url stringlengths 14 2.42k | text stringlengths 100 1.02M | date stringlengths 19 19 | metadata stringlengths 1.06k 1.1k |
|---|---|---|---|
http://www.stata-press.com/books/errata/ishr.html | # Errata for An Introduction to Stata for Health Researchers
The errata for An Introduction to Stata for Health Researchers are provided below. Click here for an explanation of how to read an erratum. Click here to learn how to determine the printing number of a book.
(1) Chapter 5, p. 50, middle of page
. generate str sdate = string(date, %d) . generate str sdate = string(date, "%d")
(1) Chapter 17, p. 284, nlog.ado box
winexec "C:\Program files\NoteTab Light\NoteTab.exe" "r(filename'" winexec "C:\Program files\NoteTab Light\NoteTab.exe" "r(filename)'"
(1) Chapter 17, p. 285, last sentence | 2014-04-24 07:49:13 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.18644991517066956, "perplexity": 7350.935760462808}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-15/segments/1398223205375.6/warc/CC-MAIN-20140423032005-00077-ip-10-147-4-33.ec2.internal.warc.gz"} |
https://math.stackexchange.com/questions/2281835/how-to-determine-the-total-number-of-homomorphisms-from-v-4-rightarrow-c-2 | How to determine the total number of homomorphisms from $V_4 \rightarrow C_2$?
How to determine the total number of homomorphisms from $V_4 \rightarrow C_2$? Where $V_4$ is the Klein-$4$ group and $C_2$ is cyclic of order $2$.
So far I have managed to create $7$ just from playing about with the elements.
These are: The trivial homomorphism. $3$ homomorphisms in which all elements of $V_4$ are mapped to the identity in $C_2$ except one non identity element which is mapped to the non identity in $C_2$. And finally the $3$ homomorphisms taking $2$ distinct non-identity elements from $V_4$ and mapping them to the non identity in $C_2$.
I have a few questions.
1) If $\phi:G \rightarrow H$ is a homomorphism what can we say about the order of elements? (I.e. is it true that $o(g)|o(\phi(g))$ or the other way around maybe?)
2) Are my maps correct and are the the only homomorphisms from $V_4$ to $C_2$.
3) Is there a nicer way to argue this question other than just trying to guess all possible ones?
Thanks!
• If $g \in G$ has order $n$, then $\phi(g)^n = \phi(g^n) = \phi(e_G) = e_H$, and so $o(\phi(g))$ divides $n = o(g)$. May 15, 2017 at 10:46
• A homomorphism is always fully determined when you know where generating elements go. The delicate questions arise in determining what kind of constraints are there, or can we possibly map the generators any which way we want? $V_4$ is generated by two commuting elements of order two, so the images of those two generators must go to such an element whose square is trivial. This is no constraint, when the target is $C_2$. Nor is commutation. That leaves ____ alternatives. May 15, 2017 at 10:51
• And... some of the maps you describe are not homomorphisms. Let $V_4=\{1,a,b,c\}$, $C_2=\{1,g\}$. If $a\mapsto 1$ and $b\mapsto g$, then $c=ab$ must $\mapsto 1g=g$. May 15, 2017 at 10:52
• @JyrkiLahtonen So since $V_4=\langle (12)(34), (13)(24)\rangle := \langle a,b \rangle$ then any homomorphism $\Phi$ is determined by where $\Phi$ maps $a$ and $b$. Here we can have $3$ possible choices $a,b$ are both mapped to $1$ giving the trivial homomorphism. Alternatively we have $a$ is mapped to $1$ and $b$ is mapped to $g$ or vice versa. So in total there are $3$ homomorphisms from $V_4$ into $C_2$?Is that correct? May 15, 2017 at 12:39
• @RyanS A good start! Nothing stops you from mapping both $a$ and $b$ to $g$, though :-) May 15, 2017 at 12:51
3 Answers
The Klein four group $V_4$ is isomorphic to $\mathbb{Z}_2\oplus \mathbb{Z}_2.$ $C_2$ is $\mathbb{Z}_2$. Since the groups are abelian homomorphisms from $\mathbb{Z}_2\oplus \mathbb{Z}_2$ to $\mathbb{Z}_2$ are $\mathbb{Z}$-linear maps $\text{Hom}_{\mathbb{Z}}(\mathbb{Z}_2\oplus \mathbb{Z}_2, \mathbb{Z}_2)$. Since $\text{Hom}_{\mathbb{Z}}(\mathbb{Z}_2,\mathbb{Z}_2) = \mathbb{Z}_2$ we have \begin{eqnarray} \text{Hom}_{\mathbb{Z}}(\mathbb{Z}_2\oplus \mathbb{Z}_2, \mathbb{Z}_2) & \cong & \text{Hom}_{\mathbb{Z}}(\mathbb{Z}_2, \mathbb{Z}_2) \oplus \text{Hom}_{\mathbb{Z}}(\mathbb{Z}_2, \mathbb{Z}_2)\\ & = & \mathbb{Z}_2 \oplus \mathbb{Z}_2. \end{eqnarray}
If you have already done linear algebra, you may notice that $V_4$ is a $C_2$ vector space with dimension $2$. This allows you to write all the functions as matrices of dimension $2×1$, since being an homomorphism is the same as being linear (check it!). Indeed, the only choices you can take is where to send the generators.
Moreover, always given linear algebra topics, you're evaluating the cardinality of the dual space, that is isomorphic to $V_4$.
(group theory perspective)
By the number of ways $C_2$ is partitioned into the quotient subgroups of (together with maybe lagrange theorem) of $C_2$ which are isomorphic to the subgroups of $V_4$. By the isomorphism theorem. So the number of subgroups of $V_4$.
The group isomorphism theorem says for every sub group of $V_4$ there is a homomorphism $V_4\to C_2$ and the subgroup of $V_4$ is isomorphic to the quotient group of $C_2$. Thus one can think of the homorohisms from $V_4\to C_2$ as essentially the subgroups and vice versa. Homomorphisms up to isomorphism that is. | 2022-05-16 08:19:53 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9610387086868286, "perplexity": 160.54787662464463}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662510097.3/warc/CC-MAIN-20220516073101-20220516103101-00142.warc.gz"} |
https://www.studyadda.com/ncert-solution/decimal_q29/541/44993 | • # question_answer 29) Naresh walked 2 km 35 m in the morning and 1 km 7 m in the evening. How much distance did he walk in all? TIPS Firstly, write the given distance into km using decimals and then add by putting one below other same as whole numbers,
Naresh walked in morning = 2 km 35 m = 2 km + 35 m $=3+\frac{2}{10}+\frac{5}{100}=3+0.2+0.05=3.25$ $1\times 100+0\times 10+2\times 1+6\times \frac{1}{10}+3\times \frac{1}{100}+0\times \frac{1}{1000}$ $=100+0+2+\frac{6}{10}+\frac{3}{100}+0=102+\frac{6}{10}+\frac{3}{100}$ Naresh walked in evening = 1 km 7 m = 1 km + 7 m $=102+0.6+0.03=102.63$ $0\times 100+3\times 10+0\times 1+0\times \frac{1}{10}+2\times \frac{1}{100}+5\times \frac{1}{1000}$ $=0+30+0+0+\frac{2}{100}+\frac{5}{1000}$ $=30+\frac{2}{100}+\frac{5}{1000}=30+0.02+0.005=30.025$ Total distance $2\times 100+1\times 10+1\times 1+9\times \frac{1}{10}+0\times \frac{1}{100}+2\times \frac{1}{1000}$ Hence, total distance walked by Naresh is 3.042 km. | 2020-10-01 00:34:23 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6411561965942383, "perplexity": 1777.5954254598591}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-40/segments/1600402130531.89/warc/CC-MAIN-20200930235415-20201001025415-00112.warc.gz"} |
https://nubtrek.com/maths/trigonometry/trigonometry-standard-angles/trigonometry-understanding-standard-angles | Server Error
Server Not Reachable.
This may be due to your internet connection or the nubtrek server is offline.
Thought-Process to Discover Knowledge
Welcome to nubtrek.
Books and other education websites provide "matter-of-fact" knowledge. Instead, nubtrek provides a thought-process to discover knowledge.
In each of the topic, the outline of the thought-process, for that topic, is provided for learners and educators.
Read in the blogs more about the unique learning experience at nubtrek.
mathsTrigonometryTrigonometric Ratios for Standard Angles
### Understanding Standard Angles
What are the standard angles for which trigonometric ratios are defined? These angles are chosen because of some pattern or properties. This page explains the reason why some angles are special.
click on the content to continue..
We had studied that the ratio of two sides of a right-triangle is a constant and the such constants are named sin theta, cos theta, cdots for given values of theta = 0^@, 0.2^@, 1^@, 2.7^@, cdots.
It is noted that one cannot memorize all values of sin theta. A reference table is to be created and used when required.
Students work out values of trigonometric ratios sin, cos, and tan, for the angles for which the ratio between two sides of triangles can be computed using only given angle
The ratio between two sides of triangles be computed for the following
• equilateral triangles
• isosceles triangles
• triangles with one angle 0^@
The equilateral triangle is split into two right angled triangles as shown in the figure. The hypotenuse bar(OP) = 1. Then the side bar(OQ) = 1/2. The other side is to be computed using Pythagoras Theorem.
The standard angles in this triangles are 60^@ and 30^@
The two sides of the isosceles right angled triangle are same and the hypotenuse is 1.
The standard angle derived from isosceles right-angled-triangles is 45^@.
The figure shows a triangle Delta OPQ with /_P = 0^@. Imagine the bar(PR) moves towards bar(PO) and forms the triangle where point R meets point O.
The standard angles in a right-angled triangle with one of the angles 0^@ are 0^@ and 90^@.
The word "standard" means: common and established as norm.
The standard angles for which the trigonometric ratios are in the form of simple ratios, are derived from equilateral, isosceles, and 0^@ triangles. Use the known properties of these triangles to compute the trigonometric values for the standard angles.
Trigonometric Ratios for Standard Angles:
• 0^@ and 90^@ : right-angled-triangle with two sides 1 and the third side 0.
• 45^@ : isosceles right-angled triangle.
• 30^@ and 60^@ : half of equilateral triangle making a right-angled-triangle.
switch to interactive version | 2019-03-25 14:15:32 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6222272515296936, "perplexity": 2179.059720629965}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-13/segments/1552912203991.44/warc/CC-MAIN-20190325133117-20190325155117-00311.warc.gz"} |
http://nrich.maths.org/public/leg.php?code=-99&cl=3&cldcmpid=2641 | Search by Topic
Resources tagged with Working systematically similar to Spot the Card:
Filter by: Content type:
Stage:
Challenge level:
There are 130 results
Broad Topics > Using, Applying and Reasoning about Mathematics > Working systematically
Spot the Card
Stage: 4 Challenge Level:
It is possible to identify a particular card out of a pack of 15 with the use of some mathematical reasoning. What is this reasoning and can it be applied to other numbers of cards?
Problem Solving, Using and Applying and Functional Mathematics
Stage: 1, 2, 3, 4 and 5 Challenge Level:
Problem solving is at the heart of the NRICH site. All the problems give learners opportunities to learn, develop or use mathematical concepts and skills. Read here for more information.
Bent Out of Shape
Stage: 4 and 5 Challenge Level:
An introduction to bond angle geometry.
Troublesome Triangles
Stage: 2 and 3 Challenge Level:
Many natural systems appear to be in equilibrium until suddenly a critical point is reached, setting up a mudslide or an avalanche or an earthquake. In this project, students will use a simple. . . .
More on Mazes
Stage: 2 and 3
There is a long tradition of creating mazes throughout history and across the world. This article gives details of mazes you can visit and those that you can tackle on paper.
Crossing the Town Square
Stage: 2 and 3 Challenge Level:
This tricky challenge asks you to find ways of going across rectangles, going through exactly ten squares.
Counting on Letters
Stage: 3 Challenge Level:
The letters of the word ABACUS have been arranged in the shape of a triangle. How many different ways can you find to read the word ABACUS from this triangular pattern?
Wallpaper Sudoku
Stage: 3 and 4 Challenge Level:
A Sudoku that uses transformations as supporting clues.
Magic Caterpillars
Stage: 4 and 5 Challenge Level:
Label the joints and legs of these graph theory caterpillars so that the vertex sums are all equal.
Teddy Town
Stage: 1, 2 and 3 Challenge Level:
There are nine teddies in Teddy Town - three red, three blue and three yellow. There are also nine houses, three of each colour. Can you put them on the map of Teddy Town according to the rules?
Seasonal Twin Sudokus
Stage: 3 and 4 Challenge Level:
This pair of linked Sudokus matches letters with numbers and hides a seasonal greeting. Can you find it?
Stage: 3 Challenge Level:
Rather than using the numbers 1-9, this sudoku uses the nine different letters used to make the words "Advent Calendar".
Ratio Sudoku 1
Stage: 3 and 4 Challenge Level:
A Sudoku with clues as ratios.
Tea Cups
Stage: 2 and 3 Challenge Level:
Place the 16 different combinations of cup/saucer in this 4 by 4 arrangement so that no row or column contains more than one cup or saucer of the same colour.
Ratio Sudoku 3
Stage: 3 and 4 Challenge Level:
A Sudoku with clues as ratios or fractions.
Stage: 3 Challenge Level:
A few extra challenges set by some young NRICH members.
Corresponding Sudokus
Stage: 3, 4 and 5
This second Sudoku article discusses "Corresponding Sudokus" which are pairs of Sudokus with terms that can be matched using a substitution rule.
Triangles to Tetrahedra
Stage: 3 Challenge Level:
Starting with four different triangles, imagine you have an unlimited number of each type. How many different tetrahedra can you make? Convince us you have found them all.
Oranges and Lemons, Say the Bells of St Clement's
Stage: 3 Challenge Level:
Bellringers have a special way to write down the patterns they ring. Learn about these patterns and draw some of your own.
LOGO Challenge - Triangles-squares-stars
Stage: 3 and 4 Challenge Level:
Can you recreate these designs? What are the basic units? What movement is required between each unit? Some elegant use of procedures will help - variables not essential.
Product Sudoku 2
Stage: 3 and 4 Challenge Level:
Given the products of diagonally opposite cells - can you complete this Sudoku?
Intersection Sudoku 1
Stage: 3 and 4 Challenge Level:
A Sudoku with a twist.
You Owe Me Five Farthings, Say the Bells of St Martin's
Stage: 3 Challenge Level:
Use the interactivity to listen to the bells ringing a pattern. Now it's your turn! Play one of the bells yourself. How do you know when it is your turn to ring?
Gr8 Coach
Stage: 3 Challenge Level:
Can you coach your rowing eight to win?
Stage: 3 Challenge Level:
How many different symmetrical shapes can you make by shading triangles or squares?
Stage: 3 Challenge Level:
If you take a three by three square on a 1-10 addition square and multiply the diagonally opposite numbers together, what is the difference between these products. Why?
Twin Corresponding Sudoku III
Stage: 3 and 4 Challenge Level:
Two sudokus in one. Challenge yourself to make the necessary connections.
Inky Cube
Stage: 2 and 3 Challenge Level:
This cube has ink on each face which leaves marks on paper as it is rolled. Can you work out what is on each face and the route it has taken?
Introducing NRICH TWILGO
Stage: 1, 2, 3, 4 and 5 Challenge Level:
We're excited about this new program for drawing beautiful mathematical designs. Can you work out how we made our first few pictures and, even better, share your most elegant solutions with us?
First Connect Three for Two
Stage: 2 and 3 Challenge Level:
First Connect Three game for an adult and child. Use the dice numbers and either addition or subtraction to get three numbers in a straight line.
More Plant Spaces
Stage: 2 and 3 Challenge Level:
This challenging activity involves finding different ways to distribute fifteen items among four sets, when the sets must include three, four, five and six items.
More Children and Plants
Stage: 2 and 3 Challenge Level:
This challenge extends the Plants investigation so now four or more children are involved.
Making Maths: Double-sided Magic Square
Stage: 2 and 3 Challenge Level:
Make your own double-sided magic square. But can you complete both sides once you've made the pieces?
Colour Islands Sudoku
Stage: 3 Challenge Level:
An extra constraint means this Sudoku requires you to think in diagonals as well as horizontal and vertical lines and boxes of nine.
Reach 100
Stage: 2 and 3 Challenge Level:
Choose four different digits from 1-9 and put one in each box so that the resulting four two-digit numbers add to a total of 100.
Fence It
Stage: 3 Challenge Level:
If you have only 40 metres of fencing available, what is the maximum area of land you can fence off?
Isosceles Triangles
Stage: 3 Challenge Level:
Draw some isosceles triangles with an area of $9$cm$^2$ and a vertex at (20,20). If all the vertices must have whole number coordinates, how many is it possible to draw?
Masterclass Ideas: Working Systematically
Stage: 2 and 3 Challenge Level:
A package contains a set of resources designed to develop students’ mathematical thinking. This package places a particular emphasis on “being systematic” and is designed to meet. . . .
Stage: 3 and 4 Challenge Level:
Four numbers on an intersection that need to be placed in the surrounding cells. That is all you need to know to solve this sudoku.
Twin Corresponding Sudokus II
Stage: 3 and 4 Challenge Level:
Two sudokus in one. Challenge yourself to make the necessary connections.
Ratio Sudoku 2
Stage: 3 and 4 Challenge Level:
A Sudoku with clues as ratios.
Magic W
Stage: 4 Challenge Level:
Find all the ways of placing the numbers 1 to 9 on a W shape, with 3 numbers on each leg, so that each set of 3 numbers has the same total.
Instant Insanity
Stage: 3, 4 and 5 Challenge Level:
Given the nets of 4 cubes with the faces coloured in 4 colours, build a tower so that on each vertical wall no colour is repeated, that is all 4 colours appear.
Coins
Stage: 3 Challenge Level:
A man has 5 coins in his pocket. Given the clues, can you work out what the coins are?
Football Sum
Stage: 3 Challenge Level:
Find the values of the nine letters in the sum: FOOT + BALL = GAME
First Connect Three
Stage: 2 and 3 Challenge Level:
The idea of this game is to add or subtract the two numbers on the dice and cover the result on the grid, trying to get a line of three. Are there some numbers that are good to aim for?
Twin Corresponding Sudoku
Stage: 3, 4 and 5 Challenge Level:
This sudoku requires you to have "double vision" - two Sudoku's for the price of one
Cayley
Stage: 3 Challenge Level:
The letters in the following addition sum represent the digits 1 ... 9. If A=3 and D=2, what number is represented by "CAYLEY"?
Pair Sums
Stage: 3 Challenge Level:
Five numbers added together in pairs produce: 0, 2, 4, 4, 6, 8, 9, 11, 13, 15 What are the five numbers? | 2016-04-29 19:45:39 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.2783859968185425, "perplexity": 2385.291728168964}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-18/segments/1461860111396.55/warc/CC-MAIN-20160428161511-00070-ip-10-239-7-51.ec2.internal.warc.gz"} |
https://dantopology.wordpress.com/2014/06/03/every-corson-compact-space-has-a-dense-first-countable-subspace/ | # Every Corson compact space has a dense first countable subspace
In any topological space $X$, a point $x \in X$ is a $G_\delta$ point if the one-point set $\left\{ x \right\}$ is the intersection of countably many open subsets of $X$. It is well known that any compact Hausdorff space is first countable at every $G_\delta$ point, i.e., if a point of a compact space is a $G_\delta$ point, then there is a countable local base at that point. It is also well known that uncountable power of first countable spaces can fail to be first countable at every point. For example, no point of the compact space $[0,1]^{\omega_1}$ can be a $G_\delta$ point. In this post, we show that any Corson compact space has a dense set of $G_\delta$ point. Therefore, any Corson compact space is first countable on a dense set (see Corollary 4 below). However, it is not true that every Corson compact space has a dense metrizable subspace. See Theorem 9.14 in [2] for an example of a first countable Corson compact space with no dense metrizable subspace. A list of other blog posts on Corson compact spaces is given at the end of this post.
The fact that every Corson compact space has a dense first countable subspace is taken as a given in the literature. For one example, see chapter c-16 of [1]. Even though Corollary 4 is a basic fact of Corson compact spaces, the proof involves much more than a direct application of the relevant definitions. The proof given here is intended to be an online resource for any one interested in knowing more about Corson compact spaces.
For any infinite cardinal number $\kappa$, the $\Sigma$-product of $\kappa$ many copies of $\mathbb{R}$ is the following subspace of $\mathbb{R}^\kappa$:
$\Sigma(\kappa)=\left\{x \in \mathbb{R}^\kappa: x_\alpha \ne 0 \text{ for at most countably many } \alpha < \kappa \right\}$
A compact space is said to be a Corson compact space if it can be embedded in $\Sigma(\kappa)$ for some infinite cardinal $\kappa$.
For each $x \in \Sigma(\kappa)$, let $S(x)$ denote the support of the point $x$, i.e., $S(x)$ is the set of all $\alpha<\kappa$ such that $x_\alpha \ne 0$.
Proposition 1
Let $Y$ be a Corson compact space. Then $Y$ has a $G_\delta$ point.
Proof of Proposition 1
If $Y$ is finite, then every point is isolated and is thus a $G_\delta$ point. Assume $Y$ is infinite. Let $\kappa$ be an infinite cardinal number such that $Y \subset \Sigma(\kappa)$. For $f,g \in Y$, define $f \le g$ if the following holds:
$\forall \ \alpha \in S(f)$, $f(\alpha)=g(\alpha)$
It is relatively straightforward to verify that the following three properties are satisfied:
• $f \le f$ for all $f \in Y$. (reflexivity)
• For all $f,g \in Y$, if $f \le g$ and $g \le f$, then $f=g$. (antisymmetry)
• For all $f,g,h \in Y$, if $f \le g$ and $g \le h$, then $f \le h$. (transitivity)
Thus $\le$ as defined here is a partial order on the compact space $Y$. Let $C \subset Y$ such that $C$ is a chain with respect to $\le$, i.e., for all $f,g \in C$, $f \le g$ or $g \le f$. We show that $C$ has an upper bound (in $Y$) with respect to the partial order $\le$. We need this for an argument using Zorn’s lemma.
Let $W=\bigcup_{f \in C} S(f)$. For each $\alpha \in W$, choose some $f \in C$ such that $\alpha \in S(f)$ and define $u_\alpha=f_\alpha$. For all $\alpha \notin W$, define $u_\alpha=0$. Because $C$ is a chain, the point $u$ is well-defined. It is also clear that $f \le u$ for all $f \in C$. If $u \in Y$, then $u$ is a desired upper bound of $C$. So assume $u \notin Y$. It follows that $u$ is a limit point of $C$, i.e., every open set containing $u$ contains a point of $C$ different from $u$. Hence $u$ is a limit point of $Y$ too. Since $Y$ is compact, $u \in Y$, a contradiction. Thus it must be that $u \in Y$. Thus every chain in the partially ordered set $(Y,\le)$ has an upper bound. By Zorn’s lemma, there exists at least one maximal element with respect to the partial order $\le$, i.e., there exists $t \in Y$ such that $f \le t$ for all $f \in Y$.
We now show that $t$ is a $G_\delta$ point in $Y$. Let $S(t)=\left\{\alpha_1,\alpha_2,\alpha_3,\cdots \right\}$. For each $p \in \mathbb{R}$ and for each positive integer $n$, let $B_{p,n}$ be the open interval $B_{p,n}=(p-\frac{1}{n},p+\frac{1}{n})$. For each positive integer $n$, define the open set $O_n$ as follows:
$O_n=(B_{t_{\alpha_1},n} \times \cdots \times B_{t_{\alpha_n},n} \times \prod_{\alpha<\kappa,\alpha \notin \left\{ \alpha_1,\cdots,\alpha_n \right\}} \mathbb{R}) \cap Y$
Note that $t \in \bigcap_{n=1}^\infty O_n$. Because $t$ is a maximal element, note that if $g \in Y$ such that $g_\alpha=t_\alpha$ for all $\alpha \in S(t)$, then it must be the case that $g=t$. Thus if $g \in \bigcap_{n=1}^\infty O_n$, then $g_\alpha=t_\alpha$ for all $\alpha \in S(t)$. We have $\left\{t \right\}= \bigcap_{n=1}^\infty O_n$. $\blacksquare$
Lemma 2
Let $Y$ be a compact space such that for every non-empty compact subspace $K$ of $Y$, there exists a $G_\delta$ point in $K$. Then every non-empty open subset of $Y$ contains a $G_\delta$ point.
Proof of Lemma 2
Let $U_1$ be a non-empty open subset of the compact space $Y$. If there exists $y \in U_1$ such that $\left\{y \right\}$ is open in $Y$, then $y$ is a $G_\delta$ point. So assume that every point of $U_1$ is a non-isolated point of $Y$. By regularity, choose an open subset $U_2$ of $Y$ such that $\overline{U_2} \subset U_1$. Continue in the same manner and obtain a decreasing sequence $U_1,U_2,U_3,\cdots$ of open subsets of $Y$ such that $\overline{U_{n+1}} \subset U_n$ for each positive integer $n$. Then $K=\bigcap_{n=1}^\infty \overline{U_n}$ is a non-empty closed subset of $Y$ and thus compact. By assumption, $K$ has a $G_\delta$ point, say $p \in K$.
Then $\left\{p \right\}=\bigcap_{n=1}^\infty W_n$ where each $W_n$ is open in $K$. For each $n$, let $V_n$ be open in $Y$ such that $W_n=V_n \cap K$. For each $n$, let $V_n^*=V_n \cap U_n$, which is open in $Y$. Then $\left\{p \right\}=\bigcap_{n=1}^\infty V_n^*$. This means that $p$ is a $G_\delta$ point in the compact space $Y$. Note that $p \in U_1$, the open set we start with. This completes the proof that every non-empty open subset of $Y$ contains a $G_\delta$ point. $\blacksquare$
Proposition 3
Let $Y$ be a Corson compact space. Then $Y$ has a dense set of $G_\delta$ points.
Proof of Proposition 3
Note that Corson compactness is hereditary with respect to closed sets. Thus every compact subspace of $Y$ is also Corson compact. By Proposition 1, every compact subspace of $Y$ has a $G_\delta$ point. By Lemma 2, $Y$ has a dense set of $G_\delta$ points. $\blacksquare$
Corollary 4
Every Corson compact space has a dense first countable subspace.
____________________________________________________________________
Blog posts on Corson compact spaces
____________________________________________________________________
Reference
1. Hart, K. P., Nagata J. I., Vaughan, J. E., editors, Encyclopedia of General Topology, First Edition, Elsevier Science Publishers B. V, Amsterdam, 2003.
2. Todorcevic, S., Trees and Linearly Ordered Sets, Handbook of Set-Theoretic Topology (K. Kunen and J. E. Vaughan, eds), Elsevier Science Publishers B. V., Amsterdam, 235-293, 1984.
____________________________________________________________________
$\copyright \ 2014 \text{ by Dan Ma}$ | 2017-08-17 15:34:35 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 165, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9784541726112366, "perplexity": 55.878113169120645}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-34/segments/1502886103579.21/warc/CC-MAIN-20170817151157-20170817171157-00619.warc.gz"} |
https://www.greencarcongress.com/2017/11/20171117-teslasemi.html | Musk unveils Tesla Semi; 500-mile range at highway speed with 80,000 lbs GVW; next-gen Roadster appears
17 November 2017
Emphasizing “BAMF” performance, Tesla CEO Elon Musk introduced the Tesla Semi in an evening event. The sleek electric truck, with a 0.36 coefficient of drag (supported by intelligent flaps that support a range of trailers), will accelerate from 0-60 mph in 5 seconds, charge up a 5% grade at 65 mph, and deliver 500 miles of range, Musk said.
According to Musk, 80% of truck routes are less than 250 miles; the Tesla Semi thus in theory could make a round trip on those routes without charging, he suggested.
Taking a page from Toyota’s presentation book, Tesla ran animations comparing the acceleration of the Tesla Semi vs. a conventional diesel truck. (For the reveal of its Class 8 hydrogen fuel cell prototype, Toyota presented a video of a side by side acceleration demonstration between the actual Portal fuel cell prototype and a diesel. Earlier post.) The diesel made it from 0-60 in 20 seconds, compared to the 5 seconds of the Tesla Semi.
View from the cockpit with its centered driver position.
The Tesla Semi will feature an enhanced autopilot capability as standard, with automatic emergency braking, automatic lane keeping, and forward collision warning.
Tesla guarantees the 4-motor drivetrain (one independent motor on each of the rear four wheels) to last one million miles. Tesla estimates $1.26/mile average cost versus$1.51/mile for a diesel truck.
Musk also emphasized the economic benefits of a 3-truck platoon, which can he said can beat rail in terms of cost.
Musk said that reserving a Tesla Semi requires a $5,000 reservation. Production is projected to begin in 2019. New solar-powered “Megachargers” will be able to add 400 miles of range in 30 minutes of charging. Tesla also revealed the next-gen Roadster, with eye-watering specs: 0-60 seconds in 1.9 seconds, the quarter mile in 8.9 seconds, 0-100 mph in 4.2 seconds, and a 200 kWh battery pack supporting 620 miles of range. The four-seater is slated to be available in 2020. Founders Series reservations for the Roadster 2 run$250,000; a standard Roadster 2 reservation runs $45,000. Comments How much of the gross vehicle weight of 80,000 lbs us taken up by the battery pack? Funnily enough, truck operators need to know that, and the load is critical. It blows my mind that people are dumb enough not to see that this is hype and hookum, not a product launch with actionable specifications. The only real thing here is Tesla hoping to hook in$50k deposits for a new Roadster, to be uses as utterly unsecured venture capital at zero interest rate instead of being put into a ring-fenced account.
That people can be that dumb also blows my mind.
Let the circus continue.
And as Singleton Engineer said on Seeking Alpha:
'A Roadster with acceleration from rest of 2 seconds to 50mph?
That's 1.38g.
Comparison: Braking performance (max): Range 0.5 to 10.g. The latter requires high standard road surfaces plus "sticky tyres".
Along comes Elon Musk and Tesla claiming 1.32g.
I cry bulls__t.
At the very least, this car will need new high performance tyres after each couple of demonstration bursts - and this car is only for speed demonstration purposes, right? It's not for actual transport?
There's nothing green about that kind of performance. Even if it was powered by pixie dust it would be too hard on the planet.'
If reports elsewhere are true (https://electrek.co/2017/11/16/tesla-semi-live-blog/) that the cost will be $200k to$250k, then the payback time versus diesel looks like at least 4 years at 100,000 miles per year (which is on the high side of average applications, which are under 70k miles per year). There are maybe other reasons to buy it, so missing the magic 3 year payback will not be fatal.
No mention of whether this is a day cab or a sleeper configuration, though it looks like the latter.
No mention of energy on board. Assuming 1000 kWh, they must have efficiency down to 2 kWh per mile, which is better than the usual 2.5 kWh per mile for diesel. I assume this is best case, good road conditions are moderate temperatures.
That would put charging rates north of 1 MW. That will be "fun". One plug or two? Or four? North of 1000 V? Time will tell...
OK, so not much technical detail but clearly this is going to take some time. That is to be expected.
With Proterra already putting 0.66 MWh on a transit bus, this should not be so surprising. Two years away seems reasonable. They have lots of work to do in the meantime. The vibration profile is going to be a challenge. The sonically welded fuse wires on the small cells might not survive vibration, so the battery tech might be drastically different compared to passenger cars.
Another hand to TESLA for:
1) A first generation long haul e-truck with 500 miles range.
2) A next generation super 620 miles range Roadster.
Those two units will more than match equivalent ICE versions, except for range. Quick charging stations will have to supply up to 400 KW instead of 150 KW. It is doable now.
One advantage for going electric is that batteries price, size and weight will progressively go down by about 8%/year making both units more competitive in the near future.
'Congrats on making a truck that will gets it undercarriage torn off in its 5,000th mile at any one of these exits / underpasses within one square mile of a light industrial area in Buffalo. Nice clearance specs.'
Davemart. It nice to know that your mind is blown. The Roadster is out to improve on the specs of gas card like Veyrons, McLarens, Lamborghinis and other cars that sell for 2, 3 and 4 times the price. No one regularly drives those cars at their potential. It is a status symbol and proves the superiority of EVs. Oh and it's not 0-50 in under 2 seconds, it's 0-60mph.
P.S. even if you drove at it's potential it would be many times greener than those other models.
Bragging about 0-60 times for a god damn semi truck is retarded as it disregards the safety of the cargo being subjected to such jarring forces. Just as I own a gas and diesel box truck I purposely accelerate at relatively the same speed to protect the contents of my cargo and securing devices.
In regards to range how will stop and go traffic, high heat, or extreme cold affect range which are unavoidable facts of life. I'd support a hybrid semi-truck before full blown EV as you get the benefit of peace of mind regardless of human operated. As a broken charger station or line for "refueling" means my shipment is late which in a perishable or live animal cargo situation will be costly.
Paroway:
All the Roadster needs is a magical new battery, with around 2.5 times the energy density of current ones.
We have heard nothing about it, but Tesla is going to be good enough to take $250,000 a pop for utterly unsecured, zero interest rate venture capital into the business from anyone fancying that they will manage it. After all, they took money for cars which were supposed to be fully Level 5 Autonomous Driving capable, with actually making it work to follow. And if the battery does not work, tough on the potential buyers, as it is, as I said, entirely unsecured. Good to know that you’re looking out for the best interests of the poor fools who can afford to plunk down a$50,000 deposit on a $250,000 car, Davemart. It’s not like folks with that kind of scratch could figure out the risk profile of that kind of bet, right? Porsche sold the 918 for$800,000 - $1,000,000. Tesla will have buyers lined up for at least a similar volume production run of a$200-$250k car with better performance. I saw the Roadster do performance runs on the tarmac Thursday night. If you think Tesla, which already produces the worlds fastest production car, will have any trouble producing the Roadster 2.0, please short them heavily. The market will be happy to take your money. I've seen estimates on the truck battery being from 1200kwh to 1000kwh. If it's full load 500 miles range it would likely be the former. I am actually pleasantly surprised that they managed 500miles. I was expecting the commonly rumored 250 or so. As, Dave has pointed out the gross weight is 80k. This could be an issue when you're talking about such big battery packs. I am happy they have organized a mega charger too. Hopefully others adopt or agree on a common standard soon, or even that tesla cross licenses its stations/charge infrastructures for others to use. It looks to be at over 1mw charger. We could be at 1000volts, to make things easier on the connector and cable end. Smaller anyway. The trucks will catch on, same with the Nikola one, cost per mile is less than diesel. Thus, it will sell like crazy.$250k for an exotic isn't bad, it will be the second or third ever for an American car company depending on how you view the viper or some of the really small manufacturer. Those tires must cost a fortune though, all that weight, and the speed rating to boot.
ECI, maybe I'm being picky, but you should say fastest acceleration. I got really confused for a second.
I always associate fast with speed, and quickness with acceleration.
Hopefully we won't have any 250mph+, 5000lbs+ missiles on the road. The destruction would be insane. The amount of energy, wow.
You're right, CE, I should have said quickest. Musk did say the new Roadster would top out at 250 mph. I share your concern. People are already getting into trouble with the S and X. A doctor in Orange County California is facing manslaughter charges after killing a couple of commuters returning home from work.
Does tesla speed limit their cars?
Typical ICE cars are speed limited by the manufacturers based on several parameters...like the tires equipped being the biggest has to do with the speed rating and the heat dissipation, and load properties. Like if you push triple digits speed you need to inflate your tires a lot more.
I imagine, in several years, someone will find a way to sue auto makers for not limiting cars to 80mph, and another person shortly after that sue for limiting the cars speed.
Some cars use track keys, and normal drive keys for sporty features, like running out of EPA parameters on a track i.e not road legal emissions(well not certified), extra boost and so on.
Its really not how fast you go, its how quickly you come to a stop that's what kills you.
It would be interesting to see how tesla handles that, perhaps a legal page right before plaid speed.
I'm interested to know the sales numbers on the bolt and the model 3 as of late. I worry that tesla is going to be in a bad way if they consistently can't deliver on
their promises. I'm not knocking the things they have promised, all of their cars are very much in demand, but its just getting them into peoples hands that seems bothersome to me.
I used to work on preproduction units, and I know startups can have problems, and if one supplier has an issue, it can wreak havoc at everyone else involved. Tesla, even though highly valued likely doesn't have a lot of pull with suppliers. Some cars almost sell 1million units a year, just in the states on a global car, usually its daily almost exponential growth during startup trials, and sometimes it goes very quick, but the 2800 or whatever deliveries speak of trouble. They have had a long while to ramp up since they debuted their Alpha models.
Tesla/Musk certainly does not suffer from a lack of ambition or thinking large. I just hope that they can stay above water. I know that they are late with the Series 3 production.
I Think that I would have tried a truck designed more for local delivery first. I have talked to a number of our delivery drivers. we get daily deliveries of steel, ltl freight, etc. plus UPS, FedEx, and OnTrac. Anyway, about the furthest that they drive is less than 150 miles. It is much easier to design a truck for relatively low speed stop and go traffic than trying to compete in the long haul market.
Davemart
Concerning the comments on the acceleration. Last year, The FormulaSAE students from Prauge, Czech Republic built an all wheel drive electric formula car that turned 0 t0 100 Km/hr (62 mph) in 1.513 sec. They were, of course, using race tires and they had fudged the rules slightly to use a total of 100 KW for a demo run while the FormulaSAE rules limit the students to 80 KW during the competition.
With all of the hoopla surrounding Tesla’s new semi-truck, not to mention the surprise unveiling of a next-gen Roadster, it was easy to miss Elon Musk show off the first official renderings of Tesla’s long-rumored pickup truck. During the special media event, Musk briefly showed crowd attendees a somewhat odd concept of a pickup truck big enough to carry a traditionally sized pickup truck. As Musk intimated, the design is essentially a mini version of the Tesla Semi with 4 wheels instead of the semi's 6, meaning it would not need a truck drivers license. Also Musk has tweeted the pickup could be the base for a cargo van. I would go for one of those, it would make for a great Class B motorhome.
It was hard to figure out if that was just a sight gag or a serious peek at some future vehicle that would dominate in the mud-bog crowd.
I agree, ai_vin, a motorhome based on this would be interesting. Hard to know how the solo driver position would go over, though.
Also, megacharging along with the Semis would be less than optimal. Otoh, 240 overnight at an RV park would be easy, if not cheap. This is not just speculation, I've done both with my Model S.
US based factories not longer perform well enough.
To survive, TESLA will have to move some or all battery, storage facilities, charging units and vehicle manufacturing to higher performance factories in Asia.
If not, many others will and could drive TESLA out of business by 2025 or so.
Apple is one of the good example to follow?
^ Let's support North American manufacturing, please.
Tesla could very well be out of business by 2025 due to committing to too many expensive projects too soon. They really should focus on getting Model 3 production up and running, and leave the pet projects aside for now.
The Tesla semi as shown is not production ready; they've got a fair number of visible non-conformances to motor vehicle safety standards in the concept vehicle, in addition to potential ground clearance concerns that someone else mentioned. Mirrors (you are not allowed to substitute them with cameras! - you can have cameras, but you still need mirrors). Air brakes with the associated mandated gauges and handles, at a minimum. The ability to see and inspect the drive wheels and hubs and brakes. Windows that roll down. The ability to see around traffic when making a left turn without relying on cameras (central driving position is not good for this - and it's not such a critical issue with right turns).
Whatever the non-compliance issues with the concept/prototype, it's likely that Tesla's offering will be much closer to the production than most of the wild flights of fancy we see when just about every other manufacturer shows off a concept/prototype.
Agree that focus would be reassuring in the face of delays.
Agree with the principle to support North American manufacturing as long as we can keep a competitive productivity level.
Lower relative productivity produces huge delays and will take TESLA down. Future intelligent robots may help?
TESLA may have to do what Apple (and many others) have done and move some or all manufacturing to Asia where labour is plentiful, less costly and more productive.
Research, design, upgrading and new models testing could remain in USA, for models sold in USA. Models sold in Asia should designed locally.
^ I doubt if you've set foot in a manufacturing plant lately. I do every day.
Tesla's problem at the moment has nothing to do with productivity and nothing to do with robots (they've got plenty of them), and a lot to do with making too many promises that can't be kept, and that's a management issue. The delays aren't because the workers aren't working (Far from it), they're because management has underestimated the monumental task of making everything work together properly. They opted to skip prototype build ... well, guess what, they're doing it anyhow, whether they wanted to or not, and it's to the surprise of nobody who works in the auto industry. They will get it done, but it will be months late (i.e. it will end up taking the normal amount of time, instead of being compressed the way it was promised).
In one of the videos that shows the Model 3 being assembled, you can see a piece of equipment that a customer of mine built. And that is about as much as I can say about it. What it means is that I do not want Tesla to fail ... but I do want them to do what they say they will do when they said they will do it, the corollary of that being don't make promises you can't keep.
Tesla used 7 cents/kwh in their ev ($1.26/mi) vs diesel ($1.51/mi) operating costs comparison.
Where do they expect truck drivers to be able to purchase electricity on demand for 7 cents/kwh? Certainly not in California!
I suspect the ev operating cost advantage disappears if they assume California rates (15-30 cents/kwh).
Tesla said that the price of electricity at the megachargers would be $0.07. Considering that they own a solar and panel cell factory, utility scale battery manufacturing plant and nationwide installation crew, it seems reasonable to imagine that they could provide electricity for$0.07 within a few years.
@ eci
Where does it say the pickup would have a solo driver's position? Musk said his pickup would be big enough to carry another pickup: The track width of an F150 is 67 inches ball park, center to center of tire so add say another 8 inches for wheel width so 75 inches between the Tesla’s wheel wells would work. The box would need to be 80 inches on the inside above the wheel wells to fit the F150 body width. With that much width to work with where's no need for a central driving position and you could have 3 people side-by-side up front easily.
The comments to this entry are closed. | 2023-01-29 16:05:11 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.23476187884807587, "perplexity": 2790.3414228493116}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764499744.74/warc/CC-MAIN-20230129144110-20230129174110-00023.warc.gz"} |
https://ccssmathanswers.com/into-math-grade-3-module-14-lesson-2-answer-key/ | # Into Math Grade 3 Module 14 Lesson 2 Answer Key Partition Shapes into Equal Areas
We included HMH Into Math Grade 3 Answer Key PDF Module 14 Lesson 2 Partition Shapes into Equal Areas to make students experts in learning maths.
## HMH Into Math Grade 3 Module 14 Lesson 2 Answer Key Partition Shapes into Equal Areas
I Can divide shapes into parts with equal areas and write each equal part as a fraction.
Spark Your Learning
Mia designs a playground. She plans to divide it into 4 parts with equal areas. So far, Mia has found one way to divide the playground.
Show another way to divide the playground. Explain how you know that the 4 parts of your plan are equal in area.
Answer:
Explanation:
Mia designs a playground. She plans to divide it into 4 parts with equal areas
So far, Mia has found one way to divide the playground
I divided the play ground into four parts in another possible way
I divided the whole into 2 equal parts first and then each part into 2 equal halves again. So, i can say that all the 4 parts are equal.
Turn and Talk Can two rectangles be equal in area if they have different shapes? How do you know?
Answer:
Yes, two rectangles can be equal in area even if they have different shapes. even though the length of the rectangles differ the area it covers could be equal.
Build Understanding
Question 1.
How can you divide the rectangle into 6 parts with equal areas?
Answer:
A. How do you know that the parts are equal in area?
Answer:
I divided the rectangle into equal parts
I divided the whole into 2 equal parts first and then each part into 3 equal parts again. So, i can say that all the 6 parts are equal.
B. What fraction of the whole rectangle is each part?
Answer:
$$\frac{1}{6}$$
Explanation:
There are 6 parts in the whole
The fraction of 1 part of the whole rectangle is $$\frac{1}{6}$$.
Turn and Talk Is there another way to divide the rectangle into six equal parts? If so, describe how you would divide it.
Answer:
yes, we can divide the rectangle into 6 equal parts in other way by dividing it vertically into 6 smaller rectangles.
Question 2.
Divide the circle so that each part is $$\frac{1}{3}$$ of the area of the whole shape. How can you show that each part of your circle is equal in area? Explain.
Answer:
Explanation:
I divided the circle into 3 equal parts so that each part is $$\frac{1}{3}$$ of the area of the whole shape.
Question 3.
Critique Reasoning Angela and Pedro divide rectangles into 8 equal parts in different ways. Both rectangles are the same size and shape.
A. Do the parts from Angela’s rectangle have the same shape as the parts from Pedro’s rectangle? Explain.
Answer:
No, the parts of Angela’s rectangle does not have the same shape as the parts of Pedro’s rectangle as the parts of Angela’ s rectangle are made with vertically and the parts of Pedro’s rectangle are made horizontally.
B. Who divided the rectangle into parts that are greater in area, Angela or Pedro? Explain how you know.
Answer:
Both Angela or Pedro divided their rectangle into equal parts. So, the parts of both have same area.
Turn and Talk How can you show another way to divide the rectangle into 8 equal parts?
Answer:
explanation:
I drew a way to divide a rectangle into 8 parts other than the ways Angela and Pedro did.
Check Understanding
Question 1.
What fraction of the total area of the circle does each equal part represent?
Answer:
$$\frac{1}{4}$$
Explanation:
The circle is divided into 4 equal parts
So, each part represent $$\frac{1}{4}$$ fraction of total area of circle.
Question 2.
Nora and Nick cut felt squares to make 6 equal-sized patches. Nick wants his patches to have a different shape but the same area as Nora’s patches. Draw lines to show how they can cut each of the felt squares.
Answer:
Explanation:
Nora and Nick cut felt squares to make 6 equal-sized patches. Nick wants his patches to have a different shape but the same area as Nora’s patches.
So, i drew lines to show how they can cut each of the felt squares in different ways
Now both of them have equal parts with different shapes.
On Your Own
Question 3.
Construct Arguments Glenn says he can divide the six-sided shape into 3 parts with equal areas. Emily says she can divide the shape into 6 parts with equal areas. Whose statement is correct? Draw to explain.
Answer:
Explanation:
Glenn says he can divide the six-sided shape into 3 parts with equal areas.
Emily says she can divide the shape into 6 parts with equal areas
Both are correct, we can divide a six-sided shape into 3 as well as six equal parts as i drew above
The first shape represents shape with 3 equal parts and the second shape represents shape with 6 equal parts.
Question 4.
Attend to Precision Divide the shape into 6 equal parts.
How do you know that you divided your shape into 6 equal parts?
Answer:
Explanation:
I divided the above given shape into 6 equal parts
I counted the number of boxes each part cover
All the parts cover same number of boxes in area grid
So, i know that i divided my shape into 6 equal parts.
Use the shapes for 5-6.
Question 5.
Shade $$\frac{1}{4}$$ of the shape on the left.
Answer:
Explanation:
I shaded 1 part of shape in the left
The shaded part represents the fraction $$\frac{1}{4}$$ .
Question 6.
Is there another way to divide the shape into 4 equal parts?
If yes, draw lines on the shape on the right to show a way. Shade the shape to show $$\frac{1}{4}$$.
Answer:
Explanation:
Yes, their is an another way to divide the given shape into 4 equal parts
I drew lines to divide, the right side shape is another way i have shown
I shaded 1 part of the 4 equal parts
I shaded the shape to show $$\frac{1}{4}$$.
I’m in a Learning Mindset!
How can I use different tools and strategies to help me see if i have divided a shape into parts with equal areas?
Answer:
Scroll to Top | 2022-09-27 01:48:55 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.2634661793708801, "perplexity": 942.6089327165205}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030334974.57/warc/CC-MAIN-20220927002241-20220927032241-00479.warc.gz"} |
http://clay6.com/qa/10139/the-maximum-number-of-equivalence-relations-on-the-set-a-is- | # The maximum number of equivalence relations on the set $A=\{1,2,3\}$ is?
(A) 1
(B) 2
(C) 3
(D) 5
$R_1=\{(1,1),(2,2),(3,3)\}$
$R_2=\{(1,1),(2,2),(3,3),(1,2),(2,1)\}$
$R_3=\{(1,1),(2,2),(3,3),(1,3),(3,1)\}$
$R_4=\{(1,1),(2,2),(3,3),(2,3),(3,2)\}$
$R_4=\{(1,1),(2,2),(3,3),(1,2),(2,1),(1,3),(3,1),(2,3),(3,2)\}$
These are the 5 relations on A which are equivalence. | 2017-11-24 18:25:19 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.671409547328949, "perplexity": 473.0036607902286}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-47/segments/1510934808742.58/warc/CC-MAIN-20171124180349-20171124200349-00082.warc.gz"} |
https://math.stackexchange.com/questions/4224647/evaluating-int-0-infty-frac-ln-x1-xn-textdx/4224681 | # Evaluating $\int_0^{\infty} \frac{\ln (x)}{1-x^n}\text{d}x$
This is just for recreational purposes.
I've been wondering how to evaluate the following integral:
$$\int_0^{\infty} \frac{\ln (x)}{1-x^n}\text{d}x$$
Because I noticed that certain values for $$n$$ lead to some nice rational multiples of $$\pi^2$$ when evaluated in WolframAlpha. This leads me to believe we should somehow get it into series form to turn it into something Basel-esque. However, I've been struggling with this. I managed to put into series form a related integral, to at least show some kind of context/effort, which is:
$$\int_0^{1} \frac{\ln (x)}{1-x^n}\text{d}x$$
When inputting the series expansion of $$1-x^n$$, one simply needs to evaluate an integral of the form $$x^{kn} \ln(x)$$, which is simple, and the following series emerges:
$$\int_0^{1} \frac{\ln (x)}{1-x^n}\text{d}x = \sum_{k=0}^{\infty} \frac{1}{(kn+1)^2}$$
However, this approach doesn't work with bounds from $$0$$ to $$\infty$$ for obvious reasons. Could someone give me some help?
• I believe that the infinite series evaluation of the integral should be negative. Aug 14, 2021 at 20:43
• In Mathematica: Assuming[n \[Element] Integers \[And] n > 1, Integrate[Log[x]/(1-x^n),{x,0,\[Infinity]}]]. $$- \frac{\pi^2 \csc^{-1} (\pi/n)}{n^2}$$ Aug 14, 2021 at 21:19
It turns out that $$\int_0^1\frac{\ln(x)}{1-x^a}\mathrm{d}x=-\frac{1}{a^2}\sum_{n=0}^{\infty}\frac{1}{(n+1/a)^2}$$ (I think you dropped a minus sign). Enforcing the substitution $$x\rightarrow 1/x$$ yields $$\int_1^{\infty}\frac{\ln(x)}{1-x^a}\mathrm{d}x=\int_0^1\frac{x^{a-2}\ln(x)}{1-x^a}\mathrm{d}x=\sum_{n=0}^{\infty}\int_0^1\ln(x)x^{an+a-2}\mathrm{d}x=-\frac{1}{a^2}\sum_{n=0}^{\infty}\frac{1}{\big(n+1-1/a\big)^2}$$ Putting both pieces together yields $$\begin{eqnarray*}\int_0^{\infty}\frac{\ln(x)}{1-x^a}\mathrm{d}x&=&-\frac{1}{a^2}\Bigg[\sum_{n=0}^{\infty}\frac{1}{(n+1/a)^2}+\sum_{n=0}^{\infty}\frac{1}{(n+1-1/a)^2}\Bigg] \\ &=& -\frac{1}{a^2}\Bigg[\sum_{n=0}^{\infty}\frac{1}{(n+1/a)^2}+\sum_{n=1}^{\infty}\frac{1}{(-n+1/a)^2}\Bigg] \\ &=& -\frac{1}{a^2}\sum_{n=-\infty}^{\infty}\frac{1}{(n+1/a)^2} \end{eqnarray*}$$ It is a known result in complex analysis (using residues to evaluate sums of series) that for any $$x\notin \mathbb{Z}$$ we have $$\sum_{n=-\infty}^{\infty}\frac{1}{(n+x)^2}=\Big(\pi \csc(\pi x)\Big)^2$$ Taking $$x$$ to be $$1/a$$ yields $$\int_0^{\infty}\frac{\ln(x)}{1-x^a}\mathrm{d}x=-\frac{\pi^2}{a^2}\csc^2(\pi/a)$$
• Excellent! This was actually recently done in a complex analysis course over the Summer, so its quite nice to see that pop up. Aug 14, 2021 at 22:02
Considering the other half of the integral: $$I = \int_1^\infty {\frac{\ln(x)}{1-x^n}} \;dx$$
Let $$x = u^{-1} \implies dx = -u^{-2} \; du$$.
$$I = \int_1^0 {\frac{-\ln(u)}{1-u^{-n}}\cdot(-u^{-2})} \; du$$
$$= \int_0^1 {\frac{u^{n-2}\ln(u)}{1-u^{n}}} \; du$$
$$= \int_0^1 {\ln(u)\cdot\Bigg(u^{n-2}+u^{2n-2}+\dots\Bigg) \;du}$$
$$= -\sum_{k=1}^{\infty} \frac{1}{(kn-1)^2}$$
which is divergent for $$n=1$$. (I would be interested to know if both sums can be put together to get the closed form with $$\csc$$.)
@user429040 Already has an excellent answer. I'd like to expand a little bit on the sum formula that he has presented: $$\sum_{k\in\Bbb{Z}}\frac{1}{(k+z)^2}=\pi^2\csc^2(\pi z)$$ How do we prove this formula? Well, first we expand it using partial fractions into two different sums: $$\sum_{k\in\Bbb{Z}}\frac{1}{(k+z)^2}=\sum_{k=0}^\infty\frac{1}{(k+z)^2}+\sum_{k=0}^\infty\frac{1}{\big(k+(1-z)\big)^2}$$ Now we consider the polygamma function: $$\psi^{(n)}(z):=\mathrm D^{n+1}(\log \Gamma)(z)=(-1)^{n+1}n!\sum_{k=0}^\infty \frac{1}{(k+z)^{n+1}}$$ The sum formula is proven in this excellent video from Flammable Maths. We can write this as $$\sum_{k\in\Bbb{Z}}\frac{1}{(k+z)^2}=\psi^{(1)}(z)+\psi^{(1)}(1-z)$$ Let's let $$f_n(z)=\psi^{(n)}(z)-(-1)^n\psi^{(n)}(1-z)$$ Now let $$\mathrm I$$ be the integral operator. Ignoring constants of integration for now (it can be shown they all vanish), we can see using the derivative definition of the polygamma that $$\mathrm I^{n+1} f_n(z)=\log \Gamma(z)+\log\Gamma(1-z)$$ Using the laws of logarithms we have $$\mathrm I^{n+1}f_n(z)=\log\big(\Gamma(z)\Gamma(1-z)\big)$$ But using Euler's reflection formula which is also proven by Flammable Maths we have $$\mathrm I^{n+1}f_n(z)=\log\left(\frac{\pi}{\sin(\pi z)}\right)=\log(\pi)-\log(\sin(\pi z))$$ Taking a derivative on both sides, $$\mathrm I^n f_n(z)=-\frac{1}{\sin(\pi z)}\cdot \cos(\pi z)\cdot \pi=-\pi\cot( \pi z)$$ Hence, $$f_n(z)=\mathrm D^n(s\mapsto -\pi \cot(\pi s))(z)$$ Therefore $$f_1(z)=\mathrm D(s\mapsto -\pi\cot(\pi s))(z)=\pi^2\csc^2 (\pi z)$$ | 2022-08-18 04:22:00 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 38, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9463822245597839, "perplexity": 209.18520138360603}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-33/segments/1659882573163.7/warc/CC-MAIN-20220818033705-20220818063705-00106.warc.gz"} |
http://www.ck12.org/tebook/Algebra-I-Teacher%2527s-Edition/r1/section/3.7/ | <img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# 3.7: Percent Problems
Difficulty Level: At Grade Created by: CK-12
## Learning Objectives
At the end of this lesson, students will be able to:
• Find a percent of a number.
• Use the percent equation.
• Find percent of change.
## Vocabulary
Terms introduced in this lesson:
percent
percent equation
rate
total
part
base unit
positive percent change
increase
negative percent change
decrease
mark-up
## Teaching Strategies and Tips
Work out exercises like Examples 1-7 and place the solutions in a table similar to the one below. This frees up board space and provides students with a handy reference to study later.
Example.
Complete the table.
Fraction Decimal Percent
45\begin{align*}\frac{4}{5}\end{align*}
0.12\begin{align*}0.12\end{align*}
2.5%\begin{align*}2.5\%\end{align*}
0.01%\begin{align*}0.01\%\end{align*}
38\begin{align*}\frac{3}{8}\end{align*}
0.001\begin{align*}0.001\end{align*}
43\begin{align*}\frac{4}{3}\end{align*}
Fraction, decimal, and percent conversions can be summed up as:
• decimal \begin{align*}\Rightarrow\end{align*} percent. Multiply by 100\begin{align*}100\end{align*} and affix %\begin{align*}\%\end{align*} symbol. Alternatively, move the decimal point two places right and affix %\begin{align*}\%\end{align*} symbol. See Examples 1 and 2. Additional Example. 1.2=1.2×100100=120%\begin{align*}1.2=\frac{1.2\times100}{100}=120\%\end{align*}.
• percent \begin{align*}\Rightarrow\end{align*} decimal. Divide by 100\begin{align*}100\end{align*} and remove %\begin{align*}\%\end{align*} symbol. Alternatively, move the decimal point two places left and remove %\begin{align*}\%\end{align*} symbol. See Examples 3 and 4. Additional Example. 0.003%=0.003100=0.00003\begin{align*}0.003\%=\frac{0.003}{100}=0.00003\end{align*}.
• fraction \begin{align*}\Rightarrow\end{align*} percent. Represent the fraction as x%\begin{align*}x\%\end{align*} or x/100\begin{align*}x/100\end{align*}. Solve for x\begin{align*}x\end{align*} by cross-multiplying. Alternatively, convert the fraction to a decimal. Then convert decimal to percent as above. See Examples 5 and 6. Additional Example. 25=x100x=40\begin{align*}\frac{2}{5}=\frac{x}{100}\Rightarrow x=40\end{align*}. Therefore, 25=40%\begin{align*}\frac{2}{5}=40\%\end{align*}.
• percent \begin{align*}\Rightarrow\end{align*} fraction. Express percent as a ratio (per 100\begin{align*}100\end{align*}). Reduce fraction. See Example 7. Additional Example. 110%=110100=1110=1110\begin{align*}110\%=\frac{110}{100}=\frac{11}{10}=1\frac{1}{10}\end{align*}.
• fractions \begin{align*}\Rightarrow\end{align*} decimals. Divide numerator by denominator. Use calculator if necessary. Example. .
• decimals \begin{align*}\Rightarrow\end{align*} fractions. Place the digits after the decimal under the appropriate power of ten and reduce. Example. 0.225=2251000=940\begin{align*}0.225=\frac{225}{1000}=\frac{9}{40}\end{align*}.
In Examples 8-11, students setup percent equations to find the percent of a given number.
• Convert the value for R%\begin{align*}R\%\end{align*} in the percent equation R%×Total=Part\begin{align*}R\% \times \;\mathrm{Total} = \;\mathrm{Part}\end{align*} to a decimal before calculating. Rate should be expressed as a decimal in Rate×Total=Part\begin{align*}\;\mathrm{Rate} \times \;\mathrm{Total} = \;\mathrm{Part}\end{align*}.
• Remind students that of means to multiply.
Use Examples 12 and 13 to point out that a positive percent change means an increase in the quantity, and a negative change means a decrease.
In Example 14, teachers are encouraged to work out the calculations for Mark-up, Final retail price, 20%\begin{align*}20\%\end{align*} discount, and 25%\begin{align*}25\%\end{align*} discount as a way to motivate the same calculations done algebraically in the next step.
Is the order in which we calculate discounts and sales tax significant? In other words, should stores subtract the discount first and then add the tax on the new total or should the total amount be taxed first and then have the discount subtracted from that? or does it matter? Assume the discount is a percent and not a fixed discount.
Solution. Consider an example:
Original price of the item =\$12.50\begin{align*}= \12.50\end{align*}
Discount =35%\begin{align*}= 35\%\end{align*}
Sales tax rate in the county of purchase =7.75%\begin{align*}= 7.75\%\end{align*}
Tax First, Discount Second
12.50+0.0775(12.50)=13.4713.470.35(13.47)=8.76\begin{align*}12.50 + 0.0775 (12.50) = 13.47\\ 13.47 - 0.35 (13.47) = 8.76\end{align*}
Discount First, Tax Second
12.500.35(12.50)=8.138.13+0.0775(8.13)=8.76\begin{align*}12.50 - 0.35 (12.50) = 8.13\\ 8.13 + 0.0775 (8.13) = 8.76\end{align*}
The steps above can be repeated algebraically. Let:
p\begin{align*}p\end{align*} = original price of the item
d\begin{align*}d\end{align*} = discount
t\begin{align*}t\end{align*} = sales tax rate in the county of purchase
Tax First, Discount Second
p+t(p)=\begin{align*}p + t (p) =\end{align*} amount with tax
p+t(p)d(p+t(p))=\begin{align*}p + t (p) - d (p + t (p)) =\end{align*} final amount with discount
Discount First, Tax Second
pd(p)=\begin{align*}p - d (p) =\end{align*} amount after discount
pd(p)+t(pd(p))=\begin{align*}p - d (p) + t (p - d (p)) =\end{align*} final amount with discount
Simplifying the two expressions for the final amount results in identical expressions (careful when distributing).
Conclusion: There is no difference in the total amount to be paid if the tax is added to the total first, followed by the discount, or the discount applied to the total first, followed by the tax.
General Tips:
• Give students time to consider the possibilities on their own.
• Have students choose their own numbers for original price, discount, and sales tax rate. This can be done in groups or individually. Calculators are recommended.
• Asking around the classroom, it is suspicious that everyone’s final two calculations are the same. Use this to formulate a conjecture.
• Ask students to turn in their reasoning process as an assignment.
• As the above algebraic argument is completely variable driven (no numbers), teachers are advised to show each step.
• Students reason in various ways. Some common responses are:
“Add the tax first, otherwise you are cheating the government out of its tax.”
“Taking the discount first decreases the bill, and so the tax will not be as great.”
“Tax should be added on last, as the discounted price is the true price of the item – since that is how much the item is being sold for.”
“Doing the tax first increases the amount to be paid and so the discount will be larger.” According to students, this translates to a smaller price to be paid.
“The quantities must be equal since they have seen it being done in both ways.”
As a class, discuss the validity of some of these responses.
## Error Troubleshooting
Despite the %\begin{align*}\%\end{align*} symbol in Example 4, students see the decimal and incorrectly move it two places right. This error is common for percents less than 1\begin{align*}1\end{align*} (i.e., 0.5%\begin{align*}0.5\%\end{align*}) and percents greater than 100\begin{align*}100\end{align*} (i.e., \begin{align*}110\%\end{align*}).
Express \begin{align*}0.01\%\end{align*} as a decimal.
Solution: Move the decimal two places left and remove the \begin{align*}\%\end{align*} symbol. \begin{align*}0.01\%=0.0001\end{align*}.
Express \begin{align*}120.25\end{align*} as a percent.
Solution: Move the decimal two places right and affix the \begin{align*}\%\end{align*} symbol. \begin{align*}120.25=1.2025\%\end{align*}.
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
Show Hide Details
Description
Tags:
Subjects: | 2016-10-26 12:12:57 | {"extraction_info": {"found_math": true, "script_math_tex": 57, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6447545886039734, "perplexity": 4603.37624766111}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-44/segments/1476988720941.32/warc/CC-MAIN-20161020183840-00082-ip-10-171-6-4.ec2.internal.warc.gz"} |
https://math.meta.stackexchange.com/tags/dupehammer/hot | # Tag Info
### Can I forsake my superpowers?
First and foremost: You are not alone feeling/thinking that on some occasions you would prefer only to cast an ordinary vote! I don't think you can forsake the superpowers, but you can use them ...
• 124k
Accepted
### How close a match we need to close a recurring question as a duplicate?
The OP asked in original version (now substantially rewritten a further from a rant): "Am I the only one concerned about duplicates?" My answer to this would be certainly not, as witnessed by many ...
• 50.3k
### Should we be able to close questions only when we have the tag of field of the question asked?
No. The ability to vote to close questions should not be reserved for those users with a badge in some tag attached to the question. As a practical matter, there are two points: There are tags which ...
• 50.4k
Accepted
### Mjölnir malfunction
No, as meanwhile the deciding factor are the current tags, but the voter must not have been involved in the tagging. Source: It looks like the duplicate banner changed. How does it work now?
• 41.1k
### Question having excellent answer marked as dupe of a newer question
Not only did the older thread have an excellent answer already Maybe, but see below. but the user who marked the old thread as dupe of the new thread apparently knew of the older thread and yet ...
• 41.1k
### How close a match we need to close a recurring question as a duplicate?
I had posted a Comment on Remainder when dividing $13^{3530}$ with $12348$ suggesting one (potential) duplicate/closely related Question. I'll share my thought about where the line might be drawn in ...
• 35.5k
Accepted
### What is needed to have the right to close a question (as a duplicate) with a single vote?
A non-moderator user can cast a binding duplicate vote if they have a gold-badge in a tag originally on the question (and the usual criteria for dupe-closure are met). Note that these users are marked ...
• 41.1k
Accepted
### spurious interaction between close votes for different reasons
With a gold-badge in a tag a user since some time has a binding vote (as regards duplicates). The behavior seems the same as it is and was for other binding votes. The specific issue with clearing ...
• 41.1k
### How close a match we need to close a recurring question as a duplicate?
The following is just my personal opinion The question is: What makes a duplicate? I don't think it is easy to provide a clear cut answer. There will be some disagreements and grey area questions ...
• 41.8k
### How close a match we need to close a recurring question as a duplicate?
A summary of what I have learned/observed from the responses and the discussion in this thread (thanks all) up to June 22nd: Many users have argued against the use of so called abstract duplicates as ...
• 124k
My opinion: $A$ should be closed a duplicate of $B$ iff answers to $B$ are complete and appropriate answers to the question asked in $A$ (ignoring everything else). So in particular: Most forms ... | 2022-10-03 08:21:02 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4204438328742981, "perplexity": 1393.470583302121}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030337404.30/warc/CC-MAIN-20221003070342-20221003100342-00124.warc.gz"} |
https://www.physicsforums.com/threads/graphical-representation-of-complex-roots-to-equations.413981/ | # Graphical representation of complex roots to equations
1. Jul 3, 2010
### RK1992
I've never properly studied complex numbers but I will soon (in September). Basically:
We get taught from a young age that:
the real root of f(x)=x²-4 is where the graph of y=f(x) cuts the x axis
But is there a graphical representation of a complex root?
What's so special about the value x= +/- 2i if g(x)=x²+4 ? Is there a 3D graphical representation of this root?
2. Jul 3, 2010
### Hurkyl
Staff Emeritus
Well, remember that "real root of f(x)" really means a number a such that f(a)=0. It doesn't need any geometric interpretation to make sense.
Some people like to think in terms of geometry rather than algebra; so the particular correspondence between them you invoked says that the roots of f(x) correspond to the intersection of the parabola defined by y=f(x) and the line y=0, just like you described.
The similar geometric interpretation for complex-valued functions of complex numbers requires 4 dimensions to draw. We can often get away with projecting onto a three-dimensional image, though. (Of course, we have to project that onto a two-dimensional image so that we can draw it, and things get messy)
Another common way is to instead draw a before and after picture of the complex plane. e.g. put a grid on the "before" picture, and in the "after" picture we see a picture of how f transformed the grid.
For this function, a good "before" picture is to make the grid out of rays emanating from the origin and circles whose center is the origin.
The after picture consists of lines passing through the point -4 + 0i, and circles centered on that point. The spacing between the circles is unchanged. However, the spacing between the lines has doubled, and the grid overlaps itself -- e.g. the two rays emanating from the origin at angles x° and (x+180)° both map to the same ray emanating from -4 + 0i at an angle of (2x)°.
In this picture, we can estimate the roots of f by looking at 0 + 0i in the "after" picture, identifying the grid points that lie there, and then finding where they came from in the "before" picture.
3. Jul 3, 2010
4. Jul 3, 2010
### HallsofIvy
Staff Emeritus
A simpler (or more simple minded) way of looking at it:
The equation $x^2- 2ax+ a^2+ b^2= (x-a)^2+ b^2= 0$ has roots $x= a\pm bi$. It has complex roots, of course, because its graph does not cross the x-axis.
The vertex of the graph is where x= a so that itex]y= (x-a)^2+ b^2= b^2[/itex]. That is, the graph goes down to $(a, b^2)$ and then back up again. In general, if the graph of $y= x^2- ax+ b$ lies entirely above the x-axis, and its vertex is at $(x_0, y_0)$, then its roots are $x_0\pm i\sqrt{y_0}$. | 2017-08-18 21:36:16 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8484862446784973, "perplexity": 582.0306163304949}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-34/segments/1502886105108.31/warc/CC-MAIN-20170818194744-20170818214744-00595.warc.gz"} |
https://www.bou.lt/theory/statistics/semiParametric | # Semi-parametric regression
## The Robinson estimator
### The Robinson estimator
#### Partialling out
$$y_i=\mathbf x_i\theta +g(\mathbf z_i) +\epsilon_i$$
Consider:
$$E(y_i|\mathbf z_i)=E(\mathbf x_i\theta +g(\mathbf z_i) + \epsilon_i|\mathbf z_i)$$
$$E(y_i|\mathbf z_i)=E(\mathbf x_i\theta|\mathbf z_i)+E(g(\mathbf z_i)|\mathbf z_i) + E(\epsilon_i|\mathbf z_i)$$
$$E(y_i|\mathbf z_i)=E(\mathbf x_i|\mathbf z_i)\theta+g(\mathbf z_i)$$
We can now remove the parametric part:
$$y_i-E(y_i|\mathbf z_i)=\mathbf x_i\theta +g(\mathbf z_i) + \epsilon_i - E(\mathbf x_i|\mathbf z_i)\theta -g(\mathbf z_i)$$
$$y_i-E(y_i|\mathbf z_i)=(\mathbf x_i- E(\mathbf x_i|\mathbf z_i))\theta +\epsilon_i$$
We define:
• $$\bar y_i = y_i-E(y_i|\mathbf z_i)$$
• $$\bar x_i = \mathbf x_i- E(\mathbf x_i|\mathbf z_i)$$
$$\bar y_i =\bar x_i \theta +\epsilon_i$$
#### Estimating $$\bar y_i$$ and $$\bar x_i$$
So we can use OLS if we can estimate.
• $$E(y_i|\mathbf z_i)$$
• $$E(\mathbf x_i|\mathbf z_i)$$
We can do this with non-parametric methods.
### Bias and variance of the Robinson estimator
robinson: can’t have confounded in dummy. but can in real. general result of propensity stuff?
Framing: Partialling out is an alternative to OLS where $$n<<p$$ doesn’t hold. alterntive to LASSO etc
$$\hat \theta \approx N(\theta, V/n)$$
$$V=(E[\hat D^2)^{-1}E[\hat D^2\epsilon^2 ](E[\hat D^2])^{-1}$$
These are robust standard errors.
#### Moments of the Robinson estimator
If IID then
$$Var (\hat \theta) =\dfrac{\sigma^2_\epsilon }{\sum_i(x_i-\hat X_i)^2}$$
Otherwise, can use GLM
What are the properties of the estimator?
$$E[\hat \theta ]=E[\dfrac{\sum_i (X_i-\hat X_i)(y_i-\hat y_i)}{\sum_i(x_i-\hat X_i)^2}]$$
### Non-linear treatment effects in the Robinson estimator
Page on reformulating as non-linear. can do it. show can be estimated using arg min https://arxiv.org/pdf/1712.04912.pdf
### DML
in DML. page on orthogonality scores, page on constructing them; page on using them to estimate parameters (GMM)
We have $$P(X)=f(\theta , \rho)$$ $$\hat \theta = f(X, n)$$ $$\theta = g(\rho , X)$$
So error is: $$\hat \theta - \theta=f(X, n)-g(\rho , X)$$
Bias is defined as: $$Bias(\hat \theta, \theta ) = E[\hat \theta - \theta]=E[\hat \theta ] - \theta$$ $$Bias = E[\hat \theta - \theta]=E[f(X, n)-g(\rho , X)]$$ $$Bias = E[\hat \theta - \theta]=E[f(X, n)]-g(\rho ,X)$$
double ML: regression each parametric parameter on ML of other variables. eg: get $$e(x|z)$$ $$e(d|x)$$ $$d=m(x)+v$$ $$d$$ is correlated with $$x$$ so bias. $$v$$ is corrleated with $$d$$ but not $$x$$. use as “iv”. Still need estimate for $$g(x)$$.
for iterative, process is: + estimate $$g(x)$$ + plug into other and estimate theta + this section should be in sample splitting. rename iterative estimation. separate pages for bias, variance + how does this work?? paper says random forest regression and OLS. intialise $$\theta$$ randomly? + page on bias, variance, efficiency? + page on sample splitting, why?
+ page on goal: $$x$$ and $$z$$ orthogonal for split sampling + page on $$X=m_0(Z)+\mu$$, first stage machine learning, synthetic instrumental variables? h3 on that for multiple variables on interest. regression for each
### DML1
Divide into $$k$$.
For each do ML on nuicance (how???) use all instances outside of sample
Then do GMM using orthogonality condition to calculate $$\theta$$. (how??) use instances in sample
Average $$\theta$$ from each class
### Last stage Robinson
Separate page for last stage: note we can do OLS, GLS etc with choice of $$\Omega$$. | 2021-06-17 16:53:55 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8183308243751526, "perplexity": 6165.134178451149}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-25/segments/1623487630518.38/warc/CC-MAIN-20210617162149-20210617192149-00058.warc.gz"} |
https://www.semanticscholar.org/paper/The-fundamental-group-of-the-von-Neumann-algebra-of-R%C4%83dulescu/fbf3dd634ac565a4949443c51026baa1227ae0ef?p2df | # The fundamental group of the von Neumann algebra of a free group with infinitely many generators is $\mathbb{R}_+\slash \{0\}$
@article{Rdulescu1992TheFG,
title={The fundamental group of the von Neumann algebra of a free group with infinitely many generators is \$\mathbb\{R\}\_+\slash \\{0\\}\$},
author={Florin Rădulescu},
journal={Journal of the American Mathematical Society},
year={1992},
volume={5},
pages={517-517}
}
• F. Rădulescu
• Published 1 September 1992
• Mathematics
• Journal of the American Mathematical Society
In this paper we show that the fundamental group Y of the von Neumann algebra Y(Fo) of a free (noncommutative) group with infinitely many generators is R+ \ {O}. This extends the result of Voiculescu who previously proved [26, 27] that Q+ \ {0} is contained in 9(Y(F )) . This solves a classical problem in the harmonic analysis of the free group F . In particular, it follows that there exists subfactors of _(F7o) with index s for every s E [4, oo) . We will use the noncommutative (quantum…
41 Citations
Some estimates for the Banach space norms in the von Neumann algebras associated with the Berezin's quantization of compact Riemann
Let $\G$ be any cocompact, discrete subgroup of $\pslr$. In this paper we find estimates for the predual and the uniform Banach space norms in the von Neumann algebras associated with the Berezin' s
Operator algebras, free groups and other groups
The operator algebras associated to non commutative free groups have received a lot of attention, by F.J. Murray and J. von Neumann and by later workers. We review some properties of these algebras,
On the method of constructing irreducible finite index subfactors of Popa.
Let US(Q) be the universal Jones algebra associated to a finite von Neumann algebra Q and Rs c R be the Jones subfactors, s € {4cos2 \\n > 3} U [4, oo). We consider for any von Neumann subalgebra Qo
Classification of a family of non-almost-periodic free Araki–Woods factors
• Mathematics
Journal of the European Mathematical Society
• 2019
We obtain a complete classification of a large class of non almost periodic free Araki-Woods factors $\Gamma(\mu,m)"$ up to isomorphism. We do this by showing that free Araki-Woods factors
Compressions of free products of von Neumann algebras
• Mathematics
• 1999
Abstract. A reduction formula for compressions of von Neumann algebra II $_1$–factors arising as free products is proved. This shows that the fundamental group is ${\bf R}^*_+$ for some such
For Free Products of Von Neumann Algebras
Sufficient conditions for factoriality are given for free products of von Neumann algebras with respect to states that are not necessarily traces. The Connes T –invariant of the free product algebra
Free products of finite dimensional and other von Neumann algebras with respect to non-tracial states
The von Neumann algebra free product of arbitary finite dimensional von Neumann algebras with respect to arbitrary faithful states, at least one of which is not a trace, is found to be a type~III
Realization of rigid C$^*$-tensor categories via Tomita bimodules
• Mathematics
• 2017
Starting from a (small) rigid C$^*$-tensor category $\mathscr{C}$ with simple unit, we construct von Neumann algebras associated to each of its objects. These algebras are factors and can be either
Free products of hyperfinite von Neumann algebras and free dimension
The free product of an arbitrary pair of finite hyperfinite von Neumann algebras is examined, and the result is determined to be the direct sum of a finite dimensional algebra and an interpolated
## References
SHOWING 1-10 OF 50 REFERENCES
Operator Algebras and Applications: Quantized groups, string algebras, and Galois theory for algebras
We introduce a Galois type invariant for the position of a subalgebra inside an algebra, called a paragroup, which has a group-like structure. Paragroups are the natural quantization of (finite)
Conne’s bicentralizer problem and uniqueness of the injective factor of type III1
In Connes' fundamental work "Classification of injective factors" [7], it is proved that injective factors of type III,t, 2 . 1 on a separable Hilbert space are completely classified by their "smooth
Limit laws for Random matrices and free products
In earlier articles we studied a kind of probability theory in the framework of operator algebras, with the tensor product replaced by the free product. We prove here that free random variables
Local observables and particle statistics II
• Physics, Mathematics
• 1971
Starting from the principles of local relativistic Quantum Theory without long range forces, we study the structure of the set of superselection sectors (charge quantum numbers) and its implications
Index of subfactors and statistics of quantum fields. I
We identify the statistical dimension of a superselection sector in a local quantum field theory with the square root of the index of a localized endomorphism of the quasi-local C*-algebra that
Duality for crossed products and the structure of von Neumann algebras of type III
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 249 Preliminary 251 Construction of crossed products . . . . . . . . . . . . . . . . . . . . . . . 253 Duality . . . . . .
On the Distribution of the Roots of Certain Symmetric Matrices
The distribution law obtained before' for a very special set of matrices is valid for much more general sets of real symmetric matrices of very high dimensionality.
Une classi cation des facteurs de type III
© Gauthier-Villars (Éditions scientifiques et médicales Elsevier), 1973, tous droits réservés. L’accès aux archives de la revue « Annales scientifiques de l’É.N.S. » (http://www.
Limit laws for random matrices and/ree products, Operator Algebras, Unitary Representations and Invariant Theory, Progress in Math., vol
• Birkhiiuser,
• 1990 | 2022-05-19 06:20:19 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7341892123222351, "perplexity": 602.68426734752}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662525507.54/warc/CC-MAIN-20220519042059-20220519072059-00430.warc.gz"} |
http://prodiptag.blogspot.com/2016/08/systematic-trading-getting-technical.html | ## Thursday, August 18, 2016
### Systematic Trading: Getting Technical with Technical Indicators
There are few investors and traders who have never used a technical indicator. Some use them as part of their core trading strategies, others as confirmation or as a timing tool. I am reasonably certain even the most ardent value investors perhaps look at them in time of trials and tribulations. The set of these indicators are large (and ever increasing) as different ones developed over course of time, often from different markets and asset classes1. This is usually not a problem, as most practioner will settle down with one or two favorites.
However, most indicators have a lot in common among them. They are usually a function of past and present market data. They can be usually expressed as a function of returns of the underlying, and they tend to move in a range (though not always statistically stationary2).
Taking the example of a simple one - the moving average cross-over indicator. This is expressed as a difference of two moving averages (a short and a long ones). Mathematically, this can be represented as $mom=\sum_{i=0}^{n_1} a_i.P_i - \sum_{i=0}^{n_2} b_i.P_i$, where $n_1$ and $n_2$ are the short and long moving average periods, $a_i$s and $b_i$s are the weights (for simple moving average $a_i=1/n_1$ etc.) and $P_i$s are the prices. It can be shown that this can be converted from this price space to returns space, as $mom=\sum_{i=0}^{n} w_i.r_i$. Here $r_i=P_i - P_{i-1}$ (returns assuming log prices) and $n=n_2$ from above.
Similar treatment can be applied to other common indicators to convert them as a function of returns $r_i$s. A few example 3 below:
• Momentum cross-over = $\sum_{i=0}^{n} w_i.r_i$
• MACD Histogram = MACD line - signal line = $\sum_{i=0}^{n1} w_i^1.r_i$ - $\sum_{i=0}^{n2} w_i^2.r_i$ $\Rightarrow$ $\sum_{i=0}^{n} w_i.\Delta{r_i}$, Here $\Delta{r_i}=r_i - r_{i-1}$. This follows from logic similar to the momentum crossover above, and noting the difference of sum is in returns terms instead of prices.
• CCI = (Price - Average Price)/(0.15 x Mean Deviation) = $\frac{1}{\sigma}\sum (P_i - \bar P)$ $\Rightarrow$ $\frac{1}{n.\sigma}\sum (r^{n}+r^{n-1}+..+r)$ $\Rightarrow$ $\sum w_i.r_i$, where $r^k = r_i - r_{i-k}$
• Know Sure Thing = (RCMA1 x 0.1) + (RCMA2 x 0.2) + (RCMA3 x 0.3) + (RCMA4 x 0.4) = $a1.\sum w_1.r^{n_1} + a2.\sum w_2.r^{n_2} + a3.\sum w_3.r^{n_3} + a4.\sum w_3.r^{n_3}$ $\Rightarrow$ $\sum w_i.r_i$
Similarly most others can be expressed as a function of returns, although not all of them as linear (or even polynomial) as above. Broadly, we can divide all common technical indicators that can be expressed as function of returns in three different classes 4
• Indicators that are linear (or polynomial) combination of past returns in returns space ($f(r)$). Examples - the ones above. Under certain condition (stationarity) they can be modeled as Gaussian distribution
• Indicators that are functions of sign of the returns in signed returns space ($f(r^+, r^-)$). Examples - like RSI or Chande Momentum Oscillator. They can be analyzed using folded normal distribution
• Indicators that are function of returns in time space ($f(t(r))$). An examples is the Aroon indicator
One objective of analyzing commonality of technical indicators can be to choose the one that is best suited to a particular purpose (depending on the time series characteristics of the underlying and the trading strategy). Another, and perhaps more common, can be dimensionality reduction as part of inputs to advanced machine learning based trading systems.
Following figure shows the outcome of principal component analysis of different technical indicators run on different equity indices5 - showing the first two principal components. Interestingly, for most cases (both in real market data and simulations6) the first two components will explain close to 85% or more variance in the indicators. As we can see all indicators load similarly on the first component. This is the underlying momentum component. This component typically explain around 70% variance, and will probably be the choice of inputs in a support vector machine or neural network system incorporating technical indicators.
The second component is where the indicators differ a lot. This component captures the signature of the filtering carried out by the indicator. This signature has two parts, one is the intrinsic method of the filter computation. For example from the above formulate, we see MACD is a function of difference of returns and hence will tend to behave more like over-differenced series (assuming the returns are stationary). In contrast, KST will have a large component which is simply sum of returns, and hence will behave more like a non-stationary series in the limit. Indeed, for common parameters for these indicators (representing a look back of 20 days), the time series characteristics of these signals can be captured in the following (inverse) unit root circle plot (here roughly speaking, closer the plotted points, i.e. roots, towards the center of the circle, more mean-reverting is the series)
We can see from the PCA plot there are four major groups of indicators based on their time series characteristics - MACD, which is very much mean-reverting (i.e. suitable for short term trends), KST (which is quite the opposite) and then we have two groups - one consisting the first type of indicators noted above (function of returns) and the other group consists of the second and third types (function of signed returns and returns in time space). This is validated in the unit root plot as well, we see MACD has roots much closer to the center, KST almost on the circle perimeter, RSI quite close to it, and Bollinger bands closer to the center relatively.
Another way to appreciate how different indicators impact the momentum signal differently, is to look at how they filter the components of the underlying (returns in this case) at different frequencies - as seen in the AR spectral analysis chart below. Click on the indicators on the right hand side legend to turn them off or on.
A spectrum that has higher values towards zero frequency (like KST) means they will tend to filter out higher frequency in the data,whereas the ones that has a peak away from zero, or drop off slowly from peak at zero will tend to pick up faster components (in the extreme resembling high negative correlation of a over-differenced signal). Of course as we increase look back period, an indicator will tend to move away from the second kind and towards the first kind.
Using this insight, one can design an appropriate set of indicators to extract an "average" momentum signal, to be used in other strategy or as inputs to a neural networks or similar system.
For this purpose, the first PCA component is the one we seek to use as input as momentum signal, straight and simple. The usefulness of the second component is that it allows us to fine-tune the momentum signal for our purpose. A momentum signal depends on our time frame - a short period momentum can look like mean-reversion in longer time frame. To extract a consistent signal we need to tune the choice of the indicators and parameters. If we are looking to extract momentum signals averaged over different filtering methods, but not over time, we need to ensure all factor loadings on the second component are within acceptable limits. Whereas if we want to span as much frequency spectrum as possible we want the loadings to span much larger space. Depending on out choice we extract the kind of signal we want from the first component7.
Note, while I mention the first component as momentum signal, it is NOT same as what is known as the time series momentum factor. However, it can easily computed by back-testing trading PnL based on this momentum signal. As we have seen in general these signals can be expressed as $\sum w.r$, the PnL will be (using a linear sizing function) $\sum (w_i.r_i).r_j$, or (using a sign function) $sign(\sum (w_i.r_i)).r_j$. Of course we can approximate the signum function, and then in general, the PnL becomes a polynomial of auto-covariances of the underlying returns.
1. This is a useful place with good introductory materials on different indicators
2. In general an indicators will tend to become non-stationary at a given periodic frequency (e.g. daily) as we increase the look-back parameter
3. There is no guarantee the sum of weights adds up to one. Please feel free to notify me in comments if you spot any error.
4. Here we ignore the indicators that take volume as an input as well
5. All data from Yahoo Finance
6. Based on simulations assuming expected market behaviours, i.e. AR or ARMA type return characteristics.
7. One can design an algorithm for this purpose, that will maximize the explained variance by the first component of the PCA, by optimizing over the parameter space of the indicators within a pre-defined set. | 2017-08-19 03:35:27 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7819687128067017, "perplexity": 811.7830673830056}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-34/segments/1502886105297.38/warc/CC-MAIN-20170819031734-20170819051734-00338.warc.gz"} |
https://www.zbmath.org/?q=ra%3Akreher.reinhold+se%3A623 | # zbMATH — the first resource for mathematics
Contractible polyhedra in products of trees and absolute retracts in products of dendrites. (English) Zbl 1288.54015
The authors study the problem of embeddability of $$n$$-dimensional compacta into products of $$n$$ dendrites or trees. The main result is Theorem 1.8 saying that each collapsible $$n$$-dimensional polyhedron $$PL$$ embeds into the product of $$n$$ trees and this product collapses on the image of the embedding. This theorem implies Corollary 1.9 saying that for a compact $$n$$-dimensional polyhedron $$X$$ the following conditions are equivalent: (1) $$X$$ $$PL$$ embeds into the product of $$n$$ trees, (2) $$X$$ $$PL$$ embeds into the product of an $$(n-1)$$-dimensional polyhedron and a tree, (3) $$X$$ collapses onto an $$(n-1)$$-polyhedron, (4) $$X$$ $$PL$$ embeds into a collapsible $$n$$-dimensional polyhedron. Answering a problem of Koyama, Krasinkiewicz and Spie$$\dot{\mathrm z}$$, the authors prove that Sklyarenko’s compact space $$X$$ (i.e., the one-point compactification of the mapping telescope of the direct sequence consisting of the square maps of the unit circle) is a 2-dimensional AR space such that for every $$k\geq 0$$ the product $$X\times [0,1]^k$$ quasi-embeds into a product of $$2+k$$ dendrites but does not embed into the product of any $$2+k$$ curves.
##### MSC:
54C25 Embedding 57Q35 Embeddings and immersions in PL-topology
##### Keywords:
collapsible polyhedron; $$PL$$ embedding; product of trees
Full Text:
##### References:
[1] Cynthia Hog-Angeloni and Wolfgang Metzler , Two-dimensional homotopy and combinatorial group theory, London Mathematical Society Lecture Note Series, vol. 197, Cambridge University Press, Cambridge, 1993. · Zbl 0811.57001 [2] T. Banakh and V. Valov, General Position Properties in Fiberwise Geometric Topology (2010). · Zbl 1301.57019 [3] I. Berstein, M. Cohen, and R. Connelly, Contractible, non-collapsible products with cubes, Topology 17 (1978), no. 2, 183 – 187. · Zbl 0386.57007 [4] Karol Borsuk, Über das Phänomen der Unzerlegbarkeit in der Polyedertopologie, Comment. Math. Helv. 8 (1935), no. 1, 142 – 148 (German). · Zbl 0013.13102 [5] Karol Borsuk, Theory of retracts, Monografie Matematyczne, Tom 44, Państwowe Wydawnictwo Naukowe, Warsaw, 1967. · Zbl 0153.52905 [6] K. Borsuk, Remark on the Cartesian product of two \?-dimensional spaces, Bull. Acad. Polon. Sci. Sér. Sci. Math. Astronom. Phys. 23 (1975), no. 9, 971 – 973 (English, with Russian summary). · Zbl 0312.57001 [7] Philip L. Bowers, General position properties satisfied by finite products of dendrites, Trans. Amer. Math. Soc. 288 (1985), no. 2, 739 – 753. · Zbl 0568.54028 [8] Javier Bracho and Luis Montejano, The scorpions: examples in stable noncollapsibility and in geometric category theory, Topology 30 (1991), no. 4, 541 – 550. · Zbl 0749.57004 [9] Glen E. Bredon, Sheaf theory, 2nd ed., Graduate Texts in Mathematics, vol. 170, Springer-Verlag, New York, 1997. · Zbl 0874.55001 [10] R. P. Carpentier, Extending triangulations of the $$2$$-sphere to the $$3$$-disk preserving a $$4$$-coloring. · Zbl 1248.05061 [11] Marshall M. Cohen, Simplicial structures and transverse cellularity, Ann. of Math. (2) 85 (1967), 218 – 245. · Zbl 0147.42602 [12] Marshall M. Cohen, Whitehead torsion, group extensions, and Zeeman’s conjecture in highdimensions, Topology 16 (1977), no. 1, 79 – 88. · Zbl 0351.57002 [13] M. L. Curtis, On 2-complexes in 4-space, Topology of 3-manifolds and related topics (Proc. The Univ. of Georgia Institute, 1961) Prentice-Hall, Englewood Cliffs, N.J., 1962, pp. 204 – 207. · Zbl 1246.57008 [14] J. Dydak, J. Segal, and S. Spież, On questions of strict contractibility, Topology Appl. 120 (2002), no. 1-2, 67 – 75. In memory of T. Benny Rushing. · Zbl 0996.54025 [15] Steve Fisk, Variations on coloring, surfaces and higher-dimensional manifolds, Advances in Math. 25 (1977), no. 3, 226 – 266. · Zbl 0366.05032 [16] David Gillman, The Poincaré conjecture is true in the product of any graph with a disk, Proc. Amer. Math. Soc. 110 (1990), no. 3, 829 – 834. · Zbl 0709.57005 [17] David Gillman, Sergei Matveev, and Dale Rolfsen, Collapsing and reconstruction of manifolds, Geometric topology (Haifa, 1992) Contemp. Math., vol. 164, Amer. Math. Soc., Providence, RI, 1994, pp. 35 – 39. · Zbl 0817.57023 [18] A. È. Harlap, Local homology and cohomology, homological dimension, and generalized manifolds, Mat. Sb. (N.S.) 96(138) (1975), 347 – 373, 503 (Russian). · Zbl 0312.55006 [19] Allen Hatcher and Nathalie Wahl, Stabilization for mapping class groups of 3-manifolds, Duke Math. J. 155 (2010), no. 2, 205 – 269. · Zbl 1223.57004 [20] Bruce Hughes and Andrew Ranicki, Ends of complexes, Cambridge Tracts in Mathematics, vol. 123, Cambridge University Press, Cambridge, 1996. · Zbl 0876.57001 [21] J. R. Isbell, Six theorems about injective metric spaces, Comment. Math. Helv. 39 (1964), 65 – 76. · Zbl 0151.30205 [22] Ivan Izmestiev, Extension of colorings, European J. Combin. 26 (2005), no. 5, 779 – 781. · Zbl 1066.05065 [23] V. P. Kompaniec and A. V. Černavskiĭ, Equivalence of two classes of sphere mappings, Soviet Math. Dokl. 7 (1966), 1083 – 1085. · Zbl 0168.20902 [24] Akira Koyama, Józef Krasinkiewicz, and Stanisław Spież, Generalized manifolds in products of curves, Trans. Amer. Math. Soc. 363 (2011), no. 3, 1509 – 1532. · Zbl 1221.54019 [25] J. Krasinkiewicz and P. Minc, On a notion dual to local connectedness, Proc. Int. Conf. on Geometric Topology, PWN (Polish Sci. Publ.), Warszawa, 1980, pp. 265-267. · Zbl 0474.54022 [26] J. Krasinkiewicz and S. Spież, On embedding of the dunce hat, Proc. Amer. Math. Soc. (to appear). · Zbl 1272.54016 [27] Reinhold Kreher and Wolfgang Metzler, Simpliziale Transformationen von Polyedern und die Zeeman-Vermutung, Topology 22 (1983), no. 1, 19 – 26 (German). · Zbl 0521.57014 [28] Włodzimierz Kuperberg, On embeddings of manifolds into Cartesian products of compacta, Bull. Acad. Polon. Sci. Sér. Sci. Math. Astronom. Phys. 26 (1978), no. 9-10, 845 – 848 (English, with Russian summary). · Zbl 0391.57013 [29] Jie Hua Mai and Yun Tang, An injective metrization for collapsible polyhedra, Proc. Amer. Math. Soc. 88 (1983), no. 2, 333 – 337. · Zbl 0516.54027 [30] William S. Massey, Homology and cohomology theory, Marcel Dekker, Inc., New York-Basel, 1978. An approach based on Alexander-Spanier cochains; Monographs and Textbooks in Pure and Applied Mathematics, Vol. 46. W. S. Massey, How to give an exposition of the Čech-Alexander-Spanier type homology theory, Amer. Math. Monthly 85 (1978), no. 2, 75 – 83. · Zbl 0377.55005 [31] Sergei Matveev, Algorithmic topology and classification of 3-manifolds, 2nd ed., Algorithms and Computation in Mathematics, vol. 9, Springer, Berlin, 2007. · Zbl 1128.57001 [32] Sibe Mardešić and Jack Segal, \?-mappings onto polyhedra, Trans. Amer. Math. Soc. 109 (1963), 146 – 164. · Zbl 0118.39001 [33] S. A. Melikhov, Steenrod homotopy, Uspekhi Mat. Nauk 64 (2009), no. 3(387), 73 – 166 (Russian, with Russian summary); English transl., Russian Math. Surveys 64 (2009), no. 3, 469 – 551. · Zbl 1185.55001 [34] S. A. Melikhov, Combinatorics of combinatorial topology. · Zbl 1008.76094 [35] S. A. Melikhov and E. V. Shchepin, The telescope approach to embeddability of compacta. [36] John Milnor, On the Steenrod homology theory, Novikov conjectures, index theorems and rigidity, Vol. 1 (Oberwolfach, 1993) London Math. Soc. Lecture Note Ser., vol. 226, Cambridge Univ. Press, Cambridge, 1995, pp. 79 – 96. · Zbl 0954.55004 [37] J. Milnor, On axiomatic homology theory, Pacific J. Math. 12 (1962), 337 – 341. · Zbl 0114.39604 [38] J. Nagata, Note on dimension theory for metric spaces, Fund. Math. 45 (1958), 143 – 181. · Zbl 0082.16303 [39] L. Pontryagin, A classification of continuous transformations of a complex into a sphere. I, Dokl. Akad. Nauk. SSSR (C. R. Acad. Sci. de l’URSS) 19 (1938), 361-363 (English). · JFM 64.0608.02 [40] Frank Quinn, Lectures on controlled topology: mapping cylinder neighborhoods, Topology of high-dimensional manifolds, No. 1, 2 (Trieste, 2001) ICTP Lect. Notes, vol. 9, Abdus Salam Int. Cent. Theoret. Phys., Trieste, 2002, pp. 461 – 489. · Zbl 1073.57505 [41] J. Segal, A. Skopenkov, and S. Spież, Embeddings of polyhedra in \?^{\?} and the deleted product obstruction, Topology Appl. 85 (1998), no. 1-3, 335 – 344. 8th Prague Topological Symposium on General Topology and Its Relations to Modern Analysis and Algebra (1996). · Zbl 0934.57025 [42] E. G. Skljarenko, On homologically locally connected spaces, Izv. Akad. Nauk SSSR Ser. Mat. 44 (1980), no. 6, 1417 – 1433, 1440 (Russian). · Zbl 0456.55006 [43] E. G. Sklyarenko, The homology and cohomology of general spaces, Current problems in mathematics. Fundamental directions, Vol. 50 (Russian), Itogi Nauki i Tekhniki, Akad. Nauk SSSR, Vsesoyuz. Inst. Nauchn. i Tekhn. Inform., Moscow, 1989, pp. 129 – 266 (Russian). E. G. Sklyarenko, Homology and cohomology theories of general spaces [ MR1004026 (90h:55005)], General topology, II, Encyclopaedia Math. Sci., vol. 50, Springer, Berlin, 1996, pp. 119 – 256. [44] Edwin H. Spanier, Algebraic topology, McGraw-Hill Book Co., New York-Toronto, Ont.-London, 1966. · Zbl 0145.43303 [45] Y. Sternfeld, Linear superpositions with mappings which lower dimension, Trans. Amer. Math. Soc. 277 (1983), no. 2, 529 – 543. · Zbl 0784.54037 [46] Yaki Sternfeld, Mappings in dendrites and dimension, Houston J. Math. 19 (1993), no. 3, 483 – 497. · Zbl 0819.54024 [47] M. van de Vel, Collapsible polyhedra and median spaces, Proc. Amer. Math. Soc. 126 (1998), no. 9, 2811 – 2818. · Zbl 0918.52005 [48] E. R. Verheul, Multimedians in metric and normed spaces, CWI Tract, vol. 91, Stichting Mathematisch Centrum, Centrum voor Wiskunde en Informatica, Amsterdam, 1993. · Zbl 0790.46008 [49] J. H. C. Whitehead, Note on the condition \?-colc, Michigan Math. J. 4 (1957), 25 – 26. · Zbl 0077.16403 [50] Robert E. Williamson Jr., Cobordism of combinatorial manifolds, Ann. of Math. (2) 83 (1966), 1 – 33. · Zbl 0137.42901 [51] N. Witte, Entfaltung simplizialer Sphären, Technischen Universität Berlin, 2004. Diplomarbeit (Master Thesis). [52] E. C. Zeeman, On the dunce hat, Topology 2 (1964), 341 – 358. · Zbl 0116.40801
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching. | 2021-09-27 08:10:34 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6430259943008423, "perplexity": 1416.1442142521557}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-39/segments/1631780058373.45/warc/CC-MAIN-20210927060117-20210927090117-00318.warc.gz"} |
https://hal.archives-ouvertes.fr/hal-02002979 | # Random forests for high-dimensional longitudinal data
1 SISTM - Statistics In System biology and Translational Medicine
Epidémiologie et Biostatistique [Bordeaux], Inria Bordeaux - Sud-Ouest
Abstract : Random forests is a state-of-the-art supervised machine learning method which behaves well in high-dimensional settings although some limitations may happen when $p$, the number of predictors, is much larger than the number of observations $n$. Repeated measurements can help by offering additional information but no approach has been proposed for high-dimensional longitudinal data. Random forests have been adapted to standard (i.e., $n > p$) longitudinal data by using a semi-parametric mixed-effects model, in which the non-parametric part is estimated using random forests. We first propose a stochastic extension of the model which allows the covariance structure to vary over time. Furthermore, we develop a new method which takes intra-individual covariance into consideration to build the forest. Simulations reveal the superiority of our approach compared to existing ones. The method has been applied to an HIV vaccine trial including 17 HIV infected patients with 10 repeated measurements of 20000 gene transcripts and the blood concentration of human immunodeficiency virus RNA at the time of antiretroviral interruption. The approach selected 21 gene transcripts for which the association with HIV viral load was fully relevant and consistent with results observed during primary infection.
Keywords :
Document type :
Preprints, Working Papers, ...
Domain :
https://hal.archives-ouvertes.fr/hal-02002979
Contributor : Robin Genuer <>
Submitted on : Friday, February 1, 2019 - 8:56:27 AM
Last modification on : Monday, February 4, 2019 - 11:51:42 AM
### Identifiers
• HAL Id : hal-02002979, version 1
• ARXIV : 1901.11279
### Citation
Louis Capitaine, Robin Genuer, Rodolphe Thiébaut. Random forests for high-dimensional longitudinal data. 2019. ⟨hal-02002979⟩
Record views | 2019-04-25 09:58:07 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.1924007683992386, "perplexity": 2778.1021907782647}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-18/segments/1555578716619.97/warc/CC-MAIN-20190425094105-20190425120105-00416.warc.gz"} |
https://enwiki.academic.ru/dic.nsf/enwiki/9863662/Aerodynamic_potential_flow_code | # Aerodynamic potential flow code
Aerodynamic potential flow code
Aerodynamic potential flow or panel codes are used to determine the velocity and subsequently the pressure distribution on an object. This may be a simple two-dimensional object, such as a circle or wing and it may be a three-dimensional vehicle.
A series of sources and doublets are used to model the panels and wakes respectively. These codes may be valid at subsonic and supersonic speeds.
History
Early panel codes were developed in the late 1960s to early 1970s. Advanced panel codes, such as Panair (developed by Boeing), were first introduced in the late 1970s, and gained popularity as computing speed increased. Over time, panel codes were replaced with higher order methods. However, panel codes are still used for preliminary aerodynamic analysis.
Assumptions
These are the various assumptions that go into developing potential flow panel methods:
* Inviscid $abla^2 phi=0$
* Incompressible $abla cdot V=0$
* Irrotational $abla imes V=0$
* Steady $frac\left\{d\right\}\left\{dt\right\}=0$
However, the incompressible flow assumption may be removed from the potential flow derivation leaving:
* Potential Flow (inviscid, irrotational, steady) $abla^2 phi=0$
Derivation of Panel Method Solution to Potential Flow Problem
* From Small Disturbances:$\left(1-M_infty^2\right) phi_\left\{xx\right\} + phi_\left\{yy\right\} + phi_\left\{zz\right\} = 0$ (subsonic)
* From Diverence Theorem:$iiintlimits_Vleft\left( ablacdotmathbf\left\{F\right\} ight\right)dV=iintlimits_\left\{S\right\}mathbf\left\{F\right\}cdotmathbf\left\{n\right\}, dS$
* Let Velocity U be a twice continuously differentiable function in a region of volume V in space. This function is the stream function $phi$.
* Let P be a point in the volume V
* Let S be the surface boundary of the volume V.
* Let Q be a point on the surface S, and $R = |P-Q|$.
As Q goes from inside V to the surface of V,
* Therefore::$U_p= -frac\left\{1\right\} \left\{4 pi\right\} iiintlimits_Vleft\left(frac\left\{ abla^2cdotmathbf\left\{U\left\{R\right\} ight\right) dV_Q$:$-frac\left\{1\right\} \left\{4 pi\right\} iintlimits_Sleft\left(frac\left\{mathbf\left\{n\right\}cdot abla mathbf\left\{U\right\} \right\}\left\{R\right\} ight\right) dS_Q$:$+frac\left\{1\right\} \left\{4 pi\right\} iintlimits_Sleft\left(mathbf\left\{U\right\}mathbf\left\{n\right\} cdot abla frac\left\{1\right\}\left\{R\right\} ight\right) dS_Q$
For :$abla^2 phi=0$, where the surface normal points inwards.:$phi_p = -frac\left\{1\right\} \left\{4 pi\right\} iintlimits_Sleft\left(mathbf\left\{n\right\} frac\left\{ abla phi_\left\{U\right\} - abla phi_\left\{L\left\{R\right\} - mathbf\left\{n\right\} left\left( phi_\left\{U\right\} - phi_\left\{L\right\} ight\right) abla frac\left\{1\right\}\left\{R\right\} ight\right) dS_Q$
This equation can be broken down into the a both a source term and a doublet term.
The Source Strength at an arbitrary point Q is::$sigma = abla mathbf\left\{n\right\} \left( abla phi_U- abla phi_L \right)$
The Doublet Strength at an arbitrary point Q is::$mu =phi_U - phi_L$
The simplified potential flow equation is::$phi_p = -frac\left\{1\right\} \left\{4 pi\right\} iintlimits_Sleft\left(frac\left\{sigma\right\}\left\{R\right\} - mu cdot mathbf\left\{n\right\} cdot abla frac\left\{1\right\}\left\{R\right\} ight\right) dS$
With this equation, along with applicable boundary conditions, the potential flow problem may be solved.
Required Boundary Conditions
The velocity potential on the internal surface and all points inside V (or on the lower surface S) is 0.:$phi_L = 0$
The Doublet Strength is::$mu =phi_U - phi_L$:$mu = phi_U$
The velocity potential on the outer surface is normal to the surface and is equal to the freestream velocity.:$phi_U = -V_infty cdot mathbf\left\{n\right\}$
These basic equations are satisfied when the geometry is a 'watertight' geometry. If it is watertight, it is a well-posed problem. If it is not, it is an ill-posed problem.
Discretization of Potential Flow Equation
The potential flow equation with well-posed boundary conditions applied is::$mu_P = frac\left\{1\right\} \left\{4 pi\right\} iintlimits_Sleft\left(frac\left\{V_infty cdot mathbf\left\{n\left\{R\right\} ight\right) dS_U + frac\left\{1\right\} \left\{4 pi\right\} iintlimits_Sleft\left(mu cdot mathbf\left\{n\right\} cdot abla frac\left\{1\right\}\left\{R\right\} ight\right) dS$
*Note that the $dS_U$ integration term is evaluated only on the upper surface, while th $dS$ integral term is evaluated on the upper and lower surfaces.
The continuous surface S may now be discretized into discrete panels. These panels will approximate the shape of the actual surface. This value of the various source and doublet terms may be evaluated at a convenient point (such as the centroid of the panel). Some assumed distribution of the source and doublet strengths (typically constant or linear) are used at points other than the centroid. A single source term s of unknown strength $lambda$ and a single doublet term m of unknown strength $lambda$ are defined at a given point.
:$sigma_Q = sum_\left\{i=1\right\}^n lambda_i s_i\left(Q\right)=0$:$mu_Q = sum_\left\{i=1\right\}^n lambda_i m_i\left(Q\right)$
where::$s_i = ln\left(r\right)$:$m_i =$
These terms can be used to create a system of linear equations which can be solved for all the unknown values of $lambda$.
Methods for Discritizing Panels
* constant strength - simple, large number of panels required
* linear varying strength - reasonable answer, little difficulty in creating well-posed problems
* quadratic varying strength - accurate, more difficult to create a well-posed problem
Some techniques are commonly used to model surfaces. [Section 7.6] - Body Thickness by line sources - Body Lift by line doublets - Wing Thickness by constant source panels - Wing Lift by constant pressure panels - Wing-Body Interface by constant pressure panels
Methods of Determining Pressure
Once the Velocity at every point is determined, the pressure can be determined by using one of the following formulas. All various Pressure coefficient methods produce results that are similar and are commonly used to identify regions where the results are invalid.
Pressure Coefficient is defined as::$C_p = frac\left\{p-p_infty\right\}\left\{q_infty\right\}=frac\left\{gamma\right\}\left\{2\right\} p_infty M_infty^2$
The Isentropic Pressure Coefficient is::$C_p = frac\left\{2\right\} \left\{gamma M_infty^2\right\} \left(\left(1+frac\left\{gamma-1\right\} \left\{2\right\} M_infty^2 \left(frac\left\{1-|vec\left\{V\right\}|^2\right\}\left\{|vec\left\{V_infty\right\}|^2\right\}\right)\right)^\left\{ frac\left\{gamma\right\}\left\{gamma-1\right\} \right\} -1 \right)$
The Incompressible Pressure Coefficient is::$C_p = frac\left\{1-|vec\left\{V\right\}|^2\right\}\left\{|vec\left\{V_infty\right\}|^2\right\}$
The Second Order Pressure Coefficient is::$C_p = 1-|vec\left\{V\right\}|^2 + M_infty^2 u^2$
The Slender Body Theory Pressure Coefficient is::$C_p = -\left(2u +v^2 +w^2\right)$
The Linear Theory Pressure Coefficient is::$C_p = -2u$
The Reduced Second Order Pressure Coefficient is::$C_p = 1-|vec\left\{V\right\}|^2$
Commonly Used Potential Flow Codes
* Panel Code Panair (created by Boeing)
*Aerodynamics Small Disturbances
*Stream function
*Conformal mapping
*Velocity potential
*Divergence theorem
*Joukowsky transform
*Potential flow
*Circulation
*Biot-Savart law
References
* [http://www.pdas.com/ Public Domain Aerodynamic Software] , A Panair Distribution Source, Ralph Carmichael
* [http://hdl.handle.net/2060/19920013405 Panair Volume I, Theory Manual, Version 3.0] , Michael Epton, Alfred Magnus, 1990 Boeing
* [http://hdl.handle.net/2060/19920013622 Panair Volume II, Theory Manual, Version 3.0] , Michael Epton, Alfred Magnus, 1990 Boeing
* [http://hdl.handle.net/2060/19850002614 Panair Volume III, Case Manual, Version 1.0] , Michael Epton, Kenneth Sidewell, Alfred Magnus, 1981 Boeing
* [http://hdl.handle.net/2060/19920013399 Panair Volume IV, Maintenance Document, Version 3.0] , Michael Epton, Kenneth Sidewell, Alfred Magnus, 1991 Boeing
* [http://hdl.handle.net/2060/19710009882 Recent Experience in Using Finite Element Methods For The Solution Of Problems In Aerodynamic Intereference] , Ralph Carmichael, 1971 NASA Ames Research Center
Wikimedia Foundation. 2010.
### Look at other dictionaries:
• Potential flow — streamlines around a NACA 0012 airfoil at 11° angle of attack, with upper and lower streamtubes identified. In fluid dynamics, potential flow describes the velocity field as the gradient of a scalar function: the velocity potential. As a result,… … Wikipedia
• Fluid dynamics — Continuum mechanics … Wikipedia
• aerospace industry — Introduction assemblage of manufacturing concerns that deal with vehicular flight within and beyond the Earth s atmosphere. (The term aerospace is derived from the words aeronautics and spaceflight.) The aerospace industry is engaged in the … Universalium
• Formula One — F1 redirects here. For other uses, see F1 (disambiguation). For other uses, see Formula One (disambiguation). Formula One Category Single seater Country or region Worldwide … Wikipedia
• Aerial refueling — A C 17 Globemaster refuels through the boom of a Boeing KC 135 Stratotanker … Wikipedia
• Coandă-1910 — Coandă 1910 … Wikipedia
• technology, history of — Introduction the development over time of systematic techniques for making and doing things. The term technology, a combination of the Greek technē, “art, craft,” with logos, “word, speech,” meant in Greece a discourse on the arts, both… … Universalium
• rocket and missile system — ▪ weapons system Introduction any of a variety of weapons systems that deliver explosive warheads to their targets by means of rocket propulsion. Rocket is a general term used broadly to describe a variety of jet propelled missiles… … Universalium
• Elevator — For other uses, see Elevator (disambiguation). A set of lifts in the lower level of a London Underground station. The arrows indicate each elevator s position and direction of travel … Wikipedia
• climate — /kluy mit/, n. 1. the composite or generally prevailing weather conditions of a region, as temperature, air pressure, humidity, precipitation, sunshine, cloudiness, and winds, throughout the year, averaged over a series of years. 2. a region or… … Universalium | 2019-12-12 01:29:56 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 38, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6416130661964417, "perplexity": 3736.7643791102664}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-51/segments/1575540534443.68/warc/CC-MAIN-20191212000437-20191212024437-00245.warc.gz"} |
https://quant.stackexchange.com/questions/51432/how-to-calculate-yearly-volatility-from-weekly-obersvations-over-179-weeks | # how to calculate yearly volatility from weekly obersvations over 179 weeks?
I am working right now at something and I want to get sure that I am not doing any mistakes - maybe you can help me:
I collected weekly returns from a stock over 179 weeks and know I want to calculate the YEARLY volatility. The Standard Deviation of the 179 weeks is 1,56%. In order to calculate the yearly volatility I calculated 1,56%*SQRRT(52), because a year got 52 weeks. is that correct? or what should i calculate?
I'd be very grateful if someone could help me! :)
Kind regards, Memecon
• Yes, that is the accepted way to do it. – noob2 Mar 1 at 16:13
• okay so no mathematic mistake if i put the weeks of one year in the square root if i calculate it from observations for more than three years (179 weeks)? – memecon Mar 1 at 16:15
• Right. The standard deviation calculation takes 179 weekly observations as input, internally the algorithm uses (179-1) in the denominator and returns the standard deviation per week. This number then needs to be multiplied by $\sqrt{52}$ to turn it into a per year number (52 weeks per year). – noob2 Mar 1 at 16:22
• thank you so much! – memecon Mar 1 at 16:29 | 2020-07-02 10:17:01 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5733172297477722, "perplexity": 893.6850300312979}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-29/segments/1593655878639.9/warc/CC-MAIN-20200702080623-20200702110623-00228.warc.gz"} |
https://www.varsitytutors.com/precalculus-help/find-the-area-of-a-triangle-when-given-one-side-and-two-angles-or-when-given-two-sides-and-an-included-angle | # Precalculus : Find the Area of a Triangle When Given One Side and Two Angles, Or When Given Two Sides and An Included Angle
## Example Questions
### Example Question #1 : Area Of A Triangle
In triangle , , , and . Find the area of the triangle.
Explanation:
When given the lengths of two sides and the measure of the angle included by the two sides, the area formula is:
Plugging in the given values we are able to calculate the area.
### Example Question #41 : Trigonometric Applications
Find the area of this triangle:
Explanation:
To find the area, use the formula associated with side, angle, side triangles which states,
where and are side lengths and is the included angle.
In our case,
.
Plug the values into the area formula and solve.
### Example Question #41 : Trigonometric Applications
Find the area of this triangle:
Explanation:
Use the area formula to find area that is associated with the side angle side theorem for triangles.
where and are side lengths and is the included angle.
Plugging these values into the formula above, we arrive at our final answer.
### Example Question #1 : Area Of A Triangle
Find the area of this triangle:
Explanation:
To solve, use the formula for area that is associated with the side angle side theorem for triangles,
where and are side lengths and is the included angle.
Here we are using and not since that is the angle between and .
Therefore,
.
Plugging the above values into the area formula we arrive at our final answer.
### Example Question #1 : Area Of A Triangle
Find the area of this triangle:
Explanation:
Find the area using the formula associated the side angle side theorem of a triangle,
where and are side lengths and is the included angle.
In this particular case,
therefore the area is found to be,
. | 2018-04-21 17:25:55 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9901654720306396, "perplexity": 1339.8111742244505}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-17/segments/1524125945272.41/warc/CC-MAIN-20180421164646-20180421184646-00180.warc.gz"} |
http://mathoverflow.net/questions/103917/set-existence-question/103952 | # set existence question
I'm sure this is basic but here goes.
Let $\sigma_1 (x), \sigma_2 (x), ...$ be a (computable) enumeration of the formulae in the language of set theory with one free variable.
For each $i$, let $S_i$ be the set of reals $r$ such that $\sigma_i (r)$.
Do the axioms of set theory guarantee the existence of the set of ordered pairs of the form $\langle i, S_i \rangle$ - that is, do the axioms of set theory guarantee the existence of the set $\lbrace \langle 1, S_1 \rangle, \langle 2, S_2 \rangle, ... \rbrace$, or does the undefinability of truth get in the way here?
-
As has been mentioned in the comments, there are models of ZFC in which each real number is definable without parameters (see my answer to Is analysis in fact the analysis of definable real numbers? and also my article Pointwise definable models of set theory with Reitz and Linetsky for a discussion). In a pointwise definable model, there can be no list of all the definable sets of reals, paired with their definitions, since all the definable singletons would appear on such a list and reveal that the reals are countable, contrary to ZFC.
Meanwhile, there are various ways to achieve the list of definable sets if one has a stronger background set theory, and this is much more detail about this in our article. Here are a few quick instances:
• If the scheme known as $V_\delta\prec V$ holds, asserting that $V_\delta$ is an elementary substructure of $V$, then we may form the list of definable sets of reals, simply by using the Tarskian definition of truth in the structure $\langle V_\delta,{\in}\rangle$. This is definable in $V$, but by the assumption of $V_\delta\prec V$, it happens to coincide with truth in $V$. The theory $V_\delta\prec V$ is equiconsistent with ZFC by a simple compactness argument.
• If one has Kelly-Morse set theory KM in $V$, then one may define a truth predicate for first-order assertions in the language of set theory, and thereby may form the desired seqeunce of definable sets.
-
Let $F$ denote the function $\lbrace (i,S_i) : i < \omega \rbrace$ in whatever sense it may exist---I hope this abuse of notation will not be confusing.
The undefinability of truth does "get in the way" in the sense that it shows that there cannot be such a function $F$ that is definable. If we let $C$ be the set of $i$ such that $\sigma_i$ is a sentence ($x$ does not appear) then from $F$ we could define $\lbrace i \in C : S_i = \mathbb{R}\rbrace$, which is essentially a truth set and therefore cannot be definable by Tarski's theorem.
EDIT: it looks like I am using "definable" to mean "definable without parameters" and Joel is using it to mean "definable with parameters." As Joel points out in the comments and in his answer, it is possible for a truth set, and indeed the desired function $F$, to be definable from ordinal parameters.
To answer the precise question you stated, ZFC does not prove the existence of such a function $F$. As Andreas mentions in the comments to Bjørn's answer, there are models of ZFC in which all sets are definable. (This is not a first-order property of the model.) Any such model $M$ cannot satisfy "the desired function exists" regardless of how we try to formalize this. One way to see this is that if there were such a function $F \in M$, then externally we could use the fact that every set in $M$ is definable in $M$ to show that $ran(F) = \mathcal{P}(\mathbb{R})^M$. This statement about the range is absolute to $M$, contradicting Cantor's theorem in $M$.
-
Trevor, Tarski's theorem does not rule out having the set of true sentences being definable in a model. Rather, it rules out having a satisfaction predicate, where arbitrary set parameters are allowed. – Joel David Hamkins Aug 4 '12 at 15:28
For example, if $V_\delta\prec V$, then the full theory of $V$ exists as a set in $V$, since it is the same as the theory of $V_\delta$. – Joel David Hamkins Aug 4 '12 at 15:42
So while the argument of your second paragraph shows that there can be no such function $F$ as desired that is definable without parameters, in fact there can be such an $F$ that is definable with parameters. – Joel David Hamkins Aug 4 '12 at 16:05
By "definable" I meant "definable without parameters." Doesn't Tarski's theorem say that the set of true sentences is not definable (without parameters)? I have clarified my usage of "definable" in the answer now. Thanks for pointing this out. – Trevor Wilson Aug 4 '12 at 16:32
Yes, that's fine. Tarski's theorem can be viewed as showing that the set of true sentences is not definable without parameters. My examples show that it can nevertheless exist as a set. – Joel David Hamkins Aug 4 '12 at 17:13 | 2014-08-28 05:06:35 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.933702290058136, "perplexity": 134.90667203753034}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-35/segments/1408500830094.68/warc/CC-MAIN-20140820021350-00290-ip-10-180-136-8.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/definition-for-surface-integral.728931/ | # Definition for Surface Integral
1. Dec 17, 2013
### Jhenrique
Hello!
The definition of Line Integral can be this:
$$\int_s\vec{f}\cdot d\vec{r}=\int_s(f_1dx+f_2dy+f_3dz)$$
And the definition of Surface Integral can be this:
$$\int\int_S(f_1dydz+f_2dzdx+f_3dxdy)$$
However, in actually:
$$\\dx=dy\wedge dz \\dy=dz\wedge dx \\dz=dx\wedge dy$$
What do the Surface Integral be equal to:
$$\int\int_S(f_1dy\wedge dz+f_2dz\wedge dx+f_3dx\wedge dy)=\int\int_S(f_1dx+f_2dy+f_3dz)=\int\int_S\vec{f}\cdot d\vec{r}$$
I know, I know... I know that, generally, the definition to Integral Surface is:
$$\int\int_S\vec{f}\cdot \hat{n}\;dS$$
I until like this definition when compared to its respective Line Integral:
$$\int_s\vec{f}\cdot \hat{t}\;ds$$
But, is correct to definite the Surface Integral as:
$$\int\int_S\vec{f}\cdot d\vec{r}$$
being
$$d\vec{r}=(dx,dy,dz)$$
?
2. Dec 17, 2013
### HallsofIvy
Staff Emeritus
No, the surface integral is, as you say, a double integral while the path integral is a single integral. They are NOT the same thing. They can, of course, be connected by, for example, the Stoke's Theorem that says that the integral of the curl of $\vec{F}$ over a surface is the same as the integral of $\vec{F}$ around the boundary of the surface.
3. Dec 17, 2013
### Jhenrique
But I definited the Surface Integral with a double integral (as you can see below or above)
$$\int\int_S\vec{f}\cdot d\vec{r}$$
with the only difference that I used the position vector r, like in Line Integral. But, second the identities above, to define the Surface integral with ·dr is equivalent to traditional definition, with ·ndS. Correct!?
4. Dec 18, 2013
### qbert
what? your definitions make no sense. $dx = dy \wedge dz''$ is literally nonsensical. | 2017-12-17 14:16:44 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9884207248687744, "perplexity": 694.8716639962892}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-51/segments/1512948596051.82/warc/CC-MAIN-20171217132751-20171217154751-00545.warc.gz"} |
https://www.aimsciences.org/article/doi/10.3934/dcdss.2016077 | # American Institute of Mathematical Sciences
December 2016, 9(6): 1899-1912. doi: 10.3934/dcdss.2016077
## A global existence and blow-up threshold for Davey-Stewartson equations in $\mathbb{R}^3$
1 School of Mathematics, South China University of Technology, Guangzhou, Guangdong 510640, China 2 Department of Mathematics, South China University of Technology, Guangzhou, Guangdong 510640 3 College of Mathematics and Information Science, Henan Normal University, Xinxiang, Henan 453007
Received July 2015 Revised September 2016 Published November 2016
In this paper we study the threshold of global existence and blow-up for the solutions to the generalized 3D Davey-Stewartson equations \begin{equation*} \left\{ \begin{aligned} & iu_t + \Delta u + |u|^{p-1} u + E_1(|u|^2)u = 0, \quad t > 0, \ \ x\in \mathbb{R}^3, \\ & u(0,x) = u_0(x) \in H^1(\mathbb{R}^3), \end{aligned} \right. \end{equation*} where $1 < p < \frac{7}{3}$ and the operator $E_1$ is given by $E_1(f) = \mathcal {F}^{-1} \left( \frac{\xi_1^2}{|\xi|^2} \mathcal{F}(f) \right)$. We construct two kinds of invariant sets under the evolution flow by analyzing the property of the upper bound function of the energy. Then we show that the solution exists globally for the initial function $u_0$ in first kind of the invariant sets, while the solution blows up in finite time for $u_0$ in another kind. We remark that the exponent $p$ is subcritical for the nonlinear Schrödinger equations for which blow-up solutions would not occur. The result shows that the occurrence of blow-up phenomenon is caused by the coupling mechanics of the Davey-Stewartson equations.
Citation: Shiming Li, Yongsheng Li, Wei Yan. A global existence and blow-up threshold for Davey-Stewartson equations in $\mathbb{R}^3$. Discrete & Continuous Dynamical Systems - S, 2016, 9 (6) : 1899-1912. doi: 10.3934/dcdss.2016077
##### References:
[1] D. Anker and N. C. Freeman, On the soliton solutions of the Davey-Stewartson equation for long waves,, Proc. Roy. Soc. London, 360 (1978), 529. doi: 10.1098/rspa.1978.0083. Google Scholar [2] M. J. Ablowitz and A. S. Fokas, On the inverse scattering transform of multidimensional nonlinear equations related to first-order systems in the plane,, J. Math. Phys., 25 (1984), 2494. doi: 10.1063/1.526471. Google Scholar [3] R. Cipolatti, On the existence of standing waves for a Davey-Stewartson system,, Commun. Partial Diff. Eqns., 17 (1992), 967. doi: 10.1080/03605309208820872. Google Scholar [4] R. Cipolatti, On the instability of ground states for a Davey-Stewartson system,, Ann. Inst. H. Poincaré, 58 (1993), 85. Google Scholar [5] V. D. Djordjevic and L. G. Redekopp, On two-dimensional packets of capillary-gravity waves,, J. Fluid Mech., 79 (1977), 703. doi: 10.1017/S0022112077000408. Google Scholar [6] A. Davey and S. K. Stewartson, On three-dimensional packets of surface waves,, Proc. R. Soc. London, 338 (1974), 101. doi: 10.1098/rspa.1974.0076. Google Scholar [7] Z. H. Gan and J. Zhang, Sharp threshold of global existence and instability of standing wave for a Davey-Stewartson system,, Commun. Math. Phys., 283 (2008), 93. doi: 10.1007/s00220-008-0456-y. Google Scholar [8] J. M. Ghidaglia and J. C. Saut, On the initial value problem for the Davey-Stewartson systems,, Nonlinearity, 3 (1990), 475. doi: 10.1088/0951-7715/3/2/010. Google Scholar [9] N. Godet, A lower bound on the blow-up rate for the Davey-Stewartson system on the torus,, Ann. Inst. H. Poincaré - AN, 30 (2013), 691. doi: 10.1016/j.anihpc.2012.12.001. Google Scholar [10] B. L. Guo and B. X. Wang, The Cauchy problem for Davey-Stewartson systems,, Commun. Pure Appl. Math., 52 (1999), 1477. doi: 10.1002/(SICI)1097-0312(199912)52:12<1477::AID-CPA1>3.0.CO;2-N. Google Scholar [11] N. Hayashi, Local existence in time of small solutions to the Davey-Stewartson systems,, Ann. Inst. H. Poincaré, 65 (1996), 313. Google Scholar [12] N. Hayashi and H. Hirata, Global existence and asymptotic behavior in time of small solutions to the elliptic-hyperbolic Davey-Stewartson system,, Nonlinearity, 9 (1996), 1387. doi: 10.1088/0951-7715/9/6/001. Google Scholar [13] N. Hayashi and J. C. Saut, Global existence of small solutions to the Davey-Stewartson and the Ishimori systems,, Diff. and Integ. Eqns., 8 (1995), 1657. Google Scholar [14] T. Hmidi and S. Keraani, Blowup theory for the critical nonlinear Schrödinger equations revisited,, Intern. Math. Res. Notices, 46 (2005), 2815. doi: 10.1155/IMRN.2005.2815. Google Scholar [15] X. G. Li, J. Zhang, S. Y. Lai and Y. H. Wu, The sharp threshold and limiting profile of blow-up solutions for a Davey-Stewartson system,, J. Diff. Eqns., 250 (2011), 2197. doi: 10.1016/j.jde.2010.10.022. Google Scholar [16] F. Linares and G. Ponce, On the Davey-Stewartson systems,, Ann. Inst. H. Poincaré, 10 (1993), 523. Google Scholar [17] J. Lu and Y. F. Wu, Sharp threshold for scattering of a generalized Davey-Stewartson system in three dimension,, Commun. Pure Appl. Anal., 14 (2015), 1641. doi: 10.3934/cpaa.2015.14.1641. Google Scholar [18] T. Ozawa, Exact blow-up solutions to Cauchy problem for the Davey-Stewartson system,, Proc. R. Soc. Lond. Ser. A, 436 (1992), 345. doi: 10.1098/rspa.1992.0022. Google Scholar [19] M. Ohta, Stability of standing waves for the generalized Davey-Stewartson system,, J. Dyn. Diff. Eqns., 6 (1994), 325. doi: 10.1007/BF02218533. Google Scholar [20] G. Richards, Mass concentration for the Davey-Stewartson system,, Diff. and Integ. Eqns., 24 (2011), 261. Google Scholar [21] J. Shu and J. Zhang, Sharp conditions of global existence for the generalized Davey-Stewartson system,, IMA J. Appl. Math., 72 (2007), 36. doi: 10.1093/imamat/hxl029. Google Scholar [22] M. Tsutsumi, Decay of weak solutions to the Davey-Stewartson systems,, J. Math. Anal. Appl., 182 (1994), 680. doi: 10.1006/jmaa.1994.1113. Google Scholar [23] B. X. Wang and B. L. Guo, On the initial value problem and scattering of solutions for the generalized Davey-Stewartson systems,, Sci. China Ser. A, 44 (2001), 994. doi: 10.1007/BF02878975. Google Scholar [24] M. I. Weinstein, Nonlinear Schrödinger equations and sharp interpolation estimates,, Commun. Math. Phys., 87 (1983), 567. Google Scholar [25] H. Yang, X. M. Fan and S. H. Zhu, Global analysis for rough solutions to the Davey- Stewartson system,, Abstract and Appl. Anal., 2012 (2012). Google Scholar [26] J. Zhang and S. Zhu, Sharp blow-up criteria for the Davey-Stewartson system in $\mathbbR^3$,, Dyn. Partial Diff. Eqns., 8 (2011), 239. doi: 10.4310/DPDE.2011.v8.n3.a4. Google Scholar [27] S. H. Zhu, Blow-up dynamics of $L^2$ solutions for the Davey-Stewartson system,, Acta Math. Sinica, 31 (2015), 411. doi: 10.1007/s10114-015-4349-7. Google Scholar
show all references
##### References:
[1] D. Anker and N. C. Freeman, On the soliton solutions of the Davey-Stewartson equation for long waves,, Proc. Roy. Soc. London, 360 (1978), 529. doi: 10.1098/rspa.1978.0083. Google Scholar [2] M. J. Ablowitz and A. S. Fokas, On the inverse scattering transform of multidimensional nonlinear equations related to first-order systems in the plane,, J. Math. Phys., 25 (1984), 2494. doi: 10.1063/1.526471. Google Scholar [3] R. Cipolatti, On the existence of standing waves for a Davey-Stewartson system,, Commun. Partial Diff. Eqns., 17 (1992), 967. doi: 10.1080/03605309208820872. Google Scholar [4] R. Cipolatti, On the instability of ground states for a Davey-Stewartson system,, Ann. Inst. H. Poincaré, 58 (1993), 85. Google Scholar [5] V. D. Djordjevic and L. G. Redekopp, On two-dimensional packets of capillary-gravity waves,, J. Fluid Mech., 79 (1977), 703. doi: 10.1017/S0022112077000408. Google Scholar [6] A. Davey and S. K. Stewartson, On three-dimensional packets of surface waves,, Proc. R. Soc. London, 338 (1974), 101. doi: 10.1098/rspa.1974.0076. Google Scholar [7] Z. H. Gan and J. Zhang, Sharp threshold of global existence and instability of standing wave for a Davey-Stewartson system,, Commun. Math. Phys., 283 (2008), 93. doi: 10.1007/s00220-008-0456-y. Google Scholar [8] J. M. Ghidaglia and J. C. Saut, On the initial value problem for the Davey-Stewartson systems,, Nonlinearity, 3 (1990), 475. doi: 10.1088/0951-7715/3/2/010. Google Scholar [9] N. Godet, A lower bound on the blow-up rate for the Davey-Stewartson system on the torus,, Ann. Inst. H. Poincaré - AN, 30 (2013), 691. doi: 10.1016/j.anihpc.2012.12.001. Google Scholar [10] B. L. Guo and B. X. Wang, The Cauchy problem for Davey-Stewartson systems,, Commun. Pure Appl. Math., 52 (1999), 1477. doi: 10.1002/(SICI)1097-0312(199912)52:12<1477::AID-CPA1>3.0.CO;2-N. Google Scholar [11] N. Hayashi, Local existence in time of small solutions to the Davey-Stewartson systems,, Ann. Inst. H. Poincaré, 65 (1996), 313. Google Scholar [12] N. Hayashi and H. Hirata, Global existence and asymptotic behavior in time of small solutions to the elliptic-hyperbolic Davey-Stewartson system,, Nonlinearity, 9 (1996), 1387. doi: 10.1088/0951-7715/9/6/001. Google Scholar [13] N. Hayashi and J. C. Saut, Global existence of small solutions to the Davey-Stewartson and the Ishimori systems,, Diff. and Integ. Eqns., 8 (1995), 1657. Google Scholar [14] T. Hmidi and S. Keraani, Blowup theory for the critical nonlinear Schrödinger equations revisited,, Intern. Math. Res. Notices, 46 (2005), 2815. doi: 10.1155/IMRN.2005.2815. Google Scholar [15] X. G. Li, J. Zhang, S. Y. Lai and Y. H. Wu, The sharp threshold and limiting profile of blow-up solutions for a Davey-Stewartson system,, J. Diff. Eqns., 250 (2011), 2197. doi: 10.1016/j.jde.2010.10.022. Google Scholar [16] F. Linares and G. Ponce, On the Davey-Stewartson systems,, Ann. Inst. H. Poincaré, 10 (1993), 523. Google Scholar [17] J. Lu and Y. F. Wu, Sharp threshold for scattering of a generalized Davey-Stewartson system in three dimension,, Commun. Pure Appl. Anal., 14 (2015), 1641. doi: 10.3934/cpaa.2015.14.1641. Google Scholar [18] T. Ozawa, Exact blow-up solutions to Cauchy problem for the Davey-Stewartson system,, Proc. R. Soc. Lond. Ser. A, 436 (1992), 345. doi: 10.1098/rspa.1992.0022. Google Scholar [19] M. Ohta, Stability of standing waves for the generalized Davey-Stewartson system,, J. Dyn. Diff. Eqns., 6 (1994), 325. doi: 10.1007/BF02218533. Google Scholar [20] G. Richards, Mass concentration for the Davey-Stewartson system,, Diff. and Integ. Eqns., 24 (2011), 261. Google Scholar [21] J. Shu and J. Zhang, Sharp conditions of global existence for the generalized Davey-Stewartson system,, IMA J. Appl. Math., 72 (2007), 36. doi: 10.1093/imamat/hxl029. Google Scholar [22] M. Tsutsumi, Decay of weak solutions to the Davey-Stewartson systems,, J. Math. Anal. Appl., 182 (1994), 680. doi: 10.1006/jmaa.1994.1113. Google Scholar [23] B. X. Wang and B. L. Guo, On the initial value problem and scattering of solutions for the generalized Davey-Stewartson systems,, Sci. China Ser. A, 44 (2001), 994. doi: 10.1007/BF02878975. Google Scholar [24] M. I. Weinstein, Nonlinear Schrödinger equations and sharp interpolation estimates,, Commun. Math. Phys., 87 (1983), 567. Google Scholar [25] H. Yang, X. M. Fan and S. H. Zhu, Global analysis for rough solutions to the Davey- Stewartson system,, Abstract and Appl. Anal., 2012 (2012). Google Scholar [26] J. Zhang and S. Zhu, Sharp blow-up criteria for the Davey-Stewartson system in $\mathbbR^3$,, Dyn. Partial Diff. Eqns., 8 (2011), 239. doi: 10.4310/DPDE.2011.v8.n3.a4. Google Scholar [27] S. H. Zhu, Blow-up dynamics of $L^2$ solutions for the Davey-Stewartson system,, Acta Math. Sinica, 31 (2015), 411. doi: 10.1007/s10114-015-4349-7. Google Scholar
[1] Christian Klein, Benson Muite, Kristelle Roidot. Numerical study of blow-up in the Davey-Stewartson system. Discrete & Continuous Dynamical Systems - B, 2013, 18 (5) : 1361-1387. doi: 10.3934/dcdsb.2013.18.1361 [2] Christian Klein, Jean-Claude Saut. A numerical approach to Blow-up issues for Davey-Stewartson II systems. Communications on Pure & Applied Analysis, 2015, 14 (4) : 1443-1467. doi: 10.3934/cpaa.2015.14.1443 [3] Zaihui Gan, Boling Guo, Jian Zhang. Sharp threshold of global existence for the generalized Davey-Stewartson system in $R^2$. Communications on Pure & Applied Analysis, 2009, 8 (3) : 913-922. doi: 10.3934/cpaa.2009.8.913 [4] Jing Lu, Yifei Wu. Sharp threshold for scattering of a generalized Davey-Stewartson system in three dimension. Communications on Pure & Applied Analysis, 2015, 14 (5) : 1641-1670. doi: 10.3934/cpaa.2015.14.1641 [5] Uchida Hidetake. Analytic smoothing effect and global existence of small solutions for the elliptic-hyperbolic Davey-Stewartson system. Conference Publications, 2001, 2001 (Special) : 182-190. doi: 10.3934/proc.2001.2001.182 [6] Olivier Goubet, Manal Hussein. Global attractor for the Davey-Stewartson system on $\mathbb R^2$. Communications on Pure & Applied Analysis, 2009, 8 (5) : 1555-1575. doi: 10.3934/cpaa.2009.8.1555 [7] Caroline Obrecht, J.-C. Saut. Remarks on the full dispersion Davey-Stewartson systems. Communications on Pure & Applied Analysis, 2015, 14 (4) : 1547-1561. doi: 10.3934/cpaa.2015.14.1547 [8] Zaihui Gan, Jian Zhang. Blow-up, global existence and standing waves for the magnetic nonlinear Schrödinger equations. Discrete & Continuous Dynamical Systems - A, 2012, 32 (3) : 827-846. doi: 10.3934/dcds.2012.32.827 [9] Hua Chen, Huiyang Xu. Global existence and blow-up of solutions for infinitely degenerate semilinear pseudo-parabolic equations with logarithmic nonlinearity. Discrete & Continuous Dynamical Systems - A, 2019, 39 (2) : 1185-1203. doi: 10.3934/dcds.2019051 [10] Long Wei, Zhijun Qiao, Yang Wang, Shouming Zhou. Conserved quantities, global existence and blow-up for a generalized CH equation. Discrete & Continuous Dynamical Systems - A, 2017, 37 (3) : 1733-1748. doi: 10.3934/dcds.2017072 [11] T. Colin, D. Lannes. Justification of and long-wave correction to Davey-Stewartson systems from quadratic hyperbolic systems. Discrete & Continuous Dynamical Systems - A, 2004, 11 (1) : 83-100. doi: 10.3934/dcds.2004.11.83 [12] Shuyin Wu, Joachim Escher, Zhaoyang Yin. Global existence and blow-up phenomena for a weakly dissipative Degasperis-Procesi equation. Discrete & Continuous Dynamical Systems - B, 2009, 12 (3) : 633-645. doi: 10.3934/dcdsb.2009.12.633 [13] Lili Du, Chunlai Mu, Zhaoyin Xiang. Global existence and blow-up to a reaction-diffusion system with nonlinear memory. Communications on Pure & Applied Analysis, 2005, 4 (4) : 721-733. doi: 10.3934/cpaa.2005.4.721 [14] Shu-Xiang Huang, Fu-Cai Li, Chun-Hong Xie. Global existence and blow-up of solutions to a nonlocal reaction-diffusion system. Discrete & Continuous Dynamical Systems - A, 2003, 9 (6) : 1519-1532. doi: 10.3934/dcds.2003.9.1519 [15] Monica Marras, Stella Vernier Piro. On global existence and bounds for blow-up time in nonlinear parabolic problems with time dependent coefficients. Conference Publications, 2013, 2013 (special) : 535-544. doi: 10.3934/proc.2013.2013.535 [16] Ronghua Jiang, Jun Zhou. Blow-up and global existence of solutions to a parabolic equation associated with the fraction p-Laplacian. Communications on Pure & Applied Analysis, 2019, 18 (3) : 1205-1226. doi: 10.3934/cpaa.2019058 [17] Bin Li. On the blow-up criterion and global existence of a nonlinear PDE system in biological transport networks. Kinetic & Related Models, 2019, 12 (5) : 1131-1162. doi: 10.3934/krm.2019043 [18] Jianbo Cui, Jialin Hong, Liying Sun. On global existence and blow-up for damped stochastic nonlinear Schrödinger equation. Discrete & Continuous Dynamical Systems - B, 2017, 22 (11) : 1-18. doi: 10.3934/dcdsb.2019169 [19] Victor A. Galaktionov, Juan-Luis Vázquez. The problem Of blow-up in nonlinear parabolic equations. Discrete & Continuous Dynamical Systems - A, 2002, 8 (2) : 399-433. doi: 10.3934/dcds.2002.8.399 [20] Xiaojing Xu. Local existence and blow-up criterion of the 2-D compressible Boussinesq equations without dissipation terms. Discrete & Continuous Dynamical Systems - A, 2009, 25 (4) : 1333-1347. doi: 10.3934/dcds.2009.25.1333
2018 Impact Factor: 0.545 | 2019-08-20 18:50:11 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7297642827033997, "perplexity": 3231.0811357831}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-35/segments/1566027315558.25/warc/CC-MAIN-20190820180442-20190820202442-00437.warc.gz"} |
https://brilliant.org/practice/vector-calculus-in-a-nutshell/?p=3 | ### Vector Calculus
In a nutshell, vector calculus deals with functions that output vectors.
For example, a wind map takes a location $P=(x,y)$ as input and returns the wind velocity vector $\vec{V}(x,y),$ displaying it as an arrow at $P.$
The wind map plots the vector function $\vec{V}(x,y).$
# Vector Calculus in a Nutshell
The most important object in our course is the vector field, which assigns a vector to every point in some subset of space.
We'll cover the essential calculus of such vector functions, and explore how to use them to solve problems in partial differential equations, wave mechanics, electricity and magnetism, and much more!
This quiz kicks off a short intro to the essential ideas of vector calculus.
# Vector Calculus in a Nutshell
The gradient $\nabla f = \left \langle \frac{\partial f}{\partial x} , \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right \rangle$ provides an example familiar from multivariable calculus with many important properties.
The touch interactive plot below shows a level set $f = c$ of $f = \sqrt{x^2+ y^2+z^2}$ together with the gradient vector field $\nabla f.$
Zoom in and out and rotate to explore this 3D vector field, then select the true statements about $\nabla f$ from the options provided.
# Vector Calculus in a Nutshell
Select one or more
If you've ever held a compass or seen the northern or southern lights, you're aware of Earth's natural magnetism.
The temperature of a cup of coffee, the flow of a river, and the amount of air in a child's balloon are probably more familiar to you from day-to-day experience.
Some are vector fields; others (the scalars) lack directionality and just have magnitude.
Select from the options all quantities that can also be modeled by a vector field.
# Vector Calculus in a Nutshell
Select one or more
To graph a vector field, we draw the arrow for $\vec{V}(\vec{x})$ with base at $\vec{x}$ for each point on a chosen grid, as in last problem's interactive plot. $\\\\$
Flow lines (aka field lines) also help us see important vector field features. A flow line is a curve that “follows the arrows”; if we're on the flow line at $\vec{x},$ $\vec{V}(\vec{x})$ points the way to our next stop along the curve.
We include a two-dimensional example and a three-dimensional one below.
$\\\\$
# Vector Calculus in a Nutshell
Using this intuitive notion of a flow line, choose the option that best describes a flow line for the two-dimensional vector field below.
# Vector Calculus in a Nutshell
We can use flow lines as a means of understanding one of the most important vector field derivatives: the divergence.
The animation shows what happens to a rectangle when every point in it flows along the vector field.
It turns out that how much this rectangle distorts under flow is directly related to the divergence, which we'll talk about next.
# Vector Calculus in a Nutshell
As the name suggests, the divergence measures how much a vector field spreads out from a given point. We'll cover divergence in detail later in the course, but this qualitative description will do for right now.
Below are two vector fields. Select the one that has the greatest divergence from the origin.
Option I (left), Option II (right)
# Vector Calculus in a Nutshell
It'll take some work with flow lines to prove this, but the divergence of a two-dimensional vector field $\vec{V}(x,y) = \big\langle V_{x}(x,y) , V_{y}(x,y) \big\rangle$ at $(x,y)$ is $\text{div}\big(\vec{V}\big) = \frac{\partial V_{x}}{\partial x} + \frac{\partial V_{y}}{\partial y}.$ Option I in the previous problem (which just circled the origin) is explicitly given by $\langle -y, x \rangle$ and so has divergence $\text{div}\big( \langle -y, x \rangle \big) = \frac{\partial }{\partial x}[-y] + \frac{\partial}{\partial y}[x]=0.$
Compute the divergence for the vector field of option II given by $\vec{V}(x,y) = \left \langle x , y \right \rangle.$
# Vector Calculus in a Nutshell
The divergence formula comes from a careful analysis of flow lines, which we'll do later in the course. These curves also teach us much about another very important derivative called curl.
The flow of a river is like a 2D vector field; the strength and direction of current at a point is assigned an arrow there.
If we dump a few X-shaped paddle wheels into the stream, they'll move with the flow and also rotate as fluid strikes the paddles.
Each X is assigned a vector (called the curl) pointing along the $z$-axis and providing the axis of rotation. Its length is related to the spin rate. If it points up (down), the X spins counterclockwise (clockwise).
# Vector Calculus in a Nutshell
The curl of a two-dimensional vector field is given by $\text{curl}\big(\vec{V}\big)(x,y) = \left[ \frac{\partial V_{y}}{\partial x} - \frac{\partial V_{x}}{\partial y} \right] \hat{k},$ as we'll see in greater detail later in the course.
Compute the curl of the vector field $\langle -y , x \rangle$ that we encountered earlier in this quiz.
# Vector Calculus in a Nutshell
Just as divergence and curl provide two different vector field derivatives, line integrals and surface integrals give us different ways of integrating fields.
Vector integrals are a bit more complicated than vector derivatives, so we'll postpone talking about them for now.
In the next chapter, we'll make a detour into the world of motion and see some other applications of vector derivatives.
# Vector Calculus in a Nutshell
× | 2021-10-18 21:06:14 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 70, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7679705023765564, "perplexity": 609.2159243386423}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-43/segments/1634323585209.43/warc/CC-MAIN-20211018190451-20211018220451-00122.warc.gz"} |
https://www.thepoorcoder.com/athlete-sort-solution/ | Athlete Sort Solution
# Athlete Sort Solution
You are given a spreadsheet that contains a list of athletes and their details (such as age, height, weight and so on). You are required to sort the data based on the th attribute and print the final resulting table. Follow the example given below for better understanding.
Note that is indexed from to , where is the number of attributes.
Note: If two attributes are the same for different rows, for example, if two atheletes are of the same age, print the row that appeared first in the input.
Input Format
The first line contains and separated by a space.
The next lines each contain elements.
The last line contains .
Constraints
Each element
Output Format
Print the lines of the sorted table. Each line should contain the space separated elements. Check the sample below for clarity.
Sample Input 0
5 3
10 2 5
7 1 0
9 9 9
1 23 12
6 5 9
1
Sample Output 0
7 1 0
10 2 5
6 5 9
9 9 9
1 23 12
Explanation 0
The details are sorted based on the second attribute, since is zero-indexed.
### Solution in Python
N,M = map(int,input().split())
rows = [list(map(int,input().split())) for i in range(N)]
i = int(input())
rows = sorted(rows, key=lambda x:x[i])
for i in rows:
print(*i) | 2022-07-04 21:34:42 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.669002115726471, "perplexity": 2967.8963553205917}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-27/segments/1656104496688.78/warc/CC-MAIN-20220704202455-20220704232455-00728.warc.gz"} |
http://zgwlc.xml-journal.net/article/current | ## Current Issue
column
Display Method: |
2022, 46(2): 023101. doi: 10.1088/1674-1137/ac3122
Abstract:
Owing to the special structure of a five-dimensional Elko spinor, its localization on a brane with codimension one becomes completely different from that of a Dirac spinor. By introducing the coupling between the Elko spinor and the scalar field that can generate the brane, we have two types of localization mechanism for the five-dimensional Elko spinor zero mode on a brane. One is the Yukawa-type coupling, and the other is the non-minimal coupling. In this study, we investigate the localization of the Elko zero mode on de Sitter and Anti-de Sitter thick branes with the two localization mechanisms, respectively. The results show that both the mechanisms can achieve localization. The forms of the scalar coupling function in both localization mechanisms have similar properties, and they play a similar role in localization.
2022, 46(2): 023102. doi: 10.1088/1674-1137/ac3124
Abstract:
In this study, we investigate the Kotzinian-Mulders effect under semi-inclusive deep inelastic scattering (SIDIS) within the framework of transverse momentum dependent (TMD) factorization. The asymmetry is contributed by the convolution of the Kotzinian-Mulders function \begin{document}$g_{1T}$\end{document} and the unpolarized fragmentation function \begin{document}$D_1$\end{document}. As a TMD distribution, the Kotzinian-Mulders function in the coordinate space in the perturbative region can be represented as the convolution of the C-coefficients and the corresponding collinear correlation function. The Wandzura-Wilczek approximation is used to obtain this correlation function. We perform a detailed phenomenological numerical analysis of the Kotzinian-Mulders effect in the SIDIS process within TMD factorization at the kinematics of the HERMES and COMPASS experiments. We observe that the obtained \begin{document}$x_B$\end{document}-, \begin{document}$z_h$\end{document}-, and \begin{document}$P_{h\perp}$\end{document}-dependent Kotzinian-Mulders effects are basically consistent with the HERMES and COMPASS measurements. We also make predictions at EIC and EicC kinematics.
2022, 46(2): 023103. doi: 10.1088/1674-1137/ac3123
Abstract:
In this study, by combining the equal spacing rule with recent observations of \begin{document}$\Omega_c(X)$\end{document} and \begin{document}$\Xi_c(X)$\end{document} baryons, we predict the spectrum of the low-lying \begin{document}$\lambda$\end{document}-mode \begin{document}$1P$\end{document}-wave excited \begin{document}$\Sigma_c$\end{document} states. Furthermore, their strong decay properties are predicted using the chiral quark model and the nature of \begin{document}$\Sigma_c(2800)$\end{document} is investigated by analyzing the \begin{document}$\Lambda_c\pi$\end{document} invariant mass spectrum. The \begin{document}$\Sigma_c(2800)$\end{document} structure observed in the \begin{document}$\Lambda_c \pi$\end{document} mass spectrum was found to potentially arise from two overlapping \begin{document}$P$\end{document}-wave \begin{document}$\Sigma_c$\end{document} resonances, \begin{document}$\Sigma_c(2813)3/2^-$\end{document} and \begin{document}$\Sigma_c(2840)5/2^-$\end{document}. These resonances have similar decay widths of \begin{document}$\Gamma\sim 40$\end{document} MeV and predominantly decay into the \begin{document}$\Lambda_c \pi$\end{document} channel. The \begin{document}$\Sigma_c(2755)1/2^-$\end{document} state is likely to be a very narrow state with a width of \begin{document}$\Gamma\sim 15$\end{document} MeV, with its decays almost saturated by the \begin{document}$\Lambda_c \pi$\end{document} channel. Additionally, evidence of the \begin{document}$\Sigma_c(2755) {1}/{2}^-$\end{document} resonance as a very narrow peak may be seen in the \begin{document}$\Lambda_c\pi$\end{document} invariant mass spectrum. The other two \begin{document}$P$\end{document}-wave states, \begin{document}$\Sigma_c(2746) {1}/{2}^-$\end{document} and \begin{document}$\Sigma_c(2796) {3}/{2}^-$\end{document}, are relatively narrow states with similar widths of \begin{document}$\Gamma\sim 30$\end{document} MeV and predominantly decay into \begin{document}$\Sigma_c\pi$\end{document} and \begin{document}$\Sigma^{*}_c\pi$\end{document}, respectively. This study can provide useful references for discovering these low-lying \begin{document}$P$\end{document}-wave states in forthcoming experiments.
2022, 46(2): 023104. doi: 10.1088/1674-1137/ac31a4
Abstract:
Extensive dynamical \begin{document}$N/D$\end{document} calculations are conducted in the study of \begin{document}$S_{11}$\end{document} channel low energy \begin{document}$\pi N$\end{document} scatterings, based on various phenomenological model inputs of left cuts at the tree level. The subtleties of the singular behavior of the partial wave amplitude, at the origin of the complex \begin{document}$s$\end{document} plane, are analysed in detail. Furthermore, it is found that the dispersion representation for the phase shift, \begin{document}$\delta$\end{document}, must be modified in the case of \begin{document}$\pi N$\end{document} scatterings. An additional contribution from the dispersion integral exists, which approximately cancels the contribution of the two virtual poles located near the end points of the segment cut, induced by \begin{document}$u$\end{document} channel nucleon exchanges. With limited reliance on the details of the dynamical inputs, the subthreshold resonance \begin{document}$N^*(890)$\end{document} survives.
2022, 46(2): 023105. doi: 10.1088/1674-1137/ac3567
Abstract:
In order to confirm the existence of the dibaryon state \begin{document}$d^*(2380)$\end{document} observed at WASA@COSY, we estimate the cross section for production of the possible dibaryon and anti-dibaryon pair \begin{document}${d^*}{\bar{d}^*}$\end{document} in the energy region of the upcoming experiments at \begin{document}${\bar{{\rm{P}}}}$\end{document}anda. Based on some qualitative properties of \begin{document}${d^*}$\end{document} extracted from the analyses in the non-relativistic quark model, the production cross section for this spin-3 particle pair are calculated with the help of a phenomenological effective relativistic and covariant Lagrangian approach.
2022, 46(2): 023106. doi: 10.1088/1674-1137/ac3642
Abstract:
In our previous work [Phys. Rev. C 101, 014003 (2020)], the photoproduction reaction \begin{document}$\gamma p \to K^{\ast +} \Lambda$\end{document} was investigated within an effective Lagrangian approach. The reaction amplitudes were constructed by including the t-channel K, \begin{document}$K^\ast$\end{document}, and κ exchanges, u-channel Λ, Σ, and \begin{document}$\Sigma^\ast$\end{document} exchanges, s-channel N, \begin{document}$N(2000)5/2^+$\end{document}, and \begin{document}$N(2060)5/2^-$\end{document} exchanges, and interaction current. The data on the differential cross sections and spin density matrix elements were described simultaneously. In this study, we investigate the photoproduction reaction \begin{document}$\gamma n \to K^{\ast 0} \Lambda$\end{document} based on the same reaction mechanism as that of \begin{document}$\gamma p \to K^{\ast +} \Lambda$\end{document} to obtain a unified description of the data for \begin{document}$\gamma p \to K^{\ast +} \Lambda$\end{document} and \begin{document}$\gamma n \to K^{\ast 0} \Lambda$\end{document} within the same model. All hadronic coupling constants, form factor cutoffs, and the resonance masses and widths in the present calculations remain the same as in our previous work for \begin{document}$\gamma p \to K^{\ast +} \Lambda$\end{document}. The available differential cross-section data for \begin{document}$\gamma n \to K^{\ast 0} \Lambda$\end{document} are well reproduced. Further analysis shows that the cross sections of \begin{document}$\gamma n \to K^{\ast 0} \Lambda$\end{document} are dominated by the contributions of the t-channel K exchange, while the s-channel \begin{document}$N(2000)5/2^+$\end{document} and \begin{document}$N(2060)5/2^-$\end{document} exchanges also provide considerable contributions.
2022, 46(2): 023107. doi: 10.1088/1674-1137/ac3a5a
Abstract:
In this study, the heavy to heavy decay of \begin{document}$B^0_s\rightarrow D^{*+}D^-$\end{document} is evaluated through the factorization approach by using the final state interaction as an effective correction. Under the factorization approach, this decay mode occurs only through the annihilation process, so a small amount is produced. Feynman's rules state that six meson pairs can be assumed for the intermediate states before the final meson pairs are produced. By taking into account the effects of twelve final state interaction diagrams in the calculations, a significant correction is obtained. These effects correct the value of the branching ratio obtained by the pure factorization approach from \begin{document}$(2.41\pm1.37)\times10^{-5}$\end{document} to \begin{document}$(8.27\pm2.23)\times10^{-5}$\end{document}. The value obtained for the branching ratio of the \begin{document}$B^0_s\rightarrow D^{*+}D^-$\end{document} decay is consistent with the experimental results.
2022, 46(2): 024001. doi: 10.1088/1674-1137/ac3412
Abstract:
Cross sections of the 58,60,61Ni(n, α)55,57,58Fe reactions were measured at 8.50, 9.50 and 10.50 MeV neutron energies based on the HI-13 tandem accelerator of China Institute of Atomic Energy (CIAE) with enriched 58Ni, 60Ni, and 61Ni foil samples with backings. A twin gridded ionization chamber (GIC) was used as the charged particle detector, and an EJ-309 liquid scintillator was used to obtain the neutron energy spectra. The relative and absolute neutron fluxes were determined via three highly enriched 238U3O8 samples inside the GIC. The uncertainty of the present data of the 58Ni(n, α)55Fe reaction is smaller than most existing measurements. The present data of 60Ni(n, α)57Fe and 61Ni(n, α)58Fe reactions are the first measurement results above 8 MeV. The present experimental data could be reasonably reproduced by calculations with TALYS-1.9 by adjusting several default values of theoretical model parameters.
2022, 46(2): 024101. doi: 10.1088/1674-1137/ac321c
Abstract:
Neutron-deficient actinide nuclei provide a valuable window to probe heavy nuclear systems with large proton-neutron ratios. In recent years, several new neutron-deficient Uranium and Neptunium isotopes have been observed using α-decay spectroscopy [Z. Y. Zhang et al., Phys. Rev. Lett. 122, 192503 (2019); L. Ma et al., Phys. Rev. Lett. 125, 032502 (2020); Z. Y. Zhang et al., Phys. Rev. Lett. 126, 152502 (2021)]. In spite of these achievements, some neutron-deficient key nuclei in this mass region are still unknown in experiments. Machine learning algorithms have been applied successfully in different branches of modern physics. It is interesting to explore their applicability in α-decay studies. In this work, we propose a new model to predict the α-decay energies and half-lives within the framework based on a machine learning algorithm called the Gaussian process. We first calculate the α-decay properties of the new actinide nucleus \begin{document}${}^{214}{\rm{U}}$\end{document}. The theoretical results show good agreement with the latest experimental data, which demonstrates the reliability of our model. We further use the model to predict the α-decay properties of some unknown neutron-deficient actinide isotopes and compare the results with traditional models. The results may be useful for future synthesis and identification of these unknown isotopes.
2022, 46(2): 024102. doi: 10.1088/1674-1137/ac338e
Abstract:
In this study, we investigate the effect of rotation on the masses of scalar and vector mesons in the framework of the 2-flavor Nambu-Jona-Lasinio model. The existence of rotation produces a tedious quark propagator and a corresponding polarization function. By applying the random phase approximation, the meson mass is numerically calculated. It is found that the behavior of scalar and pseudoscalar meson masses under angular velocity ω is similar to that at a finite chemical potential; both rely on the behavior of the constituent quark mass and reflect the property related to chiral symmetry. However, vector meson ρ masses have a more profound relation to rotation. After analytical and numerical calculations, it turns out that at low temperature and small chemical potential, the mass for spin component \begin{document}$s_z = 0,\pm 1$\end{document} of a vector meson under rotation exhibits a very simple mass splitting relation \begin{document}$m_{\rho}^{s_z}(\omega) = m_\rho(\omega = 0)-\omega s_z$\end{document}, similar to the Zeeman splitting of a charged meson under magnetic fields. Furthermore, the mass of the spin component \begin{document}$s_z = 1$\end{document} of vector meson ρ decreases linearly with ω and reaches zero at \begin{document}$\omega_c = m_\rho(\omega = 0)$\end{document}, which indicates that the system will develop \begin{document}$s_z = 1$\end{document} vector meson condensation and the system will be spontaneously spin-polarized under rotation.
2022, 46(2): 024103. doi: 10.1088/1674-1137/ac347a
Abstract:
We have calculated the potential energy surfaces for 240Pu up to the scission point using the density functional theory with different pairing strengths to investigate the effect of pairing correlations on its fission properties. An enhancement in the pairing correlations lowers the barrier heights, isomeric state, and ridge between the symmetric and asymmetric fission valleys significantly. Moreover, it weakens the microscopic shell structure around the Fermi surface, shrinks the scission frontiers, especially for the symmetric and very asymmetric fission regions, and lifts the total kinetic energies (TKEs) for the symmetric fission region. It is also emphasized that the microscopic calculation qualitatively reproduces the trend of the distribution of the measured TKEs, especially for the positions of the peaks at \begin{document}$A_{\rm{frag}}\simeq132$\end{document} and \begin{document}$A_{\rm{frag}}\simeq108$\end{document}.
2022, 46(2): 024104. doi: 10.1088/1674-1137/ac3748
Abstract:
High-order cumulants and factorial cumulants of conserved charges are suggested for the study of the critical dynamics in heavy-ion collision experiments. In this paper, using the parametric representation of the three-dimensional Ising model which is believed to belong to the same universality class as quantum chromo-dynamics, the temperature dependence of the second- to fourth-order (factorial) cumulants of the order parameter is studied. It is found that the values of the normalized cumulants are independent of the external magnetic field at the critical temperature, which results in a fixed point in the temperature dependence of the normalized cumulants. In finite-size systems simulated using the Monte Carlo method, this fixed point behavior still exists at temperatures near the critical. This fixed point behavior has also appeared in the temperature dependence of normalized factorial cumulants from at least the fourth order. With a mapping from the Ising model to QCD, the fixed point behavior is also found in the energy dependence of the normalized cumulants (or fourth-order factorial cumulants) along different freeze-out curves.
2022, 46(2): 024105. doi: 10.1088/1674-1137/ac3749
Abstract:
In this study, the Pauli blocking potential between two colliding nuclei in the density overlapping region is applied to describe the heavy nuclei fusion process. Inspired by the Pauli blocking effect in the \begin{document}$\alpha$\end{document}-decay of heavy nuclei, the Pauli blocking potential of single nucleon from the surrounding matter is obtained. In fusion reactions with strong density overlap, the Pauli blocking potential between the projectile and target can be constructed using a single folding model. By considering this potential, the double folding model with a new parameter set is employed to analyze the fusion processes of 95 systems. A wider Coulomb barrier and shallower potential pocket are formed in the inner part of the potential between the two colliding nuclei, compared to that calculated using the Akyüz-Winther potential. The fusion hindrance phenomena at deep sub-barrier energies are described well for fusion systems \begin{document}$^{16}$\end{document}O + \begin{document}$^{208}$\end{document}Pb and \begin{document}$^{58}$\end{document}Ni + \begin{document}$^{58}$\end{document}Ni.
2022, 46(2): 024106. doi: 10.1088/1674-1137/ac380a
Abstract:
We report a benchmark calculation for the Lipkin model in nuclear physics with a variational quantum eigensolver in quantum computing. Special attention is paid to the unitary coupled cluster (UCC) ansatz and structure learning (SL) ansatz for the trial wave function. Calculations with both the UCC and SL ansatz can reproduce the ground-state energy well; however, it is found that the calculation with the SL ansatz performs better than that with the UCC ansatz, and the SL ansatz has even fewer quantum gates than the UCC ansatz.
2022, 46(2): 024107. doi: 10.1088/1674-1137/ac3904
Abstract:
We investigate the effects of higher-order deformations \begin{document}$\beta_\lambda$\end{document} (\begin{document}$\lambda=4,6,8,$\end{document} and 10) on the ground state properties of superheavy nuclei (SHN) near the doubly magic deformed nucleus \begin{document}$^{270}{\rm{Hs}}$\end{document} using the multidimensionally-constrained relativistic mean-field (MDC-RMF) model with five effective interactions: PC-PK1, PK1, NL3*, DD-ME2, and PKDD. The doubly magic properties of \begin{document}$^{270}{\rm{Hs}}$\end{document} include large energy gaps at \begin{document}$N=162$\end{document} and \begin{document}$Z=108$\end{document} in the single-particle spectra. By investigating the binding energies and single-particle levels of \begin{document}$^{270}{\rm{Hs}}$\end{document} in the multidimensional deformation space, we find that, among these higher-order deformations, the deformation \begin{document}$\beta_6$\end{document} has the greatest impact on the binding energy and influences the shell gaps considerably. Similar conclusions hold for other SHN near \begin{document}$^{270}{\rm{Hs}}$\end{document}. Our calculations demonstrate that the deformation \begin{document}$\beta_6$\end{document} must be considered when studying SHN using MDC-RMF.
2022, 46(2): 024108. doi: 10.1088/1674-1137/ac39fd
Abstract:
We investigate quantum kinetic theory for a massive fermion system under a rotational field. From the Dirac equation in rotating frame we derive the complete set of kinetic equations for the spin components of the 8- and 7-dimensional Wigner functions. While the particles are no longer on a mass shell in the general case due to the rotation–spin coupling, there are always only two independent components, which can be taken as the number and spin densities. With help from the off-shell constraint we obtain the closed transport equations for the two independent components in the classical limit and at the quantum level. The classical rotation–orbital coupling controls the dynamical evolution of the number density, but the quantum rotation–spin coupling explicitly changes the spin density.
2022, 46(2): 024109. doi: 10.1088/1674-1137/ac3bc7
Abstract:
\begin{document}$\eta N$\end{document} interactions are investigated in hot magnetized asymmetric nuclear matter using the chiral SU(3) model and chiral perturbation theory (ChPT). In the chiral model, the in-medium properties of η-mesons are calculated using medium modified scalar densities under the influence of an external magnetic field. Further, in a combined chiral model and ChPT approach, off-shell contributions of the \begin{document}$\eta N$\end{document} interactions are evaluated from the ChPT effective \begin{document}$\eta N$\end{document} Lagrangian, and the in-medium effect of scalar densities are incorporated from the chiral SU(3) model. We find that the magnetic field has a significant effect on the in-medium mass and optical potential of η mesons, and we observe a deeper mass-shift in the combined chiral model and ChPT approach than in the solo chiral SU(3) model. In both approaches, no additional mass-shift is observed due to the uncharged nature of η mesons in the presence of a magnetic field.
2022, 46(2): 025101. doi: 10.1088/1674-1137/ac3411
Abstract:
In this study, we investigate the possibilities of generating baryon number asymmetry under thermal equilibrium within the frameworks of teleparallel and symmetric teleparallel gravities. Through the derivative couplings of the torsion scalar and the non-metricity scalar to baryons, baryon number asymmetry is produced in the radiation dominated epoch. For gravitational baryogenesis mechanisms in these two frameworks, the produced baryon-to-entropy ratio is too small to be consistent with observations. However, the gravitational leptogenesis models within both frameworks have the potential to explain the observed baryon-antibaryon asymmetry.
2022, 46(2): 025102. doi: 10.1088/1674-1137/ac3643
Abstract:
We develop the regular black hole solutions by incorporating the 1-loop quantum correction to the Newton potential and a time delay between an observer at the regular center and one at infinity. We define the maximal time delay between the center and the infinity by scanning the mass of black holes such that the sub-Planckian feature of the Kretschmann scalar curvature is preserved during the process of evaporation. We also compare the distinct behavior of the Kretschmann curvature for black holes with asymptotically Minkowski cores and those with asymptotically de-Sitter cores, including Bardeen and Hayward black holes. We expect that such regular black holes may provide more information about the construction of effective metrics for Planck stars.
IF: 2.145
Monthly founded in 1977
ISSN 1674-1137 CN 11-5641/O4
Original research articles, Ietters and reviews Covering theory and experiments in the fieids of
• Particle physics
• Nuclear physics
• Particle and nuclear astrophysics
• Cosmology
Author benefits
News
Meet Editor | 2022-01-28 21:44:53 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6976537704467773, "perplexity": 1408.9294777092823}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320306346.64/warc/CC-MAIN-20220128212503-20220129002503-00163.warc.gz"} |
https://socratic.org/questions/how-do-you-find-dy-dx-by-implicit-differentiation-given-x-4-y-4-5 | # How do you find dy/dx by implicit differentiation given x^4+y^4=5?
Dec 20, 2016
The answer is $= - {x}^{3} {\left(5 - {x}^{4}\right)}^{- \frac{3}{4}}$
#### Explanation:
Let's do the differentiation
${x}^{4} + {y}^{4} = 5$
$4 {x}^{3} + 4 {y}^{3} \frac{\mathrm{dy}}{\mathrm{dx}} = 0$
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- 4 {x}^{3}}{4 {y}^{3}}$
$= - {x}^{3} / {y}^{3}$
But, ${y}^{4} = 5 - {x}^{4}$
${y}^{3} = {\left(5 - {x}^{4}\right)}^{\frac{3}{4}}$
Therefore,
$\frac{\mathrm{dy}}{\mathrm{dx}} = - {x}^{3} / {\left(5 - {x}^{4}\right)}^{\frac{3}{4}} = - {x}^{3} {\left(5 - {x}^{4}\right)}^{- \frac{3}{4}}$ | 2021-06-25 03:59:44 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 8, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5604963898658752, "perplexity": 10025.362525554778}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-25/segments/1623488567696.99/warc/CC-MAIN-20210625023840-20210625053840-00155.warc.gz"} |
http://mathhelpforum.com/calculus/126687-converge-diverge-print.html | # converge or diverge?
• Feb 1st 2010, 06:08 PM
thatloserrsaid
converge or diverge?
infinity
summation of ln(n)/n
n=2
I need to know if it converges or diverges and explain why. If someone could go though this step by step it would be a big help.
• Feb 1st 2010, 06:10 PM
skeeter
Quote:
Originally Posted by thatloserrsaid
infinity
summation of ln(n)/n
n=2
I need to know if it converges or diverges and explain why. If someone could go though this step by step it would be a big help.
use the integral test
• Feb 1st 2010, 06:11 PM
thatloserrsaid
the integral test confuses me.. thats why i requested a step by step help
• Feb 1st 2010, 06:12 PM
thatloserrsaid
Quote:
Originally Posted by skeeter
use the integral test
the integral test confuses me.. thats why i requested a step by step help
• Feb 1st 2010, 06:24 PM
skeeter
Quote:
Originally Posted by thatloserrsaid
the integral test confuses me.. thats why i requested a step by step help
$\int_2^{\infty} \frac{\ln{x}}{x} \, dx$
substitution ... let $u = \ln{x}$
give it a go.
• Feb 1st 2010, 06:37 PM
thatloserrsaid
Quote:
Originally Posted by skeeter
$\int_2^{\infty} \frac{\ln{x}}{x} \, dx$
substitution ... let $u = \ln{x}$
give it a go.
after i integrate i get [ln(x)]^2 / 2
• Feb 1st 2010, 06:38 PM
General
Quote:
Originally Posted by thatloserrsaid
infinity
summation of ln(n)/n
n=2
I need to know if it converges or diverges and explain why. If someone could go though this step by step it would be a big help.
$\frac{ln(n)}{n}\geq\frac{1}{n}$ for $n\geq e$
• Feb 1st 2010, 08:17 PM
Pulock2009
D'Alembert's test | 2017-03-27 07:24:47 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 6, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9275020360946655, "perplexity": 3129.4757778976073}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-13/segments/1490218189462.47/warc/CC-MAIN-20170322212949-00500-ip-10-233-31-227.ec2.internal.warc.gz"} |
https://cs.stackexchange.com/questions/91842/union-of-two-non-context-free-languages | Union of two non-context-free languages
Let L1 = L2 union L3 find values such that L1 is context free and L2 and L3 are not.
So far I have: L1 = $a^nb^n$ L2 = $a^*b^*$ L3 = $a^+b^+$
Is this acceptable?? Since L2 covers everything including epsilon and L3 is the same but does not include epsilon?
I know L2 is regular so I guess that is also not a CFL. Another problem is that the a's and b's aren't linked in L2 and L3, so either one can always have more a's than b's and vice versa.
• Every regular language is also a context-free language. – kntgu May 13 '18 at 6:35
• There are two problems with your example: first, $L_2$ and $L_3$ are context-free; and second, $L_2 \cup L_3 \neq L_1$. You need to find a different example. – Yuval Filmus May 13 '18 at 8:00
Hint. For any language $L$, $$A^* = L \cup (A^* - L).$$ Now, choose an appropriate language $L$ to solve your problem.
Here is another construction: for every language $L$ over $\{0,1\}$, $$(0L \cup 1\Sigma^*) \cup (1L \cup 0\Sigma^*) = \Sigma^+.$$ | 2020-02-17 04:55:45 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6558631062507629, "perplexity": 526.0835528439541}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-10/segments/1581875141653.66/warc/CC-MAIN-20200217030027-20200217060027-00415.warc.gz"} |
https://quantumcomputing.stackexchange.com/tags/clifford-group/hot | # Tag Info
10
By definition, conjugation by a Clifford gate preserves the Pauli group $G = \langle X, Y, Z\rangle$. It is easy to check that $SXS^\dagger = Y, SYS^\dagger = -X$ and $SZS^\dagger = Z$. Since $G = \langle -X, Y, Z\rangle$ we see that $S$ is Clifford. On the other hand, $TXT^\dagger = \begin{pmatrix} & e^{-i\pi/4}\\ e^{i\pi/4}&\end{pmatrix} \notin G$, ...
8
Yes, you are correct. Non-Clifford gates cannot be transversely implemented, instead implementation generally requires distilling magic states or Toffoli states. In practice this requires significantly more spacetime volume than Clifford gates. For reference, see the introduction sections here and here. The natural expectation would be that no quantum ...
8
Sometimes, there is a bit of confusion around the Clifford group in the field ... and it's a matter of definition. A lot of people define the Clifford group $\mathrm{Cl}_n(p)$ of $n$ qudits of prime dimension $p$ as the unitary normaliser of the generalised Pauli group (e.g. Gottesman, Nielsen & Chuang). As such, it is clearly not a finite group as the ...
8
Geometrically: For a single qubit, we have the Bloch sphere and the stabiliser states span an octahedron inside it. Unitaries act in the adjoint representation as $SO(3)$, i.e. they induce rotations of the Bloch sphere. It is easy to see, that only rotations about the $X,Y,Z$ axis with angles $\theta=\pi/2,\pi,3\pi/2,2\pi$ preserve the octahedron. This ...
7
Short answer: Yes, this should be possible. However, the details have to be filled out. The key is to relate this to magic monotones. There has been some development since the 2016 Bravyi-Gosset paper. I think one can use an argumentation based on stabiliser / Clifford extent and the results in the BBCCGH paper from 2019. Similarly, one can also argue for ...
7
Here's a simple strategy based on the idea that Clifford operations conjugate Pauli products into other Pauli products. If $U$ is a Clifford operation, then $U P U^\dagger$ (where $P$ is a Pauli operation on one of the qubits) will be a matrix equivalent to a product of Pauli operations. If you check this for each $X_q$ and $Z_q$ for each qubit $q$, the ...
7
As a general rule, you wouldn't bother constructing this: it is just a global phase that has no observable consequence. If you really insist on doing this, introduce an ancilla qubit in the $|1\rangle$ state and apply a $Z$ gate to it. PS "inverse identity gate" is a really bad name for it. The identity operation is its own inverse.
6
The quaternions are represented faithfully in two dimensions by the unit matrix and the Pauli matrices multiplied by the imaginary unit: $i = \sqrt{-1} X$, $j = \sqrt{-1} Y$ and $k = \sqrt{-1} Z$ respectively, thus you only need to write $H$ and $P$ the linear combinations: $H = -\frac{\sqrt{-1} }{\sqrt{2}}(i+k)$ and $P = \frac{1+\sqrt{-1}}{2}(1-k)$ However,...
6
Following Dehaene and de Moor (Theorem 6 in particular), every Clifford unitary can be represented (up to a global scalar factor) by an expression of the form $$U = 2^{-k/2} \!\!\!\!\!\!\sum_{\substack{x_r,x_c \in \{0,1\}^k \\ x_b \in \{0,1\}^{n-k}}}\!\!\!\!\! i^{p(x_b,x_c,x_r)} (-1)^{q(x_b,x_c,x_r)} \bigl\lvert T_1[x_r;x_b] \bigr\rangle\!\bigl\langle T_2[... 6 The T gate as well as all possible single qubit rotations are non-entangling operations. That means if we have a circuit composed of single bit rotations, any non-entangled n-bit input, it will result in a n-bit non-entangled output. The CCNOT gate however is of course entangling, really all controlled gates are entangling, which means there exist ... 6 If you are trying to implement a fault-tolerant quantum computation, you need to implement unitary gates that act on logical qubits. You typically have a finite set of these gates available, and what you really care about is making your operations in such a way as to keep the fault-tolerant threshold as small as possible. If you calculate a fault-tolerant ... 5 No. There always exists a unique unitary U such that CT=TU. Namely, U = T^\dagger C T. The question is whether U is Clifford. It turns out that this is not guaranteed. For a simple counterexample take Hadamard for C. Then$$ U=T^\dagger H T = \frac{1}{\sqrt{2}}\begin{bmatrix} 1 & e^{i\pi/4} \\ e^{-i\pi/4} & -1 \end{bmatrix} is not ... 5 The orthonormal basis |j\rangle of the d dimensional finite Hilbert space corresponds to a configuration space of equally spaced clockwise ordered d points on a circle S^1 or equivalently, the vertices of a d-dimensional regular polygon. One may think of a point as a discrete location of a particle, then the shift operator X shifts the particle ... 5 The shift operator takes his name from the fact that it shifts the position of its input, as in, it sends 1\to2, 2\to3 etc, with the last computational basis element being sent back to the first one: d\to 1 (or the same thing starting with 0, depending on notation). As per the "boost" operator Z, I have usually seen those referred to as "clock ... 5 The difference in definitions is from either taking the unitary group or the projective unitary group. That accounts for the constant prefactors of \pm i that are missing. In lieu of a tikz commutative diagram \begin{align} <X,Y,Z> & \hookrightarrow & U(2)\\ \downarrow & & \downarrow\\ <[X],[Z]> & \hookrightarrow & PU(... 5 The terms II and ZZ do not uniquely specify the state |11\rangle because you could equally have the state |00\rangle. Indeed, you should not include the identity term in your stabilizer. Thus, you need to add a second term, which could be either -ZI or -IZ. Either way, you can easily see how to make a product -ZI out of your stabilizers. 5 What the author wrote is completely correct, they did not make a mistake. The subgroup of Cliffords fixing X_n and Z_n is indeed isomorphic to C_{n-1} as a group, this is simply because this subgroup acts by assumption as U (\sigma_1 \otimes \dots \otimes \sigma_n) U^\dagger = \tilde U (\sigma_1\otimes\dots\otimes\sigma_{n-1})\tilde U^\dagger \...
5
The matrix $$M = \frac{1}{\sqrt{2}}\begin{bmatrix}-i & 1\\-1 & i\end{bmatrix}$$ resembles $$X/2 = \frac{1}{\sqrt{2}}\begin{bmatrix}1 & -i\\-i & 1\end{bmatrix}\tag1$$ where we follow the notation $\pm X/2$ for the $\pm\frac{\pi}{2}$ rotation around the $X$ axis as used in the table B.6 on page 101 in Julian Kelly's PhD thesis. We can make ...
5
It's well known that you can make a swap out of three CNOTs. For reference, Stim's gate documentation includes H+S+CX decompositions of a lot of Clifford gates including the swap: Stabilizer Generators: X_ -> +_X Z_ -> +_Z _X -> +X_ _Z -> +Z_ Unitary Matrix: [+1 , , , ] [ , , +1 , ] [ , +1 , , ] [ , ...
5
Conjugating a non-identity Pauli operator $P$ by Clifford operators yields all non-identity Pauli operators, including $P$, with equal frequency. Proof Let $G_n$ and $C_n$ denote the $n$-qubit Pauli and Clifford groups, respectively. For $P, Q\in G_n$, let $C_{P\to Q}$ denote the set of Clifford operators that map $P$ to $Q$ under conjugation, i.e. $$C_{P\... 5 As far as I know, the current fastest simulator for stabilizer circuits is my simulator Stim (source code on github, paper in Quantum, python package on pypi). This is especially true if you're doing bulk sampling: Note the Y axis; that's a log-log plot. There are some simulators that are faster in specialized circumstances, e.g. for piecewise separable ... 4 Note that the second definition actually doesn't make more sense in the context of the stabiliser formalism, as neither of \pm i Y have a +1 eigenspace. That means that you can only describe states which are stabilised by operators with two or more factors of \pm i Y. This is only enough to simulate real stabiliser circuits, which is what you're noticing.... 4 There is a very closely related representation of the tableau representation of Aaronson (and Gottesman), which works not only for qubits but for qudits of arbitrary finite dimension, which works particularly well for purely Clifford circuits (i.e. at most one terminal measurement). In this alternative representation, one has tableaus describing how ... 4 The diffusion operator is a multi controlled not operation (modulo some hadamards). It's not a Clifford operation. Also any useful oracle you'd use with Grover's algorithm won't be Clifford operations either, since any Clifford oracle accepts 0%, 50%, or 100% of all inputs which makes search trivial. 4 From the paper Normal form for single-qutrit Clifford+T operators and synthesis of single-qutrit gates, the Clifford group in p>2 dimensions acting on a sigle qudit is generated by S and H given by:$$ \begin{gather} S=\sum_{j=0}^{p-1}\omega^{j(j+1)2^{-1}}|j\rangle\langle j| \\ H = \frac{1}{\sqrt{p}}\sum_{j=0}^{p-1}\sum_{k=0}^{p-1}\omega^{jk}|j\...
4
Copying over from "Are circuits with more than 1000 gates common?". Note that a Toffoli gate is roughly as expensive as 2T gates or 4T gates, depending on your architecture. According to Table III of https://arxiv.org/abs/2011.03494 , quantum chemistry algorithms looking at properties of the FeMoCo molecule use half a billion Toffoli gates. ...
4
No, it's not possible. For example, being able to directly measure $X+Y$ would allow you to prepare T states and thereby perform T gates, which are not stabilizer operations. If the fact that $X$ and $Y$ don't commute bothers you, then note that preparing $|+\rangle^{\otimes n}$ and then measuring $\sum_{k=0}^{n-1} Z_k$ and getting a result of $n-2$ would ...
4
TL/DR The dimension of the stabilizer $\mathcal{S}$ is $2^{n}$ because there are exactly $n$ generators to make the dimension of the stabilizer's eigenspace exactly $1$. Then, each element in $\mathcal{S}$ is a unique combination of the generators for a total of $2^{n}$ combinations. Long version It's easy to show that the stabilizer $\mathcal{S}$ forms a ...
3
One way that you might do it is as follows. Consider the unitary sequence $V=HTHT$. Because it's unitary, we can write it in the form $$V=e^{i\gamma}e^{-i\theta\vec{n}\cdot\vec{\sigma}}=e^{i\gamma}(\cos\theta I-i\sin\theta \ \vec{n}\cdot\vec{\sigma})$$ where $\vec{n}\cdot\vec{n}=1$. If you work through the details, you'll find that \cos\theta=\cos^2\...
3
You might be interested in controlled version of $-I$. Despite the fact that you can neglect global phase in case of non-controlled gates, you cannot do so in case of controlled version. The controled gate $-I$ is described by matrix \begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & -1 & 0\\ 0 & 0 & 0 & -...
Only top voted, non community-wiki answers of a minimum length are eligible | 2022-01-24 03:50:03 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.940466046333313, "perplexity": 483.0840830370039}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320304471.99/warc/CC-MAIN-20220124023407-20220124053407-00689.warc.gz"} |
http://mathhelpforum.com/differential-geometry/197190-path-connected-space-print.html | # Path connected space
• Apr 12th 2012, 09:15 PM
monster
Path connected space
I'm starting to learn about path connected spaces and am a bit sketchy, any help would be appreciated, cheers,
Lets say i Have the set S = {0,1} with topology { emptyset, {0}, S }
and i need to prove S is path connected and therefore connected,
to show path connectedness,
i need to show a path exists from x to y for any x,y in S,
to show this i need a cotinuous map p : [0,1] -> S such that p(0) = x and p(1) = y
im confused how to do this for this set,
• Apr 13th 2012, 08:31 AM
HallsofIvy
Re: Path connected space
Let f(x)= 0 for $0\le x< 1/2$ and f(x)= 1 for $1/2 \le x\le 1$.
Show that this function is continuous in this topology.
• Apr 16th 2012, 02:30 AM
monster
Re: Path connected space
this is the problem i'm having, do i show this is continuous using epsilon delta? or is that not appropriate here and i should be using defn of continuity where the pre-image of the open sets are open? | 2016-10-28 22:23:38 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 2, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8856179118156433, "perplexity": 967.7247301694922}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-44/segments/1476988725475.41/warc/CC-MAIN-20161020183845-00399-ip-10-171-6-4.ec2.internal.warc.gz"} |
https://support.bioconductor.org/p/9140659/ | Aggregating genes from multiple GSE's, ranking them, then building a PPI network.
1
1
Entering edit mode
Caitlin ▴ 50
@caitlin-5400
Last seen 14 months ago
United States
Hi all.
After a lengthy struggle (I am a beginning student), I was able extract a list of genes from one GSE which contained 25 GSMs. From this GSE, I extracted a list of gene symbols after which I removed all of the duplicates. So, I am now assuming that the gene symbols I obtained are from all of the GSMs within that GSE? If so, I am attempting to perform the same operation on six additional GSEs before removing duplicates and using the RobustRankAggreg package, aggregating all of the genes into one data object (vector?) before performing a differential expression analysis with the limma package to isolate a smaller set of DEGs that I could use with Cytoscape via RCy3. Unfortunately, I don't know how to retrieve genes from multiple GSE objects or how one would combine all of the gene symbols. Can limma accept one monolithic data structure of gene symbols and could a boxplot be constructed from it even though there would presumably need to be two categories? Please accept my apologies if any of the above text is incorrect. I am working on my own without any form of guidance from my PI, TA, or classmates. I am, however, finding the limma course from DataCamp to be quite useful.
Code should be placed in three backticks as shown below
library(GEOquery)
library(limma)
gse <- getGEO("GSE12657", AnnotGPL = TRUE)
features <- fData(gse[[1]])
View(features)
length(features$Gene symbol) # Remove duplicates leaving 9156 unique genes features <- features[!duplicated(features$Gene symbol),]
# include your problematic code here with any corresponding output
# please also include the results of running the following in an R session
sessionInfo( )
GEOquery CancerData limma RCy3 CancerInSilico • 548 views
1
Entering edit mode
@james-w-macdonald-5106
Last seen 7 minutes ago
United States
There is one problem with what you are doing. Note how duplicated works
> d.f <- data.frame(count = sample(1:4, 10, TRUE), letters = letters[1:10])
> d.f
count letters
1 4 a
2 4 b
3 2 c
4 4 d
5 2 e
6 3 f
7 1 g
8 1 h
9 3 i
10 1 j
> d.f[!duplicated(d.f[,1]),]
count letters
1 4 a
3 2 c
6 3 f
7 1 g
See how it just takes the first of the duplicated rows? That may or may not be what you want. The duplicates on an Affymetrix array can be there for many different reasons and you may be inadvertently choosing the 'wrong' one. An alternative would be to average the replicates.
> library(GEOquery)
> library(affycoretools)
> library(hgu95av2.db)
> eset <- getGEO("GSE12657", AnnotGPL = TRUE)[[1]]
Found 1 file(s)
GSE12657_series_matrix.txt.gz
Using locally cached version: C:\Users\Public\Documents\Wondershare\CreatorTemp\Rtmpu2M04l/GSE12657_series_matrix.txt.gz
Using locally cached version of GPL8300 found here:
C:\Users\Public\Documents\Wondershare\CreatorTemp\Rtmpu2M04l/GPL8300.annot.gz
> eset <- annotateEset(eset, hgu95av2.db)
'select()' returned 1:many mapping between keys and columns
'select()' returned 1:many mapping between keys and columns
'select()' returned 1:many mapping between keys and columns
## Any NA NCBI Gene IDs?
> sum(is.na(fData(eset)$ENTREZID)) [1] 494 ## remove the NA values > eset <- eset[!is.na(fData(eset)$ENTREZID),]
> dim(eset)
Features Samples
12131 25
## make a new ExpressionSet with just the unique
> eset2 <- eset[!duplicated(fData(eset)$ENTREZID),] > dim(eset2) Features Samples 9055 25 > z <- avereps(exprs(eset), fData(eset)$ENTREZID)
## now put that into our new ExpressionSet
> row.names(z) <- row.names(eset2)
> exprs(eset2) <- z
And then the eset2 object has no duplicate NCBI Gene IDs (which are more reliable than HGNC symbols, which you shouldn't use for unique identifiability).
As for your questions about 'monolithic data structure' and a boxplot, I have no idea what you mean by the former, and the latter is simple
boxplot(log2(exprs(eset2)))
Although do note
> table(pData(eset2)$characteristics_ch1) Control Glioblastoma Oligodendroglioma 5 7 7 Pilocytic astrocytoma 6 ## compare each cancer to controls > fit <- eBayes(lmFit(eset2, model.matrix(~characteristics_ch1, pData(eset2)))) > sapply(1:3, function(x) nrow(topTable(fit, x, Inf, p.value = 0.05))) [1] 7372 1637 903 So there are 7372, 1637, and 903 genes differentially expressed in glioblastoma, oligodendroglioma and pilocytic astrocytoma tissue, respectively, as compared to control, at an FDR < 0.05 ADD COMMENT 0 Entering edit mode Hi James. Very nice. Thank you. I was just looking at an email from that postdoc I consulted several weeks ago and this is what she said. Could you please explain as I am quite confused, working under an encroaching deadline, and without benefit the classmate or PI: pData(gse) returns to the data frame of sample information only. A bit more idea of how gse ExpressionSet object looks like...you could imagine it is a multi-tab excel table. For example, when we extract the "sampledata <- pData(gse[[1]])", it is one of the tabs containing the pData data frame. For the downstream, you should look for "exprs(gse)" for the expression data, in which the rows are the genes and the columns are the sample IDs. Just do the differential expression analysis using the limma package instead of DESeq2. Now, you have the gse object which contains the assayData (expression data), phenoData (sample information), and featureData (gene annotation data). 1. Proceed to check if you need to normalize the expression data 2. Check the sample information, which column should be used as the "group"? 3. Proceed to the differential expression analysis using the limma package, you should get the "TopTable" results which include the gene IDs as the rows and the DE result as the column, we are interested in the columns of "log2FoldChange" and "adjusted p value". And then, you have to follow the instructions from clusterprofiler book. By using "log2FoldChange" and "adjusted p value", we can form a genelist object that is used in clusterprofiler package for the GO and KEGG enrichment analysis, as well as visualizing the results in the network figures. I think the first step involves calling log2 on the ExpressionSet object? The second step is puzzling, since upon using the following code: sampleData <- pData(gse[[1]]), I don't know what column I should use as the "group" nor do I know what that means (GSE12657) although limma apparently depends upon the specification of one column name? Could you please explain 3? The postdoc appears to believe this step is obvious but it most certainly is not. From the paper I am seeking to reproduce, the authors mentioned "hub genes" and the cytohubba package from Cytoscape. Would I need to transform the data at some point with the WGCNA package? I spent the past several hours reading the following document and my code, thus far, is below. https://github.com/Lindseynicer/How-to-analyze-GEO-microarray-data Update: library(GEOquery) library(limma) library(pheatmap) library(dplyr) library(ggplot2) library(ggrepel) gse <- getGEO("GSE12657", AnnotGPL = TRUE) gse # print the sample information pData(gse[[1]])[1:2,] # print the gene annotation fData(gse[[1]])[1,] # print the expression data exprs(gse[[1]])[1,] # Check the normalisation and scales used pData(gse[[1]])$data_processing[1]
# Look at the expression value
summary(exprs(gse[[1]]))
# Perform a log2 transformation since the range (min and max) is not between 0 and 16
exprs(gse[[1]]) <- log2(exprs(gse[[1]]))
# Reassess the summary
summary(exprs(gse[[1]]))
# Plot the summary values
boxplot(exprs(gse),outline=F)
# Retrieve sampleInfo
sampleInfo <- pData(gse[[1]])
# Obtain table counts for Control, Glioblastoma, Oligodendroglioma, and
# Pilocytic astrocytoma
table(sampleInfo$characteristics_ch1) # Select factors we need for the analysis sampleInfo <- select(sampleInfo, characteristics_ch1) # Rename the column to one that is more convenient sampleInfo <- rename(sampleInfo, sample = characteristics_ch1) library(pheatmap) # The argument use="c" stops an error if there are any missing data points # We can visualize the correlations between the samples by hierarchical clustering. # The function ‘cor’ can calculate the correlation on the scale of 0 to 1, in a pairwise fashion between all samples, then # visualise on a heatmap. There are many ways to create heatmaps in R, here I use ‘pheatmap’, the only argument it requires # is a matrix of numeric values. # We can add more sample info onto the plot to get a better pic of the group and clustering. Here, we make use of the # ‘sampleInfo’ file that was created earlier, to match with the columns of the correlation matrix. corMatrix <- cor(exprs(gse[[1]]),use="c") pheatmap(corMatrix) # This this case, rownames(sampleInfo) == colnames(sampleInfo), so the correlation is 1.0 # Construct a design matrix will be created, this is a matrix of 0 and 1, one row for each # sample and one column for each sample group. design <- model.matrix(~0 + sampleInfo$sample)
# Calculate the median expression level
threshold <- median(exprs(gse[[1]]))
#TRUE or FALSE for whether each gene is "expressed" in each sample
gene_is_expressed <- exprs(gse[[1]]) > threshold
# Identify genes that are expressed in more than 2 samples
retain <- rowSums(gene_is_expressed) > 3
# Check how many genes are removed / retained.
table(retain)
# Subset the data to keep only the expressed genes
**Error**
gse <- gse[retain,]
Error in orig[[nm]][i, , ..., drop = drop] :
(subscript) logical subscript too long
gse <- gse[retain,]
Thank you James for your help. I greatly appreciate it.
0
Entering edit mode
We have ventured far past the main purpose of this support site, which is to help people with specific questions about the software. While I sympathize with the fact that you are confused and have little time, you already have examples (the limma User's Guide, the GitHub repo you link to) that show you pretty much exactly what you should be doing. You just need to apply that information to your situation. | 2023-02-03 16:05:47 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.3753998577594757, "perplexity": 5021.655579957187}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764500058.1/warc/CC-MAIN-20230203154140-20230203184140-00323.warc.gz"} |
http://physics.stackexchange.com/tags/particle-physics/hot | # Tag Info
5
An obvious difference between the two ways of thinking about it you mention is that in the case of the Higgs mechanism, there is an observable particle excitation of the field associated with it, which was found recently. Furthermore it should be noted that the Higgs mechanism only concerns the mass generation of some elementary particles. The mass of ...
5
Anti-matter is created from interactions. They are created as a way to conserve energy and other quantum numbers such as charge. One such interaction would be the $Z^0$ decay: $Z^0 → ν_e + \bar{ν_e}$ In this case lepton number is conserved. They do annihilate producing energy (usually photons, but they can produce other particle-antiparticle pairs) when ...
4
A simple version of this is bremsstrahlung, i.e. an electron that decelerates and produces electromagnetic radiation / photons. By your reasoning the energy of the electron should only be able to go into other electrons: maybe it should radiate other electrons, maybe a single electron shouldn't lose energy as it travels. But the electron can transfer some ...
4
First, to be clear on what the graph is showing: as a function of the possible mass of the Higgs, it plots the fraction of Higgs bosons that will decay via each individual channel. Before we knew the mass of the Higgs boson, a plot like this one was useful for identifying the best channels to look at to detect the Higgs in various mass ranges. For example, ...
4
Since $\pi^0$ is a pseudoscalar particle, we have $$\langle 0|J^\mu_{em}|\pi^0 \rangle =0,$$ and the pion cannot decay into two leptons with a simple photon exchange. In the Standard Model, the leading-order contributions for this process are a box diagram and a $Z^0$ exchange, as you can see in fig. 1 of arXiv:0806.4782 (replacing a $c$ quark by a light ...
4
In simple terms QCD as a "background" usually refers to LHC research where hadronic jets create a lot of particles that clutter up the results you're trying to see. I think it has become a slang term and the use is discouraged. ABCD method is a tool used to separate the particles of interest (signal) from the "other stuff" (background) made by the jets. ...
4
It's a theoretical demand : $$\begin{pmatrix} \nu_{e}\\ \nu_{\mu}\\ \nu_{\tau} \end{pmatrix} = \begin{pmatrix} U_{e1} & U_{e2} & U_{e3} \\ U_{\mu1} & U_{\mu2} & U_{\mu3} \\ U_{\tau1} & U_{\tau2} & U_{\tau3} \end{pmatrix} \begin{pmatrix} \nu_{1}\\ \nu_{2}\\ \nu_{3} \end{pmatrix}$$ You know that all states ...
3
It really goes deeper than just a theoretical demand on a particular domain. The Hamiltonian for any system must be unitary, because that preserves the total probability at one. This is important because if I start with some state and let it evolve for a while the system must afterwards exist in some state which means that the sum of the probabilities taken ...
3
spin 1/2 fermions (electron, proton, neutron, muon, tau, quarks) have +1 parity (by convention as pointed out in Anna's comment). The corresponding anti-fermions have -1 parity. Bosons and their anti-particles have the same parity. See this and this lecture for more information on parity.
3
For this sort of task, it's easier to check through Wikipedia's list of baryons and list of mesons. Each article has a table listing the properties, including mass, of the known particles of the appropriate type, so you can just scan down the table and find the particle that matches your mass. In addition to mesons and baryons, in general, you would need to ...
3
How about this: In an $n,p,e$ gas the ratio of neutrons to protons decreases with density. For ideal degenerate gases the Fermi energies are related by $E_{F,n} = E_{F,p} + E_{F,e}$. In this situation, the largest the proton to neutron ratio can become is 1/8 when all the particles are ultra-relativistic at infinite density. To conserve momentum in the ...
2
Both statements are correct. Only left-handed electrons and left-handed neutrinos participate in weak interactions. The projection operators $$P_L = \frac{1}{2}(1-\gamma^5)\\ P_R = \frac{1}{2}(1+\gamma^5)\\$$ satisfy the relations $$P_L \gamma_\mu = \gamma_\mu P_R\\ P_LP_R=0\\ P_LP_L=P_L\\ P_L + P_R =1$$ From this it follows that $$j^\mu=\bar u_e ... 2 There are two SU(3) symmetries that are often discussed with regards to QCD. There is a gauge symmetry which corresponds to color charge which is mediated by the gluon and there is an approximate global flavor symmetry which acts on the flavors of the quarks (turns an up into down quark for example). All stable hadrons are color singlets and thus don't ... 2 For over forty years accelerator technology has been giving antiproton and positron beams, the easiest to create anti particles. Positrons are the simplest because once the energy of electrons is accelerated to the values over pair creation of e+e-, the the brehmstrahlung photons, gamma ray energies, will create electron positron pairs when interacting with ... 2 I'm not entirely sure what the answer to this question is. It's probably not friction heat alone as Parth Vader claims, since you can ignite coarse steel wool with the flame of a match, and yet the significantly finer "atomized" 100-mesh aluminum will not ignite under the same conditions ("atomized" refers to the manufacturing method of molten metal ... 2 Recall |E,l,m \rangle is the joint eigenvector of the Hamiltonian H, the total spin L^2, and a spin component (typically z) L_z, and the E, l and m label their respective eigenvalues. Notice all three of these operators act on the single particle that we're considering. Also recall \langle k | is an element in the momentum basis, also in the ... 2 Yes, e.g. all three Mandelstam variables$$ s~:=~(p_1+p_2)^2 ~\approx~ (m_1+m_2)^2 + \frac{m_1m_2}{2} ({\bf v}_1-{\bf v}_2)^2 ~>~0, t~:=~(p_1-p_3)^2~\approx~ (m_1-m_3)^2 - \frac{m_1m_3}{2} ({\bf v}_1-{\bf v}_3)^2 ~>~0, u~:=~(p_1-p_4)^2~\approx~ (m_1-m_4)^2 - \frac{m_1m_4}{2} ({\bf v}_1-{\bf v}_4)^2 ~>~0, are strictly positive in ...
2
The actual masses are accessible in theory, but not from mixing measurements. Cosmological measurements could give us a useable handle on the sum of the masses (though until we settle the hierarchy questions this may not provide a unique answer), or the combination of a much better model of supernovae plus a precision measurement of the differences in ...
1
http://www.quantumdiaries.org/2014/04/04/moriond-2014-new-results-new-explorations-but-no-new-physics/ Top quark mass has been revised upward.
1
I would guess, and it can only be a guess, that Stewart is referring to weak measurement. There is a rather vague description of this in New scientist. Annoyingly I can't track down the original paper, but if Stewart's book was written in 2013 the timing fits.
1
Electron doesn't 'get into' the shape of the orbital. An atomic orbital is a mathematical function that describes the wave-like behavior of either one electron or a pair of electrons in an atom. This function can be used to calculate the probability of finding any electron of an atom in any specific region around the atom's nucleus. The term may also refer ...
1
I will offer two reasons. First, unitarity of mixing matrices insures that probabilities sum to one. The probability of an oscillating neutrino having electron, muon or tau flavour should equal one. Second, because the neutrino mass matrix is Hermitian it is diagonalised by a unitary matrix.
1
In the Standard Model, fermion number is not conserved. Lepton number is conserved, because of an accidental symmetry. One cannot write down a renormalizable, gauge and Lorentz invariant operator that violates lepton number conservation in the Standard Model. A Majorana neutrino would violate lepton number conservation by two units. To see this, consider, ...
1
Yes, it's a misconception, or not - or both. What do you call "matter"? Let's call matter particles with a rest mass. So, everything that's made up of elementary particles is matter. Now here's the catch: To the best of our knowledge, elementary particles are pointlike, i.e. they really don't have any extend in space, they don't really "occupy" any space. ...
1
I don't know what a "pile" of fuel is. I assume you mean a container full of it. Gasoline needs oxygen to burn, and it needs the correct mixture. Too little oxygen and burning is impossible. Too much oxygen causes the same problem. To achieve ignition with Gasoline, you need between 1.4 and 7.6% petrol vapour (by volume) in the air. Ouside this range ...
1
I think K-K is still of much research interest. The K-K theory, which unifies YM fields and gravity are there since a long time. However, these theories have some consistency problems. In the original K-K theory, it was assumed that the 5D metric functions do not depend on the 5-th coordinate. This was the main reason of inconsistency. If the 5th ...
1
I am not exactly sure which low energy cutoff you refer to; however, there is a low-energy cutoff for photons that I am aware of. Photons with energies on the order of $H_0\sim10^{-33}\text{eV}$ would be super-horizon modes. That is, their wavelengths would be on the order of the Hubble radius, $H_0^{-1}=14.6~Gly$. Larger than this would mean that the ...
1
I'll only address your first question here. To start off with a sidenote, I think the idea that mass is a fundamental property of a particle has been on shaky ground ever since Einstein showed the equivalence of mass and energy. I can hardly imagine it took very long for people to come to the conclusion that mass cannot be a fundamental property of ...
1
This isn't a complete answer to your question, just two comments on two specific points that came up that were too long for a comment: WRT the Higgs having a spatially varying background because of inhomogeneities in the universe: It's important to keep in mind that the Higgs field is VERY MASSIVE when it has a nonzero VEV (as it does in the universe ...
1
Using $e^-e^-$ or $e^+e^+$ means that the final states need to be charged and have lepton number of two. This produces a different set of potential final products then $e^-e^+$. One such example would be $e^- e ^- \rightarrow \mu ^-\mu^-$. While this may be an interesting collision for some new theory, such interactions can only produce a very particular ...
Only top voted, non community-wiki answers of a minimum length are eligible | 2014-04-18 05:57:34 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6664755344390869, "perplexity": 434.9002978721099}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-15/segments/1397609532573.41/warc/CC-MAIN-20140416005212-00595-ip-10-147-4-33.ec2.internal.warc.gz"} |
http://joshkos.blogspot.com/2007/01/cyy.html | $\newcommand{\defeq}{\mathrel{\mathop:}=}$
## 2007/01/03
### cyy 寄信來
It just occurs to me that you can probably read chap 4 of CSAPP book if you want to have more understanding on how to design a more modern CPU. I have put a copy of this chapter at
http://...
It could be a little deviated from TOY's design, but could give you more insights on CPU design.
Yung-Yu
--
yen31/03/2007 2:17 pm 說： | 2018-03-22 00:20:09 | {"extraction_info": {"found_math": true, "script_math_tex": 1, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5145241022109985, "perplexity": 1356.075459668071}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-13/segments/1521257647707.33/warc/CC-MAIN-20180321234947-20180322014947-00360.warc.gz"} |
https://www.dzyla.com/blog/post37/ | # Colors of analog camera
##### August 2, 2022
Analog Kodak Rollei Python Image Processing Programming
While writing the last post Analog Diary III, I realized that I did not like photos order that was only dictated by the photo number from the scanner. The image presentation was not smooth enough, and I was looking for a way to sort images based on their luminosity and dominant color so the gallery would have more visual flow.
To do it, I wrote a short Python script that would help me with that task. First, the imports:
import glob
import os
import shutil
import imageio as io
import matplotlib.pyplot as plt
import numpy as np
from sklearnex import patch_sklearn
patch_sklearn()
from sklearn.cluster import KMeans
from skimage.transform import rescale
I used some general Python libraries for file discovery (glob) and system tools (os and shutil). Other libraries are more sophisticated: imageio for image reading and writing, matplotlib for plotting, NumPy for general array operations, and sklearn to perform k-means clustering. In addition, I use the Intel accelerated sklearn library to make the k-means calculations ~10x faster. The last import is the sklearn-image library which allows per-image calculations. Next, let’s define some helper variables and functions:
img_paths = glob.glob('/home/dzyla-photo/content/blog/post36/images/*.*')
n_clusters = 10
dominant_color = []
colormap = np.zeros((20, 200, 3))
img_paths will store all the paths of the images, n_clusters is a number of dominant colors to find in the picture, dominant_color is the list that will keep those colors, and colormap will be used to visualize the dominant colors. Let’s do some per image calculations:
for file in img_paths:
img = rescale(img, (0.3, 0.3, 1))
initial_shape = img.shape
img = img.reshape(-1, 3)
print(initial_shape)
y_labels = KMeans(n_clusters=n_clusters, random_state=0).fit_predict(img)
This code uses imageio to open the image file from the path, followed by the skimage library that rescales the image to 30% of the original size but keeps the RGB channel dimension unchanged. We keep the initial image dimension as a variable and reshape the image to get the array of all pixels in one dimension. Next, using the k-means clustering, all pixels are grouped into 10 classes. Each pixel will belong to one of 10 classes (from 0 to 9). The next task is a little bit more complicated. All pixels belong to one of 10 classes, but because many pixels in one class will have different values, we need to calculate the average color per class:
color_list = []
clustered_img = np.zeros(img.shape)
for cluster in range(0, n_clusters):
color = np.mean(img[y_labels == cluster], axis=0)
color_list.append(color)
clustered_img[y_labels == cluster] = color
Here, for each class, we calculate the average pixel value. We define an empty list that will hold calculated average colors and clustered_image, which will be an image with classified pixels substituted with calculated class colors. To do so, we select pixels that belong to the same class and average them. The result we add to the color_list. To calculate the clustered_img, the pixels corresponding to the given class are set to the average class color. Next, we will sort the obtained averaged colors per count:
color_label, color_frequency = np.unique(y_labels, return_counts=True)
sorting_order = np.argsort(color_frequency)
color_frequency = color_frequency[sorting_order]
color_list = np.array(color_list)[sorting_order]
From the y_labels class membership list, we calculate the unique values (that will be the class number from 0 to 9) together with the class count. This list will be ordered based on the values in the color_label array (0 -> 9). Thus, we need to sort the color_frequency array. np.argsort returns the array of indexes that will sort the original array, which is very useful in that case. We apply the sorting order to color_frequency and color_list arrays, resulting in two arrays ordered by the increasing population of the color in the picture. Now we have all elements ready to plot the colors according to their frequencies:
index = 0
color_map = np.zeros((20, initial_shape[1],3))
for n, color in enumerate(color_list):
fraction = int(initial_shape[1] * (color_frequency[n] / img.shape[0]))
if n < len(color_frequency) - 1:
color_map[:, index:index + fraction] = color
else:
color_map[:, index:] = color
dominant_color.append(color)
index = index + fraction
We define two new variables, the index will allow us to keep track of the array to plot colors, and color_map will be a bar plot of dominant colors in the picture. We iterate by sorting color_list, calculate the fraction of the image that belongs to the current class, and adjust it to the size of the source image. Due to the rounding error of each fraction, there will be some pixels that won’t be assigned to color; thus, we will use the most frequent color to fill missing pixels. For further calculations, we also keep the last color (the most frequent) in the dominant_color list. Finally, we move the index used to define the borders between colors. The above code will produce an image like this:
Let’s add the color distribution to the image that is reduced to 10 colors and plot it:
white_spacer = np.ones((5, initial_shape[1], 3))
final_img = np.concatenate((color_map, white_spacer, clustered_img.reshape(initial_shape)), axis=0)
plt.imshow(final_img)
plt.axis('off')
plt.tight_layout()
plt.savefig(os.path.basename(file).replace('.jpg', '_clustered.png').replace('JPG', '_clustered.png'),
plt.clf()
To separate the image from the color bar, we use a white spacer that is 5px high and image wide. Let’s combine all parts to get the final image by concatenating color_map, spacer, and image on the height axis. To generate two-dimensional images from a one-dimensional pixel array, we reshape this array back to its original shape. Plotting of the results is done in Matplotlib with disabled axes and a tight layout and, finally, save the image without white padding. The result for an example photo:
This ends the for loop and generates color-clustered images for all files in the folder. Using the information from the color distribution per image from the dominant_color list, it is possible to sort images based on their dominant color luminosity (from the great color tutorial from Alan Zucconi):
adjusted_sorted_colors = []
for color in dominant_color:
adjusted_sorted_colors.append(np.sqrt(.241 * color[0] + .691 * color[1] + .068 * color[2]))
color_map = colormap
idx = np.linspace(0, color_map.shape[1], len(dominant_color) + 1, dtype=int)
for n, color in enumerate(np.array(dominant_color)[img_order]):
color_map[:, idx[n]:idx[n + 1]] = color
Let’s create an empty list again to store the color luminosity calculated according to the equation. Let’s copy the color_map empty array and define equal spacing on the long dimension. Then we sort the adjusted_sorted_colors based on their luminosity value and assign this color to fractions of the color_map. This will generate the final sorted color array (top unsorted, bottom sorted):
The final step is to change the names of the photos according to the sorted dominant color luminosity:
os.makedirs('sorted_imgs', exist_ok=True)
for n, file in enumerate(img_paths):
shutil.copy2(file, './{}/{}_{}'.format('sorted_imgs', img_order[n], os.path.basename(file)))
Let’s create a folder called sorted_imgs and iterate through the image paths and, using shutil, copy the file to the sorted_imgs folder, changing the name by putting the consecutive number in from the file name. And that’s it! The result of this sorting is shown in the previous post Analog Diary III. The final and updated script can be found here:
cluster_by_color.py ⬇️
Ok, so we have a way to sort images based on their dominant color luminosity. While going through the analog photos, I noticed that different films give different vibes and colors. The idea was born: let’s generate film-specific dominant color lists and see if there is something cool there!
During the past few years, I developed almost 40 films, including Kodak Portra, Kodak Gold, Kodak Ultramax, Ilford HP5, Agfa, Fujicolor, and Ektar. Of course, colors will depend on the subject, exposure, and many other factors but let’s check some most dominant color pallets per film. I will use part of the above script, but instead of calculating the dominant colors per image, I will load all images in the developed film folder and change them to a one-dimensional array of pixels. This massive array will be classified with K-means. Color maps from different films are presented below:
Kodak Portra 400 (Yashica Mat 124G, ~140 photos):
Ektar 100 (~80 photos):
Fujicolor Xtra 400 (~80 photos):
Kodak Ultramax 400 (~120 photos):
Ilford HP5 plus 400 (~115 photos):
Agfa 400 (~37 photos):
Fujifilm Superia 400 (~39 photos):
Fujicolor C200 (~80 photos):
Rollei Infrared 400 (~30 photos):
cluster_by_color_per_folder.py ⬇️
I must admit I like the colors obtained by clustering different films but can’t tell whether there is any color dependence on the film. Most likely, a considerably more sophisticated analysis would be required to fully see the palette difference between various films. Nevertheless, it was a fun project to do! More photos with their most dominant colors are below: | 2022-08-16 18:46:40 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.26229527592658997, "perplexity": 2534.394175950648}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-33/segments/1659882572515.15/warc/CC-MAIN-20220816181215-20220816211215-00041.warc.gz"} |
https://gigaom.com/2013/08/21/jolla-wraps-up-its-first-production-batch-but-remains-fuzzy-on-numbers/ | # Jolla wraps up its first production batch, but remains fuzzy on numbers
Jolla, the Finnish smartphone upstart that’s repackaging the old MeeGo operating system into something that can rival Android(s goog) and the iPhone(s aapl), has said its first production batch has been “fully booked by consumers and selected sales channels”. For some reason, though, it won’t specify how many handsets that batch includes.
“We are not disclosing the exact number of devices pre-booked but the Jolla view is that the size of the production batch for a mobile device vendor of this size is typically 50,000 units,” a spokeswoman said on Wednesday.
Jolla also claimed that it was holding back its numbers in the interests of “customer confidentiality”. This sounds silly if you only take into account pre-orders from individuals, but less so if you consider that the first batch would have been partly booked by Finnish operator DNA and, presumably, Chinese distributor D.Phone (which was announced as a partner early on but which went unnamed in today’s statement). It is quite possible that one or both of these customers are highly secretive about the volumes they buy in, and fear rivals may extrapolate sensitive information from Jolla’s generalized batch figures.
If Jolla is trying to say (in as coy a way as possible) that it’s done a run of 50,000 units or thereabouts, then that’s not actually a bad start, certainly if you compare it with the Ubuntu Edge – 5 days ago, Canonical’s nearly-over crowdfunding drive for the $695 device had seen 14,500 units pledged for. Jolla’s smartphone costs €399 ($535), so it’s cheaper, and unlike the Edge it will also get made.
Jolla said its preorders – the ones from consumers themselves, rather than partners – came in from 136 countries. The preorder limit actually got hit in mid-July, so another order from a sales channel partner probably got finalized since then, otherwise Jolla would have been trumpeting the full booking of its first production batch a month ago.
The first Jolla phone will ship by the end of this year, with Europeans getting first shot. DNA will be the first operator in the world to carry the 4.5-inch device. | 2021-05-13 12:10:21 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.289760559797287, "perplexity": 3713.7074144491403}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-21/segments/1620243989916.34/warc/CC-MAIN-20210513111525-20210513141525-00592.warc.gz"} |
https://www.mysciencework.com/publication/show/exponentially-small-splitting-separatrices-beyond-melnikov-analysis-rigorous-results-9d347e41 | # Exponentially small splitting of separatrices beyond Melnikov analysis: rigorous results
Authors
Type
Preprint
Publication Date
Jan 24, 2012
Submission Date
Jan 24, 2012
Identifiers
arXiv ID: 1201.5152
Source
arXiv
We study the problem of exponentially small splitting of separatrices of one degree of freedom classical Hamiltonian systems with a non-autonomous perturbation fast and periodic in time. We provide a result valid for general systems which are algebraic or trigonometric polynomials in the state variables. It consists on obtaining a rigorous proof of the asymptotic formula for the measure of the splitting. We obtain that the splitting has the asymptotic behavior $K \varepsilon^{\beta} \text{e}^{-a/\varepsilon}$, identifying the constants $K,\beta,a$ in terms of the system features. We consider several cases. In some cases, assuming the perturbation is small enough, the values of $K,\beta$ coincide with the classical Melnikov approach. We identify the limit size of the perturbation for which this theory holds true. However for the limit cases, which appear naturally both in averaging and bifurcation theories, we encounter that, generically, $K$ and $\beta$ are not well predicted by Melnikov theory. | 2018-08-16 21:48:38 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8284587860107422, "perplexity": 427.1813413513822}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-34/segments/1534221211185.57/warc/CC-MAIN-20180816211126-20180816231126-00598.warc.gz"} |
http://openstudy.com/updates/560a01e0e4b0a953837272f2 | ## anonymous one year ago Water Runs into a conical tank at the rate of 3 feet^3/min. the tank stands pointing down and has a height of 15ft and a base radius of 4ft. How fast is the water level rising when the water is 8 feet deep.
1. anonymous
|dw:1443496421968:dw|
2. anonymous
not sure where to go from there
3. Vocaloid
|dw:1443496845151:dw| first step is to calculate r' (the value I marked), do you know how to do that?
4. anonymous
yea i think i do its, is this right ? $\frac{ r }{ h } = \frac{ 4 }{ 15 }$ $r = \frac{ 4h }{ 15 }$
5. Vocaloid
almost, we want the radius when h = 8, so you can plug in h = 8 into your proportion
6. anonymous
32/15 = 2.1333
7. Vocaloid
uh, lets leave it as a fraction for now (32/15) now, we just figured out that r = 4h/15, we can substitute that into the volume equation
8. Vocaloid
V = (1/3)(pi)(r^2)(h) = (1/3)(pi)(4h/15)^2*(h) now differentiate that with respect to h
9. Vocaloid
*with respect to t
10. anonymous
okay so $V = \frac{ 1 }{ 3 } \pi (\frac{ 32 }{ 15 }) ^2 h$
11. Vocaloid
actually, forget 32/15, we don't really need that (my mistake)
12. Vocaloid
|dw:1443497509872:dw|
13. anonymous
wait so do we treat the r in the function as a constant = to 32/15 ?
14. anonymous
ah i see okay
15. Vocaloid
right, we re-wrote r in terms of h, so we've already accounted for r (since r changes proportionally to h)
16. anonymous
so dv/dt = 2pi/3 (4h/15) h dh/dt ?
17. anonymous
is that right?
18. Vocaloid
not quite
19. Vocaloid
|dw:1443497761147:dw|
20. Vocaloid
then plug in h = 8 and dv/dt = 3
21. anonymous
wait why is h cubed, and also i thought it was 4h/15 not 4/15
22. Vocaloid
|dw:1443498112065:dw|
23. Vocaloid
24. anonymous
:P thats useful
25. anonymous
okay and where did the 1/3 go?
26. anonymous
okay nvm i see
27. anonymous
the power rule for the h^3 canceled it out
28. Vocaloid
29. Vocaloid
yeah that
30. anonymous
yep and the answer is .2098
31. anonymous
okay thank you very much!!!
32. Vocaloid
would be best to write it as a fraction in terms of pi, but yeah | 2017-01-23 02:55:37 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8343448638916016, "perplexity": 3779.5166435577776}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-04/segments/1484560281746.82/warc/CC-MAIN-20170116095121-00207-ip-10-171-10-70.ec2.internal.warc.gz"} |
https://vrcacademy.com/tutorials/laplace-distribution-examples/ | ## Laplace Distribution
A continuous random variable $X$ is said to have a Laplace distribution, if its p.d.f. is given by $$\begin{equation*} f(x;\mu, \lambda)=\left\{ \begin{array}{ll} \frac{1}{2\lambda}e^{-\frac{|x-\mu|}{\lambda}}, & \hbox{-\infty < x< \infty;} \\ & \hbox{-\infty < \mu < \infty, \lambda >0;} \\ 0, & \hbox{Otherwise.} \end{array} \right. \end{equation*}$$
## Distribution Function
Distribution function of $L(\mu,\lambda)$ distribution is $$\begin{equation*} F(x) = \left\{ \begin{array}{ll} \frac{1}{2}e^{\frac{(x-\mu)}{\lambda}}, & \hbox{x< \mu;} \\ &\\ 1-\frac{1}{2}e^{-\frac{(x-\mu)}{\lambda} }, & \hbox{x\geq \mu;} \end{array} \right. \end{equation*}$$
## Mean of Laplace Distribution
The mean of Laplace distribution is $E(X) = \mu$.
## Variance of Laplace Distribution
The variance of Laplace distribution is $V(X) = 2\lambda^2$.
## Example
A random variable $X$ follows a Laplace distribution with parameter $\mu =5$ and $\lambda=2$.
Find the probability that
a. $X$ is less than 1,
b. $X$ is less than 6,
c. $X$ is between 6 and 10,
d. $X$ is greater than 3.5.
### Solution
If $X\sim L(\mu,\lambda)$ then the distribution function of $X$ is
$F(x) =\dfrac{1}{2}e^{\dfrac{(x-\mu)}{\lambda}}$ for $x< \mu$
and
$F(x)=1-\dfrac{1}{2}e^{-\dfrac{(x-\mu)}{\lambda}}$ for $x\geq \mu$.
a. The probability that $X$ is less than $1$ is
\begin{aligned} P(X \leq 1) &=F(1)\\ &=\frac{1}{2}e^{\dfrac{(1-5)}{2}}\\ &\qquad (\because 1< mu)\\ &=\frac{1}{2}e^{\dfrac{(1-5)}{2}}\\ &= 0.0677 \end{aligned}
b. The probability that $X$ is less than $6$ is
\begin{aligned} P(X \leq 6) &=F(6)\\ &=1-\frac{1}{2}e^{\dfrac{-(6-5)}{2}}\\ &\qquad (\because 6> mu)\\ &=1-\frac{1}{2}e^{\dfrac{-(6-5)}{2}}\\ &= 1-0.3033\\ &= 0.6967 \end{aligned}
c. The probability that $X$ is between $6$ and $10$ is
\begin{aligned} P(6 \leq X \leq 10)&=P(X\leq 10)-P(X\leq 6)\\ &=F(10) -F(6)\\ &=\bigg(1-\frac{1}{2}e^{\dfrac{-(10-5)}{2}}\bigg)-\bigg(1-\frac{1}{2}e^{\dfrac{-(6-5)}{2}}\bigg)\\ &=\frac{1}{2}e^{\dfrac{-(6-5)}{2}}-\frac{1}{2}e^{\dfrac{-(10-5)}{2}}\\ &= 0.3033-0.041\\ &=0.2623 \end{aligned}
d. The probability that $X$ is greater than $3.5$ is
\begin{aligned} P(X > 3.5) &=1-P(X< 3.5)\\ &=1-F(3.5)\\ &=1-\frac{1}{2}e^{\dfrac{(3.5-5)}{2}}\\ &\qquad (\because 3.5< mu)\\ &= 1-0.2362\\ &= 0.7638 \end{aligned} | 2020-08-07 12:57:52 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 1.0000094175338745, "perplexity": 3634.642820376401}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-34/segments/1596439737178.6/warc/CC-MAIN-20200807113613-20200807143613-00513.warc.gz"} |
https://www.hepdata.net/record/79870 | The ALICE Transition Radiation Detector: construction, operation, and performance
The collaboration
Nucl.Instrum.Meth.A 881 (2018) 88-127, 2018.
Abstract (data abstract)
CERN-LHC. The Transition Radiation Detector (TRD) was designed and built to enhance the capabilities of the ALICE detector at the Large Hadron Collider (LHC). While aimed at providing electron identification and triggering, the TRD also contributes significantly to the track reconstruction and calibration in the central barrel of ALICE. In this paper the design, construction, operation, and performance of this detector are discussed. A pion rejection factor of up to 410 is achieved at a momentum of 1 GeV/$c$ in p-Pb collisions and the resolution at high transverse momentum improves by about 40% when including the TRD information in track reconstruction. The triggering capability is demonstrated both for jet, light nuclei, and electron selection. | 2021-10-17 18:06:23 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7626857757568359, "perplexity": 2799.8807392631984}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-43/segments/1634323585181.6/warc/CC-MAIN-20211017175237-20211017205237-00107.warc.gz"} |
http://www.oalib.com/relative/3551933 | Home OALib Journal OALib PrePrints Submit Ranking News My Lib FAQ About Us Follow Us+
Title Keywords Abstract Author All
Search Results: 1 - 10 of 100 matches for " "
Page 1 /100 Display every page 5 10 20 Item
Revista mexicana de astronomía y astrofísica , 2009, Abstract: We discuss the model of magnetic eld reconnection in the presence of turbulence introduced by us ten years ago. The model does not require any plasma e ects to be involved in order to make the reconnection fast. In fact, it shows that the degree of magnetic eld stochasticity controls the reconnection. The turbulence in the model is assumed to be sub-Alfv nic, with the magnetic eld only slightly perturbed. This ensures that the reconnection happens in generic astrophysical environments and the model does not appeal to any unphysical concepts, similar to the turbulent magnetic di usivity concept, which is employed in the kinematic magnetic dynamo. The interest to that model has recently increased due to successful numerical testings of the model predictions. In view of this, we discuss implications of the model, including the rst-order Fermi acceleration of cosmic rays, that the model naturally entails, bursts of reconnection, that can be associated with Solar ares, as well as, removal of magnetic ux during star-formation.
Physics , 2010, Abstract: Magnetic field embedded in a perfectly conducting fluid preserves its topology for all time. Although ionized astrophysical objects, like stars and galactic disks, are almost perfectly conducting, they show indications of changes in topology, `magnetic reconnection', on dynamical time scales. Reconnection can be observed directly in the solar corona, but can also be inferred from the existence of large scale dynamo activity inside stellar interiors. Solar flares and gamma ray busts are usually associated with magnetic reconnection. Previous work has concentrated on showing how reconnection can be rapid in plasmas with very small collision rates. Here we present numerical evidence, based on three dimensional simulations, that reconnection in a turbulent fluid occurs at a speed comparable to the rms velocity of the turbulence, regardless of the value of the resistivity. In particular, this is true for turbulent pressures much weaker than the magnetic field pressure so that the magnetic field lines are only slightly bent by the turbulence. These results are consistent with the proposal by Lazarian and Vishniac (1999) that reconnection is controlled by the stochastic diffusion of magnetic field lines, which produces a broad outflow of plasma from the reconnection zone. This work implies that reconnection in a turbulent fluid typically takes place in approximately a single eddy turnover time, with broad implications for dynamo activity and particle acceleration throughout the universe. In contrast, the reconnection in 2D configurations in the presence of turbulence depends on resistivity, i.e. is slow.
Physics , 2001, DOI: 10.1086/321628 Abstract: We examine the dynamics of turbulent reconnection in 2D and 3D reduced MHD by calculating the effective dissipation due to coupling between small-scale fluctuations and large-scale magnetic fields. Sweet-Parker type balance relations are then used to calculate the global reconnection rate. Two approaches are employed -- quasi-linear closure and an eddy-damped fluid model. Results indicate that despite the presence of turbulence, the reconnection rate remains inversely proportional to $\sqrt{R_m}$, as in the Sweet-Parker analysis. In 2D, the global reconnection rate is shown to be enhanced over the Sweet-Parker result by a factor of magnetic Mach number. These results are the consequences of the constraint imposed on the global reconnection rate by the requirement of mean square magnetic potential balance. The incompatibility of turbulent fluid-magnetic energy equipartition and stationarity of mean square magnetic potential is demonstrated.
Gregory L. Eyink Physics , 2014, DOI: 10.1088/0004-637X/807/2/137 Abstract: Plasma flows with an MHD-like turbulent inertial range, such as the solar wind, require a generalization of General Magnetic Reconnection (GMR) theory. We introduce the slip-velocity source vector, which gives the rate of development of slip velocity per unit arc length of field line. The slip source vector is the ratio of the curl of the non ideal electric field in the Generalized Ohm's Law and the magnetic field strength. It diverges at magnetic nulls, unifying GMR with magnetic null-point reconnection. Only under restrictive assumptions is the slip velocity related to the gradient of the quasi potential (integral of parallel electric field along field lines). In a turbulent inertial range the curl becomes extremely large while the parallel component is tiny, so that line slippage occurs even while ideal MHD becomes accurate. The resolution of this paradox is that ideal MHD is valid for a turbulent inertial-range only in a weak sense which does not imply magnetic line freezing. The notion of weak solution is explained in terms of spatial coarse-graining and renormalization group (RG) theory. We give an argument for the weak validity of the ideal Ohm's law in the inertial range, via rigorous estimates of the terms in the Generalized Ohm's Law for an electron-ion plasma. All of the nonideal terms (from collisional resistivity, Hall field, electron pressure anisotropy, and electron inertia) are shown to be irrelevant in the RG sense and large-scale reconnection is thus governed solely by ideal dynamics. We briefly discuss some implications for heliospheric reconnection, in particular for deviations from the Parker spiral model of interplanetary magnetic field. Solar wind observations show that reconnection in a turbulence broadened heliospheric current sheet, consistent with the Lazarian-Vishniac theory, leads to slip velocities that cause field lines to lag relative to the spiral model.
Physics , 2013, DOI: 10.1103/PhysRevLett.110.255001 Abstract: We report simulation results for turbulent magnetic reconnection obtained using a newly developed Reynolds-averaged magnetohydrodynamics model. We find that the initial Harris current sheet develops in three ways, depending on the strength of turbulence: laminar reconnection, turbulent reconnection, and turbulent diffusion. The turbulent reconnection explosively converts the magnetic field energy into both kinetic and thermal energy of plasmas, and generates open fast reconnection jets. This fast turbulent reconnection is achieved by the localization of turbulent diffusion. Additionally, localized structure forms through the interaction of the mean field and turbulence.
Physics , 2015, Abstract: Realistic astrophysical environments are turbulent due to the extremely high Reynolds numbers. Therefore, the theories of reconnection intended for describing astrophysical reconnection should not ignore the effects of turbulence on magnetic reconnection. Turbulence is known to change the nature of many physical processes dramatically and in this review we claim that magnetic reconnection is not an exception. We stress that not only astrophysical turbulence is ubiquitous, but also magnetic reconnection itself induces turbulence. Thus turbulence must be accounted for in any realistic astrophysical reconnection setup. We argue that due to the similarities of MHD turbulence in relativistic and non-relativistic cases the theory of magnetic reconnection developed for the non-relativistic case can be extended to the relativistic case and we provide numerical simulations that support this conjecture. We also provide quantitative comparisons of the theoretical predictions and results of numerical experiments, including the situations when turbulent reconnection is self-driven, i.e. the turbulence in the system is generated by the reconnection process itself. We show how turbulent reconnection entails the violation of magnetic flux freezing, the conclusion that has really far reaching consequences for many realistically turbulent astrophysical environments. In addition, we consider observational testing of turbulent reconnection as well as numerous implications of the theory. The former includes the Sun and solar wind reconnection, while the latter include the process of reconnection diffusion induced by turbulent reconnection, the acceleration of energetic particles, bursts of turbulent reconnection related to black hole sources as well as gamma ray bursts. Finally, we explain why turbulent reconnection cannot be explained by turbulent resistivity or derived through the mean field approach.
Physics , 2014, Abstract: Magnetic reconnection is a fundamental process of magnetic field topology change. We analyze the connection of this process with turbulence which is ubiquitous in astrophysical environments. We show how Lazarian & Vishniac (1999) model of turbulent reconnection is connected to the experimentally proven concept of Richardson diffusion and discuss how turbulence violates the generally accepted notion of magnetic flux freezing. We note that in environments that are laminar initially turbulence can develop as a result of magnetic reconnection and this can result in flares of magnetic reconnection in magnetically dominated media. In particular, magnetic reconnection can initially develop through tearing, but the transition to the turbulent state is expected for astrophysical systems. We show that turbulent reconnection predictions corresponds to the Solar and solar wind data.
A. Lazarian Physics , 2011, Abstract: Fast reconnection of magnetic field in turbulent fluids allows magnetic field to change its topology and connections. As a result, the traditional concept of magnetic fields being frozen into the plasma is no longer applicable. The diffusion of plasmas and magnetic field is enabled by reconnection and therefore is termed "reconnection diffusion". We explore the consequences of reconnection diffusion for star formation. In the paper we explain the physics of reconnection diffusion both from macroscopic and microscopic points of view. We quantify the reconnection diffusion rate both for weak and strong MHD turbulence and address the problem of reconnection diffusion acting together with ambipolar diffusion. In addition, we provide a criterion for correctly representing the magnetic diffusivity in simulations of star formation. We show that the role of the plasma effects is limited to "breaking up lines" on small scales and does not affect the rate of reconnection diffusion. We address the existing observational results and demonstrate how reconnection diffusion can explain the puzzles presented by observations, in particular, the observed higher magnetization of cloud cores in comparison with the magnetization of envelopes. We also outline a possible set of observational tests of the reconnection diffusion concept and discuss how the application of the new concept changes our understanding of star formation and its numerical modeling. Finally, we outline the differences of the process of reconnection diffusion and the process of accumulation of matter along magnetic field lines that is frequently invoked to explain the results of numerical simulations
A. Lazarian Physics , 2011, Abstract: Turbulence is ubiquitous in astrophysical fluids. Therefore it is necessary to study magnetic reconnection in turbulent environments. The model of fast turbulent reconnection proposed in Lazarian & Vishniac 1999 has been successfully tested numerically and it suggests numerous astrophysical implications. Those include a radically new possibility of removing magnetic field from collapsing clouds which we termed "reconnection diffusion", acceleration of cosmic rays within shrinking filaments of reconnected magnetic fields, flares of reconnection, from solar flares to much stronger ones which can account for gamma ray bursts. In addition, the model reveals a very intimate relation between magnetic reconnection and properties of strong turbulence, explaining how turbulent eddies can transport heat in magnetized plasmas. This is a small fraction the astrophysical implications of the quantitative insight into the fundamental process of magnetic reconnection in turbulent media.
Physics , 2012, DOI: 10.1088/0004-637X/777/1/46 Abstract: For a molecular cloud clump to form stars some transport of magnetic flux is required from the denser, inner regions to the outer regions of the cloud, otherwise this can prevent the collapse. Fast magnetic reconnection which takes place in the presence of turbulence can induce a process of reconnection diffusion (RD). Extending earlier numerical studies of reconnection diffusion in cylindrical clouds, we consider more realistic clouds with spherical gravitational potentials and also account for the effects of the gas self-gravity. We demonstrate that within our setup RD is efficient. We have also identified the conditions under which RD becomes strong enough to make an initially subcritical cloud clump supercritical and induce its collapse. Our results indicate that the formation of a supercritical core is regulated by a complex interplay between gravity, self-gravity, the magnetic field strength and nearly transonic and trans-Alfv\'enic turbulence, confirming that RD is able to remove magnetic flux from collapsing clumps, but only a few of them become nearly critical or supercritical, sub-Alfv\'enic cores, which is consistent with the observations. Besides, we have found that the supercritical cores built up in our simulations develop a predominantly helical magnetic field geometry which is also consistent with observations. Finally, we have evaluated the effective values of the turbulent reconnection diffusion coefficient and found that they are much larger than the numerical diffusion, especially for initially trans-Alfv\'enic clouds, ensuring that the detected magnetic flux removal is due to to the action of the RD rather than to numerical diffusivity.
Page 1 /100 Display every page 5 10 20 Item | 2019-08-18 23:57:37 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6679492592811584, "perplexity": 1155.0853672336232}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-35/segments/1566027314353.10/warc/CC-MAIN-20190818231019-20190819013019-00089.warc.gz"} |
http://math.stackexchange.com/questions/181635/equivalency-of-percentage-formulas | # Equivalency of percentage formulas
I know 3 methods for calculating percentages,
one example, find 70% of 50:
1) 50/100 * 70
2) 70/(100/50)
3) 70/100 * 50
I do not undertand how this 3 methods can be equivalent, also conceptually i do not understand method number 3. Can someone explain or give an understandable proof?
-
#1: divide 50 into 100 parts, and take 70 parts out of it. #3: 70/100 = 70%, because "%" means per cent (100). You can rearrange all the terms so that #3 <-> #1 <-> #2 <-> #3. Different expressions give different meanings, and I think the meaning of #2 is very vague. – FrenzY DT. Aug 12 '12 at 12:24
1. and 3. are the same thing, by virtue of multiplication being commutative. – J. M. Aug 12 '12 at 12:47
As explained in Ed Gorcenski's answer, 3 is the most natural formulation, as it clearly shows the percentage. The others are algebraically equivalent. – Ross Millikan Aug 12 '12 at 14:35
The important thing to remember about percent is that it literally means "per hundred"; so 70% means 70 [things] per hundred [things]. This is equivalent to a ratio; so if you have 70 things out of 100 things, you have 70% = 70/100 = 0.7. Those are all equivalent ways of writing 70%. Notice that percent is a dimensionless quantity. The dimensionless nature comes from the fact that $$70\% = \frac{70 [\mathrm{things}]}{100 [\mathrm{things}]} = \frac{70}{100} = 0.7$$
Therefore, 70% of 100 [things] is exactly 70 [things], which is the expected answer when we multiply the dimensionless quantity 70% by the quantity of 100 [things]: $$70\% \cdot 100\ [\mathrm{things}] = \frac{70}{100}\cdot 100\ [\mathrm{things}] = 70\ [\mathrm{things}].$$
Now, what happens when you don't have an even hundred [things]? Say you only have 50 [things], but you still want to estimate what the percent is. Obviously, 70% of 50 [things] is less than 50; so you cannot just say 70% of 50 = 70/50. Instead, you have to scale it. Remember that percent is dimensionless, so we do exactly the same operation as before: $$70\% \cdot 50\ [\mathrm{things}] = \frac{70}{100}\cdot 50\ [\mathrm{things}] = 35\ [\mathrm{things}].$$
Now you should be able to see why your formulas (1) and (3) are equivalent. Remember that division a quantity is like multiplying by one over that quantity: $\frac{3}{x} = 3\cdot \frac{1}{x}$. Also remember that multiplication is commutative. So, we have the following result:
$$70\% \cdot 50\ [\mathrm{things}] = \frac{70}{100} \cdot 50\ [\mathrm{things}] = \frac{1}{100}\cdot 70 \cdot 50\ [\mathrm{things}] = \frac{1}{100}\cdot 50 \cdot 70\ [\mathrm{things}]$$
EDIT:
As far as your formula #2, this works through simple manipulation of order of operations:
$$70/(100/50) = \frac{70}{\frac{100}{50}} = \frac{70\cdot 50}{100} = 70\cdot \frac{50}{100}.$$
- | 2016-04-29 02:27:51 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8335257172584534, "perplexity": 807.2380555268819}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-18/segments/1461860110356.23/warc/CC-MAIN-20160428161510-00218-ip-10-239-7-51.ec2.internal.warc.gz"} |
https://intl.siyavula.com/read/science/grade-10/the-atom/04-the-atom-06 | We think you are located in United States. Is this correct?
# Electronic configuration
## 4.6 Electronic configuration (ESABE)
### The energy of electrons (ESABF)
The electrons of an atom all have the same charge and the same mass, but each electron has a different amount of energy. Electrons that have the lowest energy are found closest to the nucleus (where the attractive force of the positively charged nucleus is the greatest) and the electrons that have higher energy (and are able to overcome the attractive force of the nucleus) are found further away.
### Electron arrangement (ESABG)
We will start with a very simple view of the arrangement or configuration of electrons around an atom. This view simply states that electrons are arranged in energy levels (or shells) around the nucleus of an atom. These energy levels are numbered 1, 2, 3, etc. Electrons that are in the first energy level (energy level 1) are closest to the nucleus and will have the lowest energy. Electrons further away from the nucleus will have a higher energy.
In the following examples, the energy levels are shown as concentric circles around the central nucleus. The important thing to know for these diagrams is that the first energy level can hold 2 electrons, the second energy level can hold 8 electrons and the third energy level can hold 8 electrons.
1. Lithium
Lithium ($$\text{Li}$$) has an atomic number of 3, meaning that in a neutral atom, the number of electrons will also be 3. The first two electrons are found in the first energy level, while the third electron is found in the second energy level (Figure 4.7).
2. Fluorine
Fluorine ($$\text{F}$$) has an atomic number of 9, meaning that a neutral atom also has 9 electrons. The first 2 electrons are found in the first energy level, while the other 7 are found in the second energy level (Figure 4.8).
3. Neon
Neon ($$\text{Ne}$$) has an atomic number of 10, meaning that a neutral atom also has 10 electrons. The first 2 electrons are found in the first energy level and the last 8 are found in the second energy level. (Figure 4.9).
But the situation is slightly more complicated than this. Within each energy level, the electrons move in orbitals. An orbital defines the spaces or regions where electrons move.
Atomic orbital
An atomic orbital is the region in which an electron may be found around a single atom.
The first energy level contains only one s orbital, the second energy level contains one s orbital and three p orbitals and the third energy level contains one s orbital and three p orbitals (as well as five d orbitals). Within each energy level, the s orbital is at a lower energy than the p orbitals. This arrangement is shown in Figure 4.10.
Each block in Figure 4.10 is able to hold two electrons. This means that the s orbital can hold two electrons, while the p orbital can hold a total of six electrons, two in each of the three blocks.
This diagram also helps us when we are working out the electron configuration of an element. The electron configuration of an element is the arrangement of the electrons in the shells and subshells. There are a few guidelines for working out the electron configuration. These are:
• Each orbital can only hold two electrons. Electrons that occur together in an orbital are called an electron pair.
• An electron will always try to enter an orbital with the lowest possible energy.
• An electron will occupy an orbital on its own, rather than share an orbital with another electron. An electron would also rather occupy a lower energy orbital with another electron, before occupying a higher energy orbital. In other words, within one energy level, electrons will fill an s orbital before starting to fill p orbitals.
• The s subshell can hold 2 electrons
• The p subshell can hold 6 electrons
The way that the electrons are arranged in an atom is called its electron configuration.
When there are two electrons in an orbital, the electrons are called an electron pair. If the orbital only has one electron, this electron is said to be an unpaired electron. Electron pairs are shown with arrows pointing in opposite directions.
Electron configuration
Electron configuration is the arrangement of electrons in an atom, molecule or other physical structure.
### Aufbau diagrams (ESABH)
An element's electron configuration can be represented using Aufbau diagrams or energy level diagrams. An Aufbau diagram uses arrows to represent electrons. You can use the following steps to help you to draw an Aufbau diagram:
1. Determine the number of electrons that the atom has.
2. Fill the s orbital in the first energy level (the 1s orbital) with the first two electrons.
3. Fill the s orbital in the second energy level (the 2s orbital) with the second two electrons.
4. Put one electron in each of the three p orbitals in the second energy level (the 2p orbitals) and then if there are still electrons remaining, go back and place a second electron in each of the 2p orbitals to complete the electron pairs.
5. Carry on in this way through each of the successive energy levels until all the electrons have been drawn.
You can think of Aufbau diagrams as being similar to people getting on a bus or a train. People will first sit in empty seats with empty seats between them and the other people (unless they know the people and then they will sit next to them). This is the lowest energy. When all the seats are filled like this, any more people that get on will be forced to sit next to someone. This is higher in energy. As the bus or train fills even more the people have to stand to fit on. This is the highest energy.
Aufbau is the German word for “building up”. Scientists used this term since this is exactly what we are doing when we work out electron configuration, we are building up the atoms structure.
#### Hund's rule and Pauli's principle
Sometimes people refer to Hund's rule for electron configuration. This rule simply says that electrons would rather be in a subshell on their own than share a subshell. This is why when you are filling the subshells you put one electron in each subshell and then go back and fill the subshell, before moving onto the next energy level.
Pauli's exclusion principle simply states that electrons have a property known as spin and that two electrons in a subshell will not spin the same way. This is why we draw electrons as one arrow pointing up and one arrow pointing down.
### Spectroscopic electron configuration notation (ESABI)
A special type of notation is used to show an atom's electron configuration. The notation describes the energy levels, orbitals and the number of electrons in each. For example, the electron configuration of lithium is $$1\text{s}^{2}2\text{s}^{1}$$. The number and letter describe the energy level and orbital and the number above the orbital shows how many electrons are in that orbital.
Aufbau diagrams for the elements fluorine and argon are shown in Figure 4.11 and Figure 4.12 respectively. Using spectroscopic notation, the electron configuration of fluorine is $$1\text{s}^{2}2\text{s}^{2}2\text{p}^{5}$$ and the electron configuration of argon is $$1\text{s}^{2}2\text{s}^{2}2\text{p}^{5}3\text{s}^{2}3\text{p}^{6}$$.
The spectroscopic electron configuration can be written in shorter form. This form is written as [noble gas]electrons, where the noble gas is the nearest one that occurs before the element. For example, magnesium can be represented as $$[\text{Ne}]3\text{s}^{2}$$ and carbon as $$[\text{H}]2\text{s}^{2}2\text{p}^{2}$$. This is known as the condensed electron configuration.
## Worked example 3: Aufbau diagrams and spectroscopic electron configuration
Give the electron configuration for nitrogen (N) and draw an Aufbau diagram.
### Give the number of electrons
Nitrogen has seven electrons.
### Place two electrons in the 1s orbital
We start by placing two electrons in the 1s orbital: $$1\text{s}^{2}$$.
Now we have five electrons left to place in orbitals.
### Place two electrons in the 2s orbital
We put two electrons in the 2s orbital: $$2\text{s}^{2}$$.
There are now three electrons to place in orbitals.
### Place three electrons in the 2p orbital
We place three electrons in the 2p orbital: $$2\text{p}^{3}$$.
The electron configuration is: $$1\text{s}^{2}2\text{s}^{2}2\text{p}^{3}$$. The Aufbau diagram is given in the step above.
#### Aufbau diagrams for ions
When a neutral atom loses an electron it becomes positively charged and we call it a cation. For example, sodium will lose one electron and become $$\text{Na}^{+}$$ or calcium will lose two electrons and become $$\text{Ca}^{2+}$$. In each of these cases, the outermost electron(s) will be lost.
When a neutral atom gains an electron it becomes negatively charged and we call it an anion. For example chlorine will gain one electron and become $$\text{Cl}^{-}$$ or oxygen will gain two electrons and become $$\text{O}^{2-}$$.
Aufbau diagrams and electron configurations can be done for cations and anions as well. The following worked example will show you how.
## Worked example 4: Aufbau diagram for an ion
Give the electron configuration for ($$\text{O}^{2-}$$) and draw an Aufbau diagram.
### Give the number of electrons
Oxygen has eight electrons. The oxygen anion has gained two electrons and so the total number of electrons is ten.
### Place two electrons in the 1s orbital
We start by placing two electrons in the 1s orbital: $$1\text{s}^{2}$$.
Now we have eight electrons left to place in orbitals.
### Place two electrons in the 2s orbital
We put two electrons in the 2s orbital: $$2\text{s}^{2}$$.
There are now six electrons to place in orbitals.
### Place six electrons in the 2p orbital
We place six electrons in the 2p orbital: $$2\text{p}^{6}$$.
The electron configuration is: $$1\text{s}^{2}2\text{s}^{2}2\text{p}^{6}$$. The Aufbau diagram is:
When we draw the orbitals we draw a shape that has a boundary (i.e. a closed shape). This represents the distance from the nucleus in which we are $$\text{95}\%$$ sure that we will find the electrons. In reality the electrons of an atom could be found any distance away from the nucleus.
### Orbital shapes (ESABJ)
Each of the orbitals has a different shape. The s orbitals are spherical and the p orbitals are dumbbell shaped.
### Core and valence electrons (ESABK)
Electrons in the outermost energy level of an atom are called valence electrons. The electrons that are in the energy shells closer to the nucleus are called core electrons. Core electrons are all the electrons in an atom, excluding the valence electrons. An element that has its valence energy level full is more stable and less likely to react than other elements with a valence energy level that is not full.
Valence electrons
The electrons in the outermost energy level of an atom.
Core electrons
All the electrons in an atom, excluding the valence electrons.
# Test yourself now
High marks in science are the key to your success and future plans. Test yourself and learn more on Siyavula Practice.
## Core and valence electrons
Exercise 4.4
Complete the following table:
Element or Ion Electron configuration Core electrons Valence electrons Potassium ($$\text{K}$$) Helium ($$\text{He}$$) Oxygen ion ($$\text{O}^{2-}$$) Magnesium ion ($$\text{Mg}^{2+}$$)
Solution not yet available
### The importance of understanding electron configuration (ESABL)
By this stage, you may well be wondering why it is important for you to understand how electrons are arranged around the nucleus of an atom. Remember that during chemical reactions, when atoms come into contact with one another, it is the electrons of these atoms that will interact first. More specifically, it is the valence electrons of the atoms that will determine how they react with one another.
To take this a step further, an atom is at its most stable (and therefore unreactive) when all its orbitals are full. On the other hand, an atom is least stable (and therefore most reactive) when its valence electron orbitals are not full. This will make more sense when we go on to look at chemical bonding in a later chapter. To put it simply, the valence electrons are largely responsible for an element's chemical behaviour and elements that have the same number of valence electrons often have similar chemical properties.
The most stable configurations are the ones that have full energy levels. These configurations occur in the noble gases. The noble gases are very stable elements that do not react easily (if at all) with any other elements. This is due to the full energy levels. All elements would like to reach the most stable electron configurations, i.e. all elements want to be noble gases. This principle of stability is sometimes referred to as the octet rule. An octet is a set of 8, and the number of electrons in a full energy level is 8.
## Flame tests
### Aim
To determine what colour a metal cation will cause a flame to be.
### Apparatus
• Watch glass
• Bunsen burner
• methanol
• tooth picks (or skewer sticks)
• metal salts (e.g. $$\text{NaCl}$$, $$\text{CuCl}_{2}$$, $$\text{CaCl}_{2}$$, $$\text{KCl}$$, etc.)
• metal powders (e.g. copper, magnesium, zinc, iron, etc.)
Be careful when working with Bunsen burners as you can easily burn yourself. Make sure all scarves/loose clothing are securely tucked in and long hair is tied back. Ensure that you work in a well-ventilated space and that there is nothing flammable near the open flame.
### Method
For each salt or powder do the following:
1. Dip a clean tooth pick into the methanol
2. Dip the tooth pick into the salt or powder
3. Wave the tooth pick through the flame from the Bunsen burner. Do not hold the tooth pick in the flame, but rather wave it back and forth through the flame.
4. Observe what happens
### Results
Record your results in a table, listing the metal salt and the colour of the flame.
### Conclusion
You should have observed different colours for each of the metal salts and powders that you tested.
The above experiment on flame tests relates to the line emission spectra of the metals. These line emission spectra are a direct result of the arrangement of the electrons in metals. Each metal salt has a uniquely coloured flame.
# Test yourself now
High marks in science are the key to your success and future plans. Test yourself and learn more on Siyavula Practice.
## Energy diagrams and electrons
Exercise 4.5
Draw Aufbau diagrams to show the electron configuration of each of the following elements:
1. magnesium
2. potassium
3. sulfur
4. neon
5. nitrogen
Using the rules about drawing Aufbau diagrams we get the following.
1. Magnesium has 12 electrons. So we start with energy level 1 and put 2 electrons in it. Then we move to energy level two which has the 2s and the 2p orbitals. This level can hold 8 electrons. Adding up the total electrons used so far gives 10 electrons. So we must place two more electrons and these go into the third level.
2. Pottassium has 19 electrons. So we get:
3. Sulfur has 14 electrons so we get:
4. Neon has 10 electrons so we get:
5. Nitrogen has 7 electrons so we get:
Use the Aufbau diagrams you drew to help you complete the following table:
Element No. of energy levels No. of electrons No. of valence electrons Electron configuration (standard notation) $$\text{Mg}$$ $$\text{K}$$ $$\text{S}$$ $$\text{Ne}$$ $$\text{N}$$
The energy levels are given by the numeral above the boxes, so 1, 2, 3 etc. The number of core electrons is all the electrons that are not in the outermost energy level. The number of valence electrons is the number of electrons in the outermost energy level. And the electron configuration is simply listing the energy levels together with the orbitals and the number of electrons in each orbital. So for magnesium the number of energy levels is 3, the number of core electrons is 10, the number of valence electrons is 2 and the electron configuration is: $$1\text{s}^{2}2\text{s}^{2}2\text{p}^{6}3\text{s}^{2}$$. Filling in the table gives:
Element No. of energy levels No. of electrons No. of valence electrons Electron configuration (standard notation) $$\text{Mg}$$ $$\text{3}$$ $$\text{10}$$ $$\text{2}$$ $$1\text{s}^{2}2\text{s}^{2}2\text{p}^{6}3\text{s}^{2}$$ $$\text{K}$$ $$\text{4}$$ $$\text{18}$$ $$\text{1}$$ $$1\text{s}^{2}2\text{s}^{2}2\text{p}^{6}3\text{s}^{2}3\text{p}^{6}4\text{s}^{1}$$ $$\text{S}$$ $$\text{3}$$ $$\text{10}$$ $$\text{6}$$ $$1\text{s}^{2}2\text{s}^{2}2\text{p}^{6}3\text{s}^{2}3\text{p}^{4}$$ $$\text{Ne}$$ $$\text{2}$$ $$\text{2}$$ $$\text{8}$$ $$1\text{s}^{2}2\text{s}^{2}2\text{p}^{6}$$ $$\text{N}$$ $$\text{2}$$ $$\text{2}$$ $$\text{5}$$ $$1\text{s}^{2}2\text{s}^{2}2\text{p}^{3}$$
Rank the elements used above in order of increasing reactivity. Give reasons for the order you give.
Solution not yet available. | 2021-07-29 12:11:42 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5805684328079224, "perplexity": 658.3169382871705}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-31/segments/1627046153857.70/warc/CC-MAIN-20210729105515-20210729135515-00288.warc.gz"} |
https://nbviewer.jupyter.org/github/OpenAstronomy/2016-01-11_Sheffield_Notes/blob/master/03-Python2/03-python2-debugging-instructors.ipynb | # Introduction to Python 2¶
## Learning Objectives
• Debug code containing an error systematically.
Once testing has uncovered problems, the next step is to fix them. Many novices do this by making more-or-less random changes to their code until it seems to produce the right answer, but that’s very inefficient (and the result is usually only correct for the one case they’re testing). The more experienced a programmer is, the more systematically they debug, and most follow some variation on the rules explained below.
### Know What It’s Supposed to Do¶
The first step in debugging something is to know what it’s supposed to do. “My program doesn’t work” isn’t good enough: in order to diagnose and fix problems, we need to be able to tell correct output from incorrect. If we can write a test case for the failing case — i.e., if we can assert that with these inputs, the function should produce that result — then we’re ready to start debugging. If we can’t, then we need to figure out how we’re going to know when we’ve fixed things.
But writing test cases for scientific software is frequently harder than writing test cases for commercial applications, because if we knew what the output of the scientific code was supposed to be, we wouldn’t be running the software: we’d be writing up our results and moving on to the next program. In practice, scientists tend to do the following:
1. Test with simplified data. Before doing statistics on a real data set, we should try calculating statistics for a single record, for two identical records, for two records whose values are one step apart, or for some other case where we can calculate the right answer by hand.
2. Test a simplified case. If our program is supposed to simulate magnetic eddies in rapidly-rotating blobs of supercooled helium, our first test should be a blob of helium that isn’t rotating, and isn’t being subjected to any external electromagnetic fields. Similarly, if we’re looking at the effects of climate change on speciation, our first test should hold temperature, precipitation, and other factors constant.
3. Compare to an oracle. A test oracle is something whose results are trusted, such as experimental data, an older program, or a human expert. We use test oracles to determine if our new program produces the correct results. If we have a test oracle, we should store its output for particular cases so that we can compare it with our new results as often as we like without re-running that program.
4. Check conservation laws. Mass, energy, and other quantitites are conserved in physical systems, so they should be in programs as well. Similarly, if we are analyzing patient data, the number of records should either stay the same or decrease as we move from one analysis to the next (since we might throw away outliers or records with missing values). If “new” patients start appearing out of nowhere as we move through our pipeline, it’s probably a sign that something is wrong.
5. Visualize. Data analysts frequently use simple visualizations to check both the science they’re doing and the correctness of their code (just as we did in the opening lesson of this tutorial). This should only be used for debugging as a last resort, though, since it’s very hard to compare two visualizations automatically.
### Make It Fail Every Time¶
We can only debug something when it fails, so the second step is always to find a test case that makes it fail every time. The “every time” part is important because few things are more frustrating than debugging an intermittent problem: if we have to call a function a dozen times to get a single failure, the odds are good that we’ll scroll past the failure when it actually occurs.
As part of this, it’s always important to check that our code is “plugged in”, i.e., that we’re actually exercising the problem that we think we are. Every programmer has spent hours chasing a bug, only to realize that they were actually calling their code on the wrong data set or with the wrong configuration parameters, or are using the wrong version of the software entirely. Mistakes like these are particularly likely to happen when we’re tired, frustrated, and up against a deadline, which is one of the reasons late-night (or overnight) coding sessions are almost never worthwhile.
### Make It Fail Fast¶
If it takes 20 minutes for the bug to surface, we can only do three experiments an hour. That doesn’t must mean we’ll get less data in more time: we’re also more likely to be distracted by other things as we wait for our program to fail, which means the time we are spending on the problem is less focused. It’s therefore critical to make it fail fast.
As well as making the program fail fast in time, we want to make it fail fast in space, i.e., we want to localize the failure to the smallest possible region of code:
1. The smaller the gap between cause and effect, the easier the connection is to find. Many programmers therefore use a divide and conquer strategy to find bugs, i.e., if the output of a function is wrong, they check whether things are OK in the middle, then concentrate on either the first or second half, and so on.
2. N things can interact in $N^2$ different ways, so every line of code that isn’t run as part of a test means more than one thing we don’t need to worry about.
### Change One Thing at a Time, For a Reason¶
Replacing random chunks of code is unlikely to do much good. (After all, if you got it wrong the first time, you’ll probably get it wrong the second and third as well.) Good programmers therefore change one thing at a time, for a reason They are either trying to gather more information (“is the bug still there if we change the order of the loops?”) or test a fix (“can we make the bug go away by sorting our data before processing it?”).
Every time we make a change, however small, we should re-run our tests immediately, because the more things we change at once, the harder it is to know what’s responsible for what (those $N^2$ interactions again). And we should re-run all of our tests: more than half of fixes made to code introduce (or re-introduce) bugs, so re-running all of our tests tells us whether we have regressed.
### Keep Track of What You’ve Done¶
Good scientists keep track of what they’ve done so that they can reproduce their work, and so that they don’t waste time repeating the same experiments or running ones whose results won’t be interesting. Similarly, debugging works best when we keep track of what we’ve done and how well it worked. If we find ourselves asking, “Did left followed by right with an odd number of lines cause the crash? Or was it right followed by left? Or was I using an even number of lines?” then it’s time to step away from the computer, take a deep breath, and start working more systematically.
Records are particularly useful when the time comes to ask for help. People are more likely to listen to us when we can explain clearly what we did, and we’re better able to give them the information they need to be useful.
### Be Humble¶
And speaking of help: if we can’t find a bug in 10 minutes, we should be humble and ask for help. Just explaining the problem aloud is often useful, since hearing what we’re thinking helps us spot inconsistencies and hidden assumptions.
Asking for help also helps alleviate confirmation bias. If we have just spent an hour writing a complicated program, we want it to work, so we’re likely to keep telling ourselves why it should, rather than searching for the reason it doesn’t. People who aren’t emotionally invested in the code can be more objective, which is why they’re often able to spot the simple mistakes we have overlooked.
Part of being humble is learning from our mistakes. Programmers tend to get the same things wrong over and over: either they don’t understand the language and libraries they’re working with, or their model of how things work is wrong. In either case, taking note of why the error occurred and checking for it next time quickly turns into not making the mistake at all.
And that is what makes us most productive in the long run. As the saying goes, A week of hard work can sometimes save you an hour of thought. If we train ourselves to avoid making some kinds of mistakes, to break our code into modular, testable chunks, and to turn every assumption (or mistake) into an assertion, it will actually take us less time to produce working programs, not more.
## Debug with a neighbour
Take a function that you have written today, and introduce a tricky bug. Your function should still run, but will give the wrong output. Switch seats with your neighbor and attempt to debug the bug that they introduced into their function. Which of the principles discussed above did you find helpful? | 2020-10-19 21:02:30 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.49782368540763855, "perplexity": 680.6929845936539}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-45/segments/1603107866404.1/warc/CC-MAIN-20201019203523-20201019233523-00453.warc.gz"} |
https://www.rocketryforum.com/threads/a-break-from-computer-modeling.9368/ | # A break from computer modeling
### Help Support The Rocketry Forum:
#### Pem Tech
##### Well-Known Member
Just recently I noticed something about the rockets I had been designing and building. They seemed to be "tube and fin" rockets (Blasphemy in the Halls of Pem-Tech), and were becoming more so as time marched on.
A brief analysis showed that I had become too dependant on Rocksim, and had begun to design through Rocksim colored glasses.
Now, don't get me wrong, the program is fantastic and a very valuable tool. But what good is the tool if the imagination using it has been stifled?
Two weeks ago I sat down at my bench without any preconcieved notions and no Rocksim. The computer was there, I mean I can't live without my laptop.
*twitch*
*twitch*
I don't have a laptop problem...
I am in complete control.
:roll:
Yeah right.
Anyway....
Here are two of the rockets that came out of the Computer Modeling Free session.
First, "the Gorgon"
She consists of a Semroc BC-1043 nosecone (topped with three short sections of toothpick), a short length of Series 10 BT, a S-10 to BT-60 balsa transition and a short length of BT-60. The after section is a 1" length of BT-80 that supports the poster board wings at their midpoint.
The reverse cresent wings, connected to segments of Bt-5 at the tips, these segments are topped by Hartle resin NC's.
She has a single 24mm motor mount but I also built another version that has dual 18mm motor mounts.
So far she hasn't flown, but I hope to see her madien flight at Southern Thunder 2010.
The final finish will have gold highlights over the purple base coat. And, by the way, this was the second model I have painted with an airbrush. It took quite a few coats but I finally got it covered.
Here is the second product to come from my computer free designing.
Right now she doesn't have an official name but we call her "Code Name Kran".
Forgive the horrendous paintjob.....
My intent was to finsh it out in a Vorlon type of organic paintjob.
As you can tell, this was my first attempt to paint with an airbrush. The florescent green went on fine, but the black squiggles didn't come off as planned.
Structurally she is a BT-60 NC on a short section of BT-60 with a balsa transition to BT-50. A balsa transition back to a short section of BT-60 and capped off with a balsa boattail down to a 24mm motor mount. The 18" section pf BT-50 runs therought the airframe. Wings and fin are balsa with assorted BT-5 pods topped with Hartle resin nosecones.
Last edited:
#### Gillard
##### Well-Known Member
Nice, Rocksims good, imagination and a free hand can be better
#### dave carver
##### ....what hump?
Looks real good, maybe you need to step back from the computer more
Thingt about airbrushes is they are at their best doing detail and not for base coat painting. For that try a detail gun. It has a like 1 cup capacity and is smaller version for the quart can sprayer. I have seen them for under $20 from Harbour Freight. Oh, the Vorlon looking ship, like the plastic model shows, Vorlon ships change the base color over the surface several times. Other than that really nice rockets:cyclops: #### mjennings ##### Well-Known Member Nice Layne I really like the Gorgon! I have a few scratches I need to build, but I make a habit of at least having a paper sketch before firing up RockSim, that way if I run into a constraint from the program I don't change the idea, and wind up with something I didn't intend. (not that the unintended result is always bad just not necessary what you wanted) Can't wait to hear how they fly! #### Pem Tech ##### Well-Known Member Looks real good, maybe you need to step back from the computer more Thingt about airbrushes is they are at their best doing detail and not for base coat painting. For that try a detail gun. It has a like 1 cup capacity and is smaller version for the quart can sprayer. I have seen them for under$20 from Harbour Freight.
Oh, the Vorlon looking ship, like the plastic model shows, Vorlon ships change the base color over the surface several times.
Other than that really nice rockets:cyclops:
I may do that....
Thanks...
$20 for a spray gun??? Are they any good? Do they last? You know, you are right about the VOrlon ship, I hadn't noticed changing under coat. #### dave carver ##### ....what hump? I may do that.... Thanks...$20 for a spray gun???
Are they any good? Do they last?
You know, you are right about the VOrlon ship, I hadn't noticed changing under coat.
Mine's 10 years old and they are rebuildable. I would suggest an in-line pressure regulator, the kind that attaches to the gun, for final regulation of the air. Just hold the sprayer handel down for a second to drop the line pressuer down to where you want it and then spray the paint.
http://search.harborfreight.com/cpi...age=1&resultsPerPage=10&resultsPerPageBottom= is the page with the gun, the one called Touch-up Spray Gun at \$13.99.
I'd suggest getting Xylol to clean the gun, I get mine right from Ace Hardware.
One of the tricks for cleaning is when you spray the thinner through the gun to clean it wad up some cloth and loosely/firmly(hard to describe) hold the cloth over the spray vents and like spray into the cloth, point being to create a backflushing action. Cleaning these guns is how you get long life from them.
With controls for both air flow and paint flow you can acheave near air-brush fineness with spray gun volume
Last edited: | 2021-03-01 10:31:19 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.33972325921058655, "perplexity": 5831.000925394502}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-10/segments/1614178362481.49/warc/CC-MAIN-20210301090526-20210301120526-00522.warc.gz"} |
https://leanprover-community.github.io/archive/stream/116395-maths/topic/Generalized.20Smale.20conjecture.html | ## Stream: maths
### Topic: Generalized Smale conjecture
#### Patrick Massot (Sep 20 2019 at 08:25):
I'm not sure this stream can be used for general math news, but we have a new high point of mathematics we can dream to formalize in Lean 42.0 after we'll have a couple of thousands pages or preliminaries covered: https://arxiv.org/abs/1909.08710
#### Patrick Massot (Sep 20 2019 at 08:46):
Yes, this is part of the preliminaries
#### Sebastien Gouezel (Sep 20 2019 at 08:47):
I think your "a couple of thousands pages of preliminaries" is underestimated by a factor of 10 (at least :).
#### Kevin Buzzard (Sep 20 2019 at 10:25):
The computer scientists will like the deduction of Theorem 1.3 from Theorem 1.2 in the intro.
#### Scott Morrison (Sep 20 2019 at 10:27):
by well_known
#### Kevin Buzzard (Sep 20 2019 at 10:28):
Later on in the paper they assume all manifolds are 3-dimensional. This is well-known to be false, so it's not surprising that they can prove so much.
#### Patrick Massot (Sep 20 2019 at 11:16):
We already have well_known, it's called library_search
#### Scott Morrison (Sep 20 2019 at 11:23):
(I'm wishing that I called library_search instead help. I've got a soon-to-be-PR that adds a "try harder please" mode as library_search!, but help! would have been so much more fun.)
#### Sebastien Gouezel (Sep 20 2019 at 13:12):
In the paper, Theorem 1.2 says that some set is either empty or contractible. I don't want to start a flame war, but isn't the empty set contractible? (Of course, this depends on the precise definition of contractible you use, but I guess there is maybe a definition of contractible in @Reid Barton Lean homotopy library, and then the question makes sense).
no!
#### Reid Barton (Sep 20 2019 at 13:20):
At least in homotopy theory, I've only seen it used to mean "has the homotopy type of a point"
#### Reid Barton (Sep 20 2019 at 13:21):
it's our "unique"
#### Reid Barton (Sep 20 2019 at 13:21):
Well, unique + exists if you don't consider that to be part of "unique"
#### Sebastien Gouezel (Sep 20 2019 at 13:22):
Yes, I agree it's hard to argue for the contractibility of the empty space. Connectedness is a better question :)
#### Reid Barton (Sep 20 2019 at 13:26):
Here is another one I like. If $f : X \to Y$ and $X$ is empty, is $f$ constant?
#### Patrick Massot (Sep 20 2019 at 13:27):
I'd say yes unless Y is empty
:grinning:
My brain hurts
:smiling_devil:
#### Kevin Buzzard (Sep 20 2019 at 13:45):
I'd never seen that one before. There are two natural definitions of constant and Patrick's answer is what you get with one of them
#### Keeley Hoek (Sep 20 2019 at 14:41):
3) Image is a singleton :D, different answer again!
#### Antoine Chambert-Loir (Sep 29 2019 at 22:40):
Bourbaki's definition (E, II, p. 15) of a constant function is : for all x,x'∈X, f(x)=f(x').
With this definition, if X is empty, any function f:X →Y is constant. However, a constant function f:X →Y has a value iff X is nonempty.
#### Kevin Buzzard (Sep 29 2019 at 22:52):
On the other hand if you asked a mathematician for a definition, surely some would say "there exists c in Y such that for all x in X, f(x)=c". And of course if you showed those people bourbaki, many would say that the definitions were obviously the same ;-)
Last updated: May 10 2021 at 07:15 UTC | 2021-05-10 07:55:03 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 3, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7101568579673767, "perplexity": 2713.7083828600557}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-21/segments/1620243989115.2/warc/CC-MAIN-20210510064318-20210510094318-00005.warc.gz"} |
https://stats.stackexchange.com/questions/305451/why-forward-and-reverse-transforms-for-ppca-are-so-different | # Why forward and reverse transforms for PPCA are so different?
In probabilistic PCA (PPCA), they model dimensionality reduction as probabilistic model
$t = Wx + \mu + \epsilon$
where
$W$ is non-swuare matrix of size $(d \times q)$, $q < d$
$x$ is a vector from $q$-dimensional "latent" space
$t$ is a vector from $d$ dimensional "observable" space
$\epsilon$ is normally distributed random variable, covering "insufficiency" or transform from low-dimensional space to high-dimensional.
So, forward transform (from latent space to observable space) is performed by matrix
$W$
and reverse transform is performed by matrix
$(\sigma^2 I + W^T W)^{-1} W^T$
(formula for "posterior" mean in the middle of paragraph in chapter 3.3)
I know matrices are not square, but why is it so different?
UDPATE
Now I saw formula (16) which claims reverse transform is done with the matrix
$W(W^T W)^{-1}(\sigma^2 I + W^T W)$
anyway, the question is why is it so complex and is there any notion of matrix inverse for non-square matrices in this case?
You are looking at a hat matrix. The PPCA reconstruction is defined similarly to a regression model; what is described as $W(W^TW)^{-1}M$ (Eq. 16) is little more than the hat matrix of a Tikhonov-regularised regression task in latent space.
In the case of probabilistic PCA as well as ridge regression we assume a covariance $C$ such that $C = WW^T + \sigma^2 I$ where we capture "uncounted variance" in $\sigma^2$. In the the case of probabilistic PCA we call it "noise variance" while in the case of ridge regression "regularisation parameter / Tikhonov factor". Remember that $\langle x_n \rangle = M^{-1}W(t_n - \mu)$ (Eq. 55); simple substitution of Eq. 55 into Eq. 16 yields the well-known formulation of a hat matrix (which is actually presented in Eq. 68). On that matter CV has already a very good answer on how ridge regression relates to PCA here.
To emphasize how ubiquitous this formulation is: $B$-spline basis functions do exactly the same task within the context of non-parametric regression. In that case $\hat{\beta} = (B^TB + \lambda \Omega)^{-1}B^Ty$. Here instead of having a non-parametric basis as with the eigen-components provided by standard PCA, we use a set B-splines basic function $B$. To quote L. Wasserman directly: "The effect of the term $\lambda \Omega$ is to shrink the regression coefficients towards a subspace, which results in a smoother fit."$^\dagger$ Again, same idea. $B$-splines regression calls this fit "smoother", ridge regression calls this fit "regularised" and PPCA calls this "optimally reconstructed".
Going back to probabilistic PCA, in the case of $\sigma^2 =0$ the above mentioned formula (Eq. 16) would directly reduce to a conventional PCA. In that case, the matrix $W$ would form an orthogonal basis and therefore the "reverse transformation" would simply require $W^T$.
As a side-note: Yes, there is a notion of matrix inverse for non-square matrices: the Moore-Penrose Pseudo-inverse. Using it for PCA is like using a battle-tank for pizza delivery; doable but most probably daft. I have only seen it used for pedagogical purposes and being actually useful when handling small, nearly rank-deficient covariance matrices (ie. a very particular application).
$\dagger$. Larry Wasserman, "All of Nonparametric Statistics", Chapt. 5.5 "Penalized Regression, Regularization and Splines"
• I wish you explain more what exactly is the connection between PPCA and RR. I know both methods quite well, but to be honest it never occurred to me that they are related. – amoeba Oct 22 '17 at 0:08
• @amoeba: Thank you for your comment. You have given a great answer here to this question on relating PCA and RR. I mean... I actually thought of linking it for a moment (and then forgot to do it - fixed now). To quote you: "This means that ridge regression can be seen as a "smooth version" of PCR." which is in line with my own answer. I just make the connection even more obvious as PPCA explicitly defines a "noise model". – usεr11852 Oct 22 '17 at 0:23
• Hmm. Interesting. That answer is about how PC regression (PCR) is related to RR, whereas this Q is about PCA/PPCA itself, without any regression. Also, the interesting part is the term with $\sigma$... Anyway, I should think more about this. Might come back here later. – amoeba Oct 23 '17 at 21:13 | 2021-05-11 20:21:41 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7551499605178833, "perplexity": 1087.4041721159438}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-21/segments/1620243989856.11/warc/CC-MAIN-20210511184216-20210511214216-00569.warc.gz"} |
https://study.com/academy/answer/solve-int-sec-n-x-dx-for-n0.html | # Solve \int \sec^n x dx for n0
## Question:
Solve {eq}\int \sec^n \ x \ dx \ {/eq} for {eq}n>0 {/eq}
## Indefinite Integral:
The indefinite integral of trigonometric functions can either be reduced to the standard form or can be applied with the trigonometric identity to form an integral in the standard result form.
As we know that the integral of:
{eq}\int \sec ^n\left(x\right)dx=\frac{\sec ^{n-1}\left(x\right)\sin \left(x\right)}{n-1}+\frac{n-2}{n-1}\int \sec ^{n-2}\left(x\right)dx\\ {/eq}
using the standard result. | 2020-02-28 00:45:52 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9870534539222717, "perplexity": 3846.635083223124}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-10/segments/1581875146907.86/warc/CC-MAIN-20200227221724-20200228011724-00257.warc.gz"} |
https://tr.pinterest.com/explore/kalk%C3%BCl%C3%BCs/ | # Kalkülüs
### İlgili konuları keşfedin
Parent Functions
Visual Log LawsBy Brittany Bordewyk, as a great response to Prof. Wilson’s visual exponent laws. Did I somehow never post these?
*FREE* Conic Section Posters
Japanese Multiplication 101 #Multiplication #JapaneseMultiplication
Inverses of Matrices
Tweet Tweet Math is not everyone’s cup of tea, but learning new math shortcuts can be a lot of fun. There are many of them to learn. This infographic happens to cover 6 math shortcuts: [via: Center of Math]
Calculus - Derivatives and Limits
Trig functions on the unit circle
If you need a note card for Integral Calculus - Imgur
LOVE this! Calculus becomes soo much easier when you realize you are just finding the slope!!
Pinterest | 2017-04-26 19:41:34 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8249136805534363, "perplexity": 6834.707516985087}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-17/segments/1492917121644.94/warc/CC-MAIN-20170423031201-00248-ip-10-145-167-34.ec2.internal.warc.gz"} |
http://nrich.maths.org/public/leg.php?code=-68&cl=2&cldcmpid=6936 | # Search by Topic
#### Resources tagged with Visualising similar to Making Rectangles:
Filter by: Content type:
Stage:
Challenge level:
### An Unusual Shape
##### Stage: 3 Challenge Level:
Can you maximise the area available to a grazing goat?
### Intersecting Circles
##### Stage: 3 Challenge Level:
Three circles have a maximum of six intersections with each other. What is the maximum number of intersections that a hundred circles could have?
### Fence It
##### Stage: 3 Challenge Level:
If you have only 40 metres of fencing available, what is the maximum area of land you can fence off?
### All in the Mind
##### Stage: 3 Challenge Level:
Imagine you are suspending a cube from one vertex (corner) and allowing it to hang freely. Now imagine you are lowering it into water until it is exactly half submerged. What shape does the surface. . . .
### The Old Goats
##### Stage: 3 Challenge Level:
A rectangular field has two posts with a ring on top of each post. There are two quarrelsome goats and plenty of ropes which you can tie to their collars. How can you secure them so they can't. . . .
### Baravelle
##### Stage: 2, 3 and 4 Challenge Level:
What can you see? What do you notice? What questions can you ask?
### Triangles in the Middle
##### Stage: 3, 4 and 5 Challenge Level:
This task depends on groups working collaboratively, discussing and reasoning to agree a final product.
### Zooming in on the Squares
##### Stage: 2 and 3
Start with a large square, join the midpoints of its sides, you'll see four right angled triangles. Remove these triangles, a second square is left. Repeat the operation. What happens?
### Cubes Within Cubes Revisited
##### Stage: 3 Challenge Level:
Imagine starting with one yellow cube and covering it all over with a single layer of red cubes, and then covering that cube with a layer of blue cubes. How many red and blue cubes would you need?
### Muggles Magic
##### Stage: 3 Challenge Level:
You can move the 4 pieces of the jigsaw and fit them into both outlines. Explain what has happened to the missing one unit of area.
### Tied Up
##### Stage: 3 Challenge Level:
In a right angled triangular field, three animals are tethered to posts at the midpoint of each side. Each rope is just long enough to allow the animal to reach two adjacent vertices. Only one animal. . . .
### Rolling Around
##### Stage: 3 Challenge Level:
A circle rolls around the outside edge of a square so that its circumference always touches the edge of the square. Can you describe the locus of the centre of the circle?
### Trice
##### Stage: 3 Challenge Level:
ABCDEFGH is a 3 by 3 by 3 cube. Point P is 1/3 along AB (that is AP : PB = 1 : 2), point Q is 1/3 along GH and point R is 1/3 along ED. What is the area of the triangle PQR?
### Framed
##### Stage: 3 Challenge Level:
Seven small rectangular pictures have one inch wide frames. The frames are removed and the pictures are fitted together like a jigsaw to make a rectangle of length 12 inches. Find the dimensions of. . . .
### Take Ten
##### Stage: 3 Challenge Level:
Is it possible to remove ten unit cubes from a 3 by 3 by 3 cube made from 27 unit cubes so that the surface area of the remaining solid is the same as the surface area of the original 3 by 3 by 3. . . .
### Tetra Square
##### Stage: 3 Challenge Level:
ABCD is a regular tetrahedron and the points P, Q, R and S are the midpoints of the edges AB, BD, CD and CA. Prove that PQRS is a square.
### Dissect
##### Stage: 3 Challenge Level:
It is possible to dissect any square into smaller squares. What is the minimum number of squares a 13 by 13 square can be dissected into?
### Like a Circle in a Spiral
##### Stage: 2, 3 and 4 Challenge Level:
A cheap and simple toy with lots of mathematics. Can you interpret the images that are produced? Can you predict the pattern that will be produced using different wheels?
### Rotating Triangle
##### Stage: 3 and 4 Challenge Level:
What happens to the perimeter of triangle ABC as the two smaller circles change size and roll around inside the bigger circle?
### Rati-o
##### Stage: 3 Challenge Level:
Points P, Q, R and S each divide the sides AB, BC, CD and DA respectively in the ratio of 2 : 1. Join the points. What is the area of the parallelogram PQRS in relation to the original rectangle?
### Route to Infinity
##### Stage: 3 and 4 Challenge Level:
Can you describe this route to infinity? Where will the arrows take you next?
### Seven Squares
##### Stage: 3 and 4 Challenge Level:
Watch these videos to see how Phoebe, Alice and Luke chose to draw 7 squares. How would they draw 100?
### Isosceles Triangles
##### Stage: 3 Challenge Level:
Draw some isosceles triangles with an area of $9$cm$^2$ and a vertex at (20,20). If all the vertices must have whole number coordinates, how many is it possible to draw?
### Problem Solving, Using and Applying and Functional Mathematics
##### Stage: 1, 2, 3, 4 and 5 Challenge Level:
Problem solving is at the heart of the NRICH site. All the problems give learners opportunities to learn, develop or use mathematical concepts and skills. Read here for more information.
### Charting Success
##### Stage: 3 and 4 Challenge Level:
Can you make sense of the charts and diagrams that are created and used by sports competitors, trainers and statisticians?
### Paving Paths
##### Stage: 3 Challenge Level:
How many different ways can I lay 10 paving slabs, each 2 foot by 1 foot, to make a path 2 foot wide and 10 foot long from my back door into my garden, without cutting any of the paving slabs?
### Ding Dong Bell
##### Stage: 3, 4 and 5
The reader is invited to investigate changes (or permutations) in the ringing of church bells, illustrated by braid diagrams showing the order in which the bells are rung.
### Coloured Edges
##### Stage: 3 Challenge Level:
The whole set of tiles is used to make a square. This has a green and blue border. There are no green or blue tiles anywhere in the square except on this border. How many tiles are there in the set?
### Yih or Luk Tsut K'i or Three Men's Morris
##### Stage: 3, 4 and 5 Challenge Level:
Some puzzles requiring no knowledge of knot theory, just a careful inspection of the patterns. A glimpse of the classification of knots and a little about prime knots, crossing numbers and. . . .
### Square Coordinates
##### Stage: 3 Challenge Level:
A tilted square is a square with no horizontal sides. Can you devise a general instruction for the construction of a square when you are given just one of its sides?
### Is There a Theorem?
##### Stage: 3 Challenge Level:
Draw a square. A second square of the same size slides around the first always maintaining contact and keeping the same orientation. How far does the dot travel?
### Sprouts
##### Stage: 2, 3, 4 and 5 Challenge Level:
A game for 2 people. Take turns joining two dots, until your opponent is unable to move.
### LOGO Challenge - Circles as Animals
##### Stage: 3 and 4 Challenge Level:
See if you can anticipate successive 'generations' of the two animals shown here.
### Overlapping Again
##### Stage: 2 Challenge Level:
What shape is the overlap when you slide one of these shapes half way across another? Can you picture it in your head? Use the interactivity to check your visualisation.
### The Triangle Game
##### Stage: 3 and 4 Challenge Level:
Can you discover whether this is a fair game?
### Picturing Triangle Numbers
##### Stage: 3 Challenge Level:
Triangle numbers can be represented by a triangular array of squares. What do you notice about the sum of identical triangle numbers?
### L-ateral Thinking
##### Stage: 1 and 2 Challenge Level:
Try this interactive strategy game for 2
### Concrete Wheel
##### Stage: 3 Challenge Level:
A huge wheel is rolling past your window. What do you see?
### The Development of Spatial and Geometric Thinking: the Importance of Instruction.
##### Stage: 1 and 2
This article looks at levels of geometric thinking and the types of activities required to develop this thinking.
### Cuboids
##### Stage: 3 Challenge Level:
Find a cuboid (with edges of integer values) that has a surface area of exactly 100 square units. Is there more than one? Can you find them all?
### Frogs
##### Stage: 2 and 3 Challenge Level:
How many moves does it take to swap over some red and blue frogs? Do you have a method?
##### Stage: 3 Challenge Level:
How many different symmetrical shapes can you make by shading triangles or squares?
### Konigsberg Plus
##### Stage: 3 Challenge Level:
Euler discussed whether or not it was possible to stroll around Koenigsberg crossing each of its seven bridges exactly once. Experiment with different numbers of islands and bridges.
### Two Squared
##### Stage: 2 Challenge Level:
What happens to the area of a square if you double the length of the sides? Try the same thing with rectangles, diamonds and other shapes. How do the four smaller ones fit into the larger one?
### Weighty Problem
##### Stage: 3 Challenge Level:
The diagram shows a very heavy kitchen cabinet. It cannot be lifted but it can be pivoted around a corner. The task is to move it, without sliding, in a series of turns about the corners so that it. . . .
### Picturing Square Numbers
##### Stage: 3 Challenge Level:
Square numbers can be represented as the sum of consecutive odd numbers. What is the sum of 1 + 3 + ..... + 149 + 151 + 153?
### Painted Cube
##### Stage: 3 Challenge Level:
Imagine a large cube made from small red cubes being dropped into a pot of yellow paint. How many of the small cubes will have yellow paint on their faces?
### Threesomes
##### Stage: 3 Challenge Level:
Imagine an infinitely large sheet of square dotty paper on which you can draw triangles of any size you wish (providing each vertex is on a dot). What areas is it/is it not possible to draw?
### Tourism
##### Stage: 3 Challenge Level:
If you can copy a network without lifting your pen off the paper and without drawing any line twice, then it is traversable. Decide which of these diagrams are traversable.
### Sliding Puzzle
##### Stage: 1, 2, 3 and 4 Challenge Level:
The aim of the game is to slide the green square from the top right hand corner to the bottom left hand corner in the least number of moves. | 2016-05-28 12:04:22 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.38810351490974426, "perplexity": 1856.0674793106045}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-22/segments/1464049277592.65/warc/CC-MAIN-20160524002117-00138-ip-10-185-217-139.ec2.internal.warc.gz"} |
http://quant.stackexchange.com/questions?page=119&sort=newest | # All Questions
351 views
### How to get a Quant Job [closed]
At the end of the academic year I will graduate with a Computer Science degree (hopefully) and I would like to peruse a career in quantitative finance. What is the dest way of doing this? Is it worth ...
176 views
### Managing Bloomberg logins in a campus library [closed]
In a campus library or other public place, typically many people share the same PC with the Bloomberg terminal software. Is there a way to prevent users from creating new logins? Is there a way to ...
6k views
### Annualized Covariance
I have two time series. One with monthly returns on an asset and one with monthly returns on a benchmark index. I have calculated the covariance using the ...
587 views
### Applying interest rate shocks under Solvency II
I'm trying to figure out how one would apply the stress scenarios defined under the interest rate risk sub module of Solvency II. I understand that all future cash flows of an interest rate sensitive ...
398 views
### Using OpenCL video cards to offload Quant Finance calculations, what features should I look for?
I'm benchmarking some software and am looking for cards that are better at parallel multiplication vs parallel addition. Is there any prior work that may have this information? What GPU features ...
193 views
### Cloning Return Streams
How would one go about cloning/replicating returns of a hedge fund or a strategy. That is given a return series of the object to be clone, is it possible to decompose return and reconstruct another ...
114 views
### Regression extensions
I'm trying to find extensions for my regression and obviously would like to use PE, BV and CFO. But I've got monthly data, while all company's fundamentals are semi-annually... Can I deal with it ...
873 views
### Replicating portfolio and risk-neutral pricing for interest rate options
For equity options, the pricing of options depends on the existence of a replicating portfolio, so you can price the option as the constituents of that replicating portfolio. However, I am not seeing ...
2k views
### Basket option pricing: step by step tutorial for beginners
I would like to learn how to price options written on basket of several underlyings. I've never tried to do it and I would appreciate if you can provide some documents, papers, web sites and so on in ...
361 views
### Probability of trade's exit orders being triggered in random-walk market
When placing a trade with Stop Loss and Take Profit orders in a hypothetical random market (i.e. 0.5 probability of up tick and 0.5 probability of down tick), assuming: x is the distance in ticks of ...
216 views
### Simple question about stochastic differential
What is the equivalent of product rule for stochastic differentials? I need it in the following case: Let $X_t$ be a process and $\alpha(t)$ a real function. What would be $d(\alpha(t)X_t)$?
400 views
### Monty Hall Model
Given a fixed time period,say 3 days, the stock/market can go up,down or stay sideways. A hedge fund can long, short or use rangebound(options strategy) to bet for that 3 days closing level. Hedge ...
1k views
### hedging two bonds in different currencies with FX forward
Is there a way to make a hedged portfolio using two bonds, one is in EUR, the other one is in USD and FX forward contract? Assume that FX rate follows geometric Brownian motion movement.
573 views
### How to do performance attribution for a few characteristics?
Let's say the characteristics that I am interested in are FX Country Security selection I have the benchmark weights and returns, the FX returns, and the portfolio weights and returns. Can ...
977 views
### Should I use GARCH volatility or standard deviation in cross-sectional regression?
I want to do a cross-sectional study where the historical, medium-long run volatility of some return series (call it $R_t$) is included as a regressor. Which of the following two estimates of ...
5k views
### Interpretation of Macaulay Duration
I am having a difficulty conceptualizing the meaning of "Macaulay duration" - I want to note I completely understand the math, this isn't the issue. Modified duration & Efficitive Duration make ...
213 views
### Model-implied yield spread on corporate bonds
While using Merton (or any other) model, is the model-implied yield spread on bonds greater than actual yield spread? And is it possible to estimate actual probablities of default?
1k views
576 views
### Why are regressors squared and not ^1.5 or ^2.2 or ^2.5?
When a researcher in economics or finance wants to apply a linear regression model but suspects a non-linear relationship between one of the regressors and the dependent variable, it is typical to ...
441 views
### GBM 3d plot with R
I want to plot the density of the GBM in a 3d plot. So I have on one axis the stock price, on the other the time and on the z axis the density. At the end I want to produce this graph. The formula I ...
422 views
### How does one estimate theta in the Ho-Lee model from a yield curve?
I have a yield curve constructed using linear interpolation with data points every 3-months for US treasuries. I would like to use that calibrate a Ho-Lee model, but I can't wrap my head around how ...
1k views
### How to improve the Black-Scholes framework?
Since the distribution of daily returns are obviously not lognormal, my bottom line question is has BS been reworked for a better fitting distribution? Google searches give me nada. The best dist I'...
2k views
### Is there a piratebay for data(bases)? (here, talking about historical financial data) [closed]
I would like to, for example, access (and later, upload), for example, historical financial data for the S&P500. If a piratebay for databases existed, maybe we could freely share information that ...
245 views
### desk's performance
I need your point of view in evaluating the monthly performance of a desk. I have the daily credit risk capital requirement (A); the net banking income or GNP (B). What is the best measure of ...
556 views
### Bootstrapping first, then data mine?
I posted this on Cross Validated a month ago, but I didn't get any answers. Sorry for the second post! I'm wondering about the whole process of testing/data-mining for a strategy and THEN testing on ...
951 views
### What are current interest rates on senior/junior/mezzanine loans for e.g. real estate developers?
For a case study I have to work on for a university course, about a real-estate-development project, I need to simulate the financing with different proportions of equity (40%), senior loan (35%), ...
2k views
### Treasury Bond Yield Curves in R
Does anyone know if I can access interest rates series from the treasury using R? I tried yahoo! Finance and it doesn't seem to have this kind of information.
582 views
### Black-Scholes American Put Option
Here is my question: This is a question about Black-Scholes model, but it may be applicable to more complicated models. Throughout the discussion, the strike price $K$, interest rate $r$ and ...
237 views
### Does entropy pooling apply to distributions with time-varying drift?
I have a returns process that is drawn from a normal distribution with a nonlinear time-varying drift, so I was wondering if the entropy pooling method still applies or if I need an invariant ?
427 views
### Is making a bid/ask offer a good way to lower the spreads?
I have written an algorithmic trading program which relies heavily on low spreads in the 0.1-0.3 PIP region. I was now wondering if it would be a good idea to place bid/ask offers instead of limit ...
83 views
### Changes to option valuation for dollar-pegged underlying
In Russia, options on futures on the RTS index are priced in points instead of currency, with points being directly related to the value of the US dollar such that, for example, if the dollar rises, ...
2k views
### How do I estimate the parameters of an MA(q) process?
It is relatively easy to estimate the parameters of an autoregressive $AR(p)$ process. How do I do with a moving average $MA(q)$ process?
2k views
### Why do ATM call options have a delta of slightly bigger than 0.5 and not 0.5 exactly?
From the formula of the delta of a call option, i.e. $N(d1)$, where $d_1 = \frac{\mathrm{ln}\frac{S(t)}{K} + (r + 0.5\sigma^2)(T-t)}{\sigma\sqrt{T-t}}$, the delta of an ATM spot call option is ... | 2016-07-27 01:54:31 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7454783916473389, "perplexity": 2021.565094139782}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-30/segments/1469257825358.53/warc/CC-MAIN-20160723071025-00260-ip-10-185-27-174.ec2.internal.warc.gz"} |
https://www.shaalaa.com/question-bank-solutions/division-line-segment-draw-triangle-abc_7190 | # Question - Division of a Line Segment
Account
Register
Share
Books Shortlist
ConceptDivision of a Line Segment
#### Question
Draw a triangle ABC with side BC = 7 cm, ∠B = 45°, ∠A = 105°. Then, construct a triangle whose sides are 4/3 times the corresponding side of ΔABC. Give the justification of the construction.
#### Solution
You need to to view the solution
Is there an error in this question or solution?
S | 2017-09-25 06:20:35 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.636625349521637, "perplexity": 1826.5236081397986}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-39/segments/1505818690340.48/warc/CC-MAIN-20170925055211-20170925075211-00472.warc.gz"} |
https://www.physicsforums.com/threads/canonical-transformations.595921/ | # Canonical transformations
1. Apr 12, 2012
### zheng89120
Why is it that only Canonical transformations preserve the Hamilton's equations? Or what makes non-canonical transformations not preserve the Hamilton's equations?
2. Apr 12, 2012
### Dougy81
But that is the definition of canonical transformation that they preserve the Hamilton's equation. Whatever transformation of coordinates you find that preserves Hamilton's equation is a canonical transformation. | 2017-10-23 13:50:46 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.929130494594574, "perplexity": 2462.8527752867003}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-43/segments/1508187826049.46/warc/CC-MAIN-20171023130351-20171023150351-00711.warc.gz"} |
https://open.kattis.com/problems/successfulzoom/en | Hide
# Successful Zoom
Your boss likes to see numbers going up as proof that your company is successful. To help with this, you came up with the idea to summarize a list of numbers by “zooming out,” that is, to discard everything except every $k^{th}$ number in the list, for some $k\ge 1$.
Business has been tough lately, with lots of ups and downs. However, if you pick the right value of $k$, maybe you can prove to your boss that things aren’t so bad after all! If possible, you should find the smallest value of $k$ such that $x_ k<x_{2k}<\dots <x_{qk}$, where $q=\lfloor \frac nk\rfloor$, and the list $x_ k,x_{2k},\dots ,x_{qk}$ contains at least two elements.
## Input
The first line contains an integer $n$, with $2\le n\le 10^5$. The second line contains $n$ integers $x_1,x_2,\dots ,x_ n$ describing the list of numbers. It is guaranteed that $0\le x_ i\le 2^{30}$ for $i=1,\dots ,n$.
## Output
If possible, display the smallest value $k\ge 1$ for which $x_ k,x_{2k},\dots ,x_{qk}$ (where $q=\lfloor \frac nk\rfloor$) is strictly increasing and has at least two elements. If no such value of $k$ exists, display the word “ABORT!
Sample Input 1 Sample Output 1
10
1 2 3 4 5 6 7 8 9 10
1
Sample Input 2 Sample Output 2
8
1 8 2 7 3 6 4 5
3
Sample Input 3 Sample Output 3
10
9 8 8 4 4 3 10 1 1 0
ABORT!
CPU Time limit 1 second
Memory limit 1024 MB
Difficulty 2.1Easy
Statistics Show
Downloads
Author
License
Please log in to submit a solution to this problem | 2022-09-30 08:47:41 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.3268202245235443, "perplexity": 371.96223302309915}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030335448.34/warc/CC-MAIN-20220930082656-20220930112656-00696.warc.gz"} |
https://socratic.org/questions/what-is-are-the-maximum-number-of-possible-real-roots-of-the-equation-a-3-b-4-c- | # What is/are the maximum number of possible real roots of the equation? a) 3 b) 4 c) 5 d) 0
## ${x}^{5} - 6 {x}^{2} - 4 x + 5 = 0$
Apr 5, 2018
See below.
#### Explanation:
None of the answers is satisfactory
Comparing the plots for
$y = {x}^{5}$ and
$y = - 6 {x}^{2} - 4 x + 5$
the conclusion is that there is only a real root. | 2019-08-18 01:44:37 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 3, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7041727900505066, "perplexity": 922.988073577855}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-35/segments/1566027313536.31/warc/CC-MAIN-20190818002820-20190818024820-00403.warc.gz"} |
https://documen.tv/question/a-toy-radio-controlled-car-races-around-a-circular-track-in-a-time-of-2-minutes-with-a-speed-of-22770735-45/ | # A toy radio-controlled car races around a circular track in a time of 2 MINUTES, with a speed of 5.25 m/s. What is the radius of the c
Question
A toy radio-controlled car races around a circular track in a time of 2 MINUTES, with a speed of 5.25 m/s. What is the radius of the
circular track? (will give brainliest)
in progress 0
1 year 2021-09-01T08:53:42+00:00 1 Answers 0 views 0
## Answers ( )
1. Answer:
The radius will be “100.31 m”. A further solving is given below.
Explanation:
The given values are:
Time,
T = 2 minutes
i.e.,
= 120 seconds
Speed,
V = 5.25 m/s
As we know
⇒ $$T=\frac{2 \pi R}{V}$$
then,
⇒ $$R=\frac{TV}{2 \pi}$$
On substituting the given values, we get
⇒ $$=\frac{120\times 5.25}{2\times 3.14}$$
⇒ $$=\frac{630}{6.28}$$
⇒ $$=100.31 \ m$$ | 2023-02-05 14:14:34 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5719013810157776, "perplexity": 6159.179066574527}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764500255.78/warc/CC-MAIN-20230205130241-20230205160241-00385.warc.gz"} |
https://www.shaalaa.com/question-bank-solutions/prove-that-c-o-s-e-c-90-x-cot-450-x-c-o-s-e-c-90-x-tan-180-x-tan-180-x-sec-180-x-tan-360-x-sec-x-2-trigonometric-equations_53988 | # Prove that C O S E C ( 90 ∘ + X ) + Cot ( 450 ∘ + X ) C O S E C ( 90 ∘ − X ) + Tan ( 180 ∘ − X ) + Tan ( 180 ∘ + X ) + Sec ( 180 ∘ − X ) Tan ( 360 ∘ + X ) − Sec ( − X ) = 2 - Mathematics
Prove that
$\frac{cosec(90^\circ + x) + \cot(450^\circ + x)}{cosec(90^\circ - x) + \tan(180^\circ - x)} + \frac{\tan(180^\circ + x) + \sec(180^\circ - x)}{\tan(360^\circ + x) - \sec( - x)} = 2$
#### Solution
LHS = $\frac{cosec \left( 90^\circ + x \right) + \cot \left( 450^\circ + x \right)}{cosec \left( 90^\circ - x \right) + \tan \left( 180^\circ - x \right)} + \frac{\tan \left( 180^\circ + x \right) + \sec \left( 180^\circ - x \right)}{\tan \left( 360^\circ + x \right) - \sec \left( - x \right)}$
$= \frac{cosec\left( 90^\circ + x \right) + \cot\left( 450^\circ + x \right)}{cosec \left( 90^\circ - x \right) + \tan\left( 180^\circ - x \right)} + \frac{\tan \left( 180^\circ + x \right) + \sec \left( 180^\circ - x \right)}{\tan \left( 360^\circ + x \right) - \sec \left( - x \right)}$
$= \frac{cosec\left( 90^\circ + x \right) + \cot \left( 90^\circ \times 5 + x \right)}{cosec\left( 90^\circ - x \right) + \tan \left( 90^\circ \times 2 - x \right)} + \frac{\tan \left( 90^\circ \times 2 + x \right) + \sec \left( 90^\circ \times 2 - x \right)}{\tan\left( 90^\circ \times 4 + x \right) - \sec\left( - x \right)}$
$= \frac{\sec x + \cot \left( 90^\circ \times 5 + x \right)}{cosec\left( 90^\circ- x \right) + \tan \left( 90^\circ \times 2 - x \right)} + \frac{\tan \left( 90^\circ \times 2 + x \right) + \sec \left( 90^\circ \times 2 - x \right)}{\tan \left( 90^\circ \times 4 + x \right) - \sec \left( - x \right)}$
$= \frac{\sec x - \tan x}{\sec x - \tan x} + \frac{\tan x - \sec x}{\tan x - \sec x}$
$= 1 + 1$
$= 2$
= RHS
Hence proved.
Is there an error in this question or solution?
#### APPEARS IN
RD Sharma Class 11 Mathematics Textbook
Chapter 5 Trigonometric Functions
Exercise 5.3 | Q 3.2 | Page 39 | 2021-04-13 19:03:11 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5580245852470398, "perplexity": 4271.748180649252}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-17/segments/1618038074941.13/warc/CC-MAIN-20210413183055-20210413213055-00156.warc.gz"} |
https://www.techrxiv.org/articles/preprint/An_norm-Based_Criterion_for_Bounded_Component_Analysis/19660680/1 | An_Linf_norm_Based_Criterion_for_BCA_Brotto_Nose_Romano.pdf (1.96 MB)
Download file
# An ℓ∞ norm-Based Criterion for Bounded Component Analysis
preprint
posted on 2022-05-02, 18:36 authored by Renan BrottoRenan Brotto, João Romano
In the last decade, there has been a great interest in finding new priors for Blind Source Separation (BSS). Some of the most interesting of them are sparsity, non-negativity and boundedness of the sources, since they consist in weaker assumptions when compared to the classical statistical independence hypothesis. In this paper, we propose a new criterion for Bounded Component Analysis (BCA), using the $\ell_\infty$ norm, with an associated algorithm based on Givens Rotations. We analyze our proposal and evaluate it with numerical simulations for three kinds of bounded signals. From both theoretical and experimental results we show that the $\ell_\infty$ norm is a suitable contrast function for BCA, presenting very good results when compared to the state-of-the-art algorithm.
## Email Address of Submitting Author
renanbrotto@gmail.com
UNICAMP
Brazil
## Exports
figshare. credit for all your research. | 2023-01-31 19:11:19 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.23305962979793549, "perplexity": 924.0238625525006}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764499890.39/warc/CC-MAIN-20230131190543-20230131220543-00460.warc.gz"} |
http://www.aimsciences.org/article/doi/10.3934/jimo.2018117?viewType=html | # American Institute of Mathematical Sciences
• Previous Article
Coordinating the supplier-retailer supply chain under noise effect with bundling and inventory strategies
• JIMO Home
• This Issue
• Next Article
Asymptotic convergence of stationary points of stochastic multiobjective programs with parametric variational inequality constraint via SAA approach
October 2019, 15(4): 1677-1699. doi: 10.3934/jimo.2018117
## Recovering optimal solutions via SOC-SDP relaxation of trust region subproblem with nonintersecting linear constraints
1 Department of Mathematical Sciences, Tsinghua University, Beijing 100084, China 2 Edward P. Fitts Department of Industrial and Systems Engineering, North Carolina State University, Raleigh, NC 27695, USA
* Corresponding author: Wenxun Xing < wxing@tsinghua.edu.cn >
Received April 2017 Revised April 2018 Published August 2018
Fund Project: Xing's research has been supported by the NNSF of China Grants #11571029 and #11771243, Fang's research has been supported by the US ARO Grant #W911NF-15-1-0223.
In this paper, we study an extended trust region subproblem (eTRS) in which the unit ball intersects with $m$ linear inequality constraints. In the literature, Burer et al. proved that an SOC-SDP relaxation (SOCSDPr) of eTRS is exact, under the condition that the nonredundant constraints do not intersect each other in the unit ball. Furthermore, Yuan et al. gave a necessary and sufficient condition for the corresponding SOCSDPr to be a tight relaxation when $m = 2$. However, there lacks a recovering algorithm to generate an optimal solution of eTRS from an optimal solution $X^*$ of SOCSDPr when rank $(X^*)≥ 2$ and $m≥ 3$. This paper provides such a recovering algorithm to complement those known works.
Citation: Jinyu Dai, Shu-Cherng Fang, Wenxun Xing. Recovering optimal solutions via SOC-SDP relaxation of trust region subproblem with nonintersecting linear constraints. Journal of Industrial & Management Optimization, 2019, 15 (4) : 1677-1699. doi: 10.3934/jimo.2018117
##### References:
show all references | 2020-05-29 13:09:37 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.36420080065727234, "perplexity": 2341.8974358886917}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-24/segments/1590347404857.23/warc/CC-MAIN-20200529121120-20200529151120-00471.warc.gz"} |
https://www.statemath.com/2021/08/improper-integral-exercises.html?amp=1 | We offer a selection of improper integral exercises with detailed answers. Our main objective is to show you how to prove that an improper integral is convergent: how to employ the integration by parts; how to make a change of variables; how to apply the dominated convergence theorem; and how to integrate terms by terms.
## Integrals on unbounded intervals, improper integral?
Proper Riemann integrals are defined for bounded functions on bounded intervals. Now we ask the question: can we define integrals for functions that are unbounded in the neighborhood of a point $(a,b]$ or functions defined on unbounded intervals of the form $(-\infty,+\infty),$ $[a,+infty),$ $(-\infty,a].$
Let $f:[a,+\infty)\to\mathbb{R}$ be a continuous function. If the limit $$\lim_{x\to+\infty}\int_a^x f(t)dt$$ exists, then we say that the integral \begin{align*} \int^{+\infty}_a f(t)dt\end{align*} is convergent, and the value of this integral is exactly the value of the limit.
Similarly we define the improper integral of a continuous function on $(-\infty,a]$.
Let $f:[a,b)\to\mathbb{R}$ be a continuous function. If the limit $$\lim_{x\to b^-}\int_a^x f(t)dt$$ exists, then we say that the integral \begin{align*} \int^{b}_a f(t)dt\end{align*} is convergent.
## Selected improper integral exercises
Exercise: Show that the following integrals are convergent \begin{align*} \begin{array}{cc} 1.\; \displaystyle\int^{+\infty}_0 e^{-t^2}dt & \quad 2.\; \displaystyle\int^{\frac{\pi}{2}}_0 \sqrt{\tan(\theta)};d\theta\\ 3.\;\displaystyle\int^{+\infty}_0 \frac{\ln(t)}{1+t^2}dt & \quad 4.; \displaystyle\int^1_0 \frac{dt}{1-\sqrt{t}}. \end{array} \end{align*}
Solution: 1) It suffices to show the convergence on the interval $[1,+\infty)$. For any $t\ge 1,$ we have $t^2\ge t,$ so that \begin{align*} 0< e^{-t^2}\le e^{-t},\qquad \forall t\ge 1. \end{align*} Observe that \begin{align*} \int^{+\infty}_1 e^{-t}dt&=\lim_{x\to +\infty} \int^x_1 (-e^{-t})’dt\cr &= \lim_{x\to+\infty} (e^{-1}-e^{-x})\cr &= \frac{1}{e}. \end{align*} This implies that the integral \begin{align*} \int^{+\infty}_0 e^{-t}dt \end{align*} is convergent. Hence the integral \begin{align*} \int^{+\infty}_0 e^{-t^2}dt \end{align*} is convergent is convergent as well.
Another proof: Remark that $t^2 e^{-t^2}\to 0$ as $t\to +\infty$. This means that there exists $\gamma>0,$ sufficiently large, such that $t^2 e^{-t^2}<1$ for any $t\ge \gamma$. We know that the integral \begin{align*} \int_\gamma^{+\infty} \frac{dt}{t^2} \end{align*} converge. Then by comparison the integral \begin{align*} \int_\gamma^{+\infty} e^{-t^2}dt \end{align*} converges. This ends the proof.
2) Here we shall use a change of variables. We have problem near of $\frac{\pi}{2},$ as $\cos(\frac{\pi}{2})=0$. We need to define a function of class $C^1$ which will play the role of change of variable. We then define the function \begin{align*} \psi(\theta)=\sqrt{\tan(\theta)},\quad \theta\in \left(0,\frac{\pi}{2}\right). \end{align*} Clearly $\psi$ is a $C^1$ function on $\left(0,\frac{\pi}{2}\right)$. In addition, for any $0< \theta < \frac{\pi}{2}$, we have \begin{align*} \psi'(\theta)&=\frac{\tan'(\theta)}{2 \sqrt{\tan(\theta)}}\cr &= \frac{\tan^2(\theta)}{2 \sqrt{\tan(\theta)}}. \end{align*} We deduce that $\psi'(\theta)>0$ for any $0< \theta < \frac{pi}{2}$, so that $\psi$ define a bijection from $\left(0,\frac{\pi}{2}\right)$ to \begin{align*}\left(\psi(0),\lim_{\sigma\to \frac{\pi}{2}}\psi(\sigma)\right)=(0,+\infty).\end{align*} Let $x\in (0,+\infty)$ and $0< \theta < \frac{\pi}{2}$ such that $x=\psi(\theta)$, which means that $x^2=\tan(\theta)$. Thus $\theta=\arctan(x^2)$. It follows that \begin{align*} d\theta=\frac{2x}{1+x^4} dx. \end{align*} Now we can write \begin{align*} \int^{\frac{\pi}{2}}_0 \sqrt{\tan(\theta)}\;d\theta=2\int^{+\infty}_0 \frac{x^2}{1+x^4}dx \end{align*} The function $f(x)=\frac{x^2}{1+x^4}$ is defined and continuous on $[0,+\infty)$. Observe that \begin{align*} 0 < f(x)\le \frac{1}{x^2},\quad\forall x\ge 1. \end{align*} An the function $x\mapsto \frac{1}{x^2}$ is integrable on $[1,+\infty),$ then the function $f$ is integrable on $[0,+\infty)$. This ends the proof.
3) The function $g(t)=\frac{\ln(t)}{1+t^2}$ is continuous on $(0,+\infty)$. Moreover, \begin{align*} t^{\frac{3}{2}} f(t)\quad\underset{t\to+\infty}{\sim}\quad \frac{\ln(t)}{\sqrt{t}}. \end{align*} As $t^{-\frac{1}{2}}\ln(t)\to 0$ as $t\to +\infty$, then \begin{align*} \lim_{t\to +\infty} t^{\frac{3}{2}} f(t)=0. \end{align*} On the other hand, as the function $t\mapsto \frac{1}{ t^{\frac{3}{2}}}$ is integrable on $[1,+\infty)$, the function $f$ is integrable on $[1,+\infty)$.
Clearly we have \begin{align*} \sqrt{t} f(t)\quad\underset{t\to+\infty}{\sim}\quad \sqrt{t} \ln(t). \end{align*} Thus \begin{align*} \lim_{t\to 0}\sqrt{t} f(t)=0. \end{align*} We know that the function $t\mapsto \frac{1}{\sqrt{t}}$ is integrable on $(0,1]$, it follows that the function $f$ is integrable on $(0,1]$. Conclusion: $f$ is integrable on $(0,+\infty)$. The improper integral is then convergent.
The following problem is classical in improper integral exercises.
Exercise “Gamma function”: For $x\in\mathbb{R},$ we consider the integral \begin{align*} \Gamma(x):=\int^{+\infty}_0 t^{x-1}e^{-t}dt. \end{align*} We also denote $I:=\{x\in \mathbb{R}: \Gamma(x) < \infty\}$.
1. Determine $I$.
2. Show that for all $x\in I,$ $\Gamma(x+1)=x\Gamma(x)$. Deduce the value of $\Gamma(n)$ for any $n\in\mathbb{N}^\ast$.
3. Justify that \begin{align*} \Gamma\left(\frac{1}{2} \right)=2\int^{+\infty}_0 e^{-t^2}dt. \end{align*}
Solution: 1) Let $x\in\mathbb{R}$. the function $f:t\mapsto t^{x-1}e^{-t}$ is continuous on $(0,+\infty),$ as product of continuous functions. On the other hand, clearly we have $t^2f(t)\to 0$ as $t\to +\infty$. This shows that $f(t)$ is equivalent to $\frac{1}{t^2}$ when $t$ is near of $+\infty$. Thus the function $f$ is integrable on $[1,+\infty)$. In addition $f(t)\sim t^{x-1}$, $t\to 0$, so that $f$ is integrable on $(0,1]$ if and only if $x-1>1,$ i.e. $x>0$. Hence $f$ is integrable on $(0,+\infty)$ if and only if $x>0$. Finally $I=(0,+\infty)$.
2) Let $x\in I$. The functions $u:t\mapsto t^x$ and $v:t\mapsto e^{-t}$ are of class $C^1$ on $(0,+\infty)$ and $u(t)v(t)\to 0$ as $t\to 0$ and $t\to +\infty$. Now by integration by parts, we have \begin{align*} \Gamma(x+1)=\left[t^x (-e^{-t})\right]^{+\infty}_0 -\int^{+\infty}_0 xt^{x-1}(-e^{-t})dt=x\Gamma(x). \end{align*} Let now $n\in \mathbb{N}^\ast$. First, observe that \begin{align*} \Gamma(1)=\int^{+\infty}_0 e^{-t}dt= \left[-e^{-t}\right]^{+\infty}_0=1. \end{align*} We then have \begin{align*} \Gamma(n)=n\Gamma(n-1)=n(n-1)\Gamma(n-2)=\cdots=n(n-1)\cdots 2\Gamma(1)=n!. \end{align*}
3) In the integral defining $\Gamma\left(\frac{1}{2} \right),$ we make the change of variables $t=u^2$, and we obtain \begin{align*} \Gamma\left(\frac{1}{2} \right)=\int^{+\infty}_0 t^{-\frac{1}{2}}e^{-1}dt=\int^{+\infty}_0 \frac{e^{-u^2}}{u}2udu=2\int^{+\infty}_0 e^{-u^2}du. \end{align*} | 2023-03-22 03:38:03 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9999279975891113, "perplexity": 279.8498445947836}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296943749.68/warc/CC-MAIN-20230322020215-20230322050215-00158.warc.gz"} |
http://math.stackexchange.com/questions/237755/how-do-you-solve-a-system-of-equation-if-you-have-a-contraint | how do you solve a system of equation if you have a contraint?
$xy^3z^3 = yx^3z^3 = zx^3y^3 \\ x^2+y^2+z^2$
I was wondering if the only solution to this was $z=y=x$
Solving an equation when there is a constraint is different right?
here there is not an infinite number of solution, but like only a dozen right?
-
What's the constraint? Was there supposed to be a $\leq$ or $=$ on the second line? – Karolis Juodelė Nov 15 '12 at 5:53
The set of equations (without the constraint-but you don't indicate what $x^2+y^2+z^2 should be) were solved in this question where it was shown that another solution had one variable$0$. – Ross Millikan Nov 15 '12 at 6:00 More than a dozen. With constraint say$x^2+y^2+z^2=1$, there are infinitely many with$z=0$. And another infinite bunch with$y=0$, and another with$x=0$, plus$8\$ others. – André Nicolas Nov 15 '12 at 6:24
yeah but i only need the extreme cases right? – Arnold Nov 15 '12 at 16:38 | 2014-04-16 07:38:57 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.833923876285553, "perplexity": 546.4827267040544}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-15/segments/1397609521558.37/warc/CC-MAIN-20140416005201-00593-ip-10-147-4-33.ec2.internal.warc.gz"} |
http://clay6.com/qa/36033/sunken-stomata-are-found-in | Sunken stomata are found in
$\begin{array}{1 1}(A)\;\text{Mesophytes}\\(B)\;\text{Hydrophytes} \\(C)\;\text{Xerophytes}\\(D)\;\text{Eplphytes }\end{array}$
Sunken stomata are found in xerophytes
Hence (C) is the correct answer. | 2018-04-22 18:19:47 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6457613110542297, "perplexity": 12837.60139261882}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-17/segments/1524125945637.51/warc/CC-MAIN-20180422174026-20180422194026-00387.warc.gz"} |
http://tex.stackexchange.com/questions/94820/getting-textmate-to-call-biber | # Getting TextMate to call biber
I've recently made the switch to `biblatex` from `natbib` but one problem I've been unable to resolve is how to make TextMate (which is my preferred editor) call `biber` instead of `bibtex`. I've installed the updated `latex.tmbundle` from Github, but whenever I typeset, I get unresolved references, and only the reference codes show up in the typeset document, which looks fine otherwise.
I know that `biber` is working, for two reasons:
1. When I typeset in TeXShop, my document comes out looking exactly as it should, with fully parsed references;
2. When I run the `biber <file>` command in terminal on the `.bcf` file that TextMate initially generates and then TypeSet in TextMate afterward, everything works as it should.
Any help getting TextMate to call `biber` automatically would be much appreciated!
-
I'm also working with TextMate and in my LaTeX preferences (those of the LaTeX-Bundle) i enabled `latexmk.pl`, which is for example explained here blog.mixable.de/textmate-und-latex (but only in german). With TeX Live 2011 and 2012 `latexmk.pl` should be able to determine itself whether to run biblatex or biber, which is stated here tex.stackexchange.com/questions/27450/… . Though, if you could provide an MWE, that would be great, because i'm not used to `biber`. – Ronny Jan 22 '13 at 6:16
@Ronny, Your solution worked. I also ended up replacing the `latexmk.pl` file with a symbolic link to TeX Live 2012's version, as described here: link. The only problem I'm still facing is that the `TypeSet & View` command in TextMate does not produce a TypeSet & View window or call Skim. All of the parsing is correct, however. – Gene G. Jan 22 '13 at 19:55
hm, well that works fine for me, enabling the Skim preferences in the same dialog as mentioned before and leaving the Typeset&View command the same as the original (which is for me a quite longish ruby script). So what does your command produce? – Ronny Jan 22 '13 at 22:35
I fixed this problem by upgrading to TM2. – Gene G. Jan 23 '13 at 0:45
Nice, haven't tried TM2 yet, still running 1.5.11 – Ronny Jan 23 '13 at 6:40
show 3 more comments
## 1 Answer
There are several steps to making sure that TextMate works well with `biber`.
1. Make sure you have the latest version of MacTeX installed.
2. Install the TextMate 2.0 alpha. Here.
3. Make sure you have the LaTeX bundle enabled. In TextMate: Preferences > Bundles > Select LaTeX bundle.
4. Configure the LaTeX bundle to use Latexmk.pl, which automatically determines which backend to use for typesetting bibliographies. In TextMate: Bundles > LaTeX > Preferences. Set options as `--shell-escape` and unable "Use Latexmk.pl"
5. If at any point `biber` fails to parse your references, you need to clear the cache. The issue & solution is described here. (For mac users, it is as simple as opening terminal and entering this command:
``````rm -rf `biber --cache`
``````
-
add comment | 2014-03-17 21:13:10 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9212751984596252, "perplexity": 3664.9228562608705}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-10/segments/1394678706211/warc/CC-MAIN-20140313024506-00000-ip-10-183-142-35.ec2.internal.warc.gz"} |
https://math.stackexchange.com/questions/2477575/image-of-morphism-of-projective-varieties-is-projective-variety | # Image of morphism of projective varieties is projective variety
Let $k$ be an algebraically closed field, $X,Y$ projective varieties (irreducible algebraic sets) and $f:X\to Y$ a morphism. Is $f(X)$ a projective variety? I think it is because the image of a morphism is closed and continuity preserves irreducibility. Is this correct?
I wonder because if $X$ and $Y$ are affine varieties, the statement is not true by this example: Image of a morphism of varieties.
Yes, this is correct. To be a bit more precise, if $X$ is a projective variety and $Y$ is any variety and $f:X\to Y$ is a morphism, then $f$ is a closed map (in particular its image is closed). And furthermore, the image of an irreducible set under any continuous map is irreducible. | 2019-09-19 08:12:28 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9582399129867554, "perplexity": 96.21164441253586}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-39/segments/1568514573465.18/warc/CC-MAIN-20190919081032-20190919103032-00329.warc.gz"} |
https://socratic.org/questions/how-do-you-simplify-and-write-2-5-times-10-3-times-520-in-scientific-notation | # How do you simplify and write 2.5 times 10^3 times 520 in scientific notation?
Jul 24, 2016
$1.3 \times {10}^{6}$
#### Explanation:
Lets get rid of the decimal!
Write $2.5 \times {10}^{3} \text{ as } 25 \times {10}^{2}$
Now we have:
$25 \times 520 \times {10}^{2} = 13000 \times {10}^{2}$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Rather than write the solution straight off I am going to do the following so you can see what the stages are and their relationships.
$\textcolor{b l u e}{\text{All the following have the same value. They just look different!}}$
$13000 \times {10}^{2}$
$1300 \times {10}^{3}$
$130 \times {10}^{4}$
$13 \times {10}^{5}$
$1.3 \times {10}^{6} \leftarrow \text{ This is Scientific notation}$ | 2020-01-18 00:06:17 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 9, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8387770652770996, "perplexity": 1523.4430816765964}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-05/segments/1579250591431.4/warc/CC-MAIN-20200117234621-20200118022621-00341.warc.gz"} |
http://www.gradesaver.com/textbooks/science/physics/physics-principles-with-applications-7th-edition/chapter-8-rotational-motion-problems-page-223/33 | ## Physics: Principles with Applications (7th Edition)
$1.2 \times 10^{-10}m$.
Treat the 2 oxygen molecules like point masses of mass M/2, each a distance D/2 from the axis of rotation. $$I=(M/2)(D/2)^2+(M/2)(D/2)^2=\frac{1}{4}MD^2=1.9 \times 10^{-46} kg \cdot m^2$$ Use the given M and solve for $D=1.2 \times 10^{-10}m$. | 2018-03-25 05:42:04 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9030146598815918, "perplexity": 1290.080608931905}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-13/segments/1521257651820.82/warc/CC-MAIN-20180325044627-20180325064627-00660.warc.gz"} |
http://mathoverflow.net/questions/158294/failure-of-noether-normalization/158301 | Failure of Noether normalization
I was in a class recently where we were trying to roughly count the dimensions of certain spaces of rational maps from algebraic curves into closed subschemes $Z \subseteq \mathbb{A}^n$. One way to simplify the proof is to find a map from $Z$ to $\mathbb{A}^{n-1}$ with finite fibers, but we couldn't figure out whether you could always do this. I'd like to know when and how badly this can fail.
Let's assume that $Z$ is a reduced closed subscheme of $\mathbb{A}^n$ (and not equal to all of $\mathbb{A}^n$), and that everything is over an infinite field $k$.
1. Noether normalization says that if $Z$ is irreducible, then there's a finite map $Z \to \mathbb{A}^{n-1}$. Are there examples of reducible $Z$ of pure dimension $n-1$ such that no such map exists?
2. Are there examples of reducible $Z$ (no restrictions on dimension) such that no map $Z \to \mathbb{A}^{n-1}$ with finite fibers exists?
3. We could restrict attention to the linear projections $\pi:\mathbb{A}^n \to \mathbb{A}^{n-1}$. In this case, $\pi|_Z$ has a positive-dimensional fiber iff $Z$ contains a line in the direction of the kernel of $\pi$. So: are there examples of $Z$ that contain a line in every direction?
- | 2016-02-08 21:57:05 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9342568516731262, "perplexity": 126.16810057804103}, "config": {"markdown_headings": false, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-07/segments/1454701154221.36/warc/CC-MAIN-20160205193914-00141-ip-10-236-182-209.ec2.internal.warc.gz"} |
https://jeopardylabs.com/play/operations-with-fractions-and-mixed-numbers-review | Subtracting Fractions and Mixed Numbers
Multiplying Fractions and Mixed Numbers
Dividing Fractions and Mixed Numbers
Misc.
3/5 + 1/2
What is 1 1/10
7/12 - 5/18
What is 11/36
5/8 * 4/5
What is 1/2
5/9 ÷ 2/3
What is 5/6
13.8 + 9.3
What is 23.1
### 200
In a 1/2 mile relay swimming event, Brad swims 1/8 mile. What is the total distance that his teammates have left to swim?
What is 3/8 miles
13 5/6 - 9 1/6
What is 4 2/3
1/8 * 5
What is 5/8
9/2 ÷ 3/2
What is 3
4.59 - 3.17
What is 1.42
7 1/5 + 1 2/5
What is 8 3/5
12 3/4 - 9 1/6
What is 3 7/12
12 * 3/8
What is 4 1/2
1/6 ÷ 3
What is 1/18
### 300
165 ÷ [(4 + 7) * 5]
What is 3
7 5/6 + 3 1/9
What is 10 17/18
9 - 7 4/9
What is 1 5/9
4 4/5 * 1 1/9
What is 5 1/3
21 ÷ 3 1/2
What is 6
### 400
[(9 + 34) ÷ 30] ∙ 18
What is 54
6 2/5 + 11 1/6
What is 17 17/30
9 1/3 - 7 4/9
What is 1 8/9
2 3/4 * 3 1/6
What is 8 17/24
9 4/5 ÷ 1 1/13
What is 9 1/10
10.5 ÷ 0.37
What is 28.38 | 2019-09-22 13:01:09 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8004457354545593, "perplexity": 12871.83666357397}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-39/segments/1568514575513.97/warc/CC-MAIN-20190922114839-20190922140839-00396.warc.gz"} |
https://www.vedantu.com/question-answer/a-machine-was-sold-on-a-hirepurchase-system-on-class-9-maths-icse-5f5cf99a9427543f91faffb2 | # A machine was sold on a hire-purchase system on ${{1}^{st}}$ March, 2015, Rs. 9000 was paid at spot and rest was paid by four equal quarterly installments of 11000 each. The cash price of the machine was Rs. 50000. Find out the amount of interest included in each installment?
Verified
149.1k+ views
Hint: We first find the total amount to purchase the machine by adding all installments and spot cash. We then find the total interest by subtracting cash price from total amount to purchase. We use the fact that the ratio of amount of interest included in ‘n’ each installment (if all the installments are equal) is n:n-1:….:3:2:1.
Given that we have a machine that was sold on a hire-purchase system on ${{1}^{st}}$ March, 2015. The cash price of the machine is given as Rs. 50000. Rs. 9000 was paid at spot and rest was paid by four quarterly installments of 11000 each. We need to find the amount of interest included in each installment.
Let us first find the purchase amount on the hire-purchase system.
Purchase amount = total amount paid to buy a machine.
So, purchase amount = $RS.\left( 9000+11000+11000+11000+11000 \right)$.
Purchase amount = Rs. 53000.
We can calculate the interest by subtracting cash price from purchase amount.
Interest = purchase amount – cash price.
Interest = Rs. 53000 – Rs. 50000.
Interest = Rs. 3000.
This interest is to be paid in 4 installments. It will be paid in the ratio of 4:3:2:1 for the ${{1}^{st}}$, ${{2}^{nd}}$, ${{3}^{rd}}$ and ${{4}^{th}}$ installments.
Share of the ${{1}^{st}}$ installment in interest = $3000\times \dfrac{4}{4+3+2+1}$.
Share of the ${{1}^{st}}$ installment in interest = $3000\times \dfrac{4}{10}$.
Share of the ${{1}^{st}}$ installment in interest = Rs. 1200.
Share of the ${{2}^{nd}}$ installment in interest = $3000\times \dfrac{3}{4+3+2+1}$.
Share of the ${{2}^{nd}}$ installment in interest = $3000\times \dfrac{3}{10}$.
Share of the ${{2}^{nd}}$ installment in interest = Rs. 900.
Share of the ${{3}^{rd}}$ installment in interest = $3000\times \dfrac{2}{4+3+2+1}$.
Share of the ${{3}^{rd}}$ installment in interest = $3000\times \dfrac{2}{10}$.
Share of the ${{3}^{rd}}$ installment in interest = Rs. 600.
Share of the ${{4}^{th}}$ installment in interest = $3000\times \dfrac{1}{4+3+2+1}$.
Share of the ${{4}^{th}}$ installment in interest = $3000\times \dfrac{1}{10}$.
Share of the ${{4}^{th}}$ installment in interest = Rs. 300.
∴ The amount included in ${{1}^{st}}$, ${{2}^{nd}}$, ${{3}^{rd}}$ and ${{4}^{th}}$ installments are Rs. 1200, Rs. 900, Rs. 600, Rs. 300.
Note: We can solve amount of interest included alternatively as follows:
Unpaid amount at ${{1}^{st}}$ installment = Rs. 53000 – Rs. 9000.
Unpaid amount at ${{1}^{st}}$ installment = Rs. 44000.
Unpaid amount at ${{2}^{nd}}$ installment = Rs. 44000 – Rs. 11000.
Unpaid amount at ${{2}^{nd}}$ installment = Rs. 33000.
Unpaid amount at ${{3}^{rd}}$ installment = Rs. 33000 – Rs. 11000.
Unpaid amount at ${{3}^{rd}}$ installment = Rs. 22000.
Unpaid amount at ${{4}^{th}}$ installment = Rs. 22000 – Rs. 11000.
Unpaid amount at ${{4}^{th}}$ installment = Rs. 11000.
Sum of unpaid amounts in all installments = Rs. 44000 + Rs. 33000 + Rs. 22000+ Rs. 11000.
Sum of unpaid amounts in all installments = Rs. 110000.
Total interest = Rs. 3000.
We know that share of interest = $\dfrac{\text{Total interest }\!\!\times\!\!\text{ unpaid amount at that installment}}{\text{total sum of unpaid amount in installments}}$.
Share of the ${{1}^{st}}$ installment in interest = $\dfrac{3000\times 44000}{110000}$.
Share of the ${{1}^{st}}$ installment in interest = $\dfrac{3000\times 4}{10}$.
Share of the ${{1}^{st}}$ installment in interest = Rs. 1200.
Share of the ${{2}^{nd}}$ installment in interest = $\dfrac{3000\times 33000}{110000}$.
Share of the ${{2}^{nd}}$ installment in interest = $3000\times \dfrac{3}{10}$.
Share of the ${{2}^{nd}}$ installment in interest = Rs. 900.
Share of the ${{3}^{rd}}$ installment in interest = $\dfrac{3000\times 22000}{110000}$.
Share of the ${{3}^{rd}}$ installment in interest = $3000\times \dfrac{2}{10}$.
Share of the ${{3}^{rd}}$ installment in interest = Rs. 600.
Share of the ${{4}^{th}}$ installment in interest = $\dfrac{3000\times 11000}{110000}$.
Share of the ${{4}^{th}}$ installment in interest = $3000\times \dfrac{1}{10}$.
Share of the ${{4}^{th}}$ installment in interest = Rs. 300. | 2022-01-27 15:29:10 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8307406902313232, "perplexity": 5146.077326288845}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320305266.34/warc/CC-MAIN-20220127133107-20220127163107-00622.warc.gz"} |
https://datascience.stackexchange.com/questions/69338/building-a-tag-based-recommendation-engine-given-a-set-of-user-tags | # Building a tag-based recommendation engine given a set of user tags?
Basically, the idea is to have users following tags on the site, so each users has a set of tags they are following. And then there is a document collection where each document in the collection has a Title, Description, and a set of tags which are of relevance to the topic being discussed in the document as determined by the author. What is the best way of recommending documents to the user, given the information we have, which would also take into consideration the semantic relevance of the title and description of a document to the user's tags, whether that be a word embeddings solution or a tf-idf solution, or a mix, do tell. I still don't know what I'm going to do about tag synonyms, it might have to be a collaborative effort like on stackoverflow, but if there is a solution to this or a pseudo-solution, and I'm writing this in C# using the Lucene.NET library, if that is of any relevance to you.
If I got your problem description correct, you are looking for a recommender system like for example used by Netflix or Amazon. State of the art solution would be to use Latent Dirichlet Allocation topic modeling to make recommendations based on topics (in your case, topics would be the tags). Here is a very good video tutorial on this topic: https://youtu.be/3mHy4OSyRf0
In the case of the standard version of LDA, you don't even have to define the tags, you just define a value of different tags among all your documents. If you have for example 10000 documents and you want to use 100 different tags, the method will transform your words/documents matrix into a topics/documents matrix.
The entries of the words/documents matrix are simply all documents as columns and all words (from all your documents) as rows, then for each document you have the counts of each word.
The entries of the topics/documents matrix are all documents as columns and all possible topics as rows, then for each document you have entries like 78% topic1, 12.5% topic95, 0% topic99 on each topic.
Once you have this data and you want to recommend a new document to a user based on his interests(tags), or in other words you have a user_interests vector $$\vec{u}$$ with 100 entries which have values between 0 and 1, and you have topics/documents matrix $$M_{{topics}\times{documents}}$$ you calculate a new matrix by multiplaying $$M_{{topics}\times{documents}}*\vec{u}$$, from this matrix you calculate the sum from each row and recommend those documents with the highest sum.
If you just want to use predefined tags, you can skip the step where you use the LDA method to calculate the topics/documents matrix and simply use your data to represent your documents as tags/documents matrix and your users as tag_vectors, proceeding from here the same as above: multiplying the matrix with a user_vector, calculating the sum from each row and recommending the documents with the highest sum.
• One option as you say is to generate topics from the document collection, in which case each document consists of a Title and Description, then generating some user_interests vector based on likes and/or past history. This way also allows for the possibility of recommending users topics to follow much like pinterest. Also since LDA doesn’t give you back a topic but rather a group of words corresponding to a certain topic, would it be wise to train a word2vec model in order to get some average vector of all the words and call that the topic? also how does LDA deal with synonymous topics? – Wasiim Ouro-sama Mar 11 '20 at 22:14
• also what do you think about lda2vec? – Wasiim Ouro-sama Mar 12 '20 at 18:29
• There are many modified versions of LDA available as of today, I'm only familar with the base method as described in the David M. Blei, Andrew Y. Ng, Michael I. Jordan; 3(Jan):993-1022, 2003 paper here: web.archive.org/web/20120501152722/http://jmlr.csail.mit.edu/… I saw a paper once of a modified version which was able to use a data base for finding exact tags for each topic, depending on the words constellations of the words represented by the topic. – Eugen Mar 12 '20 at 23:26
• I really like LDA2Vec and the kind of statistics I can capture with it, but my only worry is having to retrain these word embeddings because of my corpus constantly changing(new chats being added), word embeddings seem to work well for more static corpuses i.e netflix movies or spotify songs, but I'm not sure how I would go about it in a situation like this. Any ideas? – Wasiim Ouro-sama Mar 25 '20 at 15:25
One solution could be to train a single embedding space, StarSpace is one such implementation. That single embedding space would contain all users, documents, and tags. Then it is a nearest neighbor search to recommend any combination. Given a user, find the nearest documents. Given a tag, find the nearest tags …
For new entities (i.e., users, documents, or tags), split the individual entity into parts. For example, a document will have tokens or tag will be associated with documents that have tokens. Then find the average embedding of all the parts. That location in the embedding space is the approximately semantic meaning of new entities.
Overall, this is a complex, open-end problem, thus there are many possible ways to create useful solutions. The next best solutions depend on what has already been done and what the next feature that would add the most value to end-users. | 2021-04-19 21:54:38 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 3, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4459046423435211, "perplexity": 792.4395841835869}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-17/segments/1618038917413.71/warc/CC-MAIN-20210419204416-20210419234416-00484.warc.gz"} |
https://www.homeworklib.com/question_710411_98-select-the-descriptor-a-b-c-d-that-best-describes-the-relationship-between-the | # (9-8) Select the descriptor (A, B, C, D): that best describes the relationship between the...
Select the descriptor (A, B, C, D): that best describes the relationship between the "reactants" and "products" below:
IOption D
is the descriptor
Option C
Concepts and reason
Descriptors: Descriptors are the symbols used to show the transformations from one to another. Descriptors are usually used between reactants and products or between two different structures of the same compound.
Fundamentals
Different types of descriptors are shown below:
A. Double headed arrow: Double headed arrow is shown between the two canonical structures (resonance structures).
B. Equilibrium symbol: Equilibrium symbol, which favors a forward reaction.
C. Equilibrium symbol: Equilibrium symbol which favors a reverse reaction.
D. Equilibrium symbol: Equilibrium symbol, which favors neither forward nor reverse reaction.
Equilibrium nature: Equilibrium shifts towards less acidic substances that are from lower ${\rm{P}}{{\rm{k}}_{\rm{a}}}$ to higher ${\rm{P}}{{\rm{k}}_{\rm{a}}}$of the substances.
Option C
## General guidance
Concepts and reason
Descriptors: Descriptors are the symbols used to show the transformations from one to another. Descriptors are usually used between reactants and products or between two different structures of the same compound.
Fundamentals
Different types of descriptors are shown below:
A. Double headed arrow: Double headed arrow is shown between the two canonical structures (resonance structures).
B. Equilibrium symbol: Equilibrium symbol, which favors a forward reaction.
C. Equilibrium symbol: Equilibrium symbol which favors a reverse reaction.
D. Equilibrium symbol: Equilibrium symbol, which favors neither forward nor reverse reaction.
Equilibrium nature: Equilibrium shifts towards less acidic substances that are from lower ${\\rm{P}}{{\\rm{k}}_{\\rm{a}}}$ to higher ${\\rm{P}}{{\\rm{k}}_{\\rm{a}}}$of the substances.
First Step | All Steps | Answer Only
## Step-by-step
### Step 1 of 2
(A)
The given reaction condition is drawn below:
The reaction is drawn below with the best descriptor which represents the reaction.
Part A
Explanation | Common mistakes | Hint for next step
It is clear that the given reaction represents acid-base reaction where Pyridinium ion acts as acid while pipyridine acts as base. In this reaction, Pyridinium ion gives proton to pipyridine and produces stable pyridine and protonated pipyridine. This reaction equilibrium favor a forward reaction due to the less acidic nature of products than reactants.
### Step 2 of 2
(B)
The given reaction condition is drawn below:
The reaction is drawn below with the best descriptor which represents the reaction.
Part B
Explanation | Common mistakes
It is clear that the given reaction represents an acid base reaction where alkyne derivative acts as acid while n-BuLi acts as base. In the reaction, alkyne derivative gives proton to n-BuLi and produces stable butane. This reaction equilibrium favors a forward reaction due to the less acidic nature of products than reactants. Butane has higher${\\rm{P}}{{\\rm{k}}_{\\rm{a}}}$than terminal alkyne derivative.
Part A
Part B
##### Add Answer of: (9-8) Select the descriptor (A, B, C, D): that best describes the relationship between the...
More Homework Help Questions Additional questions in this topic.
• #### Study the following sentence, then select the answer that best describes it
Need Online Homework Help? | 2019-06-17 06:54:45 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 5, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5218703746795654, "perplexity": 5392.745489617025}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-26/segments/1560627998440.47/warc/CC-MAIN-20190617063049-20190617085049-00510.warc.gz"} |
http://gmatclub.com/forum/are-the-positive-integers-x-and-y-consecutive-145327.html?fl=similar | Find all School-related info fast with the new School-Specific MBA Forum
It is currently 25 Apr 2015, 19:25
### 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 June
Open Detailed Calendar
# Are the positive integers x and y consecutive?
Author Message
TAGS:
Manager
Status: Never ever give up on yourself.Period.
Joined: 23 Aug 2012
Posts: 147
Location: India
Concentration: Finance, Human Resources
Schools: MBS '17 (A)
GMAT 1: 570 Q47 V21
GMAT 2: 690 Q50 V33
GPA: 3.5
WE: Information Technology (Investment Banking)
Followers: 8
Kudos [?]: 145 [0], given: 35
Are the positive integers x and y consecutive? [#permalink] 06 Jan 2013, 07:04
4
This post was
BOOKMARKED
00:00
Difficulty:
45% (medium)
Question Stats:
45% (02:33) correct 55% (01:08) wrong based on 58 sessions
Are the positive integers x and y consecutive?
(1) x^2 - y^2 = 2y + 1
(2) x^2 - xy - x = 0
[Reveal] Spoiler: OA
_________________
Don't give up on yourself ever. Period.
Beat it, no one wants to be defeated (My journey from 570 to 690) : beat-it-no-one-wants-to-be-defeated-journey-570-to-149968.html
Verbal Forum Moderator
Status: Preparing for the another shot...!
Joined: 03 Feb 2011
Posts: 1425
Location: India
Concentration: Finance, Marketing
GPA: 3.75
Followers: 142
Kudos [?]: 765 [0], given: 62
Re: Are the positive integers x and y consecutive? [#permalink] 06 Jan 2013, 10:00
Expert's post
daviesj wrote:
Are the positive integers x and y consecutive?
(1)$$x^2 - y^2 = 2y + 1$$
(2) $$x^2 - xy - x = 0$$
The question is basically asking whether $$x=y+1$$
Statement 1)
$$x^2 - y^2=2y+1$$ can be written as $$(x+y)(x-y)=2y+1$$.----equation 1
If we put $$x=y+1$$, then LHS must be equal to RHS.
Equation 1 can be written, after substituting x=y+1, as $$(2y+1)(1)=2y+1$$. They are equal. Hence x and y are consecutive.
Statement 2)
$$x(x-y-1)=0$$
The above equation can be equal to 0 only when $$x-y=1$$ because x is given to be positive.
Hope I am correct.
+1D
_________________
Current Student
Joined: 27 Jun 2012
Posts: 418
Concentration: Strategy, Finance
Followers: 56
Kudos [?]: 442 [2] , given: 182
Re: Are the positive integers x and y consecutive? [#permalink] 06 Jan 2013, 22:24
2
KUDOS
You can simplify the first statement in following way:
(1) $$x^2-Y^2=2y+1$$
i.e. $$x^2 = Y^2 + 2y+1$$
i.e. $$x^2 = (y+1)^2$$
i.e. $$x=|y+1| = y + 1$$(taking positive root as both x & y are positive integers)
Hence (1) is SUFFICIENT as x & y are consecutive
(2) $$x^2-xy-x=0$$
i.e. $$x(x-y-1)=0$$
i.e. $$x=0$$ or $$x=y+1$$
As x is positive integer x<>0, thus $$x=y+1$$
Hence (2) is SUFFICIENT.
_________________
Thanks,
PraPon
Tough 700+ Level RCs: Passage1 | Passage2 | Passage3 | Passage4 | Passage5 | Passage6 | Passage7
VOTE: vote-best-gmat-practice-tests-excluding-gmatprep-144859.html
PowerScore CR Bible - Official Guide 13 Questions Set Mapped: Click here
Math Expert
Joined: 02 Sep 2009
Posts: 27056
Followers: 4184
Kudos [?]: 40359 [0], given: 5420
Re: Are the positive integers x and y consecutive? [#permalink] 07 Jan 2013, 02:55
Expert's post
daviesj wrote:
Are the positive integers x and y consecutive?
(1) x^2 - y^2 = 2y + 1
(2) x^2 - xy - x = 0
Similar question to practice: if-x-y-and-z-are-positive-integers-where-x-y-and-z-x-126859.html
_________________
GMAT Club Legend
Joined: 09 Sep 2013
Posts: 4667
Followers: 290
Kudos [?]: 52 [0], given: 0
Re: Are the positive integers x and y consecutive? [#permalink] 30 Sep 2014, 23:21
Hello from the GMAT Club BumpBot!
Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
Re: Are the positive integers x and y consecutive? [#permalink] 30 Sep 2014, 23:21
Similar topics Replies Last post
Similar
Topics:
1 x and y are consecutive positive integers and: \left{ 7 25 Oct 2008, 20:16
x and y are consecutive positive integers and x>y. X^2 - 9 06 Nov 2007, 21:49
x and y are consecutive positive integers and x>y . Which 9 12 Oct 2007, 23:23
x and y are consecutive positive integers and x>y x^2 5 24 Nov 2006, 22:28
If x, y, and z are 3 positive consecutive integers such that 3 23 Sep 2006, 04:50
Display posts from previous: Sort by | 2015-04-26 03:25:23 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5191707015037537, "perplexity": 7012.76676749005}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-18/segments/1429246652296.40/warc/CC-MAIN-20150417045732-00160-ip-10-235-10-82.ec2.internal.warc.gz"} |
https://www.flyingcoloursmaths.co.uk/ask-uncle-colin-troublesome-triangle/ | Dear Uncle Colin,
I couldn’t make head nor tail of this geometry problem: “If $a:b=12:7$, $c=3$, and $B\hat{A}C = 2 B\hat{C}A$, find the length of the sides $a$ and $b$.”
- Totally Rubbish In Geometry
Hi, TRIG, and thank you for your message! (And don’t put yourself down like that, it doesn’t help.)
I would probably start by letting $a=12x$ and $b=7x$, and seeing what came out of the equations. (I’d also let angle $B\hat{A}C=2\alpha$ and $B\hat{C}A=\alpha$, meaning $A\hat{B}C=\pi - 3\alpha$ - if I can avoid using that, I will.)
We have $\frac{\sin\br{\alpha}}{3}=\frac{\sin\br{2\alpha}}{12x}$
Multiplying around, that becomes $12x \sin(\alpha) = 3\sin(2\alpha)$, or $2x=\cos(\alpha)$ after using the double-angle formula. (For completeness: I’ve discarded the possibility that $\sin(\alpha)=0$ as that wouldn’t make for much of a triangle.)
### How about the cosine rule?
We could probably bring in the other angle with the sine rule, but I don’t really fancy messing around with $\sin(3\alpha)$ right now.
Instead, let’s look at the cosine rule. I reckon $3^2 = (12x)^2 + (7x)^2 - 2(12x)(7x)\cos(\alpha)$.
Simplify that a bit to get $9 = 193x^2 - 168x^2 \cos(\alpha)$.
However, we know $\cos(\alpha)=2x$, so that’s $9 = 193x^2 - 336x^3$.
It turns out that $336x^3 - 193x^2 + 9$ factorises as $(16x+3)(7x-3)(3x-1)$, which gives $x = \frac{-3}{16}$, $x=\frac{3}{7}$ or $x=\frac{1}{3}$.
### Three possibilities? Let’s check them!
Since $12x$ is a distance, the first solution is invalid.
The second solution gives $a=\frac{36}{7}$ and $b=3$, which seems superficially plausible but in fact doesn’t work: it gives an equilateral triangle, with the apex double the size of the two base angles. However, the only triangle that fits that description is (don’t tell the Ninja) the 90-45-45 triangle. This triangle clearly isn’t that! ((What’s happened here? I think the apex angle is very large, and is such that its sine is equal to the sine of double each base angle - but the curse of many-to-one functions is that that doesn’t mean the apex angle is double the base angle!))
So we’re left with the last one, with $a=4$ and $b=\frac{7}{3}$.
Hope that helps!
- Uncle Colin | 2021-08-02 21:32:27 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9460846185684204, "perplexity": 539.8514700450799}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-31/segments/1627046154385.24/warc/CC-MAIN-20210802203434-20210802233434-00084.warc.gz"} |
https://vivekg.dev/posts/spectral-markov/index.html | Vivek Gopalakrishnan
# Some Spectral Properties of Markov Chains
2021-11-10 | The power of Gershgorin and his Circles
Markov chains are stoachastic models for describing discrete sequences of random behavior. While I normally dislike probability theory because I find it incredibly confusing, I like Markov chains because they can be expressed using linear algebra (which I find much less confusing!). This post explores the spectral properties of Markov chains[1] and relates this to the limiting probabilistic behaviors of Markov chains.
Three main results will be proved in this post (still need to prove #4):[2]
1. The eigenvalues of a Markov chain are bounded within the unit circle
2. The number of stationary distributions a Markov chain has is related to the geometric multiplicity of $\lambda=1$
3. A large class of Markov chains have a unique stationary distribution
4. If a Markov chain is not diagonalizable, its stationary distributions are calculable using the Jordan Canonical Form
[1] The spectrum of a matrix is the set of its eigenvalues. Spectral properties are then the properties of the eigenvalues that give rise to special behavior.
[2] Said more precisely, Result 2 really characterizes the dimension of the space of stationary distributions.
## Gershgorin and His Circles
A Markov chain describes the probability of transitioning from one state to another. These transition probabilities can be summarized using a Markov matrix:
Definition 1 (Markov matrix). A matrix $M \in M_n([0, 1])$ is a Markov matrix if the entries of each column sum to 1. Said another way, if $i$ and $j$ are states in a Markov chain, then $\mathbb P( j \mapsto i ) = M_{ij}$
The eigenvalues of a Markov matrix have a remarkable property: they're never bigger than 1 in magnitude! To prove this, we need an underrated result from linear algebra:
Theorem 1 (Gershgorin disc theorem). For the $i$-th row of a matrix $A$,
let the row sum $R_i = \sum_{j \neq i} A_{ij}$ be the radius of the disc centered at $A_{ii}$: $G_i(A) = \{ z \in \mathbb C : | z - A_{ii} | \leq R_i \}$. Let the union of these discs be $G(A) = \cup_{i=1}^n G_i(A) \,.$ Then, $\sigma(A) \subseteq G(A) \,.$[3]
Proof: A good explanation is avilable on Wikipedia.
[3] Note that $\sigma(A) = \{\lambda \in \mathbb C : Ax = \lambda x\}$ represents the set of all eigenvalues of $A$.
We can augment the Gershgorin disc theorem with a quick lemma connecting the eigenvalues of $A$ and $A^*$:
Lemma 1. Given a matrix $A \in M_n$, $\sigma(A) = \sigma(A^*)$.
Proof: The lemma holds since $A$ and $A^*$ have the same characteristic polynomial: $\det(A^*- \lambda I) = \det((A - \lambda I)^*) = \det(A - \lambda I) \,.$
Thanks to Lemma 1, we can equivalently define $R_i$ in the Gershgorin disc theorem to be the column sum, $\sum_{i \neq j} A_{ij}$, instead of the row sum. Now, the first main result follows:
Corollary 1 (Spectrum of Markov matrices). The eigenvalues of a Markov matrix $M \in M_n([0, 1]) \,,$ are bounded within the unit circle.[4]
Proof: Since the columns of $M$ must sum to $1$, this means that for every column $i$, $M_{ii} + R_i = 1$. Therefore, the disc $G_i(M)$ is a subset of the unit circle that also intersects the unit circle at $(1, 0)$.
Therefore, by Theorem 1, $\sigma(M) \subseteq G(M) \subseteq \{ z \in \mathbb C : |z| \leq 1 \}$.
Here is an illustration of this proof for a Markov matrix in $M_4([0,1])$. Interestingly, this picture also shows us that the only way we can have an complex eigenvalue with magnitude 1 is if one of the $M_{ii} = 0$... more on this later.
[4] Said mathematically, $\sigma(M) \subseteq \{ z \in \mathbb C : |z| \leq 1 \} \,.$
Next, we ask what is the largest eigenvalue of a Markov matrix?
Corollary 2 (Spectral radius of Markov matrices). The spectral radius of a Markov matrix is $\rho(M) = 1$ and $\rho(M) \in \sigma(M)$.
Proof: Let $\vec{1} \in \mathbb R^n$ be a column vector of all 1s. We can then write the column sum constraint as $M^*\vec{1} = \vec{1}$. Note that this equation shows that $\lambda = 1$ is an eigenvalue of $M^*$ (and by Lemma 1, also an eigenvalue of $M$). Because the eigenvalues of $M$ are constrained within the unit circle by Corollary 1, $\lambda = 1$ achieves the maximum possible modulus.
Therefore, $\rho(M) = 1$ and $\rho(M) \in \sigma(M)$.
## Enumerating the Stationary Distributions
Remark: Gershgorin gives us a very powerful sets of results! They show that if our Markov matrix is diagonalizable (i.e., $M = SDS^{-1}$ for $S \in M_n$ invertible and $D \in M_n$ diagonal), then $M^k = SD^kS^{-1}$ will converge to some matrix as $k \to \infty$, and the steady state for any initial condition will be determined by the eigenvectors of $M$ whose eigenvalues equal the spectral radius.[5]
[5] What happens if $M$ is not diagonalizable? In this case, we have to use the Jordan Canonical Form (JCF) of $M$ (more on this later).
For a cool application of this convergence idea, let's consider a symmetric Markov chain (i.e., forwards and backwards probability are equal). The spectral theorem tells us that the eigenvectors of this matrix are orthogonal. Additionally, assume that the geometric multiplicity of the eigenvector $1$ is $k$ for this example. Then,
\begin{aligned} \lim_{k \rightarrow \infty} M^k &= \lim_{k \rightarrow \infty} SD^kS^{-1} \\ &= S (I_k \oplus \mathbf 0_{n-k}) S^{-1} \\ &= \begin{pmatrix} s_1 \cdots s_k & \vec 0 \cdots \vec 0 \end{pmatrix} \begin{pmatrix} s_1 & \cdots & s_n \end{pmatrix}^* \\ &= \sum_{i=1}^k s_i s_i^* \,. \end{aligned}
Since the eigenvectors of $M$ are orthonormal, note that $s_i s_i^*$ is a matrix representing the orthogonal projection onto $s_i$. Thus, $P_{S_k} = \sum_{i=1}^k s_i s_i^*$ is a projection matrix onto the span of $\{s_1, \dots, s_k\}$. Finally, we can see that for some initial condition $x_0$,
$\lim_{k \rightarrow \infty} M^k x_0 = P_{S_k} x_0 \,,$
meaning the stationary distributions of $M$ are an orthogonal projection onto the eigenspace of $\lambda = 1$. Said another way, any convex combination of the eigenvectors of $M$ with eigenvalue $1$ is a valid stationary distribution.
So, in general, when does $M$ have a unique stationary distribution (i.e., when is the geometric multiplicity of $1$ equal to $1$)?
## The Perron–Frobenius Theorem
Definition 3 (Positive and non-negative matrices). A matrix $A \in M_n(\mathbb R)$ is positive (written $A > 0$) if all entries of $A$ are positive. Similarly, if all the entries of $A$ are non-negative, then $A$ is non-negative as well (i.e., $A \geq 0$).
The Perron–Frobenius Theorem gives many results about positive matrices. To study Markov matrices, we focus on one of them:
Theorem 2 (Perron–Frobenius). If $A \in M_n(\mathbb R)$ is positive, then $\rho(A)$ is an eigenvalue of $A$ with algebraic multiplicity 1.
Corollary 3. Theorem 2 still applies if $A$ is non-negative and irreducible.[6]
[6] Irreducibility is related to the concept of the connectedness of a matrix. Specifically, matrices can be represented as entry digraphs (take the matrix, binarize it, and treat that as the adjacency matrix of a directed graph). If the graph is strongly connected (i.e., each vertex is connected to every other vertex by a directed edge), then it is irreducible.
By definition, Markov matrices are non-negative. If a Markov matrix is irreducible, then the Perron–Frobenius theorem says that $\rho(M)=1$ is a simple eigenvalue of $M$. Since the geometric multiplicity of an eigenvalue is bounded above by its algebraic multiplicity, that means that there is only a single eigenvector $x$ such that $Ax = x$. This vector $x$, scaled so that its entries sum to 1, is the unique steady state distribution of $M$.
The defintion of irreducibility also helps us understand the result proved in the previous section. If a matrix is reducible, there are multiple strongly connected components. These are communities in the graph that, once you walk into, you cannot walk out of.[7] Such a community would have an eigenvector with non-zero entry for each vertex in the strongly connected component and $0$'s for every other vertex, and this eigenvector would be a stationary distribution. Each strongly connected component would have its own such eigenvector, and their convex hull would be the space of all possible stationary distributions. This is exactly what we proved in the previous section, but thanks to this spectral graph theory perspective, we now we don't need $M$ to be symmetric to prove it!
[7] You can check out any time you like, but you can never leave.
## Diagonalization with the Jordan Canonical Form
The past results have assumed that $M$ is diagonalizable. What if this is not true?
I'm not sure how the math works out really, but I'll endeavor to prove it some day :)
These will slowly be addressed in future versions of this post.
• How do we characterize stationary oscillations with spectral methods?
• How does this connect to deterministic discrete linear dynamic systems? | 2022-10-04 15:49:00 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 89, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9862117171287537, "perplexity": 203.91670195812804}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030337516.13/warc/CC-MAIN-20221004152839-20221004182839-00591.warc.gz"} |
http://snakesonabrain.com/docs/groundhog/piles/cavityexpansion.html | # Cavity expansion methods for cast in-situ piles¶
groundhog.deepfoundations.boreholestability.cavityexpansion.expansion_cylinder_tresca(insitu_pressure, borehole_pressure, diameter, undrained_shear_strength, shear_modulus, poissons_ratio=0.5, max_radius_multiplier=10.0, number_radii=250, **kwargs)[source]
Calculates the cavity expansion for a cylinder in Tresca material. The relation between borehole radius + plastic radius and pressure differential is calculated first.
This relation is then used to evaluate where the imposed pressure lies, whether it causes any plasticity around the borehole and whether it does not cause overall borehole failure.
The stresses for the given pressure are then calculated.
Parameters
• insitu_pressure – Isotropic horizontal stress in the soil mass before borehole excavation ($$p_0$$) [$$kPa$$] - Suggested range: insitu_pressure >= 0.0
• borehole_pressure – Pressure on the borehole wall due to drilling fluids or concrete ($$p$$) [$$kPa$$] - Suggested range: borehole_pressure >= 0.0
• diameter – Borehole initial diameter (equations are formulated in terms of radius but diameter is more convenient as input) ($$2 \cdot a_0$$) [$$m$$] - Suggested range: diameter >= 0.0
• undrained_shear_strength – Undrained shear strength of the material surrounding the borehole ($$S_u$$) [$$kPa$$] - Suggested range: undrained_shear_strength >= 0.0
• shear_modulus – Shear modulus of the material surrounding the borehole ($$G$$) [$$kPa$$] - Suggested range: shear_modulus >= 0.0
• poissons_ratio – Poissons ratio of the material surrounding the borehole (default for undrained material)) ($$\nu$$) [$$-$$] - Suggested range: 0.0 <= poissons_ratio <= 0.5 (optional, default= 0.5)
• max_radius_multiplier – Multiplier on borehole radius for determining the maximum extent of the calculation (:math:) [$$-$$] - Suggested range: max_radius_multiplier >= 1.0 (optional, default= 10.0)
• number_radii – Number of radii considered for the calculation (:math:) [$$-$$] (optional, default= 250)
\begin{align}\begin{aligned}\text{Elastic properties}\\n = \frac{4 \cdot S_u \cdot (1 - \nu^2) }{E}\\E = 2 \cdot G \cdot (1 + \nu)\\\text{Plastic radius and wall radius expansion during plastic deformation}\\\left( \frac{c}{a} \right)^2 = \left( \frac{a_0}{a} \right)^2 + \frac{1}{n} \cdot \left[ 1 - \left( \frac{a_0}{a} \right)^2 \right]\\\frac{p - p_0}{2 \cdot S_u} = \frac{1}{2} + \frac{1}{2} \cdot \ln \left[ \frac{G}{S_u} \cdot \left( 1 - \left( \frac{a_0}{a} \right)^2 \right) + \left( \frac{a_0}{a} \right)^2 \right]\\\text{Elastic stresses and displacements can be calculated by taking the limit for the outer radius going to infinity}\\\sigma_r = - p_0 - (p - p_0) \cdot \frac{\frac{b_0^2}{r^2} - 1}{\frac{b_0^2}{a_0^2} - 1} \implies \sigma_r = -p_0 - (p - p_0) \cdot \frac{a_0^2}{r^2}\\\sigma_{\theta} = -p_0 + (p - p_0) \cdot \frac{\frac{b_0^2}{r^2} + 1}{\frac{b_0^2}{a_0^2} - 1} \implies \sigma_{\theta} = -p_0 + (p - p_0) \cdot \frac{a_0^2}{r^2}\\u = \frac{(1 + \nu) \cdot (p - p_0)}{E} \cdot \frac{a_0^2}{b_0^2 - a_0^2} \cdot \left[ (1 - 2 \cdot \nu) \cdot r + \frac{b_0^2}{r} \right] \implies u = \frac{(1 + \nu) \cdot (p - p_0)}{E} \cdot \frac{a_0^2}{r}\\\text{The plasticity critertion can be expressed as:}\\\sigma_{r,r=a_0} - \sigma_{theta,r=a_0} = 2 \cdot S_u \implies -(p - p_0) = S_u\\\text{The elastic stresses and displacements outside of the plastic zone can be written as:}\\\sigma_r = -\frac{S_u \cdot c^2}{b_0^2} \cdot \left( \frac{b_0^2}{r^2} - 1 \right) - p_0 \implies \sigma_r = -\frac{S_u \cdot c^2}{r^2} - p_0\\\sigma_{\theta} = \frac{S_u \cdot c^2}{b_0^2} \cdot \left( \frac{b_0^2}{r^2} + 1 \right) - p_0 \implies \sigma_{\theta} = \frac{S_u \cdot c^2}{r^2} - p_0\\u = \frac{(1 + \nu) \cdot S_u \cdot c^2}{E \cdot b_0^2} \cdot \left[ (1 - 2 \cdot \nu) \cdot r + \frac{b_0^2}{r} \right] \implies u = \frac{(1 + \nu) \cdot S_u \cdot c^2}{E} \cdot \frac{1}{r}\\\text{Stresses in the plastic zone:}\\\sigma_r = -p_0 - S_u - 2 \cdot S_u \cdot \ln \left( \frac{c}{r} \right)\\\sigma_{\theta} = -p_0 + S_u - 2 \cdot S_u \cdot \ln \left( \frac{c}{r} \right)\end{aligned}\end{align}
Returns
Dictionary with the following keys:
• ’yielding’: Boolean determining whether plastic deformation is taking place or not
• ’pressure expansion function’: Dictionary with the pressure-expansion relation with keys expansion [m] and pressure difference [kPa]
• ’yielding pressure [kPa]’: Borehole pressure at which yield occurs [$$kPa$$]
• ’radii [m]’: Numpy array with radii used for the stress and displacement calculation [$$m$$]
• ’radial stresses [kPa]’: Numpy array with the radial stresses around the borehole [$$kPa$$]
• ’tangential stresses [kPa]’: Numpy array with the tangential stresses around the borehole [$$kPa$$]
• ’elastic wall expansion [m]’: Borehole elastic wall expansion [$$m$$]
• ’plastic wall expansion [m]’: Borehole plastic wall expansion ($$a - a_0$$) [$$m$$]
• ’plastic radius [m]’: Radius of the plastic zone ($$c$$) [$$m$$]
Reference - Yu, H.-S., 2000. Cavity Expansion Methods in Geomechanics. Springer-Science+Business Media, B.V.
groundhog.deepfoundations.boreholestability.cavityexpansion.expansion_tresca_thicksphere(undrained_shear_strength, internal_radius, external_radius, internal_pressure, external_pressure, youngs_modulus, poissons_ratio, seed=100, **kwargs)[source]
Calculates the stresses for cavity expansion around a thick-walled sphere in Tresca material.
The plastic radius is first calculated from the pressure boundary conditions. Using this plastic radius, the stresses in the elastic and plastic region are calculated.
Parameters
• undrained_shear_strength – Undrained shear strength of the material surrounding the spherical cavity ($$S_u$$) [$$kPa$$] - Suggested range: undrained_shear_strength >= 0.0
• internal_radius – Initial internal radius of the spherical cavity ($$a_0$$) [$$m$$] - Suggested range: internal_radius >= 0.0
• external_radius – Initial external radius of the region ($$b_0$$) [$$m$$] - Suggested range: external_radius >= 0.0
• internal_pressure – Internal pressure applied on the inside of the sphere ($$p$$) [$$kPa$$] - Suggested range: internal_pressure >= 0.0
• external_pressure – External pressure on the outside of the sphere ($$p_0$$) [$$kPa$$] - Suggested range: external_pressure >= 0.0
• youngs_modulus – Young’s modulus of the material ($$E$$) [$$kPa$$] - Suggested range: youngs_modulus >= 0.0
• poissons_ratio – Poisson’s ratio of the material ($$\nu$$) [$$-$$] - Suggested range: 0 <= poissons_ratio <= 0.5
• seed – Number of radii at which stresses and displacements are calculated (:math:) [$$-$$] (optional, default= 100)
\begin{align}\begin{aligned}\text{Elastic solutions}\\\sigma_r = -p_0 - (p - p_0) \frac{\left( \frac{b_0}{r} \right)^3 - 1}{\left( \frac{b_0}{a_0} \right)^3 - 1}\\\sigma_{\theta} = \sigma_{\phi} = -p_0 + (p - p_0) \frac{\frac{1}{2} \left( \frac{b_0}{r} \right)^3 - 1}{\left( \frac{b_0}{a_0} \right)^3 - 1}\\u = r - r_0 = \frac{p - p_0}{E} \frac{(1 - 2 \nu) r + \frac{(1 + \nu) b_0^3}{2 r^2}}{\left( \frac{b_0}{a_0} \right)^3 - 1}\\\text{Yield criterion}\\\sigma_1 - \sigma_3 = 2 \cdot S_u\\\text{Internal pressure for yielding } (\sigma_1 = \sigma_{\theta}, \sigma_3 = \sigma_r)\\p = p_{1y} = p_0 + \frac{4 \cdot S_u}{3} \left[ 1 - \left( \frac{a_0}{b_0} \right)^3 \right]\\\text{Displacement at internal and external boundaries}\\u |_{r=a_0} = \frac{2 \cdot S_u \cdot a_0}{E} \left[ \frac{2 \cdot (1 - 2 \nu) \cdot a_0^3 }{3 \cdot b_0^3} + \frac{1 + \nu}{3} \right]\\u |_{r=b_0} = \frac{2 \cdot S_u \cdot (1 - \nu) \cdot a_0 }{E \cdot b_0^2}\\\text{Stresses and displacements in the elastic region after yielding}\\\sigma_r = - \frac{4 \cdot S_u \cdot c^3}{3 \cdot b_0^3} \left[ \left( \frac{b_0}{r} \right)^3 - 1 \right] - p_0\\\sigma_{\theta} = \sigma_{\phi} = \frac{4 \cdot S_u \cdot c^3}{3 \cdot b_0^3} \left[ \frac{1}{2} \left( \frac{b_0}{r} \right)^3 + 1 \right] - p_0\\u = \frac{4 \cdot S_u \cdot c^3}{3 \cdot E \cdot b_0^3} \left[ (1 - 2 \nu) \cdot r + \frac{(1 + \nu) b_0^3}{2 \cdot r^2} \right]\\\text{Stresses in the plastic region}\\\sigma_r = - 4 \cdot S_u \cdot \ln \left( \frac{c}{r} \right) - \frac{4 \cdot S_u}{3} \left[ 1 - \left( \frac{c}{b_0} \right)^3 \right] - p_0\\\sigma_{\theta} = 2 \cdot S_u - 4 \cdot S_u \cdot \ln \left( \frac{c}{r} \right) - \frac{4 \cdot S_u}{3} \cdot \left[ 1 - \left( \frac{c}{b_0} \right)^3 \right] - p_0\\\text{The pressure required to generate a plastic radius is thus}\\p = 4 \cdot S_u \cdot \ln \left( \frac{c}{a} \right) + \frac{4 \cdot S_u}{3} \left[ 1 - \left( \frac{c}{b_0} \right)^3 \right] + p_0\\\text{Expansion of the boundary}\\\left( \frac{a}{a_0} \right)^3 = 1 + \frac{6 (1 - \nu) S_u c^3}{E \cdot a_0^3} - \frac{4 (1 - 2 \nu) S_u}{E} \left[ 3 \ln \left( \frac{c}{a_0} \right) + 1 - \left( \frac{c}{b_0} \right)^3 \right]\end{aligned}\end{align}
Returns
Dictionary with the following keys:
• ’elastic_radial_stress [kPa]’: Radial stresses for purely elastic deformation ($$\sigma_{r,elastic}$$) [$$kPa$$]
• ’elastic_tangential_stress [kPa]’: Tangential stresses for purely elastic deformation ($$\sigma_{\theta,elastic}$$) [$$kPa$$]
• ’elastic_radial_displacement [m]’: Radial displacement for purely elastic deformation ($$u_{elastic}$$) [$$m$$]
• ’yielding_pressure [kPa]’: Internal pressure which initiates yield ($$p_{1y}$$) [$$kPa$$]
• ’plastic_radius [m]’: Radius of the plastic zone for the given internal and external pressures ($$c$$) [$$m$$]
• ’elastoplastic_radii [m]’: Radii at which elastoplastic stresses are calculated (equal amount of point inside and outside the plastic radius)
• ’elastoplastic_radial_stress [kPa]’: Radial stresses for Tresca soil ($$\sigma_r$$) [$$kPa$$]
• ’elastoplastic_tangential_stress [kPa]’: Tangential stresses for Tresca soil ($$\sigma_{\theta}$$) [$$kPa$$]
• ’expanded_radius [m]’: Radius of the internal wall after expansion ($$a$$) [$$m$$]
Reference - Yu, H.-S., 2000. Cavity Expansion Methods in Geomechanics. Springer-Science+Business Media, B.V.
groundhog.deepfoundations.boreholestability.cavityexpansion.stress_cylinder_elastic_isotropic(radius, internal_pressure, farfield_pressure, borehole_radius, shear_modulus=nan, **kwargs)[source]
Calculates the radial and tangential stress around a cylindrical borehole under internal pressure, in a soil mass with isotropic virgin stress conditions in the given plane
Parameters
• radius – Radius or radii at which to calculate the stresses (float or NumPy array) ($$r$$) [$$m$$] - Suggested range: radius >= 0.0
• internal_pressure – Internal pressure on the borehole ($$p$$) [$$kPa$$] - Suggested range: internal_pressure >= 0.0
• farfield_pressure – Far-field pressure, equal to the virgin horizontal stress ($$p_0$$) [$$kPa$$] - Suggested range: farfield_pressure >= 0.0
• borehole_radius – Radius of the borehole ($$a$$) [$$m$$] - Suggested range: borehole_radius >= 0.0
• shear_modulus – Shear modulus used to calculate the radial displacement ($$G$$) [$$kPa$$] - Suggested range: shear_modulus >= 0.0 (optional, default=np.nan). If unspecified, radial displacements are not calculated
\begin{align}\begin{aligned}\frac{d \sigma_{r}}{dr} + \frac{\left(\sigma_{r} - \sigma_{\theta} \right)}{r} = 0\\\begin{split}\sigma_r | _{r=a} = p \\ \sigma_r | _{r=a} = p_0\end{split}\\\sigma_{r} = p_0 + \left( p - p_0 \right) \cdot \left( \frac{a}{r} \right)^2\\\sigma_{\theta} = p_0 - \left( p - p_0 \right) \cdot \left( \frac{a}{r} \right)^2\end{aligned}\end{align}
Returns
Dictionary with the following keys:
• ’radial stress [kPa]’: Radial stress at the specified radii ($$\sigma_r$$) [$$kPa$$]
• ’tangential stress [kPa]’: Tangential stress at the specified radii ($$\sigma_{\theta}$$) [$$kPa$$]
• ’radial displacement [m]’: Radial displacement (calculated if shear modulus is specified) ($$u$$) [$$m$$]
Basic sketch of the cavity expansion problem
Reference - Yu, H.-S., 2000. Cavity Expansion Methods in Geomechanics. Springer-Science+Business Media, B.V. | 2021-06-20 12:43:23 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 3, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 1.0000030994415283, "perplexity": 13059.29026555308}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-25/segments/1623487662882.61/warc/CC-MAIN-20210620114611-20210620144611-00164.warc.gz"} |
https://tex.stackexchange.com/questions/231738/example-images-in-latex?noredirect=1 | Example images in LaTeX?
How to make an example image in LaTeX, a dummy image, a spot holder? I have seen it before, but I can't remember how they would made. It would be nice to be able to use example images - if you should help anyone or you need to question in this community. You might say I could easily Google it, but I tried - and are obviously too stupid to find anything.
\documentclass[12pt,a4paper]{article}
\usepackage[utf8]{inputenc}
\usepackage[ngerman, danish]{babel}
\usepackage[T1]{fontenc}
%\usepackage[]{}
\begin{document}
\iffalse
\begin{figure}
\includegraphics[scale=1]{•}
\end{figure}
\fi
\end{document}
Is it also possible to resize the example image? So I need a usepackage to make an image without including an immage. An placeholder so anyone can try my code without needing images - just needing my tex-code. I hope you understand?
Kind regards!
• Related: New support package for MWEs – Werner Mar 6 '15 at 19:04
• If I google "dummy images latex", I get the mwe package documentation as the 4th result. – 1010011010 Mar 6 '15 at 19:09
• Sounds like you want placeholder images, not examples, right? – Cole Johnson Mar 7 '15 at 2:15
Thanks to Martin Scharrer, modern LaTeX systems offer you some ready-to-use images; the documentation for the mwe package (Section 4 Provided Images) describes all available images (initially you had to load to package to use the images, but then the images were made usable without the package):
\documentclass[12pt,a4paper]{article}
\usepackage{graphicx}
\begin{document}
\includegraphics[width=3cm]{example-grid-100x100pt}
\noindent\includegraphics[height=5cm]{example-image-b}
\noindent\includegraphics[scale=0.5]{example-image-c}
\noindent\includegraphics[width=3cm]{example-image}
\end{document}
• It seams example-image stops at c. Is it possible to get example-image-d? – Viesturs Oct 5 '18 at 11:45
• @Viesturs There is example-image-duck nowadays :) – TeXnician Oct 9 '18 at 15:33
The documentation folder of PStricks contains some images in a separate folder. Most notably it holds tiger (in both EPS and PDF formats):
\documentclass{article}
\usepackage{graphicx}
\begin{document}
\includegraphics[width=150pt]{c:/texlive/2014/texmf-dist/doc/generic/pstricks/images/tiger}
\end{document}
As can be seen above (under TeX Live 2014 in Windows) it's not that readily accessible. | 2019-08-19 00:22:13 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6936270594596863, "perplexity": 2513.0030036237445}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-35/segments/1566027314353.10/warc/CC-MAIN-20190818231019-20190819013019-00381.warc.gz"} |
https://questioncove.com/updates/4f8f9090e4b000310fadd945 | OpenStudy (anonymous):
Susan has a jar that contains 6 pink marbles and 7 purple marbles.She randomly draws another marbles.What is the probability of drawing 2 purple marbles if the marbles we drawn. A. Without replacement? B.With replacement?
5 years ago
OpenStudy (anonymous):
I am not sure if I remember this, its been a while. so it maybe a bit rusty a) $7/15 * 6/14$ b) $7/15 * 7/15$
5 years ago
OpenStudy (anonymous):
Thank you
5 years ago | 2017-11-18 00:32:28 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.793808102607727, "perplexity": 3892.4695129909064}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-47/segments/1510934804125.49/warc/CC-MAIN-20171118002717-20171118022717-00008.warc.gz"} |
https://www.gradesaver.com/textbooks/math/calculus/university-calculus-early-transcendentals-3rd-edition/chapter-6-practice-exercises-page-390/2 | ## University Calculus: Early Transcendentals (3rd Edition)
$\dfrac{ 8 \sqrt 3}{15}$
We need to integrate the integral as shown below: Area of each cross-section triangle, $Area=\dfrac{(2 \sqrt x-x)^2 \sin 60^{\circ}}{y}=\dfrac{\sqrt 3 }{4} (2 \sqrt x-x)^2$ $I= \int_{0}^{4} [\dfrac{\sqrt 3 }{4} (2 \sqrt x-x)^2] dx \\= \dfrac{\sqrt 3}{4}\int_{0}^{4} [4x-4x \sqrt x+x^2] dx \\=\dfrac{ 8 \sqrt 3}{15}$ | 2019-12-07 04:04:44 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9936474561691284, "perplexity": 1839.6570500343812}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-51/segments/1575540495263.57/warc/CC-MAIN-20191207032404-20191207060404-00382.warc.gz"} |
https://socratic.org/questions/how-do-you-simplify-3sqrt7-12 | # How do you simplify (3sqrt7)/12?
Jun 22, 2018
$\frac{\sqrt{7}}{4}$ (see explanation)
#### Explanation:
You just need to find the common factor of the whole numbers in the numerator and denominator.
In the numerator, we have 3 and 1 as the factors of 3, since 3 multiplied to 1 is equal to 3
.
While the denominator have the factors 4 and 3.
$\frac{3 \cdot 1 \sqrt{7}}{4 \cdot 3}$
As we can observe, both whole numbers had 3 as their common factor. Because of this, the 3 must be cancelled in order to obtain the simplified answer, which is: $\frac{\sqrt{7}}{4}$ | 2021-06-24 12:26:33 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 3, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9507662653923035, "perplexity": 380.67794156329694}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-25/segments/1623488553635.87/warc/CC-MAIN-20210624110458-20210624140458-00006.warc.gz"} |
https://stats.stackexchange.com/questions/459190/are-all-estimators-biased-is-the-unbiasedness-only-a-theoretical-or-approximati | # Are all estimators biased? Is the unbiasedness only a theoretical or approximation case?
The definition of unbiased estimator says that it's expected value has no difference comparing to a true value. So can we say that all estimators are biased (even slightly)? I thought that only in terms of infinite sampling we can converge to a true value, hence in this conditions estimators are not biased. Is the term unbiased estimator just a generalization or some real scenario issue? Why I see many times that people are calling estimators used in a statistical inference - unbiased. Is it just a approximation isue? What about convergence to a true value and infinite variance of population? Any help would highly appreciated!
As you said, unbiasedness is a theoretical property of an estimator because the expected value operation is theoretical, which means it doesn't depend on the sample size. So, an estimator is either biased or unbiased and doesn't change its state according to $$n$$.
I thought that only in terms of infinite sampling we can converge to a true value
I'd also like to comment on this one:
For example, let $$\hat \theta_n$$ be the n-sample estimator of parameter $$\theta$$. If $$\mathbb E[\hat \theta_n]=\theta$$, the estimator is unbiased and this means
• as $$n$$ increases, you get close to true $$\theta$$
• when averaged among all n-sample estimates, the estimator will converge to the true $$\theta$$ (this statement is the implication of the expected value)
Note that, the first one can also be satisfied by unbiased estimators, but the second cannot. Take for example, $$\mathbb E[\hat\theta_n]=\frac{n}{n-1}\theta$$. Then,
$$\lim_{n\rightarrow\infty}\mathbb E[\hat\theta_n]=\lim_{n\rightarrow\infty}\frac{n}{n-1}\theta=\theta$$
So, as the sample size, $$n$$, increases, the unbiased estimator converges to the true parameter.
• Thank you for the response. So can ever the estimator's expected value meet the "true" true value (except when the sample is complete)? Or we just call the estimator unbiased when it's arbitraly close to the true value? For example does the BLUE estimator really gives us the "true" true value or it just close enough to the true value so we can call it unbiased? – Tom Apr 8 at 12:06
• an unbiased estimator's expected value is always equal to the true parameter, however you won't be able to measure it because $\mathbb E[.]$ is not an operation performed over the data. It's purely theoretical. – gunes Apr 8 at 12:14
• So, an estimator is either biased or unbiased and doesn't change its state according to n This seems very strong, might be true in some cases, but certainly the bias can depend on $n$. – kjetil b halvorsen Apr 8 at 14:08
• Indeed, @kjetilbhalvorsen , the bias in my second example is a function of $n$. – gunes Apr 8 at 14:16 | 2020-07-12 07:50:20 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 10, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7373251914978027, "perplexity": 507.00920499387354}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-29/segments/1593657131734.89/warc/CC-MAIN-20200712051058-20200712081058-00117.warc.gz"} |
https://plainmath.net/16428/related-functions-described-chapter-describe-sequence-transformations | Question
# h is related to one of the parent functions described in this chapter. Describe the sequence of transformations from f to h.h(x)=−√x+4
Transformations of functions
h is related to one of the parent functions described in this chapter. Describe the sequence of transformations from f to h.
$$\displaystyle{h}{\left({x}\right)}=−\sqrt{x+4}$$
We are starting with the parent function $$f(x)=\sqrt{x}$$
STEP 1: Reflect the graph across x-axis, to get $$y=-\sqrt{x}$$
STEP 2: Shift the graph by 4 units upwards, to get $$y=-\sqrt{x+4}$$, which is the required function h(x) | 2021-08-03 17:54:39 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7538996338844299, "perplexity": 457.2386382921443}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-31/segments/1627046154466.61/warc/CC-MAIN-20210803155731-20210803185731-00512.warc.gz"} |
http://physics.stackexchange.com/tags/general-physics/new | # Tag Info
0
$$F/(ma) = 1$$ $$E/(mc^2) = 1$$ $$U/(RI) = 1$$ etc, I think you get the point :) Probably you ask this, because many equations are preferred in the form "= 0". Why this is advantageous is probably what you wanted... or should have wanted to ask, right?
2
Sure. For example: $$\varepsilon_0\cdot \mu_0 \cdot c^2 = 1$$ with the electric and magnetic field constants and the speed of light.
-1
Many of the answers given above are wrong. The word "causal" is not subjective. One of the answers includes the analogy: if I (the charge) throw a ball (the electric field) at a lamp, was the cause of the lamp breaking (the change in the magnetic field) the ball or me? But this is not a good analogy because the two cases are not at all similar. If a ...
0
If we define some event which we will call "p" to be the cause and "q" to be its effect, then p and q should satisfy the following rules, 1.p implies q but q doesn't imply p. 2.in the absense of p, q shouldn't exist either. 3.p and q shouldn't be simultaneous events as viewed from any inertial frame of reference.
0
Ambiguous is the direction of the magnetic field lines. Somehow ambiguous is the direction of the electric current too because there are a technical direction from plus to minus or the real direction of the negative charged electrons from minus to plus. So one has to use the right hand rule or the left hand rule. More important is the fact that - if the ...
1
The convention is that the direction of the magnetic dipole is determined from the direction of the current using the right hand rule. For example, looking down on a loop with a clockwise current means that the magnetic moment is downwards.
Top 50 recent answers are included | 2016-05-04 11:59:26 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7704693078994751, "perplexity": 349.153522045195}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-18/segments/1461860123023.37/warc/CC-MAIN-20160428161523-00140-ip-10-239-7-51.ec2.internal.warc.gz"} |
https://stacks.math.columbia.edu/tag/03JX | Lemma 67.5.1. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. Consider the following conditions on $X$:
• $(\alpha )$ For every $x \in |X|$, the equivalent conditions of Lemma 67.4.2 hold.
• $(\beta )$ For every $x \in |X|$, the equivalent conditions of Lemma 67.4.3 hold.
• $(\gamma )$ For every $x \in |X|$, the equivalent conditions of Lemma 67.4.5 hold.
• $(\delta )$ The equivalent conditions of Lemma 67.4.6 hold.
• $(\epsilon )$ The equivalent conditions of Lemma 67.4.7 hold.
• $(\zeta )$ The space $X$ is Zariski locally quasi-separated.
• $(\eta )$ The space $X$ is quasi-separated
• $(\theta )$ The space $X$ is representable, i.e., $X$ is a scheme.
• $(\iota )$ The space $X$ is a quasi-separated scheme.
We have
$\xymatrix{ & (\theta ) \ar@{=>}[rd] & & & & \\ (\iota ) \ar@{=>}[ru] \ar@{=>}[rd] & & (\zeta ) \ar@{=>}[r] & (\epsilon ) \ar@{=>}[r] & (\delta ) \ar@{=>}[r] & (\gamma ) \ar@{<=>}[r] & (\alpha ) + (\beta ) \\ & (\eta ) \ar@{=>}[ru] & & & & }$
Proof. The implication $(\gamma ) \Leftrightarrow (\alpha ) + (\beta )$ is immediate. The implications in the diamond on the left are clear from the definitions.
Assume $(\zeta )$, i.e., that $X$ is Zariski locally quasi-separated. Then $(\epsilon )$ holds by Properties of Spaces, Lemma 65.6.6.
Assume $(\epsilon )$. By Lemma 67.4.7 there exists a Zariski open covering $X = \bigcup X_ i$ such that for each $i$ there exists a scheme $U_ i$ and a quasi-compact surjective étale morphism $U_ i \to X_ i$. Choose an $i$ and an affine open subscheme $W \subset U_ i$. It suffices to show that $W \to X$ has universally bounded fibres, since then the family of all these morphisms $W \to X$ covers $X$. To do this we consider the diagram
$\xymatrix{ W \times _ X U_ i \ar[r]_-p \ar[d]_ q & U_ i \ar[d] \\ W \ar[r] & X }$
Since $W \to X$ factors through $X_ i$ we see that $W \times _ X U_ i = W \times _{X_ i} U_ i$, and hence $q$ is quasi-compact. Since $W$ is affine this implies that the scheme $W \times _ X U_ i$ is quasi-compact. Thus we may apply Morphisms, Lemma 29.56.9 and we conclude that $p$ has universally bounded fibres. From Lemma 67.3.4 we conclude that $W \to X$ has universally bounded fibres as well.
Assume $(\delta )$. Let $U$ be an affine scheme, and let $U \to X$ be an étale morphism. By assumption the fibres of the morphism $U \to X$ are universally bounded. Thus also the fibres of both projections $R = U \times _ X U \to U$ are universally bounded, see Lemma 67.3.3. And by Lemma 67.3.2 also the fibres of $R \to X$ are universally bounded. Hence for any $x \in X$ the fibres of $|U| \to |X|$ and $|R| \to |X|$ over $x$ are finite, see Lemma 67.3.6. In other words, the equivalent conditions of Lemma 67.4.5 hold. This proves that $(\delta ) \Rightarrow (\gamma )$. $\square$
Comment #696 by Simon Pepin Lehalleur on
There are two numberings in the lemma, one 1-9 and one $\alpha-\ldots$
Comment #697 by on
That is actually a bug of the Stacks website code, see also https://github.com/stacks/stacks-website/issues/59. I will take care of this soon.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.
In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 03JX. Beware of the difference between the letter 'O' and the digit '0'. | 2023-02-07 00:58:16 | {"extraction_info": {"found_math": true, "script_math_tex": 1, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 2, "x-ck12": 0, "texerror": 0, "math_score": 0.9659504294395447, "perplexity": 312.1757341100153}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764500368.7/warc/CC-MAIN-20230207004322-20230207034322-00376.warc.gz"} |
https://questions.examside.com/past-years/jee/question/pif-the-equation-of-the-parabola-whose-vertex-is-at-5-4-jee-main-mathematics-trigonometric-functions-and-equations-cs2pmvxozldrwvjy | 1
JEE Main 2022 (Online) 27th June Evening Shift
+4
-1
If the equation of the parabola, whose vertex is at (5, 4) and the directrix is $$3x + y - 29 = 0$$, is $${x^2} + a{y^2} + bxy + cx + dy + k = 0$$, then $$a + b + c + d + k$$ is equal to
A
575
B
$$-$$575
C
576
D
$$-$$576
2
JEE Main 2022 (Online) 27th June Morning Shift
+4
-1
Let the eccentricity of an ellipse $${{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1$$, $$a > b$$, be $${1 \over 4}$$. If this ellipse passes through the point $$\left( { - 4\sqrt {{2 \over 5}} ,3} \right)$$, then $${a^2} + {b^2}$$ is equal to :
A
29
B
31
C
32
D
34
3
JEE Main 2022 (Online) 26th June Evening Shift
+4
-1
If m is the slope of a common tangent to the curves $${{{x^2}} \over {16}} + {{{y^2}} \over 9} = 1$$ and $${x^2} + {y^2} = 12$$, then $$12{m^2}$$ is equal to :
A
6
B
9
C
10
D
12
4
JEE Main 2022 (Online) 26th June Evening Shift
+4
-1
The locus of the mid point of the line segment joining the point (4, 3) and the points on the ellipse $${x^2} + 2{y^2} = 4$$ is an ellipse with eccentricity :
A
$${{\sqrt 3 } \over 2}$$
B
$${1 \over {2\sqrt 2 }}$$
C
$${1 \over {\sqrt 2 }}$$
D
$${1 \over 2}$$
JEE Main Subjects
Physics
Mechanics
Electricity
Optics
Modern Physics
Chemistry
Physical Chemistry
Inorganic Chemistry
Organic Chemistry
Mathematics
Algebra
Trigonometry
Coordinate Geometry
Calculus
EXAM MAP
Joint Entrance Examination | 2023-03-23 04:42:44 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.898947536945343, "perplexity": 915.4537304990852}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296944996.49/warc/CC-MAIN-20230323034459-20230323064459-00320.warc.gz"} |
https://mitchellkember.com/mcv4u/cross-product.html | ## Cross product
The cross product is an operation that takes two nonzero vectors and produces a vector (not a scalar) perpendicular to both of them. Geometrically, we define the magnitude of the cross product with
|vec u xx vec v| = |vec u||vec v|sin theta,
where theta is the angle between the vectors in standard position. To find the direction of the vector, we use the right-hand rule: point in the direction of vec u with your fingers, and then curl them (naturally, not backwards) towards the direction of vec v. Your thumb will then point in the direction of vec u xx vec v.
Algebraically, we define the cross product in RR^3 with
[u_1,u_2,u_3] xx [v_1,v_2,v_3] = [w_1,w_2,w_3],
where the components are given by
Component Value Remember
w_1 u_2v_3 - u_3v_2 2, 3
w_2 u_3v_1 - u_1v_3 3, 1
w_3 u_1v_2 - u_2v_1 1, 2
The pattern shouldn’t be too hard to see. Each part has a product of two terms, and then you just subtract the same two terms with the subscripts swapped. You essentially need to remember the sequence 233112, and even that has some repetition. You do not need to remember anything for RR^2, since the cross product exists only in RR^3 and in RR^7.
Like the dot product, the cross product has some important properties. First, it can produce the zero vector:
vec v xx vec 0 = vec 0 qquad and qquad vec v xx vec v = vec 0.
Unlike the dot product, is is anticommutative—order matters:
vec u xx vec v = -(vec u xx vec v).
vec u xx (vec v + vec w) = vec u xx vec v + vec u xx vec w.
It is associative over scalar multiplication:
kvec u xx vec v = k(vec u xx vec v) = vec u xx kvec v.
We can use the dot product and cross product together to make a test for coplanarity. Here it is: vectors vec u, vec v, and vec w are coplanar if and only if
vec u * vec v xx vec w = 0.
It does not matter which vectors are placed where in that equation. As long as crossing two of the vectors and dotting the result with the third yields zero, they are coplanar. Why does this work? When we cross vec v and vec w, we get a vector perpendicular to both of them. If vec u is on the same plane, it should also be perpendicular to the cross product. We test for this using the dot product—two vectors are perpendicular if their dot product is zero.
The notation vec u * vec v xx vec w is unambiguous; it is the same as vec u * (vec v xx vec w). If you dotted first, you would then be crossing a scalar with a vector, and that is not possible. | 2020-01-27 06:59:24 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.45967432856559753, "perplexity": 1392.794469985575}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-05/segments/1579251694908.82/warc/CC-MAIN-20200127051112-20200127081112-00141.warc.gz"} |
https://nanosync.eu/5m54gq/2e20ce-harmonic-motion-meaning-in-urdu | Kitchenaid Refrigerator Water Filter Reset, Barstow Ca Mugshots, Evolution Fresh Orange Juice Review, Ronseal Laminate Floor Repair, Zeus Arc Gt Best Price, Red Lotus Chapel Hill, American Board Of Endodontics Verification, Online Jobs For College Students At Home, " />
1 Krzemowa Street, 62-002 Złotniki, Suchy Las, Poland +48 734 188 729
# harmonic motion meaning in urdu
Latin words for harmonic include conveniens, congruens, unanimans, unanimus, concors, consentiens, canorus, consonus and harmonicus. We also provide student loans to assist in the education of our members. Definition of subharmonic in the Definitions.net dictionary. Here, you can check सुसंगत is also written as Susaṃgat in Roman. Need to translate "simple harmonic motion" to Hindi? Simple Harmonic Motion, Mass Spring System - Amplitude, Frequency, ... SHM and Uniform Circular Motion in Urdu Sabaq Foundation 131 - Duration: 10:50. सुसंगत Meaning in English is Harmonic. It results in an oscillation which, if uninhibited by friction or any other dissipation of energy, continues indefinitely. harmonic mean sa Urdu Ingles - diksyonaryo Urdu. This word is written in Roman Urdu. #simple harmonic motion full detail video in Urdu. Here, you can check Musiqi Ka Ek Sar موسیقیاتی حرکت. meaning in different languages. Class 10 Physics Notes - Chapter 10 - Simple Harmonic Motion and Waves - Important MCQs for FBISE Papersthat contain important MCQs of the chapter. Harmonic How to say simple harmonic motion in English? You will find all the latest relevant news and update about the Alipore Muslim Association of South Africa. Submit the origin and/or meaning of Colt to us below. If you are looking for Susaṃgat Meaning in English, you are at the right place. Harmonic motion refers to the motion an oscillating mass experiences when the restoring force is proportional to the displacement, but in opposite directions. Simple Harmonic Motion concept and derivation using Mass Attached to spring. People often want to translate English words or phrases into Urdu. Thanks for using this online dictionary, we have been helping millions of people improve their use of … Harmonic, Harmonic Motion and Harmonic Analysis. The oscillation of a body in which its acceleration is proportional to its displacement and is directed to a fixed point. What does subharmonic mean? What does simple harmonic motion mean? An oscillator that has the same period regardless of amplitude. You have searched the English word Damped which means “ریڈیو” radio in Urdu.Damped meaning in Urdu has been searched 597 (five … Harmonic definition: Harmonic means composed , played , or sung using two or more notes which sound right and... | Meaning, pronunciation, translations and examples Simple Harmonic Motion:- It is the simplest type of oscillatory motion. simple harmonic motion - tamil meaning of சீரிசை இயக்கம். Meaning of harmonic minor scale. All time best and simple solved numerical problem for 10th class students. You have searched the English word The guitar string is an example of simple harmonic motion, or SHM. Chapter 11 – Sound. Chapter 10 – Simple Harmonic Motion & Waves. Simple harmonic motion is any motion where a restoring force is applied that is proportional to the displacement and in the opposite direction of … Search meanings in Urdu to get the better understanding of the context. Information and translations of harmonic motion in the most comprehensive dictionary definitions resource on the web. Get Class 10th Physics Video Lectures Online Free for Pakistani Students. Harmonic Meaning in Urdu Urdu meaning of Harmonic is موسیقی کا ایک سر, it can be written as Musiqi Ka Ek Sar in Roman Urdu. Shm meaning in Urdu: ایس-ایچ-ایممُخفّف "simple harmonic motion""swedish house mafia""supporting healthy marriage" - - meaning, Definition Synonyms at English to Urdu dictionary gives you the best and accurate Urdu translation and meanings of Shm, - Meaning. In addition to it, the knowledge about the origin, pronunciation, and synonyms of a word allows them to find similar words or phrases. In mechanics and physics, simple harmonic motion is a special type of periodic motion where the restoring force on the moving object is directly proportional to the object's displacement magnitude and acts towards the object's equilibrium position. Harmonic is a adjective by form. We are a non-profit organisation which is involved in numerous welfare and social responsibility schemes in the muslim community. Ham Ahang Though I have been a tactic of the natural frequency of simple harmonic oscillators. Here, k/m = ω 2 (ω is the angular frequency of the body). (in team sports…. What does harmonic motion mean? The Official Website of the Alipore Muslim Association of South Africa. Check out Harmonic similar words like Harmonic, Harmonica and Harmonicon; Harmonic Urdu Translation is … In the age of digital communication, any person should learn and understand multiple languages for better communication. Harmonic Hz away. 1856 (one thousand eight hundred and fifty-six) Let us learn more about it. Oscillating motion. times till Foal Meaning In Urdu. Harmonic Motion The Urdu Word ہم آہنگ Meaning in English is Harmonic. Harmonic motion is periodic and can be represented by a sine wave with constant frequency and amplitude. SHM can be seen throughout nature. Here's how you say it. Meaning of simple harmonic motion. Simple Harmonic Motion. moseeqiyati harkat. Harmonic Meaning In Hindi. However, it will allow you to learn the appropriate use of Harmonic Motion in a sentence. It results in an oscillation which, if uninhibited by friction or any other dissipation of energy, continues indefinitely. ... Film Flick Motion Picture Motion-Picture Show Movie Moving Picture Moving-Picture Show Pic Picture Picture Show: متحرک فلم Mutaharik Film: a form of entertainment that enacts a story by sound and a sequence of images giving the illusion of continuous movement. times till meaning in Urdu has been searched Meaning of subharmonic. synonym words moseeqiyati harkat 10th Class Physics Solved Numerical. Translation is "Ham Ahang" harmonic mean noun + gramatika (mathematics) A type of measure of central tendency calculated as the reciprocal of the mean of the reciprocals, ie, H = { n \over {1 \over x_1} + {1 \over x_2} + \cdots {1 \over x_n} } We are a non-profit organisation which is involved in numerous welfare and social responsibility schemes in the muslim community. Harmonic motion definition: a periodic motion in which the displacement is symmetrical about a point or a periodic... | Meaning, pronunciation, translations and examples Simple harmonic motion definition is - a harmonic motion of constant amplitude in which the acceleration is proportional and oppositely directed to the displacement of the body from a position of equilibrium : the projection on any diameter of a point in uniform motion around a circle. simple harmonic motion, Marathi translation of simple harmonic motion, Marathi meaning of simple harmonic motion, what is simple harmonic motion in Marathi dictionary, simple harmonic motion related Marathi | मराठी words Definition of harmonic motion in the Definitions.net dictionary. Pronunciation of Harmonic Motion in roman Urdu is "moseeqiyati harkat" and Translation of Vibratory motion in hindi/urdu physics crash course #127 youtube. Hindi words for harmonic include सुरीला, लयबद्ध, समस्वर, अनुरूप and एकताल संबंधी. Definition of harmonic motion. The other similar words are Hum Aahang and Mawafiq. Colt meaning in Urdu has been searched 13914 (thirteen thousand nine hundred and fourteen) times till Nov 25, 2020. Definition of simple harmonic motion in the Definitions.net dictionary. this video makes you clear the basic cony about simple pendulum#simple harmonic motion,#physics#oscillation In the age of digital communication, any person should learn and understand multiple languages for better communication. For a mass-spring system undergoing simple harmonic motion, the frequency of the oscillations can be found using the equation. Motion – the physics hypertextbook. #simple harmonic motion full detail video in Urdu. Simple Harmonic Motion concept and derivation using Mass Attached to spring. You have searched the English word The synonyms and antonyms of Harmonic are listed below. In the simple harmonic motion, the displacement of the object is always in the opposite direction of the restoring force. Harmonic motion definition, periodic motion consisting of one or more vibratory motions that are symmetric about a region of equilibrium, as the motion of a … meaning in Urdu is Simple harmonic motion is any motion where a restoring force is applied that is proportional to the displacement and in the opposite direction of that displacement. HARMONIC MEANING IN URDU Harmonic meaning in Urdu is Ham Ahang - Synonyms and related Harmonic meaning is Consonant, Harmonical, Harmonized and Sympathetic. Learn the difference between Linear and Damped Simple Harmonic Motion here. Utilize the online English to Urdu dictionary to check the Urdu meaning of English word. Harmonic Meaning in Hindi is सुसंगत. Harmonic Motion A body in periodic motion is said to have undergone one cycle after passing through a series of events or positions and returning to its original state. However, it will allow you to learn the appropriate use of Harmonic in a sentence. What does harmonic minor scale mean? At least hours or two full days before the in meaning doing homework urdu last min. in Urdu writing script is Simple harmonic motion is typified by the motion of a mass on a spring when it is subject to the linear elastic restoring force given by Hooke's Law.The motion is sinusoidal in time and demonstrates a single resonant frequency. Harmonic ہم آہنگ. Harmonic distortion definition at Dictionary.com, a free online dictionary with pronunciation, synonyms and translation. Thanks for using this online dictionary, we have been helping millions of people improve their use of … Barhao Meaning from Urdu to English is Progression, and in Urdu it is written as بڑھاؤ. You can get more than one meaning for one word in Urdu. Concepts of Simple Harmonic Motion (S.H.M) Amplitude: The maximum displacement of a particle from its equilibrium position or mean position is its amplitude, and its direction is always away from the mean or equilibrium position. You will find all the latest relevant news and update about the Alipore Muslim Association of South Africa. also commonly used in daily talk like as Find more Hindi words at wordhippo.com! However, a person feels better to communicate if he/she has sufficient vocabulary. Jan 07, 2021. Simple harmonic motion, in physics, repetitive movement back and forth through an equilibrium, or central, position, so that the maximum displacement on one side of this position is equal to the maximum displacement on the other side. People often want to translate English words or phrases into Urdu. • Critical damping damped harmonic motion. damped harmonic oscillation meaning in Hindi with examples: अवंमदित सरल आवर्त दोलन अवमंदित सरल आवर्त ... click for more detailed meaning in Hindi with examples, definition, pronunciation and … Simple harmonic motion is the motion in which the object moves to and fro along a line. in Urdu.Harmonic Utilize the online English to Urdu dictionary to check the Urdu meaning of English word. Harmonic Look it up now! Harmonic Motion Chapter 13 – Electrostatics A pendulum in simple harmonic motion is called a simple pendulum. Learning Paradise 29,733 views. Harmonic Motion You can get more than one meaning for one word in Urdu. Here you can check all definitions and meanings of However, a person feels better to communicate if he/she has sufficient vocabulary. Find more Latin words at wordhippo.com! The example sentences play a good role in this regard. For people who can communicate in different languages the simplest type of oscillatory motion person better. People who can communicate in different languages talk like as Harmonic, Harmonic motion is periodic and can represented. For better communication difference between Linear and damped simple Harmonic motion refers to displacement. But in opposite directions you are at the right place words or into! And update about the Alipore Muslim Association of South Africa, if uninhibited by friction or any other dissipation energy. Definitions resource on the web oscillation of a spring have been a tactic of the object is in! Flu sentence in Urdu an oscillator that has the same of people improve their use of Harmonic minor in., k/m = ω 2 ( ω is the same period regardless of amplitude loans to assist in Definitions.net. Simple Harmonic motion in the modern world, there is a dire for. Understand multiple languages for better communication always in the most comprehensive dictionary definitions resource on the system but... Loans to assist in the most comprehensive dictionary definitions resource on the other side, you can Harmonic! Comprehensive dictionary definitions resource on the web age of digital communication, any person should learn and multiple. A sentence Urdu dictionary to check the Urdu word ہم آہنگ the latest relevant news and about... Provide student loans to assist in the most comprehensive dictionary definitions resource on the web have been tactic... Person feels better to communicate if he/she has sufficient vocabulary can get more than one meaning for one word Urdu... Helping millions of people improve their use of Harmonic in Urdu Urdu language the appropriate use …. English word motion concept and derivation using Mass Attached to spring to the... That has the same effect on the system harmonic motion meaning in urdu but it is the in... Is proportional to the equation and solve for frequency, york basic books weight bouncing on a.! Spring, etc for Pakistani Students place where something or someone is, often in relation to other:. Directed to a fixed point effect on the web plug these in harmonic motion meaning in urdu the displacement of Alipore! An oscillation which, if uninhibited by friction or any other dissipation of energy, continues indefinitely and freeman structural. The modern world, there is a dire need for people who can communicate different... Translation is … Harmonic meaning is Consonant, Harmonical, Harmonized and Sympathetic examples the! The system, but in opposite directions words like Harmonic, Harmonic motion is periodic and can be by... Angular frequency of the object is always in the most comprehensive dictionary definitions resource on the.. Harmonic oscillation damped motion How to say damped in Hindi and what is the simplest of. The equation and solve for frequency, dictionary definitions resource on the.. Meanings in Urdu spring constant ),, to be are also used in daily talk like as Harmonic Harmonic! M. Hannan and freeman, structural york basic books has been searched 13914 ( thirteen thousand hundred... With 1 audio pronunciation, synonyms and antonyms of Harmonic minor scale in the age digital... He/She has sufficient vocabulary our members on a spring similar words are also commonly used in daily talk as. Physics crash course # 127 youtube a Free online dictionary with pronunciation, synonyms and antonyms of motion! Online English to Urdu dictionary to check the Urdu meaning of colt to us below online... There is a dire need for people who can communicate in different languages something someone... Side, you can check all definitions and meanings of Harmonic motion with 1 audio pronunciation, 2 synonyms 1. Displacement of the Alipore Muslim Association of South Africa need to translate simple Harmonic motion concept and using..., the displacement of the context there is a weight bouncing on a.! Hindi words for Harmonic include सुरीला, लयबद्ध, समस्वर, अनुरूप and एकताल संबंधी using Mass Attached spring. Update about the Alipore Muslim Association of South Africa feels better to communicate if he/she has sufficient vocabulary similar of... Muslim community this regard to Hindi, but it is the angular frequency the... Simple solved Numerical problem for 10th Class Physics chapter simple Harmonic motion in the age of digital,! Disadvantages of synonyms and antonyms of Harmonic minor scale in the opposite direction of the Alipore Association! Is, often in relation to other things: 2 13 – Electrostatics the Official Website of the.. On a spring, etc is Ham Ahang '' and translation of Harmonic a! Of damped in Hindi and what is the check out Harmonic similar words like Harmonic, Harmonic motion in sentence... For using this online dictionary with pronunciation, synonyms and related Harmonic meaning is Consonant, Harmonical, and..., लयबद्ध, समस्वर, अनुरूप and एकताल संबंधी, we have been helping millions of people their. Understanding of the context energy, continues indefinitely information and translations of Harmonic refers! Online English to Urdu dictionary to check the Urdu meaning of English word meaning in Urdu,!, you are at the right place Lectures in Urdu to get the better understanding of the object is in! The difference between Linear and damped simple Harmonic motion '' to Hindi organisation. News and update about the Alipore Muslim Association of South Africa in a sentence always in the Definitions.net dictionary be., synonyms and antonyms of Harmonic minor scale in the simple Harmonic motion and Waves Video Lectures online Free Pakistani... Similar words of Harmonic in a sentence best and simple solved Numerical problem for 10th Class Physics chapter Harmonic... Class with complete clarification in Urdu as several English words are Hum and... You can get more than one meaning for one word in Urdu to get better! '' and translation of Harmonic are listed below Harmonic oscillation damped motion How to say damped in Hindi آہنگ in! With complete clarification in Urdu spring constant ),, to be 10th Physics Video Lectures in Urdu body... Welfare and social responsibility schemes in the education of our members a fixed point is to. Of simple Harmonic oscillators harkat '' and translation of Harmonic stated below the simplest of. Of a body in which its acceleration is proportional to the motion of a pendulum in simple Harmonic motion the!, pronunciation, synonyms and antonyms of Harmonic motion refers to the equation and solve for frequency, and simple... Oscillator that has the same effect on the web of digital communication, any person should and.: 1. the place where something or someone is, often in relation to other things: 2 harkat and! We were given the force constant ( or spring constant ),, to be kg! 13 – Electrostatics the Official Website of the context oscillation of a pendulum in simple Harmonic motion translation in Urdu! English to Urdu dictionary to check the Urdu word ہم آہنگ the force constant ( or spring constant ),. Pronunciation roman Urdu is Ham Ahang - synonyms and translation languages for better communication times till Nov 25,.. Or phrases into Urdu check out Harmonic similar words are Hum Aahang and Mawafiq bouncing on a spring,.! Its displacement and is directed to a fixed point people who can communicate in different languages force constant ( spring! Harmonic, Harmonic motion and Harmonic Analysis ; Harmonic Urdu translation is … Harmonic is... In Hindi English word between Linear and damped simple Harmonic motion refers to the displacement, but in directions. Fourteen ) times till Nov 25, 2020, synonyms and translation of Harmonic... 1. the place where something or someone is, often in relation to other things: 2 effect on web! Student loans to assist in the education of our members Harmonic similar words of.! Understand multiple languages for better communication also used in daily talk like as Harmonic, motion... | 2021-06-16 11:37:30 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.23554839193820953, "perplexity": 2454.1798812454945}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-25/segments/1623487623596.16/warc/CC-MAIN-20210616093937-20210616123937-00531.warc.gz"} |
https://www.itninja.com/question/msi-command-line-with-folders | Hi,
My MSI file is within a folder, within a folder and therfore I zip it up for upload into KACE. There is also other files in these folders which the MSI requires and so when creating a deployment in KACE, I need to tell it where the msi file is.
The instructions are not clear. For example if I was doing it manually I would go (this is an example): [msiexec /i "C:\installs\subfolder1\msi_file\program name.msi" /qn /l logfile.log]. But In KACE which part of that do I need to enter into the command line field? Because I've uploaded the zip file surely it doesn't need to refer to C:?
Hopefully I've given enough info. Even if someone replies and says "using the above example, you would enter it into KACE command like this this: XXX" that would be awesome.
Thanks.
Create a cmd file to trigger the msi commandline
Cancel
or Comment on this question for clarity
3
Create a CMD file with:
`msiexec /i "installs\subfolder1\msi_file\program name.msi" /qn /l logfile.log`
Then enter the name of that CMD file in your command line. Make sure to include the file in your ZIP. | 2018-05-25 22:44:53 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8501359820365906, "perplexity": 3157.719235840582}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-22/segments/1526794867220.74/warc/CC-MAIN-20180525215613-20180525235613-00038.warc.gz"} |
https://www.mysciencework.com/publication/show/sampling-probability-measures-convex-order-approximation-martingale-optimal-transport-problems-6231bb82?search=1 | # Sampling of probability measures in the convex order and approximation of Martingale Optimal Transport problems
Authors
Publication Date
Sep 18, 2017
Source
HAL-UPMC
Keywords
Language
English
Motivated by the approximation of Martingale Optimal Transport problems, westudy sampling methods preserving the convex order for two probability measures$\mu$ and $\nu$ on $\mathbb{R}^d$, with $\nu$ dominating $\mu$. When$(X_i)_{1\le i\le I}$ (resp. $(Y_j)_{1\le j\le J}$) are i.i.d. according $\mu$(resp. $\nu$), the empirical measures $\mu_I$ and $\nu_J$ are not in the convexorder. We investigate modifications of $\mu_I$ (resp. $\nu_J$) smaller than$\nu_J$ (resp. greater than $\mu_I$) in the convex order and weakly convergingto $\mu$ (resp. $\nu$) as $I,J\to\infty$. In dimension 1, according to Kertzand R\"osler (1992), the set of probability measures with a finite first ordermoment is a lattice for the increasing and the decreasing convex orders. Fromthis result, we can define $\mu\vee\nu$ (resp. $\mu\wedge\nu$) that is greaterthan $\mu$ (resp. smaller than $\nu$) in the convex order. We give efficientalgorithms permitting to compute $\mu\vee\nu$ and $\mu\wedge\nu$ when $\mu$ and$\nu$ are convex combinations of Dirac masses. In general dimension, when $\mu$and $\nu$ have finite moments of order $\rho\ge 1$, we define the projection$\mu\curlywedge_\rho \nu$ (resp. $\mu\curlyvee_\rho\nu$) of $\mu$ (resp. $\nu$)on the set of probability measures dominated by $\nu$ (resp. larger than $\mu$)in the convex order for the Wasserstein distance with index $\rho$. When$\rho=2$, $\mu_I\curlywedge_2 \nu_J$ can be computed efficiently by solving aquadratic optimization problem with linear constraints. It turns out that, indimension 1, the projections do not depend on $\rho$ and their quantilefunctions are explicit, which leads to efficient algorithms for convexcombinations of Dirac masses. Last, we illustrate by numerical experiments theresulting sampling methods that preserve the convex order and their applicationto approximate Martingale Optimal Transport problems. | 2020-05-29 14:55:56 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9255017638206482, "perplexity": 406.48884744556347}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-24/segments/1590347404857.23/warc/CC-MAIN-20200529121120-20200529151120-00472.warc.gz"} |
https://www.rdocumentation.org/packages/car/versions/2.1-6/topics/scatterplot | # scatterplot
0th
Percentile
##### Scatterplots with Boxplots
Makes enhanced scatterplots, with boxplots in the margins, a nonparametric regression smooth, smoothed conditional spread, outlier identification, and a regression line; sp is an abbreviation for scatterplot.
Keywords
hplot
##### Usage
scatterplot(x, ...)# S3 method for formula
scatterplot(formula, data, subset, xlab, ylab, legend.title, legend.coords,
labels, ...)# S3 method for default
scatterplot(x, y,
smoother=loessLine, smoother.args=list(), smooth, span,
boxplots=if (by.groups) "" else "xy",
xlab=deparse(substitute(x)), ylab=deparse(substitute(y)), las=par("las"),
lwd=1, lty=1,
labels, id.method = "mahal",
id.n = if(id.method[1]=="identify") length(x) else 0,
id.cex = 1, id.col = palette()[1], id.location="lr",
log="", jitter=list(), xlim=NULL, ylim=NULL,
cex=par("cex"), cex.axis=par("cex.axis"), cex.lab=par("cex.lab"),
cex.main=par("cex.main"), cex.sub=par("cex.sub"),
groups, by.groups=!missing(groups),
legend.title=deparse(substitute(groups)), legend.coords, legend.columns,
ellipse=FALSE, levels=c(.5, .95), robust=TRUE,
col=if (n.groups == 1) palette()[3:1] else rep(palette(), length=n.groups),
pch=1:n.groups,
legend.plot=!missing(groups), reset.par=TRUE, grid=TRUE, ...)sp(x, ...)
##### Arguments
x
vector of horizontal coordinates (or first argument of generic function).
y
vector of vertical coordinates.
formula
a model'' formula, of the form y ~ x or (to plot by groups) y ~ x | z, where z evaluates to a factor or other variable dividing the data into groups. If x is a factor, then parallel boxplots are produced using the Boxplot function.
data
data frame within which to evaluate the formula.
subset
expression defining a subset of observations.
smoother
a function to draw a nonparametric-regression smooth; the default is loessLine, which does loess smoothing. The function gamLine fits a generalized additive model and allows including a link and error function. See ScatterplotSmoothers. Setting this argument to something other than a function, e.g., FALSE suppresses the smoother.
smoother.args
a list of named values to be passed to the smoother function; the specified elements of the list depend upon the smoother (see ScatterplotSmoothers).
smooth, span
these arguments are included for backwards compatility: if smooth=TRUE then smoother is set to loessLine, and if span is specified, it is added to smoother.args.
if TRUE, estimate the (square root) of the variance function. For loessLine and for gamLine, this is done by separately smoothing the squares of the postive and negative residuals from the mean fit, and then adding the square root of the fitted values to the mean fit. For quantregLine, fit the .25 and .75 quantiles with a quantile regression additive model. The default is TRUE if by.groups=FALSE and FALSE is by.groups=TRUE.
reg.line
function to draw a regression line on the plot or FALSE not to plot a regression line.
boxplots
if "x" a boxplot for x is drawn below the plot; if "y" a boxplot for y is drawn to the left of the plot; if "xy" both boxplots are drawn; set to "" or FALSE to suppress both boxplots.
xlab
label for horizontal axis.
ylab
label for vertical axis.
las
if 0, ticks labels are drawn parallel to the axis; set to 1 for horizontal labels (see par).
lwd
width of linear-regression lines (default 1).
lty
type of linear-regression lines (default 1, solid line).
id.method,id.n,id.cex,id.col,id.location
Arguments for the labelling of points. The default is id.n=0 for labeling no points. See showLabels for details of these arguments. If the plot uses different colors for groups, then the id.col argument is ignored and label colors are determined by the col argument.
labels
a vector of point labels; if absent, the function tries to determine reasonable labels, and, failing that, will use observation numbers.
log
same as the log argument to plot, to produce log axes.
jitter
a list with elements x or y or both, specifying jitter factors for the horizontal and vertical coordinates of the points in the scatterplot. The jitter function is used to randomly perturb the points; specifying a factor of 1 produces the default jitter. Fitted lines are unaffected by the jitter.
xlim
the x limits (min, max) of the plot; if NULL, determined from the data.
ylim
the y limits (min, max) of the plot; if NULL, determined from the data.
groups
a factor or other variable dividing the data into groups; groups are plotted with different colors and plotting characters.
by.groups
if TRUE, regression lines are fit by groups.
legend.title
title for legend box; defaults to the name of the groups variable.
legend.coords
coordinates for placing legend; can be a list with components x and y to specify the coordinates of the upper-left-hand corner of the legend; or a quoted keyword, such as "topleft", recognized by legend.
legend.columns
number of columns for the legend; if absent will be supplied automatically to prefer horizontal legends when plotted above the graph.
ellipse
if TRUE data-concentration ellipses are plotted.
levels
level or levels at which concentration ellipses are plotted; the default is c(.5, .95).
robust
if TRUE (the default) use the cov.trob function in the MASS package to calculate the center and covariance matrix for the data ellipses.
col
colors for lines and points; the default is taken from the color palette, with palette()[3] for linear regression lines, palette()[2] for nonparametric regression lines, and palette()[1] for points if there are no groups, and successive colors for the groups if there are groups.
pch
plotting characters for points; default is the plotting characters in order (see par).
cex, cex.axis, cex.lab, cex.main, cex.sub
set sizes of various graphical elements; (see par).
legend.plot
if TRUE then a legend for the groups is plotted in the upper margin.
reset.par
if TRUE then plotting parameters are reset to their previous values when scatterplot exits; if FALSE then the mar and mfcol parameters are altered for the current plotting device. Set to FALSE if you want to add graphical elements (such as lines) to the plot.
other arguments passed down and to plot.
grid
If TRUE, the default, a light-gray background grid is put on the graph
##### Value
If points are identified, their labels are returned; otherwise NULL is returned invisibly.
boxplot, jitter, legend, scatterplotMatrix, dataEllipse, Boxplot, cov.trob, showLabels, ScatterplotSmoothers.
##### Aliases
• scatterplot
• scatterplot.formula
• scatterplot.default
• sp
##### Examples
# NOT RUN {
scatterplot(prestige ~ income, data=Prestige, ellipse=TRUE)
if (interactive()){
scatterplot(prestige ~ income, data=Prestige, smoother=quantregLine)
}
scatterplot(prestige ~ income|type, data=Prestige, smoother=loessLine,
smoother.args=list(span=1))
scatterplot(prestige ~ income|type, data=Prestige, legend.coords="topleft")
scatterplot(vocabulary ~ education, jitter=list(x=1, y=1),
data=Vocab, id.n=0, smoother=FALSE)
scatterplot(infant.mortality ~ gdp, log="xy", data=UN, id.n=5)
scatterplot(income ~ type, data=Prestige)
# }
# NOT RUN {
scatterplot(infant.mortality ~ gdp, id.method="identify", data=UN)
scatterplot(infant.mortality ~ gdp, id.method="identify", smoother=loessLine, data=UN)
# }
Documentation reproduced from package car, version 2.1-6, License: GPL (>= 2)
### Community examples
Looks like there are no examples yet. | 2020-08-11 13:24:59 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4367763102054596, "perplexity": 8917.670689697263}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-34/segments/1596439738777.54/warc/CC-MAIN-20200811115957-20200811145957-00502.warc.gz"} |
https://brilliant.org/problems/proof-that-21/ | # Proof that 2=1
Algebra Level 2
The following are my steps when I proved that $$2=1$$.
Step 1. Suppose that$$:$$ $$y=b$$
Step 2. Multiply both sides by $$y$$$$:$$ $$y^2=yb$$
Step 3. Subtract $$b^2$$ from both sides$$:$$ $$y^2-b^2=yb-b^2$$
Step 4. Factor both sides$$:$$ $$(y-b)(y+b)=b(y-b)$$
Step 5. Divide both sides by $$y-b$$$$:$$ $$y+b=b$$
Step 6. Since $$y=b$$, substitute$$:$$ $$b+b=b$$ or $$2b=b$$
Step 7. Divide both sides by $$b$$$$:$$ $$2=1$$
Is there something with my calculation? If your answer is $$no$$, give your answer as $$10$$. If your answer is $$yes$$, input the step number which I first committed a mistake.
× | 2018-03-22 15:30:48 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.941075325012207, "perplexity": 487.24520013695616}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-13/segments/1521257647892.89/warc/CC-MAIN-20180322151300-20180322171300-00348.warc.gz"} |
http://www.popflock.com/learn?s=Fa%C3%A0_di_Bruno's_formula | Fa%C3%A0 Di Bruno's Formula
Get Fa%C3%A0 Di Bruno's Formula essential facts below. View Videos or join the Fa%C3%A0 Di Bruno's Formula discussion. Add Fa%C3%A0 Di Bruno's Formula to your PopFlock.com topic list for future reference or share this resource on social media.
Fa%C3%A0 Di Bruno's Formula
Faà di Bruno's formula is an identity in mathematics generalizing the chain rule to higher derivatives. Though it is named after Francesco Faà di Bruno (1855, 1857), he was not the first to state or prove the formula. In 1800, more than 50 years before Faà di Bruno, the French mathematician Louis François Antoine Arbogast had stated the formula in a calculus textbook,[1] which is considered to be the first published reference on the subject.[2]
Perhaps the most well-known form of Faà di Bruno's formula says that
${\displaystyle {d^{n} \over dx^{n}}f(g(x))=\sum {\frac {n!}{m_{1}!\,1!^{m_{1}}\,m_{2}!\,2!^{m_{2}}\,\cdots \,m_{n}!\,n!^{m_{n}}}}\cdot f^{(m_{1}+\cdots +m_{n})}(g(x))\cdot \prod _{j=1}^{n}\left(g^{(j)}(x)\right)^{m_{j}},}$
where the sum is over all n-tuples of nonnegative integers (m1, ..., mn) satisfying the constraint
${\displaystyle 1\cdot m_{1}+2\cdot m_{2}+3\cdot m_{3}+\cdots +n\cdot m_{n}=n.}$
Sometimes, to give it a memorable pattern, it is written in a way in which the coefficients that have the combinatorial interpretation discussed below are less explicit:
${\displaystyle {d^{n} \over dx^{n}}f(g(x))=\sum {\frac {n!}{m_{1}!\,m_{2}!\,\cdots \,m_{n}!}}\cdot f^{(m_{1}+\cdots +m_{n})}(g(x))\cdot \prod _{j=1}^{n}\left({\frac {g^{(j)}(x)}{j!}}\right)^{m_{j}}.}$
Combining the terms with the same value of m1 + m2 + ... + mn = k and noticing that mj has to be zero for j > n − k + 1 leads to a somewhat simpler formula expressed in terms of Bell polynomials Bn,k(x1,...,xnk+1):
${\displaystyle {d^{n} \over dx^{n}}f(g(x))=\sum _{k=1}^{n}f^{(k)}(g(x))\cdot B_{n,k}\left(g'(x),g''(x),\dots ,g^{(n-k+1)}(x)\right).}$
## Combinatorial form
The formula has a "combinatorial" form:
${\displaystyle {d^{n} \over dx^{n}}f(g(x))=(f\circ g)^{(n)}(x)=\sum _{\pi \in \Pi }f^{(\left|\pi \right|)}(g(x))\cdot \prod _{B\in \pi }g^{(\left|B\right|)}(x)}$
where
• ? runs through the set ? of all partitions of the set { 1, ..., n },
• "B ? ?" means the variable B runs through the list of all of the "blocks" of the partition ?, and
• |A| denotes the cardinality of the set A (so that |?| is the number of blocks in the partition ? and |B| is the size of the block B).
## Example
The following is a concrete explanation of the combinatorial form for the n = 4 case.
{\displaystyle {\begin{aligned}(f\circ g)''''(x)={}&f''''(g(x))g'(x)^{4}+6f'''(g(x))g''(x)g'(x)^{2}\\[8pt]&{}+\;3f''(g(x))g''(x)^{2}+4f''(g(x))g'''(x)g'(x)\\[8pt]&{}+\;f'(g(x))g''''(x).\end{aligned}}}
The pattern is:
${\displaystyle {\begin{array}{cccccc}g'(x)^{4}&&\leftrightarrow &&1+1+1+1&&\leftrightarrow &&f''''(g(x))&&\leftrightarrow &&1\\[12pt]g''(x)g'(x)^{2}&&\leftrightarrow &&2+1+1&&\leftrightarrow &&f'''(g(x))&&\leftrightarrow &&6\\[12pt]g''(x)^{2}&&\leftrightarrow &&2+2&&\leftrightarrow &&f''(g(x))&&\leftrightarrow &&3\\[12pt]g'''(x)g'(x)&&\leftrightarrow &&3+1&&\leftrightarrow &&f''(g(x))&&\leftrightarrow &&4\\[12pt]g''''(x)&&\leftrightarrow &&4&&\leftrightarrow &&f'(g(x))&&\leftrightarrow &&1\end{array}}}$
The factor ${\displaystyle g''(x)g'(x)^{2}}$ corresponds to the partition 2 + 1 + 1 of the integer 4, in the obvious way. The factor ${\displaystyle f'''(g(x))}$ that goes with it corresponds to the fact that there are three summands in that partition. The coefficient 6 that goes with those factors corresponds to the fact that there are exactly six partitions of a set of four members that break it into one part of size 2 and two parts of size 1.
Similarly, the factor ${\displaystyle g''(x)^{2}}$ in the third line corresponds to the partition 2 + 2 of the integer 4, (4, because we are finding the fourth derivative), while ${\displaystyle f''(g(x))}$ corresponds to the fact that there are two summands (2 + 2) in that partition. The coefficient 3 corresponds to the fact that there are ${\displaystyle {\tfrac {1}{2}}{\tbinom {4}{2}}=3}$ ways of partitioning 4 objects into groups of 2. The same concept applies to the others.
A memorizable scheme is as follows:
{\displaystyle {\begin{aligned}&{\frac {D^{1}(f\circ {}g)}{1!}}&=\left(f^{(1)}\circ {}g\right){\frac {\frac {g^{(1)}}{1!}}{1!}}\\[8pt]&{\frac {D^{2}(f\circ g)}{2!}}&=\left(f^{(1)}\circ {}g\right){\frac {\frac {g^{(2)}}{2!}}{1!}}&{}+\left(f^{(2)}\circ {}g\right){\frac {{\frac {g^{(1)}}{1!}}{\frac {g^{(1)}}{1!}}}{2!}}\\[8pt]&{\frac {D^{3}(f\circ g)}{3!}}&=\left(f^{(1)}\circ {}g\right){\frac {\frac {g^{(3)}}{3!}}{1!}}&{}+\left(f^{(2)}\circ {}g\right){\frac {\frac {g^{(1)}}{1!}}{1!}}{\frac {\frac {g^{(2)}}{2!}}{1!}}&{}+\left(f^{(3)}\circ {}g\right){\frac {{\frac {g^{(1)}}{1!}}{\frac {g^{(1)}}{1!}}{\frac {g^{(1)}}{1!}}}{3!}}\\[8pt]&{\frac {D^{4}(f\circ g)}{4!}}&=\left(f^{(1)}\circ {}g\right){\frac {\frac {g^{(4)}}{4!}}{1!}}&{}+\left(f^{(2)}\circ {}g\right)\left({\frac {\frac {g^{(1)}}{1!}}{1!}}{\frac {\frac {g^{(3)}}{3!}}{1!}}+{\frac {{\frac {g^{(2)}}{2!}}{\frac {g^{(2)}}{2!}}}{2!}}\right)&{}+\left(f^{(3)}\circ {}g\right){\frac {{\frac {g^{(1)}}{1!}}{\frac {g^{(1)}}{1!}}}{2!}}{\frac {\frac {g^{(2)}}{2!}}{1!}}&{}+\left(f^{(4)}\circ {}g\right){\frac {{\frac {g^{(1)}}{1!}}{\frac {g^{(1)}}{1!}}{\frac {g^{(1)}}{1!}}{\frac {g^{(1)}}{1!}}}{4!}}\end{aligned}}}
## Combinatorics of the Faà di Bruno coefficients
These partition-counting Faà di Bruno coefficients have a "closed-form" expression. The number of partitions of a set of size n corresponding to the integer partition
${\displaystyle \displaystyle n=\underbrace {1+\cdots +1} _{m_{1}}\,+\,\underbrace {2+\cdots +2} _{m_{2}}\,+\,\underbrace {3+\cdots +3} _{m_{3}}+\cdots }$
of the integer n is equal to
${\displaystyle {\frac {n!}{m_{1}!\,m_{2}!\,m_{3}!\,\cdots 1!^{m_{1}}\,2!^{m_{2}}\,3!^{m_{3}}\,\cdots }}.}$
These coefficients also arise in the Bell polynomials, which are relevant to the study of cumulants.
## Variations
### Multivariate version
Let y = g(x1, ..., xn). Then the following identity holds regardless of whether the n variables are all distinct, or all identical, or partitioned into several distinguishable classes of indistinguishable variables (if it seems opaque, see the very concrete example below):[3]
${\displaystyle {\partial ^{n} \over \partial x_{1}\cdots \partial x_{n}}f(y)=\sum _{\pi \in \Pi }f^{(\left|\pi \right|)}(y)\cdot \prod _{B\in \pi }{\partial ^{\left|B\right|}y \over \prod _{j\in B}\partial x_{j}}}$
where (as above)
• ? runs through the set ? of all partitions of the set { 1, ..., n },
• "B ? ?" means the variable B runs through the list of all of the "blocks" of the partition ?, and
• |A| denotes the cardinality of the set A (so that |?| is the number of blocks in the partition ? and |B| is the size of the block B).
More general versions hold for cases where the all functions are vector- and even Banach-space-valued. In this case one needs to consider the Fréchet derivative or Gateaux derivative.
Example
The five terms in the following expression correspond in the obvious way to the five partitions of the set { 1, 2, 3 }, and in each case the order of the derivative of f is the number of parts in the partition:
{\displaystyle {\begin{aligned}{\partial ^{3} \over \partial x_{1}\,\partial x_{2}\,\partial x_{3}}f(y)={}&f'(y){\partial ^{3}y \over \partial x_{1}\,\partial x_{2}\,\partial x_{3}}\\[10pt]&{}+f''(y)\left({\partial y \over \partial x_{1}}\cdot {\partial ^{2}y \over \partial x_{2}\,\partial x_{3}}+{\partial y \over \partial x_{2}}\cdot {\partial ^{2}y \over \partial x_{1}\,\partial x_{3}}+{\partial y \over \partial x_{3}}\cdot {\partial ^{2}y \over \partial x_{1}\,\partial x_{2}}\right)\\[10pt]&{}+f'''(y){\partial y \over \partial x_{1}}\cdot {\partial y \over \partial x_{2}}\cdot {\partial y \over \partial x_{3}}.\end{aligned}}}
If the three variables are indistinguishable from each other, then three of the five terms above are also indistinguishable from each other, and then we have the classic one-variable formula.
### Formal power series version
Suppose ${\displaystyle f(x)=\sum _{n=0}^{\infty }{a_{n}}x^{n}}$ and ${\displaystyle g(x)=\sum _{n=0}^{\infty }{b_{n}}x^{n}}$ are formal power series and ${\displaystyle b_{0}=0}$.
Then the composition ${\displaystyle f\circ g}$ is again a formal power series,
${\displaystyle f(g(x))=\sum _{n=0}^{\infty }{c_{n}}x^{n},}$
where c0 = a0 and the other coefficient cn for n >= 1 can be expressed as a sum over compositions of n or as an equivalent sum over partitions of n:
${\displaystyle c_{n}=\sum _{\mathbf {i} \in {\mathcal {C}}_{n}}a_{k}b_{i_{1}}b_{i_{2}}\cdots b_{i_{k}},}$
where
${\displaystyle {\mathcal {C}}_{n}=\{(i_{1},i_{2},\dots ,i_{k})\,:\ 1\leq k\leq n,\ i_{1}+i_{2}+\cdots +i_{k}=n\}}$
is the set of compositions of n with k denoting the number of parts,
or
${\displaystyle c_{n}=\sum _{k=1}^{n}a_{k}\sum _{\mathbf {\pi } \in {\mathcal {P}}_{n,k}}{\binom {k}{\pi _{1},\pi _{2},...,\pi _{n}}}b_{1}^{\pi _{1}}b_{2}^{\pi _{2}}\cdots b_{n}^{\pi _{n}},}$
where
${\displaystyle {\mathcal {P}}_{n,k}=\{(\pi _{1},\pi _{2},\dots ,\pi _{n})\,:\ \pi _{1}+\pi _{2}+\cdots +\pi _{n}=k,\ \pi _{1}\cdot 1+\pi _{2}\cdot 2+\cdots +\pi _{n}\cdot n=n\}}$
is the set of partitions of n into k parts, in frequency-of-parts form.
The first form is obtained by picking out the coefficient of xn in ${\displaystyle (b_{1}x+b_{2}x^{2}+\cdots )^{k}}$ "by inspection", and the second form is then obtained by collecting like terms, or alternatively, by applying the multinomial theorem.
The special case f(x) = ex, g(x) = ?n >= 1an /n! xn gives the exponential formula. The special case f(x) = 1/(1 − x), g(x) = ?n >= 1 (-an) xn gives an expression for the reciprocal of the formal power series ?n >= 0anxn in the case a0 = 1.
Stanley [4] gives a version for exponential power series. In the formal power series
${\displaystyle f(x)=\sum _{n}{\frac {a_{n}}{n!}}x^{n},}$
we have the nth derivative at 0:
${\displaystyle f^{(n)}(0)=a_{n}.}$
This should not be construed as the value of a function, since these series are purely formal; there is no such thing as convergence or divergence in this context.
If
${\displaystyle g(x)=\sum _{n=0}^{\infty }{\frac {b_{n}}{n!}}x^{n}}$
and
${\displaystyle f(x)=\sum _{n=1}^{\infty }{\frac {a_{n}}{n!}}x^{n}}$
and
${\displaystyle g(f(x))=h(x)=\sum _{n=0}^{\infty }{\frac {c_{n}}{n!}}x^{n},}$
then the coefficient cn (which would be the nth derivative of h evaluated at 0 if we were dealing with convergent series rather than formal power series) is given by
${\displaystyle c_{n}=\sum _{\pi =\left\{B_{1},\ldots ,B_{k}\right\}}a_{\left|B_{1}\right|}\cdots a_{\left|B_{k}\right|}b_{k}}$
where ? runs through the set of all partitions of the set {1, ..., n} and B1, ..., Bk are the blocks of the partition ?, and | Bj | is the number of members of the jth block, for j = 1, ..., k.
This version of the formula is particularly well suited to the purposes of combinatorics.
We can also write with respect to the notation above
${\displaystyle g(f(x))=b_{0}+\sum _{n=1}^{\infty }{\frac {\sum _{k=1}^{n}b_{k}B_{n,k}(a_{1},\ldots ,a_{n-k+1})}{n!}}x^{n},}$
where Bn,k(a1,...,ank+1) are Bell polynomials.
### A special case
If f(x) = ex, then all of the derivatives of f are the same and are a factor common to every term. In case g(x) is a cumulant-generating function, then f(g(x)) is a moment-generating function, and the polynomial in various derivatives of g is the polynomial that expresses the moments as functions of the cumulants.
## Notes
1. ^
2. ^ According to Craik (2005, pp. 120-122): see also the analysis of Arbogast's work by Johnson (2002, p. 230).
3. ^ Hardy, Michael (2006). "Combinatorics of Partial Derivatives". Electronic Journal of Combinatorics. 13 (1): R1.
4. ^ See the "compositional formula" in Chapter 5 of Stanley, Richard P. (1999) [1997]. Enumerative Combinatorics. Cambridge University Press. ISBN 978-0-521-55309-4. | 2020-10-31 20:41:16 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 34, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9591109156608582, "perplexity": 764.1102134363897}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-45/segments/1603107922411.94/warc/CC-MAIN-20201031181658-20201031211658-00411.warc.gz"} |
http://math.stackexchange.com/questions/295100/echelon-form-for-a-2-times-2-matrix | # Echelon form for a $2\times 2$ matrix
So I'm to describe the possible forms of a $2\times 2$ matrix in Echelon form. I thought the only way we could have a $2\times 2$ matrix in Echelon form is like this: $$\left[\begin{matrix}1&c\\ 0&0\end{matrix}\right]$$ Where $c$ is just some constant. But I'm told that we can have it in the form $$\left[\begin{matrix}1&c\\0&1\end{matrix}\right]$$ and $$\left[\begin{matrix}0&1\\0&0\end{matrix}\right]$$But how come we can have a leading entry in the last column when this just leaves the matrix inconsistent? Is it because consistency doesn't matter when it comes to just classifying a matrix as "in Echelon form"?
-
There is "row echelon form" and "reduced row echelon form"; the stricter of the two is the last.
The linked entry contains the criteria for each, first the row echelon form, then additional criteria for reduced row echelon form. So using the term "echelon form", or even the casual use of the terms "row echelon form", can be ambiguous, depending on context, and depending on textbooks, to some degree, unless an explicit distinction is made.
Also, as an aside, consistency isn't a property of a matrix, per se, but of the system of equations that an augmented coefficient matrix may represent. For your matrices, the first reveals that there exist two linearly independent row (and column) vectors. Your second matrix reveals that the row vectors (and column vectors) are linearly dependent. We can also determine the rank of a matrix from a matrix in echelon form: the rank of a matrix is equal to the number of non-zero rows in the matrix when reduced to row echelon form.
-
+1 For a $2\times 2$ matrix it is done too fast, but for higher it takes time. – Babak S. Feb 5 '13 at 3:49
Kyle - does the link help clarify matters? Do you know what I mean when I say that a matrix itself isn't consistent/non-consistent, but that consistency is a property of the linear system it may represent? (which is usually represented by an augmented coefficient matrix, as in your more recent post?) – amWhy Feb 5 '13 at 14:41 | 2015-07-06 05:10:17 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8313981890678406, "perplexity": 227.0642453411047}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-27/segments/1435375098059.60/warc/CC-MAIN-20150627031818-00046-ip-10-179-60-89.ec2.internal.warc.gz"} |
http://christwire.org/01ooc/23ed1b-multiplying-complex-numbers-graphically | If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. Products and Quotients of Complex Numbers, 10. For example, 2 times 3 + i is just 6 + 2i. by BuBu [Solved! 11.2 The modulus and argument of the quotient. A reader challenges me to define modulus of a complex number more carefully. When you divide complex numbers, you must first multiply by the complex conjugate to eliminate any imaginary parts, and then you can divide. What complex multiplication looks like By now we know how to multiply two complex numbers, both in rectangular and polar form. Home | Is there a way to visualize the product or quotient of two complex numbers? Topic: Complex Numbers, Numbers. After calculation you can multiply the result by another matrix right there! Complex numbers are the sum of a real and an imaginary number, represented as a + bi. We can represent complex numbers in the complex plane.. We use the horizontal axis for the real part and the vertical axis for the imaginary part.. Using the complex plane, we can plot complex numbers … Think about the days before we had Smartphones and GPS. We have a fixed number, 5 + 5j, and we divide it by any complex number we choose, using the sliders. • Modulus of a Complex Number Learning Outcomes As a result of studying this topic, students will be able to • add and subtract Complex Numbers and to appreciate that the addition of a Complex Number to another Complex Number corresponds to a translation in the plane • multiply Complex Numbers and show that multiplication of a Complex Example 1 EXPRESSING THE SUM OF COMPLEX NUMBERS GRAPHICALLY Find the sum of 6 –2i and –4 –3i. In each case, you are expected to perform the indicated operations graphically on the Argand plane. The following applets demonstrate what is going on when we multiply and divide complex numbers. http://www.freemathvideos.com In this video tutorial I show you how to multiply imaginary numbers. Learn how complex number multiplication behaves when you look at its graphical effect on the complex plane. Result: square the magnitudes, double the angle.In general, a complex number like: r(cos θ + i sin θ)When squared becomes: r2(cos 2θ + i sin 2θ)(the magnitude r gets squared and the angle θ gets doubled. All numbers from the sum of complex numbers? Free Complex Numbers Calculator - Simplify complex expressions using algebraic rules step-by-step This website uses cookies to ensure you get the best experience. (This is spoken as “r at angle θ ”.) Then, we naturally extend these ideas to the complex plane and show how to multiply two complex num… 3. It will perform addition, subtraction, multiplication, division, raising to power, and also will find the polar form, conjugate, modulus and inverse of the complex number. Similarly, when you multiply a complex number z by 1/2, the result will be half way between 0 and z One way to explore a new idea is to consider a simple case. Complex numbers have a real and imaginary parts. You'll see examples of: You can also use a slider to examine the effect of multiplying by a real number. Example 7 MULTIPLYING COMPLEX NUMBERS (cont.) Let us consider two cases: a = 2 , a = 1 / 2 . A complex number is a number of the form a + bi, where a and b are real numbers, and i is an indeterminate satisfying i 2 = −1.For example, 2 + 3i is a complex number. FOIL stands for first , outer, inner, and last pairs. by M. Bourne. Graph both complex numbers and their resultant. Geometrically, when we double a complex number, we double the distance from the origin, to the point in the plane. As imaginary unit use i or j (in electrical engineering), which satisfies basic equation i 2 = −1 or j 2 = −1.The calculator also converts a complex number into angle notation (phasor notation), exponential, or polar coordinates (magnitude and angle). Geometrically, when you double a complex number, just double the distance from the origin, 0. Find the division of the following complex numbers (cos α + i sin α) 3 / (sin β + i cos β) 4. Another approach uses a radius and an angle. Free Complex Number Calculator for division, multiplication, Addition, and Subtraction Solution : In the above division, complex number in the denominator is not in polar form. Have questions? Learn how complex number multiplication behaves when you look at its graphical effect on the complex plane. So you might have said, ''I am at the crossing of Main and Elm.'' Friday math movie: Complex numbers in math class. Here are some examples of what you would type here: (3i+1)(5+2i) (-1-5i)(10+12i) i(5-2i) Type your problem here. Our mission is to provide a free, world-class education to anyone, anywhere. By moving the vector endpoints the complex numbers can be changed. Such way the division can be compounded from multiplication and reciprocation. But either part can be 0, so all Real Numbers and Imaginary Numbers are also Complex Numbers. See the previous section, Products and Quotients of Complex Numbersfor some background. Movie: complex numbers are the sum of a complex number, we double the distance from the,. Two cases: a + bi cis 2θ Home solver can solve a wide range of math problems another! Numbers is graphically presented angle θ ”. 1 EXPRESSING the sum of complex numbers in polar.... We double the distance from the origin, to the point in the complex plane multiplication, multiplying complex numbers graphically multiply. = r2 cis 2θ Home the initial case, you are multiplying complex numbers graphically to multiply together..., please make sure that the domains *.kastatic.org and *.kasandbox.org are.! I\Text { for the initial case, where we are dividing by 1 5j... With the complex plane consisting of the numbers that have a zero imaginary part expressed in polar coordinate,! Numbers geometrically multiplication with complex numbers: polar & exponential form, Visualizing complex number by a real number how... Following applets demonstrate what is going on when we multiply the result by another from the origin, to vector... For example, 2 times 3 + I is just 6 + 2i number, 5 +,... Geometrically, when you look at its graphical effect on the complex plane consisting of the crossing two. Calculator will simplify any complex number has a real and an imaginary number gives real... A new idea is to provide a free, world-class education to anyone, anywhere in each case where! [ Solved! ] the crossing of two lines to locate a point (,. Graphs to a unique point on the real axis to visualize the multiplying complex numbers graphically Or quotient two... Like by now we know how to multiply them together correctly you how to imaginary... What complex multiplication looks like by now we know how to multiply two complex numbers in math class lesson. To polar form, r ∠ θ, 0 we choose, using the sliders is.: you can perform matrix multiplication with complex numbers graphically Find the sum of a complex number choose... Point on the real number page will show you how to multiply two complex numbers both. Is not in polar form = r2 cis 2θ Home not in polar form going on when we the. Multiply both parts of the numbers that have a zero imaginary part: a = 2, a complex,. “ r at angle θ ”. the denominator is not in polar form will simplify any complex has... 'Re having trouble loading external resources on our website z1 and z2 a! Are dividing by 1 − 5j numbers and evaluates expressions in the complex plane 6... Slider to examine the effect of multiplying by a scalar external resources on our website and! Idea is to provide a free, world-class education to anyone, anywhere multiplication! The explanations using the Next '' button part and an imaginary part a. Challenges me to define modulus of a complex number by Jedothek [ Solved! ] θ ) 2 = cis. Some background to consider a simple case by now we know how to imaginary... Z 1 ⋅ z 2 a very creative way to explore a new is. Number corresponds to a unique point on the complex number by a scalar simplify any complex we... These pairs as shown below z1 and z2 in a polar form all the features of Academy! An imaginary number gives a real and an imaginary part parts of the multiplication 1... The explanation given for the initial case, you might have made reference to an intersection numbers. … Here you can also be expressed in polar form, can also use a to... To visualize the product Or quotient of two streets with steps shown section 10.3 we represented the of! | Author: Murray Bourne | about & Contact | Privacy & Cookies | IntMath feed | on. Math movie: complex numbers in polar form line in the complex.. The operation with the complex plane real result also complex numbers online for free like... ) of one complex number more carefully arithmetic with complex numbers, both in rectangular and polar form consider. You 'll see examples of: you can perform matrix multiplication with complex graphically! Know how to multiply them together correctly Velocity: Application of complex Numbersfor some background is there a to... Consider two complex numbers in math class numbers that have a fixed number, like... Can be changed the origin, 0 sure that the domains *.kastatic.org and *.kasandbox.org unblocked! Be careful with your negative signs distance from the origin, to the vector representing a complex,! Perform matrix multiplication with complex numbers can be compounded from multiplication and reciprocation ( division ) of one number. Complex plane modulus of a real and an imaginary part: a + bi is a very way... On complex numbers are the sum of 6 –2i and –4 –3i and *.kasandbox.org are unblocked | Sitemap Author! | about & Contact | Privacy & Cookies | IntMath feed | page show... Of a real and an imaginary part to the vector endpoints the complex numbers, inner, we. The Next '' button line in the denominator is not in polar coordinate form multiplying. Perform matrix multiplication with complex numbers in math class http: //www.freemathvideos.com in this lesson we review idea. The imaginary axis is the crossing of Main and Elm. applet demonstrates quotient. To anyone, anywhere polar coordinate form, multiplying and dividing complex numbers graphically as a + bi explanation for. World-Class education to anyone, anywhere but it does require you to be careful with negative. Numbers - Displaying top 8 worksheets found for this concept [ Solved! ] and divide... + 0i angle θ ”. θ ”. inner, and we it... Is basically the same, but it does require you to be careful with your negative signs!... \ ( i\text { way the division can be changed the product Or quotient of two complex numbers can compounded. The crossing of Main and Elm. supposed to multiply these pairs as shown below correspondences between numerical graphical! We divide it by any complex expression, with steps shown is on! Where we are dividing by 1 − 5j choose, using the Next '' button crossing of and..., the intersection is the line in the shorter \ '' cis\ '':! Seeing this message, it means we 're having trouble loading external resources on our website \ ( {. ) Or in the complex plane consisting of the multiplication z 1 ⋅ z 2 we,... B ) in the denominator is not in polar form, so all real numbers evaluates... Multiply and divide complex numbers graphically multiplying complex numbers graphically the sum of complex Numbersfor background... ) in the complex number, just like vectors, can also use a slider to the...: complex numbers for some background: in the set of complex numbers polar! Require you to be careful with your negative signs complex number more carefully multiply two numbers... Define modulus of a complex number in denominator to polar form Cookies | feed... Consider a simple case now we know how to multiply a complex number corresponds to a point (,! And polar form shows the result of the complex numbers Solved! ] multiplication complex. Gives a real number real numbers and imaginary numbers is going on when we multiply number., r ∠ θ part and an imaginary number, just like vectors, also! In rectangular and polar form but it does require you to be careful with your signs. Number times another imaginary number times another imaginary number gives a real result Smartphones GPS. | Privacy & Cookies | IntMath feed | this idea of the complex plane red arrow shows result! 2 times 3 + I is just 6 + 2i of math problems this graph shows how we can the! For example, 2 times 3 + I is just 6 +.. Argand plane compounded from multiplication and reciprocation features of Khan Academy, please enable JavaScript in your browser, number..., where we are dividing by 1 − 5j ∠ θ by moving the vector endpoints the complex are. Expressed in polar form double a complex number in denominator to polar form negative signs is graphically presented, the... Are unblocked the above division, complex number by a real part and imaginary... Z2 in a polar form denominator is not in polar form multiplication of complex Numbersfor some background the... How complex number has a real and an imaginary part sure that the domains *.kastatic.org and * are... 3 + I is just 6 + 2i and graphical representations of arithmetic with complex,... Behaves when you double a complex number by a scalar numbers - Displaying top 8 worksheets found for this..! Some background two complex numbers consider a simple case exponential form, and., 5 + 5j, and last pairs ∠ θ the denominator not! Basic arithmetic on complex numbers - Displaying top 8 worksheets found for this concept were! Had Smartphones and GPS reader challenges me to define modulus of a complex number corresponds to friend! & Contact | Privacy & Cookies | IntMath feed |, I am at the crossing Main!, Visualizing complex number in denominator to polar form in your browser any complex number 5. The real axis real number graphs to a friend, you can multiplying complex numbers graphically! Any complex number has a real part and an imaginary part top 8 worksheets found for this..... Said, I am at the crossing of two complex numbers is graphically presented graph... Message, it means we 're having trouble loading external resources on our website Bourne | about Contact. | 2021-04-18 09:25:32 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6904890537261963, "perplexity": 799.619345822751}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-17/segments/1618038469494.59/warc/CC-MAIN-20210418073623-20210418103623-00136.warc.gz"} |
https://dml.cz/handle/10338.dmlcz/140403 | # Article
Full entry | PDF (0.1 MB)
Keywords:
Hilbert space; orthogonality; ergodic theorem
Summary:
This note deals with the orthogonality between sequences of random variables. The main idea of the note is to apply the results on equidistant systems of points in a Hilbert space to the case of the space $L^2(\Omega ,\mathcal F,\mathbb P)$ of real square integrable random variables. The main result gives a necessary and sufficient condition for a particular sequence of random variables (elements of which are taken from sets of equidistant elements of $L^2(\Omega ,\mathcal F,\mathbb P)$) to be orthogonal to some other sequence in $L^2(\Omega ,\mathcal F,\mathbb P)$. The result obtained is interesting from the point of view of the time series analysis, since it can be applied to a class of sequences random variables that exhibit a monotonically increasing variance. An application to ergodic theorem is also provided.
References:
[1] Wermuth, E. M. E.: A remark on equidistance in Hilbert spaces. Linear Algebra Appl. 236 (1996), 105-111. MR 1375608 | Zbl 0843.46015
Partner of | 2017-12-18 03:31:03 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7562056183815002, "perplexity": 257.15666998185327}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-51/segments/1512948604248.93/warc/CC-MAIN-20171218025050-20171218051050-00286.warc.gz"} |
https://blog.i-ll.cc/archives/75/ | Queues and Priority Queues are data structures which are known to most computer scientists. The Team Queue, however, is not so well known, though it occurs often in everyday life. At lunch time the queue in front of the Mensa is a team queue, for example.
In a team queue each element belongs to a team. If an element enters the queue, it first searches the queue from head to tail to check if some of its teammates (elements of the same team) are already in the queue. If yes, it enters the queue right behind them. If not, it enters the queue at the tail and becomes the new last element (bad luck). Dequeuing is done like in normal queues: elements are processed from head to tail in the order they appear in the team queue.
Your task is to write a program that simulates such a team queue.
Input
The input will contain one or more test cases. Each test case begins with the number of teams t (1
Finally, a list of commands follows. There are three different kinds of commands:
ENQUEUE x - enter element x into the team queue
DEQUEUE - process the first element and remove it from the queue
STOP - end of test case
The input will be terminated by a value of 0 for t.
Output
For each test case, first print a line saying “Scenario #k”, where k is the number of the test case. Then, for each DEQUEUE command, print the element which is dequeued on a single line. Print a blank line after each test case, even after the last one.
Sample Input
2
3 101 102 103
3 201 202 203
ENQUEUE 101
ENQUEUE 201
ENQUEUE 102
ENQUEUE 202
ENQUEUE 103
ENQUEUE 203
DEQUEUE
DEQUEUE
DEQUEUE
DEQUEUE
DEQUEUE
DEQUEUE
STOP
2
5 259001 259002 259003 259004 259005
6 260001 260002 260003 260004 260005 260006
ENQUEUE 259001
ENQUEUE 260001
ENQUEUE 259002
ENQUEUE 259003
ENQUEUE 259004
ENQUEUE 259005
DEQUEUE
DEQUEUE
ENQUEUE 260002
ENQUEUE 260003
DEQUEUE
DEQUEUE
DEQUEUE
DEQUEUE
STOP
0
Sample Output
Scenario #1
101
102
103
201
202
203
Scenario #2
259001
259002
259003
259004
259005
260001
Solution
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 #include #include #include #include using namespace std; const int maxn = 1000000; int id[maxn]; bool f[1002]; int main() { int t, cas = 1, n, x; char s[20]; while (scanf("%d",&t)!=EOF&&t!=0) { printf("Scenario #%d\n", cas++); queue q[1002]; queue team; for (int i = 1; i | 2021-10-26 05:29:09 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.21047568321228027, "perplexity": 1317.6317169135511}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-43/segments/1634323587799.46/warc/CC-MAIN-20211026042101-20211026072101-00532.warc.gz"} |
https://monach.us/operations/rke-and-iscsi-race-condition/ | #### RKI and iSCSI Race Condition
##### Resolving a race between the kubelet and the host
###### May 29, 2018
rancher kubernetes iscsi
Pods backed by iSCSI storage fail to mount the storage after a reboot of the host. The error in kubectl describe is something like:
MountVolume.WaitForAttach failed for volume "mqtt-data" : failed to get any path for iscsi disk, last err seen:
iscsi: failed to attach disk: Error: iscsiadm: Could not login to [iface: default, target: iqn.2006-04.us.monach:nas.mqtt-data, portal: 10.68.0.11,3260].
Logging in to [iface: default, target: iqn.2006-04.us.monach:nas.mqtt-data, portal: 10.68.0.11,3260] (multiple)
(exit status 12)
Note the iSCSI driver not found error. Strange, right? It’s there - lsmod shows iSCSI modules loaded into the kernel. In order to resolve this, I have to log into the host and stop/disable the iscsid service on the host itself. This allows the kubelet container to load/execute the drivers directly.
The part that I haven’t yet figured out is why the service keeps reactivating. I disable it with systemctl disable iscsid, and the next time the host boots….it’s running again.
Update 2018/06/14: This is resolved, and the answer is…well…easy. It appears that Ubuntu automatically installs and enables open-iscsi, which I noticed during another installation on a different system. This led me to the idea that the service I need to disable is not iscsid, but in fact is open-iscsi. Disabling that service prevented the restart of the service, and the next time that I brought the RKE nodes back up, Kubernetes correctly mounted the iSCSI targets for the pods without my intervention.
Sweet.
Update 2018/06/27: I’m rolling out a new Rancher cluster, and this problem reappeared. It now seems that I have to disable both open-iscsi and iscsid services in order for this to work. | 2019-02-22 20:18:48 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.23833827674388885, "perplexity": 9290.548936028985}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-09/segments/1550247526282.78/warc/CC-MAIN-20190222200334-20190222222334-00197.warc.gz"} |
https://projecteuclid.org/euclid.aoap/1177005934 | ## The Annals of Applied Probability
### Stochastic Order for Inspection and Repair Policies
#### Abstract
Inspection and repair policies for $(n - r + 1)$-out-of-$n$ systems are compared stochastically with respect to two partial orderings $\geq^{b_1}$ and $\geq^{b_2}$ on the set of permutations of $\{1, 2, \ldots, n\}$. The partial ordering $\geq^{b_1}$ is finer than the partial ordering $\geq^{b_2}$. A given permutation $\pi$ of $\{1,2, \ldots, n\}$ determines the order in which components are visited and inspected. We assume that the reliability of the $i$th independent component is given by $P_i$, where $P_1 \leq P_2 \leq \cdots \leq P_n$. If $\pi$ and $\pi'$ are two permutations such that $\pi \geq^{b_1} \pi'$, then we show that the number of inspections necessary to achieve minimal or complete repair is stochastically smaller with $\pi$ than with $\pi'$. We also consider three policies for minimal repair when the components are each made up of $t$ "parts" assembled in parallel. It is shown that if $\pi \geq^{b_2} \pi'$, then the number of repairs necessary under $\pi$ is stochastically smaller than the number necessary under $\pi'$, but that in general this is not true for the finer ordering $\geq^{b_1}$. The results enable one to make interesting comparisons between various inspection and repair policies, as well as to understand better the relationship between the orderings $\geq^{b_1}$ and $\geq^{b_2}$ on the set of permutations.
#### Article information
Source
Ann. Appl. Probab., Volume 1, Number 2 (1991), 207-218.
Dates
First available in Project Euclid: 19 April 2007
https://projecteuclid.org/euclid.aoap/1177005934
Digital Object Identifier
doi:10.1214/aoap/1177005934
Mathematical Reviews number (MathSciNet)
MR1102317
Zentralblatt MATH identifier
0737.62089
JSTOR | 2019-10-19 02:10:49 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5441409945487976, "perplexity": 398.76870480718003}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-43/segments/1570986688674.52/warc/CC-MAIN-20191019013909-20191019041409-00393.warc.gz"} |
https://motls.blogspot.com/2007/01/lep-and-tevatron-higgs-is-probably.html | ## Tuesday, January 09, 2007 ... //
### LEP and Tevatron: Higgs is probably light
The mass of the Higgs boson must be between 114 GeV and 1000 GeV. If it were lighter than 114 GeV, it would have already been detected by LEP, the Large Electron-Positron collider at CERN that lived in the same tunnel that is currently being seized by the LHC. Unless one of those strange models is right which would allow the Higgs to be as light as 105 GeV.
The God particle must also be lighter than a TeV or so because otherwise its self-interaction would have to be too strong. Because the quartic coupling is growing stronger at higher energies, we would encounter a lethal Landau pole too soon: the interaction strength would become infinite and most field theories don't like that.
The mass of the top quark is essentially (approximately, pretty accurately) determined by the RG flow as we have repeatedly mentioned. The right value depends on the electroweak couplings i.e. essentially on the W mass as well as the Higgs mass. A precise measurement of the top quark mass and the W mass can therefore be used to indirectly derive the Higgs mass. That's what was done to create the picture above.
The allowed range from 114 GeV (left) to 1000 GeV is the green band on the picture. Older measurements from LEP 1 and SLD at SLAC have led to the interior of the red squashed ellipse as the 68 percent confidence level region. Newer combined data from LEP 2 and the Tevatron suggest a somewhat lighter top quark as well as a significantly heavier W boson - a jump from 80.35 to 80.40 GeV. The blue dotted ellipse is again the 68 percent confidence level band according to these newer measurements.
You see that the new, blue ellipse doesn't quite overlap with the green band. Of course it doesn't have to. But statistically speaking, it now seems more likely that the Higgs is quite light. If you remember, LEP at CERN has had some unconvincing Higgs signals around 115 GeV just before they shut it down. It was not enough to claim a Higgs discovery but it was suggestive. The recent high precision data can revive the speculations that the 115 GeV Higgs signal was actually a real effect caused by a real God particle.
Recall that supersymmetry tends to predict a light Higgs boson, too. Those of us who are convinced that quantum field theory is a correct description up to pretty high energies where characteristic effects of string theory take over also believe the calculation that implies that the Higgs should be lighter than 180 GeV or so, regardless of the fate of supersymmetry - a much stronger bound than those 1000 GeV above. The new precision data is consistent with this expectation.
Appendix: Confidence levels
Let me recall some basic properties of the normal distribution. For one variable "x", the infinitesimal probability to get a value between "x" and "x+dx" is
• dP = dx . exp(-x2/2) / sqrt(2 pi).
In that case, we say that "x" is described by a normal (Gaussian) distribution with mean value of zero and a standard deviation of one. Note that the normalization is chosen so that the "integral dP" from "-infinity" to "+infinity" equals one. Everyone should be able to move the distribution to a different mean value - replace "x" by "x-x0". Also, everyone should know how to change the standard deviation to "sigma" - replace "x" by "(x-x0)/sigma" and divide "dP" by "sigma" to keep the integral being equal to one.
Fine. Return to the simply normalized formula for "dP". Its integral from -1 to +1 is equal to 0.683 or so. This means that there is a 68 percent probability for the normal distribution to give a result that differs from the mean value by less than 1 standard deviation. For plus minus two standard deviations, you get about 95 percent, and for plus minus three standard deviations, you get about 99.7 percent.
You must redo this calculation of the relation between the "radius" and the "confidence level" if you have "n" variables instead of one but the results only differ quantitatively, not qualitatively.
#### snail feedback (0) :
(function(i,s,o,g,r,a,m){i['GoogleAnalyticsObject']=r;i[r]=i[r]||function(){ (i[r].q=i[r].q||[]).push(arguments)},i[r].l=1*new Date();a=s.createElement(o), m=s.getElementsByTagName(o)[0];a.async=1;a.src=g;m.parentNode.insertBefore(a,m) })(window,document,'script','//www.google-analytics.com/analytics.js','ga'); ga('create', 'UA-1828728-1', 'auto'); ga('send', 'pageview'); | 2022-01-29 10:34:57 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7277780175209045, "perplexity": 900.041388992572}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320304883.8/warc/CC-MAIN-20220129092458-20220129122458-00159.warc.gz"} |
https://chemistry.stackexchange.com/questions/114281/what-is-the-least-dense-liquid-under-normal-conditions | # What is the least dense liquid under normal conditions?
What is the least dense liquid under normal conditions, room temperature, one atmosphere of pressure, doesn't combust upon contact with air, also wouldn't kill a human just for being in the same room as it?
Isopentane $$\ce{C5H12}$$ has the density of $$0.6201~\mathrm{g\,cm^{-3}}$$ at $$20~\mathrm{^\circ C}$$ [1, p. 3-330].
• @orlp That's due to the constraint imposed by OP: the compound must be liquid at NTP conditions. Once the constraint removed, there are more interesting things such as solution of lithium metal in liquid ammonia or good old liquid hydrogen with the density of $0.0709~\mathrm{g\,cm^{-3}}$, but that's not what the question is about. – andselisk Apr 25 at 8:16 | 2019-10-17 18:39:53 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 3, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.3609273433685303, "perplexity": 663.9178964987888}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-43/segments/1570986675598.53/warc/CC-MAIN-20191017172920-20191017200420-00075.warc.gz"} |