text:
# Schedule

The schedule below shows the tentative dates for all class topics, readings, and assignments. You should complete all assigned reading before class on the day it is listed. Labs will be available shortly before the assigned lab day. There may be some revisions to the schedule during the semester, but I will make sure to announce these changes in class. If you view this page with JavaScript enabled you can jump to the current week on the schedule, and you should see the next day of class highlighted in the schedule below.

Week 0
W
Jan 22

#### Introduction to Algorithms

We begin the class by exploring the definition of computer science and by trying to write some basic algorithms.

Lab
• No lab
Assigned
Th
Jan 23

Due
F
Jan 24

#### Getting Started with Linux and Scheme

We begin to consider the content and structure of the course. We also prepare ourselves to use our laboratory environment (that is, the Linux workstations and the DrRacket Programming Environment).

Lab
Week 1
M
Jan 27

#### Basic Types

We explore some of the basic types in Scheme and the operations available for those types.

Lab
Due
• Quiz 0

W
Jan 29

#### Introducing Lists

We start to explore Scheme’s list data structure and some ways to use lists to work with collections of data.

Lab

F
Jan 31

We consider new ways to write your own procedures and why you might do so.

Assigned
Week 2
M
Feb 3

#### Pair Programming

We explore the whys and hows of working with others. We also catch up on any confusing issues from the first few days of class.

Lab
• No lab
Due
• Quiz 1

W
Feb 5

#### Documenting Programs and Procedures

We consider documentation for your programs: Why to write documention, when to write documentation, how to write documentation. We also explore the 6P style of documentation that we use in this course.

Lab
• No lab
Th
Feb 6

#### Work Due

Due
F
Feb 7

We consider testing When, why, and how you might test the procedures and programs that you write.

Assigned
Week 3
M
Feb 10

#### Tables and Compound Data

We consider how to deal with compound data, such as the title, latitude, longitude, time, and date of an event.

Due
• Quiz 2

W
Feb 12

#### Reading Lists and Tables from Files

We consider a variety of techniques for gathering lists and tables of data from files.

Th
Feb 13

Due
F
Feb 14

#### Boolean Values, Predicate Procedures, and List Filters

We consider a new type and its use in selecting elements from lists.

Assigned
Week 4
M
Feb 17

#### Local Bindings

We consider how and why to name values within procedures.

Due
• Quiz 3

W
Feb 19

#### Conditionals

We consider one of the central components of algorithms.

Lab
Th
Feb 20

Due
F
Feb 21

#### Displaying Data

We explore techniques for displaying simple kinds of data, such as coordinate pairs or counts of categorical data.

Assigned
• Assignment 4
Week 5
M
Feb 24

#### Discussion of Exam 1

We consider your work on exam 1.

Lab
• No lab
Due
• Quiz 4

W
Feb 26

#### Preconditions, revisited

We revisit preconditions. We then consider programming techniques for ensuring that preconditions are met.

Th
Feb 27

#### Work Due

Due
• Assignment 4 (by 10:30pm)
F
Feb 28

#### Recursion Basics

We begin our exploration of recursion, the most general form of repetition available in Scheme. You can use recursion to both build and iterate over different kinds of values.

Assigned
Week 6
M
Mar 2

#### Recursion Basics, continued

We continue our exploration of recursion basics.

Due
• Quiz 5

W
Mar 4

#### Recursion with Helper Procedures

We consider a different form of recursion, one based on the construction of recursive helper procedures that take additional parameters. Along the way, we consider the idea of tail recursion. We also explore how careless design of recursive procedures can inadvertently lead to slow execution.

Th
Mar 5

Due
F
Mar 6

#### Other Forms of List Recursion

We conclude our initial forays into list recursion by looking for some common patterns in the design of recursive procedures.

Assigned
Week 7
M
Mar 9

#### Numeric Recursion

We consider a slightly different kind of recursion, numeric recursion. In this technique, we once again have procedures call themselves. However, the parameter that we “simplify” at every step is a number, rather than a list.

Due
• Quiz 6

W
Mar 11

#### Naming Local Procedures

We explore how and why one writes local recursive procedures.

Th
Mar 12

Due
F
Mar 13

#### Debugging

We explore techniques for understanding and correcting flaws in our programs.

Spring Break
Week 8
M
Mar 30

#### Randomness and Simulation

We consider Scheme’s random procedure and how one might use that procedure in writing simple simulations.

W
Apr 1

#### Pairs and Pair Structures

We consider pairs, the basic data type used to build lists and other structures in Scheme. We also consider why it is useful to understand about pairs.

F
Apr 3

#### Vectors

We consider vectors, an alternative to lists for storing collections of data.

Lab
• Vectors
Assigned
• Assignment 6
Week 9
M
Apr 6

#### Files in Scheme

We revisit files, considering the lower-level operations for working with files, a technique for structuring information that permits the information to persist across invocations of Scheme. Files also let our Scheme programs share information with other programs.

Lab
• Files

W
Apr 8

#### Higher-Order Procedures, revisited

We revisit the topic of higher-order procedures, one of the most important techniques in languages like Scheme. Higher-order procedures are procedures – like map, left-section, or compose – that take other procedures as parameters, return other procedures as values, or both.

Lab
• Higher-order procedures
Th
Apr 9

#### Work Due

Due
• Assignment 6 (by 10:30pm)
F
Apr 10

#### Trees

We consider trees, structures built from pairs. Trees are somewhat like two-dimensional lists.

Lab
• Binary trees
Assigned
• Exam 3
Week 10
M
Apr 13

#### Analyzing Procedures

We explore techniques for analyzing the number of calls made in evaluating procedures, particularly recursive procedures. We consider why such analysis is useful.

Lab
• Analyzing procedures

W
Apr 15

#### Association Lists

We consider association lists, a simple, but useful, technique for organizing tables of information.

Lab
• Association lists and searching
Th
Apr 16

#### Work Due

Due
• Exam 3 (by 10:30pm)
F
Apr 17

#### Project Introduction

We introduce the project.

• Class Project
Lab
• No lab
Assigned
• Assignment 7
Week 11
M
Apr 20

#### Project Work Day 1

We provide time for groups to work on their projects.

Lab
• No lab

W
Apr 22

#### Binary Search

We consider the general problem of searching. We explore binary search, one of the most efficient algorithms for searching.

Lab
• Binary search
Th
Apr 23

#### Work Due

Due
• Assignment 7 (by 10:30pm)
F
Apr 24

#### Introduction to Sorting

We explore the problem of sorting. When you sort a list, vector, or other collection, you put the elements in order. The order of the elements usually corresponds to the type of the elements. We might sort strings alphabetically, grades numerically, colors by brightness, and so on and so forth.

Lab
• No lab
Due
• Project Proposal (by 10:30pm)
Week 12
M
Apr 27

#### Project Work Day 2

We provide additional time for groups to work on their projects.

Lab
• No lab

W
Apr 29

#### Insertion Sort

We move from our general exploration of sorting to the implementation of a particular sorting algorithm, insertion sort. We also explore how the running time for that algorithm varies based on the number of values we are sorting.

Lab
• Insertion sort

F
May 1

#### Project Work Day 3

We will have some additional time to work on projects before the project deadline.

Lab
• No lab
Assigned
• Exam 4
Week 13
M
May 4

#### Merge Sort

We continue our exploration of sorting by considering the applicability of the divide-and-conquer approach to the problem of sorting. We look at one particular divide-and-conquer algorithm, merge sort. We explore how the running time for that algorithm varies based on the number of values we are sorting.

Lab
• Merge sort
Tu
May 5

#### Work Due

Due
• Class Project (by 10:30pm)
W
May 6

Lab
• No lab
Th
May 7

#### Work Due

Due
• Exam 4 (by 10:30pm)
F
May 8

#### Wrap Up

We conclude the course by considering the topics we’ve covered, and discuss the concepts you will see in future CS courses.


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# R: H is singular

For my thesis I need to estimate BEKK GARCH models. For this I have tried several packages. I keep getting the same error: "H is singular". I have found that this can be caused by highly correlated series, or series very close to zero.

To make sure it had nothing to do with my data (I am modeling cryptocurrencies and other asset classes), I tried to run the BEKK model on different series. I have tried to run in on Google, IBM, microsoft, GOLD index and other series, but I keep getting this error.

One example of code I have tried where I got this error is:

getSymbols("MSFT", from = startDate, to = endDate)

getSymbols("GOOG", from = startDate, to = endDate)

data <- data.frame(dailyReturn(MSFT), dailyReturn(IBM))

log_returns1 <- diff(log(MSFT$MSFT.Adjusted), lag=1) log_returns2 <- diff(log(GOOG$GOOG.Adjusted), lag=1)

log_ret <- merge(log_returns1[-1,], log_returns2[-1,])

estimate <- BEKK(log_ret, order = c(1, 1), params = NULL, fixed = NULL, method = "BFGS", verbose = T)

I still obtain parameters but the results cannot be right. I get totally different parameter estimates with different packages. Also, when plotting the dynamic correlation obtained from the model it looks more like a return series that bounces between -1 and +1.

I did get "Valid" parameters for the BTC-VIX series using the MTS BEKK package, namely no significant volatility spillover. But even here the dynamic correlation obtained from the model was not correct.

I have used: MTS and mGARCH-BEKK package.

Any help would be greatly appreciated.

• I sympathise with you but I would suggest you start with simple simulated series that you know the answers for the parameters you are looking and then use the more complicated data / real-life series. As it stands you might have a variety of causes for this (e.g. extremely high volatility) and your question is unanswerable in that sense. – usεr11852 Apr 22 at 16:10
• Thank you for your answer. I have just simulated BEKK garch data using simulateBEKK function of mgarchBEKK package in R. Using the simulated series I still get the "H is singular" and "negative inverted hessian matrix element" error. Code used: simulated = simulateBEKK(2, 1000, c(1,1)) simulated <- do.call(cbind, simulated\$eps) testEst <- BEKK(simulated, order = c(1, 1), params = NULL, fixed = NULL, method = "BFGS", verbose = T) – David Daoe Apr 22 at 17:02
• Then go even simpler and see why this fails. I appreciate this is time-consuming but there are not many better alternatives. (You could also try reading the R code to see what it does in more detail.) – usεr11852 Apr 22 at 19:37

format: plain


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Acta Phys. -Chim. Sin. ›› 2018, Vol. 34 ›› Issue (5): 528-536.

• ARTICLE •

### Chemical Bonding and Interpretation of Time-Dependent Electronic Processes with Maximum Probability Domains

Andreas SAVIN*()

• Received:2017-09-01 Published:2018-01-24
• Contact: Andreas SAVIN E-mail:andreas.savin@lct.jussieu.fr

Abstract:

Tools have been designed obtain information about chemical bonds from quantum mechanical calculations. They work well for solutions of the stationary Schrödinger equation, but it is not clear whether Lewis electron pairs they aim to reproduce survive in time-dependent processes, in spite of the underlying Pauli principle being obeyed in this regime. A simple model of two same-spin non-interacting fermions in a one-dimensional box with an opaque wall, is used to study this problem, because it allows presenting the detailed structure of the wave function. It is shown that ⅰ) oscillations persisting after the Hamiltonian stopped changing produce for certain time intervals states where Lewis electron pairs are spatially separated, and ⅱ) methods (like density analysis, or the electron localization function) that are widely used for describing bonding in the stationary case, have limitations in such situations. An exception is provided by the maximum probability domain (the spatial domain that maximizes the probability to find a given number of particles in it). It is conceptually simple, and satisfactorily describes the phenomenon.


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# How to use the volumes package?

I need to split a book in two volumes, the problem is that I need the table of contents and the index of the whole book in each volume. I already tried to use the package `volumes` but I can't make it work.

``````\usepackage{volumes}
\onlyvolume
``````

And according to the documentation I should put `\volumeone` and `\volumetwo` but after installig the package these commands are not recognized.

-
Welcome to TeX.sx! You tagged your question with `biblatex`, `bibtex` and `bibliographies`, but I cannot recognize how those tags relate to your question. Therefore please elaborate your question a bit more to account for this topic or remove the tags as they would lead to a mis-categorization. –  Benedikt Bauer Mar 7 '13 at 13:18
The `\volumeone`, `\volumetwo`, …, `\volumeten` macros are only defined if the package is called with the `manual` option; however they shouldn't be necessary, unless there are problem with other packages trying to redefine `\include`, according to the documentation. –  egreg Mar 7 '13 at 13:38

As long as you publish in print, the solution is easy:

1. Keep the whole opus in one main file (maybe using \include etc.)
2. Compile the file with TOC
3. Save your TOC as a extra PDF (e.g. by sending only the TOC pages to a PDF printer and the like)
4. Now split your main file in two and set the first pagenumber of the second part to the pagenumber it had before.
5. Before this first page include the TOC-PDF, using the pdfpages package.

You know how to switch between roman pagenumbers for the frontmatter and arabic pagenumbers of the mainmatter? But this depends on the documentclass.

So you need to find out how to deal with pagenumbering. We can help you, as soon as you tell us more about the documentclass and maybe give us an MWE. Minimal working examples for large documents are easy with the blindtext package and the command `\blinddocument`.

If you publish a PDF, things are more complicated, because you'll probably wish to have hyperlinks from the TOC to the chapters and sections.

-


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# How would MnO_4^(-), and SO_2(g) interact?

Mar 2, 2017

We would assume that the permanganate ion, which is a potent oxidant...............

#### Explanation:

We would assume that the permanganate ion, which is potent oxidant, oxidizes the $S {O}_{2} \left(g\right)$ up to $S \left(V I +\right)$. Let's assume such oxidation:

$\text{Oxidation rxn (i):}$

$S {O}_{2} \left(g\right) + 2 {H}_{2} O \left(l\right) \rightarrow S {O}_{4}^{2 -} + 4 {H}^{+} + 2 {e}^{-}$

$\text{Reduction rxn (ii):}$

$M n {O}_{4}^{-} + 8 {H}^{+} + 5 {e}^{-} \rightarrow M {n}^{2 +} + 4 {H}_{2} O$

$\text{Overall rxn:}$ $=$ $5 \times \left(i\right) + 2 \times \left(i i\right)$ $=$

$2 M n {O}_{4}^{-} + 5 S {O}_{2} \left(g\right) + 2 {H}_{2} O \left(l\right) \rightarrow 2 M {n}^{2 +} + 5 S {O}_{4}^{2 -} + 4 {H}^{+}$

I would recheck this question. They should give you a bit of context (I am guessing that you are an undergrad). Namely, that the reaction proceeds with the disappearance of the deep purple colour of permanganate to give the almost colourless $M {n}^{2 +}$ ion. The redox equation is as far as I know stoichiometrically balanced; i.e. garbage out equals garbage in, including electronic charge.


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# Confusion Regarding Most Probable Kinetic Energy of an Ideal gas

## Homework Statement

Find the most probable kinetic energy of an ideal gas molecule.

## Homework Equations

v=sqrt(2RT/M)
where v= most probable velocity
k= Boltzmann's constant
T= temperature of the system
M= molar mass of the gas

Maxwell's formula of probability distribution of kinetic energy

## The Attempt at a Solution

The standard solution, as prescribed by many textbooks and reference materials, involves differentiation of the probability distribution equation w.r.t. kinetic energy to find the maximum and hence showing that the most probable kinetic energy is kT/2. This seems logical and I assume it is correct.

But my doubt arises due to a different approach that I employed. Instead of differentiating the probability distribution equation if I simply plug in the value of most probable velocity in the equation for kinetic energy (mv^2/)2 then my answer turns out to be kT. I'm confused as isn't it logical to assume that the molecules with most probable velocity have the most probable kinetic energy. Please help.

Last edited:


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# geometric realization of a simplicial complex

Let $K$ be a simplicial complex and let $|K|$ be the geometric realization of $K$. Suppose that the vertices ($0$-simplices) of $K$ are $v_0,v_1,\cdots,v_k$, $k$ finite. Let $\Delta^n$ be the standard $n$-simplex and let $$K_{n}= \text{Simplicial Map}(\Delta^n,K).$$ Then $$K_n\cong\{f\in \text{Map}(\mathbb{Z}_{n+1},\mathbb{Z}_{k+1})\mid \text{ there exists a simplex } \sigma\in K \text{ such that Im}f \text{ is the set of vertices of } \sigma\}.$$ Here we use $\mathbb{Z}_{n+1}$ to represent the set of all vertices of the standard $n$-simplex $\Delta^n$ and use $\mathbb{Z}_{k+1}$ to represent the set of all vertices of $K$. Analogous to the boundary map in singular homology, there is a boundary map $$\partial_n: \mathbb{Z} K_n\longrightarrow \mathbb{Z}K_{n-1}.$$ We have that $\{K_n,\partial_n\}_{n\geq 0}$ is a simplicial complex.

Question: Whether is the geometric realization of the simplicial complex $\{K_n,\partial_n\}_{n\geq 0}$ (weak) homotopy equivalent to (or having same homology with) $|K|$?


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# Confusion related to a convex optimization problem

I was going through Stephen Boyd's lecture on convex optimization. However, I am a bit confused about a problem

Given Minimize $f(x) = x_1^2+x_2^2$

subject to $f_1(x) = \frac{x_1}{1+x_2^2} \leq 0$

How come $f_1(x)$ is not convex?

I was going through Stephen Boyd's book related to convex optimization http://www.stanford.edu/~boyd/cvxbook/bv_cvxbook.pdf

Here is the exact screenshot of the page

-

## migrated from stats.stackexchange.comJan 10 '13 at 15:40

This question came from our site for people interested in statistics, machine learning, data analysis, data mining, and data visualization.

I cant understand you question, what do you want to know? Do you want to know if $f_1$ is convex? –  Tomás Jan 10 '13 at 16:01
Closely related: math.stackexchange.com/questions/274837. –  whuber Jan 10 '13 at 16:19
The constraint only says $x_1 \le 0$. So looks like min is $0$ when $(x_1,x_2)=(0,0)$. –  coffeemath Jan 10 '13 at 17:23
Compare $f(1,0)$ with $\frac12\big(f_1(1,1)+f_1(1,-1)\big)$. Or fix $x_1=1$ and check the sign of $\frac{\mathrm d^2}{\mathrm dx_2^2}f(1,x_2)$. –  Rahul Jan 10 '13 at 19:38
@Tomás. Yeah I want to know why $f_1(x)$ is not convex? –  user34790 Jan 10 '13 at 19:43

Look at $f_1$ along any "vertical" line $x_1=c$ where $c\neq0$. For positive $c$, you get a failure of convexity near the $x_1$-axis, and for negative $c$ you get a failure of convexity far from the $x_1$-axis.


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## A community for students. Sign up today

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## manyo 3 years ago Pythagorean Theorem Help!

• This Question is Closed
1. jiberjiber

$a ^{2} + b ^{2} = c ^{2}$

2. manyo

You have a square deck with an area of $17m^{2}$ You would like to double that area. What is the length of one side of the new deck.

3. Gerardo012

a^2 + b^2 = C^2 3x^2 +4x + 7 a=3,b=4,c=7

4. joemath314159

im a little confused. whats the question? just the one about the square? that doesnt really involve pythagorean theorem.

5. manyo

Well, just answer using square.

6. joemath314159

right. you have a square thats going to have an area of 34 square meters (because we doubled the 17 square meters). So you have:$x\cdot x=34 \Longrightarrow x^2 = 34\Longrightarrow x = \sqrt{34}$|dw:1320887833591:dw|

7. manyo

Well, just answer using square.

8. manyo

and the we square the 34, which is gong to be around 5.83m, and that is going to be one length of the square?

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# How can I call the Boost C++ odeint Runge-Kutta integrator for a system of ODEs?

I would like to use Boost C++ odeint Runge-Kutta integrator on a system that looks like this :

$$\ddot x = - \frac A{||x||^3} * x$$

$x$ is a vector in 3D space, so basicaly $x(i, j, k)$
$\ddot x$ is its second derivative
${||x||^3}$ is magnitude cubed of $x$
$A$ is a constant

I know the initial conditions of the problem, namely $\dot x(t=0)$ and $x(t=0)$ .

I have checked this example in the odeint documentation, as well as the full code here. Examples show use of odeint with a single ODE. But my problem would have to be split into 6 ODE.

Can I use odeint Runge Kutta method for such a system ( 6 ODEs ) and if yes, is there any example that I can follow to help me implement my problem ?

## 1 Answer

In theory, sure, although I'd be wary of your particular ODE system for a couple reasons: (1) the right-hand side is undefined when $x = 0$, which will trip up any integrator in the Boost C++ suite, (2) based on its similarity to functions known to be non-Lipschitz in neighborhoods of the origin, even existence of solutions is unclear.

Implementing your system should be straightforward. Transform your system into first-order form, and then model your implementation after the Lorenz system attractor example on the odeint website.

• Thank you for your answer, especially the warnings. I wouldn't worry about x=0 too much, because given what I know about the system I am solving, I believe this state cannot occur, at least not in a reasonable interval of integration that I want to observe. I would like to ask a thing about Lorenz system example. Are dxdt[0], dxdt[1] and dxdt[2] the "equivalent" for the IJK coordinates in my system ? Thank you. – James C Nov 18 '14 at 3:12
• @JamesC: You would introduce an auxiliary variable, say, $y$ and set $\dot{x} = y$, and $\ddot{x} = \dot{y} = -Ax/\|x\|^{3}$. At that point, the abstraction is basically that you have a vector of six state variables $x[i]$, $x[j]$, $x[k]$, $y[i]$, $y[j]$, $y[k]$, so if you want to think of these as being x[0] through x[5], you're representing the same quantities, just relabeled so that they're in a form convenient for use in an ODE solver. – Geoff Oxberry Nov 18 '14 at 3:17


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# Integration using parametric derivatives

In mathematics, integration by parametric derivatives is a method of integrating certain functions.

For example, suppose we want to find the integral

${\displaystyle \int _{0}^{\infty }x^{2}e^{-3x}\,dx.}$

Since this is a product of two functions that are simple to integrate separately, repeated integration by parts is certainly one way to evaluate it. However, we may also evaluate this by starting with a simpler integral and an added parameter, which in this case is t = 3:

{\displaystyle {\begin{aligned}&\int _{0}^{\infty }e^{-tx}\,dx=\left[{\frac {e^{-tx}}{-t}}\right]_{0}^{\infty }=\left(\lim _{x\to \infty }{\frac {e^{-tx}}{-t}}\right)-\left({\frac {e^{-t0}}{-t}}\right)\\&=0-\left({\frac {1}{-t}}\right)={\frac {1}{t}}.\end{aligned}}}

This converges only for t > 0, which is true of the desired integral. Now that we know

${\displaystyle \int _{0}^{\infty }e^{-tx}\,dx={\frac {1}{t}},}$

we can differentiate both sides twice with respect to t (not x) in order to add the factor of x2 in the original integral.

{\displaystyle {\begin{aligned}&{\frac {d^{2}}{dt^{2}}}\int _{0}^{\infty }e^{-tx}\,dx={\frac {d^{2}}{dt^{2}}}{\frac {1}{t}}\\[10pt]&\int _{0}^{\infty }{\frac {d^{2}}{dt^{2}}}e^{-tx}\,dx={\frac {d^{2}}{dt^{2}}}{\frac {1}{t}}\\[10pt]&\int _{0}^{\infty }{\frac {d}{dt}}\left(-xe^{-tx}\right)\,dx={\frac {d}{dt}}\left(-{\frac {1}{t^{2}}}\right)\\[10pt]&\int _{0}^{\infty }x^{2}e^{-tx}\,dx={\frac {2}{t^{3}}}.\end{aligned}}}

This is the same form as the desired integral, where t = 3. Substituting that into the above equation gives the value:

${\displaystyle \int _{0}^{\infty }x^{2}e^{-3x}\,dx={\frac {2}{3^{3}}}={\frac {2}{27}}.}$


text:
pymor.reductors.era¶

Module Contents¶

class pymor.reductors.era.ERAReductor(data, sampling_time, force_stability=True, feedthrough=None)[source]

Eigensystem Realization Algorithm reductor.

Constructs a (reduced) realization from a sequence of Markov parameters $$h_i$$, for $$i\in\{1,\,\dots,\,2s-1\}$$, $$s\in\mathbb{N}$$, by a (reduced) orthogonal factorization of the Hankel matrix of Markov parameters

$\begin{split}H = \begin{bmatrix} h_1 & h_2 & \dots & h_s \\ h_2 & h_3 & \dots & h_{s+1}\\ \vdots & \vdots & \ddots & \vdots\\ h_s & h_{s+1} & \dots & h_{2s-1} \end{bmatrix}=U\Sigma V^T\in\mathbb{R}^{ps\times ms},\end{split}$

where $$r\leq\min\{ms,ps\}$$ is the reduced order. See [Kun78].

In order for the identified model to be stable, the Markov parameters decay substantially within $$s$$ samples. Stability is enforced automatically through zero-padding and can be deactivated by setting force_stability=False.

For a large number of inputs and/or outputs, the factorization of the Hankel matrix can be accelerated by tangentially projecting the Markov parameters to reduce the dimension of the Hankel matrix, i.e.

$\hat{h}_i = W_L^T h_i W_R,$

where $$n_L \leq p$$ and $$n_R \leq m$$ are the number of left and right tangential directions and $$W_L \in \mathbb{R}^{p \times n_L}$$ an $$W_R \in \mathbb{R}^{m \times n_R}$$ are the left and right projectors, respectively. See [KG16].

data[source]

NumPy array that contains the first $$n$$ Markov parameters of an LTI system. Has to be one- or three-dimensional with either:

data.shape == (n,)

for scalar-valued Markov parameters or:

data.shape == (n, p, m)

for matrix-valued Markov parameters of dimension $$p\times m$$, where $$m$$ is the number of inputs and $$p$$ is the number of outputs of the system.

sampling_time[source]

A number that denotes the sampling time of the system (in seconds).

force_stability[source]

Whether the Markov parameters are zero-padded to double the length in order to enforce Kung’s stability assumption. See [Kun78]. Defaults to True.

feedthrough[source]

(Optional) Operator or NumPy array of shape (p, m). The zeroth Markov parameter that defines the feedthrough of the realization. Defaults to None.

Methods

 error_bounds Compute the error bounds for all possible reduction orders. input_projector Construct the right/input projector $$W_2$$. output_projector Construct the left/output projector $$W_1$$. reduce Construct a minimal realization.
cache_region = memory[source]
error_bounds(num_left=None, num_right=None)[source]

Compute the error bounds for all possible reduction orders.

Without tangential projection of the Markov parameters, the $$\mathcal{L}_2$$-error of the Markov parameters $$\epsilon$$ is bounded by

$\epsilon^2 = \sum_{i = 1}^{2 s - 1} \lVert C_r A_r^{i - 1} B_r - h_i \rVert_F^2 \leq \sigma_{r + 1}(\mathcal{H}) \sqrt{r + p + m},$

where $$(A_r,B_r,C_r)$$ is the reduced realization of order $$r$$, $$h_i\in\mathbb{R}^{p\times m}$$ is the $$i$$-th Markov parameter and $$\sigma_{r+1}(\mathcal{H})$$ is the first neglected singular value of the Hankel matrix of Markov parameters.

With tangential projection, the bound is given by

$\epsilon^2 = \sum_{i = 1}^{2 s - 1} \lVert C_r A_r^{i - 1} B_r - h_i \rVert_F^2 \leq 4 \left( \sum_{i = n_L + 1}^p \sigma_i^2(\Theta_L) + \sum_{i = n_R + 1}^m \sigma_i^2(\Theta_R) \right) + 2 \sigma_{r + 1}(\mathcal{H}) \sqrt{r + n_L + n_R},$

where $$\Theta_L,\,\Theta_R$$ is the matrix of horizontally or vertically stacked Markov parameters, respectively. See [KG16] (Thm. 3.4) for details.

input_projector(num_right)[source]

Construct the right/input projector $$W_2$$.

output_projector(num_left)[source]

Construct the left/output projector $$W_1$$.

reduce(r=None, tol=None, num_left=None, num_right=None)[source]

Construct a minimal realization.

Parameters

r

Order of the reduced model if tol is None, maximum order if tol is specified.

tol

Tolerance for the error bound.

num_left

Number of left (output) directions for tangential projection.

num_right

Number of right (input) directions for tangential projection.

Returns

rom

Reduced-order LTIModel.


text:
A Valley Without Wind 2
A Valley Without Wind 2 > Discussioni generali > Dettagli della discussione
BA|Razorflamelight 7 lug 2013, ore 18:57
How you get by here?
Visit this screenshot: http://cloud-2.steampowered.com/ugc/630785219621727373/944A3E62B1F55920A423A2C04AF4464CF535F41B/

Please let me know how you get by here, because there is a perk token there that I want to get :(

Thanks!
Visualizzazione di 1-4 commenti su 4
tigersfan  [sviluppatore] 8 lug 2013, ore 3:03
That looks like a bug, I'm sorry, I'm not sure you can get there. :(
BA|Razorflamelight 8 lug 2013, ore 12:01
Nope, this doesn't look like a bug. In fact, I think this is something that requires the second feat of the game, Minimize/Shrink in order to get by.
Ultima modifica da BA|Razorflamelight; 9 lug 2013, ore 13:43
Jokaabel 3 set 2013, ore 14:04
Messaggio originale di BA|Razorflamelight:
Nope, this doesn't look like a bug. In fact, I think this is something that requires the second feat of the game, Minimize/Shrink in order to get by.

Hi Razorflamelight

The shrink feat is probably your ticket for this issue.

hope this helps
Ultima modifica da Jokaabel; 3 set 2013, ore 14:05
BA|Razorflamelight 3 set 2013, ore 15:19
Thanks. That's what I thought.
Visualizzazione di 1-4 commenti su 4
Per pagina: 15 30 50
Data di pubblicazione: 7 lug 2013, ore 18:57
Messaggi: 4
Altre discussioni


text:
Article

# Moments of charge fluctuations, pseudo-critical temperatures and freeze-out in heavy ion collisions

(Impact Factor: 2.84). 08/2011; 38(12). DOI: 10.1088/0954-3899/38/12/124098
Source: arXiv

ABSTRACT We discuss universal properties of higher order cumulants of net baryon
number fluctuations and point out their relevance for the analysis of
freeze-out and critical conditions in heavy ion collisions at LHC and RHIC.

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##### Article: Hadron Formation in Relativistic Nuclear Collisions and the QCD Phase Diagram
[Hide abstract]
ABSTRACT: We analyze hadrochemical freeze-out in central Pb+Pb collisions at CERN SPS and LHC energies. We determine the effects of baryon and antibaryon annihilation and/or regeneration occurring during the final cascade expansion stage of the collisions by deriving "survival factors" for each hadronic species and all energies considered by employing the UrQMD hybrid model. These survival factors are shown to resemble the pattern of data deviation from the statistical equilibrium calculations with the statistical model. We apply them in the SM data analysis, obtaining new points in the $T-\mu_B$ plane which follow closely the parton-hadron phase boundary predicted by lattice QCD.
Physical Review Letters 12/2012; 111(8). DOI:10.1103/PhysRevLett.111.082302 · 7.73 Impact Factor
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##### Article: Equilibrium Statistical-Thermal Models in High-Energy Physics
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ABSTRACT: We review some recent highlights from the applications of statistical-thermal models to different experimental measurements and lattice QCD thermodynamics, that have been made during the last decade. We start with a short review of the historical milestones on the path of constructing statistical-thermal models for heavy-ion physics. We discovered that Heinz Koppe formulated in 1948 an almost complete recipe for the statistical-thermal models. In 1950, Enrico Fermi generalized this statistical approach, in which he started with a general cross-section formula and inserted into it simplifying assumptions about the matrix element of the interaction process that likely reflects many features of the high-energy reactions dominated by density in the phase space of final states. In 1964, Hagedorn systematically analysed the high-energy phenomena using all tools of statistical physics and introduced the concept of limiting temperature based on the statistical bootstrap model. It turns to be quite often that many-particle systems can be studied with the help of statistical-thermal methods. The analysis of yield multiplicities in high-energy collisions gives an overwhelming evidence for the chemical equilibrium in the final state. The strange particles might be an exception, as they are suppressed at lower beam energies. However, their relative yields fulfill statistical equilibrium, as well. We review the equilibrium statistical-thermal models for particle production, fluctuations and collective flow in heavy-ion experiments. We also review their reproduction of the lattice QCD thermodynamics at vanishing and finite chemical potential. During the last decade, five conditions have been suggested to describe the universal behavior of the chemical freeze out parameters.
International Journal of Modern Physics A 10/2014; 17(17). DOI:10.1142/S0217751X1430021X · 1.09 Impact Factor
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##### Article: Chemical Freeze-Out and Higher Order Multiplicity Moments
[Hide abstract]
ABSTRACT: We calculate the non-normalized moments of the particle multiplicity within the framework of the hadron resonance gas (HRG) model. At finite chemical potential $\mu$, a non-monotonic behavior is observed in the thermal evolution of third order moment (skewness $S$) and the higher order ones as well. Among others, this observation likely reflects dynamical fluctuations and strong correlations. The signatures of non-monotonicity in the normalized fourth order moment (kurtosis $\kappa$) and its products get very clear. Based on these findings, we introduce a novel condition characterizing the universal freeze-out curve. The chemical freeze-out parameters $T$ and $\mu$ are described by vanishing $\kappa\, \sigma^2$ or equivalently $m_4=3\,\chi^2$, where $\sigma$, $\chi$ and $m_4$ are the standard deviation, susceptibility and fourth order moment, respectively. The fact that the HRG model is not able to release information about criticality related to the confinement and chiral dynamics should not veil the observations related to the chemical freeze-out. Recent lattice QCD studies strongly advocate the main conclusion of the present paper.
Nuclear Physics A 06/2013; 922. DOI:10.1016/j.nuclphysa.2013.12.008 · 2.50 Impact Factor


text:
### 3249

Brain extraction and segmentation framework for bias field rich cranial MRI scans of rats
Jacob Daniel Kirstejn Hansen1, François Lauze1, Sune Darkner1, Kristian Nygaard Mortensen2, Simon Sanggaard2, Helene Benveniste3, and Maiken Nedergaard2,4

1Department of Computer Science, University of Copenhagen, Copenhagen, Denmark, 2Center for Translational Neuromedicine, University of Copenhagen, Copenhagen, Denmark, 3Anesthesiology, Yale School of Medicine, Yale University, New Haven, CT, United States, 4Center for Translational Neuromedicine, University of Rochester, Rochester, NY, United States

### Synopsis

This abstract presents a framework to extract brain tissue and internal Cerebrospinal fluid in cranial magnetic resonance imaging of rats with strong bias fields. Desired segments are obtained through bias field correction and several passes of segmentation. A refinement procedure is proposed to remove brain surface CSF. Promising planar and 3D visualizations of results are presented and demonstrate the capabilities of the framework.

### Introduction

Magnetic resonance imaging employing surface receive coils often faces the problem of strong image bias fields in data that makes relevant tissue extractions very challenging, as bias field correction algorithms cannot fully correct for it. We present a framework for the extraction of rat brain tissue and the cerebrospinal fluid (CSF) networks, in the case where MR volumes are affected by strong bias fields. We demonstrate that our framework obtains very satisfactory segmentation results.

Animals

Unaffected male rats of the Wistar Kyoto stain (WKY) and spontaneously hypertensive rats (SHR) were obtained from Charles River, Germany. Separate groups of rats were scanned at two age-ranges: young (7-9 weeks old; WKY7, n = 11; and SHR7, n = 8) and young adults (19-21 weeks WKY19, n = 8; and SHR19, n = 9). All treatments and imaging were performed according to protocols approved by the IACUC, and according to a protocol approved by the University of Copenhagen animal experimentation committee.

MRI

All MRI investigations was conducted on a 9.4 T magnet (Bruker Biospec 9.4/30 USR) interfaced to a Bruker Advance III console and controlled by Paravision 5.1 software (Bruker BioSpin) at the Panum Institute, University of Copenhagen. Imaging was performed with an 86 mm3 volume resonator and a surface quadrature array receiver coil. Images were acquired with the spoiled gradient FLASH3D sequence (TE: 4 ms, TR: 15 ms, NA: 3; matrix: 128 x 128 x 128, voxel size: 0.24 x 0.24 x 0.26 mm, FA: 15°, scan time: 4:05 min).

### Methods

The proposed framework consists of 4 major steps and is illustrated in Figure 1. Each step is explained below.

Bias field correction: A standard bias field correction method[] is applied. This alone does not completely remove the bias and standard segmentation methods fail to deal with it.

2-class segmentation: The local-means method proposed in [2, 3] is used and seeks to minimize the functional

$$\mathcal{E}_Q(\mathbf{c},\mathbf{v})=\frac{1}{2}\sum_{i=1}^{n}\int_{\Omega}g*\left[u-c_i(\mathbf{x}))^2\mathbf{v}_i\right]d\mathbf{x}+\frac{\mu}{2}\sum_{i=1}^{n}\int_{\Omega}|D\mathbf{v}_i|^2$$

where $\mathbf{v}_i\in[0,1]$ is the label field describing the probability of belonging to class i. $u$ is the bias corrected volume, $g$ a moving average kernel and

$$c_i(\mathbf{x})=\frac{(u\mathbf{v}_i)*g(\mathbf{x})}{\mathbf{v}_i*g(\mathbf{x})}$$

This method produces a clear contour of the brain tissue, with strong edges, that makes rough brain extractions possible.

GAC brain extraction: A full brain mask is retrieved from the skull, by initializing a Geodesic Active Contour (GAC) based model[4, 5] in the center of the segmented brain. Any ’holes’ that would occur in the resulting brain mask are filled, using mathematical morphology operations.

3-class segmentation: The retrieved brain mask is used to extract the brain tissue segment of the binary segmentation result and is in turn segmented by a tighter 3-class local-means segmentation. The three classes are background, CSF, and brain tissue. A single connected brain tissue component is expected, so only the largest is kept.

To refine the segmentation of the internal CSF network, we process as follows. A run of the GAC method on the final brain tissue class provides a more accurate brain mask. It is used to extract the internal CSF network from the final CSF class, removing surface CSF. Small objects are removed in the CSF class to highlight larger networks in the brain.

### Results

The framework is run on a sample rat cranium volume affected by significant bias. In Figure 2 the result of step 2 and 4 are shown for different brain slices.

Figure 3 presents a 3D surface view of the combined brain tissue matter (pink) and CSF (blue) for different viewing angles. In Figure 4 and 5 respectively, the 3D surfaces of the isolated brain tissue and CSF network can be seen.

### Discussion and conclusion

A framework that satisfactorily segments and extracts brain tissue and internal CSF networks in rat brain MRI scans affected by strong image bias fields has been proposed. The framework combines bias field correction with several passes of segmentation to adequately extract the relevant classes. 3D surface results for brain tissue and the major CSF network as well as axial, sagittal, and coronal planes have been presented.

To further improve on the results, a distance based feature could be incorporated, as the severity of the bias field is correlated with the distance to the coils. Additionally, strong priors on the connectivity of the CSF network will be integrated in the model. While the framework introduces more parameters, we believe the improvements in extracting high quality tissues from medical imaging affected by strong bias fields are highly beneficial.

### Acknowledgements

No acknowledgement found.

### References

[1] Tustison, N.J., Avants, B.B., Cook, P.A., Zheng, Y., Egan, A., Yushkevich, P.A. and Gee, J.C. N4ITK: improved N3 bias correction. in IEEE transactions on medical imaging (2010), pp.1310-1320.

[2] Hansen, J. D. K., and Lauze, F. Local Mean Multiphase Segmentation with HMMF Models. In International Conference on Scale Space and Variational Methods in Computer Vision (2017), Springer, pp. 396–407.

[3] Brox, T., and Cremers, D. On Local Regions Models and a Statistical Interpretation of the Piecewise Smooth Mumford-Shah Functional. Int’l J. Computer Vision 84 (2009), 184–193.

[4] Álvarez, L., Baumela, L., Henríquez, P., and Márquez-Neila, P. Morphological snakes. In Computer Vision and Pattern Recognition (CVPR), 2010 IEEE Conference on (2010), IEEE, pp. 2197–2202.

[5] Márquez-Neila, P., Baumela, L., and Alvarez, L. A morphological approach to curvature-based evolution of curves and surfaces. IEEE Trans-actions on Pattern Analysis and Machine Intelligence 36, 1 (2014), 2–17.

### Figures

The proposed framework, highlighting each step as a box inside the dotted blue box. Arrows indicate that the output of the origin box is used as input in the destination box.

The rows highlight: The original image, showing a severe bias field; the bias field corrected image, which still has quite a bit of bias left; The final segmentation and extraction of the example MRI using the proposed framework. Each column provides the axial, sagittal, and coronal planes for the same brain. Brain tissue is light gray while CSF is is dark gray.

The columns show different viewing angles of the brain by rotating 180 degrees around the corresponding axis of the resulting brain. The rows from top to bottom show the x, y, and z direction for the same brain segmentation. Brain tissue is pink, while CSF is light blue.

The columns from left to right show the negative and positive direction of the brain tissue (pink) segmentation, resulting from step 4. The rows from top to bottom show the x, y, and z direction for the same brain segmentation.

Each column shows two different angles of the extracted major CSF network (light blue) segmentation, from running our framework. The rows from top to bottom presents: The raw CSF segmentation; The internal CSF network after removing surface CSF; The internal CSF network after having removed small CSF objects. When rotating the volume in a 3D visualizer the network structure and symmetry is more apparent.

Proc. Intl. Soc. Mag. Reson. Med. 26 (2018)
3249


text:
COVID-19 information for PI Residents and Visitors

# A Gleason-type theorem for qubits based on mixtures of projective measurements

We derive Born’s rule and the density-operator formalism for quantum systems with Hilbert spaces of finite dimension. Our extension of Gleason’s theorem only relies upon the consistent assignment of probabilities to the outcomes of projective measurements and their classical mixtures. This assumption is significantly weaker than those required for existing Gleason-type theorems valid in dimension two.

COVID-19 information for PI Residents and Visitors

Collection/Series:
Event Type:
Seminar
Scientific Area(s):
Event Date:
Mardi, Mai 14, 2019 - 15:30 to 17:00
Location:
Sky Room
Room #:
394


text:
# Passing a sampled signal through a filter

I was wondering why is it wrong to use a band-pass filter on a sampled signal?

If the signal we want to sample has frequencies up to fmax, we sample it with frequency fs = 2fmax (so that Nyquist theorem condition is fulfilled), and the spectrum of the sampled signal will have scaled "copies" of the original spectrum centered around frequencies k * fs (k = +-1, +-2 ...).

What would happen if we filtered that sampled signal with a band-pass filter [fs-fmax, fs+fmax], and why is it wrong? Wouldn't we end up with the signal we started with?

• how would you construct such a filter? analog? digital? ideal filters? – user28715 Oct 12 '19 at 15:00
• Ideal analog filters, I forgot to add that. I meant what would happen purely theoretically. – Dan D. Walsh Oct 12 '19 at 16:50
• To fulfill Nyquist in finite time, you need to sample above 2*Fmax. Nyquist at 2*F only applies to sampled signals longer than the existence of the universe. – hotpaw2 Oct 12 '19 at 17:23

The Nyquist criteria simply requires you to have "two samples per Hz of bandwidth". It doesn't have to be $$[-f_{max},-f_{max}]$$, it can be any frequency range that includes at least $$2 \cdot f_{max}$$ of bandwidth. However, for real signals you need to figure out what to do with the negative frequencies.


text:
# electrochemistry previous year questions jee advanced

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CONFIRMA TER 18 ANOS OU MAIS? ATENÇÃO! ESTA PÁGINA CONTÉM CONTEÚDO INAPROPRIADO PARA MENORES DE 18 ANOS


text:
1. Feb 7, 2017

### issacnewton

1. The problem statement, all variables and given/known data
Consider the function $f(x) = x^4 \sin(\frac 1 x)$ for $x \ne 0$ and $f(x) = 0$ for $x =0$. I have to prove that $x=0$ is the critical number of this function and its derivative changes the sign indefinitely.

2. Relevant equations
Definition of the critical number

3. The attempt at a solution
$x=0$ would be the critical number if $f'(0) = 0$ or if $f'(0)$ does not exist. I have been able to show that $f'(0) = 0$ using the squeeze theorem. Now I want to show that the derivative of $f(x)$ changes sign indefinitely. I am totally stuck at this point. How would I progress from here ?

Thanks $\smallsmile$

2. Feb 7, 2017

### Staff: Mentor

As x approaches 0 from the right, look at what happens for $x = \frac 2 {\pi}, \frac 2 {3\pi}, \frac 2 {5\pi}, \dots$. What are the values of 1/x at these numbers? What are the values of $\sin(1/x)$ at these numbers?.

3. Feb 8, 2017

### issacnewton

Thanks Mark for replying. I see that the sign will be altering between positive and negative as $x$ approaches $0$ from the the right at these values. So do I just write this, or is there more formal way of writing it as a proof ?

4. Feb 8, 2017

### Staff: Mentor

No, don't write just what I said, as I was trying to steer you in the right direction by asking some questions, and showing a sequence $\{x_n \}$ whose limit is zero, and for which the sequence $\{\sin(\frac 1 {x_n}) \}$ oscillates. Have you written up something that captures this idea?

Note that a similar idea can be used to show that the sequence $\{\sin(\frac 1 {x_n}) \}$ has the same behavior as x approaches zero from the left.

5. Feb 12, 2017

### issacnewton

Ok, So I have to prove an existence of a sequence in $\mathbb{R}$, such that this sequence goes to zero as $n \rightarrow \infty$ but the sequence $\{\sin(\frac{1}{x_n})\}$ oscillates. You have already come up with the sequence. So I need to prove that $\forall~n \in \mathbb{N}~\exists k, p \in \mathbb{N}$ such that $\sin(\frac{1}{x_k}) > 0$ and $\sin(\frac{1}{x_p}) < 0$. Would this be correct ?

6. Feb 12, 2017

### Staff: Mentor

As you note, you don't have to "prove" the existence of a sequence -- I gave you one that does the trick.This sequence contains one subsequence for which $\sin(x_{n_1}) > 0$ and another subsequence for which $\sin(x_{n_2}) < 0$. IMO, you should focus first on figuring this out, and less on the fancy "mathy" formatting, such as this stuff -- $\forall~n \in \mathbb{N}~\exists k, p \in \mathbb{N}$

7. Feb 12, 2017

### issacnewton

So if $x_n = \frac{2}{\pi(2n-1)}$ is the sequence given by you, then for the subsequence $x_{n_1} = \frac{2}{\pi(4n_1 - 3)}$, we have $\sin(x_{n_1}) > 0$ and for the subsequence $x_{n_2} = \frac{2}{\pi(4n_2 - 1)}$, we have $\sin(x_{n_2}) < 0$

8. Feb 12, 2017

### Staff: Mentor

Yes, those are the subsequences I was thinking of. The whole sequence should demonstrate that f ' changes sign infinitely often as x approaches zero. "Infinitely often" as a better way to say it than "changes signs indefinitely."

9. Feb 13, 2017

### issacnewton

Ok.. so I will complete the entire proof when I get the time as I think whole process is with me. But this problem is from James Stewart's Calculus (8 ed.) and this is just a computational book. So I am surprised that they have this problem. When we start introducing subsequences, we are using machinery of real analysis.

10. Feb 13, 2017

### Staff: Mentor

Maybe you are misinterpreting what the problem is asking for. If it says "show ..." rather than "prove ..." a less rigorous explanation is called for. Can you post a picture of the problem statement?

11. Feb 13, 2017

### issacnewton

Ok, I am putting snapshot of the problem.

#### Attached Files:

• ###### Capture.PNG
File size:
15.7 KB
Views:
21
12. Feb 21, 2017

### issacnewton

So I will present the proof here. Consider a sequence $x_n = \frac{2}{\pi(2n-1)}$ where $n\in\mathbb{N}$. Then we have a subsequence $x_{n_1} = \frac{2}{\pi(4n_1 - 3)}$, such that $f'(x_{n_1}) = \frac{32}{\pi^3(4n_1-3)^3} >0$ for all $n_1 \in \mathbb{N}$ and we have another subsequence $x_{n_2} = \frac{2}{\pi(4n_2 - 1)}$ such that $f'(x_{n_2}) = \frac{-32}{\pi^3(4n_2-1)^3} <0$ for all $n_2 \in \mathbb{N}$. So this proves that, the derivative changes sign infinitely often. I hope this is correct.

13. Feb 21, 2017

### issacnewton

I also have to prove that $f$ has neither a local maximum nor a local minimum at $0$. Here I present my proof. Suppose $f$ has a local maxima at $x=0$. Then we have some open interval $(a,b)$ containing $x=0$ such that $f(x)\leqslant f(0)$, for all $x \in (a,b)$. This implies $f(x) \leqslant 0$, for all $x \in (a,b)$. Now $a<0<b$. By Archimedean Property, there exists $n_1 \in \mathbb{N}$ such that $\frac 1 n_1 < \frac{\pi b}{2}$. But we have $4n_1-3 \geqslant n_1 >0$, which implies that $\frac{1}{4n_1-3} < \frac{\pi b}{2}$. Hence we have $0< \frac{2}{\pi(4n_1-3)} < b$. So $\frac{2}{\pi(4n_1-3)} \in (a,b)$. This leads us to $f\left(\frac{2}{\pi(4n_1-3)}\right) \leqslant 0$. But $\forall n_1 \in \mathbb{N}$, we have $f\left(\frac{2}{\pi(4n_1-3)}\right) = \frac{16}{\pi^4(4n_1-3)^4} > 0$. So we reach a contradiction. Hence $f$ doesn't have a local maxima at $x=0$. Now suppose that $f$ has a local minima at $x=0$. Again, there exists an open interval $(a,b)$ containing $x=0$ such that $f(0)\leqslant f(x)$, for all $x \in (a,b)$. Since we have $a<0<b$, we can come up with some $n_2 \in \mathbb{N}$, due to Archimedean property, such that $\frac 1 n_2 < \frac{\pi b}{2}$. But $(4n_2-1) > n_2 > 0$, so we get $\frac{1}{4n_2-1} < \frac{\pi b}{2}$. So $0< \frac{2}{\pi(4n_2-1)} < b$. This means that $\frac{2}{\pi(4n_2-1)} \in (a,b)$. Because of our assumption, this means that $f\left(\frac{2}{\pi(4n_2-1)}\right) \geqslant 0$. But $\forall n_2 \in \mathbb{N}$, we have that $f\left(\frac{2}{\pi(4n_2-1)}\right) = \frac{-16}{\pi^4(4n_2-1)^4} < 0$. So again we reach a contradiction, which means that $f$ can not have a local minima at $x=0$. I hope my proof is correct.


text:
# zbMATH — the first resource for mathematics

On decision problems for probabilistic Büchi automata. (English) Zbl 1139.68030
Amadio, Roberto (ed.), Foundations of software science and computational structures. 11th international conference, FOSSACS 2008, held as part of the joint European conferences on theory and practice of software, ETAPS 2008, Budapest, Hungary, March 29–April 6, 2008. Proceedings. Berlin: Springer (ISBN 978-3-540-78497-5/pbk). Lecture Notes in Computer Science 4962, 287-301 (2008).
Summary: Probabilistic Büchi automata (PBA) are finite-state acceptors for infinite words where all choices are resolved by fixed distributions and where the accepted language is defined by the requirement that the measure of the accepting runs is positive. The main contribution of this paper is a complementation operator for PBA and a discussion on several algorithmic problems for PBA. All interesting problems, such as checking emptiness or equivalence for PBA or checking whether a finite transition system satisfies a PBA-specification, turn out to be undecidable. An important consequence of these results are several undecidability results for stochastic games with incomplete information, modelled by partially observable Markov decision processes and $$\omega$$-regular winning objectives. Furthermore, we discuss an alternative semantics for PBA where it is required that almost all runs for an accepted word are accepting, which turns out to be less powerful, but has a decidable emptiness problem.
For the entire collection see [Zbl 1133.68001].

##### MSC:
 68Q45 Formal languages and automata 03D35 Undecidability and degrees of sets of sentences
POMDP
Full Text:


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# The operation delta is defined for all nonzero x and y b x

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The operation delta is defined for all nonzero x and y b x  [#permalink]

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The operation # is defined for all nonzero x and y by x#y = x + x/y. If a>0, then 1#(1#a) =

A. a
B. a+1
C. a/(a+1)
D. (a+2)/(a+1)
E. (2a+1)/(a+1)

OPEN DISCUSSION OF THIS QUESTION IS HERE: the-operation-is-defined-for-all-nonzero-x-and-y-by-x-y-162019.html

Originally posted by nfa1rhp on 29 Jul 2007, 11:56.
Last edited by Bunuel on 03 Jan 2014, 05:03, edited 1 time in total.
Renamed the topic, edited the question, added the OA and moved to PS forum.
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Posts: 1786

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29 Jul 2007, 13:53
(E) for me

o 1 delta a
= 1 + 1/a
= (a + 1)/a

o 1 delta (1 delta a)
= 1 delta ((a + 1)/a)
= 1 + 1/((a + 1)/a)
= 1 + a/(a+1)
= (2*a + 1) / (a + 1)
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Posts: 4939
Location: Singapore

### Show Tags

30 Jul 2007, 23:39
1 delta a = 1 + 1/a
Let 1 delta a be x

Then 1 delta (1 delta a) = 1 delta x = 1 + 1/x. Subsitute in x, to give

1 + 1/(1+1/a) = 1 + a/(a+1) = a+1+a/(a+1) = 2a+1/(a+1)
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Re: The operation delta is defined for all nonzero x and y b x  [#permalink]

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02 Jan 2014, 08:26
Can someone explain the math better on this one? Brunnel, do you mind?
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Posts: 47923
Re: The operation delta is defined for all nonzero x and y b x  [#permalink]

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03 Jan 2014, 05:04
jeremyl3763 wrote:
Can someone explain the math better on this one? Brunnel, do you mind?

The operation # is defined for all nonzero x and y by x#y = x + x/y. If a>0, then 1#(1#a) =
A. a
B. a+1
C. a/(a+1)
D. (a+2)/(a+1)
E. (2a+1)/(a+1)

$$1#(1#a) = 1#(1+\frac{1}{a}) = 1#(\frac{a+1}{a})= 1+ \frac{1}{(\frac{a+1}{a})}=1+\frac{a}{a+1}=\frac{a+1+a}{a+1}=\frac{2a+1}{a+1}$$.

OPEN DISCUSSION OF THIS QUESTION IS HERE: the-operation-is-defined-for-all-nonzero-x-and-y-by-x-y-162019.html
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Re: The operation delta is defined for all nonzero x and y b x  [#permalink]

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03 Jan 2014, 05:04
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Re: The operation delta is defined for all nonzero x and y b x  [#permalink]

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04 Aug 2018, 05:39
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# A Number Theory Problem by Muhammad Rasel Parvej

$\dfrac{1}{1-2^{2}}+\dfrac{1}{1-4^{2}}+\dfrac{1}{1-6^{2}}+\dfrac{1}{1-8^{2}}+\cdots+\dfrac{1}{1-1000^{2}}\$

The value of the expression above can be expressed as $$\dfrac{a}{b},$$ where $$a$$ and $$b$$ are coprime integers.

What is the number of ways (equivalently, the number of ordered pair of integers $$(a,b)$$ ) the value of the expression above can be expressed as $$\dfrac{a}{b},$$ where $$a$$ and $$b$$ are coprime integers.

Hint: Is it at all needed to compute the sum?

×


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Researchers from 25 institutions around the country, including PNNL, are working to find out how exercise changes the molecular makeup of our cells to generate health benefits.
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A new study using proteogenomics to compare cancerous tissue with normal fallopian tube samples advances insights about the molecular machinery that underlies ovarian cancer.
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Researchers at PNNL have come up with a novel way to use silicon as an energy storage ingredient, replacing the graphite in electrodes. Silicon can hold 10 times the electrical charge per gram, but it comes with problems of its own.
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A PNNL scientist is studying the structures of the proteins on the surface of the novel coronavirus, using NMR spectroscopy to reveal information about the molecular toolkit that holds the keys to a vaccine or treatment.


text:
# Electronic – Calculate resistance of coil/inductor made of PCB traces

copperresistancetrace

This is to be a NFC coil/inductor of 1.1uH according do ST's eDesignSuite tool, with 7 turns.

The drawing above follows these parameters:

ST's tool give inductance value as 1.1uH, but also I need to know the resistance of entire track, in ohms.
I have measured the antenna total lenght on the layout software and it is 418 mm long.

So I have these parameters:

• 0.25mm width
• 418mm length
• 0.5 oz (17.5 micrometers) copper thickness

Then I entered these value of this calculator: https://www.allaboutcircuits.com/tools/trace-resistance-calculator/

And the result given by the site is 1.66 Ohms. Can I trust this value? I need the know correct value of resistance to calculate the components of RF portion of NFC filter and matching circuits.

The board will have 0.5 oz copper thickness, I can guarantee this.

Resistivity of Annealed copper: 1.72x10-8 Ωm

Resistivity is defined as:

$$\ \rho = R\cdot \frac{A}{l}\$$

therefore $$\ R = \frac{\rho \cdot l}{A} \$$

your length is 0.418m your area is $$\0.25\,\mathrm{mm}\times 17.5\,\mu \mathrm{m}\$$

therefore your resistance = $$\ \frac{ 1.72\times10^{-8} \,\Omega \mathrm{m} \,\times\, 0.418\, \mathrm{m}}{ 0.25\,\mathrm{mm} \,\times\, 17.5\,\mu \mathrm{m}} = 1.64\,\Omega\$$ at 20C


text:
# Leaving out the indeterminate in an integral

Let $f,g : \mathbb{R} \to \mathbb{R}$. Is it acceptable to write $\int_a^b fg$ instead of $\int_a^b f(x)g(x)\,dx$ (i.e., would it throw others off while reading it)? The (lack of an) indeterminate is unambiguous since the function $fg$ only accepts one argument, but I've never seen others write it like that.

-
I have seen it all too often, from students who then can't handle substitution. –  André Nicolas Nov 27 '12 at 6:49
@AndréNicolas: Can you explain? I don't see the problem with substitution: $\int_a^b (f' \circ g) g' = \int_{g(a)}^{g(b)} f$. –  Snowball Nov 27 '12 at 7:15
I believe you can handle it. Was observing from grading too many calculus final exams that there is significant correlation between leaving out $dx$ and getting the wrong answer for $\int \frac{dx}{1+3x}$. –  André Nicolas Nov 27 '12 at 7:20

$$\int_0^{\pi}fg\sin\theta\,\mathrm d\theta=\int_{-1}^1fg\,\mathrm d\cos\theta\;.$$


text:
# How to use a wetmap to distribute particles (how to distribute plants where moisture collects)

This is a follow-up to this question - Geometry nodes: How to distribute points on valleys only?

One of the approaches considered was using particles to simulate rain, then using the wet map from dynamic paint to calculate where water flows and accumulates in a landscape. The animation below shows the surface with wet and dry areas.

I'd like to use the wet maps to distribute particles or apply modifiers that rely on vertex weights. Additionally, I'd like to be able to do this as part of an animation.

Is there a way to move the wet map to a vertex map as part of an animation? A preferred solution would not involve python.

The file used to generate this image is here -

# Particle Distribution

For the particle distribution, you can bake the Dynamic Paint (the wetmap to be precise) to an image sequence. The landscape object must be UV unwrapped for this to work.

Baking Step

• create the UV map for the landscape object. It will take a minute or two because your mountain landscape has many vertices.

• in the Dynamic Paint settings of the canvas (landscape object) choose Format: Image Sequence and a higher Resolution, for example, 512.

• expand the Output section, enter a folder name for Cache Path. This is where the baked images will be saved. The press the Bake Image Sequence button.

Done this, you can deactivate the Dynamic Paint physics on the canvas and the brush with monitor and camera icons on the Dynamic Paint section in the Physics tab or alternatively, on the modifier itself.

Using the Baked Image Sequence

Use the baked image sequence for the shader to visualize it.

• choose Open on the Image Texture node and select all images with A. Make sure they are ordered by the file name to have them in the correct order.
• enable [X] Auto Refresh for the Image Texture node
• set the Number of Frames

Then create a Particle System for the landscape object.

• type is Hair, activate the [X] Advanced option
• use the cone (tree) object as rendered object (Render > Re[]nder As = Object and .. > Object > Instance Object = the cone). Increase the .. > Scale to a higher values like 0.2.
• activate the [X] Rotation (advanced option), and set Orientation Axis to Object Y to have the trees all vertically aligned.
• in the Textures section of the Particle System create a new texture in the first slot with the New button below.

Switch to the Textures tab.

• select the image sequence that you have created for the shader
• tick [X] Auto Refresh and check if the Number of Frames value is correct
• in the Influence section, tick [X] Density. And [X] Hair Length if you want to have the size affected.
• set Mapping to UV and select the correct UV map if you have more than one

(the particle system has a few old bugs. Sometimes it doesn't update and you have to do a change and revert the change immediately. Here for example are tiny trees... no idea why. Hair Length is not checked).

# Geometry Nodes - Point Distribution

This is rather easy to do in Blender 2.93.5 because you can access the vertex color that is created by the Dynamic Paint physics directly in the geometry node:

The drawback here is that it is very slow when you replay the animation.

Related question

Is it possible to get a weight paint value for a vertex group inside of a material?

• the 2nd part using Geometry Nodes was exactly what I was looking for. Thanks! Oct 18, 2021 at 1:51


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# what is the difference between modulus and absolute value

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### Modulus vs Absolute - What's the difference? WikiDiff

As nouns the difference between modulus and absolute. is that modulus is (mathematics) the base with respect to which a congruence is computed while absolute is (philosophy) that which is totally unconditioned, unrestricted, pure, perfect, or complete; that which can be thought of Absolute Value or Modulus of a Real Number An Idea of

• For A, B R, |A + B |A+ |B|For A, B R, |AB= |A||B|For A, B R, |A B= ||a |b||0 < |x A A X A + Absolute Value of a Complex Number - Varsity TutorsThe absolute value of a complex number, a + bi (also called the modulus) is defined as the distance between the origin (0, 0) and the point (a, b) in the complex plane. What are Absolute Values? Definition, Symbol and ExamplesSep 19, 2019 · The symbol of absolute value is represented by the modulus symbol, |, with the numbers between it. For example, the absolute value of 9 is denoted as |9|. The distance of any number from the origin on the number line is the absolute value of that number. It also shows the polarity of the number whether it is positive or negative. What is the difference between modulo and modulus?Feb 12, 2016 · Meanwhile, "modulus" appears to have several definitions, including "absolute value" and "constant factor" as well as referring to the "5" in "modulo 5." Is it ever correct to use these terms interchangeably in the context of computer science? Are they simply different types of words that represent the same idea (such as "run" and "runner")? Modulus Function - Graph, Properties and Solved ExamplesThe absolute value of a modulus function needs to be non-negative. Since the absolute value of a modulus function generally defines the distance between two points, therefore it can be eed as:For f(x), where x represents 0, therefore, |f(0)= 0, which isn't Java Absolute Value - JavatpointIn Java, we can find the absolute value by using the abs () method. It belongs to java.lang.Math class. It is an overloaded method. It returns the absolute value of an argument. The argument can be of any type such as int, float, double, long, and short. The signature of the method is:Where x is an argument whose absolute value is to be Modulus vs Modulo - What's the difference? WikiDiffJul 28, 2015 · In context|mathematics|lang=en terms the difference between modulus and modulo is that modulus is (mathematics) the absolute value of a complex number while modulo is (mathematics) given a specified modulus of. As nouns the difference between modulus and modulo is that modulus is (mathematics) the base with respect to which a congruence is computed while modulo is Why square the difference instead of taking the absolute Having a square as opposed to the absolute value function gives a nice continuous and differentiable function (absolute value is not differentiable at 0) - which makes it the natural choice, especially in the context of estimation and regression analysis. The squared formulation also naturally falls out of parameters of the Normal Distribution. Solving absolute value equations and inequalities (Algebra The absolute number of a number a is written as $$\left a \right |$$ And represents the distance between a and 0 on a number line. An absolute value equation is an equation that contains an absolute value eion. The equation $$\left x \right |=a$$ Has two solutions x = a and x = -a because both numbers are at the distance a from 0. Difference between symbols for absolute value? - TeX Oct 08, 2017 · 1 Answer1. You ask for the differences and here's an analysis for them. First let's get rid of the last case:\mid is a relation symbol like = or < and TeX adds spaces around it that disqualify the command from being used for the absolute value. The other two seem good, but the first one is better as it doesn't need precautions. Relative vs Absolute Change - Analysis MistakesEven when the absolute change is large, if it is a change on a larger number the relative change can be small. Lets say that the national deficit increased by 5%. This may seem small, but the actual increase or absolute change to the $20,000,000,000,000 budget is$1 trillion. Absolute changes. Absolute changes work the other way: Graphing Absolute Values & Quadratics:An Overview This Purplemath lesson provides a quick overview of graphing absolute-value and quadratic functions. This page explains how to keep track of the "V" in the absolute-value graph, and how to draw the correct shape for the quadratic equation's parabola. Highlights common student mistakes and points out how to Absolute Function in Excel - Formula, Examples, How to Use Examples of the ABSOLUTE Function in Excel. For our analysis, we want the difference between Series A and Series B as given below. Ideally, if you subtract Series A from Series B you might get negative numbers depending on the values. However, if you want absolute numbers in this scenario, we can use the ABSOLUTE function. The results returned How to Calculate Absolute Value in RSep 07, 2021 · The absolute value graphically represents the line of a vector on a number line from zero to the point the vector represents. Thus, bys. By providing the absolute difference or magnitude of the functions output, it can provide a real absolute difference in scale between different types of numbers. That is it for the R absolute value tutorial.


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# Normal distribution percentage problem.

1. Apr 11, 2012

### MACHO-WIMP

1. The problem statement, all variables and given/known data
A company pays its employees an average wage of \$3.25 an hour with a standard deviation of sixty cents. If the wages are approximately normally distributed, determine
the minimum wage of the employees who are paid the highest 5%.

2. Relevant equations
z=(x-μ)/σ

3. The attempt at a solution
I did:
.95=((x-3.25)/.6) which equals .57=x-3.25
therefore x=3.82. I feel like I didn't do this correctly and I can't finish the rest of my homework if I don't know what I'm doing. Thanks.

Last edited: Apr 11, 2012
2. Apr 11, 2012

### tal444

Isn't that equation for standardizing your values? 95% has nothing to do with z-scores.

Last edited: Apr 11, 2012
3. Apr 11, 2012

### MACHO-WIMP

Oh, shoot, you're right. Well how would I find out what 95% of the wages cause I don't have the slightest clue.

4. Apr 11, 2012

### tal444

Your z-value is +-σ from the mean. What is 95% in terms of σ on a normal distribution? Also, do you use calculators to help you in your class? If you do, this question should be quite straightforward.

5. Apr 11, 2012

### Ray Vickson

If z_95 is the 95th percentile of the standard normal (which is available in tables or on some calculators) you need $$z_{95} = \frac{x-3.25}{0.6}.$$ Note: you should divide by 0.6 not 6, since 0.6 is the standard deviation in dollars.

RGV

6. Apr 11, 2012

### MACHO-WIMP

No, my teacher wants us only using this table she gave us.


text:
Issue No. 03 - July-Sept. (2014 vol. 7)
ISSN: 1939-1412
pp: 301-310
Jian Li , School of Energy Science and Engineering, University of Electronic Science and Technology of China, Room 316, No. 2006, Xiyuan Road, High Tech Zone (West Zone), Chengdu, Sichuan, China
Mahdi Tavakoli , Department of Electrical and Computer Engineering, University of Alberta, W6-037 ECERF Building, Edmonton, Canada
Qi Huang , School of Energy Science and Engineering, University of Electronic Science and Technology of China, Room 316, No. 2006, Xiyuan Road, High Tech Zone (West Zone), Chengdu, Sichuan, China
ABSTRACT
Trilateral haptic systems can be modeled as three-port networks. We present a criterion for absolute stability of a general class of three-port networks. Traditionally, existing (i.e., Llewellyn’s) criteria have facilitated the stability analysis of bilateral haptic systems modeled as two-port networks. If the same criteria were to be used for stability analysis of a three-port network, its third port termination would need to be assumed known for it to reduce to a two-port network. This is restrictive because, for absolute stability, all three terminations of the three-port network must be allowed to be arbitrary (while passive). Extending Llewellyn’s criterion, we present closed-form necessary and sufficient conditions for absolute stability of a general class of three-port networks. We first find a symmetrization condition under which a general asymmetric impedance (or admittance) matrix $Z_{3\times 3}$ has a symmetric equivalent $Z_{eq}$ from a network stability perspective. Then, via the equivalence of passivity and absolute stability for reciprocal networks, an absolute stability condition for the original nonreciprocal network is derived. To demonstrate the convenience and utility of using this criterion for both analysis and design, it is applied to the problem of designing stabilizing controllers for dual-user haptic teleoperation systems, with simulations and experiments validating the criterion.
INDEX TERMS
Stability criteria, Ports (Computers), Haptic interfaces, Impedance, Robots, Circuit stability
CITATION

J. Li, M. Tavakoli and Q. Huang, "Absolute Stability of a Class of Trilateral Haptic Systems," in IEEE Transactions on Haptics, vol. 7, no. 3, pp. 301-310, 2014.
doi:10.1109/TOH.2014.2321616


text:
# A(-8, 0), B(0, 16) and C(0, 0) are the verticals of a triangle ABC. Point P lies on AB and Q lies on AC such that AP : PB = 3 : 5 and AQ : QC = 3 : 5 - Mathematics

Sum

A(-8, 0), B(0, 16) and C(0, 0) are the verticals of a triangle ABC. Point P lies on AB and Q lies on AC such that AP : PB = 3 : 5 and AQ : QC = 3 : 5

Show that : PQ =3/8 BC

#### Solution

Given that, point P lies on AB such that AP: PB = 3: 5.
The co-ordinates of point P are

((3xx0+5xx(-8))/(3+5),(3xx16+5xx0)/(3+5))

=(-40/8,48/8)

=(-5,6)

Also, given that, point Q lies on AB such that AQ: QC = 3: 5.
The co-ordinates of point Q are

((3xx0+5xx(-8))/(3+5),(3xx0+5xx0)/(3+5))

=(-40/8,0/8)

=(-5,0)

Using distance formula,

PQ=sqrt((-5+5)^2+(0-6)^2)=sqrt(0+36)=6

Hence, proved

BC=sqrt((0-0)^2+(0-16)^2)=sqrt(0+16^2)=16

Now `3/8BC=3/8xx16=6=PQ

Concept: Distance Formula
Is there an error in this question or solution?

#### APPEARS IN

Selina Concise Maths Class 10 ICSE
Chapter 13 Section and Mid-Point Formula
Exercise 13 (C) | Q 3 | Page 183


text:
Lemma 56.2.2. Let $A$, $\mathcal{B}$, $F$ be as in Lemma 56.2.1. Assume $\mathcal{B}$ is additive and $F$ is additive. Then $F'$ is additive and commutes with arbitrary direct sums.

Proof. To show that $F'$ is additive it suffices to show that $F'(M) \oplus F'(M') \to F'(M \oplus M')$ is an isomorphism for any $A$-modules $M$, $M'$, see Homology, Lemma 12.7.1. Write $M = \mathop{\mathrm{colim}}\nolimits _ i M_ i$ and $M' = \mathop{\mathrm{colim}}\nolimits _ j M'_ j$ as filtered colimits of finitely presented $A$-modules $M_ i$. Then $F'(M) = \mathop{\mathrm{colim}}\nolimits _ i F(M_ i)$, $F'(M') = \mathop{\mathrm{colim}}\nolimits _ j F(M'_ j)$, and

\begin{align*} F'(M \oplus M') & = F'(\mathop{\mathrm{colim}}\nolimits _{i, j} M_ i \oplus M'_ j) \\ & = \mathop{\mathrm{colim}}\nolimits _{i, j} F(M_ i \oplus M'_ j) \\ & = \mathop{\mathrm{colim}}\nolimits _{i, j} F(M_ i) \oplus F(M'_ j) \\ & = F'(M) \oplus F'(M') \end{align*}

as desired. To show that $F'$ commutes with direct sums, assume we have $M = \bigoplus _{i \in I} M_ i$. Then $M = \mathop{\mathrm{colim}}\nolimits _{I' \subset I\text{ finite}} \bigoplus _{i \in I'} M_ i$ is a filtered colimit. We obtain

\begin{align*} F'(M) & = \mathop{\mathrm{colim}}\nolimits _{I' \subset I\text{ finite}} F'(\bigoplus \nolimits _{i \in I'} M_ i) \\ & = \mathop{\mathrm{colim}}\nolimits _{I' \subset I\text{ finite}} \bigoplus \nolimits _{i \in I'} F'(M_ i) \\ & = \bigoplus \nolimits _{i \in I} F'(M_ i) \end{align*}

The second equality holds by the additivity of $F'$ already shown. $\square$

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text:
These MP Board Class 11th Chemistry Notes for Chapter 14 Environmental Chemistry help students to get a brief overview of all the concepts.

## MP Board Class 11th Chemistry Notes Chapter 14 Environmental Chemistry

→ Substances produced by natural causes or by human activities which cause harmful effect on the environment are called pollutant.

→ Such substances which when present in the environment pollute it, they are called contaminant.

→ Air pollutants are maximum in Troposphere.

→ Ozone layer situated in stratosphere prevent us from ultraviolet rays.

→ Oxides of nitrogen and sulphur cause acid rain.

→ CFC is the main cause of depletion of ozone layer.

→ The heat radiations (IR rays) coming from the sun are reflected from the earth’s surface and absorbed by gases like CO2. This increases the temperature and is called green house effect.

→ Along with CO2 CFC also cause green house effect.

→ 111 effect of green house effect is Globa! warming.

→ Pollutants are of two types :

• Degradable pollutants : Pollutants which can easily break down by micro-organisms. Example : Discarded vegetables, cow dung.
• Non-degradable pollutants : Pollutants which do not break down by micro-organisms. Example : DDT, plastic, heavy metals, chemicals etc.

→ Tropospheric pollution : It occurs due to presence of undesirable solid or gaseous particles in air. Gaseous air pollutants are oxides of S, N and C, H2S, hydrocarbons, ozone and other oxidants. Particulate pollutants are dust, mist, fumes, smoke, smog etc.

→ Vaporized material formed by minute carbon particles, obtained by combustion of a substance is called smoke.

→ Thin layer of fog, smoke and dust is called Smog. It is harmful mixture of smoke and fog.

→ Smog is of two types : (i) Classical smog, (ii) Photochemical smog.

→ Fine solid particles present in air whose size range from 01 µ to 25 µ are called dust.

→ In Green house system allow low wavelength solar radiation to enter but does not allow higher wavelength infrared radiations to go out.

→ Main green house gases are CO2, CH4, N2O and CFCs.

→ Main components of Acid rain are H2SO4 and HNO3. When pH of acid rain becomes less than 5-6, it is known as Acid rain.

→ Main source of soil pollution are : Detergent, poisonous metals, insecticides, acid, base, salt, insects, pharmaceutical waste.

→ Bio-chemical oxygen demand (BOD). Amount of oxygen required by bacteria to break down the organic matter (at 206° for 5 days) present in a certain volume of water sample is called B.O.D.

BOD = $$\frac{\text { Number of milligrams of } \mathrm{O}_{2} \text { required }}{\text { Volume of sample (in litres) }}=\frac{\text { Parts of } \mathrm{O}_{2}}{\text { Million part of sample }}$$

BOD for clean water is less than 5 ppm, whereas BOD of highly polluted water is 17 ppm or can be more than this.

→ Eutrophication : The process in which nutrient enriched water bodies support a dense plant population which kills animal life by depriving it of oxygen and results in subsequent loss of biodiversity is known as Eutrophication.

→ Green chemistry: It is a way of utilizing the existing knowledge and principles of chemistry to reduce the adverse production of pollutants.