text:
# Tag Info

6

Let's look at the case $x[n] \in \mathbb{R}$, where $x[n]$ is real. Autocorrelation is basically convolution of the signal with it's time inverse. This can be easily expressed in the frequency domain. $$\mathscr{F}\Big\{ r_{xx}[n] \Big\} = \mathscr{F}\Big\{ x[n] \Big\} \cdot \mathscr{F}\Big\{ x[-n] \Big\}$$ $$R_{xx}(\omega) = X(\omega)\cdot X^*(\... 5 Since the bulk of R’s DSP capability comes from the signal package which was ported over from the open source project Octave (itself influenced by MATLAB), there's no intrinsic limitation of R. What you have picked up on, are ecosystem preferences. We learned MATLAB in university, picked up numpy/scipy/sklearn at work, so R isn't the first weapon of choice. ... 4 There is a book by Basseville and Nikiforov called "Detection of Abrupt Changes : Theory and Application" that they released to the public as a PDF several years ago (it's out of print, now, I believe). That book looks at the basic CUSUM (cumulative sum) algorithm and how to choose appropriate thresholds for it. 4 I would try a median filter. Let your original signal be f[n]. Median filter f[n] using N pixels, where N > 2 \times S + 1, where S is the maximum number of samples in the spike. The resulting signal, lets call it g[n] should have all the spikes removed. Find the absolute of the difference between the two signals, h[n] = |f[n] - g[n]|. This ... 4 Types of signals: According to their range set (values): Real Valued, Complex valued ; According to their dimensions: Scalar, Vector ; According to their values: Continuous Amplitude, Quantized ; According to their domain set (arguments): Continuous-time, Discrete-time : According to their mappings: Deterministic, Stochastic (Random) ; According to their ... 4 The default parameters of signal.spectrogram are: nperseg = 256 noverlap = nperseg/8 = 32 This means that: The length of analysis window is 256 samples (256/250 = 1.024 second) The overlap between consecutive windows is 32 samples (32/250 = 0.128 second) The timestamps returned by signal.spectrogram correspond to the centres of a window. So in ... 4 This a very complicated question, and I would say a still open topic. The concept of stationarity is manifold, from pure statistics to applied DSP (strict, strong, wide-sense, quasi-stationarity, cyclo-stationarity, to refer to a recently closed question). The lack of access to sound models and faithful realizations renders the quest quite difficult. Non-... 4 There is in general, as @Hilmar's answer points out, no unique solution to the question of a sequence that has the given perodic autocorrelation function. In the simplest case, that a shifted version y of any sequence x (e.g. y[n] = x[n-3] for all n) has the same autocorrelation function as x. Similarly, y[n] = x[-n] for all n has the same ... 3 Q1: should the model generate a time series of length 'N=16` i.e, would the output of the above model \mathbf{y} = [y_1,y_2,\ldots,y_N] contain 16 elements where n = 1,2,\ldots,16? If one thinks about your question in terms of FIR filtering, then an input signal, x of length N filtered with an FIR filter of length p (p coefficients) will have an ... 3 A centered moving average filter is a finite impulse response (FIR) filter that affects the same weight to all the samples in the window. If you only care about time domain properties, and do not care about its relatively poor performance in the spectral domain, for a signal s that is quite stationary across the window, you can use it. It has extremely ... 3 Mathematically, shifting the frequency of a signal is pretty easy: Following @OlliNiemitalo's answer, the 0.003 frequency shift can either be done in time domain, or frequency domain. I recommend doing it in time domain, by multiplying the signal with a complex sinusoid,$$e^{j 2 \pi f_\text{shift} n},\, n \in\{1,2,\ldots\}$$that way you can get ... 3 Suppose you are given a system with transfer function$$H(z)=\frac{(1-3z^{-1})(1-7z^{-1})}{(1-4z^{-1})(1-6z^{-1})} $$Poles Poles are the values of z for which the entire function will be infinity or undefined. So, they will be the roots of the denominators, right? Look here, what values of z will turn the transfer function tend to infinity? ... 3 In addition to what others have already said, I'll try to answer it from a purely practical point of view (this is also a variant of the overlap-add technique). If your FFT length is 2048, then an overlap of 1024 (50%) means that you have twice as many analysis (FFT) frames (as compared to the number of frames without any overlapping). A 512 samples overlap ... 3 From further research I've discovered that the frequency is given by the index of the FFT multiplied by the sampling rate and divided by the size of the array. And the amplitude is the magnitude of the complex number. So the full code for such a plot would be as follows # Load ggplot library library(ggplot) # X is some set of Wait times between spikes, ... 3 ...best results come from a weighted ensemble of techniques... Maybe they do, depending on the application. But each one of the similarities mentioned, is equivalent to the other at least when we are talking about signals originating from linear systems. Cross correlation provides very good results especially if you are trying to figure out if a signal is ... 3 The product x(t)y(t) of two periodic signals with fundamental periods T_x and T_y is not a periodic signal unless T_x and T_y are rational multiples of one another; that is, T_x = aT_y where a is a rational number. Thus, except when such a relationship holds, x(t)y(t) does not have a Fourier series. When T_x is a rational multiple of T_y,... 3 To be honest, I don't think CNNs, RNNs and LSTM are useful for this kind of problem – a bandpass filter followed by a threshold would be. Now, that would have three parameters: Lower cutoff frequency Upper cutoff frequency threshold value and what is usually called "Machine Learning" is nothing but finding local minima over some (loss) function with real ... 2 If the signals are as you've drawn them (flat, with abrupt changes), then perhaps this approach might work: Take the absolute value of the differences between successive time samples. The non-zero entries are the transition times. Find these times for both signals. Subtract the two. I've made an attempt to do this in the R code below. Example output: &... 2 Yes indeed, there is a huge body of works on the topic of variational methods for signal restoration (more general than denoising), one example being total variation denoising, or proximal methods. If you can be more specific, you could get more accurate and precise answers. 2 Well, any input-output representation obviously admits a state-sapce form. for your equation in y[k] you can easily construct one as follows. Create a "shift" system (an integrator chain) as$$ \begin{aligned} x_1[k+1] &= x_2[k],\\ x_2[k+1] &= x_3[k],\\ &\vdots\\ x_n[k+1] &= y[k] \end{aligned} $$In this way indeed you have x_n[k] = y[k-1]... 2 Given the strong correlation of the downward trend in each cycle, I offer the following that would offer an improved prediction over one linear regression, but not be significantly more complicated (but still would like to see what a precise "best estimator" would be): Do a linear regression to determine the general upward trend. This would be the Income - ... 2 Since you have the x/y position for each object it should be fairly straightforward. Call the position of each p_1(t) through p_4(t). Subtract the mean from all of them since it sounds like you only care about their movement, not their position. Once you have done that it should be a fairly simple matter of cross-correlating the position functions. ... 2 First remove the DC offset (C2) from each signal by subtracting the average level from each sample, then removing the arbitrary scaling (C1) of each signal by dividing each sample by the average sample magnitude. 2 Pay attention that for Scalar Random Process the Power Spectrum Density is non negative. Namely, let  y \left[ n \right] \in \mathbb{R}  a WSS Random process with its Auto Correlation function given by:$$ {R}_{y, y} \left[ m \right] = \mathbb{E} \left[ y \left[ n \right] y \left[ n - m \right] \right] $$Then the Power Spectrum Denisty is:$$ {S}_{y, ...

2

Assuming your time series are the same length, take your data and produce spectrograms, $S_k$, for each row of the $4 \times 3000$ data matrix $D$ that you have. Since these individual time series should be similar, this means you can try to extract features by stacking these spectrograms next to each other horizontally into a large matrix. So if you have \$...

2

This is an interesting problem. I've downloaded your data and written a small program to process it. There are missing data points that don't seem to be correlated with the jumps. It seems to me the best approach is to look for matching jump values and adjust the values in between. The first step was to measure the RMS of the jumps from point to point ...

2

Technically, the most reactive filter is the all-pass filter with a gain of 1, this filter has no phase shift at all. But it is not a really useful filter. Here's what you need to take in account : 1 - IIR versur FIR An IIR filter will have less phase shift than a FIR filter for the same cut-off frequency 2 - Cut-off frequency For an IIR low-pass ...

2

LPC reduces to AR modelling only if the stochastic time process is stationary (does not change distribution parameters over time) and ergodic (average over time is equivalent to mean of ensemble average). This connection between auto-regressive coefficients and the autocovariance of the process is described by the Yule-Walker-Equations (mentioned in the ...

2

A signal is just a 1D image. So if you can make it work for images, why not for signals? Pun aside, discrete wavelets combine multiscale smoother and differential operators, so they are been used as trend and singularity detectors for a while. Wavelet coefficients can be turned into feature vectors, directly or by condensating them across subband. For ...

2

Correlation is a multiply and accumulate process. So to do what you describe, simply multiply the two signals and accumulate (integrate) the resulting output. The integration time is set based on your knowledge of how long the waveforms could be correlated; a longer integration time will give you more processing gain against background noise but will also ...

Only top voted, non community-wiki answers of a minimum length are eligible


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Spaces whose rational homology and de Rham homotopy are both finite dimensional.(English)Zbl 0546.55015

Homotopie algébrique et algèbre locale, Journ. Luminy/France 1982, Astérisque 113-114 (1984).
Summary: [For the entire collection see Zbl 0535.00017.]
If a simply connected space S satisfies the hypothese of the title let n be the top integer for which $$H^ n(S;{\mathbb{Q}})\neq 0.$$ It was known that $$\sum^{2n-1}_{i=n+1}\dim \pi_ i(S)\otimes {\mathbb{Q}}\leq 1;$$ the case when equality holds is analyzed. We also show that $$\dim H^*(S;{\mathbb{Q}})\leq 2^ n,$$ and derive a formula for the Lefschetz number of a map.

MSC:

 55P62 Rational homotopy theory 55M20 Fixed points and coincidences in algebraic topology

Zbl 0535.00017

format: plain


text:
## 简介

Shenxing Zhang张神星

2016/03 - now至今
Post-doctor
in AMSS, CAS
Beijing 100190 PRC

2010/09 - 2015/11
in School of Mathematical Sciences, USTC

2006/09 - 2010/06
in School for the Gifted Young, USTC

## 研究

My specialty is algebraic number theory, and my interest in particular lies in the theory of p-adic representations and elliptic curves.

1. Birch's lemma over global function fields, with Ouyang Y, Proc Amer Math Soc (2016).
2. Newton polygons of L-functions of polynomials $x^d+ax^{d-1}$ with $p\equiv-1\bmod d$, with Ouyang Y, Finite Fields Appl 37(2016), 285-294.
3. On second 2-descent and non-congruent numbers, with Ouyang Y, Acta Arith 170(2015), no. 4, 343-360.
4. On non-congruent numbers with 1 modulo 4 prime factors, with Ouyang Y, Sci China Math 57(2014), no. 3, 649-658.

Here are some notes.

2016/09/14 - 2016/10/26
$p$-adic abelian integrals Pierre Colmez

## 教学

2010-2015年丘成桐大学生数学竞赛，如果你有意参加可以做个参考。


text:
# How to trash files / directories?

Both DeleteFile and DeleteDirectory can only delete files / directories, and lack the option to move them to the trash. Is there an elegant (and OS-independent) way to trash items?

-
It's probably not going to be easy on Windows... JLink/.NETLink etc. might be your best bet –  rm -rf Aug 3 '13 at 20:22
If you are willing to solve it via Java. You can use FileUtil from incubator project.See the link github.com/gelosie/jdic/blob/master/src/incubator/fileutil/…. There is a method public abstract boolean recycle(File file) under class NativeFileUtil. –  s.s.o Aug 3 '13 at 20:52
Naked linux does not provide a "trash can", though specific linux window managers might, so it's hard to see how this can be universally OS independent. –  image_doctor Aug 4 '13 at 0:10
Thanks for all the comments! Using a 3rd party Java library or C++ code via JNI isn't exactly elegant, but if its the only way to get the job done, then so be it. I didn't realized that on Linux the trash implementation varies from one distribution to another, so indeed making this fully OS independent seems unlikely to do. –  Teake Nutma Aug 4 '13 at 7:14

At first I thought to withhold this answer because it is for OS X only. I thought I would look silly if someone posted a cross-platform solution. But it doesn't look like that's going to happen, so I feel I might as well show how it can be done on an OS that makes it easy. The only tricky bit here is that you have to be aware that the OS X Trash won't accept two files with the same name.

moveToTrash[fullPathName_String] :=
Module[{trash, fileName},
trash = FileNameJoin[{HomeDirectory[], ".Trash"}];
fileName = FileNameTake[fullPathName];
While[FileExistsQ[FileNameJoin[{trash, fileName}]],fileName = "#" <> fileName];
RenameFile[fullPathName, FileNameJoin[{trash, fileName}]]]

Can't show any results here; this code is only for its side-effects. However, you can play with using a piece of code that was one of my unit tests. The code should be evaluated a few times, say three, to verify that multiple files with the same name can be moved to Trash.

testFile =
Export[FileNameJoin[{HomeDirectory[], "Desktop", "test.txt"}], "The quick brown fox ..."];
moveToTrash[testFile];

Remember this is for OS X only.

-
Thanks anyhow! Two small comments though: this moves everything to ~/.Trash, but AFAIK OS X has different trash locations for each volume. So if you trash something that's not on the same volume as ~/, you won't get the native behaviour. Also, how about an \$OperatingSystem === "MacOSX" switch? –  Teake Nutma Aug 4 '13 at 18:49
@TeakeNutma. 1) I happen to prefer that all files get moved to my home .Trash. Since almost every file that goes into any .Trash gets deleted from the disk drive its on, normally it doesn't matter which particular volume it's on. It might matter if I were using portable disk drives, but I don't. 2) Why bother with an OS test when the code is OS X only? My warning is clear. If a Windows user chooses to try it, he/she must accept whatever consequences ensue. –  m_goldberg Aug 5 '13 at 2:07


text:
# Need help disproving this

## Homework Statement

Thanks for the help on the last problem. Here is the final problem set I'm stuck on:

## The Attempt at a Solution

To me it seems that there will always be a positive c so that cg(n) is greater or equal to f(n). No matter how large n is, since there's no limit to how large c can be (can even be a decimal), wouldn't that always be possible?

Thanks.

Mark44
Mentor
First off, g is not a polynomial, as it is not the sum of multiples of integral powers of n. It might be helpful for you to graph the two functions, because you would see that for n larger than about 20, the linear function dominates the other function. Although f(n) = 2n + 3 is a linear function and grows at only a constant rate, that rate is larger than that of the other function, for large enough n.

Oooh I see. Since c is fixed and chosen before the n, then there will always be an n that contradicts the statement right?

I'm guessing the same thing applies to this other problem right?

Last edited:
vela
Staff Emeritus


text:
Mathematics

# Fill in the blanks so as to make the following statements true:A ray stands on a line, then the sum of the two adjacent angles so formed is .....

$180^{\circ}$

##### SOLUTION
$\Rightarrow$  A ray stands on a line, then the sum of the two adjacent angles so formed is $180^o$.
$\Rightarrow$  The statement talks about two adjacent angles forming a linear pair and we know that sum of linear pair is $180^o.$

You're just one step away

One Word Medium Published on 09th 09, 2020
Questions 120418
Subjects 10
Chapters 88
Enrolled Students 87

#### Realted Questions

Q1 Subjective Medium
Write the supplement angle of:
$(x-30)^{o}$

Asked in: Mathematics - Lines and Angles

1 Verified Answer | Published on 09th 09, 2020

Q2 Subjective Medium
In figure, if $l||m|| n$ and $\angle 1=60^o$, find $\angle 2$.

Asked in: Mathematics - Straight Lines

1 Verified Answer | Published on 17th 08, 2020

Q3 One Word Medium
$d$ is parallel to $e$, as shown in the figure above. Find the value of $y$ in deg.

Asked in: Mathematics - Lines and Angles

1 Verified Answer | Published on 09th 09, 2020

Q4 Single Correct Medium
If angle $P$ and angle $Q$ are supplementary and the measure of angle $P$ is $60^o$, then the measure of angle $Q$ is
• A. $60^o$
• B. $30^o$
• C. $20^o$
• D. $120^o$

Asked in: Mathematics - Lines and Angles

1 Verified Answer | Published on 09th 09, 2020

Q5 Single Correct Medium
$\displaystyle\overrightarrow{DC}\perp \overline{AB}$, then $\angle CDB$ is equal to:
• A. $\displaystyle 180^{\circ}$
• B. $\displaystyle 45^{\circ}$
• C. $\displaystyle 120^{\circ}$
• D. $\displaystyle 90^{\circ}$

Asked in: Mathematics - Lines and Angles

1 Verified Answer | Published on 09th 09, 2020


text:
# how to define arbitrary plot styles (dash dash dot “--.--.” and dot dot dash “..-..-”) in mathematica [closed]

I have a graph with six curves. I have already used all the Mathematica's default plot styles (solid, dashed, dotted, dashed-dotted). So, I need two more plot styles. The natural options would be the dash dash dot "--.--." and dot dot dash "..-..-". I don't want to use symbols (like stars, squares, etc.) as a plot style since the curves represent analytical equations, not numerical simulations. Is there anyway to define these plot styles?

I appreciate any suggestion!

## closed as off-topic by Szabolcs, LCarvalho, b.gates.you.know.what, Vitaliy Kaurov, WReachNov 18 '17 at 2:49

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question arises due to a simple mistake such as a trivial syntax error, incorrect capitalization, spelling mistake, or other typographical error and is unlikely to help any future visitors, or else it is easily found in the documentation." – Szabolcs, LCarvalho, b.gates.you.know.what, Vitaliy Kaurov, WReach
If this question can be reworded to fit the rules in the help center, please edit the question.

• Read about Dashing function as an option to create several dashing types – José Antonio Díaz Navas Nov 15 '17 at 11:53
• Thanks for the suggestion, but doesn't the Dashing function just change the size of the dash? I'll take a look at it again. – Farokh Nov 15 '17 at 11:59
• But also the separation... – José Antonio Díaz Navas Nov 15 '17 at 12:00
• Thanks, it's kinda working! I have to play around a bit to make it as I wanted. – Farokh Nov 15 '17 at 12:08

Based on suggestion by José Antonio Díaz Navas, here is a minimal code to this question:

l1 = .025;
l2 = .001;
s = l1/2;

Plot[x^2, {x, 0, 2}, PlotStyle -> Dashing[{l1, s, l1, s, l2, s}]]

• DotDashed is pre-defined (as Dashing[{0, Small, Small, Small}]). – Szabolcs Nov 15 '17 at 12:41
• Is there a way to change the colors of these dashed and dotted graphs while keeping the style intact? – Samapan Bhadury Jul 21 '18 at 13:33


text:
# Fit a GARCH (1,1) - model with covariates in R

I have some experiences with time series modelling, in the form of simple ARIMA models and so on. Now I have some data that exhibits volatility clustering, and I would like to try to start with fitting a GARCH (1,1) model on the data.

I have a data series and a number of variables I think influence it. So in basic regression terms, it looks like:

$$y_t = \alpha + \beta_1 x_{t1} + \beta_2 x_{t2} + \epsilon_t .$$

But I am at a complete loss at how to implement this into a GARCH (1,1) - model? I've looked at the rugarch-package and the fGarch-package in R, but I haven't been able to do anything meaningful besides the examples one can find on the internet.

Here is an example of implementation using the rugarch package and with to some fake data. The function ugarchfit allows for the inclusion of external regressors in the mean equation (note the use of external.regressors in fit.spec in the code below).

To fix notations, the model is \begin{align*} y_t &= \lambda_0 + \lambda_1 x_{t,1} + \lambda_2 x_{t,2} + \epsilon_t, \\ \epsilon_t &= \sigma_t Z_t , \\ \sigma_t^2 &= \omega + \alpha \epsilon_{t-1}^2 + \beta \sigma_{t-1}^2 , \end{align*} where $x_{t,1}$ and $x_{t,2}$ denote the covariate at time $t$, and with the "usual" assumptions/requirements on parameters and the innovation process $Z_t$.

The parameter values used in the example are as follows.

## Model parameters
nb.period  <- 1000
omega      <- 0.00001
alpha      <- 0.12
beta       <- 0.87
lambda     <- c(0.001, 0.4, 0.2)

The image below shows the series of covariate $x_{t,1}$ and $x_{t,2}$ as well as the series $y_t$. The R code used to generate them is provided below.

## Dependencies
library(rugarch)

## Generate some covariates
set.seed(234)
ext.reg.1 <- 0.01 * (sin(2*pi*(1:nb.period)/nb.period))/2 + rnorm(nb.period, 0, 0.0001)
ext.reg.2 <- 0.05 * (sin(6*pi*(1:nb.period)/nb.period))/2 + rnorm(nb.period, 0, 0.001)
ext.reg   <- cbind(ext.reg.1, ext.reg.2)

## Generate some GARCH innovations
sim.spec    <- ugarchspec(variance.model     = list(model = "sGARCH", garchOrder = c(1,1)),
mean.model         = list(armaOrder = c(0,0), include.mean = FALSE),
distribution.model = "norm",
fixed.pars         = list(omega = omega, alpha1 = alpha, beta1 = beta))
path.sgarch <- ugarchpath(sim.spec, n.sim = nb.period, n.start = 1)
epsilon     <- as.vector(fitted(path.sgarch))

## Create the time series
y <- lambda[1] + lambda[2] * ext.reg[, 1] + lambda[3] * ext.reg[, 2] + epsilon

## Data visualization
par(mfrow = c(3,1))
plot(ext.reg[, 1], type = "l", xlab = "Time", ylab = "Covariate 1")
plot(ext.reg[, 2], type = "l", xlab = "Time", ylab = "Covariate 2")
plot(y, type = "h", xlab = "Time")
par(mfrow = c(1,1))

A fit is done with ugarchfit as follows.

## Fit
fit.spec <- ugarchspec(variance.model     = list(model = "sGARCH",
garchOrder = c(1, 1)),
mean.model         = list(armaOrder = c(0, 0),
include.mean = TRUE,
external.regressors = ext.reg),
distribution.model = "norm")
fit      <- ugarchfit(data = y, spec = fit.spec)

Parameter estimates are

## Results review
fit.val     <- coef(fit)
fit.sd      <- diag(vcov(fit))
true.val    <- c(lambda, omega, alpha, beta)
fit.conf.lb <- fit.val + qnorm(0.025) * fit.sd
fit.conf.ub <- fit.val + qnorm(0.975) * fit.sd
> print(fit.val)
#     mu       mxreg1       mxreg2        omega       alpha1        beta1
#1.724885e-03 3.942020e-01 7.342743e-02 1.451739e-05 1.022208e-01 8.769060e-01
> print(fit.sd)
#[1] 4.635344e-07 3.255819e-02 1.504019e-03 1.195897e-10 8.312088e-04 3.375684e-04

And corresponding true values are

> print(true.val)
#[1] 0.00100 0.40000 0.20000 0.00001 0.12000 0.87000

The following figure shows a parameter estimates with 95% confidence intervals, and the true values. The R code used to generate it is provided is below.

plot(c(lambda, omega, alpha, beta), pch = 1, col = "red",
ylim = range(c(fit.conf.lb, fit.conf.ub, true.val)),
xlab = "", ylab = "", axes = FALSE)
box(); axis(1, at = 1:length(fit.val), labels = names(fit.val)); axis(2)
points(coef(fit), col = "blue", pch = 4)
for (i in 1:length(fit.val)) {
lines(c(i,i), c(fit.conf.lb[i], fit.conf.ub[i]))
}
legend( "topleft", legend = c("true value", "estimate", "confidence interval"),
col = c("red", "blue", 1), pch = c(1, 4, NA), lty = c(NA, NA, 1), inset = 0.01)

• how did you estimate parameters( lambda,omega,alpha,beta)?? – chs Nov 28 '17 at 5:38
• @chs The parameter estimates were obtained with the ugarchfit function. – QuantIbex Nov 30 '17 at 20:08


text:
## Sound Design Schedules

3

3

Hi!I would like to know how do you calculate times for a sound design project? How much time (in human hours) does it take to make the sound design for a feature film (1:30hrs)? If you could specify by the different processess it would help a lot:

• Cleaning/Editing Production Sound
• Designing (recording/editing from libraries/etc) Backgrounds
• Designing (recording/editing from libraries/etc) Effects
• Recording/editing Foley
• Editing Music or
• Composing Music
• Mixing in 5.1

Am I missing something? I know times may vary depending on the films necesities, but to be a little more specific you could tell me these times for a thriller/suspense movie? With 3 or 4 action scenes (including gun fighting)? In my specific case, I'm working in a 2hr movie...

Thank you very much!

2Cronograms? Amazing. – g.a.harry – 2011-10-12T22:00:26.940

Haha sorry! I meant schedules? Or how is it said? – Manuel J Gordillo – 2011-10-13T21:45:01.647

6

Reasonable times that pop off the top of my head are about 1 reel per week for cutting DX or FX, but sometimes 2 reels per week for cutting FX on a tight time budget (did that on a recent feature - not ideal turn around time, but still doable). Figure about 8-12 hours per day of work, and you'll be able to figure out how many human hours.

A good rule of thumb I learned for cutting dialogue is to meet a minimum benchmark of 1 hour per 1 minute of dialogue. I average around 1 hour per 2 minutes of dialogue, but it has taken some time to ramp up to this speed while still maintaining quality of work.

As for the other categories, it I can't provide a definite answer because it's not something I've "timed out" really or done enough of to accurately gauge a time.

Thank you very much for your the answer Stavrosound! I'm currently working in an action movie (2hrs long)that has a lot of effects, and sequences that were cut very similar to "Requiem for a Dream" (the fast sequences when the characters take drugs, or sell drugs in the streets, etc...) which require many SFX. Actually we are 3 people working on the sound design (except for the foley recording which is being done by another teem) and we were given 2 months for editing dialogues, backgrounds design and SFX. I feel we should be given at least 1 month extra, but that's why I am asking. Thank you! – Manuel J Gordillo – 2011-10-14T14:44:17.880

4

It entirely depends on the total budget of the film, the creative and practical requirements etc.... films scale by huge amounts

Hi Tim! In this case there is no problem with the budget (It's negotiable). The film is 2 hours long, and has about 6 or 7 hard action scenes including gun fighting, and fast sequences that have been cut like "Requiem for a Dream" or "Snatch" (if you know what I mean). Could you tell me a little bit from your past experiences?

Thank you very much! – Manuel J Gordillo – 2011-10-14T14:48:15.883

total budget? some films get made for US$100k, some for us$1mill, some for us$10mill, some for us$100mill – None – 2011-10-22T18:52:34.377

1

I haven't fully worked on a feature other than just foley recording/editing or ADR, but from my experience on just doing a 20min short film to the best of my ability it has taken almost 3weeks-4weeks. This was working on my own, cleaning dialogue, editing, foley, design etc... If there is three of you and you assign tasks to eachother and they're all being done at the same time, I see no problem in it being done in a month and a half maybe?

I would stress though that I'm sure once the director watches it he'll be coming back to you for changes, which you should take into consideration. Goodluck!


text:
# Re: blindtext, unicode-math, and babel won't play nice because of setmathfont

```
```
This really looks like a LaTeX question, not a LyX question. Try some LaTeX forum, or comp.text.tex.
```
Richard

On 05/05/2014 11:08 AM, asllearner wrote:
```
```
```
I am trying to use a font for my main font, and also for math latin and greek characters, while keeping the rest as is.
```At the same time, I want to use blindtext.
```
blindtext only allows \blindmathtrue if the language has been set to english: http://texblog.org/2011/02/26/generating-dummy-textblindtext-with-latex-for-testing/ This I believe I have to do via the language package custom area in the document preferences, however using
```\usepackage[english]{babel}
breaks unicode-math.
I get
! Font \l_tmpb_font has only 8 fontdimen parameters.

```
If I use the default language package, all works, except, as stated above, blindtext english/math.
```
%This is in my preamble:
\usepackage[vargreek-shape=unicode]{unicode-math}
```
\setmathfont{Latin Modern Math} % this is obligatory or no math font will be set...
```%everything works to here. But this setmathfonts is causing the problem
```
\setmathfont[range=\mathit/{latin, Latin}]{SNsanafonmaruP} % upper and lower case latin and numbers % any font will do
```
```
\setmainfont[Mapping= tex-text]{SNsanafonmaruP} % not actually relevant....
```\usepackage{blindtext}

```
The font doesn't matter; I get the same behavior whatever font I use, even standard ones...
```
```
(NB, this is heavily redacted from a larger set of uses of unicode-math for nefarious purposes, so this a best effort to produce a MWE for what seems to be the crux of the matter).
```
PS apparently an incomplete version of this post got sent. please ignore.
```
```
```


text:
Developer FAQs

• Updated

ℹ Don't see your question here? Come to our Discord and look for the #🏗coding-chat channel!

Is there an API?

There is no API in the regular sense. Perpetual Protocol uses The Graph API to provide data, and protocol functions can be called using tools that interact directly with the on-chain smart contracts.

How do I obtain a value for liquidity?

First call

Orderbook.getOpenOrder(address trader,
int24 lowerTick,
int24 upperTick
)

and the function will return a struct:

struct Info {
uint128 liquidity;
int24 lowerTick;
int24 upperTick;
uint256 lastFeeGrowthInsideX128;
uint256 baseDebt;
uint256 quoteDebt;
}

How is Net Return calculated?

$net return = impermanent loss + pending fees for this liquidity$

To get this number programmatically, you need to get your impermanent position size

impermanentPosition = accountBalance.getTotalPositionSize() - accountBalance.getTakerPositionSize()

and the open notional

impermanentOpenNotional = accountBalance.getTotalOpenNotional() - accountBalance.getTakerOpenNotional()

then you can get

impermanentLoss = impermanentPosition * indexPrice + impermanentOpenNotional

next get pending fees

pendingFees = accountBalance.getPnlAndPendingFee()

now you can calculate net return

netReturn = pendingFees + impermanentLoss

How are base and quote calculated?

We can calculate the base/quote amount in a liquidity at any given time as long as we know the current price at that time.

p_c = current price
p_u = upper price of the liquidity
p_l = lower price of the liquidity
L = liquidity amount

There are 3 cases: 1. current price is in the range:

base amount = (1/√p_c - 1/√p_u) * L
quote amount = (√p_c - √p_l) * L

2. current price is above the upper price, only quote token in the liquidity:

quote amount = (√p_u - √p_l) * L

3. current price is below the lower price, only base toke in the liquidity:

base amount = (1/√p_l - 1/√p_u) * L


text:
Create a global map of the geoid.

## Usage

geoid = MakeGeoidGridDH (cilm, r0, gm, potref, [lmax, omega, r, order, lmax_calc, a, f, sampling])

## Returns

geoid : float, dimension (2lmax+2, sampling*2lmax+2)
A global grid of the height to the potential potref above a flattened ellipsoid of equatorial radius a and flattening f.

## Parameters

cilm : float, dimension (2, lmaxin+1, lmaxin+1)
The real spherical harmonic coefficients (geodesy normalized) of the gravitational potential referenced to a spherical interface of radius r0.
r0 : float
The reference radius of the spherical harmonic coefficients.
gm : float
The product of the gravitational constant and mass of the planet.
potref : float
The value of the potential on the chosen geoid, in SI units.
lmax : optional, integer, default = lmaxin
The maximum spherical harmonic degree of the gravitational-potential coefficients. This determines the number of latitudinal and longitudinal samples.
omega : optional, float, default = 0
The angular rotation rate of the planet.
r : optional, float, default = r0
The radius of the reference sphere that the Taylor expansion of the potential is performed on.
order : optional, integer, default = 2
The order of the Taylor series expansion of the potential about the reference radius r. This can be either 1, 2, or 3.
lmax_calc : optional, integer, default = lmax
The maximum degree used in evaluating the spherical harmonic coefficients. This must be less than or equal to lmax.
a : optional, float, default = r0
The semi-major axis of the flattened ellipsoid that the output grid geoid is referenced to. The optional parameter f must also be specified.
f : optional, float, default = 0
The flattening (R_equator-R_pole)/R_equator of the reference ellipsoid. The optional parameter a (i.e., R_equator) must be specified.
sampling : optional, integer, default = 2
If 1 (default) the output grids are equally sampled (n by n). If 2, the grids are equally spaced (n by 2n).

## Description

MakeGeoidGrid will create a global map of the geoid, accurate to either first, second, or third order, using the method described in Wieczorek (2007; equation 19-20). The algorithm expands the potential in a Taylor series on a spherical interface of radius r, and computes the height above this interface to the potential potref exactly from the linear, quadratic, or cubic equation at each grid point. If the optional parameters a and f are specified, the geoid height will be referenced to a flattened ellipsoid with semi-major axis a and flattening f. The pseudo-rotational potential is explicitly accounted for by specifying the angular rotation rate omega of the planet.

It should be noted that this geoid calculation is only strictly exact when the radius r lies above the maximum radius of the planet. Furthermore, the geoid is only strictly valid when it lies above the surface of the planet (it is necessary to know the density structure of the planet when calculating the potential below the surface).

The default is to calculate grids for use in the Driscoll and Healy routines that are equally spaced (n by 2n), but this can be changed to calculate equally sampled grids (n by n) by setting the optional argument sampling to 1. This routine uses geodesy 4-pi normalized spherical harmonics that exclude the Condon-Shortley phase.

## References

Driscoll, J.R. and D.M. Healy, Computing Fourier transforms and convolutions on the 2-sphere, Adv. Appl. Math., 15, 202-250, 1994.

Wieczorek, M. A. Gravity and topography of the terrestrial planets, Treatise on Geophysics, 10, 165-206, 2007.


text:
2019
Том 71
№ 6

# Estimation of dilatations for mappings more general than quasiregular mappings

Abstract

We consider the so-called ring $Q$-mappings, which are natural generalizations of quasiregular mappings in a sense of Väisälä’s geometric definition of moduli. It is shown that, under the condition of nondegeneracy of these mappings, their inner dilatation is majorized by a function $Q(x)$ to within a constant depending solely on the dimension of the space.

English version (Springer): Ukrainian Mathematical Journal 62 (2010), no. 11, pp 1775-1782.

Citation Example: Salimov R. R., Sevost'yanov E. A. Estimation of dilatations for mappings more general than quasiregular mappings // Ukr. Mat. Zh. - 2010. - 62, № 11. - pp. 1531–1537.

Full text


text:
# 5.13: Orbitals

The flight path of a commercial airliner is carefully regulated by the Federal Aviation Administration. Each airplane must maintain a distance of five miles from another plane flying at the same altitude and 2,000 feet above and below another aircraft (1,000 feet if the altitude is less than 29,000 feet). So, each aircraft only has certain positions it is allowed to maintain while it flies. As we explore quantum mechanics, we see that electrons have similar restrictions on their locations.

We can apply our knowledge of quantum numbers to describe the arrangement of electrons for a given atom. We do this with something called electron configurations. They are effectively a map of the electrons for a given atom. We look at the four quantum numbers for a given electron and then assign that electron to a specific orbital in the next Module.

## $$s$$ Orbitals

For any value of $$n$$, a value of $$l=0$$ places that electron in an $$s$$ orbital. This orbital is spherical in shape:

## $$p$$ Orbitals

Fro the table below, we see that we can have three possible orbitals when $$l=1$$. These are designated as $$p$$ orbitals and have dumbbell shapes. Each of the $$p$$ orbitals has a different orientation in three-dimensional space.

## $$d$$ Orbitals

When $$l=2$$, $$m_l$$ values can be $$-2, \: -1, \: 0, \: +1, \: +2$$ for a total of five $$d$$ orbitals. Note that all five of the orbitals have specific three-dimensional orientations.

## $$f$$ Orbitals

The most complex set of orbitals are the $$f$$ orbitals. When $$l=3$$, $$m_l$$ values can be $$-3, \: -2, \: -1, \: 0, \: +1, \: +2, \: +3$$ for a total of seven different orbital shapes. Again, note the specific orientations of the different $$f$$ orbitals.

Orbitals that have the same value of the principal quantum number form a shell. Orbitals within a shell are divided into subshells that have the same value of the angular quantum number. Some of the allowed combinations of quantum numbers are compared in Table$$\PageIndex{1}$$.

Table $$\PageIndex{1}$$: Electron Arrangement Within Energy Levels
Principal Quantum Number $$\left( n \right)$$ Allowable Sublevels Number of Orbitals per Sublevel Number of Orbitals per Principal Energy Level Number of Electrons per Sublevel Number of Electrons per Principal Energy Level
1 $$s$$ 1 1 2 2
2 $$s$$ 1 4 2 8
$$p$$ 3 6
3 $$s$$ 1 9 2 18
$$p$$ 3 6
$$d$$ 5 10
4 $$s$$ 1 16 2 32
$$p$$ 3 6
$$d$$ 5 10
$$f$$ 7 14

## Summary

There are four different classes of electron orbitals. These orbitals are determined by the value of the angular momentum quantum number $$l$$.

## Contributors

• CK-12 Foundation by Sharon Bewick, Richard Parsons, Therese Forsythe, Shonna Robinson, and Jean Dupon.


text:
# Multiples Sequence Maze

In this math learning exercise, students help a frog get to a pond by completing a maze puzzle. Students trace the multiples of 5 from the beginning to the end.


text:
## 過去の記録

### 2011年06月29日(水)

#### 諸分野のための数学研究会

10:30-11:30   数理科学研究科棟(駒場) 056号室

[ 講演概要 ]

また、具体例として、局所的な画像信号パターンのアピアランスモデル、およびランドマーク候補間の距離に基づいた解剖学的構造物の認識手法についても紹介する。
[ 講演参考URL ]
http://info.ms.u-tokyo.ac.jp/seminar/mathvar/future.html

#### 統計数学セミナー

15:00-16:10   数理科学研究科棟(駒場) 002号室

Statistics in genetic association studies (JAPANESE)
[ 講演概要 ]

[ 講演参考URL ]
http://www.ms.u-tokyo.ac.jp/~kengok/statseminar/2011/01.html

### 2011年06月28日(火)

#### トポロジー火曜セミナー

16:30-18:00   数理科学研究科棟(駒場) 056号室
Tea: 16:00 - 16:30 コモンルーム

On a Sebastiani-Thom theorem for directed Fukaya categories (JAPANESE)
[ 講演概要 ]

### 2011年06月27日(月)

#### 複素解析幾何セミナー

10:30-12:00   数理科学研究科棟(駒場) 128号室

Vanishing cycles for the entire functions of type $A_{1/2\infty}$ and $D_{1/2\infty}$ (JAPANESE)
[ 講演概要 ]
We introduce two elementary transcendental functions $f_{A_{1/2\infty}}$ and $f_{D_{1/2\infty}}$ of two variables. They have countably infinitely many critical points. Then, the vanishing cycles associated with the critical points form Dynkin diagrams of type $A_{1/2\infty}$ and $D_{1/2\infty}$. We calculate the spectral decomposition of the monodromy transformation by embedding the lattice of vanishing cycles into a Hilbert space. All these stories are connected with a new understanding of KP and KdV integral hierarchy. But the relationship is not yet clear.

#### 代数幾何学セミナー

16:30-18:00   数理科学研究科棟(駒場) 126号室
Vladimir Lazić 氏 (Imperial College London)
MMP revisited, II (ENGLISH)
[ 講演概要 ]
I will talk about how finite generation of certain adjoint rings implies everything we currently know about the MMP. This is joint work with A. Corti.

### 2011年06月24日(金)

#### 博士論文発表会

13:15-14:30   数理科学研究科棟(駒場) 123号室

#### 古典解析セミナー

15:00-16:30   数理科学研究科棟(駒場) 128号室

A Schwarz map of Appell's $F_2$ whose monodromy group is
related to the reflection group of type $D_4$ (JAPANESE)
[ 講演概要 ]
The system of differential equations for Appell's hypergeometric function $F_2(a,b,b',c,c';x,y)$ has four fundamental solutions.
Let $u_1,u_2,u_3,u_4$ be such solutions. If the monodromy group of the system is finite, the closure of the image of the Schwarz map $U(x,y)=(u_1(x,y),u_2(x,y),u_3(x,y),u_4(x,y))$
is a hypersurface $S$ of the 3-dimensional projective space ${\\bf P}^3$. Then $S$ is defined by $P(u_1,u_2,u_3,u_4)=0$ for a polynomial $P(t_1,t_2,t_3,t_4)$.
It is M. Kato (Univ. Ryukyus) who determined the parameter
$a,b,b',c,c'$ such that the monodromy group of the system for $F_2(a,b,b',c,c';x,y)$ is finite. It follows from his result that such a group is the semidirect product of an irreducible finite reflection group $G$ of rank four by an abelian group.
In this talk, we treat the system for $F_2(a,b,b',c,c';x,y)$ with
$(a,b,b',c,c')=(1/2,1/6,-1/6,1/3,2/3$. In this case, the monodromy group is the semidirect group of $G$ by $Z/3Z$, where $G$ is the reflection group of type $D_4$. The polynomial $P(t_1,t_2,t_3,t_4)$ in this case is of degree four. There are 16 ordinary singular points in the hypersurface $S$.
In the rest of my talk, I explain the background of the study.

#### 談話会・数理科学講演会

16:30-17:30   数理科学研究科棟(駒場) 123号室

お茶&Coffee&お菓子: 16:00～16:30 (コモンルーム)。

ｐ進微分方程式の対数的延長について (JAPANESE)
[ 講演概要 ]

### 2011年06月22日(水)

#### 統計数学セミナー

15:00-16:10   数理科学研究科棟(駒場) 002号室

[ 講演概要 ]

アファイン接続などの微分幾何学的量を用いて表される.本研究では,まず代数的な

これにより計算機代数を用いて統計的推定量の有効性を評価することが可能となる.
さらに,統計的な漸近的有効性を満たしつつ,推定値を求める数値計算に必要な計算量の

[ 講演参考URL ]
http://www.ms.u-tokyo.ac.jp/~kengok/statseminar/2011/00.html

### 2011年06月21日(火)

#### 数値解析セミナー

16:30-18:00   数理科学研究科棟(駒場) 002号室

http://www.ms.u-tokyo.ac.jp/gcoe/index.html

[ 講演概要 ]

[ 講演参考URL ]
http://www.infsup.jp/utnas/

### 2011年06月20日(月)

#### 複素解析幾何セミナー

10:30-12:00   数理科学研究科棟(駒場) 128号室

シュタイン空間内の岡・グラウエルトの原理をみたす領域 (JAPANESE)

### 2011年06月16日(木)

#### 作用素環セミナー

16:30-18:00   数理科学研究科棟(駒場) 122号室

Introduction to rigidity theory of von Neumann algebras (JAPANESE)

### 2011年06月15日(水)

#### 代数学コロキウム

17:30-18:30   数理科学研究科棟(駒場) 056号室

Product formula for $p$-adic epsilon factors (ENGLISH)
[ 講演概要 ]
I would like to talk about my recent work jointly with A. Marmora on a product formula for $p$-adic epsilon factors. In 80's Deligne conjectured that a constant appearing in the functional equation of $L$-function of $\\ell$-adic lisse sheaf can be written by means of local contributions, and proved some particular cases. This conjecture was proven later by Laumon, and was used in the Lafforgue's proof of the Langlands' program for functional filed case. In my talk, I would like to prove a $p$-adic analog of this product formula.

(本講演は「東京パリ数論幾何セミナー」として、インターネットによる東大数理とIHESとの双方向同時中継で行います。)

### 2011年06月14日(火)

#### トポロジー火曜セミナー

17:00-18:00   数理科学研究科棟(駒場) 056号室
Tea: 16:40 - 17:00 コモンルーム

Donaldson-Tian-Yau's Conjecture (JAPANESE)
[ 講演概要 ]
For polarized algebraic manifolds, the concept of K-stability
introduced by Tian and Donaldson is conjecturally strongly correlated
to the existence of constant scalar curvature metrics (or more
generally extremal K\\"ahler metrics) in the polarization class. This is
known as Donaldson-Tian-Yau's conjecture. Recently, a remarkable
progress has been made by many authors toward its solution. In this
talk, I'll discuss the topic mainly with emphasis on the existence
part of the conjecture.

### 2011年06月13日(月)

#### 複素解析幾何セミナー

10:30-12:00   数理科学研究科棟(駒場) 128号室

On the complement of effective divisors with semipositive normal bundle (JAPANESE)

#### 講演会

16:00-17:30   数理科学研究科棟(駒場) 128号室
CHEN Hua 氏 (Wuhan University)
Regularity of Solutions for a Class of Degenerate Equations (ENGLISH)
[ 講演概要 ]
In this talk, I would report some recent joint results on the Gevrey (or analytic) regularities of solutions for some degenerate partial differential equations, which including
(1) generalized Kolmogorov equations,
(2) Fokker-Planck equations,
(3) Landau equations and
(4) sub-elliptic Monge-Ampere equations.

### 2011年06月09日(木)

#### 作用素環セミナー

16:30-18:00   数理科学研究科棟(駒場) 122号室

On the free Fisher information distance and the free logarithmic Sobolev inequality (JAPANESE)

#### 応用解析セミナー

16:30-18:00   数理科学研究科棟(駒場) 128号室

Spectral representations and scattering for Schr\\"odinger operators on star graphs (JAPANESE)
[ 講演概要 ]
We consider Schr\\"odinger operators defined on star graphs with Kirchhoff boundary conditions. Under suitable decay conditions on the potential, we construct a complete set of eigenfunctions to obtain spectral representations of the operator. The results are applied to give a time dependent formulation of the scattering theory. Also we use the spectral representation to determine an integral equation of Marchenko which is fundamental to enter into the inverse scattering problems.

### 2011年06月08日(水)

#### 代数学コロキウム

16:30-17:30   数理科学研究科棟(駒場) 056号室

[ 講演概要 ]
カスプ形式とEisenstein級数のFourier係数の間の合同式からそれらに付随する L 関数の特殊値の間の合同式を導くという問題を考える。
これは保型形式の重さが 2 の場合はVatsal氏によって証明された。本講演では,重さが 2 以上の場合に一般化できた結果を紹介する。
さらに、この結果を保型形式に付随する p 進Galois表現が剰余して可約という特別な場合の岩澤主予想に応用する。これは、重さが 2 の場合のGreenberg氏及びVatsal氏の結果を部分的に一般化したものである。

### 2011年06月07日(火)

#### 代数幾何学セミナー

16:30-18:00   数理科学研究科棟(駒場) 126号室
Chenyang Xu 氏 (MIT)
Log canonical closure (ENGLISH)
[ 講演概要 ]
(joint with Christopher Hacon) In this talk, we will address the problem on given a log canonical variety, how we compactify it. Our approach is via MMP. The result has a few applications. Especially I will explain the one on the moduli of stable schemes.
If time permits, I will also talk about how a similar approach can be applied to give a proof of the existence of log canonical flips and a conjecture due to Kollár on the geometry of log centers.

#### 数値解析セミナー

16:30-18:00   数理科学研究科棟(駒場) 002号室

http://www.ms.u-tokyo.ac.jp/gcoe/index.html

)

(JAPANESE)
[ 講演概要 ]

[ 講演参考URL ]
http://www.infsup.jp/utnas/

#### Lie群論・表現論セミナー

16:30-18:00   数理科学研究科棟(駒場) 056号室
トポロジー火曜セミナーと合同で行います

Rigidity of group actions via invariant geometric structures
(JAPANESE)
[ 講演概要 ]
It is a homomorphism into a FINITE dimensional Lie group that is concerned with in the classical RIGIDITY theorems such as those of Mostow and Margulis. In the meantime, differentiable GROUP ACTIONS for which we ask rigidity problems is a homomorphism into a diffeomorphism group, which is a typical example of INFINITE dimensional Lie groups. The purpose of the present talk is exhibiting several rigidity theorems for group actions in which I have been involved for years. Although quite a few fields of mathematics, such as ergodic theory, the theory of smooth dynamical systems, representation theory and so on, have made remarkable contributions to rigidity problems, I would rather emphasis geometric aspects: I would focus on those rigidity phenomenon for group actions that are observed by showing that the actions have INVARIANT GEOMETRIC STRUCTURES.

#### トポロジー火曜セミナー

16:30-18:00   数理科学研究科棟(駒場) 056号室
Lie群論・表現論セミナーと合同 Tea: 16:00 - 16:30 コモンルーム

Rigidity of group actions via invariant geometric structures (JAPANESE)
[ 講演概要 ]
It is a homomorphism into a FINITE dimensional Lie group that is concerned with in the classical RIGIDITY theorems such as those of Mostow and Margulis. In the meantime, differentiable GROUP ACTIONS for which we ask rigidity problems is a homomorphism into a diffeomorphism group, which is a typical example of INFINITE dimensional Lie groups. The purpose of the present talk is exhibiting several rigidity theorems for group actions in which I have been involved for years. Although quite a few fields of mathematics, such as ergodic theory, the theory of smooth dynamical systems, representation theory and so on, have made remarkable contributions to rigidity problems, I would rather emphasis geometric aspects: I would focus on those rigidity phenomenon for group actions that are observed by showing that the actions have INVARIANT GEOMETRIC STRUCTURES.

### 2011年06月06日(月)

#### 複素解析幾何セミナー

10:30-12:00   数理科学研究科棟(駒場) 128号室

An invariant surface of a fixed indeterminate point for rational mappings (JAPANESE)

#### 代数幾何学セミナー

16:30-18:00   数理科学研究科棟(駒場) 126号室

Multiplier ideals via Mather discrepancies (JAPANESE)
[ 講演概要 ]
For an arbitrary variety we define a multiplier ideal by using Mather discrepancy.
This ideal coincides with the usual multiplier ideal if the variety is normal and complete intersection.
In the talk I will show a local vanishing theorem for this ideal and as corollaries we obtain restriction theorem, subadditivity theorem, Skoda type theorem, and Briancon-Skoda type theorem.


text:
Publication
Title
Measurement of transverse momentum relative to dijet systems in PbPb and pp collisions at $\sqrt{s_{NN}}$=2.76 TeV
Author
Institution/Organisation
CMS Collaboration
Abstract
 An analysis of dijet events in PbPb and pp collisions is performed to explore the properties of energy loss by partons traveling in a quark-gluon plasma. Data are collected at a nucleon-nucleon center-of-mass energy of 2.76 TeV at the LHC. The distribution of transverse momentum (p(T)) surrounding dijet systems is measured by selecting charged particles in different ranges of p(T) and at different angular cones of pseudorapidity and azimuth. The measurement is performed as a function of centrality of the PbPb collisions, the p(T) asymmetry of the jets in the dijet pair, and the distance parameter R used in the anti-k(T) jet clustering algorithm. In events with unbalanced dijets, PbPb collisions show an enhanced multiplicity in the hemisphere of the subleading jet, with the p(T) imbalance compensated by an excess of low-p(T) particles at large angles from the jet axes.
Language
English
Source (journal)
Journal of high energy physics. - Bristol
Publication
Bristol : 2016
ISSN
1126-6708
1029-8479 [online]
Volume/pages
1(2016), 52 p.
Article Reference
006
ISI
000370436500001
Medium
E-only publicatie
Full text (Publisher's DOI)
Full text (open access)
UAntwerpen
 Faculty/Department Research group Publication type Subject Affiliation Publications with a UAntwerp address


text:
#### 题目列表

In an appliance store, the ratio of the number of radios in stock to the number of television sets in stock is 0.4. Exactly 10 radios and x television sets are added to the appliances in stock, and the ratio of radios to television sets in stock remains at 0.4.

x

#### Quantity B

30

y is a positive even integer.

x+y

#### Quantity B

x+1

For all x, $x^{2}+4x-12 = (x-r)(x+s)$, where r > 0 and s > 0.

r

#### Quantity B

s

x and y are integers and 0 < x < y.

#### Quantity A

$x^{y}$

#### Quantity B

$y^{x}$

O is the center of the circle, and OA = BC.

#### Quantity A

The measure of angle ABC

#### Quantity B

60°

List L: 84, 73, 87, 88, 75, 90

The standard deviation of the numbers in list L is s. List M consists of 6 different numbers, each of which is obtained by adding 9 to one of the numbers in list L. The standard deviation of the numbers in list M is t.

s+3

#### Quantity B

t

Two sides of a right triangle have lengths of 5 and 10, respectively.

#### Quantity A

The length of the third side of the triangle

#### Quantity B

11

$\frac{2+\frac{1}{2})^{4}-2^{4}}{(2+\frac{1}{2})^{2}+2^{2}}$=
A ball is dropped onto a concrete floor from a height of 50 feet, after which it bounces n times. After each bounce, the ball rebounds to $\frac{1}{3}$ of the height from which it has just fallen. To what height does the ball rebound after the nth bounce?
If $2x -3 = 5y+ 2,$ then $x- \frac{5}{2}y$=
In the freshman class at College X, there are 400 students taking a certain French course and 500 students taking a certain English course. If 200 students in the freshman class are taking both the French course and the English course, how many students in the freshman class are taking the French course or the English course or both?
Sonia went on a 360-mile trip in her car. She drove the first 200 miles in 4 hours, stopped 45 minutes for lunch, and then drove the rest of the way at an average speed of 58 miles per hour. If the total time for the trip includes the lunch stop, what was the total time for the trip?

_____hours

For how many of the categories shown was the dollar amount of the expenditures in 1995 less than $1 billion? Approximately what was the ratio of the dollar amount of the expenditures for prescriptions in 1995 to the dollar amount of the expenditures for prescriptions in 1970? From 1970 to 1995. the dollar amount of expenditures for nursing home care increased by approximately what percent? A list consists of 24 different positive four-digit integers ordered from least to greatest. Each integer in the list consists of the four digits 6, 7, 8, and 9, without repetitions. If 8,697 is the nth integer in the list, what is the value of n? n=_____ The surface area of a sphere of radius r is equal to $4πr^{2}$. Sphere K has radius k, and sphere N has radius n. If the ratio of the surface area of sphere K to the surface area of sphere N is 9 to 4, what is the ratio of k to n? At the end of each business day, the owners of Stores X and Y calculate the total daily revenue received that day. For a five-business-day period, the average (arithmetic mean) total daily revenue was$6,000 for Store X and \$5,000 for Store Y. Which of the following statements must be true?

Indicate all such statements.
2x + 5 = 3x - 7

2x+4

#### Quantity B

3x-6

List K consists of 9 positive integers. In list K, the range of the integers is 20, the mode of the integers is 3, and the median of the integers is 7.

#### Quantity A

The average (arithmetic mean) of the integers in list K

13

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My Math Forum How do I prove this seemingly simple trigonometric identity

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 December 11th, 2014, 08:15 PM #1 Senior Member   Joined: Aug 2014 From: India Posts: 343 Thanks: 1 How do I prove this seemingly simple trigonometric identity $\displaystyle a=\sinθ+\sinϕ$ $\displaystyle b=\tanθ+\tanϕ$ $\displaystyle c=\secθ+\secϕ$ Show that, $\displaystyle 8bc=a[4b^2 + (b^2-c^2)^2]$ I tried to solve this for hours and have gotten nowhere. Here's what I've got so far: $\displaystyle \\a= 2\sin(\frac{\theta+\phi}{2})\cos(\frac{\theta-\phi}{2})$ $\displaystyle \\ b = \frac{2\sin(\theta+\phi)}{\cos(\theta+\phi)+ \cos(\theta-\phi)}$ $\displaystyle \\c=\frac{2(\cos\theta+\cos\phi)}{\cos(\theta+\phi )+\cos(\theta-\phi)}$ $\displaystyle \\a^2 = \frac{\sin^2(\theta+\phi)[\cos(\theta+\phi)+1]} {\cos(\theta+\phi)+1}\\\cos(\theta-\phi)=\frac{ca}{b}-1\\\sin^2(\frac{\theta+\phi}{2})=\frac{2a^2b}{4(ca +b)}$ Last edited by skipjack; December 12th, 2014 at 01:44 AM.
 December 12th, 2014, 03:59 AM #2 Global Moderator   Joined: Dec 2006 Posts: 19,981 Thanks: 1853 \displaystyle \begin{align*}4\cos^2(\theta)\cos^2(\phi)b^2 &= 4\cos^2(\theta)\cos^2(\phi)(\tan(\theta) + \tan(\phi))^2 \\ &= 4(\sin(\theta)\cos(\phi) + \cos(\theta)\sin(\phi))^2 \\ &= 4\sin^2(\theta + \phi) \\ &= 16\sin^2((\theta + \phi)/2)\cos^2((\theta + \phi)/2) \\ &= 16(1 - \cos^2((\theta + \phi)/2))\cos^2((\theta + \phi)/2) \\ &= 16\cos^2((\theta + \phi)/2) - 16\cos^4((\theta + \phi)/2)\end{align*} \displaystyle \begin{align*}c^2 - b^2 &= \sec^2\theta + \sec^2\phi + 2\sec(\theta)\sec(\phi) - (\tan^2\theta + \tan^2\phi + 2\tan(\theta)\tan(\phi)) \\ &= 2 + 2\sec(\theta)\sec(\phi) - 2\tan(\theta)\tan(\phi)\end{align*} \displaystyle \begin{align*}\cos(\theta)\cos(\phi)(c^2 - b^2) &= 2(\cos(\theta)\cos(\phi) + 1 - \sin(\theta)\sin(\phi)) \\ &= 2(1 + \cos(\theta + \phi)) \\ &= 4\cos^2((\theta + \phi)/2)\end{align*} $\displaystyle \cos^2(\theta)\cos^2(\phi)(b^2 - c^2)^2 = 16\cos^4((\theta + \phi)/2)$ $\displaystyle \cos^2(\theta)\cos^2(\phi)(4b^2 + (b^2 - c^2)^2) = 16\cos^2((\theta + \phi)/2)$ \displaystyle \begin{align*}8\cos^2(\theta)\cos^2(\phi)bc/a &= 8\cos^2(\theta)\cos^2(\phi)(\tan(\theta) + \tan(\phi))(\sec(\theta) + \sec(\phi))/(\sin(\theta) + \sin(\phi)) \\ &= 8(\sin(\theta)\cos(\phi) + \cos(\theta)\sin(\phi))(\cos(\theta) + \cos(\phi))/(\sin(\theta) + \sin(\phi)) \\ &= 8\sin(\theta + \phi)(\cos(\theta) + \cos(\phi))/(\sin(\theta) + \sin(\phi)) \\ &= \frac{8(2\sin((\theta + \phi)/2)\cos((\theta + \phi)/2))(2\cos((\theta + \phi)/2)\cos((\theta - \phi)/2))}{2\sin((\theta + \phi)/2)\cos((\theta - \phi)/2)} \\ &= 16\cos^2((\theta+\phi)/2)\end{align*} The desired result follows.

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# Why does a radial basis function kernel imply an infinite dimension map?

I understand that each kernel implies a particular feature map. For instance for $x,z \in R^2$ the kernel $K(x,z)=(dot(x,z))^2$ implies a feature map $φ(<x_1, x_2>)=<x_1^2 , x_1 x_2 , x_1 x_2, x_2^2>$, as though my inputs are transformed from $R^2$ to $R^4$ in the process of computing $K(x,z)$.

The polynomial kernels are clear enough to see in this way. However the radial basis function kernel in particular, i.e. $K(x,z) = exp(-||x-z||^2/c^2)$, supposedly implies an infinite dimensional feature map. Exactly why is this so?

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# a occurs when price is above market equilibrium

First published 28 Nov 2010. When two lines on a diagram cross, this intersection usually means something. Conversely, if a situation is inefficient, it becomes possible to benefit at least one party without imposing costs on others. The only thing left … Now, compare the quantity demanded and quantity supplied at this price. This results in unsold inventories and forces producers to offer reduced price. A price below equilibrium creates a shortage. Generally any time the price for a good is below the equilibrium level, incentives built into the structure of demand and supply will create pressures for the price to rise. Transactions above this price is prohibited. Let’s practice solving a few equations that you will see later in the course. In the above diagram, price (P2) is below the equilibrium. As this occurs, the shortage will decrease. Price adjusts when plans don’t match. 4. (Q2-Q1) Therefore firms would reduce price and supply less. Suppose that the price is $1.20 per gallon, as the dashed horizontal line at this price in Figure 3, below, shows. The equilibrium quantity is Q1. A market occurs where buyers and sellers meet to exchange money for goods. You can also find it in Table 1 (the numbers in bold). This would encourage more … The equilibrium point of the market is the point at which the supply curves cross each other. Cutting price encourages a movement along the demand curve (more is bought) 3. Movements from this point will cause either a shortage or a surplus in the market. D. the price floor would have no impact on the market because it is higher than equilibrium price. Governmental intervention can often create surplus as well, particularly through the utilization of a price floor if it is set at a price above the market equilibrium. The equilibrium price of soda, that is, the price where Qs = Qd will be$2. We call this a situation of excess demand (since Qd > Qs) or a shortage. Finding market equilibrium with equations, Advantages and disadvantages of monopolies, NEET – ‘Not in Employment, Education or Training’. In a perfectly competitive market, a firm cannot change the price of a product by modifying the quantity of its output. Figure 5. Efficiency in the demand and supply model has the same basic meaning: the economy is getting as much benefit as possible from its scarce resources, and all the possible gains from trade have been achieved. Question: QUESTION 19 Excess Demand Occurs: A. This happens either because there is more supply than what the market is demanding or because there is more demand than the market is supplying. If you look at either Figure 1 or Table 1, you’ll see that at most prices the amount that consumers want to buy (which we call the quantity demanded) is different from the amount that producers want to sell (which we call the quantity supplied). These price increases will stimulate the quantity supplied and reduce the quantity demanded. However, if a market is not at equilibrium, then economic pressures arise to move the market toward the equilibrium price and the equilibrium quantity. If the price of a good is above equilibrium, this means that the quantity of the good supplied exceeds the quantity of the good demanded. Quantity supplied (680) is greater than quantity demanded (500). We have equilibrium price and quantity of $3.0 and 210 units respectively. In this situation, some firms will want to cut prices, because it is better to sell at a lower price than not to sell at all. With a surplus, gasoline accumulates at gas stations, in tanker trucks, in pipelines, and at oil refineries. Also, a competitive market that is operating at equilibrium is an efficient market. Imagine, for example, that the price of a gallon of gasoline was above the equilibrium price—that is, instead of$1.40 per gallon, the price is … How much will producers supply, or what is the quantity supplied? Lesson summary: Market equilibrium, disequilibrium, and changes in equilibrium. However, if a market is not at equilibrium, then economic pressures arise to move the market toward the equilibrium price and equilibrium quantity. Imagine that the price of a gallon of gasoline were $1.80 per gallon. If there was an increase in income the demand curve would shift to the right (D1 to D2). The Supply Curve B. Finally, recall that the soda market converges to the point where supply equals demand, or, We now have a system of three equations and three unknowns (Qd, Qs, and P), which we can solve with algebra. Or, to put it in words, the amount that producers want to sell is less than the amount that consumers want to buy. There is a surplus. Because the graphs for demand and supply curves both have price on the vertical axis and quantity on the horizontal axis, the demand curve and supply curve for a particular good or service can appear on the same graph. The price mechanism refers to how supply and demand interact to set the market price and amount of goods sold. Market equilibrium can be shown using supply and demand diagrams. Refer to Table 2. $\begin{array}{l}\,16-2P=2+5P\\-2+2P=-2+2P\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,14=7P\end{array}$. If this is the case, produces will be willing to supply more than consumers demand creating a surplus. If price is above the equilibrium. Figure 3. Demonstration of the law of market equilibrium. That confirms that we’ve found the equilibrium quantity. You are welcome to ask any questions on Economics. Suppose the supply of soda is, where Qs is the amount of soda that producers will supply (i.e., quantity supplied). At P2 there is disequilibrium (excess supply) 2. This means that we did our math correctly, since. We know that a firm is in equilibrium when its profits are maximum, which relies on the cost and revenue conditions of the firm. If the market price is above or below the equilibrium price, the market is in disequilibrium. If a market is at its equilibrium price and quantity, then it has no reason to move away from that point, because it’s balancing the quantity supplied and the quantity demanded. Initially, there would be a shortage of the good. Similarly, any time the price for a good is above the equilibrium level, similar pressures will generally cause the price to fall. If price is greater than equilibrium level, there will be a surplus, which forces price down. – from Â£6.99. In order for a price ceiling to be effective, it must be set below the natural market equilibrium. This mutually desired amount is called the equilibrium quantity. Market equilibrium is determined at the intersection of the market demand and market supply. Therefore there is a shortage of (Q2 â Q1). Excess demand occurs at a price less than the equilibrium price. Equilibrium in a market occurs when the price balances the plans of buyers and sellers. This is a state of disequilibrium because there is either a shortage or surplus and firms have an incentive to change the price. This is the currently selected item. A surplus exists when the price is above equilibrium, which encourages sellers to lower their prices to eliminate the surplus. Market equilibrium is said to occur when there is no tendency for the price to change and supply is in balance with demand. This balance is a natural function of a free-market economy. Let’s consider one scenario in which the amount that producers want to sell doesn’t match the amount that consumers want to buy. Consider our gasoline market example. We will explore this important concept in detail in the next module on applications of supply and demand. If price is less than equilibrium level. and both Qd and Qs are equal to 12. At this equilibrium point, the market is efficient because the optimal amount of gasoline is being produced and consumed. Whenever The Market Is Not In Equilibrium OCwhenever The Market Is In Equilibrium с. Od.when Price Is Above The Equilibrium Price QUESTION 20 The Entire Group Of Buyers And Sellers Of A Particular Good Or Service Makes Up Oa. which forces price up. As before, the equilibrium occurs at a price of$1.40 per gallon and at a quantity of 600 gallons. If price was at P2, this is above the equilibrium of P1. The answer is: a surplus or a shortage. We call this a situation of excess supply (since Qs > Qd) or a surplus. Also as price falls, firms have less incentive to supply. At the price of P2, then supply (Q2) would be greater than demand (Q1) and therefore there is too much supply. Suppose that the demand for soda is given by the following equation: where Qd is the amount of soda that consumers want to buy (i.e., quantity demanded), and P is the price of soda. Demand and Supply for Gasoline: Equilibrium. Similarly, the law of supply says that when price decreases, producers supply a lower quantity. • Policy makers set ceiling price below the market equilibrium price which they believed is too high. This would encourage more demand and therefore the surplus will be eliminated. Let’s return to our gasoline problem. The price will rise until the shortage is eliminated and the quantity supplied equals quantity demanded. These price reductions will, in turn, stimulate a higher quantity demanded. This accumulation puts pressure on gasoline sellers. What does it mean when the quantity demanded and the quantity supplied aren’t the same? The existence of this surplus gives sellers an incentive to lower their price, thus sending the price downward toward its equilibrium level. In this situation, eager gasoline buyers mob the gas stations, only to find many stations running short of fuel. Therefore the price will rise to P1 until there is no shortage and supply = demand. A shortage occurs at a price below the equilibrium level. Watch this video for a closer look at market equilibrium: Equilibrium is important to create both a balanced market and an efficient market. Select one: a. The new market equilibrium will be at Q3 and P1. As before, the equilibrium occurs at a price of $1.40 per gallon and at a quantity of 600 gallons. A market situation in w… Economists typically define efficiency in this way: when it is impossible to improve the situation of one party without imposing a cost on another. Shortage. Step 1: Isolate the variable by adding 2P to both sides of the equation, and subtracting 2 from both sides. We can do this by plugging the equilibrium price into either the equation showing the demand for soda or the equation showing the supply of soda. Disequilibrium occurs when the quantity supplied does not equal the quantity demanded. A shortage will exist at any price below equilibrium, which leads to the price of the good increasing. When the surplus is eliminated, the quantity supplied just equals the quantity demanded—that is, the amount that producers want to sell exactly equals the amount that consumers want to buy. At most prices, planned demand does not equal planned supply. The process continues until the equilibrium price is reached. As you can see, the quantity supplied or quantity demanded in a free market will correct over time to restore balance, or equilibrium. You can also find these numbers in Table 1, above. The price will rise until the shortage is eliminated and the quantity supplied equals quantity demanded. Figure 4. Price Ceiling: is legally imposed maximum price on the market. The market-clearing price and output are determined at the equilibrium point. There is a surplus of supply. On a graph, the Therefore the price will rise to P1 until there is no shortage and supply = demand. Assume actual price is above market equilibrium price.-- the negative slope of the demand curve for buyers will mean that the quantity demanded will be less than the equilibrium quantity; -- the positive slope of the supply curve for sellers will mean that the quantity supplied will be greater When the government sets a price floor which is above the equilibrium price, a surplus will develop. constant interaction of buyers and sellers brings about a stable price for a product or service How far will the price fall? So, if the price is$2 each, consumers will purchase 12. Because the market price of $2.50 is above the equilibrium price, the quantity supplied (10 cones) exceeds the quantity demanded (4 cones). Together, demand and supply determine the price and the quantity that will be bought and sold in a market. Since. Figure 2. A market is in equilibrium when price adjusts so that quantity demanded equals quantity supplied. How far will the price rise? Our site uses cookies so that we can remember you, understand how you use our site and serve you relevant adverts and content. At this price, the quantity demanded is 700 gallons, and the quantity supplied is 550 gallons. As we will see, when supply and demand are not in balance, economic forces will work until the balance is restored. At this price, demand would be greater than the supply. At any other price, the quantity demanded does not equal the quantity supplied, so the market is not in equilibrium at that price. In other words, the optimal amount of each good and service is being produced and consumed. At any price above$3.0, the quantity supplied exceeds the quantity demanded. There is a surplus. If you have only the demand and supply schedules, and no graph, you can find the equilibrium by looking for the price level on the tables where the quantity demanded and the quantity supplied are equal (again, the numbers in bold in Table 1 indicate this point). A price ceiling occurs when the government puts a legal limit on how high the price of a product can be. If the current market price was $8.00 – there would be excess supply of 12,000 units. If price was at P2, this is above the equilibrium of P1. CC BY-NC-ND: Attribution-NonCommercial-NoDerivatives, https://cnx.org/contents/aWGdK2jw@11.346:D3bzsNhU@8/Demand-Supply-and-Equilibrium-, Define equilibrium price and quantity and identify them in a market, Define surpluses and shortages and explain how they cause the price to move towards equilibrium. In economics, these forces are supply and demand. When a price ceiling is set, a shortage occurs. In other words, the market will be in equilibrium again. 1. You can see this in Figure 2 (and Figure 1) where the supply and demand curves cross. If a price ceiling is set above the market equilibrium price, the price ceiling has no impact on the economy. Let’s use demand. True Auctions in recent years have resulted in higher prices paid for letters written by John Wilkes Booth than those written by Abraham Lincoln. Changes in equilibrium price and quantity: the four-step process. Excess demand is not linked to price but to quantity b. below c. equal to d. above A supply curve is a graphical illustration of the relationship between quantity supplied and Select one: a. demand. Be… situation where the quantity demanded in a market is greater than the quantity supplied; occurs at prices above the equilibrium surplus (or excess supply): situation where the quantity demanded in a market is less than the quantity supplied; occurs at prices below the equilibrium It should be clear from the previous discussions of surpluses and shortages, that if a market is not in equilibrium, market forces will push the market to the equilibrium. Equilibrium is the point where the amount that buyers want to buy matches the point where sellers want to sell. In order for a price ceiling to be effective, it must be set below the market equilibrium price. Recall that the law of demand says that as price decreases, consumers demand a higher quantity. Did you have an idea for improving this content? In order to understand market equilibrium, we need to start with the laws of demand and supply. Market equilibrium occurs when price is at$3 per unit: Quantity Demanded = Quantity Supplied = 30 units. At this price, the quantity demanded is 500 gallons, and the quantity of gasoline supplied is 680 gallons. Therefore, the firm can alter the quantity of its output without changing the price of the product. Price will fall until S= D and the market is in equilibrium. How far will the price rise? In this situation where demand goes up, both price and quantity are going to go up assuming we have this upwards sloping supply curve again. Note that whenever we compare supply and demand, it’s in the context of a specific price—in this case, $1.80 per gallon. If the market price is above the equilibrium, there is an excess supply in the market, and the supply exceeds the demand. As price rises, there will be a movement along the demand curve and less will be demanded. As this occurs, the shortage will decrease. We’ve just explained two ways of finding a market equilibrium: by looking at a table showing the quantity demanded and supplied at different prices, and by looking at a graph of demand and supply. In the diagram below, the equilibrium price is P1. Price regulates buying and selling plans. Explain: A price floor may guarantee a price that is above the market equilibrium. We call this equilibrium, which means “balance.” In this case, the equilibrium occurs at a price of$1.40 per gallon and at a quantity of 600 gallons. The equilibrium price in the market is $5.00 where demand and supply are equal at 12,000 units If the current market price was$3.00 – there would be excess demand for 8,000 units, creating a shortage. These conditions can vary in the long and short-term. • When the price is above the equilibrium point, a surplus exists, and inventories build up. The price that equates the quantity demanded with the quantity supplied is the equilibrium price and amount that people are willing to buy and sellers are willing to offer at the equilibrium price level is the equilibrium quantity. In other words, the market will be in equilibrium again. â A visual guide Oil companies and gas stations recognize that they have an opportunity to make higher profits by selling what gasoline they have at a higher price. Suppliers lower their price in an effort to sell the unwanted goods. We’d love your input. A price above equilibrium creates a surplus. Excess Demand Occurs When The Actual Price In Some Market Is The Equilibrium Price. This situation is referred to as a ‘ surplus ’ or ‘ producer surplus.’ Due to the high inventory holding cost, suppliers will reduce the price and offer discounts or other offers to stimulate more demand. we can set the demand and supply equations equal to each other: $\begin{array}{c}\,\,Qd=Qs\\16-2P=2+5P\end{array}$. the equilibrium Excess demand occurs when the actual price in some market is_ price. B. a surplus will occur and producers will produce less and lower prices. At the price of P2, then supply (Q2) would be greater than demand (Q1) and therefore there is too much supply. • Intention of price ceiling is keeping stuff affordable for poor people. Price Floor: A price floor ensures a minimum price is charged for a specific good, often higher than that what the previous market equilibrium determined. If the market price is above the equilibrium price, A. a shortage will occur and producers will produce more and lower prices. Therefore the price and quantity supplied will increase leading to a new equilibrium at Q2, P2. Or, to put it in words, the amount that producers want to sell is greater than the amount that consumers want to buy. (Remember, these are simple equations for lines). Step 2: Simplify the equation by dividing both sides by 7. When the market price of a good or service rises above equilibrium on its own, the number of buyers exhibiting demand for it is reduced. An increase in supply would lead to a lower price and more quantity sold. (Q2-Q1). At our new equilibrium point, this is Q2 and then this right over here is P2, our new equilibrium price or our new equilibrium quantity. Supply, and Equilibrium in Markets for Goods and Services. When at the current price level, the quantity demanded is more than quantity supplied, a situation of excess demand is said to arise in the market. If there is a shortage, firms will put up prices and supply more. When two lines on a diagram cross, this intersection usually means something. Whenever there is a surplus, the price will drop until the surplus goes away. Equilibrium is formally defined as a state of rest or balance due to the equal action of opposing forces. The supply and demand curves for gasoline. Market surplus. Suppliers try to increase sales by cutting the price of a cone, and this moves the price toward its equilibrium level. Changes in equilibrium price and quantity when supply and demand change. A market situation in which the quantity demanded exceeds the quantity supplied shows the shortage of the market. Further, the input and cost conditions are given. A surplus occurs when the actual selling price is above the market equilibrium price. There is a surplus of the good on the market. Last updated 28 Nov 2019, Cracking Economics Taking the price of $2, and plugging it into the equation for quantity supplied, we get the following: $\begin{array}{l}Qs=2+5P\\Qs=2+5(2)\\Qs=2+10\\Qs=12\end{array}$, Now, if the price is$2 each, producers will supply 12 sodas. $\begin{array}{l}\underline{14}=\underline{7P}\\\,\,\,7\,\,\,\,\,\,\,\,\,\,7\\\,\,\,\,2=P\end{array}$. Now we want to determine the quantity amount of soda. Click the OK button, to accept cookies on this website. Right now, we are only going to focus on the math. This price is illustrated by the dashed horizontal line at the price of $1.80 per gallon in Figure 2, below. There are two conditions that are a direct result of disequilibrium: a shortage and a surplus. These relationships are shown as the demand and supply curves in Figure 1, which is based on the data in Table 1, below. The equilibrium price is the point at which the quantity demanded and the quantity supplied in the market are equal. there will be a shortage. In a free market, the excess supply should encourage firms to cut price. The equilibrium quantity is the quantity bought and sold at the equilibrium price. When Price Is Below The Equilibrium Price B. Quantity supplied (550) is less than quantity demanded (700). The equilibrium price is the price at which the quantity demanded equals the quantity supplied. Therefore firms would reduce price and supply less. On a graph, the point where the supply curve (S) and the demand curve (D) intersect is the equilibrium. For example, imagine the price of dragon repellent is currently \$6 $6 A price floor creates a market surplus. A. we would expect to see a surplus of carrots If a price ceiling is set above the equilibrium price: If a surplus remains unsold, those firms involved in making and selling gasoline are not receiving enough cash to pay their workers and cover their expenses. Remember, the formula for quantity demanded is the following: Taking the price of$2, and plugging it into the demand equation, we get, $\begin{array}{l}Qd=16–2(2)\\Qd=16–4\\Qd=12\end{array}$. Once some sellers start cutting prices; others will follow to avoid losing sales. Later you’ll learn why these models work the way they do, but let’s start by focusing on solving the equations. The equilibrium price is the only price where the desires of consumers and the desires of producers agree—that is, where the amount of the product that consumers want to buy (quantity demanded) is equal to the amount producers want to sell (quantity supplied). We can also identify the equilibrium with a little algebra if we have equations for the supply and demand curves. Figure 1. 1.


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Question

Find the following for path C in Figure 3.58: (a) the total distance traveled and (b) the magnitude and direction of the displacement from start to finish. In this part of the problem, explicitly show how you follow the steps of the analytical method of vector addition.

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a) $1560 \textrm{ m}$

b) $120 \textrm{ m, East}$

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# OpenStax College Physics Solution, Chapter 3, Problem 13 (Problems & Exercises) (2:10)

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This is College Physics Answers with Shaun Dychko. To find the total distance traveled along path C, we need to add the length of each part of the path or each part of the route. So we have one block going north here and then we have one, two, three, four, five blocks going to the east and then two blocks south, one block west, one block north and then three blocks to the west. So each of those numbers are written here and then we take that sum and we multiply by 120 meters per block. This gives us 1560 meters is the total distance traveled. Now for the displacement between this position and the end point, we can see that the displacement is going to be one block to the east. But the question asks us to show how we figure that out analytically instead of graphically. So we'll take the x component of the displacement as being the x component of each of these legs of the route. So we have zero x component at that part. This part is an x component of positive five. This part is an x component of zero, negative one x component, zero and then negative three x component. Those are each of the numbers here. That makes one block positive which is what we expected because we expect one block positive here. Then for the y components we have positive one block there and then this has no y component at all. This section has negative two for y component. This section has nothing. This has positive one block for y component and this has none. So each of those numbers are written here and that adds up to zero. So, magnitude of our displacement will be the square root of the sum of the squares of the components. That's the square root of one squared plus zero which is one. Then we'll take the angle as being the inverse tangent of the y component over the x component. So it's the inverse tangent of zero which is zero. This zero represents directly east. So our displacement is 120 meters east.


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# reproducible-research

## prettypublisher

prettypublisher is an R package that aims to improve your workflow by allowing an easier transition from literate code to a paper draft ready for journal submission.

## tbinenglanddataclean

R package containing the scripts required to clean data from the Enhanced Tuberculosis Surveillance system, and the Labour Force Survey, and to then calculate Tuberculosis incidence.


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# Why doesn't the Sun produce an emission spectrum?

I have read that the reason why the Sun produces an absorption spectrum is because the temperature drops as you go away from the center, such that as the various layers of the atmosphere of the sun absorb certain wavelengths, the re-emitted light will have a smaller intensity than the absorbed one, causing a dip in the spectrum (i.e., an absorption spectrum). This is consistent with Kirchhoff's laws and Planck's law for blackbody radiation. However, I checked and found that as you go from the photosphere and into the chromosphere and corona, the temperature rises instead. So then, why doesn't the opposite happen where instead of dips in the spectrum, we get peaks in it (i.e., an emission spectrum)?

(I will note that there was a temperature drop inside of the photosphere itself, but to my understanding the photosphere is opaque to all wavelengths, meaning that it can't pick out the specific wavelengths of the absorption spectrum).

Now, I already have some doubts about the logic above applying here. It seems to me that one of the assumptions being made is that the energy that is absorbed by the various layers are first distributed among the various atoms in the layers, such that the re-emitted light that comes out is the ordinary thermal radiation/blackbody radiation. However, considering the fact that the chromosphere and corona is so dilute, it would seem that that wouldn't be the case. Is it more correct to say that the radiation is rather being absorbed and re-emitted (i.e., scattered) so many times on its journey through the Sun's atmosphere that it continually loses small "bits" of energy that once it reaches us, the intensity of the re-emitted light is much smaller than the absorbed one (causing a dip in the spectrum)? Or is the gas perhaps so dilute that even that wouldn't work? If not, then again, why do we see an absorption spectrum?

The hotter layers above the solar photosphere do have an emission spectrum. The emission spectrum is much fainter than the visible photosphere and so is not easily seen through broadband filters in the optical spectrum, though it can be observed through very narrow filters centred on the emission lines (e.g. H$$\alpha$$ from the chromosphere) in question.

The hotter gas is much less dense than the photosphere and is essentially transparent to photons from the photosphere.

The situation becomes much easier in the UV and X-ray part of the spectrum, where there is very little radiation from the photosphere, but a more significant amount from the chromosphere and corona, because their hotter temperatures excite higher energy transitions. The Sun has an obvious emission line spectrum in the UV and X-ray range. Here is an example (from del Zanna & Mason 2018).

The (optically) thin chromosphere and corona are not heated by radiation from below. That could not produce the temperature inversion that is seen. Instead the chromosphere and corona are heated by the dissipation of magnetic fields in the form of currents and the acceleration of charged particles.

A shorthand way of thinking about photospheric absorption lines is that we can see to different depths in the Sun at different wavelengths. Where there is a strong atomic transition, light is readily absorbed, we cannot see very far. Thus our sight only penetrates to relatively cool layers (note again that the chromosphere and corona are transparent). In contrast, when we look at a "continuum" wavelength we see to deeper and hotter layers in the Sun. Since the brightness is very temperature dependent this is why we get a continuous spectrum with absorption lines.

• Comments are not for extended discussion; this conversation has been moved to chat. – Connor Garcia Jun 20 at 18:36

The photosphere of the sun does produce an emission spectrum (a Planck spectrum according to its temperature of about 6000K). It is only that the atmosphere above the photosphere (the chromosphere) scatters light out of the line of sight at those frequencies where the scattering cross section is very high (i.e. at the atomic resonance frequencies of a given element). This is schematically illustrated by the following diagram

We are of course unable to see the pure continuous spectrum (bottom left) in case of the sun, as the scattering 'cloud' is a layer all around its surface, but we can see the 'bright line spectrum' emitted by the chromosphere during a total solar eclipse for instance as then the direct sunlight is blocked out.

The 'missing bits' of energy are thus due to the photons reflected back by the chromosphere into the photosphere where they are absorbed (the photosphere can be considered a black body and a black body absorbs all incoming radiations). They are not truly lost though but are merely thermalized again and re-emitted as continuum photons. The geometry may be different to that shown in the diagram but the result is the same.

• That diagram makes a lot of sense, however I have some doubts about using this principle for the sun. The thing is that the sun is (approximately) spherically symmetric (around its center), meaning that if you calculate the total power radiated out of a shell with a radius of 1 AU, with the usual absorption lines, and compare that with the total power without absorption lines, then the power with absorption lines is clearly lower. So then where did that lost energy go? It would seem that it was absorbed by certain layers in the sun (which is why my proposed explanation included "losing small – Physics2718 Jun 10 at 19:54
• bits of energy". – Physics2718 Jun 10 at 19:55
• That is because that diagram is inapplicable to explaining the spectrum of the Sun. @Physics2718 The geometry and temperature structure is quite different. – ProfRob Jun 10 at 20:13
• @Physics2718 As the chromosphere is optically thick at frequencies corresponding to the centers of spectral lines, the radiation is reflected back into the photosphere and absorbed there (the photosphere can be considered a black body and a black body absorbs all incoming radiations). So whilst the geometry is obviously different to that shown in the diagram, the physics is the same. The absorption lines arise due to light being scattered out of the beam at those frequencies and being lost (well, not actually lost in this case, but thermalized again into the continuous spectrum).. – Thomas Jun 10 at 21:09

I have read that the reason why the sun produces an absorption spectrum is because the temperature drops as you go away from the center, such that as the various layers of the atmosphere of the sun absorb certain wavelengths, the re-emitted light will have a smaller intensity than the absorbed one, causing a dip in the spectrum (ie an absorption spectrum). This is consistent with Kirchhoff's laws and Planck's law for blackbody radiation.

The further away you go from the solar surface, the photons from the stellar surface/or within have to travel longer through the solar atmosphere. Star already emits a blackbody radiation but Most of the emission and absorption spectra of a star comes from the gases in this solar atmosphere.

However, I checked and found that as you go from the photosphere and into the chromosphere and corona, the temperature rises instead. So then, why doesn't the opposite happen where instead of dips in the spectrum, we get peaks in it (ie an emission spectrum)?

The Stellar atmosphere is made of gases which are atoms in ground state, atoms in excited state and free electrons. Different wavelength/frequency of photons from the star interacts in multiple ways with different types of atoms. These are some major interactions:

1. Line Scattering: The photon is absorbed by an atom $$\rightarrow$$ excitation of electron $$\rightarrow$$ de-excitation of electrons $$\rightarrow$$ photon is released. The photon could be released in any direction and although it would be released with same energy as initial photon, doppler effect plays an important role here.
2. Emission by Recombination: The energy of photon is used to de-excite a free electron. Each de-excitation results in emission of a photon.
3. Emission by collisional or photo-excitation: An atom absorbs the photon and the electron makes a transition from excited state to ground state.
4. Pure Absorption: Photon is absorbed by an atom to excite an electron. Further de-excitation of that same electron to a lower level will cause emission.
5. Masering: Atom in excited state emit the same energy photon as the stellar photon. Such a process will result in same photons travelling in same direction.

Now, getting back to why sun or most sun-like stars don't have a strong emission spectrum. We can notice, that for an emission spectrum, we need an excited or free electron to de-excite and emit photons. Usually, this condition needs a very high temperature such as a state of plasma. For sun (and many other stars), the temperature is just not high enough to have atoms in their excited state. Most atoms in solar atmosphere are in their ground state and hence sun shows much stronger absorption spectrum. Moreover, the elements found in sun's atmosphere have more transitions that support absorption rather than emission. Some of the astronomical bodies that show emission spectrum would be massive stars, Black Hole/Neutron Star accretion disk (both have high temperatures and presence of heavy metals like Fe).

Within the stellar surface, there are free electrons, plasma and excited atoms but sun's outer surface being too cold for radiative transport and is convective zone, which is opaque to internal radiation, we are unable to see the emission spectrum from within the solar surface.

• But the Sun does have an emission spectrum, just not at visible wavelengths. – ProfRob Jun 10 at 18:02
• But i never said it doesn't! I said the absorption spectrum is much stronger due to Answered reasons. – Aryan Bansal Jun 10 at 19:34
• But the absorption spectrum isn't much stronger (whatever that means) at UV and X-ray wavelengths. – ProfRob Jun 10 at 20:09
• Hot, high mass stars, such as O-type main sequence stars and Wolf-Rayet stars, have emission spectrum, due to their strong winds (especially if they are metal rich). – Daddy Kropotkin Jun 10 at 20:43

The sun is the source of emission as well the source of absorption. The hot ball beneath the surface produces a continuous spectrum of black body radiation. Photons in the ball beneath the surface layer are continuously absorbed and reemitted (scattered). This causes a continuous distribution of photon wavelengths, i.e., the radiation conformal to a black body. Upon entering the outer gas layer, part of these photons are absorbed, depending on their wavelengths. The rest of the huge energy escapes and travels to us (in part). We see an almost continuous spectrum (as can be seen in a prism). The radiation coming from the surface gas layer is almost identical to the radiation coming from the inner ball. The difference is that a part of certain wavelength photons is absorbed. So a part of the continuous inner ball radiation is absorbed. So if you subtract the outer layer radiation from the inner ball radiation, you will be left with absorption lines (which can be said to be emission lines at the same time). Where do the absorbed photons go? They are sent back to the inside, while the non-absorbed travel to us.