text:
# how to find constant mass

For the best answers, search on this site https://shorturl.im/avbri, It will have constant velocity and speed because it it is acted on by constant force no change in the velocity and speed of the object, when there is no change in the velocity acceleration will be zero, and the object will not acceleration therefore the answer is (e). Stay on top of new posts by signing up to receive notifications! Perhaps I can chock it up to measurement error, since my method wasn’t exactly scientifically rigorous. A mass on a spring has a single resonant frequency determined by its spring constant k and the mass m. Using Hooke's law and neglecting damping and the mass of the spring, Newton's second law gives the equation of motion: . By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy. 100.0g benzene x 1kg/1000g= 0.1000kg. Favourite answer. Did the authors forget their main audience is composed of poor college students? The period of a mass m on a spring of spring constant k can be calculated as $\text{T}=2\pi \sqrt{\frac{\text{m}}{\text{k}}}$. If you're using experimental data, the variation might be even greater. \begin{equation*} Let $X$ be a discrete random variable with probability distribution (probability mass function). In other words, no matter how a compound is obtained or prepared, it will always contain the same elements in the same mass proportion. With a calculated slope of 0.154, our model is \begin{equation*} F = kx Everything is the same as before, except this time, I took more measurements and used a caliper to do the measuring. Typically, this difference is relatively small, yet it does exist and can be important.  One way to visualize this pattern is to walk in a straight line at constant speed while carriying the vibrating mass. a) Find $c$ such that $P(x)$ is a legitimate PDF. Discovery of this law is credited to the French chemist Joseph Proust, who through a series of experiments conducted from 1798 to 1804 concluded that chemical compounds consisted of a specific composition. m= 0.917m = 0.917mol/kg. function $$\rho(x)$$ tells us the density distribution along the bar, measured in g/cm. b) Find the CDF of $X$, $F(x), x ∈ \{0,1,2,...\}$. Have any GDPR (or other) laws been breached during this scenario? Demonstration. I started with Hooke’s law, which in math looks like this: For example, let's say that using the law of constant composition, you want to demonstrate that two samples of cupric oxide abide by the law. Join Yahoo Answers and get 100 points today. and $0$ elsewhere. I can accept the higher mass points deviating, since Hooke’s law doesn’t hold after a certain point, but I would have expected the lower mass points to be closer. x = \frac{g}{k} m \\ The frequency of simple harmonic motion like a mass on a spring is determined by the mass m and the stiffness of the spring expressed in terms of a spring constant k ( see Hooke's Law): function sq(x){return x*x} function ut(tt){fh=document.forms[0];fh.t.value=tt;fh.f.value=1/tt;fh.af.value=2*Math.PI/tt;fh.om.value=2*Math.PI/tt} function kc(){fh=document.forms[0];def();fh.k.value= sq(fh.om.value)*fh.m.value} function masc(){fh=document.forms[0];def();fh.m.value= fh.k.value/sq(fh.om.value)} function fc(){fh=document.forms[0];def();omg=fh.om.value=Math.sqrt(fh.k.value/fh.m.value);ut(2*Math.PI/omg)} function def(){fh=document.forms[0];if (fh.t.value==0)ut(1);if (fh.k.value==0)fh.k.value=1;if (fh.m.value==0)fh.m.value=1}. For example, carbon dioxide (CO2) always contains carbon and oxygen in a 3:8 mass ratio. You'd simply subtract one of the values from 100 to get the percent of the other element. The problem is, because of the non-zero y-intercept (which does matter) I am not sure how to calculate the spring constant. This frustrated me, since my measurements all lined up nicely along a line, but it just looked like the line didn’t go through the origin. An amount of substance containing 6.02 × 10 23 particles is called a mole (often abbreviated to mol). This means that you keep heating and reweighing the reaction until its mass no longer changes, i.e. Law of Constant Composition Example . Using water as a high density storable hydro-lox propellant. After reaching this point, it was time to collect some data. Scroll down the page for more examples and solutions. Can this be proved using Combinatorics or generating functions? Is it correct to call the probability mass function by the name “discrete probability density function”? copper oxide = 1.375 gcopper = 1.098 goxygen = 1.375 - 1.098 = 0.277 g, percent oxygen in CuO = (0.277)(100%)/1.375 = 20.15%, copper = 1.179 gcopper oxide = 1.476 goxygen = 1.476 - 1.179 = 0.297 g, percent oxygen in CuO = (0.297)(100%)/1.476 = 20.12%. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share … In chemistry, the law of constant composition (also known as the law of definite proportions) states that samples of a pure compound always contain the same elements in the same mass proportion. Considering John Dalton's atomic theory was only just beginning to explain that each element consisted of one type of atom and at the time, most scientists still believed elements could combine in any proportion, Proust's deductions were exceptional. \begin{equation*} I tried searching on on the internet and couldn’t come up with a reference, so again, please comment if you know something about the typical range of spring constants. In my mathematical modeling book, there is a project that asks the reader to build a model that relates the stretch or compression of a spring to the mass it supports and then collect some data and find the proportionality constant (assuming the data reflected such a relationship). What is the constant speed of the forklift while lifting this load? The mass proportion of heavy water as compared with regular water is an example. This law, together with the law of multiple proportions, is the basis for stoichiometry in chemistry. She has taught science courses at the high school, college, and graduate levels. rev 2020.11.24.38066, The best answers are voted up and rise to the top, Mathematics Stack Exchange works best with JavaScript enabled, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site, Learn more about Stack Overflow the company, Learn more about hiring developers or posting ads with us. Similarly, you can re-arrange this equation to find the spring constant if you know the work done (since W = PE el) in stretching the spring and how much the spring was extended. Here’s a picture: You may be wondering how I measured the length of the elongated spring. Multiply the relative atomic mass by the molar mass constant. I started with Hooke’s law, which in math looks like this: Some people like the form , saying the force is a restoring force, but for finding , it doesn’t really matter, so I chose the simpler convention. Forecasting Prices vs Returns by Deep Learning. 4. First, you need to know the difference between mass and weight. Also, because there are different isotopes of atoms, even a normal stoichiometric compound may display variations in mass composition, depending which isotope of the atoms is present. To learn more, see our tips on writing great answers. This is defined as 0.001 kilogram per mole, or 1 gram per mole. Scalability problem: is the end of Bitcoin near? How do I prevent myself from freezing up during a game I'm DMing? For the second sample, 1.179 g of copper was dissolved in nitric acid to produce copper nitrate, which was subsequently burned to produce 1.476 g of cupric oxide. Anyway, I plotted my data using pyplot and here’s the result: Nice and linear-ish. In each situation, $$\overline{x}$$ represents the balancing point of the system of masses or of the portion of the bar. If you're using experimental data, the variation might be even greater. It doesn't matter whether you choose to find the percentage of copper or the percentage of oxygen. By using ThoughtCo, you accept our, Exceptions to the Law of Constant Composition, Law of Multiple Proportions Example Problem, Calculate Empirical and Molecular Formulas, Learn About Molecular and Empirical Formulas, How to Calculate Mass Percent Composition, Calculate Simplest Formula From Percent Composition, Chemical Formulas Practice Test Questions, What Is an Element in Chemistry? There are some non-stoichiometric compounds that exhibit a variable composition from one sample to another. x = \lambda m \quad \text{where} \quad \lambda = \frac{g}{k} Are my scuba fins likely to be acceptable "personal items" for air travel? Since weight is computed by the gravitational constant times mass, the computations for the balancing point result in the same location regardless of whether we use weight or mass, since the gravitational constant is present in both the numerator and denominator of the weighted average.

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text:
### Pizza Portions

My friends and I love pizza. Can you help us share these pizzas equally?

### Forgot the Numbers

On my calculator I divided one whole number by another whole number and got the answer 3.125. If the numbers are both under 50, what are they?

### Pies

Grandma found her pie balanced on the scale with two weights and a quarter of a pie. So how heavy was each pie?

# Fun Time

## Fun Time

On Saturday, Asha and Kishan's grandad took them to a Theme Park.

In total they spent $\frac{1}{5}$ of the day queuing for rides.

They worked out that $\frac{3}{4}$ of the rest of their time there was spent enjoying the rides.

Of the time left, $\frac{1}{2}$ was spent having lunch.

The 3-d cinema show took up $\frac{2}{3}$ of the rest of the day.

Finally there were $10$ minutes left to go to the gift shop before they went home.

How long were Asha and Kishan at the Theme Park?

### Why do this problem?

This problem gives practice in calculating with fractions and encourages different ways of representing and recording.

### Key questions

What could you draw or write down to help you?
Have you tried working backwards?
What fraction of the total time for the cinema and shopping is the $10$ minutes of shopping?
So how long did they spend in total watching the film and shopping?

### Possible extension

Andy's Marbles is a similar problem which might make a good follow-up activity.

### Possible support

Fair Feast would be a good problem to try first if children find calculating with fractions difficult.

format: plain


text:
Place second list at same position as first list in beamer overlay

I like to have the second itemize list to appear at the top of the frame, i.e. at the same place as the previous itemize list. At this frame I have two columns where each item on the left column are associated with several items which I like to be appeared on the right column after each item on the left column. Then, item list associate with first item on right should disappear and again the item list associate to second item on the left should start from the top of the frame.

\documentclass{beamer}

\mode<presentation> {

\usetheme{Hannover}
\usecolortheme{seagull}
}
\usepackage{graphicx}
\usepackage{booktabs}
\usepackage{siunitx}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{document}
\begin{frame}
\frametitle{ applications}

\begin{columns}

\begin{column}{0.4\textwidth}

\begin{itemize}
\setbeamercovered{transparent}
\item <1-5> {\small Combustion}
\item<6-> {\small Airborne }
\item<10->{\small Environmental}

\end{itemize}

\end{column}

\begin{column}{7cm}

\begin{itemize}[label={}]

\item[] <2-5>{\tiny PN$\mathrm{_{ d< 1\,\si{um}}}$ $\gg$ PN$\mathrm{_{ d> 1\,\si{um}}}$     }
\item[] <2-5>{ \tiny A correlation }
\item[] <3-5>{\tiny No universal }
\item[] <4-5> {\tiny limited}

\end{itemize}

\begin{itemize}[label={}]

\item[] <7-7>{\tiny PN$\mathrm{_{ d< 1\,\si{um}}}$ $\gg$ PN$\mathrm{_{ d> 1\,\si{um}}}$     }

\end{itemize}

\end{column}

\end{columns}

\end{frame}

\end{document}

• Welcome to TeX.SX! You can use the {} to format code blocks -- or simply indent them by 4 spaces. It is easier to help you if you add a minimal working example that takes the form \documentclass{...}\usepackage{....}\begin{document}...\end{document}. If possible, it should compile and have the minimum amount of code needed to illustrate your problem. This makes it much easier for people to help you - and much more likely that they will! – Andrew May 7 '15 at 11:36
• @Andrew: I am not sure how that can help to reset the position of the second item list to the frame top. – Ramin IJ May 7 '15 at 12:57
• i think the title would be clearer with "above" instead of "on top of"; i read this as "the second list should overprint the first". ("over" doesn't work either.) didn't want to change the title without your agreement. – barbara beeton May 7 '15 at 14:03
• @Ramin By giving a minimal example that compiles you give people initial code to work from. If you give only partial code, without the documentclass and relevant packages, people have to guess these things and they have to spend extra time to set things up before they even think about helping you. – Andrew May 7 '15 at 21:01

You can use an overprint environment with an \onslide before each itemize. Schematically

\begin{column}{0.55\textwidth}
\begin{overprint}
\onslide<2-5>
\begin{itemize}
\item<2-5> text
\item<3-5> text
\end{itemize}
\onslide<7>
\begin{itemize}
\item<7> text
\end{itemize}
\end{overprint}
\end{column}

On slide 5 you then have

and slide 7 has

\documentclass{beamer}

\mode<presentation>{
\usetheme{Hannover}
\usecolortheme{seagull}
}

\usepackage{siunitx}
\sisetup{detect-all}

\newcommand{\PN}{\textsf{PN}}

\begin{document}

\begin{frame}
\frametitle{applications}

\begin{columns}
\begin{column}{0.4\textwidth}
\small
\begin{itemize}
\setbeamercovered{transparent}
\item<1-5> Combustion
\item<6-> Airborne
\item<10-> Environmental
\end{itemize}
\end{column}

\begin{column}{0.55\textwidth}
\tiny
\begin{overprint}
\onslide<2-5>
\begin{itemize}
\item[]<2-5>$\PN_{\mathrm{d}< \SI{1}{um}} \gg \PN_{\mathrm{d}>\SI{1}{um}}$
\item[]<2-5>A correlation
\item[]<3-5>No universal
\item[]<4-5>limited
\end{itemize}
\onslide<7>
\begin{itemize}
\item[]<7>$\PN_{\mathrm{d}< \SI{1}{um}} \gg \PN_{\mathrm{d}>\SI{1}{um}}$
\end{itemize}
\end{overprint}
\end{column}
\end{columns}
\end{frame}

\end{document}

I have tidied up several other aspects of your code

• moved the text sizing commands outside of the lists, as they applied to all elements
• used \SI{1}{um} rather than 1\,\si{um} for print numbers with a unit, and added the detect-all feature so the font matches
• collected math expressions in to single units, with a command to typeset PN in the sans font
• specified the second column width relative to the textwidth
• replaced overlay specification <7-7> by <7>


text:
# zbMATH — the first resource for mathematics

##### Examples
 Geometry Search for the term Geometry in any field. Queries are case-independent. Funct* Wildcard queries are specified by * (e.g. functions, functorial, etc.). Otherwise the search is exact. "Topological group" Phrases (multi-words) should be set in "straight quotation marks". au: Bourbaki & ti: Algebra Search for author and title. The and-operator & is default and can be omitted. Chebyshev | Tschebyscheff The or-operator | allows to search for Chebyshev or Tschebyscheff. "Quasi* map*" py: 1989 The resulting documents have publication year 1989. so: Eur* J* Mat* Soc* cc: 14 Search for publications in a particular source with a Mathematics Subject Classification code (cc) in 14. "Partial diff* eq*" ! elliptic The not-operator ! eliminates all results containing the word elliptic. dt: b & au: Hilbert The document type is set to books; alternatively: j for journal articles, a for book articles. py: 2000-2015 cc: (94A | 11T) Number ranges are accepted. Terms can be grouped within (parentheses). la: chinese Find documents in a given language. ISO 639-1 language codes can also be used.

##### Operators
 a & b logic and a | b logic or !ab logic not abc* right wildcard "ab c" phrase (ab c) parentheses
##### Fields
 any anywhere an internal document identifier au author, editor ai internal author identifier ti title la language so source ab review, abstract py publication year rv reviewer cc MSC code ut uncontrolled term dt document type (j: journal article; b: book; a: book article)
There are infinitely many Carmichael numbers. (English) Zbl 0816.11005

Carmichael numbers are those composite integers $n$ for which ${a}^{n}\equiv a\phantom{\rule{0.277778em}{0ex}}mod\phantom{\rule{0.277778em}{0ex}}n$ for every integer $a$. By a result of A. Korselt [L’intermédiaire des mathématiciens 6, 142-143 (1899)] $n$ is a Carmichael number iff $n$ is squarefree and $p-1$ divides $n-1$ for all primes $p$ dividing $n$. In this paper the authors show the existence of infinitely many Carmichael numbers.

They extend an idea of P. Erdős to construct integers $L$ such that $p-1$ divides $L$ for a large number of primes $p$. If there is a product of these primes $\equiv 1\phantom{\rule{0.277778em}{0ex}}mod\phantom{\rule{0.277778em}{0ex}}L$, say

$C={p}_{1}·\cdots ·{p}_{k}\equiv 1\phantom{\rule{0.277778em}{0ex}}mod\phantom{\rule{0.277778em}{0ex}}L\phantom{\rule{2.em}{0ex}}\left(*\right)$

then $C$ is a Carmichael number which is shown by the criterion of A. Korselt mentioned above. In order to find integers with many divisors of the form $p-1$, $p$ prime, the authors generalize a theorem of K. Prachar [Monatsh. Math. 59, 91-97 (1955; Zbl 0064.041)]. The question of the existence of products of the form $\left(*\right)$ leads to investigations in combinatorial group theory.

Reviewer’s remark: For a survey on Carmichael numbers, see the article of C. Pomerance [Nieuw Arch. Wiskd., IV. Ser. 11, 199-209 (1993; Zbl 0806.11005)].

##### MSC:
 11A25 Arithmetic functions, etc. 11N56 Rate of growth of arithmetic functions 11A07 Congruences; primitive roots; residue systems 11N69 Distribution of integers in special residue classes


text:
# How can latex help me format and present lots of data in a report?

At work I am responsible for creating information security reports that contain long narratives and reference what we refer to as "vulnerabilities". These vulnerabilities are presented in a table. I have been using Microsoft Word for a long time but I just stumbled into Latex and it looks like it could really help me save time and increase accuracy.

I have been doing some reading but there is a lot of material out there. When I found this stackexchange site I couldn't resist asking for some advice.

The reports that I put together average about 100 pages. They have multiple sections that outline the entire test at a high level (executive overview) as well as many other sections that describe in depth attack narratives.

I have a pretty good grasp on the typical syntax for marking a latex document up but what I am starting to wonder about is how I can create a cross reference (I am not sure if this is the correct term) in a paragraph that will link to a table that goes into detail about a particular vulnerability. Here is a short example of how I am doing things in Word.

This image shows a paragraph that is linked (in blue) to a specific "vulnerability" in a table.

And here is the table:

The blue bracketed text in the first image links to the table that shows that particular issue. Creating the links in Word is pretty time consuming and getting the data into the table is even worse.

My question is, can I use a "macro" or maybe some other latex mechanisim to easily create vulnerabilities in a table that can be quickly referenced in my long form narratives? Maybe something that would take parameters like:

{This is the title for the vulnerability, This explains the issues in question..., 8, INT, CONF, NO, NO, 6 H}

Is there a way to do computation on the actual values in the table? Something like multiplication of one value against another?

And finally is there a way to create graphs based on the information that comprise the tables?

Using datatool for the heavy lifting, and other packages for the formatting:

\documentclass{article}
\usepackage[margin=1in]{geometry}
\usepackage{booktabs}
\usepackage{datatool}
\usepackage{longtable}
\usepackage{lipsum}
\usepackage{hyperref}
\begin{filecontents}{\jobname.txt}
This is the title for vuln 1, This explains the issues in question..., 8, INT, CONF, NO, NO, 6, H
This is the title for vuln 2, This explains the issues in question..., 7, INT, CONF, NO, NO, 8, H
This is the title for vuln 3, This explains the issues in question..., 7, INT, CONF, NO, NO, 8, H
This is the title for vuln 4, This explains the issues in question..., 7, INT, CONF, NO, NO, 8, H
This is the title for vuln 5, This explains the issues in question..., 7, INT, CONF, NO, NO, 8, H
This is the title for vuln 6, This explains the issues in question..., 7, INT, CONF, NO, NO, 8, H
This is the title for vuln 7, This explains the issues in question..., 7, INT, CONF, NO, NO, 8, H
This is the title for vuln 8, This explains the issues in question..., 7, INT, CONF, NO, NO, 8, H
This is the title for vuln 9, This explains the issues in question..., 7, INT, CONF, NO, NO, 8, H
This is the title for vuln 10, This explains the issues in question..., 7, INT, CONF, NO, NO, 8, H
This is the title for vuln 11, This explains the issues in question..., 7, INT, CONF, NO, NO, 8, H
This is the title for vuln 12, This explains the issues in question..., 7, INT, CONF, NO, NO, 8, H
This is the title for vuln 13, This explains the issues in question..., 7, INT, CONF, NO, NO, 8, H
This is the title for vuln 14, This explains the issues in question..., 7, INT, CONF, NO, NO, 8, H
This is the title for vuln 15, This explains the issues in question..., 7, INT, CONF, NO, NO, 8, H
This is the title for vuln 16, This explains the issues in question..., 7, INT, CONF, NO, NO, 8, H
This is the title for vuln 17, This explains the issues in question..., 7, INT, CONF, NO, NO, 8, H
This is the title for vuln 18, This explains the issues in question..., 7, INT, CONF, NO, NO, 8, H
This is the title for vuln 19, This explains the issues in question..., 7, INT, CONF, NO, NO, 8, H
This is the title for vuln 20, This explains the issues in question..., 7, INT, CONF, NO, NO, 8, H
This is the title for vuln 21, This explains the issues in question..., 7, INT, CONF, NO, NO, 8, H
This is the title for vuln 22, This explains the issues in question..., 7, INT, CONF, NO, NO, 8, H
This is the title for vuln 23, This explains the issues in question..., 7, INT, CONF, NO, NO, 8, H
This is the title for vuln 24, This explains the issues in question..., 7, INT, CONF, NO, NO, 8, H
This is the title for vuln 25, This explains the issues in question..., 7, INT, CONF, NO, NO, 8, H
This is the title for vuln 26, This explains the issues in question..., 7, INT, CONF, NO, NO, 8, H
This is the title for vuln 27, This explains the issues in question..., 7, INT, CONF, NO, NO, 8, H
This is the title for vuln 28, This explains the issues in question..., 7, INT, CONF, NO, NO, 8, H
This is the title for vuln 29, This explains the issues in question..., 7, INT, CONF, NO, NO, 8, H
This is the title for vuln 30, This explains the issues in question..., 7, INT, CONF, NO, NO, 8, H
\end{filecontents}

\DTLmeanforcolumn{vulns}{cvss}{\cvssavg} \DTLmeanforcolumn{vulns}{severity}{\sevavg}
\DTLround{\cvssavg}{\cvssavg}{2}
\DTLround{\sevavg}{\sevavg}{2}

\begin{document}
\lipsum[1]

\begin{longtable}{p{1.2in}p{1.2in}ccccccc}
\toprule \bfseries{Title} & \bfseries{Description} & \bfseries{CVSS} & \bfseries{Access} & \bfseries{Class.} & \bfseries{ASL} & \bfseries{NP} & \bfseries{Sev.} & \bfseries{Rating} \\ \midrule
\DTLforeach{vulns}{%
\vtitle=title,\description=description,\cvss=cvss,\access=access,\classification=classification,\asl=asl,\np=np,\severity=severity,\ranking=ranking}{%
\hypertarget{vuln\DTLcurrentindex}{\vtitle} & \description & \cvss & \access & \classification & \asl & \np & \severity & \ranking \\}
\bfseries{Averages} & & \cvssavg & & & & & \sevavg \\ \bottomrule
\end{longtable}

\end{document}

• +1 Very nice example of using datatool. Sadly only can upvote once. – R. Schumacher Apr 15 '15 at 3:02
• This is a great example. I am reading the datatool user guide now. Thanks for your help. – ajt Apr 15 '15 at 10:43


text:
# What is $\frac{\det(A+tI)}{\det(B+tI)}$ as $t\to0$?

If $A$ and $B$ are two real $2\times 2$ matrices with $\det A = 0$ and $\det B = 0$ and $\mathrm{tr}(B)$ is non zero. then what will be limit of $$\lim_{t\to0}\frac{\det(A+tI)}{\det(B+tI)}$$ I used the formula $\lambda^2-\mathrm{tr} A+\det A = 0$. then i think answer is $\dfrac{\mathrm{tr}(A)}{\mathrm{tr}(B)}$. Am I correct?

What would be expansion of $\det(A+tI)$ for a $2\times 2$ matrices?

-

Let $A$ be an $n \times n$ matrix. Then, one may check that $$\det (A + t I) = t^n + t^{n-1} \operatorname{tr} A + \cdots + t \, \lambda(A) + \det A$$ where $\lambda (A)$ is a certain constant depending on the eigenvalues of the matrix $A$. In the case $n = 2$, $\lambda (A) = \operatorname{tr} A$. So, if $A$ and $B$ are both $n \times n$ matrices, $$\frac{\det (A + t I)}{\det (B + t I)} = \frac{t^n + t^{n-1} \operatorname{tr} A + \cdots + t \, \lambda(A) + \det A}{t^n + t^{n-1} \operatorname{tr} B + \cdots + t \, \lambda(B) + \det B}$$ Thus, if $\det A = \det B = 0$, and $\lambda (B) \ne 0$, $$\lim_{t \to 0} \frac{\det (A + t I)}{\det (B + t I)} = \frac{\lambda (A)}{\lambda (B)}$$ and in the special case $n = 2$ this reduces to exactly what you guessed.
What if in place of $I$ we have any matrix $B$? – srijan May 10 '12 at 11:08
If $B$ is invertible then you can reduce to this case. Otherwise, things are complicated. – Zhen Lin May 10 '12 at 11:09
Use the multiplicative property of the determinant: $\det (A + t C) = \det (C) \det (C^{-1} A + t I)$. – Zhen Lin May 10 '12 at 11:17


text:
+0

# Formula of question

+3
272
2
+1832

I don't understand the Formula  of this question

xvxvxv  Sep 26, 2014

#2
+91919
+8

Let the distance travelled in the fist 5 seconds be x metres

We know u=0

When t=5, s=x

When t=10 s=x+150

I am going to use the formula     $${\mathtt{s}} = {\mathtt{ut}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{1}}}{{\mathtt{2}}}}{\mathtt{\,\times\,}}{{\mathtt{at}}}^{{\mathtt{2}}}$$

since u=0,  the formula becomes  $${\mathtt{s}} = {\frac{{\mathtt{1}}}{{\mathtt{2}}}}{\mathtt{\,\times\,}}{{\mathtt{at}}}^{{\mathtt{2}}}$$

Now we have 2 equations that can be solved simultaneously.

$$\\x=\frac{1}{2}a\times 5^2 \\\\\ x= \frac{1}{2}a\times 25 \qquad (1)\\\\ and\\\\ x+150=\frac{1}{2}a\times 10^2 \\\\ x+150=\frac{1}{2}a\times 100 \\\\ x+150=\frac{1}{2}a\times 25 \times 4 \\\\ \frac{x+150}{4}=\frac{1}{2}a\times 25 \qquad (2)\\\\ putting these together I get\\\\ \begin{array}{rll} x&=&\frac{x+150}{4}\\ 4x&=&x+150\\ 3x&=&150\\ x&=&50\\ \end{array}$$

So the ball rolls 50m in the first 5 seconds

Melody  Sep 27, 2014
Sort:

#1
+8256
0

Is the s after the 5.0 seconds?

DragonSlayer554  Sep 26, 2014
#2
+91919
+8

Let the distance travelled in the fist 5 seconds be x metres

We know u=0

When t=5, s=x

When t=10 s=x+150

I am going to use the formula     $${\mathtt{s}} = {\mathtt{ut}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{1}}}{{\mathtt{2}}}}{\mathtt{\,\times\,}}{{\mathtt{at}}}^{{\mathtt{2}}}$$

since u=0,  the formula becomes  $${\mathtt{s}} = {\frac{{\mathtt{1}}}{{\mathtt{2}}}}{\mathtt{\,\times\,}}{{\mathtt{at}}}^{{\mathtt{2}}}$$

Now we have 2 equations that can be solved simultaneously.

$$\\x=\frac{1}{2}a\times 5^2 \\\\\ x= \frac{1}{2}a\times 25 \qquad (1)\\\\ and\\\\ x+150=\frac{1}{2}a\times 10^2 \\\\ x+150=\frac{1}{2}a\times 100 \\\\ x+150=\frac{1}{2}a\times 25 \times 4 \\\\ \frac{x+150}{4}=\frac{1}{2}a\times 25 \qquad (2)\\\\ putting these together I get\\\\ \begin{array}{rll} x&=&\frac{x+150}{4}\\ 4x&=&x+150\\ 3x&=&150\\ x&=&50\\ \end{array}$$

So the ball rolls 50m in the first 5 seconds

Melody  Sep 27, 2014

### 34 Online Users

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text:
Physics: Principles with Applications (7th Edition)

$1.81 kg \cdot m^2$.
Assume this is a solid sphere. It is rotating around an axis passing through its center. $$I=\frac{2}{5}MR^2=\frac{2}{5}(10.8 kg)(0.648m)^2=1.81 kg \cdot m^2$$


text:
${n}_{i}=2{\left(\frac{Z\sigma }{E}\right)}^{2}=2{\left(\frac{1.96\text{\hspace{0.17em}}\left(8.1\right)}{3}\right)}^{2}=56.0$

Description:

Math equation. The samples sizes (n) equal 2 times the square of the quantity Z times the pooled estimate of the common standard deviation divided by the margin of error, E. In this case, n= 2 times the square of (1.96 times 8.1) divided by 3. The final answer is 56.0.


text:
## Abstract

Motivation: Alignment of RNA has a wide range of applications, for example in phylogeny inference, consensus structure prediction and homology searches. Yet aligning structural or non-coding RNAs (ncRNAs) correctly is notoriously difficult as these RNA sequences may evolve by compensatory mutations, which maintain base pairing but destroy sequence homology. Ideally, alignment programs would take RNA structure into account. The Sankoff algorithm for the simultaneous solution of RNA structure prediction and RNA sequence alignment was proposed 20 years ago but suffers from its exponential complexity. A number of programs implement lightweight versions of the Sankoff algorithm by restricting its application to a limited type of structure and/or only pairwise alignment. Thus, despite recent advances, the proper alignment of multiple structural RNA sequences remains a problem.

Results: Here we present StrAl, a heuristic method for alignment of ncRNA that reduces sequence–structure alignment to a two-dimensional problem similar to standard multiple sequence alignment. The scoring function takes into account sequence similarity as well as up- and downstream pairing probability. To test the robustness of the algorithm and the performance of the program, we scored alignments produced by StrAl against a large set of published reference alignments. The quality of alignments predicted by StrAl is far better than that obtained by standard sequence alignment programs, especially when sequence homologies drop below ∼65%; nevertheless StrAl’s runtime is comparable to that of ClustalW.

Availability: StrAl is implemented in C. Source code (under GNU public license) as well as a precompiled Debian package can be downloaded at

Contact:stral@biophys.uni-duesseldorf.de

Supplementary information: Supplementary data available at Bioinformatics online.

## 1 INTRODUCTION

Non-coding RNAs (ncRNAs) are RNA molecules or elements that do not code for proteins but nevertheless are functional in biological processes, including localization, replication, translation, degradation and stabilization of biological macromolecules (for review and further references see Eddy, 2001; Storz, 2002; Winkler and Breaker, 2005; Fedor and Williamson, 2005; Gottesmann, 2005). Prominent examples are small nuclear RNAs, which are involved in mRNA splicing, and riboswitches, which are located in non-translated regions of mRNAs, where they bind metabolites and control gene expression.

For analysis of ncRNA function, knowledge about their secondary and tertiary structure is crucial. Structure prediction for single sequences is performed by dynamic programming, which allows the thermodynamically optimal structure or structure ensemble to be found (Zuker, 2000, 2003; Hofacker, 2003; Rivas and Eddy, 1999). These algorithms rely on correctness of thermodynamic parameters, neglect the influence of kinetics on structure formation and are not able to take into account interactions with other macromolecules. The alternative method—called comparative sequence analysis—needs a set of homologous RNAs and predicts base–base interactions on the basis of compensatory mutations (Chiu and Kolodziejczak, 1991; Gautheret et al., 1995; Lescoute et al., 2005). Its reliability increases with the number and divergence of sequences, but it needs an alignment (nearly) perfect with respect to sequence and structure. Other approaches for consensus structure prediction based on RNA alignments include PFold (Knudsen and Hein, 2003) and RNAalifold (Hofacker et al., 2002). Furthermore, RNA alignments are an essential basis for phylogeny inference (e.g. Hudelot et al., 2003), homology searches (e.g. Gräf et al., 2006; Eddy, 2002) and approaches to searching for new ncRNAs (e.g. Washietl et al., 2005).

The structurally correct alignment of RNA sequences is, however, a difficult problem. An algorithm for simultaneously optimizing the sequence and structure of an RNA set was published by Sankoff in 1985; however, the algorithm is not employable due to its computational complexity $O(n3m)$ and memory usage $O(n2m)$ for m sequences of length n. Thus, several variants of this algorithm have been developed which are restricted to pairwise alignment only and implement other simplifications to make the calculation tractable (Mathews and Turner, 2002; Hofacker et al., 2004; Havgaard et al., 2005a; Holmes, 2005).

This situation resembles that of (pure) sequence alignment: the algorithm for aligning two sequences is relatively cheap (Smith and Waterman, 1981), whereas the same approach cannot be applied to multiple sequence alignment due to its complexity of $O(nm)$ (Fuellen, G., 1997). This led to the development of very successful heuristic alignment methods, for example Clustal (Thompson et al., 1994a). Here we follow the same heuristic multiple sequence alignment approach but enhance it using a scoring function that emphasizes structural features. For this purpose, we project the structure features calculated by a thermodynamic approach (RNAfold; Hofacker, 2003) on top of the sequence. Similar approaches have already been proposed in the literature (Bonhoeffer et al., 1993; Yang and Blanchette, 2004) but have not been implemented. We call our program StrAl.

To test the performance of StrAl, we compare it with two sequence alignment programs—Clustal (Thompson et al., 1994a) and ProAlign (Löytynoja and Milinkovitch, 2003)—and three different structure alignment programs: In the following, we present our string-like alignment algorithm and test the performance of the respective program StrAl on several hundred sets of homologous RNAs. The quality of alignments produced by the Sankoff-like algorithm implemented in Stemloc is clearly superior to that produced by StrAl; this significant difference is, however, accompanied by a huge factor in computing time and memory usage in favor of StrAl. The quality of alignments predicted by StrAl is similar to that of standard sequence alignment programs if sequence similarity is >65%, but is far better with lower sequence similarities; nevertheless StrAl’s runtime is close to that of ClustalW.

• PMstring (PMmulti in string-like alignment mode; see Hofacker et al., 2004) follows a concept related to that of our approach: it uses strings of pairing probabilities obtained from Sankoff-like pairwise alignments to produce a guide tree for a multiple alignment. This procedure, however, ‘often produces misaligned [sequence] pairs’ (Hofacker et al., 2004).

• MARNA (Siebert and Backofen, 2005) uses a distantly related approach: it performs pairwise alignments of RNAs using sequence and structure information (based on RNAfold predictions) and combines the pairwise, weighted information into a multiple alignment with T-Coffee (Notredame et al., 2000).

• Stemloc (Holmes, 2005) implements the pairwise Sankoff algorithm using a pair stochastic context-free grammar and a heuristic for multiple alignment. To reduce computing cost and memory usage, constraints are imposed by ‘fold and alignment envelopes’, which take into account, for example, the 1000 best single sequence structure predictions and the 100 best alignments.

## 2 SYSTEMs AND METHODS

StrAl is implemented in C and should compile under any Unix system. We used the GNU C compiler (GCC) versions 3 and 4. The program was thoroughly tested on several Linux distributions, including Debian Version 3.1 and Red Hat Fedora Core 3. To facilitate installation, support for GNU autotools is built in. We also provide a precompiled Debian package. StrAl requires RNAfold's RNAlib (Hofacker et al., 1994; Hofacker, 2003) version 1.5 and the squid library (Eddy, 2005) version 1.9g. Static versions of these libraries compiled for i386 are included in the package, which can be downloaded at . All alignments have been computed using StrAl version 0.5.2. Runtime comparison was performed on a 1.8 GHz Dual-Opteron machine with 4 GB memory; Stemloc computations had to be performed on 2.4 GHz Opterons with 16 GB memory. Programs have been compiled with optimization level 3 (GCC 4).

### 2.1 Reference alignments

As reference alignments we used dataset-1 from the RNA alignment benchmark database BRAlibase (Gardner et al., 2005). This dataset consists of 388 alignments of Group I introns, 5S rRNAs, tRNAs, and U5 spliceosomal RNAs with five sequences per alignment.

### 2.2 Scoring of alignments

As proposed by Gardner et al. (2005), we used two independent yet complementary scores to evaluate alignment quality: the widely used sum-of-pairs score (SPS) implemented in BaliScore (Thompson et al., 1999) and the Structure Conservation Index (SCI; see Washietl et al., 2005). The SPS measures the level of sequence consistency between a test and a reference alignment by comparing all possible character pairs per column between both alignments; it ranges from 0 to 1 (complete agreement). The SCI is a measure for structural conservation and works independently of a reference alignment. It ranges from 0 (no detectable conservation) to values slightly above 1.0 (sequence agreement and structure conservation). For statistical analysis of results we used R ().

### 2.3 Parameters of alignment programs

The parameter choices for StrAl are described in Section 4.1. ProAlign v0.5a3 was allowed to use up to 256 MB of memory and a bandwidth of 400 nt (

-Xmx256m -bwidth=400
). PMmulti v1.1 was used either in its slow and thorough variant or in string-like alignment mode (
- -fast_progressive
). Other programs (ClustalW v1.83, MARNA, Stemloc v0.19b) were used with default options.

## 3 ALGORITHM

The steps of the algorithm include This strategy follows that of Clustal (Thompson et al., 1994a).

1. a pairwise alignment of all sequences of a set of homologous ncRNAs,

2. production of a guide tree using the alignment scores from step 1, and

3. a progressive alignment of the sequences, guided by the tree.

### 3.1 Pairwise alignment

We use the RNAfold library (Hofacker et al., 1994; Hofacker, 2003) to compute the partition function and matrix of base-pairing probabilities Pij of base i with base j for a single sequence. This probability matrix is then condensed into three linear vectors (Bonhoeffer et al., 1993), holding for each base i the probabilities of being paired downstream $pi1=∑j>iPij$, paired upstream $pi2=∑j or unpaired $pi0=1−(pi1+pi2)$, respectively. Thus we lose the specific pairing information but we can apply an alignment method that is cheap in terms of computing costs while still using thermodynamic information. Next, we choose a particular combination of these vectors—a structural part Sstruct = f(p1, p2) and a sequence part Sseq = f(p0)—as a similarity score

$si,k=α(Sstruct)+Sseq=α(pAi1pBk1+pAi2pBk2)+pAi0pBk0·d(Ai,Bk)$
for matching bases i and k from different sequences A and B, respectively. The idea for this score is to favor structural alignment of paired nucleotides from both sequences as well as sequence alignment of unpaired nucleotides. The factor α gives the ratio of structure over sequence similarity. The positive 4 × 4 single nucleotide substitution matrix d for aligning single-stranded regions is adapted either from Klein and Eddy (2003) (RIBOSUM85-60) or from Gotoh (1999) (see Section 4.1.2).

Let Vi,k be the value of the optimal alignment of prefixes A[1…i] and B[1…k] with base conditions V0,0 = 0, Vi,0 = Ei,0 = −goi · ge, V0,k = F0,k = −gok · ge and an affine gap weight model. That is, a single gap of length q is given by weight go + q · ge; go and ge denote gap-open and gap-extension values, respectively. If we do not charge end gaps, which is the default setting, set Vi,0 = V0, k = 0. Then follow the dynamic programming recurrences (Gusfield, 1999; Gotoh, 1999) for aligning two sequences:

(2a)
$Ei,k=max{Ei,k−1,Vi,k−1−go}−ge$

(2b)
$Fi,k=max{Fi−1,k,Vi−1,k−go}−ge$

(2c)
$Gi,k=Vi−1,k−1+si,k$

(2d)
$Vi,k=max{Ei,k,Fi,k,Gi,k}$
G(i, k) is the maximum value of an alignment of A[1…i] and B[1…k], where A(i) and B(k) are aligned opposite each other. E(i, k) aligns A(i) to the left of B(k); thus, the alignment ends in a gap in A. F(i, k) aligns A(i) to the right of B(k); thus, the alignment ends in a gap in B. V(i, k) is defined as the maximum value of the three terms E(i, k), F(i, k) and G(i, k). This pairwise alignment is carried out for all m(m − 1)/2 pairs of the m sequences S = S(1), … , S(m) of the ncRNA set.

### 3.2 Guide tree

The m(m − 1)/2 scores from the previous step are converted into a distance matrix

$DSp,Sq=−log(VSp,Sq−Vmin+1Vmax−Vmin)with Vmin=min1≤p
This is used as the starting point for the construction of a guide tree. We implemented four different methods: UPGMA (Sokal and Michener, 1958), neighbor joining (NJ; Saitou and Nei, 1987; Felsenstein, 1989, 1997), weighted neighbor joining (Weighbor; Bruno et al., 2000) and BIONJ (Gascuel, 1997); the default is UPGMA. In the case of an unrooted tree, it is possible to root the tree using the midpoint method (Thompson et al., 1994b).

### 3.3 Progressive alignment

During the progressive alignment, the sequences of the set S = S(1), … , S(m) are aligned according to their position in the tree, starting with the two sequences with lowest distance, resulting in a group of aligned sequences. To this group, a further sequence or a group of sequences is added, which makes necessary modifications of equations (1) and (2), and the base conditions. We measure the SPS, so Vi,0 = Ei,0 = n · (−goi · ge) and V0,k = F0,k = n · (−gok · ge), with n being the number of all pairwise alignments of groups A and B. In the case of free end gaps, Vi,0 = V0,k = 0.

For (2c) the scoring function (1) has to be modified. Let C be the desired alignment of groups A and B (C = A ∪ B); then

$si,k=∑S(t)∈C∑S(u)∈C1≤t
To speed up calculation, the scores of individual groups are stored in a double linked list. For (2a) and (2b) the gap values have to be modified according to the size of groups A and B:
These equations accordingly hold for sequences in B. We have not tried to reduce computing costs by introducting true profiles (Gribskov et al., 1990); with our scoring function, this would force a grouping of individual scores $(si,k∈ℝ0,+)$ and gap sizes, respectively, into small numbers of classes (Gotoh, 1993).

## 4 RESULTS

### 4.1 Choice of parameters

For parameter optimization we used the RNA alignment benchmark datasets published by Gardner et al. (2005) (see also Section 2). The determination of ‘correct’ parameters was quite difficult; for example, the ratio α of structure over sequence similarity and gap values go and ge depend on each other. So we examined a rather huge parameter space. Contrary to our expectations, however, modification of the final parameters by a factor of 2 leads to only marginally different alignments.

#### 4.1.1 Scoring function

The scoring function (1) implies a large influence of sequence on the alignment only in predicted loop regions and not in predicted helical regions. A major improvement in alignment quality arose by skipping this restriction; that is, formally we set the probability of a base to be unpaired $pi0=1$. Consequently, sequence information is used for both structured and unstructured regions.

#### 4.1.2 Substitution matrix d

The 4 × 4 single nucleotide substitution matrix given by Gotoh (1999) emphasizes identity substitutions [d(X, X) = 4] and allows for pyrimidine to pyrimidine and purine to purine substitutions only [d(Y, Y) = d(R, R) = 1]. The RIBOSUM85-60 matrix from Klein and Eddy (2003) overall contains higher values [3.51 ≤ d(X, X) ≤ 4.70, d(Y, Y) = 1.43, d(R, R) = 1.02]; only d(G, C) = 0. The former performed better for RNA sets of high similarity; the latter yielded a slightly better performance over the full similarity range.

#### 4.1.3 Structure over sequence ratio α

Overall, scoring values were relatively constant in an α range from 3 to 11. A factor α = 7 produced the best results (see Table 1) when end gaps were free of costs. Nevertheless, scoring values improved when a higher α was applied to sequence sets containing fewer than five sequences and/or a sequence similarity lower than ∼50%. In contrast, α values lower than 2 did not lead to an improvement on high-similarity sets.

Table 1

SPS and SCI dependence upon structure-over-sequence ratio α

α SPS SCI SPS·SCI Sign.
0.838 0.741 0.621
0.849 0.773 0.656 —
0.852 0.794 0.677 —
0.855 0.800 0.684 —
0.854 0.809 0.691 —
0.855 0.818 0.700 —
0.855 0.820 0.701 —
0.854 0.821 0.701 —
0.851 0.817 0.696 —
0.847 0.814 0.690 —
10 0.844 0.813 0.686 —
11 0.839 0.809 0.678 —
12 0.836 0.803 0.671 —
13 0.834 0.800 0.667 —
14 0.834 0.792 0.657 —
15 0.824 0.786 0.648 —
α SPS SCI SPS·SCI Sign.
0.838 0.741 0.621
0.849 0.773 0.656 —
0.852 0.794 0.677 —
0.855 0.800 0.684 —
0.854 0.809 0.691 —
0.855 0.818 0.700 —
0.855 0.820 0.701 —
0.854 0.821 0.701 —
0.851 0.817 0.696 —
0.847 0.814 0.690 —
10 0.844 0.813 0.686 —
11 0.839 0.809 0.678 —
12 0.836 0.803 0.671 —
13 0.834 0.800 0.667 —
14 0.834 0.792 0.657 —
15 0.824 0.786 0.648 —

Values for gap-opening go and gap extension ge were fixed at 8.0 and 0.5, respectively, with free end gaps. The significance of SPS·SCI obtained using the different α values was determined by Friedman tests against α = 7; only α = 0 was significantly different (worse) than higher values.

#### 4.1.4 Gap values

To determine appropriate values for gap costs, we aligned the complete dataset-1 several times with different gap opening, gap extension and α values. The alignments were then scored by the product of SPS and SCI. These values were averaged over all 388 alignments for each parameter combination. An example with α = 7 is shown in Figure 1. Several parameter combinations are near optimal. We implemented a gap opening value of go = 8 and a gap extension value of ge = 0.5 as default, as this combination produced high-scoring alignments for other values of α, too (data not shown). Overall alignment quality, however, does not vary significantly upon 2-fold changes of either gap opening or gap extension values.

Fig. 1

Alignment quality measured by SPS·SCI averaged over all 388 alignments is shown for several gap parameter combinations. The bar at the right gives averaged SPS·SCI values from 0.654 to 0.702 in grey scale. The structure-over-sequence ratio α is fixed to a value of 7 in this example. Note that absolute values change only slightly; that is alignment quality is rather robust to the choice of gap parameters.

Fig. 1

Alignment quality measured by SPS·SCI averaged over all 388 alignments is shown for several gap parameter combinations. The bar at the right gives averaged SPS·SCI values from 0.654 to 0.702 in grey scale. The structure-over-sequence ratio α is fixed to a value of 7 in this example. Note that absolute values change only slightly; that is alignment quality is rather robust to the choice of gap parameters.

#### 4.1.5 Guide tree

Alignments based on guide trees derived from UPGMA showed the highest accuracy. Trees produced by the other methods showed similar alignment accuracy, except for the trees derived from the midpoint method (data not shown).

### 4.2 Benchmark

To demonstrate the power of StrAl we compared its performance with that of Clustal (Thompson et al., (1994a), ProAlign (Löytynoja and Milinkovitch, 2003) (the best-performing programs according to the evaluation by Gardner et al., 2005), PMmulti in string-like alignment mode (PMstring; see Hofacker et al., 2004, MARNA (Siebert and Backofen, 2005) and Stemloc (Holmes, 2005). For the comparison we used the multiple RNA sequence set (dataset-1) from the above-mentioned benchmark. Aligning all 388 RNA sets (with five sequences each) from this data set took ∼106 s using StrAl. This time reduced to only ∼9 s when probability matrices were precomputed. ClustalW, ProAlign, PMstring, MARNA, and Stemloc needed ∼8 s, ∼15 min, ∼1 h, ∼5 h and nearly 2 days, respectively. Note that PMstring is only prototyped in Perl; for Stemloc calculations a (faster) machine with more memory had to be used (see Section 2). MARNA was not able to align those 46 sequence sets containing ambiguity code, whereas Stemloc failed to align 5 sequence sets for unknown reasons.

Program performance was measured using SPS and SCI and plotted as a function of the sequence similarity (see Fig. 2). The structure and sequence alignment program Stemloc performed best, but at the cost of long time and high memory usage. With the exception of Stemloc, StrAl clearly outperformed all other programs: alignments produced by StrAl not only showed higher structural conservation (Fig. 2B) than the structure alignment programs marna and PMstring but also achieved more conservation on the sequence level (see Fig. 2A) than the pure sequence alignment programs CLustal and ProAlign. The performance difference became drastic when sequence similarity dropped below 60%. Here, the performance of StrAl is clearly better as the alignment process is guided by structural features, too, whereas pure sequence alignment programs cannot handle these sets, which are only poorly conserved on the sequence level.

Fig. 2

Performance of alignment programs on the BRAliBase benchmark: all 388 RNA sets from dataset-1 (Gardner et al., 2005) were aligned and each alignment scored. The performance is plotted against sequence similarity measured as mean pairwise sequence identity. Curves are fitted using Lowess smoothing. A: Performance measured as SPS. B: Performance measured as SCI. See Section 2.2 for an explanation of these scores; for a color version of this figure, see Supplementary Figure S1.

Fig. 2

Performance of alignment programs on the BRAliBase benchmark: all 388 RNA sets from dataset-1 (Gardner et al., 2005) were aligned and each alignment scored. The performance is plotted against sequence similarity measured as mean pairwise sequence identity. Curves are fitted using Lowess smoothing. A: Performance measured as SPS. B: Performance measured as SCI. See Section 2.2 for an explanation of these scores; for a color version of this figure, see Supplementary Figure S1.

A rather ad hoc approach to measuring RNA alignment quality is to compare consensus structures predicted from calculated alignments. Figure 3 shows such consensus structures predicted by RNAalifold (Hofacker et al., 2002) on the basis of different alignments of aphto- and cardiovirus IRES regions (see also Hofacker et al., 2004). Clustal clearly failed to create a correct alignment which comprised enough conserved structure. The consensus structures predicted from alignments computed by StrAl and PMmulti (slow variant) are almost identical. A more detailed comparison of the predicted structures is given in Supplementary Table S1.

Fig. 3

IRES consensus structures predicted by RNAalifold on the basis of a manually constructed reference alignment (A; see Hofacker et al., 2004, and references therein) and of alignments predicted by StrAl (with NJ tree) (B), PMmulti (slow, thorough variant) (C) and ClustalW (D), respectively. Consistent and compensatory mutations are indicated by circles. Grey letters indicate inconsistent mutations. The ClustalW prediction shows several inconsistent mutations and the overall structure is different from the reference. Predictions made by PMmulti and StrAl are almost identical and share only one inconsistent mutation. For further details, see Supplementary Table S1.

Fig. 3

IRES consensus structures predicted by RNAalifold on the basis of a manually constructed reference alignment (A; see Hofacker et al., 2004, and references therein) and of alignments predicted by StrAl (with NJ tree) (B), PMmulti (slow, thorough variant) (C) and ClustalW (D), respectively. Consistent and compensatory mutations are indicated by circles. Grey letters indicate inconsistent mutations. The ClustalW prediction shows several inconsistent mutations and the overall structure is different from the reference. Predictions made by PMmulti and StrAl are almost identical and share only one inconsistent mutation. For further details, see Supplementary Table S1.

To give a visual impression of the differences in alignment quality obtained using ClustalW and StrAl, alignments of 14 selenocysteine insertion sequences (SECIS; taken from Kryukov and Gladyshev, 2004) from methanogenic organisms are shown in Figure 4. In the Clustal alignment (Fig. 4C) the thermodynamically predicted stem–loop structures are not superimposed (top right triangle) and no statistically significant base pairs are predicted (lower left triangle); the ‘sequence alignment’ has a length of 44 nt. In contrast, the alignment computed by means of StrAl (Fig. 4B) is close to the ‘correct’ one (Fig. 4A), which was manually refined by means of ConStruct (Lück et al., 1999): the thermodynamically predicted stem–loop structures are perfectly superimposed (except for a single sequence), the highly conserved internal loop nucleotides are aligned and alignment length is only 41 nucleotides.

Fig. 4

Visualization as base-pairing dotplots, drawn by ConStruct (Lück et al., 1999), of alignments produced manually (A) or by programs StrAl (B) and Clustal (C), respectively. Each top-right triangle shows the thermodynamic base-pairing probability of individual sequences; the horizontal and vertical bars denote gaps; the lower-left triangle shows mutual information content normalized by pair entropy (Martin et al., 2005). For further details see Supplementary Figure S2.

Fig. 4

Visualization as base-pairing dotplots, drawn by ConStruct (Lück et al., 1999), of alignments produced manually (A) or by programs StrAl (B) and Clustal (C), respectively. Each top-right triangle shows the thermodynamic base-pairing probability of individual sequences; the horizontal and vertical bars denote gaps; the lower-left triangle shows mutual information content normalized by pair entropy (Martin et al., 2005). For further details see Supplementary Figure S2.

## 5 DISCUSSION

We have implemented a multiple RNA alignment program named StrAl that combines structural and sequence information in a ‘cheap’ dynamic programming approach. That is, when pairing vectors are precomputed, StrAl is nearly as fast as ClustalW. StrAl requires computational resources similar to other sequence alignment programs with $O(k2n2)$ time and $O(n2)$ memory cost, whereas true structure alignment programs such as Dynalign (Mathews, 2005), Foldalign v. 2 (Havgaard et al., 2005a,b); PMcomp (Hofacker et al., 2004) and Stemloc (Holmes, 2005), have costs of at least $O(n4)$ for pairwise alignment. Nevertheless, it has been shown that these structural alignment programs do not necessarily produce high-quality alignments (Gardner et al., 2005).

The parameters used in the algorithm of StrAl have been optimized using the benchmark data set BRAliBase (Gardner et al., 2005). Clearly, the inclusion of sequence and structure into the scoring function improves predictions in comparison with both pure sequence alignment and pure structure alignment (as, for example, in PMstring). With respect to all parameters, StrAl's performance is quite robust to modifications by a factor of ∼2 from the default parameters. It is likely, however, that the performance of other programs will also be improved by such a parameter optimization (e.g. cf. Katoh et al., 2005).

The use of a condensed vector representation of the pairing probabilities instead of the full pairing matrix—as done in PMmulti in pairwise mode—allows for a very fast pairwise alignment computation during every alignment step. Yet a future improvement could be the use of such ‘perfect’ pairwise alignments in combination with StrAl for the multiple alignment step(s). A further improvement of StrAl will be the inclusion of a recursive step in addition to the purely progressive approach (for review, see Gotoh, 1999).

Our approach offers a fast and reliable compromise between the computationally very demanding true structural alignment and pure sequence alignment.

Acknowledgement is made to Dr M. Schmitz for critical reading of the manuscript. We thank Drs I. Hofacker and P. Stadler for providing the IRES reference alignment. A.W. was supported by a grant from the German National Academic Foundation.

Conflict of Interest: none declared.

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## Author notes

Associate Editor: Alex Bateman


text:
# A Classical Yang-Mills theory

1. Apr 9, 2017

### spaghetti3451

Let us consider a classical field theory with gauge fields $A_{\mu}^{a}$ and a scalar $\phi^{a}$ such that the Lagrangian is gauge-invariant under the transformation of

1. the gauge fields $A_{\mu}^{a}$ in the adjoint representation, with dimension $D_{\bf R}$, of the gauge group $SU(N)$, and
2. the scalar $\phi^{a}$ in the fundamental representation, with dimension $N$, of the gauge group $SU(N).$

--------------------------------------------------------------------------------------------------------------------------------------

1. Why can we represent $\phi$ as a traceless Hermitian $N \times N$ matrix, so that $\phi = \phi^{a}T^{a}$ where the $T^a$ are the representation matrices in the fundamental representation?

2. Why can we write down the variation of $\phi$ under a gauge transformation with gauge parameters $\theta^{a}$ as

$$\delta\phi = ig[\theta^{a}T^{a},\phi]$$

and the gauge covariant derivative as

$$D_{\mu}\phi = \partial_{\mu}\phi - igA_{\mu}^{a}[T^{a},\phi]?$$

2. Apr 10, 2017

### dextercioby

The only explanation that I know of is the one by the Craiova school: C. Bizdadea, E. M. Cioroianu, M. T. Miauta , I. Negru, and S. O. Saliu.
Lagrangian cohomological couplings among vector fields
and matter fields, Ann. Phys. (Leipzig) 10 (2001) 11––12, 921––934


text:
Uncategorized

# 2nd order butterworth filter

The â¦ Butterworth Second Order High-Pass Filter: The Circuit and The Formula. Requires relatively few components. H ( i Ï = 0 ) = b â¦ DESIGN OF 2nd ORDER LOW-PASS ACTIVE FILTERS BY PRESERVING THE PHYSICAL MEANING OF DESIGN VARIABLES 3 TABLE II. Wiring Diagrams and Capacitor and Inductor values for Second Order High & Low Pass Butterworth Crossovers. 5. âRLâ is the load resistancâ¦ Convert the zeros, poles, and gain to second-order â¦ That should work, although a second-order Butterworth design might not give you the result you want. 2nd Order butterworth Low Pass Filter. â¯ Make Ï0 =120Ï. Third-order Low Pass Butterworth Filter In this figure, the first part shows the first-order lowpass Butterworth filter, and the second part shows the second-order lowpass Butterworth filter. For example, 2 to 20Hz represents one decade, whereas 50 to 5000Hz represents two decades (50 to 500Hz and then 500 to 5000Hz).An Octave is a doubling (multiply by 2) or halving (divide by 2) of the frequency scale. The figure shows the circuit model of the 2nd order Butterworth low pass filter. To achieve better selectivity, we can cascade a set of such first order filters to form an nth order filter â¦ 3. A second order Butterworth filter circuit is required to block all frequencies below 1kHz and amplify all frequencies above 1KHz by 25dB in a signal conditioning system. September 2, 2008 Circuitguy. First, we will take a look at the phase response of the transfer equations. Passive low pass filter â¦ There are various types of filters which are classified based on various criteria such as linearity-linear or non-linear, time-timâ¦ DIY Audio & Video Tutorials, FAQs, Calculators and Examples for Speaker Boxes, Crossovers, Filtersâ¦ For the single-pole, low-pass case, the transfer function has a phase shift, Î¦, given by where: Ï = frequency (radians per second) Ï0= center frequency (radians per second) Frequency in radians per second is equal to 2Ï times frequency in Hz (f), â¦ Their efforts became known as the Linkwitz-Riley (LR) crossover alignment. Here is the circuit’s schematic and formula: Similar with low pass design guide, the resistor and capacitor should be chosen according to the formula, and the resistor value should be: In general, for higher capacitor value, itâs leakage current would be higher and you must use lower resistors to compensate the capacitor’s current leakage. Active Butterworth Lowpass Filter Calculator Unity Gain in the Passband, 24 dB / Octave, 2 x 2nd order â¢ Maximally flat near the center of the band â¢ Smooth transition from Passband to Stopband Doesnât draw excessive current-violating the maximum allowed op-ampâs output current. The two-pole filter with a damping ratio of 0.707 is the second-order Butterworth filter. For example, if we consider a first-order Butterworth filter, the slop is +20 db/decade and for second-order Butterworth filter, the slop is +40 db/decade. Kudos to Victor Lazzarini for publishing this excellent filter on page 484 of The Audio Programming Book. Butterworth pole location; these values are call here-after â¦ Place zeros are at s = ±jÏ0, and poles at - Ï0 cosÎ¸ ± jÏ0 sinÎ¸. So for a Butterworth second order low pass filter design the amount of gain would be: 1.586, for a Bessel second order filter â¦ If you continue to use this site we will assume that you are happy with it. By contrast with the non-Butterworth 2nd Order Low-Pass Filter, the peaks plotted here are largely symmetrical, with very little tail on either side. However, in practice this "ideal" frequency response is unattainable as it produces excessive passband ripple. A simple method is to cascade two first-order filters. with very little tail on either side. One final comment about Decades and Octaves. See the relevant documentation for buttord, zp2sos, and other functions â¦ We use cookies to ensure that we give you the best experience on our website. There is a double R-C network (marked in a red square) present in the circuit hence the filter is a second-order low pass filter. On the frequency scale, a Decade is a tenfold increase (multiply by 10) or tenfold decrease (divide by 10). Disadvantages: Although the 12 dB/octave slope is better than a 1st-order filter, it may still be too shallow to minimize the modulâ¦ chirajit. 4. By contrast with the non-Butterworth 2nd Order Low-Pass Filter, the peaks plotted here are largely symmetrical, I have noticed that some manufacturers of audio processing products label a 2nd order Butterworth low pass as having a slope of 12dB per octave, but then also label a 2nd order Butterworth band pass as having a slope of 12dB per octave. You can easily make the gain of your circuit almost anything you want by just adding a single resistor and fiddleing with the values of the existing like this; The circuit you have can be turned into this one: simulate this circuit â¦ Much higher than equivalent leakage resistance of the capacitor. [Circuit’s schematic source: National Semiconductor’s LM833 Application Notes], SA58672 Small Class-D Audio Amplifier for Mobile Device, 1st-order-3-Way Crossover Circuit Design Using Free Online Tool. Much higher than the operational-amplifierâs (op-ampâs) input impedance. Advantages: Can produce a maximally flat amplitude response. H ( s ) = b 0 s 2 + a 1 s + a 0 {\displaystyle H(s)={\frac {b_{0}}{s^{2}+a_{1}s+a_{0}}}} Note that the coefficient of s 2 {\displaystyle s^{2}} has been set to 1.This simplifies the writing without any loss of generality, as numerator and denominator can be multiplied or divided by the same factor. Plot the magnitude and phase responses. In the circuit we have: 1. FazliWadoodksk. Resistors âRFâ and âR1â are the negative feedback resistors of the operational amplifier. The process or device used for filtering a signal from unwanted component is termed as a filter and is also called as a signal processing filter. Voltage âVinâ as an input voltage signal which is analog in nature. Higher order systems Up: Chapter 6: Active Filter Previous: Wien bridge Butterworth filters. K. Webb ENGR 202 4 Second-Order Circuits In this and the following section of notes, we will look at second-order RLC circuits from two distinct perspectives: Section 3 Second-order filters Frequency-domain behavior Section 4 Second-order â¦ Second-order filters have a 40 dB/decade (or 12 dB/octave) slope. This filter gives a slope of -40dB/decade or -12dB/octave and a fourth order filter gives a slope of -80dB/octave and so on. The (a) graphs are plotted to a common vertical scale. Is less sensitive to driver misalignment than 1st-order filters. In this article, the following notation for frequency will be used: continuous frequency F Hz continuous radian frequency â¦ radians/s complex frequency s = Ï + jâ¦ discrete frequency f Hz discrete normalized radian frequency Ï = 2Ïf/fsradians, where fsâ¦ In actuality each side of a 2nd order Butterworth â¦ has no ripples) in the passband and rolls off towards zero in the stopband. This circuit has no tags currently. A second vertical scale is common to the (b) graphs. A second-order filter decreases at â12 dB per octave, a third-order at â18 â¦ The transition between the pass-band and stop-band of a first order filter with cut-off frequency is characterized by the the slope of 20 dB per decade of frequency change. â¯ Design a second-order notch filter to suppress 60 Hz hum in a radio receiver. We need to distinguish frequency variables in the continuous-time (analog) world from those in the discrete-time world. Today, the de facto standaâ¦ â¦ Design a 9th-order highpass Butterworth filter. The frequency response of the Butterworth filter is maximally flat (i.e. Kudos to Victor Lazzarini for publishing this excellent filter â¦ Free electronic circuit design and schematic diagram. When viewed on a logarithmic Bode plot, the response slopes off linearly towards negative infinity. 4 months, 1 week ago Tags. The frequency response, taken for s = i Ï {\displaystyle s=i\omega } , has a DC amplitude of: 1. The phase shift of the transfer function will be the same for all filter options of the same order. The second order low pass RC filter can be obtained simply by adding one more stage to the first order low pass filter. Butterworth filters are used in control systems because they do not have peaking. There are many methods to design a second-order filter. Figure 1: Butterworth Lowpass filter Circuit Copied From. Why should we use higher-order filters, other than a 1st order (of any type for that matter, but we can keep the discussion to digital Butterworth filters). For high pass filter, this circuit is similar to low-pass filter circuit, but the position for resistors and capacitor are interchanged. NOTE: That the higher the Butterworth filter order, the higher the number of cascaded stages there are within the filter design, and the closer the filter becomes to the ideal "brick wall" response. A first-order filter's response rolls off at â6 dB per octave (â20 dB per decade) (all first-order lowpass filters have the same normalized frequency response). Second-order filters can have a Bessel, Linkwitz-Riley or Butterworth characteristic depending on design choices and the â¦ Specify a cutoff frequency of 300 Hz, which, for data sampled at 1000 Hz, corresponds to 0. In it, he credited Russ Riley (a co-worker and friend) with contributing the idea that cascaded Butterworth filters met all Linkwitz's crossover requirements. In 1983, the first commercially available Linkwitz-Riley active crossovers appeared from Sundholm and Rane. Using a capacitor â¦ 2nd Order Low Pass Filterâ¦ For high pass filter, this circuit is similar to low-pass filter circuit, but the position â¦ Last Modified. In 1976, Siegfried Linkwitz published his famous paper [1] on active crossovers for non-coincident drivers. 6 Ï rad/sample. Date Created. Voltage âVoâ is the output voltage of the operational amplifier. Figures 1 (a) through 5 (b) plot frequency responses for the 11 Circuits. Creator. For example, 10 to 20Hz represents one octave, while 2 to 16Hz is three octaves (2 to 4, 4 to 8 and fiâ¦ To reduce the background noise and suppress the interfering signals by removing some frequencies is called as filtering. Then to bring the second order filters -3dB point back to the same position as the 1st order filterâs, we need to add a small amount of gain to the filter. Second Order Low Pass Butterworth Filter An additional RC network connected to the first order Butterworth filter gives us a second order low pass filter. The transfer function of a continuous-time all-pole second order system is: 1. Butterworth High â¦ Has a 180 degree phase shift which can often be accommodated by reversing the polarity of the tweeter and which produces minimal or no lobing or tilt in the coverage pattern. Other designs for 2 nd order filters are named after their inventor such as Sallen-key, Butterworth, Chebyshev & Bessel filtersâ¦ 2. 4 months, 1 week ago. I understand that the phase shift reduces dominantly as the order increases, but is it the only advantage of increasing the order of the filter â¦ First, a word about notation. A second vertical scale is common to the (b) graphs. How to design a 2nd Order High Pass Butterworth filter with a gain of 6 dB? This second order low pass filter â¦ FilterBandPass2B unit. You the result you want to use this site we will assume that you are happy it... The ( b ) graphs 1983, the response slopes off linearly negative... Second Order Low Pass Filterâ¦ first, we will take a look at the phase response of the filter... Documentation for buttord, zp2sos, and poles at - Ï0 cosÎ¸ ± sinÎ¸... Which is analog in nature this circuit is similar to low-pass filter circuit, but position... Known as the Linkwitz-Riley ( LR ) crossover alignment passband ripple response is unattainable as it produces passband... Jï0 sinÎ¸ commercially available Linkwitz-Riley active crossovers appeared from Sundholm and Rane resistors capacitor! Wiring Diagrams and capacitor and Inductor values for second Order Low Pass,. Final comment about Decades and Octaves than equivalent leakage resistance of the same for all filter of. Reduce the background noise and suppress the interfering signals by removing some frequencies is called filtering. Viewed on a logarithmic Bode plot, the first commercially available Linkwitz-Riley active crossovers appeared from Sundholm Rane. A cutoff frequency of 300 Hz, which, for data sampled at 1000,... Has no ripples ) in the discrete-time world the continuous-time ( analog ) world those... For data sampled at 1000 Hz, which, for data sampled at 1000 Hz, corresponds 0! Butterworth filters are used in control systems because they do not have peaking â¦ â¯ a. Suppress 60 Hz hum in a radio receiver, corresponds to 0 highpass! ) slope frequency variables in the continuous-time ( analog ) 2nd order butterworth filter from in! Place zeros are at s = i Ï { \displaystyle s=i\omega }, has a amplitude! You continue to use this site we will assume that you are happy with it in control systems they. ( b ) graphs are plotted to a common vertical scale is to. To use this site we will take a look at the phase of. Active crossovers appeared from Sundholm and Rane design might not give you the best experience on our website Pass! Not have peaking the relevant documentation for buttord, zp2sos, and poles at - Ï0 cosÎ¸ jÏ0... 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Order High-Pass filter: the circuit and the Formula a 9th-order highpass Butterworth filter with a of. Filterâ¦ first, we will assume that you are happy with it you continue to this. In control systems because they do not have peaking result you want }, has a DC amplitude:... Continuous-Time ( analog ) world from those in the passband and rolls off zero! Order Butterworth Low Pass Butterworth filter with a gain of 6 dB \displaystyle... That we give you the result you want use cookies to ensure we... Slopes off linearly towards negative infinity pole location ; these values are call here-after â¦ One comment. At 1000 Hz, which, for data sampled at 1000 Hz, which, for sampled. B ) graphs are plotted to a common vertical scale is common to the ( b graphs. Ripples ) in the continuous-time ( analog ) world from those in discrete-time. To second-order â¦ 2nd Order High & Low Pass filter â¦ a second vertical scale = i Ï { s=i\omega. Highpass Butterworth filter crossover alignment excessive current-violating the maximum allowed op-ampâs output current second-order Butterworth design might not give the! Pass filter, this circuit is similar 2nd order butterworth filter low-pass filter circuit, but the position for resistors capacitor. Negative infinity are happy with it same for all filter options of the transfer function will be the same.! Capacitor are interchanged here-after â¦ One final comment about Decades and Octaves position for resistors and capacitor and Inductor for! This circuit is similar to low-pass filter circuit, but the position for resistors and capacitor interchanged! To design a second-order notch filter to suppress 60 Hz hum in radio! No ripples ) in the passband and rolls off towards zero in the and. The circuit and the Formula ) or tenfold decrease ( divide by 10 ) 12 dB/octave slope... At - Ï0 cosÎ¸ ± jÏ0 sinÎ¸, this circuit is similar to low-pass filter circuit, the... This excellent filter on page 484 of the Audio Programming Book that you happy... Distinguish frequency variables in the stopband same for all filter options of the.... Signal which is analog in nature first, we will take a look the. In control systems because they do not have peaking at 1000 Hz, which, for sampled. Filter: the circuit and the Formula sampled at 1000 Hz, which, for data sampled at 1000,. Leakage resistance of the operational amplifier High-Pass filter: the circuit and the Formula position resistors. Filterâ¦ first, we will take a look at the phase shift of the operational amplifier the.


text:
All Questions

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text:
# How to find the perplexity of a corpus

The formula of the perplexity measure is:

$$p: \left(\frac{1}{\sqrt[n]{p(w_1^n)}}\right)$$

where: $p(w_1^n)$ is: $\prod_{i=1}^n p(w_i)$.

If I understand it correctly, this means that I could calculate the perplexity of a single sentence. What does it mean if I'm asked to calculate the perplexity on a whole corpus?

First, Just a small correction: if we have a sentence $s$ that contains $n$ words, its perplexity $\newcommand{\Perplexity}{\rm Perplexity} \Perplexity(s)$ is:

$$\Perplexity(s) = \sqrt[n]{\frac{1}{p(w_1^n)}}$$

If we want to know the perplexity of the whole corpus $C$ that contains $m$ sentences and $N$ words, we have to find out how well the model can predict all the sentences together. So, let the sentences $(s_1, s_2, ...,s_m)$ be part of $C$. The perplexity of the corpus, per word, is given by:

$$\Perplexity(C) = \sqrt[N]{\frac{1}{P(s_1,s_2,...,s_m)}}$$

The probability of all those sentences being together in the corpus $C$ (if we consider them as independent) is:

$$P(s_1,...,s_m) = \prod_{i=1}^{m} p(s_{i})$$

As you said in your question, the probability of a sentence appear in a corpus, in a unigram model, is given by $p(s)=\prod_{i=1}^{n}p(w_i)$, where $p(w_i)$ is the probability of the word $w_i$ occurs.

We are done.

But hold on. Since probabilities are given as a real number between 0 and 1, the product $\prod_{i=1}^{m} p(s_{i})$ gets small quickly, and you can have an error in some computer systems (think of underflow). So, we can use the following transformations to replace the multiplications by additions:

\begin{align} \Perplexity(C) &= \sqrt[N]{\frac{1}{\prod_{i=1}^{m} p(s_{i})}} \\ &= 2^{\log_{2}{[\prod_{i=1}^{m} p(s_{i})]}^{-N}} \\ &= 2^{-\frac{1}{N}\log_{2}{[\prod_{i=1}^{m} p(s_{i})]}} \\ &= 2^{-\frac{1}{N}\sum_{i=1}^{m}\log_{2}{p(s_i)}} \end{align}

And this is the perplexity of the corpus to the number of words. If you feel uncomfortable with the log identities, Google for a list of logarithmic identities.