text:
# 1.1: General purpose software

Almost every computer or smart phone has the calculator. Typically, it can do simple arithmetics, sometimes also square roots and degrees. This is enough for the basic data processing. However, to do any statistical analysis, such calculator will need statistical tables which give approximate values of statistics, special characteristics of data distribution. Exact calculation of these statistics is too complicated (for example, it might require integration) and most programs use embedded statistical tables. Calculators usually do not have these tables. Even more important disadvantage of the calculator is absence of the ability to work with sequences of numbers.

To deal with many numbers at once, spreadsheets were invented. The power of spreadsheet is in data visualization. From the spreadsheet, it is easy to estimate the main parameters of data (especially if the data is small). In addition, spreadsheets have multiple ways to help with entering and converting data. However, as spreadsheets were initially created for the accounting, they oriented still to the tasks typical to that field. If even they have statistical functions, most of them are not contemporary and are not supported well. Multivariate methods are frequently absent, realization of procedures is not optimal (and frequently hidden from the user), there is no specialized reporting system, and so on.

And thinking of data visualization in spreadsheets—what if the data do not fit the window? In that case, the spreadsheet will start to prevent the understanding of data instead of helping it.

Another example—what if you need to work simultaneously with three non-neighboring columns of data? This is also extremely complicated in spreadsheets.

This is why specialized statistical software come to the scene.


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# You are designing a rotating metal flywheel that will be used to store energy. The flywheel is to...

## Question:

You are designing a rotating metal flywheel that will be used to store energy. The flywheel is to be a uniform disk with radius {eq}29.0{/eq} {eq}cm{/eq}. Starting from rest at {eq}t=0{/eq}. The flywheel rotates with constant angular acceleration {eq}3.00{/eq} {eq}rad/s^2{/eq} about an axis perpendicular to the flywheel at its center.

1. If the flywheel density (mass per unit volume) of {eq}8600{/eq} {eq}kg/m^3{/eq}, what thickness must it have to store {eq}800{/eq} {eq}J{/eq} of kinetic energy at {eq}t=8.00{/eq} {eq}s{/eq} ?

## Rotational Kinetic Energy:

The energy stored in a rotation flywheel is in the form of rotational kinetic energy given by the formula:

\begin{align*} &E_K = \frac{1}{2} I\omega^2 & \text{[Rotational Kinetic Energy of a Flywheel]]} \end{align*}

where {eq}\omega{/eq} is the angular velocity of the flywheel.

The moment of inertia {eq}I{/eq} of the flywheel about its axis of symmetry is given by the equation:

\begin{align*} &I = \frac{1}{2}mr^2 & \text{[Moment of Inertia of a Flywheel]} \end{align*}

where {eq}m{/eq} is the mass of the flywheel and {eq}r{/eq} is its radius.

Given a rotating metal flywheel with radius {eq}r = 29 \,cm = 0.29\,m{/eq} and density {eq}\rho = 8600\,\frac{kg}{m^3}{/eq}.

The flywheel's mass {eq}M{/eq} and moment of inertia about its center {eq}I{/eq} expressed in terms of its thickness {eq}t{/eq} in meters are:

{eq}\begin{align*} M &= \rho V \\ &= \rho (\pi r^2t) \\ &= \pi (8600)(0.29)^2t \\ &= 723.26\pi t & \text{[Mass of the Flywheel]} \\ \\ I &= \frac{1}{2}mr^2 \\ &= \frac{1}{2} (723.26\pi t)(0.29)^2 \\ &= 30.413083\pi t & \text{[Moment of Inertia of the Flywheel]} \end{align*} {/eq}

The flywheel started spinning from rest at {eq}t = 0{/eq} with an angular acceleration of {eq}\alpha = 3 \,\frac{rad}{s^2}{/eq}. After {eq}t = 8\,s{/eq}, the angular velocity of the flywheel is:

{eq}\begin{align*} \omega &= \alpha t \\ &= (3)(8) \\ &= 24\, \frac{rad}{s} & \text{[Angular Velocity at } t=8 \text{ s]} \end{align*} {/eq}

Finally, to store {eq}800\,J{/eq} of kinetic energy, the flywheel must have a thickness {eq}t{/eq} such that:

{eq}\begin{align*} E_K &= \frac{1}{2} I \omega^2 & \text{[Substitute the known values]} \\ 800 &= \frac{1}{2} (30.413083\pi t) (24)^2 & \text{[Solve for } t \text{]} \\ \\ t &= \frac{2(800)}{(30.413083\pi)(24)^2} \\ &= \frac{25}{9(30.413083\pi)} \\ t &= \boxed{ 0.0291 \,m = 2.91\,cm } & \text{[Required Thickness of the Flywheel to Store } 800 \text{ J of Energy]} \end{align*} {/eq}

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# Discrepancy in estimating boundary stencil for finite difference method

I am trying to estimate the FD stencil for boundary as mentioned in this paper (section 4.1) using MATLAB. The stencil order (6th) is higher than the one mentioned in paper (4th). $$f_1' +\alpha f_2' = \frac{1}{h} (c_1f_1 + c_2f_2 + c_3f_3 +c_4f_4+ c_5f_5 + c_6 f_6 + c_7 f_7 )$$ Each term of both sides is expanded using Taylor series expansion and then derivatives are matched to estimate the coefficients [$c1,c2,c3,c4,c5,c6,c7$ and $\alpha$].

I solved the above system as following:

Above system of equation can be written as M*x=y

Note:

1) The alpha was taken to the left to form a part of unknown vector x

2) the factorial on both side were simplified and adjusted to right side.

I wrote following code for solving the above system of equation.

LM = zeros(8);
LM(1,1:7)=1;                LM(1,8)=0;
LM(2,2:7)=1:6;              LM(2,8)=-1;
LM(3,2:7)=(1:6).^2;         LM(3,8)=-2;
LM(4,2:7)=(1:6).^3;         LM(4,8)=-3;
LM(5,2:7)=(1:6).^4;         LM(5,8)=-4;
LM(6,2:7)=(1:6).^5;         LM(6,8)=-5;
LM(7,2:7)=(1:6).^6;         LM(7,8)=-6;
LM(8,2:7)=(1:6).^7;         LM(8,8)=-7;
y=[0;1;0;0;0;0;0;0];
x=LM\y;
rats(x)

which gives completely different answer than reported one. Both are given below:

Solution i.e. $x =[c1,c2,c3,c4,c5,c6,c7, \alpha$] by

MATLAB estimated solution: [-69/20, -17/10, 15/2, -10/3, 5/4, -3/10, 1/30, 6]

Solution reported in literature: [-197/60, -5/12, 5, -5/3, 5/12, -1/20, 0, 5]

Any help is appreciated.


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# question about $\omega$-starcompact space

A topological space $$X$$ is said to be star compact if whenever $$\mathscr{U}$$ is an open cover of $$X$$, there is a compact subspace $$K$$ of $$X$$ such that $$X = \operatorname{St}(K,\mathscr{U})$$.

$$St(K, \mathscr{U})=\cup\{u\in \mathscr{U}: u \cap K \neq \emptyset\}$$

$$st^{n+1} (K, \mathscr{U}) = \bigcup  \{ U \in \mathscr{U} : U \cap st^{n}(K, \mathscr{U}) \neq \emptyset \}$$

Definition: A space $$X$$ is said to be $$\omega$$-starcompact if for every open cover $$\mathscr{U}$$ of $$X$$, there is some $$n \in \mathbf{N^{+}}$$ and some finite subset $$B$$ of $$X$$ such that $$st^{n}( B, \mathscr{U}) = X$$.

Definition: A Hausdorff topological space $$(X,\tau)$$ is called H-closed or absolutely closed if it is closed in any Hausdorff space, which contains $$X$$ as a subspace.

Theorem: Let $$X$$ be a Hausdorff space. $$X$$ is H-closed if and only if every open cover $$\mathcal{C}$$ of $$X$$ contains a finite subsystem $$\mathcal{D}$$ such that $$\bigcup \{\overline{D} : D \in \mathcal{D} \}=X$$, i.e., the closures of the sets from $$\mathcal{D}$$ cover $$X$$.

According to the above theorem , I concluded that each H-closed space is $$\omega$$-starcompact.

Is it possible to conclude that any $$\omega$$-starcompact space is H-closed? Or in the Hausdorff space is any $$\omega$$-starcompact space H-closed?

No. It is well-known (see, for instance, [Mat, Theorem 3]) that a Hausdorff space $$X$$ is countably compact iff for every open cover $$\mathcal U$$ there exists a finite subset $$F$$ of $$X$$ such that $$St(F,\mathcal U)=X$$. In particular, each countably compact space is $$\omega$$-starcompact. So any proper dense countably compact space $$X$$ of a Hausdorff compact space $$Y$$ is a counterexample. For instance, we can put $$Y=[0,\omega_1]$$ endowed with the order topology and $$X=[0,\omega_1)$$ or $$Y$$ be a $$\Sigma$$-product of an uncountable family of compact spaces with at least two points each. I recall that a space $$Y$$ is $$\Sigma$$-product of a family $$\{X_\alpha:\alpha\in A\}$$ if there exists a point $$x\in\prod X_\alpha$$ such that $$Y=\{y\in \prod X_\alpha: |\{\alpha:y_\alpha\ne x_\alpha\}|\le\omega\}.$$

References

[Mat] M. Matveev, A Survey on Star Covering Properties.

• I can not downlode the reference, is it an article? – adin Jul 10 at 18:28
• @adin Yes, it is an article, available, I guess, in several formats. I have just downloaded its PostScript version. – Alex Ravsky Jul 10 at 19:12
• @adin my answer has a downloadable version of the original paper. – Henno Brandsma Jul 10 at 21:11

If $$X$$ is $$H$$-closed and $$\mathcal{U}$$ is an open cover, subset of $$\mathcal{U}$$, say $$\mathcal{U}'$$, such that $$\bigcup \{\overline{U}: U \in \mathcal{U}'\} = \overline{\bigcup \mathcal{U}'} = X$$ and if we pick one $$x$$ from each $$U \in \mathcal{U}'$$ we get a finite subset $$F$$ of $$X$$ such that $$\operatorname{st}(F,\mathcal{U})\supseteq \bigcup\mathcal{U}'$$ so $$\operatorname{st}(F,\mathcal{U})$$ is dense and so $$\operatorname{st}^2(F,\mathcal{U})= \operatorname{st}(\operatorname{st}(F,\mathcal{U}),\mathcal{U})= X$$ and $$X$$ is star-2-compact, as this is called and so certainly $$\omega$$-starcompact.

The reverse is far from true, star-1-compact is equivalent to countable compactness for Hausdorff spaces van Douwen et al.'s paper). So Alex Ravsky's examples are $$\omega$$-starcompact Hausdorff (even Tychonoff) but not $$H$$-compact.


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# cloneTopology

A Workflow Engine function that copies an existing topology to a new inactive named topology if the name is not already in use. Both topologyName and cloneName can be static or a substitution value. To substitute a value, use $(<attribute_name>). For example$(custom_info.myTopology).

Follow this action with the 'activateTopology' action if you want to activate the new topology.

This function is available for alert workflows.

## Arguments

Workflow Engine function cloneTopology takes the following arguments:

Name

Required

Type

Description

topologyName

yes

string

The 'donor' named topology or substituted value for the topology. May be active or inactive. To substitute a value, use $(<attribute_name>). For example$(custom_info.myTopology).

cloneName

yes

string

The 'clone' named topology or substituted value for the topology. To substitute a value, use $(<attribute_name>). For example$(custom_info.myTopology).

## Example

The following example demonstrates typical use of Workflow Engine function cloneTopology.

If you want to create a new topology based upon a topology in the alert workflow context, set the following:

• topologyName: $(workflowContext.myTopology) • cloneName:$(workflowContext.myTopology) copy

The UI translates your settings to the following JSON:

{"topologyName":"$(workflowContext.myTopology)","cloneName":"$(workflowContext.myTopology) copy"}

For an alert that has a workflowContext.myTopolgy = "my network", the action clones the topology to "my network copy". If you run the topologies API, you can see your new inactive topology:

curl -X GET 'https://example.com/api/v1/topologies/inactive'

Returns the following:

 [{
"name": "my network copy",
"active": false,
"description": "Automatically created topology: triggering alert # 35 @ 2020-05-08T02:14:05.680Z"
}]


text:
Measuring the Performance of Classification Models

Nearly all companies use machine learning to do classification now. When it comes to deciding which model best suits the data, we need to employ some performance measures. An inappropriate measure would lead to a poorly chosen model. Here are some general thoughts on model performance criteria selection.

1. Things to do upon getting the data

Before fitting any model, the quality of data should be checked:

• What covariates do we have?
• For a categorical covariate,
• how many categories?
• are they just categorical or ordinal?
• For a numerical covariate,
• what is the shape of the distribution?
• are there any outlying values?
• Is there any missingness? (can be checked with the mice package but I haven’t given it a try yet)
• if there is missing, is the pattern at random or not at random?
• what’s an appropriate way to deal with it? (imputation, mean/median/mode, or multiple imputation)

2. Data cleaning

The next thing is to do some data cleaning, including:

• normalizing numerical covariates so that they have a mean of 0 and sd of 1;
• combine levels of categorical variable if there are too few observations in a single level
• explore the correlation between the covariates and eliminate highly correlated features

3. Classification

I usually start with a logistic regression - the simplest and most interpretable model. Variable selection with LASSO can be done at the same time using the glmnet package in R, or sklearn.LogisticRegression in Python. When used on a large and highly imbalanced data, with say 99.5% 0’s and 0.5% 1’s, the model would be dominated by the negative class. When evaluating the model on the test data, what’s most likely to happen will be that all observations are predicted to be 0.

For unbalanced dataset, such as credit fraud, or response to direct mail marketing campaigns, the major focus is often the positive case. To build a better model, people might suggest under-sampling the negative class, or over-sampling the positive class. The problem with under sampling is, we are essentially ignoring some information, which could be very important in constructing the decision rule, on purpose. What about oversampling? We will have duplicated observations in the positive class. The same observation might appear in both the training and testing data after we do data partition. Once a model is trained on the new training data after oversampling, when we test it on the new testing data, the duplicated observations will most likely be assigned the correct label, which does not reflect the true predicting power of the model!

Towards this end, some techniques are often used to “enlarge” the positive class, including the Synthetic Minority Over-sampling Technique (SMOTE), or the Adaptive Synthetic (ADASYN) approach. Both are implemented in imblearn package in Python.

• SMOTE: use linear combinations of positive class observations to create new observations
• ADASYN: an improved version of SMOTE; after a new observation is created, a small noise is added to it, making it more realistic.

Tree-based models work well on such datasets. The sklearn package contains a large collection of tree-based models, including the simplest DecisionTree, and more complicated ensemble methods like AdaBoost, GradientBoosting, and RandomForest. The very popular XGBoost method has its own package. Microsoft’s LightGBM is also powerful. The difference between it and XGBoost is where the trees make splits. While XGBoost orders the values of a feature first and calculate the gain or loss if a split is made at each of the values, LightGBM makes a histogram of the values first, and then split at different bins. This greatly improves the speed of calculation. These two, however, need the inputs to be all numerical. CatBoost is able to work on categorical inputs directly (I haven’t tried it yet).

4. Performance Measures

These methods have been so amazing that properly tuned models often yield high prediction precisions on testing datasets. Is this simple performance measure sufficient? Do we just choose the model that has the highest accuray? No, we don’t.

For binary classification problems, specifically, the performance evaluation is done using a confusion matrix:

Predicted PositivePredicted Negative
PositiveTrue positive, TPFalse negative, FN
NegativeFalse positive, FPTrue negative, TN

The simplest performance measure would be the prediction accuracy:

$\text{Acc} = \frac{\text{TP} + \text{TN}}{\text{TP} + \text{TN} + \text{FP} + \text{TN}}$

It is not difficult to notice that here we are paying equal attention to correct predictions in both the positive and negative classes. If the data is highly unbalanced, I can get a very high accuracy by just predicting everyone to be in the majority class.

The second (set of) measure(s) is sensitivity and specificity. Intuitively,

sensitivity = proportion of true positives that are correctly predicted = $\frac{\text{TP}}{\text{TP} + \text{FN}}$;
specificity = proportion of true negatives that are correctly predicted = $\frac{\text{TN}}{\text{TN} + \text{FP}}$.

Sensitivity is also called the true positive rate, while 1 - specificity, $\frac{\text{FP}}{\text{TN} + \text{FP}}$, is called the false positive rate.

All the classifiers return a set of scores for the observations, and the decision of positive/negative is made depending on the comparison of the scores with a threshold. Depending on choice of thresholds, one is able to get different combinations of (sensitivity, specificity) values. Plotting sensitivity (Y) against 1-specificity (X) gives the Receiver Operating Characteristic (ROC) curve. The area under curve (AUC) is a number between 0 and 1, and it is the probability that the classifier will assign a higher score to an originally positive observation than an originally negative observation.

Based on the AUC, another derived measure called the Gini index is defined, but they are essentially the same thing.

For very unbalanced classification, the AUC might still not be a good measure. It essentially assigns equal importance to sensitivity and specificity. In a scenario like credit card fraud detection, our major focus is the majority class. We don’t care if the system allows a normal transaction to be processed, but we do care if it successfully stops an abnormal transaction.

The third set of measures here would be the precision and recall. Similar to sensitivity and specificity, they are defined using information from the confusion matrix:

precision = proportion of correctly made positive classifications = $\frac{\text{TP}}{\text{TP} + \text{FP}}$;
recall = sensitivity = $\frac{\text{TP}}{\text{TP} + \text{FN}}$

Plotting precision against recall gives the PR curve, and correspondingly we have the PRAUC.

Based upon the precision and recall, a measure called the F-score is defined. It is the harmonic mean of precision and recall, i.e.,

$F = \frac{2}{1 / \text{precision} + 1 / \text{recall}} = \frac{2 \cdot \text{precision} \cdot \text{recall}}{\text{precision} + \text{recall}}$.

We essentially place equal emphasis on false positive and false negatives. There are times, however, that FPs and FNs need different level of emphasis. Imagine that you are a credit card issuer doing fraud detection. When a real fraud happens and your classifier missed it, your cost would be the amount of transaction. When your classifier incorrectly flags a transaction and stopped it, your cost is only sending an email/text message to your cardholder, and ask them to confirm the transaction. It is obvious that FN is a much more serious problem than FP here. Therefore we wish to place more emphasis on the FN. The $F_2$ measure comes based on this premise:

$F_2 = (1 + 2^2) \frac{\text{precision} \cdot \text{recall}}{2^2 \text{precision} + \text{recall}}$

A more general form would be $F_\beta$:

$F_\beta = (1 + \beta^2)\cdot \frac{\text{precision} \cdot \text{recall}}{\beta^2 \text{precision} + \text{recall}}$.

While $F_2$ weighs recall higher than precision, and thus puts more attention to FN, $F_{0.5}$ weights recall lower than precision, and emphasizes FP.

Yishu Xue
Data Scientist / Coder / Novice Sprinter / Gym Enthusiast

The night is dark and full of terrors.


text:
\name{dendrogramGrob}
\alias{dendrogramGrob}
\title{
Grob for Dendrogram
}
\description{
Grob for Dendrogram
}
\usage{
dendrogramGrob(dend, facing = c("bottom", "top", "left", "right"),
order = c("normal", "reverse"), gp = gpar())
}
\arguments{

\item{facing}{Facing of the dendrogram.}
\item{order}{If it is set to \code{reverse}, the first leaf is put on the right if the dendrogram is horizontal and it is put on the top if the dendrogram is vertical.}
\item{gp}{Graphic parameters for the dendrogram segments. If any of \code{col}, \code{lwd} or \code{lty} is set in the \code{edgePar} attribute of a node, the corresponding value defined in \code{gp} will be overwritten for this node, so \code{gp} is like global graphic parameters for dendrogram segments.}

}
\details{


text:
# Recursive successions limit.

Good morning everyone, I tried to solve an exercise and I want to ask you what do you think about my solution (if legit, clear, if I could have done it better) and if you can help me going forward.

Let $$x_{n+1}=\frac{|x_n^3-x_n|}{n+x_n^2} \qquad x_0=2017$$ Compute $\lim_{n\to \infty}x_n$ and $\lim_{n\to \infty}2^nx_n$.

That's what I've done: For the first question we first notice that $x_n \geq 0 \ \forall n \in \mathbb{N}$ (easy proof by induction). Suppose $x_n$ has a limit $l \in \mathbb{R}$; taking the recurrence we have:

$$\lim_{n\to \infty} x_{n+1}=\lim_{n\to \infty} \frac{|x_n^3-x_n|}{n+x_n^2}$$

So as $x_{n+1}$ and $x_n$ have the same limit we have that it must be $0$. So the unique possibilities for the limit are $0$ and $+\infty$. Let's now prove that $x_{n+1} \leq x_{n}$ so we can say that the limit exists and is $0$ as $x_n$ is decreasing and bounded.

$$x_{n+1}\leq x_n \iff x_n(n+x_n^2)\geq |x_n^3-x_n| \iff -nx_n-x_n^3\leq x_n^3-x_n \leq x_n^3+nx_n$$

and by using the fact that $x_n \geq 0$ the conclusion is clear as $n\to \infty$. So for the first point we have that $\lim_{n\to \infty}x_n=0$.

Now I'm stuck at the second point because I actually have no ideas how to proceed. I thought that I may be searching for an asymptotic behaviour for my recurrence but I can't find any so I tried to do root test but that's what I found:

$$\lim_{n \to \infty} (\frac{2^n|x_n^3-x_n|}{n+x_n^2})^{\frac{1}{n}}= 2\lim_{n \to \infty}e^{\frac{1}{n}\log\left(\frac{|x_n^3-x_n|}{n+x_n^2}\right)}$$

but I don't know how to compute the last one. Any help? Thanks in advance. Add: What about $\lim n!x_n$? I think that the limit is $\infty$ and I already proved that is monotonically increasing. How can I say that it can't go towards a real limit?

Here is another solution, progressively improving the asymptotics of $x_n$.

• We know that $(x_n)$ is non-negative for all $n$. Also, for $n \geq 1$,

$$x_{n+1} = \frac{\lvert x_n^2 - 1\rvert}{x_n^2 + n} \cdot x_n \leq \frac{x_n^2 + 1}{x_n^2 + n} \cdot x_n \leq x_n.$$

So $(x_n)$ is monotone decreasing and hence converges.

• Since $(x_n)$ is bounded, the same is true for $\lvert 1 - x_n^2\rvert$. Let $C$ be a bound of $\lvert 1 - x_n^2\rvert$. Then

$$x_{n+1} = \frac{\lvert x_n^2 - 1\rvert}{x_n^2 + n} \cdot x_n \leq \frac{C}{n} \cdot x_n \leq \cdots \leq \frac{C^n}{n!}x_1.$$

So $2^n x_n$ converges to $0$ by the squeezing theorem. In particular, $x_n$ converges to $0$ as well.

• If $x_N = 1$ for some $N$, then $x_n = 0$ for all $n > N$ and hence $n!x_n \to 0$ as $n\to\infty$. So it suffices to consider the case where $x_n \neq 1$ for all $n$. Then by noting that

$$\log \Bigg( \frac{\left| 1 - x_n^2 \right| }{1 + \frac{x_n^2}{n}} \Bigg) = \mathcal{O}(x_n^2) = \mathcal{O}\left( \frac{C^{2n-2}}{(n-1)!^2} \right) \quad \text{as } n\to\infty,$$

it follows that $\sum_{n=1}^{\infty} \log \Big( \frac{\left| 1 - x_n^2 \right| }{1 + \frac{x_n^2}{n}} \Big)$ converges. Let $A$ denote the value of this sum. Then

$$(n-1)!x_n = x_1 \prod_{k=1}^{n-1} \frac{\left| 1 - x_k^2 \right| }{1 + \frac{x_k^2}{k}} \xrightarrow[n\to\infty]{} x_1 e^{A}$$

Consequently, $n! x_n$ diverges to $\infty$ at a linear speed.

• Great solution. Now only to consider if $x_n=1$ can happen for some $n$ – Jakobian May 26 '18 at 23:13
• Yeah, great solution and nice method (I'll use it again surely)! Surely we should avoid the case $x_n=1$ for some $n$ as the recurrence would be definitely zero. +1, answer accepted. – Alberto Andrenucci May 27 '18 at 13:34

Assume $x_n>0$. Let's suppose $\lim\limits_{n\to \infty} 2^nx_n = g$. Using Stolz theorem: $$\lim\limits_{n\to \infty} 2^nx_n = \lim\limits_{n\to \infty} \frac{2^n}{\frac{1}{x_n}} = \lim\limits_{n\to \infty} \frac{2^{n+1}-2^n}{\frac{1}{x_{n+1}}-\frac{1}{x_n}} = \lim\limits_{n\to \infty} 2^nx_n\cdot \frac{x_{n+1}}{x_n-x_{n+1}}$$ $$= g \lim\limits_{n\to \infty} \frac{|x_n-x_n^3|}{(n+x_n^2)x_n-|x_n-x_n^3|} = g \lim\limits_{n\to \infty} \frac{1-x_n^2}{n-1+2x_n^2} = 0$$ It shows that if the limit exists, it must be either $0$ or $\infty$. That can lead to assume, the sequence is monotone. It can be shown it indeed is from some point on. We know the sequence $x_n$ is decreasing to $0$, so we will assume that $n$ is large enough, so that $x_n \leq 1$ $$2^{n+1}x_{n+1}\leq 2^nx_n \iff 2x_{n+1}\leq x_n \iff 2|x_n-x_n^3|\leq(n+x_n^2)x_n$$ $$\iff 2x_n-2x_n^3\leq(n+x_n^2)x_n \iff -3x_n^3\leq (n-2)x_n$$ Which is obvious for $n\geq2$. For big enough $n$ we have:

$$\frac{(n+1)!x_{n+1}}{n!x_n}=\frac{(n+1)(1-x_n^2)}{n+x_n^2}>1$$ Let's suppose the limit is finite, and equals $h$. I also use Stolz theorem here.

$$h=\lim\limits_{n\to\infty} n!x_n = \lim\limits_{n\to\infty} (n+1)!\cdot\frac{x_n-x_n^3}{n+x_n^2} = \lim\limits_{n\to\infty} \frac{(n+2)!(x_{n+1}-x_{n+1}^3)-(n+1)!(x_n-x_n^3)}{(n+1)+x_{n+1}^2-(n+x_n^2)}$$ $$= \lim\limits_{n\to\infty} [(n+2)!x_{n+1}-(n+1)!x_n] = h\lim\limits_{n\to\infty} (n+1)[(n+2)\frac{1-x_n^2}{n+x_n^2}-1]$$ $$= h\lim\limits_{n\to\infty} (n+1)\cdot\frac{2-nx_n^2-3x_n^2}{n+x_n^2} = 2h$$ Hence $h=0$, which can't be.

• Perfect, thanks! Do you have any idea for the other question i added? – Alberto Andrenucci May 26 '18 at 21:23
• I've added some more, what i think might be solution. But you need to check if i didn't make any mistakes. – Jakobian May 26 '18 at 22:23


text:
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Topic: 8 significant digit ellipse circumference formula, simplistic
Replies: 0

 thomasinventions@yahoo.com Posts: 35 Registered: 12/23/11
8 significant digit ellipse circumference formula, simplistic
Posted: Oct 2, 2012 11:23 PM

This is a variant of the Pade form, with no use of the Hoelder mean or pi in the formula.

:$C\approx 4a+b\left ( \frac{3929x^{0.5}+1639157x+19407215x^{1.5}+24302653x^{2}+12892432x^{2.5}}{86251+1924742x^{0.5}+6612384x+7291509x^{1.5}+6436977x^{2}+3158719x^{2.5}} \right )$

::$x=\left ( \frac{b}{a} \right )$

Well, since Latex is not recognized here, please go to:

http://ellipse-circumference3.blogspot.com


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# Gordon Metric Revisited

We show that Gordon metric belongs to a larger class of geometries, which are responsible to describe the paths of accelerated bodies in moving dielectrics as geodesics in a metric $\hat q {\mu u}$ different from the background one. This map depends only on the background metric and on the motion of the bodies under consideration. As a consequence, this method describes a more general property that concerns the elimination of any kind of force acting on bodies by a suitable change of the substratum metric.

Author: M. Novello; E. Bittencourt

Source: https://archive.org/


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# If 0.0030<x-√30<0.0031, which of the following options is the approxim

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If 0.0030<x-√30<0.0031, which of the following options is the approxim  [#permalink]

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If $$0.0030<x-\sqrt{30}<0.0031$$, which of the following options is the approximation of  $$\frac{1}{\sqrt{30}}-\frac{1}{x}$$?

A. 0.001
B. 0.002
C. 0.0001
D. 0.0002
E. 0.0003

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Re: If 0.0030<x-√30<0.0031, which of the following options is the approxim  [#permalink]

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24 Feb 2017, 01:46
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AustinKL wrote:
If $$0.0030<x-\sqrt{30}<0.0031$$, which of the following options is the approximation of  $$\frac{1}{\sqrt{30}}-\frac{1}{x}$$?

A. 0.001
B. 0.002
C. 0.0001
D. 0.0002
E. 0.0003

Hi

$$0.0030<x-\sqrt{30}<0.0031$$

$$\sqrt{30} + 0.0030 < x < \sqrt{30} + 0.0031$$

$$\frac{1}{\sqrt{30}}-\frac{1}{x}$$ = $$(x - \sqrt{30}) / x * \sqrt{30}$$

$$min = \frac{0.003}{(\sqrt{30} + 0.003)*\sqrt{30}} = \frac{0.003}{30 + 0.003*\sqrt{30}} ≈ \frac{0.003}{30} = \frac{3*0.001}{3*10} = 0.0001$$

$$max = \frac{0.0031}{(\sqrt{30} + 0.0031)*\sqrt{30}} = \frac{0.0031}{30 + 0.0031*\sqrt{30}} ≈ \frac{0.0031}{30} = \frac{0.003}{30} + \frac{0.0001}{30} ≈ \frac{0.003}{30} = 0.0001$$
.

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If 0.0030<x-√30<0.0031, which of the following options is the approxim  [#permalink]

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30 Sep 2018, 12:09
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hazelnut wrote:
If $$0.0030<x-\sqrt{30}<0.0031$$, which of the following options is the approximation of  $$\frac{1}{\sqrt{30}}-\frac{1}{x}$$?

A. 0.001
B. 0.002
C. 0.0001
D. 0.0002
E. 0.0003

here is my approach

The $$\sqrt{30}$$ is about 5.4

Hence x is between $$0.0030<x-5.4<0.0031$$ add 5.4 to both sides so difference between maximum and minimum value is 0.0001

so I clicked on C
If 0.0030<x-√30<0.0031, which of the following options is the approxim &nbs [#permalink] 30 Sep 2018, 12:09
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# Is there any way to hook onto the advancement of a counter?

Question title really says it all - can you hook onto

\stepcounter{mycounter}

which probably boils down to (I may very well be wrong)

\advance \mycounter by 1

Is this a hookable thing?

### Edit:

I'm not sure if this is helpful, but answers in LaTeX3 are okay, too :)

• @Werner (who had just commented \setcounter -> \stepcounter ;) Yes in fact - that would be more useful. (In the end, I'm trying to hook onto page breaks, but atbegshi and everypage aren't doing the trick. So, by looking into hooking counters, I can hook onto the page counter (or whatever, more extensible) and do it that way.) – Sean Allred Jul 11 '13 at 16:35
• I'm not entirely sure whether this question stems from a What does “XY Problem” mean? type of situation. Perhaps it would be useful to actually focus on the problem at hand, rather than fixing a possible work-around. – Werner Jul 11 '13 at 16:40
• Note that unless you are planning to use the code of your hook inside the output routine it will execute at strange places, typically half way down the following page, as TeX will have expanded macros and typeset boxes ahead of the material that finally lands on any given page. – David Carlisle Jul 11 '13 at 16:42
• @DavidCarlisle For my purposes, as long as it updates its counter only when certain sectioning elements have been set, I'm golden. I'm just updating a l3seq stack, I'm not necessarily outputting anything. It may not work, but I had the idea. Werner - specific to me, this is a good example of the XY problem, but that's why I made it more general -- perhaps the functionality is also useful elsewhere. – Sean Allred Jul 11 '13 at 16:47
• Yes but if you look at the l3seq stack from a macro expanding in the main page rather than the output routine it will have an essentially arbitrary value, perhaps reflecting the setting from several pages back as one or more pages might by typeset before the page counter is incremented. (which is why you need \pageref to refer to a page number the value of \thepage (and the value of your stack) are not useful within the document. – David Carlisle Jul 11 '13 at 16:57

In this specific case, yes, but you need to identify the base component of this advancement. The macros involved include \refstepcounter, \stepcounter and \addtocounter. The following are their definitions, taken from latex.ltx:

\def\refstepcounter#1{\stepcounter{#1}%
\protected@edef\@currentlabel
{\csname p@#1\endcsname\csname the#1\endcsname}%
}

\def\stepcounter#1{%
\begingroup
\let\@elt\@stpelt
\csname cl@#1\endcsname
\endgroup}

\refstepcounter uses \stepcounter for advancement, and \stepcounter uses \addtocounter for advancement. So, tapping into \addtocounter is the best. You can update it to hook into arguments #1 (the counter) or #2 (the amount/number).
If you're attempting to hook into page counters, then a \label-\pageref system should be considered instead, due to the asynchronous nature of the output routine. For this, the use of refcount which provides \setcounterpageref and \getpagerefnumber, both of which might be of help.


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## Quantitative Aptitude Questions Quiz Set 10 for IBPS/SBI/SSC

Quantitative Aptitude Questions Quiz Set 10 for IBPS/SBI/SSC

1. One year ago the ratio between Laxman’s and Gopal’s salary was 3: 4. The individual ratios between their last year’s and this year’s salaries are 4 : 5 and 2 : 3 respectively. At present the total of their salary is Rs. 4160. The salary of Laxman now, is

1) Rs. 1600 2) Rs. 2560 3) Rs. 1040 4) 1800 5) None of these

2. When 35 per cent of a number is added to another number, the second number increase by its 20 percent. What is the ratio between the second number and the first number?

1) 4: 7 2) 7: 4 3) 8: 5 4) Data inadequate 5) None of these

3. An amount of money is to be distributed among P, Q and R in the ratio of 5 : 8 : 12 respectively. If the total share of Q and R is four times that of P, what is definitely P’s share?

1) Rs. 3000 2) Rs. 5000 3) Rs. 8000 4) Data inadequate 5) None of these

4. An amount of money is to be divided among P,Q and R in the ratio of 4 : 9 : 16. If R gets 4 times more than P, what is Q’s share in it?

1) Rs. 1800 2) Rs. 2700 3) Rs. 3600 4) Data inadequate 5) None of these

5. Dilip, Ram and Amar started a shop by investing Rs. 2700, Rs. 8100 and Rs. 7200 respectively. At the end of one year, the profit was distributed. If Ram’s share was Rs. 3600 their total profit was

1) Rs. 10800 2) Rs. 11600 3) Rs. 8000 4) 12000 5) None of these

6. A, B and C contract a work for Rs. 550. Together A and B are to do 7/11 of the work.

The share of C should be

1) Rs. 400 2) Rs. 300 3) Rs. 2004) Rs. 183 1/3

5) None of these

7. A sum of money is to be divided among A, B and C in the ratio 2 : 3 : 7. If the total share of A and B together is Rs. 1500 less than C, what is A’s share in it?

1) Rs. 1000 2) Rs. 1500 3) Rs. 2000 4) Data inadequate 5) None of these

8. The daily wage is decreased by 20% and a person now gets Rs. 16 per day. What was his daily wage before the decrease?

1) Rs. 20 2) Rs. 25 3) Rs. 24 4) Rs. 21 5) None of these

9. The daily wage is decreased by 18% and person now gets Rs. 41 per day. What was his daily wage before the decrease?

1) Rs. 51 2) Rs. 50 3) Rs. 454) Rs. 55 5) None of these

10. Due to fall in manpower, the production in a factory decreases by 24\$. By what percent should the working hour be increased to restore the original production?

1) 600/19 5% 2) 400/19 % 3) 650/19 % 4) 450/19% 5) None of these

11. Two numbers are respectively 25% and 20% more than a third. What percentage is the first of the second?

1) 104% 2) 104.16% 3) 104.26% 4) 105% 5) None of these

12.Two numbers are respectively 15% and 84% more than a third. What percentage is the first of the second?

1) 64 ½ % 2) 65 ½ % 3) 63 ½ % 4) 62 ½ % 5) None of these

13. Two numbers are respectively 68% and 26% more than a third. What percentage is the second of the first?

1) 75% 2) 72% 3) 85%4) 78% 5) None of these

14. Pradip spends 40 per cent of his monthly income on food items, and 50 per cent of the remaining on clothes and conveyance. He saves one-third of the remaining amount after spending on food, clothes and conveyance. If he saves Rs. 19,200 every year, what is his monthly income?

1) Rs. 24000 2) Rs. 12000 3) Rs. 16000 4) Rs. 20000 5) None of these

15. The ratio of males and females in a city is 7: 8 and the percentage of children among males and females is 25% and 20% respectively. If the number of adult females in the city is 156800 what is the total population?

1) 245000 2) 367500 3) 196000 4) 171500 5) None of these

 Question No Answers 1 1 2 2 3 4 4 4 5 3 6 3 7 2 8 1 9 2 10 1 11 2 12 4 13 1 14 3 15 2


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# Tag Info

## Hot answers tagged density

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4

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2

The notion of buoyancy really only applies to fluids or very plastic/glassy solids, so might not be something to look at with the polystyrene beads. However, if we want to model it like a liquid, there are also the factors of "surface tension" and "viscosity" to consider: polystyrene beads are often electrostatically bound to each other, creating an ...

2

If we take neutron star material at say a density of $\sim 10^{17}$ kg/m$^{3}$ the neutrons have an internal kinetic energy density of $3 \times 10^{32}$ J/m$^{3}$. So even in a teaspoonful (say 5ml), there is $1.5\times10^{27}$ J of kinetic energy (more than the Sun emits in a second, or a billion or so atom bombs) and this will be released instantaneously. ...

2

So I am guessing you have a a 2 dimensional density? say, kg/m^2? In this case, you will need to get the area of your box -> say it is 2m^2. The mass of the box is 2 * 0.5 = 1 kg. But the mass of your person is 80kg. So your new mass is 81kg. This means the new density will be 81/2 = 40.5kg/m^2. The item will only float if the item has a lower density ...

2

One should always specify whether one is talking about rest mass per unit rest frame volume, $\rho_0 = m_0/V_0$, rest mass per unit observer-frame volume, $D = m_0/(V_0/\gamma) = \gamma\rho_0$, or relativistic mass per unit observer-frame volume, $(\gamma m_0)/(V_0/\gamma) = \gamma^2\rho_0$.1 (I can't imagine the fourth case, relativistic mass per unit rest ...

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


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# Number of permutations in $S_{a+b}$ with $\operatorname{maj}(\pi)=a$ and $\operatorname{maj}(\pi^{-1})=b$

$$\DeclareMathOperator\maj{maj}\DeclareMathOperator\inv{inv}$$Let $$t_{a,b}$$ be the numbers $$t_{a,b} \mathrel{:=} \lvert\{ \pi \in S_{a+b} : \maj(\pi)=a \text{ and } \maj(\pi^{-1})=b \}\rvert.$$ Here, $$S_{a+b}$$ denotes the set of permutations of $$1,2,\dotsc,a+b$$. By a result of Foata, one can also look at the pair of statistics $$(\maj, \inv)$$, and a few other combinations — these pairs of statistics will produce the same numbers.

Now, according to the OEIS entry A090806, it is proved by Garsia–Gessel that $$$$\label{*} \tag{*} \sum_{a,b} t_{a,b} q^a t^b = \prod_{i,j \geq 1} \frac{1}{1-q^i t^j}.$$$$ I cannot see exactly where in their paper one can deduce this.

My attempt: I have tried to prove this myself (mainly by resorting to RSK, the Cauchy identity, and some symmetric function identities). This leads to the following (which appears in Stanley's EC2): $$$$\sum_{n \geq 0} \frac{z^n}{(1-q)^n[n]_q!(1-t)^n [n]_t!} \sum_{\pi \in S_n} t^{\maj(\pi)} q^{\maj(\pi^{-1})} = \prod_{i,j \geq 0} \frac{1}{1-z q^i t^j},$$$$ where $$[n]_q! \mathrel{:=} [1]_q [2]_q \dotsm [n]_q$$, and $$[n]_q = 1+q+q^2+\dotsb + q^{n-1}$$. However, I do not see some short way to deduce the above generating function from this.

Question: Is there some alternative (more recent?) reference where \eqref{} is stated and easily referenced? Alternatively, someone who can see exactly where in the paper \eqref{} is proven?

Garsia, A. M.; Gessel, I., Permutation statistics and partitions, Adv. Math. 31, 288-305 (1979). ZBL0431.05007.