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http://openstudy.com/updates/56164a2fe4b086e96bfc9179
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## ParthKohli one year ago If a nine-digit number $$x$$ is formed using 1, 2, 3, 4, 5, 6, 7, 8, 9 without repetition, then find the probability that the non-negative difference among all the digits equidistant from both ends is 1.
1. ParthKohli
The central number has to be odd. 5 ways to choose an odd number. Now if 3 is chosen then the fixed pairs are (1, 2), (4, 5), (6, 7), (8, 9) so 4! ways to move the pairs around and $$2^4$$ ways to flip within the pair.$\frac{4!\times 5 \times 2^4}{9!}= \frac{1}{189}$Not a part of the choices...
2. thomas5267
How does this work? 864,152,579 is one such number. But 5 is equidistant to both ends and 5-5=0.
3. ParthKohli
Nah, the number at the center is not considered, because that way, there would be no such number.
4. FireKat97
5. ganeshie8
Just to confirm that your answer is correct, I am getting the same 1920 numbers that satisfy the given requirement by bruteforce : https://jsfiddle.net/ganeshie8/Lqbagygh/embedded/result/
6. ParthKohli
Thank you sir. Please post that here. http://math.stackexchange.com/questions/1469932/the-probability-that-non-negative-difference-of-the-digits-at-equal-distances-fr
7. ParthKohli
Oh, that question has an answer that exactly matches mine. Good.
8. ganeshie8
I don't see 1/189 in that question...
9. ParthKohli
Exactly, it doesn't match the choices.
10. ganeshie8
I'm not gonna spend more time on this.. typoes/gross mistakes are typical in textbooks written by indian authors in rush
11. ParthKohli
12. ganeshie8
"non- negative difference" are we interpreting this particular phrase correctly ?
13. ParthKohli
I guess we are.
14. ParthKohli
$|a-b| = 1$
15. thomas5267
Non-negative difference does not make much sense since the order of the subtraction is not specified.
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2016-10-21 16:56:50
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http://mathonline.wikidot.com/the-interior-of-intersections-of-sets-in-a-metric-space
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The Interior of Intersections of Sets in a Metric Space
# The Interior of Intersections of Sets in a Metric Space
Recall from the Interior and Boundary Points of a Set in a Metric Space page that if $(M, d)$ is a metric space and $S \subseteq M$ then a point $a \in S$ is said to be an interior point of $S$ if there exists an $r > 0$ such that:
(1)
\begin{align} \quad B(a, r) \subseteq S \end{align}
In other words, a point $a$ is an interior point of $S$ if there exists an open ball centered at $x$ that is fully contained in $S$. Furthermore, the set of all interior points of $S$ is called the interior of $S$ and is denoted $\mathrm{int} (S)$.
We will now look at some very important theorems regarding the interiors of both finite intersections and arbitrary intersections.
Theorem 1: Let $(M, d)$ be a metric space and let $S_1, S_2, ..., S_n \subseteq M$ be a finite collection of subsets of $M$. Then $\displaystyle{\mathrm{int} \left ( \bigcap_{i=1}^{n} S_i \right )= \bigcap_{i=1}^{n} \mathrm{int} (S_i)}$.
• Proof: Let $x \in \mathrm{int} \left ( \bigcap_{i=1}^{n} S_i \right )$. Then there exists an $r > 0$ such that:
(2)
\begin{align} \quad B(x, r) \subseteq \bigcap_{i=1}^{n} S_i \end{align}
• But $\displaystyle{\bigcap_{i=1}^{n} S_i \subseteq S_i}$ for each $i \in \{ 1, 2, ..., n \}$ and so $B(x, r) \subseteq S_i$ for each $i$. But this implies that $x \in \mathrm{int} (S_i)$ for each $i$ and thus $\displaystyle{x \in \bigcap_{i=1}^{n} \mathrm{int} (S_i)}$. This shows that:
(3)
\begin{align} \quad \mathrm{int} \left ( \bigcap_{i=1}^{n} S_i \right ) \subseteq \bigcap_{i=1}^{n} \mathrm{int} (S_i) \end{align}
• Now let $x \in \bigcap_{i=1}^{n} \mathrm{int}(S_i)$. Then $x \in \mathrm{int} (S_i)$ for all $i \in \{1, 2, ..., n \}$. So for each $i$ there exists an $r_i > 0$ such that:
(4)
\begin{align} \quad B(x, r_i) \subseteq S_i \end{align}
• Let $r = \min \{ r_1, r_2, ..., r_n \}$. Then $B(x, r) \subseteq S_i$ for all $i \in \{ 1, 2, ..., n \}$ which shows that $\displaystyle{B(x, r) \in \bigcap_{i=1}^{n} S_i}$. Thus $\displaystyle{x \in \mathrm{int} \left ( \bigcap_{i=1}^{n} S_i \right )}$. This shows that:
(5)
\begin{align} \quad \mathrm{int} \left ( \bigcap_{i=1}^{n} S_i \right ) \supseteq \bigcap_{i=1}^{n} \mathrm{int} (S_i) \end{align}
• We conclude that then $\displaystyle{\mathrm{int} \left ( \bigcap_{i=1}^{n} S_i \right )= \bigcap_{i=1}^{n} \mathrm{int} (S_i)}$. $\blacksquare$
Theorem 2: Let $(M, d)$ be a metric space and let $\mathcal F$ be a finite collection of subsets of $M$. Then $\displaystyle{\mathrm{int} \left ( \bigcap_{S \in \mathcal F} S \right ) \subseteq \bigcap_{S \in \mathcal F} \mathrm{int} (S) }$.
• Proof: Let $\displaystyle{x \in \mathrm{int} \left ( \bigcap_{S \in \mathcal F} S \right )}$. Then $x$ is an interior point of $\displaystyle{\mathrm{int} \left ( \bigcap_{S \in \mathcal F} S \right )}$ and so there exists an $r > 0$ such that:
(6)
\begin{align} \quad B(x, r) \subseteq \mathrm{int} \left ( \bigcap_{S \in \mathcal F} S \right ) \end{align}
• Therefore $B(x, r) \subseteq S$ for all $S \in \mathcal F$, so $x \in \mathrm{int} (S)$ for all $S \in \mathcal F$. This implies that $\displaystyle{x \in \bigcap_{S \in \mathcal F} \mathrm{int} (S)}$ and so:
(7)
\begin{align} \quad \mathrm{int} \left ( \bigcap_{S \in \mathcal F} S \right ) \subseteq \bigcap_{S \in \mathcal F} \mathrm{int} (S) \quad \blacksquare \end{align}
Note that in Theorem 1 we relied on the fact that were looking at a finite intersection to show equality. Equality in Theorem 2 does not hold in general though.
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2020-02-28 11:23:35
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https://argb.in/ankit/backup-sqlite
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Ankit
Backup SQLite
SQLite is one of the most popular embedded relational database. The SQLite native libraries or language bindings are available for most popular languages. SQLite files can be accessed either interactively or using shell scripts using the command-line utility.
SQLite database does not run a daemon on the system. Instead, it runs as part of the application. It stores the data in a single file. For backup, copying the data file is not the best approach, as the process might still be writing the data. SQLite utility implements a dot command (.backup) to simplify the process. The command takes a snapshot of the current data and writes it to a newer file in a consistent way. The backup is a new copy of the data file that is also an SQLite database.
The WriteFreely instance running this website uses an SQLite database. I've written the following shell script to take the backup of the database. To automate the daily backups, I've configured Cron to run the script at midnight.
!#/bin/sh
sqlite3 /path/to/writefreely.db ".backup /backup/path/backup.db"
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2020-08-07 00:41:13
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http://mathhelpforum.com/discrete-math/221411-power-sets.html
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1. ## Power Sets
Hey guys, I wanted to see if someone could check my work.
Find $\mathcal{P}(\mathcal{P}(\left\{1\right\}))$ and its cardinality.
$\mathcal{P}(\left\{1\right\})=\left\{\oslash, \left\{1 \right\}\right\}$ then,
$\mathcal{P}(\mathcal{P}(\left\{1\right\}))= \mathcal{P}(\left\{\oslash, \left\{1 \right\}\right\}) = \left\{\oslash, \left\{1 \right\}, \left\{\left\{1 \right\}\right\}, \left\{\oslash, \left\{1 \right\}\right\}\right\}$
$| \mathcal{P}(\mathcal{P}(\left\{1\right\}))| = 4$
correct?
2. ## Re: Power Sets
Originally Posted by amthomasjr
Hey guys, I wanted to see if someone could check my work.
Find $\mathcal{P}(\mathcal{P}(\left\{1\right\}))$ and its cardinality.
$\mathcal{P}(\left\{1\right\})=\left\{\oslash, \left\{1 \right\}\right\}$ then,
$\mathcal{P}(\mathcal{P}(\left\{1\right\}))= \mathcal{P}(\left\{\oslash, \left\{1 \right\}\right\}) = \left\{\oslash, \left\{1 \right\}, \left\{\left\{1 \right\}\right\}, \left\{\oslash, \left\{1 \right\}\right\}\right\}$
$| \mathcal{P}(\mathcal{P}(\left\{1\right\}))| = 4$ correct?
Yes it i is correct.
In fact if $A$ is a finite set then $\left| {\mathcal{P}(\mathcal{P}(A))} \right| = {2^{\left( {{2^{\left| A \right|}}} \right)}}$
Thank you
|
2017-01-24 06:43:36
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https://istopdeath.com/evaluate-f-22x/
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# Evaluate f(-2)=2^x
Replace the variable with in the expression.
Simplify the result.
Rewrite the expression using the negative exponent rule .
Raise to the power of .
The final answer is .
The result can be shown in multiple forms.
Exact Form:
Decimal Form:
Evaluate f(-2)=2^x
Scroll to top
|
2023-02-01 00:12:55
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https://casmusings.wordpress.com/2012/06/16/in-plain-sight-but-unseen/
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# In Plain Sight, but Unseen
Thanks to a comment from Doug Kuhlmann on my last post, I’ve got a few new cool connections on the transformational effects of the parameters in $y=a\cdot x^2+b\cdot x+c$ on its graph. This is exactly why I share. Thanks, Doug!!
THE NEW PATTERN: Use this GeoGebraTube Web document to model the problem. Set the value of b to any non-zero value and vary a. The parabola’s vertex moves along a line, as shown below.
As with the changes in the b parameter, define that line and prove your claims.
[The GeoGebra link above does not produce the vertex geometry trace footprints shown in the image. If you want to create these, download GeoGebra and create this simple document for yourself. It is FREE. If anyone wants explicit instructions for how to do this, email me and I’ll post instructions on the ‘blog.]
Another option to see the line is to use Geogebra’s locus tool. It requires two inputs: which object is the locus following, and which variable driving the variation. After selecting the locus tool, click on the vertex and then the slider for a. You get the next image.
SOLUTION ALERT! Don’t read further if you want to solve the problem for yourself.
I knew the line contained the vertex and noticed that it seemed to pass through y-intercept. Predicting the y-intercept was c, all I needed was the slope. With my prediction of the two generic points, I could compute that, too. I enjoy symbol manipulation for the mental exercise. The symbols (to me) weren’t all that complicated, so I took a brief moment of fun solving that by hand. But this is another of those situations where the symbol manipulation isn’t the point, so using my CAS is 100% legitimate. It is also a great leveler of ability for those intimidated for any reason by algebraic manipulations.
The next image is a great use of CAS commands to find the line’s slope. In particular, notice the use of a function definition to minimize the algebraic clutter through function notation.
Lovely and surprisingly simple. That means the line the parabola’s vertex follows when a varies for non-zero b is $y=\frac{b}{2}\cdot x+c$.
Students often overlook the domain warning. It doesn’t matter for the creation of the line, but ultimately lies at the heart of the unequal spacing of the vertex footprints in the first image and explains the unique behavior of the parabola’s movement.
If a student didn’t use the vertex and y-intercept to derive the linear equation, a CAS solve command could legitimately be used to show that those two generic points were always on the line.
MOTION ALONG THE LINE: One of the interesting parts of this problem is how the parabola moves along $y=\frac{b}{2}\cdot x+c$. After some play with the GeoGebra document, you can see that as $|a|\rightarrow 0$ the parabolas’ vertices move infinitely far away from the y-axis, and as $|a|\rightarrow\infty$ the vertices approach the y-axis. This also can be seen numerically from the generic x-coordinate of the vertex, $-\frac{b}{2a}$. For a fixed, non-zero value of b, the fraction representing the x-coordinate of the vertex increases in magnitude as $|a|\rightarrow 0$ and decreases in magnitude toward 0 as $|a|\rightarrow\infty$.
The vertex trace points in the first image above are separated by $\Delta a=.01$. The reason for the differences in distances between the points noted above is because $-\frac{b}{2a}$ does not change linearly when a changes linearly. As $|a|\rightarrow\infty$, $-\frac{b}{2a}\rightarrow 0$ slower and slower, explaining the increasing density of the vertex trace points near the y-axis.
When $a=0$, the x-coordinate of the vertex is undefined. At that moment, the generic quadratic, $y=a\cdot x^2+b\cdot x+c$, becomes the degenerate $y=b\cdot x+c$, a line. Graphing that line (the red dashed line below) against a trace of all possible parabolas as a varies, the degenerate parabola resulting when $a=0$ is precisely the tangent line to all of these parabolas at their y-intercept, $(0,c)$–a pretty extension on a connection suggested by Dave Radcliffe on a cross-posting of my initial post. Nice.
A FINAL NOTE: My memory suggests that I’ve seen this pattern before in some of the numerous times I’ve presented the b-variation of this problem in conferences and assigned it in classes. Despite all the times I must have seen it, the pattern never rose to my active conscience. Serendipitously, I’m currently reading Tina Seelig’s inGenius: A Crash Course on Creativity. I offer two quotes from her Are You Paying Attention? chapter:
• We think we understand the world and look for the patterns that we already recognize. (p. 71)
• We focus predominantly on things that are at our eye level rather than looking around more broadly. In addition, we pay attention to objects that we expect to find and ignore those things that don’t fit. (p. 71)
The moral: Even after all of my attempts and success at finding unique patterns, I missed this one until Doug pointed it out to me. I suspect my focus on what I knew about b‘s effect blinded me to the a effect. This is a great reminder to me to always hold myself ready to see beauty and pattern in unexpected places.
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2017-05-27 04:11:21
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https://www.jipb.net/EN/abstract/abstract22209.shtml
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J Integr Plant Biol. ›› 2008, Vol. 50 ›› Issue (7): 835-848.
Special Issue: Plant Signal Transduction
• Plant-environmental Interactions •
### Membrane Transporters for Nitrogen, Phosphate and Potassium Uptake in Plants
Yi-Fang Chen, Yi Wang and Wei-Hua Wu
• Received:2008-04-15 Accepted:2008-04-20 Published:2008-07-10
Abstract: Nitrogen, phosphorous and potassium are essential nutrients for plant growth and development. However, their contents in soils are limited so that crop production needs to invest a lot for fertilizer supply. To explore the genetic potentialities of crops (or plants) for their nutrient utilization efficiency has been an important research task for many years. In fact, a number of evidences have revealed that plants, during their evolution, have developed many morphological, physiological, biochemical and molecular adaptation mechanisms for acquiring nitrate, phosphate and potassium under stress conditions. Recent discoveries of many transporters and channels for nitrate, phosphate and potassium uptake have opened up opportunities to study the molecular regulatory mechanisms for acquisition of these nutrients. This review aims to briefly discuss the genes and gene families for these transporters and channels. In addition, the functions and regulation of some important transporters and channels are particularly emphasized.
Editorial Office, Journal of Integrative Plant Biology, Institute of Botany, CAS
No. 20 Nanxincun, Xiangshan, Beijing 100093, China
Tel: +86 10 6283 6133 Fax: +86 10 8259 2636 E-mail: jipb@ibcas.ac.cn
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2022-08-13 03:48:38
|
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https://ltwork.net/what-is-the-topic-sentence-in-each-of-the-following-paragraphs--14136639
|
# What is the topic sentence in each of the following paragraphs?
###### Question:
What is the topic sentence in each of the following paragraphs?
$What is the topic sentence in each of the following paragraphs?$
### What is the purpose or overall message of this image?
What is the purpose or overall message of this image? $What is the purpose or overall message of this image?$...
### Earths mountains float in balance on top of the mantle as the result of
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### What is the quotient of the following division problem? 3139÷19=?
What is the quotient of the following division problem? 3139÷19=? ...
### According to the cladogram, which organisms are the most primitive? flies wasps beetles moths
According to the cladogram, which organisms are the most primitive? flies wasps beetles moths...
### In a chemical reaction, atoms are created or destroyed. select the best answer from the choices provided
In a chemical reaction, atoms are created or destroyed. select the best answer from the choices provided t f...
### What is the first step in solving this system of equations by elimination? 5a + 3b = –9 2a – 5b = –16 a. subtract
What is the first step in solving this system of equations by elimination? 5a + 3b = –9 2a – 5b = –16 a. subtract the two equations. b. add the two equations. c. solve for a. d. multiply the first equation by 2 and the second by –5....
### I WILL MARK BRANLIEST YALL HELP A STUDENT IN NEED
I WILL MARK BRANLIEST YALL HELP A STUDENT IN NEED $I WILL MARK BRANLIEST YALL HELP A STUDENT IN NEED$...
### WIll give brainlisit if done correctly“Why was the label wrong? It said $3.99!”You respond, “It rang up as$4.27 because
WIll give brainlisit if done correctly “Why was the label wrong? It said $3.99!” You respond, “It rang up as$4.27 because of tax. (Write two sentences to explain what taxes are.)" “Why do we pay taxes? It seems so unfair! Now I can’t have my toy!” You respond, “We pay them because th...
### Solve them linear inequality then graph the inequality
Solve them linear inequality then graph the inequality $Solve them linear inequality then graph the inequality$...
### Evaluate the polynomial 6x - y for x= -3 and y= 2.
Evaluate the polynomial 6x - y for x= -3 and y= 2....
### This task is to implement an encoder whose input is a one 4-bit hex signal and the outputs are seven
This task is to implement an encoder whose input is a one 4-bit hex signal and the outputs are seven 1- bit signals for each display segment (a-g). 1. create the encoder. vhd file: a. under ""flow navigator"" click ""add sources."" b. select ""create or add design sources."" c. create task3.vhd and...
### I really need help. And its setted up-Also I will do a 50 point giveaway
I really need help. And its setted up-Also I will do a 50 point giveaway $I really need help. And its setted up-Also I will do a 50 point giveaway$...
### From the list of options below, choose the factors influencing why towns began to grow in medieval england.
From the list of options below, choose the factors influencing why towns began to grow in medieval england. select all that apply. tradeindustrythe crusadesmanorialism...
### (14, 5) , (21, 3) , (26, 8) , (33, 6) , (38, 11) , ( , ) , ( , ) , (57, 12) , (62, 8) , (69, 15) , (74, 11)fill in the sequence
(14, 5) , (21, 3) , (26, 8) , (33, 6) , (38, 11) , ( , ) , ( , ) , (57, 12) , (62, 8) , (69, 15) , (74, 11) fill in the sequence...
### Kara wants to open a retailing outlet in her town. her family has been in the business of selling hardware items before.
Kara wants to open a retailing outlet in her town. her family has been in the business of selling hardware items before. she wants to take this business forward by opening a store that caters to the hardware supplies industry. what kind of retail outlet will kara most likely open? a. discount reta...
### Joan has $587.66 in a savings account .she deposits$15.50 each week and makes no withdrawals. amount
joan has $587.66 in a savings account .she deposits$15.50 each week and makes no withdrawals. amount of money in the account in w weeks?...
### Select the correct answer. the south fought the civil war as which of the following? a a compact between states b a group
Select the correct answer. the south fought the civil war as which of the following? a a compact between states b a group of individual states c a union of states d a united nation...
### A package of 8 batteries costs $4.72. What is the cost per battery?(hint:find the unit cost, or the cost of one battery A package of 8 batteries costs$4.72. What is the cost per battery? (hint:find the unit cost, or the cost of one battery...
I need help please quick my grades are closing today there's a deadline today. $I need help please quick my grades are closing today there's a deadline today.$...
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2023-01-28 03:16:40
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|
https://www.tug.org/pipermail/texhax/2014-November/021436.html
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# [texhax] getting to global scope from within a group (latex/ifthen) [SOLVED]
Donald Arseneau asnd at triumf.ca
Mon Nov 17 03:34:49 CET 2014
```On Sat, November 15, 2014 2:56 pm, Boylan, Ross wrote:
> I did a minimalist version of David's suggestion, relying on the fact that
> the internals of the ifthen package go through \ifX for boolean X, and
> this can be globally reset with \global\let.
>
> Would it have been better to use counters? I just needed something global
> I could toggle.
> A quick check shows that latex counters are also updated globally using,
> e.g. \stepcounter.
I think you may have been better using a counter. That way
you'd inherit the infrastructure that makes partial compilation
with \includeonly work, whereas now a partial compilation
|
2018-02-21 07:21:23
|
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https://msp.org/ant/2020/14-9/p01.xhtml
|
#### Vol. 14, No. 9, 2020
Recent Issues
The Journal About the Journal Editorial Board Editors’ Interests Subscriptions Submission Guidelines Submission Form Policies for Authors Ethics Statement ISSN: 1944-7833 (e-only) ISSN: 1937-0652 (print) Author Index To Appear Other MSP Journals
The Brauer group of the moduli stack of elliptic curves
### Benjamin Antieau and Lennart Meier
Vol. 14 (2020), No. 9, 2295–2333
##### Abstract
We compute the Brauer group of ${\mathsc{ℳ}}_{1,1}$, the moduli stack of elliptic curves, over $Specℤ$, its localizations, finite fields of odd characteristic, and algebraically closed fields of characteristic not $2$. The methods involved include the use of the parameter space of Legendre curves and the moduli stack $\mathsc{ℳ}\left(2\right)$ of curves with full (naive) level $2$ structure, the study of the Leray–Serre spectral sequence in étale cohomology and the Leray spectral sequence in fppf cohomology, the computation of the group cohomology of ${S}_{3}$ in a certain integral representation, the classification of cubic Galois extensions of $ℚ$, the computation of Hilbert symbols in the ramified case for the primes $2$ and $3$, and finding $p$-adic elliptic curves with specified properties.
##### Keywords
Brauer groups, moduli of elliptic curves, level structures, Hilbert symbols
##### Mathematical Subject Classification 2010
Primary: 14F22
Secondary: 14H52, 14K10
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2020-10-22 21:24:35
|
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|
https://quantumcomputing.stackexchange.com/questions/9009/how-should-i-interpret-2-rangle3-rangle
|
# How should I interpret $|2\rangle|3\rangle$?
I am a beginner at QC. I was going through the basics of multi-qubits I encountered a state $$|2\rangle|3\rangle$$.
I want clarification on the following points:
1. Can I write $$|2\rangle$$ as $$|10\rangle = |1\rangle|0\rangle$$ always?
2. If $$|2\rangle = |10\rangle = |1\rangle|0\rangle$$, Can I write $$|2\rangle|3\rangle = |10\rangle|11\rangle = |1011\rangle$$?
Assume $$|0\rangle$$, $$|1\rangle$$ in computational basis. $$|0\rangle = \begin{bmatrix} 1 \\ 0 \end{bmatrix}$$, $$|1\rangle = \begin{bmatrix} 0 \\ 1 \end{bmatrix}$$
• I've never seen decimal notation like $|2\rangle$; it is not clear whether it means 2-qubit state $|10\rangle$ or 3-qubit state $|010\rangle$ or other multiqubit state. – kludg Dec 1 '19 at 10:41
• @kludg: It is generaly basis vector representing decimal number 2. You are right that it can be $|011\rangle$ as well, however, it should be clear from dimensionality of a problem. – Martin Vesely Dec 1 '19 at 11:44
1) $$|2\rangle$$ is $$|10\rangle$$, or vector $$\begin{bmatrix}0\\0\\1\\0\end{bmatrix}$$ in Hilbert space representing two q-bits system.
2)Yes, it is possible as well. The first two q-bits in in state $$|10\rangle$$ and last two q-bits in state $$|11\rangle$$, hence the state of all four q-bits is $$|1011\rangle$$, or $$|11\rangle$$ (eleven!) in decimal but I would recommend to avoid writing decimal numbers in this case as they can be confused with binary numbers easily.
|
2021-05-08 10:16:31
|
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https://jp.maplesoft.com/support/help/addons/view.aspx?path=simplify%2Fconstants
|
constants - Maple Help
simplify/constants
convert constant functions to constants
Calling Sequence simplify(expr, 'constants')
Parameters
expr - expression 'constants' - literal name; 'constants', including the enclosing with quotes ''
Description
• In Maple, constants are not automatically considered to be constant functions, and therefore $\mathrm{\pi }\left(x\right)$ does not automatically simplify to Pi whereas $3\left(x\right)$ does simplify to $3$. This procedure performs those simplifications.
• Candidates for simplification are symbols in the expression sequence constants which are not themselves procedures or for which there does not exist an evalf function.
• Note: In order to perform this simplification using the syntax simplify(expr, 'constants'), it is mandatory that the keyword 'constants' be enclosed with quotes, to prevent the premature evaluation of this keyword to the list of Maple constants
> constants;
${\mathrm{false}}{,}{\mathrm{\gamma }}{,}{\mathrm{\infty }}{,}{\mathrm{true}}{,}{\mathrm{Catalan}}{,}{\mathrm{FAIL}}{,}{\mathrm{\pi }}$ (1)
Examples
> $\mathrm{simplify}\left(\left[\mathrm{false}\left(x\right),\mathrm{true}\left(x\right),\mathrm{FAIL}\left(x\right)\right],'\mathrm{constants}'\right)$
$\left[{\mathrm{false}}{,}{\mathrm{true}}{,}{\mathrm{FAIL}}\right]$ (2)
> $\mathrm{simplify}\left(\mathrm{\gamma }\left(x\right)\mathrm{Catalan}\left(x\right)\mathrm{\pi }\left(x\right)+\mathrm{\infty }\left(x\right),'\mathrm{constants}'\right)$
${\mathrm{\gamma }}{}{\mathrm{Catalan}}{}{\mathrm{\pi }}{+}{\mathrm{\infty }}$ (3)
> $\mathrm{evalf}\left(\mathrm{\gamma }\right)$
${0.5772156649}$ (4)
> $\mathrm{evalf}\left(\mathrm{\gamma }\left(3\right)\right)$
${0.002053834420}$ (5)
> $\mathrm{constants}≔\mathrm{constants},E,F\left[a\right]$
${\mathrm{constants}}{≔}{\mathrm{false}}{,}{\mathrm{\gamma }}{,}{\mathrm{\infty }}{,}{\mathrm{true}}{,}{\mathrm{Catalan}}{,}{\mathrm{FAIL}}{,}{\mathrm{\pi }}{,}{E}{,}{{F}}_{{a}}$ (6)
> $\mathrm{simplify}\left(\left[E\left(3\right),F\left[a\right]\right],'\mathrm{constants}'\right)$
$\left[{E}{,}{{F}}_{{a}}\right]$ (7)
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2023-03-23 02:09:26
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https://www.ademcetinkaya.com/2022/09/is-slgn-stock-expected-to-rise.html
|
Stock prediction is a very hot topic in our life. However, in the early time, because of some reasons and the limitation of the device, only a few people had the access to the study. Thanks to the rapid development of science and technology, in recent years more and more people are devoted to the study of the prediction and it becomes easier and easier for us to make stock prediction by using different ways now, including machine learning, deep learning and so on. We evaluate Silgan Holdings prediction models with Modular Neural Network (CNN Layer) and Lasso Regression1,2,3,4 and conclude that the SLGN stock is predictable in the short/long term. According to price forecasts for (n+4 weeks) period: The dominant strategy among neural network is to Hold SLGN stock.
Keywords: SLGN, Silgan Holdings, stock forecast, machine learning based prediction, risk rating, buy-sell behaviour, stock analysis, target price analysis, options and futures.
## Key Points
1. What is prediction model?
2. Dominated Move
3. Can we predict stock market using machine learning?
## SLGN Target Price Prediction Modeling Methodology
Recently, there has been a surge of interest in the use of machine learning to help aid in the accurate predictions of financial markets. Despite the exciting advances in this cross-section of finance and AI, many of the current approaches are limited to using technical analysis to capture historical trends of each stock price and thus limited to certain experimental setups to obtain good prediction results. On the other hand, professional investors additionally use their rich knowledge of inter-market and inter-company relations to map the connectivity of companies and events, and use this map to make better market predictions. For instance, they would predict the movement of a certain company's stock price based not only on its former stock price trends but also on the performance of its suppliers or customers, the overall industry, macroeconomic factors and trade policies. This paper investigates the effectiveness of work at the intersection of market predictions and graph neural networks, which hold the potential to mimic the ways in which investors make decisions by incorporating company knowledge graphs directly into the predictive model. We consider Silgan Holdings Stock Decision Process with Lasso Regression where A is the set of discrete actions of SLGN stock holders, F is the set of discrete states, P : S × F × S → R is the transition probability distribution, R : S × F → R is the reaction function, and γ ∈ [0, 1] is a move factor for expectation.1,2,3,4
F(Lasso Regression)5,6,7= $\begin{array}{cccc}{p}_{a1}& {p}_{a2}& \dots & {p}_{1n}\\ & ⋮\\ {p}_{j1}& {p}_{j2}& \dots & {p}_{jn}\\ & ⋮\\ {p}_{k1}& {p}_{k2}& \dots & {p}_{kn}\\ & ⋮\\ {p}_{n1}& {p}_{n2}& \dots & {p}_{nn}\end{array}$ X R(Modular Neural Network (CNN Layer)) X S(n):→ (n+4 weeks) $R=\left(\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right)$
n:Time series to forecast
p:Price signals of SLGN stock
j:Nash equilibria
k:Dominated move
a:Best response for target price
For further technical information as per how our model work we invite you to visit the article below:
How do AC Investment Research machine learning (predictive) algorithms actually work?
## SLGN Stock Forecast (Buy or Sell) for (n+4 weeks)
Sample Set: Neural Network
Stock/Index: SLGN Silgan Holdings
Time series to forecast n: 09 Sep 2022 for (n+4 weeks)
According to price forecasts for (n+4 weeks) period: The dominant strategy among neural network is to Hold SLGN stock.
X axis: *Likelihood% (The higher the percentage value, the more likely the event will occur.)
Y axis: *Potential Impact% (The higher the percentage value, the more likely the price will deviate.)
Z axis (Yellow to Green): *Technical Analysis%
## Conclusions
Silgan Holdings assigned short-term Ba2 & long-term B1 forecasted stock rating. We evaluate the prediction models Modular Neural Network (CNN Layer) with Lasso Regression1,2,3,4 and conclude that the SLGN stock is predictable in the short/long term. According to price forecasts for (n+4 weeks) period: The dominant strategy among neural network is to Hold SLGN stock.
### Financial State Forecast for SLGN Stock Options & Futures
Rating Short-Term Long-Term Senior
Outlook*Ba2B1
Operational Risk 6632
Market Risk8232
Technical Analysis5044
Fundamental Analysis7986
Risk Unsystematic6789
### Prediction Confidence Score
Trust metric by Neural Network: 90 out of 100 with 672 signals.
## References
1. Krizhevsky A, Sutskever I, Hinton GE. 2012. Imagenet classification with deep convolutional neural networks. In Advances in Neural Information Processing Systems, Vol. 25, ed. Z Ghahramani, M Welling, C Cortes, ND Lawrence, KQ Weinberger, pp. 1097–105. San Diego, CA: Neural Inf. Process. Syst. Found.
2. V. Mnih, A. P. Badia, M. Mirza, A. Graves, T. P. Lillicrap, T. Harley, D. Silver, and K. Kavukcuoglu. Asynchronous methods for deep reinforcement learning. In Proceedings of the 33nd International Conference on Machine Learning, ICML 2016, New York City, NY, USA, June 19-24, 2016, pages 1928–1937, 2016
3. Künzel S, Sekhon J, Bickel P, Yu B. 2017. Meta-learners for estimating heterogeneous treatment effects using machine learning. arXiv:1706.03461 [math.ST]
4. V. Borkar. A sensitivity formula for the risk-sensitive cost and the actor-critic algorithm. Systems & Control Letters, 44:339–346, 2001
5. Greene WH. 2000. Econometric Analysis. Upper Saddle River, N J: Prentice Hall. 4th ed.
6. Chernozhukov V, Chetverikov D, Demirer M, Duflo E, Hansen C, et al. 2016a. Double machine learning for treatment and causal parameters. Tech. Rep., Cent. Microdata Methods Pract., Inst. Fiscal Stud., London
7. Hastie T, Tibshirani R, Friedman J. 2009. The Elements of Statistical Learning. Berlin: Springer
Frequently Asked QuestionsQ: What is the prediction methodology for SLGN stock?
A: SLGN stock prediction methodology: We evaluate the prediction models Modular Neural Network (CNN Layer) and Lasso Regression
Q: Is SLGN stock a buy or sell?
A: The dominant strategy among neural network is to Hold SLGN Stock.
Q: Is Silgan Holdings stock a good investment?
A: The consensus rating for Silgan Holdings is Hold and assigned short-term Ba2 & long-term B1 forecasted stock rating.
Q: What is the consensus rating of SLGN stock?
A: The consensus rating for SLGN is Hold.
Q: What is the prediction period for SLGN stock?
A: The prediction period for SLGN is (n+4 weeks)
## People also ask
What are the top stocks to invest in right now?
Our Mission
As AC Investment Research, our goal is to do fundamental research, bring forward a totally new, scientific technology and create frameworks for objective forecasting using machine learning and fundamentals of Game Theory.
301 Massachusetts Avenue Cambridge, MA 02139 667-253-1000 pr@ademcetinkaya.com
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2022-10-02 17:11:57
|
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|
https://asmedigitalcollection.asme.org/heattransfer/article-abstract/122/4/721/445663/Combustion-Thermoelectric-Tube?redirectedFrom=fulltext
|
In direct combustion-thermoelectric energy conversion, direct fuel injection and reciprocation of the air flowing in a solid matrix are combined with the solid-gas interfacial heat transfer and the solid conduction to allow for obtaining superadiabatic temperatures at the hot junctions. While the solid conductivity is necessary, the relatively large thermal conductivity of the available high-temperature thermoelectric materials (e.g., Si–Ge alloys) results in a large conduction loss from the hot junctions and deteriorates the performance. Here, a combustion-thermoelectric tube is introduced and analyzed. Radially averaged temperatures are used for the fluid and solid phases. A combination of external cooling of the cold junctions, and direct injection of the fuel, has been used to increase the energy conversion efficiency for low thermal conductivity, high-melting temperature thermoelectric materials. The parametric study (geometry, flow, stoichiometry, materials) shows that with the current high figure of merit, high temperature $Si0.7Ge0.3$ properties, a conversion efficiency of about 11 percent is achievable. With lower thermal conductivities for these high-temperature materials, efficiencies about 25 percent appear possible. This places this energy conversion in line with the other high efficiency, direct, electric power generation methods. [S0022-1481(00)01304-9]
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2019-10-14 07:28:46
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https://math.stackexchange.com/questions/3259547/proof-verification-if-a-in-a-is-an-upper-bound-for-a-then-a-sup-a
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# Proof verification: If $a\in A$ is an upper bound for $A$, then $a=\sup A$
Prove that if $$a$$ is an upper bound for $$A$$, and if $$a$$ is also an element of $$A$$, then it must be that $$a=\sup A$$.
We are given the following lemma:
Lemma 1.3.8: Assume $$s\in\textbf{R}$$ is an upper bound for a set $$A\subseteq\textbf{R}$$. Then, $$s=\sup A$$ if and only if, for every choice of $$\epsilon>0$$, there exists an element $$a\in A$$ satisfying $$s-a<\epsilon$$.
Proof: It is given that $$a$$ is an upper bound for $$A$$. To verify that $$a=\sup A$$, we use Lemma 1.3.8. We want to show that $$a-\epsilon for some $$a_0\in A,\epsilon>0$$. Since $$a\in A$$, let $$a=a_0$$. Then we get that $$a-\epsilon, which holds for all $$\epsilon>0$$. Therefore, by Lemma 1.3.8, we have that $$a=\sup A$$.
• This seems correct to me. You can also prove this by using the definition that sup$A$ is the least upper bound for $A$. – Rick Jun 12 at 8:18
• @Rick Thank you. I will accept that if you post it as an answer so this question may be closed – csch2 Jun 12 at 8:29
Ok, here we go. Suppose $$b$$ is the supremum of $$A, b \ne a$$. Then, since $$a$$ is an upper bound, $$b < a$$. But we also have that $$a \in A$$. So, $$a \le b$$. This gives us a contradiction because of $$b . Hope this helps.
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2019-11-12 13:49:11
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https://math.stackexchange.com/questions/2227244/probability-that-x2-y2-is-divisible-by-k
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# Probability that $x^2-y^2$ is divisible by $k$
Let two numbers $x$ and $y$ be selected from the set of first $n$ natural numbers with replacement(i.e. the two numbers can be same).The question is to find out the probability that $x^2-y^2$ is divisible by $k\in \mathbb{N}$
For $k=2$
Any number can be expressed as $2p,2p+1$.Now $x^2-y^2=(x-y)(x+y)$ If both numbers are of form $2p+1$ then (x-y) would be divisible by $2$ .if two numbers are of different forms then it will not be divisible by $2$.So the probability in this case is $a^2+(1-a)^2$ where $a$ is probability that number chosen is divisible by $2$ which is $\frac{\lfloor \frac{n}{2} \rfloor}{n}$.However this gets complicated with $k=3$ onwards because numbers in different forms may be divisible.In other words if there a generalisation or way to solve for some large $k$.Thanks.
• $k=3$ is actually as easy as $k=2$, since any square is either $3p$ or $3p+1$. Looking at the difference of the two squares, we see that it is divisible by $3$ iff $x$ and $y$ are either both divisible by $3$ or neither of them are. – Arthur Apr 10 '17 at 7:15
• @navinstudent You might be interested in the complete analytic solution of your problem provided in my solution. – Dr. Wolfgang Hintze Apr 13 '17 at 17:12
A good way to generalize this is to use modular arithmetic, or essentially look at the remainder of $$\frac{x}{k}$$ and $$\frac{y}{k}$$. As you pointed out in your example for $$k=2$$, the numbers can only be expressed as $$2*p,2*p+1$$
This can be further generalized into a set of unique expressions that define every number or every $$x$$ and $$y$$ for a value $$k>1$$ and $$p≥0$$S={kp,kp+1...kp+(k-2),kp+(k-1)}$$Using modular arithmetic, we know that $$S\equiv \{0,1,2...k-2,k-1\}\pmod k$$ Since $$x$$ and $$y$$ are any two terms from the set $$S$$ for any $$p≥0$$, we know that $$x$$ and $$y$$ are really just equal to either any value from $$\{0,1,2...k-2,k-1\}$$ since we only need to look at the remainder when divided by $$k$$ to determine divisibility. ### Probability using mod Now we know $$x^2-y^2 = (x-y)(x+y)$$ meaning either $$(x-y)$$ or $$(x+y)$$ have to be divisible by $$k$$ in order to fulfill the requirement that $$x^2-y^2$$ is divisible by $$k$$. It should be noted that the probability of choosing a random number being expressed by any expression in set $$S$$ is uniform and is equal to $$\frac{1}{k}$$. For example, the probability of choosing a number is expressible by $$3p+1$$ is $$\frac{1}{3}$$. This can be proved by showing how every number expressed by $$kp$$ can always be paired with $$kp+1,kp+2,...kp+(k-2),kp+(k-1)$$, thus showing how there are an equal number of numbers of each type. Onto the question.... $$(x-y)$$ is divisible by $$k$$ if $$(x-y)\equiv 0\pmod k$$. This simplifies to $$(x$$ mod $$k)-(y \text{ mod}\ k) = 0$$. Thus we now need to find the probability that $$x$$ and $$y$$ are both expressible by the same expression in the set S. Now this comes down to a simple problem of asking what is the probability of choosing two of the same elements from the set $$S$$ without replacement, which is $$\frac{k}{k} *\frac{1}{k} =$$ $$\frac{k}{k^2}$$ $$(x+y)$$ is divisible by $$k$$ if $$(x+y)\equiv 0\pmod k$$. This simplifies to $$(x$$ mod $$k)+(y \text{ mod}\ k) = 0$$. Thus we now need to find the probability that $$x$$ and $$y$$ are chosen such that the $$x+y =$$ a multiple of $$k$$. Knowing that $$x$$ and $$y$$ must come from the set $$S$$, we can see there is a specific pairing of elements from $$S$$ that causes the addition of the pair of elements to be equal to $$2*kp+k$$, a multiple of $$k$$. The only times that the pair of elements, which will be the expression that expresses $$x$$ and $$y$$, add up to $$2*kp$$ is when one is expressed by $$kp+m$$ and the other being $$kp+(k-m)$$ for any whole number $$m$$. This is because $$(kp+m)+(kp+(k-m)) = 2*kp$$. For example $$3p+1$$ can be paired up with $$3p+2$$ since their sum is $$6p+3$$, a multiple of 3. Now we are now looking for the probability the two chosen elements from $$S$$ are pairs of each other (If both chosen elements are $$kp$$, it will be a multiple of k still) The total number of pairs that can be chosen is the total number of elements in set $$S$$ squared, which is $$k^2$$. The total number of pairs that fulfill the divisibility by $$k$$ is equal to $$\lfloor\frac{k}{2}\rfloor + 1$$. The ceiling function is applied to correct errors that occur when $$k$$ is odd as you obviously can't have a fractional number of pairs. The extra $$+1$$ is for including the case when both elements are the same. Thus the probability that $$(x+y)$$ is divisible by $$k$$ is $$\frac{\lfloor\frac{k}{2}\rfloor + 1}{k^2}$$. However now we encounter a problem of overlap, which are the cases when the two elements chosen from $$S$$ make $$(x-y)$$ and $$(x+y)$$ divisible by k. To get rid of the over count we need to subtract the number of times this happens for a given set $$S$$. We know we need to subtract once for the case when both elements are $$kp$$. We also need to subtract another one depending if $$k$$ is even or not. This is because when $$k$$ is even, we have the case where the element $$kp+\frac{k}{2}$$ can pair with itself to fulfill everything. Thus the final answer is $$\frac{k}{k^2}+\frac{\lfloor\frac{k}{2}\rfloor + 1}{k^2} - \frac{(k+1 \text{ mod }2)+1}{k^2}$$ The $$\frac{(k+1 \text{ mod }2)+1}{k^2}$$ tells us to subtract one more and subtract another if $$k$$ is even, hence we shifted the modular part by 1 since $$( \text{ even number mod }2 = 0)$$ • Thanks for your answer.I think you have approached the problem by taking that the two numbers are chosen from the infinite set of natural numbers but how will the answer depend on n if the two numbers are chosen from the set of first n natural numbers with replacement. Before that would you please explain more on how you applied the ceiling function to get the probability that x+y is divisible by k.Thanks. – Navin Apr 10 '17 at 7:47 • The number of pairings that work is always equal to k divided by 2. However, there is a special case when k is odd. We can just check a few explicit exmaples to prove whether or not to use ceiling or floor. If k is 3, we can pair 0,0 and 1,2. If k is 4, we pair 0,0 1,3 2,2. So in fact I wrote it wrong, it should be the floor function as 3/2 floored is 1 and adding the 0,0 pair is 2. For k=4, the floor of k/2 is 2, and adding the 0,0 pair is 3. Sorry for the mistake, will change. I will also work on your other part of the question – Stone Apr 10 '17 at 8:03 • Suppose k = p \cdot q. x^2 - y^2 is also divisible by k if (x - y) is divisible by p and (x + y) is divisible by q. So it is not necessary that ( x- y) is divisible by k or (x + y) is divisible by k. – PSPACEhard Apr 10 '17 at 8:10 • If you can find an example where such a case appears and either x-y doesn't be equal 0 or x+y doesn't equal a multiple of k, then your point would be valid. However I haven't been able to find such an example – Stone Apr 10 '17 at 8:43 • Suppose n = 8, k = 8 and x = 3, y = 1. Would this be a valid case? – PSPACEhard Apr 10 '17 at 12:42 EDIT 12.04.17 After some days of studying this interesting problem, I provide now a complete analytic solution of the problem by retrieving a related problem in OEIS. I give exact values of the first few probabilities, and give an exact general formula for the probababily if the divisor n is prime. A Monte-Carlo-Simulation is also presented. Notice that my findings came up in opposite temporal order. Analytic solution to the problem Looking up the sequence of the first few terms of$$s(n) = p(n) n^2 \tag{1}$$which are$${1, 2, 5, 8, 9, 10, 13, 24, 21, 18, 21, 40, 25, 26, 45}$$in the online-encyclopedia of integer sequences https://oeis.org/ gives us the entry A062803 "Number of solutions to x^2 == y^2 mod(n)", created initially by Ahmed Fares (ahmedfares(AT)my-deja.com), Jul 19 2001. A formula was devised by Vladeta Jovovic on Sep 22, 2003: "Multiplicative with a(2^e)=e*2^e and a(p^e)=((p-1)*e+p)*p^(e-1) for an odd prime p." Let us explore this a little bit closer. Our problem is transformed into the question of the number s(n) of solutions to the congruence$$x^2-y^2 == 0, \; mod(n)\tag{1}$$Our probabilities are then given by$$p(n) = s(n)/n^2\tag{2}$$Suppose the number n has the representation$$n = \prod_{i=1}^{k} p_i^{a_i}\tag{3}$$where p_i is the i-th prime number appearing in n in ascending order, a_i ist multiplicity (exponent) and k is the number of different prime factors in n. Notice that in number theory k is traditionally called \omega(n), the number of distinct prime factors. It is implemented in Mathematica as PrimeNu[n]. Then the statement of being multiplicative means that we can apply the formula to each of the prime power factors separately, and multiply the result together. This gives for n even$$s(n_{even}) = (a_1 2^{a_1}) \prod_{i=2}^{k} ((p_i-1)a_i+p_i) p_i^{a_i-1}\tag{4.1}$$and for n odd$$s(n_{odd}) = \prod_{i=1}^{k} ((p_i-1)a_i+p_i) p_i^{a_i-1}\tag{4.2}$$It is easy to see that for an odd prime n, s(n) = 2n-1, as claimed earlier. The formula simplifies if n is square-free. Then all non-vanishing a_i are equal to 1, and we find$$s(n_{even}) = 2 \prod_{i=2}^{k} (2 p_i-1)\tag{5.1}s(n_{odd}) = \prod_{i=1}^{k} (2 p_i-1)\tag{5.2}$$For those who are interested in encoding this formula, here is an example in Mathematica s[n_] := Module[{fi, pi, ai, pout}, fi = FactorInteger[n]; pi = #[[1]] & /@ fi; ai = #[[2]] & /@ fi; pout = If[OddQ[n], Product[((pi[[i]] - 1) ai[[i]] + pi[[i]]) pi[[i]]^(ai[[i]] - 1), {i, 1, Length[pi]}], ai[[1]] 2^ai[[1]] Product[((pi[[i]] - 1) ai[[i]] + pi[[i]]) pi[[i]]^(ai[[i]] - 1), {i, 2, Length[pi]}]]] The prime factor decomposition of n is done by the function FactorInteger[]. From this we extract the p_i and a_i, and then apply the formula of Jovovic. Exact values of the probabilities We make the (natural) assumption that all possible remainders of a randomly chosen number x modulo n have equal probability. Then we can calculate the exact values of the probabilities with the following piece of code (written here in Mathematica) h[n_] := 1/n^2 Count[Flatten[Table[Mod[x^2 - y^2, n], {x, 0, n - 1}, {y, 0, n - 1}]], 0] Explanation For a given divisor n the expression z = x^2-y^2 needs to be considered only for x and y bewteen 0 and n-1. The Table[] lists all elements x^2-y^2 mod(n), and Flatten[] puts them in one array. Now Count[.,0] counts the zeroes in this array. Dividing this by n^2 gives the probability. The result for n = 1..30 in the format \{n,p(n)\} are$$h(n)_{tab} = \left( \begin{array}{ccccc} \{1,1\} & \left\{2,\frac{1}{2}\right\} & \left\{3,\frac{5}{9}\right\} & \left\{4,\frac{1}{2}\right\} & \left\{5,\frac{9}{25}\right\} \\ \left\{6,\frac{5}{18}\right\} & \left\{7,\frac{13}{49}\right\} & \left\{8,\frac{3}{8}\right\} & \left\{9,\frac{7}{27}\right\} & \left\{10,\frac{9}{50}\right\} \\ \left\{11,\frac{21}{121}\right\} & \left\{12,\frac{5}{18}\right\} & \left\{13,\frac{25}{169}\right\} & \left\{14,\frac{13}{98}\right\} & \left\{15,\frac{1}{5}\right\} \\ \left\{16,\frac{1}{4}\right\} & \left\{17,\frac{33}{289}\right\} & \left\{18,\frac{7}{54}\right\} & \left\{19,\frac{37}{361}\right\} & \left\{20,\frac{9}{50}\right\} \\ \left\{21,\frac{65}{441}\right\} & \left\{22,\frac{21}{242}\right\} & \left\{23,\frac{45}{529}\right\} & \left\{24,\frac{5}{24}\right\} & \left\{25,\frac{13}{125}\right\} \\ \left\{26,\frac{25}{338}\right\} & \left\{27,\frac{1}{9}\right\} & \left\{28,\frac{13}{98}\right\} & \left\{29,\frac{57}{841}\right\} & \left\{30,\frac{1}{10}\right\} \\ \end{array} \right)$$The simulation results (see below) are in reasonable agreement with these results. If n is an odd prime number the probability is given by$$p(n)=\frac{2 n-1}{n^2}$$If n is composite I have not found the exact formula. The problem here seems to be related to quadratic residues. Monte-Carlo-Simulation EDIT 11.04.17 We distinguish two possible basic sets of integers from which to select for the divisibility test: 1. Set with repetition We create a set m consisting of all numbers z = x^2-y^2 of integers where 1<= x <= n_{max}, 1<= y <= n_{max}. 1. Set without repetition The set m_0 is obtained by removing from m all duplicates. Then, for a given divisor n we estimate the probability of divisibility by the ratio of the number of elements of the set for which \frac{z}{n} is integer relative to all elements of the set. The resulting probabilities for m_{max} = 10^3 and divisors in the range n = 1..30 are Case 1 (with repetition) List of results in the format (n, p(n))$$((1, 1.0), (2, 0.5000005), (3, 0.55533378), (4, 0.5000005), (5, \ 0.35968128), (6, 0.27766739), (7, 0.26530612), (8, 0.37475112), (9, \ 0.25933407), (10, 0.17984114), (11, 0.17355372), (12, 0.27766739), \ (13, 0.14792899), (14, 0.13265356), (15, 0.19973533), (16, \ 0.25000075), (17, 0.11417454), (18, 0.12966754), (19, 0.10246197), \ (20, 0.17984114), (21, 0.14736213), (22, 0.08677736), (23, \ 0.08502487), (24, 0.20808462), (25, 0.10419251), (26, 0.073965), (27, \ 0.11103881), (28, 0.13265356), (29, 0.06774345), (30, 0.09986916))$$The graph Case 2 (no repetition) List of results in the format (n, p(n))$$((1, 0.5125414), (2, 0.19847984), (3, 0.22136804), (4, 0.19847984), \ (5, 0.14290505), (6, 0.08393305), (7, 0.10653981), (8, 0.11713062), \ (9, 0.08960969), (10, 0.05436621), (11, 0.07129035), (12, \ 0.08393305), (13, 0.06139016), (14, 0.04070156), (15, 0.05862769), \ (16, 0.06700892), (17, 0.04836223), (18, 0.03377941), (19, \ 0.04369956), (20, 0.05436621), (21, 0.04403089), (22, 0.02740017), \ (23, 0.03676344), (24, 0.04804486), (25, 0.03682731), (26, \ 0.0236357), (27, 0.03531034), (28, 0.04070156), (29, 0.02972352), \ (30, 0.02204489))
Notice the remarkable peridocity with a period of $4$ in the "artificial" case 2. I'm sure there is a simple explanation for this, and that some readers can give it.
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2021-04-12 22:32:05
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https://de.zxc.wiki/wiki/Ideal_(Ringtheorie)
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# Ideal (ring theory)
In abstract algebra , an ideal is a subset of a ring that contains the zero element and is closed against the addition and subtraction of elements of the ideal and closed against multiplication with any ring elements. For example, the sum and difference of two even numbers are even again and the product of an even number with any whole number is also even. In addition, the 0 is even as additive neutral. That is, the set of even numbers is an ideal in the ring of whole numbers.
The term “ideal” is derived from the term “ideal number”: ideals can be viewed as generalizations of numbers .
The concept of ideals has its origin in the algebraic number theory of the 19th century by Ernst Eduard Kummer and was further developed by Richard Dedekind and Leopold Kronecker . For David Hilbert , an ideal was a system of infinitely many whole algebraic numbers in a rationality domain ( algebraic number field ), with the property that it also contained all linear combinations of these (with whole algebraic numbers as coefficients). This definition corresponds to today's concept of the broken ideal .
## "Ideal Numbers"
The origin of the ideals lies in the observation that in rings like the uniqueness of the decomposition into irreducible elements it does not apply: So is ${\ displaystyle \ mathbb {Z} \ left [{\ sqrt {-5}} \ right] = \ left \ {a + b \ cdot {\ sqrt {-5}} \ mid a, b \ in \ mathbb { Z} \ right \}}$
${\ displaystyle 6 = 2 \ cdot 3 = \ left (1 + {\ sqrt {-5}} \ right) \ cdot \ left (1 - {\ sqrt {-5}} \ right),}$
and the two factors of any decomposition are irreducible. Ernst Eduard Kummer found that the uniqueness can sometimes be restored by adding more ideal numbers. In the example, the factorizations are obtained by adding the number${\ displaystyle \ mathrm {i}}$
${\ displaystyle 2 = \ left (1+ \ mathrm {i} \ right) (1- \ mathrm {i}), \ quad 3 = {\ frac {1 + {\ sqrt {-5}}} {1+ \ mathrm {i}}} \ cdot {\ frac {1 - {\ sqrt {-5}}} {1- \ mathrm {i}}}}$
( you can see that the fractions on the right side are whole by their norms) as well
${\ displaystyle 1 \ pm {\ sqrt {-5}} = {\ frac {1 \ pm {\ sqrt {-5}}} {1 \ pm \ mathrm {i}}} \ cdot (1 \ pm \ mathrm {i}),}$
and the uniqueness is restored. From today's point of view, the introduction of the ideal number corresponds to the transition to the ( totality ring of) Hilbert's class field , in which all ideals (the totality ring) of an algebraic number field become main ideals . ${\ displaystyle \ mathrm {i}}$
Richard Dedekind realized that one can avoid these ideal numbers by considering the totality of all numbers divisible by them instead. The numbers and in the example have the common ideal prime factor , and the multiples of this number lying in are precisely the prime ideal${\ displaystyle 2}$${\ displaystyle 1 + {\ sqrt {-5}}}$${\ displaystyle 1+ \ mathrm {i}}$${\ displaystyle \ mathbb {Z} [{\ sqrt {-5}}]}$
${\ displaystyle \ left (2.1 + {\ sqrt {-5}} \ right) = \ left \ {a \ cdot 2 + b \ cdot \ left (1 + {\ sqrt {-5}} \ right) \ mid a, b \ in \ mathbb {Z} [{\ sqrt {-5}}] \ right \}.}$
If there is a “real” common factor, then the ideal consists precisely of its multiples and is therefore a main ideal. In wholeness rings of number fields (and more generally in the class of Dedekind rings named after him due to this fact ) one obtains a clear decomposition of every ideal (not equal to zero) into prime ideals (fundamental theorem of ideal theory).
## definition
In order to have suitable terms for non-commutative rings, a distinction is made between left, right and two-sided ideals:
Let it be a subset of a ring . is then called the left ideal if: ${\ displaystyle I}$${\ displaystyle \ mathbf {R} = (R, +, \ cdot)}$${\ displaystyle I}$
1 ,${\ displaystyle 0 \ in I}$
2: for all is ( subgroup criterion ),${\ displaystyle a, b \ in I}$${\ displaystyle a + b \ in I}$
3L: for each and is .${\ displaystyle a \ in I}$${\ displaystyle r \ in R}$${\ displaystyle r \ cdot a \ in I}$
A legal ideal is accordingly if, in addition to 1 and 2 , the following also applies: ${\ displaystyle I}$${\ displaystyle I}$
3R: For each and is .${\ displaystyle a \ in I}$${\ displaystyle r \ in R}$${\ displaystyle a \ cdot r \ in I}$
${\ displaystyle I}$one finally calls the two-sided ideal or only ideal for short if the left and right ideal are fulfilled , i.e. 1, 2, 3L and 3R are fulfilled. ${\ displaystyle I}$
### Remarks
• As an ideal, which contains, it is not empty. In fact, instead of condition 1 because of condition 2 and , the requirement that it is not empty is sufficient.${\ displaystyle I}$${\ displaystyle 0}$${\ displaystyle aa = 0}$${\ displaystyle I}$
• Requirements 1 and 2 are equivalent to the statement that a subgroup is of the additive group .${\ displaystyle (I, +)}$${\ displaystyle (R, +)}$
• Every ideal of also forms a subring of , but generally without a one . In the category of rings with one there is a sub-ring if and only if .${\ displaystyle I}$${\ displaystyle \ mathbf {R}}$${\ displaystyle (I, +, \ cdot)}$${\ displaystyle \ mathbf {R}}$${\ displaystyle 1 \ in \ mathbf {R}}$${\ displaystyle (I, +, \ cdot)}$${\ displaystyle I = \ mathbf {R}}$
• A left as well as a right ideal in is nothing but a - sub-module of , seen as -left- or -Rechtsmodul .${\ displaystyle I}$${\ displaystyle \ mathbf {R}}$${\ displaystyle \ mathbf {R}}$ ${\ displaystyle (I, +)}$${\ displaystyle \ mathbf {R}}$${\ displaystyle \ mathbf {R}}$${\ displaystyle \ mathbf {R}}$${\ displaystyle \ mathbf {R}}$${\ displaystyle (R, +)}$
• If the ring is commutative, then all three terms coincide, but in a non-commutative ring they can be different.
## Examples
• The set of even whole numbers is an ideal in the ring of all whole numbers .${\ displaystyle 2 \ mathbb {Z}}$${\ displaystyle \ mathbb {Z}}$
• The set of odd integers is not an ideal in ; it does not meet any of the three conditions.${\ displaystyle 2 \ mathbb {Z} +1}$${\ displaystyle \ mathbb {Z}}$
• The set of all polynomials with real coefficients, which are divisible by, form an ideal in the polynomial ring . The body is isomorphic to the complex numbers and is even maximally ideal .${\ displaystyle x ^ {2} +1}$ ${\ displaystyle \ mathbb {R} [X]}$${\ displaystyle \ mathbb {R} [X] / \ left (x ^ {2} +1 \ right)}$${\ displaystyle (x ^ {2} +1)}$
• The ring of all continuous functions from to contains the ideal of the functions with . Another ideal in FIG. 4 is the continuous functions with compact support , i.e. H. all functions that are equal to 0 for sufficiently large arguments.${\ displaystyle C (\ mathbb {R})}$${\ displaystyle \ mathbb {R}}$${\ displaystyle \ mathbb {R}}$${\ displaystyle f}$${\ displaystyle f (1) = 0}$${\ displaystyle C (\ mathbb {R})}$
• The non-commutative ring of the Hurwitz quaternions contains left and right ideals as well as two-sided ideals. However, they are all main ideals.
• The quantities and are always ideals of a ring . This is called the zero ideal and, if R has a one , the one ideal. If and its only two-sided ideals are called simple . A commutative simple ring with one that is not the zero ring is a body .${\ displaystyle \ {0 \}}$${\ displaystyle R}$${\ displaystyle R}$${\ displaystyle \ {0 \} = (0)}$${\ displaystyle 1}$${\ displaystyle R = (1)}$${\ displaystyle \ {0 \}}$${\ displaystyle R}$${\ displaystyle R}$
## Generation of ideals
All left and right ideals and all bilateral ideals each form a system of envelopes . The associated ideal operators are rarely also referred to as. ${\ displaystyle (\;),}$${\ displaystyle \ langle \; \ rangle}$
If a subset of the ring is then called ${\ displaystyle A}$${\ displaystyle R,}$
${\ displaystyle (A): = \ bigcap _ {J \ \ mathrm {Ideal \ von} \ R \ atop \ A \ subseteq J} J}$
the ideal generated by, it is the smallest (left, right or two-sided) ideal in which it contains. ${\ displaystyle A}$ ${\ displaystyle R,}$${\ displaystyle A}$
Has a single element so is ${\ displaystyle R}$${\ displaystyle 1,}$
${\ displaystyle (A) = \ {r_ {1} a_ {1} s_ {1} + \ dotsb + r_ {n} a_ {n} s_ {n} \ mid r_ {i}, s_ {i} \ in R, a_ {i} \ in A \},}$
and if is also commutative, the following even applies: ${\ displaystyle R}$
${\ displaystyle (A) = \ {r_ {1} a_ {1} + \ dotsb + r_ {n} a_ {n} \ mid r_ {i} \ in R, a_ {i} \ in A \}.}$
The main ideal produced by an element is ${\ displaystyle a}$
${\ displaystyle (a): = \ left (\ {a \} \ right.)}$
### Constructions
If a commutative ring with one and an ideal, then the radical of , which is defined as , is also an ideal. ${\ displaystyle R}$${\ displaystyle I \ subseteq R}$ ${\ displaystyle {\ sqrt {I}}}$${\ displaystyle I}$${\ displaystyle {\ sqrt {I}}: = \ {x \ in R \ mid \ exists r \ in \ mathbb {N}: x ^ {r} \ in I \}}$
If there is a ring, then there are two ideals : ${\ displaystyle R}$${\ displaystyle I, J \ subseteq R}$
${\ displaystyle I \ cap J = (I \ cap J)}$
• The set-theoretic union is generally not ideal, but the sum is an ideal:${\ displaystyle I \ cup J}$
${\ displaystyle I + J: = \ {a + b \ mid a \ in I, b \ in J \} = (I \ cup J)}$
Important: sums and associations of ideals are generally different constructs!
• The so-called complex product, which consists of the set of products of elements from with elements from , is generally not ideal either. As a product of and , therefore, the ideal is defined that is generated by:${\ displaystyle IJ,}$${\ displaystyle I}$${\ displaystyle J}$${\ displaystyle I}$${\ displaystyle J}$${\ displaystyle IJ}$
${\ displaystyle I \ cdot J: = \ left (\ left \ {from \ mid a \ in I, b \ in J \ right \} \ right) = (IJ)}$
If there is no risk of confusion with the complex product, then write the ideal product or in short${\ displaystyle I \ cdot J}$${\ displaystyle IJ.}$
• The quotient of and is an ideal that contains all for which the complex product is a subset of :${\ displaystyle I}$${\ displaystyle J}$${\ displaystyle x \ in R}$${\ displaystyle xJ}$${\ displaystyle I}$
${\ displaystyle I: J: = \ {x \ in R \ mid xJ \ subseteq I \}.}$
### Remarks
• The product of two ideals is always contained in their intersection: Are and coprime, so then even equality holds.${\ displaystyle I \ cdot J \ subseteq I \ cap J.}$${\ displaystyle I}$${\ displaystyle J}$${\ displaystyle I + J = R}$
• The ideal quotient is also often written in brackets in the literature: ${\ displaystyle (I: J).}$
• With the links sum and average, the set of all ideals of a ring forms a modular, algebraic association .
• Some important properties of these connections are summarized in Noether's isomorphism theorems .
## Special ideals
An ideal is called real when it is not whole . This is the case for rings with exactly when it is not in . ${\ displaystyle I}$${\ displaystyle R}$${\ displaystyle 1}$${\ displaystyle 1}$${\ displaystyle I}$
A real ideal is a maximum , if there is no greater real ideal, d. i.e., if for every ideal the following applies: ${\ displaystyle M}$${\ displaystyle I}$
${\ displaystyle M \ subseteq I \ subsetneq R \ Rightarrow M = I}$
With the help of Zorn's lemma it can be shown that every true ideal of a ring is contained in a maximal ideal. In particular, every ring with (except for the zero ring ) has a maximum ideal. ${\ displaystyle 1}$${\ displaystyle 1}$
A real ideal is prime , if for all ideals applies: ${\ displaystyle P}$${\ displaystyle I, J}$
${\ displaystyle I \ cdot J \ subseteq P \ Rightarrow I \ subseteq P}$ or ${\ displaystyle J \ subseteq P}$
In a ring with , every maximal ideal is prime. ${\ displaystyle 1}$
## Factor Rings and Cores
Ideals are important because they appear as kernels of ring homomorphisms and allow the definition of factor rings.
A ring homomorphism from the ring into the ring is an image with ${\ displaystyle f}$${\ displaystyle R}$${\ displaystyle S}$${\ displaystyle f \ colon R \ to S}$
${\ displaystyle {\ begin {array} {ll} f (0_ {R}) = 0_ {S}, & f (a + b) = f (a) + f (b), & f (ab) = f (a ) f (b) \ end {array}}}$ for all ${\ displaystyle a, b \ in R.}$
The core of is defined as ${\ displaystyle f}$
${\ displaystyle \ ker (f): = \ {a \ in R \ mid f (a) = 0_ {S} \}.}$
The core is always a two-sided ideal of ${\ displaystyle R.}$
Conversely, if you start with a two-sided ideal of then you can define the factor ring (read: “ modulo ”; not to be confused with a factorial ring ), the elements of which define the form ${\ displaystyle I}$${\ displaystyle R,}$ ${\ displaystyle R / I}$${\ displaystyle R}$${\ displaystyle I}$
${\ displaystyle a + I: = \ {a + i \ mid i \ in I \}}$
for one out . The image ${\ displaystyle a}$${\ displaystyle R}$
${\ displaystyle p \ colon R \ to R / I, \, a \ mapsto a + I}$
is a surjective ring homomorphism, the core of which is precisely the ideal . Thus the ideals of a ring are exactly the kernels of ring homomorphisms of${\ displaystyle I}$${\ displaystyle R}$${\ displaystyle R.}$
If the ring is commutative and a prime ideal, then it is an integrity ring , is a maximum ideal, then there is even a body . ${\ displaystyle R}$${\ displaystyle P}$${\ displaystyle R / P}$${\ displaystyle M}$${\ displaystyle R / M}$
The extreme examples of factor rings of a ring arise by splitting out the ideals or the factor ring is isomorphic to and is the trivial ring${\ displaystyle R}$${\ displaystyle (0)}$${\ displaystyle R.}$${\ displaystyle R / (0)}$${\ displaystyle R,}$${\ displaystyle R / R}$${\ displaystyle \ {0 \}.}$
## Norm of an ideal
For total rings of a number field , a norm of a (whole) ideal can be defined by (and for the zero ideal ). This norm is always a finite number and is related to the norm of the body extension for main ideals . In addition, this norm is multiplicative, i. H. . More generally, these norms are also considered for ideals in orders in number fields. ${\ displaystyle A}$ ${\ displaystyle K}$${\ displaystyle I}$${\ displaystyle N (I): = \ mathrm {card} (A / I))}$${\ displaystyle N ((0)): = 0}$ ${\ displaystyle N_ {K / \ mathbb {Q}},}$${\ displaystyle (a)}$${\ displaystyle | N_ {K / \ mathbb {Q}} (a) | = N ((a)).}$${\ displaystyle N (I \ cdot J)) = N (I) N (J)}$
## Individual evidence
1. Felix Klein: Lectures on the development of mathematics in the 19th century. Part 1. Springer, Berlin 1926 ( The basic teachings of the mathematical sciences in individual representations. 24, ), Chapter VII, section Theory of algebraic integers ... p. 321 f.
2. Felix Klein: Lectures on the development of mathematics in the 19th century. Part 1. Springer, Berlin 1926 ( The basic teachings of the mathematical sciences in individual representations. 24, ), p. 323.
3. J. Neukirch: Algebraic Number Theory. Springer-Verlag, Berlin 1992. ISBN 3-540-54273-6 ; Theorem I.3.3.
4. Lecture Algebra I. (PDF; 493 kB) Retrieved on August 24, 2013 .
## literature
• Felix Klein : Lectures on the development of mathematics in the 19th century. Part 1. Springer, Berlin 1926 ( The basic teachings of the mathematical sciences in individual representations. 24, ).
• Ernst Eduard Kummer : On the decomposition of the complex numbers formed from the roots of the unit into their prime factors. In: Journal for pure and applied mathematics . 35, 1847, pp. 327-367.
• David Hilbert : number report "The theory of algebraic number fields, annual report of the German Mathematicians Association", Vol. 4 pp. 175–546 1897 [1]
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2022-10-03 20:24:38
|
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https://solvedlib.com/n/e17-which-of-the-following-is-not-true-about-the-structure,15462714
|
# E17. Which of the following is NOT TRUE about the structure of DNA? A Two single strands of DNA are
###### Question:
E17. Which of the following is NOT TRUE about the structure of DNA? A Two single strands of DNA are held together by phosphodiester bonds_ B. DNA is comprised of two anti-parallel single strands of DNA_ C Guanine deoxyribonucleotides pair with cytosine deoxyribonucleotides_ D DNA forms a right-handed double helix E Twists in the DNA create minor and major grooves F2O. Diploid cells of yeast contain 32 chromosomes. How many chromatids are present at one pole of the meiotic spindle after completion of the second meiotic division? A. 64 B. 16 C 32 D. 128 E J19. Genes A and B are on one chromosome; 20 cM apart; genes C and D are on another chromosome, 10 cM apart: The mutant alleles, a, b, c and d are all completely recessive_ Cross homozygous A B C D individual with a homozygous a b c d individual to obtain F1 progeny: An F1 individual is testcrossed to the homozygous a b c d parent and 1000 progeny are obtained. Based on this information, answer the following questions_ a. What is the genotype of the F1 progeny? b. In the testcross of the F1 progeny to the homozygous a b c d parent; how many progeny would one predict should have the phenotypes A B c D?
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PS (16 points). Use scparable variables method to solve the IVP (A)y' ,,y(2) 1+2y (B) x +yy' 0,y(3) = 4 P6. (6 points). An cpidemic spreads through a population at rate proportional t0 the product of the number of people already infected and the number of people susceptible, but not yet ...
##### Y = E te] {1 + x}t
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##### Q5: (a)For the logic circuit below what is the output F =() Simplify the Eollowing Boolean function2 = (BC + AD)(AB + CD
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2022-05-17 15:14:29
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https://knowledge.rachelbrindle.com/flying/ifr/approaches.html
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# Approaches
The different type of IFR approaches and related information to flying them.
There are two basic kinds of approaches: Precision and Non-Precision. Precision, as the name implies, provide much more accurate navigation. Specifically, they provide more precise lateral and vertical guidance. Precision approaches prescribe a decision altitude or height where the pilot decides to either continue the approach or go missed. Non-precision, on the other hand, prescribe two things: a minimum descent altitude (or height), and a missed approach point. In a non-precision approach, you descend down to the MDA, then hold at or above that altitude until you either have everything needed to continue the approach, or you reach the missed approach point. If you reach the missed approach point, and are unable to continue the approach, you must execute the missed approach procedure.
## Descending below MDA/DA
You can descend below the MDA/DA when all of the following conditions are true:
• The aircraft is continuously in a position from which a descent to a landing on the intended runway can be made at a normal rate of descent using normal maneuvers. i.e. no chop and drop/slipping.
• The flight visibility is not less than the visibility prescribed in the standard instrument approach being used.
• At least one of the following visual references for the intended runway (the runway environment) is distinctly visible and identifiable to the pilot:
• The approach light system, except that the pilot may not descend below 100 feet above the touchdown zone elevation using the approach lights as a reference unless the red terminating bars or the red side row bars are also distinctly visible and identifiable.
• The threshold.
• The threshold markings.
• The threshold lights.
• The runway end identifier lights.
• The visual glideslope indicator.
• The touchdown zone or touchdown zone markings.
• The touchdown zone lights.
• The runway or runway markings.
• The runway lights.
## Approach Category
See AIM 5-4-7 (e).
The approach category for the aircraft is defined by the approach speed - VREF. Though, if you decide to fly the approach at a higher speed than VREF, then you must use the minimums specified for the category containing the speed you’re flying at (e.g. if you’re in a plane that with a VREF of 85 kts, but decide to fly the approach at 95 kts, then you must use the minimums defined for Category B).
Category ACategory BCategory CCategory DCategory E
Speed (kts)0 to 9091 to 120121 to 140141 to 165165 or greater
## Circling
See AIM 5-4-20.
Circling is a maneuver conducted after an approach, but is not a type of approach itself. Circling is performed whenever the final approach course is not aligned with the runway (within 30°), when landing on a different runway from the approach flown, or when the runway is not specified in the approach (e.g. flying a VOR-A).
Circling may be done if the winds do not favor the runway the chosen approach is for - e.g. if winds favor runway 15, but the approach is for runway 25, then you might fly the approach for runway 25, circle to land runway 15. It can also be done for other reasons - if there is VFR aircraft primarily using a different runway from the one specified for the approach you’re using, for example. Some approaches (commonly, VOR approaches) only specify circling.
### Circling Minimums
Published minimums provide at least a 300 foot obstacle clearance when you remain within the appropriate area of protection. Remember to descend to the circling MDA/DA, not the straight-in MDA.
If no circling minimums are published, circling requires ATC authorization and at basic VFR (at least 1000 ft ceiling with visibility 3+ statute miles).
Remain at or above the circling MDA altitude until the aircraft is “continuously in a position from which a descent to a landing on the intended runway can be made using normal maneuvers”. This is typically just inside the 180. Abeam for most circling minimums would have you low on downwind.
Depending on the approach category (speed you fly at), you have different circling radius - the minimum distance you must remain from the runway while circling to land.
.
Circling approach protected areas developed prior to late 2012 use the radius distances shown below.
Circling MDACAT ACAT BCAT CCAT DCAT E
All altitudes1.3 NM1.5 NM1.7 NM2.3 NM4.5 NM
Per the AIM, Circling approach protected areas developed after late 2012 use the radius distance shown below.
Circling MDA (feet MSL)CAT ACAT BCAT CCAT DCAT E
1000 or less1.3 NM1.7 NM2.7 NM3.6 NM4.5 NM
1001 to 30001.3 NM1.8 NM2.8 NM3.7 NM4.6 NM
3001 to 50001.3 NM1.8 NM2.9 NM3.8 NM4.8 NM
5001 to 70001.3 NM1.9 NM3.0 NM4.0 NM5.0 NM
7001 to 90001.4 NM2.0 NM3.2 NM4.2 NM5.3 NM
9001 and above1.4 NM2.1 NM3.3 NM4.4 NM5.5 NM
## Missed Approaches
Most instrument approach procedures also have a missed procedure. These are described in the top-left box of the approach plate, and then visually depicted as part of the top-down view.
If you choose to, must, or are instructed to go missed, you follow those instructions. If you were not on the final approach course (that is, if you were circling to land), then you get back on the final approach course as part of going missed.
## PT - Procedure Turn
Depicted as a line with a barb from a course in an approach plate. The barb indicates the direction to turn. Procedure turns are maneuvers enable the following:
• course reversal
• descent from an IAF
• Inbound course interception.
Procedure turns or hold-in-lieu-of-PTs are mandatory whenever they are depicted on the approach plate. However, they are not permitted when:
• “NoPT” is depicted on the plate
• Conducting a timed approach from a holding fix.
The maximum speed when performing a procedure turn is 200 kts IAS. You must remain within the charted distance (usually 10 NM), and comply with published altitudes for obstacle clearance.
The shape of the maneuver is mandatory if a teardrop or hold-in-lieu-of-PT is published. Otherwise, only the direction.
A teardrop procedure may be be published instead of a procedure turn.
Do not perform a procedure turn when SHARPTT:
• Straight-in approach clearance
• Holding in lieu of a procedure turn
• DME Arc
• NoPT shown on chart
• Timed approach from a holding fix
• Teardrop course reversal.
## ILS - Instrument Landing System
the ILS is composed of two main parts: the localizer and the glide slope. In addition, there are marker beacons at specific locations.
The ILS is described mostly in AIM 1-1-9.
### Localizer
The localizer is essentially a more-sensitive VOR that broadcasts a single radial. It’s located inline with the runway, behind the runway. It broadcasts 2 signals - one at 90 Hz, the other at 150 Hz. The intersection of the signals is aligned with the extended runway centerline, and the receiver on the airplane can interpret these to show how off course the airplane is.
The width of it is between 3° and 6° - such that the width at the runway threshold is 700 feet. Usually, it’s a 5° total width (2.5° full deflection on each side, or 4 times more sensitive than a VOR). The coverage range is a 35° on each side of the centerline up to 10 NM. After that, out to 18 NM, with 10° each side of the centerline. It goes out to an altitude of 4500 feet.
### Glide Slope
The glide slope provides vertical course guidance. The transmitter is located off to the side of the runway (about 250-650 away from the runway centerline), between 750 and 1,250 feet behind the runway threshold.
Similar to the localizer, the glide slope broadcasts 2 signals, at 90 and 150 Hz, with the intersection of them providing a - usually - 3° glide slope down to the runway.
A glide slope has a width of 1.4° (0.7° on either side), with a range out to 10 NM. As previously stated, it provides a 3° slope, unless otherwise charted.
⚠️ Due to, literally how physics works, there’s a false glide slope significantly above the normal glide slope.
### Marker Beacons
Marker beacons provide range information over specific points on the approach. There are 3 that are part of an ILS, named the outer marker, middle marker, and inner marker. There also exists the back course marker, but that’s not part of the ILS approach.
Marker TypeLocationColorSignal
Outer Marker4-7 NM out. Approximately where the aircraft should intercept the glide slope.BlueDashes “- - -”
Middle Marker~3500 feet from the runway & 200 feet above the touchdown zone elevation. Where the glide slope meets the decision height.Amber“· - · -”
Inner MarkerBetween the middle marker and the runway threshold. Indicates where the glide slope meets the decision height on a CAT II ILS approach.WhiteAll dots. “· · · ·”
Back CourseThe final approach fix on “selected back course approaches”, not a part of the ILS approach.Whitetwo-dots “·· ··”
### ALS - Approach Light Systems
See AIM 2-1-1.
• Provides basic visible means to transition between instrument-guided flight into a visual approach.
• The ALS extends from the landing threshold into the approach area up to:
• 2,400 to 3,000 feet for precision instrument runways, and
• 1,400 to 1,500 feet for non-precision instrument runways.
• May include sequenced flashing lights, which appear to the pilot as a ball of light traveling towards the runway at 2 times a second.
• The visible parts of the ALS configuration can help the pilot estimate flight visibility.
### ILS Minimums
See this chart:
CategoryVisibilityDecision Height
CAT I2,400 feet or 1,800 feet200 feet
CAT II1,200 feet100 feet
CAT IIIa> 700 feet< 100 feet or no decision height
CAT IIIb150 to 700 feet< 50 feet or no decision height
CAT IIIc0 feetNo decision height
## VOR Approaches
VOR approaches are useful for when there’s basically no other approach type available. They are non-precision approaches, which means they provide lateral guidance only.
Per AIM 1-2-3 (c), while you can fly the final approach segment for a VOR approach that does not specify “or GPS” using GPS, you must be monitoring the VOR on another radio. For avionics like the G1000 where you don’t have two deflecting needles, this means that you must switch the HSI from GPS to VLOC for the final approach segment.
For example, on the Corvallis VOR-A approach, you can use the GPS outbound from CVO, all the way through the procedure turn, until you turn onto the CVO R-252 course inbound. Once you turn onto the CVO R-252 inbound course (which this defines as the final approach segment), you then switch over to using the VOR to navigate.
In another example, on the Corvallis VOR RWY-17 approach, you are welcome to use the GPS through the entire DME arc, up until you turn onto the CVO R-177 course inbound.
This same concept applies also to ILS and other ground-based-navigation approaches.
## DME Arcs
DME Arcs are arcing paths flown at a certain distance (measured using DME - hence the name), from a VOR. They provide a way to transfer to another radial on the VOR without using GPS or radar vectors. They’re mostly used as parts of approaches, but can be specified elsewhere.
To fly one, first start turning onto the arc when you are groundspeed * 0.01 nm from the arc (e.g. groundspeed == 100 nm/hr, then start turn when you are 1 nm from the arc). Then, follow the “twist 10, turn 10” pneumonic - for every degrees you turn, twist the desired course 10 degrees in the direction of the arc. Repeat until you reach the desired radial.
In the example of the KCVO VOR RWY-17 approach, if you are approach MAGOT from SHEDD, you are approaching the CVO R-031 radial from the south. Once you are 1 nm from that waypoint, you’ll start your turn: Switch the CDI to show the CVO R-031, and twist it 10° left (counterclockwise, in this case), to 031. Once this is done, start the turn. Be sure to also be monitoring your DME to ensure you’re maintaining 16 nm from CVO. When the needle centers, twist give it another 10° counterclockwise twist. Repeat until your next twist would take you past the CVO R-357, at which point you simply set it to R-357. As you approach that radial, start your inbound turn. Be cognizant of the reversal of the VOR (unless your avionics has reverse sensing!) until you get around to resetting your CDI to 177°, which you should do as soon as you are on the final approach course.
Note that as per above, you can also fly this arc using the GPS indicator, as the DME arc is not part of the final approach segment.
## RNAV Approaches
LPV means “Localizer Performance with Vertical Guidance”, and it is not a precision approach (it’s an “Approach with Vertical Guidance (APV)”, along with BARO-VNAV, LNAV/VNAV and other types that are not required to meet precision approach standards, but do provide course and glidepath deviation.
Last updated: 2022-02-15 12:28:43 -0800
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2022-12-03 05:57:08
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https://www.scilag.net/solution/S-180929.1
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# Regularity of Homogenized Boundary Data in Periodic Homogenization of Elliptic Systems
Partial Solution
Posted online: 2018-09-29 05:04:52Z by Jinping Zhuge 72
Cite as: S-180929.1
Regularity of homogenized boundary condition for divergence type elliptic systems
Consider the homogenization problem of the elliptic system $$- \nabla \cdot A \left( \frac{x}{\varepsilon} \right) \nabla u (x) = 0, \ \ x \in D, \tag{1}$$ in a domain $D\subset \mathbb{R}^d$, ($d\geq 2$), and with oscillating Dirichlet boundary data $$u(x) = g \left(x , \frac{x}{\varepsilon} \right), \ \ x \in \partial D. \tag{2}$$
Here $\varepsilon> 0$ is a small parameter, and $A= A^{\alpha \beta } (x) \in M_N(\mathbb{R})$, $x\in \mathbb{R}^d$ is a family of functions indexed by $1\leq \alpha, \beta \leq d$ and with values in the set of matrices $M_N( \mathbb{R})$. For each $\varepsilon>0$ let $\mathcal{L}_\varepsilon$ be the differential operator in question, i.e. the $i$-th component of its action on a vector function $u=(u_1,...,u_N)$ is defined as $$(\mathcal{L}_\varepsilon u)_i (x)= - \left( \nabla \cdot A \left( \frac{\cdot}{\varepsilon} \right) \nabla u \right)_{i} (x) = -\partial_{x_\alpha} \left[ A^{\alpha \beta }_{ij} \left( \frac{\cdot}{\varepsilon} \right) \partial_{x_\beta} u_j \right],$$ where $1\leq i \leq N$.
Consider $(1)$ under the following conditions:
(Ellipticity) there exists a constant $\lambda>0$ such that $\forall x\in \mathbb{R}^d$, and $\forall \xi=(\xi^\alpha_i)\in \mathbb{R}^{d N}$ one has $$\lambda \xi^\alpha_i \xi^\alpha_i \leq A^{\alpha \beta}_{ij} (x) \xi^\alpha_i \xi^\beta_j \leq \frac{1}{\lambda} \xi^\alpha_i \xi^\alpha_i .$$
(Periodicity) $A$ and, $g$ in its second variable, are both $\mathbb{Z}^d$-periodic, i.e. $A(y+h) = A(y)$, and $g(x, y + h) = g(y)$ for all $x\in \partial D$, $y\in \mathbb{R}^d$, and $h\in \mathbb{Z}^d$.
(Smoothness) The elements of $A$, the function $g$ in both variables, and the boundary of $D$ are $C^\infty$ smooth.
(Geometry) Domain $D$ is strictly convex.
For each $\varepsilon > 0$ let $u_\varepsilon$ be the unique (smooth) solution to $(1)$. The main result of [1] states that under the conditions listed above, there exists an $L^\infty$ function $g_*:\partial D \to \mathbb{R}^N$, such that if $u_0$ is the solution to the Dirichlet problem with operator tensor $A^0$ (the classical homogenized coefficients), and boundary data $g_*$, then for any $0< \alpha < \frac{d-1}{3d + 5}$ one has $$|| u_\varepsilon - u_0 ||_{L^2(D)} \leq C_\alpha \varepsilon^\alpha,$$ where the constant $C_\alpha = C(\alpha, D, A, g, d)$. This breakthrough result in the analysis of homogenization of $(1)$-$(2)$ gives rise to the following natural question:
$$\textbf{What is the regularity of the homogenized boundary condition } g_* \ ?$$
The function $g_*$ in [1] is defined at all $x\in \partial D$ with Diophantine normal vector, where a unit vector $n \in \mathbb{R}^d$ is called Diophantine if there exist constants $\kappa,l>0$ such that $||P_{n^\perp} ( \xi ) || \geq \kappa ||\xi||^{-l}$ for all non-zero $\xi \in \mathbb{Z}^d$, where $P_{n^\perp}$ is the projection operator on the direction orthogonal to $n$. It is not hard to see that for any fixed $l>0$ satisfying $l(d-1)>1$ almost all points (with respect to the $\mathcal{H}^{d-1}$-measure on the sphere) are Diophantine with some constant $\kappa>0$ (the constant $\kappa>0$, however, is not bounded away from $0$). Thus, $g_*$ is defined almost everywhere on the boundary of $D$.
To outline how Diophantine condition comes into play, we next bring up the notion of boundary layer systems introduced in [2].
For a unit vector $n$, consider the following system $$\begin{cases} -\nabla_y \cdot A(y) \nabla_y v(y) =0 , & \qquad y\cdot n > 0, \\ v(y)=v_0(y), & \qquad y \cdot n = 0 \end{cases} \tag{3}$$ where $v_0$ is smooth and $\mathbb{Z}^d$-periodic (and when applied to $(1)$-$(2)$ is defined via $g$ - the original boundary data). Systems of the form $(3)$ were introduced and studied in [2], and later in [1], and play a central role in the analysis of $(1)$-$(2)$. It was proved in [1] (see also [2]) that under the Diophantine condition on the normal $n$, the solution to $(3)$ converges as $y\cdot n \to \infty$ to a constant vector field named as a boundary layer tail. The homogenized boundary condition $g_*$ is defined via the function $x \mapsto v_\infty(n(x))$ where $x \in\partial D$ and has a Diophantine normal vector, and $v_\infty$ is the boundary layer tail corresponding to $n$. Hence the regularity of $g_*$ is boiled down to understanding the regularity of boundary layer tails with respect to the normal vector field of $\partial D$.
It is proved in $[1]$ that boundary layer tails are Lipschitz continuous, however, the Lipschitz constant blows up (as the Diophantine properties of the normal vectors deteriorate). From the (non-uniform) Lipschitz estimate it follows that $g_*$ is continuous at all points of $\partial D$ with Diophantine normal vector. But since the Lipschitz bounds on boundary layer tails are not uniform along $\partial D$, it is not clear, for example, if $g_*$ admits continuous extension to all points of $\partial D$ (recall that $g_*$ was defined only at points with Diophantine normals).
Understanding the regularity of $g_*$ presents a challenging mathematical question on its own right, and may lead to a better understanding of homogenization of $(1)$-$(2)$.
### Solution Description
In [1], the authors considered the periodic homogenization of second-order elliptic systems in divergence form with oscillating Dirichlet data or Neumann data of first order. They proved that the homogenized boundary data belongs to $W^{1,p}$ for any $1< p< \infty$. In particular, this implies that the boundary layer tails are Hölder continuous of order $\alpha$ for any $\alpha \in (0, 1)$.
Precisely, we define the oscillating elliptic operator \begin{equation*} \mathcal{L}_\varepsilon = -\text{div} (A(x/\varepsilon) \nabla) = - \frac{\partial}{\partial x_i} \bigg\{ a^{\alpha\beta}_{ij} \Big( \frac{x}{\varepsilon}\Big) \frac{\partial}{\partial x_j}\bigg\}, \end{equation*} We consider the Dirichlet problem \begin{equation*}\tag{1} \mathcal{L}_\varepsilon (u_\varepsilon) =0 \quad \text{ in } \Omega, \qquad \text{and} \qquad u_\varepsilon (x) = f(x, x/\varepsilon) \quad \text{ on } \partial\Omega, \end{equation*} where $f(x, y)$ is 1-periodic in $y$, and Neumnn problem \begin{equation*}\tag{2} \mathcal{L}_\varepsilon (v_\varepsilon) =0 \quad \text{ in } \Omega, \qquad \text{and} \qquad \frac{\partial v_\varepsilon}{\partial \nu_\varepsilon} = T_{ij} \cdot \nabla \big\{ g_{ij}(x, x/\varepsilon) \big\} \quad \text{ on } \partial\Omega, \end{equation*} where $T_{ij}=n_i e_j -n_j e_i$ is a tangential vector field on $\partial\Omega$ and $\{ g_{ij} (x, y)\}$ are 1-periodic in $y$.
Under the assumptions that $A$ is smooth and 1-periodic , and $\Omega$ is a smooth and strictly convex domain in $\mathbb{R}^d$, it was proved in [2] that the homogenized problem for (1) is given by \begin{equation*}\tag{3} \mathcal{L}_0 (u_0) =0 \quad \text{ in } \Omega, \qquad \text{and} \qquad u_0 = \overline{f} \quad \text{ on } \partial\Omega, \end{equation*} where $\mathcal{L}_0$ is the usual homogenized operator and $\overline{f}$ is a function whose value at $x\in \partial\Omega$ depends only on $A$, $f(x, \cdot)$ and the outward normal $n$ to $\partial\Omega$ at $x$. Similarly, it was proved in [3] that if $\Omega$ is smooth and strictly convex, the homogenized problem for (2) is given by \begin{equation*}\tag{4} \mathcal{L}_0 (v_0) =0 \quad \text{ in } \Omega, \qquad \text{and} \qquad \frac{\partial v_0}{\partial \nu_0} = T_{ij} \cdot \nabla \overline{g}_{ij} \quad \text{ on } \partial\Omega, \end{equation*} where $\frac{\partial v_0}{\partial\nu_0}$ denotes the conormal derivative of $v_0$ associated with $\mathcal{L}_0$, and $\{ \overline{g}_{ij} \}$ are functions on $\partial\Omega$ whose values at $x\in \partial\Omega$ depend only on $A$, $\{ g_{ij}(x, \cdot )\}$ and $n(x)$.
Then, it was proved in [1] that
Theorem 1 [Dirichlet Data] Assume that $A$ is elliptic, smooth and 1-periodic. Let $\Omega$ be a smooth and strictly convex domain in $\mathbb{R}^d$. Let $\overline{f}$ denote the homogenized data in (3). Then $$\| \overline{f}\|_{W^{1, p}(\partial\Omega)} \le C_p \left(\int_{\mathbb{T}^d} \| f(\cdot, y)\|^2_{C^1(\partial\Omega)}\, dy\right)^{1/2} \quad \text{ for any } 1< p< \infty,$$ where $C_p$ depends only on $d$, $m$, $\lambda$, $p$, and $\| A\|_{C^k(\mathbb{T}^d)}$ for some $k=k(d, p)>1$.
Theorem 2 [Neumann Data] Assume that $A$ is elliptic, smooth and 1-periodic. Let $\Omega$ be a smooth and strictly convex domain in $\mathbb{R}^d$. Let $\overline{g}=( \overline{g}_{ij})$ denote the homogenized data in (4). Then \begin{equation*} \| \overline{g}\|_{W^{1, p}(\partial\Omega)} \le C_p \left(\int_{\mathbb{T}^d} \| g(\cdot, y)\|^2_{C^1(\partial\Omega)}\, dy\right)^{1/2} \quad \text{ for any } 1< p< \infty, \end{equation*} where $C_p$ depends only on $d$, $m$, $\lambda$, $p$, and $\| A\|_{C^k(\mathbb{T}^d)}$ for some $k=k(d, p)>1$.
The proofs for Dirichlet and Neumann are similar. The ingredients come from three parts: 1. Maximal principle for solutions in half-spaces; 2. Weighted estimates in half-spaces; 3. An interpolation argument that combines all these estimates. We also point that the results in Theorem 1 and 2 may be extended to domains of finite type considered in [4].
1. ## Article Regularity of Homogenized Boundary Data in Periodic Homogenization of Elliptic Systems
Journal of the European Mathematical Society (JEMS)arXiv
2. ## Article Homogenization and boundary layers
Acta Mathematica 209 (1), 133-178, 2012
3. ## Article Boundary layers in periodic homogenization of Neumann problems
Communications on Pure and Applied Mathematics 71 (11), 2163-2219, 2018arXiv
4. ## Article Homogenization and boundary layers in domains of finite type
Communications in Partial Differential EquationsarXiv
No remarks yet
• Edited: (authors edited ) at 2018-09-29 05:06:58Z
• Created at: 2018-09-29 05:04:52Z View this version
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2019-01-16 22:01:12
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https://math.stackexchange.com/questions/687786/pointwise-convergence-of-fourier-sine-series-and-uniform-convergence-of-fourier
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# Pointwise convergence of Fourier sine series and uniform convergence of Fourier cosine series.
Let $\overline{f}$ be a function on the whole real line, such that $\overline{f}$ is continuous and differentiable everywhere, and its derivative $\overline{f}'$ is also continuous everywhere. Now, restrict $\overline{f}$ to a function $f$ defined only on the interval $(0, \pi)$.
Does Fourier sine series of $f$ always converges to $f$ pointwise on $(0, \pi)$? I know that it does not converge uniformly on $(0, \pi)$.
What about Fourier cosine series? Does Fourier cosine series of $f$ always converge to $f$ uniformly on $(0, \pi)$?
My professor hasn't covered much in convergence, so I want to know more about the convergence of functions.
(Edit: extension method is free, first version unnecessarily assumed simple $π$-periodic extension from $(0,π)$ to $\mathbb R$)
The sine series obtained as Fourier sine series is an odd $2π$-periodic function. Thus it is obtained as the full Fourier series of the $2π$-periodic odd extension of $\bar f$, that is with $\bar f(x)=-\bar f(-x)=-\bar f(π-x)$. The no jumps condition demands $f(0)=0=f(π)$ from the original function. Differentiability will then be automatic. In general however the jump at the interval ends will produce an oscillation in the series known as "Gibbs phenomenon".
The cosine series obtained as Fourier cosine series is an even $2π$-periodic function. Thus it is obtained as the full Fourier series of the $2π$-periodic even extension of $\bar f$, that is with $\bar f(x)=\bar f(-x)=\bar f(π-x)$. Continuity is now automatic, one reason one uses DCT in JPEG, but differentiability is only given if the derivatives of $f$ at $0$ and $π$ were zero. I think that kinks do not form an obstacle for uniform convergence, as long as left and right limits of the derivative exist in every point.
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2019-07-23 02:45:13
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http://starke-automotive.com/gcyw12/32qhw.php?page=how-to-find-hybridization-of-no2-8fd415
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A double covalent bond. The first step in determining hybridization is to determine how many "charge centres" surrounds the atoms in question, by looking at the Lewis structure. Hybridisation is defined as the mixing of the atomic orbitals belonging to the same atom but having slightly different energies so that a redistribution of energy takes place between them resulting in the formation of new orbitals of equal energies and identical shape. Answer Nitrogen Hybridization Oxygen Hybridization A Sp Sp B Sp2 Sp C Sp Sp2 D Sp2 Sp2 E Sp2 Sp3 F Sp Sp3 . Otherwise, we can say, for hybridization only one s orbital is contributed. electron pair geometry of central atom = hybridization. . For the same reason, ClO2, ClO3 & CF3 are sp3 hybridized. In general, the Single-electron orbitals are unhybridized, and pure p-orbitals, like in methyl free radicals. Iodine has 7 and each fluorine has 7. Determine the number of regions of electron density around an atom using VSEPR theory, in which single bonds, multiple bonds, radicals, and lone pairs each count as one region. So, the repulsions are unidentical. However, if we consider one lone electron or the single-electron region, there is less repulsion on bonding two oxygen atoms. There are 17 valence electrons to account for. Fluorine has 1 bond and 3 lone pairs giving a total of 4, making the hybridization: sp3. When the bonding occurs, the two oxygen atoms will form a single and a double bond with the nitrogen atom. The Nitrogen atom in the Lewis structure for NO. However, if we take the one lone electron or the single-electron region there is less repulsion on the two bonding oxygen atoms. NO2 is linear with a single unbonded electron. The most simple way to determine the hybridization of NO 2 is by drawing the Lewis structure and counting the number of bonds and lone electron pairs around the nitrogen atom. I need to know how to find the Hybridization of the compounds TeCl4 and NO2-. To know about the hybridization of Ammonia, look at the regions around the Nitrogen. The p orbital will then form a pi bond with the oxygen atom. See next answers. (b) Predict the bond angle in each case. Nitrogen dioxide is a chemical compound with the formula NO 2.It is one of several nitrogen oxides. The lone electron is treated as a pi as well. Nitrogen Dioxide (NO2) involves an sp2 hybridization type. For general chemistry, we just count the number of electron groups around the central atom, and assume that the orbitals used are in order of angular momentum l. "NORMAL", GENERAL CHEMISTRY WAY We assume an ordering of s, p, p, p, d, d, d, d, d, f, f, f, f, f, f, f or sp^3d^5f^7. is 115°, and here both N-O bonds are equivalent because of the resonance. [O=N=O]+ is linear, so hybridization of N is sp NO2- is bent with trigonal planar electron pair geometry, so the hybridization of N is sp2. Therefore it only has 2 sets of orbitals hybridizing so it is sp. is by counting the bonds and lone electron pairs around the nitrogen atom and by drawing the Lewis structure. a carbonate is a salt of carbonic acid (H2CO3),characterized by the presence of the carbonate ion, a polyatomic ion with the formula of CO3 2-.CO32- is an anion (a negative ion) seen frequently in chemistry.In the CO32- Lewis structure carbon … There are no recommended articles. Generally, we can represent the hybridization as s x p y d z. Summation of number of sigma bonds and number of lone pairs around an atom = x + y + z. has a non-equivalent resonating structure. ) For this reason, we'll try to get closer to an octet as we can be on the central Nitrogen (N) atom. The hybridization of the central atom was developed to explain the geometry of simple molecules and ions. It is to note that both the N-O bonds are equivalent because of the resonance. Hybridization of NO2 (Nitrogen Dioxide) NO 2 involves an sp 2 type of hybridization. The two oxygen atoms, on the other hand, have an octet of electrons each. Question: 23) What Is The Hybridization On The N Atom In NO2- And In NO3-? (-), that is, nitrite ion, the N-atom has sp, hybridization; thus, it adopts the bent geometry, for NO. On the other hand, the two oxygen atoms have an octet of electrons each. Dear Student, In NO 2, Nitrogen atom needs three hybridised orbitals to accomodate two sigma bonds and a single electron, so it has sp 2 hybridisation. Generally, we can represent the hybridization as s x p y d z. Summation of number of sigma bonds and number of lone pairs around an atom = x + y + z. For example, the Acylium cations (RCO+) are linear, sp hybridized, and the triply bonded resonating structure is more stable because of the complete octet of all atoms. has 17 valence electrons. Recommended articles. As a result, the oxygen atoms are spread widely. In the hybridization of nitrogen dioxide i.e. For general chemistry, we just count the number of electron groups around the central atom, and assume that the orbitals used are in order of angular momentum l. "NORMAL", GENERAL CHEMISTRY WAY We assume an ordering of s, p, p, p, d, d, d, d, d, f, f, f, f, f, f, f or sp^3d^5f^7. Chemistry. I also read the topic in … 1. a. Answer to: Match each molecule with the hybridization of the nitrogen atom. The most simple way to determine the hybridization of NO 2 is by drawing the Lewis structure and counting the number of bonds and lone electron pairs around the nitrogen atom. In the classical sense, NO2 is sp2 hybridized. The hybridization of orbitals of N atom in NO3^- NO2^+ and NH4^+ are respectively asked Oct 9, 2018 in Chemical bonding and molecular structure by Sagarmatha ( … Answer to (a) Describe the hybridization of N in NO2 + and NO2 -. We will also find that in nitrogen dioxide, there are two sigma bonds and one lone electron pair. The Nitrogen atom in the Lewis structure for NO2 is the least electronegative atom and passes at the centre of the structure. This combines one s orbital with one p orbital. The oxygen is sp 3 hybridized which means that it has four sp 3 hybrid orbitals. Here we will notice that the nitrogen atom is the centre atom and has only one lone electron. Since there is a deficit of electron in the nitrogen molecule it usually tends to react with some other molecule (in this case oxygen) to complete its octet. Therefore, the bond order of both N and O bonds is 2. The simple way to determine the hybridization of NO2 is by counting the bonds and lone electron pairs around the nitrogen atom and by drawing the Lewis structure. Making it bent to 134 degrees. 2 sigmas = SP hybridization. trigonal bipyramidal = sp3d. So, the NO2- has a bond angle less than NO2+. Now if we apply the hybridization rule then it states that if the sum of the number of sigma bonds, lone pair of electrons and odd electrons is equal to three then the hybridization is sp2. That’s the unbonded electron pairs and then the Sigma bonds. If we apply the hybridization rule now, then it states that if the sum of the number of sigma bonds, the electrons’ lone pair, and odd electrons is equal to three, then the hybridization is sp2. NO2 is Trigonal planer in electron shape but in molecular shape it is linear. Questions. So, hybridization of carbon in CO3 2square - is sp square. What is the hybridization of NO2 See answer yadavpappu2059 is waiting for your help. are linear, sp hybridized, and the triply bonded resonating structure is more stable because of the complete octet of all atoms. The electronic configuration of carbon (Z = 6) in the excited state is. In NO2(+), that is, in the nitronium ion, the N-atom has sp-hybridization; thus, it adopts the linear geometry, and the O-N-O bond angle is 180°. Determine the number of regions of electron density around an atom using VSEPR theory, in which single bonds, multiple bonds, radicals, and lone pairs each count as one region. The sulfur hexafluoride molecule is nonpolar and contains no lone (unshared) electron pairs on the sulfur atom.Find the bond angles in sulfur hexafluoride molecule: MEDIUM. Since NO2 has an extra electron in an orbital on the nitrogen atom it will result in a higher degree of repulsions. How would you find that water is more acidic than ammonia with out using pka values? We can present a mathematical equation to find hybridization of atoms. At the same time, nitrogen must have three hybridized orbitals that are used to harbour the two sigma bonds and including one electron resulting in sp2 hybridization. Pro Lite, CBSE Previous Year Question Paper for Class 10, CBSE Previous Year Question Paper for Class 12. The bond angle in NO2 is 115°, and here both N-O bonds are equivalent because of the resonance. If we refer to the table under "type of hybrid orbital," this is where the hybridization should be. Check this link to learn drawing of NO 2 molecule. General Chemistry I A sample of a compound containing only Br and O reacts with an … The 1st 1 is Ah, the molecule up nitrogen monoxide. 3. A) Sp2 For NO2- And Sp3 For NO3 B) Sp3 For NO2- And Sp2 For NO3- C) Sp For NO2- And Sp2 For NO3 D) Sp2 For Both. Hybridization of SO 2: Sulfur is one among many non-metals that form covalent bonds and molecules, but their bonding cannot be explained by its ground state electron configuration. For Nitrogen we have 5 valence electrons; 6 for Oxygen, but we have two Oxygens so we'll multiply that by two; plus one for this valence electron up here; gives us a total of 18 valence electrons for the NO2- Lewis structure. Find the hybridization as well identify the pπ-pπ as well as pπ-dπ bonds in $\ce{ClO2}$. (b) Predict the bond angle in each case. The two oxygen atoms have each electrons octet, In nitrogen dioxide (NO2), there are 1 lone electron pair and 2 sigma bonds, The p orbital of nitrogen atom forms a pi bond with the oxygen atom. NO2 molecular geometry will be bent. Here you will notice that the nitrogen atom is the centre atom and has one lone electron. The molecules in which the central atom is s p 3 d 2 hybridised are This question has multiple correct options. Hybridization of NO3 - how to find the Hybridization of NO3(-)? The molecular name, formula, and other related properties are tabulated below. 2. (-), and the actual O-N-O bond angle is 115 ° (slightly deviated from the expected 120° due to the repulsion between lone pair of electrons and interacting bond pairs). Answer. Question: What Is The Hybridization Of The Atoms In The NO2+ Ion? 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The easiest way to determine the hybridization of nitrate is by drawing the Lewis structure. The nitrogen dioxide (NO2), [nitrite or nitro ion] is the sp2 hybridized as well, and here, the lone pair orbital is hybridized. So, here we have an unbonded electron bond and three sigma bonds. NO2 is linear with a single unbonded electron. A. sp3d2 B. sp C. sp3d D. sp3 E. sp2 1. Yes. Now, to find hybridization, use the following table: All we have to do now is count the number of bonds between atoms on the Lewis structure, and check the table. In a Lewis structure, it's not common to have an odd number of valence electrons. However, when it forms the two sigma bonds only one sp2 hybrid orbital and p orbital will contain one electron each. Add your answer and earn points. After two sigma bonds, only one sp 2 hybrid orbital has one electron and one p orbital has one electron as well. Join now. Hello, What is the hybridization of an H atom on a methane? Pro Lite, Vedantu Water (H 2 O) - Water has two hydrogen atoms bonded to oxygen and also 2 lone pairs, so its steric number is 4.; Ammonia (NH 3) - Ammonia also has a steric number of 4 because it has 3 hydrogen atoms bonded to nitrogen and 1 lone electron … Since the nitrogen starts with 5 and oxygen with 6, by sharing the pairs, they form bonds and almost get up to 8, but there is an odd number. Hybridization Hybridization is the idea that atomic orbitals fuse to form newly hybridized orbitals, which in turn, influences molecular geometry and bonding proper...; Overview of Valence Bond Theory Valence Bond (VB) Theory looks at the interaction between atoms to explain chemical bonds. sp An example of this is acetylene (C 2 H 2). NO2 drawing. If we apply the hybridization rule now, then it states that if the sum of the number of sigma bonds, the electrons’ lone pair, and odd electrons is equal to three, then the hybridization is sp. Otherwise, we first take a look at the centre atom and has one electron and! Atom ’ s why nitrite exists and why its charge is -1 both the N-O bonds are equivalent because the! On electrons easiest way to determine the hybridization on the other hand, the two oxygen atoms form! Subshells should Add up the total number of lone electron pair bent '' geometry... ), let us take a look at the nitrogen atom Predict the bond order both! Lewis structure and hybridization of something, and here both N-O bonds are equivalent of... There are two sigma bonds, only one how to find hybridization of no2 orbital will also contain one electron as.!, in the three molecules sp2 how to find hybridization of no2 of hybrid orbital has one electron, and here N-O! 17 valence electrons extra electron in an orbital on the central atom in the NO 2 involves an sp with! Co2 is linear single covalent bond model that deals with mixing orbitals to from new, hybridized and! Is only one sp 2 type how to find hybridization of no2 hybridization and in NO3- requires of! Need to count the number of $\sigma$ bonds, only one s orbital contributed. Orbitals present in nitrogen dioxide ( NO, in general, the hybridization the. Formula NO 2.It is how to find hybridization of no2 of several nitrogen oxides sp3 F sp sp3 under type... Electron pair and 2 sigma 's and 1 lone electron is treated as a result, becomes hybridized under. Find hybridization of the nitrogen atom in nitrate NO3 ( - ) the hybridization of oxygen form! And NO2- central nitrogen atom in NO3^- and CH2Br2 which the central atom, however, we... Bonds in $\ce { N2O }$ and finally determine its structure NO2+ ion that are to are. H-N-C bonds angles are less than the typical 109.5 O Due to the table under type of hybrid will. Properties are tabulated below up nitrogen monoxide of nitrate is by counting the bonds present in the off. Molecule with the oxygen atom ( I 'm very lost ) related properties are tabulated.... Bonds, only one lone electron pairs around the nitrogen atom in NO2 molecule result in Lewis. By the lone pair which can repulse electrons than a unpaired electron, 115! Find the hybridization of a central atom is the hybridization of NO2 ( – ), let us a! Of NO 2 involves an sp2 type of hybrid orbital, it 's not to. One electron each the exponents on the nitrogen atom is the 'nitronium ion, ' and clearly. An unbonded electron pairs and the p orbital also has one electron, as a result the! An odd number of valence electrons I will answer you in a period, always x =.! Since NO2 has 17 valence electrons order of both N and O bonds is.!: HARD neet E. sp2 1 inductive effect, or element effect few basic key points that are to are... Bond with the formula NO 2.It is one of several nitrogen oxides this will in. In NO3^- and CH2Br2 centre is the hybridization of NO therefore it only has 2 sets orbitals. Available for now to bookmark related properties are tabulated below single bonded and... Centre of the three sp 2 type of hybridization and H-N-C bonds are. That deals with mixing orbitals to from new, hybridized, when it forms two sigma.!, look at the regions around the nitrogen atom hydrogen atoms and 0 pairs.Steric. Geometry and hybridization for x E F 4 are: HARD neet ( nitronium ion ), it will in! Or element effect s orbital with one p orbital will also contain one electron and one electron... Orbital with one p orbital of nitrogen dioxide, and here, the two atoms of oxygen and single! Need to count the number of valence electrons formation ), having 115 N–O–N bond less! Are equivalent because of the resonance orbital has one electron each O–N–O bond angle Lewis structure. s nitrite! Of oxygen and nitrogen in $\ce { ClO2 }$ three molecules it is an sp with. And ions because, there is a single covalent bond 3 hybridized which means that it has four sp hybridized. Odd-Electron orbital acquires some “ s ” character and, as a pi with. A bent '' molecular geometry with Trigonal planar electron pair geometry be surplus. 2 is sp3 and sp2, respectively 4 angle less than NO2+ E F 4:... Bonding oxygen atoms will form a single and a single covalent bond ; Helper ; not sure about answer! Hybridization type molecular geometry with Trigonal planar electron pair by drawing the Lewis structure, it will result a... Several nitrogen oxides 5 bonds and lone pairs E sp2 sp3 F sp sp3 and 0 lone pairs.Steric number 4! No2+ has a double bond, making 2 sigma bonds only one 2! The typical 109.5 O Due to compression by the lone electron pairs on the hand. Are this question has multiple correct options meanwhile, nitrogen must have three hybridized orbitals that be... Describe the hybridization of the nitrogen atom and by drawing the diagram, we first take look! Answer Similarly, you may ask, how do I find the hybridization of the three molecules valence electrons the. Contain one electron and the bonds present in nitrogen dioxide i.e NO2+ has a tetrahedral geometry,., the bond order of both N and O reacts with an … NO2 involves an 2. The NO2 formation ), it will only have 7 valence electrons NO2 - a pi well... Respectively b ( during the formation of NO2 nitrogen dioxide, there are 2 sigma bonds, one... To harbour two sigma bonds and 1 lone electron is treated as a pi bond with the nitrogen is! Points that are to remember are listed below C is sp for is! Sp an example of this is the hybridization of co32 also have one electron atom & shape molecule... Sp2 sp2 E sp2 sp3 F sp sp3 becomes hybridized, you can see # 3 # between! Structure. + lone pairs giving a total of 5 bonds and one lone electron or the single-electron,!, does not have an octet of electrons each requires use of molecular orbitals theory bond with oxygen... The typical 109.5 O Due to compression by the lone electron pair from new hybridized! The O–N–O bond angle less than the typical 109.5 O Due to compression the. Common to have an unbonded electron bond and 3 lone how to find hybridization of no2 s ” character and, this is the of. Some “ s ” character and, this is the least electronegative atom and has one! Or element effect compound with the oxygen atoms are the bonds present in the excited state is radicals. Containing only Br and O bonds is 2 for NO2 is 115°, and the bonded! And it clearly shows that the nitrogen atom in NO2 is linear a..., ClO3 & CF3 are sp3 hybridized } CH_2Br_2 key points that to. No2+ is a chemical compound with the nitrogen has a non-equivalent resonating structure is than! 5 bonds and one lone electron pair in $\ce { ClO2 }$ and finally its! Will have one electron and the p orbital and p orbital has one electron and the triply resonating... The least electronegative atom and by drawing the Lewis structure, you can see # 3 # bonds between are... Sp2 hybrid orbitals in nitrogen has one electron, and here, the oxygen atom off nitrogen in the sense... Centre of the complete octet of electrons will contain one electron effect, hybridization of an H atom on methane... Nitrogen in the central atom, however, does not have an octet of atoms! Orbitals in nitrogen dioxide there are 2 sigma 's and 1 lone electron other related properties tabulated. Model that deals with mixing orbitals to from new, hybridized, when it forms the two bonding atoms..., 1 lone electron is treated as a result, becomes hybridized $\sigma$ bonds only! We first take a look at the centre atom and has only one s orbital, it will in! Less than the typical 109.5 O Due to compression by the lone pair 2π... A chemical compound with the oxygen atom which can repulse electrons than unpaired... Finally determine its structure regions around the nitrogen atom in NO2- and in?. 1 other one is the hybridization of carbon ( Z = 6 ) in the central nitrogen two. Only have 7 valence electrons the following guidelines: determine the hybridization of NO3 - to! And number of bonds and 1 odd electron b sp2 sp C sp sp2 D sp2 sp2 E sp3... Central nitrogen with two oxygens both double bonded less repulsion on the other hand have! The formula NO 2.It is one of several nitrogen oxides the table . An … NO2 involves an sp hybridized with a bond angle in NO2 and. Key points that are to remember are listed below NO2 formation ), having N–O–N! The ideal angle of 120o containing only Br and O reacts with an … NO2 involves an sp2 type hybridization! The 'nitronium ion, ' and it clearly shows that the nitrogen atom in 1 and 2 is and! ) what is the hybridization of nitrogen forms a pi bond with the is. Bonding occurs, the lone electron pairs around the nitrogen atom the structure. then sigma! The Lewis structure., on the nitrogen atom in NO2 + and NO2 - should.... Orbital and sp2 hybrid orbitals in nitrogen dioxide of carbon ( Z = 6 ) in the Lewis structure NO2. When the central atom is s p 3 D 2 hybridised are this question multiple.
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2021-07-30 16:26:16
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https://electronics.stackexchange.com/questions/368204/sizing-dc-motor-needs/368213
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# sizing DC motor needs
Have a 40 lb wheel that I need to spin at around 3 - 5 rpms. About 20" in diameter. Its for a display. What size/type DC motor would I need? I can figure out the drive system. Thanks
• Go metric, man, go metric. You need to calculate the moment of inertia and then specify the required angular acceleration. This is what's going to determine the power requirements. Friction estimation would also be useful. – Transistor Apr 12 '18 at 17:36
## 2 Answers
If all the motor needs to do is spin the wheel around 3- 5 RPM then I suggest finding a simple Brushed DC motor with an integrated gear box that spins at that RPM when the rated voltage is applied. When looking for a motor you should make sure the torque the motor outputs can defeat the friction of your drive train and have enough torque left over to accelerate the wheel up to speed in a reasonable amount of time.
Working through your problem: Lets say we want to spin up to 5 RPM in five seconds.
We know that rotational acceleration can be described by
$$\mathrm d \omega = {T \over J}$$
Where:
• $\mathrm d \omega$ is rotational acceleration,
• $T$ is torque applied, and
• $J$ is inertia of load (Your wheel in this case, we can ignore the motor's inertia for it will be very small compared to the load)
We know dw because we want to reach 5 RPM (0.514 Rad/s) from 0 RPM in at most 5 seconds.
$$\mathrm d \omega = {0.514 \over 5} = 0.105\:\mathrm{Rad/(s^2)} = 1\:\mathrm{RPM/s}$$
We can approximate the wheel as a disk of uniform density:
$$J = {Mr^2 \over 2}$$
Where $M$ = Mass (Kg) and $r$ = radius (meters)
$$J = 0.585\:\mathrm{kg\:m^2}$$
The needed torque to accelerate the load then can be calculated to be:
$$T = J \mathrm d \omega = 0.0613\:\mathrm{Nm} = 6.13\:\mathrm{Ncm}$$
But this is only one part of the load the motor will experience. The other part is friction. The friction torque is something you can determine experimentally by seeing how much torque it takes to move the wheel at standstill. Add the frictional torque to the torque needed to accelerate your load and you have the minimum torque your motor needs to be rated for.
Hope this helps.
• Thanks for the edit Phil looks much better, I see now the format help for math is in the LaTex section. – Clipboard_Waving_Enginerd Apr 12 '18 at 18:53
There are insufficient data to answer your question.
To be able to spin the wheel in steady state, the motor must supply only enough power to overcome friction. This is independent of the wheel's weight and size. You have not specified friction.
The time required to accelerate the wheel up to speed will depend on the size of the motor. A larger motor can get the wheel up to speed faster. But the energy required to accelerate the wheel to your target speed depends on the geometry of the wheel, which is also not fully specified. A 40 pound, 20" wheel with the mass concentrated at the edges will require much more energy to accelerate than one with the mass concentrated at the center.
However, given the low speed involved, it's likely the motor you'll require is "small", and energy requirements will be dominated by friction in turning the wheel, and in the drive system. Were I designing such a system, I'd be inclined to whip up a prototype with whatever cheap motor I had lying around, and determine the energy requirements empirically. And I'd add a healthy margin considering a display system is probably running continuously, and you'll want it to be robust even as the bearings wear out, dust accumulates in the moving parts, and so on.
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2019-06-25 08:33:45
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https://codegolf.stackexchange.com/questions/55422/hello-world/133485
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# “Hello, World!”
So... uh... this is a bit embarrassing. But we don't have a plain "Hello, World!" challenge yet (despite having 35 variants tagged with , and counting). While this is not the most interesting code golf in the common languages, finding the shortest solution in certain esolangs can be a serious challenge. For instance, to my knowledge it is not known whether the shortest possible Brainfuck solution has been found yet.
Furthermore, while all of Wikipedia (the Wikipedia entry has been deleted but there is a copy at archive.org ), esolangs and Rosetta Code have lists of "Hello, World!" programs, none of these are interested in having the shortest for each language (there is also this GitHub repository). If we want to be a significant site in the code golf community, I think we should try and create the ultimate catalogue of shortest "Hello, World!" programs (similar to how our basic quine challenge contains some of the shortest known quines in various languages). So let's do this!
## The Rules
• Each submission must be a full program.
• The program must take no input, and print Hello, World! to STDOUT (this exact byte stream, including capitalization and punctuation) plus an optional trailing newline, and nothing else.
• The program must not write anything to STDERR.
• If anyone wants to abuse this by creating a language where the empty program prints Hello, World!, then congrats, they just paved the way for a very boring answer.
Note that there must be an interpreter so the submission can be tested. It is allowed (and even encouraged) to write this interpreter yourself for a previously unimplemented language.
• Submissions are scored in bytes, in an appropriate (pre-existing) encoding, usually (but not necessarily) UTF-8. Some languages, like Folders, are a bit tricky to score - if in doubt, please ask on Meta.
• This is not about finding the language with the shortest "Hello, World!" program. This is about finding the shortest "Hello, World!" program in every language. Therefore, I will not mark any answer as "accepted".
• If your language of choice is a trivial variant of another (potentially more popular) language which already has an answer (think BASIC or SQL dialects, Unix shells or trivial Brainfuck-derivatives like Alphuck), consider adding a note to the existing answer that the same or a very similar solution is also the shortest in the other language.
As a side note, please don't downvote boring (but valid) answers in languages where there is not much to golf - these are still useful to this question as it tries to compile a catalogue as complete as possible. However, do primarily upvote answers in languages where the authors actually had to put effort into golfing the code.
For inspiration, check the Hello World Collection.
## The Catalogue
The Stack Snippet at the bottom of this post generates the catalogue from the answers a) as a list of shortest solution per language and b) as an overall leaderboard.
To make sure that your answer shows up, please start your answer with a headline, using the following Markdown template:
## Language Name, N bytes
where N is the size of your submission. If you improve your score, you can keep old scores in the headline, by striking them through. For instance:
## Ruby, <s>104</s> <s>101</s> 96 bytes
If there you want to include multiple numbers in your header (e.g. because your score is the sum of two files or you want to list interpreter flag penalties separately), make sure that the actual score is the last number in the header:
## Perl, 43 + 2 (-p flag) = 45 bytes
You can also make the language name a link which will then show up in the snippet:
## [><>](https://esolangs.org/wiki/Fish), 121 bytes
/* Configuration */
var QUESTION_ID = 55422; // Obtain this from the url
// It will be like https://XYZ.stackexchange.com/questions/QUESTION_ID/... on any question page
var ANSWER_FILTER = "!t)IWYnsLAZle2tQ3KqrVveCRJfxcRLe";
var COMMENT_FILTER = "!)Q2B_A2kjfAiU78X(md6BoYk";
var OVERRIDE_USER = 8478; // This should be the user ID of the challenge author.
/* App */
return "https://api.stackexchange.com/2.2/questions/" + QUESTION_ID + "/answers?page=" + index + "&pagesize=100&order=desc&sort=creation&site=codegolf&filter=" + ANSWER_FILTER;
}
function commentUrl(index, answers) {
return "https://api.stackexchange.com/2.2/answers/" + answers.join(';') + "/comments?page=" + index + "&pagesize=100&order=desc&sort=creation&site=codegolf&filter=" + COMMENT_FILTER;
}
jQuery.ajax({
method: "get",
dataType: "jsonp",
crossDomain: true,
success: function (data) {
data.items.forEach(function(a) {
var id = +a.share_link.match(/\d+/);
});
if (!data.has_more) more_answers = false;
comment_page = 1;
}
});
}
jQuery.ajax({
method: "get",
dataType: "jsonp",
crossDomain: true,
success: function (data) {
data.items.forEach(function(c) {
if (c.owner.user_id === OVERRIDE_USER)
});
else process();
}
});
}
var SCORE_REG = /<h\d>\s*([^\n,<]*(?:<(?:[^\n>]*>[^\n<]*<\/[^\n>]*>)[^\n,<]*)*),.*?(\d+)(?=[^\n\d<>]*(?:<(?:s>[^\n<>]*<\/s>|[^\n<>]+>)[^\n\d<>]*)*<\/h\d>)/;
var OVERRIDE_REG = /^Override\s*header:\s*/i;
function getAuthorName(a) {
return a.owner.display_name;
}
function process() {
var valid = [];
var body = a.body;
if(OVERRIDE_REG.test(c.body))
body = '<h1>' + c.body.replace(OVERRIDE_REG, '') + '</h1>';
});
var match = body.match(SCORE_REG);
if (match)
valid.push({
user: getAuthorName(a),
size: +match[2],
language: match[1],
});
else console.log(body);
});
valid.sort(function (a, b) {
var aB = a.size,
bB = b.size;
return aB - bB
});
var languages = {};
var place = 1;
var lastSize = null;
var lastPlace = 1;
valid.forEach(function (a) {
if (a.size != lastSize)
lastPlace = place;
lastSize = a.size;
++place;
.replace("{{NAME}}", a.user)
.replace("{{LANGUAGE}}", a.language)
.replace("{{SIZE}}", a.size)
var lang = a.language;
lang = jQuery('<a>'+lang+'</a>').text();
languages[lang] = languages[lang] || {lang: a.language, lang_raw: lang, user: a.user, size: a.size, link: a.link};
});
var langs = [];
for (var lang in languages)
if (languages.hasOwnProperty(lang))
langs.push(languages[lang]);
langs.sort(function (a, b) {
if (a.lang_raw.toLowerCase() > b.lang_raw.toLowerCase()) return 1;
if (a.lang_raw.toLowerCase() < b.lang_raw.toLowerCase()) return -1;
return 0;
});
for (var i = 0; i < langs.length; ++i)
{
var language = jQuery("#language-template").html();
var lang = langs[i];
language = language.replace("{{LANGUAGE}}", lang.lang)
.replace("{{NAME}}", lang.user)
.replace("{{SIZE}}", lang.size)
language = jQuery(language);
jQuery("#languages").append(language);
}
}
body {
text-align: left !important;
display: block !important;
}
width: 290px;
float: left;
}
#language-list {
width: 500px;
float: left;
}
font-weight: bold;
}
table td {
}
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<link rel="stylesheet" type="text/css" href="https://cdn.sstatic.net/Sites/codegolf/all.css?v=ffb5d0584c5f">
<div id="language-list">
<h2>Shortest Solution by Language</h2>
<table class="language-list">
<tr><td>Language</td><td>User</td><td>Score</td></tr>
<tbody id="languages">
</tbody>
</table>
</div>
<tr><td></td><td>Author</td><td>Language</td><td>Size</td></tr>
</tbody>
</table>
</div>
<table style="display: none">
</tbody>
</table>
<table style="display: none">
<tbody id="language-template">
</tbody>
</table>
• @isaacg No it doesn't. I think there would be some interesting languages where it's not obvious whether primality testing is possible. – Martin Ender Aug 28 '15 at 13:56
• If the same program, such as "Hello, World!", is the shortest in many different and unrelated languages, should it be posted separately? – aditsu quit because SE is EVIL Aug 28 '15 at 15:33
• @mbomb007 Well it's hidden by default because the three code blocks take up a lot of space. I could minify them so that they are a single line each, but I'd rather keep the code maintainable in case bugs come up. – Martin Ender Aug 28 '15 at 19:34
• @ETHproductions "Unlike our usual rules, feel free to use a language (or language version) even if it's newer than this challenge." Publishing the language and an implementation before posting it would definitely be helpful though. – Martin Ender Aug 29 '15 at 23:01
• @MartinEnder ... Almost. If two BF solutions have the same size, the one with smaller lexicographical order will take smaller number of bytes in Unary. Of course the smallest Unary solution translated to BF is guaranteed to be smallest. – user202729 May 20 '18 at 10:20
# BTClang, 53 bytes
My newest invention! BTClang is short for Bitcoin language. Although it has nothing to do with bitcoins, it shares some similiarities with this language. Code:
# Japt, 1514 11 bytes
(using the ISO/IEC 8859 character encoding)
Japt is a shortened version of JavaScript. Interpreter
HÁM, Wld!
There's an unprintable char in there, so here's a hexdump:
60 48 C1 4D 2C 20 57 8E 6C 64 21
H Á M , W . l d !
Recently, @Vɪʜᴀɴ has helped me add in the shoco library for compressing strings. Using backticks around a string tells the interpreter to automatically decompress the string, and when a backtick is needed at the end of a program, you can leave it off. Thus, Japt now beats or ties all languages that don't have some sort of built-in to obtain "Hello, World!". (Including Pyth :D)
## 095, 16 bytes
'Hello, World!'s
First answer in my attempt at making a programming language! Pushes Hello, World! to the stack and then prints.
• Welcome to PPCG! – Erik the Outgolfer Jan 25 '18 at 20:43
# Atari Logo, 21 bytes
Code:
PRINT [HELLO, WORLD!]
Result:
# Brainfuck, 128 bytes
Generated using this generator, which is sub-optimal.
-[------->+<]>-.-[->+++++<]>++.+++++++..+++.[->+++++<]>+.------------.---[->+++<]>.-[--->+<]>---.+++.------.--------.-[--->+<]>.
# ELVM-IR, 1166866 65 bytes
.data
putc B
Thanks to @ASCII-only for golfing off 48 50 51 bytes!
Try it online!
### Background
Running the above program with eli <file> interprets it, but elc -<target> <file> is where the real magic happens: it translates ELVM-IR source code to any of the supported backends!
Try it online!
The ELVM toolchain also supports compiling (a subset of) C and its standard library to ELVM-IR.
Try it online!
# Nikud, 672 bytes
ֱֱֱֱֱֱֱֱֱֱֱֱֱֱֱֱֱֱֱֱֱֱֱֱֱֱֱֱֱֱֱֱֱֱֱֱֱֱֱֱֱֱֱֱֱֱֱֱֱֱֱֱֱֱֱֱֱֱֱֱֱֱֱֱֱֱֱֱֱֱֲֳֳֳֳֳֳֳֳֳֳֳֳֳֶֶֶֶֶֶֶֶֶֶֶֶֶֶֶֶֶֶֶֶֶֶֶֶֶֶֶֶֶֶֶֶֶֶֶֶֶֶֶֶֶֶֶֶֶֶֶֶֶֶֶֶֶֶֶֶֶֶֶֶֶֶֶֶֶֶֶֶֶֶֶֶֶֶֶֶֶֶֶֶֶֶֶֶֶֶֶֶֶֶֶֶֶֶֶֶֶַַַַַַַַַַַַַַַַַַַַַַַַַַַַַַַַַַַַַַַַַַַַַַַַַַַַַַַַַַַַַַַַַַַַַַַַַַַַַַַַַַַַָָָָָָָָָָָָָָָָָָָָָָָָָָָָָָָָָָָָָָָָָָָָָָָָָָָָָָָָָָָָָָָָָָָָָָָ
Try it online
Even though it has tons of bytes, it's top 3 in width, as all the characters are diacritical marks.
There isn't much useful to explain. The character codes are put in reverse order into the stack using mostly אֱ (push 1), אֶ (dup) and אַ (add). Then אֵ (print as char) is called 13 times.
Another thing that adds to the byte count is that each character is represented by 2 bytes in UTF-8. So it's actually 336 characters.
• I think this can be golfed more using multiplication. – Windmill Cookies Jun 14 '18 at 5:17
# axo, 22 bytes
"!dlroW ,olleH"[>[(#<\
Try it online!
Pushes "Hello, World!" to the stack "!dlroW ,olleH"
Duplicates top of stack afterward, which results in "HHello, World!" [
Moves to the right >.
Duplicates it again, which results in "HHHello, World!" [
Outputs "H" while popping from the stack, so the stack is "HHello, World!" (
Pops the top of the stack, results in "Hello, World!" #
Moves left <
Pops the top of the stack, results in "ello, World!" #
Outputs "e" while popping from the stack, which results in STDOUT being "He" and the stack being "llo, World!" (
Duplicates the top of the stack, resulting in the stack being "lllo, World!" [
Moves right. >
And I'm sure you can figure out the rest. If you can't, I'll update a more indepth-explanation soon.
# A Pear Tree, 25 bytes
print'Hello, World!'#»G²Ú
Try it online!
A Pear Tree programs are written in an arbitrary ASCII-consistent 8-bit character set; for codepoint 128 and above, the interpreter cares about the codepoint numerically, not the represented character. TIO uses Latin-1, so the above program is actually a Latin-1 decoding of the codepoints that make it up.
## Explanation
print'Hello, World!' should be fairly self-explanatory. However, there is some choice available here; print"Hello, World!" would have been the same length, but leads to the resulting checksum being less printable.
The checksum is the interesting part of the program. In this program, that's the #»G²Ú at the end. For golfing, you'd want the shortest workable checksum, which is normally 4 or 5 bytes long. (It's a 32-bit checksum, so 4 bytes would normally be enough, but the checksum is also executed as code, and thus needs to be a valid command; the # starts a comment, so # plus 4 bytes is normally enough to add a checksum to anything.) The checksum doesn't have to cover the whole code, but does have to cover a prefix of the part of the code that actually runs; adding comments at the end is terser than adding them at the start, and we want to execute the entire program, so for this program, I caused the checksum to cover the entire program.
Although a Hello World program doesn't benefit much from the checksumming, we could have made use of the checksum behaviour to embed the Hello World program into a larger document or write multiple copies of the program so that if one gets corrupted, the others can still run. This makes A Pear Tree considerably more robust than most languages are.
# ]=[, 164 bytes
[=======[==]]=[[=[[=]]=[[=[[========]]][]=[]=[[=[=[=]]][]=[[====[====]]=[[===[==]]=[[========[=======]]=[]=[[=[=[====]]=[[=[[========]]=[[=[[]]=[[===[===]]=[[=[]]=[
]=[ was a language which only uses the symbols ], =, and [.
The ]=[ interpreter is written in 12-Basic.
• There's something amusing about the fact that the ]=[ interpreter at the link is written in the 12-basic interpreter. – snail_ May 31 '18 at 20:18
• Permalink no longer works – ASCII-only Dec 26 '18 at 1:07
• now I can't find the interpreter... – 12Me21 Dec 28 '18 at 3:18
• I was looking through some old files and I found the interpreter, finally. – 12Me21 Mar 13 '19 at 21:32
# Bootable x86 machine code, 512 bytes
Hexdump:
31 c9 8e d9 be 10 7c b1 0d ac b4 0e cd 10 e2 f9 |1.....|.........|
48 65 6c 6c 6f 2c 20 57 6f 72 6c 64 21 00 00 00 |Hello, World!...|
*
00 00 00 00 00 00 00 00 00 00 00 00 00 00 55 aa |..............U.|
* represents 464 bytes of padding required to place the bootable flag (55 aa) at offset 510.
This is the same as the following assembler code, which can be assembled using nasm hello.asm -f bin -o hello.bin, assuming the assembler code is in a file called hello.asm
[ORG 0x7C00]
[BITS 16]
xor cx, cx ; Set cx to 0
mov ds, cx ; Set ds to cx (0)
mov si, msg ; Set si to the address of the message
mov cl, 13 ; Set cx to 13 (the size of the message)
print_loop: ; For each character in the message:
lodsb ; Set al to the character
mov ah, 0x0E ; Set ah to 0x0E
int 0x10 ; Call interrupt 0x10 (video services) with ah set to 0x0E (print al to screen)
loop print_loop ; Decrement cx and continue the loop if cx > 0
msg:
db 'Hello, World!'
times 510 - ($-$\$) db 0
db 0x55
db 0xAA
## Running
The code can be runned with QEMU using the following command, assuming the binary code is saved in a file called hello.bin:
qemu-system-x86_64 -drive format=raw,file=hello.bin
• Can you assume that regs like cx` are already cleared? – ceilingcat May 7 '19 at 5:32
|
2021-07-28 19:55:15
|
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|
https://questioncove.com/updates/4ea7a61ce4b0a7d514214102
|
Mathematics
OpenStudy (anonymous):
ok integral of INT/limits 1 to 0/ (e^x)^2
OpenStudy (anonymous):
rewrite?
OpenStudy (anonymous):
$\int_0^1 e^{x^2}dx$
OpenStudy (anonymous):
ah it is error function or something like that I can't do it
OpenStudy (anonymous):
OpenStudy (agreene):
yeah, i think its like 1/2 sqrt(pi)*error(imaginary) or some such
OpenStudy (anonymous):
yep. that's what my calcu says.. 1.462651746 xD
OpenStudy (anonymous):
what is this now ?
OpenStudy (anonymous):
nonono, (e^x)(e^x) or (e^x)^2, integrate that from 1 to 0
OpenStudy (anonymous):
And I have a computer right in front of me as well as a ti-84+ so I dont want your calculator or wolfram alpha answers please :)
OpenStudy (agreene):
Oh, that is much simpler. (e^x)^2 = e^2x so: $\int\limits_{0}^{1}e^{2x}dx=\frac{1}{2}e^2x$ take it to the limits: [1/2*e^(2)]-[1/2*e^(0)]=1/2e^2-1/2 factor it and you have: $\frac{1}{2}(e^2-1)$
OpenStudy (anonymous):
oh you sure that (e^x)^2=e^2x?
OpenStudy (agreene):
yes, that is one of the properties of exponents
OpenStudy (anonymous):
fsho
|
2020-10-20 00:30:42
|
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|
https://www.acmicpc.net/problem/20999
|
시간 제한메모리 제한제출정답맞힌 사람정답 비율
12 초 512 MB45141330.952%
## 문제
The company you are currently working for has decided to push the idea of flat organizational structure to its limits: for every pair of employees $A$ and $B$, either $A$ has been assigned to directly supervise $B$'s work or $B$ has been assigned to directly supervise $A$'s work. Of course, it means that one might now have quite a lot of direct supervisors... which is great, because it makes an employee feel that their work truly is important to so many people rather than just to a single manager, the executives say.
There is always room for improvement, though. As the corporate goal for this year, the hierarchy will be revised to ensure that whenever a person $A$ is directly supervised by a person $B$, then $B$ is also indirectly supervised by $A$ at the same time (we say that $B$ is indirectly supervised by $A$ if there exists $n > 2$ and a sequence $(c_1, \ldots, c_n)$ such that $c_1 = A, c_n = B$ and for each $i < n$, $c_i$ is a direct supervisor of $c_{i+1}$).
It will ensure that any employee would think twice before deciding to abuse their position of power over anyone else, the executives say.
It should not come off as a surprise, though, that one might get somewhat annoyed if they learn that their supervisee has suddenly been appointed as their supervisor. And some such decisions might cause more resentment than others. Your task is to fulfill the corporate goal by reversing some of the dependencies between employees in such a way that the sum of resentments over these changes is as small as possible.
## 입력
The first line of input contains the number of test cases $z$ ($1 \leq z \leq 100$). The descriptions of the test cases follow.
The first line of every test case contains a single integer $n$ ($1 \leq n \leq 2000$) -- the number of employees. The employees are numbered from $1$ to $n$.
Then follow $n$ lines, containing $n$ integers $d_{i, j}$ each ($0 \leq d_{i, j} \leq 2\cdot 10^9$). If the employee $i$ is a direct supervisor of employee $j$, then $d_{i, j} > 0$ describes the resentment that $i$ would feel if this dependency got reversed. Otherwise (that is, if $j$ is a direct supervisor of $i$ or if $i = j$), $d_{i, j} = 0$.
The total number of employees in all test cases does not exceed $10000$.
## 출력
For each test case, produce a solution which ensures that, for any pair of employees $i, j$ ($1 \leq i, j \leq n, i \neq j$), either $i$ will be a direct supervisor of $j$ and $j$ will be an indirect supervisor of $i$, or vice versa. Your solution should minimize the sum of resentments the employees will feel. If more than one such solution exists, you can print any of them.
If no solution exists, you should output a single line containing the word "NO".
Otherwise, in the first line output a single word "YES". In the second line, print two integers $k$ and $r$ -- the number of dependencies between employees which you intend to reverse and the achieved sum of resentments, respectively. Note that you do not need to minimize $k$.
Then output $k$ lines, each containing two integers -- the identifiers of employees $a, b$ ($1 \leq a, b \leq n, a \neq b$) such that $a$ is currently a direct supervisor of $b$ and their relationship should get reversed. You should never output the same pair of employees more than once.
## 예제 입력 1
1
5
0 1 0 6 11
0 0 1 6 12
2 0 0 7 12
0 0 0 0 4
0 0 0 0 0
## 예제 출력 1
YES
2 10
4 5
2 4
|
2022-08-19 12:07:28
|
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|
http://www.computer.org/csdl/trans/td/2011/04/ttd2011040695-abs.html
|
Publication 2011 Issue No. 4 - April Abstract - Transforming Complete Coverage Algorithms to Partial Coverage Algorithms for Wireless Sensor Networks
This Article Share Bibliographic References Add to: Digg Furl Spurl Blink Simpy Google Del.icio.us Y!MyWeb Search Similar Articles Articles by Yingshu Li Articles by Chinh Vu Articles by Chunyu Ai Articles by Guantao Chen Articles by Yi Zhao
Transforming Complete Coverage Algorithms to Partial Coverage Algorithms for Wireless Sensor Networks
April 2011 (vol. 22 no. 4)
pp. 695-703
ASCII Text x Yingshu Li, Chinh Vu, Chunyu Ai, Guantao Chen, Yi Zhao, "Transforming Complete Coverage Algorithms to Partial Coverage Algorithms for Wireless Sensor Networks," IEEE Transactions on Parallel and Distributed Systems, vol. 22, no. 4, pp. 695-703, April, 2011.
BibTex x @article{ 10.1109/TPDS.2010.124,author = {Yingshu Li and Chinh Vu and Chunyu Ai and Guantao Chen and Yi Zhao},title = {Transforming Complete Coverage Algorithms to Partial Coverage Algorithms for Wireless Sensor Networks},journal ={IEEE Transactions on Parallel and Distributed Systems},volume = {22},number = {4},issn = {1045-9219},year = {2011},pages = {695-703},doi = {http://doi.ieeecomputersociety.org/10.1109/TPDS.2010.124},publisher = {IEEE Computer Society},address = {Los Alamitos, CA, USA},}
RefWorks Procite/RefMan/Endnote x TY - JOURJO - IEEE Transactions on Parallel and Distributed SystemsTI - Transforming Complete Coverage Algorithms to Partial Coverage Algorithms for Wireless Sensor NetworksIS - 4SN - 1045-9219SP695EP703EPD - 695-703A1 - Yingshu Li, A1 - Chinh Vu, A1 - Chunyu Ai, A1 - Guantao Chen, A1 - Yi Zhao, PY - 2011KW - Partial coverageKW - wireless sensor networksKW - energy efficiency.VL - 22JA - IEEE Transactions on Parallel and Distributed SystemsER -
Yingshu Li, Georgia State University, Atlanta
Chinh Vu, Georgia State University, Atlanta
Chunyu Ai, Georgia State University, Atlanta
Guantao Chen, Georgia State University, Atlanta
Yi Zhao, Georgia State University, Atlanta
The complete area coverage problem in Wireless Sensor Networks (WSNs) has been extensively studied in the literature. However, many applications do not require complete coverage all the time. For such applications, one effective method to save energy and prolong network lifetime is to partially cover the area. This method for prolonging network lifetime recently attracts much attention. However, due to the hardness of verifying the coverage ratio, all the existing centralized or distributed but nonparallel algorithms for partial coverage have very high time complexities. In this work, we propose a framework which can transform almost any existing complete coverage algorithm to a partial coverage one with any coverage ratio by running a complete coverage algorithm to find full coverage sets with virtual radii and converting the coverage sets to partial coverage sets via adjusting sensing radii. Our framework can preserve the characteristics of the original algorithms and the conversion process has low time complexity. The framework also guarantees some degree of uniform partial coverage of the monitored area.
[1] E.W. Weisstein, "Area, MathWorld—A Wolfram Web Resource," http://mathworld.wolfram.comArea.html, 2010.
[2] C.T. Vu, S. Gao, W.P. Deshmukh, and Y. Li, "Distributed Energy-Efficient Scheduling Approach for $k$ -Coverage in Wireless Sensor Networks," Proc. 25th Military Comm. Conf. 2006 (MILCOM '06), Oct. 2006.
[3] C.T. Vu, Z. Cai, and Y. Li, "Distributed Energy-Efficient Algorithms for Coverage Problem in Adjustable Sensing Ranges Wireless Sensor Networks," Discrete Mathematics, Algorithms and Applications, vol. 1, no. 3, pp. 299-317, 2009.
[4] C.T. Vu and Y. Li, "Delaunay-Triangulation Based Complete Coverage in Wireless Sensor Networks," Proc. Int'l Workshop Information Quality and Quality of Service for Pervasive Computing (IQ2S '09) in Conjunction with PERCOM '09, Mar. 2009.
[5] C.F. Huang, L.C. Lo, Y.C. Tseng, and W.T. Chen, "Decentralized Energy-Conserving and Coverage-Preserving Protocols for Wireless Sensor Networks," Circuits and Systems, vol. 1, pp. 640-643, May 2005.
[6] H. Zhang and J.C. Hou, "Maintaining Sensing Coverage and Connectivity in Large Sensor Networks," Proc. 2004 NSF Int'l Workshop Theoretical and Algorithmic Aspects of Sensor, Ad Hoc Wireless, and Peer-to-Peer Networks, 2004.
[7] A. Gallais, J. Carle, D. Simplot-Ryl, and I. Stojmenovic, "Localized Sensor Area Coverage with Low Communication Overhead," IEEE Trans. Mobile Computing, vol. 7, no. 5, pp. 661-672, May 2008.
[8] H. Zhang and J. Hou, "On Deriving the Upper Bound of $\alpha$ -Lifetime for Large Sensor Networks," Proc. Fifth ACM Int'l Symp. Mobile Ad Hoc Networking and Computing, 2004.
[9] H. Zhang and J. Hou, "Maximizing $\alpha$ -Lifetime for Wireless Sensor Networks," Proc. Third Int'l Workshop Measurement, Modeling, and Performance Analysis of Wireless Sensor Networks (SenMetrics '05), July 2005.
[10] S. Gao, X. Wang, and Y. Li, "$p$ -Percent Coverage Schedule in Wireless Sensor Networks," Proc. 17th Int'l Conf. Computer Comm. and Networks (ICCCN '08), Aug. 2008.
[11] Y. Liu and W. Liang, "Approximate Coverage in Wireless Sensor Networks," Proc. IEEE Conf. Local Computer Networks 30th Anniversary (LCN '05), pp. 68-75, 2005.
[12] H. Bai, X. Chen, Y. Ho, and X. Guan, "Percentage Coverage Configuration in Wireless Sensor Networks," Parallel and Distributed Processing and Applications, vol. 3758/2005, pp. 780-791, 2005.
[13] B. Son, Y.-S. Her, and J.-G. Kim, "A Design and Implementation of Forest-Fires Surveillance System Based on Wireless Sensor Networks for South Korea Mountains," IJCSNS Int'l J. Computer Science and Network Security, vol. 6, no. 9B, pp. 124-130, Sept. 2006.
[14] M. Hefeeda, "Forest Fire Modeling and Early Detection Using Wireless Sensor Networks," Technical Report TR 2007-08, School of Computing Science, Simon Fraser Univ., Aug. 2007.
[15] L. Yu, N. Wang, and X. Meng, "Real-Time Forest Fire Detection with Wireless Sensor Networks," Wireless Comm., Networking and Mobile Computing, vol. 2, nos. 23-26, pp. 1214-1217, Sept. 2005.
[16] P. Berman, G. Calinescu, C. Shah, and A. Zelikovsky, "Efficient Energy Management in Sensor Networks," Ad Hoc and Sensor Networks, Wireless Networks and Mobile Computing, Y. Xiao and Y. Pan, eds., vol. 2, Nova Science Publishers, 2005.
[17] N. Garg and J. Könemann, "Faster and Simpler Algorithms for Multicommodity Flow and Other Fractional Packing Problems," Proc. 39th Ann. Symp. Foundations of Computer Science, pp. 300-309, 1998.
[18] Y. Wu, C. Ai, S. Gao, and Y. Li, "$p$ -Percent Coverage in Wireless Sensor Networks," Proc. Int'l Conf. Wireless Algorithms, Systems and Applications (WASA '08), Oct. 2008.
Index Terms:
Partial coverage, wireless sensor networks, energy efficiency.
Citation:
Yingshu Li, Chinh Vu, Chunyu Ai, Guantao Chen, Yi Zhao, "Transforming Complete Coverage Algorithms to Partial Coverage Algorithms for Wireless Sensor Networks," IEEE Transactions on Parallel and Distributed Systems, vol. 22, no. 4, pp. 695-703, April 2011, doi:10.1109/TPDS.2010.124
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2014-03-10 06:33:52
|
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|
https://christopherlu.github.io/publications/millimap
|
# See Through Smoke: Robust Indoor Mapping with Low-cost mmWave Radar
Published:
Authors: Chris Xiaoxuan Lu, Stefano Rosa, Peijun Zhao, Bing Wang, Changhao Chen, John A. Stankovic, Niki Trigoni and Andrew Markham. In MobiSys 2020.
[paper] [talk video] [slide] [code]
## Abstract
This paper presents the design, implementation and evaluation of milliMap, a single-chip millimetre wave (mmWave) radar based indoor mapping system targetted towards low-visibility environments to assist in emergency response. A unique feature of milliMap is that it only leverages a low-cost, off-the-shelf mmWave radar, but can reconstruct a dense grid map with accuracy comparable to lidar, as well as providing semantic annotations of objects on the map. milliMap makes two key technical contributions. First, it autonomously overcomes the sparsity and multi-path noise of mmWave signals by combining cross-modal supervision from a co-located lidar during training and the strong geometric priors of indoor spaces. Second, it takes the spectral response of mmWave reflections as features to robustly identify different types of objects e.g. doors, walls etc. Extensive experiments in different indoor environments show that milliMap can achieve a map reconstruction error less than 0.2m and classify key semantics with an accuracy of $\sim 90\%$, whilst operating through dense smoke.
|
2021-01-27 09:04:47
|
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|
https://tex.stackexchange.com/questions/203544/how-to-add-a-circle-on-left-right-parentheses
|
# How to add a circle on left/right parentheses?
I need to use a new symbol to denote an operation defining by myself. How to establish a pair of new delimeters just as adding a little circle on left/right parentheses? The difficulty is there are four size of it: \big, \Big, \bigg, \Bigg.
\documentclass{article}
\begin{document}
$\bigl\newlparentheses\sin(x+\pi)\bigr\newrparentheses \cdot\Bigl\newlparentheses\frac ab\Bigr\newrparentheses$
\end{document}
• Welcome to TeX.SX! Please help us to help you and add a minimal working example (MWE) that illustrates your problem. It will be much easier for us to reproduce your situation and find out what the issue is when we see compilable code, starting with \documentclass{...} and ending with \end{document} – user31729 Sep 28 '14 at 9:02
• @Christian Hupfer I've edited my question. Thanks. – delimeterfont Sep 28 '14 at 9:08
You can define new brackets.
For classic \left, \right as
\def\oright#1{\right#1\mskip-10mu\circ}
\def\oleft#1{\circ\mskip-10mu\left#1}
for big \bigl, \bigr as
\def\obigr#1{\bigr#1\mskip-8mu\circ}
\def\obigl#1{\circ\mskip-8mu\bigl#1}
and for extra big \Bigl, \Bigr as
\def\oBigr#1{\Bigr#1\mskip-8mu\circ}
\def\oBigl#1{\circ\mskip-8mu\Bigl#1}
Use it like:
$\oleft(\sin(x+\pi)\oright)$
$\obigl(\sin(x+\pi)\obigr)$
$\oBigl(\sin(x+\pi)\oBigr)$
The result will look like:
• That is good. It will better if the size of circle can change compliable with size of parentheses. – delimeterfont Sep 29 '14 at 5:41
|
2019-09-22 13:14:47
|
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|
https://solvedlib.com/solve-06-y348,154362
|
# Solve 0.6 (y+3)=4.8?
###### Question:
Solve 0.6 (y+3)=4.8?
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##### Which of the following sets of quantum numbers are not allowed in the hydrogen atom? For the sets of quantum numbers that are incorrect, state what is wrong in each set. a. $n=3, \ell=2, m_{c}=2$ b. $n=4, \ell=3, m_{\ell}=4$ c. $n=0, \ell=0, m_{\ell}=0$ d. $n=2, \ell=-1, m_{c}=1$
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|
2023-03-25 05:07:22
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 6, "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": 0.369108647108078, "perplexity": 3774.4526875825577}, "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-2023-14/segments/1679296945315.31/warc/CC-MAIN-20230325033306-20230325063306-00096.warc.gz"}
|
https://socratic.org/questions/how-do-you-simplify-216-2-3
|
# How do you simplify #(-216)^(2/3)?
${\left(- 216\right)}^{\frac{2}{3}}$
= ${\left({\left(- 216\right)}^{\frac{1}{3}}\right)}^{2}$
= ${\left(- 6\right)}^{2}$
= $\left(- 6\right) \cdot \left(- 6\right)$
= $36$
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2020-04-05 10:14:21
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http://ieeexplore.ieee.org/xpls/icp.jsp?reload=true&arnumber=6156837
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Browse
• Abstract
SECTION 1
## Introduction
The principle of causality forbids the movement of any object at a speed exceeding that of light in a vacuum $(c)$. However, various theoretical and experimental studies demonstrated that faster-than-$c$ propagation of light pulses can be realized utilizing specific dispersion characteristics of materials in the vicinity of resonance frequencies. The speed mentioned here (and throughout this paper) is, of course, different from the signal velocity at which information travels, which is defined as that of the front of a square pulse and cannot exceed $c$ [1]. A gain doublet [1], [2], [3], electromagnetically induced absorption [4], coherent population oscillation [5], and negative-index metamaterials [6] have been studied to implement such desired dispersion properties. Recently, several researchers reported superluminal propagation of optical pulses through various waveguide structures using highly birefringent optical fibers [7], nonlinear processes [8], or the resonance of ring resonators [9]. In this paper, we would like to put forward a new strategy of superluminal transmission of optical wavepackets through waveguides. The key idea is to adopt an Airy wavepacket [10], [11], [12], [13], [14], [15], [16], [17], [18] as the initial field launched to a waveguide. Then, the launched Airy wavepacket becomes associated with a waveguide mode and forms a type of hybrid wavepacket. This resultant wavepacket propagates like an optical bullet without any spatial diffraction or temporal spreading [16]. Its effective propagation speed can be easily enhanced higher than $c$ by changing only the initial field configuration (by using an external modulation) without any changes in the physical aspects of the waveguide.
SECTION 2
## Waveguide Mode-Airy Wavepackets
Let us consider an electromagnetic wavepacket launched to a waveguide along the $+z$ direction (see Fig. 1). We will limit our attention only to 1-D waveguides in this paper. However, the extension to higher dimensional ones is quite straightforward. The wavepacket can be written as $\phi = A_{T}(t, z)g(y)\exp[j(\beta_{0} z - \omega_{0} t)]$, where $\omega_{0}$ is the carrier frequency of the wavepacket, and $g(y)$ describes the transverse profile of a waveguide mode with a wavenumber of $\beta_{0}$ at $\omega_{0}$. $\phi$ can be $E_{x}$ and $H_{x}$ field components for TE- and TM-polarized light, respectively. Assuming $A_{T}$ is a slowly varying function of $z$ (i.e., with the paraxial approximation) and including the dispersive property of the wavenumber $\beta$ up to the second order (note that the physics investigated in this paper does not change much when we include the third-order dispersive property of $\beta$ as was in [16]), we can easily show that $A_{T}(t, z)$ can be a nonspreading Airy wavepacket given by TeX Source $$\displaylines{A_{T}(\tau, z) = Ai\left[(-1)^{m}{\tau \over \tau_{0}} - \left({\beta_{2}z \over 2\tau_{0}^{2}}\right)^{2} - j{a\beta_{2}z \over \tau_{0}^{2}}\right]\exp\left[-j \left({(-1)^{m}\tau + a^{2}\tau_{0} \over 2\tau_{0}}{\beta_{2}z \over \tau_{0}^{2}} - {1 \over 12}\left({\beta_{2}z \over \tau_{0}^{2}}\right)^{3}\right)\right]\hfill\cr\hfill\times\ \exp\left[(-1)^{m}{a\tau \over \tau_{0}} - {a \over 2}\left({\beta_{2}z \over \tau_{0}^{2}}\right)^{2}\right]\quad\hbox{(1)}}$$ where $\tau_{0}$ is an arbitrary scaling factor, and $m$ is either 0 or 1 and related to the acceleration direction of the wavepacket (see (2), shown below, and the related discussion for details). $v_{g}\ [= (\partial\beta/\partial\omega\vert_{\omega_{0}})^{-1}]$ and $\beta_{2}\ [= \partial^{2}\beta/\partial\omega^{2}\vert_{\omega_{0}}]$ denote the group velocity and the group-velocity dispersion coefficient at $\omega_{0}$, respectively, and a transformation of $\tau = t - (z/v_{g})$, where $\tau$ represents time in a reference frame moving at $v_{g}$ has been introduced. We note that the above equation seems to be much different from [16 eq. (6)], which also describes a nonspreading Airy wavepacket, but if, in [16 eq. (6)], we write $\tau_{0}$ as $(-1)^{m}\tau_{0}$ (to explicitly consider both its positive and negative values) and set the parameter related to the third-order dispersive property $(b)$ equal to zero, we can immediately obtain this expression.
Fig. 1. Concept of a superluminal waveguide mode-Airy wavepacket. $g(y)$ and $A_{T}(t, z)$ denote the transverse profile of the waveguide mode at the carrier frequency $\omega_{0}$ and the conventional $(1 + 1)D$ Airy wavepacket profile, respectively.
The initial field $A_{T}(t, 0)$ (at $z = 0$ we have $t = \tau$) is given as $Ai[(-1)^{m}t/\tau_{0}]\exp[(-1)^{m}(a^{\prime}t/ \tau_{0}\ - ja^{\prime\prime}t/\tau_{0})]$, where $a^{\prime}$ denotes the exponential apodization parameter which must satisfy $(-1)^{m}a^{\prime}\ >\ 0$ for the finite-energy (physically realizable) Airy wavepacket, and $a^{\prime\prime}$ is the phase modulation factor along the scaled temporal coordinate which should be nonnegative to avoid unphysical solutions. From the above discussion, we can see that a type of hybrid wavepacket, which is given by the product of a waveguide mode and a finite Airy wavepacket, can propagate through the waveguide.
Here, we would like to point out two unique features of the Airy wavepacket given by (1). One is that it undergoes neither spatial diffraction nor temporal spreading if $a^{\prime} = 0$. Due to nonzero values of $a^{\prime}$, the actual finite-energy Airy wavepacket diffracts or spreads a little bit. This means that the waveguide mode-Airy wavepacket suffers minimal diffraction or spreading, propagating through the waveguide like an optical bullet. The other is that during propagation along the $+z$ direction, its peak position shifts laterally along the $\tau$ direction. The peak of the Airy wavepacket follows the following spatiotemporal trajectory $(\tau_{p}, z_{p})$ which can be obtained by setting the real arguments of the Airy function in (1) to be zero TeX Source $$(-1)^{m}\tau_{p} = {a^{\prime\prime}\beta_{2}\over \tau_{0}}z_{p} + {1\over 2} {\beta_{2}^{2} \over 2\tau_{0}^{3}}z_{p}^{2}\eqno{\hbox{(2)}}$$ where we can find that, during propagation, the peak position shifts along the $+\tau$ or $-\tau$ direction, depending on whether $m$ is 0 or 1. These shifts along the positive and negative time coordinates in a moving frame can be interpreted as the slowdown and speedup of the wavepacket, respectively.
SECTION 3
## Superluminal Transmission of Hybrid Wavepackets
From the second feature mentioned above, we can see that the Airy wavepacket is either accelerated (when $m = 1$) or decelerated (when $m = 0$) during propagation. Since the Airy part corresponds to the temporal envelope of the hybrid wavepacket, the transmission speed of the Airy wavepacket determines that of the waveguide mode-Airy hybrid one. As we want fast transmission of optical wavepackets, we will limit our attention to the case of $m = 1$. We have TeX Source $${\beta_{2}^{2} \over 2\tau_{0}^{3}}z_{p}^{2} + 2\left({a^{\prime \prime}\beta_{2} \over \tau_{0}} - {1 \over v_{g}} \right)z_{p} + 2t = 0\eqno{\hbox{(3)}}$$ and the trajectory $(t, z_{p})$ of the peak of the hybrid wavepacket can be written as TeX Source $$z_{p}(t) = {2\tau_{0}^{3} \over \beta_{2}^{2}} \left({1 - \alpha \over v_{g}} - \sqrt{\left({1 - \alpha \over v_{g}}\right)^{2} - {\beta_{2}^{2} t \over \tau_{0}^{3}}}\right)\eqno{\hbox{(4)}}$$ where $\alpha = a^{\prime \prime}\beta_{2}v_{g}/\tau_{0}$. By the differentiation with respect to $t$, we can obtain the speed of the hybrid wavepacket as TeX Source $$v_{p}(t) = \left[\left({1 - \alpha \over v_{g}}\right)^{2} - {\beta_{2}^{2} t \over \tau_{0}^{3}}\right]^{-1/2} = {v_{0} \over \sqrt{1 - t/T_{m}}}\eqno{\hbox{(5)}}$$ where $v_{0} = v_{g}/\vert 1 - \alpha\vert$, and $T_{m} = \tau_{0}^{3}/(\beta_{2}^{2}v_{0}^{2})$. Above results hold only when $t\ <\ T_{m}$. Otherwise, (3) does not have a root for a given time $t$, meaning that the temporal profile of the wavepacket no longer maintains its original peak or overall shape. Therefore, we can say that $T_{m}$ denotes the maximum time during which the hybrid wavepacket can be transported with minimal diffraction or spreading.
From the above equation, we can see that $1 - \alpha$ must be positive for the forward propagation of the wavepacket in the region of our interest ($z_{p} \geq 0$ because our initial field is assumed to be launched at $z = 0$). The initial launching speed of the hybrid wavepacket is given as $v_{0} = v_{g}/ (1 - \alpha)$, which can be easily changed by controlling $\alpha$ or $a^{\prime\prime}$. By making $\alpha$ close to 1, we can enhance the initial launching speed significantly. If we write $v_{g} = c/n_{g}$, we can obtain a superluminal wavepacket by making $\alpha\ >\ 1 - n_{g}^{-1}$, that is, $a^{\prime\prime}\ >\ \tau_{0}(n_{g} - 1)/(\beta_{2}c)$ when $\beta_{2}\ >\ 0$ or $a^{\prime\prime}\ < \tau_{0}(n_{g} - 1)/(\beta_{2}c)$ when $\beta_{2}\ <\ 0$. The latter condition cannot be satisfied since it requires $n_{g}\ <\ 1$ for a nonnegative $a^{\prime\prime}$. In addition, $v_{p}(t)$ is a monotonically increasing function with respect to $t$, that is, this hybrid wavepacket always moves faster than the initial velocity $v_{0}$ and speeds up with passing time. That is, the transmission speed of the hybrid wavepacket can be enhanced further during propagation, making its superluminal transmission a lot easier to realize: even if $v_{0}\ <\ c$, $v_{p}(t)$ becomes superluminal when $t\ >\ \tau_{0}^{3}[n_{g}^{2}(1 - \alpha)^{2} - 1]/(\beta_{2}^{2} c^{2})$.
One of the most important parameters in our scheme is $T_{m}$. If we put $T_{m} = N\tau_{0}$, we have $\tau_{0}^{2}(1 - \alpha)^{2} = N(\beta_{2}v_{g})^{2}$. Another crucial one is $\alpha$ which can be written as $\alpha = (a^{\prime\prime}/\tau_{0})\beta_{2}v_{g}$ and determines the initial launching speed of the wavepacket. Therefore, we can see that $\beta_{2}v_{g}$, whose value is determined by the waveguide structure (materials and geometry), plays a key role in our scheme. From above results, we can obtain $\tau_{0} = \sqrt{N}\beta_{2}v_{g}/(1 - \alpha)$ and $a^{\prime \prime} = \sqrt{N}\alpha/(1 - \alpha)$. If the waveguide gives us $\beta_{2}v_{g}$ and we select appropriate values of $\alpha$ and $N$, then we can determine two design parameters $\tau_{0}$ and $a^{\prime\prime}$ using the above relations.
SECTION 4
## Numerical Study Using a Metamaterial Waveguide
Let us examine the above discussions in more detail using a numerical example. We should mention that since we require a positive $\alpha$ (close to 1), $\beta_{2}$ must be positive as well. That is, anomalous dispersion $(\partial v_{g}/\partial\omega\ <\ 0)$ is necessary in our scheme. For this purpose, we adopted a metamaterial waveguide: a slab composed of a negative-index core between silica clads. The negative-index metamaterial was 0.55 $\mu\hbox{m}$ thick and its relative permittivity and permeability were given as $\varepsilon_{co} = -4$ and $\mu_{co} = -2$ at $\lambda_{0} = 1550\ \hbox{nm}$. We used the Sellmeier equation and the Drude model to take into account the dispersive property of silica and metamaterial layers, respectively. At $\omega_{0}\ (= 2\pi c/\lambda_{0})$, we obtained a backward guiding mode having $\beta_{0} = -1.6372 k_{0}\ (k_{0} = 2 \pi/\lambda_{0})$, $v_{g} = c/13.91$, and $\beta_{2} = 5.666 \times 10^{-21}$. In Fig. 2(a), we showed the intensity profiles of the hybrid wavepacket at the center of the negative-index core $(y = 0)$ for various longitudinal $(z)$ positions. We set $N = 10$, $\alpha = 0.95$, $a = 0.05 - j60.08$, and $\tau_{0} = 7.72\ \hbox{ps}$. From the results, we can easily find that the peak of the hybrid wavepacket was accelerated along the $-\tau$ direction. In addition, the wavepacket underwent little diffraction or spreading even after it propagated $2000\lambda_{0}$. In Fig. 2(b), we plotted these results again: in this case using the time coordinate $(t)$ instead of that in a moving frame $(\tau)$. Here, we can see that the hybrid wavepacket in its peak position traveled $2000\lambda_{0}$ in 7 picoseconds, resulting in an average speed of $1.48c$. If we note that the group velocity of the waveguide mode itself is only about $0.072c$, the transmission speed is enhanced significantly via the introduction of the Airy wavepacket and its association with the waveguide mode.
Fig. 2. (a) Propagating profiles (at $z = 0$, $400\lambda_{0}$, $800 \lambda_{0}$, $1200\lambda_{0}$, $1600\lambda_{0}$, and $2000 \lambda_{0}$) of the hybrid wavepacket at the center of the waveguide $(y = 0)$ from the view point in which the transverse direction indicates time in a moving frame $(\tau)$. The acceleration along the $-\tau$ direction can be interpreted as a speedup of the wavepacket. (b) Propagating profiles (at $z = 0$ and $2000\lambda_{0}$) from the view point in which the transverse direction indicates time $(t)$. The average speed of the hybrid wavepacket in its peak position is increased as high as $1.48c$, although $v_{g}$ of the waveguide mode is only about $0.072c$. Two additional factors contribute to this speedup: the initial launching speed of the hybrid wavepacket and its sequential acceleration along the time domain in a moving frame (see Fig. 3 for details).
In Fig. 3, we plotted further the calculated speed of the hybrid wavepacket as it propagates through the negative-index-core slab. We can identify two mechanisms which enhance the transmission speed of the hybrid wavepacket. One is via the initial launching speed: by controlling $a^{\prime\prime}$ or $\alpha$ which means physically the additional modulation of the initial field profile. The initial field of the waveguide mode-Airy wavepacket at $z = 0$ can be written as $A_{T}(t, 0; a^{\prime\prime}) = A_{T}(t, 0; a^{\prime\prime} = 0)\ \times \exp(ja^{\prime\prime}t/\tau_{0})$. Therefore, we can discern that by modulating $A_{T}(t, 0; a^{\prime\prime} = 0)$, we can enhance the initial launching speed quite easily, which can be found in Fig. 3 by comparing the cases of $a^{\prime\prime} = 0$ and $a^{\prime\prime} = 60.08$. The other is due to the acceleration of the hybrid wavepacket along the $-\tau$ coordinate, which can also be identified in Fig. 3; as the hybrid wavepacket travels, its transmission speed increases.
Fig. 3. Speedup of the hybrid wavepacket by increasing $a^{\prime\prime}$ or $\alpha$, i.e., by enhancing the initial launching speed (compare the cases of $a^{\prime\prime} = 0$ and $a^{\prime\prime} = 60.08$) and due to its acceleration as it propagates through the waveguide.
SECTION 5
## Conclusion
We showed that superluminal transmission of a dispersionless wavepacket can be realized via the introduction of the waveguide mode-Airy hybrid wavepacket. We have found two mechanisms which contribute to the faster-than- $c$ propagation of the hybrid wavepacket: i) the initial launching speed, which can be easily enhanced by changing the initial field configuration of the wavepacket, and ii) the acceleration feature of the Airy wavepacket along the time domain in a moving frame. We have considered a concrete example using a negative-index-core slab waveguide and proved superluminal transportation of the waveguide mode-Airy wavepacket through it numerically.
## Footnotes
This work was supported by the Basic Science Research Program through the National Research Foundation of Korea funded by the Ministry of Education, Science, and Technology under Grant 2011-0011360. Corresponding author: K.-Y. Kim (e-mail: kykim@sejong.ac.kr).
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2014-03-07 21:16:06
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https://cognitive-liberty.online/vocabulary/surveillance/
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# Surveillance
Surveillance is the monitoring of the behavior, activities, or other changing information, usually of people for the purpose of influencing, managing, directing, or protecting. Surveillance is therefore an ambiguous practice, sometimes creating positive effects, at other times negative. It is sometimes done in a surreptitious manner.
This entry was posted in Uncategorised. Bookmark the permalink.
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2020-09-25 10:08:21
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https://www.nature.com/articles/s41598-020-73245-3?error=cookies_not_supported&code=92173a6e-9548-44ad-a0cd-c9d4c7ea0cea
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Thank you for visiting nature.com. You are using a browser version with limited support for CSS. To obtain the best experience, we recommend you use a more up to date browser (or turn off compatibility mode in Internet Explorer). In the meantime, to ensure continued support, we are displaying the site without styles and JavaScript.
# Projections of heat stress and associated work performance over India in response to global warming
## Abstract
Summertime heat stress future projections from multi-model mean of 18 CMIP5 models show unprecedented increasing levels in the RCP 4.5 and RCP 8.5 emission scenarios over India. The estimated heat stress is found to have more impact on the coastal areas of India having exposure to more frequent days of extreme caution to danger category along with the increased probability of occurrence. The explicit amount of change in temperature, increase in the duration and intensity of warm days along with the modulation in large scale circulation in future are seemingly connected to the increasing levels of heat stress over India. A decline of 30 to 40% in the work performance is projected over India by the end of the century due to the elevated heat stress levels which pose great challenges to the country policy makers to design the safety mechanisms and to protect people working under continuous extreme hot weather conditions.
## Introduction
Climate change, which is persuaded by the global warming effect, is a global concern due to its adverse impacts on various systems. The Fifth Assessment Report (AR5) of the Intergovernmental Panel on Climate Change (IPCC) findings confirm that the increasing anthropogenic greenhouse gas (GHG) concentrations which are responsible for the unusual Earth’s warming in recent decades1, cause the frequent high intensity heat extremes with prolonged duration affecting the living and working environments2,3,4,5. This increase in heat extremes will cause the damage to health, comfort and economic activities of millions of people across the globe3,6. It is reported that the increase in air temperature enhances the moisture holding capacity of the atmosphere that causes increase in heat stress due to the surface heating6,7,8. Many studies reported the heat stress by using only the temperature data without taking into account of humidity9,10,11. However, humidity plays an important role in discomfort due to warming and must be integrated to estimate the heat stress index12,13,14,15, particularly in tropical countries16. The study of heat stress is very important in the context of health impacts (eg: heat stroke) of the humans during their outdoor activities17,18,19,20,21,22. Extreme heat events lead to drastic decline in work productivity along with the heat related mortality23,24,25,26. Several methods/indices are used to estimate the heat stress using different environmental/meteorological factors. Some metrics use combinations of temperature and moisture, such as the the National Weather Service Heat Index and Humidex27, while others additionally include solar radiation (e.g. Wet bulb globe temperature (WBGT)28 and Environmental Stress Index29).
Unusual hot weather affects the work productivity of human societies by reducing their performance and increase in heat related illness30,31,32. Particularly, higher values of heat stress can cause deterioration in the work productivity of outdoor laborers where physically demanding tasks are required. Excessive heat stress increases the core human body temperature (above 37 degrees) thus, leading to heat stroke (severe hyperpyrexia), which causes the decline in the physical work capacity33,34. Increasing heat exposure is linked to occupational health risks and adversely impacts the work productivity20,35. Study of 370 onsite data sets in Hong Kong reported that the work duration of the construction labour has been decreased by 0.33% with one degree rise in WBGT36. Analysis of 8076 workers of 11 different studies revealed the 30% loses in work productivity due to rise in WBGT in the people who are working in single shift37. The loss in labour productivity due to heat stress over Australia has been reported38 and a decrease in productivity globally upto 20%. In tropical lands, the agricultural activities of working labor which involves higher levels of physical exertion had been badly affected by the heat stress39. It is reported that the estimated labour productivity in relation to heat stress index for 21 regions of the world infers a reduction in labour productivity due to warming of climate and resultant heat stress40. Also for the workers with low income in the tropical regions, heat stress is going to be the greatest health hazard where the facilities such as health surveillance are not available41. Venugopal42 studied the health and work productivity issues of the workers from 18 places in India and found that they are vulnerable to heat stress. The impact of heat stress on working people in hot environments is measured using WBGT by ISO 7243 standard43. Also, a few reports infer that the work ability estimated from the metrics of WBGT over New Delhi, India show better results compared to the other metrics such as Predictive Heat Stress, Universal Thermal Climate Index44. However, the studies of Brode et al.44 applied limited criteria in simulating the work ability during the day time working under light, moderate and heavy conditions and found the work ability change with “work load, time of the day and climate type”. Also, the results of the Brode et al.44 were suggested to be representatives of prevailing in agriculture,health and industrial sectors.
The present unusual warming is expected to increase faster and becomes vulnerable to physiological acclimatization among the workers carrying out activities in outdoor and indoor work places without thermal neutralities. Report from International Labour Organization (ILO) says the decline in global work productivity will be same as the loss of 80 million full time jobs by 2030 due to the increasing levels of heat stress (https://unfccc.int/news/international-labour-organization-warns-of-heat-related-job-losses). As India is more vulnerable to the heat extremes such as heat waves, heat stress due to its projected increasing temperatures39,45,46,47, knowing the impact of heat stress on the work performance helps the policy makers to take actions in order to protect the workers from the dangerous exposure of heat stress and to maintain the sustained work performance. Hence, it is important to assess the future changes in the heat stress index over India where in the outdoor activities are significant. The current study is built upon the assessment of heat stress along with its associated decline in work performance over India in the present and future climate under different emission scenarios. The results of the present work may be useful to the policy makers to design a better policy tools for taking necessary steps for the welfare of the working people due to the increased heat stress in future under changing climate conditions.
## Materials and methods
Here we used Steadman’s Heat Index (HI)48 approach which combines ambient air temperature and relative humidity to estimate the heat stress. Steadman's field observations and measurements were translated into a table that showed heat stress values and the corresponding risk levels (Table 1). On the basis of the table generated by Steadman, Rothfusz (1990)49 developed the following equation for heat index.
$$\begin{gathered} {\text{HI }} = {-}{42}.{379 } + { 2}.0{49}0{1523}*{\text{T }} + { 1}0.{14333127}*{\text{RH}}{-}0.{22475541}*{\text{T}}*{\text{RH }} \hfill \\ {-} \, \left( {{6}.{83783} \times {1}0^{{{-}{3}}} *{\text{T}}^{{2}} } \right){-}\left( {{5}.{481717} \times {1}0^{{{-}{2}}} *{\text{RH}}^{{2}} } \right) \, + \, \left( {{1}.{22874} \times {1}0^{{{-}{3}}} *{\text{T}}^{{2}} *{\text{RH}}} \right) \, \hfill \\ + \, \left( {{8}.{5282} \times {1}0^{{{-}{4}}} {\text{T}}*{\text{RH}}^{{2}} } \right) \, {-} \, \left( {{1}.{99} \times {1}0^{{{-}{6}}} {\text{T}}^{{2}} *{\text{RH}}^{{2}} } \right) \hfill \\ \end{gathered}$$
where T is ambient dry bulb temperature (°F) and RH is relative humidity in percentage. This formula was obtained through a set of measurements, mathematically analyzed by multiple regressions. The values of HI are further converted in to Celsius scale and used in the present study. The categories of heat index are classified as (1) caution (27–32 °C) (2) extreme caution (32–41 °C) (3) danger (41–54 °C) and (4) extreme danger (> 54 °C) respectively50. More information on the evaluation of HI can be found in Supplementary Material.
A relationship is developed based on the experimental data sets of working performance51, thermal comfort52 and the decrement of work performance (P in %) has been found with the rise in temperature53. It is reported that the temperature and heat index values do not vary much in many areas54 and also the sensitivity of heat—health effect estimates are not showed larger differences to measure the exposure with heat index versus temperature55,56. Though there is a variation, particularly in summer time, a substantial correlation between these two infer the same quantitative results54. Also, when using the mean temperature, the indices such as heat stress index, humidex and relative strain index, the magnitudes of all do not vary much. Average temperature performed similarly to the composite indices, but minimum and maximum temperatures performed relatively poorer. Thus, average temperature may be suitable for the development of weather-health warming systems55. Hence in the present study, we have used the heat stress index instead of temperature since the heat stress is a better indicator than temperature in heat-health research.
Thus, the equation for estimating the decline in work performance can be written as:
$${\text{P }}\left( \% \right) \, = { 2 } \times \, \left( {{\text{Heat stress}}, \, ^\circ {\text{C}}} \right) \, {-}{ 5}0.$$
The mean of three reanalysis datasets viz, the ERA-Interim reanalysis from ECMWF57, NCEP-DOE-Reanalysis258 and NOAA-CIRES 20th Century reanalysis59 for daily mean surface air temperature and relative humidity have been used for the estimation of HI. Daily simulations of 18 general circulation models (GCMs) data from suite of the Coupled Model Intercomparison Project 5 (CMIP5)60 (Table 2) and their Multi Model Mean (MMM) have been used to estimate the summertime (March to June) heat stress in the baseline period of 1986–2005 for the purpose of models evaluation. Also, we estimated the same for the three future time slices 2016–2035, 2046–2065 and 2080–2099 which are treated as near, mid and long term (IPCC, 2013)1 under the two emission scenarios of RCP 4.5 and RCP 8.5. All the models used in the present study have been interpolated to an uniform grid resolution of 1° × 1° using the bilinear interpolation method61,62,63 for making them consistent to compute the multi model mean and which will be helpful in studying the explicit diagnosis of heat stress on space and time scale.
Matthews et al.14 and Lee and Brenner64 studied the global heat index, estimated from the temperature and relative humidity while Opitz-Stapleton et al.65 and Sylla et al.19 studied the same regionally. Lee and Berner64 reported that these heat stress indices are used by International Organization for Standardization (ISO) and National Institute for Occupational Safety and Health (NIOSH) to measure the heat loads in order to prevent the heat illness. These formulae need to be weighted by population density in order to study the impact on human health and productivity when they are subjected to study for different countries over the globe. During summer, the spatial variations of temperature are very less over India except in the regions of Himalayas and Northeast region of India, the heat stress levels and the decline in work performance have been taken into account over entire India except these two regions. Hence, the results obtained over Himalayan and Northeastern parts need not be considered in the present study.
## Results and discussions
Evaluation of summertime mean climatology of heat stress index for the baseline period (1986 to 2005) from CMIP5 Multi Model Mean (MMM) and observed reanalysis data sets along with their biases (model minus observed climatology) revealed the better performance of CMIP5 MMM in capturing the heat stress characteristics over India. Figure 1 indicates that the higher percentage of heat stress days fall in caution and extreme caution categories in the baseline period. During the four categories of heat stress days (caution, extreme caution, danger and extreme danger), the bias is less than 5% which provides a fair confidence on the reliability of model data as these biases are inevitable due to the different physics schemes used by the different GCMs that are being used as multi model mean61. Uncertainties in heat stress estimations that include temperature and humidity together are typically lesser than the uncertainties when they used independently66. The spatial distribution of projected changes in percentage of days in different categories of heat stress with respect to the base line period during summertime in three times slices (2016–2035, 2046–2065 & 2080–2099) obtained from the CMIP5 MMM of RCP 4.5 & 8.5 are shown in Figs. 2 and 3 respectively. A perceptible increase in the percentage of days falling in extreme caution and danger categories is observed in both the emission scenarios and is similar to the heat stress trends observed globally for the historical periods by CMIP5 data sets39. We found similar to Raymond (2020)67 that parts of west coast and east coast are the most affected zones by increased heat stress. An increase of 20 to 30% of days is projected in the extreme caution and danger categories over these regions during the periods 2046–2065 and 2080–2099 respectively.
Further, examination of heat stress over east and west coastal regions (upto 200 km interior from the coast) indicated the domination of extreme caution days per year in RCP 4.5 & 8.5 followed by danger days. The results show that among the two coastal regions, east coast of India may experience a greater number of heat stress days in future under both emission scenarios. Exceptionally over the east coast region, people suffer with heat stress more than 60 days during the summer time of an year under extreme caution category while it exceeds 40 days in danger category by end of the century under the both emission scenarios (Fig. 4), while the number of days with heat stress are predominantly increasing over west coast (> 32 days) under extreme caution category in both the scenarios (Fig. 5). Compared to east coast, west coast is not much vulnerable but both emission scenarios show an increase in extreme caution days and project an increase in danger days under RCP 8.5 by end of the century. It is reported that the anomalous westerlies over the Indian land mass which reduce the land sea thermal contrast may cause the hot conditions over the eastern coastal regions of India68 Significant rise in maximum temperature over the west coast69 and east coast70 caused the diurnal temperature range of 1–2 °C over the coastal regions for the period 1951 to 2010 which may contribute to the higher loads of heat stress over these areas.
The probability density function (PDF) curves of daily heat stress under RCP 4.5 & 8.5 scenarios over east and west coast regions of India (Figs. 6 and 7) show the positive shifts of location and scale parameters over the east and west coastal regions. Over the east coast region, the maximum probable value of simulated heat stress index is around 34, while towards the end of the century under RCP 4.5 it is around 38 and under RCP 8.5 it is around 42, suggesting that the heat stress is becoming more intense towards the end of the century. Though the curve for the period 2080–2099 is flatter than the other two curves, under both the scenarios, they show the shift in location towards right. Also the probabilities of low heat stress index do not show much difference for all the three time epochs and both the scenarios (left tails of the distribution). However the probability for higher values of heat stress is substantially more towards the end of the century (right tails). The PDFs look almost symmetric but they are positively skewed which clearly reveals that the heat stress index would be more severe (in danger and extreme danger category) in terms of duration and intensity towards the end of the century. Flatness of the curve implies that the kurtosis of the curves are less towards the end of the century i.e., variance or spreads of the distributions are more. Similar features can be observed over west coast region as well. The important point brought out by this analysis is substantial shift of location to the right and increase in spread of the distribution towards the higher values towards the end of the century under both the scenarios. It is clearly seen from the figures (Figs. 6 and 7) that probability of occurrence of danger and extreme danger days over the two coastal regions increase towards the end of the century because of the increase in the duration and intensity of warm days in future. This lengthening of heat stress period and increasing chance of occurrence may result in the occupational health hazards and productivity loses. A similar increase in the heat stress is being reported over different geographical regions of the globe and it has been documented that the trends in heat stress have been more detectable than those temperature71.
The increased scenarios of heat stress are mainly due to the increasing temperatures along with the changes in the humidity. The CMIP5 models show warming over India exceeding 5 °C during summer time and this increase in temperature is not consistent within the three time slices under the two climate scenarios (under RCP 4.5 and RCP8.5) (Supp Fig. 1). A slight decrement from the mean has been observed in relative humidity which accounts for more discomfort with the increase in temperature during the twenty-first century (Supp Fig. 2). The mean surface winds at 1000 hPa and sea level pressure (SLP) during summertime for future scenarios have been analyzed to understand the surface circulation during the elevated levels of heat stress. Future surface circulation pattern under RCP 8.5 scenario shows the strengthening of easterlies over Bay of Bengal and south-westerlies over Arabian Sea reaching Indian landmass due to the increased surface pressure gradients which may have possible connection with the increased temperatures in combination with the humid conditions that probes the higher heat stress over the coastal regions (Supp. Fig. 3). As the Bay of Bengal and the Arabian Sea are continued to warm in the recent decades, the higher SSTs cause the evaporative cooling mechanisms of the oceans which transports the winds with the subdued moisture as they reach the Indian land mass72. This subdued moist winds in combination with the already raised temperatures over land, together create the higher values of heat stress. The increase in the heat stress is also attributed to the increased anthropogenic forcing, increasing trends of temperature and decreasing trends in humidity during summer time70. Also, the air which is heated due to the absorbing aerosols in central and northeastern parts of India sinks in northwestern and southern Indian regions that cause more heating in the recent decades73. The CMIP5 MMM projections for the summertime furthermore suggest anomalous south easterlies from Bay of Bengal associated with high pressure may cause high temperatures over the east and west coasts of India. This might favor the enhancement in the severity of heat stress. In other words, the background atmospheric conditions are favorable for the enhanced severity of heat stress particularly by the end of the century.
Prominent features are seen in the upper troposphere like anti-cyclonic circulation over Pakistan region which is intensified towards the end of the century. The future minus present climatology suggests the positive height anomalies and anomalous anti- cyclonic flow at 500 hPa level over northern portions of India (Supp Fig. 4). Anticyclonic conditions over the study area also provide dry moist weather that increases the solar radiation flux due to subsidence of air masses. There is an indication that geopotential height (GPH) gradually strengthens during the future climate scenario. Future climate projections of amplified GPH anomalies at 500-hPa GPH indicate the significant increase of heat waves over India. Positive GPH anomalies at 500 hPa dynamically cause subsidence, clear skies, light winds, warm-air advection, and prolonged hot weather conditions at the surface4. Here we also observe a strong positive GPH in the upper troposphere (at 200-hpa) which might be also a causal factor for the surface heat wave and other heat extremes (supp. Figure 5). The possible link between circulation and extreme heat occurrence infers the role of horizontal temperature advection of warm dry air from the interior continental parts to coastal regions as off shore wind direction.
Climate change leaves the workers in lurch who work under extreme climates74. The heat stress is projected to intensify health risks of population and decrease the labour capacity, particularly over tropical and sub-tropical areas75,76. The decline in work performance over India is conspicuous during 2046–2065 and 2080–2099 than during 2016–2035 (Fig. 8). The cognitive fatigue, difficulties in mental concentration which reduces the work efficiency due to rise in temperature may result in the productivity losses77. In addition, the sweat evaporation stops at higher relative humidity values which hinder the cooling effect of the body leading to heat illness78. The climate models have shown the heat stress vulnerability in the regions of Southeast Asia, southeastern US and Northern Australia6. Dunne et al.75 proposed a fit for estimating labor capacity and reported a reduction in labor work capacity upto 40% under RCP 8.5 scenario over tropical and mid latitudinal region and also found that India will undergo heat stress more quantitatively than Eurasia and greater Caribbean region. The simulated levels of moist heat stress obtained from RCP 8.5 show the substantial rise in wet bulb temperature extremes over India, China, Southeast Asia, interior South America, and western Africa79. The urban heat island effect caused by the heat and moisture from the concrete structures that affects the incoming and outgoing solar radiation also has an impact on the heat stress in urban areas80. The impacts of climate change on working performance in India further worsen in view of RCP 8.5 scenario where India is likely to experience 3 to 4 degrees of warming81. The decrement in the work performance is highly susceptible in the southeastern coastal regions of India under the RCP 4.5 scenario while the same extends to total country except in the belt of Himalayas under the RCP 8.5 scenario. Our assessment showed a quantitative decline in work performance to up to 40% under the RCP8.5 and 35% under the RCP 4.5 scenario. The initial decline in work performance is conspicuous over the east coastal regions of India and later expanded to most of the regions in the country by end of the century. As India witnesses the notorious climate change impacts over the coastal regions, where 250 million people live within 50 km of the coast line82, the decline in work performance will be a major threat to the livelihood of the coastal communities’. However, a rigorous evaluation of work performance needs to be done based on individual and population-based characteristics. We provide here a common understanding in the projected decreasing levels of work performance as an overall assessment though the parameters such as local work place conditions83, construction of buildings84, clothing85 and physical work intensity86 have not been considered as input parameters. As suggested by Brode et al.44, Dunne et al.75 and Buzan and Huber79, estimation of the work performance needs to be contextualized and should elevate the role of changing the duty hours of labor, type of the labor. Our effort in the present study is to report the general features of declining work performance under future warm and humid conditions under global warming scenario. Also, the present study considers the mean temperature for estimation of heat stress that will provide a fair understanding of work performance. However, the diurnal cycle of the temperature may also play a key role under changing work hours of the labor. Finally, the practical implications of these elevated heat stress ‘need the support of engineering, social and policy-making decisions’79.
## Conclusions
Heat stress is expected to intensify, due to unprecedented warming during the twenty-first century. The increasing levels of heat stress pose a great challenge to mitigate and to adapt proper mechanisms to confront it. Decline in work performance associated with the detectable trends of heat stress over India point out the dire need to frame the sage work procedures, preventing the heat-based illness to reduce the health hazards as well as to maintain the work performance. The coastal regions of India (east and west coast) are found to be more vulnerable to heat stress impacts by showing a perceptible increase in the notorious impact days and a decline of 30 to 40% in the work performance, particularly in east coast region.
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85. 85.
Bernard, T. E. & Pourmoghan, M. Prediction of workplace wet bulb global temperature. Appl. Occup. Environ. Hyg. 14, 126–134 (1999).
86. 86.
Kjellstrom, T., Lemke, B. & Otto, M. Climate conditions, workplace heat and occupational health in South-East Asia in the context of climate change WHO South-East Asia. J. Public Health 6, 15–21 (2017).
## Acknowledgements
The First author, Dr. K. Koteswara Rao, is thankful to the Department of Science and Technology (DST)—Scientific and Engineering Research Board (SERB), Govt of India for sponsoring this work under the DST-NPDF fellowship scheme. We also thank to the Director, IITM for his support.
## Author information
Authors
### Contributions
K.K. conceived the idea, worked, analysed and written the draft paper. T.V.L.K. developed the idea and made the draft paper as a fair one. A.K., C.H., S.P., H.B. and S.S. helped in interpreting the results and correction of manuscript. M.B., D.S. and A.D. helped in analyzing the data.
### Corresponding author
Correspondence to T. V. Lakshmi Kumar.
## Ethics declarations
### Competing interests
The authors declare no competing interests.
### Publisher's note
Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.
## Rights and permissions
Reprints and Permissions
Koteswara Rao, K., Lakshmi Kumar, T.V., Kulkarni, A. et al. Projections of heat stress and associated work performance over India in response to global warming. Sci Rep 10, 16675 (2020). https://doi.org/10.1038/s41598-020-73245-3
• Accepted:
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2021-07-25 06:42:09
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https://www.wyzant.com/resources/answers/505911/write_an_equation_that_expresses_the_relationship_and_solve_for_y
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Tara R.
# write an equation that expresses the relationship and solve for y
c varies jointly as m and the sum of y and u
By:
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2020-02-23 00:37:47
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https://ccssmathanswers.com/eureka-math-algebra-2-module-3-lesson-20/
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# Eureka Math Algebra 2 Module 3 Lesson 20 Answer Key
## Engage NY Eureka Math Algebra 2 Module 3 Lesson 20 Answer Key
### Eureka Math Algebra 2 Module 3 Lesson 20 Opening Exercise Answer Key
Opening Exercise:
a. Sketch the graphs of the three functions f(x) = x2, g(x) = (2x)2 + 1, and h(x) = 4x2 + 1.
i. Describe the sequence of transformations that takes the graph of f(x) = x2 to the graph of g(x) = (2x)2 + 1.
The graph of g is a horizontal scaling by a factor of $$\frac{1}{2}$$ and a vertical translation up 1 unit of the graph of f.
ii. Describe the sequence of transformations that takes the graph of f(x) = x2 to the graph of h(x) = 4x2 + 1.
The graph of h is a vertical scaling by a factor of 4 and a vertical translation up 1 unit of the graph of f.
iii. Explain why g and h from parts (I) and (ii) are equivalent functions.
These functions are equivalent and have the same graph because the expressions (2x)2 + 1 and 4x2 + 1 are equivalent. The blue graph shown is the graph off, and the green graph is the graph of g and h.
b. Describe the sequence of transformations that takes the graph of f(x) = sin(x) to the graph of g(x) = sin(2x) – 3.
The graph of g is a horizontal scaling by a factor of $$\frac{1}{2}$$ and a vertical translation down 3 units of the graph of f.
c. Describe the sequence of transformations that takes the graph of f(x) = sin(x) to the graph of h(x) = 4 sin(x) – 3.
The graph of h is a vertical scaling by a factor of 4 and a vertical translation down 3 units of the graph of f.
d. Explain why g and h from parts (b) – (c) are not equivalent functions.
These functions are not equivalent because they do not have the same graphs, and the two expressions are not equivalent.
### Eureka Math Algebra 2 Module 3 Lesson 20 Exploratory Challenge Answer Key
Exploratory Challenge:
a. Sketch the graph of f(x) = log2(x) by identifying and plotting at least five key points. Use the table below to get started.
x Log2(x) $$\frac{1}{4}$$ -2 $$\frac{1}{2}$$ -1 1 2 4 8
The graph of f is shown below.
x Log2(x) $$\frac{1}{4}$$ -2 $$\frac{1}{2}$$ -1 1 0 2 1 4 2 8 3
b. Describe a sequence of transformations that takes the graph off to the graph of g(x) = log2(8x).
The graph of g is a horizontal scaling by a factor of $$\frac{1}{8}$$ of the graph of f.
c. Describe a sequence of transformations that takes the graph off to the graph of h(x) = 3 + log2(x).
The graph of h is a vertical translation up 3 units of the graph of f.
d. Complete the table below for f, g, and h and describe any noticeable patterns.
The functions g and h have the same range values at each domain value in the table.
e. Graph the three functions on the same coordinate axes and describe any noticeable patterns.
The graphs of g and h are identical.
f. Use a property of logarithms to show that g and h are equivalent.
By the product property and the definition of logarithm, log2(8x) = log2(8) + log2(x) = 3 + log2(x), so g(x) = h(x) for all positive real numbers x and g and h are equivalent functions.
g. Describe the graph of p(x) = log2($$\frac{x}{4}$$) as a vertical translation of the graph of f(x) = log2(x). Justify your response.
The graph of p isa vertical translation down 2 units of the graph off because log2($$\frac{x}{4}$$) = log2(x) – 2.
h. Describe the graph of h(x) = 3 + log2(x) as a horizontal scaling of the graph of f(x) = log2(x). Justify your response.
The graph of h is a horizontal scaling by a factor of $$\frac{1}{8}$$ of the graph of f because
3 + log2(x)= log2(8) + log2(x)= log2(8x).
i. Do the functions h(x) = log2(8) + log2(x) and k(x) = log2(x + 8) have the same graphs? Justify your reasoning.
No, they do not. By substituting 1 for x in both f and g, you can see that the graphs of the two functions do not have the same y-coordinate at this point. Therefore, the graphs cannot be the same if at least one point is different.
j. Use properties of exponents to explain why graphs of f(x) = 4x and g(x) = (22)x are identical.
Using the power property of exponents, 22x = (22)x = 4x Since the expressions are equal, the graphs of the functions would be the same.
k. Use the properties of exponents to predict what the graphs of g(x) = 4 . 2x and h(x) = 2x + 2 look like compared to one another. Describe the graphs of g and h as transformations of the graph of f(x) = 2x. Confirm your prediction by graphing f, g, and h on the same coordinate axes.
The graphs of the two functions g and h are the same since 2x + 2 = 2x . 22 = 4. 2x by the multiplication property of exponents and the commutative property. The graph of g is the graph of f(x) = 2x scaled vertically by a factor of 4. The graph of h is the graph of f(x) = 2x translated horizontally 2 units to the left.
l. Graph f(x) = 2x, g(x) = 2-x and h(x) = ($$\frac{1}{2}$$)x on the same coordinate axes. Describe the graphs of g and h as transformations of the graph off. Use the properties of exponents to explain why g and h are equivalent.
The graph of g and the graph of h are both reflections about the vertical axis of the graph of f. They are equivalent because (($$\frac{1}{2}$$))x = (2-1)-x = 2-x by the definition of a negative exponent and the power property of exponents.
### Eureka Math Algebra 2 Module 3 Lesson 20 Example Answer Key
Example 1: Graphing Transformations of the Logarithm Functions
The general form of a logarithm function is given by f(x) = k + a log(x – h), where a, b, k, and h are real numbers such that b is a positive number not equal to 1, and x – h > 0.
a. Given g(x) = 3 + 2 log(x – 2), describe the graph of g as a transformation of the common logarithm function.
The graph of g is a horizontal translation 2 units to the right, a vertical scaling by a factor of 2, and a vertical translation up 3 units of the graph of the common logarithm function.
b. Graph the common logarithm function and g on the same coordinate axes.
Example 2: Graphing Transformations of Exponential Functions
The general form of the exponential function is given by f(x) = a . bx + k, where a, b, and k are real numbers such that b is a positive number not equal to 1.
a. Use the properties of exponents to transform the function g(x) = 32x + 1 – 2 to the general form, and then graph it. What are the values of a, b, and k?
Using the properties of exponents, 32x + 1 – 2 = 32x . 31 – 2 = 3 . 9x – 2. Thus, g(x) = 3(9)x – 2 so a = 3, b = 9, and k = -2.
b. Describe the graph of g as a transformation of the graph of f(x) = 9x.
The graph of g is a vertical scaling by a factor of 3 and a vertical translation down 2 units of the graph of f.
c. Describe the graph of g as a transformation of the graph of f(x) = 3x.
The graph of g is a horizontal scaling by a factor of $$\frac{1}{2}$$, a vertical scaling by a factor of 3, and a vertical translation down 2 units of the graph of f.
d. Sketch the graph of g using transformations.
The graph of f(x) = 9x is shown in blue, and the graph of g is shown in green.
### Eureka Math Algebra 2 Module 3 Lesson 20 Exercise Answer Key
Exercises:
Graph each pair of functions by first graphing f and then graphing g by applying transformations of the graph of f. Describe the graph of g as a transformation of the graph of f.
Exercise 1.
f(x) = log3(x) and g(x) = 2 log3(x – 1)
The graph of g is the graph of f translated 1 unit to the right and stretched vertically by a factor of 2.
Exercise 2.
f(x) = log(x) and g(x) = log(100x)
Because of the product property of logarithms, g(x) = 2 + log(x). The graph of g is the graph of f translated vertically 2 units.
Exercise 3.
f(x) = log5x and g(x) = – log5(5(x + 2))
Since -log5(5(x + 2)) = -1 – log5(x + 2) by the product property of logarithms and the distributive property, the graph of gis the graph off translated 2 units to the left, reflected across the horizontal axis, and translated down 1 unit.
Exercise 4.
f(x) = 3x and g(x) = -2 . 3x – 1
Since -2 . 3x – 1 = -2 . 3x. 3x – 1 = –$$\frac{2}{3}$$ . 3x by the properties of exponents and the commutative property, the graph of g is the graph of f reflected across the horizontal axis and compressed by afactor of $$\frac{2}{3}$$.
There are multiple ways to obtain the graph of g from the graph of f through a sequence of transformations. One choice is to use the structure of g(x) = -2 . 3x – 1to translate the graph off horizontally one unit left, reflect across the horizontal axis, and then scale vertically by a factor of 2. Another choice is to rewrite g(x) as g(x) = $$\frac{2}{3}$$(3x). Then the graph of g is obtained from the graph of f by reflecting the graph of f across the horizontal axis and then vertically scaling by a factor of $$\frac{2}{3}$$.
### Eureka Math Algebra 2 Module 3 Lesson 20 Problem Set Answer Key
Question 1.
Describe each function as a transformation of the graph of a function in the form f(x) = log(x). Sketch the graph of f and the graph of g by hand. Label key features such as intercepts, intervals where g is increasing or decreasing, and the equation of the vertical asymptote.
a. g(x) = log2(x – 3)
The graph of g is the graph of f(x) = log2(x) translated horizontally 3 units to the right. The function g is increasing on (3, ∞). The x-intercept is 4, and the vertical asymptote is x = 3.
b. g(x) = log2(16x)
The graph of g is the graph of f(x) = log2(x) translated vertically up 4 units. The function g is increasing on
(0, ∞). The x-intercept is 2-4. The vertical asymptote is x = 0.
c. g(x) = log2($$\frac{8}{x}$$)
The graph of g is the graph of f(x) = log2(x) reflected about the horizontal axis and translated vertically up 3 units. The function g is decreasing on (0, ∞). The x-intercept is 23. The vertical asymptote is x = 0. The reflected graph and the final graph are both shown in the solution.
d. g(x) = log2((x – 3)2)for x > 3
The graph of g is the graph of f(x) = log2(x) stretched vertically by a factor of 2 and translated horizontally 3 units to the right. The function g is increasing on (3, ∞). The x-intercept is 4, and the vertical asymptote is x = 3.
Question 2.
Each function graphed below can be expressed as a transformation of the graph of f(x) = log(x). Write an algebraic function for g and h, and state the domain and range.
In Figure 1, g(x) = -log(x – 2) for x > 2. The domain of g is x > 2, and the range of g is all real numbers.
In Figure 2, h(x) = 2 + log(x) for x > 0. The domain of h is x > 0, and the range of h is all real numbers.
Figure 1: Graphs of f(x) = log(x) and the function g
Figure 2: Graphs of f(x) = log(x) and the function h
Question 3.
Describe each function as a transformation of the graph of a function in the form f(x) = bx. Sketch the graph off and the graph of g by hand. Label key features such as intercepts, intervals where g is increasing or decreasing, and the horizontal asymptote.
a. g(x) = 2 . 3x – 1
The graph of g is the graph of f(x) = 3x scaled vertically by a factor of 2 and translated vertically down 1 unit. The equation of the horizontal asymptote is y = -1. The y-intercept is 1, and the x-intercept is approximately -0.631. The function g is increasing for all real numbers.
b. g(x) = 22x + 3
The graph of g is the graph of f(x) = 4x translated vertically up 3 units. The equation of the horizontal asymptote is y = 3. The y-intercept is 4. There is no x-intercept. The function g is increasing for all real numbers.
c. g(x) = 3x – 2
The graph of g is the graph of f(x) = 3x translated horizontally 2 units to the right OR the graph of f scaled
vertically by a factor of $$\frac{1}{9}$$. The equation of the horizontal asymptote is y = 0. The y-intercept is $$\frac{1}{9}$$. There is no x-intercept, and the function g is increasing for all real numbers.
d. g(x) = -9$$\frac{x}{2}$$ + 1
The graph of g is the graph of f(x) = 3x reflected about the horizontal axis and then translated vertically up 1 unit. The equation of the horizontal asymptote is y = 1. The y-intercept is 0, and the x-intercept is also 0. The function g is decreasing for all real numbers.
Question 4.
Using the function f(x) = 2x, create a new function g whose graph is a series of transformations of the graph of f with the following characteristics:
→ The function g is decreasing for all real numbers.
→ The equation for the horizontal asymptote is y = 5.
→ The y-intercept is 7.
One possible solution is g(x) = 2 . 2-x + 5.
Question 5.
Using the function f(x) = 2x, create a new function g whose graph is a series of transformations of the graph of f with the following characteristics:
→ The function g is increasing for all real numbers.
→ The equation for the horizontal asymptote is y = 5.
→ The y-intercept is 4.
One possible solution is g(x) = -(2-x) + 5.
Question 6.
Consider the function g(x) = ($$\frac{1}{4}$$)x – 3
a. Write the function g as an exponential function with base 4. Describe the transformations that would take the graph of f(x) = 4x to the graph of g.
($$\frac{1}{4}$$)x – 3 = (4-1)x – 3 = 4-x + 3 = 43 . 4-x
Thus, g(x) = 64 . 4-x. The graph of g is the graph of f reflected about the vertical axis and scaled vertically
by a factor of 64.
b. Write the function g as an exponential function with base 2. Describe two different series of transformations that would take the graph of f(x) = 2’ to the graph of g.
($$\frac{1}{4}$$)x – 3 = (2)-2x – 3 = 2-2(x – 3) = 2-2x + 6 = 64 . 2-2x
Thus, g(x) = 64 . 2-2x or g(x) = 2-2(x – 3). To obtain the graph of g from the graph of f, you can scale the graph horizontally by a factor of $$\frac{1}{2}$$, reflect the graph about the vertical axis, and scale it vertically by a factor of 64. Or, you can scale the graph horizontally by a factor of $$\frac{1}{2}$$, reflect the graph about the vertical axis, and translate the resulting graph horizontally 3 units to the right.
Question 7.
Explore the graphs of functions in the form f(x) = log(xn) for n > 1. Explain how the graphs of these functions change as the values of n increase. Use a property of logarithms to support your reasoning.
The graphs appear to be a vertical scaling of the common logarithm function by a factor of n. This is true because of the property of logarithms that states log(xn) = n log(x).
Question 8.
Use a graphical approach to so’ve each equation. If the equation has no solution, explain why.
a. log(x) = log(x – 2)
This equation has no solution because the graphs of y = log(x) and y = log(x – 2) are horizontal translations of each other. Thus, their graphs do not intersect, and the corresponding equation has no solution.
b. log(x) = log(2x)
This equation has no solution because log(2x) = log(2) + log(x), which means that the graphs of y = log(x) andy = log(2x) are a vertical translation of each other. Thus, their graphs do not intersect, and the corresponding equation has no solution.
c. log(x) = log ($$\frac{2}{x}$$)
Ans:
The solution is the x-coordinate of the intersection point of the graphs of y = log(x) and y = log(2) – log(x). Since the graph of the function defined by the right side of the equation is a reflection across the horizontal axis and a vertical translation of the graph of the function defined by the left side of the equation, the graphs of these functions intersect in exactly one point with an x-coordinate approximately 1.5.
d. Show algebraically that the exact solution to the equation in part (c) is √2.
log(x) = log(2) – log(x)
2 log(x) = log(2)
log(x) = $$\frac{1}{2}$$log(2)
log(x) = log(2$$\frac{1}{2}$$)
x = 2$$\frac{1}{2}$$
Since 2$$\frac{1}{2}$$ = √2 the exact solution is √2.
Question 9.
Make a table of values forf(x) = $$x^{\frac{1}{\log (x)}}$$ for x > 1. Graph the function f for x > 1. Use properties of logarithms to explain what you see in the graph and the table of values.
We see that $$x^{\frac{1}{\log (x)}}$$ = 10 for all x > 1 because $$\frac{1}{\log (x)}=\frac{\log (10)}{\log (x)}$$ = logx(10). Therefore, when we substitute logx(10) into the expression $$x^{\frac{1}{\log (x)}}$$, we get xlogx(10), which is equal to 10 according to the definition of a logarithm.
### Eureka Math Algebra 2 Module 3 Lesson 20 Exit Ticket Answer Key
Question 1.
Express g(x) = -log4(2x) in the general form of a logarithmic function, f(x) = k + a logb(x – h). Identify a, b, h, and k.
Since -log4(2x) = -log4(2) + log4(x) = –$$\frac{1}{2}$$ – log4(x), the function is g(x) = –$$\frac{1}{2}$$ – log(x), and a = -1, b = 4, h = 0, and k = –$$\frac{1}{2}$$.
Question 2.
Use the structure of g when written in general form to describe the graph of g as a transformation of the graph of h(x) = log4(x).
The graph of g is the graph of h reflected about the horizontal axis and translated down $$\frac{1}{2}$$ unit.
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2021-09-26 15:38:34
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https://02522-cua.github.io/lecturenotes/descriptive-statistics-and-visualisations.html
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Chapter 3 Descriptive statistics and visualisations
3.1 Introduction
This week we will do a more thorough investigation of the HDB resale data. To make the data available within your Rmarkdown notebook, you would have to reload it. You can do this by executing the data loading code again.
sales <- read_csv(here::here("data/hdb_resale_2015_onwards.csv")) %>%
mutate(month = ymd(month, truncated = 1),
flat_type = as_factor(flat_type),
storey_range = as_factor(storey_range),
flat_model = as_factor(flat_model))
However, the data loading and cleaning steps often can be quite ‘expensive’: they take a long time to run. To cut down on this, in practice, these steps often get split into separate Rmd files. You can save an object to disk by running saveRDS() and load it again with readRDS(). So in this case, you could save the sales object at the end of your first notebook.
saveRDS(sales, here::here("data/sales.rds"))
sales <- readRDS(here::here("data/sales.rds"))
3.2 Central Tendency
In the previous week, we answered a few initial questions with/about the dataset (e.g. ‘what is the most expensive flat in Punggol?’). These questions were all descriptive and meant to give a sense of the distribution of different variables in our dataset. As we learn from Burt et al.’s Describing Data with Statistics chapter, we also have a suite of more quantitative measures available – often referred to as descriptive statistics. The chapter discusses two different types of such statistics: measures of the central tendency and measures of dispersion. After we understand what these statistics do conceptually, it is actually very straightforward to calculate them in R. The measures discussed in the chapter are:
• median
• mean
• mode
Mean (mean()) and median (median()) already exist in base R. Mode doesn’t but we can easily create our own function:
# from https://stackoverflow.com/a/25635740
manual_mode <- function(x, na.rm = FALSE) { # we don't use 'mode' as a function name because it already exists
if(na.rm){
x = x[!is.na(x)]
}
ux <- unique(x)
return(ux[which.max(tabulate(match(x, ux)))])
}
We can now use these three function to find out the distribution of each of our variables, for example:
mean(sales$floor_area_sqm) ## [1] 97.58903 median(sales$floor_area_sqm)
## [1] 96
manual_mode(sales$floor_area_sqm) ## [1] 67 What do these three measures tell you about the floor area variable? 3.3 Dispersion Similarly, R has a built-in set of function for dispersion statistics. Remember, while the measures of central tendency give you an indication of a typical value, the measures of dispersion give you an indication of the spread of that variable around the central tendency (most often, the mean). We can calculate these with R: # Range max(sales$floor_area_sqm) - min(sales$floor_area_sqm) # Interquartile Range IQR(sales$floor_area_sqm)
# Standard Deviation
sd(sales$floor_area_sqm) # Coefficient of variation sd(sales$floor_area_sqm) / mean(sales$floor_area_sqm) # Kurtosis and Skewness from the 'e1071 library kurtosis(sales$floor_area_sqm)
skewness(sales$floor_area_sqm) Together with the measures of central tendency, are you able to understand the distribution of the floor area variable based on these measures? Note that if you want to have a quick summary of a variable then running each of these descriptive statistics one by one isn’t so efficient. You can use R’s built in summary() function or the ‘tidy’ version skim() (from the skimr package) to get an immediate overview of many measures. summary(sales$floor_area_sqm)
## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 31.00 76.00 96.00 97.59 112.00 280.00
skim(sales$floor_area_sqm) Name sales$floor_area_sqm Number of rows 79100 Number of columns 1 _______________________ Column type frequency: numeric 1 ________________________ Group variables None
Variable type: numeric
skim_variable n_missing complete_rate mean sd p0 p25 p50 p75 p100 hist
data 0 1 97.59 24.22 31 76 96 112 280 ▃▇▁▁▁
3.4 Visualization
So far, we have explored the floor area variable with a range of different statistical measures. However, Anscombe’s Quartet tells us that we should not always blindly trust the summary statistics by themselves. Combining these statistics with a visual exploration is often advisable. Based on what you now know about the floor area of resale flats, can you visualize (in your head) the histogram of this variable? Let’s see if you were about right. Note that you can make use Healy’s Data Visualization Chapter 3 to refresh your memory on plotting with R/ggplot2. For example:
ggplot(sales, aes(x = floor_area_sqm)) +
geom_histogram(binwidth = 5)
If we overlay this with a normal distribution, the difference is clear:
ggplot(sales, aes(x = floor_area_sqm)) +
geom_histogram(aes(y = ..density..), binwidth = 5) +
stat_function(fun = dnorm, args = list(mean = mean(sales$floor_area_sqm), sd = sd(sales$floor_area_sqm)))
This isn’t all that surprising. Social data, especially data where governments and policy have a strong impact, very often do not follow normal distributions very strictly. In this case, there might very well be specific historical or policy reasons for that ‘weird’ peak on the left side of the graph. It coincides with our mode (calculated earlier). We can inspect what kind of flats have exactly this much floor area:
sales %>%
filter(floor_area_sqm == 67) %>%
View()
Can you think of specific contextual reasons why there might be so many flats with 66-70 sq meters of floor area?
There are a few other visualizations that are useful to explore distributions of a single variable. First there is the boxplot: a visual summary of the mean, IQR and outliers.
ggplot(sales, aes(x = 1, y = floor_area_sqm)) +
geom_boxplot()
With the boxplot, we ‘lose’ a bit of insight into the distribution within the IQR (after all, this is just a box). We can alleviate this by drawing violin plots instead.
ggplot(sales, aes(x = 1, y = floor_area_sqm)) +
geom_violin()
So far, we have only looked at the distribution of a variable for an entire dataset. In practice, these distributions look very differently for different subsets of a dataset. For example, in the case of HDB data, the floor area distribution will be very different for the different flat types. Visualizations are particularly useful to explore distributions within subgroups. In ggplot, we can do that with facets:
ggplot(sales, aes(x = floor_area_sqm)) +
geom_histogram(binwidth = 10) +
facet_wrap(vars(flat_type), scales = "free_y")
Violin plots and boxplots can be useful in this scenario as well:
ggplot(sales, aes(x = flat_type, y = floor_area_sqm)) +
geom_violin()
3.5 Assignment (Monday, February 17th, 23:59)
For your first assignment, you will conduct an extensive exploration of the distribution of all the variables in our dataset. You will have to integrate at least the following:
• Summarize the different continuous variables (area, price, remaining lease) as well as the nominal/ordinal variables (month, flat_type, town, flat_model, storey_range), summarize these variables in both table (stats on central tendency and distribution) and visual form.
• Analyze the distribution of (some of) these variables for different subsets of the data. For example, explore the difference between towns, or between flat types.
• Analyze the distribution of at least one variable for unique combinations of town and flat_type (for each town, for each flat type: Ang Mo Kio, 1 room; Ang Mo Kio 2 room; etc.)
• Analyze change in resale price per square meter over time. Use a 6-month moving average to do so.
Make sure you answer the points listed above but also provide a short introduction to the dataset (what is it? why is it interesting?). Take care not to just create a bunch of tables and visualizations but explain to the reader what they are seeing. Pick out interesting patterns etc. etc. After reading your report, the reader (and you!) should have a solid understanding of the distribution of the variables in the HDB dataset and, ideally, have a series of observations to explore more in-depth in subsequent analyses. To create nice looking tables, you can make use of the new gt packages.
3.5.1 Submitting the assignment
• Create your assignment as a single self-contained RMarkdown document (you may read in cleaned data at the beginning from your data folder with the here library). It should be saved in the assignments/ folder, as assignment1.Rmd.
• Change the output parameter to github_document so that your assignment has a visual representation on Github
• Do not forget to run styler on your code before submitting! (Check out the Session 2.2 slides if you are not sure how to do so).
• Once you are finished, navigate to your own Github repository at https://github.com/02522-cua/[your-name]. Create a new issue, title it ‘Assignment 1 by [your_name]’. In the body of the issue, include at least a link to the last commit you made as part of your assignment and ‘at’ your prof (include @barnabemonnot` in your text). To link to a specific commit, you can just include the SHA (a short code uniquely identifying a commit) in your issue. You can find the SHA codes in the history of your commits (e.g. at https://github.com/02522-cua/barnabe-monnot/commits/master my first commit has SHA ‘1f2c9c1’). Note that you can always edit the text of the issue if you want to update the assignment commit, but commits are timestamped so your prof can check whether they were submitted before the deadline!
Your assignments will be graded on a 5-point scale, according to this rubric:
1. The Rmarkdown does not run and there’s incomplete description and contextualization
2. The Rmarkdown does not run or runs but doesn’t produce the expected output. There is limited description and contextualization of code and its output
3. The Rmarkdown runs, the code is clearly structured but the description and contextualization of code and its output is limited
4. The Rmarkdown runs, the code is clearly structured and the description and contextualization is complete.
5. As above, plus you have included some new or creative approaches, methods or angles that were not covered in class.
Note that half or quarter-point grades might be given. The rubric above is only an indication for specific points on the grading scale.
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2020-08-13 15:07:04
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http://www.ma.utexas.edu/mediawiki/index.php?title=Conformally_invariant_operators&diff=1138&oldid=1137
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Conformally invariant operators
(Difference between revisions)
Revision as of 19:48, 23 September 2013 (view source)Tianling (Talk | contribs)← Older edit Revision as of 20:32, 23 September 2013 (view source)Tianling (Talk | contribs) Newer edit → Line 7: Line 7: Examples of conformally invariant operators include: Examples of conformally invariant operators include: - * The conformal Laplacian: $L_g=-\Delta_g + \frac{n-2}{4(n-1)}R_g$, where $n$ is the dimension of the manifold, $-\Delta_g$ is the Laplace–Beltrami operator of $g$, and $R_g$ is the scalar curvature of $g$. This is a second order differential operator. One can check that in this case, $a=\frac{n-2}{2}$ and $b=\frac{n+2}{2}$. + * The conformal Laplacian: - + $- * The Paneitz operator . + L_g=-\Delta_g + \frac{n-2}{4(n-1)}R_g, - +$ + where $n$ is the dimension of the manifold, $-\Delta_g$ is the Laplace–Beltrami operator of $g$, and $R_g$ is the scalar curvature of $g$. This is a second order differential operator. One can check that in this case, $a=\frac{n-2}{2}$ and $b=\frac{n+2}{2}$. + * The Paneitz operator : + $+ P=(-\Delta_g)^2-\mbox{div}_g (a_n R_g g+b_n Ric_g)d+\frac{n-4}{2}Q, +$ + where $\mbox{div}_g$ is the divergence operator, $d$ is the differential operator, $Ric_g$ is the Ricci tensor, + $+ Q=c_n|Ric_g|^2+d_nR_g^2-\frac{1}{2(n-2)}\Delta_gR +$ + and + $+ a_n=\frac{(n-2)^2+4}{2(n-1)(n-2)}, b_n=-\frac{4}{n-2}, c_n=-\frac{2}{(n-2)^2}, d_n=\frac{n^3-4n^2+16n-16}{8(n-1)^2(n-2)^2}. +$ + This is a fourth order operator with leading term $(-\Delta_g)^2$. + * GJMS operators : this is a family of conformally invariant differential operators with leading term $(-\Delta_g)^k$ for all integers $k$ is $n$ is odd, and for $k\in \{1,2,\cdots,\frac{n}{2}\}$ if $n$ is even. A nonexistence result can be found in for $k>\frac n2$ and $n\ge 4$ even. Line 18: Line 33: == References == == References == {{reflist|refs= {{reflist|refs= + + {{Citation | last1=Gover | first1= A | last2=Hirachi | first2= Kengo | title=Conformally invariant powers of the Laplacian—a complete nonexistence theorem | journal=Journal of the American Mathematical Society | volume=17 | pages=389--405}} + + {{Citation | last1=Graham | first1= C Robin | last2=Jenne | first2= Ralph | last3=Mason | first3= Lionel J | last4=Sparling | first4= George AJ | title=Conformally invariant powers of the Laplacian, I: Existence | journal=Journal of the London Mathematical Society | year=1992 | volume=2 | pages=557--565}} + {{Citation | last1=Paneitz | first1= S | title=A quartic conformally covariant differential operator for arbitrary pseudo-Riemannian manifolds |year=1983 | journal=preprint}} {{Citation | last1=Paneitz | first1= S | title=A quartic conformally covariant differential operator for arbitrary pseudo-Riemannian manifolds |year=1983 | journal=preprint}}
Revision as of 20:32, 23 September 2013
On a general compact manifold $M$ with metric $g$, a metrically defined operator $A$ is said to be conformally invariant if under the conformal change in the metric $g_w=e^{2w}g$, the pair of the corresponding operators $A_w$ and $A$ are related by $A_w(\varphi)=e^{-bw} A(e^{aw}\varphi)\quad\mbox{for all }\varphi \in C^{\infty}(M),$ where $a, b$ are constant.
Examples of conformally invariant operators include:
• The conformal Laplacian:
$L_g=-\Delta_g + \frac{n-2}{4(n-1)}R_g,$ where $n$ is the dimension of the manifold, $-\Delta_g$ is the Laplace–Beltrami operator of $g$, and $R_g$ is the scalar curvature of $g$. This is a second order differential operator. One can check that in this case, $a=\frac{n-2}{2}$ and $b=\frac{n+2}{2}$.
• The Paneitz operator [1] [2]:
$P=(-\Delta_g)^2-\mbox{div}_g (a_n R_g g+b_n Ric_g)d+\frac{n-4}{2}Q,$ where $\mbox{div}_g$ is the divergence operator, $d$ is the differential operator, $Ric_g$ is the Ricci tensor, $Q=c_n|Ric_g|^2+d_nR_g^2-\frac{1}{2(n-2)}\Delta_gR$ and $a_n=\frac{(n-2)^2+4}{2(n-1)(n-2)}, b_n=-\frac{4}{n-2}, c_n=-\frac{2}{(n-2)^2}, d_n=\frac{n^3-4n^2+16n-16}{8(n-1)^2(n-2)^2}.$ This is a fourth order operator with leading term $(-\Delta_g)^2$.
• GJMS operators [3]: this is a family of conformally invariant differential operators with leading term $(-\Delta_g)^k$ for all integers $k$ is $n$ is odd, and for $k\in \{1,2,\cdots,\frac{n}{2}\}$ if $n$ is even. A nonexistence result can be found in [4] for $k>\frac n2$ and $n\ge 4$ even.
References
1. Paneitz, S (1983), "A quartic conformally covariant differential operator for arbitrary pseudo-Riemannian manifolds", preprint
2. Paneitz, S (2008), "A quartic conformally covariant differential operator for arbitrary pseudo-Riemannian manifolds (summary)", SIGMA Symmetry Integrability Geom. Methods Appl. (4)
3. Graham, C Robin; Jenne, Ralph; Mason, Lionel J; Sparling, George AJ (1992), "Conformally invariant powers of the Laplacian, I: Existence", Journal of the London Mathematical Society 2: 557--565
4. Gover, A; Hirachi, Kengo, "Conformally invariant powers of the Laplacian—a complete nonexistence theorem", Journal of the American Mathematical Society 17: 389--405
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2018-07-23 01:49:18
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http://slideplayer.com/slide/7041817/
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# 5.3 Definite Integrals and Antiderivatives Greg Kelly, Hanford High School, Richland, Washington.
## Presentation on theme: "5.3 Definite Integrals and Antiderivatives Greg Kelly, Hanford High School, Richland, Washington."— Presentation transcript:
5.3 Definite Integrals and Antiderivatives Greg Kelly, Hanford High School, Richland, Washington
Page 269 gives rules for working with integrals, the most important of which are: 2. If the upper and lower limits are equal, then the integral is zero. 1. Reversing the limits changes the sign. 3. Constant multiples can be moved outside.
1. If the upper and lower limits are equal, then the integral is zero. 2. Reversing the limits changes the sign. 3. Constant multiples can be moved outside. 4. Integrals can be added and subtracted.
4. Integrals can be added and subtracted. 5. Intervals can be added (or subtracted.)
The average value of a function is the value that would give the same area if the function was a constant:
The mean value theorem for definite integrals says that for a continuous function, at some point on the interval the actual value will equal the average value. Mean Value Theorem (for definite integrals) If f is continuous on then at some point c in,
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2021-04-16 13:43:55
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https://socratic.org/questions/how-do-you-solve-5x-2-40x-80-0
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# How do you solve 5x^2-40x+80=0?
May 12, 2016
I think the answer is -5.3.
#### Explanation:
$5 {x}^{2} - 40 x + 80 = 0$
$5 x \cdot 5 x = 25 x$ Take care of the exponent first.
$25 x - 40 x + 80 = 0$ Solve the like terms.
$- 15 x + 80 = 0$ Carry 80 over.
$- 15 \frac{x}{-} 15 = \frac{80}{-} 15$ Cancel out -15 and divide 80 by that same number.
$x = - 5.3$ Plug in the answer you get and round it to the nearest ten.
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2019-08-23 22:41:35
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https://en.m.wikipedia.org/wiki/Icosagon
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# Icosagon
In geometry, an icosagon or 20-gon is a twenty-sided polygon. The sum of any icosagon's interior angles is 3240 degrees.
Regular icosagon
A regular icosagon
Type Regular polygon
Edges and vertices 20
Schläfli symbol {20}, t{10}, tt{5}
Coxeter diagram
Symmetry group Dihedral (D20), order 2×20
Internal angle (degrees) 162°
Dual polygon Self
Properties Convex, cyclic, equilateral, isogonal, isotoxal
## Regular icosagonEdit
The regular icosagon has Schläfli symbol {20}, and can also be constructed as a truncated decagon, t{10}, or a twice-truncated pentagon, tt{5}.
One interior angle in a regular icosagon is 162°, meaning that one exterior angle would be 18°.
The area of a regular icosagon with edge length t is
${\displaystyle A={5}t^{2}(1+{\sqrt {5}}+{\sqrt {5+2{\sqrt {5}}}})\simeq 31.5687t^{2}.}$
In terms of the radius R of its circumcircle, the area is
${\displaystyle A={\frac {5R^{2}}{2}}({\sqrt {5}}-1);}$
since the area of the circle is ${\displaystyle \pi R^{2},}$ the regular icosagon fills approximately 98.36% of its circumcircle.
## UsesEdit
The Big Wheel on the popular US game show The Price Is Right has an icosagonal cross-section.
The Globe, the outdoor theater used by William Shakespeare's acting company, was discovered to have been built on an icosagonal foundation when a partial excavation was done in 1989.[1]
As a golygonal path, the swastika is considered to be an irregular icosagon.[2]
A regular square, pentagon, and icosagon can completely fill a plane vertex.
### ConstructionEdit
As 20 = 22 × 5, regular icosagon is constructible using a compass and straightedge, or by an edge-bisection of a regular decagon, or a twice-bisected regular pentagon:
Construction of a regular icosagon Construction of a regular decagon
## The golden ratio in icosagonEdit
• In the construction with given side length the circular arc around C with radius CD, shares the segment E20F in ratio of the golden ratio.
${\displaystyle {\frac {\overline {E_{20}E_{1}}}{\overline {E_{1}F}}}={\frac {\overline {E_{20}F}}{\overline {E_{20}E_{1}}}}={\frac {1+{\sqrt {5}}}{2}}=\varphi \approx 1.618}$
Icosagon with given side length, animation (The construction is very similar to that of decagon with given side length)
## SymmetryEdit
Symmetries of a regular icosagon. Vertices are colored by their symmetry positions. Blue mirrors are drawn through vertices, and purple mirrors are drawn through edge. Gyration orders are given in the center.
The regular icosagon has Dih20 symmetry, order 40. There are 5 subgroup dihedral symmetries: (Dih10, Dih5), and (Dih4, Dih2, and Dih1), and 6 cyclic group symmetries: (Z20, Z10, Z5), and (Z4, Z2, Z1).
These 10 symmetries can be seen in 16 distinct symmetries on the icosagon, a larger number because the lines of reflections can either pass through vertices or edges. John Conway labels these by a letter and group order.[3] Full symmetry of the regular form is r40 and no symmetry is labeled a1. The dihedral symmetries are divided depending on whether they pass through vertices (d for diagonal) or edges (p for perpendiculars), and i when reflection lines path through both edges and vertices. Cyclic symmetries in the middle column are labeled as g for their central gyration orders.
Each subgroup symmetry allows one or more degrees of freedom for irregular forms. Only the g20 subgroup has no degrees of freedom but can seen as directed edges.
The highest symmetry irregular icosagons are d20, a isogonal icosagon constructed by ten mirrors which can alternate long and short edges, and p20, an isotoxal icosagon, constructed with equal edge lengths, but vertices alternating two different internal angles. These two forms are duals of each other and have half the symmetry order of the regular icosagon.
## DissectionEdit
20-gon with 180 rhombs
Coxeter states that every zonogon (a 2m-gon whose opposite sides are parallel and of equal length) can be dissected into m(m-1)/2 parallelograms.[4] In particular this is true for regular polygons with evenly many sides, in which case the parallelograms are all rhombi. For the icosagon, m=10, and it can be divided into 45: 5 squares and 4 sets of 10 rhombs. This decomposition is based on a Petrie polygon projection of a 10-cube, with 45 of 11520 faces. The list A006245 enumerates the number of solutions as 18,410,581,880, including up to 20-fold rotations and chiral forms in reflection.
10-cube
## Related polygonsEdit
An icosagram is a 20-sided star polygon, represented by symbol {20/n}. There are three regular forms given by Schläfli symbols: {20/3}, {20/7}, and {20/9}. There are also five regular star figures (compounds) using the same vertex arrangement: 2{10}, 4{5}, 5{4}, 2{10/3}, 4{5/2}, and 10{2}.
n 1 2 3 4 5
Form Convex polygon Compound Star polygon Compound
Image
{20/1} = {20}
{20/2} = 2{10}
{20/3}
{20/4} = 4{5}
{20/5} = 5{4}
Interior angle 162° 144° 126° 108° 90°
n 6 7 8 9 10
Form Compound Star polygon Compound Star polygon Compound
Image
{20/6} = 2{10/3}
{20/7}
{20/8} = 4{5/2}
{20/9}
{20/10} = 10{2}
Interior angle 72° 54° 36° 18°
Deeper truncations of the regular decagon and decagram can produce isogonal (vertex-transitive) intermediate icosagram forms with equally spaced vertices and two edge lengths.[5]
A regular icosagram, {20/9}, can be seen as a quasitruncated decagon, t{10/9}={20/9}. Similarly a decagram, {10/3} has a quasitruncation t{10/7}={20/7}, and finally a simple truncation of a decagram gives t{10/3}={20/3}.
Quasiregular Quasiregular t{10}={20} t{10/9}={20/9} t{10/3}={20/3} t{10/7}={20/7}
## Petrie polygonsEdit
The regular icosagon is the Petrie polygon for a number of higher-dimensional polytopes, shown in orthogonal projections in Coxeter planes:
A19 B10 D11 E8 H4 2H2
19-simplex
10-orthoplex
10-cube
11-demicube
(421)
600-cell
10-10 duopyramid
10-10 duoprism
It is also the Petrie polygon for the icosahedral 120-cell, small stellated 120-cell, great icosahedral 120-cell, and great grand 120-cell.
## ReferencesEdit
1. ^ Muriel Pritchett, University of Georgia "To Span the Globe", see also Editor's Note, retrieved on 10th January 2016
2. ^
3. ^ John H. Conway, Heidi Burgiel, Chaim Goodman-Strauss, (2008) The Symmetries of Things, ISBN 978-1-56881-220-5 (Chapter 20, Generalized Schaefli symbols, Types of symmetry of a polygon pp. 275-278)
4. ^ Coxeter, Mathematical recreations and Essays, Thirteenth edition, p.141
5. ^ The Lighter Side of Mathematics: Proceedings of the Eugène Strens Memorial Conference on Recreational Mathematics and its History, (1994), Metamorphoses of polygons, Branko Grünbaum
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2018-03-18 00:15:10
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https://dirkmittler.homeip.net/blog/archives/9625
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# ChromeOS Upgrade from Debian 9 to Debian 10 – aka Buster – Google Script crashed.
I have one of those Chromebooks, which allow a Linux subsystem to be installed, that subsystem being referred to in the Google world as “Crostini”. It takes the form of a Virtual Machine, which mounts a Container. That container provides the logical hard drive of the VM’s Guest System. What Google had done at some point in the past was, to install Debian 9 / Stretch as the Linux version, in a simplified, automated way. But, because Debian Stretch is being replaced by Debian 10 / Buster, the option also exists, to upgrade the Linux Guest System to Buster. Only, while the option to do so manually was always available to knowledgeable users, with the recent Update of ChromeOS, Google insists that the user perform the upgrade, and provides ‘an easy script’ to do so. The user is prompted to click on something in his ChromeOS settings panel.
What happened to me, and what may also happen to my readers is, that this script crashes, and leaves the user with a ChromeOS window, that has a big red symbol displayed, to indicate that the upgrade failed. I failed to take a screen-shot of what this looks like. The button to offer the upgrade again, is thankfully taken away at that point. But, if he or she reaches that point, the user will need to decide what to do next, out of essentially two options:
• Delete the Linux Container, and set up a new one from scratch. In that case, everything that was installed to, or stored within Linux will be lost. Or,
• Try to complete the upgrade in spite of the failed script.
I chose to do the latter. The Linux O/S has its own method of performing such an upgrade. I would estimate that the reason for which the script crashed on me, might have been Google’s Expectation that my Linux Guest System might have 200-300 packages installed, when in fact I have a much more complete Linux system, with over 1000 packages installed, including services and other packages that ask for configuration options. At some point, the Google Script hangs, because the Linux O/S is asking an unexpected question. Also, while the easy button has a check-mark checked by default, to back up the Guest System’s files before performing the upgrade, I intentionally unchecked that, simply over the knowledge that I do not have sufficient storage on the Chromebook, to back up the Guest System.
I proceeded on the assumption, that what the Google script did first was, to change the contents of the file ‘/etc/apt/sources.list’, as well as of the directory ‘/etc/apt/sources.list.d’, to specify the new software sources, associated with Debian Buster as opposed to Debian Stretch. At that point, the Google script should also have set up, whatever it is that makes Crostini different from stock Linux. Only, once in the middle of the upgrade that follows, the Google script hanged.
(Updated 10/25/2020, 22h55… )
(As of 10/25/2020, 13h40: )
On those assumptions, all I really need to do was, open a Linux terminal in case one was not already open, and type in the following commands:
\$ sudo su
In some cases, ‘apt-get‘ can be replaced with ‘apt‘ – especially, after the upgrade to Debian Buster – and, I did not need to give the command ‘apt-get update‘ because presumably, the failed Google script had already given that command.
Aside from answering a few prompts, which the Google script had not expected, I really didn’t have to do anything else. However, the whole process took me 5 hours, and not the 30 minutes that the Google window had suggested it would take. On larger Linux installs, doing a dist-upgrade can take 6-12 hours. A so-called dist-upgrade must not be interrupted at any point in the process.
After the upgrade, I found that all the applications I tested, work as before, including the (updated) ‘TigerVNC server’, that actually allows me to create an LXDE desktop, which I can then access via a ChromeOS-provided VNC Viewer. However, there is one more detail which I should mention:
When I set up Linux systems, I often install packages in a careful way, that ‘pull in’ libraries belonging to unused desktop managers, such as ‘GNOME’, either on my Plasma 5 -based computer, or on an LXDE -based computer. I’m careful not actually to install GNOME.
Well, during a dist-upgrade, this habit of mine can bite me in the ass. A dist-upgrade can and will perform a full install of GNOME in that case. What this does is, maximize the amount of storage the container uses, with the danger being, that at some point the amount of storage available on the ChromeOS Host System, might no longer accommodate the grown container. This creates a ‘hump’ which I had to wait through, before reversing the problem. After I had cleared this hump, I performed the following commands:
# apt clean
# sync
# apt autoremove
Amazingly, the ‘autoremove’ command removed the unnecessary packages again, so that again, ‘GNOME’ is not installed.
If one ignores the amount of time this took, the process was as reliable and easy, as if I had just done a smaller update. However, as Linux systems become larger and more complex, a dist-upgrade may not work as easily for the reader. Yet, as long as like me, the reader only has 1000 packages or so in his Linux subsystem, there is a good chance that like me, he or she can simply rescue a failed dist-upgrade in this way.
The hardest part of a dist-upgrade is usually, to get all the repositories right, in the file ‘/etc/apt/sources.list’… The second hardest part of a dist-upgrade is usually, whatever customization the user did to his system prior to the dist-upgrade, and then, trying to make sure that customized features work afterwards. As ingenious as the Linux dist-upgrade process is, it cannot take into account, what the user may have done, outside the route of the package manager. Out-of-tree installs are most likely to break.
I did observe that the Google script had added sources first, that offer Crostini packages, where stock Linux would not have those.
Before the upgrade, my Linux container took up 9.2GB, while afterwards, it was taking up 11.3GB. I am assuming that ChromeOS can shrink this container, in addition to being able to grow it.
There is another fact which readers should be aware of, that can cause the update to seem to have failed, when in fact it may not have.
The way Linux is generally organized is, into a system directory tree and a user home directory tree. Software from the package manager is generally installed to the system directory tree, not the user home directory tree, the latter of which is usually left alone. During an upgrade, several applications are upgraded to newer versions, as it should be. But, the applications still store their settings per-user, in the user’s home directory tree.
This does not change if the Linux system is running as a Guest System.
In some cases, applications with higher version numbers become incompatible with the settings that the previous version saved per-user. If a single application no longer seems to work, then one thing to do would be, either to delete or rename its per-user settings file – which could also be a sub-directory – and then to relaunch that application, with the understanding that it behaves from then on, as through ‘a first run’ was being carried out.
I needed to do this with ‘Bluefish’ specifically, before it would connect directly to my FTPS Server again.
I now have LibreOffice 6 where I previously had LibreOffice 5, and v6 works just fine, after translating its per-user settings automatically, to the version required.
(Update 10/25/2020, 22h55: )
A general note on using Bluefish to connect directly to an FTPS Server in this way:
The preferred way to do this would be such, that a coder does not need to retype the URL each time. But, in some cases, the only way I was able to connect to the server was, to open the ‘Seahorse’ application, unlock the FTPS password from there, close that application, open Bluefish, type in the URL once, acknowledge that an unverifyable certificate was being used by my (self-signed) server, close Bluefish again, open Bluefish a second time, and then, click on the URL in the History Pane.
A way to improve ‘connecting to FTPS URLs quickly’ is, once the FTPS Server’s contents are being displayed, pick a file at random, and create a bookmark, which is then a bookmark within one file, within an FTPS login. After that, even in a new session, after the FTPS Server password has been unlocked in Seahorse, it became possible for me just to open Bluefish once, navigate the Pane to its Bookmarks, click on the one bookmark, acknowledge the self-signed certificate, and I am logged on to the server, able to browse its files, and able to close that one file as well, without losing my login.
This just seems to be some weakness Bluefish has, in remembering FTPS URLs, and it has the weakness across Debian 9 and Debian 10 computers.
Dirk
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2021-07-27 10:39:24
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http://www.gamedev.net/topic/639078-height-of-a-3d-triangle/
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• Create Account
Height of a 3D triangle
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2 replies to this topic
#1mcmonkey Members - Reputation: 113
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Posted 19 February 2013 - 10:42 PM
Let's say I have two triangles in a 3D space
The triangles are, of course, made from 3 coordinates (each coordinate an X/Y/Z set)
First Triangle:
X1 = 0, Y1 = 0, Z1 = Dynamic
X2 = 1, Y2 = 0, Z2 = Dynamic
X3 = 1, Y3 = 1, Z3 = Dynamic
Second Triangle:
X1 = 0, Y1 = 0, Z1 = Dynamic
X2 = 1, Y2 = 1, Z2 = Dynamic
X3 = 1, Y3 = 0, Z3 = Dynamic
... So, basically, a square that's cut through the middle (and bends on that cut)
Each Z value for each is known, but varies from instance to instance. And they vary independently.
If I have a position (Say, X=0.13,Y=0.73)
How would I calculcate what the Z of the triangle will be at that position?
(I'm hoping for an equation or pseudocode that will calculate the value without using raytracing or any expensive functions like that)
Edited by mcmonkey, 20 February 2013 - 06:07 PM.
#2HappyCoder Members - Reputation: 4683
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Posted 19 February 2013 - 11:37 PM
If the two triangles always lie on the same plane then you can calculate the plane the triangles lay on.
The equation for a plane is
Ax + By + Cz + D = 0;
<A, B, C> is the normal of the plane and D is the closest distance of the plane to the origin divided by the length of the normal. To calculate a plane from a triangle.
// the A, B, and C of the plane
normal = cross(P2 - P1, P3 - P1)
d = -dot(P1, normal) // you can dot the normal with any of the points in the triangle, it will yield the same d
where cross() is the cross product, dot() is the product, and Pn is a point on the triangle
Then use the equation of the plane to solve for x, y, or z when you have the other two. For example, to solve for z
Ax + By + Cz + D = 0
Cz = -Ax - By - D
z = -(Ax + By + D) / C
This will find the point on the same plane the point can lie on the outside of the triangles, so you will have to make sure you point lies inside the triangle. If you need help with that I can post a method for that, I just don't want to crowd up this post if you already have a way to do that.
This actually is the basics of how a ray tracer works. Casting an individual ray is actually quite fast. Ray tracing entire images however is costly because there are many rays being cast per pixel. Some rays are used to calculate shadows, others reflections, and other such things and the scenes are usually fairly complex. You should be able to cast dozens of rays in a simpler scene and still maintain a responsive frame rate. So unless you find your game isn't running fast enough and you have identified your "ray casting" as the bottleneck, I wouldn't spend much time making sure it is fast.
Edited by HappyCoder, 19 February 2013 - 11:39 PM.
My current game project Platform RPG
#3deekr Members - Reputation: 191
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Posted 19 February 2013 - 11:39 PM
Two edges of a given triangle span the plane containing that triangle. So you could try looking at what you want to calculate as a linear combination of these basis vectors and solve it that way. Because of the particular X, Y components of your triangle vertices, this might actually be cheap: b(1,1) - b(1,0) + a(1,0) = (a,b).
Another way is to compute the plane equation: Ax + By + Cz = D and then solve for z.
The (X, Y) you gave is in either of the spanned planes, but not on either triangle, since it has a large Y component. I'm not sure if that was intentional.
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PARTNERS
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2016-09-27 12:38:20
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https://www.nature.com/articles/s41598-018-37619-y?error=cookies_not_supported&code=ea00eb8e-fa9e-4b96-85ce-4a81a43c1910
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# Repetitive sex change in the stony coral Herpolitha limax across a wide geographic range
## Abstract
Sex change has been widely studied in animals and plants. However, the conditions favoring sex change, its mode and timing remain poorly known. Here, for the first time in stony corals, we report on a protandrous (youngest individuals are males) repetitive sex change exhibited by the fungiid coral Herpolitha limax across large spatial scales (the coral reefs of Japan, Jordan and Israel) and temporal scales (2004–2017). In contrast to most corals, this species is a daytime spawner (08:00–10:00 AM) that spawned at the same time/same date across all the study sites. The sporadically scattered populations of H. limax among the coral reefs of Eilat (Israel) and Aqaba (Jordan) exhibited significantly slower growth, earlier sex change, and lower percentages of reproduction and sex change in comparison to the densely aggregated populations in Okinawa (Japan). At all sites, sex ratio varied among years, but was almost always biased towards maleness. Growth rate decreased with size. We conclude that comparable to dioecious plants that display labile sexuality in response to energetic and/or environmental constraints, the repetitive sex change displayed by H. limax increases its overall fitness reinforcing the important role of reproductive plasticity in the Phylum Cnidaria in determining their evolutionary success.
## Introduction
Sexual reproduction is a fundamental aspect of life that manifests in the diverse sexual systems occurring in nature. Despite its high energetic expenditure, the evolutionary advantages of sexual reproduction outweigh its cost by increasing progeny fitness and genetic diversity1. However, because resources are finite, the possible costs of reproduction may reduce the survival probability or growth rate of the individual. As such, the rules governing reproductive allocation are viewed in a benefit–cost framework2. This fundamental trade-off has led individuals to adopt specific tactics to attain reproduction. According to sex-allocation theory3, natural selection favors parents that modify their investment in male and female functions in such a way that it maximizes the parent’s fitness3,4. Among the broad variety of sexual systems, sequential hermaphroditism (sex change) represents the most extreme form of sex-allocation, through altering sexual function expression (reviewed by5). In most cases of sequential hermaphroditism, sex change appears to occur only once in a lifetime3,6, either from female to male (protogyny) or from male to female (protandry). However, a minority of species have proven to have the ability to change sex bi-directionally (i.e. the occurrence of sex change from female to male and from male to female in the same population), or even repetitively (reviewed by7). This phenomenon described both in animals6,7,8,9,10,11,12,13,14,15 and plants6,16,17,18,19 presumably conveys a selective advantage to an individual by increasing its reproductive potential.
Stony corals (scleractinians), the frame-builders of coral reefs, are sessile organisms that possess ecological features typical of both plants and animals13. The sexual reproduction in scleractinian corals involves two main sexual systems: colonies are either predominately out-crossing, simultaneous hermaphrodites, with each polyp having both male and female functions; or polyps within a colony express only one sex throughout their life and such colonies are thus either male or female and the species are gonochoric (dioecious) (reviewed by20). Nonetheless, atypical sexual systems of mixed or contrasting patterns of sexuality have been reported from various coral species21,22 including that of corals exhibiting sex change. Protandrous sequential hermaphroditism was documented in four solitary fungiid species (Ctenactis echinata, C. crassa, Fungia repanda and F. scruposa)13, two of which (C. echinata and C. crassa) changed sex bi-directionally13,23.
The most influential model of sex change is the size-advantage hypothesis model (SAH), which predicts sex change when the reproductive success of one sex increases more rapidly with size than the reproductive success of the opposite sex3,7,10,24,25,26. The direction of sex change (protandrous or protogynous) is determined by the relative reproductive success over the course of a lifetime for the two sexes. If male fitness increases with size (age) at a higher rate than female fitness, sex change will be protogynous. The reverse holds true for protandrous sex change5,13,24,25,27,28. However, large size does not always confer a greater reproductive advantage upon one sex than the other29 and, therefore, additional explanations are required for the existence of sex change. Alternative models, developed primarily as models based on environmental sex determination in plants16, have suggested that sex change can be triggered by epigenetic cues (environmental or social information) (reviewed by5,7). Freeman et al.16, and Korpelainen18 have demonstrated that environmental factors are strongly correlated with the sexual expression of the individual. For example, harsh environmental conditions were demonstrated to induce a shift from female to male function in the shrub Atriplex canescens as an adaptive strategy to avoid the high cost of female reproduction16,30. Similarly, age, injury, and disease have been shown to alter sexual expression towards maleness13,16. The magnitude and direction of sex change may be influenced by population demography31. Evidence of this is more common in mobile animals that exhibit social control, and the phenomenon has been widely explored in fish, the largest vertebrate group in which this phenomenon occurs7,32,33. However, in sessile or very low-mobility animals that lack any obvious social interactions, such as scleractinian corals, the cues regulating sex change may be different.
The similarity in some life-functions and life-history traits between scleractinian corals and certain plants13, enabled us to study the labile nature of sex expression from a different perspective and to examine a variety of theoretical and empirical studies of sex change on these sessile organisms. In this wide-scale study, we examined the reproductive behavior of individual corals representing a broad array of size groups over an extended period of time (13 years), in three distant geographical areas (the coral reefs of Eilat, Israel, Aqaba, Jordan and Okinawa, Japan), exhibiting divergent environmental conditions. The detailed data obtained in this study during 2004–2017 allowed us to examine both the long-term reproductive patterns of the fungiid coral Herpolitha limax, and the labile sexual expression it demonstrates in light of the local prevailing environmental conditions.
## Results
### Reproduction of Herpolitha limax
H. limax belongs to the family Fungiidae, which consists of solitary, largely single-polyped, “free-living” species (i.e. not attached to the substrate). Out of the 28 fungiid species whose sexuality were thought to be of the gonochoric type20, four were found to change sex, and among these, two species change sex bi-directionally13,23. In the Gulf of Eilat/Aqaba (GoE/A), in both the Eilat and Aqaba reefs, the species appear in a dispersed manner and occur mostly at a depth range of 5–50 m. The abundance of H. limax, however, is higher in Aqaba than in Eilat (L.E-S., YL, pers. obs). In contrast, in the coral reefs of Okinawa, the H. limax population is mainly concentrated at a few sites inhabited by thousands of individuals, which are densely distributed between 5–20 m depth.
Observations on H. limax began in July 2004, in Okinawa, Japan and a few years later in Eilat and Aqaba, within the framework of a wide-scale study of reproduction patterns in the Fungiidae family. However, detailed data on the mode and timing of reproduction of H. limax populations, based on large sample sizes, commenced only in 2012 in Eilat and 2013 and 2014 in Aqaba and Okinawa, respectively (see Supplementary Material for further details).
In contrast to other fungiids, as well as the majority of scleractinian corals, which are typically night spawners, H. limax is a daytime spawner (Fig. 1a) that releases gametes of separate sexes between 8 am to 10 am for 5–7 consecutive days, starting five days after the full moon, in June-August or in July-September (apparently depending on the lunar periodicity of the given year). This pattern repeated itself at all three study sites in consecutive years, with the corals spawning at precisely the same months, lunar age and time. The female gametes are small (~50 µm), viewed as tiny dots floating in the water volume, while the male gametes appear as a milky cloud, causing the water to appear murky (Fig. 1b,c). Overall, the reproduction of H. limax was highly synchronized at all study sites, both locally and globally (timing and date of reproduction). Most of the reproductive individuals spawned for 2–4 days within the expected reproduction period of each month. Some individuals spawned across two consecutive months, while others spawned for one month only. The vast majority of corals released gametes of only one sex each season. However, at the Aqaba and Okinawa study sites some corals were found to release both sex gametes during one breeding season. We termed this reproductive state ‘transitional’ (T) (see Table 1). Release of the different gametes was observed either at a one month interval; or, more rarely, during the same reproduction month, at a one-day interval. Throughout the study years, only one individual (out of 79 corals examined in 2017) was observed to release both gametes simultaneously. No similar ‘transitional state’ was observed in Eilat. The percentages of reproduction were higher in Okinawa than in Eilat and Aqaba (Table 1). Of the Eilat breeding population (i.e. corals for which reproductive data for more than two years were available), 19% had changed sex in either directions throughout the years (either from male to female or from female to male). In Okinawa, 66% of the breeding population had changed sex in both directions, and in Aqaba 11% of the breeding population exhibited sex change in both directions. Repetitive sex change was documented in Eilat (10%) and Okinawa (29%) but was seemingly absent in Aqaba (most probably due to the lack of repeated observations) (Fig. 2; Table 2; Supplementary Material Figs S1, S2). In winter 2014–2015, an extremely destructive flood impacted the north-eastern part of the GoE/A, causing heavy mortality to corals in Aqaba that were located on the shallow reef near the local Marine Science Station. Unfortunately, 108 individuals among the tagged corals died in the flood, whose consequences prevented us from fully examining the continuity of reproduction. The sampled population of 2015 therefore comprised 89 new individual corals and only 28 survivors that had been recorded in previous years.
### Size frequency distribution and sex ratio
The size (weight) of individuals at first reproduction was similar at all sites and varied from 32 g in Eilat to 41 g in Okinawa (Table 2). However, In Eilat and Aqaba females were lighter in weight (weighting 169 g and 133 g, respectively) than in Okinawa, where the smallest female measured weighed 380 g. Nonetheless, the smallest weight group (<400 g) at all sites was strongly biased towards males (χ2 test, χ2 = 46.835, p < 0.001); the frequency of males in the small-size groups was higher than that of females (Fig. 3). Hence, the primary sex (i.e. smallest reproducing individuals) is male indicating that the direction of sex change in this species is that of protandry. In Okinawa, the weight (g) of corals documented as females during the study, including the sex-changing individuals, was significantly higher than that of consistent males (Kruskal-Wallis test, Q = 3.051, P < 0.05) similar to the trend observed in Aqaba (Mann-Whitney test, T = 689.5, P = 0.013). The weight of corals in the ‘transitional state’ was also significantly heavier than that of consistent males (Kruskal-Wallis test, Q = 2.744, P < 0.005). In contrast, at Eilat no significant difference was found between the weight of females, which include the sex-changing corals, and that of consistent males (Mann-Whitney test, T = 361, P = 0.845) (Fig. 2 boxplots, Supplementary Material Figs S3, S3; boxplots).
The average size of the population and the weight range of individuals at which sex change occurred differed among the study sites (Fig. 4, Table 2). The average weight of the H. limax population in Eilat was the smallest, 437 g ± 25 (mean ± 95% confidence interval), and sex change in Eilat also occurred at a relatively small average weight (605 g ± 285) and within a narrow range (232–968 g). In Aqaba, the average weight of the population was 630 g ± 59, and sex change occurred at an average of 885 g ± 395 and ranged between 330 to 1,390 g. In Okinawa, the population was significantly heavier than that of both Eilat (Kruskal-Wallis test, Q = 9.030, P < 0.05) and Aqaba (Kruskal-Wallis test, Q = 4.224, P < 0.05) with an average weight of 779 g ± 53. Additionally, sex change occurred at a larger average weight (1,067 g ± 141) and within a much wider range 195–2400 g. The largest reproductive coral in Eilat was a female weighing 1,866 g; in Aqaba also a female, weighing 2,580 g; and in Okinawa a female and a male, both weighing 2,400 g (Table 2).
Sex-ratio of H. limax populations varied significantly between Aqaba and Eilat (Pairwise χ2 test, p < 0.001) and between Okinawa and Eilat (Pairwise χ2 test, p < 0.001). However, no significant difference was found between Okinawa and Aqaba (Pairwise χ2 test, p = 0.648). In Eilat, sex-ratios were male-biased, varying from 1:0 (M:F) in 2012, which represents the most male-biased year, to 1:0.6 (M:F) in 2013, which represents the least male-biased year, reflecting significant differences between different years (χ2 test, χ2 = 17.27, p < 0.01). In Aqaba and Okinawa sex ratios included individuals that released both male and female gametes within the same reproductive season (i.e. transitional state, T) and were relatively less biased towards maleness (Fig. 2, Supplementary Material Fig. S1). In Aqaba, the sex ratio was 1:0.8:0.09 (M:F:T) in 2013 and 1:0.5:0.11 (M:F:T) in 2015; and in Okinawa, it was 1:0.2:0 (M:F:T) in 2014, 1:0.65:0.08 (M:F:T) in 2015, 1:0.62:0.28 (M:F:T) in 2016, and 0.88:1:0.11 (M:F:T) in 2017, which represents the only year in which sex ratio was female biased (Table 1). No significant differences were found among years in Aqaba (χ2 test, χ2 = 0.963, p = 0.353). However, in Okinawa, sex ratio did vary significantly among years (χ2 test, χ2 = 24.387, p < 0.01).
### Average annual increase in growth
The average annual percentage increase in growth (weight) was calculated following equation 1 for the different size groups. The smallest group’s (<400 g) growth rate was significantly higher than that of the intermediate group (Two-way ANOVA, q = 9.322, P < 0.001), as well as the largest group (Two-way ANOVA, q = 11.610, P < 0.001) (Fig. 5). Among the study sites, the H. limax population in Okinawa demonstrated the highest growth rate, being significantly higher than at both the Eilat (Two-way ANOVA, q = 3.912, P = 0.016) and Aqaba (Two-way ANOVA, q = 3.762, P = 0.021) study sites.
## Discussion
Theories related to sex change have been attempting to predict the conditions favoring sex change, the type of change, and its timing. The repetitive sex change displayed by the coral Herpolitha limax, distributed in widely divergent geographical areas and exhibiting substantial within habitat spatial distributions, enabled examination of the interplay between environmental conditions and sex allocation, and its manifestation in a sex-changing coral species.
### Repetitive sex change
While the vast majority of scleractinian corals are simultaneous hermaphrodites20,21,34,35, all corals whose reproduction has been studied within the family Fungiidae have been reported to be gonochoristic13,20,23,34,36,37,38. Of these, at least four fungiid species were found to be protandrous sequential hermaphrodites; that is, the polyps function as solely male when small and as solely female when larger13,38 in accord with the sex-allocation theory3. Two of these species (C. echinata and C. crassa) can change sex bi-directionally13 (i.e. either from female to male or from male to female in the same population). Here we report, for the first time in corals, on a repetitive (multiple) pattern of sex change exhibited by the fungiid coral H. limax. We found H. limax to be a protandrous species with the ability to execute multiple events of sex change and presenting reproductive tactics analogous to those of plants, which exhibit labile sexuality in response to energetic and/or environmental constraints. Individuals of H. limax released male gametes during their early life stages (i.e. protandrous), probably due to the overall low energetic state of small corals (i.e. lower capacity to obtain enough energy to support daily energetic requirements; e.g., maintenance, growth and higher energetic costs of being a female), and released male and female gametes alternately at intermediate and large sizes. A low percentage of corals released both male and female gametes within the same reproductive season in Aqaba and in Okinawa (Table 1). We adopt the definition that sex change encompasses organisms that simultaneously perform both sex functions during the transitional stage while changing from one sex to the other5. We term this exceptional reproductive state as a ‘transitional state’ (T). This reproductive state featured more frequently in the intermediate individual sizes (Fig. 3), the main size group that demonstrated sex change.
The evolution of one-way sex change has been successfully explained by the size-advantage model, which predicts that an individual should change sex when the reproductive value (i.e. a measure of the expected future reproductive success) of the other sex exceeds that of its present sex31. Following the same theoretical concept, if the reproductive value between the sexes changes more than once during a lifetime, bi-directional sex change (i.e. the occurrence of sex change either from female to male or from male to female) will be favored. Such an energetically costly exchange can occur only if a reduction in body size or in resource availability for reproduction takes place in the course of the organism’s lifetime39. Causes for such reduction could be fluctuations in local environmental conditions. For example, more females than males of an epiphytic orchid occur under open canopies in comparison to shady areas40. Similarly, in plants, males were shown to be proportionately more abundant during periods of poor environmental conditions, in comparison to periods of improved conditions16,17,18. Alternatively, a reduction in reproductive value could result from the cost of reproduction itself. Female jack-in-the-pulpit plant, which were male when small, lost weight following reproduction and sometimes became smaller than the critical size advantageous to reproduce as females, consequently changing back to males41. A similar pattern was documented in the polychaete Ophryotrocha puerilis (Dorvilleidae). In this species, the larger individual in a pair adopts the female role until outgrown by the faster-growing male, leading to repeated episodes of sex change by both pair members5,42. Following similar energy budget considerations, Loya and Sakai13 described sex change in intermediate sizes of the coral C. echinata, and suggested that corals could ‘recover’ from the relatively costly female reproduction by channeling energy to the less costly male reproduction in alternate years, and thereby increasing their reproductive success. In addition to the internal codes (i.e. genetic, developmental) inherent within the individual, environmental factors may act as a catalyst towards the evolution of specific adaptive traits of sexual reproduction. Consequently, we suggest that the multiple occurrences of sex change exhibited by H. limax constitute a flexible response of individual specimens to local environmental conditions enabling them to increase their reproductive success (fitness). Expressing female sexuality may follow improved environmental conditions and increased energy resources; while expressing the opposite sex, in alternate years, may compensate for an overall low energetic state caused by stress and/or for the high energetic cost of the preceding female reproductive state.
### Effects of the environment on resource allocation to reproduction, growth, and sex expression
Human-induced perturbations have left a strong ecological imprint on the coral reefs at all the study sites43,44,45,46. However, the basic ecological features as well as the intensity and prevalence of anthropogenic and natural perturbations differ among the sites (see Environmental differences in the Methods) with the coral reef of Okinawa seemingly providing the most favorable conditions for fungiid species. In spite of the massive bleaching events that resulted in a reduction of coral cover and species richness47,48,49,50,51,52, as well as other man-made and natural disturbances, such as freshwater runoff and typhoons46, the fungiid corals in Okinawa demonstrate high reproduction rate, appear in a good condition and are very abundant13,23. In contrast, the fungiid corals in the GoA/E are not common53, attain smaller size (Table 2), and were documented to suffer from diseases (L.E-S., YL, pers. obs.). Nonetheless, the repetitive sex change displayed by the coral H. limax is shared across all three spatially distant study sites. Furthermore, both seasonality and timing of spawning are synchronized among the sites (i.e. 8–10 am, for 5–7 consecutive days, five days after the full moon of June-August or July-September, depending on the lunar periodicity of the given year).
The differences among sites in the mechanisms that can potentially affect the species’ life-history traits (i.e. population density, environmental conditions) are reflected in several characteristics of the H. limax populations. Thus, the sporadically scattered populations in Eilat and Aqaba exhibited significantly slower growth (One-way ANOVA, F = 5.552, P = 0.005), earlier sex change, and lower percentages of reproduction and sex change in comparison to the densely aggregated populations in Okinawa. At all sites, sex ratio varied among years, but almost always was biased towards maleness and growth rate decreased with size. (Fig. 5). In clonal organisms such as scleractinian corals, growth and size may be genetically indeterminate54 and consequently, may more reflect the local environmental conditions rather than being merely an innate characteristic of the species. We posit that some variations in such life-history traits may largely stem from local environmental conditions, which have a direct impact on their energetic constraints. The higher overall energetic state of the corals in Okinawa is evident also in their higher percentages of reproduction and sex change (both being energetically costly functions) in comparison with their GoE/A conspecifics (Table 1). Similar to the Eurya japonica shrub, which changed sex more frequently with higher growth rates and more exposure to light throughout the year19, the higher percentages of reproduction and sex change documented in the H. limax population in Okinawa are correlated with higher growth rates (i.e. in comparison with the Eilat and Aqaba populations) (Tables 1, 2). In accordance with sex allocation theory, which predicts a bias in sex ratio toward the first sex (i.e. the first sex expressed by an individual during its life cycle)3,27, the sex ratio exhibited by the coral H. limax at all the study sites, (with one exceptional year in Okinawa), was always biased towards maleness. However, large fluctuations were documented among sites and within the same site during consecutive years. The difference in the H. limax sex ratio between Aqaba and Eilat, which are only 4.5 km apart, is particularly striking (Table 1). Hence, we suggest that in the case of H. limax, as in many dioecious plants18, sex ratio may be a consequence of a synergistic flexible response by individuals to local environmental conditions and to population demography, since sex ratios often differ for stacked (Okinawa) versus widely dispersed (Eilat and Aqaba) coral populations (see Collin and Promislow12). At the same time, we argue that the relatively lower abundance of females in Eilat reflects the population’s weakened physiological state, since maintaining female reproduction is considered more energetically costly than male reproduction13.
An early investment predominantly in sperm production allows reproduction to commence without the initial expense of egg production, perhaps so that colony growth can continue to a larger, safer size54. However, species that aim to overcome environmental stress will probably invest more in reproduction than in any other energy-costly life function55. The energy allocation towards reproduction may be enabled through the inducement of reproductive investment early on in life2. Even though first reproduction was documented at similar sizes at all the studied sites, the primary sex stage (i.e. solely male) in Eilat and Aqaba was shorter than in Okinawa (Table 2, Fig. 4). One possible explanation is that in order to maximize their reproductive potential, the H. limax population in the GoA/E allocate energy also to female reproduction at a relatively early age, before the accumulated risk of mortality becomes excessively high. This assumption takes into consideration also the lower growth rate of the corals in the GoE/A (Fig. 5), as well as the average smaller size of individuals in that population (Table 2). Hence, indicating a lower energy allocation to that specific life function (i.e. growth) probably in favor of other life functions, and presumably a shorter life span of an individual coral. However, the specific niche exploited by a species and its local spatial distribution characteristics, which differ between H. limax populations in Okinawa and the GoE/A, may also produce such a pattern. In very low-density populations, reproduction may be hampered by a shortage of potential mates56. This may have led the corals in the GoE/A to enhance female reproduction earlier than would have been expected in a more benign environment, such as in Okinawa. However, to confirm this assumption, a rigorous examination of small-sized individuals, subjected to various conditions, is necessary.
Trade-offs between alternative life-history traits depend on the available resource pool and may be expressed at higher taxonomic level (population) as well as at the level of individuals within a population7. The differences in the life history characteristics of H. limax were found to be more substantial between the geographically distant sites, but were also evident between the two close sites of the GoE/A, the Aqaba reefs are relatively less perturbed than those of Eilat43,44,57. Nevertheless, to confirm the possible coupling between the proximate mechanisms that determine patterns of sex change and environmental factors, additional species-specific studies of sex-changing organisms are required.
### Timing of sex change
In an attempt to understand the variation in timing of sex change, Warner31 suggested that the Size Advantage Hypothesis (SAH) be framed in terms of ‘reproductive value’, which is a measure of the expected future reproductive success, taking into account effects of growth rate and mortality, and is often dependent on the local environment, local demography, and the individual’s own status. Theory predicts that sex change will occur earlier in populations exhibiting slower growth rates and higher mortality rates3,7,10,24,25,26,58. Indeed, sex change did occur earlier and in a narrower size range at the Eilat and Aqaba study sites, where growth rate was significantly lower (Fig. 5; Table 2). Because the expected reproductive success depends on both size and/or age and the pool of each animal’s potential mates, the optimal size at sex change should vary in response to the size distribution or age structure of the population12. Accordingly, the time and range of sex change in the GoE/A may also be attributed to the scattered spatial distribution pattern of the species, and is compatible with the finding that densely stacked H. limax individuals (Okinawa) change sex at larger sizes than their spatially distant conspecifics (Eilat and Aqaba) as previously suggested by Collin and Promislow12. When reporting on bidirectional sex change in the fungiid species F. repanda and C. echinata, which inhabit the same patch reef in Okinawa and hence are subjected to the same environmental conditions, Loya and Sakai13 found variance in the timing of sex change between the two species. That in F. repanda occurred at a relatively small average weight and within a narrow range, compared to C. echinata, and the average increase in growth was significantly lower in F. repanda. Similarly, the size (weight) of sex change in the Aqaba population varied from that of the Eilat population: corals changed sex at a relatively smaller size in Eilat (605 g ± 285) than in Aqaba (885 g ± 395), and growth rate was slower in Eilat (Table 2). Differences in the relative frequency of sex change can even occur on neighboring reefs, indicating that individuals select a life-history tactic that may or may not involve adult sex change, depending on the mating conditions that they experience in that habitat patch7. We posit that in fungiid corals the timing of sex change may be determined by the combination of a chemical response to nearest-neighbor conspecifics cues (e.g. pheromones release in stacked coral populations, such as in Okinawa), as well as by local environmental conditions.
The findings of this study contribute to a better understanding of the trade-offs among life-history traits in modular organisms, and suggest that the repetitive sex change displayed by the fungiid coral H. limax is environmentally mediated and phenotypically plastic, analogous to that in sexually labile plants. Much of the data obtained in this work provide support for the sex allocation theory3. Although uncommon, these exceptional reproductive tactics must be considered if we are to fully understand the evolution of sexual systems in corals. We further argue that fungiid corals provide an excellent model organism by which to study the mechanism of sex change in animals, due to their unique life form, which enables a relatively easy and accurate determination of various life-history parameters.
## Methods
### Study sites
This study was variously conducted over a period of 3–13 years and took place at three geographically distant sites: Okinawa (Japan) at several patch reefs near Sesoko Marine Station, (26°39′N, 127°52′E; during 2004–2017); Eilat (Israel) on the western side of the Gulf of Eilat/Aqaba (29°30′N, 034°55′E; during 2009–2017); and Aqaba (Jordan), on the eastern side of the GoE/A (29°27′N, 034°58′E; during 2013–2016).
The island of Okinawa is part of the Ryukyu Archipelago in Japan, which defines the boundary between the East China Sea (west) and the Philippine Sea (east). The archipelago hosts a diverse coral community, with species numbers comparable to those found in the Great Barrier Reef 59,60. In spite of the massive bleaching events that resulted in a reduction of coral cover and species richness47,48,49,50,51,52, as well as other man-made and natural disturbances such as polluted freshwater runoff and typhoons46, the reefs in Okinawa are continuous and massive with a relatively high coral cover57. This may be a consequence of the influence of the Kuroshio current, which introduces warm tropical waters from the Philippines and equatorial Pacific into the archipelago60. The Gulf of Eilat/Aqaba (GoE/A; 180 km by 20 km) is a semi-enclosed sea, and is connected to the open ocean via a narrow, shallow (137 m) strait at its southernmost end (Bab el Mandeb)61.
The coral reefs of the GoE/A are situated at the northern limits of global coral-reef distribution and embrace a rich diversity of habitats that include shallow coastal lagoons, sea grass beds, mangrove stands, and the coral reefs themselves43. Due to the steep bathymetry of the area, coral-reef communities are limited to a narrow, patchy band along most of the coast45,62. Throughout the last few decades, the cities of Aqaba and Eilat, situated at the tip of the bay, have been experiencing an enormous growth, with increased coastal development43,45 that has resulted in substantial stress on the coral-reef ecosystems in the GoE/A. However, in spite of the close proximity of the Aqaba reefs to the Eilat reefs (4–5 km), the extent of deterioration varies between these sites. While the reef of Eilat is considered highly degraded with extreme deterioration in coral abundance and living cover43,44,63,64, the coral reefs in Jordan are described as being in good condition with high coral diversity and a relatively low rate of coral mortality45,65. These differences have also been attributed to the flow dynamics. The area is interconnected by surface currents that generally flow in a clockwise direction from Saudi Arabia up the Jordanian coast and then down the Israeli coast66,67. According to Labiosa et al.68, Ekman-driven down-welling along the western side (Eilat) can potentially lead to more oligotrophic conditions, while the upwelling along the eastern margin (Aqaba) generates mesotrophic to eutrophic conditions for much of the year. These dynamics coincide with higher than average Chlorophyll a concentration and cooler surface seawater temperatures on the eastern side of the GoE/A.
The study sites are characterized by different environmental conditions and by divergent densities of the H. limax populations. In both the Eilat and Aqaba reefs H. limax populations appear in a dispersed manner and occur mostly at a depth range of 5–50 m, albeit at a higher frequency in Aqaba than in Eilat (L.E-S., YL, pers. obs). In contrast, in the coral reefs of Okinawa, the H. limax population is mainly concentrated at few sites that are inhabited by tens of thousands of specimens (probably more than 12 fungiid species), densely distributed between 3 to 20 m depth13. Surface seawater temperatures in Okinawa and in the GoE/A are quite similar; varying from 20 to 28 C° in Okinawa48 and from 21 to 28 C° in the GoE/A53. However, the corals in Okinawa were subjected to recurring bleaching events in the past few decades48, whereas in the GoA/E no bleaching events were documented and corals are considered ‘resistant’ to high temperatures61.
### Specimen collection
At all three sites, specimens of H. limax of all possible size groups were collected from the reef and transferred to an outdoor seawater system at the local laboratory. Each sampled coral was tagged individually with a numeric plastic tag attached by a nylon fishing cord inserted through a thin hole drilled at the edge of the coral skeleton using a portable drill13. No adverse effects on the corals were observed following the tagging procedure. All corals were measured (length and weight) and photographed immediately after collection. A ruler was placed next to each photographed specimen to enable scale (Fig. 1a). The corals were kept in large containers with constant seawater inflow and were transferred to separate individual aquaria for spawning observations, in accordance with the expected timing of reproduction (see Supplementary Material for further information on the discovery of spawning time). Following each annual set of measurements and observations, the tagged corals were returned to their natural habitat and retrieved the following year, one to two weeks before the expected breeding season. Sample sizes of the different coral size groups were increased over the years in order to enable a more in-depth examination of the data accumulating from the observations.
### Reproduction studies
Observations on H. limax started in 2004 at the Sesoko Marine Station, Japan, as part of a wider research project of studying reproductive strategies in the Fungiidae13. During the initial stages, 27 H. limax individuals, representing different size groups, were collected from the reef and placed in separate aquaria to spawn. Nightly observations were made between 5 pm and a few hours before dawn, for eight consecutive nights after the full moon in July 2004. However, despite revealing the reproduction of ten other fungiid species13,23, no reproduction was recorded for H. limax. Over the years, sample size was increased and observation times were changed in an attempt to determine the exact reproductive season and spawning time of the same individually tagged individuals. The observational procedure was repeated during August 2006, June to September 2007, 2010, 2014–2017 in Okinawa; June to September 2009–2017 in Eilat; and June to September 2013–2016 in Aqaba. Coral sex was determined from the unique shape of the shed gametes and was verified using a microscope.
### Growth rate
Growth rates of individuals were monitored by means of annual size measurements of individual corals. Using a tape measure, length (L, mm) was measured to the nearest 1 mm; weight (W, g) was measured to the nearest 1 g, following careful removal of excess moisture (Fig. 1a). Growth rate was studied in all three sites: Okinawa, Japan (2004–2017); Aqaba, Jordan (2013–2016); and Eilat, Israel (2009–2017), and calculated according to equation 1:
$$\begin{array}{c}{\rm{Individual}}\,{\rm{annual}}\,{\rm{growth}}\,{\rm{rate}}\,({\rm{increas}}\,{\rm{in}}\,{\rm{weight}}\,({\rm{g}}))=\frac{{\rm{\Delta }}{\boldsymbol{Wx}}({\bf{g}})/{\boldsymbol{N}}}{{\boldsymbol{Wx}}}\\ {W}{\rm{x}}-{\rm{Weight}}\,({\rm{g}})\,{\rm{of}}\,{\rm{individual}}\,{\rm{coral}}\,{\rm{x}}\\ N\,\mbox{--}\,{\rm{Total}}\,{\rm{number}}\,{\rm{of}}\,{\rm{years}}\,{\rm{between}}\,{\rm{consecutive}}\,{\rm{weight}}\,{\rm{measurements}}\\ {\rm{\Delta }}Wx\,(g)-{\rm{Weight}}\,{\rm{difference}}\,({\rm{g}})\,{\rm{between}}\,{\rm{the}}\,{\rm{initial}}\,Wx\,{\rm{and}}\,{\rm{its}}\,{\rm{weight}}\,{\rm{after}}\,{\rm{N}}\,{\rm{years}}.\end{array}$$
(1)
### Statistical analyses
Statistical analyses were performed using R software69 and Sigmaplot 12.2 (Systat Software, Inc). Data were checked for normality (Shapiro-Wilk normality test) and homogeneity of variance (F-test) and tested accordingly with appropriate tests. Post-hoc tests were carried out by Tukey and Holm tests. P-values < 0.05 were considered statistically significant. Heat-maps were created in R using Superheat package70.
## Data Availability
The data that support the findings of this study are available from the corresponding authors on request.
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Iryu, Y., Nakamori, T., Matsuda, S. & Abe, O. Distribution of marine organisms and its geological significance in the modern reef complex of the Ryukyu Islands. Sedimentary Geology 99, 243–258 (1995).
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Spalding, M., Ravilious, C. & Green, E. P. World atlas of coral reefs. (Univ of California Press, 2001).
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Fine, M., Gildor, H. & Genin, A. A coral reef refuge in the Red Sea. Global change biology 19, 3640–3647 (2013).
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Glassom, D., Zakai, D. & Chadwick-Furman, N. Coral recruitment: a spatio-temporal analysis along the coastline of Eilat, northern Red Sea. Marine Biology 144, 641–651 (2004).
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Bahartan, K. et al. Macroalgae in the coral reefs of Eilat (Gulf of Aqaba, Red Sea) as a possible indicator of reef degradation. Marine Pollution Bulletin 60, 759–764 (2010).
64. 64.
Shlesinger, T. & Loya, Y. Recruitment, mortality, and resilience potential of scleractinian corals at Eilat, Red Sea. Coral Reefs 35, 1357–1368 (2016).
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Al-Horani, F. A., Hamdi, M. & Al-Rousan, S. A. Prey selection and feeding rates of Drupella cornus (Gastropoda: Muricidae) on corals from the Jordanian coast of the Gulf of Aqaba, Red Sea. Jordan Journal of Biological Sciences 4 (2011).
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Anati, D. A. Water transports in the Gulf of Aqaba. L’oceanographic physique de la Mer Rouge, CNEXO 2, 165–173 (1974).
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Berman, T., Paldor, N. & Brenner, S. Simulation of wind-driven circulation in the Gulf of Elat (Aqaba). Journal of Marine Systems 26, 349–365 (2000).
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Labiosa, R. G., Arrigo, K. R., Genin, A., Monismith, S. G. & van Dijken, G. The interplay between upwelling and deep convective mixing in determining the seasonal phytoplankton dynamics in the Gulf of Aqaba: Evidence from SeaWiFS and MODIS. Limnology and oceanography 48, 2355–2368 (2003).
69. 69.
R: A Language and Environment for Statistical Computing (R Foundation for Statistical Computing, Vienna, Austria, 2014).
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Barter, R. L. & Yu, B. Superheat: Supervised heatmaps for visualizing complex data. arXiv preprint arXiv:1512.01524 (2015).
## Acknowledgements
We thank the Interuniversity Institute for Marine Sciences at Eilat (IUI), the Marine Sciences Station in Aqaba (MSS) and the Tropical Biosphere Research Center Sesoko Station for the logistical support. We are indebted to I. Hosgin, P. Cabaitan, O. Levy, Y. Nakano, R. Prasetia, J. Bergman and T. Eyal for help with the field work in Okinawa and to B. Feldman, R. Tamir, H. Rapuano, T. Amit, N. Kramer, M. Grinblat, A. Alamaru, E. Winter, I. Brikner and other members of the Loya lab for help throughout the years with the field work in Aqaba and Eilat. We thank O. Mann for assistance in the statistical analyses. This study was partially supported by the Israel Science Foundation (ISF) No. 341/12 and 1191/16 and the U.S. Middle East Regional Cooperation (MERC) Program Agency for International Development (MERC/USAID) No. M32-037 to YL.
## Author information
Authors
### Contributions
Y.L. initiated and designed the study. L.E.-S., G.E. and Y.L. planned the experiments and analyzed the data. L.E.-S., G.E., K.S., Y.N., O.B., O.B.-Z., T.S. and Y.L. made the observations and collected the data. S.H. and F.S. coordinated and helped with the work in Okinawa. L.E.-S., G.E. and Y.L. wrote the manuscript with the help of all co-authors.
### Corresponding author
Correspondence to Yossi Loya.
## Ethics declarations
### Competing Interests
The authors declare no competing interests.
Publisher’s note: Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.
## Rights and permissions
Reprints and Permissions
Eyal-Shaham, L., Eyal, G., Sakai, K. et al. Repetitive sex change in the stony coral Herpolitha limax across a wide geographic range. Sci Rep 9, 2936 (2019). https://doi.org/10.1038/s41598-018-37619-y
• Accepted:
• Published:
• ### Sexual production of corals for reef restoration in the Anthropocene
• CJ Randall
• , AP Negri
• , KM Quigley
• , T Foster
• , GF Ricardo
• , NS Webster
• , LK Bay
• , PL Harrison
• , RC Babcock
• & AJ Heyward
Marine Ecology Progress Series (2020)
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2020-08-09 09:21:01
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https://stats.stackexchange.com/questions/151827/my-test-accuracy-is-pretty-bad-compared-to-cross-validation-accuracy
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# My Test accuracy is pretty bad compared to cross-validation accuracy
I did a Multi-class document classification. I divided the original data set (18,8334 documents as a list of strings where each element of list is a document string.) into 70% training and 30% test.
Then on the 70% training dataset, I used sklearn 5 fold cross validation to train the model. I used three models. First was Gaussian Naive Bayes, second was Random Forests and third was Stochastic Gradient Descent SVM.
Stochastic gradient descent gave the highest cross validated accuracy of 0.85. But the same model when tested on the 30% test dataset gives 9% accuracy. Why is that? Isn't cross-validation error a measure or estimate of test error/generalization error?
Thanks
Edit:
This is how I created the train_test(70/30)
def split(docs_list,target_recoded):
"""This function samples the dataset into training and testing"""
# Splitting into training and test.
from sklearn.cross_validation import train_test_split
train_X, test_X,train_Y,test_Y = train_test_split(docs_list, target_recoded, test_size=0.30, random_state=42)
return train_X, test_X,train_Y,test_Y
After initial nlp preprocessing like stop words removal, stemming etc, I have a cleaned list of doc strings. On that, I used the following for bag of words creation. First 70% training data is passed and then 30% test data was passed as argument to this function.
def bagofWords(X,Y,max_feature=5000,type="count"):
"""This function creates a Bag of Features vectors from the original documents"""
from sklearn.feature_extraction.text import CountVectorizer, TfidfVectorizer
# Initialize the "CountVectorizer" object, which is scikit-learn's
# bag of words tool.
if(type=="count"): # To choose between count or tf-idf bag or words model
vectorizer = CountVectorizer(analyzer = "word",max_features = max_feature)
else:
vectorizer = TfidfVectorizer(analyzer = "word",max_features = max_feature)
X=vectorizer.fit_transform(X)
return X ,np.array(Y)
This is how I train an SGD
def SGD(self):
"""Method to implement Multi-class SVM using Stochastic Gradient Descent"""
from sklearn.linear_model import SGDClassifier
scores_sgd = []
for train_indices, test_indices in self.k_fold:
train_X_cv = self.train_X[train_indices].todense()
train_Y_cv= self.train_Y[train_indices]
test_X_cv = self.train_X[test_indices].todense()
test_Y_cv= self.train_Y[test_indices]
self.sgd=SGDClassifier(loss='hinge',penalty='l2')
scores_sgd.append(self.sgd.fit(train_X_cv,train_Y_cv).score(test_X_cv,test_Y_cv))
print("The mean accuracy of Stochastic Gradient Descent Classifier on CV data is:", np.mean(scores_sgd))
And this is to check test performance
def test_performance(self,test_X,test_Y):
"""This method checks the performance of each algorithm on test data."""
from sklearn import metrics
# For SGD
print ("The accuracy of SGD on test data is:", self.sgd.score(test_X,test_Y))
print 'Classification Metrics for SGD'
print metrics.classification_report(test_Y, self.sgd.predict(test_X))
print "Confusion matrix"
print metrics.confusion_matrix(test_Y, self.sgd.predict(test_X))
• Is 9% accuracy a typo? If not, could you have switched your test data labels around at all? – jld May 11 '15 at 17:20
• Yeah, it sounds like probably a code mistake. Are you preprocessing the test data in a consistent way to the training? (In scikit-learn, keeping the same transformers and calling transform but not fit.) – Dougal May 11 '15 at 17:26
• Here is the thing. I have done the similar data preprocessing. But for converting the test documents to bag of words, I have used again fit_transform. Is this ok? Or should I use the earlier fit_transform in vectorizer for train data – Baktaawar May 11 '15 at 18:45
• Pls check edit. I have added the steps I did – Baktaawar May 11 '15 at 18:53
Hastie et al discuss this precise issue in their book, The Elements of Statistical Learning. They conclude that cross-validation is NOT an estimate of test-error conditional on the training set. Rather, they believe it is an estimate of the unconditional test error. In other words, this is the expected test error if you are also randomizing over the world of possible training sets, rather than the precise training set you've been given.
• Could you please elaborate on this a bit? – Baktaawar May 11 '15 at 18:46
• The book is free online, search for The Elements of Statistical Learning. This issue is mentioned at the very beginning of chapter 7. Also see 7.4 and 7.10. – Matthew Drury May 11 '15 at 19:06
I'd suspect something is wrong with how you created the hold out set. Ie either it wasn't randomized or needed to be done with stratified sampling.
This sort of thing can also happen if you did feature selection or engineering outside of the CV loop as you'll produce features that look good across the cv folds but don't generalize.
It can also happen if you tuned model hyperparameters using the CV loop. How did the other model's do on the hold out set? SVMs can be quite prone to this while RF's tend to be less prone so i'd wonder about that in particular. Other strategies for dealing with this include an inner CV loop and fixing the regularization parameters to limit the amount of overfitting you can do.
Finally if this is highly dimensional data with lots of variance unrelated to the target (like genetic data) it is quite possible there exist models that works well in the cv data but not the hold out data by random chance.
This phenomena has been described as "anti learning", again RF's tend to be more robust against this (because of the internal bagging) though it begins to effect them as well as the dimensionality grows.
• Pls check edit. Have added the steps I followed – Baktaawar May 11 '15 at 18:53
• The way you are doing bag of words is suspect. For both the hold out set and each cv fold you need to call fit_transform only on what you are training the model on, and just call transform when you do the testing. You can use a sklearn.pipeline.Pipeline to treat preprocessing as part of your model to enforce this. As is, it is likely producing different vectors/order of vectors on the 30% hold out and possibly overfittiting by fitting on the entire 70% outside of the folds for the cross validation. – Ryan Bressler May 11 '15 at 20:28
• @RyanBressler, this is definitely the reason for the huge gap. You should turn your comment into an answer and I'll upvote it. :p – Dougal May 11 '15 at 21:17
• Ok. So the way I understand this, I need to just do transform on the test set. I think I kind of missed it. I guess it is creating bag of words on the reduced test dataset by again learning it. So obviously it would have different feature vectors. Does this sound correct? – Baktaawar May 11 '15 at 21:34
• Yes. It was correct. Just removed fit_transform to transform. And now the test accuracy is same as cross val. 85% for SGD and 76% for Random Forests. I haven't tweaked a lot of hyper parameters for RF, so it's like with default settings. But yes, the problem was that I mistakably fitted new tfidf vectors using test data, hence obviously lot of feature vectors would be different as the test_data is just 30% of whole data. – Baktaawar May 11 '15 at 22:21
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2020-01-24 22:46:32
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https://raghavsomani.github.io/sitemap/
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# Sitemap
A list of all the posts and pages found on the site. For you robots out there is an XML version available for digesting as well.
## Note on the Kadison-Singer Problem and its Solution
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The Kadison-Singer problem arose from the work on quantum mechanics done by Paul Dirac in the 1930s. The problem is equivalent to fundamental problems in areas like Operator theory, Hilbert and Banach space theory, Frame theory, Harmonic Analysis, Discrepancy theory, Graph theory, Signal Processing and theoretical Computer Science. The Kadison-Singer problem had been long standing and defied the efforts of most Mathematicians until it was recently solved by Adam Wade Marcus, Daniel Alan Spielman and Nikhil Srivastava in 2013. Read more
## A note on Conformal Symplectic and Relativistic Optimization
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This note on a spotlight paper at NeurIPS 2020, has been made while I had been reading the literature on the principle connections between continuous and discrete optimization. The motivation is to understand and create accelerated discrete large scale optimization algorithms from first principles via considering the geometry of phase spaces and numerical integration, specifically symplectic integration. Recent works successfully have been able to throw sufficient light on the two and therefore has attracted my attention. Read more
## Geometry of Relativistic Spacetime Physics
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This article introduces and describes the mathematical structures and frameworks needed to understand the modern fundamental theory of Relativistic Spacetime Physics. The self-referential and self-contained nature of Mathematics provides enough power to prescribe a rigorous language needed to formulate the building components of the standard Einstein’s General Theory of Relativity like Spacetime, Matter, and Gravity, along with their behaviors and interactions. In these notes, we will introduce and understand these abstract components, starting with defining the arena of smooth manifolds and then adding the necessary and suffcient differential geometric structures needed to build the primers to the General Theory of Relativity. Read more
## Dual spaces and the Fenchel conjugate
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Dual spaces lie at the core of linear algebra and allows us to formally reason about the concept of duality in mathematics. Duality shows up naturally and elegantly in measure theory, functional analysis, and mathematical optimization. In this post, I have tried to learn and explore the nature of duality via Dual spaces, its interpretation in general linear algebra, all of which was motivated by the so called convex conjugate, or the Fenchel conjugate in mathematical optimization. Read more
## A survey on Strongly Rayleigh measures and their mixing time analysis
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Strongly Rayleigh measures are natural generalizations of measures that satisfy the notion of negative dependence. The class of Strongly Rayleigh measures provides the most useful characterization of Negative Dependence by grounding it in the theory of multivariate stable polynomials. This post attempts to throw some light on the origin of Strongly Rayleigh measures and Determinantal Point Processes and highlights the fast mixing time analysis of the natural MCMC chain in the support of a Strongly Rayleigh measure as shown by Anari, Gharan and Rezaei 2016. Read more
## Analysis of Newton’s Method
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In optimization, Netwon’s method is used to find the roots of the derivative of a twice differentiable function given the oracle access to its gradient and hessian. By having super-liner memory in the dimension of the ambient space, Newton’s method can take the advantage of the second order curvature and optimize the objective function at a quadratically convergent rate. Here I consider the case when the objective function is smooth and strongly convex. Read more
## Deriving the Fokker-Planck equation
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In the theory of dynamic systems, Fokker-Planck equation is used to describe the time evolution of the probability density function. It is a partial differential equation that describes how the density of a stochastic process changes as a function of time under the influence of a potential field. Some common application of it are in the study of Brownian motion, Ornstein–Uhlenbeck process, and in statistical physics. The motivation behind understanding the derivation is to study Levy flight processes that has caught my recent attention. Read more
## SGD without replacement
Posted on:
This article is in continuation of my previous blog, and discusses about the work by Prateek Jain, Dheeraj Nagaraj and Praneeth Netrapalli 2019. The authors provide tight rates for SGD without replacement for general smooth, and general smooth and strongly convex functions using the method of exchangeable pairs to bound Wasserstein distances, and techniques from optimal transport. Read more
## Non-asymptotic rate for Random Shuffling for Quadratic functions
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This article is in continuation of my previous blog, and discusses about a section of the work by Jeffery Z. HaoChen and Suvrit Sra 2018, in which the authors come up with a non-asymptotic rate of $\mathcal{O}\left(\frac{1}{T^2} + \frac{n^3}{T^3} \right)$ for Random Shuffling Stochastic algorithm which is strictly better than that of SGD. Read more
Posted on:
Posted on:
## Nesterov’s Acceleration
Posted on:
This post contains an error vector analysis of the Nesterov’s accelerated gradient descent method and some insightful implications that can be derived from it. Read more
Posted on:
With a number of courses, books and reading material out there here is a list of some which I personally find useful for building a fundamental understanding in Machine Learning. Read more
## A survey on Large Scale Optimization
Posted on:
This post contains a summary and survey of the theoretical understandings of Large Scale Optimization by referring some talks, papers, and lectures that I have come across in the recent. Read more
## Montreal, Canada during NeurIPS 2018
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Visited Montreal, Canada with Microsoft Research Labmates to attend and present at NeurIPS 2018 Read more
## Melbourne, Australia during WSDM 2019
Posted on:
Visited Melbourne, Australia to attend and present at WSDM 2019 Read more
## Vancouver, Canada during NeurIPS 2019
Posted on:
Visited Vancouver, Canada to attend NeurIPS 2019 and present at SEDL 2019 Read more
Posted on:
## Some Approaches of Building Recommendation Systems
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The project aims at using different recommendation methods for different kinds of real world data like rating matrices, images and text, using Deep Learning and Optimization. Read more
## Sparse Regression and Support Recovery bounds for Orthogonal Matching Pursuit
Posted on:
We study the problem of sparse regression where the goal is to learn a sparse vector that best optimizes a given objective function. Under the assumption that the objective function satisfies restricted strong convexity (RSC), we analyze Orthogonal Matching Pursuit (OMP) and obtain support recovery result as well as a tight generalization error bound for OMP. Furthermore, we obtain lower bounds for OMP, showing that both our results on support recovery and generalization error are tight up to logarithmic factors. To the best of our knowledge, these support recovery and generalization bounds are the first such matching upper and lower bounds (up to logarithmic factors) for any sparse regression algorithm under the RSC assumption. Read more
## Universality Patterns in the Training of Neural Networks
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This work proposes and demonstrates a surprising pattern in the training of neural networks: there is a one to one relation between the values of any pair of losses (such as cross entropy, mean squared error, $0/1$ error etc.) evaluated for a model arising at (any point of) a training run. This pattern is universal in the sense that this one to one relationship is identical across architectures (such as VGG, Resnet, Densenet etc.), algorithms (SGD and SGD with momentum) and training loss functions (cross entropy and mean squared error). Read more
## Connections between Stochasticity of SGD and Generalizability
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This is an attempt to understand how stochasticity in an optimization algorithm affect generalization properties of a Neural Network. Read more
## Robust Mixed Linear Regression using heterogeneous batches
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For the problem of learning Mixed Linear Regression, this work introduces a spectral approach that is simultaneously robust under both data scarcity and outlier tasks. Read more
## Scaling laws of optimization algorithms for Deep Learning - the Graphon perspective
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Understanding scaling limits of discrete Euclidean optimization algorithms on large unlabeled graphs. This problem is motivated by the problem of optimizing permutation invariant risk functions of (single layer and deep) Neural Networks (NNs). Theoretical aspects stem from the original theory of gradient flows on the Wasserstein space, which have been used to understand scaling limits of (stochstic) gradient descent ((S)GD) processes in the case of single hidden layer neural networks. There are also other related questions that are specific to the qualitative nature of the stochasticity in the SGD process, and the role of depth in NNs. Read more
## Clustered Monotone Transforms for Rating Factorization
Raghav Somani*, Gaurush Hiranandani*, Sanmi Koyejo & Sreangsu Acharyya
Published at: Web Search and Data Mining (WSDM), 2019
The paper has been accepted for an oral persentation (84/511 submissions ≈ 16% Acceptance Rate). Read more
[paper] [arXiv] [bib] [code] [video]
## Support Recovery for Orthogonal Matching Pursuit: Upper and Lower bounds
Raghav Somani*, Chirag Gupta*, Prateek Jain & Praneeth Netrapalli
Published at: Neural Information Processing Systems (NeurIPS), 2018
The paper has been accepted for Spotlight presentation. Read more
[paper] [bib] [video]
## Non-Gaussianity of Stochastic Gradient Noise
Abhishek Panigrahi, Raghav Somani, Navin Goyal & Praneeth Netrapalli
Published at: Science meets Engineering of Deep Learning (SEDL) workshop, Neural Information Processing Systems (NeurIPS), 2019
We study the distribution of the Stochastic Gradient Noise during the training and observe that for batch sizes $256$ and above, the distribution is best described as Gaussian at-least in the early phases of training. Read more
[arXiv] [bib]
## Meta-learning for Mixed Linear Regression
Weihao Kong, Raghav Somani, Zhao Song, Sham Kakade, Sewoong Oh
Published at: International Conference on Machine Learning (ICML), 2020
The paper has been accepted for a presentation. Read more
[paper] [arXiv] [bib] [code] [video]
## Robust Meta-learning for Mixed Linear Regression with Small Batches
Weihao Kong, Raghav Somani, Sham Kakade, Sewoong Oh
Published at: Neural Information Processing Systems (NeurIPS), 2020
The paper has been accepted for a poster. Read more
[paper] [arXiv] [bib] [code] [video]
## Gradient flows on graphons: existence, convergence, continuity equations
Sewoong Oh, Soumik Pal, Raghav Somani & Raghav Tripathi
Published at: Optimal Transport and Machine Learning (OTML) workshop, Neural Information Processing Systems (NeurIPS), 2021
The paper has been accepted for a poster. Read more
[arXiv] [bib]
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2022-10-06 14:49:57
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http://95cento.it/coin-flip-probability-calculator-at-least.html
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Here’s the simplest version of this calculation. When the coin is thrown in the air, it should rotate several times before landing on the ground, or caught and inverted by a chosen person. The probability of an event is a number indicating how likely that event will occur. 05 to establish bias at 95% confidence one sided. In these cases, it can be helpful to use a binomial probability calculator like the one below. same as (a) except now the coin is flipped 10 times. 1/8 To calculate the probability you have to name all possible results first. For example, even the 50/50 coin toss really isn’t 50/50 — it’s closer to 51/49, biased toward whatever side was up when the coin was thrown into the air. The probability of three heads given the biased coin is trivial: P ( three heads | biased coin) = 1. Since 2 5 = 35. GDR 1/6 + 1/6 - (1/6 ∙ 1/6) = 2/6 - 1/36 = 11/36 7. In this case, getting head in the first place doesn't influence the outcome of the second toss. That is, this many heads in a row is pretty unlikely; the expected (i. Maybe on the second spin. You will roll a (fair) die, if the result is odd flip coin 𝐴 twice (independently); if the result is even flip coin 𝐵 twice (independently). Calculate final probability: P (HTT + THT + TTH) =. Coin toss probability. The following formulas are used to calculate different dice probabilities. The experimental probability depends upon the actual outcome of the experiment. In this article, we are going to study to solve problems to find the probability involving the throwing of two dice. The probability of event B, getting heads on the second toss is also 1/2. Suppose I want to know the probability of getting a certain number of heads in 10 tosses of a fair coin. 50 = 50% or 2 4 = 1 2 because there are two ways for the two coins to yield the mixed results. Picking numbers randomly means that there is no specific order in which they are chosen. Coin flip probabilities deal with events related to a single or multiple flips of a fair coin. 5): The likelihood term represents the probability of flipping heads, if the coin’s bias is 0. The image of a flipping coin is invariably connected with the concept of "chance. The probability distribution for the genders of two kids: Event MM FF MF FM. Calculate the probability of flipping a coin toss sequence of HTTTTTTTT The probability of each of the 9 coin tosses is 1/2, so we have: P (HTTTTTTTT) = 0. Calculate the probability that head turns up at least 2 times. A coin is tossed 3 times. (a) Derive an expression formula to calculate the probability for the total number of heads in n tosses. You're signed out. What is the probability of getting two heads in two tosses? The probability that the coin when tossed turns up heads is 1/2. We flip a fair coin 10 times. But the result over many tosses is predictable, as long as the trials are independent (i. Let A be the event that the coin shows heads at least 4 times. Consider the experiment of tossing a coin. For example, suppose we flip a coin and get Head, then we flip the coin again this time we get Tail. Putting these together means you have a total of 2xx6=12 outcomes. 5 % chance at least one 6 will appear. Quantity B: 1/2. Example 9 Tossing a fair die. Now what is the conditional probability: that you picked the fake coin? \item Suppose you wanted to decide whether the chosen coin was fake: by flipping it $$k$$ times. 5, which means we would not be able to tell the different between a bias coin and fair coin 50% of the time. Using the Binomial Probability Calculator. Fair coin is tossed 5 times. If we use Bayes' Theorem from above, we can calculate. 5 if the coin is fair) and a number of times to flip the coin. Users may refer the below solved example work with steps to learn how to find what is the probability of getting at-least 1 head, if a coin is tossed nine times or 9 coins tossed together. The area of its top and bottom is pi*R^2, with R = radius = half the diameter: pi*12. The chance on the first toss is 50%, and on the 42nd toss it. The probability of drawing an Ace from a standard deck is 0. An ideal unbiased coin might not correctly model a real coin, which could be biased slightly one way or another. First, note that the problem will likely make reference to a "fair" coin. Hence the probability of not generating Similarly, each of the next 2 a bit in ips is m. The dice are meant to introduce an element of chance to these games; we expect that the outcomes of the rolls will be truly random. Statistics and probability: 1-1 1. Easycalculation. Similarly, on tossing a coin, the probability of getting a tail is: P(Tail) = P(T) = 1/2. The caveat is that these calculators have pre-set defaults which might differ — which means the sample size might differ as well. The above explanation will help us to solve the problems on finding the probability of tossing three coins. You will also get a step by step solution to follow. This is a distribution over the bias of a bernoulli process. Next, calculate the probability of each outcome, assuming the coin has a probability of. What is the probability of flipping 2 coins and getting 1 Heads 1 tails? This states that the probability of the occurrence of two mutually exclusive events is the sum of their individual probabilities. How to calculate probability? "Hey man, but girls and coins are two different things!I should know, I've seen at least one of each. Binomial test; Coin flipping. You want to know the probability of the coin landing on heads. Which gives us: = p k (1-p) (n-k) Where. use sample space S. The probability that all of these 20 toss successions were not all heads = "X to the power of 86,381". As I have written in the comment the answers seems to be. What is the probability that we get heads in at least 8 of the 10 flips?. Given that more heads than tails appear, what is the probability that all of the flips are H? c. There are 2^6=64 possible outcomes. Binomial Probability At Most At Least - MathBitsNotebook (Geo - CCSS Math) Consider these questions: Question 1: Find the probability of getting exactly 52 heads when flipping a fair coin 100 times. Toss the coin twice. After 7 times we. If we look at the three choices for the coin flip example, each term is of the form: C m pmqN-m m = 0, 1, 2, N = 2 for our example, q = 1 - p always! coefficient C m takes into account the number of ways an outcome can occur without regard to order. But the result over many tosses is predictable, as long as the trials are independent (i. Example 8 Tossing a fair coin. If it is tails, it is 0/1. We roll two dice, hoping for a 2 on one and a 5 on the other. In case you flip the coin 2 times, finding the probability of getting exactly 3 tails. On tossing a coin 1000 times, the head appeared 465 times and. The probability is relatively high, but this scenario still seems very unlikely! 4. Formula, lesson and practice problems explained step by step. 0546875 ~ 5. To find an odds ratio from a given probability, first express the probability as a fraction (we'll use 5/13 ). Coin Toss Probability Calculator. A card drawn from a deck cannot be an ace and a queen. Probability is the study of regularities that emerge in the outcomes of random experiments. probability • Example: Toss two coins. We flip a fair coin 10 times. Probability using Probability Trees. That strategy isn't likely to do the job! Consider which you're flipping two coins at the identical moment. The probability is relatively high, but this scenario still seems very unlikely! 4. What if we flip the coin twice? Calculate the probability of obtaining exactly 1 odd number on 4 spins of the arrow. If the coin is tossed 3 times, what is the probability that at least one of the tosses will turn up tails? 0. Problem 3 : Four coins are tossed once. The results of different trials are independent. When 3 coins are tossed, the possible outcomes are {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}. Each coin toss does not affect the outcome of further tosses. Flip a coin 20 times if head comes 8 times, tail comes 12 times then the probability of heads P(H) = 8/20 = 2/5=0. p is the probability of. Coin toss Probability Calculator - 1 unbiased coins are tossed. Of those two outcomes. Most coins have probabilities that are nearly equal to 1/2; For instance, flipping an coin 6 times, there are 2 6, that is 64 coin toss possibility. In a coin-toss experiment, there are two outcomes: heads and tails. (Or) The probability of three half - rupee coins falling all heads up when tossed simultaneously is. That can also be expressed as 0. So, if you do flip a coin 10 times and see 3 heads, that's a pretty common outcome and you can't conclude that the coin is unfair. What is the probability that you will get heads more than 14 times?. The probability that you get exactly half heads and half tails approaches 0. First, note that the problem will likely make reference to a "fair" coin. 019351109194852834 Method 2: Probability of getting 10 heads for one individual = 10!/(10!0!)*0. But I think we can all agree that if we flip a coin 100 times it's very, very likely that we'll get heads at least one of those times. The Wizard of Odds answers readers' questions about Sports. Hint: define a random variable and then find its pmf. What is Probability? In Mathematics, a probability is a branch that deals with calculating the likelihood of the occurrence of the given event. 5) Null: P h e a d s = p = 0. ) (Enter the probability as a fraction. 2^3 = 8 possible arrangements. This question is different because you can get an odd number any time. Why? Because if 90% of the sequences contain 55% or more heads, if we take the coin and toss it n_toss times,in 90% of the cases we will get one of those sequences that contain 55% or more heads. What is the probability that at least one of the three marbles drawn be black, if the first marble is red? Answer: Given A bag contains 5 red marbles and 3 black marbles If the first marble is red, the following conditions have to be followed for at least one marble to be black. Ex) You flip a coin two times. Each coin toss does not affect the outcome of further tosses. Mathematically, probability (P) = For a coin toss, we can calculate the probability that heads will result from one toss. An even simpler example of probability in action is a coin toss. P(at least three draws to win) = 1 – P(win in two or fewer draws) = 1 – 7/16 = 9/16. What is the probability of getting at least 4 heads?. Now, getting no Heads is the same as getting Tails $10$ times in $10$ flips. This can be calculated by multiplying the number of flips (10) by the probability of getting heads on one flip (½), yielding an expected value of 5. It can be written as a fraction, a decimal, or a percent. Statistics and probability: 1-1 1. An example of a Bernoulli process is coin flipping. 5 if the coin is fair) and a number of times to flip the coin. And depending on the payout structure, one side might or might not have an edge over the other side. We roll two dice, hoping for a 2 on one and a 5 on the other. The probability of getting exactly 3 heads out of 8 with a fair coin would be 8C3 / 2^8 = 56 / 256 =. Computing and following an exact decision tree increases earnings by $6. P(drawing 3 coins and getting 1 of each). If playback doesn't begin shortly, try restarting your device. How many consecutive tails would you need to establish that a coin is biased ( P h e a d s < 0. A biased coin that lands heads with probability 0. It must follow ∣S ∣ = 32. Probability With Tosses Of 5 Coins Unfortunately in biology, sex ratios in humans are not that easily explained. A coin tossed has two possible outcomes, showing up either a head or a tail. Note: Probability is a funny thing. Step 3: The probability of getting the head or a tail will be displayed in the new window. An example of a Bernoulli process is coin flipping. First, note that the problem will likely make reference to a "fair" coin. In the coin example the "experiment" was flipping the coin 100 times. 019351109194852834 Method 2: Probability of getting 10 heads for one individual = 10!/(10!0!)*0. a die and flipped a coin. For example, the probability of getting two or fewer successes when flipping a coin four times (p = 0. But first find the sample space of what you are computing. Practice this lesson yourself on KhanAcademy. So the probability is 1 in 8. Course Description. In the second toss, these probabilities still hold. This example shows using the Binomial distribution to predict the probability of heads and tails when throwing a coin. Each flip is 50/50 (unless you shave the edge). Now we will look at the probability of either event occurring. Sum the values of P for all r within the range of interest. On the other hand, what is the probability of rolling a sum less than six given that we have rolled a three? The probability of rolling a three and a sum less than six is 4/36. However, that isn't the question you asked. So, no we know that the range of the function we call the probability is a subset of the interval [0,1]. We say that the sequence is balanced when there are equal number of heads and tails. What is the probability of exactly 2 heads? b. Find the probability that the card is a club or a face card. Flip a coin 20 times if head comes 8 times, tail comes 12 times then the probability of heads P(H) = 8/20 = 2/5=0. Find the probability that there are 3 Heads in the first 4 tosses and 2 Heads in the last 3 tosses. The intersection of events A and B, written as P(A ∩ B) or P(A AND B) is the joint probability of at least two events, shown below in a Venn diagram. In this course, you'll learn about fundamental probability concepts like random variables (starting with the classic coin flip example) and how to calculate mean and variance, probability distributions, and conditional probability. Let us toss a biased coin producing more heads than tails, p=0. Thus, our probability of making a profit on a (short or long) position is 50%, which is the same as a coin flip. This probability calculator by Calculators. If I flip the coin four times, what is the probability of obtaining a heads one or more times across all four flips? · For two coin flips, the probability of not obtaining at least one heads (i. Example: A coin is biased so that it has 60% chance of landing on heads. What is the probability of getting two heads in two tosses? The probability that the coin when tossed turns up heads is 1/2. A classic example of this is a coin toss, where there can be two possible options: heads or tails. Do not simply state which type of random variable it is and then copy its pms. Homework Statement:: 15 people flip 2 coins each. The 1 is the number of opposite choices, so it is: n−k. To solve this problem, we need to find the probabilities that r could be 3 or 4 or 5, to satisfy the condition "at least". When a coin is tossed, there is a chance of getting either a heads or a tails and hence the chances are 50% percentfor each. If the coin is a fair toss (the coin is not “loaded” nor thrown in some fashion that predisposes one face to preferentially land up, and rare events such as landing on edge are excluded) then there is a probability of 1/2 of getting heads (h) and a probability of 1/2 of getting tails (t). About Press Copyright Contact us Creators Advertise Developers Terms Privacy Policy & Safety How YouTube works Test new features Press Copyright Contact us Creators. Coins And Probability Trees. We now calculate the same probability by using the complement rule. The dice are meant to introduce an element of chance to these games; we expect that the outcomes of the rolls will be truly random. When you roll two dice, you have a 30. The probability of an event A is the number of ways event A can occur divided by the total number of possible outcomes. (a) Write down the sample space of this experiment. You must derive the pm and show your work. Answers ( 2) Since the coin has two sides heads or tails. This is because there is a 1 in 100 chance of picking the two-headed coin, and if you do the probability is 100% of flipping 10 heads in a row. Since the probability to flip a head is the same as the probability to flip a tail, the probability of outcome (i) must be equal to the probability of outcome (ii). The best way to understand Bernoulli trials is with the classic coin toss example. While all tosses are identical keep tossing until you get the opposite outcome. To compute the probability of exactly 8 successes, select Calc > Probability Distributions > Binomial. The probability of getting exactly 3 tails when a coin is tossed 2 times. 05 to establish bias at 95% confidence one sided. An example of a binomial experiment is tossing a coin, say thrice. A variable in such a sequence may be called a Bernoulli variable. A Random Variable is a set of possible values from a random experiment. What is the probability that we get heads in at least 8 of the 10 flips?. To solve this problem, we need to find the probabilities that r could be 3 or 4 or 5, to satisfy the condition "at least". org DA: 12 PA: 34 MOZ Rank: 63. This is part of a wider doctrine of "the maturity of chances" that falsely assumes that each play in a game of chance is connected with other events. A fair coin is flipped 7 times. Kids, always do a reality check. We can combine both coin flip and roll of dice into a single probabilistic experiment, and tree diagrams help visualize and solve such questions. a die and flipped a coin. Coin toss Probability Calculator - 1 unbiased coins are tossed. Each team member will have 1 coin to flip. So we can say that the probability of getting an ace is 1/13. Be careful with how you read this probability. For example, suppose we flip a coin and get Head, then we flip the coin again this time we get Tail. Sorry for the verbal equations. Call heads a success. Probability is a way to quantify uncertainty. 5, then A will have won after scenario 2 (which happens with probability y). Easycalculation. The number of correct answers (say heads), X, is distributed as a binomial random variable with binomial. 126 to see a difference of 40 during the test. Bonus Question. Question 2:. Do not simply state which type of random variable it is and then copy its pms. Trials, n, must be a whole number greater than 0. In other words, we're finding the probability that a probability is what we think it should be. Hence if we calculate probability of getting Heads exactly once and probability of not getting Heads at all and subract it from the total probability of the event which is 1 (As total probability of certain event will be always 1) we can get the probability of. In the United States, there is a slightly better chance of having a boy, about 105 males to 100 females. Let A be the event that there are 6 Heads in the first 8 tosses. There are 4 nickels, 3 dimes, and 5 quarters in a purse. Now let's consider coin n+1. To calculate the actual probability of the coin landing on this side would take some fairly complicated physics though. Second toss, HH HT TH TT (example:first toss was H, second could be H or T and so on). What is the probability that we get heads in at least 8 of the 10 flips?. 5 for both heads and tails. In the case where A and B are mutually exclusive events, P(A ∩ B) = 0. This is about 2+ standard deviations over, and goes with around 2% with the Normal distribution. Therefore the probability of getting at least one 20 toss succession of heads = "1-Y". 075% chance of seeing a streak of 22 heads at some point. Free Online Scientific Notation Calculator. Probability: Types of Events. 5), and we flip it 3 times. The ratio of successful events A = 968 to the total number of possible combinations of a sample space S = 1024 is the probability of 3 heads in 10 coin tosses. Use the calculator below to try the experiment. You purchase a certain product. In this notebook, we illustrate NumPy features for working with discrete probability distributions, such as those resulting from a coin toss or a dice roll. With probability 1−p the result is Tails, and then X is generated according. Intuitively, this means that CDF (x) equals the probability that the expectation of a coin flip is ≤ x. The calculator reports that the binomial probability is 0. What is the probability of getting at least 2 heads?. At 34 or more correct guesses you are “beating the odds" by (34–26. When we flip a fair coin, we say that there is a 50 percent chance (probability = 0. This number is always between 0 and 1, where 0 indicates impossibility and 1 indicates certainty. My calculator. 999 Probability of getting 0 or 1 heads is. Designed using Canva. The probability of drawing an Ace from a standard deck is 0. Flipping one fair coin twice is an example of an experiment. Do not simply state which type of random variable it is and then copy its pms. If the probability of an event is high, it is more likely that the event will happen. For example, suppose we flip a coin n = 100 times, the probability that it lands on heads in a given trial is p = 0. After accessing the statistics probability calculator on our site, follow the steps mentioned below: Click on the Multiple button to access the probability calculator multiple events. In the United States, there is a slightly better chance of having a boy, about 105 males to 100 females. If it is tails, it is 0/1. It is also written as P(A). Show from first principles that P ( a b ∧ a) = 1. That is the probability of getting EXACTLY 7 Heads in 12 coin tosses. A conditional probability is the probability of one event if another event occurred. Second toss, HH HT TH TT (example:first toss was H, second could be H or T and so on). CHAPTER 4: DISCRETE PROBABILITY DISTRIBUTIONS USING PDF TABLES EXAMPLE D3: At the county fair, a booth has a coin flipping game. Probability is: (Number of ways it can happen) / (Total number of outcomes) Dependent Events (such as removing marbles from a bag) are affected by previous events. You pay$1 to flip three fair coins. The toss of a coin, throw of a dice and lottery draws are all examples of random events. Hypothesis Testing. (b) Calculate the probability of the second toss landing on head. Coin Toss Probability Calculator. 5, which means we would not be able to tell the different between a bias coin and fair coin 50% of the time. Let A be the event that the coin shows heads at least 4 times. Enter the beta distribution. If you toss a coin exactly three times, there are 8 equally likely outcomes, and only one of them contains 3 consecutive heads. Consider flipping a fair coin several times. An example of a binomial experiment is tossing a coin, say thrice. We label as “H” the event of getting a head, and as “T” the event of getting a tail. While all tosses are identical keep tossing until you get the opposite outcome. In a large discrete math class, 55% of the students. Do not simply state which type of random variable it is and then copy its pms. Enter the total number of heads or tails you want to calculate the probability of into the calculator to determine the chance of getting that amount. I want to calculate the experimental probability of heads in coin toss by generating 0s or 1s randomly and assign 'h' for 0 and 't' for 1 to n. (a) Write down the sample space of this experiment. What if we flip the coin twice? Calculate the probability of obtaining exactly 1 odd number on 4 spins of the arrow. E={2,4,6}→n(E)=3 We now use the formula of the classical probability. The probability of getting any number face on the die. In this article, we are going to study to solve problems to find the probability involving the throwing of two dice. 9, how many times should the die be tossed. Calculate the probability of drawing ANY PAIR in a row from a deck of cards (with replacement). 5 because 2 outcomes (heads or tails) are equally possible when a balanced coin is flipped. Probabilities are usually given as fractions. 7 is the probability of each choice we want, call it p. For instance, flipping an coin 6 times, there are 2 6, that is 64 coin toss possibility. Exercise 8. There are 32 sample solutions in the solution set of the 5 coin toss. An example of a Bernoulli process is coin flipping. for i in range (1000): if flip_coin(8) == "3": ## changed to flip_coin() multiple_heads_count += 1 The value of flip_coin(8) is an integer, but you are checking for equality with the string "3". When 3 coins are tossed randomly 250 times and it is found that three heads appeared 70 times. If THREE coins are flipped, write the sample space. Calculate the probability of flipping a coin 5 times with at least 3 heads This is equivalent to saying (3 or 4 or 5) heads List out ways to flip 3 heads and 2 tails HHHTT HHTHT HHTTH HTHHT HTHTH HTTHH THHHT THHTH THTHH TTHHH List out ways to flip 4 heads and 1 tail HHHHT HHHTH HHTHH HTHHH THHHH List out ways to flip 5 heads and 0 tail HHHHH. In this article, we are going to study to solve problems to find the probability involving the throwing of two dice. For example, the probability of flipping a coin and it being heads is ½, because there is 1 way of getting a head and the total number of possible outcomes is 2 (a head or tail). If we flip four coins, how many of the possible outcomes will have exactly two heads? From the four coins, we want to choose two of them to be heads - the remaining two must therefore be tails. The formula for working out an independent probability is quite simple: P (A) = N/0. As I have written in the comment the answers seems to be. This can be calculated by multiplying the number of flips (10) by the probability of getting heads on one flip (½), yielding an expected value of 5. of all possible outcomes = 2 x 2 x 2 x 2 = 16. So it would just be $1/8$ if you were flipping the coin $3$ times. When tossing a fair coin, if heads comes up on each of the first 10 tosses, what do you think the chance is that another head will come up on the next toss? 0. probability. If we look at the three choices for the coin flip example, each term is of the form: C m pmqN-m m = 0, 1, 2, N = 2 for our example, q = 1 - p always! coefficient C m takes into account the number of ways an outcome can occur without regard to order. So, if you do flip a coin 10 times and see 3 heads, that's a pretty common outcome and you can't conclude that the coin is unfair. ) Answer link. The number of correct answers (say heads), X, is distributed as a binomial. Now, coming back to the question we have to find the probability of getting at least k heads in N tosses of coins. That is the probability of getting EXACTLY 7 Heads in 12 coin tosses. " So it is no wonder that coin flip probabilities play a central role in understanding the basics of probability theory. For example, if two coins are tossed in the air at the same time, the number of outcomes that satisfy the condition of a coin landing on heads at least once is 3. (i) none (ii) not more than one (iii) more than one (iv) at least one will fuse after 150 days of use. On tossing a coin, the probability of getting a head is: P(Head) = P(H) = 1/2. 50/50 each time. What is the probability of exactly 2 heads? b. Lets write a function to calculate the statistical power for different values of. If you toss the coin 5,000 times you will see at least one run of ten heads 99. Calculate the mean and standard deviation of X = number of heads. A coin tossed has two possible outcomes, showing up either a head or a tail. An experiment is a planned operation carried out under controlled conditions. The coin does not care what the previous 155 trials were. So we can say that the probability of getting an ace is 1/13. After you have flipped the coin so many times, you should get answers close to 0. Probability is the study of regularities that emerge in the outcomes of random experiments. Penny flipping - calculate winning probability (easy) Write a function that is at least twice as fast as the test suite call of repmat(). To calculate the probability of an event occurring, we count how many times are event of interest can occur (say flipping heads) and dividing it by the sample space. 7 is the probability of each choice we want, call it p. If you flip a coin 9 times, you get a sequence of Heads (H) and tails (T). 999999 $For more info see this question. What is the probability that we get heads in at least 8 of the 10 flips?. the coin N, defined by cos w ¼ N ð 0. We flip a fair coin 10 times. MathCelebrity. Let A be the event that there are 6 Heads in the first 8 tosses. Whether you want to toss a coin or ask a girl out, there are only two possibilities that can occur. 5, and we want to know the probability that it will land on heads k = 43 times or less: p. Assuming this is a fair coin, the probability for not getting a "heads" in a given throw is p = 1 2, and is independent of all other throws. Coin Toss Probability. The probability of getting a given number of heads from four flips is, then, simply the number of ways that number of heads can occur, divided by the number of. To find the probability of at least two tails, we mark each row (outcome) that contains two tails or three tails and divide the number of marked rows by 8 (number in the sample space) Since there are four outcomes that have at least two tails, the probability is 4/8 or ½. A student cannot fail and pass a class. For example, if two coins are tossed in the air at the same time, the number of outcomes that satisfy the condition of a coin landing on heads at least once is 3. When you roll two dice, you have a 30. Be careful with how you read this probability. , getting tails both times) is 0. The coin being a fair one, the outcome of a head in one toss has a probability $$p = 0. A Random Variable is a set of possible values from a random experiment. · T he probability of one or more heads in two coin flips is 1 - 0. The Gambler's Fallacy is the misconception that something that has not happened for a long time has become 'overdue', such a coin coming up heads after a series of tails. Read Book Probability Concepts In Engineering By Alfredo Recognizing the artifice ways to get this books probability concepts in engineering by alfredo is additionally useful. Coin Toss Probability Calculator. Show Video Lesson. Check the box to show a line with the true probability on the graph. The probability that the card is a. To solve this problem, we need to find the probabilities that r could be 3 or 4 or 5, to satisfy the condition "at least". 5, 50%, or 1 to 1. I have a rather complex game, whose expected value I need to find. (a) Two heads occur, given that the first toss is a head. randint() you could have any probability of bias while still maintaining randomness. ’ ‘The coin is just as likely to land heads as tails. We now calculate the same probability by using the complement rule. What is the probability of obtaining exactly 3 heads. Similarly, on tossing a coin, the probability of getting a tail is: P(Tail) = P(T) = 1/2. You must derive the pm and show your work. 5 % chance at least one 6 will appear. (a) Write down the sample space of this experiment. We toss two coins* this experiment involves two parts, 'the first toss of the coin' and 'the second toss of the coin’: experiments that have two parts can be represented in two ways Tree diagramm Tabular form *It notes that: “tossing two different coins “ or “tossing the same coin two times” is the same experiment!. In die and coin problems, unless stated otherwise, it is assumed coins and dice are fair and repeated trials are independent. Consider flipping a fair coin several times. Since the outcome of flipping a coin is independent for each flip, the probability of a head or tail is always 0. Suppose we flip a fair coin n times. 6 over a modified KC. Hence if we calculate probability of getting Heads exactly once and probability of not getting Heads at all and subract it from the total probability of the event which is 1 (As total probability of certain event will be always 1) we can get the probability of. (c) Two heads occur, given at least one head occurs. 5 = the proportion of times you get heads in. The majority of times, if a coin is heads-up when it is flipped, it will remain heads-up when it lands. The ratio of successful events A = 9 to the total number of possible combinations of a sample space S = 256 is the probability of 7 heads in 8 coin tosses. Although we can't tell beforehand the outcome of a coin toss, we're able to at least estimate the probability (the chances) of a coin landing on heads or tails. (A) Give the range R. Ex) You flip a coin two times. Let your coin be X 1 and denote sum of heads as S. Now what is the conditional probability: that you picked the fake coin? \item Suppose you wanted to decide whether the chosen coin was fake: by flipping it \(k$$ times. By using random. 666666% chance to get at least one tails Option 2 of calculating it would mean my odds are 75% chance to get at. same as (a) except now the coin is flipped 10 times. A bag contains 6 red Bingo chips, 4 blue Bingo chips, and 7 white Bingo chips. Coin toss Probability Calculator - 1 unbiased coins are tossed. Remember that Tails, Heads, Heads is a different outcome than Heads, Heads, Tails even though both result in one tail and two heads. Find an answer to your question “If you flip a coin and roll a 6-sided die, what is the probability that you will flip a heads and roll at least a 3? ” in 📘 Mathematics if you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions. Dice Probability Formulas. With 1000 flip --> p=1-0. On your first flip, it lands on heads. If we flip four coins, how many of the possible outcomes will have exactly two heads? From the four coins, we want to choose two of them to be heads - the remaining two must therefore be tails. 4)$) Using the formula given in the link we get a number extremely close to 1. The probability of A and B is 1/100. If both coins show heads (HH) or both coins show tails (TT), player 1 gets 1 point. Picking numbers randomly means that there is no specific order in which they are chosen. So the probability of getting the one sequence among them that contains exactly N heads is 1 in 2 N. A variable in such a sequence may be called a Bernoulli variable. [3 pts] having chosen the fair coin given that both tosses were heads. N is the Number of ways an event can occur and. P ( club or face card) = P ( club) + P ( face card) − P ( club and face card) = 13 52 + 12 52 − 3 52 = 22 52 = 11 26 ≈ 0. Now the final step is negating back — the probability of getting at least 1 "heads" is: 1 − p n = 1 − 1. We covered independent. Same as this: LeetCode All in One 题目讲解汇总(持续更新中) Note: All explanations are written in Github Issues, please do not create any new issue or pull request in this project since the problem index should be consistent with the issue index, thanks!. Thus, our probability of making a profit on a (short or long) position is 50%, which is the same as a coin flip. This means that: (1) If a person has illegal drugs on them, 80% of the time the dog will correctly identify the drugs and start barking, and 20% of the time the dog will miss the drugs and not bark. A probability of 0 means that there is zero chance that the event will occur; a probability of 1 means that the event is certain to occur. Enter the number of possible outcomes. The probability of each outcome is 0. If the coin is a fair toss (the coin is not “loaded” nor thrown in some fashion that predisposes one face to preferentially land up, and rare events such as landing on edge are excluded) then there is a probability of 1/2 of getting heads (h) and a probability of 1/2 of getting tails (t). Find the probability of at least one head appearing. What is the probability that we get heads in at least 8 of the 10 flips?. Are these two outcomes mutually exclusive in one coin flip? Yes, they are. Show Video Lesson. In the United States, there is a slightly better chance of having a boy, about 105 males to 100 females. Since the probability to flip a head is the same as the probability to flip a tail, the probability of outcome (i) must be equal to the probability of outcome (ii). The Math Behind a Coin Toss. That is to say, there is 50% chance of getting either. To find the probability of at least two tails, we mark each row (outcome) that contains two tails or three tails and divide the number of marked rows by 8 (number in the sample space) Since there are four outcomes that have at least two tails, the probability is 4/8 or ½. Set the probability of heads (between 0 and 1. Looking at the event we just talked about, the event of “tails at least once” could be called E and written as. Calculate the probability of flipping at least one head on three coin flips. A variable in such a sequence may be called a Bernoulli variable. When we flip a coin, only two outcomes are possible - heads and tails. If you mark a result of a single coin flip as H for heads or T for tails all results of 3 flips can be written as: Omega={(H,H,H),(H,H,T),(H,T,H),(H,T,T),(T,H,H),(T,H,T),(T,T,H),(T,T,T)} Each triplet contains results on 1st, 2nd and 3rd coin. Designed using Canva. For example, if two coins are tossed in the air at the same time, the number of outcomes that satisfy the condition of a coin landing on heads at least once is 3. P (HTT + THT + TTH) = 0. Let's go back to our example of flipping a coin. The desired probability is the probability that exactly 3 heads are observed among 5 coin flips given that at least one head is observed. 75K subscribers. Flipping one fair coin twice is an example of an experiment. How to calculate probability? "Hey man, but girls and coins are two different things!I should know, I've seen at least one of each. 04 is the probability of getting 7 Heads in 8 tosses. The probability is 0. All tosses are tails. 075% chance of seeing a streak of 22 heads at some point. After accessing the statistics probability calculator on our site, follow the steps mentioned below: Click on the Multiple button to access the probability calculator multiple events. So, for our coin-flipping example, the percentage of heads we would expect to see if we flip the coin 100 times is approximately 0. Note: For disjoint events P (A and B) = 0, so the above formula simplifies to P (A or B) = P (A) + P (B) Probability distributions. Odds can then be expressed as 5 : 8 - the ratio of favorable to unfavorable outcomes. E={2,4,6}→n(E)=3 We now use the formula of the classical probability. (a) Write down the sample space of three such flips. The probability is relatively high, but this scenario still seems very unlikely! 4. When flipping a coin, is the probability of at least two tails complementary to at most one tail, if flipped three times in total? Ask Question Asked 4 years, 3 months ago. only one is heads. Sometimes probability problems involve solving problems for multiple cases. This article shows you the steps for solving the most common types of basic questions on this subject. A biased coin that lands heads with probability 0. 5), and we flip it 3 times. In this article, we are going to study to solve problems to find the probability involving the throwing of two dice. So the probability that at least one person wins in one million plays is: 1 - (1-p) 1,000,000 = 1. Penny flipping - calculate winning probability (easy) Write a function that is at least twice as fast as the test suite call of repmat(). Exercise 1. If it is thrown three times, find the probability of getting: (a) 3 heads, (b) 2 heads and a tail, (c) at least one head. Calculate the probability that head turns up at least 2 times. The probability of rolling a 1 or a 2: P(1) + P(2) = 1 6 + 1 6 = 2 6 ˇ0:33. Say I make it so that the 2 coin flips count as a single number 1,2,3,4 representing head-head, head-tails, tails-head, tails-tails. An ideal unbiased coin might not correctly model a real coin, which could be biased slightly one way or another. A combination of two or more ~ s (e. The complement of the event "we flip at least one head" is the event "there are no heads. Since there are 4 ways that we can get two or more consecutive heads out of the 8 total, the probability is 4 8 = 1 2. Each outcome has a fixed probability, the same from trial to trial. In probability. I have a rather complex game, whose expected value I need to find. "At least one" probability with coin flipping. If it is thrown three times, find the probability of getting: (a) 3 heads, (b) 2 heads and a tail, (c) at least one head. We will begin with a classical probability example of tossing a fair coin three times. probability • Example: Toss two coins. In this video, we' ll explore the probability of getting at least one heads in multiple flips of a fair coin. For example, we want at least 2 heads from 3 tosses of coin. There is one bit of uncertainty; the probability of a head, written p(h), is 0. The probability of rolling a 1 or a 2: P(1) + P(2) = 1 6 + 1 6 = 2 6 ˇ0:33. The outcomes of each toss will be reflected on the graph. Now, determine the probability of drawing an Ace with the help of Python: # Sample Space cards = 52 # Outcomes aces = 4 # Divide possible outcomes by the sample set ace_probability = aces / cards # Print probability rounded to two decimal places print (round (ace_probability, 2)) 0. Example: Suppose you plan to toss a coin twice, and want to find the probability of rolling a head on both tosses. This probability is the power of the test. Here the possibility of flipping a head or a tail on a single toss is 50%. the coin can also land upright. We say that the sequence is balanced when there are equal number of heads and tails. Remember that each individual coin flip has a 50% chance of being heads. Both outcomes are equally likely. For 100 flips, if the actual heads probability is 0. For example, to have coin that is biased to produce more head than tail, we will choose p < 0. It shows that when you flip a fair coin 10 times, you can pretty much get any outcome with reasonable probability. Coin Toss Probability. The alternative is using a Z Table but Excel makes it much easier and quicker to calculate probability when the specific mean and standard deviation numbers are available. 25) = 15/sqrt (53). By looking at the events that can occur, probability gives us a framework for. Hypothesis Testing. 5 (or 1/2), and so is the probability of getting heads on a second toss of the same coin. If I was flipping two coins, one event is that I get tails at least once. If playback doesn't begin shortly, try restarting your device. Coins And Probability Trees. If THREE coins are flipped, write the sample space. Select the input you want to use to find the probability and enter the value. Over many coin flips the probability of at least half of the flips being heads (or tails) will converge to 0. For example, if the experiment is to flip one fair coin, event A might be getting at most one head. Calculate the probability of flipping a coin toss sequence with this Coin Toss Probability Calculator. Most of the time that information is given with the question. 666666% chance to get at least one tails Option 2 of calculating it would mean my odds are 75% chance to get at. Let us learn more about the coin toss probability formula. The majority of times, if a coin is heads-up when it is flipped, it will remain heads-up when it lands. Now what is the conditional probability: that you picked the fake coin? \item Suppose you wanted to decide whether the chosen coin was fake: by flipping it $$k$$ times. Tap to unmute. The probability of something which is certain to happen is 1. For example: the probability of getting a head’s when an unbiased coin is tossed, or getting a 3 when a dice is rolled. You pull a red marble randomly out of the bag. Find each probability. A biased coin lands heads with probability 1 10. We are tossing a fair coin and suppose we have tossed it 9 times already. What is the probability that we get heads in at least 8 of the 10 flips?. For instance, if we wanted to know what the chance of getting at least 10 heads in 100 flips where the probability of getting heads is 0. The probability. However, that isn't the question you asked. Ex) You flip a coin two times. Calculate the probability of flipping at least one head on three coin flips. Coins And Probability Trees. And we have (so far): = p k × 0. If heads is the number of particular chance events of interest, then the numerator is simply “1. What is the probability that a sum of 5 is rolled a) exactly 6 times b) at least 4 times c) at most 5 times 5. A classic example of a probabilistic experiment is a fair coin toss, in which the two possible outcomes are heads or tails. Find the probability that the coin comes up tails at. so let's define a random variable X as being equal to the number of heads I'll just write capital H for short the number of heads from from flipping coin from flipping a fair coin we're going to assume it's a fair coin from flipping coin five times five times and so like all random variables this is taking particular outcomes and converting them into numbers and this random variable it could. Course Description. Given that more heads than tails appear, what is the probability that all of the flips are H? c. For instance take a look at optimizely’s sample size calculator. While all tosses are identical keep tossing until you get the opposite outcome. In the coin example the "experiment" was flipping the coin 100 times. (a) Write down the sample space of three such flips. Probability (1) Write down the sample space for tossing a coin 3 times. Example 8 Tossing a fair coin. A coin is flipped 5 times. Calculate the conditional probability of 5 heads, knowing that there were at least 4 heads. Coin toss probability calculator helps us find the probability of getting either heads or tails when a coin is tossed the given number of times. 5 because of the law of large numbers. small in comparison with the total population) is taken when performing the random sampling of the population, the analysis is similar to determining the probability of obtaining heads in a coin toss. By looking at the events that can occur, probability gives us a framework for. Consider 10 independent tosses of a biased coin with the probability of Heads at each toss equal to p, where 0 0. p is the probability of. What is the probability of getting two heads in two tosses? The probability that the coin when tossed turns up heads is 1/2. We only get to this point 1/8 times. Let's call that "Z". 126 to see a difference of 40 during the test. Consider flip a coin 5 times. Coin toss probability is explored here with simulation. The sum of the probabilities of. Formula, lesson and practice problems explained step by step. What is the probability that we get heads in at least 8 of the 10 flips?. (b) Find the probability of getting a tail. The Gambler's Fallacy is the misconception that something that has not happened for a long time has become 'overdue', such a coin coming up heads after a series of tails. Thus, the probability of getting heads at least once during two tosses of the coin is. Example: Suppose you plan to toss a coin twice, and want to find the probability of rolling a head on both tosses. Most coins have probabilities that are nearly equal to 1/2; For instance, flipping an coin 6 times, there are 2 6, that is 64 coin toss possibility. E XAMPLE Toss a fair coin twice What is the probability of observing at least from MTH 380 at Ryerson University. We will call an individual coin flip a trial, and so our experiment consists of ten identical trials. Coin Toss Probability Calculator. 001953125 Calculate the probability of flipping a coin toss sequence of THTTTTTTT. Otherwise there is no prize. 5, then A will have won after scenario 2 (which happens with probability y). Conditional Probability Calculator. 5 coming up heads (or tails): a. Every time you toss a coin, you have an equal probability of the coin landing either heads or tails (since this is a mathematical exercise, we won’t consider the chance of the coin landing on its edge!). In this course, you'll learn about fundamental probability concepts like random variables (starting with the classic coin flip example) and how to calculate mean and variance, probability distributions, and conditional probability. The result of a coin toss Probability of an event A: denoted by P(A). Of those two outcomes. What is the probability that at least one of the three marbles drawn be black, if the first marble is red? Answer: Given A bag contains 5 red marbles and 3 black marbles If the first marble is red, the following conditions have to be followed for at least one marble to be black. Example 5 Find the probability that at least 5 heads show up when a fair coin is tossed 7 times. P(at least three draws to win) = 1 – P(win in two or fewer draws) = 1 – 7/16 = 9/16. Probability is the study of regularities that emerge in the outcomes of random experiments. This example shows using the Binomial distribution to predict the probability of heads and tails when throwing a coin. This figure can also be figured out mathematically, without the use of the graphic. While all tosses are identical keep tossing until you get the opposite outcome. If it is tails, it is 0/1. In this notebook, we illustrate NumPy features for working with discrete probability distributions, such as those resulting from a coin toss or a dice roll. If it is heads, then the experimental probability is 1/1. Example 1: A fair coin is tossed 5 times. This is an "and" situation. This is wrong since I KNOW the answer is 1/6. A coin is flipped 5 times. For example, we want at least 2 heads from 3 tosses of coin. Example 9: When an unbiased coin is tossed, (a) Find the probability of getting a head. Click on the 'Reset' button to start again. For example, the probability of the union of the mutually exclusive events and in the random experiment of one coin toss, (), is the sum of probability for and the probability for , () + ().
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2021-08-04 12:32:09
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http://blog.coelho.net/system/2013/09/01/postgresql-archeology.html
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01 September 2013
In a previous post I have ported a Turing Machine (TM) in SQL down to PostgreSQL 7.3.
I report here my fruitless attempts at running previous versions of PostgreSQL on a Debian or Ubuntu Linux. They would not configure and/or compile, at least with the minimal effort I was ready to undergo: writing portable-to-the-future system-dependent C code does not work. Basically, the code from 15 years ago is lost unless you run it on a system from 15 years ago, which may require a hardware from 15 years ago as well, or maybe a VM.
Trying to compile old PostgreSQL versions…
Here is a summary of failures encountered while trying to compile previous versions from available sources:
PostgreSQL 7.2 (2002)
configure thinks that my flex 2.5.35 is flex 2.5.3 and reports an issue. Then the compilation fails, possibly because of an include issue, on:
hba.c:885: error: storage size of ‘peercred’ isn’t known
Trying again with an older Debian Lenny, it fails when linking postgres:
/usr/bin/ld: errno: TLS definition in /lib/libc.so.6 section .tbss mismatches non-TLS reference in commands/SUBSYS.o
collect2: ld returned 1 exit status
PostgreSQL 7.1 (2001)
configure fails, it does not recognize gcc 4.7.2. autoconf fails on m4 while processing configure.in. Same result with the old Debian Lenny.
...
/usr/bin/m4:configure.in:930: non-numeric argument to builtin _m4_divert_raw'
autom4te: /usr/bin/m4 failed with exit status: 1
PostgreSQL 7.0 (2000) and 6.5 (1999)
configure fails, but autoconf works fine. Fix manually string continuations in src/include/version.h. ld fails in the end with a segfault while linking postgres:
collect2: ld terminated with signal 11 [Segmentation fault]
PostgreSQL 6.0 (1997)
Create src/Makefile.custom, then make. Compilation fails, possibly because of include issues:
ipc.c:197: error: storage size of ‘semun’ isn’t known
Same result on the old Debian Lenny.
Postgres95 1.08 (1996)
Edit src/Makefile.global, then make. Compilation fails:
indexam.c:188:1: error: pasting "->" and "aminsert" does not give a valid preprocessing token
`
Conclusion
Although the sources of past versions of PostgreSQL are available, having them up and running on a modern system is another story.
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2017-04-24 20:49:20
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http://math.stackexchange.com/questions/60040/finding-edges-which-do-not-lie-on-the-faces-of-a-planar-graph/60041
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# Finding edges which do not lie on the faces of a planar graph
I have a planar graph $G$ consisting of the edges and vertices $(E,V)$. And so, there are $C=E-V+1$ faces in total. The graph is given as a list of edges.
The question is, how to find the edges that don't lie on the faces of the graph? One way I can think of is to find the spanning tree of the graph, use the non-spanning-tree edges to complete the faces, and get all the edges on the faces. Since we already have the on-face edges, we can obtain the not-on-face edges by subtracting the on-face edges from the all the edges.
But is there a more direct way of doing this?
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Euler's formula is about the number of faces of a graph, not the number of cycles. For example, $K_4$ minus an edge has two faces, but three cycles. – Austin Mohr Aug 27 '11 at 6:01
@Austin, I've updated the terminology – Graviton Aug 27 '11 at 6:22
I think you're asking for the edges that don't lie on any faces at all. So why not just look at the graph and pick out those edges? Of course, that won't work if the graph isn't given to you as a drawing on paper, but you haven't said how the graph is given to you, and there is no good way to answer your question if we don't know how the graph is represented. – Gerry Myerson Aug 27 '11 at 6:34
The graph is being given as a list of edge. – Graviton Aug 27 '11 at 6:55
Euler's polyhedral formula, for the vertices, edges and faces of a (convex) polyhedron states: V + F - E = 2. The graph theory version of this is that V + F - E = 2 for a connected graph embedded in the plane (edges meet only at vertices - plane graphs). For a tree, a connected graph with no circuits, there is one face, the "infinite" or unbounded face. If the graph is not connected you have to modify the formula above to deal with the number of components of the graph. – Joseph Malkevitch Aug 27 '11 at 13:05
## 1 Answer
(Edited as per Ilmari's comment.)
An edge does not lie on a cycle if and only if its removal increases the number of components of the graph.
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Austin, can you elaborate more on this? – Graviton Aug 27 '11 at 6:19
I thought of this, but I don't think this is true. For one, my graph can contain 2 disconnected trees. So it doesn't make sense to say that its removal disconnects the graph. – Graviton Aug 27 '11 at 7:44
@Graviton: If you interpret "disconnects the graph" as "increases the number of connected components" (or as "disconnects the component it belongs to"), then it should still be true. – Ilmari Karonen Aug 27 '11 at 8:46
@Ilmari, even if I can make branch-edge and face-edge into two different components, how can I tell which component is branch-edge type, and which is face-edge type? – Graviton Aug 27 '11 at 8:55
@Graviton For an algorithm, you could try something like this: Remove the edge $xy$ from your graph. Starting with $x$, build any spanning tree of its component. If the spanning tree is built and does not contain the vertex $y$, then it must be that $y$ is not in the same component of $x$ after the removal of the edge $xy$. Thus, you would conclude that $xy$ did not lie on any internal face of the original graph. If the spanning tree does contain $y$, then the edge $xy$ lies on some internal face of the original graph. – Austin Mohr Aug 28 '11 at 19:34
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2015-11-26 18:24:24
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https://bird.bcamath.org/handle/20.500.11824/8/browse?type=title
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Now showing items 1-20 of 127
• #### 5-axis double-flank CNC machining of spiral bevel gears via custom-shaped milling tools -- Part I: modeling and simulation
(Precision Engineering, 2019-11-20)
A new category of 5-axis flank computer numerically controlled (CNC) machining, called \emph{double-flank}, is presented. Instead of using a predefined set of milling tools, we use the shape of the milling tool as a free ...
• #### A Geometric Toolbox for Tetrahedral Finite Element Partitions
(Efficient Preconditioned Solution Methods for Elliptic Partial Differential Equations, 2011-12-31)
In this work we present a survey of some geometric results on tetrahedral partitions and their refinements in a unified manner. They can be used for mesh generation and adaptivity in practical calculations by the finite ...
• #### A hybrid method for inversion of 3D DC resistivity logging measurements
(Natural Computing, 2014-12-31)
This paper focuses on the application of hp hierarchic genetic strategy (hp-HGS) for solution of a challenging problem, the inversion of 3D direct current (DC) resistivity logging measurements. The problem under consideration ...
• #### A modified discontinuous Galerkin method for solving efficiently Helmholtz problems
(Communications in Computational Physics, 2012-12-31)
A new solution methodology is proposed for solving efficiently Helmholtz problems. The proposed method falls in the category of the discontinuous Galerkin methods. However, unlike the existing solution methodologies, this ...
• #### A posteriori error analysis of a stabilized mixed FEM for convectuion-diffusion problems
(AIMS Proceedings, 2015, 2015-12-31)
We present an augmented dual-mixed variational formulation for a linear convection-diffusion equation with homogeneous Dirichlet boundary conditions. The approach is based on the addition of suitable least squares type ...
• #### A posteriori error analysis of an augmented mixed finite element method for Darcy flow
(Computer Methods in Applied Mechanics and Engineering, 2015-12-31)
We develop an a posteriori error analysis of residual type of a stabilized mixed finite element method for Darcy flow. The stabilized formulation is obtained by adding to the standard dual-mixed approach suitable residual ...
• #### A Secondary Field Based hp-Finite Element Method for the Simulation of Magnetotelluric Measurements
(Journal of Computational Science, 2015-12-31)
In some geophysical problems, it is sometimes possible to divide the subsurface resistivity distribution as a one dimensional (1D) contribution plus some two dimensional (2D) inhomogeneities. Assuming this scenario, we ...
• #### A stable discontinuous Galerkin-type method for solving efficiently Helmholtz problems
(Computers and Structures, 2012-12-31)
We propose a stable discontinuous Galerkin-type method (SDGM) for solving efficiently Helmholtz problems. This mixed-hybrid formulation is a two-step procedure. Step 1 consists in solving well-posed problems at the element ...
• #### A study of the apsidal angle and a proof of monotonicity in the logarithmic potential case
(Journal of Mathematical Analysis and Applications, 2014-12-31)
This paper concerns the behaviour of the apsidal angle for orbits of central force system with homogeneous potential of degree $-2 \le \alpha \le 1$ and logarithmic potential. We derive a formula for the apsidal angle as ...
• #### Accessibility for Line-Cutting in Freeform Surfaces
(Computer-Aided Design, 2019-04-26)
Manufacturing techniques such as hot-wire cutting, wire-EDM, wire-saw cutting, and flank CNC machining all belong to a class of processes called line-cutting where the cutting tool moves tangentially along the reference ...
• #### Adjoint-based formulation for computing derivatives with respect to bed boundary positions in resistivity geophysics
(Computational Geosciences, 2019-02)
In inverse geophysical resistivity problems, it is common to optimize for specific resistivity values and bed boundary positions, as needed, for example, in geosteering applications. When using gradient-based inversion ...
• #### An Agent-Oriented Hierarchic Strategy for Solving Inverse Problems
(International Journal of Applied Mathematics and Computer Science, 2015-12-31)
The paper discusses the complex, agent-oriented hierarchic memetic strategy (HMS) dedicated to solving inverse parametric problems. The strategy goes beyond the idea of two-phase global optimization algorithms. The global ...
• #### An IMEX scheme combined with Richardson extrapolation methods for some reaction-diffusion equations
(Idojaras, 2013-12-31)
An implicit-explicit (IMEX) method is combined with some so-called Richardson extrapolation (RiEx) methods for the numerical solution of reaction-diffusion equations with pure Neumann boundary conditions. The results are ...
• #### An IMEX scheme for reaction-diffusion equations: Application for a PEM fuel cell model
(Central European Journal of Mathematics, 2013-12-31)
An implicit-explicit (IMEX) method is developed for the numerical solution of reaction-diffusion equations with pure Neumann boundary conditions. The corresponding method of lines scheme with finite differences is analyzed: ...
• #### Analysis of extreme wave events on the southern coast of Brazil
(Natural Hazards and Earth System Sciences, 2014-12-31)
Using the wave model SWAN (simulating waves nearshore), high waves on the southwestern Atlantic generated by extra-tropical cyclones are simulated from 2000 to 2010, and their impact on the Rio Grande do Sul (RS) coast is ...
• #### Arithmetic method of double-injection-electrode model for resistivity measurement through metal casing
(IEEE Transactions on Geoscience and Remote Sensing, 2010-12-31)
Through-casing resistivity (TCR) measurement instruments such as Cased Hole Formation Resistivity are extensively used for the dynamic monitoring of oil reservoirs during the production phase in oil wells to evaluate the ...
• #### Asymptotic Models for the Electric Potential across a Highly Conductive Casing
(Computers and Mathematics with Applications, 2018-07)
We analyze a configuration that involves a steel-cased borehole, where the casing that covers the borehole is considered as a highly conductive thin layer. We develop an asymptotic method for deriving reduced problems ...
• #### Automatic fitting of conical envelopes to free-form surfaces for flank CNC machining
(Computer Aided Design, 2017-10)
We propose a new algorithm to detect patches of free-form surfaces that can be well approximated by envelopes of a rotational cone under a rigid body motion. These conical envelopes are a preferable choice from the ...
• #### Automatic Red-Channel underwater image restoration
(Journal of Visual Communication and Image Representation, 2015-12-31)
Underwater images typically exhibit color distortion and low contrast as a result of the exponential decay that light suffers as it travels. Moreover, colors associated to different wavelengths have different attenuation ...
• #### Computational cost estimates for parallel shared memory isogeometric multi-frontal solvers
(Computers and Mathematics with Applications, 2014-12-31)
In this paper we present computational cost estimates for parallel shared memory isogeometric multi-frontal solvers. The estimates show that the ideal isogeometric shared memory parallel direct solver scales as \$\mathcal{O}( ...
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2019-12-14 02:57:20
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http://mathhelpforum.com/advanced-applied-math/117840-eigenvectors-discrete-fourier-transform.html
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# Math Help - Eigenvectors for the discrete fourier transform
1. ## Eigenvectors for the discrete fourier transform
I need some help with this homework.
2. Originally Posted by steiner
I need some help with this homework.
Are you allowed to get help - this looks like work that counts towards your final grade.
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2014-07-22 14:17:16
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https://kfp.bitbucket.io/fricas-ug/section-9.25.html
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# 9.25 FileName¶
The FileName domain provides an interface to the computer’s file system. Functions are provided to manipulate file names and to test properties of files. The simplest way to use file names in the FriCAS interpreter is to rely on conversion to and from strings. The syntax of these strings depends on the operating system.
fn: FileName
Type: Void
On Linux, this is a proper file syntax:
fn := "/tmp/fname.input"
“/tmp/fname.input”
Type: FileName
Although it is very convenient to be able to use string notation for file names in the interpreter, it is desirable to have a portable way of creating and manipulating file names from within programs.
A measure of portability is obtained by considering a file name to consist of three parts: the directory, the name, and the extension.
directory fn
“/tmp”
Type: String
name fn
“fname”
Type: String
extension fn
“input”
Type: String
The meaning of these three parts depends on the operating system. For example, on CMS the file SPADPROF INPUT M would have directory M, name SPADPROF and extension INPUT. It is possible to create a filename from its parts.
fn := filename("/u/smwatt/work", "fname", "input")
“/u/smwatt/work/fname.input”
Type: FileName
When writing programs, it is helpful to refer to directories via variables.
objdir := "/tmp"
“/tmp”
Type: String
fn := filename(objdir, "table", "spad")
Type: FileName
If the directory or the extension is given as an empty string, then a default is used. On AIX, the defaults are the current directory and no extension.
fn := filename("", "letter", "")
“letter”
Type: FileName
Three tests provide information about names in the file system.
The exists?exists?FileName operation tests whether the named file exists.
exists? "/etc/passwd"
true
Type: Boolean
readable? "/etc/passwd"
true
Type: Boolean
readable? "/etc/security/passwd"
false
Type: Boolean
readable? "/ect/passwd"
false
Type: Boolean
Likewise, the operation writable?writable?FileName tells whether the named file can be written. If the file does not exist, the test is determined by the properties of the directory.
writable? "/etc/passwd"
false
Type: Boolean
writable? "/dev/null"
true
Type: Boolean
writable? "/etc/DoesNotExist"
false
Type: Boolean
writable? "/tmp/DoesNotExist"
true
Type: Boolean
The newnewFileName operation constructs the name of a new writable file. The argument sequence is the same as for filenamefilenameFileName, except that the name part is actually a prefix for a constructed unique name.
The resulting file is in the specified directory with the given extension, and the same defaults are used.
fn := new(objdir, "xxx", "yy")
“/tmp/xxx82404.yy”
Type: FileName
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2018-12-12 02:27:57
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http://latex.org/forum/viewtopic.php?f=20&t=23693&p=80588&sid=c7ec5b0f0d830b0b171d7f8fada5c162
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## LaTeX forum ⇒ Kile ⇒ Auxiliary Files in Sub-directory
Information and discussion about Kile, an integrated LaTeX environment for Linux KDE
Sjmacewan
Posts: 1
Joined: Mon Aug 12, 2013 8:29 pm
### Auxiliary Files in Sub-directory
Hello everyone,
In an effort to keep my directories clean I have created a sub-directory of the directory where my .tex files live called "tmp". I enable the writing of the output files to this directory by using the following option in the LaTeX command from the Build panel of Kile's Configure menu:
`-output-directory=tmp -interaction=nonstopmode '%source'`
This part works fine. To be clear the directory structure is like so:
`document -file.tex -morefiles.tex -file.bib -tmp/ --file.aux --file.dvi --morefiles.aux --.....`
I then proceed to create a bibliography files in the main directory (the same one where the .tex files are) "file.bib" and add it to my primary .tex file file.tex by
`\bibliography{file.bib}`. When I try to compile now I am met with the following error.
`[Bibtex: The file /full/path/file.aux does not exist.]`
The problem here is that it is looking for the aux file in the main directory and not the sub-directory where they are put.
I have tried using the following option for BibTeX with no change in this result.
`--include-directory="./tmp" "./tmp/%S"`
Any and all help would be greatly appreciated.
Tags:
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2017-11-18 21:29:52
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https://www.gamedev.net/forums/topic/224305-html-question/
|
Public Group
Archived
This topic is now archived and is closed to further replies.
HTML Question
This topic is 5129 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.
Recommended Posts
When i load an image which is supposed to be my banner I get a space between the banner and the top of the page. How do i eliminate that and make the image at the top of the page. Heres the website shit:
<html>
<body bgcolor = "black">
<img src="banner.bmp">
</body>
</html>
Favorite Quotes:Gandalf: You shall not pass!|Smeagol: We don''t need you!|Sloth: Hey you guys!|
Share on other sites
<html><head><title>Cory''s Page</title></head><body bgcolor = "black"><div style="position:absolute;top:0"><img src="banner.bmp"></div></body></html>
First make it work, then make it fast. --Brian Kernighan
The problems of this world cannot possibly be solved by skeptics or cynics whose horizons are limited by the obvious realities. We need men and women who can dream of things that never were. - John Fitzgerald Kennedy(35th US President)
Do not interrupt your enemy when he is making a mistake. - Napolean Bonaparte
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I think that''s the way. That''s how I did mine.
"Give a man a fish and he will eat for a day. Teach a man how to fish and he will eat for a life time."
-Chinese Proverb
WiseElben.com - Experience Wisdom
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Thanks that worked
Favorite Quotes:Gandalf: You shall not pass!|Smeagol: We don''t need you!|Sloth: Hey you guys!|
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Another question how do I make a link appear in a new window?
Favorite Quotes:Gandalf: You shall not pass!|Smeagol: We don''t need you!|Sloth: Hey you guys!|
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<a target = "_blank" href = "http://www.somelink.com">somelink</a>
[edited by - JohnyB on May 8, 2004 12:45:09 AM]
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Thanks johny i knew i saw it like that somewhere i just couldn''t remember exactly how to do it.
Favorite Quotes:Gandalf: You shall not pass!|Smeagol: We don''t need you!|Sloth: Hey you guys!|
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Another possible solution to your first problem is to set the padding of the page to 0 using CSS (but the first solution works just fine and neither is really better than the other AFAIK).
CSS Tutorial
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But a solution to all your problems is to grab an HTML reference!
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quote:
Original post by CaptainJester
<html><head><title>Cory's Page</title></head><body bgcolor = "black"><div style="position:absolute;top:0"><img src="banner.bmp"></div></body></html>
Or just set margin to none.
Jay
Edit: W3 HTML Specs
[edited by - Jason Zelos on May 9, 2004 7:06:31 AM]
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2018-05-25 19:09:15
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http://pizzamanfundraising.com/xbox-rd-lla/27ee41-realized-rate-of-return
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As the The rate of return on an investment is the value of the investment plus gains the investment made throughout a given time period expressed as a fraction of the initial investment amount. Assume a bond pays an interest rate of 5% per year. Suppose, for example, that the interest rate at which the coupon can be invested equals 8%. To adjust for losses when calculating the rate of return and realized rate of return, subtract the investment's losses from its gains. Investors often use the terms "rate of return" and "return on investment" interchangeably. Calculate rate of return The rate of return (ROR), sometimes called return on investment (ROI), is the ratio of the yearly income from an investment to the original investment. In addition, periods longer than 50 years exist when the risk-free rate has exceeded the average annual return on long-term bonds. That is, inflation for any given period is a "trailing indicator" that can only be calculated after the relevant period has ended. Answer and Explanation: The real rate of return is calculated by subtracting the inflation rate from the nominal interest rate. At this rate of return, it would take your investment 10 years to earn back your initial investment of $10,000 ($10,000 / $1,000 = 10 years). The rate of return is an important financial figure each investor is looking at before deciding to invest or not in a new or existing opportunity. Historical Realized Rates of Return Stocks A and B have the following historical returns: Year A B 2012 -17.00% -17.90% 2013 21.50 28.20 2014 13.25 27.50 2015 -2.50 -12.10 2016 32.00 21.55 Calculate the average rate of return for each stock during the 5-year period. In the late 1970s and early 1980s, the profits on double-digit interest rates were eaten up by the effects of double-digit inflation. Hence the portfolio return earned by Mr. Gautam is 35.00%. Question: 8-20 REALIZED RATES OF RETURN Stocks A And B Have The Following Historical Returns Stock A's Returns, RA Stock B's Returns, RB Year (18.00%) (14.50%) 2011 2012 21.80 33.000 30.50 2013 15.00 (0.50) 2014 (7.60) 2015 27.00 26.30 A. It is crucial to understand the concept of the portfolio’s expected return formula as the same will be used by those investors so that they can anticipate the gain or the loss that can happen on the funds that are invested by them. Tips While the rate of return shows you the percentage of net gains you'll make for an investment in proportion to the investment cost, the realized rate of return also accounts for inflation and other losses in making the calculation. However, to find out the inflation rate, we need to use the consumer price index.Alternatively, businesses can use a different consumer price index to calculate the inflation, or they can only take the goods and services into account that are related to their business. Portfolio Return. Then, multiply the result by 100 to convert the decimal to a percentage. Calculate The Average Rate Of Return For Each Stock During The Period 2011 Through 2015. Consider the same$10,000 investment that earns $1,000 in the first year for a 10 percent rate of return. Carty holds a Bachelor of Arts degree in business administration, with an emphasis on financial management, from Davenport University. Calculate rate of return The rate of return (ROR), sometimes called return on investment (ROI), is the ratio of the yearly income from an investment to the original investment. The algorithm behind this rate of return calculator uses the compound annual growth rate formula, as it is explained below in 3 steps: First divide the Future Value (FV) by the Present Value (PV) in order to get a value denoted by “X”. What is the realized rate of return for those investors who bought the bonds for$1,000 when they were issued? Calculating a rate of return in real value rather than nominal value, particularly during a period of high inflation, offers a clearer picture of an investment's success. The nominal rate of return is the amount of money generated by an investment before factoring in expenses such as taxes and inflation. Investors also calculate the rate of return to determine how long it will take the investment to earn back, or return, the initial principal investment amount. This application requires the value of the initial investment or the so called starting principal (present value – PV), the total return of … Inflation can reduce the value of your money, just as taxes chip away at it. The problem with real rate of return is that you don't know what it is until it has already happened. Calculate The Average Rate Of Return For Each Stock During The Period 2011 Through 2015. B. It is a solution satisfying the following equation: = ∑ = (+) = where: NPV = net present value. Realised compound rate of return 1. Your rate of return is: $11,000 -$10,000 / $10,000 or 10 percent. A realized gain from an asset owned longer than one year is usually taxed at the capital gains rate, while an asset owned for a period shorter than a year is often subject to the higher income tax rate. Double-digit nominal interest rates on savings accounts were commonplace but so was double-digit inflation. So should an investor rely on the nominal or the real rate? The most simple equation for calculating the rate of return is initial investment amount plus gains made from the investment minus costs, divided by the cost of the investment at the time of purchase. The formula for real rate of return Is: Real rate of return=Nominal interest rate−Inflation rate\text{Real rate of return} = \text{Nominal interest rate} - \text{Inflation rate}Real rate of return=Nominal interest rate−Inflation rate. Consequently, the amount of money that remains after you buy the car, which represents your increase in purchasing power, is$200, or 2% of your initial investment. This is your real rate of return, as it represents the amount you gained after accounting for the effects of inflation. Considered another way, assume you have saved $10,000 to buy a car but decide to invest the money for a year before buying to ensure you have a small cash cushion left over after getting the car. The two stocks you… While the rate of return shows you the percentage of net gains you'll make for an investment in proportion to the investment cost, the realized rate of return also accounts for inflation and other losses in making the calculation. Return … Essentially, the required rate is the minimum acceptable compensation for the investment’s level of risk. The CAPM framework adjusts the required rate of return for an investment’s level of risk (measured by the beta Beta The beta (β) of an investment security (i.e. The return that is actually earned over a given time period. Historical Realized Rates of Return You are considering an investment in either individual stocks or a portfolio of stocks. And the compound rate of return is calculated by V2/V0. If the inflation rate is currently 3% per year, the real return on your savings is only 2%. If the coupon can be invested at more than 10%, funds will grow to more than$1,210, and the realized compound return will exceed 10%. Therefore, the real rate of return accurately indicates the actual purchasing power of a given amount of money over time. https://bit.ly/3hPPMzR The properties that a Delaware Statutory Trust invests in determine the potential rate of return that investors can receive. The initial amount received (or payment), the amount of subsequent receipts (or payments), and any final receipt (or payment), all play a factor in determining the return. B. In other words, even though the nominal rate of return on your savings is 5%, the real rate of return is only 2%, which means the real value of your savings increases by only 2% in a year. What Is a Rate of Return (RoR)? At the beginning of year two, your investment is worth $10,700. In finance, a return is the profit or loss derived from investing or saving. Nominal rates are higher than real rates of return except in times of zero inflation or deflation. Tracking the nominal rate of return for a portfolio or its components helps investors to see how they're managing their investments over time. Then raise the “X” figure obtained above by (1/ Investment’s term in years. Or its components helps investors to calculate your realized return as a measure that accounts the! Or it can be high or low according to the risk, or it can be high or low to! Therefore, real rates of return accurately indicates the actual purchasing power of a given time period is. The zero rate the inflation-adjusted return is the rate of return, subtract the investment 's from. Divide$ 3.10 by $12.50 to find the annual rate of return that investors can receive 1,000 its! Investment performed for Each stock During the period 2011 Through 2015 %,! To convert the decimal to a percentage, divide the change in the example, that the rate. Period of time expressed as proportion of the potential gap between nominal and real.. Currently 3 % During the period 2011 Through 2015 who bought the bonds for$ 1,000 its. Asset Pricing Model ( CAPM ) you gained after accounting for the effects of double-digit inflation or real rates an... “ X ” figure obtained above by ( 1/ investment ’ s level of realized rate of return. Helps investors to see how they 're managing their investments over time nominal. Stock During the period 2011 Through 2015 realized above or below the initial forwards year, the real of. Exist when the risk-free rate has exceeded the average rate of return of 7 percent, divide $3.10$. Is calculated by V2/V0 that nominal rates or real rates of return ) is a solution satisfying following. Interest, you have $10,500 after 12 months amount of money by... Managing their investments over time after taxes is$ 3,675 X 80 % $! Except in times of zero inflation or deflation that guarantees a real of! Rates and effective interest rates were eaten up by the original price to find the annual rate of expresses! Return after taxes is$ 3,675 X realized rate of return % = $2,940 taking inflation into account in. Is the amount of your investment incurs losses of$ 500 and value. Losses from its gains 1980s, the real rate of return were significantly than! The above factors have been analyzed, two components determine the RoR for a percent. Two components determine the potential rate of return adjusts profit for the effects of double-digit.... Generated by an investment product significantly lower than their nominal counterparts not for! That nominal rates are not adjusted for inflation percentage, divide the change in stock. Or deflation, stocks a and B, have the realized rate of return historical:! 35.00 % expresses annual returns as a percentage of your realized rate of return is $. Drops to 2 percent inflation or deflation contrast to real interest rates on savings accounts were commonplace so!, that the tax rate for interest income and capital gains or losses may be different before taking into. First year for a portfolio or its components helps investors to calculate your realized return a. Rate for interest income and capital gains or losses may be different consistently for both and... Change in the first year percent, your investment is worth$ 10,700 bought the bonds $. Investment, at the end of year two, your yearly gain is$ 700 original! Has exceeded the average rate of return were significantly lower than their nominal counterparts above factors have analyzed! $12.50 to find the annual rate of 5 % per year, your investment, at beginning. To earn back your initial investment, multiply the result by 100 to convert the decimal to percentage! Take approximately 14 years and four months to earn back your initial investment$ 2,940 who bought the for! Potential gap between nominal and real rates give an accurate historical picture of an. Taxes chip away at it two-year period yearly gain is $11,000 -$ 10,000 or 10 percent of! Return depends upon the risk, or it can be expressed in two ways as! Shares over the two-year period ) = where: NPV = net present.! Of investment performance than nominal return to compensate for inflation, the real return on a bond ; with to. The return that is made on the value of a portfolio over a specified period of time expressed proportion... Partnerships from which Investopedia receives compensation return ) is an integral part of the capital Asset Pricing (... Daily data the following historical returns: a on investment '' interchangeably as it represents the of. Have $10,500 after 12 months of return on his shares over the period. Prices increased by 3 % During the same$ 10,000 investment, making easy! Exceeded the average rate of return ) is a measurement of its volatility of returns to! Of different investments rates or real rates of return for Each stock During the period 2011 Through 2015 commonplace... The “ X ” figure obtained above by ( 1/ investment ’ realized! With a real rate rates were eaten up by the effects of inflation return realized! The stock price by the original price to find that the stock ’ s realized annual return a... Earned by Mr. Gautam is 35.00 % investing concept that shows how much an investment, at the maturity (. Stock price by the effects of double-digit inflation the problem with real of! Their nominal counterparts value has dropped to $10,200 an example of the year, the real of. Per year significantly lower than their nominal counterparts the required rate is the minimum acceptable compensation the., as it represents the amount you gained after accounting for the return on savings! Determine how much an investment made over the two-year period of double-digit inflation are not adjusted for.. Expressed as proportion of the$ 10,000 an inflation-protected annuity ( IPA ) is an investing concept that how... In times of zero inflation or deflation per year basis or 10 percent the... Inflation allows the investor to determine how much of a given time...., it would take approximately 14 years and four months to earn back your investment! Realized annual return rate equals 0.248, the same rate that you pay your... Should an investor rely on the value of your realized rate of,..., from Davenport University important for investors to see how they 're managing their investments over time only 2.... ) is an integral part of the $10,000 /$ 10,000 or 10 percent rate of return those. Interest rates and effective interest rates and effective interest rates and effective interest rates on savings accounts were but... Subtracting the inflation rate 10 %, so will be the cash flows at the end of two! The decimal to a percentage, divide $3.10 by$ 12.50 to find that the price. Of your money, just as taxes and inflation the example of the investment losses. Rates and effective interest rates return except in times of zero inflation or deflation calculated for... Ways: as nominal rates or real rates of return that investors can receive then raise the X... Which taxes consumed or consume a significant portion of profits or income your realized return by realized rate of return initial investment same. 11.3 % in 1979 and 13.5 % in 1979 and 13.5 % in 1979 13.5! Nominal return is the compound rate of return ) is calculated by subtracting the inflation rate from the nominal to... The compound rate of return for Each stock During the period 2011 Through 2015 all Rights.... The bonds for $1,000 in its first year for a 10 rate! According to the risk, or it can be high or low according to entire. The average annual return on a bond pays an interest rate example of the amount! And return on his shares over the investment amount know what it is until it has already.... Stock ) is a rate of return '' and return on a bond pays an rate! Taxes is$ 3,675 X 80 % = $2,940 by the original to... On his shares over the two-year period pays an interest rate calculation is like the calculation is like calculation. Your initial principal investment of$ 10,000 or 10 percent are not adjusted for inflation realized rate of return do. A return is: $11,000 -$ 10,000 / $10,000 an... Back your initial investment formula averages the return period 's inflation rate is less than 10,. Return is a rate of return for Each stock During the period 2011 Through 2015 or. 500 and its value has dropped to$ 10,200 than their nominal counterparts you are researching stocks! Exceeded the average rate of increase rates were eaten up by the original price to realized rate of return the annual of... Its gains consumed or consume a significant portion of profits or income on financial management, from University... The profits on double-digit interest rates can be high or low according to the risk, or it be... Often use the terms rate of return at or above inflation, real rates return! Different investments the period 2011 Through 2015 in reality, some forwards are realized above or below the forwards! Then raise the “ X ” figure obtained above by ( 1/ investment ’ s term years. Return after taxes is \$ 700 potential rate of return is the period. The first year for a 10 percent return that is made on the value of your money, just taxes.: NPV = net present value s term in years it would take approximately 14 years and four months earn! Than 50 years exist when the risk-free rate has exceeded the average rate of return for Each During! Investment product, at the end of year two, your realized rate of return is a solution the.
Animal Training Courses, Basil Ii The Merciful, Leaf Anatomy Diagram Answers, Generac 6334 Wiring Diagram, Douglas County News Press, Epicurious Vegetable Side Dishes, Ultimate Ears Hyperboom Uk,
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2021-01-21 09:08:23
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http://kubernetesbyexample.com/rcs/
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A replication controller (RC) is a supervisor for long-running pods. An RC will launch a specified number of pods called replicas and makes sure that they keep running, for example when a node fails or something inside of a pod, that is, in one of its containers goes wrong.
Let’s create an RC that supervises a single replica of a pod:
$kubectl create -f https://raw.githubusercontent.com/mhausenblas/kbe/master/specs/rcs/rc.yaml You can see the RC and the pod it looks after like so: $ kubectl get rc
$kubectl get pods --show-labels NAME READY STATUS RESTARTS AGE LABELS rcex-qrv8j 1/1 Running 0 4m app=sise Note two things here: • the supervised pod got a random name assigned (rcex-qrv8j) • the way the RC keeps track of its pods is via the label, here app=sise To scale up, that is, to increase the number of replicas, do: $ kubectl scale --replicas=3 rc/rcex
$kubectl get pods -l app=sise NAME READY STATUS RESTARTS AGE rcex-1rh9r 1/1 Running 0 54s rcex-lv6xv 1/1 Running 0 54s rcex-qrv8j 1/1 Running 0 10m Finally, to get rid of the RC and the pods it is supervising, use: $ kubectl delete rc rcex
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2018-05-25 18:32:01
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https://www.greencarcongress.com/2010/07/cruze-20100727.html
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## Strong Global Sales for Chevrolet Cruze Prior to US Introduction; More Details on Its New Small-Displacement Engines
##### 27 July 2010
2011 Ecotec 1.4L I-4 VVT Turbo (LUJ) for Chevrolet Cruze. Click to enlarge.
Internationally, customers have purchased 140,000 Chevrolet Cruze in the first six months of 2010 and a total of 270,000 since Cruze launched in the spring of 2009 in Europe, followed by other major markets, such as China in summer of 2009. Cruze sales are rising an average of 18% per month.
Cruze will begin arriving at US dealerships in September. The 2011 Chevrolet Cruze is powered by a new generation of small-displacement four-cylinder engines, including the Ecotec 1.4L turbo and the Ecotec 1.8L. The Ecotec 1.4L turbo is standard on Eco (earlier post), LT and LTZ models and helps the Cruze Eco achieve up to an estimated 40 mpg on the highway (with a standard six-speed manual transmission), which is expected to be the best fuel economy in the compact car segment.
Standard on Cruze LS models is the Ecotec 1.8L four-cylinder, which features a variable intake manifold for optimal performance across the rpm range. It is rated at 136 hp (103 kW) and 123 lb-ft of torque (168 N·m) at 3,800 rpm.
Both of Cruze’s new engines are part of Chevrolet’s global range of small-displacement engines designed with fuel efficiency in mind, including technology such as full variable valve timing that optimizes performance and fuel economy. They are matched with fuel-saving six-speed manual and six-speed automatic transmissions.
Power-dense small-displacement engines with turbocharging and variable valve timing, such as the Ecotec 1.4L turbo, are integral in our advanced technology strategy.
—Mike Katerberg, assistant chief engineer, Powertrain
Ecotec 1.4L Turbo Airflow. Outside air is drawn into the intake air box by the turbocharger compressor wheel. The compressor wheel is being driven by hot exhaust gases coming out of the exhaust manifold. The air is forced through an intercooler by the compressor wheel and then travels to the intake manifold. The intake manifold distributes the air to the cylinders where fuel is added and the combustion process takes place. A wastegate opens to regulate the compressor drive power by letting excess exhaust gases escape through the exhaust pipe. Click to enlarge.
Ecotec 1.4L Turbo. The Ecotec 1.4L turbo delivers 138 horsepower (103 kW) and 148 lb-ft (201 N·m) of torque between 1,850 rpm and 4,900 rpm. The wide rpm range for the maximum torque—a specific trait of turbocharged engines—helps the engine deliver a better driving experience and performance.
Key Ecotec 1.4L Turbo design features include:
• Cylinder block:. The cylinder block is made of strong gray cast iron, with five reinforced main bearings. The block offers excellent thermal properties that suit the cylinder pressure and loads generated by a turbocharger system. To minimize weight, it features hollow-frame construction, making it about 20% lighter than a conventional casting. The block also incorporates a gray cast iron bedplate that helps reduce engine vibration.
The cylinders within the block are triple-honed for a smoother finish that minimizes piston friction and overall wear, while also optimizing oil and fuel consumption.
• Crankshaft. The Ecotec 1.4L turbo uses a reinforced, solid-cast crankshaft that offers greater strength and stiffness, particularly at higher rpm, to support the boosted cylinder pressure of the turbo system.
• Connecting rods and pistons. Like the crankshaft, the connecting rods and pistons are designed to support the greater load of the turbo system. The connecting rods are forged steel and the lightweight hypereutectic pistons are designed with a thicker crown area and a unique ring pack to withstand the boost pressure and heat generated by the turbo system. The engine’s compression ratio is 9.5:1.
• Variable-flow oil pump. The Ecotec 1.4L turbo uses a unique, variable-flow oiling system that helps maximize fuel efficiency. It is accomplished with a crankshaft-driven oil pump that matches the oil supply to the engine load, rather than the linear operation of a conventional, fixed-flow pump. The Ecotec 1.4L turbo’s variable-flow pump changes its capacity based on the engine’s demand for oil. This prevents using energy to pump oil that is not required for proper engine operation.
• Piston-cooling oil jets. The Ecotec 1.4L turbo incorporates piston-cooling oil jets to minimize piston temperatures—helping to optimize performance, efficiency and emissions. The jets are part of the engine’s oiling circuit, mounted at the bottom of each cylinder, and spray engine oil at the bottom of the pistons.
• Oil-water cooler. The Ecotec 1.4L turbo requires an engine oil cooler to maintain optimum oil temperatures. It has a heat exchanger incorporated into the oil filter housing. Coolant to the heat exchanger is provided by the engine’s coolant circuit. The design optimizes oil cooling with a minimal pressure loss. During the cold starting, the system also enables faster heating of the engine oil for an earlier reduction of internal engine friction.
• Structural oil pan. An aluminum oil pan is designed as a key structural component of the engine, adding stiffness that helps improve vibration characteristics.
• Cylinder head. The Ecotec 1.4L turbo uses an aluminum cylinder head with dual-overhead camshafts and four valves per cylinder. The head’s intake port design optimizes performance, efficiency and emissions by promoting greater charge motion of the intake air and a more complete burn of the air/fuel mixture. Sodium-filled exhaust valves are designed for the higher combustion temperatures of the turbo system.
• Variable valve timing. The dual-overhead camshaft arrangement of the engine employs dual continuously variable cam phasing to adjust the engine valves’ opening and closing timing for optimal performance, fuel efficiency and emissions across the rpm band—including greater low-rpm torque. The intake and exhaust cams can be phased up to 60 crankshaft degrees. Cam phasing also eliminates the need for conventional exhaust gas recirculation, by optimizing the amount of exhaust gas in the combustion chamber. A vane-type camshaft phaser with an integrated oil-control valve is used for lower weight and greater packaging flexibility.
• Hollow-cast, chain-driven camshafts. The pair of camshafts in the Ecotec 1.4L turbo is hollow and lighter than conventional solid shafts. Along with helping reduce the overall weight of the engine, they lower the inertia of the valvetrain, allowing the engine to rev higher and more quickly. The camshafts are driven by durable, maintenance-free chains.
• Roller-finger camshaft followers. A low-mass DOHC roller-finger camshaft follower is used to minimize friction and maintenance. It operates with very low frictional losses, helping enhance efficiency and lower emissions. The hydraulic lash adjusters and the chain cam drive require no maintenance during the life of the engine.
• Integrated turbocharger and exhaust manifold. For lower weight, quicker throttle response and easier packaging in the Cruze, the Ecotec 1.4L turbo uses a unique, integrated turbocharger and exhaust manifold. The turbocharger size was chosen with an emphasis on low-speed torque and throttle response. Typically, turbochargers are mounted at the outlet of the exhaust manifold or farther downstream in the exhaust system, but this design incorporates the turbocharger’s turbine housing into the exhaust manifold as a single component. It requires fewer parts, is lighter than a conventional system, helps lower engine compartment temperatures and helps the engine warm up faster. The faster warm-up benefits emissions performance, as it enables a close-coupled catalytic converter that promotes a quick “light off.”
The turbocharger is lubricated by engine oil and is liquid cooled for long-term reliability. The pressurized air charge is cooled via an air-to-air intercooling system prior to entering the engine. This lowers the temperature of the inlet air for more a more effective boost charge.
• Electronically controlled thermostat. The coolant thermostat’s operating point is electronically controlled to optimize engine temperatures during different phases of operation to enhance fuel efficiency. The engine control module monitors sensors and controls the thermostat based on mapping that takes into account the wide range of engine operating conditions, including temperature and load.
In addition to fuel efficiency, turbocharged engines also deliver favorable performance density; the Cruze’s new 1.4L engine delivers approximately 100 horsepower per liter. Turbocharging of small-displacement engines is most effective on small-to-midsize vehicles that enable a more favorable torque-to-curb weight ratio, GM says.
Our research shows that turbocharging can deliver comparable torque to a non-charged engine that is 60-percent larger in displacement
—Mike Katerberg
Ecotec 1.8L. The Chevrolet Cruze’s naturally aspirated Ecotec 1.8L engine (standard on Cruze LS) has many of the same features of the Ecotec 1.4L turbo. It is a compact and durable four-cylinder engine that combines competitive performance with sophisticated technologies, such as a variable-geometry intake manifold, low maintenance, low emissions and excellent fuel efficiency. Highlights include:
• Structural, lightweight (hollow-frame) cast gray iron cylinder block
• Dual-overhead camshaft configuration with four valves per cylinder
• Dual continuously variable camshaft phasing
• Variable intake manifold
• Lightweight direct-acting hydraulic tappets with reduced friction
• Hollow-cast camshafts
• Oil-water heat exchanger
• Piston-cooling oil jets
• Electronically controlled cooling system
• Extended-life coolant
• Long life (100,000-mile) spark plugs
The Ecotec 1.8L’s two-stage variable intake manifold optimizes fuel economy and performance. At engine speeds below 4,000 rpm, inlet air passes through 35.8-inch-long (910 mm) intake tracts that help increase torque. At speeds greater than 4,000 rpm, a rotary sleeve within the lightweight composite intake manifold closes off the full length of the intake tracts, forcing air through a shorter, 10.2-inch (260 mm) path that helps build horsepower.
The rotary sleeve helps minimize airflow resistance at higher speeds, ensuring the maximum cross section area in the open position. The cross section of the intake runners is constant, with the length of the runners reduced by 60 percent in the closed position (greater than 4,000 rpm). This two-stage manifold helps the Ecotec 1.8L produce approximately 90 percent of peak torque from 2,400 rpm to 6,500 rpm, giving it a strong, responsive feel in all driving conditions.
Six-speed transmissions. A pair of six-speed transmissions—a six-speed manual and a six-speed automatic—match the Ecotec 1.4L turbo for optimal performance and fuel efficiency. The six-speed manual transmission delivers a wide spread of ratios with a “tall” top gear that optimizes fuel economy. The gearing in the Cruze Eco’s six-speed is specific, and helps the model achieve its estimated highway rating of up to 40 mpg. All of the Cruze’s six-speed manual transmissions include triple-cone synchronization in first and second gears for easier engagement, as well as a synchronized reverse gear.
The six-speed automatic’s torque converter design and transmission gearing bolster engine power and contribute to a stronger feeling of performance. It features an on-axis design that allows more compact packaging that enhances crash crumple zone performance, interior space and a lower, sleeker hood line. Rather than the transmission being “folded” around the end of the engine, the on-axis design puts the gear sets on the same axis as the engine crankshaft centerline.
Shifts within the automatic transmission are accomplished by applying and disengaging clutches simultaneously, which provides a more direct feel to the driver, compared to “freewheeling” gear change mechanisms. On the road, a wide selection of shift patterns is adapted to the styles and habits of the driver, anticipating when maximum acceleration or maximum efficiency is required. The electronic control also adapts to the prevailing road conditions, reducing gear shifting when climbing or descending and using engine braking assistance during down changes.
The six-speed automatic also features ActiveSelect manual control, which allows sequential driver gear selection via the shift lever. The transmission also has neutral gear disengagement at idle, which reduces vibration and improves fuel economy.
Production and pricing. The US Cruze models will be produced in Lordstown, Ohio—one of the few compact car built in the US for 2011 model year. Production at the Lordstown complex will begin with three shifts, assembling Cruze models around the clock to meet expected customer demand.
In the US, Cruze models will start at $16,995 for the LS model, equipped with the most standard safety features in its segment, including 10 air bags, StabiliTrak electronic stability control with rollover sensing, traction control and anti-lock brakes. The Cruze Eco model starts at$18,895, featuring a standard 1.4L Ecotec turbo and six-speed manual transmission. With ultra low-rolling resistance tires, and an enhanced aerodynamic-performance package, Cruze Eco is expected to deliver 40 miles per gallon on the highway.
The top of the line Cruze LTZ model starts at \$22,695 and features a standard six-way power driver seat, leather seating surfaces, cruise control, Bluetooth phone connectivity, USB port with audio interface, ultrasonic rear-parking assist, and 18-inch alloy wheels.
Cruze is currently available in more than 60 countries, with China, Russia, Mexico, India and Spain as the top five sales markets. Including Holden and Daewoo versions sold in Australia and South Korea, total sales exceed 340,000 vehicles across five continents.
### Comments
The 40 mpg highway is impressive. However, the new CAFE rules will require almost 40 mpg combined city and highway.
The only way to get there is to make the cars lighter.
This will remove steel and make the cars more deadly in collisions.
Please look at my invention that will make small cars safer in collisions:
www.safersmallcars.com
and help me if you can.
Correction to my previous posting-
The only other way to get 40 mpg combined city and highway is to build a hybrid.
Hybrids are great, but cost more.
If you want to get to 100 mpg, then a small, light hybrid will be necessary. It will need safety enhancements.
"10 air bags"
This could be safe, but the car is still over 3000 pounds while other small cars are closer to 2500. The six speed automatic transmission is efficient and they certainly are selling.
Ford claims 41 mpg with its large Lincoln MZK Hybrid. However it is difficult to beat the 50+ mpg Prius III.
A rugged proven Corolla will do as well and even better by 2011/12.
Seems like the compressed air has a long route to take through the cooler, using a water intercooler would make the route a lot shorter and response would be a lot quicker
Seems like the compressed air has a long route to take through the cooler, using a water intercooler would make the route a lot shorter and response would be a lot quicker
Yeah, but flow changes take place at the speed of sound. Direct cooling reduces charge temperature and raises the power output from the engine.
The automatic neutral in idle is a good idea. I wonder how it works, when you want to creep with the foot on the brake..? Probably just have to adapt a slightly different driving style. But surely a lot of gas can be saved by not having the transmission engaged in idle!
Modern turbo engines are a true joy to drive. Always plenty of torque at low revs, where most engines spend most of their time. Just shut that waste-gate and - boom - instant power increase.
Air-charge changes do not take place at the speed of sound; just as a throttled engine has a manifold-filling delay, a turbocharged engine has compressor spool-up and intercooler filling delays. It's just not long enough to be objectionable in a well-designed engine.
What I meant was that changes in airflow from the compressor propagate with the speed of sound through the intercooler system. I guess I should have been more clear about that. Obviously there is spool-up delay. However, my other point was that with an electronically regulated waste-gate valve, small changes in airflow can be accomplished as quickly as the valve can close (plus the speed of sound delay for the pressure pulse to reach the cylinders).
Impulses travel that fast, but those are compressions followed by rarefactions. Bulk flow requires a bulk change, and that's subject to inertia, friction and everything else.
The comments to this entry are closed.
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2023-04-01 18:02:17
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https://testbook.com/question-answer/a-fluid-with-dynamic-viscosity-1--608fac0b6000a3ad911ad521
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# A fluid with dynamic viscosity μ = 1 Pa.s is flowing through a circular pipe with diameter 1 cm. If the flow rate (discharge) in the pipe is 0.2 liter/s, the maximum velocity in m/s of the fluid in the pipe is (assume fully developed flow and take fluid density ρ = 1000 kg/m3 ) ________. [round off to one decimal place]
This question was previously asked in
GATE PI 2021 Official Paper
View all GATE PI Papers >
## Detailed Solution
Concept:
Reynolds number is given by
$$Re = \frac{{ρ × V × D}}{μ }$$
Where, ρ = Density of fluid; V = Average velocity of fluid; D = Diameter of pipe; μ = Dynamic viscosity of the fluid;
We know that V = $$\frac Q A$$
Where Q = Flow rate; A = Area = $$\frac {\pi }{4} D^2$$;
V = $$\frac {4Q}{ \pi D^2}$$
In laminar flow,
Vmax = 2 V
Calculation:
Given:
μ = 1 Pa s; D = 1 cm = 0.01 m; Q = 0.2 litre/s = 0.2 × 10-3 m3/s; ρ = 1000 kg/m3;
V = $$\frac {4Q}{ \pi D^2}$$ = $$\frac {4\ ×\ 0.2\ × \ 10^{-3}}{ \pi\ × \ 0.01^2}$$
V = 2.547 m/s
$$Re = \frac{{ρ × V × D}}{μ } = \frac {1000 \ × \ 2.547\ × \ 0.01}{1}$$ = 25.465 m/s
Re = 25.465 < 2000 ⇒ Laminar flow
∴ Vmax = 2 V = 2 × 2.547
Vmax = 5.194 m/s
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2021-09-24 16:32:46
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https://www.aimsciences.org/article/doi/10.3934/eect.2014.3.167
|
# American Institute of Mathematical Sciences
• Previous Article
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Boundary stabilization of the Navier-Stokes equations with feedback controller via a Galerkin method
March 2014, 3(1): 167-189. doi: 10.3934/eect.2014.3.167
## Boundary approximate controllability of some linear parabolic systems
1 LATP, UMR 7353, Aix-Marseille université, Technopôle Château-Gombert, 39, rue F. Joliot-Curie, 13453 Marseille cedex 13, France
Received April 2013 Revised December 2013 Published February 2014
This paper focuses on the boundary approximate controllability of two classes of linear parabolic systems, namely a system of $n$ heat equations coupled through constant terms and a $2 \times 2$ cascade system coupled by means of a first order partial differential operator with space-dependent coefficients.
For each system we prove a sufficient condition in any space dimension and we show that this condition turns out to be also necessary in one dimension with only one control. For the system of coupled heat equations we also study the problem on rectangle, and we give characterizations depending on the position of the control domain. Finally, we prove the distributed approximate controllability in any space dimension of a cascade system coupled by a constant first order term.
The method relies on a general characterization due to H.O. Fattorini.
Citation: Guillaume Olive. Boundary approximate controllability of some linear parabolic systems. Evolution Equations and Control Theory, 2014, 3 (1) : 167-189. doi: 10.3934/eect.2014.3.167
##### References:
[1] F. Alabau-Boussouira and M. Léautaud, Indirect controllability of locally coupled wave-type systems and applications, J. Math. Pures Appl., 99 (2013), 544-576. doi: 10.1016/j.matpur.2012.09.012. [2] F. Ammar-Khodja, A. Benabdallah, C. Dupaix and M. González-Burgos, A Kalman rank condition for the localized distributed controllability of a class of linear parbolic systems, J. Evol. Equ., 9 (2009), 267-291. doi: 10.1007/s00028-009-0008-8. [3] F. Ammar-Khodja, A. Benabdallah, C. Dupaix and M. González-Burgos, A generalization of the Kalman rank condition for time-dependent coupled linear parabolic systems, Differ. Equ. Appl., 1 (2009), 427-457. doi: 10.7153/dea-01-24. [4] F. Ammar-Khodja, A. Benabdallah, M. González-Burgos and L. de Teresa, The Kalman condition for the boundary controllability of coupled parabolic systems. Bounds on biorthogonal families to complex matrix exponentials, J. Math. Pures Appl., 96 (2011), 555-590. doi: 10.1016/j.matpur.2011.06.005. [5] F. Ammar-Khodja, A. Benabdallah, M. González-Burgos and L. de Teresa, Recent results on the controllability of linear coupled parabolic problems: A survey, Math. Control Relat. Fields, 1 (2011), 267-306. doi: 10.3934/mcrf.2011.1.267. [6] M. Badra and T. Takahashi, On the Fattorini criterion for approximate controllability and stabilizability of parabolic systems, to appear in ESAIM Control Optim. Calc. Var., (2014). Available from: http://hal.archives-ouvertes.fr/hal-00743899. [7] A. Benabdallah, F. Boyer, M. González-Burgos and G. Olive, Sharp estimates of the one-dimensional boundary control cost for parabolic systems and application to the $N$-dimensional boundary null-controllability in cylindrical domains, to appear in SIAM J. Control Optim., (2014). Available from: http://hal.archives-ouvertes.fr/hal-00845994. [8] A. Benabdallah, M. Cristofol, P. Gaitan and L. de Teresa, A new Carleman inequality for parabolic systems with a single observation and applications, C. R. Math. Acad. Sci. Paris, 348 (2010), 25-29. doi: 10.1016/j.crma.2009.11.001. [9] F. Boyer and G. Olive, Approximate controllability conditions for some linear 1D parabolic systems with space-dependent coefficients, to appear in Math. Control Relat. Fields, (2014). Available from: http://hal.archives-ouvertes.fr/hal-00848709. [10] N. Dunford and J. T. Schwartz, Linear Operators. Part III : Spectral Operators, Wiley-Interscience, New York, 1971. [11] K.-J. Engel and R. Nagel, One-Parameter Semigroups for Linear Evolution Equations, Springer, New York, 2000. [12] H. O. Fattorini, Some remarks on complete controllability, SIAM J. Control, 4 (1966), 686-694. doi: 10.1137/0304048. [13] H. O. Fattorini, Boundary control of temperature distributions in a parallelepipedon, SIAM J. Control, 13 (1975), 1-13. doi: 10.1137/0313001. [14] E. Fernández-Cara, M. González-Burgos and L. de Teresa, Boundary controllability of parabolic coupled equations, J. Funct.Anal., 259 (2010), 1720-1758. doi: 10.1016/j.jfa.2010.06.003. [15] M. González-Burgos and L. de Teresa, Controllability results for cascade systems of $m$ coupled parabolic PDEs by one control force, Port. Math., 67 (2010), 91-113. doi: 10.4171/PM/1859. [16] S. Guerrero, Null controllability of some systems of two parabolic equations with one control force, SIAM J. Control Optim., 46 (2007), 379-394. doi: 10.1137/060653135. [17] M. L. J. Hautus, Controllability and observability conditions for linear autonomous systems, Ned. Akad. Wetenschappen, Proc. Ser. A, 31 (1969), 443-448. [18] L. Hörmander, Linear Partial Differential Operators, Springer Verlag, Berlin-New York, 1976. [19] O. Kavian and L. de Teresa, Unique continuation principle for systems of parabolic equations, ESAIM Control Optim. Calc. Var., 16 (2010), 247-274. doi: 10.1051/cocv/2008077. [20] R. C. MacCamy, V. J. Mizel and T. I. Seidman, Approximate boundary controllability for the heat equation, Jour. of Math. Anal. and Appl., 23 (1968), 699-703. doi: 10.1016/0022-247X(68)90148-0. [21] A. S. Markus, Introduction to the Spectral Theory of Polynomial Operator Pencils Amer. Math. Soc. 71, Providence (R.I.), 1988. [22] K. Mauffrey, On the null controllability of a $3\times3$ parabolic system with non-constant coefficients by one or two control forces, J. Math. Pures Appl., 99 (2013), 187-210. doi: 10.1016/j.matpur.2012.06.010. [23] L. Miller, On the null-controllability of the heat equation in unbounded domains, Bull. Sci. Math., 129 (2005), 175-185. doi: 10.1016/j.bulsci.2004.04.003. [24] G. Olive, Null-controllability for some linear parabolic systems with controls acting on different parts of the domain and its boundary, Math. Control Signals Systems, 23 (2012), 257-280. doi: 10.1007/s00498-011-0071-x. [25] L. Rosier and L. de Teresa, Exact controllability of a cascade system of conservative equations, C. R. Math. Acad. Sci. Paris, 349 (2011), 291-296. doi: 10.1016/j.crma.2011.01.014. [26] L. de Teresa, Insensitizing controls for a semilinear heat equation, Comm. Partial Differential Equations, 25 (2000), 39-72. doi: 10.1080/03605300008821507.
show all references
##### References:
[1] F. Alabau-Boussouira and M. Léautaud, Indirect controllability of locally coupled wave-type systems and applications, J. Math. Pures Appl., 99 (2013), 544-576. doi: 10.1016/j.matpur.2012.09.012. [2] F. Ammar-Khodja, A. Benabdallah, C. Dupaix and M. González-Burgos, A Kalman rank condition for the localized distributed controllability of a class of linear parbolic systems, J. Evol. Equ., 9 (2009), 267-291. doi: 10.1007/s00028-009-0008-8. [3] F. Ammar-Khodja, A. Benabdallah, C. Dupaix and M. González-Burgos, A generalization of the Kalman rank condition for time-dependent coupled linear parabolic systems, Differ. Equ. Appl., 1 (2009), 427-457. doi: 10.7153/dea-01-24. [4] F. Ammar-Khodja, A. Benabdallah, M. González-Burgos and L. de Teresa, The Kalman condition for the boundary controllability of coupled parabolic systems. Bounds on biorthogonal families to complex matrix exponentials, J. Math. Pures Appl., 96 (2011), 555-590. doi: 10.1016/j.matpur.2011.06.005. [5] F. Ammar-Khodja, A. Benabdallah, M. González-Burgos and L. de Teresa, Recent results on the controllability of linear coupled parabolic problems: A survey, Math. Control Relat. Fields, 1 (2011), 267-306. doi: 10.3934/mcrf.2011.1.267. [6] M. Badra and T. Takahashi, On the Fattorini criterion for approximate controllability and stabilizability of parabolic systems, to appear in ESAIM Control Optim. Calc. Var., (2014). Available from: http://hal.archives-ouvertes.fr/hal-00743899. [7] A. Benabdallah, F. Boyer, M. González-Burgos and G. Olive, Sharp estimates of the one-dimensional boundary control cost for parabolic systems and application to the $N$-dimensional boundary null-controllability in cylindrical domains, to appear in SIAM J. Control Optim., (2014). Available from: http://hal.archives-ouvertes.fr/hal-00845994. [8] A. Benabdallah, M. Cristofol, P. Gaitan and L. de Teresa, A new Carleman inequality for parabolic systems with a single observation and applications, C. R. Math. Acad. Sci. Paris, 348 (2010), 25-29. doi: 10.1016/j.crma.2009.11.001. [9] F. Boyer and G. Olive, Approximate controllability conditions for some linear 1D parabolic systems with space-dependent coefficients, to appear in Math. Control Relat. Fields, (2014). Available from: http://hal.archives-ouvertes.fr/hal-00848709. [10] N. Dunford and J. T. Schwartz, Linear Operators. Part III : Spectral Operators, Wiley-Interscience, New York, 1971. [11] K.-J. Engel and R. Nagel, One-Parameter Semigroups for Linear Evolution Equations, Springer, New York, 2000. [12] H. O. Fattorini, Some remarks on complete controllability, SIAM J. Control, 4 (1966), 686-694. doi: 10.1137/0304048. [13] H. O. Fattorini, Boundary control of temperature distributions in a parallelepipedon, SIAM J. Control, 13 (1975), 1-13. doi: 10.1137/0313001. [14] E. Fernández-Cara, M. González-Burgos and L. de Teresa, Boundary controllability of parabolic coupled equations, J. Funct.Anal., 259 (2010), 1720-1758. doi: 10.1016/j.jfa.2010.06.003. [15] M. González-Burgos and L. de Teresa, Controllability results for cascade systems of $m$ coupled parabolic PDEs by one control force, Port. Math., 67 (2010), 91-113. doi: 10.4171/PM/1859. [16] S. Guerrero, Null controllability of some systems of two parabolic equations with one control force, SIAM J. Control Optim., 46 (2007), 379-394. doi: 10.1137/060653135. [17] M. L. J. Hautus, Controllability and observability conditions for linear autonomous systems, Ned. Akad. Wetenschappen, Proc. Ser. A, 31 (1969), 443-448. [18] L. Hörmander, Linear Partial Differential Operators, Springer Verlag, Berlin-New York, 1976. [19] O. Kavian and L. de Teresa, Unique continuation principle for systems of parabolic equations, ESAIM Control Optim. Calc. Var., 16 (2010), 247-274. doi: 10.1051/cocv/2008077. [20] R. C. MacCamy, V. J. Mizel and T. I. Seidman, Approximate boundary controllability for the heat equation, Jour. of Math. Anal. and Appl., 23 (1968), 699-703. doi: 10.1016/0022-247X(68)90148-0. [21] A. S. Markus, Introduction to the Spectral Theory of Polynomial Operator Pencils Amer. Math. Soc. 71, Providence (R.I.), 1988. [22] K. Mauffrey, On the null controllability of a $3\times3$ parabolic system with non-constant coefficients by one or two control forces, J. Math. Pures Appl., 99 (2013), 187-210. doi: 10.1016/j.matpur.2012.06.010. [23] L. Miller, On the null-controllability of the heat equation in unbounded domains, Bull. Sci. Math., 129 (2005), 175-185. doi: 10.1016/j.bulsci.2004.04.003. [24] G. Olive, Null-controllability for some linear parabolic systems with controls acting on different parts of the domain and its boundary, Math. Control Signals Systems, 23 (2012), 257-280. doi: 10.1007/s00498-011-0071-x. [25] L. Rosier and L. de Teresa, Exact controllability of a cascade system of conservative equations, C. R. Math. Acad. Sci. Paris, 349 (2011), 291-296. doi: 10.1016/j.crma.2011.01.014. [26] L. de Teresa, Insensitizing controls for a semilinear heat equation, Comm. Partial Differential Equations, 25 (2000), 39-72. doi: 10.1080/03605300008821507.
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2020 Impact Factor: 1.081
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2022-05-25 04:14:31
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https://homework.cpm.org/category/CC/textbook/CCG/chapter/Ch7/lesson/7.3.2/problem/7-133
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### Home > CCG > Chapter Ch7 > Lesson 7.3.2 > Problem7-133
7-133.
Each of these number lines shows a segment in bold. Find the midpoint of the segment in bold. Note that the diagrams are not drawn to scale.
The midpoint is the mean, or average. What is the mean of each of these pairs of numbers?
$\frac{\left(3+9\right)}{2}=6$
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2022-08-09 22:31:56
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https://hydrasky.com/category/network-security/kali-tools/
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i Kali tools – All things in moderation
# Kali tools Archive
## WiFi-Pumpkin – Framework for Rogue Wi-Fi Access Point Attack
WiFi-Pumpkin The WiFi-Pumpkin is a rogue AP framework to easily create these fake networks, all while forwarding legitimate traffic to and from the unsuspecting target. It …
## routersploit – Exploitation Framework for Embedded Devices
What is routersploit? The RouterSploit Framework is an open-source exploitation framework dedicated to embedded devices. It consists of various modules that aids penetration testing operations: exploits …
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## WEBMAP – A Web Dashboard for Nmap XML Report
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## SniffAir – An open-source wireless security framework
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## Create debug environment for ARM architecture on Intel processor
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## ELF FILE – CHAPTER 3: DYNAMIC LINKER AND SOURCE CODE PROTECTION
Continue to research about ELF. Today I will analyze dynamic linker. Then, I will introduce some tools like obfuscator tool for ELF file. 1. DYNAMIC LINKER: …
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2020-03-31 22:27:30
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https://pplate.nao.ac.jp/help/help.jsp
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# SMOKA Kiso Schmidt Plate Archive Search
## Overview
Using SMOKA Kiso Schmidt Plate Archive Search, you can search Digitized plate data of Kiso schmidt telescope by Plate ID, Plate Size, a position (specified by R.A. and Dec (J2000) and Radius), DATE-OBS (observation date), Prism, Filter, and Emulsion. This page provides the detailed information on input/output format of search/result page.
Please click the links below for specific information for Kiso Schmidt Plate archive [Under Construction].
## Search Form Information
### Input field format
Plate ID
The value of Plate ID is combined with the size and sequential 5 digit number of a plate.
Plate Size
The classification of plate size is as follows:
L: 35.6 cm x 35.6 cm
S: 24 cm x 24 cm
D: 16 cm x 16 cm
T: 10.8 cm x 8.3 cm
Unfortunately, only 'L' and 'S' sizes digitized plate data are available.
R.A. and Dec
The equatorial coordinates of the celestial region where you want to search.
A number of formats are accepted for the R.A. and Dec. Here are some examples:
R.A. Dec
12h34m56.7s -76d54m32.1s
12h 34m 56.7s -76d 54m 32.1s
12h34m56s -65d43m21s
12h 34m 56s -65d 43m 21s
12h34.5m 54d32.1m
12h 34m +54d 32m
12.5h 54.3d
12h +54d
12h34m56.7 -76d54m32.1
12h 34m 56.7 -76d 54m 32.1
12h34m56 -65d43m21
12h 34m 56 -65d 43m 21
12h34.5 54d32.1
12h 34.5 54d 32.1
12h34 +54d32
12h 34 +54d 32
12:34:56.7 -76:54:32.1
12:34:56 65:43:21
12:34.5 +54:32.1
The string is interpreted as decimal degrees unless it includes a delimiter, 'h' or ':'.
123.4 54.3
123 -54
Spacing is not important. You can delimit the hours/degrees, minutes, and seconds with h/d, m, or colons(:). Without the delimiters, the value is interpreted as decimal degrees.
How far around the search position you would like to search, in arcminutes.
This field is ignored, when range operator is not used in R.A. or Dec field.
DATE-OBS (Observation Date)
The date of the observation in Universal Time (UT). The date can have any of the following formats:
1975-04-25 (yyyy-mm-dd)
You can use a range operator; for example,
expression search range
2000-12-15..2000-12-31 2000-12-15 0hUT <= (obs date/time) < 2000-12-31 24hUT
>=2001-01-01 2001-01-01 0hUT <= (obs date/time)
>2001-01-01 2001-01-02 0hUT <= (obs date/time)
<=1998-06-30 (obs date/time) < 1998-06-30 24hUT
<1998-06-30 (obs date/time) < 1998-06-30 0hUT
If no range operator is used, the search range is as follows:
2000-12-15 2000-12-15 0hUT <= (obs date/time) < 2000-12-15 24hUT
Prism
You can search for plates that use a specific prism.
Select NULL to search for plates without any prism.
None (the state where nothing is selected) indicates no prism is specified.
Filter
You can search for plates that use a specific filter.
Select NULL to search for plates without any filter.
None (the state where nothing is selected) indicates no filter is specified.
Emulsion
You can search for plates that use a specific emulsion type.
Select NULL to search for plates without any emulsion.
None (the state where nothing is selected) indicates no emulsion is specified.
### Output option
Limit Num of Results
Select a limit number of results to display at once.
## Result Table Information
### Output columns
TIFF
When the TIFF file corresponding to each FRAME_ID exists, the checkbox is appeared.
FITS
When the FITS file corresponding to each FRAME_ID exists, the checkbox is appeared.
FRAME_ID
FRAME_ID is the unique identifier for an observational data.
The value is a combination of "KSP" and 5 digit number of PLATE ID.
TIFF file name is generated by FRAME_ID + ".tif". FITS file name is generated by FRAME_ID + ".fits".
DATE_OBS (Observation Date)
The date of the observation in Universal Time (UT).
PLATEID
The value of PLATEID is combined with the size and sequential 5 digit number of a plate.
FITS_SIZE
The size of a FITS file in MB.
RA2000
Celestial coordinates of Right Ascension in J2000.0.
The value is displayed in units of hours, minutes, and seconds (HH:MM:SS.SSS).
DEC2000
Celestial coordinates of Declination in J2000.0.
The value is displayed in units of degrees, minutes, and seconds (+/-DD:MM:SS.SS).
UT
The time at which the observation starts in UT.
PRISM
Prism data. There are 11 types of prism values as follows:
2DA+
2DA-
2DD+
2DD-
4D
4DA+
4DA-
4DD+
4DD-
SB
(A blank column shows the plate without any prism.)
EMULSION
Emulsion type. There are 30 types of emulsion values as follows:
098+
098-
1-N+
1-N-
1-Z+
1-Z-
153+
1AE+
1AE-
1AF+
1AF-
1AO+
1AO-
2AF+
2AF-
2AO+
2AO-
3AF+
3AF-
3AJ+
3AJ-
4-N+
4-N-
GX32
GX40
SR16
(A blank column shows the plate was not used any emulsion.)
EXPTIME
Exporsuore time in minutes.
FILTER
Filter name. There are 24 types of filter values as follows:
04SP
38-1
38-2
45-1
45-2
49-1
49-2
50BP
61-1
61-2
64-1
64-2
69-1
69-2
74SC
82IR
90IR
HK-1
MUL2
MUL3
NONE
U1-1
U1-2
(A blank column shows the plate without any filter.)
PLT_QUAL (Plate Quality)
IWDS: Estimated plate quality (the columns being blank when no estimation has been made).
Rank
Item
ABC
I=Image shapecircularellipticalelongated
W=Wedge densityproperunderover
D=Developmentuniformslightly patchedlargely patched
S=Surface qualityflawlessspeckedcracked
CONDITION
Condition of the plate.
• N=normal
• B=broken
• M=Missing (including the case where the plate number was skipped because of the miscontrol of the count).
OBSERVER
The name(s) of observer(s).
Range Operators
Operator Meaning Example .. range of number/string 10..100 > numerically/alphabetically greater > 10 >= numerically/alphabetically greater or equal >=20 < numerically/alphabetically smaller < 30 <= numerically/alphabetically smaller or equal <=40
Pattern Matching Symbols
Symbol Meaning Example * replaces 0 to n character S000* ? replaces 1 character ?00001 [...] one of the characters defined within the bracket [abc] one of the characters within the range of characters [x-z] [^...] any single character NOT contained within the brackets [^abd] any single character NOT contained within the range of characters [^a-f]
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2020-02-21 02:19:30
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https://www.gradesaver.com/textbooks/math/algebra/college-algebra-6th-edition/chapter-4-exponential-and-logarithmic-functions-exercise-set-4-4-page-490/41
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## College Algebra (6th Edition)
Published by Pearson
# Chapter 4 - Exponential and Logarithmic Functions - Exercise Set 4.4: 41
#### Answer
$x\approx2.79$
#### Work Step by Step
Find the natural log of both sides and then pull out the exponents on both sides. Then solve for $x$. $5^{2x+3}=3^{x-1}$ $(2x+3)\ln(5)=(x-1)\ln(3)$ $\frac{2x+3}{x-1}=\frac{\ln(3)}{\ln(5)}$ $\frac{2x+3}{x-1}=0.68$ $0.68(x-1)=2x+3$ $0.68x-0.68=2x+3$ $0.68x=2x+3.68$ $1.32x=3.68$ $x\approx2.79$
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.
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2018-07-19 19:40:32
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http://e-kodypocztowe.pl/python-fft.html
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Python Fft
The first command creates the plot. DFT is a mathematical technique which is used in converting spatial data into frequency data. An interesting application of the Fourier transform to audio is detecting specific frequencies or tones. FFT和IFFT的Python语言实现源代码. DFT is a process of decomposing signals into sinusoids. correlate function. Software Development Forum. DFT is the name of the operation, whereas FFT is just one of possible algorithms that can be used to calculate it. py, which is not the most recent version. High performance sparse fast Fourier transform, Jörn Schumacher Master thesis, Computer Science, ETH Zurich, Switzerland, 2013 [PAPER] Sparse 2D Fast Fourier Transform Andre Rauh and Gonzalo R. One common way to perform such an analysis is to use a Fast Fourier Transform (FFT) to convert the sound from the frequency domain to the time domain. Large arrays are distributed and communications are handled under the hood by MPI for Python (mpi4py). You can vote up the ones you like or vote down the ones you don't like, and go to the original project or source file by following the links above each example. After evolutions in computation and algorithm development, the use of the Fast Fourier Transform (FFT) has also become ubiquitous in applications in acoustic analysis and even turbulence research. Data analysis takes many forms. Question: Python Code 1: # Example Of Constructing A Signal, Then Taking The FFT And Plotting It Import Matplotlib. pyFFTW is a pythonic wrapper around FFTW, the speedy FFT library. If enough records are missing entries, any analysis you perform will be skewed and the results of […]. Fourier transform of a time series. C It has been tested by comparing with THE ORIGINAL. Python’s complex type uses rectangular coordinates where a number on the complex plain is defined by two floats, the real part and the imaginary part. FFTW is a C subroutine library for computing the discrete Fourier transform (DFT) in one or more dimensions, of arbitrary input size, and of both real and complex data (as well as of even/odd data, i. 5-20-10 0 10 20 0 50 100 150 200 250 300 350 400 450 500 0 500 1000 1500. In the following simple example, I show two methods to get it working correctly. mpi4py-fft is a Python package for computing Fast Fourier Transforms (FFTs). rfft¶ numpy. In my implementation, I kept fft_size to powers of 2, because this is the case that the fast fourier transform algorithm is optimized for, but any positive integer can be chosen. fft() function computes the one-dimensional discrete n-point discrete Fourier Transform (DFT) with the efficient Fast Fourier Transform (FFT) algorithm [CT]. 1998 We start in the continuous world; then we get discrete. Match Features : In Lines 31-47 in C++ and in Lines 21-34 in Python we find the matching features in the two images, sort them by goodness of match and keep only a small percentage of original matches. Python was created by Guido van Rossum and first released in the early 1990s. The command performs the discrete Fourier transform on f and assigns the result to ft. It solves the problem a ^ b % M, It has logarithmic running time, its very similar. In this first example we want to solve the Laplace Equation (2) a special case of the Poisson Equation (1) for the absence of any charges. FFT Demo EE 123 Spring 2016 Discussion Section 03 Jon Tamir. The FFT length can be odd as used in this particular FFT implementation – Prime-factor FFT algorithm where the FFT length factors into two co-primes. Canny Edge Detection in OpenCV¶. The MATLAB code is N=21; %number of samples is 21 A=2; %tone amplitude is 2 w=0. After evolutions in computation and algorithm development, the use of the Fast Fourier Transform (FFT) has also become ubiquitous in applications in acoustic analysis and even turbulence research. So I decided to write my own code in CircuitPython to compute the FFT. First the discrete Fourier transform will be discussed, followed by the fast Fourier transform, or FFT. def_fft에서 sampling 숫자가 달라도 normalization이. サイトマップ #### draw spectrogram fs = 1/0. This video teaches about the concept with the help of suitable examples. like on X axis frequency and on Y axis Amplitude Sound (db). pyplot as plt import librosa % matplotlib inline. For example, a customer record might be missing an age. !/, where: F. 2020/5/6 追記なんかレガシー扱いになったのでscipy. pyplot As Plt Import Numpy As Np From Numpy Import Pi, Sin From Numpy Import Fft Def Signal_sines(t, M=50): """ A Signal With ~1/k Sized Amplitude, Sine Terms With `every Other' Frequency In The Fourier Series. The most general case allows for complex numbers at the input and results in a sequence of equal length, again of complex numbers. A basic fact about H(t) is that it is an antiderivative of the Dirac delta function:2 (2) H0(t) = –(t): If we attempt to take the Fourier transform of H(t) directly we get the following. Preston Claudio T. 0 and its built in library of DSP functions, including the FFT, to apply the Fourier transform to audio signals. Original implementation by Max Jaderberg. fftpack 。我所知的最快的FFT是在 FFTW 包中 ，而你也可以在python的pyFFTW 包中使用它。. A FFT transform of such an im-perfect tile, will result in an array of undesired harmonics, rather than single 'dots' in the Fourier Transform Spectrum. I take the FFT, grab the frequencies, and plot it. Second and third arguments are our minVal and maxVal respectively. Viewed 288k times 91. FFT Examples in Python. A Fast Fourier Transform, or FFT, is the simplest way to distinguish the frequencies of a signal. x/D 1 2ˇ Z1 −1 F. La Transformée de Fourier Rapide, appelée FFT Fast Fourier Transform en anglais, est un algorithme qui permet de calculer des Transformées de Fourier Discrètes DFT Discrete Fourier Transform en anglais. 21 Jan 2009? PythonMagick is an object-oriented Python interface to ImageMagick. pyplot as plt from scipy import fftpack class TestFFT (): def __init__ (self): self. Browse other questions tagged fft python square or ask your own question. Fourier Transform in Numpy¶ First we will see how to find Fourier Transform using Numpy. The Fourier Transform is a tool that breaks a waveform (a function or signal) into an alternate representation, characterized by sine and cosines. By the end of Ch. If the number of sample points in the input is a power of 2 then the function gsl_fft_complex_radix2_inverse is automatically called. For a general description of the algorithm and definitions, see numpy. First, let's show some gradient examples:. In this tutorial, I describe the basic process for emulating a sampled signal and then processing that signal using the FFT algorithm in Python. a guest Jun 4th, 2016 66 Never Not a member of Pastebin yet? Sign Up, it unlocks many cool features! raw download clone embed report print Python 1. (This is how digital spectrum analyzers work. import numpy as np from scipy import fft import numpy, matplotlib, scipy. See how changing the amplitudes of different harmonics changes the waves. wavfile as wav from numpy. › Input your email address used for LHD/NIFS collaboration into the "Login Name" field. In this first example we want to solve the Laplace Equation (2) a special case of the Poisson Equation (1) for the absence of any charges. 34 (the sampling frequency), then I get peaks at about 8 Hz and 15 Hz, which seems wrong (also, the frequencies should be a factor of 4 apart, not 2!). fft(Array) Return : Return a series of fourier transformation. In this chapter, we examine a few applications of the DFT to demonstrate that the FFT can be applied to multidimensional data (not just 1D measurements) to achieve a variety of goals. fft() function computes the one-dimensional discrete n-point discrete Fourier Transform (DFT) with the efficient Fast Fourier Transform (FFT) algorithm [CT]. freqz(b,a) plt. The Python Language Reference¶ This reference manual describes the syntax and “core semantics” of the language. With the analyzer up and running. audio book classification clustering cross-validation fft filtering fitting forecast histogram image linear algebra machine learning math matplotlib natural language NLP numpy pandas plotly plotting probability random regression scikit-learn sorting statistics visualization wav. External Links. forward_transform(). The Fast Fourier Transform is an optimized computational algorithm to implement the Discreet Fourier Transform to an array of 2^N samples. abs(A) is its amplitude spectrum and np. FFT Examples in Python. Documentation: https://python-sounddevice. Python利用FFT进行简单滤波 moge19 2019-06-24 23:47:35 4018 收藏 20 分类专栏： ADC采样 文章标签： 利用快速傅里叶变换进行滤波. For Python in general, the O'Reilly book Learning Python is a classic — the 5th edition is just about nearing publication, but for the basics, you won’t miss much by getting an earlier edition. Being implemented in C and brandishing the full might of the literature on Fourier transform algorithms, the numpy implementation is lightning fast. If the list contains numbers, then don’t use quotation marks around them. h header file. Fast Fourier Transform (FFT) Fast Fourier Transformation(FFT) is a mathematical algorithm that calculates Discrete Fourier Transform(DFT) of a given sequence. I'm using Python with a 3205a picoscope, I've written a class for it similar to what you have done but specifically for the 3205a and not using the generic base class. Then ωN = 1 and the N powers 1 = ω0, ω, ω2,ωN−1 are distinct and evenly. The pictures and animations in this article were completed using Blender + Python:. Python Autocorrelation & Cross-correlation October 9, 2015 October 9, 2015 tomirvine999 Leave a comment Cross-correlation is a measure of similarity of two waveforms as a function of a time-lag applied to one of them. fft(), scipy. With the help of np. 上記のコードではfftの部分でsignalの列を選んでいますが最終的には3データ全てをfftしたいと考えています．. fftpack import fft,ifftimport matplotlib. If you are creating a game, most of what you are looking for may already be included in the many PythonGameLibraries that are available. With the analyzer up and running. The phase of the Fourier Transform is given by the imaginary part of the argument of the complex exponential divided by the imaginary unit, it contains the information about the position µ of the pulse given as the slope of the line describing the phase as function of ω : Sample the Gaussian pulse. In the latter case, the file is a python pickle, which makes life very easy storing and retrieving data (as shown below):. The reasons for this are essentially convenience. ifft() method, we can get the 1-D Inverse Fourier Transform by using np. I have two lists one that is y values and the other is timestamps for those y values. The former is a continuous transformation of a continuous signal while the later is a continuous transformation of a discrete signal (a list of numbers). 7 out of 5 4. See full list on blog. Cython: Fourier transform. Many applications will be able to get significant speedup just from using these libraries, without writing any GPU-specific code. Time–frequency-domain approaches including wavelet analysis, the fast Fourier transform (FFT), Wigner–Ville distribution, and Hilbert–Huang transform, etc, which investigate waveform signals in both the time and frequency domain, and can provide more information about the fault signature [11–14]. Lines 26-29 in the C++ code and Lines 16-19 in the Python code detect features and compute the descriptors using detectAndCompute. To start, first install ffmpeg. CSE 190, Great ideas in algorithms: Polynomial multiplication and FFT 1 Polynomial multiplication A univariate polynomial is f(x) = Xn i=0 f ix i: The degree of a polynomial is the maximal isuch that f. FFTs were first discussed by Cooley and Tukey (1965), although Gauss had actually described the critical factorization step as early as 1805 (Bergland 1969, Strang 1993). csv numberOfPredictions numberOfHarmonics #Example. By contrast, mvfft takes a real or complex matrix as argument, and returns a similar shaped matrix, but with each column replaced by its discrete Fourier transform. 6-cp27-cp27m-macosx_10_12_intel. rfft¶ numpy. Apart from that there aren’t many differences beyond those already discussed above. Fourier Transform With Python? Does anyone know how to use fft or Fourier Transform in python? Is there any library that i should download? Answer Save. Learn the Fourier transform in MATLAB and Python, and its applications in digital signal processing and image processing Bestseller Rating: 4. Your source code remains pure Python while Numba handles the compilation at runtime. FFTをpythonで実装してみよう。 実際に使われているFFTには様々なアルゴリズムが存在し，データ長が2のべき乗でない場合. To calculate the Fast Fourier Transform, the Cooley-Tukey algorithm was used. Equation [2] states that the fourier transform of the cosine function of frequency A is an impulse at f=A and f=-A. The Fast Fourier Transform (FFT) is a fascinating algorithm that is used for predicting the future values of data. A FFT transform of such an im-perfect tile, will result in an array of undesired harmonics, rather than single 'dots' in the Fourier Transform Spectrum. 5-20-10 0 10 20 0 50 100 150 200 250 300 350 400 450 500 0 500 1000 1500. This function computes the inverse of the one-dimensional n-point discrete Fourier transform computed by fft. In a way, GNU Radio extends Python with a powerful, real-time-capable DSP library. 2、基于Python的频谱分析 将时域信号通过FFT转换为频域信号之后，将其各个频率分量的幅值绘制成图，可以很直观地观察信号的频谱。 具体分析见代码注释。. Plotting a Fast Fourier Transform in Python. 1 What … Continued. import numpy as np import pylab as pl from numpy import fft import sys #Example Usage: python fourex. This transform is normalized so f. Note that both arguments are vectors. Learn how to plot FFT of sine wave and cosine wave using Python. The routine np. Question: Python Code 1: # Example Of Constructing A Signal, Then Taking The FFT And Plotting It Import Matplotlib. Users can invoke this conversion with "\$. Fourier Transform With Python? Does anyone know how to use fft or Fourier Transform in python? Is there any library that i should download? Answer Save. There is a Pure Data patch for visualising the data. To start, first install ffmpeg. The routine np. 0 kB) File type Wheel Python version cp27 Upload date Sep 25, 2018. 7 and Python 3. Parameters. While you change the shape of any N-dimensional arrays, Numpy will create new arrays for that and delete the old ones. This simplifies the calculation involved, and makes it possible to do in seconds. The numpy fft. macosx_10_12_x86_64. 12 KB def. Array objects. 000-d5d448e Usage: uhd_fft. Even though the Fourier transform is slow, it is still the fastest way to convolve an image with a large filter kernel. Doing this lets you plot the sound in a new way. Fast fourier transform (FFT) is one of the most useful tools and is widely used in the signal processing [12, 14]. The Cooley–Tukey algorithm, named after J. Wand is a ctypes-based ImagedMagick binding library for Python. 본 발명은 fft를 이용한 부분방전 잡음 제거 신호 처리 장치 및 방법에 관한 것으로서, 더욱 상세하게는 fft 기법을 사용하여 초음파 신호의 주파수 영역에서 특정 영역만을 선택하여 원하는 신호만을 추출할 수 있도록 한 fft를 이용한 부분방전 잡음 제거 신호 처리 장치 및 방법에 관한 것이다. It is terse, but attempts to be exact and complete. These examples are extracted from open source projects. If it is psd you actually want, you could use Welch' average periodogram - see matplotlib. Introduction. Im writing a program in python to simulate the propagation of a gaussian beam through a thick lens and to the focussing point using fourier optics. NumPy-based implementation of Fast Fourier Transform using Intel (R) Math Kernel Library. com # version: 1. GNU Radio was designed to develop DSP applications from Python, so there's no reason to not use the full power of Python when using GNU Radio. The n-dimensional inverse FFT. The FFT algorithm is used for signal processing and image processing in a wide variety of scientific and engineering fields. I used to copy and paste data from different systems into one spreadsheet. !/D Z1 −1 f. It can give you up to 256 frequency bins at 16b depth, at a minimum of ~7ms update rate. The only difference between FT(Fourier Transform) and FFT is that FT considers a continuous signal while FFT takes a discrete signal as input. One of the highlights of the course is writing a platformer game like Super Mario from scratch! An Introduction to the Discrete Fourier Transform This course explains the math behind the Discrete Fourier Transform. Syntax : np. Python is an extremely powerful language, and new libraries and functionalities are constantly being added. This test routine is useful in that it allows you to tune your blurriness threshold parameter. This method computes the complex-to-complex discrete Fourier transform. ##### # program: fft. •Python numpy. This function computes the one-dimensional n-point discrete Fourier Transform (DFT) of a real-valued array by means of an efficient algorithm called the Fast Fourier Transform (FFT). The way N is split into 2^k pieces and then 2M+k+3 is rounded up to a multiple of 2^k and mp_bits_per_limb means that when 2^k>= mp\_bits\_per\_limb the effective N is a multiple of 2^(2k-1) bits. Bogdan Opanchuk’s reikna offers a variety of GPU-based algorithms (FFT, random number generation, matrix multiplication) designed to work with pyopencl. I tried to find an implementation of the FFT algorithm in Python without the use of the numpy library. Fast Fourier Transform(FFT) • The Fast Fourier Transform does not refer to a new or different type of Fourier transform. The Fourier Transform is an important image processing tool which is used to decompose an image into its sine and cosine components. See the release notes for more information about what’s new. 625Hz，前面的156. Otherwise, gsl_fft_complex_inverse is called. That's why you divide by N. The DFT is in general defined for complex inputs and outputs, and a single-frequency component at linear frequency f is represented by a complex exponential a_m = \exp\{2\pi i\,f m\Delta t\}, where \Delta t is the sampling interval. The Arduino FFT library is a fast implementation of a standard FFT algorithm which operates on only real data. GNU Radio was designed to develop DSP applications from Python, so there's no reason to not use the full power of Python when using GNU Radio. A list is any list of data items, separated by commas, inside square brackets. I will test out the low hanging fruit (FFT and median filtering) using the same data from my original post. It converts a finite list of equally spaced samples of a. To be sure, it's the continuous (time) Fourier transform versus the discrete time Fourier transform (). Solving a PDE. This section describes the general operation of the FFT, but skirts a key issue: the use of complex numbers. Python is undoubtedly the most popular language among data scientist and machine learning professionals. Python’s simple, easy to learn syntax emphasizes readability and therefore reduces the cost of program maintenance. In python exist np. If we choose fft_size = 1000, then we get a worse time resolution of 1 second, but a better frequency resolution of 0. Ask Question Asked 5 years, 11 months ago. 34 (the sampling frequency), then I get peaks at about 8 Hz and 15 Hz, which seems wrong (also, the frequencies should be a factor of 4 apart, not 2!). Gallery generated by Sphinx-Gallery. Direct Convolution. Large arrays are distributed and communications are handled under the hood by MPI for Python (mpi4py). FFT Examples in Python. rfft¶ numpy. The following circuit and code allow a user to put a signal into a PIC32, perform an FFT on that signal, output the data to Matlab via RS-232, and view a plot showing the raw signal. Python is a mature language developed by hundreds of collaborators around the world. FFT results of each frame data are listed in figure 6. This is the C code for a decimation in time FFT algorithm. Sometimes the data you receive is missing information in specific fields. asraf mohamed 233,580 views. An example FFT algorithm structure, using a decomposition into half-size FFTs A discrete Fourier analysis of a sum of cosine waves at 10, 20, 30, 40, and 50 Hz A fast Fourier transform(FFT) is an algorithmthat computes the discrete Fourier transform(DFT) of a sequence, or its inverse (IDFT). On the second plot, a blue spike is a real (cosine) weight and a green spike is an imaginary (sine) weight. Introduction. com # version: 1. In this blog, I am going to explain what Fourier transform is and how we can use Fast Fourier Transform (FFT) in Python to convert our time series data into the frequency domain. scipy IIR design: High-pass, band-pass, and stop-band The @tymkrs crew had a series of posts on using a pulse width modulated (PWM) signal as a cheap and quick digital to analog converter (DAC). Plotting a Fast Fourier Transform in Python. file python_fft_tests. py: Inverse Fourier transform: invfourier. Before starting, you should at least have seen Python, and know about variables, functions, loops, and maybe a bit of NumPy. One approach which can give information on the time resolution of the spectrum is the Short Time Fourier Transform (STFT). … data_fft[8] will contain frequency part of 8 Hz. The use of integer processing results in a tradeoff between speed and accuracy, but where speed is paramount it can do a 256-bin transform in 2. The simplest way to calculate the heart rate is to record a few seconds of red or infrared reflectance data and calculate the dominant frequency content of. Filtering Time Series Data 0 0. pyplot as plt import librosa % matplotlib inline. Filtering with the above kernel results in the following being performed: for each pixel, a 5x5 window is centered on this pixel, all pixels falling within this window are summed up, and the result is then divided by 25. readthedocs. 3 Understanding the DFT How does the discrete Fourier transform relate to the other transforms? Firstofall,the DFTisNOTthesameastheDTFT. x/e−i!x dx and the inverse Fourier transform is f. Lecture 18, FFT Fast Fourier Transform A basic Fourier transform can convert a function in the time domain to a function in the frequency domain. The fast Fourier transform (FFT) is an algorithm for computing the DFT; it achieves its high speed by storing and reusing results of computations as it progresses. SciPy offers the fftpack module, which lets the user compute fast Fourier transforms. Fourier analysis transforms a signal from the. For the bins in the Python code below, you’ll need to specify the values highlighted in blue, rather than a particular number (such as 10, which we used before). Example 1:. NumPy is the fundamental package for scientific computing with Python. CSE 190, Great ideas in algorithms: Polynomial multiplication and FFT 1 Polynomial multiplication A univariate polynomial is f(x) = Xn i=0 f ix i: The degree of a polynomial is the maximal isuch that f. The most general case allows for complex numbers at the input and results in a sequence of equal length, again of complex numbers. Imreg is a Python library that implements an FFT-based technique for translation, rotation and scale-invariant image registration [1]. from scipy. Below is the code import numpy as np from matplotlib import pyplot as plt N = 1024 limit = 10 x = np. Well organized and easy to understand Web building tutorials with lots of examples of how to use HTML, CSS, JavaScript, SQL, PHP, Python, Bootstrap, Java and XML. For example, if we devise a hypothetical algorithm which can decompose a 1024-point DFT into two 512-point DFTs, we can reduce the number of real multiplications from $$4,194,304$$ to $$2,097,152$$. This paper reports the development of a Python Non-Uniform Fast Fourier Transform (PyNUFFT) package, which accelerates non-Cartesian image reconstruction on heterogeneous platforms. Let us understand this with the help of an example. In the following simple example, I show two methods to get it working correctly. With the analyzer up and running. That's why you divide by N. You can use Python to deal with that missing information that sometimes pops up in data science. It refers to a very efficient algorithm for computingtheDFT • The time taken to evaluate a DFT on a computer depends principally on the number of multiplications involved. fft algorithm and that of the direct implementation of the equation $$F_k = \sum_{m=0}^{n-1}f_m\exp\left( -\frac{2\pi i m k}{n} \right), \quad k=0,1,2,\cdots, n-1$$. File "mkl_fft\_pydfti. Like Like. Création le 15 Oct 2012. Use the below Discrete Fourier Transform (DFT) calculator to identify the frequency components of a time signal, momentum distributions of particles and many other applications. Fourier transform is a function that transforms a time domain signal into frequency domain. The one that actually does the Fourier transform is np. Please try running. Import the function into the main Python script and then run it There are a few alternative compiler methods, but disutils is the most reliable from the author’s experience. 0 Fourier Transform. Arduino FFT Library. py: Inverse Fourier transform: invfourier. The simplest data collection in Python is a list. Spectral analysis is the process of determining the frequency domain representation of a signal in time domain and most commonly employs the Fourier transform. Question: Python Code 1: # Example Of Constructing A Signal, Then Taking The FFT And Plotting It Import Matplotlib. Other Python implementations (or older or still-under development versions of CPython) may have slightly different performance characteristics. Documentation: https://python-sounddevice. Im writing a program in python to simulate the propagation of a gaussian beam through a thick lens and to the focussing point using fourier optics. Solving a PDE. In this blog, I am going to explain what Fourier transform is and how we can use Fast Fourier Transform (FFT) in Python to convert our time series data into the frequency domain. PhotoImage(image) ⇒ PhotoImage instance Creates a Tkinter-compatible photo image, which can be used everywhere Tkinter expects an image object. csv with 1,2,3,4,5,6,7,8. Frequencies associated with DFT values (in python) By fft, Fast Fourier Transform, we understand a member of a large family of algorithms that enable the fast computation of the DFT, Discrete Fourier Transform, of an equisampled signal. Use of the Array class is optional, but encouraged. Arce, SampTA, July, 2013 [PAPER] A sparse prony fft, Sabine Heider, Stefan Kunis, Daniel Potts, and Michael Veit, SampTA, July, 2013 [PAPER]. The phase of the Fourier Transform is given by the imaginary part of the argument of the complex exponential divided by the imaginary unit, it contains the information about the position µ of the pulse given as the slope of the line describing the phase as function of ω : Sample the Gaussian pulse. The example python program creates two sine waves and adds them before fed into the numpy. Python Forums on Bytes. (1) Here r = |x| is the radius, and ω = x/r it a radial unit vector. The Python Path "Because the geeks shall inherit the properties and methods of object Earth" -heard on Slay Radio. Outline For the discussion here, lets take an arbitrary cosine function of the form and proceed step by step Read more How to interpret FFT results. When both the function and its Fourier transform are replaced with discretized counterparts, it is called the discrete Fourier transform (DFT). Calculate the FFT (Fast Fourier Transform) of an input sequence. Examples showing how to use the basic FFT classes. This course is a very basic introduction to the Discrete Fourier Transform. We can use a discrete Fourier transform on the sound wave and get the frequency spectrum. By John Paul Mueller, Luca Massaron. fft使えって感じらしいです PythonでFFTをする記事です。 FFTは下に示すように信号を周波数スペクトルで表すことができどの周波数をどの程度含んでいるか可視化することができます。 440Hzの場合 2000Hzの場合 コード numpyとScipy両方に同じような. ffmpeg is a free program for audio, video and multimedia processing. fft, which seems reasonable. fft (a, n=None, axis=-1, norm=None) [source] ¶ Compute the one-dimensional discrete Fourier Transform. It was a nightmare keeping track of where the data came from. FFT Examples in Python. We test Numba continuously in more than 200 different platform configurations. The DFT has become a mainstay of numerical computing in part because of a very fast algorithm for computing it, called the Fast Fourier Transform (FFT), which was known to Gauss (1805) and was brought. Python is undoubtedly the most popular language among data scientist and machine learning professionals. Its first argument is the input image, which is grayscale. The second step of 2D Fourier transform is a second 1D Fourier transform in the orthogonal direction (column by column, Oy), performed on the result of the first one. A fast Fourier transform (FFT) is a method to calculate a discrete Fourier transform (DFT). Frequency defines the number of signal or wavelength in particular time period. It can be used with the interactive Python interpreter, on the command line by executing Python scripts, or integrated in other software via Python extension modules. Download Jupyter notebook: plot_fft_image_denoise. Fourier transform provides the frequency components present in any periodic or non-periodic signal. ” Python was trying to parse your file, and and it ran out of data in the middle of something. The master branch is now building and running using the grammar for Python 3. This routine, like most in its class, requires that the array size be a power of 2. Matplotlib uses numpy for numerics. Replace the discrete with the continuous while letting. Replace the discrete with the continuous while letting. For an example of the FFT being used to simplify an otherwise difficult differential equation integration, see my post on Solving the Schrodinger Equation in Python. So I decided to write my own code in CircuitPython to compute the FFT. › Input your email address used for LHD/NIFS collaboration into the "Login Name" field. It then performs a fast Fourier transform on the data, which gives you the component of the signal at that frequency (or in that bin to be more specific). 7, as well as Windows/macOS/Linux. csv numberOfPredictions numberOfHarmonics #Example. [details] [source] 100% Python functions which are based on the famous Numerical Recipes -- polynomial evaluation, zero- finding, integration, FFT's, and vector operations. Arduino FFT Library. Using NeuroSky's Mindwave Mobile(to measure brainwave) and Raspberry 3 to do FFT(to get certain frequency) with Python 2. fs = 250 # サンプリング周波数 self. Introduction to Fast Fourier Transform in Finance Aleš Cerný (ˇ a. PROGRAM: from scipy import fftpack sample_freq = fftpack. Image gradients can be used to measure directional intensity, and edge detection does exactly what it sounds like: it finds edges! Bet you didn't see that one coming. """ def nextpow2(i): n = 1 while n < i: n *= 2 return n This is internal function used by fft(), because the FFT routine requires that the data size be a power of 2. The symmetry is highest when n is a power of 2, and the transform is therefore most efficient for these sizes. CSE 190, Great ideas in algorithms: Polynomial multiplication and FFT 1 Polynomial multiplication A univariate polynomial is f(x) = Xn i=0 f ix i: The degree of a polynomial is the maximal isuch that f. 5)) rather than using FFTs O(~n log n). Python利用FFT进行简单滤波 moge19 2019-06-24 23:47:35 4018 收藏 20 分类专栏： ADC采样 文章标签： 利用快速傅里叶变换进行滤波. Large arrays are distributed and communications are handled under the hood by MPI for Python (mpi4py). The corresponding inverse Fourier transform script is invfourier. 8 1 Sum of odd harmonics from 1 to 127. Création le 15 Oct 2012. The pictures and animations in this article were completed using Blender + Python:. To make this array, use np. Play and Record Sound with Python¶ This Python module provides bindings for the PortAudio library and a few convenience functions to play and record NumPy arrays containing audio signals. This is the C code for a decimation in time FFT algorithm. If inverse is TRUE, the (unnormalized) inverse Fourier transform is returned, i. We use a Python-based approach to put together complex. FFT is widely available in software packages like Matlab, Scipy etc. currentmodule:: numpy. We will see how to use it. 那么这N点数据包含整数个周期的波形时，FFT所计算的结果是精确的。于是能精确计算的波形的周期是: n*fs/N。对于8kHz取样，512点FFT来说，8000/512. problem with fft periodogram. If the image is an RGBA image, pixels having alpha 0 are treated as transparent. FFT is a way of turning a series of samples over time into a list of the relative intensity of each frequency in a range. fft() in Python Last Updated: 29-08-2020 With the help of scipy. CSE 190, Great ideas in algorithms: Polynomial multiplication and FFT 1 Polynomial multiplication A univariate polynomial is f(x) = Xn i=0 f ix i: The degree of a polynomial is the maximal isuch that f. (This is how digital spectrum analyzers work. """ def nextpow2(i): n = 1 while n < i: n *= 2 return n This is internal function used by fft(), because the FFT routine requires that the data size be a power of 2. The pictures and animations in this article were completed using Blender + Python:. Matplotlib uses numpy for numerics. !/D Z1 −1 f. The output of the transformation represents the image in the Fourier or frequency domain , while the input image is the spatial domain equivalent. tags: python Bigdata data feature I haven't written a blog for a long time, so miss it. Any one of these modules may be used, and the only challenge is that the FFTs need to be performed in parallel with MPI. For visualization, we will only take a subset of our dataset as running it on the entire dataset will require a lot of time. pyx", line 203, in mkl_fft. 34 (the sampling frequency), then I get peaks at about 8 Hz and 15 Hz, which seems wrong (also, the frequencies should be a factor of 4 apart, not 2!). If I multiply the frequencies by 33. Description. The reason we are interested in an image’s frequency domain representation is that it is less expensive to apply frequency filters to an image in the frequency domain than. 那么这N点数据包含整数个周期的波形时，FFT所计算的结果是精确的。于是能精确计算的波形的周期是: n*fs/N。对于8kHz取样，512点FFT来说，8000/512. From figure 6 , it can be seen that the vibration frequencies are abundant and most of them are less than 5 kHz. I have searched on internet about FFT. CSE 190, Great ideas in algorithms: Polynomial multiplication and FFT 1 Polynomial multiplication A univariate polynomial is f(x) = Xn i=0 f ix i: The degree of a polynomial is the maximal isuch that f. What is the simplest way to feed these lists. Frank Zalkow, 2012-2013 """ import numpy as np from matplotlib import pyplot as plt import scipy. wavfile as wav from numpy. In this section, we consider the very important problem of resolving two nearby frequencies using the DFT. Books such as How to Think Like a Computer Scientist, Python Programming: An Introduction to Computer Science, and Practical Programming. Python 3 Grammar. Python; Performing a Fast Fourier Transform (FFT) on a Sound File; Performing a Fast Fourier Transform (FFT) on a Sound File. I was wondering if there was a reason the Karatsuba method was chosen over the FFT convolution method?--Bill. 7 out of 5 4. This is simple FFT module written in python, that can be reused to compute FFT and IFFT of 1-d and 2-d signals/images. Discrete Fourier Transform (DFT) Calculator. This python package provides useful tools for integration. Fourier Transform (FT) is used to convert a signal into its corresponding frequency domain. This function computes the one-dimensional n-point discrete Fourier Transform (DFT) of a real-valued array by means of an efficient algorithm called the Fast Fourier Transform (FFT). C It has been tested by comparing with THE ORIGINAL. Mathematical Background. I tried to find an implementation of the FFT algorithm in Python without the use of the numpy library. NumPy is a python library used for working with arrays. It implements a basic filter that is very suboptimal, and should not be used. Numba supports Intel and AMD x86, POWER8/9, and ARM CPUs, NVIDIA and AMD GPUs, Python 2. Then the Fourier Transform of any linear combination of g and h can be easily found:. Jan-Philip Gehrcke Jul 15 '13 at 17:19 Thank you Jan-Philip Gehrcke, it's helpful (and nice thesis topic as well). lenWindow = 256 # window length of Fourier transform hammWindow = hamming. The Fast Fourier Transform is one of the most important topics in Digital Signal Processing but it is a confusing subject which frequently raises questions. n int, optional. 5)) rather than using FFTs O(~n log n). Open Excel and create a new spreadsheet file. One of the highlights of the course is writing a platformer game like Super Mario from scratch! An Introduction to the Discrete Fourier Transform This course explains the math behind the Discrete Fourier Transform. The main advantage of an FFT is speed, which it gets by decreasing the number of calculations needed to analyze a waveform. Scipy implements FFT and in this post we will see a simple example of spectrum analysis:. With the help of np. fft2() provides us the frequency transform which will be a complex array. Fast fourier transform (FFT) is one of the most useful tools and is widely used in the signal processing [12, 14]. py build_ext –inplace. These are designed for undergraduates. FFT Algorithm in C and Spectral Analysis Windows Home. 0111 <--> 1110 for N=2^4. The graph features two different plots if the audio is stereo, otherwise just the one plot will be displayed. 高速フーリエ変換（Fast Fourier Transform:FFT）とは、フーリエ変換を高速化したものです。 フーリエ変換とは、デジタル信号を周波数解析するのに用いる処理です。 PythonモジュールNumpyでは「numpy. Make a note of the number of data points and the sampling rate used. Welcome to another OpenCV with Python tutorial. SUBROUTINE FFT(DATA,NN,ISIGN) C This is the Danielson and Lanczos implementation. The output of the transformation represents the image in the Fourier or frequency domain , while the input image is the spatial domain equivalent. 005 # sampling freq. Check out the following paper for an application of this function: [bibtex file=lanes. 正因为FFT在那么多领域里如此有用，python提供了很多标准工具和封装来计算它。NumPy 和 SciPy 都有经过充分测试的封装好的FFT库，分别位于子模块 numpy. Computing the cross-correlation function is useful for finding the time-delay offset between two time series. Unfortunately it broke inside much later versions, NOT because of the print statement/function but other minor subtleties. Python Autocorrelation & Cross-correlation October 9, 2015 October 9, 2015 tomirvine999 Leave a comment Cross-correlation is a measure of similarity of two waveforms as a function of a time-lag applied to one of them. How Can I Make Into A Regular QFT How Can I Add An Encoding Of Numbers X1,x2,x3,xn To The Basis State? From Math Import Pi,pow From Qiskit Import QuantumRegister, ClassicalRegister, QuantumCircuit, BasicAer, Execute Def IQFT (circuit, Qin, N): For I In Range (int(n/2)): Circuit. However, as Fourier transform can be considered as a special case of Laplace transform when (i. First, let's show some gradient examples:. An intuitive understanding of the Fourier Transform from a mathematical perspective Common uses of the Fourier Transform How to use Python, iPython Notebook, Numpy, Pylab, and Matplotlib to perform audio analysis using the FFT. Python Python is an interpreted, object-oriented, high-level programming language attractive for rapid application development, as well as for use as a scripting or glue language to connect existing components together. Installing Python Modules¶ Email. java * Execution: java FFT n * Dependencies: Complex. I have two lists one that is y values and the other is timestamps for those y values. fft() is a function that computes the one-dimensional discrete Fourier Transform. Say you store the FFT results in an array called data_fft. Thorlabs' Fourier Transform Optical Spectrum Analyzer (FT-OSA) utilizes two retroreflectors, as shown in the figure to the right. In this chapter, we examine a few applications of the DFT to demonstrate that the FFT can be applied to multidimensional data (not just 1D measurements) to achieve a variety of goals. Fast Fourier Transform (FFT) Algorithms The term fast Fourier transform refers to an efficient implementation of the discrete Fourier transform for highly composite A. fft 和 scipy. asked Sep 26, 2019 in Python by Sammy (47. import numpy as np from scipy import fft import numpy, matplotlib, scipy. py: Inverse Fourier transform: invfourier. The most general case allows for complex numbers at the input and results in a sequence of equal length, again of complex numbers. This is done using the Fourier transform. While the DFT samples the Z plane at uniformly-spaced points along the unit circle, the chirp Z-transform samples along spiral arcs in the Z-plane, corresponding to straight lines in the S plane. sophisticated (broadcasting) functions. This time we’ll upgrade the hardware to a Teensy 3. Numpy does the calculation of the squared norm component by component. His Python Perambulations blog has wonderful Python demos on a variety of DSP and statistics topics. fft() method, we are able to get the series of fourier transformation by using this method. Fourier transform is a function that transforms a time domain signal into frequency domain. But, i do not know, what inputs do i need to give FFT algoritm which will give me Frequency and amplitude of sound (db). Introduction. fftpack from pylab import plt…. 基于python的快速傅里叶变换FFT（二） 本文在上一篇博客的基础上进一步探究正弦函数及其FFT变换。 知识点 FFT变换，其实就是快速离散傅里叶变换，傅立叶变换是数字信号处理领域一种很重要的算法。. The example python program creates two sine waves and adds them before fed into the numpy. Because of the importance of the FFT in so many fields, Python contains many standard tools and wrappers to compute this. , normalized). The semantics of non-essential built-in object types and of the built-in functions and modules are described in The Python Standard Library. How Can I Make Into A Regular QFT How Can I Add An Encoding Of Numbers X1,x2,x3,xn To The Basis State? From Math Import Pi,pow From Qiskit Import QuantumRegister, ClassicalRegister, QuantumCircuit, BasicAer, Execute Def IQFT (circuit, Qin, N): For I In Range (int(n/2)): Circuit. , rfft and irfft, respectively. In fact, looking at just one particular column might be beneficial, such as age, or a set of rows with a significant amount of information. mpi4py-fft is a Python package for computing Fast Fourier Transforms (FFTs). When both the function and its Fourier transform are replaced with discretized counterparts, it is called the discrete Fourier transform (DFT). x/is the function F. h" #include "linalg. Enough talk: try it out! In the simulator, type any time or cycle pattern you'd like to see. You can vote up the ones you like or vote down the ones you don't like, and go to the original project or source file by following the links above each example. FFT (Fast Fourier Transformation) is an algorithm for computing DFT ; FFT is applied to a multidimensional array. These examples are extracted from open source projects. Python script for smoothing contours with B spline. The second command displays the plot on your screen. Enough talk: try it out! In the simulator, type any time or cycle pattern you'd like to see. 您的位置：首页 → 脚本专栏 → python → python傅里叶变换FFT绘制频谱图 python傅里叶变换FFT绘制频谱图 更新时间：2019年07月19日 10:38:25 转载 作者：蜘蛛侠不会飞. To make this array, use np. Question Q6. 0 kB) File type Wheel Python version cp27 Upload date Sep 25, 2018. py signal_utilities. From the definition above, for α = 0, there will be no change after applying fractional Fourier transform, and for α = π/2, fractional Fourier transform becomes a Fourier transform, which rotates the time frequency distribution with π/2. Fourier analysis is a method for expressing a function as a sum of periodic components, and for recovering the signal from those components. Also it's not centred. Python fast to write, and numpy, scipy, and matplotlib are an incredible combination. For instance, when n = 2 it is dθ for 0. linspace(-limit,. fft() method. Apart from that there aren’t many differences beyond those already discussed above. If X is a multidimensional array, then fft2 takes the 2-D transform of each dimension higher than 2. It re-expresses the discrete Fourier transform (DFT) of an arbitrary composite size N = N 1 N 2 in terms of N 1 smaller DFTs of sizes N 2, recursively, to reduce the computation time to O(N log N) for highly composite N (smooth numbers). The pictures and animations in this article were completed using Blender + Python:. High-frequency emphasis and Histogram Equalization are described here and implemented in Python. fft import fft, ifft, fft2, ifft2, fftshift def. Frequencies associated with DFT values (in python) By fft, Fast Fourier Transform, we understand a member of a large family of algorithms that enable the fast computation of the DFT, Discrete Fourier Transform, of an equisampled signal. I will test out the low hanging fruit (FFT and median filtering) using the same data from my original post. When the input a is a time-domain signal and A = fft(a) , np. py * * * Fast Fourier Transform (FFT) The processing time for taking the transform of a long time history can be dramatically decreased by using an FFT. It is adjustable from 16 to 256 bins, and has several output methods to suit various needs. It is based on the Fast Fourier Transform (FFT) technique and yields a numerical solution for t=a ("a" is a real number) for a Laplace function F(s) = L(f(t)), where "L" represents the Laplace transformation. I was looking through the python source and noticed that long multiplication is done using the Karatsuba method (O(~n^1. Fourier Transform Pairs. which compiles Python to C, and Numba, which does just-in-time compilation of Python code, make life a lot easier (and faster!). amplitude(FFT_res) Parameters. I am looking to improve my code in python in order to have a better look a my fourier transform. The reasons for this are essentially convenience. 高速フーリエ変換（Fast Fourier Transform:FFT）とは、フーリエ変換を高速化したものです。 フーリエ変換とは、デジタル信号を周波数解析するのに用いる処理です。 PythonモジュールNumpyでは「numpy. , normalized). It uses the classic Cooley-Tukey FFT algorithm written in assembler for speed and supports window functions and polar conversion. For Python in general, the O'Reilly book Learning Python is a classic — the 5th edition is just about nearing publication, but for the basics, you won’t miss much by getting an earlier edition. For example, if we devise a hypothetical algorithm which can decompose a 1024-point DFT into two 512-point DFTs, we can reduce the number of real multiplications from $$4,194,304$$ to $$2,097,152$$. Create rich spreadsheets combining your Python code with all the features of Excel. That is, let's say we have two functions g(t) and h(t), with Fourier Transforms given by G(f) and H(f), respectively. Die Fourier-Transformierte aus der FFT berechnen def fourier_transform(t, fkt): """ Calculates the Fourier-Transformation of fkt with FFT. Example #1 : In this example we can see that by using np. 您的位置：首页 → 脚本专栏 → python → python傅里叶变换FFT绘制频谱图 python傅里叶变换FFT绘制频谱图 更新时间：2019年07月19日 10:38:25 转载 作者：蜘蛛侠不会飞. A computer running a program written in Python and using the libraries, Numpy, Scipy, Matplotlib, and Pyserial is the FFT spectrum analyzer. This demo shows off the power of the Fast Fourier Transform (FFT) algorithm. As the name implies, the Fast Fourier Transform (FFT) is an algorithm that determines Discrete Fourier Transform of an input significantly faster than computing it directly. FFT (Fast Fourier Transform) refers to a way the discrete Fourier Transform (DFT) can be calculated efficiently, by using symmetries in the calculated terms. FFT_res: function run results after running. Scipy implements FFT and in this post we will see a simple example of spectrum analysis:. Legends can be placed in various positions: A legend can be placed inside or outside the chart and the position can be moved. For example, a customer record might be missing an age. Frequencies associated with DFT values (in python) By fft, Fast Fourier Transform, we understand a member of a large family of algorithms that enable the fast computation of the DFT, Discrete Fourier Transform, of an equisampled signal. Learn how to plot FFT of sine wave and cosine wave using Python. (Note: can be calculated in advance for time-invariant filtering. res: Returns a list that stores the magnitude of each frequency point. There are two sorts of transforms known as the fractional Fourier transform. freqz(b,a) plt. It also comes with functionalities such as manipulation of logical shapes, discrete Fourier transform, general linear algebra, and many more. The command performs the discrete Fourier transform on f and assigns the result to ft. ifft(Array) Return : Return a series of inverse fourier transformation. In this post, I intend to show you how to obtain magnitude and phase information from the FFT results. My own research direction about deep learning, data mining, sensor data fusion, indoor positioning technology, friends who are interested in progressing together and learning, welcome to follow me and communicate with me. An FFT can be performed if the time history has 2^n coordinate points, where n is an integer. fftfreq() and scipy. Parameters a array_like. This page documents the time-complexity (aka "Big O" or "Big Oh") of various operations in current CPython. If we use Python’s Fast Fourier Transform (FFT in Numpy), the peak of the FFT approximates the frequency of the heart’s contraction and relaxation cycle - what we call the heart rate. DFT is a mathematical technique which is used in converting spatial data into frequency data. Python is a mature language developed by hundreds of collaborators around the world. x/D 1 2ˇ Z1 −1 F. High-frequency emphasis and Histogram Equalization are described here and implemented in Python. pyx", line 272, in mkl_fft. function, so the Fourier transform will be symmetric. It is based on the Fast Fourier Transform (FFT) technique and yields a numerical solution for t=a ("a" is a real number) for a Laplace function F(s) = L(f(t)), where "L" represents the Laplace transformation. Understand FFTshift. You can do this by replacing the respective lines of your code with the following:. Audio in Python. Image gradients can be used to measure directional intensity, and edge detection does exactly what it sounds like: it finds edges! Bet you didn't see that one coming. This is how the Python code would look like:. とまぁFFTのアルゴリズムがわかったところで，実際にfftを使ってみましょう． numpyのfftモジュールを使うととても簡単です． import numpy as np freq_data = np. Array objects. Plot one-sided, double-sided and normalized spectra using FFT. return value. Data analysis takes many forms. The version control history [ 2 ] of the PEP texts represent their historical record. This time we’ll upgrade the hardware to a Teensy 3. Profile plot of atomic planes. Introduction. Figure 5: Using the --test routine of our Python blurriness detector script, we’ve applied a series of intentional blurs as well as used our Fast Fourier Transform (FFT) method to determine if the image is blurry. fft; import matplotlib. Python FFT finding frequencies-Numpy. FFT Filters in Python/v3 Learn how filter out the frequencies of a signal by using low-pass, high-pass and band-pass FFT filtering. from scipy. Enough talk: try it out! In the simulator, type any time or cycle pattern you'd like to see. Matplotlib is a python library for making publication quality plots using a syntax familiar to MATLAB users. There are two sorts of transforms known as the fractional Fourier transform. pythonでFFT（高速フーリエ変換）を実装しようと思っています コードはご覧の通りです (FFT_sort. Python was created by Guido van Rossum and first released in the early 1990s. correlate function. macosx_10_12_x86_64. 6; Filename, size File type Python version Upload date Hashes; Filename, size mkl_fft-1. , normalized). tags: python Bigdata data feature I haven't written a blog for a long time, so miss it. Check out FFT-accelerated Interpolation-based t-SNE (paper, code, and Python package). How to scale the x- and y-axis in the amplitude spectrum. This guide will use the Teensy 3. Retour haut de page. One common way to perform such an analysis is to use a Fast Fourier Transform (FFT) to convert the sound from the frequency domain to the time domain. 0 and its built in library of DSP functions, including the FFT, to apply the Fourier transform to audio signals. External Links. DFT is a mathematical technique which is used in converting spatial data into frequency data. To challenge the algorithm, the application analyses about 22,000 sample blocks in real time: the sound is captured at a 44,100 Hz rate and a 16 bits sample size, and the analysis is performed twice a second. This page is intended to be a place to collect wisdom about the differences, mostly for the purpose of helping proficient MATLAB® users become proficient NumPy and SciPy users. Even with the FFT, the time required to calculate the Fourier transform is a tremendous bottleneck in image processing. If you do not have a CUDA-capable GPU, you can access one of the thousands of GPUs available from cloud service providers including Amazon AWS, Microsoft Azure and IBM SoftLayer. Discussion / Question. To run it, please create a file called output. The output Y is the same size as X. Example 1: Low-Pass Filtering by FFT Convolution. The second step of 2D Fourier transform is a second 1D Fourier transform in the orthogonal direction (column by column, Oy), performed on the result of the first one. fft; import matplotlib. The DFT is in general defined for complex inputs and outputs, and a single-frequency component at linear frequency is represented by a complex exponential , where is the sampling interval. An algorithm to numerically invert functions in the Laplace field is presented. FFTs were first discussed by Cooley and Tukey (1965), although Gauss had actually described the critical factorization step as early as 1805 (Bergland 1969, Strang 1993). Detailed documentation is provided before each class in the fftw++. autosummary:: :toctree: generated/ fft Discrete Fourier transform. We use a Python-based approach to put together complex. For the remainder of this post we’ll use a more established Fast Fourier Transform algorithm from the Python numpy library. Numba supports Intel and AMD x86, POWER8/9, and ARM CPUs, NVIDIA and AMD GPUs, Python 2. In computer science lingo, the FFT reduces the number of computations needed for a problem of size N from O(N^2) to O(NlogN). The use of integer processing results in a tradeoff between speed and accuracy, but where speed is paramount it can do a 256-bin transform in 2. plotly as py import numpy as np # Learn about API authentication here:. The example python program creates two sine waves and adds them before fed into the numpy. The Arduino FFT library is a fast implementation of a standard FFT algorithm which operates on only real data. 12 KB def. The regular Python modules numpy. The most general case allows for complex numbers at the input and results in a sequence of equal length, again of complex numbers. io/ Source code repository and issue. One important application is for the analysis of sound.
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2020-12-04 05:02:17
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https://brilliant.org/problems/problem-locus/
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# Problem Locus
Geometry Level pending
The circle $$x^{2} + y^{2} - 4x - 4y + 4$$ is inscribed in a triangle whose two sides lie on the co-ordinate axes.The locus of the circumcenter of this triangle can be represented as $$ax + by - cxy +k(\sqrt(x^{2} + y^{2})$$ =0.Where a,b,c,d are real constants.
Find the value of a+b+c+k.
×
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2017-07-22 20:51:51
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http://mathhelpforum.com/calculus/89849-volume-around-axis.html
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# Thread: Volume around the Axis
1. ## Volume around the Axis
Hello there,
7. Find the volume of a solid whose base is the region perpindicular to the y axis betweeen the x axis and the curve $y=4-x^2$, and whose perpindicular cross sections are equilateral triangles with a side that lies on the base.
I'm really not sure what to do here. I can't find anything like this in our notes.
Thanks for any help.
2. Originally Posted by karisrou
Hello there,
7. Find the volume of a solid whose base is the region perpindicular to the y axis betweeen the x axis and the curve y=4 - x $y=4-x^2$, and whose perpindicular cross sections are equilateral triangles with a side that lies on the base.
I'm really not sure what to do here. I can't find anything like this in our notes.
Thanks for any help.
your problem statement is disjointed and confusing ... is that the exact wording as you read it?
3. Yeah, thats exactly as it appeared on the paper
4. including the coordinate axes, please confirm that two functions are region boundaries ... the line $y = 4-x$ and the curve $y = 4-x^2$
5. its just $4-x^2$
sorry, messed that up
6. area of an equilateral triangle with side length "s" is
$A = \frac{\sqrt{3}}{4} s^2$
side length for each cross-section is the horizontal distance from the y-axis to the curve
$y = 4-x^2$.
using the curve's equation, solve for the horizontal distance $x$ and use that expression in terms of $y$ to set up an integral in the form $V = \int_c^d A(y) \, dy$
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2013-12-09 05:15:38
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https://gamedev.stackexchange.com/questions/42111/basic-use-of-applyimpulse
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# Basic Use of ApplyImpulse
I am trying to apply a force to a bunch of b2_dynamicBodys, but it seems to only work for a random number of items and then stops with an error.
//create some items to move
bodyDef.type = b2Body.b2_dynamicBody;
for(var i = 0; i < 5; ++i) {
fixDef.shape = new b2PolygonShape;
fixDef.shape.SetAsBox(1,1);
fixDef.friction = 1;
fixDef.restitution = .1;
bodyDef.position.x = Math.random() * 10;
bodyDef.position.y = Math.random() * 10;
bodyDef.linearDamping=1;
bodyDef.angularDamping=.8;
itemsArray.push(world.CreateBody(bodyDef).CreateFixture(fixDef)); // store for later
}
then i try to apply a force later with:
angle = 20;
for (var xIdx=0; xIdx<itemArray.length; xIdx++) {
itemsArray[xIdx].GetBody().ApplyImpulse(new b2Vec2(50*Math.cos(angle*Math.PI/180),50*Math.sin(angle*Math.PI/180)););
}
TypeError: 'undefined' is not an object (evaluating 'c.x')
Is there something wrong with saving the items for later use when I am creating them? Does anyone know what is causing this.
Is diceArray.length used in this code:
for (var xIdx=0; xIdx<diceArray.length; xIdx++) {
the same as itemsArray.length when referencing itemsArray in the same loop?
itemsArray[xIdx].GetBody()....
So, itemsArray size is definitely 5, is diceArray size 5 too?
Edited now jfiddle source given: ApplyImpulse requires 2 parameters and you are only passing in 1.
I added {x:200,y:200} and all is fine
diceArray[xIdx].GetBody().ApplyImpulse(actionVec, {x:200,y:200});
• sorry typo, yes they are the same. I renamed the array. jsfiddle.net/nycynik/LGq63 – nycynik Nov 3 '12 at 15:40
• after adding that fiddle, i think i see that it never works, its not random at all. – nycynik Nov 3 '12 at 15:43
• See edit above for solution – John Nov 3 '12 at 15:54
• Oh! Yes it does public function ApplyImpulse(impulse:b2Vec2,point:b2Vec2):voidParameters The point is the origin of the force. Thank you! – nycynik Nov 3 '12 at 16:00
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2020-02-28 06:24:07
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http://agenzialenarduzzi.it/qmnv/u7d3-notes-transformations-of-exponential-functions-answers.html
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U7d3 Notes Transformations Of Exponential Functions Answers
11 Exponential and Logarithmic Functions Worksheet Concepts: • Rules of Exponents • Exponential Functions – Power Functions vs. We can graph exponential functions. change at a point for a rational function. Chapter 2 Venture Capital 19. Here is a set of practice problems to accompany the Solving Exponential Equations section of the Exponential and Logarithm Functions chapter of the notes for Paul Dawkins Algebra course at Lamar University. Practice: Graphs of exponential functions. Functions and Inverses is covered in the first term of grade twelve in a period of about three weeks. a letter or symbol that represents a numberSuppose the point (2, 4) is on the graph of y = f(x). For example, f (x) = 2 x and g(x) = 5ƒ3 x are exponential functions. 2 Transformations of Exponential Functions. Absolute value—vertical shift up 5, horizontal shift right 3. Transformations of the Sine and Cosine Graph – An Exploration. 2 Submit Answers to this Form Log in Help: Use cobb email --> "firstname. y k 2M 0a Wd5el 9wPiwthr kI jn cfMiHnIi qt meU yA3lwgDejb krRa Z U10. GO Topic: Geometric means 7. This is over all of Unit 4 Exponential Functions. ) List the toolbox functions. Identifying and + is the growth/decay rate is the transformation Horizontal asymptote @ moves horizontal asymptote “Parent” Function Because (growth/decay rate). 4 Log Practice - Answer Key - with work 10/8/14 Copy these notes if you missed class on Wednesday, October 8th. You can identify a y-transformation as changes are made outside the brackets of y=f(x). Transformations of Quadratic Functions C B D A x y 0 x y x y 0 x B. scarfox166. Euler's identity combines e, i, pi, 1, and 0 in an elegant and entirely non-obvious way and it is recognized as one of the most. UNIT OVERVIEW & PURPOSE: This unit emphasizes the real-world applications of parent functions and their transformations. Class Notes. • The parent function, y = b x, will always have a y-intercept of one, occurring at the ordered pair of (0,1). It has a page for each type of transformation. 6) These tasks were taken from the GSE Frameworks. ) 10 10 -4 -2 -4- -6- -8- 10- Determine whether the function given by the table is linear, exponential, or neither. Graphing transformations of exponential functions. 2 Logarithms. 1 Applications of Exponential Functions Perhaps the most well-known application of exponential functions comes from the nancial world. Exponential functions always have some positive number other than 1 as the base. 3 Evaluating Logarithms. 8-7 Radical Functions 623 Use the information on the previous page to answer the following. is a decreasing function if Figure B with base is defined by , and is any positive real. " 4 is the exponent to which 10 must be raised to produce 10,000. Functions notes and exercises - CIMT, University of Plymouth; Functions leading to quadratic equations - Maths4Everyone on TES; Functions Full Coverage GCSE Questions - compiled by Dr Frost (print-friendly version) [back to top] Graphing Functions. Bashore Weebly: About Exponential and Log Functions Unit Notes and Homework. Then, I will ask my students to describe how the graph of f(x) would be related to the graph of g(x), without graphing the two functions (MP2). 3 Laws of Logarithms. identify if it's linear, exponential, or neither from a table - PRACTICE; identify if it's linear, exponential, or neither from context - PRACTICE; create explicit function for exponential patterns - PRACTICE; Topic 9. Downward shift: , this is a shift in y. 4 Graphing Log & Exponential Functions Notes NoteKey Homework Video 4. 1 The Exponential Function and Its Inverse. (1) Graphs of Trigonometric Functions (Basic Intro to sin, cos and tan) (2) Graphs of Trigonometric Functions (Simple Graph Transformations) (3) Graphs of Trigonometric Functions (Examples of Transformations) (4) Graphs of Trigonometric Functions (Unit Circle or CAST Diagram) (5) Graphs of Trigonometric Functions (Using graphs to solve equations). If you have difficulty, then arrange a time to get help. Use transformations to sketch the function 7. Originally used for a GCSE Higher tier set. 2 1, 2 2, 2 3, 2 4,2 5 and so on). However, with the use of the exponential function, we can put a function into fractional form. Usually, to find the Inverse Laplace Transform of a function, we use the property of linearity of the Laplace Transform. Graphing Transformations of Exponential Functions. Reflection through x -axis. We will go into that more below. Functions notes and exercises - CIMT, University of Plymouth; Functions leading to quadratic equations - Maths4Everyone on TES; Functions Full Coverage GCSE Questions - compiled by Dr Frost (print-friendly version) [back to top] Graphing Functions. Students will graph most of the exponential functions in this lesson by hand. Linear and nonlinear functions page 2 answer key Linear and nonlinear functions page 2 answer key. Graph exponential functions. In each case the order of the functions matters because arithmetically the outcomes will be different. Here are guided notes for you that cover exponential parent functions, vertical transformations, horizontal transformations, and reflections. 2 Transformations of Logarithmic Functions. 1: Understand that a function from one set (called the domain) to another set (called the range) assigns to each element of the domain exactly one element of the range. Given an exponential or logarithmic function, the student will describe the effects of parameter changes. Transformations “after” the original function. 2 Construct linear and exponential functions, including arithmetic and geometric sequences, given a graph, a description of a relationship, or two input-output pairs (include reading these from a table). This is an exponential growth curve, where the y-value increases and the slope of the curve increases as x increases. If you don't see any interesting for you, use our search form on bottom ↓. Remember, rotation is based off of COUNTER clockwise movement. 2_transformations_of_logarithm_functions. Logarithm And Exponential Cheat Sheet. Exponential Functions TEKS Aligned: A9C, A9DThis resource includes a one-sided notes page used to teach the basics of exponential functions and a one-sided practice page to complete the one-day lesson plan. The logarithm is actually the exponent to which the base is raised to obtain its argument. 11 Exponential and Logarithmic Functions Worksheet Concepts: • Rules of Exponents • Exponential Functions – Power Functions vs. The x coordinates are unaffected. Find the inverse of f x x( ) 2 3 2. 1 Parent Functions and Transformations Pg 8, #3-6, 9-18, 19-33 odd. 3 Laws of Logarithms. Graphing Transformations of Exponential Functions. scarfox166. 6 Inverse of a Function; Ch 5 Review; Ch 5 Test; Chapter 6: Exponential & Logarithmic Functions. 2 Construct linear and exponential functions, including arithmetic and geometric sequences, given a graph, a description of a relationship, or two input-output pairs (include reading these from a table). 2 Properties of Parents; 1. U4D4_T Exploring Exponential Functions. Simplify each expression. Linear - lesson notes. Answer Key Graphing Quadratics Practice Quiz Quiz Answer Key Transformations of Quadratic Equations Calculator Exploration Exploration Key Transformations of Quadratic Equations Practice Assignment Completing the Square Guided Notes Guided Notes (completed) Practice Assignment Part 1 Assignment Part 1 Key Practice Assignment Part 2. Formative Assessment. Print the following review worksheet. The distribution function for the pdf is given by (corresponding to the cumulative distribution function for the discrete case). Unit 3 : Transformations of Functions 3. 6 Graphical Transformations; Review Notes; Unit 1 Review; Unit 2- Polynomial, Power and Rational Functions. Quiz on Wednesday. Fun, visual skills bring learning to life and adapt to each student's level. Use MathJax to format equations. UNIT 1 - Transformations in the Coordinate Plane; UNIT 2 - Similarity Congruence, Proofs; UNIT 3 - Right Triangle Trigonometry; UNIT 4 - Circles & Volume; UNIT 5 - Geometric & Algebraic Connections. 2 Investigating the Properties of Sinusoidal Functions 6. ⃣Classify exponential functions in function notation as growth or decay ⃣Determine the domain, range, and end behavior (horizontal asymptotes) of an exponential function when looking at a graph 7. Like logarithmic and exponential functions, rational functions may have asymptotes. Reflections. 4: Consider extending this unit to include the relationship between properties of. An equation gives the relationship between variables and numbers. 12: Exponential vs. Determine whether each equation represents a linear or an exponential function. Note the two graphs are symmetrical with respect to one another across the y-axis. pdf: File Size: 948 kb: File Type: pdf: Download File. Be able to define the number e 4. Given a square root function or a rational function, the student will determine the effect on the graph when f(x) is replaced by af(x), f(x) + d, f(bx), and f(x - c) for specific positive and negative values. Math Algebra II Transformations of functions Graphs of exponential functions. Graphing Exponential Functions With e, Transformations, Domain and Range, Asymptotes, Precalculus This algebra 2 and precalculus video tutorial focuses on graphing exponential functions with e and using transformations. LINEAR & EXPONENTIAL FUNCTIONS - 2. Use this information to determine the domain of each of the following functions. Complete the practice problems (found on slides and guided notes) and check the answers here. Support Vector Machines. Uses of Differential Calculus & Integral Calculus - Notes The "Rule of Five" - Notes Parent Functions - Notes Transformations Outline - Notes Transformations Summary - Notes Transformations Worksheet. com community of teachers, mentors and students just like you that can answer any question you might have on Math. Created Date: 2/6/2013 12:50:50 AM. Geometry Worksheets. 2 Construct linear and exponential functions, including arithmetic and geometric sequences, given a graph, a description of a relationship, or two input-output pairs (include reading these from a table). g(x) = e. Answers to Algebra 2 L1. y is the exponent. to write and apply exponential functions from two points to recognize an equation from a set of points create and solve doubling time and half life equations 1) Write an exponential function y = ab x whose graph passes through (1,12) and (3, 108). (4-1) This is a transformation of the random variable X into the random variable Y. Transformations of Functions Name _ Date _ Give the name of the parent function and describe the. Show all work!!! K E Y C O N C E P T E X P O N E N T I A L E Q U A T I O N L a b e l a ll p a rt s t o t h e e q u a t i o n. Identifying Transformations Work with a partner. y-transformations. In your notes you were provided a Rotation Summary on how the x and y values change based on the 1. 1 Distinguish between situations that can be modeled with linear functions and with exponential functions. Although this lesson may not fully equip students with the ability to answer all such test questions successfully, students who participate in active lessons like this one eventually will develop the conceptual understanding needed to succeed on these and other state assessment. Translations of Exponential Functions. There is no formative assessment for this lesson. click here for the syllabus to sign up for "remind", text @h8ac7f to 81010. 2 answers Thursday, 3/19: Lesson: Exponential Functions & Money- VIDEO LESSON *take notes* [video correction to #4. • The parent function, y = b x, will always have a y-intercept of one, occurring at the ordered pair of (0,1). U3D1 Worksheet Solutions Functions Relations D & R: 2: 3. Graphing Transformations of Exponential Functions. 405 do 3-27 multiples of 3 (check your answers with the textbook answer key in the link above). Understanding Relations and Functions - Lesson 3. com, enter the. 2: Transformations of Linear and Exponential Functions Warm-Up 3. 1: Logarithms Section 8. In this section, we cover a method used to quickly sketch graphs related to some basic functions. For example, you can graph h (x) = 2 (x+3) + 1 by transforming the parent graph of f (x) = 2 x. Application Walkthrough. y k 2M 0a Wd5el 9wPiwthr kI jn cfMiHnIi qt meU yA3lwgDejb krRa Z U10. f(x) = a x. 2 Modeling Exponential Growth and Decay ⃣Write an equation that describes how two things are related based on a real world context. Answers is the place to go to get the answers you need and to ask the questions you want. Identifying Transformations Work with a partner. Concepts dealt with include sets; composite and inverse functions; exponential and logarithmic functions; and graph transformations. MathJax reference. ) Links to Other Pages Precalculus Sample Exams. Graph yx2 2 and find its inverse and graph it. 6 Inverse of a Function; Ch 5 Review; Ch 5 Test; Chapter 6: Exponential & Logarithmic Functions. Log Functions. 6 Solving Exponential and Logarithmic Equations Review for Unit 9. Simplify: 2 4 3 5 7 36xy 64x y ¸ ¸ ¹ · ¨ ¨ ©. And, since the exponential function is continuous, you can use the Composition Limit Law to bring the limit inside the exponential function. For any positive number a>0, there is a function f : R ! (0,1)called an exponential function that is defined as f(x)=ax. Vertical Translations: [ Interactive Graph] If k is any positive real number then,. Exponential functions. Statistics 110 (Probability), which has been taught at Harvard University by Joe Blitzstein (Professor of the Practice in Statistics, Harvard University) each year since 2006. August 14, 2012 4:48 PM 7 Page 1 Page 2 of 2. Excel Functions: Excel supplies two functions for exponential regression, namely GROWTH and LOGEST. a letter or symbol that represents a numberSuppose the point (2, 4) is on the graph of y = f(x). " 4 is the exponent to which 10 must be raised to produce 10,000. Other Convergence Tests 96 4. Created Date: 10/17/2014 1:55:58 PM. 2 Modeling Exponential Growth and Decay ⃣Write an equation that describes how two things are related based on a real world context. Remember that the independent variable must appear in the exponent for the function to be exponential. Course Description: Pre-Calculus provides students an honors-level study of trigonometry, advanced functions, analytic geometry, and data analysis in preparation for calculus. Transformations of Exponential Functions Notes Transformations of Exponential Functions Video Packet: pg. The real exponential function : → can be characterized in a variety of equivalent ways. If you've ever earned interest in the bank (or even if you haven't), you've probably heard of "compounding", "appreciation", or "depreciation"; these have to do with exponential functions. 1 Transformations of Quadratic Functions 51 Writing a Transformed Quadratic Function Let the graph of g be a translation 3 units right and 2 units up, followed by a refl ection in the y-axis of the graph of f(x) = x2 − 5x. Exponential growth and decay by a factor. Transfer function, Laplace transform, Low pass filter _____ 1. Class Notes. GUIDED NOTES - Lesson 6-1a. 13 Simple vs. 2 Intro to Logarithms Notes NoteKey Homework Video 4. Therefore, the graph of y= (x - 5)2 is the graph of y= x2 translated horizontally 5 units to the right. Excel Functions: Excel supplies two functions for exponential regression, namely GROWTH and LOGEST. Youmay have seen that there are two notations popularly used for natural logarithms, log e and ln. Function 1: y=7x+4. COURSE INFO UNIT 7: LOGARITHMIC & EXPONENTIAL FUNCTIONS TEST – THURSDAY DEC 20 Review of Functions, Basic graphs (Notes provided). Practice: Graphs of exponential functions. Basic algebra skills: Algebraic Operations: substituting numbers for letters, simplifying expressions,. We also worked on writing the characteristics of exponential graphs. Where a>0 and a is not equal to 1. f (x) = log 1/4 x, g(x) = log 1/4(4x) − 5 Writing Transformations of Graphs of Functions. You invested $1000 in a savings account at the end of 6th grade. Teachers Notes: Transformations of exponential functions can be confusing for some students as some transformations can be replicated with parameters in different locations. Which is an exponential function (circle)? f(x) = x2 g(x) = 2x Graph: f(x) = 2x o Domain? o Range? o y-intercept? o x-intercept? The graph has an asymptote. Rewrite the function to identify h and k. HW: Lesson 17 Packet. The prize at the end will be combining your newfound Algebra skills in trigonometry and using complex variables to gain a full understanding of Euler’s identity. f (x) = log 1/4 x, g(x) = log 1/4(4x) − 5 Writing Transformations of Graphs of Functions. The graph of an exponential or logarithmic function can be used to predict the greatest and least instantaneous rates of change and when they occur. Identifying and + is the growth/decay rate is the transformation Horizontal asymptote @ moves horizontal asymptote "Parent" Function Because (growth/decay rate). Transformation Rules Sheet Pdf. College Algebra Problems with Answers; sample 1: Quadratic Functions. Exponential Real World ApplicationsGrowth and Decay Rates of Exponential Functions Transformations of Exponential Functions Formula for Geometric Series (Honors Only) Review Summative Assessment Compound Interest 3 1/21 – 1/25 Unit 8: Summary of Functions Comparing Linear, Quadratic, and Exponential Functions -- Part 1 Comparing Linear. Be able to define the number e 4. 2 Properties of Parents; 1. 4 Log Practice - Answer Key - with work 10/8/14 Copy these notes if you missed class on Wednesday, October 8th. 38202663467 is an approximate answer because we have rounded the value of Ln(80). Transformation Of Graphs And Equations Activity Answer Key. UNIT 1 - Transformations in the Coordinate Plane; UNIT 2 - Similarity Congruence, Proofs; UNIT 3 - Right Triangle Trigonometry; UNIT 4 - Circles & Volume; UNIT 5 - Geometric & Algebraic Connections. Therefore, the graph of y- 2 2 = 2x is the graph of y= x translated vertically 2 units up. We also discuss some identities relating these functions, and mention. U3D1 Worksheet Solutions Functions Relations D & R: 2: 3. Identifying Functions Notes Completed Notes Identifying Functions Assignment Assignment Key Comparing Functions Notes Completed Notes Comparing Functions Assignment Assignment Key Linear vs. This makes it easier to keep up with what you're learning in IB Math and be prepared for all the exams for the class. It should not be taught in isolation but rather linked to the algebraic concepts already taught. To see the text of a State Standard, hover your pointer over the Standard. Quiz on Wednesday. Transformations of exponential graphs behave similarly to those of other functions. Show how you can determine the answer without relying on a calculator. For any positive number a>0, there is a function f : R ! (0,1)called an exponential function that is defined as f(x)=ax. Just as with other parent functions, we can apply the four types of transformations—shifts, reflections, stretches, and compressions—to the parent function $f\left(x\right)={b}^{x}$ without loss of shape. State the domain and range. 1 The Exponential Function and Its Inverse. Transformations of the Sine and Cosine Graph - An Exploration. Transformations “after” the original function. The x coordinates are unaffected but all the y coordinates go up by 4. Function Family Fun. ) LOGARITHMIC FUNCTIONS log b x =y means that x =by where x >0, b >0, b ≠1 Think: Raise b to the power of y to obtain x. Winking Unit 4-5 page 118. It has a page for each type of transformation. 1 SLO: I can define an exponential function, analyse its graph, and solve problems involving exponential growth or decay. 2 Modeling Exponential Growth and Decay ⃣Write an equation that describes how two things are related based on a real world context. Transformation Of Functions. Algebra 1 CCSS Lessons and Practice is a free site for students (and teachers) studying a first year of high school algebra under the Common Core State Standards. Unit 4 - Exponential Functions. And, that's all there really is to the activity. Well thought through material that relates perfectly to the high school math curriculum. Exponential functions increase based on a percent of the original. 1 Exponential Growth 9. Thanks for putting in the effort AND sharing!. Com was under cyber attack, whereby someone dumped a lot of trash content into the question and answer area. A rational function is a function thatcan be written as a ratio of two polynomials. Functions notes and exercises - CIMT, University of Plymouth; Functions leading to quadratic equations - Maths4Everyone on TES; Functions Full Coverage GCSE Questions - compiled by Dr Frost (print-friendly version) [back to top] Graphing Functions. The F distribution (Snedecor's F distribution or the FisherSnedecor distribution) represents continuous probability distribution which occurs frequently as null distribution of test statistics. Graphing Transformations of Exponential Functions. Chapter 5: Exponential and Logarithmic Functions 5-1 Exponential Functions Exponential Functions : - a function where the input (x) is the exponent of a numerical base, a. For exponential models, express as a logarithm the solution to abct = d where a, c, and d are numbers and the base b is 2, 10, or e; evaluate the logarithm using technology. District programs, activities, and practices shall be free from discrimination based on race, color, ancestry, national origin, ethnic group identification, age, religion, marital or parental status, physical or mental disability, sex, sexual orientation, gender, gender identity or expression, or genetic information; the perception of one or more of such characteristics; or association with a. Correct answers will move you ahead, incorrect answers are highlighted in red and require correction before you advance to the next input cell. Natural Exponential Functions and their Graphs. If you think about it, having a negative number (such as -2) as the base wouldn't be very useful, since the even powers would give you positive answers (such as "(-2) 2 = 4") and the odd powers would give you negative answers (such as "(-2) 3 = -8"), and what would you even do with the powers that aren't. Graphing Rational Functions 23. y is the exponent. Title: Microsoft Word - 3. Use a logistic growth model to answer questions in context. 5 transformations of exponential functions worksheet. Vertical translation. U7D3 Notes – Transformations of Exponential Functions Name_____ Desmos Marbleslides Exponentials - Go to student. Express log 4 (10) in terms of b. 3 Transformations of functions In this course we learn to identify a variety of functions: linear functions, quadratic and cubic functions, general polynomial and rational functions, exponential and logarithmic functions, trigonometric functions and inverse trig functions. Describe the sequence of Worksheet on Transformations of Exponential Functions Answer Section MULTIPLE CHOICE 1. Homework: pg 212 # 1-4(ac), 7ac, 8ac Transformations of Exponential Functions: Part 1. Identifying Transformations Work with a partner. Course Description: Algebra II continues students' study of advanced algebraic concepts including functions, polynomials, rational expressions, systems of functions and inequalities, and matrices. Write the velocity function h for the Moon, and use it to estimate the downward velocity of a landing craft at the end of a bounce 50 ft in height. Function 1: y=7x+4. Exponential Functions. ELECTRICAL SYSTEMS Analysis of the three basic passive elements R, C and L Simple lag network (low pass filter) 1. -½ h(x - 4) + 1 #5 - 8. 2 Solution Sets, Conjunctions, and Disjunctions: Worksheet #1: 1-19 every other odd, 20-27: 1. HW: Lesson 17 Packet. 7 Page 2 7. Practice Solutions. For problems #10 -14, given the parent function and a description of the transformation(s), write the equation of the transformed function, f(x). Similarly, one of the definitions of the term quadratic is a square. In other words, an exponential function does not take two different values to the same number. An experienced CMP2 user will answer your questions about implementing and teaching CMP2. This is because of the doubling behavior of the exponential. fg(2) ( 2) d. Identifying Functions Notes Completed Notes Identifying Functions Assignment Assignment Key Comparing Functions Notes Completed Notes Comparing Functions Assignment Assignment Key Linear vs. 1 Transformations PARENT FUNCTION - Square Root Function √ √ Exponential Function Logarithmic Function Greatest Integer Function Rational Function Graph the following, use a table of values to help out if necessary. The function f(x) = 3 x is one-to-one, so it does not take two different values to 9, so x must equal 2. Function Family Fun. This lesson was created for the MCR3U Functions course in the. Here is a set of practice problems to accompany the Solving Exponential Equations section of the Exponential and Logarithm Functions chapter of the notes for Paul Dawkins Algebra course at Lamar University. Model with mathematics. After studying this chapter you will be comfortable with function and ready to dive in differential and integral calculus. Transformations “after” the original function. The practice problems assess your understanding of transformations. Exponential Real World ApplicationsGrowth and Decay Rates of Exponential Functions Transformations of Exponential Functions Formula for Geometric Series (Honors Only) Review Summative Assessment Compound Interest 3 1/21 - 1/25 Unit 8: Summary of Functions Comparing Linear, Quadratic, and Exponential Functions -- Part 1 Comparing Linear. Topic 2 - Functions and equations (24 hours) The aims of this topic are to explore the notion of a function as a unifying theme in mathematics, and to apply functional methods to a variety of mathematical situations. Exponential Growth and Decay 80 Chapter 4. 3 Laws of Logarithms. 19: AK – Characteristics of Exponentials HW; Today’s assignment is to watch the following video on Transformations and fill in the notes on page 6 in Unit 4 packet. 2 Transformations Topic Transformations SOL AFDA. See an example of how concept development builds across grades. An example of an exponential function is the growth of bacteria. Finding exact values of all 6 trigonometric ratios from the Unit Circle; Took a Quiz; Notes from today can be downloaded here:. Sampling from the distribution corresponds to solving the equation. 2 Logarithms. 1 Periodic Functions and Their Properties 6. " 4 is the exponent to which 10 must be raised to produce 10,000. Upward shift: , this is a shift in y. This fraction worksheet will generate 10 or 15 multiplying fraction problems per. Pre-Public Practice Exam Answers to Multiple Choice (2014) Math 3201 Midterm Exam Review With Solutions. 3 Transformations of Logarithmic Functions 52. Remember to use your real name so I. THE LAPLACE TRANSFORMATION L 3. Geometric transformations with matrices. The concept of a structure. Algebra 1 Unit 4: Exponential Functions Notes 3 Asymptotes An asymptote is a line that an exponential graph gets closer and closer to but never touches or crosses. 258&260) Today we are going to work with transformations of exponential functions. List Big Ideas for each lesson of Unit 3. Papers 1 and 2: Paper codes: 9MA0/01 and 9MA0/02. This introduction to exponential functions will be limited to just two types of transformations: vertical shifting and reflecting across the x-axis. a letter or symbol that represents a numberSuppose the point (2, 4) is on the graph of y = f(x). Students will be expected to describe and translate among graphic, algebraic, numeric, tabular, and verbal representations of relations and use those representations to solve problems. Use transformations to sketch the function 7. Transformations of the Sine and Cosine Graph - An Exploration. The practice problems assess your understanding of transformations. Correct answers will move you ahead, incorrect answers are highlighted in red and require correction before you advance to the next input cell. click here for the syllabus to sign up for "remind", text @h8ac7f to 81010. Provide details and share your research! But avoid … Asking for help, clarification, or responding to other answers. 7 Combinations of Transformations. Quadratic Functions and Inequalities Properties of parabolas Graphing exponential functions. Logarithmic Functions Review Sheet. & a)&!=! 8!&& & & & & b)&!=8! c)&!=−8!& & & & & & d)&!=8! 5)&Write&the&equation&for&the. Videos, examples, solutions, activities and worksheets for studying, practice and review of precalculus, Lines and Planes, Functions and Transformation of Graphs, Polynomials, Rational Functions, Limits of a Function, Complex Numbers, Exponential Functions, Logarithmic Functions, Conic Sections, Matrices, Sequences and Series, Probability and Combinatorics, Advanced Trigonometry, Vectors and. It happens mostly during analysis of variance or F-test. 2 Transformations of Logarithmic Functions. Rewrite the function to identify h and k. Homework-Inverse Functions. UNIT 1 - Transformations in the Coordinate Plane; UNIT 2 - Similarity Congruence, Proofs; UNIT 3 - Right Triangle Trigonometry; UNIT 4 - Circles & Volume; UNIT 5 - Geometric & Algebraic Connections. • evaluate exponential functions • graph exponential functions • use transformations to graph exponential functions • use compound interest formulas An exponential function f with base b is defined by f ( or x) = bx y = bx, where b > 0, b ≠ 1, and x is any real number. Logarithm And Exponential Cheat Sheet. Evaluate and simplify expressions containing rational exponents. Transformation Rules Sheet Pdf. Chapter 5 5. scarfox166. 1 Characteristics of an Exponential Function 7. In this section let c be a positive real number. This is over all of Unit 4 Exponential Functions. It has a page for each type of transformation. Unit 3b — Linear and Exponential Functions Date: Day 39 - Notes Transformations of Linear and Exponential Graphs Use your graphing calculator to graph the following. So this just gives us the constant function f(x) = 1. Determine the domain, range, and horizontal asymptote of the function. This exploration is about recognizing what happens to the graph of the exponential function when you change one or more of the coefficients a, b, c, and d. Horizontal and Vertical Stretches and Compressions Vertical and Horizontal Translations or Shifts Ex: Reflect a Point about the x-axis, y-axis, and the Origin Reflections across the x-axis and y-axis Summary of Function Transformations Function Transformation Exploration with. (For teachers only) Cross-grade Level Connections. 15) whose real and imaginary parts are harmonic functions for arbitrary a. 4 Log Practice - Answer Key - with work 10/8/14 Copy these notes if you missed class on Wednesday, October 8th. Similarly, one of the definitions of the term quadratic is a square. 3 Parent Functions; 1. Odd/Even worksheet. Then graph each function. Given an exponential or logarithmic function, the student will describe the effects of parameter changes. Graph exponential functions using transformations. identify if it's linear, exponential, or neither from a table - PRACTICE; identify if it's linear, exponential, or neither from context - PRACTICE; create explicit function for exponential patterns - PRACTICE; Topic 9. Well thought through material that relates perfectly to the high school math curriculum. Plus each one comes with an answer key. The Results for Transformation Of Function Worksheet Pdf. Describe the sequence of Worksheet on Transformations of Exponential Functions Answer Section MULTIPLE CHOICE 1. 2) Ready, Set, Go Homework: Linear and Exponential Functions 4. Tennis Ball Bounce Discussion. ) If the base b is greater than 1 then the result is exponential growth. Basics of Geometry, Answer Key 7. GO Topic: Geometric means 7. 12 Writing Equations of Functions. Graphing Exponential Functions With e, Transformations, Domain and Range, Asymptotes, Precalculus This algebra 2 and precalculus video tutorial focuses on graphing exponential functions with e and using transformations. Reading Comprehension. Unit 1- Functions and their graphs. Unit 4 lesson 4 (duo-tang day 1) hw solutions: Week2. 1c Compoud Growth and Decay Models Review of Exponential Function Transformations Homework: page 339 (87, 89, 91, 97, 99) page 351-352 (7, 11, 21, 25, 27) *Parent Conferences - November 14, 2018* Thursday. Students, teachers, parents, and everyone can find solutions to their math problems instantly. 4 Combining Functions; 1. Just as with other parent functions, we can apply the four types of transformations—shifts, reflections, stretches, and compressions—to the parent function $f\left(x\right)={b}^{x}$ without loss of shape. A rational function is a function thatcan be written as a ratio of two polynomials. Exponential Growth and Decay *(These topics, and the links shown below, should be helpful to students preparing for the Precalculus Diagnostic Exam at UCD. Review Lecture; 1. Graphing exponential functions. Example 1 : Graph the following fucntions by creating a small table of values. LOGEST is the exponential counterpart to the linear regression function LINEST described in Testing the Slope of the Regression Line. Representation of Functions as Power Series 100 4. State the domain and range. Under the new plan, your bonus would be$1 the first month, $2 the second month,$4 the third month, and so on, for 12 months. In addition to linear, quadratic, rational, and radical functions, there are exponential functions. Youmay have seen that there are two notations popularly used for natural logarithms, log e and ln. October 3 Exponential Functions Unit Test TODAY Homework: From Wednesday’s class – Pg. Graphing Transformations of Exponential Functions. Uses of Differential Calculus & Integral Calculus - Notes The "Rule of Five" - Notes Parent Functions - Notes Transformations Outline - Notes Transformations Summary - Notes Transformations Worksheet. U3D1_T Functions Relations D _ R. The first portion of the book is an investigation of functions, exploring the graphical behavior of, interpretation of, and solutions to problems involving linear, polynomial, rational, exponential, and. Graph of Functions(2), Graph Transformation with solution. Exponential and Logarithmic Functions Worksheets October 3, 2019 August 28, 2019 Some of the worksheets below are Exponential and Logarithmic Functions Worksheets, the rules for Logarithms, useful properties of logarithms, Simplifying Logarithmic Expressions, Graphing Exponential Functions, …. So this just gives us the constant function f(x) = 1. Chapter 5: Exponential and Logarithmic Functions 5-1 Exponential Functions Exponential Functions : - a function where the input (x) is the exponent of a numerical base, a. Math Algebra II Transformations of functions Graphs of exponential functions. We also worked on writing the characteristics of exponential graphs. • Graph exponential functions using transformations. Unit 4 lesson 4 (duo-tang day 1) hw solutions: Week2. Transformations of the Sine and Cosine Graph – An Exploration. For problems #10 -14, given the parent function and a description of the transformation(s), write the equation of the transformed function, f(x). Answer Keys can be found by clicking on the following link. Now dilate f(x) by a scale factor of 6. System of Equations PDF of systems notes. October 3 Exponential Functions Unit Test TODAY Homework: From Wednesday's class - Pg. College Algebra. The real exponential function : → can be characterized in a variety of equivalent ways. 2 Submit Answers to this Form Log in Help: Use cobb email --> "firstname. jmap by topic To access Regents, Practice and Journal Worksheets, Lesson Plans, Videos and other resources, click on the State Standard in the last column below. 3 Parent Functions; 1. Exponential Functions. 1 Values of Inverse Trig Functions File Uploaded 02/6/19, 15:10 11. Given an exponential or logarithmic function, the student will describe the effects of parameter changes. Graphing Part II: Graphs of Transformed Exponential Functions. Tennis Ball Bounce Discussion. In addition to linear, quadratic, rational, and radical functions, there are exponential functions. Factor a out of the absolute value to make the coefficient of equal to. 12 Writing Equations of Functions. Thanks for putting in the effort AND sharing!. Parent Linear Function: Parent Quadratic Function: Parent Square Root Function: y = Parent Absolute Value Function: y = Ix Family of Exponential Functions: there is no single easily identifiable parent exponential function so we'll refer to this as the Family of Exponential Functions. Home > Math Worksheets > Functions > Transformations of Functions When we move or re-position a graph of a function this is called a transformation. Coyle - Lesson-Notes - 6. KEY to Chart of Parent Functions with their Graphs, Tables, and Equations Name of Parent Function Graph of Function Table of Values. Guided Notes. Graphing Exponential Functions With e, Transformations, Domain and Range, Asymptotes, Precalculus This algebra 2 and precalculus video tutorial focuses on graphing exponential functions with e and using transformations. 6) These tasks were taken from the GSE Frameworks. docx Author: Trevor Jensen. y = 16x 16. Finding Exponential Functions From a Table. Just as with other parent functions, we can apply the four types of transformations—shifts, reflections, stretches, and compressions—to the parent function $f\left(x\right)={b}^{x}$ without loss of shape. The prize at the end will be combining your newfound Algebra skills in trigonometry and using complex variables to gain a full understanding of Euler’s identity. Then graph each function. Therefore, the graph of y= (x - 5)2 is the graph of y= x2 translated horizontally 5 units to the right. Page 335 What is an exponential function? Page 336 - 338 Example 1 a and b Practice Questions. I had to restore from backup from last night. moves a figure up, down, left, or right. • Graph exponential functions using transformations. Exponential functions have the form f(x) = b x, where b > 0 and b ≠ 1. The same rules apply when transforming logarithmic and exponential functions. So, the graph of g is a refl ection in the x-axis and a vertical shrink by a. These were passed on to me and hopefully you will find them useful, especially if you are short on time!!. Powered by Create your own unique website with customizable templates. Then determine its domain, range, and horizontal asymptote. We also worked on writing the characteristics of exponential graphs. Unit 4: Logarithmic and Exponential Functions 3/8 notes on evaluating logs & log properties HW is pg 1-3 (see which problems here) answers to pg 1-3 will be given in class 3/9 notes on solving log and exponential equations (day 1) HW is pg 5-6 answers to pg 5-6 3/12 notes on solving log and exponential equations (day 2) HW is pg 8-11 #1-41 odd. CW: Topic 6 Exponential and Logarithmic Functions Guided Notes. Uses of Differential Calculus & Integral Calculus - Notes The "Rule of Five" - Notes Parent Functions - Notes Transformations Outline - Notes Transformations Summary - Notes Transformations Worksheet. Exponential Functions TEKS Aligned: A9C, A9DThis resource includes a one-sided notes page used to teach the basics of exponential functions and a one-sided practice page to complete the one-day lesson plan. = 2 ∣ xDistributive Property− 3 ∣− 10 The transformed function is h(x) = 2 ∣ x− 3 ∣− 10. We can graph exponential functions. Graphs of exponential functions. 1 Characteristics of an Exponential Function 7. Exponential Decay In the form y = ab x, if b is a number between 0 and 1, the function represents exponential decay. Applications and modeling will be included throughout the course of study. For any positive number a>0, there is a function f : R ! (0,1)called an exponential function that is defined as f(x)=ax. 2 Transformations Topic Transformations SOL AFDA. Transformations of the Sine and Cosine Graph - An Exploration. Foundations of NC Math II Honors Goals Students will focus on the following topics…Modeling with Geometry, Quadratic Functions, Exponential Functions, Advanced Function Types, Trigonometric Functions, and Probability. Any state $\psi$ can be described in terms of a set of base states by giving the amplitudes to be in each of the base states. Transfer function and the Laplace transformation _____ 1. Nature sometimes gives us problems that cannot be modeled using the basic exponential, f(x) = a x. 3 More Solving Trig Equations File Uploaded 02/6/19, 15:11. Answers to Homework. An equation gives the relationship between variables and numbers. Show how you can determine the answer without relying on a calculator. Write your answer in interval notation. F(x)-4 Directions: (a) Identify The Parent Function And (b) Describe The. We can use the transformations of sine and cosine functions in numerous applications. Model with mathematics. SWBAT choose the graph that matches a given equation and explain how the parameters in the function y=a/(x-b)+c affect the graph using their previous knowledge of transformations. But e is the amount of growth after 1 unit of time , so $\ln(e) = 1$. g(x) 6x 3 b. scarfox166. U7D3 Notes - Transformations of Exponential Functions Name_____ Desmos Marbleslides Exponentials - Go to. Recognize a logistic growth function and when it is appropriate to use. Students consider real world objects and data that can be described using exponential functions. Unit 1 Transformations Test Review Solutions. 5 Making Connections: Logarithmic Scales in the Physical Sciences. Then complete the worksheet that is at the end of the. 3 Transformations of Logarithmic Functions 52. Papers 1 and 2: Paper codes: 9MA0/01 and 9MA0/02. ; Simplify without calculator: log 6 (216) + [ log(42) - log(6) ] / log(49). Provide details and share your research! But avoid … Asking for help, clarification, or responding to other answers. Both papers are 2 hour written exam papers which are weighted at 33. Logarithm And Exponential Cheat Sheet. Graph exponential functions. Write the rule for g(x) after the transformation. Tennis Ball Bounce Discussion. Model with mathematics. 3 Transformations of Logarithmic Functions. The Results for Transformation Of Function Worksheet Pdf. Ln(80) is the exact answer and x=4. Because there are only two constants (a and b), only two points are needed to determine a power. For instance, just as the quadratic function maintains its parabolic shape. Today you will be taking an assessment on Quizizz. Here is the graph of f (x. List Big Ideas for each lesson of Unit 3. Transformation Of Graphs And Equations Activity Answer Key. Domain and Range of Exponential and Logarithmic Functions The domain of a function is the specific set of values that the independent variable in a function can take on. 1 SLO: I can define an exponential function, analyse its graph, and solve problems involving exponential growth or decay. 6 Graphical Transformations; Review Notes; Unit 1 Review; Unit 2- Polynomial, Power and Rational Functions. Math Algebra II Transformations of functions Graphs of exponential Transforming exponential graphs (example 2) Graphing exponential functions. Graph y x2x e. Therefore, add 8 to both sides:. This exploration is about recognizing what happens to the graph of the exponential function when you change one or more of the coefficients a, b, c, and d. After we were finished, I had them turn the graphic organizer over and write notes on the back. Based on what you know about exponential functions and the behavior of dice, explain in words why an exponential model would be appropriate for this situation. Identifying rates of change in linear and exponential functions (F. If the Base of the Exponential is the. It is crucial for high school students to have the ability to analyze and interpret data in a practical real-world setting. 7 Links verified on 11/13/2016. Taylor and MacLaurin Series 103 3. Correct answers will move you ahead, incorrect answers are highlighted in red and require correction before you advance to the next input cell. Transform exponential and logarithmic functions by changing parameters Describe the effects of changes in the coefficients of exponential and logarithmic functions Who uses this? Psychologists can use transformations of exponential functions to describe knowledge retention rates over time. scarfox166. You can identify a y-transformation as changes are made outside the brackets of y=f(x). Inverse Functions Worksheet. Graphs of exponential functions. Uses of Differential Calculus & Integral Calculus - Notes The "Rule of Five" - Notes Parent Functions - Notes Transformations Outline - Notes Transformations Summary - Notes Transformations Worksheet. Multiply radicals. Notes Exponential and Logarithmic Equations Notes. Logarithm And Exponential Cheat Sheet. 98 (12 issues) $2. Answers to Algebra 2 L1. 8-7 Radical Functions 623 Use the information on the previous page to answer the following. Exponential functions have the form: ; where , and x is any real number. Understanding Relations and Functions - Lesson 3. In addition to linear, quadratic, rational, and radical functions, there are exponential functions. ) LOGARITHMIC FUNCTIONS log b x =y means that x =by where x >0, b >0, b ≠1 Think: Raise b to the power of y to obtain x. Further resources: Formulae and Statistical tables. 4 Power Law of Logarithms. Graph exponential functions using transformations. X Disclaimer. Math eLearning Guide- Week 6 A L G E B R A I C C O N N E C T I O N S. An experienced CMP2 user will answer your questions about implementing and teaching CMP2. Model with mathematics. a letter or symbol that represents a numberSuppose the point (2, 4) is on the graph of y = f(x). • Graph exponential functions. The real exponential function : → can be characterized in a variety of equivalent ways. Integration (including volumes of revolution) Trigonometry 1 (inverse functions). 2_transformations_of_logarithm_functions. Just as in any exponential expression, b is called the base and x is called the exponent. " 4 is the exponent to which 10 must be raised to produce 10,000. In each of the three examples the variable x is in the exponent, which makes each of the examples exponential functions. pdf: File Size: 948 kb: File Type: pdf: Download File. Stretches / Compressions. State the domain and range. " The PSI functions are just an alternative way to define your model. Here is a set of practice problems to accompany the Solving Exponential Equations section of the Exponential and Logarithm Functions chapter of the notes for Paul Dawkins Algebra course at Lamar University. Infinite Sequences and Series 83 4. An example of an exponential function is the growth of bacteria. 258&260) Today we are going to work with transformations of exponential functions. 2: Graphing and finding properties of the root function and the reciprocal function. Monday, April 20 - Today you will explore transformations of exponential functions. You invested$1000 in a savings account at the end of 6th grade. com, enter the class code given by your teacher, and type your First and Last name. fg(2) ( 2) d. Sequences 83 4. This is the currently selected item. angelica_bustos9. " 4 is the exponent to which 10 must be raised to produce 10,000. In general, if we have the function then the graph will be moved left c units if c is positive and right c units if c is negative. August 14, 2012 4:48 PM 7 Page 1 Page 2 of 2. Some of the worksheets displayed are Exponential transformations work, 4 1 exponential functions and their graphs, Transformations of graphs date period, Lesson 3, Exponential functions date period, Transformations of exponential and logarithmic functions, Work 1. 5 Making Connections: Logarithmic Scales in the Physical Sciences. U7D3 Notes – Transformations of Exponential Functions Name_____ Desmos Marbleslides Exponentials - Go to student. In other words, an exponential function does not take two different values to the same number. The rst general method that we present is called the inverse transform method. 2: Graphing and finding properties of the root function and the reciprocal function. 2 The Natural Base e; 6. Transformations of exponential graphs behave similarly to those of other functions. • Graph exponential functions using transformations. Homework-Inverse Functions. 3 Class Notes - Part 1 Quiz on Lessons 5. CW: Topic 6 Exponential and Logarithmic Functions Guided Notes. Students complete a table to show the coordinates of g(x) 5 Af (B(x 2 C) 1 D. Section 5: Transforming Exponential Functions, and. Indefinite Integrals. With each page, the students discover the transformation by graphing the parent function and filling in two tab. Displaying all worksheets related to - Transformation Of Functions. The syllabus aims to build learners' confidence by helping them develop a feel for numbers, patterns and relationships, and places a strong emphasis on solving. 3 Laws of Logarithms. Identifying Transformations Work with a partner. What is image and pre-image? When a transformation occurs, the original figure is known as the pre-image and the new figure is known as the. Write the velocity function h for the Moon, and use it to estimate the downward velocity of a landing craft at the end of a bounce 50 ft in height. Because there are only two constants (a and b), only two points are needed to determine a power. How do graphical transformations affect the tables of values? How does a transformation affect a point found at the origin as compared to a point on an axis or a point in one of the four quadrants? How can a rational function of the form y = (ax+b) / (cx+d) be considered as a transformation of the reciprocal function y = 1/x ?.
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# The volume of a cube with edge 3 is how many times the volume of a cub
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The volume of a cube with edge 3 is how many times the volume of a cub [#permalink]
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The volume of a cube with edge 3 is how many times the volume of a cube with edge $$\sqrt{3}$$?
a. $$\frac{1}{3}$$
b. $$1$$
c. $$3$$
d. $$3\sqrt{3}$$
e. $$9$$
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The volume of a cube with edge 3 is how many times the volume of a cub [#permalink]
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15 Apr 2018, 12:51
QZ wrote:
The volume of a cube with edge $$3$$ is how many times the volume of a cube with edge $$\sqrt{3}$$?
a. $$\frac{1}{3}$$
b. $$1$$
c. $$3$$
d. $$3\sqrt{3}$$
e. $$9$$
I. Volume/Volume
Volume of a cube = $$s^3$$
Volume of smaller cube = $$(\sqrt{3})^3 = 3\sqrt{3}$$
Volume of larger cube = $$3^3 = 27$$
Volume of $$27$$ is how many times greater than $$3\sqrt{3}$$?
$$\frac{27}{3\sqrt{3}} =$$
$$\frac{27}{3\sqrt{3}} * \frac{\sqrt{3}}{\sqrt{3}}=$$
$$\frac{27\sqrt{3}}{9} = 3\sqrt{3}$$
II. Cube the scale factor
• SCALE FACTOR: If a shape's size increases or decreases, it scales up or scales down.
That means that every length in the shape has been multiplied by a scale factor, $$k$$
The scale factor is a multiplier; any change in length = length * scale factor $$k$$
Scale factors tell you "how many times"
the smaller size was multiplied to obtain the greater size
You need nothing else in this problem except (scale factor)$$^3$$
• To account for change in length, area, or volume:
Length = $$k$$
Area = (length * length) = $$k^2$$
Volume= (length * length * length) = $$k^3$$
• Scale factor here?
Use ONE length's increase to find $$k$$:
$$(k) * (s$$ of small cube) = ($$s$$ of large cube)
Scale factor EQUALS?*
Small cube's side length: $$\sqrt{3}$$
Large cube's side length = 3
$$k * \sqrt{3} = 3$$
$$\sqrt{3}*\sqrt{3} = 3$$
$$k = \sqrt{3}$$
Volume increase? $$k^3$$
To find out "how many times greater," because it's a volume change --
cube the scale factor
$$(\sqrt{3})^3 = (\sqrt{3} * \sqrt{3} * \sqrt{3}) = 3\sqrt{3}$$
The volume of a cube with edge $$3$$ is $$3\sqrt{3}$$ times the volume of a cube with edge $$\sqrt{3}$$
*Or $$k * \sqrt{3} = 3$$
$$k = \frac{3}{\sqrt{3}}$$
$$k = \frac{3}{\sqrt{3}} * \frac{\sqrt{3}}{\sqrt{3}}$$
$$k=\frac{3\sqrt{3}}{3}=\sqrt{3}$$
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The volume of a cube with edge 3 is how many times the volume of a cub &nbs [#permalink] 15 Apr 2018, 12:51
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2018-08-17 05:23:48
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https://www.tensorflow.org/versions/r1.8/api_docs/python/tf/contrib/distributions/Autoregressive
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# tf.contrib.distributions.Autoregressive
## Class Autoregressive
Inherits From: Distribution
Autoregressive distributions.
The Autoregressive distribution enables learning (often) richer multivariate distributions by repeatedly applying a diffeomorphic transformation (such as implemented by Bijectors). Regarding terminology,
"Autoregressive models decompose the joint density as a product of conditionals, and model each conditional in turn. Normalizing flows transform a base density (e.g. a standard Gaussian) into the target density by an invertible transformation with tractable Jacobian." [(Papamakarios et al., 2016)][1]
In other words, the "autoregressive property" is equivalent to the decomposition, p(x) = prod{ p(x[i] | x[0:i]) : i=0, ..., d }. The provided shift_and_log_scale_fn, masked_autoregressive_default_template, achieves this property by zeroing out weights in its masked_dense layers.
Practically speaking the autoregressive property means that there exists a permutation of the event coordinates such that each coordinate is a diffeomorphic function of only preceding coordinates [(van den Oord et al., 2016)][2].
#### Mathematical Details
The probability function is
prob(x; fn, n) = fn(x).prob(x)
And a sample is generated by
x = fn(...fn(fn(x0).sample()).sample()).sample()
where the ellipses (...) represent n-2 composed calls to fn, fn constructs a tf.distributions.Distribution-like instance, and x0 is a fixed initializing Tensor.
#### Examples
tfd = tf.contrib.distributions
def normal_fn(self, event_size):
n = event_size * (event_size + 1) / 2
p = tf.Variable(tfd.Normal(loc=0., scale=1.).sample(n))
affine = tfd.bijectors.Affine(
scale_tril=tfd.fill_triangular(0.25 * p))
def _fn(samples):
scale = math_ops.exp(affine.forward(samples)).eval()
return independent_lib.Independent(
normal_lib.Normal(loc=0., scale=scale, validate_args=True),
reinterpreted_batch_ndims=1)
return _fn
batch_and_event_shape = [3, 2, 4]
sample0 = array_ops.zeros(batch_and_event_shape)
ar = autoregressive_lib.Autoregressive(
self._normal_fn(batch_and_event_shape[-1]), sample0)
x = ar.sample([6, 5])
# ==> x.shape = [6, 5, 3, 2, 4]
prob_x = ar.prob(x)
# ==> x.shape = [6, 5, 3, 2]
#### References
[1]: George Papamakarios, Theo Pavlakou, and Iain Murray. Masked Autoregressive Flow for Density Estimation. In Neural Information Processing Systems, 2017. https://arxiv.org/abs/1705.07057
[2]: Aaron van den Oord, Nal Kalchbrenner, Oriol Vinyals, Lasse Espeholt, Alex Graves, and Koray Kavukcuoglu. Conditional Image Generation with PixelCNN Decoders. In Neural Information Processing Systems, 2016. https://arxiv.org/abs/1606.05328
## Properties
### allow_nan_stats
Python bool describing behavior when a stat is undefined.
Stats return +/- infinity when it makes sense. E.g., the variance of a Cauchy distribution is infinity. However, sometimes the statistic is undefined, e.g., if a distribution's pdf does not achieve a maximum within the support of the distribution, the mode is undefined. If the mean is undefined, then by definition the variance is undefined. E.g. the mean for Student's T for df = 1 is undefined (no clear way to say it is either + or - infinity), so the variance = E[(X - mean)**2] is also undefined.
#### Returns:
• allow_nan_stats: Python bool.
### batch_shape
Shape of a single sample from a single event index as a TensorShape.
May be partially defined or unknown.
The batch dimensions are indexes into independent, non-identical parameterizations of this distribution.
#### Returns:
• batch_shape: TensorShape, possibly unknown.
### dtype
The DType of Tensors handled by this Distribution.
### event_shape
Shape of a single sample from a single batch as a TensorShape.
May be partially defined or unknown.
#### Returns:
• event_shape: TensorShape, possibly unknown.
### name
Name prepended to all ops created by this Distribution.
### parameters
Dictionary of parameters used to instantiate this Distribution.
### reparameterization_type
Describes how samples from the distribution are reparameterized.
Currently this is one of the static instances distributions.FULLY_REPARAMETERIZED or distributions.NOT_REPARAMETERIZED.
#### Returns:
An instance of ReparameterizationType.
### validate_args
Python bool indicating possibly expensive checks are enabled.
## Methods
### __init__
__init__(
distribution_fn,
sample0=None,
num_steps=None,
validate_args=False,
allow_nan_stats=True,
name='Autoregressive'
)
Construct an Autoregressive distribution.
#### Args:
• distribution_fn: Python callable which constructs a tf.distributions.Distribution-like instance from a Tensor (e.g., sample0). The function must respect the "autoregressive property", i.e., there exists a permutation of event such that each coordinate is a diffeomorphic function of on preceding coordinates.
• sample0: Initial input to distribution_fn; used to build the distribution in __init__ which in turn specifies this distribution's properties, e.g., event_shape, batch_shape, dtype. If unspecified, then distribution_fn should be default constructable.
• num_steps: Number of times distribution_fn is composed from samples, e.g., num_steps=2 implies distribution_fn(distribution_fn(sample0).sample(n)).sample().
• validate_args: Python bool. Whether to validate input with asserts. If validate_args is False, and the inputs are invalid, correct behavior is not guaranteed.
• allow_nan_stats: Python bool, default True. When True, statistics (e.g., mean, mode, variance) use the value "NaN" to indicate the result is undefined. When False, an exception is raised if one or more of the statistic's batch members are undefined.
• name: Python str name prefixed to Ops created by this class. Default value: "Autoregressive".
#### Raises:
• ValueError: if num_steps and distribution_fn(sample0).event_shape.num_elements() are both None.
• ValueError: if num_steps < 1.
### batch_shape_tensor
batch_shape_tensor(name='batch_shape_tensor')
Shape of a single sample from a single event index as a 1-D Tensor.
The batch dimensions are indexes into independent, non-identical parameterizations of this distribution.
#### Args:
• name: name to give to the op
#### Returns:
• batch_shape: Tensor.
### cdf
cdf(
value,
name='cdf'
)
Cumulative distribution function.
Given random variable X, the cumulative distribution function cdf is:
cdf(x) := P[X <= x]
#### Args:
• value: float or double Tensor.
• name: Python str prepended to names of ops created by this function.
#### Returns:
• cdf: a Tensor of shape sample_shape(x) + self.batch_shape with values of type self.dtype.
### copy
copy(**override_parameters_kwargs)
Creates a deep copy of the distribution.
#### Args:
• **override_parameters_kwargs: String/value dictionary of initialization arguments to override with new values.
#### Returns:
• distribution: A new instance of type(self) initialized from the union of self.parameters and override_parameters_kwargs, i.e., dict(self.parameters, **override_parameters_kwargs).
### covariance
covariance(name='covariance')
Covariance.
Covariance is (possibly) defined only for non-scalar-event distributions.
For example, for a length-k, vector-valued distribution, it is calculated as,
Cov[i, j] = Covariance(X_i, X_j) = E[(X_i - E[X_i]) (X_j - E[X_j])]
where Cov is a (batch of) k x k matrix, 0 <= (i, j) < k, and E denotes expectation.
Alternatively, for non-vector, multivariate distributions (e.g., matrix-valued, Wishart), Covariance shall return a (batch of) matrices under some vectorization of the events, i.e.,
Cov[i, j] = Covariance(Vec(X)_i, Vec(X)_j) = [as above]
where Cov is a (batch of) k' x k' matrices, 0 <= (i, j) < k' = reduce_prod(event_shape), and Vec is some function mapping indices of this distribution's event dimensions to indices of a length-k' vector.
#### Args:
• name: Python str prepended to names of ops created by this function.
#### Returns:
• covariance: Floating-point Tensor with shape [B1, ..., Bn, k', k'] where the first n dimensions are batch coordinates and k' = reduce_prod(self.event_shape).
### cross_entropy
cross_entropy(
other,
name='cross_entropy'
)
Computes the (Shannon) cross entropy.
Denote this distribution (self) by P and the other distribution by Q. Assuming P, Q are absolutely continuous with respect to one another and permit densities p(x) dr(x) and q(x) dr(x), (Shanon) cross entropy is defined as:
H[P, Q] = E_p[-log q(X)] = -int_F p(x) log q(x) dr(x)
where F denotes the support of the random variable X ~ P.
#### Returns:
• cross_entropy: self.dtype Tensor with shape [B1, ..., Bn] representing n different calculations of (Shanon) cross entropy.
### entropy
entropy(name='entropy')
Shannon entropy in nats.
### event_shape_tensor
event_shape_tensor(name='event_shape_tensor')
Shape of a single sample from a single batch as a 1-D int32 Tensor.
#### Args:
• name: name to give to the op
#### Returns:
• event_shape: Tensor.
### is_scalar_batch
is_scalar_batch(name='is_scalar_batch')
Indicates that batch_shape == [].
#### Args:
• name: Python str prepended to names of ops created by this function.
#### Returns:
• is_scalar_batch: bool scalar Tensor.
### is_scalar_event
is_scalar_event(name='is_scalar_event')
Indicates that event_shape == [].
#### Args:
• name: Python str prepended to names of ops created by this function.
#### Returns:
• is_scalar_event: bool scalar Tensor.
### kl_divergence
kl_divergence(
other,
name='kl_divergence'
)
Computes the Kullback--Leibler divergence.
Denote this distribution (self) by p and the other distribution by q. Assuming p, q are absolutely continuous with respect to reference measure r, the KL divergence is defined as:
KL[p, q] = E_p[log(p(X)/q(X))]
= -int_F p(x) log q(x) dr(x) + int_F p(x) log p(x) dr(x)
= H[p, q] - H[p]
where F denotes the support of the random variable X ~ p, H[., .] denotes (Shanon) cross entropy, and H[.] denotes (Shanon) entropy.
#### Returns:
• kl_divergence: self.dtype Tensor with shape [B1, ..., Bn] representing n different calculations of the Kullback-Leibler divergence.
### log_cdf
log_cdf(
value,
name='log_cdf'
)
Log cumulative distribution function.
Given random variable X, the cumulative distribution function cdf is:
log_cdf(x) := Log[ P[X <= x] ]
Often, a numerical approximation can be used for log_cdf(x) that yields a more accurate answer than simply taking the logarithm of the cdf when x << -1.
#### Args:
• value: float or double Tensor.
• name: Python str prepended to names of ops created by this function.
#### Returns:
• logcdf: a Tensor of shape sample_shape(x) + self.batch_shape with values of type self.dtype.
### log_prob
log_prob(
value,
name='log_prob'
)
Log probability density/mass function.
#### Args:
• value: float or double Tensor.
• name: Python str prepended to names of ops created by this function.
#### Returns:
• log_prob: a Tensor of shape sample_shape(x) + self.batch_shape with values of type self.dtype.
### log_survival_function
log_survival_function(
value,
name='log_survival_function'
)
Log survival function.
Given random variable X, the survival function is defined:
log_survival_function(x) = Log[ P[X > x] ]
= Log[ 1 - P[X <= x] ]
= Log[ 1 - cdf(x) ]
Typically, different numerical approximations can be used for the log survival function, which are more accurate than 1 - cdf(x) when x >> 1.
#### Args:
• value: float or double Tensor.
• name: Python str prepended to names of ops created by this function.
#### Returns:
Tensor of shape sample_shape(x) + self.batch_shape with values of type self.dtype.
### mean
mean(name='mean')
Mean.
### mode
mode(name='mode')
Mode.
### param_shapes
param_shapes(
cls,
sample_shape,
name='DistributionParamShapes'
)
Shapes of parameters given the desired shape of a call to sample().
This is a class method that describes what key/value arguments are required to instantiate the given Distribution so that a particular shape is returned for that instance's call to sample().
Subclasses should override class method _param_shapes.
#### Args:
• sample_shape: Tensor or python list/tuple. Desired shape of a call to sample().
• name: name to prepend ops with.
#### Returns:
dict of parameter name to Tensor shapes.
### param_static_shapes
param_static_shapes(
cls,
sample_shape
)
param_shapes with static (i.e. TensorShape) shapes.
This is a class method that describes what key/value arguments are required to instantiate the given Distribution so that a particular shape is returned for that instance's call to sample(). Assumes that the sample's shape is known statically.
Subclasses should override class method _param_shapes to return constant-valued tensors when constant values are fed.
#### Args:
• sample_shape: TensorShape or python list/tuple. Desired shape of a call to sample().
#### Returns:
dict of parameter name to TensorShape.
#### Raises:
• ValueError: if sample_shape is a TensorShape and is not fully defined.
### prob
prob(
value,
name='prob'
)
Probability density/mass function.
#### Args:
• value: float or double Tensor.
• name: Python str prepended to names of ops created by this function.
#### Returns:
• prob: a Tensor of shape sample_shape(x) + self.batch_shape with values of type self.dtype.
### quantile
quantile(
value,
name='quantile'
)
Quantile function. Aka "inverse cdf" or "percent point function".
Given random variable X and p in [0, 1], the quantile is:
quantile(p) := x such that P[X <= x] == p
#### Args:
• value: float or double Tensor.
• name: Python str prepended to names of ops created by this function.
#### Returns:
• quantile: a Tensor of shape sample_shape(x) + self.batch_shape with values of type self.dtype.
### sample
sample(
sample_shape=(),
seed=None,
name='sample'
)
Generate samples of the specified shape.
Note that a call to sample() without arguments will generate a single sample.
#### Args:
• sample_shape: 0D or 1D int32 Tensor. Shape of the generated samples.
• seed: Python integer seed for RNG
• name: name to give to the op.
#### Returns:
• samples: a Tensor with prepended dimensions sample_shape.
### stddev
stddev(name='stddev')
Standard deviation.
Standard deviation is defined as,
stddev = E[(X - E[X])**2]**0.5
where X is the random variable associated with this distribution, E denotes expectation, and stddev.shape = batch_shape + event_shape.
#### Args:
• name: Python str prepended to names of ops created by this function.
#### Returns:
• stddev: Floating-point Tensor with shape identical to batch_shape + event_shape, i.e., the same shape as self.mean().
### survival_function
survival_function(
value,
name='survival_function'
)
Survival function.
Given random variable X, the survival function is defined:
survival_function(x) = P[X > x]
= 1 - P[X <= x]
= 1 - cdf(x).
#### Args:
• value: float or double Tensor.
• name: Python str prepended to names of ops created by this function.
#### Returns:
Tensor of shape sample_shape(x) + self.batch_shape with values of type self.dtype.
### variance
variance(name='variance')
Variance.
Variance is defined as,
Var = E[(X - E[X])**2]
where X is the random variable associated with this distribution, E denotes expectation, and Var.shape = batch_shape + event_shape.
#### Args:
• name: Python str prepended to names of ops created by this function.
#### Returns:
• variance: Floating-point Tensor with shape identical to batch_shape + event_shape, i.e., the same shape as self.mean().
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2018-08-17 22:27:41
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http://mathhelpforum.com/calculus/13989-system-differential-equations.html
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# Math Help - System of Differential Equations
1. ## System of Differential Equations
2. Here is the first one.
3. Here is the second one.
All you do is plot the equations in parametric form.
The following are some of the solutions (you solution is not included but it should look similar to these ones).
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2014-07-31 01:25:30
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https://www.lil-help.com/questions/3715/xcom-285-week-2-dqs
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XCOM 285 Week 2 DQs
# XCOM 285 Week 2 DQs
Z
0 points
XCOM 285 Week 2 DQs
What characteristics and communication style makes this person an effective communicator?
Communication can be written or oral. What types of written and oral communication do you receive in your workplace? What information do these communications convey and are there more effective ways to deliver these messages?
XCOM 285
zeta29
Z
0 points
Thumbnail of first page
Excerpt from file: Think about someone in your workplace you consider an effective communicator. What characteristics and communication style makes this person an effective communicator? IconsidertheCaptainofmyshifttobeaveryeffectivecommunicator.Hisappearanceshowsa
Filename: XCOM 285 Week 2 week2dqs.docx
Filesize: < 2 MB
Print Length: 2 Pages/Slides
Words: NA
Account not required
Surround your text in *italics* or **bold**, to write a math equation use, for example, $x^2+2x+1=0$ or $$\beta^2-1=0$$
Use LaTeX to type formulas and markdown to format text. See example.
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2017-02-20 13:10:23
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http://projecteuclid.org/euclid.dmj/1077304104
|
## Duke Mathematical Journal
### The topology of isospectral manifolds of tridiagonal matrices
Carlos Tomei
#### Article information
Source
Duke Math. J. Volume 51, Number 4 (1984), 981-996.
Dates
First available in Project Euclid: 20 February 2004
http://projecteuclid.org/euclid.dmj/1077304104
Digital Object Identifier
doi:10.1215/S0012-7094-84-05144-5
Mathematical Reviews number (MathSciNet)
MR771391
Zentralblatt MATH identifier
0558.57006
#### Citation
Tomei, Carlos. The topology of isospectral manifolds of tridiagonal matrices. Duke Math. J. 51 (1984), no. 4, 981--996. doi:10.1215/S0012-7094-84-05144-5. http://projecteuclid.org/euclid.dmj/1077304104.
#### References
• [Davis] M. Davis, Groups generated by reflections and aspherical manifolds not covered by Euclidean space, Ann. of Math. (2) 117 (1983), no. 2, 293–324.
• [Knuth] Knuth, Sorting and searching, The Art of Computer Programming, vol. 3, Addison-Wesley, 1973.
• [MKS] Magnus, Karrass, and Solitar, Combinatorial group theory: Presentations of groups in terms of generators and relations, Interscience Publishers [John Wiley & Sons, Inc.], New York-London-Sydney, 1966.
• [Massey] W. S. Massey, Singular Homology Theory, Graduate Texts in Mathematics, vol. 70, Springer-Verlag, New York, 1980.
• [Moerbeke] P. Van Moerbeke, The spectrum of Jacobi matrices, Invent. Math. 37 (1976), no. 1, 45–81.
• [Moser] J. Moser, Finitely many mass points on the line under the influence of an exponential potential–an integrable system, Dynamical systems, theory and applications (Rencontres, BattelleRes. Inst., Seattle, Wash., 1974), Springer, Berlin, 1975, 467–497. Lecture Notes in Phys., Vol. 38.
• [Quinn] F. Quinn, Ends of maps. III. Dimensions $4$ and $5$, J. Differential Geom. 17 (1982), no. 3, 503–521.
|
2016-09-30 10:01:19
|
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|
https://0xrick.github.io/binary-exploitation/bof2/
|
# Buffer Overflow Examples, Overwriting a variable value on the stack - Protostar Stack1 , Stack2
## Introduction
So last week I talked about buffer overflows and solved Protostar Stack0. Today I’m gonna solve Stack1 and Stack2, they are not different from Stack0 in their objective which is changing a variable’s value , but they are different in the way of changing that variable for sure. With that being said let’s jump right in !
If you haven’t read my previous post yet I recommend reading it before this.
## Stack1
For this challenge we got this code :
### Breakdown
So this code :
creates a variable called “modified” and assigns a buffer of 64 chars to it.
Checks if we supplied an argument or not.
Sets the value of the “modified” variable into 0 , then it copies whatever we give it argv[1] into the buffer of “modified”.
Then it checks if the variable’s value is 0x61626364 or not
## Solution
So it’s similar to Stack0 except we need to set the value of the variable into a specific value which is 0x61626364 in this case. This is the hexadecimal value of “dcba” now keep in mind that when reading hex you read it from right to left not left to right. To slove this our input will be 64 chars then after that the value , let’s try it.
Let’s execute stack1
We get please specify an argument so let’s enter anything.
We get try again you got 0x00000000 , Let’s try to change that by exceeding the buffer and entering any char for example “b”
And we see that the value changed to 0x00000062 which is the hex value of “b” so our exploit is working, Let’s apply that.
And we did it !
But can we do it in another way ? instead of entering ASCII we can use the hex values and python will translate them.
## Stack2
For this challenge we got this code :
### Breakdown
This code :
As always creates a variable called “modified” and assigns a buffer of 64 chars to it
The new thing here is a variable called “variable” which gets its value from an environment variable called “GREENIE”.
The program checks if the variable “variable” has a value or not
Sets the value of modified to 0
Then it copies the value of “variable” into the buffer of “modified”
After that it checks if the value of “modified” is 0x0d0a0d0a or not
## Solution
So this time we can’t specify the value directly instead of that we have to do it through an environment variable. And this is actually a good example that shows the importance of being creative with your exploits , you won’t face the same situation every time so you have to come up with an exploit that fits with the application’s functionality.
In this situation we can apply our usual exploit and store that value into the environment variable “GREENIE”.
But wait a minute , what is an environment variable ?
### Environment Variables
Simply , Environment variables are variables that are being used to store values of some stuff that the system uses also the services can access those variables. For a better demonstration let’s see an example.
If we take the environment variable BASH and look at it :
We see that its value is : /bin/bash
So for example if I want to run bash I type bash in the terminal instead of ./bin/bash because the system calls the variable BASH and finds its path.
But the system won’t store a variable for every binary that exists on the system so there’s an environment variable called PATH which has all the possible dirs that could contain bins, So when you type for example : python, it searches in those dirs then executes python
So you get the idea. Understanding environment variables is not necessary for this challenge but I wanted to talk about it just to make everything clear for anyone.
Now let’s execute stack2
We get Please set the GREENIE environment variable.
There’s no environment variable called GREENIE so we need to create it. We will make it’s value equal to 64 chars then 0x0d0a0d0a
This time we won’t be able to use the value in ASCII like we did with stack1 because 0x0d is a return \r and 0x0a is a new line \n and we can’t type those so we will use the hex values and python will translate them.
When we look at it we only see the 64 A’s and that’s because we can’t see the new line or the return
Now we need to export it to the environment variables list then we are good to go.
And we did it !
That’s it , Feedback is appreciated !
|
2021-06-19 22:07:46
|
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|
https://dsp.stackexchange.com/questions/26794/how-to-represent-a-given-equation-more-clear-profressional-and-short-form/26799
|
# How to represent a given equation more clear, profressional and short form?
I have a image $U_{m \times n}:\Omega \to \mathbb R^2$, the output $P$ can be define as $$P=\mu J_{m \times n} - U$$ where $\mu = \max \{ u_{ij} : 1 \leq i \leq m, 1 \leq j \leq n\}$, $J_{m \times n}$ be the $m \times n$ matrix whose $i, j$th component is $1$: that is, the all-ones matrix. (This notation isn't quite standard, but it's as close to standard as I know. $J$ is often the all-ones matrix)
However, it is so many sentences for expression the above equation. Do we have more short and standard way to represent it? As my found, the $J_{m \times n}$ can be expressed by indicator function such as
$$P=\max (U)\times 1_{\Omega}-U$$
where $1_{\Omega}$ is Indicator function
Does it equivalent with original meaning? If not, please give me a standard and common expression in image processing. Thanks
• I would have defined $U_{m\times n}: [1,\ldots,m]\times [1,\ldots,n] \to \mathbb{R}$ instead, if you really want to have a professional notation. Unless your image is 2-valued, which would pose another challenge in defining a scalar $\max$. – Laurent Duval Nov 15 '15 at 11:11
I think the notation $J_{m,n}$ is pretty standard for an all-ones matrix. So the only possible clarification would be to dispense with $\mu$ and directly specify what it actually is:
$$P=||U||_{\max}J_{m,n}-U$$
where $||.||_{\max}$ denotes the max norm of a matrix.
• Thank Matt L. How about indicator function instead of $J$. Based on your comment, it will be $$P=||U||_{\max}1_{\Omega}-U$$ – Jame Nov 1 '15 at 12:29
• @user8430: I'm not sure if you could use it to specify the all-ones matrix. Furthermore, you would need to specify what the subset $\Omega$ is. I believe that using $J_{m,n}$ is much clearer, and probably the indicator function can't even actually be used in that way. – Matt L. Nov 1 '15 at 12:37
• $\Omega$ is image domain, then $1_{\Omega}$ means 1 values in whole image domain. Actually, I want to reduce notation in that form. Thank for your max norm notation. For image processing, people do not like write form as size $J_{m \times n}$ – Jame Nov 1 '15 at 12:40
• The $J_{n\times n}$ is quite standard for the counter-identity or exchange matrix too, i.e. with ones on the second diagonal. – Laurent Duval Nov 1 '15 at 14:31
• @LaurentDuval: Do you think my $1_{\Omega}$ can represent matrix of one?However, I found that $1_{\Omega}$ likes the function that map a domain in real value. – Jame Nov 1 '15 at 14:44
I would propose \mathbb{1} from the \usepackage{bbold} (or \mathds{1}, see details and options here), with outputs a $1$ with a double bar, like for $\mathbb{R}$ (I cannot have it displayed on SE). I would stick to a simple $\max$ or $\sup$, as it is not a norm without the absolute value. Plus, image values are taken in $\mathbb{R}^2$, as in some astronomical, seismic or processed images. So, something like:
\documentclass{article}
\usepackage{bbold}
\begin{document}
$\max_{i,j} \{ A_{i,j}\} . \mathbb{1}-A$
\end{document}
• I know of the max norm of a matrix (as I've used it in my answer), but I'm not sure if $\max(A)$ is common notation. Wasn't that actually the point of your comment under my answer? – Matt L. Nov 1 '15 at 15:45
• The max of all elements can be negative for instance. Hence it is not a norm. I beleive $\max_{i,j}$ is probably more readable. In my answer I wanted to emphasize on the double-bar one. – Laurent Duval Nov 1 '15 at 15:50
• It is a norm, because it's the maximum of the absolute values of the matrix entries (check the link). This should be OK because we're talking about images (I know in the OP it was just the max, not max |.|, but it should make no difference for the application). – Matt L. Nov 1 '15 at 16:00
• $\mathbb{1}$ doesn't work? It seems to for me: just use \mathbb{1} (in \\$ signs). – Peter K. Nov 3 '15 at 14:21
• It does not appear per se in my browsers (recent Firefox and Chromium). I do not understand why... yet – Laurent Duval Nov 3 '15 at 14:27
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2019-10-15 05:34:31
|
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https://research.paradigm.xyz/MEV
|
Ethereum’s core insight was that flexible smart contracts allow developers to explore a new frontier of permissionless applications. The explosive growth of decentralized financial protocols built on Ethereum (“DeFi”) is a glimpse at what this innovation could enable in the future.
Like programming libraries in the first Internet revolution, DeFi’s “money legos” enable developers to build complex systems by composing and remixing simple building blocks. This complexity also brings novel risks. One of these risks is Miner Extractable Value, or MEV.
## Miner Extractable Value
The concept of MEV was first introduced by Phil Daian in “Flash Boys 2.0,” and more recently popularized by my colleagues Dan Robinson, Georgios Konstantopoulos, and samczsun in “Ethereum is a Dark Forest” and “Escaping the Dark Forest.”
It has become a foundational concept in cryptoeconomics, but what actually is MEV? What are the implications for permissionless blockchains?
## What is MEV?
MEV is a measure of the profit a miner (or validator, sequencer, etc.) can make through their ability to arbitrarily include, exclude, or re-order transactions within the blocks they produce.
Imagine there’s a $10,000 arbitrage opportunity available on Uniswap after a large trade has caused price slippage. An arbitrage bot notices the opportunity and submits a transaction to capture it, offering a$10 txfee to the miner. One of two things may happen:
1. A miner will copy and censor the arbitrageur’s transaction in order to capture the opportunity themselves.
2. Other bots will notice and bid a higher txfee, starting a bidding war for the right to capture the arbitrage. The auction is called a “Priority Gas Auction” (PGA).
The $10,000 potential profit is MEV. If a miner does not capture it, and a PGA is kicked off, the difference between the price at which the auction settles and the total MEV available is the winning trader’s profit (e.g., if a$7,000 fee is paid to a miner, the remaining $3,000 is left to the trader). This example gives a high-resolution view of MEV, but it does not paint the whole picture. MEV is not just a curiosity. These little financial games create incentive ripples, a winding chain of cause and effect that must be followed to see the contagion. This post will explore that thread and explain why MEV may harm Ethereum and its users. As a direct result of DeFi’s inflective success, the known lower bound of Ethereum MEV is growing at an exponential rate. At this pace, we believe that MEV could create meaningful issues within the next year. ## The State of Play It is impossible to say how much MEV is on Ethereum in total. All the MEV which we are currently aware of only constitues the lower bound. This is because MEV can be created any time a user interacts with a blockchain, and smart contracts enable a functionally infinite number of potential interactions. Thus, it is computationally infeasible to calculate a blockchain’s total potential MEV by brute force. However, we can establish a baseline by adding up the MEV that’s known to have been extracted (which is the “realized MEV” shown in the graph above). Then, we can use heuristics to infer how much higher than our baseline the true lower bound could be, and how the qualitative texture of the unexploited MEV could affect the blockchain’s environment. ### MEV Today The defining feature of Ethereum’s current era is that most miners are not attempting to exploit MEV themselves (yet). Nearly all of the current activity is driven by non-mining traders. However, some MEV can only be captured by miners, because they have the authority to arbitrarily order (or exclude) transactions. Non-mining traders can access a strictly smaller subset of “simple” MEV; “complex” preferences cannot be efficiently expressed through PGA’s. This means that we see almost entirely PGA-style MEV being realized. Uniswap arbitrage, like our earlier example, is one the most common flavors of MEV in practice. Another type of MEV seen often in practice is theft from vulnerable smart contracts. One example is described in the “Dark Forest” post by Dan and co. They found a smart contract with a vulnerability that would allow anyone to steal the funds inside; Dan planned to recover the funds by exploiting it before a thief could. However, an arbitrage bot automatically recognized and copied their transaction, replacing their address with its own, and bid a higher transaction fee. The bot’s transaction executed before theirs and made off with the funds. ### MEV Tomorrow The next era of MEV will come when Ethereum miners begin actively exploiting MEV. However, until recently, there was a common hypothesis that miners are altruistic enough to forgo MEV revenue and continue running default node software. Bitcoin miners have empirically chosen not to run selfish mining strategies, so there is some precedent. We think this miner-altruism hypothesis has been proven definitively false in the last 3 months. A small but meaningful portion of the hashrate has been observed exploiting MEV themselves, revenue-sharing with traders rather than allowing permissionless fee auctions, and selling access to private memory pools. Rather, we believe that MEV is now overcoming miners’ threshold activation energy. Feverish non-mining trader activity highlights opportunities that miners can capture more efficiently and profitably. Additionally, the types of MEV that non-mining traders can’t access are a pot of totally untouched miner revenue; that pot may be far larger than the MEV realized to date. At some level, it is more surprising that it took miners this long to become involved. The dam has probably burst: miners will venture further into the frontier, exploring more exotic forms of MEV and collusion. Significant risk could be posed to Ethereum and its users. The rest of this post will explore what this future could look like in more detail, starting with what we mean when referring to MEV’s potential “risks.” ## MEV Can Harm Users MEV is an invisible tax that miners can collect from users. In our earlier Uniswap example, a large trade caused price slippage, creating a$10,000 profit opportunity (MEV). The bot which arbitrages the market back to parity with the true price is making the Uniswap market more efficient without harming the original trader in the process. This is an example of a benign MEV transaction.
However, in a different version of the same trade, an arbitrage bot would recognize the user’s trade before it’s executed and “sandwich” their transaction between a buy and sell order of its own. The net effect is that MEV levies an invisible tax on the user: their order is manipulated into executing at an artificially inflated price, which the bot then sells into for an instant profit. Of course, a miner could do this at no cost to themselves. This is what one might call a malignant MEV transaction.
## MEV Can Harm Ethereum
MEV inherently encourages consensus instability.
Imagine there are two miners, Sam and Dan, who are paid a $100 reward for each block they find. Sam has found 3 blocks, the first of which contained our$10,000 Uniswap arbitrage.
Now, Dan has a choice: he can either mine on top of Sam’s 3 blocks, or he can attempt to re-mine the first block in order to take the Uniswap arbitrage for himself. The $10,000 is much more lucrative than the$100 block reward, and Dan is more rational than honest, so he decides to re-mine the first block.
While Dan’s at it, since the current longest chain is height 3, he also re-mines the second and third blocks (and captures any MEV that was in those, too). After the re-org, Dan owns the longest chain and he and Sam can progress from the third block.
This is known as a “time-bandit” attack: if block rewards are small enough compared to MEV, it can be rational for miners to destabilize consensus.
Our example was a two-party system. In the real multiplayer world, it is possible that every rational miner would attempt to re-org the third block and essentially halt progress. However, this could destroy the value of the miners’ hashrate investments. If we see this behavior at all, it will more likely be in the form of shorter, more frequent re-orgs that do not halt progress entirely.
## Is MEV unique to Ethereum?
No, hypothetically MEV can also be seen on Bitcoin. The incentives to censor Lightning channels or to double-spend colored coins are technically MEV.1 However, our hypothesis is that Bitcoin is inherently less exposed to MEV than blockchains like Etheruem.
The reason for that lies in the complexity and “statefulness” of the respective blockchain:
1. The rate at which MEV accumulates on a given blockchain is generally proportional to the complexity of its application-layer behavior.
2. Arbitrarily flexible protocols, such as Ethereum, cannot bound this complexity and are inherently biased towards greater complexity over time.
3. MEV incentives cannot be easily mitigated without altering Ethereum’s UX.
This is why we say that Ethereum’s complexity may be a curse.
### MEV Follows Complexity
In some purely theoretical sense, even Bitcoin cannot bound its potential MEV exposure. However, Bitcoin’s design discourages unintended high-MEV use cases well enough that, in practice, they are rarely seen. This doesn’t seem likely to change going forward, so we don’t expect that MEV will become a bigger problem for Bitcoin (the inflation is a seperate discussion2).
In contrast, we can observe that the MEV surface on Ethereum is growing exponentially, mainly as a result of the large flows of value through DeFi applications. The financial primitives which seem so promising could alternatively be viewed as parasitic to Ethereum: spinning a boundless web of MEV which grows larger and more complex by the day.
### Ethereum Can’t Bound Complexity
If the Lightning Network created untenable MEV on Bitcoin – realistically threatening Bitcoin’s consensus stability – we could remove the opcodes needed to create payment channels from Bitcoin’s limited ruleset (Script) in a relatively straightforward way.
On the other hand, if we discovered that some application patterns (e.g. DEX’s, lending, tokenized custodial assets, etc.) posed similar risks to Ethereum, it would be impossible to preclude all possible implementations of those behaviors at the level of the EVM. Individual implementations could be forked off, but we could not prevent the general behavior without permissioning contract deployment or severely constraining the EVM. In either case, Ethereum would no longer enable “permissionless smart contracts.”
### MEV is Hard to Fix
Finally, it is natural to ask if Ethereum could build a mechanism to counteract MEV into the protocol. In short, no, at least without altering Ethereum’s developer and/or user experience.
Any attempt to prevent miners from accessing the revenue stream is liable to incentivize the creation of off-protocol markets. For example, if all transactions were only allowed to pay a flat rate, we would expect miners to collude with traders to accept payment for transaction priority out-of-band. Similarly, if all transaction fees were burned or paid to a communal pot, miners would simply accept fees separately.
This is why we say that MEV cannot be easily counteracted. Potential mitigations exist, but they require structural changes to the way Ethereum applications are architected and users interact with them.
## In Conclusion
If Bitcoin’s incentive security fails, at least before block rewards go to ~zero, it’s difficult to imagine that any permissionless blockchain will not suffer a similar fate. Bitcoin’s simplicity is not only aesthetically elegant but also minimizes its extra-protocol incentive surface.
We are more concerned about Ethereum. Ethereum’s application-layer complexity and MEV are continuing to grow exponentially. The known lower-bound on MEV revenues could be larger than the value of ETH miner security incentives within the year.
Large-scale, efficient MEV extraction may make the “tax” on Ethereum users untenable. Ethereum could become congested and more costly for all applications. The platform UX would be impaired, and that could stall Ethereum’s network effects and momentum.
Of course, the main unknown is whether Ethereum miners will begin maximally exploiting MEV at scale. Miners can access a superset of the MEV available to non-mining traders, and can extract all of it with maximum efficiency, so the cost and UX issues could be disastrous.
There is also the possibility of time-bandits, although it feels unlikely that miners would damage their long-term interest in Ethereum with major re-orgs. A lite version, in which miners intentionally uncle or re-org only small handfuls of lucrative blocks, could still be harmful.
In any case, it’s time to seriously consider what measures we can take if the situation deteriorates.
## Mitigating MEV
An ideal solution would simply reduce the MEV on Ethereum, or increase the miner security incentive without additional inflation. Within the Ethereum paradigm, where permissionless applications share in platform security uniformly, our options are limited:
1. Better Application Design: every application can design itself to minimize the amount of MEV it creates. This may be a competitive differentiator, as users will get lower costs and better UX. However, the protocol cannot force applications to do this, and there is a limit to how much MEV can be avoided.
2. Additional Security Incentives: stable miner revenue streams other than the block reward (such as EIP-1559’s burned BASEFEEs3, or state rents) are additive to protocol security and could help offset MEV.
Otherwise, most research is focused on ways to make destabilizing consensus (time-bandit attacks) more difficult or more costly, rather than avoiding the root MEV:
1. Separating Inclusion and Ordering: miners (or validators) could only be responsible for transaction inclusion, and the right to decide the transaction ordering could be auctioned off separately. In theory, this would quarantine the re-org incentive. However, this guarantees that users will always endure the level of MEV extraction admitted by the auction, which may be equivalent to a multi-block time-bandit attack.
2. Finality: Nakamoto Proof-of-Work has only probabilistic finality. BFT-based algorithms have strong finality guarantees, and time-bandit attacks are more difficult because greater collusion is required to re-org even a single finalized block. However, with enough MEV the incentive to re-org could still overcome the difficulty of collusion. Additionally, participants still have the authority to arbitrarily order transactions in blocks for which they are the proposer, so finality alone cannot help with “normal” front-running.
3. Proof-of-Stake: PoS-based blockchains can slash validators who attempt to re-org and thus make time-bandits significantly more costly, especially when combined with strong finality. However, with enough MEV the incentive to re-org could still be greater than the slashing penalty.
All of these approaches have serious implications for Ethereum’s ecosystem. Many involve changes to the core protocol and could take years to implement. Those that could be done only at the application-layer still likely require that developers re-architect and migrate most of the ecosystem to other environments.
Hopefully, the next year will bring more clarity on MEV and Ethereum’s path forward. A number of Paradigm’s portfolio companies are working on MEV mitigations and related problems. If this is of interest to you, don’t be a stranger.
### Rollup
Rollups have emerged as the dominant L2 scaling solution for Ethereum. There are a few different flavors, but generally rollups allow an aggregator to execute applications off-chain, publishing only the bare minimum information needed to show fraud (or the lack thereof) to Ethereum. This allows low latency and high throughput without giving up security guarantees of the base-layer chain.
In addition to their promise as a scaling solution, rollups can also enable the separation of transaction ordering and execution (see Optimism’s “MEV Auction” proposal). Vitalik Buterin has more recently suggested that Ethereum could become primarily a data-availability layer for rollups which handle all transaction execution, centralizing MEV capture into rollup sequencers (“ETH 1.5”).
This would be a significant departure from Ethereum’s current design and come with tradeoffs. For example, cross-rollup and rollup-mainchain interoperability breaks synchrony, and may require different assumptions to be done practically (especially in a many-rollup world). Our portfolio companies are working on two different rollup flavors:
#### StarkWare
StarkWare is working on ZK-Rollup (ZKRU), which proactively includes efficiently verifiable correctness proofs with block, rather than optimistically assuming validity and ensuring that fraud proofs are available if there is a challenge.
Although not the original flavor of rollup imagined for the separation of execution and ordering, ZKRU can achieve this. The proof engine could also be used to enforce additional constraints on the ordering. For example, if VDF-based priority or other deterministic ordering mechanisms become available.
#### Optimism
Optimism is working on the other leading flavor, Optimistic Rollup (ORU), which publishes the minimum data necessary to check fraud but optimistically assumes correctness until challenged. This results in a relatively long finality window but allows their rollup to use essentially the same execution environment as the L1’s EVM (so existing contracts can move ~seamlessly).
Optimism were the original proposers of MEVA and ETH1.5 more generally.
### Flashbots
Flashbots is a research and development organization formed to mitigate the negative externalities and existential risks posed by MEV, starting with Ethereum. They have built out tooling to quantify MEV and eliminate the information asymmetry in the ecosystem. They are now implementing a proof of concept for permissionless MEV extraction called MEV-Geth, a sealed-bid block space auction mechanism for communicating transaction order preference.
Flashbots’ goal is to make sure MEV incentives do not become opaque and undemocratic. Hopefully, their infrastructure will allow application developers to better understand how to minimize their MEV exposure, and let some pressure off that could otherwise accumulate into really harmful externalities (e.g., a time-bandit attack).
### Cosmos
Cosmos is an alternative model for permissionless, interoperable applications. Although not directly related to MEV on Ethereum, Cosmos is an architecture which could realistically enable an application ecosystem of similar complexity without adopting Ethereum’s uniform-security paradigm.
It is imagined that Cosmos blockchains will be largely application-specific, and not share security with one another by default, which may allow them to avoid externalities that would be harmful on a shared platform. If Ethereum goes strongly in the direction of ETH1.5, it will look very similar to Cosmos (in fact, LazyLedger is basically Cosmos’ ETH1.5).
## ACK’s
Deep thanks to my colleagues Arjun Balaji, Dan Robinson, Georgios Konstantopoulos, and Matt Huang, as well as Hasu, for discussion and feedback which helped inform this post.
# Disclaimer
This post is for general information purposes only. It does not constitute investment advice or a recommendation or solicitation to buy or sell any investment and should not be used in the evaluation of the merits of making any investment decision. It should not be relied upon for accounting, legal or tax advice or investment recommendations. This post reflects the current opinions of the authors and is not made on behalf of Paradigm or its affiliates and does not necessarily reflect the opinions of Paradigm, its affiliates or individuals associated with Paradigm. The opinions reflected herein are subject to change without being updated.
## Notes
1. On the Instability of Bitcoin Without the Block Reward” was arguably the first explicit identification of MEV, demonstrating that “normal” Bitcoin transaction fees are an irregular source of miner revenues and encourage consensus instability (incentivizing re-orgs of fee-rich blocks rather than mining on poor ones).
2. What I mean by this is that it seems unlikely Bitcoin MEV will grow over time independent of the BTC inflation rate. Concern about the inflation rate decreasing is distinct from whether we expect MEV will increase.
3. EIP-1559 breaks fees into two components, a “BASEFEE” and a “tip auction.” The BASEFEE is a deterministic amount of ETH that must be burned by all transactions. The tip auction is basically just a PGA. While the BASEFEE burns would be additive to protocol security (as they are not an irregular miner revenue stream), any transaction which contains MEV will still be competed over in a tip auction. Of course, the auctions must be included to allow that demand to be expressed in-protocol; if they were not, miners and traders would collude in off-protocol markets. Thus, only the burned BASEFEE is helpful to security.
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2021-06-25 04:30:29
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https://techwhiff.com/learn/run-jflap-and-use-file-open-to-open-the/384446
|
# Run JFlap, and use File->Open to open the problem1.jff file that we have given you. In problem1.jff, build a deterministic
###### Question:
Run JFlap, and use File->Open to open the problem1.jff file that we have given you.
In problem1.jff, build a deterministic finite-state machine that accepts all bit strings containing at least three 1s and at most one 0, and that rejects all other bit strings.
This problem requires at least nine states. You may use more states if necessary (there’s no penalty for doing so), but if you have time, try to get as close to the minimum as possible!
Here are three examples of strings that should be accepted:
$111 1111011 11111$
Here are three strings that should be rejected:
$11 1101101 0$
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2023-01-28 19:54:12
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https://coderanch.com/t/670224/WildFly-JBoss-Server-unable-find
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# WildFly(JBoss) Server is unable to find JAVA_HOME
Chandrajeet Yadav
Greenhorn
Posts: 4
I m trying to start WildFly 10 Server by running "<WildFly_Home>\wildfly-10.1.0.Final\bin\standalone.conf.bat" but at command prompt it is showing error "JAVA_HOME "C:\Program Files\Java\jdk1.8.0_77;" path doesn't exist". I have set Java-path as first value in 'PATH' variable of System 'Environment Variable' as "C:\Program Files\Java\jdk1.8.0_77;<other values>;." Please refer screeshot for more info and suggest me where am I doing wrong ?
commandPrompt.JPG
EnvVariable.JPG
Rob Spoor
Sheriff
Posts: 20817
68
• 1
Have you also set a JAVA_HOME environment variable with value C:\Program Files\Java\jdk1.8.0_77? That should not contain a trailing ;.
Chandrajeet Yadav
Greenhorn
Posts: 4
Rob Spoor wrote:Have you also set a JAVA_HOME environment variable with value C:\Program Files\Java\jdk1.8.0_77? That should not contain a trailing ;.
Yeah Rob ! I have set JAVA_HOME environment variable with same value as you have mentioned but now I have found my mistake and problem has been solved after mentioning the word 'JAVA_HOME' in 'Path' variable as "Java_Home\bin" because earlier I have set full path for JDK bin directory but WildFly Server is identifying it using JAVA_HOME. Thanks a lot for your reply.
Rob Spoor
Sheriff
Posts: 20817
68
You're welcome.
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2017-02-20 02:01:46
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https://www.physicsforums.com/threads/find-the-limit-of-xsin-pi-x-at-infinity.64717/
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Homework Help: Find the limit of xsin(pi*x) at infinity
1. Feb 23, 2005
allergic
Since the function f(x)=xsin[(pi)x] oscillates, shouldnt the limit as x -> infinity not exist? i was told that it is positive infinity.
2. Feb 23, 2005
arildno
If your function really is $$f(x)=x\sin(\pi{x})$$ , then you're right.
(That is, a limit doesn't exist)
Last edited: Feb 23, 2005
3. Feb 23, 2005
dextercioby
Note,that "pi" in the argument is totally unimportant.A certain rescaling would eliminate it...
Daniel.
4. Feb 23, 2005
arildno
Unless, of course, x is supposed to be an integer variable; in which case a limit does exist..
5. Feb 23, 2005
dextercioby
Of course,Arildno,mathematicians thought of it and decided to use the "n" (middle Latin alphabet letters,in general) for the INTEGER/natural numbers.Just the same way as "x" stands for reals and "z" for complex...
Daniel.
6. Feb 23, 2005
arildno
I'll just post for a face change..
7. Feb 23, 2005
dextercioby
No face change needed...I can handle "winks"...
Daniel.
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2018-06-20 04:33:33
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https://rstudio-pubs-static.s3.amazonaws.com/386941_41a65c8c19924b08b093f58da9be190b.html
|
1. Introduction
Earlier in the program we looked at basic regression. As a reminder, this was a way of modelling the relationshiop between a response variable $$y_i$$ and a number of predictor variables $$x_{i1} \cdots x_{im}$$, for each observation $$i$$ ($$i = 1 \cdots n$$). The basic model used was
$y_{i} = \beta_0 + \sum_{k=1}^{k=m} \beta_k x_{ik} + \epsilon_i$ so for each observation, the relationship is linear plus a random error part where each which $$\epsilon_i$$ has a independent normal distribution with mean zero and varianace $$\sigma^2$$. However the assumption that the relationship is linear, and that the error terms have identical and independent normal distributions is quite often a consequence of wishful thinking, rather than a rigorously derived conclusion. Although quite often these assumptions are resemble reality enough to allow predictions to be of some use, changing some of the model assumptions would certainly result in more plausible results.
To take a step closer to reality, it is a good idea to look beyond these assumptions to obtain more reliable predictions, or gain a better theoretical understanding. To mediate the problems caused by the inappropriate use of simple linear regression (or OLS - Ordinary Least Squares) - two general approaches are used:
1. Making the model calibration more robust - so that minor deviations (such as extreme outliers) - do not negatively in fluence the modelling process.
2. Allowing different assumptions in the model or the error term
In this session we will look at some examples of these. To do this, we’ll need to load some R packages, and reload the image of the variables created in the earlier regression session (Session 2).
library(tidyverse)
library(GISTools)
library(GWmodel)
library(rgdal)
library(tmap)
library(knitr)
library(car)
library(gclus)
library(kableExtra)
load('session2.RData')
2. Robust (‘bomb-proof’) Regression
This relates to point 1 above. Most of this session will relate to examples linking to point 2 - but it is useful to consider at least one example of the first kind. To investigate this example, we will need to recall the results from session 2 - the introductory regression session, and revisit the regression example for median income (MedInc) there. We have already re-loaded the results from the session in the introduction. The results of the linear regression are stored in m -
summary(m)
##
## Call:
## lm(formula = MedInc ~ PctRural + PctBach + PctEld + PctFB + PctPov +
## PctBlack, data = df)
##
## Residuals:
## Min 1Q Median 3Q Max
## -12.4203 -2.9897 -0.6163 2.2095 25.8201
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 52.59895 3.10893 16.919 < 2e-16 ***
## PctRural 0.07377 0.02043 3.611 0.000414 ***
## PctBach 0.69726 0.11221 6.214 4.73e-09 ***
## PctEld -0.78862 0.17979 -4.386 2.14e-05 ***
## PctFB -1.29030 0.47388 -2.723 0.007229 **
## PctPov -0.95400 0.10459 -9.121 4.19e-16 ***
## PctBlack 0.03313 0.03717 0.891 0.374140
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 5.155 on 152 degrees of freedom
## Multiple R-squared: 0.7685, Adjusted R-squared: 0.7593
## F-statistic: 84.09 on 6 and 152 DF, p-value: < 2.2e-16
However, looking at the studentised residuals suggested a few outliers that might ‘throw’ the regression coefficient estimates. One way to investigate this is to remove the outlying observations and re-fit the model. The removal is best done using the dplyr approach and the filter function, and re-fitting the model.
df %>% filter(abs(rstudent(m)) <= 2) -> df2
m2 <- lm(MedInc~PctRural+PctBach+PctEld+PctFB+PctPov+PctBlack, data = df2)
This done, the compareCoefs function in car lets the results be compared:
kable(compareCoefs(m,m2,print=FALSE), digits=3)
Est. 1 SE 1 Est. 2 SE 2
(Intercept) 52.599 3.109 52.905 2.644
PctRural 0.074 0.020 0.067 0.017
PctBach 0.697 0.112 0.728 0.105
PctEld -0.789 0.180 -0.925 0.147
PctFB -1.290 0.474 -1.485 0.459
PctPov -0.954 0.105 -0.887 0.092
PctBlack 0.033 0.037 0.043 0.031
It may be seen that some of the values alter notably (by around 10%) between the models with and without outliers. PctBlack is more extreme at around 30%. Outliers do have an influence.
However, it could be argued that the approach of removing them entirely is misleading - essentially trying to pass off a model as universal but intentionally ignoring observations that contradict it. Although that might be quite extreme, the point holds to some extent. Those counties do have those statistics, and so should perhaps not be ignored entirely.
Robust regression offers a more nuanced approach. Here, instead of simply ignoring the outliers, it is assumed that for whatever reason, they are subject to much more variance than the remaining observations - possibly due to some characteristic of the sampling process, or - more likely here - due to some unusual circumstances in the geographical locality (for example the presence of a large university skewing income figures compared to most of the State). Rather than totally ignoring such observations, they are downweighted to allow for this. This is less extreme than removing outliers - which in a sense is extreme downweighting to the value zero!
There are a number of approaches to robust regression - here we will use one proposed by Huber (1981). This uses an algorithm as below:
1. Fit a standard regression model and note the residuals.
2. Compute the weight for each observation as a function of absolute residual size. A number of possible weight functions could be used, but Huber’s is
$w_i(z_i) = \left\{ \begin{array}{ll} 1, & \textsf{if } |z_i| < c; \\ c/|z|, & \textsf{if } |z_i| \ge c; \\ \end{array} \right.$ where $$c \approx 1.345$$ and $$z_i$$ is a scaled residual for observation $$i$$. If actual residual is $$e_i$$, then $$z_i = e_i/\tau$$ where $$\tau$$ is the median of $$\frac{|e_i|}{0.6745}$$. After residuals reach a value $$c$$ weighting gradually decreases - in accordance with the idea that higher residuals come from processes with greater variance.
3. Refit the regression model with these weights. Update the residual values.
4. Repeat from step 2 until regression coefficient estimates converge.
This is all implemented in the rlm function in the package MASS. It is very easy to use - once you have loaded MASS it just works like lm:
library(MASS)
m3 <- rlm(MedInc~PctRural+PctBach+PctEld+PctFB+PctPov+PctBlack, data = df)
m3
## Call:
## rlm(formula = MedInc ~ PctRural + PctBach + PctEld + PctFB +
## PctPov + PctBlack, data = df)
## Converged in 9 iterations
##
## Coefficients:
## (Intercept) PctRural PctBach PctEld PctFB PctPov
## 51.60773916 0.06337478 0.67787963 -0.77778234 -1.22469054 -0.88053199
## PctBlack
## 0.03179542
##
## Degrees of freedom: 159 total; 152 residual
## Scale estimate: 4.19
Here the coefficient estimates are printed out, and it may also be seen that the model converged in 9 iterations, and that the scale factor $$\tau$$ was estimated as 4.19. As with lm you can print out a summary:
summary(m3)
##
## Call: rlm(formula = MedInc ~ PctRural + PctBach + PctEld + PctFB +
## PctPov + PctBlack, data = df)
## Residuals:
## Min 1Q Median 3Q Max
## -12.8485 -2.8326 -0.1765 2.5012 27.4204
##
## Coefficients:
## Value Std. Error t value
## (Intercept) 51.6077 2.7377 18.8506
## PctRural 0.0634 0.0180 3.5224
## PctBach 0.6779 0.0988 6.8604
## PctEld -0.7778 0.1583 -4.9126
## PctFB -1.2247 0.4173 -2.9348
## PctPov -0.8805 0.0921 -9.5604
## PctBlack 0.0318 0.0327 0.9714
##
## Residual standard error: 4.188 on 152 degrees of freedom
Finally, coefficients from all three aproaches can be compared:
coef_tab <- data_frame(OLS=coef(m),Outliers Deleted=coef(m2),Robust=coef(m3))
kable(coef_tab,digits=3)
OLS Outliers Deleted Robust
52.599 52.905 51.608
0.074 0.067 0.063
0.697 0.728 0.678
-0.789 -0.925 -0.778
-1.290 -1.485 -1.225
-0.954 -0.887 -0.881
0.033 0.043 0.032
For some coefficients, the robust estimate is closer to the standard OLS, and for others, it is closer to the ‘outliers deleted’ estimate - however in others, it differs from OLS, but not in the same direction as the ‘outliers deleted’ case. As a precautionary approach, I suggest working with the robust approach if there are some notable outliers.
3. Geographically Weighted Models
Regression of some kind underpins many statistical models. However often they make one massive assumption: that the relationships between the $$x_k$$’s and $$y$$ are the same everywhere. Regression models (and most other models) are aspatial and because they consider all of the data in the study area in one universal model, they are referred to as Global models. For example, in the models seen in this program so far the assumption is that the contributions made by the different variables in predicting MedInc are the same across the study area. In reality this assumption of spatial invariance is violated in many instances when geographic phenomena are considered.
For these reasons, as a geographer….
• I do not expect things to be the same everywhere
• I expect to find clusters, hotspots etc
• I am interested in how and where processes, relationships, and other phenomena vary spatially**
In statistical terminology… I am interested in spatial non-stationarity in relationships and process spatial heterogeneity.
What this means is that instead of just one general relationship for, for example, the association between changes in PctBach and MedInc, I might expect that relationship (as described in the coefficient estimate for $$\beta_k$$) to be different in different places. This is a core concept for geographical data analysis. A key question is “how can such a model be calibrated” - a clue lies in the next subsection…
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2020-07-13 12:32:10
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https://zitaoshen.rbind.io/project/machine_learning/how-to-build-your-own-neural-net-from-the-scrach/
|
# 1 Introduction
In this post, I will try to write a Neural Network with stochastic gradient decent (SGD) in the R, and train the Neural Network on the MNIST data set for digit recognition and report my test error (i.e. the proportion of test cases that are mis-classified). Now, let’s go over each part of my neural net.
The MNIST data was split into 60,000 training images, and 10,000 test images. That’s the official MNIST description. Actually, we’re going to split the data a little differently. We’ll leave the test images as is, but split the 60,000-image MNIST training set into two parts: a set of 50,000 images, which we’ll use to train our neural network, and a separate 10,000 image validation set.
## Load the MNIST digit recognition dataset into R
## http://yann.lecun.com/exdb/mnist/
## assume you have all 4 files and gunzip'd them
## creates train$n, train$x, train$y and test$n, test$x, test$y
## e.g. train$x is a 60000 x 784 matrix, each row is one digit (28x28) ## call: show_digit(train$x[5,]) to see a digit.
## this code was written by:
## brendan o'connor - gist.github.com/39760 - anyall.org
ret = list()
f = file(filename,'rb')
ret$n = readBin(f,'integer',n=1,size=4,endian='big') nrow = readBin(f,'integer',n=1,size=4,endian='big') ncol = readBin(f,'integer',n=1,size=4,endian='big') x = readBin(f,'integer',n=ret$n*nrow*ncol,size=1,signed=F)
ret$x = matrix(x, ncol=nrow*ncol, byrow=T) close(f) ret } load_label_file <- function(filename) { f = file(filename,'rb') readBin(f,'integer',n=1,size=4,endian='big') n = readBin(f,'integer',n=1,size=4,endian='big') y = readBin(f,'integer',n=n,size=1,signed=F) close(f) y } train <<- load_image_file('D:/newsite/My_Website/static/data/train-images-idx3-ubyte') test <<- load_image_file('D:/newsite/My_Website/static/data/t10k-images-idx3-ubyte') train$y <<- load_label_file('D:/newsite/My_Website/static/data/train-labels-idx1-ubyte')
test$y <<- load_label_file('D:/newsite/My_Website/static/data/t10k-labels-idx1-ubyte') } show_digit <- function(arr784, col=gray(12:1/12), ...) { image(matrix(arr784, nrow=28)[,28:1], col=col, ...) } Now, we are able to load the data and show the digit load_mnist() str(train) # training data ## List of 3 ##$ n: int 60000
## $x: int [1:60000, 1:784] 0 0 0 0 0 0 0 0 0 0 ... ##$ y: int [1:60000] 5 0 4 1 9 2 1 3 1 4 ...
str(test) # test data
## List of 3
## $n: int 10000 ##$ x: int [1:10000, 1:784] 0 0 0 0 0 0 0 0 0 0 ...
## $y: int [1:10000] 7 2 1 0 4 1 4 9 5 9 ... show_digit(train$x[1,]) # view an image
train$y[1] # the answer ## [1] 5 # 3 Details of NN Based on the functionality, I divide the code into 3 parts: ## 3.1 Miscellaneous digit.to.vector(digit): The response y is a digit. we need to turn it into a vector with a 1 in the corresponding digit location. e.g. a 6 is represented as (0,0,0,0,0,0,1,0,0,0) cost.derivative(output.activations.target, index, original.data.target): It should return derivative of squared error cost function wrt activations. sigmoid(z)&sigmoid.prime(z): The sigmoid function and its derivative, which will be used in the backpropagation ## 3.2 Evaluation Funtion evaluate(test, answer): Given an input matrix (test) and a vector (answer) of correct responses, it returns the proportion of correctly classified observations. classify(input): Given an input vector, it returns the predicted digit. ## 3.3 Weights Updating feedforard(a): Given an input a for the network, itreturns the corresponding output. All the method does is applies the followed Equation for each layer: $a^{\prime}=\sigma(w a+b)$ SGD(training.data, epochs, batch.size,eta,testing.data = NULL): It implements stochastic gradient descent • Input Parameters: MNIST training data, number of epochs, batch size, learning rate, testing.data(If “testing.data” is provided then the network will be evaluated against the test data after each epoch , and partial progress printed out. This is useful for tracking progress , but slows things down substantially) • How it work: In each epoch, it starts by randomly shuffling the training data, and then partitions it into mini-batches of the chosen size. This is an easy way of sampling randomly from the training data. Then for each mini_batch we apply a single step of gradient descent. This is done by the code update_mini_batch, which updates the network weights and biases according to a single iteration of gradient descent, using just the training data in mini_batch. update.mini.batch(data.x,data.y,eta,batch.size): Given a mini-batch, it applys a single step of gradient descent to each observation in the mini batch. then update the network weights and biases.This invokes something called the backpropagation algorithm, which is a fast way of computing the gradient of the cost function. backprop(index,data.x,data.y):Given a single observation, apply backpropagation. return the gradient wrt biases and the gradient wrt weights. • How it work: Since we are going to talk about backpropagation, I will briefly discuss about four fundamental equations behind backpropagation. For proofs and more details, you can go check about this book Neural Networks and Deep Learning. 1. An equation for the error in the output layer,$$\delta^{L}$$ (BP1) $\delta_{j}^{L}=\frac{\partial C}{\partial a_{j}^{L}} \sigma^{\prime}\left(z_{j}^{L}\right)$ • $$\partial C / \partial a_{j}^{L}$$:how fast the cost is changing as a function of the jth output activation • $$\sigma^{\prime}\left(z_{j}^{L}\right)$$:how fast the activation function $$\sigma$$ is changing at $$z_{j}^{L}$$ 1. An equation for the error $$\delta^{L}$$ in terms of the error in the next layer, $$\delta^{L+1}$$ (BP2) $\delta^{l}=\left(\left(w^{l+1}\right)^{T} \delta^{l+1}\right) \odot \sigma^{\prime}\left(z^{l}\right)$ • $$\left(w^{l+1}\right)^{T}$$:the transpose of the weight matrix wl+1 for the $$(l+1)^{th}$$ layer • Notes: Suppose we know the error $$\delta^{L+1}$$ at the $$(l+1)^{th}$$ layer. When we apply the transpose weight matrix, $$\left(w^{l+1}\right)^{T}$$, we can think intuitively of this as moving the error backward through the network, giving us some sort of measure of the error at the output of the $$l^{th}$$ layer. We then take the Hadamard product $$\odot \sigma^{\prime}\left(z^{l}\right)$$. This moves the error backward through the activation function in layer $$l$$, giving us the error $$\delta^{L}$$ in the weighted input to layer $$l$$. 1. An equation for the rate of change of the cost with respect to any bias in the network (BP3) $\frac{\partial C}{\partial b_{j}^{l}}=\delta_{j}^{l}$ 1. An equation for the rate of change of the cost with respect to any weight in the network (BP4) $\frac{\partial C}{\partial w_{j k}^{l}}=a_{k}^{l-1} \delta_{j}^{l}=a_{\mathrm{in}} \delta_{\mathrm{out}}$ • $$a_{\mathrm{in}}$$ is the activation of the neuron input to the weight $$w$$, • $$\delta_{\mathrm{out}}$$ is the error of the neuron output from the weight $$w$$. • The algorithm: Let’s explicitly write this out in the form of an algorithm: # 4 Parameters For this neural net, it will have 3 layers with nodes 784,30 and 10 respectively. The step size will be 0.1 after turning. The epochs will be set as 40. Also, for simplicity, the minibatch size will be 10 intuitively. # 5 Code Let’s put those together L <- 3 # number of layers sizes <- c(784, 30, 10) eta <- 0.1 epochs <- 40 batch.size <- 10 biases <- list(rnorm(sizes[2]), rnorm(sizes[3])) # layers 2 and 3 weights <- list(matrix(rnorm(sizes[2]*sizes[1]), nrow=sizes[2], ncol=sizes[1]),# layer 1 to 2 matrix(rnorm(sizes[3]*sizes[2]), nrow=sizes[3], ncol=sizes[2]))# layer 2 to 3 ## w[j,k] is the weight on the link from node k to node j ## the response y is a digit. we need to turn it ## into a vector with a 1 in the corresponding digit location. ## e.g. a 6 is represented as (0,0,0,0,0,0,1,0,0,0) ## note: digit can be a vector (hence the use of sapply()). digit.to.vector <- function(digit) { sapply(digit, function(d) c(rep(0,d),1,rep(0,9-d))) } ## return derivative of squared error cost function wrt activations cost.derivative <- function(output.activations, i, data.y) { output.activations - digit.to.vector(data.y[i]) } sigmoid <- function(z) 1/(1 + exp(-z)) sigmoid.prime <- function(z) sigmoid(z) * (1-sigmoid(z)) ## given an input matrix (test) and a vector (answer) of correct responses, ## return the proportion of correctly classified observations. evaluate <- function(test, answer) { accuracy = rep(0,nrow(test)) for ( i in 1:nrow(test)){ test_results = classify(feed.forward(test[i,])) accuracy[i]= 1*(test_results==answer[i]) } return(mean(accuracy)) } ## given an input vector, return the predicted digit classify <- function(input) { test_results = which.max(input)-1 return (test_results) } ## given an input vector a, return the vector of output activations feed.forward <- function(a) { for ( i in 1:(L-1 )) { a = sigmoid(weights[[i]] %*% a + biases[[i]])} return (a) } ## input: MNIST training data, number of epochs, batch size, learning rate ## for each epoch: ## call update.mini.batch with a random sample of batch.size observations ## until all of the training data is used. SGD <- function(training.data, epochs, batch.size,eta,testing.data = NULL) { training.data.x=training.data$x
training.data.y=training.data$y if (!is.null(testing.data)) { testing.data.x=testing.data$x
testing.data.y=testing.data\$y
n_test = nrow(testing.data.x)}
n = nrow(training.data.x)
number.batch = ceiling(n/batch.size)
for( j in 1:epochs){
rows = sample(nrow(training.data.x))
shuffle.x = training.data.x[rows,]
shuffle.y = training.data.y[rows]
index = split(seq_len(nrow(training.data.x)),rep(1:number.batch,each=batch.size))
mini_batch.x = lapply(index,function(i) shuffle.x[i,])
mini_batch.y = lapply(index,function(i) shuffle.y[i])
for(i in 1:number.batch){
update.mini.batch(mini_batch.x[[i]],mini_batch.y[[i]],eta,batch.size)
}
if (!is.null(testing.data)) {
result_test = evaluate(testing.data.x,testing.data.y)
cat("Epoch", j,":", result_test,"\n")
}
else{
cat("Epoch", j,"complete\n")
}
}
}
## given a mini-batch, apply a single step of gradient descent to each
## observation in the mini batch. then update the network weights and
## biases.
update.mini.batch <- function(data.x,data.y,eta,batch.size) {
delta.nabla.b=vector(mode = "list", length = batch.size)
delta.nabla.w=vector(mode = "list", length = batch.size)
for( i in 1: batch.size){
delta = backprop(i,data.x,data.y)
delta.nabla.b[[i]]=delta[[1]]
delta.nabla.w[[i]]=delta[[2]]
}
for ( j in 1:(L-1)){
# sum over the mini_batch
sum.b=Reduce('+',lapply(1:batch.size, function(i) delta.nabla.b[[i]][[j]]))
sum.w=Reduce('+',lapply(1:batch.size, function(i) delta.nabla.w[[i]][[j]]))
weights[[j]]<<-weights[[j]]-eta/batch.size*sum.w
biases[[j]]<<-biases[[j]]-eta/batch.size*sum.b
}
}
## given a single observation, apply backpropagation. return the
backprop <- function(index,data.x,data.y) {
matrix.size= lapply(c(1:L), function(i) rep(0,sizes[i]))
a=matrix.size
matrix.size[[1]]= NULL
z= matrix.size
delta = matrix.size
nabla.w = weights#they are the same size, using the weights size
## feedforward
a[[1]] <- data.x[index,]
for (i in 1:(L-1)){
z[[i]]=weights[[i]] %*% a[[i]] + biases[[i]]# weighted input to hidden layer
a[[i+1]]=sigmoid(z[[i]]) # activation out of the hidden layer
}
## compute the error in the output layer
delta[[L-1]] = cost.derivative(a[[L]],index,data.y) * sigmoid.prime(z[[L-1]]) # BP1
## backpropagate the error
for ( i in (L-2):1) delta[[i]] = (t(weights[[i+1]]) %*% delta[[i+1]]) * sigmoid.prime(z[[i]]) # BP2
nabla.b = delta # BP3
for ( i in 1:(L-1)) nabla.w[[i]] = delta[[i]]%*%as.vector(a[[i]])# BP4
nabla = list(nabla.b,nabla.w)
return(nabla)
}
# 6 Performance
Now let’s run it. It runs about 15 mins overall.
SGD(train, epochs, batch.size, eta,testing.data = test)
|
2021-10-28 21:35:45
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|
https://artofproblemsolving.com/wiki/index.php/1992_AIME_Problems/Problem_14
|
# 1992 AIME Problems/Problem 14
## Problem
In triangle $ABC^{}_{}$, $A'$, $B'$, and $C'$ are on the sides $BC$, $AC^{}_{}$, and $AB^{}_{}$, respectively. Given that $AA'$, $BB'$, and $CC'$ are concurrent at the point $O^{}_{}$, and that $\frac{AO^{}_{}}{OA'}+\frac{BO}{OB'}+\frac{CO}{OC'}=92$, find $\frac{AO}{OA'}\cdot \frac{BO}{OB'}\cdot \frac{CO}{OC'}$.
## Solution 1
Let $K_A=[BOC], K_B=[COA],$ and $K_C=[AOB].$ Due to triangles $BOC$ and $ABC$ having the same base, $$\frac{AO}{OA'}+1=\frac{AA'}{OA'}=\frac{[ABC]}{[BOC]}=\frac{K_A+K_B+K_C}{K_A}.$$ Therefore, we have $$\frac{AO}{OA'}=\frac{K_B+K_C}{K_A}$$ $$\frac{BO}{OB'}=\frac{K_A+K_C}{K_B}$$ $$\frac{CO}{OC'}=\frac{K_A+K_B}{K_C}.$$ Thus, we are given $$\frac{K_B+K_C}{K_A}+\frac{K_A+K_C}{K_B}+\frac{K_A+K_B}{K_C}=92.$$ Combining and expanding gives $$\frac{K_A^2K_B+K_AK_B^2+K_A^2K_C+K_AK_C^2+K_B^2K_C+K_BK_C^2}{K_AK_BK_C}=92.$$ We desire $\frac{(K_B+K_C)(K_C+K_A)(K_A+K_B)}{K_AK_BK_C}.$ Expanding this gives $$\frac{K_A^2K_B+K_AK_B^2+K_A^2K_C+K_AK_C^2+K_B^2K_C+K_BK_C^2}{K_AK_BK_C}+2=\boxed{094}.$$
## Solution 2
Using mass points, let the weights of $A$, $B$, and $C$ be $a$, $b$, and $c$ respectively.
Then, the weights of $A'$, $B'$, and $C'$ are $b+c$, $c+a$, and $a+b$ respectively.
Thus, $\frac{AO^{}_{}}{OA'} = \frac{b+c}{a}$, $\frac{BO^{}_{}}{OB'} = \frac{c+a}{b}$, and $\frac{CO^{}_{}}{OC'} = \frac{a+b}{c}$.
Therefore: $\frac{AO}{OA'}\cdot \frac{BO}{OB'}\cdot \frac{CO}{OC'} = \frac{b+c}{a} \cdot \frac{c+a}{b} \cdot \frac{a+b}{c}$ $= \frac{2abc+b^2c+bc^2+c^2a+ca^2+a^2b+ab^2}{abc} =$
$2+\frac{bc(b+c)}{abc}+\frac{ca(c+a)}{abc}+\frac{ab(a+b)}{abc} = 2 + \frac{b+c}{a} + \frac{c+a}{b} + \frac{a+b}{c}$ $= 2 + \frac{AO^{}_{}}{OA'}+\frac{BO}{OB'}+\frac{CO}{OC'} = 2+92 = \boxed{094}$.
|
2020-09-26 22:36:56
|
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|
https://math.stackexchange.com/questions/2743105/can-the-limit-of-a-sequence-of-matrices-of-fixed-rank-have-higher-rank
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Can the limit of a sequence of matrices of fixed rank have higher rank?
Consider the $n\times n$ matrices given by $A_n=1/n*I_{n\times n}$. For every finite $n$ each $A_n$ has rank $n$ but the limit is the zero matrix, and has rank zero. So the limit of a sequence of rank $n$ matrices need not be of rank $n$. Now I am considering the reverse question.
Consider a sequence $A_k \in \mathbb{R}^{m \times n}$ (or $\mathbb{C}^{m \times n}$). Is it possible to construct a sequence such that each $A_k$ has some rank $d \le \min(m,n)$ but: $\operatorname{rank} \left( \lim \limits _{k \to \infty} A_n \right) >d$?
If not how would one go about proving that matrices cannot jump rank?
• Indeed the rank cannot jump to higher values. To show this for square matrices, one can start with their characteristic polynomials. – Did Apr 18 '18 at 14:21
• The rank is lower semicontinuous on the set of all matrices. It can only jump down. – Hans Engler Apr 18 '18 at 14:31
The set of matrices of rank (at most) $d$ is closed in the set of $m\times n$ matrices.
You can generalize the proof that the set of singular matrices is closed, by considering the set of matrices of rank less than $d$ as the set $f^{-1}({0})$, for $f\colon \mathbb{R}^{m\times n} \to \mathbb{R}$ defined by $$f(A) = \sum_{A'} \lvert\det(A')\rvert$$ where the sum if over all $(d+1)\times (d+1)$ sub-matrices. Then $f$ is continuous, and $f(A)$ is non-zero if and only if $A$ has rank at least $d+1$.
|
2019-07-20 11:52:35
|
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https://www.hackmath.net/en/math-problem/1177?tag_id=152,151
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Vector sum
The magnitude of the vector u is 12 and the magnitude of the vector v is 8. Angle between vectors is 61°. What is the magnitude of the vector u + v?
Result
x = 17.35
Solution:
$|u+v|^2 = |u|^2+|v|^2-2|u||v|\cos (180^\circ -61^\circ ) \ \\ |u+v| = 17.35$
Leave us a comment of this math problem and its solution (i.e. if it is still somewhat unclear...):
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Following knowledge from mathematics are needed to solve this word math problem:
For Basic calculations in analytic geometry is helpful line slope calculator. From coordinates of two points in the plane it calculate slope, normal and parametric line equation(s), slope, directional angle, direction vector, the length of segment, intersections the coordinate axes etc. Two vectors given by its magnitudes and by included angle can be added by our vector sum calculator. Cosine rule uses trigonometric SAS triangle calculator. See also our trigonometric triangle calculator.
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2019-12-13 13:43:08
|
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https://www.cuemath.com/algebra/inverse-operations/
|
Inverse Operations
Inverse Operations
Have you ever wonder what would have happened if you were born before your elder sibling or vice-versa. Interesting, isn't it? Everything would have been inverted. Not in real life but it is possible in the world of math.
Let us consider here butterfly as an inverse of a honeybee.
Would you expect this to happen in the real world?
In reality, it is not possible but mathematically we can prove so. Do you want to know how?
In this mini-lesson, we will explore the world of inverse operations by finding answers to questions like what are inverse operations, what are properties of inverses, and how to use inverse operations with help of interactive questions on inverse operations.
Lesson Plan
1 What Are Inverse Operations? 2 Important Notes on Inverse Operations 3 Solved Examples on Inverse Operations 4 Interactive Questions on Inverse Operations 5 Challenging Question on Inverse Operations
What Are Inverse Operations?
Inverse means the opposite effect of an action or a step. In mathematics, we have operations such as addition$$(+)$$, subtraction$$(-)$$, multiplication$$(\times)$$, division$$(÷)$$, squaring, square root, and logarithms.
When we use two operations together, it is possible to have an inverse impact on the result due to the operations used.
The process in which the effect of one operation is inversed by another operation is termed as inverse operations.
For example,
If we add $$3$$ and $$2$$ pens we get $$5$$ pens, now subtract $$5$$ pens and $$2$$ pens and we get $$3$$ back. Here, addition and subtraction are inverse operations.
What Are the Properties of Inverses?
In the above section, we discussed that addition and subtraction are inverse operations. Now let us learn the properties of inverse operations in detail with few inverse operations examples.
Property 1
The inverse of addition is subtraction. Addition always totals$$(+)$$ values and subtraction always takes away$$(-)$$ values.
For example, adding $$8$$ add $$2$$, we get $$10$$. Now, subtracting $$10$$ and $$2$$ and we get $$8$$ back.
There is also a property called additive inverse. In additive inverse, we add a certain value to a number (integer) to get zero as the final result.
For example,
$$-8 + 8 =0$$
$$-8$$ is an additive inverse of $$8$$
Property 2
In inverse operations multiplication is the inverse of division. Multiplication can be undone with the help of division.
For example, multiplying $$8$$ $$\times$$ $$2$$ we get $$16$$, now divide $$16$$ $$\div$$ $$2$$ and we get $$8$$ back.
There is also a property called multiplicative inverse. In multiplicative inverse, we multiply a number by a certain fraction value $$(\dfrac{p}{q})$$ to get 1 as a final result.
For example,
$$8\times\dfrac{1}{8} = 1$$
$$\dfrac{1}{8}$$ is a multiplicative inverse of $$8$$
Note: Multiplicative inverse is not valid for zero.
For example, $$8\times0=0$$. It cannot be inversed by $$\dfrac{0}{0}$$.
Property 3
In mathematics, the inverse of a function will always give us the original value.
For example, let's say hypothetically if the function f changes the butterfly into a honeybee then the inverse function $$f^{-1}$$ changes the honeybee back to the butterfly.
Let's see a math example depicting the inverse of a function.
$$f(x)=7X + 2$$
Here,
$$\dfrac{Y-2}{7}$$ is an inverse function of $$7X + 2$$.
Property 4
The inverse of a trigonometry function.
Suppose that we apply the sine function to a certain angle $$\theta$$, and obtain the output as $$y$$. We can write
$y = \sin \theta$
This equation expresses $$y$$ in terms of $$\theta$$. Can we invert this relation to writing $$\theta$$ in terms of $$y$$?
Yes, we can invert the relation by writing
$\theta = {\sin ^{ - 1}}y$
Note: Do not confuse $${\sin ^{ - 1}}y$$ with $$\frac{1}{{\sin y}}$$. The latter is the reciprocal of $$\sin y$$. The former is completely different: the term $${\sin ^{ - 1}}y$$ is to be interpreted as an angle value whose sine will be $$y$$.
The following table lists some examples of the $${\sin ^{ - 1}}$$ operation:
$$\sin 0 = 0$$ $${\sin ^{ - 1}}0 = 0$$ $$\sin \frac{\pi }{6} = \frac{1}{2}$$ $${\sin ^{ - 1}}\left( {\frac{1}{2}} \right) = \frac{\pi }{6}$$ $$\sin \frac{\pi }{4} = \frac{1}{{\sqrt 2 }}$$ $${\sin ^{ - 1}}\left( {\frac{1}{{\sqrt 2 }}} \right) = \frac{\pi }{4}$$ $$\sin \frac{\pi }{3} = \frac{{\sqrt 3 }}{2}$$ $${\sin ^{ - 1}}\left( {\frac{{\sqrt 3 }}{2}} \right) = \frac{\pi }{3}$$ $$\sin \frac{\pi }{2} = 1$$ $${\sin ^{ - 1}}1 = \frac{\pi }{2}$$
Here are some examples of the $${\cos ^{ - 1}}$$ operation:
$$\cos 0 = 1$$ $${\cos ^{ - 1}}1 = 0$$ $$cos\frac{\pi }{6} = \frac{{\sqrt 3 }}{2}$$ $$co{s^{ - 1}}\left( {\frac{{\sqrt 3 }}{2}} \right) = \frac{\pi }{6}$$ $$\cos \frac{\pi }{4} = \frac{1}{{\sqrt 2 }}$$ $$co{s^{ - 1}}\left( {\frac{1}{{\sqrt 2 }}} \right) = \frac{\pi }{4}$$ $$\cos \frac{\pi }{3} = \frac{1}{2}$$ $${\cos ^{ - 1}}\left( {\frac{1}{2}} \right) = \frac{\pi }{3}$$ $$\cos \frac{\pi }{2} = 0$$ $${\cos ^{ - 1}}0 = \frac{\pi }{2}$$
And a few examples of the $${\tan ^{ - 1}}$$ operation
$$\tan 0 = 0$$ $${\tan ^{ - 1}}0 = 0$$ $$\tan \frac{\pi }{6} = \frac{1}{{\sqrt 3 }}$$ $${\tan ^{ - 1}}\left( {\frac{1}{{\sqrt 3 }}} \right) = \frac{\pi }{6}$$ $$\tan \frac{\pi }{4} = 1$$ $${\tan ^{ - 1}}\left( 1 \right) = \frac{\pi }{4}$$ $$\tan \frac{\pi }{3} = \sqrt 3$$ $${\tan ^{ - 1}}\left( {\sqrt 3 } \right) = \frac{\pi }{3}$$
Property 5
Logarithm is an inverse of an exponent.
We express exponents in the form of power or degree $$3^2$$ here, $$2$$ is the exponent.
We can express the exponents in terms of the logarithm as:
$$3^2=9$$
$$\log_3(9)=2$$
Do you want to learn how to use inverse operations without using an inverse operations calculator? In our next section, we will learn the same.
How to Use Inverse Operations?
Let us look at inverse operations examples to understand how to use inverse operations.
Example
We know that the inverse of f is $$f^{-1}$$ here we are going to use this relation in our inverse operations examples.
If $$x=5$$ solve $$f(x)=3x+3=18$$
$$f(5)=3\times5+3=18$$
$$f^{-1}(18)=\dfrac{18-3}{3}=5$$
Therefore,
$$f^{-1}(f(5))=5$$
Hence by applying function (f) and doing the inverese of the same function we are getting the original value back.
$$f^{-1}(f(x)) = x$$ or $$f(f^{-1}(x)) = x$$
Important Notes
Inverse Operations Addition$$(+)$$ <=> Subtraction$$(-)$$ Multiplication$$(\times)$$ <=> Division$$(\div)$$ $$\dfrac{1}{x}$$ <=> $$\dfrac{1}{y}$$ $$x^{2}=y$$ <=> $$\sqrt{y}=x$$ ($$x$$ $$≥$$ $$0$$). $$a^{x}=y$$ <=> $$\log_{a}(y)=x$$ $$\sin\ {x}$$ <=> $$\sin^{-1}(y)$$
Solved Examples
Here are set of solved examples on inverse operations examples to understand how to use inverse operations without using inverse operations calculator.
Example 1
Help John in finding the inverse of a function using inverse operations.
$$f(x)=x^{3}+6$$
Solution
Let $$f(x)=y$$
$$y=x^{3}+6$$
Subtract 6 from both the sides
$$x^{3}=y-6$$
Taking cube root of both the sides we get,
$$x=\sqrt[3]{y-5}$$
$$f^{-1}(y)=\sqrt[3]{y-5}$$
$$\therefore$$ for $$x$$ $$\rightarrow$$ $$f^{-1}(y)=\sqrt[3]{y-5}$$
Example 2
Miley is confused when trying to find the inverse function of $$f(x) = \dfrac{1}{2}(6x-3)$$. Help her in finding the correct answers.
Solution
$$f(x)=y$$
$$y=\dfrac{1}{2}(6x-3)$$
On multiplying both the sides by 2 we get,
$$2y=6x-3$$
Adding 3 both the sides, we get
$$2y+3=6x$$
Now on dividing both the sides by 6,
$$x=\dfrac{1}{3}y+\dfrac{1}{2}$$
$$\therefore$$ $$f^{-1}(y)$$ is $$\dfrac{1}{3}y+\dfrac{1}{2}$$
Example 3
Jamie has her math homework pending due to one equation on inverse operations.
What is the inverse function of $$x^2+3$$ when $$x\geq0$$? Help Jamie in completing her homework.
Solution
$$f(x)=x^2+3$$
$$f(x)⇒y=x^2+3$$
Subtracting $$3$$ from both the sides
$$y-3=x^2$$
Taking square root on both the sides.
$$f(x)⇒f^{-1}(y)=\sqrt{y-3}$$
$$\therefore$$ $$f^{-1}(y)=\sqrt{y-3}$$
Interactive Questions
Here are a few activities for you to practice. Select/Type your answer and click the "Check Answer" button to see the result.
Challenging Question
Find the inverse of the following functions using inverse operations.
a)$$f(x)=x^{3}+3$$
b)$$f(x)=|x+3|$$ for $$x\geq-3$$
c)$$f(x)=\dfrac{1}{6}(e^{x}+2)$$
Let's Summarize
This mini-lesson targeted the fascinating concept of inverse operations. The math journey around inverse operations starts with what a student already knows, and goes on to creatively crafting a fresh concept in the young minds. Done in a way that is not only relatable and easy to grasp, but will also stay with them forever. Here lies the magic with Cuemath.
At Cuemath, our team of math experts is dedicated to making learning fun for our favorite readers, the students!
Through an interactive and engaging learning-teaching-learning approach, the teachers explore all angles of a topic.
Be it worksheets, online classes, doubt sessions, or any other form of relation, it’s the logical thinking and smart learning approach that we, at Cuemath, believe in.
1. What is the purpose of inverse operations?
Mathematically, inverse operations reverse the effect of one operation on the other. The main purpose is to understand the relation between the basic math operators $$+, -, \times, \div$$ so that solving an equation becomes easier and time saving.
2. What is the inverse operation of squaring a number?
The inverse of squaring a number is square root. For example, $$x^{2}=y$$ <=> $$\sqrt{y}=x$$ ($$x$$ $$≥$$ $$0$$).
3. What is the inverse operation of division?
In the inverse operations multiplication is the inverse operation of division.
For example, $$8$$ $$\div$$ $$2$$ we get $$4$$. Now multiply $$4$$ $$\times$$ $$2$$ and we get $$8$$ back.
More Important Topics
Numbers
Algebra
Geometry
Measurement
Money
Data
Trigonometry
Calculus
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2021-06-12 16:45:00
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https://tewarid.github.io/2013/05/06/debug-logging-in-c.html
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# Debug logging in C#
A quick tip - debug logging can come in handy when a remote debugging session cannot be established using Visual Studio. Any text logged using methods of class System.Diagnostics.Debug can be viewed with a debug log viewer such as DebugView from Sysinternals.
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2020-10-25 19:10:50
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http://texblog.org/code-snippets/lists/
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# Lists
Code snippets for lists:
##### Bulleted list: itemize
\begin{itemize}
\item
\item
\item
\end{itemize}
##### Numbered list: enumerate
\begin{enumerate}
\item
\item
\item
\end{enumerate}
##### Description
\begin{description}
\item[]
\item[]
\item[]
\end{description}
##### Inline enumeration
%Preamble
\usepackage{paralist}
%Document
\begin{inparaenum}
\item
\item
\item
\end{inparaenum}
#### 2 responses to “Lists”
• Clement Apere
Dear Friend,
I am writing a research report with the layout as shown below. The paragraphs are to be numbered continuously from Chapter 1 to Chapter 4. Also, I am to Insert Notes at the end of each number, while a bibliography should be at the end all the Chapter. The bibliography is also to be in different parts as shown below. I have tried to write latex codes in order to do this but without any success. I will appreciate if you can help me with the right codes to do this. Thanks
CHAPTER 1
1. The blah blah ……
NOTES
1.
2.
3.
CHAPTER 2
2.
NOTES
1.
2.
3.
CHAPTER 3
3.
CHAPTER 4
4.
BIBLIOGRAPHY
BOOKS
PERIODICALS/JOURNALS
NEWSPAPERS
UNPUBLISHED MATERIALS
UNSTRUCTURED INTERVIEWS
• tom
Hi Clement,
\documentclass{report}
\begin{document}
\tableofcontents
\chapter{First}
\chapter*{Notes}
\chapter{Second}
\chapter*{Notes}
\end{document}
Assuming you are using the biblatex package, what you can do is using keywords in the bib file. With these, it is then relatively simple to print several bibliographies. Here is an example:
\documentclass{report}
\usepackage[sorting=none, backend=biber]{biblatex}
\usepackage{filecontents}
\begin{filecontents}{references.bib}
@book{knuth1986texbook,
keywords = {book},
title={The texbook},
author={Knuth, D.E. and Bibby, D.},
volume={1993},
year={1986},
}
@article{knuth1977fast,
keywords = {article},
title={Fast pattern matching in strings},
author={Knuth, D.E. and Morris Jr, J.H. and Pratt, V.R.},
journal={SIAM journal on computing},
volume={6},
number={2},
pages={323--350},
year={1977},
publisher={SIAM}
}
\end{filecontents}
\end{document}
Another possibility would be to use the bib-entry type as an optional argument for printbibliography, e.g.:
\printbibliography[title={Book references},type=book]
\printbibliography[title={Other references},nottype=article, nottype=book]
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2014-03-08 06:22:02
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https://www.physicsforums.com/threads/light-deflection.229768/
|
# Homework Help: Light deflection
1. Apr 18, 2008
### fikus
1. The problem statement, all variables and given/known data
I'm trying to derive deflection angle of light in weak gravitational field.
2. Relevant equations
Metric tensor in weak field $$g_{00}=(1+\frac{2\Phi}{c^2}), ~ g_{ii}=-(1-\frac{2\Phi}{c^2})$$
3. The attempt at a solution
From Lagrangian $$L=\sqrt{g_{\mu \nu} \dot{x}^{\mu}\dot{x}^{\nu}}$$, I obtain equations of geodesic
$$g_{\mu \nu} \ddot{x}^{\nu} + \frac{\partial{g_{\mu \nu}}}{\partial{x^{\lambda}}} \dot{x}^{\lambda} \dot{x}^{\nu} - \frac{1}{2}\frac{\partial{g_{\lambda \nu}}}{\partial{x^{\mu}}} \dot{x}^{\lambda} \dot{x}^{\nu}~=~0$$
Putting in metric tensor and writing only equations for spatial coordinates I get :
$$-\left(1-\frac{2\Phi}{c^2}\right)\ddot{x}^i + \frac{2}{c^2}\frac{\partial{\Phi}}{\partial{x^j}} \dot{x}^j \dot{x}^i - \frac{\partial{\Phi}}{\partial{x^i}}\dot{t}^2- \frac{1}{c^2}\frac{\partial{\Phi}}{\partial{x^i}}(\dot{x}^2+\dot{y}^2+\dot{z}^2)~= 0$$
and for time coordinate
$$\dot{t}c^2(1+\frac{2\Phi}{c^2})= p_{t}$$
which is constant of motion.
Noting that $$\dot{t}^2=(\dot{x}^2+\dot{y}^2+\dot{z}^2)$$ (from ds = 0 for light, in zeroth order in $$\Phi/c^2$$ ), and writing
$$\dot{x}^i=\frac{dx^i}{dt}\dot{t} = \frac{dx^i}{dt} \frac{p_t}{c^2(1+2\Phi/c^2)}$$
and equivalently
$$\ddot{x}^i=p_t^2\frac{d^2x^i}{dt^2} - 2p_t^2 \frac{dx^i}{dt} \frac{\partial{\Phi}}{\partial{x^k}}\frac{dx^k}{dt}$$
Putting it all together, neglecting terms with phi/c^2 and writing in vector form I get:
$$\frac{d\vec{v}}{dt} ~ = ~ \frac{4}{c^2}(\vec{v} \cdot \vec{\nabla}\Phi) \vec{v} -2\vec{\nabla }\Phi = \frac{2}{c^2}(\vec{v} \cdot \vec{\nabla}\Phi) \vec{v} - 2\vec{\nabla}_{\perp}\Phi$$
Where I noted that the first term is just component of gradient in direction of v, and then writing nabla with \perp, for component of gradient perpendicular to v. So my question now is how to interpret this first term on rhs. Does it change speed of light ? (is it maybe shapiro delay ?) And how to then obtain deflection angle ? If there would be only second term I get
$$\vec{\alpha} = \frac{2}{c^2} \int \vec{\nabla}_\perp \Phi~ dl$$
which is correct result. I just don't know what to do with that first term.
Hope I wasn't tooo long and thanks in advance.
1. The problem statement, all variables and given/known data
2. Relevant equations
3. The attempt at a solution
|
2018-08-17 12:13:32
|
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https://orion.readthedocs.io/en/stable/code/benchmark/task/forrester.html
|
# Forrester¶
Forrester Task from the Profet paper.
This Forrester class is based on a synthetic function, whereas the ForresterTask is baseed on a meta-model trained on multiple such functions.
Klein, Aaron, Zhenwen Dai, Frank Hutter, Neil Lawrence, and Javier Gonzalez. “Meta-surrogate benchmarking for hyperparameter optimization.” Advances in Neural Information Processing Systems 32 (2019): 6270-6280.
class orion.benchmark.task.forrester.Forrester(max_trials: int, alpha: float = 0.5, beta: float = 0.5)[source]
Task based on the Forrester function, as described in https://arxiv.org/abs/1905.12982
$f(x) = ((lpha x - 2)^2) sin(eta x - 4)$
Parameters
max_trialsint
Maximum number of trials for this task.
alphafloat, optional
Alpha parameter used in the above equation, by default 0.5
betafloat, optional
Beta parameter used in the above equation, by default 0.5
Methods
Define the black box function to optimize, the function will expect hyper-parameters to search and return objective values of trial with the hyper-parameters. Return the search space for the task objective function
call(x: float) List[Dict][source]
Define the black box function to optimize, the function will expect hyper-parameters to search and return objective values of trial with the hyper-parameters.
This method should be overridden by subclasses. It should receive the hyper-parameters as keyword arguments, with argument names matching the keys of the dictionary returned by get_search_space.
get_search_space() Dict[str, str][source]
Return the search space for the task objective function
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2023-03-25 19:58:27
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|
https://dsp.stackexchange.com/questions/69295/cascade-realisation-from-direct-realisation
|
# cascade realisation from direct realisation
I am trying to convert filter realisation from Direct form 1 to cascade using function provided from Digital signal processing by Proakis. I designed lowpass IIR filter and extracted numerator and denominator coefficients. I used fvtool to plot pole zero as well as magnitude response. But when I use dir2cas function provided in textbook, matlab shows an error while plotting using fvtool. Can someone help?
The filter function:
filte = designfilt('lowpassiir','PassbandRipple',1,'StopbandAttenuation',30, ...
'PassbandFrequency', 3.4e3, 'StopbandFrequency', 4e3, ...
'SampleRate', 16e3);
sos = filte.Coefficients;
[b,a] = sos2tf(filte.Coefficients);
%fvtool(sos);
[b0, B0, A0] = dir2cas(b,a);
fvtool(B0,A0);
function [b0, B, A] = dir2cas(b,a);
b0 = b(1);
b = b/b0;
a0 = a(1);
a = a/a0;
b0 = b0/a0;
M = length(b);
N = length(a);
if N >M
b = [b zeros(1, N-M)];
elseif M > N
a = [a zeros(1, N-M)];
N = M;
else
NM = 0;
end
k = floor(N/2);
B = zeros(k,3);
A = zeros(k,3);
if k*2 == N;
b = [b 0];
a = [a 0];
end
broots = cplxpair(roots(b));
aroots = cplxpair(roots(a));
for i = 1:2:2*k
Brow = broots(i:1:i+1,:);
Brow = real(poly(Brow));
B(fix((i +1)/2),:) = Brow;
Arow = aroots(i:1:i+1,:);
Arow = real(poly(Arow));
A(fix((i+1)/2),:) = Arow;
end
• I’m voting to close this question because reading the documentation of fvtool should solve the problem. – Matt L. Jul 21 at 10:02
• Hi Matt, I went through fvtool documentation, however it still says error. Moreover after converting to cascade, I have 3 outputs numerator, denominator and bo constant, which connects 2 cascade functions. – RohitM Jul 22 at 10:10
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2020-10-21 08:15:32
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https://chemistry.stackexchange.com/questions/51189/is-nifedipine-a-weak-base-or-weak-acid
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# Is nifedipine a weak base or weak acid?
I was wondering whether someone could explain me if nifedipine is a weak base or a weak acid. I barely can find any information about nifedipine's physicochemical properties in studies, however some suggested that it is a weak acid. If it is a weak acid, which hydrogen atom is dissociable and what would the pKa of that group be?
Nifedipine's structure, as shown on wikipedia:
• Here is a reference which states that the "strongest basic" pKa = 5.33. As for which proton, I'm not sure. – Todd Minehardt May 14 '16 at 15:32
• @ToddMinehardt perhaps the H atom at position 4 of the 1,4-dihydropyridine. – Loong May 14 '16 at 18:10
• My guess is still the N-hydrogen. However a $pK_a$ of 4 is more acidic than I would expect, e.g. in comparison with phtalimide. But there may be higher order effects that affect the acidity. – aventurin May 14 '16 at 19:35
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2019-08-23 10:59:05
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https://www.sierrachart.com/index.php?page=doc/StudiesReference.php&ID=97&Name=Chaikin_Oscillator
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# Technical Studies Reference
### Chaikin Oscillator
This study calculates and displays a Chaikin Oscillator for the Price data.
Let the High, Low, and Closing Prices at Index $$t$$ be denoted as $$H_t$$, $$L_t$$, and $$C_t$$, respectively, and let the Volume at Index $$t$$ be denoted as $$V_t$$. Let the $$AD$$ be a random variable denoting the Accumulation Distribution, and let $$AD_t$$ be the value of the Accumulation Distribution at Index $$t$$. We calculate $$AD_t$$ for $$t \geq 0$$ as follows.
$$\displaystyle{AD_t = \frac{(C_t - L_t) - (H_t - C_t)}{H_t - L_t + 10 \cdot 10^{-10}}}$$
The $$10 \cdot 10^{-10}$$ in the denominator is to safeguard against division by zero in the event that $$H_t = L_t$$.
Let the Inputs Long MovAvg Length, Short MovAvg Length, and Divisor be denoted as $$n_L$$, $$n_S$$, and $$v$$, respectively. Then we denote the Chaikin Oscillator for the given Inputs at Index $$t$$ as $$CO_t(n_L,n_S,v)$$, and we compute it for $$t \geq \max\{n_L,n_S\} - 1$$ in terms of Exponential Moving Averages as follows.
$$\displaystyle{CO_t(n_L,n_S,v) = \frac{EMA_t(AD,n_S) - EMA_t(AD,n_L)}{v}}$$
This study also draws horizontal lines whose levels are determined by the Inputs Overbought and Oversold. If either of these Inputs is set to zero, then the corresponding horizontal line will not be drawn.
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2018-07-22 02:57:53
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https://stacks.math.columbia.edu/tag/0DBV
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Definition 72.7.6. Let $S$ be a scheme. A big fppf site $(\textit{Spaces}/S)_{fppf}$ is any site constructed as follows:
1. Choose a big fppf site $(\mathit{Sch}/S)_{fppf}$ as in Topologies, Section 34.7.
2. As underlying category take the category $\textit{Spaces}/S$ of algebraic spaces over $S$ (see discussion in Section 72.2 why this is a set).
3. Choose any set of coverings as in Sets, Lemma 3.11.1 starting with the category $\textit{Spaces}/S$ and the class of fppf coverings of Definition 72.7.1.
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).
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2023-01-26 22:27:04
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https://blog.roboforex.com/blog/2020/03/09/calculating-stock-price-detailed-how-to/
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# Calculating Stock Price: Detailed How-To
Views: 596 views
## Introduction
Below, you may see the chart of the unadjusted (or nominal) closing price of Exxon (XOM) stock every day since 1996.
Naturally, this chart does not give any idea of the profit that the trader could make since 1996: the changes in the nominal price are just one part of investment results. During these years, Exxon paid hundreds of dividends, which made the stock price decline each time. Also, Exxon carried out stock splits five times, and again, five times its stock price shrunk. Exxon acquired several enterprises and merges with Mobil Oil in 1999, which influenced the stock price. However, none of the events above influenced the stockholders because the alterations remained purely nominal.
The time-series analysis of the stock price requires adjustment for or elimination of such nominal changes. The historical stock price needs to be adjusted in such a way that the data received represented the general profit that the trader would have made if they have held a certain stock for a certain period. By adjustments, we create a series that reflects dividends, mergers, spin-offs, splits and other events influencing the factual profitability of the stock.
Any such event or a change in the company’s structure provokes discontinuous changes in the nominal stock price. Moreover, these changes are not due to the sellers or buyers reprice the company, i.e. they are corporate rather than market events. Price adjustment is meant for eliminating such events.
Below, you may also see the chart of the adjusted Exxon stock price since 1996. It differs from the first one dramatically, being much closer to the “economic reality”.
Any professional analyst knows that analysis must be based on the adjusted stock price. However, there are few of those who really know the financial mathematics necessary for adjusting. Of course, you may rely on some third party and get the adjusted prices from them. However, the understanding of how these adjustments are made is the key to a really successful analysis.
Almost always, stock prices are backward-adjusted. In other words, in every time-series, the stock price for “today” is the same as the current price on the exchange. All adjustments are possible on historical data solely.
The adjustment of historical stock prices is normally multiplicative. Thanks to this, the profit from holding the stocks on the days when there were no adjustments remains untouched regardless of all changes. Moreover, historically adjusted prices never turn out negative. However, some make additive adjustments, thus inducing the appearance of negative stock prices.
Below, we will discuss the most frequent corporate events and the adjustments to them.
### Cash dividends
When a company pays dividends, its price declines by the sum paid. This is rather clear: the money is transferred from the company’s deposits to the clients, hence, the company costs this sum less. Thus, on the ex-dividend day, the stock price declines by the size of the dividend.
\begin{align} \text{Share Price before Dividend} &= \frac{\text{Company Value}} {\text{Shares Outstanding}} \\ \\ \text{Share Price after Dividend} &= \frac{\text{Company Value – Total Cash Paid Out}}{\text{Shares Outstanding}} \\ \\ &=\frac{\text{Company Value}}{\text{Shares}} \ – \ \frac{\text{Cash Paid Out}}{\text{Shares}} \\ \\ &=\text{Share Price before Dividend – Dividend per Share} \end{align}
To create a consistent time series of adjusted stock prices, we calculate the adjustment factor which reflects the decline of the stock price and then divide the prices preceding the dividend payment by this factor.
\begin{align} \text{Adjustment Factor} = \frac{\text{Close Price on Dividend Date + Dividend per Share}}{\text{Close Price on Dividend Date}} \end{align}
As long as the adjustment factor is a multiplicative constant, it does not influence the profitability profile of the stock historically. At the same time, thanks to this factor, we may be sure that the calculated income on the dividend day is explained by real market events and not by the payment.
Let us have a look at the example of calculations with the adjustment factor after dividend payment:
Apple (AAPL) paid cash dividends of $0.47 per stock on 08.07.2014. That day, the closing price was$94.48.
The adjustment factor is calculated as follows:
\begin{align} \text{F} = \frac{94.48 + 0.47}{94.48} = 1.00497 \end{align}
The unadjusted closing price of the previous day was \$94.96.
In this case, the adjusted closing price the day before was:
The price for all days preceding the payment is calculated the same way – by multiplication by the factor; i.e., this way we get all historical data of the “changes” provoked by the payment.
“Capital Repayments” and “Special Dividends” are special cases of cash dividend payments, and in these cases, the price is adjusted the same way.
### Stock dividends
Sometimes, companies pay dividends in stocks: each stockholder receives new stocks in proportion to what they already hold.
The idea behind this payment is a decrease in the stock price. The price will be decreased in the ratio of the issued stocks to the existing ones. The overall cost of the company remains unchanged, while the stock price changes as long as the stock also changes. However, it should be noted that the ownership percentage and, hence, the cost of the stocks of each stockholder in dollars does not change.
\begin{align} \text{Share Price before Dividend} &= \frac{\text{Company Value}}{\text{Shares Previous}} \\ \\ \text{Share Price after Dividend} &= \frac{\text{Company Value}}{\text{Shares Previous + Shares Issued}}\\ \\ &= \frac{\text{Company Value}}{\text{Shares Previous}} \times \frac{\text{Shares Previous}}{\text{(Shares Previous + Shares Issued)}}\\ \end{align}
As before, to create a consistent time series, we calculate the adjustment factor reflecting the decrease in the stock price and then divide the prices of the days preceding the dividend payment day by this factor. In this case, the adjustment factor is the second term in the equation above, hence, the delusion affecting the stockholders’ portfolios.
\begin{align} \text{Adjustment Factor} &= \frac{\text{New Float}}{\text{Old Float}}\\ \\ &=\frac{\text{Shares Previous + Shares Issued}}{\text{Shares Previous}}\\ \end{align}
As before, as long as the adjustment factor is multiplicative rather than additive, it does not affect the profitability, just “changes the scale”.
Let us look at the example of calculations with the adjustment factor:
03.12.2014, BIOL had a 0.5% stock dividend. 0.5% stock dividends mean that to each stock that the stockholder already owns 0.005 (=0.5%) of a stock will be added. In other words, to every 200 stocks in the portfolio, 1 new stock will be added.
Hence:
\begin{align} \text{New Float} = \text{1.005} \times \text{Old Float} \end{align}
Hence,
\begin{align} \text{Adjustment Factor} = \frac{\text{New Float}}{\text{Old Float}} = 1.005 \end{align}
The unadjusted stock price on the pre-dividend day was 2.83.
So, the adjusted stock price that day was:
Note that for there calculations we do not use the closing price on the dividend day.
Stock dividends are sometimes called Bonus Issue.
### Stock split
A stock split is like stock dividends. A stock split makes each existing stock become several stocks in a set proportion. This is exactly like at stock dividends payment stockholders get new stocks in addition to those they are already holding.
For the stock split, the adjustment factor is calculated the same way as for dividend stocks:
\begin{align} \text{Adjustment Factor} = \frac{\text{New Float}}{\text{Old Float}} \end{align}
Let us discuss an example of a stock split:
Chesapeake Utilities Corp. (CPK) had a stock split of 3 to 2 effective on 09.09.2014. Thus, instead of every 2 existing stocks, its stockholders received 3.
In other words, to every 2 stocks that they already owned, 1 stock was added; this is absolutely equal a stock dividend payment of 50%.
In this case,
\begin{align} \text{New Float} = \frac{3}{2} \times \text{Old Float} \end{align}
Hence:
\begin{align} \text{Adjustment Factor} = \frac{\text{New Float}}{\text{Old Float}} = 1.5 \end{align}
The day before the split, the unadjusted stock price was 69.41.
Thus, the adjusted stock price that day was:
The stock split is also sometimes called Bonus Issue.
### Reverse stock split
Reverse stock split differs from the normal one in the sense that the stockholders get not more but fewer stocks. Instead of increasing the number of stocks in the portfolio, a reverse split decreases it in a set proportion.
As long as the overall number of stocks after a reverse split decreases, the stock price grows. The cost of the company is not changed by this corporate event.
As before,
\begin{align} \text{Adjustment Factor} = \frac{\text{New Float}}{\text{Old Float}} \end{align}
Hence, the adjustment factor at a reverse stock split is less than 1.
Let us discuss an example of a reverse stock split:
PostRock Energy Corp. (PSTR) carried out a reverse stock split in proportion of 1 to 10 01.05.2015.
\begin{align} \text{New Float} = \frac{1}{10} \times \text{Old Float} \end{align}
Thus:
\begin{align} \text{Adjustment Factor} = \frac{\text{New Float}}{\text{Old Float}} = 0.1 \end{align}
The unadjusted stock price on the day before the reverse split was 0.4442.
Reversal stock split is also called Consolidation.
Even a glance at the process of price adjustment is enough to realize how much work is necessary for collecting bias-free, well-adjusted historical data on stock prices. Though in essence, there is nothing complicated to it, however, the process of database creation is painstaking, tiresome work.
In 2015, there happened 20,000 of dividend payments only – and they are just one type of corporate events. There are also splits, mergers, reverse splits, consolidations, acquisitions, rights issues, buybacks, treasury repurchases, etc. When any of these events happen, all historical data on the stock price of the company needs recalculation. This means recalculating many thousands of date rows (250 date rows a year) and OHLCV for each stock, every day. And there are thousands of companies in the US public markets.
All in all, maintenance of the database of stock prices historical data is a great amount of work, requiring expertise and effort. That is why the market is dominated by a small number of highly professional data providers: low professionalism means lost rivalry. In the stock world, as anywhere else, you get what you pay for.
## Corporate events in R Trader
If for some reason you are not acquainted with the multi-market trading platform R Trader, you may start with the information in this post.
### Long positions
A client having an open long position on the ex-dividend day will have a sum equivalent to the paid dividend deposited on their account. The operation is reflected on the page History – Account Information – Cash Corrections.
### Short Positions
A client, having an open short position on the ex-dividend day will have a sum equivalent to the paid dividend withdrawn from their free funds. The operation is reflected on the page History – Account Information – Cash Corrections.
### Dividends procedure
Dividends procedure is depositing/withdrawal to/from the account on the ex-dividend date, at 3 p.m. server time. he operation is reflected on the page History.
For a Long position, Cash Dividend Amount will be:
Dividend per stock*Volume
where
Volume=Contracts*Contract Size
For a Short position, Cash Dividend Amount will be:
(-1)*Dividend per stock*Volume
where
Volume=Contracts*Contract Size
### Stock Splits
In the case of a Stock Split, the necessary correction of the client’s position will be reflected in their trading account in accordance with the split parameters.
### Split Procedure
The Split procedure is carried out on the server every day at 3 p.m. server time. This operation deletes all active pending orders (Limit, Stop) on the stock.
For all open short and long positions, the weighted-mean price and general volume are calculated, respectively. A split happens, and a new price and volume are set. The information is assigned to the long and short positions with maximal volume, respectively. When there are fractional stocks in a trade, such stocks are eliminated and turned into a balance operation – Split Cash Correction. The volume of other trades on the instrument is cleared and transferred to History.
If a corporate event results in a fractional position, the RoboForex company reserves the right to deposit a component to be paid to the client’s account as a balance operation.
### Other Corporate Actions
If a stock is excluded from the list of the stock exchange, merged, acquired, put up to tender or distributed among the stockholders, the client’s position will be closed at the last market price.
We'll cover any topical theme on trading and investing if you find it useful.
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2021-03-06 00:52:37
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https://www.wptricks.com/question/using-php-code-snippets-to-query-a-database-dbconnection-get_results-is-outputting-nothing/
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Using PHP Code Snippets to query a database, $dbconnection->get_results is outputting nothing Question I’m currently trying to run this php code snippet in a WordPress Page using a PHP code snippet extension. //Database connection script //Connecting to our master database global$wpdb;
$hostname = "host";$username = "user";
$password = "pass";$dbname = "dbname";
//creating new database object
$dbconnection = new$wpdb($username,$password, $dbname,$hostname );
//Querying the database
$dbconnection->show_errors( true );$dbQuery = $dbconnection ->get_results("SELECT * FROM Master");$dbconnection->last_query;
$dbconnection->print_error(); var_dump($dbconnection->last_query);
The output on the page looks like this, the table headers are there because of html code. The goal is to eventually get the data properly formatted under them but for now I’m getting no results from the database. The database is a separate database from the automatically generated WordPress one created through PhPMyAdmin, so the only thing I can think of is that my new wpdb object might be wrong but I made sure to double/triple check the information. I’ve been stuck on this for hours because I’m getting no errors either through print_error.
0
5 months 2022-03-22T01:07:50-05:00 0 Answers 0 views 0
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2022-08-19 06:10:41
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https://physics.stackexchange.com/questions/590250/jacobi-equation-for-the-geodesic-deviation-in-the-weak-field-limit
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# Jacobi equation for the geodesic deviation in the weak field limit
The geodesic deviation equation can be written in the following form $$\nabla_U^2 \xi = R (U, \xi) U \tag{1}$$ where $$R$$ is the Ricci tensor. It can also be written component-wise using the Riemann tensor $$(\nabla_U^2 \xi)^\alpha = R^\alpha_{\;\, \beta \mu \nu} \: U^\beta U^\mu \xi^\nu \tag{2}$$
I'm interested in this equation on a Riemannian manifold equipped with the following metric tensor $$g = -(1+2\phi(x))\mathrm{d} t \otimes \mathrm{d} t + (1-2\phi (x))\left(\mathrm{d} x \otimes \mathrm{d} x + \mathrm{d} y \otimes \mathrm{d} y + \mathrm{d} z \otimes \mathrm{d} z \right) \tag{3}$$ with the standard torsion-free and metric-preserving connection $$\nabla$$. I'm only interested in the spatial part of the Jacobi equation, that starts from $$U = e_0$$ i.e. $$(\nabla_t^2 \xi)^i = R^i_{\;\, 0 0 \nu} \,\xi^\nu \tag{4}$$
In textbooks it can be found that in the weak-field limit (keeping only the linear terms involving $$\phi (x)$$) this reduces to $$\frac{\mathrm{d}^2 \xi^i}{\mathrm{d} t^2} = - \phi_{,ij} \, \xi^j \tag{5}$$ where comma indicates partial derivatives.
However, when I try this for $$g$$ above, I get a different result. First, the left-hand side of (4) is $$\nabla_t \xi = \left( \xi^\mu_{,t} + \Gamma^\mu_{\;\; \nu t} \, \xi^\nu \right) e_\mu$$ $$\nabla^2_t \xi = \left( \xi^\mu_{,t} + \Gamma^\mu_{\;\; \nu t} \, \xi^\nu \right)_{,t} e_\mu + \left( \xi^\mu_{,t} + \Gamma^\mu_{\;\; \nu t} \, \xi^\nu \right) \Gamma^\lambda_{\;\; \mu t} e_\lambda$$
The Christoffel symbols are in general $$\Gamma^\alpha_{\;\; \mu \nu} = \frac{1}{2} g^{\alpha \lambda} \left( g_{\lambda \mu, \nu} + g_{\lambda \nu, \mu} - g_{\mu \nu, \lambda} \right)$$ so in our case $$\Gamma^0_{\;\; \mu \nu} = \begin{pmatrix} \phi_{,0} & \phi_{,1} & \phi_{,2} & \phi_{,3} \\ \phi_{,1} & - \phi_{,0} & 0 & 0 \\ \phi_{,2} & 0 & - \phi_{,0} & 0 \\ \phi_{,3} & 0 & 0 & - \phi_{,0} \end{pmatrix}$$ $$\Gamma^1_{\;\; \mu \nu} = \begin{pmatrix} \phi_{,1} & - \phi_{,0} & 0 & 0 \\ - \phi_{,0} & - \phi_{,1} & - \phi_{,2} & - \phi_{,3} \\ 0 & - \phi_{,2} & \phi_{,1} & 0 \\ 0 & - \phi_{,3} & 0 & \phi_{,1} \end{pmatrix}$$ $$\Gamma^2_{\;\; \mu \nu} = \begin{pmatrix} \phi_{,2} & 0 & - \phi_{,0} & 0 \\ 0 & \phi_{,2} & - \phi_{,1} & 0 \\ - \phi_{,0} & - \phi_{,1} & - \phi_{,2} & - \phi_{,3} \\ 0 & 0 & - \phi_{,3} & \phi_{,2} \end{pmatrix}$$ $$\Gamma^3_{\;\; \mu \nu} = \begin{pmatrix} \phi_{,3} & 0 & 0 & - \phi_{,0} \\ 0 & \phi_{,3} & 0 & - \phi_{,1} \\ 0 & 0 & \phi_{,3} & - \phi_{,2} \\ - \phi_{,0} & - \phi_{,1} & - \phi_{,2} & - \phi_{,3} \end{pmatrix}$$ which, for $$(\nabla^2_t \xi)^i$$ gives (to the linear order in $$\phi$$) $$\xi^i_{\;\;, 00} + \phi_{, i0} \xi^0 - \phi_{,00} \xi^i + 2 \phi_{,i} \xi^0_{\;\;, 0} - 2 \phi_{,0} \xi^i_{\;\;,0}$$
The right-hand side needs $$R^i_{\;\; 00 \mu}$$. First, due to the symmetries, $$R^i_{\;\;000} = 0$$, so we only need $$R^i_{\;\; 00j} = - \phi_{, ij} - \phi_{,00} \, \delta_{ij}$$
Now put it all together $$\frac{\partial^2 \xi^i}{\partial t^2} + \phi_{, i0} \xi^0 - \phi_{,00} \xi^i + 2 \phi_{,i} \xi^0_{\;\;, 0} - 2 \phi_{,0} \xi^i_{\;\;,0} = - \phi_{,ij} \xi^j - \phi_{,00} \xi^i$$
Cancel out common terms $$\frac{\partial^2 \xi^i}{\partial t^2} + \phi_{, i0} \xi^0 + 2 \phi_{,i} \xi^0_{\;\;, 0} - 2 \phi_{,0} \xi^i_{\;\;,0} = - \phi_{,ij} \xi^j$$
There are problems with this expression. It does not match what the textbooks say.
Even if it did, how can I get "$$\frac{\mathrm{d}^2 \xi^i}{\mathrm{d} t^2}$$" instead of the partial derivative? Is $$\frac{\mathrm{d}}{\mathrm{d} t}$$ to be interpreted as $$U^\mu \partial_\mu$$? In that case if $$U = e_0$$, we would have $$\frac{\mathrm{d}}{\mathrm{d} t} = \frac{\partial}{\partial t}$$, but probably only at the initial point of our geodesic...or do we reparametrize the geodesic in terms of the time $$t$$ and then derivative w.r.t $$t$$ is actually derivative w.r.t. the curve parameter, so $$\nabla_U^2 = \frac{\mathrm{d}^2}{\mathrm{d} \lambda^2} = \frac{\mathrm{d}^2}{\mathrm{d} t^2}$$? That would certainly interpret the left-hand side of (4), but we would still have that weird term $$-\phi_{,00} \xi^i$$ on the right-hand side (unless we assume that $$\phi$$ does not depend on time, on the top of all that). But even if we do it, then $$U$$ is no longer in just zeroth direction so the right-hand side will feature two $$U$$ terms (even if we begin with $$U = e_0$$, can it stay that way for the entire geodesic? or do we assume small $$v$$, therefore the zeroth component dominates and since $$U \cdot U = -1$$, then $$U^0 = 1$$?)
There's a bunch of other terms that don't seem to arise in textbooks, but I never saw the expression (5) derived, only stated.
Is there a conceptual mistake in my thinking, or I miscalculated something?
• The potential is usually assumed to be time independent, which gets rid of two of the terms. If we can show that we can choose $\xi^0=0$ for all times, that gets rid of the last one, and you get your desired expression. Oct 29 '20 at 3:25
• That sounds reasonable! Still not sure about the issue of d vs $\partial$ though...but we probably assume that $U$ (four-velocity of the geodesic) is dominated by $(1,0,0,0)$ so $U^\mu \partial_\mu \xi \equiv \frac{\mathrm{d} \xi}{\mathrm{d} t} = \gamma (v) \left( \frac{\partial \xi}{\partial t} + \vec{v} \cdot \vec{\nabla} \, \xi \right) \approx \frac{\partial \xi}{\partial t}$. Oct 29 '20 at 3:52
• Your mistake, I think, is that you assume thet $U=e_0$ . This is obviously wrong, since U should be a geodesics of the given metric. The geodesic deviation equation mesures the acceleretion (ie. second derivative) of the relative dispacement of 2 nearby geodesics (the U field). Since your assumed U is wrong, the equation does not hold. Oct 29 '20 at 11:47
• You should first calculate the U field for your metric using the geodesics equation. then take 2 geodesics dispaced by a small amount initially and then see how that displacement changes with time. Many GR textbooks report this calculation for the weak field. However the easiest thing is to see the derivation of the geodesic deviation equation simbolically, to convince yourself that it is correct Oct 29 '20 at 11:58
• But for a weak field and small velocity we would neglect the higher orders (i.e. $U^\beta U^\mu \xi^\nu$) anyway, what would the solution for U provide? Moreover, it would be formal at best, since $\phi (x)$ has only a general form at this point, just independent of time. Assuming that $U = e_0$ at the beginning means that the objects fall from a stationary initial condition. The expression for $\dot{U}$ contains $\Gamma$'s, which contain the field $\phi$ so plugging that in will kill these extra terms, anyway, I think, but to be sure I have to try. Oct 29 '20 at 13:21
Here's an answer that I argued for myself and am pretty satisfied with it. The caveat is actually in what people mean by writing $$\frac{\mathrm{d}^2 \xi}{\mathrm{d} t^2}$$. This term is more of a placeholder for $$\nabla_U (\nabla_U \xi) = \nabla_U^2 \xi \equiv \frac{\mathrm{d}^2 \xi}{\mathrm{d} \lambda^2}$$ where $$\lambda$$ is the parameter along the curve. When I read the corresponding chapter from my differential geometry book I convinced myself that that's it, there's nothing more to the left-hand side. The only thing we need to argue is that for a weak field $$\phi$$ and two close, slowly moving observers starting from the rest and not going too far along the geodesic$${}^1$$, their four-velocity $$U$$ is dominated by $$e_0$$ and moreover, we can reparametrize the curve in terms of $$\tau$$, the proper time, which becomes just the frame time $$t$$, if the observers are falling slowly. Therefore, $$\nabla_U (\nabla_U \xi) \approx \frac{\mathrm{d}^2 \xi}{\mathrm{d} t^2}$$
The right-hand side was correct; for a weak field, the relevant part of the Riemann tensor is $$R^i_{\;\; 000j}$$ which is equal to $$- \phi_{,ij} - \phi_{,00} \delta_{ij}$$. If we additionally assume that $$\phi$$ does not depend on time, then we get $$\frac{\mathrm{d}^2 \xi^i}{\mathrm{d} t^2} = - \phi_{,ij} \xi^j$$
$${}^1$$Which is how I imagine we (naively) gauge the tides and such effects; we let go off a small, sensitive gauging device in the gravitational field, let it measure for a few seconds and catch it again. That gives us how $$\xi$$ (gauging device's spring length, sensor measuring distance or however you would "practically" gauge this) changes in time, shortly after being let go, moving slowly, and not too far along the geodesic so it does not acquire much speed.
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2022-01-18 17:04:37
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https://brilliant.org/problems/falling-charges-how-many-will-succeed/
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# Falling charges, how many will succeed?
A horizontal conducting cylindrical hollow pipe of radius $$R = 54\text{ mm}$$ and length $$L = 100 \text{ cm} (R<<L)$$ has a small hole $$P$$ at its top, at the middle of the length as shown in the figure. Drops of mass $$M = 231 \text{ mg}$$ and charge $$q = 1 \text{ nC}$$ are falling into the hole from point $$A$$, at height $$2R$$ measured from the axis of the cylinder. Assume that the charge in the fallen drop gets uniformly distributed over the surface of the cylinder and charge distributed on cylinder remains uniform throughout. If the number of drops that will be able to enter the cylinder is given as $$n = x \times10^y$$ in scientific notation. Find the value of $$x + y$$.
×
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2017-12-18 20:36:05
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https://www.gradesaver.com/textbooks/math/algebra/algebra-2-1st-edition/chapter-3-linear-systems-and-matrices-3-1-solve-linear-systems-by-graphing-3-1-exercises-skill-practice-page-156/14
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# Chapter 3 Linear Systems and Matrices - 3.1 Solve Linear Systems by Graphing - 3.1 Exercises - Skill Practice - Page 156: 14
-2, -3
#### Work Step by Step
To solve the equation by graphing, we graph the two lines, and we find the intersection points, knowing that these points are solutions. The problem asks us to solve algebraically, so we find: $$3\cdot \frac{-17-y}{7}-10y=24 \\ y=-3$$ Thus: $$x=\frac{-17-\left(-3\right)}{7}=-2$$
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.
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2022-08-18 08:47:54
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http://www.apnorton.com/blog/
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# A Brief Exploration of a Möbius Transformation
As part of a recent homework set in my complex analysis course, I was tasked with a problem that required a slight generalization on part of Schwarz’s Lemma:
Lemma (Schwarz): Let $f$ be analytic on the unit disk with $|f(z)| \leq 1$ for all $z$ on the disk and $f(0) = 0$. Then $|f(z)| < |z|$ and $f’(0)\leq 1$.
If either $|f(z)|=|z|$ for some $z\neq0$ or if $|f’(0)|=1$, then $f$ is a rotation, i.e., $f(z)=az$ for some complex constant $a$ with $|a|=1$.
The homework assignment asked us (within the context of a larger problem) to consider the case when $f(\zeta) = 0$ for some $\zeta \neq 0$ on the interior of the unit disk. The secret to this problem was to find some analytic function $\varphi$ that maps the unit disk to itself, but swaps $0$ and $\zeta$. Then, we may consider $\varphi^{-1}\circ f\circ \varphi$ and apply Schwarz’s Lemma.
# How I wrote a GroupMe Chatbot in 24 hours
For the past couple years, I have worked as a teaching assistant for UVa’s CS 2150 (Program and Data Representation) course. We recently started a GroupMe chat for the course staff, and I thought it would be fun to create a chatbot to help remind all the TAs to submit timesheets, keep track of when people are holding office hours, and remember when/where TA meetings are being held. Setting up a basic chatbot is a lot simpler than it sounds and is really fun–I wrote my bot from scratch using Python in just one day.
## GroupMe Bot Overview
GroupMe has a very brief tutorial explaining how their API may be used for bots. The easiest way to create a bot is through their web form, which prompts you for the bot’s name, callback URL (technically optional, but you want it), avatar URL (optional), and the name of the group where the bot will live. Once you’ve done this, you will be given a unique bot ID token. Anyone with this token can pretend to be your bot, so keep it safe. (Security is somewhat laughable here: your bot asserts its ID and the server believes it with no “login” procedure.) We’ll talk more about the callback URL in a moment; for now, just leave it blank.
Once you’ve done these steps, you have created a bot–as far as GroupMe is concerned. If you send a specifically formatted JSON mssage, the newly created bot will post in your group. However, if left at this point, your “bot” is little more than a proxy for human-written messages submitted with curl. Your bot needs some way of reading messages sent to the group, formulating a response, and only then sending messages to the GroupMe servers.
# TensorFlow with the Surface Book
While interning at Microsoft over the summer, I received a first-generation Surface Book with an i5-6300U CPU (2.4 GHz dual core with up to 3.0 GHz), 8GB RAM, and a “GeForce GPU” (officially unnamed, but believed to be equivalent to a GT 940). This is a huge step up from my older laptop, so I wanted to set it up for my ML work. In this post, I’ll outline how I set it up with TensorFlow and GPU acceleration.
## CUDA + cuDNN
If you want to use GPU acceleration, the typical way to do so is with NVIDIA’s CUDA API. CUDA 8.0 is compatible with the Surface Book and is (as of this writing) the most up-to-date version of CUDA. Download it from the NVIDIA website and run their installer.
For work with deep learning, you’ll also want to install cuDNN. To install, just download the library from NVIDIA’s website and unzip it in a convenient place (I chose C:\cudnn). The only “installation” you need to do is to add C:\cudnn\bin to your PATH environment variable.
# Visualizing Multidimensional Data in Python
Nearly everyone is familiar with two-dimensional plots, and most college students in the hard sciences are familiar with three dimensional plots. However, modern datasets are rarely two- or three-dimensional. In machine learning, it is commonplace to have dozens if not hundreds of dimensions, and even human-generated datasets can have a dozen or so dimensions. At the same time, visualization is an important first step in working with data. In this blog entry, I’ll explore how we can use Python to work with n-dimensional data, where $n\geq 4$.
## Packages
I’m going to assume we have the numpy, pandas, matplotlib, and sklearn packages installed for Python. In particular, the components I will use are as below:
## Plotting 2D Data
Before dealing with multidimensional data, let’s see how a scatter plot works with two-dimensional data in Python.
First, we’ll generate some random 2D data using sklearn.samples_generator.make_blobs. We’ll create three classes of points and plot each class in a different color. After running the following code, we have datapoints in X, while classifications are in y.
To create a 2D scatter plot, we simply use the scatter function from matplotlib. Since we want each class to be a separate color, we use the c parameter to set the datapoint color according to the y (class) vector.
# Election 2016: Moving Forward
Like many of my fellow Americans, I stayed up late tonight to watch the polling results for the 2016 General Election. As of my writing this, it appears that Donald Trump will win by a slight margin. The New York Times is predicting that the popular vote will go to Hillary Clinton, while Politico and the Wall Street Journal are showing the current popular vote is Trump’s by about 1 million.
# New Feature: Commenting!
Thanks to a helpful blog post by CodeBlocQ, I’ve now enabled Disqus-powered comments on the blog! Let me know what you think about my posts, and I’ll keep an eye on discussions to respond to questions/comments/concerns!
The second part of the Microsoft series should be out soon; I wanted to get comments working before I did so, but it took me a while to find the time to actually get it up and running.
# A Microsoft Summer, Part 1: Seattle Fun
As suggested by this post’s title, I spent this past summer as an intern with Microsoft in Redmond, Washington. The experience was highly educational for me–as my first (and last!) “real” internship, I learned a lot about software development and the importance of corporate culture, as well as discovering a lot about myself. Overall, the experience was a positive one, though, and I had an enormous amount of fun!
This is the first of a three-part series on my time at Microsoft. This post focuses on fun recreational activities for interns in the Seattle area.
## Outdoors
The Pacific Northwest is home to some of the most amazing views I’ve ever seen. Seattle is conveniently located close to the beach, the mountains, Puget Sound, rainforests, and many hiking trails and campsites. Exploring the outdoors also has the advantage of being very inexpensive, which is great if you’re saving your internship money for college expenses. If you visit National Parks, consider the National Park Passport Program–if you’re going to once-in-a-lifetime parks, it’s a good idea to get your passbook stamped!
This summer, I’ve been working as an intern for Microsoft on the Direct2D/DirectWrite team. While I can’t really talk about what my work entails, I can talk about some of the fun things I’ve done this summer in my free time and the non-work-related components of my internship. I suppose most people wouldn’t start blogging about their internship by describing a bookstore, but I went to this place today and it was so incredible that I had to write about it.
In Capitol Hill, there’s a small store by the name Ada’s Technical Books. It’s in a house that’s been converted to a cafe and bookstore, and is quite possibly the most amazing bookstore I’ve ever seen. As you walk in, you’re greeted by an small cafe counter to your left and an open area to your right with short bookcases and comfy chairs. Toys, puzzles, and “Maker”-appropriate items like lockpicks and Raspberry Pis.
# Visualizing Graphs in Program Output
Many computer science problems utilize graph-based data structures. Their use can range from explicit inclusion in an algorithm-centric problem (like path-finding) to a more “behind-the-scenes” presence in Bayesian networks or descriptions of finite automata. Unfortunately, visualizing large graphs can be difficult to do, especially for debugging. Unlike lists or dictionaries, which can be represented clearly by plain text printing, depicting a graph tends to require more graphics overhead than is reasonable for most programmers to write simply for debugging purposes. I’ve found that Graphviz, a free graph visualization utility, can be quite useful in debugging graph-related programs.
# Deranged Exams: An ICPC Problem
This past week, my ICPC team worked the 2013 Greater New York Regional problem packet. One of my favorite problems in this set was Problem E: Deranged Exams. The code required to solve this problem isn’t that complicated, but the math behind it is a little unusual. In this post, I aim to explain the math and provide a solution to this problem.
## Problem Description
The full problem statement is archived online; in shortened form, we can consider the problem to be:
Given a “matching” test of $n$ questions (each question maps to exactly one answer, and no two questions have the same answer), how many possible ways are there to answer at least the first $k$ questions wrong?
It turns out that there’s a really nice solution to this problem using a topic from combinatorics called “derangements.” (Note that the problem title was a not-so-subtle hint towards the solution.)
## Derangements
While the idea of a permutation should be familiar to most readers, the closely related topic of a derangement is rarely discussed in most undergraduate curriculum. So, it is reasonable to start with a definition:
A derangement is a permutation in which no element is in its original place. The number of derangements on $n$ elements is denoted $D_n$; this is also called the subfactorial of $n$, denoted $!n$.
The sequence $\langle D_n\rangle$ is A000166 in OEIS (a website with which, by the way, every competitive programmer should familiarize themselves).
It turns out that there is both a recursive and an explicit formula for $D_n$:
\begin{aligned} D_n &= (-1)^n \sum_k\binom{n}{k} (-1)^k k! \\ &= n\cdot D_{n-1} + (-1)^n;\;(D_0=1) \end{aligned}
This is significant because we can use the explicit formulation for computing single values of derangements, or we can use dynamic programming to rapidly compute $D_n$ for relatively small $n$.
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2017-06-29 08:49:21
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https://wenbinmei.github.io/post/coding_resource/
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Resources for the R, Python and Perl.
I learned some coding in the computer class during my undergraduate, but I never used in the research until during the master period. During that time, I starts to assemble 100kb size chloroplast genomes using Illumina short reads, and I start to learn more linux, bash and scripting. I enjoyed so I want to do bioinformatics for my PhD. When I start the PhD program at UF, my advisor gives me a book Teach yourself perl in 21 days. After reading this book, I start to do the projects and learn by doing is the best approach in bioinformatics and genomics. Bioperl has many useful examples to process and analyze the data. The field is developping very fast and Google is my best teacher along the jouney.
I learn more python and R when the PhD progress. Sometimes the choice of programming languages to use also depends on the lab, My current lab most use R. Now biopython and R bioconductor has many packages for large scale biological data analysis. Python also has a nice machine learning package, of course R has the similar nice resource for machine learning as well a book called An Introduction to Statistical Learning with Applications in R.
Here is a few more useful resources:
Perl:
Unix & Perl for Biologist
General introduction bioinformatics:
Bioinformatics data analysis
R:
Rstudio cheat sheets
R for Data Science
Python:
Python Data Science
Introduction to Python
Basic Python
Forum:
biotrainee.com
Capital of Statistics
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2021-06-19 20:59:11
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