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# Tag Info

## Hot answers tagged hyperparameter

72

It's common to find code snippets that treat $T$ as a hyper-parameter, and attempt to optimize over it in the same way as any other hyper-parameter. This is just wasting computational power: when all other hyper-parameters are fixed, the model’s loss stochastically decreases as the number of trees increases. Intuitive explanation Each tree in a random ...

52

Random search has a probability of 95% of finding a combination of parameters within the 5% optima with only 60 iterations. Also compared to other methods it doesn't bog down in local optima. Check this great blog post at Dato by Alice Zheng, specifically the section Hyperparameter tuning algorithms. I love movies where the underdog wins, and I love machine ...

26

This is how I have trained a xgboost classifier with a 5-fold cross-validation to optimize the F1 score using randomized search for hyperparameter optimization. Note that X and y here should be pandas dataframes. from scipy import stats from xgboost import XGBClassifier from sklearn.model_selection import RandomizedSearchCV, KFold from sklearn.metrics ...

26

As the lead developer of Optunity I'll add my two cents. We have done extensive benchmarks comparing Optunity with the most popular Bayesian solvers (e.g., hyperopt, SMAC, bayesopt) on real-world problems, and the results indicate that PSO is in fact not less efficient in many practical cases. In our benchmark, which consists of tuning SVM classifiers on ...

26

Like you already observed yourself, your choice of features (feature selection) may have an impact on which hyperparameters for your algorithm are optimal, and which hyperparameters you select for your algorithm may have an impact on which choice of features would be optimal. So, yes, if you really really care about squeezing every single percent of ...

25

There are basically four methods: Manual Search: Using knowledge you have about the problem guess parameters and observe the result. Based on that result tweak the parameters. Repeat this process until you find parameters that work well or you run out of time. Grid Search: Using knowledge you have about the problem identify ranges for the hyperparameters. ...

21

I suspect what is meant by hyper-parameter depends on the context, but here goes: I would say that the parameters of a model are those that are directly fitted to the data, and the hyper-parameters are those parameters that are set by the user or which are indirectly fitted to the data. For instance in ridge regression, the parameters are the regression ...

20

The effects of this bias can be very great. A good demonstration of this is given by the open machine learning competitions that feature in some machine learning conferences. These generally have a training set, a validation set and a test set. The competitors don't get to see the labels for either the validation set or the test set (obviously). The ...

20

Unfortunately, it doesn't work that way. Hyperparameters cooperate in hard-to-predict ways. For example, a bit extreme to make the point. You have no hidden layers, in other words, you are fitting a logistic regression. A logistic regression will usually not really overfit. So you use a relatively big learning rate and a lot of epochs, and find that that ...

18

The term hyperparameter is pretty vague. I will use it to refer to a parameter that is in a higher level of the hierarchy than the other parameters. For an example, consider a regression model with a known variance (1 in this case) $$y \sim N(X\beta,I)$$ and then a prior on the parameters, e.g. $$\beta \sim N(0,\lambda I)$$ Here $\lambda$ ...

18

Number of trees is not a parameter that should be tuned, but just set large enough usually. There is no risk of overfitting in random forest with growing number of trees, as they are trained independently from each other. See our paper for more information about this: https://arxiv.org/abs/1705.05654 Max depth is a parameter that most of the times should be ...

17

Nested cross validation explained without nesting Here's how I see (nested) cross validation and model building. Note that I'm chemist and like you look from the application side to the model building process (see below). My main point here is from my point of view I don't need a dedicated nested variety of cross validation. I need a validation method (e.g. ...

17

Is hyperparameter tuning on sample of dataset a bad idea? A: Yes, because you risk overfitting (the hyperparameters) on that specific test set resulting from your chosen train-test split. Do I limit my classification accuracy? A: Yes, but common machine learning wisdom is: with your optimal hyperparameters, say $\lambda^*$, refit your model(s) on the ...

17

In addition to Jim's (+1) answer: For some classifiers, the hyper-parameter values are dependent on the number of training examples, for instance for a linear SVM, the primal optimization problem is $\mathrm{min} \frac12\|w\|^2 + C\sum_{i=1}^\ell \xi_i$ subject to $y_i(x_i\cdot w _ b) \geq 1 - \xi_i, \quad \mathrm{and} \quad \xi_i \geq 0 \quad \forall i$ ...

15

5x2cv as far as I have seen in the literature, always refer to a 5 repetition of a 2-fold. There is no nesting at all. do a 2-fold (50/50 split between train and test), repeat it 4 more times. The 5x2cv was popularised by the paper Approximate statistical tests for comparing supervised classification learning algorithms by Dietterich as a way of obtaining ...

15

Here's the "default" nested cross-validation procedure to compare between a fixed set of models (e.g. grid search): Randomly split the dataset into $K$ folds. For $i$ from 1 to $K$: Let test be fold $i$. Let trainval be all the data except that which is in test. Randomly split trainval into $L$ subfolds $(i, 1), (i, 2), \dots, (i, L)$ . So, each subfold $(... 14 A wide variety of methods exist. They can be largely partitioned in random/undirected search methods (like grid search or random search) and direct methods. Be aware, though, that they all require testing a considerable amount of hyperparameter settings unless you get lucky (hundreds at least, depends on the number of parameters). In the class of direct ... 14 results are sensitive to parameters of the surrogate model, which are typically fixed at some value; this underestimates uncertainty; or else you have to be fully Bayesian and marginalize over hyper parameter distributions, which can be expensive and unwieldy. it takes a dozen or so samples to get a good surrogate surface in 2 or 3 dimensions of search space;... 14 As a result of doing that you will also overfit the validation set (the more so the more you tuned the hyperparameters - if you tried two or three configurations, the effect is less than if you did some systematic search e.g. using the Gaussian process approach). The standard solution to this would be to not just have a training and validation set, but a ... 14 The decision threshold creates a trade-off between the number of positives that you predict and the number of negatives that you predict -- because, tautologically, increasing the decision threshold will decrease the number of positives that you predict and increase the number of negatives that you predict. The decision threshold is not a hyper-parameter in ... 13 Look again at the graphic from the paper (Figure 1). Say that you have two parameters, with 3x3 grid search you check only three different parameter values from each of the parameters (three rows and three columns on the plot on the left), while with random search you check nine (!) different parameter values of each of the parameters (nine distinct rows and ... 13 In regards to the evidence in regards to the claim, I believe the only evidence supporting the claim can be found on figure 4 in their paper. They show the final results under a range of different values for$\beta_1$,$\beta_2$and$\alpha$. Personally, I don't find their argument convincing, in particular because they do not present results across a ... 12 I think that the answer here is the same as everywhere in data science: it depends on the data :-) It might happen that one method outperforms another (here https://arimo.com/data-science/2016/bayesian-optimization-hyperparameter-tuning/ people compare Bayesian hyperparameter optimization and achieve a better result on the San Francisco crime kaggle ... 12 Information is leaked because you're using the validation data to make hyper-parameter choices. Essentially, you're creating a complicated optimization problem: minimize the loss over hyper-parameters$\phi$as evaluated against the validation data, where these hyper-parameters regularize a neural network model that has parameters$\theta$trained by use of ... 12 But varying the threshold will change the predicted classifications. Does this mean the threshold is a hyperparameter? Yup, it does, sorta. It's a hyperparameter of you decision rule, but not the underlying regression. If so, why is it (for example) not possible to easily search over a grid of thresholds using scikit-learn's GridSearchCV method (as you ... 11 Adam learns the learning rates itself, on a per-parameter basis. The parameters$\beta_1$and$\beta_2\$ don't directly define the learning rate, just the timescales over which the learned learning rates decay. If they decay really fast, then the learning rates will jump about all over the place. If they decay slowly, it will take ages for the learning rates ...

11

As you noted, tol is the tolerance for the stopping criteria. This tells scikit to stop searching for a minimum (or maximum) once some tolerance is achieved, i.e. once you're close enough. tol will change depending on the objective function being minimized and the algorithm they use to find the minimum, and thus will depend on the model you are fitting. ...

10

Look no further! Yoshua Bengio published one of my favorite applied papers, one that I recommend to all new machine learning engineers when they start training neural nets: Practical recommendations for gradient-based training of deep architectures. To get his perspective on hyperparameter turning: including learning rate, learning rate schedule, early ...

10

A hyperparameter is simply a parameter that impacts, completely or partly, other parameters. They do not directly solve the optimization problem you face, but rather optimize parameters that can solve the problem (hence the hyper, because they are not part of the optimization problem, but rather are "addons"). For what I've seen, but I have no reference, ...

10

"Why don't we just learn the hyper parameters?" It's a great question! I'll try to provide a more general answer. The TL;DR answer is that you can definitely learn hyperparameters, just not from the same data. Read on for a slightly more detailed reply. A hyperparameter typically corresponds to a setting of the learning algorithm, rather than one of its ...

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


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# Evaluate $\int_0^1\ln(1-x)\ln x\ln(1+x) \mathrm{dx}$

Evaluate
$$\int_0^1\ln(1-x)\ln x\ln(1+x) \mathrm{dx}$$

#### Solutions Collecting From Web of "Evaluate $\int_0^1\ln(1-x)\ln x\ln(1+x) \mathrm{dx}$"

I get (and verified by Mathematica)
$$\int_0^1 \log(1-x) \log x \log(1+x) \, dx = -6 + 4 \log 2 – \log^2 2 + \frac{5}{2} \zeta(2) – 3\zeta(2) \log 2 + \frac{21}{8} \zeta(3).$$

Transforming to a double sum.

\begin{align*}
\log(1+x)\log(1-x) & = \left(\sum_{k=1}^{\infty} \dfrac{(-x)^k}k \right)\left(\sum_{k=1}^{\infty} \dfrac{x^k}k \right)\\
& = \sum_{k=1}^{\infty}\sum_{m=1}^{\infty} \dfrac{(-1)^k x^{k+m}}{km}.
\end{align*}
Using the integral $$\int_0^1 x^r \log x \, dx = – \frac{1}{(1+r)^2},$$
we have
$$\int_0^1 \log(1-x) \log x \log(1+x) \, dx = \sum_{k=1}^{\infty} \sum_{m=1}^{\infty} \frac{(-1)^{k+1}}{km(k+m+1)^2}.$$

Evaluating the inner sum.

Partial fractions decomposition on the summand in $m$ yields
\begin{align*}
\sum_{m=1}^{\infty} \frac{1}{m(k+m+1)^2} &= \sum_{m=1}^{\infty} \left(\frac{1}{(k+1)^2 m} – \frac{1}{(k+1)^2 (m+k+1)} – \frac{1}{(k+1) (m+k+1)^2}\right) \\
&= \sum_{m=1}^{k+1} \frac{1}{(k+1)^2 m} – \sum_{m=k+2}^{\infty} \frac{1}{(k+1)m^2} \\
&= \frac{H_{k+1}}{(k+1)^2} – \frac{\zeta(2)}{k+1} + \frac{H^{(2)}_{k+1}}{k+1},
\end{align*}
where $H^{(r)}_n = \sum_{i=1}^n i^{-r}$, the $n$th $r$-harmonic number.

Now we’re left with evaluating
$$\sum_{k=1}^{\infty} \frac{(-1)^{k+1}H_{k+1}}{k(k+1)^2} – \sum_{k=1}^{\infty} \frac{(-1)^{k+1} \zeta(2)}{k(k+1)} + \sum_{k=1}^{\infty} \frac{(-1)^{k+1} H^{(2)}_{k+1}}{k(k+1)}. \tag{1}$$
We take the three sums in Eq. (1) in turn.

The first sum in Eq. (1).

For the first sum, applying partial fractions decomposition yields
\begin{align*}
\sum_{k=1}^{\infty} \frac{(-1)^{k+1}H_{k+1}}{k(k+1)^2} &= \sum_{k=1}^{\infty} \frac{(-1)^{k+1}H_{k+1}}{k} – \sum_{k=1}^{\infty} \frac{(-1)^{k+1}H_{k+1}}{(k+1)} – \sum_{k=1}^{\infty} \frac{(-1)^{k+1}H_{k+1}}{(k+1)^2} \\
&= \sum_{k=1}^{\infty} \frac{(-1)^{k+1}H_k}{k} + \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k(k+1)} + \sum_{k=1}^{\infty} \frac{(-1)^{k+1}H_k}{k} – 1 + \sum_{k=1}^{\infty} \frac{(-1)^{k+1}H_k}{k^2} – 1 \\
&= -2 + \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k(k+1)} + 2\sum_{k=1}^{\infty} \frac{(-1)^{k+1}H_k}{k} + \sum_{k=1}^{\infty} \frac{(-1)^{k+1}H_k}{k^2} \\
&= -3 + 2 \log 2 + 2\sum_{k=1}^{\infty} \frac{(-1)^{k+1}H_k}{k} + \sum_{k=1}^{\infty} \frac{(-1)^{k+1}H_k}{k^2},
\end{align*}
where in the last step we used the evaluation for the second sum we’re about to do.

The second sum in Eq. (1).

For the second sum in Equation (1), applying partial fractions decomposition yields
\begin{align*}
\zeta(2) \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k(k+1)} &= \zeta(2) \left(\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k} + \sum_{k=1}^{\infty}\frac{(-1)^{k+2}}{k+1}\right) \\
&= \zeta(2) \left(2\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k} -1\right) \\
&= 2 \zeta(2) \log 2 – \zeta(2).
\end{align*}

The third sum in Eq. (1).

For the third sum in Equation (1), applying partial fractions decomposition yields
\begin{align*}
\sum_{k=1}^{\infty} \frac{(-1)^{k+1} H^{(2)}_{k+1}}{k(k+1)} &= \sum_{k=1}^{\infty} \frac{(-1)^{k+1} H^{(2)}_{k+1}}{k} – \sum_{k=1}^{\infty} \frac{(-1)^{k+1} H^{(2)}_{k+1}}{(k+1)} \\
&= \sum_{k=1}^{\infty} \frac{(-1)^{k+1} H^{(2)}_k}{k} + \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k(k+1)^2} + \sum_{k=1}^{\infty} \frac{(-1)^{k+1} H^{(2)}_k}{k} – 1 \\
&= -1 + 2\sum_{k=1}^{\infty} \frac{(-1)^{k+1} H^{(2)}_k}{k} + \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k} – \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k+1} – \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{(k+1)^2} \\
&= -1 + 2\sum_{k=1}^{\infty} \frac{(-1)^{k+1} H^{(2)}_k}{k} + 2 \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k} – 1 + \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k^2} – 1\\
&= -3 + 2 \log 2 + \frac{1}{2} \zeta(2) + 2\sum_{k=1}^{\infty} \frac{(-1)^{k+1} H^{(2)}_k}{k} ,\\
\end{align*}
where in the last step we use the identity, for $p > 1$, $$\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k^p} = (1 – 2^{1-p}) \zeta(p)$$

Combining the results.

Putting all this together, we have that the integral evaluates to
$$-6 + 4 \log 2 + \frac{3}{2} \zeta(2) – 2 \zeta(2) \log 2 + 2\sum_{k=1}^{\infty} \frac{(-1)^{k+1}H_k}{k} + \sum_{k=1}^{\infty} \frac{(-1)^{k+1}H_k}{k^2} + 2\sum_{k=1}^{\infty} \frac{(-1)^{k+1} H^{(2)}_k}{k}.$$

Evaluating the alternating Euler sums.

The three remaining sums all have similar forms. Letting $$A(p,q) = \sum_{k=1}^{\infty} \frac{(-1)^{k+1}H^{(p)}_k}{k^q},$$ we have $A(1,1)$, $A(1,2)$, and $A(2,1)$ left to evaluate. Sums of the form $A(p,q)$ are known as alternating Euler sums. Euler sums and their close variants are an ongoing research area. These three, though, seem like the three simplest alternating Euler sums, and one would think that there would be relatively simple ways to evaluate them. However, despite a decent amount of work today I could not find simple proofs for any of them, either on my own or in the literature.

At any rate, we have $A(1,1) = \frac{1}{2} \zeta(2) – \frac{1}{2} \log^2 2$, $A(1,2) = \frac{5}{8} \zeta(3)$, and $A(2,1) = \zeta(3) – \frac{1}{2}\zeta(2) \log 2$.

(Here’s a link to my question where I ask for a proof for the $A(1,1)$, $A(1,2)$, and $A(2,1)$ evaluations. There are multiple nice answers there. In particular, Robjohn’s three answers show how to do the evaluations by manipulating the sums, the same approach I was trying here.)

Finally.

Therefore,
$$\int_0^1 \log(1-x) \log x \log(1+x) \, dx = -6 + 4 \log 2 – \log^2 2 + \frac{5}{2} \zeta(2) – 3\zeta(2) \log 2 + \frac{21}{8} \zeta(3).$$

\begin{align}
\log(1+x)\log(1-x) & = \left(\sum_{k=1}^{\infty} \dfrac{(-x)^k}k \right)\left(\sum_{k=1}^{\infty} \dfrac{x^k}k \right)\\
& = \sum_{k=1}^{\infty}\sum_{m=1}^{\infty} \dfrac{(-1)^k x^{k+m}}{km}\\
& = \sum_{l=2}^{\infty} \left(\sum_{k=1}^{l-1} \dfrac{(-1)^k}{k(l-k)} \right) x^l \\
& = \sum_{l=1}^{\infty} \left(\sum_{k=1}^{2l-1} \dfrac{(-1)^k}{k(2l-k)} \right) x^{2l}
\end{align}
Now $$\int_0^1 x^{2l} \log(x) dx = -\dfrac1{(2l+1)^2}$$
Hence, the integral is
$$\sum_{l=2}^{\infty} \left(\sum_{k=1}^{l-1} \dfrac{(-1)^k}{k(l-k)} \right) x^l = \sum_{l=1}^{\infty} \left(\sum_{k=1}^{2l-1} \dfrac{(-1)^{k-1}}{k(2l-k)} \right)\dfrac1{(2l+1)^2}$$

P.S: Bit too long for a comment.

Mathematica gives for the final result:
$$\frac{21 \zeta (3)}{8}-6-\log ^2(2)+\log (16)-\frac{1}{12} \pi ^2 (\log(64)-5)$$
The indefinite integral is really really long..

Regarding a series what’s wrong with a Maclaurin series around, say, $x=1/e$?

format: plain


text:
# HSB

Color model that describes colors in the same way as the human eye perceives them, using hue, saturation (also known as chroma) and brightness (or luminance). The hue is defined by its position in a color circle and is specified by an angle lying between 0 and 360?. The saturation corresponds to the amount of gray in the color mixture and has values of between 0 percent for gray and 100 percent for pure color. The values for brightness also range from 0 percent for black and 100 percent for white.


text:
We share R tutorials from scientists at academic and scientific institutions with a goal to give everyone in the world access to a free knowledge. Our tutorials cover different topics including statistics, data manipulation and visualization!
Introduction

# Streamline your analyses linking R to SAS: the workfloweR experiment

We all know R is the first choice for statistical analysis and data visualisation, but what about big data munging? tidyverse (or we’d better say hadleyverse) has been doing a lot in this field, nevertheless it is often the case this kind of activities being handled from some other coding language. Moreover, sometimes you get as an input pieces of analyses performed with other kind of languages or, what is worst, piece of databases packed in proprietary format (like .dta .xpt and other). So let’s assume you are an R enthusiast like I am, and you do with R all of your work, reporting included, wouldn’t be great to have some nitty gritty way to merge together all these languages in a streamlined workflow?
Yes, we all know great products like microsoft Azure and sas viya, but you know what? They don’t come free, and this can sometime become an obstacle. Moreover all of them involve some kind of sophisticated setup to go trough. But what if we could reach some useful results just leveraging a useful r package and a cleaver setup? We actually can do this and I’ll show you how within coming paragraphs.

## The main character: rio package

I met rio package some years ago, and since then, I never stopped using it. What rio package basically does is automagically guessing the file type you are trying to import and subsequently calling the proper function to effectively import within your R workspace. All you have to do is running the import() function, enclosing in brackets the complete path to your data file, or relative path if within your working directory. be sure to include file extension within the path string.
As import(), rio also comes with an export() function, which does exactly what you are guessing: exporting your r object into a user-defined file. to complete the suite we find convert() which takes a file as input and convert it into a user-defined output file.
How all this comes in help for our purposes? this is actually our main block: we are going to use rio to transform output from one given language into the input for R scripts or any other language. so what about the second block, our cleaver setup?

## The ancillary character: a cleaver setup

So at the moment we know how to take non-R output as R-input and how to export R output to non-R languages, but how do we structure this in an ordered and clear manner? I came up with the following logical flow:

### The data step

As you can see we first have a data step. this logical step involves keeping all data pieces in one physical location which all languages script will have to reference to get and release data files. We are going to teach to every language to point to that location when data loading or production is involved.

### The coding step

The second step is represented by some languages logos. it stands for the logical step involving actual code production from different languages. within this step we consider different scripts, each of them performing its tasks pointing to the common data location. For sure we have to recognise the R script prominence, since within this script we convert, if needed, data coming from for instance from sas to data eatable from spss. it could even be the case that a proper R script is set aside only to perform this kind of task, without impacting our “real” R script where real analyses are performed.

Why do we have arrows coming and going from first and second step? simply because every language can take and put files into data location.

### The reporting step

last step is the reporting one: your analyses are finalised and you want to share them with your stakeholder, what are you going to use for it? If you are a real R enthusiast you will for sure fire up an Rmarkdown file, and our logical flow is here to help.
Rmarkdown is a powerful tool which combines main advantages of markdown with powerful features of R languages. You can actually dynamically embeed R code results within you markdown file, having the R code recompiled every time you compile the main file. This means that if your analyses changes your report will change as well, and everything will always be in sync without requiring you to get trapped within the copy and paste swamp.

Where your report will take the needed data from? from data location, of course. and this complete our workflow. But now let’s have a look to a real example:

## A real example: linking togheter R and sas

We start our analysis munging some heavy data within sas environment. We read our data from a txt/csv file into sas, creating two sas7bdat tables which we need to merge togheter in order to associate dimensional data to some kind of customer list. so we run our two or three data step and we end up having one complete table. This table, in was language will be referenced as data.complete_data. We now want to apply some cool and trending statistical models on this data, just to show we are real data scientists, where should we do this if not in R? So here we go, creating a new R script and loading our data in. How do we do it? Leveraging rio! we will have just to run the following line of code, leaving all the hard job to the package:

complete_data <- import('data/complete_data.sas7bdat')

This line of code will create a data.frame object ready to be analysed with our favourite language. and now it’s time to go for the last step: reporting analyses results. this is actually an easy step, since we could also merge it with the r script one: once we can read our sas-munged data into R why not readying them directly within the Rmarkdown file? Anyway, we can handle the last step in different complementary ways:

• producing all final results within a separate R script and saving them to an Rdata object to be loaded within the Rmarkdown file
• producing pieces of results in sas and R joining them only within the Rmarkdown file through the import() function.
• performing the whole analyses from munged data within the Rmarkdown file.

The trade-off you have to consider when choosing within this continuum is between weight and flexibility: loading final results into rmarkdown will may results in lighter objects then raw datasets but will also generally reduce the possibility to let the user interact with your data, for instance with interactive components like the ones from shiny framework. as is often the case, the experience will teach you how to move from one side to the other.

Ok, so everything seems finalised, doesn’t it? not quite, since we have a bonus track: the workfloweR experiment.

## Bonus track: the workfloweR experiment

Since I often use more than one programming language when performing my job and I am a lazy programmer, I decided to automize a bit all the workflow described within previous paragraph, that is where workfloweR comes from.

What it does is asking you ( through a simple shiny app) to define a path where you would like to initialize the analysis workspace and select languages you are going to use for your analyses. Once the selection is done and the usual initialize button is pressed, workfloweR takes care of creating within submitted path all needed data folders and languages scripts. It morover adds proper references to data folder within different scripts, for instance employing the libname statement to create a sas library within the sas script. finally, worflower creates an rmarkdown report ready to be employed to share results of your wonderful work.

## How to try workflower

An updated version of workflower is currently hosted on Github an you can try it following this steps:

Clone or download the repository from Github

source ‘app.R’

specify where to initialize your analysis workspace

specify the languages you are going to use for your analyses

initialize the workspace

enjoy !

Just open up your pre-initialised R or sas scripts and start performing you analyses. When you will be done with your work an Rmarkdown file will be waiting for you within the deliverable folder.

## May I ask for your help?

If you enjoy the idea of workflower, would you mind give me an help to further expand it? As mentioned you can find the project on github, feel free to clone it and work on it. I can see two main developement paths:

• Move the all stuff into an R package, which will make its use really easy directly from Github via devtools
• Add more languages beside R and sas.

Languages integration is a powerfull way to let best feature of each one shine, that’s why I thought it was a good idea to share this pragmating and simple approach to exchange data an sum up results from analyses performed with R and other languages. Let me know if you find it useful or if you have got any other similar trick.


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Textbook by: Denny Burzynski, Wade Ellis. E-mail the authors

# Subtraction of Whole Numbers

Module by: Wade Ellis, Denny Burzynski. E-mail the authorsEdited By: Math Editors

Summary: This module is from Fundamentals of Mathematics by Denny Burzynski and Wade Ellis, Jr. This module discusses how to subtract whole numbers. By the end of this module, students should be able to understand the subtraction process, subtract whole numbers, and use a calculator to subtract one whole number from another whole number.

## Section Overview

• Subtraction
• Subtraction as the Opposite of Addition
• The Subtraction Process
• Subtraction Involving Borrowing
• Borrowing From Zero
• Calculators

## Subtraction

### Subtraction

Subtraction is the process of determining the remainder when part of the total is removed.

Suppose the sum of two whole numbers is 11, and from 11 we remove 4. Using the number line to help our visualization, we see that if we are located at 11 and move 4 units to the left, and thus remove 4 units, we will be located at 7. Thus, 7 units remain when we remove 4 units from 11 units.

### The Minus Symbol

The minus symbol (-) is used to indicate subtraction. For example, 114114 size 12{"11" - 4} {} indicates that 4 is to be subtracted from 11.

### Minuend

The number immediately in front of or the minus symbol is called the minuend, and it represents the original number of units.

### Subtrahend

The number immediately following or below the minus symbol is called the subtrahend, and it represents the number of units to be removed.

### Difference

The result of the subtraction is called the difference of the two numbers. For example, in 114=7114=7 size 12{"11" - 4=7} {}, 11 is the minuend, 4 is the subtrahend, and 7 is the difference.

## Subtraction as the Opposite of Addition

Subtraction can be thought of as the opposite of addition. We show this in the problems in Sample Set A.

### Sample Set A

#### Example 1

85=385=3 size 12{8 - 5=3} {} since 3+5=83+5=8 size 12{3+5=8} {}.

#### Example 2

93=693=6 size 12{9 - 3=6} {} since 6+3=96+3=9 size 12{6+3=9} {}.

### Practice Set A

Complete the following statements.

#### Exercise 1

75=75= size 12{7 - 5={}} {}

since

+5=7+5=7 size 12{+5=7} {}.

##### Solution

7-5=27-5=2 since 2+5=72+5=7

#### Exercise 2

91=91= size 12{9 - 1={}} {}

since

+1=9+1=9 size 12{+1=9} {}.

##### Solution

9-1=89-1=8 since 8+1=98+1=9

#### Exercise 3

178=178= size 12{"17" - 8={}} {}

since

+8=17+8=17 size 12{+8="17"} {}.

##### Solution

17-8=917-8=9 since 9+8=179+8=17

## The Subtraction Process

We'll study the process of the subtraction of two whole numbers by considering the difference between 48 and 35.

which we write as 13.

### Example 3: The Process of Subtracting Whole Numbers

To subtract two whole numbers,

The process

1. Write the numbers vertically, placing corresponding positions in the same column.

48 - 35 ̲ 48 - 35 ̲

2. Subtract the digits in each column. Start at the right, in the ones position, and move to the left, placing the difference at the bottom.

48 - 35 ̲ 13 48 - 35 ̲ 13

### Sample Set B

Perform the following subtractions.

#### Example 4

275 - 142 ̲ 133 275 - 142 ̲ 133

5 2 = 3 . 7 4 = 3 . 2 1 = 1 . 5 2 = 3 . 7 4 = 3 . 2 1 = 1 . alignr { stack { size 12{5 - 2=3 "." } {} # size 12{7 - 4=3 "." } {} # size 12{2 - 1=1 "." } {} } } {}

#### Example 5

46,042 -   1,031 ̲ 45,011 46,042 -   1,031 ̲ 45,011

2 1 = 1 . 4 3 = 1 . 0 0 = 0 . 6 1 = 5 . 4 0 = 4 . 2 1 = 1 . 4 3 = 1 . 0 0 = 0 . 6 1 = 5 . 4 0 = 4 . alignr { stack { size 12{2 - 1=1 "." } {} # size 12{4 - 3=1 "." } {} # size 12{0 - 0=0 "." } {} # size 12{6 - 1=5 "." } {} # size 12{4 - 0=4 "." } {} } } {}

#### Example 6

Find the difference between 977 and 235.

Write the numbers vertically, placing the larger number on top. Line up the columns properly.

977 - 235 ̲ 742 977 - 235 ̲ 742

The difference between 977 and 235 is 742.

#### Example 7

In Keys County in 1987, there were 809 cable television installations. In Flags County in 1987, there were 1,159 cable television installations. How many more cable television installations were there in Flags County than in Keys County in 1987?

We need to determine the difference between 1,159 and 809.

There were 350 more cable television installations in Flags County than in Keys County in 1987.

### Practice Set B

Perform the following subtractions.

#### Exercise 4

534 - 203 ̲ 534 - 203 ̲

331

#### Exercise 5

857 -   43 ̲ 857 -   43 ̲

814

#### Exercise 6

95,628 - 34,510 ̲ 95,628 - 34,510 ̲

61,118

#### Exercise 7

11,005 -   1,005 ̲ 11,005 -   1,005 ̲

10,000

#### Exercise 8

Find the difference between 88,526 and 26,412.

##### Solution

62,114

In each of these problems, each bottom digit is less than the corresponding top digit. This may not always be the case. We will examine the case where the bottom digit is greater than the corresponding top digit in the next section.

## Subtraction Involving Borrowing

### Minuend and Subtrahend

It often happens in the subtraction of two whole numbers that a digit in the minuend (top number) will be less than the digit in the same position in the subtrahend (bottom number). This happens when we subtract 27 from 84.

84 - 27 ̲ 84 - 27 ̲

We do not have a name for 4747 size 12{4 - 7} {}. We need to rename 84 in order to continue. We'll do so as follows:

Our new name for 84 is 7 tens + 14 ones.

= 57 = 57

Notice that we converted 8 tens to 7 tens + 1 ten, and then we converted the 1 ten to 10 ones. We then had 14 ones and were able to perform the subtraction.

### Borrowing

The process of borrowing (converting) is illustrated in the problems of Sample Set C.

### Sample Set C

#### Example 8

1. Borrow 1 ten from the 8 tens. This leaves 7 tens.
2. Convert the 1 ten to 10 ones.
3. Add 10 ones to 4 ones to get 14 ones.

#### Example 9

1. Borrow 1 hundred from the 6 hundreds. This leaves 5 hundreds.
2. Convert the 1 hundred to 10 tens.
3. Add 10 tens to 7 tens to get 17 tens.

### Practice Set C

Perform the following subtractions. Show the expanded form for the first three problems.

#### Exercise 9

53 - 35 ̲ 53 - 35 ̲

#### Exercise 10

76 - 28 ̲ 76 - 28 ̲

#### Exercise 11

872 - 565 ̲ 872 - 565 ̲

#### Exercise 12

441 - 356 ̲ 441 - 356 ̲

85

#### Exercise 13

775 -   66 ̲ 775 -   66 ̲

709

#### Exercise 14

5,663 - 2,559 ̲ 5,663 - 2,559 ̲

3,104

### Borrowing More Than Once

Sometimes it is necessary to borrow more than once. This is shown in the problems in Section 12.

### Sample Set D

Perform the Subtractions. Borrowing more than once if necessary

#### Example 10

1. Borrow 1 ten from the 4 tens. This leaves 3 tens.
2. Convert the 1 ten to 10 ones.
3. Add 10 ones to 1 one to get 11 ones. We can now perform 118118 size 12{"11" - 8} {}.
4. Borrow 1 hundred from the 6 hundreds. This leaves 5 hundreds.
5. Convert the 1 hundred to 10 tens.
6. Add 10 tens to 3 tens to get 13 tens.
7. Now 135=8135=8 size 12{"13" - 5=8} {}.
8. 53=253=2 size 12{5 - 3=2} {}.

#### Example 11

1. Borrow 1 ten from the 3 tens. This leaves 2 tens.
2. Convert the 1 ten to 10 ones.
3. Add 10 ones to 4 ones to get 14 ones. We can now perform 145145 size 12{"14" - 5} {}.
4. Borrow 1 hundred from the 5 hundreds. This leaves 4 hundreds.
5. Convert the 1 hundred to 10 tens.
6. Add 10 tens to 2 tens to get 12 tens. We can now perform 128=4128=4 size 12{"12" - 8=4} {}.
7. Finally, 40=440=4 size 12{4 - 0=4} {}.

#### Example 12

71529 - 6952 ̲ 71529 - 6952 ̲ alignr { stack { size 12{"71529"} {} # size 12{ {underline { - "6952"}} } {} } } {}

After borrowing, we have

### Practice Set D

Perform the following subtractions.

#### Exercise 15

526 - 358 ̲ 526 - 358 ̲

168

#### Exercise 16

63,419 -   7,779 ̲ 63,419 -   7,779 ̲

55,640

#### Exercise 17

4,312 - 3,123 ̲ 4,312 - 3,123 ̲

1,189

## Borrowing from Zero

It often happens in a subtraction problem that we have to borrow from one or more zeros. This occurs in problems such as

1. 503 -   37 ̲ 503 -   37 ̲

and
2. 5000 -     37 ̲ 5000 -     37 ̲

We'll examine each case.

### Example 13: Borrowing from a single zero.

Consider the problem 503 -   37 ̲ 503 -   37 ̲

Since we do not have a name for 3737 size 12{3 - 7} {}, we must borrow from 0.

Since there are no tens to borrow, we must borrow 1 hundred. One hundred = 10 tens.

We can now borrow 1 ten from 10 tens (leaving 9 tens). One ten = 10 ones and 10 ones + 3 ones = 13 ones.

Now we can suggest the following method for borrowing from a single zero.

### Borrowing from a Single Zero

To borrow from a single zero,

1. Decrease the digit to the immediate left of zero by one.
2. Draw a line through the zero and make it a 10.
3. Proceed to subtract as usual.

### Sample Set E

#### Example 14

Perform this subtraction.

503 -   37 ̲ 503 -   37 ̲

The number 503 contains a single zero

1. The number to the immediate left of 0 is 5. Decrease 5 by 1.

5 1 = 4 5 1 = 4 size 12{5 - 1=4} {}

2. Draw a line through the zero and make it a 10.
3. Borrow from the 10 and proceed.

1 ten+10 ones1 ten+10 ones

10 ones+ 3 ones=13 ones10 ones+3 ones=13 ones

### Practice Set E

Perform each subtraction.

#### Exercise 18

906 -   18 ̲ 906 -   18 ̲

888

#### Exercise 19

5102 -   559 ̲ 5102 -   559 ̲

4,543

#### Exercise 20

9055 -   386 ̲ 9055 -   386 ̲

8,669

### Example 15: Borrowing from a group of zeros

Consider the problem 5000 -     37 ̲ 5000 -     37 ̲

In this case, we have a group of zeros.

Since we cannot borrow any tens or hundreds, we must borrow 1 thousand. One thousand = 10 hundreds.

We can now borrow 1 hundred from 10 hundreds. One hundred = 10 tens.

We can now borrow 1 ten from 10 tens. One ten = 10 ones.

From observations made in this procedure we can suggest the following method for borrowing from a group of zeros.

### Borrowing from a Group of zeros

To borrow from a group of zeros,

1. Decrease the digit to the immediate left of the group of zeros by one.
2. Draw a line through each zero in the group and make it a 9, except the rightmost zero, make it 10.
3. Proceed to subtract as usual.

### Sample Set F

Perform each subtraction.

#### Example 16

40,000 -     125 ̲ 40,000 -     125 ̲

The number 40,000 contains a group of zeros.

1. The number to the immediate left of the group is 4. Decrease 4 by 1.

4 1 = 3 4 1 = 3 size 12{4 - 1=3} {}

2. Make each 0, except the rightmost one, 9. Make the rightmost 0 a 10.

3. Subtract as usual.

#### Example 17

8,000,006 -       41,107 ̲ 8,000,006 -       41,107 ̲

The number 8,000,006 contains a group of zeros.

1. The number to the immediate left of the group is 8. Decrease 8 by 1.

8 1 = 7 8 1 = 7 size 12{8 - 1=7} {}

2. Make each zero, except the rightmost one, 9. Make the rightmost 0 a 10.

3. To perform the subtraction, we’ll need to borrow from the ten.

1 ten = 10 ones 10 ones + 6 ones = 16 ones 1 ten = 10 ones 10 ones + 6 ones = 16 ones

### Practice Set F

Perform each subtraction.

#### Exercise 21

21,007 -   4,873 ̲ 21,007 -   4,873 ̲

16,134

#### Exercise 22

10,004 -   5,165 ̲ 10,004 -   5,165 ̲

4,839

#### Exercise 23

16,000,000 -      201,060 ̲ 16,000,000 -      201,060 ̲

15,789,940

## Calculators

In practice, calculators are used to find the difference between two whole numbers.

### Sample Set G

Find the difference between 1006 and 284.

 Display Reads Type 1006 1006 Press − − size 12{ - } {} 1006 Type 284 284 Press = 722

The difference between 1006 and 284 is 722.

(What happens if you type 284 first and then 1006? We'll study such numbers in (Reference)Chapter 10.)

### Practice Set G

#### Exercise 24

Use a calculator to find the difference between 7338 and 2809.

4,529

#### Exercise 25

Use a calculator to find the difference between 31,060,001 and 8,591,774.

22,468,227

## Exercises

For the following problems, perform the subtractions. You may check each difference with a calculator.

### Exercise 26

15 -   8 ̲ 15 -   8 ̲

7

### Exercise 27

19 -   8 ̲ 19 -   8 ̲

### Exercise 28

11 -   5 ̲ 11 -   5 ̲

6

### Exercise 29

14 -   6 ̲ 14 -   6 ̲

### Exercise 30

12 -   9 ̲ 12 -   9 ̲

3

### Exercise 31

56 - 12 ̲ 56 - 12 ̲

### Exercise 32

74 - 33 ̲ 74 - 33 ̲

41

### Exercise 33

80 - 61 ̲ 80 - 61 ̲

### Exercise 34

350 - 141 ̲ 350 - 141 ̲

209

### Exercise 35

800 - 650 ̲ 800 - 650 ̲

### Exercise 36

35,002 - 14,001 ̲ 35,002 - 14,001 ̲

21,001

### Exercise 37

5,000,566 - 2,441,326 ̲ 5,000,566 - 2,441,326 ̲

### Exercise 38

400,605 - 121,352 ̲ 400,605 - 121,352 ̲

279,253

### Exercise 39

46,400 -   2,012 ̲ 46,400 -   2,012 ̲

### Exercise 40

77,893 -      421 ̲ 77,893 -      421 ̲

77,472

### Exercise 41

42 - 18 ̲ 42 - 18 ̲

### Exercise 42

51 - 27 ̲ 51 - 27 ̲

24

### Exercise 43

622 -   88 ̲ 622 -   88 ̲

### Exercise 44

261 -   73 ̲ 261 -   73 ̲

188

### Exercise 45

242 - 158 ̲ 242 - 158 ̲

### Exercise 46

3,422 - 1,045 ̲ 3,422 - 1,045 ̲

2,377

### Exercise 47

5,565 - 3,985 ̲ 5,565 - 3,985 ̲

### Exercise 48

42,041 - 15,355 ̲ 42,041 - 15,355 ̲

26,686

### Exercise 49

304,056 -   20,008 ̲ 304,056 -   20,008 ̲

### Exercise 50

64,000,002 -     856,743 ̲ 64,000,002 -     856,743 ̲

63,143,259

### Exercise 51

4,109 -   856 ̲ 4,109 -   856 ̲

### Exercise 52

10,113 -   2,079 ̲ 10,113 -   2,079 ̲

8,034

### Exercise 53

605 -   77 ̲ 605 -   77 ̲

### Exercise 54

59 - 26 ̲ 59 - 26 ̲

33

### Exercise 55

36,107 -   8,314 ̲ 36,107 -   8,314 ̲

### Exercise 56

92,526,441,820 - 59,914,805,253 ̲ 92,526,441,820 - 59,914,805,253 ̲

32,611,636,567

### Exercise 57

1,605 -   881 ̲ 1,605 -   881 ̲

### Exercise 58

30,000 - 26,062 ̲ 30,000 - 26,062 ̲

3,938

### Exercise 59

600 - 216 ̲ 600 - 216 ̲

### Exercise 60

9,000,003 -    726,048 ̲ 9,000,003 -    726,048 ̲

#### Solution

8,273,955

For the following problems, perform each subtraction.

### Exercise 61

Subtract 63 from 92.

#### Hint:

The word "from" means "beginning at." Thus, 63 from 92 means beginning at 92, or 92 63 92 63 size 12{"92" - "63"} {} .

### Exercise 62

Subtract 35 from 86.

51

### Exercise 63

Subtract 382 from 541.

### Exercise 64

Subtract 1,841 from 5,246.

3,405

### Exercise 65

Subtract 26,082 from 35,040.

### Exercise 66

Find the difference between 47 and 21.

26

### Exercise 67

Find the difference between 1,005 and 314.

### Exercise 68

Find the difference between 72,085 and 16.

72,069

### Exercise 69

Find the difference between 7,214 and 2,049.

### Exercise 70

Find the difference between 56,108 and 52,911.

3,197

### Exercise 71

How much bigger is 92 than 47?

### Exercise 72

How much bigger is 114 than 85?

29

### Exercise 73

How much bigger is 3,006 than 1,918?

### Exercise 74

How much bigger is 11,201 than 816?

10,385

### Exercise 75

How much bigger is 3,080,020 than 1,814,161?

### Exercise 76

In Wichita, Kansas, the sun shines about 74% of the time in July and about 59% of the time in November. How much more of the time (in percent) does the sun shine in July than in November?

15%

### Exercise 77

The lowest temperature on record in Concord, New Hampshire in May is 21°F, and in July it is 35°F. What is the difference in these lowest temperatures?

### Exercise 78

In 1980, there were 83,000 people arrested for prostitution and commercialized vice and 11,330,000 people arrested for driving while intoxicated. How many more people were arrested for drunk driving than for prostitution?

11,247,000

### Exercise 79

In 1980, a person with a bachelor's degree in accounting received a monthly salary offer of $1,293, and a person with a marketing degree a monthly salary offer of$1,145. How much more was offered to the person with an accounting degree than the person with a marketing degree?

### Exercise 80

In 1970, there were about 793 people per square mile living in Puerto Rico, and 357 people per square mile living in Guam. How many more people per square mile were there in Puerto Rico than Guam?

436

### Exercise 81

The 1980 population of Singapore was 2,414,000 and the 1980 population of Sri Lanka was 14,850,000. How many more people lived in Sri Lanka than in Singapore in 1980?

### Exercise 82

In 1977, there were 7,234,000 hospitals in the United States and 64,421,000 in Mainland China. How many more hospitals were there in Mainland China than in the United States in 1977?

57,187,000

### Exercise 83

In 1978, there were 3,095,000 telephones in use in Poland and 4,292,000 in Switzerland. How many more telephones were in use in Switzerland than in Poland in 1978?

For the following problems, use the corresponding graphs to solve the problems.

### Exercise 84

How many more life scientists were there in 1974 than mathematicians? (Media 29)

165,000

### Exercise 85

How many more social, psychological, mathematical, and environmental scientists were there than life, physical, and computer scientists? (Media 29)

### Exercise 86

How many more prosecutions were there in 1978 than in 1974? (Media 30)

74

### Exercise 87

How many more prosecutions were there in 1976-1980 than in 1970-1975? (Media 30)

### Exercise 88

How many more dry holes were drilled in 1960 than in 1975? (Media 31)

4,547

### Exercise 89

How many more dry holes were drilled in 1960, 1965, and 1970 than in 1975, 1978 and 1979? (Media 31)

For the following problems, replace the ☐ with the whole number that will make the subtraction true.

### Exercise 90

14 - ̲ 3 14 - ̲ 3

11

### Exercise 91

21 - ̲ 14 21 - ̲ 14

### Exercise 92

35 - ̲ 25 35 - ̲ 25

10

### Exercise 93

16 - ̲ 9 16 - ̲ 9

### Exercise 94

28 - ̲ 16 28 - ̲ 16

#### Solution

12

For the following problems, find the solutions.

### Exercise 95

Subtract 42 from the sum of 16 and 56.

### Exercise 96

Subtract 105 from the sum of 92 and 89.

76

### Exercise 97

Subtract 1,127 from the sum of 2,161 and 387.

### Exercise 98

Subtract 37 from the difference between 263 and 175.

51

### Exercise 99

Subtract 1,109 from the difference between 3,046 and 920.

### Exercise 100

Add the difference between 63 and 47 to the difference between 55 and 11.

60

### Exercise 101

Add the difference between 815 and 298 to the difference between 2,204 and 1,016.

### Exercise 102

Subtract the difference between 78 and 43 from the sum of 111 and 89.

165

### Exercise 103

Subtract the difference between 18 and 7 from the sum of the differences between 42 and 13, and 81 and 16.

### Exercise 104

Find the difference between the differences of 343 and 96, and 521 and 488.

214

### Exercises for Review

#### Exercise 105

((Reference)) In the number 21,206, how many hundreds are there?

#### Exercise 106

((Reference)) Write a three-digit number that has a zero in the ones position.

#### Exercise 107

((Reference)) How many three-digit whole numbers are there?

#### Exercise 108

((Reference)) Round 26,524,016 to the nearest million.

27,000,000

#### Exercise 109

((Reference)) Find the sum of 846+221+116846+221+116 size 12{"846"+"221"+"116"} {}.

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 Title: Analysis of rotationally resolved spectra to non-degenerate (a′′1) upper-state vibronic levels in the Ã 2E′′−X̃2A′2 electronic transition of NO3 Author(s): Miller, Terry A. Contributor(s): Stanton, John F.; Melnik, Dmitry G.; Tran, Henry; Chen, Ming-Wei; Codd, Terrance Joseph; Roudjane, Mourad Subject(s): Radicals Abstract: The vibronic structure of the $tilde{A}-tilde{X}$ electronic spectrum of NO$_3$ has been observed using both room-temperature and jet-cooled samples. A recent analysis of this structure is consistent with the Jahn-Teller effect (JTE) in the $e^prime$ $nu_3$ vibrational mode (N-O stretch) being quite strong while the JTE in the $e^prime$ $nu_4$ mode (O-N-O) bend) is rather weak. Electronic structure calculations qualitatively predict these results but the calculated magnitude of the JTE is quantitatively inconsistent with the spectral analysis. Rotationally resolved spectra have been obtained for over a dozen vibronic bands of the $tilde{A}-tilde{X}$ electronic transition in NO$_3$. An analysis of these spectra should provide considerably more experimental information about the JTE in the $tilde{A}$ state of NO$_3$ as the rotational structure should be quite sensitive to the geometric distortion of the molecule due to the JTE. This talk will focus upon the parallel bands, which terminate on $tilde{A}$ state levels of $a^{primeprime}_1$ vibronic symmetry, which were the subject of a preliminary analysis reported at this meeting in 2014. We have now recorded the rotational structure of over a half-dozen parallel bands and have completed analysis on the 3$^1_0$ and 3$^1_0 4^1_0$ transitions with several other bands being reasonably well understood. Two general conclusions emerge from this work. (i) All the spectral bands show evidence of perturbations which can reasonably be assumed to result from interactions of the observed $tilde{A}$ state levels with high vibrational levels of the $tilde{X}$ state. The perturbations range from severe in some bands to quite modest in others. (ii) Analyses of observed spectra, insofar as the perturbations permit, have all been performed with an oblate symmetric top model including only additional spin-rotation effects. This result is, of course, consistent with an effective, undistorted geometry for NO$_3$ of D$_{3h}$ symmetry on the rotational timescale. Issue Date: 23-Jun-15 Publisher: International Symposium on Molecular Spectroscopy Citation Info: ACS Genre: CONFERENCE PAPER/PRESENTATION Type: Text Language: English URI: http://hdl.handle.net/2142/79137 Date Available in IDEALS: 2016-01-05


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# Processing a detergent powder formulation: Direct compression, and high shear wet granulation followed by compression

Abstract : In this paper, detergent tablets processing for industrial multi-purpose uses, i.e., hard surfaces cleaning, has been investigated. The detergent formulation is a complex blend of over 95% of powders: inorganic salts of alkaline metal called builders, and a system of surfactants composed of anionic and nonionic organic molecules. Two processes were examined: wet granulation prior to compression (processing 1), and direct compression (processing 2). Wet granulation is performed in a laboratory high shear mixer with a blend of powdered raw materials excluding one of the salts, the metasilicate. This blend is called initial blend for granulation.'' The granulation parameters studied are the intensity of shear related to the impeller speed, and the amount of liquid added, water being used as liquid binder. Powdered and granulated formulations are characterized in terms of particle size distribution, flowability, and compressibility. Compression is performed on a uniaxial laboratory press, either with the granules mixed with sodium pentahydrated metasilicate for processing 1, or with a mix of powdered and pre-granulated raw materials, called here initial blend for direct compression,'' for processing 2. The following features of the compacts are considered: mechanical strength, ftiability, density, solubility in cold water, and microscope observation of fractured surfaces. Comparison of compacts showed that prior granulation leads to increased mechanical properties due to higher plasticity and stronger binding with water, while it slightly deactivates the alkaline product. Compression of the material significantly increases dissolution times in both cases.
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Marie-Élise Chateau, Laurence Galet, Yannick Soudais, Jacques Fages. Processing a detergent powder formulation: Direct compression, and high shear wet granulation followed by compression. Powder Technology, Elsevier, 2005, 157 (1-3), pp.191-198. ⟨10.1016/j.powtec.2005.05.026⟩. ⟨hal-01593338⟩

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# A Difficult partial differential Problem

1. Dec 22, 2017

### big dream

Problem:
$${\frac {\partial }{\partial t}}A\left( y,t \right) +6\,\Lambda\,\Omega\, \left( {y}^{2}-y \right) \sin \left( t \right) ={\frac {\partial ^{2}}{\partial {y}^{2}}}A \left( y,t \right)$$
$${\frac{\partial }{\partial y}}A \left( t,0 \right) ={\frac {\partial }{\partial y}}A \left( t,1 \right) =0$$
Boundary condition
$${\frac{\partial }{\partial y}}A \left( t,0 \right) ={\frac {\partial }{\partial y}}A \left( t,1 \right) =0$$
$$A \left( y,t \right) =6\,\Lambda\, \left( \Im \right) \, \left\{ [{\frac {i\sinh \left( \alpha\,y \right) }{\alpha}}-{\frac {i \left( 1-\cosh \left( \alpha \right) \right) \cosh \left( \alpha\,y \right) }{\alpha\,\sinh \left( \alpha \right) }}+i{y}^{2}-iy+2\,{\Omega}^{-1}]{{\rm e}^{it}} \right\}$$
Where, $$\alpha=1/2\, \left( 1+i \right) \sqrt {2}\sqrt {\Omega}$$
attempt at a solution

Maple didn't give an answer. I don't know how to get this kind of solution.

2. Dec 23, 2017

### bigfooted

3. Dec 24, 2017

### big dream

Thank you, sir. I got it. This thread can be closed now.


text:
# Proof for (Abelian) Group - or not?

• November 4th 2013, 10:06 AM
MageKnight
Proof for (Abelian) Group - or not?
Hello together!

I need to show whether the following is a(n abelian) group or not:

$G = \{w, f\}; a*b := (a \leftrightarrow b)$

I try to show the axioms.
1) Closure
Is given, since
w*f = f
f*w = f
w*w = w
f*f = f

2.) Associativity
???

3.) Identity
I need an element e with e*a = a*e = a. (for all elements of G)
So..

a*e = a <-> e = w.
because w*w = w and f*f = f.

4.) Inverse element;
I need an element i with
a*i = i*a = e.

the inverse for w is w, since w*w=w=e.
the inverse for f is f, since f*f=w=e.

5.) Commutativity:

Is a*b=b*a ?
This can be shown be the few examples:
w*f = f
f*w = f
• November 4th 2013, 10:29 AM
emakarov
Re: Proof for (Abelian) Group - or not?
Ha-ha, you forgot to translate w into English!

Quote:

Originally Posted by MageKnight
2.) Associativity
???

Wikipedia says that associativity holds. In the worst case, you have 8 cases to check.

Quote:

Originally Posted by MageKnight
3.) Identity
I need an element e with e*a = a*e = a. (for all elements of G)
So..

a*e = a <-> e = w.
because w*w = w and f*f = f.

You mean f * w = f.

The rest seems OK.
• November 4th 2013, 10:38 AM
SlipEternal
Re: Proof for (Abelian) Group - or not?
I am not sure I understand your definition for $a \leftrightarrow b$. To me, that reads $a$ if and only if $b$. But, the multiplication you give in part (1) suggests otherwise. Then, for part (4), your multiplication changes: f*f = w in part (4) but f*f = f in part (1) and part (3). I think you are correct in part (4) since that multiplication is correct for the biconditional.

For part (3), you say w*w = w and f*f=f. I think you want to show w*w = w and f*w = w*f = f. So, w is the identity element.

For part (4), I agree with you.

For part (5), you have f*f = f*f = w, f*w = w*f = f, and w*w = w. Those are the three cases.

Anyway, for associativity, you need to show that $a*(b*c) = (a*b)*c$ for all $a,b,c\in G$. So, try it out. There are only eight possible combinations for $a,b,c \in G$.


text:
I've got an url coming in as:

/index.php?/location/&gclid=287ejek22kj

This is going to a 404 page because of the gclid...

I need it to go to:

/index.php?/location

I've tried this, with no success:

RewriteEngine On
RewriteBase /
RewriteRule ^(.*)&gclid=(.*)1 [L]

All help is appreciated. Thanks.

-

## migrated from serverfault.comMay 1 '11 at 7:08

This question came from our site for system and network administrators.

is there a reason you're leaving in the /index.php? I assume that's your controller script, but using rewrites, you could easily make it simply make it /location. The main issue with what you're trying to do is that you're trying to capture the QUERY_STRING (everything after the ?). You need to treat this specially in RewriteRules –  djdrey May 1 '11 at 2:26
in the next set of rules i'm removing the index.php... but i'm having the issue with simply ignoring the gclid... on those adword pages only of course –  Jeffrey May 1 '11 at 2:50

This is pretty much answered in concept at least here: http://stackoverflow.com/questions/2111182/rewriting-url-with-selected-query-string-parameters-in-htaccess

-
thanks, but i need to see it with my conditions –  Jeffrey May 1 '11 at 2:48

I'm not sure how you're ending up with AdWords auto tagging appended to a URL without the proper structure but if you don't want the gclid parameter at all you could alway just switch off autotagging in AdWords. You shouldn't need to remove it using a rewrite rule as it's just an option that you can choose not to use.

-

Try this (you :

<IfModule mod_rewrite.c>

RewriteEngine On
RewriteBase /

RewriteCond %{REQUEST_URI} ^/$RewriteCond %{QUERY_STRING} ^(gclid=.*) RewriteRule ^(.*)$ /index.php/ [L,PT]


text:
# Equivalence of Hölder norms [closed]

Let $0 <\alpha \leq 1$. Define $$|f|_1= \sup_{x\in\mathbb{R}}|f(x)|+ \sup_{x,h\in\mathbb{R}}\left|\frac{f(x+h)-f(x)}{|h|^{\alpha}}\right|$$ $$|f|_2= \sup_{x\in\mathbb{R}}|f(x)|+ \sup_{x,h\in\mathbb{R}}\left|\frac{f(x+h)-2f(x)+f(x-h)}{|h|^{\alpha}}\right|$$ for $f\in C(\mathbb{R}^d)$.

Prove that for every $\alpha\in (0,1)$ and $f\in C_c^{\infty}(\mathbb{R}^d)$ there exists a constant $c \geq 1$ s.t. $$c^{-1}|f|_2\leq |f|_1\leq c|f|_2$$ Discuss the same question for $\alpha = 1$.

Can anyone help me and give me a hint...?

• $x \in \mathbb{R}$ in the supremum but $f \in \mathcal{C}(\mathbb{R}^d)$ so $d=1$? – Rafael Wagner Oct 23 '17 at 17:54
• oh, I didnt notice that before. I think $x\in \mathbb{R}^d$ is meant in the supremum. – user494831 Oct 23 '17 at 18:00

## 1 Answer

First it is easy to see that, $$\|f\|_{\infty} +\left|\frac{f(x+h)-2f(x)+f(x-h)}{|h|^{\alpha}}\right| =\|f\|_\infty+ \left|\frac{f(x+h)-f(x) -(f(x)-f(x-h))}{|h|^{\alpha}}\right|\\\le \|f\|_\infty+\left|\frac{f(x+h)-f(x)}{h^\alpha} \right|+\left|\frac{(f(x)-f(x-h))}{|h|^{\alpha}}\right| \le 2|f|_1$$

Which implies, by taking the supremum that, $$|f|_2 \le 2|f|_1$$ this means that, if we define the embedding

Next we consider $z= x+k$ and $h=2k$ hence, $x+h =z+k$ and $x =z-k$

Then, for $0<\alpha <1,$ $$\left|\frac{f(x+h)-f(x)}{|h|^{\alpha}}\right| = \left|\frac{f(z+k)-2f(z)+f(z-k) +\color{red}{2 [f(z) -f(z-k)]}}{|2k|^{\alpha}}\right|\\ \le 2^{-\alpha} \left|\frac{f(z+k)-2f(z)+f(z-k)}{|k|^{\alpha}}\right| + 2^{1-\alpha} \left|\frac{\color{red}{ [f(z) -f(z-k)]}}{|k|^{\alpha}}\right|$$

passing through the supremum yields, $$|f|_1 \le 2^{-\alpha}|f|_2 +2^{1-\alpha} |f|_1 \implies (1-2^{1-\alpha})|f|_1 \le 2^{-\alpha}|f|_2$$ that is

$$|f|_1 \le \frac{2^{-\alpha}}{(1-2^{1-\alpha})}|f|_2$$

Let $$\color{blue}{c=\max(2,\frac{2^{-\alpha}}{(1-2^{1-\alpha})})}.$$ Thus, $$c^{-1}|f|_2\le |f|_1 \le c |f|_2~~~for ~~all ~~f$$

If $\alpha =1$ then for $f\in C^\infty_c(\Bbb R)$ we that $$f(x+h)-f(x) =\int_0^1 f'(x+th) h dt$$ that is $$\left|\frac{f(x+h)-f(x)}{|h|} \right| = \left|\int_0^1 f'(x+th) dt.\right|$$

Unlike, for the second case applying the above trick twice we have,$$\left|\frac{f(x+h)-2f(x)+f(x-h)}{|h|}\right| =\left|\int_0^1 f'(x+th)- f'(x-th)dt.\right|\\ = \left|\int_0^1t\int_0^1 f''(x-th+2sth)h\right| \le |h|\|f''\|_\infty$$

If we assume that $|f|_1 \le c |f|_2$ then this leads to $$\|f\|_\infty + \left|\frac{f(x+h)-f(x)}{|h|} \right| \le |f|_1 \le c(\|f\|_\infty + |h|\|f''\|_\infty)~~~~\forall ~~h$$

• you shouldn't prove that $c=2$ as the question of OP? – Rafael Wagner Oct 23 '17 at 18:58
• But in the question he says to prove that $c^{-1}|f|_2\le |f|_1 \le c |f|_2$. – Rafael Wagner Oct 23 '17 at 19:06
• It is straight forward but I don't see it as irrelevant. Also made your question much more elegant in my opinion. – Rafael Wagner Oct 23 '17 at 19:22
• Ok sorry it was not appropriate as word to use – Guy Fsone Oct 23 '17 at 19:25
• Thanks for your solution. Two questions: 1) Is there any reason why $f\in C_c^{\infty}$ in the statement? You didnt use it or? 2) Is there a problem for $\alpha =1$? – user494831 Oct 24 '17 at 9:49


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# Quick start 2: How to setup iterations for a simulation¶

In the following quick start tutorial, we will setup a basic simulation that runs multiple iterations (each time changing the input parameters for a functional block).

Part 1: Drag and drop the required components onto the design layout

1. Launch a new application of SystemLab|Design by double left-clicking on the SystemLab-Design.exe executable file.

2. Go to the Functional block library (left panel of GUI) and hover over the triangle in front of the Waveform generators group (under Electrical).

3. Left-click mouse to expand the menu.

4. Left-click select and hold over the Sine Generator component and start moving your mouse towards the design scene for Project_1.

5. Release the mouse button anywhere over the design layout. [An icon representing a sinusoidal signal generator should appear on the layout].

Part 2: Setup the functional block script to update its parameters for each iteration

1. On the design layout (Project_1), double left-click on the Sine functional block. [The Functional block properties dialog will open].

2. Next to the Script module name click on the Edit Script button. [The Functional block script editor associated with the Sine Generator will open].

1. On line 27 of the Sine_Generator script, comment out the current freq parameter definition and add the following new lines of code as shown below.

[The variable “iteration” is linked to the current iteration for the simulator (an integer value of 1, 2, 3, …n where n is the total number of iterations). For each iteration, the “freq” parameter of the Sine Generator will be increased based on the iteration value times the base frequency of 10 GHz]

'''==INPUT PARAMETERS=================='''
#freq = float(parameters_input[0][1]) #Hz
freq = iteration*10e9
signal_amp = float(parameters_input[1][1]) #Peak amplitude (crest-to-crest = 2*Peak amplitude)
bias = float(parameters_input[2][1])

2. On line 46 of the Sine_Generator script, add two new lines of code just below signal_gen_results = [ ] as follows:

'''==RESULTS===================================================='''
signal_gen_results = []
freq_result = ['Generator frequency', freq*1e-9, 'GHz', ' ', False]
signal_gen_results = [freq_result]

3. Click on the Save icon in the Tool Bar to save all the changes.

4. Close the SciTE dialog (File/Exit)

5. Select OK to close the Functional block properties dialog.

Part 3: Setup the desired number of iterations and run a simulation

1. Select the Settings icon on the on the Tool bar, and within the Simulation settings tab of the Project settings dialog, change the Number of iterations from 1 to 5 (as shown below).

1. Select OK to save and close the Project settings dialog.

2. From the Tool bar, click on the blue Start button (blue triangle) to launch the simulator. [The Simulation status dialog should indicate that all iterations (in this case five) were successfully completed!].

3. Double left-click on the output port of the Sine generator to open the Electrical signal data analyzer and select the Frequency data tab.

4. Within the Iterations group box you will see that the Current iteration spin box is set to “1”. To change the current iteration, click on upper arrow of the spin box to increase the iteration index to “2” and continue until reaching “5”.

As shown in the images below (showing iteration 1 and 5) the signal data will be refreshed to reflect the current saved iteration for the simulation. As designed, we see that the frequency increases from its start point of 10 GHz all the way to 50 GHz. After a simulation, the signal data for all calculated ports in a design are saved to data dictionaries and can be accessed through via the Iterations spin boxes located within each tab of a port data viewer.

1. It is also possible to change iterations by clicking on the spin box located on the upper right of the main application window. Close the Electrical Signal Data Analyzer and set the current index of the main application Iteration spin box to 3 as shown below:

Note

The Iteration spin box in the main application window is linked to data contained in the functional blocks and data panels - the iteration spin boxes of the signal data viewers operate independently.

1. Hover your mouse over the Sine functional block to view the component status and results.

As shown below, the functional block result we created in step 9 is shown for iteration 3 (displaying a frequency of 30 GHz). Using the iteration spin box in the upper right of the main application window, results from functional blocks and data panels can be analyzed quickly over all iterations. Similarly, by hovering over signal ports, signal metrics data can be viewed for the selected iteration (see second figure below)


text:
# How heavy-tailed is the t distribution? (Part 2)

In this previous post, we explored how heavy-tailed the $t$ distribution is through the question: “What is the probability that the random variable is at least x standard deviations (SDs) away from the mean?” For the most part, the smaller the degrees of freedom, the larger this probability was (more “heavy-tailed”), until we realized that the trend reversed for really small degrees of freedom (2.1 in the post). In fact, for $1 < df \leq 2$, the variance of the $t$ distribution is infinite, and so the random variable is always within 1 SD of the mean!

We need another way to think about heavy-tailedness. (The code to produce the figures in this post is available here.)

A first approach that doesn’t work

You might be wondering, why didn’t I just plot $P\{ T > x \}$ with $T \sim t_{df}$ against $x$, for various values of $x$ and $df$? If I did that, I would have ended up with the plot below (for the log of the probabilities):

That seems to be exactly what we want: the smaller the degrees of freedom, the slower this probability decays…

The problem is that the comparison above ignores the scale of the random variables. Imagine if we tried to make the plot above, but instead of plotting lines for the $t$ distribution with different degrees of freedom, let’s plot it for the normal distribution with different standard deviations. This is what we would get:

That seems to give the same trend as the plot before! Can we then conclude that the $\mathcal{N}(0, 10^2)$ distribution is more heavy-tailed than the $\mathcal{N}(0, 1)$ distribution??

One way to incorporate scale

The discussion above illustrates the need to take scale into account. We tried to do this in the previous post by scaling each distribution by its own SD, but that idea broke down for small degrees of freedom.

Here’s an idea: Pick some threshold $x'$. For each random variable $T$, find the scale factor $k$ such that $P \{ kT > x' \} = P \{ \mathcal{N}(0, 1) > x' \}$. For this value of $k$, $kT$ and $\mathcal{N}(0, 1)$ are on the same scale w.r.t. this threshold. We then compare the tail probabilities of $kT$ and $\mathcal{N}(0, 1)$ (instead of $T$ and $\mathcal{N}(0, 1)$).

Finding $k$ is not hard: here’s a three-line function that does it for the $t$ distribution in R:

getScaleFactor <- function(df, threshold) {
tailProb <- pnorm(threshold, lower.tail = FALSE)
tQuantile <- qt(tailProb, df = df, lower.tail = FALSE)
return(threshold / tQuantile)
}

Let’s plot the log10 of the tail probability $P \{ kT > x \}$ with $T \sim t_{df}$ against $x$ for various values of $x$ and $df$, with the scale factor $k$ computed as above:

By definition, the tail probabilities will coincide when $x$ is equal to the threshold used to compute the scale factors. We now see a clear trend with no breakdown: for smaller values of $df$, the tail probability $P \{ kT > x \}$ is larger.

Another side benefit of this way to looking at tail probabilities is that we can now compare distributions which have infinite variance, or even an undefined mean (like the Cauchy distribution, which is the t distribution with one degree of freedom)! Here is the same plot as above but for smaller degrees of freedom:


text:
## Precalculus (10th Edition)

$1$
I know that $\cos$ has a period of $360^\circ$, hence $\cos{\theta}=\cos{(\theta-360^\circ)}.$ Hence: $\cos{400^\circ}\sec{40^\circ}=\cos{(400^\circ-360^\circ)}\sec{40^\circ}=\cos{40^\circ}\sec{40^\circ}$. We also know that $\cos{\theta}\sec{\theta}=1$. Thus: $\cos{40^\circ}\sec{40^\circ}=1$.


text:
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text:
In finite-dimensional spaces, the matrix representation (with respect to an orthonormal basis) of an orthogonal transformation is an orthogonal matrix. orthonormal matrix polynomials are unique up to a multiplication on the left by an orthogonal matrix. ~u j = ( 1 if i = j , 0 otherwise; which implies A T A = I n . Non-square matrices with orthonormal columns. So if a column has norm different from one, the corresponding vector of the base (which has norm 1) changes its norm. The rows of U are Given U is an orthc ationship between U and U-1? This means that the orthonormal … So, a column of 1's is impossible. Bilden diese All identity matrices are an orthogonal matrix. Matrix Proof: Show that Qx = 0 implies x = 0.Since Q Would a square matrix with orthogonal columns, but not orthonormal, change the norm of a vector? Both routines compute the decomposition using Householder reflections. Here, the term ‘vector’ is used in the sense that it is an element of a vector space – an algebraic structure used in linear algebra. My confusion comes when the columns of $Q$ are orthogonal, but not orthonormal, i.e., if the columns are weighted by weights $w_1,\dots,w_N$, the dot product of any two different columns would still be zero, but $Q^H Q \neq I$ anymore. Let v 1, v 2 v 1, v 2 It will be an orthonormal matrix only when norm(k)==1 (which implies k=1/sqrt(3) in your examples, as the others have noted). and from the first property, we know that, so we can conclude from both the facts that. Adding lists to specific elements in a list. Bei der Bestimmung einer Orthonormalbasis aus Eigenvektoren ist die folgende Erkenntnis nützlich: ist die reelle Matrix symmetrisch, so sind ihre Eigenvektoren zu verschiedenen Eigenwerten orthogonal zueinander. The literature always refers to matrices with orthonormal columns as orthogonal, however I think that's not quite accurate. You will then explore another application of orthogonal projections: creating a matrix factorization widely used in practical applications of linear algebra. $$replacing G by an orthogonal matrix, since the norms of the rows of G follow the -distribution, while rows of an orthogonal matrix have the unit norm. (2) Orthonormal Matrix If in addition, all the vectors are unit vectors if, OpenCV is designed for detecting, it is not meant for quantification. 278 Appendix C Vectors, Matrices, Orthogonal Functions When we take the scalar product of two vectors with complex components, we can write (w, V) = <й| V) = (и, «2 u 3) \v 3J = щ v, + u 2 v 2 + «3 3. It is then desired to ﬁnd the “nearest’’ orthonormal matrix. Orthogonal vs Orthonormal In mathematics, the two words orthogonal and orthonormal are frequently used along with a set of vectors. Adventure cards and Feather, the Redeemed? # Of a square matrix: such that its transpose is equal to its inverse. Changes in orientation are given by an orthogonal matrix Q, which includes rotations and reflections, leaving intact the essential distance and angular properties of the configuration. A p×q orthonormal matrix T=P with q columns represents a projection from p to q dimensions. As adjectives the difference between orthonormal and orthogonal is that orthonormal is (mathematics) of a set of vectors, both orthogonal and normalized while orthogonal is (geometry) of two objects, at right angles; perpendicular to each other. a vector enters an expression on the right side and comes out on the left. their dot product is 0. Hence Q is the product of a unitary matrix U with a diagonal matrix D. Calculate the orthonormal basis for the range of A using orth. Making statements based on opinion; back them up with references or personal experience. This paper describes an algorithm for simultaneously diagonalizing by orthogonal transformations the blocks of a partitioned matrix having orthonormal columns. Die Länge der Vektoren und der Winkel zwischen den Vektoren bleiben dabei erhalten. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Two vectors are orthogonal to each other if their inner product with each other is 0. rev 2020.12.3.38123, The best answers are voted up and rise to the top, Mathematics Stack Exchange works best with JavaScript enabled, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site, Learn more about Stack Overflow the company, Learn more about hiring developers or posting ads with us. Let T be a linear transformation from R^2 to R^2 given by the rotation matrix. ~u j = (1 if i = j, 0 otherwise; which implies A T A = I n. Conversely, if A T A = I A T A = I This is a preview of subscription content, log in to check access. Check if rows and columns of matrices have more than one non-zero element? Proof — part 2 (optional) For an n × n symmetric matrix, we can always find n independent orthonormal eigenvectors. If Q=(x_1,\ldots,x_n) is a matrix with orthogonal columns (x_i^Hx_j=0), then provided that its columns x_1,\ldots,x_n are nonzero, we have Orthogonal Matrix :- Whereas A … Folglich müsste man die hier beschriebene Matrix orthonormale Matrix nennen. The normal vector and tangent vector at a given point are orthogonal . Hence a matrix is orthogonal iff the image of the standard orthonormall basis is an orthonormal basis again. But we might be dealing with some subspace, and not need an orthonormal basis for the entire space. Det kan ses, at en ortogonalmatrix har determinant 1 eller − 1, og en ortogonal matrix med determinant 1 kaldes en speciel ortogonal matrix. … Thanks for watching!! Since U is an orthogonal matrix, U=U'. do u mean uncorrelated matrix? If vector x and vector y are also unit vectors then they are orthonormal. The set of rows of Q forms a bases in Rd. Changes in orientation are given by an orthogonal matrix Q, which includes rotations and reflections, leaving intact the essential distance and angular properties of the configuration. The linear transformation matrix of ORF has the following form W ORF = 1 SQ, (2) where Q is a uniformly distributed random orthogonal matrix1. Are the natural weapon attacks of a druid in Wild Shape magical? There are many possible ways to produce an orthonormal matrix from the resulting matrix e.g. Thank you. Such matrices are usually denoted by the letter Q. (C.2) MATRIX MANIPULATION IN EXCEL Microsoft Excel is set up to manipulate matrices5 for the following operations: Matrix Inversion Two elements u and v of a vector space with bilinear form B are orthogonal when B(u, v) = 0. Thanks for contributing an answer to Mathematics Stack Exchange! Note that Active 7 years, 8 months ago. Similarly, a matrix Q is orthogonal if its transpose is equal to its inverse. Also, learn how to identify the given matrix is an orthogonal matrix with solved examples at BYJU'S. The determinant of an orthogonal matrix is equal to 1 or -1. An orthogonal matrix is a square matrix with real entries whose columns and rows are orthogonal unit vectors or orthonormal vectors. Why don't you try a simple 2\times 2 example and see … Use MathJax to format equations. Damit ist die Inverse einer orthogonalen Matrix gleichzeitig ihre Transponierte. Checking for finite fibers in hash functions. What can be deduced from the fact that U' is an orthogonal matrix? Thank you, although I don't know what you meant by "image of canonical base". 2는 어디서 많이 본 그림일 것이다. > > Cheers > ZAB > Start a New Thread. Why does this movie say a witness can't present a jury with testimony which would assist in making a determination of guilt or innocence? Orthogonal matrices are the most beautiful of all matrices. Basis vectors. Why is the matrix product of 2 orthogonal matrices also an orthogonal matrix? One way to think about a 3x3 orthogonal matrix is, instead of a 3x3 array of scalars, as 3 vectors. Two vector x and y are orthogonal if they are perpendicular to each other i.e. Fig. Unitary Matrix:- A Complex Square matrix U is a Unitary Matrix if its Conjugate transpose (U*) is its inverse.$$ This can be generalized and extended to 'n' dimensions as described in group theory. MAINTENANCE WARNING: Possible downtime early morning Dec 2, 4, and 9 UTC…. Solved exercises. Then. What are these matrices called? But if matrix A is orthogonal and we multiply transpose of matrix A on both sides we get. What is the difference between orthogonal and orthonormal in terms of vectors and vector space? What is Orthogonal Matrix? Eine orthogonale Matrix ist in der linearen Algebra eine quadratische, reelle Matrix, deren Zeilen- und Spaltenvektoren orthonormal bezüglich des Standardskalarprodukts sind. Bilden die Spalten einer quadratischen Matrix ein System zueinander orthogonaler Einheitsvektoren, so heißt diese Matrix orthogonale Matrix. This preview shows page 1 - 2 out of 2 pages.. 2. should the first row of A be [0 0 0 0 0 -1] instead of [0 0 0 0 -1 0] ? The collection of the orthogonal matrix of order n x n, in a group, is called an orthogonal group and is denoted by ‘O’. That puts A into a nice looking form and allows a solution like you are talking about. Orthonormal (orthogonal) matrices are matrices in which the columns vectors form an orthonormal set (each column vector has length one and is orthogonal to all the other colum vectors). They should be mutually perpendicular to each other (subtended at an angle of 90 degrees with each other). This is true even if Q is not square. Definition: if the columns of a matrix are orthonormal, the matrix itself is called orthogonal. Why put a big rock into orbit around Ceres? One way to express this is Q T Q = Q Q T = I , {\displaystyle Q^{\mathrm {T} }Q=QQ^{\mathrm {T} }=I,} Its rows are mutually orthogonal vectors with unit norm, so that the rows constitute an orthonormal basis of V. The columns of the matrix form another orthonormal basis of V. if so. If matrix Q has n rows then it is an orthogonal matrix (as vectors q1, q2, q3, …, qn are assumed to be orthonormal earlier)Here we are using the property of orthonormal … In linear algebra, a semi-orthogonal matrix is a non-square matrix with real entries where: if the number of columns exceeds the number of rows, then the rows are orthonormal vectors; but if the number of rows exceeds the number of columns, then the columns are orthonormal vectors. Exercise 1. Why does the FAA require special authorization to act as PIC in the North American T-28 Trojan? Hadamard matrices are used in signal processing and statistics. Prove that this linear transformation is an orthogonal transformation. Show that the rows of U form an orthonormal basis of R". Since det(A) = det(Aᵀ) and the determinant of product is the product of determinants when A is an orthogonal matrix. Then the transformation matrix of the basis-changer from P to Q is orthogonal. An orthogonal matrix is a square matrix and satisfies the following condition: Inveniturne participium futuri activi in ablativo absoluto? For what purpose does "read" exit 1 when EOF is encountered? In this tutorial, we will dicuss what it is and how to create a random orthogonal matrix … A matrix P is orthogonal if P T P = I, or the inverse of P is its transpose. This is because the singular values of A are all nonzero. In this tutorial, we will dicuss what it is and how to create a random orthogonal matrix with pyhton. To summarize, for a set of vectors to be orthogonal : Assuming vectors q1, q2, q3, ……., qn are orthonormal vectors. Instead of performing Gaussian elimination you can just multiply transpose of coefficient matrix with constant matrix and get the solution. Exercises on orthogonal matrices and Gram-Schmidt Problem 17.1: (4.4 #10.b Introduction to Linear Algebra: Strang) Orthonormal vectors are automatically linearly independent. Then the transformation matrix of the basis-changer from P to Q is orthogonal. gives you a square matrix with mutually orthogonal columns, no matter what's the vector kk. Let $Q$ be an $N \times N$ unitary matrix (its columns are orthonormal). Questions about Orthogonal vs Perspective in OpenGL [closed] Ask Question Asked 7 years, 8 months ago. We are given a matrix, we need to check whether it is an orthogonal matrix or not. The rows of u are the same as the columns of UT. Orthogonal matrix. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. An orthogonal set of vectors is said to be orthonormal if .Clearly, given an orthogonal set of vectors , one can orthonormalize it by setting for each .Orthonormal bases in “look” like the standard basis, up to rotation of some type.. We call an matrix orthogonal if the columns of form an orthonormal … Linear Algebra - Definition of Orthogonal Matrix What is Orthogonal Matrix? The orthogonal matrix has all real elements in it. An interesting property of an orthogonal matrix P is that det P = ± 1. Orthogonal matrices are the most beautiful of all matrices. They can be written as A = ZQR[Z.sup.T], where Z is an orthonormal matrix, R is upper triangular, and Q is an orthonormal matrix such that parts of some columns and of some rows are zero, depending at which iterations the stagnation of the residual norms happens. (statistics) Statistically independent, with reference to variates. It's difficult to tell what is being asked here. "Would a square matrix with orthogonal columns, but not orthonormal, change the norm of a vector?" O D. Since U is the inverse of U, U is the inverse of UT. Orthogonal In mathematics, orthogonality is the generalization of the notion of perpendicularity to the linear algebra of bilinear forms. An orthogonal matrix is a matrix whose column vectors form an orthonormal set. Maybe you mean that the column should be [1;1;1;1;1;1] /sqrt(6). To learn more, see our tips on writing great answers. Subject: [matlab] Orthogonal random matrix > Hi, > > I want to generate orthogonal random matrix in matlab. whose product with the transpose A′ gives the identity matrix, that is, AA′ = E and A′ A = E. The elements of an orthogonal matrix satisfy the relations. The product of two orthogonal matrices is also an orthogonal matrix. The set of all linearly independent orthonormal vectors is an orthonormal basis. or the equivalent relations. An orthonormal basis for the range of matrix A is matrix B, such that: B'*B = I , where I is the identity matrix. By using our site, you acknowledge that you have read and understand our Cookie Policy, Privacy Policy, and our Terms of Service. For the second question: yes, the columns of the matrix are the image of the canonical base. Learn the orthogonal matrix definition and its properties. 2 직교행렬(orthogonal matrix)이면서 정방행렬(square matrix)인 단위행렬(identity matrix)의 시각화 Fig. Code Samples from TFCO — TensorFlow Constrained Optimization. Alternatively, a matrix is orthogonal if and only if its columns are orthonormal, meaning they are P Difference between orthogonal and orthonormal matrices. Solution: We know that a square matrix with orthonormal columns satisfies Q-1 = Q T, so QQ T = I. Conversely, if A T A = I In case Q is square, ofQ–1 Q A matrix P is orthogonal if P T P = I, or the inverse of P is its transpose. In other words, a square matrix whose column vectors (and row vectors) are mutually perpendicular (and have magnitude equal to 1) will be an orthogonal matrix. Transpose and the inverse of an orthonormal matrix are equal. The determinant of an orthogonal matrix has value +1 or -1. The concept of orthogonality for a matrix is defined for just one matrix: A matrix is orthogonal if each of its column vectors is orthogonal to all other column vectors and has norm 1. The columns of B span the same space as the columns of A . site design / logo © 2020 Stack Exchange Inc; user contributions licensed under cc by-sa. Gram-Schmidt orthogonalization, SVD decomposition and replace the diagonal matrix with identity matrix, QR decomposition and discard R, etc. I matrixteori er en reel ortogonal matrix (eller en reel ortogonalmatrix) en kvadratisk matrix Q hvis transponerede er dens inverse: = =. In case Q is square, of course this means that Q–1 = QT. > How can I do it? (n/> 0) are orthogonal matrices, i.e., The largest eigenvalue is Vektoren, die nicht nur orthogonal zueinander stehen sondern auch normiert sind, bezeichnet man als orthonormale Vektoren. By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy. Name for multiples of orthogonal matrices. There appears to be no established name for them. Show that if Q is a square matrix with orthonormal columns, then Q also has orthonormal rows. Is it illegal to carry someone else's ID or credit card? Is it more efficient to send a fleet of generation ships or one massive one? An orthogonal matrix of order n is a matrix. In order to understand the definition of a unitary matrix, we need to remember the following things. Proof that if Q is an n x n orthogonal matrix, then det(Q) = + - 1. Are there ideal opamps that exist in the real world? Orthogonal matrix is an important matrix in linear algebra, it is also widely used in machine learning. By taking the subspace to be the column space of a matrix, you will develop a method for producing approximate (“least-squares”) solutions for inconsistent systems. Given the eigenvector of an orthogonal matrix, x, it follows that the product of the transpose of x and x is zero. geometry), an estimate M of an orthonormal matrix Rrepresenting rotation is recovered. Since both generating a d d orthogonal matrix (O(d3) time and O(d2) space) and computing the transformation (O(d 2 ) time and space) are prohibitively expensive for high-dimensional data, we further propose Structured Orthogonal Random Features (SORF) in Section 4. 바로 표준 기저(standard basis) 이다. Let P and Q be orthonormal bases of U and V respectively. # Of grid graphs, board games and polyominoes: vertical or horizontal but not diagonal. Q=\left(\frac{x_1}{\|x_1\|},\ldots,\frac{x_n}{\|x_n\|}\right)\begin{pmatrix}\|x_1\|\\ &\ddots\\ &&\|x_n\|\end{pmatrix}=UD. The norm of the columns (and the rows) of an orthogonal matrix must be one. Let U be an nxn orthogonal matrix. Alternatively, a matrix is orthogonal if and only if its columns are orthonormal, meaning they are orthogonal and of unit length. O A. Notice that QTQ = I. Such matrices are usually denoted by the letter Q. Since $Q$ is unitary, it would preserve the norm of any vector $X$, i.e., $\|QX\|^2 = \|X\|^2$. The columns of ut form an orthonormal set. Is there any way that a creature could "telepathically" communicate with other members of it's own species? Viewed 1k times -2. The $k$-th column of the matrix $A$ is $A e_k$ where $e_k$ is the $k$-th element of the canonical basis. A p×q orthonormal matrix T=P with q p to q Orthogonal matrix multiplication can be used to represent rotation, there is an equivalence with quaternion multiplication as described here.
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