text:
# A Lego Toolbox for Flexible Bayesian Regression (and Beyond)

library("bamlss") prefix <- "http://www.bamlss.org/articles/" ## "" prefix2 <- "http://www.bamlss.org/reference/" ## ""

## Overview

The R package bamlss provides a modular computational framework for flexible Bayesian regression models (and beyond). The implementation follows the conceptional framework presented in @bamlss:Umlauf+Klein+Zeileis:2017 and provides a modular "Lego toolbox" for setting up regression models. In this setting not only the response distribution or the regression terms are "Lego bricks" but also the estimation algorithm or the MCMC sampler.

The highlights of the package are:

• The usual R "look & feel" for regression modeling.
• Estimation of classic (GAM-type) regression models (Bayesian or frequentist).
• Estimation of flexible (GAMLSS-type) distributional regression models.
• An extensible "plug & play" approach for regression terms.
• Modular combinations of fitting algorithms and samplers.

Especially the last item is notable because the models in bamlss are not limited to a specific estimation algorithm but different engines can be plugged in without necessitating changes in other aspects of the model specification.

More detailed overviews and examples are provided in the articles:

## Installation

The stable release version of bamlss is hosted on the Comprehensive R Archive Network (CRAN) at https://CRAN.R-project.org/package=bamlss and can be installed via

install.packages("bamlss")

The development version of bamlss is hosted on R-Forge at https://R-Forge.R-project.org/projects/bayesr/ in a Subversion (SVN) repository. It can be installed via

install.packages("bamlss", repos = "http://R-Forge.R-project.org")

## Basic Bayesian regression

This section gives a first quick overview of the functionality of the package and demonstrates that the usual "look & feel" when using well-established model fitting functions like glm() is an elementary part of bamlss, i.e., first steps and basic handling of the package should be relatively simple. We illustrate the first steps with bamlss using a data set taken from the Regression Book [@bamlss:Fahrmeir+Kneib+Lang+Marx:2013] which is about prices of used VW Golf cars. The data is loaded with

data("Golf", package = "bamlss") head(Golf)

In this example the aim is to model the price in 1000 Euro. Using bamlss a first Bayesian linear model could be set up by first specifying a model formula

f <- price ~ age + kilometer + tia + abs + sunroof

afterwards the fully Bayesian model using MCMC simulation is estimated by

library("bamlss") set.seed(111) b1 <- bamlss(f, family = "gaussian", data = Golf)

Note that the default number of iterations for the MCMC sampler is 1200, the burnin-phase is 200 and thinning is 1 (see the manual of the default MCMC sampler GMCMC()). The reason is that during the modeling process, users usually want to obtain first results rather quickly. Afterwards, if a final model is estimated the number of iterations of the sampler is usually set much higher to get close to i.i.d. samples from the posterior distribution. To obtain reasonable starting values for the MCMC sampler we run a backfitting algorithm that optimizes the posterior mode. The bamlss package uses its own family objects, which can be specified as characters using the bamlss() wrapper, in this case family = "gaussian" (see also BAMLSS Families). In addition, the package also supports all families provided from the gamlss families.

The model summary gives

summary(b1)

indicating high acceptance rates as reported by the alpha parameter in the linear model output, which is a sign of good mixing of the MCMC chains. The mixing can also be inspected graphically by

plot(b1, which = "samples")
bsp <- b1 bsp$samples <- bsp$samples[, c("mu.p.(Intercept)", "sigma.p.(Intercept)")] plot(bsp, which = "samples")

Note, for convenience we only show the traceplots of the intercepts. Considering significance of the estimated effects, only variables tia and sunroof seem to have no effect on price since the credible intervals of estimated parameters contain zero. This information can also be extracted using the implemented confint() method.

confint(b1, prob = c(0.025, 0.975))

Since the prices cannot be negative, a possible consideration is to use a logarithmic transformation of the response price

set.seed(111) f <- log(price) ~ age + kilometer + tia + abs + sunroof b2 <- bamlss(f, family = "gaussian", data = Golf)

and compare the models using the DIC()

DIC(b1, b2)

indicating that the transformation seems to improve the model fit.

Instead of using linear effects, another option would be to use polynomial regression for covariates age, kilometer and tia. A polynomial model using polynomials of order 3 is estimated with

set.seed(222) f <- log(price) ~ poly(age, 3) + poly(kilometer, 3) + poly(tia, 3) + abs + sunroof b3 <- bamlss(f, family = "gaussian", data = Golf)

Comparing the models using the DIC() function

DIC(b1, b2, b3)

suggests that the polynomial model is not really an improvement, however, this is probably caused by an insignificant tia effect. The effects can be inspected graphically, to better understand their influence on price. Using the polynomial model, graphical inspections can be done using the predict() method.

One major difference compared to other regression model implementations is that predictions can be made for single variables, only, where the user does not have to create a new data frame containing all variables. For example, posterior mean estimates and 95% credible intervals for variable age can be obtained by

nd <- data.frame("age" = seq(min(Golf$age), max(Golf$age), length = 100)) nd$page <- predict(b3, newdata = nd, model = "mu", term = "age", FUN = c95, intercept = FALSE) head(nd) Note that the prediction does not include the model intercept. Similarly for variables kilometer and tia nd$kilometer <- seq(min(Golf$kilometer), max(Golf$kilometer), length = 100) nd$tia <- seq(min(Golf$tia), max(Golf$tia), length = 100) nd$pkilometer <- predict(b3, newdata = nd, model = "mu", term = "kilometer", FUN = c95, intercept = FALSE) nd$ptia <- predict(b3, newdata = nd, model = "mu", term = "tia", FUN = c95, intercept = FALSE) Here, we need to specify for which model predictions should be calculated, and if predictions only for variable age are created, argument term needs also be specified. Argument FUN can be any function that should be applied on the samples of the linear predictor. For more examples see the documentation of the predict.bamlss() method. Then, the estimated effects can be visualized with par(mfrow = c(1, 3)) ylim <- range(c(nd$page, nd$pkilometer, nd$ptia)) plot2d(page ~ age, data = nd, ylim = ylim) plot2d(pkilometer ~ kilometer, data = nd, ylim = ylim) plot2d(ptia ~ tia, data = nd, ylim = ylim)
par(mfrow = c(1, 3), mar = c(4.1, 4.1, 0.1, 1.1)) ylim <- range(c(nd$page, nd$pkilometer, nd$ptia)) plot2d(page ~ age, data = nd, ylim = ylim) plot2d(pkilometer ~ kilometer, data = nd, ylim = ylim) plot2d(ptia ~ tia, data = nd, ylim = ylim) The figure clearly shows the negative effect on the logarithmic price for variable age and kilometer. The effect of tia is not significant according the 95% credible intervals, since the interval always contains the zero horizontal line. ## Location-scale model data("mcycle", package = "MASS") f <- list(accel ~ s(times, k = 20), sigma ~ s(times, k = 20)) set.seed(123) b <- bamlss(f, data = mcycle, family = "gaussian") As a second startup on how to use bamlss for full distributional regression, we illustrate the basic steps on a small textbook example using the well-known simulated motorcycle accident data [@bamlss:Silverman:1985]. The data contain measurements of the head acceleration (in$g$, variable accel) in a simulated motorcycle accident, recorded in milliseconds after impact (variable times). data("mcycle", package = "MASS") head(mcycle) To estimate a Gaussian location-scale model with $$\texttt{accel} \sim \mathcal{N}(\mu = f(\texttt{times}), \log(\sigma) = f(\texttt{times}))$$ we use the following model formula f <- list(accel ~ s(times, k = 20), sigma ~ s(times, k = 20)) where s() is the smooth term constructor from the mgcv [@bamlss:Wood:2018]. Note, that formulae are provided as lists of formulae, i.e., each list entry represents one parameter of the response distribution. Also note that all smooth terms, i.e., te(), ti(), etc., are supported by bamlss. This way, it is also possible to incorporate user defined model terms. A fully Bayesian model is the estimated with set.seed(123) b <- bamlss(f, data = mcycle, family = "gaussian") using the default of 1200 iterations of the MCMC sampler to obtain first results quickly (see the documentation GMCMC() for further details on tuning parameters). Note that per default bamlss() uses a backfitting algorithm to compute posterior mode estimates, afterwards these estimates are used as starting values for the MCMC chains. The returned object is of class "bamlss" for which generic extractor functions like summary(), plot(), predict(), etc., are provided. For example, the estimated effects for distribution parameters mu and sigma can be visualized by plot(b, model = c("mu", "sigma")) par(mar = c(4.1, 4.1, 1.1, 1.1), mfrow = c(1, 2)) plot(b, pages = 1, spar = FALSE, scheme = 2, grid = 100) The model summary gives summary(b) showing, e.g., the acceptance probabilities of the MCMC chains (alpha), the estimated degrees of freedom of the optimizer and the successive sampler (edf), the final AIC and DIC as well as parametric model coefficients (in this case only the intercepts). As mentioned in the first example, using MCMC involves convergence checks of the sampled parameters. The easiest diagnostics are traceplots plot(b, which = "samples") bsp <- b bsp$samples <- bsp$samples[, c("mu.p.(Intercept)", "sigma.p.(Intercept)")] plot(bsp, which = "samples") Note again that this call would show all traceplots, for convenience we only show the plots for the intercepts. In this case, the traceplots do not indicate convergence of the Markov chains for parameter "mu". To fix this, the number of iterations can be increased and also the burnin and thinning parameters can be adapted (see GMCMC()). Further inspections are the maximum autocorrelation of all parameters, using plot.bamlss() setting argument which = "max-acf", besides other convergence diagnostics, e.g., diagnostics that are part of the coda package [@bamlss:Plummer+Best+Cowles+Vines:2006]. Inspecting randomized quantile residuals [@bamlss:Dunn+Gordon:1996] is useful for judging how well the model fits to the data plot(b, which = c("hist-resid", "qq-resid")) par(mfrow = c(1, 2)) plot(b, which = c("hist-resid", "qq-resid"), spar = FALSE) Randomized quantile residuals are the default method in bamlss, which are computed using the CDF function of the corresponding family object. The posterior mean including 95% credible intervals for new data based on MCMC samples for parameter$\mu$can be computed by nd <- data.frame("times" = seq(2.4, 57.6, length = 100)) nd$ptimes <- predict(b, newdata = nd, model = "mu", FUN = c95) plot2d(ptimes ~ times, data = nd)
par(mar = c(4.1, 4.1, 1.1, 1.1)) nd <- data.frame("times" = seq(2.4, 57.6, length = 100)) nd$ptimes <- predict(b, newdata = nd, model = "mu", FUN = c95) plot2d(ptimes ~ times, data = nd) and as above in the first example, argument FUN can be any function, e.g., the identity() function could be used to calculate other statistics of the distribution, e.g., plot the estimated densities for each iteration of the MCMC sampler for times = 10 and times = 40: ## Predict for the two scenarios. nd <- data.frame("times" = c(10, 40)) ptimes <- predict(b, newdata = nd, FUN = identity, type = "parameter") ## Extract the family object. fam <- family(b) ## Compute densities. dens <- list("t10" = NULL, "t40" = NULL) for(i in 1:ncol(ptimes$mu)) { ## Densities for times = 10. par <- list( "mu" = ptimes$mu[1, i, drop = TRUE], "sigma" = ptimes$sigma[1, i, drop = TRUE] ) dens$t10 <- cbind(dens$t10, fam$d(mcycle$accel, par)) ## Densities for times = 40. par <- list( "mu" = ptimes$mu[2, i, drop = TRUE], "sigma" = ptimes$sigma[2, i, drop = TRUE] ) dens$t40 <- cbind(dens$t40, fam$d(mcycle$accel, par)) } ## Visualize. par(mar = c(4.1, 4.1, 0.1, 0.1)) col <- rainbow_hcl(2, alpha = 0.01) plot2d(dens$t10 ~ accel, data = mcycle, col.lines = col[1], ylab = "Density") plot2d(dens$t40 ~ accel, data = mcycle, col.lines = col[2], add = TRUE)


text:
### Home > CALC > Chapter 2 > Lesson 2.1.3 > Problem2-40

2-40.
1. For the given function, write an expression in summation notation that will approximate A(f, 3 ≤ x ≤ 7) using eight rectangles. Specify if you use left, right, or midpoint rectangles. Then enter this summation expression into your graphing calculator and evaluate the approximate area. Homework Help ✎

Recall problem 2-17:

i. For left-endpoint and midpoint rectangles, let the index start at 0 and end at n-1.
For right-endpoint rectangles, let the index start at 1 and end at n.
n = number of rectangles.

$\text{ii. The }\frac{1}{2}\text{ represents the WIDTH/BASE of all rectangles.}$

To determine the width of each rectangle, divide the domain by the number of rectangles:

$\frac{7-3}{1}=\frac{1}{2}.$

$\text{Each rectangle has a width, }\Delta x \text{ is } \frac{1}{2}.$

iii. The transformed function represents the HEIGHT of each rectangle.
For left- and right-endpoint rectangles, the heights are represented by f(a + Δxi).
For midpoint rectangles, the heights are represented by

$f \left(a+\frac{\Delta x}{2}i\right).$

a = starting value
Δx = width of each rectangle.

format: plain


text:
# Math Help - Finding x & y intercepts (found the y-intercept)

1. ## Finding x & y intercepts (found the y-intercept)

I need help please with finding the x-intercept for the following:

$y = -2x^2 +4$

I got the y-intercept (0,4) by setting x = 0

However, when I set y = 0 and attempt to solve I get stuck

$0 = -2x^2 + 4$

What are the next steps for finding the x-intercept? Thanks in advance!

2. $x= \pm \sqrt 2$?

3. Originally Posted by lilrhino
I need help please with finding the x-intercept for the following:

$y = -2x^2 +4$

I got the y-intercept (0,4) by setting x = 0

However, when I set y = 0 and attempt to solve I get stuck

$0 = -2x^2 + 4$

What are the next steps for finding the x-intercept? Thanks in advance!
$0 = -2x^2 + 4$ .........subtract 4 from both sides

$\Rightarrow -4 = -2x^2$ .........Now divide through by -2

$\Rightarrow 2 = x^2$ .............Now take the squareroot of both sides

$\Rightarrow \pm \sqrt {2} = x$

So our x-intercepts are: $\left( \sqrt {2} , 0 \right) \mbox { and } \left( - \sqrt {2} , 0 \right)$


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# beamer to show nested-subitem when all (non-nested) items covered

I am trying to get an alternating highlighting effect between \itemize items and subitems (which are just nested itemize items), but there is an issue in that when all of the non-nested items are hidden the nested subitem fails to show. The desired effect is that:

• Overlay #1: the non-nested items will be initially shown in full opacity while the nested subitem will be initially shown as transparent
• Overlay #2: the non-nested items are hidden as transparent items, while the nested sub-item is shown in full opacity
• Overlay #3: all items are shown in full opacity

A minimal working example to reproduce the issue (using code that reflects what I'm trying to achieve):

\documentclass{beamer}
\begin{document}
\setbeamercovered{transparent}
\begin{frame}
\begin{itemize}
\item<1,3>{outside}
\begin{itemize}
\item<2->{inside}
\end{itemize}
\item<1,3>{outside}
\end{itemize}
\end{frame}
\end{document}

Note that the issue seems to be specifically at the last non-nested item before the nested itemize environment. I.e., the issue above is that the sub-item fails perform the overlay spec of appearing at slide #2, but this spec works in the example below when the non-nested item above it is also displayed during slide #2 (instead of above where it is hidden during slide #2):

\documentclass{beamer}
\begin{document}
\setbeamercovered{transparent}
\begin{frame}
\begin{itemize}
\item<1->{outside}  % <---- this is the difference from above
\begin{itemize}
\item<2->{inside}
\end{itemize}
\item<1,3>{outside}
\end{itemize}
\end{frame}
\end{document}

Include \visible with your \items

\documentclass{beamer}
\begin{document}
\setbeamercovered{transparent}
\begin{frame}
\begin{itemize}
\item\visible<1,3>{outside}
\begin{itemize}
\item\visible<2->{inside}
\end{itemize}
\item\visible<1,3>{outside}
\end{itemize}
\end{frame}
\end{document}

• Thanks, this worked (although I use \uncover instead of \visible). – Daniel Naftalovich May 4 '16 at 0:51
• @DanielNaftalovich You're welcome. I wasn't sure if transparent meant covered or invisible, but I'm glad it worked for you. – erik May 4 '16 at 0:53

Here is a hack that almost accomplishes the intention, but non-ideal because the bullet does not become transparent (and also is a hassle to code)

\setbeamercovered{transparent}
\begin{frame}
\begin{itemize}
\item{\temporal<2>{outside}{\color{black!15}{outside}}{outside}}
\begin{itemize}
\item<->{inside}
\end{itemize}
\item<1,3>{later}
\end{itemize}
\end{frame}


text:
## Thursday, September 04, 2014 ... /////

### Brian Cox's incompetence

Like Sean Carroll, Brian Cox pretends to be a scientist but in reality, he is confused about some very rudimentary facts about modern physics and science in general.

It's not just the lunar phases or locality or the exclusion principle that he totally misunderstands (be sure that I haven't discussed every misconception of his that has made me very angry). He actually doesn't build on science; he builds on licking the rectums of the powerful and those who are brainwashed by currently fashionable political deviations. Cox is a kitsch for the least demanding audiences.

Yesterday, The Guardian published a diatribe under the title

Brian Cox: scientists giving false sense of debate on climate change
I agree with this title. Genuine science doesn't have significant doubts about the fact that the climate panic is a pile of rubbish and it's pathetic for the media hosts and others to keep on inviting assorted alarmist loons and fraudsters whenever the topic is related to the climate or the energy policy.

But you surely know that the message that should have been conveyed by the title was upside down.

The subtitle ends with words "is manipulated by nonsensical sceptics". No comments about that are needed. The following paragraph says
Scientists are doing the public a disservice in their attempts to communicate certainty in climate change science, often giving a “false sense of debate” by being overly precise, says broadcaster and physicist Professor Brian Cox.
Wow, so this Gentleman urges everyone to be "less precise". This is particularly shocking recommendation in the case of the climate debate where imprecise and misleading statements are the ultimate cause of most mistakes and irrational reactions. Both skeptics and alarmists are sometimes "too imprecise" but Cox's statement should erase your doubt which side actually loves the imprecision – and downright demagogy – and benefits from it.

But cheap Mr Cocks, most of the public has already outgrown you. People who care about this question at all already know something and if any conversation has any meaning for them, they surely want the exchanges to be more precise, not less precise, than before. I fully understand why precision is your enemy but you can't do anything about it. Lies have short legs and at some moment, everyone will understand that the likes of you are crooks.

In his diatribe, sentences like the following are repeated approximately 100 times:
He said scientists could say with total confidence that climate science was uncontroversial and the current predictions for warming were the best advice available. “The scientific view at the time is the best, there’s nothing you can do that’s better than that. So there’s an absolutism. It’s absolutely the best advice,” he said.
Because of the defining properties of science (which is something else than a fanatical environmentalist religion), conclusions are never absolute, especially not in cases when they are known to be almost certainly wrong. Cox must believe that a lie repeated 100 times becomes the truth but it doesn't. It's enough to say once that the advises by the climate alarmists are among the most untrue, ill-conceived, and counterproductive claims one could listen to in the present world.

But there is a gem in the next paragraph:
Cox, a physicist who works on the Large Hadron Collider where the Higgs boson was discovered, said that 95% certainty in science is effectively total.
Wow. If he actually were a competent particle physicist rather than a pathetic environmentalist clown pretending to be a caricature of a scientist, he would know that the 95% confidence – also known as 2-sigma evidence – is almost equivalent to no evidence at all. Moreover, even this figure 95% is vastly exaggerated and all papers that claim this level of certainty (there are not too many such papers) are fraudulent or they are not scientific papers at all (like the IPCC reports' summaries where politicians just made these numbers up).

As all actual particle physicists would agree, 95% is just an insignificant hint. The hints only become evidence worth discussing when the confidence approaches the 99.7% or 3-sigma level and an effect is only claimed to be discovered when the confidence reaches 99.9999% or 5-sigma. This convention of hard science is no excessively demanding dream. If a new effect is actually there, collecting the 5-sigma evidence takes just slightly (6.25 times) more data than accumulating the 2-sigma evidence. If a 2-sigma hint refuses to grow to 5 sigma for a decade or two, you may be pretty sure that it was a fluke and the effect doesn't exist. And indeed, most effects that are, at one point, observed to be there at the 95% confidence level ultimately go away.

To be sure that you don't interpret the totally wrong comments about the 95% confidence above as an isolated typo and to be sure that Brian Cox really doesn't have the slightest clue what probabilities of various magnitude mean in science, they add the following paragraph that made me laugh out loud:
“We had it with the Large Hadron Collider and people were saying: “Is it going to destroy the world?” Well of course it bloody isn’t. But [in scientific terms] we’re putting a confidence level on that statement … at the 95% confidence level, but you don’t want to go there,” he said.
Wow, so the LHC physicists only had 95% confidence that the collider wouldn't destroy the world? There was a 5% risk of the world destruction? That would be pretty awful: the world would survive 20 or so experiments like that.

Of course that I would have become a big fighter for the stopping of the LHC if the certainty that the LHC wouldn't destroy the world were lower than 99.9999%. Any certainty lower than that would be unacceptable because similar experiments would become by far the largest known threat for the Earth (or the world).

What were the actual odds that scientists estimated that the collider experiments wouldn't destroy the world? Open e.g. this 1999 paper by Jaffe, Busza, Sandweiss, and Wilczek on page 25. At the end of the next-to-last paragraph, you will read:
So the bound on dangerous strangelet production during the RHIC lifetime is more nearly ${\mathfrak p} \lt 10^{-21}$.
The risk was at most ten to the minus twenty-first power! The certainty that the RHIC wouldn't destroy the world (and the same argument actually applies to the LHC as well) was at least 99.9999999999999999999%. There are twenty-one digits nine over there. 95% would really, really not be enough to soothe a rational person.

Cox's next paragraph says:
“What I think about climate change actually is it’s obviously true and clearly true to all of us who look at the debate that goes on.”
Wow. In reality, one can't get any information – let alone certainty – from looking at a "debate" at all. One can only get information and insights about the behavior of the physical system by looking at the actual scientific evidence. The scientific evidence overwhelmingly indicates that there is no reason to think that a "significant enough to be dangerous" change of the climate is going to occur in the coming decades or centuries.
“You’re allowed to say, well I think we should do nothing. That’s a policy choice. But what you’re not allowed to do is to claim there’s a better estimate of the way that the climate will change, other than the one that comes out of the computer models. It’s nonsensical to say ‘we know better’, you can’t know better.”
Thankfully, I live in the Czech Republic where we are "allowed" to say that the existing climate models are absolutely worthless for the prediction of the future climate. They have been shown to fail so many times – to predict quantities of a wrong sign, empirically invalid regional patterns and vertical profiles, quadrupled values of warming rates over 20 years, and so on – that a person who is not completely blind or a hopeless moron must know very well that they are not at the level that could actually help us with a rational planning.

Their number of adjustable (and often adjusted with an intentional bias) components is so high and, in fact, so much higher than the number of checks that were used to adjust them that there is also absolutely no theoretical reason to think that their predictions will be correlated with the reality.

Every other sentence he writes is outrageous but they are so repetitive that I don't want to comment on all of them. Let me just reproduce the last paragraph:
“Don’t undermine the science just because you don’t like the economics. That’s a dangerous slope, because the problem of course is you’re not undermining just that, you’re undermining the basis of rational decision-making in society.”
Because Cox's understanding of all relevant issues is upside down, geologist Bob Carter had to repair this sentence. The corrected version addressed to those who remain silent about the indefensibility of the climate panic reads:
Don’t undermine the science just because you don’t like being politically incorrect in environmental terms. That’s a dangerous slope, because the problem of course is you’re not undermining just that, you’re undermining the basis of rational decision-making in society.”
At any rate, I am very angry that the likes of Brian Cox are sometimes being presented as if they were spokesmen for science. What they say is so wrong and dangerous not just for policymaking but for the public's understanding of and attitude to many fundamental aspects of science. So please, the editors in The Guardian, kindly notice that Brian Cox is full of šit. Thank you very much.


text:
/ hep-ex CERN-PH-EP-2012-326

Search for a heavy narrow resonance decaying to e mu, e tau, or mu tau with the ATLAS detector in sqrt(s) = 7 TeV pp collisions at the LHC

Pages: 11

Abstract: This letter presents the results of a search for a heavy particle decaying into an e mu, e tau, or mu tau final state in pp collisions at sqrt(s) = 7 TeV. The data were recorded with the ATLAS detector at the LHC during 2011 and correspond to an integrated luminosity of 4.6/fb. No significant excess above the Standard Model expectation is observed, and exclusions at 95% confidence level are placed on the cross section times branching ratio for the production of an R-parity-violating supersymmetric tau sneutrino. These results considerably extend constraints from Tevatron experiments.

Note: 11 pages plus author list (32 pages total), 3 figures, 1 table, submitted to Physics Letters B, All figures including auxiliary figures are available at http://atlas.web.cern.ch/Atlas/GROUPS/PHYSICS/PAPERS/SUSY-2012-20/

Total numbers of views: 2253
Numbers of unique views: 903


text:
# How do you evaluate \frac { 2( x ^ { 2} + 32) } { ( x + 8) ( x - 8) } - \frac { x - 4} { x + 8}?

Feb 1, 2017

$\frac{x + 4}{x - 8}$

#### Explanation:

Take out $\frac{1}{x + 8}$ as a common factor

$\frac{1}{x + 8} \left[\frac{2 \left({x}^{2} + 32\right)}{x - 8} - \left(x - 4\right)\right]$

$\frac{1}{x + 8} \left[\frac{2 \left({x}^{2} + 32\right) - \left(x - 4\right) \left(x - 8\right)}{x - 8}\right]$

$\frac{1}{x + 8} \left[\frac{2 {x}^{2} + 64 - {x}^{2} + 12 x - 32}{x - 8}\right]$

$\frac{{x}^{2} + 12 x + 32}{\left(x + 8\right) \left(x - 8\right)}$

Now factorise the numerator

$\frac{\left(x + 8\right) \left(x + 4\right)}{\left(x + 8\right) \left(x - 8\right)}$

$\frac{x + 4}{x - 8}$

Feb 1, 2017

$= \frac{x + 4}{x - 8}$

#### Explanation:

$\frac{2 \left({x}^{2} + 32\right)}{\left(x + 8\right) \left(x - 8\right)} - \frac{x - 4}{x + 8}$

$= \frac{2 \left({x}^{2} + 32\right) - \left(x - 4\right) \left(x - 8\right)}{\left(x + 8\right) \left(x - 8\right)}$

$= \frac{2 \left({x}^{2} + 32\right) - \left({x}^{2} - 8 x - 4 x + 32\right)}{\left(x + 8\right) \left(x - 8\right)}$

$= \frac{2 {x}^{2} + 64 - {x}^{2} + 12 x - 32}{\left(x + 8\right) \left(x - 8\right)}$

$= \frac{{x}^{2} + 12 x + 32}{\left(x + 8\right) \left(x - 8\right)}$

$= \frac{{x}^{2} + 8 x + 4 x + 32}{\left(x + 8\right) \left(x - 8\right)}$

$= \frac{x \left(x + 8\right) + 4 \left(x + 8\right)}{\left(x + 8\right) \left(x - 8\right)}$

$= \frac{\left(x + 8\right) \left(x + 4\right)}{\left(x + 8\right) \left(x - 8\right)}$

$= \frac{x + 4}{x - 8}$


text:
TITLE

# The Compact Quantum Group U q (2) ( II)

AUTHOR(S)
Xiao Zhang
PUB. DATE
July 2006
SOURCE
Acta Mathematica Sinica;Jul2006, Vol. 22 Issue 4, p1221
SOURCE TYPE
DOC. TYPE
Article
ABSTRACT
In this paper, we first prove that the Î¸-deformation U Î¸ (2) of U(2) constructed by Connes and Violette is our special case of the quantum group U q (2) constructed in our previous paper. Then we will show that the set of traces on the C*â€“algebra U Î¸ , Î¸ irrational, is determined by the set of the traces on a subalgebra of U Î¸ .
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The sequence $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$ is such that $a_{n} = a_{n-1} + 5$ for $2 \leq n \leq 5$. If $a_{5} = 31$, what is the value of $a_{1}$?
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TTP05-10 One-loop weak corrections to hadronic production of Z bosons at large transverse momenta

TTP05-10 One-loop weak corrections to hadronic production of Z bosons at large transverse momenta

TTP05-10 One-loop weak corrections to hadronic production of Z bosons at large transverse momenta

To match the precision of present and future measurements of Z-boson production at hadron colliders, electroweak radiative corrections must be included in the theory predictions. In this paper we consider their effect on the transverse momentum ($p_T$) distribution of Z bosons, with emphasis on large $p_T$. We evaluate, analytically and numerically, the full one-loop corrections for the parton scattering reaction $q\bar q \to Z g$ and its crossed variants. In addition we derive compact approximate expressions which are valid in the high-energy region, where the weak corrections are strongly enhanced by logarithms of $\hat s/M_W^2$. These expressions include quadratic and single logarithms as well as those terms that are not logarithmically enhanced. This approximation, which confirms and extends earlier results obtained to next-to-leading logarithmic accuracy, permits to reproduce the exact one-loop corrections with high precision. Numerical results are presented for proton-proton and proton-antiproton collisions. The corrections are negative and their size increases with $p_T$. For the Tevatron they amount up to -7\% at 300 GeV. For the LHC, where transverse momenta of 2 TeV or more can be reached, corrections up to -40\% are observed. We also include the dominant two-loop effects of up to 8\% in our final LHC predictions.

 J. H. K\“uhn, A. Kulesza, S. Pozzorini, M. Schulze Nucl. Phys. B 727 368-394 2005 PDF PostScript arXiv


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# Difference between revisions of "2019 AMC 10A Problems/Problem 9"

## Problem

What is the greatest three-digit positive integer $n$ for which the sum of the first $n$ positive integers is $\underline{not}$ a divisor of the product of the first $n$ positive integers?

$\textbf{(A) } 995 \qquad\textbf{(B) } 996 \qquad\textbf{(C) } 997 \qquad\textbf{(D) } 998 \qquad\textbf{(E) } 999$

### Solution 1

Because the sum of $n$ positive integers is $\frac{(n)(n+1)}{2}$, and we want this to not be a divisor of the $n!$, $n+1$ must be prime. The greatest three-digit integer that is prime is $997$. Subtract $1$ to get $\boxed{\textbf{(B) } 996}$.

### Solution 2

We can use the fact that $n+1$ must be prime to eliminate answer choices as possible values of $n$. $A$, $C$, and $E$ don't work because $n+1$ is even, and $D$ does not work since $999$ is divisible by $9$. Thus, the only correct answer is $\boxed{\textbf{(B) } 996}$.


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Set and get number of threads to be used in data.table functions that are parallelized with OpenMP. The number of threads is initialized when data.table is first loaded in the R session using optional environment variables. Thereafter, the number of threads may be changed by calling setDTthreads. If you change an environment variable using Sys.setenv you will need to call setDTthreads again to reread the environment variables.

setDTthreads(threads = NULL, restore_after_fork = NULL, percent = NULL, throttle = NULL)
getDTthreads(verbose = getOption("datatable.verbose"))

## Arguments

threads NULL (default) rereads environment variables. 0 means to use all logical CPUs available. Otherwise a number >= 1 Should data.table be multi-threaded after a fork has completed? NULL leaves the current setting unchanged which by default is TRUE. See details below. If provided it should be a number between 2 and 100; the percentage of logical CPUs to use. By default on startup, 50%. 1024 (default) means that, roughly speaking, a single thread will be used when nrow(DT)<=1024, 2 threads when nrow(DT)<=2048, etc. The throttle is to speed up small data tasks (especially when repeated many times) by not incurring the overhead of managing multiple threads. Hence the number of threads is throttled (restricted) for small tasks. Display the value of relevant OpenMP settings plus the restore_after_fork internal option.

## Value

A length 1 integer. The old value is returned by setDTthreads so you can store that prior value and pass it to setDTthreads() again after the section of your code where you control the number of threads.

## Details

data.table automatically switches to single threaded mode upon fork (the mechanism used by parallel::mclapply and the foreach package). Otherwise, nested parallelism would very likely overload your CPUs and result in much slower execution. As data.table becomes more parallel internally, we expect explicit user parallelism to be needed less often. The restore_after_fork option controls what happens after the explicit fork parallelism completes. It needs to be at C level so it is not a regular R option using options(). By default data.table will be multi-threaded again; restoring the prior setting of getDTthreads(). But problems have been reported in the past on Mac with Intel OpenMP libraries whereas success has been reported on Linux. If you experience problems after fork, start a new R session and change the default behaviour by calling setDTthreads(restore_after_fork=FALSE) before retrying. Please raise issues on the data.table GitHub issues page.

The number of logical CPUs is determined by the OpenMP function omp_get_num_procs() whose meaning may vary across platforms and OpenMP implementations. setDTthreads() will not allow more than this limit. Neither will it allow more than omp_get_thread_limit() nor the current value of Sys.getenv("OMP_THREAD_LIMIT"). Note that CRAN's daily test system (results for data.table here) sets OMP_THREAD_LIMIT to 2 and should always be respected; e.g., if you have written a package that uses data.table and your package is to be released on CRAN, you should not change OMP_THREAD_LIMIT in your package to a value greater than 2.

Some hardware allows CPUs to be removed and/or replaced while the server is running. If this happens, our understanding is that omp_get_num_procs() will reflect the new number of processors available. But if this happens after data.table started, setDTthreads(...) will need to be called again by you before data.table will reflect the change. If you have such hardware, please let us know your experience via GitHub issues / feature requests.

Use getDTthreads(verbose=TRUE) to see the relevant environment variables, their values and the current number of threads data.table is using. For example, the environment variable R_DATATABLE_NUM_PROCS_PERCENT can be used to change the default number of logical CPUs from 50% to another value between 2 and 100. If you change these environment variables using Sys.setenv() after data.table and/or OpenMP has initialized then you will need to call setDTthreads(threads=NULL) to reread their current values. getDTthreads() merely retrieves the internal value that was set by the last call to setDTthreads(). setDTthreads(threads=NULL) is called when data.table is first loaded and is not called again unless you call it.

setDTthreads() affects data.table only and does not change R itself or other packages using OpenMP. We have followed the advice of section 1.2.1.1 in the R-exts manual: "... or, better, for the regions in your code as part of their specification... num_threads(nthreads)... That way you only control your own code and not that of other OpenMP users." Every parallel region in data.table contain a num_threads(getDTthreads()) directive. This is mandated by a grep in data.table's quality control script.

setDTthreads(0) is the same as setDTthreads(percent=100); i.e. use all logical CPUs, subject to Sys.getenv("OMP_THREAD_LIMIT"). Please note again that CRAN's daily test system sets OMP_THREAD_LIMIT to 2, so developers of CRAN packages should never change OMP_THREAD_LIMIT inside their package to a value greater than 2.

Internally parallelized code is used in the following places:

• between.c - between()

• cj.c - CJ()

• coalesce.c - fcoalesce()

• fifelse.c - fifelse()

• fread.c - fread()

• forder.c, fsort.c, and reorder.c - forder() and related

• froll.c, frolladaptive.c, and frollR.c - froll() and family

• fwrite.c - fwrite()

• gsumm.c - GForce in various places, see GForce

• nafill.c - nafill()

• subset.c - Used in [.data.table subsetting

• types.c - Internal testing usage

## Examples

  getDTthreads(verbose=TRUE)
#>   OpenMP version (_OPENMP)       201511
#>   omp_get_num_procs()            1
#>   R_DATATABLE_NUM_PROCS_PERCENT  unset (default 50)
#>   data.table is using 1 threads with throttle==1024. See ?setDTthreads.#> [1] 1


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Repozitorij Univerze v Novi Gorici

# Izpis gradiva A+ | A- | SLO | ENG

Naslov: Method of Regularized Sources for Stokes Flow Problems with Improved Calculation of Velocity Derivatives at the Boundary Šarler, Božidar (Lastnik avtorskih pravic)Šarler, Božidar (Avtor)shiting, wen (Avtor)ming, li (Avtor) Gradivo nima datotek. Gradivo je morda fizično dosegljivo v knjižnici fakultete, zalogo lahko preverite v COBISS-u. Angleški jezik Delo ni kategorizirano (r6) 1.12 - Objavljeni povzetek znanstvenega prispevka na konferenci UNG - Univerza v Novi Gorici The solution of Stokes flow problems with Dirichlet and Neumann boundary conditions is performed by a non-singular Method of Fundamental Solutions which does not require artificial boundary, i.e. source points of fundamental solution coincide with the collocation points on the boundary. Instead of Dirac delta force, an exponential function, called blob, with a free parameter epsilon is employed, which limits to Dirac delta function when epsilon limits to zero. The solution of the problem is sought as a linear combination of the fields due to the regularized sources that coincide with the boundary and with their intensities chosen in such a way that the solution complies with the boundary conditions. A two-dimensional flow between parallel plates is chosen to assess the properties of the method. The results of the method are accurate except for the derivatives at the boundary. A correction of the method is proposed which can be used to properly assess also the derivatives at the boundary stokes flow, method of regularized sources, meshless method 28 4417531 URN:SI:UNG:REP:LIMZLHNP 2722 0 Gradivo ni uvrščeno v področja.

Skupna ocena: (0 glasov) Ocenjevanje je dovoljeno samo prijavljenim uporabnikom.

Postavite miškin kazalec na naslov za izpis povzetka. Klik na naslov izpiše podrobnosti ali sproži prenos.

Nazaj


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# Tag Info

12

Yes, steel (even some stainless) may be cheaper than aluminum, but the material cost of an item is seldom the majority of the total cost, especially a small item such as a potato peeler. Making aluminum parts with complex, curved shapes can be fairly easily done by casting. Aluminum pours at around 1500 °F, which can be achieved with low-cost furnaces and ...

10

A very important property of steel used in reinforced concrete is that it has similar coefficient of thermal expansion as concrete: Concrete: $14.5 \cdot 10^{-6} \frac{m}{m\,K}$ Steel: $12.0 \cdot 10^{-6} \frac{m}{m\,K}$ Compare this with: Aluminum: $22.2 \cdot 10^{-6} \frac{m}{m\,K}$ So a likely result of putting aluminum inside a concrete beam would ...

9

The work history and temperature history of any metal can make a big difference in the final shape after machining. If the material was cold worked (e.g. rolled) there may be significant residual stresses within the material. When you start selectively removing material, those stresses may cause the part to warp into a new shape. Cast metals and ones that ...

8

I'm going to run with this assuming the arm looks like the following diagram (I'm ignoring the 11.4 pounds of the bar for now to make the concepts easier to explain - that can be added later by assuming all 11.4 pounds run through the center of the bar.): This is a pretty simple setup. To figure out the loading on the shaft, we have to move the load from ...

7

Most soda cans in the US are on the order of 0.1 mm thick according to the sources I've seen. So 0.25 mm seems plausible to me, assuming that you made no errors in your calculations.

7

In short: you can't use aluminium to reinforce concrete. You shouldn't even embed uncoated aluminium in concrete. Quoting Corrosion of Non Ferrous Metals in Contact with Concrete, "Aluminium suffers attack when embedded in concrete". Corrosion of aluminium embedded in concrete causes total destruction of aluminium bars, therefore it's dangerous and ...

6

You can use a sort of "double diameter washer". I've made that name up, but I am pretty sure such a device exists: That's a cross section of one of your linkages. The white things are the aluminum pieces, the grey one is the screw, the yellow is the nut while the red is the "double diameter washer". If you can found some washer that is ...

6

Summary The general term for changing tube and pipe diameter is swaging. The tools you are looking for exist but vary widely in price and complexity. It may be more straightforward to find a local metal supplier who can do the swaging for you based on your specifications. Flaring vs. Swaging I am unsure if there is a specific name for the process which is ...

6

Using your plate as boat hull implies a complex load profile probably involving compression, shear and bending, possibly even torsion. In such cases, it is effectively impossible to give a simple answer to your question. @RainerJ's answer is perfectly acceptable for tensile and shear loads: if you're replacing material X with material Y and Y's yield ...

5

We use this equation for a simply supported beam loaded at its center with 200lbs. M= P x L/4 =200 x 3 /4 = 150 lbs.ft for calculating the tube stress we convert this to 150 x 12 = 1800 lbs.inch For a hollow round tube we calculate section module which is $Z = 0.78 (R_o^4 -R_i^4)/R_o$ assuming you have 16 gauge tube with 0.065 inch walls, Z=0.1838 . ...

5

I looked up 6000 grade aluminum to be about \$2/kg, while 304 stainless steel is about \$3.40/kg. This means stainless steel is about 70% more expensive than aluminum, which is consistent with my experience in doing mechanical design. I'm not sure where your numbers came from that aluminum is 4x more expensive than stainless steel. I also have a ...

5

The Wikipedia page puts tin foil thickness at between 0.2 mm and 0.006 mm. That means that the upper end of the tin foil is easily comparable with the thickness of my box. Ooook. Let's look at this, from an engineering perspective. First, scales. The smallest measurement from Wikipedia vs the largest is a 33x factor difference. This is perhaps hard to ...

4

A shoulder bolt is a good approach in applications with light loading, but it needs to be specified such that it can be fully tightened while still allowing the necessary clearance for movement without binding. Use shims where needed. You can reduce some of the play in the joint by having the bolt threaded into one of the aluminum pieces (but still backed ...

4

The dross is mostly composed of aluminium oxide which is already present on the surface of the metal before it even goes into the furnace, as the aluminium melts the oxides separate and float to the top. Recycled aluminium will tend to be made up of relatively thin material with a high surface area to volume ratio and so a relatively high proportion of ...

4

Yes it is but there a a few things you need to watch out for. You will need to allow enough excess material for any surfaces which need to be machined subsequent to casting. All castings shrink somewhat so the pattern needs to be a some percentage larger than the finished casting needs to be. Its not uncommon for castings to have internal voids which ...

4

Aluminum sections are extruded from billets heated to 800-925F under high pressure and then pushed through a die. During the entire process temperature is carefully controlled, because depending on the aluminum alloy and performance expected heat stages at the beginning and exit out of die are crucial. here is the diagram of the extrusion machine.

4

The tube is sealed in the following manner: The aluminum from which the tube is formed is laminated or otherwise coated with a thin layer of plastic on its inside, which is easily heat-fused together and which thereby seals off the material inside from oxygen on the outside. The outer surface is also typically laminated or otherwise coated with another type ...

3

For a one-off structure swaging or flaring tube is probably a bit of a long way round to do it. There are actually quite a few off the shelf solutions for joining tube, an example at random Something else to consider is that, for something which is disassembled and moved about a lot, if your system relies on one tube fitting inside another then and damage ...

3

Having spent several years in product development at Kitchen-Aid's parent company, I can guarantee that if there were a less expensive way to provide the same function to the consumer, that is how it would be done. They have whole teams of engineers tasked only with cost reduction. The reason that items are marked as "hand wash only" often have more to do ...

3

In your situation (as long as you don't actually intend to use the thing!) 6061-T6 will be just fine. The main reason to pick it is that many more product forms are available in it than most other alloys, and it is relatively inexpensive. It is one of the strongest among the weldable aluminum alloys and very easy to machine (as most aluminum is.) The one ...

3

Re-tooling and re-equipping a factory to use a different metal can be very expensive. If the required rate of return from re-tooling/re-equipping cannot be achieved be it simply isn't done. The production rate from some utensil manufacturers may be low and the profit margins small resulting in no major new investing in changing to other metals and re-...

3

I'm guessing that shipping cost could be a reason. For a pallet (or several) of utensils, shaving a few percent weight off of each one could amount to a significant savings in shipping cost, especially if the trade route is between China and Europe or North/South America.

3

This is unlikely to work very well. The biggest issue is that you won't get good adhesion between the two surfaces. aluminium has a particularly resilient and inert surface oxide layer so molten aluminium won't just stick to the surface of solid aluminium. Welding it requires an inert atmosphere and reverse polarity current or alternating current to clean ...

3

I used to build structures using the 80/20 aluminum extrusions , I used a mitre saw with a good aluminum cutting blade lubricated with stick wax (or any good cutting fluid). Makes beautiful square cuts with minimal burring. Be careful applying the stick wax to the rotating blade, otherwise your mitre saw will become a finger removing device by default :-(

3

(Short answer below) A phase diagram like the one you posted is to be intended true for a solution(either solid or liquid) of the two components(i.e. Al and Cu). Adding Cu to molten Al does not properly mean to have a solution(the two components must be dispersed in thw whole bath in a uniform way). Also a phase diagram is drawn as a "succession of ...

3

My reputation is too low to comment so I'll give some pointers in an answer :) Looking at your design your structures weak points are the areas where the plate and the sensors interface, especially the ones in the corners. This is because you have a very long lever from your load to the feet of your sensors and a very small contact area between the sensor ...

2

Steel or aluminum can be made equally strong. Generally aluminum can more easily be designed to have a higher strength to weight ratio because it has the a larger area moment of inertia for a given mass. Like Olin Lanthrop mentioned in the comments, the weakest link will be your connection between the table top and the legs. You can't count on the friction ...

2

Normally the production of specific alloys is a fairly specialised process and they would be bought in as ingots or other stock types from commercial manufacturer rather than being made as part of a manufacturing process. This is as much about the quality control and analysis required to produce a given alloy to a specification as the actual mechanics of ...

2

Beside very cheap products, the only aluminum parts I have found in my kitchen tools are either injected or molded. Why this? Because the fusion point of aluminum is much lower than steel, plain steel molds can be easily used. The opposite is not true and for steel casting, it requires disposable sand molds. Thus, the process of steel casting is more ...

2

I am going to start out by saying that aluminum is a mess. There are too many designations and things that you can do to it after it solidifies. To directly answer your question, what might necessitate an additional round of heat treating is directly affected by the desired final mechanical properties. Each of the designations that you listed is a complete ...

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


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1. ## Linear Programming - Help please!

An orchadist has 10 hectares of land available for growing Fuji apples and Nashi pears. She estimates that each hectare of apples will cost $5000 to establish, and each hectare of pears will cost$20 000 to establish. She is not able to spend more than $100 000 on establishing the orchard. It is also estimated that the profit per hectare from apples will be$4000 and the profit per hectare from pears will be $8000. What is the maximum profit the orchardist can expect? How many hectares of Fuji apples and Nashi pears should she plant to achieve this? Please show your working clearly, with an explanation for the reason behind the answer you give. Thank you for your help. 2. Originally Posted by LilDragonfly An orchadist has 10 hectares of land available for growing Fuji apples and Nashi pears. She estimates that each hectare of apples will cost$5000 to establish, and each hectare of pears will cost $20 000 to establish. She is not able to spend more than$100 000 on establishing the orchard.

It is also estimated that the profit per hectare from apples will be $4000 and the profit per hectare from pears will be$8000. What is the maximum profit the orchardist can expect? How many hectares of Fuji apples and Nashi pears should she plant to achieve this? Please show your working clearly, with an explanation for the reason behind the answer you give.

There are two decision variables:
hectares of apples, A
hectares of pears, P

The constraints are:
A +P <= 10 ----------------------(1)

5000A +20,000P <= 100,000
Divide both sides by 5000,
A +4P <= 20 ----------------------(2)

Non-negative, non-zero constraints:
A > 0 ------------------------------(3)
P > 0 ---------------------------(4)

The objective function is
Profit = 4000A +8000P -----------***

Now, plot the inequalities (1), (2), (3) and (4) on the same (A,P) cartesian or rectangular coordinate axes.
Get the intersection points of the 4 inequalities. Since A or P cannot be zero, then the intersection of (1) and (2) , being the only intersection point left, should determine the optimum (maximum here) profit.

The intersection of (1) and (2):
Consider them as equations only.
(2) minus (1),
3P = 10
So, P = 10/3
And, using (1), A = 10 -P = 10 -(10/3) = 20/3
Therefore, the intersection of (1) and (2), and at the same time the only effective corner of the feasible/solution region, is (20/3, 10/3).
Plugging that into the objective function,
Profit = 4000A +8000P
Maximum profit = 4000(20/3) +8000(10/3) = 160,000/3 = $53,333.33 ----------answer. To achieve that, the orchardist should plant (6 and 2/3) hectares of apples and (3 and 1/3) hectares of pears. ------------------answer. 3. Originally Posted by ticbol Non-negative, non-zero constraints: A > 0 ------------------------------(3) P > 0 ---------------------------(4) Non-negativity constraints are (there is no stipulation that they are non-zero as far as I can see):$\displaystyle
A \ge 0$,$\displaystyle
P \ge 0$Get the intersection points of the 4 inequalities. Since A or P cannot be zero, then the intersection of (1) and (2) , being the only intersection point left, should determine the optimum (maximum here) profit. We will need to consider the casses where$\displaystyle A$and/or$\displaystyle P$are zero, unless you determine that these cannot correspond to the optimum by other means. (this will not change the answer as the point you give is in fact the minimum as we know ) RonL 4. Originally Posted by CaptainBlack Non-negativity constraints are (there is no stipulation that they are non-zero as far as I can see):$\displaystyle
A \ge 0$,$\displaystyle
P \ge 0$We will need to consider the casses where$\displaystyle A$and/or$\displaystyle P$are zero, unless you determine that these cannot correspond to the optimum by other means. (this will not change the answer as the point you give is in fact the minimum as we know ) RonL The question says Apples and Pears are to be raised, not Apples or Pears, or not Apples and/or Pears. That was or is very easy to understand. So, A or P cannot be zero. 5. Originally Posted by ticbol The question says Apples and Pears are to be raised, not Apples or Pears, or not Apples and/or Pears. That was or is very easy to understand. So, A or P cannot be zero. No it actually says "An orchadist has 10 hectares of land available for growing Fuji apples and Nashi pears" Which if you wish to interpret it as both apples and pears must be grown you are welcome. We can overlook the fact that this is an exercise in linear programming, which usually requires the feasible region to be closed (among other things) to guarantee that a solution exists. RonL 6. Maybe this will help, 7. Originally Posted by CaptainBlack No it actually says "An orchadist has 10 hectares of land available for growing Fuji apples and Nashi pears" Which if you wish to interpret it as both apples and pears must be grown you are welcome. We can overlook the fact that this is an exercise in linear programming, which usually requires the feasible region to be closed (among other things) to guarantee that a solution exists. RonL No, I do not wish that. That is what it really is. That was very EASY to understand. The feasible region must be closed to guarantee that a solution exists? That is new to me, to say it mildly. 8. Originally Posted by ticbol The feasible region must be closed to guarantee that a solution exists? I believe so. If the feasible region is not closed, then IF the optimized solution exists then it must be on the endpoints. This does not means that it does indeed exists. 9. Originally Posted by ThePerfectHacker I believe so. If the feasible region is not closed, then IF the optimized solution exists then it must be on the endpoints. This does not means that it does indeed exists. Then I can safely say, if you wish to believe that. Now, a question, what if the feasible region extends to infinity, and that the optimum is a minimum? Have you not encountered those in linear programming yet? There are garozillions of them, I wish to believe. Wait, did I understand really your, "If the feasible region is not closed, then IF the optimized solution exists then it must be on the endpoints. This does not means that it does indeed exists." You mean the optimum is at the endpoints? At the points on the axes? Not at corners of the feasible region away from the axes? An optimum does not exist? Now, you lost me there. 10. Originally Posted by ticbol Then I can safely say, if you wish to believe that. Now, a question, what if the feasible region extends to infinity, and that the optimum is a minimum? Have you not encountered those in linear programming yet? There are garozillions of them, I wish to believe. I belive max/min follow the same rules in linear programming. I am not an expert, a noob rather in this field. I am sure CaptainBlack will have the answers. 11. Standard formulation of the Linear Programming problem from Wikipedia attached. The constraints are usualy expressed as$\displaystyle \le$or$\displaystyle \ge$to ensure that the solution algorithms can find a solution in the feasible region. An example of a psedo LP problem without a solution (if I have done this right) is: Maximise:$\displaystyle
P(x,y)=4000x+1000y
$Subject to the constraints:$\displaystyle
x+y\le10
\displaystyle
x+4y \le 20
\displaystyle
x>0, y>0
$RonL 12. Originally Posted by ThePerfectHacker I belive max/min follow the same rules in linear programming. I am not an expert, a noob rather in this field. I am sure CaptainBlack will have the answers. A minimization problem can always be rewritten as a maximization problem by replacing the objective function$\displaystyle O(\rm{X})$by a new objective function$\displaystyle O'(\mathrm{X}) =- O (\rm{X})\$.

RonL

13. Originally Posted by CaptainBlack
No it actually says "An orchadist has 10 hectares of land available for
growing Fuji apples and Nashi pears"
a solution exists.

RonL
The orchadist has 10 hectares available for growing apples and pears. This
means in this context that apples and/or pears can be grown on this land,
but the orchadist is not obliged to grow any of either. The only thing that
make the orchadist grow anything on this land is the desire to maximize
profit.

RonL

14. Originally Posted by CaptainBlack
The orchadist has 10 hectares available for growing apples and pears. This
means in this context that apples and/or pears can be grown on this land,
but the orchadist is not obliged to grow any of either. The only thing that
make the orchadist grow anything on this land is the desire to maximize
profit.

RonL
That is another problem, not the original problem.


text:
One sample Wilcoxon test for each boxplot ggplot2
1
1
Entering edit mode
11 months ago

Hi folks! I need your help with a ggplot2 representation.

I have a data set that has the following structure:

log2FoldChange      Sequence_biotype      Knockdown
-1.40               LTR                   A
-1.11               DNA                   B
-3.46               Protein               A
-1.25               Protein               C
1.03               DNA                   B
...                ...                   ...

I am plotting the foldChanges as boxplots, one for each Sequence_biotype. I have the knockdown variable faceted. What I'm trying to do is to do a one sample Wilcoxon test for each boxplot, comparing the log2FoldChange to 0 (to see if there is a significant change). That can be achieved with the following code:

wilcox.test(x = data, mu = 0)

when the data is grouped by Sequence_biotype and Knockdown.

My question is, how could I introduce the results of the wilcox test to the plot as labels or how could I compute it directly in the plot (I have been trying with stat_compare_means(method = "wilcox", paired = FALSE) from ggpubr package but all the pvalues are piling up to the same spot)

Thank you before hand, any help is appreciated!

Best,

Jordi

R ggplot2 • 1.3k views
0
Entering edit mode

Hi,

Can you provide the code that you've tried as well the figure that you got?

António

0
Entering edit mode

Sure! This is the code and the plot I currently have

com = ggboxplot(data  = com_sig.df, x = "Sequence_biotype", y = "log2FoldChange", fill = "Sequence_biotype",
facet.by = "Knockdown") +
geom_hline(yintercept = 0, linetype = "dashed", color = "red") +
theme_bw() +
theme(legend.position = "none",
axis.title.x = element_text(hjust = .975, vjust = -.6),
plot.title = element_text(hjust = .5),
axis.text.x = element_blank(),
strip.background = element_blank(),
strip.text.x = element_blank()) +
stat_compare_means(method = "wilcox.test")
com

plot_generated

0
Entering edit mode

Hi! An expansion of y axis may help to avoid piling up stat_compare_means() results. Alternatively, results of Wilcoxon test may be added on the plot using ggpubr::stat_pvalue_manual(). In such cases, I compute Wilcoxon using ggpubr::compare_means() - maybe because it is directly compatible with ggpubr::stat_pvalue_manual(), I don't recall why)))

0
Entering edit mode

I have tried the compare_means approach and it works, I obtain the kind of data that I am looking for. However, I still don't know how to embed the data into the plot, I tried with stat_pvalue_manual() but I have the same problem as with stat_compare_means(), that the values are piled up on top of each other. Any clue why?

0
Entering edit mode

I guess that the problem arises because stat_compare_means compares all groups against all groups - there are too many values to plot. If you specify ref.group or comparisons, p-values will move to the respective bars. As far as I know, it will require an explicit plotting of compared groups on x-axis.

0
Entering edit mode

Yes indeed, if I specify a reference group all the labels move to the right place! The problem is that I don't want any reference group, but to compare it to mu=0... It is a wilcoxon one sample test what I would like to do

0
Entering edit mode

There should be one p-value for each facet, right?

0
Entering edit mode

One p-value for each biotype (or boxplot) in each facet. In other words, if I have 6 boxplots per facet, I should get 6 p-values per facet

1
Entering edit mode

If so, I can suggest making a table with p-values calculated for the corresponding Sequence_biotype and Knockdown type, then add arbitrary y-axis (log2FoldChange) value higher than its actual values to place p-values on top of the bars (say, 5) and finally add this table to the plot with geom_text(data = your_table, aes(label = p_values)). The presence of biotypes, facetting variable and y-axis values will take care of proper p-value positioning. Since so many p-values will overlap, you can 1) rotate p-values or a whole plot 2) incrementally increase y-axis value for each next p-value. For 6 Sequence_biotypes and 3 Knockdown types sorted in accordance with your plot, it will be simply like rep(seq(5, 7, length.out = 6), 3).

0
Entering edit mode

Thank you so much! This work smoothly! I am pasting the code with the answer and closing the question

5
Entering edit mode
11 months ago

In case anyone faces the same problem, I will paste the code I used to get the plot.

# Subset the dataset into different knockdowns
A = com_sig.df[com_sig.df$Knockdown=="A",] B = com_sig.df[com_sig.df$Knockdown=="B",]
C = com_sig.df[com_sig.df$Knockdown=="C",] # Compute the test for the first knockdown and the position in the plot stat.test_A = compare_means(log2FoldChange ~ 1, paired = FALSE, data = A, method = "wilcox.test", group.by = "Sequence_biotype", mu = 0) %>% mutate(y.position = 4.5) # Add a label to identify the knockdown stat.test_A$Knockdown = "A"

# Same for second knockdown
stat.test_B = compare_means(log2FoldChange ~ 1, paired = FALSE, data = B, method = "wilcox.test",
group.by = "Sequence_biotype", mu = 0) %>%
mutate(y.position = 4.5)
stat.test_B$Knockdown = "B" # Same for the third one stat.test_C = compare_means(log2FoldChange ~ 1, paired = FALSE, data = C, method = "wilcox.test", group.by = "Sequence_biotype", mu = 0) %>% mutate(y.position = 4.5) stat.test_C$Knockdown = "C"

# Combine all 3 datasets into 1
stat.test = do.call(rbind, list(stat.test_A, stat.test_B, stat.test_C))

# change column name of the variable used to compute the wilcox test
names(stat.test)[names(stat.test)==".y."] = "log2FoldChange"

com = ggboxplot(data  = com_sig.df, x = "Sequence_biotype", y = "log2FoldChange", fill = "Sequence_biotype",
facet.by = "Knockdown") +
geom_hline(yintercept = 0, linetype = "dashed", color = "red") +
theme_bw() +
theme(legend.position = "none",
axis.title.x = element_text(hjust = .975, vjust = -.6),
plot.title = element_text(hjust = .5),
axis.text.x = element_blank(),
strip.background = element_blank(),
strip.text.x = element_blank()) +
geom_text(data = stat.test, aes(y=y.position,label = p.signif))
com

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text:
# msp430¶

## Testing¶

To test the code for msp430, the openmsp430 project is used.

https://opencores.org/project,openmsp430

## Module¶

To flash the msp430 board, the following program can be used:

http://www.ti.com/tool/msp430-flasher

class ppci.arch.msp430.Msp430Arch(options=None)

Texas Instruments msp430 target architecture

determine_arg_locations(arg_types)

Given a set of argument types, determine location for argument ABI: param1 = r12 param2 = r13 param3 = r14 param4 = r15 further parameters are put on stack. retval = r12

determine_rv_location(ret_type)

Determine the location of a return value of a function given the type of return value

gen_call(frame, label, args, rv)

Generate instructions for a function call.

gen_epilogue(frame)

Return epilogue sequence for a frame. Adjust frame pointer and add constant pool

gen_function_enter(args)

Generate code to extract arguments from the proper locations

The default implementation tries to use registers and move instructions.

Parameters: args – an iterable of virtual registers in which the arguments must be placed.
gen_prologue(frame)

Returns prologue instruction sequence

get_runtime()

Compiles the runtime support for msp430. It takes some c3 code and some assembly helpers.

litpool(frame)

Generate instruction for the current literals

move(dst, src)

Generate a move from src to dst

static round_upwards(v)

Round value upwards to multiple of 2


text:
A triangle has sides A, B, and C. Sides A and B have lengths of 9 and 4, respectively. The angle between A and C is (pi)/12 and the angle between B and C is  (2pi)/3. What is the area of the triangle?

Oct 2, 2016

Let angle beween A and C $= \beta = \frac{\pi}{12}$

Angle beween B and C $= \alpha = \frac{2 \pi}{3}$

So angle beween A and B $= \gamma = \pi - \frac{\pi}{12} - \frac{2 \pi}{12} = \frac{\pi}{4}$

Again sides $A = 9 \mathmr{and} B = 4$

The area of the triangle

$\Delta = \frac{1}{2} \times A \times B \times \sin \gamma$

$\implies \Delta = \frac{1}{2} \times 9 \times 4 \times \sin \left(\frac{\pi}{4}\right)$

$= 18 \times \frac{1}{\sqrt{2}} = 9 \sqrt{2} s q u n i t$


text:
# complex table with split and merged cells

I would like to make a table like the one in the following screenshot. I could use \multicolumn to create the second row with S1,S2 etc. In the next stage i couldn't split the cells for 'V' and 'I'.

Here is how I would do it (with suggestions from egreg incorporated):

\documentclass{article}

\usepackage[hmargin = 3cm]{geometry} % to avoid overfull \hbox' warning
\usepackage{booktabs,multirow,makecell}

\def\mc#1#2{\multicolumn{#1}{c}{#2}}
\def\mr[#1]#2#3{\multirowcell{#2}[#1]{#3}}

\begin{document}

\begin{table}
\setlength\tabcolsep{9pt}
\centering
\begin{tabular}{*{16}{l}}
\toprule
\mc{2}{Data1}
& \mc{14}{Data2} \\
\cmidrule{3-16}
&& \mc{2}{S1}
& \mc{2}{S2}
& \mc{2}{S3}
& \mc{2}{S4}
& \mc{2}{S5}
& \mc{2}{S6}
& \mc{2}{S7} \\
\cmidrule(lr){3-4}
\cmidrule(lr){5-6}
\cmidrule(lr){7-8}
\cmidrule(lr){9-10}
\cmidrule(lr){11-12}
\cmidrule(lr){13-14}
\cmidrule(lr){15-16}
&& V & I & V & I & V & I & V & I & V & I & V & I & V & I \\
\midrule
\mr[-0.5ex]{2}{Subdata1} & Try1 &&&&&&&&&&&&&& \\
\cmidrule(l){2-16}
& Try2 &&&&&&&&&&&&&& \\
\midrule
\mr[-0.5ex]{2}{Subdata2} & Try1 &&&&&&&&&&&&&& \\
\cmidrule(l){2-16}
& Try2 &&&&&&&&&&&&&& \\
\midrule
\mr[-0.5ex]{2}{Subdata3} & Try1 &&&&&&&&&&&&&& \\
\cmidrule(l){2-16}
& Try2 &&&&&&&&&&&&&& \\
\midrule
&      &&&&&&&&&&&&&& \\
\cmidrule(l){2-16}
&      &&&&&&&&&&&&&& \\
\bottomrule
\end{tabular}
\end{table}

\end{document}

Notice the absence of any vertical lines; they often disturb more than they help.

• I'd align “Data1” with ”Data2”; also there's no need of vertically centering “SubdataX” if you use a \cmidrule, which however should be trimmed on the left with \cmidrule(l){2-16}. Depending on the data, the \cmidrule could be unneeded. – egreg Jan 1 '15 at 16:46
• @egreg I've change the alignment and added (l) but I don't understand why I don't need to vertically center “SubdataX” if I use \cmidrule(l){2-16}. If you will change the code (and possibly the screenshot) regarding this, that would be nice. – Svend Tveskæg Jan 1 '15 at 17:02
• Here is a picture of how I'd do it. There's no need to vertically center any cell; it's clear that “Subdata1” refers to both rows. Note also the seven \cmidrule commands in the header. – egreg Jan 1 '15 at 17:12

The less vertical rules in a table, the better. If we accept this axiom, then zero is the right number of vertical rules. They serve no purpose and are simply a hindrance to reading.

Also horizontal rules should be used sparingly, mainly to separate chunks of rows that are linked together. Shifting a cell's content vertically is not needed: a blank cell in the table body means “repeat the value above”.

So here's my proposal for the table, that's quite similar to Svend Tveskæg, but differs from it in the placement of the row headers.

Note that \cmidrule(lr) is used for making clear how the groups of columns should be interpreted.

\documentclass{article}

\usepackage{booktabs}

\begin{document}

\begin{table}
\centering
\begin{tabular}{*{16}{l}}
\toprule
\multicolumn{2}{c}{Data1} & \multicolumn{14}{c}{Data2} \\
\cmidrule(lr){3-16}
&& \multicolumn{2}{c}{S1} & \multicolumn{2}{c}{S2}
& \multicolumn{2}{c}{S3} & \multicolumn{2}{c}{S4}
& \multicolumn{2}{c}{S5} & \multicolumn{2}{c}{S6}
& \multicolumn{2}{c}{S7} \\
\cmidrule(lr){3-4}\cmidrule(lr){5-6}\cmidrule(lr){7-8}\cmidrule(lr){9-10}
\cmidrule(lr){11-12}\cmidrule(lr){13-14}\cmidrule(lr){15-16}
&& V & I      & V & I      & V & I      & V & I      & V & I      & V & I      & V & I      \\
\midrule
Subdata1 & Try1 &&&&&&&&&&&&&& \\
& Try2 &&&&&&&&&&&&&& \\
\midrule
Subdata2 & Try1 &&&&&&&&&&&&&& \\
& Try2 &&&&&&&&&&&&&& \\
\midrule
Subdata3 & Try1 &&&&&&&&&&&&&& \\
& Try2 &&&&&&&&&&&&&& \\
\bottomrule
\end{tabular}
\end{table}

\end{document}

You don't need to think of split cells, but could merge different amounts of cells. As you already know about \multicolumn and its syntax is pretty basic, here is the way for producing the first three rows with different merging.

• I generate a row with all simple cells
• I made a row with merged 2-cells
• I inserted the row with the merged 14-cell
• I took care to preserve the desired lines by also adding them to the multicolumn format specifications

\documentclass{article}
\renewcommand*{\familydefault}{\sfdefault}
\renewcommand*{\arraystretch}{1.5}
\begin{document}
\begin{tabular}{|l|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|}
\hline
\multicolumn{2}{|l|}{Data1} & \multicolumn{14}{c|}{Data 2} \\
\cline{3-16}
\multicolumn{2}{|c|}{} & \multicolumn{2}{l|}{S1} & \multicolumn{2}{l|}{S2}
& \multicolumn{2}{l|}{S3} & \multicolumn{2}{l|}{S4}
& \multicolumn{2}{l|}{S5} & \multicolumn{2}{l|}{S5}
& \multicolumn{2}{l|}{S7} \\
\cline{3-16}
\multicolumn{2}{|c|}{} & V & I & V & I & & & & & & & & & & \\
\hline
\end{tabular}
\end{document}
`


text:
http://iet.metastore.ingenta.com
1887

## Input shaping control

• DOI:

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## Thank you

This chapter is on input shaping, also known as preview control. This is a feedforward control approach that can be applied to reference and disturbance inputs that are known in advance. A discrete-time closed-loop controlled system is considered and its approximate inverse filter that is not causal is designed to achieve desirable inputto-output behaviour. The external input is assumed to be known in advance. Thus, the approximate inverse filter also known as the preview filter as it is not causal. It is applied to the known input offline, resulting in a new input which is applied online to the closed-loop system (hence the name input shaping). This chapter starts with a characterisation of non-minimum phase zeros and approximate inversion methods used to treat them as they cannot be inverted directly. The ZP, ZPG, ZPGE and ZPGO and TSA methods are presented in this chapter as different possible approximate inverse filters for feedforward control.

Chapter Contents:

• 4.1 Introduction to input shaping control
• 4.2 Discrete-time NMP zeros
• 4.3 Different feed forward controller designs
• 4.4 Zero phase compensation
• 4.5 Zero phase gain compensation
• 4.6 Characterisation of complex non-minimum phase zeros
• 4.7 Zero phase extended gain compensation
• 4.8 Zero phase optimal gain compensation
• 4.9 Truncated series approximation compensation
• 4.10 Case study: electrohydraulic testing system
• 4.11 Robustness analysis
• 4.12 Chapter summary and concluding remarks
• References

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Input shaping control, Page 1 of 2

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text:
# Lipschitz flows on Banach spaces

I have a simple question on a Lipschitz flow. Recall that a flow on a non-empty set $X$ is a map $\Phi: \mathbb R \times X \to X$ satisfying

$$\Phi(0,x) = x \text{ and } \Phi(s, \Phi(t, x)) = \Phi(s + t, x)$$

Now let $X$ be a Banach space and $F$ a Lipschitz-function on $X$. Now it is well known that there exists for every $x \in X$ a unique function $u_x:\mathbb R \to X$ which is continuously strongly differentiable and satisfies:

$$u_x'(t) = F(u_x(t)) \text{ where x \in \mathbb R and } u_x(0) = x.$$ Where $u'$ is the strong derivative in the Banach space.

Now we can set $\Phi(t,x) = u_x(t)$. This should be a flow, why is that? The first property is easy to verify, but the second one requires us to show that $u_{u_x(t)}(s) = u_x(s + t)$. Can someone give a hint how I obtain this? I'm probably missing something simple.

-

$$v_x'(t) = F(v_x(t)),\qquad\qquad v_x(0) = u_x(s).$$
For every $s\in \mathbb R$, $u_{u_x(t)}(s)$ and $u_x(s + t)$ are both solutions with the same initial data at $t=0$.


text:
# Upon heating, calcium carbonate decomposes to produce calcium oxide and carbon dioxide. What is the percent yield of CO_2 if 97.5 g CO_2 is collected?

Mar 24, 2016

$C a C {O}_{3} \left(s\right) + \Delta \rightarrow C a O \left(s\right) + C {O}_{2} \left(g\right) \uparrow$
$\frac{97.5 \cdot g}{44.01 \cdot g \cdot m o {l}^{-} 1 \text{ carbon dioxide}}$ $\cong$ $2$ $m o l$.
But we don't know with how much calcium carbonate you started. You started with over $2$ $m o l$ because you collected over $2$ $m o l$ carbon dioxide, and the stoichiometry shows that the moles of calcium carbonate to carbon dioxide were $1 : 1$.


text:
# hyperbola and cricle

Level pending

If a variable line has its intercepts on the coordinate axes $$e,e_{1}$$ , where $$\frac {e}{2} , \frac{e_{1}}{2}$$ are the eccentricities of a hyperbola and its conjugate hyperbola respectively, the the line always touches the circle $$x^2+y^2=r^2$$. What is the value of $$r$$ ?

×


text:
# RS Aggarwal Class 6 Solutions Ratio Proportion And Unitary Method

RS Aggarwal class 6 solutions chapter 10 ratio, proportion, and unitary method. This chapter is divided into three parts. In the first part, the students will learn about the ratio, what is a ratio, how to calculate the ratio. In the second part, the students will learn about proportions and unitary method in the third part. To master the concepts of ratio, proportion and unitary method students are advised to solve questions from the RS Aggarwal for class 6.

Q1: Find each of the following ratios in the simplest form:

(i) 24 to 56

(ii) 84 paise to Rs. 3

(iii) 4 kg to 750 g

(iv) 1.8 kg to 6 kg

(v) 48 minutes to 1 hour

(vi) 2.4 km to 900 m

Sol:

(i) 24:56 = $\frac{24}{56} = \frac{24 \div 8}{56 \div 8} = \frac{3}{7}$

As the H.C.F of 3 and 7 is 1, the simplest form of 24 : 56 = 3 : 7

(ii) 84 paise to Rs. 3 = Rs. 0.84 to  Rs. 3

$\frac{0.84}{3} = \frac{0.84 \div 3}{3 \div 3} = \frac{0.28}{1} = \frac{28}{100} = \frac{28 \div 4}{100 \div 4} = \frac{7}{25}$

As the H.C.F. of 7 and 25 is 1, the simplest form of 0.84 : 3 is 7 : 25.

(iii) 4 kg : 750 g = 4000g : 750 g = $\frac{4000 \div 250}{750 \div 250} = \frac{16}{3}$

As the H.C.F of 16 and 3 is 1, the simplest form of 4000:750 is 16 : 3.

(iv) 1.8 kg : 6 kg = $\frac{1.8}{6} = \frac{18}{60} = \frac{18 \div 6}{60 \div 6} = \frac{3}{10}$

As the H.C.F. of 3 and 10 is 1, the simplest form of 1.8 : 6 = 3 : 10.

(v) 48 minutes to 1 hour = 48 minutes to 60 minutes = 48 : 60 = $\frac{48 \div 12}{60 \div 12} = \frac{4}{5}$

As the H.C.F. of 4 and 5 is 1, the simplest form of 48 : 60 = 4 : 5.

(vi) 2.4 km to 900 m = 2400 m to 900 m = $\frac{2400}{900} = \frac{24}{9} = \frac{24 \div 3}{9 \div 3} = \frac{8}{3}$

As the H.C.F. of 8 and 3 is 1, the simplest form of 2400 : 900 = 8 : 3.

Q2: Express each of the following ratios in the simplest form:

(i) 36 : 90

(ii) 324 : 144

(iii) 85 : 561

(iv) 480 : 384

(v) 186 : 403

(vi) 777 : 1147

Sol:

(i) 36 : 90 = $\frac{36}{90} = \frac{36 \div 18}{90 \div 18} = \frac{2}{5}$    (As the H.C.F. of 36 and 90 is 18)

Since the H.C.F. of 2 and 5 is 1, the simplest form of 36 : 90 is 2 : 5.

(ii) 324 : 144 = $\frac{324}{144} = \frac{324 \div 36}{144 \div 36} = \frac{9}{4}$    (As the H.C.F. of 324 and 144 is 36)

Since the H.C.F. of 9 and 4 is 1, the simplest form of 324 : 144 is 9 : 4.

(iii) 85 : 561 = $\frac{85}{561} = \frac{85 \div 17}{561 \div 17} = \frac{5}{33}$    (As the H.C.F. of 85 and 561 is 17)

Since the H.C.F. of 5 and 33 is 1, the simplest form of 85 : 561 is 5 : 33.

(iv) 480 : 384 = $\frac{480}{384} = \frac{480 \div 96}{384 \div 96} = \frac{5}{4}$    (As the H.C.F. of 480 and 384 is 96)

Since the H.C.F. of 5 and 4 is 1, the simplest form of 480 : 384 is 5 : 4

(v) 186 : 403 = $\frac{186}{403} = \frac{186 \div 31}{403 \div 31} = \frac{6}{13}$    (As the H.C.F. of 186 and 403 is 31)

Since the H.C.F. of 6 and 13 is 1, the simplest form of 186 : 403 is 6 : 13

(vi) 777 : 1147 = $\frac{777}{1147} = \frac{777 \div 37}{1147 \div 37} = \frac{21}{31}$    (As the H.C.F. of 777 and 1147 is 37)

Since the H.C.F. of 21 and 31 is 1, the simplest form of 777 : 1147 is 21 : 31.

Q3: Write each of the following ratios in the simplest form:

(i) Rs. 6.30 to Rs. 16.80

(ii) 3 weeks : 30 days

(iii) 3m 5 cm : 35 cm

(iv) 48 min : 2 hours 40 min

(v) 1 L 35 mL : 270 mL

(vi) 4 kg : 2kg 500g

Sol:

(i) Rs. 6.30 to Rs. 16.80

$\frac{6.30}{16.80} = \frac{63}{168} = \frac{63 \div 21}{168 \div 21} = \frac{3}{8}$    (H.C.F. of 63 and 168 is 21.)

Ratio = 3 : 8

(ii) 3 weeks : 30 days = 21 days : 30 days

$\frac{21}{30} = \frac{21 \div 3}{30 \div 3} = \frac{7}{10}$    (H.C.F. of 21 and 30 is 3.)

Ratio = 7 : 10.

(iii) 3m 5 cm : 35 cm = 305 cm : 25 cm       (as 1 m = 100 cm)

$\frac{305}{35} = \frac{305 \div 5}{35 \div 5} = \frac{61}{7}$    (H.C.F. of 305 and 35 is 5.)

Ratio = 61 : 7.

(iv) 48 min : 2 hours 40 min = 48 min : 160 min          (as 1 hour = 60 min)

$\frac{48}{160} = \frac{48 \div 16}{160 \div 16} = \frac{3}{10}$    (H.C.F. of 48 and 160 is 16.)

Ratio = 3 : 10.

(v) 1 L 35 mL : 270 mL = 1035 mL : 270 mL     ( as 1 L = 1000 mL)

$\frac{1035}{270} = \frac{1035 \div 45}{270 \div 45} = \frac{23}{6}$    (H.C.F. of 1035 and 270 is 45.)

Ratio = 23 : 6.

(vi) 4 kg : 2kg 500g = 4000 g : 2500 g   (as 1 kg = 1000 g)

$\frac{4000}{2500} = \frac{4000 \div 500}{2500 \div 500} = \frac{8}{5}$    (H.C.F. of 4000 and 2500 is 500.)

Ratio = 8 : 5.

Q4: Mr Sahai and his wife are both school teachers and earn Rs. 16800 and Rs. 10500 per months respectively. Find the ratio of

(i) Mr. Sahil’s income to his wife’s income.

(ii) Mrs. Sahil’s income to her husband’s income.

(iii) Mr. Sahil’s income to the total income of the two.

Sol:

Mr Sahai’s earning = Rs. 16800

Mrs. Sahai’s earning = Rs. 10500

(i) Ratio = 16800 : 10500 = 168 : 105 = $\frac{168 \div 21}{105 \div 21} = \frac{8}{5}$ (H.C.F. of 168 and 105 is 21.)

Mr Sahai’s income : Mrs. Sahai’s income = 8 : 5

(ii) Ratio = 10500 : 16800 = 105 : 168 = $\frac{105 \div 21}{168 \div 21} = \frac{5}{8}$   (H.C.F. of 168 and 105 is 21.)

Mrs. Sahai’s income : Mr Sahai’s income = 5 : 8

(iii) Total income = 16800 + 10500 = Rs. 27300

Ratio = 16800 : 27300 = 168 : 273 = $\frac{168 \div 21}{273 \div 21} = \frac{8}{13}$     (H.C.F. of 168 and 273 is 21.)

Mrs. Sahai’s income : Total income : 8 : 13

Q5: Rohit earns Rs. 15300 and saves Rs. 1224 per month. Find the ratio of

(i) his income and savings

(ii) his income and expenditure

(iii) his expenditure and saving

Sol:

Rohit’s income = Rs. 15300

Rohit’s saving = Rs. 1224

(i) Income : Savings = 15300 : 1224 = $\frac{15300 \div 612}{1224 \div 612} = \frac{25}{2}$   (H.C.F. of 15300 and 1224 is 612.)

Income : Savings = 25 : 2

(ii) Monthly expenditure = Rs. (15300 – 1224) = Rs. 14076

Income : Expenditure = 14076 : 1224 = $\frac{15300 \div 612}{14076 \div 612} = \frac{25}{23}$        (H.C.F. of 15300 and 14076 is 612.)

Income : Expenditure = 25 : 23

(iii) Expenditure : Savings = 14076 : 1224 = $\frac{14076 \div 612}{1224 \div 612} = \frac{23}{2}$         (H.C.F. of 14076 and 1224 is 612.)

Expenditure : Savings = 23 : 2

Q6: The ratio of the number of males and females workers in a textile mill is 5 : 3. If there are 115 male worker, what is the number of female workers in the mill?

Sol:

Number of males : Number of females = 5 : 3

Let the number be x.

Number of males = 5x

Number of females = 3x

Number of male workers = 115

Now 5x = 115

$\Rightarrow$ x = 23

Number of female workers in the mill = 3x = 3 $\times$ 23 = 69

Q7: The boys and the girls in a school are in the ratio 9 : 5. If the total strength of the school is 448, find the number of girls.

Sol:

Boys : Girls = 9 : 5

Let the number of boys = 9x

Let the number of girls = 5x

Total strength of the school = 448

According to given condition, we have:

9x + 5x = 448

$\Rightarrow$ 14x = 448

$\Rightarrow$ x = 35

Number of boys = 9x = 9 $\times$ 35 = 288

Number of girls = 5x = 5 $\times$ 35 = 160

Q8: Divide Rs. 1575 between Kamal and Madhu in the ratio 7 : 2.

Sol:

Kamal : Madhu = 7 : 2

Sum of the ratio terms = 7 + 2 = 9

Kamal’s share = $\frac{7}{9} \times 1575 = \frac{11025}{9}$ = Rs. 1225

Madhu’s share = $\frac{2}{9} \times 1575 = \frac{3150}{9}$ = Rs. 350

Q9: Divide Rs. 3450 among A, B and C in the ratio 3 : 5 : 7.

Sol:

A: B : C = 3 : 5 : 7

Sum of the ratio terms = 3 + 5 + 7 = 15

A’s share = $\frac{3}{15} \times 3450 = \frac{10350}{15}$ = Rs. 690

B’s share = $\frac{5}{15} \times 3450 = \frac{17250}{15}$ = Rs. 1150

C’s share = $\frac{7}{15} \times 3450 = \frac{24150}{15}$ = Rs. 1610

Q10: Two numbers are in the ratio 11 : 12 and their sum is 460. Find the numbers.

Sol:

Two numbers are in the ratio 11 : 12

Let the numbers be 11x and 12x.

Given: 11x + 12x = 460

$\Rightarrow$ 23x = 460

$\Rightarrow$ x = 20

First number = 11x = 11 $\times$ 20 = 220

Second number = 12x = 12 $\times$ 20 = 240

Hence, the numbers are 220 and 240.

Q11: A 35 cm line  segment is divided into two parts in the ratio 4 : 3. Find the length of each part.

Sol:

Ratio of the two parts of line segment = 4 : 3

Sum of the ratio terms = 4 + 3 = 7

First part = $\frac{4}{7} \times 35 = 4 \times 5$ = 20 cm

Second part = $\frac{3}{7} \times 35 = 3 \times 5$ = 15 cm

Q12: A factory produces electric bulbs. If 1 out of every 10 bulbs is defective and the factory produces 630 bulbs per day, find the number of defective bulbs produced each day.

Sol:

Number of bulbs produced each day = 630

Out of 10 bulbs, 1 is defective.

Number of defective bulbs = $\frac{630}{10} = 63$

Therefore, Number of defective bulbs produced each day = 63

Q13: Find the ratio of the prices of a pencil to that of the ball pen if pencil cost Rs. 96 per score and ball pens cost Rs. 50.40 per dozen.

Sol:

Price of pencil = Rs. 96 per score

Price of ball pen = Rs. 50.40 per dozen

Price per unit of pencil = $\frac{96}{20} = 4.8$

Price per unit of ball pen = $\frac{50.40}{12} = 4.2$

Ratio = $\frac{4.8}{4.2} = \frac{48}{42} = \frac{48 \div 6}{42 \div 6} = \frac{8}{7}$

Price of pencil : Price of ball pen = 8 : 7

Q14: The ratio of the length of a field to its width is 5 : 3. Find its length if the width is 42 metres.

Sol:

Length : Width = 5 : 3

Let the length and the width of the field be 5x and 3x m respectively.

Width = 42 m

3x = 42

x = $\frac{42}{3} = 14$

Therefore,5x = 5 $\times$ 14 = 70 m

Q15: The ratio of income to savings of a family is 11 : 2. Find the expenditure if the savings is Rs. 1520.

Sol:

Income : Savings = 11 : 2

Let the income and the savings be Rs. 11x and Rs. 2x respectively.

Savings = Rs. 1520

2x = 1520

x = $\frac{1520}{2} = 760$

Therefore,Income = Rs 11x = Rs (11 $\times$ 760) = Rs. 8360

Expenditure = Income – Saving

= Rs (8360 – 1520)

Rs. 6840

Q16: The ratio of income to expenditure of a family is 7 : 6. Find the savings if the income is Rs. 14000.

Sol:

Income : Expenditure = 7 : 6

Let the income and the expenditure be Rs. 7x and 6x, respectively.

Income = Rs 14000

7x = 14000

X = 2000

Expenditure = Rs 6x = Rs 6 $\times$ 2000 = Rs. 12000

Therefore,Saving = Income – Expenditure

= Rs. (14000 – 12000)

= Rs. 2000

Q17: The ratio of zinc and copper in an alloy is is 7 : 9. If the weight of copper in the alloy is 11.7 kg, find the weight of zinc in it.

Sol:

Let the weight of zinc be x kg.

Ratio of zinc and copper = 7.9

Weight of copper in the alloy = 11.7 kg

$\frac{7}{9} = \frac{x}{11.7}$

$\Rightarrow x = \frac{7 \times 11.7}{9} = \frac{81.9}{9} = 9.1$

Weight of zinc = 9.1 kg

Q18: A bus covers 128 km in 2 hours and a train covers 240 km in 3 hours. Find the ratio of their speeds.

Sol:

A bus covers 128 km in 2 hours.

Speed of the bus = $\frac{Distance}{time} = \frac{128}{2} = 64$ km/h

A train covers 240 km in 3 hours = $\frac{Distance}{time} = \frac{240}{3} = 80$ km/h

Ratio of their speeds = 64 : 80 = $\frac{64}{80} = \frac{64 \div 16}{80 \div 16} = \frac{4}{5}$

Therefore,Ratio of the speeds of the bus and the train = 4 : 5

Q19: From each of the given pairs, find which ratio is larger:

(i) (3 : 4) or (9 : 16)

(ii) (5 : 12 ) or (17 : 30)

(iii) (3 : 7) or (4 : 9)

(iv) (1 : 2) or (13 : 27)

Sol:

(i) (3 : 4) or (9 : 16)

Making the denominator equal

$\frac{3 \times 4}{4 \times 4} = \frac{12}{16}$ and

$\frac{12}{16} > \frac{9}{16}$

Therefore, 3 : 4 > 9 : 16

(ii) (5 : 12 ) or (17 : 30)

Making the denominator equal

$\frac{5 \times 5}{12 \times 5} = \frac{25}{60}$ and

$\frac{17 \times 2}{30 \times 2} = \frac{34}{60}$

$\Rightarrow \frac{25}{60} < \frac{34}{60}$

Therefore, 5 : 12 < 17 : 30

(iii) (3 : 7) or (4 : 9)

Making the denominator equal

$\frac{3 \times 9}{7 \times 9} = \frac{27}{54}$ and

$\frac{4 \times 7}{9 \times 7} = \frac{28}{63}$

$\Rightarrow \frac{27}{63} < \frac{28}{63}$

Therefore, 3 : 7 < 4 : 9

(iv) (1 : 2) or (13 : 27)

Making the denominator equal

$\frac{1 \times 27}{2 \times 27} = \frac{27}{54}$ and

$\frac{13 \times 2}{27 \times 2} = \frac{26}{54}$

$\Rightarrow \frac{27}{54} > \frac{26}{54}$

Therefore, 1 : 2 > 13 : 27

Q20: Fill in the blank places:

(i) $\frac{24}{40} = \frac{}{5} = \frac{12}{}$

(ii) $\frac{36}{63} = \frac{4}{} = \frac{}{21}$

(iii) $\frac{5}{7} = \frac{}{28} = \frac{35}{}$

Sol:

(i) $\frac{24}{40} = \frac{24 \div 8}{40 \div 8} = \frac{3}{5} = \frac{3 \times 4}{5 \times 5} = \frac{12}{20}$

(ii) $\frac{36}{63} = \frac{36 \div 9}{63 \div 9} = \frac{4}{7} = \frac{4 \times 3}{7 \times 3} = \frac{12}{21}$

(iii) $\frac{5}{7} = \frac{5 \times 4}{7 \times 4} = \frac{20}{28} = \frac{5 \times 7}{7 \times 7} = \frac{35}{49}$

#### Practise This Question

Draw a line AB. At A, draw an arc of length 3cm using compass such that it intersects AB at O. With the same spread of compass, put the compass pointer at O and make an arc that intersects the previous arc at P. With the same spread again, put the compass pointer at P and draw an arc that intersects the first arc at Q. Join A and Q. Using the protractor, measure QAB. What is the value of QAB.


text:
# Math Help - Groups exercise questions

1. ## Groups exercise questions

I was wondering if someone could help with following exercise (image attached):

2. To show something is a subgroup, you need to show it contains the identity, $I_2$. It is closed under multiplication and inverses.

You are capable of doing this, it is very good practice.

1) if b and c are 0 you are just talking about diagonal matrices, but also note since they are in GL(2,R) neither of the diagonal entries can be 0 or else they will not be invertible since the determinant is the product of the diagonal entries.
Certainly the identity is in there. Write out multiplication of diagonal matrices, is it diagonal? What about inverses? is the inverse of a diagonal matrix diagonal the answer should be clear from what you got when you multiplied two arbitrary diagonal matrices (it is important here that neither entry is 0). See why?

3) is clearly not since it is not invertible if the entries must be integers. consider the one with both diagonal entries 2. It's inverse is the one with both diagonal entries 1/2, but that is not an integer, so it is not in the set, so its not closed under inverse.

4) Hint this is called the special linear GROUP. The chances are it is a group. When checking it recall $det(AB)=det(A)det(B)=1*1=1$
and $1=det(I_2)=det(AA^{-1})=det(A)det(A^{-1}) \Rightarrow det(A^{-1})=\frac{1}{det(A)}=\frac{1}{1}=1$

5) no what is the determinant of a matrix with first row all 0s? Are empty sets groups?

2) I leave to you, it will make you stronger.