text:
Update October 31, 2016: I received an announcement that CS 294-112 will be taught again next semester! That sounds exciting, and while I won’t be enrolling in the course, I will be following its progress and staying in touch on the concepts taught.

And by the way, today I finally published my reinforcement learning post that I said I would write in my July update. You can see it here.

Update July 18, 2016: This post seems to have gotten a considerable amount of attention, at least compared to my normal blog posts, so I would like to answer some of the common questions that I’ve received in either the comments or by private email.

1. If you’re looking for homework assignments, I first want to warn you that, as I emphasize in the my review, the assignments are probably not going to be as educational as you would want them to be. If you’re still interested, our TA for the class posted a github repository on the Berkeley RLL page with the homework for this class. The homeworks are iPython notebooks (now called Jupyter notebooks, I think). If there’s code in the “YOUR CODE HERE” sections, then you’re probably reading the solutions; I’m not sure if there’s a clean version of the assignments there.

2. Unfortunately, we did not have any video lectures, slides, or readings outside of what you can see on the class website. A note for those who are reading the comments after this update: the class website was originally pulled down due to “some tyrants” (according to the course staff), but it’s happily now up.

3. If you’re looking for other resources to learn about deep reinforcement learning, I have several recommendations. In terms of courses, check out David Silver’s reinforcement learning course and the recent Machine Learning Summer School; the latter had our class instructor as part of the course staff, so the material is probably going to be similar to what we covered. (Coming up in a few weeks is the Deep Learning Summer School, something you might also want to check out.) I have all these courses bookmarked and am trying to carve out some time to read the slides. In terms of code, I would strongly recommend starting with either the deep_q_rl library or OpenAI Gym. The former is a super easy-to-read Python library that allows you to replicate DeepMind’s results in their 2013 and 2015 papers on Atari games. The latter was recently launched, and I don’t have experience with it, but it sounds really cool as we can compare our reinforcement learning implementations.

4. This is more of a comment than an answer, but I thought I’d mention it anyway: my blog’s comments are handled by Disqus, and in the moderation panel I can see the emails of the commenters. Thus, there is no need to post your email publicly as I can see it regardless.

Thanks everyone, and that’s all! After this paragraph is the original post as I had written it. But one more thing: after rereading this post, I think I was a little too harsh on the class. Furthermore, even though people have said they liked this post, I don’t think I gave reinforcement learning its due. So to rectify my regrets, I’m planning on launching a new series of deep reinforcement learning posts on this blog, similar to the style of Andrej Karpathy’s excellent blog post. I’ve already written a post on basic reinforcement learning, so I’m hoping to progress towards more advanced topics. My goal is to have the first post up sometime in August. Hopefully those will be a good resource for some enthusiasts out there.

At the start of the semester, Pieter Abbeel told me, “If you want to know more about the work we do, take the deep reinforcement learning class.”

“Which course is that? The one taught by John?” I asked.

“Yes, take that.” Pieter said. “Hope to see you there.”

I somewhat reluctantly agreed to join late, knowing that I had missed the first two lectures, and furthermore, that I would have to sit through the first one I did attend without sign language interpreting. In retrospect, however, interpreting services provided the least benefit for me in this class compared to any other class I had ever taken. Given my history of taking highly technical, difficult-to-interpret courses, that’s saying a lot!

So what was this course? It was a new two-credit class called Deep Reinforcement Learning (CS 294-112) and taught by Pieter’s graduate student, John Schulman. It seemed like a cross between a research seminar and a normal lecture course. The former tend to be one or two credits and are principally about relevant research results; the latter tend to be three or four credits and have lectures, homeworks, exams, and projects.

In AI and robotics, reinforcement learning is a standard way of framing a problem. For example, if a robot needs to learn how to play a game, it must engage in “reinforcement learning” to try out different actions, get rewards, and then modify its policy. The word “deep” refers to how deep neural networks have recently become the workhorse of state-of-the-art reinforcement learning. (This is why the class wasn’t taught until now.) The broader category of deep learning involves the use of deep neural networks in other applications, such as image classification and speech recognition. Deep learning has become so popular that Google even paid \$400 million to buy a deep learning company, DeepMind.

The class had about eighty students, so to avoid getting into trouble with the building managers about stuffing too many people in one room, John gave two identical lectures for each class day. I remained in the afternoon session to make it easy on the interpreters’ schedules, but unfortunately, most of the other students picked the afternoon session, but hey, they don’t have my excuse … perhaps they can’t wake up early? So once again, a graduate level class had some of its students sitting on the floor. Seems like that’s a common problem here, huh?

Anyway, back to the class discussion. The first few lectures were about Markov Decision Processes and neural networks, so if there were any classes to miss, it would be those because I already knew the material.

The remaining lectures were, to be frank, difficult, and I often felt mentally stressed in class. Most of the content was pure math, and the derivations were a long sequence of sums, expectations, and other terms, each of which were more sums and expectations. For instance, look at the formula for policy gradients:

To understand1 this, one has to process lots of material, such as what it means to take the gradient of the log of a policy, and that $\Psi_t$ isn’t just a simple scalar but can represent concepts like the advantage function, which involves another sequence of expectations and sums of rewards. Connecting this material is challenging in real time, and I felt that the lectures did not provide sufficient intuition. My sign language interpreters tried to repeat the exact words John uttered, but despite this, I could not translate this process into clear mathematical comprehension.

Given that the lectures were difficult for me to follow, I hoped that homeworks would be more useful. The homeworks in this class were provided as IPython/Jupyter notebooks. We had starter code and needed to fill in the “YOUR CODE HERE” sections.

The first homework was nearly trivial for people who knew about the basics of Markov Decision Process, Value Iteration, Policy Iteration, and Q-Learning. I wrote about thirty lines of Python code for the entire assignment.

The second homework, on policy gradients, was more interesting, but the release date kept getting postponed. It soon became a running joke in class whenever John said: “Oh, and about that second homework, we plan to release it in a few days…”. It was finally released on October 11. (John on Piazza: “You may have given up hope that this day would ever come, but behold, HW2 is finally here.”) To put this in perspective, the first homework assignment was due on September 7.

Fortunately, the second assignment was more challenging than the first, and I had to be careful in implementing formulas since math from research papers doesn’t always translate neatly into code. I was pleased to see that the homework was designed so naive implementations of formulas would take too long to test. (I believe AI assignments should require code to be reasonably optimized.)

We were going to have homeworks on approximate dynamic programming and supervised learning, but since the second homework got delayed so much and the third one would have taken too long to create, the staff canceled all future assignments.

To be honest, the main deep reinforcement learning material I learned this semester didn’t even come from this class. In Pieter’s Advanced Robotics (CS 287) class, which I also took this semester, my final project was about deep learning for Atari games. I had time to sufficiently read and absorb the Atari deep learning research papers, which helped me to better understand some of the material in this class (CS 294-112). Consequently, my recommendation for someone who wants to take this class in the future is to, if possible, take CS 287 concurrently and do a project that uses neural networks. That way, one gets to do deep reinforcement learning.

To recap, here are some of the positive aspects of the class:

• It covers a popular and interesting research area.
• It presents many relevant research papers, including those from Berkeley students.
• For a class that is almost like a research seminar, there are many online resources one can consult for additional background. Unfortunately, a lot of the written references are also hard to understand.
• It is easy to obtain homework help on Piazza.

Here are some of the negative ones:

• The lectures were not polished and involved lots of math without intuition. This issue is understandable because it was a first time course taught by a graduate student.
• There did not seem to be much advance preparation for the course in terms of lecture material. The course website had a brief outline of lectures, but we had to change some of that on short notice.
• It did not provide sufficiently many or sufficiently difficult homework assignments. Having more in-depth assignments would let me deeply reinforce my understanding (pun fully intended).

Ultimately, this course allowed me to scratch the surface of deep reinforcement learning, though it was immensely frustrating for me to try and understand the material directly from the lecture, and the haphazard nature of the course did not help. I suspect that future iterations of the course will proceed more smoothly, and yes, even though no one’s told me personally, this class will be offered again (in some form) so long as deep learning remains the king of machine learning.

1. Update November 3, 2016: After studying more about policy gradients, I now feel like I truly understand this formula.


text:
# Typeset one citation with all authors

I'm using the authoryear style in biblatex with the following options name length options:

maxcitenames=1, mincitenames=1, maxbibnames=999, minbibnames=999

So citations (usually issued with \cite) are properly typeset with only one author like

Shakespeare et al, The...

However I want to typeset a single citation (via \fullcite) with all authors listed, eg.

W. Shakespeare, J. Austen and D. Adams, The...

How can I temporarily increase the values of maxcitenames and mincitenames? Or is there another way to typeset only the \fullcite citatation with all names?

If you want to be able to make full use of all of biblatex's features for \fullcite, namely pre- and postnotes, you might want to try this version.

We temporarily set maxcitenames to maxbibnames; apparently, there is no need to re-set the counter, because everything is wrapped in a group, so the assignment is locally.

A simpler solution than the below wrapper is

\makeatletter
\DeclareCiteCommand{\longfullcite}
{\usebibmacro{prenote}}
{\usedriver
{\c@maxnames\blx@maxbibnames\relax
\DeclareNameAlias{sortname}{default}}
{\thefield{entrytype}}}
{\multicitedelim}
{\usebibmacro{postnote}}
\makeatother

\makeatletter
\newcommand{\tempmaxup}[1]{\def\blx@maxcitenames{\blx@maxbibnames}#1}
\makeatother

\DeclareCiteCommand{\longfullcite}[\tempmaxup]
{\usebibmacro{prenote}}
{\usedriver
{\DeclareNameAlias{sortname}{default}}
{\thefield{entrytype}}}
{\multicitedelim}
{\usebibmacro{postnote}}

Unfortunately, I have found no better way of doing this than via that wrapper command.

The MWE

\begin{filecontents}{\jobname.bib}
@Article{Reference:1994,
author           =  {First I. Last and Second Y. Author and Third Z. Author and Fourth Q. Author},
title            = {This is the article title},
journal          = {T Journal T},
journallongtitle = {The Journal Title},
year             = 1994,
volume           = 50,
number           = 6,
pages            = {30--40},
}
\end{filecontents}

\documentclass{article}
\usepackage[style=authoryear, maxcitenames=1, maxbibnames=999]{biblatex}

\makeatletter
\newcommand{\tempmaxup}[1]{\def\blx@maxcitenames{\blx@maxbibnames}#1}
\makeatother

\DeclareCiteCommand{\longfullcite}[\tempmaxup]
{\usebibmacro{prenote}}
{\usedriver
{\DeclareNameAlias{sortname}{default}}
{\thefield{entrytype}}}
{\multicitedelim}
{\usebibmacro{postnote}}

\begin{document}
\begin{tabular}{ l p{8cm} }
\verb|\autocite| & \autocite{Reference:1994}\\
\verb|\longfullcite| &\longfullcite{Reference:1994}\\
\verb|\longfullcite| & \longfullcite[see][14]{Reference:1994}\\
\verb|\fullcite| & \fullcite{Reference:1994}\\
\verb|\autocite| &\autocite{Reference:1994}\\
\end{tabular}

\printbibliography
\end{document}
% end of file comment

yields

• Nice. Why does \tmpmaxup have to be defined as taking an argument? – Alan Munn Nov 4 '13 at 11:48
• @AlanMunn Well, I could not get the redefinition of the counter to work in the pre- or post-code fields, so it had to be the wrapper command. The wrapper command must take one argument, which is the citation. (That is, if you take \mkbibparens as wrapper, the \cite command will we wrapped in parentheses.) – moewe Nov 4 '13 at 16:35
• Ok. That makes a bit more sense now. I hadn't thought of the analogy to \mkbibparens. (I also tried to put the redefinition in the pre-code without success.) – Alan Munn Nov 4 '13 at 16:43
• +1. Very nice solution. I've tried it and it works as advertised. – halloleo Nov 4 '13 at 23:08
• @moewe: I am having trouble using your \longfullcite command with certain Biber/BibTeX entries. The problem is too complicated to describe in a comment so I've opened a separate question for it. – Psychonaut Jan 27 '17 at 20:44

To change maxnames for one citation only you may use the counter:

\AtNextCite{\defcounter{maxnames}{99}}\fullcite{Reference:1994}

This sets both maxcitenames and maxbibnames but in your case this is not probably the problem. Unfortunately there is no maxcitenames counter in biblatex.

You may also define a dedicated command for such citations:

\newcommand{\longfullcite}{%
\AtNextCite{\defcounter{maxnames}{99}}%
\fullcite}

• +1 It seems that it will take me longer to learn biblatex than it did to learn LaTeX. – Alan Munn Nov 4 '13 at 14:02
• I adore biblatex for the things like this :) – Oleg Domanov Nov 4 '13 at 17:14

Here's a kind of hacky way to do this. Because of the complexity of the arguments of the \cite commands, it's not easy to wrap code around them. So the following solution assumes that you will just use \fullcite with a single argument and none of its optional arguments. I'd be interested in seeing a more elegant way to do this.

\begin{filecontents}{\jobname.bib}
@Article{Reference:1994,
author =   {First I. Last and Second Y. Author and Third Z. Author and Fourth Q. Author },
title =    {This is the article title},
journal =  {T Journal T},
journallongtitle =     {The Journal Title},
year =     1994,
volume =   50,
number = 6,
pages =    {30--40}
}
\end{filecontents}

\documentclass{article}
\usepackage[style=authoryear,maxcitenames=1, mincitenames=1, maxbibnames=999,
minbibnames=999]{biblatex}

\makeatletter
\newcommand{\upmax}{\def\blx@maxcitenames{99}}
\newcommand{\dnmax}{\def\blx@maxcitenames{1}}
\makeatother

\newcommand\myfullcite[1]{\upmax\fullcite{#1}\dnmax} % does not allow pre or postnotes

\begin{document}
\autocite{Reference:1994}

\myfullcite{Reference:1994}

\autocite{Reference:1994}

\printbibliography
\end{document}

format: plain


text:
# How do you simplify (2x^2y^4*4x^2y^4*3x)/(3x^-3y^2) and write it using only positive exponents?

Feb 17, 2017

See the entire simplification process below:

#### Explanation:

First, to simplify the numerator we will rewrite this expression as:

$\frac{\left(2 \cdot 3 \cdot 4\right) \left({x}^{2} \cdot {x}^{2} \cdot x\right) \left({y}^{4} \cdot {y}^{4}\right)}{3 {x}^{-} 3 {y}^{2}} \to \frac{24 \left({x}^{2} \cdot {x}^{2} \cdot x\right) \left({y}^{4} \cdot {y}^{4}\right)}{3 {x}^{-} 3 {y}^{2}}$

We can now use these two rules for exponents to simplify the numerator:

$a = {a}^{\textcolor{red}{1}}$ and ${x}^{\textcolor{red}{a}} \times {x}^{\textcolor{b l u e}{b}} = {x}^{\textcolor{red}{a} + \textcolor{b l u e}{b}}$

$\frac{24 \left({x}^{2} \cdot {x}^{2} \cdot {x}^{\textcolor{red}{1}}\right) \left({y}^{4} \cdot {y}^{4}\right)}{3 {x}^{-} 3 {y}^{2}} \to \frac{24 {x}^{2 + 2 + 1} {y}^{4 + 4}}{3 {x}^{-} 3 {y}^{2}} \to \frac{24 {x}^{5} {y}^{8}}{3 {x}^{-} 3 {y}^{2}} \to$

$\frac{8 {x}^{5} {y}^{8}}{{x}^{-} 3 {y}^{2}}$

We can now use this rule of exponents to complete the simplification:

${x}^{\textcolor{red}{a}} / {x}^{\textcolor{b l u e}{b}} = {x}^{\textcolor{red}{a} - \textcolor{b l u e}{b}}$

$\frac{8 {x}^{\textcolor{red}{5}} {y}^{\textcolor{red}{8}}}{{x}^{\textcolor{b l u e}{- 3}} {y}^{\textcolor{b l u e}{2}}} \to 8 {x}^{\textcolor{red}{5} - \textcolor{b l u e}{- 3}} {y}^{\textcolor{red}{8} - \textcolor{b l u e}{2}} \to 8 {x}^{\textcolor{red}{5} + \textcolor{b l u e}{3}} {y}^{\textcolor{red}{8} - \textcolor{b l u e}{2}} \to$

$8 {x}^{8} {y}^{6}$


text:
# Finite Mathematics and Calculus with Applications

All new and solved questions in Finite Mathematics and Calculus with Applications category

##### Actual Sales=75000 and Break even sales =25000 .Then Margin of safetywill be ?
Actual Sales=75000 and Break even sales =25000 .Then Margin of safety will be ?...
##### To compute the construction cost per square foot of a building
To compute the construction cost per square foot of a building...
##### Determine the average sale of chocolate bars at a mini mart where the sold bars are 4 pcs. at PHP 25.50 each ,6 pcs.at PHP 26.25 each ,5 pcs.at PHP 29.00 each ,and 7 pcs. for PHP 32.50 each.
Determine the average sale of chocolate bars at a mini mart where the sold bars are 4 pcs. at PHP 25.50 each ,6 pcs.at PHP 26.25 each ,5 pcs.at PHP 29.00 each ,and 7 pcs. for PHP 32.50 each....
##### In a lottery, 4 distinct numbers are chosen from 1 to 49. Players pay $10 to guess 4 numbers and win$30 for each number that matches the winning set. Let X be the net money gained by playing the lottery.
In a lottery, 4 distinct numbers are chosen from 1 to 49. Players pay $10 to guess 4 numbers and win$30 for each number that matches the winning set. Let X be the net money gained by playing the lottery....
##### In a lottery, 4 distinct numbers are chosen prom 1 to 49. Players pay $10 to guess 4 numbers and win$30 for each number that matches the winning set. Let X be the net moneygained by playing the lottery. Determine the probobility clistributienpor X. 7 decimal places .
In a lottery, 4 distinct numbers are chosen prom 1 to 49. Players pay $10 to guess 4 numbers and win$30 for each number that matches the winning set. Let X be the net moneygained by playing the lottery. Determine the probobility clistributienpor X. 7 decimal places ....
##### In a lottery, 4 distinct numbers are chosen prom 1 to 49. Players pay $10 to guess 4 numbers and win$30 for each number that matches the winning set. Let X be the net moneygained by playing the lottery. Determine the probability distribution for X. 7 decimal places .
In a lottery, 4 distinct numbers are chosen prom 1 to 49. Players pay $10 to guess 4 numbers and win$30 for each number that matches the winning set. Let X be the net moneygained by playing the lottery. Determine the probability distribution for X. 7 decimal places ....
suppose that...
##### Suppose there are four points {(8, 1),(4, 1),(0, 1),(?2, ?1)} in the (x, y)-space, where x isthe input variable and y is the label variable. We will attempt to linearly separate thesefour points.(a) Suppose the initialization parameters are ?1 = 1 and ?0 = ?1. Write out theequation of the initial linear separator in terms of x.(b) Identify the misclassified point(s) in the set of four points using this initiallinear separator.(c) We will perform one step of the perceptron algorithm. Choose one
Suppose there are four points {(8, 1),(4, 1),(0, 1),(?2, ?1)} in the (x, y)-space, where x is the input variable and y is the label variable. We will attempt to linearly separate these four points. (a) Suppose the initialization parameters are ?1 = 1 and ?0 = ?1. Write out the equation of the in...


text:
# How do you simplify (2-3i)(4+i)?

Feb 5, 2016

To simplify this expression, multiply top and bottom by the complex conjugate, $\left(4 - i\right)$. This yields $\frac{11 - 4 i}{17}$

#### Explanation:

To simplify an expression like this, we multiply top and bottom of the fraction (denominator and numerator) by the complex conjugate of the bottom (numerator). For a complex expression $\left(a + b i\right)$, the complex conjugate is $\left(a - b i\right)$. When we do the calculation, it will become clear why this works so well.

Note that $\frac{a - b i}{a - b i}$ is just the same as 1, so when we do this multiplication the result is the same number we started with.

$\frac{2 - 3 i}{4 + i} \cdot \frac{4 - i}{4 - i} = \frac{8 - 12 i - 2 i + 3 {i}^{2}}{16 + 4 i - 4 i - {i}^{2}}$

But since $i = \sqrt{-} 1$, ${i}^{2} = - 1$. Using this and collecting like terms:

$\frac{8 - 14 i - 3}{16 + 1} = \frac{11 - 4 i}{17}$


text:
Does an oriented $S^3$ fiber bundle admit the structure of a principal $SU(2)$-bundle?

Let $S \to X$ be an $S^3$-fiber bundle over a smooth manifold $X$. If $S$ is an oriented manifold does this fiber bundle admit the structure of an $SU(2)$-principal bundle?

There is a similar theorem for the case of circle bundles and is proved in Morita's book on differential forms. Unfortunately, I do not see a way to extend his argument to this case and he does not discuss $SU(2)$-principal bundles in detail.

• A nontrivial result of A. Hatcher says that the diffeomorphism group of $S^3$ is homotopic with $O(4)$. If the bundle is orientable, then the bundle has an $SO(4)$-structure, i.e., it is the unit sphere bundle of an oriented rank $4$ real vector bundle. – Liviu Nicolaescu Sep 29 '17 at 8:47
• (I'm assuming you mean that the bundle is orientable not just the total space). From there maybe use that SO(4)/SU(2) \cong SO(3). This is topologically split, and SO(3) has nontrivial pi_4. Using the splitting gets you an oriented vector bundle on S^5 that doesn't have an SU(2) structure. Of course, with the counterexample in hand, we didn't need the Smale conjecture :) – Dylan Wilson Sep 29 '17 at 9:07
• Dylan, maybe you meant $\pi_3$, so a bundle over $S^4$? – Tom Goodwillie Sep 29 '17 at 23:22

No. (The main idea here is present in Dylan Wilson's comment.)

Every principal $SU(2)$-bundle over $S^2$ is trivial, because $\pi_1 SU(2)$ is trivial. But there is a nontrivial oriented bundle over $S^2$ with fiber $S^3$, namely the unit sphere bundle of the nontrivial rank $4$ vector bundle. (There are precisely two rank $r$ oriented vector bundles over $S^2$ if $r\ge 3$, because $\pi_1 SO(r)$ has order two.)

I should explain why the nontrivial vector bundle has nontrivial unit sphere bundle, that is, why $\pi_1 SO(4)$ injects into $\pi_1 SDiff(S^3)$.

You can use the big theorem of Hatcher ("Smale Conjecture"), which says that $\pi_k SO(4)$ maps isomorphically to $\pi_k SDiff(S^3)$ for all $k$.

Alternatively, you can use that the bundle is detected by the Stiefel-Whitney class $w_2$, and that Stiefel-Whitney classes of vector bundles are invariant under fiber homotopy equivalence of unit sphere bundle.

Or you can use $\pi_3$ instead of $\pi_1$ as suggested by Dylan Wilson, making a bundle over $S^4$; there are elements of $\pi_3 SO(4)\cong \mathbb Z\times \mathbb Z$ not coming from $\pi_3 SU(2)\cong\mathbb Z$. Again, the resulting sphere bundle is nontrivial either by Hatcher's theorem or by using characteristic classes. I'm not sure what the most elementary version of the characteristic-class argument would be.

One example of a rank $4$ real vector bundle over $S^4$ that does not admit a complex structure is the tangent bundle!


text:
1. Exploratory Data Analysis
1.3. EDA Techniques
1.3.5. Quantitative Techniques
1.3.5.18. Yates Algorithm

## Important Factors

Identify Important Factors We want to select the most appropriate model to represent our data. This requires balancing the following two goals.
1. We want the model to include all important factors.
2. We want the model to be parsimonious. That is, the model should be as simple as possible.
In short, we want our model to include all the important factors and interactions and to omit the unimportant factors and interactions.

Seven criteria are utilized to define important factors. These seven criteria are not all equally important, nor will they yield identical subsets, in which case a consensus subset or a weighted consensus subset must be extracted. In practice, some of these criteria may not apply in all situations.

These criteria will be examined in the context of the eddy current data set. The parameter estimates computed using least-squares analysis are shown below.

     Parameter    Estimate
---------    --------
Mean          2.65875
X1            1.55125
X2           -0.43375
X1*X2         0.06375
X3            0.10625
X1*X3         0.12375
X2*X3         0.14875
X1*X2*X3      0.07125

In practice, not all of these criteria will be used with every analysis (and some analysts may have additional criteria). These critierion are given as useful guidelines. Most analysts will focus on those criteria that they find most useful.

Criteria for Including Terms in the Model The seven criteria that we can use in determining whether to keep a factor in the model can be summarized as follows. The first four criteria focus on parameter estimates with three numeric criteria and one graphical criteria. The fifth criteria focuses on effects, which are twice the parameter estimates. The last two criteria focus on the residual standard deviation of the model. We discuss each of these seven criteria in detail in the sections that following.
Parameters: Engineering Significance The minimum engineering significant difference is defined as
$$|\hat{\beta_{i}}| > \Delta$$
where $$|\hat{\beta_{i}}|$$ is the absolute value of the parameter estimate and Δ is the minimum engineering significant difference.

That is, declare a factor as "important" if the parameter estimate is greater than some a priori declared engineering difference. This implies that the engineering staff have in fact stated what a minimum difference will be. Oftentimes this is not the case. In the absence of an a priori difference, a good rough rule for the minimum engineering significant Δ is to keep only those factors whose parameter estimate is greater than, say, 10% of the current production average. In this case, let's say that the average detector has a sensitivity of 2.5 ohms. This would suggest that we would declare all factors whose parameter is greater than 10 % of 2.5 ohms = 0.25 ohm to be significant (from an engineering point of view).

Based on this minimum engineering significant difference criterion, we conclude that we should keep two terms: X1 and X2.

Parameters: Order of Magnitude The order of magnitude criterion is defined as
$$|\hat{\beta_{i}}| < 0.10 \cdot \max |\hat{\beta_{i}}|$$
That is, exclude any factor that is less than 10 % of the maximum parameter size. We may or may not keep the other factors. This criterion is neither engineering nor statistical, but it does offer some additional numerical insight. For the current example, the largest parameter is from X1 (1.55125 ohms), and so 10 % of that is 0.155 ohms, which suggests keeping all factors whose parameters exceed 0.155 ohms.

Based on the order-of-magnitude criterion, we thus conclude that we should keep two terms: X1 and X2. A third term, X2*X3 (0.14875), is just slightly under the cutoff level, so we may consider keeping it based on the other criterion.

Parameters: Statistical Significance Statistical significance is defined as
$$|\hat{\beta_i}| > 2 \cdot \mbox{s.e.}(\hat{\beta_i})$$
That is, declare a factor as important if its parameter is more than 2 standard deviations away from 0 (0, by definition, meaning "no effect").

The "2" comes from normal theory (more specifically, a value of 1.96 yields a 95 % confidence interval). More precise values would come from t-distribution theory.

The difficulty with this is that in order to invoke this criterion we need the standard deviation, σ of an observation. This is problematic because

1. the engineer may not know σ;
2. the experiment might not have replication, and so a model-free estimate of σ is not obtainable;
3. obtaining an estimate of σ by assuming the sometimes employed assumption of ignoring 3-term interactions and higher may be incorrect from an engineering point of view.
For the eddy current example:
1. the engineer did not know σ;
2. the design (a 23 full factorial) did not have replication;
3. ignoring 3-term interactions and higher interactions leads to an estimate of σ based on omitting only a single term: the X1*X2*X3 interaction.
For the eddy current example, if one assumes that the 3-term interaction is nil and hence represents a single drawing from a population centered at zero, then an estimate of the standard deviation of a parameter is simply the estimate of the 3-factor interaction (0.07125). Two standard deviations is thus 0.1425. For this example, the rule is thus to keep all $$|\hat{\beta_{i}}|$$

This results in keeping three terms: X1 (1.55125), X2 (-0.43375), and X1*X2 (0.14875).

Parameters: Probability Plots Probability plots can be used in the following manner.
1. Normal Probability Plot: Keep a factor as "important" if it is well off the line through zero on a normal probability plot of the parameter estimates.

2. Half-Normal Probability Plot: Keep a factor as "important" if it is well off the line near zero on a half-normal probability plot of the absolute value of parameter estimates.
Both of these methods are based on the fact that the least-squares estimates of parameters for these two-level orthogonal designs are simply half the difference of averages and so the central limit theorem, loosely applied, suggests that (if no factor were important) the parameter estimates should have approximately a normal distribution with mean zero and the absolute value of the estimates should have a half-normal distribution.

Since the half-normal probability plot is only concerned with parmeter magnitudes as opposed to signed parameters (which are subject to the vagaries of how the initial factor codings +1 and -1 were assigned), the half-normal probability plot is preferred by some over the normal probability plot.

Normal Probablity Plot of Parameters The following normal probability plot shows the parameter estimates for the eddy current data.

For the example at hand, the probability plot clearly shows two factors (X1 and X2) displaced off the line. All of the remaining five parameters are behaving like random drawings from a normal distribution centered at zero, and so are deemed to be statistically non-significant. In conclusion, this rule keeps two factors: X1 (1.55125) and X2 (-0.43375).

Averages: Youden Plot A Youden plot can be used in the following way. Keep a factor as "important" if it is displaced away from the central-tendancy "bunch" in a Youden plot of high and low averages. By definition, a factor is important when its average response for the low (-1) setting is significantly different from its average response for the high (+1) setting. (Note that effects are twice the parameter estimates.) Conversely, if the low and high averages are about the same, then what difference does it make which setting to use and so why would such a factor be considered important? This fact in combination with the intrinsic benefits of the Youden plot for comparing pairs of items leads to the technique of generating a Youden plot of the low and high averages.
Youden Plot of Effect Estimates The following is the Youden plot of the effect estimatess for the eddy current data.

For the example at hand, the Youden plot clearly shows a cluster of points near the grand average (2.65875) with two displaced points above (factor 1) and below (factor 2). Based on the Youden plot, we conclude to keep two factors: X1 (1.55125) and X2 (-0.43375).

Residual Standard Deviation: Engineering Significance This criterion is defined as
Residual Standard Deviation > Cutoff
That is, declare a factor as "important" if the cumulative model that includes the factor (and all larger factors) has a residual standard deviation smaller than an a priori engineering-specified minimum residual standard deviation.

This criterion is different from the others in that it is model focused. In practice, this criterion states that starting with the largest parameter, we cumulatively keep adding terms to the model and monitor how the residual standard deviation for each progressively more complicated model becomes smaller. At some point, the cumulative model will become complicated enough and comprehensive enough that the resulting residual standard deviation will drop below the pre-specified engineering cutoff for the residual standard deviation. At that point, we stop adding terms and declare all of the model-included terms to be "important" and everything not in the model to be "unimportant".

This approach implies that the engineer has considered what a minimum residual standard deviation should be. In effect, this relates to what the engineer can tolerate for the magnitude of the typical residual (the difference between the raw data and the predicted value from the model). In other words, how good does the engineer want the prediction equation to be. Unfortunately, this engineering specification has not always been formulated and so this criterion can become moot.

In the absence of a prior specified cutoff, a good rough rule for the minimum engineering residual standard deviation is to keep adding terms until the residual standard deviation just dips below, say, 5 % of the current production average. For the eddy current data, let's say that the average detector has a sensitivity of 2.5 ohms. Then this would suggest that we would keep adding terms to the model until the residual standard deviation falls below 5 % of 2.5 ohms = 0.125 ohms.

                                                            Residual
Model                                                      Std. Dev.
-----------------------------------------------------      ---------
Mean + X1                                                    0.57272
Mean + X1 + X2                                               0.30429
Mean + X1 + X2 + X2*X3                                       0.26737
Mean + X1 + X2 + X2*X3 + X1*X3                               0.23341
Mean + X1 + X2 + X2*X3 + X1*X3 + X3                          0.19121
Mean + X1 + X2 + X2*X3 + X1*X3 + X3 + X1*X2*X3               0.18031
Mean + X1 + X2 + X2*X3 + X1*X3 + X3 + X1*X2*X3 + X1*X2            NA

Based on the minimum residual standard deviation criteria, and we would include all terms in order to drive the residual standard deviation below 0.125. Again, the 5 % rule is a rough-and-ready rule that has no basis in engineering or statistics, but is simply a "numerics". Ideally, the engineer has a better cutoff for the residual standard deviation that is based on how well he/she wants the equation to peform in practice. If such a number were available, then for this criterion and data set we would select something less than the entire collection of terms.
Residual Standard Deviation: Statistical Significance This criterion is defined as
Residual Standard Deviation > σ
where σ is the standard deviation of an observation under replicated conditions.

That is, declare a term as "important" until the cumulative model that includes the term has a residual standard deviation smaller than σ. In essence, we are allowing that we cannot demand a model fit any better than what we would obtain if we had replicated data; that is, we cannot demand that the residual standard deviation from any fitted model be any smaller than the (theoretical or actual) replication standard deviation. We can drive the fitted standard deviation down (by adding terms) until it achieves a value close to σ, but to attempt to drive it down further means that we are, in effect, trying to fit noise.

In practice, this criterion may be difficult to apply because

1. the engineer may not know σ;
2. the experiment might not have replication, and so a model-free estimate of σ is not obtainable.

For the current case study:

1. the engineer did not know σ;
2. the design (a 23 full factorial) did not have replication. The most common way of having replication in such designs is to have replicated center points at the center of the cube ((X1,X2,X3) = (0,0,0)).

Thus for this current case, this criteria could not be used to yield a subset of "important" factors.

Conclusions In summary, the seven criteria for specifying "important" factors yielded the following for the eddy current data:

 Parameters, Engineering Significance: X1, X2 Parameters, Numerically Significant: X1, X2 Parameters, Statistically Significant: X1, X2, X2*X3 Parameters, Probability Plots: X1, X2 Effects, Youden Plot: X1, X2 Residual SD, Engineering Significance: all 7 terms Residual SD, Statistical Significance: not applicable

Such conflicting results are common. Arguably, the three most important criteria (listed in order of most important) are:

 Parameters, Probability Plots: X1, X2 Parameters, Engineering Significance: X1, X2 Residual SD, Engineering Significance: all 7 terms

Scanning all of the above, we thus declare the following consensus for the eddy current data:

1. Important Factors: X1 and X2
2. Parsimonious Prediction Equation:

$$\hat{Y} = 2.65875 + 1.55125 \cdot X_1 - 0.43375 \cdot X_2$$

(with a residual standard deviation of 0.30429 ohms)

Note that this is the initial model selection. We still need to perform model validation with a residual analysis.


text:
# What fraction is equivalent to 15

Answer: 15% is expressed as 3/20 as a fraction.

## What fraction is equivalent to 1.15?

The fraction 2/30 is equal to 1/15 when reduced to lowest terms. To find equivalent fractions, just multiply the numerator and denominator of that reduced fraction (1/15) by any interger number, ie, multiply by 2, 3, 10, 30 … and so on …

## How do you change a percent into a fraction?

Convert Percents to Fractions To convert a Percent to a Fraction follow these steps: Step 1: Write down the percent divided by 100 like this: percent 100 Step 2: If the percent is not a whole number, then multiply both top and bottom by 10 for every number after the decimal point. (For example, if there is one number after the decimal, then use 10, if there are two then use 100, etc.)

## How do you convert a decimal into a fraction?

• First write the given decimal in the form of a ratio (p/q), where the denominator is equal to 1.
• Now multiply the numerator and denominator by multiples of 10, for every decimal point, such that the decimal in the numerator becomes a whole number. …
• Simply the resulting fraction

## How do you divided a fraction?

Dividing Fractions. Turn the second fraction upside down, then multiply. There are 3 Simple Steps to Divide Fractions: Step 1. Turn the second fraction (the one you want to divide by) upside down (this is now a reciprocal). Step 2. Multiply the first fraction by that reciprocal Step 3.

## What is the equal of 15?

15 = 1 x 15 or 3 x 5. Factors of 15: 1, 3, 5, 15.

## What is 15 in its simplest form?

1 Answer. The simplified fraction form of 15% is 320 .

## What is an equivalent fraction of 15 20?

2 equivalent fractions of 15/20=30/40 & 45/60.

## What is the equivalent fraction of 12 by 15?

:4/5. 4/5×3/3=12/15 so equivalent fraction is 4/5 ….

## What is 15 as a decimal?

0.15Convert from percent to decimal: For example, 15% is equivalent to the decimal 0.15.

## What is fraction equivalent?

Equivalent fractions are fractions that represent the same value, even though they look different. For example, if you have a cake, cut it into two equal pieces, and eat one of them, you will have eaten half the cake.

## What is 1/3 the same as?

Answer: The fractions equivalent to 1/3 are 2/6, 3/9, 4/12, etc. Equivalent fractions have the same value in the reduced form. Explanation: Equivalent fractions can be written by multiplying or dividing both the numerator and the denominator by the same number.

## What is 3/5 equivalent to as a fraction?

So, 3/5 = 6/10 = 9/15 = 12/20.

## What does 2 5 equal as a fraction?

Answer: The fractions equivalent to 2/5 are 4/10, 6/15, 8/20, etc.

## What is the decimal for 12 15?

Let’s look at a quick example of this by using the fraction 1215 and converting it to decimal using a calculator. As you can see, in one quick calculation, we’ve converted the fraction 1215 into it’s decimal expression, 0.8.

## What grade is a 12 15?

80 %Your grade for 12 of 15 is 80 % / B-

## What is 5/6 equal to as a fraction?

10/12Decimal and Fraction Conversion ChartFractionEquivalent Fractions5/610/1260/721/72/1412/842/74/1424/843/76/1436/8423 more rows

## What is in simplified form?

1:184:10What is the Simplest Form of a Fraction? | Don’t Memorise – YouTubeYouTubeStart of suggested clipEnd of suggested clip12 by 6 is 2 and 18 by 6 is 3 12 by 18 and 2 by 3 are the same and this 2 by 3 here is called theMore12 by 6 is 2 and 18 by 6 is 3 12 by 18 and 2 by 3 are the same and this 2 by 3 here is called the simplest form it means that this fraction cannot be reduced further.

## Can you simplify 3 15?

0:391:25How to simplify 3/15 – YouTubeYouTubeStart of suggested clipEnd of suggested clip5 right so that is actually our final answer our original 13 over 15 becomes 1 over 5 and that isn’tMore5 right so that is actually our final answer our original 13 over 15 becomes 1 over 5 and that isn’t divisible any. Further. So that is our final simplified or reduced fraction.

## What is the simplified fraction of 5 15?

1 Answer. Aviv S. The simplified form is 13 .

## What is the simplest form of 6 15?

2/5Answer: The fraction 6/15 in simplest form is 2/5. The given fraction 6/15, is further simplified by dividing both numerator and denominator by 3.

## What are Equivalent Fractions?

Equivalent fractions are fractions with different numbers representing the same part of a whole. They have different numerators and denominators, but their fractional values are the same.

## How to make a fraction equivalent?

Multiply both the numerator and denominator of a fraction by the same whole number. As long as you multiply both top and bottom of the fraction by the same number, you won’t change the value of the fraction , and you’ll create an equivalent fraction.

## What is half of a fraction?

For example, think about the fraction 1/2. It means half of something. You can also say that 6/12 is half, and that 50/100 is half. They represent the same part of the whole. These equivalent fractions contain different numbers but they mean the same thing: 1/2 = 6/12 = 50/100

## What are Equivalent Fractions?

Equivalent fractions state that two or more than two fractions are said to be equal if both results are the same fraction after simplification. Let us say, a/b and c/d are two fractions, after the simplification of these fractions, both result in equivalent fractions, say e/f, then they are equal to each other.

## Why are fractions the same?

These fractions are actually the same because when we multiply or divide both the numerator and the denominator by the same number, the value of the fraction actually doesn’t change. Therefore, equivalent fractions, when reduced to their simplified value, will all be the same. For example, consider the fraction 1/5.

## What happens when two fractions are given?

If two fractions are given, we can simply find their decimals to check if they are equivalent fractions.

## How to make the denominator equal to 9?

Multiply 2/3 by 3/3 to make the denominator equal to 9.

## Is a fraction equivalent to a decimal?

Since, both the fractions results in the same decimal, thus they are equivalent.

## Do denominators have a common multiple?

The answer to this question is that, as the numerator and denominator are not co-prime numbers, therefore they have a common multiple, which on division gives exactly the same value.

## Is 10/12 a fraction?

But, ⅚ and 10/12 are equivalent fractions because of 10/12 = ⅚.

## How to find equivalent fractions?

To find equivalent fractions, you just need to multiply the numerator and denominator of that reduced fraction ( 1 15) by the same natural number, ie, multiply by 2, 3, 4, 5, 6

## Is 1 15 a fraction?

Important: 1 15 looks like a fraction, but it is actually an improper fraction.

## Can you convert fractions to decimals?

This Equivalent Fractions Table/Chart contains common practical fractions. You can easily convert from fraction to decimal, as well as, from fractions of inches to millimeters.

## How to find equivalent fractions?

To find equivalent fractions, you just need to multiply the numerator and denominator of that reduced fraction ( 56) by the same natural number, ie, multiply by 2, 3, 4, 5, 6

## Is 15 18 a fraction?

Important: 15 18 looks like a fraction, but it is actually an improper fraction.

## Can you convert fractions to decimals?

This Equivalent Fractions Table/Chart contains common practical fractions. You can easily convert from fraction to decimal, as well as, from fractions of inches to millimeters.

## How to find equivalent fractions?

To find equivalent fractions, you just need to multiply the numerator and denominator of that reduced fraction ( 215) by the same natural number, ie, multiply by 2, 3, 4, 5, 6

## Is 2 15 a fraction?

Important: 2 15 looks like a fraction, but it is actually an improper fraction.

## Can you convert fractions to decimals?

This Equivalent Fractions Table/Chart contains common practical fractions. You can easily convert from fraction to decimal, as well as, from fractions of inches to millimeters.

## How to find equivalent fractions?

Two fractions are equivalent when they are both equal when written in lowest terms. The fraction 24 30 is equal to 4 5 when reduced to lowest terms. To find equivalent fractions, you just need to multiply the numerator and denominator of that reduced fraction ( 4 5) by the same natural number, ie, multiply by 2, 3, 4, 5, 6 …

## Equivalent Fractions – Samples

While every effort is made to ensure the accuracy of the information provided on this website, neither this website nor its authors are responsible for any errors or omissions. Therefore, the contents of this site are not suitable for any use involving risk to health, finances or property.


text:
## Divide 180 into two part which are in the ratio 3:2

Question

Divide 180 into two part which are in the ratio 3:2

in progress 0
3 days 2021-10-12T16:19:20+00:00 2 Answers 0 views 0

let the ratio be 3x&2x

3x+2x=180 since we have to divide 180 in the ratio of 3:2

5x=180x

x=180/5

x=36

2. As given, we need to find two numbers that satisfy the ratio 2:3 and their sum must me 180. So we can take 2x + 3x = 180. So we have got the equation, Let’s get started


text:
#### Authors

David Eriksson, Kun Dong, Eric Lee, David Bindel, Andrew G. Wilson

#### Abstract

Gaussian processes (GPs) with derivatives are useful in many applications, including Bayesian optimization, implicit surface reconstruction, and terrain reconstruction. Fitting a GP to function values and derivatives at $n$ points in $d$ dimensions requires linear solves and log determinants with an ${n(d+1) \times n(d+1)}$ positive definite matrix-- leading to prohibitive $\mathcal{O}(n^3d^3)$ computations for standard direct methods. We propose iterative solvers using fast $\mathcal{O}(nd)$ matrix-vector multiplications (MVMs), together with pivoted Cholesky preconditioning that cuts the iterations to convergence by several orders of magnitude, allowing for fast kernel learning and prediction. Our approaches, together with dimensionality reduction, allows us to scale Bayesian optimization with derivatives to high-dimensional problems and large evaluation budgets.


text:
# Difference between revisions of "Pootle User Guide/Translation Built-in Help"

 [[{{{PrevPage}}}|< Previous Page ]] [[{{{NextPage}}}|Next Page >]]

## Basics about the Built-in Help Format

The format of the help is derived from XML and specified in the Understanding and Authoring OpenOffice 2.0 Online Help (pdf) document. Translators need not know all details, but should have a basic understanding of some terms.

Some content here is taken from Localizing Documentation (pdf) and adapted. The book itself refers to another translation tool, which was used for OpenOffice.org 2 and cannot be used directly.

Element
An XML element is a logical xml fragment that is surrounded by tags, e.g. <paragraph role="heading">Paragraph Content</paragraph>
Tag
A tag is the delimiter of an element and is enclosed in brackets:<>. Most elements have an opening tag at the beginning and a closing tag at the end. The opening tag may contain attributes, for example <paragraph role="heading"> in the example above. The closing tag always starts with a slash / and does not contain attributes, for example </paragraph> in the example above. Some tags don't have any content, for example

. These are sometimes also written as one standalone tag, for example
.
Attribute
An attribute might appear in the opening or standalone tag of an element and is used to further specify the element, e.g. role in the example above. An attribute always is associated with a value, e.g. "heading" in the example above.
Child element
Element content can contain other (nested) elements. Such is called child element. You must copy the opening and closing tag of the child element exactly and translate the content of the child element.

You must not translate anything inside a tag but only the element content. The only exception is the value of the name attribute of a link element.

Example File: helpcontent2/source/text/swriter/guide.po, Location: words_count.xhp#par_id3149821.5.help.text, String 12958811

Choose <emph>Tools - Word Count</emph>.

<emph> is the opening tag, </emph> is the closing tag. Do not translate them. You have only to translate the parts Choose and Tools - Word Count.

In the source text in Pootle you see

Notice, that Pootle uses different colors for the tags and for the actual text.

The tag pair <emph> … </emph> causes, that the content is emphasized. This is done by showing the content in bold.

Sometimes Pootle shows you only the content, but in other cases you see the tags as well.

## Line Breaks and Whitespace

Sometimes you see red wave-lines, see picture below. They belong to line breaks in the source file. Such line breaks do not influence how the help text is shown, but are used by the developers to make the source text easier to read. You can ignore these red wave-lines.

The line breaks you see in Pootle depend on the width of the browser window and are irrelevant too. Write your text in the input field without any explicit line break. Texts which have to go to different paragraphs are separate records in Pootle.

## Strings Used in the Index of the Help

Look at the example File:helpcontent2/source/text/swriter/guide.po, Locations: words_count.xhp#bm_id3149686.help.text, string 12958808.

The help file has the content

<bookmark xml-lang="en-US" branch="index" id="bm_id3149686">
<bookmark_value>words; counting in text</bookmark_value>
<bookmark_value>number of words</bookmark_value>
<bookmark_value>documents; number of words/characters</bookmark_value>
<bookmark_value>text; number of words/characters</bookmark_value>
<bookmark_value>characters; counting</bookmark_value>
<bookmark_value>number of characters</bookmark_value>
<bookmark_value>counting words</bookmark_value>
<bookmark_value>word counts</bookmark_value>
</bookmark>

You see in Pootle

The bookmark element in connection with the attribute branch="index" means, that this is content to be shown in the Index tab of the help. Each of the bookmark_value elements determines an item for the index. If their is a semicolon, the left part is the main item and the right part is a subitem. Items in the index are sorted automatically, so you need not care of alphabetical order.

## Embedded Content

The help format allows to embed content from one help file into another. When you read text in the help window, you will not notice this. Use the help manipulation described in the section Transforming the Style of the Built-in Help to make such embedding visible.

Embedding is done by the elements embed and embedvar, for example <embedvar href="text/shared/guide/paintbrush.xhp#formatpaintbrush"/>. Such elements refer to a paragraph, section or variable in another file by the filename and the id value of the referenced part, formatpaintbrush in the example. The id is shown in the manipulated help, so that you can search for it in Pootle by searching for id="an id" in the source text. For example search for id="formatpaintbrush" to find the origin of the above mentioned embedded text fragment.

## Images

Images are included by an image element. Inside of it you see an alt child element. It specifies an alternative text for the image. At a first glance it is confusing. But look at the colors in the source text. In the example you see, that only the word Icon has to be translated.

The image element may have a caption child element, but that is currently not used.

## Extended Tooltips

The text for the extended tips are stored in the built-in help and therefore their text is in the project Apache OpenOffice 4.x Help although the text is shown in the UI.

Extended tips are provided by an ahelp element. It has a hid attribute to specify the help identifier, which connects the text to the UI element.

If you have set the environment variable HELP_DEBUG=true, you see the extended tip itself in the upper part and the help identifier in the lower part of the tip.

If you know the help identifier of an UI element, you can search for the related text in Pootle. But in contrast to the project Apache OpenOffice 4.x you do not search in location but in source text.

Unfortunately, if the attribute is hid=".", the help identifier is given earlier in the help file, in a branch attribute of a bookmark element for example. And such bookmark element has no content to translate and therefore no entry in the Pootle database.

Example: Edit a chart and call the context menu. It has the item Data Ranges. The extended tooltip shows Opens the Data Ranges dialog where you can edit Data Range and Data Series. and the help identifier uno.DataRanges. Imagine the text has an error and you need to find the string in Pootle to correct it. Your search for DataRanges gets no result.

In such cases you should search for the extended tip itself. If you use not only a single word but a longer part of the tip, the number of search results will be small.

Sometimes the attribute visibility="hidden" is used. Translate the content nevertheless. The value "hidden" causes, that the text is only shown as extended tooltip and not in the help window.

## Switches

Switches are used to show only one of several alternative strings in the help window. Which content is shown is determined at runtime based on a condition. The switch element chooses a paragraph or section, the switchinline element chooses a fragment inside a paragraph.

The type of condition is given by a select attribute. Possible values are:

sys
Operating system. Runtime conditions are WIN UNIX MAC
appl
Application. Runtime conditions are WRITER CALC DRAW IMPRESS MATH BASIC CHART
distrib, target, lang, ver
These conditions are currently not used.

The different contents are given in case or caseinline elements. They have a select attribute too. It has one of the possible runtime condition as value. The default or defaultinline element is used, when the current runtime condition does not fulfill any of the other case or caseinline elements.

In a usual help window you see only those texts, where the condition is fulfilled by your current runtime environment.

On a Windows or Linux system the user sees hold down Ctrl and, on a MAC system the user sees hold down Command and.

The user on a Windows system sees The scanner must support the TWAIN standard. The user on a Linux system sees The scanner must support the SANE standard. The user on a MAC system sees nothing.

Calling the help from Impress the user sees Rename Slide, calling it from Draw or another application but Impress the user sees Rename Page.

Being a translator you likely want to see each possible text in the help window. For that purpose you have to manipulate your installation as described in section Transforming the Style of the Built-in Help.

The link element creates a hyperlink, similar to those known from webpages. The target URL is given in the href attribute. The content of the link element provides the visible text of the link.

According to the specification, a link element should have a name attribute to fulfill accessibility requirements. And in contrast to all other elements, the value of the name attribute has to be translated, although it is inside the opening tag.

## Items

The item element is used to characterize a part of the text. It is mostly used to provide a unique formatting for a class of objects. The class itself is given in the type attribute. For translators the case type="literal" is relevant. It indicates, that the content should perhaps not be translated, for example because it is a book title or a Basic statement. In other cases it may be translated, for example for localized function names in Calc. So you have to examine the content carefully to decide whether a translation is needed.

## Placeholders

The help contains some placeholders. You will see current placeholders %PRODUCTNAME and %PRODUCTVERSION, and the deprecated placeholders $[officename] and $[officeversion]. When the help is displayed at runtime, the Main Transformation Style Sheet main_transform.xsl replaces the placeholder with the information from the configuration file.

Do not translate placeholders, but transfer them unchanged. Do not use the product name directly but always a placeholder. Otherwise the help will be wrong for different branding.

## What must not be Translated

ToDo

<There are some examples in LibreOffice Wiki and on http://openoffice.apache.org/translate.html>

<show examples for each item>

<tag>

<placeholder>

<StarBasic source code>

## Finding items

ToDo

<Find an item by filename and paragraph ID., link to Pootle_User_Guide/Technical_Helpers#Transforming_the_Style_of_the_Built-in_Help >

### Finding errors from GSIcheck in POOTLE

After translating you might have checked the translated strings with GSIcheck. Which could have given you a list of ERROR and/or WARNING to be corrected, before the files are merged into POOTLE. By slightly adjusting the filename and paragraph identifier from the errorreport from GSIcheck, you can search for the offensive string in POOTLE.

You may find that GSIcheck gives you a number of errors like the one below:

Error: Translation, Line 45107, UniqueID
helpcontent2/source\text\sdraw\guide\eyedropper.xhp/help/par_id3150202/30//:
Expected Symbol: Close tag for \<emph\> at Position 35  ""

You can find these in POOTLE by using the search on Location

The location consist of the name of the file and the paragraph-ID (par_id)
There is a catch though.

GSIcheck delivers in the format:
helpcontent2/source\text\sdraw\guide\eyedropper.xhp/help/par_id3150202/30//
Searching in POOTLE uses the format :
eyedropper.xhp#par_id3150202.30

By partly adjusting the string, delivered by GSIcheck, you can search in POOTLE on Location

So, replace /help/ from the string of GSIcheck to # and a possible / (slash) in the par_id between digits by a .' (dot)

Also, be sure to include the last part of the name of the file to avoid possible double par_ids

From the above example this will give you:

eyedropper.xhp#par_id3150202.30

which is a valid search for Location in POOTLE and gives you:

(The last </emph> was indeed missing the / at first, but that is already corrected in the image)


text:
# Which geometry does not allow the existence of matter?

I have seen these lectures by Fredric Schuller that discuss the obstruction theory and the role of global geometric properties in admitting a spin structure.

See the video at 01:27:52

https://youtu.be/Way8FfcMpf0?t=5279

For example, it is said that electrons can live on a curved space like $S^2$ because, using the bundle language, it is a compact and orient-able base manifold and in particular its second class Stiefel–Whitney vanishes; which is a necessary and sufficient condition.

Now I am trying to imagine some physical intuition about the role of topology invariant constraints on the matter (physical) fields.

Is there any easy to understand geometry that does not allow the existence of electrons? I say that a Möbius shaped piece of metal should meet the Stiefel–Whitney condition, because it exists, but are there any shape to suggest which is not too much exotic to imagine and does not allow the existence of electrons on it due to the obstruction theory results?

Or the put it another way, which geometry does not allow the existence of matter?

Can you suggest a low dimensional and easy to understand geometry in which second class Stiefel–Whitney is not zero?

• Have asked there too. But appreciate a purely math answer also. – user56963 Jun 27 '18 at 7:57
• I don't think you will get a purely mathematical answer because concepts like electron or matter are not purely mathematical afaik. – freakish Jun 27 '18 at 7:58
• How about suggesting an easy to understand geometry in which second class Stiefel–Whitney is not zero. – user56963 Jun 27 '18 at 8:00
• How about $\mathbb{C}P^2$? – Tyrone Jun 27 '18 at 9:16
• Thanks, Where is its definition? – user56963 Jun 27 '18 at 9:19


text:
# When to aggregate for time series forecasting?

I have a some historical sales data for various product SKUs, including category information ("department" "category", "subcategory"). I want to use this to generate sales curve (a baseline forecast for future demand), no doubt using an appropriate exponential smoothing algorithm.

The trouble is that the data for a particular SKU will often be too sparse to be able to infer a sales curve. I'm thinking that when that is the case, the sales data for the SKU could be aggregated with other SKUs for the same product and I could then use the sales curve for the product rather than the SKU (which would presumably be more reliable). If the product doesn't have enough sales history then I'd use the sales curve for the subcategory, and so on and so forth… potentially even having to resort to a sales curve for the top level category (a department in this case) if insufficient sales data is available to generate reliable sales curves for lower level categories.

What I'm wondering is, how do I determine when I have enough data to be able to trust a sales curve? How many sales "data points" do I need for a SKU or a subcategory in order to be able to trust its sales curve?

Perhaps I'm approaching this the wrong way though. An alternative that I'd considered was to use data from all levels of the hierarchy in the forecast with appropriate weightings for each. So a product's future sales would be predicted by W1 * Forecast(Product) + W2 * Forecast(Subcategory) + W3 * Forecast(Category) + W4 * Forecast(Department).... Working out the weightings might be a bit complex but not impossible (and might be done using a regression analysis).

I'm sure this kind of problem has been solved before but I'm a bit of a stats newbie. Are there "off the shelf" algorithms/techniques that could use for this problem?

• You can never absolutely trust a prediction - you need confidence intervals. Have you tried plotting a projection with confidence intervals? Your second prediction algorithm doesn't make much sense to me. Isn't a given product always going to be in the same subcategory, category, and department? If so, isn't the forcast for the product already going to include all the information that the categories could add anyway? What would you be regressing against? – naught101 May 17 '13 at 2:02
• ah... enter keys. Imagine I have three products SKU1, SKU2 and SKU3: SKU1 Stock Levels: 0 0 0 10 9 8 SKU1 Sales Levels: 0 0 0 1 1 2 SKU2 Stock Levels: 10 8 4 10 9 8 SKU2 Sales Levels: 2 4 0 1 1 2 SKU3 Stock Levels: 10 8 7 0 0 0 SKU3 Sales Levels: 2 1 7 0 0 0 SKU1 doesn't have "sales" data for t=1, t=2 and t=3. SKU3 lacks data for t=4..6. However if all of the SKUs are part of the same subcategory, we could perhaps "guess" at what SKU1 and SKU3 sales would have in those periods by borrowing the seasonal and trend components for those SKUs in something like a Holt-Winters algorithm). – James Crosswell May 17 '13 at 11:01
• More generally, the historical sales for an individual SKU might not be the best predictor for future sales of that SKU. Perhaps seasonal, cyclical and trend data for the subcategory are better predictors. I'm thinking of an algorithm that includes the trend data for all levels in the hierarchy (subcategory, category etc.), with different weights to each. The best values for the weightings is going to be tricky to guess at though... so this is what I was thinking of using a regression for. – James Crosswell May 17 '13 at 11:02
• Ah, I see the regression idea was stupid now... not very appropriate for one-step-foreward prediction algorithms 8-/ – James Crosswell May 17 '13 at 13:31


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# Desmos

Desmos is a free online graphing calculator. Some classes at NCSSM require its usage, but it is a tool that any student of mathematics will appreciate. Here are enumerated some of the simple and complex functions of the software you might find useful. If this page is not sufficient, see the Desmos user guide for more information.

Below is a brief description of how to effectively use the Desmos online graphing calculator. Click the images to enlarge them.

## Graphing

### Functions

Simple functions of one variable can be graphed in three ways:

• Typing the equation (${\displaystyle x^{2}}$)
• Typing ${\displaystyle y=}$ and then the equation (${\displaystyle y=x^{3}}$)
• Typing ${\displaystyle f(x)=}$ and then the equation (${\displaystyle f(x)={\frac {x^{2}}{7}}}$)

Functions can be hidden from the graph by clicking the colored circle to the left of their equation.

Functions can be functions of other functions. For example:

• ${\displaystyle f(x)=x^{2}}$
• ${\displaystyle g(x)=f(x+1)}$

Functions can be graphed with x in terms of y: ${\displaystyle x={\sqrt {1-y^{2}}}}$

Functions can be of multiple variables. For example, ${\displaystyle g(a,b)=a^{2}+b}$ and ${\displaystyle h(a)=g(a,3)}$.

#### Parametric Equations

In Desmos you can graph parametric equations. The standard form of this is ${\displaystyle (f(t),g(t))}$. For example, ${\displaystyle (sin(2t),cos(3t))}$. It is necessary to use the variable ${\displaystyle t}$ in order for the equation to be interpreted as a parametric equation.

#### Implicit Functions

Solutions of equations involving x and y can be plotted without solving for x or y.

### Points and Tables

Points can be graphed in two ways. One way is by clicking the plus in the upper left, and then selecting table. Points can be added manually, or pasted in from an outside source, such as Logger Pro. The points will appear on the graph, but the viewport will not scale to accommodate them. To do this, see Resizing the viewport.

The other way to graph points is by adding them in the form ${\displaystyle (1,4)}$.

Movable points can be graphed by setting one or both of the parameters to variables: ${\displaystyle (1,g)}$. Movable points can be moved either with the sliders or by clicking and dragging the point. You can restrict a movable point to being on a function by graphing it in the form ${\displaystyle (a,f(a))}$.

You can add multiple points by separating them with commas, like so: ${\displaystyle (2,5),(3,10)}$. Points added this way cannot be accessed as a list of x and y values, unlike points in a table.

### Variables and Sliders

Sometimes, for example, when looking for the value of a constant or seeing how a function reacts to a change in one of its parts, it is useful to use a slider. Sliders represent a named variable.

To add a slider, write a function with a free variable, such as ${\displaystyle y=mx}$, then click on "add slider: m".

Another way to add variables is to type the variable letter, "=", and then a starting value. For example, ${\displaystyle d=100.42}$.

Clicking on either of the constraints of a slider allows you to set the minimum, maximum, and step values.

Performing regression analysis of data in Desmos is simple. First, enter the data points in a table as explained in Graphing Points.

Next, decide on the type of regression. If you want to do a linear regression, the equation would take the form ${\displaystyle y=mx+b}$. If you want a quadratic regression, the equation would look like ${\displaystyle y=ax^{2}+bx+c}$. The enterprising student can extrapolate this to any form they would like, such as ${\displaystyle y=e^{m}x+bx}$.

After you have decided the form your equation should take, substitute "${\displaystyle y}$" with "${\displaystyle y_{1}}$", "${\displaystyle x}$" with "${\displaystyle x_{1}}$", and "${\displaystyle =}$" with ${\displaystyle \sim }$. For example, to do a linear regression, you would enter ${\displaystyle y_{1}\sim mx_{1}+b}$. This is saying: "find the constant values m and b that best satisfy ${\displaystyle y=mx+b}$ for every ${\displaystyle x}$ and ${\displaystyle y}$ in my table, where ${\displaystyle x=x_{1}}$ and ${\displaystyle y=y_{1}}$". See the examples below for a demonstration.

Desmos will find the best constants that fit the points and function you put in. The value of each constant is shown under your function entry. As you can see, the function it finds is not always in the correct form. (This is because you provide the form, Desmos only finds the best values for the free variables that it can.) Above, the same points were used, but a line and a parabola were both found. The equation you get out is only as good as the equation form you put in.

One way to check whether the form of the function is correct is to plot the residuals, and see if there is a pattern. If you see a clear trend, such as the residuals moving in a straight line, parabola, or sinusoidal curve, you know you chose the wrong function to model your data, even if the r-value is small.

See Desmos regressions tour for an interactive demonstration for plotting regressions.

### Restrictions

It can be useful to restrict the domain or range of a function. To do this, add a restriction of the form ${\displaystyle \left\{2 directly onto the end of a function. This works with any variable: for a function of ${\displaystyle c}$ where c should be greater than 4, type ${\displaystyle \left\{c>4\right\}}$.

### Inequalities

Desmos also has the capability to graph inequalities. Simple inequalities are easy: type an expression followed by a comparison sign (e.g. ${\displaystyle <}$, ${\displaystyle >=}$) and then a value. For example, ${\displaystyle 3x<4}$. A slightly more complicated example might be ${\displaystyle x^{2}>y>2}$

You can also use this on functions you define. For example, say you're modeling compound interest, and want to set up an inequality for the space under the curve. To do this, define a function for amount given time, ${\displaystyle A(t)}$, and an inequality ${\displaystyle 0. Notice the use of ${\displaystyle x}$ instead of ${\displaystyle t}$. Desmos is picky about variables, and for inequalities it implicitly defines x and y as the input and output.

We can use this powerful tool in tandem with restrictions. To graph the area between functions ${\displaystyle f(x)}$ and ${\displaystyle g(x)}$, all we have to do is type ${\displaystyle f\left(x\right)y>f\left(x\right)\right\}}$.

## Built-in Functions and Symbols

Desmos comes with many built in functions and symbols. Here are some of the documented and undocumented ones. Most of these can be accessed by opening the keyboard and clicking "functions".

### Variables

Input Result Explanation
theta ${\displaystyle \theta }$ A variable, like ${\displaystyle x}$.

### Constants

Input Result Explanation
pi ${\displaystyle \pi }$ The constant 3.14...
tau ${\displaystyle \tau }$ The constant 6.28...
e ${\displaystyle e}$ The constant 2.71...

### Exponent and Log Functions

Input Result Explanation
exp(x) exp(x) ${\displaystyle e^{x}}$
ln(x) ln(x) The natural log of x
log(x) log(x) The log (base 10) of x
log_n(x) ${\displaystyle log_{n}(x)}$ The log (base n) of x
x^n ${\displaystyle x^{n}}$ x to the nth power
sqrtx ${\displaystyle {\sqrt {x}}}$ The square root of x
nthrootx ${\displaystyle {\sqrt[{n}]{x}}}$ The generalized root function.

### Precalculus and Calculus Functions

Input Result Explanation
sum ${\displaystyle \sum _{n=}}$ The summation operator
prod ${\displaystyle \prod _{n=}}$ The product operator.
int ${\displaystyle \int }$ The integral operator.
d/dx ${\displaystyle {\frac {d}{dx}}}$ The differential operator. Can be used on functions.

### Trig Functions

 sin(x) arcsin(x) or ${\displaystyle sin^{-1}(x)}$ sinh(x) cos(x) arccos(x) or ${\displaystyle cos^{-1}(x)}$ cosh(x) tan(x) arctan(x) or ${\displaystyle tan^{-1}(x)}$ tanh(x) sec(x) arcsec(x) or ${\displaystyle sec^{-1}(x)}$ sech(x) csc(x) arccsc(x) or ${\displaystyle csc^{-1}(x)}$ csch(x) cot(x) arccot(x) or ${\displaystyle cot^{-1}(x)}$ coth(x)

### Stats and Probability Functions

Function Explanation
total(${\displaystyle x_{1}}$) Sum of all elements in a list
length(${\displaystyle x_{1}}$) Number of elements in a list
mean(${\displaystyle x_{1}}$) Mean of a list
median(${\displaystyle x_{1}}$) Median element in a list
min(${\displaystyle x_{1}}$) Minimum element in a list
max(${\displaystyle x_{1}}$) Maximum element in a list
quantile(${\displaystyle x_{1},p}$) The quantile function
mad(${\displaystyle x_{1}}$) Mean absolute deviation
stdev(${\displaystyle x_{1}}$) Sample standard deviation
stdevp(${\displaystyle x_{1}}$) Population standard deviation
var(${\displaystyle x_{1}}$) Variance
cov(${\displaystyle x_{1},x_{2}}$) Covariance
corr(${\displaystyle x_{1},x_{2}}$) Pearson correlation coefficient of two lists
nCr(${\displaystyle n,r}$) Number of combinations
nPr(${\displaystyle n,r}$) Number of permutations
${\displaystyle x!}$ Factorial

### Miscellaneous Functions

Input Explanation
lcm(a,b) The least common multiple of a and b
gcd(a,b) The greatest common denominator of a and b
mod(a, b) The modulo function
ceil(x) The closest integer above x
floor(x) The closest integer below x
round(x) The closest integer to x
sign(x) The sign of x
abs(x) The absolute value function
${\displaystyle |x|}$ The absolute value operator
a%b a percent of b

## Making graphs easier to read

### Resizing the viewport

The viewport (the section of the coordinate plane visible onscreen) can be resized to give a better view of your data and functions. One way to resize it is to scroll down to zoom out, and up to zoom in, and to click and drag to move. This method, however, keeps the ratio of the domain and range constant. Sometimes it is useful to be able to zoom to an arbitrary window.

To do this, click the wrench in the top right corner (graph settings). After this, a window will appear. In this window, there are inequalities for ${\displaystyle x}$ and ${\displaystyle y}$ (${\displaystyle a\leq x\leq b}$ and ${\displaystyle c\leq y\leq d}$) which can be edited to resize the viewport with the domain set by ${\displaystyle x}$ and range set by ${\displaystyle y}$.

### Changing line colors/styles

If you have many functions graphed, it can be useful to change their appearance. To do this, perform a long left click on the icon to the left of the function definition. From here, you may change the style to dashed, dotted, or unbroken, and the color to one of six options.

### Projector mode

When saving images for reports or projects, it can be useful to have thicker, fuller lines and larger text. To enable this, go to graph settings and enable "Projector Mode".


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Exact random sums

Sometimes, you need a list of random numbers that sum to a known constant. There's a known algorithm to provide this list of numbers with the proper distribution, but a straight-forward implementation may give a list that doesn't sum exactly to the desired constant because of rounding error.

This article describes the basic algorithm, why the rounding error happens, and the solution.

In some sense, random numbers uniformly-distributed in the range $$[0, 1)$$ are the easiest class of random number to generate.


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# Sentinel Company is considering an investment in technology to improve its operations. The invest...

Sentinel Company is considering an investment in technology to improve its operations. The investment will require an initial outlay of $257,000 and will yield the following expected cash flows. Management requires investments to have a payback period of 3 years, and it requires a 8% return on investments. (PV of$1, FV of $1, PVA of$1 and FMA of \$1) (Use appropriate factor(s) from the table provided.) Period Cash Flow 48,500 53,100 76,500 95,200 125,300 Required: 1. Determine the payback period for this investment. (Enter cash outflows with a minus sign. Round your Payback Period answer to 1 decimal place.) Cumulative Net Cash inflow Year Cash inflow (outflow) (outflow) (257,000) Payback period


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## A group of friends went out to eat at a restaurant. Their bill came to $57.89. They want to leave an 18% tip. How much should they leave for Question A group of friends went out to eat at a restaurant. Their bill came to$57.89. They want to leave an 18% tip. How much should they leave for the tip?

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6 months 2021-08-10T02:02:11+00:00 1 Answers 6 views 0