text:
# Tangent lines and rates of change

How tangent lines are a limit of secant lines, and where the derivative and rate of change fit into all this.

## Introduction

The position of a car driving down the street, the value of currency adjusted for inflation, the number of bacteria in a culture, and the AC voltage of an electric signal are all examples of quantities that change with time. In this section, we will study the rate of change of a quantity and how is it related geometrically to secant and tangent lines.

## Secant and tangent lines

If two distinct points P, left parenthesis, x, start subscript, 0, end subscript, comma, y, start subscript, 0, end subscript, right parenthesis and Q, left parenthesis, x, start subscript, 1, end subscript, comma, y, start subscript, 1, end subscript, right parenthesis lie on the curve y, equals, f, left parenthesis, x, right parenthesis, the slope of the secant line connecting the two points is
$\displaystyle m_{\sec} = \frac{y_1 - y_0} {x_1 - x_0} = \frac{f(x_1) - f(x_0)} {x_1 - x_0}\,$.
If we let the point x, start subscript, 1, end subscript approach x, start subscript, 0, end subscript, then Q will approach P along the graph f. The slope of the secant line through points P and Q will gradually approach the slope of the tangent line through P as x, start subscript, 1, end subscript approaches x, start subscript, 0, end subscript. In the limit, the previous equation becomes
$\displaystyle m_{\tan} =\lim_{x_1 \to x_0} \frac{f(x_1) - f(x_0)} {x_1 - x_0}\,$.
If we let h, equals, x, start subscript, 1, end subscript, minus, x, start subscript, 0, end subscript, then x, start subscript, 1, end subscript, equals, x, start subscript, 0, end subscript, plus, h and h, right arrow, 0 as x, start subscript, 1, end subscript, right arrow, x, start subscript, 0, end subscript. We can rewrite the limit as
$\displaystyle m_{\tan} =\lim_{h \to 0} \frac{f(x_0 + h) - f(x_0)} {h}$.
When the limit exists, its value m, start subscript, tangent, end subscript is the slope of the tangent line to the graph of f at the point P, left parenthesis, x, start subscript, 0, end subscript, comma, y, start subscript, 0, end subscript, right parenthesis.

## Example 1

Find the slope of the tangent line to the graph of the function f, left parenthesis, x, right parenthesis, equals, x, start superscript, 3, end superscript at the point left parenthesis, 2, comma, 8, right parenthesis.

## Solution

Since left parenthesis, x, start subscript, 0, end subscript, comma, y, start subscript, 0, end subscript, right parenthesis, equals, left parenthesis, 2, comma, 8, right parenthesis, using the slope of the tangent line formula
$\displaystyle m_{\tan} =\lim_{h \to 0} \frac{f(x_0 + h) - f(x_0)} {h}$
we get
\begin{aligned} m_{\tan} &=\lim_{h \to 0} \frac{f(2 + h) - f(2)} {h}\\ \\ &=\lim_{h \to 0} \frac{\left(h^3 + 6h^2 + 12h + 8\right) - 8} {h}\\ \\ &=\lim_{h \to 0} \frac{h^3 + 6h^2 + 12h} {h}\\ \\ &=\lim_{h \to 0} \left(h^2 + 6h + 12\right)\\ \\ &= 12. \end{aligned}
Thus, the slope of the tangent line is 12. Recall from algebra that the point-slope form of the equation of the tangent line is
y, minus, y, start subscript, 0, end subscript, equals, m, start subscript, tangent, end subscript, dot, left parenthesis, x, minus, x, start subscript, 0, end subscript, right parenthesis.
The point-slope formula gives us the equation
y, minus, 8, equals, 12, dot, left parenthesis, x, minus, 2, right parenthesis
which we can rewrite as
y, equals, 12, x, minus, 16.

## Finding the slope at any point

Next we are interested in finding a formula for the slope of the tangent line at any point on the graph of f. Such a formula would be the same formula that we are using except we replace the constant x, start subscript, 0, end subscript by the variable x. This yields
$\displaystyle m_{\tan} =\lim_{h \to 0} \frac{f(x + h) - f(x)} {h}\,$.
We denote this formula by
$\displaystyle f\,'(x) =\lim_{h \to 0} \frac{f(x + h) - f(x)} {h}\,$,
where f, space, prime, left parenthesis, x, right parenthesis is read "f prime of x." The next example illustrates its usefulness.

## Example 2

If f, left parenthesis, x, right parenthesis, equals, x, start superscript, 2, end superscript, minus, 3, find f, space, prime, left parenthesis, x, right parenthesis and use the result to find the slopes of the tangent lines at x, equals, 2 and x, equals, minus, 1.
Tangent lines to f, left parenthesis, x, right parenthesis, equals, x, start superscript, 2, end superscript, minus, 3 at the points x, equals, minus, 1 and x, equals, 2.

## Solution

Since
$\displaystyle f\,'(x) =\lim_{h \to 0} \frac{f(x + h) - f(x)} {h}$,
then
\begin{aligned} f\,'(x) &=\lim_{h \to 0} \frac{\left[(x + h)^2 - 3\right] - \left[x^2 - 3\right]} {h}\\ \\ &=\lim_{h \to 0} \frac{x^2 + 2xh + h^2 - 3 - x^2 + 3} {h}\\ \\ &=\lim_{h \to 0} \frac{2xh + h^2} {h}\\ \\ &=\lim_{h \to 0} (2x + h)\\ \\ &= 2x. \end{aligned}
To find the slope, we substitute x, equals, 2 and x, equals, minus, 1 into the result f, space, prime, left parenthesis, x, right parenthesis, We get
f, space, prime, left parenthesis, 2, right parenthesis, equals, 2, left parenthesis, 2, right parenthesis, equals, 4
and
f, space, prime, left parenthesis, minus, 1, right parenthesis, equals, 2, left parenthesis, minus, 1, right parenthesis, equals, minus, 2.
Thus, slopes of the tangent lines at x, equals, 2 and x, equals, minus, 1 are 4 and minus, 2, respectively.

## Example 3

Find the slope of the tangent line to the curve y, equals, 1, slash, x at the point left parenthesis, 1, comma, 1, right parenthesis.
Tangent line to the curve f, left parenthesis, x, right parenthesis, equals, 1, slash, x at the point x, equals, 1

## Solution

Using the slope of the tangent line formula
$\displaystyle f\,'(x) =\lim_{h \to 0}\frac{f(x + h) - f(x)} {h}$
and substituting y, equals, 1, slash, x gives us
\begin{aligned} y\,' &=\lim_{h \to 0} \dfrac{\left (\dfrac{1} {x+h} \right) - \dfrac{1} {x}} {h}\\ \\ &= \lim_{h \to 0} \frac{\dfrac{x - (x + h)} {x(x + h)}} {h}\\ \\ &= \lim_{h \to 0} \frac{x - x - h} {hx(x + h)}\\ \\ &= \lim_{h \to 0} \frac{-h} {hx(x + h)}\\ \\ &= \lim_{h \to 0} \frac{-1} {x(x + h)}\\ \\ &= \frac{-1} {x^2}. \end{aligned}
Substituting x, equals, 1 yields
y, space, prime, equals, start fraction, minus, 1, divided by, left parenthesis, 1, right parenthesis, start superscript, 2, end superscript, end fraction, equals, minus, 1.
Thus, the slope of the tangent line at x, equals, 1 for the curve y, equals, 1, slash, x is m, equals, minus, 1. To find the equation of the tangent line, we use the point-slope formula,
y, minus, y, start subscript, 0, end subscript, equals, m, dot, left parenthesis, x, minus, x, start subscript, 0, end subscript, right parenthesis,
where left parenthesis, x, start subscript, 0, end subscript, comma, y, start subscript, 0, end subscript, right parenthesis, equals, left parenthesis, 1, comma, 1, right parenthesis. The equation of the tangent line is
\begin{aligned} y - 1 &= -1 \cdot ( x - 1)\\ y&= -x + 1 + 1\\ y&= -x + 2. \end{aligned}

## Average speed

The primary concept of differential is calculating the rate of change of one quantity with respect to another. For example, speed is defined as the rate of change of the distance traveled with respect to time. If a car travels 120 miles in 4 hours, his speed is
start fraction, 120, space, m, i, l, e, s, divided by, 4, space, h, o, u, r, s, end fraction, equals, 30, space, m, i, slash, h, r.
This speed is called the average speed or the average rate of change of distance with respect to time. Of course, a car that travels 120 miles at an average rate of 30 miles per hour for 4 hours does not necessarily do so at constant speed. It may have slowed down or sped up during the 4 hour period.
However, if the car hits a tree, it would not be its average speed that determines the resulting damage but its speed at the instant of the collision. So here we have two distinct kinds of speeds, average speed and instantaneous speed.
The average speed of an object is defined as the object’s displacement triangle, x divided by the time interval triangle, t during which the displacement occurs:
v, equals, start fraction, triangle, x, divided by, triangle, t, end fraction, equals, start fraction, x, start subscript, 1, end subscript, minus, x, start subscript, 0, end subscript, divided by, t, start subscript, 1, end subscript, minus, t, start subscript, 0, end subscript, end fraction.
The average speed is also the expression for the slope of a secant line connecting the two points. Figure 1 shows the start color purple, s, e, c, a, n, t, space, l, i, n, e, end color purple through the points start color pink, left parenthesis, t, start subscript, 0, end subscript, comma, x, start subscript, 0, end subscript, right parenthesis, end color pink and start color pink, left parenthesis, t, start subscript, 1, end subscript, comma, x, start subscript, 1, end subscript, right parenthesis, end color pink on the start color blue, p, o, s, i, t, i, o, n, negative, v, e, r, s, u, s, negative, t, i, m, e, space, c, u, r, v, e, end color blue.
Figure 1. The slope of the secant line through two points on the position curve gives the average speed.
Thus we conclude that the average speed of an object between time t, start subscript, 0, end subscript and t, start subscript, 1, end subscript is represented geometrically by the slope of the secant line connecting the two points left parenthesis, t, start subscript, 0, end subscript, comma, x, start subscript, 0, end subscript, right parenthesis and left parenthesis, t, start subscript, 1, end subscript, comma, x, start subscript, 1, end subscript, right parenthesis. If we choose t, start subscript, 1, end subscript close to t, start subscript, 0, end subscript, then the average speed will closely approximate the instantaneous speed at time t, start subscript, 0, end subscript.

## Rates of change

The average rate of change of an arbitrary function f on an interval is represented geometrically by the slope of the secant line to the graph of f. The instantaneous rate of change of f at a particular point is represented by the slope of the tangent line to the graph of f at that point. Let's consider each case in more detail.

### Average rate of change

The average rate of change of the function f over the interval open bracket, x, start subscript, 0, end subscript, comma, x, start subscript, 1, end subscript, close bracket is
$\displaystyle m_{\sec} = \frac{f(x_1) - f(x_0)} {x_1 - x_0}$.
Figure 2 shows the start color purple, s, e, c, a, n, t, space, l, i, n, e, end color purple through the points start color pink, left parenthesis, x, start subscript, 0, end subscript, comma, f, left parenthesis, x, start subscript, 0, end subscript, right parenthesis, right parenthesis, end color pink and start color pink, left parenthesis, x, start subscript, 1, end subscript, comma, f, left parenthesis, x, start subscript, 1, end subscript, right parenthesis, right parenthesis, end color pink on the start color blue, g, r, a, p, h, space, o, f, space, f, end color blue. The slope of the secant line is the average rate of change $m_{\sec}$.
Figure 2. The slope of the secant line through two points on the graph of a function gives the function's average rate of change over the interval.

### Instantaneous rate of change

The instantaneous rate of change of the function f at the point x, start subscript, 0, end subscript is
$\displaystyle m_{\tan} = f\,'(x_0) = \lim_{x_1 \to x_0} \frac{f(x_1) - f(x_0)} {x_1 - x_0}$.
Figure 3 shows start color purple, t, a, n, g, e, n, t, space, l, i, n, e, end color purple through the point start color pink, left parenthesis, x, start subscript, 0, end subscript, comma, f, left parenthesis, x, start subscript, 0, end subscript, right parenthesis, right parenthesis, end color pink on the start color blue, g, r, a, p, h, space, o, f, space, f, end color blue. The slope of the tangent line is the instantaneous rate of change m, start subscript, tangent, end subscript.
Figure 3. The slope of the tangent line through a point on the graph of a function gives the function's instantaneous rate of change at that point.

## Example 4

Suppose that y, equals, x, start superscript, 2, end superscript, minus, 3.
(a) Find the average rate of change of y with respect to x over the interval open bracket, 0, comma, 2, close bracket.
(b) Find the instantaneous rate of change of y with respect to x at the point x, equals, minus, 1.

## Solution

(a) Applying the formula for average rate of change with f, left parenthesis, x, right parenthesis, equals, x, start superscript, 2, end superscript, minus, 3 and x, start subscript, 0, end subscript, equals, 0 and x, start subscript, 1, end subscript, equals, 2 yields
\begin{aligned} m_{\sec} &= \frac{f(x_1) - f(x_0)} {x_1 - x_0}\\ \\ &= \frac{f(2) - f(0)} {2 - 0}\\ \\ &= \frac{1 - (-3)} {2}\\ \\ &= 2 \end{aligned}
This means the average rate of change over the interval open bracket, 0, comma, 2, close bracket is 2 units of increase in y for each unit of increase in x.
(b) From Example 2 above, we found that f, space, prime, left parenthesis, x, right parenthesis, equals, 2, x, so
\begin{aligned} m_{\tan} &= f\,'(x_0)\\ &= f\,'(-1)\\ &= 2(-1)\\ &= -2. \end{aligned}
This means that the instantaneous rate of change is negative. That is, y is decreasing at x, equals, minus, 1. It is decreasing at a rate of 2 units in y for each unit of increase in x.


text:
## Introduction to Probability Models

Lecture 24

Qi Wang, Department of Statistics

Oct 20, 2017

### Important Concepts

• PDF: $f_X(x)$ is a non-negative function of $x$ which integrates to 1, $$f_X(x) \ge 0, \int_{-\infty}^{\infty}f_X(x) dx = 1$$
• CDF: The area under the PDF to the left of a point x, $$F_X(x) = P(X \le x) = \int_{-\infty}^{x}f_X(x) dx$$
• Generally, $P(a \le X \le b) = \int_{a}^{b} f_X(x) dx = F_X(b) - F_X(a)$

### Expected Value and Variance for a Continuous Random Variable

X is a continuous random variable with PDF $f_X(x)$

• Expected Value: $$E[X] = \int_{-\infty}^{+\infty} x f_X(x) dx$$
• Variance: \begin{align} Var(X) &= E[(E - E[X])^2] \\ & = E[X^2] - E[X]^2 \\ &= \int_{-\infty}^{+\infty} x^2f_X(x) dx - (\int_{-\infty}^{+\infty} xf_X(x) dx)^2 \end{align}

### Percentiles

X is a continuous random variable with PDF $f_X(x)$ and CDF $F_X(x)$

• Definition:$p_{th}$ percentile of X means a value k at which $P(X < k) = p\%$ or $F_X(k) = p\%$
• For example, $10_{th}$ percentile of X means a value k at which $P(X < k) = 0.10$ or $F_X(k) = 0.10$
• Quantile:
• $1_{st}$ quantile: $25_{th}$ percentile
• $2_{nd}$ quantile, Median: $50_{th}$ percentile
• $3_{rd}$ quantile: $75_{th}$ percentile

### Example 1

Let X represent the diameter in inches of a circular disk cut by a machine. Let $f_X(x) = c(4x - x^2)$ for $1 \le x \le 4$ and $0$ otherwise.

1. Find the value of c that makes this a valid PDF.
2. Find the expected value and variance of $X$
3. Find the CDF $F_X(x)$
4. Find the probability that X is within 0.5 inches of the average diameter.
5. What is the third quartile of X?

### Example 2

Suppose that a continuous random variable, X, has the probability density function (PDF) given below: $f_X(x)=\left\{ \begin{array}{ll} \frac{3}{2}x, 0 \le x \le 1\\ \frac{1}{4}, 5 \le x \le 6\\ 0, otherwise \end{array} \right.$

1. What is the probability that $X$ is equal to 4?
2. What is the probability that $X$ is more than 4?
3. Find $F_X(5.6)$
4. Knowing that $X$ is more than $0.8$, what is the probability that $X$ is less than 5.6?
5. What is the $85_{th}$ percentile of $X$?

format: plain


text: despite that, it somehow fails to evaluate which is probably an issue with how octave is embedded. I suggest you to open up a terminal and run octave directly. there it works fine.


text:
proof that a domain is Dedekind if its ideals are products of maximals

Let $R$ be an integral domain. We show that it is a Dedekind domain if and only if every nonzero proper ideal can be expressed as a product of maximal ideals. To do this, we make use of the characterization of Dedekind domains as integral domains in which every nonzero integral ideal is invertible (proof that a domain is Dedekind if its ideals are invertible).

First, let us suppose that every nonzero proper ideal in $R$ is a product of maximal ideals. Let $\mathfrak{m}$ be a maximal ideal and choose a nonzero $x\in\mathfrak{m}$. Then, by assumption,

 $(x)=\mathfrak{m}_{1}\cdots\mathfrak{m_{n}}$

for some $n\geq 0$ and maximal ideals $\mathfrak{m}_{k}$. As $(x)$ is a principal ideal, each of the factors $\mathfrak{m}_{k}$ is invertible. Also,

 $\mathfrak{m}_{1}\cdots\mathfrak{m_{n}}\subseteq\mathfrak{m}.$

As $\mathfrak{m}$ is prime, this gives $\mathfrak{m}_{k}\subseteq\mathfrak{m}$ for some $k$. However, $\mathfrak{m}_{k}$ is maximal so must equal $\mathfrak{m}$, showing that $\mathfrak{m}$ is indeed invertible. Then, every nonzero proper ideal is a product of maximal, and hence invertible, ideals and so is invertible, and it follows that $R$ is Dedekind.

We now show the reverse direction, so suppose that $R$ is Dedekind. Proof by contradiction will be used to show that every nonzero ideal is a product of maximals, so suppose that this is not the case. Then, as $R$ is defined to be Noetherian (http://planetmath.org/Noetherian), there is an ideal $\mathfrak{a}$ maximal (http://planetmath.org/MaximalElement) (w.r.t. the partial order of set inclusion) among those proper ideals which are not a product of maximal ideals. Then $\mathfrak{a}$ cannot be a maximal ideal itself, so is strictly contained in a maximal ideal $\mathfrak{m}$ and, as $\mathfrak{m}$ is invertible, we can write $\mathfrak{a}=\mathfrak{mb}$ for an ideal $\mathfrak{b}$.

Therefore $\mathfrak{a}\subseteq\mathfrak{b}$ and we cannot have equality, otherwise cancelling $\mathfrak{a}$ from $\mathfrak{a}=\mathfrak{ma}$ would give $\mathfrak{m}=R$. So, $\mathfrak{b}$ is strictly larger than $\mathfrak{a}$ and, by the choice of $\mathfrak{a}$, is therefore a product of maximal ideals. Finally, $\mathfrak{a}=\mathfrak{mb}$ is then also a product of maximal ideals.

Title proof that a domain is Dedekind if its ideals are products of maximals ProofThatADomainIsDedekindIfItsIdealsAreProductsOfMaximals 2013-03-22 18:35:04 2013-03-22 18:35:04 gel (22282) gel (22282) 5 gel (22282) Proof msc 13F05 msc 13A15 DedekindDomain MaximalIdeal FractionalIdeal


text:
Canadian Mathematical Society www.cms.math.ca
 location:  Publications → journals → CJM
Abstract view

# On the Tempered Spectrum of Quasi-Split Classical Groups II

Published:2001-04-01
Printed: Apr 2001
• David Goldberg
• Freydoon Shahidi
Features coming soon:
Citations   (via CrossRef) Tools: Search Google Scholar:
 Format: HTML LaTeX MathJax PDF PostScript

## Abstract

We determine the poles of the standard intertwining operators for a maximal parabolic subgroup of the quasi-split unitary group defined by a quadratic extension $E/F$ of $p$-adic fields of characteristic zero. We study the case where the Levi component $M \simeq \GL_n (E) \times U_m (F)$, with $n \equiv m$ $(\mod 2)$. This, along with earlier work, determines the poles of the local Rankin-Selberg product $L$-function $L(s, \tau' \times \tau)$, with $\tau'$ an irreducible unitary supercuspidal representation of $\GL_n (E)$ and $\tau$ a generic irreducible unitary supercuspidal representation of $U_m (F)$. The results are interpreted using the theory of twisted endoscopy.
 MSC Classifications: 22E50 - Representations of Lie and linear algebraic groups over local fields [See also 20G05] 11S70 - $K$-theory of local fields [See also 19Fxx]

© Canadian Mathematical Society, 2014 : http://www.cms.math.ca/


text:
You can format your text by using wiki markup. This consists of normal characters like asterisks, single quotes or equal signs which have a special function in the wiki, sometimes depending on their position. For example, to format a word in italic, you include it in two pairs of single quotes like ''this''.

## Paragraphs

MediaWiki ignores single line breaks. To start a new paragraph, leave an empty line. You can force a line break within a paragraph with the HTML tag
.

## Text formatting markup

Description You type You get
character (inline) formatting – applies anywhere
Italic text
''italic''

italic

Bold text
'''bold'''

bold

Bold and italic
'''''bold & italic'''''

bold & italic

Strike text
 strike text

strike text
Escape wiki markup
<nowiki>no ''markup''</nowiki>

no ''markup''

Escape wiki markup once
[[API]]<nowiki/>extension

APIextension

section formatting – only at the beginning of the line
Headings of different levels

== Level 2 ==

=== Level 3 ===

==== Level 4 ====

===== Level 5 =====

====== Level 6 ======

• Level 1 is the page name level so we avoid using it in content formatting.
• An article with 4 or more headings automatically creates a table of contents.

## Level 2

### Level 3

#### Level 4

##### Level 5
###### Level 6
Horizontal rule
Text above
----
Text below

Text above

Text below

Bullet list
* Start each line
* with an [[Wikipedia:asterisk|asterisk]] (*).
** More asterisks gives deeper
*** and deeper levels.
* Line breaks
don't break levels.
*** But jumping levels creates empty space.
Any other start ends the list.

• Start each line
• with an asterisk (*).
• More asterisks gives deeper
• and deeper levels.
• Line breaks
don't break levels.
• But jumping levels creates empty space.

Any other start ends the list.

Numbered list
# Start each line
# with a [[Wikipedia:Number_sign|number sign]] (#).
## More number signs gives deeper
### and deeper
### levels.
# Line breaks
don't break levels.
### But jumping levels creates empty space.
# Blank lines

# end the list and start another.
Any other start also
ends the list.

1. Start each line
2. with a number sign (#).
1. More number signs gives deeper
1. and deeper
2. levels.
3. Line breaks
don't break levels.
1. But jumping levels creates empty space.
4. Blank lines
1. end the list and start another.

Any other start also ends the list.

Definition list
;item 1
: definition 1
;item 2
: definition 2-1
: definition 2-2

item 1
definition 1
item 2
definition 2-1
definition 2-2
Indent text
: Single indent
:: Double indent
::::: Multiple indent

This workaround may be controversial from the viewpoint of accessibility.
Single indent
Double indent
Multiple indent

## HTML tags

Some HTML tags are allowed in MediaWiki, for example ,
,  and . These apply anywhere you insert them.

Description You type You get
Inserted
(Displays as underline in most browsers.)
Inserted

or

Underline

Inserted

or

Underline

Deleted
(Displays as strikethrough in most browsers.)
<s>Struck out</s>

or

Deleted

Struck out

or

Deleted

Fixed width text
Source code

or

<tt>Fixed width text</tt>

Source code

or

Fixed width text

Blockquotes
text above
text above

blockquote

text below
text below

text above text above

blockquote

text below text below

Comment

Comments are only visible
in the edit window.

Comments are only visible in the edit window.

## LaTeX equations

This wiki is equipped with LaTeX support, meaning contributors can include robust math equations with their edits by entering a formula in LaTeX format.

If you know LaTeX, simply add your equation code to any page with any of the methods outlined below. If you do not, one of the easiest ways to generate LaTeX is via a Chrome extension called the Daum Equation Editor.

### Daum Equation Editor

To install Daum, click here and then click the blue ADD TO CHROME button.

To open Daum, open a new tab in Chrome and click Apps from the lower center of the window. Then, select Daum Equation Editor from the options displayed. The app will open in the current tab with an interface as described below:

• The upper, dark grey area of the application provides some of the basic equation forms and options to get you started.
• The center, white, area is where you build your equation.
• The lower, grey area is where your LaTeX code will be generated.

### Adding LaTeX to a page

Once you've built your equation, select and copy the text from the lower, grey area and paste it into the wiki page you are working with surrounded by the appropriate characters from the list below:

The following math environments are defined for inline style math:

• $...$ (can be turned off, even per page),
• $$...$$ and
• $...$.

And the following math environments are defined for block display style math:

• $$...$$ (can be turned off, even per page),
• $...$,
• \begin{...}...\end{...} and
• :$...$.

### Examples of Equations

Using the methods above, can result in equations that look like this:

$$x = {-b \pm \sqrt{b^2-4ac} \over 2a}.$$

$$x=1a \over 2b$$

$${ T }_{ ij }=\sum _{ ij }^{ }{ { e }^{ ij } } \left\lceil \ln { { X }_{ ij } } \right\rceil$$

$$sum_(i=1)^n i^3=((n(n+1))/2)^2$$

If you click on Edit next to the heading Examples of Equations, you can see exactly how to specify the equation in your wiki article.


text:
# Direction of magnetic force

1. Dec 6, 2004

### agirlsrepublic

how do you find the direction of the magnetic force on a current loop..a loop that experiences a torque?
i know you use some kind of right hand rule but im confused

also..the direction of a charged particle in a magnetic field

Last edited: Dec 6, 2004
2. Dec 6, 2004

### Phymath

its the right hand rule, where ur fingers are in the direction of the current flow...im pretty sure

3. Dec 7, 2004

### James R

If you curl the fingers of your right hand in the direction the current flows around the loop, then your thumb points in the direction of the loop's magnetic dipole moment. The force on the loop will tend to align the dipole moment vector to be in the same direction as the external magnetic field the loop is sitting in.

The direction of the force on a charged particle in a magnetic field is:

$$\vec{F} = \vec{v} \times \vec{B}$$

where [itex]\vec{v}[\itex} is the velocity and [itex]\vec{B}[\itex] is the field.

Do you know how to apply the right hand rule to cross products such as this one?


text:
## Get the Most Out of Your AA – Achieve the Productivity and Speed of Sequential ICP-OES with Agilent’s AA Systems

Agilent’s AA range is productive, user-friendly, and exceptionally reliable. The instruments deliver the high performance that analysts require, while being equally at home in routine laboratories where reliability and simple operation are vital.

In this webinar we take a look at maximizing productivity and speed fo Sequential ICP-OES with the use of Agilent’s AA Systems.


text:
# Lesson 3

Which One?

These materials, when encountered before Algebra 1, Unit 1, Lesson 3 support success in that lesson.

## 3.1: Math Talk: What Was the Final Temperature? (5 minutes)

### Warm-up

The purpose of this Math Talk is to elicit strategies and understandings students have for mental subtraction. These understandings help students develop fluency and will be helpful later in this lesson when students will need to be able to compute interquartile range.

### Launch

Display one problem at a time. Give students quiet think time for each problem and ask them to give a signal when they have an answer and a strategy. Keep all problems displayed throughout the talk. Follow with a whole-class discussion.

### Student Facing

Mentally evaluate the final temperature in each scenario.

The temperature was 20 degrees Celsius and it dropped 18 degrees.

The temperature was 20 degrees Celsius and it dropped 20 degrees.

The temperature was 20 degrees Celsius and it dropped 25 degrees.

The temperature was 20 degrees Celsius and it dropped 33 degrees.

### Student Response

For access, consult one of our IM Certified Partners.

### Activity Synthesis

Ask students to share their strategies for each problem. Record and display their responses for all to see. To involve more students in the conversation, consider asking:

• “Who can restate $$\underline{\hspace{.5in}}$$’s reasoning in a different way?”
• “Did anyone have the same strategy but would explain it differently?”
• “Did anyone solve the problem in a different way?”
• “Does anyone want to add on to $$\underline{\hspace{.5in}}$$’s strategy?”
• “Do you agree or disagree? Why?”

## 3.2: Best Representation (20 minutes)

### Activity

Previously, students learned and reviewed how to construct box plots and dot plots. In this activity, students will practice constructing these. If they recall how to construct histograms from previous instruction, this is an opportunity to practice. If necessary, demonstrate how to construct a histogram. They will extend their thinking about data representations to understanding when and how each one is useful in analyzing data.

### Launch

Either measure each student’s height, or ask students to report their height in inches. Poll the class and display the list of heights for all to see. (If you use a different question other than asking about height, ensure it is a question with a range of responses similar to student height in inches, something like 48–76.)

If student data is unavailable or if there are not enough students to create a useful display of data, use the data about the height of students in a small class:

• 60
• 60
• 60
• 63
• 63
• 65
• 65
• 65
• 65
• 67
• 67
• 70
• 70
• 70
• 70
• 70
• 72
• 72
• 72

Arrange students in groups of 3 or 4. Allow students to work together to create the different representations. If necessary, demonstrate how to create one of the histograms and allow students to practice by creating the others. If time is limited, encourage students to assign a graph to each group member and to take a look at each other’s graphs when they have completed them.

### Student Facing

Use the class data to create a dot plot, box plot, and three histograms, each with different bin sizes.

1. Create a dot plot.

2. Create a box plot.

3. Create a histogram using intervals of length 20.

4. Create a histogram using intervals of length 10.

5. Create a histogram using intervals of length 5.

6. Which of these representations would you use to summarize your class’ data: the dot plot, the box plot, or one of the histograms? Explain your reasoning.

### Student Response

For access, consult one of our IM Certified Partners.

### Anticipated Misconceptions

Students sometimes forget the characteristics of each graph. Encourage them to ask people in their group for assistance and provide previously constructed box plots, dot plots, and histograms as references.

Students often confuse histograms with bar graphs. Histograms are used for representing a distribution of numerical data along a continuum, while bar graphs are used for displaying categorical data, but it's not crucial that students can explain this difference. However, ensure that students are creating proper histograms (with the intervals touching).

When constructing the intervals, students may not know which data points to include in a given range. In these materials, the lower bound of each interval is included and the upper bound is not. For example, if the intervals are 60–70 and 70–80, then 70 would be included in the second interval. Ensure that this rule is applied consistently.

### Activity Synthesis

Select several students or groups to share their preferred representations and their explanations. If not already mentioned by students, discuss the different insights that each display offers, or different challenges it poses. Ensure that students notice how each graph provides different information even when they are all displaying the same data set. Here are sample questions to prompt class discussion:

• “What information can we get from using a dot plot to display the data?” (It displays everyone’s height in the class. This gives us more insight in to the center of the data than a box plot.)
• "What information can we get from using a box plot to display the data? (It provides a summary of the data as well as where the median is located.)
• “Does the histogram give meaningful snapshot of the distribution?” (A histogram gives meaningful insight into the spread of everyone’s height in the class because we can see how much the numbers range just by looking.)
• “How did your graphs differ from one another?” (The histograms showed different shapes of the data sets. With each different interval size, the shape changes.)
• “Did all three types of graphs display the same information?” (No, the box plot showed the median, and the dot plot and histograms do not.)
• “How were all three histograms alike or different?” (They were alike because the same data points were included in each. They were different because they each used different intervals. Using a smaller bin size makes the distribution look more like the dot plot.)
• "To know the range and median at a glance, which would you use?" (box plot)

## 3.3: Which One? (15 minutes)

### Activity

Previously, students practiced constructing and interpreting data representations. In the first activity in this lesson, students were introduced to the idea that each representation can be used for different reasons. In this activity, students will further practice interpreting data by agreeing or disagreeing with statements about the information presented in data displays.

### Launch

Ensure students understand that each question has two parts: agreeing or disagreeing with the claim, then explaining a reasoning, including identifying which data display was useful.

### Student Facing

There are several baskets on a table, and each basket contains a certain number of strawberries. Here are three data displays showing the number of strawberries in each basket.

1. Kiran makes these claims. For each claim, decide whether you agree or disagree. Explain your reasoning using at least one of the data displays.
1. There are 4 baskets that contain 11 strawberries.
2. The range of the number of strawberries in baskets can be found using any of the three data displays.
3. The number of baskets of strawberries can only be found using the dot plot.
4. The interquartile range can be found using the dot plot or box plot, but is easiest with the box plot.
5. The total number of strawberries in all the baskets can only be determined from the dot plot.
2. Complete the table to show the frequency of baskets containing strawberries in each range. Which representation did you use?
number of strawberries   frequency
0–6
6–12
12–18

### Student Response

For access, consult one of our IM Certified Partners.

### Activity Synthesis

Students should understand that each display shows certain information, and they can choose a type of display to help them answer specific questions. Some of the displays cannot answer certain questions, and some questions can be answered using multiple displays, but may be more easily seen in one type than in another. Select students to share their solutions. Some questions for discussion:

• “What information can be easily found in both dot plots and box plots?” (The range of the data and clusters of data can be seen in both dot plots and box plots.)
• “Overall, what types of questions do box plots help answer? Histograms?”(Box plots help answer questions specifically about quartiles, interquartile range, range, median, lowest value, and highest value. Histograms help answer questions about ranges of data points and the shape of the distribution.)
• “If the histogram had bin sizes of 5, how would the graph’s display change? Would this graph be more useful for filling in the frequency table than the current histogram?” (If the bin sizes were 5, then the histogram would have fewer bins than it currently does. No, it would not be more useful for filling in the frequency table, because the frequency table has intervals of 6, and the histogram would have intervals of 5. It would be difficult to determine which data points to include from an interval of 5–10 in a frequency table that has 6–12.)

The goal of the lesson is for students to be able to determine which data representation will be most useful, given what information they need from a data set. Discuss how students knew which data representation is most useful in answering the questions. Here are sample questions to promote a class discussion:

• “Were there any patterns in the questions you chose to use a dot plot (or histogram and box plot) to answer?” (I noticed all the questions that I used a dot plot to answer were questions about frequencies or accounting for each individual data point.)
• “How did you know which data representation would be most useful in finding your answers?” (I thought about what the question asked, then connected that to the characteristics of each graph.)


text:
# Doppler effect to measure angle, not speed

1. Jul 17, 2007

### goodspeeler

DISCLAIMER: i may very well have exactly zero idea what i'm talking about. please feel free to berate me if i am way off base...

so i know the doppler effect is normally used to determine an unknown radial velocity, but i'm assuming that if i know the velocity of the source, i can use the frequency shift to determine the angle of the velocity vector relative to the receiver.

What i'm wondering about is the accuracy of this determination. I have heard that the standard formula for EM doppler effect is only an approximation. Does one of those assumptions require that you are parallel with the velocity vector?

Consider the following setup:

* receiver (arbitrary position)

*-------->v
source moving at a known speed |v| in a known direction (angle = 0)

if $$\theta$$ is the angle between the source velocity and the source->receiver direction, can I determine this angle by simply saying the following:

$$|v|cos(\theta) = v_{rad}$$ (where $$v_{rad}$$ is the source velocity's component toward the receiver)

and then say that
$$f_{shift} = \frac{f_{orig}*v_{rad}}{c}$$ (where $$f_{orig}$$ is the original EM frequency)

rearranging and plugging back in for $$v_{rad}$$ i get:

$$\theta = cos^{-1}\left(\frac{f_{shift}*c}{f_{orig}*|v|}\right)$$

so that's the setup. What i'm wondering is
1. does the doppler effect work this way, or have i violated too many of its assumptions?
2. if it works, how accurate can my angle determination be? put differently, to what precision can one measure a frequency shift? from some quick calculations with plausible $$|v|$$ and $$f_{orig}$$ values, it seemed like determining frequency to within 1-3 Hz would suffice, but i have zero idea how accurate such measurements can be. I would like to use the RF range (say, 100Mhz - 100GHz). is there some other way to get 1-3Hz accuracy in this range other than an FFT sampling at ~1THz?

hopefully the setup and my questions are relatively clear. just hoping for some input with some experience working with these sorts of things.

thanks in advance.

-gs

2. Jul 18, 2007

### bernhard.rothenstein

doppler effect and aberration

The Doppler formula in the case when the light signal propagates along a direction which makes a given angle A(A') with the line which joins source and receiver reads
f=f'g(V)(1+vcosA'/c).
Solved for v it gives the answer to your question.
I think that it is simpler to use the aberration of light formula
cosA=(cosA'+v/c)/[(1+(v/c)cosA')]
in order to solve the problem you propose.

3. Jul 18, 2007

### mgb_phys

You would have to know (and fix) the speed and direction of the source to the same accuracy which could be tricky.
Determining the frequency to 1 part in 1e9 is possible with a long enough sample time but building a source with an output this stable might be interesting.

4. Jul 18, 2007

### goodspeeler

thanks for the responses so far.

bernhard: i'm not sure i followed your explanation, though what i gathered was your agreement that it was possible?

mgb_phys: in my half-baked plan, the source speed and direction would be fixed. I was imagining several boxes sitting at defined positions around a room, each containing a microwave transmitter mounted on some kind of actuator that would move the transmitter side to side (or perhaps around in a vertically-oriented circle) at a defined speed and at a defined orientation. the receiver would then be the only thing moving around the room, and would interpret the frequency shifts from each of the transmitters (each on a slightly different frequency of course) to determine the angle with respect to each and then, in effect, triangulate the position.

after posting last night i came across http://www.freesoft.org/radar/Guardian/" [Broken] fun little link that mentioned that if you mix the shifted signals with the unshifted source signal, what you get out is actually the beat pattern that is at a frequency of exactly that shifted amount. so that seems to eliminate the needs for insanely high frequency sampling, as long as i can generate a reasonable facsimile of the known source frequency on my receiver to mix it with. that is, telling the difference between a 3Hz and a 6Hz shift now just requires telling the difference between a 3Hz and 6Hz signal. obviously, as you mentioned, I would still need the the source frequency and speeds to be stable. I don't have any clear notion of how difficult that would be. stable velocity doesn't seems like it would be prohibitively difficult, but I really have no idea what the typical frequency tolerances for microwave emitters are.

the one remaining factor would be the motion of the receiver itself. it would be small in magnitude relative to the actuator motion of the sources, but would still need to be accounted for. i haven't worked it out yet, but i imagine this can be done by coordinating between the various sources.

in any case, what i'm really hoping to do here is determine position using RF-frequency signals with millimeter or submillimeter accuracy. often people just use stationary transmitters and detect the drop in intensity and triangulate thusly, but for such accuracy this requires energy with suitably short wavelength which ends up being in the visible light range. problem with visible light range is that it will be blocked by anything sitting between the transmitters and the receiver. i was hoping that by relying on changes in frequency rather than intensity, i could eliminate the need to use such short wavelengths and thereby use signals like microwave that would penetrate obstacles and still reach the receiver.

anybody have any insight into whether using frequency in this way rather than intensity would in fact solve the accuracy problem? or is there some fundamental property of waves/energy i'm overlooking?

thanks again

-gs

Last edited by a moderator: May 3, 2017
5. Jul 18, 2007

### mgb_phys

A collegue of mine invented a system based on this 20years ago.
It uses a mobile receiver that constantly mixes the signal received from a number of fixed radio transmitters to calculate the relative distance from each. As the position of the transmitters is fixed you can work out the position of the mobile station. A big advantage was that it used commercial FM broadcast transmitters and could use phase methods to determine it's position to cm.

For various reasons to do with atmosphere to get very high accuracy you also had to a fixed receiver listening to the same signals and broadcast a small periodic correction to the mobile station - a little like Differential GPS.

The system was called "Cursor" from the company Cambridge Positioning Systems.

6. Jul 18, 2007

### BobG

Yes.

In fact, if you knew the direction and the velocity component directly towards you or away from you (the range rage), waited a little bit, took the direction of the signal and the range rate again, waited a little bit, and took the direction of the signal and the range rate again, you could compute the trajectory without even knowing the velocity beforehand.

In fact, tracking stations for satellites pass TEARR data (time, elevation, azimuth, range, and range rate) to the orbital analysts as soon as a newly launched satellite comes over the horizon, somewhere near mid pass, and right before the satellite goes over the horizon. It's not a terribly accurate trajectory, but accurate enough for the next tracking station to pick it up and it only takes a few widely separated observations to calculate the orbit for a newly launched satellite (which is never exactly what was planned since the rockets that boost the satellite into orbit never function perfectly).

Of course, even knowing the speed and the orbit, the satellite's location can't be predicted with millimeter or submilliter accuracy. For one thing, environmental perturbations are constantly changing the orbit. You'd run into the same problem, but even more so, for an object travelling on Earth. The object isn't going to maintain a constant speed, so, at best, you're only going to get an approximation for average speed, which means there's no way you're going to get the accuracy you want no matter what wavelength you use.

Last edited: Jul 18, 2007
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook


text:
Copied to
clipboard

## G = Dic18⋊2C6order 432 = 24·33

### 2nd semidirect product of Dic18 and C6 acting faithfully

Series: Derived Chief Lower central Upper central

 Derived series C1 — C18 — Dic18⋊2C6
 Chief series C1 — C3 — C9 — C18 — C2×3- 1+2 — C2×C9⋊C6 — C4×C9⋊C6 — Dic18⋊2C6
 Lower central C9 — C18 — Dic18⋊2C6
 Upper central C1 — C2 — D4

Generators and relations for Dic182C6
G = < a,b,c | a36=c6=1, b2=a18, bab-1=a-1, cac-1=a7, cbc-1=a18b >

Subgroups: 438 in 128 conjugacy classes, 52 normal (30 characteristic)
C1, C2, C2, C3, C3, C4, C4, C22, C22, S3, C6, C6, C2×C4, D4, D4, Q8, C9, C9, C32, Dic3, C12, C12, D6, C2×C6, C2×C6, C4○D4, D9, C18, C18, C3×S3, C3×C6, C3×C6, Dic6, C4×S3, C2×Dic3, C3⋊D4, C2×C12, C3×D4, C3×D4, C3×Q8, 3- 1+2, Dic9, Dic9, C36, C36, D18, C2×C18, C2×C18, C3×Dic3, C3×C12, S3×C6, C62, D42S3, C3×C4○D4, C9⋊C6, C2×3- 1+2, C2×3- 1+2, Dic18, C4×D9, C2×Dic9, C9⋊D4, D4×C9, D4×C9, C3×Dic6, S3×C12, C6×Dic3, C3×C3⋊D4, D4×C32, C9⋊C12, C9⋊C12, C4×3- 1+2, C2×C9⋊C6, C22×3- 1+2, D42D9, C3×D42S3, C36.C6, C4×C9⋊C6, C2×C9⋊C12, Dic9⋊C6, D4×3- 1+2, Dic182C6
Quotients: C1, C2, C3, C22, S3, C6, C23, D6, C2×C6, C4○D4, C3×S3, C22×S3, C22×C6, S3×C6, D42S3, C3×C4○D4, C9⋊C6, S3×C2×C6, C2×C9⋊C6, C3×D42S3, C22×C9⋊C6, Dic182C6

Smallest permutation representation of Dic182C6
On 72 points
Generators in S72
(1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36)(37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72)
(1 37 19 55)(2 72 20 54)(3 71 21 53)(4 70 22 52)(5 69 23 51)(6 68 24 50)(7 67 25 49)(8 66 26 48)(9 65 27 47)(10 64 28 46)(11 63 29 45)(12 62 30 44)(13 61 31 43)(14 60 32 42)(15 59 33 41)(16 58 34 40)(17 57 35 39)(18 56 36 38)
(2 32 26 20 14 8)(3 27 15)(4 22)(5 17 29)(6 12 18 24 30 36)(9 33 21)(10 28)(11 23 35)(16 34)(37 55)(38 50 62)(39 45 51 57 63 69)(41 71 65 59 53 47)(42 66 54)(43 61)(44 56 68)(48 72 60)(49 67)

G:=sub<Sym(72)| (1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36)(37,38,39,40,41,42,43,44,45,46,47,48,49,50,51,52,53,54,55,56,57,58,59,60,61,62,63,64,65,66,67,68,69,70,71,72), (1,37,19,55)(2,72,20,54)(3,71,21,53)(4,70,22,52)(5,69,23,51)(6,68,24,50)(7,67,25,49)(8,66,26,48)(9,65,27,47)(10,64,28,46)(11,63,29,45)(12,62,30,44)(13,61,31,43)(14,60,32,42)(15,59,33,41)(16,58,34,40)(17,57,35,39)(18,56,36,38), (2,32,26,20,14,8)(3,27,15)(4,22)(5,17,29)(6,12,18,24,30,36)(9,33,21)(10,28)(11,23,35)(16,34)(37,55)(38,50,62)(39,45,51,57,63,69)(41,71,65,59,53,47)(42,66,54)(43,61)(44,56,68)(48,72,60)(49,67)>;

G:=Group( (1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36)(37,38,39,40,41,42,43,44,45,46,47,48,49,50,51,52,53,54,55,56,57,58,59,60,61,62,63,64,65,66,67,68,69,70,71,72), (1,37,19,55)(2,72,20,54)(3,71,21,53)(4,70,22,52)(5,69,23,51)(6,68,24,50)(7,67,25,49)(8,66,26,48)(9,65,27,47)(10,64,28,46)(11,63,29,45)(12,62,30,44)(13,61,31,43)(14,60,32,42)(15,59,33,41)(16,58,34,40)(17,57,35,39)(18,56,36,38), (2,32,26,20,14,8)(3,27,15)(4,22)(5,17,29)(6,12,18,24,30,36)(9,33,21)(10,28)(11,23,35)(16,34)(37,55)(38,50,62)(39,45,51,57,63,69)(41,71,65,59,53,47)(42,66,54)(43,61)(44,56,68)(48,72,60)(49,67) );

G=PermutationGroup([[(1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36),(37,38,39,40,41,42,43,44,45,46,47,48,49,50,51,52,53,54,55,56,57,58,59,60,61,62,63,64,65,66,67,68,69,70,71,72)], [(1,37,19,55),(2,72,20,54),(3,71,21,53),(4,70,22,52),(5,69,23,51),(6,68,24,50),(7,67,25,49),(8,66,26,48),(9,65,27,47),(10,64,28,46),(11,63,29,45),(12,62,30,44),(13,61,31,43),(14,60,32,42),(15,59,33,41),(16,58,34,40),(17,57,35,39),(18,56,36,38)], [(2,32,26,20,14,8),(3,27,15),(4,22),(5,17,29),(6,12,18,24,30,36),(9,33,21),(10,28),(11,23,35),(16,34),(37,55),(38,50,62),(39,45,51,57,63,69),(41,71,65,59,53,47),(42,66,54),(43,61),(44,56,68),(48,72,60),(49,67)]])

50 conjugacy classes

 class 1 2A 2B 2C 2D 3A 3B 3C 4A 4B 4C 4D 4E 6A 6B 6C 6D 6E 6F 6G 6H 6I 6J 6K 9A 9B 9C 12A 12B 12C 12D 12E 12F 12G 12H 12I 12J 12K 18A 18B 18C 18D ··· 18I 36A 36B 36C order 1 2 2 2 2 3 3 3 4 4 4 4 4 6 6 6 6 6 6 6 6 6 6 6 9 9 9 12 12 12 12 12 12 12 12 12 12 12 18 18 18 18 ··· 18 36 36 36 size 1 1 2 2 18 2 3 3 2 9 9 18 18 2 3 3 4 4 6 6 6 6 18 18 6 6 6 4 6 6 9 9 9 9 18 18 18 18 6 6 6 12 ··· 12 12 12 12

50 irreducible representations

 dim 1 1 1 1 1 1 1 1 1 1 1 1 12 2 2 2 2 2 2 2 2 4 4 6 6 6 type + + + + + + - + + + - + + + image C1 C2 C2 C2 C2 C2 C3 C6 C6 C6 C6 C6 Dic18⋊2C6 S3 D6 D6 C4○D4 C3×S3 S3×C6 S3×C6 C3×C4○D4 D4⋊2S3 C3×D4⋊2S3 C9⋊C6 C2×C9⋊C6 C2×C9⋊C6 kernel Dic18⋊2C6 C36.C6 C4×C9⋊C6 C2×C9⋊C12 Dic9⋊C6 D4×3- 1+2 D4⋊2D9 Dic18 C4×D9 C2×Dic9 C9⋊D4 D4×C9 C1 D4×C32 C3×C12 C62 3- 1+2 C3×D4 C12 C2×C6 C9 C32 C3 D4 C4 C22 # reps 1 1 1 2 2 1 2 2 2 4 4 2 1 1 1 2 2 2 2 4 4 1 2 1 1 2

Matrix representation of Dic182C6 in GL10(𝔽37)

 0 0 0 1 0 0 0 0 0 0 0 0 36 36 0 0 0 0 0 0 0 36 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 36 0 0 0 0 0 0 0 0 1 36 0 0 0 0 0 0 0 0 0 0 0 36 0 0 0 0 0 0 0 0 1 36 0 0 0 0 36 1 0 0 0 0 0 0 0 0 36 0 0 0 0 0
,
 0 0 31 0 0 0 0 0 0 0 0 0 6 6 0 0 0 0 0 0 31 0 0 0 0 0 0 0 0 0 6 6 0 0 0 0 0 0 0 0 0 0 0 0 33 29 0 0 0 0 0 0 0 0 25 4 0 0 0 0 0 0 0 0 0 0 0 0 12 33 0 0 0 0 0 0 0 0 8 25 0 0 0 0 0 0 12 33 0 0 0 0 0 0 0 0 8 25 0 0
,
 26 0 0 0 0 0 0 0 0 0 0 26 0 0 0 0 0 0 0 0 0 0 11 0 0 0 0 0 0 0 0 0 0 11 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 36 1 0 0 0 0 0 0 0 0 36 0 0 0 0 0 0 0 0 0 0 0 0 36 0 0 0 0 0 0 0 0 1 36

G:=sub<GL(10,GF(37))| [0,0,0,1,0,0,0,0,0,0,0,0,36,1,0,0,0,0,0,0,0,36,0,0,0,0,0,0,0,0,1,36,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,36,36,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,36,36,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,36,36,0,0],[0,0,31,6,0,0,0,0,0,0,0,0,0,6,0,0,0,0,0,0,31,6,0,0,0,0,0,0,0,0,0,6,0,0,0,0,0,0,0,0,0,0,0,0,33,25,0,0,0,0,0,0,0,0,29,4,0,0,0,0,0,0,0,0,0,0,0,0,12,8,0,0,0,0,0,0,0,0,33,25,0,0,0,0,0,0,12,8,0,0,0,0,0,0,0,0,33,25,0,0],[26,0,0,0,0,0,0,0,0,0,0,26,0,0,0,0,0,0,0,0,0,0,11,0,0,0,0,0,0,0,0,0,0,11,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,36,36,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,36,36] >;

Dic182C6 in GAP, Magma, Sage, TeX

{\rm Dic}_{18}\rtimes_2C_6
% in TeX

G:=Group("Dic18:2C6");
// GroupNames label

G:=SmallGroup(432,363);
// by ID

G=gap.SmallGroup(432,363);
# by ID

G:=PCGroup([7,-2,-2,-2,-3,-2,-3,-3,176,590,303,10085,1034,292,14118]);
// Polycyclic

G:=Group<a,b,c|a^36=c^6=1,b^2=a^18,b*a*b^-1=a^-1,c*a*c^-1=a^7,c*b*c^-1=a^18*b>;
// generators/relations

׿
×
𝔽


text:
A Comparison of the Weak and Weak* Topologies

# A Comparison of the Weak and Weak* Topologies

We will now compare the weak and weak* topologies.

Topology Alternative Names Defined on Description
The Weak Topology The $X^*$-Weak Topology $X$ The weakest topology which makes all the linear functionals in $X^*$ continuous with respect to the topology.
The Weak* Topology The $J(X)$-Weak Topology $X^*$ The weakest topology which makes all the linear functionals in $J(X)$ continuous with respect to the topology. Each linear functional in $J(X)$ is of the form $J_x : X^* \to \mathbb{C}$ defined for all $\varphi \in X^*$ by $J_x(\varphi) = \varphi(x)$.

Observe that weak can also define the weak topology on $X*$. In that case, the weak topology on $X*$ is the weaest topology which makes all linear functionals in $X^{**}$ continuous with respect to the topology. Since every linear functional in $J(X)$ is already continuous with respect to this topology, we see that:

(1)
\begin{align} \quad \mathrm{weak* \: topology \: on \:} X^* \subseteq \mathrm{weak \: topology \: on \:} X \end{align}


text:
# Phototransistor transimpedance amplifier

I have a typical NPN phototransistor. I have it working in a common-collector configuration; see figure 2 of this app note.

Increasing Re will increase the sensitivity, but decrease the speed. I have been studying phototransistors for a few days now, and I think that a transimpedance amplifier can give me additional sensitivity without sacrificing speed, since I won't have Re loading the emitter anymore.

However, I can't seem to find straightforward implementations. The vast majority of app notes describe photodiodes. Unlike a photodiode, a phototransistor needs to be biased, and the few application notes that discuss using phototransistors assume the presence of a negative biasing voltage in their transimpedance amplifiers. I need a solution that works with a single-supply op-amp.

Would a virtual ground on the non-inverting input of the transimpedance amplifier correctly bias the phototransistor? Usually virtual ground is half-way between VCC and GND, but I don't think it has to be. My phototransistor saturation voltage is 0.15V; given VCC=3.3V, does that mean my virtual ground could be at ~3V?

Is there a better way to design this circuit? I would like the output to get as close to GND as possible, because there will probably be a second-stage amplifier.

EDIT:

More details on the application. I'm sensing light levels; low, very low, and off. There are no problems with ambient light, so I'd rather not focus on the phototransistor aspect of this question too much. The bandwidth of interest is around 1-10 kHz. The common-collector almost works; I've raised Re as high as it can go while maintaining the bandwidth I want, but I'd still like Re about 2x larger which results in a signal that's too slow.

• Check out the receiver designed for the RONJA project: ronja.twibright.com/schematics – Optimal Cynic Jul 13 '11 at 6:11
• There's no such thing as a "typical NPN phototransistor"? Why don't you just tell us what type it is? Also, like I wrote in my edited answer: what's the application? Light levels, Code reception? Visible light, IR? etc. etc. – stevenvh Jul 13 '11 at 6:40
• Sure there is such a thing. Search digikey for "npn photo transistor", average all the values in the datasheets together (e.g. Vcesat, Vceo, Ic, etc), and I would call that "typical". Ultimately, it's an indication that the amplifier circuit is more important – ajs410 Jul 13 '11 at 15:18

I've been trying to do a very low light level project myself the past 2 days with photodiodes and phototransistors. This is for people like myself and the original poster who are pushing light detection without a photomultiplier to the limit (below 0.1 mW/cm^2).

I looked at the first receiver module and its minimum irradiance detection was 0.2 mW/m^2 which is about 10,000 times more (less capable) than what discrete photodiodes and phototransistors can do (maybe they meant cm^2 instead of m^2?). Neither are good for really low light levels according to "Art of Electronics" (1 uA per uW of light page 996), totally not capable of getting near what the human eye can do due to leakage current and noise. He describes using photomultipliers which may be required if your light levels are too low. However, in shining light through my fingers in a well-lit room, I am able to see what my eye can't detect on an oscilliscope (with either PhotoDiode or PhotoTransistor).

Assuming his 1 uA per uW is correct, here is an example: a 5 mm photodiodes and phototransistors have an area of 20 micro m^2. So 1 uW/m^2 (1/1000th of noon sunlight) would generate 20 uA (according to Art of Electr) . [[ 1/1000th of noon sunlight is 1 W/m^2 which is about twice as strong as a 20W incadescent light at 1 meter (6W light output into 12 m^2 surface area of a surrounding sphere). ]]

However, my 880nm phototransistor datasheet indicates 600 uA at 1W/m^2 (0.1 mW/cm^2), wich is 30 times more. This assumes all the light is within the active range of the diode's junction.

Sharp has a much better application note, but it seems to be lacking in explaining which design is best for which situations. Figure 13 is most applicable to what original poster and I need, and figure 10B is very interesting but I don't know what they mean by "improves response". http://physlab.lums.edu.pk/images/1/10/Photodiode_circuit.pdf

When used with an op amp, a phototransistor may not capabale of getting as good of a gain as a photodiode for very low light levels because it to uses a "cheap" method of getting its initial gain (transistor instead of op amp). I suspect a photodiode with a JFET op amp (very low input current) would ultimately provide a higher gain with less noise. In any event, the photodiode or phototransistor with the largest optical receiving area might have the best ability to detect low light levels, but that might also increase noise and leakage by a proportional amount and they are usually the underlying problem. So there is a limit to this type of light detection and ideally efficient phototransistors and photodiodes may ultimately be equally good when used with an op amp, but theoretically I suspect photodiode is a little better. It will be the op amp design that matters and I think figure 13 with a JFET op amp and a well-chosen shunt capacitor across the feedback will be best for phototransistor.

For the dual supply op amp, you can use a "lowish" valued resistor pair (two 1k for 10V Vcc to get 5 mA bias) to split the voltage to create a false ground for the +Vin.

I found R=1M for the feedback resistor much better than R=4.7M. Forrest Mimms in his simple opto book used a 10 M with a parallel 0.002uF and a solar cell instead of a phototransistor or photodiode for "extrememly" low light levels" (maybe a solar cell would be better for your application) It seems all PN junctions seem to operate as a solar cell to some extent, as I've read of using clear-cased small signal diodes to detect light. I'm using a regular 830 nm LED as my "photodiode".

The lens angle of whichever 5mm optical diode you use makes a big difference. +/-10 degrees is roughly 4 times more sensitive than +/-20 degrees....if the light source is coming in from less than +/-10 degrees. If the light source is a big area that is +/-20 degrees in front, then it doesn't matter.

I tested the two circuits below. I could detect 0.3V, 5 ms pulses on the phototransistor's Vo which means 0.3 uA which means 0.05 uW/cm^2 if my reading of the datasheet is correct and if it remained linear (big ifs) all the way down to 0.3uA. Maybe it was 5 uW/cm^2. If 0.05 uW/cm^2 is correct, then the off-the-shelf 830 LED was reading down to 0.5 uW/cm^2. I was shining 10 mW 830 nm light through 1 cm of tissue (my finger). I know that if the light levels I was working with were red, it would have been barely visible. The link below shows using 500 M ohm feedback with a photodiode, indicating much lower light levels. Notice the orientation of their photodiode, which is the same as my LED (backwards from most internet links). I got better results this way.

http://www.optics.arizona.edu/palmer/OPTI400/SuppDocs/pd_char.pdf

• I posted this before editing and immediately got 3 negative votes. Hopefully the edited version will not be disliked so much. – scott roberts Aug 15 '12 at 17:56
• It's a bit wordy, and you didn't address the whole trans-impedance opamp thing the original poster was asking about. But you're new here, so I'll give you the +1 for effort. – user3624 Aug 15 '12 at 18:09
• He says raising Re decreases bandwidth. I thought it would have no effect. Looking at his subsequent comments, figure 13 of the link I provided is an answer to his request for a specific design. Also figure 10B is an interesting idea that "improves the response" whatever that means (gain, BW, or both?). If he can't get it to work from light levels that are too low, "photomultiplier", "lens", or "increase light source" are possible answers. – scott roberts Aug 15 '12 at 18:56
• +1 for actually addressing the question, which was about single-supply amplifier circuits for phototransistors. – ajs410 Aug 24 '12 at 18:48

I was also thinking about an inverting opamp amplifier. The nicest thing would be a dual power supply so that you don't have to bias the inputs to create a virtual ground. The picture shows the schematic. You'll have a positive ground-referenced signal: more light = higher output voltage.

$V_{OUT} = I_{PHOTOTRANSISTOR} \times R_{FEEDBACK}$

Since you only have a single supply you'll have to use a virtual ground, which I would place at half $V_{CC}$. You have to bias both inputs, the output will have this bias as well.

edit dd. 2012-08-15
In this answer Alfred showed that a photodiode will also sink current without a voltage drop across it. That means that we don't need the negative supply, and a single supply is possible:

Make sure that's an RRIO (Rail-to-Rail I/O) opamp.

edit
In the above I presumed that you want to measure light levels, i.e. analog values. But rereading your question it nowhere says you do. The mention of speed suggests pulse code reception. If that's what you want, what does the signal look like? What's the wavelength (IR or visible light?) Can't you use an IR receiver module?

• I've updated the question with extra application information. Unfortunately, I need to use a single-supply op-amp. Also, from what I've read, the transimpedance amp will probably need a feedback capacitor in parallel with the resistor to compensate. Fortunately, the bandwidth reduction of this feedback cap probably won't be a problem for me. – ajs410 Jul 13 '11 at 15:51

If you really need flexibility, consider using a photodiode instead of a phototransistor-- you're already building a transimpedance amp, so why not go all the way?

Also, there's a great book on the subject, with many detailed circuit examples, for low noise and/or high speed.
Building Electro-Optical Systems: Making It All Work, by Hobbs.

• +1 for "why not use a photodiode". – Optimal Cynic Jul 13 '11 at 6:09
• Photodiodes are too fast. Photodarlingtons are too slow. Phototransistors are juuuust right - with a good amplifier. Photodiodes also require more gain in a circuit with an already large gain; they would probably require a transistor somewhere, and at that point a phototransistor has the transistor on-die, rather than at the end of some traces. – ajs410 Jul 13 '11 at 15:21
• "Too fast?" You can always slow things down with cheap, tiny capacitors. Or use a larger, slower photodiode-- not all photodiodes are fast. And, op-amp circuits can give you ludicrously high gain-- in normal cases, you can use an op-amp OR a transistor. – Windell Oskay Jul 13 '11 at 22:51
• Why use a discrete transistor on my PCB when the phototransistor has it integrated on-die? And with ludicrously high gain comes a requirement for ludicrously low offset voltage. I'm looking for an amplifier circuit to go with a phototransistor. I appreciate the book suggestion, but it does not appear to contain such examples. – ajs410 Jul 14 '11 at 16:08


text:
# rank of block triangular matrix and its relation to the rank of its diagonal blocks

Prove that $$rank\begin{pmatrix}A & X \\ 0 & B \\ \end{pmatrix}\ge rank(A) + rank(B)$$ where $$A,B\in \mathbb C^{m \times m}$$.

I know the intuition behind it (i.e. maximal independent rows, etc.), but I am looking for a formal proof. I have tried QR decomposition of A and B, then broke the block triangular matrix up into a matrix product and used the inequality rank(AB)<= min{rank(A), rank(B)}, but have not made any great progress. One fact I thought may help was, if A=QR, where Q is unitary, then rank(A)=rank(R). Again I thought this would help but it did not. Any help on this problem would be much appreciated.

Let $a = \text{rank}(A)$ and $b=\text{rank}(B)$. Consider a set of $a+b$ columns $C_1, \ldots, C_{a+b}$ of your matrix, where $C_1, \ldots, C_a$ correspond to linearly independent columns of $A$ and $C_{a+1}, \ldots, C_{a+b}$ correspond to linearly independent columns of $B$. Suppose $\sum_{i=1}^{a+b} s_i C_i = 0$ where $s_1, \ldots, s_{a+b}$ are scalars. Because those $b$ columns of $B$ are linearly independent, we must have $s_{a+1} = \ldots = s_{a+b} = 0$. And then because those $a$ columns of $A$ are linearly independent, $s_1 = \ldots = s_a = 0$.


text:
GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 21 Nov 2018, 04:19

GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Events & Promotions

Events & Promotions in November
PrevNext
SuMoTuWeThFrSa
28293031123
45678910
11121314151617
18192021222324
2526272829301
Open Detailed Calendar
• All GMAT Club Tests are Free and open on November 22nd in celebration of Thanksgiving Day!

November 22, 2018

November 22, 2018

10:00 PM PST

11:00 PM PST

Mark your calendars - All GMAT Club Tests are free and open November 22nd to celebrate Thanksgiving Day! Access will be available from 0:01 AM to 11:59 PM, Pacific Time (USA)
• Free lesson on number properties

November 23, 2018

November 23, 2018

10:00 PM PST

11:00 PM PST

Practice the one most important Quant section - Integer properties, and rapidly improve your skills.

A study based on a random sample revealed that, on average, 2 out of 5

Author Message
TAGS:

Hide Tags

Senior PS Moderator
Joined: 26 Feb 2016
Posts: 3311
Location: India
GPA: 3.12
A study based on a random sample revealed that, on average, 2 out of 5  [#permalink]

Show Tags

Updated on: 02 Nov 2018, 01:16
00:00

Difficulty:

5% (low)

Question Stats:

100% (00:43) correct 0% (00:00) wrong based on 20 sessions

HideShow timer Statistics

A study based on a random sample revealed that, on average, 2 out of 5 adults have high blood pressure. If these results hold true for the 580,000 adults in City A, approximately how many adults in City A have high blood pressure?

(A) 116,000
(B) 145,000
(C) 232,000
(D) 250,000
(E) 290,000

_________________

You've got what it takes, but it will take everything you've got

Originally posted by pushpitkc on 02 Nov 2018, 01:14.
Last edited by Bunuel on 02 Nov 2018, 01:16, edited 1 time in total.
Edited the question.
Manager
Status: Quant Expert Q51
Joined: 02 Aug 2014
Posts: 93
Re: A study based on a random sample revealed that, on average, 2 out of 5  [#permalink]

Show Tags

03 Nov 2018, 02:05
Easy,

$$\frac{2}{5}*580000 = 232000$$

_________________
Re: A study based on a random sample revealed that, on average, 2 out of 5 &nbs [#permalink] 03 Nov 2018, 02:05
Display posts from previous: Sort by


text:
QUESTION

# Shatabdi express has a capacity of 500 seats of which 10% are in Executive class and the rest being in chair cars. During one journey the train was booked 85% of its capacity. If Executive class was booked to 96% of its capacity, then how many chair car seats were empty during that journey?${\text{A}}{\text{. 113}}${\text{B}}{\text{. 93}}${\text{C}}{\text{. 73}}$${\text{D}}{\text{. 53}}$

Hint: In such types of questions find the number of seats in the executive class by solving the percentage. Then using simple subtraction find seats in the chair car and proceed to solve the question.

We have been given in the question that-
Total number of seats $= {\text{ }}500$
So, as per the question, Number of Executive class seats = 10% of 500 i.e.,
Number of Executive class seats$= \dfrac{{10}}{{100}} \times 500$
Now, Number of seats in chair cars = total number of seats – number of seats in Executive class
Number of seats in chair cars $= {\text{ }}500{\text{ }}-{\text{ }}50$
Number of seats in chair cars$\; = {\text{ }}450$
It is given that during a journey the train was booked 85% of its capacity. So
Number of seats booked in the complete train = 85% of 500
Number of seats booked in the complete train $\dfrac{{85}}{{100}} \times 500 = 85 \times 5 = 425$
It is also given that the Executive class is booked 96% of its capacity. So
Number of seats booked in Executive class = 96% of 50
Number of seats booked in Executive class $\dfrac{{96}}{{100}} \times 50 = \dfrac{{96}}{2} = 48$
Number of chair car seats booked = number of seats booked in the train – number of seats booked in Executive class
Number of chair car seats booked $= {\text{ }}425{\text{ }}-{\text{ }}48$
Number of chair car seats booked $= {\text{ }}377$
So,
Total seats empty in chair car = number of seats in chair car – number of seats booked in chair car
Total seats empty in chair car $= {\text{ }}450{\text{ }}-{\text{ }}377{\text{ }} = {\text{ }}73$
Hence option (C) is correct.

Note: Whenever solving such types of questions, use the basic principle of addition, subtraction and percentage calculation. Always note down the things given in the question to reduce complexity. Step by step solve the equations and reach the answer. As mentioned in the solution first find 10% of the total seats and then proceed with further steps of finding the no. of seats.


text:
To give slightly more context, it was Exercise 3.3 from this[1] paper by Hutchings and Taubes

I've just realised that I spent several hours attempting to prove, ∀ v, ∀ u, (u = v) ⇔ f(u) = f(v) (v, u and f are all specific things but it's the shape of the problem that's important).

I should have been proving ∀ v, (∃ u, u = v) ⇔ (∃ u, f(u) = f(v)) (which is, at least, possible).

quantum boosted

The pseudorhomicuboctahedron can never be an Archimedean solid.
Boost if you agree

The Hodge decomposition (in particular the fact that every de Rham cohom. class has a harmonic representative) probably doesn't extend to pseudo-Riemannian mflds. because the pR Laplacian isn't elliptic.

Is there a weaker result that does extend to pR mflds. or some sort of pR variant of Hodge theory?

quantum boosted

I will go kicking and screaming into category theory

quantum boosted

Boosts appreciated for sample size :)

quantum boosted

People in 2021 pining for Google Reader feels like standing next to a clean mountain spring and wishing you had an old Dasani bottle.

So I'm reasonably sure it's $$\mathbf{Z}^{\ast n}$$, I'm just having trouble figuring out how to Seifert-Van Kampen it together,

quantum boosted

rabid tech vent

why the bloody damned fucked fuckety fuck the tabs need to all reload every time I switch to another app with this so called web browser for Android?

can someone fucking invent multitasking for fucks sake? can we get to 1960 I fucking beg you it's not too much to fucking ask is it??

I guess this is a problem for the morning

Five minutes of work have determined that this is too tedious a problem to solve tonight

> brain: What is $$\pi_1(X_n)$$ where $$X_n$$ is the complement of the $$n$$-fold wedge product of $$S^1$$ with itself in $$\mathbf{R}^3$$?
> me (tired): _Why?_

quantum boosted

> We explore the ability for the CPU to “double speculate”, where a second stalled indirect jump while the CPU is already speculating causes it to predict the target and speculate a second time.

> [..] It appears Haswell chips do not speculate while already speculating [...] Both Sandy Bridge (which preceded Haswell) and Kaby Lake (which followed Haswell) support nested speculation. We find that when a CPU does support nested speculation, there appears to be no limit to nested depth besides the speculative instruction limit.

We heard you like speculative execution. So we put speculative execution in your speculative execution so you can do speculative execution when you do speculative execution.

quantum boosted

Does anyone know a good book to explain gender stuff to a boomer with relatively no background but an interest in learning?

quantum boosted

and then god created enbies and he was like "oh shit thats tight these people are actually in my image, way moreso than male and female"
- genesis 1:27.5

quantum boosted

the safest way to use computers is still abstinence

quantum boosted
quantum boosted

Here's a question: is it OK to omit parentheses around the argument to a non-standard function?

Like, "sin x = sin(x)" and "log z = log(z)" are common, but what about "f x = f(x)"?

Does the function name need to be more than one letter?

"flop x = flop(x)"

quantum boosted

@jordyd category theorists will literally reverse an arrow instead of going to therapy

Show older

The social network of the future: No ads, no corporate surveillance, ethical design, and decentralization! Own your data with Mastodon!


text:
# Solving a system of linear equations stored in a table

I have a system of $n$ linear equations with an equal or smaller number of unknowns stored in a table. The variables are of the form $f[x_1,x_2,x_3]$. For example, the equations are:

f[1,1,1]= 2f[3,2,1] + 4f[2,1,1]
f[1,4,2] = 3f[1,3,1] + 3f[2,4,1] + 3ab
.....

I need to solve those equations for f[1,1,1], f[1,1,2],... in terms of a, b, ... and other constants.

Solve[f[i,j,k]==0, {f[i,j,k], {i,m}, {j,m}, {k,m}}] doesn't work.

Any help is appreciated.

ClearAll[f]

• Thanks. Is there a way of storing the solution as the variables' values? For example, f[5] -> 8 so f[5] will have a value of 8 and so on? – Bran Nov 18 '17 at 22:54
• @Subzero-273K Yes, you can ReplaceAll[solns, {Rule -> Set}], but generally it is better not to do this. – Alan Nov 19 '17 at 16:04


text:
## Acamedic

Liz was happy it wasn’t a physicist this time.

Physicists were usually bad cases. Not the way you’d expect — no weird glowing gadgets and radiation burns. Not even hazardous lasers. No; instead they were hollow-eyed quantum folks, human husks that lay curled on the floor muttering things a mind without a Ph.D. couldn’t understand. Basically the “Walls wouldn’t stand — planes wouldn’t fly” stuff with gradients and contour integrals thrown in, or the glassy-eyed drooling “I’m n-furcating into $10^6 \pm 10^4$ worlds every second!” stuff, which was frankly creepy. Happened all the time once they really got into the quantum mindset; her guess was that it was three weeks of heavy speculative work and then, bing! a psychosis.

Or many years of calm, productive work, and then all of a sudden, bing! bing! bing! and a person flinging around hamsters, saying it would be a groundbreaking results if he could get macroscopic objects to quantum tunnel through the walls.

Biologists were almost as bad; not because of any animals (not since the incident of the chancellor and the hidden incontinent rabbit), but because things tended to escalate with them. She had contemplated putting up a sign in their lounge, saying “YOU DO NOT KNOW HOW TO POP YOUR OR THE PERSON NEXT DOOR’S SHOULDER BACK INTO PLACE SO PLEASE DO NOT TRY”; but that might be seen as disrespectful.

It was a mathematician this time; by Liz’s experience this could be a very bad thing, or then not. Could be someone who had gotten trapped under a falling bookshelf, and not called for help until all working pens were out of reach (one with a laptop had lain in place for two days before someone happened in); could as well be someone whose complaint was “my hand doesn’t open the door anymore”, and whose reason was (largely) unmedical, being a mix-up with the keys.

She knocked, entered, and took in the room. Shelves full of papers and some books; a plastic skull atop one; a few faded conference announcement posters decorating the walls; a battered wooden desk, an office chair so beat-up it looked like a gas-driven projectile enema just waiting to happen, and sitting on it, a shriveled man.

It was difficult to see his age behind the thick glasses and the chalk dust-covered skin; late thirties was her guess by the yellowing of the papers all around them.

“Academic medic Liz Riddell, hi. You called?”

The man frowned, blinked, then opened his mouth with almost childish glee. “Oh yes! That bothersome practical thing.”

She sucked in a breath — oh, no. Let it be just a coincidence that those were the very words that had began her introduction into the bothersome case of “I think I saw someone sitting in a room quite dead this morning; blood running out of nose and all. Is that, er, the sort of a thing I should tell someone like you? If this doesn’t concern you I’m sorry to have wasted your — er, are you all right?” Then, a case of spatial dementia and ninety-six offices later, the passed-out professor of statistics with hemophilia.

“Practical thing?”

“Ah yes. It is this, this swine flu thing. Matter. Event cluster.”

She took an involuntary step back.

“Or rather this” — the mathematician waved a sheet of paper — “this instruction about it. To quote verbatim, ‘Everyone is expected to wash their hands for at least 15 seconds for each toilet visit.'”

“Er, what of it?”

“By a conservative estimate of mine, during my childhood I visited the toilet roughly 6570 times, plus-minus 50 as I did not keep exact count; on roughly 10% of those visits, plus-minus 2%, I was in too much of a hurry to wash my hands. In addition to these, an estimated 1210 plus-minus 120 times I visited to pick up —”

“Wh-what does this have to do with…”

“By largely similar estimates I have calculated that to retroactively achieve the desired result of 15 seconds per each toilet visit I should have to wash my hands for at least eight hours, nine minutes and thirty seconds, plus-minus fifteen minutes and fifteen.”

“Erm…”

“Given, of course, certain trivial assumptions on the definition of a toilet and a toilet visit, and the exclusion from consideration of diapers, since I could not find a published, peer-reviewed paper laying out the necessary terminology; but I have a graduate student working on that. I am honestly surprised there hasn’t been a wider outcry about this frankly nonsensical and easily misinterpreted policy. I am a busy man, and I do not have the time to spare.”


text:
### Improved Approximation Guarantees for Weighted Matching in the Semi-Streaming Model

Leah Epstein, Asaf Levin, JuliÃ¡N Mestre & Danny Segev
We study the maximum weight matching problem in the semi-streaming model, and improve on the currently best one-pass algorithm due to Zelke (Proc.\ STACS~'08, pages 669--680) by devising a deterministic approach whose performance guarantee is $4.91 + \eps$. In addition, we study {\em preemptive} online algorithms, a sub-class of one-pass algorithms where we are only allowed to maintain a feasible matching in memory at any point in time. All known results prior to Zelke's belong...


text:
# [NTG-context] add more blank space when using \starteffect[hidden]

Hans Hagen j.hagen at xs4all.nl
Sat Mar 13 10:18:36 CET 2021

On 3/12/2021 10:08 PM, Jeong Dal wrote:
> Hi,
>
> Sometimes, I need to hide the solution of a problem.
> So I use \doifmode{} and \starteffect[hidden] as in the following sample
> file.
> Since I want to set more blank space if it is hidden, I add \blank[5cm]
> when it is in the \enablemode[hideSolution].
> It works well except the figures and bars are not hidden. Only texts,
> even  in the mp part, are hidden.
> It is OK for me.
>
> However, it would be much better to set the height of blank space
> depends on the height of hidden texts.
> Is there a way to set the variable height  using “overlay” stuff instead
> of \blank[5cm]?
>
>
> Best regards,
>
> Dalyoung
>
>
> \startbuffer[circles]
> draw fullcircle scaled 1cm;
> draw fullcircle scaled 1cm shifted (2cm,0);
> dotlabel.top(“O”,origin);
> \stopbuffer
>
> \def\StartHide{\doifmode{hideSolution}{\starteffect[hidden]}}
> \def\StopHide{\doifmode{hideSolution}{\stopeffect\blank[5cm]}}
>
> %\enablemode[hideSolution]
>
> \starttext
> \framed[]{This is what I want:}
> \framedtext[]{
> \StartHide
> \samplefile{knuth}
> Overbar is not \overbar{hidden}.
> {\processMPbuffer[circles]}
> \StopHide
> }
> \stoptext
In your case it's easier than you mightt think:

\framedtext[empty=yes]

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


text:
# single precision floating point example

Range of … and the minimum positive (subnormal) value is Significand is Fraction with the “1.” restored. In double precision, 52 bits are used for mantissa. Decimal to IEEE 754 Floating Point Representation - YouTube In floating point representation, each number (0 or 1) is considered a “bit”. The IEEE 754 standard specifies a binary32 as having: This gives from 6 to 9 significant decimal digits precision. We see that O and 1. 2 By default, 1/3 rounds up, instead of down like double precision, because of the even number of bits in the significand. Hence after determining a representation of 0.375 as The mantissa is part of a number in scientific notation or a floating-point number, consisting of its significant digits. Single precision (32-bit) Double precision (64-bit) IEEE FLOATING-POINTFORMAT. By using our site, you In the IEEE 754-2008 standard, the 32-bit base-2 format is officially referred to as binary32; it was called single in IEEE 754-1985. 10 0.25 Encodings of qNaN and sNaN are not specified in IEEE 754 and implemented differently on different processors. The mantissa is part of a number in scientific notation or a floating-point number, consisting of its significant digits. × From these we can form the resulting 32-bit IEEE 754 binary32 format representation of real number 0.25: Example 3: In general, refer to the IEEE 754 standard itself for the strict conversion (including the rounding behaviour) of a real number into its equivalent binary32 format. Example 1: As shown in Example 1.36, we can convert the numbers into floating point as follows: 0.510 = 0 01110110 (1)000 0000 0000 0000 0000 0000. similarly, 0.7510 = 0 01110110 (1)100 0000 0000 0000 0000 0000. If an IEEE 754 single-precision number is converted to a decimal string with at least 9 significant digits, and then converted back to single-precision representation, the final result must match the original number.[5]. Bias number is 127. {\displaystyle 2^{-126}\approx 1.18\times 10^{-38}} 1 − ) 1. The exponent is an 8-bit unsigned integer from 0 to 255, in biased form: an exponent value of 127 represents the actual zero. {\displaystyle (0.375)_{10}} 2 {\displaystyle (0.011)_{2}} The x86 family and the ARM family processors use the most significant bit of the significand field to indicate a quiet NaN. In single precision, 23 bits are used for mantissa. The minimum positive normal value is 12.375 An IEEE 754 standard floating point binary word consists of a sign bit, exponent, and a mantissa as shown in the figure below. 42883EF9 Note that exponent is encoded using an offset-binary representation, which means it's always off by 127. Integer arithmetic and bit-shifting can yield an approximation to reciprocal square root (fast inverse square root), commonly required in computer graphics. 1.100011 Understanding “volatile” qualifier in C | Set 2 (Examples), Random Access Memory (RAM) and Read Only Memory (ROM), floating-point number is represented in two ways, Difference Between Single and Double Quotes in Shell Script and Linux, Difference between Single Bus Structure and Double Bus Structure, Difference between float and double in C/C++, Difference between Single and Multiple Inheritance in C++, Difference between Single User and Multi User Database Systems, Differences between Single Datapath and Pilpeline Datapath, Differences between Single Cycle and Multiple Cycle Datapath, Single pass, Two pass, and Multi pass Compilers, Introduction of Single Accumulator based CPU organization, Working of 8085-based Single board microcomputer, Difference between Stop and Wait, GoBackN and Selective Repeat, Difference between Stop and Wait protocol and Sliding Window protocol, Similarities and Difference between Java and C++, Difference and Similarities between PHP and C, Difference between Time Tracking and Time and Attendance Software, Difference between Input and Output devices, Computer Organization | Instruction Formats (Zero, One, Two and Three Address Instruction), Logical and Physical Address in Operating System, Difference between == and .equals() method in Java, Differences between Black Box Testing vs White Box Testing, Write Interview − The bits of 1/3 beyond the rounding point are 1010... which is more than 1/2 of a unit in the last place. − 10 There are two floating-point data representations on the C67x processor: single precision (SP) and double precision (DP). Consider a value of 0.375. The bits are laid out as follows: The real value assumed by a given 32-bit binary32 data with a given sign, biased exponent e (the 8-bit unsigned integer), and a 23-bit fraction is. It is used for minimization of approximation. O and 1. − 2 ) Therefore single precision has 32 bits total that are divided into 3 different subjects. Single Precision is a format proposed by IEEE for representation of floating-point number. With that methodology, I came up with an average decimal precision for single-precision floating-point: 7.09 digits. can be exactly represented in binary as These subjects consist of a sign (1 bit), an exponent (8 bits), and a mantissa or fraction (23 bits). It is widely used in games and programs requiring less precision and wide representation. Single Precision: Then we need to multiply with the base, 2, to the power of the exponent, to get the final result: where s is the sign bit, x is the exponent, and m is the significand. Single-precision floating-point format (sometimes called FP32 or float32) is a computer number format, usually occupying 32 bits in computer memory; it represents a wide dynamic range of numeric values by using a floating radix point. It uses 8 bits for exponent. We immediately see that $x$ is a negative number and so the sign is $\sigma = 1$. We can see that: 0.011 ( For example, the rational number 9÷2 can be converted to single precision float format as following, 9 (10) ÷ 2 (10) = 4.5 (10) = 100.1 (2) × ( 2 Consider a value 0.25. − 1.100011 It … {\displaystyle {(1.1)_{2}}\times 2^{-2}} Since the exponents of both numbers are the same, there is … Two representations. It is used in simple programs like games. 2 Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below. 2 Before the widespread adoption of IEEE 754-1985, the representation and properties of floating-point data types depended on the computer manufacturer and computer model, and upon decisions made by programming-language designers.


text:
# udfGetElmWDetJ()

Return the weighted determinant of the element Jacobian at the quadrature point.

## Syntax

wghtDetJac = udfGetElmWDetJ( udfHd ) ;

## Type

User Defined Element

## Parameters

udfHd
The opaque handle (pointer) which was passed to the user function.

## Return Value

wghtDetJac (Real*)
Pointer to one dimensional real array of the weighted determinant of the Jacobian at the element quadrature point. The array dimension is the number of elements, nItems.

## Description

This routine returns the weighted determinant of the element Jacobian at the quadrature point. The unit of this quantity is volume. The sum over the quadrature points is the volume of the element. For example,
Real* wghtDetJac ;
Real wDetJ ;
Integer elem ;
...
wghtDetJac = udfGetElmWDetJ( udfHd ) ;
for ( elem = 0 ; elem < nItems ; elem++ ) {
wDetJ = wghtDetJac[elem] ;
...
}

## Errors

• This routine expects a valid udfHd.
• This routine may only be called within a Body Force, Material Model or Component Model user function.


text:
CBSE Class 10CBSE
Account
Register

Share

Books Shortlist

Solution - Prove the Following Identities, Where the Angles Involved Are Acute Angles for Which the Expressions Are Defined. - CBSE Class 10 - Mathematics

ConceptTrigonometric Identities

Question

Prove the following identities, where the angles involved are acute angles for which the expressions are defined.

(sintheta - 2sin^3theta)/(2costheta - costheta) =tan theta

Solution

You need to to view the solution
Is there an error in this question or solution?

Reference Material

Solution for question: Prove the Following Identities, Where the Angles Involved Are Acute Angles for Which the Expressions Are Defined. concept: Trigonometric Identities. For the course CBSE
S