text:
# Choosing a Desktop Product Dataset

This document will help you find the best data source for a given analysis. It focuses on descriptive datasets and does not cover anything attempting to explain why something is observed. This guide will help if you need to answer questions like:

• How many Firefox users are active in Germany?
• How many crashes occur each day?
• How many users have installed a specific add-on?

If you want to know whether a causal link occurs between two events, you can learn more at tools for experimentation.

## Raw Pings

We receive data from our users via pings. There are several types of pings, each containing different measurements and sent for different purposes. To review a complete list of ping types and their schemata, see this section of the Mozilla Source Tree Docs.

Pings are also described by a JSONSchema specification which can be found in the mozilla-pipeline-schemas repository.

There are a few pings that are central to delivering our core data collection primitives (Histograms, Events, Scalars) and for keeping an eye on Firefox behaviour (Environment, New Profiles, Updates, Crashes).

For instance, a user's first session in Firefox might have four pings like this:

### "main" ping

The "main" ping is the workhorse of the Firefox Telemetry system. It delivers the Telemetry Environment as well as Histograms and Scalars for all process types that collect data in Firefox. It has several variants each with specific delivery characteristics:

ReasonSent whenNotes
shutdownFirefox session ends cleanlyAccounts for about 80% of all "main" pings. Sent by Pingsender immediately after Firefox shuts down, subject to conditions: Firefox 55+, if the OS isn't also shutting down, and if this isn't the client's first session. If Pingsender fails or isn't used, the ping is sent by Firefox at the beginning of the next Firefox session.
dailyIt has been more than 24 hours since the last "main" ping, and it is around local midnightIn long-lived Firefox sessions we might go days without receiving a "shutdown" ping. Thus the "daily" ping is sent to ensure we occasionally hear from long-lived sessions.
environment-changeTelemetry Environment changesIs sent immediately when triggered by Firefox (Installing or removing an addon or changing a monitored user preference are common ways for the Telemetry Environment to change)
aborted-sessionFirefox session doesn't end cleanlySent by Firefox at the beginning of the next Firefox session.

It was introduced in Firefox 38.

### "first-shutdown" ping

The "first-shutdown" ping is identical to the "main" ping with reason "shutdown" created at the end of the user's first session, but sent with a different ping type. This was introduced when we started using Pingsender to send shutdown pings as there would be a lot of first-session "shutdown" pings that we'd start receiving all of a sudden.

It is sent using Pingsender.

It was introduced in Firefox 57.

### "event" ping

The "event" ping provides low-latency eventing support to Firefox Telemetry. It delivers the Telemetry Environment, Telemetry Events from all Firefox processes, and some diagnostic information about Event Telemetry. It is sent every hour if there have been events recorded, and up to once every 10 minutes (governed by a preference) if the maximum event limit for the ping (default to 1000 per process, governed by a preference) is reached before the hour is up.

It was introduced in Firefox 62.

### "update" ping

Firefox Update is the most important means we have of reaching our users with the latest fixes and features. The "update" ping notifies us when an update is downloaded and ready to be applied (reason: "ready") and when the update has been successfully applied (reason: "success"). It contains the Telemetry Environment and information about the update.

It was introduced in Firefox 56.

### "new-profile" ping

When a user starts up Firefox for the first time, a profile is created. Telemetry marks the occasion with the "new-profile" ping which sends the Telemetry Environment. It is sent either 30 minutes after Firefox starts running for the first time in this profile (reason: "startup") or at the end of the profile's first session (reason: "shutdown"), whichever comes first. "new-profile" pings are sent immediately when triggered. Those with reason "startup" are sent by Firefox. Those with reason "shutdown" are sent by Pingsender.

It was introduced in Firefox 55.

### "crash" ping

The "crash" ping provides diagnostic information whenever a Firefox process exits abnormally. Unlike the "main" ping with reason "aborted-session", this ping does not contain Histograms or Scalars. It contains a Telemetry Environment, Crash Annotations, and Stack Traces.

It was introduced in Firefox 40.

### "deletion-request" ping

In the event a user opts out of Telemetry, we send one final "deletion-request" ping to let us know. It contains only the common ping data and an empty payload.

It was introduced in Firefox 72, replacing the "optout" ping (which was in turn introduced in Firefox 63).

### "coverage" ping

The coverage ping (announcement) is a periodic census intended to estimate telemetry opt-out rates.

We estimate that 93% of release channel profiles have telemetry enabled (and are therefore included in DAU).

### Pingsender

Pingsender is a small application shipped with Firefox which attempts to send pings even if Firefox is not running. If Firefox has crashed or has already shut down we would otherwise have to wait for the next Firefox session to begin to send pings.

Pingsender was introduced in Firefox 54 to send "crash" pings. It was expanded to send "main" pings of reason "shutdown" in Firefox 55 (excepting the first session). It sends the "first-shutdown" ping since its introduction in Firefox 57.

### Analysis

The large majority of analyses can be completed using only the main ping. This ping includes histograms, scalars, and other performance and diagnostic data.

Few analyses actually rely directly on any raw ping data. Instead, we provide derived datasets which are processed versions of these data, made to be:

• Easier and faster to query
• Organized to make the data easier to analyze
• Cleaned of erroneous or misleading data

Before analyzing raw ping data, check to make sure there isn't already a derived dataset made for your purpose. If you do need to work with raw ping data, be aware that the volume of data can be high. Try to limit the size of your data by controlling the date range, and start off using a sample.

#### Accessing the Data

Ping data lives in BigQuery and is accessible in STMO; see the BigQuery cookbook section for more information.

You can find the complete ping documentation. To augment our data collection, see Collecting New Data and the Data Collection Policy.

## Main Ping Derived Datasets

The main ping includes most of the measurements that track the performance and health of Firefox in the wild. This ping includes histograms, scalars, and events.

In its raw form, the main ping can be a bit difficult to work with. To make analyzing data easier, some datasets have been provided that simplify and aggregate information provided by the main ping.

### Clients Daily

Many questions about Firefox take the form "What did clients with characteristics X, Y, and Z do during the period S to E?" The clients_daily table aims to answer these questions. Each row in the table is a (client_id, submission_date) and contains a number of aggregates about that day's activity.

See the clients_daily reference for more information.

### Clients Last Seen

The clients_last_seen dataset is useful for efficiently determining exact user counts such as DAU and MAU. It can also allow efficient calculation of other windowed usage metrics like retention via its bit pattern fields. It includes the most recent values in a 28 day window for all columns in the clients_daily dataset.

See the clients_last_seen reference for more information.


text:
[OS X TeX] extrange error

Peter Dyballa Peter_Dyballa at Web.DE
Sun Feb 7 04:44:51 EST 2010

Am 07.02.2010 um 01:01 schrieb Sr Sur:

> The reason is, that, if I use all
> the imputs in my master file I have the error described before...
> but if I
> throw away all except some of the imputs, the file compiles with no
> problem.

So the problem lies in the use of these "imputs in my master file."

> The same file compiled in my laptop with the 2.29 version compiles
> without
> problems I'm Using MACTex-2009 distribution in both cases
> (MacBookpro and
> iMac...) and TEX+DVI script.

TeXShop is just a TeX front-end, like a (TeX) shell, or whatever. It
has no influence on how any TeX compiler converts the source file(s)
to whatever other format. You better check what TeX software you have
installed on both computers! One simple option is to use the
\listfiles statement in the preamble.

--
Greetings

Pete

Life is the only flaw in an otherwise perfect nonexistence
– Schopenhauer

format: plain


text:
• NCERT Chapter
Question 2

Check the injectivity and surjectivity of the following functions:(i) f : N → N given by f(x) = x2(ii) f : Z → Z given by f(x) = x2(iii) f : R → R given by f(x) = x2(iv) f : N → N given by f(x) = x3(v) f : Z → Z given by f(x) = x3

(i) fN → N is given by,

f(x) = x2

It is seen that for xy ∈Nf(x) = f(y) ⇒ x2 = y2 ⇒ x = y.

f is injective.

Now, 2 ∈ N. But, there does not exist any x in N such that f(x) = x2 = 2.

∴ f is not surjective.

Hence, function f is injective but not surjective.

(ii) fZ → Z is given by,

f(x) = x2

It is seen that f(-1) = f(1) = 1, but -1 ≠ 1.

∴ f is not injective.

Now,-2 ∈ Z. But, there does not exist any element x ∈Z such that f(x) = x2 = -2.

∴ f is not surjective.

Hence, function f is neither injective nor surjective.

(iii) fR → R is given by,

f(x) = x2

It is seen that f(-1) = f(1) = 1, but -1 ≠ 1.

∴ f is not injective.

Now,-2 ∈ R. But, there does not exist any element x ∈ R such that f(x) = x2 = -2.

∴ f is not surjective.

Hence, function f is neither injective nor surjective.

(iv) fN → N given by,

f(x) = x3

It is seen that for xy ∈Nf(x) = f(y) ⇒ x3 = y3 ⇒ x = y.

f is injective.

Now, 2 ∈ N. But, there does not exist any element x in domain N such that f(x) = x3 = 2.

∴ f is not surjective

Hence, function f is injective but not surjective.

(v) fZ → Z is given by,

f(x) = x3

It is seen that for xy ∈ Zf(x) = f(y) ⇒ x3 = y3 ⇒ x = y.

∴ f is injective.

Now, 2 ∈ Z. But, there does not exist any element x in domain Z such that f(x) = x3 = 2.

∴ f is not surjective.

Hence, function f is injective but not surjective.

">

Let A = R – {3} and B = R – {1}. Consider the function  f : A → B defined by

. Is f one-one and onto? Justify your answer.

• Q:-

Prove that the Greatest Integer Function f : R → R, given by f(x) = [x], is neither one-one nor onto, where [x] denotes the greatest integer less than or equal to x.

• Q:- Show that the relation R in the set R of real numbers, defined as R = {(a, b): a ≤ b2} is neither reflexive nor symmetric nor transitive.
• Q:-

Show that the Modulus Function f : R → R, given by f(x) = |x|, is neither oneone nor onto, where | x | is x, if x is positive or 0 and |x| is – x, if x is negative.

• Q:- Show that the relation R in the set A = {1, 2, 3, 4, 5} given by R = { (a,b) ; |a - b| is even}, is an equivalence relation. Show that all the elements of {1, 3, 5} are related to each other and all the elements of {2, 4} are related to each other. But no element of {1, 3, 5} is related to any element of {2, 4}.
• Q:- Show that the relation R in R defined as R = {(a, b): a ≤ b}, is reflexive and transitive but not symmetric.
• Recently Viewed Questions of Class 12th mathematics

">

Let A = R – {3} and B = R – {1}. Consider the function  f : A → B defined by

. Is f one-one and onto? Justify your answer.

• Q:-

Show that the function f : R* → R* defined by f(x) = 1/x is one-one and onto,where R* is the set of all non-zero real numbers. Is the result true, if the domain R* is replaced by N with co-domain being same as R

• Q:-

Answer the following as true or false.

\begin{align}(i) \overrightarrow{a}\;  and\; \overrightarrow{-a}\; are\; collinear.\end{align}

(ii) Two collinear vectors are always equal in magnitude.

(iii) Two vectors having same magnitude are collinear.

(iv) Two collinear vectors having the same magnitude are equal.

• Q:- Let L be the set of all lines in XY plane and R be the relation in L defined as R = {(L1, L2): L1 is parallel to L2}. Show that R is an equivalence relation. Find the set of all lines related to the line y = 2x + 4.


text:
# Advanced Signals and Systems - Stochastic Processes I

### 3. LAPLACE's distribution

Let $$v$$ be a random variable with the probability-density function $$f_{v}(V)=\frac{\alpha}{2}e^{-\alpha |V|}$$ (density of LAPLACE's distribution).

1. Determine the probability function $$F_v(V)$$.
2. Determine the mean value and the variance of $$v$$.
3. Determine the characteristic function of LAPLACE's distribution.
4. Verify the results of (2) with the characteristic function.

## Amount and difficulty

• Working time: approx. xx minutes
• Difficulty: easy

## Solution

1. In general: \begin{equation*} F_v(V) = \int \limits_{x=-\infty}^V f_v(x) \, dx \end{equation*} For solving the integral a case differentiation for $$V<0$$ and $$V\leq0$$ is done and the following result can be derived: \begin{equation*} F_v(V) = \frac{1}{2} \cdot e^{-\alpha |V|} + \left[ 1 - e^{-\alpha |V|}\right] \delta_{-1}(V)\end{equation*}
2. Determining the mean in general: \begin{equation*} \text{E}\left\{ v\right\} = \int \limits_{V=-\infty}^\infty V f_v(V) \, dV = \mu_v \end{equation*} Here we can conclude \begin{equation*} \text{E}\left\{ v\right\} = \int \limits_{V=-\infty}^\infty \underbrace{\underbrace{V}_{\text{odd}} \cdot \underbrace{\frac{\alpha}{2} \cdot e^{-\alpha |V|}}_{\text{even}}}_{\text{odd}} \, dV = 0 \end{equation*} because the integral of an odd function equals to zero.

Determining the variance in general: \begin{equation*} \sigma_v^2 = \text{E}\{v^2\} - {\text{E}\left\{v\right\}}^2 \end{equation*} Here we can conclude \begin{align*} \sigma_v^2 &= \text{E}\{v^2\} - \underbrace{\text{E}\{v\}^2}_{=0} = \text{E}\{v^2\}\\ &= \int \limits_{-\infty}^\infty V^2 f_v(V) \, dV = \int \limits_{-\infty}^\infty \underbrace{V^2 \cdot \frac{\alpha}{2} e^{-\alpha |V|}}_{\text{even}} \, dV\\ &= \alpha \int \limits_{0}^\infty V^2 e^{-\alpha V} dV \end{align*} After substitution with $$t=\alpha \cdot V$$ and applying the gamma function $$\Gamma(x) = \int \limits_0^\infty t^{x-1} e^{-t} dt$$ the variance is defined by: \begin{equation*} \sigma_v^2 = \frac{1}{\alpha^2} \cdot \Gamma(3) = \frac{2}{\alpha^2} \end{equation*}
3. \begin{align*} C(\omega) &= \text{E} \{ e^{j\omega V} \} = \int \limits_{V=-\infty}^\infty e^{j\omega V} f_v(V) \, dV \\ &= \frac{\alpha^2}{\omega^2 + \alpha^2} \end{align*}
4. Moment theorem: \begin{align*} \text{E} \left\{x^k \right\} &= \left. \frac{1}{j^k} \frac{d^k C(\omega)}{d \omega ^k} \right| _{\omega =0}\\ \Rightarrow \mu_v &= \text{E} \{x \} = \left. \frac{1}{j} \frac{d C(\omega)}{d \omega }\right| _{\omega =0} = \cdots = 0 \\ \Rightarrow \sigma_v^2 &= \text{E} \{x ^2\} = \left. \frac{1}{j^2} \frac{d^2 C(\omega)}{d \omega ^2}\right| _{\omega =0} = \cdots = \frac{2}{\alpha^2} \end{align*}The same results as in (b) can be achieved.

Laplace distribution:

Example speech signal:

### 4. Properties of ACF/CCF.

1. Show that for the CCF of the complex random signals $$v_1(n)$$ and $$v_2(n)$$ holds: \begin{equation*}s_{v_1v_2}(\kappa) = s^\ast_{v_2v_1}(-\kappa) \end{equation*} What does this imply for the CCF of real signals $$v_1(n)$$ and $$v_2(n)$$?
How can this relationship be used to determine the ACF of a real signal?
2. Show that the CAUCHY-SCHWARZ inequality for real random variables holds: \begin{equation*} \left[E\{xy\}\right]^2 \leq E\{x^2\}\cdot E\{y^2\} \end{equation*} What does this imply for the ACF of a real signal?
3. How can the above results for discrete signals be applied to the CCF and ACF of continuous signals?

## Amount and difficulty

• Working time: approx. xx minutes
• Difficulty: easy

## Solution

1. Stationarity has to be assumed. According to the lecture, the cross-correlation is defined by \begin{equation*} s_{v_1v_2}(\kappa) = \text{E} \left\{ v_1^*(n) \cdot v_2(n+\kappa) \right\} \end{equation*} Further it can be shown: \begin{align*}s_{v_2v_1}^*(-\kappa) &= \left[ \text{E}\left\{ v_2^*(n) \cdot v_1(n-\kappa) \right\} \right]^*\\ &= \left[ \text{E}\left\{ v_2^*(n+\kappa) \cdot v_1(n) \right\} \right]^*\\ &= \text{E}\left\{ v_2(n+\kappa) \cdot v_1^*(n) \right\}\\ &= s_{v_1v_2}(\kappa)\end{align*} Special case: $$v_1(n), v_2(n) \in \mathbb{R}$$ \begin{equation*}s_{v_1v_2}(\kappa) = \text{E} \{ v_1(n) \cdot v_2(n+\kappa)\} = \cdots = s_{v_2v_1}(-\kappa);\end{equation*} Special case ACF: $$v_1(n) = v_2(n) =v(n)$$ \begin{align*} v(n) \in \mathbb{C} &\rightarrow s_{vv}(\kappa) = s_{vv}^* (-\kappa)\\ v(n) \in \mathbb{R} &\rightarrow s_{vv}(\kappa) = s_{vv} (-\kappa) \end{align*}
2. Let $$x,y$$ be real random variables and let $$a \in \mathbb{R}$$. \begin{align*} \text{E} \{ (ax-y)^2 \} \geq 0 \\ a^2 \, \text{E} \{ x^2 \} +2a\,\text{E} \{ xy \} +\text{E} \{ y^2\}\geq 0 \end{align*} By solving this equation and inserting $$a=\frac{\text{E} \{ xy\}}{\text{E} \{ x^2\}}$$ it follows that \begin{equation*}\text{E}^2 \{ xy \} \leq \text{E}\{x^2\} \cdot \text{E}\{y^2\}\end{equation*}
3. The above results hold true also for continuous signals.

Example 1:

Stationary?

Ergodic?

Example 2:

Stationary?

Ergodic?

Example 3:

Stationary?

Ergodic?

### Website News

01.10.2017: Started with a Tips and Tricks section for KiRAT.

01.10.2017: Talks from Jonas Sauter (Nuance) and Vasudev Kandade Rajan (Harman/Samsung) added.

13.08.2017: New Gas e.V. sections (e.g. pictures or prices) added.

05.08.2017: The first "slide carousel" added.

### Recent Publications

J. Reermann, P. Durdaut, S. Salzer, T. Demming,A. Piorra, E. Quandt, N. Frey, M. Höft, and G. Schmidt: Evaluation of Magnetoelectric Sensor Systems for Cardiological Applications, Measurement (Elsevier), ISSN 0263-2241, https://doi.org/10.1016/j.measurement.2017.09.047, 2017

S. Graf, T. Herbig, M. Buck, G. Schmidt: Low-Complexity Pitch Estimation Based on Phase Differences Between Low-Resolution Spectra, Proc. Interspeech, pp. 2316 -2320, 2017

### Contact

Prof. Dr.-Ing. Gerhard Schmidt

E-Mail: gus@tf.uni-kiel.de

Christian-Albrechts-Universität zu Kiel
Faculty of Engineering
Institute for Electrical Engineering and Information Engineering
Digital Signal Processing and System Theory

Kaiserstr. 2
24143 Kiel, Germany

## Recent News

Jens Reermann Defended his Dissertation with Distinction

On Friday, 21st of June, Jens Reermann defended his research on signals processing for magnetoelectric sensor systems very successfully. After 90 minutes of talk and question time he finished his PhD with distinction. Congratulations, Jens, from the entire DSS team.

Jens worked for about three and a half years - as part of the collaborative research center (SFB) 1261 - on all kinds of signal ...


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# accelerated failure time model pdf

Background and Purpose: The goal of this study is application of the propor tional hazards model (PH) and accelerated failure time model (AFT) , with consideration Weibull distribution, to determine the level of effectiveness of the fact ors affecting on the level of disease-free survival (DFS) of the patients with breast cancer. to failure time. Rank estimators have been studied by Prentice (1978), Tsi As a result of its direct physical interpretation, this model provides Keywords: Insurance attrition, Survival analysis, Accelerated failure time model, Proportional hazards model. In the presence of a nonsusceptible population, Li and Taylor (2002) and Zhang and Peng (2007) considered the accelerated failure time mix-ture cure model and proposed … The presence of censoring poses major challenges in the semiparametric analysis of the accelerated failure time model. 5.1 The Accelerated Failure Time Model Before talking about parametric regression models for survival data, let us introduce the ac-celerated failure time (AFT) Model. We address the issue of performing hypothesis testing in accelerated failure time models for non-censored and censored samples. The accelerated failure time (AFT) model is specified by logT = +µ σε with location and scale parameters µ, σ, respectively. 32–4; Cox & Oakes, 1984, pp. Accelerated Failure Time (AFT) model is one of the most commonly used models in survival analysis. The model is of the following form: The model is of the following form: $\ln{Y} = \langle \mathbf{w}, \mathbf{x} \rangle + \sigma Z$ 1 Introduction The growing need to include covariates in the analysis of time-to-event data has brought forth the two popular regression models: the Cox proportional hazards model (PH model) and the accelerated failure time (AFT) model. 64–5). Several complications arise when the covariates are measured Komarek and Lesa re, 2008). “Bayesian Accelerated Failure Time Model with Multivariate Doubly-Interval-Censored Data and Flexible Distributional Assumptions” Arnoˇst Kom ´arek and Emmanuel Lesaffre Biostatistical Centre, Katholieke Universiteit Leuven, Kapucijnenvoer 35, B–3000, Leuven, Belgium E-mail: Arnost.Komarek@med.kuleuven.be Emmanuel.Lesaffre@med.kuleuven.be The accelerated failure time model or accelerated life model relates the logarithm of the failure time linearly to the covariates (Kalbﬂeisch & Prentice, 1980, pp. PARAMETRIC MODELS-ACCELERATED FAILURE TIME MODEL Procedures LIFEREG and RELIABILITY can be used for inference from survival data that have a combination of left, right and interval censored observations. native to the proportional hazards model due to its direct physical interpretation (Reid (1994)). The accelerated failure time (AFT) model is an attractive alternative to the Cox model when the proportionality assumption fails to capture the relation between the survival time and longitudinal covariates. proportional hazards model is the accelerated failure time (AFT) model, which relates the logarithm or a known transformation of the failure time to its covariates. In some situations, the AFT model could be preferred over the proportional hazards model due to its quite direct physical interpretation (see, e.g. II. Denote by S1(t)andS2(t) the survival functions of two populations. The AFT models says that there is … Using Weibull accelerated failure time regression model to predict survival time and life expectancy Enwu Liu1,2* 1 Musculoskeletal Health and Ageing Research Program, Mary MacKillop Institute for Health Research, Australian Catholic University, Melbourne, Victoria, Australia The performances of the likelihood ratio test and a recently proposed test, the gradient test, are compared through This model may provide more accurate or more concise summarization of the data than the proportional hazards model in certain applications.


text:
# Simple VS Compound

Lesson

Now that we've had a good look at both simple interest and compound interest, let's compare them side by side.

##### example

Let's say we have $\$10000$$10000 to invest and we have two options. Option 1: Invest at 5%5% per annum simple interest Option 2: Invest at 4.5%4.5% per annum compound interest, compounded annually At first glance it appears that the higher interest rate in Option 1 would naturally be the better choice, but let's see what happens. Year Simple Interest Compound Interest Simple Interest Balance Compound Interest Balance 11 500500 450450 1050010500 1045010450 22 500500 470.25470.25 1100011000 10920.2510920.25 33 500500 491.41491.41 1150011500 11411.6611411.66 44 500500 513.52513.52 1200012000 11925.1811925.18 55 500500 536.63536.63 1250012500 12641.8112641.81 66 500500 560.78560.78 1300013000 13022.5913022.59 We can see that the interest remains constant each year for simple interest but due to the nature of compounding, the interest is increasing each year. At the 6 year mark, the balance for the compound interest option yields a greater return. Let's look at the same situation, but this time let's look at what happens if Option 2 was compounded monthly. Simple Interest Balance after 6 years: \13000$$13000

Compound Interest Balance after 6 years, compounded annually: $\$13022.59$$13022.59 Compound Interest Balance after 6 years, compounded monthly: 10000\left(1+\frac{4.5}{100\times12}\right)^{\left(12\times6\right)}10000(1+4.5100×12)(12×6)=\13093.03$$13093.03

We can see here that the more often the investment is compounded, the greater the amount of interest accrued.


text:
### Binary Relations in Mathematical Economics: On the Continuity, Additivity and Monotonicity Postulates in Eilenberg, Villegas and DeGroot

This chapter examines how positivity and order play out in two important questions in mathematical economics, and in so doing, subjects the postulates of continuity, additivity and monotonicity to closer scrutiny. Two sets of results are offered: the first departs from Eilenberg's (1941) necessary and sufficient conditions on the topology under which an anti-symmetric, complete, transitive and continuous binary relation exists on a topologically connected space; and the second, from DeGroot's (1970) result concerning an additivity postulate that ensures a complete binary relation on a {\sigma}-algebra to be transitive. These results are framed in the registers of order, topology, algebra and measure-theory; and also beyond mathematics in economics: the exploitation of Villegas' notion of monotonic continuity by Arrow-Chichilnisky in the context of Savage's theorem in decision theory, and the extension of Diamond's impossibility result in social choice theory by Basu-Mitra. As such, this chapter has a synthetic and expository motivation, and can be read as a plea for inter-disciplinary conversations, connections and collaboration.

### Forecasting with Bayesian Grouped Random Effects in Panel Data

In this paper, we estimate and leverage latent constant group structure to generate the point, set, and density forecasts for short dynamic panel data. We implement a nonparametric Bayesian approach to simultaneously identify coefficients and group membership in the random effects which are heterogeneous across groups but fixed within a group. This method allows us to incorporate subjective prior knowledge on the group structure that potentially improves the predictive accuracy. In Monte Carlo experiments, we demonstrate that our Bayesian grouped random effects (BGRE) estimators produce accurate estimates and score predictive gains over standard panel data estimators. With a data-driven group structure, the BGRE estimators exhibit comparable accuracy of clustering with the nonsupervised machine learning algorithm Kmeans and outperform Kmeans in a two-step procedure. In the empirical analysis, we apply our method to forecast the investment rate across a broad range of firms and illustrate that the estimated latent group structure facilitate forecasts relative to standard panel data estimators.

### Spectral Targeting Estimation of $λ$-GARCH models

This paper presents a novel estimator of orthogonal GARCH models, which combines (eigenvalue and -vector) targeting estimation with stepwise (univariate) estimation. We denote this the spectral targeting estimator. This two-step estimator is consistent under finite second order moments, while asymptotic normality holds under finite fourth order moments. The estimator is especially well suited for modelling larger portfolios: we compare the empirical performance of the spectral targeting estimator to that of the quasi maximum likelihood estimator for five portfolios of 25 assets. The spectral targeting estimator dominates in terms of computational complexity, being up to 57 times faster in estimation, while both estimators produce similar out-of-sample forecasts, indicating that the spectral targeting estimator is well suited for high-dimensional empirical applications.

### Teacher-to-classroom assignment and student achievement

We study the effects of counterfactual teacher-to-classroom assignments on average student achievement in elementary and middle schools in the US. We use the Measures of Effective Teaching (MET) experiment to semiparametrically identify the average reallocation effects (AREs) of such assignments. Our findings suggest that changes in within-district teacher assignments could have appreciable effects on student achievement. Unlike policies which require hiring additional teachers (e.g., class-size reduction measures), or those aimed at changing the stock of teachers (e.g., VAM-guided teacher tenure policies), alternative teacher-to-classroom assignments are resource neutral; they raise student achievement through a more efficient deployment of existing teachers.

### Semi-nonparametric Latent Class Choice Model with a Flexible Class Membership Component: A Mixture Model Approach

This study presents a semi-nonparametric Latent Class Choice Model (LCCM) with a flexible class membership component. The proposed model formulates the latent classes using mixture models as an alternative approach to the traditional random utility specification with the aim of comparing the two approaches on various measures including prediction accuracy and representation of heterogeneity in the choice process. Mixture models are parametric model-based clustering techniques that have been widely used in areas such as machine learning, data mining and patter recognition for clustering and classification problems. An Expectation-Maximization (EM) algorithm is derived for the estimation of the proposed model. Using two different case studies on travel mode choice behavior, the proposed model is compared to traditional discrete choice models on the basis of parameter estimates' signs, value of time, statistical goodness-of-fit measures, and cross-validation tests. Results show that mixture models improve the overall performance of latent class choice models by providing better out-of-sample prediction accuracy in addition to better representations of heterogeneity without weakening the behavioral and economic interpretability of the choice models.

### Spatial Iterated Prisoner's Dilemma as a Transformation Semigroup

The prisoner's dilemma (PD) is a game-theoretic model studied in a wide array of fields to understand the emergence of cooperation between rational self-interested agents. In this work, we formulate a spatial iterated PD as a discrete-event dynamical system where agents play the game in each time-step and analyse it algebraically using Krohn-Rhodes algebraic automata theory using a computational implementation of the holonomy decomposition of transformation semigroups. In each iteration all players adopt the most profitable strategy in their immediate neighbourhood. Perturbations resetting the strategy of a given player provide additional generating events for the dynamics. Our initial study shows that the algebraic structure, including how natural subsystems comprising permutation groups acting on the spatial distributions of strategies, arise in certain parameter regimes for the pay-off matrix, and are absent for other parameter regimes. Differences in the number of group levels in the holonomy decomposition (an upper bound for Krohn-Rhodes complexity) are revealed as more pools of reversibility appear when the temptation to defect is at an intermediate level. Algebraic structure uncovered by this analysis can be interpreted to shed light on the dynamics of the spatial iterated PD.

### Off-Policy Exploitability-Evaluation and Equilibrium-Learning in Two-Player Zero-Sum Markov Games

Off-policy evaluation (OPE) is the problem of evaluating new policies using historical data obtained from a different policy. Off-policy learning (OPL), on the other hand, is the problem of finding an optimal policy using historical data. In recent OPE and OPL contexts, most of the studies have focused on one-player cases, and not on more than two-player cases. In this study, we propose methods for OPE and OPL in two-player zero-sum Markov games. For OPE, we estimate exploitability that is often used as a metric for determining how close a strategy profile is to a Nash equilibrium in two-player zero-sum games. For OPL, we calculate maximin policies as Nash equilibrium strategies over the historical data. We prove the exploitability estimation error bounds for OPE and regret bounds for OPL based on the doubly robust and double reinforcement learning estimators. Finally, we demonstrate the effectiveness and performance of the proposed methods through experiments.

### Robust Causal Inference Under Covariate Shift via Worst-Case Subpopulation Treatment Effects

We propose the worst-case treatment effect (WTE) across all subpopulations of a given size, a conservative notion of topline treatment effect. Compared to the average treatment effect (ATE) that solely relies on the covariate distribution of collected data, WTE is robust to unanticipated covariate shifts, and ensures positive findings guarantee uniformly valid treatment effects over underrepresented minority groups. We develop a semiparametrically efficient estimator for the WTE, leveraging machine learning-based estimates of heterogenous treatment effects and propensity scores. By virtue of satisfying a key (Neyman) orthogonality property, our estimator enjoys central limit behavior---oracle rates with true nuisance parameters---even when estimates of nuisance parameters converge at slower rates. For both observational and randomized studies, we prove that our estimator achieves the optimal asymptotic variance, by establishing a semiparametric efficiency lower bound. On real datasets where robustness to covariate shift is of core concern, we illustrate the non-robustness of ATE under even mild distributional shift, and demonstrate that the WTE guards against brittle findings that are invalidated by unanticipated covariate shifts.

### Bridging the COVID-19 Data and the Epidemiological Model using Time Varying Parameter SIRD Model

This paper extends the canonical model of epidemiology, SIRD model, to allow for time varying parameters for real-time measurement of the stance of the COVID-19 pandemic. Time variation in model parameters is captured using the generalized autoregressive score modelling structure designed for the typically daily count data related to pandemic. The resulting specification permits a flexible yet parsimonious model structure with a very low computational cost. This is especially crucial at the onset of the pandemic when the data is scarce and the uncertainty is abundant. Full sample results show that countries including US, Brazil and Russia are still not able to contain the pandemic with the US having the worst performance. Furthermore, Iran and South Korea are likely to experience the second wave of the pandemic. A real-time exercise show that the proposed structure delivers timely and precise information on the current stance of the pandemic ahead of the competitors that use rolling window. This, in turn, transforms into accurate short-term predictions of the active cases. We further modify the model to allow for unreported cases. Results suggest that the effects of the presence of these cases on the estimation results diminish towards the end of sample with the increasing number of testing.

### Dynamic Awareness

We investigate how to model the beliefs of an agent who becomes more aware. We use the framework of Halpern and Rego (2013) by adding probability, and define a notion of a model transition that describes constraints on how, if an agent becomes aware of a new formula $\phi$ in state $s$ of a model $M$, she transitions to state $s^*$ in a model $M^*$. We then discuss how such a model can be applied to information disclosure.


text:
Code in a setup() block is run immediately in a clean environment. Code in a teardown() block is run upon completion of a test file, even if it exits with an error. Multiple calls to teardown() will be executed in the order they were created.

teardown(code, env = parent.frame())

setup(code, env = parent.frame())

## Arguments

code Code to evaluate Environment in which code will be evaluted. For expert use only.

## Examples

if (FALSE) {

tmp <- tempfile()
setup(writeLines(tmp, "some test data"))
}


text:
# How to compute $I_n=\int_{-\infty}^{+\infty}\mathrm{d}x\frac{x^{2n}}{\cosh^2 x}$?

I'd like to compute: $$I_n = \int_{-\infty}^{+\infty}\mathrm{d}x\frac{x^{2n}}{\cosh^2 x}.$$ We have, quite easily: $$I_0 = \int_{-\infty}^{+\infty}\mathrm{d}x\frac{1}{\cosh^2 x}=\left[\tanh x\right]_{-\infty}^{+\infty}=2.$$ So to begin with, I gave a try at $$I_1 = \int_{-\infty}^{+\infty}\mathrm{d}x\frac{x^{2}}{\cosh^2 x}.$$ My idea was to use the following contour $\Gamma_{\varepsilon, M}$ to integrate on in the complex plane: the rectangle with vertices $M$, $M+i\frac{\pi}{2}$, $-M+i\frac{\pi}{2}$, $-M$, and an indent of radius $\varepsilon$ at $i\frac{\pi}{2}$, where $\cosh z=0$. Then by Residues Theorem: $$\oint_{\Gamma_{\varepsilon, M}} \mathrm{d}z\frac{z^{2}}{\cosh^2 z} = 0.$$

Problem is, when I make the calculations the integrand is still singular $O(\frac{1}{\varepsilon})$ on my semicircle. More explicitly: $$\int_{0}^{-\pi} \mathrm{d}\theta i\varepsilon e^{i\theta}\frac{(i\frac{\pi}{2}+\varepsilon e^{i\theta})^{2}}{\cosh^2 (i\frac{\pi}{2}+\varepsilon e^{i\theta})} = \int_{0}^{-\pi} \mathrm{d}\theta i\varepsilon e^{i\theta} \frac{-\frac{\pi^2}{4} + O(\varepsilon)}{\varepsilon^2e^{2i\theta}+O(\varepsilon^3)}.$$ To avoid this problem I tried: $$\oint_{\Gamma_{\varepsilon, M}} \mathrm{d}z\frac{(z-i\frac{\pi}{2})^{2}}{\cosh^2 z}=0.$$

But then I only get: $$\int \mathrm{d}x\frac{(x-i\frac{\pi}{2})^{2}}{\cosh^2x}-\int \mathrm{d}x\frac{x^{2}}{\cosh^2(x-i\frac{\pi}{2})}=0$$ $$\int \mathrm{d}x\left(\frac{x^2}{\cosh^2x}+\frac{x^2}{\sinh^2x}\right) = \frac{\pi^2}{2}$$

since $\cosh(x+i\frac{\pi}{2}) = i\sinh x$.

Why not simply go up to $\Im{z} = \pi$ so that the rectangle $C$ has vertices $\pm R$ and $\pm R+i \pi$? Then we consider

$$\oint_C dz \frac{z^{2 n+1}}{\cosh^2{z}}$$

which is equal to

$$\int_{-R}^R dx \frac{x^{2 n+1}}{\cosh^2{x}} + i \int_0^{\pi} dy \frac{(R+i y)^{2 n+1}}{\cosh^2{(R + i y)}} \\ + \int_R^{-R} dx \frac{(x+i \pi)^{2 n+1}}{\cosh^2{x}} + i \int_{\pi}^0 dy \frac{(-R+i y)^{2 n+1}}{\cosh^2{(-R + i y)}}$$

As $R \to \infty$, the second and fourth integrals vanish. Thus the contour integral is equal to

$$\int_{-\infty}^{\infty} dx \frac{x^{2 n+1}-(x+i \pi)^{2 n+1}}{\cosh^2{x}}$$

Note that the highest power in the numerator is $x^{2 n}$, and that all odd powers vanish upon integration. Thus the contour integral is expressible in terms of integrals of lower powers:

$$-i \pi (2 n+1)\int_{-\infty}^{\infty} dx \frac{x^{2 n}}{\cosh^2{x}} +i \sum_{k=0}^{n-1} (-1)^k \binom{2 n+1}{2 k} \pi^{2 (n-k)+1}\int_{-\infty}^{\infty} dx \frac{x^{2 k}}{\cosh^2{x}}$$

By the residue theorem, the contour integral is also equal to $i 2 \pi$ times the residue of the integrand at the pole $z=i \pi/2$, which is a double pole.

$$\operatorname*{Res}_{z=i \pi/2} \frac{z^{2 n+1}}{\cosh^2{z}} = \lim_{z\to i \pi/2} \frac{d}{dz}\frac{(z-i \pi/2)^2 z^{2 n+1}}{\cosh^2{z}} = (-1)^{n+1} (2 n+1) \left ( \frac{\pi}{2}\right )^{2 n}$$

In order to verify this you can proceed as follows: $$\lim_{z\to i \pi/2} \frac{d}{dz}\frac{(z-i \pi/2)^2 z^{2 n+1}}{\cosh^2{z}} = \lim_{u\to 0} \frac{d}{dz}\frac{u^2 (u+i\pi/2)^{2 n+1}}{-\sinh^2{u}} = \lim_{u\to 0} (u+i\pi/2)^n\ \frac{-\left((2n+1)u^2+2(u+i\pi/2)u\right)2\sinh u+2(u+i\pi/2)u^2\cosh u}{\sinh^3u} = \lim_{u\to 0} (u+i\pi/2)^n\ \frac{-(2n+3)u^3-i\pi u +2u^3 + i\pi u + O(u^4)}{u^3+O(u^4)}= (-1)^{n+1}(2n+1)\left(\frac{\pi}{2}\right) ^{2n}.$$

You then need to find the integrals of lower powers. Actually, you can get around this by setting up a system of equations for each $k$ from $0$ to $n$. Let

$$I_k = \int_{-\infty}^{\infty} dx \frac{x^{2 k}}{\cosh^2{x}}$$

and

$$R_k = (-1)^{k} (2 k+1) \left ( \frac{\pi}{2}\right )^{2 k}$$

Then we have

$$(2 n+1) I_n - \sum_{k=0}^{n-1} (-1)^k \binom{2 n+1}{2 k} \pi^{2 (n-k)} I_k = 2 R_n$$ $$(2 n-1) I_{n-1} - \sum_{k=0}^{n-2} (-1)^k \binom{2 n-1}{2 k} \pi^{2 (n-1-k)} I_k = 2 R_{n-1}$$ $$\cdots$$ $$3 I_1 - \pi^2 I_0 = 2 R_1$$ $$I_0 = 2$$

• Thanks Ron. I'm having problems with the computation of the residue at the double pole: shouldn't it be second derivative? And if so: I've done the calculations a couple of times but it doesn't turn out correct. Thanks in advance! – Brightsun Feb 17 '14 at 10:44
• @Brightsun: Double pole is a derivative; in general for an $n$-tuple pole, it is the $n-1$th derivative. The calculation is messy. – Ron Gordon Feb 17 '14 at 13:33
• Totally right dude, thanks. – Brightsun Feb 17 '14 at 13:37

Note that: $$\int _{-\infty }^{\infty }{\frac {{x}^{2\,n}}{ \cosh^2 \left( x \right) }}\,{\mathrm{d}x}=2\,\int _{0}^{\infty }\!{\frac {{x}^{2\,n }}{\cosh^2 \left( x \right) }}\,{\mathrm{d}x}\tag{1}$$ then, consider the following integral for $\Re(s)>1$: \begin{aligned} \int _{0}^{\infty }{\frac {{x}^{s}}{ \cosh^2 \left( x \right) }}{dx}&=-2\,\int _{0}^{\infty }\!{x}^{s}{\frac {\mathrm{d}}{\mathrm{d}x}} \left( \frac{1}{ 1+{{\rm e}^{2\,x}}}\right)\, {\mathrm{d}x},\\ \mbox{integration by parts...}\quad&=2\,s\int _{0}^{\infty }\!{\frac {{x}^{s-1}}{1+{{\rm e}^{2\,x}}}}\,{\mathrm{d}x},\\ \mbox{partial fractions...}\quad&=2\,s\int _{0}^{\infty }\!2\,{\frac {{x}^{s-1}}{1-{{\rm e}^{4\,x}}}}\,{\mathrm{d}x}-2\,s\int _{0}^{\infty }\!{\frac {{x}^{s-1}}{1-{{\rm e}^{2\,x}}}}\,{\mathrm{d}x},\\ \mbox{rescale the variables...}\quad&=-s \left( {2}^{-2\,s+2}-{2}^{1-s} \right) \int _{0}^{\infty }\!{\frac {{x}^{s-1}}{-1+{{\rm e}^{x}}}}\,{\mathrm{d}x},\\ \mbox{geometric series...}\quad&=-s \left( {2}^{-2\,s+2}-{2}^{1-s} \right) \sum _{n=1}^{\infty } \left( \int _{0}^{\infty }\!{x}^{s-1}{{\rm e}^{-xn}}\,{\mathrm{d}x} \right),\\ \mbox{rescale the variable...}\quad&=-s \left( {2}^{-2\,s+2}-{2}^{1-s} \right) \left(\sum _{n=1}^{\infty } \frac{1}{n^s}\right)\left(\int _{0}^{\infty }\!{x}^{s- 1}{{\rm e}^{-x}}\,{\mathrm{d}x}\right),\\ \mbox{function definitions...}\quad&=-s \left( {2}^{-2\,s+2}-{2}^{1-s} \right) \Gamma \left( s \right) \zeta \left( s \right),\\ \mbox{absorb the s into}\,\, \Gamma \mbox{...}\quad&=\left( -{2}^{-2\,s+2}+{2}^{1-s} \right) \zeta \left( s \right) \Gamma \left( s+1 \right) \end{aligned} \tag{2} where $\zeta$ is the Riemann zeta function. Then, if $s=2n,\, n\in \mathbb{Z}$: $$\Gamma \left( 2\,n+1 \right) = \left( 2\,n \right) !,\quad\zeta \left( 2\,n \right) ={\frac { \left( -1 \right) ^{n+1}B_{2n} \left( 2\,\pi \right) ^{2\,n}}{2\left( 2\,n \right) !}} \tag{3}$$ where $B$ denotes the Bernoulli number (or polynomial later) and thus: $$\int _{-\infty}^{\infty }\!{\frac {{x}^{2\,n}}{ \cosh^2 \left( x \right) }}{dx}= \left( 2-2^{2-2n} \right)\left( -1 \right) ^{n+1}B_{2n}\, {\pi }^{2\,n} \tag{4}$$ or if you prefer: $$\int _{-\infty}^{\infty }\!{\frac {{x}^{2\,n}}{ \cosh^2 \left( x \right) }}{dx}=2B_{2n}\left(\frac{1}{2} \right) \left( i\pi \right) ^{2\,n} \tag{5}$$

• +1 . Nice job. It's well known in Statistical Physics with Bose-Einstein distribution. – Felix Marin Feb 16 '14 at 22:36
• Oh cool, cheers! :) – Graham Hesketh Feb 16 '14 at 22:54
• Actually I had those integrals jumping out of Sommerfeld's Expansions for highly degenerate metals! – Brightsun Feb 17 '14 at 18:27

\begin{align} \color{#c00000}{I_{n}}&=8\int_{0}^{\infty}{x^{2n}\expo{2x} \over \pars{\expo{2x} + 1}^{2}}\,\dd x =-4\int_{x\ =\ 0}^{x\ \to\ \infty}x^{2n}\,\dd\pars{1 \over \expo{2x} + 1} \\[3mm]&=2\delta_{n0} + 8n \color{#00f}{\int_{0}^{\infty}{x^{2n - 1}\expo{-2x} \over 1 + \expo{-2x}}\,\dd x} \end{align}

\begin{align} &\color{#00f}{\int_{0}^{\infty}{x^{2n - 1}\expo{-2x} \over 1 + \expo{-2x}}\,\dd x} =\sum_{k = 0}^{\infty}\pars{-1}^{k}\int_{0}^{\infty} x^{2n - 1}\expo{-\pars{2k + 2}x}\,\dd x \\[3mm]&=\sum_{k = 0}^{\infty}{\pars{-1}^{k} \over \pars{2k + 2}^{2n}} \int_{0}^{\infty}x^{2n - 1}\expo{-x}\,\dd x =-\,{\Gamma\pars{2n} \over 2^{2n}} \sum_{k = 1}^{\infty}{\pars{-1}^{k} \over k^{2n}} \\[3mm]&=-{\Gamma\pars{2n} \over 2^{2n}} \bracks{{1 \over 2^{2n}}\sum_{k = 1}^{\infty}{1 \over k^{2n}} -\sum_{k = 1}^{\infty}{1 \over \pars{2k - 1}^{2n}}} \\[3mm]&=-{\Gamma\pars{2n} \over 2^{2n}} \bracks{{1 \over 2^{2n}}\sum_{k = 1}^{\infty}{1 \over k^{2n}} -\sum_{k = 1}^{\infty}{1 \over k^{2n}} + \sum_{k = 1}^{\infty}{1 \over \pars{2k}^{2n}}} \\[3mm]&=-{\Gamma\pars{2n} \over 2^{2n}}\pars{{1 \over 2^{2n - 1}} - 1} \sum_{k = 1}^{\infty}{1 \over k^{2n}} ={\Gamma\pars{2n} \over 2^{4n - 1}}\pars{2^{2n - 1} - 1}\zeta\pars{2n} \end{align}

$$\color{#66f}{\large I_{n}} \equiv\int_{-\infty}^{\infty}{x^{2n} \over \cosh^{2}\pars{x}}\,\dd x =\color{#66f}{\large2\delta_{n0} + n{2^{2n - 1} - 1 \over 2^{4n - 4}}\,\Gamma\pars{2n} \zeta\pars{2n}}$$

I would like to post an answer based on a slight variation of @Graham Hesketh's method, both because I find it a bit simpler and because it reflects the steps leading to the integrals temselves in the context where they appear.

For $t>0$ and $s>0$, we consider $$\int_0^\infty \frac{x^{s-1}}{e^{tx}-1}dx=\sum_{n=1}^\infty\int_0^\infty x^{s-1}e^{-ntx}dx=t^{-s}\sum_{n=1}^\infty n^{-s} \int_0^\infty x^{s-1}e^{-x}dx=\frac{\zeta(s)\Gamma(s)}{t^s}\,,$$ and note that $$\int_0^\infty \frac{x^{s-1}}{e^{tx}-1}dx -\int_0^\infty\frac{x^{s-1}}{e^{tx}+1}dx=2\int_0^\infty \frac{x^{s-1}}{e^{2tx}-1}dx = 2^{1-s}\int_0^\infty\frac{x^{s-1}}{e^{tx}-1}dx$$ so, by comparison, $$\int_0^\infty \frac{x^{s-1}}{e^{tx}+1}dx=\frac{1-2^{1-s}}{t^s}\zeta(s)\Gamma(s)\,.$$ The derivative of both sides of this equation, with respect to $t$, yields $$\int_0^\infty \frac{x^s}{\cosh^2(\frac{tx}{2})}dx = \frac{4s(1-2^{1-s})}{t^{s+1}}\zeta(s)\Gamma(s)\,.$$ In particular, for $s=2n$ with $n\in\mathbb N$, and $t=2$ we retrieve the integrals we wanted $$I_{n}=\int_{-\infty}^{+\infty} \frac{x^{2n}}{\cosh^2 x}dx = 2^{1-2n}(1-2^{1-2n})\,4n\,\zeta(2n)\Gamma(2n)\,.$$ E.g. $$I_0=2\,,\qquad I_1=\frac{\pi^2}{6}\,,\qquad I_2=\frac{7\pi^4}{120}\,.$$ (The case $n=0$ requires formally an analytic continuation $\epsilon \Gamma(\epsilon)=1+\mathcal O(\epsilon)$ and $\zeta(0)=-1/2$).


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题目列表

At a furniture store on Monday, the retail price of a table was n dollars. On Wednesday that week, the retail price of the table was reduced to $56. On Friday that week, the price of table was reduced again to 25 percent less than the price on Wednesday. If the ratio of n to the price of the table on Friday is 32 to 21, what is the value of n? Last year Leo bought two paintings. This year he sold them for$21,000 each. On one, he made a 25% profit, and on the other he had a 25% loss. What was his net loss or profit?

$_____ On the first day, the ticket price of a concert is 15% less than the original price; on the second day, the ticket price of the concert continued to drop by 20% than the first day. What percent is the price of the ticket on the second day less than the original price? Give your answer as a percent. _____% For each of the years 1993 and 1994, the population of City M increased by 20 percent during the year. If the population was 280,000 on December 31, 1994, approximately what was the population on January 1, 1993? y=2x+1, where x > 0. Quantity A The percent increase in the value of y when the value of x is increased by 10% Quantity B 10% Three variables, x, y, and z, in a scientific experiment are related by the equation $z=x^{2}y$. In the experiment, if the value of x decreases by 40 percent while the value of y increases by 50 percent, what is the percent decrease in the value of z? The total revenue that firm P receives from the sale of a particular item is determined by multiplying the number of the items it sells by the price of the item. If the price of the item is to be decreased by 20 percent, by what percent must the number of the items sold increase to keep the total revenue unchanged? Evelyn owns two different properties with areas of 1.5 acres and 0.5 acre, respectively. What is the total area, in square feet of the two properties combined? (1 acre = 4,840 square yards, 1 yard = 3 feet) _____square feet 1 yard=3 feet 1 acre=4840 square yard Quantity A One acre when transformed in terms of square feet Quantity B 40,000 The area of a certain floor is 150 square feet. (1 yard=3 feet) Quantity A The area of the floor, in square yards Quantity B 30 1 cup=8 ounces 1 pint=2 cups 1 quart=2 pints A large coffee jug contains 3 quarts, 1 pint, and 1.5 cups of coffee. What is the greatest number of 12-ounces mugs of coffee that can be filled from the jug? A certain store sells circular rugs at a constant price per square foot. If a circular rug with diameter 5 feet costs$250, what is the cost of a circular rug with diameter 9 feet?
After a bird feeder was filled with x pounds of birdseed, 12 ounces of birdseed were consumed each day for the next 8 days. (1 pound=16 ounces)

Quantity A

At the end of the 8 days, the weight of the birdseed in the bird feeder that was not consumed

Quantity B

x-5 pounds

The length of a model car is 8.2 centimeters, and the ratio of the length of the model car to the length of the actual car that it represents is 1 to 64. The length of the actual car is how many meters? (1 centimeter=0.01 meter)

_____meters
To fill a cube, machine A needs 5 hours. If machine A and B together fill the cube, it will take 1/3 of the time if machine B fills the cube alone. How many hours will it take if B fill the cube alone?
It takes machine A 2 minutes to fill a cube whose volume is 50m3 with water. If the cube holds 10m3 of water at first, and then machine A works for 1.5 minutes. By then, what percent of the cube is filled with water?

_____%
By draining 40 gallons of water from a tank, the amount of water in the tank was decreased from $\frac{1}{5}$ of the tank 's full capacity to $\frac{2}{11}$ of the tanks full capacity. Water was then added to the tank until the tank was full. How many gallons of water were added to the tank?
The overall volume of a water tank is g. If the inlet valve is open for a minute, x volume of water will be added to the tank; If the discharge valve is open for a minute, y volume of water will be removed from the tank (y>x). If the tank is filled with water at first, then in how many minutes will the water volume in the tank drop to a half if the inlet valve and discharge valve are both open?

Water started running into an empty tank and continued running until it was filled to its capacity. Water ran at a constant rate of 200 kiloliters per minute for the first 20 minutes and then at a constant rate of 100 kiloliters per minute until the tank was filled. If it took a total to 80 minutes to fill the empty tank to its capacity. how many minutes did it take to fill the tank to $\frac{1}{2}$ of its capacity?
A pump delivered water to fill an empty swimming pool. The pump delivered the water at a constant rate of 450 liters per minute until the pool was $\frac{1}{2}$ full. Then the pump became partially clogged and delivered the water at a slower constant rate until the pool was full. For the whole time during which the pump delivered water to fill the empty pool, its average rate was 360 liters per minute. What was the pump`s slower constant rate, in liters per minute?


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tinfoQ

2021-09-29

The set $B=\left\{1+{t}^{2},t+{t}^{2},1+2t+{t}^{2}\right\}$ is a basis for ${\mathbb{P}}_{2}$. Find the coordinate vector of $p\left(t\right)=1+4t+7{t}^{2}$ relative to B.

Nathanael Webber

Given basis can be written in matrix form (first row - second power, second row - first power, third row - free coefficient):
$B=\left[\begin{array}{ccc}1& 1& 1\\ 0& 1& 2\\ 1& 0& 1\end{array}\right]$
Solve $Bx=p\left(t\right)$ using augmented matrix:
$\left[\begin{array}{ccc}1& 1& 1|7\\ 0& 1& 2|4\\ 1& 0& 1|1\end{array}\right]$
, multiply first row with -1 and add to third
$\sim \left[\begin{array}{ccc}1& 1& 1|7\\ 0& 1& 2|4\\ 0& -1& 0|-6\end{array}\right]$
$-{x}_{2}=-6\to {x}_{2}=6$
${x}_{2}+2{x}_{3}=4$
$6+2{x}_{3}=4$
$2{x}_{3}=4-6=-2$
${x}_{3}=-1$
${x}_{1}+{x}_{2}+{x}_{3}=7$
${x}_{1}+6-1=7$
${x}_{1}=7-5=2$
${p}_{B}=\left[\begin{array}{c}2\\ 6\\ -1\end{array}\right]$
Results:
${p}_{B}=\left[\begin{array}{c}2\\ 6\\ -1\end{array}\right]$

Do you have a similar question?


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# Prove that jump functions are measurable

This question comes from the exercises of Stein and Shakarchi's Real analysis Ex. 5.14.

Define

$$j_n(x)= \begin{cases} 0& \text{if } x< x_n\\ \theta_n & \text{if } x=x_n\\ 1 &\text{if } x>x_n \end{cases}$$

where $\{x_n\}$ is a sequence of real number and $0\le\theta_n\le 1$.

Further define

$$J(x)=\sum^{\infty}_{n=1}\alpha_nj_n(x)$$

where $\alpha_n>0$ and $\sum\alpha_n$ converges.

Prove that for any $\epsilon$, the set of $x$ satisfying $$\limsup_{h\to0}\frac{J(x+h)-J(x)}{h}>\epsilon$$

is measurable.

So far what I can prove is that the sum in $J(x)$ can converge absolutely. And I know the limit of measurable functions is measurable. However, Since $J$ is not continuous, I cannot assume $h\to0$ as $\frac1n;n\to\infty$, then I cannot figure out whether the above set is measurable.

Edit: There is a hint in the book: consider $$F^N_{k,m}(x)=\sup_{1/k\le|h|\le1/m}\left|\frac{J_N(x+h)-J_N(x)}{h}\right|$$ where $J_N$ is the $N^{\text{th}}$ partial sum of $J$

Show that $F^N_{k,m}$ is measurable (which I couldn't tell why) and so does $$\lim_{m\to\infty}\lim_{k\to\infty}\lim_{N\to\infty}F^N_{k,m}(x)$$

(Which I also cannot relate it to the original set because I am confused of the multiple limit and supremum)

• Just to be clear, the exercise asks you to show that the map $x\mapsto \limsup_{h\to0}{J(x+h)-J(x)\over h}$ is measurable. The measurability of the set follows from this.
– user940
Aug 8, 2015 at 22:19

For brevity, write

$$\Delta(x,h) := \frac{J(x+h)-J(x)}{h} \qquad \text{and} \qquad \Delta_n(x,h) := \frac{J_n(x+h)-J_n(x)}{h}$$

Step 1: By definition of limsup, we have

\begin{align*} \limsup_{h \to 0} \Delta(x,h) > \epsilon &\iff \forall m \in \mathbb{N} \, \exists h \in \left[- \frac{1}{m}, \frac{1}{m} \right] \backslash \{0\}: \Delta(x,h) > \epsilon \\ &\iff \forall m \in \mathbb{N} \, \exists k \geq m \, \exists h \in \left[- \frac{1}{m}, \frac{1}{m} \right] \backslash \left(- \frac{1}{k}, \frac{1}{k} \right): \Delta(x,h)>\epsilon \\ &\iff \forall m \in \mathbb{N} \, \exists k \geq m: \sup_{\frac{1}{k} \leq |h| \leq \frac{1}{m}} \Delta(x,h)>\epsilon. \end{align*}

This shows that

$$\{x; \limsup_{h \to 0} \Delta(x,h)>\epsilon\} = \bigcap_{m \in \mathbb{N}} \bigcup_{k \geq m} \bigg\{x; \sup_{\frac{1}{k} \leq |h| \leq \frac{1}{m}} \Delta(x,h)>\epsilon \bigg\}.$$

Consequently, it suffices to prove that for each fixed $k,m \in \mathbb{N}$ and $\epsilon>0$ the set

$$A := A_{k,m,\epsilon} := \bigg\{x; \sup_{\frac{1}{k} \leq |h| \leq \frac{1}{m}} \Delta(x,h)>\epsilon \bigg\} \tag{1}$$

is measurable.

Step 2: We claim that $$A = \bigcup_{\ell \in \mathbb{N}} \bigcup_{N \in \mathbb{N}} \bigcap_{n \geq N} \bigg\{x; \sup_{\frac{1}{k} \leq |h| \leq \frac{1}{m}} \Delta_n(x,h)> \epsilon + \frac{1}{\ell} \bigg\}. \tag{2}$$

For brevity of notation, we set $$I := I_{k,m} := \left[- \frac{1}{m}, \frac{1}{m} \right] \backslash \left(- \frac{1}{k}, \frac{1}{k} \right).$$

Proof: Let $x \in A$, then $\sup_{\frac{1}{k} \leq |h| \leq \frac{1}{m}} \Delta(x,h)>\epsilon$. By the definition of $\sup$, we can find $h \in I$ such that $\Delta(x,h)>\epsilon$. In particular, there exists $l \in \mathbb{N}$ such that $\Delta(x,h) > \epsilon + \frac{2}{\ell}$. As $\Delta(x,h) = \lim_{n \to \infty} \Delta_n(x,h)$, we have $\Delta_n(x,h) \geq \epsilon+ \frac{1}{\ell}$ for all $n$ sufficiently large. This proves "$\subseteq$".

Now let $x$ be an element of the right-hand side of $(2)$. Then there exist $\ell ,N \in \mathbb{N}$ such that $$\sup_{\frac{1}{k} \leq |h| \leq \frac{1}{m}} \Delta_n(x,h)> \epsilon + \frac{1}{\ell}$$ for all $n \geq N$. By the definition of $\sup$, we can find $h(n) \in I$ such that $$\Delta_n(x,h(n))> \epsilon + \frac{1}{\ell}. \tag{3}$$ Since $I$ is compact, there exists a convergent subsequence $h_n \to h \in I$. Choose $n=n(x) \geq N$ sufficiently large such that $$|J_n(x)-J(x)| \leq \frac{1}{2k \ell}.$$ This implies $$\frac{|J_n(x)-J(x)|}{h} \leq \frac{1}{2\ell}$$ for all $h \in I$; hence in particular for $h=h_n$. Combining this with $(3)$, we get

\begin{align*} \frac{J_n(x+h_n)-J(x)}{h_n} &= \frac{J_n(x+h_n)-J_n(x)}{h_n} + \frac{J_n(x)-J(x)}{h_n}\\ &\geq \epsilon + \frac{1}{\ell} - \frac{1}{2\ell} = \epsilon + \frac{1}{2\ell} > \epsilon. \end{align*} \tag{4}

Finally, we note that $J_n(y) \leq J(y)$ implies

$$\epsilon \stackrel{(4)}{<} \frac{J_n(x+h_n)-J(x)}{h_n} \leq \frac{J(x+h_n)-J(x)}{h_n} = \Delta(x,h_n).$$

Consequently, $$\sup_{\frac{1}{k} \leq |h| \leq \frac{1}{m}} \Delta(x,h) \geq \Delta(x,h_n) > \epsilon,$$ i.e. $x \in A$.

Step 3: Recall that we actually want to show that $A$ defined in $(1)$ is measurable. By the second part, it suffices to show that sets of the form $$B := B_{N,n,k,m,\epsilon} := \left\{x; \sup_{\frac{1}{k} \leq |h| \leq \frac{1}{m}} \Delta_n(x,h)> \epsilon \right\}$$ are measurable.

To show this, we use the following elementary result:

Lemma: Let $f: (a,b) \to \mathbb{R}$ be of the form $f = \frac{f_1}{f_2}$ where $f_1$ is increasing and $f_2 \neq 0$ continuous. Then $$\limsup_{\mathbb{Q} \ni q \downarrow x} f(q) \geq f(x)$$ for any $x \in (a,b)$.

Since the mapping $h \mapsto J_n(x+h)-J_n(x)$ is increasing, we may apply the previous result to obtain $$\sup_{\frac{1}{k} \leq |h| \leq \frac{1}{m}} \Delta_n(x,h) = \sup_{(\mathbb{Q} \cap I) \cup \{\frac{1}{m}\}} \Delta_n(x,h) \tag{5}$$ where $I$ is defined as in step 2. (The inequality "$\geq$" holds in any case, the lemma above gives "$\leq$".) Since $\Delta_n(\cdot,h)$ is measurable, this proves that the left-hand side of $(5)$ is measurable and so is $B$.

• (If anyone else comes up with a better idea, I'll remove my answer. Might well be that my proof is much too complicated.)
– saz
Aug 8, 2015 at 10:36
• Wow! It is a great proof. I just have one question, which is why we can assume that $x+h\ne x_k$ in your second part? Aug 8, 2015 at 13:35
• @UnemChan I have changed the second part; there was some mistake in it. Should be fine now...
– saz
Aug 8, 2015 at 15:26
• @saz It seems to me that the set $I$ in equation (3) might depend on $x$.
– user940
Aug 8, 2015 at 22:32
• @ByronSchmuland I don't think so, but to get rid of this hand-waving argumentation, I have also changed this part of my answer (...which is now pretty close to yours, but, in fact, I had already thought of this before - I simply didn't realize that this can be used to show the measurability directly, without this approximation-stuff. +1 for your answer :)) [And thanks for correcting my typos...]
– saz
Aug 9, 2015 at 9:49

$J$ is a non-decreasing function, so for any fixed $h\neq 0$ if $q_n\to h$ with $|q_n|\geq |h|$, then $$\lim_n {J(x+q_n)-J(x)\over q_n} \geq {J(x+h)-J(x)\over h}\geq 0.$$ (Note that the absolute value bars are not necessary.) Hence for $m\geq 1$ we have $$\sup_{q \in\mathbb{Q},\, 0<|q|<1/m}{J(x+q)-J(x)\over q}= \sup_{0<|h|<1/m} {J(x+h)-J(x)\over h},\tag1$$ where $\mathbb{Q}$ means the rational numbers. Since $\mathbb{Q}$ is a countable set, equation (1) shows that $$\Psi_{m}(x):= \sup_{0<|h|<1/m}{J(x+h)-J(x)\over h}$$ is a measurable map into $[0,\infty].$ Therefore
$$\limsup_{h\to0}{J(x+h)-J(x)\over h} =\inf_{m\geq 1} \Psi_{m}(x)$$ is also measurable into $[0,\infty].$


text:
zbMATH — the first resource for mathematics

A fast Las Vegas algorithm for triangulating a simple polygon. (English) Zbl 0681.68061
The authors present in this revised and expanded version of a conference paper their randomized algorithm that triangulates a simple polygon of n vertices in 0(nlog*n) expected time.
Reviewer: H.D.Hecker

MSC:
 68Q25 Analysis of algorithms and problem complexity
Full Text:
References:
 [1] M. J. Atallah, R. Cole, and M. T. Goodrich, Cascading divide-and-conquer: a technique for designing parallel algorithms,Proceedings of the 28th Annual Symposium on Foundations of Computer Science, 1987, pp. 151-160. · Zbl 0677.68022 [2] Chazelle, B.; Incerpi, J., Triangulation and shape complexity, ACM Transactions on Graphics, 3, 135-152, (1984) · Zbl 0592.68083 [3] Clarkson, K. L., New applications of random sampling in computational geometry, Discrete and Computational Geometry, 2, 195-222, (1987) · Zbl 0615.68037 [4] K. L. Clarkson, Applications of random sampling in computational geometry, II,Proceedings of the Fourth Annual Symposium on Computational Geometry, 1988, pp. 1-11. [5] K. L. Clarkson, R. Cole, and R. E. Tarjan, private communication. [6] K. L. Clarkson and P. W. Shor, Applications of random sampling in computational geometry, II,Discrete and Computational Geometry, this issue, 387-421. · Zbl 0681.68060 [7] K. L. Clarkson, R. E. Tarjan, and C. J. Van Wyk, A fast Las Vegas algorithm for triangulating a simple polygon,Proceedings of the Fourth Annual Symposium on Computational Geometry, 1988, pp. 18-22. · Zbl 0681.68061 [8] P. Erdos and J. Spencer,Probabilistic Methods in Combinatorics, Academic Press, New York, 1974. · Zbl 0308.05001 [9] Fournier, A.; Montuno, D. Y., Triangulating simple polygons and equivalent problems, ACM Transactions on Graphics, 3, 153-174, (1984) · Zbl 0592.68084 [10] K. Y. Fung, T. M. Nicholl, R. E. Tarjan, and C. J. Van Wyk, Simplified linear-time Jordan sorting and polygon clipping,ACM Transactions on Graphics, submitted. · Zbl 0697.68036 [11] Garey, M. R.; Johnson, D. S.; Preparata, F. P.; Tarjan, R. E., Triangulating a simple polygon, Information Processing Letters, 7, 175-180, (1978) · Zbl 0384.68040 [12] Haussler, D.; Welzl, E.$$, ɛ$$-nets and simplex range queries, Discrete and Computational Geometry, 2, 127-151, (1987) · Zbl 0619.68056 [13] Hoffman, K.; Mehlhorn, K.; Rosenstiehl, P.; Tarjan, R., Sorting Jordan sequences in linear time using level-linked search trees, Information and Control, 68, 170-184, (1986) · Zbl 0614.68051 [14] F. P. Preparata and M. I. Shamos,Computational Geometry: An Introduction, Springer-Verlag, New York, 1985. · Zbl 0759.68037 [15] Tarjan, R. E.; Wyk, C. J., An $$O(n$$ log log $$n)$$-time algorithm for triangulating a simple polygon, SIAM Journal on Computing, 17, 143-178, (1988) · Zbl 0637.68044
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching.


text:
# Riemann surface

In mathematics, particularly in complex analysis, a Riemann surface is a one-dimensional complex manifold. Riemann surfaces can be thought of as "deformed versions" of the complex plane: locally near every point they look like patches of the complex plane, but the global topology can be quite different. For example, they can look like a sphere or a torus or a couple of sheets glued together.

The main point of Riemann surfaces is that holomorphic functions may be defined between them. Riemann surfaces are nowadays considered the natural setting for studying the global behavior of these functions, especially multi-valued functions such as the square root or the logarithm.

Every Riemann surface is a two-dimensional real analytic manifold (i.e. a surface), but it contains more structure (specifically a complex structure) which is needed for the unambiguous definition of holomorphic functions. A two-dimensional real manifold can be turned into a Riemann surface (usually in several inequivalent ways) if and only if it is orientable. So the sphere and torus admit complex structures, but the Möbius strip, Klein bottle and projective plane do not.

Geometrical facts about Riemann surfaces are as "nice" as possible, and they often provide the intuition and motivation for generalizations to other curves, manifolds or varieties. The Riemann-Roch theorem is a prime example of this influence.

 Contents

## Formal definition

Let X be a Hausdorff space. A homeomorphism from an open subset UX to a subset of C is called a chart. Two charts f and g whose domains intersect are said to be compatible if the maps f o g-1 and g o f-1 are holomorphic over their domains. If A is a collection of compatible charts and if any x in X is in the domain of some f in A, then we say that A is an atlas. When we endow X with an atlas A, we say that (X, A) is a Riemann surface. If the atlas is understood, we simply say that X is a Riemann surface.

Different atlases can give rise to essentially the same Riemann surface structure on X; to avoid this ambiguity, one sometimes demands that the given atlas on X be maximal, in the sense that it is not contained in any other atlas. Every atlas A is contained in a unique maximal one by Zorn's lemma.

## Examples

• The complex plane C is perhaps the most trivial Riemann surface. The map f(z) = z (the identity map) defines a chart for C, and {f} is an atlas for C. The map g(z) = z* (the conjugate map) also defines a chart on C and {g} is an atlas for C. The charts f and g are not compatible, so this endows C with two distinct Riemann surface structures. In fact, given a Riemann surface X and its atlas A, the conjugate atlas B = {f* : f ∈ A} is never compatible with A, and endows X with a distinct, incompatible Riemann structure.
• In an analogous fashion, every open subset of the complex plane can be viewed as a Riemann surface in a natural way. More generally, every open subset of a Riemann surface is a Riemann surface.
• Let S = C ∪ {∞} and let f(z) = z where z is in S \ {∞} and g(z) = 1 / z where z is in S \ {0} and 1/∞ is defined to be 0. Then f and g are charts, they are compatible, and { f, g } is an atlas for S, making S into a Riemann surface. This particular surface is called the Riemann sphere because it can be interpreted as wrapping the complex plane around the sphere. Unlike the complex plane, it is compact.
• The theory of compact Riemann surfaces can be shown to be equivalent to that of algebraic curves that are defined over the complex numbers and non-singular. Important examples of non-compact Riemann surfaces are provided by analytic continuation (see below.)

## Properties and further definitions

A function f : MN between two Riemann surfaces M and N is called holomorphic if for every chart g in the atlas of M and every chart h in the atlas of N, the map h o f o g-1 is holomorphic (as a function from C to C) wherever it is defined. The composition of two holomorphic maps is holomorphic. The two Riemann surfaces M and N are called conformally equivalent if there exists a bijective holomorphic function from M to N whose inverse is also holomorphic (it turns out that the latter condition is automatic and can therefore be omitted). Two conformally equivalent Riemann surfaces are for all practical purposes identical.

Every simply connected Riemann surface is conformally equivalent to C or to the Riemann sphere C ∪ {∞} or to the open disk {zC : |z| < 1}. This statement is known as the uniformization theorem.

Every connected Riemann surface can be turned into a complete 2-dimensional real Riemannian manifold with constant curvature -1, 0 or 1. This Riemann structure is unique up to scalings of the metric. The Riemann surfaces with curvature -1 are called hyperbolic; the open disk is the canonical example. The Riemann surfaces with curvature 0 are called parabolic; C is a typical parabolic Riemann surface. Finally, the surfaces with curvature +1 are known as elliptic; the Riemann sphere C ∪ {∞} is an example.

For every closed parabolic Riemann surface, the fundamental group is isomorphic to a rank 2 lattice group, and thus the surface can be constructed as C/Γ, where C is the complex plane and Γ is the lattice group. The set of representatives of the cosets are called fundamental domains.

Similarly, for every hyperbolic Riemann surface, the fundamental group is isomorphic to a Fuchsian group, and thus the surface can be modelled by a Fuchsian model H/Γ where H is the upper half-plane and Γ is the Fuchsian group. The set of representatives of the cosets of H/Γ are free regular sets and can be fashioned into metric fundamental polygons.

When a hyperbolic surface is compact, then the total area of the surface is [itex]4\pi(g-1)[itex], where g is the genus of the surface; the area is obtained by applying the Gauss-Bonnet theorem to the area of the fundamental polygon.

We noted in the preamble that all Riemann surfaces, like all complex manifolds, are orientable as a real manifold. The reason is that for complex charts f and g with transition function h = f(g-1(z)) we can consider h as a map from an open set of R2 to R2 whose Jacobian in a point z is just the real linear map given by multiplication by the complex number h'(z). However, the real determinant of multiplication by a complex number α equals |α|^2, so the Jacobian of h has positive determinant. Consequently the complex atlas is an oriented atlas.

## History

Riemann surfaces were first studied by Bernhard Riemann and were named after him.

## References

• Art and Cultures
• Countries of the World (http://www.academickids.com/encyclopedia/index.php/Countries)
• Space and Astronomy


text:
Arrange business cards with crop marks

I have a single business card as PDF (paper size is 85mm x 54mm). Does anyone have a solution to arrange copies along an A4 paper with crop marks?

Here's an example:

With pdfpages, arranging works fine, but I can't get cropmarks. One could probably achieve something like this using Tikz, but I'd prefer something more out-of-the-box.

-
Please post a Minimal Working Example. For example, post the code you use to arrange the cards using pdfpages so people don't have to start from scratch! Maybe ctan.org/pkg/ncccropbox? –  cfr Mar 23 '14 at 22:30
@cfr I did not provide a minimal example because I know that pdfpages does not support crop marks (according to the manual), so a minimal example using pdfpages would not help. I did not try solutions using TIkz yet, as I'd hope for a solution as automated as possible ;-) –  Marcus Bitzl Mar 24 '14 at 17:23

\documentclass[a4paper]{article}
\usepackage[scale=.95,landscape]{geometry}
\usepackage{ncccropbox,graphicx,multido}
\begin{document}
\noindent
\multido{}{3}{%
\multido{}{3}{%
\cropbox{\includegraphics[angle=90]{54x85}}}\\}
\end{document}

will produce

Note: The text is only the wrong way around because I used a 85x54 image rather than a 54x85 one so turned it.

-

Not very dynamic, but this it can be achieved with \rule.

\documentclass{article}

\usepackage[landscape,textwidth=275mm,textheight=182mm]{geometry}
\usepackage{xcolor}
\usepackage{mathtools}
\pagestyle{empty}

\begin{document}

\noindent
\smash{\rule[-0.5pt]{8mm}{1pt}}%
\hspace{2mm}%
\clap{\rule[2mm]{1pt}{8mm}}%
\hspace{85mm}%
\clap{\rule[2mm]{1pt}{8mm}}%
\hspace{85mm}%
\clap{\rule[2mm]{1pt}{8mm}}%
\hspace{85mm}%
\clap{\rule[2mm]{1pt}{8mm}}%
\hspace{2mm}%
\smash{\rule[-0.5pt]{8mm}{1pt}}\\[-1pt]
\smash{\rule[-0.5pt]{8mm}{1pt}}%
\hspace{2mm}%
\textcolor{lightgray}{\rule{85mm}{54mm}}%
\textcolor{gray}{\rule{85mm}{54mm}}%
\textcolor{lightgray}{\rule{85mm}{54mm}}%
\hspace{2mm}%
\smash{\rule[-0.5pt]{8mm}{1pt}}\\[-1pt]
\smash{\rule[-0.5pt]{8mm}{1pt}}%
\hspace{2mm}%
\textcolor{gray}{\rule{85mm}{54mm}}%
\textcolor{lightgray}{\rule{85mm}{54mm}}%
\textcolor{gray}{\rule{85mm}{54mm}}%
\hspace{2mm}%
\smash{\rule[-0.5pt]{8mm}{1pt}}\\[-1pt]
\smash{\rule[-0.5pt]{8mm}{1pt}}%
\hspace{2mm}%
\textcolor{lightgray}{\rule{85mm}{54mm}}%
\textcolor{gray}{\rule{85mm}{54mm}}%
\textcolor{lightgray}{\rule{85mm}{54mm}}%
\hspace{2mm}%
\smash{\rule[-0.5pt]{8mm}{1pt}}\\[-1pt]
\hspace*{1cm}%
\rule{0pt}{1cm}%
\clap{\rule{1pt}{8mm}}%
\hspace{85mm}%
\clap{\rule{1pt}{8mm}}%
\hspace{85mm}%
\clap{\rule{1pt}{8mm}}%
\hspace{85mm}%
\clap{\rule{1pt}{8mm}}%

\end{document}

-
This doesn't really answer the question, though, does it? It does reproduces the image posted to illustrate the desired layout. It doesn't show how to include the business card or produce crop marks. –  cfr Mar 23 '14 at 22:54
I understood the question as that he knows how to add the business cards, but not the crop marks. The code adds crop marks around the edge like his example. –  Jonas Granholm Mar 24 '14 at 1:50
Yes but the OP is using pdfpages to add the card. That won't work straightforwardly with this solution because pdfpages will put the card on a new page. It is possible to address that, but it is not trivial to do so. And I don't understand why the checks at all. –  cfr Mar 24 '14 at 1:53
Hi, maybe I was a bit unclear: I statet that pdfpages does not support crop marks. A solution using \includegraphics (or anything else) like this is as fine as well. –  Marcus Bitzl Mar 24 '14 at 17:20


text:
## Geometry: Common Core (15th Edition)

Published by Prentice Hall

# Chapter 1 - Tools of Geometry - 1-7 Midpoint and Distance in the Coordinate Plane - Lesson Check - Page 53: 3

8.5

#### Work Step by Step

Let $K(-9,8)$ be $(x_1,y_1)$. Let $L(-6,0)$ be $(x_2,y_2)$. Use the distance formula. $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$ $d=\sqrt{(-6-(-9))^2+(0-8)^2}$ $d=\sqrt{(3)^2+(-8)^2}$ $d=\sqrt{9+64}$ $d=\sqrt{73}$ $d\approx8.5$

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.


text:
# Properties

 Label 6930.g Number of curves 2 Conductor 6930 CM no Rank 0 Graph

# Related objects

Show commands for: SageMath
sage: E = EllipticCurve("6930.g1")
sage: E.isogeny_class()

## Elliptic curves in class 6930.g

sage: E.isogeny_class().curves
LMFDB label Cremona label Weierstrass coefficients Torsion order Modular degree Optimality
6930.g1 6930j1 [1, -1, 0, -44955, 3679825] 2 23040 $$\Gamma_0(N)$$-optimal
6930.g2 6930j2 [1, -1, 0, -42525, 4093411] 2 46080

## Rank

sage: E.rank()

The elliptic curves in class 6930.g have rank $$0$$.

## Modular form6930.2.a.g

sage: E.q_eigenform(10)
$$q - q^{2} + q^{4} - q^{5} + q^{7} - q^{8} + q^{10} + q^{11} + 4q^{13} - q^{14} + q^{16} + 4q^{17} + 8q^{19} + O(q^{20})$$

## Isogeny matrix

sage: E.isogeny_class().matrix()

The $$i,j$$ entry is the smallest degree of a cyclic isogeny between the $$i$$-th and $$j$$-th curve in the isogeny class, in the LMFDB numbering.

$$\left(\begin{array}{rr} 1 & 2 \\ 2 & 1 \end{array}\right)$$

## Isogeny graph

sage: E.isogeny_graph().plot(edge_labels=True)

The vertices are labelled with LMFDB labels.


text:
# Start a new discussion

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## Site Tag Cloud

Vanilla 1.1.10 is a product of Lussumo. More Information: Documentation, Community Support.

• Added a section on terminology.

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Anonymous

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Jon Awbrey

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Jan Głowacki

• just a stub for the moment, in order to make links work

• some minimum

• The entry unit of an adjunction had a big chunk of mixed itex+svg code at the beginning to display an adjunction. On my machine though the output of that code was ill typeset. So I have removed the code and replaced it by plain iTex encoding of an adjunction.

(Just in case anyone deeply cares about the svg that was there. It’s still in the history. If it is preferred by anyone, it needs to be fixed first.)

• starting something. Not done yet, but need to save

• a bare list of fundamental scales (speed of light, Planck scale, etc. ), to be !include-ed into the “Related entries”-section of relevant entries

• splitting this off from Schwinger effect, to go with a list of entries on fundamental scales

• needed a minimum here, for satisfying links

• a stub, to satisfy links

• Felipe Rosso, Global aspects of conformal symmetry and the ANEC in dS and AdS (arXiv:1912.08897)
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• starting something

• Todd,

when you see this here and have a minute, would you mind having a look at monoidal category to see if you can remove the query-box discussion there and maybe replace it by some crisp statement?

Thanks!

• created cohesive topos.

• wrote an Idea-section that is meant to explain why the concept is very natural, trying to provide some of the chat that one cannot find in the terse (but beautiful in its own way) article by Lawvere

• spelled out the definition in some detail, here, too, trying to fill in things that Lawvere is glossing over, making it all very explicit;

• started an Examples-section:

• copied over the discussion that $Sh(CartSp)$ is a connected topos. checking the remaining axioms for cohesive topos are easy, but i have not typed that yet

• included a little discussion of how diffeological spaces fit in, following our conversation in another thread

• started an analogous section for $\infty Sh(CartSp)$, but just a stub so far

• but added in a section that goes rgrough the various items in Lawvere’s definition and discusses their meaning in a cohesive oo-topos

• I created the page

with some discussion and references on the various triangulated categories associated to a scheme (e.g. bounded derived category of coherent sheaves, triangulated category of perfect complexes).

• removed typo

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• Clarified relationship between polycategories, properads and PROPs.

• I noticed that some Anonymous Coward had created a very stubby stub at two-out-of-six property, and used it as a place to put proofs of two theorems about when the 2-out-of-6 property is equivalent to saturation of the class of weak equivalences.

• Page created, but author did not leave any comments.


text:
# How to use -For parameter in Restart-Computer cmdlet in PowerShell?

PowerShellMicrosoft TechnologiesSoftware & Coding

In the –Wait parameter 3 main checks are validated − PowerShell, WMI, and WINRM connectivity. So whenever you specify the –Wait parameter, you have to wait until all three checks pass, however, if any of the checks fail then script freezes and can not further execute. If you know which specific check you want to perform then you can specify that value with the –For parameter as shown below.

Here, for the example, we will run a single WMI check for the remote server.

Restart-Computer Test1-Win2k12 -Wait -For Wmi -Force

-For parameter must be used with –Wait parameter. Here, again if something goes wrong while validation, it may take longer and longer time for the validation to pass or fail. To prevent the deadlock situation, you can use the -Timeout parameter.

Published on 24-Jul-2020 10:34:27