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# Andreas von Ettingshausen

Andreas von Ettingshausen
Andreas Freiherr von Ettingshausen (1796–1878)
Born 25 November 1796
Died 25 May 1878 (aged 81)
Residence Austria
Nationality Austrian
Alma mater University of Vienna
Known for Electric machines
Scientific career
Fields Physicist and mathematician
Institutions University of Innsbruck
University of Vienna
Vienna Polytechnic Institute
Doctoral students Ernst Mach
Jožef Stefan
Viktor von Lang

Andreas Freiherr von Ettingshausen (25 November 1796 – 25 May 1878) was an Austrian mathematician and physicist.

## Biography

Ettingshausen studied philosophy and jurisprudence at the University of Vienna. In 1817, he joined the University of Vienna and taught mathematics and physics as an adjunct professor. In 1819 he became professor of physics at the University of Innsbruck and 1821 professor of higher mathematics at the University of Vienna. His lectures of that time marked a new era for the University of Vienna, and they were published in 1827 in 2 volumes. In 1834 Ettingshausen became the chair of physics.

Ettingshausen was the first to design an electromagnetic machine, which used the electrical induction for power generation. He promoted optics and wrote a textbook of physics. His method of lecturing was widely influential. In addition he wrote a book on combinatorial analysis (Vienna 1826). In 1866, he retired.

Among his lasting impacts in mathematics is the introduction of the notation ${\displaystyle {\tbinom {n}{k}}}$ for the binomial coefficient, which is the coefficient of xk in the expansion of the binomial (x+1)n and, more generally, the number of k-element subsets of an n-element set.[2][3]

## References

1. ^ Andreas von Ettingshausen, Vorlesungen über die höhere Mathematik: Vorlesungen über die Analysis, Volume 1, Gerold, 1827, p. v.
2. ^ Ettingshausen, Andreas von (1826). Die combinatorische Analysis als Vorbereitungslehre zum Studium der theoretischen höhern Mathematik [Combinatorial analysis as preparatory instruction for the study of theoretical higher mathematics] (in German). Vienna, Austria: J.B. Wallishauffer. pp. 30, 31. From p. 30: "Da wir im Folgenden sehr häufig Gelegenheit haben werden, von dem numerischen Ausdrucke dieser Menge Gebrauch zu machen, so wollen wir dafür das Zeichen (n k) wählen, und es mit dem Worten n über k ausprechen, wobei die obere Zahl stets die Anzahl der combinirten Elemente, die untere aber den Rang der Combinationsklasse angibt." (Since we will very frequently have occasion in the following to make use of numerical expressions of these quantities, then we will choose for that purpose the symbol (n k) and will express it with the words "n over k", whereby the upper number always specifies the number of combined elements whereas the lower [number] specifies the rank of the classes of combinations.) Page 31 shows that (n k) = n! / k! (n-k)! .
3. ^ Nicholas J. Higham. Handbook of Writing for the Mathematical Sciences. Society for Industrial and Applied Mathematics. p. 25. ISBN 0-89871-420-6.


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# liminf and limsup of a sequence (II)

I came across the following problem about limit supremums and infimums of sequences:

Let $(a_n)$ be a bounded sequence in $\mathbb{R}$.

1. Prove that $b = \text{lim sup} \ a_n \implies (\forall \epsilon >0) \ a_n < b+\epsilon$ ultimately and $a_n > b-\epsilon$ frequently.

2. Show that the condition first condition characterizes the limit superior, in the sense that if $b \in \mathbb{R}$ satisfies the condition then necessarily $b = \text{lim sup} \ a_n$.

Here is my attempted solution:

1. Proof. Let $(a_n)$ is a bounded sequence in $\mathbb{R}$. Suppose $b = \text{lim sup} \ a_n$. By definition, $$b = \text{lim sup} \ a_n = \inf\limits_{n \geq 1} \left(\sup\limits_{k \geq n} \ a_k \right)$$ Let $A = \left\{\sup\limits_{k \geq n} \ a_k \right\}$. Then by definition of infimum, $b+ \epsilon >a_n$ for every $a_n \in A$. So $(\exists N) \ \ni n \geq N \implies a_n < b+ \epsilon$. Thus $a_n < b+ \epsilon$ frequently. Since $b$ is a lower bound of $A$, $b \leq a_n$ for all $n$. Thus $b- \epsilon < a_n$ for all $n$. Hence $a_n > b- \epsilon$ frequently.

2. Proof. Suppose $(\forall \epsilon >0) \ a_n < b+ \epsilon$ ultimately and $a_n > b- \epsilon$ frequently. Then $b \leq \text{lim sup} \ a_n$ and $b \geq \text{lim sup} \ a_n$. Hence $b = \text{lim sup} \ a_n$ by antisymmetry.

Are these on the right track?

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In your attempted proof, I am confused between $a_n$ being a member of the original sequence and $a_n$ being an element of $A$; the suprema do not need to be in the original sequence. –  Henry Jun 26 '11 at 17:43
What do you mean by "ultimately" and "frequently"? –  omar Jun 26 '11 at 18:42
@Omar: "Ultimately" means that there exists $N$ such that the condition holds for all $n\geq N$. "Frequently" means that for every $N$ there exists (at least one) $k\geq N$ such that the condition holds for $k$. That is, a condition on a sequence holds "ultimately" if it holds for a tail, and holds "frequently" if it holds for a subsequence. –  Arturo Magidin Jun 26 '11 at 18:55
For the first condition in part 1 you need to take a given $\epsilon$ and show that there is an N beyond which the condition holds - eventually the sequence stays below $b+ \epsilon$. For the second condition, given $\epsilon$ consider what would happen if there were only finitely many n for which $a_n > b- \epsilon$ - what would that do to your lim sup (hint - there would be an N beyond which the condition would never hold)? Frequently here should mean infinitely often, but not necessarily for all n or for all n beyond a certain point. –  Mark Bennet Jun 26 '11 at 19:02
Damien, I think it will serve you well if you come to understand these concepts (after which the proofs will become obvious). GH Hardy "Pure Mathematics" has a short section (no 82 in 10th edition, p156, also read sect 81) on the limits of indetermination of a bounded function. Amongst other things it has narrative definitions, and a short narrative proof of what you are asked and a number of examples which illustrate the various possibilities which may arise. Hardy may help you to understand why lim sup is a useful concept. –  Mark Bennet Jun 26 '11 at 20:05

As soon as you write "Then by the definition of infimum", you are saying incorrect things.

First: if $b$ is the infimum of $A$, then for every $\epsilon\gt 0$ there exists at least one $a\in A$ such that $a\lt b+\epsilon$. However, you claim this holds for all elements of $A$, which is false. Take $A = (0,1)$. Then $\inf A = 0$; and it is indeed true that for every $\epsilon\gt0$ there exists at least one $a\in A$ with $a\lt\epsilon$; but if $\epsilon=\frac{1}{10}$, it is certainly false that every $a\in A$ is smaller than $\epsilon$.

Second: the elements of $A$ are not $a_n$s! They are suprema of infinite sequences of $a_n$s, and as such cannot be assumed to be $a_n$s. For example, if $a_n = 1-\frac{1}{n}$, then $A=\{1\}$, and no $a_n$ is equal to any element of $A$.

So that sentence is not just wrong, it's doubly wrong. The rest of course is now nonsense.

The second part does not seem to be proving anything; you are just asserting things. Why do the conditions imply the inequalities? What properties of the limit superior are you using? It's a mystery.

Rather: let $A_n = \mathop{\sup}\limits_{k\geq n}(a_k)$. Prove that $A_n$ is a decreasing sequence: $A_{n+1}\leq A_n$ for all $n\in \mathbb{N}$. Your set $A$ is precisely the set of $A_n$s.

Now let $b= \inf A = \inf\{ A_n\}$. By the definition of infimum, for every $\epsilon\gt 0$ there exists $N$ such that $b\leq A_N\lt b+\epsilon$. Since the sequence of $A_n$s is decreasing, then for all $n\geq N$ we have $b\leq A_n\leq A_N\lt b+\epsilon$, so in fact we have that $A_n\lt b+\epsilon$ ultimately. Moreover, since $a_n\leq A_m$ for all $n\geq m$, this implies that $a_n\lt b+\epsilon$ ultimately, as required.

For the second clause of the first part, let $\epsilon\gt 0$. Then $b-\epsilon\lt A_n$ for all $n$. Now remember what $A_n$ is. $A_n = \mathop{\sup}\limits_{k\geq n}(a_k)$; since $b-\epsilon\lt A_n$, there exists $k\geq n$ such that $b-\epsilon\lt a_k\leq A_n$. That is: for all $n$, there exists $k\geq n$ such that $a_k\gt b-\epsilon$, so $a_k\gt b-\epsilon$ frequently.

For part (2), let $b$ be a real number that satisfies the given properties. Since for every $\epsilon\gt 0$ we have that $b-\epsilon \lt a_n$ frequently, that means that $b-\epsilon$ is not an upper bound for $\{a_k\mid k\geq n\}$ for any $n$. Therefore, $\sup\{a_k\mid k\geq n\} = A_n\gt b-\epsilon$. This holds for all $A_n$, so $\liminf a_n = \inf\{A_n\mid n\in\mathbb{N}\} \geq b-\epsilon$. This holds for all $\epsilon\gt 0$, so $\liminf a_n \geq b$.

Now, since $b+\epsilon\gt a_n$ ultimately, then $b+\epsilon$ is an upper bound for $A_m$ all sufficiently large $m$; since the $A_m$ are decreasing, that means that $\inf A_m \lt b+\epsilon$, hence $\liminf a_n\lt b+\epsilon$; this holds for all $\epsilon\gt 0$, so $\liminf a_n \leq b$.

Now that we have established the inequalities (rather than merely asserting them), we have $b\leq \liminf a_n \leq b$, hence $b=\liminf a_n$, as claimed. QED

-
@Arturo: I think it should be for all $n \geq N$ we have that $b \leq a_n \leq A_{N} < b+ \epsilon$. –  Damien Jun 26 '11 at 19:25
@Damien: Where? Which line of which paragraph? I have a lot of inequalities, I don't know which one you are refering to. –  Arturo Magidin Jun 26 '11 at 19:27
@Arturo: It would be the third line of the seventh "paragraph." Also in the 9th paragraph, fourth line: $\text{lim inf} \ a_n = \sup \{A_n: n \in \mathbb{N} \} \geq b- \epsilon$. –  Damien Jun 26 '11 at 19:31
@Damien: No, in the third line of the seventh paragraph, I wrote what I meant to write. Since the $A_n$s form a decreasing sequence, and $b\leq A_n$ for all $n$, you have that if $A_N\lt b+\epsilon$, then for all $n\geq N$ you have $b\leq A_n\leq A_N\lt b+\epsilon$. There is also no error in the 9th paragraph fourth line. What makes you think those $A_n$'s are supposed to be $a_n$? $\inf\{A_n\mid n\in\mathbb{N}\}$ is the same as $\mathop{\inf}\limits_{n\in\mathbb{N}}A_n$. –  Arturo Magidin Jun 26 '11 at 19:35
@Arturo: In the third line of the seventh paragraph you have $b_{\epsilon}$ instead of $b+\epsilon$. And in the 9th paragraph fourth line, shouldnt $\text{lim inf} \ a_n = \sup \{A_n: n \in \mathbb{N} \}$ instead of $\text{lim inf} \ a_n = \inf\{A_n: n \in \mathbb{N} \}$? –  Damien Jun 26 '11 at 19:38
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## Stats NZ has a new website.

For new releases go to

www.stats.govt.nz

As we transition to our new site, you'll still find some Stats NZ information here on this archive site.

Comparing revisions in time series data: A report on seasonally adjusted versus trend series

In this paper, we describe our investigation into whether using the trend rather than the seasonally adjusted series could improve the stability of the figures in Statistics New Zealand publications.

### Summary of our methodology and results

• We analysed 15 Statistics NZ time series from the business and social areas, using absolute total revisions and Levene’s test for equality of variances.
• We calculated and analysed the absolute total revisions for the seasonally adjusted and trend estimators using various statistical tests, including Levene’s test.
• We used Levene’s test to identify if there is a statistically significant difference between the variances of the seasonally adjusted and trend series in terms of total revisions.
• Levene’s test for equality of variances plus other tests confirmed that the difference between the variances of the seasonally adjusted and trend series was statistically significant for all 15 time series.
• The absolute total revision graphs showed that the trend series had larger revisions than the seasonally adjusted series in all 15 series we analysed.
• The plots of the seasonally adjusted revisions and trend revisions showed the trend series to be less stable than the seasonally adjusted series. This was more pronounced towards the recent end of the series due to the end-point problem.
• The seasonally adjusted series showed consistent significantly smaller variance than the trend cycle series.
• We found the seasonally adjusted series consistently outperformed the trend series in the relative size of revisions.
• For greater stability in revisions in Statistics NZ’s official estimates, we recommend leading with the seasonally adjusted series.

To read this publication, download or print the PDF from 'Available files' above. If you have problems viewing the files, see opening files and PDFs.

Citation
Buchanan, R, and MacCormick, C (2014). Comparing revisions in time series data: A report on seasonally adjusted versus trend series. Available from www.stats.govt.nz.

ISBN

978-0-478-40877-5 (online)
Published 7 July 2014

Top


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# Why double error correction and quadruple error detection cannot occur at the same time?

What does [ we cannot both correct double errors and detect quadruple errors because this would require us to interpret a received codeword in two different ways] means? I understand how double error correction works but I cannot imagine quadruple error detection. Can anybody show me example why double error correction and quadruple error detection cannot occur at the same time because it requires different interpretation?

For example, if double error correction occur,

This is an excerpt from Computer Network Book by Andrew S. Tanenbaum

The data link layer(chapter 3), Page 208, Fifth edition.

• Where is the snippet taken from? Please attribute the author. May 22 '20 at 14:48
• This is an excerpt from Computer Network Book by Andrew S. Tanenbaum. Thank you for reminding me to attribute the author. I totally had forgotten. May 22 '20 at 15:18
• Can you update your question with this information, also adding the chapter and page numbers, as well as which edition of the book you are using? May 22 '20 at 15:20
• The data link layer(chapter 3), Page 208, Fifth edition. May 22 '20 at 15:23

Your interpretation is mistaken. Both properties are satisfied at the same time, since they are properties of the code itself.

In more detail, using the notation $$d(x,y)$$ for the Hamming distance between $$x,y$$:

• If $$x$$ is the sent codeword, $$y$$ is the received codeword, and $$0 \leq d(x,y) \leq 4$$, then we would be able to detect that $$y \neq x$$. This is quadruple error detection.

• If $$x$$ is the send codeword, $$y$$ is the received codeword, and we are promised that $$0 \leq d(x,y) \leq 2$$, then we can recover $$x$$ from $$y$$. This is double error correction.

Rephrasing:

• If there were errors in the channel but at most 4, we will be able to detect that errors occurred.
• If there were errors in the channel but at most 2, we can not only detect that errors occurred, but also determine their locations.

In the second case, we both detect that errors occurred, and are able to undo them. In that sense, we can both detect and correct errors at the same time.

Your quote is trying to make a different point. Given a codeword $$y$$, what can you do with it?

• You can try to detect whether errors occurred. This detection is guaranteed to be successful if up to 4 errors have occurred.
• You can try to determine the sent codeword $$x$$. This will be successful if up to 2 errors have occurred.

In practice, this is used in the following way:

• It is highly unlikely that more than 4 errors occur (or this is taken care of in some other way). Your error-handling strategy is to detect whether any errors occurred, and if so, ask the sender to resend the message.
• It is highly unlikely that more than 2 errors occur (or this is taken care of in some other way). Your error-handling strategy is to silently correct the errors, without communicating with the sender.

The two strategies are mutually exclusive, since the error detection mechanism doesn't differentiate between the case in which the number of errors is 0,1,2 and the case in which the number of errors is 3,4. So if you assume that there are at most 4 errors, you have no way of knowing whether the input can be corrected (there were up to 2 errors) or not.

The author doesn't mention a third approach, list decoding, which allows you to handle more errors as part of a longer communication protocol. In list decoding, you get a small "list" of possible sent messages, which you winnow down in some other way. I don't know whether this is actually used in practice, but it has been very influential in coding theory and theoretical computer science.

With a hamming distance of five, you can make one of two assumptions:

A. No word has more than four errors. Since you need five changes to get from one valid word to another valid word, and five errors don’t happen, every invalid word has 1 to 4 errors.

B. No word has more than two errors. If you find an invalid word, then it has at most two changes from a valid word. To reach any different valid word would take at least 3 more changes, so there is only one valid word within two changes, so you can fix the errors.

But you can’t assume both at the same time. If you find an invalid word, it could have two or four errors, and trying error correction would give you a totally wrong result.

If you have a hamming-distance of 7, you can indeed detect up to four errors and fix up to two errors at the same time. A word with four or three errors cannot be made valid with 1 or 2 changes, so you can detect 3 or 4 errors without being able to fix them, and you can fix 1 or 2 errors.


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# Legendre chi function

(Redirected from Legendre's chi-function)

In mathematics, the Legendre chi function is a special function whose Taylor series is also a Dirichlet series, given by

${\displaystyle \chi _{\nu }(z)=\sum _{k=0}^{\infty }{\frac {z^{2k+1}}{(2k+1)^{\nu }}}.}$

As such, it resembles the Dirichlet series for the polylogarithm, and, indeed, is trivially expressible in terms of the polylogarithm as

${\displaystyle \chi _{\nu }(z)={\frac {1}{2}}\left[\operatorname {Li} _{\nu }(z)-\operatorname {Li} _{\nu }(-z)\right].}$

The Legendre chi function appears as the discrete Fourier transform, with respect to the order ν, of the Hurwitz zeta function, and also of the Euler polynomials, with the explicit relationships given in those articles.

The Legendre chi function is a special case of the Lerch transcendent, and is given by

${\displaystyle \chi _{\nu }(z)=2^{-\nu }z\,\Phi (z^{2},\nu ,1/2).}$

## Identities

${\displaystyle \chi _{2}(x)+\chi _{2}(1/x)={\frac {\pi ^{2}}{4}}-{\frac {i\pi }{2}}|\ln x|\qquad (x>0).}$
${\displaystyle {\frac {d}{dx}}\chi _{2}(x)={\frac {{\rm {arctanh\,}}x}{x}}.}$

## Integral relations

${\displaystyle \int _{0}^{\pi /2}\arcsin(r\sin \theta )d\theta =\chi _{2}\left(r\right)}$
${\displaystyle \int _{0}^{\pi /2}\arctan(r\sin \theta )d\theta =-{\frac {1}{2}}\int _{0}^{\pi }{\frac {r\theta \cos \theta }{1+r^{2}\sin ^{2}\theta }}d\theta =2\chi _{2}\left({\frac {{\sqrt {1+r^{2}}}-1}{r}}\right)}$
${\displaystyle \int _{0}^{\pi /2}\arctan(p\sin \theta )\arctan(q\sin \theta )d\theta =\pi \chi _{2}\left({\frac {{\sqrt {1+p^{2}}}-1}{p}}\cdot {\frac {{\sqrt {1+q^{2}}}-1}{q}}\right)}$
${\displaystyle \int _{0}^{\alpha }\int _{0}^{\beta }{\frac {dxdy}{1-x^{2}y^{2}}}=\chi _{2}(\alpha \beta )\qquad {\rm {if}}~~|\alpha \beta |\leq 1}$


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• Monisa Rajkhowa

Articles written in Pramana – Journal of Physics

• Deviation from tri-bimaximal mixings through ﬂavour twisters in inverted and normal hierarchical neutrino mass models

We explore a novel possibility for lowering the solar mixing angle ($\theta_{12}$) from tri-bimaximal mixings, without sacriﬁcing the predictions of maximal atmospheric mixing angle ($\theta_{23} = 45^{\circ}$) and zero reactor angle ($\theta_{13} = 0^{\circ}$) in the inverted and normal hierarchical neutrino mass models having 2-3 symmetry. This can be done through the identiﬁcation of a ﬂavour twister term in the texture of neutrino mass matrix and the variation of such term leads to lowering of solar mixing angle. For the observed ranges of $\Delta m_{21}^{2}$ and $\Delta m_{23}^{2}$, we calculate the predictions on tan2 $\theta_{12} = 0.5, 0.45, 0.35$ for different input values of the parameters in the neutrino mass matrix. We also observe a possible transition from inverted hierarchical model having even CP parity (Type-IHA) to inverted hierarchical model having odd CP parity (Type-IHB) in the ﬁrst two mass eigenvalues, when there is a change in input values of parameters in the same mass matrix. The present work differs from the conventional approaches for the deviations from tri-bimaximal mixing, where the 2-3 symmetry is broken, leading to $\theta_{23} \neq 45^{\circ}$ and $\theta_{13} \neq 0^{\circ}$.

• # Pramana – Journal of Physics

Volume 95, 2021
All articles
Continuous Article Publishing mode

• # Editorial Note on Continuous Article Publication

Posted on July 25, 2019


text:
"Fire and Ice" is a nine-line poem written by Robert Frost. It was first published in December $$1920$$ in Harper's Magazine and later in $$1923$$ in his collection of poems called New Hampshire. It is one of his well-known and most anthologised poems.

In the poem, the speaker raises the age old question as to how the world would end. The poet makes two prophecies about the end of the world: fire or ice. According to the speaker, "fire" symbolises desire, greed, and ambition. They expand the more you make an effort to appease them. It goes on forever. They engulf your entire existence and spread quickly like fire. One develops a nasty and occasionally egotistical nature. On the other hand, the speaker claims that "ice" represents hatred, rigidity, and coldness. One develops indifference and a lack of conacern for the feelings of others. According to the poet, the planet would soon perish in either fire or ice because both are spreading quickly.


text:
Site search: .ya-page_js_yes .ya-site-form_inited_no { display: none; }
Results of the Baikal experiment of observations of macroscopic nonlocal correlations in reverse time
Горохов Ю.В. (Gorohov Yu.V.), Буднев Н.М., Киктенко Е.О. (Kiktenko E.O.), Сердюк В.О. (Serdyuk V.O.), Коротаев С.М. (Korotaev S.M.) Results of the Baikal experiment of observations of macroscopic nonlocal correlations in reverse time // Unified Field Mechanics. London: World Scientific. 2015. P. 366-373.

Категории: Исследование, Авторский указатель

## Results of the Baikal experiment of observations of macroscopic nonlocal correlations in reverse time 0.0/5 rating (0 votes)/*<![CDATA[*/jQuery(function($) {$('#c4aab8cf-bd11-4dce-bfe9-a2fdc567666a-614e5290ee7c9').ElementRating({ url: '/en/eelectropublications?task=callelement&format=raw&item_id=7860&element=c4aab8cf-bd11-4dce-bfe9-a2fdc567666a' }); });/*]]>*/

Abstract

Although the general theory macroscopic quantum entanglement of is still in its infancy, consideration of the matter in the framework of action-at-a distance electrodynamics predicts for the random dissipative processes observability of the advanced nonlocal correlations (time reversal causality). These correlations were really revealed in our previous experiments with some large-scale heliogeophysical processes as the source ones and the lab detectors as the probe ones. Recently a new experiment has been performing on the base of Baikal Deep Water Neutrino Observatory. The thick water layer is an excellent shield against any local impacts on the detectors. The first annual series 2012/2013 has demonstrated that detector signals respond to the heliogeophysical (external) processes and causal connection of the signals directed downwards: from the Earth surface to the Baikal floor. But this nonlocal connection proved to be in reverse time. In addition advanced nonlocal correlation of the detector signal with the regional source-process: the random component of hydrological activity in the upper layer was revealed and the possibility of its forecast on nonlocal correlations was demonstrated. But the strongest macroscopic nonlocal correlations are observed at extremely low frequencies, that is at periods of several months. Therefore the above results should be verified in a longer experiment. We verify them by data of the second annual series 2013/2014 of the Baikal experiment. All the results have been confirmed, although some quantitative parameters of correlations and time reversal causal links turned out different due to nonstationarity of the source-processes. A new result is displaying of the advanced response of nonlocal correlation detector to the earthquake. This opens up the prospect of the earthquake forecast on the new physical principle, although further confirmation in the next events is certainly needed. The continuation of the Baikal experiment with expanded program is burning.

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###### Question:

Can someone help me with 1-5 plz I need to get this done today and I need help on this.

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WILL GIVE BRAINLIEST Which of these is the human body's first line of defense against invading pathogens such as bacteria and viruses? a. macrophages b. skin cells c .vitamins...

### Alan launches a ball of clay into the air using a catapult. He calculates its kinetic energy to be 9.0 J. Then he tears it in half and forms a ball that has half the mass. He launches the smaller ball into the air at the same speed. What is the kinetic energy of the smaller ball

Alan launches a ball of clay into the air using a catapult. He calculates its kinetic energy to be 9.0 J. Then he tears it in half and forms a ball that has half the mass. He launches the smaller ball into the air at the same speed. What is the kinetic energy of the smaller ball...


text:
In this post, I will explain how to design a keypad by using an internal ADC of the microcontroller or microprocessor system.

ADC will be used to determine which any key is pressed in the state. In principle, there is a circuit that can generate a differrent voltages at each place of push button pressing. A simple circuit is by applying a voltage divider circuit using some  series resitors, which can lead to different voltages for each keystrokes.

For example, it is necessary to distinguish the five key functions in the program the microcontroller, the UP, DOWN, RIGHT, LEFT, and SELECT button. There are required 6 pieces of resistor: 2K, 220, 330, 1K, 3k3 and 10K, in order to generate the  rising voltage at ADC of microcontroller by successive key presses: UP, DOWN, LEFT, RIGHT, SELECT.

1. The circuit will work as follow:

1. If  the “UP” button is pressed, the voltage to ADC will be 0 Volt, because the 2K resistor will be pulled down to ground.
2. If the “DOWN” button is pressed, the voltage to ADC will be:
$\frac{330\Omega&space;}{330\Omega+2K\Omega}.5V=0.7082V$
3. If the “LEFT” button is pressed, the voltage to ADC will be:
$\frac{330\Omega+620\Omega&space;}{330\Omega+620\Omega+2K\Omega}.5V=1.6102V$
4. If the “RIGHT” button is pressed, the voltage to ADC will be:
$\frac{330\Omega+620\Omega+1K\Omega&space;}{330\Omega+620\Omega+1K\Omega+2K\Omega}.5V=2.4684V$
5. If the “SELECT” button is pressed, the voltage to ADC will be:
$\frac{330\Omega+620\Omega+1K\Omega+3K3\Omega&space;}{330\Omega+620\Omega+1K\Omega+3K3\Omega+2K\Omega}.5V=3.6207V$
6. If no button is pressed, the voltage to ADC will be:
$\frac{330\Omega+620\Omega+1K\Omega+3K3\Omega+10K\Omega&space;}{330\Omega+620\Omega+1K\Omega+3K3\Omega+10K\Omega+2K\Omega}.5V=4.4203V$

If the output of the voltage divider circuit (“to ADC”) fed to the 10bit ADC input, then any key presses will produce a decimal:

1. UP button: 0V will be 0 decimal.
2. DOWN button: 0.7082 Volt will be 145 decimal.
3. LEFT button: 1.6102 Volt will be 329 decimal.
4. RIGHT button: 2.4684 Volt will be 505 decimal.
5. SELECT button: 3.6207 Volt will be 741 decimal.
6. No Button: 4.4203 Volt  will be 905 decimal.

3. How to programming tricks

In fact every keystrokes are not exactly generate a voltage as described in point 1. Means a range slightly above or below the voltage. Thus, the voltage value of the middle is taken, for example: the “UP” from 0V to 0.7082 / 2 = 0.3541V, the “DOWN” from 0.7082V to 0.7082 + (1.6102-0.7082) / 2 = 1.1592 V, and so on.
More easily calculated from point 2, the decimal after being converted by the ADC.
1. The button of  “UP” on pressed state when the conversion result of ADC is defined below value of 145/2 or 72 decimal.
2. The button of  “DOWN” on pressed state when the conversion result of ADC is defined below value of 145+(329-145)/2 or 237 decimal.
3. The button of  “LEFT” on pressed state when the conversion result of ADC is defined below value of 329+(505-329)/2 or 417  decimal.
4. The button of  “RIGHT” on pressed state when the conversion result of ADC is defined below value of 505+(741-505)/2 or 623 decimal.
5. The button of  “SELECT” on pressed state when the conversion result of ADC is defined below value of 741+(905-741)/2 or 823 decimal.
6. No button on pressed state when the conversion result of ADC is defined below value of 905+(1024-905)/2 or 964 decimal.

C  Programming

/*********************************************************/
// created by: Bambang Siswoyo
// Electronic department - University of Brawijaya
/*********************************************************/

int adc_button_table[6] = { 72, 237, 417, 623, 823, 964 };
int NUM_KEYS = 6;
#define UP      0
#define DOWN    1
#define LEFT    2
#define RIGHT   3
#define SELECT  4
#define NOKEY   5

{
............................................
............................................
}

int get_key()
{
int k_num;            // number of key

// Convert voltage from voltage devider
for (k_num = 0; k_num < NUM_KEYS; k_num++)
{
{ return k_num }
}
}

void main()
{ // main program for testing only
int btn;

btn = get_key();
// action
switch(btn)
{
case UP:
//do something when var equals 0
break;
case DOWN:
//do something when var equals 1
break;
case LEFT:
//do something when var equals 2
break;
case RIGHT:
//do something when var equals 3
break;
case SELECT:
//do something when var equals 4
break;
case NOKEY:
//do something when var equals 5
break;
}

Last updated by at .


text:
# teaching machines

## CS 330 Lecture 31 – Maps, Filters, and Folds

### Agenda

• what ?s
• a gallery of maps, filters, and folds
• pairs
• lambdas
• composition

### TODO

• On a 1/4 sheet, write these functions using map, filter, or fold:
•  lengths, which accepts a list of strings and yields a list of all their lengths. For example, lengths ["a", "bcd", "efghi"] yields [1, 3, 5].
•  tweens, which accepts a low, a high, and a list of numbers. It yields a list of all numbers between low and high, inclusive. For example, tweens 13 19 [5, 13, 18, 25] yields [13, 18].
• seconds, which accepts a list of pairs and returns a list of only the second elements. For example, seconds [("chris", 0), ("matt", 17), ("jacob", 10), ("caitlin", 23)] yields [0, 17, 10, 23].

### Problems

• Write a function resort that accepts a list and yields the list sorted in reverse. Use sort and reverse.
• Write a function expand that accepts a list of string-n pairs. It yields a list in which each pair has been expanded into a list comprised of n instances of its string. For example: expand [("dog", 2), ("cat", 3")] yields [["dog", "dog"], ["cat", "cat", "cat"]].
• Write a function proportionalize that accepts a list of integer weights. It returns a list of the weights, proportionalized. For example, proportionalize [1, 3, 4] yields [0.125, 0.375, 0.5].
• Write a function scalebias that accepts a scale, a bias, and a list of numbers. It yields a list where all elements have been scaled by the scale factor and offset by the bias.
• Write a function nonzeroes that accepts a list of numbers and yields a list containing only the nonzero elements. For example, nonzero [0, 1, -3, 0] yields [1, -3].
• Write a function overs that accepts an n and a list of strings and yields a list containing only those strings whose length is greater than n. For example, overs 3 ["the", "quick", "brown", "fox"] yields ["quick", "brown"].
• Write a function all that accepts a list of booleans and yields true if all elements are true.
• Write a function any that accepts a list of booleans and yields true if any element is true.
• Write a function sum that accepts a list of numbers and yields their sum.
• Write a function product that accepts a list of numbers and yields their product.
• Write a function length that accepts a list and yields its length.
• Write a function join that accepts a list of strings and yields their concatenation.
• Write a function max that accepts a list of numbers and yields their maximum.

### Code

#### Hof.hs

import Data.List

resort :: Ord a => [a] -> [a]
resort = (reverse . sort)

-- pairexpander :: (String, Int) -> [String]
-- pairexpander (s, n) = replicate n s

expand :: [(String, Int)] -> [[String]]
-- expand list = map pairexpander list
-- expand list = map (\(s, n) -> replicate n s) list
expand = map (\(s, n) -> replicate n s)

proportionalize :: [Double] -> [Double]
-- proportionalize list = map (\n -> n / total) list
-- where total = sum list
proportionalize list = map (/ total) list
where total = sum list

all' :: [Bool] -> Bool
all' [] = True
all' (first:rest) = first && all' rest

any' :: [Bool] -> Bool
any' [] = False
any' (first:rest) = first || any' rest

sum' :: [Double] -> Double
sum' [] = 0
sum' (first:rest) = first + sum' rest

foldr' mix accum [] = accum
foldr' mix accum (first:rest) = foldr' mix (mix accum first) rest

all'' = foldr' (&&) True
any'' = foldr' (||) False
sum'' = foldr' (+) 0
product'' = foldr' (*) 1
join'' = foldr' (++) ""

### Haiku

on productivity
I outsourced my job
To map and filter and fold


text:
# User:Guy vandegrift/Quizbank/Archive1/Calculus Physics/T4study

## CalcIIT4_Study

If you are reading this as a Wikiversity page, proper pagebreaks should result if printed using your browser's print option. On Chrome, Explorer, and Firefox, this option is available in the upper right hand corner of your screen. But, pagebreaks do not render properly if you use "Printable version" on Wikiversity's Print/export option on the left-hand sidebar.

This document contains either a study guide OR pairs of exams taken from the same exam bank
If two exams have the same s-number, then v1 and v2 have the same questions, presented in different (random) order.
Exams with different s-numbers have different questions and may not have the same difficulty.
Click items in the table of contents and appropriate page should be reached. This feature should allow you to print only those pages that you need.
At the end of this document
Attribution for the quizzes identifies where the questions were obtained

### CalcIIT4_Study-v1s1

1. Two orbiting satellites are orbiting at a speed of 52 km/s perpendicular to a magnetic field of 41 μT. They are connected by a cable that is 33 km long. A voltmeter is attached between a satellite and one end of the cable. The voltmeter's internal impedance far exceeds the net resistance through the ionosphere that completes the circuit. What is the measured voltage?

___a) 4.79 x 104 volts.
___b) 5.81 x 104 volts.
___c) 7.04 x 104 volts.
___d) 8.52 x 104 volts.
___e) 1.03 x 105 volts.

2. An loop of wire with 43 turns has a radius of 0.27 meters, and is oriented with its axis parallel to a magetic field of 0.68 Tesla. What is the induced voltage if this field is reduced to 36% of its original value in 3.8 seconds?

___a) 6.34 x 10-1 volts
___b) 7.68 x 10-1 volts
___c) 9.31 x 10-1 volts
___d) 1.13 x 100 volts
___e) 1.37 x 100 volts

3. A circlular capactitor of radius 4.2 m has a gap of 12 mm, and a charge of 94 μC. What is the electric field between the plates?

___a) 1.92E+05 N/C (or V/m)
___b) 2.32E+05 N/C (or V/m)
___c) 2.81E+05 N/C (or V/m)
___d) 3.41E+05 N/C (or V/m)
___e) 4.13E+05 N/C (or V/m)

4. A circlular capactitor of radius 4.1 m has a gap of 8 mm, and a charge of 24 μC. Compute the surface integral ${\displaystyle c^{-2}\oint {\vec {E}}\cdot d{\vec {A}}}$ over an inner face of the capacitor.

___a) 2.05E-11 Vs2m-1
___b) 2.49E-11 Vs2m-1
___c) 3.02E-11 Vs2m-1
___d) 3.65E-11 Vs2m-1
___e) 4.43E-11 Vs2m-1

5. A circlular capactitor of radius 4.7 m has a gap of 19 mm, and a charge of 27 μC. The capacitor is discharged through a 6 kΩ resistor. What is the decay time?

___a) 1.60E-04 s
___b) 1.94E-04 s
___c) 2.35E-04 s
___d) 2.85E-04 s
___e) 3.45E-04 s

6. A circlular capactitor of radius 3.1 m has a gap of 9 mm, and a charge of 85 μC. The capacitor is discharged through a 5 kΩ resistor. What is what is the maximum magnetic field at the edge of the capacitor? (There are two ways to do this; you should know both.)

___a) 2.33E-08 Tesla
___b) 2.93E-08 Tesla
___c) 3.69E-08 Tesla
___d) 4.65E-08 Tesla
___e) 5.85E-08 Tesla

7.
Shown is a corrective lens by a person who needs glasses. This ray diagram illustrates
___ a) how a nearsighted person might see a distant object
___ b) how a farsighted person might see an object that is too close for comfort
___ c) how a farsighted person might see a distant object
___ d) how a nearsighted person might see an object that is too close for comfort

8.
Shown is a corrective lens by a person who needs glasses. This ray diagram illustrates
___ a) how a farsighted person might see an object that is too close for comfort
___ b) how a nearsighted person might see a distant object
___ c) how a farsighted person might see a distant object
___ d) how a nearsighted person might see an object that is too close for comfort

9. In optics, normal means

___ a) to the right of the optical axis
___ b) perpendicular to the surface
___ c) to the left of the optical axis
___ d) parallel to the surface

10. The law of reflection applies to

___ a) flat surfaces
___ b) telescopes but not microscopes
___ c) curved surfaces
___ d) both flat and curved surfaces
___ e) only light in a vacuum

11. When light passes from air to glass

___ a) it bends towards the normal
___ b) the frequency increases
___ c) it does not bend
___ d) the frequency decreases
___ e) it bends away from the normal

12. When light passes from glass to air

___ a) it bends towards the normal
___ b) it does not bend
___ c) the frequency increases
___ d) the frequency decreases
___ e) it bends away from the normal

13. An important principle that allows fiber optics to work is

___ a) the invariance of the speed of light
___ b) total external refraction
___ c) partial internal absorption
___ d) the Doppler shift
___ e) total internal reflection

14. The focal point is where

___ a) rays meet whenever they are forming an image
___ b) the center of the lens
___ c) rays meet if they were parallel to the optical axis before striking a lens
___ d) rays meet whenever they pass through a lens
___ e) rays meet if they are parallel to each other

15. An object is placed 6.3 cm to the left of a diverging lens with a focal length of 8.9 cm. How far is the image from the lens?

___a) 1.17 x 100 cm
___b) 2.07 x 100 cm
___c) 3.69 x 100 cm
___d) 6.56 x 100 cm
___e) 1.17 x 101 cm

16. An object is placed 4.15 cm to the left of a converging lens with a focal length of 3.6 cm. How far is the image from the lens?

___a) 8.59 x 100 cm
___b) 1.53 x 101 cm
___c) 2.72 x 101 cm
___d) 4.83 x 101 cm
___e) 8.59 x 101 cm

17. An object of height 0.75 cm is placed 147 cm behind a diverging lens with a focal length of 86 cm. What is the height of the image?

___a) 2.77 x 10-1 cm
___b) 3.32 x 10-1 cm
___c) 3.99 x 10-1 cm
___d) 4.78 x 10-1 cm
___e) 5.74 x 10-1 cm

18. An object is placed 13.7 cm to the left of a diverging lens with a focal length of 17.7 cm. On the side, at a distance of 5.5 cm from the diverging lens is a converging lens with focal length equal to 4 cm. How far is the final image from the converging lens?

___a) 5.73 x 10-2 cm
___b) 1.81 x 10-1 cm
___c) 5.73 x 10-1 cm
___d) 1.81 x 100 cm
___e) 5.73 x 100 cm

19. Which lens has the shorter focal length?

___ a) They have the same focal lengh.
___ b)
___ c)

20. If this represents the eye looking at an object, where is this object?

___ a) at infinity
___ b) very far away
___ c) One focal length in front of the eye
___ d) directly in front of the eye (almost touching)
___ e) Two (of the other answers) are true

21. After passing through a the lens of a camera or the eye, the focal point is defined as where the rays meet.

___ a) true
___ b) false

22. Mr. Smith is gazing at something as shown in the figure to the left. Suppose he does not refocus, but attempts to stare at the star shown in the figures below. Which diagram depicts how the rays from the star would travel if he does not refocus?

___ a)
___ b)
___ c)

#### Key to CalcIIT4_Study-v1s1

1. Two orbiting satellites are orbiting at a speed of 52 km/s perpendicular to a magnetic field of 41 μT. They are connected by a cable that is 33 km long. A voltmeter is attached between a satellite and one end of the cable. The voltmeter's internal impedance far exceeds the net resistance through the ionosphere that completes the circuit. What is the measured voltage?

-a) 4.79 x 104 volts.
-b) 5.81 x 104 volts.
+c) 7.04 x 104 volts.
-d) 8.52 x 104 volts.
-e) 1.03 x 105 volts.

2. An loop of wire with 43 turns has a radius of 0.27 meters, and is oriented with its axis parallel to a magetic field of 0.68 Tesla. What is the induced voltage if this field is reduced to 36% of its original value in 3.8 seconds?

-a) 6.34 x 10-1 volts
-b) 7.68 x 10-1 volts
-c) 9.31 x 10-1 volts
+d) 1.13 x 100 volts
-e) 1.37 x 100 volts

3. A circlular capactitor of radius 4.2 m has a gap of 12 mm, and a charge of 94 μC. What is the electric field between the plates?

+a) 1.92E+05 N/C (or V/m)
-b) 2.32E+05 N/C (or V/m)
-c) 2.81E+05 N/C (or V/m)
-d) 3.41E+05 N/C (or V/m)
-e) 4.13E+05 N/C (or V/m)

4. A circlular capactitor of radius 4.1 m has a gap of 8 mm, and a charge of 24 μC. Compute the surface integral ${\displaystyle c^{-2}\oint {\vec {E}}\cdot d{\vec {A}}}$ over an inner face of the capacitor.

-a) 2.05E-11 Vs2m-1
-b) 2.49E-11 Vs2m-1
+c) 3.02E-11 Vs2m-1
-d) 3.65E-11 Vs2m-1
-e) 4.43E-11 Vs2m-1

5. A circlular capactitor of radius 4.7 m has a gap of 19 mm, and a charge of 27 μC. The capacitor is discharged through a 6 kΩ resistor. What is the decay time?

-a) 1.60E-04 s
+b) 1.94E-04 s
-c) 2.35E-04 s
-d) 2.85E-04 s
-e) 3.45E-04 s

6. A circlular capactitor of radius 3.1 m has a gap of 9 mm, and a charge of 85 μC. The capacitor is discharged through a 5 kΩ resistor. What is what is the maximum magnetic field at the edge of the capacitor? (There are two ways to do this; you should know both.)

-a) 2.33E-08 Tesla
-b) 2.93E-08 Tesla
+c) 3.69E-08 Tesla
-d) 4.65E-08 Tesla
-e) 5.85E-08 Tesla

7.
Shown is a corrective lens by a person who needs glasses. This ray diagram illustrates
+ a) how a nearsighted person might see a distant object
- b) how a farsighted person might see an object that is too close for comfort
- c) how a farsighted person might see a distant object
- d) how a nearsighted person might see an object that is too close for comfort

8.
Shown is a corrective lens by a person who needs glasses. This ray diagram illustrates
+ a) how a farsighted person might see an object that is too close for comfort
- b) how a nearsighted person might see a distant object
- c) how a farsighted person might see a distant object
- d) how a nearsighted person might see an object that is too close for comfort

9. In optics, normal means

- a) to the right of the optical axis
+ b) perpendicular to the surface
- c) to the left of the optical axis
- d) parallel to the surface

10. The law of reflection applies to

- a) flat surfaces
- b) telescopes but not microscopes
- c) curved surfaces
+ d) both flat and curved surfaces
- e) only light in a vacuum

11. When light passes from air to glass

+ a) it bends towards the normal
- b) the frequency increases
- c) it does not bend
- d) the frequency decreases
- e) it bends away from the normal

12. When light passes from glass to air

- a) it bends towards the normal
- b) it does not bend
- c) the frequency increases
- d) the frequency decreases
+ e) it bends away from the normal

13. An important principle that allows fiber optics to work is

- a) the invariance of the speed of light
- b) total external refraction
- c) partial internal absorption
- d) the Doppler shift
+ e) total internal reflection

14. The focal point is where

- a) rays meet whenever they are forming an image
- b) the center of the lens
+ c) rays meet if they were parallel to the optical axis before striking a lens
- d) rays meet whenever they pass through a lens
- e) rays meet if they are parallel to each other

15. An object is placed 6.3 cm to the left of a diverging lens with a focal length of 8.9 cm. How far is the image from the lens?

-a) 1.17 x 100 cm
-b) 2.07 x 100 cm
+c) 3.69 x 100 cm
-d) 6.56 x 100 cm
-e) 1.17 x 101 cm

16. An object is placed 4.15 cm to the left of a converging lens with a focal length of 3.6 cm. How far is the image from the lens?

-a) 8.59 x 100 cm
-b) 1.53 x 101 cm
+c) 2.72 x 101 cm
-d) 4.83 x 101 cm
-e) 8.59 x 101 cm

17. An object of height 0.75 cm is placed 147 cm behind a diverging lens with a focal length of 86 cm. What is the height of the image?

+a) 2.77 x 10-1 cm
-b) 3.32 x 10-1 cm
-c) 3.99 x 10-1 cm
-d) 4.78 x 10-1 cm
-e) 5.74 x 10-1 cm

18. An object is placed 13.7 cm to the left of a diverging lens with a focal length of 17.7 cm. On the side, at a distance of 5.5 cm from the diverging lens is a converging lens with focal length equal to 4 cm. How far is the final image from the converging lens?

-a) 5.73 x 10-2 cm
-b) 1.81 x 10-1 cm
-c) 5.73 x 10-1 cm
-d) 1.81 x 100 cm
+e) 5.73 x 100 cm

19. Which lens has the shorter focal length?

- a) They have the same focal lengh.
+ b)
- c)

20. If this represents the eye looking at an object, where is this object?

- a) at infinity
- b) very far away
- c) One focal length in front of the eye
- d) directly in front of the eye (almost touching)
+ e) Two (of the other answers) are true

21. After passing through a the lens of a camera or the eye, the focal point is defined as where the rays meet.

- a) true
+ b) false

22. Mr. Smith is gazing at something as shown in the figure to the left. Suppose he does not refocus, but attempts to stare at the star shown in the figures below. Which diagram depicts how the rays from the star would travel if he does not refocus?

+ a)
- b)
- c)


text:
# highlight text between selector without highlighting nor deleting the selectors [duplicate]

So I'm trying to render something beautiful for my PHP code using the \lstdefinestyle command.

I think the basic configuration is good, but I would like to highlight the function name to render it in the color normal.

Here is my code:

\documentclass[10pt,a4paper]{article}
\usepackage[T1]{fontenc}
\usepackage[utf8]{inputenc}
\usepackage[english]{babel}
\usepackage{listings}             % to insert code
\usepackage{xcolor}
\definecolor{background}{RGB}{40, 44, 52}           % gris foncé
\definecolor{comment}{RGB}{92, 99, 112}             % gris clair
\definecolor{keyworld}{RGB}{86, 182, 194}           % cyan
\definecolor{string}{RGB}{152, 195, 121}            % vert
\definecolor{identifier}{RGB}{198, 120, 221}        % mauve
\definecolor{classic}{RGB}{224, 108, 117}           % genre de rouge
\definecolor{normal}{RGB}{171, 178, 191}            % blanc
\definecolor{specialkeyworld}{RGB}{229, 192, 123}   % jaune orange
\definecolor{veridad}{RGB}{209, 154, 102}           % orange jaune

\lstdefinestyle{myPHP}{
language=PHP,
backgroundcolor=\color{background},
basicstyle=\footnotesize\ttfamily\color{normal} ,
keywordstyle=\color{keyworld},
numberstyle=\color{background},
identifierstyle=\color{classic},
stringstyle=\color{string},
emph        =[1]{function,try,return,catch, new},
emphstyle   =[1]\color{identifier},
emph        =[2]{Exception, PDO},
emphstyle   =[2]\color{specialkeyworld},
emph        =[3]{true, false},
emph        =[4]{connectAccess},
emphstyle   =[4]\color{normal},
stringstyle=\color{string},
breakatwhitespace=false,
breaklines=false,
keepspaces=true,
numbers=left,
numbersep=4pt,
showspaces=false,
showstringspaces=false,
}
\begin{document}
\begin{lstlisting}[style=myPHP]
/**
*  Connect an Access database
*  @param  string  $path Name of the database you want to connect to * @return PDO object initialise */ function connectAccess($path){
try {
$db = new PDO("odbc:DRIVER={Microsoft Access Driver (*.mdb)}; Dbq=$path;");
return $db; } catch (Exception$e) {
echo 'Connection failed: ' . \$e->getMessage();
return false;
}
}
\end{lstlisting}
\end{document}

and the result :

I would like to replace the two lines

  emph        =[4]{connectAccess},
emphstyle   =[4]\color{normal},

with something that will highlight what is between function and ( without selecting those two, because I will be displaying a lot of code in my document and I dont want to escape the function name each time.

I've tried moredelim=[is][\color{normal}\ttfamily]{function}{(}, but it render like this :

with moredelim=[s][\color{normal}\ttfamily]{function}{(}, the 'function' selector is also rendered in the \color{normal} font...

and I have pretty much the same results with morecomment.

What am I doing wrong? aren't moredelim and morecomment the right parameters to use? if not, what are the original purpose of these two?

Is there a(nother) way to do it or will I have to escape the function name each time?

## marked as duplicate by Stefan Pinnow, Paul Gaborit, Romain Picot, egreg, user13907 May 24 '16 at 9:13

• Welcome to TeX.SX Guillaume ! – ebosi May 9 '16 at 7:40
• I've seen these type of comments for improper questions, do you think I have to edit my question for something specific or is it just a way to say hello? – guillaume latour May 9 '16 at 8:16
• @guillaumelatour Hello :-) – touhami May 9 '16 at 8:18
• That's quite a rainbow you show here. Imho you are overdoing with the colors. Beside this I don't think tha there is a better solution than the one you found. Delimiters are printed in the same style as the part between them. So you can't use them here. Side remark: You named your color "keyworld" instead of "keyword". – Ulrike Fischer May 9 '16 at 9:13
• Possible duplicate: Listings - Only highlight between delimiters – jkuczm May 22 '16 at 9:15

\newcommand{\functionDefHighlight}[1]{\textcolor{identifier}{function} \textcolor{normal}{ #1}(}

moredelim = [is][\functionDefHighlight]{function\ }{(},


text:
Which quantity contains the LEAST number of atoms?

Aug 2, 2017

I would go for $\text{option b}$.

Explanation:

You gots.......

$a .$ $\text{Two moles of NaOH;}$ $2 \times N {a}^{+} + 2 H {O}^{-}$ $\text{moles}$

$b .$ $\text{One mole of ammonium; }$ $1 \times N {H}_{3} \cdot m o l$

$c .$ $\text{Two moles of calcium chloride; }$ $2 \times C {a}^{2 +} + 4 \times C {l}^{-} \cdot m o l$

$d .$ $\text{One mole of alumina; }$ $2 \times A {l}^{3 +} + 3 \times {O}^{2 -} \cdot m o l$

Is this clear?


text:
##### Context sensitive language
• A context sensitive grammar (CSG) is a grammar where all productions are of the form αAβ → αγβ.
• During derivation non-terminal A will be changed to γ only when it is present in context of α and β.
• CSG is a Noncontracting grammar.
• The language generated by Context free grammar is known as Context sensitive language.

Created:


text:
# Variational calculus … Prove that a linear functional $\varphi [h]$ cannot have an extremum unless $\varphi [h] \equiv 0$

From definitions in the book calculus of variations - Gelfand and fomin http://web.cs.iastate.edu/~cs577/handouts/variations.pdf

I try prove that a linear functional $$\varphi[h]$$ cannot have an extremum unless $$\varphi[h] \equiv 0$$.

I tried the following 1- prove that $$\varphi[h]$$ is differentiable ann use that theorem 2 pag 13 in the book

2- I tried the use that a $$J[y]$$ has an extremum in $$t$$ if $$J[y] -J[t]$$ does not change its sing in some neighborhood, but for me it is not clear the not change of sign of $$\varphi[h]$$

Any help is good. Thanks.

• Use $f(ax) = af(x)$ for suitable choices of $a$ to find points close to $x$ with a larger/smaller value. – Winther Jun 8 at 17:12


text:
# Workshop on Immersed surfaces in 3-manifolds Mar 26-30, 2012

Organizers: Danny Calegari, Bill Goldman, Vlad Markovic, Jean-Marc Schlenker, Alan Reid.

### Invited speakers:

• Ian Agol (UC Berkeley)
• Nicolas Bergeron (Paris 6)
• Martin Bridgeman (Boston College)
• Jeff Brock (Brown)
• Danny Calegari (Cambridge)
• Will Cavendish (Princeton)
• François Guéritaud (Lille)
• Jeremy Kahn (Brown)
• Sang-hyun (Sam) Kim (KAIST)
• François Labourie (Orsay)
• Tao Li (Boston College)
• Brice Loustau (Orsay)
• Jason Manning (Buffalo)
• Jessica Purcell (Brigham Young)
• Claire Renard (ENS Cachan)
• Alden Walker (Caltech)

## Titles and Abstracts

### Ian Agol

Title: The virtual Haken conjecture.

Abstract: We prove that cubulated hyperbolic groups are virtually special. Work of Haglund and Wise on special cube complexes implies that they are therefore linear groups, and quasi-convex subgroups are separable. A consequence is that closed hyperbolic 3-manifolds have finite-sheeted Haken covers, which resolves the virtual Haken question of Waldhausen and Thurston's virtual fibering question. The results depend on a recent result of Wise, the malnormal special quotient theorem; the cubulation of closed hyperbolic 3-manifolds by Bergeron-Wise using the existence of nearly geodesic surfaces by Kahn-Markovic; and a generalization of previous work with Groves and Manning to the case of torsion (which is joint with Groves and Manning).

### Nicolas Bergeron

Title: Cubulation of hyperbolic 3-manifolds.

Abstract: I will describe a simple criterion in terms of the boundary for the existence of a proper cocompact action of a hyperbolic group on a CAT(0) cube complex. Thanks to the proof, by Kahn and Markovic, of the surface subgroup conjecture, the criterion is satisfied by all compact hyperbolic 3-manifolds.

### Martin Bridgeman

Title: Pleated Surfaces, Convex Hulls and Domains.

Abstract:

### Jeff Brock

Title: Fat, exhaustive integer homology 3-spheres.

Abstract:

### Danny Calegari

Title: Stable commutator length - statistics, concentration and compression.

Abstract: We discuss the statistical distribution of stable commutator length in various classes of groups, and some applications. For certain classes of groups (e.g. central extensions of lattices in $$\mathrm{Sp}(2n, \mathbf{R})$$) stable commutator length is distributed like distance to the origin for a random walk in a finite dimensional Euclidean space. For other classes of groups (e.g. hyperbolic groups, braid groups) there is a concentration of values, clustered around some fixed scale $$Cn/\log(n)$$ where the constant $$C$$ should conjecturally be derived in a simple manner from the (growth) entropy. This concentration should be thought of as a random analogue of the phenomenon of Mostow rigidity for hyperbolic manifolds. Finally, the growth rate of stable commutator length is an obstruction to the existence of (nonelementary) homomorphisms to hyperbolic groups, or actions on certain hyperbolic spaces. Some of this is joint work with Koji Fujiwara, Joseph Maher and Alden Walker.

### Will Cavendish

Title: Towers of Covering Spaces of 3-manifolds and Mapping Solenoids.

Abstract: This talk concerns the following question: given a map $$f:X\toY$$ between compact CW complexes and a collection $$C$$ of finite-sheeted covering spaces of $$Y$$, which covering spaces of $$X$$ appear as pull-backs of covers in the collection $$C$$? We will begin by introducing a construction called the mapping solenoid of $$f$$, and to show how the cohomology groups of this object can be used to give qualitative answers to this question. To demonstrate some uses of this construction we will study $$\pi_1$$-injective immersions $$f$$ from surfaces into 3-manifolds. In this setting, we will show that certain cohomology classes on the mapping solenoid of $$f$$ can be viewed as obstructions to solving lifting problems in the tower $$C$$. We will go on to show that if $$C$$ is a tower of non-Haken 3-manifolds, then the cohomology of the mapping solenoid can often be computed exactly, and gives rise to $$\pi_1(M)$$-modules with curious algebraic properties.

### François Guéritaud

Title: On Lorentzian spacetimes with constant curvature.

Abstract: I will discuss a characterization of Lorentzian spacetimes with constant negative curvature (i.e. complete AdS manifolds) in terms of contracting equivariant self-maps of the hyperbolic plane. I will also describe an infinitesimal version (corresponding to complete flat Lorentz manifolds), and describe how geodesic laminations arise in the boundary of the deformation space of such objects. Joint work with F. Kassel.

### Jeremy Kahn

Title: Essential Immersed Surfaces in Closed Hyperbolic 3-Manifolds.

Abstract: We prove that every closed hyperbolic 3-manifold has an essential immersed surfaces, and we present some results on the number of these surfaces of a given genus. In particular, we will sketch a proof that in certain cases, a typical closed immersed surface of large genus will not be evenly distributed in the given 3-manifold.

### Sang-hyun (Sam) Kim

Title: Doubles of free groups and hyperbolic surface subgroups.

Abstract: Let G be the double of a rank-two free group. We prove that if G is one-ended, then G has a hyperbolic surface subgroup. The proof is done by a partial resolution of a stronger graph theoretical conjecture, and is based upon a previous joint work with Henry Wilton. We also discuss a generalization to a graph of free groups with cyclic edge groups. This is a joint work with Sang-il Oum.

### François Labourie

Title: Random complex projective structures.

Abstract: In this talk, I will present a way to describe as a dynamical system the space of all complex projective structure on surfaces.

Quotienting by the action of a closed 3-manifold group yields a compact space sharing many properties of the usual geodesic flow. This space which is foliated by Riemann surfaces have the usual chaotic properties: genericity of dense leaves, density of the space of closed leaves, stability, many invariant ergodic measures of full support (associated to complex cross ratios). These results are not new but they present a framework to ask questions suscited by Kahn-Markovic theory.

### Tao Li

Title: Rank and genus of 3-manifolds.

Abstract: We construct a counterexample to the Rank versus Genus Conjecture, i.e. a closed orientable hyperbolic 3-manifold with rank of its fundamental group smaller than its Heegaard genus. Moreover, we show that the discrepancy between rank and Heegaard genus can be arbitrarily large for hyperbolic 3-manifolds. We also construct toroidal such examples containing hyperbolic JSJ pieces.

### Brice Loustau

Title: Minimal surfaces in almost-Fuchsian manifolds and renormalized volume.

Abstract: Minimal surfaces in quasifuchsian 3-manifolds provide a parametrization of the set of almost-Fuchsian structures by holomorphic quadratic differentials. This parametrization is different from the classical Schwarzian parametrization. We use the notion of renormalized volume to compare the symplectic structures induced on the set of almost-Fuchsian structures.

### Jason Manning

Title: Height, multiplicity, and Dehn filling.

Abstract: We show that given a virtually special quasiconvex subgroup $$H$$ of a hyperbolic group $$G$$, and some $$g \notin H$$, there is a hyperbolic quotient of $$G$$ in which $$H$$ projects to a finite subgroup, disjoint from the image of the element $$g$$. This statement is used in Ian Agol's recent proof of the Virtual Haken Conjecture. The method of proof is an induction on the height of the quasiconvex subgroup, using relatively hyperbolic Dehn filling to reduce height. Along the way, we give a new characterization of height as the multiplicity of a certain map, and a new proof (which does not depend on torsion-freeness) that Dehn filling reduces height of quasi-convex subgroups. This is joint work with Ian Agol and Daniel Groves.

### Jessica Purcell

Title: State surfaces in knot complements: fibers, guts, and the Jones polynomial.

Abstract: In the last decade, there have been several breakthroughs in hyperbolic geometry and 3-manifold topology. One goal in the research community is to use these new results to shed light on related problems, for example problems in knot and link theory. In this talk, we will describe some of our work in this direction. We determine geometric information on incompressible spanning surfaces in a large class of knot and link diagrams. Study of the geometric properties of these surfaces allows us to detect fibers and bound volumes, and relate them to certain coefficients of the Jones polynomial. This is joint with David Futer and Efstratia Kalfagianni.

### Claire Renard

Title: Finite covers of a hyperbolic 3-manifold and virtual fibrations.

Abstract: One can ask if there are some 'nice' conditions for a finite cover of a hyperbolic 3-manifold $$M$$ to fiber over the circle, or at least to contain a virtual fiber. In this talk, we provide an inequality on the Heegaard genus $$g$$ of a finite cover $$M'$$ of a hyperbolic 3-manifold $$M$$ and the degree $$d$$ of the cover such that if this inequality is satisfied, $$M'$$ contains a fiber which is an embedded surface of genus at most $$g$$. This inequality involves an explicit constant $$k$$ depending only on the volume and the injectivity radius of $$M$$. This machinery also applies to the setting of a circular decomposition associated to a non trivial homology class, to obtain sufficient conditions for a non trivial homology class of $$M$$ to correspond to a fibration over the circle. Similar methods lead also to a sufficient condition for an incompressible embedded surface in $$M$$ to be a fiber.

### Alden Walker

Title: Ziggurats and rotation numbers.

Abstract: (Joint with Danny Calegari) We establish the existence of new rigidity and rationality phenomena in the theory of nonabelian group actions on the circle, and introduce tools to translate questions about the existence of actions with prescribed dynamics into finite combinatorics. A special case of our theory gives a very short new proof of Naimi's theorem (i.e. the conjecture of Jankins--Neumann) which was the last step in the classification of taut foliations of Seifert fibered spaces.


text:
Thread: Using the Binomial Theory to determine the approx. percentage change in the area?

thank you

2. Re: Using the Binomial Theory to determine the approx. percentage change in the area?

You are "stuck" implies that you have started the problem but then ran into a problem. Okay, what have you done and exactly where did you hit the "problem"?
If you have not done anything, perhaps the problem is with basic formulae or definitions. Do you know how to find the area of a circle? Do you know what the "binomial theorem" is?

3. Re: Using the Binomial Theory to determine the approx. percentage change in the area?

When i said stuck i meant im working on a college assignment and came to this question. the reason im stuck on it is because our teacher hasnt gone through this with us yet. ive tried google'ing binomial theorem and i cant find anything that relates to what i need to do. is there any possibility that you could show me step by step how id work this out??

Thank you

4. Re: Using the Binomial Theory to determine the approx. percentage change in the area?

But yes i know how to work out the area of a circle.

Originally Posted by HallsofIvy
You are "stuck" implies that you have started the problem but then ran into a problem. Okay, what have you done and exactly where did you hit the "problem"?
If you have not done anything, perhaps the problem is with basic formulae or definitions. Do you know how to find the area of a circle? Do you know what the "binomial theorem" is?

binomial series for calculting the approximate percentage error

Click on a term to search for related topics.


text:
# An Introduction to Proportionality

## Introduction

The bioinformatic evaluation of gene co-expression often begins with correlation-based analyses. However, correlation lacks validity when applied to relative data. This includes data generated by high-throughput RNA-sequencing, chromatin immunoprecipitation (ChIP), ChIP-sequencing, Methyl-Capture sequencing, and more. This package implements three measures of proportionality to measure dependence between relative features using compositional data analysis: $$\phi$$ (Lovell 2015), $$\rho$$ (Lovell 2015; Erb 2016), and $$\phi_s$$ (Erb 2016; Quinn 2017). Unlike correlation, proportionality gives the same result for both relative and absolute data. Meanwhile, pairs that are strongly proportional in relative space are also strongly correlated in absolute space. Proportionality helps avoid the pitfall of spurious correlation.

## Theory

Let $$A_i$$ and $$A_j$$ each represent a log-ratio transformed feature vector measured across $$N$$ samples. We then define the metrics $$\phi$$, $$\rho$$, and $$\phi_s$$:

$\phi(A_i, A_j) = \frac{var(A_i - A_j)}{var(A_i)}$

$\rho(A_i, A_j) = 1 - \frac{var(A_i - A_j)}{var(A_i) + var(A_j)}$

$\phi_s(A_i, A_j) = \frac{var(A_i - A_j)}{var(A_i + A_j)}$

Log-ratio transformation is applied to each subject vector, $$x$$, containing $$D$$ features (e.g., genes). In this vignette, we consider the centered log-ratio (clr) transformation and the additive log-ratio (alr) transformation. In clr-transformation, sample vectors undergo a transformation based on the logarithm of the ratio between the individual elements and the geometric mean of the vector, $$g(\textrm{x}) = \sqrt[D]{x_i...x_D}$$. In alr-transformation, sample vectors undergo a transformation based on the logarithm of the ratio between the individual elements and chosen reference feature, $$x_D$$. We define these accordingly:

$\textrm{clr(x)} = \left[\ln\frac{x_i}{g(\textrm{x})};...;\ln\frac{x_D}{g(\textrm{x})}\right]$

$\textrm{alr(x)} = \left[\ln\frac{x_i}{x_D};...;\ln\frac{x_{D-1}}{x_D}\right]$

Note that this package also implements the interquartile log-ratio transformation (iqlr) as used by the ALDEx2 package. For more information on integrating propr with ALDEx2, we refer the reader to the “Frequently Asked Questions” vignette.

## Calculating proportionality

We provide three principal functions for calculating proportionality. The first, phit, calculates $$\phi$$. The second, perb, calculates $$\rho$$. The third, phis, calculates $$\phi_s$$.

The measures $$\phi$$ and $$\rho$$ differ in three ways. First, the values of $$\phi$$ range from $$[0, \infty)$$ (with lower $$\phi$$ values indicating more proportionality) while the values of $$\rho$$ range from $$[-1, 1]$$ (with greater $$|\rho|$$ values indicating more proportionality and negative $$\rho$$ values indicating inverse proportionality). Second, $$\phi$$ lacks symmetry, although one can force symmetry by reflecting the lower left triangle of the matrix across the diagonal (toggled by the argument symmetrize = TRUE). Third, $$\rho$$ corrects for the individual variance of each feature in the pair, rather than for just one of the features. On the other hand, $$\phi_s$$ is a naturally symmetric variant of $$\phi$$ that also corrects for the individual variance of each feature in the pair.

Let us begin by building an arbitrary dataset of 4 features (e.g., genes) measured across 100 subjects. In this example, the feature pairs “a” and “b” will show proportional change as will the feature pairs “c” and “d”.

set.seed(12345)
N <- 100
X <- data.frame(a=(1:N), b=(1:N) * rnorm(N, 10, 0.1),
c=(N:1), d=(N:1) * rnorm(N, 10, 1.0))

Let $$D$$ represent any number of features measured across $$N$$ observations exposed to a binary or continuous event $$E$$. For example, $$E$$ could represent differences in case-control status, treatment status, treatment dose, or time. These functions convert a “count matrix” with $$N$$ rows and $$D$$ columns into a proportionality matrix of $$D$$ rows and $$D$$ columns measuring proportionality for each feature pair. One can think of this matrix as analogous to a dissimilarity matrix (in the case of $$\phi$$) or a correlation matrix (in the case of $$\rho$$). Both functions return the proportionality matrix bundled within an object of the class propr. This object contains four key slots:

• @counts A matrix. Stores the original “count matrix” input.
• @logratio A matrix. Stores the log-ratio transformed “count matrix”.
• @matrix A matrix. Stores the proportionality metrics.
• @pairs A vector. Indexes proportionality of interest.
library(propr)
phi <- phit(X, symmetrize = TRUE)
rho <- perb(X, ivar = 0)
phs <- phis(X, ivar = 0)

Note that the log-ratio transformation, by its nature, fails if the input data contain any zero values. To avoid an error in this case, these functions automatically replace all zero values with 1. However, the topic of zero replacement is controversial. Proceed carefully when analyzing data that contain zero values.

## Subsetting propr objects

We have provided methods for indexing and subsetting objects belonging to the propr class. Using the familiar [ method, we can efficiently index the proportionality matrix (i.e., the @matrix slot) based on an inequality operator and a reference value.

In this first example, we use [ to index the matrix by $$\rho > .95$$. This indexes the location of all values satisfying that inequality (i.e., in the lower left triangle of the matrix), and saves those indices to the @pairs slot. Indexing helps guide some of the bundled visualization methods in lieu of copy-on-modify subsetting. Note that indexing an already indexed object appends the new index to the previous index (via a union merge).

rho99 <- rho[">", .95]
rho99@pairs
## [1]  2 12

Alternatively, using the subset method, we can subset an entire propr object based on a vector of feature indices or names. However, this method does copy-on-modify the proportionality matrix, making it potentially unsuitable for large datasets.

In this second example, we subset by the feature names “a” and “b”.

rhoab <- subset(rho, select = c("a", "b"))
rhoab@matrix
##           a         b
## a 1.0000000 0.9999151
## b 0.9999151 1.0000000

The convenience function, simplify, can subset an entire propr object based on the index saved in its @pairs slot. This function converts the saved index into a paired list of coordinates and passes them along to the subset method. As such, this method does copy-on-modify the proportionality matrix, making it potentially unsuitable for large datasets. Unlike subset, simplify returns an object with the @pairs slot updated. Most users will find simplify preferable to subset.

simplify(rho99)
## @counts summary: 100 subjects by 4 features
## @logratio summary: 100 subjects by 4 features
## @matrix summary: 4 features by 4 features
## @pairs summary: 2 feature pairs

## Visualizing pairs

Each feature belonging to a highly proportional data pair should have about the same log-ratio transformed abundance across all subjects. The method plot (or, equivalently, smear) provides a means by which to visually inspect whether this holds true. Since this function will plot all pairs unless indexed with the [ method, we recommend the user first index or subset the propr object before plotting. A “noisy” relationship between some feature pairs could suggest that the proportionality cutoff is too lenient. We include this plot as a handy “sanity check” when working with high-dimensional datasets.

plot(rho99)
## Alert: Generating plot using indexed feature pairs.

## Computational burden

High-throughput genomic sequencing has the ability to measure tens of thousands of features for each subject. Since calculating proportionality generates a matrix sized $$D^2$$, this method uses a lot of RAM when applied to real biological datasets. To address this, propr harnesses the power of C++ (via the Rcpp package) to minimize the run-time and RAM overhead. Below, we provide a table that estimates the approximate amount of RAM needed to render a proportionality matrix based on the number of features studied. The user should account for up to 25% more RAM for subsequent [ indexing and visualization.

Features Peak RAM (MiB)
1000 8
2000 31
4000 123
8000 491
16000 1959
24000 4405
32000 7829
64000 31301
100000 76406

## Limitations

Although we developed this package with biological count data in mind, compositional count data is not truly compositional in that count data contain integer values only. As such, measuring “Gene A” as $$1$$ in one subject and “Gene B” as $$2$$ in another subject (i.e., the feature vector $$[1, 2]$$) does not carry the same information as measuring “Gene A” as $$1000$$ in one subject and “Gene B” as $$2000$$ in another subject (i.e., the feature vector $$[1000, 2000]$$) due to how additive variation affects the relative abundance of small counts more than large counts (Quinn 2017). We advise the investigator to proceed with caution when analyzing low counts.

## References

1. Erb, Ionas, and Cedric Notredame. “How Should We Measure Proportionality on Relative Gene Expression Data?” Theory in Biosciences = Theorie in Den Biowissenschaften 135, no. 1-2 (June 2016): 21-36. http://dx.doi.org/10.1007/s12064-015-0220-8.

2. Lovell, David, Vera Pawlowsky-Glahn, Juan José Egozcue, Samuel Marguerat, and Jürg Bähler. “Proportionality: A Valid Alternative to Correlation for Relative Data.” PLoS Computational Biology 11, no. 3 (March 2015): e1004075. http://dx.doi.org/10.1371/journal.pcbi.1004075.

3. Quinn, Thomas, Mark F. Richardson, David Lovell, and Tamsyn Crowley. “Propr: An R-Package for Identifying Proportionally Abundant Features Using Compositional Data Analysis.” bioRxiv, February 1, 2017, 104935. http://dx.doi.org/10.1101/104935.


text:
# Math Help - residue

1. ## residue

I have an integral to evaluate via contour integration:
$({1 \over 1+cos^2\theta}) \ wrt d\theta$
Using the substitution $z =e^{i\theta}$I find the integral becomes,with $d\theta \ becoming \ {dz \over iz}$
${4z^2 \over z^4 + 6z^2 + 1}$ is that correct?
Also how would I evaluate this via the residue theorem?

2. That depends what you're integrating. For example, suppose:

$\int_0^{2\pi}\frac{1}{1+\cos^2(t)}dt$

and I let $z=e^{it}$, then I'd get:

$-i\mathop\oint\limits_{|z|=1}\frac{4z}{z^4+6z^2+1}d z=(-i) 2\pi i \sum\mathop\text{Res}\frac{4z}{z^4+6z^2+1}$

where the sum is over the residues of the enclosed poles.

3. I've written it differently:

${1\over i} \oint ({3\over2z}+{z\over4}+{1\over4z^3})^{-1}dz$

how can I pick the relevant term (the cofefficient of 1/z) to give the residue? The inverse is complicating matters....

4. I don't see why you'd want to write it that way. Just factor the denominator of the expression I got, get the enclosed zeros, bingo-bango.

5. Originally Posted by shawsend
I don't see why you'd want to write it that way. Just factor the denominator of the expression I got, get the enclosed zeros, bingo-bango.
Thanks.
So solve $z^4 + 6z^2 +1 =0$ to begin with...? But how would I?

6. Originally Posted by bigdoggy
Thanks.
So solve $z^4 + 6z^2 +1 =0$ to begin with...? But how would I?
Let $z^2=k$..."bingo-bango"!

7. Originally Posted by Drexel28
Let $z^2=k$..."bingo-bango"!
Hi, which gives $k=-3 \pm2\sqrt2 =z^2$
Am I doing what you're implying...!? But how can I get the poles here...?

8. Originally Posted by bigdoggy
Hi, which gives $k=-3 \pm2\sqrt2 =z^2$
Am I doing what you're implying...!? But how can I get the poles here...?
The poles are merely the zeros of the denominator. You know that $z^2=-3\pm \sqrt{2}\implies\text{Denominator equals zero}$ therefore $z=?\implies\text{Denominator equals zero}$?

9. Originally Posted by Drexel28
The poles are merely the zeros of the denominator. You know that $z^2=-3\pm \sqrt{2}\implies\text{Denominator equals zero}$ therefore $z=?\implies\text{Denominator equals zero}$?
I can't see it being $z= \pm\sqrt{-3\pm \sqrt{2}}$
I'm missing something here ...

10. Originally Posted by bigdoggy
I can't see it being $z= \pm\sqrt{-3\pm \sqrt{2}}$
I'm missing something here ...
You screwed up the quadratic equation.

- Wolfram|Alpha -Step 1

- Wolfram|Alpha -Step 2...validattion.

11. Thankyou.

So if the integrand is around |z|=1, the solution is $z=\pm i\sqrt{3 -2\sqrt2}$
I can't see how this would give a 'nice' (multiple of pi) answer (which I was told it should!)....

I find the residue = $1 \over \sqrt2$ correct??

12. The zeros are $z_0=i\sqrt{3+2\sqrt{2}},z_1=i\sqrt{3-2\sqrt{2}}, z_2=-i\sqrt{3+2\sqrt{2}},z_3=-i\sqrt{3-2\sqrt{2}}$. Now, which ones are in the unit circle? Isn't it the $3-2\sqrt{2}$ ones? So then we have:

$\int_0^{2\pi} \frac{1}{1+\cos^2(t)}dt=2\pi \mathop\text{Res}\limits_{z_1,z_3} \left\{\frac{4z}{z^4+6z^2+1}\right\}$.

I'll do one of the residues:

$r_1=\frac{4z}{(z-i\sqrt{3+2\sqrt{2}})(z+i\sqrt{3+2\sqrt{2}})(z+i\sq rt{3-2\sqrt{2}})}\biggr|_{z=i\sqrt{3-2\sqrt{2}}}=\frac{1}{2\sqrt{2}}$

Edit: Also, forgot to mention, if it's a simple pole and the derivative of the denominator is not zero there then:

$\mathop\text{Res}\limits_{z=z_0} \frac{f}{g}=\frac{f(z_0)}{g'(z_0)}$

May be easier to calculate these residues this way.


text:
# Multiple linear regression model selection. Individually, X's and Y not correlated, but X's are significant predictors of Y in multiple regression?

My study is largely exploratory and looks at how environmental factors such as (physical characteristics, water quality and pesticide type and concentration) affect relative antioxidant enzyme mRNA expression. I have categorized my x variables into the 3 aforementioned categories and each category includes a 8-10 variables. Some of these variables were correlated with each other which required removal from the model. I am not 100% sure that this approach is correct in the sense that I included all variables that may have an effect on y "in theory" but not statistically (i.e. I included variables that are not independently correlated with y). This may be wrong but there is mixed info on this topic and my supervisor suggested this approach. In the end, each gene I tested did not have sig relationships with more than one variable regardless. This entire model approach was done for completeness to include all data collected since the results end up being pretty lean.

I am using a backward stepwise model selection approach (every variable is included to begin with) to analyze if explanatory (x) variables are explaining my response variable. I have dealt with multicollinearity already and am using a p value (≤ 0.05) exclusion method for the remaining variables. So I am going one by one, starting with the highest p value and dropping each non significant variable as in Chowdhury & Turin 2020.

I encountered an issue where several individually non-correlated (x and y) variables are showing to be significant when in the model together. Clearly these results are spurious. I read a few posts that suggest either multicollinearity (not the case here) or confounding variables but I can't seem to find out how to deal with this issue in a statically sound way. Do I report this issue? But how do i decide what to do next to generate the final model (even if the model is non significant in the end).

What I have done so far is just used VIF to determine which of the "sig" variables maybe were correlating with the others in a way that I was unable to detect and deleted the variable with the highest VIF. When I do this, there is no longer any significance which further convinces me that this was just an artifact. Any help would be greatly appreciated as I am trying to write up my PhD dissertation in 6 months and am very strapped for time!

Here is an example of a full model for one category (physical characteristics)

Here is an an example of the issue for context

Here is an example of an individual relationship between one x and my y

Here are the VIF

I removed Velocity

• " Clearly these results are spurious" - what does spurious mean here? If you know that if a variable is not correlated with Y it cannot be statistically significant, why would you add it to the model? (A: You don't, so you do.) Jan 27 at 17:43
• Merely removing individual covariates based on their correlation with the dependent variable can be problematic. Rather, why not include all then apply regularization for instance? Jan 27 at 17:43
• Although the question itself is not clear, due to incorrect assumptions, its nature appears to be one that has been discussed before: see stats.stackexchange.com/….
– whuber
Jan 27 at 17:47
• If these are complete data, then the real problem seems to be overfitting. You only have a dozen observations yet are trying to fit models with 4-5 predictors. That can lead to a model that fits your data set very well but fails to extend to new data samples. Please edit the question to say more about the study at a higher level: the hypothesis you're trying to test, details of what you've done so far to "deal with multicollinearity," and the number of variables that you started with before you imposed your "exclusion method." There might be better solutions at those early stages of analysis.
– EdM
Jan 27 at 18:15
• Thanks so much for your feedback so far guys! I just added more details for context and my apologies for the lack of clarity. I am pretty new at complex stats (this is quite complex for me !)
– Pria
Jan 27 at 18:35

If you only have 12 observations in a biological/ecological study, as your plot and regression summaries indicate, it's really hard to get reliable results from regression. The main problem here is overfitting, not multicollinearity among predictors.

The usual rule of thumb for regression analysis in such studies is that you shouldn't try to evaluate more than 1 predictor for every 10-20 observations.* Otherwise you will tend to get an overfitted model that doesn't generalize well to new data samples. You only show data from 12 observations.

That's the problem you face. Your "full model" evaluated 10 predictors on 12 observations. With the necessary addition of an intercept to the model, that left only 1 "degree of freedom" to allow for variability in outcomes not accounted for by the predictors. The very high $$R^2$$ value in that model was because there was almost no way for the 12 data points to deviate from the 10-predictor model.

You then removed enough predictors to get to a 5-predictor model. But as the selection of which to remove was driven by how well things fit your data set, you ended up with a model in which the remaining 5 predictors still fit your data set almost perfectly. It almost certainly would not work on a new data set. Note that neither your "full model" (p = 0.12) nor your 4-predictor model (p = 0.26) showed overall statistical significance by the usual p < 0.05 criterion.

Also, breaking up your predictors into classes and then doing separate regressions on each of them is generally not a good idea. Then you can't examine how different types of predictors work together.

There might be ways to get something useful here, but that will probably require working closely with an experienced local statistician to whom you can explain details of your study and what you are trying to accomplish scientifically.

For example, you can try to combine multiple predictors of the same type intelligently--but without looking at the outcome values--into a single predictor, based on your understanding of the subject matter. I'd recommend that you look at Frank Harrell's course notes on regression modeling, in particular Chapter 4, with careful attention to section 4.7 on data reduction. That might give you some ideas about how to combine multiple predictors usefully.

To get around the overfitting problem with too many predictors even after you've combined many of them, you could try ridge regression. That so-called "penalized" or "regularized" method allows you to use more predictors without overfitting than with standard regression, because it reduces the magnitudes of the regression coefficients to compensate. That is implemented in the R glmnet package, with an example worked through in Section 6.6.1 of An Introduction to Statistical Learning. I'd advise against using the variable-selection methods described in that chapter of ISL, like forward-stepwise, best-subset, or LASSO regressions, as the particular predictors selected typically don't carry over well to new samples.

*It won't help even if you have mRNA values for multiple enzymes at each of 12 observation locations, as you will effectively be doing regressions for each type of mRNA.


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# If x is the product of the positive integers from 1 to 8, in

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If x is the product of the positive integers from 1 to 8, in  [#permalink]

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20 Jun 2010, 22:52
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If x is the product of the positive integers from 1 to 8, inclusive, and if i, k, m, and p are positive integers such that $$x = 2^i*3^k*5^m*7^p$$, then i + k + m + p =

(A) 4
(B) 7
(C) 8
(D) 11
(E) 12

OPEN DISCUSSION OF THIS QUESTION IS HERE: https://gmatclub.com/forum/if-x-is-the- ... 46157.html
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Re: If x is the product of the positive integers from 1 to 8, in  [#permalink]

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21 Jun 2010, 01:15
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divakarbio7 wrote:
If x is the product of the positive integers from 1 to 8,
inclusive, and if i, k, m, and p are positive integers
such that x = 2i3k5m7p, then i + k + m + p =

(A) 4
(B) 7
(C) 8
(D) 11
(E) 12

I am not understanding the reasoning behind this problem.

It should be "if $$x=2^i*3^k*5^m*7^p$$".

$$x=8!=2*3*4*5*6*7*8=2*3*(2^2)*5*(2*3)*7*(2^3)=2^7*3^2*5^1*7^1=2^i*3^k*5^m*7^p$$ --> $$i=7$$, $$k=2$$, $$m=1$$, and $$p=1$$ --> $$i+k+m+p=7+2+1+1=11$$.

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Re: If x is the product of the positive integers from 1 to 8, in  [#permalink]

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30 Nov 2014, 16:54
Here the idea is to go expressing each factor from one to eight, decomposed into its primes:

1 * 2 * 3 * 4 * 5 * 6 * 7 * 8 = 1 * 2 * 3 *(2*2) * 5 +(2*3) * 7 *(2*2*2) = 2 ^ 7 * 3 ^ 2 * 5 ^ 1 * 7 ^ 1

Then

i = 7

k = 2

m = 1

p = 1

i + k + m + p = 11

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Re: If x is the product of the positive integers from 1 to 8, in  [#permalink]

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30 Nov 2014, 21:59
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Incidentally 2,3,5,7 are prime factors of x = 8!

$$x = 2^i * 3^k * 5^m * 7^p$$ , we require to find (i + k + m + p)

There is no need to find individual values of i,k,m,p

Just write all prime numbers in there powers & add

$$x = 2^1 * 3^1 * (2^2) * 5^1 * (2^1 3^1) * 7^1 * (2^3)$$

i+k+m+p = 1+1+2+1+1+1+1+3 = 11

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Re: If x is the product of the positive integers from 1 to 8, in  [#permalink]

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03 Dec 2014, 07:27
Following the logic of
PareshGmat

There would be no need to write products between cousins, it is enough to write directly i + k + m + p = 11, that if you can do it in the CAT, obviously, but for people who can't do it directly would require the previous steps, to have crystal clarity and understand how to deal with similar situations, on the understanding that this is the purpose of some learning.

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Re: If x is the product of the positive integers from 1 to 8, in  [#permalink]

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16 Jan 2018, 06:59
divakarbio7 wrote:
If x is the product of the positive integers from 1 to 8, inclusive, and if i, k, m, and p are positive integers such that $$x = 2^i*3^k*5^m*7^p$$, then i + k + m + p =

(A) 4
(B) 7
(C) 8
(D) 11
(E) 12

OPEN DISCUSSION OF THIS QUESTION IS HERE: https://gmatclub.com/forum/if-x-is-the- ... 46157.html
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Re: If x is the product of the positive integers from 1 to 8, in &nbs [#permalink] 16 Jan 2018, 06:59
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