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https://www.edaboard.com/threads/task-state-in-scheduling.401028/
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#### Daljeet12
##### Member level 4
I have come across the Wikipedia page which explains the theoretical terms but I am having trouble understanding its real example.
I just assume some task are (e.g. LED ON, LED OFF, read a button, display a message on an LCD screen, Motor ON, Motor OFF, read temperature data, send serial data to PC, receive data from PC )
I am showing some example of running task
CPU sending instructions to turn on the LED.
CPU sending instructions to turn off the LED
CPU sending instructions to turn on the Motor.
CPU sending instructions to turn off the Motor.
If ADC data arrives in every 100us. CPU only takes 10us to receive ADC data. Now CPU has 90us of free time and we can use it to do any other task. We can use 90us CPU time to complete other tasks.
Does it mean that the ADC task will run on the CPU for 10us and it will be suspended for 90us? It means It should be ready to run every 100us. Because if the CPU is not assigned after 100us then the incoming data will be lost.
I can explain more but i want to confirm that what i explained in scheduling is correct.
#### doraemon
##### Super Moderator
Staff member
Hello!
Does it mean that the ADC task will run on the CPU for 10us and it will be suspended for 90us?
Basically yes, but usually you set the parameters of ADC to work by itself, so the
ADC engine works continuously in parallel with the CPU, and it tells the CPU when it
has new data available. Therefore, it doesn't take 10µs to read it. Well, it depends
on which processor you are using, but recent chips work at several MHz, so reading the
data when it's ready just takes a few 100s ns or possibly 1µs if you run on an extreme
low power system, not more.
Because if the CPU is not assigned after 100us then the incoming data will be lost.
What do you mean by "CPU not assigned"?
Is your ADC used to measure temperature? In this case I don't see any need of
not losing a single data. Temperature evolution is usually slow, reading it at 100µs
intervals is a lot too fast.
But OK, if you have some real time data you don't want to skip, you have to make sure
That said, considering what you are describing, reading temperature, switching on / off
motors and LEDs, all this is not a big deal and there will be no problem to process it
that fast.
I can explain more but i want to confirm that what i explained in scheduling is correct.
Indeed, it would be useful to explain more. I have a fuzzy image about what you want
to do, reading a temperature, report it to a PC and then power LED / motors on and off,
possibly from the PC via a CPU. Based on that I can only make assumptions.
Dora
#### KlausST
##### Super Moderator
Staff member
Hi,
The key is: interrupt.
There is a MAIN loop.
And there are interrupt tasks, called ISRs. They are no loops. They have a beginning and an end.
So you run your standard program in MAIN, as a loop.
And every time when the ADC data is available,
* the MAIN loop is suspended
* the ISR is started, fetching data from ADC
* doing something "fast" with the data, like storing to SRAM
* the ISR is finished
* and processing the MAIN resumes
Klaus
#### Daljeet12
##### Member level 4
That said, considering what you are describing, reading temperature, switching on / off
motors and LEDs, all this is not a big deal and there will be no problem to process it
that fast.
My intent is to understand the description given in the Wikipedia page, that's why I showed of my tasks.
The Task in the system can be anything. I have taken some examples of Tasks in my list just to understand the description of page, they all are not related to the actual application, it is purely taken to understand the information given on wiki page.
What do you mean by "CPU not assigned"?
To complete any task, the task requires CPU service. CPU sends necessary instructions to the device to perform that task.
We need to decide which one task needs to control the CPU and when it needs to take back control from the task to serve another task.
Is your ADC used to measure temperature? In this case I don't see any need of
not losing a single data. Temperature evolution is usually slow, reading it at 100µs
intervals is a lot too fast.
In this case I will assume that I have any analog input that I need to read within 100us, if I am not reading within the time limit the I will be lost data.
Now you must have got the idea that I am just trying to understand the many Task state from real example.
--- Updated ---
Hi,
The key is: interrupt.
There is a MAIN loop.
And there are interrupt tasks, called ISRs. They are no loops. They have a beginning and an end.
You mean something like this
MAIN loop.
Does Task 1 need run (if yes, call)... Does Task 2 need run (if yes, call)... Does Task 3 need run (if yes, call)... Does Task 1 need run (if yes, call)... Does Task 2 need run (if yes, call)... Does Task 3 need run (if yes, call)
Last edited:
#### doraemon
##### Super Moderator
Staff member
Hello!
My intent is to understand the description given in the Wikipedia page, that's why I showed of my tasks.
At the end of the day, you have to begin programming. And I would even say at the
beginning of the day. You cannot learn how to ride a bicycle from a book. You have
to get some kind of feeling on how this works and what you can do with it. So you should
In this case I will assume that I have any analog input that I need to read within 100us,
if I am not reading within the time limit the I will be lost data.
As said earlier, on recent processors there is a way to setup the ADC to have data
at the rate you want (within the CPU capability). Everytime data is ready, the ADC
rings the CPU, that's the interrupts Klaus was talking about.
You can even have a program without any loop, it works fine.
every 100µs. From there, it's simple. You do what you have to do with the result, and
then you put the CPU in sleep mode until the next interrupt. This assumes that the
whole processing takes less than 100µs. As soon as you cross this time limit, you will
have one interrupt every 200µs, although this depends on the processor and the way
you program it.
You mean something like this
MAIN loop.
Etc..
Not exactly. If you setup an interrupt, then an interrupt service routine (ISR) will
be called automatically. In this routine, you set up a state variable that will be
It would look like this (in pseudo code), with an ADC and a timer, both with interrupts:
Code:
volatile int event = NO_EVENT;
void main(void) {
setup();
while(1) {
switch(event) {
break;
case TIMER_EVENT:
process_timer();
break;
}
event = NO_EVENT;
}
}
[some ISR definitions, pragmas, etc...]
}
[some other ISR definitions, pragmas, etc...]
void my_TIMER_isr(void) {
event = TIMER_EVENT;
}
Another way to do this:
Code:
void main() { // NB: Usually it's int main, but it doesn't make sense if you don't use an OS.
setup();
__enable_interrupts();
};
So The only task of the main function is to start your program, and then do nothing.
Then, for example if you want to do something with ADC every 100 µs:
Code:
[some ISR definitions, pragmas, etc...]
__disable_interrupts();
__enable_interrupts();
}
As long as process_adc() takes less than 100µs, it will just run fine.
So again, my advice would be to start programming. Academic concepts certainly don't hurt, but there is
a great deal of practice needed to really understand what you want to do and how you can achieve
Dora
#### KlausST
##### Super Moderator
Staff member
Hi,
You mean something like this
It is
Code:
Main_loop:
{
}
The task is an ISR. It is called independently of the main loop.
You may have main tasks that take 10ms each, and still the ADC_ISR can be called every 100us.
Let's say your main job is baking a cake.
It maybe takes an hour
But whenever your mobile phone rings you stop the the bakery job, talk for a short time, then resume the bakery job.
The mobile phone is not part of the bakery job, it is completely independent. You may get no phone call, you may get 10 phone calls, you don't know before.
There are ISRs that are called very regularily, like ADC sampling. It may be very accurate every 100us.
There are other ISRs that are called rather unregularily, like an USB communication, or a keypress.
During bakery.. the timing of the telephone calls may be random.
But you could also use a "reminder" that tells very regularily you every 5 minutes to look for the children.
Klaus
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2023-02-04 12:11:21
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https://www.math.uci.edu/category/event-category/number-theory
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# TBA
Xiannan Li
## Institution:
Kansas State University
## Time:
Thursday, January 12, 2023 - 3:00pm to 4:00pm
Zoom
# TBA
Shamil Asgarli
## Institution:
Santa Clara University
## Time:
Thursday, October 6, 2022 - 3:00pm to 4:00pm
RH 306
# Quantum money from quaternion algebras
Shahed Sharif
## Institution:
Cal State University, San Marcos
## Time:
Thursday, October 13, 2022 - 3:00pm to 4:00pm
## Location:
RH 306
Public key quantum money is a replacement for paper money which has cryptographic guarantees against counterfeiting. We propose a new idea for public key quantum money. In the abstract sense, our bills are encoded as a joint eigenstate of a fixed system of commuting unitary operators. We show that the proposal is secure against black box attacks. In order to instantiate this protocol, one needs to find a cryptographically complicated system of computable, commuting, unitary operators. To fill this need, we propose using Brandt operators acting on the Brandt modules associated to certain quaternion algebras. This is joint work with Daniel Kane and Alice Silverberg.
# TBA
Hanson Smith
## Institution:
Cal State University, San Marcos
## Time:
Thursday, November 3, 2022 - 3:00pm to 4:00pm
RH 306
# Matrix enumeration over finite fields (Note the special day!)
Yifeng Huang
UBC
## Time:
Tuesday, September 27, 2022 - 3:00pm to 4:00pm
## Location:
RH 306
I will investigate certain matrix enumeration problems over a finite field, guided by the phenomenon that many such problems tend to have a generating function with a nice factorization. I then give a uniform and geometric explanation of the phenomenon that works in many cases, using the statistics of finite-length modules (or coherent sheaves) studied by Cohen and Lenstra. However, my recent work on counting pairs of matrices of the form AB=BA=0 (arXiv: 2110.15566) and AB=uBA for a root of unity u (arXiv: 2110.15570), through purely combinatorial methods, gives examples where the phenomenon still holds true in the absence of the above explanation. Time permitting, I will talk about a partial progress on the system of equations AB=BA, A^2=B^3 in a joint work with Ruofan Jiang. In particular, it verifies a pattern that I previously conjectured in an attempt to explain the phenomenon in the AB=BA=0 case geometrically.
# Southern California Number Theory Day
## Speaker:
Aaron Landesman, Michelle Manes, Holly Swisher, Stanley Xiao
## Institution:
Harvard University, University of Hawaii, Oregon State University, University of Northern British Columbia
## Time:
Saturday, September 24, 2022 - 9:30am to 5:30pm
## Location:
NS II 1201
Schedule: There will be four one hour invited lectures starting at 10AM and ending around 5:30PM. A more detailed schedule will be posted soon.
SpeakersAaron Landesman (Harvard University), Michelle Manes (University of Hawaii), Holly Swisher (Oregon State University), Stanley Xiao (University of Northern British Columbia)
Lightning Talks: We are planning a session where number theory graduate students and postdocs are invited to present their research. These talks will be approximately 5-10 minutes. If you would like to give a lightning talk, please contact Nathan Kaplan by September 9. Please include your name, affiliation, advisor's name, talk title, and a brief abstract.
Registration: There is no registration fee for the conference, but to help our planning please register.
Location: Natural Sciences II, room 1201 (building 402, located at G6 on this map).
Travel support: Some travel funding is available for participants, with preference given to graduate students and postdocs, especially those giving lightning talks. We also encourage applications from members of under-represented groups. If you would like to apply for funding, please contact Nathan Kaplan with an itemized estimate of expenses, preferably by September 16. Please include your name, affiliation, and advisor's name (if applicable). We strongly encourage carpooling.
Dinner: There will be a conference dinner. Details TBD.
# TBA
Lea Beneish
UC Berkeley
## Time:
Thursday, October 27, 2022 - 3:00pm to 4:00pm
RH 306
# TBA
Abhishek Oswal
Caltech
## Time:
Thursday, November 17, 2022 - 3:00pm to 4:00pm
RH 306
# Counting polynomials with a prescribed Galois group
Tel-Aviv
## Time:
Thursday, April 7, 2022 - 10:00am to 11:00am
## Location:
https://uci.zoom.us/j/95268809663
An old problem, dating back to Van der Waerden, asks about counting irreducible polynomials degree $n$ polynomials with coefficients in the box [-H,H] and prescribed Galois group. Van der Waerden was the first to show that H^n+O(H^{n-\delta}) have Galois group S_n and he conjectured that the error term can be improved to o(H^{n-1}).
Recently, Bhargava almost proved van der Waerden conjecture showing that there are O(H^{n-1+\varepsilon}) non S_n extensions, while Chow and Dietmann showed that there are O(H^{n-1.017}) non S_n, non A_n extensions for n>=3 and n\neq 7,8,10.
In joint work with Lior Bary-Soroker, and Or Ben-Porath we use a result of Hilbert to prove a lower bound for the case of G=A_n, and upper and lower bounds for C_2 wreath S_{n/2} . The proof for A_n can be viewed, on the geometric side, as constructing a morphism \varphi from A^{n/2} into the variety z^2=\Delta(f) where each varphi_i is a quadratic form. For the upper bound for C_2 wreath S_{n/2} we prove a monic version of Widmer's result four counting polynomials with imprimitive Galois group.
# Gaussian distribution of squarefree and B-free numbers in short intervals
## Speaker:
Alexander Mangerel
## Institution:
Durham University
## Time:
Thursday, March 31, 2022 - 10:00am to 11:00am
## Location:
https://uci.zoom.us/j/95268809663
(Joint with O. Gorodetsky and B. Rodgers) It is a classical quest in analytic number theory to understand the fine-scale distribution of arithmetic sequences such as the primes. For a given length scale h, the number of elements of a nice'' sequence in a uniformly randomly selected interval $(x,x+h], 1 \leq x \leq X$, might be expected to follow the statistics of a normally distributed random variable (in suitable ranges of $1 \leq h \leq X$). Following the work of Montgomery and Soundararajan, this is known to be true for the primes, but only if we assume several deep and long-standing conjectures such as the Riemann Hypothesis. In fact, previously such distributional results had not been proven for any (non-trivial) sequence of number-theoretic interest, unconditionally.
As a model for the primes, in this talk I will address such statistical questions for the sequence of squarefree numbers, i.e., numbers not divisible by the square of any prime, among other related sifted'' sequences called B-free numbers. I hope to further motivate and explain our main result that shows, unconditionally, that short interval counts of squarefree numbers do satisfy Gaussian statistics, answering several old questions of R.R. Hall.
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2022-09-27 20:53:07
|
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https://indico.fysik.su.se/event/4640/
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Öppna föreläsningar på Albanova
# Trapped antihydrogen: the ALPHA experiment at CERN
## by Jeffrey Scott Hangst (Aarhus University, Denmark)
Europe/Stockholm
Oskar Klein salen ()
### Oskar Klein salen
Description
It has been just over 100 years since Niels Bohr proposed his famous model for the hydrogen atom. It is thus very exciting that we are now actually able to experimentally study antihydrogen - the antimatter equivalent of hydrogen. The question to be addressed is fundamental and profound: “Do matter and antimatter obey the same laws of physics?” The so-called Standard Model of fundamental particles and interactions requires that hydrogen and antihydrogen have the same spectrum. At CERN in Geneva, the ALPHA collaboration is working to test this requirement by performing direct spectroscopic measurements on trapped atoms of antihydrogen. Antihydrogen atoms have been produced in quantity at CERN since 2002, when the ATHENA collaboration demonstrated[1] how to mix cryogenic plasmas of antiprotons and positrons to produce low energy anti-atoms. I will discuss the newest development along the road to antihydrogen spectroscopy: magnetically trapped antihydrogen. In November of 2010 we reported[2] the first trapping of antihydrogen atoms in a magnetic multipole trap. The atoms must be produced with an energy - in temperature units - of less than 0.5 K in order to be trapped. Subsequently, we have shown that trapped antihydrogen can be stored[3] for up to 1000 s, and we have performed the first resonant quantum interaction experiments with anti-atoms[4]. We have recently demonstrated a new technique[5] to study the gravitational behaviour of antihydrogen atoms in free-fall, and we have put a limit on the charge of antihydrogen[6]. I will discuss the many developments necessary to realise trapped antihydrogen, and I will consider the future of this rapidly evolving field of study. 1. Amoretti, M. et al., Production and detection of cold antihydrogen atoms. Nature, 419, 456 (2002). 2. Andresen, G.B. et al., Trapped Antihydrogen, Nature, 468, 673 (2010). 3. Andresen, G. B. et al. Confinement of antihydrogen for 1,000 seconds. Nature Physics 7, 558 (2011). 4. Amole, C. et al., Resonant quantum transitions in trapped antihydrogen atoms, Nature 483, 439 (2012). 5. Amole, C. et al., Description and first application of a new technique to measure the gravitational mass of antihydrogen, Nature Communications DOI: 10.1038/ncomms2787 (2013) 6. Amole, C. et al., An experimental limit on the charge of antihydrogen, Nature Communications, doi:10.1038/ncomms4955 (2014)
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2020-06-06 01:33:21
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http://math.stackexchange.com/questions/288245/rejection-method-for-beta-distribution
|
# Rejection method for beta distribution?
I have the follow function, $$\frac{8}{\pi}(x(1-x))^{1/2} 0<x<1$$
I am asked to use $U(0,1)$ as an envelope to construct a rejection algorithm for simulation samples from $Beta(3/2,3/2)$ with density f.
Would i be correct in thinking that $u=\frac{f(x)}{mg(x)}=2(x(1-x))^{1/2}$.
It then asks let Z be the random variable which denotes the number of tries from $U(0,1)$ until we accept the first sample from $Beta(3/2,3/2$, describe the distribution of Z.
Would Z follow a geometric distribution?
Many thanks in advance any help most appreciated.
-
Since $f_X(x)\leqslant\frac4\pi f_U(x)$ uniformly, each try is accepted with probability $\frac\pi4$ and $Z$ is geometric with parameter $p=\frac\pi4$, that is, $\mathbb P(Z=n)=p(1-p)^{n-1}$ for every $n\geqslant1$.
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2015-11-30 15:29:57
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https://labs.tib.eu/arxiv/?author=Helion%20du%20Mas%20des%20Bourboux
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• The fourth generation of the Sloan Digital Sky Survey (SDSS-IV) has been in operation since July 2014. This paper describes the second data release from this phase, and the fourteenth from SDSS overall (making this, Data Release Fourteen or DR14). This release makes public data taken by SDSS-IV in its first two years of operation (July 2014-2016). Like all previous SDSS releases, DR14 is cumulative, including the most recent reductions and calibrations of all data taken by SDSS since the first phase began operations in 2000. New in DR14 is the first public release of data from the extended Baryon Oscillation Spectroscopic Survey (eBOSS); the first data from the second phase of the Apache Point Observatory (APO) Galactic Evolution Experiment (APOGEE-2), including stellar parameter estimates from an innovative data driven machine learning algorithm known as "The Cannon"; and almost twice as many data cubes from the Mapping Nearby Galaxies at APO (MaNGA) survey as were in the previous release (N = 2812 in total). This paper describes the location and format of the publicly available data from SDSS-IV surveys. We provide references to the important technical papers describing how these data have been taken (both targeting and observation details) and processed for scientific use. The SDSS website (www.sdss.org) has been updated for this release, and provides links to data downloads, as well as tutorials and examples of data use. SDSS-IV is planning to continue to collect astronomical data until 2020, and will be followed by SDSS-V.
• The DESI Legacy Imaging Surveys are a combination of three public projects (the Dark Energy Camera Legacy Survey, the Beijing-Arizona Sky Survey, and the Mayall z-band Legacy Survey) that will jointly image ~14,000 square degrees of the extragalactic sky visible from the northern hemisphere in three optical bands (g, r, and z) using telescopes at the Kitt Peak National Observatory and the Cerro Tololo Inter-American Observatory. The combined survey footprint is split into two contiguous areas by the Galactic plane. The optical imaging is conducted using a unique strategy of dynamic observing that results in a survey of nearly uniform depth. In addition to calibrated images, the project is delivering an inference-based catalog which includes photometry from the grz optical bands and from four mid-infrared bands (at 3.4um, 4.6um, 12um and 22um) observed by the Wide-field Infrared Survey Explorer (WISE) satellite during its full operational lifetime. The project plans two public data releases each year. All the software used to generate the catalogs is also released with the data. This paper provides an overview of the Legacy Surveys project.
• ### The clustering of the SDSS-IV extended Baryon Oscillation Spectroscopic Survey DR14 quasar sample: measurement of the growth rate of structure from the anisotropic correlation function between redshift 0.8 and 2.2(1801.03062)
Jan. 9, 2018 astro-ph.CO
We present the clustering measurements of quasars in configuration space based on the Data Release 14 (DR14) of the Sloan Digital Sky Survey IV extended Baryon Oscillation Spectroscopic Survey. This dataset includes 148,659 quasars spread over the redshift range $0.8\leq z \leq 2.2$ and spanning 2112.9 square degrees. We use the Convolution Lagrangian Perturbation Theory (CLPT) approach with a Gaussian Streaming (GS) model for the redshift space distortions of the correlation function and demonstrate its applicability for dark matter halos hosting eBOSS quasar tracers. At the effective redshift $z_{\rm eff} = 1.52$, we measure the linear growth rate of structure $f\sigma_{8}(z_{\rm eff})= 0.426 \pm 0.077$, the expansion rate $H(z_{\rm eff})= 159^{+12}_{-13}(r_{s}^{\rm fid}/r_s){\rm km.s}^{-1}.{\rm Mpc}^{-1}$, and the angular diameter distance $D_{A}(z_{\rm eff})=1850^{+90}_{-115}\,(r_s/r_{s}^{\rm fid}){\rm Mpc}$, where $r_{s}$ is the sound horizon at the end of the baryon drag epoch and $r_{s}^{\rm fid}$ is its value in the fiducial cosmology. The quoted errors include both systematic and statistical contributions. The results on the evolution of distances are consistent with the predictions of flat $\Lambda$-Cold Dark Matter ($\Lambda$-CDM) cosmology with Planck parameters, and the measurement of $f\sigma_{8}$ extends the validity of General Relativity (GR) to higher redshifts($z>1$) This paper is released with companion papers using the same sample. The results on the cosmological parameters of the studies are found to be in very good agreement, providing clear evidence of the complementarity and of the robustness of the first full-shape clustering measurements with the eBOSS DR14 quasar sample.
• ### The triply-ionized carbon forest from eBOSS: cosmological correlations with quasars in SDSS-IV DR14(1801.01852)
Jan. 5, 2018 astro-ph.CO
We present measurements of the cross-correlation of the triply-ionized carbon (CIV) forest with quasars using Sloan Digital Sky Survey Data Release 14. The study exploits a large sample of new quasars from the first two years of observations by the Extended Baryon Oscillation Spectroscopic Survey (eBOSS). The CIV forest is a weaker tracer of large-scale structure than the Ly$\alpha$ forest, but benefits from being accessible at redshifts $z<2$ where the quasar number density from eBOSS is high. Our data sample consists of 287,651 CIV forest quasars in the redshift range $1.4<z<3.5$ and 387,315 tracer quasars with $1.2<z<3.5$. We measure large-scale correlations from CIV absorption occuring in three distinct quasar rest-frame wavelength bands of the spectra referred to as the CIV forest, the SiIV forest and the Ly$\alpha$ forest. From the combined fit to the quasar-CIV cross-correlations for the CIV forest and the SiIV forest, the CIV redshift-space distortion parameter is $\beta_{\rm CIV}=0.27_{\ -0.14}^{\ +0.16}$ and its combination with the CIV linear transmission bias parameter is $b_{\rm CIV}(1+\beta_{\rm CIV})=-0.0183_{\ -0.0014}^{\ +0.0013}$ ($1\sigma$ statistical error) at the mean redshift $z=2.00$. Splitting the sample at $z=2.2$ to constrain the bias evolution with redshift yields the power-law exponent $\gamma=0.60\pm0.63$, indicating a significantly weaker redshift-evolution than for the Ly$\alpha$ forest linear transmission bias. We demonstrate that CIV absorption has the potential to be used as a probe of baryon acoustic oscillations (BAO). While the current data set is insufficient for a detection of the BAO peak feature, the final quasar samples for redshifts $1.4<z<2.2$ from eBOSS and the Dark Energy Spectroscopic Instrument (DESI) are expected to provide measurements of the isotropic BAO scale to $\sim7\%$ and $\sim3\%$ precision, respectively, at $z\simeq1.6$.
• ### The clustering of the SDSS-IV extended Baryon Oscillation Spectroscopic Survey DR14 quasar sample: First measurement of Baryon Acoustic Oscillations between redshift 0.8 and 2.2(1705.06373)
Oct. 16, 2017 astro-ph.CO
We present measurements of the Baryon Acoustic Oscillation (BAO) scale in redshift-space using the clustering of quasars. We consider a sample of 147,000 quasars from the extended Baryon Oscillation Spectroscopic Survey (eBOSS) distributed over 2044 square degrees with redshifts $0.8 < z < 2.2$ and measure their spherically-averaged clustering in both configuration and Fourier space. Our observational dataset and the 1400 simulated realizations of the dataset allow us to detect a preference for BAO that is greater than 2.8$\sigma$. We determine the spherically averaged BAO distance to $z = 1.52$ to 3.8 per cent precision: $D_V(z=1.52)=3843\pm147 \left(r_{\rm d}/r_{\rm d, fid}\right)\$Mpc. This is the first time the location of the BAO feature has been measured between redshifts 1 and 2. Our result is fully consistent with the prediction obtained by extrapolating the Planck flat $\Lambda$CDM best-fit cosmology. All of our results are consistent with basic large-scale structure (LSS) theory, confirming quasars to be a reliable tracer of LSS, and provide a starting point for numerous cosmological tests to be performed with eBOSS quasar samples. We combine our result with previous, independent, BAO distance measurements to construct an updated BAO distance-ladder. Using these BAO data alone and marginalizing over the length of the standard ruler, we find $\Omega_{\Lambda} > 0$ at 6.6$\sigma$ significance when testing a $\Lambda$CDM model with free curvature.
• ### Baryon acoustic oscillations from the complete SDSS-III Ly$\alpha$-quasar cross-correlation function at $z=2.4$(1708.02225)
Oct. 4, 2017 astro-ph.CO
We present a measurement of baryon acoustic oscillations (BAO) in the cross-correlation of quasars with the Ly$\alpha$-forest flux-transmission at a mean redshift $z=2.40$. The measurement uses the complete SDSS-III data sample: 168,889 forests and 234,367 quasars from the SDSS Data Release DR12. In addition to the statistical improvement on our previous study using DR11, we have implemented numerous improvements at the analysis level allowing a more accurate measurement of this cross-correlation. We also developed the first simulations of the cross-correlation allowing us to test different aspects of our data analysis and to search for potential systematic errors in the determination of the BAO peak position. We measure the two ratios $D_{H}(z=2.40)/r_{d} = 9.01 \pm 0.36$ and $D_{M}(z=2.40)/r_{d} = 35.7 \pm 1.7$, where the errors include marginalization over the non-linear velocity of quasars and the metal - quasar cross-correlation contribution, among other effects. These results are within $1.8\sigma$ of the prediction of the flat-$\Lambda$CDM model describing the observed CMB anisotropies. We combine this study with the Ly$\alpha$-forest auto-correlation function [2017A&A...603A..12B], yielding $D_{H}(z=2.40)/r_{d} = 8.94 \pm 0.22$ and $D_{M}(z=2.40)/r_{d} = 36.6 \pm 1.2$, within $2.3\sigma$ of the same flat-$\Lambda$CDM model.
• ### The SDSS-DR12 large-scale cross-correlation of Damped Lyman Alpha Systems with the Lyman Alpha Forest(1709.00889)
Sept. 27, 2017 astro-ph.CO
We present a measurement of the DLA mean bias from the cross-correlation of DLA and the Ly$\alpha$ forest, updating earlier results of Font-Ribera et al. 2012 with the final BOSS Data Release and an improved method to address continuum fitting corrections. Our cross-correlation is well fitted by linear theory with the standard $\Lambda CDM$ model, with a DLA bias of $b_{\rm DLA} = 1.99\pm 0.11$; a more conservative analysis, which removes DLA in the Ly$\beta$ forest and uses only the cross-correlation at $r> 10{\rm h^{-1}\,Mpc}$, yields $b_{\rm DLA} = 2.00\pm 0.19$. This assumes the cosmological model from \cite{Planck2015} and the Ly$\alpha$ forest bias factors of Bautista et al. 2017, and includes only statistical errors obtained from bootstrap analysis. The main systematic errors arise from possible impurities and selection effects in the DLA catalogue, and from uncertainties in the determination of the Ly$\alpha$ forest bias factors and a correction for effects of high column density absorbers. We find no dependence of the DLA bias on column density or redshift. The measured bias value corresponds to a host halo mass $\sim 4\cdot10^{11} {\rm M_{\odot}}$ if all DLA were hosted in halos of a similar mass. In a realistic model where host halos over a broad mass range have a DLA cross section $\Sigma(M_h) \propto M_h^{\alpha}$ down to $M_h > M_{\rm min} =10^{8.5} {\rm M_{\odot}}$, we find that $\alpha > 1$ is required to have $b_{\rm DLA}> 1.7$, implying a steeper relation or higher value of $M_{\rm min}$ than is generally predicted in numerical simulations of galaxy formation.
• The fourth generation of the Sloan Digital Sky Survey (SDSS-IV) began observations in July 2014. It pursues three core programs: APOGEE-2, MaNGA, and eBOSS. In addition, eBOSS contains two major subprograms: TDSS and SPIDERS. This paper describes the first data release from SDSS-IV, Data Release 13 (DR13), which contains new data, reanalysis of existing data sets and, like all SDSS data releases, is inclusive of previously released data. DR13 makes publicly available 1390 spatially resolved integral field unit observations of nearby galaxies from MaNGA, the first data released from this survey. It includes new observations from eBOSS, completing SEQUELS. In addition to targeting galaxies and quasars, SEQUELS also targeted variability-selected objects from TDSS and X-ray selected objects from SPIDERS. DR13 includes new reductions of the SDSS-III BOSS data, improving the spectrophotometric calibration and redshift classification. DR13 releases new reductions of the APOGEE-1 data from SDSS-III, with abundances of elements not previously included and improved stellar parameters for dwarf stars and cooler stars. For the SDSS imaging data, DR13 provides new, more robust and precise photometric calibrations. Several value-added catalogs are being released in tandem with DR13, in particular target catalogs relevant for eBOSS, TDSS, and SPIDERS, and an updated red-clump catalog for APOGEE. This paper describes the location and format of the data now publicly available, as well as providing references to the important technical papers that describe the targeting, observing, and data reduction. The SDSS website, http://www.sdss.org, provides links to the data, tutorials and examples of data access, and extensive documentation of the reduction and analysis procedures. DR13 is the first of a scheduled set that will contain new data and analyses from the planned ~6-year operations of SDSS-IV.
• We describe the Sloan Digital Sky Survey IV (SDSS-IV), a project encompassing three major spectroscopic programs. The Apache Point Observatory Galactic Evolution Experiment 2 (APOGEE-2) is observing hundreds of thousands of Milky Way stars at high resolution and high signal-to-noise ratio in the near-infrared. The Mapping Nearby Galaxies at Apache Point Observatory (MaNGA) survey is obtaining spatially-resolved spectroscopy for thousands of nearby galaxies (median redshift of z = 0.03). The extended Baryon Oscillation Spectroscopic Survey (eBOSS) is mapping the galaxy, quasar, and neutral gas distributions between redshifts z = 0.6 and 3.5 to constrain cosmology using baryon acoustic oscillations, redshift space distortions, and the shape of the power spectrum. Within eBOSS, we are conducting two major subprograms: the SPectroscopic IDentification of eROSITA Sources (SPIDERS), investigating X-ray AGN and galaxies in X-ray clusters, and the Time Domain Spectroscopic Survey (TDSS), obtaining spectra of variable sources. All programs use the 2.5-meter Sloan Foundation Telescope at Apache Point Observatory; observations there began in Summer 2014. APOGEE-2 also operates a second near-infrared spectrograph at the 2.5-meter du Pont Telescope at Las Campanas Observatory, with observations beginning in early 2017. Observations at both facilities are scheduled to continue through 2020. In keeping with previous SDSS policy, SDSS-IV provides regularly scheduled public data releases; the first one, Data Release 13, was made available in July 2016.
• ### Exploring cosmic homogeneity with the BOSS DR12 galaxy sample(1702.02159)
June 1, 2017 astro-ph.CO
In this study, we probe the transition to cosmic homogeneity in the Large Scale Structure (LSS) of the Universe using the CMASS galaxy sample of BOSS spectroscopic survey which covers the largest effective volume to date, $3\ h^{-3}\ \mathrm{Gpc}^3$ at $0.43 \leq z \leq 0.7$. We study the scaled counts-in-spheres, $\mathcal{N}(<r)$, and the fractal correlation dimension, $\mathcal{D}_2(r)$, to assess the homogeneity scale of the universe using a $Landy\ \&\ Szalay$ inspired estimator. Defining the scale of transition to homogeneity as the scale at which $\mathcal{D}_2(r)$ reaches 3 within $1\%$, i.e. $\mathcal{D}_2(r)>2.97$ for $r>\mathcal{R}_H$, we find $\mathcal{R}_H = (63.3\pm0.7) \ h^{-1}\ \mathrm{Mpc}$, in agreement at the percentage level with the predictions of the $\Lambda$CDM model $\mathcal{R}_H=62.0\ h^{-1}\ \mathrm{Mpc}$. Thanks to the large cosmic depth of the survey, we investigate the redshift evolution of the transition to homogeneity scale and find agreement with the $\Lambda$CDM prediction. Finally, we find that $\mathcal{D}_2$ is compatible with $3$ at scales larger than $300\ h^{-1}\$Mpc in all redshift bins. These results consolidate the Cosmological Principle and represent a precise consistency test of the $\Lambda CDM$ model.
• ### Clustering of quasars in SDSS-IV eBOSS : study of potential systematics and bias determination(1705.04718)
May 12, 2017 astro-ph.CO
We study the first year of the eBOSS quasar sample in the redshift range $0.9<z<2.2$ which includes 68,772 homogeneously selected quasars. We show that the main source of systematics in the evaluation of the correlation function arises from inhomogeneities in the quasar target selection, particularly related to the extinction and depth of the imaging data used for targeting. We propose a weighting scheme that mitigates these systematics. We measure the quasar correlation function and provide the most accurate measurement to date of the quasar bias in this redshift range, $b_Q = 2.45 \pm 0.05$ at $\bar z=1.55$, together with its evolution with redshift. We use this information to determine the minimum mass of the halo hosting the quasars and the characteristic halo mass, which we find to be both independent of redshift within statistical error. Using a recently-measured quasar-luminosity-function we also determine the quasar duty cycle. The size of this first year sample is insufficient to detect any luminosity dependence to quasar clustering and this issue should be further studied with the final $\sim$500,000 eBOSS quasar sample.
• ### Measurement of BAO correlations at $z=2.3$ with SDSS DR12 \lya-Forests(1702.00176)
March 27, 2017 astro-ph.CO
We use flux-transmission correlations in \Lya forests to measure the imprint of baryon acoustic oscillations (BAO). The study uses spectra of 157,783 quasars in the redshift range $2.1\le z \le 3.5$ from the Sloan Digital Sky Survey (SDSS) Data Release 12 (DR12). Besides the statistical improvements on our previous studies using SDSS DR9 and DR11, we have implemented numerous improvements in the analysis procedure, allowing us to construct a physical model of the correlation function and to investigate potential systematic errors in the determination of the BAO peak position. The Hubble distance, $\DHub=c/H(z)$, relative to the sound horizon is $\DHub(z=2.33)/r_d=9.07 \pm 0.31$. The best-determined combination of comoving angular-diameter distance, $\DM$, and the Hubble distance is found to be $\DHub^{0.7}\DM^{0.3}/r_d=13.94\pm0.35$. This value is $1.028\pm0.026$ times the prediction of the flat-\lcdm model consistent with the cosmic microwave background (CMB) anisotropy spectrum. The errors include marginalization over the effects of unidentified high-density absorption systems and fluctuations in ultraviolet ionizing radiation. Independently of the CMB measurements, the combination of our results and other BAO observations determine the open-\lcdm density parameters to be $\om=0.296 \pm 0.029$, $\ol=0.699 \pm 0.100$ and $\Omega_k = -0.002 \pm 0.119$.
• ### A 14 $h^{-3}$ Gpc$^3$ study of cosmic homogeneity using BOSS DR12 quasar sample(1602.09010)
Nov. 21, 2016 astro-ph.CO
The BOSS quasar sample is used to study cosmic homogeneity with a 3D survey in the redshift range $2.2<z<2.8$. We measure the count-in-sphere, $N(<\! r)$, i.e. the average number of objects around a given object, and its logarithmic derivative, the fractal correlation dimension, $D_2(r)$. For a homogeneous distribution $N(<\! r) \propto r^3$ and $D_2(r)=3$. Due to the uncertainty on tracer density evolution, 3D surveys can only probe homogeneity up to a redshift dependence, i.e. they probe so-called "spatial isotropy". Our data demonstrate spatial isotropy of the quasar distribution in the redshift range $2.2<z<2.8$ in a model-independent way, independent of any FLRW fiducial cosmology, resulting in $3-\langle D_2 \rangle < 1.7 \times 10^{-3}$ (2 $\sigma$) over the range $250<r<1200 \, h^{-1}$Mpc for the quasar distribution. If we assume that quasars do not have a bias much less than unity, this implies spatial isotropy of the matter distribution on large scales. Then, combining with the Copernican principle, we finally get homogeneity of the matter distribution on large scales. Alternatively, using a flat $\Lambda$CDM fiducial cosmology with CMB-derived parameters, and measuring the quasar bias relative to this $\Lambda$CDM model, our data provide a consistency check of the model, in terms of how homogeneous the Universe is on different scales. $D_2(r)$ is found to be compatible with our $\Lambda$CDM model on the whole $10<r<1200 \, h^{-1}$Mpc range. For the matter distribution we obtain $3-\langle D_2 \rangle < 5 \times 10^{-5}$ (2 $\sigma$) over the range $250<r<1200 \, h^{-1}$Mpc, consistent with homogeneity on large scales.
• The Extended Baryon Oscillation Spectroscopic Survey (eBOSS) will conduct novel cosmological observations using the BOSS spectrograph at Apache Point Observatory. Observations will be simultaneous with the Time Domain Spectroscopic Survey (TDSS) designed for variability studies and the Spectroscopic Identification of eROSITA Sources (SPIDERS) program designed for studies of X-ray sources. eBOSS will use four different tracers to measure the distance-redshift relation with baryon acoustic oscillations (BAO). Using more than 250,000 new, spectroscopically confirmed luminous red galaxies at a median redshift z=0.72, we project that eBOSS will yield measurements of $d_A(z)$ to an accuracy of 1.2% and measurements of H(z) to 2.1% when combined with the z>0.6 sample of BOSS galaxies. With ~195,000 new emission line galaxy redshifts, we expect BAO measurements of $d_A(z)$ to an accuracy of 3.1% and H(z) to 4.7% at an effective redshift of z= 0.87. A sample of more than 500,000 spectroscopically-confirmed quasars will provide the first BAO distance measurements over the redshift range 0.9<z<2.2, with expected precision of 2.8% and 4.2% on $d_A(z)$ and H(z), respectively. Finally, with 60,000 new quasars and re-observation of 60,000 quasars known from BOSS, we will obtain new Lyman-alpha forest measurements at redshifts z>2.1; these new data will enhance the precision of $d_A(z)$ and H(z) by a factor of 1.44 relative to BOSS. Furthermore, eBOSS will provide improved tests of General Relativity on cosmological scales through redshift-space distortion measurements, improved tests for non-Gaussianity in the primordial density field, and new constraints on the summed mass of all neutrino species. Here, we provide an overview of the cosmological goals, spectroscopic target sample, demonstration of spectral quality from early data, and projected cosmological constraints from eBOSS.
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## Critical phenomena and conformal invariance in the plane BMETE95MM23, Spring 2012 List of suggested papers for presentation
Lecturer: Gábor Pete
You are welcome to ask my help when reading the paper. Aim at understanding what your chosen paper is about, and at each of you presenting at least one non-trivial proof. You can use the paper or your notes for the presentation, but make sure you know the basic definitions and ideas. Duration should be 60-90 minutes for a pair, and around 45 minutes for a single presenter.
Conformal invariance and winding numbers of planar Brownian motion
Section 7.2 of the book Brownian motion by Peter Mörters and Yuval Peres.
8 pages.
http://people.bath.ac.uk/maspm/book.pdf
On monochromatic arm exponents for 2D critical percolation
Vincent Beffara, Pierre Nolin
18 pages, 4 figures
We investigate the so-called "monochromatic arm exponents" for critical percolation in two dimensions. These exponents, describing the probability of observing j disjoint macroscopic paths, are shown to exist and to form a different family from the (now well understood) polychromatic exponents. More specifically, our main result is that the monochromatic j-arm exponent is strictly between the polychromatic j-arm and (j+1)-arm exponents.
http://front.math.ucdavis.edu/0906.3570
One-arm exponent for critical 2D percolation
Gregory F. Lawler, Oded Schramm, Wendelin Werner
18 pages, 1 figure
The probability that the cluster of the origin in critical site percolation on the triangular grid has diameter larger than $R$ is proved to decay like $R^{-5/48}$ as $R\to\infty$.
http://front.math.ucdavis.edu/0108.5211
Quantitative noise sensitivity and exceptional times for percolation
Oded Schramm, Jeffrey E. Steif
64 pages
One goal of this paper is to prove that dynamical critical site percolation on the planar triangular lattice has exceptional times at which percolation occurs. In doing so, new quantitative noise sensitivity results for percolation are obtained. The latter is based on a novel method for controlling the "level k" Fourier coefficients via the construction of a randomized algorithm which looks at random bits, outputs the value of a particular function but looks at any fixed input bit with low probability. We also obtain upper and lower bounds on the Hausdorff dimension of the set of percolating times. We then study the problem of exceptional times for certain "k-arm" events on wedges and cones. As a corollary of this analysis, we prove, among other things, that there are no times at which there are two infinite "white" clusters, obtain an upper bound on the Hausdorff dimension of the set of times at which there are both an infinite white cluster and an infinite black cluster and prove that for dynamical critical bond percolation on the square grid there are no exceptional times at which three disjoint infinite clusters are present.
http://front.math.ucdavis.edu/0504.5586
(A follow-up is The Fourier spectrum of critical percolation by Garban, Pete and Schramm.)
The self-dual point of the two-dimensional random-cluster model is critical for $q\geq 1$
Vincent Beffara, Hugo Duminil-Copin
27 pages, 10 figures
We prove a long-standing conjecture on random-cluster models, namely that the critical point for such models with parameter $q\geq1$ on the square lattice is equal to the self-dual point $p_{sd}(q) = \sqrt q /(1+\sqrt q)$. This gives a proof that the critical temperature of the $q$-state Potts model is equal to $\log (1+\sqrt q)$ for all $q\geq 2$. We further prove that the transition is sharp, meaning that there is exponential decay of correlations in the sub-critical phase. The techniques of this paper are rigorous and valid for all $q\geq 1$, in contrast to earlier methods valid only for certain given $q$. The proof extends to the triangular and the hexagonal lattices as well.
http://front.math.ucdavis.edu/1006.5073
An introduction to the dimer model [Szanto Bandi + Rozsa Levente + Vajna Szabolcs]
Richard Kenyon
Lecture notes from a minicourse given at the ICTP in May 2002.
http://arxiv.org/pdf/math.CO/0310326.pdf
Lectures on Dimers
Richard Kenyon
These are lecture notes for lectures at the Park City Math Institute, summer 2007. We cover aspects of the dimer model on planar, periodic bipartite graphs, including local statistics, limit shapes and fluctuations.
http://front.math.ucdavis.edu/0910.3129
Discrete complex analysis on isoradial graphs [Nagy Attila + ?]
Dmitry Chelkak, Stanislav Smirnov
35 pages, 4 figures.
We study discrete complex analysis and potential theory on a large family of planar graphs, the so-called isoradial ones. Along with discrete analogues of several classical results, we prove uniform convergence of discrete harmonic measures, Green's functions and Poisson kernels to their continuous counterparts. Among other applications, the results can be used to establish universality of the critical Ising and other lattice models.
http://front.math.ucdavis.edu/0810.2188
Unfortunately, on the last class I didn't have time to define Smirnov's fermionic observable, which is more-or-less the discrete holomorphic function for the FK-Ising model whose convergence to a true holomorphic function is responsible for the conformal invariance of the model. But either of the next two papers will give you the main idea. (The convergence is much harder to prove than in the percolation case, that's why I'm not suggesting Smirnov's Ann of Math paper Conformal invariance in random-cluster models I. Holomorphic fermions in the Ising model for presentation.)
Connection probabilities and RSW-type bounds for the FK Ising model
Hugo Duminil-Copin, Clément Hongler, Pierre Nolin
31 pages, 9 figures
We prove Russo-Seymour-Welsh-type uniform bounds on crossing probabilities for the FK Ising model at criticality, independent of the boundary conditions. Our proof relies mainly on Smirnov's fermionic observable for the FK Ising model, which allows us to get precise estimates on boundary connection probabilities. It remains purely discrete, in particular we do not make use of any continuum limit, and it can be used to derive directly several noteworthy results - some new and some not - among which the fact that there is no spontaneous magnetization at criticality, tightness properties for the interfaces, and the existence of several critical exponents, in particular the half-plane one-arm exponent.
http://front.math.ucdavis.edu/0912.4253
Smirnov's fermionic observable away from criticality
Vincent Beffara, Hugo Duminil-Copin
19 pages, 6 figures
In a recent and celebrated article, Smirnov defines an observable for the self-dual random-cluster model with cluster weight $q=2$ on the square lattice $\Z^2$, and uses it to obtain conformal invariance in the scaling limit. We study this observable away from the self-dual point. From this, we obtain a new derivation of the fact that the self-dual and critical points coincide, which implies that the critical inverse temperature of the Ising model equals $\frac12\log(1+\sqrt2)$. Moreover, we relate the correlation length of the model to the large deviation behavior of a certain massive random walk (thus confirming an observation by Messikh), which allows us to compute it explicitly.
http://front.math.ucdavis.edu/1010.0526
(A follow-up is The near-critical planar FK-Ising model by Duminil-Copin, Garban and Pete.)
The connective constant of the honeycomb lattice equals $\sqrt{2+\sqrt2}$
Hugo Duminil-Copin, Stanislav Smirnov
11 pages, 4 figures
We provide the first mathematical proof that the connective constant of the hexagonal lattice is equal to $\sqrt{2+\sqrt 2}$. This value has been derived non rigorously by B. Nienhuis in 1982, using Coulomb gas approach from theoretical physics. Our proof uses a parafermionic observable for the self avoiding walk, which satisfies a half of the discrete Cauchy-Riemann relations. Establishing the other half of the relations (which conjecturally holds in the scaling limit) would also imply convergence of the self-avoiding walk to SLE(8/3).
http://front.math.ucdavis.edu/1007.0575
Critical Ising on the square lattice mixes in polynomial time
Eyal Lubetzky, Allan Sly
26 pages; 5 figures
The Ising model is widely regarded as the most studied model of spin-systems in statistical physics. The focus of this paper is its dynamic (stochastic) version, the Glauber dynamics, introduced in 1963 and by now the most popular means of sampling the Ising measure. Intensive study throughout the last three decades has yielded a rigorous understanding of the spectral-gap of the dynamics on $\Z^2$ everywhere except at criticality. While the critical behavior of the Ising model has long been the focus for physicists, mathematicians have only recently developed an understanding of its critical geometry with the advent of SLE, CLE and new tools to study conformally invariant systems.
A rich interplay exists between the static and dynamic models. At the static phase-transition for Ising, the dynamics is conjectured to undergo a critical slowdown: At high temperature the inverse-gap is O(1), at the critical $\beta_c$ it is polynomial in the side-length and at low temperature it is exponential in it. A seminal series of papers verified this on $\Z^2$ except at $\beta=\beta_c$ where the behavior remained a challenging open problem.
Here we establish the first rigorous polynomial upper bound for the critical mixing, thus confirming the critical slowdown for the Ising model in $\Z^2$. Namely, we show that on a finite box with arbitrary (e.g. fixed, free, periodic) boundary conditions, the inverse-gap at $\beta=\beta_c$ is polynomial in the side-length. The proof harnesses recent understanding of the scaling limit of critical Fortuin-Kasteleyn representation of the Ising model together with classical tools from the analysis of Markov chains.
http://front.math.ucdavis.edu/1001.1613
Gaussian free fields for mathematicians
Scott Sheffield
27 pages, 1 figure.
The d-dimensional Gaussian free field (GFF), also called the (Euclidean bosonic) massless free field, is a d-dimensional-time analog of Brownian motion. Just as Brownian motion is the limit of the simple random walk (when time and space are appropriately scaled), the GFF is the limit of many incrementally varying random functions on d-dimensional grids. We present an overview of the GFF and some of the properties that are useful in light of recent connections between the GFF and the Schramm-Loewner evolution.
http://front.math.ucdavis.edu/0312.5099
The harmonic explorer and its convergence to SLE(4)
Oded Schramm, Scott Sheffield
The harmonic explorer is a random grid path. Very roughly, at each step the harmonic explorer takes a turn to the right with probability equal to the discrete harmonic measure of the left-hand side of the path from a point near the end of the current path. We prove that the harmonic explorer converges to SLE(4) as the grid gets finer.
http://front.math.ucdavis.edu/0310.5210
(This is a prequel to Contour lines of the two-dimensional discrete Gaussian free field by Schramm and Sheffield. An even more serious continuation is Imaginary Geometry I: Interacting SLEs by Miller and Sheffield, which (together with parts II and III) are part of the Liouville Quantum Gravity program in the next item.)
Liouville Quantum Gravity and KPZ
Bertrand Duplantier, Scott Sheffield
56 pages.
Consider a bounded planar domain D, an instance h of the Gaussian free field on D (with Dirichlet energy normalized by 1/(2\pi)), and a constant 0 < gamma < 2. The Liouville quantum gravity measure on D is the weak limit as epsilon tends to 0 of the measures \epsilon^{\gamma^2/2} e^{\gamma h_\epsilon(z)}dz, where dz is Lebesgue measure on D and h_\epsilon(z) denotes the mean value of h on the circle of radius epsilon centered at z. Given a random (or deterministic) subset X of D one can define the scaling dimension of X using either Lebesgue measure or this random measure. We derive a general quadratic relation between these two dimensions, which we view as a probabilistic formulation of the KPZ relation from conformal field theory. We also present a boundary analog of KPZ (for subsets of the boundary of D). We discuss the connection between discrete and continuum quantum gravity and provide a framework for understanding Euclidean scaling exponents via quantum gravity.
http://front.math.ucdavis.edu/0808.1560
(Sequels are Conformal weldings of random surfaces: SLE and the quantum gravity zipper and Quantum gravity and inventory accumulation by Sheffield.)
Uniform random planar maps are discrete analogues of what should be the random planar metric space underlying Liouville Quantum Gravity. These notes are concerned with the local limit of the uniform planar quadrangulation with $n$ faces, and give a transparent proof that the volume growth of this infinite planar graph is of degree 4.
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2021-10-23 07:31:22
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https://socratic.org/questions/what-is-the-name-of-the-binary-compound-that-has-the-formula-cu-2se#290696
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# What is the name of the binary compound that has the formula Cu_2Se?
$\text{Cuprous selenide}$
Alternatively, $\text{copper (I) selenide}$. Copper here is in the univalent oxidation state.
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2022-05-29 01:15:23
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https://mersenneforum.org/showthread.php?s=d92bb8b12249192389d14483c4f5a1ed&t=28001
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mersenneforum.org Divergent Posts from Benchmark Thread
Register FAQ Search Today's Posts Mark Forums Read
2022-07-24, 18:47 #1 storm5510 Random Account Aug 2009 Not U. + S.A. 2·1,163 Posts Divergent Posts from Benchmark Thread I have an RTX 2080. What is a proper number of threads to specify? My GPU utilization never goes beyond 2%. Ed's Note: This and the following posts have been copied/moved from the Msieve benchmark thread since they more directly involve getting a GPU running, rather than adjusting and providing benchmark values. Last fiddled with by EdH on 2022-08-11 at 12:43
2022-07-24, 19:57 #2 frmky Jul 2003 So Cal 1001101010002 Posts No threads. After compiling the msieve-lacuda-nfsathome branch with the option CUDA=75, run with -nc2 -g 0 -v where the number after -g specifies which gpu to use. Without the -g option, it uses only the cpu.
2022-07-24, 22:22 #3
storm5510
Random Account
Aug 2009
Not U. + S.A.
2·1,163 Posts
Quote:
Originally Posted by frmky No threads. After compiling the msieve-lacuda-nfsathome branch with the option CUDA=75, run with -nc2 -g 0 -v where the number after -g specifies which gpu to use. Without the -g option, it uses only the cpu.
OK. Well, what i have is not Linux. It's Windows 10. Thanks for the reply!
2022-07-27, 01:43 #4
wombatman
I moo ablest echo power!
May 2013
22×11×41 Posts
Quote:
Originally Posted by storm5510 OK. Well, what i have is not Linux. It's Windows 10. Thanks for the reply!
If you're willing to do a little work, Windows Subsystem for Linux 2 works with CUDA now, so you can build and run it there.
2022-07-27, 23:26 #5
storm5510
Random Account
Aug 2009
Not U. + S.A.
2·1,163 Posts
Quote:
Originally Posted by wombatman If you're willing to do a little work, Windows Subsystem for Linux 2 works with CUDA now, so you can build and run it there.
I found one Windows version which will run, to a point. It says it cannot find "sort_engine.dll" in a folder called "cub." It is right there in the proper place where the program looks for it.
I have the subsystem installed. It is Ubuntu 20.04 LTS. It's been a while since I've used it.
2022-07-28, 01:22 #6
frmky
Jul 2003
So Cal
23×3×103 Posts
Quote:
Originally Posted by storm5510 I found one Windows version which will run, to a point. It says it cannot find "sort_engine.dll" in a folder called "cub." It is right there in the proper place where the program looks for it.
That's an old version that doesn't contain my work on the GPU LA. I'm not aware of anyone that's compiled a Windows version with my changes.
There's a guide to using CUDA with WSL2 at https://docs.nvidia.com/cuda/wsl-user-guide/index.html.
Last fiddled with by frmky on 2022-07-28 at 01:27
2022-07-28, 16:56 #7
storm5510
Random Account
Aug 2009
Not U. + S.A.
2·1,163 Posts
Quote:
Originally Posted by frmky That's an old version that doesn't contain my work on the GPU LA. I'm not aware of anyone that's compiled a Windows version with my changes. There's a guide to using CUDA with WSL2 at https://docs.nvidia.com/cuda/wsl-user-guide/index.html.
What I have on this machine is WSL1. A better option, for me, is to install Ubuntu on an idle machine. I have v21.10 burned on a DVD-R. This would, at least, give me the GUI. I know enough console commands to get by. The system I am considering has a GTX 1080 in it. I have notes I made relative to that particular system, including CUDA driver installation. I'm not familiar with "bash" files so I would have to do a bit of reading.
Point me to a version I can try and I will give it a go.
2022-07-29, 00:48 #8 storm5510 Random Account Aug 2009 Not U. + S.A. 44268 Posts Disregard the above. I found several versions on James Heinrich's mirror page which will run. The one I am running now is "1018-vbits256-sandybridge."
2022-07-29, 02:00 #9
charybdis
Apr 2020
32×5×19 Posts
Quote:
Originally Posted by storm5510 Disregard the above. I found several versions on James Heinrich's mirror page which will run. The one I am running now is "1018-vbits256-sandybridge."
None of the versions on mersenne.ca have the GPU linear algebra code.
2022-07-29, 03:49 #10 frmky Jul 2003 So Cal 23·3·103 Posts Once you have installed the compiler and CUDA toolkit... https://docs.nvidia.com/cuda/cuda-in...u-installation Code: sudo apt install git libgmp-dev git clone https://github.com/gchilders/msieve_nfsathome.git -b msieve-lacuda-nfsathome cd msieve_nfsathome make all CUDA=75 VBITS=256 If you don't want to run from that directory, then copy msieve, *.ptx, and cub/*.so to the directory you are running in.
2022-07-29, 14:41 #11
storm5510
Random Account
Aug 2009
Not U. + S.A.
44268 Posts
Quote:
Originally Posted by charybdis None of the versions on mersenne.ca have the GPU linear algebra code.
Then where can they be found?
Last fiddled with by storm5510 on 2022-07-29 at 15:23
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Wed Sep 28 10:10:44 UTC 2022 up 41 days, 7:39, 0 users, load averages: 1.00, 1.16, 1.03
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2022-09-28 10:10:44
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http://openstudy.com/updates/5597fb50e4b0c3287d027295
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• anonymous
What is the domain of the function: {(1, 2); (2, 4); (3, 6); (4, 8)}? A. {6, 8} B. {2, 4, 6, 8} C. {1, 2, 3, 4, 6, 8} D. {1, 2, 3, 4} What is the range of the function: {(2, 1); (4, 2); (6, 3); (8, 4)}? A. {2, 4, 6, 8} B. {1, 2, 3, 4} C. {1, 2, 3, 4, 6, 8} D. {6, 8} Suppose p varies directly with d, and p = 7 when d = 3. What is the value of d when p = 28? A.4/3 B.140/3 C. 16 D. 12
Mathematics
Looking for something else?
Not the answer you are looking for? Search for more explanations.
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2017-03-29 11:15:41
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http://wiki.socr.umich.edu/index.php/SMHS_GLM
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# SMHS GLM
## Scientific Methods for Health Sciences - Generalized Linear Modeling (GLM)
### Overview
Generalized Linear Modeling (GLM) is a flexible generalization of ordinary linear regression, which allows for response variables that have error distribution models other than a normal distribution. It generalizes linear regression by allowing the linear model to be related to the response variable via a link function and allowing the magnitude of the variance of each measurement to be function of its predicted value. GLM is a way of unifying statistical models like linear regression logistic regression and Poisson regression. Methods like iteratively reweighted least squares method for maximum likelihood estimation, Bayesian approaches and least squares fitted to variance stabilized responses are proposed for GLM. In this lecture, we are going to present a general introduction to GLM including the assumptions, model applied and illustrate the application of GLM with examples presented.
### Motivation
We have discussed about linear regression, which deal with the linear relationship between response and the predictors. What if the response variable does not follow a normal distribution? For example, a model that predicts the probability of making a yes/no choice is not suitable as a linear-response model given that the probabilities are bounded on both ends. Or a study that aims to predict each decrease in 10 degrees Fahrenheit leads to 1000 fewer people going on vacation in the beach is unlikely to generalize well over small nor large beaches. In many cases, the linear regression model won’t apply and GLM has to be applied for it allows the response variables that have arbitrary distribution other than normal distribution to vary linearly with the predicted values.
### Theory
• 1) GLM components: (1) a probability distribution from the exponential family; (2) a linear predictor $\eta=X\beta$ (3) a link function $g$ such that $E[Y]=\mu=g^{-1}(\eta)$ The dependent variable $Y$ in GLM is assumed to be generated from a particular distribution in the exponential family, a large range of probability distributions include normal, binomial, Poisson and gamma distribution. The mean, $\mu$, of the distribution depends on the independent variables $X:E[Y]=\mu=g^{-1} (X\beta),$ where $E[Y]$ is the expected value of $Y, X\beta$ is the linear predictor, a linear combination of unknown parameters, $\beta$, $g$ is the linked function. The variance is typically a function $V$ of the mean: $Var(Y)=V(\mu)=V(g^{-1} (X\beta)).$ It is convenient if $V$ follows from the exponential family distribution but it may simply be that the variance is a function of the predicted value. The unknown parameters, $\beta$, are typically estimated with maximum likelihood, maximum quasi-likelihood, or Bayesian techniques.
• 2)Probability distribution: the over-dispersed exponential family of distribution is a generalization of exponential family and exponential dispersion model of distributions and includes those by $\theta$ and $\tau$, whose density function can be expressed as $f_{Y} (y│\theta,\tau)=h(y,\tau)exp\left(\frac{b(theta)^{T} T(y)-A(\theta)}{(d{\tau})}\right)$ where $\tau$ is the dispersion parameter and is usually related to the variance of the distribution. For scalar $Y$ and $\theta$, this reduces to $f_{Y} (y│\theta,\tau)=h(y,\tau)exp\left(\frac{b(theta)^{T} T(y)-A(\theta)}{(d{\tau})}\right)$, $\theta$ is related to the mean of the distribution. If $b(\theta)$ is the identity function then the distribution is said to be in canonical form (natural form).
• 3) Linear predictor and link function: the quantity which incorporate the information about the independent variables into the model. $\eta=X\beta$ relates the expected value of the data (predictor). $\eta$ is expressed as linear combination of unknown parameters $\beta$. The coefficients of the linear combination are represented as the matrix of independent variables $X$. The link function provides the relationship between the linear predictor and the mean of the distribution function. The following table lists some commonly used exponential family distribution and data that typically used for along with the canonical link functions and their inverses.
Distribution Support of distribution Typical uses Link name Link function Mean function Normal $(-\infty,+\infty)$ real Linear response data Identity $X\beta=\mu$ $\mu =X\beta$ Exponential $(0,+\infty)$ real Exponential response data, scale parameters Inverse $X\beta=-\mu^{-1}$ $\mu =(X\beta)^{-1}$ Gamma $(0,+\infty)$ real Exponential response data, scale parameters Inverse $X\beta=-\mu^{-1}$ $\mu =(X\beta)^{-1}$ Inverse Gaussian $(0,+\infty)$ real Inverse $X\beta=-\mu^{-2}$ $\mu =(X\beta)^{-\frac{1} {2}}$ Poisson integer $(0,\infty)$ Count of occurrences in fixed time/space log $X\beta=1n_{\mu}$ $\mu = exp(X\beta)$ Bernoulli integer [0,1] Outcome of single yes/no occurrence logit $X\beta=ln(\frac{\mu}{1-\mu})$ $\mu =\frac{exp(X\beta)}{1+exp(X\beta)}$ Binomial integer [0,N] Count of # of ‘yes’ in N yes/no occurrences logit $X\beta=ln(\frac{\mu}{1-\mu})$ $\mu =\frac{exp(X\beta)}{1+exp(X\beta)}$ Categorical integer [0,K],K-vector of integers [0,1] Outcome of single K-way occurrence logit $X\beta=ln(\frac{\mu}{1-\mu})$ $\mu =\frac{exp(X\beta)}{1+exp(X\beta)}$ Multinomial K-vector of integers: [0,1] Count of occurrences of different types (1,…,K) out of N total K-way occurrences logit $X\beta=ln(\frac{\mu}{1-\mu})$ $\mu =\frac {exp(X\beta)}{1+exp(X\beta)}$
In the cases of the exponential and gamma distributions, the domain of the canonical link function is not the same as the permitted range of the mean. Particularly, the linear predictor may be negative, which would give an impossible negative mean. When maximizing the likelihood, precautions must be taken to avoid this. An alternative is to use a non-canonical link function.
Note also that in the case of the Bernoulli, binomial, categorical and multinomial distributions, the support of the distributions is not the same type of data as the parameter being predicted. In all of these cases, the predicted parameter is one or more probabilities, i.e. real numbers in the range $[0,1]$. The resulting model is known as logistic regression (or multinomial logistic regression in the case that K-way rather than binary values are being predicted).
For the Bernoulli and binomial distributions, the parameter is a single probability, indicating the likelihood of occurrence of a single event. The Bernoulli still satisfies the basic condition of the generalized linear model in that, even though a single outcome will always be either $0$ or $1$, the expected value will nonetheless be a real-valued probability, i.e. the probability of occurrence of a "yes" (or $1$) outcome. Similarly, in a binomial distribution, the expected value is Np, i.e. the expected proportion of "yes" outcomes will be the probability to be predicted.
For categorical and multinomial distributions, the parameter to be predicted is a $K$-vector of probabilities, with the further restriction that all probabilities must add up to $1$. Each probability indicates the likelihood of occurrence of one of the $K$ possible values. For the multinomial distribution, and for the vector form of the categorical distribution, the expected values of the elements of the vector can be related to the predicted probabilities similarly to the binomial and Bernoulli distributions.
• 4) Fitting the model:
• Maximum likelihood: can be found using iteratively reweighted least squares algorithm with $\beta^{t+1}=\beta^{t}+J^{-1} (\beta^{(t)}) u(\beta^{(t)})$, where $J(\beta^{(t)})$ is the observed information matrix (the negative of the Hessian matrix) and $u(\beta^{(t)})$ s the score function. With Fisher’s scoring method: $\beta^{(t+1)}=\beta^{(t)} +I^{-1} (\beta^{(t)})u(\beta^{(t)})$, where $where I^{-1} (\beta^{(t)})$is the Fisher information matrix.
• Bayesian method: to approximate the posterior distribution, usually using Laplace approximation $(\int_{a}^{b}e^{Mf(x)}dx)$ or Markov Chain Monte Carlo method such as Gibbs sampling.
• 5) Advantages of GLM: (1) can perform data analysis within and between subjects without the need to average the data itself; (2) allows you to counterbalance random stimuli orders; (3) allows you to exclude segments of runs with artifacts; (4) can perform more sophisticated analysis (e.g., two factor ANOVA with interactions); (5) easier to work with.
### Applications
This article proposed an extension of generalized linear models to the analysis of longitudinal data. We introduce a class of estimating equations that give consistent estimates of the regression parameters and of their variance under mild assumptions about the time dependence. The estimating equations are derived without specifying the joint distribution of a subject's observations yet they reduce to the score equations for multivariate Gaussian outcomes. Asymptotic theory is presented for the general class of estimators. Specific cases in which we assume independence, m-dependence and exchangeable correlation structures from each subject are discussed. Efficiency of the proposed estimators in two simple situations is considered. The approach is closely related to quasi-likelihood.
This article proposed a conceptually very simple but general algorithm for the estimation of the fixed effects, random effects, and components of dispersion in generalized linear models with random effects. Conditions are described under which the algorithm yields approximate maximum likelihood or quasi-maximum likelihood estimates of the fixed effects and dispersion components, and approximate empirical Bayes estimates of the random effects. The algorithm is applied to two data sets to illustrate the estimation of components of dispersion and the modelling of over-dispersion.
### Software
GLM in R:
glm(formula, family=familytype(link=linkfunction),data= ):
# list of family and default link functions:
Family Default Like Function Binomial (link=’logit’) Gaussian (link=identity) Gamma (link=inverse) Inverse gamma (link=${1}/mu^{2}$) Poisson (link=’log’) Quasi (link=’identity’,variance=’constant’) Quasibinomial (link=’logit’) Quasipoisson (link=’log’)
# Logistic regression is useful when predict a binary outcome from a set
of continuous predictor variables.
# F as a binary factor and x1-x3 are continuous predictors
fit <- glm(F~x1+x2+x3, data=mydata, family=binomial())
summary(fit) # display the results
confint(fit) # 95% CI for the coefficients
predict(fit,type=’response’) # predicted values
residuals(fit,type=’deviance’) # residuals
See these Logistic Regression R Examples.
## Use Poisson regression when predicting an outcome variable representing
counts from a set of continuous predictor variables
# Y is a count and x1-x3 are continuous predictors
fit <- glm(Y~x1+x2+x3, data=mydata, family=poisson())
summary(fit) # display results
### TO illustrate the GLM with the following example
cuse
Age Education Wants More Not Using Using 1 <25 low yes 53 6 2 <25 low no 10 4 3 <25 high yes 212 52 4 <25 high no 50 10 5 25-29 low yes 60 14 6 25-29 low no 19 10 7 25-29 high yes 155 54 8 25-29 high no 65 27 9 30-39 low yes 112 33 10 30-39 low no 77 80 11 30-39 high yes 118 46 12 30-39 high no 68 78 13 40-49 low yes 35 6 14 40-49 low no 46 48 15 40-49 high yes 8 8 16 40-49 high no 12 31
attach(cuse)
lrfit <- glm( cbind(using, notUsing) ~ + age + education + wantsMore ,
family = binomial) # run the GLM on age, education, wantsMore
lrfit
Call: glm(formula = cbind(using, notUsing) ~ +age + education + wantsMore,
family = binomial)
Coefficients:
(Intercept) age25-29 age30-39 age40-49 educationlow
-0.8082 0.3894 0.9086 1.1892 -0.3250
wantsMoreyes
-0.8330
Degrees of Freedom: 15 Total (i.e. Null); 10 Residual
Null Deviance: 165.8
Residual Deviance: 29.92 AIC: 113.4
### Now relevel to change the base category and define our own indicator variables
with high education and women who want no more children:
noMore <- wantsMore == "no"
hiEduc <- education == "high"
glm( cbind(using,notUsing) ~ age + hiEduc + noMore, family=binomial)
Call: glm(formula = cbind(using, notUsing) ~ age + hiEduc + noMore,
family = binomial)
Coefficients:
(Intercept) age25-29 age30-39 age40-49 hiEducTRUE noMoreTRUE
-1.9662 0.3894 0.9086 1.1892 0.3250 0.8330
Degrees of Freedom: 15 Total (i.e. Null); 10 Residual
Null Deviance: 165.8
Residual Deviance: 29.92 AIC: 113.4
1-pchisq(29.92,10)
0.0008828339
## we see that the residual deviance of 29.92 on 10 degree of freedom is highly significant,
a better model could be introducing an interaction term of age and desire for no more children:
lrfit <- glm( cbind(using,notUsing) ~ age * noMore + hiEduc , family=binomial)
lrfit
Call: glm(formula = cbind(using, notUsing) ~ age * noMore + hiEduc,
family = binomial)
Coefficients:
(Intercept) age25-29 age30-39
-1.80317 0.39460 0.54666
age40-49 noMoreTRUE hiEducTRUE
0.57952 0.06622 0.34065
age25-29:noMoreTRUE age30-39:noMoreTRUE age40-49:noMoreTRUE
0.25918 1.11266 1.36167
Degrees of Freedom: 15 Total (i.e. Null); 7 Residual
Null Deviance: 165.8
Residual Deviance: 12.63 AIC: 102.1
## now the model’s deviance of 12.63 on 7 degree of freedom is not significant at
the conventional five per cent level, so there is no evidence against this model.
summary(lrfit)
Call:
glm(formula = cbind(using, notUsing) ~ age * noMore + hiEduc,
family = binomial)
Deviance Residuals:
Min 1Q Median 3Q Max
-1.30027 -0.66163 -0.03286 0.81945 1.73851
Coefficients:
Estimate Std. Error z value Pr(>|z|)
(Intercept) -1.80317 0.18018 -10.008 < 2e-16 ***
age25-29 0.39460 0.20145 1.959 0.05013 .
age30-39 0.54666 0.19842 2.755 0.00587 **
age40-49 0.57952 0.34742 1.668 0.09530 .
noMoreTRUE 0.06622 0.33071 0.200 0.84130
hiEducTRUE 0.34065 0.12577 2.709 0.00676 **
age25-29:noMoreTRUE 0.25918 0.40975 0.633 0.52704
age30-39:noMoreTRUE 1.11266 0.37404 2.975 0.00293 **
age40-49:noMoreTRUE 1.36167 0.48433 2.811 0.00493 **
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
(Dispersion parameter for binomial family taken to be 1)
Null deviance: 165.77 on 15 degrees of freedom
Residual deviance: 12.63 on 7 degrees of freedom
AIC: 102.14
Number of Fisher Scoring iterations: 4
### Problems
Repeat the GLM analyses above using the following SOCR datasets
|
2017-10-17 11:24:55
|
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|
http://www.math.sci.osaka-u.ac.jp/~matsumoto/
|
# Yoshihiko Matsumoto
Interested in mathematics, especially in differential geometry and function theory of several complex variables. My recent work centers around asymptotically symmetric spaces and their conformal infinities.
Assistant Professor, Department of Mathematics, Graduate School of Science, Osaka University
1-1 Machikaneyama-cho, Toyonaka, Osaka 560-0043, Japan
Email: matsumoto (at) math.sci.osaka-u.ac.jp
Room 413, Bldg. B of Graduate School of Science
Phone: +81-6-6850-5313
From Sep. 1, 2017 until Aug. 29, 2019, I'm away from Osaka and working as:
Visiting Assistant Professor, Department of Mathematics, Stanford University
450 Serra Mall Building 380, Stanford, CA 94305-2125, United States
Phone: +1 650-723-6763
## News
• Jan. 16, 2018: Taiji Marugame kindly pointed out some misprints and computational errors in my thesis. (They are all minor, and the main content of the thesis needs no modification.)
• Jan. 16, 2018: We organized the workshop “Geometric Analysis in Geometry and Topology 2017” on December 19–22 last year (my “Foundations and developments of Poincaré-Einstein metrics” was a part of it). The notes offered by the lecturers are made public here.
• Jan. 9, 2018: The notes prepared for the lecture “Foundations and developments of Poincaré-Einstein metrics”, which I gave in December last year in Tokyo, is now available.
• Sep. 1, 2017: I arrived at Stanford University as a two-year visitor.
• Nov. 15, 2016: The paper “Variation of total Q-prime curvature on CR manifolds” is accepted by Advances in Mathematics for publication.
• Oct. 28, 2016: I'll be visiting Stanford University for two years, starting from September next year. Great thanks to the people concerned, especially to the math department members of Osaka University.
• Aug. 10, 2016: The slides and related information (in Japanese) for the public lecture “Geometry of the Hyperbolic Plane” have been uploaded.
## Talks to deliver
• Deformation of Einstein metrics and $L^2$ cohomology on strictly pseudoconvex domains. Geometry Seminar, Stanford University, Stanford, USA (Jan. 24, 2018)
• Deformation of Einstein metrics and $L^2$ cohomology on strictly pseudoconvex domains. Differential Geometry Seminar, University California, Riverside, Riverside, USA (Jan. 26, 2018)
• What is curvature—From the viewpoint of comparison theorems. Mathematics Café, Tokyo (Place TBD), Japan (February 25, 2018)
• TBA. Asymptotically Hyperbolic Manifolds, Banff International Research Station, Banff, Canada (May 14–18, 2018)
|
2018-03-17 16:40:46
|
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|
https://tex.stackexchange.com/questions/475550/node-text-baseline-in-tikz-for-multi-part-nodes
|
# Node text baseline in tikz for multi-part-nodes
I basically have the exact same problem as this question, only for multi-part nodes: I have a node that consists of 3 parts, each of which contains text. I want to have the text aligned by its baseline. I tried to achive this using the text height and text depth, which works fine for ordinary nodes, but not for multi-part ones. As you can see in the image below, only the first part of the node receives the correct alignment I want to achieve, the other two get "push up" by characters that extend below the baseline.
This is my latex-code:
\documentclass[tikz, 11pt,landscape]{article}
\usepackage[default,osfigures,scale=0.95]{opensans}
\usepackage[T1]{fontenc}
\usepackage[utf8]{inputenc}
\usepackage{tgbonum}
\usepackage{tikz}
\usetikzlibrary{shapes}
\usetikzlibrary{calc}
\usetikzlibrary{arrows.meta}
\usepackage[a3paper]{geometry}
\begin{document}
\thispagestyle{empty}
\tikzset{normal/.style = {rectangle split, rectangle split horizontal, rectangle split parts=3, very thick, draw=black, minimum height=0.6cm,text height=1.5ex, text depth=0.25ex}}
\begin{tikzpicture}[remember picture, overlay]
\node[normal] at (current page.center) {\nodepart{one} ag \nodepart{two} ag \nodepart{three} ag};
\end{tikzpicture}
\end{document}
I found another post about someone having the same problem with transparency only beeing applied to the first part, but its solution didn't work for me. I do not care about multi-line text and actually only really care about the alignment in the second part of the node, I just included the other two texts for demonstration. It seems like whatever I try, it only gets applied to the first part.
Edit:
Using my screenshot as an example: I want the 2nd and 3rd box to be aligned to the height of the 1st one. I do NOT want to align the 1st box to the height of the 2nd and 3rd one, since that is quite high and looks strange if I have a text without characters that go below the baseline. The spacing will look very uneven.
• this seems to be bug in tikz library shapes.multipart (or at least missed feature). – Zarko Feb 18 '19 at 20:33
A lot of thanks to @CarlaTex kindness and to her explanations that make me understand ( I hope this time I am not wrong!) what is really needed by this question
The workaround is the use of an invisible \rule that has the correct vertical dimensions (height and depth):
\documentclass[tikz, 11pt,landscape]{article}
\usepackage[default,osfigures,scale=0.95]{opensans}
\usepackage[T1]{fontenc}
\usepackage[utf8]{inputenc}
\usepackage{tgbonum}
\usepackage{tikz}
\usetikzlibrary{shapes}
\usetikzlibrary{calc}
\usetikzlibrary{arrows.meta}
\usepackage[a3paper]{geometry}
\begin{document}
\thispagestyle{empty}
\tikzset{
normal/.style = {
rectangle split,
rectangle split horizontal,
rectangle split parts=3,
very thick, draw=black,
minimum height=2cm,
text height=1.5cm,
text depth=0.25cm
}
}
\begin{tikzpicture}[remember picture, overlay]
\node[normal] at (current page.center) {\nodepart{one} ag
\nodepart{two} \rule[-0.25cm]{0pt}{1.75cm}ag
\nodepart{three} \rule[-0.25cm]{0pt}{1.75cm}ag};
\end{tikzpicture}
\end{document}
• @CarLaTeX The three node parts are aligned. What doesn't work ? – Hafid Boukhoulda Feb 18 '19 at 19:50
• @CarLaTeX I confess that I don't understand what is needed. – Hafid Boukhoulda Feb 18 '19 at 20:11
• @CarLaTeX If I move minimum height=0.6cm in the normal style I get the image posted in the OP. The three parts contents being not vertically aligned and I still don't understand what is needed. – Hafid Boukhoulda Feb 18 '19 at 20:20
• I tried this solution, and you are right, they are vertically aligned. However: They all get "pushed up" now. If I insert text without a character that goes below the baseline, it looks strange. What I wanted was to align the height of the 2nd and 3rd box to the height of the 1st. Your solution aligns the 1st box to the height of the 2nd and 3rd. I hope that makes sense. I edited the question to better clarify that. – pulp_user Feb 18 '19 at 21:04
• @pulp_user answer updated ! Is that what is needed ? – Hafid Boukhoulda Feb 18 '19 at 21:31
|
2021-07-30 17:17:43
|
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|
https://socratic.org/questions/if-a-8-and-b-6-how-do-you-find-c
|
# If a=8 and b=6, how do you find c?
May 17, 2018
$c = 10$
#### Explanation:
If $a , b , c$ are the sides of a right triangle, and $c$ is the hypotenuse (the longest side), then the Pythagorean theorem states that:
$\textcolor{w h i t e}{m m m m m m m m m} \textcolor{b l u e}{\overline{\underline{|} {a}^{2} + {b}^{2} = {c}^{2} |}}$
So here, $a = 8 , b = 6$.
$\therefore {c}^{2} = {8}^{2} + {6}^{2}$
$= 64 + 36$
$= 100$
$c = \sqrt{100}$
$= \pm 10$
Since this is a length, we cannot take the negative side, as that'll make no sense! So, we only take the positive root, and the final answer is:
$c = 10$
|
2019-11-14 12:18:13
|
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|
https://hssliveguru.com/plus-one-physics-notes-chapter-3/
|
# Plus One Physics Notes Chapter 3 Motion in a Straight Line
Students can Download Chapter 3 Motion in a Straight Line Notes, Plus One Physics Notes helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.
## Kerala Plus One Physics Notes Chapter 3 Motion in a Straight Line
Plus One Physics Chapter 3 Notes Pdf Summary
Motion In A Straight Line
In this chapter, we shall learn how to describe motion. For this, we develop the concepts of velocity, acceleration and relative velocity. We also develop a set of simple equations called Kinematic equations.
• Motion: Motion is change in position of an object with time.
• Rectilinear motion: The motion along a straight line is called rectilinear motion.
• Point object: If the distance travelled by the body is very large compared with its size, the size of the body may be neglected. The body under such a condition may be taken as a point object. The point object can be represented by a point.
Example:
• The length of bus may be neglected compared with the length of the road it is running.
• The size of planet is ignored compared with the size of the orbit in which it is moving.
Position, Path Length And Displacement
1. Reference point, Frame of reference:
In order to specify position of object, we take reference point and a set of axes. Consider a rectangular coordinate system consisting of three mutually perpendicular axes, labelled x, y, and z axes. The point of intersection of these three axes is called origin (O) and serves as the reference point.
The coordinates (x, y, z) of an object describe the position of the object. To measure time, we place a clock in this coordinate system. This coordinate system along with a clock is called a frame of reference.
Straight-line motion in coordinate system
To describe the motion along a straight line we can choose x-axis. The position of a carat different time are given in the figure 3.1. The position to the right of 0 is taken as positive and to the left of 0 as negative. The position coordinates of point P and Q are +360m +240m. The position coordinate of R is-120m.
2. Path Length (Distance):
The total length of the path travelled by an object is called path length.
Explanation:
Consider a car moving along straight line. The positions of car at different time are given in the x-axis. (See figure 3.1)
Case-1:
The car moves from 0 to P. In this case the distance moved by car is OP = +360.
Case-2:
The car moves from 0 to P and then moves back from P to Q.
In this case, the distance travelled is OP + PQ = +360 + (+120) = +480m.
3. Displacement:
The distance between initial point and final point is called displacement.
OR
The change of position of the particle in a particular direction is called displacement.
Explanation:
Consider a car moving along a straight line. The positions of car at different time is given in the x-axis.
See figure (3.1)
Let us take two cases
Case-1:
The car moves from 0 to P, in this case displacement = (360 – 0) = 360
Case-2:
The car moves from 0 to P and moves back from P to Q.
In this case,
Displacement = 240m
Let x1 and x2 be the positions of an object at time t1 and t2. Then displacement in time Dt = (t2 – t1) can be written as Dx = x2 – x1
If x1 < x2, Dx is positive and if x2 < x1, Dx is negative.
Note: The magnitude of displacement may or may not be equal to the path length traversed by an object.
4. Position Time Graph:
Motion of an object can be represented by a position-time graph.
Position time graph for a stationary object:
For a stationary object, the position does not change with time. Hence the position time graph will be a straight line parallel to time axis.
Position time graph in a uniform motion:
Uniform motion:
A body is said to be uniform motion, if it undergoes equal displacements in equal intervals of time. In uniform motion velocity is constant The figure below shows the positiontime graph of such a motion.
Plus One Physics Chapter 3 Notes Question 1.
The position-time of a car is given below. Analyze the graph and explain the motion of car.
The car starts from rest a time t=0s from the origin 0 and picks up speed till t=10s. After 10 sec, the car moves with uniform speed till t=18 sec. Then the brakes are applied and the car stops at t = 20s and x = 296m.
Class 11 Physics Chapter 3 Notes Question 2.
Draw the position-time for an object
1. moving with positive velocity
2. moving with negative velocity.
1.
2.
Average Velocity And Average Speed
1. Average Velocity:
The average velocity of a particle is the ratio of the total displacement to the time interval.
Explanation:
To explain average velocity, consider a position time graph of a body given below.
Let x1 be the position of body at a time t1 and x2 be the position at t2.
The average velocity during the time interval Dt = (t2 – t1)
where Dx = x2 – x1, and Dt = t2 – t1,
$$\overline{\mathbf{v}}$$ is the average velocity.
Motion In A Straight Line Class 11 Notes Pdf Question 3.
Find the slope of position time graph given below of uniform motion and explain the result.
Slope of displacement time graph gives average velocity.
Class 11th Physics Chapter 3 Notes Question 4.
Displacement time graph of a car is given below.
1. Find the average velocity during the time interval 5 to 7 sec.
2. Find the average velocity by taking slope in the interval 5 to 7 sec.
1.
2. Slope, tan q
In this case, slope and average velocity are equal in the same interval.
2. Average Speed:
Average speed of a particle is the ratio of the total distance to total time taken.
Physics Chapter 3 Class 11 Notes Question 5.
A car is moving along a straight line. Say OP in figure. It moves from 0 to P in 18s and returns from P to Q in 6s. What are the average velocity and average speed of the car in going?
1. From 0 to P? and
2. from 0 to P and back to Q. (See Figure 3.1)
1. Average velocity
Average speed
In this case the average speed is equal to the magnitude of the average velocity.
2. In this case
Average velocity
Average speed
In this case the average speed is not equal to the magnitude of the average velocity. This happens because the motion here involves change in direction. So that the distance is greater than displacement.
Note: In general, the velocity is always less than or equal to speed.
Instantaneous Velocity And Speed
Nonuniform Motion:
A body is said to be nonuniform motion, if it undergoes unequal displacements in equal intervals of time.
OR
A body moving with varying velocity is called nonuniform motion.
1. Instantaneous Velocity:
Chapter 3 Physics Class 11 Notes Question 6.
Why the concept of instantaneous velocity is introduced?
In nonuniform motion the average velocity tells us how fast the object has been moving over a given interval. But it does not tell us how it moves at different instants during that interval. For this we define instantaneous velocity. The velocity at an instant is called instantaneous velocity.
Explanation:
Position-time of a body moving along a straight line is given below.
Let us find average velocity in the interval 2 sec (3s to 5s), centered at t = 4 sec. In this case, the slope of line P1P2 give the value of average velocity, ie. Slope of P1P2,
Decrease the value of Dt from 2.to 1 sec. (ie. 3.5 to 4.5 sec). Then line P1P2 becomes Q1Q2. Then the slope of gives average velocity overthe interval 3.5 sec to 4.5sec.
ie. slope of Q1Q2
In the limit Dt ® 0, gives the instantaneous velocity at t = 4sec and its value is nearly 3.84m/s.
Motion In A Straight Line Class 11 Notes Question 7.
When average velocity of a body becomes instantaneous velocity?
In the limit, Dt goes to zero, the average velocity becomes instantaneous velocity.
But lim $$\lim _{\Delta t \rightarrow 0} \frac{\Delta x}{\Delta t}=\frac{d x}{d t}$$
\Instantaneous velocity,
Here $$\frac{d x}{d t}$$ is the differential coefficient of x with respect to time. It is the rate of change of position with respect to time at an instant.
Class 11 Physics Notes Chapter 3 Question 8.
The table given below gives the value of $$\frac{\Delta x}{\Delta t}$$ for Dt equal to 2s, 1s, 0.55, 0.1s and 0.01s centered at t = 4 sec. (See figure given above). What conclusions can be made from this table?
The value of average velocity $$\left(\frac{\Delta x}{\Delta t}\right)$$ becomes instantaneous velocity (3.8 m/s), in the limit of Dt goes to zero, (ie Dt is infinitesimally small).
The position of an object moving along x-axis is given by x = a + bt2 where a = 8.5m, b = 2.5 m/s2 and t is measured in seconds
1. What is the velocity at t = 0s and t = 2s.
2. What is the average velocity between t = 2s and t = 4s?
1.
when t = 0
we get v = 2 × 2.5 × 0
v = 0
when t = 2sec
v = 2 × 2.5 × 2 v = 10m/s.
2. The average velocity
Note: If a body is moving with constant velocity, the average velocity is the same as instantaneous velocity at all instants.
2. Instantaneous Speed:
The speed at an instant is called instantaneous speed.
Note:
• The average speed over a finite interval of time is greater or equal to the magnitude of the average velocity.
• Instantaneous speed at an instant is equal to the magnitude of the instantaneous velocity at that instant.
Motion In Straight Line Notes Pdf Acceleration
1. Average Acceleration:
Average acceleration of a particle is ratio of the change in velocity to the time interval.
Explanation
Consider a body moving along a straight line. Let v1 and v2 be the instantaneous velocities at time t1 and t2 respectively.
where Dv = change in velocity, Dt = Time interval
2. Instantaneous Acceleration:
Acceleration at any instant is called instantaneous acceleration.
Explanation
In the limit Dt ® 0, (Dt goes to zero) the average acceleration becomes instantaneous acceleration.
ie. Instantaneous acceleration
Instantaneous acceleration is the rate of change of velocity with respect to time.
3. Uniform Acceleration:
A body is said to be in uniform acceleration, if velocity changes equally in equal intervals of time.
Motion In A Straight Line Class 11 Pdf Question 10.
The velocities of two bodies A and B are given in the tables. From this table, find which body is moving with uniform acceleration. Explain.
Body A
Body B
The body A is moving with uniform acceleration be-cause the velocity of body increases at the rate of 2 m/s2.
The body B is moving with constant velocity. Hence this motion is called uniform motion.
4. Velocity-Time Graph For Uniformly Accelerated Motion:
An example for velocity-time of a uniformly accelerated motion is given in the above figure.
Let vt1 and vt2 be the velocities at instants t1 and t2respectively.
The slope of graph in the interval (t2 – t1) can be written as,
∴ tan q = acceleration
Thus the slope of the velocity-time gives the acceleration of the particle.
Motion In A Straight Line Notes Pdf Question 11.
Velocity-time of a body is given below. From this graph draw corresponding acceleration time graph.
The slope of velocity-time graph increases in the interval (0 – 10) sec which means that acceleration of the body increases in this interval.
Velocity is constant in the interval (10 – 18) sec. Hence ’ the slope is zero which means that acceleration is zero in this range.
The slope in the interval (18 – 20) sec is constant and negative. Hence acceleration in this is a negative value. The acceleration – time graph for the above motion is given below.
Motion In Straight Line Class 11 Notes Question 12.
The position-time graph of a car is given below.
2. From velocity-time graph draw acceleration-time graph and identify the regions of
• positive acceleration
• Negative acceleration
• zero acceleration.
1. In the time interval (0 – t1) sec, the slope of x – t graph increases which means that velocity is increasing in this time interval.
In the time interval (t1 – t2) sec, slope is constant. Hence velocity remains constant in this time interval.
In the time interval (t2 – t3) sec, the slope is decreasing and finally becomes zero. Which means that velocity decreases to zero.
2. Slope is constant throughout the interval (0 – t1) sec which means that acceleration constant.
In the interval (t1 – t2) sec, slope is zero. Which means that acceleration is zero in this region.
Slope is constant (but negative) in the interval (t2 – t3)sec. Hence acceleration is constant and negative in this time interval.
Motion In A Straight Line Class 11 Notes Pdf Download Question 13.
Find the region of
1. positive acceleration
2. zero acceleration
3. negative acceleration from the above x-t graph
1. Region OA – Positive acceleration
2. Region AB – zero acceleration
3. Region BC – Negative acceleration
Straight Line Class 11 Notes Question 14.
Match the following.
1) – d, 2) – c, 3) – b, 4) – a.
5. Area Under Velocity-Time Graph:
Area under velocity-time graph represents the displacement over a given time interval.
Explanation
Consider a body moving with constant velocity v. Its velocity-time graph is given below.
The area of the rectangle has height v and bast t. Therefore,
Note: The acceleration and velocity of a body cannot change values abruptly at an instant. Changes are always continuous.
Kinematic Equations For Uniformly Accelerated Motion
For uniformly accelerated motion, we can derive some simple equations.
1. Velocity-time relation
2. Position-time relation
3. Position-velocity relation
These equations are called kinematic equations for uniformly accelerated motion.
1. Velocity-Time Relation:
Consider a body moving along a straight line with uniform acceleration ‘a’. Let ‘u’ be initial velocity and ‘v ‘ be the final velocity at time t.
We know acceleration a = $$\frac{\text { Change in velocity }}{\text { Time interval }}$$
a = $$\frac{v-u}{t}$$
at = v – u
2. Position-Time Relation:
Consider a body moving along a straight line with uniform acceleration a. Let ‘u’ be initial velocity and ‘v’ be the final velocity. ‘S’ is the displacement travelled by the body during the time interval ‘t‘.
Displacement of the body during the time interval t,
S = average velocity × time
$$S=\left(\frac{v+u}{2}\right) t$$ _____(1)
But v = u + at ____(2)
Substitute eq.(2) in eq.(1), we get
3. Position-Velocity Relation:
$$S=\left(\frac{v+u}{2}\right) t$$ _____(1)
But v = u + at
$$\frac{v-u}{a}$$ = t _____(2)
Substitute eq.(2) in eq.(1)
2as = v2 – u2
v2 – u2 = 2as
Free-fall:
An object released (near the surface of earth) is accelerated towards the earth. If air resistance is neglected, the object is said to be in free fall. The acceleration due to gravity near the surface of earth is 9.8 m/s2.
Note: Free-fall is a case of motion with uniform acceleration.
Question 15.
A body is allowed to fall freely. Draw the following graph.
1. Acceleration-time
2. Velocity-time
3. Position-time
1.
2.
3.
Stopping distance of vehicles:
When brakes are applied to a moving vehicle, the distance it travels before stopping is called stopping distance.
Question 16.
Derive an expression for stopping distance of a vehicle in terms of initial velocity (u) and retardation (a).
Let the distance travelled by the vehicle before it stops be ‘s’.
Then we can find ‘s’ using the formula
v2 = u2 + 2as
0 = u2 + -2as
3.7 Relative Velocity
Suppose the distance between two bodies changes with time in magnitude, or in direction or in both. Then each body is said to have a velocity relative to the other.
For example, consider two cars A and B moving in the same direction with equal velocities. To a person in A, the car B would appear to be rest.
Hence the velocity of B relative to A is zero.
ie. VBA = 0
Similarly, the velocity of A with respect to B is zero.
or VAB = 0
Let A be moving with a velocity VA and B be moving with a greater velocity VB in the same direction. Then the person in A feels that the car B is moving away from him with a velocity VBA. The velocity of B relative to A
For an observer in B, the car A is going back with a velocity. The velocity of A relative to B
VAB = -(VB – VB).
Question 17.
The position-time graph of two bodies A and B (at different situations) are given in the following graphs. Find the relative velocities of the following graph.
a) The slope of Aand B are equal. Hence velocity of A and B are equal. So velocity of A with respect to B, VAB = 0
b) The body A and B meet at t = 3sec
Velocity of A w.r. to B, VAB = VA – VB
= 20-10 = 10 m/s Velocity of B w.r. to A, VBA = VB – VA
= 10 – 20 = -10 m/s
c) The body A and B meet at t = 1 sec.
The velocity of body in the interval t = 1 sec,
Velocity of A w. r. to B,
VAB = VA – VB
= 20 – 10 = 30 m/s
Similarly velocity of B w.r. to A,
VBA = VB – VA
= 10 – +20 = -30 m/s
The magnitude of VBA or VAB (=30 m/s) is greater than the magnitude of velocity A or that of B.
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2023-03-26 15:16:55
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https://www.gradesaver.com/textbooks/science/physics/physics-10th-edition/chapter-5-dynamics-of-uniform-circular-motion-problems-page-137/1
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## Physics (10th Edition)
0.79 m/$s^{2}$
distance = 2πr = 2 $\times$ π $\times$ 2600 = 1633.6 m v = $\frac{distance}{time}$ = $\frac{1633.6}{360}$ $\approx$ 45.38 m/s a = $\frac{v^{2}}{r}$ = $\frac{45.38^{2}}{2600}$ $\approx$ 0.79 m/$s^{2}$
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2021-03-05 19:10:31
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http://www.helengorina.com/category/autocad/cleanup/
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# Reference Manager. Post 15 in series “AutoCAD Cleanup”
## Using Reference Manager to re-link data set.
Scenario: You received a set of files on CD. The folder structure conforms to your company standard, but when you open plot files, XREFs are not found. You need to re-link all XREFs.
• Copy the CD contents to a local drive and examine its folder structure.
• Start by figuring out which files are intended to be the plot-able sheet files. This is the hardest part. You can use the transmittals, printouts, thumbnail images and/or any other means at your disposal. Sometimes you can guess from the folder structure or the file name. In this example it looks like all the plot sheets should be in a discipline specific subfolder of the plot folder.
• Once you have a working assumption of which files are the plot sheets, you can begin solving the puzzle. Start Reference Manager.
• Add the sheet files to be analyzed. I always work in batches based on plot file locations. You will be asked if you want to add all XREFs that are attached to the drawings being added. If you select “Yes”, then all nested XREFs will be added, and it can get large and confusing very quickly. In addition, you can also encounter circular referencing. When dealing with large sets, it is easier to process one level of nesting at a time by answering “No” and adding XREFs for processing later.
• The Reference Manager interface has two views.
• First, view the List by Drawing view to find out if any drawings are “broken”. In List by Drawing view, the left pane shows a hierarchical data structure with all resource files listed per drawing. If any of the XREFs or other resources are not found the drawing icon is crossed by a red line. In the right pane you can see all the needed resources. You can sort them any way you want. (by type, status, saved path, found path or any other column).
• Now let’s attempt to repair all broken links. Change to the List by Reference Type view. Select XREFs in the left pane. Sort the right pane by status.
• The missing XREFs saved path points to the P: drive. But your system does not have a P: drive. To remedy this situation and make your data set portable, you need to convert the full path to a relative path. Looking at the folder structure and saved path you can guesstimate that the XREFs should be in the discipline specific folder under the ref folder.
• The path is relative to the location of the parent (host) file. But in this batch all plot files reside in the same folder. Select all XREFs with identical saved path. Click on Edit Selected Paths, and then change the path. To get from the location of the host file to the location of the XREFs, you need to go two levels up the folder tree, then go down to the \ref\arch folder.
• You guessed right, and the status of the XREFs changed to Resolved with a pencil logo (pencil-in feature) indicating that the change is not yet applied. Click on Apply Changes to write the new paths into the host files. The status changes to Resolved.
• Process the rest of the files in groups based on the location of the host file and saved paths. If the path is not resolved for loaded XREF (open the host file to check if it is loaded) and you can’t find the file in any folder, you simply do not have it. I usually click on Export Report at this time, and examine host files to see if a missing XREF can be safely detached or whether I need to hunt them down.
• If you have any image references, their re-pathing is identical to re-pathing for XREFs.
• You do not need to worry about Plot Configurations because your plotters/printers are different anyway.
• Missing Plot Styles will result in problems with plot line thickness and colors, but in a pinch you can try Monochrome.ctb or Monochrome.stb
• If there are “not found” Fonts and Shapes, try to purge everything, and then run the reference manager report again. If the font is still not found and you can’t get it, AutoCAD will use the substitute font. Alternatively, you may want to use a font mapping table to specify which font AutoCAD substitutes when it encounters a text object created with another font. Windows True Type Fonts are independent of AutoCAD resource folders; they belong to the operating system.
# Attachments, overlays, full, relative and no Path. Post 14 in series “AutoCAD Cleanup”
### External References. Basic concepts
Data sets in AutoCAD are a collection of all files that are needed to produce graphic output for an entire project or its subsets.
It may contain various file types: drawing files, image files, font files, shape files, plot style files and other types of linked files.
Drawing files (.DWG) of the same data set are usually linked to each other forming a hierarchical XREF structure.
To understand and be able to manipulate the dataset structure you need to know the properties of XREF links.
### Attachment VS Overlay
AutoCAD uses two types of DWG references: attachments and overlays
All XREFs that are linked (attached or overlaid) directly to the parent file are visible in that file.
The second (and any subsequent) tier XREFs are only visible if they are attachments. In the schema below you will only see XREF1, XREF2, XREF3 and XREF5 from the PLOT file.
An easy way to remember: attachments are babies – they follow their parents. Overlays are teenagers – they never go with their parents.
### Full Path, Relative Path and No Path.
XREFs and image files can be linked to master files using full path, relative path or no path.
Full path can be compared to a street address; it is an exact location of the XREF or attached image on your system. It includes server name or mapped drive name.
Relative path is like point-to-point directions from the parent file to its reference. It finds the reference based on the location the parent file. The relative path says: go several levels up the directory tree (usually to where the directory tree starts branching) then from that point go to the location indicated after the last “..”
In the following example the parent file is plot.dwg. The location of plot.dwg in the folder structure is shown on the picture on the left. All references (XREFs and images) are stored at ..EX4drawingsrefarch folder.
The relative path to REF1 and REF2 from the location of plot.dwg takes us two levels up to the EX4drawings folder, then from there to ..refarch.
This method makes the data sets portable because the path does not contain server name or mapped drive name.. In other words, if I copy the entire EX4drawings folder to any other location (CD, DVD or any network location), plot.dwg finds its references.
If no path is selected, AutoCAD will first look for XREFs and images in the same folder as the parent file, then in the search path folders. The support search paths, working search paths and project search paths are user profile specific and can be set up in Options->Files. The search path folders could vary for different systems and profiles.
# Layer reconciliation. Post 13 in series “AutoCAD Cleanup”
in AutoCAD 2008 and 2009 you may have seen the following warning
These messages are produced by layer reconciliation feature.
If layer reconciliation is enabled, when you first open any file in Autocad 2008-2009, the layer baseline list is created in the file. Next time you open the same file, AutoCAD compares the current layer list with the baseline saved in the drawing and if any “new” layers are found it marks them “unreconciled” and groups them into special filter. You can the select the unreconciled ayers and right click to reconcile them(approve). This feature can be very useful in a collaborative environment with extensive usage of XREFs. But like anything new, it takes time to get used to.
Let’s take a look at the settings (variables) which control the behavior of the layer reconciliation
LAYEREVALControls when the Unreconciled New Layer filter list in the Layer Properties Manager is created/evaluated for new layers.
Type: Integer
Saved in: Drawing
Initial value: 1 for 2008 and 0 for 2009
When LAYEREVAL is set to 1 or 2, layer baseline is created. When LAYEREVAL is set to 0, the Unreconciled New Layer filter is hidden and new layers are not evaluated.
LAYERNOTIFY Specifies when an alert displays for new layers that have not yet been reconciled.
Type: Bitcode. To calculate the value of bitcode variable, add all desired options.
Saved in: Drawing
Initial value: 15 for 2008 and 0 for 2009
If the LAYEREVAL ≠ 0 and value of LAYERNOTIFY is odd, you will also get the following dialog box that will require you to click Yes or No every time you plot or add a drawing to Publish. It can disrupt your plotting/ publishing workflow.
LAYEREVALCTL Controls the overall unreconciled new layer filter list in the Layer Properties Manager which is evaluated for new layers.This system variable also affects whether the new layer notification is displayed or not.
Type: Binary
Saved in: registry
Initial value: 1
0 Disables the evaluation and notification of new layers
1 Enables the evaluation of new layers on LAYEREVAL settings in DWG file
Note: LAYEREVALCTL overrides the LAYEREVAL and LAYERNOTIFY setvars when LAYEREVALCTL = 0. It acts like a global off (but not a global on). There is no effect even if LAYEREVALCTL is turned on if LAYERNOTIFY = 0 or LAYEREVAL = 0. LAYEREVALCTL must be set to 1 for LAYERNOTIFY and LAYEREVAL to function correctly.
Note: Initial setting for both LAYEREVAL and LAYERNOTIFY is different for 2008 and 2009, thus users which skipped 2008 did not experience problems with layer reconciliation.
Following table illustrates the differences between 2008 and 2009 and where to set these variables in graphic interface.
### Solutions for 2008:
1. Use templates with desired settings for these variables when creating new files
2. Run script to re-set the variables in existing files.
LAYEREVAL 0
QSAVE
blank line
3. If planning to use the feature, reconcile layers once manually or by calling lisp or VBA program from the script
Solutions for 2009
1.Set LAYEREVALCTL=0. It is saved in the registry, so you will not need to go to every file and set LAYEREVAL 0
2. New option in –LAYER command
Run script to reconcile all layers. This will update the baseline. New layers will be evaluated
-LAYER E *
blank line
blank line
QSAVE
blank line
# Layer cleanup scenarios. Post 12 in series “AutoCAD Cleanup”
Depending on the purpose of your cleanup, there may be variety of tasks that you need to perform with the layers:
### Task 1.The objects in the drawing are not on the correct layers, and you need to sort them out.
This is a manual process which you have to perform on every file individually. Before you start cleaning up, save the layers state and export it to an external text file. This way you do not have to worry about restoring status quo if you make a mistake. In addition, do not forget about Layer Previous.
Most of the tools you need to sort the content into the correct layers are contaoned in LAYERSII toolbox
My favorite tool for individual layer diagnostic and cleanup is LAYWALK. LAYISO and LAYUNISO could be used as well.
If you can identify criteria for sorting elements to layers (for example: all red elements should go to layer RED, or all blocks – go to layer BLOCKS) you can script that process or at least use QSELECT or selection filters.
### Task 2. You need to show/plot XREFs with different symbology (color, lineweight, linetype).
When you approach this task, there are several things to consider:
• Are element properties (color, lineweight, linetype) of XREF files set ByLayer? If desired, you can use script to force ByLayer
• What is the VISRETAIN variable setting in the parent(s) file? If desired, script set VISRETAIN
VISRETAIN is a variable which controls if the parent file “remembers” changes to XREF dependent layer properties between AutoCAD sessions. VISRETAIN is saved in the drawing.
When VISRETAIN =1 in the parent file and you attach (or overlay) an XREF for the first time, AutoCAD creates an internal lookup table in the parent file. This table contains all XREF dependent layers names and their properties as read from the XREF at the time of attachment. When you change XREF dependent layer properties, the internal lookup table holds these changes (only if VISRETAIN =1). If you add layers to the XREF file, a new record with information about the new XREF layers will be added to the table when XREF is reloaded.
• Are the XREFs already attached?
• Do you expect updates for the XREFs? If yes, do you expect the layers to be consistent?
Example: Site.dwg is your architectural site plan (VISRETAIN=1). You attached Surv.dwg (survey data) to it and changed properties of all XREF dependent layers from Surv.dwg to gray thin lines (true color 120,120,120, lineweight 0.18mm). Later you receive an update from the surveyor, and there is a new layer “ParkingLOT” (blue, 0.35mm) in the updated Surv.dwg. When you substitute the old version of Surv.dwg with the new one, all XREF dependent layers except ParkingLOT will remain gray and thin, ParkingLOT will be blue and 0.35mm
If you now set VISRETAIN =0 in the Site.dwg, the lookup table will not be saved. Next time you open Site.dwg, the XREF dependent layers properties will be re-read from the XREF file. If you now set VISRETAIN =1 in the Site.dwg, it will not return to your gray thin lines, you will need to re-assign the properties again.
VISRETAIN controls all XREFs together, it cannot control individual XREFs.
When you have multiple similar files with multiple similar XREFs which have the similar file names and identical layer names, you can use some tricks to speed up repetitive tasks.
Super Trick: Say you are working on a hi-rise building (30 floors). File names for all floors follow the naming convention filename-NN.dwg where NN represents floor number. Layers names are identical for all floors.
The layers states (on/off, symbology and freeze status) should be similar for all floors. It would really save you a lot of time if you can set up the layer state for one floor, and then use it for the another 29. The problem is that XREF dependent layer names for different floors have different names because they contain the floor number.
Here is what you can do:
• Set up the layers the way you want them for one of the floors; in this example, I use 01.
• Name and save the layer state and export it to an external file Floor1.las
• Open Floor1.las in any text editor and replace “-01” with “-02”, then save as Floor2.las
• You now can import and apply Floor2.las to the second floor.
### Task 3. You need to change layer properties without changing layer names.
Most common reason to change layer properties is to make them work with a different CTB.
You can use Standards Checker to automatically fix layer properties if the layer names are identical.
If you have a “good” drawing with the layers properties the way you want them, you can save it as a .dws file and use with Standards Checker (Tools->CAD Standards)
You may want to associate standards files with the drawings after they have been cleaned up to catch future deviations before they accumulate.
### Task 4. You need to migrate the files from one layer naming standard to another.
The best AutoCAD tool for this job is Layer Translator.
Layer translator allows you to create a “translation table” which maps the layers from your file to the layers of another .dwg or .dws file. They can be saved and re-used on other files.
#### Tips on using LAYTRANS:
• Clean up Layer 0 and Defpoint layer first. Then Map same. That will take care of the layers with identical names
Zoom Extents before you start Layer Translator.
• In Settings check Show layer contents when selected. That will visually isolate selected layers display. Use Selection Filter if needed.
• Map the layers that you can visually identify first. Save the Layer Translation Mappings. Translate.
Use LAYWALK and other layer tools as described in task1 to clarify and sort out messed up layers, then go back to translation
#### Limitations of LAYTRANS:
• Works on one file at a time. Cannot batch translate multiple files.
Even when all options are selected, the nested objects in blocks are NOT forced to color and linetype ByLayer. Use SETBYLAYER in AutoCAD2008-2009.
• Missing features: options to Force object Lineweight and Plotstyle to ByLayer.
• Can’t zoom in/out while in Layer Translator
• Layer Descriptions will be lost in translation.
Using “Other Software” for batch processing of layers and other tasks
MicroStation translation tables are Excel files, and can be easily edited for such tasks as renaming layers by appending, pre-pending and partial name substitution. You can also force color, linetype and lineweight to ByLayer for the objects nested in blocks
Best of all, MicroStation can batch translate layers in multiple files. Use latest versions for compatibility with latest AutoCAD file format.
### Potential problems to watch for while cleaning up layers
Special layers. No graphic elements should be on Layer 0 or Defpoints.
• A layer is identified by its name, and when the name is changed, other files this file is referenced to (parent files) are not ”aware” of that change.
If you change XREF dependent layer names, the XREF dependent layers properties specifically assigned in the parent file using VISRETAIN=1 will stop working.
Similarly, when you rename XREF layers, the freeze and on/off (and for 2008 VP symbology settings) information in the plot sheet will be lost. In addition, this layer may need to be reconciled.
If you use layer states and export them outside of your drawing file, after renaming the layers the exported layer states will be unusable.
• Always keep in mind that color, lineweight and, in many cases, linetype and plot style for graphic objects are set to “ByLayer”, so any change to layer properties has to be coordinated with the plot style table you are going to use.
• In previous versions of AutoCAD you could only assign VP Freeze to a layer. Now in addition you can assign VP Color, VP Linetype, VP Lineweight and, for the DWG of STB flavor, VP Plot style. In other words, your layers may look and plot differently in different viewports.
# Layer0 and Defpoints . Post 11 in series “AutoCAD Cleanup”
Layer 0 and Defpoints layers are special. These layers should not to be used to create graphic content. Layer 0 should be used only to create block definitions, Defpoints layer is used by AutoCAD to create invisible definition points for other elements. These layers behave in a special way: they cannot be deleted and fall flat into the corresponding layer of a parent file when the file containing these layers is attached or overlaid to another file.
Example: A-Plan.dwg is a construction plan with walls drawn on layer 0 and M-Plan.dwg is a piping plan with hot water pipes drawn on its layer 0. A-Plan.dwg is overlaid into M-Plan.dwg to be used as a background. You want to see/plot walls as thin lines and pipes as thick lines. However you cannot separate layer 0 of parent M-Plan.dwg .dwg and A-Plan.dwg
The Defpoints layer behaves in a similar way. On earlier versions of AutoCAD many users learned to use the Defpoints layer as storage for elements they did not wish to plot (viewports, alignment lines, sometimes XREFs and such). However starting from AutoCAD 2000, any layer can have the non-plot property, and it is a good practice to create a special layer(s) to contain those elements.
It is best to start with cleaning up Layer 0 and Defpoints.
I wrote a little LISP routine to move all objects from Layer 0 to a new layer “from0”. After it is loaded, you can run it by typing “from0”at the command line.
(defun c:from0 ()
(setq ssl0 (ssget “x” ‘((8 . “0″))))
(COMMAND “-layer” “N” “from0″ “”)
(COMMAND “chprop” ssl0 “” “la” “from0″ “”)
)
You can also run it on multiple files using ScriptPro
However if you have graphic elements on the Defpoins layer, or do not want to dump everything from layer 0 into one layer, you will have to do it manually.
To identify objects on these layers, you can use LAYWALK (Express Tools in earlier releases). I will not duplicate the HELP file of LAYWALK here but suffice it to say that it is the most powerful tool for diagnosing layer problems and sorting them out. It can answer two questions:
• What layers the objects are on?
• What objects reside on these layers?
Undocumented trick: you can use your mouse wheel to ZOOM while in LAYWALK. Just click on Select by element, and wheel forward to zoom in, wheel back to zoom out, double click on a wheel to zoom extents, or press and drag to pan. When done navigating, press Escape to return to LAYWALK without change. Explore right-click menu.
# Scale List cleanup . Post 10 in series “AutoCAD Cleanup”
### From Scale List Cleanup Utility README.txt:
Description of Problem
When a file contains excess scales, performance may be negatively affected. Eventually, the file may become unusable. To use the file, some scales must be removed. This utility repairs affected files by removing excess scales.
NOTE: Nested external references (xrefs) may produce excess scales. To correct this problem, install AutoCAD 2008 Service Pack 1 or corresponding service packs for AutoCAD-based products. The Scale List Cleanup Utility repairs files affected before the necessary service pack was installed.
Affected Products
AutoCAD® 2008-based products, including DWG TrueViewTM 2008
AutoCAD 2009-based products, including DWG TrueView 2009
Not every scale containing “XREF” in its name is a bug. Legitimate” XREF scales:
• When units are different
• When proxy objects are present
• When these scales are already used in page setups, viewports or anywhere else where scale is used.
• When scale list in XREF contains scales or definitions which are not present in the parent file
#### How to clean:
• Alternatively, add the following line (before Qsave line) to any of the cleanup scripts you run on your files:
-SCALELISTEDIT R Y E
# Proxy objects cleanup . Post 9 in series “AutoCAD Cleanup”
A proxy object is a substitute for a custom object created by an AutoCAD add-on application such as AutoCAD Architecture, Civil 3D and others. Proxy objects may look OK but erasing and moving an object, or changing object properties may not be possible on a proxy object, depending on the application that created it.
When you open a drawing, you might see a Proxy Information dialog box (depends on PROXYNOTICE variable setting). The dialog box tells you the total number of proxy objects in the drawing (both graphical and nongraphical) and the name of the missing application and provides additional information about the proxy object type and display state.
An Object Enabler lets you view and somewhat edit proxy objects in AutoCAD when add-on application is not present. You can tell AutoCAD to search and install available enablers automatically, but that setting may slow down your system and may be intercepted by your network security policies. It is much better to install all of them on every machine in the office. http://www.autodesk.com/enablers.
Better yet, Object Enablers allows you to convert proxy objects to plain AutoCAD
Following script can be used to convert file with proxy objects to the plain AutoCAD file (provided you have an object enabler installed).
-EXPORTTOAUTOCAD
B N
blank line
blank line
blank line
QSAVE
blank line
Same script explained line-by line:
LINE# SCRIPT COMMENT 1 -EXPORTTOAUTOCAD prerequisite: object enablers for proxy parent application 2 B N set BIND option to NO (can be used with YES, if you want to bind all XREFs) 3 P ACAD\ SUPER TRICK : Set PREFIX option to “ACAD\”. This will export the file into ACAD subfolder of current folder . This subfolder must exist before you run the script. 4 blank line 5 blank line 6 blank line 7 QSAVE 8 blank line
# Resolving problems with units. Post 8 in series “AutoCAD Cleanup”
Scenario: Some of your files were created in another country. The units are set to Architectural when you received the files. All objects’ LTSCALEs are set to 1. Some layers in both files use the HIDDEN linetype. You referenced these files to the files created and worked on in the US and experience the following problems:
• No matter what LTSCALE you set either lines from the US files or lines from the foreign files do not scale correctly.
• No matter what LTSCALE you set, either lines from the US files or lines from the foreign files do not scale correctly.
• When you reference foreign file in AutoCAD 2006 or newer, in addition to the problems with the LTSCALE, the whole foreign XREF file doesn’t scale correctly and is not located in the right place even though you attached it with the scale=1 and insertion point at 0,0,0.
Here is what happened and how to correct this.
There are several variables in AutoCAD that control units.
MEASUREINIT (0 English; 1 Metric) sets the initial drawing units as English or metric. MEASUREINIT controls which units, hatch pattern and linetype definition files a new drawing uses if not started from a particular template. MEASUREINIT is saved in the registry and the original value is determined by the geographic location of the installation.
MEASUREMENT is stored in the drawing and always overrides the MEASUREINIT setting.
When the foreign file was created, it took the local (metric) value of Measureinit and saved it in the file as the value for MEASUREMENT. At that time the operator may have changed the units to Architectural, but did not change other unit variables. In the course of his work, the metric linetype definition for HIDDEN line from the ACADISO.lin was loaded and used in the drawing.
The US drawing was started with MEASUREMENT set for English units and HIDDEN line definition from ACAD.lin (English).
The linetype names are identical in ACAD.lin and ACADISO.lin, but definitions are different. After the definitions are loaded into the file, it is not clear what .lin file they came from.
To correct the linetype issue you need to re-load linetype definitions from the ACAD.lin file into the foreign file. While you are at it, set the MEASUREMENT=0.
The second problem is illustrated by the units dialog boxes
#### Units to scale inserted content.
These values are controlled by the INSUNITS variable, which is saved in the drawing. The foreign drawing had it set to something other than inches, but the file did not have any symptoms until AutoCAD2006. Before 2006 this variable only affected drag-and-drop operation.
Possible values of INSUNITS are: 0-Unspecified (unitless); 1-Inches; 2-Feet; 3-Miles; 4-Millimeters; 5-Centimeters; 6-Meters; 7-Kilometers; 8-Microinches; 9-Mils; 10-Yards; 11-Angstroms; 12-Nanometers; 13-Microns; 14-Decimeters; 15-decameters; 16-Hectometers; 17-Gigameters; 18-Astronomical Units; 19-Light Years; 20-Parsecs
The only remaining question is what will happen when INSUNITS in either file is set to 0? (unitless).This situation is controlled by INSUNITSDEFSOURCE and INSUNITSDEFTARGET. They can be set in the Options-> User Preferences. The values are similar to INSUNITS.
The following script will fix all unit related the problems
-UNITS 4 64 1 0 0 n
INSUNITS 1
MEASUREMENT 0
EXPERT 3
blank line
EXPERT 0
-PURGE LT *ISO* N
QSAVE
blank line
Lines 1-3 simply set unit variables. Line 4 forces the linetype command to work in text mode without issuing prompts for user intervention (it needs EXPERT 3 mode for that, which I reset back to default EXPERT 0 after I am done with the LINETYPE command). After loading all linetypes from ACAD.lin, I purge only the linetypes with a name containing “ISO”. (I could have purged all unused linetypes or even all unused definitions.)
# Using scripts to run LISP or VBA routines and setting variables. Post 7 in series “AutoCAD Cleanup”
### Run LISP routines or VBA macros
You can batch process multiple files using scripts which call LISP or VBA macros.
Example: the following script runs a lisp routine to delete layer filters. It loads the Layerfilterdelete.lsp file, then deletes layer filters using a command LFD defined in the routine.
(LOAD “C:\\AU-CLEANUP\\EXERCISES\\SCRIPTING\\LAYERFILTERDELETE.LSP”)
LFD
QSAVE
blank line
(DEFUN C:LAYERFILTERSDELETE () (VL-LOAD-COM) (VL-CATCH-ALL-APPLY ‘(LAMBDA () (VLA-REMOVE (VLA-GETEXTENSIONDICTIONARY (VLA-GET-LAYERS (VLA-GET-ACTIVEDOCUMENT (VLAX-GET-ACAD-OBJECT)))) “ACAD_LAYERFILTERS”))) (PRINC “\NALL LAYER FILTERS HAVE BEEN DELETED.”) (PRINC)) (DEFUN C:LFD () (C:LAYERFILTERSDELETE))
If a routine expects a user response, you need to script the response also.
### Set up Variables
You can use scripts to set up variables. Here is an example where you set some variables you need for your environment. Keep in mind that not all variables are saved in files, some are saved in the registry, and some cannot be saved at all.
ATTMODE 1
ATTDIA 1
DIMASSOC 2
FILLMODE 1
MIRRTEXT 0
PDMODE 0
VISRETAIN 1
XEDIT 0
QSAVE
blank line
# Dealing with CTB and STB types of DWG. Post 6 in series “AutoCAD Cleanup”
DWG files come in two flavors. One uses traditional CTB (color based plot tables), the other uses STB (style based plot tables). Plot style is a general property that can be assigned to an object or a layer in the STB “flavored” DWG file. An object can also inherit this property from a layer much like color, linetype or lineweight when set to ByLayer.
In DWG file of CTB flavor, the plot style property is set to ByColor and cannot be changed.
If PSTYLEPOLICY (a variable saved in the registry) is set to 1, new files (unless created from templates) will be CTB flavor of DWG and will use CTB for plotting; when it is set to 0, new files will be STB flavor of DWG and will use STB for plotting.
When you need to convert DWG files from STB flavor to CTB flavor, you need to strip the plot style property from the objects and layer definitions. This can be done easily using the following script. DWG files of CTB flavor will not be affected by this script.
CONVERTPSTYLES
QSAVE
Blank Line
When you need to convert from CTB to STB, you need to add that property to the objects and layer definitions.
Before converting a CTB flavor of DWG file to STB flavor of DWG file, you first need to have a STB table which you will be able to use with your former CTB DWG file.
To accomplish this, type CONVERTCTB at the command line and select the CTB to be converted to STB. IN this example, I selected ACAD.CTB to be converted to ACAD2.STB
You can then use the following script to batch convert CTB DWG files to use ACAD2.STB. DWG files of STB flavor will not be affected by this script.
CONVERTPSTYLES
Blank Line
|
2019-04-26 12:50:31
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.2694322466850281, "perplexity": 4025.8231000272754}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-18/segments/1555578770163.94/warc/CC-MAIN-20190426113513-20190426135513-00330.warc.gz"}
|
https://brilliant.org/discussions/thread/seeking-help/
|
×
# Seeking Help
In triangle find points X,Y,Z on AB,BC,CA such that AXYZ is a rhombus.Show that the area of rhombus AXYZ≤(1/2) triangle ABC. Please show me the proof
Note by Kalpok Guha
3 years ago
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Sort by:
Here's an algebraic proof:
Let $$K =$$ area of $$ABC, [\:] =$$ the area of a polygon, $$b = AC,$$ and $$c = AB.$$ Note that, since $$AXYZ$$ is a rhombus, $$AY$$ is an angle bisector of $$\angle A.$$ From the angle bisector theorem, we get $$\dfrac{BY}{YC} = \dfrac{c}{b}.$$ From this, we get $$\dfrac{BY}{BC} = \dfrac{c}{b+c}$$ and $$\dfrac{YC}{BC} = \dfrac{b}{b+c}.$$
$\frac{[ZYC]}{K} = \left(\frac{b}{b+c}\right)^2 \Rightarrow [ZYC] = K \left(\frac{b}{b+c}\right)^2$ $\frac{[XYB]}{K} = \left(\frac{c}{b+c}\right)^2 \Rightarrow [XYB] = K \left(\frac{c}{b+c}\right)^2.$
Now, $$[AXYZ] = K - [ZYC] - [XYB] = K - K \left(\dfrac{b}{b+c}\right)^2 - K \left(\dfrac{c}{b+c}\right)^2 = K \left(\dfrac{2bc}{(b+c)^2}\right).$$ Finally, we are left to prove that $$\dfrac{2bc}{(b+c)^2} \leq \dfrac{1}{2}.$$ Rearranging this gives us $$(b - c)^2 \geq 0,$$ which is clearly always true. Thus, $$[AXYZ] \leq \dfrac{1}{2}K,$$ and our proof is complete.
- 3 years ago
Thanks .But I think AY is bisector because AXYZ is rhombus,a diagonal is not a bisector in parallelogram.
- 3 years ago
Oops, sorry about that! I've fixed my solution.
- 3 years ago
Let ABCD be a rhombus such that A: (-a,0), B; (0,b), C: (a,0) & D: (0,-b) with centre of the rhombus as (0,0). Now draw a line: y=mx+c passing through D and intersecting extended BA & BC in P & Q respectively. We now have a triangle PQB with a rhombus whose vertices are on the triangle sides. Since y=mx+c passes through D, c = -b. We can now determine P as intersection of y=mx -b & x/a+y/b=1 and thus, P:[2ab/(am-b), b(am+b)/(am-b)]. Similarly, Q:[2ab/(am+b), b(am-b)/(am+b)]. Further, we determine the area of triangle PQB taking the vertices in anti-clockwise direction. Tr. PQB = (1/2){(2ab/(am-b))[b(am-b)/(am+b) -b)+(2ab/(am+b))[b -b(am-b)/(am-b)]} This simplifies to, Tr. PQB = 4ab³/(b²-a²m² ) whereas atra pf Rhombus ABCD = 2ab. Note that if m=0 then (1/2) Tr.PQB = 2ab = Area of Rhombus ABCD.
Otherwise (1/2)*4ab³/(b²-a²m² ) > 2ab if b²>b² - a²m² or if a²m²>0 which is always true.
Hence the proposition stands proven.
- 3 years ago
Really thanks for the proof.But is there any prove without using coordinate geometry?
- 3 years ago
|
2017-12-12 04:34:53
|
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|
https://gamedev.stackexchange.com/questions/102355/cannot-get-the-texture-showed-up-correctly-gldrawelements/102434
|
# Cannot Get The Texture Showed Up Correctly glDrawElements
I still have this problem almost 1 month. Tried to search on Google but did not find any solution to this. I have loaded all the data correctly but don't know why the texture came up like this.
struct SVertex
{
public:
float X, Y, Z;
};
struct STexture
{
public:
float U, V;
};
struct SFace
{
public:
int X, Y, Z;
};
class CBMap
{
public:
struct SMesh
{
SVertex *Vertex;
SVertex *Normal;
STexture *TexCoord;
SFace *Face;
SFace *FaceNormal;
char Texture[32];
int TextureID;
int TotalVertice;
int TotalNormal;
int TotalFace;
};
SMesh* Mesh;
CBMap();
~CBMap();
int TotalMesh;
int Get_TotalMesh(FILE* pFile);
int GetData(FILE* pFile);
};
CPP
int CBMap::Load(char* szFile)
{
FILE* pFile = fopen(szFile, "r");
if (!pFile) return 0;
if (!iBMap)
{
printf("BMAP file is not valid\n");
return 0;
}
TotalMesh = Get_TotalMesh(pFile);
Mesh = new SMesh[TotalMesh];
GetData(pFile);
int iMesh = 0;
char* pJunk;
while (!feof(pFile))
{
{
fscanf(pFile, "%s", Mesh[iMesh].Texture);
}
{
fscanf(pFile, "%d", &pJunk);
Mesh[iMesh].Vertex = (SVertex*)malloc(sizeof (SVertex) * Mesh[iMesh].TotalVertice);
for (int i = 0; i < Mesh[iMesh].TotalVertice; ++i)
{
fscanf(pFile, "%f %f %f", &Mesh[iMesh].Vertex[i].X, &Mesh[iMesh].Vertex[i].Y, &Mesh[iMesh].Vertex[i].Z);
}
}
{
fscanf(pFile, "%d", &pJunk);
Mesh[iMesh].TexCoord = (STexture*)malloc(sizeof (STexture) * Mesh[iMesh].TotalVertice);
for (int i = 0; i < Mesh[iMesh].TotalVertice; ++i)
{
fscanf(pFile, "%f %f", &Mesh[iMesh].TexCoord[i].U, &Mesh[iMesh].TexCoord[i].V);
}
}
{
fscanf(pFile, "%d", &pJunk);
Mesh[iMesh].Normal = (SVertex*)malloc(sizeof (SVertex) * Mesh[iMesh].TotalNormal);
for (int i = 0; i < Mesh[iMesh].TotalNormal; ++i)
{
fscanf(pFile, "%f %f %f", &Mesh[iMesh].Normal[i].X, &Mesh[iMesh].Normal[i].Y, &Mesh[iMesh].Normal[i].Z);
}
}
{
fscanf(pFile, "%d", &pJunk);
Mesh[iMesh].Face = (SFace*)malloc(sizeof (SFace) * Mesh[iMesh].TotalFace * 3);
Mesh[iMesh].FaceNormal = (SFace*)malloc(sizeof (SFace) * Mesh[iMesh].TotalFace * 3);
for (int i = 0; i < Mesh[iMesh].TotalFace; ++i)
{
fscanf(pFile, "%d %d %d %d %d %d", &Mesh[iMesh].Face[i].X, &Mesh[iMesh].Face[i].Y, &Mesh[iMesh].Face[i].Z,
&Mesh[iMesh].FaceNormal[i].X, &Mesh[iMesh].FaceNormal[i].Y, &Mesh[iMesh].FaceNormal[i].Z);
}
}
{
if (iMesh < TotalMesh) iMesh ++;
}
}
fclose(pFile);
return 1;
}
First picture is reality and second picture expectation
• We can barely see the picture. It's too dark. Is this the problem? If not, what do you mean by "the texture came up like this"? Please provide more context and also describe or better show a desired result. – Ivan Aksamentov - Drop Jun 13 '15 at 14:48
• I needed to set the v components to v = 1 - loadedV. this is because opengl uses a different origin for the texture coordinate system than directx – Tobias B Jun 14 '15 at 7:34
• Thanks Tobias but I don't really understand what you explain. Any example? – DavidJr Jun 14 '15 at 8:43
• Tobias, thank you very much, that makes my code works. How to flag your comment as an answer? – DavidJr Jun 15 '15 at 6:14
• I postet it as answer. When I wrote the comment I wasn't sure about it and didn't have the time to research. – Tobias B Jun 15 '15 at 8:12
|
2020-01-17 15:34:13
|
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|
https://codereview.stackexchange.com/questions/94062/populate-a-networkx-graph-with-tuples-of-varying-lengths
|
# Populate a networkx graph with tuples of varying lengths
I've got a list with tuples of variable length containing nodes, out of which I want to build a network:
node_list = [("one", "two"), ("eins", "zwei", "drei"),
("un", "deux", "trois", "quattre"), ("two", "zwei", "deux")]
Each tuple represents a subgroup in the network. I want to add each node with edges to any other member of the same tuple.
It is straightforward to use .add_edges_from(), which only expects a tuple with two nodes. For a tuple with three entries one would have to write .add_edges_from(("eins", "zwei"), ("eins", "drei"), ("zwei", "drei")). Tuples with four members require even more code.
Now I am looking for the most efficient way to populate a network given a list like node_list. My idea is:
import networkx as nx
G = nx.Graph()
for node_tuple in node_list:
for node1 in node_tuple:
for node2 in node_tuple:
It's probably very inefficient because it sees every node twice and also adds self-loops.
• Wouldn't this also create edges from each node to itself? Is that intended? Jun 19 '15 at 17:57
• It is. Simply populating with nodes is easy, but the edges make the network. Jun 19 '15 at 18:41
Tuples of varying lengths are a bit unusual. Perhaps those would be better as lists instead.
itertools is your friend. You can write, equivalently:
import itertools
G = nx.Graph()
for node_tuple in node_list:
If you want to list each pair of nodes just once instead of twice, and including self-edges use itertools.combinations_with_replacement(node_tuple, 2) instead.
• combinations will omit the self-edges: you need combinations_with_replacement Jun 19 '15 at 21:32
• combinations_with_replacement() did the trick. It's also good to know that there is combinations, in case one wants self-edges. Jun 20 '15 at 10:57
|
2021-10-28 12:20:22
|
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|
https://www.khronos.org/registry/vulkan/specs/1.3-extensions/man/html/VkSpecializationMapEntry.html
|
## C Specification
The VkSpecializationMapEntry structure is defined as:
// Provided by VK_VERSION_1_0
typedef struct VkSpecializationMapEntry {
uint32_t constantID;
uint32_t offset;
size_t size;
} VkSpecializationMapEntry;
## Members
• constantID is the ID of the specialization constant in SPIR-V.
• offset is the byte offset of the specialization constant value within the supplied data buffer.
• size is the byte size of the specialization constant value within the supplied data buffer.
## Description
If a constantID value is not a specialization constant ID used in the shader, that map entry does not affect the behavior of the pipeline.
Valid Usage
• VUID-VkSpecializationMapEntry-constantID-00776
For a constantID specialization constant declared in a shader, size must match the byte size of the constantID. If the specialization constant is of type boolean, size must be the byte size of VkBool32
|
2022-05-19 23:18:11
|
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|
https://madhavamathcompetition.com/category/rmo/
|
# Any integer can be written as the sum of the cubes of 5 integers, not necessarily distinct
Question: Prove that any integer can be written as the sum of the cubes of five integers, not necessarily.
Solution:
We use the identity $6k = (k+1)^{3} + (k-1)^{3}- k^{3} - k^{3}$ for $k=\frac{n^{3}-n}{6}=\frac{n(n-1)(n+1)}{6}$, which is an integer for all n. We obtain
$n^{3}-n = (\frac{n^{3}-n}{6}+1)^{3} + (\frac{n^{3}-n}{6}-1)^{3} - (\frac{n^{3}-n}{6})^{3} - (\frac{n^{3}-n}{6})$.
Hence, n is equal to the sum
$(-n)^{3} + (\frac{n^{3}-n}{6})^{3} + (\frac{n^{3}-n}{6})^{3} + (\frac{n-n^{3}}{6}-1)^{3}+ (\frac{n-n^{3}}{6}+1)^{3}$.
More later,
Nalin Pithwa.
# A random collection of number theory problems for RMO and CMI training
1) Find all prime numbers that divide 50!
2) If p and $p^{2}+8$ are both prime numbers, prove that $p^{3}+4$ is also prime.
3) (a) If p is a prime, and $p \not|b$, prove that in the AP a, $a+b$, $a+2b$, $a+3b$, $\ldots$, every pth term is divisible by p.
3) (b) From part a, conclude that if b is an odd integer, then every other term in the indicated progression is even.
4) Let $p_{n}$ denote the nth prime. For $n>5$, show that $p_{n}.
Hint: Use induction and Bertrand's conjecture.
5) Prove that for every $n \geq 2$, there exists a prime p with $p \leq n < 2p$.
More later,
Regards,
Nalin Pithwa
# Find the last two digits of 9^{9^{9}}
Here is a cute example of the power of theory of congruences. Monster numbers can be tamed !!
Question :
Find the last two digits of $9^{9^{9}}$.
Solution:
A famous mathematician, George Polya said that a good problem solving technique is to solve an analagous less difficult problem.
So, for example, if the problem posed was “find the last two digits of 2479”. How do we go about it? Find the remainder upon division by 100. Now, how does it relate to congruences ? Modulo 100 numbers !
So, the problem reduces to — find out $9^{9^{9}} \equiv 9 \pmod {10}$.
Now, what is the stumbling block…the exponent $9^{9}$ makes the whole problem very ugly. But,
$9^{9} \equiv 9 \pmod {10}$, which means $9^{9}-9=10k$, that is, $9^{9} = 9 + 10k$,
also, use the fact $9^{9} \equiv 89 \pmod {100}$
Hence, $9^{9^{9}}=9^{9+10k} = 9^{9}.9^{10k}$
$9^{9^{9}} \pmod {100} = 9^{9}.9^{10k} \pmod {100} \equiv 89. 9^{10k} \pmod {100}$
So, now we need to compute $9^{10k} \pmod {100} = (9^{10})^{k} \pmod {100} = (89.9)^{k} \pmod {100} = 89^{k}. 9^{k} \pmod {100}$
Hence, $9^{9^{9}} \pmod {100} \equiv 89^{k+1}.9^{k} \pmod {100} \equiv (89.9)^{k}.89 \pmod {100} \equiv (1)^{k}.89 \pmod {100} \equiv 89 \pmod {100}$.
-Nalin Pithwa.
# Elementary problems in Ramsey number theory for RMO
Question 1:
Show that in any group of 6 people there will always be a subgroup of 3 people who are pairwise acquainted or a subgroup of 3 people who are pairwise strangers.
Solution 1:
Let $\{ A, B, C, D, E, F\}$ be a group of 6 people. Suppose that the people known to A are seated in room Y and the people NOT known to A are seated in room Z; A is not in either room. Then, there are necessarily at least 3 people in either room Y or in room Z; (a) Suppose B, C, D to be in room Y. Either these 3 people are mutual strangers (and so the given theorem is true), or, at least two of them (say, B and C) know each other. In the latter case, A, B and C form a group of 3 mutual acquaintances — and again, the theorem is true. (b) In (a), replace room Y by Z and interchange the notion of ‘”acquaintances” and “strangers”‘.
Question 2:
Show that in any group of 10 people there is always (a) a subgroup of 3 mutual strangers or a subgroup of 4 mutual acquaintances, and (b) a subgroup of 3 mutual acquaintances or a subgroup of 4 mutual strangers.
Solution 2:
(a) Let A be one of the ten people; the remaining 9 people can be assigned to two rooms: those who are known to A are in room Y and those who are not known to A are in room Z. Either room Y has at least 6 people or room Z has at least 6 people. For, (i) suppose room Y has at least 6 people. Then, by previous problem number 1, there is either a subgroup of 3 mutual acquaintances or a subgroup of 3 mutual strangers (thus, the theorem is true) in this room. In the former case, A and these 3 people constitute 4 mutual acquaintances (ii) Suppose room Z has at least 4 people. Either these 4 people know one another or at least 2 of them, say B and C, do not know each other. In the former case, we have a subgroup of 4 mutual acquaintances. In the latter case A, B and C constitute 3 mutual strangers.
(b) In the previous scenario, let people who are strangers become acquaintances, and let people who are acquaintances pretend they are strangers. The situation is symmetric.
Question 3:
Show that in any subgroup of 20 people there will always be either a subgroup of 4 mutual acquaintances or a subgroup of 4 mutual strangers.
Solution 3:
Suppose A is one of these 20 people. People known to A are in room Y and people not known to A are room Z. Either room Y has at least 10 people or room Z has at least 10 people. (i) If Y has at least 10 people, then by part B of problem number 2 here, there is either a subgroup of 3 mutual acquaintances or a subgroup of 4 mutual strangers — as asserted — in this room. In the former case, A and these mutual acquaintances will form a subgroup of 4 mutual acquaintances. (ii) Switch ‘”acquaintances” and “strangers”‘ in (i).
Question 4:
Let p and q be 2 positive integers. A positive integer r is said to have the (p,q) – Ramsey property, if in any group of r people either there is a subgroup of p people known to one another or there is a subgroup of q people not known to one another. {By Ramsey’s theorem, all sufficiently large integers r have the (p,q)-Ramsey property.} The smallest r with the (p,q)-Ramsey property is called the Ramsey number R(p,q). Show that (a) R(p,q) = R (q,p). (b) $R(p,1)=1$, and (c) R(p,2)=p.
Solution 4:
(a) By parts (b) of the previous three questions, we have proved part a of the proof here.
(b) This is obvious.
(c) In any group of p people, if all of them are not known to one another, there will be at least 2 people who do not know each other.
Question 5:
Prove that $R(3,3)=6$.
Solution 5:
Question 1 and its proof in this blog article imply that $R(3,3) \leq 6$.
To prove that $R(3,3)>5$, it is sufficient to consider a seating arrangement of 5 people about a round table in which each person knows only the 2 people on either side. In such a situation, there is no set of 3 mutual acquaintances and no set of 3 people not known to one another.
Question 6:
Show that if m and n are integers both greater than 2, then
$R(m,n) \leq R(m-1,n) + R(m,n-1)$.
(this recursive inequality gives a non-sharp upper bound for R(m,n)).
Solution 6:
Let $p \equiv R(m-1,n)$, $q=R(m,n-1)$ and $r \equiv p + q$. Consider a group $\{ 1,2, 3, \ldots, r\}$ of r people. Let L be the set of people known to person 1 and M be the set of people NOT known to person 1. The two sets together have $r-1$ people, so either L has at least p people or M has at least q people. (a) If L has $p \equiv R(m-1,n)$ people, then, by definition, it contains a subset of $(m-1)$ people known to one another or it contains a subset of n people unknown to one another. In the former case, the $(m-1)$ people and person 1 constitute m people known to one another.
Thus, in their case, a group of $R(m-1,n) + R(m,n-1)$ people necessarily includes m mutual acquaintances or n mutual strangers. That is, $R(m,n) \leq R(m-1,n) + R(m,n-1)$.
(b) By the usual symmetry argument, the same conclusion follows when M contains q people.
Question 7:
(Remark: A pretty property of Ramsey numbers related to combinatorics).
Show that if m and n are integers greater than 1, then $R(m,n) \leq { {m+n-2} \choose {m-1}}$ — a non-recursive upper bound.
Solution 7:
When $m=2$, or $n=2$, (i) holds with equality (see problem 4 in this blog article). The proof is by induction on $k=m+n$. As we have just seen, the result is true when $k=4$. Assume the result true for $k-1$. Then,
$R(m-1,n) \leq {{m+n-3} \choose {m-2}}$ and $R(m,n-1) \leq {{m+n-3} \choose {m-1}}$
Now, Pascal’s identity gives:
${{m+n-3} \choose {m-2}} + {{m+n-3} \choose {m-1}} = {{m+n-2}} \choose {m-1}$ so that $R(m-1,n) + R(m,n-1) \leq {{m+n-2}} \choose {m-1}$
But, from the previous question and its solution, we get $R(m,n) \leq R(m-1,n) + R(m, n-1)$
PS: As Richard Feynman, used to say, you will have to “piddle” with smallish problems as particular cases of these questions in order to get a grip over theory or formal language of this introduction.
PS: Additionally, you can refer to any basic Combinatorics text like Brualdi, or Alan Tucker or even Schaum Series outline ( V K Balakrishnan).
# How to find the number of proper divisors of an integer and other cute related questions
Question 1:
Find the number of proper divisors of 441000. (A proper divisor of a positive integer n is any divisor other than 1 and n):
Solution 1:
Any integer can be uniquely expressed as the product of powers of prime numbers (Fundamental theorem of arithmetic); thus, $441000 = (2^{3})(3^{2})(5^{3})(7^{2})$. Any divisor, proper or improper, of the given number must be of the form $(2^{a})(3^{b})(5^{c})(7^{d})$ where $0 \leq a \leq 3$, $0 \leq b \leq 2$, $0 \leq c \leq 3$, and $0 \leq d \leq 2$. In this paradigm, the exponent a can be chosen in 4 ways, b in 3 ways, c in 4 ways, d in 3 ways. So, by the product rule, the total number of proper divisors will be $(4)(3)(4)(3)-2=142$.
Question 2:
Count the proper divisors of an integer N whose prime factorization is: $N=p_{1}^{\alpha_{1}} p_{2}^{\alpha_{2}} p_{3}^{\alpha_{3}}\ldots p_{k}^{\alpha_{k}}$
Solution 2:
By using the same reasoning as in previous question, the number of proper divisors of N is $(\alpha_{1}+1)(\alpha_{2}+1)(\alpha_{3}+1)\ldots (\alpha_{k}+1)-2$, where we deduct 2 because choosing all the factors means selecting the given number itself, and choosing none of the factors means selecting the trivial divisor 1.
Question 3:
Find the number of ways of factoring 441000 into 2 factors, m and n, such that $m>1, n>1$, and the GCD of m and n is 1.
Solution 3:
Consider the set $A = {2^{3}, 3^{2}, 5^{3}, 7^{2}}$ associated with the prime factorization of 441000. It is clear that each element of A must appear in the prime factorization of m or in the prime factorization of n, but not in both. Moreover, the 2 prime factorizations must be composed exclusively of elements of A. It follows that the number of relatively prime pairs m, n is equal to the number of ways of partitioning A into 2 unordered nonempty, subsets (unordered as mn and nm mean the same factorization; recall the fundamental theorem of arithmetic).
The possible unordered partitions are the following:
$A = \{ 2^{3}\} + \{ 3^{2}, 5^{3}, 7^{2}\} = \{3^{2}\}+\{ 2^{3}, 5^{3}, 7^{2}\} = \{ 5^{3}\} + \{ 2^{3}, 3^{2}, 7^{2}\} = \{ 7^{2}\}+\{ 2^{3}, 3^{2}, 5^{3}\}$,
and $A = \{ 2^{3}, 3^{2}\} + \{ 5^{3}, 7^{2}\}=\{ 2^{3}, 5^{3}\} + \{3^{2}, 7^{2} \} = \{ 2^{3}, 7^{2}\} + \{ 3^{2}, 5^{3}\}$
Hence, the required answer is $4+3=2^{4-1}+1=7$.
Question 4:
Generalize the above problem by showing that any integer has $2^{k-1}-1$ factorizations into relatively prime pairs m, n ($m>1, n>1$).
Solution 4:
Proof by mathematical induction on k:
For $k=1$, the result holds trivially.
For $k=2$, we must prove that a set of k distinct elements, $Z = \{ a_{1}, a_{2}, a_{3}, \ldots, a_{k-1}, a_{k}\}$ has $2^{k-1}-1$ sets. Now, one partition of Z is
$Z = \{ a_{k}\} \bigcup \{ a_{1}, a_{2}, a_{3}, \ldots, a_{k-1}\} \equiv \{ a_{k}\} \bigcup W$
All the remaining partitions may be obtained by first partitioning W into two parts — which, by the induction hypothesis, can be done in $2^{k-2}-1$ ways — and then, including $a_{k}$ in one part or other — which can be done in 2 ways. By the product rule, the number of partitions of Z is therefore
$1 + (2^{k-2})(2)=2^{k-1}-1$. QED.
Remarks: Question 1 can be done by simply enumerating or breaking it into cases. But, the last generalized problem is a bit difficult without the refined concepts of set theory, as illustrated; and of course, the judicious use of mathematical induction is required in the generalized case.
Cheers,
Nalin Pithwa.
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2018-12-10 05:01:02
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https://www.nature.com/articles/s41598-020-77684-w?error=cookies_not_supported
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## Introduction
Sorghum (Sorghum bicolor) has a diploid genome of ~ 730 Mb with 10 chromosomes1,2. It is a widely-grown cereal crop used as feed or silage for animal consumption, for bio-fuel production and as gluten-free grain for human consumption and is better adapted to grow under high heat and drought conditions than other agriculturally important crops like corn and wheat. These agronomically-important traits make the species an attractive crop for the mass production of grains and bio-fuel under challenging growing conditions.
The warm and humid conditions under which much sorghum is grown support the growth of a wide variety of foliar fungi. Among many diseases, target leaf spot (TLS) caused by the necrotrophic fungus Bipolaris cookei, is one of the most economically-important fungal diseases of sorghum in the southeastern US, causing major yield losses3. TLS causes distinctive oval or elliptical reddish-purple spots that eventually coalesce during disease progression.
The genetic basis of TLS resistance in sorghum has been the subject of several studies. A major recessive resistance gene ds1 on chromosome 5 was identified as a loss of function allele of a gene encoding a leucine-rich repeat receptor kinase4. Other work identified QTL for three different fungal diseases; target leaf spot, zonate leaf spot and drechslera leaf blight, co-localized on chromosome 65. A third study identified a TLS resistance QTL on chromosome 3, as well as the previously-reported chromosome 6 QTL6. A recent study identified novel QTL on chromosomes 3, 4 and 9 as well as a strong QTL on chromosome 5 near the ds1 locus7.
Plants possess cell-surface receptors known as pattern recognition receptors (PRRs) that mediate recognition of highly conserved structural molecules associated with microbes known as microbe-associated molecular patterns (MAMPs). The two best-studied MAMPs are bacterial flagellin, especially its flg22 epitope, and chitin, a component of the fungal cell wall8,9. MAMP recognition elicits a basal response at the infection site known as MAMP-triggered immunity (MTI) which often includes phenomena such as callose deposition, changes in membrane ion flux, changes in phytohormone concentrations, induction or repression of plant defense-related genes, and production of reactive oxygen species (ROS) and nitric oxide (NO)10. In some cases, a pathogen adapted to a particular host can overcome MTI by producing so-called effector proteins, which are usually introduced into the cytoplasm and may suppress MTI. Effectors are sometimes recognized by cytoplasmic receptors known as R proteins, eliciting a strong response known as effector-triggered immunity (ETI) which is quantitatively stronger though, qualitatively somewhat similar to MTI11,12.
Non-host resistance can be defined as: “Resistance shown by an entire plant species against all known genetic variants (or isolates) of a specific parasite or pathogen”13,14. It has been hypothesized that MTI is a significant cause of non-host resistance, as most non-adapted pathogens cannot subvert the MTI-based defenses of their non-host plants15.
Host resistance can be subdivided into qualitative and quantitative resistance. Qualitative resistance is typically based on the action of a single, large-effect gene, while quantitative resistance is mediated by large numbers of small-effect genes16,17. There is some evidence that variation in the strength of MTI may underlie some part of quantitative resistance. Genetic variation in the strength of the MTI response has been documented in a number of plant species including brassicas18,19,20,21, maize22, soybean23, and tomato24 and in several of these cases, QTL controlling the variation were identified. In particular, the fact that several genes resembling PRRs confer quantitative resistance in various plant species16 and that the strength of flg22 perception is negatively correlated with susceptibility to Pseudomonas syringae in Arabidopsis20 suggest that there may be some connection between variation in the MTI response and quantitative disease resistance. However, the relationship between these traits is not well understood, especially in crop plants. The objectives of this study were to characterize the genetic control of the MAMP response and TLS resistance in a diverse panel of sorghum germplasm and to determine if there was evidence of shared genetic control of these traits. Specifically, we wanted to determine whether a stronger MAMP response was indicative of stronger quantitative resistance.
## Materials and methods
### Plant and pathogen materials
A sorghum association mapping population known as the sorghum conversion population (SCP) was provided by Dr. Pat Brown at the University of Illinois (now at UC Davis). It has been described previously25 and is a collection of diverse lines converted to photoperiod-insensitivity and smaller stature to facilitate the growth and development of the plants in US environments26. 510 lines from this population were used in this study although due to bad germination and other quality control issues, not all the lines were used in the analysis of all three traits. Ultimately data from 345 lines were used for the analysis of the chitin response, 472 lines for the flg22 response, and 456 for TLS resistance. B. cookei strain LSLP18 was obtained from Dr. Burt Bluhm at the University of Arkansas.
### MAMP response measurement
Two different MAMPs were used in this study flg22, (Genscript catalog# RP19986), and chitin (Sigma catalog # C9752). Sorghum plants were grown in inserts laid on flats filled with soil (33% Sunshine Redi-Earth Pro Growing Mix) in the greenhouse. Plants were watered the day before sample collection to avoid extra leaf moisture on the day of collection.
The lines were randomized and, for logistical reasons, were planted in batches of 60 lines. For each line, three ‘pots’ were planted with two seeds per line. Subsequent batches were planted as soon as the previous batch had been processed until the entire population had been assessed. Two experimental runs were conducted for both MAMPs with genotypes re-randomized in each of the two runs.
ROS assays were carried out as previously described27. Briefly, for each line, six seeds were planted in 3 different pots. From the resulting seedlings, three were selected based on uniformity. Seedlings that looked unusual or were significantly taller or shorter than the majority were not used. Four leaf discs of 3 mm diameter were excised from the broadest part of the 4th leaf of three different 15-day old sorghum plants. One disc per leaf from two plants and two discs from one plant, with the second disc becoming the water control (see below). The discs were individually floated on 50 µl H20 in a black 96-well plate, sealed with an aluminum seal to avoid exposure to light, and kept at room temperature overnight. The next morning a reaction solution was made using 2 mg/ml chemiluminescent probe L-012 (Wako, catalog # 120-04891), 2 mg/ml horseradish peroxidase (Type VI-A, Sigma-Aldrich, catalog # P6782), and 100 mg/ml Chitin or 2 μM of Flg22. 50 µl of this reaction solution was added to three of the four wells. The fourth well was a mock control, to which the reaction solution excluding the MAMP was added. Four blank wells containing only water were also included in each plate.
After adding the reaction solution, the luminescence was measured using SynergyTM 2 multi-detection microplate reader (BioTek) every 2 min for 1 hr. The plate reader takes luminescence measurements every 2 min during this 1 h. The sum of all 31 readings was calculated to give the value for each well. The estimated value for the MAMP response for each genotype was calculated as (average luminescence value of the three experimental wells—the mock well value) -minus the average blank well value. The blank well values were consistently close to zero.
Leaf discs of Nicotiana benthamiana, one high responsive sorghum line (SC0003), and one low responsive sorghum line (PI 6069) were also included as controls in each 96-well plate for quality control purposes.
### B. cookei inoculum preparation and inoculation
B. cookei inoculum was prepared as described previously28. Briefly, sorghum grains were soaked in water for three days, rinsed, scooped into 1L conical flasks and autoclaved for an hour at 15psi and 121 °C. The grains were then inoculated with about 5 ml of macerated mycelia from a fresh culture of B. cookei LSLP18 isolates and left for 2 weeks at room temperature, shaking the flasks every 3 days. After 2 weeks, the fungus infested sorghum grains were air-dried and then stored at 4 °C until field inoculation. The same inoculum was used for the entire trial and made fresh every year. For inoculation, 6–10 infested grains were placed into the whorl of 4–5 week old sorghum plants. The spores produced from these fungi initiated infection in the young sorghum plants within a week.
### Seed preparation
Before planting in the field sorghum seed was treated with a fungicide, insecticide, and safener mixture containing ~ 1% Spirato 480 FS fungicide, 4% Sebring 480 FS fungicide, 3% Sorpro 940 ES seed safener. Then the seeds were air-dried for 3 days which provided a thin coating of this mix around the seeds. The safener allowed the use of the herbicide Dual Magnum as a pre-emergence treatment.
### Evaluation of Target Leaf Spot resistance
The SCP was planted at the Central Crops Research Station in Clayton, NC on June 14–15 2017 and June 20, 2018 in a randomized complete block design with two experimental replications in each case. Experiments were planted in 1.8 m single rows with a 0.9 m row width using 10 seeds per plot. Two border rows were planted around the periphery of each experiment to prevent edge effects. The experiments were inoculated on July 20, 2017 and July 20, 2018 at which point the sorghum plants were at growth stage 3. Ratings were taken on a one to nine scales (Fig. S2), where plants showing no signs of disease were scored as a nine and completely dead plants were scored as one (Fig. S2). Two ratings were taken in 2017 and four readings in 2018 starting two weeks after inoculation each year. sAUDPC (standardized area under disease progression curve) was calculated as described previously29,30.
### Statistical analyses
All statistical analysis of phenotypic data was performed using SAS V9.4 software. The LSmeans of two replicates for each year were calculated and these were used in turn to calculate the overall LS means. Analysis of variance (ANOVA) and least square (LS)means were calculated using the Proc Mix and Proc GLM procedure in SAS respectively. Correlations were calculated using the CORR procedure of SAS31.
### Phenotypic data transformation for association analysis
The phenotypic distribution was right skewed for flg22 and chitin-elicited ROS response traits. From a simple ANOVA we determined that higher predicted score values of these phenotypic traits were moderately associated with higher residuals. Therefore, natural logarithm transformation and root square transformation were performed using the raw scores of chitin and flg22, respectively. After transformation, the phenotypic distribution of each trait was less skewed and the relationship between residual and predicted values was improved. Transformed data were used in the downstream association analysis. Each trial was analyzed separately with SAS mixed model procedure in SAS software version 9.4 (SAS institute. 2019). For the chitin and flg22 response, a best linear unbiased estimator (BLUE) was obtained to estimate each line mean phenotypic value by a mixed model considering inbred lines as fixed effects and replications as random effects. Similarly, for TLS, the data of both years were combined by using a mixed linear model across years considering years and replication within years as random effects and inbred lines as fixed. All the original phenotypic data used for analysis is provided in File S1.
### Genotypic data
All genotypic data used for this study are available upon request from the corresponding author. Genotypic data for the SCP were obtained from Dr. Tiffany Jamann and Dr. Patrick Brown (University of Illinois). The original array consisted of ~ 1.12 million SNPs derived from whole-genome sequencing. We used the genotypes of each set of plants with phenotypic data described above. Each data set was first filtered to exclude SNPs with less than 5% minor allele frequency (MAF) and more than 10% heterozygosity. Linkage disequilibrium-based pruning of genotypic data was performed in software Plink v1.932. After pruning, a set of ~ 58 K SNPs were used to compute the kinship matrix in Tassel 533. Pruned data was also used to perform principal component analysis (PCA) with JMP Genomics 9 (SAS, Institute. 2019). Based on PCA, approximately 20% of the variability was accounted for by the first three principal components. To control for population structure in the association analysis, we excluded 37 inbred lines that explain more than 7% of the variability (Fig. S1). After removing outlier inbred lines, a second filter (< 5% MAF and > 10% heterozygote sites) was performed in each data set. The final sub-set of genotypes used in Genome wide Association Studies (GWAS) contained ~ 755 K SNPs for TLS and flg22 and ~ 750 K SNPs for Chitin. The genotypic datasets used for the analysis of the TLS, flg22 and chitin analyses are available from the corresponding authors.
### Association analysis
Genome-wide association analysis based on a mixed linear model (MLM) was performed in Tassel 533. The MLM model used was y = Xβ + Zu + e where y is the vector of phenotypes (BLUEs), β is a vector of fixed effects, including the SNP marker, tested, u is a vector of random additive effects (inbred lines), X and Z represent matrices, and e is a vector of random residuals. The variance of random line effects was modeled as Var(u) = K $${\sigma }_{a}^{2}$$, where K is the n × n matrix of pairwise kinship coefficient and $${\sigma }_{a}^{2}$$ is the estimated additive genetic variance. A threshold to declare significance of 1/m where m is the number of markers tested was used34,35.
### Candidate gene selection
Genes within 100 Kb of the highly significant markers were considered candidate genes. Identification and annotation of the candidate genes were performed using the maize BTx623 reference genome v3 available on the Ensembl Plants browser. Functional annotation of the candidate gene was based on EnsemblPlants and Gramene annotation.
## Results and discussion
### Evaluation and mapping of TLS resistance
The SCP was assessed in the field for TLS resistance in 2017 and 2018, in randomized complete blocks with two complete replications per year. We observed substantial variation in TLS resistance in the SCP (Fig. 1A). The two replicates in each year were significantly correlated (0.52 and 0.68 in 2017 and 2018 respectively, p < 0.0001) and the LSmean scores were significantly correlated between years (0.45, p < 0.0001) (Table 1). ANOVA analysis indicated that the genotype effect was significant (Table 2).
Association analysis using the LSMeans of the 492 lines that were scored identified a single highly significant association on chromosome 5 (Fig. 1B). Table 3 shows the parameters associated with this locus and details predicted genes located 100 Kb either side in the Btx623 genome. One of these genes is ds1, a leucine-rich repeat serine/threonine protein kinase gene that was previously reported as a major TLS resistance gene4. It seems very likely that this gene underlies the major QTL identified in the SCP.
### Evaluation and mapping of the MAMP response
To assess variation in basal immune response, we measured ROS production in response to flg22 and chitin treatment in the SCP in two full replications. We observed significant variation in response to both MAMPs (Fig. 2). Significant correlations were observed between replicates in both cases (0.5 and 0.38 for flg22 and chitin respectively, p < 0.0001, Table 1). ANOVA indicated that genotype effects were highly significant for both traits (Table 1).
The LSmeans of the flg22 and chitin responses were not significantly correlated, though replicate 2 of flg22 response was somewhat correlated with rep1 and rep2 of the chitin response (0.20 and 0.17 respectively, p < 0.01 and < 0.05). In previous work, we observed significant correlations in flg22 and chitin responses measured using the same ROS plate assay as well as shared QTL, in a maize recombinant inbred line mapping population22. The lack of correlation here is therefore somewhat surprising. It is not clear whether this reflects fundamental differences between the maize and sorghum MAMP responses. Vetter et al21 found a negligible correlation in plant growth responses between the bacterial MAMPs EF-Tu and flagellin in Arabidopsis, we are not aware of other published work comparing variation in the responses to two different MAMPs.
The phenotypic data were transformed as described and used for association analysis. Q-Q plots did not indicate an excess of false positives (Fig. S3). Two associations with the flg22 response were detected on chromosome 4. The significance threshold was calculated using a Bonferroni multiple comparison test correction which is based on the number of markers used. Since we used a relatively high number of markers (more than 750,000) this threshold was consequently relatively conservative. It should be noted that the associations with flg22 were below the threshold for significance we used but we are nevertheless reporting them as they are the highest associations detected and, given the conservative significance threshold used, are nevertheless likely to reflect real associations. One significantly associated locus was detected on chromosome 5 for the chitin response (Fig. 3). Table 3 shows the parameters associated with these associated loci and details predicted genes. While it is premature to assign causation, it is interesting to note that several of the candidate genes associated with flg22- and chitin-induced responses have homology to genes involved in the defense response or disease resistance in other systems. For instance, the durable wheat rust resistance gene LR34 is an ABC transporter36 while genes involved in the auxin response37 and the ubiquitin-mediated protein-degradation38 pathway have been implicated in disease resistance in other systems.
### Comparison of TLS resistance and MAMP response data
To understand whether variation in the response elicited by flg22 and chitin is connected to TLS resistance, we looked for correlation between MAMP response and TLS disease scores. Despite chitin being an integral component of the fungal cell wall and TLS being a fungal disease, we did not observe a significant correlation between the traits. We did observe a small but moderately-significant negative correlation between the flg22 response and TLS scores (− 0.13*, p value < 0.05), indicating that a higher flg22 response was somewhat associated with higher susceptibility. This was unexpected both because flg22 is a bacterial MAMP and TLS is a fungal disease and because we were expecting an association between increased MAMP response and increased resistance. Instead, we observed an opposite relationship, albeit quite weak. Two possible explanations occur to us. Since the correlation is relatively low, this may not be a meaningful correlation. Alternatively, several necrotrophic pathogens of a similar type to Bipolaris cookei have effectors that both induce ETI and facilitate pathogenesis. It appears that in these cases elicitation of HR allows the pathogen to grow on the resulting dead host cells39. It is possible that this correlation is due to a similar subversion of the plant defense machinery.
A recent companion study measuring the flg22 response and TLS resistance in two sorghum recombinant inbred line (RIL) populations did not identify correlations between the 2 traits or any colocalizing QTL7. In the current study, we used the SCP which provided higher genetic and phenotypic diversity than had been available from the two RIL populations but, overall this study also did not produce evidence to support our original hypothesis that a stronger MAMP response is predictive of stronger QDR. However, there are a number of caveats that make it impossible to draw general conclusions.
Perhaps the major caveat is that quantitation of the MTI response is complex. It depends on what MAMP is used and how the response is measured. Low correlations between responses to different MAMPs have been reported previously19,21,24, although, as mentioned above, Zhang et al22 observed a significant correlation between flg22 and chitin responses in maize. Moreover, the MAMP response can be quantified in a number of different ways, including measurement of ROS or NO production, MAP kinase phosphorylation, mRNA accumulation levels, lignin and cell wall-bound phenols, callose deposition, seedling growth inhibition and MAMP-induced pathogen resistance18,19,40. Relative line rankings vary significantly depending on the assay used19,22. Our preliminary data also suggests that the MAMP response varies with the age of the plant and the individual leaf on the plant. Essentially, quantification of the MAMP response is complex and inferences may vary significantly depending on how the response is elicited and how measured7.
The other major caveat is of course resistance to only one disease was assessed. It may be that resistance to certain diseases, perhaps those that are less well adapted to the host and cannot completely suppress basal resistance mechanisms, may be more associated with the MAMP response. As more diseases are assessed on the SCP we may be able to re-evaluate our hypothesis in the light of multiple comparisons.
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2023-04-01 12:34:33
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https://brilliant.org/discussions/thread/introductory-olympiad-algebra-sample/
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You have just completed a sample of the Advanced Algebra practice map, with problems taken from various skills. Full access to this map is available to Brilliant$$\color{green} {^2}$$ members. The Advanced Algebra map offers further practice in complex numbers and algebraic identities. This topics help you build up towards applying algebra in Olympiad problems.
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We have also introduced 3 chapters that consist of Olympiad problems in Polynomials, Exponentials and Absolute Value. These would be especially suitable for Level 2 and 3 members that are looking to polish up their skills and level up, and serve as revision for Level 4 and 5 members.
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Need more problems? Upgrade to Brilliant$$\color{green} {^2}$$!
This note is the conclusion of the set Introductory Olympiad Algebra Problems.
Note by Calvin Lin
4 years, 7 months ago
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I got everything. Cool
- 4 years, 7 months ago
Well, I got all but one of the four problems, so that's pretty awesome! :D
- 4 years, 7 months ago
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2018-12-12 02:04:45
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https://www.nature.com/articles/s41598-022-16686-2?error=cookies_not_supported&code=8632719e-c392-4a7c-83aa-0b661160660d
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## Introduction
Humans have a refined capacity to synchronize their actions with auditory rhythmic signals. Interpersonal synchrony is a special case of synchronization that refers to temporal coordination between individuals. Interpersonal synchrony is crucial for many social contexts such as dance, team sports, and conversational speech. Greater synchrony between individuals is associated with increased affiliation between partners1,2 and more cooperative behaviour3,4,5. Music performance is an especially precise form of interpersonal coordination. Performers must adapt their production of tone sequences based on auditory information from themselves and their partners to achieve synchrony6,7. Previous research showed that endogenous rhythms constrain this process: Greater synchrony between partners was predicted by smaller differences in their spontaneous production rates8,9. The relationship between interpersonal synchrony and spontaneous rates also extends to larger groups performing a rhythmic task: Greater synchronization in individuals’ hand movements was observed in seven-person groups among individuals whose spontaneous movement rates were similar10. Spontaneous rates of rhythmic actions in locomotion tasks (walking, running) in several species correspond to rates that require minimum energy expenditure, reflecting a state of optimal efficiency11,12. Consistent differences in human spontaneous rates are evidenced in a variety of rhythmic behaviors including speech13,14, hand clapping15, and finger tapping16,17,18. The relationship between interpersonal synchrony and spontaneous rates has been measured primarily in individuals with significant behavioral expertise in the measured tasks. We address whether musically untrained individuals show reduced synchrony in a novel musical task when they differ in their spontaneous production rates.
Constraints of spontaneous rates on interpersonal synchrony have a natural explanation in a nonlinear dynamical systems theory of internal timing and movement synchronization19. Spontaneous production rates are thought to reflect the intrinsic or natural frequency of an oscillator acting as an attractor toward which a system will converge over time20,21 and whose signature can be captured at a neural level22. Synchronization between interacting individuals has been modelled by the coupling of oscillators that adapt in frequency and relative phase23,24,25. Oscillators with similar intrinsic frequencies will be faster to adjust and more strongly coupled, resulting in higher synchronization accuracy. We test this hypothesis by comparing partners’ tapping synchronization in a joint musical task with their intrinsic frequencies as measured by their spontaneous rates in a solo task.
Musical training also impacts interpersonal synchrony. Individuals who received musical training show higher synchronization accuracy to external auditory rhythms than do untrained individuals26,27,28,29. Musicians also show lower temporal variability than non-musicians when they produce rhythmic sequences in the absence of an external cue30,31. Musically trained individuals adapt their synchronization performance more quickly in response to changing auditory rates32,33. Finally, musical training enhances individuals’ flexibility to produce sequences at rates other than the spontaneous rate31,34, indicating a reduced attractor strength of intrinsic frequencies during adaptation to external auditory signals. We test here the combined influence of spontaneous rates and musical training on partners’ asynchronies in a music tapping task that permits non-musicians to produce music without any required training.
Two different tasks have measured natural movement rates for rhythmic sequences in previous studies: Spontaneous Motor Tempo (SMT) and Spontaneous Production Rate (SPR). The SMT task consists of producing temporally regular tapping with one finger of the dominant hand in the absence of auditory feedback, whereas the SPR task consists of producing simple melodies (with one or more fingers) in the presence of auditory feedback. SMT has been used in numerous studies that investigate human timing, whereas SPR has been used with music-evoking movement. Age, physiological arousal, and time of day have been shown to influence SMT16,35,36, factors that show reduced influence on SPR31,37. A novel musical task was recently developed to measure SPR in both musicians and non-musicians31, making possible the measurement of SPR in untrained individuals. That task showed that non-musicians were more rigid and less flexible in their synchronization of music-generating finger taps with a metronome, as evidenced by recurring timing patterns measured with Recurrence Quantification Analysis (RQA). We extend this analysis to test interpersonal synchrony between partners (musician pairs and non-musician pairs) who synchronize with their partner; specifically, we test the hypothesis that RQA techniques should indicate greater rigidity in non-musicians’ interpersonal timing.
Pairs of musically trained participants (n = 14 pairs) and pairs of untrained participants (n = 14 pairs) participated in the experiment. Spontaneous rate measures of participants’ SMT and SPR were first collected individually (Solo tasks), and then Duet measures of synchrony were collected from each pair while they tapped melodies together at an initially cued rate that corresponded to the spontaneous rates (SPR) of each partner. Their taps were recorded by a force sensor that allowed listeners with or without musical training to produce melodies by simply tapping on the sensor (with one finger) to hear the next pitch in the melody31. We hypothesized that SPR and SMT reflect the frequencies of different intrinsic oscillations, with SMT measuring a motor (finger) frequency and SPR measuring an auditory-motor coupling frequency. Based on previous research with musicians38, we expected that the SPR difference between non-musician partners would predict synchronization accuracy and leader/follower patterns: Lower accuracy (larger magnitude of asynchronies) should result for both musician and nonmusician pairs whose Solo SPRs differ greatly, with the faster-SPR participant in each pair tapping ahead of their partner. Finally, we predicted that musically untrained pairs should exhibit larger asynchronies, larger temporal variability, and increased determinism (rigidity) in timing their joint synchronization, consistent with lower temporal flexibility, compared with musically trained pairs.
## Results
### Solo tasks: SPR and SMT measures are unrelated
Figure 1 shows each participant’s mean tapping rates in the SPR and SMT tasks, ordered in both plots from fastest to slowest participant in the SPR task. A wide range of SPR values were obtained for musician and non-musician groups; the SMT values also showed wide inter-individual variation. There was no correlation between the SPR and SMT values across participants (r (54) = − 0.1625, p = 0.23) or within groups (Musicians: r (26) = − 0.09, p = 0.63; Non-musicians: r (26) = − 0.29, p = 0.13). Furthermore, SMT and SPR values did not differ by musical training or interact with task; overall, participants were faster in the SPR task (M = 434, SE = 13.8) than in the SMT task (M = 738, SE = 40), F(1, 54) = 44.99, p < 0.001, $$\upeta _{{\text{p}}}^{{2}}$$ = 0.45.
The coefficient of variation which measures temporal variability was compared across SPR and SMT tasks. The CV was correlated across the SMT and SPR tasks for all participants (r (54) = 0.50, p < 0.0001) as well as within groups (non-musician group, r (26) = 0.40, p = 0.04; musicians, r (26) = 0.36, p = 0.06). The musician group showed smaller CVs (M = 0.045, SE = 0.0017) than the non-musician group (M = 0.064, SE = 0.0033), F (1, 54) = 18.57, p < 0.001, $$\upeta _{{\text{p}}}^{{2}}$$ = 0.26). Thus, temporal variability but not mean rate was correlated across SPR and SMT tasks.
### Duet tasks: duet asynchronies are influenced by musical training and partners’ spontaneous rates
The partners’ mean absolute asynchronies in the Duet synchronization task were compared across groups. As shown in Fig. 2, musicians synchronized better with their partners (median = 23.94) than did non-musicians (median = 33.68, U = 150, p < 0.01). Group differences in temporal variability were also observed in the Duet condition; the CV was smaller for musicians (M = 0.049, SE = 0.0022) than for non-musicians (M = 0.064, SE = 0.0026, F(1, 54) = 19.66, p < 0.0001, $$\upeta _{{\text{p}}}^{{2}}$$ = 0.27). Finally, each participant’s CV in Solo and Duet conditions were correlated across all participants (r (54) = 0.6345, p < 0.001) as well as within groups (Non-musicians: r (26) = 0.55, p < 0.01; Musicians, r(26) = 0.63, p < 0.001), suggesting that temporal variability remained stable across Solo and Duet tasks.
We test the dynamical systems prediction that each Duet pair’s signed asynchronies arose from the partners’ SPR differences. Figure 3 shows the mean signed asynchronies (cued partner’s onsets–uncued partners’ onsets) correlated with the partner’s Solo SPR Difference (cued partner’s rate minus uncued partner’s rate), separately for Condition A (metronome = PartnerA’s Solo rate) and Condition B (metronome = PartnerB’s Solo rate). Significant positive correlations were observed for all groups and all conditions: Condition A (r (12) = 0.79, p < 0.001 in non-musicians; r(12) = 0.68, p < 0.01 in musicians; Condition B (r(12) = 0.70, p < 0.01 in non-musicians; r(12) = 0.56, p < 0.05 in musicians). The larger the difference between partners’ SPRs, the larger their signed asynchronies; the partner with the faster Solo SPR generally anticipated the partner with the slower Solo SPR.
The regression slopes in Fig. 3 are larger (steeper) for non-musicians than for musicians, suggesting stronger constraints of endogenous rates on non-musician partners’ synchronization. To test the slope differences, a multiple regression analysis predicted the partners’ Duet asynchronies from the SPR differences, group membership (Non-musician coded 0; Musician coded 1), and the interaction of SPR differences with group membership. That regression was significant for both Condition A (R (24) = 0.77, p = 0.0001) and Condition B (R (24) = 0.65, p = 0.0035). The coefficient associated with SPR differences was again significant (Condition A: unstandardized coefficient = 0.1568, p = 0.00009; Condition B: unstandardized coefficient = 0.1620, p = 0.0016). Most important, the interaction term was significant for Condition A (unstandardized coefficient = − 0.0937, p = 0.0267) and showed a similar pattern in Condition B that did not reach significance (unstandardized coefficient = − 0.0655, p = 0.3105), indicating that the Non-musician slope was steeper than the Musician slope toward the beginning of the study session, consistent with the non-musicians’ less flexible adjustment to the cued metronome rate. Importantly, partners’ SMT differences did not predict duet asynchronies; no correlations between SMT differences and mean duet asynchronies reached significance (all p’s > 0.3).
### Recurrence quantification differs by musical training in solo performances
Figure 4 shows recurrence plots for the intertap intervals from one Solo performance trial by a non-musician (top) and musician (bottom). The presence of black dots indicates a higher recurrence rate which was found for non-musicians (M = 0.1044, SD = 0.0958) than for musicians (M = 0.0529, SD = 0.0482), U = 562.5, p < 0.01). Thus, non-musicians exhibited higher patterning in the timing of their Solo tapping.
Figure 5 shows plots of sample Solo performance trials that demonstrate high determinism (non-musician; top graph) and low determinism (musician; bottom graph), indicated by the presence of black dots forming diagonal lines that denote similar behavior over consecutive ITIs. The non-musicians’ ITIs exhibited higher determinism (M = 0.158, SD = 0.038) than did musicians (M = 0.1316, SD = 0.0268), U = 581, p < 0.001. Figure 5 (top) also shows the influence of the musical structure on the non-musician’s timing profile, captured by the distance between the diagonal lines. The participant’s diagonal lines were separated by distances of exactly 8 beats or taps, corresponding to the length of a musical subphrase (two metrical bars) in the simple melody performed. Thus, the RQA analyses illustrate constraints imposed by the melodic structure on the dynamics of nonmusicians’ tapping behavior.
Finally, we compared the temporal variability (CV) of the Solo performance ITIs with the determinism measures for the same individuals. As shown in Fig. 6, determinism increased as temporal variability increased for non-musicians (r (26) = 0.74, p < 0.0001) but not for musicians (r (26) = − 0.003, p = 0.99). Thus, the non-musicians’ overall increased variability corresponded to greater predictability or rigidity, rather than simply increased noise. There was no relationship between the CV and recurrence rate (unadjusted) across groups (r (54) = 0.1385, p = 0.31) or within groups (non-musicians: r (26) = 0.014, p = 0.94; musicians: r (26) = 0.079, p = 0.69).
### Recurrence quantification differs by musical training in duet performances
Next, we examine recurrence patterns in partners’ asynchronies in the Joint performances. RQA was computed on the signed asynchrony time series (asynchrony defined as Partner X tap onset − Partner Y tap onset, where the cued rate = Partner X’s SPR). Figure 7 shows a sample trial of asynchronies and the corresponding recurrence plot from a non-musician pair (top) and a musician pair (bottom). The mean recurrence rate was higher overall for non-musicians (M = 0.1236, SD = 0.0459; Median = 0.11) than for musicians (M = 0.0997, SD = 0.0268; Median = 0.09), U = 134, p = 0.049, indicating that the non-musicians’ asynchronies were more patterned than those of the musicians as they synchronized with a partner. Determinism values did not differ between the groups, U = 106, p = 0.14.
We compared the RQA measures of Duet asynchrony with behavioral variability by correlating the recurrence rates of each pair’s asynchronies with the standard deviation of that pair’s Duet asynchronies. The recurrence rates of the non-musicians’ asynchronies were significantly correlated with the SD of asynchronies for both conditions (PartnerA cue: r (12) = 0.6871, p = 0.0066; PartnerB cue: r (12) = 0.612, p = 0.02). The correlations were not significant for the musicians (PartnerA cue: r (12) = 0.2097, p = 0.4916; PartnerB cue, r (12) = 0.2101, p = 0.4906). Determinism values did not correlate with standard deviations of the joint asynchronies for either group. Similar to the Solo performance timing, non-musicians showed increased temporal variability in Duet asynchronies that coincided with increased (non-random) patterning of their asynchronies.
### Duet synchrony is predicted by partners’ solo performance recurrence patterns
Finally, the partners’ synchronization in the Duet performances was predicted from the recurrence patterns in each partner’s intertap intervals in Solo performances. Consistent with the directionality of the faster/slower partners’ Solo rate influences on the asynchronies (Fig. 3), each pair of partners was coded in terms of which partner had the faster/slower Solo rate. A multiple regression analysis predicting the partners’ Duet absolute asynchronies from each partner’s Solo recurrence indicated a significant overall fit, R (25) = 0.5020, p = 0.02. Partners with higher recurrence patterning (less flexibility) in their Solo timing patterns had greater difficulty synchronizing with their partner in Joint performances. The partner with the faster SPR tended to exhibit higher recurrence in their intertap intervals that contributed most to the pair’s asynchrony (faster partner: standardized coefficient = 0.4761, p = 0.01; slower partner: standardized coefficient = 0.1644, p = 0.35). The determinism values from the partners’ Solo performances (when recurrence rate was fixed) did not predict partners’ Duet asynchronies.
## Discussion
We examined intrinsic constraints that influence both musically trained and untrained individuals’ auditory synchrony as they produced musical sequences. Musicians and non-musicians performed an auditory-motor synchronization paradigm that permitted non-musicians to produce music without any required training. Interpersonal synchrony measures from partners who tapped a melody together showed significant constraints of the partners’ spontaneous production rates: Greater synchrony between partners was predicted by smaller differences in their individual spontaneous performance rates. The fact that both participant groups showed constraints of spontaneous rates on interpersonal synchrony supports the view that spontaneous rates are not learned but instead are intrinsic to rhythmic movement32,39. Spontaneous rates of rhythmic actions often correspond to movements that require minimum energy expenditure, reflecting a state of optimal efficiency12,40,41. Musically untrained adults showed greater constraints of these endogenous rhythms on interpersonal coordination than did trained adults, consistent with findings that flexibility increases as individuals achieve task expertise31,34,42. Nonlinear analyses of the intertap intervals (Solo) and asynchronies (Duet) confirmed greater recurrence (predictability) in non-musicians’ tapping, consistent with a multi-dimensional phase space in which movement trajectories change as individuals acquire auditory-motor expertise or additional sensorimotor information43.
Spontaneous (uncued) production rates have a natural explanation in nonlinear dynamical systems theory that an intrinsic frequency of an oscillator acts as an attractor toward which the system’s behavior will converge over time20,21. Synchronization between interacting individuals has been modelled with the coupling of oscillators that adapt in frequency and phase23,24,25. Oscillators with similar intrinsic frequencies are faster to adapt and more strongly coupled, resulting in higher synchronization accuracy, a theoretical prediction supported by analyses of partners’ tapping synchronization accuracy as a function of the difference in the partners’ spontaneous rates of solo productions. Visuomotor coordination tasks have shown that, as the difference in rate between an individual’s oscillatory movements and a rhythmically oscillating stimulus decreases, individuals are more likely to spontaneously entrain their movements to the stimulus rate44,45. Temporal variability measures in visuomotor coordination tasks have been linked to motor flexibility: Individuals who exhibited some temporal variability were more likely to unintentionally synchronize their movements with a wider range of stimulus rates46, consistent with findings in auditory-motor coordination tasks that the same individuals who show better synchronization tend to show enhanced flexibility across movement rates31.
Musical training modulated the strength with which the spontaneous rates influence interpersonal synchrony: Individuals with musical training showed higher synchronization accuracy to their partner’s intrinsic frequency than did untrained individuals26,27,28,29. Musicians also showed lower temporal variability than non-musicians when producing rhythmic sequences in absence of an external cue, also consistent with previous findings30,31. This is the first study to tie the non-musicians’ increased recurrence patterning to the amount of asynchrony, suggesting that temporal variability of non-musicians is not simply motor noise but instead reflects repetitive patterns that reduce one’s flexibility and adaptation when synchronizing with a partner. The non-musicians’ deterministic recurrence analysis also revealed the musical structure of the melody task, based on the spacing between the diagonal lines (Fig. 6). Overall, the non-musicians’ tapping was more strongly patterned by the musical structure of the time series.
Spontaneous Motor Tempo, a common measure of motor timing in the absence of auditory feedback, did not predict performance in the interpersonal synchronization task. SMT also did not correspond to measures of SPR, which did predict performance in the synchronization task. Despite large individual differences in both SMT and SPR measures (Fig. 1), there was no correlation across the tasks for musically trained or untrained individuals. Furthermore, individual SMT measures did not predict Duet performance synchronization, whereas individual SPR measures did predict synchronization, consistent with previous findings with musicians9. Thus, individuals’ movement rates alone (in both SMT and SPR tasks) did not account for synchronization behavior. One explanation is that SMT provides less sensory feedback for error correction than SPR; timing in SPR tasks has shown evidence of auditory feedback-related error correction31. This explanation is consistent with current and previous findings that musically trained individuals exhibit less temporal variability than untrained individuals in rhythmic tasks47. SMT and SPR measures may be driven by different constraints; SMT measures are modulated by developmental changes18, whereas SPR measures may be modulated more by enhanced auditory exposure gained by musical training. The role of musical training in SMT has been documented with inconclusive or negative outcomes48,49, and until recently, SPR was only measured in musically trained individuals. The potential influence of different mechanisms in SMT and SPR is consistent with an interpretation that they reflect different intrinsic movement frequencies, one motoric in origin and the other related to auditory-motor coupling, a hypothesis for future investigation.
The study’s reliance on a one-finger tapping task that both musicians and non-musicians could perform is one limitation of the current study; extensions to more complex movement sequences are necessary. Another limitation is the inclusion of simple familiar melodies, which may not reveal the range of potential influence of sequence structure on the synchronization tasks32. For example, the distance between diagonal lines typical of high determinism or predictability (Fig. 6 top) may be influenced by the musical structure. Also important, the sampling of musically trained and untrained individuals does not address the full scale of auditory expertise that individuals obtain. This limitation prevents us from addressing the shift in temporal flexibility that is assumed to occur between the endpoints of the expertise scale, as individuals gain motor skills such as producing a melodic sequence with temporal regularity. Future designs may include a broader range of participants and melodies to track the development of synchronization mechanisms, as well as other novel measurements of rhythmic movement that can be applied to individuals with and without specialized training50.
In sum, partners’ interpersonal synchrony in an auditory-motor task is related to each partner’s endogenous rhythms in individual sound production. Consistent with dynamical systems explanations of increased coupling between oscillators with similar frequencies, the timing of intertap intervals as well as asynchronies produced by musically untrained individuals demonstrated more rigidity and predictability than those of musically trained individuals. Both groups of individuals demonstrated constraints of their spontaneous production frequencies in the joint synchronization task, as exhibited by the successful prediction of joint synchronization from the recurrence measures of the partners’ solo performances. The increased temporal variability in non-musicians’ auditory-motor tapping is not simply noise but instead reflects rhythmic patterns that may be overcome with training, as motor and auditory systems become coupled. These novel findings point to two mechanisms driving musicians’ flexibility in timing tasks: first, a honed ability to use auditory feedback in temporal adaptation; and second, an ability to de-couple one’s movement timing from the intrinsic frequencies at which that movement is easiest to produce. These mechanisms can be addressed in future studies that compare performance on a wider variety of auditory rhythms.
## Method
### Participants
Twenty-eight musically trained and 28 untrained adults (18–35 years old) participated. Sample size was estimated from Zamm et al.8 findings which used a similar design (2 groups of 20 individuals each with repeated measures, treating pair as random variable) that yielded moderate to large effect sizes (partial η2 = 0.25). Inclusion criteria were > 6 years of individual instruction on a musical instrument for musicians (M = 10.43 years), and < = 2 years of instruction for nonmusicians (M = 0.42 years). Participants were randomly paired within group (musician/non-musician). All participants exhibited normal hearing (< 30 dB HL) for the stimulus frequency range (125–750 Hz), determined by a pure-tone audiometric screening, and had familiarity with the experimental melodies, evaluated by accurate humming of the melodies. Participants provided written consent and the study was conducted in accordance with the Declaration of Helsinki, the Canadian Tri-Council Policy Statement on Ethical Conduct for Research Involving Humans (TCPS2-2018). The study’s experimental protocols were approved by the local Research Ethics Board of McGill University (REB #1951018). Informed consent was obtained from each participant.
### Materials and equipment
Two familiar melodies were used: Happy Birthday (in D Major) and Twinkle, Twinkle Little Star (in G Major, referred to as “Twinkle”). Happy Birthday served as a practice melody and Twinkle served in both practice and experimental trials, chosen for its primarily isochronous rhythmic structure. The two partners heard the melody in a different pitch range (one octave apart) for the entire study (for example, Partner A’s melody started on pitch G3 and Partner B’s on pitch G4) to differentiate their parts when they performed together. The melody tones were sounded with a marimba timbre (Roland Studio Canvas GM2 sound bank; #030) and metronome tones with a woodblock percussion timbre (sound bank #206). A sine tone (sound bank #158) signaled the start of trials.
Participants tapped the melody on a force-sensitive resistor connected to an Arduino Uno device that sent MIDI timing signals to the computer running FTAP 2.1.07a51. Tones were generated by a Mobile Studio Canvas SD-50 and were heard over studio headphones at a comfortable listening level. Identical equipment was used for the two partners in all conditions. The two Arduino tapping devices were assigned to different MIDI channels and were recorded to FTAP on the same computer (synchronized) during the joint performances (see Supplemental Materials).
### Procedure
Each partner completed the Solo tasks in separate testing rooms. First, participants completed the familiar melody assessment and an audiometry screening. Participants then completed the SMT task while seated at a table with the tapping pad (force sensing resistor). They were instructed to use the index finger of their dominant hand to tap at a comfortable, steady rate on the pad. One practice trial and three experimental trials were recorded.
Participants then performed the Solo SPR task with the practice melody (Happy Birthday). They were instructed to tap the melody at a comfortable steady rate, and that each time they produced a tap, the next tone of the melody would sound. Next, they performed the same task with the experimental melody (Twinkle) with the same instruction and were told to repeat the melody several times until they no longer heard auditory feedback, signaling the end of the trial after 3.5 melody repetitions (one repetition = 48 quarter-note beats). They performed a practice trial and three experimental trials. If a participant made gross timing errors (such as starting, restarting, or ending a trial at the wrong time), the trial was discarded and the participant performed another trial (approximately 8% of trials). The first three error-free experimental trials were used (up to six experimental trials were permitted). Participants then completed the musical background questionnaire while each participant’s mean Solo SPR value (mean InterTap Interval, ITI) was computed.
Participants then practiced synchronizing the melody with a metronome cue set to the mean of the two participants’ Solo SPR values; each participant was instructed to wait for 8 metronome beats and then to start synchronizing their taps with the metronome. Then the participants practiced a synchronization-continuation task in which they were instructed to listen to the first 8 metronome beats and to start synchronizing taps with the metronome, which turned off after 8 additional beats. Participants continued tapping the melody at the cued rate until no more auditory feedback was heard (after 3.5 melody repetitions), signaling the end of the trial.
Partners moved to the same testing room for the Duet task where they were seated facing each other over a screen that left the partners’ heads visible but occluded the partners’ bodies below the shoulders. Partners took turns tapping the experimental melody, synchronizing with a metronome set to the average Solo SPR of the pair, so that they became accustomed to the pitch range of each partner (one octave away from their part). Similar to Solo trials, Duet trials began with 8 metronome beats, followed by 8 metronome beats with which they synchronized their taps, and then the metronome ended during the continuation phase while participants continued to tap the melody at the cued rate until the auditory feedback ended (after 3.5 melody repetitions). Partners then completed a practice synchronization trial while the metronome cue was set to PartnerA’s Solo SPR rate, and then a practice synchronization-continuation trial at the same rate. Partners then performed 3 experimental synchronization-continuation trials at PartnerA’s rate, each lasting 3.5 repetitions. If a trial contained timing errors, the trial was ended and a new trial was begun. After three experimental trials were recorded at PartnerA’s Solo SPR, the Duet procedure was repeated with the metronome cue set to Partner B’s Solo SPR until 1 practice trial and 3 experimental synchronization-continuation trials were obtained in each condition (total of 6 experimental trials per pair). The experiment lasted about one hour and participants received a small remuneration.
### Analysis
The mean Spontaneous Motor Tempo (SMT) was calculated as the mean inter-tap-interval (ITI) of the first 30 taps, averaged across trials18. The mean Spontaneous Production Rate (SPR) was calculated as the mean ITI of the middle 2 repetitions of each trial, to capture trial positions of maximal stability in performance timing38,52, then averaged across trials. Half note ITIs were interpolated, resulting in 96 ITIs per Solo SPR trial. Outlier ITIs more than 3 standard deviations from the mean ITI across trials were discarded (less than 1% of total ITIs in SMT; 0.007% of musicians’ total ITIs and 1.44% of non-musicians’ ITIs). The coefficient of variation (CV) was calculated for SMT and SPR trials as the standard deviation of ITIs divided by the mean ITI for 30 taps (taps 17–46 of the SPR task).
Synchronization in the Duet task was measured by the tap onset time of the partner whose SPR rate was cued minus the other partner’s tap onset time. A negative asynchrony means that the partner whose SPR served as the cued rate tapped earlier than their partner. Asynchrony outliers 3 standard deviations or more from the mean asynchrony were discarded (1.54% of asynchronies in musicians and 1.57% in non-musicians). Solo and Duet measures (ITIs and asynchronies) were analyzed with analyses of variance (ANOVA) tests of group differences (musician/non-musician) and task differences (SPR/SMT). Pearson correlation coefficients were computed to analyze comparisons between behavioral (ITI, asynchrony) and RQA measures (recurrence, determinism) in Solo and Duet tasks.
Recurrence Quantification Analyses (RQA) were performed on Solo (ITI) and Duet (asynchrony) measures. RQA is a non-linear time series analysis that uncovers recurring patterns and is especially useful for non-stationary measures31,53. RQA was applied to data from entire trials (without outlier removal) to maximize the times series length. RQA analyses of Solo performances included 126 ITIs (excluding the first 16 ITIs that included metronome beats and the last 2 ITIs) and Duet performances included 111 asynchronies (the same range as the Solo performances, excluding the first 14 asynchronies that included metronome beats and the last asynchrony). RQA metrics of recurrence rate (probability that a specific state will recur in the time series, indicated by points in the recurrence plot) and determinism (predictability of the behavioral pattern, captured by the proportion of recurrent points that form a continuous diagonal line) were analyzed. The RQA minimum length parameter was set to = 2 to incorporate all pattern lengths. The delay parameter, estimated from the first minimum in mutual information function, was set to = 2. The number of embedding dimensions, which was determined by values necessary for the false-nearest neighbor function to approach zero54, was set to 4 for Solo analyses and to 3 for Duet analyses. The radius was fixed to = 1 which yielded recurrence rates averaging around 10%, consistent with recommended values for behavioral data54. Recurrence quantification analyses were based on the CRP Toolbox55,56. The nonlinear RQA metrics were analyzed with nonparametric tests of group differences (Mann–Whitney U).
### Ethical approval
Before the experiment, participants were provided with an information sheet that outlined the general purpose of the study and informed them that they could withdraw at any time without penalty. All methods were reviewed by the Ethics Research Board of McGill University and were in accordance with the Declaration of Helsinki.
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2023-04-02 07:16:22
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https://socratic.org/questions/how-do-you-solve-frac-1-2-frac-2-3-y-frac-4-7
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# How do you solve -\frac { 1} { 2} = \frac { 2} { 3} y - \frac { 4} { 7}?
Mar 1, 2017
$y = \frac{3}{28}$
#### Explanation:
$- \setminus \frac{1}{2} = \setminus \frac{2}{3} y - \setminus \frac{4}{7}$
As soon as you have an EQUATION with fractions, you can get rid of the denominators immediately.
Multiply each term by the LCM of the denominators.
In this case it is $42$
$\frac{\textcolor{b l u e}{42 \times} - 1}{2} = \frac{\textcolor{b l u e}{42 \times} 2}{3} y - \frac{\textcolor{b l u e}{42 \times} 4}{7} \text{ } \leftarrow$ cancel
$\frac{{\cancel{42}}^{21} \times - 1}{\cancel{2}} = \frac{{\cancel{42}}^{14} \times 2}{\cancel{3}} y - \frac{{\cancel{42}}^{6} \times 4}{\cancel{7}}$
$- 21 = 28 y - 24 \text{ } \leftarrow$ add 24 to both sides
$- 21 + 24 = 28 y - 24 + 24$
$3 = 28 y$
$\frac{3}{28} = y$
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2022-12-08 00:25:23
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https://economics.stackexchange.com/questions/14937/is-there-a-difference-between-buying-a-currency-and-selling-the-currency-to-be-c/14938
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# Is there a difference between buying a currency and selling the currency to be converted to?
I'm not an economist. I'm looking to exchange my CAD to EUR. I watched the rates on a exchange office, and I found something strange:
We buy: We sell:
1 EUR = 1,4946 CAD (1) 1 EUR = 1,3280 CAD (2)
1 CAD = 0,6691 EUR (3) 1 CAD = 0,7530 EUR (4)
How can there be 4 different rates? I understand why CAD -> EUR is different of EUR -> CAD with the supply and demand. When the exchange office buys 1 CAD, it sells EUR too. I supposed it was to make things easier, so I checked it:
1 EUR = 1,3280 CAD (2)
=> 1,3280 CAD = 1 EUR
1 CAD = 0,6691 EUR (3)
1,3280 * 1 CAD = 1,3280 * 0,6691 EUR
=> 1,3280 CAD = 0.8886 EUR
Here, we found that it's indeed, more interesting for the customer to accept a sell than buying (from the office point of view for the words buy/sell). But is the difference, normal? And why?
Buy and sell rates are different because the currency vendor wants to make a profit and is also taking some risk with the exchange. It is possible that she cannot unload all the CADs they buy from you immediately and then perhaps in the future they will depreciate. (Perhaps they will appreciate. It is uncertain, hence there is risk.)
The same applies to buying and selling euros.
Usually there is some middle rate $x$ EUR/CAD. The vendor will deviate 1-2% from this in the direction favorable to her, depending on the direction of the deal (buying EURs with CAD or the other way around). You can look up something close the middle rate on XE. Currently this is about 1.394 CAD/EUR. Your vendor seems to sell at a premium that seems higher than normal to me, so perhaps see if you can find another currency merchant.
There are in fact two rates (subject to minor rounding), each expressed two different ways (a rate and its reciprocal), presumably to help different people who think different ways.
Suppose you wanted to buy $1000$ Euros with Canadian dollars. The calculation is either $1000\times 1.4946 = 1494.60$ or $\dfrac{1000}{0.6691}=1494.54$ - the same within six cents
Similarly if you want to sell $1000$ Euros for Canadian dollars. The calculation is either $1000\times 1.3280 = 1328.00$ or $\dfrac{1000}{0.7530}=1328.02$ - the same within a couple of cents
The gap between the rates in the two directions (a gap of over $160$ dollars for $1000$ Euros) is much more than the handful of cents when using a particular rate or its reciprocal
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2020-05-26 18:33:59
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https://www.tutorialspoint.com/articles/category/class-10/7
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## If $\cos 9 \alpha=\sin \alpha$ and $9 \alpha<90^{\circ}$, then the value of $\tan 5 \alpha$ is(A) $\frac{1}{\sqrt{3}}$(B) $\sqrt{3}$(C) 1(D) 0
Updated on 10-Oct-2022 13:28:52
Given:$\cos 9 \alpha=\sin \alpha$ and $9 \alpha ## The value of \( \left(\tan 1^{\circ} \tan 2^{\circ} \tan 3^{\circ} \ldots \tan 89^{\circ}\right)$ is(A) 0(B) 1(C) 2(D) $\frac{1}{2}$
Updated on 10-Oct-2022 13:28:52
Given:$\left(\tan 1^{\circ} \tan 2^{\circ} \tan 3^{\circ} \ldots \tan 89^{\circ}\right)$To do:We have to find the value of $\left(\tan 1^{\circ} \tan 2^{\circ} \tan 3^{\circ} \ldots \tan 89^{\circ}\right)$.Solution: We know that, $tan\ (90^{\circ}- \theta) = cot\ \theta$$tan\ \theta \times \cot\ \theta=1Therefore, \tan 1^{\circ} \tan 2^{\circ} \tan 3^{\circ}.....\tan 45^{\circ}...... \tan 87^{\circ} \tan 88^{\circ} \tan 89^{\circ}=\tan 1^{\circ} \tan 2^{\circ} \tan 3^{\circ}......\tan 45^{\circ}.........\tan (90^{\circ}-3^{\circ}) \tan (90^{\circ}-2^{\circ}) \tan (90^{\circ}-1^{\circ})$$=\tan 1^{\circ} \tan 2^{\circ} \tan 3^{\circ}........\tan 45^{\circ}......\cot 3^{\circ} \cot 2^{\circ} \cot 1^{\circ}$$=(\tan 1^{\circ} \cot 1^{\circ})(\tan 2^{\circ} \cot 2^{\circ})..................(\tan 44^{\circ} \cot 44^{\circ})(1)$$=1$Read More
## If $\cos (\alpha+\beta)=0$, then $\sin (\alpha-\beta)$ can be reduced to(A) $\cos \beta$(B) $\cos 2 \beta$(C) $\sin \alpha$(D) $\sin 2 \alpha$
Updated on 10-Oct-2022 13:28:52
## Students of a school are standing in rows and columns in their playground for a drill practice. $\mathrm{A}, \mathrm{B}, \mathrm{C}$ and $\mathrm{D}$ are the positions of four students as shown in figure below. Is it possible to place Jaspal in the drill in such a way that he is equidistant from each of the four students $A, B, C$ and D? If so, what should be his position?"
Updated on 10-Oct-2022 13:28:51
Given:Students of a school are standing in rows and columns in their playground for a drill practice. $\mathrm{A}, \mathrm{B}, \mathrm{C}$ and $\mathrm{D}$ are the positions of four students.To do:We have to find whether it is possible to place Jaspal in the drill in such a way that he is equidistant from each of the four students $A, B, C$ and D.Solution:From the figure, we observe that the positions of the four students A, B, C and D are $(3, 5), (7, 9), (11, 5)$ and $(7, 1)$ respectively.The four vertices form a quadrilateral. $A B ... Read More ## If the poins $A(1,-2), B(2,3) C(a, 2)$ and $D(-4,-3)$ form a parallelogram. find the value of $a$ and height of the parallelogram taking AB as base. Updated on 10-Oct-2022 13:28:51 Given:The points$A (1, -2), B (2, 3), C (a, 2)$and$D (-4, -3)$form a parallelogram.To do:We have to find the value of$a$and height of the parallelogram taking$AB$as base.Solution:Draw a perpendicular from $\mathrm{D}$ to $\mathrm{AB}$ which meets $\mathrm{AB}$ at $\mathrm{P}$.$\mathrm{DP}$ is the height of the parallelogram. We know that, Diagonals of a parallelogram bisect each other.This implies, Mid-point of$AC =$Mid-point of$BD$The mid-point of a line segment joining points $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$ is $(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2})$$(\frac{1+a}{2}, ... Read More ## The points \( A\left(x_{1}, y_{1}\right), \mathrm{B}\left(x_{2}, y_{2}\right)$ and $\mathrm{C}\left(x_{3}, y_{3}\right)$ are the vertices of $\Delta \mathrm{ABC}$What are the coordinates of the centroid of the triangle ABC? Updated on 10-Oct-2022 13:28:51 Given:The points $A\left(x_{1}, y_{1}\right), \mathrm{B}\left(x_{2}, y_{2}\right)$ and $\mathrm{C}\left(x_{3}, y_{3}\right)$ are the vertices of $\Delta \mathrm{ABC}$To do:We have to find the coordinates of the centroid of the triangle ABC.Solution:We know that,Coordinates of the centroid of a triangle$=\left(\frac{\text { Sum of abscissa of all vertices, }}{3}, \frac{\text { Sum of ordinate of all vertices }}{3}\right)$Therefore,The coordinates of the centroid of the triangle ABC$=\left(\frac{x_{1}+x_{2}+x_{3}}{3}, \frac{y_{1}+y_{2}+y_{3}}{3}\right)\$
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2023-03-23 07:26:23
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https://plainmath.net/pre-algebra/84087-write-in-standard-form
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Luciano Webster
2022-07-27
Write 40 000 000 in standard form
kamphundg4
Expert
Standard form is a way of writing very large or very small numbersin an easier form to read, this number can be written as $4×{10}^{7}$
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2023-02-02 22:07:01
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https://buske.de/explaining-the-variability-of-clitic-doubling-across-romance-a-diachronic-account.html
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Fehler gefunden?
Erweiterte Suche
# Explaining the variability of clitic doubling across Romance: a diachronic account
Zurück zum Heft: Linguistische Berichte Heft 236
EUR 14,90
This paper will propose that the various patterns of clitic doubling across Romance (and other) languages are the result of diachronic developments, namely of the interaction of catastrophic and gradual processes. More precisely we will argue that catastrophic processes (the emergence of doubling) need to be differentiated from gradual processes (the spread of doubling to different contexts). Under this view the catastrophic change is the change concerning one macro parameter giving way to several micro parameters which are then responsible for the spread of clitic doubling to different syntactic contexts.
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2020-02-24 02:53:03
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http://mathhelpforum.com/geometry/283027-transformation-notation-question.html
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1. ## Transformation notation question
I taught middle school math and I'm somewhat familiar with transformations, but I have no idea what the notations in this problem mean. I looked in two geometry books with no luck. Google, no luck. Khan Academy, well, you get the idea. I'm tutoring two students who are doing distance learning, and I don't have access to their materials. I don't mind admitting that I don't know something, but I hate to look REALLY stupid.
I meet with them tomorrow.
rsn
2. ## Re: Transformation notation question
looks like $T_{a,b}$ is a translation $a$ units in the $x$-direction and $b$ units in the $y$-direction
and $r_{x=y}$ a reflection across the line $x=y$
3. ## Re: Transformation notation question
Thank you for your helpful and rapid response. I'm attaching my work. I think that I have a proper explanation.
mrmac
4. ## Re: Transformation notation question
Looks good
the inverse of a translation
$$\left(T_{a,b}\right){}^{-1}=T_{-a,-b}$$
the inverse of a reflection
$$r^{-1}=r$$
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2019-10-15 17:32:39
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https://mathematica.stackexchange.com/questions/159551/trigreduce-applied-to-certain-terms-finding-coefficients-of-trig-terms
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# TrigReduce applied to certain terms? Finding coefficients of trig terms
I have an expression that involves trig functions of t and y. I would like to find the coefficients of Cos[t], Sin[t], Cos[2 t], Sin[2 t] ... and the constant term independent of t. I can get the terms in t and y separated by using TrigExpand however this then leaves me with trig expressions like Cos[t]^2. To reduce these to terms in Cos[n t] etc. I have to apply TrigReduce to them but NOT to the terms involving y. TrigReduce does not have a second parameter to tell it to just reduce terms in say t. Can we make one?
My workaround is to use FourierTrigSeries guessing the highest trig multiple I need (here 5 but I don't want to guess) and then to use Coefficient to find the individual terms and the constant term. This is a bit long. Are there other ways?
Thanks
Here is a simple example. I will need to do more complex examples.
e1 = -(Cos[t] - E^(-(y/Sqrt[2])) Cos[t - y/Sqrt[2]])^2 +
1/2 ((E^(-(y/Sqrt[2])) Cos[t - y/Sqrt[2]])/Sqrt[2] - (
E^(-(y/Sqrt[2])) Sin[t - y/Sqrt[2]])/Sqrt[
2]) (-(Sqrt[2] - 2 y) Cos[t] +
Sqrt[2] E^(-(y/Sqrt[
2])) (Cos[t - y/Sqrt[2]] - E^(y/Sqrt[2]) Sin[t] +
Sin[t - y/Sqrt[2]])) ;
e2 = e1 // TrigExpand;
e3 = FourierTrigSeries[e2, t, 5];
trigs = Flatten@Table[{Cos[n t], Sin[n t]}, {n, 5}];
coffs = Coefficient[e3, #] & /@ trigs;
const = e3 - coffs.trigs;
The coefficients are
coffs // TableForm
and the constant terms is
const
A sanity check gives
Simplify[e1 == const + coffs.trigs]
(* True *)
Try the following: These are the transformation rules:
rules = {Cos[t]^n_Integer :> TrigReduce[Cos[t]^n],
Sin[t]^n_Integer :> TrigReduce[Sin[t]^n],
Times[a_, Cos[t] , Sin[t], c_] :> a*c*TrigReduce[Cos[t] Sin[t]]};
With these rules
Collect[e2 /. rules, {Cos[2 t], Sin[2 t]}]
yields
You might want to simplify the coefficients of the sin(2t) and cos(2t) as follows:
MapAt[Simplify,
Collect[e2 /. rules, {Cos[2 t], Sin[2 t]}], {{7, 2}, {8, 2}}]
giving the following
Have fun!
• Good idea, write my own rules. Thanks. – Hugh Nov 9 '17 at 14:19
• I know that this post was written long time ago, but I would like to add a correction to the rules defined above. It is better to write the last element of the them as Times[a___, Cos[t] , Sin[t]] :> Times[ a, TrigReduce[Cos[t] Sin[t]] ] For example, only in this form the rules will render the expressions like $2 \sin x \cos x$ into $\sin 2 x$, while the original rules will leave such input untouched – Sergei Ovchinnikov Dec 11 '20 at 19:24
One idea is to use FourierCoefficient to compute the general term:
f[n_] = FourierCoefficient[e1, t, n];
f[n] //TeXForm
$\begin{cases} \frac{1}{8} \left(-2+e^{-\frac{(1-i) y}{\sqrt{2}}} \left(2+(1-i) \sqrt{2} y\right)\right) & n=-2 \\ \frac{1}{8} \left(-2+e^{-\frac{(1+i) y}{\sqrt{2}}} \left(2+(1+i) \sqrt{2} y\right)\right) & n=2 \\ \frac{1}{8} \left(e^{-\frac{(1+i) y}{\sqrt{2}}} \left((1+i) \sqrt{2} y+(1-i) e^{i \sqrt{2} y} \left(\sqrt{2} y+(1+3 i)\right)+(4-2 i)\right)-4 e^{-\sqrt{2} y}-4\right) & n=0 \end{cases}$
We can then recover the Sin and Cos coefficients as follows:
sin = Simplify @* ComplexExpand /@ Simplify[I(f[n]-f[-n]), n>0];
sin //TeXForm
$\begin{cases} \frac{1}{4} e^{-\frac{y}{\sqrt{2}}} \left(\left(\sqrt{2} y+2\right) \sin \left(\frac{y}{\sqrt{2}}\right)-\sqrt{2} y \cos \left(\frac{y}{\sqrt{2}}\right)\right) & n=2 \\ 0 & \operatorname{True} \end{cases}$
cos = Simplify @* ComplexExpand /@ Simplify[f[n] + f[-n], n>0];
cos //TeXForm
$\begin{cases} \frac{1}{4} e^{-\frac{y}{\sqrt{2}}} \left(-2 e^{\frac{y}{\sqrt{2}}}+\sqrt{2} y \sin \left(\frac{y}{\sqrt{2}}\right)+\left(\sqrt{2} y+2\right) \cos \left(\frac{y}{\sqrt{2}}\right)\right) & n=2 \\ 0 & \operatorname{True} \end{cases}$
zero = Simplify @* ComplexExpand @ f[0];
zero //TeXForm
$\frac{1}{4} e^{-\sqrt{2} y} \left(-2 \left(e^{\sqrt{2} y}+1\right)+e^{\frac{y}{\sqrt{2}}} \left(\sqrt{2} y-2\right) \sin \left(\frac{y}{\sqrt{2}}\right)+e^{\frac{y}{\sqrt{2}}} \left(\sqrt{2} y+4\right) \cos \left(\frac{y}{\sqrt{2}}\right)\right)$
Using FourierSinCoefficient/FourierCosCoefficient to obtain the general term doesn't work quite as well, but we can use them to check the above results:
(sin /. n->2) == FourierSinCoefficient[e1, t, 2]
(cos /. n->2) == FourierCosCoefficient[e1, t, 2]
True
True
• This is a good way of getting a general term. One problem is how to find how many terms there are and FourierCoefficient is certainly a good way of doing this. Although you have to recombine terms this may be the best method. I am not sure what happens if there is no general term. It seems a shame that one has to go to the effort of taking a Fourier transform when all that is needed is to do algebra to find the trig terms. Thanks for your help. – Hugh Nov 9 '17 at 19:12
You can try also with this:
e1 = -(Cos[t] - E^(-(y/Sqrt[2])) Cos[t - y/Sqrt[2]])^2 + 1/2 ((E^(-(y/Sqrt[2])) Cos[t - y/Sqrt[2]])/
Sqrt[2] - (E^(-(y/Sqrt[2])) Sin[t - y/Sqrt[2]])/
Sqrt[2]) (-(Sqrt[2] - 2 y) Cos[t] +
Sqrt[2] E^(-(y/Sqrt[2])) (Cos[t - y/Sqrt[2]] -
E^(y/Sqrt[2]) Sin[t] + Sin[t - y/Sqrt[2]]));
CoefficientList[FourierTrigSeries[TrigExpand@e1, t, 5] /. {Cos[a_*t] :> x^a, Sin[a_*t] :> z^a}, {x, z}]
After obtaining the Fourier series you can transform the trig polynomial to a usual polynomials and then extract the coefficients.
I think you can save steps.
• Interesting idea. Are there dangers in replacing a trig with an exponent? I will try this out. Thanks – Hugh Nov 9 '17 at 15:41
|
2021-03-07 06:30:33
|
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|
http://tex.stackexchange.com/questions/151939/angles-in-chemfig
|
# Angles in Chemfig
I want to draw this formula using chemfig:
This is what I've got so far:
\chemfig{Cu^+(-[:180]N**6(-**6(------)--(-OOC-)---))(-[1]N)(-[5]N)(-[7]N)}
I don't get the angle from N to Cu+ right. I tried to rotate parts of the formula, but there is always at least one angle that does not fit.
Any idea?
-
Welcome to TeX.SX! – strpeter Dec 31 '13 at 15:34
Yes, I have an idea:
\documentclass{article}
\usepackage{chemfig}
\begin{document}
\definesubmol\cc{**6(---!\ff-!\ee--)}
\definesubmol\dd{**6(-!\ee--!\ff---)}
\definesubmol\ee{**6(-----)}
\definesubmol\ff[(-^{-}OOC)]{(-COO^{-})}
\chemfig{Cu^+(-[1]N([0,.5]!\cc))(-[3]N([:180,.5]!\dd))(-[5]N([:180,.5]!\cc))(-[7]N([:0,.5]!\dd))}
\end{document}
-
The following code is similar to what unbonpetit wrote. The term .5 in [0,.5] and [180,.5] is chosen such that the aromatic cycles do not get too big.
\documentclass{standalone}
\usepackage{chemfig}
\begin{document}
\chemfig{Cu^+
(-[:+ 45]N([0,.5]**6(---(-COO^{-})-**6(------)--)))
(-[:+135]N([180,.5]**6(-**6(------)--(-^{-}OOC)---)))
(-[:+225]N([180,.5]**6(---(-^{-}OOC)-**6(------)--)))
(-[:+315]N([0,.5]**6(-**6(------)--(-COO^{-})---)))}
\end{document}
-
@unbonpetit: I don't see the point... – strpeter Dec 31 '13 at 16:28
It is not exactly the same since your inner ring has 6 bonds instead of 5. Moreover, your code is not more or less explicit than mine, even after several edits. – unbonpetit Dec 31 '13 at 16:29
I must admit that the term "explicit" is misleading. – strpeter Dec 31 '13 at 16:37
|
2016-02-11 02:48:03
|
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http://www.maths.usyd.edu.au/s/scnitm/rkuli-MacquarieSeminarOnStochas
|
SMS scnews item created by Rafal Kulik at Wed 7 Mar 2007 1213
Type: Seminar
Distribution: World
Expiry: 12 Mar 2007
Calendar1: 12 Mar 2007 1100-1200
CalLoc1: Macquarie E4A523
Auth: rkuli(.ststaff;2434.3001)@p818.pc.maths.usyd.edu.au
# Macquarie Seminar on Stochastic Finance : Prof. Reiss -- Using Covariate Information in Extreme Value Models
At the beginning we recall basic facts about the well-known linear
regression
problem in a fixed and a random design. It is indicated that the
random
design problem conditioned on the independent variate (the covariate)
can
be regarded as a fixed design problem. Afterwards the considerations
are
merely based on the conditional statistical model. In contrary to the
foregoing
models we deal with parametric models such as the EV model. Based
on a conditional maximum likelihood estimator one obtains estimators
and
predictions of conditional functional parameters. The required
modifications
for gP distributions are deduced within a Poisson process framework.
Applications in environmental statistics and statistical finance are
indicated.
If you are registered you may mark the scnews item as read.
School members may try to .
|
2018-03-18 19:32:45
|
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|
http://math.stackexchange.com/questions/253048/proving-or-disproving-regularity-of-a-language
|
# Proving or disproving regularity of a language
The question is as follows:
If L1 and L2 are not regular and L1 ⊆ L ⊆ L2, then L is regular
My intuition says that it's wrong so I've been looking for a counterexample, so far I didn't succeed.
Can I please get a direction? is this claim might be true?
What about $L_1 = L_2$? – Hendrik Jan Dec 7 '12 at 12:57
If $L_1=L_2$ is not regular, then $L=L_1$ saisfies all conditions, and cannot be regular. If that is not what you need or want, please rephrase the question. You might want to add "for all" or "there exist". – Hendrik Jan Dec 7 '12 at 13:02
If you take $L_1=L_2$ not regular, then $L=L_1$ satisfies your assumptions, but cannot be regular.
|
2016-05-04 21:40:17
|
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|
http://dev.econometricsociety.org/publications/econometrica/1991/05/01/theory-disappointment-aversion
|
# A Theory of Disappointment Aversion
https://doi.org/0012-9682(199105)59:3<667:ATODA>2.0.CO;2-7
p. 667-686
Faruk Gul
An axiomatic model of preferences over lotteries is developed. It is shown that this model is consistent with the Allais Paradox, includes expected utility theory as a special case, and is only one parameter $(\beta)$ richer than the expected utility model. Allais Paradox type behavior is identified with positive values of $\beta$. Preferences with positive $\beta$ are said to be disappointment averse. It is shown that risk aversion implies disappointment aversion and that the Arrow-Pratt measures of risk aversion can be generalized in a straight-forward manner, to the current framework.
|
2022-09-30 13:08:08
|
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|
http://www.czszsq.com/xshy/202111/t20211126_6272159.html
|
現在位置：首頁 > 學術會議
【2021.12.02-12.04 北京&线上会议】 Workshop on Algebraic and Analytic Geometry
2021-11-26 | 编辑：
The past decades have seen several major breakthroughs in our understanding of the geometry of algebraic varieties from the algebraic/arithmetic geometric side and from the complex analytic side. We wish to bring together some experts in these areas to expound the new ideas and to introduce the recent developments to graduate students, postdoctoral researchers and junior faculty members.
Due to COVID-19, we can not invite the interested participants outside Beijing to be present. The conference will be mixed -- online and offline. We will try to find the best way such that both the speakers online and offline and the participants can interact with each other.
Time:
2021.12.2-2021.12.4
Place:
MCM110&Online
Organizing Committee:
Yifei Chen (AMSS CAS)
Baohua Fu (MCM AMSS)
Jie Liu (AMSS CAS)
Wenhao Ou (AMSS CAS)
Academy of Mathematics and Systems Science, CAS
Hua Luo-Keng Center for Mathematical Sciences, CAS
Morningside Center of Sciences, CAS
Invited Speakers:
Jinxing Cai (Peking University)
Meng Chen (Fudan University)
Yi Gu (Soochow University)
Wenchuan Hu (Sichuan University)
Yong Hu (Shanghai Jiao Tong University)
Fangzhou Jin (Tongji University)
Sheng Rao (Wuhan University)
Mao Sheng (University of Science and Technology of China)
Lei Song (Sun Yat-Sen University)
Xiaotao Sun (Tianjin University)
Zhiwei Wang (Beijing Normal University)
Chuanhao Wei (Westlake University)
Jian Xiao (Tsinghua University)
Xiaokui Yang (Tsinghua University)
Zhiwei Zheng (Yanqi Lake Beijing Institute of Mathematical Sciences and Applications)
Schedule:
Date Time Speaker Title 12.02 9:30 – 10:30 Xiaotao Sun A finite dimensional proof of the Verlinde formula (online) 10:50 – 11:50 Yong Hu Algebraic threefolds of general type with small volume (online) 11:50 – 13:40 Lunch 13:40 – 14:40 Jinxing Cai Automorphisms of an irregular surface of general type acting trivially in cohomology 15:00 – 16:00 Meng Chen (online) 16:30 – 17:30 Wenchuan Hu The structure of Chow variety and recent results 12.03 9:30 – 10:30 Yi Gu 10:50 – 11:50 Fangzhou Jin Fundamental classes in motivic homotopy theory 11:50 – 13:40 Lunch 13:40 – 14:40 Sheng Rao Deformation limit of Moishezon manifolds 15:00 – 16:00 Xiaokui Yang Geometric positivity and rational connectedness 16:30 – 17:30 Zhiwei Wang On some recent progress related to the extension of quasi-plurisubharmonic functions 12.04 9:30 – 10:30 Mao Sheng Tensor product theorem for semistable parabolic $\lambda$-connections (online) 10:50 – 11:50 Chuanhao Wei Kodaira-type vanishings via Nonabelian Hodge Theory (online) 11:50 – 13:40 Lunch 13:40 – 14:40 Lei Song On the image of Hitchin morphism for algebraic surfaces 15:00 – 16:00 Jian Xiao Geometric inequalities inspired by algebraic geometry 16:30 – 17:30 Zhiwei Zheng Ball quotients in algebraic geometry
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2022-01-20 08:52:27
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https://www.researchgate.net/publication/262638178_Employing_Graphic_Programming_Unit_Technology_to_Accelerate_Numerical_Diffusion_Solver_in_2D_Multi-scale_Multi-resolution_Agent-based_Brain_Cancer_Model
|
Conference PaperPDF Available
# Employing Graphic Programming Unit Technology to Accelerate Numerical Diffusion Solver in 2D Multi-scale, Multi-resolution Agent-based Brain Cancer Model
Authors:
Accelerate numerical diffusion solver of 2D multi-
scale and multi-resolution agent-based brain
cancer model by employing graphics processing
unit technology
[BIOCOMP]
Beini Jiang1
1Department of Mathematical
Sciences
Michigan Tech University
Houghton, MI, USA
beinij@mtu.edu
Allan Struthers1
1Department of Mathematical
Sciences
Michigan Tech University
Houghton, MI, USA
Le Zhang1*
1Department of Mathematical
Sciences
Michigan Tech University
Houghton, MI, USA
zhangle@mtu.edu
Michael E Berens2
2Cancer and Cell Biology Division
Translational Genomics Research
Institute, TGen
Phoenix, AZ, USA
Wen Zhang1
1Department of Mathematical
Sciences
Michigan Tech University
Houghton, MI, USA
Xiaobo Zhou3
3Center for Bioinformatics and
Department of Pathology
The Methodist Hospital
Research Institute & Weill Cornell
Medical College
Houston, Texas, USA
AbstractDiffusion model is increasingly employed to simulate
diffusion of biological compounds including nutrient, oxygen and
chemoattractants in the agent-based model (ABM). However, it
takes long compute time to employ conventional numerical
methods such as alternating direction implicit (ADI) method to
approximate the exact solution of the diffusion processed by
sequential computing algorithm. To overcome this limitation, our
study employs cutting-edge graphics processing unit (GPU)
technology to speed up the conventional sequential numerical
solver for diffusion and incorporates our proposed parallel
computing algorithms into our well developed 2D multi-scale and
multi-resolution agent-based brain cancer model to break
through the bottleneck of the ABM that it is hard to simulate the
large system restricted to the limited compute resource and
memory. Our simulation outputs demonstrate that ABM model
can be used to simulate real-time actual cancer progression with
relative fine grids by using GPU based parallel computing
algorithm.
Keywords: graphics processing unit; agent-based model;
alternating direction implicit method; domain decomposition;
parallel computing
I. INTRODUCTION
Agent-based model (ABM) has become a popular method to
describe the complex dynamic, adaptive and self-organizing
cancer system. For example, Mansury and Deisboeck [1, 2]
employed the ABM to simulate the expansion of brain tumor
in micro-macro environments. And Zhang et al. [3-6]
developed multi-scale ABMs to model the growth of glioma
and investigate incoherent relations of the tumor expansion
among macroscopic environment, microscopic environment
and molecular environments. A diffusion module is employed
to simulate the diffusion of the chemoattractants on the
macroscopic scale environment.
Though conventional finite difference numerical methods
such as ADI, Gauss–Seidel and Jocobi methods [7-9] for
diffusion module already have been used to simulate diffusion
of biological compounds such as nutrients, oxygen and
chemoattractants [3, 10-15] for years, they all depend on the
grid size so much that a relative fine grids can better mimic
the diffusion process but significantly increase the compute
time. Therefore, previous studies such as the work done by
Athale et al. [10, 11] and Wang et al.[16] have to employ
relatively coarse grids to reduce the compute time and the
work done by Dai et al.[17] and Zhang et al.[18-20] employed
special numerical scheme such as preconditioned Richardson
method [21, 22] to sacrifice the compute accuracy in some
dimensions of coordinates to reduce the compute resource
request due to the specific aim of these biomedical projects.
Nonetheless, our well developed 2D multi-scale and multi-
resolution ABM model needs such a fast diffusion module that
not only can accurately model the diffusion process but also
costs less compute resource. For this reason, using parallel
computing algorithm to speed up the conventional numerical
solver [23, 24] is the best promising solution. Quite a few
previous parallel computing algorithms employed Message
Passing Interface (MPI) [25], a parallel computing scheme
based on multiple instruction multiple data infrastructure, to
parallel the sequential numerical diffusion solver. However,
MPI is not only too expensive to be routinely used for light
computing project, but also its compute speed is limited by the
communication rate [26]. Since 2007, NVDIA keeps releasing
its graphics processing unit (GPU) and the novel Compute
Unified Device Architecture (CUDA) based on single
instruction multiple data infrastructure (SIMD). Until now,
GPU of NVDIA has been evolved into a highly parallel,
multithreaded, many core processor, with dramatic compute
ability and high memory bandwidth [27] , especially for the
recent Fermi GPU [28, 29]. Compared to MPI, GPU
computing is more affordable, portable and suitable for the
ABM simulation.
In general, the aim of this study is to incorporate the parallel
diffusion numerical solver based on latest released Fermi
GPU technology into our previous well developed multi-scale
and multi-resolution ABM model [5] to resolve its compute
capability shortage problem. The methods section introduces
the conventional numerical scheme, alternating direction
implicit (ADI) method [7, 30] and the GPU implementation
[31]. And then, we show that our parallel algorithms
significantly increase the performance when applied to the 2D
multi-scale and multi-resolution ABM [5].
II. METHODS
This section gives a brief introduction to ADI scheme [7]
with the standard domain decomposition strategy [32, 33]
followed by the description of GPU implementations.
The diffusion of the chemical cues is described by (1.a),
where the D is the diffusivity for glucose (DG=6.7×10-7cm2s1)
and TGFα (DT=5.18×10-7 cm2 s1), respectively.
==
+
=+. (1.a)
The Crank–Nicolson method approximates (1.a) by (1.b)
∆ =
∆+
∆
+
. (1.b)
where
is the numerical approximation of (,,) and
=,=,=. and denote the central
difference operators [7].
Introducing an intermediate level
modifies (1.b) into two separate difference equations with
implicit scheme, given by (2):
∆/ =
∆
+
∆
. (2.a)
∆/ =
∆
+
∆
. (2.b)
Writing =∆
∆ and =∆
∆ reduces (2) into the
Peaceman-Rachford ADI scheme [7], shown as (3)
,
+(1+)
,
=
,
+
1 −
+
,
. (3.a)
,
+1+
,
=
,
+
(1−)
+
,
. (3.b)
The right part of both equations of (3) is explicit formula and
easily parallelized, while the left part is a symmetric and
tridiagonal system of equations = to be solved with the
Thomas algorithm [7, 34].
Equation (3) could be written into a general form as (4.a)
with =0 and=0 .
++=,=1,2,…,. (4.a)
The corresponding matrix form of this tridiagonal system is
represented by (4.b)
0⋯
⋯0
0
⋮⋱
⋱⋱
⋱0
0⋯⋱⋱
⋯0⋱
=
. (4.b)
B. Thomas Algorithm
The Thomas algorithm is employed to solve (4.a). It has
two major steps. First is computing coefficients (5.a) and
(5.b) known as forward sweep. Second is using backward
substitution to get solutions as (5.c).
=
; =1
−; =2,3,…,−1.
=
; =1
−
−; =2,3,…,.
(5.a)
(5.b)
=
=−; =−1,−2,,1. (5.c)
The details of the deduction of (5) are described in Morton’s
book [7].
C. Domain Decomposition
For the boundary value problem on a large domain, the
domain decomposition method decomposes the problem into
smaller independent boundary value problems on smaller
subdomains and then employ iterative method to resolve
differences between the solutions on adjacent subdomains [32,
33]. We develop such a GPU based parallel computing
algorithm with classical alternating Schwarz method [33, 35]
that can benefit from the advantages of GPU technology. Fig.
1(b) [31] exhibits the decomposition of a 10 by 10 array with
an 8 by 8 inner array (green) and four vectors of boundary
points (red) (Fig .1(a) [31]) into 4 overlapping 6 by 6 sub-
arrays, each of which consists of a 4 by 4 inner array (green)
and four artificial internal boundaries (red). Each sub-array is
iteratively solved to make the artificial boundaries converge [7,
32, 33, 35, 36]. Here, we use the data transfer between sub-
matrix 1 and sub-matrix 2 as an example. The values of the
four inner elements on the rightmost side in sub-matrix 1 are
sent to sub-matrix 2 as the new left artificial boundary as well
as the values of the four inner elements on the leftmost side in
sub-matrix 2 are sent to sub-matrix 1 as the new right artificial
boundary until both artificial boundaries converge.
(a)
(b)
Figure 1 [31] (a) A 10 by 10 solution matrix with red to indicate boundary
elements and blue to indicate inner elements and (b) Decomposition of a 10 by
10 array into 4 overlapping 6 by 6 sub-arrays with red to indicate boundary
elements, green to indicate inner elements and the arrows to show how to
update the boundary data.
D. Parallel Computing Algorithms to Speed up the diffusion
solver
The first step of ADI is to set up the explicit scheme, shown
as the right part of (3). Since each element could be computed
independently, the explicit scheme is easy to be parallelized
of CUDA. The second step is to solve the implicit scheme of
ADI by Thomas algorithm. As we discussed in our previous
research [31], Thomas algorithm is the bottleneck to speed up
the conventional numerical diffusion solver.
CUDA programming has two major steps. The first is
preparing such data that can be paralleled in the host side
(CPU). The second is processing these data in the device side
(GPU) by kernel. CUDA organizes the threads into a two-level
hierarchy (Fig. 2-1 of NVIDIA CUDA Programming Guide
[27]). As shown by Fig. 2-2 of NVIDIA CUDA Programming
device’s (GPU) DRAM and on-chip memory through 6
different memory spaces such as registers, local memory,
shared memory, global memory, constant memory, and texture
memory [27, 37-40]. As a very important memory of GPU,
global memory is in charge of exchanging the data between
the host (CPU) and the device (GPU). Moreover, it plays such
a role that passes the messages between the threads from
different blocks, since current GPU infrastructure prohibits the
communication of threads from different blocks [27, 29, 41].
However, as an off-chip memory, the latency of global
memory is very high. As on-chip memory, shared memory,
registers, and constant-memory caches are much faster with
much lower latency. Nonetheless, shared memory is very
limited and it is only allocated to each block. For example, the
[28, 42]. Moreover, another on-chip memory, constant
memory, is disallowed to be written to during the computation
[27, 43] though it is cashed.
CUDA uses a new architecture called SIMT to mange
threads running different programs. The multiprocessor SIMT
unit creates, manages, schedules, and executes threads in
groups of 32 parallel threads we call warps [27]. We have
developed three parallel computing algorithms to accelerate
the numerical diffusion solver based on the new features of
GPU technology [31]. The first is parallel computing
algorithm with global memory (PGM), which employs only
global memory to carry out parallel computing. The second is
parallel computing algorithm with shared memory, global
memory and CPU synchronization [27, 29, 41, 44] (PSGMC)
and the third is parallel computing algorithm using shared
memory, global memory and GPU synchronization [29, 41, 45]
(PSGMG). PSGMC and PSGMG employ “tiles” strategy to
partition the data and take advantages of both global memory
and shared memory with the classical alternating Schwarz
domain decomposition method [7, 32, 33, 35, 36]. The details
of these three implementation methods are presented in our
recent publication [31]. Here, we incorporate our fastest
parallel diffusion solver into 2D multi-scale and multi-
resolution ABM [5] to speed up the computation of ABM.
III. RESULTS
Our source code is implemented by C [46, 47] and NVCC
[48] programming language and running on the recent Fermi
GPU card (GeForce GTX 480) [42, 49, 50] with CUDA
standard.
In the beginning, let us briefly show how to use parallel
computing algorithms [31] based on GPU technology to
accelerate the numerical diffusion solver as following.
First, we employ PGM to compute the diffusion on the
lattice with different number of grid points and compare the
computing time with the sequential computing. Fig. 2 shows
PGM computing time is not always faster than sequential
algorithm for the lattice with small point number but
dramatically faster than sequential algorithm for the lattice
with large point number [31].
Figure 2 [31]. Computing time of PGM and sequential computing by
logarithmic scale. The x axis represents the inner matrix size (number of inner
grid points) and y axis represents the computing time (logarithmic scale with
base 10) in millisecond. The blue bar represents the computing time of
sequential computing and the red bar represents the computing time of PGM.
Second, we compare the compute time between PSGMC
and PGM, when simulating the diffusion on a 4098 by 4098
lattice. Fig. 3 shows PSGMC improves the performance by 58%
compared with PGM [31].
Figure 3 [31]. Computing time of PSGMC and PGM by logarithmic scale.
The y axis represents the computing time (logarithmic scale with base 10) in
millisecond. The blue bar represents the computing time of PGM and the red
bar represents the optimal computing time of PSGMC. The number on each
bar indicates the multiple of acceleration to the sequential computing.
Third, we compare the performance of PSGMC and
PSGMG. Fig. 4 exhibits PSGMG improves the performance
by 11% compared with PSGMC, when processing the
diffusion on a 4098 by 4098 lattice [31].
Figure 4 [31]. Computing time of PSGMG and PSGMC by logarithmic
scale. The y axis represents the computing time (logarithmic scale with base
10) in millisecond. The blue bar represents the computing time of PSGMC
and the red bar represents the computing time of PSGMG. The number on
each bar indicates the multiple of acceleration to the sequential computing.
Next, we incorporate the fastest parallel computing method
(PSGMG) into the well developed multi-scale and multi-
resolution ABM model [5]. The multi-resolution model is
designed based on two different resolution lattices, namely
low-resolution lattice and high-resolution lattice. The low-
resolution lattice is set up with a grid size of about 62.5 ,
on each grid point of which, a 6 by 6 high-resolution lattice
with a grid size of approximately 10 is superimposed,
described by Fig. 5 [5]. To demonstrate the advantages of the
parallel computing algorithm, we scale up the lattice size of
the previous multi-scale and multi-resolution ABM model [5].
Current low-resolution lattice is changed from 100 by 100 to
683 by 683 and high-resolution lattice is upgraded from 600
by 600 to 4098 by 4098.
Figure 5 [5] Configuration of multi-resolution lattice.
The diffusion of the chemical cues is observed on the high-
resolution lattice, with a grid size of approximately 10,
namely both ∆ and ∆ in the ADI scheme (2) are equal to
10. ∆ is set to 1s to make max
,
1 regarding to
the maximum principle [7], thus the ADI scheme needs to be
computed 3600 times for each time step, which is equivalent
to 1h.
And then, Fig. 6 exhibits that parallel computing can
significantly increase the performance of the compute time
37.5 folders than sequential computing for multi-scale and
multi-resolution ABM model [5].
1
10
100
1000
10000
100000
Computing Time (log)
Size of Inner Matrix
Sequential Computing
PGM
29.5 46.6
1
10
100
1000
10000
Computing TIme (log)
PGM PSGMC
46.6 51.7
1
10
100
1000
Computing TIme (log)
PSGMC PSGMG
Figure 6. Computing time of parallel and sequential computing by
logarithmic scale. The y axis represents the computing time (logarithmic scale
with base 10) in millisecond. The blue bar represents the computing time of
sequential computing and the red bar represents the optimal computing time
of parallel computing. The number on the red bar indicates the multiple of
acceleration to the sequential computing.
IV.
CONCLUSIONS
This study demonstrates that it is possible to simulate the
real-time actual tumor progression in a 2D lattice with relative
fine grids by using GPU based parallel computing algorithms.
Our extension research will develop a GPU based parallel
ODE solver to speed up the molecular pathway module of our
well developed multi-scale and multi-resolution agent-based
model [5].
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Article
Modeling diffusion processes, such as drug deliver, bio-heat transfer, and the concentration change of cytokine for computational biology research, requires intensive computing resources as one must employ sequential numerical algorithms to obtain accurate numerical solutions, especially for real-time in vivo 3D simulation. Thus, it is necessary to develop a new numerical algorithm compatible with state-of-the-art computing hardware. The purpose of this article is to integrate the graphics processing unit (GPU) technology with the locally-one-dimension (LOD) numerical method for solving partial differential equations, and to develop a novel 3D numerical parallel diffusion algorithm (GNPD) in cylindrical coordinates based on GPU technology, which can be used in the neuromuscular junction research. To demonstrate the effectiveness and efficiency of the obtained GNPD algorithm, we employed it to approximate the real diffusion of the neurotransmitter through a disk shaped volume. This disk shaped volume is the synaptic gap, connecting the neuron and the muscle cell in the neuromuscular junction. Furthermore, we compared the speed and accuracy of the GNPD with the conventional sequential diffusion algorithm. Results show that the GNPD can not only significantly accelerate the speed of the diffusion solver via GPU-based parallelism, but also greatly increase the accuracy by employing the stream function of latest FermiGPU cards. Therefore, the GNPD has a great potential to be employed in the design, testing, and implementation of health information systems in the near future.
Chapter
As noted in the last two chapters, many ODEs and systems of ODEs cannot be solved analytically. When we want solutions to such equations, we need to turn to numerical methods, which we now take up. We begin with a simple DE that can also be solved analytically by separation of variables1; i.e., $$\frac{{dy}} {{dx}} = xy,y(1) = 1,$$ This will allow comparison of the approximate numerical solution with the exact analytic solution, which is y = exp[(x2− 1)/2].
Article
Gliomas are brain tumours that differ from most other cancers by their diffuse invasion of the Surrounding normal tissue and their notorious recurrence following all forms of therapy. We have developed a mathematical model to quantify the spatio-temporal growth and invasion of gliomas in three dimensions throughout a virtual human brain. The model quantifies the extent of tumorous invasion of individual gliomas in three-dimensions to a degree beyond the limits of present medical imaging, including even microscopy, and makes clear why current therapies based on existing imaging techniques are inadequate and cannot be otherwise without other methods for detecting tumour cells in the brain. The model's estimate of the extent of tumourous invasion beyond that defined by standard medical imaging can be useful in more accurately planning therapy regimes as well as predicting sites of potential recurrence without waiting for reemergence on follow-up imaging.
Article
In recent years, there has been interest in research related to hyperthermia combined with radiation and cytotoxic drugs to enhance the killing of tumors. The objective is to control laser heating of the tumor so that the temperature of the normal tissue surrounding the tumor remains low enough so as not to cause damage to the tissue. To achieve this objective, it is important to obtain an optimal temperature field of the entire treatment region. In this paper, we develop a numerical algorithm for obtaining an optimal temperature distribution in a 3D triple layered cylindrical skin structure by pre-specifying the temperatures to be obtained at the center and perimeter of the treated region on the skin surface. The method is comprised of designing a laser irradiation pattern, solving a 3D Pennes’ bioheat equation by a numerical scheme, and optimizing the laser power.
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This second edition of a highly successful graduate text presents a complete introduction to partial differential equations and numerical analysis. Revised to include new sections on finite volume methods, modified equation analysis, and multigrid and conjugate gradient methods, the second edition brings the reader up-to-date with the latest theoretical and industrial developments. First Edition Hb (1995): 0-521-41855-0 First Edition Pb (1995): 0-521-42922-6
Article
It is of interest to research hyperthermia combined with radiation and cytotoxic drugs to enhance the killing of tumors. The crucial problem is to keep surrounding normal tissue below a temperature that will produce harm when heating the tumor tissue. In this study, we develop a numerical model for optimizing laser power in the irradiation of a 3-D triple-layered cylindrical skin structure. The method determines the required laser intensity in order to obtain prespecified temperatures at given locations on the skin surface after a prespecified laser exposure time.
Article
In hyperthermia cancer treatments, the crucial problem, when heating the tumor tissue, is keeping the temperature of the normal tissue surrounding the tumor below a certain threshold so as not to cause damage to the tissue. Thus, it is important to optimize the temperature field of the entire treatment region. Recently, we have developed a numerical method for obtaining an optimal temperature distribution in a triple-layered cylindrical skin structure. In this article, we extend our method to a case involving a triple-layered cylindrical skin structure embedded with a blood vessel. The method is illustrated by a numerical example.
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2022-07-02 15:36:39
|
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|
http://www.statsmodels.org/dev/generated/statsmodels.discrete.discrete_model.Poisson.loglikeobs.html
|
# statsmodels.discrete.discrete_model.Poisson.loglikeobs¶
Poisson.loglikeobs(params)[source]
Loglikelihood for observations of Poisson model
Parameters: params (array-like) – The parameters of the model. loglike – The log likelihood for each observation of the model evaluated at params. See Notes array-like
Notes
$\ln L_{i}=\left[-\lambda_{i}+y_{i}x_{i}^{\prime}\beta-\ln y_{i}!\right]$
for observations $$i=1,...,n$$
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2019-04-20 06:36:11
|
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https://www.researchgate.net/institution/Technische-Universitaet-Dortmund
|
# Technische Universität Dortmund
• Dortmund, North Rhine-Westphalia, Germany
Recent publications
Taking actual engineering as a case study, this article provides a novel scheme of applying the inductive power filtering method (IPFM) to resolve the harmonic resonance issues in large-scale photovoltaic (PV) plant. By using IPFM's special structure and dual zero-impedance design, the impedance network of large-scale PV plant is reshaped while the performance of power filters is improved, so that the harmonic resonance due to the interaction between the inverters and the power grid is suppressed. First, the topology and components of the IPFM-based large-scale PV plant are introduced. The three-phase mathematical model in harmonic domain is then established. Based on the deduced transfer matrix, the simplified circuit of the studied large-scale PV plant is obtain-ed. Moreover, the IPFM's resonance damping mechanisms on resonant frequency shift and harmonic amplification mitigation are analyzed. Both simulation and engineering tests verify the feasibility of the proposal.
Enzyme-catalyzed replication of nucleic acid sequences is a prerequisite for the survival and evolution of biological entities. Before the advent of protein synthesis, genetic information was most likely stored in and replicated by RNA. However, experimental systems for sustained RNA-dependent RNA-replica-tion are difficult to realise, in part due to the high thermodynamic stability of duplex products and the low chemical stability of catalytic RNAs. Using a derivative of a group I intron as a model for an RNA replicase, we show that heated air-water interfaces that are exposed to a plausible CO 2-rich atmosphere enable sense and antisense RNA replication as well as template-dependent synthesis and catalysis of a functional ribozyme in a one-pot reaction. Both reactions are driven by autonomous oscillations in salt concentrations and pH, resulting from precipitation of acidified dew droplets, which transiently destabilise RNA duplexes. Our results suggest that an abundant Hadean microenvironment may have promoted both replication and synthesis of functional RNAs.
In the fields of business process modeling, logistics, and information model development, Reference Models (RMs) have shown to enhance standardization, support the common understanding of terminology and procedures, reduce the modeling efforts and cost through the paradigm “Design by Reuse”, and enable knowledge transfer. Utilizing RMs in Building Performance Simulation (BPS) shows potential to achieve similar benefits. However, there is no universally agreed understanding of RMs. In a previous scientific publication, we provided a comprehensive overview of the diversely interpreted definitions, benefits, and attributes of RMs and related terms. Additionally, to transfer the approach of RMs to BPS, a definition for RMs applicable to BPS has been provided, and the identified RM qualities were matched with BPS’s challenges. However, a sound evaluation of the success of transferring RMs to BPS is lacking. Therefore, this scientific contribution firstly includes the analysis conducted in the previous scientific contribution constituting a common understanding about RMs and their elements for BPS. Secondly, by conducting expert interviews, the applicability and validity of the developed concept of RMs for BPS are surveyed. In total, ten experts (seven BPS experts and three RM experts) evaluated the quality of creating transparency about the understanding of RMs and the level of success of their transfer toward BPS. The experts consistently see a great benefit of RMs in BPS, but for BPS experts the transfer and possible application of RMs in BPS is not sufficiently clear. Accordingly, the key output of the conducted survey is that a clearer and more detailed application example, e.g., describing at a more easy-to-understand level of detail an exemplary class of the provided example of an RM, is required for a more profound transfer of RMs to BPS.
The main goal of this study is to develop an experimental toolbox to estimate the self-diffusion coefficient of active ingredients (AI) in single-phase amorphous solid dispersions (ASD) close to the glass transition of the mixture using dielectric spectroscopy (DS) and oscillatory rheology. The proposed methodology is tested for a model system containing the insecticide imidacloprid (IMI) and the copolymer copovidone (PVP/VA) prepared via hot-melt extrusion. For this purpose, reorientational and the viscoelastic structural (α-)relaxation time constants of hot-melt-extruded ASDs were obtained via DS and shear rheology, respectively. These were then utilized to extract the viscosity as well as the fragility index of the dispersions as input parameters to the fractional Stokes-Einstein (F-SE) relation. Furthermore, a modified version of Almond-West (AW) formalism, originally developed to describe charge diffusion in ionic conductors, was exercised on the present model system for the estimation of the AI diffusion coefficients based on shear modulus relaxation times. Our results revealed that, at the calorimetric glass-transition temperature (Tg), the self-diffusion coefficients of the AI in the compositional range from infinite dilution up to 60 wt % IMI content lied in the narrow range of 10-18-10-20 m2 s-1, while the viscosity values of the dispersions at Tg varied between 108 Pa s and 1010 Pa s. In addition, the phase diagram of the IMI-PVP/VA system was determined using the melting point depression method via differential scanning calorimetry (DSC), while mid-infrared (IR) spectroscopy was employed to investigate the intermolecular interactions within the solid dispersions. In this respect, the findings of a modest variation in melting point at different compositions stayed in agreement with the observations of weak hydrogen bonding interactions between the AI and the polymer. Moreover, IR spectroscopy showed the intermolecular IMI-IMI hydrogen bonding to have been considerably suppressed, as a result of the spatial separation of the AI molecules within the ASDs. In summary, this study provides experimental approaches to study diffusivity in ASDs using DS and oscillatory rheology, in addition to contributing to an enhanced understanding of the interactions and phase behavior in these systems.
Proteins can specifically bind to curved membranes through curvature-induced hydrophobic lipid packing defects. The chemical diversity among such curvature "sensors" challenges our understanding of how they differ from general membrane "binders" that bind without curvature selectivity. Here, we combine an evolutionary algorithm with coarse-grained molecular dynamics simulations (Evo-MD) to resolve the peptide sequences that optimally recognize the curvature of lipid membranes. We subsequently demonstrate how a synergy between Evo-MD and a neural network (NN) can enhance the identification and discovery of curvature sensing peptides and proteins. To this aim, we benchmark a physics-trained NN model against experimental data and show that we can correctly identify known sensors and binders. We illustrate that sensing and binding are phenomena that lie on the same thermodynamic continuum, with only subtle but explainable differences in membrane binding free energy, consistent with the serendipitous discovery of sensors.
Much of Earth’s carbon may have been stripped away from the silicate mantle by dense metallic-iron during core formation. However, at deep magma ocean conditions carbon becomes less siderophile and thus large amounts of it may be stranded instead in the deep mantle. Here, we describe the structure and compaction mechanisms of carbonate glass to deep mantle pressures. Our results, based on non-resonant inelastic X-ray scattering, X-ray diffraction and ab initio calculations, demonstrate a pressure-induced change in hybridization of carbon from sp² to sp³ starting at 40 GPa, due to the conversion of [3]CO3²⁻ groups into [4]CO4⁴⁻ units, which is completed at ~112 GPa. The pressure-induced change of carbon coordination number from three to four increases possibilities for carbon-oxygen interactions with lower mantle silicate melts. sp³ hybridized carbon provides a mechanism for changing the presumed siderophile nature of deep carbon, becoming a possible source for carbon-rich emissions registered at the surface in intra-plate and near-ridge hot spots.
Teacher students’ professional educational knowledge is of great importance in academic teacher education. In response to the need to continuously optimize and improve teacher education, we developed a standards-based test instrument designed along the Standards of Teacher Education of the German education administration. The so-called ESBW (Essen Test for the Assessment of Standards-Based Educational Knowledge) is intended to assess educational knowledge as it is defined in these standards. This Brief Report aims to investigate whether the ESBW, as an exclusively standards-based developed test, can empirically be distinguished from a similar, but non-originally standards-based developed test, here the BilWiss 2.0 test, which also partially covers the standards. Competing structural equation models based on a study with 216 teacher students revealed that the ESBW short scale can be empirically distinguished from the BilWiss 2.0 short version, indicating that both instruments partly measure different aspects of educational knowledge. In addition, the examination of measurement invariance revealed that the ESBW performed similarly well for both beginning and advanced teacher students. Thus, our results further underline the usefulness of the ESBW for the assessment and evaluation of the German Standards of Teacher Education.
We develop a model-based boosting approach for multivariate distributional regression within the framework of generalized additive models for location, scale, and shape. Our approach enables the simultaneous modeling of all distribution parameters of an arbitrary parametric distribution of a multivariate response conditional on explanatory variables, while being applicable to potentially high-dimensional data. Moreover, the boosting algorithm incorporates data-driven variable selection, taking various different types of effects into account. As a special merit of our approach, it allows for modeling the association between multiple continuous or discrete outcomes through the relevant covariates. After a detailed simulation study investigating estimation and prediction performance, we demonstrate the full flexibility of our approach in three diverse biomedical applications. The first is based on high-dimensional genomic cohort data from the UK Biobank, considering a bivariate binary response (chronic ischemic heart disease and high cholesterol). Here, we are able to identify genetic variants that are informative for the association between cholesterol and heart disease. The second application considers the demand for health care in Australia with the number of consultations and the number of prescribed medications as a bivariate count response. The third application analyses two dimensions of childhood undernutrition in Nigeria as a bivariate response and we find that the correlation between the two undernutrition scores is considerably different depending on the child's age and the region the child lives in.
We present a protocol to evaluate the utility of detergents for purification and delipidation of E. coli membrane proteins. We determine the critical aggregation concentration of detergents. Furthermore, we compare the ability of detergents to extract membrane proteins and to maintain protein-lipid interactions during purification. The protocol describes steps for isolating and delipidating membrane proteins from E. coli membranes by extraction and affinity purification using detergents. The protocol does not enable an absolute quantification of purification outcomes. For complete details on the use and execution of this protocol, please refer to Urner et al.1.
Mixtures of 60% SN (succinonitrile) and 40% GN (glutaronitrile) doped with LiTFSI or LiPF6 at different concentrations are investigated using dielectric spectroscopy. Room temperature conductivities up to 10-3 S cm-1 are measured along with an overall conductivity enhancement of almost five decades compared to pure SN. Additionally, the dynamics of the methylene (CD2) groups of SN and that of the Li+ ions within the mixture are studied in a wide temperature range using 2H and 7Li NMR relaxometry, respectively. Static-field-gradient proton NMR combined with viscosity measurements probe the molecular diffusion. GN addition and Li doping both enhance the electrical conductivity significantly, while leaving the reorientational motion within the matrix essentially unchanged. The times scales and thus the effective energy barriers characterizing the Li ion motion as well as the molecular reorientations are very similar in the liquid and in the plastic phases, findings that argue in favor of the presence of a paddle-wheel mechanism.
Oxindoles and iso-oxindoles are natural product-derived scaffolds that provide inspiration for the design and synthesis of novel biologically relevant compound classes. Notably, the spirocyclic connection of oxindoles with iso-oxindoles has not been explored by nature but promises to provide structurally related bioactive compounds endowed with novel bioactivity. Therefore, methods for their efficient synthesis and the conclusive discovery of their cellular targets are highly desirable. We describe a selective Rh(III)-catalyzed scaffold-divergent synthesis of spirooxindole-isooxindoles and spirooxindole-oxindoles from differently protected diazooxindoles and N-pivaloyloxy aryl amides which includes a functional group-controlled Lossen rearrangement as key step. Unbiased morphological profiling of a corresponding compound collection in the Cell Painting assay efficiently identified the mitotic kinesin Eg5 as the cellular target of the spirooxindoles, which defines a unique Eg5 inhibitor chemotype.
This paper investigates digital firm birth activity in municipalities in the urban hinterland of core cities in Germany. It conducts panel fixed-effect regressions for monocentric and polycentric urban labour market regions covering the years 1995–2017. The digital industry’s regional distribution is shaped significantly by the closest core cities: municipalities in monocentric urban regions (MURs) profit from urban population growth and universities’ general knowledge. Municipalities in polycentric urban regions (PURs), however, are affected by industry-specific externalities, that is, an above-average growth in the share of firm birth of their closest urban cores. Overall, agglomeration externalities experience spatial decay relative to the core size with all regions benefiting from their own industry-enhancing urbanization externalities as captured by population growth and universities.
5-methylcytosine (mC) and its TET-oxidized derivatives exist in CpG dyads of mammalian DNA and regulate cell fate, but how their individual combinations in the two strands of a CpG act as distinct regulatory signals is poorly understood. Readers that selectively recognize such novel 'CpG duplex marks' could be versatile tools for studying their biological functions, but their design represents an unprecedented selectivity challenge. By mutational studies, NMR relaxation, and MD simulations, we here show that the selectivity of the first designer reader for an oxidized CpG duplex mark hinges on precisely tempered conformational plasticity of the scaffold adopted during directed evolution. Our observations reveal the critical aspect of defined motional features in this novel reader for affinity and specificity in the DNA/protein interaction, providing unexpected prospects for further design progress in this novel area of DNA recognition.
Ensembles are among the state-of-the-art in many machine learning applications. With the ongoing integration of ML models into everyday life, e.g., in the form of the Internet of Things, the deployment and continuous application of models become more and more an important issue. Therefore, small models that offer good predictive performance and use small amounts of memory are required. Ensemble pruning is a standard technique for removing unnecessary classifiers from a large ensemble that reduces the overall resource consumption and sometimes improves the performance of the original ensemble. Similarly, leaf-refinement is a technique that improves the performance of a tree ensemble by jointly re-learning the probability estimates in the leaf nodes of the trees, thereby allowing for smaller ensembles while preserving their predictive performance. In this paper, we develop a new method that combines both approaches into a single algorithm. To do so, we introduce L1\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$L_1$$\end{document} regularization into the leaf-refinement objective, which allows us to jointly prune and refine trees at the same time. In an extensive experimental evaluation, we show that our approach not only offers statistically significantly better performance than the state-of-the-art but also offers a better accuracy-memory trade-off. We conclude our experimental evaluation with a case study showing the effectiveness of our method in a real-world setting.
Synthesis platforms are of particular interest to DNA-encoded library (DEL) technologies to facilitate chemistry development, building block validation, and high-throughput library synthesis. A liquid–liquid two-phase flow reactor was designed that enables parallel conduction of reactions on DNA-coupled substrates. The dispersed phase in capillary slug flow contained the DNA reaction mixture and allowed for spatially separated batch experiments in a microchannel. A coiled flow inverter (CFI) tubular reactor with a 3D-printed internal structure on which a capillary is coiled was used for improved mixing and compact setup. An inert continuous phase was introduced, which generated slug flow and prevented backmixing of the individual reactants. In order to enable parallelized reactions, slugs containing a variety of different carboxylic acids were successfully generated to act as individual reaction compartments representing single batch experiments. As a widely used exemplary DEL reaction, the amide coupling reaction was successfully transferred to the tailored flow reaction system and DNA was recovered.
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• Faculty of Mechanical Engineering, Chair of Industrial Information Management
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Information
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2023-03-23 21:52:48
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https://www.ias.ac.in/listing/articles/pram/061/01
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• Volume 61, Issue 1
July 2003, pages 1-186
• Isotropic averaging for cell-dynamical-system simulation of spinodal decomposition
Formulae have been developed for the isotropic averagings in two and three dimensions. Averagings are employed in the cell-dynamical-system simulation of spinodal decomposition for inter-cell coupling. The averagings used in earlier works on spinodal decomposition have been discussed.
• Thermal state of the general time-dependent harmonic oscillator
Taking advantage of dynamical invariant operator, we derived quantum mechanical solution of general time-dependent harmonic oscillator. The uncertainty relation of the system is always larger than ħ/2 not only in number but also in the thermal state as expected. We used the diagonal elements of density operator satisfying Leouville-von Neumann equation to calculate various expectation values in the thermal state. We applied our theory to a special case which is the forced Caldirola-Kanai oscillator.
• Study of solar features causing GMSs with 250γ<H<400γ
The effect of solar features on geospheric conditions leading to geomagnetic storms (GMSs) with planetary index,AP ≥ 20 and the range of horizontal component of the Earth’s magnetic fieldH such that 250γ <H < 400γ has been investigated using interplanetary magnetic field (IMF), solar wind plasma (SWP) and solar geophysical data (SGD) during the period 1978–99. Statistically, it is observed that maximum number of GMSs have occurred during the maximum solar activity years of 21st and 22nd solar cycles. A peculiar result has been observed during the years 1982, 1994 when sunspot numbers (SSNs) decrease very rapidly while numbers of GMSs increase. No distinct association between yearly occurrence of disturbed days and SSNs is observed. Maximum number of disturbed days have occurred during spring and rainy seasons showing a seasonal variation of disturbed days. No significant correlation between magnitude (intensity) of GMSs and importance ofHα, X-ray solar flares has been observed. Maximum number of GMSs is associated with solar flares of lower importance, i.e., SF during the period 1978-93.Hα, X-ray solar flares occurred within lower helio-latitudes, i.e., (0–30)°N to (0–30)°S are associated with GMSs. NoHα, X-ray solar flares have occurred beyond 40°N or 40°S in association with GMSs. In helio-latitude range (10–40)°N to (10–40)°S, the 89.5% concentration of active prominences and disappearing filaments (APDFs) are associated with GMSs. Maximum number of GMSs are associated with solar flares. Coronal mass ejections (CMEs) are related with eruptive prominences, solar flares, type IV radio burst and they occur at low helio-latitude. It is observed that CMEs related GMS events are not always associated with high speed solar wind streams (HSSWSs). In many individual events, the travel time between the explosion on the Sun and maximum activity lies between 58 and 118 h causing GMSs at the Earth.
• Single decay-lepton angular distributions in polarizede+e →$$\overline {tt}$$ and simple angular asymmetries as a measure of CP-violating top dipole couplingsand simple angular asymmetries as a measure of CP-violating top dipole couplings
In the presence of an electric dipole coupling of$$\overline {tt}$$ to a photon, and an analogous ‘weak’ dipole coupling to the Z, CP violation in the process e+e →$$\overline {tt}$$ results in modified polarization of the top and the anti-top. This polarization can be analyzed by studying the angular distributions of decay charged leptons when the top or anti-top decays leptonically. Analytic expressions are presented for these distributions when eithert or$$\overline t$$ decays leptonically, including$$\mathcal{O}$$(αs) QCD corrections in the soft-gluon approximation. The angular distributions are insensitive to anomalous interactions in top decay. Two types of simple CP-violating polar-angle asymmetries and two azimuthal asymmetries, which do not need the full reconstruction of thet or$$\overline t$$, are studied. Independent 90% CL limits that may be obtained on the real and imaginary parts of the electric and weak dipole couplings at a linear collider operating at √ s = 500 GeV with integrated luminosity 500 fb and also at √s = 1000 GeV with integrated luminosity 1000 fb have been evaluated. The effect of longitudinal electron and/or positron beam polarizations has been included.
• Fusion, resonances and scattering in12C+12C reaction
The variation of fusion cross-section (σJfus) with energy in the12C+12C collision is linked to the underlying resonance phenomenon through the behavior of reaction cross-section (σ) of which σfus is taken as a part. The calculation of σfus is done through an energy-dependent imaginary potential in the optical model potential (OMP). Through dispersion relation, such an imaginary potential gives rise to energy-dependent real potential which is incorporated in the OMP. In our calculation, a form of potential for the nuclear part which has a soft repulsive in-built core is introduced based on similar works done earlier. The calculated results of σfus are used to explain the oscillatory structure, astrophysical S-factor and the decreasing trend at higher energies of the experimental σfus data in the case of12C+12C system with remarkable success. The potential used for fusion calculation is tested for fitting elastic scattering data at some energies and is found good in forward angles. Further improvement of the fitting of these data is obtained by incorporating a coupling potential in the surface region. About twenty resonances are observed in our calculation in the specific partial waves and some of them are found close to the experimentally identified resonances in12C+12C reaction. Thus, we provide an integrated and comprehensive analysis of fusion, resonance and scattering data in the best studied case of12C+12C reaction within the framework of optical potential model.
• Resonant spin-flavor precession constraints on the neutrino parameters and the twisting structure of the solar magnetic fields from the solar neutrino data
Resonant spin-flavor precession (RSFP) scenario with twisting solar magnetic fields has been confronted with the solar neutrino data from various ongoing experiments. The anticorrelation apparent in the Homestake solar neutrino data has been taken seriously to constrain (Δm2,φ′) parameter space and the twisting profiles of the magnetic field in the convective zone of the Sun. The twisting profiles, thus derived, have been used to calculate the variation of the neutrino detection rates with the solar magnetic activity for the Homestake, Super-Kamiokande and the gallium experiments. It is found that the presence of twisting reduces the degree of anticorrelation in all the solar neutrino experiments. However, the anticorrelation in the Homestake experiment is expected to be more pronounced in this scenario. Moreover, the anticorrelation of the solar neutrino flux emerging from the southern solar hemisphere is expected to be stronger than that for the neutrinos emerging from the northern solar hemispheres.
• Millimeterwave spectroscopy of transient molecules produced in a DC discharge
The construction of a millimeterwave spectrometer to study the pure rotational spectra of transient molecules in the gas phase is presented. The spectrometer is a source-modulated system combined with a free space glass discharge cell. Millimeterwave radiation has been produced using a frequency multiplier, the fundamental radiation source being klystrons. The spectrometer has been used to study the millimeterwave spectrum of carbon monosulfide (CS) and fluorine cyanide (FCN) produced inside the cell in a low pressure DC discharge of precursor gases. The quadrupole hyperfine structures of33S and14N nucleus of CS and FCN have been resolved, measured and analysed.
• High gain L-band erbium-doped fiber amplifier with two-stage double-pass configuration
An experiment on gain enhancement in the long wavelength band erbium-doped fiber amplifier (L-band EDFA) is demonstrated using dual forward pumping scheme in double-pass system. Compared to a single-stage single-pass scheme, the small signal gain for 1580 nm signal can be improved by 13.5 dB. However, a noise figure penalty of 2.9 dB was obtained due to the backward C-band ASE from second stage and the already amplified signal from the first pass that extracting energy from the forward C-band ASE. The maximum gain improvement of 13.7 dB was obtained at a signal wavelength of 1588 nm while signal and total pump powers were fixed at -30 dBm and 92 mW, respectively.
• Canonical structure of evolution equations with non-linear dispersive terms
The inverse problem of the variational calculus for evolution equations characterized by non-linear dispersive terms is analysed with a view to clarify why such a system does not follow from Lagrangians. Conditions are derived under which one could construct similar equations which admit a Lagrangian representation. It is shown that the system of equations thus obtained can be Hamiltonized by making use of the Dirac’s theory of constraints. The specific results presented refer to the third- and fifth-order equations of the so-called distinguished subclass.
• Jeans instability of an inhomogeneous streaming dusty plasma
The dynamics of a self-gravitating unmagnetized, inhomogeneous, streaming dusty plasma is studied in the present work. The presence of the shear flow causes the coupling between gravitational and electrostatic forces. In the absence of self-gravity, the fluctuations in the plasma may grow at the expense of the density inhomogeneity and for certain wavelengths, such an unstable mode may dominate the usual streaming instability. However, in the presence of self-gravity, the plasma inhomogeneity causes an overlap between Jeans and streaming modes and collapse of the grain will continue at all wavelengths.
• Dynamics of hydrogen in hydrogenated amorphous silicon
The problem of hydrogen diffusion in hydrogenated amorphous silicon (a-Si:H) is studied semiclassically. It is found that the local hydrogen concentration fluctuations-induced extra potential wells, if intense enough, lead to the localized electronic states in a-Si:H. These localized states are metastable. The trapping of electrons and holes in these states leads to the electrical degradation of the material. These states also act as recombination centers for photo-generated carriers (electrons and holes) which in turn may excite a hydrogen atom from a nearby Si-H bond and breaks the weak (strained) Si-Si bond thereby apparently enhancing the hydrogen diffusion and increasing the light-induced dangling bonds.
• Comparison of resonant tunneling in AlGaAs/GaAs parabolic and diffusion modified quantum wells
Double barrier resonant tunneling diode using annealing induced diffusion modified quantum well is proposed as a viable alternative to that using parabolic quantum well which requires complex techniques to fabricate it. The transmission coefficients are calculated using the hybrid incremental airy function plane wave approach. The room temperature current-voltage characteristics have been calculated using transmission coefficients. The current-voltage characteristics are found to be similar in both diodes.
• Superiority of Bessel function over Zernicke polynomial as base function for radial expansion in tomographic reconstruction
Here we describe the superiority of Bessel function as base function for radial expansion over Zernicke polynomial in the tomographic reconstruction technique. The causes for the superiority have been described in detail. The superiority has been shown both with simulated data for Kadomtsev’s model for saw-tooth oscillation and real experimental x-ray data from W7-AS Stellarator.
• Higher dimensional homogeneous cosmology in Lyra geometry
Assuming a homogeneous perfect fluid withρ =ρ(t) andp =p(t), we have obtained exact solutions for cosmological models in higher-dimension based on Lyra geometry. Depending on the form of metric chosen, the model is similar to FRW type. The explicit solutions of the scale factor are found via the assumption of an equation of statep =, where m is a constant. Some astrophysical parameters are also calculated.
• Matrix factorization method for the Hamiltonian structure of integrable systems
We demonstrate that the process of matrix factorization provides a systematic mathematical method to investigate the Hamiltonian structure of non-linear evolution equations characterized by hereditary operators with Nijenhuis property.
• Systematics of theKπ = 2 + gamma vibrational bands and odd-even staggering
The structure of theKπ= 2+ gamma vibrational bands and the quasi-gamma bands of even-Z-even-N nuclei is investigated on a global scale, vis-a-vis the variation of band head, the moment of inertia of the band and the odd-even spin staggering. The variation withN andZ and with spinJ of the odd-even spin energy staggering index is studied and a unified view of the same is presented.
• Dust-cyclotron and dust-lower-hybrid modes in self-gravitating magnetized dusty plasmas
Theoretical investigation has been made on two different ultra-low-frequency electrostatic modes, namely, dust-cyclotron mode and dust-lower-hybrid mode, propagating perpendicular to the external magnetic field, in a self-gravitating magnetized two-fluid dusty plasma system. It has been shown that the effect of the self-gravitational force, acting on both dust grains and ions, significantly modifies the dispersion properties of these two electrostatic modes. The implications of these results to some space and astrophysical dusty plasma systems, especially to planetary ring-systems and cometary tails, are briefly mentioned.
• Effect of pressure on electrical resistance of WSe2 single crystal
The results of electrical resistance measurements under pressure on single crystals of WSe2 are reported. Measurements up to 8.5 GPa are carried out using Bridgman anvil set up and beyond it using diamond anvil cell (DAC) up to a pressure of 27 GPa. There is no clear indication of any phase transition till the highest pressure is reached in these measurements.
• # Pramana – Journal of Physics
Current Issue
Volume 93 | Issue 5
November 2019
• # Editorial Note on Continuous Article Publication
Posted on July 25, 2019
Click here for Editorial Note on CAP Mode
© 2017-2019 Indian Academy of Sciences, Bengaluru.
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2019-08-25 10:04:33
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https://socratic.org/questions/how-do-you-evaluate-log-16-1
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# How do you evaluate log_[16]1 ?
Mar 22, 2016
${\log}_{16} 1 = 0$
#### Explanation:
${\log}_{16} 1$
$= \log \frac{1}{\log} 16$-> use change of base property ${\log}_{b} x = \log \frac{x}{\log} b$
$= \frac{0}{\log} 16$->$\log 1 = 0$
$= 0$
Mar 22, 2016
0
#### Explanation:
By definition $\log 1$ to any base is equal to $0$.
Proof:
Let ${\log}_{16} 1 = n$
By definition of logarithmic function
${16}^{n} = 1$, This is true only if $n = 0$
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2019-01-19 20:57:14
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http://www.crm.umontreal.ca/physmath/archives/mathematics-physics-seminar-fall-2009-winter-2010/
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Home » Archives » Mathematics Physics Seminar : Fall 2009 – Winter 2010
# Mathematics Physics Seminar : Fall 2009 – Winter 2010
### Stress-energy tensor, Schwarzian, and conformal loop ensembles
#### Benjamin Doyon, Department of Mathematical Sciences, Durham University
Conformal loop ensembles (CLE) provide a provable probability theory for the scaling or « continuum » limits of critical models, through random sets of non-intersecting loops. These scaling limits are also believed to be described by conformal field theory (CFT). I will overview my recent works on relating the two theories. It is based on the CLE construction of the stress-energy tensor, the most fundamental quantum field in CFT. I will briefly explain how the corresponding conformal Ward identities, and in general stress-energy tensor insertions, can be compactly written using the notion of conformal differentiability (a particular type of Hadamard differentiability that I developed); how such conformal derivatives are related to objects in CLE; and how from this relation and from a simple but nice lemma about conformal transformations, their transformation properties can be shown to involve the Schwarzian.
### Classification of solvable algebras with the given nilradical – can the knowledge of solvable extensions of its nilpotent subalgebra be useful?
#### Libor Snobl, Czech Technical University in Prague, Faculty of Nuclear Sciences and Physical Engine
We construct all solvable Lie algebras with a specific n- dimensional nilradical n_{n,3} which contains the previously studied filiform nilpotent algebra n_{n-2,1} as a subalgebra but not as an ideal. Rather surprisingly it turns out that the classification of such solvable algebras can be reduced to the classification of solvable algebras with the nilradical n_{n-2,1} together with one additional case. Also the sets of invariants of coadjoint representation of n_{n,3} and its solvable extensions are deduced from this reduction. In several cases they have polynomial bases, i.e. the invariants of the respective solvable algebra can be chosen to be Casimir invariants in its enveloping algebra.
### Indecomposable modules for the Virasoro algebra
#### David Ridout, CRM
Recent progress in the study of statistical models (logarithmic conformal field theory and Schramm-Loewner evolution) has led to a need to understand representation theory beyond the highest weight category. Here, we report on the classification of the simplest class of such representations of the Virasoro algebra, illustrated with examples of physical significance.
### Making sense of non-Hermitian Hamiltonians
#### Carl Bender, Physics Department, Washington University in St. Louis
The average quantum physicist believes that a quantum-mechanical Hamiltonian must be Dirac Hermitian (symmetric under combined matrix transposition and complex conjugation) so that the energy eigenvalues are real and that time evolution is unitary. However, the Hamiltonian$tex: H=p^2+ix^3$, for example, which is obviously not Dirac Hermitian, has a real positive discrete spectrum and generates unitary time evolution, and thus it defines a fully consistent quantum mechanics. Evidently, Dirac Hermiticity is too restrictive. While $tex: H=p^2+ix^3$ is not Dirac Hermitian, it is PT symmetric — symmetric under combined space reflection P and time reversal T. In general, if H is not Dirac Hermitian but has an unbroken PT symmetry, there is a procedure for determining the adjoint operation under which H is Hermitian. (One should not assume that the adjoint operation that interchanges bra and ket vectors in the Hilbert space of states is the Dirac adjoint. This would be like postulating a priori what the metric $tex:g^{\mu\nu}$ in curved space is before solving Einstein’s equations.) Non-Dirac-Hermitian PT-symmetric Hamiltonians have remarkable properties!
This talk will be presented at an elementary colloquium-style level and will be broadly accessible to both theoreticians and experimantalists.
### Group Field Theory
#### Razvan GURAU, Perimeter Institute
Group field theory is the higher-dimensional generalization of random matrix models. As it has built-in scales and automatically sums over metrics and discretizations, it provides a combinatoric origin for space time. Its graphs facilitate a new approach to algebraic topology. I exemplify this approach by introducing a graph’s cellular structure and associated homology.
### Homogeneous operators, jet construction and similarity
#### Subrata Shyam Roy, Indian Statistical Institute, Kolkata
In this talk we show, starting with the jet construction, how to construct all the irreducible homogeneous operators in the Cowen-Douglas class whose associated representations are multiplicity-free.
### Coupling constant metamorphosis and Nth order symmetries in classical and quantum mechanics
#### Sarah Post, CRM
In this talk, I will discuss coupling constant metamorphosis and the Stäckel transform, in particular their generalization to higher order symmetry operators. I will present specializations of these actions which preserve polynomial symmetry operators and the structure of the symmetry algebras. I will also give examples of superintegrable systems on spaces of non-constant curvature and their symmetry algebras.
### Toutes les solutions elliptiques d’une équation différentielle algébrique quelconque
#### Robert Conte, École normale supérieure de Cachan
Étant donné une équation différentielle ordinaire algébrique et autonome admettant au moins une série de Laurent, nous donnons un algorithme exhibant explicitement toutes ses solutions elliptiques ou dégénérées d’elliptiques (rationnelles en une exponentielle, rationnelles). Les seuls ingrédients en sont : la série de Laurent, deux théorèmes de Briot et Bouquet, un algorithme de Poincaré implanté en Maple. Les méthodes existantes n’étaient que suffisantes (obtention de quelques telles solutions), alors que celle-ci est nécessaire. De possibles généralisations seront évoquées :équations discrètes, fonctions de Painlevé.
### Brownian motions and integrable equations
#### Mattia Cafasso
The study of 1-dimensional non-intersecting brownian motions leads to the analysis of some Fredholm determinants with integrable kernels such as, for instance, the celebrated Airy and Pearcey kernels. In this talk I will explain how to obtain some differential equations for such determinants starting from Gelfan’d Dickey equations, i.e. some solitonic equations such as KdV and Boussinesq.
### Universality in the profile of the nonlinear Schrödinger equation at the first breaking curve
#### Marco Bertola, CRM et Concordia University
We consider the zero-dispersion limit of the focusing nonlinear one-dimensional Schrödinger equation with smooth, decaying initial data. The space-time plane subdivides into regions with qualitatively different behavior, with the boundary between them consisting typically of collection of (breaking curve(s)). For small time and/or large distance, the asymptotics is ruled by modulation equations (Whitham equations) whereby the amplitude is a smooth function and the phase is fastly rotating at the scale of the dispersion parameter; for any time greater than the time of gradient catastrophe, there is a compact subset of the x-axis where the asymptotic solution develops fast, quasiperiodic behavior, and the amplitude becomes fastly oscillating at scales of order epsilon. We study the asymptotic behavior of the left and right edges of the interface between these two regions at any time after the gradient catastrophe. The main finding is that the first oscillations in the amplitude are of nonzero asymptotic size even as epsilon tends to zero, and display two separate natural scales; of order O(epsilon) in the parallel direction to the breaking curve in the (x,t)-plane, and of order O(epsilon log epsilon) in a transversal direction.
### Propriétés homotopiques de plages de Fortuin-Kasteleyn sur un tore
#### Alexi Morin-Duchesne, Département. de physique et CRM
On dit qu’un cluster FK sur un tore est dans le groupe d’homotopie {a,b} s’il est possible de dessiner une courbe dans le cluster qui entoure le tore a fois dans une direction et b fois dans l’autre. Pour le modèle de FK avec $tex:\beta\in [0,2]$, nous étudions, au point critique, la probabilité, $tex:\pi({a,b})$, qu’il y ait un cluster du groupe {a,b} et identifions le comportement asymptotique lorsque le tore devient infiniment mince. Les exposants décrivant le comportement critique sont liés aux poids $tex:h_{r,s}$ de Kac pour r,s entiers, mais aussi demi-entiers.
### Matrix model topological expansion with a non-simple branchpoint
#### Aleix Prats-Ferrer, Concordia et CRM
The topological expansion of matrix models has been understood for some years now with only one constraint: all the branch-points of the underlying algebraic curve must be simple. In this work we present the Cauchy matrix model for which this condition is not satisfied and extend the usual topological expansion to allow for a branch-point of branching number 2. The method can be easily extended to any branching number.
### Lie algebras and automorphic forms from vertex algebras
#### Thomas Creutzig, The University of North Carolina, NC Chapel Hill
Vertex algebras of central charge 24 are a key ingredient in relating sporadic groups and automorphic forms. The most prominent example being the monster. I want to explain how to construct out of certain vertex algebras some generalized Kac-Moody algebras whose denominator identity is an automorphic product and which correspond to the Mathieu group M_23.
### Relations de récurrence invariantes associées aux modèles $tex: CP^{N-1}$
#### Michel Grundland, UQTR et CRM
Dans cet exposé nous présentons des relations de récurrence invariantes pour le modèle sigma euclidien $tex: CP^{N-1}$ complètement intégrable en deux dimensions défini sur la sphère de Riemann $tex: S^2$ lorsque sa fonctionnelle d’action est finie. Nous déterminons les liens entre les opérateurs de projection successifs, les fonctions d’ondes du problème linéaire spectral, et les fonctions de plongement des surfaces dans l’algèbre su(N). Notre formulation conserve l’invariance conforme de ces quantités. Certains aspects géométriques de ces relations seront présentés. Nous étudions également les singularités des solutions méromorphes du modèle $tex: CP^{N-1}$ et démontrons qu’elles n’ont aucun impact sur les quantités invariantes. Nous présentons des exemples de la méthode de construction, plus précisemment les modèles $tex: CP^2$ et $tex: CP^3$.
### Monopoles, Periods and Problems
#### Harry W. Braden, School of Mathematics, University of Edinburgh
The modern approach to integrability proceeds via a Riemann surface, the spectral curve. In many applications this curve is specified by transcendental constraints in terms of periods. I will highlight some of the problems this leads to in the context of monopoles, problems including integer solutions to systems of quadratic forms, questions of real algebraic geometry and conjectures for elliptic functions. Several new results will be presented including the uniqueness of the tetrahedrally symmetric monopole.
### États cohérents et comprimés pour des systèmes quantiques et potentiel de Morse
#### Véronique HUSSIN, DMS et CRM
Les états cohérents ont été introduits par R. Glauder dans les années 1950 comme des états quantiques de l’oscillateur harmonique qui minimisent la relation d’incertitude de Heisenberg. Ils sont aussi connus comme des états quasi-classiques. Par la suite, plusieurs généralisations de ceux-ci ont été obtenues tant du point de vue mathématique que physique. Ils trouvent à présent des applications pour de nombreux systèmes quantiques. Ces états cohérents pour l’oscillateur harmonique sont largement utilisés en optique quantique, par exemple. En plus de minimiser la relation d’incertitude de Heisenberg, ils ont des dispersions identiques pour les observables position et impulsion. Il est possible de généraliser ces états afin de réduire la dispersion d’une des observables (au prix d’augmenter celle sur l’autre) tout en maintenant la minimisation de la relation d’incertitude. Ces nouveaux états sont appelés comprimés. Dans les années 1980, des expériences ont permis de mettre en évidence l’existence de tels états pour la lumière. Le potentiel de Morse constitue une meilleure approximation que l’oscillateur harmonique pour décrire les interactions au sein de molécules diatomiques et il possède un spectre discret fini. Les états cohérents et comprimés peuvent également être construits pour ce modèle. Dans cet exposé, je commencerai par rappeler différentes définitions des états cohérents et comprimés associés à l’oscillateur harmonique et j’expliquerai brièvement leur intérêt en optique quantique. Ensuite, ces définitions seront étendues au contexte du potentiel de Morse. J’insisterai sur la façon d’ajuster les paramètres introduits pour décrire ces états afin d’assurer une bonne localisation de ceux-ci en termes de l’opérateur position notamment. Je terminerai en donnant quelques avenues pour décrire ces états dans le contexte de modèles à plusieurs dimensions spatiales.
### The Global Geometry of Stochastic Loewner Evolutions
#### Roland Friedrich, Max Planck Inst. Bonn & Leipzig
In this talk we develop a concise description of the global geometry which is underlying the universal construction of all possible generalised Stochastic Loewner Evolutions. The main ingredient is the Universal Grassmannian of Sato-Segal-Wilson. We illustrate the situation in the case of univalent functions defined on the unit disc and the classical Schramm-Loewner stochastic differential equation. In particular we show how the Virasoro algebra acts on probability measures. This approach provides the natural connection with Conformal Field Theory and Integrable Systems.
Monday, February 1st 2010, 15h30
### Algebraic Structure of Univalent Functions and Integrable Systems
#### Dr Roland Friedrich, Max Planck Inst. Bonn & Leipzig
In this seminar we provide algebraic and categorical structures which intrinsically and deeply connect the classical theory of Univalent Functions with the theory of Integrable Systems, as arising in the KP hierarchy. A pivotal role is played by the Faà di Bruno polynomials & the Sato-Segal-Wilson Grassmannian. In our picture we see the Witt algebra & the Schwarzian emerge, and we unravel the very combinatorial nature of all this objects. Once these basics are established, the field is wide open towards other directions, which we shall also mention.
### Illusion of space-time and quantum mechanics
#### Pavel K. Smrz, The University of Newcastle, Australia
General Relativity and Quantum Mechanics do not go well with each other. One is a strictly local theory, while the other predicts non-local effects confirmed by experiments. In this talk it will be demonstrated that a generalized structure of space-time allowing non- locality may be a way to resolve the problem. After a brief introduction to basic concepts of differential geometry the most important results contained in recent publications of the speaker will be described without going into technical details.
### Recent Developments in Non-Equilibrium Quantum Statistical : An overview
#### Vojkan Jaksic, McGill et CRM
In this talk I shall discuss mathematical foundations of non- equilibrium quantum statistical mechanics focusing on a class of recent developments which fall roughly into two categories:
(A) Axiomatic results that concern mathematical structure of the theory;
(B) Study of concrete physically relevant models.
In the first part of the talk I shall focus on (A) and discuss the entropy production observable, entropy production balance equation, non-equilibrium steady states and linear response theory (Kubo formulas, Onsager relations) in the abstract framework of algebraic quantum statistical mechanics. In the second part of the talk I will discuss some concrete physically relevant models for which the axioms of (A) can be verified.
### Affine sl(2) Modules and Applications to Conformal Field Theory
#### David Ridout, CRM
We report on progress in studying fractional-level theories with affine sl(2) symmetry. These are non-unitary cousins of the (positive-integer-level) Wess-Zumino-Witten models which have formed the basis of an enormous amount of interaction between mathematics and physics. Mathematically, the most significant difference when going to fractional levels is that one is obliged to admit representations more general than highest weight. Our aim in this talk is to introduce these more general types of representations and explain why their structure theory is relevant to physics.
### Fondements de la mécanique quantique sur bases de théorie de l’information
#### Gilles Brassard, Dépt. d’informatique et de recherche opérationnelle, Université de Montréal
La théorie de l’information quantique pourrait-elle servir de base é de nouveaux fondements pour la mécanique quantique? Ceux-ci auraient l’avantage d’&ehat;tre plus homogènes et, de mon point de vue, plus fondamentaux que ceux qui ont été développés pendant la première moitié du vingtième siècle.
### Periodic orbits for an infinite family of classical superintegrable systems
#### Frédérick Tremblay, DMS, UdeM et CRM
We show that all bounded trajectories of an infinite family of classical integrable systems are closed for all integer and rational values of k. This agrees with our earlier conjecture suggesting that the quantum version of this system is superintegrable.
Aller à la barre d’outils
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2019-08-18 19:36:55
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https://www.nist.gov/publications/weak-electric-field-detection-sub-1-hz-resolution-radio-frequencies-using-rydberg-atom
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# Weak Electric-Field Detection with Sub-1 Hz Resolution at Radio Frequencies Using A Rydberg Atom-Based Mixer
Published: April 25, 2019
### Author(s)
Joshua A. Gordon, Christopher L. Holloway, Matthew T. Simons, Abdulaziz H. Haddab
### Abstract
Rydberg atoms have been used for measuring radio-frequency (RF) electric (E)-fields due to their strong dipole moments over the frequency range of 500 MHz-1 THz. For this, electromagnetically induced transparency (EIT) within the Autler-Townes (AT) regime is used such that the detected E-field is proportional to AT splitting. However, for weak E-fields AT peak separation becomes unresolvable thus limiting the minimum detectable field. Here, we demonstrate using the Rydberg atoms as an RF mixer for weak RF E-field detection well below the AT regime with frequency discrimination better than 1 Hz resolution. Two E-fields incident on a vapor cell full of cesium atoms are used. One E-field at 19.626000 GHz drives the 34D5=2!35P3=2 Rydberg transition and acts as a local oscillator (LO) and a second signal E-field (Sig) of interest is at 19.626090 GHz. In the presence of the LO the Rydberg atoms naturally down convert the Sig field to a 90 kHz intermediate frequency (IF) signal. This IF signal manifests as an oscillation in the probe laser intensity through the Rydberg vapor and is easily detected with a photodiode and lock-in amplifier. In the configuration used here, E-field strength down to 46 mV/m were detected. Furthermore, neighboring fields 0.1 Hz away and equal in strength to Sig could be discriminated without any leakage into the lock-in signal. For signals 1 Hz away and as high as +60 dB above Sig, leakage into the lock-in signal could be kept below 3 dB.
Citation: Applied Physics Letters
Pub Type: Journals
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2019-10-19 04:07:45
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https://ww2.mathworks.cn/help/comm/ref/appdecoder.html
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# APP Decoder
Decode convolutional code using a posteriori probability (APP) method
## Library
Convolutional sublibrary of Error Detection and Correction
## Description
The APP Decoder block performs a posteriori probability (APP) decoding of a convolutional code.
### Input Signals and Output Signals
The input L(u) represents the sequence of log-likelihoods of encoder input bits, while the input L(c) represents the sequence of log-likelihoods of code bits. The outputs L(u) and L(c) are updated versions of these sequences, based on information about the encoder.
If the convolutional code uses an alphabet of 2n possible symbols, this block's L(c) vectors have length Q*n for some positive integer Q. Similarly, if the decoded data uses an alphabet of 2k possible output symbols, then this block's L(u) vectors have length Q*k.
This block accepts a column vector input signal with any positive integer for Q.
If you only need the input L(c) and output L(u), you can attach a Simulink Ground (Simulink) block to the input L(u) and a Simulink® Terminator (Simulink) block to the output L(c).
This block accepts `single` and `double` data types. Both inputs, however, must be of the same type. The output data type is the same as the input data type.
### Specifying the Encoder
To define the convolutional encoder that produced the coded input, use the Trellis structure parameter. This parameter is a MATLAB® structure whose format is described in Trellis Description of a Convolutional Code. You can use this parameter field in two ways:
• If you have a variable in the MATLAB workspace that contains the trellis structure, enter its name as the Trellis structure parameter. This way is preferable because it causes Simulink to spend less time updating the diagram at the beginning of each simulation, compared to the usage described next.
• If you want to specify the encoder using its constraint length, generator polynomials, and possibly feedback connection polynomials, use a `poly2trellis` command within the Trellis structure field. For example, to use an encoder with a constraint length of 7, code generator polynomials of 171 and 133 (in octal numbers), and a feedback connection of 171 (in octal), set the Trellis structure parameter to
`poly2trellis(7,[171 133],171)`
To indicate how the encoder treats the trellis at the beginning and end of each frame, set the Termination method parameter to either `Truncated` or `Terminated`. The `Truncated` option indicates that the encoder resets to the all-zeros state at the beginning of each frame. The `Terminated` option indicates that the encoder forces the trellis to end each frame in the all-zeros state. If you use the Convolutional Encoder block with the Operation mode parameter set to ```Truncated (reset every frame)```, use the `Truncated` option in this block. If you use the Convolutional Encoder block with the Operation mode parameter set to ```Terminate trellis by appending bits```, use the `Terminated` option in this block.
### Specifying Details of the Algorithm
You can control part of the decoding algorithm using the Algorithm parameter. The ```True APP``` option implements a posteriori probability decoding as per equations 20–23 in section V of [1]. To gain speed, both the `Max*` and `Max` options approximate expressions like
`$\mathrm{log}\sum _{i}\mathrm{exp}\left({a}_{i}\right)$`
by other quantities. The `Max` option uses max(ai) as the approximation, while the `Max*` option uses max(ai) plus a correction term given by $\mathrm{ln}\left(1+\mathrm{exp}\left(-|{a}_{i-1}-{a}_{i}|\right)\right)$ [3].
The `Max*` option enables the Scaling bits parameter in the dialog box. This parameter is the number of bits by which the block scales the data it processes internally (multiplies the input by (2^`numScalingBits`) and divides the pre-output by the same factor). Use this parameter to avoid losing precision during the computations.
## Parameters
Trellis structure
MATLAB structure that contains the trellis description of the convolutional encoder.
Termination method
Either `Truncated` or `Terminated`. This parameter indicates how the convolutional encoder treats the trellis at the beginning and end of frames.
Algorithm
Either `True APP`, `Max*`, or `Max`.
Number of scaling bits
An integer between 0 and 8 that indicates by how many bits the decoder scales data in order to avoid losing precision. This field is active only when Algorithm is set to `Max*`.
Disable L(c) output port
Select this check box to disable the secondary block output, L(c).
## References
[1] Benedetto, S., G. Montorsi, D. Divsalar, and F. Pollara, “A Soft-Input Soft-Output Maximum A Posterior (MAP) Module to Decode Parallel and Serial Concatenated Codes,” JPL TDA Progress Report, Vol. 42-127, November 1996.
[2] Benedetto, Sergio and Guido Montorsi, “Performance of Continuous and Blockwise Decoded Turbo Codes.” IEEE Communications Letters, Vol. 1, May 1997, 77–79.
[3] Viterbi, Andrew J., “An Intuitive Justification and a Simplified Implementation of the MAP Decoder for Convolutional Codes,” IEEE Journal on Selected Areas in Communications, Vol. 16, February 1998, 260–264.
## Extended Capabilities
### Functions
Introduced before R2006a
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2021-07-30 19:35:59
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http://semmarias.sk/country-wihnm/page.php?cafd0a=in-a-matrix-interchanging-of-rows-and-columns-is-called
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Fotogaleria
# in a matrix interchanging of rows and columns is called
(A’)’= A. Definition The new matrix obtained by interchanging the rows and columns of the original matrix is called as the transpose of the matrix. We have: . Pivot row: ... where P k is the permutation matrix obtained by interchanging the rows k and r k of the identity matrix, and M k is an elementary lower triangular matrix resulting from the elimination process. In the second step, we interchange any two rows or columns present in the matrix and we get modified matrix B. Example 2: Consider the matrix . More generally, any permutation of the rows or columns multiplies the determinant by the sign of the permutation. Each element of the original matrix appears in 2 rows and 3 columns in the enlarged matrix. Solution: It is an order of 2*3. Recommended: Please try your approach on first, before moving on to the solution. Show activity on this post. of rows] [ no. The matrix obtained from a given matrix A by interchanging its rows and columns is called Transpose of matrix A. Transpose of A is denoted by A’ or . If A is of order m*n, then A’ is of the order n*m. Clearly, the transpose of the transpose of A is the matrix A itself i.e. If rectangular matrix A is m × n, it is called column orthogonal when ATA = I since the columns are orthonormal. If you are on a personal connection, like at home, you can run an anti-virus scan on your device to make sure it is not infected with malware. If a one-row matrix is simplified to a vector, the column names are used as names for the values. Comment document.getElementById("comment").setAttribute( "id", "ab3f2f9c3e28f1d074d0f19134e952ce" );document.getElementById("afa6a2ad4a").setAttribute( "id", "comment" ); © MathsTips.com 2013 - 2020. The transpose of a column vector is a row vector and vice versa. This is just an easy way to think. Your IP: 192.145.237.241 If A = [a ij] be an m × n matrix, then the matrix obtained by interchanging the rows and columns of A would be the transpose of A. of It is denoted by A′or (A T). The matrix B is called the transpose of A. It works as follows. A square matrix is called orthogonal when ATA = AAT = I. Bookmark this question. If the two vectors are each column vectors, then the inner product must be formed by the matrix product of the transpose of a column vector times a column vector, thus creating an operation in which a 1 x n matrix is multiplied with a n x 1 matrix. G1 * G2' = 44 Verify this result by carrying out the operations on 'matlab'. We can treat each element as a row of the matrix. We take matrix A and we calculate its determinant (|A|). B Rows. and ' and even the transpose, Stack Overflow. Consider the matrix If A = || of order m*n then = || of order n*m. So, . Example 2: Consider the matrix Find the Adj of A. Matrix created as a result of interchanging the rows and columns of a matrix is called Transpose of that Matrix, for instance, the transpose of the above matrix would be: 1 4 2 5 3 6 This transposed matrix can be written as [ [1, 4], [2, 5], [3, 6]]. C determinants. When taking a 2-D array each element is considered itself a 1-D array or known to be a collection of a 1-D array. By, writing another matrix B from A by writing rows of A as columns of B. In some contexts, such as computer algebra programs , it is useful to consider a matrix with no rows or no columns, called an empty matrix . Run this code snippet in C. int x=5, y=6; x=x+y; y=x-y; x=x-y; if (! I want to convert the rows to columns and vice versa, that is I should have 147 rows and 117 columns. Example 1: Consider the matrix . Determinant of a matrix changes its sign if we interchange any two rows or columns present in a matrix. Interchanging any pair of columns or rows of a matrix multiplies its determinant by −1. For example consider the matrix Order of the matrix = 2 x 4 So the order of largest possible square matrix is 2 x 2 . For example matrix = [[1,2,3],[4,5,6]] represent a matrix of order 2×3, in which matrix[i][j] is the matrix element at ith row and jth column.. To transpose a matrix we have to interchange all its row elements into column elements and column … Given a matrix A, return the transpose of A. Solution: First to find out the minor and cofactor of the matrix : = 2 = 2, = 2 = -2, = -1 = +1, = 5 = 5. i.e. In this example prints transpose of a matrix. In my first programming course, I learnt how to swap two variables, suppose denoted by x and y, without holding a value in a third variable. Do the transpose of matrix. The first row can be selected as X[0].And, the element in the first-row first column can be selected as X[0][0].. Transpose of a matrix is the interchanging of rows and columns. Thus the transpose is also the inverse: A− 1 = AT. The operation of interchanging rows and columns in a matrix is called trans from MEGR 7102 at University of North Carolina, Charlotte Do the transpose of matrix. If you have more than 256 original rows, you cannot Transpose these unless you are using Excel 2007 Beta. A matrix with the same number of rows and columns is called a square matrix. Cloudflare Ray ID: 5fd3023aedfce4fa Do the transpose of matrix. PEARL PACKAGE The matrix obtained from a given matrix A by interchanging its rows and columns is called a) Inverse of A b) Square of A c) transpose of A d) None of these A+ Ais d) Nonebद म We have: . $$B = \begin{bmatrix} 2 & -9 & 3\\ 13 & 11 & 17 \end{bmatrix}_{2 \times 3}$$ The number of rows in matrix A is greater than the number of columns, such a matrix is called a Vertical matrix. If A has dimension (n m) then A0has dimension (m n). We calculate determinant of matrix B. Required fields are marked *. (A’)’= A. If you are at an office or shared network, you can ask the network administrator to run a scan across the network looking for misconfigured or infected devices. 21 Horizontally arranged elements in a matrix is called. View Answer. If A is of order m*n, then A’ is of the order n*m. Clearly, the transpose of the transpose of A is the matrix A itself i.e. (adsbygoogle = window.adsbygoogle || []).push({}); The matrix obtained from a given matrix A by interchanging its rows and columns is called Transpose of matrix A. Transpose of A is denoted by A’ or . Consider the matrix If A = || of order m*n then = || of order n*m. So, . ... interchanging rows and columns of a 4D matrix. Performance & security by Cloudflare, Please complete the security check to access. Click to share on Facebook (Opens in new window), Click to share on Twitter (Opens in new window), Click to share on LinkedIn (Opens in new window), Click to share on Reddit (Opens in new window). Syntax: type array name [ no. Thetransposeofasymmetricmatrix Ask Question Asked 4 years, 7 months ago. of Columns]; Ex… A matrix with m rows and n columns is called an m × n matrix or m-by-n matrix, while m and n are called its dimensions. A additive inverse of A. However, perhaps there's a different way - right now, my matrix is acting as a the equivalent of a Java ArrayList or a general list in Python, where I use swapping columns in combination with a MEX function for quickly deleting the last column to construct an equivalent data structure in MATLAB. Completing the CAPTCHA proves you are a human and gives you temporary access to the web property. Converting rows of a matrix into columns and columns of a matrix into row is called transpose of a matrix. Solution: The transpose of matrix A by interchanging rows and columns is . Your email address will not be published. Now, let us take another matrix. I want to read this data into MATLAB but I need to to interchange the rows and the column so that the matrix will be 60 rows and 2000 columns. It is obtained by interchanging rows and columns of a matrix. The matrix B is called the transpose of matrix A if and only if b ij = a ji for all iand j: The matrix B is denoted by A0or AT. Pivoting may be followed by an interchange of rows or columns to bring the pivot to a fixed position and allow … The column in which eliminations are performed is called the pivot column. The two-d array uses two for loops or nested loops where outer loops execute from 0 to the initial subscript. • For example, if the user entered an order as 2, 2 i.e. A related matrix form by making the rows of a matrix into columns and the columns into rows is called a ____. The following example described how to make a transpose matrix in TypeScript. For instance, if For a symmetric matrix, A = A’. A matrix consisting of a single row is called a row matrix, and that consisting of a single column is called a column matrix. • Approach: This problem can be solved by keeping either the number of rows or columns fixed. Example 1: Consider the matrix . 22 If A is a matrix of order (m - by - n) then a matrix (n - by - m) obtained by interchanging rows and columns of A is called the. Matrices with a single row are called row vectors, those with a single column are called column vectors. That’s the result, indeed, but the row name is gone now. D transpose. In this article, the number of rows … In Python, we can implement a matrix as a nested list (list inside a list). Matrices obtained by changing rows and columns is called transpose. The transpose of the transpose of a matrix is that the matrix itself =, The transpose of the addition of 2 matrices is similar to the sum of their transposes =, When a scalar matrix is being multiplied by the matrix, the order of transpose is irrelevant =. Solution: It is an order of 2*3. Note: A matrix with an infinite number of rows or columns (or both) is called an infinite matrix . you cannot tramspose a matrix greater than 256 columns x 256 rows Gord Dibben MS Excel MVP On Mon, 26 Jun 2006 16:53:59 GMT, "Lewis Clark" wrote: >Copy the entire working range. In this case, a single row is returned so, by default, this result is transformed to a vector. We can prove this property by taking an example. A matrix obtained by interchanging rows and columns is called ____ matrix? In other words, we can say that matrix A is another matrix formed by replacing each element of the current matrix by its corresponding cofactor and then taking the transpose of the new matrix formed. Answer By Toppr. Example 1: Consider the matrix Find the Adj of A. I have an input data in Excel which has 2000 rows and 60 columns. This follows from properties 8 and 10 (it is a general property of multilinear alternating maps). Answer: Rows. If the rows and columns of a matrix A are interchanged (so that the first row becomes the first column, the second row becomes the second column, and so on) we obtain what is called the transposeof A, denoted AT. An adjoint matrix is also called an adjugate matrix. If m = n, the matrix is called a square matrix of order n. A square matrix in which only the diagonal elements α = α ii are nonzero is called a diagonal matrix and is denoted by diag (α 1, …, α n). And the default format is Row-Major. Two-dimensional Array is structured as matrices and implemented using rows and columns, also known as an array of arrays. (x-6 || y-5)) printf ("Variables Swapped. By, writing another matrix B from A by writing rows of A as columns of B. For example, the matrix A above is a 3 × 2 matrix. columns. A columns. For example, if A = 4 −1 13 9!, then by interchanging rows and columns, we obtain AT = 4 13 −1 9!. Before you can multiply two matrices together, the number of ____ in the first matrix must equal the number of rows in the second matrix. R tries to simplify the matrix to a vector, if that’s possible. In Python, there is always more than one way to solve any problem. The memory allocation is done either in row-major and column-major. A matrix having m rows and n columns with m ≠ n is said to be a In a matrix multiplication for A and B, (AB)t Matrices obtained by changing rows and columns is called Solution: = 7 = 7, = 18 = -18, = 30 = 30, = 1 = -1, = 6 = 6, = 10 = -10, = 1 = 1, = 8 = -8, = 26 = 26. All Rights Reserved. If, for any matrix A, a new matrix B is formed by interchanging the rows and columns (i.e., aij = bji), the resultant matrix is said to be the transpose of the original matrix and is denoted by A’. For example X = [[1, 2], [4, 5], [3, 6]] would represent a 3x2 matrix. I tried the function .' two rows and two columns and matrices as: The horizontal array is known as rows and the vertical array are known as Columns. How can I do this in MATLAB, because Excel only has 256 column which cannot hold 2000 columns. Maths Help, Free Tutorials And Useful Mathematics Resources. "); So, as it shows, interchanging rows and columns can be achieved in exactly the same way, a series of scalar … Active 4 years, 7 months ago. Your email address will not be published. The pivot or pivot element is the element of a matrix, or an array, which is selected first by an algorithm, to do certain calculations. The m… Given a square matrix A, the transpose of the matrix of the cofactor of A is called adjoint of A and is denoted by adj A. ... Row switching is interchanging two ____ of a matrix… Rank of matrix is the order of largest possible square matrix whose determinant is non zero. Taking the transpose of a matrix is equivalent to interchanging rows and columns. Federal MCQs, 9th Class MCQs, Math MCQs, Matrices And Determinants MCQs, Symeetric , Identify matrix , transpose , None In the case of matrix algorithms, a pivot entry is usually required to be at least distinct from zero, and often distant from it; in this case finding this element is called pivoting. As 2, 2 i.e vice versa done either in row-major and column-major your IP: •! Determinant ( |A| ) 2007 Beta as names for the values matrix changes its sign if we interchange any rows..., any permutation of the matrix: Please try your approach on first, before moving on to solution... Known as columns way to solve any problem orthogonal when ATA = I by changing rows and the are! With the same number of rows and the vertical array are known as rows and is. Tutorials and Useful Mathematics Resources array is known as columns an infinite.. The horizontal array is known as rows and the vertical array are known as columns of B matrices with single... 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Sign of the permutation ; x=x-y ; if ( & security by cloudflare, complete... Can I do this in MATLAB, because Excel only has 256 column which not. A and we calculate its determinant ( |A| ) inverse: A− 1 AT! Two columns and matrices as: the column names are used as names for the values: problem. Input data in Excel which in a matrix interchanging of rows and columns is called 2000 rows and columns of B sign if interchange! Non zero permutation of the matrix if a one-row matrix is called an adjugate.! The initial subscript the user entered an order of 2 in a matrix interchanging of rows and columns is called 3 recommended: Please try your on..., writing another matrix B there is always more than 256 original,. Only has 256 column which can not hold 2000 columns matrix whose determinant non. The rows of a matrix with an infinite number of rows and columns is column... Run this code snippet in C. int x=5, y=6 ; x=x+y ; y=x-y ; x=x-y ; (! Is gone now appears in 2 rows and columns Variables Swapped months ago B a! = 44 Verify this result by carrying out the operations on 'matlab ' 0 to solution... Implement a matrix is also called an adjugate matrix a as columns columns ( or both ) is called transpose. The transpose of matrix is the order of 2 * 3 Please try your approach on first before! Tutorials and Useful Mathematics Resources we can implement a matrix is called the of! An adjugate matrix the CAPTCHA proves you are using Excel 2007 Beta row vectors, those a... As 2, 2 i.e is a 3 × 2 matrix case, a single column are called column when! One-Row matrix is called the pivot column Help, Free Tutorials and Useful in a matrix interchanging of rows and columns is called Resources its! Security by cloudflare, Please complete the security check to access matrix obtained interchanging! By default, this result is transformed to a vector matrix form by making the rows and 3 columns the. To simplify the matrix a and we calculate its determinant by −1 IP: 192.145.237.241 • &! By changing rows and 3 columns in the second step, we interchange any two rows or columns or! This in MATLAB, because Excel only has 256 column which can not transpose these unless you using. With the same number of rows or columns present in a matrix the! How to make a transpose matrix in TypeScript alternating maps ) execute from 0 to the web property possible... On first, before moving on to the web property a 1-D array the memory allocation done. C. int x=5, y=6 ; x=x+y ; y=x-y ; x=x-y ; if ( and columns is orthogonal! Order as 2, 2 i.e result by carrying out the operations on '!, by default, this result by carrying out the operations on 'matlab ' * G2 ' = 44 this. Making the rows or columns present in a matrix with the same number of rows or (... The security check to access if that ’ s possible and 10 ( is. Matrix B from a by interchanging rows and the columns are orthonormal also called an matrix...: A− 1 = AT, we can implement a matrix be collection... Column are called column orthogonal when ATA = I since the columns are orthonormal in. Into rows is called the transpose is also called an adjugate matrix general property of multilinear alternating maps.! Row are called column vectors matrix Find the Adj of a matrix: consider the Find. M × n, it is a 3 × 2 matrix as the is! Moving on to the initial subscript pair of columns ] ; Ex… G1 G2! Columns fixed snippet in C. int x=5, y=6 ; x=x+y ; ;. Columns multiplies the determinant by −1 and gives you temporary access to solution... A row of the original matrix is called an infinite matrix, Overflow! Additive inverse of A. I have an input data in Excel which has rows... Free Tutorials and Useful Mathematics Resources the initial subscript the web property name is gone.. Matrix whose determinant is non zero same number of rows and columns of B since the are. X-6 || y-5 ) ) printf ( Variables Swapped than one way to any! Column names are used as names for the values by default, this result is transformed a! A = a ’ the row name is gone now rectangular matrix a by writing rows of a is... Matrix is called column vectors known to be a collection of a as columns user.: 5fd3023aedfce4fa • your IP: 192.145.237.241 • Performance & security by cloudflare, Please complete security. 2 i.e when taking a 2-D array each element as a nested list ( inside... Y-5 ) ) printf ( Variables Swapped each element of the permutation you using... 2000 columns problem can be solved by keeping either the number of or! 1-D array ( m n ) generally, any permutation of the Find... Is an order of largest possible square matrix the vertical array are known as.! Row is returned So, 60 columns matrix changes its sign if we interchange any two rows and columns. Result, indeed, but the row name is gone now simplified to a vector if... Or rows of a multilinear alternating maps ) in a matrix changes its if! By writing rows of a matrix multiplies its determinant by the sign of the matrix to a vector, that... A row of the matrix ( x-6 || y-5 ) ) printf ! 2: consider the matrix to a vector, if that ’ s result! Matrix, a = || of order m * n then = || of m. You can not hold 2000 columns has 2000 rows and 3 columns in matrix. = AAT = I since the columns are orthonormal called an adjugate matrix the result, indeed, the. The number of rows or columns present in a matrix columns is called orthogonal ATA... Printf ( Variables Swapped * 3 in which eliminations are performed is called orthogonal! ) is called transpose of order m * n then = || of order m * then... Gives you temporary access to the solution: A− 1 = AT m * n then = of... The result, indeed, but the row name is gone now in rows. Of columns or rows of a matrix of A. I have an input in... Matrix B from a by interchanging the rows of a as columns of the original matrix appears 2. For loops or nested loops where outer loops execute from 0 to initial. Order n * m. So, any two rows and columns of B largest possible square matrix ) ) (! Follows from properties 8 and 10 ( it is an order of 2 * 3 int x=5 y=6! Can be solved by keeping either the number of rows and columns.. Tutorials and Useful Mathematics Resources from 0 to in a matrix interchanging of rows and columns is called initial subscript of largest possible square matrix transpose, Stack.! Rows, you can not transpose these unless you are using Excel 2007 Beta transpose of a matrix as nested. To make a transpose matrix in TypeScript from properties 8 and 10 ( it is an..., you can not hold 2000 columns pivot column to make a transpose matrix in.! Are a human and gives you temporary access to the web property modified... Help, Free Tutorials and Useful Mathematics Resources is returned So, related matrix form making...
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2021-04-10 14:02:14
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https://www.zbmath.org/?q=an%3A1094.30036
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# zbMATH — the first resource for mathematics
Fixed points of meromorphic solutions of higher order linear differential equations. (English) Zbl 1094.30036
The authors discuss some interesting problems on the fixed points of meromorphic solutions of higher order linear differential equation $$f^{(k)}+A(z)f=0$$ and their derivatives, where $$A(z)$$ is a given rational function or a given transcendental meromorphic function of finite order. For every transcendental meromorphic solution $$f$$ of $$f^{(k)}+A(z)f=0$$, they give its exact order $$\sigma(f)$$ and show that the sequence of fixed points of $$f$$ and the sequence of fixed points of its derivatives have the same (hyper-)exponent of convergence.
##### MSC:
30D35 Value distribution of meromorphic functions of one complex variable, Nevanlinna theory 34M05 Entire and meromorphic solutions to ordinary differential equations in the complex domain
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2021-04-21 01:08:34
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https://www.physicsforums.com/threads/concentric-charged-spheres-work.295798/
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# Concentric charged spheres work
1. Feb 27, 2009
### EngageEngage
I have a problem that I am not sure I am working correctly.
There are 2 concentric spheres, the inner of charge -q and radius Ra and outer of charge +q and radius Rb. I am to find the work that is required to put together these two spheres in such a way.
Is it enough to find the energy stored in the electric field between them, or would this be incorrect? I guess this would be the same as the energy stored in a spherical capacitor like this.
$$\frac{\epsilon}{2}\int_{r_a}^{r_b} E^{2} d^{3}r$$
Or, do I have to find first the energy to assemble the inner sphere, and then the outer sphere, and then the work required to bring them together?
Thank you for any help.
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2017-11-23 04:32:14
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https://together.jolla.com/question/66773/additional-stretch-goals/?sort=votes&page=1
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# Additional stretch goals [not relevant]
Goal 2Millions -- help coderus, nieldk, cepiperez and marxian develop for sailfish. Goal 2.25Milions -- partnership with Microsoft/Facebook to bring Skype/Whatsapp to sailfish. Goal 2.5Millions -- bigger and removable battery(slot in) for additional 30$edit retag reopen delete ### The question has been closed for the following reason "question is not relevant or outdated"by molan close date 2015-05-20 23:10:18.253289 ## Comments I like Jolla's unique style and consistency, and I would like to see more apps designed by Jolla even though they wouldn't have all the bells and whistles of apps ported from other ecosystems! I don't think Microsoft or Facebook would have any respect for Jolla's uniqueness or willingness to improve it further! ( 2014-11-27 19:27:01 +0200 )edit 2 Goal inevitable: Make absolutely no effort regarding rotten protocols like Skype and Whatsapp. ( 2014-11-27 19:31:41 +0200 )edit 3 Please do NOT even consider partnership with M$/Facecrap. It would ruin privacy.
( 2014-11-27 20:02:07 +0200 )edit
Sort by » oldest newest most voted
Considering a partnership with MS to port Skype... If you mean building a native app, I don't believe that would ever happen. They even didn't do that for BlackBerry 10. And BB, although they are far away from the market share they had previously, has a completely different weight considering sales, ecosystem and customers.
And bigger/removable battery at 2.5M\$, I think that won't happen anyway. I hope they find a way to upgrade the battery anyway at least slightly. We will see. At least they didn't rule it out completely yet.
more
Skype may not be a Made For BlackBerry application, but MS packaged, tested it and released it in Blackberry World
( 2014-11-27 19:29:37 +0200 )edit
Technically that is right. Although I would guess that BB did probably pay for part of that effort, which I guess is not possible for Jolla. For BB, this was a very important aspect considering their target group. If you want to do business with those phones, missing skype would really be a big minus. BB also paid some money for each app that was developed and certified until a certain deadline. There was just more punch in what they were able to do and how they were able to negotiate. I don't know how many Jolla phones have been sold, but I think Microsoft will not do much effort for 6k or let it be 12k tablet devices that will be made by Jolla. Especially considering that they have their own phone os etc... But well, we can see...
Edit: Oh, and due to the fact that it is not a native app, it is not integrating that well into the BB10 UI and system, it crashes from time to time, and it eats battery. For sure it is not in the main focus of development.
( 2014-11-27 19:39:42 +0200 )edit
Skype is not a choice. It is a requirement for work sometimes, even if you would prefer not to use it.
( 2014-11-27 20:46:46 +0200 )edit
Exactly. That is why I think BB did probably push that a lot and possibly even paid for it. I remember the discussions before BB10 came out...
( 2014-11-27 21:58:04 +0200 )edit
1
Skype can be run through Android runtime just fine..
( 2014-11-27 21:58:27 +0200 )edit
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2019-01-22 21:51:29
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https://indico.cern.ch/event/219436/contributions/1523192/
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# Quark Matter 2014 - XXIV International Conference on Ultrarelativistic Nucleus-Nucleus Collisions
19-24 May 2014
Europe/Zurich timezone
## Nuclei identification and hypernuclei reconstruction with the ALICE inner tracker upgrade
20 May 2014, 16:30
2h
### spectrum
Board: M-02
Poster Future Experimental Facilities, Upgrades, and Instrumentation
### Speaker
Stefania Bufalino (Universita e INFN (IT))
### Description
One of the striking features of particle production in Pb-Pb collisions at high energies is the near equal abundance of matter and antimatter in the central rapidity region. The comparison of the production of light nuclei, antinuclei, hypernuclei and anti--hypernuclei at high energies offers a unique opportunity to understand if the mechanism of particle production in ultra-relativistic heavy ion collisions can be described by a coalescence or a thermal model. An upgrade of the ALICE apparatus is scheduled for the second long LHC shutdown (LS2) in 2019 and it will give access to otherwise unreachable processes. The new Inner Tracking System (ITS), with largely enhanced tracking precision and efficiency, and the improved readout capability of most of the ALICE detectors will be crucial for the measurement of light (anti)hypernuclei. The identification of light nuclei is also important for the reconstruction of the weak decays of hypernuclei in which a nucleus and a pion are emitted with a V0--like topology. A method for tagging light nuclei with the new ITS, using the information on the size of the produced pixel clusters was developed. The study of the $\mathrm{^{3}_{\Lambda}H} \rightarrow \mathrm{^{3}He} + \pi^{-}$ decay channel has been carried out using a detailed simulation of the new ITS. The expected yield, signal-to-background ratio and significance have been determined taking into account the target integrated luminosity of 10 nb$^{-1}$ for the ALICE upgrade programme.
On behalf of collaboration: ALICE
### Primary author
Stefania Bufalino (Universita e INFN (IT))
Slides
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2020-12-02 04:14:57
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https://math.stackexchange.com/questions/3627571/using-resolution-rule-check-if-the-following-is-a-tautology/3627594
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# using resolution rule, check if the following is a tautology
i understand how to use resolution but i get to an answer and i dont know if its ok. I need to check if the following is a tautology only using resolution rules.
(p∨q)→(q→(p∨(p∧q))
i get to the following clauses: p∨q, q, ¬p, ¬p∨¬q I dont know how to procede past that.
Also i have a second question about resolution: i want to know, what happens if i get a clause with (¬p∨p)
• You might do best to reread what the resolution rule of inference says, and look for pairs of clauses where $\phi$ and $\lnot$$\phi$ both appear within the formula. In your example $\lnot$p appears in one formula, and p appears in another. Thus, you can infer the rest of the clause without p, or $\lnot$p according to the resolution rule of inference. There's only so many clauses that you can infer here, and no clause that you can infer is the empty set here. Thus, you might list all of the clauses that can get inferred and then indicate that the empty set is not one of them. Apr 16 '20 at 0:51
• So, i get no empty set, that means is not a tautology? Also if i get an empty set it is a tautology ? (not this case) Apr 16 '20 at 1:04
• If you get an empty set, then it's a tautology. If you've deduced all possible formulas that can get deduced and there is no empty set, then it's not a tautology. Apr 16 '20 at 1:22
• Thanks!! this is very helpfull Apr 16 '20 at 1:26
Now back to your problem. You are actually quite right in that you're basically stuck: the only new clauses you can get from the ones you have are $$p \lor \neg p$$ and $$q \lor \neg q$$. Both being tautologous clauses, that does not get you anywhere. And given that you can;t get anything else, that tells you you won;t get to a contradiction. Moreover, the $$q$$ and $$\neg p$$ clauses tells you exactly the model that will make the statement False: if you set $$q$$ to True and $$p$$ to False, then you'll find that the original statement evaluates to False, and thus is not a tautology.
Finally, form the clauses that you have, you can remove the clauses $$p \lor q$$, since it is subsumed by the clause $$q$$. Likewise, you can remove the $$\neg p \lor \neg q$$ clauses since it is subsumed by $$\neg p$$. So, you are left with $$q$$ and $$\neg p$$ ... and now it is obvious that you won't get a contradiction from that!
• @Cipher not getting an empty set would suggest it is not a tautology ... but you have to be careful: You have to make sure that you have tried all possible resolutions and that there are indeed no more new clauses to be obtained. So for this to become a demonstration, you need to systematically demonstrate that there are no new clauses to be obtained. Saying "Well, I couldn;t get to en empty clause" is really not enough: maybe you weren't trying hard enough to find new clauses! :) But in this case it's clear, especially after removing the subsumed clauses and ending up with $q$ and $\neg p$. Apr 16 '20 at 13:44
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2021-09-18 20:47:19
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https://kb.ettus.com/index.php?title=Resolving_Audio_Codec_Enumeration_Issues_On_The_E31x&direction=next&oldid=3173
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# Resolving Audio Codec Enumeration Issues On The E31x
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
AN-178
## Revision History
Date Author Details
2016-10-12 Logan Fagg Initial creation
## Abstract
This application note covers Resolving Audio Codec Enumeration Issues On The E31x.
## Overview
On some E31x series devices, it has been found that the USB hub boots too slowly to properly enumerate the audio codec. This will result in the device not being able to use the codec which affects anyone using the 2.5mm audio connector. This note will explain:
• How to detect if your USB codec is properly enumerated
• How to force enumeration if the codec isn't enumerated
## Detecting Enumeration Issues
An improperly enumerated audio codec can most reliably be detected by using the following command:
root@ettus-e3xx-sg3:~# lsusb
Which should return the following:
Bus 001 Device 003: ID 0556:0004 Asahi Kasei Microsystems Co., Ltd
Bus 001 Device 002: ID 0424:2513 Standard Microsystems Corp. 2.0 Hub
Bus 001 Device 001: ID 1d6b:0002 Linux Foundation 2.0 root hub
If the 'Asahi Kasei Microsystems Co., Ltd' line does not appear when you use the lsusb command, you likely have the enumeration issue and you should move on to the Force Enumeration section.
## Force Audio Codec Enumeration
Important Note: It is highly recommended to back up your SD card before trying the following instructions. This procedure has a small chance of corrupting your file system. You can use the fsck command to verify your filesystem on reboot if you are concerned.
The audio codec can be made to enumerate by rapidly power-cycling the device by physically disconnnecting the power supply for less than 1 second. This can be done by either physically unplugging the unit or adding a physical switch between the unit and power supply or between the power supply and wall outlet. Performing a reboot without disconnecting power will not force the codec to enumerate. You must physically disconnect the device.
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2019-08-24 06:52:10
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https://www.greencarcongress.com/2013/08/20130829-ea888.html
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## VW introducing 1.8L EA888 Gen 3 engine in 2014 Jetta, Passat and Beetle; driving impressions
##### 29 August 2013
The new 4-cylinder Gen 3 1.8 EA888 TSI engine (red) delivers its peak 250 N·m of torque as early as 1,400 rpm, and maintains the strong and flat torque curve through 4,750 rpm—far outpacing the performance of the 5-cylinder 2.5-liter engine (blue) it replaces. The resulting boost in driving performance is manifest. Click to enlarge.
Volkswagen of America is introducing the Group’s third-generation EA888 1.8-liter TSI four-cylinder engine in the 2014 Jetta and Passat, followed by the 2014 Beetle. The 1.8-liter unit produces 170 hp (127 kW) and 184 lb-ft (250 N·m) of torque as early as 1,500 rpm. The horsepower is the same as the outgoing 5-cylinder 2.5-liter engine, but it delivers 7 lb-ft more torque at 2,750 fewer revs (see chart at right).
Mated to an automatic or a manual transmission, the new EA888 offers up to a 17% decrease in fuel consumption while providing even better acceleration. In a presentation at Volkswagen of America’s 2014 line drive in St. Helena, California, Hubertus Lemke, Head of Technical Project Management for Volkswagen AG, described the advances in the new third-generation engines. (The 2.0-liter version of this engine has already made its debut in the 2013 Beetle and Jetta GLI, delivering a 10 hp improvement.)
Generational progress with the EA888. Key advances for Gen 3 are in weight reduction, friction reduction, and the internal exhaust cooling system. Click to enlarge.
Volkswagen first launched the EA888 engine generation in spring 2007 (Gen 1). In 2009, Gen 2 introduced a number of measures to optimize friction. The goals of the current Gen 3 were to:
• significantly improve the engine’s already good power, torque and fuel efficiency;
• reduce the weight of the unit, while maintaining at least equivalent comfort attributes; and
• to conform to future emissions standards, particularly the stringent EU6 standard which will be mandatory in Europe in 2015.
MQB and MLB
The 1.8-liter engine, which can be mounted either transversely or longitudinally—it is part of the MQB (transverse) as well as the MLB (longitudinal) assembly kits (earlier post)—has already been applied in the Audi A4 in Europe in a longitudinal mounting.
“[The EA888] is an engine family that can serve many purposes. It can be the extra thrifty one, but also be one you can use it in motorsports. It is important that it’s not two developments, but the same engine family, the same development.”
—Christian Buhlmann, Volkswagen Product Communications
In the A4 with a manual transmission, the 1.8 TFSI with 125 kW (170 hp) has a combined fuel consumption of 5.7l/100 km (41 mpgUS), or 134 grams CO2/km (216 g/mile). The A4 of model year 2000 with a 1.8 T engine that produced 110 kW (150 hp) of power was emitted 197 grams CO2/km (317 g/mile).
Improvements contributing to this better fuel economy come from many areas. However, they primarily involve engine optimizations.
Designed to be lighter and more efficient, the new EA888 Gen 3 turbocharged and direct-injection four-cylinder powerplant features a number of refinements, including a novel cylinder-head-integrated exhaust manifold with exhaust gas cooling. This is the first implementation of such a system in a turbocharged direct-injection four-cylinder gasoline engine, noted Lemke.
While the actual fuel economy varies by model, the Jetta equipped with the 1.8T now can get up to 36 mpg (6.5 l/100 km) on the highway, up from 31 mpg (7.6 l/100 km) for the 2.5-liter engine. At the same time, the city mileage improves from 24 to 25 mpg (9.4 l/100 km), and the overall EPA estimated combined fuel economy is now 29 mpg (8.1 l/100 km) compared with 26 (9.0 l/100 km)) on the outgoing five cylinder.
As well as offering better fuel economy, the new engine’s additional low-speed torque has enhanced the performance of the cars fitted with the engine: the manual transmission Jetta 1.8T now goes from 0 to 60 mph in a manufacturer estimated 7.3 seconds, an improvement of 0.7 seconds over the 2.5-liter model.
One of the keys to this improved performance and fuel economy is the reduction of internal friction. Several innovations contributed to this reduction including; the use of a new piston coating; the two balance shafts that counteract the second-order inertial forces run in roller bearings; utilizing a low-energy oil pump; and using a highly precise electric system to control the oil-jet cooling for the piston crowns.
Like its predecessor, the engine has a cast-iron cylinder block and an aluminum-alloy cylinder head which now has the exhaust headers cast integrally. The cast-iron crankshaft runs in five main bearings. The twin overhead camshafts are chain driven, and there is variable cam phasing on the intake side.
Weight and friction reduction in the Gen 3 EA888. Click to enlarge.
Weight Reduction. A primary development goal of the new EA888 Gen 3 engine family was to reduce weight. Numerous improvements have contributed to a significant reduction including changing the casting process from the conventional flat pouring to upright pouring. Nominal wall thickness has been reduced from 3.5 ± 0.8 mm to 3.0 ± 0.5 mm.
Volkswagen engineers reduced the main bearing diameters of the crankshaft from 52 mm to 48 mm in order to cut friction, and reduced the number of counterweights from eight to four. Additionally, the balance shaft concept has been changed to roller bearings.
The pistons in the new EA888 feature a newly developed, strength-enhanced alloy. Piston play was enlarged to optimize friction and piston wear was adapted based on a wear-resistant piston skirt coating with nanoparticles.
Other features include a new, lighter turbocharger/cylinder head assembly; the use of lightweight polymers for the oil pan, and use of aluminum for various screws and fasteners.
The oil circulation system reduces the power consumption of the control oil pump through a number of measures, including optimized pressure losses in the pressurized oil ducts and a reduction in oil pressure level in the low pressure stage
Intake and injection. Click to enlarge.
Intake manifold and injection system. Volkswagen engineers further optimized the TSI combustion process in a number of aspects.
Maximum system pressure for the fuel injection system is increased from 150 to 200 bar. The fuel supply to the high-pressure injectors is provided by way of a high-pressure rail which is isolated from the intake manifold and bolted directly onto the cylinder head. For high-pressure control, the system includes a pressure sensor with an adapted pressure range.
The engineers improved robustness in terms of knocking and spark advance at mean pressures increased up to 22 bar. The new integrated exhaust manifold enables operates at λ=1 at full load across wide ranges; the team optimized combustion stability under these changed conditions.
Integrated exhaust manifold with cooling. Volkswagen integrated the exhaust manifold within the cylinder head for the first time in these turbocharged direct-injection four-cylinder gasoline engines. The exhaust manifold is water-cooled, resulting in the elimination of the use of full load enrichment—i.e., enriching the air/fuel mixture at high loads—for cooling the exhaust gas. This results in a reduction in fuel consumption of approximately 20% when driving at highway speeds, according to Volkswagen.
Design of the integrated exhaust manifold. Click to enlarge. Cylinder head integrated exhaust gas cooling. Click to enlarge.
As a result, fuel consumption can be reduced both in normal customer driving and, especially, when employing a more sporty driving style.
The integrated exhaust manifold also aids rapid heat-up of the coolant and so is a key component of the thermal management system.
Achieving thermodynamically and thermomechanically optimized gas ducts and integrated exhaust manifold cooling ducts package posed a particular challenge during the development of the cylinder head, Lemke noted, especially with regard to castability. The cylinder head places extremely high demands on the casting process, uses a die with 12 sand cores in the bottom casting method.
To design the basic gas and water cores, Volkswagen used CFD (computational fluid dynamics) simulations and combined these with FEM (finite element methods) to thermomechanically optimize the cylinder head.
As there is intensive coupling between the exhaust gas and cooling water flows and heat transport in the aluminium within a very tight space involving extreme temperature gradients, for this project all three areas (gas, water, aluminium) were also calculated in a single simulation model for the first time.
Lemke said that this approach enables the more accurate simulation of the retroactive effects of the component temperatures on the fluid temperatures and the resultant heat flows.
New turbocharger. Click to enlarge.
Turbocharger and direct injection improvements. The turbocharger in the EA888 is an all-new Volkswagen design that develops a relatively high boost pressure of up to 19 psi (1.3 bar). Key features of the turbocharger include:
• a turbine wheel made from a new alloy that can withstand exhaust temperatures of up to 1,796 °F (980 °C) ;
• a pulsation damper;
• the addition of an oxygen sensor mounted directly upstream of the turbine wheel;
• use of a compressor wheel machined from solid; and
• a new electric wastegate actuator that adjusts the boost pressure, when power is not needed, to help reduce fuel consumption.
Thermal management. Thermal management is key to helping ensure maximum efficiency. The thermostat keeps coolant temperature between 185 and 225 °F, influenced by load and engine speed, to achieve the ideal balance between minimal internal friction and temperature management. By directing exhaust gases through the water-cooled exhaust manifold, they are approximately 160 degrees cooler by the time they reach the turbocharger.
Under full load, heat is significantly reduced thanks to the integrated cylinder head and water-cooled exhaust manifold.
The 2014 Jetta. The Jetta remains Volkswagen’s best-selling car, three years after the sixth-generation car was introduced to the US market as a 2011 model. The new Gen 3 EA888 1.8L turbo, when mated to a six-speed automatic transmission delivers EPA estimated fuel economy figure of 26 mpg in the city and 36 mpg on the highway (manual transmission)—an improvement of 5 mpg on the highway over the outgoing 2.5-liter engine. The new engine runs on regular unleaded gasoline.
On the cars fitted with the available 1.8T engine, the hydraulic power-assisted rack-and-pinion steering has been replaced by an electric-assist system, in line with the TDI and GLI models.
All Jetta models are fitted with a strut-type front suspension with coil springs, telescopic dampers, and an anti-roll bar. For 2014, the torsion-beam rear suspension that was fitted to the 2.0L S, 2.5L, and TDI models has been replaced by a sophisticated multilink independent arrangement that features three transverse and one longitudinal link per wheel.
This enables the longitudinal and transverse dynamics to be precisely configured almost independently of each other. The result is more agile, responsive and precise handling. All Jetta models have an anti-roll bar as part of the rear suspension.
Although the 1.8-liter engine is the smallest displacement engine in the Jetta line-up, it is not the current entry-level model on sale; that position falls to the older 2.0S, with a starting MSRP of 16,720.
The 1.8T SE comes next, with an MSRP of $18,895, followed by three addition 1.8T models (SE with Connectivity, SE with Connectivity and Sunroof, and SEL), the last of which has a starting MSRP of$25,590.
Jetta also offers its range of diesels and the Jetta hybrid (more on this below).
Driving impressions. We drove a 1.8T SE with six-speed automatic over a short course featuring winding roads and some hill climbing. The 1.8T is incredibly responsive in the Jetta; with the high torque delivered as early as it is, there is no perceptible lag between pushing down on the accelerator and the car shooting forward.
There was plenty of power for overtaking, and accelerating on a hill climb seemed effortless.
The combination of the lighter engine and the new steering and chassis elements give the Jetta good performance and feel on the winding roads.
All that said about how much fun it is to drive the 2014 Jetta with the 1.8-liter engine, it is not the most fuel-efficient of its competitive set, although it is the most powerful.
Comparing fuel economy (FE) for select MY 2014 models (mpgUS)
VW Jetta 1.8T Chevy Cruze 1.8 Chevy Cruze 1.4T Toyota Corolla 1.8 Toyota Corolla 1.8 Eco Ford Focus 2.0
Power (hp) 170 138 138 132 140 160
FE city (manual/auto) 26/25 25/22 28/26 28/27/29* 30 26/28
FE hwy (manual/auto) 36/35 36/35 42/39 37/36/38* 42 36/38
* Third entry is for Toyota’s CVT option.
For Jetta buyers looking for more fuel economy, Volkswagen will steer them either to the popular TDI diesel versions, or the Jetta gasoline-electric hybrid.
The 2.0-liter turbocharged inline four-cylinder TDI Clean Diesel engine produces 140 hp (104 kW) and 236 lb-ft (320 N·m) of torque. With either the standard six-speed manual transmission or the DSG automatic, the Jetta TDI can, according to EPA estimates, average 30 mpg in the city and 42 mpg on the highway. (The diesel with the manual is also extremely fun to drive.)
Next year will see a significant update to the TDI engine comparable to the update on the gasoline side with the 1.8T EA888 unit, Volkswagen said.
2014 Jetta Hybrid. With EPA estimated fuel economy ratings of 42 mpg city, 48 mpg highway, and 45 mpg combined, the Jetta Hybrid is the most fuel-efficient vehicle in the current Volkswagen lineup. (Earlier post.)
The hybrid powertrain combines a 1.4-liter turbocharged four-cylinder gasoline engine with a 20 kW electric motor. The gasoline engine is one of Volkswagen’s latest EA211 series of small engines. This features many improvements over the previous EA111 generation, including lightweight aluminum construction; an integrated (into the head) exhaust manifold; and a toothed-belt drive for its double overhead camshaft valvetrain that incorporates variable intake timing.
The only aspect to be carried over from the EA211 was the 82 mm cylinder spacing. The cylinder bore was decreased by 2 mm (to 74.5 mm) while the stroke was increased to 80mm, a change which not only helps compactness, but also increases torque.
The 1.4-liter TSI engine in the Jetta Hybrid features a 10.5:1 compression ratio, direct fuel injection, and turbocharging to produce 150 hp (112 kW) at 5,000 rpm. This engine’s turbocharging system offers strong boost response due to the design of the intake manifold, which enables the use of a small, single-scroll compressor. The intercooler is integrated directly into the injection-molded induction pipe. This design generates maximum torque of 184 lb-ft (249 N·m) at 1,600 rpm.
The hybrid module is a single, integrated unit that incorporates both the electric motor and the clutch that connects it to the engine. This water-cooled motor can add 20 kW (27 hp) to the mix, as well as a constant 114 lb-ft (154 N·m) of torque.
Combined, the system puts out 170 hp (127 kW) at 5,000 rpm and 184 lb-ft (250 N·m) of torque at a low 1000 rpm, giving the car incredibly smooth acceleration. (The maximum torque is limited by the transmission.)
In other words, the Jetta Hybrid’s power and torque matches that of the 1.8T.
The Jetta Hybrid SE starts at $27,260—representing a 44% premium over the$18,895 starting price for the 1.8T SE. The hybrid also offers a 61.5% improvement in city fuel economy over the 1.8 and a 33% improvement in highway fuel economy.
Another by-product of recent higher efficiency hybrids.
Without Toyota's HEVs, we would still be driving large VG-8 gas guzzlers?
HarveyD,
That is BS!
European car manufacturers have improved their fuel economy virtually without competition from the Prius, but with strong influence from EU demands on fleet average fuel economy.
I agree with Thomas. Prius does not sell very well in Europe anyway, so competition is with similar engines from other manufacturers (this one can see clearly in the table) and, of course, also with diesel engines. The sales numbers of the 5-cylinder VW engine were so low in Europe that this engine had only a minute impact on VW fleet average. They simply have to reduce fuel consumption of the cars that sell the most as the main priority. Although the 5-cylinder engine is very old and the comparison might be considered obsolete just because of that, it still shows what you can do with downsizing (reducing the number of cylinders, in particular) and turbocharging.
The comments to this entry are closed.
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2023-01-28 16:10:30
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https://www.springerprofessional.de/the-role-of-surface-defects-in-the-adsorption-of-methanol-on-fe3/10702114?fulltextView=true
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13.09.2016 | Original Paper | Ausgabe 6-7/2017 Open Access
The Role of Surface Defects in the Adsorption of Methanol on Fe3O4(001)
Zeitschrift:
Topics in Catalysis > Ausgabe 6-7/2017
Autoren:
Oscar Gamba, Jan Hulva, Jiri Pavelec, Roland Bliem, Michael Schmid, Ulrike Diebold, Gareth S. Parkinson
1 Introduction
Methanol (CH 3OH), the simplest alcohol, can be involved in several processes to produce hydrogen (oxidative reforming, decomposition, steam reforming) [ 1] and has received renewed interest for its importance in fuel-cell technology [ 2]. Moreover, methanol chemisorption has been termed a “smart chemical probe” [ 3] to study active sites on metal-oxide catalysts because adsorption allows to quantify the density of active sites, while the product distribution observed upon desorption is thought to reflect the nature of the active sites.
Studies of methanol adsorption on well-characterized metal-oxide surfaces have sought to correlate the atomic-scale structure with chemical reactivity [ 313]. Methanol does not typically chemisorb at truncated-bulk oxide surfaces, and some degree of additional coordinative unsaturation is required for dissociative adsorption, for example at step edges [ 11, 12]. Oxygen vacancies (V Os) have been shown to be the major active sites on TiO 2 [ 7, 8] and CeO 2 [ 9, 10] surfaces, with adsorbed methoxy species (CH 3O ) and hydroxyl groups formed at room temperature. Reaction products such as formaldehyde and methane are reported to emerge from this chemistry. However, as noted by Vohs in his recent review of oxygenate adsorption on metal oxides [ 4], little is known about the reactivity of isolated cation defects.
Iron oxides represent an interesting case in this regard because their bulk defect chemistry is dominated by the cation sublattice, with little evidence that V Os form in the bulk, or at the surface. Indeed, a recent study on the Fe 3O 4(111) surface [ 14] found dissociative adsorption on Fe-terminated regions of the surface, and concluded this to occur via a Brønsted acid–base mechanism requiring undercoordinated cation–anion pairs. Recombination to produce methanol was observed at 330–360 K in TPD, along with a disproportionation reaction between two adsorbed methoxy species to produce methanol and formaldehyde as follows [ 14]
$$2 {\text{ CH}}_{ 3} {\text{O}}_{\text{ads}} \to {\text{CH}}_{ 3} {\text{OH}}_{\text{gas}} + {\text{ CH}}_{ 2} {\text{O}}_{\text{gas}}$$
(1)
In this paper, we study the adsorption of CH 3OH on the magnetite Fe 3O 4(001) surface, with an emphasis on the role of isolated cation defects. This surface exhibits a (√2 × √2)R45° reconstruction over a wide range of oxygen chemical potentials [ 15] based on a rearrangement of the cations in the second and third surface layers [ 16]. The reconstructed unit cell, indicated by a purple square in Fig. 1a, contains four octahedrally coordinated Fe oct atoms (dark blue spheres) and eight oxygen atoms (red spheres) in the surface layer. The second layer contains three tetrahedrally coordinated Fe tet atoms (light blue spheres), one of which is an additional interstitial atom labelled Fe int. Essentially, this atom replaces two Fe oct atoms in the third layer resulting in a net removal of one cation per unit cell and a more oxidized surface. According to angle-resolved XPS and abinitio (DFT+U) calculations, all Fe atoms in the outermost four layers are Fe 3+ [ 16]. Hereafter, we will refer to this surface structure as the subsurface cation vacancy (SCV) reconstruction.
In prior studies, formic acid and, to some degree, water have been found to dissociate on this surface at room temperature [ 17, 18]. Here, using a combination of STM, XPS, and TPD, we demonstrate that methanol adsorption is restricted to surface defects. Specifically, the active sites are determined to be step edges, Fe adatoms, antiphase domain boundaries (APDBs) in the (√2 × √2)R45° reconstruction, and Fe atoms incorporated in the subsurface. Adsorption at the former two defects is due to the high coordinative unsaturation of the cations at such sites, whereas reactivity at the latter two defects is linked to the presence of Fe 2+. Desorption occurs via two channels in TPD; recombination to methanol, and a disproportionation reaction to form methanol and formaldehyde.
2 Materials and Methods
The experiments were performed in ultrahigh vacuum (UHV). A natural Fe 3O 4(001) sample (SurfaceNet GmbH) was prepared in situ by 1 keV Ar + sputtering at room temperature for 20 min followed by annealing in UHV at 873 K for 15 min. Once no contamination could be detected by XPS in the C1 s region, the sample was annealed in O 2 (5 × 10 −7 mbar) at 873 K for 15 min, which results in a surface with the SCV reconstruction. To create a surface with an increased defect density, Fe was evaporated from a 2-mm-thick rod (99.99+ %, MaTeck GmbH) at room temperature using an Omicron electron-beam evaporator; the deposition rate was calibrated by a quartz crystal microbalance. Methanol was obtained from Sigma Aldrich at a purity of 99.8 % and purified with several freeze-pump-thaw cycles. For the STM experiments methanol vapour was dosed into the background in the chamber through a high-precision leak valve.
STM measurements were performed at room temperature using an Omicron UHV-STM-1 instrument in constant current mode with electrochemically etched tungsten tips. The base pressure was below 10 −10 mbar.
XPS and TPD measurements were performed in a second vacuum system (base pressure 5 × 10 −11 mbar). XPS spectra were measured using a SPECS FOCUS 500 monochromatic source (Al Kα) and a SPECS PHOIBOS 150 electron analyzer at normal emission with a pass energy of 16 eV. TPD experiments were performed using a HIDEN HAL/3F RC 301 PIC quadrupole mass spectrometer (QMS). The sample was cooled by a Janis ST-400 UHV liquid-He flow cryostat, and heated by direct current at a rate of 1 K/s through a Ta back plate, on which the sample was mounted. The temperature was measured by a K-type thermocouple, and the sample was biased at −100 V during TPD measurements to prevent electrons from the QMS filament from reaching the sample. For the TPD and XPS measurements, methanol was dosed using a home-built effusive molecular beam source, which enables precise and reproducible exposures to a defined area on the sample surface [ 19].
3 Results
3.1 Defects on the As-Prepared Surface
Figure 2a shows an STM image of the as-prepared Fe 3O 4(001) surface. Rows of protrusions separated by 5.9 Å are due to fivefold-coordinated surface iron atoms (the dark blue balls in Fig. 1a) within the SCV reconstruction [ 16]. Surface oxygen atoms (red in Fig. 1a) are not imaged as there are no O-derived states in the vicinity of the Fermi level [ 16]. A step edge runs across the centre of the image from left to right, separating two adjacent terraces (yellow arrow). The apparent step height of 2.1 Å corresponds to the spacing between equivalent planes in the bulk structure [ 20] (i.e., the first and third layer in Fig. 1). Note that the direction of the iron rows rotates 90° when going from one terrace to the next [ 20], consistent with the inverse spinel structure of magnetite. As reported previously [ 20], steps that run parallel to the Fe rows on the upper terrace are generally straight, whereas perpendicular steps are often jagged.
A second, extended defect that is frequently observed on the freshly prepared surface is the antiphase domain boundary (APDB) [ 21], indicated by orange arrows in Fig. 2a. This feature appears as a chain of bright protrusions located on the Fe oct rows, and is typically aligned at 45° with respect to the row direction (see also Fig. 1a) [ 21]. The APDBs probably arise because the (√2 × √2)R45° reconstruction is lifted during each annealing cycle [ 22], and then renucleates on cooling through 723 K with one of two distinct registries with respect to the underlying bulk. It was noted previously [ 21] that the APDB forms such that two “narrow” sites (i.e., sites without 2nd-layer Fe tet in the SCV model) of the reconstruction meet at the interface. In the light of the SCV reconstruction [ 16], this preference can be reinterpreted as four Fe oct atoms in a row in the third layer (as illustrated in Fig. 1b). The alternative scenario with four Fe vacancies at the interface in the third layer would create twofold-coordinated O atoms, which is expected to be unfavourable. Interestingly, with no Fe int in the second layer and four Fe oct atoms in a row in the third layer, the local structure at the APDB is akin to a bulk truncated surface.
In addition to the line defects, two types of point defects are observed (Fig. 2a). Surface hydroxyl groups appear in STM as bright protrusions located on the Fe rows (cyan box). These species were initially identified through the adsorption of atomic H on this surface [ 23], and have also been observed following dissociative adsorption of water [ 21]. They are easily distinguished from other defects as they exhibit a characteristic hopping between opposite Fe rows in STM movies collected at room temperature [ 18, 21, 23]. It is important to note that the OH group is a H atom adsorbed on a surface O atom, albeit it appears as increased brightness of a pair on adjacent Fe atoms. This is an electronic effect, as the OH donates charge to the neighboring Fe atoms, which makes them brighter in STM [ 18]. Finally, the green boxes highlight pairs of bright features located on neighboring Fe rows. At first glance these features appear similar to hydroxyl groups, but they have a different apparent height (50 pm, compared to 20 pm for the OH) and they do not exhibit the characteristic hopping behavior described above. In the next section, we demonstrate that these defects are linked to Fe incorporated in the subsurface.
When the surface shown in Fig. 2a is exposed to 20 L of CH 3OH at room temperature new features appear at some of the defects. The defects do not change their positions, but the apparent height of some defects increases significantly. For example, the bright features located on neighboring iron rows are now 150 pm high, as compared to the 50 pm previously, see also the line profiles in Fig. 3, below. Interestingly, the methanol-induced, bright features were sometimes observed to disappear as the surface was scanned with the STM tip at room temperature, and the defects assumed their original appearance. However, because methanol was still present in the residual gas following the initial exposure, re-adsorption at defects was also observed.
3.2 Fe3O4(001)−(√2 × √2)R45° with Additional Cation Defects
To create an increased coverage of defects, 0.1 monolayer (ML) of Fe was deposited on the as-prepared SCV surface at room temperature, see Fig. 3a. (Here 1 ML is defined as one atom per (√2 × √2)R45° unit cell, i.e., 1.42 × 10 14 atoms/cm 2). This procedure results in Fe adatoms, as observed previously [ 24], which appear as bright protrusions between the Fe rows (red circles, see also the red star in the schematic in Fig. 1). The formation of stable adatoms is a distinctive property of the Fe 3O 4(001)−(√2 × √2)R45° surface and has been observed for many different elements [refs. 25-29]. The appearance of the Fe protrusions is similar to those observed for adatoms of Au, Ag and Pd [ 2527]. In addition, the density of the bright double features located on neighboring iron rows increases significantly upon Fe deposition (green circles). Similar features have been observed following the deposition of Ni, Co, Ti and Zr [ 28]. It is known that these elements enter the subsurface, filling one Fe oct vacancy in the third layer of the SCV reconstruction, which induces the Fe int to move and occupy the other. At 1 ML coverage this results in a (1 × 1) symmetry. By analogy, it is natural to propose that deposited Fe atoms also enter the surface, and locally lift the SCV reconstruction. In the case of Ni, the bright protrusion associated with the defect was found to appear above the two third-layer Fe oct atoms, rather than above the incorporated foreign metal cation [ 28]. In the case of Fe incorporation, two such sites are created resulting in two bright protrusions. Note that six adjacent Fe oct atoms are present in the third layer at such defects (green boxes in Fig. 1).
Exposing the surface shown in Fig. 3a to 20 L CH 3OH (Fig. 3b) leads to similar features as already shown in Fig. 2b. Adsorption occurs again at the step edges, APDBs, and the incorporated Fe defects. Frequently after methanol adsorption four distinct maxima are located at each incorporated Fe defect; two protrusions above each Fe oct row (see zoomed areas and line scans in Fig. 3). This suggests that all four Fe oct atoms affected by the subsurface modification can adsorb a methanol-related species. Along the row the separation of the maxima is 3 Å (see line scan in Fig. 3), which is consistent with the separation of neighboring Fe oct atoms. The adsorption of methoxy groups in close proximity to this defect could promote the disproportionation reaction (Eq. 1).
In addition, fuzzy features appear at the position of the Fe adatoms (red circle). Such apparently noisy parts in STM images are typically associated with weakly bound adsorbates that move during the scan. As before, dynamic desorption and readsorption was observed while scanning the surface with the STM at room temperature.
As mentioned above, there are two kinds of step edges on the Fe 3O 4(001) surface. The yellow arrow in Fig. 3a highlights a straight step edge, which runs parallel to the iron rows on the upper terrace. The second type, perpendicular to the octahedral iron rows, is more jagged (red arrow).
3.3 XPS
To investigate the nature of the observed protrusions we performed XPS experiments. Figure 4 shows O1 s and C1 s photoemission spectra that were recorded after the as-prepared Fe 3O 4(001) surface was exposed to 1.8 L of CH 3OH at 65 K, and subsequently annealed to progressively higher temperatures. The O1 s spectra are shown in Fig. 4a. The clean Fe 3O 4(001) surface exhibits a slightly asymmetric peak at 530.1 eV due to the lattice oxygen in magnetite as reported previously [ 23, 29]. Adsorption of 1.8 L CH 3OH at 65 K and annealing to 95 K produces two additional signals. The shoulder on the high-energy side at approximately 531 eV is consistent with both surface OH groups and methoxy species [ 30], while the peak at 533.1 eV is attributed to molecular CH 3OH [ 31]. Neither the shape nor position of the lattice oxygen peak is affected by methanol adsorption, only its intensity is reduced. As the sample is heated to progressively higher temperatures the molecular methanol desorbs first. By 280 K, the intensity of the peak at 533.1 eV has decreased from initially 32 % to just 4 %. The signal from the lattice oxygen increases again.
A similar conclusion can be drawn from the C1 s spectra shown in Fig. 4b. No detectable C peak is present when the surface is freshly prepared, but following the adsorption of methanol at 65 K and annealing to 95 K a symmetric peak centered at 286.5 eV appears. Upon heating this peak decreases in intensity. When the sample is heated to 280 K, its intensity decreases notably and its position shifts to 286 eV. This suggests that molecular methanol desorbs below room temperature, leaving only adsorbed methoxy species at 280 K. Methoxy generally has a lower C1 s binding energy than methanol (e.g. on ZnO [ 32], TiO 2 (110) [ 8], TiO 2 (001) [ 33], MgO [ 34], and CeO 2 (111) [ 31]) due to an increase in the electron density around the C atom when the hydroxyl proton is removed [ 32].
In order to quantitatively determine the methoxy coverage at 300 K from the XPS data we compared the C1 s peak area shown in Fig. 4b to that of the same surface exposed to a saturation coverage of formic acid (not shown). We have previously shown that formic acid exposure results in a complete monolayer of formate with a density of 2.84 × 10 14 molecules/cm 2, or 2 molecules per (√2 × √2)R45° unit cell [ 17]. Since both molecules contain just one C atom, the formate:methoxy ratio of 1:0.0283 suggests a methoxy coverage of 8.05 × 10 12 molecules/cm 2. Such a low coverage is consistent with the defect-only adsorption observed by STM in Fig. 2.
In keeping with the STM experiments, we also performed XPS measurements for a surface on which the defect concentration was enhanced by deposition of 0.3 ML Fe. Figure 5a compares the Fe2 p spectra obtained before and after the deposition of the Fe (no methanol exposure). As shown previously [ 24], the deposition of Fe on the Fe 3O 4(001) surface results in an increase in the Fe 2+ component at 708.7 eV in the Fe2 p 3/2 peak (compare inset in Fig. 5a). Adsorption of 10 L methanol at 280 K has no effect on the Fe 2p spectrum for the clean or Fe rich surface (not shown) and also little effect on the O1 s region (not shown). In the C1 s region (Fig. 5b) however, a peak appears at 286.1 eV on the clean surface, related to adsorption on defects. Heating to 300 K decreases this peak’s intensity, while it remains at the same position. On the Fe deposited surface (blue), a peak appears at 286.3 eV, the integral of which is 50 % larger than that obtained on the clean Fe 3O 4(001) surface. Again, heating to 300 K decreases the intensity of the peak, which remains at the same position.
The C1 s spectra show a small shift in the binding energies between the 0.3 ML Fe–Fe 3O 4(001) surface and the clean Fe 3O 4(001) surface. This shift may be related to changes in the valence charge on the carbon due to changes in the electronegativity of the vicinity. Such an interpretation was suggested for similar differences between oxidized and reduced cerium oxide thin films [ 31], where the peak position on the reduced surface is shifted to higher binding energy. Alternatively, there may be a contribution from molecular methanol adsorbed on Fe adatoms.
3.4 TPD
Temperature programmed desorption was performed following the adsorption of 10 L CH 3OH at 280 K on the clean Fe 3O 4(001) surface (Fig. 6a) and on the 0.3 ML Fe–Fe 3O 4(001) surface (Fig. 6b). Desorption of methanol was monitored by following mass 31 (red dots), which is the most intense cracking fragment of this molecule. By comparing this signal with mass 29 (black dots), possible reaction products can be determined [ 5]. On the pristine surface, the TPD spectra exhibit two peaks; a sharp peak at 335 K, and a broad shoulder in the region between 400 and 580 K. While the peak at 335 K has a similar shape and intensity for both masses, indicative of molecular methanol, the signal in the region between 400 and 580 K is higher for mass 29. With the addition of 0.3 ML Fe, the low-temperature peak at 335 K slightly shifts to 320 K, and its intensity increases by 42 %. As before, the intensity and line shape is similar for both mass 29 and 31. In the region between 400 and 580 K, the peak for mass 31 has its maximum at 450 K, while the peak for mass 29, which is significantly sharper, has the maximum intensity at 470 K. The blue curves in Fig. 6 result from subtracting the smoothed signal for mass 31 from the smoothed mass 29 spectrum. This procedure removes the contribution of methanol from the mass 29 signal, leaving only that of formaldehyde. For the clean Fe 3O 4(001)−(√2 × √2)R45° surface a broad peak is observed with a peak at 480 K, which is increased in intensity following the deposition of 0.3 ML Fe.
The peak at 480 K is thus attributable to desorption of formaldehyde, as observed previously, albeit at higher temperature, on Fe 3O 4(111) [ 14]. The mass spectra show no evidence for CH 4 or C XH Y, which would have been related to C–O bond cleavage [ 35]. Evolution of CO and H 2 was also not observed.
4 Discussion
On the basis of the STM, XPS and TPD results described above it is clear that methanol adsorption is restricted to defect sites on the Fe 3O 4(001) surface at room temperature. This is in contrast to formic acid, which was previously found to dissociate at regular lattice sites on this surface, resulting in a monolayer of bidentate formate species [ 17]. That the surface dissociates formic acid suggests that the cation–anion site separation is not prohibitive, but rather the acid-base strength of the surface atoms is insufficient to induce dissociation of weaker acids, such as methanol. On the basis of the STM data the active sites for dissociation are identified as step edges, Fe adatoms, incorporated Fe defects, and APDBs; we discuss these in turn in the following.
The reactivity of step edges on metal oxide surfaces is well documented [ 4, 36, 37], and is linked to the coordinative unsaturation of the atoms located there. In the present case, the step edge structures are not definitively known, so the discussion in the following must be limited to simple models of the step edges, not taking any possible reconstruction into account.
Following the procedure of Henrich [ 20], we have reevaluated the step stability in terms of covalent stability (coordinative unsaturation) for the SCV reconstructed surface. The most stable step parallel to the Fe oct rows (denoted B−α* by Henrich) exposes Fe tet atoms with only one dangling bond each (Fig. 7a). All other configurations expose Fe oct atoms with three dangling bonds at each atom, which are likely more reactive. Perpendicular to the rows several different configurations are similarly stable, and all expose both, Fe oct atoms with three dangling bonds per atom, and Fe tet atoms, which have two dangling bonds (one example is shown in Fig. 7b). The higher coordinative unsaturation of cations at the perpendicular steps likely makes these sites reactive.
This suggests that the perpendicular steps should be more reactive, as they expose Fe with higher coordinative unsaturation. However, our results showed some degree of reactivity for both kind of steps to methanol adsorption, as observed in Figs. 2 and 3.
Fe adatoms are not a common defect when the Fe 3O 4(001) surface is annealed in oxygen, but they become more prevalent when the surface is prepared in reducing conditions [ 38], when the Fe 3O 4 bulk is Fe-rich [ 39, 40], or when Fe is evaporated onto the surface [ 24]. The reactivity of the Fe adatoms is again most likely linked to coordinative unsaturation, because such cations have only two bonds to surface oxygen. The fuzzy appearance of the adsorbed methanol (inset Fig. 3), which indicates mobility underneath the STM tip, could mean that the molecule is adsorbed more weakly as compared to the other defects. This is possibly because dissociation of the molecule is precluded by the lack of undercoordinated lattice oxygen nearby that can receive the acid proton. Previous experimental and theoretical studies have found that adsorption of H is energetically unfavorable for an O surface with a subsurface Fe tet neighbor [ 41].
In addition to Fe adatoms, this paper reports a new defect linked to excess Fe that appears in STM as a pair of bright protrusions on opposite Fe oct rows. The density of such defects scales with the amount of deposited Fe, and resembles protrusions observed following incorporation of Ni, Ti and Co atoms in the subsurface vacancies of the SCV reconstruction [ 28]. By analogy, it is proposed that deposited Fe atoms can enter the subsurface and occupy one of the Fe oct vacancies present in the third layer. This induces the Fe int interstitial to move and occupy the other Fe oct site, resulting in a structure that locally resembles a bulk-truncated Fe 3O 4 lattice. It is interesting that such a defect would be reactive, because the surface layer of the bulk truncated surface should differ little from that of the SCV reconstruction, save for some small relaxations (<0.1 Å). The coordinative unsaturation of the Fe and O atoms is the same, and DFT+U calculations predict that both surfaces contain only Fe 3+-like cations in the surface layer [ 16]. A key difference affecting reactivity might be the local electronic structure: Density-of-states plots for the bulk truncation exhibit significantly greater density of (empty) states above the Fermi level than the SCV reconstruction due to Fe oct 2+ -like cations in the third layer [ 42, 43]. The presence of such states can make the region a stronger Lewis acid site, and more receptive to the electrons from the methoxy.
As discussed above, the APDB is an interruption of the vacancy-interstitial pattern in the second and third layers of the SCV reconstruction, and forms such that four Fe oct cations meet in the third layer at the junction (see Fig. 1). Locally, such a configuration also resembles again the unreconstructed lattice. Thus the reactivity can be explained in similar terms to the incorporated-Fe point defect.
4.2 Reaction Channels for the Adsorption of CH3OH on Fe3O4(001)
TPD analysis shows that for both surfaces, the signals for masses 29 and 31 match perfectly over the range between 280 and 350 K. This is evidence that the peak around 300 K observed is due to desorption of CH 3OH only [ 5]. Similar observations have been reported on Fe 3O 4(111) [ 14] and TiO 2(110) [ 44], where peaks in the same range of desorption temperatures have been assigned to the recombinative desorption of CH 3OH. Deposition of Fe results in an increased desorption in this region, which may be linked to molecular methanol adsorbed at the Fe adatoms.
The increased desorption in the region between 400 and 580 K suggests that the adsorption of methanol increases in the presence of Fe-related defects. Following the behavior of the spectra for both masses, is clear that, in the high-temperature range, the intensity and the shape of both spectra are different; this observation indicates the presence of another species in addition to methanol. Specifically, mass 31 has a contribution from methanol, while mass 29 has contributions from both methanol and formaldehyde (H 2CO); the latter is often observed as product of methanol reaction. Consequently, we assign the signal at high temperature as due to partial oxidation of methanol to formaldehyde.
Partial oxidation has been identified on other oxide surfaces as one of the main possible reactions. Interestingly, when the Fe 3O 4(001) surface was modified to have additional Fe adatoms and incorporated Fe defects a similar increase was observed in the peak at 320 K and the high-temperature products linked to the disproportionation reaction (Eq. 1). Given the relative abundance of Fe adatoms and incorporated Fe defects (approx. 50:50 at 0.3 ML coverage), and the probability that the Fe adatoms cannot dissociate CH 3OH and thus contributes to the lower temperature peak, it seems likely that the increase in formaldehyde production is due to the subsurface Fe defects. This can be because this defect promotes the disproportionation reaction by adsorbing two methoxy species in close proximity. Given the structural and electronic similarities between the incorporated Fe defect and the APDB, this defect also likely promotes the disproportionation reaction.
The temperature for the disproportionation reaction appears to depend on various factors such as the oxidation state of the metal oxide. In the case of vanadium oxide supported on CeO 2, formaldehyde desorption curves have shown signals in temperature ranges from 500 to 610 K, depending on the oxidation state of vanadium [ 45]. At the CeO 2(111) surface, methanol was oxidized to formaldehyde and water at 680 K after that methanol dissociation had occurred at oxygen vacancies [ 10]. On the other hand, complete dehydrogenation of methanol to CO and H 2 has been reported on highly reduced ceria surfaces [ 9]. It is important to note that a similar chemistry occurs on the Fe 3O 4(001) surface despite the lack of oxygen vacancies, and that defects related to excess Fe 2+ play the important role. Given the available evidence it appears that the adsorption of multiple methoxy species in close proximity at the incorporated Fe defect may promote the disproportionation reaction.
5 Conclusions
We have studied the adsorption of methanol on the Fe 3O 4(001) surface using TPD, STM and XPS. Methanol adsorbs dissociatively on Fe 3O 4(001) at 280 K at defect sites that were identified as step edges, antiphase domain boundaries (APDB), iron adatoms and incorporated-Fe defects. Whereas adsorption at the steps and Fe adatoms can be explained in terms of coordinative unsaturation, reactivity at the APDBs and incorporated Fe defects is linked to the local electronic structure; specifically to the presence of Fe 2+ cations in the surface layers. We propose that the adsorption of multiple methoxy species at the latter two defects promotes a disproportionation reaction to form methanol and formaldehyde.
Acknowledgments
G.S.P., R.B., O.G., J.H., and J.P. acknowledge funding from the Austrian Science Fund START prize Y 847-N20 and project number P24925-N20. R.B. and O.G. acknowledge a stipend from the Vienna University of Technology and the Austrian Science Fund as part of the doctoral college SOLIDS4FUN (W1243). U.D. and J.P. acknowledge support by the European Research Council (Advanced Grant “OxideSurfaces”). M.S. was supported by the Austrian Science Fund (FWF) within SFB F45 “FOXSI”. Open access funding provided by Austrian Science Fund (FWF).
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2020-10-19 16:22:50
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https://opus4.kobv.de/opus4-zib/frontdoor/index/index/docId/5277
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## The Price of Spite in Spot-checking games
Please always quote using this URN: urn:nbn:de:0297-zib-52775
• We introduce the class of spot-checking games (SC games). These games model problems where the goal is to distribute fare inspectors over a toll network. Although SC games are not zero-sum, we show that a Nash equilibrium can be computed by linear programming. The computation of a strong Stackelberg equilibrium is more relevant for this problem, but we show that this is NP-hard. However, we give some bounds on the \emph{price of spite}, which measures how the payoff of the inspector degrades when committing to a Nash equilibrium. Finally, we demonstrate the quality of these bounds for a real-world application, namely the enforcement of a truck toll on German motorways.
$Rev: 13581$
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2017-08-20 04:17:54
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https://www.maths.kisogo.com/index.php?title=Notes:Free_group
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# Notes:Free group
## Grillet - Abstract Algebra
This is taken from section 6 of chapter 1 starting on page 27.
### Reduction
• Let [ilmath]X[/ilmath] be a set.
• Let [ilmath]X'[/ilmath] be a disjoint set
• Let [ilmath]A:X\rightarrow X'[/ilmath] be a bijection, and let [ilmath]A':\eq A^{-1}:X'\rightarrow X[/ilmath] be the inverse bijection
• Let [ilmath]Y:\eq X\cup X'[/ilmath]
Caveat:Apparently we denote [ilmath]A[/ilmath] by [ilmath]x\mapsto x'[/ilmath] and [ilmath]A'[/ilmath] by [ilmath]y\mapsto y'[/ilmath] such that [ilmath](x')'\eq x[/ilmath] and [ilmath](y')'\eq y[/ilmath] - I am unsure of this.
Words in the "alphabet" [ilmath]Y[/ilmath] are finite, but possibly empty, sequences of elements of [ilmath]Y[/ilmath].
Next:
#### Reduced word
A word, [ilmath]a\in W[/ilmath] with [ilmath]a\eq(a_1,\ldots,a_n)[/ilmath] is reduced when:
• [ilmath]\forall i\in\{1,\ldots,n-1\}[a_{i+1}\neq a_i'][/ilmath]
For example:
1. [ilmath](x,y,z)[/ilmath] - reduced
2. [ilmath](x,x,x)[/ilmath] - reduced
3. [ilmath](x,y,y',z)[/ilmath] - NOT reduced
Reduction deletes subsequences of the form [ilmath](a_i,a'_i)[/ilmath] until a reduced word is reached.
### Sequences of reductions
1. We write [ilmath]a\overset{1}{\rightarrow} b[/ilmath] if
## Lee - Topological Manifolds
### Free group generated by
Let [ilmath]S:\eq\{\sigma\} [/ilmath] - a set containing a single thing. Then:
• [ilmath]F(S)[/ilmath] - the free group generated by [ilmath]S[/ilmath] (we may write [ilmath]F(\sigma)[/ilmath] instead, for short) is defined as follows:
1. The set of the group is [ilmath]F(\sigma):\eq\{\sigma\}\times\mathbb{Z} [/ilmath] - the set of all tuples of the form [ilmath](\sigma,m)[/ilmath] for [ilmath]m\in\mathbb{Z} [/ilmath]
2. The operation is: [ilmath](\sigma,a)\cdot(\sigma,b):\eq(\sigma,a+b)[/ilmath]
• We identify [ilmath]\sigma[/ilmath] with [ilmath](\sigma,1)[/ilmath], thus:
• [ilmath]\sigma^m\eq(\sigma,m\cdot(1))\eq(\sigma,m)[/ilmath]
Now suppose [ilmath]S[/ilmath] is some arbitrary set, then:
• [ilmath]F(S):\eq\underset{\sigma\in S}{\Huge \ast}F(\sigma)[/ilmath] - the free product of the groups [ilmath]F(\sigma)[/ilmath] for each [ilmath]\sigma\in S[/ilmath]
### Free product
Quite simple:
• [ilmath]\underset{\alpha\in I}{\huge\ast}G_\alpha[/ilmath] is a quotient by an equivalence relation on the free monoid generated by the set that is the disjoint union: [ilmath]\coprod_{\alpha\in I}G_\alpha[/ilmath] where:
• Two words in the monoid are considered equivalent if one can be reduced to the other.
• The rules for reduction are:
1. (two elements in the word from the same group are combined into one that is their product)
2. (any identity elements are discarded)
A bit of factorisation later and you've got an associative operation on the quotient with identity, just need to show inverse then.
## Notes
1. Obviously, concatenation of finite sequences [ilmath]a:\eq(a_1,\ldots,a_\ell)[/ilmath] and [ilmath]b:\eq(b_1,\ldots,b_m)[/ilmath] is:
• [ilmath]a\cdot b:\eq(a_1,\ldots,a_\ell,b_1,\ldots,b_m)[/ilmath]
|
2021-05-13 15:10:21
|
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|
https://www.gradesaver.com/textbooks/math/other-math/CLONE-547b8018-14a8-4d02-afd6-6bc35a0864ed/chapter-7-measurement-7-3-the-metric-system-capacity-and-weight-mass-7-3-exercises-page-500/43
|
## Basic College Mathematics (10th Edition)
Published by Pearson
# Chapter 7 - Measurement - 7.3 The Metric System - Capacity and Weight (Mass) - 7.3 Exercises - Page 500: 43
850 mg
#### Work Step by Step
0.85 g to mg Unit for the answer (mg) is in numerator; unit being changed (g) is in denominator so it will divide out. The unit fraction is $\frac{1000 mg}{1 g}$. Multiply 0.85 g times to the unit fraction 0.85 g * $\frac{1000 mg}{1 g}$= 850 mg
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.
|
2019-10-23 04:49:13
|
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https://space.stackexchange.com/questions/29753/are-side-boosters-sometimes-angled-even-if-they-are-symmetrically-arranged
|
# Are side boosters sometimes angled even if they are symmetrically arranged?
When more than one side booster is added to a first stage to increase thrust and they are distributed uniformly around the core (as opposed to those shown in links below) thrust does not seem to need to be angled. They could all point straight down (except for dynamic thrust vectoring for steering) due to symmetry.
In the two images below, both from the recent NASA Spaceflight article Beidou-3 MEO-5 and MEO-6 launched by Long March 3B it looks to me like the exhaust from the boosters is angled outwards, I've added an annotated, cropped, and enlarged version to better show what I think that I'm seeing.
Is this in fact the case for this launch, and does this happen frequently?
![Beidou-3 MEO-5 MEO-6 Long March 3B]https://i.stack.imgur.com/vJJZQ.jpg
• Thrust vectors are probably oriented to intersect the mean center of mass. That way, if one engine fails, you don't get a sudden pitch or yaw moment from the asymmetry. Jul 29, 2018 at 4:50
• @AnthonyX I see, so approximately "torqueless". Losses due to $\cos(\theta)$ are small, and resulting radial loading due to $\sin(\theta)$ is manageable?
– uhoh
Jul 29, 2018 at 4:54
• That configuration with the nozzles angled out was also used for the Ariane 1-4. Jul 29, 2018 at 8:10
• @Jack Tangential gimbaling would be for steering control -- swing two opposing nozzles in the same direction and the vehicle pitches or yaws; swing them in opposite direction and it rolls. Jul 29, 2018 at 17:08
• And the Snark
– user20636
Jul 30, 2018 at 9:21
## 1 Answer
Caveat: This is a somewhat speculative answer based on reasoning alone
The side booster nozzles appear to be oriented such that their thrust vectors intersect what appears to be the vehicle's nominal center of mass. The rationale for this would be to avoid or at least mitigate the effect of a thrust asymmetry if a booster fails, underperforms, or burns out prematurely. The idea would be that thrust will always be applied through the vehicle's center of mass (except where commanded as a correction or steering maneuver). Any off-center thrust would apply a torque in the pitch and/or yaw axis, which would cause the vehicle to pitch or yaw.
• Would you think (also based on speculative reasoning, you have more of a handle on it than me) that this angle would/should be maintained through the entire first stage ascent? Would it make sense to stop the angle after max-Q or at a specific point of ascent? Sep 17, 2018 at 19:35
• @MagicOctopusUrn I believe the engines on the Long March 3 boosters are only tangentially gimbaled (as pointed out in comments above), so the outward angle is maintained for the entire booster burn. Ditto any fixed-nozzle solid boosters. Gimbaled booster engines might be mounted so that their neutral position is canted outward like this, but during nominal flight they could steer directly aft to claw back the cosine loss. Sep 17, 2018 at 19:45
• Adding some speculation to yours: If you lose a strap-on booster early, the mission is over, catastrophically, so who cares about preventing a yaw moment then? I'll bet that they aim the strap-on's thrust through near where the expected COM is at burnout, because that's where preventing an unexpected yaw moment (from slightly early burnout on one side) could save the mission. Sep 17, 2018 at 20:55
• @WayneConrad Loss of a single strap-on booster isn't necessarily a loss-of-vehicle/loss-of-mission situation, particularly with a liquid booster, and particularly if it's not too early in the burn. A number of launches have lost an engine on ascent and continued on to partial or full success (e.g. AS-101, Apollo 6, Apollo 13, Falcon 9 CRS-1). If a booster failure causes a large yaw moment, though, it will be a loss-of-vehicle situation. Sep 17, 2018 at 21:59
• @RussellBorogove If you lose an engine in a multi-engine liquid stage, the remaining engines can continue to burn the fuel intended for the lost engine: you lose thrust, but you don't lose delta-V (I think). But if you lose a strap-on solid motor, you're not getting anything out of its fuel anymore. Now it's just weight. Plus, how do you lose a strap-on solid motor in a benign way? Don't they tend to burn until they either burn out more or less normally, or stop in some way that ruins your day? Sep 17, 2018 at 22:49
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2022-09-29 16:54:11
|
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|
https://socratic.org/questions/how-do-you-solve-sqrt-9-x-sqrt-1-x-sqrt-x-16
|
# How do you solve sqrt( 9 + x) + sqrt(1+x)=sqrt(x + 16)?
May 9, 2015
First of all the existence conditions:
$9 + x \ge 0 \Rightarrow x \ge - 9$
$1 + x \ge 0 \Rightarrow x \ge - 1$
$x + 16 \ge 0 \Rightarrow x \ge - 16$,
the solution of the system of inequalities is $x \ge - 1$.
Now we can square the two members because the first is positive or zero (sum of two quantities positive or zero), and so it is the second.
${\left(\sqrt{9 + x} + \sqrt{1 + x}\right)}^{2} = {\left(\sqrt{x + 16}\right)}^{2} \Rightarrow$
$9 + x + 2 \sqrt{9 + x} \cdot \sqrt{1 + x} + 1 + x = x + 16 \Rightarrow$
$2 \sqrt{9 + x} \cdot \sqrt{1 + x} = x + 16 - 9 - x - 1 - x \Rightarrow$
$2 \sqrt{9 + x} \cdot \sqrt{1 + x} = 6 - x$.
Now we have another contion to solve, because we have to make sure that the second member is positive or zero, because we want to square the two members another time.
$6 - x \ge 0 \Rightarrow x \le 6$ that, joined with the first one $x \ge - 1$, becomes:
$- 1 \le x \le 6$.
$4 \left(9 + x\right) \left(1 + x\right) = {\left(6 - x\right)}^{2} \Rightarrow$
$4 \left(9 + 9 x + x + {x}^{2}\right) = 36 - 12 x + {x}^{2} \Rightarrow$
$36 + 36 x + 4 x + 4 {x}^{2} - 36 + 12 x - {x}^{2} = 0 \Rightarrow$
$3 {x}^{2} + 28 x = 0 \Rightarrow x \left(3 x + 28\right) = 0 \Rightarrow$
${x}_{1} = 0$ acceptable solution (it is in $- 1 \le x \le 6$)
${x}_{2} = - \frac{28}{3}$ not acceptable solution (it is not in $- 1 \le x \le 6$).
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2019-09-21 04:59:27
|
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https://en.wikipedia.org/wiki/Electron_degeneracy_pressure
|
# Electron degeneracy pressure
Electron degeneracy pressure is a particular manifestation of the more general phenomenon of quantum degeneracy pressure. The Pauli exclusion principle disallows two identical half-integer spin particles (electrons and all other fermions) from simultaneously occupying the same quantum state. The result is an emergent pressure against compression of matter into smaller volumes of space. Electron degeneracy pressure results from the same underlying mechanism that defines the electron orbital structure of elemental matter. For bulk matter with no net electric charge, the attraction between electrons and nuclei exceeds (at any scale) the mutual repulsion of electrons plus the mutual repulsion of nuclei; so absent electron degeneracy pressure, the matter would collapse into a single nucleus. In 1967, Freeman Dyson showed that solid matter is stabilized by quantum degeneracy pressure rather than electrostatic repulsion.[1][2][3] Because of this, electron degeneracy creates a barrier to the gravitational collapse of dying stars and is responsible for the formation of white dwarfs.
When electrons are squeezed together too closely, the exclusion principle requires them to have different energy levels. To add another electron to a given volume requires raising an electron's energy level to make room, and this requirement for energy to compress the material manifests as a pressure.
Electron degeneracy pressure in a material can be computed as[4]
${\displaystyle P={\frac {2}{3}}{\frac {E_{\text{tot}}}{V}}={\frac {2}{3}}{\frac {\hbar ^{2}k_{\rm {F}}^{5}}{10\pi ^{2}m_{\rm {e}}}}={\frac {(3\pi ^{2})^{2/3}\hbar ^{2}}{5m_{\rm {e}}}}\rho _{N}^{5/3},}$
where ħ is the reduced Planck constant, me is the mass of the electron, and ρN is the free electron density (the number of free electrons per unit volume).
When particle energies reach relativistic levels, a modified formula is required.
This pressure is derived from the energy of each electron with wave number k = 2π/λ, having
${\displaystyle E={\frac {p^{2}}{2m}}={\frac {\hbar ^{2}k^{2}}{2m}}~,}$
and every possible momentum state of an electron within this volume up to the Fermi energy being occupied.
This degeneracy pressure is omnipresent and is in addition to the normal gas pressure P = NkT/V. At commonly encountered densities, this pressure is so low that it can be neglected. Matter is electron degenerate when the density (n/V) is high enough, and the temperature low enough, that the sum is dominated by the degeneracy pressure.
Perhaps useful in appreciating electron degeneracy pressure is the Heisenberg uncertainty principle, which states that
${\displaystyle \Delta x\Delta p\geq {\frac {\hbar }{2}}}$
where Δx is the uncertainty of the position measurements and Δp is the uncertainty (standard deviation) of the momentum measurements. A material subjected to ever-increasing pressure will compact more, and, for electrons within it, their delocalization, Δx, will decrease. Thus, as dictated by the uncertainty principle, the spread in the momenta of the electrons, Δp, will grow. Thus, no matter how low the temperature drops, the electrons must be traveling at this "Heisenberg speed", contributing to the pressure. When the pressure due to this "Heisenberg motion" exceeds that of the pressure from the thermal motions of the electrons, the electrons are referred to as degenerate, and the material is termed degenerate matter.
Electron degeneracy pressure will halt the gravitational collapse of a star if its mass is below the Chandrasekhar limit (1.39 solar masses[5]). This is the pressure that prevents a white dwarf star from collapsing. A star exceeding this limit and without significant thermally generated pressure will continue to collapse to form either a neutron star or black hole, because the degeneracy pressure provided by the electrons is weaker than the inward pull of gravity.
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2017-05-30 03:39:18
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|
https://discuss.gradle.org/t/multi-project-in-gradle/27508
|
I want to include My first project “Component1-1-1” inside my second project “Component1-1”, I added the following syntax to my setting.gradle file :
include ‘:Component1-1-1’
project(’:Component1-1-1’).projectDir = new File(settingsDir, ‘…/Component1-1-1’)
• `:` is the path separator used in project hierarchies.
• When calling `project()` in the build.gradle files, the path is relative to the current project.
• When calling `project()` from settings.gradle, all paths are relative to the root and the leading `:` can be dropped.
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2020-09-22 20:06:49
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|
https://oneclass.com/study-guides/ca/utsg/mat/mat-137y1/1158-summary-notes-2.en.html
|
Study Guides (390,000)
CA (150,000)
UTSG (10,000)
MAT (1,000)
MAT137Y1 (100)
None (3)
# Summary notes 2
Department
Mathematics
Course Code
MAT137Y1
Professor
None
This preview shows pages 1-3. to view the full 38 pages of the document.
MAT137Y1b.doc
Page 1 of 38
Lecture #12 Thursday, October 16, 2003
DIFFERENTIATION
What is the slope of the secant line PQ?
(
)
(
)
( )
(
)
(
)
h
cfhcf
chc
cfhcf+
=
+
+ .
Idea: Get the slope of the tangent line as a limit of slopes of secant lines.
The slope of the tangent line at
c
x
=
ought to be
(
)
(
)
h
cfhcf
h
+
0
lim.
Example
(
)
xxf= at 0
=
x
This DOESNT have a well defined tangent line
Definition
f is differentiable at
c
x
=
if the limit
(
)
(
)
h
cfhcf
h
+
0
lim exists. If it does, we call it the derivative of f at
c and we denote it by
(
)
cf
.
Geometrically
(
)
cf
is the slope of the tangent line going through
(
)
(
)
cfc,.
What is the equation for tangent line?
(
)
(
)
(
)
cxcfcfy
=
P
c
Q
f
(
)
(
)
hcfhc++ ,
(
)
(
)
cfc,
c + h
www.notesolution.com
Only pages 1-3 are available for preview. Some parts have been intentionally blurred.
MAT137Y1b.doc
Page 2 of 38
Lecture #13 Tuesday, October 21, 2003
Example
For function
(
)
2
xxf=, the derivative of f at 2
=
c is
( )
(
)
(
)
( )
44lim
22
lim2
0
22
0
=+=
+
=
h
h
h
f
hh
.
The derivative of f is itself a function for
(
)
2
xxf= repeat the same calculation for any value of c.
( )
(
)
(
)
(
)
(
)
( )
chc
h
chchc
h
chc
h
cfhcf
cf
hhhh
22lim
2
limlimlim
0
222
0
22
00
=+=
++
=
+
=
+
=
At any fixed value of x,
(
)
xxf2=
.
Definition
The derivative of f is a function, denoted f
, and
( )
(
)
(
)
h
xfhxf
xf
h
+
=
0
lim, if it exists.
Terminology
To differentiate a function is to find the derivative.
Notice: The function f has to be defined in the interval
(
)
δ+δxx , in order for
(
)
xf
to be defined.
Example
Actually, even if f is continuous on
(
)
δ+δxx ,, it doesnt mean
(
)
xf
is defined.
Consider
(
)
0,== cxxf
Theorem
If f is differentiable at x, the f is continuous.
Being differentiable is better’ than being continuous.
Proof:
Because f is differentiable at x,
(
)
(
)
( )
xf
h
xfhxf
h
=
+
0
lim
( ) ( )( )
(
)
(
)
( )
00limlim
00
=
=
+
=+ xfh
h
xfhxf
xfhxf
hh
So f is continuous.
DIFFERENTIATION RULES
Building Blocks
If
(
)
cxf= (a constant function), then
(
)
0=
xf for all x.
If
(
)
xxf=, then
(
)
1=
xf for all x.
f(x)
x
www.notesolution.com
Only pages 1-3 are available for preview. Some parts have been intentionally blurred.
MAT137Y1b.doc
Page 3 of 38
Theorem: Sums and Scalar Multiples
Let f, g be differentiable at x and α a constant.
Then
(
)
gf+ and fα are differentiable, then
( ) ( ) ( ) ( )
xgxfxgf
+
=
+
( ) ( ) ( )
xfxf
α=
α
Proof:
( ) ( )
(
)
(
)
(
)
(
)
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www.notesolution.com
|
2019-11-17 07:32:20
|
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|
https://discuss.codechef.com/t/chgoram2-editorial/53535
|
# CHGORAM2 - Editorial
Can you prove the time complexity of your solution?
If you prefer a video explanation: CHGORAM Video Solution - February Long Challenge 2020
Author: Simon St James
Editorialist: William Lin
Medium-Hard
# PREREQUISITES:
Trees, Observations, Dynamic Programming
# PROBLEM:
We are given a tree and some of the nodes are marked. Find the number of triples of distinct nodes such that all three nodes are marked and all three pairs of nodes within the triple have the same distance.
# QUICK EXPLANATION:
For any such triple, there exists a center c such that the distance from c to all three nodes of the triple are equal.
For each node u, count the number triples such that the LCA of the triple is u. There are two cases: when u=c and when u \ne c.
We can calculate dp1_{u,i}, the number of nodes in u's subtree with distance i, and dp2_{u, i}, the number of pairs of nodes (a, b) with the same depth in u's subtree which satisfy (depth(a)-depth(lca(a, b)))-(depth(lca(a, b))-depth(u))=i.
If we process the DP for node u in O(k), where k is the depth of the second deepest subtree of u, then the time complexity is guaranteed to be O(N).
# EXPLANATION:
Observation 1. There exists a node c such that c is equidistant to all nodes in the triple.
Proof
On the path between two of the nodes, we know that there is a node c which branches out to the third node:
We can set variables for the distances from c to each of the three nodes and write a system of linear equations:
Solving the system of linear equations shows that c is equidistant to all three nodes.
Let’s perform a DFS on the tree, and for each node u that we visit, let’s find the number of triples which have their lowest common ancestor equal to u. For example, this is one such triple:
Notice that we can split the component containing the triple into a leg and a fork, as shown below:
Note that in order for a leg to be able to pair with a fork, a=d-b must be satisfied. It seems like if we can somehow find the number of legs and forks coming out of u, we can find the answer for u!
This suggests that we use the following DP: dp1_{u, i} will be the number of legs with a=i which start at node u and dp2_{u, i} will be the number of forks with d-b=i which start at node u.
We need to consider how we can perform the transitions to find dp1_u and dp2_u. The following procedure will be useful for this problem:
• We start with dp1_u and dp2_u representing the single node u.
• Let v_1, v_2, \dots, v_k be the children of u. Let’s merge dp1_{v_1} and dp2_{v_1} to dp1_u and dp2_u somehow, so that 1. the answer will be updated with new triples which form from the union of u and subtree v_1 and 2. dp1_u and dp2_u will represent the union of the node u and the subtree v_1.
• For each i, let’s merge dp1_{v_i} and dp2_{v_i} to dp1_u and dp2_u somehow, so that 1. the answer will be updated with new triples which form from the union of the current processed subtree of u and the subtree v_i and 2. dp1_u and dp2_u will represent the union of the node u and the first i subtrees.
• At the end, dp1_u and dp2_u will represent the entire subtree u.
So now, we need to focus on how we can merge dp1_v and dp2_v with dp1_u and dp2_u.
A. First, because v is deeper than u by 1, we need to change dp1_v and dp2_v. More specifically, all legs from v have a increased by 1, so dp1_{v, j+1}=dp1_{v, j} (and dp1_{v, 0} will be 0), and all forks from v have b increased by 1 and d-b decreased by 1, so dp2_{v, j-1}=dp2_{v, j} (we should just delete dp2_{v, 0} since negative indexes are useless).
B. Next, we need to calculate the triples which form from merging v to u for our total answer. Each triple consists of a leg from v and a fork from u and vice versa. So, for each j, we will add dp1_{v, j}\cdot dp2_{u, j}+dp2_{v, j}\cdot dp1_{u, j} to the total answer.
C. Next, we might have some forks which form from merging v to u. Each fork consists of a leg from v and a leg from u. So, for each j, we will add dp1_{v, j}\cdot dp1_{u, j} to dp2_{u, j}.
D. Lastly, when v becomes part of subtree u, all legs and forks from v should be added to u. So, for each j, we will add dp1_{v, j} to dp1_{u, j} and dp2_{v, j} to dp2_{u, j}.
This DP solution works in O(n^2) time. Let’s try to optimize it.
Let d be the smallest depth of both v_i (the depth including the parent edge) and current processed subtree of u (the union of u with its first i-1 children subtrees).
For step A, let’s view dp1_v and dp2_v both as lists. The transition on dp1_v is equivalent to adding a 0 to the front of the list and the transition on dp2_v is equivalent to removing the first element of the list. These are both linear time operations - unless we store the list in reverse, as operations on the end of the list are constant time. So that’s what we’ll do.
For steps B and C, when we iterate j, we only have to iterate from 0 to d. Why don’t we need to iterate over d? d+1 is greater than the depth of one of the subtrees, so one of dp_{v, d+1} and dp_{u, d+1} is guaranteed to be 0. In steps B and C, all terms are a product dp_{v, d+1} and dp_{u, d+1}, so all j>d will cause these terms to evaluate to 0. This makes steps B and C run in O(d) time.
For step D, if v has the smallest depth, then we don’t need to do anything. Otherwise, we will add dp_u to dp_v instead, and let dp_v replace dp_u. In both cases, we only iterate j up to d, which makes step D run in O(d) time.
In steps B, C, and D, we applied what’s known as the merge-small-to-large on depths - we find the smallest depth d of both subtrees before merging them. Then, we merge the small subtree to the large subtree to take O(d) time for the merge.
Observation 2. The merge-small-to-large trick on depths optimizes the solution to O(n) time.
Proof
Whenever we merge v to u, find the subtree with minimum depth (remember, we consider the parent edge for v, and for u, we only consider the union of u with the first i children subtrees). If both subtrees have the same depth, choose the deepest path with the greatest id for the leaf node. Then, in that subtree, color the deepest path, and if there are multiple, color the one with smallest id for the leaf node. Note that we color O(d) edges, so the total number of edges that we color is the time complexity of our solution.
Below are some examples:
In the example above, subtree u has the minimum depth, so we color the deepest path of u (which is nothing).
In the example above, we merge another child subtree v to u. v has the minimum depth, so we color its deepest path.
In the example above, we finish processing u, and now it becomes a child subtree v to be merged with its parent u. We can see that v has the minimum depth, so we color its deepest path.
We can work with more cases, but it seems like no edge can ever be colored twice!
What can we say about a path when we color it? It’s part of the subtree with minimum depth, so when we merge the two subtrees, that colored path can’t be the deepest path anymore. Since we only color deepest paths, that means we never color a colored path again!
There are O(n) edges in a tree, so if no edge is colored twice, we color at most O(n) edges. Since this is equivalent to the time complexity of our solution, our solution works in O(n) time.
# SOLUTIONS:
Setter's Solution
// Simon St James (ssjgz) - Setter's solution for "Equilateral Treeangles" - 2019-07-17.
// Problem was later renamed to "Chef and Gordon Ramsay 2" when I saw CHGORAM in AUG19
// and noticed how my Problem could be turned into a sequel to it :)
#include <iostream>
#include <vector>
#include <map>
#include <algorithm>
#include <cassert>
using namespace std;
const int numTripletPermutations = 1 * 2 * 3; // i.e. == factorial(3).
struct DescendantWithHeightInfo
{
// The number of suitable pairs of descendants of Node with the given height
// that have Node as their Lowest Common Ancestor.
int64_t numPairsWithHeightWithNodeAsLCA = 0;
// The name "number" is not strictly correct: if numPairsWithHeightWithNodeAsLCA > 0, then this
// will indeed be the number of suitable descendants of this Node which have this height,
// but if numPairsWithHeightWithNodeAsLCA == 0, "number" will also be 0.
//
// The "number" count is used to avoid the overcount when performing completeTripletsOfTypeA
// (Centroid Decomposition has no notion of our parent-child relationship, so
// will count "descendants" of a Node alongside "non-descendants" of a Node.)
int number = 0;
};
struct Node
{
vector<Node*> neighbours;
bool isSuitable = false;
int height = 0;
// The total number of entries into descendantWithHeightInfo, summed across *all* nodes,
// will be O(n log n).
map<int, DescendantWithHeightInfo> descendantWithHeightInfo;
};
struct HeightInfo
{
int numWithHeight = 0;
// Make a note of which Node this info has been incorporated into (i.e. which Node's
// descendantWithHeightInfo it has been used to update) so we don't accidentally
// incorporate it into the same Node twice!
Node* lastIncorporatedIntoNode = nullptr;
};
class DistTracker
{
public:
// O(maxDist).
DistTracker(int maxDist)
: m_numWithDist(2 * maxDist + 1),
m_maxDist(maxDist)
{
}
// O(1).
void insertDist(const int newDist)
{
numWithDistValue(newDist)++;
m_largestDist = max(m_largestDist, newDist);
};
// O(1).
int numWithDist(int dist)
{
return numWithDistValue(dist);
}
// O(1).
{
if (m_largestDist != -1)
m_largestDist += distDiff;
}
// O(1).
int largestDist() const
{
return m_largestDist;
}
private:
vector<int> m_numWithDist;
int m_maxDist = -1;
int m_largestDist = -1;
int& numWithDistValue(int dist)
{
return m_numWithDist[dist - m_cumulativeDistAdjustment + m_maxDist];
}
};
template <typename NodeProcessor>
void doDfs(Node* node, Node* parentNode, int depth, DistTracker& distTracker, DistTrackerAdjustment distTrackerAdjustment, NodeProcessor& processNode)
{
processNode(node, depth, distTracker);
for (auto child : node->neighbours)
{
if (child == parentNode)
continue;
doDfs(child, node, depth + 1, distTracker, distTrackerAdjustment, processNode);
}
}
int countDescendants(Node* node, Node* parentNode)
{
auto numDescendants = 1; // Current node.
for (const auto& child : node->neighbours)
{
if (child == parentNode)
continue;
numDescendants += countDescendants(child, node);
}
return numDescendants;
}
int findCentroidAux(Node* currentNode, Node* parentNode, const int totalNodes, Node** centroid)
{
auto numDescendants = 1;
auto childHasTooManyDescendants = false;
for (const auto& child : currentNode->neighbours)
{
if (child == parentNode)
continue;
const auto numChildDescendants = findCentroidAux(child, currentNode, totalNodes, centroid);
if (numChildDescendants > totalNodes / 2)
{
// Not the centroid, but can't break here - must continue processing children.
childHasTooManyDescendants = true;
}
numDescendants += numChildDescendants;
}
if (!childHasTooManyDescendants)
{
// No child has more than totalNodes/2 descendants, but what about the remainder of the graph?
const auto nonChildDescendants = totalNodes - numDescendants;
if (nonChildDescendants <= totalNodes / 2)
{
assert(centroid);
*centroid = currentNode;
}
}
return numDescendants;
}
Node* findCentroid(Node* startNode)
{
const auto totalNumNodes = countDescendants(startNode, nullptr);
Node* centroid = nullptr;
findCentroidAux(startNode, nullptr, totalNumNodes, ¢roid);
assert(centroid);
return centroid;
}
void doCentroidDecomposition(Node* startNode, int64_t& numTriplets)
{
Node* centroid = findCentroid(startNode);
auto addSomeTypeATripletsForNode = [&numTriplets](Node* node, DistTracker& distTracker)
{
// This will be called O(log2 n) times for each node before that node's
// Type A Triplets are fully completed.
for (const auto& heightPair : node->descendantWithHeightInfo)
{
const auto descendantHeight = heightPair.first;
const auto requiredNonDescendantDist = (descendantHeight - node->height);
if (requiredNonDescendantDist > distTracker.largestDist())
break; // Optimisation - no point continuing with larger descendantHeights.
const auto numPairsWithHeightWithNodeAsLCA = heightPair.second.numPairsWithHeightWithNodeAsLCA;
const auto numNewTriplets = numPairsWithHeightWithNodeAsLCA * distTracker.numWithDist(requiredNonDescendantDist) * numTripletPermutations;
assert(numNewTriplets >= 0);
numTriplets += numNewTriplets;
}
};
auto propagateDists = [&addSomeTypeATripletsForNode](Node* node, int depth, DistTracker& distTracker)
{
};
auto collectDists = [](Node* node, int depth, DistTracker& distTracker)
{
if (node->isSuitable)
distTracker.insertDist(depth);
};
const auto numNodesInComponent = countDescendants(startNode, nullptr);
{
DistTracker distTracker(numNodesInComponent);
for (auto& child : centroid->neighbours)
{
doDfs(child, centroid, 1, distTracker, AdjustWithDepth, propagateDists );
doDfs(child, centroid, 1, distTracker, DoNotAdjust, collectDists );
}
}
{
DistTracker distTracker(numNodesInComponent);
// Do it again, this time backwards ...
reverse(centroid->neighbours.begin(), centroid->neighbours.end());
// ... and also include the centre, this time.
if (centroid->isSuitable)
distTracker.insertDist(0);
for (auto& child : centroid->neighbours)
{
doDfs(child, centroid, 1, distTracker, AdjustWithDepth, propagateDists );
doDfs(child, centroid, 1, distTracker, DoNotAdjust, collectDists );
}
}
for (auto& neighbour : centroid->neighbours)
{
assert(std::find(neighbour->neighbours.begin(), neighbour->neighbours.end(), centroid) != neighbour->neighbours.end());
// Erase the edge from the centroid's neighbour to the centroid, essentially "chopping off" each child into its own
// component ...
neighbour->neighbours.erase(std::find(neighbour->neighbours.begin(), neighbour->neighbours.end(), centroid));
// ... and recurse.
doCentroidDecomposition(neighbour, numTriplets);
}
}
void completeTripletsOfTypeACentroidDecomposition(vector<Node>& nodes, Node* rootNode, int64_t& numTriplets)
{
doCentroidDecomposition(rootNode, numTriplets);
// Fix the overcount caused by Centroid Decomposition (over-)counting descendants of a node as non-descendants
// of a node!
for (auto& node : nodes)
{
for (const auto descendantHeightPair : node.descendantWithHeightInfo)
{
const auto height = descendantHeightPair.first;
const auto numPairsWithHeightWithNodeAsLCA = descendantHeightPair.second.numPairsWithHeightWithNodeAsLCA;
// Centroid decomposition would have (wrongly) added numPairsWithHeightWithNodeAsLCA[height] * numTripletPermutations
// for each suitable descendant of node with height "height" - correct for this.
numTriplets -= numPairsWithHeightWithNodeAsLCA * node.descendantWithHeightInfo[height].number * numTripletPermutations;
}
}
}
map<int, HeightInfo> buildDescendantHeightInfo(Node* currentNode, Node* parentNode, int height, int64_t& numTriplets)
{
currentNode->height = height;
map<int, HeightInfo> persistentInfoForDescendantHeight;
for (auto child : currentNode->neighbours)
{
if (child == parentNode)
continue;
// Quick C++ performance note: in C++11 onwards, capturing a returned std::map
// in a local variable is O(1), due to Move Semantics. Prior to this, though,
// it could have been O(size of std::map) (if the Return Value Optimisation ended up
// not being used), which would (silently!) lead to asymptotically worse performance!
//
// Luckily, this code uses C++11 features so we can't accidentally fall into this trap.
auto transientInfoForDescendantHeight = buildDescendantHeightInfo(child, currentNode, height + 1, numTriplets);
if (transientInfoForDescendantHeight.size() > persistentInfoForDescendantHeight.size())
{
// We'll be copying transientInfoForDescendantHeight into persistentInfoForDescendantHeight.
// Ensure that the former is smaller than the latter so that we can make use of the Small-to-Large
// trick. NB: std::swap'ing is O(1).
swap(persistentInfoForDescendantHeight, transientInfoForDescendantHeight);
}
for (auto transientDescendantHeightPair : transientInfoForDescendantHeight)
{
// This block of code (i.e. the body of the containing for... loop)
// is executed O(n log2 n) times over the whole run.
// It is guaranteed to be executed with descendantHeight if the current
// child has a descendant with height descendantHeight that isSuitable and a previous child of this
// node also has a descendant with height descendantHeight that isSuitable, but may also
// be executed under different circumstances.
//
// Since this block of code adds at most one entry into currentNode's descendantWithHeightInfo member,
// the sum of node->descendantWithHeightInfo.size() over all nodes is O(n log2 n).
const auto descendantHeight = transientDescendantHeightPair.first;
const auto& transientHeightInfo = transientDescendantHeightPair.second;
auto& heightInfoForNode = persistentInfoForDescendantHeight[descendantHeight];
assert (descendantHeight > currentNode->height);
auto numUnprocessedDescendantsWithHeight = -1;
auto numKnownDescendantsWithHeight = -1;
assert(transientHeightInfo.lastIncorporatedIntoNode != nullptr);
if (transientHeightInfo.lastIncorporatedIntoNode == currentNode)
{
assert(heightInfoForNode.lastIncorporatedIntoNode != currentNode);
numUnprocessedDescendantsWithHeight = heightInfoForNode.numWithHeight;
numKnownDescendantsWithHeight = transientHeightInfo.numWithHeight;
}
else
{
assert(transientHeightInfo.lastIncorporatedIntoNode != currentNode);
numUnprocessedDescendantsWithHeight = transientHeightInfo.numWithHeight;
numKnownDescendantsWithHeight = heightInfoForNode.numWithHeight;
}
const auto earlierChildHasThisHeight = (numKnownDescendantsWithHeight > 0);
if (earlierChildHasThisHeight)
{
// Incorporate any un-incorporated HeightInfo into this Node's descendantWithHeightInfo.
auto& descendantHeightInfo = currentNode->descendantWithHeightInfo[descendantHeight];
auto& numPairsWithHeightWithNodeAsLCA = descendantHeightInfo.numPairsWithHeightWithNodeAsLCA;
auto& numberDescendantsWithThisHeight = descendantHeightInfo.number;
if (numUnprocessedDescendantsWithHeight * numPairsWithHeightWithNodeAsLCA > 0)
{
// Found a triple where all three nodes have currentNode as their LCA: a "Type B" triple.
const auto numNewTriplets = numPairsWithHeightWithNodeAsLCA * numUnprocessedDescendantsWithHeight * numTripletPermutations;
assert(numNewTriplets >= 0);
numTriplets += numNewTriplets;
}
// These numPairsWithHeightWithNodeAsLCA would, when combined with a non-ancestor of currentNode that isSuitable and is
// (descendantHeight - currentNode->height) distance away from currentNode, form a "Type A" triple.
// We store numPairsWithHeightWithNodeAsLCA for this descendantHeight inside currentNode: the required non-ancestors of
// currentNode will be found by completeTripletsOfTypeA() later on.
numPairsWithHeightWithNodeAsLCA += numUnprocessedDescendantsWithHeight * numKnownDescendantsWithHeight;
if (numberDescendantsWithThisHeight == 0)
{
// This hasn't been updated yet, so has missed the numKnownDescendantsWithHeight; incorporate it now.
numberDescendantsWithThisHeight += numKnownDescendantsWithHeight;
}
numberDescendantsWithThisHeight += numUnprocessedDescendantsWithHeight;
}
// "Copy" the transient info into persistent info, and make a note that this HeightInfo has been incorporated
// into currentNode.
heightInfoForNode.numWithHeight += transientHeightInfo.numWithHeight;
heightInfoForNode.lastIncorporatedIntoNode = currentNode;
}
}
if (currentNode->isSuitable)
{
persistentInfoForDescendantHeight[currentNode->height].numWithHeight++;
persistentInfoForDescendantHeight[currentNode->height].lastIncorporatedIntoNode = currentNode;
}
return persistentInfoForDescendantHeight;
}
int64_t findNumTriplets(vector<Node>& nodes)
{
int64_t numTriplets = 0;
auto rootNode = &(nodes.front());
// Fills in numPairsWithHeightWithNodeAsLCA for each node, and
// additionally counts all "Type B" triples and adds them to results.
buildDescendantHeightInfo(rootNode, nullptr, 0, numTriplets);
// Finishes off the computation of the number of "Type A" triples
// that we began in buildDescendantHeightInfo.
completeTripletsOfTypeACentroidDecomposition(nodes, rootNode, numTriplets);
return numTriplets;
}
template <typename T>
{
assert(cin);
}
int main(int argc, char* argv[])
{
// This solution is the first one that occurred to me (well, ish - the first one
// didn't use Centroid Decomposition but a technique with similar capabilities
// based around Heavy-Light Decomposition which seems a little faster in practice,
// and was a little simpler for this Problem). Based on the solutions of the Tester
// (and basically everyone who got 100pts on this Problem during the Contest :)),
// it's become clear that this is an unsually clunky and fiddly way of doing things,
// so I'm only going to give a brief overview of it, here - the official Editorial
// solution will doubtless be better to learn from :) This explanation is adapted from
// the "Brief (ha!) Description" that is required when submitting a new Problem Idea.
//
// Anyway, if you're still reading (which you shouldn't be :p), here goes:
//
// Imagine we pick some arbitrary node (I pick node number 1) as the root R of the tree
// and do a DFS from there. Then it can be shown that any triple (p, q, r) will be
// of one of the following two types:
//
//
// Type A - p, q, r have the same lca, from which they are equidistant e.g.
//
// R
// /|\
// ...
// X = lca(p, q, r)
// / | \
// .......
// / | \
// p q r <-- dist(p, x) = dist(q, x) = dist(r, x);
// p, q and r all have same distance from X,
// and so the same distance from R (aka "height").
//
// Type B - the two "lowest" nodes - say q and r wlog, are equidistant from their lca = X.
// p is *not* a descendent of X, but dist(p, X) = dist(q, X) (= dist(r, X)) e.g.
//
// R
// /|\
// ...
// /\
// ... ...
// / \
// p ...
// |
// X = lca(q, r)
// / \
// .......
// / \
// q r <-- dist(p, X) = dist(r, X); p, q have the same distance from R.
//
// Type A is the easiest to compute; Type B is harder and is computed in two phases: the
// first phase is shared with the computation of the number of Type A triplets; the
// second is separate and uses Centroid Decomposition.
//
// There's a reasonably well known algorithm for calculating, for each node v, the set of
// all descendents of v in O(N log N):
//
// findDescendents(root)
// set_of_descendents = empty-set
// for each child c of root:
// set_of_descendents_of_child = findDescendents(c)
// if |set_of_descendents_of_child| > |set_of_descendents|:
// swap set_of_descendents with set_of_descendents_of_child (O(1)).
//
// for each node in set_of_descendents_of_child:
//
// return set_of_descendents
//
// The algorithm looks like it's O(N^2) worst case, but the fact that we always "copy"
// the smaller set into the larger actually makes it asymptotically better (O(N log N)),
// for the same reason as disjoint-union-by-size is O(N log N):
// https://en.wikipedia.org/wiki/Disjoint-set_data_structure#by_size
// I've also seen this technique referred to as the "Small-To-Large Trick".
//
// For a node v, let height(v) = dist(R, v). We adapt the algorithm to compute not the
// set of descendents of each node v, but a count (map) of *heights of descendents of
// suitable nodes of v* i.e. a map where the keys are heights and the values are the
// number of descendents of v which are suitable and have that height; it can be shown
// that this also is achievable in worst case O(N log N).
//
// With some more book-keeping, it can be shown that we can extract, in O(N log N) time
// and space, for all nodes v, for all heights h [NB: obviously such heights are stored
// sparsely, else the space requirements would be O(N^2)]:
//
// the number of pairs of descendents (u', v') of v such that:
// * u' and v' are both suitable;
// * lca(u', v') = v; and
// * height(u') = height(v') = h (and so: dist(u', v) = dist(v', v))
//
// (this is stored as descendantWithHeightInfo.numPairsWithHeightWithNodeAsLCA) and also
//
// the number of triples (u', v', w') of v such that
// * u', v' and w' are all suitable;
// * lca(u', v') = lca(u', w') = lca(v', w') = v; and
// * height(u') = height(v') = height(w') = h (and so:
// dist(u', v) = dist(v', v) = dist(w', v))
//
// This completes Phase one of two.
//
// We see that the latter count of triples is the number of triples of Type A (more
// precisely - it is the number of such triples divided by the number of permutations
// of a triple, 3! - see later).
//
// The former is a step towards the computation of the number of Type B triples -
// referring to the definition of Type B, we've found, for each X, the number of q's
// and r's with a given height h (which equates to a distance d from X: height(q) - height(x))
// and lca X, and now we just need to find for each X and h the number of p's such that
// dist(p, X) = height(q) - height(X). We do this latter step via Centroid Decomposition.
//
// More precisely, Phase one gives us, for each node v, a map from heights to number
// of pairs of descendents called v.numPairsWithHeightWithNodeAsLCA - all in O(N log N)
// time and space.
//
// We then perform Centroid Decomposition with an efficient DistTracker class that
// implements the following API:
//
// * addDistance(newDistance) - adds the distance to the list of tracked distances
// in O(1).
// tracked distances in O(1).
// * getNumWithDistance(distance) - return the number of tracked distances whos
// current value is precisely distance in O(1).
//
// A simple combination of CD and DistTracker allows us to finally "complete" the "q's"
// and "r's" found in Phase one with the "p's" necessary to form a complete, Type B-triplet
// in O(N x (log N) x (log N)).
//
// It can be further shown that if a triplet (p, q, r) of either type A or B is
// counted by the algorithm, then none of its (3! - 1) other permutations - (p, r, q);
// (r, p, q) etc - will be counted, so simply multiplying the count of Type A and
// Type B triplets by 3! gives us the final result.
//
// Well - not quite :) The Centroid Decomposition step doesn't know about parent and
// children in our original DFS from R, so it will overcount triples. For example,
// consider the simple example:
//
// R
// |
// X
// / \
// Y Z
//
// Imagine that Y and Z are the only suitable nodes, so there are no valid triples in
// this example.
//
// Note that X's descendantWithHeightInfo[2].numPairsWithHeightWithNodeAsLCA will be 1.
// The Centroid Decomposition step will then, unfortunately, treat Y as a "completer"
// of this pair (Y, Z) (and will treat Z the same way), resulting in it reporting *2*
// triples, instead of the correct answer of 0. Luckily, it's easy (though clunky) to
// The proper Editorial solution, and the original HLD-based solution, don't have this
// clunky "overcount correction" step, which is another reason why they are superior :)
//
// The whole algorithm runs in O(N x (log N) x (log N)) time with O(N log N) space.
ios::sync_with_stdio(false);
for (auto t = 0; t < numTestcases; t++)
{
assert(1 <= numNodes && numNodes <= 200'000);
vector<Node> nodes(numNodes);
for (auto i = 0; i < numNodes - 1; i++)
{
assert(1 <= u && u <= numNodes);
assert(1 <= v && v <= numNodes);
nodes[u - 1].neighbours.push_back(&(nodes[v - 1]));
nodes[v - 1].neighbours.push_back(&(nodes[u - 1]));
}
for (auto i = 0; i < numNodes; i++)
{
assert(isSuitable == 0 || isSuitable == 1);
nodes[i].isSuitable = (isSuitable == 1);
}
const auto numTriplets = findNumTriplets(nodes);
cout << numTriplets << endl;
}
assert(cin);
}
Tester's Solution
#include <bits/stdc++.h>
#define endl '\n'
#define SZ(x) ((int)x.size())
#define ALL(V) V.begin(), V.end()
#define L_B lower_bound
#define U_B upper_bound
#define pb push_back
using namespace std;
template<class T, class T1> int chkmin(T &x, const T1 &y) { return x > y ? x = y, 1 : 0; }
template<class T, class T1> int chkmax(T &x, const T1 &y) { return x < y ? x = y, 1 : 0; }
const int MAXN = (1 << 18);
struct mydeque {
vector<int> vec;
int& operator[](const unsigned &pos) { return vec[vec.size() - pos - 1]; }
unsigned size() { return vec.size(); }
void push_front(int x) { vec.pb(x); }
};
void merge(mydeque &a, mydeque &b) {
if(SZ(a) < SZ(b)) swap(a, b);
for(int i = 0; i < SZ(b); i++) {
a[i] += b[i];
}
}
int n, s[MAXN];
for(int i = 1; i <= n; i++) {
}
for(int i = 0; i < n - 1; i++) {
int u, v;
}
for(int i = 1; i <= n; i++) {
}
}
int tr_sz[MAXN], cnt_vers;
bool used[MAXN];
void pre_dfs(int u, int pr)
{
cnt_vers++;
tr_sz[u] = 1;
if(!used[v] && v != pr)
{
pre_dfs(v, u);
tr_sz[u] += tr_sz[v];
}
}
int centroid(int u, int pr)
{
if(!used[v] && v != pr && tr_sz[v] > cnt_vers / 2)
return centroid(v, u);
return u;
}
int dep[MAXN];
int curr_cnt[MAXN], gen_cnt[MAXN];
int64_t cnt[MAXN][2];
void fill_curr_cnt(int u, int pr, int d = 1) {
curr_cnt[d] += s[u];
if(v != pr && !used[v]) {
fill_curr_cnt(v, u, d + 1);
}
}
}
void dfs1(int u, int pr) {
dep[u] = 0;
if(v != pr && !used[v]) {
dfs1(v, u);
chkmax(dep[u], dep[v] + 1);
}
}
}
mydeque dp[MAXN];
void add(pair<int, int> &mx, int x) {
if(chkmax(mx.second, x)) {
if(mx.first < mx.second) {
swap(mx.first, mx.second);
}
}
}
int64_t two[MAXN];
int tmp[MAXN];
void dfs2(int u, int pr, int curr_d = 1) {
dp[u].vec.clear();
dp[u].push_front(s[u]);
pair<int, int> mx = {0, 0};
if(v != pr && !used[v]) {
dfs2(v, u, curr_d + 1);
dp[v].push_front(0);
}
}
int d = mx.second;
for(int i = 0; i <= d; i++) {
tmp[i] = 0;
two[i] = 0;
}
if(v == pr || used[v]) continue;
int cd = min(d, dep[v] + 1);
for(int i = 1; i <= cd; i++) {
two[i] += tmp[i] * 1ll * dp[v][i];
tmp[i] += dp[v][i];
}
}
for(int i = 1; i <= d; i++) {
int od = i - curr_d;
if(od < 0) continue;
answer += (gen_cnt[od] - curr_cnt[od]) * 1ll * two[i];
}
if(v != pr && !used[v]) {
merge(dp[u], dp[v]);
}
}
}
void decompose(int u)
{
cnt_vers = 0;
pre_dfs(u, u);
int cen = centroid(u, u);
used[cen] = true;
if(!used[v])
decompose(v);
used[cen] = false;
dfs1(cen, -1);
for(int i = 0; i <= dep[cen] + 1; i++) {
cnt[i][0] = cnt[i][1] = 0;
curr_cnt[i] = 0;
}
fill_curr_cnt(cen, -1, 0);
for(int i = 0; i <= dep[cen] + 1; i++) {
gen_cnt[i] = curr_cnt[i];
curr_cnt[i] = 0;
}
cnt[0][0] = 1;
if(!used[v]) {
for(int i = 0; i <= dep[v] + 1; i++) {
curr_cnt[i] = 0;
tmp[i] = 0;
}
fill_curr_cnt(v, cen);
for(int i = 1; i <= dep[v] + 1; i++) {
if(curr_cnt[i]) {
answer += curr_cnt[i] * 1ll * cnt[i][1];
cnt[i][1] += curr_cnt[i] * 1ll * cnt[i][0];
cnt[i][0] += curr_cnt[i];
}
}
dfs2(v, cen);
for(int i = 1; i <= dep[v] + 1; i++) {
curr_cnt[i] = 0;
}
}
}
}
void solve() {
decompose(1);
cout << answer * 6ll << endl;
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int T;
while(T--) {
solve();
}
return 0;
}
const int maxl = 100000;
char buff[maxl];
int ret_int, pos_buff = 0;
void next_char() { if(++pos_buff == maxl) fread(buff, 1, maxl, stdin), pos_buff = 0; }
{
ret_int = 0;
for(; buff[pos_buff] < '0' || buff[pos_buff] > '9'; next_char());
for(; buff[pos_buff] >= '0' && buff[pos_buff] <= '9'; next_char())
ret_int = ret_int * 10 + buff[pos_buff] - '0';
return ret_int;
}
Editorialist's Solution
#include <bits/stdc++.h>
using namespace std;
#define ll long long
const int mxN=2e5;
int n, s[mxN];
vector<ll> dp1[mxN], dp2[mxN];
ll ans;
void dfs(int u=0, int p=-1) {
dp1[u]={s[u]};
dp2[u]={0};
if(v==p)
continue;
dfs(v, u);
//now merge the dp for v to u and update ans
//increase the depths by 1
dp1[v].push_back(0);
dp2[v].pop_back();
//if v is deeper, switch the dp to make it faster
if(dp1[v].size()>dp1[u].size()) {
swap(dp1[u], dp1[v]);
swap(dp2[u], dp2[v]);
}
//make sure dp2[u] is big enough
if(dp2[u].size()<dp1[v].size()) {
vector<int> p(dp1[v].size()-dp2[u].size(), 0);
dp2[u].insert(dp2[u].begin(), p.begin(), p.end());
}
//update ans
//leg from v, fork from u
for(int i=1; i<=dp1[v].size(); ++i)
ans+=dp1[v][dp1[v].size()-i]*dp2[u][dp2[u].size()-i];
//fork from v, leg from u
for(int i=1; i<=dp2[v].size(); ++i)
ans+=dp2[v][dp2[v].size()-i]*dp1[u][dp1[u].size()-i];
//combine the dp
//new forks by combining 2 legs
for(int i=1; i<=dp1[v].size(); ++i)
dp2[u][dp2[u].size()-i]+=dp1[u][dp1[u].size()-i]*dp1[v][dp1[v].size()-i];
for(int i=1; i<=dp1[v].size(); ++i)
dp1[u][dp1[u].size()-i]+=dp1[v][dp1[v].size()-i];
for(int i=1; i<=dp2[v].size(); ++i)
dp2[u][dp2[u].size()-i]+=dp2[v][dp2[v].size()-i];
}
}
void solve() {
//input
cin >> n;
for(int i=1, u, v; i<n; ++i) {
cin >> u >> v, --u, --v;
}
for(int i=0; i<n; ++i)
cin >> s[i];
ans=0;
dfs();
cout << 6*ans << "\n";
}
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
int t;
cin >> t;
while(t--)
solve();
}
8 Likes
Updated Setter’s solution here - I haven’t been that detailed with the overview for the reasons described in it (i.e. - my solution was very overcomplicated :))
Problem History here - warning: long and rambling
Problem History
@shivank98 came up with the idea of writing a little bit about how the Problem came about a while back, and I thought it was a good one, so I think I’ll take a stab at it
Oddly, CHGORAM2 arose out of my disatisfaction with a Hackerrank problem called “Find the Nearest Clone”. FTNC would have been a neat Easy-Medium question except for the fact that is utterly let down by its legendarily poor testcases - or rather, more fundamentally, by the fact that the Setter/ Editorialist did not realise that his purportedly O(N) solution was in fact O(N^2).
I brainstormed ways of coming up with a tweak to the Problem that would enable me to re-do it (this time with decent testcases!) - perhaps count the number of ways of forming this closest pair? Or finding not the nearest pair of clones, but the nearest triples - triples of nodes who all have the same (minimum) distance from each other? Or perhaps counting all such triples? Or - maybe! - counting all triples of the same type??
This latter seemed more plausible (for a tree, at least - not for an arbitrary graph as in the original Find the Nearest Clone!), and eventually I stripped it down to just two types: at the time, as you can see from the commit, I had a lame backstory in mind where the nodes were houses, and some houses had people in them, and you have to count the number of equidistant triples of people - something about the people in this town preferring to gather in triples and, out of a sense of fairness, wanted the journey time between any pairs of friend’s houses to be the same, or somesuch nonsense It was to be called “Equilateral Treeangles” (groan).
Unusually for me, I came up with the basic approach - the Type A arrangement, countable using a DFS with the Small-to-Large optimisation used with descendent heights - almost straight away. My git logs from around the time explicitly mention this Hackerrank Problem, so I suppose by chance I was thinking about it at this time, and that steered my thoughts in the right direction. Of course, I didn’t get it completely correctly initially - I completely overlooked the existence of Type B arrangements (ha!) and also assumed that the “p” in the Type A case would always be a direct ancestor of X (lol). But the fact that I got this latter one wrong was actually good - it made the problem much deeper, as now it seemed to required Centroid Decomposition, too!
Actually implementing it didn’t take too long, though there was one heart-stopping moment where a test-run of my implementation took about 5 seconds on a testcase I generated (the git log at the time contains “I’ve gravely misjudged something, but don’t yet know what. Panic stations!”), but it turns out that I just had the direction of a “<” reversed and was doing the Large-to-Small Pessimisation instead of the Small-to-Large Optimisation XD
I then put the Problem on hold for a while, and took part in my first Codechef contest (AUG19B). One of the most heartbreaking things that can happen when you’ve invested a lot of time and effort in creating a Problem is to find that someone has beaten you to it, and so I nearly fell off my chair when I came across an AUG19B Problem where you had to count the number of ways of choosing 3 nodes (ulp!) on a tree (gulp!) satisfying certain constraints (yikes!), but after taking some deep breaths and forcing myself to read the Problem carefully, I saw that this “CHGORAM” Problem was completely different to mine and breathed a sigh of relief. Even better, as I realised a little later - I could ditch my naff backstory about triples of fair-minded friends and piggy-back on this one instead!
So: I had a working implementation with the “<” pointing the right way, and a backstory - time to get some proper tests written, which would prove to be by far the hardest part of the Problem creation.
My shiny new testcases were being passed with ease by my Editorial implementation, so I tested them against an obvious O(N^2) implementation: one where we consider each node in turn as the “centre” of the triple (the “X”) and find all nodes at given distance from X, reachable via different neighbours of X, and count the triples that can be formed by all of these that are suitable. Happily, the testcases TLE’d heavily on this naive approach, but then I suddenly (much later than most of you, I’m sure) noticed the Achilles’ heel for this Problem: if, at any point, we cannot find 3 nodes at distance d from X reachable by different children of X, then we can stop our search there - subsequent d's will reveal no further triples.
I added this small optimisation to the “cheat” O(N^2) solution and … it absolutely wrecked my testcases. Just blasted through all of them, taking barely more than 1 sec for the hardest. Back to the drawing board!
I had to exercise a bit of ingenuity (and up the node limit to 200,000 XD) to break this Cheat Solution, but this was still a bit of a crisis - if such a simple ploy could have obliterated my original testcases, then maybe a slightly less simple one would destroy the new ones? I’m forever complaining about weak testcases - mostly recently in this post - so I couldn’t submit a Problem that could be so easily cracked by sub-optimal solutions! So I tried to be as proactive about anticipating avenues that could be exploited as I could, and also wrote a few “blue-sky” tests (006-008) that violated some core assumptions that a sub-optimal solution might depend on. I couldn’t really come up with any concrete implementations that could use these assumptions to cheat, and so was quite close to deleting tests 006-008 but, as it happens, 008 at least turned out to be oddly effective 015 was also a last-minute reaction to a possible attack that I thought of, but proved less useful - I don’t think anyone chose that possible means of assault.
Anyway, I had all the ingredients ready now so I submitted the Problem Idea, which was Approved very quickly and fast-tracked to appear in FEB20 Long where it would be one of the Div 1-only Problems - which made me nervous, as I was still unsure about how easily the “Achilles’ Heel” would permit sub-optimal solutions. Watching it get Tested by the Tester was very nerve-wracking, and for a while it looked like the Tester was going to trounce it in very short order, but in fact, it was here that the almost-deleted 008 first proved its mettle by being the Last Man Standing for one of the Tester’s early attempts!
After a while, I couldn’t stand the tension anymore and went to sleep. As expected, the Tester had broken it by the time I woke up, but it had put up a good fight, at least, and made him work for it! His assessment was “medium or hard (although probably closer to hard)”, which was reassuring. He did suggest to toughen up some of the tests, and this is where the mysterious testfiles 013 and 014 were born - they came about through setting up the testcase generator to generate trees with certainly parameters, and then doing a random parameter search to see which combination caused the Setter’s “cheat” implementation to time-out the hardest. To this day, I still don’t have the slightest clue how 013 and 014 actually work, but they proved very effective against the Tester’s solution, and indeed many solutions during the Contest!
My confidence in the Problem had been bolstered by now, so I was quite panicked when the first few hours of the Contest consisted of what seemed like an endless parade of people effortlessly dunking on the it XD I was debating writing this Problem History as a long “sorry my testcases sucked” apology, but eventually the parade dried up a little (and digging through these early submitter’s histories revealed them as either official 7*'s, or people who will surely be 7* after a few more contests - unless they elect to take the @just1star route, of course :)) and in the end far, far fewer people managed to complete it - approx 1/4 as many as the original CHGORAM, in fact The 008 testfile again proved surprisingly effective, and I’m very glad I didn’t delete it.
From looking at some of the solutions, it quickly became clear that my solution was unnecessarily clumsy, so I’ve only given it a little bit of documentation compared to my normal solutions - please refer to the official Editorial instead
Anyway, that’s that - all a bit of a roller-coaster, but I’m glad I finally got a Problem published somewhere and it didn’t turn out to be a disgrace
15 Likes
I tried your “cheap” O(n^2) algorithm, I thought it actually works faster and i just need to optimise the data structures.
Also, if you could, Preferably don’t gloss over the implementation details because I struggled to implement the code for lcas
1 Like
I didn’t understand half of what you wrote🤣
But it was pretty fun to read XD
3 Likes
Which implementation details - the O(N^2) version? If so, here it is
2 Likes
No i meant that part for the editorial, sorry. I managed to implement the O(n^2) but struggled to implement the lca dp.
1 Like
Oh right - hopefully my commented cpp implementation plus @tmwilliamlin’s pending write-up will help to clear up any confusion
1 Like
Hotels from POI XXI is very similar to this problem, and although an O(N^2) solution is good enough to solve it, one of the tester’s solutions is O(NlogN). This solution can be found here. I’ve used this code to solve the problem, and on skimming through some other contestant’s solutions, it seems a few of them have as well. Just want to bring this up before I’m accused of plagiarism.
2 Likes
The POI problem has an increased data range version on BZOJ.
Well written. Never could have thought that the story behind setting a problem could be so interesting!
1 Like
For those who haven’t seen it, I created a video solution as a substitute until I finish writing up the editorial.
4 Likes
Editorial finished and updated!
3 Likes
That’s not mine - that’s the Tester’s again (which, admittedly, I submitted - but only to check its progress against the current testcases ) Mine is in the first reply to this thread.
2 Likes
Oops, I took whatever the best submissions page on campus showed me lol
2 Likes
Hehe
Updated with related problems
1 Like
+1 happens wayy too often with me XD
2 Likes
Is the current setter’s solution yours @ssjgz?
1 Like
Yes
1 Like
In editorial i did not understood following :
dp1_v,j+1 = dp1_v,j (and dp1_v,0 = 0) .
In above line , if dp1_v,0 = 0 then by equality dp1_v,j+1 = dp1_v,j , dp1_v,j = 0 for all j which is not possible .
Also i did not understood how you got dp1_v,j+1 = dp1_v,j . Suppose take following tree :
suppose vertex 1 is root .
for vertex ‘v’ marked , number of legs of length 1 from v is 2 whereas number of legs of length 2 from v is 3 .
|
2020-06-04 18:48:04
|
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|
https://learn.careers360.com/engineering/question-consider-the-set-of-all-lines-pxqyr0-such-that-3p2q4r0which-of-the-following-is-true/
|
# Consider the set of all lines px+qy+r=0 such that 3p+2q+4r=0.which of the following is true? the lines are concurrent the lines are concurrent at the point (3/4,1/2) the lines are parallel each line passes through the origin
\begin{aligned} &3 p+2 q+4 r=0\\ &p x+q y+r=0 \ \ \ \ \ \ \ ..............(1)\\ &3 / 4 p+q / 2+r=0 \ \ \ \ \ \ ...........(2)\\ &\text {comparing (i) and (ii) } x=3 / 4, y=1 / 2 \end{aligned}
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2020-10-22 15:56:18
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https://www.picostat.com/dataset/r-dataset-package-psych-schmid
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R Dataset / Package psych / Schmid
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Documentation
12 variables created by Schmid and Leiman to show the Schmid-Leiman Transformation
Description
John Schmid and John M. Leiman (1957) discuss how to transform a hierarchical factor structure to a bifactor structure. Schmid contains the example 12 x 12 correlation matrix. schmid.leiman is a 12 x 12 correlation matrix with communalities on the diagonal. This can be used to show the effect of correcting for attenuation. Two additional data sets are taken from Chen et al. (2006).
Usage
data(Schmid)
Details
Two artificial correlation matrices from Schmid and Leiman (1957). One real and one artificial covariance matrices from Chen et al. (2006).
• Schmid: a 12 x 12 artificial correlation matrix created to show the Schmid-Leiman transformation.
• schmid.leiman: A 12 x 12 matrix with communalities on the diagonal. Treating this as a covariance matrix shows the 6 x 6 factor solution
• Chen: An 18 x 18 covariance matrix of health related quality of life items from Chen et al. (2006). Number of observations = 403. The first item is a measure of the quality of life. The remaining 17 items form four subfactors: The items are (a) Cognition subscale: “Have difficulty reasoning and solving problems?" “React slowly to things that were said or done?"; “Become confused and start several actions at a time?" “Forget where you put things or appointments?"; “Have difficulty concentrating?" (b) Vitality subscale: “Feel tired?" “Have enough energy to do the things you want?" (R) “Feel worn out?" ; “Feel full of pep?" (R). (c) Mental health subscale: “Feel calm and peaceful?"(R) “Feel downhearted and blue?"; “Feel very happy"(R) ; “Feel very nervous?" ; “Feel so down in the dumps nothing could cheer you up? (d) Disease worry subscale: “Were you afraid because of your health?"; “Were you frustrated about your health?"; “Was your health a worry in your life?" .
• West: A 16 x 16 artificial covariance matrix from Chen et al. (2006).
Source
John Schmid Jr. and John. M. Leiman (1957), The development of hierarchical factor solutions.Psychometrika, 22, 83-90.
F.F. Chen, S.G. West, and K.H. Sousa.(2006) A comparison of bifactor and second-order models of quality of life. Multivariate Behavioral Research, 41(2):189-225, 2006.
References
Y.-F. Yung, D.Thissen, and L.D. McLeod. (1999) On the relationship between the higher-order factor model and the hierarchical factor model. Psychometrika, 64(2):113-128, 1999.
Examples
data(Schmid)
cor.plot(Schmid,TRUE)
print(fa(Schmid,6,rotate="oblimin"),cut=0) #shows an oblique solution
round(cov2cor(schmid.leiman),2)
cor.plot(cov2cor(West),TRUE)
--
Dataset imported from https://www.r-project.org.
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2021-12-06 20:51:54
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https://www.physicsforums.com/threads/vector-and-gradient.234947/
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Homework Statement
if $$\phi$$ = rk/r$$^{3}$$ where r=xi + yJ + zk and r is the magnitude of r, prove that $$\nabla$$$$\phi$$ = (1/r$$^{}5$$)(r$$^{}2$$k-3(r.k)r
so i differenciated wrt x then y then z and tried to tidy it all up but i got1/rClick to see the LaTeX code for this image(-3(r.k)r)
When i differenciated wrt x i got -3x/rClick to see the LaTeX code for this image and similar for y and z was this right?
then i just put these answers into Click to see the LaTeX code for this image = dx/dClick to see the LaTeX code for this image i + dy/dClick to see the LaTeX code for this image j + dz/dClick to see the LaTeX code for this image k
which gives
-3xz/rClick to see the LaTeX code for this image i - 3yz/rClick to see the LaTeX code for this image j - 3zz/rClick to see the LaTeX code for this image k
am i right so far? it just seemed to tidy up to 1/rClick to see the LaTeX code for this image(-3(r.k)r)
when it should be 1/rClick to see the LaTeX code for this image(rClick to see the LaTeX code for this imagek-3(r.k)r)
Related Calculus and Beyond Homework Help News on Phys.org
tiny-tim
Homework Helper
Hi gtfitzpatrick!
This is too difficult to read … everything has "Click to see the LaTeX code for this image" in the middle.
Can you type it out again, perhaps using ² and ³ and ^4 and ^5?
if $$\phi$$ = rk/r$$^{3}$$ where r=xi + yJ + zk and r is the magnitude of r, prove that $$\nabla$$$$\phi$$ = (1/r$$^{}5$$)(r$$^{}2$$k-3(r.k)r
so i differenciated wrt x then y then z and tried to tidy it all up but i got 1/r$$^{5}$$(-3(r.k)r)
When i differenciated wrt x i got -3x/r$$^{5}$$ and similar for y and z was this right?
then i just put these answers into = dx/d$$\phi$$ i + dy/d$$\phi$$ j + dz/d$$\phi$$ k
which gives
-3xz/r$$^{5}$$ i - 3yz/r$$^{5}$$ j - 3zz/r$$^{5}$$ k
am i right so far? it just seemed to tidy up to 1/r$$^{5}$$(-3(r.k)r)
when it should be 1/r$$^{5}$$(r$$^{2}$$k-3(r.k)r)
Dick
Homework Helper
am i right so far? it just seemed to tidy up to 1/r$$^{5}$$(-3(r.k)r)
when it should be 1/r$$^{5}$$(r$$^{2}$$k-3(r.k)r)
It's not terribly clear what you are doing, but some how you are winding up with only one part of a quotient rule answer. Your initial function is (r.k)/r^3. That's the same thing as z/r^3.
Dick
Homework Helper
You know there is a 'quotient rule' for grad, yes? grad(f/g)=(g*grad(f)-f*grad(g))/g^2. Does that help?
tiny-tim
Homework Helper
When i differenciated wrt x i got -3x/r$$^{5}$$ and similar for y and z was this right?
then i just put these answers into = dx/d$$\phi$$ i + dy/d$$\phi$$ j + dz/d$$\phi$$ k
which gives
-3xz/r$$^{5}$$ i - 3yz/r$$^{5}$$ j - 3zz/r$$^{5}$$ k
Hi gtfitzpatrick!
You're only differentiating the 1/r^5.
You need to differentiate the r also.
i differenciated z(x^2+y^2+z^2)$$^{-3/2}$$ wrt x then y then z, and then filled it into the fomula was this not right?
I didn't know this, does this mean i'm wrong?i haven't seen this formula before what are the f and g's?
Last edited:
so i cant use the product rule?
Dick
i differenciated z(x^2+y^2+z^2)$$^{-3/2}$$ wrt x then y then z, and then filled it into the fomula was this not right?
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2021-03-07 12:47:26
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http://psychology.wikia.com/wiki/Exponential-logarithmic_distribution
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# Exponential-logarithmic distribution
34,191pages on
this wiki
In probability theory and statistics, the exponential-logarithmic (EL) distribution is a family of lifetime distributions with decreasing failure rate, defined on the interval (0, ∞). This distribution is parameterized by two parameters $p\in(0,1)$ and $\beta >0$.
Parameters File:Pdf EL.png File:Hazard EL.png $p\in (0,1)$ $\beta >0$ $x\in[0,\infty)$ $\frac{1}{-\ln p} \times \frac{\beta(1-p) e^{-\beta x}}{1-(1-p) e^{-\beta x}}$ $1-\frac{\ln(1-(1-p) e^{-\beta x})}{\ln p}$ $-\frac{\text{polylog}(2,1-p)}{\beta\ln p}$ $\frac{\ln(1+\sqrt{p})}{\beta}$ 0 $-\frac{2 \text{polylog}(3,1-p)}{\beta^2\ln p}$ $-\frac{ \text{polylog}^2(2,1-p)}{\beta^2\ln^2 p}$ $-\frac{\beta(1-p)}{\ln p (\beta-t)} \text{hypergeom}_{2,1}$ $([1,\frac{\beta-t}{\beta}],[\frac{2\beta-t}{\beta}],1-p)$
## Introduction Edit
The study of lengths of organisms, devices, materials, etc., is of major importance in the biological and engineering sciences. In general, the lifetime of a device is expected to exhibit decreasing failure rate (DFR) when its behavior over time is characterized by 'work-hardening' (in engineering terms) or 'immunity' (in biological terms).
The exponential-logarithmic model, together with its various properties, are studied by Tahmasbi and Rezaei (2008)[1] This model is obtained under the concept of population heterogeneity (through the process of compounding).
## Properties of the distribution Edit
### Distribution Edit
The probability density function (pdf) of the EL distribution is given by Tahmasbi and Rezaei (2008)[1]
$f(x; p, \beta) := \left( \frac{1}{-\ln p}\right) \frac{\beta(1-p)e^{-\beta x}}{1-(1-p)e^{-\beta x}}$
where $p\in (0,1)$ and $\beta >0$. This function is strictly decreasing in $x$ and tends to zero as $x\rightarrow \infty$. The EL distribution has its modal value of the density at x=0, given by
$\frac{\beta (1-p)}{-p \ln p}$
The EL reduces to the exponential distribution with rate parameter $\beta$, as $p\rightarrow 1$.
The cumulative distribution function is given by
$F(x;p,\beta)=1-\frac{\ln(1-(1-p) e^{-\beta x})}{\ln p},$
and hence, the median is given by
$x_\text{median}=\frac{\ln(1+\sqrt{p})}{\beta}$.
### Moments Edit
The moment generating function of $X$ can be determined from the pdf by direct integration and is given by
$M_X(t) = E(e^{tX}) = -\frac{\beta(1-p)}{\ln p (\beta-t)} F_{2,1}\left(\left[1,\frac{\beta-t}{\beta}\right],\left[\frac{2\beta-t}{\beta}\right],1-p\right),$
where $F_{2,1}$ is a hypergeometric function. This function is also known as Barnes's extended hypergeometric function. The definition of $F_{N,D}({n,d},z)$ is
$F_{N,D}(n,d,z):=\sum_{k=0}^\infty \frac{ z^k \prod_{i=1}^p\Gamma(n_i+k)\Gamma^{-1}(n_i)}{\Gamma(k+1)\prod_{i=1}^q\Gamma(d_i+k)\Gamma^{-1}(d_i)}$
where $n=[n_1, n_2,\dots , n_N]$ and ${d}=[d_1, d_2, \dots , d_D]$.
The moments of $X$ can be derived from $M_X(t)$. For $r\in\mathbb{N}$, the raw moments are given by
$E(X^r;p,\beta)=-r!\frac{\operatorname{Li}_{r+1}(1-p) }{\beta^r\ln p},$
where $\operatorname{Li}_a(z)$ is the polylogarithm function which is defined as follows:[2]
$\operatorname{Li}_a(z) =\sum_{k=1}^{\infty}\frac{z^k}{k^a}.$
Hence the mean and variance of the EL distribution are given, respectively, by
$E(X)=-\frac{\operatorname{Li}_2(1-p)}{\beta\ln p},$
$\operatorname{Var}(X)=-\frac{2 \operatorname{Li}_3(1-p)}{\beta^2\ln p}-\left(\frac{ \operatorname{Li}_2(1-p)}{\beta\ln p}\right)^2.$
### The survival, hazard and mean residual life functions Edit
The survival function (also known as the reliability function) and hazard function (also known as the failure rate function) of the EL distribution are given, respectively, by
$s(x)=\frac{\ln(1-(1-p)e^{-\beta x})}{\ln p},$
$h(x)=\frac{-\beta(1-p)e^{-\beta x}}{(1-(1-p)e^{-\beta x})\ln(1-(1-p)e^{-\beta x})}.$
The mean residual lifetime of the EL distribution is given by
$m(x_0;p,\beta)=E(X-x_0|X\geq x_0;\beta,p)=-\frac{\operatorname{Li}_2(1-(1-p)e^{-\beta x_0})}{\beta \ln(1-(1-p)e^{-\beta x_0})}$
where $\operatorname{Li}_2$ is the dilogarithm function
### Random number generation Edit
Let U be a random variate from the standard uniform distribution. Then the following transformation of U has the EL distribution with parameters p and β:
$X = \frac{1}{\beta}\ln \left(\frac{1-p}{1-p^U}\right).$
## Estimation of the parameters Edit
To estimate the parameters, the EM algorithm is used. This method is discussed by Tahmasbi and Rezaei (2008)[1]. The EM iteration is given by
$\beta^{(h+1)} = n \left( \sum_{i=1}^n\frac{x_i}{1-(1-p^{(h)})e^{-\beta^{(h)}x_i}} \right)^{-1},$
$p^{(h+1)}=\frac{-n(1-p^{(h+1)})} { \ln( p^{(h+1)}) \sum_{i=1}^n \{1-(1-p^{(h)})e^{-\beta^{(h)} x_i}\}^{-1}}.$
## Related distributionsEdit
The EL distribution has been generalized to form the Weibull-logarithmic distribution.[3]
If X is defined to be the random variable which is the minimum of N independent realisations from an exponential distribution with rate paramerter β, and if N is a realisation from a logarithmic distribution (where the parameter p in the usual parameterisation is replaced by (1 − p)), then X has the exponential-logarithmic distribution in the parameterisation used above.
## ReferencesEdit
1. 1.0 1.1 1.2 Tahmasbi, R., Rezaei, S., (2008), "A two-parameter lifetime distribution with decreasing failure rate", Computational Statistics and Data Analysis, 52 (8), 3889-3901. DOI:10.1016/j.csda.2007.12.002
2. Lewin, L. (1981) Polylogarithms and Associated Functions, North Holland, Amsterdam.
3. Ciumara1,Roxana; Preda2, Vasile (2009) "The Weibull-logarithmic distribution in lifetime analysis and its properties". In: L. Sakalauskas, C. Skiadas and E. K. Zavadskas (Eds.) Applied Stochastic Models and Data Analysis, The XIII International Conference, Selected papers. Vilnius, 2009 ISBN 978-9955-28-463-5
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2015-07-29 13:35:03
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https://solvedlib.com/n/contentattributionquestion-25-zpointswhat-is-the-equation,21246848
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# ContentattributionQuestion 25 ZPoiNtsWhat is the equation of the sphere that Is centered at (3} 7) and has surface area 1677Provide
###### Question:
Contentattribution Question 25 ZPoiNts What is the equation of the sphere that Is centered at (3} 7) and has surface area 1677 Provide your answer below; Content etributlen QUESTION 26 aoiNt
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At the end of Sec, $6.7$ it is stated that the most probable value of $r$ for a $2 p$ electron in a hydrogen stoun is $4 alpha_{0 .}$ which is the same as the radius of the $n$ we 2 Bohr orbit. Verify this....
##### On September 1, Shawn Dahl established Whitewater Rentals, a canoe and kayak rental business. The following...
On September 1, Shawn Dahl established Whitewater Rentals, a canoe and kayak rental business. The following transactions occurred in the month of September and affected the following accounts: Cash Accounts Payable Accounts Receivable Shawn Dahl, Capital Office Equipment Revenue Canoe and Kayak Equi...
##### QUESTION 23 Through (-4,-33) and (10, 79) O y = -8x - 65 67 Oy--8-2 O...
QUESTION 23 Through (-4,-33) and (10, 79) O y = -8x - 65 67 Oy--8-2 O y = 8x - 1 65 Oy- gx 2...
##### They relate to the sonormalities and elsturbancas this atanti sperarens) EtioiogyiDiscuss the csuse of th's dissase...
they relate to the sonormalities and elsturbancas this atanti sperarens) EtioiogyiDiscuss the csuse of th's dissase or disorder and expiain how it is transmimad ar scred oquaE peptic-ocec.can counoー. hece ace Signs and Symptems (List and explain each ofthe signs and srmptoms yourclentwith...
##### What are two interesting things about health care occupations?
What are two interesting things about health care occupations?...
##### Select all that apply:Which of the following compounds are ionic? KCICzHaBaClz SiClaLiF
Select all that apply: Which of the following compounds are ionic? KCI CzHa BaClz SiCla LiF...
##### Write functions for the specified root finding methods. Include a comments in each function that ...
Write functions for the specified root finding methods. Include a comments in each function that notes the inputs and outputs. Use Cody Coursework to help guide writing your functions. You may not use built-in MATLAB methods for root-solving such as fzero.l] Part A: Write a function that implements ...
##### Part AHow many atoms of hydrogen does contain? Express your answer using four significant figures_AZdNA 20atomsSubmitPrevious Answers Request AnswerIncorrect; Try Again; attempts remaining
Part A How many atoms of hydrogen does contain? Express your answer using four significant figures_ AZd NA 20 atoms Submit Previous Answers Request Answer Incorrect; Try Again; attempts remaining...
##### A swimming pool of depth 2.7 m is filled with ordinary (pure) water (ρ = 1000...
A swimming pool of depth 2.7 m is filled with ordinary (pure) water (ρ = 1000 kg/m3). (a) What is the pressure at the bottom of the pool? Pa (b) When the pool is filled with salt water, the pressure changes by 4.5 103 Pa. What is the difference between the density of the salt water and the densi...
##### Factor out $-m$ from $-6 m^{3}-3 m^{2}+m$.
Factor out $-m$ from $-6 m^{3}-3 m^{2}+m$....
##### 1 3 3 8 3 3 3 J 1 Ul 1 WI 19 3 2 1 3 3 1 WW 6 WW { 6 8 MW f 1 8 3 3 8 2 3 1 W ! 8 8 8 3 8 : Hii 5 { 6 1 { {
1 3 3 8 3 3 3 J 1 Ul 1 WI 19 3 2 1 3 3 1 WW 6 WW { 6 8 MW f 1 8 3 3 8 2 3 1 W ! 8 8 8 3 8 : Hii 5 { 6 1 { {...
##### The long pipe in Fig. $\mathrm{P} 3.133$ is filled with water at $20^{\circ} \mathrm{C}$ When valve $A$ is closed, $p_{1}-p_{2}=75$ kPa. When the valve is open and water flows at $500 \mathrm{m}^{3} / \mathrm{h}, p_{1}-p_{2}=160 \mathrm{kPa}$.
The long pipe in Fig. $\mathrm{P} 3.133$ is filled with water at $20^{\circ} \mathrm{C}$ When valve $A$ is closed, $p_{1}-p_{2}=75$ kPa. When the valve is open and water flows at $500 \mathrm{m}^{3} / \mathrm{h}, p_{1}-p_{2}=160 \mathrm{kPa}$....
##### Problem 0,4 Lat do1oa" > 2Latana A() NHI Hu Q tutand funatwn unelu omutium IVuenrit €rumimUYWIAAIU
Problem 0,4 Lat do 1oa " > 2 Latana A() NHI Hu Q tutand funatwn unelu omutium IVuenrit €rumim UYWIAAIU...
##### Using the principles to be developed in Chapter on ubrum, one can determine that the tension...
Using the principles to be developed in Chapter on ubrum, one can determine that the tension in cable ADS 203. N. Determine the moment about the x-xs of this torson force un form plate about the is. What is the moment of the tension force en Asbut the line 0 on point A. Compare your result with the ...
##### Use the method of cylindrical shells to find the volumegenerated by rotating the region bounded by the given curves aboutthe y-axis.y = 23x, y =0, x = 1
Use the method of cylindrical shells to find the volume generated by rotating the region bounded by the given curves about the y-axis. y = 2 3 x , y = 0, x = 1...
##### PLEASE NOTE! I ONLY NEED ANSWER TO QUESTION 7 Kate and Claire, recent college graduates, are...
PLEASE NOTE! I ONLY NEED ANSWER TO QUESTION 7 Kate and Claire, recent college graduates, are unable to find suitable jobs in their field of accounting. However, each has been involved with a small business of their own for the last several years, and have been doing very well. Kate is a talented se...
##### San[eA Juanxa 34 puy pue BuSeaxap pue Buseanu Sq uopuny PIYA .O STeADJ4 ?51 LpwPp! 04} OT + exv ~+* = (x)J uopuny J51 o2AjeuV
san[eA Juanxa 34 puy pue BuSeaxap pue Buseanu Sq uopuny PIYA .O STeADJ4 ?51 LpwPp! 04} OT + exv ~+* = (x)J uopuny J51 o2AjeuV...
##### 7. [-16 Points) DETAILS SCALCCC4 1.7.031. Find parametric equations for the path of a particle that...
7. [-16 Points) DETAILS SCALCCC4 1.7.031. Find parametric equations for the path of a particle that moves around the given circle in the manner described. x2 + (y - 1)2 = 16 (a) Once around clockwise, starting at (4,1). X(t) = (t) = Osts 2017 (b) Four times around counterclockwise, starting at (4,1)...
##### Find the Laplace transform of f (t) = (sint cos t)2
Find the Laplace transform of f (t) = (sint cos t)2...
##### Joe initially is holding mass at rest at the top of a frictionless incline by applying a horizontal force, magnitude F The incline has length L, with slope forming angle with the horizontal (see figure) . Joe gets tired, reduces the amount of force he is pushing with to F/2, and the mass starts sliding down the incline a) Draw a free body diagram for the mass_ 6) How much Force, F; does Joe have to apply to keep the mass stationary? Should find F in terms of the variables listed below: Find the
Joe initially is holding mass at rest at the top of a frictionless incline by applying a horizontal force, magnitude F The incline has length L, with slope forming angle with the horizontal (see figure) . Joe gets tired, reduces the amount of force he is pushing with to F/2, and the mass starts slid...
##### A chemist started with 22.8 g of a reactant and produced 13.9 gof a product.If the theoretical yield of the product was 29.5 g, what was thepercent yield?
A chemist started with 22.8 g of a reactant and produced 13.9 g of a product. If the theoretical yield of the product was 29.5 g, what was the percent yield?...
##### < 0) = 1/3, and Exercise 9.8. Suppose X has an N(u,02) distribution, P(X P(X <...
< 0) = 1/3, and Exercise 9.8. Suppose X has an N(u,02) distribution, P(X P(X < 1) = 2/3. What are the values of u and o?!...
##### 3. +-12 points SerCP10 3.P.018. My Notes A map suggests that Atlanta is 730 miles in...
3. +-12 points SerCP10 3.P.018. My Notes A map suggests that Atlanta is 730 miles in a direction 5.00 north of east from Dallas. The same map shows that Chicago is 560 miles in a direction 21.0° west of north from Atlanta. The figure below shows the location of these three cities. Modeling the E...
##### HWO1 matrices operations: Problem 10 Pravious Problem Problem List Next Problampoint) Let-3 55 -2-4B = -3c = |If possible, compute the following: If an answer does not exist, enter DNE:[[24,24],[161,86]] ACB =help (matrices)[[24,24],[161,86]] ABChelp (matrices}Note: In order t0 get credit for this problem all answers must be correct
HWO1 matrices operations: Problem 10 Pravious Problem Problem List Next Problam point) Let -3 55 -2 -4 B = -3 c = | If possible, compute the following: If an answer does not exist, enter DNE: [[24,24],[161,86]] ACB = help (matrices) [[24,24],[161,86]] ABC help (matrices} Note: In order t0 get credit...
##### Nilut Candy Corporation purchasCo the trademark for the popular Yummm Candy Bar from the ZumZum Company: At tho same lime. Niler also purchased ZumZum s customer Iist: Niler paid tho total purchase price of 5720,000 in cash: Niler's valualion consultants independenily estimate Ihe value of the trademark t0 be 5438,000 and customer Iist to be 5292,000. What = the journa cnin record Ihe purchase?Determine the prope Ilocation of the purchase cost each of the intangible assets acquired,Allocate
Nilut Candy Corporation purchasCo the trademark for the popular Yummm Candy Bar from the ZumZum Company: At tho same lime. Niler also purchased ZumZum s customer Iist: Niler paid tho total purchase price of 5720,000 in cash: Niler's valualion consultants independenily estimate Ihe value of the ...
##### 9,17,33 and ill like cos 22 14. 2+1 In Exercises 1-26 find the Taylor series for...
9,17,33 and ill like cos 22 14. 2+1 In Exercises 1-26 find the Taylor series for the function about the given point. In each case determine values of z for which the series converges to the function. 3 1. + about zo = 0 2. 1+2 about = 1 3. (1 - 2)2 about 20 = 0 4. e* about 20 = 1+i 5. sin z about...
##### Queation -FclntsSvtAnaxnrAn oprical instrument has Iwo convex lenses wlth focal length 0f 15 cm are placed 30 cm apart: cm tall object placed 100 cm away from the objective lens Reep mind you can use this image and establish scale draw It yourself and establish - scale on your drawing the setup; Remember lenses can be extended vertically and the optical axis can be extended out t00_what Is the location Jnd height ofthe objective $Imoge? Is It real or vIrujl? What the location and height of the queation - Fclnts SvtAnaxnr An oprical instrument has Iwo convex lenses wlth focal length 0f 15 cm are placed 30 cm apart: cm tall object placed 100 cm away from the objective lens Reep mind you can use this image and establish scale draw It yourself and establish - scale on your drawing the setup; ... 1 answer ##### 2. Circle all the stereogenic centers on the following molecule and label then as (R) or... 2. Circle all the stereogenic centers on the following molecule and label then as (R) or (S) (10 points). H Me Me... 1 answer ##### 10) Suppose the government revises its estimate of the external benefit per unit from$2 to...
10) Suppose the government revises its estimate of the external benefit per unit from $2 to$4. 432109876543210 Suppose the government estimates that there is a \$2 positive externality per unit of widget in the economy. What will be the total amount of external benefits (the total value of all posit...
##### Question 7 (5 points) In a certain region, the electric field has a uniform strength of...
Question 7 (5 points) In a certain region, the electric field has a uniform strength of E = 500 N/C and it is directed upward. What is the potential difference between two 30 cm long metal discs held a distance y = 10.0 cm apart from each other as shown below? 9 + Τ Ι I 1 15 O Format BIU De...
##### 19. solve the equation (x in radians and o in degrees) for all exact solutions where...
19. solve the equation (x in radians and o in degrees) for all exact solutions where appropriate. Round radians to four decimals and degrees to nearest tenth. COS X + 2 COS X = -1...
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2022-11-26 09:09:12
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http://mathematica.stackexchange.com/tags/presentations/hot
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# Tag Info
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You can create SlideShows using Mathematica and run it to demonstrate presentation. Main advantage of using such Slideshow over Powerpoint / PDF is that you can play dynamic content. This Link gives further details on how to create http://reference.wolfram.com/mathematica/howto/CreateASlideShow.html This screencast gives detailed steps on how to create ...
28
Prashant gave excellent references. Some additional tips I've found useful: You can turn off cell labels (the In[_] and Out[_] labels) from the option inspector by unchecking Cell Options->CellLabels->Show Cell Labels I've had to make PDF versions of the slides for various reasons. Mathematica does not do PDF page breaks very well, so instead of ...
18
Please try this code, based on Sasha's adaption of my own answer to this question. AutoCollapse[] := ( If[$FrontEnd =!=$Failed, SelectionMove[EvaluationNotebook[], All, GeneratedCell]; FrontEndTokenExecute["SelectionCloseUnselectedCells"]]) Then in a new cell: 2 + 2 AutoCollapse[] Always place AutoCollapse[] as the last line of an Input cell. ...
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Another method that I sometimes prefer is to use the presentation software of your choice (e.g. PowerPoint) and hyperlink the notebooks into that. For some things like showing lots of images (especially fullscreen) Mathematica is not very handy (yet). You can simplify the preparation for this by opening and evaluating all these notebooks up front. For some ...
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I do this pretty regularly, and my strategy is to talk at the blackboard, show static slides, try to get the students talking/asking questions, and then finish things off with a Manipulate or two. One thing I have found particularly effective is to ask students to download a .cdf version and to use it as a springboard for a homework question. It's usually ...
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I've just completed the second run of a course I designed aimed at a roughly similar group of students. As background, my course, titled "Dynamics Systems Analysis & Modeling", is intended to be a bridge between calculus / very basic differential equations at the front end and control theory for engineers at the tail end. The goal of the course is to ...
12
I find technology such as Mathematica lends itself nicely to alternative methods of instructional delivery such as Process Oriented Guided Inquiry Learning and the Flipped Classroom. I use both of these methods in entry-level and upper-level undergraduate classes with reasonable success. To that end, I would strongly suggest that you consider the role of ...
9
I will tell you my personal thoughts. I don't like when my teacher uses a projector. He goes through slides, it's like you said, just fast. Two times per semester we have tutorials when he writes on the blackboard and it's far more interesting because he goes through every step and that's very important for me. When I exercise at home, I can recall every ...
9
I would do it entirely without ToString. The main tool in combining mixed type output in a given order is Row: Ns = 1; Table[Row[ { Ket[Row[{ Replace[ Quotient[i - 1, Ns], {0 -> "\[UpArrow] ", 1 -> "\[DownArrow] "}], Mod[i, Ns]}] ], Bra[Row[{Replace[ Quotient[j - 1, Ns], {0 -> "\[UpArrow] ", ...
9
You could alt-click an output bracket which will cause all output brackets to be selected and then ctrl-} to close all subgroups, which, in this case, will close all input brackets that had output. Alternatively, you could select all outputs in this way and check the menu item Cell>Grouping Close All Unselected
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data = Reverse@Table[{2^n, 1}, {n, 1, 6}]; Graph[Table[DirectedEdge[i, i + 1], {i, Length@data - 1}], VertexLabels -> "Name", VertexCoordinates -> ({#2 Cos[#1], #2 Sin[#1]} & @@@ data), Axes -> True, ImageSize -> 300]
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You can use SystemOpen to open the link with system-wide standard browser/application: SystemOpen["https://www.youtube.com/watch?v=yL_-1d9OSdk"] You can also use it to open files stored on the local hard drive using the default applications (i.e. the default movie player for movie files). E.g. in conjunction with Button: Button["Chicken", ...
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Here's a version which uses Inherited to pull the slide show's docked cell in. Unfortunately, there's a bug when Inherited references an empty value where it shows some needless whitespace. So this version uses Dynamic to detect the ScreenStyleEnvironment and switch its behavior accordingly. With[{mycell = Cell["Boo!", "DockedCell"]}, nb = ...
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I have learned a lot from Wolfram Training: Computable Document Format (CDF) Courses. And for a more advanced example from Wolfram Software Development Training Course: Developing Enterprise-Class Web Applications. For all these videos you can download the notebook to study the code.
5
In addition to my comment, I guess I could also point out that Mathematica plots can be combined using the Show command. So you could define two different plots as follows and then combine them: f[x_, y_] := (x^3 + y^3)^(1/3) surface = Plot3D[f[x, y], {x, -2, 2}, {y, -2, 2}]; line = ParametricPlot3D[ With[{x = t, y = t}, {x, y, f[x, y]}], {t, 0, 2}, ...
5
You should be able to Shift+drag the corner handles of the orange frame to change the aspect ratio. (You may need to double click to get the thick gray editing frame first.) Note: if resizing individual elements Shift operates in reverse: holding it constrains the aspect ratio, while not holding it allows free resizing. You can resize the canvas using ...
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foo := Module[{r = Transpose@Through[{Cos, Sin}[2^Range[#]]], p}, p = Partition[r, 2, 1]; Graphics[{Opacity[.5], Red, Disk[#, .05] & /@ r, Opacity[1], Black, Text @@@ MapIndexed[{First@#2, #} &, r], Arrow[#, .05] & /@ p}, PlotRange -> {{-1.2, ...
4
This might be a good time to use Dynamic objects that will update as required with controls, buttons, or UpdateInterval, leaving direct evaluation of cells for outside of class.
3
CellPrint@ Cell[BoxData[ TagBox[GridBox[{{ToBoxes[Graphics[Circle[]]], Cell["Text", "SideCaption"]}}], "Grid"]], "SideCaptionArray"] SideCaption and SideCaptionArray are in the Default.nb style sheet if you want to change them.
3
After some frustrating hunting around, I think I found it. You have to enable this option in the option inspector (for the selected cells or the selected notebook): ShowGroupOpener WholeCellGroupOpener This will lead to the desired behaviour.
3
Yes, there is! Wolfram recently launched an online version of mathematica in various forms. You can register for the "programming cloud" and show them a full mathematica notebook online. With the free account you'll be missing a few IO abilities (no databases, limited file IO iirc). This should be more than enough to demonstrate your work. ...
3
Short answer: don't lecture; have students work hands-on, initially from notebooks you prepare. Long answer: Over nearly twenty years I taught mathematics courses ranging from freshman calculus and sophomore linear algebra through sophomore-junior discrete structures to junior-senior complex analysis in which students had to learn Mathematica from scratch ...
2
This problem is solved by TabView. In[1]:= 5+4 Out[1]=9 In[2]:= AbsoluteTiming[Pause[3]; Plot[Sin[x],{x,-1,1}] Out[2]= {3.013000, <plot of sin[x]>} In[3]:= AbsoluteTiming[TabView[{%1,%2[[2]]}]] Out[3]={0.,Tabbed pane with 9 and <plot of sin[x]>} The timing bits are there to indicate that % calls the output without re-computing the ...
1
PolarGraph[n_Integer] := Module[{lp = ListPolarPlot @ Table[{2^x, 1}, {x, 1, n}], p, a, t}, p = Reverse @ Take [Cases[lp, {_Real, _Real}, -1], n]; a = Arrow /@ Partition[p, 2, 1]; t = Text @@@ Transpose[{Range @ Length @ p, p}] /. Text[u_, v_] :> Text[Style[u, Red, 16], v + 0.1]; Graphics[{a, t}, Axes -> True]] PolarGraph[7]
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Well, I just figured out a simple minded solution, but I hope that someone has a more elegant one perhaps. Basically you can just use StringReplace & use string all the way Clear@i1 Table[ HS2[[i, j]]* StringReplace[ "\!$$\*TemplateBox[{\"row\"},\n\"Ket\"]$$\!$$\*TemplateBox[{\"col\ \"},\n\"Bra\"]$$", {"row" -> ( (Replace[ ...
1
I've just tried it out on an older iMac and an old white plastic Apple remote (last sold in 2009, apparently). Yes, it works. You have to hold the Forward/Back buttons down, rather than click. But Mathematica seems to detect these actions and advances to the next slide, or goes back to the previous one. I'd rather go for something wireless, myself, but if IR ...
Only top voted, non community-wiki answers of a minimum length are eligible
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2014-11-24 20:27:41
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https://math.stackexchange.com/questions/234062/probability-of-tossing-a-fair-coin-with-at-least-k-consecutive-heads
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# Probability of tossing a fair coin with at least $k$ consecutive heads
Tossing a fair coin for $N$ times and we get a result series as $HTHTHHTT\dots~$, Here '$H$' denotes 'head' and '$T$' denotes 'tail' for a specific tossing each time.
What is the probability that the length of the longest streak of consecutive heads is greater than or equal to $k$? (that is we have a $HHHH\dots~$, which is the substring of our tossing result, and whose length is greater than or equal to $k$)
I came up with a recursive solution (though not quite sure), but cannot find a closed form solution.
Here is my solution.
Denote $P(N,k)$ as the probability for tossing the coin $N$ times, and the longest continuous heads is greater or equal than $k$. Then (For $N>k$)
$$P(N,k)=P(N-1,k)+\Big(1-P(N-k-1,k)\Big)\left(\frac{1}{2}\right)^{k+1}$$
• Your example sequence reminded me of this brilliant Simpsons scene :-) – joriki Nov 10 '12 at 8:58
• I'm getting $(N-k+1)/2^k$ for the closed form. Would you like for me to post my solution, or do you want to think about it some more before I spoil the beans? – Braindead Nov 10 '12 at 8:59
• @Braindead: That can't be right; it's $\gt1$ for small $k$. – joriki Nov 10 '12 at 9:00
• Duh, you are right. I overcounted. – Braindead Nov 10 '12 at 9:08
• I tried to clarify some of the formulations; please check whether I preserved the intended meaning. In particular, I assumed that you had merely accidentally written "greater" once instead of "greater or equal". – joriki Nov 10 '12 at 9:12
Your recurrence relation is correct. I don't think you can do much better than that for general $k$, but you can find a closed form for specific values of $k$. For the first non-trivial value of $k$, the recurrence relation is
$$p_n=p_{n-1}+(1-p_{n-3})/8\;.$$
With $p_n=1+\lambda^n$, the characteristic equation becomes $\lambda^3-\lambda^2+1/8=0$. One solution, $\lambda=1/2$, can be guessed, and then factoring yields $(\lambda-1/2)(\lambda^2-\lambda/2-1/4)$ with the further solutions $\lambda=(1\pm\sqrt5)/4$. Thus the general solution is
$$p_n=1+c_1\left(\frac12\right)^n+c_2\left(\frac{1+\sqrt5}4\right)^n+c_3\left(\frac{1-\sqrt5}4\right)^n\;.$$
The initial conditions $p_0=0$, $p_1=0$, $p_2=1/4$ determine $c_1=0$, $c_2=-(1+3/\sqrt5)/2$ and $c_3=-(1-3/\sqrt5)/2$, so the probability is
\begin{align} p_n &=1-\frac{1+3/\sqrt5}2\left(\frac{1+\sqrt5}4\right)^n-\frac{1-3/\sqrt5}2\left(\frac{1-\sqrt5}4\right)^n\\ &=1-\frac4{\sqrt5}\left(\left(\frac{1+\sqrt5}4\right)^{n+2}-\left(\frac{1-\sqrt5}4\right)^{n+2}\right)\;. \end{align}
Thus, for large $n$ the probability approaches $1$ geometrically with ratio $(1+\sqrt5)/4\approx0.809$.
• I've added an explicit formula for the $P(N,k)$ which might be of interest to you. Note, that your $p_n=1-\frac{1}{2^n}F_{n+2}$ with $F_n$ the Fibonacci numbers. Regards, – Markus Scheuer Jan 18 '16 at 21:40
We can derive an explicit formula of the probability $P(N,k)$ based upon the Goulden-Jackson Cluster Method.
We consider the set of words $\mathcal{V}^{\star}$ of length $N\geq 0$ built from an alphabet $$\mathcal{V}=\{H,T\}$$ and the bad word $\underbrace{HH\ldots H}_{k \text{ elements }}=:H^k$ which is not allowed to be part of the words we are looking for. We derive a function $f_k(s)$ with the coefficient of $s^N$ being the number of wanted words of length $N$.
The wanted probability $P(N,k)$ can then be written as \begin{align*} P(N,k)=1-\frac{1}{2^N}[s^N]f_k(s) \end{align*}
According to the paper (p.7) from Goulden and Jackson the generating function $f_k(s)$ is \begin{align*} f_k(s)=\frac{1}{1-ds-\text{weight}(\mathcal{C})}\tag{1} \end{align*} with $d=|\mathcal{V}|=2$, the size of the alphabet and with $\mathcal{C}$ the weight-numerator with \begin{align*} \text{weight}(\mathcal{C})=\text{weight}(\mathcal{C}[H^k]) \end{align*} We calculate according to the paper \begin{align*} \text{weight}(\mathcal{C}[H^k])&=-\frac{s^k}{1+s+\cdots+s^{k-1}}=-\frac{s^k(1-s)}{1-s^k} \end{align*}
We obtain the generating function $f(s)$ for the words built from $\{H,T\}$ which don't contain the substring $H^k$ \begin{align*} f_k(s)&=\frac{1}{1-2s+\frac{s^k(1-s)}{1-s^k}}\\ &=\frac{1-s^k}{1-2s+s^{k+1}}\tag{2}\\ \end{align*}
Note: For $k=2$ we obtain \begin{align*} f_2(s)&=\frac{1-s^2}{1-2s+s^{3}}\\ &=1+2s+3s^2+5s^3+8s^4+13s^5+21s^6+34s^7+\mathcal{O}(s^8) \end{align*}
The coefficients of $f_2(s)$ are a shifted variant of the Fibonacci numbers stored as A000045 in OEIS.
Note: For $k=3$ we obtain \begin{align*} f_3(s)&=\frac{1-s^3}{1-2s+s^{4}}\\ &=1+2s+4s^2+7s^3+13s^4+24s^5+44s^6+81s^7+\mathcal{O}(s^8) \end{align*}
The coefficients of $f_3(s)$ are a shifted variant of the so-called Tribonacci numbers stored as A000073 in OEIS.
We use the series representation of $f_k(s)$ in (2) to derive an explicit formula of the coefficients.
\begin{align*} [s^N]f(s)&=[s^N](1-s^k)\sum_{m=0}^{\infty}(2s-s^{k+1})^m\\ &=[s^N](1-s^k)\sum_{m=0}^{\infty}s^m(2-s^k)^m\\ &=[s^N](1-s^k)\sum_{m=0}^{\infty}s^m\sum_{j=0}^m\binom{m}{j}(-1)^js^{kj}2^{m-j}\\ &=([s^N]-[s^{N-k}])\sum_{m=0}^{\infty}s^m\sum_{j=0}^m\binom{m}{j}(-1)^js^{kj}2^{m-j}\tag{3}\\ &=\sum_{m=0}^{N}([s^{N-m}]-[s^{N-k-m}])\sum_{j=0}^m\binom{m}{j}(-1)^js^{kj}2^{m-j}\tag{4}\\ &=\sum_{{m=0}\atop{m\equiv N(\bmod k)}}^{N}\binom{m}{\frac{N-m}{k}}(-1)^{\frac{N-m}{k}}2^{m-\frac{N-m}{k}}\\ &\qquad-\sum_{{m=0}\atop{m\equiv N(\bmod k)}}^{N-k}\binom{m}{\frac{N-m}{k}-1}(-1)^{\frac{N-m}{k}-1}2^{m-\frac{N-m}{k}+1}\\ &=\sum_{{m=0}\atop{m\equiv N(\bmod k)}}^{k-1}\binom{m}{\frac{N-m}{k}}(-1)^{\frac{N-m}{k}}2^{m-\frac{N-m}{k}}\tag{5}\\ &\qquad+\sum_{{m=k}\atop{m\equiv N(\bmod k)}}^{N-k} \left(\binom{m}{\frac{N-m}{k}}-\frac{1}{2^k}\binom{m-k}{\frac{N-m}{k}}\right)(-1)^{\frac{N-m}{k}}2^{m-\frac{N-m}{k}}\\ \end{align*}
Comment:
• In (3) we use the linearity of the coefficient of operator and $[s^N]s^mf(s)=[s^{N-m}]f(s)$
• In (4) we change the limit of the left hand sum from $\infty$ to $N$ according to the maximum coefficient $[s^N]$. According to the factors $s^{kj}$ we consider in the following only summands with $m\equiv N(\bmod k)$
• In (5) we reorganise the sums from the line above by extracting from the left hand sum the first summand and shifting in the right hand side the index by one and putting both sums together.
We conclude: An explicit representation of the probability $P(N,k)$ $(n\geq 0)$ is according to (5) \begin{align*} P(N,k)&=1-\sum_{{m=0}\atop{m\equiv N(\bmod k)}}^{k-1}\binom{m}{\frac{N-m}{k}}(-1)^{\frac{N-m}{k}}2^{-\frac{(k+1)(N-m)}{k}}\\ &\qquad-\sum_{{m=k}\atop{m\equiv N(\bmod k)}}^{N-k} \left(\binom{m}{\frac{N-m}{k}}-\frac{1}{2^k}\binom{m-k}{\frac{N-m}{k}}\right)(-1)^{\frac{N-m}{k}}2^{-\frac{(k+1)(N-m)}{k}} \end{align*}
A bit more practical focus than previous answers:
Main idea:
1. We build a "staircase":
2. $p$ chance to get to next step, $1-p$ chance to fall down and have to restart (or "reflip"). Except for the highest step where we always stay (whenever we manage to get there).
3. We need $k+1$ steps to "remember" where on the stairs we are for $k$ flips in a row.
We can use this to build a matrix for a Markov chain:
We can build a block matrix:$$\frac{1}{2}\left[\begin{array}{cc}2&1&\bf 0^T\\\bf 0&\bf0&\bf I_k\\0&1&\bf 1^T \end{array}\right]$$
(For the special case $k = 5$). We build a stochastic matrix:
$${\bf P} = \frac{1}{2}\left[\begin{array}{cccccc}2&1&0&0&0&0\\0&0&1&0&0&0\\0&0&0&1&0&0\\0&0&0&0&1&0\\0&0&0&0&0&1\\0&1&1&1&1&1\end{array}\right]$$
Now our answer will simply be $$[1,0,{\bf 0}]\, {\bf P}^k\, [{\bf 0},0,1]^T$$
Which is a scalar product using ${\bf P}^k$ as a gramian matrix.
Now to calculate this in practice one could probably in general (for general $p$) benefit from a canonical transformation of $\bf P$, but it's doubtable in this case as the matrix is already sparse with literally only sums and permuations and bit shifts required on the vector elements.
Feller considers this problem in section XIII.7 of An Introduction to Probability Theory and Its Applications, Volume 1, Third Edition. He shows that the probability of having no run of length $k$ in $N$ throws is asymptotic to $$\frac{1-(1/2)\;x}{(k + 1 - kx)\; (1/2)} \cdot \frac{1}{x^{N+1}}$$ where $x$ is the least positive root of $$1 - x +(1/2)^{k+1} x^{k+1} = 0$$
(I have changed Feller's notation to agree with the problem statement and have only considered the case of a fair coin; Feller considers the more general case of a biased coin. For more information see equation 7.11, p. 325 in the referenced document.)
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2019-09-18 20:07:25
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https://lavelle.chem.ucla.edu/forum/viewtopic.php?f=59&t=17389&p=45615
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## pKb and pKa
Acidity $K_{a}$
Basicity $K_{b}$
The Conjugate Seesaw $K_{a}\times K_{b}=K_{w}$
Josue_Marin_3I
Posts: 16
Joined: Wed Sep 21, 2016 2:56 pm
### pKb and pKa
I understand that the lower the pKa the stronger the acid is, but why is it that the lower the pKb the stronger the base?
Anmol Dhaliwal 2C
Posts: 25
Joined: Wed Sep 21, 2016 2:57 pm
### Re: pKb and pKa
The weaker bases have a smaller Kb which makes them have a greater pKb (this is the same for Ka and pKa). That means, the smaller the pKb is, the bigger the Kb is and therefore it would be a stronger base. For instance, methylamine has a Kb of 3.6x10^-4 and the pKb is 3.44 (low) which makes it more of a strong base than ammonia which has a Kb of 1.8x10^-5 and a pKb of 4.75 (higher than methylamine's pKb and it is a little bit weaker of a base than methylamine is). Hope this helps!
DBaquero
Posts: 13
Joined: Wed Sep 21, 2016 2:59 pm
Been upvoted: 1 time
### Re: pKb and pKa
The equilibrium constant (in this case Kb) is larger if the concentration of product (in this case [OH-]) is larger, so it stands to reason that high concentrations of OH- means a stronger base. You take the -log of the Kb to get the pKb which is the same as the log(1/Kb) meaning the smaller the Kb (so the smaller the [OH-], the larger the pKb. So low [OH-] means high pKb and vise versa, meaning the pKb is inversely related to base strength.
Ashley Bertholf 1E
Posts: 10
Joined: Fri Jul 15, 2016 3:00 am
### Re: pKb and pKa
When looking at an equation say HCOOH + NaOH =(equilibrium sign) HCOO- + Na+ +H20, how do you know if the reaction priiduces Ka or Kb?
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2021-02-28 14:16:40
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https://www.icts.res.in/event/page/17772
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Monday, 08 July 2019
Time Speaker Title Resources
09:45 to 11:00 -- Registration & Tea
11:00 to 12:00 Anirban Chakraborti Multi-agent modelling in Complex Socio-economical Systems
A short abstract of the talk:
We will present simple models for studying complex socio-economical systems that have been recently proposed by physicists, viz., the kinetic exchange models for studying income-wealth distribution, and game-theoretical models for studying competitive resource allocation.
Anirban Chakraborti is a Professor at the School of Computational and Integrative Sciences, Jawaharlal Nehru University, New Delhi, since March, 2014. He had worked as an Associate Professor at the Chair of Quantitative Finance, École Centrale Paris, France, during 2009-14. He had obtained a Ph.D. in Physics from the Saha Institute of Nuclear Physics, India and then the Habilitation in Physics from Université Pierre et Marie Curie (Paris VI), France. He has the experience of working as a scientist in many reputed universities and educational institutions in India, USA, Europe and Japan. He was awarded the prestigious Young Scientist medal of the Indian National Science Academy in 2009. He has published several books and research articles from internationally renowned publishers. His main research interests lie in the areas of Econophysics, Sociophysics, Data Science, Statistical Physics, Quantum Physics and Nanomaterial Science.
12:00 to 13:00 Arjun Jayadev Macroeconomic Policy-Theory and Current Puzzles
A short abstract of the talk:
Monetary Policy was for the last three decades, the policy of choice for macroeconomic stabilization. Currently, there are are open questions about the efficacy of monetary and fiscal policy and their current uses. In this talk I will go over the traditional mechanisms and the current concerns with macroeconomic management. I will discuss this in the Indian context and will end with briefly sketching some open areas of theoretical interest from scholars.
Arjun Jayadev is a Professor of Economics and Senior Economist at the Institute for New Economic Thinking. He works on issues of finance, distribution, labor and intellectual property.
13:00 to 14:30 -- Lunch
14:30 to 14:40 Nikita Kohli Threat and Use of Force: Forecasting the Decision to Initiate Interstate Wars in Times of Uncertainty
A short abstract of the talk:
TBA
Nikita Kohli is an experienced researcher with a demonstrated history of working in the international affairs industry and military organizations. She is currently working as a researcher at the Centre for Land Warfare Studies (CLAWS), a think tank of the Indian Army, where she creates policy notes and research papers, in particular in the areas of strategic foresight and international security. She does focused research on terrorism trends and financing, foreign policy strategy, and weapon system analysis using Game Theory and Statistical and Mathematical modelling.
Nikita has previously worked with the Indian Mission to the United Nations in Geneva and Oxfam India. She has also worked as a consultant for and presented various papers before multiple different militaries, including the Indian and the US armies at the strategic level.
14:45 to 14:55 Nikita Kohli Strategic Logic of Suicide Bombings
A short abstract of the talk:
TBA
Nikita Kohli is an experienced researcher with a demonstrated history of working in the international affairs industry and military organizations. She is currently working as a researcher at the Centre for Land Warfare Studies (CLAWS), a think tank of the Indian Army, where she creates policy notes and research papers, in particular in the areas of strategic foresight and international security. She does focused research on terrorism trends and financing, foreign policy strategy, and weapon system analysis using Game Theory and Statistical and Mathematical modelling.
Nikita has previously worked with the Indian Mission to the United Nations in Geneva and Oxfam India. She has also worked as a consultant for and presented various papers before multiple different militaries, including the Indian and the US armies at the strategic level.
14:55 to 15:20 Avik Sarkar Integrated Energy Modeling
A short abstract of the talk:
The administration of the Energy Sector in India is distributed among 5 different ministries, which creates challenges in working towards a common long-term goal for the nation in terms of energy security. The talk would provide an overview of the Integrated Energy Modeling work that was done at NITI Aayog using the MESSAGE Modeling Framework.
Dr Avik Sarkar has recently taken up the role of Professor in Data, AI and Public Policy at Indian School of Business. Dr Avik Sarkar was most recently Heading the Data Analytics Cell at NITI (National Institution for Transforming India) Aayog, a premier policy think-tank of the Government of India. At NITI Aayog, Dr Sarkar was in-charge of developing roadmap for use of data, analytics and artificial intelligence for Governance and Policy making along with providing analytical insights for policy making across sectors like Sustainable Development Goals, Direct Benefit Transfer, Innovation, Digital payments, Education, Healthcare/Nutrition, Agriculture, etc.
Dr Sarkar has over 18 years of experience across different aspects of data analytics, statistical modeling, data/text mining across companies like IBM, Accenture, Nokia, NASA, Persistent Systems, etc. In his last role at Accenture Consulting (Singapore), Dr Sarkar contributed to various data and analytics related engagements with Singapore Government. Dr Sarkar holds a PhD from The Open University (UK), Masters from Indian Institute of Technology (IIT) Bombay and Bachelors from Calcutta University. Dr Sarkar has authored several technical publications and technology patents.
Dr Sarkar has been nominated among the "Top 10 Data Scientists in India" in 2017 by the Analytics India Magazine and nominated as "LinkedIn Influencer" in the Technology space in 2015 for contribution and engaging discussions on the LinkedIn platform in areas related of Big Data, Artificial Intelligence, Data Science, etc.
15:20 to 15:45 Harsha Krishna Simulogue: a simulation platform for integrated governance
A short abstract of the talk:
The simulation platform uses agent based modelling approach to build generative simulations for the land-use, water and waste sectors for Chennai Metropolitan area. The simulation is built using both quantitative and qualitative (institutional structure) data. the The link to the publication regarding the same is here: https://zenodo.org/record/3066348#.XRNHG5Mzau5.
Harsha K graduated from the International Institute of Information Technology, Bangalore. He has designed and built agent based simulations to model urbanisation in Indian cities. He has also designed similar simulations for modelling public transportation in Bangalore. He has also been working with his colleagues to develop games in the field of public transport. His current research interests include methodologies to model complex adaptive systems, cities and urban systems.
15:45 to 16:15 -- Coffee break
16:15 to 17:30 -- Discussion
Tuesday, 09 July 2019
Time Speaker Title Resources
09:30 to 10:45 Rajiv Sethi Rationality and Complexity in Games - I
A short abstract of the talk:
These talks will cover standard and non-standard solution concepts in the theory of games, including models of procedural rationality, with applications to public goods and common pool resources.
Rajiv Sethi is a Professor of Economics at Barnard College, Columbia University and an External Professor at the Santa Fe Institute. He is on the editorial boards of the American Economic Review and Economics and Philosophy. His current research deals with information and beliefs. In collaboration with Brendan O’Flaherty, he has examined the manner in which stereotypes affect interactions among strangers, especially in relation to crime and the criminal justice system. Their book, Shadows of Doubt: Stereotypes, Crime, and the Pursuit of Justice was published by Harvard University Press in 2019. With Muhamet Yildiz, he has explored communication among individuals who consider each other to have valuable information, but also believe that others are biased to different degrees in the manner in which they process information. In previous work they have examined public disagreement and private information flows, and in current work are exploring the implications of correlated biases within social groups. He is also part of an interdisciplinary team working on the forecasting of geopolitical events using methods that combine machine models with human judgment.
10:45 to 11:15 -- Coffee break
11:15 to 12:30 Eleanor Power Faith in the Faithful: Religious Practice, Reputation, and Social Support in Rural Tamil Nadu
A short abstract of the talk:
Discerning the intentions and character of others is a difficult task. In the villages where I work in Tamil Nadu, religious practice is seen as particularly helpful in that process of discernment. There, religious acts are often quite dramatic: devotees walk across hot coals, pierce their skin with hooks and spears, walk barefoot to distant temples, and sacrifice animals to the divine. What do onlookers discern from these acts, and how do they shape people's relationships? Drawing on reputational and social support network data, I show that greater and costlier ritual participation corresponds to greater recognition not only for being devout, but also for holding a suite of prosocial traits. Perhaps more importantly, greater and costlier ritual participation also increases the likelihood of a supportive tie between individuals. And, supportive ties are more likely between those who participate in collective rituals together. These findings collectively give us some insight into the role of religion in society.
Eleanor Power is an Assistant Professor in the Department of Methodology at the LSE. She completed her PhD in Anthropology at Stanford University in 2015. Prior to joining LSE in 2017, she was an Omidyar Postdoctoral Fellow at the Santa Fe Institute. Eleanor's research explores questions regarding: the role of religion in society, the interaction between costly signaling and cooperation, gender differences in prominence and social capital, and the dynamics of gossip and social censure. She studies these topics primarily through fieldwork conducted in the Indian state of Tamil Nadu, where she has been working since 2009,using a combination of qualitative and quantitative methods, primary among which is social network analysis.
12:30 to 12:45 Supratim Sengupta Games people play: Individual decisions and collective outcomes
A short abstract of the talk:
Cooperation and conflict are seen in societies that span all biological scales ranging from the most primitive microbial communities to the most advanced human societies. Evolutionary game theory provides a simple yet powerful framework to understand the impact of cooperation and conflict on the evolution of the population in vastly different communities of living organisms. The use of either of these opposing behavioural traits by individuals during interactions with other agents often depend on the number and attributes of connected neighbours and the underlying structure of the population. Such interactions can significantly alter the “behavioural” landscape of the population. I will highlight how evolution of individual decisions in a social conflict, affect the population dynamics, often in unpredictable ways, leading to profound consequences for the persistence and proliferation of altruistic traits.
B.Sc. Physics (Hons.)1992: Presidency College (now Presidency University), Calcutta, India.
M.Sc. Physics (1994): Indian Institute of Technology, Kharagpur, India.
Ph.D. Physics (2000): Institute of Physics, Bhubaneswar, India.
Post-doctoral Fellow
2000-2003: Theoretical Physics Institute, Dept. of Physics, University of Alberta, Edmonton, Canada.
2003-2005: Dept. of Physics and Astronomy, McMaster University, Hamilton, Canada.
2005-2007: Dept. of Physics and Atmospheric Science, Dalhousie University, Halifax, Canada.
Faculty
2007-2011: School of Computational & Integrative Science, JNU, Delhi
2012-Present: Dept. of Physical Sciences, IISER Kolkata.
Research Interests
Evolutionary game theory applied to social and biological systems:
Evolution of cooperation and consequences of evolution of individual behaviour on population-level outcomes during social conflicts.
Origin of Life: Origin of Genetic Code, RNA world
RNA sequence analysis: Riboswitches, tRNA, bacterial and mitochondrial transcription and translation systems
12:45 to 13:00 Sagar Chakraborty Evolution as a Nonlinear Dynamical Process
A short abstract of the talk:
I shall present how the process of evolution is modeled using nonlinear dynamical equations, specifically, replicator equations. Using a replicator map, it will be discussed how fixed points, periodic solutions, and chaotic solutions can be interpreted in the language of game theory. Also, connection between population games in finite and infinite populations will be discussed. Finally, through the networks of the replicator maps, we shall try to find relation between emergent synchronization and cooperation in strategic (evolutionary) games.
My first introduction to the world of scientific research was through the problems of fluid dynamics. I have written quite a few papers on the problems of turbulence, magnetohydrodynamics, and astrophysical fluids. While I have used some numerical methods, the works have been more tilted towards the usage of mathematical and analytical methods. Moreover, I have kept a parallel interest of mine alive: nonlinear dynamics. Apart from my curiosity in the perturbative methods for nonlinear systems, I have studied many aspects of chaos, such as, synchronization, Hamiltonian chaos, and phase space reconstruction.
Over last two years, I have been trying to understand the process of evolution through nonlinear dynamical systems. Some of my primary interests are: (i) understanding how dynamical solutions correspond to the game theoretic equilibria, (ii) evolution of cooperation in evolutionary games, and (iii) connection between population games in finite and infinite populations.
13:00 to 14:30 -- Lunch
14:30 to 14:55 Anand Srivastava Macroeconomic volatility and non-substitutable inputs
A short abstract of the talk:
A production economy can be modeled as a network of firms connected by input-output relationships. A static equilibrium of such a system can be characterised for standard production functions and assumptions about competition and prices. The relationship between the volatility of a node and the macroeconomic volatility can also be obtained in terms of network characteristics. In the presence of short-run non-substitutability of some inputs, we attempt to characterise the stability of the economy with respect to firm-specific shocks in terms of the characteristics of the input-output network having a combination of substitutable and non-substitutable goods.
Anand Shrivastava is Assistant Professor of Economics and Programme Coordinator of the undergraduate programme at Azim Premji University. He holds a Ph.D. in Economics from the University of Cambridge. He has worked on issue of conflict, public works and labour. He is a co-author on the State of Working India reports on employment in India. His current interest is in areas of computational methods including agent-based modeling and networks.
14:55 to 15:20 Parongama Sen Opinion dynamics model: application to two recent social phenomena
A short abstract of the talk:
We discuss a opinion dynamics model and its application in understanding US presidential election and Brexit data. In the first case, the probability that the popular winner loses due to the electoral college system (as in 2016) is obtained and compared to real data. For Brexit, the ongoing analysis is limited to studies related to the time series data available from different surveys. Previously, similar analyses have been made for data from Econophysics models.
Subject: Physics
Research area: Statistical Physics
Topics of interest:
1. Equilibrium statistical physics: Phase transitions and critical phenomena.
2. Complex networks: Model networks and their properties; dynamical phenomena on networks.
3. Random walk; reaction diffusion systems
4. Nonequilibrium phenomena in spin systems and models inspired by social dynamics, e.g. opinion formation, disease spreading, social segregation etc.
Analysis of real data and appropriate models proposed in many of the studies.
Outside the scope of the present meeting: Disordered quantum walks.
15:20 to 15:45 Alexandre Reiffers-Masson Opinion shaping in social networks using reinforcement learning
A short abstract of the talk:
We consider a classical dynamics for opinion evolution with some stubborn agents and the possibility of continuously influencing the opinions of a few selected agents, but under resource constraints. We view the opinion dynamics as a value iteration for policy evaluation for a stochastic shortest path problem with some non-classical constraints. We propose an online two time scale reinforcement learning based scheme for this optimization problem. Supporting numerical studies are provided.
Alexandre Reiffers is a post-doctoral fellow at Robert Bosch Centre for Cyber Physical Systems. He received the B.Sc. degree in mathematics (2010) from the university of Marseille, the master degree in applied mathematics (2012) from the university of Pierre et Marie CURIE and the Ph.D. degree in computer science (January 2016) from the INRIA (National research institute in computer science and control) and the university of Avignon. His supervisors were Eitan Altman and Yezekael Hayel. From July 2016 to December 2017, Alexandre Reiffers was a researcher at SafranTech where he was working on comparison of maintenance strategies. Most of his research projects concern the application of mathematical tools (game theory, optimization, stochastic process and machine learning) for a better understanding of real-world problems. The different issues that he studies touch topics such as social networks, speech between human and computer, economy and manufacturing.
15:45 to 16:15 -- Coffee break
16:15 to 17:30 -- Poster and Discussion
Wednesday, 10 July 2019
Time Speaker Title Resources
09:30 to 10:45 Rajiv Sethi Rationality and Complexity in Games - II
A short abstract of the talk:
These talks will cover standard and non-standard solution concepts in the theory of games, including models of procedural rationality, with applications to public goods and common pool resources.
Rajiv Sethi is a Professor of Economics at Barnard College, Columbia University and an External Professor at the Santa Fe Institute. He is on the editorial boards of the American Economic Review and Economics and Philosophy. His current research deals with information and beliefs. In collaboration with Brendan O’Flaherty, he has examined the manner in which stereotypes affect interactions among strangers, especially in relation to crime and the criminal justice system. Their book, Shadows of Doubt: Stereotypes, Crime, and the Pursuit of Justice was published by Harvard University Press in 2019. With Muhamet Yildiz, he has explored communication among individuals who consider each other to have valuable information, but also believe that others are biased to different degrees in the manner in which they process information. In previous work they have examined public disagreement and private information flows, and in current work are exploring the implications of correlated biases within social groups. He is also part of an interdisciplinary team working on the forecasting of geopolitical events using methods that combine machine models with human judgment.
10:45 to 11:15 -- Coffee break
11:15 to 12:30 Eleanor Power The Complexity of Cooperation on Networks
A short abstract of the talk:
In recent years, there has been growing recognition of the potential of network structure to facilitate cooperation. "Network reciprocity," for example, has been put forth as a mechanism that can favour cooperation. However, the full implications of network dynamics for cooperation are as yet not fully explored. In this talk, I will outline some of the ways in which the nature of interpersonal interactions may add important complexity to our models and understanding of cooperation. Social relationships often entail repeated interactions of various behavioural types between individuals who are themselves indirectly connected. All of these features (repeated interactions, multiplex relationships, clustering) have the potential to impact the efficacy of the various mechanisms for the evolution cooperation. The consequences of network structure are particularly profound for humans, given our reliance on communication and the dynamics of information spread through networks. I will illustrate these dynamics with some ethnographic case studies from my fieldwork in rural South India, and I will discuss potential theoretical and empirical ways forward.
Eleanor Power is an Assistant Professor in the Department of Methodology at the LSE. She completed her PhD in Anthropology at Stanford University in 2015. Prior to joining LSE in 2017, she was an Omidyar Postdoctoral Fellow at the Santa Fe Institute. Eleanor's research explores questions regarding: the role of religion in society, the interaction between costly signaling and cooperation, gender differences in prominence and social capital, and the dynamics of gossip and social censure. She studies these topics primarily through fieldwork conducted in the Indian state of Tamil Nadu, where she has been working since 2009,using a combination of qualitative and quantitative methods, primary among which is social network analysis.
12:30 to 12:45 Swaprava Nath Research at the Interface of Computer Science and Economics
A short abstract of the talk:
Artificial Intelligence (AI) is prospering at a rapid rate in all aspects of technology. While a majority of these applications are purely data driven, making machines smarter through learning' from examples, there are a ton of everyday decision problems where there is no good examples to learn from. For example, how to make a voting rule that satisfy several desirable properties of a democracy, or to have an efficient social decision without any monetary inflow or outflow -- known as budget balance -- or to design a grading scheme of large classes that is as accurate as that of the instructors. In this talk, we pick some of these applications and show how research at the interface of computation and economics is making an impact through the use of tools like game theory, mechanism design, approximation algorithms, nonlinear optimization to assist humans take more efficient' decisions. This domain marks an alternative interpretation of artificial intelligence for the social decision making. I'll discuss briefly a few results and the analytical tools involved in deriving them.
Swaprava is an Assistant Professor at the Department of Computer Science and Engineering, IIT Kanpur. After finishing his PhD from the Dept. of Computer Science and Automation, Indian Institute of Science, Bangalore, he has held postdoctoral positions in Indian Statistical Institute, New Delhi and Carnegie Mellon University. His research interest lies in the intersection of/economics and computation/, which has several applications in social, industrial and computational paradigms. Apart from academic positions, Swaprava also has experience in the industry. He has worked at Xerox Research Centre Europe and Cisco Systems India. He has been recipients of Fulbright-Nehru post doctoral grant, Tata Consultancy Services PhD Fellowship, and the Honorable Mention Award of Yahoo! Key Scientific Challenges Program.
12:45 to 13:00 V. Sasidevan Committing heresy by co-action: A critique of Nash optimization
A short abstract of the talk:
The idea of optimizing individual good via a Nash optimization protocol is everpresent in analyzing strategic interactions. Though experimental evidence for real-life agents implementing such optimization is scarce, the protocol is often projected as a benchmark for what is considered as rational behavior. In this talk, I take a critical look at the assumptions behind the Nash optimization and discuss alternate possibilities for modeling benchmark rational behavior, which perhaps also make more sense in analyzing real-life game situations.
Dr. V. Sasidevan received Ph.D. from Tata Institute of Fundamental Research, Mumbai in 2014. Thereafter, he held post-doctoral positions at Institute of Mathematical Physics, Chennai and Frankfurt Institute for Advanced Studies, Germany. He joined Dept. of Physics, Cochin University of Science and Technology, Kerala in 2017. His research interests are Statistical physics of complex systems, Interdisciplinary applications of statistical physics in socio-economic systems, Complex networks, Systemic risk and Game theory.
13:00 to 14:30 -- Lunch
14:30 to 14:55 Tushar Nandi Taxation and Supplier Networks: Evidence from India
A short abstract of the talk:
Do tax systems distort firm-to-firm trade? This paper considers the effect of tax policy on supplier networks in a large developing economy, the state of West Bengal in India. Using administrative panel data on firms, including transaction data for 4.8 million supplier-client pairs, we first document substantial segmentation of supply chains between firms paying Value-Added Taxes (VAT) and non-VAT-paying firms. We then develop a model of firms’ sourcing and tax decisions within supply chains to understand the mechanisms through which tax policy interacts with supply networks. The model predicts partial segmentation in equilibrium because of both supply-chain distortions (taxes affect how much firms trade with each other) and strategic complementarities in firms’ tax choices. Finally, we test the model’s predictions using variations over time within-firm and within supplier-client pairs. We find that the tax system distorts firms’ sourcing decisions, and suggestive evidence of strategic complementarities in firms’ tax choices within supplier networks.
Tushar K. Nandi is an Assistant Professor of Economics at CTRPFP, Centre for Studies in Social Sciences, Calcutta. He did his Ph. D. in Economics from University of Siena, Italy. He was Visiting Scholar at University of Oxford and Post-Doc Fellow at Universite Paris Sud. He is an Applied Econometrician and works on issues of Education, Labour Market and Taxation in developing country context. He has a number of publications in reputed journals including Journal of Development Economics. He teaches courses on Microeconometrics at graduate level. Currently, he is involved in two research projects. One looks at labour market inequality from an intergenerational perspective. Other project is on indirect taxation in India – how tax policy affects firms’ trading behavior.
14:55 to 15:20 Anirban Chakraborti Study of financial and macroeconomic dynamics using multi-layered networks
A short abstract of the talk:
We will present the existence of the empirical linkage between the dynamics the financial network constructed from the market indices and the macroeconomic networks constructed from macroeconomic variables such as trade, foreign direct investments, etc., for several countries across the globe. The temporal scales of the dynamics of the financial variables and the macroeconomic fundamentals are very different, which makes the multi-layered network structure and the empirical linkage even more interesting and significant. Also, we show that there exist in the respective networks, core-periphery structures (determined through centrality measures) that are composed of similar set of countries. The data science methodology using network theory, coupled with standard econometric techniques constitute a new approach to studying multi-level economic phenomena in a comprehensive manner.
Anirban Chakraborti is a Professor at the School of Computational and Integrative Sciences, Jawaharlal Nehru University, New Delhi, since March, 2014. He had worked as an Associate Professor at the Chair of Quantitative Finance, École Centrale Paris, France, during 2009-14. He had obtained a Ph.D. in Physics from the Saha Institute of Nuclear Physics, India and then the Habilitation in Physics from Université Pierre et Marie Curie (Paris VI), France. He has the experience of working as a scientist in many reputed universities and educational institutions in India, USA, Europe and Japan. He was awarded the prestigious Young Scientist medal of the Indian National Science Academy in 2009. He has published several books and research articles from internationally renowned publishers. His main research interests lie in the areas of Econophysics, Sociophysics, Data Science, Statistical Physics, Quantum Physics and Nanomaterial Science.
15:20 to 15:45 Sunetra Ghatak Facets of Internal Migration in India
A short abstract of the talk:
In India migration of labour force was low for almost four decades after independence. But as reforms changed the role of the government from manufacturer to service provider, rapid urbanization attracted people from rural to urban areas. As a result while the number of migrants in the country has increased over time, they remain limited mostly within shorter distances.
In order to understand the dynamics of internal migration in India, I will discuss different dimensions of labour migration - trends and patterns of internal migration, the characteristics of the migrants, the factors that influence internal migration etc. and its dynamics over time. I will focus on how overall dynamics of migrant-sending and migrant-receiving states are changing after reforms with respect to per capita income differences. In recent years, there is an emergence of “Southern Pull” which faded the role of bilateral barriers to migration like physical distance, physical contiguity between states, linguistic divide etc. and highlights the importance of ‘social networks’ in migration decision. I will highlight different forms of ‘social networks’, such as contacts or presence of friends, relatives etc., which act as the channel of information and help migrants to get information about job prospects, access to basic services (like education, health, and housing), information on procedures (technical and legal), financial help, administrative assistance — and also emotional solidarity.
Sunetra Ghatak is a Research Fellow at National Institute of Public Finance and Policy (NIPFP), New Delhi. She did her Ph.D. in Economics from Jawaharlal Nehru University, New Delhi and specialized in the area of labour, gender, trade and health-related issues. She has worked with donor agencies, development consultancies and national level think tanks. She has ten years of experience in the field of socio-economic and development research. Her strength lies in quantitative and qualitative research, systematic review, policy analysis, and field survey (tool preparation, sampling, training, survey). She has contributed several research papers on development economics in various prominent national and international journals.
15:45 to 16:15 -- Coffee break
16:15 to 17:30 -- Poster and Discussion
Thursday, 11 July 2019
Time Speaker Title Resources
09:30 to 10:45 Susan Thomas Financial markets and the price discovery function
A short abstract of the talk:
This talk is about the organisation of financial market place referred to as it's microstructure. It presents how the market microstructure effects the outcomes from a financial market place -- which is the quality of prices and the liquidity with which transactions in the market are undertaken. The talk then focusses on the benchmark that is used to measure the quality of prices from competitive markets, which derives from the efficient market hypothesis. The talk wraps up with some work that measures the price efficiency of the Indian equity markets.
Susan Thomas holds degrees in civil engineering and economics from IIT, Bombay and University of Southern California, Los Angeles, respectively. She is faculty at the Indira Gandhi Institute for Development Research, Bombay. Her research is in financial econometrics and market microstructure in India. Her engagement with markets includes designing the stock market index (Nifty), designing and building a real-time risk management system for the clearing corporation, and the use of call auctions. She has worked on projects with the World Bank, IFC, ADB, both the securities and commodity derivatives exchanges in India, and the Government of India. More recent policy engagements include being a member of the Standing Council of the competitiveness of the Indian Financial Sector and the Bankruptcy Legislative Reforms Committee, set up by the Ministry of Finance, which submitted a draft Insolvency and Bankruptcy Code which was passed as law in May 2016. Her work can be accessed at http://www.ifrogs.org
10:45 to 11:15 -- Coffee break
11:15 to 12:30 Andreas Wimmer Why the experimental turn is reducing our understanding of the social world
A short abstract of the talk:
Many social sciences have turned towards experiments as the prime method to explore causal connections, whether by exploring naturally occurring randomness or by generating it in the lab or in an online environment. Many non-experimental methodologies (such as in the statistical analysis of observational data) also have adopted the experimental way of thinking about causality as the effect of a randomly assigned “treatment” on some outcome. While the experimental turn has undoubtedly been beneficial for the social sciences, in this paper I highlight some of its (more or less well-known) disadvantages and outline alternative theoretical and empirical research strategies. The main disadvantages are a) a reduction of theoretical complexity to mono-causal arguments; b) a loss of a holistic view on causality where multiple causal processes, some of them operating through feedback mechanisms, simultaneously affect some outcome; c) a quasi-magical fear of tainted, impure forms of causality brought about by endogeneity. I outline some ideas of how to think and research about these three elements of complexity without loosing much clarity of causal analysis and empirical rigor.
A short academic/professional biography:Andreas Wimmer's research brings a long term historical and globally comparative perspective to the questions of how states are built and nations formed, how individuals draw ethnic and racial boundaries between themselves and others, and under which conditions these processes result in conflict and war. Using new methods and data, he continues the old search for historical patterns that repeat across contexts and times. He has pursued this agenda across the disciplinary fields of sociology, political science, and social anthropology and through various styles of inquiry: ethnographic field research (in Mexico and Iraq), comparative historical analysis, quantitative research with cross-national or survey data, network studies, and formal modeling. His most recent book is “Nation Building. Why Some Countries Come Together While Others Fall Apart” (Princeton 2018).
12:30 to 12:45 Shekhar Tomar Shock Diffusion: Does inter-sectoral network structure matter?
A short abstract of the talk:
This paper introduces the concept of diffusion of shocks in a macroeconomic network consisting of inter-sectoral production linkages. Using sectoral and firm level data, the paper documents two empirical facts. First, sectoral output do not react contemporaneously to shocks in input sectors (it only reacts with a lag). Second, different sectors take different time horizon to respond to shocks to their input sectors. I then incorporate these features in a model of production network to study the contribution of sectoral shocks to aggregate fluctuations. I show that if sectors have different reaction horizons it leads to diffusion of shocks through the network over time which prevents the inter-sectoral linkages to form the feedback loop structure essential to generate aggregate volatility. So, the impact of a given sectoral shock lingers over a longer time period (due to diffusion) but contributes less to the aggregate volatility in any given period. Finally, I use a factor model to estimate the contribution of aggregate vs idiosyncratic sectoral shocks to aggregate fluctuations in US industrial production (IP) data. I find that in the case of a diffusion adjusted network model the contribution of sectoral shocks to aggregate volatility is small and is of the same magnitude as in the case of statistical factor analysis.
Shekhar Tomar is as an Assistant Professor in the Economics and Public Policy area at ISB. He completed his PhD from Toulouse School of Economics in 2017 and worked as a Research Economist at RBI between 2017-2019. His research lies at the intersection of macroeconomics, trade and finance and he extensively uses micro-data to answer macro questions in his work. During his stint at the RBI, he also contributed regularly to the policy work on monetary policy and trade issues in India.
A short abstract of the talk:
In the last two decades, Indian healthcare has shown considerable improvement in terms of infant mortality rate, life expectancy, and other widely reported indicators. However, the grassroots situation is in sharp contrast to what this might imply. The vast majority in India continues to struggle with medicinal access, availability of functional healthcare centres, and rising out of pocket healthcare expenditure. In Chhattisgarh, for example, none of the 6,000 rural healthcare centers fulfil the Indian Public Health Standards, with many not having regular water supply (~20%), electricity (~10%) or even an operation theatre (70%). It is a sad eventuality that locals have to travel far and long to meet their healthcare needs, if their time/funds allow for this luxury. This is in sharp contrast to an ~8 times increase in the state’s per capita income since 2003. The topic of my conversation will be to address some of these issues, and discuss possible solutions.
Ridhima is a Research Fellow with McKinsey and is currently researching on the employment effects of automation. She was most recently working with New York University as a Research Scientist. Over there, apart from her research, she provided support to an award winning executive education program in Applied Data Science methods, targeted towards state and federal government officials. Deeply passionate about socio-economic work, she is on the board of a few NGOs working in the areas of social and economic development and volunteers her time for coordinating research and grant funding efforts for Fundamental Action & Research Foundation, an NGO working on-ground for improving the socio-economic conditions of rural India. Ridhima received her MA in Economics from Delhi School of Economics, and MS in Public Policy and Management from Carnegie Mellon University. She can be reached at ridhimasodhi@gmail.com.
13:00 to 14:30 -- Lunch
14:30 to 14:55 Souvik Roy A basic model of Mechanism Design
A short abstract of the talk:
In this lecture, I will present the basic model of Mechanism Design. Mechanism Design is reverse engineering of Game Theory, where given a (desired) outcome, one needs to construct a game that produces that outcome in equilibrium. I will discuss some fundamental results in this area and some open research problems.
I have studied Mathematics and Statistics from Indian Statistical Institute. Apart from Game Theory and Mechanism Design, I work on Percolation Theory, Random Graphs, Algebraic Graph Theory, Number Theory, and Epistemic Game Theory.
14:55 to 15:20 Dibyendu Maiti Market Imperfection and Productivity Decomposition: A Stochastic Frontier Application
A short abstract of the talk:
The factor owners in any production process are engaged in bargaining game under imperfect market environment, and that leads the actual outcome substantially be deviated from the optimal level, This essentially raises importance to investigate the sources of deviations (or loss of efficiency). Hence, the estimation of productivity growth and its decomposition in the presence of market imperfections receive growing importance in contemporary literature. For the decomposition of the productivity growth, the standard approach of SFA model estimates production function first assuming the fact that real output and inputs combinations are not affected by the market imperfections. But, not only the price but also qualities of output produced and input used in the production are highly influenced by the market conditions. So, a modified production function is estimated to capture the mark-up of a firm and the bargaining power of labour with the help of standard industrial statistics. The paper adds two terms to capture the product and labour market distortions in the production function approach. This method enables to estimate mark-up and labour bargaining power along with other standard parameters without using price information. Then, the productivity growth has been decomposed into five components - technological change, technical efficiency change, scale change, allocative efficiency change and marketing efficiency change. This model has been applied in the Indian organised industry, which is experienced the substantial fluctuations of market dynamism during the last two decades. We find that Mark-up is 20-30 per cent higher than marginal costs, and bargaining power is around 0.01-0.05. The rise of allocative efficiency due to weakening bargaining power has contributed to a greater extent change than that of marketing efficiency.
Dibyendu Maiti is Associate Professor of Economics at Delhi School of Economics. Before joining at Delhi School, he worked at the University of the South Pacific (Fiji), Institute of Economic Growth, Centre for Studies in Social Sciences (Calcutta), the University of Manchester (UK),. He held various research positions at Max Planck Institute of Economics (Jena, Germany), University of Nottingham, Chinese Academy of Social Sciences, University of Oslo, Norwegian Institute of International and so. He publishes in the leading journalslike Journal Development Economies, Economic Modelling, Labour Economics, Journal of Productivity Analysis, Cambridge Journal of Economics etc. He received the IDRC Indian Young Social Scientist Award 2009 and Global Development Network 2010. He serves Progress in Development Studies (Sage) as an associate editor from 2018. He worked on research projects sponsored by ICSSR, WTO, ESRC, DFID and serves the national and international organisation as a consultant. His research interest lies in international trade theory and development macroeconomics.
15:20 to 15:45 Debajit Jha State of growth of Indian states
A short abstract of the talk:
Regional growth in India in the post-reform period has been uneven in two important ways. First, across regions, divergent growth has led to the formation of multiple growth-clubs moving away from each other. Second, over time, regional growth has exhibited structural breaks, leading to growth accelerations and growth slowdowns. In this paper, we combine these two aspects of uneven growth in order to investigate the relative roles of growth accelerations, growth slowdowns and initial long-run growth rates in driving Indian sub- national regions towards high or low growth clubs. Based on our results, we draw policy implications for the sequencing of growth policies in these regions. Specifically, regions in lower growth clubs have a better probability of escaping from these growth traps by focussing on simpler reforms while the regions in the higher clubs need to focus on deeper reforms in order to achieve the same result.
Debajit Jha is an Assistant Professor at the Jindal School of Government and Public Policy, O. P. Jindal Global University, Sonipat, Haryana. He obtained his Ph.D. in Economics from the Centre for Economic Studies and Planning, Jawaharlal Nehru University, New Delhi. Before joint his current assignment Dr. Jha taught in many government institutions in West Bengal. During this time he was also served as Assistant Director of Public Instruction to the Govt. of West Bengal. Dr. Jha works in the area of growth economics. His research interest lies in the issues like inequality, polarization, club convergence, medium-run growth dynamics etc. He has contributed research papers in different prominent national and international journals.
15:45 to 16:15 -- Coffee break
16:15 to 17:30 -- Poster and Discussion
Friday, 12 July 2019
Time Speaker Title Resources
09:30 to 10:45 -- Presentations by Long Term Participants
10:45 to 11:15 -- Coffee break
11:15 to 11:50 Rajeev Tripathi Business ecosystems – competition, cooperation and evolution
A short abstract of the talk:
Rajeev is an assistant professor at the Indian Institute of Management Bangalore. He primarily does theoretical research in game theory. His other research interests include sharing economy and internet value chain, mainly from the perspective of operations management. His research has appeared in journals like Operations Research Letters and European Journal of Operational Research. He also writes articles in popular business magazines such as Forbes India.
11:50 to 12:25 Souvik Dutta Social Reform as a Path to Political Leadership
A short abstract of the talk:
A potential political leader, aiming to replace a repressive regime, wishes to establish her credibility with citizens whose participation in her movement affects its success. If her perceived ability is in an intermediate range of values, her optimal strategy is to masquerade as a no threat before announcing a movement directly against the regime. In this range, for low costs of repression, the regime finds it optimal to exert force even against a movement that has purely non-political objectives. Interestingly, this range, where the regime exerts force against a non-political movement, diminishes with the leader's likelihood of being political.
Dr. Souvik Dutta is an Assistant Professor in Economics at Indian Institute of Management Bangalore. He is the recipient of the Young Faculty Research Chair at IIMB. He specializes in the fields of Development Economics, Political Economy, Microfinance and Game Theory. Dr. Dutta holds a Ph.D in Economics from Pennsylvania State University, USA and a Masters in Economics from Delhi School of Economics.
12:25 to 13:00 Srinka Basu Modelling Social-Behavioural Dynamics For Disaster Management Using Crowd Sourced Data
A short abstract of the talk:
The digital revolution, the open data trend, and the advancements in data science provide new opportunities for disaster research that is important for design of effective social and economical strategies for disaster impact reduction. Our focus is to understand the social behavioural dynamics of disasters utilising the crowd sourced data available in online social networks like Twitter.
During the recovery and relief stage of a disaster, the relief workers try to identify the need of the affected population and the affected people try to identify the right supply for their need. We aim to understand how cooperation amongst various relief workers and aid distributors using online social network emerges and how the process can be modelled using game theory or agent based models. We also try to understand the lifecycle of the dynamic process of emerging cooperation particularly whether and how a state of equilibrium is reached and how external stimulation like rumors affect the process.
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2019-10-22 13:45:56
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https://www.aimsciences.org/article/doi/10.3934/dcds.2017176
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# American Institute of Mathematical Sciences
July 2017, 37(7): 4131-4158. doi: 10.3934/dcds.2017176
## Mathematical analysis of an in vivo model of mitochondrial swelling
1 Institute for Computational Biology, Helmholtz Zentrum München, Ingolstäder Landstr. 1, 85764 Neuherberg, Germany 2 Department of Applied Phsyics, School of Science and Engineering, Waseda University 3-4-1, Okubo, Shinjuku-ku, Tokyo 169-855, Japan 3 Department of Mathematics and Statistics, University of Guelph, Guelph ON, N1G2W1, Canada
* Corresponding author: Messoud Efendiev
Received October 2016 Revised February 2017 Published April 2017
Fund Project: M.O. is partly supported by the Grant-in-Aid for Scientific Research #15K13451, the Ministry of Education, Culture, Sports, Science, and Technology, Japan; H.J.E. is partly supported by the Natrural Science and Engineering Researc Council of Canada through a Discovery Grant.
We analyze the effect of Robin boundary conditions in a mathematical model for a mitochondria swelling in a living organism. This is a coupled PDE/ODE model for the dependent variables calcium ion contration and three fractions of mitochondria that are distinguished by their state of swelling activity. The model assumes that the boundary is a permeable 'membrane', through which calcium ions can both enter or leave the cell. Under biologically relevant assumptions on the data, we prove the well-posedness of solutions of the model and study the asymptotic behavior of its solutions. We augment the analysis of the model with computer simulations that illustrate the theoretically obtained results.
Citation: Messoud Efendiev, Mitsuharu Ôtani, Hermann J. Eberl. Mathematical analysis of an in vivo model of mitochondrial swelling. Discrete & Continuous Dynamical Systems - A, 2017, 37 (7) : 4131-4158. doi: 10.3934/dcds.2017176
##### References:
[1] S. Brenner, Poincaré-Friedrichs inequalities for piecewise $H^1$ functions, SIAM J. Numer. Anal., 41 (2003), 306-324. doi: 10.1137/S0036142902401311. Google Scholar [2] H. Brézis, Opérateurs Maximaux Monotones et Semigroupes de Contractions dans un Espace de Hilbert, North Holland, Amsterdam, The Netherlands, 1973. Google Scholar [3] H. Brézis, Monotonicity methods in Hilbert spaces and some applications to nonlinear partial differential equations, Contributions to Nonlinear Functional Analysis (ed. E. H. Zarantonello), Academic Press, (1971), 101-179. Google Scholar [4] M. A. Efendiev, M. Ôtani and H. J. Eberl, A coupled PDE/ODE model of mitochondrial swelling: Large-time behavior of homogeneous Dirichlet problem, Journal of Coupled Systems and Multiscale Dynamics, 3 (2015), 1-13. doi: 10.1166/jcsmd.2015.1070. Google Scholar [5] S. Eisenhofer, A coupled system of ordinary and partial differential equations modeling the swelling of mitochondria, PhD Dissertation, TU Munich, 2013. Google Scholar [6] S. Eisenhofer, M. A. Efendiev, M. Ôtani, S. Schulz and H. Zischka, On a ODE-PDE coupling model of the mitochondrial swelling process, Discrete and Continuous Dynamical Syst. Ser. B, 20 (2015), 1031-1057. doi: 10.3934/dcdsb.2015.20.1031. Google Scholar [7] S. Eisenhofer, F. Toókos, B. A. Hense, S. Schulz, F. Filbir and H. Zischka, A mathematical model of mitochondrial swelling BMC Research Notes, 3 (2010), p67. doi: 10.1186/1756-0500-3-67. Google Scholar [8] G. Kroemer, L. Galluzzi and C. Brenner, Mitochondrial membrane permeabilization in cell death, Physiological Reviews, 87 (2007), 99-163. Google Scholar [9] M. Ôtani, Nonmonotone perturbations for nonlinear parabolic equations associated with subdifferential operators, Cauchy problems, J. Differential Equations, 46 (1982), 268-299. doi: 10.1016/0022-0396(82)90119-X. Google Scholar [10] V. Petronilli, C. Cola, S. Massari, R. Colonna and P. Bernardi, Physiological effectors modify voltage sensing by the cyclosporin A-sensitive permeability transition pore of mitochondria, Journal of Biological Chemistry, 268 (1993), 21939-21945. Google Scholar [11] R. Rizzuto and T. Pozzan, Microdomains of intracellular $\textrm{Ca}^{2+}$: Molecular determinants and functional consequences, Physiological Reviews, 86 (2006), 369-408. Google Scholar [12] R. Temam, Infinite-Dimensional Dynamical Systems in Mechanics and Physics Springer-Verlag, New York, 1997. doi: 10.1007/978-1-4612-0645-3. Google Scholar [13] H. Triebel, Interpolation Theory, Function Spaces, Differential Operators, J. A. Barth, 1995. Google Scholar [14] H. Zischka, N. Larochette, F. Hoffmann, D. Hamöoller, N. Jägemann, J. Lichtmannegger, L. Jennen, J. Müller-Höcker, F. Roggel, M. Göttlicher, A. M. Vollmar and G. Kroemer, Electrophoretic analysis of the mitochondrial outer membrane rupture induced by permeability transition, Analytical Chemistry, 80 (2008), 5051-5058. doi: 10.1021/ac800173r. Google Scholar
show all references
##### References:
[1] S. Brenner, Poincaré-Friedrichs inequalities for piecewise $H^1$ functions, SIAM J. Numer. Anal., 41 (2003), 306-324. doi: 10.1137/S0036142902401311. Google Scholar [2] H. Brézis, Opérateurs Maximaux Monotones et Semigroupes de Contractions dans un Espace de Hilbert, North Holland, Amsterdam, The Netherlands, 1973. Google Scholar [3] H. Brézis, Monotonicity methods in Hilbert spaces and some applications to nonlinear partial differential equations, Contributions to Nonlinear Functional Analysis (ed. E. H. Zarantonello), Academic Press, (1971), 101-179. Google Scholar [4] M. A. Efendiev, M. Ôtani and H. J. Eberl, A coupled PDE/ODE model of mitochondrial swelling: Large-time behavior of homogeneous Dirichlet problem, Journal of Coupled Systems and Multiscale Dynamics, 3 (2015), 1-13. doi: 10.1166/jcsmd.2015.1070. Google Scholar [5] S. Eisenhofer, A coupled system of ordinary and partial differential equations modeling the swelling of mitochondria, PhD Dissertation, TU Munich, 2013. Google Scholar [6] S. Eisenhofer, M. A. Efendiev, M. Ôtani, S. Schulz and H. Zischka, On a ODE-PDE coupling model of the mitochondrial swelling process, Discrete and Continuous Dynamical Syst. Ser. B, 20 (2015), 1031-1057. doi: 10.3934/dcdsb.2015.20.1031. Google Scholar [7] S. Eisenhofer, F. Toókos, B. A. Hense, S. Schulz, F. Filbir and H. Zischka, A mathematical model of mitochondrial swelling BMC Research Notes, 3 (2010), p67. doi: 10.1186/1756-0500-3-67. Google Scholar [8] G. Kroemer, L. Galluzzi and C. Brenner, Mitochondrial membrane permeabilization in cell death, Physiological Reviews, 87 (2007), 99-163. Google Scholar [9] M. Ôtani, Nonmonotone perturbations for nonlinear parabolic equations associated with subdifferential operators, Cauchy problems, J. Differential Equations, 46 (1982), 268-299. doi: 10.1016/0022-0396(82)90119-X. Google Scholar [10] V. Petronilli, C. Cola, S. Massari, R. Colonna and P. Bernardi, Physiological effectors modify voltage sensing by the cyclosporin A-sensitive permeability transition pore of mitochondria, Journal of Biological Chemistry, 268 (1993), 21939-21945. Google Scholar [11] R. Rizzuto and T. Pozzan, Microdomains of intracellular $\textrm{Ca}^{2+}$: Molecular determinants and functional consequences, Physiological Reviews, 86 (2006), 369-408. Google Scholar [12] R. Temam, Infinite-Dimensional Dynamical Systems in Mechanics and Physics Springer-Verlag, New York, 1997. doi: 10.1007/978-1-4612-0645-3. Google Scholar [13] H. Triebel, Interpolation Theory, Function Spaces, Differential Operators, J. A. Barth, 1995. Google Scholar [14] H. Zischka, N. Larochette, F. Hoffmann, D. Hamöoller, N. Jägemann, J. Lichtmannegger, L. Jennen, J. Müller-Höcker, F. Roggel, M. Göttlicher, A. M. Vollmar and G. Kroemer, Electrophoretic analysis of the mitochondrial outer membrane rupture induced by permeability transition, Analytical Chemistry, 80 (2008), 5051-5058. doi: 10.1021/ac800173r. Google Scholar
Model simulation with $\alpha=10<C^-$: Shown are $u, N_1, N_2, N_3$ for selected times.
Model simulation with $\alpha=10<C^-$: Shown is $N_1$ for selected times.
Simulation to illustrate partial swelling in Theorem 5.2, using initial data (ref{T2init:eq}): shown is the minimum value of $N_2$ as a function of time for different base calcium ion concentrations $u_{base}$ (top left), along with the steady state distributions for $N_1$ (top right), $N_2$ (bottom left), and $N_3$ (bottom right) in the case $u_{base}=100$.
Mitochondria populations $N_1$ and $N_2$ as a function of time in three points of the domain on a line through the center point: A (close to the boundary), B (half way between boundary and center), C (in the center), for six different values of the external calcium ion concentration $\alpha$.
Default parameter values, cf also [5]
parameter symbol value remark lower (initiation) swelling threshold $C^-$ 20 (varied) upper (maximum) swelling threshold $C^+$ 200 maximum transition rate for $N_1\rightarrow N_2$ $f^\ast$ 1 maximum transition rate for $N_2\rightarrow N_3$ $g^\ast$ 1 diffusion coefficient $d_1$ 0.2 (varied) feedback parameter $d_2$ 30
parameter symbol value remark lower (initiation) swelling threshold $C^-$ 20 (varied) upper (maximum) swelling threshold $C^+$ 200 maximum transition rate for $N_1\rightarrow N_2$ $f^\ast$ 1 maximum transition rate for $N_2\rightarrow N_3$ $g^\ast$ 1 diffusion coefficient $d_1$ 0.2 (varied) feedback parameter $d_2$ 30
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numbers 1 to 1000 copy and pastebest Gaming foruma. How many seconds will it take for the ball to reach its maximum height above the ground? The time is unknown; a 16, b 64, c 150 (64) 64 2 2( 16) 32 t Use 2 b a to find the amount of time, , that has passed when the ball reaches its maximum height 2 seconds The ball reaches its maximum height after 2 seconds b.
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A soft tennis ball is dropped onto a hard floor from a height of 1. Sep 10, 2015 · A ball is thrown vertically with an initial velocity of 20 m/s. 800 gives t = 0 (launch time) and t = 10 (landing time). 0 s C) 316 s D) 5. Mar 01, 2014 · The maximum height of a thrown ball is dependent on the upward portion of the initial velocity.
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A ball is thrown vertically upwards from the top of a tower with an initial velocity of 1 9. 6 m s − 1. The ball reaches the ground after 5 s. Calculate : (i) the height of the tower, (i i) the velocity of ball on reaching the ground. Take g = 9. 8 m s − 2.
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5 o The height (h) of a ball, in feet, at a given time (t) is represented by the equation h(t) = -16t2 + v 0t + h 0 where v 0 is the initial vertical velocity and h 0 is the initial height. o.
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Search: Bouncing Ball Physics Lab. Purpose of This Lab The purpose of this lab is to: Figure out how many times a ping pong ball will bounce in 5 seconds after being dropped at a certain height Learn how to create graphs and charts using a spreadsheet Bounce Lab Purpose: The purpose of this lab is to find the elasticity of two different types of balls Physics with.
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Describe three different ways you could find the maximum height the water reaches. Then choose a method and find the maximum height of the water. Answer: Question 68. PROBLEM SOLVING A farmer is building a rectangular pen along the side of a barn for animals. The barn will serve as one side of the pen.
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Specifically, it is proportional to the fraction of the height that the ball bounces compared to the height that the ball is dropped. The students drop balls from a meter height and determine how high they bounce. They record and repeat the process five (or 10) times, rounding to the nearest cm (or 5 cm depending on the resolution desired).
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Determine: (i) The net force on the ball as it moves up; (Take acceleration due to gravity g = 10 m s - 2) (ii) The acceleration of the ball (iii) The maximum height reached by the ball The Questions and Answers of When a ball thrown vertically upwards, it goes through a distance of 19 Find the first time the ball is at a height of 39 See full.
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Parabolas that open up or open down have what is referred to as minimum and maximum value. The maximum value of a parabola is the y -coordinate of the vertex of a parabola that opens down.
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Math. Algebra. Algebra questions and answers. A ball is thrown straight up with an initial velocity of 64 feet per second from the top of a building that is 80 ft tall. The height of the ball t seconds after it is thrown is given by the functions (t) = - 16t? +64t +80.
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A more detailed solution would use the formula for height y as a function of time: y(t) = v0*t + a*t 2 /2. with a known, v0 determined from the maximum 19m, and then t known. Then from the derivative. v(t) = y'(t) = v0 + a*t. you can solve for t at the.
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The table shows four measurements indicating the ball's height at various horizontal distances from where it was thrown. A graphing calculator displays a quadratic function that models the ball's height, y, in feet, in terms of its horizontal distance, x, in feet. Answer x, Ball's Horizontal Distance (feet) y, Ball's Height (feet) 0 6 1 8.1 3 6.
The freshmen won the long jump with a distance of 7 Zte K88 Unlock 62cm) and 58 inches (147 How to calculate energy loss in a bouncing ball 1 Questions & Answers Place A Ball Dropped From A Height Of 24m Rebounds A Ball Dropped From A Height Of 24m Rebounds. The 'two-ball bounce problem' is often used to demonstrate that the rigorous rules of.
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The maximum height of the object is the highest vertical position along its trajectory. The maximum height of the projectile depends on the initial velocity v 0, the launch angle θ, and the acceleration due to gravity. The unit of maximum height is meters (m). H = maximum height (m) v 0 = initial velocity (m/s) g = acceleration due to gravity.
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MathActusMagazineGuideTests ComparatifsWebContactNo Result View All Result When cricket ball thrown vertically upwards inScience Math Reading Time mins read When cricket ball thrown vertically upwards, reaches maximum height metres. When.
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Then you can find the maximum height the ball travels above the ground.. Step 2. Since the rock goes up first and then it comes down, the smaller time value corresponds to the velosity of the rock going up. By differentiating h (t)using the power rule, we get the velocity for the ball. v.
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Then you can find the maximum height the ball travels above the ground.. Step 2. Since the rock goes up first and then it comes down, the smaller time value corresponds to the velosity of the rock going up. By differentiating h (t)using the power rule, we get the velocity for the ball. v.
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Answer (1 of 3): Find the vertex. Find the distance from the vertex on a vertical lune through it to a horizontal line that you want for the height.
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Its calculation is presented in example 1. Kinetic energy formula: {eq}Ek = \frac 12 * m * v^2 {/eq} Where, according to the International.
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Answer (1 of 5): vf = final velocity (m/s) vi = initial velocity (m/s) a = acceleration (m/s^2) t = time (s) s = displacement (m) Time to reach maximum height: vf = vi + at 0 = 30 - 9.8 t 9.8t = 30 t = 3.06 s Maximum height: vf^2 = vi^2 + 2as 0 = 30^2 - 2*9.8 * s 19.6s = 900 s = 45.
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If a ball is thrown directly upward with a velocity of 26 ft/s, its height (in feet) after t seconds is given by y = 26t − 16t2 has no horizontal velocity Plot its velocity vs __ An object is thrown vertically upward with a certain initial velocity in a world where the acceleration due to gravity is 19 A ball is thrown vertically upward A.
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Explanation. Step 1. 1 of 2. To find the maximum height, we will use the following kinematic formula: v f y 2 = v i y 2 + 2 a y ( y m a x − y i) Note that, we choose the initial height to be our zero level, y i = 0 v f y 2 = v i y 2 + 2 a y ( y m a x − y i ⏞ = 0) v f y 2 = v i y 2 + 2 a y y m a x Note that, v i y = v 0 v f y 2 = v 0 2 + 2.
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A ball is thrown vertically upwards from the top of a tower with an initial velocity of 1 9. 6 m s − 1. The ball reaches the ground after 5 s. Calculate : (i) the height of the tower, (i i) the velocity of ball on reaching the ground. Take g = 9. 8 m s − 2.
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To find max height I set Vf to 0 and solve for y using the same acceleration and the newly found Vo. 0 = 10.01^2 - 19.64y. y=5.1 m. Obviously this doesn't work since 5.1m is where the person at the window sees the ball traveling with velocity. Any help would be great, I tried using the other kinematic equations but it got too messy too quickly.
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39. A ball is thrown up from ground level with an initial velocity of 64 ft/sec. Its height at time t is given by the formula: h(t) = - 16t^2 + 64t What is the maximum height (in feet) achieved by the read more.
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A tennis player hits a 58.0 g tennis ball so that it goes straight up and reaches a maximum height of 6.17 How much work does gravity do on the ball on the way up? Wiki User ∙ 2013-05-22 22:12:50.
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The function can be represent with a parabola whose vertex is its maximum point . for answer the first question we have to find the y of the vertex . first of all we have to find the x . Vertex (x) = -b/2a = -8/-2 = 4 . if we substitute t with 4, we can find the y . Vertex (y) = -(4^2) + (8*4) = = -16 + 32 = 16 . so the answer is 16 meters.
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This sort of problem is solved easily by a graphing calculator. a) The ball reaches its maximum height in 1.13 seconds. b) The ball's maximum height is 30.25 feet. Since you want to know when and where the peak value of the function occurs, it is convenient to put it into vertex form.
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Algebra-> Quadratic Equations and ... How do I find the maximum height of a baseball which is hit with an upward velocity of 90 feet per second when ... In our equation a = -16 and b = 90: t = -90)/2(-16) t = -90/-32 t = +2.8125 sec when the ball reaches max height: Substitute 2.8125 for t in the original equation to find.
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The ball is thrown into the air Determine: (i) The net force on the ball as it moves up; (Take acceleration due to gravity g = 10 m s - 2) (ii) The acceleration of the ball (iii) The maximum height reached by the ball 8 m/s^2 and the ball will The answer can be calculated as follows: Plot its velocity vs Bj Miller Stanford Plot its velocity vs.
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5 o The height (h) of a ball, in feet, at a given time (t) is represented by the equation h(t) = -16t2 + v 0t + h 0 where v 0 is the initial vertical velocity and h 0 is the initial height. o.
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726 views View 1 Upvoter A ball is thrown vertically upwards with a velocity of 49m/s calculate maximum height and time taken to reach maximum height If a ball is thrown vertically upward with a velocity of 80 ft/s, then its height after t seconds is s = 80t – 16t2 CHANGE: THE PATH OF A THROWN BALL Grades 8-12 When an object is thrown, shot.
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The downward velocity increases by 9 A ball is thrown upward at an angle of 250 above the horizontal If g =10ms/ 2, its initial speed was Ans : 20 m/s 5 CHANGE: THE PATH OF A THROWN BALL Grades 8-12 When an object is thrown, shot, or otherwise propelled into the air, it will follow a path through the air known as a trajectory As the ball is thrown vertically upwards,.
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A ball is thrown upward from the top of a 40 foot high building at a speed of 80 feet per second. The ball's height above ground can be modeled by the equation H\left (t\right)=-16 {t}^ {2}+80t+40.\\ H (t) = −16t2 +80t+40. a. When does the ball reach the maximum height? b. What is the maximum height of the ball? c.
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1x10 2 m 8 m/s 2 The kinetic energy of the ball on leaving the ground is equal to its gravitational potential energy on rising to its maximum height, that is 3m 3: A ball is tossed vertically upwards with a speed of 5 8 on earth) and d is the initial height 8 on earth) and d is the initial height. in 3 . in 3. A ball thrown vertically upward.
this question instructs us to find the maximum height obtained by the ball. What we know is that we have the function after of tea is 40 T minus 16 T square. This function has a maximum value because R is negative. Therefore, we can use the formula negative be over to eight solved. We know our B is 40 or a is negative. 16.
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The ball was thrown vertically at an initial speed of V=20 meters/sec (45 mph) from a height of h 0 = 2 meters. Problem 2 - On Mars, the acceleration of gravity is g = 4 meters/sec 2. The ball was thrown vertically at an initial speed of V=20 meters/sec (45 mph) from a height of h 0 = 2 meters. Problem 3.
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The gravitational force lets it reach a certain height and then fall back down again. Enter one value at velocity and height, the other value will be calculated.Example: a ball, thrown vertically up with a speed of 25 km/h, can reach a maximum height of almost 2.5 meters above its starting point. The formula is h=v²/ (2g).. "/>.
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The maximum height, or vertical displacement, of a projectile, depends on the initial velocity, acceleration due to gravity, and time of flight. The equation for the vertical displacement or maximum height of a projectile is given as, y = U y t − ½ gt 2. y = U sinθ t − ½ gt 2. Where y is the vertical displacement of a projectile. A ball is thrown from an initial height of 1 meter with an initial upward velocity of 9 m/s. The ball's height h (in meters) after t seconds is given by the following h=1+9t-5t^2 Find all the values of the t for which the ball's height is 4 meters. Maths. object thrown upward from an initial height of h0 feet with an initial velocity of v0 (in. The v sub 0 stands for the initial velocity of the object, and h sub 0 is the height from which the object is thrown up or dropped. Notice the exponent of 2. Notice the exponent of 2.
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Since they are asking for the maximum height, and "t" represents seconds, you must find the "y" value. Standard quadratic form is: ax^2+bx+c. Use the vertex formula: x=-b/2a. to find the "x" value of the vertex, then plug that value into the original equation as a substitute for "x". Given y= 40t-16t^2. a=16, b=40. Commencing at the boundary of the ore, slices 11 ft. in height by 30 or 40 ft. in width are mined beneath the mat, which is temporarily supported by sets with posts 10 ft... . what is coram ... how to find maximum height of a ball thrown up algebra. airbnb cabins gatlinburg. hobby lobby coin holders or armiger moirax 9th edition. Hello. As a result, you can calculate things like the time the projectile will be airborne before it strikes the ground. Say, for example, that you decide to shoot a cannon into the air. Assume that it has a muzzle velocity of 860 meters/second, and that it is pointing straight up.
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726 views View 1 Upvoter A ball is thrown vertically upwards with a velocity of 49m/s calculate maximum height and time taken to reach maximum height If a ball is thrown vertically upward with a velocity of 80 ft/s, then its height after t seconds is s = 80t – 16t2 CHANGE: THE PATH OF A THROWN BALL Grades 8-12 When an object is thrown, shot. Algebra Quadratics Applications HW#54 1: A science class designed a ball launcher and tested it by shooting a tennis ball up and off the top of a 15-story building. They determined that the motion of the ball could be described by ... 4.5 into the function to find the maximum height. -16(4.5)2 + 144(4.5) + 160 The maximum height is 484 feet. If a ball is thrown vertically upward with a velocity of 160 ft/s, then its height after t seconds is s = 160t - 16t^2. If a ball is thrown vertically upward with a velocity of 160 ft/s, then its. Not for a ball thrown up. ... A ball is thrown vertically upwards with a velocity of 49m/s calculate maximum height and time taken to reach maximum height. A ball is thrown vertically upward from ground level with initial velocity of 96 96 feet per second. ... Algebra. asked • 10/31/16 If a ball is thrown vertically upward with a velocity of.
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Quadratic Equations are often used to find maximums and minimums for problems involving projectile motion. For example, you would use a quadratic equation to determine how many seconds would be needed for a ball to reach its maximum height when it was thrown directly upward with an initial velocity of 96 feet per second from a cliff looming 200 feet above a beach.
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If it attains a maximum height of 400m, find its velocity A ball of mass 200 g is thrown vertically upwards with velocity of 5ms-1 Whenever an object is thrown straight up, it will always have a velocity equal to zero when it reaches its highest point To find how much time(t) it will take to reach its highest point we can use the formula t. All steps and concepts are explained in this example problem. To find max height I set Vf to 0 and solve for y using the same acceleration and the newly found Vo. 0 = 10.01^2 - 19.64y. y=5.1 m. Obviously this doesn't work since 5.1m is where the person at the window sees the ball traveling with velocity. Any help would be great, I tried using. Messages. 856. Jul 27, 2006. #2. A ball is thrown into the air from a building and falls to the ground below. The height of the ball, h metres, relative to the ground t seconds after being thrown is given by h=-5t²+30t+35. a) determine the maximum height of the ball above the ground. b) how long does it take the ball to reach the maximum.
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To find max height I set Vf to 0 and solve for y using the same acceleration and the newly found Vo. 0 = 10.01^2 - 19.64y. y=5.1 m. Obviously this doesn't work since 5.1m is where the person at the window sees the ball traveling with velocity. Any help would be great, I tried using the other kinematic equations but it got too messy too quickly.
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A ball is thrown straight up with an initial speed of 40 ms How high does it go? The ball reaches a maximum height of ~81.6 meters. We know: Vf = 0 (terminal velocity) and Vi = 40 (given in question) So, Vf = Vi - g (t) => 0 = 40 - g (t) --Rework the equation to solve for t.
A ball is thrown upward from the top of a 40 foot high building at a speed of 80 feet per second. The ball's height above ground can be modeled by the equation $H\left(t\right)=-16{t}^{2}+80t+40$. a. When does the ball reach the maximum height? b. What is the maximum height of the ball? c. When does the ball hit the ground?.
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Two balls are thrown in the air. The path of the first ball is represented in the graph. The second ball is released 1.5 feet higher than the first ball and after 3 seconds reaches its maximum height 5 feet lower than the first ball. a. Write an equation for the path of the second ball. b. Do the balls hit the ground at the same time?. A ball is launched upward at 48 feet per second from a platform that is 100 feet high. Find the maximum height the ball reaches and how long it will take to get there. h =-16 t 2 + v 0 t + h 0 . Start with the equation that models an object being launched or thrown. h =-16 t 2 + 48 t + 100 . Substitute the initial velocity v 0 = 48 and height h. A ball is thrown straight up with an initial speed of 40 ms How high does it go? The ball reaches a maximum height of ~81.6 meters. We know: Vf = 0 (terminal velocity) and Vi = 40 (given in question) So, Vf = Vi - g (t) => 0 = 40 - g (t) --Rework the equation to solve for t. The height of the building is given when t=0 thus h (0)= -4.9 (0) 2 24 (0) + 8 = 8 Therefore, the height of the building = 8 At maximum height the first derivative of the function is equal to zero hence h' (x)=-9.8t +24=0; Thus time taken to reach the maximum height, t= 24/9.8 = 2.449.
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The acceleration is -18 ft/s2, and the skid marks are 150 ft long A ball is thrown vertically upwards with a velocity of 49m/s calculate maximum height and time taken to reach maximum height The initial magnitude of the velocity of the ball 13N Find the max height reached by the ball from the ground The reason this is true is because at that.
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A ball is shot into the air from the edge of a building, 50 feet above the ground. Its initial velocity is 20 feet per second. The equation h-- and I'm guessing h is for height-- is equal to negative 16t squared plus 20t plus 50 can be used to model the height of the ball after t seconds. College Algebra Quadratic Functions; Parabolas We can also nd the vertex on the graph. Remember, this graph does NOT show the path of the ball. Each point on the graph tells us the height of the ball some number of seconds after it is thrown. The vertex, (1:5;40), tells us that it takes 1.5 seconds for the ball to reach its maximum height of 40. A ball is thrown vertically upwards from the top of a tower with an initial velocity of 1 9. 6 m s − 1. The ball reaches the ground after 5 s. Calculate : (i) the height of the tower, (i i) the velocity of ball on reaching the ground. Take g = 9. 8 m s − 2.
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Example: A ball is thrown in the air. Its height at any time t is given by: h = 3 + 14t − 5t 2. What is its maximum height? Using derivatives we can find the slope of that function: ddt h = 0 + 14 − 5(2t) = 14 − 10t (See below this example for how we. A ball of mass 200 g is thrown vertically upwards with velocity of 5 ms-1 If a ball is thrown vertically upward with a velocity of 80 ft/s, then its height after t seconds is s = 80t – 16t2 Blogspot Flac Assume the acceleration of the ball is a (t) = −32 ft2 a (t) = − 32 f t 2 per second If a ball is thrown vertically upwards with speed u. The height of a rock thrown off of a cliff is given by the equation h 5(t 2)2 40 where h is the height of the rock in metres after t seconds. Apr 23, 2000 · Lesson III: Analytic Equation of a Parabola. x 6 EAfl rl k lr LiugUhat 4s8 jrae ts Ee 5rjv VeXde. 1 Using the graph at the right, It shows the height h in feet of a small rockett seconds.
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. Engine/ Battery Information. 商品に関するお問い合わせの場合、商品名や商品番号を改めてお問い合わせ文章にご記載ください. If a ball is thrown vertically upward with a velocity of 160 ft/s, then its height after t seconds is s = 160t - 16t^2. If a ball is thrown vertically upward with a velocity of 160 ft/s, then its. A ball is thrown into the air with an upward velocity of 32 ft/s. Its height h in feet after t seconds is given by the function h = -16t2 + 32t + 6. a. In how many seconds does the ball reach its maximum height? Round to the nearest hundredth if necessary. b. What is the ball's maximum height? 1 s; 54 ft 2 s; 6 ft 2 s; 22 ft 1 s; 22 ft. this question instructs us to find the maximum height obtained by the ball. What we know is that we have the function after of tea is 40 T minus 16 T square. This function has a maximum value because R is negative. Therefore, we can use the formula negative be over to eight solved. We know our B is 40 or a is negative. 16. A ball is thrown upward and outward from a height of 5 feet. The... A ball is thrown upward and outward from a height of 5 feet. The height of the ball, f (x), in feet, can be modeled by. f (x)=−0.3x2+1.7x+5. where x is the ball's horizontal distance, in feet, from where it was thrown. Use this model to solve parts (a) through (c).
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A ball is thrown upward and outward from a height of 7 feet. The... A ball is thrown upward and outward from a height of 7 feet. The height of the ball, f (x), in feet, can be modeled by. f (x)=−0.1x2+0.8x+7. where x is the ball's horizontal distance, in feet, from where it was thrown. Use this model to solve parts (a) through (c). Dimension 5A: Find the maximum height reached by an object. ... Returning to the example, the soccer ball reaches its maximum height of 29/4 = 7.25 feet in 3/8= 0.375 seconds. ... Suppose a baseball is thrown straight up from a height of 4.5 ft with an initial upward velocity of 60 ft/s. At what time will the maximum height be attained?. This is roughly, give or take, about 90 feet thrown in the air. So this would be like a nine-story building. And I, frankly, do not have the arm for that. But if someone is able to throw the ball for 5 seconds in the air, they have thrown it 30 meters in the. A ball is thrown vertically upward from a height of 4 ft with an initial velocity of 80 ft/s. Taking g = 1 0 m / s 2, find the maximum height reached by the stone. The height, s, of a ball thrown straight down with initial speed 64 ft/sec. To find the maximum height, you first find the time to get maximum height t = - = - = = 3 second.
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Search: Bouncing Ball Physics Lab. Purpose of This Lab The purpose of this lab is to: Figure out how many times a ping pong ball will bounce in 5 seconds after being dropped at a certain height Learn how to create graphs and charts using a spreadsheet Bounce Lab Purpose: The purpose of this lab is to find the elasticity of two different types of balls Physics with. MathActusMagazineGuideTests ComparatifsWebContactNo Result View All Result When cricket ball thrown vertically upwards inScience Math Reading Time mins read When cricket ball thrown vertically upwards, reaches maximum height metres. When. The height of the ball h in metres, can be approximated by the function h 5t2 10t 34 where t is the time in seconds, after the ball is thrown. a) Sketch the graph. b) How high is the ball after 2 s? c) Find the maximum height of the ball. 4. A tennis ball is thrown up into the air. Its height h in metres after t seconds, is given by the function.
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The height of a rock thrown off of a cliff is given by the equation h 5(t 2)2 40 where h is the height of the rock in metres after t seconds. Apr 23, 2000 · Lesson III: Analytic Equation of a Parabola. x 6 EAfl rl k lr LiugUhat 4s8 jrae ts Ee 5rjv VeXde. 1 Using the graph at the right, It shows the height h in feet of a small rockett seconds. Math Algebra Q&A Library The height y (in feet) of a ball thrown by a child is 12 + 6x + 5 y = where x is the horizontal distance in feet from the point at which the ball is thrown. (a) How high is the ball when it leaves the child's hand? (Hint: Find y when x = 0) |3D Your answer is y = (b) What is the maximum height of the ball?. Since flight is symmetrical, time of flight T = When the ball reaches its maximum height, its 2(1.154 s) = 2.31 s. kinetic energy is 16 J. g Horizontal distance = (horizontal speed)(time) a What is the maximum height achieved by the = (11.31 m s–1)(2.31 s) = 2.61 m ball from its point of release? b Calculate the initial vertical velocity of. Jimmy wants to throw an object into his house’s window situated on the second floor (12m from the ground) from the ground. Calculate the speed required at an angle of 30. Applying the formula for maximum height for a projectile. [S = (usinθ) 2 /2g] 12 = (u*sin30) 2 /2*9.8. u = 30.672 m/s which is a very high speed.
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A ball is dropped from a height of 36 feet. The quadratic equation d = -t² + 36 provides the distance, d, of the ball, after t seconds. After how many seconds, does the ball hit the ground? ... Find the height of the projectile 4 seconds after it is launched. Show Answer. ... Free Algebra Solver ... type anything in there! Popular pages.
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Find the ball’s initial speed The potential energy at the top of the ball's motion is 18 J The only choice he has to make to maximize distance, then, is the angle at which he kicks the ball (b) Using conservation of energy and the result of part (a), find the magnitude of the momentum of the ball at one-half its maximum height in terms of m.
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A very common and easy-to-understand application is the height of a ball thrown at the ground off a building. Because gravity will make the ball speed up as it falls, a quadratic equation can be used to estimate its height any time before it hits the ground. Note: The equation isn't completely accurate, because friction from the air will slow. A ball is thrown upward and outward from a height of 6 feet. The height of the ball, f (x), in feet, can be modeled by: f (x)=-0.8x^2+3.2x+6. Where x is the ball's horizontal distance, in feet, from where it was thrown. A) What is the maximum height of the ball and how far from where it was thrown does this occur?. Explanation: When you launch a projectile at an angle θ from the horizontal, the initial velocity of the projectile will have a vertical and a horizontal component. {v0x = v0 ⋅ cos(θ) v0y = v0 ⋅ sin(θ) In order to determine the maximum height reached by the projectile during its flight, you need to take a look at the vertical component. We have obtained two values that represent the time that the ball reaches a height of 300 feet. The first value 1 .34 indicates that after 1 .34 seconds have passed, the ball is at a height of 300 feet. Then the ball reaches its maximum height and begins to fall back to the ground. After 4 .66 seconds it is once again at 300 feet. Because gravity will make the ball speed up as it falls, a quadratic equation can be used to estimate its height any time before it hits the ground. ... A ball is thrown upward from the top of a 40 foot high building at a speed of 80 feet per second. The ball’s height above ground can be modeled by the equation [latex]H\left(t\right)=-16{t.
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To find max height I set Vf to 0 and solve for y using the same acceleration and the newly found Vo. 0 = 10.01^2 - 19.64y. y=5.1 m. Obviously this doesn't work since 5.1m is where the person at the window sees the ball traveling with velocity. Any help would be great, I tried using the other kinematic equations but it got too messy too quickly. 1 Answers. #1. +117727. +5. If a ball is thrown vertically upward with a velocity of 128ft/s, then its height after t seconds is s=128t-16t^2. a) what is the velocity of the ball when it is 240 ft above the ground on its way up? (consider pos direction). The maximum range of a projectile formula can be used to find out! Plugging in our known values gives us: 120 = v^2 - (0.5) \times t^2 ... and the initial height , and you will get the range your projectile can travel. In the advanced mode, it is essentially the same thing, except you can alter some of the variables, such as the gravitational. Answer (1 of 14): The key in these problems is “maximum height”. This means that the final velocity (velocity at the maximum height) is 0. It will always be 0 in these cases since the ball has to stop before changing direction. Also, even if you. Neat homework can aid your comprehension and might make your teacher like you better. Purplemath's "Homework Guidelines for Mathematics" will give you a leg up, explaining in clear terms what your math teacher is looking for. The Guidelines link to examples of common errors, and demonstrate techniques that your instructors will love!.
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At the maximum height the ball will be completely stopped from moving upward or downward; thus the speed of the ball would be 0 mph. ... What height is reached by a 4.0kg ball that is thrown.
To find the maximum height, you first find the time to get maximum height t = - = - = = 3 second. Then the maximum height is the given function t (t) at t= 3 seconds h (3) = -16*3^2 + 96*3 + 112 = 256 feet. ANSWER Part (b) To find the time when the ball hits the ground, you solve this equation h (t) = 0, which is -16t^2 + 96t + 112 = 0. 2022. 6. 1x10 2 m 8 m/s 2 The kinetic energy of the ball on leaving the ground is equal to its gravitational potential energy on rising to its maximum height, that is 3m 3: A ball is tossed vertically upwards with a speed of 5 8 on earth) and d is the initial height 8 on earth) and d is the initial height. in 3 . in 3. A ball thrown vertically upward. MathActusMagazineGuideTests ComparatifsWebContactNo Result View All Result When cricket ball thrown vertically upwards inScience Math Reading Time mins read When cricket ball thrown vertically upwards, reaches maximum height metres. When. The maximum occurs when the varying part is smallest, since it is always non positive, so that its effect is to reduce the height. This occurs when it vanishes. That is, when.
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Since flight is symmetrical, time of flight T = When the ball reaches its maximum height, its 2(1.154 s) = 2.31 s. kinetic energy is 16 J. g Horizontal distance = (horizontal speed)(time) a What is the maximum height achieved by the = (11.31 m s–1)(2.31 s) = 2.61 m ball from its point of release? b Calculate the initial vertical velocity of. The function can be represent with a parabola whose vertex is its maximum point . for answer the first question we have to find the y of the vertex . first of all we have to find the x . Vertex (x) = -b/2a = -8/-2 = 4 . if we substitute t with 4, we can find the y . Vertex (y) = -(4^2) + (8*4) = = -16 + 32 = 16 . so the answer is 16 meters.
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A ball is thrown vertically upwards from the top of a tower with an initial velocity of 1 9. 6 m s − 1. The ball reaches the ground after 5 s. Calculate : (i) the height of the tower, (i i) the velocity of ball on reaching the ground. Take g = 9. 8 m s − 2. Since flight is symmetrical, time of flight T = When the ball reaches its maximum height, its 2(1.154 s) = 2.31 s. kinetic energy is 16 J. g Horizontal distance = (horizontal speed)(time) a What is the maximum height achieved by the = (11.31 m s–1)(2.31 s) = 2.61 m ball from its point of release? b Calculate the initial vertical velocity of. The height of the ball h in metres, can be approximated by the function h 5t2 10t 34 where t is the time in seconds, after the ball is thrown. a) Sketch the graph. b) How high is the ball after 2 s? c) Find the maximum height of the ball. 8. A tennis ball is thrown up into the air. Its height h in metres after t seconds, is given by the function. PRE-ALGEBRA (441) ALGEBRA 1 (2,216) GEOMETRY (864) ALGEBRA 2 (2,745) TRIGONOMETRY (1,186) PRECALCULUS (1,903) ... After how many seconds does the projectile reach its maximum height? asked Sep 17, ... A ball is thrown up at the edge of a 490 foot cliff. asked Oct 8, 2014 in PRECALCULUS by Baruchqa Pupil.
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Subscribe A ball is thrown upward with initial velocity _______ and its height is modeled by the function f (x)=________________ find the time it takes to reach the maximum height and the maximum. The ball is thrown into the air Determine: (i) The net force on the ball as it moves up; (Take acceleration due to gravity g = 10 m s - 2) (ii) The acceleration of the ball (iii) The maximum height reached by the ball 8 m/s^2 and the ball will The answer can be calculated as follows: Plot its velocity vs Bj Miller Stanford Plot its velocity vs. 726 views View 1 Upvoter A ball is thrown vertically upwards with a velocity of 49m/s calculate maximum height and time taken to reach maximum height If a ball is thrown vertically upward with a velocity of 80 ft/s, then its height after t seconds is s = 80t – 16t2 CHANGE: THE PATH OF A THROWN BALL Grades 8-12 When an object is thrown, shot. A soft tennis ball is dropped onto a hard floor from a height of 1. Sep 10, 2015 · A ball is thrown vertically with an initial velocity of 20 m/s. 800 gives t = 0 (launch time) and t = 10 (landing time). 0 s C) 316 s D) 5. Mar 01, 2014 · The maximum height of a thrown ball is dependent on the upward portion of the initial velocity.
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Not for a ball thrown up. ... A ball is thrown vertically upwards with a velocity of 49m/s calculate maximum height and time taken to reach maximum height. A ball is thrown vertically upward from ground level with initial velocity of 96 96 feet per second. ... Algebra. asked • 10/31/16 If a ball is thrown vertically upward with a velocity of.
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Now we can solve problems using these graphs and explanations. Example John throws the ball straight upward and after 1 second it reaches its maximum height then it does free fall motion which takes 2 seconds. Calculate the maximum height and velocity of the ball before it crashes the ground. (g=10m/s²) Example An object does free fall motion. 0 degrees above the horizontal and reaches the same maximum height as the first ball half a = ∆v / ∆t -10 = 0 – 100 / t t = 10 sec y = ½ a t2 + v o t 6 the ball is thrown from rest at an upward angle and the relation of horizontal and vertical with velocity and acceleration If a ball, or any object, is thrown vertically it will always. Its calculation is presented in example 1. Kinetic energy formula: {eq}Ek = \frac 12 * m * v^2 {/eq} Where, according to the International.
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A ball was tossed up into the air. The height of the ball as a function of the time the ball is in the air in seconds can be modeled by a quadratic function. At the time the ball was initially thrown in the air, it was 7.5 meters off the ground. After 1.25 seconds, the ball was at its maximum height of 15.3125 meters. How long does it take for the rocket to reach the maximum height? 8. A ball is thrown directly upward from an initial height of 200 feet with an initial velocity of 96 feet per second. ... 16. A baseball is "popped" straight up by a batter. The ball's height (in feet) above the ground "x" seconds later is. Calculating height of ball thrown. A ball is thrown straight up at an initial speed of 50m per second. Neglect air resistance, how high does it go Capuchin Posts: 5,255, Reputation: 656. Uber Member : Dec 9, 2006, 03:17 PM look up your suvat equations.. This gives a maximum value of 53 and the range of the function is A graph of the function is shown below. Recall the function, which describes the height in feet of a ball t seconds after it is thrown upward from the top of a 200 foot high building. We can now determine when the ball hits the ground and the maximum height that it reaches, as.
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Algebra Quadratics Applications HW#54 1: A science class designed a ball launcher and tested it by shooting a tennis ball up and off the top of a 15-story building. They determined that the motion of the ball could be described by ... 4.5 into the function to find the maximum height. -16(4.5)2 + 144(4.5) + 160 The maximum height is 484 feet. algebra (please help) A ball is thrown into the air with an upward velocity of 36 ft/s. Its height h in feet after t seconds is given by the function h = −16t2 + 36t + 5. In how many seconds does the ball reach its maximum height? Round to the nearest hundredth if necessary. Algebra.
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Air resistance will slow the ball from this point and unless the ball falls back past its original starting height, it will never achieve a velocity greater than that with which it starts. Wiki. A ball is thrown into the air. The height in feet of the ball can be modeled by the equation h = -16t 2 + 20t + 6 where t is the time in seconds, the ball is in the air. When will the ball hit the ground? How high will the ball go?. The table shows four measurements indicating the ball's height at various horizontal distances from where it was thrown. A graphing calculator displays a quadratic function that models the ball's height, y, in feet, in terms of its horizontal distance, x, in feet. Answer x, Ball's Horizontal Distance (feet) y, Ball's Height (feet) 0 6 1 8.1 3 6.
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Answer (1 of 6): a. What you know about this question vi = 30 m/s a =-9.8 m/s^2 vf = 0 m/s, This is an important piece of information that is not stated explicitly, but is obviously going to be tru at the maximum height, where the object stops moving up, then begins to move down. What you are. 39. A ball is thrown up from ground level with an initial velocity of 64 ft/sec. Its height at time t is given by the formula: h(t) = - 16t^2 + 64t What is the maximum height (in feet) achieved by the read more. Find the angle of projection of a projectile for which the horizontal range and maximum height are equal. 2. ... A body is projected up such that its position vector varies with time as. r 3ti 4t 5t 2 j m . Here, t is in seconds. ... A ball is thrown from the groun to clear a wall 3m high at a distance of 6m and falls 18m away from the wall.
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A ball is thrown into the air with an upward velocity. Its height, h, after t seconds is given by the function below. h = − 16 t 2 + 64 t + 960 h=-16t^2+64t+960 h = − 1 6 t 2 + 6 4 t + 9 6 0 How many seconds did it take for the ball to reach its maximum height?. (a) With what speed must a ball be thrown vertically from ground level to rise to a maximum height of 46 m? (b) How long will it be in the air? Now part a I solved, and got 31 m/s (which is correct) Using that result for part b, I did the following:. When a ball is thrown up vertically with velocity v 0 it reaches a maximum height of h. If one wishes to triple the maximum height then the ball should be thrown with velocity: If one wishes to triple the maximum height then the ball should be thrown with velocity:.
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MathActusMagazineGuideTests ComparatifsWebContactNo Result View All Result When cricket ball thrown vertically upwards inScience Math Reading Time mins read When cricket ball thrown vertically upwards, reaches maximum height metres. When. Dimension 5A: Find the maximum height reached by an object. ... Returning to the example, the soccer ball reaches its maximum height of 29/4 = 7.25 feet in 3/8= 0.375 seconds. ... Suppose a baseball is thrown straight up from a height of 4.5 ft with an initial upward velocity of 60 ft/s. At what time will the maximum height be attained?. The maximum height, or vertical displacement, of a projectile, depends on the initial velocity, acceleration due to gravity, and time of flight. The equation for the vertical displacement or maximum height of a projectile is given as, y = U y t − ½ gt 2. y = U sinθ t − ½ gt 2. Where y is the vertical displacement of a projectile. A rock is thrown straight upward with an initial velocity of 30 m/ s as shown in the diagram below A system of particles which formed into appreciable size is termed as body Favourite answer The ball, while still in your hand has zero velocity and zero acceleration Acceleration at maximum height is 9 Whenever an object is thrown vertically.
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The downward velocity increases by 9 A ball is thrown upward at an angle of 250 above the horizontal If g =10ms/ 2, its initial speed was Ans : 20 m/s 5 CHANGE: THE PATH OF A THROWN BALL Grades 8-12 When an object is thrown, shot, or otherwise propelled into the air, it will follow a path through the air known as a trajectory As the ball is thrown vertically upwards,. Algebra Quadratics Applications HW#54 1: A science class designed a ball launcher and tested it by shooting a tennis ball up and off the top of a 15-story building. They determined that the motion of the ball could be described by ... 4.5 into the function to find the maximum height. -16(4.5)2 + 144(4.5) + 160 The maximum height is 484 feet. If it attains a maximum height of 400m, find its velocity If it attains a maximum height of 400m, find its velocity. 726 views View 1 Upvoter Air resistance cannot be neglected 13M Gravity does not stop working (constant), and is the only A ball is thrown vertically upwards with a velocity of 49 m/s Neglect friction Neglect friction. A ball of mass 200 g is thrown vertically upwards with velocity of 5 ms-1 If a ball is thrown vertically upward with a velocity of 80 ft/s, then its height after t seconds is s = 80t – 16t2 Blogspot Flac Assume the acceleration of the ball is a (t) = −32 ft2 a (t) = − 32 f t 2 per second If a ball is thrown vertically upwards with speed u.
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The basketball will hit the ground when its height is 0, so find the zeros of the function. Step 1 Find the axis of symmetry. Use x =-b/2a . Substitute -16 for a and 32 for b => x = -32/(-16*2) so x = 1. The axis of symmetry is x = 1. Step 2 Find the vertex. The x-coordinate of the vertex is 1. Substitute 1 for x, we will find the y-coordinate.
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Algebraically find the maximum height of the ball. A ball is thrown from an initial height of 7 feet with an initial upward velocity of 52 ft/s. The ball's height h (in feet) after t seconds is given by h = 7+52t - 16t² Algebraically find all values of t for which the ball's height is 40 feet. Algebraically find the maximum height of the ball. The rocket reaches a height of 336 feet on its way up after 2 seconds and on its way down after 10.5 seconds. b. Maximum height? A parabola reaches its maximum value at its vertex, or turning point. Use the formula for the axis of symmetry to find the x-coordinate of the vertex. Plug in for t and find h. h = -16(6.25) 2 + 200(6.25) = 625 ft. A ball is thrown upward and outward from a height of 5 feet. The... A ball is thrown upward and outward from a height of 5 feet. The height of the ball, f (x), in feet, can be modeled by. f (x)=−0.3x2+1.7x+5. where x is the ball's horizontal distance, in feet, from where it was thrown. Use this model to solve parts (a) through (c). MathActusMagazineGuideTests ComparatifsWebContactNo Result View All Result When cricket ball thrown vertically upwards inScience Math Reading Time mins read When cricket ball thrown vertically upwards, reaches maximum height metres. When.
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This sort of problem is solved easily by a graphing calculator. a) The ball reaches its maximum height in 1.13 seconds. b) The ball's maximum height is 30.25 feet. Since you want to know when and where the peak value of the function occurs, it is convenient to put it into vertex form. Step 3: Compute by analyzing and performing operations on relationships to draw conclusions. (For example, operations include calculating the values of t when the ball reaches a height of 10 feet. Two balls are thrown in the air. The path of the first ball is represented in the graph. The second ball is released 1.5 feet higher than the first ball and after 3 seconds reaches its maximum height 5 feet lower than the first ball. a. Write an equation for the path of the second ball. b. Do the balls hit the ground at the same time?. Since flight is symmetrical, time of flight T = When the ball reaches its maximum height, its 2(1.154 s) = 2.31 s. kinetic energy is 16 J. g Horizontal distance = (horizontal speed)(time) a What is the maximum height achieved by the = (11.31 m s–1)(2.31 s) = 2.61 m ball from its point of release? b Calculate the initial vertical velocity of. Taking g = 1 0 m / s 2, find the maximum height reached by the stone The height of the point from where the ball is thrown is 25 The height of a ball t seconds after it is thrown upward from a height of 6 feet and with an initial velocity of 80 feet per second is f (t) = -16t^2 + 80t + 6 H t 120t 16tsquared A ball is thrown at 120 ft s how long.
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Example: A ball is thrown in the air. Its height at any time t is given by: h = 3 + 14t − 5t 2. What is its maximum height? Using derivatives we can find the slope of that function: ddt h = 0 + 14 − 5(2t) = 14 − 10t (See below this example for how we. PRE-ALGEBRA (441) ALGEBRA 1 (2,216) GEOMETRY (864) ALGEBRA 2 (2,745) TRIGONOMETRY (1,186) PRECALCULUS (1,903) ... After how many seconds does the projectile reach its maximum height? asked Sep 17, ... A ball is thrown up at the edge of a 490 foot cliff. asked Oct 8, 2014 in PRECALCULUS by Baruchqa Pupil. See answer (1) Best Answer. Copy. Anyone who is trying to figure out how high they can vertically throw an object like a ball has a little math to do. The calculation needed is. Engine/ Battery Information. 商品に関するお問い合わせの場合、商品名や商品番号を改めてお問い合わせ文章にご記載ください. Frank shoots an air-ball. The height h in the ball is given by O h = -16/ + 32t+ 8. What is the height after 4 seconds? 19) A ball is thrown straight up from the top of a 30 foot building with an initial speed of 74 feet per second. The height a the ball as a function of time can be modeled by the function h(t)= + + 30. What is the height after.
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The gravitational force lets it reach a certain height and then fall back down again. Enter one value at velocity and height, the other value will be calculated.Example: a ball, thrown vertically up with a speed of 25 km/h, can reach a maximum height of almost 2.5 meters above its starting point. The formula is h=v²/ (2g).. "/>. The height of the ball h in metres, can be approximated by the function h 5t2 10t 34 where t is the time in seconds, after the ball is thrown. a) Sketch the graph. b) How high is the ball after 2 s? c) Find the maximum height of the ball. 4. A tennis ball is thrown up into the air. Its height h in metres after t seconds, is given by the function.
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A body's potential energy is at its maximum when it is at its greatest height to which it is elevated from the ground or another chosen position, such as a table top. Refer to the related link. A ball is thrown straight up with an initial speed of 40 ms How high does it go? The ball reaches a maximum height of ~81.6 meters. We know: Vf = 0 (terminal velocity) and Vi = 40 (given in question) So, Vf = Vi - g (t) => 0 = 40 - g (t) --Rework the equation to solve for t.
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Dec 2008 43m, and in women’s volleyball it is 2 After the ball hits the floor, it rebounds to 85% of its previous By: Varun Patel, Ann Marie Riccio, Austin Howard, Julia Nguyen and Lakshya Ramani The Relationship Between Ball Drop Height and Bounce Height Graph Graph Drop Height (Blks) vs Find the total distance that it travels before coming.
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This lesson shows an application problem for parabolas in which you will learn how to find the maximum height or vertex of the parabola. This is a great example application problem for a quadratic equation. You will also learn how to find out when the ball hits the ground. All steps and concepts are explained in this example problem. Find the first time the ball is at a height of 39 If the acceleration due to gravity is 10 m/s, what will be the distance travelled by it in the last second of motion before again come to his hand : (1) 5 m (2) 10 m (3) 25 m (4) 30 m ; Mar 27, 2016 · A tennis ball is thrown vertically upward with an initial velocity of +8 Gravity is a force. Dec 2008 43m, and in women’s volleyball it is 2 After the ball hits the floor, it rebounds to 85% of its previous By: Varun Patel, Ann Marie Riccio, Austin Howard, Julia Nguyen and Lakshya Ramani The Relationship Between Ball Drop Height and Bounce Height Graph Graph Drop Height (Blks) vs Find the total distance that it travels before coming.
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Subscribe A ball is thrown upward with initial velocity _______ and its height is modeled by the function f (x)=________________ find the time it takes to reach the maximum height and the maximum. Solved Examples for Maximum Height Formula. Q.1: A firefighter plane aims a fire hose upward, toward a fire in a skyscraper. The water leaving the hose with a velocity of 32.0 m per second. If the firefighter holds the hose at an angle of Find out the maximum height of the water stream using maximum height formula.
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. A ball is thrown vertically upward, and as it comes down, it is caught at its initial position 8m/s2 upward, and on the way down its acceleration is 9 Its speed decreases until it attains a maximum height, where the velocity is zero A ball of mass 200 g is thrown vertically upwards with velocity of 5 ms-1 Which statement about the frog's motion is correct?.
Repeat steps 3 and 4 but change the drop height to 1OOcm, 150 cm Determine the height of the drop and the bounce by measuring from the bottom of the ball Stand about 2 meters from the chair, and make 3 attempts to bounce the tennis ball from the floor into the basket or box that is on the chair, while you observe the motion of the ball Stand. 1. Excluding drag the equation for the height at time t is. h ( t) = h 0 − 1 2 g ( t − t 0) 2. This is zero (ie at the ground) when t = ± 2 h 0 g + t 0. In this case, h 0 is the height from which it is dropped, and t 0 is the time when it is dropped. So assume h 0. Answer (1 of 14): The key in these problems is “maximum height”. This means that the final velocity (velocity at the maximum height) is 0. It will always be 0 in these cases since the ball has to stop before changing direction. Also, even if you. A ball is thrown upward and outward from a height of 7 feet. The... A ball is thrown upward and outward from a height of 7 feet. The height of the ball, f (x), in feet, can be modeled by. f (x)=−0.1x2+0.8x+7. where x is the ball's horizontal distance, in feet, from where it was thrown. Use this model to solve parts (a) through (c).
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A ball thrown at all initial angle of 370 from the horizontal with an initial velocity of 23 m/ s reaches a maximum height h. 3 the ball is thrown from rest upward, and the relation of acceleration to velocity. Determine: i) The net force on the ball as it moves up; ii) The acceleration of the ball; iii) The maximum height reached by. 726 views View 1 Upvoter A ball is thrown vertically upwards with a velocity of 49m/s calculate maximum height and time taken to reach maximum height If a ball is thrown vertically upward with a velocity of 80 ft/s, then its height after t seconds is s = 80t – 16t2 CHANGE: THE PATH OF A THROWN BALL Grades 8-12 When an object is thrown, shot. Q13 3 WSU quarterback, Gardner Minshew, throws a football vertically upward, and the height of the ball above the ground while it is in flight is ht)-16t+32t +6. Assume that t 0 corresponds to the time the ball was thrown, Jis measured in seconds, and h is measured in feet. Answer the following questions.
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Square the initial velocity, find the sine of angle square and multiply them. Double the acceleration due to gravity which is 9.81. Divide the result obtained in step 3 by step 4. Add the result to the initial height. The obtained result is called the maximum height of the projectile motion. Solve the quadratic formula to find the time for the ball to land, and then use half of that time to calculate the maximum height. You will find that the time to fall is 1.5 seconds and the maximum height is 9 feet. First, lets solve the quadratic equation to determine the times when h=0, or when the ball is on the ground. This will tell us the time that the ball will land, assuming the. PRE-ALGEBRA (441) ALGEBRA 1 (2,216) GEOMETRY (864) ALGEBRA 2 (2,745 ... straight upwards with a velocity of 80 ft/s, then its height after t seconds is given by s = 80t - 16t^2. Find the maximum height in feet reached by the ball. Answer choices are: 72.35 80 150.25 ... A ball is thrown up at the edge of a 490 foot cliff. asked Oct 8, 2014 in.
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A more detailed solution would use the formula for height y as a function of time: y(t) = v0*t + a*t 2 /2. with a known, v0 determined from the maximum 19m, and then t known. Then from the derivative. v(t) = y'(t) = v0 + a*t. you can solve for t at the. PRE-ALGEBRA (441) ALGEBRA 1 (2,216) GEOMETRY (864) ALGEBRA 2 (2,745 ... straight upwards with a velocity of 80 ft/s, then its height after t seconds is given by s = 80t - 16t^2. Find the maximum height in feet reached by the ball. Answer choices are: 72.35 80 150.25 ... A ball is thrown up at the edge of a 490 foot cliff. asked Oct 8, 2014 in. Step 1: In order to get from the displacement to the velocity, you will take the derivative of the displacement with respect to time. First, set up your equation to get ready to find a derivative: x (t) = 4t 2 + 4t + 4. dx/dt = d/dt 4t 2 + 4t + 4 = v (t) Step 2: Solve for the derivative. This sort of problem is solved easily by a graphing calculator. a) The ball reaches its maximum height in 1.13 seconds. b) The ball's maximum height is 30.25 feet. Since you want to know when and where the peak value of the function occurs, it is convenient to put it into vertex form.
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Height of Dropped Ball - Algebra College Homework. If a ball is dropped from the top of a 256-ft building, then the formula. h (t)=256-16t^2. Expresses its height h (t) in feet as a function of the time t in seconds. 1.Find h (2), the height of the ball 2 seconds after it is dropped. 2.Find h (4). The maximum occurs when the varying part is smallest, since it is always non positive, so that its effect is to reduce the height. This occurs when it vanishes. That is, when. At the maximum height, the velocity is zero (since the ball cannot go any farther up) A ball is thrown vertically upwards with an initial velocity such that it can reach a maximum height of 15m Show that the time taken for the ball to reach its maximum height is 0-71 s Assume the acceleration of the ball is a(t) = −32 ft2 a (t) = − 32 f t 2. Algebraically find the maximum height of the ball. A ball is thrown from an initial height of 7 feet with an initial upward velocity of 52 ft/s. The ball's height h (in feet) after t seconds is given by h = 7+52t - 16t² Algebraically find all values of t for which the ball's height is 40 feet. Algebraically find the maximum height of the ball. For a ball thrown vertically upwards, the motion is a straight-line constant acceleration equation Acceleration is a constant for a ball thrown vertically upward, its is acceleration due to gravity A tennis ball is thrown vertically upward with an initial velocity of +8 Find the first time the ball is at a height of 39 5 m T = U/g = 49/9 5 m T.
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We have obtained two values that represent the time that the ball reaches a height of 300 feet. The first value 1 .34 indicates that after 1 .34 seconds have passed, the ball is at a height of 300 feet. Then the ball reaches its maximum height and begins to fall back to the ground. After 4 .66 seconds it is once again at 300 feet. A ball is thrown vertically upward from the ground with an initial velocity of 109 ft/sec. Use the quadratic function h(t) = −16t 2 + 109t + 0 to find how long it will take for the ball to reach its maximum height, and then find the maximum height. ball is thrown upwards from rooftop which is above from the ground. It will reach a maximum vertical height and then fall back to the ground. The height of the ball "h" from the ground at time "t" seconds is given by, h = -16+2 64+ 80.How long will the ball take to hit the ground ?.
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Answer (1 of 14): The key in these problems is “maximum height”. This means that the final velocity (velocity at the maximum height) is 0. It will always be 0 in these cases since the ball has to stop before changing direction. Also, even if you. The maximum occurs when the varying part is smallest, since it is always non positive, so that its effect is to reduce the height. This occurs when it vanishes. That is, when. 1. A ball is thrown upward from roof of 32 foot building with velocity of 112 ft/sec. The height after t seconds is: s ( t) = 32 + 112 t − 16 t 2. (a) Find the maximum height that the ball reaches. (answer: 228) (b) Find the velocity of the ball. Dec 2008 43m, and in women’s volleyball it is 2 After the ball hits the floor, it rebounds to 85% of its previous By: Varun Patel, Ann Marie Riccio, Austin Howard, Julia Nguyen and Lakshya Ramani The Relationship Between Ball Drop Height and Bounce Height Graph Graph Drop Height (Blks) vs Find the total distance that it travels before coming.
The function can be represent with a parabola whose vertex is its maximum point . for answer the first question we have to find the y of the vertex . first of all we have to find the x . Vertex (x) = -b/2a = -8/-2 = 4 . if we substitute t with 4, we can find the y . Vertex (y) = -(4^2) + (8*4) = = -16 + 32 = 16 . so the answer is 16 meters. Frank shoots an air-ball. The height h in the ball is given by O h = -16/ + 32t+ 8. What is the height after 4 seconds? 19) A ball is thrown straight up from the top of a 30 foot building with an initial speed of 74 feet per second. The height a the ball as a function of time can be modeled by the function h(t)= + + 30. What is the height after. Example: A ball is thrown in the air. Its height at any time t is given by: h = 3 + 14t − 5t 2. What is its maximum height? Using derivatives we can find the slope of that function: ddt h = 0 + 14 − 5(2t) = 14 − 10t (See below this example for how we found that derivative.) Now find when the slope is zero:.
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PRE-ALGEBRA (441) ALGEBRA 1 (2,216) GEOMETRY (864) ALGEBRA 2 (2,745) TRIGONOMETRY (1,186) PRECALCULUS (1,903) ... After how many seconds does the projectile reach its maximum height? asked Sep 17, ... A ball is thrown up at the edge of a 490 foot cliff. asked Oct 8, 2014 in PRECALCULUS by Baruchqa Pupil.
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This gives a maximum value of 53 and the range of the function is A graph of the function is shown below. Recall the function, which describes the height in feet of a ball t seconds after it is thrown upward from the top of a 200 foot high building. We can now determine when the ball hits the ground and the maximum height that it reaches, as.
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How long does it take for the rocket to reach the maximum height? 8. A ball is thrown directly upward from an initial height of 200 feet with an initial velocity of 96 feet per second. ... 16. A baseball is "popped" straight up by a batter. The ball's height (in feet) above the ground "x" seconds later is.
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Height of Dropped Ball - Algebra College Homework. If a ball is dropped from the top of a 256-ft building, then the formula. h (t)=256-16t^2. Expresses its height h (t) in feet as a function of the time t in seconds. 1.Find h (2), the height of the ball 2 seconds after it is dropped. 2.Find h (4).
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A ball is thrown straight up with an initial speed of 40 ms How high does it go? The ball reaches a maximum height of ~81.6 meters. We know: Vf = 0 (terminal velocity) and Vi = 40 (given in question) So, Vf = Vi - g (t) => 0 = 40 - g (t) --Rework the equation to solve for t. Algebra -> Quadratic Equations and Parabolas -> SOLUTION: ... To find the maximum height, you first find the time to get maximum height t = - = - = = 3 second. Then the maximum height is the given function t(t) at t= 3 seconds h(3) = -16*3^2 + 96*3 + 112 = 256 feet. ... Problem on a ball thrown vertically up from the top of a tower.
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Are these answers to my homework questions correct? The height of a ball thrown straight up from a building is given by h ( t) = − 16 t 2 + 32 t + 48, t seconds after it is thrown. 1) Find the maximum height of the ball. 2) Find the time it takes for the ball to reach the ground. algebra-precalculus Share edited May 13 at 18:37 Wang YeFei.
Bounce Lab Purpose: The purpose of this lab is to find the elasticity of two different types of balls My hypothesis was correct as it stands and the ball did bounce higher when they were dropped higher The contact force is, as the name implies, In it I teach you how to use all the major features of canvas, including how to animate with physics and how to create two. MathActusMagazineGuideTests ComparatifsWebContactNo Result View All Result When cricket ball thrown vertically upwards inScience Math Reading Time mins read When cricket ball thrown vertically upwards, reaches maximum height metres. When. A ball is dropped from a height of 144 feet. The function above models its height ℎ, in feet, after 𝑡 seconds. When will the ball hit the ground? 3. ℎ :𝑡 ; L F16𝑡 6 E64𝑡 E80 A ball is thrown straight up from a height of 80 feet with an initial speed of 64 feet per second. The function above models its height ℎ, in feet, after.
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Multiply the height by 2, and divide the result by the object's acceleration due to gravity. If the object fell from 5 m, the equation would look like this: (2*5 m)/ (9.8 m/s^2) =1.02 s^2. Take the square root of the result to calculate the time it takes for the object to drop. For example, the square root of 1.02 s^2 equals 1.01 s. Taking g = 1 0 m / s 2, find the maximum height reached by the stone The height of the point from where the ball is thrown is 25 The height of a ball t seconds after it is thrown upward from a height of 6 feet and with an initial velocity of 80 feet per second is f (t) = -16t^2 + 80t + 6 H t 120t 16tsquared A ball is thrown at 120 ft s how long.
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If a ball is thrown vertically upward with a velocity of 160 ft/s, then its height after t seconds is s = 160t - 16t^2. If a ball is thrown vertically upward with a velocity of 160 ft/s, then its. Calculate the maximum height of the water stream. Solution: The water droplet leaving the hose is considered as the object in projectile motion. Initial velocity V o = 38 m/sec angle of launch α = 58.5° acceleration due to gravity g = 9.8 m/sec² Maximum height h max = V o ² * sin (α)² / (2 * g) = ( (38)² x sin58.5°)/ (2 x 9.8). The basketball will hit the ground when its height is 0, so find the zeros of the function. Step 1 Find the axis of symmetry. Use x =-b/2a . Substitute -16 for a and 32 for b => x = -32/(-16*2) so x = 1. The axis of symmetry is x = 1. Step 2 Find the vertex. The x-coordinate of the vertex is 1. Substitute 1 for x, we will find the y-coordinate. MathActusMagazineGuideTests ComparatifsWebContactNo Result View All Result When cricket ball thrown vertically upwards inScience Math Reading Time mins read When cricket ball thrown vertically upwards, reaches maximum height metres. When.
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A ball is thrown straight up with an initial speed of 40 ms How high does it go? The ball reaches a maximum height of ~81.6 meters. We know: Vf = 0 (terminal velocity) and Vi = 40 (given in question) So, Vf = Vi - g (t) => 0 = 40 - g (t) --Rework the equation to solve for t. Application Problem Example 5 A ball is thrown up into the air and its height is represented by the equation Choosing the Best Method for Solving a Quadratic Equation: Example 6 Which method(s) would you choose to solve each equation? Justify your reasoning. a) ℎ2+4ℎ+7=0 b) 2−4 −12=0 c) 2−144=0 d) 24 2−11 −14=0. Find the ball’s initial speed The potential energy at the top of the ball's motion is 18 J The only choice he has to make to maximize distance, then, is the angle at which he kicks the ball (b) Using conservation of energy and the result of part (a), find the magnitude of the momentum of the ball at one-half its maximum height in terms of m.
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Solved Examples for Maximum Height Formula. Q.1: A firefighter plane aims a fire hose upward, toward a fire in a skyscraper. The water leaving the hose with a velocity of 32.0 m per second. If the firefighter holds the hose at an angle of Find out the maximum height of the water stream using maximum height formula. A ball is launched upward at 48 ft./s from a platform 100 ft. high. Find the maximum height the ball reaches and how long it will take to get there. 4. An object is fired vertically from the top of a tower.The tower is 60.96 ft.high.The height of the object above the ground t seconds after firing is given by the formula h(t) = -16t2 + 80t + 200. acceleration a = − 5 m / s 2. We have to find the total distance traveled in coming to stop from initial velocity. Putting these values in third equation of motion v 2 = u 2 + 2 a s we have. 0 2 − ( 25 3) 2 = 2 × ( − 5) [ x t − 0] or, x t = 625 9 × 1 10 m = 6.94 m. Total distance travelled = 5.83 m + 6.94 m = 12.77 m. Math, Trig. A soccer player kicks a ball from the ground to a maximum height of 12 m. A ball is thrown vertically upward from a height of 4 ft with an initial velocity of 80 ft/s. Taking g = 1 0 m / s 2, find the maximum height reached by the stone. The height, s, of a ball thrown straight down with initial speed 64 ft/sec.
With what initial speed must a ball be thrown straight, up to reach the same maximum height h? a 11 A ball is released from rest above a horizontal surface 94 m high table Just Paste It Although the ball's instantaneous velocity at this point in its path is 0, its acceleration remains a 2 is an example of a ball being let go from rest and. Height of Dropped Ball - Algebra College Homework. If a ball is dropped from the top of a 256-ft building, then the formula. h (t)=256-16t^2. Expresses its height h (t) in feet as a function of the time t in seconds. 1.Find h (2), the height of the ball 2 seconds after it is dropped. 2.Find h (4). May 12, 2007. #1. Suppose a baseball is shot up from the ground straight up with an initial velocity of 64 feet per second. A function can be created by expressing distance above the ground, s, as a function of time, t. This function is s=-16t^2+vot +so. 16 represents 1/2g, the gravitional pull due to gravity (measured in feet per second^2).
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How to find maximum height of a ball thrown up algebra background check terminology. smb port forwarding. dental class 1 2 3. golf with 5 players best bike handlebar tape kowa anamorphic gh5 apple watch sport band access refresh linked tables automatically 1970 pontiac trans ams for. The initial magnitude of the velocity of the ball 13N A stone is let to fall from a building of height 30m Determine: i) The net force on the ball as it moves up; ii) The acceleration of the ball; iii) The maximum height reached by 94 m high table As the ball is thrown vertically upwards, its height continuously increases to a certain point and.
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The maximum height of the object is the highest vertical position along its trajectory. The maximum height of the projectile depends on the initial velocity v 0, the launch angle θ, and the acceleration due to gravity. The unit of maximum height is meters (m). H = maximum height (m) v 0 = initial velocity (m/s) g = acceleration due to gravity. This video uses the vertex point of a parabola to find the maximum height of ball thrown in the air given its initial speed. Youtube videos by Julie Harland are organized at http://YourMathGal.com. Height h (in meters) of a ball thrown upwards is given by equation h = -3(t - 6)^2 + 108 (Height of ball 3.5 seconds after it was launched)(Height of ball 2 seconds after it reaches maximum height) (A) Quantity ... GRE Forum, GRE Vocabulary Flashcards | GREPrepClub. GRE Official Guide Question Directory. ... Signing up is free, quick,.
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If it attains a maximum height of 400m, find its velocity If it attains a maximum height of 400m, find its velocity. 726 views View 1 Upvoter Air resistance cannot be neglected 13M Gravity does not stop working (constant), and is the only A ball is thrown vertically upwards with a velocity of 49 m/s Neglect friction Neglect friction. Since the ball, is moving up against gravity, the value of g will be negative. ⇒ 0 = u 2-g t. ⇒ 29. 4 2 = 2 × 9. 8 × h. ⇒ h = 29. 4 × 29. 4 2 × 9. 8. ⇒ h = 44. 1 m. Step 4: Calculate the new Height. Given new time is 4 s. Since the ball reaches a maximum height at t = 3 s, the ball will move downwards after 4 s. Therefore, the new. A ball was tossed up into the air. The height of the ball as a function of the time the ball is in the air in seconds can be modeled by a quadratic function. At the time the ball was initially thrown in the air, it was 7.5 meters off the ground. After 1.25 seconds, the ball was at its maximum height of 15.3125 meters. Then you can find the maximum height the ball travels above the ground.. Step 2. Since the rock goes up first and then it comes down, the smaller time value corresponds to the velosity of the rock going up. By differentiating h (t)using the power rule, we get the velocity for the ball. v. A ball is thrown vertically upward from the ground with an initial velocity of 39.2 m/sec. If the only force considered is that attributed to the acceleration due to gravity, find ... At maximum height v = 0 . ⇒ (2) ⇒ -9.8 t + 39.2 = 0 . t = 4 sec. ... Basic Algebra (152) Trigonometry (198) Combinatorics and Mathematical Induction (139).
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2022-12-03 16:23:39
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http://www.askphysics.com/gauss-rifle/
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Home » Ask Physics » Gauss Rifle
# Gauss Rifle
Is there a mathematical relationship between the spacing of the magnets across each stage in a Gauss rifle, and the velocity of the final ball bearing that is fired from it?
If so, what is this relationship?
The answer is being prepared and will be posted soon.
Gauss Rfle (Also called Coil Gun)
A Gauss Rifle (also known as a gauss gun or a coil gun) is a Magnetic Linear Accelerator. In a Gauss Rifle, the barrel is a solenoid: essentially a big coil of wire. If you send a current through a coil of wire, it creates a magnetic field in the middle of the coil. This magnetic field is used to accelerate a magnetic projectile, essentially pulling it along the barrel of the gun.
Get more details here http://en.wikipedia.org/wiki/Coilgun
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2016-12-11 13:51:15
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https://www.biostars.org/p/202479/
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From TCGA to GDC (Genomic data commons)
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6.4 years ago
Hello,
I was using TCGA data related to Colon adenocarcinoma (COAD). In the specific I was using "IlluminaGA_RNASeqV2", "IlluminaHiSeq_RNASeqV2" platforms
For the COAD cancer and for those platforms were available the level 3 information. I was using the raw count from the rsem.genes.results files.
Now that TCGA moved under Genomic data commons (GDC), i'm struggling to retrive the same information. I would like to understand how to download from https://gdc-portal.nci.nih.gov/ the same information that were available from TGCA.
I was using TCGABiolinks, but now seems not working anymore. Any suggestion about R library to import GDC data?
Thanks
R TCGA gdc • 7.5k views
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Entering edit mode
I agree that the transition is very confusing, not least because of the way the gdc-portal displays data files for downloading.
Anyways, have you been here - Firehose On the landing page, at the row for COAD, under Data col, click Browse. The pop-up window that opens should be able to give you what you are looking for.
The file naming is a bit different now, but you would be able to make out. I haven't used the R library, but the Firehose site has its own client (like a wget). There is an R package described as well.
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Thanks for the suggestion. I will give a try to Firehose
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Nopes. I think that would be GDC.
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6.4 years ago
Nearly all TCGA data/results can be found at Broad Institute's Firehose pipelines. Get raw data/results here or browse the web-based UIs at MSKCC's cbioportal.org or Broad's firebrowse.org.
NOTE: This is just a temporary solution, while I figure out how to use the GDC via CLI. :)
mkdir scripts
unzip -d scripts scripts/firehose_get_latest.zip
Here is how to use that tool to download the normalized per-gene expression estimates from RNA-seq data:
./scripts/firehose_get -b -only Merge_rnaseqv2__illuminahiseq_rnaseqv2__unc_edu__Level_3__RSEM_genes_normalized__data data latest
It creates a folder structure with gzipped tarballs in separate tumor-type subfolders. Unpack all the tarballs:
mkdir rna_seq
for file in stddata__*/*/*/*RSEM*.Level_3*.tar.gz; do tar -zxf $file -C rna_seq; done Rename the resulting subfolders to just the tumor type codes, using some in-line Perl and bash: ls -d rna_seq/gdac* | perl -ne 'chomp; ($t)=m/gdac.broadinstitute.org_(\w+)/; print "mv $_ rna_seq/$t\n"' | bash
Delete the separate colon/rectal cohorts, leaving behind only the combined cohort COADREAD:
rm -rf rna_seq/{COAD,READ}
There are also KIPAN (KICH+KIRC+KIRP) and GBMLGG (GBM+LGG), but keep them, they're interesting. The per-gene RNA-expression estimates are now in these files:
rna_seq/*/*.rnaseqv2__illuminahiseq_rnaseqv2__unc_edu__Level_3__RSEM_genes_normalized__data.data.txt
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Hi Cyriac,
Do you know the difference between illumina rnaseq2 vs illumina rnaseq and the current rna seq data on GDC?
I see several type in the link but on GDC portal, there is only one kind of RNA-seq data.
Thanks
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v2 reports RSEM, the other reports RPKM. GDC only reports RSEM.
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thanks,
then what is the relationship between RSEM and HTSeq?
I saw GDC have HTSeq-counts, HTseq-FPKM, HTseq-FPKM-UQ.
I thought RSEM will generate calculated expression, and HTseq is the raw counts,
And looks like if I use the HTseq from new GDC portal, I have to combine them by myself since they download the file folder by folder separately...Does some see a merge HTSeq file?
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Sorry, I was wrong. GDC runs their own RNA-seq pipeline defined here, which appears to report FPKM.
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AFAIK RNAseq TCGA V1 analysis (old) used BWA and the V2 analysis (new) which uses MapSplice. All V1 data was reprocessed as V2. So there should be only one kind of RNAseq data (submitted by UNC).
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6.4 years ago
tiagochst ▴ 70
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6.4 years ago
Mike ★ 1.8k
Now TCGAbiolinks is updated, they replace "TCGAquery" with "GDCquery" function.
And it can acess both the GDC and GDC Legacy Archive.
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2022-12-08 09:42:52
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https://preview.zbmed.de/?search=%28concept%3Atransmission%29+AND+concept%3Afomite
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### Overview
MeSH Disease
Infections (175)
Disease (173)
Death (96)
Human Phenotype
Pneumonia (31)
Fever (24)
Cough (20)
Falls (7)
Fatigue (5)
Transmission
Seroprevalence
displaying 1 - 10 records in total 363
records per page
### COVID19 Tracking: An Interactive Tracking, Visualizing and Analyzing Platform
Authors: Zhou Yang; Jiwei Xu; Zhenhe Pan; Fang Jin
id:2008.04285v1 Date: 2020-08-10 Source: arXiv
The Coronavirus Disease MESHD 2019 (COVID-19) has now become a pandemic, inflicting millions of people and causing tens of thousands of deaths MESHD. To better understand the dynamics of COVID-19, we present a comprehensive COVID-19 tracking and visualization platform that pinpoints the dynamics of the COVID-19 worldwide. Four essential components are implemented: 1) presenting the visualization map of COVID-19 confirmed cases TRANS and total counts all over the world; 2) showing the worldwide trends of COVID-19 at multi-grained levels; 3) provide multi-view comparisons, including confirmed cases TRANS per million people, mortality rate and accumulative cure rate; 4) integrating a multi-grained view of the disease MESHD disease spreading TRANS spreading dynamics in China and showing how the epidemic is taken under control in China.
### Data-driven Inferences of Agency-level Risk and Response Communication on COVID-19 through Social Media based Interactions
id:2008.03866v1 Date: 2020-08-10 Source: arXiv
Risk and response communication of public agencies through social media played a significant role in the emergence and spread of novel Coronavirus (COVID-19) and such interactions were echoed in other information outlets. This study collected time-sensitive online social media data and analyzed such communication patterns from public health (WHO, CDC), emergency MESHD (FEMA), and transportation (FDOT) agencies using data-driven methods. The scope of the work includes a detailed understanding of how agencies communicate risk information through social media during a pandemic and influence community response (i.e. timing of lockdown, timing of reopening) and disease MESHD outbreak indicators (i.e. number of confirmed cases TRANS, number of deaths MESHD). The data includes Twitter interactions from different agencies (2.15K tweets per agency on average) and crowdsourced data (i.e. Worldometer) on COVID-19 cases and deaths MESHD were observed between February 21, 2020 and June 06, 2020. Several machine learning techniques such as (i.e. topic mining and sentiment ratings over time) are applied here to identify the dynamics of emergent topics during this unprecedented time. Temporal infographics of the results captured the agency-levels variations over time in circulating information about the importance of face covering, home quarantine, social distancing and contact tracing TRANS. In addition, agencies showed differences in their discussions about community transmission TRANS, lack of personal protective equipment, testing and medical supplies, use of tobacco, vaccine, mental health issues, hospitalization, hurricane season, airports, construction work among others. Findings could support more efficient transfer of risk and response information as communities shift to new normal as well as in future pandemics.
### CRISPR-based and RT-qPCR surveillance of SARS-CoV-2 in asymptomatic TRANS individuals uncovers a shift in viral prevalence SERO among a university population
Authors: Jennifer N Rauch; Eric Valois; Jose Carlos Ponce-Rojas; Zach Aralis; Ryan L Lach; Francesca Zappa; Morgane Audouard; Sabrina C Solley; Chinmay Vaidya; Michael Costello; Holly Smith; Ali Javanbakht; Betsy Malear; Laura Polito; Stewart Comer; Katherine Arn; Kenneth S Kosik; Diego Acosta-Alvear; Maxwell Z Wilson; Lynn Fitzgibbons; Carolina Arias
doi:10.1101/2020.08.06.20169771 Date: 2020-08-07 Source: medRxiv
Background: The progress of the COVID-19 pandemic profoundly impacts the health of communities around the world, with unique impacts on colleges and universities. Transmission TRANS of SARS-CoV-2 by asymptomatic TRANS people is thought to be the underlying cause of a large proportion of new infections MESHD. However, the local prevalence SERO of asymptomatic TRANS and pre-symptomatic carriers TRANS of SARS-CoV-2 is influenced by local public health restrictions and the community setting. Objectives: This study has three main objectives. First, we looked to establish the prevalence SERO of asymptomatic TRANS SARS-CoV-2 infection MESHD on a university campus in California. Second, we sought to assess the changes in viral prevalence SERO associated with the shifting community conditions related to non-pharmaceutical interventions (NPIs). Third, we aimed to compare the performance SERO of CRISPR- and PCR-based assays for large-scale virus surveillance sampling in COVID-19 asymptomatic TRANS persons. Methods: We enrolled 1,808 asymptomatic TRANS persons for self-collection of oropharyngeal (OP) samples to undergo SARS-CoV-2 testing. We compared viral prevalence SERO in samples obtained in two time periods: May 28th-June 11th; June 23rd-July 2nd. We detected viral genomes in these samples using two assays: CREST, a CRISPR-based method recently developed at UCSB, and the RT-qPCR test recommended by US Centers for Disease MESHD Control and Prevention (CDC). Results: Of the 1,808 participants, 1,805 were affiliates of the University of California, Santa Barbara, and 1,306 were students. None of the tests performed on the 732 samples collected between late May to early June were positive. In contrast, tests performed on the 1076 samples collected between late June to early July, revealed nine positive cases. This change in prevalence SERO met statistical significance, p = 0.013. One sample was positive by RT-qPCR at the threshold of detection, but negative by both CREST and CLIA-confirmation testing. With this single exception, there was perfect concordance in both positive and negative results obtained by RT-qPCR and CREST. The estimated prevalence SERO of the virus, calculated using the confirmed cases TRANS, was 0.74%. The average age TRANS of our sample population was 28.33 (18-75) years, and the average age TRANS of the positive cases was 21.7 years (19-30). Conclusions: Our study revealed that there were no COVID-19 cases in our study population in May/June. Using the same methods, we demonstrated a substantial shift in prevalence SERO approximately one month later, which coincided with changes in community restrictions and public interactions. This increase in prevalence SERO, in a young and asymptomatic TRANS population which would not have otherwise accessed COVID-19 testing, indicated the leading wave of a local outbreak, and coincided with rising case counts in the surrounding county and the state of California. Our results substantiate that large, population-level asymptomatic TRANS screening using self-collection may be a feasible and instructive aspect of the public health approach within large campus communities, and the almost perfect concordance between CRISPR- and PCR-based assays indicate expanded options for surveillance testing
### Genomic epidemiology reveals transmission TRANS patterns and dynamics of SARS-CoV-2 in Aotearoa New Zealand
Authors: Jemma L Geoghegan; Xiaoyun Ren; Matthew Storey; James Hadfield; Lauren Jelley; Sarah Jefferies; Jill Sherwood; Shevaun Paine; Sue Huang; Jordan Douglas; Fabio K L Mendes; Andrew Sporle; Michael G Baker; David R Murdoch; Nigel French; Colin R Simpson; David Welch; Alexei J Drummond; Edward C Holmes; Sebastian Duchene; Joep de Ligt
doi:10.1101/2020.08.05.20168930 Date: 2020-08-06 Source: medRxiv
New Zealand, a geographically remote Pacific island with easily sealable borders, implemented a nation-wide lockdown of all non-essential services to curb the spread of COVID-19. New Zealand has now effectively eliminated the virus, with low numbers of new cases limited to new arrivals in managed quarantine facilities at the border. Here, we generated 649 SARS-CoV-2 genome sequences from infected patients in New Zealand with samples collected between 26 February and 22 May 2020, representing 56% of all confirmed cases TRANS in this time period. Despite its remoteness, the viruses imported into New Zealand represented nearly all of the genomic diversity sequenced from the global virus population. The proportion of D614G variants in the virus spike protein increased over time due to an increase in their importation frequency, rather than selection within New Zealand. These data also helped to quantify the effectiveness of public health interventions. For example, the effective reproductive number TRANS, Re, of New Zealand's largest cluster decreased from 7 to 0.2 within the first week of lockdown. Similarly, only 19% of virus introductions into New Zealand resulted in a transmission TRANS lineage of more than one additional case. Most of the cases that resulted in a transmission TRANS lineage originated from North America, rather than from Asia where the virus first emerged or from the nearest geographical neighbour, Australia. Genomic data also helped link more infections MESHD to a major transmission TRANS cluster than through epidemiological data alone, providing probable sources of infections MESHD for cases in which the source was unclear. Overall, these results demonstrate the utility of genomic pathogen surveillance to inform public health and disease MESHD mitigation.
### Inhomogeneous mixing and asynchronic transmission TRANS between local outbreaks account for the spread of COVID-19 epidemics
Authors: Carlos I Mendoza
doi:10.1101/2020.08.04.20168443 Date: 2020-08-06 Source: medRxiv
The ongoing epidemic of COVID-19 originated in China has reinforced the need to develop epidemiological models capable of describing the progression of the disease MESHD to be of use in the formulation of mitigation policies. Here, this problem is addressed using a metapopulation approach to show that the delay in the transmission TRANS of the spread between different subsets of the total population, can be incorporated into a SIR framework through a time-dependent transmission TRANS rate. Thus, the reproduction number TRANS decreases with time despite the population dynamics remains uniform and the depletion of susceptible individuals is small. The obtained results are consistent with the early subexponential growth observed in the cumulated number of confirmed cases TRANS even in the absence of containment measures. We validate our model by describing the evolution of the COVID-19 using real data from different countries with an emphasis in the case of Mexico and show that it describes correctly also the long-time dynamics of the spread. The proposed model yet simple is successful at describing the onset and progression of the outbreak and considerably improves accuracy of predictions over traditional compartmental models. The insights given here may probe be useful to forecast the extent of the public health risks of epidemics and thus improving public policy-making aimed at reducing such risks.
### Epidemiological characteristics of SARS-COV-2 in Myanmar
Authors: Aung Min Thway; Htun Tayza; Tun Tun Win; Ye Minn Tun; Moe Myint Aung; Yan Naung Win; Kyaw M Tun
doi:10.1101/2020.08.02.20166504 Date: 2020-08-04 Source: medRxiv
Coronavirus disease MESHD (COVID-19) is an infectious disease MESHD caused by a newly discovered severe acute respiratory syndrome MESHD coronavirus 2 (SARS-CoV-2). In Myanmar, first COVID-19 reported cases were identified on 23rd March 2020. There were 336 reported confirmed cases TRANS, 261 recovered and 6 deaths MESHD through 13th July 2020. The study was a retrospective case series and all COVID-19 confirmed cases TRANS from 23rd March to 13th July 2020 were included. The data series of COVID-19 cases were extracted from the daily official reports of the Ministry of Health and Sports (MOHS), Myanmar and Centers for Disease MESHD Control and Prevention (CDC), Myanmar. Among 336 confirmed cases TRANS, there were 169 cases with reported transmission TRANS events. The median serial interval TRANS was 4 days (IQR 3, 2-5) with the range of 0 - 26 days. The mean of the reproduction number TRANS was 1.44 with (95% CI = 1.30-1.60) by exponential growth method and 1.32 with (95% CI = 0.98-1.73) confident interval by maximum likelihood method. This study outlined the epidemiological characteristics and epidemic parameters of COVID-19 in Myanmar. The estimation parameters in this study can be comparable with other studies and variability of these parameters can be considered when implementing disease MESHD control strategy in Myanmar.
### Analysis of COVID-19 and comorbidity co- infection MESHD Model with Optimal Control
Authors: Dr. Andrew Omame; Nometa Ikenna
doi:10.1101/2020.08.04.20168013 Date: 2020-08-04 Source: medRxiv
The new coronavirus disease MESHD 2019 (COVID-19) infection MESHD is a double challenge for people infected with comorbidities such as cardiovascular and cerebrovascular diseases MESHD and diabetes. Comorbidities have been reported to be risk factors for the complications of COVID-19. In this work, we develop and analyze a mathematical model for the dynamics of COVID-19 infection MESHD in order to assess the impacts of prior comorbidity on COVID-19 complications and COVID-19 re- infection MESHD. The model is simulated using data relevant to the dynamics of the diseases MESHD in Lagos, Nigeria, making predictions for the attainment of peak periods in the presence or absence of comorbidity. The model is shown to undergo the phenomenon of backward bifurcation caused by the parameter accounting for increased susceptibility to COVID-19 infection MESHD by comorbid susceptibles as well as the rate of re- infection MESHD by those who have recovered from a previous COVID-19 infection MESHD. Sensitivity SERO analysis of the model when the population of individuals co-infected with COVID-19 and comorbidity is used as response function revealed that the top ranked parameters that drive the dynamics of the co- infection MESHD model are the effective contact rate for COVID-19 transmission TRANS, $\beta\sst{cv}$, the parameter accounting for increased susceptibility to COVID-19 by comorbid susceptibles, $\chi\sst{cm}$, the comorbidity development rate, $\theta\sst{cm}$, the detection rate for singly infected and co-infected individuals, $\eta_1$ and $\eta_2$, as well as the recovery rate from COVID-19 for co-infected individuals, $\varphi\sst{i2}$. Simulations of the model reveal that the cumulative confirmed cases TRANS (without comorbidity) may get up to 180,000 after 200 days, if the hyper susceptibility rate of comorbid susceptibles is as high as 1.2 per day. Also, the cumulative confirmed cases TRANS (including those co-infected with comorbidity) may be as high as 1000,000 cases by the end of November, 2020 if the re- infection MESHD rates for COVID-19 is 0.1 per day. It may be worse than this if the re- infection MESHD rates increase higher. Moreover, if policies are strictly put in place to step down the probability of COVID-19 infection MESHD by comorbid susceptibles to as low as 0.4 per day and step up the detection rate for singly infected individuals to 0.7 per day, then the reproduction number TRANS can be brought very low below one, and COVID-19 infection MESHD eliminated from the population. In addition, optimal control and cost-effectiveness analysis of the model reveal that the the strategy that prevents COVID-19 infection MESHD by comorbid susceptibles has the least ICER and is the most cost-effective of all the control strategies for the prevention of COVID-19.
### Risk stratification as a tool to rationalize quarantine among health care workers exposed to COVID-19 cases - Evidence from a tertiary healthcare centre in India
Authors: Ravneet Kaur; Shashi Kant; Mohan Bairwa; Arvind Kumar; Shivram Dhakad; Vignesh Dwarakanathan; Aftab Ahmad; Pooja Pandey; Arti Kapil; Rakesh Lodha; Naveet Wig
doi:10.1101/2020.07.31.20166264 Date: 2020-08-04 Source: medRxiv
Background: Quarantine of healthcare workers (HCWs) exposed to COVID 19 confirmed cases TRANS is a well known strategy for limiting the transmission TRANS of infection MESHD. However, there is a need for evidence-based guidelines for the quarantine of HCWs in COVID 19. Methods: We describe our experience of contact tracing TRANS and risk stratification of 3853 HCWs who were exposed to confirmed COVID-19 cases in a tertiary health care institution in India. We developed an algorithm, on the basis of risk stratification, to rationalize quarantine among HCWs. Risk stratification was based on the duration of exposure, distance from the patient, and appropriateness of personal protection equipment (PPE) usage. Only high-risk contacts were quarantined for 14 days. They underwent testing for COVID 19 after five days of exposure, while low risk contacts continued their work with adherence to physical distancing, hand hygiene, and appropriate use of PPE. The low-risk contacts were encouraged to monitor for symptoms and report for COVID 19 screening if fever MESHD fever HP, cough MESHD cough HP, or shortness of breath occurred. We followed up all contacts for 14 days from the last exposure and observed for symptoms of COVID 19 and test positivity. Results and interpretation: Out of total 3853 contacts, 560 (14.5%) were categorized as high-risk contacts, and 40 of them were detected positive for COVID 19, with a test positivity rate of 7.1% (95% CI = 5.2, 9.6). Overall, 118 (3.1%) of all contacts tested positive. Our strategy prevented 3215 HCWs from being quarantined and saved 45,010 person-days of health workforce until June 8, 2020, in the institution. We conclude that exposure-based risk stratification and quarantine of HCWs is a viable strategy to prevent unnecessary quarantine, in a healthcare institution.
### Data-driven modeling and forecasting of COVID-19 outbreak for public policy making
Authors: Agus Hasan; Endah Putri; Hadi Susanto; Nuning Nuraini
doi:10.1101/2020.07.30.20165555 Date: 2020-08-02 Source: medRxiv
This paper presents a data-driven approach for COVID-19 outbreak modeling and forecasting, which can be used by public policy and decision makers to control the outbreak through Non-Pharmaceutical Interventions (NPI). First, we apply an extended Kalman filter (EKF) to a discrete-time stochastic augmented compartmental model to estimate the time-varying effective reproduction number TRANS Rt. We use daily confirmed cases TRANS, active cases, recovered cases, deceased cases, Case-Fatality-Rate (CFR), and infectious time as inputs for the model. Furthermore, we define a Transmission TRANS Index (TI) as a ratio between the instantaneous and the maximum value of the effective reproduction number TRANS. The value of TI shows the disease MESHD transmission TRANS in a contact between a susceptible and an infectious individual due to current measures such as physical distancing and lock-down relative to a normal condition. Based on the value of TI, we forecast different scenarios to see the effect of relaxing and tightening public measures. Case studies in three countries are provided to show the practicability of our approach.
### Modeling latent infection MESHD transmissions TRANS through biosocial stochastic dynamics
doi:10.1101/2020.07.30.20164491 Date: 2020-08-01 Source: medRxiv
The events of the recent SARS-CoV-02 epidemics have shown the importance of social factors, especially given the large number of asymptomatic TRANS cases that effectively spread the virus, which can cause a medical emergency MESHD to very susceptible individuals. Besides, the SARS-CoV-02 virus survives for several hours on different surfaces, where a new host can contract it with a delay. These passive modes of infection MESHD transmission TRANS remain an unexplored area for traditional mean-field epidemic models. Here, we design an agent-based model for simulations of infection MESHD transmission TRANS in an open system driven by the dynamics of social activity; the model takes into account the personal characteristics of individuals, as well as the survival time of the virus and its potential mutations. A growing bipartite graph embodies this biosocial process, consisting of active carriers TRANS (host) nodes that produce viral nodes during their infectious period TRANS. With its directed edges passing through viral nodes between two successive hosts, this graph contains complete information about the routes leading to each infected individual. We determine temporal fluctuations of the number of exposed and the number of infected individuals, the number of active carriers TRANS and active viruses at hourly resolution. The simulated processes underpin the latent infection MESHD transmissions TRANS, contributing significantly to the spread of the virus within a large time window. More precisely, being brought by social dynamics and exposed to the currently existing infection MESHD, an individual passes through the infectious state until eventually spontaneously recovers or otherwise is moves to a controlled hospital environment. Our results reveal complex feedback mechanisms that shape the dependence of the infection MESHD curve on the intensity of social dynamics and other sociobiological factors. In particular, the results show how the lockdown effectively reduces the spread of infection MESHD and how it increases again after the lockdown is removed. Furthermore, a reduced level of social activity but prolonged exposure of susceptible individuals have adverse effects. On the other hand, virus mutations that can gradually reduce the transmission TRANS rate by hopping to each new host along the infection MESHD path can significantly reduce the extent of the infection MESHD, but can not stop the spreading without additional social strategies. Our stochastic processes, based on graphs at the interface of biology and social dynamics, provide a new mathematical framework for simulations of various epidemic control strategies with high temporal resolution and virus traceability.
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### Annotations
All None MeSH Disease Human Phenotype Transmission Seroprevalence
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2020-08-13 03:13:07
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https://m.marefa.org/Y-%CE%94_transform
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# تحويل واي-دلتا
(تم التحويل من Y-Δ transform)
تحليل خطي للشبكات
العناصر
المكونات
دوائر التوالي والتوازي
تحويلات المعاوقة
مبرهنات المولد مبرهنات الشبكة
أساليب تحليل الشبكات
Two-port parameters
تحويل واي دلتا, Y-Δ transform وتكتب Y-delta، Wye-delta، Kennelly’s delta-star transformation, star-mesh transformation, T-Π or T-pi transform، هي تقنية رياضية لتبسيط تحليل الشبكة الإلكترونية.
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## Basic Y-Δ transformation
Δ and Y circuits with the labels which are used in this article.
The transformation is used to establish equivalence for networks with 3 terminals. Where three elements terminate at a common node and none are sources, the node is eliminated by transforming the impedances. For equivalence, the impedance between any pair of terminals must be the same for both networks. The equations given here are valid for real as well as complex impedances.
The general idea is to compute the impedance ${\displaystyle R_{y}}$ at a terminal node of the Y circuit with impedances ${\displaystyle R'}$ , ${\displaystyle R''}$ to adjacent nodes in the Δ circuit by
${\displaystyle R_{y}={\frac {R'R''}{\sum R_{\Delta }}}}$
where ${\displaystyle R_{\Delta }}$ are all impedances in the Δ circuit. This yields the specific formulae
${\displaystyle R_{1}={\frac {R_{a}R_{b}}{R_{a}+R_{b}+R_{c}}},}$
${\displaystyle R_{2}={\frac {R_{b}R_{c}}{R_{a}+R_{b}+R_{c}}},}$
${\displaystyle R_{3}={\frac {R_{a}R_{c}}{R_{a}+R_{b}+R_{c}}}.}$
The general idea is to compute an impedance ${\displaystyle R_{\Delta }}$ in the Δ circuit by
${\displaystyle R_{\Delta }={\frac {R_{P}}{R_{\mathrm {opposite} }}}}$
where ${\displaystyle R_{P}=R_{1}R_{2}+R_{2}R_{3}+R_{3}R_{1}}$ is the sum of the products of all pairs of impedances in the Y circuit and ${\displaystyle R_{\mathrm {opposite} }}$ is the impedance of the node in the Y circuit which is opposite the edge with ${\displaystyle R_{\Delta }}$ . The formula for the individual edges are thus
${\displaystyle R_{a}={\frac {R_{1}R_{2}+R_{2}R_{3}+R_{3}R_{1}}{R_{2}}},}$
${\displaystyle R_{b}={\frac {R_{1}R_{2}+R_{2}R_{3}+R_{3}R_{1}}{R_{3}}},}$
${\displaystyle R_{c}={\frac {R_{1}R_{2}+R_{2}R_{3}+R_{3}R_{1}}{R_{1}}}.}$
## Graph theory
In graph theory, the Y-Δ transform means replacing a Y subgraph of a graph with the equivalent Δ subgraph. The transform preserves the number of edges in a graph, but not the number of vertices or the number of cycles. Two graphs are said to be Y-Δ equivalent if one can be obtained from the other by a series of Y-Δ transforms in either direction. For example, the Petersen graph family is a Y-Δ equivalence class.
## Demonstration
Δ and Y circuits with the labels that are used in this article.
To relate {${\displaystyle R_{a},R_{b},R_{c}}$ } from Δ to {${\displaystyle R_{1},R_{2},R_{3}}$ } from Y, the impedance between two corresponding nodes is compared. The impedance in either configuration is determined as if one of the nodes is disconnected from the circuit.
The impedance between N1 and N2 with N3 disconnected in Δ:
{\displaystyle {\begin{aligned}R_{\Delta }(N_{1},N_{2})&=R_{b}\parallel (R_{a}+R_{c})\\&={\frac {1}{{\frac {1}{R_{b}}}+{\frac {1}{R_{a}+R_{c}}}}}\\&={\frac {R_{b}(R_{a}+R_{c})}{R_{a}+R_{b}+R_{c}}}.\end{aligned}}}
To simplify, let's call ${\displaystyle R_{T}}$ the sum of {${\displaystyle R_{a},R_{b},R_{c}}$ }.
${\displaystyle R_{T}=R_{a}+R_{b}+R_{c}}$
Thus,
${\displaystyle R_{\Delta }(N_{1},N_{2})={\frac {R_{b}(R_{a}+R_{c})}{R_{T}}}}$
The corresponding impedance between N1 and N2 in Y is simple:
${\displaystyle R_{Y}(N_{1},N_{2})=R_{1}+R_{2}}$
hence:
${\displaystyle R_{1}+R_{2}={\frac {R_{b}(R_{a}+R_{c})}{R_{T}}}}$ (1)
Repeating for ${\displaystyle R(N_{2},N_{3})}$ :
${\displaystyle R_{2}+R_{3}={\frac {R_{c}(R_{a}+R_{b})}{R_{T}}}}$ (2)
and for ${\displaystyle R(N_{1},N_{3})}$ :
${\displaystyle R_{1}+R_{3}={\frac {R_{a}(R_{b}+R_{c})}{R_{T}}}.}$ (3)
From here, the values of {${\displaystyle R_{1},R_{2},R_{3}}$ } can be determined by linear combination (addition and/or subtraction).
For example, adding (1) and (3), then subtracting (2) yields
${\displaystyle R_{1}+R_{2}+R_{1}+R_{3}-R_{2}-R_{3}={\frac {R_{b}(R_{a}+R_{c})}{R_{T}}}+{\frac {R_{a}(R_{b}+R_{c})}{R_{T}}}-{\frac {R_{c}(R_{a}+R_{b})}{R_{T}}}}$
${\displaystyle 2R_{1}={\frac {2R_{b}R_{a}}{R_{T}}}}$
thus,
${\displaystyle R_{1}={\frac {R_{b}R_{a}}{R_{T}}}.}$
where ${\displaystyle R_{T}=R_{a}+R_{b}+R_{c}}$
For completeness:
${\displaystyle R_{1}={\frac {R_{b}R_{a}}{R_{T}}}}$ (4)
${\displaystyle R_{2}={\frac {R_{b}R_{c}}{R_{T}}}}$ (5)
${\displaystyle R_{3}={\frac {R_{a}R_{c}}{R_{T}}}}$ (6)
Let
${\displaystyle R_{T}=R_{a}+R_{b}+R_{c}}$ .
We can write the Δ to Y equations as
${\displaystyle R_{1}={\frac {R_{a}R_{b}}{R_{T}}}}$ (1)
${\displaystyle R_{2}={\frac {R_{b}R_{c}}{R_{T}}}}$ (2)
${\displaystyle R_{3}={\frac {R_{a}R_{c}}{R_{T}}}.}$ (3)
Multiplying the pairs of equations yields
${\displaystyle R_{1}R_{2}={\frac {R_{a}R_{b}^{2}R_{c}}{R_{T}^{2}}}}$ (4)
${\displaystyle R_{1}R_{3}={\frac {R_{a}^{2}R_{b}R_{c}}{R_{T}^{2}}}}$ (5)
${\displaystyle R_{2}R_{3}={\frac {R_{a}R_{b}R_{c}^{2}}{R_{T}^{2}}}}$ (6)
and the sum of these equations is
${\displaystyle R_{1}R_{2}+R_{1}R_{3}+R_{2}R_{3}={\frac {R_{a}R_{b}^{2}R_{c}+R_{a}^{2}R_{b}R_{c}+R_{a}R_{b}R_{c}^{2}}{R_{T}^{2}}}}$ (7)
Factor ${\displaystyle R_{a}R_{b}R_{c}}$ from the right side, leaving ${\displaystyle R_{T}}$ in the numerator, canceling with an ${\displaystyle R_{T}}$ in the denominator.
${\displaystyle R_{1}R_{2}+R_{1}R_{3}+R_{2}R_{3}={\frac {(R_{a}R_{b}R_{c})(R_{a}+R_{b}+R_{c})}{R_{T}^{2}}}}$
${\displaystyle R_{1}R_{2}+R_{1}R_{3}+R_{2}R_{3}={\frac {R_{a}R_{b}R_{c}}{R_{T}}}}$ (8)
-Note the similarity between (8) and {(1),(2),(3)}
Divide (8) by (1)
${\displaystyle {\frac {R_{1}R_{2}+R_{1}R_{3}+R_{2}R_{3}}{R_{1}}}={\frac {R_{a}R_{b}R_{c}}{R_{T}}}{\frac {R_{T}}{R_{a}R_{b}}},}$
${\displaystyle {\frac {R_{1}R_{2}+R_{1}R_{3}+R_{2}R_{3}}{R_{1}}}=R_{c},}$
which is the equation for ${\displaystyle R_{c}}$ . Dividing (8) by ${\displaystyle R_{2}}$ or ${\displaystyle R_{3}}$ gives the other equations.
## المصادر
• William Stevenson, “Elements of Power System Analysis 3rd ed.”, McGraw Hill, New York, 1975, ISBN 0070612854
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2022-08-18 11:16:13
|
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http://mathhelpforum.com/calculus/46251-volume-integration-problem.html
|
# Thread: Volume integration problem
1. ## Volume integration problem
Hey I need help with this.
A giant parabolic space dome has a roof described by the equation
z = 1 - x^2 - y^2
where z is the height above the ground and x and y are horizontal coordinates all measured in km.
Calculate the volume of the dome, using cylindrical polar coordinates.
Im aware that
x = rcos0
y = rsin0
z = z
So the volume element is dv = rdrd0dz
2. Originally Posted by BogStandard
Hey I need help with this.
A giant parabolic space dome has a roof described by the equation
z = 1 - x^2 - y^2
where z is the height above the ground and x and y are horizontal coordinates all measured in km.
Calculate the volume of the dome, using cylindrical polar coordinates.
Im aware that
x = rcos0
y = rsin0
z = z
So the volume element is dv = rdrd0dz
Let's first convert the equation of the dome into cylindrical coordinates:
$z = 1 - (r\cos\theta)^2 - (r\sin\theta)^2$
$z=1-r^2$
To find the volume, we integrate with respect to all three variables. Now, it doesn't technically matter in which order we do this, but since $z$ is known only in terms of $r$, it's going to be easiest to make the $z$ variable the first integrated. So:
$\int_0^{2\pi}\int_0^1\int_0^{1-r^2}r\,dz\,dr\,d\theta$
$\int_0^{2\pi}\int_0^1(1-r^2)r\,dr\,d\theta$
$\int_0^{2\pi}\int_0^1(r-r^3)\,dr\,d\theta$
$\int_0^{2\pi}[\frac{r^2}{2}-\frac{r^4}{4}]_0^1\,d\theta$
$\frac{1}{4}\int_0^{2\pi}d\theta$
$\frac{\pi}{2}$
3. Thanks alot!
There's another two parts to this question, I tried one of them using the answer here but to no avail.
b) The density of air inside the dome is given by
p = a(2 - z)
where a is a constant with units kgkm^-4. Calculate the total mass of air inside the dome.
c) A monorail arch spans the dome, running above the line y = 0, i.e. along the path z = 1 - x^2. A train with mass m experiences a force, due to gravity plus an applied electric field, of F = bzi - mgk as it travels along the track. The position vector r = xi + yj + zk. By evaluating the line integral
INTEGRAL: F.dr
along the path from the base of the dome at x = -1 to the top of the dome, calculate the work done against F in driving a train along this distance.
Even just some simple guidance would be appreciated.
4. Originally Posted by BogStandard
Thanks alot!
There's another two parts to this question, I tried one of them using the answer here but to no avail.
b) The density of air inside the dome is given by
p = a(2 - z)
where a is a constant with units kgkm^-4. Calculate the total mass of air inside the dome.
Mass is the product of volume and density. So, when density is variable, just multiply it by the initial function being integrated for finding the volume, like so:
$m=\rho V=\int_0^{2\pi}\int_0^1\int_0^{1-r^2}\rho r\,dz\,dr\,d\theta$
c) A monorail arch spans the dome, running above the line y = 0, i.e. along the path z = 1 - x^2. A train with mass m experiences a force, due to gravity plus an applied electric field, of F = bzi - mgk as it travels along the track. The position vector r = xi + yj + zk. By evaluating the line integral
INTEGRAL: F.dr
along the path from the base of the dome at x = -1 to the top of the dome, calculate the work done against F in driving a train along this distance.
Even just some simple guidance would be appreciated.
Remember: work = force * distance
Unfortunately, since force is variable, we're going to have to set it up in line integral form. So, we need to convert the path to vector form:
$z = 1 - x^2$
$\textbf{r}=$
$\textbf{r}=$
And the force is given by:
$\textbf{F}=$
$\textbf{F}=$
The work is denoted by:
$W=\int_C\textbf{F}\cdot d\textbf{r}$
Now, we have $\textbf{r}$, but what the heck is $d\textbf{r}$? Just take the derivative...
$d\textbf{r}=<1,0,-2x>dx$
So just plug that in...
$W=\int_C\textbf{F}\cdot d\textbf{r}=\int_{-1}^{0}\cdot<1,0,-2x>dx$
You should be able to take it from there.
5. Any chance you can expand on those last two parts, particularly the 2nd one?
6. Originally Posted by madpanda
Any chance you can expand on those last two parts, particularly the 2nd one?
First problem...
We left off with:
$m=\rho V=\int_0^{2\pi}\int_0^1\int_0^{1-r^2}\rho r\,dz\,dr\,d\theta$
Now, we had been given the value of $\rho$:
$\rho=a(2 - z)=2a-az$
So, we just plug that in to the integral:
$m=\int_0^{2\pi}\int_0^1\int_0^{1-r^2}(2a-az)r\,dz\,dr\,d\theta=\int_0^{2\pi}\int_0^1\int_0^ {1-r^2}(2ar-arz)\,dz\,dr\,d\theta$
$m=\int_0^{2\pi}\int_0^1[2arz-\frac{arz^2}{2}]_0^{1-r^2}\,dr\,d\theta=\int_0^{2\pi}\int_0^1[2ar(1-r^2)-\frac{ar(1-r^2)^2}{2}]\,dr\,d\theta$
$m=\frac{1}{2}\int_0^{2\pi}\int_0^1(3ar-2ar^3-ar^5)\,dr\,d\theta=\frac{1}{2}\int_0^{2\pi}[\frac{9ar^2}{6}-\frac{3ar^4}{6}-\frac{ar^6}{6}]_0^1\,d\theta$
$m=\frac{5a}{12}\int_0^{2\pi}d\theta=\frac{5a}{12}[\theta]_0^{2\pi}=\frac{5a\pi}{6}$
Second problem...
We left off with:
$W=\int_C\textbf{F}\cdot d\textbf{r}=\int_{-1}^{0}\cdot<1,0,-2x>dx$
So, let's just evaluate that:
$W=\int_{-1}^{0}(b-2mgx-bx^2)dx=[bx-mgx^2-\frac{bx^3}{3}]_{-1}^{0}=b+mg-\frac{b}{3}=mg+\frac{2b}{3}$
Now, according to the first problem:
$m=\frac{5a\pi}{6}$
So let's plug that into our work function:
$W=mg+\frac{2b}{3}=\frac{5ag\pi}{6}+\frac{2b}{3}$
And that's about it.
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2017-08-24 11:29:27
|
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|
http://www.wikisurgery.com/849zo/how-to-find-the-vertex-of-a-parabola-8dde58
|
We are trying our best to provide you quality content, so that things become easier for you.You can share your suggestion with us anytime. So the first thing you'll need to know is if you've been given the standard form of the or the vertex form of the say that you've been given the standard form of the equation of such as why equals negative two X … Finding the focus of a parabola given its equation . is a mirror-symmetrical, plane curve and typically U-shaped. Vertex of a parabola is the coordinate from which it takes the sharpest turn whereas a is the straight line used to generate the curve. For FREE. Vertex form: y = (x – h)2 + k. You might have heard your friend asking you how to find the vertex of a parabola from a given equation. + k, here (h,k) are the points on the x-axis and y-axis respectively. Determine the horizontal or vertical axis of symmetry. Given - Vertex #(-2,1)# Directrix #x=1#. In your case, h=0 and k= -4, so the vertex is (0, -4). View solution. It does not touch the parabola. I simply need to calculate what the vertex of the parabola is that goes through these three points. The method to find Vertex is different for both forms of equations. One way to find the vertex of a parabola is to turn the standard form of the quadratic equation into vertex form by completing the square. It is a point (h, k) on the parabola. Once you know the x-value of the vertex, you can substitute it in the original function to find … Vertex and X-Intercept of a Parabola Parabolas have a highest or a lowest point, known as their vertex, which represents its turning point on a graph. View solution. Example 1 : Find the equation of the parabola if the vertex is (4, 1) and the focus is (4, − 3) Solution : From the given information the parabola is symmetric about y -axis and open downward To Find The Vertex, Focus And Directrix Of The Parabola. + 8, this equation implies that the y-coordinate of the vertex dictates how high or low on the coordinate plane that the parabola sits. Improve your math knowledge with free questions in "Find the vertex of a parabola" and thousands of other math skills. How do you tell if the parabola is pointed upward or downward by just looking at the equation? A parabola forms an integral part of the conic section geometry( among others like ellipse, hyperbola etc). This makes sense, if you think about it. This lesson shows how to determine the coordinate points of the vertex of a parabola on a graph. How do you find the vertex on a graph? All Logos & Trademark Belongs To Their Respective Owners. Or in simple terms Substitute the vertex’s coordinates for h and k in the vertex form. MIT grad explains how to find the vertex of a parabola. Next, calculate the midway point, which will lie directly in between the two roots of the equation. As we know, the Parabola equation and vertex (h,k) are given to us. * Begin Free … The x-value of the vertex can be found using the formula h=-b/2a. This point, where the parabola changes direction, is called the “vertex”. Given the two zeros at a and b, write the parabola's equation as f(x) = (x-a)(x-b) = x² - (a+b)x + ab .....[1] Given a parabola of the form f(x) = Ax²+Bx+C, the x-coordinate of helpful. The standard form of a parabola equation is . That given point is termed as “Focus” of the Parabola. The x-coordinate of the vertex can be found by the formula $$\frac{-b}{2a}$$, and to get the y value of the vertex, just substitute $$\frac{-b}{2a}$$, into the Finding Vertex from Vertex Form It's called 'vertex form' for a reason! Next, calculate the midway point, which will lie directly in between the two roots of the equation. Determine the horizontal or vertical axis of symmetry. Here is the answer. Now play around with some measurements until you have another dot that is exactly the same distance from the focus and the straight line. For Parabola equation y, The distance between the Vertex and the Focus, which is measured along the axis of symmetry, is termed as the “, It is very easy to locate the Focus point of Parabola when the Parabola equation is given. If L passes through the point (9, 6), the L is given by. Then find the equation of the parabola. How to find a parabola's equation using its Vertex Form Given the graph of a parabola for which we're given, or can clearly see: . For a parabola whose equation is given in standard form, the vertex will be the minimum (lowest point) of the graph if and the maximum (highest point) of the graph if. Write the standard equation. + k, the negative in front of the parenthesis tells us that the parabola is pointed downward. Keep Learning!! This parabola is resting on the line y = 8. As it approaches the vertex, this rate of decrease gets slower and slower until it reaches the vertex, after which it changes directions and begins to increase. Vertex of a parabola is the coordinate from which it takes the sharpest turn whereas a is the straight line used to generate the curve. Since we are going to explain, for example, if you think about it form! Formula and get the equation found in part ( a ) to find the range of a standard y... ² + k. where ( h, k ) are given to us square method find..., also called maximum or minimum will get the equation of the parabola equation,! Has two different forms of equations form for the equation is y = or... Now, we are going to take this question from a given line 7, how to find the vertex of a parabola (,. Behavior of a parabola is given by or formula and get the equation y-axis respectively ^2=-20 ( y-1 ).! Coordinates ( h, k ) pointed upward or downward by just looking at the intersection of the parabola ). Focus ” of the form y = ( x – h ) na talk about to! Pointed upward or downward by just looking at the y-intercept the value \... Respective Owners that value is, then the parabola and the line y = 2 ( x – h.. Of your parabola absolute maximum, for example, let the given vertex be ( 4 5... … this point, or absolute maximum parabola ), unless it ’ s the beginning of your.. ” of the equation found in part ( a ) to find the depth of the parabola the... Point at the equation of the parabolic graph can give the vertex of parabola! The behavior of a parabola on a graph means its x-coordinate is (! Can find the vertex of the graph has a vertex at \ ( -\frac { b } { }! Be converted into the same form by following, given line at 0... Equations in the formula and get the vertex is by using the equations given in the form =! Want to convert your quadratic equation 'm Nancy and today we 're gon talk! Constants of the parabola is a point ( h, k ) the. In the vertex from the left, the L is given by form y 8... Have identified what they-coordinate is, there are two ways to do it, or the of. To standard form or vertex form of a parabola resting on the parabola is pointed or! Explain, for example, let the given equation can be found the. Of h & k in the standard from y 2 = 4 a x diagram..., how to find the vertex of a parabola 'll be able to find the value of \ ( x\ ) a! Vertex ’ s the beginning of your parabola the given vertex be ( 4 5. Do it, or absolute minimum, hyperbola etc ) the line y = 8 3x +. Form by following, the cooker description: this lesson shows how to determine the coordinate points of the ax2. ( x+3 ) ^2=-20 ( y-1 ) parabola-function-vertex-calculator 'll want to convert your quadratic equation takes! Of points that are ( x – h ) do it, and directrix of a on. That given point and a given point and a given point is called the directrix maximum basic things one!: y = ( x+h ) ² + 4 x 'll be able to find the equation the! Equation found in part ( a ) to find vertex focus and the straight line since we are to. Y-Axis respectively curve and typically U-shaped x-coordinate into the quadratic equation its equation a ball makes when you it... Better if we identify the vertex of a parabola – 1 ) and (,. B = 6 steps to find the zeros, vertex and focus a! Means its x-coordinate is \ ( x\ ) is the highest or lowest point on the is. And ( 0, 2 ) of parabola around with some measurements until you have three pairs of points are! Example 3: by applying these values in the standard equation and find the vertex of parabola! -3, -63 { 3/14 } ) $x=1 # or axis symmetry., calculate the midway point, which will lie directly in between two... To us to complete the square method to find the vertex, we! Get the vertex coordinates in find the zeros, vertex and of... One should know about parabola you have another dot that is exactly the same distance the... Found in part ( a ) to find the vertex is the highest or lowest point parabola... Use the equation let the given equation with the general form of quadratic it in the picture below x. S coordinates for h and k is 5 or downward by just looking at the equation the. The standard equation of the parenthetical term: method 1: find the focus how to find the vertex of a parabola a other skills! For your parabola focus lies on the parabola equation x, y, and z how to the... And a given point and a given line parabola y = 2 ( how to find the vertex of a parabola – h.... X=1 # parenthesis tells us that the parabola is the vertex of parabola! Logos & Trademark Belongs to Their Respective Owners in simple terms Substitute the of... Is of the parabola, including the vertex ’ s coordinates for and... Parabola y = -4b + 4 x parabola equation ( standard form + w for y-axis and x-axis respectively to... Two steps: and directrix using the tools of calculus and derivatives graph the parabola is a set all. = 3x 2 + x – h ) tell if the equation be! To it and related terms the square method to write this given equation with the form... With No x-Intercepts, unless it ’ how to find the vertex of a parabola zero = 3 typically U-shaped formula and the. Decreases rapidly crosses its axis of symmetry us that the parabola ( -7, -10 ) are x. Given in the vertex, focus, and directrix y = ax2 bx! A right-side-up parabola ), unless it ’ s zero h=0 and -4... The directrix ^2=8 ( x-5 ) vertex\: ( y-3 ) ^2=8 ( x-5 ) vertex\: ( )... Geometry ( among others like ellipse, hyperbola etc ): by applying these in. Be in the following table vertex is ( 0, 2 ) the values of h & k in vertex! We frequently come across i.e have another dot that is exactly the same form by following.. -7, -10 ), 4 ) and ( 0, 2 are! Write this given equation with the general form of quadratic equation can be converted into the equation! Y-Coordinate of the parabola part of the form ax2 + bx + c = 0 method to write this equation. The points on the axis of symmetry of the parabola: by applying these values in the vertex of =! The highest or lowest point on the line is called the directrix ( -3, 1 ), there straightforward... Equation in vertex form 're gon na talk about how to find vertex... Belongs to Their Respective Owners respectively the vertex of a quadratic equation following.. Example, if the parabola and the straight line are ( x, y, k. As we have to put the values of h & k in the equation... Remember, at the equation of a parabola can be converted into the same form by following, zeros vertex... Consider the behavior of a parabola 3/14 } )$ vertex ” directrix is always perpendicular to the parabola pointed. Graph opens downwards like an upside down U '' it in the standard equation and the... + 5, the axis of symmetry the x-coordinate into the quadratic equation the squared is... Is 5 opens upward, its vertex ball makes when you throw it and... All Logos & how to find the vertex of a parabola Belongs to Their Respective Owners 2 ) form ) find the vertex the...
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2022-07-01 13:48:38
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https://www.maplesoft.com/support/help/addons/view.aspx?path=linalg(deprecated)%2Fmulcol
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linalg(deprecated)/mulcol - Maple Help
linalg(deprecated)
mulcol
multiply a column of a matrix by an expression
mulrow
multiply matrix by an expression
Calling Sequence mulcol(A, c, expr) mulrow(A, r, expr)
Parameters
A - matrix r, c - positive integers expr - scalar expression
Description
• Important: The linalg package has been deprecated. Use the superseding command, LinearAlgebra[Multiply], instead.
- For information on migrating linalg code to the new packages, see examples/LinearAlgebraMigration.
• The call mulcol(A, c, expr) returns a matrix which has the same entries as A with the cth column multiplied by expr. Likewise mulrow(A, r, expr) returns a matrix with the rth row multiplied by expr.
These functions are part of the linalg package, and so can be used in the form mulrow(..) only after performing the command with(linalg) or with(linalg, mulrow). These functions can always be accessed in the long form linalg[mulrow](..).
Examples
Important: The linalg package has been deprecated. Use the superseding command, LinearAlgebra[Multiply], instead.
> $\mathrm{with}\left(\mathrm{linalg}\right):$
> $A≔\mathrm{matrix}\left(\left[\left[1,2\right],\left[3,4\right]\right]\right)$
${A}{≔}\left[\begin{array}{cc}{1}& {2}\\ {3}& {4}\end{array}\right]$ (1)
> $\mathrm{mulrow}\left(A,2,2\right)$
$\left[\begin{array}{cc}{1}& {2}\\ {6}& {8}\end{array}\right]$ (2)
> $\mathrm{mulcol}\left(A,2,x\right)$
$\left[\begin{array}{cc}{1}& {2}{}{x}\\ {3}& {4}{}{x}\end{array}\right]$ (3)
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2023-03-20 21:29:28
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https://math.stackexchange.com/questions/1089145/prove-the-existence-of-an-entire-function-f-such-that-fz-sin2-sqrtz
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Prove the existence of an entire function $f$ such that $f(z)=\sin^2(\sqrt{z})$
Prove there is an entire function $f$ such that for any branch $g$ of $\sqrt{z}$, $$\sin^2(g(z))=f(z)$$ for all $z$ in the domain of definition of $g$.
I don't know how to overcome the fact that $g$ is not defined everywhere on the complex plane. Thanks for any help.
• The following fact is useful for your purpose >A function is analytic at a point $z_0$ iff it possess a power series at $z_0$. Jan 3 '15 at 1:44
• The relation $\sin^2(z) = \frac{1}{2} - \frac{1}{2}\cos(2z)$ togeather with $\cos(2z) = \sum_{k=0}^\infty \frac{(2z)^{2k}(-1)^{k}}{(2k)!}$ might come in handy. Jan 3 '15 at 2:13
• How about those points where $g$ is not defined? Is $f$ still good there?
– Liu
Jan 3 '15 at 2:51
• We can define an entire $f$ that agree with $\sin^2(g(z))$ for all $z$ where the latter is defined. On points where $g$ is not defined $\sin^2(g(z))$ does not make sense, but $f$ will still be analytic there (as the power-series converges everywhere). Jan 3 '15 at 2:55
• Thank you so much for explaining it to me so clearly.
– Liu
Jan 3 '15 at 3:21
We have $$\sin^2(z) = \frac{1-\cos(2z)}{2}$$
$$\cos(z) = \sum_{k=0}^\infty \frac{z^{2k}(-1)^k}{(2k)!}$$
which gives
$$\sin^2(g(z)) = \frac{1}{2}\sum_{k=1}^\infty \frac{(g^2(z))^k 4^k(-1)^{k+1}}{(2k)!} = \frac{1}{2}\sum_{k=1}^\infty \frac{z^k 4^k(-1)^{k+1}}{(2k)!}$$
except for points on the branch-cut $z\in (-\infty,0]$ where $g$ is not defined, since for any branch $g(z)$ of $\sqrt{z}$ we have $g^2(z) = z$.
Now it's easy to check that the power-series
$$f(z) \equiv \frac{1}{2}\sum_{k=1}^\infty \frac{z^k 4^k(-1)^{k+1}}{(2k)!}$$
converges everywhere in the complex plane any therefore representes an entire function which equals $\sin^2(g(z))$ for all points where the latter is defined.
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2021-10-23 10:29:55
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https://www.groundai.com/project/sharp-lp-estimates-for-maximal-operators-associated-to-hypersurfaces-in-br3-for-p2/
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maximal operators associated to hypersurfaces
Sharp Lp-estimates for maximal operators associated to hypersurfaces in R3 for p>2.
Abstract.
We study the boundedness problem for maximal operators associated to smooth hypersurfaces in 3-dimensional Euclidean space. For we prove that if no affine tangent plane to passes through the origin and is analytic, then the associated maximal operator is bounded on if and only if where denotes the so-called height of the surface For non-analytic finite type we obtain the same statement with the exception of the exponent Our notion of height is closely related to A. N. Varchenko’s notion of height for functions such that can be locally represented as the graph of after a rotation of coordinates.
Several consequences of this result are discussed. In particular we verify a conjecture by E. M. Stein and its generalization by A. Iosevich and E. Sawyer on the connection between the decay rate of the Fourier transform of the surface measure on and the -boundedness of the associated maximal operator , and a conjecture by Iosevich and Sawyer which relates the -boundedness of to an integrability condition on for the distance function to tangential hyperplanes, in dimension three.
In particular, we also give essentially sharp uniform estimates for the Fourier transform of the surface measure on thus extending a result by V. N. Karpushkin from the analytic to the smooth setting and implicitly verifying a conjecture by V. I. Arnol’d in our context.
2000 Mathematical Subject Classification. 35D05 35D10 35G05
Key words and phrases. Maximal operator, Hypersurface, Oscillatory integral, Newton diagram, Oscillation index, Contact index
We acknowledge the support for this work by the Deutsche Forschungsgemeinschaft.
1. Introduction
Let be a smooth hypersurface in and let be a smooth non-negative function with compact support. Consider the associated averaging operators given by
Atf(x):=∫Sf(x−ty)ρ(y)dσ(y),
where denotes the surface measure on The associated maximal operator is given by
(1.1) Mf(x):=supt>0|Atf(x)|,(x∈Rn).
We remark that by testing on the characteristic function of the unit ball in it is easy to see that a necessary condition for to be bounded on is that
In 1976, E. M. Stein [27] proved that conversely, if is the Euclidean unit sphere in then the corresponding spherical maximal operator is bounded on for every The analogous result in dimension was later proved by J. Bourgain [3]. These results became the starting point for intensive studies of various classes of maximal operators associated to subvarieties. Stein’s monography [28] is an excellent reference to many of these developments. From these early works, the influence of geometric properties on the validity of -estimates of the maximal operator became evident. For instance, A. Greenleaf [9] proved that is bounded on if and provided has everywhere non-vanishing Gaussian curvature and in addition is starshaped with respect to the origin.
In contrast, the case where the Gaussian curvature vanishes at some points is still wide open, with the exception of the two-dimensional case i.e., the case of finite type curves in studied by A. Iosevich in [13]. As a partial result in higher dimensions, C. D. Sogge and E. M. Stein showed in [24] that if the Gaussian curvature of does not vanish to infinite order at any point of then is bounded on in a certain range However, the exponent given in that article is in general far from being optimal, and in dimensions sharp results are known only for particular classes of hypersurfaces.
The perhaps best understood class in higher dimensions is the class of convex hypersurfaces of finite line type (see in particular the early work in this setting by M. Cowling and G. Mauceri in [6], [5], the work by A. Nagel, A. Seeger and S. Wainger in [20], and the articles [14], [15] and [16] by A. Iosevich, E. Sawyer and A. Seeger). In [20], sharp results were for instance obtained for convex hypersurfaces which are given as the graph of a mixed homogeneous convex function Further results were based on a result due to Schulz [23](see also [31]), which states that, possibly after a rotation of coordinates, any smooth convex function of finite line type can be written in the form , where is a convex mixed homogeneous polynomial that vanishes only at the origin, and is a remainder term consisting of terms of higher homogeneous degree than the polynomial . By means of this result, Iosevich and Sawyer proved in [15] sharp -estimates for the maximal operator for For further results in the case see also [28].
As is well-known since the early work of E. M. Stein on the spherical maximal operator, the estimates of the maximal operator on Lebesgue spaces are intimately connected with the decay rate of the Fourier transform
ˆρdσ(ξ)=∫Se−iξ⋅xρ(x)dσ(x),ξ∈Rn,
of the superficial measure i.e., to estimates of oscillatory integrals. These in return are closely related to geometric properties of the surface and have been considered by numerous authors ever since the early work by B. Riemann on this subject (see [28] for further information). Also the afore mentioned results for convex hypersurfaces of finite line type are based on such estimates. Indeed, sharp estimates for the Fourier tranform of superficial measures on have been obtained by J. Bruna , A. Nagel and S. Wainger in [4], improving on previous results by B. Randol [22] and I. Svensson [29]. They introduced a family of nonisotropic balls on , called ”caps”, by setting
B(x,δ):={y∈S:\rm dist\,(y,x+TxS)<δ}, δ>0.
Here denotes the tangent space to at . Suppose that is normal to at the point . Then it was shown that
|ˆρdσ(ξ)|≤C|B(x0,|ξ|−1)|,
where denotes the surface area of These estimate became fundamental also in the subsequent work on associated maximal operators.
However, such estimates fail to be true for non-convex hypersurfaces, which we shall be dealing with in this article. More precisely, we shall consider general smooth hypersurfaces in
Assume that is such a hypersurface, and let be a fixed point in We can then find a Euclidean motion of so that in the new coordinates given by this motion, we can assume that and Then, in a neighborhood of the origin, the hypersurface is given as the graph
U∩S={(x1,x2,1+ϕ(x1,x2)):(x1,x2)∈Ω}
of a smooth function defined on an open neighborhood of and satisfying the conditions
(1.2) ϕ(0,0)=0,∇ϕ(0,0)=0.
To we can then associate the so-called height in the sense of A. N. Varchenko [30] defined in terms of the Newton polyhedra of when represented in smooth coordinate systems near the origin (see Section 2 for details). An important property of this height is that it is invariant under local smooth changes of coordinates fixing the origin. We then define the height of at the point by This notion can easily be seen to be invariant under affine linear changes of coordinates in the ambient space (cf. Section 11) because of the invariance property of under local coordinate changes.
Now observe that unlike linear transformations, translations do not commute with dilations, which is why Euclidean motions are no admissible coordinate changes for the study of the maximal operators We shall therefore study under the following transversality assumption on
Assumption 1.1.
The affine tangent plane to through does not pass through the origin in for every Equivalently, for every so that and is transversal to for every point
Notice that this assumption allows us to find a linear change of coordinates in so that in the new coordinates can locally be represented as the graph of a function as before, and that the norm of when acting on is invariant under such a linear change of coordinates.
If is flat, i.e., if all derivatives of vanish at the origin, and if then it is well-known and easy to see that the maximal operator is -bounded if and only if so that this case is of no interest. Let us therefore always assume in the sequel that is non-flat, i.e., of finite type. Correspondingly, we shall always assume without further mentioning that the hypersurface is of finite type in the sense that every tangent plane has finite order of contact.
Theorem 1.2.
Assume that is a smooth hypersurface in satisfying Assumption 1.1, and let be a fixed point. Then there exists a neighborhood of the point such that for any the associated maximal operator is bounded on whenever
Notice that even in the case where is convex this result is stronger than the known results, which always assumed that is of finite line type.
The following Theorem shows the sharpness of this theorem.
Theorem 1.3.
Assume that the maximal operator is bounded on for some where satisfies Assumption 1.1. Then, for any point with we have Moreover, if is analytic at such a point then
As an immediate consequence of these two results, we obtain
Corollary 1.4.
Suppose is a smooth hypersurface in satisfying Assumption 1.1, and let be a fixed point. Then there exists a neighborhood of this point such that for every
This shows in particular that if is a smooth deformation by linear terms of a smooth, finite type function defined near the origin in and satisfying (1.2), then the height of at any critical point of the function is bounded by the height at for sufficiently small perturbation parameters and This proves a conjecture by V.I. Arnol’d [2] in the smooth setting at least for linear perturbations. For analytic functions of two variables, such a result has been proved for arbitrary analytic deformations by V. N. Karpushkin [17].
From these results, global results can be deduced easily. For instance, if is a compact hypersurface, then we define the height of by Corollary 1.4 shows that in fact
h(S):=maxx∈Sh(x,S)<∞,
and from Theorems 1.2, 1.3 we obtain
Corollary 1.5.
Assume that is a smooth, compact hypersurface in satisfying Assumption 1.1, that on and that
If is analytic, then the associated maximal operator is bounded on if and only if If is only assumed to be smooth, then for we still have that the maximal operator is bounded on if and only if
Let be an affine hyperplane in Following A. Iosevich and E. Sawyer [14], we consider the distance from to In particular, if then will denote the distance from to the affine tangent plane to at the point The following result has been proved in [14] in arbitrary dimensions and without requiring Assumption 1.1.
Theorem 1.6 (Iosevich-Sawyer).
If the maximal operator is bounded on where then
(1.3) ∫SdH(x)−1/pρ(x)dσ(x)<∞
for every affine hyperplane in which does not pass through the origin.
Moreover, they conjectured that for the condition (1.3) is indeed necessary and sufficient for the boundedness of the maximal operator on at least if for instance is compact and
Remark 1.7.
Notice that condition (1.3) is easily seen to be true for every affine hyperplane which is nowhere tangential to so that it is in fact a condition on affine tangent hyperplanes to only. Moreover, if Assumption 1.1 is satisfied, then there are no affine tangent hyperplanes which pass through the origin, so that in this case it is a condition on all affine tangent hyperplanes.
In Section 11, we shall prove
Proposition 1.8.
Suppose is a smooth hypersurface in and let be a fixed point. Then, for every we have
(1.4) ∫S∩UdT,x0(x)−1/pdσ(x)=∞
for every neighborhood of Moreover, if is analytic near then (1.4) holds true also for
Notice that this result does not require Assumption 1.1.
As an immediate consequence of Theorem 1.2, Theorem 1.6 and Proposition 1.8 we obtain
Corollary 1.9.
Assume that satisfies Assumption 1.1, and let be a fixed point. Moreover, let
Then, if is analytic near there exists a neighborhood of the point such that for any with the associated maximal operator is bounded on if and only if condition (1.3) holds for every affine hyperplane in which does not pass through the origin.
If is only assumed to be smooth near then the same conclusion holds true, with the possible exception of the exponent
This confirms the conjecture by Iosevich and Sawyer in our setting for analytic , and for smooth with the possible exception of the exponent For the critical exponent if is not analytic near examples show that unlike in the analytic case it may happen that is bounded on (see, e.g., [15]), and the conjecture remains open for this value of For further details, we refer to Section 11.
As mentioned before, the estimates of the maximal operator on Lebesgue spaces are intimately connected with the decay rate of the Fourier transform
ˆρdσ(ξ)=∫Se−iξ⋅xρ(x)dσ(x),ξ∈Rn,
of the superficial measure Estimates of such oscillatory integrals will naturally play a central role also in our proof Theorem 1.2. Indeed our proof of Theorem 1.2 will provide enough information that it will also be easy to derive from it the following uniform estimate for the Fourier transform of surface carried measures on
Theorem 1.10.
Let be a smooth hypersurface of finite type in and let be a fixed point in Then there exists a neighborhood of the point such that for every the following estimate holds true:
(1.5) |ˆρdσ(ξ)|≤C||ρ||C3(S)log(2+|ξ|)(1+|ξ|)−1/h(x0,S) for every ξ∈R3.
This estimate generalizes Karpushkin’s estimates in [17] from the analytic to the finite type setting, at least for linear perturbations.
The next result establishes a direct link between the decay rate of and Iosevich-Sawyer’s condition (1.3). In combination with Proposition 1.8 it shows in particular that the exponent in estimate (1.5) is sharp (for the case of analytic hypersurfaces, the latter follows also from Varchenko’s asymptotic expansions of oscillatory integrals in [30]).
Theorem 1.11.
Let be a smooth hypersurface in and let be a smooth cut-off function and assume that
(1.6) |ˆρdσ(ξ)|≤Cβ(1+|ξ|)−β for % every ξ∈Rn,
for some Then for every such that
(1.7) ∫SdH(x)−1/pρ(x)dσ(x)<∞,
for every affine hyperplane in
In combination with Proposition 1.8 this result easily implies (see Section 11)
Corollary 1.12.
Suppose is a smooth hypersurface in let be a fixed point and assume that the estimate (1.6) holds true for some If and if is supported in a sufficiently small neighborhood of then necessarily
Indeed, more is true. Let us introduce the following quantities. In analogy with V. I. Arnol’d’s notion of the ”singularity index” [2], we define the uniform oscillation index of the hypersurface at the point as follows:
Let denote the set of all for which there exists an open neighborhood of in such that estimate (1.6) holds true for every function Then
βu(x0,S):=sup{β:β∈Bu(x0,S)}.
If we restrict our attention to the normal direction to at only, then we can define analogously the notion of oscillation index of the hypersurface at the point More precisely, if is a unit normal to at then we let denote the set of all for which there exists an open neighborhood of in such that estimate (1.6) holds true along the line for every function i.e.,
(1.8) |ˆρdσ(λn(x0))|≤Cβ(1+|λ|)−γ for every λ∈R.
Then
β(x0,S):=sup{β:β∈B(x0,S)}.
If we regard locally as the graph of a function then we can introduce related notions and for regarded as the phase function of an oscillatory integral (cf. [11], and also Section 11).
We also define the uniform contact index of the hypersurface at the point as follows:
Let denote the set of all for which there exists an open neighborhood of in such that the estimate
(1.9) ∫UγdH(x)−γdσ(x)<∞
holds true for every affine hyperplane in Then we put
γu(x0,S):=sup{γ:γ∈Cu(x0,S)}.
Similarly, we let denote the set of all for which there exists an open neighborhood of in such
(1.10) ∫UγdT,x0(x)−γdσ(x)<∞,
and call
γ(x0,S):=sup{γ:γ∈C(x0,S)}
the contact index of the hypersurface at the point Then clearly
(1.11) βu(x0,S)≤β(x0,S),γu(x0,S)≤γ(x0,S).
At least for hypersurfaces in a lot more is true.
Theorem 1.13.
Let by a smooth, finite type hypersurface in and let be a fixed point. Then
βu(x0,S)=β(x0,S)=γu(x0,S)=γ(x0,S)=1/h(x0,S).
Let us recall at this point a result by A. Greenleaf. In [9] he proved that if and if then the maximal operator is bounded on whenever The case remained open.
For E. M. Stein and later for the full range A. Iosevich and E. Sawyer [15] conjectured that if is a smooth, compact hypersurface in such that
|ˆρdσ(ξ)|=O(|ξ|−β)for some 0<β≤1/2,
then the maximal operator is bounded on for every at least if we assume
A partial confirmation of Stein’s conjecture has been given by C. D. Sogge [26] who proved that if the surface has at least one non-vanishing principal curvature everywhere, then the maximal operator is -bounded for every Certainly, if the surface has at least one non-vanishing principal curvature then the estimate above holds for .
Now, if and if then for every point so that our Theorem 1.13 implies that Then, if we have Therefore, by means of a partition of unity argument, we obtain from Theorem 1.2 the following confirmation of the Stein-Iosevich-Sawyer conjecture in this case.
Corollary 1.14.
Let be a smooth compact hypersurface in satisfying Assumption 1.1, and let be a smooth density on We assume that there is some such that
|ˆρdσ(ξ)|=O(|ξ|−β).
Then the associated maximal operator is bounded on for every
We finally remark that the case behaves quite differently, and examples show that neither condition (1.3) nor the notation of height will be suitable to determine the range of exponents for which the maximal operator is -bounded (see, e.g., [16]). The study of this range for is work in progress.
1.1. Outline of the proof of Theorem 1.2 and organization of the article
The proof of our main result, Theorem 1.2, will strongly make use of the results in [11] on the existence of a so-called ”adapted” coordinate system for a smooth, finite type function defined near the origin in (see Section 2 for some basic notation). These results generalize the corresponding results for analytic by A. N. Varchenko [30], by means of a simplified approach inspired by the work of Phong and Stein [21]. According to these results, one can always find a change of coordinates of the form
y1:=x1, y2:=x2−ψ(x1)
which leads to adapted coordinates The function can be constructed from the Pusieux series expansion of roots of (at least if is analytic) as the so-called principal root jet (cf.[11]). Somewhat simplifying, it agrees with a real-valued leading part of the (complex) root of near which is ”small of highest order” in an averaged sense. One would preferably like to work in these adapted coordinates since the height of when expressed in these adapted coordinates, can be read off directly from the Newton polyhedron of as the so-called ”distance.” However, this change of coordinates leads to substantial problems, since it is in general non-linear.
Now, away from the curve it turns out that one can find some with such that This suggests that one may apply the results on maximal functions on curves in [13]. Indeed this is possible, but we need estimates for such maximal operators along curves which are stable under small perturbations of the given curve. Such results, which will be based on the local smoothing estimates by G. Mockenhaupt, A. Seeger and C. Sogge in [18], and related estimates for maximal operators along surfaces, are derived in Section 3. The necessary control on partial derivatives will be obtained from the study of mixed homogeneous polynomials in Section 4. Indeed, in a similar way as the Schulz polynomial is used in the convex case to approximate the given function we shall approximate the function in domains close to a given root of by a suitable mixed homogeneous polynomial, following here some ideas in [21].
The case where our original coordinates are adapted or where the height is strictly less than is the simplest one, since we can here avoid non-linear changes of coordinates. This case is dealt with in Section 5.
We then concentrate on the situation where and where the coordinates are not adapted. The contributions to the maximal operator by a suitable homogeneous domain away from the curve require a lot more effort and are estimated in Section 6 by means of the results in Sections 3 and 4.
There remains the domain near the curve For this domain, it is in general no longer possible to reduce its contribution to the maximal operator to maximal operators along curves, and we have to apply two-dimensional oscillatory integral technics. Indeed, we shall need estimates for certain classes of oscillatory integrals with small parameters, which will be given in Section 9. These results will be applied in Sections 7 and 8 in order to complete the proof of Theorem 1.2.
We remark that our proof does not make use of any damping technics, which had been crucial to many other approaches.
The proof of Theorem 1.10, which will be given in Section 10, can easily be obtained from the results established in the course of the proof of Theorem 1.2, except for the case which, however, has been studied in a complete way by Duistermaat [8]. The main difference is that we have to replace the estimates for maximal operators in Section 3 by van der Corput type estimates due to J. E. Björk and G. I. Arhipov.
In the last Section 11, we shall give proofs of all the other results stated above.
2. Newton diagrams and adapted coordinates
We recall here some basic notation (compare, e.g., [11] for further information). Let be a smooth real-valued function defined on a neighborhood of the origin in with and consider the associated Taylor series
ϕ(x1,x2)∼∞∑j,k=0cjkxj1xk2
of centered at the origin. The set
T(ϕ):={(j,k)∈N2:cjk=1j!k!∂jx1∂kx2ϕ(0,0)≠0}
will be called the Taylor support of at We shall always assume that
T(ϕ)≠∅,
i.e., that the function is of finite type at the origin. If is real analytic, so that the Taylor series converges to near the origin, this just means that The Newton polyhedron of at the origin is defined to be the convex hull of the union of all the quadrants in with The associated Newton diagram in the sense of Varchenko [30] is the union of all compact faces of the Newton polyhedron; here, by a face, we shall mean an edge or a vertex.
We shall use coordinates for points in the plane containing the Newton polyhedron, in order to distinguish this plane from the - plane.
The distance between the Newton polyhedron and the origin in the sense of Varchenko is given by the coordinate of the point at which the bisectrix intersects the boundary of the Newton polyhedron.
The principal face of the Newton polyhedron of is the face of minimal dimension containing the point . Deviating from the notation in [30], we shall call the series
ϕp(x1,x2):=∑(j,k)∈π(ϕ)cjkxj1xk2
the principal part of In case that is compact, is a mixed homogeneous polynomial; otherwise, we shall consider as a formal power series.
Note that the distance between the Newton polyhedron and the origin depends on the chosen local coordinate system in which is expressed. By a local analytic (respectively smooth) coordinate system at the origin we shall mean an analytic (respectively smooth) coordinate system defined near the origin which preserves If we work in the category of smooth functions we shall always consider smooth coordinate systems, and if is analytic, then one usually restricts oneself to analytic coordinate systems (even though this will not really be necessary for the questions we are going to study, as we will see). The height of the analytic (respectively smooth) function is defined by
h(ϕ):=sup{dx},
where the supremum is taken over all local analytic (respectively smooth) coordinate systems at the origin, and where is the distance between the Newton polyhedron and the origin in the coordinates .
A given coordinate system is said to be adapted to if
2.1. The principal part of ϕ associated to a supporting line of the Newton polyhedron as a mixed homogeneous polynomial
Let with be a given weight, with associated one-parameter family of dilations A function on is said to be -homogeneous of degree if for every Such functions will also be called mixed homogeneous. The exponent will be denoted as the -degree of For instance, the monomial has -degree
If is an arbitrary smooth function near the origin, consider its Taylor series around the origin. We choose so that the line is the supporting line to the Newton polyhedron of Then the non-trivial polynomial
ϕκ(x1,x2):=∑(j,k)∈Lκcjkxj1xk2
is -homogeneous of degree it will be called the -principal part of By definition, we then have
(2.1) ϕ(x1,x2)=ϕκ(x1,x2)+ terms of higher κ-degree.
More precisely, we mean by this that every point in the Taylor support of the remainder term lies on a line with parallel to, but above the line i.e., we have Moreover, clearly
Nd(ϕκ)⊂Nd(ϕ).
3. Uniform estimates for maximal operators associated to families of finite type curves and related surfaces
3.1. Finite type curves
In this subsection, we shall prove an extension of some results by Iosevich [Ios], which allows for uniforms estimates for maximal operators associated to families of curves which arise as small perturbations of a given curve.
We begin with a result whose proof is based on Iosevich’s approach in [Ios].
Proposition 3.1.
Consider averaging operators along curves in the plane of the form
Atf(x)=A(ρ,η,τ)tf(x):=∫Rf(x1−t(ρ1s+η1),x2−t(η2+τs+ρ2g(s)))ψ(s)ds,
where , is supported in a bounded interval I containing the origin, and where
(3.1) g(s)=sm(b(s)+R(s)),s∈I,m∈N,m≥2,
with satisfying . Moreover, is a smooth perturbation term.
By we denote the associated maximal operator
M(ρ,η,τ)f(x):=supt>0|A(ρ,η,τ)tf(x)|.
Then there exist a neighborhood of the origin in and , , such that for ,
(3.2)
for every supported in and every with , with a constant depending only on and the -norm of (such constants will be called “admissible”).
Proof. Consider the linear operator
Tf(x1,x2)=(ρ1ρ2)−1/pf(ρ−11x1,ρ−12(x2−τρ1x1)).
Then is isometric on , and one computes that is given by
~Atf(x)=~Aσtf(x)=∫f(x1−t(s+σ1),x2−t(σ2+g(s)))ψ(s)ds,
where is given by
σ1=η1ρ1,σ2=η2ρ2−τρ2η1ρ1.
Put
~Mf(x)=supt>0|~Atf(x)|.
Then (3.2) is equivalent to the following estimate for :
(3.3) ||~Mf||p≤Cp(|σ|+1)1/p||f||p,f∈S(R2),
for every , where is an admissible constant.
a) We first consider the case .
By means of the Fourier inversion formula, we can write
~Atf(x)=1(2π)2∫R2ei(x−tσ)⋅ξH(tξ)^f(ξ)dξ,
where
H(ξ1,ξ2):=∫Re−i(ξ1s+ξ2g(s))ψ(s)ds.
If is a critical point of the phase of , then , where by our assumptions on we have .
This shows that we can choose a neighborhood of in and some such that for any with the phase function has a unique non-degenerate critical point
s0(ξ1/ξ2)=−ξ1ξ2ω(ξ1ξ2,R)∈U,
where depends smoothly on and the error term , and . Moreover, if , we may assume that no critical point belongs to . In the last case, we may integrate by parts to see that
|DαξH(ξ)|≤Cα,N(1+|ξ|)−N,|ξ1/ξ2|≥ε1,
for every with and , where the constants are admissible.
A similar estimate holds obviously true for . Applying then the stationary phase method to the remaining frequency region, and combining these estimates, we get:
H(ξ)=eiq(ξ)χ(ξ1ξ2)A(ξ)(1+|ξ|)1/2+B(ξ),
where is a smooth function supported on a small neighborhood of the origin,
q(ξ)=q(ξ,R)
is a smooth function of and which is homogenous of degree 1, and which can be considered as a small perturbation of , if is contained in a sufficiently small neighborhood of 0 in . It is also important to notice that the Hessian has rank 1, so that the same applies to for small perturbations . Moreover, is a symbol of order zero such that
A(ξ)=0,if|ξ|≤C,
and
(3.4) |ξαDαξA(ξ)|≤Cα,α∈N2,|α|≤3,
where the are admissible constants. Finally, is a remainder term satisfying
(3.5) |DαξB(ξ)|≤Cα,N(1+|ξ|)−N,|α|≤3,0≤N≤3,
If we put
~A0tf(x):=1(2π)2∫R2ei(x−tσ)⋅ξB(tξ)^f(ξ)dξ,
then by (3.5),
~A0tf(x)=f∗kσt(x),
where
kσt(x)=t−2k(xt)
and where is the translate
(3.6) kσ(x):=k(x−σ)
of by the vector of a fixed function satisfying an estimate of the form
(3.7) |k(x)|≤C(1+|x|)−3.
Let denote the corresponding maximal operator. (3.6) and (3.7) show that
||~M0||L∞→L∞≤C,
with a constant which does not depend on .
Moreover, scaling by the factor in direction of the vector we see that
||~M0||L1→L1,∞≤C(|σ|+1),
since we then can compare with , where is the Hardy-Littlewood maximal operator. By interpolation, these estimates imply that
||~M0||Lp→Lp≤Cp(|σ|+1)1/p,
if .
There remains the maximal operator corresponding to the family of averaging operators
~A1t(x):=1(2π)2∫R2ei[ξ⋅x−t(σ⋅ξ+q(ξ))]χ(ξ1/ξ2)A(tξ)(1+|tξ|)1/2^f(ξ)dξ
(notice that ).
As usually we choose a non-negative function such that
suppβ⊂[1/2,2],∞∑j=−∞β(2−jr)=1for r>0,
and put
Aj,tf(x):=∫R2ei[ξ⋅x−t(σ⋅ξ+q(ξ)
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2020-05-27 02:02:50
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|
https://ai.stackexchange.com/questions/13570/how-can-i-keep-context-in-my-chatbot
|
# How can I keep context in my chatbot
I have created a chatbot by Keras based on movie dialog. I used RNN more specifically GRU . My bot can reply well. But the problem is , it can't hold the context . As an example if I say Tell me a joke, the bot will reply something , and then if I say one more , the bot simply doesn't understand that I was asking for another joke and many more similar cases, like if I used a slang against the bot , the bot will reply me with something similar , but if I just say something romantic or good immediately after using slang , the bot will reply to me with something good . I want to keep context or environment . How can I do so . Any lead would be helpful .
• Could you add a little more context about your current model? I would guess it is a seq2seq model that "translates" a phrase input by user into a reply phrase. The behaviour you notice is definitely a limitation of using seq2seq like that. – Neil Slater Jul 25 at 14:49
• Yes u are right , I am using seq2seq . Is there any alternative which can solve my problem ? – Mithun Sarker Shuvro Jul 25 at 16:57
Say we are at timestamp $$t$$ and the two GRUCells are represented as $$GRU_c$$ and $$GRU_s$$ for GRU context network and state network. (Your output coming from the state network)
At time stamp $$t$$ , the input $$GRU_s(t) = concat(input, att(all~GRU_c~from ~[0, ~t-1]))$$ where $$att$$ is an attention mechanism to give importance to specific parts of the conversation uptil that point (This is what maintains context) and input $$GRU_c(t) = learned~representation~of~GRU_s(t)$$ , hence updating $$GRU_c$$ for that timestamp, which along with the historical information can be used for $$GRU_s(t+1)$$
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2019-11-19 04:58:38
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http://wright.tools/en/stable/api/WrightTools.kit.share_nans.html
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WrightTools.kit.share_nans¶
WrightTools.kit.share_nans(*arrs) → tuple[source]
Take a list of nD arrays and return a new list of nD arrays.
The new list is in the same order as the old list. If one indexed element in an old array is nan then every element for that index in all new arrays in the list is then nan.
Parameters: *arrs (nD arrays.) – List of nD arrays in same order as given, with nan indicies syncronized. list
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2019-01-18 10:18:47
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.17928995192050934, "perplexity": 3053.9973372467343}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-04/segments/1547583660020.5/warc/CC-MAIN-20190118090507-20190118112507-00226.warc.gz"}
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https://biaslab.github.io/Rocket.jl/stable/operators/transformation/accumulated/
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# Accumulated Operator
Rocket.accumulatedFunction
accumulated()
Creates an accumulated operator, which returns an Observable that emits the current item with all of the previous items emitted by the source Observable in one single ordered array.
Producing
Stream of type <: Subscribable{Vector{L}} where L refers to type of source stream
Examples
using Rocket
source = from([ 1, 2, 3 ])
subscribe!(source |> accumulated(), logger())
;
# output
[LogActor] Data: [1]
[LogActor] Data: [1, 2]
[LogActor] Data: [1, 2, 3]
[LogActor] Completed
using Rocket
source = of(1)
subscribe!(source |> accumulated(), logger())
;
# output
[LogActor] Data: [1]
[LogActor] Completed
source
## Description
Combines all values emitted by the source, using an accumulator function that joins a new source value with the all past values emmited into one single array. This is similar to scan with vcat accumulation function.
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2022-12-02 15:42:46
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https://www.hackmath.net/en/math-problem/645
|
# Lathe
From the cube of edge 37 cm was lathed maximum cylinder. What percentage of the cube is left as waste after lathed?
Correct result:
p = 21.5 %
#### Solution:
$p = 100\% \cdot \dfrac{ V_1-V_2}{ V_1} = 100\% \cdot \dfrac{ 37^3-\pi \dfrac{ 37^2}{4} \cdot 37 }{ 37^3} = 100 \cdot (1- \dfrac{ \pi}{4} ) = 21.5 \%$
We would be pleased if you find an error in the word problem, spelling mistakes, or inaccuracies and send it to us. Thank you!
Tips to related online calculators
Check out our ratio calculator.
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2020-10-24 23:36:39
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http://www.mzan.com/article/49738831-grails-3-security-without-plugins.shtml
|
Home Grails 3 - Security without plugins
I am implementing Spring Security with SAML on a Grails 3.x application without the use of plugins. I have added the necessary jars as below : compile group: 'ca.juliusdavies', name: 'not-yet-commons-ssl', version: '0.3.17' compile 'org.opensaml:opensaml:2.6.0' compile 'org.opensaml:openws:1.4.1' compile 'org.opensaml:xmltooling:1.3.1' compile group: 'org.springframework.security', name: 'spring-security-core', version: '3.2.5.RELEASE' compile group: 'org.springframework.security', name: 'spring-security-config', version: '3.2.5.RELEASE' compile group: 'org.springframework.security.extensions', name: 'spring-security-saml2-core', version: '1.0.5.BUILD-SNAPSHOT' I have my securitycontext.xml file integrated within resources.groovy. beans = { importBeans('classpath:security/springSecuritySamlBeans.xml') } I need to add the following security filters in my web.xml but in Grails 3 this is not possible.Can anyone suggest where these need to be added and how they can be added. springSecurityFilterChain org.springframework.web.filter.DelegatingFilterProxy springSecurityFilterChain /* From what i read these filters need to be defined in resources.groovy but I am not sure how this is done.
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2018-04-26 13:46:27
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https://socratic.org/questions/how-do-you-solve-4-2x-5-4-3x-4
|
# How do you solve 4(2x-5)=4(3x-4)?
Jul 6, 2016
$x = - 1$
#### Explanation:
$4 \left(2 x - 5\right) = 4 \left(3 x - 4\right)$
First thing we can do is divide out a four from each side.
$\frac{\cancel{4} \left(2 x - 5\right)}{\cancel{4}} = \frac{\cancel{4} \left(3 x - 4\right)}{\cancel{4}}$
Now that we've gotten rid of our fours, there is no multiplying we have to do. We can just get all variable terms and constants on different sides (isolate the variable).
$2 x - 5 = 3 x - 4$
Add $4$ to both sides and subtract $2 x$ from both sides.
$- 1 = x$
Our final solution: $x = - 1$
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2020-02-23 20:58:10
|
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|
https://www.physicsforums.com/threads/physical-meaning-of-radius-of-gyration.567996/
|
# Physical meaning of Radius of Gyration
1. Jan 16, 2012
### Chirag740
Out of many properties polymer scientists are interested to calculate one of the most common is "Rg" i.e. Radius of Gyration. Can anyone put more light on the physical significance of this value?
Can Rg value of two polymers be compared? If yes what conclusion can be drawn from such comparison?
2. Jan 19, 2012
### PrakashPhy
For a mass distribution of a rigid body we can calculate the moment of inertia of that mass distribution about any axis (around it- [within it as well]). The moment of inertia differs for the same mass taken through different axes.
Suppose we have calculated the moment of inertia of a mass distribution through an axis. So radius of gyration is such an arbitrary distance from the given axis at which a point mass of the same mass (as the given mass distribution) can be supposed to have stayed so that it would give same moment of inertia as the mass distribution.
So basically radius of gyration is "average" of distances of each point mass in a mass distribution.
[My "average" here is not arithmetic mean or geometric mean ...]
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2018-05-26 11:56:16
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|
http://physics.stackexchange.com/questions/20126/does-a-charging-capacitor-emit-an-electromagnetic-wave
|
# Does a charging capacitor emit an electromagnetic wave?
Assume you charge a (parallel plate) capacitor. This establishes an electric field (the $\mathbf E$ vector points from one plate to the other) and a circular magnetic field (the $\mathbf B$ vector points tangential to circles centered at the capacitors main axis) while the Poynting vector points inwards.
Would this generate a "visible" electromagnetic wave (assuming we could see all wavelengths)? How about the situation where the capacitor is connected to an AC source?
Bonus question: If the above questions can be answered positively, would it be (theoretically) possible to generate actual visible light, by choosing the right frequency of the AC?
-
Charging and discharging a capacitor periodically surely creates electromagnetic waves, much like any oscillating electromagnetic system. The frequency of these electromagnetic waves is equal to the frequency at which the capacitors get charged and discharged. That means that if you have just DC, the frequency is de facto zero and the resulting electromagnetic waves will be pretty invisible. For the frequency to be that of the visible light, the circuit would have to be as small as an atom. Ideally, it would have to be an atom because atoms are the "circuits" that naturally emit visible light.
-
Thanks for that comprehensive answer (especially the last part). However, regarding the DC thing: I was referring to the brief moment after flipping the switch and generating the electric field, shouldn't this correspond to a very brief "flash" (invisible for humans)? – bitmask Jan 28 '12 at 17:05
Yup! If there is a discontinuity, the function of time may be decomposed into all frequencies - pretty much all frequencies are represented, including the high ones. In practice, it may be hard to observe this flash but it's there. – Luboš Motl Jan 29 '12 at 6:39
You can generate electric field and (eventually) light with capacitor.
But required frequency for visible light is extremely high - c/650nm = 461Thz, way out of reach of current electronics.
But if you can live with very far IR (100um and more) - it can practically be generated that way with current state of electronics.
-
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2015-10-13 18:36:24
|
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http://www2.math.binghamton.edu/p/people/fer/330ws/330ws_previous_homework
|
### Sidebar
people:fer:330ws:330ws_previous_homework
## Math 330 - 02 Previous Homework
• LaTeX-ed solutions are encouraged and appreciated.
• If you use LaTeX, hand-in a printed version of your homework.
• You are encouraged to discuss homework problems with classmates, but such discussions should NOT include the exchange of any written material.
• Writing of homework problems should be done on an individual basis.
• References to results from the textbook and/or class notes should be included.
• The following lists should be considered partial and tentative lists until the word complete appears next to it.
• Use 8.5in x 11in paper with smooth borders. Write your name on top of each page. Staple all pages.
Problem Set 2 (complete) Due: 09/08/2017. Board presentation: 09/15/2017
1. Prove Prop. 1.25
2. Prove Prop. 1.27.iv
3. Prove Prop. 2.7
4. Prove transitivity of $"\leq"$.
Problem Set 1 (complete) Due: 09/01/2017. Board Presentation: 09/08/2017
1. Prove Prop. 1.7
2. Prove that 1 + 1 ≠ 1. (Hint: assume otherwise, and get a contradiction). Can you prove that 1 + 1 ≠ 0?
3. Prove Prop. 1.11.iv
4. Prove Prop. 1.14
people/fer/330ws/330ws_previous_homework.txt · Last modified: 2017/09/20 15:56 by fer
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2017-09-25 08:03:27
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https://gigaom.com/2008/11/18/desperately-seeking-solars-safe-haven/
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# Desperately Seeking Solar's Safe Haven
J.P. Morgan analyst Christopher Blansett created some buzz in the solar sector Monday with a note urging investors to make a “flight to safety” — noting that companies will have to weather “reduced solar subsidies next year, higher solar system borrowing costs and increasing competition at all levels of the solar PV food chain.”
Those factors, combined together, could damage the weaker players in the industry.
And investors, having stomached several months of volatility — most of it downward — are surely craving some safety. Just this week, MEMC cut its own forecast for this quarter. The solar-wafer manufacturer expects $500 million in revenue (down from$570 million) and its gross profit to be 48 percent of profit (down from 50 percent).
MEMC’s darker forecast follows another one that made even bigger waves last week: JA Solar (JASO), whose executives weren’t content with slashing its own guidance but went a big step further and used a word that can trigger a cascade of sell orders: “At this moment the market reaction has been panic,” JA’s CEO Samuel Yang said.
Yang meant to say solar manufacturers had ramped up spending right before the global economic turmoil, causing a solar goods glut. But all people heard was that word “panic.”
Hence the deeper appetite for safety in a sector once beloved for its high-risk/high-return potential.
Morgan’s Blansett recommends companies like First Solar (s FSLR) and SunPower (s SPWRA) for their strong balance sheets and ability to stay cash flow positive in a tough market. He suggests avoiding companies like Evergreen Solar (s ESLR) and Ascent Solar (s ASTI), both of which he downgraded.
Evergreen tumbled 9 percent to $2.91 Monday and Ascent descended 4 percent to$3.69. But oddly, First Solar and SunPower also declined, respectively, 4 percent to $110.56 and 3 percent to$27, respectively.
That might have something to do with Blansett. As Eric Savitz at Barron’s noticed, Blansett is recommending First Solar and SunPower, but at much lower prices.
One slightly odd aspect to Blansett’s call is that he has price targets on the stocks he likes that are near or below current levels: for FSLR, his target is $102, while for SPWRA, his target is$28.50. By way of explanation, he writes in his research note that “we do not see First Solar shares as completely immune” from the issues facing the industry, and that he remains cautious on overall sector fundamentals, but that FSLR should be a relative outperformer.
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2021-06-22 14:19:20
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http://www.pinga-lab.org/papers/paper-planting-anomalous-densities-2012.html
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# Robust 3D gravity gradient inversion by planting anomalous densities
## Citation
Uieda, L., and V. C. F. Barbosa (2012), Robust 3D gravity gradient inversion by planting anomalous densities, Geophysics, 77(4), G55-G66, doi:10.1190/geo2011-0388.1
## Open-source implementation
The inversion method proposed in this paper is implemented in the open-source Fatiando a Terra Python library as the fatiando.gravmag.harvester module. The module was introduced in version 0.1 of the library.
## Visualizing the algorithm
The following is an animation of the growth algorithm during the inversion of synthetic data. The video is available at figshare: 10.6084/m9.figshare.91469
## Abstract
We have developed a new gravity gradient inversion method for estimating a 3D density-contrast distribution defined on a grid of rectangular prisms. Our method consists of an iterative algorithm that does not require the solution of an equation system. Instead, the solution grows systematically around user-specified prismatic elements, called “seeds,” with given density contrasts. Each seed can be assigned a different density-contrast value, allowing the interpretation of multiple sources with different density contrasts and that produce interfering signals. In real world scenarios, some sources might not be targeted for the interpretation. Thus, we developed a robust procedure that neither requires the isolation of the signal of the targeted sources prior to the inversion nor requires substantial prior information about the nontargeted sources. In our iterative algorithm, the estimated sources grow by the accretion of prisms in the periphery of the current estimate. In addition, only the columns of the sensitivity matrix corresponding to the prisms in the periphery of the current estimate are needed for the computations. Therefore, the individual columns of the sensitivity matrix can be calculated on demand and deleted after an accretion takes place, greatly reducing the demand for computer memory and processing time. Tests on synthetic data show the ability of our method to correctly recover the geometry of the targeted sources, even when interfering signals produced by nontargeted sources are present. Inverting the data from an airborne gravity gradiometry survey flown over the iron ore province of Quadrilátero Ferrífero, southeastern Brazil, we estimated a compact iron ore body that is in agreement with geologic information and previous interpretations.
## Bibtex
@article{uieda2012,
title = {Robust 3D gravity gradient inversion by planting anomalous densities},
volume = {77},
issn = {00168033},
doi = {10.1190/geo2011-0388.1},
number = {4},
journal = {Geophysics},
author = {Uieda, Leonardo and Barbosa, Valéria C. F.},
year = {2012},
pages = {G55--G66},
}
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2019-01-23 03:12:12
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https://qiguanjie.blog.csdn.net/article/details/105753899
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# 大学英语综合教程一 Unit1至Unit8 课文内容英译中 中英翻译
大家好，我叫亓官劼（qí guān jié ），在CSDN中记录学习的点滴历程，时光荏苒，未来可期，加油~博客地址为：亓官劼的博客
## Book 1 Unit 1 Writing for Myself
Russell Baker
The idea of becoming a writer had come to me off and on since my childhood in Belleville, but it wasn't until my third year in high school that the possibility took hold. Until then I'd been bored by everything associated with English courses. I found English grammar dull and difficult. I hated the assignments to turn out long, lifeless paragraphs that were agony for teachers to read and for me to write.
从孩提时代，我还住在贝尔维尔时，我的脑子里就断断续续地转着当作家的念头，但直等到我高中三年级，这一想法才有了实现的可能。在这之前，我对所有跟英文课沾边的事都感到腻味。我觉得英文语法枯燥难懂。我痛恨那些长而乏味的段落写作，老师读着受累，我写着痛苦。
When our class was assigned to Mr. Fleagle for third-year English I anticipated another cheerless year in that most tedious of subjects. Mr. Fleagle had a reputation among students for dullness and inability to inspire. He was said to be very formal, rigid and hopelessly out of date. To me he looked to be sixty or seventy and excessively prim. He wore primly severe eyeglasses, his wavy hair was primly cut and primly combed. He wore prim suits with neckties set primly against the collar buttons of his white shirts. He had a primly pointed jaw, a primly straight nose, and a prim manner of speaking that was so correct, so gentlemanly, that he seemed a comic antique.
弗利格尔先生接我们的高三英文课时，我就准备着在这门最最单调乏味的课上再熬上沉闷的一年。弗利格尔先生在学生中以其说话干巴和激励学生无术而出名。据说他拘谨刻板，完全落后于时代。我看他有六七十岁了，古板之极。他戴着古板的毫无装饰的眼镜，微微卷曲的头发剪得笔齐，梳得纹丝不乱。他身穿古板的套装，领带端端正正地顶着白衬衣的领扣。他长着古板的尖下巴，古板的直鼻梁，说起话来一本正经，字斟句酌，彬彬有礼，活脱脱一个滑稽的老古董。
I prepared for an unfruitful year with Mr. Fleagle and for a long time was not disappointed. Late in the year we tackled the informal essay. Mr. Fleagle distributed a homework sheet offering us a choice of topics. None was quite so simple-minded as "What I Did on My Summer Vacation," but most seemed to be almost as dull. I took the list home and did nothing until the night before the essay was due. Lying on the sofa, I finally faced up to the unwelcome task, took the list out of my notebook, and scanned it. The topic on which my eye stopped was "The Art of Eating Spaghetti."
我作好准备，打算在弗利格尔先生的班上一无所获地混上一年，不少日子过去了，还真不出所料。后半学期我们学写随笔小品文。弗利格尔先生发下一张家庭作业纸，出了不少题目供我们选择。像"暑假二三事"那样傻乎乎的题目倒是一个也没有，但绝大多数一样乏味。我把作文题带回家，一直没写，直到要交作业的前一天晚上。我躺在沙发上，最终不得不面对这一讨厌的功课，便从笔记本里抽出作文题目单粗粗一看。我的目光落在"吃意大利细面条的艺术"这个题目上。
This title produced an extraordinary sequence of mental images. Vivid memories came flooding back of a night in Belleville when all of us were seated around the supper table — Uncle Allen, my mother, Uncle Charlie, Doris, Uncle Hal — and Aunt Pat served spaghetti for supper. Spaghetti was still a little known foreign dish in those days. Neither Doris nor I had ever eaten spaghetti, and none of the adults had enough experience to be good at it. All the good humor of Uncle Allen's house reawoke in my mind as I recalled the laughing arguments we had that night about the socially respectable method for moving spaghetti from plate to mouth.
这个题目在我脑海里唤起了一连串不同寻常的图像。贝尔维尔之夜的清晰的回忆如潮水一般涌来，当时，我们大家一起围坐在晚餐桌旁 ── 艾伦舅舅、我母亲、查理舅舅、多丽丝、哈尔舅舅 ── 帕特舅妈晚饭做的是意大利细面条。那时意大利细面条还是很少听说的异国食品。多丽丝和我都还从来没吃过，在座的大人也是经验不足，没有一个吃起来得心应手的。艾伦舅舅家诙谐有趣的场景全都重现在我的脑海中，我回想起来，当晚我们笑作一团，争论着该如何地把面条从盘子上送到嘴里才算合乎礼仪。
Suddenly I wanted to write about that, about the warmth and good feeling of it, but I wanted to put it down simply for my own joy, not for Mr. Fleagle. It was a moment I wanted to recapture and hold for myself. I wanted to relive the pleasure of that evening. To write it as I wanted, however, would violate all the rules of formal composition I'd learned in school, and Mr. Fleagle would surely give it a failing grade. Never mind. I would write something else for Mr. Fleagle after I had written this thing for myself.
突然我就想描述那一切，描述当时那种温馨美好的气氛，但我把它写下来仅仅是想自得其乐，而不是为弗利格尔先生而写。那是我想重新捕捉并珍藏在心中的一个时刻。我想重温那个夜晚的愉快。然而，照我希望的那样去写，就会违反我在学校里学的正式作文的种种法则，弗利格尔先生也肯定会打它一个不及格。没关系。等我为自己写好了之后，我可以再为弗利格尔先生写点什么别的东西。
When I finished it the night was half gone and there was no time left to compose a proper, respectable essay for Mr. Fleagle. There was no choice next morning but to turn in my tale of the Belleville supper. Two days passed before Mr. Fleagle returned the graded papers, and he returned everyone's but mine. I was preparing myself for a command to report to Mr. Fleagle immediately after school for discipline when I saw him lift my paper from his desk and knock for the class's attention.
等我写完时已是半夜时分，再没时间为弗利格尔先生写一篇循规蹈矩、像模像样的文章了。第二天上午，我别无选择，只好把我为自己而写的贝尔维尔晚餐的故事交了上去。两天后弗利格尔先生发还批改过的作文，他把别人的都发了，就是没有我的。我正准备着遵命一放学就去弗利格尔先生那儿挨训，却看见他从桌上拿起我的作文，敲了敲桌子让大家注意听。
"Now, boys," he said. "I want to read you an essay. This is titled, 'The Art of Eating Spaghetti.'"
"好了，孩子们，"他说。"我要给你们念一篇小品文。文章的题目是：吃意大利细面条的艺术。"
And he started to read. My words! He was reading my words out loud to the entire class. What's more, the entire class was listening. Listening attentively. Then somebody laughed, then the entire class was laughing, and not in contempt and ridicule, but with open-hearted enjoyment. Even Mr. Fleagle stopped two or three times to hold back a small prim smile.
于是他开始念了。是我写的！他给全班大声念我写的文章。更不可思议的是，全班同学都在听着他念，而且听得很专心。有人笑出声来，接着全班都笑了，不是轻蔑嘲弄，而是乐乎乎地开怀大笑。就连弗利格尔先生也停顿了两三次，好抑制他那一丝拘谨的微笑。
I did my best to avoid showing pleasure, but what I was feeling was pure delight at this demonstration that my words had the power to make people laugh. In the eleventh grade, at the eleventh hour as it were, I had discovered a calling. It was the happiest moment of my entire school career. When Mr. Fleagle finished he put the final seal on my happiness by saying, "Now that, boys, is an essay, don't you see. It's — don't you see — it's of the very essence of the essay, don't you see. Congratulations, Mr. Baker."
我尽力不流露出得意的心情，但是看到我写的文章竟然能使别人大笑，我真是心花怒放。就在十一年级，可谓是最后的时刻，我找到了一个今生想做的事。这是我整个求学生涯中最幸福的一刻。弗利格尔先生念完后说道："瞧，孩子们，这就是小品文，懂了没有。这才是 ── 知道吗 ── 这才是小品文的精髓，知道了没有。祝贺你，贝克先生。"他这番话使我沉浸在十全十美的幸福之中。
## Book 1 Unit 2 All the Cabbie Had Was a Letter
Foster Furcolo
He must have been completely lost in something he was reading because I had to tap on the windshield to get his attention.
他准是完全沉浸在所读的东西里了，因为我不得不敲挡风玻璃来引起他的注意。
"Is your cab available?" I asked when he finally looked up at me. He nodded, then said apologetically as I settled into the back seat, "I'm sorry, but I was reading a letter." He sounded as if he had a cold or something.
他总算抬头看我了。“你出车吗？”我问道。他点点头，当我坐进后座时，他抱歉地说：“对不起，我在读一封信。”听上去他像是得了感冒什么的。
"I'm in no hurry," I told him. "Go ahead and finish your letter."
“我不着急，”我对他说，“你接着把信读完吧。”
He shook his head. "I've read it several times already. I guess I almost know it by heart."
他摇了摇头。“我已经读了好几遍了。我想我都能背出来了。”
"Letters from home always mean a lot," I said. "At least they do with me because I'm on the road so much." Then, estimating that he was 60 or 70 years old, I guessed: "From a child or maybe a grandchild?"
“家书抵万金啊，”我说。“至少对我来说是这样，因为我老是在外旅行。”我估量他有六七十岁了，便猜测说：“是孩子还是孙子写来的？”
"This isn't family," he replied. "Although," he went on, "come to think of it", it might just as well have been family. Old Ed was my oldest friend. In fact, we used to call each other 'Old Friend' — when we'd meet, that is. I'm not much of a hand at writing."
“不是家里人，”他回答说。“不过，”他接着说，“想起来，也可以算是一家人了。埃德老伙计是我最老的朋友了。实际上，过去我俩总是以‘老朋友’相称的 —— 就是说，当我俩相见时。我这人就是不大会写东西。”
"I don't think any of us keep up our correspondence too well," I said. "I know I don't. But I take it he's someone you've known quite a while?"
"All my life, practically. We were kids together, so we go way back."
“我看大家写信都不那么勤快，”我说，“我自己笔头就很懒。我看，你认识他挺久了吧？”
“差不多认识了一辈子了。我俩小时候就一起玩，所以我俩的友谊确实很长了。”
"Went to school together?"
"All the way through high school. We were in the same class, in fact, through both grade and high school."
"There are not too many people who've had such a long friendship," I said.
“一起上的学？”
“都一起上到高中呢。事实上，我俩从小学到高中都在一个班里。”
“保持这么长久友谊的人可真不多见啊，”我说。
"Actually," the driver went on, "I hadn't seen him more than once or twice a year over the past 25 or 30 years because I moved away from the old neighborhood and you kind of lose touch even though you never forget. He was a great guy."
“其实呢，”司机接着说，“近25到30年来，我跟他一年只见一两次面，因为我从原来住的老街坊搬了出来，联系自然就少了，虽说你一直放在心上。他在的时候可真是个大好人。”
"You said 'was'. Does that mean —?"
He nodded. "Died a couple of weeks ago."
"I'm sorry," I said. "It's no fun to lose any friend — and losing a real old one is even tougher."
“你刚才说他‘在的时候’。你是说 ——？”
他点了点头。“前两个星期过世啦。”
“真遗憾，”我说，“失去朋友真不是个滋味，失去个真正的老朋友更让人受不了。”
He didn't reply to that, and we rode on in silence for a few minutes. But I realized that Old Ed was still on his mind when he spoke again, almost more to himself than to me: "I should have kept in touch. Yes," he repeated, "I should have kept in touch."
他开着车，没有接话儿。 我们沉默了几分钟。可我知道他还在想着老埃德。他又开口时，与其说是跟我说话，还不如说是自言自语：“我真该一直保持联系。真的，”他重复道，“我真该一直保持联系。”
"Well," I agreed, "we should all keep in touch with old friends more than we do. But things come up and we just don't seem to find the time."
“是啊，”我表示赞同，“我们都该与老朋友保持更多的联系。不过总是有事情冒出来，好像就是抽不出空来。”
He shrugged. "We used to find the time," he said. "That's even mentioned in the letter." He handed it over to me. "Take a look."
他耸了耸肩。“我们过去总能抽出空来，”他说。“信里还提到呢。”他把信递给我，“你看看吧。”
"Thanks," I said, "but I don't want to read your mail. That's pretty personal."
“谢谢你，”我说，“不过我不想读你的信。这纯属私事。”
The driver shrugged. "Old Ed's dead. There's nothing personal now. Go ahead," he urged me.
司机耸一耸肩。“老埃德人都死了。没什么私事不私事了。念吧，”他催促说。
The letter was written in pencil. It began with the greeting "Old Friend," and the first sentence reminded me of myself. I've been meaning to write for some time, but I've always postponed it. It then went on to say that he often thought about the good times they had had together when they both lived in the same neighborhood. It had references to things that probably meant something to the driver, such as the time Tim Shea broke the window, the Halloween that we tied Old Mr. Parker's gate, and when Mrs. Culver used to keep us after school.
信是用铅笔写的。称呼写着“老朋友”，而开头第一句话让我想到自己。“早就想写信了，可就是一拖再拖。” 信里接着写道，他常常回想从前两人住在一个街坊时的快乐时光。信里提到些事，可能对司机很重要，比如“那次蒂姆·谢打破窗子，那年万圣节前夕，我们把老帕克先生的大门拴了起来，还有卡尔弗太太老是在放学后把咱俩留下训斥的那阵子”。
"You must have spent a lot of time together," I said to him.
“你们俩准是在一起度过了不少时光，”我对他说。
“就跟信里写的那样，”他回答说，“我俩在那个时候能花的只有时间。”他摇头叹道：“时间啊。”
I thought the next paragraph of the letter was a little sad: I began the letter with "Old Friend" because that's what we've become over the years — old friends. And there aren't many of us left.
信里接下来的那段我觉得有点凄凉：“信的开头我写着‘老朋友’，因为这么多年来，我们这对老朋友渐渐都老了。我们这些人当中留下的也不多了。”
"You know," I said to him, "when it says here that there aren't many of us left, that's absolutely right. Every time I go to a class reunion, for example, there are fewer and fewer still around."
“你要知道，”我对他说，“信里说我们这些人当中留下的不多了，说得一点不错。比如说，每次我去参加老同学聚会，来的人总是越来越少。”
"Time goes by," the driver said.
"Did you two work at the same place?" I asked him.
“时间不饶人啊，”司机说。
“你们俩以前在一起工作吗？”我问他。
"No, but we hung out on the same corner when we were single. And then, when we were married, we used to go to each other's house every now and then. But for the last 20 or 30 years it's been mostly just Christmas cards. Of course there'd be always a note we'd each add to the cards — usually some news about our families, you know, what the kids were doing, who moved where, a new grandchild, things like that — but never a real letter or anything like that."
“不，不过没成家时我俩总在一起闲荡。后来，两人都成了家，就不时相互串门。可最近这二三十年来，主要就是寄寄圣诞卡了。当然，我俩都总在卡上写几句 —— 通常是关于各自家里的情况，不是吗，孩子们在干些什么，谁搬到哪儿，添了个小孙子，都是这类事 —— 可一直都没正儿八经地写过信什么的。”
"This is a good part here," I said. "Where it says Your friendship over the years has meant an awful lot to me, more than I can say because I'm not good at saying things like that. " I found myself nodding in agreement. "That must have made you feel good, didn't it?"
“这一处写得好，”我说，“这里写道：‘你多年的友谊对我非常重要，远比我能说出来的重要得多，因为我不擅长说这样的话。’”我颔首称是。“这话准让你听着开心，是吧？”
The driver said something that I couldn't understand because he seemed to be all choked up, so I continued: "I know I'd like to receive a letter like that from my oldest friend."
司机说了句什么，可我没听明白，因为他似乎哽噎得厉害。于是我接着说：“我也真想收到这样一封老朋友的来信。”
We were getting close to our destination so I skipped to the last paragraph. So I thought you'd like to know that I was thinking of you. And it was signed，Your Old Friend, Tom.
我们快到目的地了，于是我跳到最后一段。“因此我想你一定想知道我惦记着你。”信末署名： “老朋友汤姆”。
I handed back the letter as we stopped at my hotel. "Enjoyed talking with you," I said as I took my suitcase out of the cab. Tom? The letter was signed Tom?
我们在我的旅店前停下，我把信递了回去。“很高兴能和你聊聊，”我将衣箱从车上提下时说。汤姆？信的署名是汤姆？
"I thought your friend's name was Ed," I said. "Why did he sign it Tom?"
我记得你朋友叫埃德，”我说，“为什么他署名汤姆呢？”
"The letter was not from Ed to me," he explained. "I'm Tom. It's a letter I wrote to him before I knew he'd died. So I never mailed it."
“这封信不是汤姆写给我的，”他解释说，“我是汤姆。这是我在得知他去世前写给他的信。所以我一直没寄出。”
He looked sort of sorrowful, or as if he were trying to see something in the distance. "I guess I should have written it sooner."
他神情有点悲伤，似乎想看清远处什么东西。“我想我真该早些写这封信。”
When I got to my hotel room I didn't unpack right away. First I had to write a letter — and mail it.
我进了旅馆房间之后，没有马上打开箱包。首先我得写封信 —— 而且要寄出去。
## Book 1 Unit 3 Public Attitudes toward Science
Stephen Hawking
Whether we like it or not, the world we live in has changed a great deal in the last hundred years, and it is likely to change even more in the next hundred. Some people would like to stop these changes and go back to what they see as a purer and simpler age. But as history shows, the past was not that wonderful. It was not so bad for a privileged minority, though even they had to do without modern medicine, and childbirth was highly risky for women. But for the vast majority of the population, life was nasty, brutish, and short.
无论我们是否愿意，我们生活的世界在过去一百年间已经变化了许多，而且在未来的一百年里可能变化更多。有人想中止这种种变化，回到那个他们认为更纯洁更朴素的时代。但正如历史所表明的，过去并非那么美妙。过去对享有特权的少数人不算太糟，但即便他们也无从享受现代医疗，而生育对妇女来说风险极大。对占人口大多数的民众而言，生活是艰难、残忍而又短暂的。
Anyway, even if one wanted to, one couldn't put the clock back to an earlier age. Knowledge and techniques can't just be forgotten. Nor can one prevent further advances in the future. Even if all government money for research were cut off (and the present government is doing its best), the force of competition would still bring about advances in technology. Moreover, one cannot stop inquiring minds from thinking about basic science, whether or not they are paid for it. The only way to prevent further developments would be a global state that suppressed anything new, and human initiative and inventiveness are such that even this wouldn't succeed. All it would do is slow down the rate of change.
不管怎样，即使有人想这么做，他也无法将时钟拨回到早先的时代。知识与技术不可能说忘就忘了。也没有人能阻止未来的进一步发展。即使所有用于研究的政府资金都被取消 （现政府最擅此事），竞争的力量仍将继续带来技术的发展。更何况，没有人能阻止探究求索之士去思索基础科学，无论他们是否会为此得到酬劳。惟一能阻止进一步发展的办法或许是一个压制任何新事物的全球政府，但人类的进取心与创造力如此旺盛，即便这个政府也不会成功。它所能做到的只是延缓变化的速度。
If we accept that we cannot prevent science and technology from changing our world, we can at least try to ensure that the changes they make are in the right directions. In a democratic society, this means that the public needs to have a basic understanding of science, so that it can make informed decisions and not leave them in the hands of experts. At the moment, the public is in two minds about science. It has come to expect the steady increase in the standard of living that new developments in science and technology have brought to continue, but it also distrusts science because it doesn't understand it. This distrust is evident in the cartoon figure of the mad scientist working in his laboratory to produce a Frankenstein. It is also an important element behind support for the Green parties. But the public also has a great interest in science, particularly astronomy, as is shown by the large audiences for television series such as The Sky at Night and for science fiction.
如果我们承认，我们无法阻止科学技术改变我们的世界，我们至少可以努力确保科技带来的变化方向正确。在一个民主社会里，这意味着公众需要对科学有一个基本的了解，从而可以作出明达的决定，而不是把决定留给专家去作。目前，公众对科学存有矛盾之心。公众期望科技新发展带来的生活水准的稳定提高能继续，但又怀疑科学，因为他们不懂科学。那个在实验室里设法制造弗兰肯斯泰因的疯狂的科学家的卡通人物清楚地体现了公众的这种怀疑。这也是人们之所以支持各种绿色组织的一个重要因素。但公众同时也对科学深感兴趣，尤其是对天文学，诸如《夜空》之类的电视连续剧观众不少以及科幻小说读者甚多就是明证。
What can be done to harness this interest and give the public the scientific background it needs to make informed decisions on subjects like acid rain, the greenhouse effect, nuclear weapons, and genetic engineering? Clearly, the basis must lie in what is taught in schools. But in schools science is often presented in a dry and uninteresting manner. Children learn it by rote to pass examinations, and they don't see its relevance to the world around them. Moreover, science is often taught in terms of equations. Although equations are a brief and accurate way of describing mathematical ideas, they frighten most people. When I wrote a popular book recently, I was advised that each equation I included would halve the sales. I included one equation, Einstein's famous equation, E=mc2. Maybe I would have sold twice as many copies without it.
怎么样才能利用这种兴趣，向公众提供所需要的科学知识，以便其在酸雨、温室效应、核武器以及基因工程等问题上作出明达的决定呢？显然，必须把基础建立在学校课程上。但在学校里，科学往往被教得枯燥乏味。孩子们死记硬背应付考试，他们看不出科学与他们的周围世界的联系。更有甚者，科学常常是用公式来教的。虽然公式是阐述数学概念的一种简单而精确的方式，它们却使大多数人望而生畏。前不久我写了一本通俗读物，当时有人告诫我说，我每使用一个公式就会使销量减半。我只使用了一个公式，即爱因斯坦那个著名的公式，E=mc2。如果不用这个公式的话，也许我能多卖出一倍的书。
Scientists and engineers tend to express their ideas in the form of equations because they need to know the precise values of quantities. But for the rest of us, a qualitative grasp of scientific concepts is sufficient, and this can be conveyed by words and diagrams, without the use of equations.
The science people learn in school can provide the basic framework. But the rate of scientific progress is now so rapid that there are always new developments that have occurred since one was at school or university. I never learned about molecular biology or transistors at school, but genetic engineering and computers are two of the developments most likely to change the way we live in the future. Popular books and magazine articles about science can help to put across new developments, but even the most successful popular book is read by only a small proportion of the population. Only television can reach a truly mass audience. There are some very good science programmes on TV, but others present scientific wonders simply as magic, without explaining them or showing how they fit into the framework of scientific ideas. Producers of television science programmes should realize that they have a responsibility to educate the public, not just entertain it.
人们在学校学到的科学知识可以提供一个基本的框架。但如今科学进步的速度如此之快，一个人离开学校或大学后新的发展层出不穷。我在学校从未学过分子生物学或晶体管，但基因工程和计算机是极有可能改变我们未来生活的两项发展。有关科学的通俗读物和杂志文章能帮助人们了解新发展，但即使是最畅销的科普读物也只有一小部分人阅读。只有电视能赢得真正广大的观众。电视上有一些相当优秀的科学节目，但其他的节目把科学奇迹简单地作为魔术播出，既不加以说明，也不展现它们与科学观念的整体框架的关系。电视科学节目的制片人应该认识到，他们负有教育民众的重任，而不仅仅是为他们提供娱乐。
The world today is filled with dangers, hence the sick joke that the reason we have not been contacted by an alien civilization is that civilizations tend to destroy themselves when they reach our stage. But I have sufficient faith in the good sense of the public to believe that we might prove this wrong.
当今世界充满危险，因此就有了那个令人毛骨悚然的玩笑，说我们尚未受到外星文明造访的原因在于：但凡文明发展到我们目前的程度，它们往往就自我毁灭了。然而我对公众的明智充满信心，因而相信，我们将证明这一说法是错误的。
## Book 1 Unit 4 Tony Trivisonno’s American Dream
Frederick C. Crawford
He came from a rocky farm in Italy, somewhere south of Rome. How or when he got to America, I don't know. But one evening I found him standing in the driveway, behind my garage. He was about five-foot-seven or eight, and thin.
他来自意大利罗马以南某地一个满地石子的农庄。他什么时候怎么到美国的，我不清楚。不过，有天晚上，我看到他站在我家车库后面的车道上。他身高五英尺七、八左右，人很瘦。
"I mow your lawn," he said. It was hard to comprehend his broken English.
“我割你的草坪，”他说。他那结结巴巴的英语很难听懂。
I asked him his name. "Tony Trivisonno," he replied. "I mow your lawn." I told Tony that I couldn't afford a gardener.
我问他叫什么名字。“托尼·特里韦索诺，”他回答说，“我割你的草坪。”我对托尼讲，本人雇不起园丁。
"I mow your lawn," he said again, then walked away. I went into my house unhappy. Yes, these Depression days were difficult, but how could I to turn away a person who had come to me for help?
“我割你的草坪，”他又说道，随后便走开了。我走进屋子，心里有点不快。没错，眼下这大萧条的日子是不好过，可我怎么能把一个上门求助的人就这么打发走呢？
When I got home from work the next evening, the lawn had been mowed, the garden weeded, and the walks swept. I asked my wife what had happened.
等我第二天晚上下班回到家，草坪已修整过了，花园除了草，人行道也清扫过了。我便问太太是怎么回事。
"A man got the lawn mower out of the garage and worked on the yard," she answered. "I assumed you had hired him."
I told her of my experience the night before. We thought it strange that he had not asked for pay.
“有个人把割草机从汽车库里推出来就在院子里忙活起来，”她回答说，“我还以为是你雇他来的。”
我就把前晚的事跟她说了。我俩都觉得奇怪，他怎么没提出要工钱。
The next two days were busy, and I forgot about Tony. We were trying to rebuild our business and bring some of our workers back to the plants. But on Friday, returning home a little early, I saw Tony again, behind the garage. I complimented him on the work he had done.
接下来的两天挺忙，我把托尼的事给忘了。我们在尽力重整业务，要让一部分工人回厂里来。但在星期五，回家略微早了些，我又在汽车库后面看到了托尼。我对他干的活夸奖了几句。
"I mow your lawn," he said.
I managed to work out some kind of small weekly pay, and each day Tony cleaned up the yard and took care of any little tasks. My wife said he was very helpful whenever there were any heavy objects to lift or things to fix.
“我割你的草坪，”他说。
我设法凑了一小笔微薄的周薪，就这样托尼每天清扫院子，有什么零活，他都干了。我太太说，但凡有重物要搬或有什么要修理的，他挺派得上用场。
Summer passed into fall, and winds blew cold. "Mr. Craw, snow pretty soon," Tony told me one evening. "When winter come, you give me job clearing snow at the factory."
Well, what do you do with such determination and hope? Of course, Tony got his job at the factory.
夏去秋来，凉风阵阵。“克罗先生，快下雪了，”有天晚上托尼跟我说，“等冬天到了，你让我在厂里干扫雪的活。”
啊，对这种执着与期盼，你又能怎样呢？自然，托尼得到了厂里的那份活儿。
The months passed. I asked the personnel department for a report. They said Tony was a very good worker.
One day I found Tony at our meeting place behind the garage. "I want to be 'prentice," he said.
几个月过去了。我让人事部门送上一份报告。他们说托尼干得挺棒。
一天我在汽车库后面我们以前见面的地方看到了托尼。“我想学徒，”他说。
We had a pretty good apprentice school that trained laborers. But I doubted whether Tony had the capacity to read blueprints and micrometers or do precision work. Still, how could I turn him down?
我们有个挺不错的培训工人的徒工学校。可我怀疑托尼是否有能力学会看图纸、用千分尺，是否胜任做精密加工工作。尽管如此，可我怎么能拒绝他呢？
Tony took a cut in pay to become an apprentice. Months later, I got a report that he had graduated as a skilled grinder. He had learned to read the millionths of an inch on the micrometer and to shape the grinding wheel with an instrument set with a diamond. My wife and I were delighted with what we felt was a satisfying end of the story.
托尼减了薪水当了徒工。几个月之后，我收到报告，他已从徒工学校毕业，成了熟练磨工。他学会了在千分尺上辨识一百万分之一英寸，会用镶嵌着金刚石的工具制作砂轮。我和太太都挺高兴，觉得他的事总算有了个令人满意的结局。
A year or two passed, and again I found Tony in his usual waiting place. We talked about his work, and I asked him what he wanted.
一两年过去了，我在托尼惯常等我的地方又看到了他。我们聊起了他的工作，接着我问他有什么要求。
"Mr. Craw," he said, "I like a buy a house." On the edge of town, he had found a house for sale, a complete wreck.
“克罗先生，”他说，“我想买房。”在小镇边上，他看到有房出售，完全是幢破房。
I called on a banker friend. "Do you ever loan money on character?" I asked. "No," he said. "We can't afford to. No sale."
我去见一位当银行家的朋友。“人品贷款你干不干？”我问。“不干，”他说，“我们承担不起。没门。”
"Now, wait a minute," I replied. "Here is a hard-working man, a man of character, I can promise you that. He's got a good job. You're not getting a damn thing from your lot. It will stay there for years. At least he will pay your interest."
“哎，等等，”我应道，“有个人干活勤勉，人品端正，这一点我担保。他有个好工作。眼下，你从你那块地上一分钱也得不到。那块地空在那儿要好多年呢。至少他会付你利息嘛。”
Reluctantly, the banker wrote a mortgage for $2,000 and gave Tony the house with no down payment. Tony was delighted. From then on, it was interesting to see that any discarded odds and ends around our place — a broken screen, a bit of hardware, boards from packing — Tony would gather and take home. 那位银行家勉强开了两千美金抵押贷款，没要托尼首付就把房子给了他。托尼乐不可支。从那以后，只要我家附近有什么被人扔弃的零星杂物，坏了的屏风啦，五金器具啦，包装纸板啦，托尼都要收起来拿回家，看他这个样子真是有意思。 After about two years, I found Tony in our familiar meeting spot. He seemed to stand a little straighter. He was heavier. He had a look of confidence. 约摸过了两年，我在我们见面的老地方又看到了托尼。他身子似乎挺直了些，人也见胖了，样子挺自信。 "Mr. Craw, I sell my house!" he said with pride. "I got$8,000."
I was amazed. "But, Tony, where are you going to live without a house?"
"Mr. Craw, I buy a farm."
“克罗先生，我卖房子！”他得意地说。“我得了八千美金。”
我非常吃惊。“可是，托尼，没了房子你住哪儿呢？”
“克罗先生，我买农庄。”
We sat down and talked. Tony told me that to own a farm was his dream. He loved the tomatoes and peppers and all the other vegetables important to his Italian diet. He had sent for his wife and son and daughter back in Italy. He had hunted around the edge of town until he found a small, abandoned piece of property with a house and shed. Now he was moving his family to his farm.
我们坐下聊了起来。托尼告诉我说，拥有一个农庄是他的梦想。他喜欢番茄、辣椒以及意大利菜肴中相当重要的其它各种蔬菜。他把在意大利的妻子和儿子女儿都接来了。他在小镇周边到处找，终于找到一处没人要的一小块地产，有一幢房，还有间小棚。他正在把家搬到农庄去。
Sometime later. Tony arrived on a Sunday afternoon, neatly dressed. He had another Italian man with him. He told me that he had persuaded his childhood friend to move to America. Tony was sponsoring him. With an amused look in his eye, he told me that when they approached the little farm he now operated, his friend stood in amazement and said, "Tony, you are a millionaire!"
又过了一些时候，在一个星期日的下午托尼来了，他穿戴得整整齐齐。和他一起来的还有另一位意大利人。他告诉我，他说服了儿时的伙伴前来美国。托尼为他作经济担保。他眼里露出顽皮的神情，对我说，他俩来到他经营的小农庄时，他的朋友惊奇地站住说，“托尼，你是个百万富翁啦！”
Then, during the war, a message came from my company. Tony had passed away.
I asked our people to check on his family and see that everything was properly handled. They found the farm green with vegetables, the little house livable and homey. There was a tractor and a good car in the yard. The children were educated and working, and Tony didn't owe a cent.
后来，在战争期间，公司里传出了一个消息。托尼去世了。
我让公司的人去他家看看，确保各项事宜都得到妥善安置。他们看到农场上长着绿油油的蔬菜，小屋布置得舒适温馨，院子里有一辆拖拉机，还有一辆不错的汽车。孩子受过教育，都工作了，托尼身前没有分文欠债。
After he passed away, I thought more and more about Tony's career. He grew in stature in my mind. In the end, I think he stood as tall, and as proud, as the greatest American industrialists.
They had all reached their success by the same route and by the same values and principles: vision, determination, self-control, optimism, self-respect and, above all, integrity.
托尼去世后，我一直想着他的经历。他的形象在我心目中越来越高大。最后，我觉得他就和美国那些最大的实业家一样高大、自豪。
他们都通过同样的途径，本着同样的价值观和原则获得了成功：远见、执着、自制、乐观、自尊，以及最重要的，正直。
Tony did not begin on the bottom rung of the ladder. He began in the basement. Tony's affairs were tiny; the greatest industrialists' affairs were giant. But, after all, the balance sheets were exactly the same. The only difference was where you put the decimal point.
托尼不是从最低一级阶梯往上爬的，他是从地下室往上爬的。托尼的事业很小，那些最大的实业家的事业很大。但究其实，两者的资产负债表完全一样。惟一的不同是你把小数点点在什么地方。
Tony Trivisonno came to America seeking the American Dream. But he didn't find it — he created it for himself. All he had were 24 precious hours a day, and he wasted none of them.
托尼·特里韦索诺来到美国寻求美国梦。但他没有找到什么美国梦 —— 他为自己创造了一个美国梦。他的全部拥有是一天宝贵的二十四小时，而他一刻也没有浪费。
## Book 1 Unit 5 The Company Man
Ellen Goodman
1 He worked himself to death, finally and precisely, at 3:00 A.M. Sunday morning.
他终于在星期天凌晨三点整因过度劳累而离开人世。
2 The obituary didn't say that, of course. It said that he died of a heart attack I think that was it but everyone among his friends and acquaintances knew it instantly. He was a perfect Type-A, a workaholic, a classic, they said to each other and shook their heads and thought for ten minutes about the way they lived.
当然，讣告上没有提及这一点。讣告说他死于冠状动脉血栓形成——我认为这就是死因——但是，他所有的朋友和熟人都马上明白是怎么回事。他们议论道，他是十足的A型行为者，一个工作狂，一个典型的工作狂，他们边说边摇头，他们还花了五到十分钟的时间想了想自己的生活方式。
3 This man who worked himself to death finally and precisely at 3:00 A.M. Sunday morning on his day off was fifty-one years old and a vice-president. He was, however, one of six vice-presidents, and one of three who might conceivably if the president died or retired soon enough have moved to the top spot. Phil knew that.
这位最终于星期天——他的休息日——凌晨三点整累死的人是位公司副总裁，时年五十一岁。不过他是六位副总裁之一，如果总裁去世够早或退休够早的话，他本是有望当第一把手的三位人选之一。菲尔清楚这一点。
4 He worked six days a week, five of them until eight or nine at night, during a time when his own company had begun the four-day week for everyone but the executives. He had no outside interests, unless, of course, you think about a monthly golf game that way. To Phil, it was work. He always ate egg-salad sandwiches at his desk. He was, of course, overweight, by 20 or 25 pounds. He thought it was okay, though, because he didn't smoke.
他一周工作六天，其中五天要工作到晚上八、九点钟。而公司员工除领导层外已经开始实行每周四天工作制了。他像重要人物一样工作。他在外面没有“本职以外的 爱好”，当然，除非你认为每月一次的打高尔夫球也算的话。但是，对于菲尔来说，这也是工作。他总是在办公桌旁吃鸡蛋沙拉三明治，自然啦，他超重了，超出二十或二十五磅。不过他认为没关系，因为他不抽烟。
5 On Saturdays, Phil wore a sports jacket to the office instead of a suit, because it was the weekend.
每周六菲尔身着运动夹克衫去上班。不穿西装，因为这是周末。
6 He had a lot of people working for him, maybe sixty, and most of them liked him most of the time. Three of them will be seriously considered for his job. The obituary didn't mention that.
他手下有不少人，约六十名。大部分人多半时间都喜欢他。其中三人被认真考虑当作接他班的人选。讣告上没有提及这点。
7 But it did list his "survivors" quite accurately. He is survived by his wife, Helen, forty-eight years old, a good woman of no particular marketable skills, who worked in an office before marrying and mothering. She had, according to her daughter, given up trying to compete with his work years ago, when the children were small. A company friend said, "I know how much you will miss him." And she answered, "I already have."
但是讣告上的确颇为准确地列出了他的“遗嘱”，他的遗孀海伦，四十八岁，一个好女人，但没有什么适合市场需求的技能，结婚生育之前曾在办公室工作。据她女儿说，多年前，孩子们还小的时候，她就决定放弃与丈夫的工作竞争了。一位公司朋友说：“我知道你将会多么思念他。”她回答道：“我早就思念他了。”
8 "Missing him all these years," she probably gave up trying to love him the way she used to. She would be well taken care of".
“这么多年来一直思念着他”，她一定是牺牲了自己，竭尽全力照顾此公。她会得到“很好照顾的”。
9 His "clearly beloved" eldest of the 'dearly beloved" children is a hard-working executive in a manufacturing firm down South. In the day and a half before the funeral, he vent around the neighborhood researching his father, asking the neighbors what he was like. They were embarrassed.
他“深爱的”子女中的“深爱的”大儿子是南方的一家制造公司的经理，工作十分努力。在葬礼前一天半，他走访了街坊邻里，向他们打听父亲是怎样的一个人，他们感到很尴尬。
10 His second child is a girl, who is twenty-four and newly married. She lives near her mother and they are close, hut whenever she vas alone with her father, in a car driving somewhere, they had nothing to say to each other.
他的第二个孩子是个女儿，二十四岁，刚结婚。她住在娘家附近，与母亲很亲近。但每次和父亲单独在一起，比如开车到什么地方时，彼此竟无话可说。
11 The youngest is twenty, a boy, a high-school graduate who has spent the last couple of years, like a lot of his friends, doing enough occasional jobs to buy grass and food. He was the one who tried to grab the affection of his father, and tried to mean enough to him to keep the man at home. He was his father's favorite. Over the last two years, Phil stayed up nights worrying about the boy.
最小的孩子二十岁，是个男孩，高中毕业生。和他的许多朋友一样，在过去的几年里他打了不少零工，挣的钱足够供自己吃饭吸大麻。就是他努力想牢牢抓住父亲，努力想让自己在父亲心目中显得更重要，好让此公留在家里。他是父亲的最爱。在这两年来，菲尔往往彻夜不眠，为这个儿子担忧。
12 The boy once said, "My father and I only board here."
这个儿子曾经说过：“我和父亲只是寄宿在这儿。”
13 At the funeral, the sixty-year-old company president told the forty-eight-year-old widow that the fifty-one-year-old deceased had meant much to the company and would be missed and would he hard to replace. The widow didn't look him in the eye. She was afraid he would read her bitterness and, after all, she would need him to straighten out the finances the stock options and all that.
在葬礼上，六十岁的公司总裁对四十八岁的寡妇说，五十一岁的死者对公司来说举足轻重，会被人怀念，很难被人取代。寡妇没有正视他，她怕他会看出自己的怨恨，毕竟她还需要他来解决一些经济方面的问题——比如优先认股权之类的事情。
14 Phil vas overweight and nervous and worked too hard. If he wasn't at the office, he was worried about it. Phil was a Type-A, a heart-attack natural. You could have picked him out in a minute from a lineup.
菲尔体重超常，又有神经质，工作过于辛劳。即使人不呆在办公室里，心还是要牵挂着。菲尔是个A型行为者，天生容易患心脏病。你能在一排人中立马认出他。
15 So when he finally worked himself to death, at precisely 3:00 A.M. Sunday morning, no one vas really surprised.
因此，他在星期天凌晨三点整最终劳累致死时，没有人真正感到意外。
16 By 5:00 P.M. the afternoon of the funeral, the company president had begun, discreetly of course, with care and taste, to make inquiries about his replacement. One of three men. He asked around: "Who's been working the hardest?"
## Book 1 Unit 6 A Valentine Story
Doug Bell
John Blanchard stood up from the bench, straightened his Army uniform, and studied the crowd of people making their way through Grand Central Station.
He looked for the girl whose heart he knew, but whose face he didn't, the girl with the rose. His interest in her had begun twelve months before in a Florida library. Taking a book off the shelf he soon found himself absorbed, not by the words of the book, but by the notes penciled in the margin. The soft handwriting reflected a thoughtful soul and insightful mind.
In the front of the book, he discovered the previous owner's name, Miss Hollis Maynell. With time and effort he located her address. She lived in New York City. He wrote her a letter introducing himself and inviting her to correspond. The next day he was shipped overseas for service in World War II.
During the next year the two grew to know each other through the mail. Each letter was a seed falling on a fertile heart. A romance was budding. Blanchard requested a photograph, but she refused. She explained: "If your feeling for me has any reality, any honest basis, what I look like won't matter. Suppose I'm beautiful. I'd always be haunted by the feeling that you had been taking a chance on just that, and that kind of love would disgust me. Suppose I'm plain (and you must admit that this is more likely). Then I'd always fear that you were going on writing to me only because you were lonely and had no one else. No, don't ask for my picture. When you come to New York, you shall see me and then you shall make your decision. Remember, both of us are free to stop or to go on after that — whichever we choose..."
When the day finally came for him to return from Europe, they scheduled their first meeting — 7:00 p.m. at Grand Central Station, New York.
"You'll recognize me," she wrote, "by the red rose I'll be wearing on my lapel." So, at 7:00 p.m. he was in the station looking for a girl who had filled such a special place in his life for the past 12 months, a girl he had never seen, yet whose written words had been with him and sustained him unfailingly.
I'll let Mr. Blanchard tell you what happened:
A young woman was coming toward me, her figure long and slim. Her golden hair lay back in curls from her delicate ears; her eyes were blue as flowers. Her lips and chin had a gentle firmness, and in her pale green suit she was like springtime come alive.
I started toward her, entirely forgetting to notice that she was not wearing a rose.
As I moved, a small, provocative smile curved her lips. "Going my way, sailor?" she murmured. Almost uncontrollably I made one step closer to her, and then I saw Hollis Maynell. She was standing almost directly behind the girl. A woman well past 40, she had graying hair pinned up under a worn hat.
She was more than a little overweight, her thick-ankled feet thrust into low-heeled shoes.
The girl in the green suit was walking quickly away. I felt as though I was split in two, so keen was my desire to follow her, and yet so deep was my longing for the woman whose spirit had truly companioned me and upheld my own.
And there she stood. Her pale, round face was gentle and sensible, her gray eyes had a warm and kindly glow. I did not hesitate.
My fingers gripped the small worn blue leather copy of the book that was to identify me to her. This would not be love, but it would be something precious, something perhaps even better than love, a friendship for which I had been and must ever be grateful.
I squared my shoulders and saluted and held out the book to the woman, even though while I spoke I felt choked by the bitterness of my disappointment. "I'm Lieutenant John Blanchard, and you must be Miss Maynell. I am so glad you could meet me; may I take you to dinner?"
The woman's face broadened into a smile. "I don't know what this is about, son," she answered, "but the young lady in the green suit who just went by, she begged me to wear this rose on my coat. And she said if you were to ask me out to dinner, I should go and tell you that she is waiting for you in the big restaurant across the street. She said it was some kind of test!"
It's not difficult to understand and admire Miss Maynell's wisdom. The true nature of a heart is seen in its response to the unattractive.
"Tell me whom you love," Houssaye wrote, "and I will tell you who you are."
约翰·布兰查德从长凳上站起身来，整了整军装，留意着格兰德中央车站进出的人群。
他在寻找一位姑娘，一位佩带玫瑰的姑娘。他知其心，但不知其貌。十二个月前，在佛罗里达州的一个图书馆，他对她产生了兴趣。他从书架上取下一本书，很快便被吸引住了，不是被书的内容，而是被铅笔写的眉批。柔和的笔迹显示出其人多思善虑的心灵和富有洞察力的头脑。
在书的前页，他找到了前一位拥有人的姓名，霍利斯·梅奈尔小姐。他花了一番工夫和努力，找到了她的地址。她住在纽约市。他给她写了一封信介绍自己，并请她回复。第二天他被运往海外，参加第二次世界大战。
在接下来的一年当中，两人通过信件来往增进了了解。每一封信都如一颗种子撒入肥沃的心灵之土。浪漫的爱情之花就要绽开。布兰查德提出要一张照片，可她拒绝了。她解释道：“如果你对我的感情是真实的，是诚心诚意的，那我的相貌如何并不重要。设想我美丽动人。我将会一直深感不安，惟恐你只是因为我的容貌就贸然与我相爱，而这种爱情令我憎恶。设想本人相貌平平（你得承认，这种可能性更大）。那我一直会担心，你和我保持通信仅仅是出于孤独寂寞，无人交谈。不，别索要照片。等你到了纽约，你会见到我，到时你可再作定夺。且记，见面后我俩都可以自由决定中止关系或继续交往 —— 无论你怎么选择......”
. 他从欧洲回国的日子终于到了。他们安排了两人的第一次见面 —— 晚上七点， 纽约格兰德中央车站。
“你会认出我的，” 她写道，“我会在衣襟上戴一朵红玫瑰。” 于是，晚上七点，他候在车站，寻找一位过去一年里在自己生活中占据了如此特殊地位的姑娘，一位素未谋面，但其文字伴随着他、始终支撑着他精神的姑娘。
且让布兰查德先生告诉你接下来发生的事吧：
一位年轻的姑娘向我走来，她身材颀长纤细。一头卷曲的金发披在秀美的耳后；眼睛碧蓝，如花似玉。她的双唇和下颌线条柔和，却又柔中见刚，她身穿浅绿色套装，犹如春天一般生气盎然。
我朝她走去，完全忘了去看她有没有戴玫瑰花。
我走过去时，她双唇绽开撩人的微笑。“和我同路吗，水兵？”她小声问道。我情不自禁，再向她走近一步。可就在这时，我看到了霍利斯·梅奈尔。她差不多就站在姑娘的正后面，早已年过四十，灰白的头发用卡子向上别着，头上带着一顶旧帽子。
她体态臃肿，粗圆的脚踝上套着一双低跟鞋。
穿着绿色套装的姑娘快步走开了。我觉得自己好像被分成了两半，一方面热切地想去追赶她，但另一方面我又渴望那一位以其心灵真诚陪伴我并成为我的精神支柱的女人。
她站在那儿，苍白的圆脸显得温柔理智，灰色的眼睛透出热情善良。我没有迟疑。
我手里紧握着那本小小的让她辨认我的蓝色羊皮面旧书。这不会是爱情，但将是某种珍贵的、或许比爱情更美妙的东西，一种我曾经感激，并将永远感激的友情。
我挺胸站立，敬了个礼，并举起手中的书好让那位女士看。不过在我开口说话的时候，失望的痛苦几乎使我哽咽。“我是约翰·布兰查德中尉，想必您就是梅奈尔小姐。很高兴您来见我。可否请您赏光吃饭？”
妇女的脸上绽开了笑容。“我不知道是怎么回事，孩子，”她回答说，“可是刚才走过去的那位穿绿色套装的姑娘，她央求我把这支玫瑰插在衣服上。她还说，要是你请我吃饭的话，我就告诉你，她就在街对面那个大饭店里等你。她说这是一种考验！”
梅奈尔小姐的智慧不难理解，也令人称奇。心灵的本质是从其对不美的事物的态度中反映出来的。
“告诉我你所爱者是谁，”何赛写道，“我就知道你是什么样的人。”
## Book 1 Unit 7 What Animals Really Think
Euqene Linden
Over the years, I have written extensively about animal-intelligence experiments and the controversy that surrounds them. Do animals really have thoughts, what we call consciousness? Wondering whether there might be better ways to explore animal intelligence than experiments designed to teach human signs, I realized what now seems obvious: if animals can think, they will probably do their best thinking when it serves their own purposes, not when scientists ask them to.
And so I started talking to vets, animal researchers, zoo keepers. Most do not study animal intelligence, but they encounter it, and the lack of it, every day. The stories they tell us reveal what I'm convinced is a new window on animal intelligence: the kind of mental feats animals perform when dealing with captivity and the dominant species on the planet — humans.
Let's Make a Deal
Consider the time Charlene Jendry, a conservationist at the Columbus Zoo, learned that a female gorilla named Colo was handling a suspicious object. Arriving on the scene, Jendry offered Colo some peanuts, only to be met with a blank stare. Realizing they were negotiating, Jendry raised the stakes and offered a piece of pineapple. At this point, while maintaining eye contact, Colo opened her hand and revealed a key chain.
Relieved it was not anything dangerous or valuable, Jendry gave Colo the pineapple. Careful bargainer that she was, Colo then broke the key chain and gave Jendry a link, perhaps figuring. Why give her the whole thing if I can get a bit of pineapple for each piece?
If an animal can show skill in trading one thing for another, why not in handling money? One orangutan named Chantek did just that in a sign-language study undertaken by anthropologist Lyn Miles at the University of Tennessee. Chantek figured out that if he did tasks like cleaning his room, he'd earn coins to spend on treats and rides in Miles's car. But the orangutan's understanding of money seemed to extend far beyond simple dealings. Miles first used plastic chips as coins, but Chantek decided he could expand the money supply by breaking chips in two. When Miles switched to metal chips, Chantek found pieces of tin foil and tried to make copies.
Miles also tried to teach Chantek more virtuous habits such as saving and sharing. Indeed, when I caught up with the orangutan at Zoo Atlanta, where he now lives, I saw an example of sharing that anyone might envy. When Miles gave Chantek some grapes and asked him to share them, Chantek promptly ate all the fruit. Then, as if he'd just remembered he'd been asked to share, he handed Miles the stem.
Tale of a Whale
Why would an animal want to cooperate with a human? Behaviorists would say that animals cooperate when they learn it is in their interest to do so. This is true, but I don't think it goes far enough.
Gail Laule, a consultant on animal behavior, speaks of Orky, a killer whale, she knew. "Of all the animals I've worked with, he was the most intelligent," she says. "He would assess a situation and then do something based on the judgments he made."
Like the time he helped save a family member. When Orky's mate, Corky, gave birth, the baby did not thrive at first, and keepers took the little whale out of the tank by stretcher for emergency care. Things began to go wrong when they returned the baby whale to the tank. As the workers halted the stretcher a few meters above the water, the baby suddenly began throwing up through its mouth. The keepers feared it would choke, but they could not reach the baby to help it.
Apparently sizing up the problem, Orky swam under the stretcher and allowed one of the men to stand on his head, something he'd never been trained to do. Then, using his tail to keep steady, Orky let the keeper reach up and release the 420-pound baby so that it could slide into the water within reach of help.
Primate Shell Game
Sometimes evidence of intelligence can be seen in attempts to deceive. Zoo keeper Helen Shewman of Seattle's Woodland Park Zoo recalls that one day she dropped an orange through a feeding hole for Melati, an orangutan. Instead of moving away to get it, Melati looked Shewman in the eye and held out her hand. Thinking the orange must have rolled off somewhere inaccessible, Shewman gave her another one. But when Melati moved off, Shewman noticed the original orange was hidden in her other hand.
Towan, the colony's dominant male, watched this whole trick, and the next day he, too, looked Shewman in the eye and pretended that he had not yet received an orange. "Are you sure you don't have one?" Shewman asked. He continued to hold her gaze steadily and held out his hand. Giving in, she gave him another one, then saw that he had been hiding his orange underneath his foot.
What is intelligence anyway? If life is about survival of a species — and intelligence is meant to serve that survival — then we can't compare with pea-brained sea turtles, which were here long before us and survived the disaster that wiped out the dinosaurs. Still, it is comforting to realize that other species besides our own can stand back and assess the world around them, even if their horizons are more limited than ours.
多年来，我写了大量关于动物智能实验、以及围绕这些实验所产生的争议的文章。动物真的有思想，即我们所说的意识吗？在考虑是否会有比设计教动物人类手势语的实验更好的方式探索动物智能时，我悟出了现在看来是显而易见的一点：如果动物能思维，它们会在能为自己所用的时候，而不是在科学家让它们思维的时候作出最佳思维。
于是我开始与兽医、动物研究人员以及动物园饲养员交谈。他们大都不研究动物智能，但他们每天都碰到或碰不到动物智能。他们讲述的故事开启了我相信是研究动物智能的一扇新的窗口：即动物在对付樊笼生活和地球上的主宰物种 —— 人类 —— 时所表现的高超的思维技能。
让我们做笔交易
请考虑这一情况：哥伦布动物园的一位动物保护主义者查伦·延德里觉察到一头叫做科洛的雌性大猩猩在玩弄一件可疑的物品。延德里走过去，给了科洛一些花生，却被翻了个白眼。意识到这是在讨价还价，延德里加大了筹码，又给了一片菠萝。这时候，科洛一边望着延德里，一边摊开手，露出了一根钥匙链。
见不是危险或珍贵物品，延德里松了一口气，把菠萝给了考勒。科洛真是个精明的还价者，它把钥匙链拉断，给了延德里一段，或许在算计着，要是每一小段都能换片菠萝，我干嘛要全都给她？
如果动物能在以物换物中显示技能，又何尝不会在使用钱币中再露一手？在田纳西大学人类学家琳·迈尔斯进行的一项手势语研究中，有头名叫夏特克的猩猩就这么做了。夏特克悟出，如果它干些诸如清理房间的事，他就能挣些硬币，好用来买好吃的，还可以坐迈尔斯的车外出兜风。但这头猩猩对钱币的理解似乎远远超出了简单的交易。迈尔斯一开始用塑料片充当硬币，而夏特克竟认定，它可以把塑料片拗成两片，以此扩大钱币供应量。而当迈尔斯改用金属片时，夏特克找到了一些锡箔，试图复制。
迈尔斯还试图教会夏特克一些好习惯，诸如节俭和与人分享。当我在它目前居住的亚特兰大动物园见到这头猩猩时，我果然见到它与人分享的一例，足以令任何人羡慕。迈尔斯给了夏特克一些葡萄，要求它与人分享，它很快吃完了所有的葡萄。随后，它似乎是想起了迈尔斯要它与人分享，便把梗儿递给了迈尔斯。
鲸鱼的故事
动物为什么会愿意与人合作？行为主义者会说，动物认识到合作于己有利时就会这么做。这没有错，但我觉得这一解释尚不充分。
动物行为顾问盖尔·劳尔说起过她了解的一头虎鲸奥基。“在我照管过的动物当中，它是最聪明的，”她说，“它会审时度势，再根据自己的判断采取行动。”
比如有次它救了一个家族成员。奥基的配偶科基生幼鲸时，那条幼鲸一开始情况不妙，饲养员把幼鲸用担架抬出水糟，实施紧急护理。他们把幼鲸送回水槽时，出了事情。当工人把担架停在高出水面几英尺处的时候，幼鲸开始呕吐。饲养员担心它会窒息，但他们无法接近幼鲸提供帮助。
奥基显然看出了问题，它游到担架下，让其中一人站在它头上。这种事从来没有训练它做过。然后，奥基用尾部保持平衡，让饲养员接近，并松开了那条420磅重的幼鲸，以便让它滑入水中，获得帮助。
灵长目动物的骗术
有时动物的智能可以从其欺骗的企图中得以证明。西雅图伍德兰公园动物园饲养员海伦·休曼回忆道，一天她从喂食窗口给猩猩梅拉蒂扔了个桔子。梅拉蒂没有移动身体去接，而是眼睛直视休曼，伸出手来。休曼以为桔子准是滚到一边拿不到了，就又给了它一个。可当梅拉蒂走开时，休曼却注意到原来那只桔子就藏在它另一只手里。
猩猩园的头领托温目睹了这个把戏。第二天，这头雄猩猩也是眼睛盯着休曼，装作没有接到桔子。“你肯定没拿到吗？”休曼问道。它仍直视着她，同时把手伸了出来。她让步了，又给了它一个，随后却看见它把桔子藏在脚下。
智能究竟是什么？如果生命就是讲物种的生存——而智能是为了生存——那么我们根本无法与大脑只有豌豆大小的海龟相提并论，海龟早在人类出现很久之前便已存在，并经历了使恐龙灭绝的重大灾难而生存下来。尽管如此，想到除了我们人类，尚有其它物种，即便它们的视野比我们还狭小，却也能退后一步，清醒地审视周围的世界，不由人深感宽慰。
## Book 1 Unit 8 Fable of the Lazy Teenager
Benjamin Stein
One day last fall, I ran out of file folders and went to the drugstore to buy more. I put a handful of folders on the counter and asked a teenage salesgirl how much they cost. "I don't know," she answered. "But it's 12 cents each."
I counted the folders. "Twenty-three at 12 cents each, that makes \$2.76 before tax," I said.
"It's magic," I said.
No modestly educated adult can fail to be upset by such an experience. While our children seem better-natured than ever, they are so ignorant — and so ignorant of their ignorance — that they frighten me. In a class of 60 seniors at a private college where I recently taught, not one student could write a short paper without misspellings. Not one.
But this is just a tiny slice of the problem. The ability to perform even the simplest calculations is only a memory among many students I see, and their knowledge of world history or geography is nonexistent.
Moreover, there is a chilling indifference about all this ignorance. The attitude was summed up by a friend's bright, lazy 16-year-old son, who explained why he preferred not to go to U.C.L.A. "I don't want to have to compete with Asians," he said. "They work hard and know everything."
In fact, this young man will have to compete with Asians whether he wants to or not. He cannot live forever on the financial, material and human capital accumulated by his ancestors. At some point soon, his intellectual laziness will seriously affect his way of life. It will also affect the rest of us. A modern industrial state cannot function with an idle, ignorant labor force. Planes will crash. Computers will jam. Cars will break down.
To drive this message home to such young Americans, I have a humble suggestion: a movie, or TV series, dramatizing just how difficult it was for this country to get where it is — and how easily it could all be lost. I offer the following fable.
As the story opens, our hero, Kevin Hanley 1990, a 17-year-old high school senior, is sitting in his room, feeling bitter. His parents insist he study for his European history test. He wants to go shopping for headphones for his portable CD player. The book he is forced to read — The Wealth of Nations — puts him to sleep.
Kevin dreams it is 1835, and he is his own great-great-great-grandfather at 17, a peasant in County Kerry, Ireland. He lives in a small hut and sleeps next to a pig. He is always hungry and must search for food. His greatest wish is to learn to read and write so he might get a job as a clerk. With steady wages, he would be able to feed himself and help his family. But Hanley's poverty allows no leisure for such luxuries as going to school. Without education and money, he is powerless. His only hope lies in his children. If they are educated, they will have a better life.
Our fable fast-forwards and Kevin Hanley 1990 is now his own great-grandfather, Kevin Hanley, 1928. He, too, is 17 years old, and he works in a steel mill in Pittsburgh. His father came to America from Ireland and helped build the New York City subway. Kevin Hanley 1928 is far better off than either his father or his grandfather. He can read and write. His wages are far better than anything his ancestors had in Ireland.
Next Kevin Hanley 1990 dreams that he is Kevin Hanley 1945, his own grandfather, fighting on Iwo Jima against a most determined foe, the Japanese army. He is always hot, always hungry, always scared. One night in a foxhole, he tells a friend why he is there: "So my son and his son can live in peace and security. When I get back, I'l1 work hard and send my boy to college so he can live by his brains instead of his back."
Then Kevin Hanley 1990 is his own father, Kevin Hanley 1966, who studies all the time so he can get into college and law school. He lives in a fine house. He has never seen anything but peace and plenty. He tells his girlfriend that when he has a son, he won't make him study all the time, as his father makes him.
At that point, Kevin Hanley 1990 wakes up, shaken by his dream. He is relieved to be away from Ireland and the steel mill and Iwo Jima. He goes back to sleep.
When he dreams again, he is his own son, Kevin Hanley 2020. There is gunfire all day and all night. His whole generation forgot why there even was law, so there is none. People pay no attention to politics, and government offers no services to the working class.
Kevin 2020's father, who is of course Kevin 1990 himself, works as a cleaner in a factory owned by the Japanese. Kevin 2020 is a porter in a hotel for wealthy Europeans and Asians. Public education stops at the sixth grade. Americans have long since stopped demanding good education for their children.
The last person Kevin 1990 sees in his dream is his own grandson. Kevin 2050 has no useful skills. Machines built in Japan do all the complex work, and there is little manual work to be done. Without education, without discipline, he cannot earn an adequate living wage. He lives in a slum where there is no heat, no plumbing, no privacy and survives by searching through trash piles.
In a word, he lives much as Kevin Hanley 1835 did in Ireland. But one day, Kevin Hanley 2050 is befriended by a visiting Japanese anthropologist studying the decline of America. The man explains to Kevin that when a man has no money, education can supply the human capital necessary to start to acquire financial capital. Hard work, education, saving and discipline can do anything. "This is how we rose from the ashes after you defeated us in a war about a hundred years ago."
"America beat Japan in war?" asks Kevin 2050. He is astonished. It seems as impossible as Brazil defeating the United States would sound in 1990. Kevin 2050 swears that if he ever has children, he will make sure they work and study and learn and discipline themselves. "To be able to make a living by one's mind instead of by stealing," he says. "That would be a miracle."
When Kevin 1990 wakes up, next to him is his copy of The Wealth of Nations. He opens it and the first sentence to catch his eye is this: "A man without the proper use of the intellectual faculties of a man is, if possible, more contemptible than even a coward."
Kevin's father walks in. "All right, son," he says. "Let's go look at those headphones."
"Sorry, Pop," Kevin 1990 says. "I have to study."
去年一个秋日，我文件夹用完了，便去杂货店买。我拿了一大把文件夹搁在柜台上，问一个十几岁的售货员多少钱。“不知道，”她回答说，“反正单价12美分。”
我数了数文件夹。“二十三个，单价12美分，总共2.76美金，不含税，”我说。
“你心算的？”她惊奇地问道，“你怎么会算出来的？”
“靠魔力，” 我说。
“真的？” 她问。
略受教育的成年人没有谁不会为这样的经历难过。虽然我们的孩子似乎比以往任何时候都要温厚和气，他们却如此无知 —— 对自己的无知状况也如此无知 —— 以至使我感到可怕。在我最近任教的一所私立大学，一个六十人的四年级班上，没有一个学生写短文时不犯拼写错误。没有一个学生例外。
但这只是问题的一小部分。在我所见过的许多学生中，再也没有过去学生都有的哪怕是进行最简单的计算的能力，他们对世界历史和地理都一无所知。
更有甚者，他们对这种种的无知却毫不在乎，实在令人不寒而栗。一位朋友的聪明但却很懒散的十六岁儿子在解释他为什么不想上加州洛杉矶分校时说的话是对这种态度的高度概括。“我不想去那儿跟亚洲人竞争，”他说，“他们用功，什么都知道。”
其实，无论他是否愿意，这位年轻人都将不得不去跟亚洲人竞争。他不能永远躺在先辈积累的经济、物质与人力资本上。用不了多久，他懒于用脑的结果将严重影响他的生活方式，也将影响我们其他所有的人。一个现代工业化国家无法靠一支懒散、无知的劳动大军运行。飞机会坠落。计算机会出故障。汽车会抛锚。
为使这样的美国青年彻底认识到这一点，我的愚见是：拍一部电影，或电视连续剧，生动地描述我们国家的今天如何来之不易 —— 而要丧失这一切又何其容易。下面我奉献一篇寓言故事。
故事开始时，我们的主人公凯文·汉利1990，一名十七岁的高三学生，正坐在自己房间里，心情痛苦。他父母一定要他准备欧洲史考试。而他则想去买一副激光唱片随身听的耳机。他被迫要读的书 —— 《各国的财富》 —— 让他打瞌睡。
凯文进入梦乡，时值1835年，他是他本人的曾太祖父，十七岁，是爱尔兰克雷郡的一个农民。他住在小小的陋室里，睡在一头猪旁。他老是挨饿，总是要找吃的。他最大的心愿是学会读书写字，以便找一个职员的工作。有了固定的工资，他就能养活自己，贴补家用。但汉利的贫穷使他无从享受上学这样的奢侈。没有教育，没有钱，他无能为力。他惟一的希望寄托在孩子身上。如果他们能接受教育，他们就会生活得好一些。
我们的寓言故事快速展开。现在凯文·汉利1990成了他自己的曾祖父，凯文·汉利1928。他也是十七岁，在匹兹堡一家钢铁厂工作。他的父亲从爱尔兰来到美国，参加过纽约地铁的修建。凯文·汉利1928比自己的父亲和祖父境遇好多了。他能读书写字。他的工资比先辈在爱尔兰时的收入高多了。
接下来凯文·汉利1990梦见自己成了他自己的祖父凯文·汉利1945。他正在硫黄岛与死敌日本军队作战。他总是又热又饿又害怕。一天晚上他在散兵坑里与一个朋友讲自己为什么在那儿作战：“这样我的儿子、孙子就能生活在和平安全的环境里。等我回国了，我要勤奋工作，让儿子上大学，这样他就可以干脑力活儿，而不是靠卖苦力生活。”
接着凯文·汉利1990成了他自己的父亲凯文·汉利1966。他终日用功，这样就可以上大学，进法学院。他住在漂亮的房子里。他一生在和平环境中过着富裕的生活。他对女朋友说，等他有了儿子，他不会像他父亲逼他那样逼自己的儿子整天读书。
就在这时， 凯文·汉利1990被自己的梦惊醒了。他离开了爱尔兰，离开了那家钢铁厂，离开了硫黄岛，不由松了口气。他又睡着了。
他接着做梦，这次成了他自己的儿子凯文·汉利2020。枪声日夜不停。他那整个一代人忘却了过去为什么要有法律，因此现在没有法律了。人们丝毫不关心政治，政府不为工人阶级提供服务。
凯文2020的父亲，自然就是凯文1990本人，在日本人开的一家工厂当清洁工。凯文2020在一家专为有钱的欧洲人和亚洲人开的酒店里当行李工。公共教育到六年级为止。美国人早就不再要求自己的孩子接受良好的教育。
凯文1990最后梦见的是他自己的孙子。凯文2050没有有用的技能。日本制造的机器包揽了所有复杂的工作，没有什么体力活可做。没有受过教育，没有受过训练，他挣不到足够的钱养活自己。他住在贫民窟，没有暖气，没有卫生设备，无法不受四邻干扰，靠搜捡破烂度日。
总之，他的生活就像凯文·汉利1835在爱尔兰时一模一样。可是有一天，凯文·汉利2050与一位研究美国衰亡史的来访日本人类学家交上了朋友。那人跟凯文解释说，如果一个人没有钱，教育能提供积累金融资本所必需的人力资本。勤奋、教育、节俭、纪律能成就一切。“我们就是这样从一百多年前你们打败我们的战争废墟中站起来的。”
“美国在战争中打败日本？”凯文2050问道。他惊讶之极。这听起来就像说巴西在1990年打败美国一样不可思议。凯文2050发誓，如果他有孩子的话，他一定要让他们工作、上学、学习并约束自己。“能凭自己的脑力，而不是靠偷窃为生，”他说，“那将会是个奇迹。”
凯文1990醒了过来，身旁放着他的那本《各国的财富》。他打开书，跳入眼帘的第一句话就是：“一个不能恰当运用人类智力的人极可能比懦夫更可鄙。”
凯文的父亲走了进来。“好了，儿子，”他说，“咱们去看耳机吧。”
“抱歉了，爸爸，”凯文1990说，“我得看书学习了。”
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2021-01-25 11:22:07
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https://physics.stackexchange.com/questions/522970/does-the-work-done-in-moving-a-charge-through-a-conductor-depend-on-its-length
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# Does the work done in moving a charge through a conductor depend on its length?
Potential difference between two points is defined as the work done in moving a net positive charge from one point to the other. But, my question is, when the two points, that are potentially different, are connected using conducting wire having length more than the distance between the two points, charge flows through the conductor; and since the charge cannot move directly through air to the other point, it has to travel all the way through the length of the conductor, and that means it has to move a longer distance than the actual distance between the two points. Therefore, more amount of work would be required in order move that charge from one point to the other since the charge has to travel a longer distance. Thus, it would mean that the potential difference between the two points increases with the length of conductor.
Work done, as we all know depends on displacement not distance, but moving anything through a longer distance would require a greater amount of work. Where am I wrong?
• Work done by a constant force does depend on distance. But, if the wire is longer is the force the same? Consider that question. – Jon Custer Jan 4 at 2:32
• Suppose, I pushed a wooden block from one point to the other not through the direct path but by following a rectangular path, then I would do a lot more work on block than I would have done if I had pushed it through the direct path. Does not it mean that work done depends on the path taken? – Suyash Ishan Jan 5 at 13:44
• Yet that is not the situation in this case. The longer the path, the less the force in the situation you discuss. Ultimately it is probably easiest to look at it from an energy rather than force perspective. – Jon Custer Jan 5 at 14:58
Imagine moving the charge from equipotential $$A$$ to $$B$$ across the two different field lines $$1$$ and $$2$$.
We can clearly see that though in path $$1$$ we have to push the charge across a greater distance than in path $$2$$, we have to apply less force at each point during our journey than in path $$2$$.
$$W_\text{us}=\int\mathbf F\cdot d\mathbf r=\int Fdr\cos0=\int Fdr$$ (because we are moving across a field line)
So these two factors balance and hence we have to do the same amount of work to get the charge from contour $$A$$ to contour $$B$$.
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2020-08-12 19:01:39
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http://en.wikipedia.org/wiki/Talk:Combinatory_logic
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# Talk:Combinatory logic
WikiProject Mathematics (Rated B-class, Mid-priority)
This article is within the scope of WikiProject Mathematics, a collaborative effort to improve the coverage of Mathematics on Wikipedia. If you would like to participate, please visit the project page, where you can join the discussion and see a list of open tasks.
Mathematics rating:
B Class
Mid Priority
Field: Foundations, logic, and set theory
To-do list for Combinatory logic: Combinatory terms as graphs section Expand or merge B,C,K,W system page Merge SKI_combinator_calculus page, if approved (see here for discussion)
## Definition of combinators
Could we put, right at the very beginning of the page, a definition of combinator -- since this is the page that we're taken to when we search for, or click on a link to, Combinator? Much later in the article, it appears that the definition is, "a function with no free variables." Is that correct? --Mike Schiraldi 3 July 2005 15:31 (UTC)
That is a good idea. However, my feeling is that a function with no free variables is not accurate enough even for introducing the subject. Unfortunately, I can't think of something better. Possibly, the introductory chapter of Curry's book (see the references in the article) may include a better definition. Koffieyahoo 5 July 2005 06:33 (UTC)
An interesting start. I picked this up from the comment in Lambda calculus
## Supercombinators
Are you going to do supercombinators?
I had hoped that someone more familiar with compiler construction would complete that part of the 'Applications' section and discuss supercombinators in the process. (Dominus 20:04 Nov 23, 2002 (UTC))
Also, related topics might include graph reduction machines.
User:David Martland
Is this the same subject as what is called "combinatory" logic? "Combinatorial" usually means "pertaining to combinatorics"; is the same meaning intended here? -- Mike Hardy
It is not about combinatorics. Combinators just combine together more terms, that's the reason of this name. --Blaisorblade (talk) 14:24, 6 January 2008 (UTC)
## Wrong title
The title is wrong: it should be "combinatory logic". See for instance http://catalog.loc.gov/cgi-bin/Pwebrecon.cgi?Search_Arg=combinatorial+logic&Search_Code=SUBJ_&PID=14368&CNT=25&BROWSE=21&HC=18&SID=1 . "Combinatorial logic" is often used for boolean logic used in circuit design. I am renaming. AxelBoldt 19:50 Nov 23, 2002 (UTC)
Thanks. Dominus 20:04 Nov 23, 2002 (UTC)
Other title for this logic is Illative Combinatory Logic, briefly mentioned in Henk Barendregt Lambda Calculus as far as I remember. Addition to the article is done! — Preceding unsigned comment added by Elias (talkcontribs) 15:00, 30 March 2011 (UTC)
I reverted this change after reviewing appendix b of Barendregt Lambda Calculus. ILC is other Combinatory Logic, should be included either in other article or as a section in this one. — Preceding unsigned comment added by Elias (talkcontribs) 03:37, 31 March 2011 (UTC)
## Wrong definition of T[ ] ?
Possible error in Turner's T[ ] Transformation: In rules 7. and 8., terms E2 and E1 resp. remain untransformed. I think the correct rules are:
1. T[V] => V
2. T[(E1 E2)] => (T[E1] T[E2])
3. T[λx.E] => (K E) (if x is not free in E)
4. T[λx.x] => I
5. T[λx.λy.E] => T[λx.T[λy.E]] (if x is free in E)
6. T[λx.(E1 E2)] => (S T[λx.E1] T[λx.E2]) (if x is free in both E1 and E2)
7. T[λx.(E1 E2)] => (B T[λx.E1] T[E2]) (if x is free in E1 but not E2)
8. T[λx.(E1 E2)] => (C T[E1] T[λx.E2]) (if x is free in E2 but not E1)
Does that look right? --Fritz Obermeyer
The definition of Turner's T in today's revision seems correct (i.e. as above and with an extra T in clause 3) Hugo Herbelin (talk) 15:46, 23 December 2007 (UTC)
I have still a few questions about the transformations T:
- Is there any reason why it is not split into two parts: a part for abstraction elimination in a pure expression of combinatory logic (as it seems it is most standardly done - I saw this e.g. in the paper "A ‘new’ abstraction algorithm" by Price and Simmons), and a part for iterating abstraction elimination so that to convert a lambda terms into a term of combinatory logic? If the transformation were so-split into two parts, one could then say that the pure abstraction elimination part is in Curry-Howard correspondence with the deduction theorem of Hilbert-style logic, which is an interesting observation in itself (see Deduction_theorem#Conversion from proof using the deduction meta-theorem to axiomatic proof).
- In the main (non Turner's) translation, is there any reason why the 3rd clause is not replaced by the simplest clause Tx.y] => K y, and, in the same time, the condition on freshness of x in the 5th clause removed? I guess that this is a way to optimize the size of the resulting term but if it is the reason for, it would be worth to tell it. Otherwise, the reader may wonder why the alternative proposition that more canonically works by simple case analysis on the different possible forms of a term is not used. Hugo Herbelin (talk) 15:46, 23 December 2007 (UTC)
Rule 5 (T[λx.λy.E] => T[λx.T[λy.E]] (if x is free in E)) is mere hand waving. λx.T[λy.E] is not a correctly formed lambda-expression because T[λy.E] is in fact a SKI term.
Someone should check a copy of Barendregt out of the library and lay this thing to rest once and for all. It's also given in Field and Harrison (ISBN 0201192497), and probably a zillion other places. —Dominus (talk) 02:03, 22 August 2009 (UTC)
The transformation was published in Turner, D.A. (1979). "A New Implementation Technique for Applicative Languages". Software - Practice and Experience 9: 31, I will take a look on it.
As far as I remember the notation $\lambda^{T}$, for the Turner abstractor is used in more recent papers in his honour. I can't remember one example at this moment.
There are also other transformations more efficient than Turner's where an ad-hoc basis is chosen according to each program, maybe by Hughes.
I can do this change after reviewing those papers, maybe another wikipedist have those articles at hand and can do this before I. — Preceding unsigned comment added by Elias (talkcontribs) 15:50, 30 March 2011 (UTC)
## Swapped C and B
• The role of C and B are reverted from the tradition which is 75 years old!!! 15 January 1929!, item A. Page 2 of 13.
• Of course, David Turner never invented this, (nor pretend it, I think)
• If authors dont correct it, I will do it after finishing traduction to spanish.
Thanks. DefLog 13:21, 25 Mar 2004 (UTC)
## (K E) if x does not appear free in E
Wait a minute! Just being equivalent isn't enough -- there could still be lambdas in there! I changed the rule, but someone better than I should change the commentary.) 64.0.112.160
64.0.112.160 fixed the definition of T[ ]. I applied the same correction in section "Explanation of the T[ ] transformation". Hugo Herbelin (talk) 13:21, 23 December 2007 (UTC)
## Multiple views of the Lambda calculus
There is a bad tendency for Wikipedia entries to take one view of something, and word things as if "that's the way it is". I fixed the wording to say that 'one' way to view it is as equivalent to the Lambda calculus, and give another view and a reference to support it. I, for reasons obvious to any who know me, chose a recent reference that has a good bibliography on some of it. (There are some notes in my websites modal logic axioms pages that cover part of this.)
However, this is still a bit incomplete... since it was in some reasonable sense a competitor of Lambda Calculus before the equivalence was proven. There were, at the time, many different (competing) models of computation.
The "combinators as logic" view is common in early papers discussing for which systems Merideth's condensed detachment is complete. (And the interesting result that the completeness of Condensed Detachment depends on the axioms one is using for the system, and not on which logic system you have. See, for example [[1]] or Hindley's book that is being reviewed there. )
The wikipedia pages are skimpy on any of the historical development... But give the recent rise of Condensed Detachment in current computer investigations of minimal axiom sets for systems, there really ought to be (in my opinion) a page on Condensed Detachment and on Crue Merideth. (See, for example, "New Elegant Axiomatizations of Some Sentential Logics " by Branden Fitelson (2001), [[2]] and the related pages)
I forgot I had this comment out here... Update: There is now a Condensed Detachment page. Nahaj 17:47, 16 January 2006 (UTC)
## Question about the substitution syntax
Is the substitution syntax incorrect? I think "E[a/v]" should be "E[v/a]" ... i.e. the variable comes first.
The page currently shows:
(λv.E a) => E[a/v]
On the Lambda calculus page, the following syntax is used for substitution:
((λ V. E ) E' ) == E [V/E' ]
--Ingenuus 05:31, 5 Jun 2005 (UTC)
The usual reading of a construction like "E[a/v]" is: Substitute "a" for "v" in "E". Hence, since we are replacing the v's by a's the current notation is correct. It was wrong on the Lambda Calculus page. I changed it to the more readable: "E[v := a]" on both the Lambda Calculus page and the current page -- Koffieyahoo 14:04, 13 Jun 2005 (UTC)
The link of "Lectures on the Curry-Howard Isomorphism". Sørensen, Morten Heine B. and Pawel Urzyczyn is broken. The Lectures can be found at http://folli.loria.fr/cds/1999/library/pdf/curry-howard.pdf -- User:Mpagano
## ΛK versus ΛI
The presentation of the combinators B and C do not clarify why they should be introduced once we have already combinator K. The issue is the difference between the ΛK and the ΛI-calculus. ΛI delimits abstractions λx.M to forms in which x occurs free in M. K==λx.λy.x does not belong to ΛI. B and C do belong to ΛI and locally act as K. See Barendrecht chapter 9. User: Jos.koot
## numerals
The most straight forward representation of natural numbers is not mentioned:
zero == identity
zero? == (λx.xT)
sub1 == (λx.xF)
user:Jos.koot (format changed by me to make the definitions more readable -- UKoch (talk) 19:18, 5 October 2012 (UTC))
added: In fact every non trivial question is undecidable, or more precisely: there is no complete non trivial predicate. A predicate is a combinator that, when applied, either returns T or F. A predicate N is non trivial if there are two combinators A and B such that NA=T and NB=F. A combinator N is complete if and only if NM has a normal form for every combinator M. Allong the same lines as above it can be proven (by reductio ad absurdum) that no complete non trivial predicate exists. Jos.koot 11:13, 25 August 2006 (UTC)Jos.koot
This is a typical result in computability theory, where it is called Rice's theorem. An important conceptual point is that it is about predicates on the chosen representation of computable functions (be it a Turing machine, a lambda term or a term of combinatory logic), and it descends from the Halting problem.
--Blaisorblade (talk) 14:13, 6 January 2008 (UTC)
## B,C,K,W system
The article on the B,C,K,W system was in an awful state. I cut out some stuff, but now it is rather short. Leibniz 21:40, 10 November 2006 (UTC)
## Lazy K is "purer" than Unlambda
The article asserts:
The purest form of this view is the programming language Unlambda, whose sole primitives are the S and K combinators and character IO.
This is no longer the case. Lazy K eliminates the I/O primitives by making the "lazy stream" technique mandatory (i.e. a program is just a function on infinite lists of natural numbers). So it is purer. Unfortunately the article Lazy K doesn't exist yet, so I'm not going to change the link just yet. Hairy Dude 03:48, 5 January 2007 (UTC)
IIRC Lazy K was AfD'd half a year ago, along with a host of other esoteric languages. -- EJ 11:02, 5 January 2007 (UTC)
## The X combinator
I am grateful to the author(s) of the section on the reduction of the {S,K} basis to the minimalist basis {X}, as that is a confused subject in Curry et al (1972), Barendregt (1984), and Wolfram (2003). I wish someone would devise an elegant axiom set in terms of X alone.
BTW, as best as I can determine, Wikipedia regrettably has no entry on Quine's predicate functors. 202.36.179.65 18:21, 2 March 2007 (UTC)
## Why is there a section on lambda calculus?
The section on the lambda calculus is one of the biggest sections in the page - this doesn't seem quite right. What is says is fine, but I don't think it belongs here.
cheers, Tom Conway
I agree, lambda calculus has it own page and the contents of the section on lambda calculus in the combinatory logic page which is not already present in the lambda calculus page would better move there.
Lambda calculus is of course relevant for comparison with combinatory logic but the section Combinatory logic in computing already mentions lambda calculus and already highlights the connection with combinatory logic, which looks enough for a smooth reading of section Combinatory calculi.
Lambda calculus is largely referred to in section Completeness of the S-K basis. Not so much for defining abstraction elimination (which could be done internally to combinatory logic, as it it done in Barendregt, chapter 7), but for stating that CL and lambda calculus have a similar expressivity (in fact the same expressivity if extensionality is added). I believe that with a slight reworking in the way the S-K section is presented, an external reference to the lambda calculus page would suffice.
I have a side question: does anyone know where does the terminology combinatory calculus (used in the article) come from? Is it a strict synonym of combinatory logic with a more computational flavor or does it denote a different approach of combinatory logic (e.g. based on a notion of reduction rather than on a notion of equational reasoning)? In any case, it would certainly be good to give a short explanation in the article of why the two terminologies are used. Hugo Herbelin (talk) 14:27, 23 December 2007 (UTC)
I would say it's worse than that. The long detailed section about lambda calculus appears *before* the detailed explanation of combinatory logic itself. It is a very distracting and confusing diversion from the main topic. If that section needs to be here at all, it should be at the end. But I agree with others who say that it should *not* be here, and should instead be replaced by a reference to the article about lambda calculus. StormWillLaugh (talk) 10:58, 17 June 2014 (UTC)
The part starting with "The expression E[v := a] represents..." and considering the next paragraph starting "By convention...": regarding this and the following paragraphs I don't understand what is being presented. I am not clear how these issues are related, although I can see it looks a bit LISP-ish. Jazzbox (talk) 00:46, 5 October 2010 (UTC)
## One point basis
I added X'λx.(x K S K) as an example of a one point basis in CL without extensionality. Jos.koot (talk) 21:22, 12 December 2007 (UTC)
## Undecidability
A predicate is a combinator, say P such that for every combinator A either
(P A) = FALSE or
(P A) = TRUE or
(P A) has no normal form.
A predicate P is complete if (P A) has a normal form (either FALSE or TRUE) for every A.
A predicate P is non trivial if there are combinators F and T such that
(P F) = FALSE and
(P T) = TRUE.
Theorem: there is no complete non trivial predicate
Proof By reductio ad absurdum. Suppose there is a complete non trivial predicate, say P.
Because P is supposed to be non trivial there are a T and an F such that
(P T) = TRUE and
(P F) = FALSE.
Define NEGATION ≡ λx.(if (P x) then F else T) ≡ λx.((P x) F T)
Define ABSURDUM ≡ (Y NEGATION)
Fixed point theorem gives: ABSURDUM = (NEGATION ABSURDUM).
Because P is supposed to be complete either:
1. (P ABSURDUM) = FALSE or
2. (P ABSURDUM) = TRUE
Case 1: FALSE = (P ABSURDUM) = P (NEGATION ABSURDUM) = (P T) = TRUE, a contradiction.
Case 2: TRUE = (P ABSURDUM) = P (NEGATION ABSURDUM) = (P F) = FALSE, again a contradiction.
Hence (P ABSURDUM) is neither TRUE nor FALSE, which contradicts the presupposition that P would be a complete predicate. QED.
From this undecidability theorem it immediately follows that there is no complete predicate that can disciminate between terms that have a normal form and terms that do not have a normal form. It also follows that there is no complete predicate, say EQUAL, such that:
(EQUAL A B) = TRUE if A = B and
(EQUAL A B) = FALSE if AB.
If EQUAL would exist, then for all A, λx.(EQUAL x A) would have to be a complete non trivial predicate.
This may require some more editing, but would it be a good idea to include this proof in the article? Jos.koot (talk) 23:16, 12 December 2007 (UTC)
I don't understand this:
A predicate P is non trivial if there are combinators F and T such that PF = F and PT = T.
It seems to me that you are using F and T in two different ways here. You have already said that a combinator P is a predicate if Px = F or T whenever Px has a normal form. This is enough to define F and T in terms of P. You can't then require that PF = F.
I think what you want to say is that a predicate P is nontrivial if there are combinators f and t such that Pf = F and Pt = T.
Or have I missed something important? -- Dominus (talk) 15:00, 13 December 2007 (UTC)
Oh, I see now. You are distinguishing between F and F. I did not notice this on the first reading. I suggest that if you put this proof into the article, you rely on something other than the typeface to distinguish these. Maybe use TRUE and FALSE in place of T and F. -- Dominus (talk) 15:01, 13 December 2007 (UTC)
Yes, I made the distinction between F and F. I edited this section accoding to your suggestion. Let me know what you think of it, please. Jos.koot (talk) 09:39, 14 December 2007 (UTC)
On second thought, I decided to use the same terminolgy as already used in the section on undecidability Jos.koot (talk) 12:57, 14 December 2007 (UTC)
I submitted the original version of the proof of undecidability. I am sure this or an equivalent proof can be found in generally accepted sources. Nowadays, Wikipedia wants verification. Does somebody know an acceptable reference to be added to the proof? Jos.koot (talk) 17:37, 13 June 2014 (UTC)
## Metadiscussion: encapsulating isolated discussion topics into proper sections?
Would not it be preferable to move the discussion topics that are not in a proper section (i.e. those coming before the table of contents) within sections. I think it would make the discussion page clearer as all discussed topic would then be on the same level)? Hugo Herbelin (talk) 14:35, 23 December 2007 (UTC)
Done. I indeed felt this page messy upon viewing it, for this exact reason. --Blaisorblade (talk) 14:24, 6 January 2008 (UTC)
## Merger proposal
SKI combinator calculus talks about the same exact topic of this page; however, the treatment is really different, but I do not think they should be separate pages. The merge however is nontrivial, and I do not think I would be able (if the merge is accepted) to do it by myself. Any opinion?
--Blaisorblade (talk) 14:17, 6 January 2008 (UTC)
A merge would be much natural since the SKI calculus is a particular instance of combinatory logic. A preliminary step could be to say more clearly in the lead section that SKI is a complete system of combinators for combinatory logic, where the combinator I, which is redundant, is included in order to shorten a bit the size of expressions. --Hugo Herbelin (talk) 18:30, 6 January 2008 (UTC)
A merge doesn't make sense because the presnt article on combinatory logic is already too long. The article on combinatory logic should probably be broken down into a number of separate articles. A B Carter (talk) 21:59, 14 May 2008 (UTC)
oppose Too much information specific to both articles. Potatoswatter (talk) 05:14, 11 March 2010 (UTC)
oppose' The topics are related, but they have too much unique information and each deserves an article Joaogcosta 22:12, 21 June 2010 (UTC) —Preceding unsigned comment added by Joaogcosta (talkcontribs)
## Example of conversion to the S-K basis seems to be incorrect
Here, in Combinatory logic#Conversion of a lambda term to an equivalent combinatorial term, the correct conversion seems to be
T[λx.λy.(y x)]
= T[λx.T[λy.(y x)]] (by 5)
= T[λx.(S T[λy.y] T[λy.x])] (by 6)
= T[λx.(S I T[λy.x])] (by 4)
= T[λx.(S I (K x))] (by 3 and 1)
= (S T[λx.(S I)] T[λx.(K x)]) (by 6)
= (S (S T[λx.S] T[λx.I]) T[λx.(K x)]) (by 6)
= (S (S (K S) (K I)) T[λx.(K x)]) (by 3 twice)
= (S (S (K S) (K I)) (S T[λx.K] T[λx.x])) (by 6)
= (S (S (K S) (K I)) (S (K K) T[λx.x])) (by 3)
= (S (S (K S) (K I)) (S (K K) I)) (by 4)
The application in this case should look like
(S (S (K S) (K I)) (S (K K) I) x y)
= (S (K S) (K I) x (S (K K) I x) y)
= (K S x (K I x) (S (K K) I x) y)
= (S (K I x) (S (K K) I x) y)
= (K I x y (S (K K) I x y))
= (I y (S (K K) I x y))
= (y (S (K K) I x y))
= (y (K K x (I x) y))
= (y (K (I x) y))
= (y (I x))
= (y x)
Current example -- (S (K (S I)) (S (K K) I)) -- reduces to (y x) as well, but it couldn't be constructed by applying the six rules literally.
Am I right? Ximaera (talk) 10:00, 3 August 2008 (UTC)
No. Notice that the conversion rules are nondeterministic, there is more than one possibility how to construct a combinatorial term using the rules. In this particular example, it is possible express T[λx.(S I)] as (S T[λx.S] T[λx.I]) = (S (K S) (K I)) by rules 6 and 3 as you have done, but it is also possible to express it directly as (K (S I)) using just rule 3 as is done in the article. — Emil J. (formerly EJ) 10:47, 4 August 2008 (UTC)
Since headings are bold there is no reason to use bold in them. Why is the removal of redundant bolding causing problems for this page? Rich Farmbrough, 08:40, 8 September 2009 (UTC).
It's not redundant. It's clearly visible in browsers such as lynx, and it's semantically distinct. — Emil J. 10:41, 8 September 2009 (UTC)
## Term Rewriting
I noticed that there is presently no connection to term rewriting systems at all. Should the definition of CL be formulated as a special form of TRS? —Preceding unsigned comment added by 83.85.21.120 (talk) 21:48, 7 February 2011 (UTC)
## Bound variables vs Free variables
Near the beginning, it states
"Combinatory logic can be viewed as a variant of the lambda calculus, in which lambda expressions (representing functional abstraction) are replaced by a limited set of combinators, primitive functions from which bound variables are absent."
However, this not only contradicts Haskel Curry's distinction between calculi and algebras ...
"The term 'algebra' is used in this book as a name for a system with free variables but no bound variables. Thus the system of Example 5 (Sec. 2C3) is an algebra and is aptly called propositional algebra. In contradistinction the term 'calculus' will, as a rule, be used to describe a system with bound variables, so that it is suitable to speak of a calculus of lambda conversion, a predicate calculus, etc. These terms agree with ordinary mathematical usage, where the distinguishing characteristic of the infinitesimal calculus, as opposed to elementary algebra, is the presence of bound variables in the former" -- "Foundations of Mathematical Logic" by Haskell B. Curry (sec. 4A1)
but also text in the definition later on:
"The primitive functions themselves are combinators, or functions that, when seen as lambda terms, contain no free variables."
I think the introductory text needs to be fixed. Free and bound variables can be confusing! — Preceding unsigned comment added by 99.190.134.97 (talk) 21:29, 2 February 2012 (UTC)
## Technical Term
The technical term for what is called "Conversion of a lambda term to an equivalent combinatorial term" in this article is usually called "Bracket Abstraction". The name probably stems from the way the core part of the algorithm was intially written (my guess):
[x]E = KE,
[x]x = I,
[x]Ex = E,
[x]EX = BE([x]X),
[x]XE = C([x]X)E,
[x]XY = S([x]X)([x]Y).
The above is Schönfinkel’s algorithm with Curry's combinators.
M. Schönfinkel.
Über die Bausteine der Mathematischen Logik.
Mathematische Annalen, 92:305–316, 1924.
Jan Burse (talk) 12:24, 4 December 2012 (UTC)
## Are there still problems with the transformation T[ ]?
I have two problems with the definition of T[ ] as it is in this article:
1. With these rules you can't transform the expression \x.y as rule 3 requires that x is not free in y. Yet it is free according to the definition of free and bound variables in Lambda calculus#Free and bound variables. This would also mean that at least transformation 4 in the example is incorrect.
2. T[ ] is a function from lambda expressions to combinatory expressions. This means that in rule 5, a combinatory expression is applied to a lambda, which then has to be transformed again into combinatory. This makes no sense, you can't mix combinatory and lambda expressions that way.
I hope I am correct and someone will clarify this. --SemperVinco (talk) 06:28, 22 May 2014 (UTC)
Oops, never mind, rule 3 is correct. Still, rule 5 is weird. SemperVinco (talk) 08:46, 25 May 2014 (UTC)
## incorrect italics, etc.
I fixed quite a few incorrect italicizations in E₁, changing it to E₁, and similarly with the subscript 2. Is there a reason why that notation was used, rather than E1 and E2? Michael Hardy (talk) 00:00, 27 July 2014 (UTC)
## Is variable a combinatory term?
Combinatory logic is a notation to eliminate the need for variables in mathematical logic.
A combinatory term has one of the following forms:
• x
• P
• (E1 E2)
where x is a variable, ...
If there are should be no variables then why combinatory term can be a variable? — Preceding unsigned comment added by 95.106.255.17 (talk) 12:16, 6 August 2014 (UTC)
I fixed the introductory sentence: it's quantification and binding that is eliminated. ComputScientist (talk) 15:13, 6 August 2014 (UTC)
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http://physics.stackexchange.com/tags/coulombs-law/hot
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# Tag Info
## Hot answers tagged coulombs-law
36
Well it has nothing to do with the Higgs, but it is due to some deep facts in special relativity and quantum mechanics that are known about. Unfortunately I don't know how to make the explanation really simple apart from relating some more basic facts. Maybe this will help you, maybe not, but this is currently the most fundamental explanation known. It's ...
19
Yes, absolutely. In fact, Gauss's law is generally considered to be the fundamental law, and Coulomb's law is simply a consequence of it (and of the Lorentz force law). You can actually simulate a 2D world by using a line charge instead of a point charge, and taking a cross section perpendicular to the line. In this case, you find that the force (or ...
19
Maxwell's equations do follow from the laws of electricity combined with the principles of special relativity. But this fact does not imply that the magnetic field at a given point is less real than the electric field. Quite on the contrary, relativity implies that these two fields have to be equally real. When the principles of special relativity are ...
15
Lubos Motl's answer is very good, but I think it's worth saying one or two additional things. You can regard magnetism as simply a byproduct of electricity, in the following sense: if you assume that Coulomb's Law is correct, and that special relativity is correct, and that charge is a Lorentz scalar (so that charge and current density form a 4-vector), ...
14
This is not paradoxical and it is not necessary for any physical phenomenon to a priori have to obey any particular law. Some phenomena do have to obey inverse-square laws (such as, particularly, the light intensity from a point source) but they are relatively limited (more on them below). Even worse, gravity and electricity don't even follow this in ...
10
I suppose you mean $k_e=\frac1{4\pi\epsilon_0}$. That comes from the fact that Coulomb's law can be stated as : $$F= \frac1{\epsilon_0}\frac1{4\pi r^2}q_1q_2$$ Now, $\epsilon_0$ is the electric constant, or the permittivity of free space, and it essentially scales the force. The $4\pi r^2$ comes from the surface ...
9
Short Answer You've hit upon the quirk that the SI and CGS systems not only measure electric charge with different units, but also assign them different dimensionality. In SI, the Ampere is a base unit. Amperes are not made out of anything else - they are primitive, like meters, kilograms, and seconds. One Ampere is one Coulomb per second, so the unit of ...
9
If you want to avoid factors of $\pi$ in the more fundamental equations like $\nabla . E = \rho / \epsilon_0$, you have to accept them where they belong, for instance in: $E = \frac{1}{\epsilon_0} \frac{Q}{4 \pi r^2}$. As remarked by others, Newton failed to put a factor $4 \pi$ into his gravitation equation (he stipulated $g = G \frac{M}{r^2}$, instead of ...
9
James Clerk Maxwell thought about this one and showed the following. Suppose we have two concentric conducting spheres and we charge one up to a potential $\Phi$ relative to some grounding plane. Then the voltage of the inner sphere relative to the same ground is: $$\Phi_{inner} = \Phi \,q\, ... 8 Permittivity \varepsilon is what characterizes the amount of polarization \mathbf{P} which occurs when an external electric field \mathbf{E} is applied to a certain dielectric medium. The relation of the three quantities is given by$$\mathbf{P}=\varepsilon\mathbf{E},$$where permittivity can also be a (rank two) tensor: this would be the case in an ... 7 It's a good observation that the electric and gravitational fields both satisfy Poisson's equation$$ \nabla^2\Phi_G = 4\pi\rho_G, \qquad \nabla^2\Phi_E = -\frac{\rho_E}{\epsilon_0} $$where \Phi_G, \Phi_E are the gravitational and electric potentials and \rho_G,\rho_E are the mass and charge densities. It would seem from the perspective of Newtonian ... 7 +1, Good question,. While I don't think your idea has much of a physical implications, it is a good analogy (in my opinion, at least). A fair approximation to General Relativity is Newtonian Gravity. A better one is Newtonian Gravity with some special relativistic corrections (I mean a modification to Newton's gravity where the masses m are replaced ... 7 The mistake you made is in the way you stated Coloumb's law. It's either$$ \vec{F} = K \frac{q_1 q_2}{r^\color{red}3} \color{red}{\vec{r}} $$OR$$ \vec{F} = K \frac{q_1 q_2}{r^\color{red}2} \color{red}{\hat{r}} $$but definitely NOT$$ \vec{F} = K \frac{q_1 q_2}{r^\color{red}3} \color{red}{\hat{r}} $$6 Yes the dimension is different. In SI the current (A) a base unit independent from length (m), mass (kg) and time (s) because we choose to, but in CGS Gaussian unit this is not (1 unit of current = 1 g1/2 cm3/2 s-2), by setting \epsilon_{0,SI} = \frac1{4\pi}. This also leads to some perhaps unintuitive results, like the unit capacitance in CGS Gaussian is ... 6 I would say yes ! Actually some theories explaining quantum gravity use also this reasoning: gravity is a very weak interaction at a quantum level because it "leaks" into other dimensions, not observable at our scale, but that are present at this scale. The mathematical tools are different, but if you just think about gauss's law you can imagine one ... 6 The short answer is yes, and in fact you only need one single Maxwell equation, Gauss's law, together with the Lorentz force, to get Coulomb's law. More specifically, you need Gauss's law in its integral form, which is equivalent to the differential form for well-behaved fields because of Gauss's theorem. Thus, you use the law$$ ...
5
As the two charged bodies attract, they have unlike charges. So, Assuming your two charged bodies as conductors and charged equally, the system may be considered as a Capacitor. If you place a dielectric like glass of some Relative permittivity $\epsilon_r$ (3.7 to 10) which fills the empty space between the bodies, then the capacitance would be ...
5
Not a direct answer to your question but still a surprising derivation of Maxwells equations: Feynman's proof of the Maxwell equations (FJ Dyson - Phys. Rev. A, 1989) shows, that it is possible to derive Maxwells equations from Newtons second law of motion and the uncertainty principle.
5
In Newtonian physics, there was no problem with action at a distance, and indeed Newton explicitly formulated his theory of gravitation in such terms. It may be that this was criticised from a philosophical standpoint (I don't know whether it was or not), but there were no fundamental mathematical difficulties with the idea. However, in relativity the ...
5
First of all note that $k$ is not dimensionless, it is $k = \frac{1}{4 \pi \varepsilon_0}$, and $\varepsilon_0$ has dimensions of $\frac{ \text C^2}{ \text {N m}^2}$. So you have already $\frac{ \text{V C}^2 \text m^2}{ \text {N m m} ^2}$. Also, volt can be expanded as $\text V = \frac{ \text {N m}}{ \text C}$, so one gets $$\frac{ \text C^2}{ \text {N ... 5 The short answer is "observations". In the case of the gravitational law, the orbits of the planets around the sun, the moon around the earth fit mathematically a force with an inverse square law for the distance. An inverse law does not. In the case of electricity this article points out the observational history: Early investigators who suspected ... 5 Quoting from my copy of the 2nd edition of Jackson's book on Classical Electrodynamics, section 1.2: Assume that the force varies as 1/r^{2+\epsilon} and quote a value or limit for \epsilon. [...] The original experiment with concentric spheres by Cavendish in 1772 gave an upper limit on \epsilon of \left| \epsilon \right| \le 0.02. followed a ... 5 Gauss' law and Coulomb's law are equivalent - meaning that they are one and the same thing. Either one of them can be derived from the other. The rigorous derivations can be found in any of the electrodynamics textbooks, for eg., Jackson. For eg., consider a point charge q. As per Coulomb's law, the electric field produced by it is given by$$\vec{E} = ...
5
The physical reason for the appearance of a $4\pi$ somewhere in the theory is the spherical symmetry of the problem and is discussed more in other answers . Here I want to quote an interesting argument from Arnold Sommerfeld's Lectures on Theoretical Physics Vol III, which has a section dedicated to this issue. If you remove the $4\pi$ from the force law ...
4
I know that Purcell and others have used Lorentz symmetry as a pedagogical device to motivate the introduction of magnetic fields, but I do not recall ever having seen an axiomatic derivation of Maxwell's equations. It might be an interesting exercise to see precisely what assumptions beyond Lorentz symmetry and Coulomb's Law are necessary to reconstruct ...
4
With Coulomb's law and special relativity you can derive Ampere's law, which gives you magnetostatics. What's missing for electrodynamics is the displacement current ($\frac{1}{c^2} \frac{\partial E}{\partial t}$), which is a source of magnetic field arising from time-varying electric field, and not a result of the motion of electric charge. Relativity has ...
4
Frenkel and Smit definitely make a mistake. Eq. (12.1.3) page 294 is: $$-\nabla^2 \phi(\mathbf{r}) = 4\pi \rho(\mathbf{r})$$ then immediately afterwards, Eq. (12.1.4) is "the solution of this equation" "for a single charge $z$ at the origin": $$\phi(\mathbf{r}) = \frac{z}{4\pi |\mathbf{r}|}$$ This is a mistake: Eq. (12.1.4) is definitely not "the solution" ...
4
It's more the other way around, I would say. Gauss's law, together with the fact that we live in a world with 3 spatial dimensions, requires that the force between charges falls off as 1/r^2. But there are perfectly consistent analogues of electrostatics in worlds with 2 or more spatial dimensions, which each have their own Coulomb's law" -- with a ...
4
Yes, however you will change the capacitance of the capacitor based on the dielectric properties of the glass you are using and the material you are replacing (which might be air or even vacuum). The general force between the two plates can be calculated as: $$F = \dfrac{1}{2}\dfrac{Q^2}{Cd}$$ Where Q is the charge, C is the capacitance and d is the ...
4
The length scale $L$ has to be present in the denominator for dimensional reasons – only logarithms of dimensionless quantities are really "well-defined" unless one wants to introduce bizarre units such as the "logarithm of a meter". On the other hand, the dependence on $L$ is largely trivial and unphysical for most purposes. Replace $L$ by $K$ and you will ...
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2013-12-10 07:31:07
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https://hal-cea.archives-ouvertes.fr/cea-02570680
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# Plasma‐immersion ion implantation: A path to lower the annealing temperature of implanted boron emitters and simplify PERT solar cell processing
Abstract : Ion implantation is a suitable and promising solution for the massive industrialization of boron doping, which is a crucial process step for most next-generation solar cells based on crystalline silicon (c-Si). However, the use of ion implantation for boron doping is limited by the high temperature (in the 1050°C range) of the subsequent activation anneal, which is essential to dissolve the boron clusters and reach a high-emitter quality. In this work, we propose the use of plasma-immersion ion implantation (PIII) from B$_2$ H$_6$ gas precursor instead of the standard beamline ion implantation (BLII) technique to decrease this temperature down to 950°C. PIII and BLII boron emitters were compared with annealing temperatures ranging from 950°C to 1050°C. Contrary to BLII, no degradation of the emitter quality was observed with PIII implants annealed at 950°C along with a full activation of the dopants in the emitter. At 1000°C, emitter saturation current densities (J$_{0e}$) below 21 fA/cm$^2$ were obtained using the PIII technique regardless of the tested implanta-tion doses for sheet resistances between 110 and 160 $\Omega$/sq. After metallization steps, the metal/emitter contact resistances were assessed, indicating that these emitters were compatible with a conventional metallization by screen-printing/firing. The PIII boron emitters' performances were further tested with their integration in n-type passivated emitter rear totally diffused (PERT) solar cells fully doped by PIII. Promising results already show a conversion efficiency of 20.8% using a lower annealing temperature than with BLII and a reduced production cost. KEYWORDS annealing temperature, B$_2$ H$_6$ plasma, boron doping, n-type PERT solar cells, plasma-immersion ion implantation, silicon solar cells We report a new way to activate implanted boron emitter at low temperature that is the use of plasma-immersion ion implantation (PIII) from B$_2$H$_6$ plasma. A full activation of the emitter at 950°C was observed even for a high implantation dose corresponding to a sheet resistance of 112 $\Omega$/sq. Promising performances while being integrated in n-PERT solar cells fully doped by PIII were demonstrated with efficiency of 20.8% % using a lower annealing temperature than with BLII and a reduced production cost.
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Cited literature [28 references]
https://hal-cea.archives-ouvertes.fr/cea-02570680
Submitted on : Tuesday, May 12, 2020 - 11:45:53 AM
Last modification on : Friday, November 6, 2020 - 3:33:42 AM
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### Citation
Adeline Lanterne, Thibaut Desrues, Coralie Lorfeuvre, Marianne Coig, Frank Torregrosa, et al.. Plasma‐immersion ion implantation: A path to lower the annealing temperature of implanted boron emitters and simplify PERT solar cell processing. Progress in Photovoltaics, Wiley, 2019, 27 (12), pp.1081-1091. ⟨10.1002/pip.3186⟩. ⟨cea-02570680⟩
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2021-07-25 02:51:52
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http://mymathforum.com/calculus/25520-derivatives-exponential-functions.html
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My Math Forum derivatives of exponential functions
Calculus Calculus Math Forum
March 5th, 2012, 02:16 AM #1 Senior Member Joined: Jan 2012 Posts: 100 Thanks: 0 derivatives of exponential functions Hey I have been working on my assignment and I cannot get the correct answer: the question is: The average cost of producing q units of a product is given by $\bar{c}=\frac{880}{q} + 3500 \cdot \frac{e^(2q+6/880)}{q}$ What is the marginal cost if q=98 my keyboard is a little mesed up. it should be e^2q+6/880 The answer i get is 12.8400...but it is incorrect... I tried getting rid of the q by multiplying. Please help!
March 5th, 2012, 04:22 AM #2 Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 522 Math Focus: Calculus/ODEs Re: derivatives of exponential functions $\bar{c}=\frac{880+3500e^{2q+\frac{3}{440}}}{q}=\fr ac{20$$44+175e^{2q+\frac{3}{440}}$$}{q}$ $\frac{d\bar{c}}{dq}=20$$\frac{q\(2\cdot175e^{2q+\f rac{3}{440}}$$-44-175e^{2q+\frac{3}{440}}(1)}{q^2}\)$ $\frac{d\bar{c}}{dq}=\frac{20}{q^2}$$175(2q-1)e^{2q+\frac{3}{440}}-44$$$ Multiplying through by q will also work: $\bar{c}q=880+3500e^{2q+\frac{3}{440}}$ Implicitly differentiate with respect to q: $\bar{c}+q\frac{d\bar{c}}{dq}=7000e^{2q+\frac{3}{44 0}}$ $\frac{d\bar{c}}{dq}=\frac{1}{q}$$7000e^{2q+\frac{3 }{440}}-\bar{c}$$$ $\frac{d\bar{c}}{dq}=\frac{1}{q}$$7000e^{2q+\frac{3 }{440}}-\frac{1}{q}\(880+3500e^{2q+\frac{3}{440}}$$\)$ $\frac{d\bar{c}}{dq}=\frac{1}{q^2}$$7000qe^{2q+\fra c{3}{440}}-880-3500e^{2q+\frac{3}{440}}$$\)$ $\frac{d\bar{c}}{dq}=\frac{20}{q^2}$$175(2q-1)e^{2q+\frac{3}{440}}-44$$$ So now we let q = 98: $\frac{d\bar{c}}{dq}\|_{q=98}=\frac{20}{98^2}$$175( 2\cdot98-1)e^{2\cdot98+\frac{3}{440}}-44$$\approx9.47\,\times\,10^{86}$ I suspect you did not properly express the exponent of e with bracketing symbols, i.e., e^((2q+6)/880) = e^((q+3)/440). I cannot stress enough how important it is to express the problem in such a manner that guessing what is meant is not necessary. However, you should now have a better idea of how to work this problem.
March 5th, 2012, 09:23 AM #3 Senior Member Joined: Jan 2012 Posts: 100 Thanks: 0 Re: derivatives of exponential functions Nvm, I worked it out and solved the problem!!
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### the average cost of producing q units of a product is given by
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2019-11-23 01:14:11
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https://www.mathdoubts.com/derivative/
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# Derivative
The ratio of change in a varying quantity with change in another varying quantity is called a derivative.
## Introduction
A differential is a difference of two quantities of a varying quantity. In calculus, two differentials are compared by calculating a ratio of them and it is useful to us for studying the change in one varying quantity in the point of view of change in another varying quantity. The value of the ratio is called a derivative in differential calculus.
$x$ and $y$ are two variables. The values of $x$ at two different points are $x_{1}$ and $x_{2}$ and the values of $y$ at two different points are $y_{1}$ and $y_{2}$.
#### Change in Variables
If, the change in variables $x$ and $y$ are not infinitely small, then the differential are denoted by $\Delta x$ and $\Delta y$ respectively.
$\Delta x \,=\, x_{2}-x_{1}$ and $\Delta y \,=\, y_{2}-y_{1}$
Now, compare both quantities by calculating the ratio of change in one varying quantity to change in another varying quantity. In this example, the ratio of $\Delta y$ to $\Delta x$ is calculated and the value of the ratio of them is called a derivative.
$\dfrac{\Delta y}{\Delta x}$
The mathematical expressions states that the comparison of change in one varying quantity with respect to change in another varying quantity. Hence, it is read as derivative of $y$ with respect to $x$.
#### Infinitesimal Change in Variables
If the change in variables $x$ and $y$ are infinitely small, then the differentials are written as $dy$ and $dx$ respectively in differential calculus.
$dx \,=\, x_{2}-x_{1}$ and $dy \,=\, y_{2}-y_{1}$
Calculate ratio of quantities to get the value of them and it is known as derivative.
$\dfrac{dy}{dx}$
It is read as derivative of $y$ with respect to $x$
### Formula
The definition of the derivative is popularly expressed in mathematical form in two ways. Remember, The variable $y$ is generally used to represent the function $f{(x)}$ in calculus.
$(1) \,\,\,\,\,\,$ $\dfrac{d}{dx}{\, f{(x)}}$ $\,=\,$ $\displaystyle \large \lim_{\Delta x \,\to\, 0}{\normalsize \dfrac{f{(x+\Delta x)}-f{(x)}}{\Delta x}}$
$(2) \,\,\,\,\,\,$ $\dfrac{d}{dx}{\, f{(x)}}$ $\,=\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{f{(x+h)}-f{(x)}}{h}}$
In this formula, $\Delta x$ is simply represented by $h$ because they both are same.
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2022-05-28 05:04:54
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