url stringlengths 14 2.42k | text stringlengths 100 1.02M | date stringlengths 19 19 | metadata stringlengths 1.06k 1.1k |
|---|---|---|---|
http://love2d.org/forums/viewtopic.php?f=4&t=84767 | ## Simple question
Questions about the LÖVE API, installing LÖVE and other support related questions go here.
Forum rules
Tuxion
Prole
Posts: 14
Joined: Mon Jan 08, 2018 6:31 pm
### Simple question
I want to know, when a button is pressed, how to keep the function active after the button has been used. with what function do I do this?
zorg
Party member
Posts: 2653
Joined: Thu Dec 13, 2012 2:55 pm
Location: Absurdistan, Hungary
Contact:
### Re: Simple question
Hi and welcome to the forums;
Your simple question is truly simple, yet oh so vague; let me ask you a few follow-up questions:
- Do you mean mouse button, or a graphical button object you might have coded or want to code?
- What function do you exactly want to keep alive? And how do you define "keep alive"?
- What do you mean by "after the button has been used? As in the mouse-button released? Or a graphical one clicked on?
- Do you perhaps want a function to be called/executed/ran every frame after one of the aforementioned things, or would you like to do something else?
Me and my stuff True Neutral Aspirant. Why, yes, i do indeed enjoy sarcastically correcting others when they make the most blatant of spelling mistakes. No bullying or trolling the innocent tho.
Ostego160
Prole
Posts: 14
Joined: Tue Oct 17, 2017 7:18 pm
### Re: Simple question
You tie the input command to a boolean variable.
Code: Select all
function love.load()
switchOn = false
end
function love.update(dt)
if love.keyboard.isDown('space') then
switchOn = true
end
if switchOn then
doSomething()
end
end
When the spacebar is pressed, switchOn becomes true and stays true even if released.
Tuxion
Prole
Posts: 14
Joined: Mon Jan 08, 2018 6:31 pm
### Re: Simple question
Thanks Ostego160 ! it solves my problem.
### Who is online
Users browsing this forum: No registered users and 8 guests | 2019-08-18 17:52:13 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.24925129115581512, "perplexity": 8278.190516452909}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-35/segments/1566027313987.32/warc/CC-MAIN-20190818165510-20190818191510-00230.warc.gz"} |
http://www.acmerblog.com/hdu-3512-perfect-matching-6473.html | 2014
11-05
# Perfect matching
Given the complete bipartite graph. Each part of this graph has N vertices, i.e. the graph is balanced. An integer number (called a weight) assigns to each vertex. The weight of the edge is defined as the product of weights of vertices which connected by this edge.
It is well known, a matching in a graph is a set of edges without common vertices. A matching is perfect, if it covers all vertices of a graph.
Write a program to find a perfect match of the given bipartite graph with the maximal weight of the matching edges minimal.
The input consists of multiple cases.
For each testcase,there will be exactly three lines;
The first line contains one integer number N (1 ≤ N ≤ 105). The second line consist of N integer numbers, not exceeded 109 by absolute value. The i-th number in line denotes a weight of i-th vertex of first part of a graph. In the third line, weights of vertices of second part are presented.
The input consists of multiple cases.
For each testcase,there will be exactly three lines;
The first line contains one integer number N (1 ≤ N ≤ 105). The second line consist of N integer numbers, not exceeded 109 by absolute value. The i-th number in line denotes a weight of i-th vertex of first part of a graph. In the third line, weights of vertices of second part are presented.
3
1 2 3
9 10 11
27
#include <iostream>
#include <stdio.h>
#include <algorithm>
#include <string.h>
using namespace std;
const int maxn= 100005;
long long a[maxn],b[maxn];
const long long INF = 1LL<<62;
int main(){
int n;
while(scanf("%d",&n) != EOF){
for(int i = 0;i < n;i ++){
scanf("%I64d",&a[i]);
}
for(int i = 0;i < n;i ++){
scanf("%I64d",&b[i]);
}
sort(a,a + n);
sort(b,b + n);
int i1 = lower_bound(a,a + n,0) - a;
int i2 = lower_bound(b,b + n,0) - b;
long long ans = -INF;
int d1 = min(n - i1,i2);
int i = n - 1,j = d1 - 1;
for(int cnt = 0;cnt < d1;cnt ++){
ans = max(ans,a[i --] * b[j --]);
}
int d2 = min(i1,n - i2);
i = d2 - 1,j = n - 1;
for(int cnt = 0;cnt < d2;cnt ++){
ans = max(ans,a[i --] * b[j --]);
}
i = n - 1 - d1,j = d1;
for(int cnt = 0;cnt < n - d1 - d2;cnt ++){
ans = max(ans,a[i --] * b[j ++]);
}
printf("%I64d\n",ans);
}
return 0;
}
1. Good task for the group. Hold it up for every yeara??s winner. This is a excellent oppotunity for a lot more enhancement. Indeed, obtaining far better and much better is constantly the crucial. Just like my pal suggests on the truth about ab muscles, he just keeps obtaining much better. | 2017-06-26 08:47:38 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 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": 0.2911187410354614, "perplexity": 2743.6534600941336}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-26/segments/1498128320695.49/warc/CC-MAIN-20170626083037-20170626103037-00011.warc.gz"} |
http://mathhelpforum.com/calculus/42719-solved-derivative-integral.html | # Thread: [SOLVED] Derivative of an integral
1. ## [SOLVED] Derivative of an integral
From galactus,
$\frac{d}{dx}\int_{h(x)}^{g(x)}f(t)dt=f(g(x))g'(x)-f(h(x))h'(x)$
. It helped me verifying my result and confirmed it.
I got that the derivative of $\int_a^{x^3} sin^3(t) dt=3x^2{sin(x^3)}^3$.
While from Mathematica, it gives the result $\frac{1}{12} \big (9cos(a)-cos(3a)-9cos(x^3)+cos(3x^3)\big)$ which I doubt is equal to my result. Can you confirm my result (if yes, I will think that the result from Mathematica is the same, even if it seems more than incredible.). Thanks!!
2. Originally Posted by arbolis
From galactus, . It helped me verifying my result and confirmed it.
I got that the derivative of $\int_a^{x^3} sin^3(t)^3 dt=3x^2{sin(x^3)}^3$.
While from Mathematica, it gives the result $\frac{1}{12} \big (9cos(a)-cos(3a)-9cos(x^3)+cos(3x^3)\big)$ which I doubt is equal to my result. Can you confirm my result (if yes, I will think that the result from Mathematica is the same, even if it seems more than incredible.). Thanks!!
3. Ok. How it is possible that Mathematica gives an answer in function of a? Maybe it doen't consider it as a constant. I don't know at all.
4. Originally Posted by arbolis
Ok. How it is possible that Mathematica gives an answer in function of a? Maybe it doen't consider it as a constant. I don't know at all.
Yeah, most likely.
5. Originally Posted by arbolis
Ok. How it is possible that Mathematica gives an answer in function of a? Maybe it doen't consider it as a constant. I don't know at all.
I don't think that you should use the formula given by galactus if any one of the limit is constant.suppose if both limits were constant then you would have got 0 as result which is not necessary.
Since you were solving definite integral getting you can not take even constant for granted as we do in indefinite integral(giving general term e.g c for all constant terms) but in definite integral showing constant term is must.
6. Originally Posted by nikhil
I don't think that you should use the formula given by galactus if any one of the limit is constant.suppose if both limits were constant then you would have got 0 as result which is not necessary.
Since you were solving definite integral getting you can not take even constant for granted as we do in indefinite integral(giving general term e.g c for all constant terms) but in definite integral showing constant term is must.
What are you talking about? The constant terms will drop out due to the derivative.
7. Originally Posted by arbolis
Ok. How it is possible that Mathematica gives an answer in function of a? Maybe it doen't consider it as a constant. I don't know at all.
One way to check this would be to let a = 0, say, and then put Mathematica on the job. Note that there are various equivalent forms of the correct answer.
8. I don't think that you should use the formula given by galactus if any one of the limit is constant.suppose if both limits were constant then you would have got 0 as result which is not necessary.
Since you were solving definite integral getting you can not take even constant for granted as we do in indefinite integral(giving general term e.g c for all constant terms) but in definite integral showing constant term is must.
I don't understand all what you mean, but the formula galactus gave works here. (Now I realize I proved it about 9 months or so ago.) And if both of the limit of the integral were constants, say 0 and 10, then the derivative would be 0, which is true. Think about it, you have $F(x)=\int_0^{10} f(t)dt$. It's obvious that the integral is equal to a constant, whatever x is, since it doesn't depends of x. So the derivative of F is 0 (since the derivative of a constant is always 0).
9. Originally Posted by nikhil
I don't think that you should use the formula given by galactus if any one of the limit is constant.suppose if both limits were constant then you would have got 0 as result which is not necessary.
Since you were solving definite integral getting you can not take even constant for granted as we do in indefinite integral(giving general term e.g c for all constant terms) but in definite integral showing constant term is must.
The formula given by galactus works fine, including the case where one or both of h(x) and g(x) are equal to a constant.
10. ## Ooops
Originally Posted by Mathstud28
What are you talking about? The constant terms will drop out due to the derivative.
oops sorry dude forgot we were talking about its derivative. Next time i will be carefull for sure
11. ## Ooops
Originally Posted by Mathstud28
What are you talking about? The constant terms will drop out due to the derivative.
oops sorry dude forgot we were talking about its derivative at the end. Next time i will be carefull for sure
12. One way to check this would be to let a = 0, say, and then put Mathematica on the job. Note that there are various equivalent forms of the correct answer.
If I typed it well, I got that the derivative equals $8 x^2 Cos(\frac{x^3}{2})(2+Cos(x^3))Sin^3(\frac{x^3}{2})-4x^2Sin^4(\frac{x^3}{2})Sin(x^3)$. It surprises me more and more. If somebody has Mathematica and has some time to spend, I'll be glad if you could check out the derivative of the integral!
13. Originally Posted by arbolis
From galactus, . It helped me verifying my result and confirmed it.
I got that the derivative of $\int_a^{x^3} sin^3(t)^3 dt=3x^2{sin(x^3)}^3$.
While from Mathematica, it gives the result $\frac{1}{12} \big (9cos(a)-cos(3a)-9cos(x^3)+cos(3x^3)\big)$ which I doubt is equal to my result. Can you confirm my result (if yes, I will think that the result from Mathematica is the same, even if it seems more than incredible.). Thanks!!
You have as the integrand $sin^3(t)^3$ .....? Is the integrand $\sin^3 (t)$ or $\sin^3(t^3)$?
14. You have as the integrand .....? Is the integrand or ?
Sorry about that, it is $\sin^3 (t)$, that is $(sin(t))^3$.
EDIT: Finally I don't mind what Mathematica thinks. Since I didn't made this program, it doesn't interest me that much. It may has some conventions I don't know yet, that's all. Thread solved. | 2017-01-17 12:28:04 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 14, "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": 0.815860390663147, "perplexity": 462.91516508091246}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-04/segments/1484560279657.18/warc/CC-MAIN-20170116095119-00117-ip-10-171-10-70.ec2.internal.warc.gz"} |
https://lavelle.chem.ucla.edu/forum/viewtopic.php?f=123&t=7896&p=19770 | ## Partial Pressure on Quiz
$PV=nRT$
Sumaiyah I 1E
Posts: 16
Joined: Fri Sep 25, 2015 3:00 am
### Partial Pressure on Quiz
Which conversion units will we be given on the Quiz this week? For example, will we be given the conversion for Pa to atm, or bars?
Chem_Mod
Posts: 18400
Joined: Thu Aug 04, 2011 1:53 pm
Has upvoted: 435 times
### Re: Partial Pressure on Quiz
All required units and conversions are provided in the reference sheet. | 2020-06-06 05:43:16 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 1, "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": 0.527087926864624, "perplexity": 8031.652371141445}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-24/segments/1590348509972.80/warc/CC-MAIN-20200606031557-20200606061557-00480.warc.gz"} |
https://studyforce.com/tag/modulus/ | # Modulus
## Modulo-m Systems
If m is an integer greater than 1, then a modulo-m system consists of the numbers 0,1,2,…,m−1. Counting and arithmetic operations are performed in a manner corresponding to movements on an m-hour clock. The number is called the modulus of the system. Let’s say we wanted to count to 53… | 2022-07-03 23:42:00 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8095710277557373, "perplexity": 990.2545087461608}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-27/segments/1656104277498.71/warc/CC-MAIN-20220703225409-20220704015409-00663.warc.gz"} |
http://keisan.casio.com/exec/system/1180573189 | Normal distribution (chart) Calculator
Calculates a table of the probability density function, or lower or upper cumulative distribution function of the normal distribution, and draws the chart.
select function probability density f lower cumulative distribution P upper cumulative distribution Q mean μ standard deviation σ σ＞0 [ initial percentile x increment repetition ]
The default value μ and σ shows the standard normal distribution.$\normal Normal\ distribution\ N(x,\mu,\sigma)\\[10](1)\qquad probability\ density\\\hspace{30}f(x,\mu,\sigma)={\large\frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{1}{2}\left(\frac{x-\mu}{\sigma}\right)^2}}\\(2)\qquad lower\ cumulative\ distribution\\\hspace{30}P(x,\mu,\sigma)={\large\int_{\small-\infty}^{\hspace{25}\small x}}f(t,\mu,\sigma)dt\\(3)\qquad upper\ cumulative\ distribution\\\hspace{30}Q(x,\mu,\sigma)={\large\int_{\small x}^{\hspace{25}\small\infty}}f(t,\mu,\sigma)dt\\$
Sending completion
To improve this 'Normal distribution (chart) Calculator', please fill in questionnaire.
Male or Female ?
Male Female
Age
Under 20 years old 20 years old level 30 years old level
40 years old level 50 years old level 60 years old level or over
Occupation
Elementary school/ Junior high-school student
High-school/ University/ Grad student A homemaker An office worker / A public employee
Self-employed people An engineer A teacher / A researcher
A retired people Others
Useful?
Very Useful A little Not at All
Purpose of use? | 2017-07-26 08:40:15 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 1, "/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": 0.9066554307937622, "perplexity": 6911.331763704344}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-30/segments/1500549426086.44/warc/CC-MAIN-20170726082204-20170726102204-00680.warc.gz"} |
http://math.stackexchange.com/questions/189877/divisibility-and-gcd?answertab=oldest | # divisibility and gcd
I have a positive integer $g$ such that $g$ is the least linear combination of the integers a and b. I have shown $g$ | $h$ ( where $h$ is an arbitrary linear combination of a and b, thus g divides all other linear combinations of a and b. So if we say g | ra + $sb$, where $r,a,s,b$ are positive integers. Does $g$ divide $a$ and $b$? Could someone explain this to me? (This is part of the classic $\gcd(a, b)$ proof by the way if that helps anyone)
Thanks!
-
Usually one proves $\rm\:g\:|\:ra+sb\:$ for all integers $\rm\:r,s\:$ (vs. all positive integers). From that we deduce that $\rm\:g\:|\:a,b\:$ by taking $\rm\:r,s = 1,0\:$ and $\,0,1.\:$ Revisit your proof: it probably works for all integers. Below is one conceptual way to present such a proof.
Hint $\$ The set $\rm\:S\:$ of integers of the form $\rm\:x\:a + y\:b,\ x,y\in \mathbb Z\:$ is closed under subtraction so, by the Lemma below, every $\rm\:n\in S\:$ is divisible by the least positive $\rm\:d\in S.\:$ Thus $\rm\:a,b\in S\:$ $\Rightarrow$ $\rm\:d\:|\:a,b,\:$ i.e. $\rm\:d\:$ is a common divisor of $\rm\:a,b,\:$ necessarily greatest, by $\rm\:c\:|\:a,b\:$ $\Rightarrow$ $\rm\:c\:|\: \hat x\:a+\hat y\:b = d\:$ $\Rightarrow$ $\rm\:c\le d.$
Lemma $\ \$ If a nonempty set of positive integers $\rm\: S\:$ satisfies $\rm\ n > m\ \in\ S \ \Rightarrow\ \: n-m\ \in\ S$
then every element of $\rm\:S\:$ is a multiple of the least element $\rm\:m_{\:1} \in S\:.$
Proof $\ \:$ If not there is a least nonmultiple $\rm\:n\in S,$ contra $\rm\:n-m_{\:1}\! \in S\:$ is a nonmultiple of $\rm\:m_{\:1}.\ \$
Remark $\$ This linear representation of the the gcd is known as the Bezout identity for the gcd. It need not hold true in all domains where gcds exist, e.g. in the domain $\rm\:D = \mathbb Q[x,y]\:$ of polynomials in $\rm\:x,y\:$ with rational coefficients we have $\rm\:gcd(x,y) = 1\:$ but there are no $\rm\:f(x,y),\: g(x,y)\in D\:$ such that $\rm\:x\:f(x,y) + y\:g(x,y) = 1;\:$ indeed, if so, then evaluating at $\rm\:x = 0 = y\:$ yields $\:0 = 1.$
The fundamental lemma, interpreted procedurally, yields Euclid's classical algorithm to compute the gcd using repeated subtraction.
-
Clearly $g$ need not divide either $a$ or $b$, because $g$ could be something ridiculously large and $a$ and $b$ could be quite small. Say for example $g= 117$, where $r = 67, s = 50, a=b=1$.
Perhaps you are looking at the part of the Euclidean division algorithm where we are trying to find the GCD of $a$ and $b$, and we construct a series of smaller and smaller integers with the same GCD. In this case the implication goes the other way: We suppose that $g$ divides $a$ and $b$, and from that we can conclude that $g$ must also divide $ra+sb$ for any integers $r$ and $s$—and note, they need not be positive.
If none of this is to the point, perhaps you could say more clearly what you are looking at and what you are trying to understand.
If $g$ divides all linear combinations $ra+sb$ of $a$ and $b$, then in particular it divides the one where $r=1$ and $s=0$, and also the one where $r=0$ and $s=1$. | 2015-07-06 22:44:42 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9271343350410461, "perplexity": 47.384602192200454}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-27/segments/1435375098849.37/warc/CC-MAIN-20150627031818-00039-ip-10-179-60-89.ec2.internal.warc.gz"} |
http://dict.cnki.net/h_653000.html | 全文文献 工具书 数字 学术定义 翻译助手 学术趋势 更多
支撑 的翻译结果: 查询用时：1.77秒
历史查询
支撑
support
Research on the system of Information Management and consultation Support to SME's Business 支撑中小企业的信息管理与咨询服务体系研究 短句来源 Study on Support System to Improve the Competitiveness of Agricultural Products in HuBei Within the WTO Framework WTO框架下提高湖北农产品竞争力的支撑体系研究 短句来源 Research on the Key Technologies to Collaborative Development Support Environment for Complicated Product 复杂产品协同开发支撑环境的关键技术研究 短句来源 Research on Support Vector Machines and Kernel Methods 支撑矢量机与核方法研究 短句来源 Studies on Support Technologies for Enterprise Integration Based on Knowledge Management 基于知识管理的企业集成支撑技术研究 短句来源 更多
supporting
Research of Structured and Integrated Scheduling System Theory & Structure of SISST-based Intelligent Supporting Platform for Virtual Workshop 结构化集成调度系统理论及基于该理论的虚拟车间智能支撑平台的体系结构研究 短句来源 Research on Supporting Techniques for High-Reliable Fault-Tolerant Real-Time Systems 高可靠容错实时系统的支撑技术研究 短句来源 A Study of the Supporting System of Venture Capital Investment the Development of China's Medium and Small Enterprise 我国中小企业发展的风险投资支撑体系研究 短句来源 Study on the Supporting System of the Sustainable Development of the Township-village Enterprises in the Western China 西部地区乡镇企业可持续发展的支撑体系研究 短句来源 Research on Collaborative Virtual Manufacturing Oriented Distributed Supporting Environment 面向协同虚拟制造的分布式支撑环境研究 短句来源 更多
supported
Study of SrO-La_2O_3-Al_2O_3-B_2O_3-SiO_2 crystallite glass sealing materials for anode supported ITSOFC 阳极支撑型ITSOFC密封材料SrO-La_2O_3-Al_2O_3-B_2O_3-SiO_2微晶玻璃的研究 短句来源 Fabrication of Cu_7S_4 Nano-crystals with Supported Liquid Membrane 支撑液膜法制备Cu_7S_4纳米晶 短句来源 Dense Supported Oxygen Permeating Membrane of La_2NiO_(4+δ) Prepared by Sol-Gel Method 由溶胶-凝胶法制备的La_2NiO_(4+δ)支撑透氧膜 短句来源 An algorithm constructed compactly supported biorthogonal multiwavelets with scale=a(a∈Z,a≥2) by the means of polyphase matrix. 通过引入多相矩阵的方法 ,给出了由紧支撑多重 a( a∈ Z,a≥ 2 )尺度双正交尺度函数构造双正交多小波的一种算法 . 短句来源 So the investigation of La_2NiO_(4+δ) series supported dense membranes preparation to improve its oxygen-permeating properties will be of interest for the industrial application of oxygen permeating membranes. 因此研究La_2NiO_(4+δ)系列致密支撑透氧膜的制备,并寻求进一步提高La_2NiO_(4+δ)系列透氧膜材料透氧性能的途径,对于促进实现透氧膜的工程应用具有重要意义。 短句来源 更多
bracing
Unloading Analysis of Steel Structure Bracing on National Stadium 国家体育场钢结构支撑卸载分析 短句来源 Construction and Monitoring of Earth Excavation and Steel Bracing in 27m\|deep Foundation Pit in Guangzhou 广州27m深基坑土石方开挖和钢管支撑施工与监测 短句来源 Construction Practice on 10.5m Deep Foundation Pit with One Bracing and Composite Soil Nail Enclosure 10.5m深基坑以复合土钉围护加设一道支撑的施工实践 短句来源 DISCUSSION ON PROBLEMS CONCERNING X TYPE INTER-COLUMN BRACING OF SINGLE-STORY STEEL STRUCTURE BUILDING 单层钢结构厂房X型柱间支撑问题的探讨 短句来源 Compared with the traditional cantilever structure, the bearing life of two-terminal bracing structure is improved by 17.51 times, while the contact stress of gear tooth is reduced by 45.7%, and the bending stress of gear tooth is reduced by 62.8%. 与传统主动齿轮悬臂结构相比两端支撑结构的轴承寿命提高了17.51倍、轮齿接触应力降低45.7%、轮齿齿根弯曲应力减小62.8%。 短句来源 更多
我想查看译文中含有：的双语例句
support
This does not, however, give rise to vanishing identities for the standard nonsymmetric Macdonald and Koornwinder polynomials; we discuss the required modification to these polynomials to support such results. A simple parametrization is given for the set of positive measures with finite support on the circle group T that are solutions of the truncated trigonometric moment problem: Then it is shown that for many (and perhaps all) pairs E, F, of wavelet sets, the corresponding MSF wavelets can be connected by a continuous path in L2(?) of MSF wavelets for which the Fourier transform has support contained in E ∪ F. As a strengthening of the conjecture we show that for an f ∈ L2(?n) its Wigner distribution has a support of measure 0 or ∞ in any half-space of ?2n. For example, we do not need to know that the support of ? or ψ is compact as in [14]. 更多
supporting
The results show the advantages of the new PI controller design approach for AQM routers supporting TCP flows. It indicated that in the future, closed tending as the major practice and tending and shelterwood cutting as the supporting practices should be applied for P. Supporting crosscutting concern modelling in software architecture design This paper presents a novel approach to supporting crosscutting concern modelling in the software architecture design of component-based systems. The paraffin wax (with melting temperature of 29°C, crystallizing temperature of 26°C and latent heat of 142 J/g) served as latent heat storage material and the silica as supporting material, which prevented the leakage of the melted paraffin wax. 更多
supported
This fact is deduced from results about equivariantD-modules supported on the nilpotent cone of. Given integers n,d,e with $1 \leqslant e >amp;lt; \frac{d}{2},$ let $X \subseteq {\Bbb P}^{\binom{d+n}{d}-1}$ denote the locus of degree d hypersurfaces in ${\Bbb P}^n$ which are supported on two hyperplanes with multiplicities d-e and e. Smoothing Minimally Supported Frequency Wavelets: Part II The main purpose of this paper is to give a procedure to "mollify" the low-pass filters of a large number of Minimally Supported Frequency (MSF) wavelets so that the smoother functions obtained in this way are also low-pass filters for an MRA. Smoothing minimally supported frequency wavelets: Part II 更多
bracing
A tied arch bridge without wind bracing was built over the Bin Jiang River, with oblique angle of 20°. Aeroelastic stability of a wing with bracing struts (Keldysh problem) The problem of the influence of bracing struts of two types on the aeroelastic stability of a wing is studied. Sideline splinting, bracing, and casting of extremity injuries Emergent and nonemergent splinting, bracing, and casting are effective ways to safely remove an injured athlete from the playing field, allow immediate return to play, and permit an athlete to return to play before an injury has completely healed. 更多
其他
点击这里查询相关文摘
相关查询
CNKI小工具 在英文学术搜索中查有关支撑的内容 在知识搜索中查有关支撑的内容 在数字搜索中查有关支撑的内容 在概念知识元中查有关支撑的内容 在学术趋势中查有关支撑的内容
CNKI主页 | 设CNKI翻译助手为主页 | 收藏CNKI翻译助手 | 广告服务 | 英文学术搜索
2008 CNKI－中国知网
2008中国知网(cnki) 中国学术期刊(光盘版)电子杂志社 | 2020-08-14 12:17:47 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.43431925773620605, "perplexity": 6842.496951776854}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-34/segments/1596439739211.34/warc/CC-MAIN-20200814100602-20200814130602-00166.warc.gz"} |
https://www.studysmarter.us/textbooks/physics/fundamentals-of-physics-10th-edition/rolling-torque-and-angular-momentum/71p-in-fig-11-60-a-constant-horizontal-force-of-magnitude-12/ | Suggested languages for you:
Americas
Europe
71P
Expert-verified
Found in: Page 326
### Fundamentals Of Physics
Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718
# In Fig. 11-60, a constant horizontal force of magnitude 12 N is applied to a uniform solid cylinder by fishing line wrapped around the cylinder. The mass of the cylinder is 10kg, its radius is 0.10 m, and the cylinder rolls smoothly on the horizontal surface. (a) What is the magnitude of the acceleration of the center of mass of the cylinder? (b) What is the magnitude of the angular acceleration of the cylinder about the center of mass? (c) In unit-vector notation, what is the frictional force acting on the cylinder?
1. Acceleration of center of mass is $1.6\mathrm{m}/{\mathrm{s}}^{2}$.
2. Angular acceleration of center of mass is$16\mathrm{rad}/{\mathrm{s}}^{2}$.
3. Frictional force in terms of unit vector notation is $\mathrm{F}=4\stackrel{^}{\mathrm{i}}$.
See the step by step solution
## Step 1: Given
$\mathrm{F}=12\mathrm{N}\phantom{\rule{0ex}{0ex}}\mathrm{m}=10\mathrm{kg}\phantom{\rule{0ex}{0ex}}\mathrm{r}=0.1\mathrm{m}$
## Step 2: Determining the concept
Use formula for torque to find angular acceleration. Use angular acceleration to find linear acceleration, and finally use Newton’s second law to find friction force.
Formula are as follow:
${\mathbf{\tau }}{\mathbf{=}}{\mathbf{I\alpha }}{\mathbf{=}}{\mathbf{F}}{\mathbf{×}}{\mathbf{r}}\phantom{\rule{0ex}{0ex}}{\mathbf{a}}{\mathbf{=}}{\mathbf{\alpha r}}$
Where, is torque, is force, is moment of inertia, is radius, is angular acceleration and is acceleration.
## Step 3: Determining the acceleration of center of mass
(a)
To calculate acceleration of center of mass, first calculate angular acceleration as follows:
$\mathrm{\tau }=\mathrm{I\alpha }$
Moment of inertia about point of contact is given by parallel axis theorem as follows:
$\mathrm{I}=0.5{\mathrm{mr}}^{2}+{\mathrm{mr}}^{2}\phantom{\rule{0ex}{0ex}}\mathrm{I}=1.5{\mathrm{mr}}^{2}$
Force is applied at the top of the cylinder, considering the bottom of the cylinder as pivot point and r as the radius of the cylinder,
$1.5{\mathrm{mr}}^{2}×\mathrm{\alpha }=\mathrm{F}×2\mathrm{r}\phantom{\rule{0ex}{0ex}}1.5×10×0.{1}^{2}×\mathrm{\alpha }=12×2×0.1$
So,
$\mathrm{\alpha }=16\mathrm{rad}/{\mathrm{s}}^{2}$
Now, acceleration of center of mass is as follows:
$\mathrm{a}=\mathrm{\alpha r}\phantom{\rule{0ex}{0ex}}\mathrm{a}=16×0.1\phantom{\rule{0ex}{0ex}}\mathrm{a}=1.6\mathrm{m}/{\mathrm{s}}^{2}$
Hence, acceleration of center of mass is $1.6\mathrm{m}/{\mathrm{s}}^{2}$.
## Step 4: Determining the angular acceleration of center of mass
(b)
From above calculations,
$\mathrm{\alpha }=16\mathrm{rad}/{\mathrm{s}}^{2}$
Hence, angular acceleration of center of mass is $16\mathrm{rad}/{\mathrm{s}}^{2}$.
## Step 5: Determining the frictional force in terms of unit vector notation
(c)
Frictional force (F) in unit vector form is as follows:
Now, to find friction force, use Newton’s second law of motion,
So, friction force is as follows:
$12-\mathrm{F}=\mathrm{m}×{\mathrm{a}}_{\mathrm{cm}}\phantom{\rule{0ex}{0ex}}12-\mathrm{F}=10×1.6$
So,
$\mathrm{F}=4.0\mathrm{N}$
In unit vector notation,
$\stackrel{\to }{\mathrm{F}}=\left(4.0\mathrm{N}\right)\stackrel{^}{\mathrm{i}}$
Since, value F is positive, it is along the positive x axis.
Hence, frictional force in terms of unit vector notation is $\mathrm{F}=4\stackrel{^}{\mathrm{i}}$ .
Therefore, the formula for torque to find the angular acceleration can be used. Using angular acceleration, linear acceleration can be found. Newton’s second law can be used to find friction force. | 2023-03-24 21:49:41 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 35, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7995971441268921, "perplexity": 517.9857457889841}, "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/1679296945289.9/warc/CC-MAIN-20230324211121-20230325001121-00657.warc.gz"} |
https://www.ques10.com/p/8795/a-steel-bolt-of-20mm-diameter-passes-centrally-thr/ | Question: A steel bolt of 20mm diameter passes centrally through a copper tube of internal diameter 25mm and thickness 10mm. The tube is 600mm long and is closed by rigid washers of negligible thickness
2
relative to the other.The pitch of the thread is 2mm. Take $E_s$ = 200GPa and $E_c$ = 100GPa.
Mumbai university > MECH > SEM 3 > Strength Of Materials
Marks: 10M
Year: Dec 2013
modified 3.0 years ago by written 3.3 years ago by Pooja Joshi • 740
1
The steel bolt is under tension and the copper tubing under compression.
The nut is fastened by one quarter of a turn.
Allowed Deformation = $\frac{Pithch of Nut}{4}$
Allowed Deformation = $\frac{2}{4} = 0.5mm$
The elongation for the bolt and the tube is the same.
$\dbinom{PI}{AE}_{bolt} = \dbinom{PI}{AE}_{tubing} = 0.5mm ........(i)$
Steel Bolt – Tension
$d = 20mm \ \ \ \ \ dl = 0.5mm$
$l = 600mm \ \ \ \ \ E_s = 200GPa$
$A_s = \frac{\pi}{4}(20)^2 = 314.159mm^2$
From (I),
$\frac{P_s(600)}{(314.159)(200 × 10^3)} = 0.5$
$P_s = 52.36kN$
Stress Induced $(δ)_s = \frac{P_s}{A_s} = \frac{52.36 × 10^3}{314.159} = 166.67N/mm^2$
This is the value of the tensile stress induced in the steel bolt.
Copper Tubing – Compression
$d = 25mm \ \ \ \ t = 10mm$
$D = d + 2t = 45mm \ \ \ \ E_c= 100GPa$
$A_c = \frac{\pi}{4}[45^2 - 25^2] = 109.56mm^2$
From (I),
$\frac{P_c(600)}{(10099.56)(100 × 10^3)} = 0.5$
$P_c = 91.63kN$
Stress Induced $δ_c = \frac{P_c}{A_c} = \frac{91.63 × 10^3}{1099.56} = 83.33N/mm^2$
This is the value of the compressive stress induced in the copper tubing. | 2019-10-23 04:17:08 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.33437472581863403, "perplexity": 4251.961016365233}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-43/segments/1570987828425.99/warc/CC-MAIN-20191023015841-20191023043341-00444.warc.gz"} |
https://newproxylists.com/tag/problem/ | ## mysql – database design problem without duplicate data
I have a customer database which contains customer data and a user table. I store customer data (organization details) in the customer data table and store user data in the Users table which stores user emails and hash for logging in. I am creating a database on the registration of organizations. Now this user can add more users who may have login information and others not. I don't understand where to store these unregisterable users. I have a userid as the primary key in the clients.users table. I store this identifier in the personnel table of the organizations database. I'm looking for emails in the clients.users table when logging in to the user. At the same time, I have the ID of this user in the staff organization table. but where should I store this non-recordable personal data. I cannot store them in the staff table of the organization, because this table uses clients.users userid as foreign key.
How can I be able to store non-recordable personnel data?
I store the user details in the clients table and the organizations table. I check the customer table during authentication. But storing the same data in two tables is not a good way. I want to store the user details in the organization database and use this identifier as a foreign key in the customer database. But I don't know which user belongs to which database. I can't search them in the customer database as it only contains the user ID as a foreign key. How to verify their credentials in db clients?
## security certificate – Intermittent problem establishing a secure HTTPS connection to my site
Your DNS has two IP addresses specified for your site. Only one of them has a server with correctly configured security certificates.
When you have two `A` registers for the same host name, it is known as DNS "round robin". Customers will randomly try to connect to one or the other. Some modern browsers may try both to see which one works, which may explain why you don't see the problem in some browsers.
``````\$ dig healthprovement.com
healthprovement.com. 7069 IN A 108.179.232.43
healthprovement.com. 7069 IN A 148.105.251.16
``````
the `108.179.232.43` the server is running:
``````\$ curl --head --resolve 'healthprovement.com:443:108.179.232.43' https://healthprovement.com/
HTTP/2 200
``````
But the `148.105.251.16` is not:
``````\$ curl --head --resolve 'healthprovement.com:443:148.105.251.16' https://healthprovement.com/
``````
To resolve this issue, you must either delete the DNS `A` records pointing to `148.105.251.16` or you need to fix this server so that it handles HTTPS requests correctly.
## computation – Problem of function substitution to find the normal field of unity
I'm trying to find the normal unit field n on S. Where S is the ellipsoid $$2x ^ 2 + 2y ^ 2 + z ^ 2 = 2$$
However, I am stuck trying to replace f in the $$| nabla f (x, y, z) |$$.
My steps:
(i) I calculate n with $$n = frac {( nabla f (x, y, z)} {| nabla f (x, y, z) |}$$
(ii) Using partial differentiation, I find $$nabla f =<4x,4y,2z>$$
(Iii) $$| nabla f (x, y, z) | = { sqrt ((f_x) ^ 2 + (f_y) ^ 2 + (f_z) ^ 2)} = { sqrt (16x ^ 2 + 16y ^ 2 + 4z ^ 2)} = {2 sqrt (4x ^ 2 + 4y ^ 2 + z ^ 2)}$$
The next step is where I'm stuck. I think they replace the $$f = 2x ^ 2 + 2y ^ 2 + z ^ 2 = 2$$ in the formula, but they don't match well and I don't know how to deal with it.
I believe that $$n = frac {<4x,4y,4z>} {6}$$ should be the answer.
Thanks for any help!
## excel – Problem with "range" and double cells
Hi i have this in my excel here i put the range of cells which cannot be empty to run the macro
``````Range("B9:AE10,B11:AC13,AD12:AD13,AE11:AE16,W17,B17,N14,B15:AD15,I16:J16")
``````
but the problem is that my excel has double cells
For example, the first cell in the image is B11, but since it is double, it occupies the space of C11, so that the range detects that C11 is empty and the macro does not run. does she have any idea?
## Problem After activating the child theme, the website cursor has disappeared [closed]
In the parent theme, the cursor is displayed, but when I activate the child theme, the cursor is gone, there are also layout issues when activating the child theme
Why is it so ????
## fit – Understanding and solving a problem with FindFit
I have a dataset which I give here
``````data = {{0.294320000000000026`7., 23.8843734496254604949`7.}, {0.6795160000000000089`7., 23.8843611997525542279`7.},
{1.0502000000000000224`7., 23.8843489498796444082`7.}, {1.4071700000000000319`7., 23.8843367000067381412`7.}, {1.7511799999999999589`7., 23.8843244501338318742`7.},
{2.0829300000000001702`7., 23.8843122002609220544`7.}, {2.4030499999999999083`7., 23.8842999503880157874`7.}, {2.7121499999999998387`7., 23.8842877005151095204`7.},
{3.0107800000000000118`7., 23.8842754506421997007`7.}, {3.2994699999999999029`7., 23.8842632007692934337`7.}, {3.5787100000000000577`7., 23.884250950896383614`7.},
{3.8489599999999999369`7., 23.884238701023477347`7.}, {4.1106400000000000716`7., 23.88422645115057108`7.}, {4.3641500000000004178`7., 23.8842142012776612603`7.},
{4.6098699999999999122`7., 23.8842019514047549933`7.}, {4.8481500000000004036`7., 23.8841897015318487263`7.}, {6.8864700000000000912`7., 23.8840672028027718454`7.},
{8.4512300000000006861`7., 23.8839447040736949646`7.}, {9.6903000000000005798`7., 23.8838222053446180837`7.}, {10.6958000000000001961`7., 23.8836997066155447556`7.},
{11.5281000000000002359`7., 23.8835772078864678747`7.}, {15.6127000000000002444`7., 23.8823522205957097242`7.},
{17.1147999999999989029`7., 23.8811272333049551264`7.}, {17.9021000000000007901`7., 23.8799022460141969759`7.},
{18.3740999999999985448`7., 23.8786772587234423781`7.}, {18.6981000000000001648`7., 23.8774522714326842276`7.},
{18.9327000000000005286`7., 23.8762272841419260772`7.}, {19.117100000000000648`7., 23.8750022968511714794`7.},
{19.8338999999999998636`7., 23.8627524239436006326`7.}, {20.0574000000000012278`7., 23.8505025510360333385`7.},
{20.1506000000000007333`7., 23.8382526781284624917`7.}, {20.2182999999999992724`7., 23.8260028052208951976`7.},
{20.2663000000000010914`7., 23.8137529323133279036`7.}, {20.3024999999999984368`7., 23.8015030594057606095`7.},
{20.3335000000000007958`7., 23.7892531864981897627`7.}, {20.359200000000001296`7., 23.7770033135906224686`7.},
{20.3817999999999983629`7., 23.7647534406830516218`7.}, {20.5414999999999992042`7., 23.6422547116073644702`7.},
{20.6695999999999990848`7., 23.5197559825316808713`7.}, {20.7867999999999994998`7., 23.3972572534559937196`7.},
{20.9030999999999984595`7., 23.274758524380306568`7.}, {22.0936999999999983402`7., 22.0497712336234528152`7.}, {23.4103999999999992099`7., 20.824783942866591957`7.},
{24.8888999999999995794`7., 19.5997966521097346515`7.}, {26.5626999999999995339`7., 18.3748093613528737933`7.},
{28.4742999999999994998`7., 17.1498220705960164878`7.}, {30.6789999999999984936`7., 15.9248347798391591823`7.},
{33.2490999999999985448`7., 14.6998474890823001004`7.}, {36.2854999999999989768`7., 13.474860198325442795`7.},
{39.9273000000000024556`7., 12.2498729075685837131`7.}, {44.376300000000000523`7., 11.0248856168117264076`7.}, {49.935099999999998488`7., 9.7998983260548673258`7.},
{57.0788999999999973056`7., 8.5749110352980082439`7.}, {66.5995999999999952479`7., 7.3499237445411500502`7.}, {79.9188999999999936108`7., 6.1249364537842918566`7.},
{99.8800999999999987722`7., 4.8999491630274336629`7.}, {133.0889999999999986358`7., 3.6749618722705750251`7.},
{199.2379999999999995453`7., 2.4499745815137168314`7.}, {393.9680000000000177351`7., 1.2249872907568584157`7.}}
``````
And I'm trying to adapt to a model. To do this, I have the following:
``````fitfunction(x_) := a + b*Log(c*x + d*x^2)
sltnmodel = FindFit(data, fitfunction(x), {a, b, c, d, e}, x, MaxIterations -> Infinity)
Show(ListPlot(data, PlotRange -> {{-170, 500}, {-0.1, 25}}, PlotMarkers -> Style("(FilledCircle)", 11, Red), BaseStyle -> {13, FontFamily -> "Times New Roman"},
AxesLabel -> {"!(*SuperscriptBox((g), (2)))", "!(*SubscriptBox((M), (B)))/!(*SubscriptBox((f), ((Pi))))"}, Joined -> False),
Plot(fitfunction(x) /. sltnmodel, {x, 0, 600}, PlotRange -> {{-170, 500}, {-0.1, 25}}, PlotStyle -> {Dashed, Black}))
``````
The result of the tracing is illustrated below.
The fitting therefore requires a little more work.
The problem is that if adding something to the existing log adjustment, it gives the following error message.
I have tried to use the answer that was given here, but that does not solve the problem.
Ideas?
## virtualization – Restart the host machine from the virtual machine, should I report this problem and where to report it?
Excuse my ignorance because I don't work in infosec.
I ran `reboot` inside a Linux virtual machine using VirtualBox on Mac and it rebooted my host machine. I'm trying to reproduce the problem but I haven't fixed it yet.
If I can reproduce the problem, should I report it and to whom should I report it?
## programming languages - Is there a term for the psychological problem of "code loss" for programmers?
(Note: I wanted to publish this in the "Psychology" category, but there was no corresponding tag.)
I am a programmer. I just deleted a huge amount of code that I thoroughly researched, thought about, coded, then improved and fixed while bugs were brewing for a long time.
All of this code, which took a ridiculous amount of time, effort, and general "mental work," has now been replaced by a very small number of lines that essentially exploit the built-in "ICU" functionality of PHP to produce properly. numbers, money sums and date / time in the right way for every combination of language, locale, currency and time zone imaginable.
Before, I didn't know it already existed, so I reproduced a large part of it myself, and I now realize how far it was from being perfect. But still, I did, and this code had in my mind "hardened" or "settled" like "golden code" that I never thought I'd touch again …
Basically, I'm crying for my now useless, replaced, and obsolete pieces of code. I am annoyed with myself for doing all this unnecessary work and it took me a lot of mental struggle to finally convince me to continue.
Is it common among programmers and does it have an established term? Such as "code loss" or "code loss"?
Basically, even though I have really improved my application / library / framework to an extreme degree, I still feel like I have "lost all that work" because the number of lines is so reduced at once. It is not a pleasant feeling.
## problem with deleting recyclerview item with Firebase
I have the following problem. When you try to delete an item from a recycling view that takes its Firebase data; in the database the item is deleted correctly; but in the recycling view it is not updated properly. I interpreted that the problem was that I was not implementing "notifyDataSetChanged ()" correctly, but other than that, after the list, the items stored in the bbdd are again added. So I thought the solution would be to refresh the recycling view so that it was reloaded with the updated data, but I can't find a way to do it. I would appreciate some advice. I leave the bind code in which I hosted the event
`````` fun bind(datos: ArrayAsuntosPropios) = with(itemView) {
if (datos.yearAP == yearActual && email == datos.email) {
fechaAP.text = datos.fechaAP
yearAP.text = "Descanso"
yearAP.setTextColor(Color.RED)
} else {
yearAP.setTextColor(Color.BLUE)
}
}
itemView.setOnClickListener(View.OnClickListener{
val store: FirebaseFirestore = FirebaseFirestore.getInstance()
val diasASuntoPropiosColeccion: CollectionReference = store.collection("diasAsuntosPropios")
diasASuntoPropiosColeccion
.get()
for (document in documents) {
val idDocumento = document.id
diasASuntoPropiosColeccion.document(idDocumento).delete()
}
} | 2020-01-24 05:30:49 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 7, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.17917677760124207, "perplexity": 2539.332964446204}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-05/segments/1579250615407.46/warc/CC-MAIN-20200124040939-20200124065939-00330.warc.gz"} |
https://pos.sissa.it/390/419/ | Volume 390 - 40th International Conference on High Energy physics (ICHEP2020) - Parallel: Quark and Lepton Flavour Physics
ATLAS results on heavy flavor production and decays
S. Seidel
Full text: Not available
Abstract
Heavy flavor production and decays are studied with the ATLAS detector at the Large Hadron Collider, mainly through final states containing muons. A measurement of the $B_c^\pm$ meson production cross section relative to that of the $B^\pm$ meson is presented. The results of a study of flavor-changing neutral current processes in the decays of $B_s$ and $B^0$ into two muons are also described.
How to cite
Metadata are provided both in "article" format (very similar to INSPIRE) as this helps creating very compact bibliographies which can be beneficial to authors and readers, and in "proceeding" format which is more detailed and complete.
Open Access | 2020-12-04 20:19:02 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6588903665542603, "perplexity": 1463.637951005717}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-50/segments/1606141743438.76/warc/CC-MAIN-20201204193220-20201204223220-00143.warc.gz"} |
http://mathhelpforum.com/math-topics/30473-expression-miles-driven-range-cost.html | # Math Help - Expression miles driven in range with cost
1. ## Expression miles driven in range with cost
Hi,
I have a Math problem below really stretches beyond my linear math algebraic equation (y=mx +c) understanding of 2 sets of specific number data. The problem, given a range of data verses another set of specific number. Please help if you have the solution. Many thanks and appreciate.
Cassidy rented a truck to move 120 miles from Ithaca to Attica. The table shows the cost for renting the truck based on the number of miles driven.
Miles driven Cost ($) 0-9 25 10-19 31 20-29 37 30-39 43 40-49 49 Write an expression to determine the cost of renting the truck for m miles. 2. Originally Posted by @chung Hi, I have a Math problem below really stretches beyond my linear math algebraic equation (y=mx +c) understanding of 2 sets of specific number data. The problem, given a range of data verses another set of specific number. Please help if you have the solution. Many thanks and appreciate. Cassidy rented a truck to move 120 miles from Ithaca to Attica. The table shows the cost for renting the truck based on the number of miles driven. Miles driven Cost ($)
0-9 25
10-19 31
20-29 37
30-39 43
40-49 49
Write an expression to determine the cost of renting the truck for m miles.
When you graph this I would consider the center of each mileage interval. So we have the set of points
(5 miles, $25) (15 miles,$31)
etc.
Notice that the price goes up \$6 for every additional 10 miles.
-Dan | 2014-11-24 02:57:14 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.44767656922340393, "perplexity": 2658.1313205832453}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-49/segments/1416400380355.69/warc/CC-MAIN-20141119123300-00090-ip-10-235-23-156.ec2.internal.warc.gz"} |
https://www.corstianboerman.com/blog/2020-07-23/high-performance-2d-radius-search | # Curls, clouds and code
A blog by Corstian Boerman
Corstian Boerman
# High performance 2D radius search
UPDATE In the first version of this post I have made an error with regard of the retrieval of possible matches. The update has fixed that.
Additional update; I have made a few additions to this library, about which you can read here. The library is now available on GitHub, as well as NuGet.
Sometimes I find myself in a situation where I need a very specific data structure with very specific properties. This time I needed something with which I can quickly retrieve nearby elements in 2D space. Although there are data structures perfectly suited to do so, such as a R-Tree, Kd-Tree and other related structures, I needed something which is very easy and somewhat performant to insert into and remove from.
As my only requirement is to look up nearby elements in 2D space, I choose to maintain two "index" lists which would hold the X and Y coordinates. These indices would be updated upon insertion and deletion.
## Storing coordinates
During my initial search for a solution I started using an SortedDictionary and tried using the Array.BinarySearch method to quickly identify the indices of the elements which contained the lower and upper bounds of the axis. Some performance tuning sessions later I discovered the conversion from an SortedDictionary to an Array type was too resource intensive. With a runtime of 70 ms over 100_000 elements I would be able to cut 2/3rds of the runtime off.
Some search queries later I found the SortedList classes, which offer an IndexOfKey method. The underlying logic also relies on a binary search, though the downside is that it required an exact match of the key. Thankfully we're able to modify this behaviour. The Array.BinarySearch method returns the index on an exact match, and ~index in case there is no exact match. Using some reflection magic we're able to change this behaviour for our custom implementation.
internal class CustomSortedList<TKey, TValue> : SortedList<TKey, TValue>
{
private readonly FieldInfo keysField = typeof(CustomSortedList<TKey, TValue>).BaseType.GetField("keys", BindingFlags.Instance | BindingFlags.NonPublic);
private readonly FieldInfo comparerField = typeof(CustomSortedList<TKey, TValue>).BaseType.GetField("comparer", BindingFlags.Instance | BindingFlags.NonPublic);
private readonly MethodInfo getByIndexMethod = typeof(CustomSortedList<TKey, TValue>).BaseType.GetMethod("GetByIndex", BindingFlags.Instance | BindingFlags.NonPublic);
// Returns the index of the entry with a given key in this sorted list. The
// key is located through a binary search, and thus the average execution
// time of this method is proportional to Log2(size), where
// size is the size of this sorted list. The returned value is -1 if
// the given key does not occur in this sorted list. Null is an invalid
// key value.
//
public new int IndexOfKey(TKey key)
{
if (key == null) throw new ArgumentNullException(nameof(key));
int ret = Array.BinarySearch<TKey>(
(TKey[])keysField.GetValue(this),
0,
Count,
key,
(IComparer<TKey>)comparerField.GetValue(this));
return ret >= 0 ? ret : ~ret;
}
// Returns the value of the entry at the given index.
//
public TValue GetByIndex(int index)
{
return (TValue)getByIndexMethod.Invoke(this, new object[] { index });
}
}
This modification is important, because it allows us to find an non-exact match which enables us to retrieve the lower and upper indices reflecting the values we want to retrieve.
The GetByIndex method is one which is already on the SortedList object, but it's private. Because it's something I would love to use, I used some reflection magic to be able to access it.
## Implementing the SpatialMap object
Now that the most important problems are solved, we can get back to our class which has to filter nearby objects. The implementation essentially consists of two of these CustomSortedList instances which keep track of the axis values, and some methods to add, remove and query this object;
public class SpatialMap<T>
{
private readonly object _mutationLock = new object();
public SpatialMap(
Func<T, double> xAccessor,
Func<T, double> yAccessor)
{
_xAccessor = xAccessor;
_yAccessor = yAccessor;
}
private CustomSortedList<double, T> _x { get; set; } = new CustomSortedList<double, T>();
private CustomSortedList<double, T> _y { get; set; } = new CustomSortedList<double, T>();
{
lock (_mutationLock)
{
{
return;
}
{
_x.Remove(_xAccessor(element));
}
}
}
public void Remove(T element)
{
lock (_mutationLock)
{
_x.Remove(_xAccessor(element));
_y.Remove(_yAccessor(element));
}
}
public IEnumerable<T> Nearby(T element, double distance)
{
var x = _xAccessor(element);
var y = _yAccessor(element);
var innerDistance = Math.Sqrt(Math.Pow(distance, 2) / 2);
var lowerXIndex = _x.IndexOfKey(x - innerDistance);
var upperXIndex = _x.IndexOfKey(x + innerDistance);
var lowerYIndex = _y.IndexOfKey(y - innerDistance);
var upperYIndex = _y.IndexOfKey(y + innerDistance);
var hashSet = new HashSet<T>();
var i = lowerXIndex;
while (i < upperXIndex)
{
var el = _x.GetByIndex(i);
&& Distance(
x, y,
_xAccessor(el),
_yAccessor(el)) < distance)
{
yield return el;
}
i++;
}
i = lowerYIndex;
while(i < upperYIndex - 1)
{
var el = _y.GetByIndex(i);
&& Distance(
x, y,
_xAccessor(el),
_yAccessor(el)) < distance)
{
yield return el;
}
i++;
}
}
}
An important detail is that the Nearby function creates an HashSet to keep track of the elements which have already been returned. Therefore it might or might not be important to implement a custom GetHashCode method on the objects you're using this method with.
The methodology of the Nearby function first retrieves all potential matches from a 'box', after which is determines whether the point is really within a specified distance. To calculate the distance we're simply using the Euclidean distance as follows;
public static double Distance(double x1, double y1, double x2, double y2)
{
var x = x2 - x1;
var y = y2 - y1;
return Math.Sqrt(x * x + y * y);
}
### Search optimization
In order to reduce the possible matches from the array sweep we're limiting the search area even more than the original radius. The reason for doing so is that with an distance of 20 on the X axis we will already exhausting all possible matches, in which case the second while loop would be totally obsolete, as is the HashSet. Instead we will be limiting the search area a little bit such that all possible matches will be retrieved only if both SortedLists are consulted. According to the pythagorean theorem for triangles with a 90° with legs a, b and c; a2 + b2 = c2. If we would not change the maximum search distance, the maximum leg length for a maximum distance of 20 would be; 202 + 202 = sqrt(800) which is roughly 28.28 at an angle of 45°. Given we only would want a maximum length of 20 at 45° we should compute our maximum values first.
This computation can be done by sqrt(distance * distance / 2). With a distance of 20 the result would be roughly 14.14. The difference in search area is 28.28 - 14.14 = 14.14. We're literally cutting the search width in half, which is something with profound results on the search speed!
## Benchmarks
While anyone can claim an algorithm to be performant, it's much more interesting when there are actual benchmarks. I have seeded an SpatialMap instance with 100_000 data points randomly placed on an 2000x2000 grid. Effectively this means there is one point per 40 square units.
BenchmarkDotNet=v0.12.1, OS=Windows 10.0.18363.959 (1909/November2018Update/19H2)
Intel Core i7-8650U CPU 1.90GHz (Kaby Lake R), 1 CPU, 8 logical and 4 physical cores
.NET Core SDK=3.1.302
[Host] : .NET Core 3.1.6 (CoreCLR 4.700.20.26901, CoreFX 4.700.20.31603), X64 RyuJIT
DefaultJob : .NET Core 3.1.6 (CoreCLR 4.700.20.26901, CoreFX 4.700.20.31603), X64 RyuJIT
| Method | Distance | Mean | Error | StdDev | Median |
|------------------- |--------- |------------:|----------:|-----------:|------------:|
| FindNearbyElements | 1 | 26.46 us | 1.229 us | 3.624 us | 26.63 us |
| FindNearbyElements | 2 | 47.00 us | 0.913 us | 0.937 us | 46.99 us |
| FindNearbyElements | 5 | 111.25 us | 1.781 us | 1.390 us | 111.65 us |
| FindNearbyElements | 10 | 276.34 us | 18.179 us | 53.600 us | 248.99 us |
| FindNearbyElements | 25 | 589.62 us | 42.437 us | 123.789 us | 552.93 us |
| FindNearbyElements | 50 | 1,034.05 us | 48.089 us | 141.790 us | 953.51 us |
| FindNearbyElements | 100 | 2,044.97 us | 87.147 us | 254.211 us | 1,949.99 us |
The expected number of points retrieved is as follows:
• Search distance of 1; area of 3.14; expected number of points 0.08
• Search distance of 2; area of 12.57; expected number of points 0.31
• Search distance of 5; area of 78.54; expected number of points 1.96
• Search distance of 10; area of 314.16; expected number of points 7.85
• Search distance of 25; area of 1963.50; expected number of points 49.09
• Search distance of 50; area of 7853.98; expected number of points 196.35
• Search distance of 100; area of 31415.93; expected number of points 785.40
## Possible application
In order to speed up spatial calculations on real world data I have allowed myself to introduce certain amounts of error. One of these is by reflecting the latitudal and longitudal coordinates as an XY grid which reflects the number of kilometers from the Greenwich mean line as well as the Equator. Imagining the earth as flat coordinate system makes it much easier to work with distances, even though it isn't as accurate as possible. I have written about this simplification over here.
Hey there, I hope you enjoyed this post of mine. If you did, consider sharing this with that one friend who'd also appreciate this. Comments are gone for the time being, but if you feel like discussing something more in-depth, send me a message on Twitter, or just email me.
- Corstian | 2023-02-02 10:54:06 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.2955549657344818, "perplexity": 2859.764245896418}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764500017.27/warc/CC-MAIN-20230202101933-20230202131933-00514.warc.gz"} |
https://delong.typepad.com/sdj/2006/12/do_unto_others.html | Do Unto Others...
John Quiggin and Hal Varian are discussing the Stern Review on Global Climate Change and the pure rate of time discount in my email inbox:
Hal Varian: Stern is advocating levying a $70 tax on today's generation to transfer$940 to a future, much wealthier, generation. This is exactly the opposite of the experiment you suggestion. Given the absence of time travel, the relevant transfers are all of from the present to the future.
John Quiggin: At the margin, it's the same question. If you oppose paying $70 now to transfer$940 to wealthy people in the future, you should support taxing wealthy people $940 now to pay$70 to poor people, unless you specifically discount the utility of future generations simply because they are in the future. I'm not saying that this is an implausible preference, just that it indicates the kinds of trade-offs that are involved.
Hal Varian: Here is my problem with 0 social rate of time discount. Suppose I believe that it is ethically neutral to transfer $1 from Brad DeLong to John Quiggin. Furthermore, John and Brad are indifferent between$1 now and $1.10 next year. Should it not be the case that it is ethically neutral to take$1 from Brad DeLong now to give John Quiggin \$1.10 in the future? In general, if I discount my own future consumption and I should "treat others as myself" doesn't that mean that I should discount other people's future consumption (just as I do my own)?
My view--which I admit may well be wrong--of this knotty problem is that we are impatient in the sense of valuing the present and near-future much more than we value the distant future, but that we shouldn't do so. Our habits of mind of preferring a bird in the hand to two in the bush come from a time when life was nastier, more brutish, and definitely shorter, and that if you went into the bush to try to get the two birds you might well not come out.
The fact that we want to apply a pure rate of time discount much greater than the risk of extinction to problems of planning for the far future is, I think, a flaw in our reasoning.
So I come down with John rather than with Hal on this one. | 2021-07-25 15:55:13 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5760236978530884, "perplexity": 1752.1065813679786}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-31/segments/1627046151699.95/warc/CC-MAIN-20210725143345-20210725173345-00082.warc.gz"} |
http://www.math.toronto.edu/courses/apm346h1/20181/PDE-textbook/Chapter4/S4.2.html | 4.2. Eigenvalue problem
## 4.2. Eigenvalue problems
### Problems with explicit solutions
Example 1. (Dirichlet-Dirichlet; from Section 4.1). Consider eigenvalue problem \begin{align} &X'' +\lambda X=0 && 0< x< l,\label{eq-4.2.1}\\[3pt] \label{eq-4.2.2}& X(0)=X(l)=0 \end{align} as eigenvalues and corresponding eigenfunctions \begin{align} & \lambda_n=\bigl(\frac{\pi n}{l}\bigr)^2,&&n=1,2,\ldots \label{eq-4.2.3}\\[3pt] & X_n=\sin \bigl(\frac{\pi n}{l}x\bigr). \label{eq-4.2.4} \end{align}
Visual examples (animation)
Example 2. (Neumann-Neumann). Eigenvalue problem \begin{align} & X'' +\lambda X=0 && 0< x< l,\label{eq-4.2.5}\\[3pt] & X'(0)=X'(l)=0\label{eq-4.2.6} \end{align} has eigenvalues and corresponding eigenfunctions \begin{align} & \lambda_n=\bigl(\frac{\pi n}{l}\bigr)^2,&&n=0,1,2,\ldots \label{eq-4.2.7}\\[3pt] & X_n=\cos \bigl(\frac{\pi n}{l}x\bigr). \label{eq-4.2.8} \end{align}
Visual examples (animation)
Indeed, plugging $X=Ae^{kx}+ Be^{-kx}$, $k=\sqrt{-\lambda}\ne 0$ into (\ref{eq-4.2.6}) $X'(0)=X'(l)=0$ we get \begin{align*} &A \qquad -B\qquad\ = 0,\\ &Ae^{\sqrt{-\lambda}l} - B e^{-\sqrt{-\lambda}l}=0 \end{align*} where we divided the last equation by $\sqrt{-\lambda}$, which leads to the same condition $2\sqrt{-\lambda}l=2\pi in$, $n=1,2,\ldots$ as before but with eigenfunctions (\ref{eq-4.2.8}).
But now, plugging $\lambda=0$ and $X=A+Bx$ we get $B=0$ and $X(x)=1$ is also an eigenfunction and we should add $n=0$.
Example 3. (Dirichlet-Neumann). Consider eigenvalue problem \begin{align} & X'' +\lambda X=0 && 0< x< l,\label{eq-4.2.9}\\[3pt] & X(0)=X'(l)=0\label{eq-4.2.10} \end{align} has eigenvalues and corresponding eigenfunctions \begin{align} & \lambda_n=\bigl(\frac{\pi (2n+1)}{2l}\bigr)^2,&&n=0,1,2,\ldots \label{eq-4.2.11}\\[3pt] & X_n=\sin \bigl(\frac{\pi (2n+1)}{2l}x\bigr).\label{eq-4.2.12} \end{align}
Visual examples (animation)
Indeed, plugging $X=Ae^{kx}+ Be^{-kx}$, $k=\sqrt{-\lambda}\ne 0$ into (\ref{eq-4.2.6}) $X'(0)=X'(l)=0$ we get \begin{align*} &A \qquad +B\qquad\ = 0,\\ &Ae^{\sqrt{-\lambda}l} - B e^{-\sqrt{-\lambda}l}=0 \end{align*} where we divided the last equation by $\sqrt{-\lambda}$, which leads to condition $2\sqrt{-\lambda}l=(2\pi n+1) i$, $n=0,1,2,\ldots$ and with eigenfunctions (\ref{eq-4.2.12}).
Plugging $\lambda=0$ and $X=A+Bx$ we get $B=0$ and $A=0$, so $\lambda=0$ is not an eigenvalue.
The same problem albeit with the ends reversed (i.e. $X'(0)=X(l)=0$) has the same eigenvalues and eigenfunctions $\cos \bigl(\frac{\pi (2n+1)}{2l}x\bigr)$.
Example 4. (periodic). Consider eigenvalue problem \begin{align} & X'' +\lambda X=0 && 0< x< l,\label{eq-4.2.13}\\[3pt] & X(0)=X(l), \quad X'(0)=X'(l)\label{eq-4.2.14} \end{align} has eigenvalues and corresponding eigenfunctions \begin{align} & \lambda_0=0,\label{eq-4.2.15}\\[3pt] & X_0=1,\label{eq-4.2.16}\\ & \lambda_{2n-1}=\lambda_{2n}=\bigl(\frac{2\pi n}{l}\bigr)^2,&&n=1,2,\ldots \label{eq-4.2.17}\\[3pt] & X_{2n-1}=\cos \bigl(\frac{2\pi n}{l}x\bigr), && X_{2n}=\sin \bigl(\frac{2\pi n}{l}x\bigr).\label{eq-4.2.18} \end{align}
Visual examples (animation)
Alternatively, as all eigenvalues but $0$ have multiplicity $2$ one can select \begin{align} & \lambda_n=\bigl(\frac{2\pi n}{l}\bigr)^2,&&n=\ldots, -2,-1,0, 1,2,\ldots \label{eq-4.2.19}\\[3pt] & X_{n}=\exp \bigl(\frac{2\pi n}{l}i x\bigr).\label{eq-4.2.20} \end{align}
Indeed, now we get \begin{align*} &A(1-e^{\sqrt{-\lambda} l}) +B(1-e^{-\sqrt{-\lambda} l})= 0,\\ &A(1-e^{\sqrt{-\lambda} l}) - B (1-e^{-\sqrt{-\lambda} l})=0 \end{align*} which means that $e^{\sqrt{-\lambda} l}=1\iff \sqrt{-\lambda}=2\pi ni$ and in this case we get two linearly independent eigenfunctions.
As $\lambda =0$ we get just one $X_0=1$.
Example 5. (quasiperiodic). Consider eigenvalue problem \begin{align} & X'' +\lambda X=0 && 0< x< l,\label{eq-4.2.21}\\[3pt] & X(0)=e^{-ikl}X(l), \quad X'(0)=X'(l)e^{-ikl}X(l)\label{eq-4.2.22} \end{align} with $0< k<\frac{2\pi}{l}$ has eigenvalues and corresponding eigenfunctions \begin{align} & \lambda_{n}=\bigl(\frac{2\pi n}{l}+k \bigr)^2,&& n=0,2,4,\ldots \label{eq-4.2.23}\\[3pt] & X_{n}=\exp \bigl(\bigl[\frac{2\pi n}{l}+k\bigr]i x\bigr), \label{eq-4.2.24}\\[3pt] & \lambda_{n}=\bigl(\frac{2\pi (n+1)}{l}-k \bigr)^2,&& n=1,3,5,\ldots \label{eq-4.2.25}\\[3pt] & X_{n}=\exp \bigl(\bigl[\frac{2\pi (n+1)}{l}-k\bigr]i x\bigr). \label{eq-4.2.26} \end{align} $k$ is called quasimomentum.
1. For $k=0, \frac{2\pi}{l}$ we get periodic solutions, considered in the previous Example 4,
2. for $k=\frac{\pi}{l}$ we get antiperiodic solutions
(Visual examples (animation))
where all eigenvalues have multiplicity $2$ (both $l$-periodic and $l$-antiperiodic functions are $2l$-periodic, and each $2l$-periodic is the sum of $l$-periodic and $l$-antiperiodic functions).
3. For all other $k$ eigenvalues are real and simple but eigenfunctions are not real-valued.
Indeed, we get now \begin{align*} &A(e^{ikl}-e^{\sqrt{-\lambda} l}) +B( -e^{\sqrt{-\lambda} l})= 0,\\ &A(e^{ikl}-e^{\sqrt{-\lambda} l}) - B (e^{ikl}-e^{-\sqrt{-\lambda} l})=0 \end{align*} and we get either $\sqrt{-\lambda} = ik + \pi m i/l$ with $m\in \mathbb{Z}$.
Remark 1. This is the simplest example of problems appearing in the description of free electrons in the crystals; much more complicated and realistic example would be Schrödinger equation \begin{equation*} X'' +\bigl(\lambda-V(x)\bigr)X=0 \end{equation*} or its $3D$-analog.
### Problems with "almost" explicit solutions
Example 6. (Robin boundary conditions). Consider eigenevalue problem \begin{align} & X'' +\lambda X=0 && 0< x< l,\label{eq-4.2.27}\\[3pt] & X'(0)=\alpha X(0), \quad X'(l)=-\beta X(l)\label{eq-4.2.28} \end{align} with $\alpha\ge 0$, $\beta\ge 0$ ($\alpha+\beta>0$). Then \begin{align} \lambda \int_0^l X^2\,dx=&-\int_0^l X''X\,dx\notag \\ =&\int_0^l X'^2\,dx-X'(l)X(l) +X'(0)X(0)\notag\\ =&\int_0^l X'^2\,dx+\beta X(l)^2 +\alpha X(0)^2\qquad \label{eq-4.2.29} \end{align} and $\lambda_n=\omega_n^2$ where $\omega_n>0$ are roots of \begin{align} & \tan (\omega l)= \frac{(\alpha+\beta)\omega}{\omega^2-\alpha\beta}; \label{eq-4.2.30}\\ & X_n= \omega \cos (\omega_n x) +\alpha \sin (\omega_n x);\label{eq-4.2.31} \end{align} ($n=1,2,\ldots$).
Indeed, looking for $X=A\cos(\omega x) + B\sin(\omega x)$ with $\omega=\sqrt{\lambda }$ (we are smarter now!) we find from the first equation of (\ref{eq-4.2.28}) that $\omega B= \alpha A$ and we can take $A=\omega$, $B= \alpha$, $X= \omega \cos(\omega x) -\alpha\sin( \omega x)$.
Then plugging to the second equation we get \begin{gather*} -\omega ^2\sin(\omega l)+\alpha \omega \cos(\omega l) =-\beta \omega \cos(\omega l) - \alpha \beta \sin(\omega l) \end{gather*} which leads us to (\ref{eq-4.2.30})--(\ref{eq-4.2.31}).
Observe that
1. $\alpha,\beta\to 0^+\implies \omega_n \to \frac{\pi (n-1)}{l}$.
2. $\alpha,\beta\to +\infty \implies \omega_n \to \frac{\pi n}{l}$.
3. $\alpha\to 0^+,\beta\to +\infty\implies \omega_n \to \frac{\pi (n-\frac{1}{2})}{l}$.
Example 7. (Robin boundary conditions (negative eigenvalues)). However if $\alpha$ and/or $\beta$ are negative, one or two negative eigenvalues $\lambda=-\gamma^2$ can also appear where \begin{align} & \tanh (\gamma l )= {-\frac{(\alpha + \beta)\gamma }{\gamma ^2 + \alpha\beta}}, \label{eq-4.2.32}\\ & X(x) = \gamma \cosh (\gamma x) + \alpha \sinh (\gamma x). \label{eq-4.2.33} \end{align}
Indeed, looking for $X=A\cosh(\gamma x) + B\sinh(\gamma x)$ with $\gamma=\sqrt{-\lambda }$ we find from the first equation of (\ref{eq-4.2.28}) that $\gamma B= \alpha A$ and we can take $A=\gamma$, $B= \alpha$, $X= \gamma \cosh(\gamma x) -\alpha \sinh( \gamma x)$.
Then plugging to the second equation we get \begin{gather*} \gamma ^2\sinh(\gamma l)+\alpha \gamma \cosh(\gamma l) =-\beta \gamma \cosh(\gamma l) - \alpha \beta \sinh(\gamma l) \end{gather*} which leads us to (\ref{eq-4.2.32})--(\ref{eq-4.2.33}).
To investigate when it happens, consider the threshold case of eigenvalue $\lambda=0$: then $X=cx+d$ and plugging into boundary conditions we have $c=\alpha d$ and $c=-\beta (d+lc)$; this system has non-trivial solution $(c,d)\ne 0$ iff $\alpha+\beta+\alpha\beta l =0$. This hyperbola divides $(\alpha,\beta)$-plane into three zones:
To calculate the number of negative eigenvalues one can either apply the general variational principle or analyze the case of $\alpha=\beta$; for both approaches see Appendix 4.A.
Example 8. (Oscillations of the rod (beam)) Small oscillations of the rod are described by
$$u_{tt} + Ku_{xxxx}=0 \label{eq-4.2.34}$$ with the boundary conditions \begin{align} &u|_{x=0}=u_x|_{x=0}=0\label{eq-4.2.35},\\ &u|_{x=l}=u_x|_{x=l}=0\label{eq-4.2.36} \end{align} (both ends are clamped) or \begin{align} &u|_{x=0}=u_{x}|_{x=0}=0\label{eq-4.2.37},\\ &u_{xx}|_{x=l}=u_{xxx}|_{x=l}=0\label{eq-4.2.38} \end{align} (left end is clamped and the right one is free); one can consider different and more general boundary conditions.
Separating variables we get \begin{align} &X^{IV}-\lambda X=0, \label{eq-4.2.39}\\ &T'' +\omega^2 T=0, && \omega=\sqrt[4]{K\lambda} \label{eq-4.2.40} \end{align} with the boundary conditions \begin{align} &X(0)=X'(0)=0\label{eq-4.2.41},\\ &X(l)\,=X'(l)\,=0\label{eq-4.2.42} \end{align} or \begin{align} &X(0)\, =X'(0)\,=0\tag{41},\\ &X''(l)=X'''(l)=0.\label{eq-4.2.43} \end{align} Eigenvalues of the both problems are positive (we skip the proof). Then \begin{gather} X(x)= A\cosh(kx) + B\sinh (kx) +C\cos(kx) +D\sin(kx). \label{eq-4.2.44} \end{gather} and (\ref{eq-4.2.41}) implies that $C=-A$, $D=-B$ and $$X(x)= A\bigl(\cosh(kx) -\cos(k x)\bigr)+ B\bigl(\sinh (kx) -\sin(kx)\bigr). \label{eq-4.2.45}$$ Plugging into (\ref{eq-4.2.43}) we get \begin{align*} &A(\cosh(kl) -\cos(kl))+ B(\sinh (kl) -\sin(kl))=0,\\ &A(\sinh(kl)\ +\sin (kl))+ B(\cosh (kl) -\cos(kl))=0. \end{align*} Then determinant must be $0$: \begin{gather*} (\cosh(kl) -\cos(kl))^2 -(\sinh^2 (kl) -\sin^2(kl))=0 \end{gather*} which is equivalent to \begin{gather} \cosh(kl)\cdot\cos(kl)=1. \label{eq-4.2.46} \end{gather}
On the other hand, plugging \begin{equation*} X(x)= A\bigl(\cosh(kx) -\cos(k x)\bigr)+ B\bigl(\sinh (kx) -\sin(kx)\bigr). \tag{45} \end{equation*} into (\ref{eq-4.2.43}) $X''(l)=X'''(l)=0$ leads us to \begin{align*} &A(\cosh(kl) +\cos(kl))+ B(\sinh (kl) +\sin(kl))=0,\\ &A(\sinh(kl)\ -\sin (kl))+ B(\cosh (kl) + \cos(kl))=0. \end{align*} Then determinant must be $0$: \begin{gather*} (\cosh(kl) +\cos(kl))^2 -(\sinh^2 (kl) -\sin^2(kl))=0 \end{gather*} which is equivalent to \begin{gather} \cosh(kl)\cdot\cos(kl)=-1. \label{eq-4.2.47} \end{gather}
We solve (\ref{eq-4.2.46}) and (\ref{eq-4.2.47}) graphically:
Case of both ends free, results in the same eigenvalues $\lambda_n =k_n^4$ as when both ends are clumped, but with eigenfunctions \begin{equation*} X(x)= A\bigl(\cosh(kx) +\cos(k x)\bigr)+ B\bigl(\sinh (kx) +\sin(kx)\bigr). \end{equation*} and also in double eigenvalue $\lambda=0$ and eigenfunctions $1$ and $X$. | 2021-01-22 22:13:40 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 20, "equation": 2, "x-ck12": 0, "texerror": 0, "math_score": 0.9991034269332886, "perplexity": 7552.946296066806}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-04/segments/1610703531429.49/warc/CC-MAIN-20210122210653-20210123000653-00162.warc.gz"} |
http://lib.physcon.ru/doc?id=da77b037eb25 | # IPACS Electronic library
## PERTURBATION ANALYSIS OF SIMPLE EIGENVALUES OF SINGULAR LINEAR SYSTEMS
M. Isabel Garcia-Planas, Sonia Tarragona
In this work a study of
the behavior of a simple eigenvalue of singular linear system family $E(p)\dot x=A(p)x+B(p)u$, $y=C(p)x$ smoothly dependent on a vector of real parameters $p=(p_{1},\ldots ,p_{n})$ is presented. | 2018-05-27 01:12:40 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8472756147384644, "perplexity": 2765.5722388339086}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-22/segments/1526794867977.85/warc/CC-MAIN-20180527004958-20180527024958-00054.warc.gz"} |
https://www.semanticscholar.org/paper/Solvability-of-a-Lie-algebra-of-vector-fields-their-Cari%C3%B1ena-Falceto/4eea5e79251be17a311b7e4de76c3fe85cf1402b | # Solvability of a Lie algebra of vector fields implies their integrability by quadratures
@article{Cariena2016SolvabilityOA,
title={Solvability of a Lie algebra of vector fields implies their integrability by quadratures},
author={Jos{\'e} F. Cari{\~n}ena and Fernando Falceto and Janusz Grabowski},
journal={Journal of Physics A: Mathematical and Theoretical},
year={2016},
volume={49}
}
• Published 8 June 2016
• Mathematics
• Journal of Physics A: Mathematical and Theoretical
We present a substantial generalisation of a classical result by Lie on integrability by quadratures. Namely, we prove that all vector fields in a finite-dimensional transitive and solvable Lie algebra of vector fields on a manifold can be integrated by quadratures.
10 Citations
• Mathematics
• 2020
We present a local and constructive differential geometric description of finite-dimensional solvable and transitive Lie algebras of vector fields. We show that it implies a Lie's conjecture for such
• Mathematics
Journal of Physics A: Mathematical and Theoretical
• 2023
Motivated by the time-dependent Hamiltonian dynamics, we extend the notion of Arnold–Liouville and noncommutative integrability of Hamiltonian systems on symplectic manifolds to that on cosymplectic
• Mathematics
Symmetry
• 2021
The general theory of the Jacobi last multipliers in geometric terms is reviewed and the theory is applied to different problems in integrability and the inverse problem for one-dimensional mechanical systems.
A geometric approach to integrability and reduction of dynamical system is developed from a modern perspective. The main ingredients in such analysis are the infinitesimal symmetries and the tensor
The Gauss-Manin equations are solved for a class of flat-metrics defined by Novikov algebras, this generalizing a result of Balinskii and Novikov who solved this problem in the case of commutative
• S. Grillo
• Mathematics
Analysis and Mathematical Physics
• 2021
The non-commutative integrability (NCI) is a property fulfilled by some Hamiltonian systems that ensures, among other things, the exact solvability of their corresponding equations of motion. The
• Mathematics
Journal of Physics A: Mathematical and Theoretical
• 2022
A stratified Lie system is a nonautonomous system of first-order ordinary differential equations on a manifold M described by a t-dependent vector field X=∑α=1rgαXα , where X 1, …, X r are vector
• Mathematics
• 2019
A $\mathcal{F}$- foliated Lie system is a first-order system of ordinary differential equations whose particular solutions are contained in the leaves of the foliation $\mathcal{F}$ and all
• Mathematics
International Journal of Geometric Methods in Modern Physics
• 2019
The theory of quasi-Lie systems, i.e. systems of first-order ordinary differential equations that can be related via a generalized flow to Lie systems, is extended to systems of partial differential
• Mathematics
• 2020
Transitive local Lie algebras of vector fields can be easily constructed from dilations of $\mathbb{R}^n$ associating with coordinates positive weights (give me a sequence of $n$ positive integers
## References
SHOWING 1-10 OF 14 REFERENCES
• Mathematics
• 2014
In this paper, we extend the Lie theory of integration by quadratures of systems of ordinary differential equations in two different ways. First, we consider a finite-dimensional Lie algebra of
This paper addresses a class of problems associated with the conditions for exact integrability of systems of ordinary differential equations expressed in terms of the properties of tensor
Preface.- Basic Concepts.- Semisimple Lie Algebras.- Root Systems.- Isomorphism and Conjugacy Theorems.- Existence Theorem.- Representation Theory.- Chevalley Algebras and Groups.- References.-
• Mathematics
• 1978
In this paper we shall show that the equations of motion of a solid, and also Liouville's method of integration of Hamiltonian systems, appear in a natural manner when we study the geometry of level
is integrable by quadratures [1] (for details, see also [2, 3]). More precisely, all of its solutions can be found by “algebraic operations” (including inversion of functions) and “quadratures,” that
In [1] a local description of analytic vector fields finitely generating a transitive nilpotent Lie algebra L on a manifold is given. Our aim is to generalize this result by (i) omitting the
Let M be a real analytic manifold, and let L be a transitive Lie algebra of real analytic vector fields on M. A concept of completeness is introduced for such Lie algebras. Roughly speaking, L is
We study finite-dimensional Lie algebras of polynomial vector fields in $n$ variables that contain the vector fields ${\partial}/{\partial x_i} \; (i=1,\ldots, n)$ and \$x_1{\partial}/{\partial x_1}+
Part 1 Newtonian mechanics: experimental facts investigation of the equations of motion. Part 2 Lagrangian mechanics: variational principles Lagrangian mechanics on manifolds oscillations rigid | 2023-01-27 08:42:03 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 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": 0.8045046329498291, "perplexity": 695.7261629129925}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764494974.98/warc/CC-MAIN-20230127065356-20230127095356-00013.warc.gz"} |
https://mouse-connectivity-models.readthedocs.io/en/latest/auto_examples/plot_nadaraya_watson.html | Note
This example is in part a copy of plot_kernel_ridge_regressions by Jan Hendrik Metzen found in the package Scikit-Learn.
Nadaraya-Watos (NW) regression learns a non-linear function by using a kernel- weighted average of the data. Fitting NW can be done in closed-form and is typically very fast. However, the learned model is non-sparse and thus suffers at prediction-time.
This example illustrates NW on an artificial dataset, which consists of a sinusoidal target function and strong noise added to every fifth datapoint.
Out:
GridSearchCV 5 fold cross validation fitted in 0.20 s
optimal bandwidth found: 3.16
NadarayaWatsonCV leave-one-out cross validation fitted in 0.02 s
optimal bandwidth found: 3.162
# Authors: Joseph Knox <josephk@alleninstitute.org>
# NOTE: modified from plot_kernel_ridge_regression.py by Jan Hendrik Metzen
# from the package Scikit-Learn licensed under the 3 clause BSD License
# reproduced below:
#
#
# Copyright (c) 2007–2018 The scikit-learn developers.
#
#
# Redistribution and use in source and binary forms, with or without
# modification, are permitted provided that the following conditions are met:
#
# a. Redistributions of source code must retain the above copyright notice,
# this list of conditions and the following disclaimer.
# b. Redistributions in binary form must reproduce the above copyright
# notice, this list of conditions and the following disclaimer in the
# documentation and/or other materials provided with the distribution.
# c. Neither the name of the Scikit-learn Developers nor the names of
# its contributors may be used to endorse or promote products
# derived from this software without specific prior written
# permission.
#
#
# THIS SOFTWARE IS PROVIDED BY THE COPYRIGHT HOLDERS AND CONTRIBUTORS "AS IS"
# AND ANY EXPRESS OR IMPLIED WARRANTIES, INCLUDING, BUT NOT LIMITED TO, THE
# IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR PURPOSE
# ARE DISCLAIMED. IN NO EVENT SHALL THE REGENTS OR CONTRIBUTORS BE LIABLE FOR
# ANY DIRECT, INDIRECT, INCIDENTAL, SPECIAL, EXEMPLARY, OR CONSEQUENTIAL
# DAMAGES (INCLUDING, BUT NOT LIMITED TO, PROCUREMENT OF SUBSTITUTE GOODS OR
# SERVICES; LOSS OF USE, DATA, OR PROFITS; OR BUSINESS INTERRUPTION) HOWEVER
# CAUSED AND ON ANY THEORY OF LIABILITY, WHETHER IN CONTRACT, STRICT
# LIABILITY, OR TORT (INCLUDING NEGLIGENCE OR OTHERWISE) ARISING IN ANY WAY
# OUT OF THE USE OF THIS SOFTWARE, EVEN IF ADVISED OF THE POSSIBILITY OF SUCH
# DAMAGE.
from __future__ import division, print_function
import time
import numpy as np
import matplotlib.pyplot as plt
from sklearn.model_selection import GridSearchCV
print(__doc__)
rng = np.random.RandomState(0)
# #############################################################################
# Generate sample data
X = 5 * rng.rand(10000, 1)
y = np.sin(X).ravel()
y[::5] += 3 * (0.5 - rng.rand(X.shape[0] // 5))
X_plot = np.linspace(0, 5, 1e3)[:, None]
# #############################################################################
# Fit regression model
train_size = 100
param_grid=dict(kernel=["rbf"], gamma=np.logspace(-2, 2, 25))
# fit 5-fold using GridSearch
t0 = time.time()
nw_gs.fit(X[:train_size], y[:train_size])
gs_fit = time.time() - t0
print("GridSearchCV 5 fold cross validation fitted in %.2f s" % gs_fit)
print("\toptimal bandwidth found: %.2f" % nw_gs.best_estimator_.gamma)
t0 = time.time()
nw_cv.fit(X[:train_size], y[:train_size])
cv_fit = time.time() - t0
print("NadarayaWatsonCV leave-one-out cross validation fitted in %.2f s" % cv_fit)
print("\toptimal bandwidth found: %.3f" % nw_cv.gamma)
# predict
y_gs = nw_gs.predict(X_plot)
y_cv = nw_cv.predict(X_plot)
# #############################################################################
# Look at the results
plt.scatter(X[:100], y[:100], c='k', label='data', zorder=1,
edgecolors=(0, 0, 0))
plt.plot(X_plot, y_gs, 'c-', lw=3, label='5-Fold GridSearchCV')
plt.plot(X_plot, y_cv, 'r:', lw=3, label='LOO NadarayaWatsonCV')
plt.xlabel('data')
plt.ylabel('target') | 2023-02-04 12:01:49 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.37106552720069885, "perplexity": 2396.9663423340116}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764500126.0/warc/CC-MAIN-20230204110651-20230204140651-00580.warc.gz"} |
https://math.stackexchange.com/questions/2233231/prove-the-unit-ball-in-the-space-ell-2-of-infinite-sequences-is-not-totally-b | # Prove the unit ball in the space $\ell_2$ of infinite sequences is not totally bounded.
Prove that the unit ball $B$ in the space $\ell_2$ of infinite sequences $\{x_n\}$ satisfying the condition: $\sum^\infty_{n=1}x^2_n \leq 1$ is not totally bounded.
To begin, I would like to say I know the concept of what a totally bounded metric space $M$ means. I can see that the set of all points $x=(x_1,x_2,\ldots,x_n,\ldots)$ satisfying the above condition is a convex body (a non empty convex set with a non empty interior). It follows that the interior is the set of all points $x=(x_1,x_2,\ldots,x_n,\ldots)$ satisfying $\sum^\infty_{n=1}x^2_n < 1$. But how am I to show that the ball is not totally bounded in a formal proof? Help/hints greatly appreciated, thanks.
Consider the set $\{v_n : n=1,2,3,\ldots\}$ in $\ell_2$ where $v_n$ is the point $(x_1,x_2,x_3,\ldots)$ for which $x_n=1$ and $x_k=0$ for all $k\ne n.$ What happens if you try to cover it with balls of radius $1/10\text{?}$ Each ball can cover only one of these points. All of these points are on the boundary of the closed unit ball. Thus the closed unit ball cannot be covered by only finitely many open balls of radius $1/10.$
• Interesting, thanks! So you considered a set of sequences such that it was a set of discontinuous sequences in the sense that sequences are isolated from each other? And all of these isolated points are on the boundary of the closed unit ball because you made $x_n=1$? The one thing I do not follow is why each ball can cover only one of these points, sorry if I am missing something trivial. – Rick Owens Apr 14 '17 at 2:10
• Must you first define the metric of the space $\ell _2$? – Rick Owens Apr 14 '17 at 2:12
• The metric is the one following from the norm -- you have the norm $\ell_2$, so the metric follows from that: $d(x,y) = \lVert x-y\rVert_2$ – Clement C. Apr 14 '17 at 2:13
• @ReginaldDick : We have, for example, $$v_3 = (0,0,1,0,0,0,0,\ldots)$$ where all entries are $0$ except the third one, and $$v_3 = (0,0,0,1,0,0,0,0,\ldots,$$ etc. The norm is $\|(x_1,x_2,x_3,\ldots)\| = \sqrt{x_1^2+x_2^2+x_3^2+\cdots}. \qquad$ – Michael Hardy Apr 14 '17 at 2:25 | 2020-01-18 06:31:03 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9314727187156677, "perplexity": 94.72353960145999}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-05/segments/1579250592261.1/warc/CC-MAIN-20200118052321-20200118080321-00219.warc.gz"} |
https://crutattoo.com/michael-van-zlfqawh/69e8ba-orbital-structure-of-ethane | of new orbitals of equal energies and identical shape. In the ethane molecule, the bonding picture according to valence orbital theory is very similar to that of methane. Ethane is structurally the simplest hydrocarbon that contains a single carbonâcarbon bond. Get detailed, expert explanations on sp3 hybrid orbitals and the structure of ethane that can improve your comprehension and help with homework. Each C â H bond length is 109 pm (1.09 Å). One s-orbital and three p-orbitals (px, py, pz) undergo Sp 3-hybridization to produce four Sp 3-hybrid orbitals for ⦠However, in ethane an sp 3 orbital of one carbon atom overlaps end to end with an sp 3 orbital of a second carbon atom to form a Ï bond between the two carbon atoms. Chapter 4.25 - Structure of Methane and Ethane I n the previous section 4.24, we saw the properties of hybrid orbitals. Log in. (b) These ⦠The six hydrogen atoms will each share their 1 electron with carbon to form six carbon - ⦠Composition of ethane molecule: Ethane molecule consists of two carbon atoms and six H-atoms (C 2 H 6 ). What is the Structure of an Ethane Molecule?. The structure ⦠The structure and overall outline of the bonding orbitals of ethane are shown in Figure 12. Both carbons are sp 3-hybridized, meaning that both have four bonds arranged with tetrahedral geometry. At standard temperature and pressure, ethane is a colorless, odorless gas. A carbon atom has 4 electrons in its outer shell. There are two ways for the orbital combination to occur-an additive way and a subtractive way. Like an atomic orbital, a molecular orbital has a specific size, shape, and energy. For symmetry reasons, the MOs of ethane, CH 3 CH 3, are rather different to those of methane as the Ï-bonds and Ï*-antibonds are present.. Ethane possesses 8 atoms, 14 (valence) electrons and 7 MOs (four bonding and three antibonding). Hydrogen can only form 1 bond.. sagarparise 21.06.2018 Chemistry Secondary School +5 pts. 5 years ago. The ethane molecule has fourteen valence electrons occupying seven bonding molecular orbitals. Its chemical formula is C2H6. Learn all about sp3 hybrid orbitals and the structure of ethane. p-orbitals (px, py, pz) undergo Sp3-hybridization to produce four Sp3-hybrid orbitals for each carbon atom. Discuss the hybridization of carbon atoms in alkene C3H4 and show the Ï- orbital overlaps. An orbital view of the bonding in ethene. Answered What is the structure of ethane 2 See answers The orientation of the two CH 3 groups is not fixed relative to each other. Structure o Ethane C2H6 In ethane molecule both the carbon atoms are sp3 hybridised. Ethene is built from hydrogen atoms (1s 1) and carbon atoms (1s 2 2s 2 2p x 1 2p y 1). The carbon atom doesn't have enough unpaired electrons to form the required number of bonds, so it needs to promote one of the 2s 2 pair into the empty 2p z orbital. 2 0. Sp³ Hybridization - Ethane Structure Definition Hybridization is defined as a phenomenon where the mixing of pure atomic orbital takes place but with slightly different energies, resulting in the formation of equal no. structure of ethane. In picture 1 we show the molecular orbital structure of F2. Answered Molecular orbital and atomic orbital picture of ethane 1 See answer In the case of Ethene, there is a difference from methane or ethane, because each carbon is only joining to three other atoms rather than four. What if a molecule can be described by multiple good (major) Also draw the orbital diagram? Fig 7: Structure of Ethene. Note that we denote ions with brackets around the structure indicating the charge outside the brackets When several arrangements of atoms are possible as for CHO 2 sp3 orbital. Ask your question. The structure of Ethane is CH3-CH3 1. ⢠One 2s orbital and two 2p orbitals form three sp2 hybrid orbitals. The dashed lines show the remaining p orbitals which do not take part in the bonding. Carbon is in group 4 of the periodic table. MEDVID. Ethane. Ethane is a hydrocarbon compound that exists in nature as an odorless and colorless gas at standard temperature and pressure. Join now. Ethane is a colorless, odorless, and flammable gas with a chemical formula of C2H6; it has two carbon (C) atoms and six hydrogen (H) atoms. Figure 12. Log in. Each of the remaining sp 3 hybrid orbitals overlaps with an s orbital of a hydrogen atom to form carbonâhydrogen Ï bonds. (a) In the ethane molecule, C 2 H 6, each carbon has four sp 3 orbitals. Join now. Join now. The out-of-phase combination the anti-bonding orbital. asked Jun 1, 2019 in Chemistry by Ruksar (68.7k points) hydrocarbons; class-11 +1 vote. 36 Structure and Bonding Hybridization and Bonding in Organic Molecules: Making a model of ethane illustrates one additional feature about its structure. **The bonding Ï orbital is the lower energy orbital and contains both p electrons (with opposite spins) in the ground state of the molecule. Later in this section, we will also see the basic details about sigma bonds Join now. the carbons in ethane are sp3 hybridized, the carbons in ethene are sp2 hybridized (double bonded C's) the carbons in ethyne are sp hybridized (triple bonded C's) Source(s): full year of gen chem, currently in organic chem. sp2. Follow. ⢠One 2s orbital and one 2p orbital form two sp hybrid orbitals. Orbital Structures are divided into two types: Logistics Structures and Tactical Structures. Ethene sp2 hybridization with a pi bond. Learn vocabulary, terms, and more with ⦠Start studying CH 1.6 & 1.7: sp3 Hybrid Orbitals and the Structure of Methane & Ethane. eg. Valence-Bond (Orbital Hybridization) provides more insight than Lewis model ability to connect structure and reactivity to hybridization develops with practice Molecular Orbital potentially the most powerful method but is the most abstract requires the most experience to use effectively Ask your question. In the $\mathrm{H}_{2}$ molecule, for example, two singly occupied $1 \mathrm{s}$ atomic orbitals combine to form two molecular orbitals. 1 Logistics Structures 1.1 TEC Structures 1.2 Advent Structures 1.3 Vasari Structures 2 Tactical Structures 2.1 TEC Structures 2.2 Advent Structures 2.3 Vasari Structures 2.4 Pirate Structures Logistics structures will help you ⦠Experimental evidence shows that rotation around Ï bonds occurs easily. Anonymous. Like in methane - the molecular orbitals of ethane show increasing nodal structure with increasing orbital energy. A hydrogen atom has 1 electron in its outer shell. Sigma bond: A covalent bond resulting from the formation of a molecular orbital by the end-to-end overlap of atomic orbitals, ... Ethane, C 2 H 6. As can be seen from the energy diagram - four of the molecular orbitals occur as degenerate pairs. In picture 2 we show the overlapping p orbitals, which form the bond between the two fl uorine atoms, in red and green gradients. The same kind of orbital hybridization that accounts for the methane structure also accounts for the bonding together of carbon atoms into chains and rings to make possible many millions of organic compounds. This preview shows page 15 - 21 out of 26 pages.. Orbitals and the Structure of Methane Carbon has 4 valence electrons (2 s 2 2 p 2) In CH 4, all CâH bonds are identical (tetrahedral) sp 3 Methane Carbon has 4 valence electrons (2 s 2 2 p 2) In CH 4, all CâH bonds are identical (tetrahedral) sp 3 Explain the structure of (C2H6) Ethane using hybridization concept. For ethene, the Ï framework is created by the interaction of the sp 2 hybrid orbitals of the C atoms and H1s orbitals. The Lewis structure of ethane is 7, which shows that there are one carbon-carbon bond and six carbon-hydrogen bonds in the ethane molecule. Example: What is hybridization on nitrogen atoms? Ethane is the simplest hydrocarbon since it contains only one carbonâcarbon bond in its structure. The region of greatest probability of finding the electrons in the bonding Ï orbital is a region generally situated above and below the plane of the Ï-bond framework between the two ⦠1. The carbon-carbon bond, with a bond length of 1.54 Å, is formed by overlap of one sp 3 orbital ⦠Polyatomic Species Molecular Orbital Theory Chemogenesis. shaukat8981217 4 minutes ago Chemistry Secondary School +5 pts. Each C has a p orbital unused by the hybrids and it is these on the adjacent C atoms that interact to form the C-C Ï bond. The second most important constituent of natural gas, Log in. Ethane molecule consists of two carbon atoms and six H-atoms (C2H6 ). 1. Due to the presence of a single bond in its structure, it is classified as an alkane hydrocarbon and is included in the first four primary alkanes: methane, ethane, propane, and butane.Ethane ⦠2: ethane. 6 years ago | 144 views. Nature of Hybridization: In ethane each C-atom is Sp 3-hybridized containing four Sp 3-hybrid orbitals. In its formation, one hybrid orbital of one carbon atom overlaps with one sp3 hybrid orbital of a second carbon atom along the internuclear axis to form a sigma (Ï) C â C bond. sp3? This is exactly the same as happens ⦠Ethane, a colourless, odourless, gaseous hydrocarbon (compound of hydrogen and carbon), belonging to the paraffin series; its chemical formula is C2H6. RE: Ethane is a colourless, odourless, gaseous hydrocarbon (compound containing only hydrogen and carbon) which belongs to the paraffin subcategory. 35 Hybridization & Bonding in Organic Molecules. Staggered Ethane â D 3d. Molecular orbital and atomic orbital picture of ethane - 18345601 1. Products from Oil. 1 answer. The structure ⦠Including Resonance in Geometry For each atom, the lowest hybridization state observed in major resonance structures is the correct one. Ethane is a chemical compound with chemical formula CâHâ. Ethane is isolated on an industrial scale from natural gas, and as a byproduct of petroleum refining. Log in. Like methane, ethane a covalent compound. Orbital-orbital Interactions and Symmetry Adapted Linear Combinations; ... Home / Structure and Bonding / Symmetry / Staggered Ethane â D3d. (ii) Molecular orbital picture of ethane: In ethane molecule, both carbon atoms are in the sp3 hybrid state. Bonding in Ethane. In this section, we will see how those orbitals can be used to make a satisfactory model of CH 4. Ethane, C2H6, is the simplest molecule containing a carbonâcarbon bond. Ï z y x Ï* x y z Construct the molecular orbital diagram ⦠Experimentally, ethane contains two elements, carbon and hydrogen, and the molecular formula of ethane is C 2 H 6. This Site Might Help You. Improve your comprehension and help with homework molecule containing a carbonâcarbon bond in its outer shell take! The properties of hybrid orbitals 1 electron in its structure atoms are sp3 hybridised Products Oil... Containing a carbonâcarbon bond its structure the Ï framework is created by the of. Illustrates one additional feature about its structure explanations on sp3 hybrid orbitals both have four arranged... H 6, each carbon atom formula of ethane is a hydrocarbon compound that in. Pressure, ethane contains two elements, carbon and hydrogen, and the molecular orbitals of is. Show the remaining sp 3 hybrid orbitals of equal energies and identical shape each C-atom is sp 3-hybridized meaning... 68.7K points ) hydrocarbons ; class-11 +1 vote to make a satisfactory model of ethane illustrates additional... The dashed lines show the Ï- orbital overlaps each carbon has four sp 3-hybrid.. Be seen from the energy diagram - four of the sp 2 hybrid orbitals not fixed to. Do not take part in the ethane molecule, the bonding picture according to orbital! To produce four Sp3-hybrid orbitals for each carbon has four sp 3-hybrid orbitals arranged with tetrahedral geometry one carbonâcarbon.! Increasing orbital energy about its structure / structure and bonding Hybridization and bonding Hybridization bonding! Length is 109 pm ( 1.09 Å ) orbital form two sp hybrid overlaps! Are sp 3-hybridized containing four sp 3 hybrid orbitals of ethane show increasing nodal structure with increasing orbital.. Atoms in alkene C3H4 and show the remaining p orbitals which do not take part the... Two elements, carbon and hydrogen, and as a byproduct of petroleum refining length 109... ) ethane using Hybridization concept the remaining p orbitals which do not take part in the hybrid. Form carbonâhydrogen Ï bonds occurs easily 4 minutes ago Chemistry Secondary School +5 pts colorless. Is sp 3-hybridized containing four sp 3-hybrid orbitals carbon-carbon bond and six carbon-hydrogen bonds the. Structurally the simplest hydrocarbon that contains a single carbonâcarbon bond, C 2 H 6, carbon! Molecular formula of ethane show increasing nodal structure with increasing orbital energy subtractive.. Electron in its outer shell in this section, we will See how those orbitals can be seen from energy! Valence orbital theory is very similar to that of methane rotation around bonds... Energies and identical shape is very similar to that of methane form Ï! Ethane illustrates one additional feature about its structure 4 electrons in its outer shell periodic... About its structure form carbonâhydrogen Ï bonds occurs easily carbon atoms in alkene C3H4 and show remaining. Ways for the orbital combination to occur-an additive way and a subtractive way those orbitals can described! Of hybrid orbitals of equal energies and identical shape form carbonâhydrogen Ï.. Structure with increasing orbital energy to valence orbital theory is very similar to that of methane to other... Ethane I n the previous section 4.24, we saw the properties of hybrid.. To produce four Sp3-hybrid orbitals for each carbon atom as a byproduct of petroleum.... Hybridization of carbon atoms are in the ethane molecule, the bonding in Organic:! O ethane C2H6 in ethane molecule, C 2 H 6 has fourteen valence electrons seven... / structure and bonding in Organic Molecules: Making a model of ethane - 18345601 1 H.! Staggered ethane â D3d a subtractive way, and as a byproduct of refining... Evidence shows that there are one carbon-carbon bond and six carbon-hydrogen bonds in the hybrid... Orbital energy sp hybrid orbitals overlaps with an s orbital of a atom! Exists in nature as an odorless and colorless gas at standard temperature and,! With chemical formula CâHâ picture according to valence orbital theory is very similar to that of methane C-atom... With chemical formula CâHâ will See how those orbitals can be described by multiple good ( major an! The previous section 4.24, we saw the properties of hybrid orbitals like in methane the. A molecule can be used to make a satisfactory model of ethane illustrates one feature!, each carbon has four sp 3 hybrid orbitals overlaps with an s orbital of a hydrogen atom to carbonâhydrogen! Carbons are sp 3-hybridized, meaning that both have four bonds arranged with tetrahedral geometry be seen from energy. Both have four bonds arranged with tetrahedral geometry, carbon and hydrogen and. To produce four Sp3-hybrid orbitals for each carbon atom and a subtractive way this is the... Sp3 hybrid orbitals overlaps with an s orbital of a hydrogen atom has 1 electron in outer... Ï- orbital overlaps used to make a satisfactory model of CH 4 that rotation around bonds!, ethane is 7, which shows that there are two ways for the orbital combination to occur-an way! Is C 2 H 6 of an ethane molecule has fourteen valence electrons occupying seven bonding molecular orbitals occupying bonding. Orbital Structures are divided into two types: Logistics Structures and Tactical Structures ethene, the bonding â H length... +1 vote are in the ethane molecule, C 2 H 6, carbon. 3-Hybridized, meaning that both have four bonds arranged with tetrahedral geometry, that! Not take part in the sp3 hybrid orbitals and the molecular orbitals occur degenerate... With tetrahedral geometry the carbon atoms are sp3 hybridised view of the two CH 3 groups is not relative! Fixed relative to each other px, py, pz ) undergo Sp3-hybridization to produce orbital structure of ethane Sp3-hybrid orbitals for carbon! ) undergo Sp3-hybridization to produce four Sp3-hybrid orbitals for each carbon atom has 4 electrons in its outer.... ( a ) in the ethane molecule, the Ï framework is created by the interaction the. Orbital view of the bonding in ethene exactly the same as happens ⦠Products Oil., and the molecular formula of ethane: in ethane each C-atom sp. Molecules: Making a model of CH 4 Sp3-hybridization to produce four Sp3-hybrid orbitals for each carbon atom that! Molecule containing a carbonâcarbon bond in its outer shell experimental evidence shows that are! Of petroleum refining a ) in the ethane molecule, both carbon atoms are the. View of the C atoms and H1s orbitals structure o ethane C2H6 in each. C-Atom is sp 3-hybridized, meaning that both have four bonds arranged with tetrahedral geometry 18345601 1 form... Framework is created orbital structure of ethane the interaction of the two CH 3 groups is not fixed relative to other! Will See how those orbitals can be described by multiple good ( major ) orbital! In methane - the molecular orbital picture of ethane 1 See answer the ethane molecule, 2... Tetrahedral geometry orbital of a hydrogen atom has 1 electron in its outer shell that of methane of CH.. Answers the out-of-phase combination the anti-bonding orbital of an ethane molecule has fourteen valence electrons occupying seven bonding molecular.. Which shows that rotation around Ï bonds to each other has fourteen electrons... The sp 2 hybrid orbitals sp 3-hybridized, meaning that both have bonds! Feature about its structure what if a molecule can be seen from the energy -. Of the sp 2 hybrid orbitals CH 4 Logistics Structures and Tactical Structures that rotation around Ï bonds in 1. Six carbon-hydrogen bonds in the ethane molecule, both carbon atoms are in the sp3 hybrid of... Orbital picture of ethane 2 See answers the out-of-phase combination the anti-bonding orbital ( px, py, ). +1 vote the molecular orbital picture of ethane that can improve your comprehension and help with homework orbitals! And a subtractive way degenerate pairs on sp3 hybrid state overlaps with an s orbital of a hydrogen atom 1... 1 we show the molecular formula of ethane is C 2 H 6 ) an view... In nature as an odorless and colorless gas at standard temperature and.. Hybrid orbitals of equal energies and identical shape C-atom is sp 3-hybridized meaning! Orbitals which do not take part in the sp3 hybrid orbitals overlaps with s! And one 2p orbital form two sp hybrid orbitals of the bonding Chemistry Secondary School +5 pts this exactly! As degenerate pairs group 4 of the molecular orbitals like in methane the. Molecule both the carbon atoms are in the ethane molecule, both carbon atoms are in ethane! Each carbon atom bond length is 109 pm ( 1.09 Å ) feature about its.!, carbon and hydrogen, and as a byproduct of petroleum refining has 1 electron in its shell... Energy diagram - four of the sp 2 hybrid orbitals and the molecular orbitals as byproduct..., ethane contains two elements, carbon and hydrogen, and as a byproduct petroleum... Of hybrid orbitals like in methane - the molecular orbitals ( 68.7k points ) hydrocarbons ; class-11 +1...., C 2 H 6, each carbon atom bonding / Symmetry / Staggered â! Are sp 3-hybridized, meaning that both have four bonds arranged with tetrahedral geometry asked Jun,... For each carbon atom has 4 electrons in its outer shell be by. Ethane molecule and help with homework Hybridization of carbon atoms are sp3 hybridised detailed, expert on. With chemical formula CâHâ six carbon-hydrogen bonds in the ethane molecule has fourteen valence electrons occupying seven molecular... Ch 4 with tetrahedral geometry not take part in the sp3 hybrid orbitals undergo Sp3-hybridization to four., both carbon atoms in alkene C3H4 and show the remaining p orbitals which do take... Ethene, the bonding bond and six carbon-hydrogen bonds in the ethane molecule, carbon. / Staggered ethane â D3d combination the anti-bonding orbital relative to each other, the bonding picture according to orbital...
0 | 2021-02-25 18:52:59 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5550451874732971, "perplexity": 3974.3154455116523}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-10/segments/1614178351454.16/warc/CC-MAIN-20210225182552-20210225212552-00398.warc.gz"} |
http://math.stackexchange.com/questions/865926/computing-complex-number | # Computing complex number [duplicate]
"Compute $(1 + i)^{1000}$.
So far I have: $(1+i)^{4 (2^2 5^3)}$ but I am not sure how to proceed. Ideas?
-
## marked as duplicate by apnorton, RecklessReckoner, daw, Normal Human, Steven Taschuk Jul 13 '14 at 17:56
Have you tried computing $(1+i)^2$? – Mark Bennet Jul 13 '14 at 13:11
HINT:
${(i+1)}^{1000} = {({(i+1)}^{2})}^{500} = {(2i)}^{500} = {2}^{500}{i}^{500} = {2}^{500}{(i^2)}^{250} = {2}^{500}{(-1)}^{250}$
-
why can you make the jump from $((1+i)^2)^{500} = (2i)^{500}$ ? – Ozera Jul 13 '14 at 13:18
@Ozera: ${(1+i)}^{2} = (1+i)(1+i) = 1^2+2i+i^2 = 1+2i-1 = 2i$ – barak manos Jul 13 '14 at 13:20
Shouldn't that $255$ be $250$? – Mark Dickinson Jul 13 '14 at 13:36
@MarkDickinson: Ooops :) Thanks – barak manos Jul 13 '14 at 13:40
@Ozera: Please note that in the original answer I wrote $255$ instead of $250$ (which essentially changes the answer from positive to negative). – barak manos Jul 13 '14 at 13:41
Powers of complex numbers are often easier in polar coordinates. The absolute value of $(1+i)$ is $\sqrt{2}$, and its angle is $\pi/4$. So the question becomes $$(\sqrt{2}e^{i\pi/4})^{1000}$$
-
Here, $\exp(z)$ stands for $e^z$
\begin{align}(1+i)^{1000}&=(\sqrt{2}(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4}))^{1000} \\ &= \sqrt2^{1000}\exp(1000\cdot\frac{\pi}{4}i)\\ &=\sqrt{2}^{1000}\exp(250\pi i) \\ &=2^{500}(\cos(250\pi)+ i\sin(250\pi)) \\ &\text{For even multiples of \pi, \cos \theta =1 , \sin \theta = 0} \\ &=2^{500}(1+0)=2^{500} \end{align}
- | 2015-12-01 10:07:29 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 1.0000066757202148, "perplexity": 1338.2422726858117}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-48/segments/1448398466178.24/warc/CC-MAIN-20151124205426-00161-ip-10-71-132-137.ec2.internal.warc.gz"} |
https://vapedev.ru/solving-ratios-word-problems-6742.html | # Solving Ratios Word Problems
Stack Exchange network consists of 175 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Question: how many gallons of yellow paint, and how many gallons or red paint, to make two gallons of orange paint? However, in this section of the text book, I'm not sure that it's "allowed" to do any of that.Visit Stack Exchange Four gallons of yellow paint plus two gallons of red paint make orange paint. Is it possible to solve this just with cross-multiplying a ratio?"Suzy has eight pairs of red socks and six pairs of blue socks. If her little sister owns nine pairs of purple socks and loses two of Suzy's pairs, how many pairs of socks do the sisters have left?
Tags: Gelato Business PlanEssay About Art NouveauWrite Portfolio EssaySmoking Persuasive EssaysAsk HomeworkPurpose Of Imprisonment EssayAerospace Engineer Research PaperWaste Disposal Research PaperEssay Indentation
Subtract the two missing pairs for a final equation of (8 6 9) - 2 = n, where n is the number of pairs of socks the sisters have left.
Using the equation, solve the problem by plugging in the values and solving for the unknown variable.
Each word problem may require a different format, but a visual representation of the necessary information makes it easier to work with.
In the example, the question asks how many socks the sisters own together, so you can disregard the information about Mark. This eliminates much of the information and leaves you with only the total number of socks that the sisters started with and how many the little sister lost. For example, the words and phrases "sum," "more than," "increased" and "in addition to" all mean to add, so write in the " " symbol over these words.
In the example, you know by adding up all the numbers for the sisters that you have a maximum of 23 socks.
Since the problem mentions that the little sister lost two pairs, the final answer must be less than 23.You can answer even the most complex word problems, provided you understand the mathematical concepts addressed.While the degree of difficulty may change, the way to solve word problems involves a planned approach that requires identifying the problem, gathering the relevant information, creating the equation, solving and checking your work.Exponents and roots come first, then multiplication and division, and finally addition and subtraction.Check if your answer makes sense with what you know.Now, multiply by $n$ which is the number of gallons of orange paint you want to make. So, making $n$ gallons of orange paint require mixing $\frac$ gallons of yellow paint and $\frac$ gallons of red paint. Question 3: Given ratio are- a:b = 2:3 b:c = 5:2 c:d = 1:4 Find a:b:c.Word problems often confuse students simply because the question does not present itself in a ready-to-solve mathematical equation.Divide these numbers by 6 in order to come back to one gallon of orange paint.Then, one gallon of orange paint is made mixing $\frac=\frac$ gallons of yellow paint and $\frac=\frac$ gallons of red paint.
## Comments Solving Ratios Word Problems
• ###### How do you solve ratio's word problems
The rules for solving word problems are read the problem, decide what you need to do, solve the problem, and check your the ratio and make that the denominator of the proportion and cross multiply. A proportion will help you solve problems like the one below.…
• ###### Solving Math Word Problems explanation and exercises
There are two steps to solving math word problems Translate the wording into a numeric equation that combines smaller "expressions"the problem entirely Get a feel for the whole problem. List information and the variables you identify Attach units of measure to the variables gallons, miles.…
• ###### Steps to Word Problem Solving Sciencing
Word problems often confuse students simply because the question does not present itself in a ready-to- solve mathematical equation. You can answer even the most complex word problems, provided you understand the mathematical concepts addressed. While the degree of difficulty may change, the.…
• ###### Solving Word Problems with Ratios and Proportions
Want to know something exciting? I taught ratios and proportions in Algebra 1 without once mentioning cross multiplying. Two kids brought it up as an option, but I quickly told them that there was a way to do these problems without cross multiplication.… | 2021-09-27 06:59:05 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5438926219940186, "perplexity": 818.3778720747395}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-39/segments/1631780058373.45/warc/CC-MAIN-20210927060117-20210927090117-00467.warc.gz"} |
https://physics.stackexchange.com/questions/514262/if-a-photon-truly-goes-through-both-slits-at-the-same-time-then-why-cant-we/514677 | # If a photon truly goes through both slits (at the same time), then why can't we detect it at both slits (at the same time)?
I am not asking about whether the photon goes through both slits, or why. I am not asking whether the photon is delocalized as it travels in space, or why.
Do we really know which slit the photon passed through in Afshar's experiment?
Which theory explains the path of a photon in Young's double-slit experiment?
Shooting a single photon through a double slit
Where John Rennie says:
The photons do not have a well defined trajectory. The diagram shows them as if they were little balls travelling along a well defined path, however the photons are delocalised and don't have a specific position or direction of motion. The photon is basically a fuzzy sphere expanding away from the source and overlapping both slits. That's why it goes through both slits. The photon position is only well defined when we interact with it and collapse its wave function. This interaction would normally be with the detector.
Lasers, Why doesn’t a photon go through the same slit every time?
Where ThePhoton says:
for example, if you put a detector after a two-slit aperture, the detector only tells you the photon got to the detector, it doesn't tell you which slit it went through to get there. And in fact there is no way to tell, nor does it even really make sense to say the photon went through one slit or the other.
In classical terms, this question might be obvious, because a classical billiard ball cannot be at two places in space at the same time. But this is not a billiard ball, this is a photon, a QM phenomenon. And this is not classical terms, but QM.
And if we truly accept that the photon travels through both slits, then it basically must exist in space at both places (both slits) at the same time.
But as soon as we interact with it (the wave function collapses), the photon becomes spatially localized, but only at a single location (at a certain time).
What is not obvious from QM, is how we can have these two things at the same time:
1. the photon pass through both slits
2. but we can only interact with it at one slit (not both)
What is that basic thing in QM, that will disallow for the photon to pass through both slits and be interacted with at both slits too? Somehow the QM world underneath will change to classical as soon as we measure, and interact with the photon. This change from QM to classical is where the possibility of the photon being at both places (both slits) at the same time gets disallowed somehow. This could be decoherence, as the QM entity gets information from the environment (because of the measurement), or just the fact that the wavefunction collapses and that has to have a single spatial location for the photon when measured.
So basically the photon goes through both slits, thus, it in some form exists at both slits at the same time. But when we try to interact with it, it will only be spatially localizable at one of the slits, not both at the same time.
Question:
1. If the photon truly goes through both slits (at the same time), then why can't we detect it at both slits (at the same time)?
• What's wrong with a good ol' EM wave interfering through the multiple slits? – user192234 Nov 16 '19 at 19:34
• John is correct in saying there is not a defined trajectory. So why do you ask a question assuming there is a trajectory? – BioPhysicist Nov 16 '19 at 20:11
• @AaronStevens I am not assuming a trajectory, just the opposite, I am assuming the photon goes through both slits. The question is, if it goes through both (thus in some form exists at both slits), then why can't we somehow measure this existence of the photon at both slits (at the same time)? What is the reason that when measured, from QM, we switch (maybe because of decoherence) to classical, where it is obvious that the photon cannot be (measured) at both slits at the same time? – Árpád Szendrei Nov 16 '19 at 20:39
• This seems a question related to the measurement problem: en.wikipedia.org/wiki/Measurement_problem so I would say there is not a definitive answer to it – Claudio P Nov 17 '19 at 22:02
• There is no proof or reason that a single photon goes through more than one slit at a time. Place a detector at both openings and only one will register. There is no reason it can’t be classical all the way and still produce QM predictions. – Bill Alsept Nov 18 '19 at 22:24
If the photon truly goes through both slits (at the same time), then why can't we detect it at both slits (at the same time)?
Alright, let's play some word games:
This isn't a well-defined question. "Detect a particle" doesn't mean anything in quantum mechanics. Quantum mechanical measurements are always measurements of specific observables. There is no holistic act of "observing all properties of a system at once" like there is in classical mechanics - a measurement is always specific to the one observable it measures, and the measurement irrevocably alters the state of the system being measured.
People often use "detect a particle" as shorthand for "perform a position measurement of a particle". By definition, a measurement of position has as its outcome a single position, and interacts with the state of the particle being measured such that it now really is in the state in which it is at that single position and nowhere else. So if you could perform position measurements that yielded both slits as the position of the particle, this would mean you have performed an impossible feat - there are now two particles, each in the state of being at one slit and that slit only. Quantum mechanics may be weird, but it is hopefully clear it is not this weird - we cannot duplicate a particle out of thin air just by measuring it.
If you don't insist on "detect" meaning "performing a position measurement", then of course the standard double slit setup is a "detection" of the photon at both slits - the pattern on the screen is only explainable by the particle's wavefunction passing through both slits and interfering with itself. This is of course just indirect reasoning - there simply is no observable whose eigenstates would naively correspond to "we have detected the photon at both slits at once".
Lastly, you seem to confuse "interacting" with "measuring" or "detecting". Of course we can interact with the particle at both slits - we just cannot perform position measurements (or other "which-way" measurements) at both slits and expect them to yield the impossible result of the particle split in two. But if you look at more sophisticated setups like the quantum erasers, there certainly is interaction with the particle at both slits - just carefully set up to not destroy the interference pattern, and hence no obtaining useable which-way information.
• Thank you i appreciate your answer. Do you mean that if there are detectors at both slits that do not collapse the wavefunction (I guess just elastic scattering) then it is possible to interact with the photon at both slits (this just does not qualify as measurement)? – Árpád Szendrei Nov 17 '19 at 2:10
• "The measurement irrevocably alters the state of the system being measured" is false. Rather, the measurement constrains the possibilities of other measurements to ones consistent with it, as determined by the underlying probability model. – R.. GitHub STOP HELPING ICE Nov 17 '19 at 20:09
• @R.. I fail to see the difference between changing the probabilities of future measurements and changing the state. A quantum state is essentially defined by the potential results of measurements on it, when you change these, you change the state. – ACuriousMind Nov 17 '19 at 21:29
• @ÁrpádSzendrei: Yes! Put a vertical polarizer at one slit and a horizontal polarizer at the other. Then the particle still passes through both slits but no longer in a manner that results in the interference pattern. Changing the angle of one of the polarizers gradually results in a gradual reappearance of the intereference pattern. This clearly shows interaction can happen at both slits while not completely collapsing the wavefunction. More simply, just put a glass piece after one slit, which would shift the interference pattern due to the higher refractive index of glass than air. – user21820 Nov 19 '19 at 7:15
• Despite my comment above, I may be closer to understanding the double-slit experiment with individual photons than I ever have been before, thanks to this answer. – T.J. Crowder Nov 19 '19 at 17:58
Think of it this way: A photon is the detection event. When there is only one photon, there is only one detection event. The probability distribution of detection events is associated with the photon's wavefunction.
• I really like this answer. The photon itself is the manifestation of the measurement that we make on the excitation of the field, and that can only be at one place. – Árpád Szendrei Nov 17 '19 at 0:40
• While a deeper explanation might demonstrate truth to this statement, the current short version does not sit right with me. This answer feels very "hand wavy." – Aaron Nov 18 '19 at 15:13
• Surely "photon" is the name we use to talk about a degree of excitation of a field, whereas a detection event is an avalanche of electrons in a detector, or a chemical reaction on a film, or a flicker of a needle. Furthermore, we can easily design detection events which indicate unambiguously that the field had non-zero excitation at two places at once. – Andrew Steane Nov 18 '19 at 19:08
• Can that be done for single photons? If so, how? – S. McGrew Nov 18 '19 at 19:32
• Please provide a reference. – S. McGrew Nov 19 '19 at 14:16
We've had a lot of answers already (because this problem invites them), but let me offer one more way to think about it. (As best I can tell, this is the interpretation of quantum mechanics closest to the point I'll make. As @PedroA notes below, what follows is interpretation-dependent.)
If the photon truly goes through both slits (at the same time), then why can't we detect it at both slits (at the same time)?
I think you're imagining we, as the scientists with our detector, are a classical system studying a separate quantum-mechanical one. But the entire experiment, including the detector and whoever inspects it, is also part of the quantum-mechanical setup. Our superposition isn't just of the photon passing through slit $$1$$ and its passing through slit $$2$$; it's of us detecting one and us detecting the other.
From a God's-eye point of view (if there is such a thing), we are superposed between announcing one result and announcing the other. We're not outside a quantum-mechanical system with such a God's-eye view, and therefore don't see the whole of the superposition. Hence we only see one result, not a bit of both.
• This is an interesting answer after careful read, +1. However, even though you added fantastic links to wikipedia, I think you should emphasize that this is only one of the various interpretations, and this matter is mostly philosophical than physical. When I first read your answer, the tone seemed to imply that what you're stating is scientifically true. – Pedro A Nov 19 '19 at 23:00
• This is the over-all approach of what is called Everettian Interpretation. It is focused on the Schrodinger equation, the dynamics of states and their time evolution. A lot of postulates can actually be derived if you obey the equation and study it. However you are not telling about some very surprising ontological facts that emerge if you pay close attention to this piece of work :) – user192234 Nov 27 '19 at 20:03
Quantum mechanics was not designed to make sense. It was designed to get correct answers. You can't expect it to make sense. That isn't what it's for.
If you want a story that makes sense (but might be wrong) here's one: Light traveling through space behaves exactly like a wave. There is no problem whatsoever about a wave going through two slits at the same time. That just vanishes.
Our methods to detect light are all quantized methods. Light changes a crystal on a photographic film. Or it sets off a photomultipler tube. Etc. They all give quantized detection. If you want a detector to tell you the amplitude of the wave, you need something that will take so many quantized measurements that they average out to something that seems continuous.
Since the measurements are quantized, of course QM will predict quantized results. That's what it ought to do if it is going to get correct answers. It will get answers that are compatible with the data.
There might be some weirdities in how light interacts with atoms. Those will affect the data. But there are no known weirdnesses about light traveling through space, it is all entirely compatible with light traveling as a wave.
QED is partly about describing light as quantum particles that behave exactly like waves. There's a lot of handwaving about probability functions etc. It's simpler and easier to just describe it as a wave, but QED gets the right measured answers too.
• Do you mean "You can't expect it to make intuitive sense?" QM tends to make a lot of sense for people who accept that it is not intuitive in the sense that CM is... – Marius Ladegård Meyer Nov 17 '19 at 0:39
• Yes, that's what I mean. – J Thomas Nov 17 '19 at 5:04
• I dont think this answer is true. Its possible to place a detector at both slits, get no result and yet the diffraction pattern will be altered. – thermomagnetic condensed boson Nov 18 '19 at 12:14
• If light is acting as a wave, why would you expect that a detector at one or both slits would not affect the diffraction pattern? – J Thomas Nov 18 '19 at 14:04
• @JiK, touché. The point I was trying to make was that the phrase "does not make sense" implies that it is wrong. Of course intuition in most fields improves with experience and expertise, but I think I will stand by the statement that QM is not intuitive in the sense that CM is. At least the professors and postdocs I've worked with seem to feel that way. – Marius Ladegård Meyer Nov 19 '19 at 16:50
Yes we can but the detectors should not completely destroy coherence. If not the interference pattern will be gone. For example two parallel polarisation filters should not destroy interference.
• Can you elaborate? (Really interested) – user192234 Nov 16 '19 at 20:31
• Can you please tell me, are you saying that we can have two detectors on the two slits, and for a single photon that is shot, both detectors will interact with the photon, just this should be elastic scattering? The polarization filters should be set up how? – Árpád Szendrei Nov 16 '19 at 20:45
• I somewhat remember from undergrad days, that the experiment has been done (with electrons). Turn on the detector for either slit, and the interference pattern vanishes because you know which slit each electron went through. (This for the regime where the electrons are microseconds apart, and the (non-)interference pattern is accumulated over seconds to minutes of "exposure"). – nigel222 Nov 19 '19 at 14:04
Can the photon be detected at both slits, of course not, it can not even be detected at one slit ... it is only detected when the EM field energy collapses and excites an electron .... science today cannot detect when a photon passes close to an electron (in a slit) and maybe disturbs it somehow. So why do you even care whether a photon passes thru one slit or the other? ... you care because because you are trying to explain this mysterious pattern that appears on the screen and you have been told it is due to "interference". Historically it has been described as an "interference" pattern because the pattern looked much like water wave interference. (And of course this is the basis for the described wave nature of light.) You believe this explanation but it requires that energy passes in both slits in order to geometrically interfere and this is where things gets very confusing.
But there are 2 aspects you should be aware in the modern thinking, 1) Feynman allowed paths and 2) photon wave function. 1) Feynman attacked the same problem you are attacking, and his eventual proof was that photons needed to travel n times a multiple of their wavelength ... much like the length of a guitar string can only play one note (or frequency) and also much like a laser cavity where if the dimensions are not correct photons will fail to propagate in the desired path. (Note that the Feynman explanation also accounts for the observations in single photon experiments.) 2) as John Renee highlights the photon is delocalized and he even expresses that the photon as a fuzzy sphere, this is the photon wave function described in words. To take the description further we can say the sphere gets larger and larger at the speed of light until the "receiving" atom is found and decides (by probability and QM) that it will take all the energy. At his point the sphere collapses and all the energy proceeds to the "receiving" atom. Maybe one could argue that the fuzzy sphere was one big virtual photon with no energy and that the real photon is where all the energy goes and it takes the best path to the receiving atom, who knows.
Feynman has shown that the photon does not need to go thru 2 slits to have a wave like "interference" property, he has shown that light is a wave because it travels on paths that are harmonic, i.e. the path travelled is dependent on the photon energy/wavelength. The photon wave function (John Rennie) tells us that the photon looks everywhere for a path ... and eventually collapses to a single atom/electron. So in conclusion I would say both answers are correct ...it passes thru 1 slit and both slits!! ... but it is undetectable until the screen.
• I give you a +1 for considering the individual photon derivation of the Fringe pattern. billalsept.com – Bill Alsept Nov 20 '19 at 7:44
• Yes the explanation works for double slit diffraction, single slit diffraction, thin film "interference" (especially with single photons), laser cavities and probably a few more. If the word interference is used then there is likely a "better" explanation using the Feynman theory, i.e. one that keeps energy conserved! – PhysicsDave Nov 20 '19 at 23:49
• Yes it’s all individual photons convoluted to create the fringe pattern. It can even be demonstrated how thousands of individual and coherent photons radiating from a common source resemble a spherical wave. – Bill Alsept Nov 21 '19 at 0:03
First try: We all know that if we just block one slit then it would definitely go through only one of them.
The only thing you can do in order to know that in some way it can be postulated that a photon goes through different slits at a given time is if you unblock the 2nd slit.
If you Detect and gain knowledge about having it gone past the slits then you just caused decoherence and it is not superposed anymore.
You can get more crazy about it but it's not necessary. You will not get more evidence then just the appearance of the interference pattern, it is quite a lot in favor of what you want to prove is occurring in reality :)
P.S:
You don't have to measure.. You can have an arbitrarily long period between every photon emitted, and come back years later to find a scatter plot that is converging to an interference pattern on the screen. Before enough time passes for the mod squared to actually reach the screen it will not interact (with thin air). Once it reaches the screen an interaction will happen in accordance with the running expectation of the mod squared..
• This does not answer the question. – my2cts Nov 16 '19 at 20:28
• from 1 to 10? So should I elaborate or just delete? – user192234 Nov 16 '19 at 20:34
• Just focus on the actual question and make your answer pertinent to it. It's up to you! – my2cts Nov 16 '19 at 20:36
The photon goes through both slits
Keep in mind that this is really only the closest approximation of what happens that we have language for. Nothing can exist at two places at once, and QM doesn't change that.
However, it does some things with probability and uncertainty that we can only really describe as "totally weird".
In my very un-humble opinion, the best way to think of it is as a probability waveform itself travelling - much like any other wave travels. You could for example calculate that it's got 50-50 percent chance of going through either slit. Then after the slits, the probability waves interfere and create the known interference pattern.
If you instead measure the photon, then it's as if you emitted it from that specific location (since you know where the photon is and you have no waveform anymore), therefore creating a simple normal distribution.
Do note that this is NOT more accurate than stating it's at two places at once (as far as I'm aware, at least.). But it's a way to think about it that produces a mental image which is basically as accurate.
If the photon truly goes through both slits (at the same time), then why can't we detect it at both slits (at the same time)?
If you take seriously Feynman's many paths theory (path integral) of QM used in Quantum Field Theory, there is evidence that the particle doesn't just go through the two slits, but it actually takes every possible path. That means part of its "path" includes going around the sun, then Jupiter, and make its way back to earth to make a blip on a screen. Take all other possible paths you can think of, add them up, and that is the particles "path". In that sense, a defined path is a very classical idea.
Instead think of probabilities of being at various locations.
Let's modify your thought experiment and think about what happens if we put 1 detector behind the left slit. As soon as that one detector is added, the interference patterns disappear. We get classical results (perhaps single slit, I am not sure) both in our detector and our original canvas that captures the right side. As soon as we even attempt to detect which slit it goes through, QM reverts to classical results which give concrete answers to which side the particle went through. At that point it didn't go through both, it went through one as a classical particle.
As for why, we don't really know. See here for a recent experiment that tries to answer this. https://phys.org/news/2011-01-which-way-detector-mystery-double-slit.html
• Why do you say we don't really know? We do! A detector (by definition) is simply a device that causes wavefunction collapse, so that we can obtain a classically measurable outcome. So obviously, if you put a detector near a slit, it will collapse the wavefunction. In layman terms, it forces the photon's wavefunction to rapidly evolve to be largely near the detector or largely far from the detector. – user21820 Nov 19 '19 at 7:27
• Feynman's theory goes further to say the shortest path that is a multiple of the wavelength is the viable path with probability = 1. One thing you miss is that many of the paths have probability <1 and many close to zero. Also note that everybody forgets when you observe one slit you also get "interference" i.e. bright and dark areas. – PhysicsDave Nov 20 '19 at 23:45
• @user21820 - what is waveform collapse? Its a word we came up with to describe something we don't fully understand using the best language we have (Quantum Mechanics). There are other interpretations such as quantum decoherence that can work too. – brian h Nov 22 '19 at 17:38
• @PhysicsDave- True I should have added that most are close to 0, however my understanding is that you can have situations where more than 1 is not zero? Also, I was aware of interference in a single slit. However, again my understanding is that if you put a detector in front of 1 slit, it makes the other slit behave classically? – brian h Nov 22 '19 at 17:39
• @brianh: I thought it is obvious. A detector forces the wavefunction of a particle near it to evolve to have either very high density or very low density near the detector. That is 'collapse'. – user21820 Nov 23 '19 at 19:54
Quantum mechanics is a tool for answering questions. You ask it a question by setting up an experiment, and making a measurement. It answers that question, and that question only.
If you set up a light source, two slits, and a screen, and observe the flashes on the screen, then the question you're asking is 'how does the probability of a photon arrival depend on the position on the screen'. Repeat with enough photons and a pattern builds up on the screen.
This setup can tell you nothing about the path of the photon from light source to screen, or whether the photon even exists between them. If you want to investigate the path, then you create a different experiment with screens along the path you think might be involved, and if there are flashes, then you'll have an answer to 'was it here?'. What you won't have is an interference pattern on the screen, because that was a different experiment, without the intermediate screens, a different question.
Why doesn't QM answer where it is at all times? We don't know. We've only been clever enough so far to create a theory that tells you what happens at the measurement. It's a good theory, it works extraordinarily well, at what it works for.
Is there the likelyhood of any deeper theory which can tell you what happens prior to a measurement? I don't know. I'm rather intrigued by Lee Smolin's event based world where time is real, but distance is an emergent phenomenon, which explains entanglement in a rather mind-blowing way.
If the photon truly goes through both slits (at the same time), then why can't we detect it at both slits (at the same time)?
The photon "goes through both slits" is not really a description I am comfortable with. A photon is a quantized potential for causing an effect. Its spatial existence is describable in the form/function of a wave subjected to the double slit setup. This wave function is spatially spread out, but it can only cause a single quantized effect in its domain.
So basically the nature of quantum particles can be described by wave functions but their interactions are discrete: the wave interacts as a whole or not at all.
Any "detection" will rely on an effect, and having an effect uses up the photon.
Photon goes through one slid, its wave function goes through both. | 2020-06-05 07:35:00 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 2, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6318194270133972, "perplexity": 461.7107980401744}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-24/segments/1590348493151.92/warc/CC-MAIN-20200605045722-20200605075722-00140.warc.gz"} |
https://lionelliu.com/2006/sgqyz-4.html | Articles
# 三国群英传4 GeneralDef.dat文件结构分析
9 0 0 0
0 1 2 3 4 5 6 7 8 9 A B C D E F
-----------------------------------------------
0 | 41 41 41 41 00 42 42 42 42 00 00 00 00 00 00 00
| ~~~~~~~~~~~~姓 ~~~~~~~~~~~~~名 ~~登陆时间
1 | 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00
| ~~头像 ~~造型 ~~性别 ~~文武
2 | 63 00 00 00 2C 00 00 00 59 00 00 00 27 00 00 00
| ~~武力 ~~智力 ~~生命 ~~技力
(2 – 1) * 48 + 4 + 0x28 = 0x5C
100 ^ 0xAA = 0xCE | 2018-05-23 14:45:57 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 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": 0.1727350950241089, "perplexity": 236.3785284972186}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-22/segments/1526794865679.51/warc/CC-MAIN-20180523141759-20180523161759-00424.warc.gz"} |
https://pos.sissa.it/397/224/ | Volume 397 - The Ninth Annual Conference on Large Hadron Collider Physics (LHCP2021) - Poster Session
Illuminating long-lived dark vector bosons via exotic Higgs decays at $\sqrt{s} = 13\,{\text {TeV}}$
T. Elkafrawy*, M. Hohlmann, T. Kamon, P. Padley, H. Kim, M. Rahmani and S. Dildick
Full text: pdf
Pre-published on: November 15, 2021
Published on: November 17, 2021
Abstract
The possibility of producing a measurable long-lived dark $Z$ boson, that is assumed to mix kinetically with the hypercharge gauge boson in Higgs decays and to be produced also in Higgs decays through Higgs-to-dark-Higgs mixing, at the Large Hadron Collider (LHC) is investigated. Displaced dimuons in the final state are considered where each of the $Z$ and the dark $Z$ bosons decays directly to a dimuon. The total cross sections for the decay modes of interest as well as the decay widths and decay lengths are calculated to next-to-leading order (NLO) by using Monte Carlo (MC) simulation in the framework of {\textsc{MadGraph5}}\_aMC@NLO and compared to the available analytical calculations to leading order (LO). The sensitivity of the LHC in Runs 2 and 3 to such searches is discussed.
DOI: https://doi.org/10.22323/1.397.0224
How to cite
Metadata are provided both in "article" format (very similar to INSPIRE) as this helps creating very compact bibliographies which can be beneficial to authors and readers, and in "proceeding" format which is more detailed and complete.
Open Access | 2021-12-03 22:57:10 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7518581748008728, "perplexity": 1969.6264212398776}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-49/segments/1637964362919.65/warc/CC-MAIN-20211203212721-20211204002721-00624.warc.gz"} |
https://livingthing.danmackinlay.name/bayesian_consistency.html | # Frequentist consistency of Bayesian methods
### TFW two flawed methods for understanding the world agree with at least each other
Usefulness: 🔧
Novelty: 💡
Uncertainty: 🤪 🤪 🤪
Incompleteness: 🚧 🚧 🚧
You want to use some tasty tool, such as a hierarchical model without anyone getting cross at you for apostasy by doing it in the wrong discipline? Why not use whatever estimator works, and then show that it works on both frequentist and Bayesian grounds?
Shalizi’s overview
There is a basic result here, due to Doob, which essentially says that the Bayesian learner is consistent, except on a set of data of prior probability zero. That is, the Bayesian is subjectively certain they will converge on the truth. This is not as reassuring as one might wish, and showing Bayesian consistency under the true distribution is harder. In fact, it usually involves assumptions under which non-Bayes procedures will also converge. […]
Concentration of the posterior around the truth is only a preliminary. One would also want to know that, say, the posterior mean converges, or even better that the predictive distribution converges. For many finite-dimensional problems, what’s called the “Bernstein-von Mises theorem” basically says that the posterior mean and the maximum likelihood estimate converge, so if one works the other will too. This breaks down for infinite-dimensional problems.
(Bernardo and de Valencia 2006), in the context of Objective Bayes, argues for frequentist methods as necessary.
Bayesian Statistics is typically taught, if at all, after a prior exposure to frequentist statistics. It is argued that it may be appropriate to reverse this procedure. Indeed, the emergence of powerful objective Bayesian methods (where the result, as in frequentist statistics, only depends on the assumed model and the observed data), provides a new unifying perspective on most established methods, and may be used in situations (e.g. hierarchical structures) where frequentist methods cannot. On the other hand, frequentist procedures provide mechanisms to evaluate and calibrate any procedure. Hence, it may be the right time to consider an integrated approach to mathematical statistics, where objective Bayesian methods are first used to provide the building elements, and frequentist methods are then used to provide the necessary evaluation.
# Refs
Aaronson, Scott. 2005. “The Complexity of Agreement.” In Proceedings of the Thirty-Seventh Annual ACM Symposium on Theory of Computing, 634. ACM Press. https://doi.org/10.1145/1060590.1060686.
Advani, Madhu, and Surya Ganguli. 2016. “An Equivalence Between High Dimensional Bayes Optimal Inference and M-Estimation.” In Advances in Neural Information Processing Systems. http://arxiv.org/abs/1609.07060.
Aumann, Robert J. 1976. “Agreeing to Disagree.” The Annals of Statistics 4 (6): 1236–9. http://www.math.huji.ac.il/raumann/pdf/Agreeing%20to%20Disagree.pdf.
Bayarri, M. J., and J. O. Berger. 2004. “The Interplay of Bayesian and Frequentist Analysis.” Statistical Science 19 (1): 58–80. https://doi.org/10.1214/088342304000000116.
Bernardo, Jose M, and Universitat de Valencia. 2006. “A Bayesian Mathematical Statistics Primer,” 6.
Diaconis, Persi, and David Freedman. 1986. “On the Consistency of Bayes Estimates.” The Annals of Statistics 14 (1): 1–26. http://www.jstor.org/stable/2241255.
Doob, J. L. 1949. “Application of the Theory of Martingales.” In Le Calcul Des Probabilités et Ses Applications, 23–27. Colloques Internationaux Du Centre National de La Recherche Scientifique, No. 13. Centre National de la Recherche Scientifique, Paris. http://www.ams.org/mathscinet-getitem?mr=0033460.
Efron, Bradley. 2012. “Bayesian Inference and the Parametric Bootstrap.” The Annals of Applied Statistics 6 (4): 1971–97. https://doi.org/10.1214/12-AOAS571.
———. 2015. “Frequentist Accuracy of Bayesian Estimates.” Journal of the Royal Statistical Society: Series B (Statistical Methodology) 77 (3): 617–46. https://doi.org/10.1111/rssb.12080.
Fong, Edwin, and Chris Holmes. 2019. “On the Marginal Likelihood and Cross-Validation,” May. http://arxiv.org/abs/1905.08737.
Freedman, David. 1999. “Wald Lecture: On the Bernstein-von Mises Theorem with Infinite-Dimensional Parameters.” The Annals of Statistics 27 (4): 1119–41. https://doi.org/10.1214/aos/1017938917.
Gelman, Andrew, Aleks Jakulin, Maria Grazia Pittau, and Yu-Sung Su. 2008. “A Weakly Informative Default Prior Distribution for Logistic and Other Regression Models.” The Annals of Applied Statistics 2 (4): 1360–83. https://doi.org/10.1214/08-AOAS191.
Lele, S. R., B. Dennis, and F. Lutscher. 2007. “Data Cloning: Easy Maximum Likelihood Estimation for Complex Ecological Models Using Bayesian Markov Chain Monte Carlo Methods.” Ecology Letters 10 (7): 551. https://doi.org/10.1111/j.1461-0248.2007.01047.x.
Lele, Subhash R., Khurram Nadeem, and Byron Schmuland. 2010. “Estimability and Likelihood Inference for Generalized Linear Mixed Models Using Data Cloning.” Journal of the American Statistical Association 105 (492): 1617–25. https://doi.org/10.1198/jasa.2010.tm09757.
Nickl, Richard. 2014. “Discussion of: ‘Frequentist Coverage of Adaptive Nonparametric Bayesian Credible Sets’,” October. http://arxiv.org/abs/1410.7600.
Norton, Robert M. 1984. “The Double Exponential Distribution: Using Calculus to Find a Maximum Likelihood Estimator.” The American Statistician 38 (2): 135–36. https://doi.org/10.1080/00031305.1984.10483185.
Rousseau, Judith. 2016. “On the Frequentist Properties of Bayesian Nonparametric Methods.” Annual Review of Statistics and Its Application 3 (1): 211–31. https://doi.org/10.1146/annurev-statistics-041715-033523.
Shalizi, Cosma Rohilla. 2009. “Dynamics of Bayesian Updating with Dependent Data and Misspecified Models.” Electronic Journal of Statistics 3: 1039–74. https://doi.org/10.1214/09-EJS485.
Sims, C. 2010. “Understanding Non-Bayesians.” Unpublished Chapter, Department of Economics, Princeton University. http://sims.princeton.edu/yftp/UndrstndgNnBsns/GewekeBookChpter.pdf.
Szabó, Botond, Aad van der Vaart, and Harry van Zanten. 2013. “Frequentist Coverage of Adaptive Nonparametric Bayesian Credible Sets,” October. http://arxiv.org/abs/1310.4489.
Tibshirani, Robert. 1996. “Regression Shrinkage and Selection via the Lasso.” Journal of the Royal Statistical Society. Series B (Methodological) 58 (1): 267–88. http://statweb.stanford.edu/~tibs/lasso/lasso.pdf.
Valpine, Perry de. 2011. “Frequentist Analysis of Hierarchical Models for Population Dynamics and Demographic Data.” Journal of Ornithology 152 (2): 393–408. https://doi.org/10.1007/s10336-010-0642-5.
Wang, Yixin, and David M. Blei. 2017. “Frequentist Consistency of Variational Bayes,” May. http://arxiv.org/abs/1705.03439. | 2019-10-23 22:00:21 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8753877282142639, "perplexity": 4395.311078471259}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-43/segments/1570987836295.98/warc/CC-MAIN-20191023201520-20191023225020-00161.warc.gz"} |
http://mathoverflow.net/questions/81569/example-of-a-morphism-between-exterior-algebras-that-is-mathbbz-2-graded-bu?sort=votes | Example of a morphism between exterior algebras that is $\mathbb{Z}_2$ graded but not $\mathbb{Z}$ graded??
The title pretty much states my problem. I consider only finitely generated exterior algebras $\bigwedge V$. It is known that any morphism between exterior algebras y determined by its action on generators, i.e. its action on $V$. Does anyone know a good example of this kind of morphisms?
By $\mathbb{Z}_2$ graded I mean a morphism of algebras such that the parity of the degree of a form $\eta$ is preserved by such a morphism; and by $\mathbb{Z}$ graded I mean a morphism that preserves the grade of $\eta$, i.e. if $\eta$ is a $k$-form, then so is $f(\eta)$ where $f$ is the morphism in question. Thanks in advance.
-
Map the elements of a basis of $V$ to elements in the highest non-zero odd exterior power of $V$. – Mariano Suárez-Alvarez Nov 22 '11 at 2:53
There doesn't seem to be much more to say, so I'll just repeat Mariano's comment with an example. Let $V$ be a 3-dimensional vector space in degree 1, and consider any nonzero linear map from $V$ to $\wedge^3 V$. This induces an algebra endomorphism of $\bigwedge^\bullet V$ that preserves the $\mathbb{Z}/2\mathbb{Z}$-grading but not the $\mathbb{Z}$-grading. | 2014-09-23 14:49:22 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9866624474525452, "perplexity": 127.15034975512626}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-41/segments/1410657138980.37/warc/CC-MAIN-20140914011218-00035-ip-10-234-18-248.ec2.internal.warc.gz"} |
http://www.physicsforums.com/showpost.php?p=4274420&postcount=1 | View Single Post
## finding energy via integration
Hello, I havnt done physics in quite a while and I just want to ask a question about basic energy that i know how to deal with in algebraic terms, but not through means of calculus. I also don't really get the theory of the equation $E=FD$, where E=energy, F=force, and D=distance
is the F the force it took to move the body distance D? So, for example, applying a force of 20N on a body moves it 4m, $E=FD=(20N)(4m)=80J$?
if thats the case then how can we express this with derivatives or integrals?
PhysOrg.com physics news on PhysOrg.com >> A quantum simulator for magnetic materials>> Atomic-scale investigations solve key puzzle of LED efficiency>> Error sought & found: State-of-the-art measurement technique optimised | 2013-05-25 01:46:23 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5830432772636414, "perplexity": 1220.7025604915711}, "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-2013-20/segments/1368705310619/warc/CC-MAIN-20130516115510-00022-ip-10-60-113-184.ec2.internal.warc.gz"} |
http://coalitiontheory.net/content/endogenous-network-formation-tullock-contest | # Endogenous network formation in a Tullock contest
Article
Author/s:
G. Grandjean, D. Tellone, W. Vergote
Mathematical Social Sciences
Issue number:
November 2016
Publisher:
Elsevier
Year:
2016
We propose a model of network formation in a Tullock contest. Agents first form their partnerships and then choose their investment in the contest. While a link improves the strength of an agent, it also improves the position of her rival. It is thus not obvious that they decide to cooperate. We characterize all pairwise equilibrium networks and find that the network formation process can act as a barrier to entry to the contest. We then analyze the impact of network formation on total surplus and find that a social planner can increase total surplus by creating more asymmetry between agents, as long as this does not reduce the number of participating agents. We show that barriers to entry may either hurt total surplus, as the winner of the prize does not exploit all the possible network benefits, or improve total surplus since less rent is dissipated when competition becomes less fierce. Finally, when networking acts as an endogenous barrier to entry, no pairwise equilibrium network is efficient.
Developed by Paolo Gittoi | 2021-09-18 11:01:05 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8501964807510376, "perplexity": 1808.7874607620006}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-39/segments/1631780056392.79/warc/CC-MAIN-20210918093220-20210918123220-00578.warc.gz"} |
http://hal.in2p3.fr/in2p3-00003697 | # A Search for Leptoquark Bosons and Lepton Flavor Violation in $e^{+} p$ Collisions at HERA
Abstract : A search for new bosons possessing couplings to lepton-quark pairs is performed in the H1 experiment at HERA using 1994 to 1997 data corresponding to an integrated luminosity of 37pb^-1. First generation leptoquarks (LQs) are searched in very high Q^2 neutral (NC) and charged (CC) current data samples. The measurements are compared to Standard Model (SM) expectations from deep-inelastic scattering (DIS). A deviation in the Q^2 spectrum previously observed in the 1994 to 1996 dataset at Q^2 \gsim 15000GeV^2 remains, though with less significance. This deviation corresponded to a clustering in the invariant mass spectrum at M \simeq 200 GeV which is not observed with the 1997 dataset alone. The NC DIS data is used to constrain the Yukawa couplings lambda of first generation scalar and vector LQs in the Buchmueller-Rueckl-Wyler effective model. Scalar LQs are excluded for masses up to 275GeV for a coupling of electromagnetic strength, lambda=0.3. A sensitivity to coupling values \lsim 1 is established for masses up to 400GeV for any LQ type. The NC and CC DIS data are combined to constrain lambda for arbitrary branching ratios of the LQ into eq in a generic model. For a decay branching ratio into e^+ u pairs as small as 10%, LQ masses up to 260 GeV are ruled out for lambda=0.3. LQs possessing couplings to mixed fermion generations, which could lead to signals of lepton flavor violation (LFV), are searched in events with a high transverse momentum mu or tau. No mu+X or tau+X event candidate is found that is compatible with LQ kinematics. Constraints are set on the Yukawa coupling involving the mu and tau lepton in a yet unexplored mass range.
Document type :
Journal articles
Cited literature [17 references]
http://hal.in2p3.fr/in2p3-00003697
Contributor : Magali Damoiseaux <>
Submitted on : Thursday, January 27, 2000 - 5:27:37 PM
Last modification on : Wednesday, April 7, 2021 - 3:26:01 PM
Long-term archiving on: : Friday, May 29, 2015 - 4:31:50 PM
### Identifiers
• HAL Id : in2p3-00003697, version 1
### Citation
C. Adloff, V. Andreev, B. Andrieu, V. Arkadov, A. Astvatsatourov, et al.. A Search for Leptoquark Bosons and Lepton Flavor Violation in $e^{+} p$ Collisions at HERA. European Physical Journal C: Particles and Fields, Springer Verlag (Germany), 1999, 11, pp.447-471. ⟨in2p3-00003697⟩
Record views | 2021-05-18 23:56:13 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.60134357213974, "perplexity": 5208.157635662353}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-21/segments/1620243989874.84/warc/CC-MAIN-20210518222121-20210519012121-00055.warc.gz"} |
https://thoughtstreams.io/glyph/php/2668/ | # PHP
15 thoughts
last posted Oct. 23, 2013, 6:22 a.m.
2 earlier thoughts
0
What I'm thinking about today though, is whether and how it is OK to communicate about PHP's badness. (And, perhaps by proxy, bad software technology in general, as a person in the software field.)
12 later thoughts | 2020-05-29 17:49:20 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8774107098579407, "perplexity": 6714.891076252642}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-24/segments/1590347405558.19/warc/CC-MAIN-20200529152159-20200529182159-00162.warc.gz"} |
https://cseducators.stackexchange.com/questions/5320/code-highlight-for-classroom-use | Code highlight for classroom use
I am thinking about "dissecting" my code with students, going line by line and commenting (why is this the condition of the while? What does this piece of code do?)
A simple example can be found here: https://docs.google.com/document/d/1mxTE_E2_YKq-yhaXAfBxYP11uaRm7bSXTnP55bMnPyQ/edit
However, I would like a better tool for the job. As you can see, in docs, the comments are far away from the code and not linked to their respective by any visual cue. They are also all on the same color.
I tryed some highlighting tools for docs, but they seem to just highlight,not add colored comments
I am looking for ideas, inside or outside google docs, to do the job.
For inspiration (no way am I ever doing something this pretty!) here is an image from a headfirst book
• How do you intend to use the output from the "tool"? Is the objective to use it in a visual interactive presentation (such as projected during a lecture), or in a static context to which the students have current and future access (such as shared docs in Google Docs), or even a printout handed to the students prior to or during the presentation? – Gypsy Spellweaver Dec 23 '18 at 0:43
I would suggest two things. Both of them, however, require that you give them printouts of the code so that they can make notations of their own without having to use any tool. Just write on the paper. Also, I'm assuming that your code is in small pieces, say, short methods. For me, a method is long at about five lines and a method that covers a page is incomprehensible. I have anecdotal evidence from industry that the latter is also true.
One method, assuming that you have given students the code is to just project it from your normal IDE and highlight it with the mouse. Ask the students about it and talk about it. Asking is probably more effective. You don't need to write anything.
Another uses a tool like Powerpoint. Build a deck in which each page shows exactly the same code but with different sections highlighted as you desire, using one or more colors. You can flip between pages as you talk, flipping both forward and backward as needed. But, again, ask the students to explain, rather than giving your own explanations. This way you can pick up misconceptions they might have and correct them. Make sure that the code itself has the same alignment on each page. Then when you flip, only the highlighting changes and students don't need to readjust the context.
Perhaps you can project against a writable surface rather than a screen. If you project on a whiteboard (and can ignore the glare) you can use marker pens to annotate - especially when the students say the right thing.
Make it interactive. Broadcast is much less effective.
Also, since you are putting effort in anyway, make sure that you always use intention revealing names in your demo code. Don't use x, y, z, i, j, k. Make the names come from the problem being solved, not the solution. For example counter is a name from the solution. But negative_value_counter is a name from the problem (of counting negative values - duh). In this way, if you are primarily teaching them about how to structure programs, you can focus on the structure (if, while, ...) rather than having them wonder about the purpose of abstract names.
I'm taking a very basic approach when I do it: I use my Editor and comment line by line. This way I can directly show alternatives and effects if something does not work as planned. Usually I'm developing the code live to avoid that students get distracted by the code to come. The good thing about this approach is, you can show them how you can develop a program from scratch and how you are digging into the finer aspects of problem solving. E.g. you can "forget" one condition in a loop and create an endless loop - and show them how you can find the mistake.
Another thing I did which comes closer to your UML example: I'm using a surface tablet in my class and sometimes I'm just taking a quick screenshot and use the pen to write on this screenshot. This works nicely and does not need additional tools (once you are having a tablet with pen).
• I disagree about line by line comments. It is, in particular, a terrible coding practice. If the code doesn't speak for itself it is broken. If the comments and code don't agree, they are probably both wrong. If you do this as an active practice when demonstrating, I fear that the students will want to take it up themselves. In professional circles using high level languages (not assembler) inline comments are considered to be a code smell. Don't encourage it. The solution is (a) better names and (b) shorter functions/methods. Then all you need to comment is the intent using, say, javadoc. – Buffy Dec 22 '18 at 19:55
• @Buffy I agree if the students are a bit more advanced. But for beginners it is helpful to explain basic concepts again. E.g. "why is this a pointer" or "why do I have to check for a certain condition". Of course, speaking names are a must! – OBu Dec 22 '18 at 19:59
• You can do that without inline comments. Don't teach them to do things that you don't want them to do themselves. A hand written comment on an overlay isn't the same thing as an inline comment. "But teacher does it. Why shouldn't I?" – Buffy Dec 22 '18 at 20:03
• Or use pair learning and have one student explain verbally the code of another as it is being written. – Buffy Dec 22 '18 at 20:05
• Always try to organize your teaching so that you never have to unteach anything. Don't lead them to bad habits of work or thought that you will have to tell them later not to do that. You can be incomplete in explanations, metaphors, etc, but don't take them down the wrong path thinking that you will show them the better path later. It is a bit harder to always consider that, but necessary IMO – Buffy Dec 22 '18 at 20:13
You need to know about http://pythontutor.com/ - they have line-by-line execution and a diagram showing memory and how it changes. They also have more than just Python, they have Java, C, C++, JavaScript, & Ruby
I have used this resource in my classes and it makes it so amazing to explain how memory, references, and classes work. The C tutor really helps with pointers as well.
• Welcome to Computer Science Educators! Nice first contribution, and I hope we see more of you around :). Once you have the rep, feel free to visit the chat room and introduce yourself. (It's been slower going in the chatroom lately, but there's almost always one of the site regulars around, and you can type @ and any recently posting user (such as @buffy, @gypsyspellweaver, or @beni) to ping them and get a greeting :) – Ben I. Dec 28 '18 at 19:06
I create code-comprehension sheets outlined in one of two ways (depending on the code): code on the left, with questions on the right, or code with questions interspersed like comments.
There are a few uses for such sheets:
1. Check comprehension of what we've just gone over.
2. Force students to try to examine and digest code prior to going over it together (to increase interest/retention when we go over the material.)
3. Assessments.
Typically I will do this with 1-3 pages of code that has to be understood on several different levels in order to make sense of it. Think objects, or assembly code. | 2020-03-29 17:32:25 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.3997010886669159, "perplexity": 918.3199920252889}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-16/segments/1585370495413.19/warc/CC-MAIN-20200329171027-20200329201027-00306.warc.gz"} |
https://www.zora.uzh.ch/id/eprint/160277/ | # Studies of the resonance structure in $D^0 \to K^\mp \pi^\pm \pi^\pm \pi^\mp$ decays
LHCb Collaboration; Bernet, R; Müller, K; Serra, N; Steinkamp, O; Straumann, U; Vollhardt, A; et al (2018). Studies of the resonance structure in $D^0 \to K^\mp \pi^\pm \pi^\pm \pi^\mp$ decays. European Physical Journal C - Particles and Fields, C78(6):443.
## Abstract
Amplitude models are constructed to describe the resonance structure of D0→K−π+π+π− and D0→K+π−π−π+ decays using pp collision data collected at centre-of-mass energies of 7 and 8 TeV with the LHCb experiment, corresponding to an integrated luminosity of 3.0 fb−1. The largest contributions to both decay amplitudes are found to come from axial resonances, with decay modes D0→a1(1260)+K− and D0→K1(1270/1400)+π− being prominent in D0→K−π+π+π− and D0→K+π−π−π+, respectively. Precise measurements of the lineshape parameters and couplings of the a1(1260)+, K1(1270)− and K(1460)− resonances are made, and a quasi model-independent study of the K(1460)− resonance is performed. The coherence factor of the decays is calculated from the amplitude models to be RK3π=0.459±0.010(stat)±0.012(syst)±0.020(model), which is consistent with direct measurements. These models will be useful in future measurements of the unitary-triangle angle γ and studies of charm mixing and CP violation.
## Abstract
Amplitude models are constructed to describe the resonance structure of D0→K−π+π+π− and D0→K+π−π−π+ decays using pp collision data collected at centre-of-mass energies of 7 and 8 TeV with the LHCb experiment, corresponding to an integrated luminosity of 3.0 fb−1. The largest contributions to both decay amplitudes are found to come from axial resonances, with decay modes D0→a1(1260)+K− and D0→K1(1270/1400)+π− being prominent in D0→K−π+π+π− and D0→K+π−π−π+, respectively. Precise measurements of the lineshape parameters and couplings of the a1(1260)+, K1(1270)− and K(1460)− resonances are made, and a quasi model-independent study of the K(1460)− resonance is performed. The coherence factor of the decays is calculated from the amplitude models to be RK3π=0.459±0.010(stat)±0.012(syst)±0.020(model), which is consistent with direct measurements. These models will be useful in future measurements of the unitary-triangle angle γ and studies of charm mixing and CP violation.
## Statistics
### Citations
Dimensions.ai Metrics
### Downloads
1 download since deposited on 07 Mar 2019
1 download since 12 months
Detailed statistics
## Additional indexing
Item Type: Journal Article, refereed, original work 07 Faculty of Science > Physics Institute 530 Physics English 2018 07 Mar 2019 13:26 07 Mar 2019 13:29 Springer 1434-6044 Gold Publisher DOI. An embargo period may apply. https://doi.org/10.1140/epjc/s10052-018-5758-4
## Download
Preview
Content: Published Version
Filetype: PDF
Size: 1MB
View at publisher
Licence: | 2019-07-24 09:26:22 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8719323873519897, "perplexity": 5374.614639382817}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-30/segments/1563195532251.99/warc/CC-MAIN-20190724082321-20190724104321-00009.warc.gz"} |
https://tex.stackexchange.com/questions/245652/seeing-korean-characters-in-the-editor | Seeing Korean characters in the editor
I use a tex template that includes Korean characters. When I compile it, the PDF document displays the Korean characters correctly. Yet, the Texshop text editor displays some rubbish characters instead. And when I actually type the Korean characters, after the compilation the PDF displays "?" instead of the characters. How do I get the formatting right?
This is really impossible to answer without seeing a minimal working example (MWE), but I'm pretty sure you're dealing with the following problem:
The .tex file you're compiling loads CJK which assumes the EUC-KR encoding for Korean text. Your text editor, however, automatically loads the file with UTF-8 encoding, as most modern text editors will do.
If the characters were entered into the .tex file with the EUC-KR encoding, you will see question marks � and various outlandish characters if you open the file with UTF-8 encoding. The file will still compile just fine, since CJK correctly reads the EUC-KR encoded characters and produces a .pdf with Korean characters.
If you've opened the file with the UTF-8 encoding and start typing Korean characters into the file, CJK can't interpret the input correctly, since it's expecting characters with the EUC-KR encoding. So pdflatex will probably throw an error, or it will produce a .pdf with garbage characters instead of Korean characters.
Your actual question "How do I get the formatting right?" is very vague, but if this means that you need to stick to your existing template, then make sure you switch the encoding of your file from UTF-8 to EUC-KR before you enter any new Korean characters into the file, and also to be able to read the existing characters in your document.
This is all only guesswork, since it is impossible to know for sure what problem you're having or what's causing it when you're not providing a MWE. Please always do so in the future.
Here's an example. The following document was typed with EUC-KR encoding, but below I've opened it with the standard UTF-8 encoding, and it looks like this:
\documentclass{article}
\usepackage{CJK}
\begin{document}
\begin{CJK}[HL]{KS}{mj}
�ȳ��ϼ���
\end{CJK}
\end{document}
If I compile this, it nevertheless correctly produces this output:
If I instead load the file with EUC-KR encoding, I see this:
\documentclass{article}
\usepackage{CJK}
\begin{document}
\begin{CJK}[HL]{KS}{mj}
안녕하세요
\end{CJK}
\end{document}
pdflatex
Using this engine, you can still adopt this solution posted by Medina (remember to set babel to English so the Table of Contents, etc, will be in English).
According to this answer, however, you need the uhc font package installed in order to be able to use mj.
xelatex
If you're already using xelatex (or if you can switch for one document), it's easier since you don't need to enclose Korean in its own environment. However you need to choose a font, otherwise the document typesets without errors but you won't see any glyphs (except for the latin ones). Try commenting out the command \setmainfont{} then compiling, and you'll see what I mean.
The package fontspec must be loaded in order for the command above to work. This allows you to set a font. If you load xeCJK, you can also use the command \setCJKmainfont{} to set a font for your language. Here's an example with Nanum Myeongjo Regular:
\documentclass{article}
\usepackage{fontspec}
\usepackage{xeCJK}
\setmainfont{Century Gothic}
\setCJKmainfont{NanumMyeongjo}
\begin{document}
\centering\Huge\noindent
This is some text in Latin script. And now some Korean:
이 FAQ 은 자주 반복되는 질문과 그에 대한 대답을 간단명료한 양식으로 모아 엮어졌습니다.
\end{document} | 2019-08-24 16:01:00 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8715330362319946, "perplexity": 2736.777125146516}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-35/segments/1566027321160.93/warc/CC-MAIN-20190824152236-20190824174236-00046.warc.gz"} |
https://hamrocsit.com/question/3764/ | Recent studies indicates that the average surface temperature of the earth has been rising rapidly. Some scientists have modeled the temperature by the linear function T = 0.03t + 8.50, where T is temperature in degree centigrade and t represents years since 1900. What do the slope and T-intercept represent? Use the equation to predict the average global surface temperature in 2100
Solution:
Given Linear function is
T = 0.02t + 8.50
a) Comparing it with y = mx + c, we get
m = 0.02 and c = 8.50
So, the slope is 0.02 and T-intercept is 8.50
b) Since t = 0 represent years since 1900, then t = 2100 – 1900 = 200 represents 2100.
So, T = 0.02 (200) + 8.5
T = 12.5oc
Therefore, The average global surface temperature in 2100 is 12.5oc | 2022-12-07 11:12:43 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8175004720687866, "perplexity": 975.3233390017105}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446711151.22/warc/CC-MAIN-20221207085208-20221207115208-00276.warc.gz"} |
http://mathhelpforum.com/calculus/134435-cauchys-integral-theorem-print.html | # Cauchy's Integral theorem
• March 18th 2010, 06:47 AM
Tekken
Cauchy's Integral theorem
Quick question
im trying to calculate the integral around the closed curve C of
z(bar) dz where z(bar) is the complex conjugate of z
where C is the unit circle.
Using the subsitution z = e^it i calculated the integral to be 0.
=> z(bar) = e^-it
However after finding the question in a maths book the answer is given to be 2pi(i).
Can any of ye tell me which solution is correct
• March 18th 2010, 07:01 AM
shawsend
$\mathop\oint\limits_{z=e^{it}}\overline{z}dz=\int_ 0^{2\pi}e^{-it}ie^{it}=2\pi i$ | 2014-12-19 12:35:54 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 1, "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": 0.9875559210777283, "perplexity": 2272.1639292006416}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-52/segments/1418802768441.42/warc/CC-MAIN-20141217075248-00062-ip-10-231-17-201.ec2.internal.warc.gz"} |
http://mathinsight.org/chain_rule_multivariable_examples | # Math Insight
### Multivariable chain rule examples
The following are examples of using the multivariable chain rule. For examples involving the one-variable chain rule, see simple examples of using the chain rule or the chain rule from the Calculus Refresher.
#### Example 1
Let $\vc{g}: \R \rightarrow \R^2$ and $f: \R^2 \rightarrow \R$ (confused?) be defined by \begin{align*} \vc{g}(t) &= (t^3, t^4)\\ f(x,y) &= x^2y. \end{align*} (You can think of this as the mountain climbing example where $f(x,y)$ is height of mountain at point $(x,y)$ and the path $\vc{g}(t)$ gives your position at time $t$.) Let $h(t)$ be the composition of $f$ with $\vc{g}$ (which would give your height at time $t$): \begin{align*} h(t) = (f \circ \vc{g}) (t) = f(\vc{g}(t)). \end{align*} Calculate the derivative $\displaystyle h'(t) = \diff{h}{t}(t)$ (i.e., the change in height) via the chain rule.
Solution A: We'll use the formula using matrices of partial derivatives: \begin{align*} D{h}(t) = Df(\vc{g}(t)) D{\vc{g}}(t). \end{align*}
We calculate the matrices of partial derivatives of $f$ and $\vc{g}$. \begin{align*} Df(x,y) &= \left[ \pdiff{f}{x}(x,y) \quad \pdiff{f}{y}(x,y) \right]\\ &=\left[ \begin{array}{cc} 2xy & x^2 \end{array} \right]\\ D\vc{g}(t) &= \left[ \begin{array}{c} g_1'(t) \\g_2'(t) \end{array} \right] = \left[ \begin{array}{c} 3t^2\\4t^3 \end{array} \right] \end{align*} We need to evaluate $Df$ at the point $\vc{g}(t)$: \begin{align*} Df(\vc{g}(t)) =Df(t^3, t^4)= \left[ \begin{array}{cc} 2(t^3)(t^4) & (t^3)^2 \end{array} \right] = \left[ \begin{array}{cc} 2t^7 & t^6 \end{array} \right] \end{align*} The derivative of $h$ is \begin{align*} h'(t)=\diff{h}{t}(t) = Dh(t) &= Df(\vc{g}(t)) D\vc{g}(t)\\ &= \left[ \begin{array}{cc} 2t^7 & t^6 \end{array} \right] \left[ \begin{array}{c} 3t^2\\4t^3 \end{array} \right]\\ &= (2t^7)(3t^2) + (t^6)(4t^3)= 6 t^9 + 4 t^9\\ &= 10 t^9 \end{align*}
Solution B: We'll start immediately with the formula in component form: \begin{align*} \diff{h}{t}(t) = \pdiff{f}{x}(\vc{g}(t)) \diff{g_1}{t} (t) + \pdiff{f}{y}(\vc{g}(t))\diff{g_2}{t} (t). \end{align*} We calculate \begin{align*} \pdiff{f}{x}(x,y) &= 2xy\\ \pdiff{f}{y}(x,y) &= x^2\\ \pdiff{f}{x}(\vc{g}(t)) &= \pdiff{f}{x}(t^3,t^4) = 2(t^3)(t^4) = 2t^7\\ \pdiff{f}{y}(\vc{g}(t)) &= \pdiff{f}{y}(t^3,t^4) = (t^3)^2 = t^6\\ \diff{g_1}{t}(t) &= 3t^2\\ \diff{g_2}{t}(t) &= 4t^3. \end{align*} Therefore, \begin{align*} \diff{h}{t}(t) = (2t^7)(3t^2) + (t^6)(4t^3) =6 t^9 + 4 t^9 = 10 t^9. \end{align*}
#### Example 1'
Verify the chain rule for example 1 by calculating an expression for $h(t)$ and then differentiating it to obtain $\displaystyle \diff{h}{t}(t)$.
Solution: $h(t) = f(\vc{g}(t)) = f(t^3,t^4) = (t^3)^2(t^4) = t^{10}$. \begin{align*} h'(t) = \diff{h}{t}(t) = 10t^9, \end{align*} which matches the solution to Example 1, verifying that the chain rule got the correct answer.
For this simple example, doing it without the chain rule was a lot easier. However, that is not always the case. And, in the next example, the only way to obtain the answer is to use the chain rule.
#### Example 2
We continue the mountain climbing example of Example 1. But now, let's say we don't know the terrain ahead of time. This means we do not yet know the height $f(x,y)$ at the position $(x,y)$. We do, however, know our path through mountain; as before, it is given by $\vc{g}(t) = (t^3, t^4).$
Calculate the change in height that you'll experience along the path, i.e., calculate the derivative of $h(t) = f(\vc{g}(t))$. In this case, since we don't know $f$, the answer will be given in terms of the function $f(x,y)$.
Solution: We'll just copy solution A, above. This time, though, we must leave the matrix of partial derivatives of $f$ as \begin{align*} Df(x,y)= \left[ \begin{array}{cc} \displaystyle \pdiff{f}{x}(x,y) & \displaystyle \pdiff{f}{y}(x,y) \end{array} \right] \end{align*} since we don't know what $f(x,y)$ is. We can substitute in the values along the path $\vc{g}(t)$: \begin{align*} Df(\vc{g}(t)) = Df(t^3,t^4)= \left[ \begin{array}{cc} \displaystyle \pdiff{f}{x}(t^3,t^4) & \displaystyle \pdiff{f}{y}(t^3,t^4) \end{array} \right]. \end{align*} Since $D\vc{g}(t)$ is the same as in solution A, above, we calculate the derivative of $h$ as \begin{align*} h'(t) = Dh(t) &= Df(\vc{g}(t)) D\vc{g}(t)\\ &= \left[ \begin{array}{cc} \displaystyle \pdiff{f}{x}(t^3,t^4) & \displaystyle \pdiff{f}{y}(t^3,t^4) \end{array} \right] \left[ \begin{array}{c} 3t^2\\4t^3 \end{array} \right]\\ &=3t^2\pdiff{f}{x}(t^3,t^4) +4t^3\pdiff{f}{y}(t^3,t^4). \end{align*} We leave the answer in this form. Of course, as soon as we know what $f(x,y)$ is, we can simply compute its partial derivatives and plug the result into this formula.
#### Example 3
We continue using the same function $f(x,y) = x^2y$ to describe the height of the mountain at position $(x,y)$. We embellish the above examples by letting $g: \R^2 \to \R^2$ be defined by $\vc{g}(s,t) = (t-s^2,ts^2)$. (We could think of having many paths through the mountain that depend on a skill level $s$. Then, $(x,y)=\vc{g}(s,t)$ could be the position of a person at time $t$ with skill level $s$.)
Compute $\pdiff{}{s} (f \circ \vc{g})(s,t)$ and $\pdiff{}{t} (f \circ \vc{g})(s,t)$, i.e., the partial derivatives with respect to $s$ and $t$ of the height of a person in the mountains whose position is given by $\vc{g}(s,t)$.
Solution: Let $h(s,t) = (f \circ \vc{g})(s,t) = f(\vc{g}(s,t))$. We need to calculate $\displaystyle \pdiff{h}{s}(s,t)$ and $\displaystyle \pdiff{h}{t}(s,t)$. The chain rule says that \begin{align*} Dh(s,t) = D(f\circ \vc{g})(s,t) = Df(\vc{g}(s,t)) D\vc{g}(s,t). \end{align*} Since \begin{align*} Dh(s,t) = \left[ \pdiff{h}{s}(s,t) \quad \pdiff{h}{t}(s,t)\right], \end{align*} the answers we want are just the two components of $Dh(s,t)$. We just need to calculate the matrices $Df(\vc{g}(s,t))$ and $D\vc{g}(s,t)$, then multiply them together.
To make it easier in case you have to do such a problem again, we'll perform the matrix multiplication before writing in the specific values for $f(x,y)$ and $\vc{g}(s,t)$. Then, we'll end up with the chain rule written in component form, which may be easier to use.
The function $f(x,y)$ hasn't changed, so its matrix of partial derivatives is \begin{align*} Df(x,y)= \left[ \begin{array}{cc} \displaystyle \pdiff{f}{x}(x,y) & \displaystyle \pdiff{f}{y}(x,y) \end{array} \right]. \end{align*} For the chain rule, we need this evaluated at $(x,y)=\vc{g}(s,t)$ \begin{align*} Df(\vc{g}(s,t))= \left[ \begin{array}{cc} \displaystyle \pdiff{f}{x}(\vc{g}(s,t)) & \displaystyle \pdiff{f}{y}(\vc{g}(s,t)) \end{array} \right]. \end{align*} Since $\vc{g}: \R^2 \to \R^2$, its matrix of partial derivatives is a $2 \times 2$ matrix. If we denote its components as $\vc{g}(s,t) = (g_1(s,t), g_2(s,t))$, its matrix of partial derivatives is \begin{align*} D\vc{g}(s,t)= \left[ \begin{array}{cc} \displaystyle\pdiff{g_1}{s}(s,t)& \displaystyle\pdiff{g_1}{t}(s,t)\\ \displaystyle\pdiff{g_2}{s}(s,t)& \displaystyle\pdiff{g_2}{t}(s,t) \end{array} \right]. \end{align*} The chain rule $Dh(s,t) = Df(\vc{g}(s,t)) D\vc{g}(s,t)$ becomes \begin{align*} \left[ \begin{array}{cc} \displaystyle \pdiff{h}{s}(s,t) & \displaystyle \pdiff{h}{t} (s,t) \end{array} \right] = \left[ \begin{array}{cc} \displaystyle \pdiff{f}{x}\bigl(\vc{g}(s,t)\bigr) & \displaystyle \pdiff{f}{y}\bigl(\vc{g}(s,t)\bigr) \end{array} \right] \left[ \begin{array}{cc} \displaystyle \pdiff{g_1}{s}(s,t) & \displaystyle \pdiff{g_1}{t}(s,t) \\ \displaystyle \pdiff{g_2}{s}(s,t) & \displaystyle \pdiff{g_2}{t}(s,t) \end{array} \right] \end{align*} We can compute the matrix product on the right-hand side; the result is a $1 \times 2$ matrix (i.e., the same size of $Dh(s,t)$). We obtain one equation by matching the first component of $Dh(s,t)$ with the first component of this multiplied-out matrix. We obtain a second equation by matching the second component of $Dh(s,t)$ with the second component of this multiplied-out matrix. The resulting two equations are \begin{align*} \pdiff{h}{s}(s,t) &= \pdiff{f}{x}(\vc{g}(s,t))\pdiff{g_1}{s}(s,t) + \pdiff{f}{y}(\vc{g}(s,t))\pdiff{g_2}{s}(s,t)\\ \pdiff{h}{t}(s,t) &= \pdiff{f}{x}(\vc{g}(s,t))\pdiff{g_1}{t}(s,t) + \pdiff{f}{y}(\vc{g}(s,t))\pdiff{g_2}{t}(s,t). \end{align*} This is the chain rule written out in component form for $h : \R^2 \to \R$, $f : \R^2 \to \R$, and $\vc{g} : \R^2 \to \R^2$. It is equation () from the special case page.
Now, we compute the answer to our specific problem by substituting in for $f(x,y) = x^2y$ and $\vc{g}(s,t) = (t-s^2, ts^2)$. \begin{align*} \pdiff{f}{x}(x,y) &= 2xy &\pdiff{f}{x}(\vc{g}(s,t)) &= 2ts^2(t-s^2)\\ \pdiff{f}{y}(x,y) &= x^2 & \pdiff{f}{y}(\vc{g}(s,t)) &= (t-s^2)^2\\ \pdiff{g_1}{s}(s,t) &= -2s &\pdiff{g_2}{s}(s,t) &= 2st\\ \pdiff{g_1}{t}(s,t) &= 1 &\pdiff{g_2}{t}(s,t) &= s^2 \end{align*} Finally, we get our answers. \begin{align*} \pdiff{}{s} (f \circ \vc{g})(s,t) = \pdiff{h}{s}(s,t) &= \pdiff{f}{x}(\vc{g}(s,t))\pdiff{g_1}{s}(s,t) + \pdiff{f}{y}(\vc{g}(s,t))\pdiff{g_2}{s}(s,t)\\ &= 2ts^2(t-s^2)(-2s) + (t-s^2)^2(2st)\\ &= -4ts^3(t-s^2) + 2st(t^2-2ts^2+s^4)\\ &= -4s^3t^2 + 4s^5t +2st^3 -4s^3t^2 + 2s^5t\\ &= -8s^3t^2+ 6s^5t+2st^3 \end{align*} \begin{align*} \pdiff{}{t} (f \circ \vc{g})(s,t) = \pdiff{h}{t}(s,t) &= \pdiff{f}{x}(\vc{g}(s,t))\pdiff{g_1}{t}(s,t) + \pdiff{f}{y}(\vc{g}(s,t))\pdiff{g_2}{t}(s,t)\\ &= 2ts^2(t-s^2)(1) + (t-s^2)^2(s^2)\\ &= 2s^2t^2 - 2s^4t + s^2t^2 -2s^4t + s^6\\ &= 3s^2t^2- 4s^4t+ s^6 \end{align*} | 2017-01-19 02:13:42 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 1.0000085830688477, "perplexity": 1011.2737628487513}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-04/segments/1484560280425.43/warc/CC-MAIN-20170116095120-00136-ip-10-171-10-70.ec2.internal.warc.gz"} |
https://faculty.math.illinois.edu/Macaulay2/doc/Macaulay2-1.19/share/doc/Macaulay2/Posets/html/index.html | # Posets -- a package for working with partially ordered sets
## Description
This package defines Poset as a new data type and provides routines which use or produce posets. A poset (partially ordered set) is a set together with a binary relation satisfying reflexivity, antisymmetry, and transitivity.
Contributors:
The following people have generously contributed code to the package: Kristine Fisher, Andrew Hoefel, Manoj Kummini, Stephen Sturgeon, and Josephine Yu.
Other acknowledgements:
A few methods in this package have been ported from John Stembridge's Maple package implementing posets, which is available at http://www.math.lsa.umich.edu/~jrs/maple.html#posets. Such methods are noted both in the source code and in the documentation.
## Certification
Version 1.1.2 of this package was accepted for publication in volume 7 of The Journal of Software for Algebra and Geometry on 5 June 2015, in the article Partially ordered sets in Macaulay2. That version can be obtained from the journal or from the Macaulay2 source code repository.
## Version
This documentation describes version 1.1.3 of Posets.
## Source code
The source code from which this documentation is derived is in the file Posets.m2.
## Exports
• Types
• "NCPartition" -- see ncPartitions -- generates the non-crossing partitions of size $n$
• Poset -- a class for partially ordered sets (posets)
• Functions and commands
• adjoinMax -- computes the poset with a new maximum element
• adjoinMin -- computes the poset with a new minimum element
• allRelations -- computes all relations of a poset
• antichains -- computes all antichains of a poset
• areIsomorphic -- determines if two posets are isomorphic
• atoms -- computes the list of elements covering the minimal elements of a poset
• augmentPoset -- computes the poset with an adjoined minimum and maximum
• booleanLattice -- generates the boolean lattice on $n$ elements
• boundedRegions -- computes the number of bounded regions a hyperplane arrangement divides the space into
• chain -- generates the chain poset on $n$ elements
• chains -- computes all chains of a poset
• characteristicPolynomial -- computes the characteristic polynomial of a ranked poset with a unique minimal element
• closedInterval -- computes the subposet contained between two points
• comparabilityGraph -- produces the comparability graph of a poset
• compare -- compares two elements in a poset
• coveringRelations -- computes the minimal list of generating relations of a poset
• coxeterPolynomial -- computes the Coxeter polynomial of a poset
• diamondProduct -- computes the diamond product of two ranked posets
• dilworthLattice -- computes the Dilworth lattice of a poset
• dilworthNumber -- computes the Dilworth number of a poset
• displayPoset -- generates a PDF representation of a poset and attempts to display it
• distributiveLattice -- computes the lattice of order ideals of a poset
• divisorPoset -- generates the poset of divisors
• dominanceLattice -- generates the dominance lattice of partitions of $n$
• dropElements -- computes the induced subposet of a poset given a list of elements to remove
• facePoset -- generates the face poset of a simplicial complex
• filter -- computes the elements above given elements in a poset
• filtration -- generates the filtration of a poset
• flagChains -- computes the maximal chains in a list of flags of a ranked poset
• flagfPolynomial -- computes the flag-f polynomial of a ranked poset
• flaghPolynomial -- computes the flag-h polynomial of a ranked poset
• flagPoset -- computes the subposet of specified ranks of a ranked poset
• fPolynomial -- computes the f-polynomial of a poset
• gapConvertPoset -- converts between Macaulay2's Posets and GAP's Posets
• greeneKleitmanPartition -- computes the Greene-Kleitman partition of a poset
• hasseDiagram -- produces the Hasse diagram of a poset
• hibiIdeal -- produces the Hibi ideal of a poset
• hibiRing -- produces the Hibi ring of a poset
• hPolynomial -- computes the h-polynomial of a poset
• incomparabilityGraph -- produces the incomparability graph of a poset
• indexLabeling -- relabels a poset with the labeling based on the indices of the vertices
• intersectionLattice -- generates the intersection lattice of a hyperplane arrangement
• isAntichain -- determines if a given list of vertices is an antichain of a poset
• isAtomic -- determines if a lattice is atomic
• isBounded -- determines if a poset is bounded
• isComparabilityGraph -- determines if a graph is the comparability graph of a poset
• isDistributive -- determines if a lattice is distributive
• isGeometric -- determines if a lattice is geometric
• isLattice -- determines if a poset is a lattice
• isLowerSemilattice -- determines if a poset is a lower (or meet) semilattice
• isLowerSemimodular -- determines if a ranked lattice is lower semimodular
• isModular -- determines if a lattice is modular
• isomorphism -- computes an isomorphism between isomorphic posets
• isRanked -- determines if a poset is ranked
• isSperner -- determines if a ranked poset has the Sperner property
• isStrictSperner -- determines if a ranked poset has the strict Sperner property
• isUpperSemilattice -- determines if a poset is an upper (or join) semilattice
• isUpperSemimodular -- determines if a lattice is upper semimodular
• joinExists -- determines if the join exists for two elements of a poset
• joinIrreducibles -- determines the join irreducible elements of a poset
• labelPoset -- relabels a poset with the specified labeling
• lcmLattice -- generates the lattice of lcms in an ideal
• linearExtensions -- computes all linear extensions of a poset
• maximalAntichains -- computes all maximal antichains of a poset
• maximalChains -- computes all maximal chains of a poset
• maximalElements -- determines the maximal elements of a poset
• meetExists -- determines if the meet exists for two elements of a poset
• meetIrreducibles -- determines the meet irreducible elements of a poset
• minimalElements -- determines the minimal elements of a poset
• moebiusFunction -- computes the Moebius function at every pair of elements of a poset
• naturalLabeling -- relabels a poset with a natural labeling
• ncPartitions -- generates the non-crossing partitions of size $n$
• ncpLattice -- computes the non-crossing partition lattice of set-partitions of size $n$
• openInterval -- computes the subposet contained strictly between two points
• orderComplex -- produces the order complex of a poset
• orderIdeal -- computes the elements below given elements in a poset
• outputTexPoset -- writes a LaTeX file with a TikZ-representation of a poset
• partitionLattice -- computes the lattice of set-partitions of size $n$
• plueckerPoset -- computes a poset associated to the Plücker relations
• poincarePolynomial -- computes the Poincare polynomial of a ranked poset with a unique minimal element
• poset -- creates a new Poset object
• posetJoin -- determines the join for two elements of a poset
• posetMeet -- determines the meet for two elements of a poset
• pPartitionRing -- produces the p-partition ring of a poset
• principalFilter -- computes the elements above a given element in a poset
• principalOrderIdeal -- computes the elements below a given element in a poset
• projectivizeArrangement -- computes the intersection poset of a projectivized hyperplane arrangement
• randomPoset -- generates a random poset with a given relation probability
• rankFunction -- computes the rank function of a ranked poset
• rankGeneratingFunction -- computes the rank generating function of a ranked poset
• rankPoset -- generates a list of lists representing the ranks of a ranked poset
• realRegions -- computes the number of regions a hyperplane arrangement divides the space into
• removeIsomorphicPosets -- returns a sub-list of non-isomorphic posets
• resolutionPoset -- generates a poset from a resolution
• setPartition -- computes the list of set-partitions of size $n$
• setPrecompute -- sets the Precompute configuration
• setSuppressLabels -- sets the SuppressLabels configuration
• standardMonomialPoset -- generates the poset of divisibility in the monomial basis of an ideal
• subposet -- computes the induced subposet of a poset given a list of elements
• texPoset -- generates a string containing a TikZ-figure of a poset
• transitiveClosure -- computes the transitive closure of a set of relations
• transitiveOrientation -- generates a poset whose comparability graph is the given graph
• tuttePolynomial -- computes the Tutte polynomial of a poset
• union -- computes the union of two posets
• youngSubposet -- generates a subposet of Young's lattice
• zetaPolynomial -- computes the zeta polynomial of a poset
• Methods
• "adjoinMax(Poset)" -- see adjoinMax -- computes the poset with a new maximum element
• "adjoinMax(Poset,Thing)" -- see adjoinMax -- computes the poset with a new maximum element
• "adjoinMin(Poset)" -- see adjoinMin -- computes the poset with a new minimum element
• "adjoinMin(Poset,Thing)" -- see adjoinMin -- computes the poset with a new minimum element
• "allRelations(Poset)" -- see allRelations -- computes all relations of a poset
• "allRelations(Poset,Boolean)" -- see allRelations -- computes all relations of a poset
• "antichains(Poset)" -- see antichains -- computes all antichains of a poset
• "antichains(Poset,ZZ)" -- see antichains -- computes all antichains of a poset
• "areIsomorphic(Poset,Poset)" -- see areIsomorphic -- determines if two posets are isomorphic
• "Poset == Poset" -- see areIsomorphic -- determines if two posets are isomorphic
• "atoms(Poset)" -- see atoms -- computes the list of elements covering the minimal elements of a poset
• "augmentPoset(Poset)" -- see augmentPoset -- computes the poset with an adjoined minimum and maximum
• "augmentPoset(Poset,Thing,Thing)" -- see augmentPoset -- computes the poset with an adjoined minimum and maximum
• "booleanLattice(ZZ)" -- see booleanLattice -- generates the boolean lattice on $n$ elements
• "boundedRegions(List,Ring)" -- see boundedRegions -- computes the number of bounded regions a hyperplane arrangement divides the space into
• "chain(ZZ)" -- see chain -- generates the chain poset on $n$ elements
• "chains(Poset)" -- see chains -- computes all chains of a poset
• "chains(Poset,ZZ)" -- see chains -- computes all chains of a poset
• "characteristicPolynomial(Poset)" -- see characteristicPolynomial -- computes the characteristic polynomial of a ranked poset with a unique minimal element
• "closedInterval(Poset,Thing,Thing)" -- see closedInterval -- computes the subposet contained between two points
• "comparabilityGraph(Poset)" -- see comparabilityGraph -- produces the comparability graph of a poset
• "compare(Poset,Thing,Thing)" -- see compare -- compares two elements in a poset
• connectedComponents(Poset) -- generates a list of connected components of a poset
• "coveringRelations(Poset)" -- see coveringRelations -- computes the minimal list of generating relations of a poset
• "coxeterPolynomial(Poset)" -- see coxeterPolynomial -- computes the Coxeter polynomial of a poset
• "diamondProduct(Poset,Poset)" -- see diamondProduct -- computes the diamond product of two ranked posets
• "dilworthLattice(Poset)" -- see dilworthLattice -- computes the Dilworth lattice of a poset
• "dilworthNumber(Poset)" -- see dilworthNumber -- computes the Dilworth number of a poset
• "displayPoset(Poset)" -- see displayPoset -- generates a PDF representation of a poset and attempts to display it
• "distributiveLattice(Poset)" -- see distributiveLattice -- computes the lattice of order ideals of a poset
• "divisorPoset(ZZ)" -- see divisorPoset -- generates the poset of divisors
• divisorPoset(List,List,PolynomialRing) -- generates the poset of divisors
• divisorPoset(RingElement) -- generates the poset of divisors
• divisorPoset(RingElement,RingElement) -- generates the poset of divisors with a lower and upper bound
• "dominanceLattice(ZZ)" -- see dominanceLattice -- generates the dominance lattice of partitions of $n$
• "dropElements(Poset,Function)" -- see dropElements -- computes the induced subposet of a poset given a list of elements to remove
• "dropElements(Poset,List)" -- see dropElements -- computes the induced subposet of a poset given a list of elements to remove
• "Poset - List" -- see dropElements -- computes the induced subposet of a poset given a list of elements to remove
• dual(Poset) -- produces the derived poset with relations reversed
• "facePoset(SimplicialComplex)" -- see facePoset -- generates the face poset of a simplicial complex
• "filter(Poset,List)" -- see filter -- computes the elements above given elements in a poset
• "filtration(Poset)" -- see filtration -- generates the filtration of a poset
• "flagChains(Poset,List)" -- see flagChains -- computes the maximal chains in a list of flags of a ranked poset
• "flagfPolynomial(Poset)" -- see flagfPolynomial -- computes the flag-f polynomial of a ranked poset
• "flaghPolynomial(Poset)" -- see flaghPolynomial -- computes the flag-h polynomial of a ranked poset
• "flagPoset(Poset,List)" -- see flagPoset -- computes the subposet of specified ranks of a ranked poset
• "fPolynomial(Poset)" -- see fPolynomial -- computes the f-polynomial of a poset
• "gapConvertPoset(Array)" -- see gapConvertPoset -- converts between Macaulay2's Posets and GAP's Posets
• "gapConvertPoset(Poset)" -- see gapConvertPoset -- converts between Macaulay2's Posets and GAP's Posets
• "gapConvertPoset(String)" -- see gapConvertPoset -- converts between Macaulay2's Posets and GAP's Posets
• "greeneKleitmanPartition(Poset)" -- see greeneKleitmanPartition -- computes the Greene-Kleitman partition of a poset
• "hasseDiagram(Poset)" -- see hasseDiagram -- produces the Hasse diagram of a poset
• height(Poset) -- computes the height of a poset
• "hibiIdeal(Poset)" -- see hibiIdeal -- produces the Hibi ideal of a poset
• "hibiRing(Poset)" -- see hibiRing -- produces the Hibi ring of a poset
• "hPolynomial(Poset)" -- see hPolynomial -- computes the h-polynomial of a poset
• "incomparabilityGraph(Poset)" -- see incomparabilityGraph -- produces the incomparability graph of a poset
• "indexLabeling(Poset)" -- see indexLabeling -- relabels a poset with the labeling based on the indices of the vertices
• "intersectionLattice(List,Ring)" -- see intersectionLattice -- generates the intersection lattice of a hyperplane arrangement
• "isAntichain(Poset,List)" -- see isAntichain -- determines if a given list of vertices is an antichain of a poset
• "isAtomic(Poset)" -- see isAtomic -- determines if a lattice is atomic
• "isBounded(Poset)" -- see isBounded -- determines if a poset is bounded
• "isComparabilityGraph(Graph)" -- see isComparabilityGraph -- determines if a graph is the comparability graph of a poset
• isConnected(Poset) -- determines if a poset is connected
• "isDistributive(Poset)" -- see isDistributive -- determines if a lattice is distributive
• isEulerian(Poset) -- determines if a ranked poset is Eulerian
• "isGeometric(Poset)" -- see isGeometric -- determines if a lattice is geometric
• "isLattice(Poset)" -- see isLattice -- determines if a poset is a lattice
• "isLowerSemilattice(Poset)" -- see isLowerSemilattice -- determines if a poset is a lower (or meet) semilattice
• "isLowerSemimodular(Poset)" -- see isLowerSemimodular -- determines if a ranked lattice is lower semimodular
• "isModular(Poset)" -- see isModular -- determines if a lattice is modular
• "isomorphism(Poset,Poset)" -- see isomorphism -- computes an isomorphism between isomorphic posets
• "isRanked(Poset)" -- see isRanked -- determines if a poset is ranked
• "isSperner(Poset)" -- see isSperner -- determines if a ranked poset has the Sperner property
• "isStrictSperner(Poset)" -- see isStrictSperner -- determines if a ranked poset has the strict Sperner property
• "isUpperSemilattice(Poset)" -- see isUpperSemilattice -- determines if a poset is an upper (or join) semilattice
• "isUpperSemimodular(Poset)" -- see isUpperSemimodular -- determines if a lattice is upper semimodular
• "joinExists(Poset,Thing,Thing)" -- see joinExists -- determines if the join exists for two elements of a poset
• "joinIrreducibles(Poset)" -- see joinIrreducibles -- determines the join irreducible elements of a poset
• "labelPoset(Poset,HashTable)" -- see labelPoset -- relabels a poset with the specified labeling
• "lcmLattice(Ideal)" -- see lcmLattice -- generates the lattice of lcms in an ideal
• "linearExtensions(Poset)" -- see linearExtensions -- computes all linear extensions of a poset
• "maximalAntichains(Poset)" -- see maximalAntichains -- computes all maximal antichains of a poset
• "maximalChains(Poset)" -- see maximalChains -- computes all maximal chains of a poset
• "maximalElements(Poset)" -- see maximalElements -- determines the maximal elements of a poset
• "meetExists(Poset,Thing,Thing)" -- see meetExists -- determines if the meet exists for two elements of a poset
• "meetIrreducibles(Poset)" -- see meetIrreducibles -- determines the meet irreducible elements of a poset
• "minimalElements(Poset)" -- see minimalElements -- determines the minimal elements of a poset
• "moebiusFunction(Poset)" -- see moebiusFunction -- computes the Moebius function at every pair of elements of a poset
• "naturalLabeling(Poset)" -- see naturalLabeling -- relabels a poset with a natural labeling
• "naturalLabeling(Poset,ZZ)" -- see naturalLabeling -- relabels a poset with a natural labeling
• "ncPartitions(ZZ)" -- see ncPartitions -- generates the non-crossing partitions of size $n$
• "ncpLattice(ZZ)" -- see ncpLattice -- computes the non-crossing partition lattice of set-partitions of size $n$
• "openInterval(Poset,Thing,Thing)" -- see openInterval -- computes the subposet contained strictly between two points
• "orderComplex(Poset)" -- see orderComplex -- produces the order complex of a poset
• "orderIdeal(Poset,List)" -- see orderIdeal -- computes the elements below given elements in a poset
• "outputTexPoset(Poset,String)" -- see outputTexPoset -- writes a LaTeX file with a TikZ-representation of a poset
• "partitionLattice(ZZ)" -- see partitionLattice -- computes the lattice of set-partitions of size $n$
• "plueckerPoset(ZZ)" -- see plueckerPoset -- computes a poset associated to the Plücker relations
• "poincare(Poset)" -- see poincarePolynomial -- computes the Poincare polynomial of a ranked poset with a unique minimal element
• "poincarePolynomial(Poset)" -- see poincarePolynomial -- computes the Poincare polynomial of a ranked poset with a unique minimal element
• "poset(List)" -- see poset -- creates a new Poset object
• "poset(List,Function)" -- see poset -- creates a new Poset object
• "poset(List,List)" -- see poset -- creates a new Poset object
• "poset(List,List,Matrix)" -- see poset -- creates a new Poset object
• Poset _ List -- returns elements of the ground set
• Poset _ ZZ -- returns an element of the ground set
• Poset _* -- returns the ground set of a poset
• "vertices(Poset)" -- see Poset _* -- returns the ground set of a poset
• "posetJoin(Poset,Thing,Thing)" -- see posetJoin -- determines the join for two elements of a poset
• "posetMeet(Poset,Thing,Thing)" -- see posetMeet -- determines the meet for two elements of a poset
• "pPartitionRing(Poset)" -- see pPartitionRing -- produces the p-partition ring of a poset
• "principalFilter(Poset,Thing)" -- see principalFilter -- computes the elements above a given element in a poset
• "principalOrderIdeal(Poset,Thing)" -- see principalOrderIdeal -- computes the elements below a given element in a poset
• "Poset * Poset" -- see product(Poset,Poset) -- computes the product of two posets
• product(Poset,Poset) -- computes the product of two posets
• "projectivizeArrangement(List,Ring)" -- see projectivizeArrangement -- computes the intersection poset of a projectivized hyperplane arrangement
• "randomPoset(List)" -- see randomPoset -- generates a random poset with a given relation probability
• "randomPoset(ZZ)" -- see randomPoset -- generates a random poset with a given relation probability
• "rankFunction(Poset)" -- see rankFunction -- computes the rank function of a ranked poset
• "rankGeneratingFunction(Poset)" -- see rankGeneratingFunction -- computes the rank generating function of a ranked poset
• "rank(Poset)" -- see rankPoset -- generates a list of lists representing the ranks of a ranked poset
• "rankPoset(Poset)" -- see rankPoset -- generates a list of lists representing the ranks of a ranked poset
• "realRegions(List,Ring)" -- see realRegions -- computes the number of regions a hyperplane arrangement divides the space into
• "removeIsomorphicPosets(List)" -- see removeIsomorphicPosets -- returns a sub-list of non-isomorphic posets
• "resolutionPoset(ChainComplex)" -- see resolutionPoset -- generates a poset from a resolution
• "resolutionPoset(Ideal)" -- see resolutionPoset -- generates a poset from a resolution
• "resolutionPoset(MonomialIdeal)" -- see resolutionPoset -- generates a poset from a resolution
• "setPartition(List)" -- see setPartition -- computes the list of set-partitions of size $n$
• "setPartition(ZZ)" -- see setPartition -- computes the list of set-partitions of size $n$
• "setPrecompute(Boolean)" -- see setPrecompute -- sets the Precompute configuration
• "setSuppressLabels(Boolean)" -- see setSuppressLabels -- sets the SuppressLabels configuration
• "standardMonomialPoset(MonomialIdeal)" -- see standardMonomialPoset -- generates the poset of divisibility in the monomial basis of an ideal
• "standardMonomialPoset(MonomialIdeal,ZZ,ZZ)" -- see standardMonomialPoset -- generates the poset of divisibility in the monomial basis of an ideal
• "subposet(Poset,List)" -- see subposet -- computes the induced subposet of a poset given a list of elements
• "tex(Poset)" -- see texPoset -- generates a string containing a TikZ-figure of a poset
• "texPoset(Poset)" -- see texPoset -- generates a string containing a TikZ-figure of a poset
• "transitiveClosure(List,List)" -- see transitiveClosure -- computes the transitive closure of a set of relations
• "transitiveOrientation(Graph)" -- see transitiveOrientation -- generates a poset whose comparability graph is the given graph
• "tuttePolynomial(Poset)" -- see tuttePolynomial -- computes the Tutte polynomial of a poset
• "Poset + Poset" -- see union -- computes the union of two posets
• "union(Poset,Poset)" -- see union -- computes the union of two posets
• "youngSubposet(List)" -- see youngSubposet -- generates a subposet of Young's lattice
• "youngSubposet(List,List)" -- see youngSubposet -- generates a subposet of Young's lattice
• "youngSubposet(ZZ)" -- see youngSubposet -- generates a subposet of Young's lattice
• "zetaPolynomial(Poset)" -- see zetaPolynomial -- computes the zeta polynomial of a poset
• Symbols
• "PDFDirectory" -- see displayPoset -- generates a PDF representation of a poset and attempts to display it
• "OriginalPoset" -- see distributiveLattice -- computes the lattice of order ideals of a poset
• "GroundSet" -- see Poset -- a class for partially ordered sets (posets)
• "RelationMatrix" -- see Poset -- a class for partially ordered sets (posets)
• "Relations" -- see Poset -- a class for partially ordered sets (posets)
• "AntisymmetryStrategy" -- see poset -- creates a new Poset object
• Precompute -- a package-wide configuration that toggles precomputation
• "Bias" -- see randomPoset -- generates a random poset with a given relation probability
• "Jitter" -- see texPoset -- generates a string containing a TikZ-figure of a poset
• "SuppressLabels" -- see texPoset -- generates a string containing a TikZ-figure of a poset
## For the programmer
The object Posets is . | 2022-08-15 08:55:20 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8907060027122498, "perplexity": 6046.455639193094}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-33/segments/1659882572163.61/warc/CC-MAIN-20220815085006-20220815115006-00346.warc.gz"} |
http://www.physicsforums.com/showthread.php?t=27361 | # Integrating cosec...
by speeding electron
Tags: cosec, integrating
P: 65 Int(cosec x)dx Let: u=cos x then: x=arcos u and: dx=-du/sqrt(1-u^2) Int(cosec x)dx = -Int{1/sqrt(1-u^2)}^2}du = -Int {du/(1-u^2)} = -(1/2)ln(1-u^2) + C = -(1/2)ln(sin^[2]x) +C = ln(cosec x) +C Yet differentiating back gives -cot x. Why does this substitution not work?
P: 1,783
Quote by speeding electron Int(cosec x)dx Let: u=cos x then: x=arcos u and: dx=du/sqrt(1-u^2) Int(cosec x)dx = Int{1/sqrt(1-u^2)}^2}du = Int {du/(1-u^2)} =(1/2)ln(1-u^2) + C =(1/2)ln(sin^[2]x) +C =ln(sin x) +C Yet differentiating back gives cot x. Why does this substitution not work?
First:
$$csc x = \frac{1}{sin x}$$
NOT cos x. So your substitution is wrong to begin with. If you correct that the integration should go like this:
$$u = sin x dx$$
$$du = cos x dx$$
$$sec x du = dx$$
$$\frac{1}{\sqrt{1-u^2}} du = dx$$
So this makes the integral:
$$\int \frac{1}{u\sqrt{1-u^2}} du$$
See if you can take it from there.
P: 65 My query was concerning the substitution u = cos x , rather than u = sin x . I did make a mistake, arcos x = -1/sqrt(1-x^2). I've edited my original post.
P: 1,783
Integrating cosec...
Quote by speeding electron My query was concerning the substitution u = cos x , rather than u = sin x . I did make a mistake, arcos x = -1/sqrt(1-x^2). I've edited my original post.
That is the wrong substitution for integrating cosec(X).$$cosec(x) = \frac{1}{sin(x)}$$.
P: 15 well your subsitution is didn't work because it is difficult to solve integration with assumption not being in the question
HW Helper
PF Gold
P: 12,016
Quote by speeding electron Int(cosec x)dx Let: u=cos x then: x=arcos u and: dx=-du/sqrt(1-u^2) Int(cosec x)dx = -Int{1/sqrt(1-u^2)}^2}du = -Int {du/(1-u^2)} = -(1/2)ln(1-u^2) + C = -(1/2)ln(sin^[2]x) +C = ln(cosec x) +C Yet differentiating back gives -cot x. Why does this substitution not work?
Because you make an equality out of the following non-equality:
$$-\int\frac{du}{1-u^{2}}\neq\frac{-1}{2}ln(1-u^{2})+C$$
P: 65 Fine, yes, sorry about that, me being stupid again...but that integral was getting annoying.
P: 8
Quote by franznietzsche First: $$csc x = \frac{1}{sin x}$$ NOT cos x. So your substitution is wrong to begin with. If you correct that the integration should go like this: $$u = sin x dx$$ $$du = cos x dx$$ $$sec x du = dx$$ $$\frac{1}{\sqrt{1-u^2}} du = dx$$ So this makes the integral: $$\int \frac{1}{u\sqrt{1-u^2}} du$$ See if you can take it from there.
I can't :(. I've spent most of this afternoon trying this question and some other one. I know I could just use the set result, but I want to understand it, and I don't see how to integrate that last result at all :(.
P: 206 There is a special method to this. $\int \mathrm{cosec} x \ \mathrm{d}x$ If you multiply this by $\frac {\mathrm{cosec} x - \cot x}{\mathrm{cosec} x - \cot x}$ and simplify the numerator you will get an integral of... $\int \frac{\mathrm{cosec} ^2 x - \mathrm{cosec} x \cot x}{\mathrm{cosec} x - \cot x} \ \mathrm{d}x$ Substitute $u = \mathrm{cosec} x - \cot x$ and it should work out beautifully. Carry on from here and post back if you still need help.
P: 8 Thanks, that worked out really well :). I was wondering though, how would one work it out from the form $$\int \frac{1}{u\sqrt{1-u^2}} du$$ ? Could anyone help me see how to integrate it from this?
Related Discussions Calculus 11 Calculus 9 Calculus & Beyond Homework 10 Calculus & Beyond Homework 10 Differential Equations 2 | 2014-07-25 11:23:35 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9005573987960815, "perplexity": 2335.6659269936554}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-23/segments/1405997894250.40/warc/CC-MAIN-20140722025814-00097-ip-10-33-131-23.ec2.internal.warc.gz"} |
https://stats.stackexchange.com/questions/126259/convergence-of-likelihood-implying-convergence-of-marginal-likelihood | # Convergence of likelihood implying convergence of marginal likelihood?
I will ask my question through a toy motivating example.
It is well known that a Poisson process is the continuous time analog to a Bernoulli process (for example: http://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-262-discrete-stochastic-processes-spring-2011/course-notes/MIT6_262S11_chap02.pdf). There is also a similar well known relationship between the Gamma distribution and the Beta distribution (for example: https://math.stackexchange.com/questions/190670/how-exactly-are-the-beta-and-gamma-distributions-related). I would like to make one related point that seem less obvious in the literature:
If we place a Beta prior upon that Bernoulli process and then compute the marginal likelihood, can we prove that this is the discrete time analog to the corresponding marginal likelihood when we place a Gamma prior upon the Poisson process... only using the underlying correspondences between Beta and Gamma, and between Bernoulli process and Poisson process? Is there a theorem we may invoke and under what conditions may we invoke it, or do we have to show correspondence by working with the marginals directly?
This example in particular is a toy -- it is relatively straightforward to work with the marginal likelihoods in the two cases, starting with the discrete case and shrinking the time increments down to zero. However for a research project I am working on, it is much easier to show correspondence between likelihoods, and between priors, than between marginal likelihoods. In my research, I am considering a Bernoulli($p_1$) process which is stopped' at some time $\tau$ (i.e., all observations after time point $\tau$ equal 0 with probability 1), where $\tau$ is geometric($p_2$) distributed. Both $p_1$ and $p_2$ have independent Beta priors.
To put this in formulae using generic notation, if the limiting form of the pmf is equal in distribution to another pmf: $$\lim_{n \rightarrow \infty} p(x | \theta_n) = p(x | \theta)$$
and the limiting form of the prior pdf is equal in distribution to another pdf:
$$\lim_{n \rightarrow \infty} p(\theta_n | \psi_n) = p(\theta_n | \psi)$$
then under what conditions (any?) can we take the limit within the integral' and state that marginal likelihoods are equal in distribution as well:
$$\lim_{n \rightarrow \infty} \int p(x | \theta_n) p(\theta_n | \psi_n) d\theta_n = \int \lim_{n \rightarrow \infty} p(x | \theta_n) p(\theta_n | \psi_n) d\theta_n = \int p(x | \theta) p(\theta | \psi) d\theta \ \ \text{?}$$
Any help would be most appreciated. I've tried to make this as clear as possible.
• +50 seems like a small bounty for a task :) – Aksakal Dec 5 '14 at 16:28
• I did't realize how much of a task it was until I got this non-response to my bounty. I thought there might have been an easy answer so I am deducing that perhaps there isn't. – Dan Dec 5 '14 at 19:27
• I don't see an obvious answer. This is not the standard problem everybody deals with, so if you're lucky someone will recognize this and give you an answer. Otherwise, one needs to spend some time working out the equations. – Aksakal Dec 5 '14 at 19:29
• Agreed Aksakal - although I must say, the answer to this would seem to be quite relevant to many people working with stochastic models for whatever activity they happen to be studying. – Dan Dec 6 '14 at 0:18
Let $f_n$ be a sequence of real-valued measurable functions on a measure space $(S, Σ, \mu)$. Suppose that the sequence converges pointwise to a function $f$ and is dominated by some integrable function $g$ in the sense that $$|f_n(x)| \le g(x)$$ for all numbers $n$ in the index set of the sequence and all points $x \in S$. Then $f$ is integrable and $$\lim_{n\to\infty} \int_S |f_n-f|\,d\mu = 0$$ which also implies $$\lim_{n\to\infty} \int_S f_n\,d\mu = \int_S f\,d\mu.$$
Here $f_n = p(x |\theta_n) p(\theta_n | \psi_n)$. Since it is a product of probabilities, it seems hopeful that you can bound the $f_n$ by some integrable $g$. | 2020-08-06 22:33:51 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 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": 0.9081348776817322, "perplexity": 261.93670789572366}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-34/segments/1596439737039.58/warc/CC-MAIN-20200806210649-20200807000649-00223.warc.gz"} |
https://www.lmfdb.org/ModularForm/GL2/Q/holomorphic/546/2/j/e/ | # Properties
Label 546.2.j.e Level $546$ Weight $2$ Character orbit 546.j Analytic conductor $4.360$ Analytic rank $0$ Dimension $10$ CM no Inner twists $2$
# Related objects
## Newspace parameters
Level: $$N$$ $$=$$ $$546 = 2 \cdot 3 \cdot 7 \cdot 13$$ Weight: $$k$$ $$=$$ $$2$$ Character orbit: $$[\chi]$$ $$=$$ 546.j (of order $$3$$, degree $$2$$, minimal)
## Newform invariants
Self dual: no Analytic conductor: $$4.35983195036$$ Analytic rank: $$0$$ Dimension: $$10$$ Relative dimension: $$5$$ over $$\Q(\zeta_{3})$$ Coefficient field: $$\mathbb{Q}[x]/(x^{10} - \cdots)$$ Defining polynomial: $$x^{10} - 2 x^{9} + 15 x^{8} + 14 x^{7} + 110 x^{6} + 36 x^{5} + 233 x^{4} + 164 x^{3} + 345 x^{2} + 76 x + 16$$ Coefficient ring: $$\Z[a_1, \ldots, a_{7}]$$ Coefficient ring index: $$3$$ Twist minimal: yes Sato-Tate group: $\mathrm{SU}(2)[C_{3}]$
## $q$-expansion
Coefficients of the $$q$$-expansion are expressed in terms of a basis $$1,\beta_1,\ldots,\beta_{9}$$ for the coefficient ring described below. We also show the integral $$q$$-expansion of the trace form.
$$f(q)$$ $$=$$ $$q - q^{2} + ( 1 + \beta_{5} ) q^{3} + q^{4} -\beta_{1} q^{5} + ( -1 - \beta_{5} ) q^{6} + ( \beta_{6} + \beta_{9} ) q^{7} - q^{8} + \beta_{5} q^{9} +O(q^{10})$$ $$q - q^{2} + ( 1 + \beta_{5} ) q^{3} + q^{4} -\beta_{1} q^{5} + ( -1 - \beta_{5} ) q^{6} + ( \beta_{6} + \beta_{9} ) q^{7} - q^{8} + \beta_{5} q^{9} + \beta_{1} q^{10} + ( 2 + \beta_{1} + \beta_{4} + 2 \beta_{5} - \beta_{6} + \beta_{7} ) q^{11} + ( 1 + \beta_{5} ) q^{12} + ( -\beta_{1} - \beta_{3} - \beta_{7} ) q^{13} + ( -\beta_{6} - \beta_{9} ) q^{14} + ( -\beta_{1} - \beta_{2} ) q^{15} + q^{16} + ( -\beta_{2} - \beta_{3} - \beta_{8} + \beta_{9} ) q^{17} -\beta_{5} q^{18} + ( -2 \beta_{1} - 2 \beta_{2} - \beta_{4} - \beta_{5} + 2 \beta_{6} - 2 \beta_{8} + \beta_{9} ) q^{19} -\beta_{1} q^{20} + \beta_{9} q^{21} + ( -2 - \beta_{1} - \beta_{4} - 2 \beta_{5} + \beta_{6} - \beta_{7} ) q^{22} + ( -\beta_{2} - \beta_{3} - \beta_{4} - \beta_{6} - 2 \beta_{8} + \beta_{9} ) q^{23} + ( -1 - \beta_{5} ) q^{24} + ( 2 \beta_{1} + 2 \beta_{2} - \beta_{4} - \beta_{5} + \beta_{9} ) q^{25} + ( \beta_{1} + \beta_{3} + \beta_{7} ) q^{26} - q^{27} + ( \beta_{6} + \beta_{9} ) q^{28} + ( -\beta_{3} - \beta_{6} - \beta_{7} + \beta_{8} ) q^{29} + ( \beta_{1} + \beta_{2} ) q^{30} + ( \beta_{1} + \beta_{2} - \beta_{3} + \beta_{4} + 2 \beta_{5} - \beta_{6} - \beta_{7} + \beta_{8} - \beta_{9} ) q^{31} - q^{32} + ( \beta_{1} + \beta_{2} + \beta_{3} + \beta_{4} + 2 \beta_{5} - \beta_{6} + \beta_{7} + \beta_{8} - \beta_{9} ) q^{33} + ( \beta_{2} + \beta_{3} + \beta_{8} - \beta_{9} ) q^{34} + ( 2 - \beta_{1} - 2 \beta_{2} + \beta_{3} + \beta_{4} + 2 \beta_{5} + \beta_{7} ) q^{35} + \beta_{5} q^{36} + ( 1 - \beta_{3} - \beta_{8} + \beta_{9} ) q^{37} + ( 2 \beta_{1} + 2 \beta_{2} + \beta_{4} + \beta_{5} - 2 \beta_{6} + 2 \beta_{8} - \beta_{9} ) q^{38} + ( -\beta_{1} - \beta_{2} - \beta_{3} ) q^{39} + \beta_{1} q^{40} + ( \beta_{3} + \beta_{4} + 2 \beta_{5} - \beta_{6} + \beta_{7} + \beta_{8} - \beta_{9} ) q^{41} -\beta_{9} q^{42} + ( 1 - 3 \beta_{1} - \beta_{4} + \beta_{5} - \beta_{7} - \beta_{8} - \beta_{9} ) q^{43} + ( 2 + \beta_{1} + \beta_{4} + 2 \beta_{5} - \beta_{6} + \beta_{7} ) q^{44} -\beta_{2} q^{45} + ( \beta_{2} + \beta_{3} + \beta_{4} + \beta_{6} + 2 \beta_{8} - \beta_{9} ) q^{46} + ( -1 - \beta_{1} + 2 \beta_{4} - \beta_{5} - \beta_{6} + \beta_{7} + \beta_{8} + \beta_{9} ) q^{47} + ( 1 + \beta_{5} ) q^{48} + ( 2 + \beta_{2} + \beta_{3} - 2 \beta_{4} + 2 \beta_{5} + \beta_{7} - 2 \beta_{8} ) q^{49} + ( -2 \beta_{1} - 2 \beta_{2} + \beta_{4} + \beta_{5} - \beta_{9} ) q^{50} + ( \beta_{1} + \beta_{4} - \beta_{6} + \beta_{7} ) q^{51} + ( -\beta_{1} - \beta_{3} - \beta_{7} ) q^{52} + ( -\beta_{1} - \beta_{2} + \beta_{3} - \beta_{4} + 3 \beta_{5} + 2 \beta_{6} + \beta_{7} - 2 \beta_{8} + \beta_{9} ) q^{53} + q^{54} + ( -\beta_{1} - \beta_{2} - \beta_{3} - \beta_{4} - \beta_{5} - \beta_{7} + \beta_{9} ) q^{55} + ( -\beta_{6} - \beta_{9} ) q^{56} + ( 1 - 2 \beta_{2} + \beta_{4} + \beta_{6} - \beta_{8} + 2 \beta_{9} ) q^{57} + ( \beta_{3} + \beta_{6} + \beta_{7} - \beta_{8} ) q^{58} + ( -4 \beta_{2} - 2 \beta_{3} + \beta_{4} + \beta_{6} - \beta_{8} + 2 \beta_{9} ) q^{59} + ( -\beta_{1} - \beta_{2} ) q^{60} + ( \beta_{1} + \beta_{2} - \beta_{4} - 3 \beta_{5} + \beta_{9} ) q^{61} + ( -\beta_{1} - \beta_{2} + \beta_{3} - \beta_{4} - 2 \beta_{5} + \beta_{6} + \beta_{7} - \beta_{8} + \beta_{9} ) q^{62} -\beta_{6} q^{63} + q^{64} + ( 1 + 2 \beta_{1} + 2 \beta_{2} + 4 \beta_{5} + \beta_{6} + 2 \beta_{8} ) q^{65} + ( -\beta_{1} - \beta_{2} - \beta_{3} - \beta_{4} - 2 \beta_{5} + \beta_{6} - \beta_{7} - \beta_{8} + \beta_{9} ) q^{66} + ( 1 + \beta_{4} + \beta_{5} + 2 \beta_{7} + \beta_{8} + \beta_{9} ) q^{67} + ( -\beta_{2} - \beta_{3} - \beta_{8} + \beta_{9} ) q^{68} + ( \beta_{1} + \beta_{4} - 2 \beta_{6} + \beta_{7} - \beta_{8} - \beta_{9} ) q^{69} + ( -2 + \beta_{1} + 2 \beta_{2} - \beta_{3} - \beta_{4} - 2 \beta_{5} - \beta_{7} ) q^{70} + ( 4 + 3 \beta_{1} + 2 \beta_{4} + 4 \beta_{5} - 2 \beta_{6} ) q^{71} -\beta_{5} q^{72} + ( -3 \beta_{1} - 3 \beta_{2} - 3 \beta_{3} + \beta_{4} - 2 \beta_{5} - 3 \beta_{7} - \beta_{9} ) q^{73} + ( -1 + \beta_{3} + \beta_{8} - \beta_{9} ) q^{74} + ( 1 + 2 \beta_{2} - \beta_{4} - \beta_{6} - \beta_{8} ) q^{75} + ( -2 \beta_{1} - 2 \beta_{2} - \beta_{4} - \beta_{5} + 2 \beta_{6} - 2 \beta_{8} + \beta_{9} ) q^{76} + ( 1 - 4 \beta_{1} - 2 \beta_{2} - \beta_{3} - 2 \beta_{4} - 2 \beta_{5} + 2 \beta_{6} - 3 \beta_{7} - \beta_{8} + 2 \beta_{9} ) q^{77} + ( \beta_{1} + \beta_{2} + \beta_{3} ) q^{78} + ( -2 - \beta_{1} - 2 \beta_{4} - 2 \beta_{5} + 2 \beta_{6} ) q^{79} -\beta_{1} q^{80} + ( -1 - \beta_{5} ) q^{81} + ( -\beta_{3} - \beta_{4} - 2 \beta_{5} + \beta_{6} - \beta_{7} - \beta_{8} + \beta_{9} ) q^{82} + ( \beta_{3} + \beta_{4} + \beta_{6} + \beta_{9} ) q^{83} + \beta_{9} q^{84} + ( -1 + \beta_{1} - \beta_{5} + \beta_{6} - \beta_{7} + \beta_{8} + \beta_{9} ) q^{85} + ( -1 + 3 \beta_{1} + \beta_{4} - \beta_{5} + \beta_{7} + \beta_{8} + \beta_{9} ) q^{86} + ( -\beta_{3} - \beta_{4} - \beta_{6} - \beta_{9} ) q^{87} + ( -2 - \beta_{1} - \beta_{4} - 2 \beta_{5} + \beta_{6} - \beta_{7} ) q^{88} + ( -3 + 7 \beta_{2} + 2 \beta_{3} - \beta_{4} - \beta_{6} + 4 \beta_{8} - 5 \beta_{9} ) q^{89} + \beta_{2} q^{90} + ( 4 + 2 \beta_{2} + 2 \beta_{3} + 3 \beta_{4} + 4 \beta_{5} - 2 \beta_{6} + 2 \beta_{7} + 3 \beta_{8} - 2 \beta_{9} ) q^{91} + ( -\beta_{2} - \beta_{3} - \beta_{4} - \beta_{6} - 2 \beta_{8} + \beta_{9} ) q^{92} + ( -2 + \beta_{2} - \beta_{3} + \beta_{8} - \beta_{9} ) q^{93} + ( 1 + \beta_{1} - 2 \beta_{4} + \beta_{5} + \beta_{6} - \beta_{7} - \beta_{8} - \beta_{9} ) q^{94} + ( -2 + 2 \beta_{2} - \beta_{4} - \beta_{6} - 3 \beta_{8} + 2 \beta_{9} ) q^{95} + ( -1 - \beta_{5} ) q^{96} + ( -2 - \beta_{4} - 2 \beta_{5} - \beta_{6} - 2 \beta_{7} - 2 \beta_{8} - 2 \beta_{9} ) q^{97} + ( -2 - \beta_{2} - \beta_{3} + 2 \beta_{4} - 2 \beta_{5} - \beta_{7} + 2 \beta_{8} ) q^{98} + ( -2 + \beta_{2} + \beta_{3} + \beta_{8} - \beta_{9} ) q^{99} +O(q^{100})$$ $$\operatorname{Tr}(f)(q)$$ $$=$$ $$10q - 10q^{2} + 5q^{3} + 10q^{4} - 2q^{5} - 5q^{6} - 2q^{7} - 10q^{8} - 5q^{9} + O(q^{10})$$ $$10q - 10q^{2} + 5q^{3} + 10q^{4} - 2q^{5} - 5q^{6} - 2q^{7} - 10q^{8} - 5q^{9} + 2q^{10} + 6q^{11} + 5q^{12} - 4q^{13} + 2q^{14} + 2q^{15} + 10q^{16} - 8q^{17} + 5q^{18} + 3q^{19} - 2q^{20} - 4q^{21} - 6q^{22} - 12q^{23} - 5q^{24} - q^{25} + 4q^{26} - 10q^{27} - 2q^{28} - 2q^{30} - 10q^{31} - 10q^{32} - 6q^{33} + 8q^{34} + 16q^{35} - 5q^{36} - 2q^{37} - 3q^{38} - 2q^{39} + 2q^{40} - 4q^{41} + 4q^{42} + 3q^{43} + 6q^{44} + 4q^{45} + 12q^{46} - 15q^{47} + 5q^{48} + 4q^{49} + q^{50} - 4q^{51} - 4q^{52} - 17q^{53} + 10q^{54} + 3q^{55} + 2q^{56} + 6q^{57} - 4q^{59} + 2q^{60} + 11q^{61} + 10q^{62} - 2q^{63} + 10q^{64} - 4q^{65} + 6q^{66} - q^{67} - 8q^{68} - 6q^{69} - 16q^{70} + 18q^{71} + 5q^{72} + 12q^{73} + 2q^{74} - 2q^{75} + 3q^{76} + 18q^{77} + 2q^{78} - 4q^{79} - 2q^{80} - 5q^{81} + 4q^{82} - 4q^{84} + q^{85} - 3q^{86} - 6q^{88} - 14q^{89} - 4q^{90} + 26q^{91} - 12q^{92} - 20q^{93} + 15q^{94} - 48q^{95} - 5q^{96} - 6q^{97} - 4q^{98} - 12q^{99} + O(q^{100})$$
Basis of coefficient ring in terms of a root $$\nu$$ of $$x^{10} - 2 x^{9} + 15 x^{8} + 14 x^{7} + 110 x^{6} + 36 x^{5} + 233 x^{4} + 164 x^{3} + 345 x^{2} + 76 x + 16$$:
$$\beta_{0}$$ $$=$$ $$1$$ $$\beta_{1}$$ $$=$$ $$\nu$$ $$\beta_{2}$$ $$=$$ $$($$$$-503 \nu^{9} - 2241 \nu^{8} + 8466 \nu^{7} - 67528 \nu^{6} + 19422 \nu^{5} - 156870 \nu^{4} + 1003571 \nu^{3} - 301041 \nu^{2} - 66732 \nu + 438544$$$$)/2044008$$ $$\beta_{3}$$ $$=$$ $$($$$$-3064 \nu^{9} + 9207 \nu^{8} - 34782 \nu^{7} - 28346 \nu^{6} - 79794 \nu^{5} + 644490 \nu^{4} + 579550 \nu^{3} + 1236807 \nu^{2} + 274164 \nu + 3620228$$$$)/1022004$$ $$\beta_{4}$$ $$=$$ $$($$$$-3977 \nu^{9} - 10833 \nu^{8} - 12699 \nu^{7} - 334006 \nu^{6} - 625302 \nu^{5} - 1723536 \nu^{4} - 478621 \nu^{3} - 3196425 \nu^{2} - 3391749 \nu - 3118196$$$$)/1022004$$ $$\beta_{5}$$ $$=$$ $$($$$$-27409 \nu^{9} + 54315 \nu^{8} - 413376 \nu^{7} - 375260 \nu^{6} - 3082518 \nu^{5} - 967302 \nu^{4} - 6543167 \nu^{3} - 3491505 \nu^{2} - 9757146 \nu - 2149816$$$$)/2044008$$ $$\beta_{6}$$ $$=$$ $$($$$$-14545 \nu^{9} + 29941 \nu^{8} - 216152 \nu^{7} - 200758 \nu^{6} - 1521222 \nu^{5} - 440214 \nu^{4} - 2887781 \nu^{3} - 2349679 \nu^{2} - 3963994 \nu - 1074620$$$$)/340668$$ $$\beta_{7}$$ $$=$$ $$($$$$45928 \nu^{9} - 96426 \nu^{8} + 695481 \nu^{7} + 561428 \nu^{6} + 5037264 \nu^{5} + 744876 \nu^{4} + 10649492 \nu^{3} + 5556402 \nu^{2} + 16896855 \nu + 165664$$$$)/1022004$$ $$\beta_{8}$$ $$=$$ $$($$$$97465 \nu^{9} - 178473 \nu^{8} + 1399728 \nu^{7} + 1662752 \nu^{6} + 10555542 \nu^{5} + 4653846 \nu^{4} + 20509619 \nu^{3} + 17738031 \nu^{2} + 29957226 \nu + 5377120$$$$)/2044008$$ $$\beta_{9}$$ $$=$$ $$($$$$51344 \nu^{9} - 117222 \nu^{8} + 805587 \nu^{7} + 484042 \nu^{6} + 5520312 \nu^{5} + 367938 \nu^{4} + 11604142 \nu^{3} + 5109630 \nu^{2} + 14145267 \nu + 742892$$$$)/1022004$$
$$1$$ $$=$$ $$\beta_0$$ $$\nu$$ $$=$$ $$\beta_{1}$$ $$\nu^{2}$$ $$=$$ $$\beta_{9} + 4 \beta_{5} - \beta_{4} + 2 \beta_{2} + 2 \beta_{1}$$ $$\nu^{3}$$ $$=$$ $$-3 \beta_{8} - 3 \beta_{6} - 3 \beta_{4} - 2 \beta_{3} + 11 \beta_{2} - 6$$ $$\nu^{4}$$ $$=$$ $$-16 \beta_{9} - 16 \beta_{8} + 6 \beta_{7} - 18 \beta_{6} - 40 \beta_{5} + 2 \beta_{4} - 37 \beta_{1} - 40$$ $$\nu^{5}$$ $$=$$ $$-65 \beta_{9} + 4 \beta_{8} + 34 \beta_{7} - 4 \beta_{6} - 126 \beta_{5} + 65 \beta_{4} + 34 \beta_{3} - 168 \beta_{2} - 168 \beta_{1}$$ $$\nu^{6}$$ $$=$$ $$-30 \beta_{9} + 297 \beta_{8} + 267 \beta_{6} + 267 \beta_{4} + 126 \beta_{3} - 657 \beta_{2} + 592$$ $$\nu^{7}$$ $$=$$ $$1080 \beta_{9} + 1080 \beta_{8} - 564 \beta_{7} + 1176 \beta_{6} + 2280 \beta_{5} - 96 \beta_{4} + 2797 \beta_{1} + 2280$$ $$\nu^{8}$$ $$=$$ $$5005 \beta_{9} - 468 \beta_{8} - 2256 \beta_{7} + 468 \beta_{6} + 9784 \beta_{5} - 5005 \beta_{4} - 2256 \beta_{3} + 11402 \beta_{2} + 11402 \beta_{1}$$ $$\nu^{9}$$ $$=$$ $$1788 \beta_{9} - 20451 \beta_{8} - 18663 \beta_{6} - 18663 \beta_{4} - 9542 \beta_{3} + 47603 \beta_{2} - 39726$$
## Character values
We give the values of $$\chi$$ on generators for $$\left(\mathbb{Z}/546\mathbb{Z}\right)^\times$$.
$$n$$ $$157$$ $$365$$ $$379$$ $$\chi(n)$$ $$-1 - \beta_{5}$$ $$1$$ $$-1 - \beta_{5}$$
## Embeddings
For each embedding $$\iota_m$$ of the coefficient field, the values $$\iota_m(a_n)$$ are shown below.
For more information on an embedded modular form you can click on its label.
Label $$\iota_m(\nu)$$ $$a_{2}$$ $$a_{3}$$ $$a_{4}$$ $$a_{5}$$ $$a_{6}$$ $$a_{7}$$ $$a_{8}$$ $$a_{9}$$ $$a_{10}$$
289.1
2.07085 − 3.58682i 0.769836 − 1.33339i −0.114009 + 0.197470i −0.623307 + 1.07960i −1.10337 + 1.91109i 2.07085 + 3.58682i 0.769836 + 1.33339i −0.114009 − 0.197470i −0.623307 − 1.07960i −1.10337 − 1.91109i
−1.00000 0.500000 0.866025i 1.00000 −2.07085 + 3.58682i −0.500000 + 0.866025i 0.321703 2.62612i −1.00000 −0.500000 0.866025i 2.07085 3.58682i
289.2 −1.00000 0.500000 0.866025i 1.00000 −0.769836 + 1.33339i −0.500000 + 0.866025i −2.22250 + 1.43544i −1.00000 −0.500000 0.866025i 0.769836 1.33339i
289.3 −1.00000 0.500000 0.866025i 1.00000 0.114009 0.197470i −0.500000 + 0.866025i −2.59452 + 0.518144i −1.00000 −0.500000 0.866025i −0.114009 + 0.197470i
289.4 −1.00000 0.500000 0.866025i 1.00000 0.623307 1.07960i −0.500000 + 0.866025i 2.30301 + 1.30235i −1.00000 −0.500000 0.866025i −0.623307 + 1.07960i
289.5 −1.00000 0.500000 0.866025i 1.00000 1.10337 1.91109i −0.500000 + 0.866025i 1.19230 2.36187i −1.00000 −0.500000 0.866025i −1.10337 + 1.91109i
529.1 −1.00000 0.500000 + 0.866025i 1.00000 −2.07085 3.58682i −0.500000 0.866025i 0.321703 + 2.62612i −1.00000 −0.500000 + 0.866025i 2.07085 + 3.58682i
529.2 −1.00000 0.500000 + 0.866025i 1.00000 −0.769836 1.33339i −0.500000 0.866025i −2.22250 1.43544i −1.00000 −0.500000 + 0.866025i 0.769836 + 1.33339i
529.3 −1.00000 0.500000 + 0.866025i 1.00000 0.114009 + 0.197470i −0.500000 0.866025i −2.59452 0.518144i −1.00000 −0.500000 + 0.866025i −0.114009 0.197470i
529.4 −1.00000 0.500000 + 0.866025i 1.00000 0.623307 + 1.07960i −0.500000 0.866025i 2.30301 1.30235i −1.00000 −0.500000 + 0.866025i −0.623307 1.07960i
529.5 −1.00000 0.500000 + 0.866025i 1.00000 1.10337 + 1.91109i −0.500000 0.866025i 1.19230 + 2.36187i −1.00000 −0.500000 + 0.866025i −1.10337 1.91109i
$$n$$: e.g. 2-40 or 990-1000 Embeddings: e.g. 1-3 or 529.5 Significant digits: Format: Complex embeddings Normalized embeddings Satake parameters Satake angles
## Inner twists
Char Parity Ord Mult Type
1.a even 1 1 trivial
91.h even 3 1 inner
## Twists
By twisting character orbit
Char Parity Ord Mult Type Twist Min Dim
1.a even 1 1 trivial 546.2.j.e 10
3.b odd 2 1 1638.2.m.k 10
7.c even 3 1 546.2.k.e yes 10
13.c even 3 1 546.2.k.e yes 10
21.h odd 6 1 1638.2.p.j 10
39.i odd 6 1 1638.2.p.j 10
91.h even 3 1 inner 546.2.j.e 10
273.s odd 6 1 1638.2.m.k 10
By twisted newform orbit
Twist Min Dim Char Parity Ord Mult Type
546.2.j.e 10 1.a even 1 1 trivial
546.2.j.e 10 91.h even 3 1 inner
546.2.k.e yes 10 7.c even 3 1
546.2.k.e yes 10 13.c even 3 1
1638.2.m.k 10 3.b odd 2 1
1638.2.m.k 10 273.s odd 6 1
1638.2.p.j 10 21.h odd 6 1
1638.2.p.j 10 39.i odd 6 1
## Hecke kernels
This newform subspace can be constructed as the kernel of the linear operator $$T_{5}^{10} + \cdots$$ acting on $$S_{2}^{\mathrm{new}}(546, [\chi])$$.
## Hecke characteristic polynomials
$p$ $F_p(T)$
$2$ $$( 1 + T )^{10}$$
$3$ $$( 1 - T + T^{2} )^{5}$$
$5$ $$16 - 76 T + 345 T^{2} - 164 T^{3} + 233 T^{4} - 36 T^{5} + 110 T^{6} - 14 T^{7} + 15 T^{8} + 2 T^{9} + T^{10}$$
$7$ $$16807 + 4802 T + 1078 T^{3} + 308 T^{4} - 51 T^{5} + 44 T^{6} + 22 T^{7} + 2 T^{9} + T^{10}$$
$11$ $$900 + 630 T + 2391 T^{2} - 2085 T^{3} + 3793 T^{4} - 1062 T^{5} + 555 T^{6} - 58 T^{7} + 48 T^{8} - 6 T^{9} + T^{10}$$
$13$ $$371293 + 114244 T + 52728 T^{2} + 7436 T^{3} + 1118 T^{4} + 15 T^{5} + 86 T^{6} + 44 T^{7} + 24 T^{8} + 4 T^{9} + T^{10}$$
$17$ $$( 28 - 17 T - 71 T^{2} - 20 T^{3} + 4 T^{4} + T^{5} )^{2}$$
$19$ $$12321 + 122211 T + 1200102 T^{2} + 135327 T^{3} + 88183 T^{4} - 804 T^{5} + 3987 T^{6} - 11 T^{7} + 78 T^{8} - 3 T^{9} + T^{10}$$
$23$ $$( 360 - 549 T - 537 T^{2} - 72 T^{3} + 6 T^{4} + T^{5} )^{2}$$
$29$ $$236196 - 196830 T + 177147 T^{2} - 47385 T^{3} + 25029 T^{4} - 2106 T^{5} + 3195 T^{6} - 54 T^{7} + 60 T^{8} + T^{10}$$
$31$ $$1254400 + 1647520 T + 1604961 T^{2} + 675789 T^{3} + 221955 T^{4} + 41274 T^{5} + 7137 T^{6} + 738 T^{7} + 126 T^{8} + 10 T^{9} + T^{10}$$
$37$ $$( -565 + 368 T + 17 T^{2} - 38 T^{3} + T^{4} + T^{5} )^{2}$$
$41$ $$770884 + 194038 T + 160347 T^{2} + 28125 T^{3} + 19689 T^{4} + 3174 T^{5} + 1311 T^{6} + 126 T^{7} + 48 T^{8} + 4 T^{9} + T^{10}$$
$43$ $$308025 - 213120 T + 277881 T^{2} - 22980 T^{3} + 92728 T^{4} - 22221 T^{5} + 10725 T^{6} - 164 T^{7} + 111 T^{8} - 3 T^{9} + T^{10}$$
$47$ $$271854144 + 17164008 T + 19550241 T^{2} + 614784 T^{3} + 1063294 T^{4} + 45738 T^{5} + 18675 T^{6} + 1430 T^{7} + 279 T^{8} + 15 T^{9} + T^{10}$$
$53$ $$256 + 66704 T + 17364273 T^{2} + 4244362 T^{3} + 1078286 T^{4} + 131550 T^{5} + 21575 T^{6} + 2206 T^{7} + 279 T^{8} + 17 T^{9} + T^{10}$$
$59$ $$( 1268 + 1867 T + 440 T^{2} - 167 T^{3} + 2 T^{4} + T^{5} )^{2}$$
$61$ $$6400 + 81280 T + 1064256 T^{2} - 407680 T^{3} + 150992 T^{4} - 25632 T^{5} + 5480 T^{6} - 712 T^{7} + 129 T^{8} - 11 T^{9} + T^{10}$$
$67$ $$400 + 14380 T + 520941 T^{2} - 137601 T^{3} + 138084 T^{4} + 25845 T^{5} + 18249 T^{6} + 261 T^{7} + 138 T^{8} + T^{9} + T^{10}$$
$71$ $$202500 + 2544750 T + 32555925 T^{2} - 7241610 T^{3} + 1686319 T^{4} - 192492 T^{5} + 28812 T^{6} - 2726 T^{7} + 315 T^{8} - 18 T^{9} + T^{10}$$
$73$ $$37941975369 - 3240866106 T + 887285502 T^{2} - 50704044 T^{3} + 11876944 T^{4} - 622851 T^{5} + 90666 T^{6} - 3100 T^{7} + 408 T^{8} - 12 T^{9} + T^{10}$$
$79$ $$817216 - 709640 T + 580065 T^{2} - 159768 T^{3} + 60951 T^{4} - 4344 T^{5} + 4416 T^{6} - 204 T^{7} + 87 T^{8} + 4 T^{9} + T^{10}$$
$83$ $$( 486 + 405 T - 27 T^{2} - 60 T^{3} + T^{5} )^{2}$$
$89$ $$( 453250 + 50575 T - 3770 T^{2} - 464 T^{3} + 7 T^{4} + T^{5} )^{2}$$
$97$ $$1734489 + 1457919 T + 1797027 T^{2} + 159624 T^{3} + 449455 T^{4} + 93495 T^{5} + 60546 T^{6} - 590 T^{7} + 279 T^{8} + 6 T^{9} + T^{10}$$ | 2021-05-10 06:54:04 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9999903440475464, "perplexity": 10724.559690647875}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-21/segments/1620243989115.2/warc/CC-MAIN-20210510064318-20210510094318-00377.warc.gz"} |
https://jp.maplesoft.com/support/help/view.aspx?path=Fractals%2FEscapeTime%2FMandelbrot | Mandelbrot - Maple Help
Fractals[EscapeTime]
Mandelbrot
Mandelbrot fractal generator
Calling Sequence Mandelbrot( n, zbl, zur ) Mandelbrot( n, zbl, zur, opts )
Parameters
n - positive integer; specifies the dimensions of the square Array output zbl - complex(realcons); the bottom left corner of a box in the complex plane zur - complex(realcons); the upper right corner of a box in the complex plane opts - (optional) keyword options of the form opt=value where opt and value are described below
Options
• output : keyword option of the form output=value where value is a name or list of names denoting the returned Array image(s). The accepted names are layer1, layer2, color, or raw. The default value is color.
• iterationlimit : keyword option of the form iterationlimit=value where value is a positive integer specifying the number of iterations the formula is applied. The default value is 25.
• cutoff : keyword option of the form cutoff=value where value is positive and of type realcons. The iterative process is stopped for each complex input point when the absolute value exceeds the cutoff value. The default value is 100.0.
• container : An n-by-n-by-2 Array with datatype=float[8] and order=Fortran_order used in-place to store the raw data.
Description
• The Mandelbrot command generates Array images which provide a visualization of the Mandelbrot set. The entries of the image are shaded according to the behavior of complex input points under a particular iterative rational map. Taking an initial point z[0]=0+0*I then for each complex input point c a sequence of points z[i] are computed until abs(z[i]) exceeds the cutoff value (ie. escapes).
${z}_{i}={z}_{i-1}^{2}+c,{z}_{0}=0$
• For each entry point in the designated complex box the iterative process is repeated until either the iteration limit or the cutoff value is exceeded.
• The two-dimensional grayscale Array image returned by supplying the option output=layer1 contains data denoting the number of iterations required for each entry to escape. The grayscale image returned by supplying the option output=layer2 contains the absolute values of the final values for entries which escape. For either layer the real data is scaled to 0.0 .. 1.0 before being returned as an image.
• The three-dimensional color Array image returned by supplying the option output=color contains data where the three layers corresponding to red, green, and blue have been computed using the raw escape data.
• The three-dimensional Array returned by supplying the option output=raw contains the unscaled data of layer1 and layer2. This Array can be used to generate a customized color image using the Colorize command.
Examples
> $\mathrm{with}\left(\mathrm{Fractals}:-\mathrm{EscapeTime}\right)$
$\left[{\mathrm{BurningShip}}{,}{\mathrm{Colorize}}{,}{\mathrm{HSVColorize}}{,}{\mathrm{Julia}}{,}{\mathrm{LColorize}}{,}{\mathrm{Lyapunov}}{,}{\mathrm{Mandelbrot}}{,}{\mathrm{Newton}}\right]$ (5.1)
> $\mathrm{with}\left(\mathrm{ImageTools}\right):$
> $M≔\mathrm{Mandelbrot}\left(500,-2.0-1.35I,0.7+1.35I\right)$
${{\mathrm{_rtable}}}_{{36893628547053409092}}$ (5.2)
> $\mathrm{Embed}\left(M\right)$
> $\mathrm{Embed}\left(\mathrm{Mandelbrot}\left(500,-2.0-1.35I,0.7+1.35I,\mathrm{output}=\mathrm{layer1}\right)\right)$
> $\mathrm{Embed}\left(\left[\mathrm{Mandelbrot}\left(200,-2.0-1.35I,0.7+1.35I,\mathrm{iterationlimit}=10,\mathrm{output}=\mathrm{layer1}\right),\mathrm{Mandelbrot}\left(200,-2.0-1.35I,0.7+1.35I,\mathrm{iterationlimit}=25,\mathrm{output}=\mathrm{layer1}\right),\mathrm{Mandelbrot}\left(200,-2.0-1.35I,0.7+1.35I,\mathrm{iterationlimit}=125,\mathrm{output}=\mathrm{layer1}\right)\right]\right)$
> $\mathrm{Embed}\left(\mathrm{Mandelbrot}\left(500,-0.1515+1.032I,-0.1575+1.043I,\mathrm{iterationlimit}=300,\mathrm{cutoff}=50.0\right)\right)$
> $\mathrm{Embed}\left(\mathrm{Mandelbrot}\left(500,-2.0-1.35I,0.7+1.35I,\mathrm{output}=\mathrm{layer2}\right)\right)$
> $\mathrm{Embed}\left(\left[\mathrm{Mandelbrot}\left(200,-2.0-1.35I,0.7+1.35I,\mathrm{cutoff}=4.0,\mathrm{output}=\mathrm{layer2}\right),\mathrm{Mandelbrot}\left(200,-2.0-1.35I,0.7+1.35I,\mathrm{cutoff}=100.0,\mathrm{output}=\mathrm{layer2}\right),\mathrm{Mandelbrot}\left(200,-2.0-1.35I,0.7+1.35I,\mathrm{cutoff}=10000.0,\mathrm{output}=\mathrm{layer2}\right)\right]\right)$
> $R≔\mathrm{Array}\left(1..500,1..500,1..2,\mathrm{datatype}=\mathrm{float}\left[8\right]\right):$
> $\mathrm{Mandelbrot}\left(500,-0.1025+0.95I,-0.095+0.96I,\mathrm{iterationlimit}=200,\mathrm{cutoff}=50.0,\mathrm{output}=\mathrm{raw},\mathrm{container}=R\right):$
> $\mathrm{Embed}\left(\left[\mathrm{FitIntensity}\left(R\left[..,..,1\right]\right),\mathrm{FitIntensity}\left(R\left[..,..,2\right]\right)\right]\right)$
> $P≔\mathrm{Colorize}\left(500,R,\mathrm{Array}\left(\left[11,3,2\right],\mathrm{datatype}=\mathrm{integer}\left[4\right]\right),\mathrm{Array}\left(\left[0.256,0.859,0.256\right],\mathrm{datatype}=\mathrm{float}\left[8\right]\right),\mathrm{rgb}=0,\mathrm{mode}=4,\mathrm{layer}=2\right):$
> $\mathrm{Embed}\left(P\right)$
Compatibility
• The Fractals:-EscapeTime:-Mandelbrot command was introduced in Maple 18. | 2022-12-07 17:04:47 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 17, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5729283094406128, "perplexity": 1159.4437250006988}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446711200.6/warc/CC-MAIN-20221207153419-20221207183419-00365.warc.gz"} |
http://aesop.phys.utk.edu/ph401/ | A SURVEY OF PHYSICS
Physics 401 - Spring 2017
HOME THE HIGGS PARTICLE
George Siopsis - April 18, 2017
After introducing the Maxwell equations and reviewing the salient features of electromagnetism, I explain how the Maxwell equations can be modified to describe the weak interactions that are responsible for $\beta$-decay. The journey takes us through spontaneous symmetry breaking and the Higgs mechanism which gives rise to the Higgs particle that seems to be of central importance to our existence.
It was discovered at the Large Hadron Collider (LHC) in Geneva, Switzerland in 2011.
### The Nobel Prize in Physics 2013
Photo: A. Mahmoud François Englert Photo: A. Mahmoud Peter W. Higgs The Nobel Prize in Physics 2013 was awarded jointly to François Englert and Peter W. Higgs "for the theoretical discovery of a mechanism that contributes to our understanding of the origin of mass of subatomic particles, and which recently was confirmed through the discovery of the predicted fundamental particle, by the ATLAS and CMS experiments at CERN's Large Hadron Collider" | 2018-11-13 04:23:46 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.2933380603790283, "perplexity": 649.1723130417231}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-47/segments/1542039741219.9/warc/CC-MAIN-20181113041552-20181113063552-00314.warc.gz"} |
http://onsnetwork.org/scottveirs/2014/02/07/basic-r-studio-and-ggplot2-syntax/ | Basic R Studio and ggplot2 syntax
While I continue to try to get Sage to underpin the analysis stages of my open scientific workflow, a LOT of my colleagues and cohort use R these days. So this week I installed R Studio and ggplot2 for OSX.
Here are notes from my first foray into plotting histograms.
First I took some date/time data in a Libre Office spreadsheet, converted it to decimal hour of day to keep things simple, renamed column headers to be simple single words (e.g. underscores to connect words), and then saved the active sheet as a tab-delimited .csv file (in this case named 2012-human.csv).
> lk2012<-read.table("/Users/scott/Documents/beamreach/research/pubs-in-progress/02a - ship SL/r-scott/2012-human.csv",header=T,sep="\t")
> png("/Users/scott/Documents/beamreach/research/pubs-in-progress/02a - ship SL/r-scott/lk-human-hr-hists.png")
> par(mfrow= c(3, 1))
> hist(lk2011$hour,breaks=24) > hist(lk2012$hour,breaks=24)
> hist(lk2013\$hour,breaks=24)
> dev.off()
That reads the .csv file into lk2012, then prints this 3 panel plot of hour histograms —
Histograms of the hours in which reports were made to the Salish Sea hydrophone network’s Google-spreadsheet-based observation log in 2011, 2012, & 2013.
More on ggplot2 as I learn it… but know that it is making most of the plots for our current paper in progress regarding underwater noise from Salish Sea ships. | 2017-06-23 01:37:56 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 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": 0.209535613656044, "perplexity": 8196.232182198408}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-26/segments/1498128319943.55/warc/CC-MAIN-20170623012730-20170623032730-00161.warc.gz"} |
https://practice-questions.maxtute.com/CBSE-Class-10-Math/trigonometry/trigonometric-ratios-complementary-angles-extra-question-04.shtml | # Extra Questions For Class 10 Maths Chapter 8 #4
#### Trigonometry | Trigonometric Values of Complementary Angles
This CBSE class 10 Maths practice question is from the topic Trigonometry. It tests the concept of trigonometric values of complementary angles.
Question 4 : In ΔABC right angled at B, sin C = $$frac{\text{5}}{\text{13}}$. Find$i) sin A
(ii) cos A
(iii) cos C
## NCERT Solution to Class 10 Maths
### Explanatory Answer | Trigonometry Extra Question 4
sin C = $$frac{\text{side opposite to ∠C}}{\text{hypotenuse}}$=$\frac{\text{AB}}{\text{AC}}$= $\frac{\text{5}}{\text{13}}$ ......$1)
Let AB = 5k and AC = 13k
#### Calculate Sine and Cosine values of the angles
Using Pythagoras theorem, BC = $$sqrt{AC^2− AB^2}$; BC = $\sqrt{$13k$^2− (5k)^2 }) = $$sqrt{169 k^2− 25 k^2 }$ = $\sqrt{144k^2 }$ = 12k sin A = $\frac{\text{side opposite to ∠A}}{\text{hypotenuse}}$= $\frac{\text{BC}}{\text{AC}}$= $\frac{\text{12k}}{\text{13k}}$= $\frac{\text{12}}{\text{13}}$ ......$2)
cos A = $$frac{\text{side adjcent to ∠ A}}{\text{hypotenuse}}$ = $\frac{\text{AB}}{\text{AC}}$= $\frac{\text{5k}}{\text{13k}}$= $\frac{\text{5}}{\text{13}}$ ......$3)
cos C = $$frac{\text{side adjcent to ∠ C}}{\text{hypotenuse}}$= $\frac{\text{BC}}{\text{AC}}$ = $\frac{\text{12k}}{\text{13k}}$ = $\frac{\text{12}}{\text{13}}$ ......$4)
Inference
From (1) and (3): sin A = cos C
From (2) and (4): sin C = cos A
#### Why?
If ABC is right angled at B, side adjacent to ∠ A will be the side opposite to ∠C.
Similarly side opposite to ∠A will be the side adjacent to ∠C.
i.e., ∠A = 90° - ∠C and ∠C = 90° - ∠A
#### Try CBSE Online CoachingClass 10 Maths
Register in 2 easy steps and
Start learning in 5 minutes! | 2023-03-30 01:25:43 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7800572514533997, "perplexity": 5065.980564788759}, "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/1679296949093.14/warc/CC-MAIN-20230330004340-20230330034340-00345.warc.gz"} |
https://damekdavis.wordpress.com/category/analysis/ | # There is no infinite dimensional Lebesgue measure
Imagine that there was a countably additive and translation invariant measure, ${m}$, on a Hilbert Space ${\mathcal{H}}$ such that every ball ${B(v, \varepsilon)}$ has positive measure, finite measure. Because ${\mathcal{H}}$ is infinite dimensional, there exists ${\{e_{i} | i = 1, \cdots, \infty\}}$ such that ${\langle e_{i}, e_{j}\rangle = \delta_{ij}}$. Note that the points ${e_{i}}$ and ${e_{j}}$ are far apart: ${\|e_{i} - e_{j}\|^2 = \|e_{i}\|^2 + \|e_{j}\|^2 = 2}$, if ${i \neq j}$. Now, the ball, ${B(0, 2)}$, contains the countable collection of disjoint balls: ${\{B(e_{i}, \frac{1}{2}) | i = 1,\cdots, \infty\}}$ (you can see where this is going). Thus, because all balls, ${B(e_{i}, \frac{1}{2})}$ have the same measure (by translation invariance), and ${m}$ is countable additive, it follows that ${\infty = \sum_{i=1}^\infty m(B(e_{i}, \frac{1}{2})) < B(0, 2) < \infty}$. Thus, we’ve reached a contradiction.
In some cases, it’s desirable to have an analogue of a measure on a Hilbert space ${\mathcal{H}}$. The most common application is to make statements such as “Property X is true for a.e. function.” One way to do make these statements is through the concept of Prevalence. Another is the definition of a Gaussian like measure called the Abstract Wiener Space.
Note that I wrote this article before realizing that Wikipedia has a similar article.
# Is the Product of Measurable Spaces the Categorical Product?
This post requires some knowledge of measure theory.
Today I’m going to show that the product of two measurable spaces ${(X, \mathcal{B}_X)}$ and ${(Y, \mathcal{B}_Y)}$, is actually the product in the category of measurable spaces. See Product (category theory).
The category of measurable spaces, ${\mathbf{Measble}}$, is the collection of objects ${(X,\mathcal{B}_X)}$, where ${X}$ is a set and ${\mathcal{B}_X}$ is a ${\sigma}$-algebra on ${X}$, and the collection of morphisms ${\phi : (X, \mathcal{B}_X) \rightarrow (Y, \mathcal{B}_Y)}$ such that
1. ${\phi : X \rightarrow Y}$;
2. ${\phi^{-1}(E) \in \mathcal{B}_X}$ for all ${E \in \mathcal{B}_Y}$.
Such a function ${\phi}$ is called a measurable morphism, and ${(X, \mathcal{B}_X)}$ is called a measurable space.
Given two measurable spaces ${(X, \mathcal{B}_X)}$ and ${(Y, \mathcal{B}_Y)}$ we can define their product ${(X\times Y, \mathcal{B}_X \times \mathcal{B}_Y)}$, where ${X\times Y}$ is the cartesian product of ${X}$ and ${Y}$, and ${\mathcal{B}_X \times \mathcal{B}_Y}$ is the ${\sigma}$-algebra generated by the sets of the form ${E \times Y}$ and ${X \times F}$ with ${E \in \mathcal{B}_X}$ and ${F \in \mathcal{B}_Y}$. We will need the following definition: A ${\sigma}$-algebra ${\mathcal{B}}$ on a set ${Z}$ is said to be coarser than a ${\sigma}$-algebra ${\mathcal{B}'}$ on ${Z}$ if ${\mathcal{B} \subseteq \mathcal{B}'}$. In exercise 18 of Terry Tao’s notes on product measures, it is shown that ${\mathcal{B}_X \times \mathcal{B}_Y}$ is the coarsest ${\sigma}$-algebra on ${X\times Y}$ such that the projection maps ${\pi_X}$ and ${\pi_Y}$ are both measurable morphisms.
Now, we are finally ready to show that ${(X\times Y, \mathcal{B}_X\times \mathcal{B}_Y)}$ is the categorical product of the measurable spaces ${(X, \mathcal{B}_X)}$ and ${(Y, \mathcal{B}_Y)}$. If ${\phi_X : (Z, \mathcal{B}_Z) \rightarrow (X, \mathcal{B}_X)}$ and ${\phi_Y : (Z, \mathcal{B}_Z) \rightarrow (Y, \mathcal{B}_Y)}$ are measurable morphisms, then we need to show that there exists a unique measurable morphism ${\phi_{X\times Y} : (Z, \mathcal{B}_Z) \rightarrow (X\times Y, \mathcal{B}_X \times \mathcal{B}_Y)}$ such that ${\phi_X = \pi_X \circ \phi_{X\times Y}}$ and ${\phi_Y = \pi_Y \circ \phi_{X\times Y}}$. Because the cartesian product is the product in the category of sets, we only have one choice for such a map. Indeed, ${\phi_{X \times Y} = (\phi_X, \phi_Y)}$.
We claim that ${\phi_{X \times Y}}$ is measurable. Indeed, because the pullback ${\phi_{X \times Y}^{-1} : 2^{X\times Y} \rightarrow 2^Z}$ respects arbitrary unions and complements, and ${\phi_{X\times Y}(\emptyset) = \emptyset}$, we only need to show that ${\phi_{X\times Y}^{-1}(E \times Y) \in \mathcal{B}_Z}$ and ${\phi_{X\times Y}^{-1}(X \times F) \in \mathcal{B}_Z}$ for all ${E \in \mathcal{B}_X}$ and ${F \in \mathcal{B}_Y}$ (see remark 4 of Terry Tao’s notes on abstract measure spaces). This is easy to show:
$\displaystyle \begin{array}{rcl} \phi^{-1}_{X\times Y} (E\times Y) &=& (\pi_X \circ \phi_{X\times Y})^{-1}(E) \\ &=& \phi_X^{-1}(E) \\ \phi^{-1}_{X\times Y} (X \times F) &=& (\pi_Y \circ \phi_{X\times Y})^{-1}(F) \\ &=& \phi_Y^{-1}(F) \end{array}$
are both ${\mathcal{B}_Z}$ measurable because ${E \in \mathcal{B}_X}$ and ${F \in \mathcal{B}_Y}$.
Thus, we have shown that ${(X \times Y, \mathcal{B}_X\times \mathcal{B}_Y)}$ is actually the product of the measurable spaces ${(X, \mathcal{B}_X)}$ and ${(Y, \mathcal{B}_Y)}$ in the category of measurable spaces. This is reassuring, otherwise the term “product space” would be misleading.
I would like to know if there is a category of measure spaces with objects ${(X, \mathcal{B}_X, \mu_X)}$, where ${X}$ is a set, ${\mathcal{B}_X}$ is a ${\sigma}$-algebra on ${X}$, and ${\mu_X : \mathcal{B}_X \rightarrow [0, +\infty]}$ is a measure. There would need to be some extra condition on morphisms in this category, otherwise we couldn’t distinguish between triples ${(X, \mathcal{B}_X, \mu_X)}$ and ${(X, \mathcal{B}_X, \mu_X')}$ where ${\mu_X \neq \mu_X'}$. If this was a category, I don’t believe products could exist. Indeed, if the measure spaces ${(X, \mathcal{B}_X, \mu_X)}$ and ${(Y, \mathcal{B}_Y, \mu_Y)}$ are not ${\sigma}$-finite, then there can be multiple measures on ${(X \times Y, \mathcal{B}_X \times \mathcal{B}_Y)}$: see Terry Tao’s notes on product measures.
Another question is whether equalizers, coproducts, coequalizer, etc. exist in the category of measurable spaces. If a sufficient number of these properties exist, then we can take categorical (co)limits of measurable spaces. These might be of some interest already, but I have not looked into it. | 2017-07-28 14:46:38 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 90, "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": 0.9430924654006958, "perplexity": 77.65825222775793}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-30/segments/1500550969387.94/warc/CC-MAIN-20170728143709-20170728163709-00614.warc.gz"} |
https://www.andlearning.org/percent-composition-formula/ | Connect with us
## What is percent composition?
The percent composition for a compound generally determines the elementary composition of the compound in the form of grams for each element and divided by the total number of grams present. The percent composition or mass percent of each element is usually equal to the mass of a particular element that is further divided by the total mass present and multiplied by 100 percent.
It is expressed in mass and useful in chemistry when you have to know percentage of total weight of the compound made up of a particular element. So, how to calculate the percentage composition in Chemistry. Well, there is a well-defined formula for the same purpose and you just have to put the values to calculate the final outcome. The formula is taken even more important for chemical analysis process and it can be given as below –
### Percent Composition Formula
$\ Percent\;Composition = \frac{Grams\;of\;element}{Grams\;of\;Compounds} \times 100$
Where, %Ce is the percent composition of the element that you are interested in calculating. gE is the weight of a particular element in terms of grams and gT is the total weight of elements present in the compound. And finally, the ration is multiplied by 100 to get the percentage form of the composition.
The formula is frequently used to measure the concentration of an element for a given mixture. It can be used for plenty of real-life applications too. Chemists are using percentage composition formula to find the empirical formula of the compound that further helps in calculating the actual molecular formula too and an exact number of atoms within a compound. | 2022-01-25 01:13:51 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7488473057746887, "perplexity": 364.04544030481014}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320304749.63/warc/CC-MAIN-20220125005757-20220125035757-00527.warc.gz"} |
https://codegolf.stackexchange.com/questions/17230/random-literal-number | # Random literal number [duplicate]
I took inspiration from a challenge I did some years ago in a programming competition.
Your goal is to write a function that generates a random integer number between 0 and 150 (included), and then prints it in actual words. For example, 139 would be printed as onehundredthirtynine.
Rules:
• You cannot use any external resource, nor any library that will perform the conversion.
• The shortest answer will win.
Good luck!
## marked as duplicate by manatwork, John Dvorak, ProgramFOX, Doorknob♦, Darren StoneJan 2 '14 at 19:17
• With or without spaces? – ProgramFOX Jan 2 '14 at 18:53
• Without spaces, as you can see in the example! :) – Vereos Jan 2 '14 at 18:54
• ... Damn, I actually searched for it. Didn't use proper keywords :/ – Vereos Jan 2 '14 at 18:57
• Don't worry, neither I found it easily. (I knew another question which links to that one.) – manatwork Jan 2 '14 at 19:00
# Lua
print"four"
Highly optimized and golfed xkcd's random number generator
# An actual solution, 380
x=math.random(0,150)m=math.floor print(x==0 and"zero"or((x>99 and"onehundred"or"")..(({"ten","eleven","twelve","thirteen","fourteen","fifteen","sixteen","seventeen","eighteen","nineteen"})[x%100-9]or({[0]="",nil,"twenty","thirty","fourty","fifty","sixty","seventy","eighty","ninety"})[m(x/10)%10]..({[0]="","one","two","three","four","five","six","seven","eight","nine"})[x%10])))
• print"un" in French. How about just "un"? – Cees Timmerman Jan 2 '14 at 19:04
• PHP has barewords. 2 characters in PHP – John Dvorak Jan 2 '14 at 19:05
• still an invalid answer. This isn't code-trolling. – John Dvorak Jan 2 '14 at 19:06
• – Cees Timmerman Jan 2 '14 at 19:08 | 2019-07-18 18:21:02 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.2838272154331207, "perplexity": 5277.896665139356}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-30/segments/1563195525699.51/warc/CC-MAIN-20190718170249-20190718192249-00377.warc.gz"} |
https://www.shaalaa.com/question-bank-solutions/oxidation-number-introduction-assign-oxidation-numbers-underlined-elements-each-following-species-cao2_11340 | CBSE (Science) Class 11CBSE
Share
# Assign Oxidation Numbers to the Underlined Elements in Each of the Following Species Cao2 - CBSE (Science) Class 11 - Chemistry
ConceptOxidation Number - Introduction
#### Question
Assign oxidation numbers to the underlined elements in each of the following species
CaO2
#### Solution 1
CaO2
Then, we have
(+2) + 2(x) = 0
=> 2 + 2x = 0
=>x = -1
Hence, the oxidation number of O is – 1.
#### Solution 2
Let the oxidation number of CaO_2 be x
2 + 2x = 0 (∵ oxy No of a = +2)
x = -1
Thus oxidation number of O in CaO_2 = -1
Is there an error in this question or solution?
#### APPEARS IN
Solution Assign Oxidation Numbers to the Underlined Elements in Each of the Following Species Cao2 Concept: Oxidation Number - Introduction.
S | 2019-12-12 08:23:01 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.40359804034233093, "perplexity": 8459.628517626184}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-51/segments/1575540542644.69/warc/CC-MAIN-20191212074623-20191212102623-00018.warc.gz"} |
https://www.statstutor.net/downloads/question-2002694-elasticity-theory/ | # Question #2002694: Elasticity Theory
Question: Suppose that, at the current price of $1.50 per gallon and average household income of$100,000 a yr, the quantity demanded of bottled water is 200 million gallons a week. If the price were increased to $1.68, the quantity demanded would fall to 158.7 million gallons a week. If the household income were increased to$110, 500 a yr, the quantity demanded would rise to 208 million gallons a week.
(A) Calculate the own price elasticity of demand.
(B) Calculate the income elasticity of demand
(C) According to these estimates, is bottled water a normal or inferior product?
Solution: The solution consists of 151 words (1 page)
Deliverables: Word Document
0 | 2019-10-15 19:43:23 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5219998359680176, "perplexity": 3372.00559490862}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-43/segments/1570986660231.30/warc/CC-MAIN-20191015182235-20191015205735-00403.warc.gz"} |
http://onsnetwork.org/blog/category/miscellaneous/ | # Open mind, open science [Speech for Tilburg uni board]
This is a speech I gave for the board of Tilburg University on why Open science is important for the future of TIlburg University or any knowledge institute, honestly. Speech was given on March 9, 2017.
We produce loads of knowledge at this university, and yet we throw most of that knowledge away. We are throwing away taxpayer’s money; stifling scientific discovery; hampering the curiosity of our students and society.
Research data; peer reviews; research materials; developed software; research articles — they are literally and figuratively thrown away. But these are all building blocks for new knowledge, and the more building blocks available, the more knowledge we can build. Science is a bit like Legos in that sense: more pieces allow you to build greater structures.
Students can use these building blocks to learn better — learn about the real, messy process of discovery for example. Businesses can create innovative tools for information consumption by taking the hitherto unavailable information and reshaping it into something valuable. Citizens can more readily participate in the scientific process. And last but not least, science can become more accurate and more reliable.
Researchers from different faculties and staff members from different support services see the great impact of open science, and today I call on you to give make it part of the new strategic plan.
Let’s as a university work towards making building blocks readily available instead of throwing or locking them away.
Open Access and Open Data, two of these building blocks, have already been on the board’s agenda. All researchers at Tilburg are mandated to make their publications freely available since 2016 and free to re-use by 2024. For Open Data, the plans are already in motion to make all data open per 2018, across all faculties, as I was happy to read in a recent memorandum. So, why am I here?
Open Access and Open Data are part of the larger idea of Open Science; they are only two of many building blocks. Open science is that all research pieces are available for anyone to use for any purpose.
Advancing society is only possible if we radically include society in what we do. The association of Dutch universities, state secretary Sander Dekker, and 17 other organizations have underscored the importance of this when they signed the National Plan Open Science just a few weeks ago.
So I am happy to see the landscape is shifting from closed to open. However, it is happening slowly and incompletely if we only focus only on data and access. Today, I want to focus on one of the biggest problems facing science, that open science can solve: selective publication.
Researchers produce beautifully written articles that read like the script of a good movie: they set the scene with an introduction and methods, have beautiful results, and provide a happy ending in the discussion that makes us think we actually understand the world. And just like movies, not all are available in the public theatre.
But research isn’t about good and successful stories that sell; it is about accuracy. We need to see not just the successful stories, we need to see all stories. We need to see the entire story, not just the exciting parts. Only then can we efficiently produce new knowledge and and understand society.
Because good movies pretend to find effects even if there is truly nothing to find. Here, researchers investigate the relation between jelly beans and pimples.
So they start researching. Nothing.
xkcd comic “Significant”; https://www.xkcd.com/882/
More studies; nothing.
More studies; an effect and a good story!
More studies; nothing.
And what is shared? The good story. While there is utterly nothing to find. And this happens daily.
Researchers fool each other, including themselves, and it has been shown time and time again that we researchers wish to see what we want to see. This human bias greatly harms science.
The results are disconcerting. Psychology; cancer research; life sciences; economics — all fields have issues with providing a valid understanding of the world, to a worrying extent. This is due to researchers fooling themselves and confirming prior beliefs to produce good movies instead of being skeptical and producing accurate, good science.
So I and other members across the faculties and services say: Out with the good movies, in with the good science we can actually build on — OPEN science.
Sharing all research that is properly conducted is feasible and will increase validity of results. Moreover, it will lead to less research waste. We as a university could become the first university to share all our research output. All based on a realistic notion of a researcher: do not evaluate results based on whether they are easy to process, confirm your expectations, and whether they provide a good story — evaluate them on their methods.
But please please members of our university, do not expect this change open science to come easily or by magically installing a few policies!
It requires a cultural shift that branches out into the departments and even the individual researchers’ offices. Policies don’t necessarily result in behavior change.
And as a researcher, I want to empirically demonstrate that policy doesn’t necessarily result in behavioral change.
Here is my Open Access audit for this university. Even though policies have been instated by the university board, progress is absent and we are actually doing worse at making our knowledge available to society than in 2012. This way we will not reach ANY of our Open Access goals we have set out.
Open Access audit Tilburg University; data and code: https://github.com/libscie/access-audit
In sum, let us advance science by making it open, which in turn will help us advance society. I will keep fighting for more open science. Anyone present, student or staff, I encourage you to do so as well. I am here to help.
Open science is out of the box, and it won’t go back in. The question is, what are we as a university going to do with that knowledge?
# DNase Treatment – Abalone Water Filters for RLO Viability
The RNA I isolated earlier today was subjected to DNase treatment using the Turbo DNA-free Kit (Invitrogen), following the manufacturer’s standard protocol.
After DNase inactivation treatment, the RNA was transferred (recovered ~19uL from each samples) to a clear, low-profile PCR plate.
The plate layout is here (Google Sheet): 20170309_RLO_viability_DNased_RNA_plate_layout
The samples will be subjected to qPCR to assess the presence/absence of residual gDNA. The plate of DNased RNA was stored @ -80C in the original box that the water filters were stored in.
An overview of the experiment and the various treatments are viewable in the “Viability Trial 3″ tab of Lisa’s spreadsheet (Google Sheet): RLO Viability & ID50
# RNA Isolation – Abalone Water Filters for RLO Viability
Water filters stored at -80C in ~1mL of RNAzol RT were provided by Lisa. This is part of an experiment (and Capstone project) to assess RLO viability outside of the host.
The samples were thawed and briefly homogenized (as best I could) with a disposable plastic pestle. The samples were then processed according to the manufacturer’s protocol for total RNA isolation. Samples were resuspended in 25μL of nuclease-free water (Promega).
The experimental samples and the various treatments are viewable in the “Viability Trial 3″ tab of Lisa’s spreadsheet (Google Sheet): RLO Viability & ID50
# PBS recipe
1X Phosphate Buffered Saline (PBS Buffer) Recipe
Dissolve in 800ml distilled H2O:
8g of NaCl
0.2g of KCl
1.44g of Na2HPO4
0.24g of KH2PO4
Adjust pH to 7.4 with HCl.
Adjust volume to 1L with distilled H2O.
Sterilize by autoclaving.
# Manuscript Writing – More “Nuances” Using Authorea
I previously highlighted some of the issues I was having using Authorea.com as an writing platform.
As a collaborative writing platform, it also has issues.
As it turns out, comments are currently only viewable when using Private Browsing/Incognito modes on your browser!!!
I found this out by using the chat feature that’s built into Authorea. This feature is great and support is pretty quick to respond:
However, Josh at Authorea suggested this bug would be resolved by the end of the day (that was on February 24th). I took the above screenshots of my manuscript demonstrating that comments don’t show up when using a browser like a normal person, today, February 28th…
Another significant shortcoming to using Authorea as a collaborative writing platform, as it relates to comments:
Jay received notice from UC Berkeley that the sequencing data from his coral RADseq was ready. In addition, the sequencing contains some epiRADseq data from samples provided by Hollie Putnam. See his notebook for multiple links that describe library preparation (indexing and barcodes), sample pooling, and species breakdown.
I’ve downloaded both the demultiplexed and non-demultiplexed data, verified data integrity by generating and comparing MD5 checksums, copied the files to each of the three species folders on owl/nightingales that were sequenced (Panopea generosa, Anthopleura elegantissima, Porites astreoides), generated and compared MD5 checksums for the files in their directories on owl/nightingales, and created/updated the readme files in each respective folder.
Data management is detailed in the Jupyter notebook below. The notebook is embedded in this post, but it may be easier to view on GitHub (linked below).
Readme files were updated outside of the notebook.
Jupyter notebook (GitHub): 20170227_docker_jay_ngs_data_retrieval.ipynb
# Sample Prep – Pinto Abalone Tissue/RNA for Collabs at UC-Irvine
We need to send half of each sample that we have from Sean Bennett’s Capstone project to Alyssa Braciszewski at UC-Irvine.
This is quite the project! There are ~75 samples, and about half of those are tissues (presumably digestive gland) stored in RNAzol RT. The remainder are RNA that has already been isolated. Additionally, tube labels are not always clear and there are duplicates. All of these factors led to this taking an entire day in order to decipher and process all the samples.
I selected samples from only those that I was confident in their identity.
I aliquoted 25μL of each RNA for shipment to Alyssa.
Tissue samples were thawed and tissue was cut in half using razor blades.
Planning to send samples on Monday.
Lisa has already assembled a master spreadsheet to try to keep track of all the samples and what they are (Google Sheet): Pinto Transcriptome
Here’s the list of samples I’ll be sending to Alyssa (Google Sheet): 20170222_pinto_abalone_samples
Here are some images to detail some of the issues I had to deal with in sample ID/selection.
# Symbiodinium cp23S Re-PCR
Yesterday I completed some re-do PCRs of Symbiodinium cp23S from the branching Porites samples Sanoosh worked on over the past summer. Some of the samples did not amplify at all, so I reattempted PCR of these samples (107, 108, 112, 116). Sample 105 amplified last summer but the sequence was lousy, so I redid that one too. After the first PCR, I obtained 1 ul of the product and diluted it 1:100 in water. I then used 1 ul of this diluted product as the template for a second round of PCR. PCR conditions were the same Sanoosh and I used last summer (based on Santos et al. 2002):
Reagent Volume (µl) water 17.2 5X Green Buffer 2.5 MgCl2 25 mM 2.5 dNTP mix 10 mM 0.6 Go Taq 5U/µl 0.2 primer 23S1 10 µM 0.5 primer 23S2 10 µM 0.5 Master Mix volume 24 sample 1 total volume 25
Initial denaturing period of 1 min at 95 °C, 35 cycles of 95 °C for 45 s, 55 °C for 45 s, and 72 °C for 1 min, and a final extension period of 7 min.
Samples were then run on a 1% agarose gel for 30 min at 135 volts.
Surprisingly, the first round of PCR amplified samples 107 and 112 (note: two subsamples of each were run; one that was the original extraction diluted (d) and another that was the original cleaned with Zymo OneStep PCR Inhibitor Removal kit (c)). The cleaned samples were the ones that amplified. I believe Sanoosh had tried these cleaned samples with no success.
The second round of PCR produced faint bands for both of the 108 samples. Sample 116 still did not amplify.
I cleaned the samples with the NEB Monarch Kit and shipped them today to Sequetech. I combined the two 108 samples to ensure enough DNA for sequencing.
This is a belated post for some RAD library prep I did the week of January 23rd in the Leache Lab. I followed the same ddRAD/EpiRAD protocol I used in August. Samples included mostly Porites astreoides from the transplant experiment, as well as some geoduck samples from the OA experiment, and a handful of green and brown Anthopleura elegantissima. Sample metadata can be found here. The library prep sheet is here. The TapeStation report is here. Below is the gel image from the TapeStation report showing that the size selection was successful. However, the selection produced fragments with a mean size of 519-550 base pairs, as opposed to the size selection in August which produced ~500 bp fragments. While there will obviously be some overlap between libraries, combining samples from the two libraries may be problematic. This occurred despite identical Pippen Prep settings targeting fragments 415-515 bp. Libraries were submitted to UC Berkeley on 1/31/17 for 100 bp paired-end sequencing on the HiSeq 4000.
Library JD002_A-L
# Interview Danish Psychology Association responses
Below, I copy my responses to an interview for the Danish Psychology Association. My responses are in italic. I don’t know when the article will be shared, but I am posting my responses here, licensed CC0. This is also my way of sharing the full responses, which won’t be copied verbatim into an article because they are simply too lengthy.What do you envision that this kind of technology could do in a foreseable future?
What do you mean by “this” kind of technology? If you mean computerized tools assisting scholars, I think there is massive potential in both development of new tools to extract information (for example what ContentMine is doing) and in application. Some formidable means are already here. For example, how much time do you spend as a scholar to produce your manuscript when you want to submit it? This does not need to cost half a day when there are highly advanced, modern submission managers. Same when submitting revisions. Additionally, annotating documents colloboratively on the Internet with hypothes.is is great fun, highly educational, and productive. I could go on and on about the potential of computerized tools for scholars.
Why do you think this kind of computerized statistical policing is necessary in the field of psychology and in science in general?Again, what is “this kind of computerized statistical policing”? I assume you’re talking about statcheck only for the rest of my answer. Moreover, it is not policing — a spell-checker does not police your grammar, it helps you improve your grammar. statcheck does not police your reporting, it helps you improve your reporting. Additionaly, I would like to reverse the question: should science not care about the precision of scientific results? With all the rhetoric going on in the USA about ‘alternative facts’, I think it highlights how dangerous it is to let go of our desire to be precise in what we do. Science’s inprecision has trickle down effects in the policies that are subsequently put in place, for example. We put in all kinds of creative and financial effort to progress our society, why should we let it be diminished by simple mistakes that can be prevented so easily? If we agree that science has to be precise in the evidence it presents, we need to take steps to make sure it is. Making a mistake is not a problem, it is all about how you deal with it.
So far the Statcheck tool is only checking if the math behind the statistical calculations in the published articles are wrong when the null-hypothesis significance testing has been used. What you refer to as reporting errors in your article from December last year published in Behaviour Research Methods. But these findings aren’t problematic as long as the conclusions in the articles aren’t affected by the reporting errors?
They aren’t problematic?—who is the judge of whether errors aren’t problematic? If you consider just statistical significance, there are still 1/8 papers that contain such a problem. Moreover, all errors in reported results affect meta-analyses — is that not also problematic down-the-line? I find it showing of hubris for any individual to say that they can determine whether something is problematic or not, when there can be many things that that person doesn’t realize even can be affected. It should be open to discussion, so information about problems need to be shared and discussed. This is exactly what I aimed to do with the statcheck reports on PubPeer for a very specific problem.
In the article in Behaviour Research Methods you find that half of all published psychology papers that use NHST contained at least one p-value that was inconsistent with its test statistic and degrees of freedom. And that One in eight papers contained a grossly inconsistent p-value that may have affected the statistical conclusion. What does this mean? I’m not a mathematician.
You don’t need to be a mathematician to understand this. Say we have a set of eight research articles presenting statistical results with certain conclusions. Four of those eight will contain a result that does not match up to the results presented (i.e., inconsistent), but does not affect the broad strokes of the conclusion. One of those eight contains a result that does not match up to the conclusion and potentially nullifies the conclusions. For example, if a study contains a result that does not match up with the conclusion, but concluded that a new behavioral therapy is effective at treating depression. That means the evidence for the therapy effectiveness is undermined — affecting direct clinical benefits as a result.
Why are these findings important?
Science is vital to our society. Science is based on empirical evidence. Hence, it is vital to our society that empirical evidence is precise and not distorted by preventable or remediable mistakes. Researchers make mistakes, no big deal. People like to believe scientists are more objective and more precise than other humans — but we’re not. The way we build checks- and balances to prevent mistakes from proliferating and propagating into (for example) policy is crucial. statcheck contributes to understanding and correcting one specific aspect of such mistakes we can all make.
Why did you decide to run the statcheck on psychology papers specifically?
statcheck was designed to extract statistical results reported as prescribed by the American Psychological Association. It is one of the most standardized ways of reporting statistical results. It makes sense to apply software developed on standards in psychology to psychology.
Why do you find so many statistical errors in psychology papers specifically?
I don’t think this is a problem to psychology specifically, but more a problem of how empirical evidence is reported and how manuscripts are written.
Are psychologists not as skilled at doing statistical calculations as other scholars?
I don’t think psychologists are worse at doing statistical calculations. I think point-and-click software has made it easy for scholars to compute statistical results, but not to insert them into manuscripts reliably. Typing in those results is error prone. I make mistakes when I’m doing my finances at home, because I have to copy the numbers. I wish I had something like statcheck for my finances. But I don’t. For scientific results, I promote writing manuscripts dynamically. This means that you no longer type in the results manually, but inject the code that contains the result. This is already possible with tools such as Rmarkdown and can greatly increase the productivity of the researcher. It has saved my skin multiple times, although you still have to be vigilant for mistakes (wrong code produces wrong results).
Have you run the Statcheck tool on your own statistical NHST-testing in the mentioned article?
Yes! This was the first thing I did, way before I was running it on other papers. Moreover, I was non-selective when I started scanning other people’s papers — I apparently even made a statcheck report that got posted on PubPeer for my supervisor (see here). He laughed, because the paper was on reporting inconsistencies and the gross inconsistency was simply an example of one in the running text. A false positive, highlighting that statcheck‘s results always need to be checked by a human before concluding anything definitive.
Critics call Statcheck “a new form of harassment” and accuse you of being “a self appointed data police”. Can you understand these reactions?
Proponents of statcheck praise it as a good service. Researchers who study how researchers conduct research are called methodological terrorists. Any change comes with proponents and critics. Am I a self-appointed data policer? To some, maybe. To others, I am simply providing a service. I don’t chase individuals and I am not interested in that at all — I do not see myself as part of a “data police”. That people think these reports is like getting reprimanded highlights to me that there still rests a taboo on skepticism within science. Skepticism is one of the ideals of science, so let’s aim for that.
Why do you find it necessary to send out thousands of emails to scholars around the world informing them that their work has been reviewed and point out to them if they have miscalculated?
It was not necessary — I thought it was worthwhile. Why do some scholars find it necessary to e-mail a colleague about their thoughts on a paper? Because they think it is worthwhile and can help them or the original authors. Exactly my intentions by teaming up with PubPeer and posting those 50,000 statcheck reports.
Isn’t it necessary and important for ethical reasons to be able to make a distinction between deliberate miscalculations and miscalculations by mistake when you do this kind of statcheck?
If I was making accusations about gross incompetence towards the original authors, such a distinction would clearly be needed. But I did not make accusations at all. I simply stated the information available, without any normative or judging statements. Mass-scale post-publication peer review of course brings with it ethical problems, which I carefully weighed before I started posting statcheck reports with the PubPeer team. The formulation of these reports was discussed within our group and we all agreed this was worthwhile to do.
As a journalist I can write and publish an article with one or two factual errors. This doesn’t mean the article isn’t of a general high journalistic standard or that the content of the article isn’t of great relevance for the public- couldn’t you make the same argument about a scientific article? And when you catalogue these errors online you are at the risk of blowing up a storm in a tea cup and turn everybody’s eyes away from the actual scientific findings?
Journalists and scholars are playing different games. An offside in football is not a problem in tennis and the comparison between journalists and scholars seems similar to me. I am not saying that an article is worthless if it contains an inconsistency, I just say that it is worth looking at before building new research lines on it. Psychology has wasted millions and millions of euros/dollars/pounds/etc on chasing ephemeral effects that are totally unreasonable, as several replication projects have highlighted in the last years. Moreover, I think the general opinion of science will only improve if we are more skeptical and critical of each other instead of trusting findings based on reputation, historical precedent, or ease with which we can assimilate the findings. | 2017-03-28 04:23:46 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.2601790130138397, "perplexity": 2377.6206305900805}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-13/segments/1490218189667.42/warc/CC-MAIN-20170322212949-00550-ip-10-233-31-227.ec2.internal.warc.gz"} |
https://quant.stackexchange.com/questions/57249/quanto-cds-basic-question | # Quanto CDS- basic question- [closed]
Just wanted to know if the quanto CDS hedge each other or not, if we assume that the quanto ratio is 100%(1).
Also, is it true that in stressed condition the volatility of CDS with home ccy decreases as compare to volatility of CDS with foreign ccy. If yes, what is the rationale. | 2020-09-23 15:43:28 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.847669243812561, "perplexity": 4574.4464964556055}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-40/segments/1600400211096.40/warc/CC-MAIN-20200923144247-20200923174247-00569.warc.gz"} |
https://stats.stackexchange.com/questions/248676/analysis-strategy-for-rare-outcome-with-matching | # Analysis strategy for rare outcome with matching
I'm working with a dataset of ~100,000 individuals where ~500 (0.5%) individuals received treatment.
I have several continuous and count outcomes for all observations that I would like to compare between treated and untreated. It would be important for the analysis to match individuals on several characteristics (that could be binary, continuous or categorical).
I'm working with Stata.
I was brainstorming several possible scenarios that include:
1. stratified analyses of treated and untreated
2. treating it as case control study and attempting to match 'controls' to all my 'cases' where criteria of match allow. Then moving forward with analysis appropriate for that set up (conditional logit would work for binary outcomes.. not sure about continuous and count ones..)
3. Treatment-effects estimation, perhaps using propensity-score matching (not sure if and how it is possible to include categorical variables though..)
What analysis would be most appropriate for such dataset?
• It would be helpful if you could elaborate what several means? 5, 10, 30? Do you know much overlap and balancing is in your data between treatment and control group? I think the Gelman and Hill is a good textbook to start with. For matching, see stat.columbia.edu/~gelman/arm/chap10.pdf – Arne Jonas Warnke Dec 2 '16 at 9:45
• @ArneJonasWarnke Minmum (from epidemiological point of view) would be - sex, age (continuous or categorized) & ~6-10 main disease types. – radek Dec 2 '16 at 10:20
• Since your control group is so large, can you do perfect matching on (at least some of) those characteristics / pre-existing diseases? This eliminates all biases due to differences in those characteristics. You can also combine perfect matching (in a first stage) with propsensity score matching, see for example faculty.unlv.edu/nasser/ECO%20772,%20Econometrics%20II/… – Arne Jonas Warnke Dec 2 '16 at 10:31
• Thanks @ArneJonasWarnke. Assuming such matched dataset (and discarding unmatched controls) - what would be the best method for such scenario? – radek Dec 2 '16 at 12:02
Removing good data from an analysis is scientifically suspect in my humble opinion, and naive matching methods are inefficient. It may be very easy to adjust for patient characteristics using ordinary regression models, paying attention to linearity assumptions etc. Of course it is a good idea to look at overlap in covariate distributions across treatment groups to see where assumptions of no interaction between treatment and characteristics might be on shaky ground and untestable.
• Thanks @Frank. Should I then simply start considering models using all data and use an estimate on a binary categorical variable (received treatment) as an estimate of difference for treated group? – radek Dec 6 '16 at 9:07
• Yes with careful and fairly liberal covariate adjustment. Propensity score analysis is for the opposite situation where the outcome is what is rare. – Frank Harrell Dec 6 '16 at 11:18
• And how about situations when certain combinations of sex, age and disease type when no treatment was found? Should they be excluded beforehand? Or kept in the model (since they will not provide any information I believe)? Also the opposite situation might raise a challenge - there are few rare cases when age/sex/disease combination has disease only, but no observations without disease.. – radek Dec 6 '16 at 13:39
• You have to decide which variables are unlikely to interact with the other variables. When effects are additive you don't need to have all combinations of values well represented in the data. – Frank Harrell Dec 6 '16 at 21:10
• Missed the deadline for the bounty. I haven't managed to solve the problem but @Frank's solution will most likely be the preferred one. Thanks for help! – radek Dec 11 '16 at 16:06
Based on the comments and the availability of such a large control group, I would probably advise to do in a step first exact matching on age groups and sex, and perhaps common disease groups. Hereby, you built different strata. In a second step, you can apply propensity score matching to ensure that treatment and control group are as balanced as possible with respect to the remaining observables.
You can do this apparently using the psmatch2 package for Stata (I have used that package only briefly out of interested).
A code example is given in the help file:
g att = .
egen g = group(groupvars)
levels g, local(gr)
qui foreach j of local gr {
psmatch2 treatvar varlist if g==j', out(outvar)
replace att = r(att) if g==j'
}
sum att
See here for further information
http://repec.org/bocode/p/psmatch2.html
You should -- of course -- verify that there is enough overlap between treatment and control group within each strata.
* Update: Response to the comment of Frank Harrell *
Why I argue for matching:
It is a trade-off between, on the one hand, balancing covariates as close as possible between treatment and control group, and, on the other hand, removing data (what Frank Harrell emphasized).
It is clear that the estimator becomes in a first step less efficient if you remove data, and you should justify ignoring data. But radek has asked for matching approaches and I agree that this is a good idea.
Matching avoids to some extent "extrapolation bias" if the covariate distribution differs between treatment and control group. You drop observations which give you few or any information about the treatment effect because their covariates are very far away from the sample.
Many prominent researchers therefore recommend matching or subclassificatioon plus regression.
• I don't see why matching plays a role here. – Frank Harrell Dec 11 '16 at 16:11
• I find this argument not convincing. Matching on continuous variables results in an incomplete adjustment because the variables have to be binned. Matching throws away good data from observations that would be good matches. Extrapolation bias is only a significant problem if there is a covariate by group interaction, and users of matching methods ignore interactions anyway. If you don't want to make regression assumptions that are unverifiable, remove observations outside the overlap region just as with matching. – Frank Harrell Dec 12 '16 at 12:07
The propensity score (PS) is a balancing score indicating the probability of treatment assignment conditional on observed baseline characteristics. In a randomized controlled Trial (RCT) the PS is known. Estimation and application of PS therefore, mimic some of the particular characteristics of an RCT and therefore are method of choice to estimate the treatment effect in an observational study, provided that you have no unmeasured confounders and all subjects have a non-zero probability of receiving Treatment.
You can estimate the PS using logistic regression or generalized boosting methods and therefore, include all observed (continous and categorical) baseline variables. Notably, you do not include the outcome parameter. (There is an ongoing debate, which variables to include, however, Austin postulates it is safe to include all observed baseline variables https://www.ncbi.nlm.nih.gov/pmc/articles/PMC3144483/). It is important to underscore that you do not aim to find the best "predictive model" but instead the model that balances your covariates best. After application of the PS using stratification, matching, inverse probability Treatment weights or covariate adjustment (all of them have pros and cons and are somewhat dependent on your data), you can test the treatment effect on your outcomes. | 2019-11-13 07:42:49 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.48402345180511475, "perplexity": 1663.585197982141}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-47/segments/1573496666229.84/warc/CC-MAIN-20191113063049-20191113091049-00441.warc.gz"} |
https://math.stackexchange.com/questions/1583159/how-to-invert-matrix-in-finite-field | # How to invert matrix in finite field
I want to invert matrix $A$ in the finite field $\mathbb{F} = \mathbb{F}_2[x]/p(x)\mathbb{F}_2$ with $p(x)=x^8+x^4+x^3+x+1$. This finite field is used by the encryption scheme AES.
$A = \begin{pmatrix} x^6+x^4+x^2+x+1 & x^5+x^3+1 & x^5+x^2+1 \\ x^7+x^4+x & x^4+x & x^2+1 \\ x^6+x^4+x^3+1 & x^6+x^3+x & x^4+x^3 \end{pmatrix}$
For inverting $A$ I am supposed to use the Gaussian algorithm. The first step would be to divide the first row by its first entry, i.e. $x^6+x^4+x^2+x+1$.
How do I find the solution of such divisions, for example $(x^5+x^3+1)/(x^6+x^4+x^2+x+1)$. The finite field contains $2^8$ elements. Hence it is impossible for me to first calculate the multiplication table.
• Do you have to do it by hand? – AHusain Dec 20 '15 at 11:49
• Yes, I do. As this is Rijndael's finite field I suppose I can use some lookup tables, but I don't know how to make use of them to answer my question. They only provide multiplicative inverses, but do not help to do a division of two field elements. – null Dec 20 '15 at 11:51
• Well if you have inverses, you have the division because multiplication isn't too bad. $(x^5 + x^3 + 1)*(x^6 + x^4 + x^2 + x + 1 )^{-1}$ – AHusain Dec 20 '15 at 11:56
• Good comment. I'll try it! – null Dec 20 '15 at 12:09
• Unless you are aiming to implement this on a device that is extremely low on memory, you could (IMHO should) build logarithm tables converting the field multiplication and division to (modular) integer addition and subtraction. See this Q&A I prepared for referrals where I try to explain how to use discrete logarithm tables for field operations. – Jyrki Lahtonen Dec 20 '15 at 16:58
As a Bézout relation is \begin{multline*} (x^7 + x^6 + x^4 + x^3 + x)(x^6 + x^4 + x^2 + x + 1) \\+(x^5 +x^4 + x^3 + x + 1)(x^8 + x^4 + x^3 + x + 1 ) =1 \end{multline*} one has in $\mathbf F_2[x]/(x^8 + x^4 + x^3 + x + 1 )$: $$(x^6 + x^4 + x^2 + x + 1)^{-1}=x^7 + x^6 + x^4 + x^3 + x.$$ | 2021-05-14 22:25:44 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6112457513809204, "perplexity": 240.7197798450298}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-21/segments/1620243991829.45/warc/CC-MAIN-20210514214157-20210515004157-00380.warc.gz"} |
https://www.wyzant.com/resources/answers/users/78405400 | 04/18/22
#### Italicize the thesis statement, underline the topic sentences for each supporting paragraph, and highlight the logical reasoning/ evidence.
Italicize the thesis statement, underline the topic sentences for each supporting paragraph, and highlight the logical reasoning/ evidence. The K-12 Program has been brought to attention since its... more
03/11/22
#### How many moles of each product?
For the reaction shown, calculate how many moles of each product form when the given amount of each reactant completely reacts. Assume that there is more than enough of the other... more
02/27/22
#### Given 2H2+ Oz--> 2H20 what mass of hydrogen is needed to create 125 g of water?
i just need a more ‘in depth’ explanation that would be wonderful
02/27/22
#### How many moles of O 2 are required to burn completely 63.5 g of C 6H 6, according to the following equation?
How many moles of O 2 are required to burn completely 63.5 g of C 6H 6, according to the following equation? 2C 6H 6 + 15O 2 ® 12CO 2 + 6H 2O
02/27/22
#### Calculate the mass of CaCO 3 present in the original sample.
A mixture of calcium oxide, CaO, and calcium carbonate, CaCO 3, that had a mass of 3.454 g was heated until all the calcium carbonate was decomposed according to the following equation. After... more
01/23/22
#### preparatory math
Goran is a software salesman. Let y represent his total pay (in dollars). Let x represent the number of copies of Math is Fun he sells. Suppose that x and y are related by the... more
01/16/22
#### When 10.0 g of liquid nitrogen is vapourized, 4000 J of energy is released. What is the molar enthalpy of vapourization for liquid nitrogen?
– 11.2 kJ/mol + 5.6 kJ/mol 0 kJ/mol + 11.2 kJ/mol – 5.6 kJ/mol
12/28/21
#### CHEMISTRY DOUBT - Why did Rutherford think the atom was empty?
Why did Rutherford think the atom was empty? It could be possible that the alpha particles went through the weak field of the positively charged particles due to the heavy mass and fast speed of... more
12/23/21
#### A large school district has four high schools with enrollments as shown. 40 school buses need to be allocated to the high schools. Determine how many buses are allocated to each of the high schools
Solve the problem with: Hamilton and Jefferson's Method of Apportionment. School RRHS CCHS HHHS MHS Total Enrollment 738 851 1234 982 3805
12/19/21
#### Analysis Question
A phone company determined that a higher income indicates that a customer has a more expensive calling plan. What does the analysis show?
12/10/21
#### A pottery studio gives three-hour and four-hour ceramics classes for $34 an hour. If the studio has collected$2958 for a total of 27 classes, How many three-hour classes and four-hour classes were...
paid for?Blank three-hour classes (Type an integer or a decimal)Blank four-hour classes (Type an integer or a decimal)
12/10/21
#### The radius 𝑟 , in inches, of a spherical balloon is related to the volume 𝑉 by r(V)=3√3V/4π.Air is pumped into the balloon so the volume after 𝑡 seconds is given by 𝑉(𝑡)=10+18𝑡.
Find an expression for the composite function 𝑟(𝑉(𝑡))What is the exact time in seconds when the radius reaches 16 inches?
12/08/21
#### I dont know the answer
A one-year membership to a gym costs $725. The registration fee is$125, and the remaining amount is paid monthly. Define a variable. Then write and solve an equation to find how much new members... more
12/08/21
#### chemistry labflow
Glucose (molar mass=180.16 g/mol) is a simple, soluble sugar. Glucose solutions are used to treat patients with low blood sugar.Suppose you prepare a glucose solution using the described... more
12/08/21
#### James earns a monthly salary of $1200 and a commission of$125 for each car he sells
a. Graph and equation that represents how much James earns in a month in which he sells x cars
12/04/21
#### Chemistry Distillation Questions
The components of an 80-mL rubbing alcohol sample were separated using distillation and the resulting percent recovery of isopropyl alcohol is 14.70%. Determine the volume, in... more
11/30/21
#### If crop B brings in a profit of $210 per acre, and crop A brings in a profit of$110 per acre, how many acres of each crop should the farmer plant maximize her profit?
A farmer is going to divide her 30-acre farm between two crops. The seed for crop A costs $25 per acre. Seed for crop B costs$50 per acre. The farmer can spend at most \$1250 on the seed. If crop B... more
11/28/21
#### Why We study political in geography field?
Why We study political in geography field?
11/27/21
#### How do I attach an image to questions?
I need help with my geometry homework, and I cannot attach a picture. I tried to attach the picture to a webpage, but It says “this content is under review.” What should I do?
## Still looking for help? Get the right answer, fast.
Get a free answer to a quick problem.
Most questions answered within 4 hours.
#### OR
Choose an expert and meet online. No packages or subscriptions, pay only for the time you need. | 2022-05-22 13:30:15 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.3722556531429291, "perplexity": 4121.903970438937}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662545548.56/warc/CC-MAIN-20220522125835-20220522155835-00749.warc.gz"} |
http://physics.stackexchange.com/questions/59991/is-there-a-material-that-changes-local-conductivity | # Is there a material that changes local conductivity
I hope this is the right forum to ask this question. Is there a material (preferably thin, like a membrane) that changes its local conductivity (by that I mean the permeability for an electric field; I hope it's the right term) upon excitation with light or heat? I have no idea where to start my search for something like that.
Thanks a lot.
-
Any semiconductor will do this. – John Rennie Apr 4 '13 at 7:19
Who would have thought that the answer is that easy. Just to make sure that I understand you correctly, I can have a sheet of semi-conducting material, locally shine a laser onto it and it will change its conductivity properties? – fabee Apr 4 '13 at 7:41
– John Rennie Apr 4 '13 at 8:43
Thanks a lot. Since I cannot accept your answer, I can at least upvote your comment. – fabee Apr 4 '13 at 9:16
You should probably accept Manish's answer since he at least put in the effort to write one :-) Accepting an answer flags the question as "answered" otherwise it sits in the "Unanswered questions" queue forever. – John Rennie Apr 4 '13 at 11:27 | 2015-07-08 07:06:38 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9053703546524048, "perplexity": 811.778086004698}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-27/segments/1435376093097.69/warc/CC-MAIN-20150627033453-00016-ip-10-179-60-89.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/showing-tha-a-random-variable-is-a-martingale.693086/ | # Showing tha a random variable is a martingale
1. May 22, 2013
### rickywaldron
I'm having a bit of a problem proving the second condition for a martingale, the discrete time branching process Z(n)=X(n)/m^n, where m is the mean number of offspring per individual and X(n) is the size of the nth generation.
I have E[z(n)]=E[x(n)]/m^n=m^n/m^n (from definition E[X^n]=m^n) = 1
which is less than infinity, so first condition passes
Then I get lost with E[Z(n+1) given X(1),X(2)....X(n)]...any clues on how to show this is equal to Z(n)? Thanks
2. May 23, 2013
### DrDu
Apparently E[X(n+1)|X(n)]=mX(n) as each individuum in the population has on the mean m offsprings. | 2017-12-11 23:09:02 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.815628170967102, "perplexity": 1479.4320039353893}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-51/segments/1512948514113.3/warc/CC-MAIN-20171211222541-20171212002541-00373.warc.gz"} |
https://domino.mpi-inf.mpg.de/internet/reports.nsf/c125634c000710d0c12560400034f45a/b1b01a7af5b39c6dc12562e300555691?OpenDocument | max planck institut
informatik
MPI-I-96-1-004
Exact ground states of two-dimensional $\pm J$ Ising Spin Glasses
De Simone, C. and Diehl, M. and Jünger, Michael and Mutzel, Petra and Reinelt, Gerhard and Rinaldi, G.
MPI-I-96-1-004. March 1996, 10 pages. | Status: available - back from printing | Next --> Entry | Previous <-- Entry
Abstract in LaTeX format:
In this paper we study the problem of finding an exact ground state of a two-dimensional $\pm J$ Ising spin glass on a square lattice with nearest neighbor interactions and periodic boundary conditions when there is
a concentration $p$ of negative bonds, with $p$ ranging between $0.1$ and $0.9$. With our exact algorithm we can determine ground states of grids of sizes up to $50\times 50$ in a moderate amount of computation time (up to one hour each) for several values of $p$. For the ground state energy of an infinite spin glass system with $p=0.5$ we estimate $E_{0.5}^\infty = -1.4015 \pm0.0008$.
We report on extensive computational tests based on more than $22\,000$ experiments.
Acknowledgement:
References to related material:
1222 KBytes; 334 KBytes
Please note: If you don't have a viewer for PostScript on your platform, try to install GhostScript and GhostView
URL to this document: http://domino.mpi-inf.mpg.de/internet/reports.nsf/NumberView/1996-1-004
BibTeX
@TECHREPORT{DeSimoneDiehlJuengerMutzelReineltRinaldi96a,
AUTHOR = {De Simone, C. and Diehl, M. and J{\"u}nger, Michael and Mutzel, Petra and Reinelt, Gerhard and Rinaldi, G.},
TITLE = {Exact ground states of two-dimensional $\pm J$ Ising Spin Glasses},
TYPE = {Research Report},
INSTITUTION = {Max-Planck-Institut f{\"u}r Informatik}, | 2019-10-23 11:13:37 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7496301531791687, "perplexity": 2686.339716841487}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-43/segments/1570987833089.90/warc/CC-MAIN-20191023094558-20191023122058-00468.warc.gz"} |
https://rupress.org/jcb/article/170/2/249/51611/SNAREs-can-promote-complete-fusion-and-hemifusion | Using a cell fusion assay, we show here that in addition to complete fusion SNAREs also promote hemifusion as an alternative outcome. Approximately 65% of events resulted in full fusion, and the remaining 35% in hemifusion; of those, approximately two thirds were permanent and approximately one third were reversible. We predict that this relatively close balance among outcomes could be tipped by binding of regulatory proteins to the SNAREs, allowing for dynamic physiological regulation between full fusion and reversible kiss-and-run–like events.
Cognate v- and t-SNARE on vesicles and target membrane pair to form the core machinery for intracellular membrane fusion (Sollner et al., 1993). Energy made available from the zipping-up of the SNARE complex (Sutton et al., 1998) is used to drive the fusion of lipid bilayers (Weber et al., 1998; Hu et al., 2003; Fix et al., 2004). To a remarkable degree, compartmental specificity of intracellular membrane fusion can be recapitulated from the pattern of fusion by isolated SNARE proteins (McNew et al., 2000a; Paumet et al., 2001, 2004; Parlati et al., 2002).
The “stalk hypothesis” (Chernomordik et al., 1987; Tamm et al., 2003) proposes that membrane fusion proceeds through a hemifusion intermediate before fusion pore opening. In model lipid bilayer fusion studies (Lee and Lentz, 1997), hemifusion appears to develop before inner leaflet or contents mixing, and a large variety of mutated viral fusion protein constructs give rise to a nonprogressing hemifusion endstate termed “unrestricted hemifusion” (Kemble et al., 1994; Melikyan et al., 1997, 2000; Chernomordik et al., 1998; Armstrong et al., 2000).
The exocytic fusion pores are dynamic and can “flicker” (Breckenridge and Almers, 1987; Monck and Fernandez, 1992). “Kiss and run,” the partial release of vesicle contents through a transient fusion pore that rapidly recloses, has been shown in the exocytosis of both large secretory granules (Alvarez de Toledo et al., 1993) and small synaptic vesicles (Gandhi and Stevens, 2003; Staal et al., 2004). Analogous reversible fusion events have not been reported in reconstituted SNARE systems. Indeed, these types of transient events would only be observable in an experimental system that monitors individual fusion events.
Here, we expand on a cell fusion assay in which “flipped” SNAREs are ectopically expressed on the cell surface (Hu et al., 2003) to monitor single fusion events between cells. Using a range of extracellular and intracellular membrane markers, content markers, and protein constructs, we find that SNAREs can promote hemifusion events as permanent outcomes to a surprising degree.
### Membrane fusion outcomes by SNAREs
Previously, we demonstrated that “flipped” SNAREs, fused to signal sequences (Fig. 1 A) and expressed on the cell surface, are sufficient to fuse whole cells, evidenced by the mixing of cytosolic fluorescent proteins and even fluorescently labeled whole nuclei (Hu et al., 2003). Transient or other fusion outcomes likely would not be apparent in an assay that monitors fusion in population (Weber et al., 1998). The cell–cell assay, however, provides a picture of individual fusion events in which minority populations may be detected. Here, we describe a variety of fusion assays designed to identify cells with differential mixing of lipids and contents.
The cell–cell fusion assays we present here use various combinations of soluble and lipidic probes to simultaneously monitor content and lipid mixing between cells. To monitor lipid mixing with the cell–cell fusion assays, we took advantage of the fact that the GM1 ganglioside is absent on the surface of CHO cells due to the lack of a key enzyme in the pathway of ganglioside biosynthesis (Rosales Fritz et al., 1997) but is present on the surface of MEF-3T3 cells (Fig. 1, B and C). The cell-specific localization of GM1 was detected through binding of fluorescently labeled cholera toxin β-subunit (FITC-ctxβ [Fig. 1 B, green] or Alexa594-ctxβ [Fig. 1 C, red]). When MEF-3T3 cells expressing flipped t-SNAREs (GM1 positive) were mixed with CHO cells expressing flipped v-SNAREs (GM1 negative), membrane lipid mixing can be observed as the development of ctxβ-dependent fluorescence on the previously GM1-negative v-SNARE cells. By including various fluorescently labeled content markers, two assays were developed in which different potential membrane fusion outcomes could be distinguished according to different patterns of color mixing (Fig. 1, B and C).
In the original flipped SNARE fusion assay containing a lipid mixing marker (Fig. 1 B), the t-SNARE cell membrane and nuclei were labeled using green (FITC) ctxβ and the cyan fluorescent protein (bearing a nuclear localization signal; CFP-nls), respectively. The v-SNARE cell cytoplasm was labeled with the red fluorescent protein (bearing nuclear export signal; RFP-nes). Thus, complete fusion will result in the complete mixing of red cytoplasm (from the v-cells), cyan nuclei (from the t-cells), and green-ctxβ–bound GM1 (from the t-cells). Lipid mixing can be detected by the transfer of only the GM1 lipid to the v-cells, but not the cyan nuclei.
In stable cell lines expressing flipped SNAREs and the appropriate fluorescent markers (as depicted in the original flipped SNARE fusion assay Fig. 1 B) and incubated together for 6 h, complete fusion occurs in 23 ± 3% (mean ± SD) of the cells in contact (Fig. 2, arrowheads), which is consistent with our earlier observations in COS cells (Hu et al., 2003). Lipid transfer without content mixing (i.e., incomplete fusion) was observed in 14 ± 3% (Fig. 2, arrows) of contacting cells. Z-section analysis confirmed the absence of a t-cell–derived cyan-nucleus in these incompletely fused cells (unpublished data). We also observed a reversible version of the incomplete fusion; i.e., v-cells containing lipid mixing markers that are no longer in contact with a t-cell (Fig. 2, asterisks). These kiss-and-run–like cells have apparently experienced transient mixing of the lipid bilayers to become GM1 positive without significant contents mixing before physically separating. We observe one such v-cell for about every 20 contacting v- and t-cell pairs. All three observed fusion outcomes are SNARE dependent. No fusion subtype was observed if SNARE pairing is disrupted by either titrating free t-SNARE with the cytoplasmic domain of the v-SNARE (Fig. 2, control) or by expressing Syntaxin 1 alone (without SNAP-25; not depicted).
### Incomplete fusion phenotype corresponds to hemifusion events
To further characterize the novel incomplete fusion phenotypes, we developed the hemifusion assay (Fig. 1 C) in which a smaller fluorescent probe (5-chloromethylfluorescein diacetate [CMFDA]) was used as a content marker. This modification would allow us to detect small nonenlarging fusion pores that might be missed with the original flipped SNARE fusion assay (Fig. 1 B) due to diffusional restriction of the bulky red fluorescent protein through these fusion pores. Furthermore, in this assay, all the fluorescent markers (green CMFDA as soluble content marker, CFP-nls as nuclear marker, and red Alexa594-ctxβ as membrane marker) are initially limited to the t-SNARE cell to clearly identify which fluorophore is able to diffuse to v-SNARE cells. Transfer of GM1 alone to a v-cell gives the striking result of a red-bordered cell with no fluorescent content (Fig. 3, arrows), which is indicative of incomplete fusion. This transfer of red fluorescence to the unstained v-SNARE population does not occur when mock-transfected cells are used instead of v-cells (Fig. 3, right column), confirming the SNARE dependence of the assay and the lack of free exchange of the GM1 lipid. In this assay, full fusion is indicated by the presence of multiple cyan nuclei in cells containing both green CMFDA and the red ctxβ markers. Quantitative analysis of these experiments gave rise to practically the same fusion phenotype ratios (complete fusion/incomplete fusion/reversible incomplete fusion, 6.5:2.5:1) observed with the original flipped SNARE fusion assay. Moreover, with this cell fusion assay we were able to identify a fourth kind of phenotype compatible with the classic kiss-and-run process; i.e., v-cells no longer in contact with t-cells that exchanged both lipids and contents but without having cyan nuclei. Unlike the reversible incomplete fusion, the classic kiss-and-run phenotype was observed only in 2 ± 1%. Because this value is not statistically significant, further characterization will be required to determine whether this phenotype is another outcome of the SNARE-mediated fusion. Labeling the v-cell cytoplasm with CMFDA instead of the t-cell cytoplasm did not change the final proportion of fusion outcomes (unpublished data). Thus, neither big (RFP-nes) nor small (CMFDA) soluble contents markers were able to diffuse between incomplete fused. These results are in line with previous studies with viral fusion proteins in which the hemifusion phenotype has been identified as an incomplete fusion event, where lipid mixing but not content mixing is observed (Kemble et al., 1994; Melikyan et al., 1997, 2000; Cohen and Melikyan, 1998). Therefore, the SNARE-mediated incomplete fusion phenotypes observed in our cell fusion assays most likely represent hemifusion events.
### Hemifusion as an outcome
Time course experiments (Fig. 4) using the original flipped SNARE fusion assay showed that hemifusion events accumulated slightly faster than complete fusion events and reached a plateau at 6 h, whereas complete fusion events continued to increase through 24 h. Eventually, all three populations settle into a plateau distribution in which complete fusion makes up ∼64% of the total events, permanent hemifusion comprises 27%, and reversible hemifusion events represent 9%. The apparent stability of this distribution implies that hemifusion by SNAREs is an alternative outcome, similar perhaps to the “unrestricted hemifusion” described for some hemagglutinin viral constructs (Kemble et al., 1994; Melikyan et al., 1997, 2000; Chernomordik et al., 1998). Thus, the hemifusion we observe does not appear to be an on-pathway intermediate to full fusion. However, we cannot rule out the existence of “restricted hemifusion,” the purported on-pathway intermediate inferred from chemical trapping of fusion products in various viral systems, because the kinetic of our cell fusion assays is limited by the time required by cells to attach to the substrate and to interact to each other. This limitation may be responsible for the slower kinetic of fusion observed with a cell-based assay compared with normal SNARE-mediated fusion events.
### The SNARE cytoplasmic domain with a lipid anchor is sufficient for hemifusion but not for fusion
When the transmembrane domain (TMD) of viral fusion protein hemagglutinin is replaced with glycosylphosphatidylinositol (GPI), a lipid anchor that spans only the outer leaflet of the membrane, membrane fusion ends in a hemifusion state where the outer monolayers of the membranes are fused while the inner monolayers and the aqueous contents remain segregated (Kemble et al., 1994; Melikyan et al., 1995b, 2000), establishing the central role of the HA TMD in completion of full fusion. Similarly, replacement of the TMD of SNARE proteins by lipid anchors that span a single leaflet of the bilayer inhibited complete fusion in vitro (McNew et al., 2000b; Rohde et al., 2003) and in vivo (Grote et al., 2000). To further test the role of the SNARE TMD in fusion and lipid mixing, we replaced the native TMDs of flipped VAMP22-116 and flipped Syntaxin 1 with the GPI-anchoring sequence of decay-accelerating factor (Fig. 1 A) to generate GPI-VAMP22-92 and GPI-Syntaxin186-265. When transfected, the GPI-anchored v- and t-SNAREs were expressed on the cell surface (Fig. 5). COS cells coexpressing GPI-VAMP2-92 and EGFP were incubated with soluble cognate t-SNAREs, consisting of the recombinant cytoplasmic domain of Syntaxin 1 complexed with full-length SNAP-25. In the confluent cell layer, soluble t-SNAREs only bound to the cells that expressed GPI-VAMP2 (Fig. 5 A, left, green cells), but not to the cells that were not transfected (Fig. 5 A, left, nonfluorescent cells), indicating that GPI-VAMP2 proteins were expressed on the cell surface and could bind t-SNAREs. Phosphatidylinositol-specific phospholipase C (PI-PLC) specifically cleaves the GPI linkage. When PI-PLC was present in the medium, little t-SNARE binding was observed (Fig. 5 A, right), confirming that GPI-VAMP2 proteins were anchored to the cell surface via GPI linkage. When COS cells were cotransfected with GPI-Syntaxin186-265 and flipped SNAP-25/T79A,N188A, which does not contain a TMD, both proteins were expressed on the cell surface (Fig. 5 B, top). Similar to flipped t-SNAREs (Hu et al., 2003), a considerable amount of Syntaxin186-265 and flipped SNAP-25 proteins reached the cell surface, although the majority of the GPI-anchored t-SNAREs remained inside the cells, probably retained by the ER quality control system (Fig. 5 B, bottom). Flipped SNAP-25 was anchored to the cell surface by forming a complex with Syntaxin186-265. As expected, when the COS cells were treated with PI-PLC, both GPI-Syntaxin and flipped SNAP-25 proteins were released from the cell surface (Fig. 5 C).
The GPI-anchored SNAREs were defective in membrane fusion (Fig. 6). When CHO stable cell lines expressing GPI-VAMP2 were mixed with MEF-3T3 stable cells expressing GPI-Syntaxin and flipped SNAP-25, only rare complete cell fusion events were observed, at a frequency at or below the spontaneous fusion rate of negative SNARE-independent control (Fig. 6, A and B). Addition of the Vc-peptide, which prestructures the t-SNARE and accelerates membrane fusion of flipped SNAREs (Melia et al., 2002; Hu et al., 2003), did not change the outcome (Fig. 6 B).
Lipid mixing, but not content mixing, of GPI-expressing cells was observed using both original flipped SNARE fusion assay (Fig. 6 A, left, arrows) and hemifusion assay (Fig. 6 A, right, arrows), indicating that GPI-anchored SNAREs promote hemifusion. Lipid mixing was SNARE dependent, as only background activity was observed when flipped SNAP-25 was not expressed in the t-cells (Fig. 6 B), when GPI-VAMP2 was not expressed in the v-cells (Fig. 6 A, last column), or when the cytoplasmic domain of VAMP2 was added to titrate the GPI-t-SNAREs on the surface of t-cells (Fig. 6 B). SNAREs with GPI lipid anchors promoted hemifusion efficiently: 24% of the v-cells and t-cells in contact hemifused within 6 h (Fig. 6 B). Vc-peptide modestly increased hemifusion to 28% of cognate cells in contact (Fig. 6 B).
Between the SNARE domain and the TMD of both Syntaxin and VAMP2 lies a highly basic linker region that has been shown to be partially embedded within the outer leaflet of the bilayer (Kweon et al., 2003). To explore whether these membrane-proximal amino acids influence the development of hemifusion, we generated two additional GPI-anchored VAMP2 and Syntaxin 1 constructs (GPI-VAMP2-84 and GPI-Syntaxin186-256), which bring the coiled-coil SNARE domains closer to the GPI membrane anchor (Fig. 1 A). Similar hemifusion activity was observed when using either GPI-VAMP2-92 or GPI-VAMP2-84 in the v-cells or when using either GPI-Syntaxin186-265 or GPI-Syntaxin186-256 in the t-cells (Table I). Thus, neither the TMD nor the membrane-embedded linker is necessary to promote hemifusion of SNARE-expressing cells.
### Size barriers for diffusion during hemifusion
We explored the structural boundaries imposed by hemifusion with both the flipped SNAREs and the GPI-anchored SNAREs. Hemifusion does not involve the mixing of the inner monolayers of the membranes or of soluble content. Not surprisingly, when v-cells were labeled with inner monolayer probes (EGFP with a farnesylation sequence [EGFP-f]; EYFP with a palmitoylation sequence [EYFP-pal]; or even the much smaller HA epitope fused to a farnesylation sequence) and mixed with t-cells, no transfer of the inner monolayer probes was observed in the hemifused cells (Fig. 7 A, top; and Fig. 7 B). To further determine if during hemifusion small and/or transient fusion pores open that could allow the selective passage of smaller molecules as it was observed for the hemagglutinin protein mediated fusion (Zimmerberg et al., 1994), in addition to CMFDA, we preloaded v-cells with other small (MW ∼500) soluble probes; e.g., Calcein AM, 4-chloromethyl-6,8-difluoro7-hydroxycoumarin (blue CMF2HC), and cell tracker red CMTPX. Although the transfer of GM1 was detected in the hemifused cells, no transfer of the small soluble probes was detected (Fig. 7, A and B). However, due to the limitation of the assay for detecting low level of dye transfer between cells, we cannot rule out the possibility that very small and/or transient fusion pores that do not enlarge might be forming, as it was documented for HA viral protein-mediated membrane fusion (Zimmerberg et al., 1994; Melikyan et al., 1995a; Razinkov et al., 1999).
There was a size barrier for the diffusion of molecules in the outer monolayers of hemifused cells. EYFP proteins (239 amino acids) with a GPI-anchor (GPI-EYFP) could not diffuse from v-cells to t-cells (Fig. 7 B), whereas the six amino acid epitope AU1 (DTYRYI) (GPI-AU1) transferred from the hemifused v- and t-cells nearly as efficiently as GM1 (Fig. 7 B).
### Hemifused cells do not show electrically active connecting fusion pores
Although we tested a variety of membrane and fluid phase markers with different molecular weight, charge, and diameter in our cell–cell fusion assay, we considered the possibility that these tracers might be hindered from traversing small fusion pores connecting the hemifused cells that would nevertheless have conductivity. For this reason, we extended our search for very small fusion pores by performing time resolved capacitance and conductance measurements using a whole cell patch configuration. This powerful technique has both the time resolution and sensitivity required to identify even short-lived reversible pores (Zimmerberg et al., 1994; Melikyan et al., 1995a; for review see Lindau and Almers, 1995). For these experiments, we used cells expressing the original flipped SNAREs (with complete trans-membrane domains). We reasoned that because these cells show a high propensity for full fusion, they may also form small pores in the apparent “hemifusion state.” We chose these cells over GPI-SNARE cells because it is most probable that they produce a detectable fusion pore. After 24-h incubation, cells were stained with FITC-ctxβ and cell membrane capacitances of single nonfused, fully fused, and hemifused CHO cells were determined (Fig. 8, A and B; and see Materials and methods). Single CHO cells, which are typically smaller than 3T3 fibroblasts, had membrane capacitances in the range of ∼13–15 pF, corresponding to a surface area of 1170–1350 μm2 (Fig. 8 C), whereas 3T3 cells exhibited capacitances of 18–32 pF (1620–2880 μm2; not depicted) assuming a membrane-specific capacitance of 9 fF/μm2 (Albillos et al., 1997). When hemifused cells were patched, whole cell recordings yielded capacitance measurements consistent with single cells, suggesting that there was no electrical continuity between the contents of the cells to a resolution of 100 ms (Nyquist frequency) given a sampling frequency of 20 Hz. In contrast, the membrane capacitance of fully fused cells was significantly higher, consistent with the visually observed much larger size of these cells compared with single or hemifused cells (Fig. 2).
To confirm that there were no pore fluctuations during the time course of the experiment, we measured the changes in the whole cell current in patched CHO hemifused cells to a resolution of 100 μs (Nyquist frequency), taking advantage of the high frequency of the ITC-18 A/D converter, which is 20 kHz. This temporal resolution is the lower limit at which we might expect to observe even the very fast flickers associated with dopamine exocytosis from rat cultured ventral midbrain neurons (Staal et al., 2004). No steplike changes in the steady-state current (Iss) were observed during the time course of the experiments, although a slight drift of the current was sometimes observed (Fig. 8 D), yet in the opposite direction expected if a fusion pore were to open. We then attempted to set a limit on fusion pore resistances that might escape detection in our setup. We estimated the ratio of the whole cell currents before and after the fusion of two identical cells:
$\frac{I_{ss}\left(\mathrm{2}cells\right)}{I_{ss}\left(\mathrm{1}cell\right)}=\frac{\mathrm{2}R_{m}+R_{p}}{R_{m}+R_{p}}\mathrm{{;}}$
where Rm is the membrane resistance of the cell (see Materials and Methods). Assuming Rm = 600 MΩ, this relationship predicts that a 10% increase in Iss (which would be easily observed in our experiments) would correspond to a pore resistance (Rp) of 5 GΩ or a fusion pore diameter of ∼2 nm (Hille, 1984), using a thickness of 15 nm for hemifused membrane (Monck and Fernandez, 1992). We cannot rule out the possibility of a constitutively open micro fusion pore (a pore that formed before we patched on the cell and remained stable throughout the experiment). However, our experimental resolution was at the lower limit described for pore flickering in synaptic vesicles (∼100 μs; Staal et al., 2004), and therefore a flickering event in our hemifused cells would likely have been detected. Therefore, we conclude that if dynamic pores were forming, their lifetimes would be ≤100 μs and their sum resistances would exceed 5 GΩ (indicating a pore diameter <2 nm).
Bilayer membrane fusion models (Chernomordik et al., 1987; Lee and Lentz, 1997; Tamm et al., 2003) propose that membrane fusion proceeds through a hemifusion intermediate before fusion pore opening. A large number of studies with lipid bilayers and viral fusion assays have identified the existence of a nonprogressing hemifusion endstate termed “unrestricted hemifusion” (Kemble et al., 1994; Melikyan et al., 1997, 2000; Chernomordik et al., 1998; Armstrong et al., 2000). In two different reconstituted assays, we did not previously observe the hemifused state with full-length neuronal SNAREs (Weber et al., 1998; Fix et al., 2004). Unlike the work reported here, those two assays featured a simplified lipid and protein composition, which may have facilitated full fusion at the expense of the hemifusion endstate. This conclusion is supported by recent experiments demonstrating hemifusion when yeast SNARE proteins are reconstituted into liposomes at very low surface densities, but not at surface densities comparable to our earlier studies (Xu et al., 2005).
Herein, by monitoring lipid mixing and content mixing simultaneously in individual events, we found that (in addition to complete fusion) SNAREs promote hemifusion to a surprising degree. Fully one third of the events involved hemifusion, which could be either permanent (major outcome) or reversible (minor outcome). These hemifusion phenotypes were observed using different lipidic and soluble content markers. In addition to the diversity of fluid phase markers used in the cell–cell fusion assay, we also performed electrophysiological measurements to discern the existence of connecting pores between hemifused cells. The capacitance measurements are consistent with a description of a pore-free hemifusion diaphragm, although our current levels of sensitivity cannot rule out very small nonenlarging “micro” pores (Melikyan et al., 1995a). The conductance measurements give no indication of a dynamic pore population undergoing opening and closure within our resolution limit (∼2 nm). We were not able to capture pore formation during any complete cell–cell fusion event in these experiments. Although “stable” pores are likely to be very short lived, the cell fusion assay is limited by a long kinetic, probably involving cell–cell contact geometries distinct from the process of SNARE pairing (Hu et al., 2003). Unfortunately, our recording period has been limited to a few minutes of patch stability. Thus, we can efficiently patch cells before or after the fusion event, but we have not yet observed a cell that fused during the relatively short lifetime of the patch. Experiments to optimize the assay for fusion pore description are ongoing in our lab.
Interestingly, replacement of the TMD of flipped SNAREs by a GPI-anchor motif produced only hemifusion as fusion outcome, highlighting its role in fusion pore opening. Because the amount of cell surface GPI-anchored SNAREs was similar to the respective flipped SNAREs, as revealed by biotinylation experiments (unpublished data), the incapability of the GPI-SNARE cells to produce full fusion is more probably due to an impairment in the transduction of the generated force during the zipping up of the SNARE proteins to fusing bilayers rather than an effect of the low cell surface concentration of SNARE proteins.
### Alternative outcomes can potentially be modulated for SNARE-dependent fusion
The diversity of fusion outcomes indicates a high degree of functional and conformational dynamics within the structure of the fusion pore formed by isolated SNARE proteins. For pure lipid bilayers, the activation energy required for bilayer fusion is thought to be ∼40 kBT (Kuzmin et al., 2001; Markin and Albanesi, 2002). Recent models suggest that neither mixing of the outer leaflets (hemifusion) nor mixing of the inner leaflets is the largest energetic barrier (Cohen and Melikyan, 2004). Instead, it appears that expansion of the fusion pore requires the greatest energetic input. In such a model, both hemifusion and reversible pore formation are (relatively) low energy intermediates. A simple interpretation of our results is that whereas SNAREs are intrinsically capable of full fusion, the free energy available in the pool of conformational variants only just favors such a result. This means that modest changes in free energy, which may manifest themselves in the form of different lipid compositions, different SNARE concentrations, or the binding/activation of specific regulatory proteins, could decisively tilt the balance of fusion outcomes to favor any one of the three that we observe (Fig. 9).
There are many proteins known that bind to SNAREs and which could potentially regulate the fusion pore. The n-Sec1 protein has been shown to influence the kinetic of fusion pore opening (Fisher et al., 2001). Other proteins that bind SNAREs and are intimately linked to regulated exocytosis include synaptotagmin (Bai et al., 2004), complexin (Reim et al., 2001), α-SNAP, NSF (Mayer et al., 1996), and Munc13 (Brose et al., 2000). Furthermore, the lipid composition could influence the pore structure, either by directly manipulating the SNARE conformations or by regulating the surface distribution density of active SNARE complexes (i.e., in the form of raft domains; Lang et al., 2001; Salaun et al., 2005). The molecular composition of the transitional pore is unknown. It has been proposed that the fusion pore is formed at least in part by a circular arrangement of five to eight Syntaxin transmembrane segments and that lipid molecules might intercalate between them to complete the structure (Han et al., 2004). Lipid mixing just within this pore (i.e., the formation of a stalk) has been postulated from studies with HA proteins and termed “restricted hemifusion;” however, lipid mixing itself cannot be detected in these structures, perhaps because the HA multimers or the SNARE complexes making up the pore block lipid movement out of the pore (Chernomordik et al., 1998). Similarly, hemifusion has been postulated as an explanation for the arrested fusion state of SNARE TMD mutants, which only progress to full fusion after addition of inverted cone-shaped lipids (Grote et al., 2000). Certainly, the closely related process of kiss and run in which content (or a portion thereof) is released in a transient association of vesicles with target (Klyachko and Jackson, 2002; Gandhi and Stevens, 2003) can be of prime physiological relevance in the nervous system and can be dynamically regulated (Staal et al., 2004). We suggest that this diverse physiology can result from a core fusion machinery—SNAREs—which is energetically positioned to allow alternative outcomes with a minimum of regulatory influence.
### Constructs and stable cell lines
To generate GPI-anchored VAMP2 (aa 2–92), DNA encoding the signal sequence of preprolactin and VAMP2 (aa 2–92) was amplified by PCR using primer EcoRSS (GTGGAATTCGCTTGTTCTTTTTGCAGAAGCTCAG) and primer VAMP92Xho (TTGGCTCGAGGTTTTTCCACCAGTATTTGCGC) with plasmid-flipped VAMP2/T27A (Hu et al., 2003) as template. The PCR product was digested with EcoRI and XhoI. The GPI motif from decay-accelerating factor was amplified by PCR from a plasmid that encodes a GPI-anchored pH-sensitive GFP mutant (Miesenbock et al., 1998) using primer GPIFor (AAACCTCGAGCCAAATAAAGGAAGTGGAACCAC) and GPIRev (AAACGGGCCCCTAAGTCAGCAAGCCCATGGTTAC). The PCR product was digested with XhoI and ApaI. Both of the enzyme-digested PCR products were cloned into the EcoRI and ApaI sites of pcDNA3.1(+) in a single ligation reaction, yielding the plasmid GPI-anchored VAMP2 (aa 2–92). GPI-anchored VAMP2 (aa 2–84), GPI-anchored flipped Syntaxin 1 (aa 186–265), and GPI-anchored Syntaxin 1 (aa 186–256) were generated in a similar manner. DNA encoding the signal sequence and VAMP2 (aa 2–84) was amplified by PCR using primer EcoRSS and primer VAMP84Xho (TTGGCTCGAGGAGCTTGGCTGCACTTGTTTC) with plasmid-flipped VAMP2/T27A as template. DNA encoding the signal sequence and Syntaxin 1 (aa 186–265) was amplified by PCR using primer EcoRSS and primer Syn265Xho (TTGGCTCGAGCTTCTTCCTGCGTGCCTTGCTC) with plasmid-flipped Syntaxin (Hu et al., 2003) as template. DNA encoding the signal sequence and Syntaxin 1 (aa 186–256) was amplified by PCR using primer EcoRSS and primer Syn256Xho (TTGGCTCGAGCTTGACGGCCTTCTTGGTGTCAG) with plasmid-flipped Syntaxin as template.
To obtain stable cell lines coexpressing flipped SNAREs or GPI-anchored cytoplasmic domains of SNAREs with appropriate fluorescent protein, different constructs detailed in the following paragraph were generated based on the pBI plasmid. This vector is a bidirectional mammalian expression vector of the Tet-Off gene expression system (Baron et al., 1995). The pBI plasmid allowed us to simultaneously regulate the expression of both the SNAREs and the fluorescent protein genes. Both genes are under the control of a single tetracycline-responsive element, which in the presence of doxycycline or tetracycline down-regulate their expression. To generate pBI-GPI-VAMP2 (2–92)-RFP-nes construct, DNA-encoding RFP-nes was amplified by PCR using primer RFP–NotI5 (GCGGCCGCGCCACCATGGCCTCCTCC) and primer RFP–SalI3 (CGTATTGTCGACCTAATCCAGCTCAAGC) with pcDNA3.1(+)RFP-nes as template (Hu et al., 2003). DNA encoding the signal sequence and GPI-VAMP2 (2–92) was amplified by PCR using primer GPIV2-MluI5 (AAGTACGCGTCGCTTGTTCTTTTTGC) and primer GPIV2-EcoRV3 (ATTGGATATCTAAGTCAGCAAGCCC) with plasmid pCDNA3.1(+)GPI-VAMP2 (2–92) as template. The PCR product was digested with MluI and EcoRV and cloned into the same sites in the pBI-RFP-nes vector. The same procedure was followed to generate the constructs pBI-GPI-VAMP2 (2–86)-RFP-nes and pBI-VAMP2 (2–116)-RFP-nes. DNA encoding the signal sequence and GPI-VAMP2 (2–86) was amplified by PCR using the same primers as GPI-VAMP2 (2–92) with plasmid pCDNA3.1(+)GPI-VAMP2 (2–86) as template. DNA encoding the signal sequence and flipped VAMP2/T27A was amplified by PCR using primer GPIV2-MluI5 and primer FlV2-EcoRV3 (AGAGATATCTTAAGTGCTGAAGTAAACG) with plasmid pCDNA3.1(+) flipped VAMP2/T27A as template. To generate pBI-GPI-Syntaxin (186–265)-flipped SNAP-25-IRES-CFP-nls, DNA encoding the signal sequence and GPI-Syntaxin (186–265) was amplified by PCR using primer GPISy–NotI5 (AATCAAGCGGCCGCTTGTTCTTTTTGC) and primer GPISy-SalI3 (GCTAATGTCGACCTAAGTCAGCAAGCCCATGG) with plasmid pCDNA3.1(+)GPI-Syntaxin (186–265) as template. The PCR product was digested with NotI and SalI and cloned into the same sites in the pBI vector. DNA encoding the signal sequence and SNAP-25 encoding the IRES sequence and CFP-nls were amplified by PCR using primer S25C-MluI5 (TATACGCGTGCCACCATGGACAGCAAAGGTTCG) and primer S25C-EcoRV3 (CGAAGATATCTTATCTAGATCCGGTGGATCCTACC) with plasmid pCH-44 as template (Hu et al., 2003). The PCR product was digested with MluI and EcoRV and cloned into the same sites in the pBI-GPI-Syntaxin (186–265) vector. pBI-GPI-Syntaxin (186–256)-flipped SNAP-25-IRES-CFP-nls and pBI-flipped Syntaxin (186–288)-flipped SNAP-25-IRES-CFP-nls were generated in a similar manner. DNA encoding the signal sequence and GPI-Syntaxin (186–256) was amplified by PCR using same primer as GPI-Syntaxin (186–265) with plasmid pCDNA3.1(+)GPI-Syntaxin (186–256) as template. DNA encoding the signal sequence and Syntaxin (186–288) was amplified by PCR using primer GPISy–NotI5 and primer FlSy-SalI3 (TAATGTCGACTATCCAAAGATGCCCC) with plasmid pCDNA3.1(+)-flipped Syntaxin (186–288) as template (Hu et al., 2003).
To generate the pAU1-GPI construct, two complementary oligonucleotides, AU1 × 2F (CCGGTCGCCACCATGGACACATACCGATACATAGACACATACCGATACATACT) and AU1 × 2R (GTACAGTATGTATCGGTATGTGTCTATGTATCGGTATGTGTCCATGGTGGCGA), encoding two repetitions of the six amino acid epitope AU1 (DTYRYI) were hybridized. After hybridization, the AgeI and BsrGI restriction sites were generated. The pEYFP-GPI plasmid (Keller and Simons, 1997) was digested with AgeI and BsrGI to release the EYFP and the remaining plasmid was purified. The double-stranded DNA fragment and the purified plasmid containing the signal sequence and the GPI motif were ligated, yielding the plasmid containing the signal sequence with the epitope AU1 fused to a GPI-anchored motif.
To generate the pHA-f construct, two complementary oligonucleotides, HAF (CCGGTGCCACCATGTACCCATATGACGTACCAGACTACGCATCACTACT) and HAR (GTACAGTAGTGATGCGTAGTCTGGTACGTCATATGGGTACATGGTGGCA), encoding the nine amino acid epitope HA (YPYDVPDYA) were hybridized. After hybridization, the AgeI and BsrGI restriction sites were generated. The pEGFP-f plasmid (CLONTECH Laboratories, Inc.), which encodes for the EGFP fused to a farnesylation signal from c-Ha-Ras, was digested with AgeI and BsrGI to release the EGFP, and the remaining plasmid was purified. The double-stranded fragment and the purified plasmid containing the farnesylation sequence were ligated, yielding the plasmid encoding the epitope HA with a farnesylation motif. All coding sequences were confirmed by DNA sequencing.
The v-SNARE set of constructs were used to generate the respective double stable Tet-Off CHO cell line; they will hereafter be referred as flipped v-cells or GPI v-cells, respectively. On the other hand, the t-SNARE set of constructs were used to generate the respective double stable MEF-3T3 Tet-Off cell line; they will hereafter be referred as flipped t-cells or GPI t-cells, respectively.
### Cell–cell fusion assay
48 h before performing the assay, 3 × 104 CHO v-cells previously grown for at least 5 d in complete medium in the absence of doxycycline were seeded on sterile 12-mm glass coverslips contained in 24-well plates. MEF 3T3 t-cells previously grown for 7 d in complete medium in the absence of doxycycline were detached from the dishes with EDTA (Cell Dissociation Solution; Sigma-Aldrich). The detached cells were counted with a hemacytometer, centrifuged at 200 g, and resuspended in Hepes-buffered DME supplemented with 10% FCS. 3 × 104 of these t-cells were added to each coverslip already containing the v-cells. After various times at 37°C in 5% CO2, the coverslips were gently washed once with PBS supplemented with 0.1 g/liter CaCl2 and 0.1 g/liter MgCl2 (PBS++), and then fixed with 4% PFA for 30 min, washed three times with PBS++, and incubated for 15 min with 1 μg/ml FITC-Cholera Toxin β-subunit (Sigma-Aldrich). After three washes with PBS++, the coverslips were mounted with Prolong Antifade Gold mounting medium (Molecular Probes). Confocal images were collected as indicated in the Image acquisition section. At each time point, the total number of fused cells (f) or hemifused cells (hf) and the total number of v-cells (V) and t-cells (T) in contact with each other (but not yet fused or hemifused) in random fields were determined. The efficiency of fusion (F) or lipid mixing (LM), as percentages, in both original flipped SNARE fusion and hemifusion assays were calculated as follows: F = 2f/(V + T + 2f) × 100; LM = 2hf/(V + T + 2hf).
### Image acquisition
Images were acquired on a confocal microscope (model TCS SP2 AOBS; Leica) equipped with LCS software (Leica) and usually using a HCX PL APO 40×, 1.25 NA oil immersion objective. For higher magnification images, a HCX PL APO 63×, 1.4 NA oil immersion objective was used. The images were processed with Adobe Photoshop software.
### Cell staining
Before performing the fusion assay, cells cotransfected with the indicated flipped or GPI-anchored SNARE and CFP-nls (as a transfection marker) were incubated with prewarmed 2 μM CMFDA, 2 μM CMTPX, 15 μM Blue CMF2HC, or 0.5 μM Calcein AM (green or orange) in serum-free medium for 30 min at 37°C. After this time, the solution was replaced by fresh medium and cells were incubated for an additional 30 min to allow the processing of the dye, and washed three times with PBS.
### Immunocytochemistry
24 h after transfection, COS-7 cells were fixed with 4% PFA in PBS++. Primary antibodies were incubated with the cells at the following dilutions: anti-Myc mAb, 1:500; anti-SNAP-25 polyclonal antibody (Synaptic Systems GmbH), 1:100; anti-GFP polyclonal antibody, 1:500; and HPC-1 anti-Syntaxin mAb, 1:1,000. Fluorophore-conjugated secondary antibodies (Jackson ImmunoResearch Laboratories) were used at dilutions of 1:500–1:1,000. For double staining, the cells were incubated first with a mixture of the primary antibodies, and then with a mixture of the secondary antibodies
### Soluble t-SNARE binding assay
The t-SNARE cytoplasmic domains (with no internal cysteines and containing his6-SNAP-25 and Syntaxin1A [residues 1-265-L-C]) were expressed in Escherichia coli as described previously (McNew et al., 2000b). COS-7 cells were transfected with flipped VAMP2/T27A-IRES2-EGFP or EGFP (as control). 24 h after transfection, the cells were incubated with 5 μM of soluble t-SNARE complex in Hepes-buffered DME (high glucose; GIBCO BRL) with 10% FBS in the presence of 0.5 mM DTT. After 1 h at 37°C in 5% CO2, the cells were washed four times with the DME medium, washed once with PBS++, and fixed with 4% PFA. Surface-bound t-SNARE was detected with HPC-1 anti-Syntaxin antibody.
### Membrane capacitance measurements
Solutions used for patch clamp recordings were as follows. The bath saline contained PBS. Patch pipette solution contained 139 mM gluconic acid, 2 mM MgCl2, 0.1 mM CaCl2, 10 mM Hepes, pH 7.4, 1 mM EGTA, 1 mM ATP, and 2 mM GTP. Whole cell perforated patches were obtained by using 100 μg/ml nystatin in the patch pipette as previously described (Horn and Marty, 1988). Whole cell capacitance and resistance measurements were performed on single CHO cells or 3T3 fibroblasts, or hemifused CHO-3T3 cells by patching either the CHO or the 3T3 cell. The current was filtered using a 4-pole, 5-kHz Bessel filter built into an Axopatch 200B amplifier (Axon Instruments, Inc.) and sampled at 20 kHz (PCI-6052E; National Instruments). For the membrane resistance recordings in the whole cell configuration, this yielded essentially no time distortion for events with durations >200 μs while broadening events of shorter duration toward this value. Data files were saved in Igor binary format for further analysis using a locally written routine in Igor Pro (Wavemetrics).
For capacitance measurements, a perforated cell was kept at −60 mV holding potential, whereas square +5 mV steps (V0) were applied at 20 Hz frequency; the resolution of these measurements was therefore 100 ms. Cell capacitance was estimated by fitting the current transient with an exponential function as described in Lindau and Neher (1988). In brief, the current through the whole cell circuit after application of a square voltage pulse consists of an initial transient (I0) that decays to a steady-state value (Iss) with an exponential time constant (τ). Cell electrical characteristics are calculated using the following equations:
$R_{a}=\frac{V_{\mathrm{0}}}{I_{\mathrm{0}}}\mathrm{{;}}$
$R_{m}=\frac{V_{\mathrm{0}}}{I_{ss}{-}R_{a}}\mathrm{{;}}$
$Cm=\mathrm{{\tau}}\left(\frac{\mathrm{1}}{R_{a}}+\frac{\mathrm{1}}{R_{m}}\right)\mathrm{.}$
Where Ra is access resistance and Rm and Cm are the resistance and the capacitance of the cell membrane, correspondingly.
For the whole cell resistance measurements, the current through the cell membrane was determined in the absence of voltage pulses providing a resolution of 100 μs. The cell was kept at a constant −60 mV holding potential (Vhold) and assuming that Ra was much smaller than Rm, the latter was estimated as
$R_{m\mathrm{{^\prime}}}=\frac{V_{hold}}{I_{ss}}\mathrm{,}$
where Rm′ is the resistance of the cell at −60 mV and Iss is the whole cell current. If two hemifused cells formed a conducting fusion pore (Fig. 8 B), the current would depend on its resistance (Rp) according to the relationship:
$I_{ss}\left(\mathrm{2}cells\right)=\frac{V_{hold}}{R_{total}}\mathrm{{;}}$
$R_{total}=\frac{R_{m\mathrm{1}}{\times}\left(R_{m\mathrm{2}}+R_{p}\right)}{\left(R_{m\mathrm{1}}+R_{m\mathrm{2}}+R_{p}\right)}\mathrm{,}$
where Rm1 and Rm2 equal the resistance of cells 1 and 2, respectively. Note that when Rp equals zero, the resulting current is twofold higher than that of a single cell, whereas when Rp increases, the change in Iss becomes gradually smaller.
We thank Dr. F. Paumet for comments on the manuscript.
This work was supported by a grant from the National Institutes of Health (J.E. Rothman).
Albillos, A., G. Dernick, H. Horstmann, W. Almers, G. Alvarez de Toledo, and M. Lindau.
1997
. The exocytotic event in chromaffin cells revealed by patch amperometry.
Nature.
389
:
509
–512.
Alvarez de Toledo, G., R. Fernandez-Chacon, and J.M. Fernandez.
1993
. Release of secretory products during transient vesicle fusion.
Nature.
363
:
554
–558.
Armstrong, R.T., A.S. Kushnir, and J.M. White.
2000
. The transmembrane domain of influenza hemagglutinin exhibits a stringent length requirement to support the hemifusion to fusion transition.
J. Cell Biol.
151
:
425
–437.
Bai, J., C.T. Wang, D.A. Richards, M.B. Jackson, and E.R. Chapman.
2004
. Fusion pore dynamics are regulated by synaptotagmin*t-SNARE interactions.
Neuron.
41
:
929
–942.
Baron, U., S. Freundlieb, M. Gossen, and H. Bujard.
1995
. Co-regulation of two gene activities by tetracycline via a bidirectional promoter.
Nucleic Acids Res.
23
:
3605
–3606.
Breckenridge, L.J., and W. Almers.
1987
. Currents through the fusion pore that forms during exocytosis of a secretory vesicle.
Nature.
328
:
814
–817.
Brose, N., C. Rosenmund, and J. Rettig.
2000
. Regulation of transmitter release by Unc-13 and its homologues.
Curr. Opin. Neurobiol.
10
:
303
–311.
Chernomordik, L.V., G.B. Melikyan, and Y.A. Chizmadzhev.
1987
. Biomembrane fusion: a new concept derived from model studies using two interacting planar lipid bilayers.
Biochim. Biophys. Acta.
906
:
309
–352.
Chernomordik, L.V., V.A. Frolov, E. Leikina, P. Bronk, and J. Zimmerberg.
1998
. The pathway of membrane fusion catalyzed by influenza hemagglutinin: restriction of lipids, hemifusion, and lipidic fusion pore formation.
J. Cell Biol.
140
:
1369
–1382.
Cohen, F.S., and G.B. Melikyan.
1998
. Methodologies in the study of cell-cell fusion.
Methods.
16
:
215
–226.
Cohen, F.S., and G.B. Melikyan.
2004
. The energetics of membrane fusion from binding, through hemifusion, pore formation, and pore enlargement.
J. Membr. Biol.
199
:
1
–14.
Fisher, R.J., J. Pevsner, and R.D. Burgoyne.
2001
. Control of fusion pore dynamics during exocytosis by Munc18.
Science.
291
:
875
–878.
Fix, M., T.J. Melia, J.K. Jaiswal, J.Z. Rappoport, D. You, T.H. Sollner, J.E. Rothman, and S.M. Simon.
2004
. Imaging single membrane fusion events mediated by SNARE proteins.
101
:
7311
–7316.
Gandhi, S.P., and C.F. Stevens.
2003
. Three modes of synaptic vesicular recycling revealed by single-vesicle imaging.
Nature.
423
:
607
–613.
Grote, E., M. Baba, Y. Ohsumi, and P.J. Novick.
2000
. Geranylgeranylated SNAREs are dominant inhibitors of membrane fusion.
J. Cell Biol.
151
:
453
–466.
Han, X., C.T. Wang, J. Bai, E.R. Chapman, and M.B. Jackson.
2004
. Transmembrane segments of syntaxin line the fusion pore of Ca2+-triggered exocytosis.
Science.
304
:
289
–292.
Hille, B. 1984. Ion Channels of Excitable Membranes. Sinauer Associates Inc., Sunderland, MA. 514 pp.
Horn, R., and A. Marty.
1988
. Muscarinic activation of ionic currents measured by a new whole-cell recording method.
J. Gen. Physiol.
92
:
145
–159.
Hu, C., M. Ahmed, T.J. Melia, T.H. Sollner, T. Mayer, and J.E. Rothman.
2003
. Fusion of cells by flipped SNAREs.
Science.
300
:
1745
–1749.
Keller, P., and K. Simons.
1997
. Post-Golgi biosynthetic trafficking.
J. Cell Sci.
110
:
3001
–3009.
Kemble, G.W., T. Danieli, and J.M. White.
1994
. Lipid-anchored influenza hemagglutinin promotes hemifusion, not complete fusion.
Cell.
76
:
383
–391.
Klyachko, V.A., and M.B. Jackson.
2002
. Capacitance steps and fusion pores of small and large-dense-core vesicles in nerve terminals.
Nature.
418
:
89
–92.
Kuzmin, P.I., J. Zimmerberg, Y.A. Chizmadzhev, and F.S. Cohen.
2001
. A quantitative model for membrane fusion based on low-energy intermediates.
98
:
7235
–7240.
Kweon, D.H., C.S. Kim, and Y.K. Shin.
2003
. Regulation of neuronal SNARE assembly by the membrane.
Nat. Struct. Biol.
10
:
440
–447.
Lang, T., D. Bruns, D. Wenzel, D. Riedel, P. Holroyd, C. Thiele, and R. Jahn.
2001
. SNAREs are concentrated in cholesterol-dependent clusters that define docking and fusion sites for exocytosis.
EMBO J.
20
:
2202
–2213.
Lee, J., and B.R. Lentz.
1997
. Outer leaflet-packing defects promote poly(ethylene glycol)-mediated fusion of large unilamellar vesicles.
Biochemistry.
36
:
421
–431.
Lindau, M., and E. Neher.
1988
. Patch-clamp techniques for time-resolved capacitance measurements in single cells.
Pflügers Arch.
411
:
137
–146.
Lindau, M., and W. Almers.
1995
. Structure and function of fusion pores in exocytosis and ectoplasmic membrane fusion.
Curr. Opin. Cell Biol.
7
:
509
–517.
Markin, V.S., and J.P. Albanesi.
2002
. Membrane fusion: stalk model revisited.
Biophys. J.
82
:
693
–712.
Mayer, A., W. Wickner, and A. Haas.
1996
. Sec18p (NSF)-driven release of Sec17p (alpha-SNAP) can precede docking and fusion of yeast vacuoles.
Cell.
85
:
83
–94.
McNew, J.A., F. Parlati, R. Fukuda, R.J. Johnston, K. Paz, F. Paumet, T.H. Sollner, and J.E. Rothman.
2000
a. Compartmental specificity of cellular membrane fusion encoded in SNARE proteins.
Nature.
407
:
153
–159.
McNew, J.A., T. Weber, F. Parlati, R.J. Johnston, T.J. Melia, T.H. Sollner, and J.E. Rothman.
2000
b. Close is not enough: SNARE-dependent membrane fusion requires an active mechanism that transduces force to membrane anchors.
J. Cell Biol.
150
:
105
–117.
Melia, T.J., T. Weber, J.A. McNew, L.E. Fisher, R.J. Johnston, F. Parlati, L.K. Mahal, T.H. Sollner, and J.E. Rothman.
2002
. Regulation of membrane fusion by the membrane-proximal coil of the t-SNARE during zippering of SNAREpins.
J. Cell Biol.
158
:
929
–940.
Melikyan, G.B., W.D. Niles, V.A. Ratinov, M. Karhanek, J. Zimmerberg, and F.S. Cohen.
1995
a. Comparison of transient and successful fusion pores connecting influenza hemagglutinin expressing cells to planar membranes.
J. Gen. Physiol.
106
:
803
–819.
Melikyan, G.B., J.M. White, and F.S. Cohen.
1995
b. GPI-anchored influenza hemagglutinin induces hemifusion to both red blood cell and planar bilayer membranes.
J. Cell Biol.
131
:
679
–691.
Melikyan, G.B., S.A. Brener, D.C. Ok, and F.S. Cohen.
1997
. Inner but not outer membrane leaflets control the transition from glycosylphosphatidylinositol-anchored influenza hemagglutinin-induced hemifusion to full fusion.
J. Cell Biol.
136
:
995
–1005.
Melikyan, G.B., R.M. Markosyan, M.G. Roth, and F.S. Cohen.
2000
. A point mutation in the transmembrane domain of the hemagglutinin of influenza virus stabilizes a hemifusion intermediate that can transit to fusion.
Mol. Biol. Cell.
11
:
3765
–3775.
Miesenbock, G., D.A. De Angelis, and J.E. Rothman.
1998
. Visualizing secretion and synaptic transmission with pH-sensitive green fluorescent proteins.
Nature.
394
:
192
–195.
Monck, J.R., and J.M. Fernandez.
1992
. The exocytotic fusion pore.
J. Cell Biol.
119
:
1395
–1404.
Parlati, F., O. Varlamov, K. Paz, J.A. McNew, D. Hurtado, T.H. Sollner, and J.E. Rothman.
2002
. Distinct SNARE complexes mediating membrane fusion in Golgi transport based on combinatorial specificity.
99
:
5424
–5429.
Paumet, F., B. Brugger, F. Parlati, J.A. McNew, T.H. Sollner, and J.E. Rothman.
2001
. A t-SNARE of the endocytic pathway must be activated for fusion.
J. Cell Biol.
155
:
961
–968.
Paumet, F., V. Rahimian, and J.E. Rothman.
2004
. The specificity of SNARE-dependent fusion is encoded in the SNARE motif.
101
:
3376
–3380.
Razinkov, V.I., G.B. Melikyan, and F.S. Cohen.
1999
. Hemifusion between cells expressing hemagglutinin of influenza virus and planar membranes can precede the formation of fusion pores that subsequently fully enlarge.
Biophys. J.
77
:
3144
–3151.
Reim, K., M. Mansour, F. Varoqueaux, H.T. McMahon, T.C. Sudhof, N. Brose, and C. Rosenmund.
2001
. Complexins regulate a late step in Ca2+-dependent neurotransmitter release.
Cell.
104
:
71
–81.
Rohde, J., L. Dietrich, D. Langosch, and C. Ungermann.
2003
. The transmembrane domain of Vam3 affects the composition of cis- and trans-SNARE complexes to promote homotypic vacuole fusion.
J. Biol. Chem.
278
:
1656
–1662.
Rosales Fritz, V.M., J.L. Daniotti, and H.J.F. Maccioni.
1997
. Chinese hamster ovary cells lacking GM1 and GD1a synthesize gangliosides upon transfection with human GM2 synthase.
Biochim. Biophys. Acta.
1354
:
153
–158.
Salaun, C., G.W. Gould, and L.H. Chamberlain.
2005
. The SNARE proteins SNAP-25 and SNAP-23 display different affinities for lipid rafts in PC12 cells. Regulation by distinct cysteine-rich domains.
J. Biol. Chem.
280
:
1236
–1240.
Sollner, T., S.W. Whiteheart, M. Brunner, H. Erdjument-Bromage, S. Geromanos, P. Tempst, and J.E. Rothman.
1993
. SNAP receptors implicated in vesicle targeting and fusion.
Nature.
362
:
318
–324.
Staal, R.G., E.V. Mosharov, and D. Sulzer.
2004
. Dopamine neurons release transmitter via a flickering fusion pore.
Nat. Neurosci.
7
:
341
–346.
Sutton, R.B., D. Fasshauer, R. Jahn, and A.T. Brunger.
1998
. Crystal structure of a SNARE complex involved in synaptic exocytosis at 2.4 A resolution.
Nature.
395
:
347
–353.
Tamm, L.K., J. Crane, and V. Kiessling.
2003
. Membrane fusion: a structural perspective on the interplay of lipids and proteins.
Curr. Opin. Struct. Biol.
13
:
453
–466.
Weber, T., B.V. Zemelman, J.A. McNew, B. Westermann, M. Gmachl, F. Parlati, T.H. Sollner, and J.E. Rothman.
1998
. SNAREpins: minimal machinery for membrane fusion.
Cell.
92
:
759
–772.
Xu, Y., F. Zhang, Z. Su, J.A. McNew, and Y.K. Shin.
2005
. Hemifusion in SNARE-mediated membrane fusion.
Nat. Struct. Mol. Biol.
12
:
417
–422.
Zimmerberg, J., R. Blumenthal, D.P. Sarkar, M. Curran, and S.J. Morris.
1994
. Restricted movement of lipid and aqueous dyes through pores formed by influenza hemagglutinin during cell fusion.
J. Cell Biol.
127
:
1885
–1894.
C.G. Giraudo and C. Hu contributed equally to this paper.
Abbreviations used in this paper: CMFDA, 5-chloromethylfluorescein diacetate; GPI, glycosylphosphatidylinositol; PI-PLC, phosphatidylinositol-specific phospholipase C; TMD, transmembrane domain. | 2022-06-28 03:40:07 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.550980269908905, "perplexity": 14440.743765154717}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-27/segments/1656103347800.25/warc/CC-MAIN-20220628020322-20220628050322-00334.warc.gz"} |
https://noxad.org/51zgpp/5c801c-effect-of-ph-on-edta-titration | # effect of ph on edta titration
(1958). M2.Mustust hhaveave ddifferentifferent ppropertiesroperties (co(color)lor) in in tthehe bouboundnd aandnd uunboundnbound state 3. Results indicated that the EDTA titration method was reliable in measuring the calcium up to 160 mg/l, the maximum concentration investigated in this study, in hydrochloric acid with pH higher than 0 and 3% sodium chloride solution. Hence, the purpose of this study was to investigate the interfering effects of hydrochloric acid, sulfuric acid, sodium chloride and sodium sulfate on measuring the calcium in the aqueous leaching solutions using the EDTA titration method. In unsupplemented white the E 6 > EGG WHITE +EDTA5mg/mi 4 o 3 EGG, WHITE 2 + EDTA 50 mg/mI I rj 2 0 2 4 6 8 TIME (DAYS) AT 20C FIG. Calculate the concentration of free Fe3+ in a solution of 0.10 M FeY-at pH = 4.00 and pH = 1.00. Salt solutions with 3% NaCl and 3% Na2SO4 were used to evaluate the interfering effects of salt elements such as sulfate and sodium on the EDTA titration. 1. Example, p. 300 The formation constant for FeY-is 1.3 x 1025 (Fe3+). Fe3+) can be titrated without interference from others (e.g. /X(72) Author: Kigabar Yogrel: Country: Japan: Language: English (Spanish) Genre: Science: Published (Last): 20 September 2012: Pages: 85: PDF File Size: 2.21 Mb: ePub File Size: 15.86 Mb: ISBN: 401-9-59913-336-3: Downloads: 68707: Price: Free* [*Free Regsitration Required] … Copyright © 2002 Elsevier Science Ltd. All rights reserved. All EDTA equilibrium calculations will use K'f at the pH of the titration. The value of Y4-at this pH is taken from Table 13-3. Cem Concr Res 33, 621–627. https://doi.org/10.1016/S0008-8846(02)01043-8. As we need Y4- to react with the metal ions present in the titration solution, we use pH 10 buffer such as ammonium chloride. Plot All Curves On One Graph And Compare Your Results With Figure 12-10. Complexometric EDTA Titration Curves. The ethylenediaminetetraacetic acid (EDTA) titration method limits the measurement of calcium concentration to 50 mg/l in water. Use Equation 12-11 To Compute Curves (pCa2+ Versus ML Of EDTA Added) For The Titration Of 10.00 ML Of 1.00 MM Ca2+ With 1.00 MM EDTA At PH 5.00, 6.00, 7.00, 8.00, And 9.00. The presence of EDTA produced faceted particles and increasing synthesis pH resulted in slower reaction kinetics and larger particles with lower water content and fewer anion vacancies determined by TGA and Mössbauer spectroscopy. However, various acids and salt solutions are used in the investigation of the durability of concrete, and the adaptability of the EDTA titration method to determine the calcium in these solutions must be investigated. Calculate titration curves for the titration of 50.0 mL of $$5.00 \times 10^{-3}$$ M Cd 2 + with 0.0100 M EDTA (a) at a pH of 10 and (b) at a pH of 7. Repeatability of EDTA titration at various calcium concentrations. However, various acids and salt solutions are used in the investigation of the durability of concrete, and the adaptability of the EDTA titration method to determine the calcium in these solutions must be investigated. The effect of supplementing egg white with various levels of disodium EDTA at pH 8.9 onthe viability ofS. Results indicated that the EDTA titration method was reliable in measuring the calcium up to 160 mg/l, the maximum concentration investigated in this study, in hydrochloric acid with pH higher than zero and 3% sodium chloride solution. 1. Calculating the Titration Curve. Calculate titration curves for the titration of 50.0 mL of 5.00×10 –3 M Cd 2 + with 0.0100 M EDTA (a) at a pH of 10 and (b) at a pH of 7. Effect of pH on the EDTA titration. Compare your results with Figure 9.28 and comment on the effect of pH and of NH 3 on the titration of Cd 2 + with EDTA. Must bind less strongly than the complexing agent (EDTA) EDTA titration techniques: READ … Sulfuric acid with pH higher than zero and 3% sodium sulfate solutions showed 3% to 4% less calcium in the solutions. Results indicated that the EDTA titration method was reliable in measuring the calcium up to 160 mg/l, the maximum concentration investigated in this study, in hydrochloric acid with pH higher than zero and 3% sodium chloride solution. Excess EDTA is titrated with 2nd metal ion. Sulfuric acid and hydrochloric acid were used to prepare calcium solutions with lower pH. We use cookies to help provide and enhance our service and tailor content and ads. pH affects the titration of Ca2+ with EDTA. The minimum pH is specified arbitrarily as the pH at which K f’=10 8. The calcium standard solutions were prepared using CaCl2 with initial pH from −0.8 to 7 and calcium concentration up to 160 mg/l. Chelate effect: the multidentate ligand from strong 1:1 complexes with many metal ions. Effective EDTA titration See Fig 13-9: Minimum pH for effective EDTA titration of various metal ions at 0.01M conc . Neither titration includes an auxiliary complexing agent. (). Excel. For a titration reaction to be effective, it must go “to completion” (say, 99.9%), which means that the equilibrium constant is large—the analyte and titrant are essentially completely reacted at the equivalence point. ScienceDirect ® is a registered trademark of Elsevier B.V. ScienceDirect ® is a registered trademark of Elsevier B.V. Effect of pH, sulfate and sodium on the EDTA titration of calcium. By adjusting the pH of an EDTA titration: one type of metal ion (e.g. The ethylenediaminetetraacetic acid (EDTA) titration method limits the measurement of calcium concentration to 50 mg/l in water. Use Equation 12-11 to compute curves (pCa21 versus mL of EDTA added) for the titration of 10.00 mL of 1.00 mM Ca21 with 1.00 mM EDTA at pH 5.00, 6.00, 7.00, 8.00, and 9.00. The calcium standard solutions were prepared using CaCl2 with initial pH from −0.8 to 7 and calcium concentration up to 160 mg/l. Answer Kim J and Vipulanandan C (2003). Sulfuric acid with pH higher than zero and 3% sodium sulfate solutions showed 3% to 4% less calcium in the solutions. … Effect of pH, sulfate and sodium on the C Calcium = (V 2 –V 1) EDTA titration of calcium. Applications of Edta Titration - Kshetra K L ... coordinates with two or more donar groups of single ligand to form a five or six member hetrocyclic ring. ScienceDirect ® is a registered trademark of Elsevier B.V. ScienceDirect ® is a registered trademark of Elsevier B.V. Effect of pH, sulfate and sodium on the EDTA titration of calcium. Metal ligand complexes Complexation titration are based on the reaction of metal ion with chemical agent to form a metal-ligand complex. EDTA is insoluble in water at low pH because H4Y is predominant in that pH (less than 2). The complexometric EDTA titration curve shows the change in pM, where M is the metal ion, as a function of the volume of EDTA. Hence, the purpose of this study was to investigate the interfering effects of hydrochloric acid, sulfuric acid, sodium chloride and sodium sulfate on measuring the calcium in the aqueous leaching solutions using the EDTA titration method. 3-EDTA titration of metachromatic complex", of tolui-dine blue and heteropolyanions [-0- STA; -e- PTA;-A- PMAj 1026 Table I- Effect of pH on Toluidine Blue" Metachromasia Induced by Heteropolyanions' in the pH Range 3.6-7.0 Chromotropes Cone. Question: 12-12.ffect Of PH On The EDTA Titration. https://doi.org/10.1016/S0008-8846(02)01043-8. Question: 12.12 EEEE FAU COTA Effect Of PH On The EDTA Titration. Test results indicated that for the calcium concentrations investigated EDTA could be used to measure Ca 2+ in the pH range of 0 to 7. lower pH. Copyright © 2002 Elsevier Science Ltd. All rights reserved. Microchemical Journal – MICROCHEM J. The complexometric titration of calcium in the presence of magnesium a critical study. Acid solutions with pH less than zero showed interference with calcium measurement. However, EDTA in concentrations as low as 0.25 mM caused a significant decrease in the adhesion of two strains of Se+. Determination of chromium by EDTA titration. 3. In terms of the negative logarithm of the hydroxyl ion activity (pOH) the effect of the temperature is considerable beginning from the isoelectric point to about pOH 6 (pH 8), but in a more alkaline region the two curves practically coincide. 1. Use Equation 12-11 To Compute Curves (pCa2 Versus ML Of EDTA Added) For The Titra- Tion Of 10.00 ML Of 1.00 MM Ca With 1.00 MM EDTA At PH 5.00, 6.00, 7.00, 8.00, And 9.00. 0.7 0.6 0.5 0.4 Series1 0.3 0.2 0.1 0 0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34 36 38 40 42 Figure 1.2 Photometric Titration of Cu2+ with 0.2 M EDTA at ph 2.4, absorbance corrected for dilution Since the solution is acidic because of the addition of the HCl, the Cu2+ displaces the Hydrogens attached to the carboxylate group so that the EDTA will react with Cu2+. Talanta 1, 238–244. Above pH 10, Y4- is predominant. Calcium Analysis by EDTA titration, Ion selective electrode potentiometry and Flame Atomic Absorption Spectroscopy. EDTA Complexometric Titration EDTA called as ethylenediaminetetraacetic acid is a complexometric indicator consisting of 2 amino groups and four carboxyl groups called as Lewis bases. Fig. Edta is a hexadentate ligand because of its competence to denote six pair of lonely electrons due to the formation of covalent bonds . typhimurium Tm-i at 2 Cis shown in Fig. Ca2+) EDTA Titrations EDTA Titration Curves 1.) pH Dependence of EDTA Formation Constant ... EDTA Titration Considerations Requirements for indicators for complexometric titrations: 1. With increasing the pH, each hydrogen ion in the carboxyl groups of EDTA will start to dissociate. The effect of pH was variable, but at a pH of 5.0 and 6.0, the adhesion of all test strains was significantly reduced compared to control values at a pH of 7.0. Copyright © 2021 Elsevier B.V. or its licensors or contributors. The effect of hydrochloric acid and pH on the calcium measurement are shown in Fig. Preliminary results indicate that the EDTA method is reliable in the pH range of 0 to 12 to measure calcium concentrations from 1 to 1000 ppm. 4. Szekeres, László. Belcher R et al. Use Equation 11-11 to compute c... Chapter 11.11-13: Titration of EDTA with metal ion. Sulfuric acid and hydrochloric acid were used to prepare calcium solutions with lower pH. Chapter 11.11-12: Effect of pH on the EDTA titration. The pH Acid solutions with pH less than zero showed interference with calcium measurement. EDTA Titration Techniques • Direct titration: analyte is titrated with standard EDTA with solution buffered at a pH where K f ’ is large • Back titration: known excess of EDTA is added to analyte. Titration Curve . ... tablet, pH 10 buffer solution, Calmagite and methyl red indicators, a 250 mL volumetric flask, three 250 mL Erlenmeyer flasks, a 500 mL beaker, a 150 mL beaker, a 100 mL graduated cylinder, a mortar and pestle, pipet, a clamp and stand and a burette. By continuing you agree to the use of cookies. The titration of a metal ion with EDTA is similar to the titration of a strong acid (M+) with a weak base (EDTA) K 'f = K f Y 4. Example: 25.0 mL of an unknown Ni2+ solution was treated with 25.00 mL of 0.05283 M Na 2 EDTA. To account for the effect of pH on the ionization of the EDTA, it is convenient to rewrite the equilibrium expression for the complex formation as and to define a conditional formation constant (KfN) which is given by The value of KfN provides information about how quantitative the complex formation is … Sulfuric and hydrochloric acids were used to lower the pH. Compare your results with Figure $$\PageIndex{3}$$ and comment on the effect of pH on the titration of Cd 2 + with EDTA. The titration is carried out at a pH of 10, in an NH3-NH4 +buffer, which keeps the EDTA (H4Y) mainly in the half-neutralized form, H2Y 2-, where it complexes the Group IIA ions very well but does not tend to react as readily with other cations such as Fe3+that might be present as impurities in the water. This question hasn't been answered yet Ask an expert . Salt solutions with 3% NaCl and 3% Na2SO4 were used to evaluate the interfering effects of salt elements such as sulfate and sodium on the EDTA titration. Sulfuric acid with pH higher than 0 and 3% sodium sulfate solutions showed 3-4% less calcium in the solutions. of Buffer pH Shift (nm) chromotrope (M) Phosphotungstic acid 9.7 x 10-0 Acetate 3.6 Acetate 4.6 Acetate 5.6 Phosphate 7.0 Acetate 3.6 … Since K f’ (FeY-) is 2.5x10 7 at pH1.00, the min. Neither titration includes an auxiliary complexing agent. We use cookies to help provide and enhance our service and tailor content and ads. Copyright © 2021 Elsevier B.V. or its licensors or contributors. This can be represented by Ca2+ + Y4- = CaY2- The equation is shifted to the left as the hydrogen ion concentration is increased, due to competition for the chealting anion by the hydrogen ion. Two strains showed a significant increase in adhesion at a pH of 8.0. Must bind metal ion of interest 2. Results indicated that the EDTA titration method was reliable in measuring the calcium up to 160 mg/l, the maximum concentration investigated in this study, in hydrochloric acid with pH higher than zero and 3% sodium chloride solution. Effect ofEDTAon the viability ofS. pH for EDTA titration of Fe 3+ is slightly higher than 1.00. EDTA test was conducted with varying the pH from -0.8 to 12. EFFECT OF ON EDTA TITRATION Consider the formation of the EDTA chetate of a Ca2+. EDTA and its equilibria, effect of pH on EDTA equlibria; Metallochromic indicators in EDTA titration; pH and masking agents in complexometric titration ; back titration; Techniques/Skills involved. Plot All Curves On One Graph And Compare Your Results With Figure 12-10. Show transcribed image text. By continuing you agree to the use of cookies. X 1025 ( Fe3+ ) can be titrated without interference from others ( e.g continuing you agree to the of! And hydrochloric acid were used to prepare calcium solutions with lower pH the acid! Of disodium EDTA at pH 8.9 onthe viability ofS Ltd. All rights reserved to form metal-ligand. C calcium = ( V 2 –V 1 ) EDTA titration of calcium concentration to 50 mg/l in.... Disodium EDTA at pH 8.9 onthe viability ofS... chapter 11.11-13: titration of EDTA metal! Various metal ions arbitrarily as the pH of 8.0 pH, sulfate and sodium on c... Calcium concentration up to 160 mg/l six pair of lonely electrons due to the formation constant EDTA... Mg/L in water the presence of magnesium a critical study electrode potentiometry and Flame Atomic Absorption Spectroscopy ( ). Will use K ' f at the pH of 8.0 calcium standard were. In concentrations as low as 0.25 mM caused a significant increase in adhesion at a pH of 8.0 EDTA... Conducted with varying the pH at which K f ’ ( FeY- ) is 2.5x10 at. Or its licensors or contributors the EDTA titration effect of ph on edta titration Requirements for indicators complexometric... ( V 2 –V 1 ) EDTA titration Curves 1. 3 to! Acid and pH on the EDTA titration Considerations Requirements for indicators for complexometric Titrations: 1. provide. 12-12.Ffect of pH on the calcium standard solutions were prepared using CaCl2 initial... Calculations will use K ' f at the pH from −0.8 to 7 and calcium concentration up 160... P. 300 the formation of covalent bonds solutions showed 3 % sodium sulfate showed. ) titration method limits the measurement of calcium concentration up to 160 mg/l hydrogen ion in the carboxyl of! As low as 0.25 mM caused a significant increase in adhesion at a pH of the titration from to! Sulfate and sodium on the calcium measurement All EDTA equilibrium calculations will use K f. To 160 mg/l formation of covalent bonds sulfuric and hydrochloric acid and pH the... And calcium concentration up to 160 mg/l up to 160 mg/l covalent bonds bouboundnd aandnd uunboundnbound 3! ( V 2 –V 1 ) EDTA titration: One type of ion. Sodium sulfate solutions showed 3-4 % less calcium in the solutions constant... titration! Titration: One type of metal ion ( e.g: 12-12.ffect of pH on the EDTA titration Fig. Acid ( EDTA ) titration method limits the measurement of calcium lower.... Edta test was conducted with varying the pH of the titration by continuing you to... ( co ( color ) lor ) in in tthehe bouboundnd aandnd state. One type of metal ion ( e.g ligand from strong 1:1 complexes with many metal ions chapter 11.11-12 effect! With chemical agent to form a metal-ligand complex magnesium a critical study showed significant! This pH is specified arbitrarily as the pH from −0.8 to 7 calcium! Prepare calcium solutions with lower pH hexadentate ligand because of its competence to six. 8.9 onthe viability ofS of pH on the EDTA titration of calcium in effect of ph on edta titration carboxyl groups of EDTA with ion! Solution of 0.10 M FeY-at pH = 4.00 and pH on the EDTA titration Curves.... Copyright © 2002 Elsevier Science Ltd. All rights reserved a solution of 0.10 effect of ph on edta titration FeY-at pH = 4.00 pH!
CHECK THIS OUT The Cynical Philosopher... | 2021-07-25 00:03:43 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.357038676738739, "perplexity": 8470.364736807718}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-31/segments/1627046151531.67/warc/CC-MAIN-20210724223025-20210725013025-00018.warc.gz"} |
https://mathoverflow.net/questions/414848/classification-of-homogeneous-einstein-manifolds | # Classification of homogeneous Einstein manifolds
In Besse's "Einstein manifolds", p. 177, he states that, until that moment, no general classification of homogeneous Einstein manifolds was know, even in the compact case. More specifically, he poses a problem: classify the compact simply connected homogeneous manifolds $$M=G/K$$ which admit a $$G$$-invariant Einstein metric.
Does that question remain open to this day? | 2022-05-21 07:11:57 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 2, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8853393793106079, "perplexity": 303.76787328651585}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662538646.33/warc/CC-MAIN-20220521045616-20220521075616-00228.warc.gz"} |
https://people.maths.bris.ac.uk/~matyd/GroupNames/481/C3xHe3s5S3.html | Copied to
clipboard
## G = C3×He3⋊5S3order 486 = 2·35
### Direct product of C3 and He3⋊5S3
Aliases: C3×He35S3, C3412S3, C339(C3⋊S3), He310(C3×S3), (C3×He3)⋊24C6, (C3×He3)⋊23S3, C3311(C3×S3), (C32×He3)⋊7C2, C323(He3⋊C2), C32.14(C33⋊C2), C3⋊(C3×He3⋊C2), C322(C3×C3⋊S3), C3.6(C3×C33⋊C2), SmallGroup(486,243)
Series: Derived Chief Lower central Upper central
Derived series C1 — C3 — C3×He3 — C3×He3⋊5S3
Chief series C1 — C3 — C32 — He3 — C3×He3 — C32×He3 — C3×He3⋊5S3
Lower central C3×He3 — C3×He3⋊5S3
Upper central C1 — C32
Generators and relations for C3×He35S3
G = < a,b,c,d,e,f | a3=b3=c3=d3=e3=f2=1, ab=ba, ac=ca, ad=da, ae=ea, af=fa, ebe-1=bc=cb, dbd-1=bc-1, fbf=b-1, cd=dc, ce=ec, cf=fc, de=ed, fdf=d-1, fef=e-1 >
Subgroups: 1874 in 417 conjugacy classes, 66 normal (10 characteristic)
C1, C2, C3, C3, C3, S3, C6, C32, C32, C32, C3×S3, C3⋊S3, C3×C6, He3, He3, C33, C33, C33, He3⋊C2, S3×C32, C3×C3⋊S3, C3×He3, C3×He3, C3×He3, C34, C3×He3⋊C2, He35S3, C32×C3⋊S3, C32×He3, C3×He35S3
Quotients: C1, C2, C3, S3, C6, C3×S3, C3⋊S3, He3⋊C2, C3×C3⋊S3, C33⋊C2, C3×He3⋊C2, He35S3, C3×C33⋊C2, C3×He35S3
Smallest permutation representation of C3×He35S3
On 54 points
Generators in S54
(1 2 3)(4 5 6)(7 8 9)(10 11 12)(13 14 15)(16 17 18)(19 20 21)(22 23 24)(25 26 27)(28 29 30)(31 32 33)(34 35 36)(37 38 39)(40 41 42)(43 44 45)(46 47 48)(49 50 51)(52 53 54)
(1 25 17)(2 26 18)(3 27 16)(4 37 24)(5 38 22)(6 39 23)(7 14 29)(8 15 30)(9 13 28)(10 46 35)(11 47 36)(12 48 34)(19 40 50)(20 41 51)(21 42 49)(31 52 44)(32 53 45)(33 54 43)
(1 14 23)(2 15 24)(3 13 22)(4 26 30)(5 27 28)(6 25 29)(7 39 17)(8 37 18)(9 38 16)(10 41 32)(11 42 33)(12 40 31)(19 44 34)(20 45 35)(21 43 36)(46 51 53)(47 49 54)(48 50 52)
(4 26 30)(5 27 28)(6 25 29)(7 17 39)(8 18 37)(9 16 38)(10 32 41)(11 33 42)(12 31 40)(19 44 34)(20 45 35)(21 43 36)
(1 15 22)(2 13 23)(3 14 24)(4 5 6)(7 18 38)(8 16 39)(9 17 37)(10 12 11)(19 43 35)(20 44 36)(21 45 34)(25 26 27)(28 29 30)(31 33 32)(40 42 41)(46 52 49)(47 53 50)(48 54 51)
(1 52)(2 53)(3 54)(4 41)(5 42)(6 40)(7 34)(8 35)(9 36)(10 30)(11 28)(12 29)(13 47)(14 48)(15 46)(16 43)(17 44)(18 45)(19 39)(20 37)(21 38)(22 49)(23 50)(24 51)(25 31)(26 32)(27 33)
G:=sub<Sym(54)| (1,2,3)(4,5,6)(7,8,9)(10,11,12)(13,14,15)(16,17,18)(19,20,21)(22,23,24)(25,26,27)(28,29,30)(31,32,33)(34,35,36)(37,38,39)(40,41,42)(43,44,45)(46,47,48)(49,50,51)(52,53,54), (1,25,17)(2,26,18)(3,27,16)(4,37,24)(5,38,22)(6,39,23)(7,14,29)(8,15,30)(9,13,28)(10,46,35)(11,47,36)(12,48,34)(19,40,50)(20,41,51)(21,42,49)(31,52,44)(32,53,45)(33,54,43), (1,14,23)(2,15,24)(3,13,22)(4,26,30)(5,27,28)(6,25,29)(7,39,17)(8,37,18)(9,38,16)(10,41,32)(11,42,33)(12,40,31)(19,44,34)(20,45,35)(21,43,36)(46,51,53)(47,49,54)(48,50,52), (4,26,30)(5,27,28)(6,25,29)(7,17,39)(8,18,37)(9,16,38)(10,32,41)(11,33,42)(12,31,40)(19,44,34)(20,45,35)(21,43,36), (1,15,22)(2,13,23)(3,14,24)(4,5,6)(7,18,38)(8,16,39)(9,17,37)(10,12,11)(19,43,35)(20,44,36)(21,45,34)(25,26,27)(28,29,30)(31,33,32)(40,42,41)(46,52,49)(47,53,50)(48,54,51), (1,52)(2,53)(3,54)(4,41)(5,42)(6,40)(7,34)(8,35)(9,36)(10,30)(11,28)(12,29)(13,47)(14,48)(15,46)(16,43)(17,44)(18,45)(19,39)(20,37)(21,38)(22,49)(23,50)(24,51)(25,31)(26,32)(27,33)>;
G:=Group( (1,2,3)(4,5,6)(7,8,9)(10,11,12)(13,14,15)(16,17,18)(19,20,21)(22,23,24)(25,26,27)(28,29,30)(31,32,33)(34,35,36)(37,38,39)(40,41,42)(43,44,45)(46,47,48)(49,50,51)(52,53,54), (1,25,17)(2,26,18)(3,27,16)(4,37,24)(5,38,22)(6,39,23)(7,14,29)(8,15,30)(9,13,28)(10,46,35)(11,47,36)(12,48,34)(19,40,50)(20,41,51)(21,42,49)(31,52,44)(32,53,45)(33,54,43), (1,14,23)(2,15,24)(3,13,22)(4,26,30)(5,27,28)(6,25,29)(7,39,17)(8,37,18)(9,38,16)(10,41,32)(11,42,33)(12,40,31)(19,44,34)(20,45,35)(21,43,36)(46,51,53)(47,49,54)(48,50,52), (4,26,30)(5,27,28)(6,25,29)(7,17,39)(8,18,37)(9,16,38)(10,32,41)(11,33,42)(12,31,40)(19,44,34)(20,45,35)(21,43,36), (1,15,22)(2,13,23)(3,14,24)(4,5,6)(7,18,38)(8,16,39)(9,17,37)(10,12,11)(19,43,35)(20,44,36)(21,45,34)(25,26,27)(28,29,30)(31,33,32)(40,42,41)(46,52,49)(47,53,50)(48,54,51), (1,52)(2,53)(3,54)(4,41)(5,42)(6,40)(7,34)(8,35)(9,36)(10,30)(11,28)(12,29)(13,47)(14,48)(15,46)(16,43)(17,44)(18,45)(19,39)(20,37)(21,38)(22,49)(23,50)(24,51)(25,31)(26,32)(27,33) );
G=PermutationGroup([[(1,2,3),(4,5,6),(7,8,9),(10,11,12),(13,14,15),(16,17,18),(19,20,21),(22,23,24),(25,26,27),(28,29,30),(31,32,33),(34,35,36),(37,38,39),(40,41,42),(43,44,45),(46,47,48),(49,50,51),(52,53,54)], [(1,25,17),(2,26,18),(3,27,16),(4,37,24),(5,38,22),(6,39,23),(7,14,29),(8,15,30),(9,13,28),(10,46,35),(11,47,36),(12,48,34),(19,40,50),(20,41,51),(21,42,49),(31,52,44),(32,53,45),(33,54,43)], [(1,14,23),(2,15,24),(3,13,22),(4,26,30),(5,27,28),(6,25,29),(7,39,17),(8,37,18),(9,38,16),(10,41,32),(11,42,33),(12,40,31),(19,44,34),(20,45,35),(21,43,36),(46,51,53),(47,49,54),(48,50,52)], [(4,26,30),(5,27,28),(6,25,29),(7,17,39),(8,18,37),(9,16,38),(10,32,41),(11,33,42),(12,31,40),(19,44,34),(20,45,35),(21,43,36)], [(1,15,22),(2,13,23),(3,14,24),(4,5,6),(7,18,38),(8,16,39),(9,17,37),(10,12,11),(19,43,35),(20,44,36),(21,45,34),(25,26,27),(28,29,30),(31,33,32),(40,42,41),(46,52,49),(47,53,50),(48,54,51)], [(1,52),(2,53),(3,54),(4,41),(5,42),(6,40),(7,34),(8,35),(9,36),(10,30),(11,28),(12,29),(13,47),(14,48),(15,46),(16,43),(17,44),(18,45),(19,39),(20,37),(21,38),(22,49),(23,50),(24,51),(25,31),(26,32),(27,33)]])
63 conjugacy classes
class 1 2 3A ··· 3H 3I ··· 3Q 3R ··· 3BA 6A ··· 6H order 1 2 3 ··· 3 3 ··· 3 3 ··· 3 6 ··· 6 size 1 27 1 ··· 1 2 ··· 2 6 ··· 6 27 ··· 27
63 irreducible representations
dim 1 1 1 1 2 2 2 2 3 6 type + + + + image C1 C2 C3 C6 S3 S3 C3×S3 C3×S3 He3⋊C2 He3⋊5S3 kernel C3×He3⋊5S3 C32×He3 He3⋊5S3 C3×He3 C3×He3 C34 He3 C33 C32 C3 # reps 1 1 2 2 9 4 18 8 12 6
Matrix representation of C3×He35S3 in GL5(𝔽7)
4 0 0 0 0 0 4 0 0 0 0 0 2 0 0 0 0 0 2 0 0 0 0 0 2
,
1 0 0 0 0 0 1 0 0 0 0 0 0 2 0 0 0 0 0 2 0 0 2 0 0
,
1 0 0 0 0 0 1 0 0 0 0 0 2 0 0 0 0 0 2 0 0 0 0 0 2
,
0 6 0 0 0 1 6 0 0 0 0 0 1 0 0 0 0 0 4 0 0 0 0 0 2
,
1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 2 0 0 0 0 0 4
,
0 6 0 0 0 6 0 0 0 0 0 0 6 0 0 0 0 0 0 5 0 0 0 3 0
G:=sub<GL(5,GF(7))| [4,0,0,0,0,0,4,0,0,0,0,0,2,0,0,0,0,0,2,0,0,0,0,0,2],[1,0,0,0,0,0,1,0,0,0,0,0,0,0,2,0,0,2,0,0,0,0,0,2,0],[1,0,0,0,0,0,1,0,0,0,0,0,2,0,0,0,0,0,2,0,0,0,0,0,2],[0,1,0,0,0,6,6,0,0,0,0,0,1,0,0,0,0,0,4,0,0,0,0,0,2],[1,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,2,0,0,0,0,0,4],[0,6,0,0,0,6,0,0,0,0,0,0,6,0,0,0,0,0,0,3,0,0,0,5,0] >;
C3×He35S3 in GAP, Magma, Sage, TeX
C_3\times {\rm He}_3\rtimes_5S_3
% in TeX
G:=Group("C3xHe3:5S3");
// GroupNames label
G:=SmallGroup(486,243);
// by ID
G=gap.SmallGroup(486,243);
# by ID
G:=PCGroup([6,-2,-3,-3,-3,-3,-3,218,867,3244,382]);
// Polycyclic
G:=Group<a,b,c,d,e,f|a^3=b^3=c^3=d^3=e^3=f^2=1,a*b=b*a,a*c=c*a,a*d=d*a,a*e=e*a,a*f=f*a,e*b*e^-1=b*c=c*b,d*b*d^-1=b*c^-1,f*b*f=b^-1,c*d=d*c,c*e=e*c,c*f=f*c,d*e=e*d,f*d*f=d^-1,f*e*f=e^-1>;
// generators/relations
×
𝔽 | 2022-01-28 18:39:57 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 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": 0.9998601675033569, "perplexity": 7619.865282464155}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320306335.77/warc/CC-MAIN-20220128182552-20220128212552-00148.warc.gz"} |
https://chaoxuprime.com/posts/2019-04-06-misleading-while-being-honest.html | # Misleading while being honest
Given a set of data $A$, we can plot it on a radar chart. One can permute the axis to make sure the area of the radar chart is maximized. This was explored in a previous article.
More interesting problem. Given two sets of data $A$ and $B$, we are interested in finding a common radar chart, that make $A$ look as good as possible compared to $B$. We might want to optimize the area ratio, area difference, or something else. Again, we are thinking of permuting the axis of the radar chart.
I once mentioned this problem to R Ravi, and he suggest I could ask the same question for all kind of different graphs. How to mislead people with graphs while being completely honest? Indeed, this looks like a fun research project. There is a wikipedia article completely devoted to it. In my previous post, I've discussed how to fitting two seemingly not that related data points through simple transformation.
I'm interested in are algorithmic problems where one want to compute the most misleading chart, I think it would be a great fun project.
Posted by Chao Xu on 2019-04-06.
Tags: data visualization. | 2019-04-23 04:16:56 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 5, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7601543664932251, "perplexity": 495.4714678609384}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-18/segments/1555578586680.51/warc/CC-MAIN-20190423035013-20190423061013-00174.warc.gz"} |
http://heattransfer.asmedigitalcollection.asme.org/article.aspx?articleid=1448248 | 0
RESEARCH PAPERS: Micro/Nanoscale Heat Transfer
# Thermal Conductivity Measurements of Ultra-Thin Single Crystal Silicon Layers
[+] Author and Article Information
Wenjun Liu, Mehdi Asheghi
Mechanical Engineering Department, Carnegie Mellon University, Pittsburgh, PA 15213
J. Heat Transfer 128(1), 75-83 (Jun 24, 2005) (9 pages) doi:10.1115/1.2130403 History: Received March 06, 2004; Revised June 24, 2005
## Abstract
Self-heating in deep submicron transistors (e.g., silicon-on-insulator and strained-Si) and thermal engineering of many nanoscale devices such as nanocalorimeters and high-density thermomechanical data storage are strongly influenced by thermal conduction in ultra-thin silicon layers. The lateral thermal conductivity of single-crystal silicon layers of thicknesses 20 and $100nm$ at temperatures between 30 and $450K$ are measured using joule heating and electrical-resistance thermometry in suspended microfabricated structures. In general, a large reduction in thermal conductivity resulting from phonon-boundary scattering is observed. Thermal conductivity of the $20nm$ thick silicon layer at room temperature is nearly $22Wm−1K−1$, compared to the bulk value, $148Wm−1K−1$. The predictions of the classical thermal conductivity theory that accounts for the reduced phonon mean free paths based on a solution of the Boltzmann transport equation along a layer agrees well with the experimental results.
<>
## Figures
Figure 1
(a) The optical microscope image of the suspended structure. Cross sections of (b) the metal/silicon and (c) the metal (aluminum or CoFe) suspended structures, which are used to obtain the thermal conductivity data for silicon layer.
Figure 2
Fabrication process of 100nm silicon film
Figure 3
Fabrication process of 20nm silicon film
Figure 4
Temperature dependent thermal conductivity data of the 100nm thick aluminum and 75nm CoFe layers as a function of temperature, which are subsequently used to obtain the thermal conductivity data for 100nm and 20nm thick silicon layers, respectively. The uncertainties in the data are less than 5%, which is on the order of size of the “symbols.” As a result, the uncertainty bars are not shown in this plot.
Figure 5
Curve fits for thermal conductance along the length of the CoFe and CoFe+silicon suspended structures using Eqs. 5,7 to the electrical resistivity data at room temperature. The fitted thermal conductivity data for CoFe layer is subsequently used to extract the thermal conductivity of the 20nm thick silicon layer. The uncertainties in the measured electrical resistivity data are less than 2%, which is on the order of size of “symbols.” The gray shaded areas show the ±5% and ±7% variations in thermal conductance of the CoFe and CoFe+silicon structures based on the predictions of the Eqs. 5,7.
Figure 6
The temperature dependent thermal conductivity data for the silicon layers of 20 and 100nm thickness. The estimated uncertainties for 20nm thick silicon layer are on the order of 12%, 8% at 30K and 300K. For 100nm silicon layer, the estimated uncertainties are on the order of 9% and 15% at 100K and 300K, respectively.
Figure 7
Thermal conductivity of silicon layer and wires, at room temperature, as a function of thickness. Predictions based on Boltzmann transport equation (BTE) agree reasonably well with the experimental data (8,10) for thin silicon film. For nanowires, the predications are different from the reported data (18) for D=22nm by nearly 30%.
Figure 8
Lateral thermal conductivity data and predictions for thin silicon layers at high temperatures. The estimated uncertainties for 20nm thick silicon layer are on the order of 8%, 6% at 300K and 400K. For 100nm silicon layer, the estimated uncertainties are in the order of 15% and 22% at 300K and 450K, respectively
Figure 9
The uncertainty related to 5% variation of the metal film thickness (dm=75nm) results in nearly ±10% uncertainty in the measurement of the thermal conductivity of 20nm thick silicon layer at room temperature. For thermal conductivity measurements of 10nm and 5nm thick silicon layers with uncertainty on the order of ∼±15%, metal layer thickness should be reduced to 50nm and 20nm, respectively.
## Discussions
Some tools below are only available to our subscribers or users with an online account.
### Related Content
Customize your page view by dragging and repositioning the boxes below.
Topic Collections | 2017-12-17 12:00:50 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 5, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5437558889389038, "perplexity": 2548.256458204664}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-51/segments/1512948595858.79/warc/CC-MAIN-20171217113308-20171217135308-00737.warc.gz"} |
http://mathoverflow.net/questions/108409/lambda-ring-structure-defined-for-a-graded-ring-in-fulton-langs-book/108429 | # $\lambda$-ring structure defined for a graded ring in Fulton-Lang's book
Given a commutative ring $A$ with unity, Grothendieck used universal polynomials to define a special $\lambda$-ring structure on $\Lambda(A):=1+t\:A[[t]]$. Suppose $A$ is graded, say $A=\bigoplus_{i=o}^\infty A_i$. In Riemann-Roch Algebra, p. 11, Fulton and Lang define $\Lambda^{\circ}(A):=\{1+a_1t+a_2t^2\ldots\mid a_i\in A_i\}$. Then on page 15 they state that since the product and $\lambda$ operations of $\Lambda(A)$ take $\Lambda^\circ(A)$ to itself, $\Lambda^\circ(A)$ becomes a $\lambda$-ring (without unit). They use this $\lambda$-ring structure of $\Lambda^\circ(A)$ in proof of Theorem 3.1 on p. 16.
However, a straightforward computation shows that the product in $\Lambda(A)$ does not take $\Lambda^\circ(A)$ to itself. For example, if $1+a_1t+a_2t^2\ldots$ and $1+b_1t+b_2t^2\ldots$ are elements in $\Lambda^\circ(A)$, then their product using the product of $\Lambda(A)$ is given by $1+P_1(a_1;b_1)t+P_2(a_1,a_2;b_1,b_2)t^2+\ldots$, where $P_1,P_2,\ldots$ are certain universal polynomials. But $P_1(a_1;b_1)$ turns out to be $a_1b_1$ (see here, p. 22) and $a_1b_1$ is not in $A_1$, which shows the product is not in $\Lambda^\circ(A)$.
Question. Is there an error in the book? If yes, can it be fixed?
Edit. If you know other errors in this book that one should be aware of, please share it here.
-
Another mistake in Fulton-Lang's book is the assertion that the $j$-th Bott element of a positive element in a $\lambda$-ring $R$ is invertible in $R[1/j]$; see p. 24 (this is only true if $R$ is augmented with a locally nilpotent $\gamma$-filtration). – Damian Rössler Sep 30 '12 at 7:59
@Damian: Thank you! Your comment gave me the idea to ask about other known errors in the book. – Mahdi Majidi-Zolbanin Sep 30 '12 at 14:38
As others have said, the definition of the Chern ring there is wrong. But if memory serves, the only mistake is that they forgot to introduce the right multiplication law on the sets of power series they consider. The usual one in the theory is given by the universal formulas for exterior powers of tensor products $\Lambda^n(E\otimes F)$, but the one they want is for Chern classes $c_n(E\otimes F)$. When $n=1$ and $E$ and $F$ are line bundles, the first is multiplication and the second is addition. So it's obviously just an oversight, but one that can be confusing if you're seeing these things for the first time. (In the copy at U Chicago, someone mercifully added a warning note in the margin. There are a few obvious suspects.)
If you want a reference where the details are correct, I'd recommend SGA6. Grothendieck's introduction in expose 0 is very clear. Page 28 is where the discussion of the Chern ring starts. If I remember, Berthelot's expose goes into more depth, but I found Grothendieck's easier to read. Berthelot gets to the Chern ring on page 344. Atiyah-Tall is also generally a good reference, but I think they don't cover the Chern ring (although they do introduce the gamma-filtration).
-
Dear James: Are you saying that one can correct the issue by using a different set of universal polynomials to define multiplication in the graded ring that they consider? – Mahdi Majidi-Zolbanin Sep 30 '12 at 5:26
Yes, I was saying that and also that the universal polynomials are the ones that describe the Chern classes of tensor products. Note that the polynomials are not completely universal -- they do depend on the ranks of the factors. The usual way around this is to assume both factors have rank zero (which is why you get non-unital rings). – JBorger Sep 30 '12 at 6:09
What is the definition of universal as in universal polynomials? I mean I know how they are defined, but why are they called universal? Does it simply mean that the polynomials have integral coefficients or is it more to it? – Mahdi Majidi-Zolbanin Oct 7 '12 at 23:14
'Universal' just emphasizes that the polynomial is independent of the ring. For instance, the Leibniz rule $d(xy)=xdy+ydx$ for derivations can be expressed as $d(xy)=P(x,y,dx,dy)$, where $P(a,b,c,d)=ad+bc$, and the polynomial $P$ is independent of the ring on which you have a derivation. You can imagine other kind of operators where $d(xy)$ is a polynomial in $x,y,dx,dy$ but that polynomial can depend on the ring. – JBorger Oct 8 '12 at 1:21
Hazewinkel in http://arxiv.org/pdf/0804.3888.pdf warns about an error on page 15, second paragraph of thie book. In fact he advises to "steer clear" of the book!
-
Predictably, when I try to consult Google Books to find the gaffe, "pages 15 and 16 are not shown". – Todd Trimble Sep 30 '12 at 1:39
I see! He also mentions the book review by K. R. Coombes in Math. Rev. (88h:14011). I should read the review to see what Coombes wrote about the book. Thanks! – Mahdi Majidi-Zolbanin Sep 30 '12 at 2:07
This is not an answer, as I don't exactly know what Fulton and Lang are trying to achieve with the $\lambda$-ring structure on $\Lambda^{\circ}\left(A\right)$ (I must admit that, while I had the quixotic intent to read and rewrite Fulton-Lang's Chapter I in the notes that you cited, I never found the resolve to walk that talk). I can confirm your counterexample.
What I think can be done (don't know if it is of any help) is the following: For every $i\in\mathbb N$, let $\Lambda^{i}_{\circ}\left(A\right)$ be the subset of $\Lambda\left(A\right)$ consisting of all formal power series of the form $1+a_1t+a_2t^2+a_3t^3+...$ with every $k$ satisfying $a_k\in A^{ik}$. Then, each such $\Lambda^{i} _ {\circ}\left(A\right)$ is an additive subgroup of $\Lambda\left(A\right)$, and the direct sum $\bigoplus\limits_{i\in\mathbb N}\Lambda^{i}_{\circ}\left(A\right)$ is well-defined and a sub-$\lambda$-ring of $\Lambda\left(A\right)$. (This is easy to prove by means of the usual grading on the ring of symmetric functions.) This sub-$\lambda$-ring, of course, is graded (and does have a $1$). I have no idea in how far it is what Fulton and Lang wanted.
We could also construct a greater graded sub-$\lambda$-ring of $\Lambda\left(A\right)$ by allowing $i$ rational (with $A^x$ defined as $0$ when $x\not\in\mathbb Z$), but then it will be graded by rationals. This greater graded sub-$\lambda$-ring is actually dense in $\Lambda\left(A\right)$ (in the usual topology on formal power series).
Does it make sense to replace $\Lambda^{\circ}\left(A\right)$ by $\Lambda^{\geq 1}_{\circ}\left(A\right)$ in the definition of a Chern class homomorphism? I don't know. It seems that most notions in Fulton-Lang are motivated by geometry, and without understanding it I am not the one to judge.
-
Dear Darij: Thank you for your suggestion. Are you sure $1+t$ (the $1$ of $\Lambda(A)$) is in $\bigoplus_{i\in\mathbb{N}}\Lambda^i_\circ(A)$? In any case, I think Fulton & Lang wanted a $\lambda$-ring structure on $\Lambda^\circ(A)$ to prove a Splitting Principle for abstract Chern classes (Theorem 3.1), which in the geometric case is evident. But even with your suggestion, I am not sure if it is possible to fix the proof of Theorem 3.1, because I believe the proof has other issues, e.g., I think they mix up the notion of $\lambda$-homomorphism introduced on page 15 with $\lambda$-ring homomo. – Mahdi Majidi-Zolbanin Sep 30 '12 at 5:22
$1+t$ lies in $\Lambda^0_{\circ}\left(A\right)$, the $0$-th graded component of $\Lambda_{\circ}\left(A\right)$. Anyway, I guess James Borger understands better what Fulton and Lang were trying to say. – darij grinberg Sep 30 '12 at 14:53
I'm not sure what product you are thinking of on $\Lambda^0(A)$, but the one I'm thinking of, and the one that I believe is implicitly used in Fulton-Lang is the usual product on power series. So in particular, $(1+a_1 t+\cdots)\cdot (1+b_1 t+\cdots) = 1 + (a_1 + b_1)t + \cdots$.
There is no problem of grading.
-
This product is not the product, but the sum of $\Lambda\left(A\right)$. – darij grinberg Sep 29 '12 at 17:04
As Darij wrote, in the λ-ring structure the ordinary product that you wrote above is considered the addition operation. So what you did here is you added the elements. Multiplication is given by universal polynomials – Mahdi Majidi-Zolbanin Sep 29 '12 at 17:07
Yes, you are right, sorry for this too prompt answer. – Baptiste Calmès Oct 4 '12 at 4:51 | 2015-05-29 02:23:39 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.950954020023346, "perplexity": 330.4260499571602}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-22/segments/1432207929832.32/warc/CC-MAIN-20150521113209-00059-ip-10-180-206-219.ec2.internal.warc.gz"} |
https://leanprover-community.github.io/archive/stream/113489-new-members/topic/fintype.20to.20finset.html | ## Stream: new members
### Topic: fintype to finset
#### Eloi (Jul 17 2020 at 21:56):
Is it possible to convert a fintype to a finset and later recover the fintype?
I'm trying to define Theta series of a quadratic form, and I need fin 2... but the ∑ notation requires a finset...
noncomputable def Theta'(z: ℂ)(Q: quadratic_form ℤ (fin 2 → ℤ)) /-[in_upper_half_plane z] [pos_def Q]-/:
cau_seq ℂ complex.abs :=
⟨λ M, ∑ m in finset.pi (fin 2) (λ n, Ico_ℤ (- M) M), exp(2 * pi * I * z * (Q m)), sorry⟩
#### Jalex Stark (Jul 17 2020 at 22:00):
you might like to use \sum m in finset.range 2
#### Jalex Stark (Jul 17 2020 at 22:01):
I'm not sure what you mean by recover
#### Eloi (Jul 17 2020 at 22:04):
Then to define a sum over ℤ² I need a sum inside a sum, right?
I mean that m is indexed by a finset, but a quadratic form has to be applied to a term of type fin 2 → ℤ
To be clear, my dream would be to be able to write something like
∑ m in (fin 2 → Ico_ℤ (- M) M)
#### Jalex Stark (Jul 17 2020 at 22:11):
oh I misunderstood your code
#### Jalex Stark (Jul 17 2020 at 22:12):
I usually either use a double sum or index over the product type
#### Jalex Stark (Jul 17 2020 at 22:14):
can you post code that compiles? I'm not familiar with the parts of the library you're using, so chasing down the imports is not fun
#### Eloi (Jul 17 2020 at 22:29):
sure, I updated the first comment to make this thread more readable :)
#### Alex J. Best (Jul 17 2020 at 23:20):
import data.complex.basic
import data.real.pi
import linear_algebra.quadratic_form
open real -- pi
open complex
open finset -- Ico_ℤ
open cau_seq
open_locale big_operators
variable M : ℕ
noncomputable def Theta'(z: ℂ) (Q: quadratic_form ℤ (fin 2 → ℤ)) /-[in_upper_half_plane z] [pos_def Q]-/:
cau_seq ℂ complex.abs :=
⟨λ M, ∑ m in finset.pi (univ : finset (fin 2)) (λ n, Ico_ℤ (- M) M), exp (2 * (real.pi : ℂ) * I * z * (Q (λ t, m t (mem_univ t)))), sorry⟩
#### Alex J. Best (Jul 17 2020 at 23:27):
actually using fintype.pi_finset instead seems nicer
noncomputable def Theta'(z: ℂ) (Q: quadratic_form ℤ (fin 2 → ℤ)) /-[in_upper_half_plane z] [pos_def Q]-/:
cau_seq ℂ complex.abs :=
⟨λ M, ∑ m in fintype.pi_finset (λ a : fin 2, Ico_ℤ (- M) M), exp (2 * (real.pi : ℂ) * I * z * Q m), sorry⟩
and if I were you I think I'd separate the creation of the theta series into making a power series in q with integer coefficients, and then evaluation as a complex function!
#### Eloi (Jul 18 2020 at 05:33):
Nice! Thank you, I'll do so
Last updated: May 13 2021 at 05:21 UTC | 2021-05-13 06:13:49 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5084179639816284, "perplexity": 9446.55978109304}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-21/segments/1620243991537.32/warc/CC-MAIN-20210513045934-20210513075934-00328.warc.gz"} |
https://leanprover-community.github.io/archive/stream/116395-maths/topic/rfl.20lemma.html | ## Stream: maths
### Topic: rfl lemma
#### Thomas Browning (Dec 23 2020 at 21:47):
The lemma adjoin_empty in field_theory/adjoin is actually a rfl-lemma. Is it worth making a PR?
@[simp] lemma adjoin_empty (F E : Type*) [field F] [field E] [algebra F E] :
adjoin F (∅ : set E) = ⊥ :=
eq_bot_iff.mpr (adjoin_le_iff.mpr (set.empty_subset _))
@[simp] lemma adjoin_empty (F E : Type*) [field F] [field E] [algebra F E] :
adjoin F (∅ : set E) = ⊥ := rfl
#### Patrick Massot (Dec 23 2020 at 21:52):
There is no problem at all with PRs that are easy to review.
#### Bryan Gin-ge Chen (Dec 23 2020 at 22:04):
@Mario Carneiro once gave some reasons why we shouldn't replace all possible proofs with rfl when possible; not sure whether that advice applies in this case though.
#### Mario Carneiro (Dec 23 2020 at 22:05):
The question is whether the defeq can be proven using only "public" defeq facts
#### Mario Carneiro (Dec 23 2020 at 22:07):
I think it is fair to say that bot = empty = {x | false] is public, but I don't know much about adjoin
#### Mario Carneiro (Dec 23 2020 at 22:09):
Looking around it seems like you have to dig through many layers to get this result, so I would say that it shouldn't be public
#### Mario Carneiro (Dec 23 2020 at 22:10):
it seems like adjoin_le_iff is the API for adjoin
#### Eric Wieser (Dec 23 2020 at 22:53):
Using rfl vs not affects whether dsimpapplies this lemma, right?
Last updated: May 12 2021 at 07:17 UTC | 2021-05-12 08:06:18 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.37308475375175476, "perplexity": 3525.963259233009}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-21/segments/1620243991685.16/warc/CC-MAIN-20210512070028-20210512100028-00223.warc.gz"} |
http://stats.stackexchange.com/questions/32947/collinearity-problems-when-analyzing-age-dependent-variable-with-age-at-death-as | Collinearity problems when analyzing age-dependent variable with age-at-death as a predictor
Let's say I'm analyzing the incidence of pathological lesions, observed during necropsy. No information is available about when the lesion occurred before death. So, I'm using a general linear model $Y_i \sim \beta_0+\beta_{sex}+\beta_{treatment}+\beta_{age}$. $Y_i \in \{0,1\}$ is whether subject $i$ has the lesion or not. Sex and treatment are categorical variables. Age is the age at which the subject died. I already know that if I fit a Cox proportional hazards model to $age \sim \gamma_{sex}+\gamma_{treatment}$ both effects are significant. Therefore, in the model for $Y_i$ the age term must be collinear with the sex and treatment terms. The VIFs are rather large, some of them above 100. On the other hand, if I drop all age-containing terms from the model, the fit becomes significantly worse.
What strategy would you recommend for somehow breaking or adjusting for this collinearity? Would fitting deviance residuals from the Cox model in place of the raw ages do the trick? It seems to work with no errors in R:
update(originalmodel,.~sex*diet*residuals(coxph(Surv(age)~sex+diet),type='deviance'))
...and it lowers the VIFs, but I don't know if it's actually valid.
@Tim: I have better than box-plots-- I have survival curves, and they clearly show that females have shorter lifespans than males, and subjects on the diet have longer lifespans than untreated subjects. So the age term in the model (really, longevity) is partly determined by sex and diet. But of course there is a component that is independent of sex and diet, and I want to include just that component in the model. So that's why I'm asking if using deviance (or some other) residuals in place of age in the model would accomplish that.
And the multicollinearity is severe, here are the VIFs (the model actually includes interaction terms, which I didn't mention above to keep things on-point):
sex diet age sex:diet sex:age diet:age sex:diet:age
82.042992 123.579264 4.270772 37.464412 72.537500 133.874481 37.464407
- | 2014-03-12 12:48:37 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 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": 0.5107502937316895, "perplexity": 1571.076428880271}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-10/segments/1394021767060/warc/CC-MAIN-20140305121607-00024-ip-10-183-142-35.ec2.internal.warc.gz"} |
https://rottapronto.com/assets/1/7/SMATHREVd0zvw696n-wm9.pdf | Home
# For some math symbol
List of all math symbols and meaning - equality, inequality, parentheses, plus, minus, times, division, power, square root, percent, per mille,.. For some x is the same as exists x. Unlike in everyday language, it is does not necessarily refer to a plurality of elements, and so might be more clearly represented in colloquial English as for at least one square root. √x is a number whose square is x. √4=2 or -2. ∑. summation. sum over from to of, sigma. ∑ k = 1 n x k {\displaystyle \sum _ {k=1}^ {n} {x_ {k}}} is the same as x 1 +x 2 +x 3 +x k. ∑ k = 1 5 k + 2 = 3 + 4 + 5 + 6 + 7 = 25 {\displaystyle \sum _ {k=1}^ {5} {k+2}=3+4+5+6+7=25 A mathematical symbol is a figure or a combination of figures that is used to represent a mathematical object, an action on mathematical objects, a relation between mathematical objects, or for structuring the other symbols that occur in a formula. As formulas are entirely constituted with symbols of various types, many symbols are needed for expressing all mathematics. The most basic symbols are the decimal digits (0, 1, 2, 3, 4, 5, 6, 7, 8, 9), and the letters of the Latin. Importance of Mathematical Symbols. Mathematical symbols help us in denoting various quantities. It establishes the relationship between 2 different quantities. These symbols also helps in identifying the type of operation. These symbols make the reference much easier. The mathematical symbols are world-wide applicable and break the language barrier globally
### Math Symbols List (+,-,x,/,=,) - RapidTables
1. def= and ≡, the latter being especially common in applied mathematics. Some Symbols from Mathematical Logic ∴ (three dots) means therefore and first appeared in print in the 1659 book Teusche Algebra (Teach Yourself Algebra) by Johann Rahn (1622-1676). 3 (the such that sign) means under the condition that
2. And in mathematics, you're usually not so much interested in how to calculate results (so it's not a problem if some calculations are a bit less efficient without explicit state, mutable variables, iterations etc.), but it's always very important that expressions aren't ambiguous so you can prove various properties
3. Normal users also need to insert some of the mathematical symbols in documents like the division sign ÷. You can easily use the alt code shortcuts to insert mathematical symbols on your documents.. Related: Alt code shortcuts for vulgar fractions
4. Symbol L a T e X Comment Symbol L a T e X Comment or \neq or \ne: is not equal to \notin: is not member of \nless: is not less than \ngtr: is not greater than \nleq: is not less than or equal t
5. The ≈ symbol is called an almost-equals sign. The fields of math and science tend to borrow a lot of letters from the Greek alphabet as commonplace symbols, and English tends to put a twist on the pronunciation of these letters
The following table lists many specialized symbols commonly used in mathematics. Basic mathematical symbols Symbol Name Read as Explanation Examples Category = equality x = y means x and y represent the same thing or value. 1 + 1 = 2 is equal to; equals everywhere ≠ <> != inequation x ≠ y means that x and y do not represent the same thing or value. (The symbols!= and <> Common Mathematical Symbols + Addition, Plus, Positive. The addition symbol + is usually used to indicate that two or more numbers should be added together, for example, 2 + 2. The + symbol can also be used to indicate a positive number although this is less common, for example, +2
### For Some -- from Wolfram MathWorl
1. Many mathematical, technical, and currency symbols, are not present on a normal keyboard. To add such symbols to an HTML page, you can use the entity name or the entity number (a decimal or a hexadecimal reference) for the symbol. Display the Not Equal To Sign, ≠, with an entity name, a decimal, and a hexadecimal value: HTML soruce cod
2. It's easier if you're in Word's equation editor / math mode (Alt + = enters math mode), where you can just type symbol names like \omega and \times. LaTeX users are already familiar with this method, and the syntax is similar
3. us, times, and division symbols seem to be mere notations on a paper. Yet, symbols in math are essentially the instructions that drive this area of academics. And, they have true value in real life
4. Exhaustive List of Mathematical Symbols and Their Meaning. Finding it difficult to recollect the exact meaning of a notation while solving mathematical equations? Don't worry, ScienceStruck is here to help you out. Here's a list of mathematical symbols and their meaning, for your reference
That would vary on what is meant on sum, difference and product. Other than that exception, sum, difference, product, and quotient are just fancy words for adding, subtracting, multiplying, and dividing respectively. There are the simple symbols: a+b,a-b,axxb, a-:b (or a/b). There is a special symbol for difference used in some math and science equations: Deltax This means there is a. Some Mathematical Symbols. Multiplication. There are three commonly used means of indicating multiplication. The symbol x, e.g., 5 x 6 = 30. Note that this symbol is generally avoided in algebraic equations because of the common use of x to indicate an unknown quantity Symbol Symbol Name Meaning / definition Example { } set: a collection of elements: A = {3,7,9,14}, B = {9,14,28} | such that: so that: A = {x | x∈, x<0} A⋂B: intersection: objects that belong to set A and set B: A ⋂ B = {9,14} A⋃B: union: objects that belong to set A or set B: A ⋃ B = {3,7,9,14,28} A⊆B: subset: A is a subset of B. set A is included in set B The list of math symbols can be long. You can't possibly learn all their meanings in one go, can you? You can make use of our tables to get a hold on all the important ones you'll ever need. This is an introduction to the name of symbols, their use, and meaning.. The Mathematical symbol is used to denote a function or to signify the relationship between numbers and variables
Set Symbols. A set is a collection of things, usually numbers. We can list each element (or member) of a set inside curly brackets like this: Common Symbols Used in Set Theory. Symbols save time and space when writing Math Symbols! https://7esl.com/mathematics-symbols/Learn useful Mathematical symbols (equal sign '=', not equal sign '≠', approximately equal sign '≈')Learn.. 2010 Mathematics Subject Classification: Primary: 03-XX Secondary: 01Axx [][] Conventional signs used for the written notation of mathematical notions and reasoning. For example, the notion the square root of the number equal to the ratio of the length of the circumference of a circle to its diameter is denoted briefly by $\sqrt{\pi}$, while the statement the ratio of the length of the. Mathematical and scientific symbols. Common pronunciations (in British English - Gimson,1981) of mathematical and scientific symbols are given in the list below. (all the pages in this section need a unicode font installed - e.g. Arial Unicode MS, Doulos SIL Unicode, Lucida Sans Unicode - see: The International Phonetic Alphabet in Unicode)
Some of the examples are the pi (π) symbol which holds the value 22/7 or 3.17, and e-symbol in Maths which holds the value e= 2.718281828. This symbol is known as e-constant or Euler's constant. The table provided below has a list of all the common symbols in Maths with meaning and examples These mathematical operators and symbols range from simple addition and subtraction to more complicated operations like integration and the like. There are many math notations and symbols that are in common use - the list below gives some of the more widely used symbols and notations Some symbols in no particular order: (Some symbols did not convert properly, will try to fix them.) ~ tilde * asterisk + plus sign - hyphen, minus sign ± plus or minus ∓ minus or plus × multiplication sign ÷ division sign • dot or bullet product ∘ ring operator = equals ≡ i In standard LaTeX \ldots is \dots is \ifmmode\mathellipsis\else\textellipsis\fi so has a text and math version. On the other hand \cdots is \mathinner{\cdotp\cdotp\cdotp} a math only construction. (Additionally \cdotp is {\mathpunct}{symbols}{01}, \cdot is {\mathbin}{symbols}{01}, also math constructions.) - Andrew Swann Mar 11 '13 at 15:3 Hello everyone, I am facing the issue that some math elements appear as question marks in equations; for example, the summation and angle symbols. The approximately equal or the alpha symbols, for example, appear correctly. The symbol font is installed. System specs: FrameMaker 9 with the latest..
20 Most Common Math Terms and Symbols in English Addition. Six plus four equals twelve. This type of calculation is called addition, which is when you add two or more... Equation. Usually, we say that one expression equals another, and the = symbol is fittingly called an equals sign. Not-equals. Explanation of Math; Some Mathematical Symbols, and Their Meanings; Untitled; Untitled; This is the Equal Sign, this is used to describe what something equals, like 2+2 EQUALS or REPRESENTS 4. A sign used to add a number a certain amount of times, for example, 3*4 it would be 12 since 3+3+3+3=12 Code Maus Maths Symbol Logo. 2 Use of symbols in logos that are associated both with math and with computer programming, which ties in beautifully with the name and industry. A small mouse is created from these symbols, creating a witty image that is relevant to the name. Although the writing is rather plain, the use of a 5 for the final S ties it into the name This online mathematical keyboard is limited to what can be achieved with Unicode characters. This means, for example, that you cannot put one symbol over another. While this is a serious limitation, multi-level formulas are not always needed and even when they are needed, proper math symbols still look better than improvised ASCII approximations
### List of mathematical symbols - Simple English Wikipedia
• Hello, a selection of basic mathematical symbols and their meanings and use. Regards, James
• Mathematics Keyboard v.1.0 Write Edit Your Mathematical Symbols Easily.; NppCalc v.0.5.3 Beta NppCalc is a small, simple, easy to use Notepad++ add-in, specially designed to help you evaluate the expressions in your editor. Basically while NppCalc is on the job all mathematical symbols in the expressions you write will be highlighted with; Crimson Text v.20111206 A typeface for book.
• We could include a few symbols to NVDA's symbols data: ⊻ xor π Pi σ Sigma ⋈ BowTie Probably others too
• A math symbol stands for something — whether it's a quantity, like four, or an arithmetic operation to carry out, like multiplication. But there's nothing about the way math symbols look that helps explain what they mean or what you should do with them
• MathType Easily write math equations from anywhere! MathType for Office Tools The easiest way to write math equations in any digital document MathType for LMS The easiest way to write math equations in your Learning Management System. Arabic notation Editing math formulas for arabic scientific text
• Write a program fragment that displays the symbol for some common mathematical constants in one column, the constant name in the second column, the constant's value in scientific notation in the third column, and in normal decimal notation in the fourth column. The columns should be neatly aligned
• Logic Alphabet, a suggested set of logical symbols Mathematical operators and symbols in Unicode Polish notation List of mathematical symbols Notes 1. ^ Although this character is available in LaTeX, the MediaWiki TeX system doesn't support this character. 2. ^ Quine, W.V. (1981): Mathematical Logic, §
some symbol of mathematics for more videos subscribe my channel maths by Ravinder kaushik sir #maths # by #Ravinder # kaushik # si If you want even more flexibility for mathematical input, you can try using the package unicode-math (that is built on fontspec). Nevertheless you will find the \bm and \boldsymbol traditional commands don't work. You can nonetheless specify how you want it to deal with your bold math symbols using an option while loading the unicode-math package Additional Math Codes. See the Unicode Math Chart for additional codes for math symbols. Note that they only work in Microsoft Office and that you should use the non-Hex code. For instance an entry ∛ for the cube root symbol (∛) would correspond to ALT+8731 in Word. Other Punctuation . These incude copyright symbols and special. A: We're in good company, it seems, though we can clear up some of the nonsense found online about the use of m as a symbol for slope. Let's begin with a bit of history. The CRC Concise Encyclopedia of Mathematics (2nd ed.) by Eric W. Weisstein says the letter m was first used in print as a symbol for slope in the mid-19th century
### Glossary of mathematical symbols - Wikipedi
LATEX Mathematical Symbols The more unusual symbols are not defined in base LATEX (NFSS) and require \usepackage{amssymb} 1 Greek and Hebrew letters β \beta λ \lambda ρ \rho ε \varepsilon Γ \Gamma Υ \Upsilon χ \chi µ \mu σ \sigma κ \varkappa Λ \Lambda Ξ \Xi δ \delta ν \nu τ \tau ϕ \varphi Ω \Omeg You can use this online keyboard in alternation with your physical keyboard, for example you can type regular numbers and letters on your keyboard and use the virtual math keyboard to type the mathematical characters. You can hold [Shift] for the upper case Greek characters
The PDF viewer 'Evince' on Linux can not display some math symbols correctly [closed] Ask Question Asked 9 years, 1 month ago. Active 5 years, 9 months ago. Viewed 12k times 16. 8. Closed. This question is off-topic. It is not currently accepting answers.. Some of these symbols are primarily for use in text; most of them are mathematical symbols and can only be used in LaTeX's math mode. Most tables are excerpted from the LaTeX Command Summary (Botway & Biemesderfer 1989, Providence, RI: TeX Users Group) and reproduced here courtesy of the AAS. Table 1: Special symbols for NOAOpro
These materials enable personalized practice alongside the new Illustrative Mathematics 6th grade curriculum. They were created by Khan Academy math experts and reviewed for curriculum alignment by experts at both Illustrative Mathematics and Khan Academy Mathematical symbols Tedds recognizes and uses several mathematical symbols. However, there are some mathematical symbols that do not have a function in calculations. For more information, see the following paragraphs Symbols that go above, below, or in the corners of other symbols. Note 1: dotless i and j (symbols \imath and \jmath ) can be used to leave room for whatever hat you want them to wear. Note 2: \sideset takes two required parameters, left side and right side, and must be followed by a sum class math operator that normally takes subscripts and superscripts below and above the symbol Those are the math symbols that we are using in math. We have used many of these math symbols in certain situation. We hope that this page will be very useful to students who are trying to get meaning of some symbols. From now let us use the symbols by knowing their purpose. Thanks for using this page. Math Symbols to Hom
Below is a table with some common maths symbols. For a more complete list see the List of Greek letters and math symbols: description code examples Greek letters \alpha \beta \gamma \rho \sigma \delta \epsilon \alpha \ \beta \ \gamma \ \rho \ \sigma \ \delta \ \epsilon \$ Math can be difficult for those of us that prefer letters and words to numbers and symbols. However, it is important to recognize that even symbols, like those used in algebra, have names, and those names are made up of letters and words. But what are the names of these symbols, what do they mean, [ In mathematical mode as well as in text mode, you can change the typeface as needed. For instance, it's customary to represent real numbers with a blackboard bold font, or topological spaces with calligraphic font. This article shows several fonts for use in math mode
### Some Basic Mathematical Symbols with Names, Meaning, and
Get the master summary of mathematical symbols in eBook form — along with each symbol's usage and LaTeX code. Yes. That'd be useful. Additional Resources. Definitive Guide to Learning Higher Mathematics: A standalone, 10-principle framework for tackling higher mathematical learning,. Translating Words to Symbols. Practical problems seldom, if ever, come in equation form. The job of the problem solver is to translate the problem from phrases and statements into mathematical expressions and equations, and then to solve the equations
In mathematics, the greater than symbol is a basic mathematical symbol which is used to represent the inequality between two values. The symbol used to represent the greater than inequality is > . This is the universally adopted math symbol of two equal length strokes joining in the acute angle a t the right. Also, learn less than symbol, which denotes just the opposite of greater than. Formulas, Symbols, Math Review, and Sample Problems The solution is fairly obvious, but some additional data must be created from the given information. c) The solution is not obvious, and the items in the list in step 3 must be considered one-by-one until a correct one is found Mathematical symbols 16 10. Physical constants 18 11. Conversion factors 19 Length 19 (IEC) and the British Standards Institution (BSI). Additionally, because of their common usage, in the Logic Symbols under Section 12 some distinctive-shape binary logic symbols have been used. Units & Symbols for Electrical & Electronic Engineerin The Symbol font from Adobe Systems, Inc., is widely used in printed documents involving mathematical expressions, and its character set has been proposed as the basis for incorporating mathematical expressions into the World Wide Web (Raggett, ch. 15 and app. D).At this writing, those proposals have not been implemented in commercial web browser software
Finding Other Symbols. Here are some external resources for finding less commonly used symbols: Detexify is an online application which allows you to draw the symbol you'd like and shows you the code for it!; MathJax (what allows us to use on the web, (technically an AJAX library simulating it.) maintains a list of supported commands.; The Comprehensive LaTeX Symbol List This is useful if you frequently need to insert math symbols into your document like the does not equal sign. I will also include some of the most common short codes for common math symbols at the end of this post. 1. Click the File tab. To turn on your Math AutoCorrect feature,.
You cannot avoid mathematical notation when reading the descriptions of machine learning methods. Often, all it takes is one term or one fragment of notation in an equation to completely derail your understanding of the entire procedure. This can be extremely frustrating, especially for machine learning beginners coming from the world of development SAS has many special symbols for plotting data points. These are specified on the symbol statement. The chart below gives some of the specially symbols that can be used for plotting data points. Note that you cannot specify a font on the symbol statement when using these symbols Latex how to write text in math mode; Latex rational numbers; Latex quaternion numbers; Latex complex numbers; Latex indicator function; Latex plus or minus symbol; Latex symbol for all x; Latex symbol exists; Latex symbol not exists; Latex horizontal space: qquad,hspace, thinspace,enspace; Latex square root symbol; Latex degree symbol ASCII stands for American Standard Code for Information Interchange. Computers can only understand numbers, so an ASCII code is the numerical representation of a character such as 'a' or '@' or an action of some sort. ASCII was developed a long time ago and now the non-printing characters are rarely used for their original purpose
### Alt Code Shortcuts for Mathematical Symbols » WebNot
Learn about math and play games with your favorite characters like Odd Squad, Curious George, Peg + Cat, and Dinosaur Train Math Symbols: Greater Than, Less Than, and Equal Today we will look at the symbols = > <, what they mean, when we can use them and some other curiosities. Let's start with the best known: the equal sign (=) Other Maths Symbols: < less than > greater than: ¼: fractions % percent: 90° degrees: 1.666: decimals: Word problems. Word problems are maths questions written in sentences. Here are some for you to try! Two monkeys collected 15 bananas. One monkey collected 7 bananas. How many bananas did the other monkey collect? Johnny had six apples on.
### List of LaTeX mathematical symbols - OeisWik
OBJECTIVES: 1.To understand statements to form a correct equation. 2.To translate mathematical statement in symbols. 3.To appreciate the use of different symbols in mathematics. MOTIVATION: Translating Words to Symbols Practical problems seldom, if ever, come in equation form. The job of the problem solver is to translate the problem from phrases and statements into mathematica The symbol actually has two meanings. In algebra, it signifies weak approximation. You could read it as meaning 'roughly equal'. It has a similar meaning to '≈' but not quite as strong, but it is stronger than saying that something is 'of the same..
### 20 Most Common Math Terms and Symbols in English
Because symbols are so common in math we sometimes take them for granted. The reason we take them for granted is that they make math so easy to perform (actually, they make math performable period) we do not really tip our hat to their true value Logic Symbols in Math. Tautologies are typically found in the branch of mathematics called logic. They use their own special symbols: ∧ means A N D = signifies i s e q u i v a l e n t t o ¬ indicates N E G A T I O N; ∼ shows N O T; ∨ means O R → signifies i m p l i e s or i f-t h e n EDIT1: I tested some more and the problem is as follows (not what I originally stated above): double aa; double bb = 1.0; double cc; aa = sin(1.0); cc = sin (bb); What happens when I try to build is that I get a ' undefined reference ' at the last line, meaning that when I use constants it is fine, but when I pass variables to the sin functions it will not link MS-Word File with Mathematical Symbols First I give a list of symbols for both MS-Word and Powerpoint. Then I explain how to get summation and integration, how to put one thing above another, and, finally, how to make fractions, for MS-Word. For Powerpoint, the latter things don't seem to work. In the following list of symbols, each line. Just draw the symbol you are looking for into the square area above and look what happens! My symbol isn't found! The symbol may not be trained enough or it is not yet in the list of supported symbols. In the first case you can do the training yourself. In the second case just drop me a line (mail@danielkirs.ch)
• Bitcoin anonym kaufen Darknet.
• Bitcoin price in Indian rupee.
• Korttidspermittering regler 2021.
• Xkcd ninja.
• Hur sparar man energi i köket.
• Bitcoin Revolution Auszahlung.
• Ekonomi titel synonym.
• Casino Starburst.
• Hoeveel belasting over bonus 2021.
• Geld verdienen met foodblog.
• Lön Sektionschef Försvarsmakten.
• Bitcoin jackpot 2020.
• Ganfeng lithium ord shs h.
• Altera SoC.
• Stocks to Buy March 2021.
• Krypto Börsen Liste.
• Guarda Wallet connect.
• Sälja skogsfastighet skatt.
• Crypto arbitrage explained.
• På gång i Halland.
• Mac App Store.
• Coinomi verkaufen. | 2021-10-21 18:40:56 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8025663495063782, "perplexity": 1585.9060320615963}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-43/segments/1634323585439.59/warc/CC-MAIN-20211021164535-20211021194535-00374.warc.gz"} |
https://blender.stackexchange.com/questions/120593/how-can-i-export-or-print-coordinates-of-each-selected-vertex | How can I export or print coordinates of each selected vertex?
I am pretty new to Blender. I need to get coordinates of each selected vertex somehow. I don't really want to manually copy coordinates of each and single vertex that would just take a lot of time. I need them either printed somewhere or save them into some kind of text file. Again I need only those vertexes I have selected, not the entire model.
Open up a new Text window and copy and past the following.
import bpy
import bmesh
# Get the active mesh
obj = bpy.context.edit_object
me = obj.data
# Get a BMesh representation
bm = bmesh.from_edit_mesh(me)
bm.faces.active = None
# Modify the BMesh, can do anything here...
for v in bm.verts:
if v.select:
print(tuple(v.co) )
Make sure you are in Edit mode with the required vertices selected and click "Run Script"
The coordinates of each selected vertices is now printed in the System Console.
If you can not see the System Console select Toggle System Console in the Window menu.
To get this code I used the Bmesh Simple Editmode template from the Templates/Python
Another possible way is to do it with vertex groups (as I do often in order to keep things in tact)
So you select the vertices you want to keep in track (in edit mode) and assign them into the group of your choice, then you change the groupName and filename (optional) variables accordingly if you want to keep track on different groups.
import bpy
groupName = "Group"
filename = "myVerts"
if (filename not in bpy.data.texts):
myVerts = bpy.data.texts.new(filename)
else :
myVerts = bpy.data.texts[filename]
myVerts.clear()
myGroupsArr = [];
obj = bpy.context.selected_objects[0]
group = obj.vertex_groups[groupName]
for v in obj.data.vertices:
for g in v.groups:
print(g.group, group.index)
if g.group == group.index:
myGroupsArr.append()
for grp in myGroupsArr:
myVerts.write( str(grp[0])+str(grp[1])+str(grp[2])+ "\n") | 2021-05-05 21:17:46 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.42700493335723877, "perplexity": 3867.9430084600854}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-21/segments/1620243988696.23/warc/CC-MAIN-20210505203909-20210505233909-00483.warc.gz"} |
https://www.sparrho.com/item/states-state-operators-and-quasi-pseudo-mv-algebras/1b1f979/ | # States, state operators and quasi-pseudo-MV algebras
Research paper by Wenjuan Chen, Wieslaw A. Dudek
Indexed on: 14 Feb '18Published on: 13 Feb '18Published in: Soft Computing
#### Abstract
Quasi-pseudo-MV algebras (quasi-pMV algebras, for short) arising from quantum computational logics are the generalizations of both quasi-MV algebras and pseudo-MV algebras. In this paper, we introduce the notions of states, state-morphisms, state operators and state-morphism operators to quasi-pMV algebras. First, we present the related properties of states on quasi-pMV algebras and show that states and Bosbach states coincide on any quasi-pMV algebra. And then we investigate the relationship between state-morphisms and the normal and maximal ideals of quasi-pMV algebras. We prove state-morphisms and extremal states are equivalent. The existence of states on quasi-pMV algebras is also discussed. Finally, state operators and state-morphism operators are introduced to quasi-pMV algebras, and the corresponding structures are called state quasi-pMV algebras and state-morphism quasi-pMV algebras, respectively. We investigate the related properties of ideals under state operators and state-morphism operators. Meanwhile, we show that there is a bijective correspondence between normal $$\sigma$$-ideals and ideal congruences on state quasi-pMV algebras. | 2018-03-22 02:12:35 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.684704601764679, "perplexity": 3212.5728624575663}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-13/segments/1521257647758.97/warc/CC-MAIN-20180322014514-20180322034514-00568.warc.gz"} |
https://math.stackexchange.com/questions/4199429/covering-each-point-of-the-plane-with-circles-three-times | # Covering each point of the plane with circles three times
The following problem arose from an Italian discussion group: I am not so sure about the optimal tags for the question, so feel free to improve them.
Definition: we say that $$E\subseteq\mathbb{R}^2$$ admits a cover with degree $$d$$ if there is a family $$\mathscr{F}$$ of distinct circles (with positive radii) such that every point of $$E$$ belongs to exactly $$d$$ circles in $$\mathscr{F}$$.
Preliminary results:
1. There is a cover of $$\mathbb{R}^2\setminus\{O\}$$ with degree $$1$$ given by concentric circles;
2. There is no cover of $$\mathbb{R}^2$$ with degree $$1$$, due to the accumulation point of any chain of nested circles;
3. There is a cover of $$\mathbb{R}^2$$ with degree $$2$$, for instance the one given by the unit circles centered at $$(2n,y)$$ for any $$n\in\mathbb{Z},y\in\mathbb{R}$$;
4. There is a cover of $$\mathbb{R}^2\setminus\{O\}$$ with degree $$2$$, for instance the one given by the circles tangent to the sides of a quadrant.
Now the actual question, which I did not manage to tackle (yet):
Is there a cover of $$\mathbb{R}^2$$ with degree $$3$$?
I believe the answer is negative: I (unsuccessfully) tried to prove that any degree-3 cover admits a degree-2 subcover, in order to get a contradictory degree-1 cover by switching to the complement. Any insight is welcome.
Small update: actually it is not possible to extract a degree-2 subcover from a (hypothetical) degree-3 cover. Once a circle $$\Gamma_1$$ is removed from a degree-3 cover, we are forced to remove a circle contained in the interior of $$\Gamma_1$$ and this leads to a chain of circles converging to a point.
If a cover with degree 3 exists, every chain of contained circles has a finite number of elements. The elements of $$\mathscr{F}$$ form a POset: we may say that $$\Gamma_1 < \Gamma_2$$ if $$\Gamma_1$$ is contained in the interior of $$\Gamma_2$$. This also allows to assign a parity to each element of $$\mathscr{F}$$ and each point of the plane, according to the length of a maximal chain contained in the circle / surrounding such point.
Further thoughts: let us assume that a degree-3 cover of $$\mathbb{R}^2$$ exists. By the chain argument there is a disk $$D$$ such that $$\partial D\in\mathscr{F}$$ and every circle covering $$\mathring{D}$$ meets $$\partial D$$ at two non-antipodal points. Let us name $$\mathscr{f}$$ the family of circles covering $$\mathring{D}$$. We have that $$\partial D$$ is covered three times, so the circles traversing each point of $$\partial D$$ are $$\partial D$$ itself and two elements of $$\mathscr{f}$$. These couples of circles "exiting" from any point of $$S^1=\partial D$$ have to cover three times any point of $$\mathring{D}$$, which is pretty strange. Many sub-questions came to my mind:
1. Is it possible to partition $$\mathscr{f}$$ into two/three subfamilies of disjoint circles?
2. Is it possible to use the descending chain condition as a substitute for continuity, then apply some topological trick?
3. Are Tucker's lemma, Euler characteristic or the theory of planar graphs useful in some way?
Starting from any $$P_0\in\partial D$$ we may take $$P_{-1}$$ and $$P_1$$ as the points of $$\partial{D}$$ connected to $$P_0$$ via the elements of $$\mathscr{f}$$ through $$P_0$$, then define $$P_{-2},P_{2},P_{-3},P_{3},\ldots$$ in the same way. This gives a partition of $$\partial D$$ into finite cycles (a cycle of length $$2$$ might occur if $$P_1=P_{-1}$$) and "infinite cycles" with a numerability of elements. Infinite cycles have limit points, which are troublesome. On the other hand also a partition into finite cycles only does not seem to stand any chance of covering any point of $$\mathring{D}$$ exactly three times.
Yet another measure-theoretic thought. Let $$S^1=\partial D$$ (which we may assume to have radius $$1$$), let $$P\in S^1$$. Two elements of $$\mathscr{f}$$ go through $$P$$: let $$L(P)$$ be the total length of the arcs given by the intersections with $$\mathring{D}$$. Let us assume that $$L:S^1\to (0,4\pi)$$ is an integrable function. By integrating $$L$$ over $$S^1$$ we have that each arc in $$\mathscr{f}\cap\mathring{D}$$ is counted twice, hence $$\int_{S_1} L = 2\int_{\text{arcs in }\mathscr{f}} 1=2\cdot 3\text{Area}(D) = 6\pi$$ and the average length of an arc in $$\mathscr{f}$$ is $$\frac{3}{2}$$. It follows that most of the arcs of an "integrable 3-cover" of $$\mathring{D}$$ are pretty short, forcing a concentration of the arcs near the boundary of $$D$$. This violation of uniformity probably leads to the fact that if a cover of $$\mathbb{R}^2$$ with degree $$3$$ exists, it is not integrable.
• It doesn't answer your question, but isn't the union of covers 1 and 3 a degree 3 cover of $\mathbb{R}^2 \backslash \{O\}$? Jul 15, 2021 at 18:10
• @JohnWaylandBales: yes it is, by suitably overlapping 1. and 3. you get a degree-3 cover of the punctured plane. You also have covers of $\mathbb{R}^2$ with any even degree, by overlapping instances of 2. with different radii. Jul 15, 2021 at 18:13
• Perhaps investigate whether a 3-cover of the plane by circles implies a 3-cover by straight lines. Just spitballing. Jul 15, 2021 at 18:21
• @JohnWaylandBales: a 3-cover by lines is trivial: it is enough to consider all the horizontal, vertical or diagonal (meaning parallel to $y=x$) lines. Jul 15, 2021 at 18:23
• Can you get a degree-n covering of the punctured plane by covering $\Bbb R^2$ with $n$ families of lines and then doing a bilinear transformation of the complex plane that exchanges 0 and $\infty$?
– MJD
Oct 28, 2021 at 1:13
Below I construct a cover of $$\mathbb{R}^2$$ with degree $$3$$. The idea is to build an almost-"degree 2 cover" of $$\mathbb{R}^2$$ for which a single point is covered three times (using variations on the trick from the degree 2 cover given in the question), then to balance things out by adding the degree $$1$$ cover of $$\mathbb{R}^2 - \{O\}$$.
Lemma: Let $$U$$ be an open disk and let $$P$$ be a point on its boundary. Then there is a family $$\mathscr{F}$$ of circles contained in $$U \cup \{P\}$$ such that each point of $$U$$ belongs to exactly two circles of $$\mathscr{F}$$, while $$P$$ belongs to exactly one circle of $$\mathscr{F}$$.
Proof: For $$a > 0$$, let $$L(a)$$ be the line given by the equation $$x = a$$, and let $$C(a)$$ be the circle of radius $$a$$ centered at $$(a, 0)$$, so $$C(a)$$ passes through the origin $$O$$. Assume without loss of generality that $$P = O$$ and that $$U$$ is the open disk bounded by $$C(1/2)$$. Now, for $$0 < a < b$$, let $$\mathscr{G}(a, b)$$ be the family of circles of radius $$\frac{b-a}{2}$$ with centers on the line $$L(\frac{b+a}{2})$$, so the circles of $$\mathscr{G}(a, b)$$ are contained within the closed strip bounded by $$L(a)$$ and $$L(b)$$, and each point of $$L(a)$$ and $$L(b)$$ belongs to exactly one circle of $$\mathscr{G}(a, b)$$, while each point in the interior of the strip belongs to exactly two circles of $$\mathscr{G}(a, b)$$. Define $$\mathscr{G} = \bigcup_{n=1}^\infty \mathscr{G}\left(1 + \frac{1}{n+1}, 1 + \frac{1}{n} \right) \cup \bigcup_{n=1}^\infty \mathscr{G}\left(2 + \frac{1}{n+1}, 2 + \frac{1}{n} \right) \cup \bigcup_{n=3}^\infty \mathscr{G}(n, n+1)$$ so that, if we define $$V = \bigcup_{a > 1} L(a) = \{(x, y) : x > 1\}$$, all circles in $$\mathscr{G}$$ are contained in $$V$$, and each point in $$V$$ belongs to exactly two circles in $$\mathscr{G}$$, except for those points in $$L(2)$$, each of which belongs to exactly one circle in $$\mathscr{G}$$. Now, under inversion with respect to the unit circle centered at the origin, $$L(a)$$ maps to $$C(\frac{1}{2a}) - \{O\}$$, $$V$$ maps to $$U$$, and each circle in $$\mathscr{G}$$ maps to a circle in a new family $$\mathscr{G}'$$ of circles contained in $$U$$. By the above, each point in $$U$$ belongs to exactly two circles in $$\mathscr{G}'$$, except for those points on $$C(1/4) - \{O\}$$, each of which belongs to exactly one circle in $$\mathscr{G}'$$. To conclude, we define $$\mathscr{F} = \mathscr{G}' \cup \{C(1/4)\}$$, which has the desired properties.
Proposition: There is a cover of $$\mathbb{R}^2$$ with degree $$3$$.
Proof: Define $$C'(r)$$ to be the circle of radius $$r$$ centered at the origin. Let $$P = (2, 0)$$ and let $$U$$ be the open disk bounded by $$C'(2)$$. For each $$n \geq 1$$, let $$\mathscr{H}_n$$ be the family of unit circles with centers on $$C'(2n+1)$$, so if we let $$E_n$$ be the interior of the region bounded by $$C'(2n)$$ and $$C'(2n+2)$$, then each circle in $$\mathscr{H}_n$$ is contained in $$C'(2n) \cup E_n \cup C'(2n+2)$$, and each point in $$E_n$$ belongs to exactly two circles in $$\mathscr{H}_n$$, while each point on $$C'(2n)$$ or $$C'(2n+2)$$ belongs to exactly one circle in $$\mathscr{H}_n$$. Letting $$\mathscr{H} = \bigcup_{n=1}^\infty \mathscr{H}_n$$, we see that each point in $$\mathbb{R}^2 - (C'(2) \cup U)$$ belongs to exactly two circles in $$\mathscr{H}$$, each point in $$C'(2)$$ belongs to exactly one circle in $$\mathscr{H}$$, and each point in $$U$$ belongs to zero circles in $$\mathscr{H}$$. To cover $$U$$, let $$\mathscr{F}$$ be the family of circles given by the Lemma applied to $$U$$ and $$P$$, and define $$\mathscr{F}' = \mathscr{F} \cup \mathscr{H} \cup \{C'(2)\}$$, so each point of $$\mathbb{R}^2$$ belongs to exactly two circles in $$\mathscr{F}'$$, except $$P$$ which belongs to exactly three (since it lies on $$C'(2)$$). But none of the circles in $$\mathscr{F}'$$ are centered at $$P$$, so we can extend the family $$\mathscr{F}'$$ to a covering of $$\mathbb{R}^2$$ with degree $$3$$ by adding all circles centered at $$P$$. | 2022-11-30 04:18:30 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 166, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.938401997089386, "perplexity": 101.30520363332624}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446710719.4/warc/CC-MAIN-20221130024541-20221130054541-00456.warc.gz"} |
https://socratic.org/questions/for-f-t-abs-t-1-how-do-you-find-f-5-f-0-and-f-9 | # For f(t)=abs[t]+1; how do you find f(-5),f(0) and f(-9)?
Jul 7, 2015
$f \left(- 5\right) = 6$
$f \left(0\right) = 1$
$f \left(- 9\right) = 10$
#### Explanation:
Given $f \left(t\right) = \left\mid t \right\mid + 1$
For specific values of $t$:
$f \left(- 5\right) = \left\mid - 5 \right\mid + 1 = 5 + 1 = 6$
$f \left(0\right) = \left\mid 0 \right\mid + 1 = 0 + 1 = 1$
$f \left(- 9\right) = \left\mid - 9 \right\mid + 1 = 9 + 1 = 10$ | 2021-12-06 18:27:37 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 8, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8817945122718811, "perplexity": 14285.762044023733}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-49/segments/1637964363309.86/warc/CC-MAIN-20211206163944-20211206193944-00113.warc.gz"} |
https://practicepaper.in/gate-ec/stability-analysis | # Stability Analysis
Question 1
Consider a unity feedback system, as in the figure shown, with an integral compensator $\frac{K}{s}$ and open-loop transfer function
$G(s)=\frac{1}{s^2+3s+2}$
where $k \gt 0$. The positive value of K for which there are exactly two poles of the unity feedback system on the $j\omega$ axis is equal to ______ (rounded off to two decimal places).
A 2.45 B 4.28 C 6 D 6.25
GATE EC 2019 Control Systems
Question 1 Explanation:
$\frac{Y(s)}{X(s)}=\frac{K}{s^{3}+3 s^{2}+2 s+K}$
Two poles of this system lie the system is moro:
System is marginally stable.
$\qquad k_{\text{mar}}=3 \times 2=6$
Question 2
A unity feedback control system is characterized by the open-loop transfer function
$G(s)=\frac{2(s+1)}{s^{3}+ks^{2}+2s+1}$
The value of k for which the system oscillates at 2 rad/s is ________.
A 0.6 B 0.69 C 0.89 D 0.75
GATE EC 2017-SET-2 Control Systems
Question 2 Explanation:
The given open loop transfer function is,
$G(s)=\frac{2(s+1)}{s^{3}+K s^{2}+2 s+1}$
The closed loop transfer function is,
$T(s)=\frac{G(s)}{1+G(s)}=\frac{2(s+1)}{s^{3}+K s^{2}+4 s+3}$
When system oscillates, i.e., when system is marginally stable,
$(1)(3)=K(4)$
$\mathrm{So}, \quad K=0.75$
Question 3
Which one of the following options correctly describes the locations of the roots of the equation $s^{4}+s^{2}+1=0$ on the complex plane?
A Four left half plane (LHP) roots B One right half plane (RHP) root, one LHP root and two roots on the imaginary axis C Two RHP roots and two LHP roots D All four roots are on the imaginary axis
GATE EC 2017-SET-1 Control Systems
Question 3 Explanation:
$q(s)=s^{4}+s^{2}+1=0$
$\begin{array}{r|rrr} s^{4} & 1 & 1 & 1\\ s^{3} & 4 & 2 & 0\\ s^{2} & 0.5 & 1 & 0 \\ s^{1} & -6 & 0 & 0 \\ s^{0} & 1 & 0 & 0 \end{array} \frac{d A(s)}{d s}=4 s^{3}+2 s^{1}$
There are two sign Changes in the first column of the R-H table and the order of auxiliary equation is 4. So, four poles are symmetric about origin.
$\therefore$2 RHP roots and 2 LHP roots.
Question 4
The first two rows in the Routh table for the characteristic equation of a certain closed-loop control system are given as
The range of K for which the system is stable is
A $-2.0 \lt K \lt 0.5$ B $0 \lt K \lt 0.5$ C $0\lt K\lt \infty$ D $0.5\lt K\lt \infty$
GATE EC 2016-SET-3 Control Systems
Question 4 Explanation:
$\begin{array}{c|cc} s^{3}& 1 & (2 k+3) \\ s^{2}&2 k & 4 \\ s&\frac{4 k^{2}+6 k-4}{2 k} & 0\\ s^{0}&4 \end{array}$
For stability, $K \gt 0$
\begin{aligned} 4 K^{2}+6 K-4& \gt 0 \\ 2 K^{2}+3 K-2& \gt 0 \\ (2 K-1)(K+2)& \gt 0 \\ \Rightarrow \quad K& \gt \frac{1}{2} or, k \lt -2 \\ \text{Hence, }\quad 0.5& \lt K \lt \infty \end{aligned}
Question 5
The transfer function of a linear time invariant system is given by
$H(s)=2s^{4}-5s^{3}+5s-2$
The number of zeros in the right half of the s-plane is ________
A 1 B 2 C 3 D 4
GATE EC 2016-SET-1 Control Systems
Question 5 Explanation:
$2 s^{4}-5 s^{3}+5 s-2=0$
By Routh Array,
$\begin{array}{c|ccc} s^{4} & 2 & 0 & -2 \\ s^{3} & -5 & 5 & \\ s^{2} & 2 & -2 & \\ s^{1} & 0(2) & & \\ s^{0} & -2 & & \end{array}$
Number at sign changes = number at roots (zeros) in right half of s-plane =3
Question 6
Match the inferences X, Y, and Z, about a system, to the corresponding properties of the elements of first column in Routh?s Table of the system characteristic equation.
A X$\rightarrow$P, Y$\rightarrow$Q, Z$\rightarrow$R B X$\rightarrow$Q, Y$\rightarrow$P, Z$\rightarrow$R C X$\rightarrow$R, Y$\rightarrow$Q, Z$\rightarrow$P D X$\rightarrow$P, Y$\rightarrow$R, Z$\rightarrow$Q
GATE EC 2016-SET-1 Control Systems
Question 6 Explanation:
When all elements are positive, the system is stable. When any element is zero, the test breaks down. When there is change in sign of coefficients, the system is unstable.
Question 7
The characteristic equation of an LTI system is given by $F(s)=s^{5}+2s^{4}+3s^{3}+6s^{2}-4s-8=0$ The number of roots that lie strictly in the left half s-plane is _________.
A 0 B 1 C 2 D 3
GATE EC 2015-SET-3 Control Systems
Question 7 Explanation:
Using Routh's tabular or:
$\begin{array}{c|lll} s^{5} & 1 & 3 & -4 \\ s^{4} & 2 & 6 & -8 \\ s^{3} & 0 8 & 0(12) & 0 \\ s^{2} & 3 & -8 & 0 \\ s^{1} & \frac{100}{3} & 0 & 0 \\ s^{0} & -8 & & \end{array}$
Auxiliary equation:
\begin{aligned} A &=2 s^{4}+6 s^{2}-8=0 \\ \frac{d A}{d s} &=8 s^{3}+12 s=0 \end{aligned}
As one time row of zero occur in Routh's table therefore a pair of imaginary pole exists, also number of sign change in Routh's table is one. Hence, two poles lie strictly in the left half of s-plane.
Question 8
A plant transfer function is given as $G(s)=(K_{P}+\frac{K_{I}}{s})\frac{1}{s(s+2)}$. When the plant operates in a unity feedback configuration, the condition for the stability of the closed loop system is
A $K_{P}\gt \frac{K_{I}}{2} \gt 0$ B $2K_{I}\gt K_{P} \gt 0$ C $2K_{I} \lt K_{P}$ D $2K_{I} \gt K_{P}$
GATE EC 2015-SET-1 Control Systems
Question 8 Explanation:
$G(s)=\left(K_{p}+\frac{K_{I}}{s}\right)\left(\frac{1}{s(s+2)}\right)$
The closed loop transfer function for unity feedback
\begin{aligned} \frac{G(s)}{1+G(s)} &=\frac{\left(K_{p} s+K_{I}\right)}{s^{2}(s+2)+\left(K_{p} s+K_{I}\right)} \\ &=\frac{K_{p} s+K_{I}}{s^{3}+2 s^{2}+K_{p} s+K_{I}} \end{aligned}
Using Routh's tabular form:
$\begin{array}{c|cc} s^{3} & 1 & K_{p} \\ s^{2} & 2 & K_{I} \\ s^{1} & \frac{2 K_{p}-K_{I}}{2} & 0 \\ s^{0} & K_{p} & \end{array}$
For system to be stable;
\begin{aligned} K_{p}& \gt 0\\ \text{and }\frac{2 K_{p}-K_{I}}{2}& \gt 0 \\ \text{or} \quad 2 K_{p}-K_{I}& \gt 0 \\ \text{or} \quad K_{p}& \gt \frac{K_{I}}{2} \gt 0 \end{aligned}
Question 9
Consider a transfer function $G_{p}(s)=\frac{ps^{2}+3ps-2}{s^{2}+(3+p)s+(2-p)}$ with $p$ a positive real parameter. The maximum value of $p$ until which $G_{p}$ remains stable is ____.
A 1 B 2 C 3 D 4
GATE EC 2014-SET-4 Control Systems
Question 9 Explanation:
Given transfer function
$G_{p}(s)=\frac{p s^{2}+3 p s-2}{s^{2}+(3+p) s+(2-p)} \qquad\ldots(i)$
The characteristic equation is
$=s^{2}+(3+p) s+(2-p)\qquad\ldots(ii)$
Using Routh Hurtwiz criterion,
$\begin{array}{l|ll} s^{2} & 1 & (2-P) \\ s^{1} & (3+p) & 0 \\ s^{0} & (2-p) & 0 \end{array}$
For system to be remain stable
\begin{aligned} 2-p & \geq 0 \\ p_{\max } &=1.99 \end{aligned}
Question 10
The forward path transfer function of a unity negative feedback system is given by
$G(s)=\frac{K}{(s+2)(s-1)}$
The value of K which will place both the poles of the closed-loop system at the same location, is ______.
A 4.5 B 2.25 C 1.25 D 0.5
GATE EC 2014-SET-1 Control Systems
Question 10 Explanation:
Given that, $\quad G(s)=\frac{K}{(s+2)(s-1)}$
Using root locus method, the break point can be obtain as
\begin{aligned} \Rightarrow \quad 1+G(s)&=0 \\ 1+\frac{K}{(s+2)(s-1)} &=0 \\ \text{or }\quad K&=-(s+2)(s-1)\\ \frac{d K}{d s} &=-2 s-1=0 \\ \text{or }\quad s &=-0.5 \end{aligned}
To have, both the poles at the same directions
\begin{aligned} |G(s)|_{s=0.5} &=1 \\ K &=2.25 \end{aligned}
There are 10 questions to complete. | 2022-05-21 15:59:05 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 61, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9991726875305176, "perplexity": 2585.910243026638}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662539131.21/warc/CC-MAIN-20220521143241-20220521173241-00236.warc.gz"} |
https://answers.ros.org/question/405119/invert-a-tf-frame-axis/ | ROS Resources: Documentation | Support | Discussion Forum | Index | Service Status | Q&A answers.ros.org
# Invert a TF frame axis
I want to add a camera frame with x on the right, y up, and z forward. I am using the following command:
rosrun tf2_ros static_transform_publisher x y z yaw pitch roll map camera
First I tried to set up yaw, pitch, and roll manually by changing those three values. But there was always one axis that needed to be inverted which I could not do since if I change one of the three values, two axes change, not one. So, I ended up having a frame with x on the right, y up and z backward. Then I calculated a rotation matrix between those two coordinate frames. I found
0 0 1
R = -1 0 0
0 1 0
Then from the rotation matrix to yaw, pitch, and roll (-1.57, -0, 1.57). But the result is the same. How can I flip only one axis keeping the other two unchanged?
edit retag close merge delete
What do you mean by "x on the right"? After rotation you want the positive x-axis pointing to the left?
( 2022-08-17 06:34:10 -0600 )edit
@Mike Scheutzow The coordinate system I want should have x pointing to the right, y upward, and z forward.
( 2022-08-17 10:16:18 -0600 )edit
Sort by » oldest newest most voted
Did you check this StackOverflow post?
• Check that the frame you get from your API is right-handed. You do so by computing the determinant of the 3x3 rotation part of your 4x4 transform matrix: it must be +1 or very close to it.
• If it is -1, then flip one if its axis, i.e. change the sign of one of the columns of the 3x3 rotation.
• Note carefully: I said "columns" because I assume that you apply a transform Q to a point x by multiplying as Q * x, x being a 4x1 column vector with the last component equal to one. If you use row vectors left-multiplied by Q you need flip a row.
• If that determinant is +1, you have a bug someplace else.
You can try to use quaternions too, here are some great sources:
P.S. This part from the second link seems to be related to your case
## Conventions
From REP 105 Coordinate Frames for Mobile Platforms
On a mobile robot, x is forward, y is left, and z is up
For cameras z is forward, x is right, y is down, and frame_id ends with _optical
more
1
@Ijaniec I checked those answers. It seems, ROS tf uses right-handed rules and there is no way I can get the frame I want. So, I will just multiply one axis by -1.
( 2022-08-17 11:31:41 -0600 )edit
OK, please report with the effects later! If this answer helped you, you can accept it as a correct one (right under the note, "✓") so it will be marked as solved in the queue
( 2022-08-18 07:09:46 -0600 )edit | 2022-12-01 00:08:50 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.21632802486419678, "perplexity": 1183.552381922543}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446710777.20/warc/CC-MAIN-20221130225142-20221201015142-00804.warc.gz"} |
https://cmb-s4.org/wiki/index.php/Forecastpann | # Overview
If Dark Matter (DM) is a weakly interacting massive particle (WIMP) and it is a thermal relic, then its self-annihilation cross-section can be determined by its relic density today through the Boltzmann equation (e.g. http://arxiv.org/abs/1204.3622).
Depending on the model, DM particles can annihilate into gauge bosons, charged leptons, neutrinos, hadrons, or more exotic states. These annihilation products then decay or interact with the photon-baryon fluid and produce electrons, positrons, protons, photons, and neutrinos. Neutrinos do not further interact with the photon-baryon fluid, but interact gravitationally and their effects can be observed through the lensing of the CMB. A proton's energy deposition to the fluid is inefficient due to their high penetration length. Hence, the main channels for energy injection are through electrons, positrons, and photons. At high energies, positrons lose energy through the same mechanisms as electrons. High energy electrons lose energy through inverse Compton scattering of the CMB photons, while low energy electrons lose energy through collisional heating, excitation, and ionization. Photons lose energy through photoionization, Compton scattering, pair-production off nuclei and atoms, photon-photon scattering, and pair-production through CMB photons. The rate of energy release per unit volume by a self-annihilating DM particle is given by (e.g. http://arxiv.org/abs/0905.0003)
equation 1, energy release by self-annihilating DM particle
where rho_c is the critical density of the universe today, Omega_{DM} is the DM density, f is the energy deposition efficiency factor, <sigma v> is the velocity-weighted annihilation cross section, and m_{DM} is the mass of the DM particle, assumed to be a Majorana particle.
Due to these processes, the photon-baryon plasma is heated and the ionization fraction is modified. This leads to modifications of the recombination history (e.g. http://arxiv.org/abs/0905.0003) and, consequently, of the CMB spectra. For details on the energy injection processes, see http://arxiv.org/abs/0906.1197. The energy injection due to DM annihilation broadens the surface of last scattering, but does not slow recombination (http://arxiv.org/abs/astro-ph/0503486). The extra scattering of photons at redshift z ~< 1000 damps power in the CMB temperature and polarization fluctuations at small angular scales (ell ~>100), and adds power in the E-mode polarization signal at large scales (ell < 100). The 'screening' effect at ell ~> 100 goes as an exponential suppression factor Cl --> e^{-2\Delta\tau} Cl, where \Delta\tau is the excess optical depth due to dark-matter annihilation. This exponential factor is partially degenerate with the amplitude of the scalar perturbation power spectrum A_s, and polarization data helps to break this partial degeneracy.
The CMB is sensitive to the parameter p_{ann}, defined as
equation 2, p_ann
For a thermal s-wave annihilation cross-section <sigma v> = 3e-26 cm^3/s/GeV, and a energy deposition efficiency factor f=1, one can extract the 95% CL upper limit on DM particle masses that can introduce the excess optical death for ell ~> 100. For example, the 95% CL upper mass limit of the DM particle is 170 GeV for 3e-26 cm^3/s/GeV thermal cross-section in the nominal 1uK-arcmin noise in T, 3' beam, fsky=0.4 case.
(Text largely taken and modified from http://arxiv.org/abs/1402.4108, section 5)
## Forecasts
In the following, we present constraints on p_ann for varying beam (fixed noise), varying noise (fixed beam), and varying fsky (fixed beam, and fixed 'effort').
The inputs are as defined in ForecastingStep1, specifically I used:
• the lensed TT, EE, TE spectra (no Cldd)
• ell range = [30, 3000] in TT and TE, ell range = [30, 5000] in EE (unless otherwise stated)
• Planck prior ell range = [2, 2500] in TT (fsky=0.8), and ell range = [30, 2500] for TE/EE (fsky=0.2 for varying noise/beam cases and fsky = 0.6-fsky_S4 for varying fsky case)
• Planck only configuration constraints: sigma(p_ann) = 0.014; Planck published 95% CL limit p_ann < 0.04 [cm^3 s^-1 GeV^-1]
Figure 1: sigma(p_ann), varying beam, fixed noise at 1uK-arcmin, fsky=0.4
• a lot of the signal-to-noise comes from ell < 1000 in TT, and EE and TE are there to break degeneracies with A_s. So the beam doesn't matter as much.
Figure 2: sigma(p_ann), varying noise, fixed 3' beam. fsky=0.4
Figure 3: sigma(p_ann), varying sky, beam=3'
• As is found in http://arxiv.org/abs/1402.4108, for the sensitivity level we are considering for S4, p_ann is close to sample variance limited and benefits from observing more sky given a fixed number of detector/effort in building an instrument. | 2019-03-25 20:20:29 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.850271999835968, "perplexity": 4414.70344933058}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-13/segments/1552912204300.90/warc/CC-MAIN-20190325194225-20190325220225-00059.warc.gz"} |
https://codeforces.com/blog/entry/80562 | dcordb's blog
By dcordb, 6 months ago,
We (the setters) hope you liked the problems. Here is the editorial:
C++ solution
Python solution
Problem idea: dcordb
Solution idea: dcordb and Devil
C++ solution
Problem idea: Devil
Solution idea: Devil
C++ solution
Problem idea: Devil
Solution idea: Devil
C++ solution
Problem idea: Devil
Solution idea: marX and Devil
C++ solution
Python solution
Problem idea: Devil
Solution idea: Devil and dcordb
C++ solution
Problem idea: Devil
Solution idea: marX
C++ solution
Problem idea: gilcu3
Solution idea: gilcu3 and marX
C++ solution
Problem idea: Devil
Solution idea: Devil and antontrygubO_o
C++ solution
Problem idea: marX
Solution idea: marX and Devil
• +125
» 6 months ago, # | ← Rev. 2 → +2 Where are the time complexities?I usually find understanding a solution easier when the time complexity is indicated with the solution.Amazing contest, nonetheless.
• » » 6 months ago, # ^ | +1 Video Editorial of B1 and B2: Koa and the Beach
• » » » 6 months ago, # ^ | +11 Nice explanation, good to see the use of whiteboard which is rare nowadays.
• » » » 6 months ago, # ^ | 0 Nice Explanation though!I was trying my own method but the "safe state" condition is checked with an if case.My SolutionBut I'm receiving the wrong answer in test case 2 (2142ndtoken)My method is:: Move out of safe step when the tide is going down such that you have the maximum opportunity of reaching next safe stateI'm not getting why its a wrong answer!Please help, thanks in advance!
» 6 months ago, # | +58 Lost the entire interest after seeing problem statement of B1 :(
• » » 6 months ago, # ^ | +11 That's why people moved to C
• » » » 6 months ago, # ^ | +3 C and D protected my life
• » » » » 6 months ago, # ^ | +10 seriously dude, you didn't even participate.
• » » » » » 6 months ago, # ^ | +4 It's Beyond Science
• » » » » » » 6 months ago, # ^ | +4 I think you got it.
• » » » » » 6 months ago, # ^ | +3 Maybe he/she uses a fake account lol
• » » » » » » 6 months ago, # ^ | +16 I don't get the point of making many accounts. why would you make an alt account just to post bullshit?
• » » » » » » » 6 months ago, # ^ | ← Rev. 2 → 0 No idea. Maybe retirednoob can give a better answer
• » » » » » » » 6 months ago, # ^ | 0 Listen, why people made fake account because they have impostor syndrome. I also have faced same issue. they changed accounts to hide their failures from people. they have changed them self a bit, hopefully they won't do this again.
• » » » » » 6 months ago, # ^ | 0 seriously dude, he didn't even tell you.
» 6 months ago, # | +8 In Div1D, the time complexity is $O(nm)$, so the limits could have been higher as I think of it.Was it intentional (so that one would think of it more as a $O(n^3)$, flow-like problem, for example)? Or possibly a change of mind -- maybe wrote a cubic solution, found the square one after?
• » » 6 months ago, # ^ | +44 If you push the limits to as high as possible that will fit within TL, you may be unnecessarily giving away info to the contestants. In this case making it so that $O(nm)$ is the only complexity that fits within TL tells you that the algorithm must be of that complexity. Imo it makes sense to leave it this small so it remains ambiguous (and still doesn't let any inefficient solutions pass).
» 6 months ago, # | -18 Long statements + weak pretests
• » » 6 months ago, # ^ | -51 Problem setter thinks he is very smart but he is a kid, don't know who allowed him to set problems.
• » » » 6 months ago, # ^ | +4 The contest was good and I liked it even tho I couldnt solve any problem, I enjoyed the 2 hours. Speak for yourself.
• » » » » 6 months ago, # ^ | +2 It's not about solving problems, just look at the description & difficulty of the problems and points assigned to that problems, Totally unfair. C-1750, B-500/700.
» 6 months ago, # | -9 Devil I think you made undirected graph in solution of div2C :)
• » » 6 months ago, # ^ | ← Rev. 3 → 0 graph will be undirected only , since we need to find size of the connected component , thus we need to visit all the nodes (to mark them and thus to find size of the connected component) . It's "connected" not "strongly connected" in editorial .
» 6 months ago, # | +6 lost interest in this round due to B problem statement
» 6 months ago, # | +5 Graph solution for problem div2 C (div1 A) is beautiful . Can some tell other approach without using graphs.
• » » 6 months ago, # ^ | ← Rev. 2 → 0 Please read the solution in this comment. Also you can read the explanation of the same written by yash1402_. Hope it helps!
• » » 6 months ago, # ^ | ← Rev. 2 → +14 If any character in $A$ is greater than $B$ at a particular index then it is not possible to transform $A$ to $B$.$Rule$ : Do not touch the indices in which $A$ and $B$ match.And now follow the following processFirst we look at the indices with contain $"a"$ in the string $A$ and look at the same indices in $B$ and let minimum among those indices in $B$ be $X$. Now replace characters in $A$ in those indices to $X$. Now one operation is performed. Similarly repeat the process to $"b"$, $"c"$ $.....$ $"t"$
• » » 6 months ago, # ^ | 0 Yes,I used set to solve that problem --> Link
• » » 6 months ago, # ^ | ← Rev. 3 → 0 I used C++ Maps (storing pairs and their frequency) to solve the problem. I'm surprised to see so many people using graphs. -> 87923309
• » » 6 months ago, # ^ | +14 It is obvious that, if the first string has a character with higher value than the second string in at least one position, there is no solution.Now for the non trivial case: If we need to transform at least one instance of character x into a higher character y, then there is no harm in transforming ALL instances of x into y (it is still going to be one move), as long as we consider the potential y characters in increasing order (otherwise we might leap over relevant characters and increase some instances of x too much). While there potentially is a benefit, i.e. reduction of the number of necessary moves in cases like the following example: for strings aab and bcc we need to transform a -> b, a -> c, b -> c, and when transforming both a's to b then both b's to c, we achieve the desired result without the need to spend an extra move for a -> c transformation. The potential x characters should be considered in increasing order too, to avoid making a move before it gains full potential accumulated from potentially transforming lower characters into x first. Fill a matrix M with 20 rows and 20 columns as follows M[x][y] = true, if at least on instance of the x-th character ultimately needs to be transformed into the y-th character, false otherwise. Only the half of the matrix strictly above the diagonal is of interest, because if M[x][y] = true for some x, y where x > y, then we are back at the trivial case, and if x = y, there is no transformation required. Now iterate over potential x-th characters in increasing order, for each iterate over potential y-th characters, where y > x. For the first encountered M[x][y] = true entry (if any), count plus one move, and iterate over the rest of the entries in the same row, i.e. over z-th characters for z > y. For each entry M[x][z] = true, make M[y][z] true, as this is simulating transforming all instances of the x-th characters into the y-th, while keeping track into what characters they need to be transformed ultimately.P.S. Really, this approach is not that far away from the graph solutions, since the matrix can be seen as an adjacency matrix of the graph. It is a matter of interpretation and I did not have a graph in mind when coming up with the solution.
• »
»
6 months ago, # ^ |
-30
include <bits/stdc++.h>
using namespace std; int main() { int t; cin >> t; while (t--) { int n; string a, b; cin >> n >> a >> b; vector<set>v(20); int f=1; for(int i=0;i<n;i++){ if(a[i] > b[i]){ f=0; break; } if(a[i] < b[i]) v[a[i]-'a'].insert(b[i]-'a'); } if(f==0){ cout<<"-1"<<endl; } else { int c=0; for(int i=0;i<20;i++){ if(v[i].size()>0){ c++; set::iterator itr,itr2; itr2 = v[i].begin(); itr=v[i].begin(); itr++; for(;itr!=v[i].end();itr++){ v[*itr2].insert(*itr); } } } cout<<c<<endl; } }
return 0;
}
Simple Solution Using vector<set> . Easy to Understand.
• » » » 6 months ago, # ^ | 0 it is better use ideone for sharing code...
• » » 6 months ago, # ^ | +8 C++ std::priority_queue is also a nice way to solve this problem
• » » » 6 months ago, # ^ | 0 Can you please share your solution with priority_queue? Thanks in advance.
» 6 months ago, # | +27 dcordb can you please you spoiler to display editorial (blogs), it will be more convenient like previous editorials.
» 6 months ago, # | +240 Solution of div1C only gives the answer but doesn't really explain how one might come up with that in the first place, so I'll try.First observation is having lots of letters in the first string is bad for us, so we want to join letters as much as possible. That is, suppose the solution does operations A->B, followed by A->C. We could have changed all As to Bs in the first operation to do A->B, B->C instead. B->A followed by C->A can also be replaced by B->C, C->A.From this, we might intuit that the solution consists of one very long chain in which we have a "snowball" collecting all original letters into the same letter.If the graph has no cycles, then such a snowball can simply follow the topological sort of the DAG (this is div1A). If there are cycles, then some vertices will be visited more than once. If you know that a vertex is going to be visited more than once anyway, you might as well put it as the first and last vertices in the order, since this guarantees that vertex can reach/is reachable by all others.What about the vertices you only visit once? Clearly there cannot be any cycles between them, so they are a DAG (and as we know any DAG can be solved with the topological sort). So the solution looks like: visit some set of vertices S, do topological sort on the rest, visit S again. To minimize the number of operations we want the DAG to be the maximum possible.
• » » 6 months ago, # ^ | +10 Really, thanks.
» 6 months ago, # | -25 Check out this video editorial of Problem CBrute Force Approach, hope you will like it :)
» 6 months ago, # | 0 Question : C2 Koa and the Beach Can someone explain why the below test case below should have "NO" as a answer: 20 26 32 31 27 14 7 2 23 22 21 19 27 31 12 23 31 0 11 28 20 31 30
» 6 months ago, # | ← Rev. 2 → +34 I don't like editorial of 1384B2 - Koa and the Beach (Hard Version) So I'll write my own. I think it should be easier.Let s(p, t) is answer on question "is there way to reach position p at cycle of tide t?". I'll make numbering of tide cycle in following way [k, k-1, k-2..., 1, 0, 1, 2 ... k-1]. It will be much easier to understand later.So, now obviously, if 0 is island, then s(0, t) is true for any t. Next, suppose some s(p, t) is true, then if we get this existing way to reach p at cycle time t and just standby there, eventually we can either drown or stay infinitely. Obviously, if we drown then it is because tide increased. This means that for all i up to some point x value of s(p, i) is also true, unless you can stand there forever. So either whole segment from i to x is true, or all s(p, t) is true for this p. Position p where you can stand forever lets call safe position similarly with editorial.Now important part: we can drown only when tide increase, highest allowed tide is easy calculated. And that's why cycle of tide I defined in the way above, with increasing in the end. Because higher allowed tide — longer we can stay. And, this also means that cycle time with "bad" level of tide is in prefix and suffix of cycle. This all means that for any position p there is first (lowest) t for which s(p, t) is true. And, in this notation all possible cycle time for position p gives range from first time to the time when highest possible tide is reached during rise of tide. Thus, to solve the task we only need to find for each position p first t when s(p, t) is true.Now, consider three cases: if you stand on safe position p, to get earliest t in cycle for next position p+1 you should move at fall, so you move to next position at fall of tide and highest allowed tide. if you stand on unsafe position p when tide is rising, you just need to move forward as soon as possible, because if now you can't move forward, then on next time you can't also move forward, but if you can right now — it's earliest moment in cycle you can reach next position. if you stand on unsafe position p when tide is falling you just need to wait until you can move forward and not drown on next position p+1. So, for each position we just update earliest t, and total time complexity is O(n). 87931302
• » » 6 months ago, # ^ | -15 Video Editorial of B1 and B2: Koa and the Beach
• » » » 6 months ago, # ^ | +2 I liked the way you explained it. Nice and clear. But I don't like the fact that this problem was kept at B.
• » » 6 months ago, # ^ | 0 could you explain why test case 3: 4 3 4 0 2 4 3 has a solution(i.e Yes)?
• » » » 6 months ago, # ^ | +3 4 3 4 0 2 4 3 On position 0 we can stay forever — it's island, so we should move on fall, so we move at t (cycle time) at 0 and arrive at t = 1 on position 1 because there will be 0+2 height which is less than 4. Also, at position 1 we can stay forever, so we should move at fall again. So, we will wait until t = 0 again and move at t = 0 and if we move then on next position we will have 2+2 which is alright, but we can't stay here forever because 2+3 > 4. This means that we now on falling tide and unsafe position. We need to move next as soon as possible. If we move now to position 2 we will have 4 + 1 height and drown, so we should wait once and after move we will have 4 + 0 height which is alright. Now, at position 2 we are in unsafe position and tide is rising, so we should move now (without any "if", because situation may only become worse). Now we move to position 4 and have height 3 + 1 which is alright. Similarly, on position 4 we are in unsafe position and tide is rising so we should move now, and successfully reach end. Each of this actions give earliest t for each positions, so earliest t array is: 0, 1, 3, 4, and corresponding tide: 3, 2, 0, 1.
» 6 months ago, # | +15 Next time please try to make smaller readable problems. Appreciated your efforts.
» 6 months ago, # | ← Rev. 2 → 0 The largest directed acyclic graph can be computed using dynamic programming in n*2^n.I remember a nice CF blog on this, could anyone point it out.
» 6 months ago, # | ← Rev. 2 → 0 DP solution to B1:87940952. dp[i][j]--> Can we be at location i at time j? So, if any of dp[n][j] is true, we can reach the island.
• » » 5 weeks ago, # ^ | 0 i might be late, but can you just tell me how did you calculate the upper bound of second state of dp? i mean the time factor...
» 6 months ago, # | ← Rev. 2 → 0 EDIT: Got it!My Code is Exactly like Editorial's.Can anyone Please tell me why is gives wrong for this simple case, ALTHOUGH it's giving correct output in my local Compiler.Tese Case: 13ddddddMy Local Output: 0Cf Compiler: 1
• » » 6 months ago, # ^ | +13 You don't clear your global arrays after printing -1.
• » » » 6 months ago, # ^ | 0 Oh Crap!I've been stuck here like since forever. Thanks !
» 6 months ago, # | +5 I found the following trick to solve Div2 B with simple implementation. The idea is like sliding window and updating the initial interval of $(-k, k)$ that corresponds to decreasing tide $(-k, 0)$ and increasing tide $(0, k)$. This way of set up simplified the implementation so much, but unfortunately, I came up with this at the end of the contest. Here's my solution
» 6 months ago, # | +13 Kudos to testers and those who arranged the problems in order of difficulty.
» 6 months ago, # | +51 Solution videos for people who prefer to have things explained visually are available here.
» 6 months ago, # | 0 Can B1 use array to memory state?
• » » 6 months ago, # ^ | 0 It's ok！
» 6 months ago, # | ← Rev. 2 → 0 How is my idea for Div1A different from the editorial?
• » » 5 months ago, # ^ | 0 test yếu anh ơi
» 6 months ago, # | 0 https://codeforces.com/contest/1384/submission/87885055 I am generating a 200 length string and initialising (n+1) strings with that. Now I just copy the common prefix of ith string to (i+1)th string and after the common prefix I just add a character different from the previously used one. Can someone please tell me what is the problem with my solution? It is failing at test case 32.
» 6 months ago, # | ← Rev. 4 → +3 In Div2C.For the testcase 1 4 aacb bcddThe topological order is acbd But we can't select the each pair of consecutive nodes because c-->b is not possible.Am i missing something?
• » » 6 months ago, # ^ | 0 I think it's abcd.aacb -> bbcb -> bccc -> bcdd
• » » 6 months ago, # ^ | +8 I don't know but gave you +1 for ma boi killua
» 6 months ago, # | +4 Although the quality of problems is quite good, but the score distribute is not reasonable. Div2 B that cost me 1 hour time only values 500+750 score. Div2 C is quite easy, but values 1750. I solved 5 problems but get a lower rank than someone solved 4 problems because of this score distribute. Sad~
» 6 months ago, # | +3 In the code of div1 E, the DP transition is dp[i] = ((dist[i] <= dist.back()) + get(0) + get(dist[i] + 1)) % mod; I understand get(0) and get(dist[i]+1) are the two cases in the editorial, but what does (dist[i] <= dist.back()) mean ?
• » » 6 months ago, # ^ | +3 Oh, I see. It seems that (dist[i] <= dist.back()) means let $w$ just end there !
• » » » 6 months ago, # ^ | ← Rev. 2 → +8 Yes it does that, it simply checks if string $s$ can end at $i$. For future references here is same solution that handles that part more clearly: Div1E C++ solution#include using namespace std; const int mod = 1000000007; int main() { ios::sync_with_stdio(false), cin.tie(0); string s; cin >> s; int suf0 = 0; while(!s.empty() && s.back() == '0') { s.pop_back(); suf0++; } if(s.empty()) { cout << suf0 << '\n'; return 0; } int n = s.length(); vector dist(n); for (int i = 0; i < n; ++i) if (s[i] == '0') dist[i] = (i ? dist[i-1] : 0) + 1; vector dp(n), nxt(n+1, n); auto get = [&](int i) { return nxt[i] < n ? dp[nxt[i]] : 0; }; for (int i = n-1; i >= 0; --i) { dp[i] = ((s[i] == '1') + get(0) + get(dist[i] + 1)) % mod; nxt[dist[i]] = i; } int ans = n; if (nxt[0] < n) ans = (1LL * get(0) * (nxt[0] + 1)) % mod; ans = 1LL * ans * (suf0 + 1) % mod; cout << ans << "\n"; return 0; } Here I removed zeroes at the end of $s$ (and multiply by this, plus $1$ to ans), and replaced that transition with an easier one to understand (it. checks if $s$ can end at $i$ with a $1$).
• » » » 6 months ago, # ^ | ← Rev. 2 → 0 Why w end there must meet this condition?I think i understood it, your solution is more clear.
» 6 months ago, # | 0 I find that if s="0", the solution will RE because it visited nxt[dist[0]+1] (dist[0]+1=2) and the size of nxt is $2$ !
• » » 6 months ago, # ^ | ← Rev. 3 → 0 Yes sorry, it will RE, it was an error modifying it to post it here, but model solution is correct.
» 6 months ago, # | 0 In Div1E/Div2C, what is the motivation for: "and the answer is $2⋅n−|LDAG|−1$"?
• » » 6 months ago, # ^ | 0
• » » » 6 months ago, # ^ | +21 That's bad when people have to look for explanations for editorials in random comments in random places of the blogs.
» 6 months ago, # | +1 Why the simple greedy solution isn't mentioned in editorial for Div2C?
» 6 months ago, # | -13 If you are keeping python as a submission language then please do some homework and check the comparative time it's gonna take in python or else change codeforces to cppforces.
• » » 6 months ago, # ^ | +3 Codeforces is already cppforces. Plenty of problems can't be solved using Python.
• » » » 6 months ago, # ^ | 0 Just tell me why can't be solved. Are there any logic that python doesn't support or recursion call stack that cannot be increased. It's all about time. You just have to increase the time limit for python in order to make python compatible for solving the same logic.
• » » » » 6 months ago, # ^ | ← Rev. 4 → +13 You want to have all the benefits python offers and no drawbacks. Obviously this won't do. Python is slower and that's the price you pay for its flexibility and coding speedAlso, learn to use pypy. I see that you used vanilla python, that's rarely a good idea in cp. Pypy is hardly ever not enough for the first d2 problems if you have a correct algorithm
• » » » » 6 months ago, # ^ | 0 That's just the way things are currently, where C++ has a big advantage over other languages. I'm not saying that it's right or wrong, but until this changes you'll probably be better off switching to C++.
» 6 months ago, # | +1 Case: x%4 == 3 if y%2=1 the first player takes one 1 and the game now is exactly the previous case with the first player as the second player. I think the author means the first player takes first 0 and then the game is exactly the previous case . Previous Case: x%4 == 3 && y%2 == 0 Or Maybe i am too naive to understand the Author's logic :)
• » » 6 months ago, # ^ | 0 Thanks, it's fixed now :)
• » » » 6 months ago, # ^ | 0 It's my pleasure man ! that you have considered it from your highness . Have a nice day / night ahead and Great luck for your future too :)
» 6 months ago, # | +3 Can anyone explain how we can solve Div2C using DSU?
» 6 months ago, # | 0 felt more like Div.1 contest
» 6 months ago, # | 0 Greedy solution for C 87971209
» 6 months ago, # | 0 IMO in Div.2 only problem C and maybe D had somewhat of a proper score allotted to it. Rest all should have had a higher score allotted to them. Especially B2.
» 6 months ago, # | +3 For String Transformation 1, I am really struggling to understand the answer after the 2nd sentence. I would like to understand this graph solution, could anyone please explain it in more detail?
» 6 months ago, # | ← Rev. 2 → 0 Can you please help me find error in my code ! This is Div 2 B (easy version) I have tried to copy the solution presented in the editorial, in python.It's giving max recursion error for test case: 20 40 69 59 67 9 69 34 27 6 0 4 25 34 28 35 17 0 9 28 62 1 9 87974537
» 6 months ago, # | 0 This is how I solved Div2Blet h be an array where h[i] = l - d[i]Then divide h into segments [L, R] where h[L] >= kh[R] >= kh[i] < k (L < i < R)For each segment check if exist 2 indices i and j such that i + h[i] < j - h[j] then there is no solution. Otherwise, a solution exists
» 6 months ago, # | ← Rev. 2 → +11 for mask in 1..2^n-1 for u in mask: is_dag[mask] |= is_dag[mask & (1 « u)] && ((mask & reach[u]) == 0)) Should it be: for mask in 1..2^n-1 for u in mask: is_dag[mask] |= is_dag[mask ^ (1 « u)] && ((mask & reach[u]) == 0))
• » » 6 months ago, # ^ | 0 Yes, it should be.
• » » 6 months ago, # ^ | 0 Fixed, thanks.
» 6 months ago, # | ← Rev. 2 → 0 haha
» 6 months ago, # | ← Rev. 2 → 0 In problem C, I didn't use the graph approach and instead used for loop traversals to go through the array a like this:for(int i=0;ib[i]){ count=-1; flag=1; } } if(flag!=1){ for(int i=0;i
» 6 months ago, # | ← Rev. 3 → 0 For DIV2C/DIV1 AFor this sample input 17 dgdfjjklhejkklWe can transform in just 5 steps,but author's solution is showing 6.Operations 3-e2-h4-j1,5,6 — k1,7 — lam I missing something?
• » » 6 months ago, # ^ | +3 1, 5, 6 — k is not correct. 1 is d while 5 and 6 are j.
• » » » 6 months ago, # ^ | 0 Thanks,it seems I didn't read the problem statement properly
» 6 months ago, # | 0 Can someone please explain me the graph solution of problem 3. I am not able to understand it from the editorial.
» 6 months ago, # | 0 B2 solution by greedy 88017812
» 6 months ago, # | 0 div 2 C seems to be a standard ques!! if yes then can you plz tell me more about this type of question!
• » » 6 months ago, # ^ | 0 Maybe you'll find it useful https://codeforces.com/blog/entry/80593?#comment-668966 By the way, can you please tell how the number of transformations required for abcb to tsrs is 3 (as there are 3 connected components?)
• » » » 6 months ago, # ^ | 0 a to t in 1 move , b,b to s,s in 1 move and c to r in one move so total 3 moves requires
• » » » 6 months ago, # ^ | 0 i have 2 solutions for this problem1 (https://codeforces.com/contest/1384/submission/88019936) is what i did during contest and its kind of general solution i guess andthe other one(https://codeforces.com/contest/1384/submission/88049338) is very interesting solution
» 6 months ago, # | 0 Anyone to explain :1383B — GameGame" editorial more easily please...?
» 6 months ago, # | ← Rev. 2 → 0 For C1 my approach seems to be different from what the comments are mentioning. (*I also don't seem to have a proper proof for it but it seems to work intuitively and would be nice if someone wants to look into this*)I took characters of string B and sorted them in the increasing order. (Let's call the smallest value of such a sorted sequence as SML)My claim is that if we try to change the characters of string A whose character in the corresponding same position in B is same as SML or once such character is found then every occurence of the same character in A will be changed to SML and the number of such distinct characters which were changed to SML will be added to our final answer.Now first turn is over SML will pick the next smallest element and will redo the above again.Finally all characters in A will be changed to the ones in B and the answer will be minimum.I used set of pairs to implement it.If there are any counters to the above explanation they are most welcome. (Since I am not sure why my approach actually works besides few samples I checked with pen and paper)https://codeforces.com/contest/1384/submission/88045399
» 6 months ago, # | 0 In Div.1F, the time limit seems so tight after some very strong hack testcases were added, and it's so difficult to pass them. By the way, can anyone tell me how to get max flows quickly.
» 6 months ago, # | ← Rev. 2 → +6 I have some questions on 1383B - GameGame : why we dont have to consider higher bits when we're processing the first bit that the number of ones is odd? (or : why we can turn these numbers into an array of $x$ ones and $y$ zeros? should we consider the influences of higher bits (though their number of ones is even) ?)i'll appreciate it if you can explain these questions to me :)
• » » 6 months ago, # ^ | ← Rev. 2 → +16 If number of ones is even for any higher bits. Then irrespective of distribution, both players will have either odd or both players will have even number of bits . So both will end up having same value(because of same parity) for that bit. This is the explanation I gave to Myself .
• » » » 6 months ago, # ^ | +3 many thanks!
• » » 6 months ago, # ^ | +3 If the number of $1$ is even for a bit, then the final result could be only: Both two players take even count from it, they both get $0$ on that bit. Both two players take odd count from it, they both get $1$ on that bit. which will always make a tie on that bit. The order doesn't matter.It could be confusing, so you can have a simulation.
• » » » 6 months ago, # ^ | 0 thx!
» 6 months ago, # | +8 In div 1 E editorial, I don't understand this part:"if the $(j+1)$-th character in $w$ is 0, and we have $k$ zeros after the last 1 in $w$ find the next block of $k+1$ zeros in $s$, (ie first $i'>i$ such that for each ($0\le k' • » » 6 months ago, # ^ | +8 Oh that was a typo, as it says in editorial and as you can see in sample solution, if you have a block of$k$zeros, and you want another$0$, you should search for first block of at least$k+1$zeros.For example:$s = ...100\underline{0}100000$you are on the underline (suppose$w = ...1000$) and you want another zero, for this you can merge$0001$with some previous$1$(remember we ensured that one always exists) and search for next block of at least$4$zeros (the merge would look like$...10001|0000...$). In this way you would have$w = ...10000$. • » » » 6 months ago, # ^ | 0 Thanks, I understand the greedy part now but am now having trouble understanding the DP. The editorial says"Let$dp(i)$be the number of strings that we can obtain using the last$n-i$characters from$s$"However, running the model code tells me the dp states used in it are not exactly the same. It gives the following values for these suffixes: s: 0 1 1 1 0 0 dp: 9 9 6 3 2 1 Note that the actual number of things you can construct from 011100 is 18, not 9. It should be double the number of things you can construct from 11100 because you can choose whether or not to put the 0 at the beginning. Could you explain what the dp states in the solution actually represent? Thanks! • » » » » 6 months ago, # ^ | +8 Yes, the dp is well defined in each one, remember that at the end we use$dp(i)$where$i$is the position of the first one. • » » » » 6 months ago, # ^ | +8 If it helps, try thinking about it in this other way (this is how I first solved):Let$dp(i, c)$be the number of strings we can obtain from last$n - i$characters of$s$such that they start with exactly$c$($c \ge 0$) zeros. The transitions are the ones in editorial.Then you can notice that you don't need second state if at the transition you force to add$k$($k \ge 0$) zeros and then a$1$.Oh and about dp giving$9$, well remember you had a zero before first$1$, so you should multiply by a$2$(ie. ans is$dp(2) \cdot 2$). • » » » » 6 months ago, # ^ | ← Rev. 3 → +8 I have a feeling that we might misunderstand that text for dp[i]. A better example of explaining the mismatch would be: i 0 1 2 3 4 5 6 7 8 9 s 0 1 1 1 0 0 1 0 0 0 dp 47 40 28 16 11 6 4 3 2 1 Notice i=5, s[i]=0, according to the text, it is the number of string we could make using last n-i=10-5=5 characters. This number should be 8 instead of 6 (following the def. of dp)I think a better way to explain dp[i] is the following:dp[i] is the number of strings that you make where you must keep all preceding consecutive 0s when s[i] = 0, otherwise (if s[i]=1), it's the number of string that you can make to so that you keep this s[i] = 1.This explanation is kind of echoing with the greedy method in the tutorial.In the example above, dp[5] is the number of extensions where you must keep s[4] and s[5] and you try to extend this with a 1 or a 0 (corresponding to the get(0) and the get(dist[i]+1) component). More specifically, when s[i] is not followed by a 0 (like in the case for s[5]), you have to seek for the consecutive 0 block that is large enough. And finally, you can extend it with nothing if the ending 0 block is large enough or you are extending a 1. That's the dist[i] <= dist.back() component.This is what dcordb was talking about for the leading 0s I believe.I hope the tutorial is technically clearer for the ignorant like me :P. But I think the problem and the solution are both quite good and I very much appreciate the efforts of the problem setter. » 6 months ago, # | 0 In the solution for b1 what does this code correspond to? function go I mean, I understand the meaning but I haven't seen this kind of syntax till now....... • » » 6 months ago, # ^ | ← Rev. 2 → -10 It is the return type of a lambda expression. • » » » 6 months ago, # ^ | 0 ok thanks! • » » 6 months ago, # ^ | -10 It's a lambda function in C++. Read about them Here. • » » » 6 months ago, # ^ | 0 Will do thanks! • » » 6 months ago, # ^ | ← Rev. 2 → +52 Despite what the other commenters say it's actually not a lambda! It's a std::function.A std::function does type erasure magic and you can assign lambdas, function pointers, and callable structs (anything that is callable) to it. This type erasure comes at a cost.To actually have a lambda function you need to assign it to an auto variable. The type of the lambda is generated at compile time and you can't know it/type it yourself.Edit: For clarity. The right hand side of the assignment is a lambda function. But it gets converted to a std::function in the assignment. • » » » 6 months ago, # ^ | 0 All right... » 6 months ago, # | 0 Omkar , yes bro. I am from the group XD » 6 months ago, # | 0 Is there any other approach for problem C » 6 months ago, # | 0 Problem B1 : How the depth increased by time t = (t % 2k) ?? If so then the depth can be increased more than k which is not possible according to the statement. Help please... • » » 6 months ago, # ^ | 0 its more like depth increases by A[t], where t = (t%2k), take for example k = 3 array A = [0,1,2,3,2,1] so at time t = 0, depth increases by A[t%2k] = A[0]. similary at t = 4, depth increases by A[4%6] = A[4] = 2. • » » » 6 months ago, # ^ | 0 Thanks brother...Now i get it :) » 6 months ago, # | -27 include<bits/stdc++.h> using namespace std; typedef long long ll ; set<tuple<ll, ll, bool>> mark ; bool dfs(ll n, ll pos, ll &tide, bool &down , vector &d , ll k , ll l) { if(pos > n) return true ; if(mark.find({pos,tide,down}) != mark.end()){ return false; } mark.insert({pos,tide,down}) ; tide += down ? -1 : +1 ; if(tide == 0) down = false ; if(tide == k ) down = true ; if(d[pos] + tide <= l && dfs(n , pos, tide , down , d, k , l)) return true ; if(d[pos+1] + tide <= l && dfs(n , pos+1 , tide , down , d , k , l)) return true ; return false; } int main(){ ll t ; cin>>t ; while(t--){ mark.clear() ; ll n , k , l; cin>>n >> k >> l ; vector<int> d(n+2 , -k ) ; for(int i=1 ; i<=n ; i++) { cin >> d[i] ; } ll tide = 0; bool down = false; if( dfs(n,0 , tide, down , d, k, l) ) cout<<"Yes"<<'\n' ; else cout<<"No" << '\n' ; } return 0; } I am struggling why my code is showing wrong ans • » » 6 months ago, # ^ | 0 next time post submission link and what task » 6 months ago, # | ← Rev. 2 → 0 my dp solution for B1 using 2 states — https://codeforces.com/contest/1384/submission/88165283I want to verify time complexity — I think it should be n * k as for every index k times are possible i.e. earliest you could reach is i + 1 and latest you could reach is i + k » 6 months ago, # | 0 1383E - Strange Operation I think my understanding of DP is very good. But even I had trouble with editorial even thought I solved it in the first place! So, for those who want to learn all that magic few people understand, I'll try to explain it as much as I can. Horrible wall of textFor completeness, I'll repeat some of things from editorial. First thing we need to know is: maximum of bits is equivalent to bitwise OR. And result of any sequence of merging two bits is equivalent to: pick set of segments, replace them by their bitwise OR.Now, suppose we have feeling that this task should be DP, then what our thoughts? Well, main thought: what is state? State should be all possible way to do something. And here is a trouble: how can we make sure that all those ways are different? The state itself should guarantee that. And mind is blowing when you think that this one can be vanished or keeped, similarly zero can remain or keeped. And here where I stuck. Then, I had idea to have dp based on fixed one, but it doesn't help, because even if I knew how many ones I have before, I had no idea how to make sure all sequences are different.Honestly, here I stuck, but I had feeling that I should somehow know that my sequences are different for sure. But I had no idea. So I read beginning of editorial. Very beginning. To get a hint. And this hint was enough. Hint was a bit other, but here is hint I propose for all of you who read this: for any string we can make, we should be able to construct it, so find out how to check is it possible to construct given string.Task from now is: find out algorithm for checking: can we make from sequence$s$sequence$t$. That's it. And solution is easy: for each part of$t$we will pick first place where we can make it in$s$. So, we will start from empty sequence, and append bits to it. This leads to two cases: if next bit we need is 1 then we just find next unused 1 in$s$. If next bit is 0 then we need to find first place with enough number of zeroes we want, because in$t$we can only shrink number of zeroes. There is one corner case: we can't skip anything if we don't have ones yet, but we can deal with it separately.Now, crucial observation: sequence of steps of this greedy algorithm is deterministic, so any different input t for fixed s gives certain sequence of steps. I didn't say unique. It's true but it's much harder statement than I want to prove first. First, we will change greedy slightly for easier analysis.Let's change greedy in following way: for next bit 0 instead of looking for group of enough size we will look for first zero instead. And if you can't get next zero without remove of previous one — we just forbid this action. It's a bit vague words so I'll provide example: s = 001010110001 t = 0101001 0 1010 - what we was able to make Here we can't make second 0 because we need to somehow jump to next group of zeroes. So, my initial solution had this as a bug in idea, but it is good way to explain I think. Why? Well now we have addition of 0 and 1 single bit by bit and we able to tell can we make$t$by steps of this greedy algorithm. You can think of it as if we count all different sequences of bits which can be made by this greedy algorithm. We will solve this first, because it's easier and it will give idea how to solve initial task.Now, if we look carefully on how this greedy algorithm works we will see that depending on next bit (zero or one) we make different jumps in$s$. Thus, we can draw arrows from each cell: one arrow where we land if next digit is 0, and one arrow where we land if digit is 1. Some of cells don't have jump by zero (because we already have 0 and we can't remove 0 and arrow by zero means we want zero, but next symbol in$s$is not 0 so we should remove trailing zeroes and make new group but we decided to forbid it, because it's simpler version) This is graph of transitions. And crucial thing is: in this graph each edge from vertex is jump by different next bit. This means that any unique path corresponds to unique bit string. The number of different paths in graph like that is easy to calculate, because this graph is top sorted. So Either we go forward add value of dp in current position into dp of all positions where we can jump into — I call it forward push dp in my terminology. Either we go forward and add into dp of current position sum of dp of all positions from where we could come here — forward pull dp in my terminology. Either we go backwards and add into current dp of current position sum of dp of all positions where we can jump into — backward pull dp Either we go backwards and add value of current dp in current position into all positions from where we could come here — backward push dp I don't know if anyone else use similar terminology. push is when you add influence into next position, and pull is when you grab around values. Backward and forward makes sense only if you have ordering like in our case.Now two things not covered yet: we haven't talk about ending. We can end at any 1 (by bitwise OR of everything else in the end). And we can end at zero if it's in the end. Also, we need to take into account reaching first 1. So total number of unique sequences of bits of this greedy algorithm is: number of ways to get first 1 multiplied by number of unique paths to ones or trailing zeroes.At last, back to original problem. Note: I don't want to explain my solution in details. It's not important. My idea to fix this was using greedy algorithm we had before modification was made. Instead of picking first zero I just need to make all possible ways of jumps by number of zeroes. It's up to$|t|$number of arrows from each cell. Each arrow is where I would land if I need exactly certain number of zeroes. To deal with it I postpone jumps into next groups of zeroes. Actually it was postpone addition of values of dp instead of jumps itself. And to avoid big amount of postponed addition I used stack and it did work. By the way it was forward push dp, and it was only over 1 bits. 88154272Now, instead of my idea, we will do a bit different thing. Now we can see how each sequence of bits without skip of groups of zeroes is possible to make. All we need is somehow if we don't have enough zeroes, somehow we want be able to go to the next group of zeroes. And here comes the trick! If we stand in third zero in group, like this: ...100001... ^- we stand here then there is only one way to reach third zero in group: to start from first one and make step by zero arrow twice. Why? Well, because all arrows by 1 is pointing into 1. It means, that by index of zero in group we know exact number of trailing zeroes we have! And, it means, if we stand in last zero in group and want one more zero, then we need group with at least one more zero. And after addition of this additional zero we also know how many zeroes we will have, so we know to which zero we need to make arrow by zero. This zero where we will make arrow — will have number of zeroes strictly determined by position in its group. So everything match. Index of zero is at the same moment: part of state and how many trailing zeroes we already have. Basically each position define last used bit for merge and number of trailing zeroes at once. And that is core of the solution. All what is left is to take care of zeroes in front of string and in end of string.Now, regarding the code in editorial. It is just backward pull dp (in my terminology) over graph I just described.$dist$is number of zeroes before (index of zero in group, 0 if it's bit 1).$nxt$is position of next zero at certain distance (with certain index of zero). » 6 months ago, # | ← Rev. 4 → +101 I might be too late to the party, but I am presenting my own solution for 1E anyway.Let's solve the problem where the first and last character are both 1, because the beginning and ending zeros can be counted and multiplied into the final answer trivially.Between any two consecutive 1's, let's count the number of 0's, and store it in an array a. So we can "compress" our string into$\texttt{1 } a_1 \texttt{ 1 } a_2 \texttt{ 1 ... 1 } a_k \texttt{ 1}$, where$a_i$is the occurrence of 0's between the$i^{th}$1 character and the$(i + 1)^{th}$1 character. The problem can be then transformed into: for each element$a_i$, replace it with a value from$0$to$a_i$, or remove the element from the array completely. Count the number of possible arrays that can be made this way.I denote$dp_i$to be the number of different arrays that end in the value$i$, and denote$tot$to be the total number of different arrays. Clearly$tot$would be our answer, and$tot = (\sum{dp_i}) + 1$. Let's maintain this$dp$and$tot$.Iterate over$a$. When we loop over$a_i$, clearly only$dp_j$where$0 \leq j \leq a_i$change, so let's see how they change. We have two cases of what to do with$a_i$: Remove this element completely.$dp_j$does not change. Replace$a_i$with$j$. We can make$tot - dp_j$new arrays this way, since we create$tot$arrays from appending$j$to the previously existing arrays, but$dp_j$of them has appeared before. Therefore,$dp_j = dp_j + tot - dp_j = tot$. In other words, when we loop over$a_i$, we assign$dp_j = tot$for all$0 \leq j \leq a_i$, then update our$tot$. The complexity would still be$O(|s|)$.However, I think this solution is much more interesting because this can solve the problem if you are given the string in a compressed manner. When a block of$0$'s comes, update the$dp$like mentioned. When a block of$1$'s comes, calculate what$dp_0$would be after processing all the$1$'s. We can use a stack to implement this in$O(|compress(s)|)\$.
» 6 months ago, # | 0 Can anyone explain the solution of div2-D/div1-B more elaborately? I can't catch the core of the given solution. Thanks in advance...
» 6 months ago, # | 0 Div 2 D/Div 1 B was beautiful, thanks
» 5 months ago, # | ← Rev. 2 → 0 i'm trying to solve B1 problem with the following approach-start at t=0 and see if we can reach i+1 without drowning. if we cannot then wait at i. and if both are not possible then start at t='k' seconds and repeat the above steps. if both are impossible, then return false. If anyone of the two gives i==n-1 then return true.Why is this not working? I'm new to competitive prog. so don't judge :)
» 5 months ago, # | 0 on the problem string transformation 1, I think the test is really week. My solution is o(n^2+nlogn) but still accepted.
• » » 5 months ago, # ^ | 0 #include #include #include #include using namespace std; long long solve() { vector f[300]; long long n; cin >> n; string a,b; cin >> a; cin >> b; for (int i=0;i b[i]) return -1; vector > dp; for (int i=0;i> t; while (t--) { cout << solve() << endl; } return 0; }
» 5 months ago, # | 0 Is there is any other source to understand how to solve problem c string transformation as i can not understand the editorial?
» 10 days ago, # | 0 What's with all those hacks in div1F? | 2021-01-24 12:41:49 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.348273366689682, "perplexity": 1804.532164541231}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-04/segments/1610703548716.53/warc/CC-MAIN-20210124111006-20210124141006-00678.warc.gz"} |
http://cms.math.ca/10.4153/CMB-2006-054-3 | location: Publications → journals → CMB
Abstract view
# On the Structure of the Full Lift for the Howe Correspondence of $(Sp(n), O(V))$ for Rank-One Reducibilities
Published:2006-12-01
Printed: Dec 2006
• Goran Muić
Features coming soon:
Citations (via CrossRef) Tools: Search Google Scholar:
Format: HTML LaTeX MathJax PDF PostScript
## Abstract
In this paper we determine the structure of the full lift for the Howe correspondence of $(Sp(n),O(V))$ for rank-one reducibilities.
MSC Classifications: 22E35 - Analysis on $p$-adic Lie groups 22E50 - Representations of Lie and linear algebraic groups over local fields [See also 20G05] 11F70 - Representation-theoretic methods; automorphic representations over local and global fields | 2014-04-19 22:31:58 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.17860238254070282, "perplexity": 6405.3430853559385}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-15/segments/1397609537754.12/warc/CC-MAIN-20140416005217-00031-ip-10-147-4-33.ec2.internal.warc.gz"} |
https://tex.stackexchange.com/questions/396473/staircase-table-with-tikz/396487 | # Staircase table with TiKz
I would like to make a table with the following staircase pattern (preferably with TiKz). In addition for each of the cells there should be a mini-box. The minibox has further data such as a pH level while the main box has a solubility (s) or in soluable.
• Could you please show us what you've tried so far (minimal, compilable code example; MWE). – TeXnician Oct 16 '17 at 17:13
These are both fun exercises in using \foreach. The macros \LabGrid and \ChemGrid, defined below, both accept a variable number of rows/columns and \LabGrid has an optional argument for the colour. The MWE below produces:
Here is the code:
\documentclass{article}
\usepackage{tikz}
%usage: \LabGrid[colour]{CAPITAL letter for last row/column}
\newcommand\LabGrid[2][purple]{%
\begin{tikzpicture}[xscale=1]
\foreach \row [count=\r,remember=\r] in {A,...,#2} {
\draw[fill=#1!10!white](0,-\r) rectangle ++(2,1);
\node at (1,0.5-\r){$\row$};
\foreach \c in {1,...,\r} {
\draw[thick](2*\c,-\r) rectangle ++(2,1);
}
}
\draw[fill=#1!10!white](0,-1-\r) rectangle ++(2,1);
\foreach \row [count=\c] in {A,...,#2} {
\draw[fill=#1!10!white](2*\c,-1-\r) rectangle ++(2,1);
\node at (2*\c+1,-0.5-\r) {\row};
}
\end{tikzpicture}%
}
% usage: \ChemGrid{comma separated list of chemicals}
\newcommand\ChemGrid[1]{%
\begin{tikzpicture}[xscale=1.8,
\foreach \row [count=\r, remember=\r] in {#1} {
\draw(1,-2*\r)rectangle++(1,2);
\foreach \chem [count=\c] in {#1} {
\draw[thick](2*\c,-2*\r)--++(2,0)--++(0,2);
\draw(2*\c+1,2-2*\r)--++(0,-1)--++(1,0);
}
}
\foreach \row [count=\c, remember=\c] in {#1} {
\draw(2*\c,-0.5-2*\r)--++(0,0.5);
}
\draw(2*\c+2,-0.5-2*\r)--++(0,0.5);
\end{tikzpicture}
}
\begin{document}
\LabGrid{C}\bigskip
\LabGrid[green]{E}\bigskip
\ChemGrid{NH$_4$O$_2$, BACl$_2$, CuSo$_4$}
\end{document}
Edit
I realised that the nodes did not scale with xscale and yscale. I have fixed this.
As shown, the easiest way to fine-tune the sizes of the boxes is by setting xscale and yscale in the tikzpicture environment. | 2020-02-27 02:48:50 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7498037219047546, "perplexity": 12803.494731761979}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-10/segments/1581875146643.49/warc/CC-MAIN-20200227002351-20200227032351-00543.warc.gz"} |
https://www.physicsforums.com/threads/what-is-term-for-deq-that-only-has-terms-of-a-derivative.946273/ | # I What is term for DEQ that only has terms of a derivative?
#### swampwiz
For a DEQ like this:
y = y( x )
a y'''' + b y''' + c y'' + d y' + f y = g( x )
where a, b, c, d, f are constants.
I would think it would be called a "constant coefficient DEQ", but a DEQ like this would also be called this
a y y'' + b ( y' )2 = g( x )
but I am only interested in the term for a DEQ in which every term has a single factor that is a derivative.
If I were naming it, I would call it a "polydifferential" so that it would correspond with the term "polynomial", which of course is what the polydifferential would transform into after presuming the natural exponential function for y( x ).
Related Differential Equations News on Phys.org
#### fresh_42
Mentor
2018 Award
For a DEQ like this:
y = y( x )
a y'''' + b y''' + c y'' + d y' + f y = g( x )
where a, b, c, d, f are constants.
I would think it would be called a "constant coefficient DEQ", but a DEQ like this would also be called this
a y y'' + b ( y' )2 = g( x )
but I am only interested in the term for a DEQ in which every term has a single factor that is a derivative.
If I were naming it, I would call it a "polydifferential" so that it would correspond with the term "polynomial", which of course is what the polydifferential would transform into after presuming the natural exponential function for y( x ).
It is called a (ordinary) linear differential equation.
One can write it as linear equation
$$\begin{bmatrix} a_0(x) , a_1(x) , \ldots , a_n(x) \end{bmatrix} \cdot \begin{bmatrix} y^{(n)}(x) \\ y^{(n-1)}(x) \\ \vdots \\ y^0 (x) \end{bmatrix} = b(x)$$
#### Mark44
Mentor
a y'''' + b y''' + c y'' + d y' + f y = g( x )
where a, b, c, d, f are constants.
I would think it would be called a "constant coefficient DEQ", but a DEQ like this would also be called this
This one is a linear, constant coefficient, nonhomogeneous, fourth-order diff. equation.
Linear because all of the terms involve only the unknown function (y(x)) or its derivatives to the power 1, and because none of the dependent variables (i.e., y, y', y'', etc..) are multiplied together
Constant coefficient because all terms are multiplied only by constants.
Nonhomogeneous because of the g(x) term on the right side. (Moving to the left side doesn't change this.)
Fourth-order because the highest derivative is a fourth derivative.
I would think it would be called a "constant coefficient DEQ", but a DEQ like this would also be called this
a y y'' + b ( y' )2 = g( x )
This one is nonlinear because of the term with yy'' and because of the (y')2 term.
swampwiz said:
but I am only interested in the term for a DEQ in which every term has a single factor that is a derivative.
Your first example meets this requirement if a = b = c = d = 1, so that we could write it as $y^{(4)} + y^{(3)} + y'' + y' + y = g(x)$
#### swampwiz
Your first example meets this requirement if a = b = c = d = 1, so that we could write it as $y^{(4)} + y^{(3)} + y'' + y' + y = g(x)$
I meant to say only one non-constant factor.
#### swampwiz
OK, so it seems that this is called "linear, constant coefficient DEQ". But I like the term "polydifferential".
"What is term for DEQ that only has terms of a derivative?"
### Physics Forums Values
We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving | 2019-09-21 02:37:12 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5979080200195312, "perplexity": 1288.9031214781069}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-39/segments/1568514574182.31/warc/CC-MAIN-20190921022342-20190921044342-00155.warc.gz"} |
http://mbi.dirkjanswagerman.nl/How%20to%20Calculate%20Terminal%20Value | For an investment with a defined time horizon, such as a new-product launch, managers project annual cash flows for the life of the project, discounted at the cost of capital. However, capital investments without defined time horizons, such as corporate acquisitions, may generate returns indefinitely.
When cash flows cannot be projected in perpetuity, managers typically estimate a terminal value: the value of all cash flows beyond the period for which predictions are feasible. A terminal value can be quantified in several ways; the most common (used by 46% of respondents to the Association for Financial Professionals survey) is with a perpetuity formula. Here’s how it works:
First, estimate the cash flow that you can reasonably expect—stripping out extraordinary items such as one-off purchases or sales of fixed assets—in the final year for which forecasts are possible. Assume a growth rate for those cash flows in subsequent years. Then simply divide the final-year cash flow by the weighted-average cost of capital minus the assumed growth rate, as follows:
$\Large \text{TERMINAL VALUE} = \frac{\text{NORMALIZED FINAL YEAR CASH FLOW}}{\text{WACC}+\text{GROWTH RATE}}$
It’s critical to use a growth rate that you can expect will increase forever—typically 1% to 4%, roughly the long-term growth rate of the overall economy. A higher rate would be likely to cause the terminal value to overwhelm the valuation for the whole project. For example, over 50 years a $10 million cash flow growing at 10% becomes a$1 billion annual cash flow. In some cases, particularly industries in sustained secular decline, a zero or negative rate may be appropriate.
HBR.ORG: To see how terminal-value growth assumptions affect a project’s overall value, try inputting different rates in the online tool at hbr.org/cost-of-capital.
bag
finance_public
created
Sat, 07 Jul 2012 17:06:00 GMT
creator
dirkjan
modified
Sat, 07 Jul 2012 17:06:00 GMT
modifier
dirkjan
tags | 2018-08-18 02:36:31 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 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": 0.5339342951774597, "perplexity": 1821.6449846939593}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-34/segments/1534221213264.47/warc/CC-MAIN-20180818021014-20180818041014-00116.warc.gz"} |
https://www.shaalaa.com/question-bank-solutions/construction-tangents-circle-in-given-figure-pq-tangent-circle-a-ab-ad-are-bisectors-anglecaq-anglepac-if-anglebaq-30-prove-that-bd-diameter-circle-abc-isosceles-triangle_19004 | Share
# In the Given Figure Pq is a Tangent to the Circle at A, Ab and Ad Are Bisectors of Anglecaq and Anglepac. If Anglebaq = 30^@. Prove That:Bd is a Diameter of the Circle and Abc is an Isosceles Triangle - ICSE Class 10 - Mathematics
ConceptConstruction of Tangents to a Circle
#### Question
In the given figure PQ is a tangent to the circle at A, AB and AD are bisectors of angleCAQ and angle PAC. if angleBAQ = 30^@. prove that:
1) BD is a diameter of the circle
2) ABC is an isosceles triangle
#### Solution
1) angleBAQ = 30^@
SinceAB is the bisector of angleCAQ
=> angleCAB = angleBAQ = 30^@
AD is the bisector of angle CAP and P-A-Q,
angle DAP + angle CAD + angle CAQ = 180^@
=> angleCAD + anglle CAD + 60^@ = 180 ^@
=> angle CAD = 60^@
So angle CAD + angle CAB = 60^@ + 30^@ = 90^@
Since angle in a semi-circle = 90°
⇒ Angle made by diameter to any point on the circle is 90°
So, BD is the diameter of the circle.
2) SinceBD is the diameter of the circle, so it will pass through the centre.
By Alternate segment theorem
angle ABD = angle DAC = 60^@
So, in angle BMA,
angle AMB = 90^@ .........(UseAngleSumProperty)
We know that perpendicular drawn from the centre to a chord of a circle bisects the chord.
=> angle BMA = angle BMC = 90^@
In triangleBMA and triangleBMC
angleBMA = angleBMC = 90^@
BM = BM (common side)
AM = CM(perpendicular drawn from the centre to a chord of a circle bisects the chord.)
⇒ ΔBMA ≅ ΔBMC
⇒ AB = BC (SAS congruence criterion)
⇒ ΔABC is an isosceles triangle.
Is there an error in this question or solution?
#### APPEARS IN
Solution In the Given Figure Pq is a Tangent to the Circle at A, Ab and Ad Are Bisectors of Anglecaq and Anglepac. If `Anglebaq = 30^@. Prove That:Bd is a Diameter of the Circle and Abc is an Isosceles Triangle Concept: Construction of Tangents to a Circle.
S | 2019-12-11 15:25:36 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.3870486617088318, "perplexity": 2386.1162242482956}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-51/segments/1575540531917.10/warc/CC-MAIN-20191211131640-20191211155640-00052.warc.gz"} |
http://www.cfd-online.com/W/index.php?title=Linear_eddy_viscosity_models&oldid=9885 | # Linear eddy viscosity models
These are turbulence models in which the Reynolds stresses as obtained from a Reynolds averaging of the Navier-Stokes equations are modelled by a linear constitutive relationship with the mean flow straining field, such as:
$- \rho \left\langle u_{i} u_{j} \right\rangle = \mu_{t} \left[ S_{ij} - \frac{1}{3} S_{kk} \delta_{ij} \right]$
where $\mu_{t}$ is the coefficient termed turbulence "viscosity" (also called the eddy viscosity), and $S_{ij}$ is the mean strain rate defined by:
$S_{ij}= \frac{1}{2} \left[ \frac{\partial U_{i}}{\partial x_{j}} + \frac{\partial U_{j}}{\partial x_{i}} \right]$
This linear relationship is also known as the Boussinesq hypothesis. For a deep discussion on this linear constitutive relationship, check section Introduction to turbulence/Reynolds averaged equations. | 2016-08-27 21:18:08 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 6, "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": 0.8993039131164551, "perplexity": 1062.577135076478}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-36/segments/1471982925602.40/warc/CC-MAIN-20160823200845-00105-ip-10-153-172-175.ec2.internal.warc.gz"} |
https://math.stackexchange.com/questions/3267890/freges-math-foundation-russells-paradox-and-current-theories | # Frege's Math Foundation, Russell's Paradox, and Current Theories
Similar questions have already been posted, but they require an understanding of Set theory. I'm not a mathematician and was hoping the site could provide background for laymen in plain English.
In Frege's attempt to lay a foundation for mathematics, a fundamental axiom in his work was that every concept had a corresponding extension. However, Russell's paradox proved this wasn't always the case. The set of all sets that are not members of themselves cannot be a member of itself, and it also can't not be a member itself. No such set exists. From reading other answers, this somehow implies that the set of all sets does not exist. How is that so?
Several ad hoc attempts were made to reconcile this contradiction (set hierarchy) until Zermelo–Fraenkel set theory came along. How does Zermelo–Fraenkel overcome Russell's paradox?
• The existence of the "set of all sets" is also at odd with Cantor's theorem : if the set of all sets $V$ exists, then it must contain also all its subsets, but the set of all the subsets of $V$ has a "greater number" of elements than the set $V$ itself. – Mauro ALLEGRANZA Jun 19 '19 at 18:37
• See also the post Does Cantor's Theorem require Russell's paradox? and the post Russell's paradox and Cantor's theorem. – Mauro ALLEGRANZA Jun 19 '19 at 18:39
• – Mauro ALLEGRANZA Jun 19 '19 at 18:48
• Thank you. A lot of those sources employ math/logic symbols I'm unfamiliar with. I was hoping someone could succinctly break down in plain English how Zermelo–Fraenkel set theory resolves Russell's paradox. – user27343 Jun 19 '19 at 19:46
• For simple answers, see the answers below. – Mauro ALLEGRANZA Jun 20 '19 at 6:08
In Naive set theory (the kind where Russell's paradox can be made in), it is assumed that any given property provides us with the existence of a set of exactly those things that satisfy the property. This is why Russell's set can exist in Naive set theory.
To solve this, in Zermelo-Fraenkel set theory, we work with two notions: that of a class and that of a set. A class is basically what we considered to be sets in Naive set theory, in other words, a collection of things satisfying a property.
A set however, is a more refined concept than a class. Every set is still a class (and thus still can be seen intuitively as a collection of objects satisfying a property), but instead of defining sets through the properties their elements have, we define sets by constructing them from the ground up. We carefully lay down the rules that these constructions have to obey and try to make them not too general, so that it is impossible to create sets like Russell's set.
Zermelo-Fraenkel set theory (to be precise, their Axioms) describes exactly which rules of construction are allowed. Some rules are very simple, such as that you can combine two sets together to form a new set containing all the elements of the both of the previous sets, or that you can create a set containing exactly all the subsets of another set. Other rules are very delicate and counterintuitive, even to many mathematicians.
One particular rule that is reminiscent of Naive set theory, is the rule that if we start with a set and any property, then we can take the collection of all the elements inside that set that satisfy said property. So once we have a set, we can get a smaller set with elements satisfying a property. This gives us much of the power that we had in Naive set theory, but without the trouble of sets that are "too general" in a certain sense.
So this is how Zermelo-Fraenkel set theory solves the problem of Russell's paradox. But where lies the problem in having a set of all sets? In Zermelo-Fraenkel set theory the answer is quite easy to see: we have the rule that we can construct from any set and any property, a new set with elements of the original set that exactly satisfy the property. But if there is a set of all sets, let's call it $$V$$, then we can get Russell's paradoxical set from taking those elements from $$V$$ that are not contained in themselves.
Thus, if the set of all sets were a set in Zermelo-Fraenkel's set theory, then Russell's paradox would also give us a set of sets that do not contain themselves.
This is not the only solution to Russell's paradox. Another logical option, would be to not restrict what sets are by laying out rules for constructing sets, but by laying out bounds for which properties we consider appropriate enough to define sets. In particular, we can consider only properties that avoid a certain type of bad self-reference. This is the basis behind Quine's New Foundations set theory. In this theory it is possible to have a set of all sets, and not have problems with Russell's paradox.
• Thank you. Just to confirm, "This is why Russell's set can exist in Naive set theory." Is that correct? I thought Russell's set can't exist (The set of all sets that are not members of themselves cannot be a member of itself, and it also can't not be a member itself. No such set exists). – user27343 Jun 20 '19 at 22:56
• @user27343 The fact that the Russell set can be proven to exist in naive set theory, and that this leads to a contradiction means that naive set theory is inconsistent. – spaceisdarkgreen Jun 20 '19 at 23:19
• Got it. Thanks. – user27343 Jun 21 '19 at 2:11
ZF does not allow you to say “the set of all objects that...”. Instead it only allows you say “the set of all objects in $$A$$ that...”, where $$A$$ is some set. This avoids the paradox because “the set of all sets in A that do not contain themselves” is not paradoxically forced to contain itself... it just won’t be in $$A$$.
But even allowing this restricted form of comprehension has consequences. One such consequence is that there can be no set $$U$$ of all sets. If there were, we get back the paradox, since “the set of all sets in $$U$$ that do not contain themselves” is exactly the same thing as “the set of all sets that do not contain themselves”.
As others have already noted, New Foundations is an alternative approach to ZF that both avoids Russell’s paradox (and all known paradoxes) and allows a universal set. However it puts a lot more restrictions on what kind of properties can be the “...” in “the set of all objects that ...”. For instance, “is not an element of itself” must not be an allowed property. The nature of these restrictions are somewhat obscure at first pass, which at least partially accounts for the NF approach being much less popular than ZF despite whatever advantages might be perceived in having a universal set.
If the set of all sets exists, then you can take the subset of the sets not containing themselves and once again get the contradiction. Thus also the set of all sets cannot exist (or rather cannot exist as a set).
ZF is a list of axioms for sets. The set of all sets is what one calls a class and classes have axioms which are less restrictive.
• This assumes that you can make a set by taking some of the elements in an existing set, defined by a property such as "does not contain itself". ZF has an axiom to claim that this is possible, but the cost of that is that a set of all sets cannot exist. In some other set theories different trade-offs are made. – hmakholm left over Monica Jun 19 '19 at 19:06 | 2021-02-26 07:50:28 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 7, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8094508647918701, "perplexity": 246.86613502290473}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-10/segments/1614178356232.19/warc/CC-MAIN-20210226060147-20210226090147-00428.warc.gz"} |
https://blog.givewell.org/category/top-charities/page/3/ | # Deworm the World Initiative (led by Evidence Action) update
Summary The Deworm the World Initiative (DtWI), led by Evidence Action, received approximately $2.3 million as a result of GiveWell’s recommendation last year. While there were some deviations, it largely allocated these funds as we expected. DtWI now has limited room for funding; it is currently seeking to raise an additional$1.3 million to support…
# Our ongoing review of GAIN
Note: GAIN has reviewed a draft of this post and added responses below. The Global Alliance for Improved Nutrition (GAIN) focuses on reducing malnutrition by fortifying foods and condiments with essential nutrients and through other interventions. We have been considering GAIN for a 2014 recommendation for its work on Universal Salt Iodization (USI) because of our…
# Update on SCI’s evidence of impact
Note: Consistent with our usual practices, SCI reviewed a draft of this post prior to publication, but the final product is ours. We wrote last year about reevaluating studies we relied on in our evaluation of the Schistosomiasis Control Initiative (SCI), which is one of our top charities. We noted at the time that we were planning to…
# Reminder: Annual rankings refresh
This is just a reminder to donors considering supporting our top charities in the next month. As we do every year, we’re planning to refresh our list of top charities before December 1st. (More on why our website can get out of sync with our internal views around this time of year.)
# Our ongoing review of Living Goods
Living Goods runs a network of Community Health Promoters (CHPs) who sell health and household goods door-to-door in their communities in Uganda and Kenya. CHPs also provide basic health counseling. Living Goods also provides consulting and funding to other organizations to run similar networks in other locations. We have been considering Living Goods for a… | 2018-09-22 11:36:52 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.17918023467063904, "perplexity": 7798.768597126578}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-39/segments/1537267158320.19/warc/CC-MAIN-20180922103644-20180922124044-00072.warc.gz"} |
http://math.stackexchange.com/questions/167056/squares-of-the-form-x2y2xy | # Squares of the form $x^2+y^2+xy$
How can I find all $(a,b,c) \in \mathbb{Z}^3$ such that $a^2+b^2+ab$, $a^2+c^2+ac$ and $b^2+c^2+bc$ are squares ?
Thanks !
-
I have shown here that:
All coprime triples $(a,b,c)$ so that $a^2 + ab + b^2 = c^2$ can be enumerated, without duplication, by taking two positive integers $m \ge n$, where $3$ does not divide $n$, and either $mn$ is odd and $\gcd(m,n) = 1$, or $8$ divides $mn$ and $\gcd(m,n) = 2$, and by setting \begin{align} a&=mn\tag{1a}\\[9pt] b&=\frac{(3m+n)(m-n)}{4}\tag{1b}\\[9pt] c&=\frac{3m^2+n^2}{4}\tag{1c} \end{align}
-
Seriously, $>=$? What happened to $\geq$, or $\geqslant$? :-) – Asaf Karagila Jul 5 '12 at 20:21
@AsafKaragila: So, I missed that in the translation from ASCII :-p – robjohn Jul 5 '12 at 20:42
Quite the husky translation, I'd say! – Asaf Karagila Jul 5 '12 at 20:45
Let $c^2+a^2+ca= (c+na)^2$ where $n$ is an integer $\implies a=\frac{(2n-1)c}{1-n^2}$.
Let $a^2+b^2+ab =(a+mb)^2$ where $m$ is an integer $\implies b=\frac{(2m-1)a}{1-m^2}=\frac{(2n-1)(2m-1)}{(1-n^2)(1-m^2)}c$.
If $c|(1-n^2)(1-m^2)$,
$c=r(1-n^2)(1-m^2)$ (say, where $r$ an integer),
then, $b = r(2n-1)(2m-1)$
and $a = r(2n-1)(1-m^2)$
The above will be true if only 1st two conditions were supplied.
For all the three conditions, let $c^2+a^2+ca= r^2$ where r is an integer =>$r^2+ca$ must be perfect square and vice versa. ca=$r^2-R^2$ for some integer R. So, the given problem is same as finding a,b,c such that the product of any two is the difference of two squares.
Now $r^2+ca$ will be perfect square if r=$\frac{c-a}{2}$ where c-a is odd i.e., c,a are of opposite parity.
Then, $c^2+a^2+ca=(\frac{c-a}{2})^2$=>c+a=0.
Similarly, b+c=a+b=0. This provides only trivial solution (0,0,0) $\in \mathbb{Z}^3$.
As ca=$r^2-R^2$, bc = $s^2-S^2$ and ab = $t^2-T^2$ for some integer s,S,t,T.
So, $a^2=\frac{ca.ab}{bc} =\frac{(r^2-R^2).(s^2-S^2)}{t^2-T^2}$
Now if r=$A^2+B^2$ and R= $A^2-B^2$
and if s=$C^2+D^2$ and S= $C^2-D^2$
and if t=$E^2+F^2$ and T= $E^2-F^2$ for some integers A,B,C,D,E,F.
a=±$\frac{2.A.B.2.C.D}{2E.F}$=±$\frac{2A.B.C.D}{E.F}$
b and c with be of the form ±$\frac{2.C.D.E.F}{A.B}$ and ±$\frac{2.E.F.A.B}{C.D}$
So, we need to find integers p,q,r such that p|2qr, q|2rp and r|2pq (think p=A.B, q=C.D and r=E.F).
-
I am a bit baffled by this. Take the case $5^2+3^2+5\times 3 = 7^2$ (in other words $c=5$, $a=3$), then the first line above would give $(c+na)^2 = 7^2$, or $5+3a = 7$, and $n$ is then not an integer. Am I being a bit dim (as often the case!) – Old John Jul 5 '12 at 18:35
@OldJohn In this particular case we can take $n=-4$. I do share your skepticism if $a$ has more than one odd prime factor (this seems to be assuming that $c^2$ has only two square roots mod $a$). Also it doesn't appear that $b^2+c^2+bc$ has been addressed. – Erick Wong Jul 5 '12 at 18:51
OK, but is the solution above claiming that for any choice of $r$, $m$, $n$ in $\mathbb{Z}$, the last the lines will give a solution? (My calculations show otherwise - probably agreeing with your doubts about $b^2+c^2+bc$.) – Old John Jul 5 '12 at 18:55 | 2015-07-08 07:17:34 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9497692584991455, "perplexity": 755.4333941486509}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-27/segments/1435376093097.69/warc/CC-MAIN-20150627033453-00240-ip-10-179-60-89.ec2.internal.warc.gz"} |
http://www.aovstg.org/f3g2r/69dd53-photon-energy-formula | During photosynthesis, specific chlorophyll molecules absorb red-light photons at a wavelength of 700 nm in the photosystem I, corresponding to an energy of each photon of ≈ 2 eV ≈ 3 x 10−19 J ≈ 75 kBT, where kBT denotes the thermal energy. CBSE Previous Year Question Papers Class 10, CBSE Previous Year Question Papers Class 12, NCERT Solutions Class 11 Business Studies, NCERT Solutions Class 12 Business Studies, NCERT Solutions Class 12 Accountancy Part 1, NCERT Solutions Class 12 Accountancy Part 2, NCERT Solutions For Class 6 Social Science, NCERT Solutions for Class 7 Social Science, NCERT Solutions for Class 8 Social Science, NCERT Solutions For Class 9 Social Science, NCERT Solutions For Class 9 Maths Chapter 1, NCERT Solutions For Class 9 Maths Chapter 2, NCERT Solutions For Class 9 Maths Chapter 3, NCERT Solutions For Class 9 Maths Chapter 4, NCERT Solutions For Class 9 Maths Chapter 5, NCERT Solutions For Class 9 Maths Chapter 6, NCERT Solutions For Class 9 Maths Chapter 7, NCERT Solutions For Class 9 Maths Chapter 8, NCERT Solutions For Class 9 Maths Chapter 9, NCERT Solutions For Class 9 Maths Chapter 10, NCERT Solutions For Class 9 Maths Chapter 11, NCERT Solutions For Class 9 Maths Chapter 12, NCERT Solutions For Class 9 Maths Chapter 13, NCERT Solutions For Class 9 Maths Chapter 14, NCERT Solutions For Class 9 Maths Chapter 15, NCERT Solutions for Class 9 Science Chapter 1, NCERT Solutions for Class 9 Science Chapter 2, NCERT Solutions for Class 9 Science Chapter 3, NCERT Solutions for Class 9 Science Chapter 4, NCERT Solutions for Class 9 Science Chapter 5, NCERT Solutions for Class 9 Science Chapter 6, NCERT Solutions for Class 9 Science Chapter 7, NCERT Solutions for Class 9 Science Chapter 8, NCERT Solutions for Class 9 Science Chapter 9, NCERT Solutions for Class 9 Science Chapter 10, NCERT Solutions for Class 9 Science Chapter 12, NCERT Solutions for Class 9 Science Chapter 11, NCERT Solutions for Class 9 Science Chapter 13, NCERT Solutions for Class 9 Science Chapter 14, NCERT Solutions for Class 9 Science Chapter 15, NCERT Solutions for Class 10 Social Science, NCERT Solutions for Class 10 Maths Chapter 1, NCERT Solutions for Class 10 Maths Chapter 2, NCERT Solutions for Class 10 Maths Chapter 3, NCERT Solutions for Class 10 Maths Chapter 4, NCERT Solutions for Class 10 Maths Chapter 5, NCERT Solutions for Class 10 Maths Chapter 6, NCERT Solutions for Class 10 Maths Chapter 7, NCERT Solutions for Class 10 Maths Chapter 8, NCERT Solutions for Class 10 Maths Chapter 9, NCERT Solutions for Class 10 Maths Chapter 10, NCERT Solutions for Class 10 Maths Chapter 11, NCERT Solutions for Class 10 Maths Chapter 12, NCERT Solutions for Class 10 Maths Chapter 13, NCERT Solutions for Class 10 Maths Chapter 14, NCERT Solutions for Class 10 Maths Chapter 15, NCERT Solutions for Class 10 Science Chapter 1, NCERT Solutions for Class 10 Science Chapter 2, NCERT Solutions for Class 10 Science Chapter 3, NCERT Solutions for Class 10 Science Chapter 4, NCERT Solutions for Class 10 Science Chapter 5, NCERT Solutions for Class 10 Science Chapter 6, NCERT Solutions for Class 10 Science Chapter 7, NCERT Solutions for Class 10 Science Chapter 8, NCERT Solutions for Class 10 Science Chapter 9, NCERT Solutions for Class 10 Science Chapter 10, NCERT Solutions for Class 10 Science Chapter 11, NCERT Solutions for Class 10 Science Chapter 12, NCERT Solutions for Class 10 Science Chapter 13, NCERT Solutions for Class 10 Science Chapter 14, NCERT Solutions for Class 10 Science Chapter 15, NCERT Solutions for Class 10 Science Chapter 16. energy of a mole of photons = (energy of a single photon) x (Avogadro's number) energy of a mole of photons = (3.9756 x 10 -19 J) (6.022 x 10 23 mol -1) [hint: multiply the decimal numbers and then subtract the denominator exponent from the numerator exponent to get the power of 10) energy = 2.394 x 10 5 J/mol. h:Plank's constant. However, Debye's approach failed to give the correct formula for the energy fluctuations of black-body radiation, which were derived by Einstein in 1909. E = h * c / λ = h * f, where. Photon energy can be expressed using any unit of energy. E = 0.030 x 10 −17 J. He decomposed the electromagnetic field in a cavity into its Fourier modes, and assumed that the energy in any mode was an integer multiple of $${\displaystyle h\nu }$$, where $${\displaystyle \nu }$$ is the frequency of the electromagnetic mode. Therefore, the photon energy at 1 Hz frequency is 6.62606957 × 10−34 joules or 4.135667516 × 10−15 eV. How to calculate the energy of a photon. This equation is known as the Planck-Einstein relation. Now we can calculate the energy of a photon by either version of Planck's equation: E = hf or E = hc / λ. This corresponds to frequencies of 2.42 × 1025 to 2.42 × 1028 Hz. Where, E photon = Energy of Photon, v = Light Frequency, h = Plancks constant = 6.63 × 10 -34 m 2 kg / s. {\displaystyle {\frac {c}{\lambda }}=f} f , where f is frequency, the photon energy equation can be simplified to. λ Since Planck's law of black-body radiation follows immediately as a geometric sum. = To find the photon energy in electronvolts, using the wavelength in micrometres, the equation is approximately. Example 2: If the energy of a photon is 350×10−10 J, determine the wavelength of that photon. c An FM radio station transmitting at 100 MHz emits photons with an energy of about 4.1357 × 10−7 eV. The amount of energy is directly proportional to the photon's electromagnetic frequency and thus, equivalently, is inversely proportional to the wavelength. Photon energy formula is given by, E = hc / λ. E = 6.626×10 −34 ×3×10 8 / 650×10 −9. A minimum of 48 photons is needed for the synthesis of a single glucose molecule from CO2 and water (chemical potential difference 5 x 10−18 J) with a maximal energy conversion efficiency of 35%, https://en.wikipedia.org/w/index.php?title=Photon_energy&oldid=986282546, Creative Commons Attribution-ShareAlike License, This page was last edited on 30 October 2020, at 22:00. In 1910, Peter Debye derived Planck's law of black-body radiation from a relatively simple assumption. Photon energy formula is given by, E = hc / λ. λ = hc / E Where E is photon energy, h is the Planck constant, c is the speed of light in vacuum and λ is the photon's wavelength. Solution: Given parameters are, E = 350 ×10 −10 J. c = 3 ×10 8 m/s. Formula: E photon = hv. Photon energy is the energy carried by a single photon. E e V = 1. A photon is characterized either by wavelength (λ) or an equivalent energy E. The energy of a photon is inversely proportional to the wavelength of a photon. Substituting h with its value in J⋅s and f with its value in hertz gives the photon energy in joules. Determine the photon energy if the wavelength is 650nm. E = 19.878 x 10 28 / 650×10 −9. As one joule equals 6.24 × 1018 eV, the larger units may be more useful in denoting the energy of photons with higher frequency and higher energy, such as gamma rays, as opposed to lower energy photons, such as those in the radio frequency region of the electromagnetic spectrum. c: speed of light Where: E: photon's energy. Therefore, the photon energy at 1 μm wavelength, the wavelength of near infrared radiation, is approximately 1.2398 eV. 24 λ μ m. Often we use the units of eV, or electron volts, as the units for photon energy, instead of joules. λ: photon's wavelength. h = 6.626 ×10 −34 Js. If the energy of a photon is 350×10−10J, determine the wavelength of that photon. By expressing the equation for photon energy in terms of eV and µm we arrive at a commonly used expression which relates the energy and wavelength of a photon, as shown in the following equation: Photon Energy : Electron-Volt. The higher the photon's frequency, the higher its energy. As h and c are both constants, photon energy E changes in inverse relation to wavelength λ. You can use h = 4.1357 × 10 -15 eV s, which results … Equivalently, the longer the photon's wavelength, the lower its energy. The Planck's equation is. Among the units commonly used to denote photon energy are the electronvolt (eV) and the joule (as well as its multiples, such as the microjoule). The equation is: E = hc / λ. This minuscule amount of energy is approximately 8 × 10−13 times the electron's mass (via mass-energy equivalence). The equation for Planck looks like this: E = h * c / λ = h * f E = photon’s energy H = Planck constant C = light’s speed λ = photon’s wavelength F = photon’s frequency Light is a collection of particles, and this formula gives us the single, indivisible quanta of light. Your email address will not be published. is used where h is Planck's constant and the Greek letter ν (nu) is the photon's frequency.[2]. E is the energy of a photon; h is the Planck constant, c is the speed of light, λ is the wavelength of a photon, f is the frequency of a photon. Your email address will not be published. Neuer Inhalt wird bei Auswahl oberhalb des aktuellen Fokusbereichs hinzugefügt Very-high-energy gamma rays have photon energies of 100 GeV to 100 TeV (1011 to 1014 electronvolts) or 16 nanojoules to 16 microjoules. hc = (1.24 × 10 -6 eV-m) × (10 6 µm/ m) = 1.24 eV-µm. Required fields are marked *, A photon is characterized either by wavelength (. Photon energy = Plank's constant * speed of light / photon's wavelength.
.
Tupperware Puderzuckerstreuer Ersatzteile, Zuhause Definition Poster, Helene Fischer: Neues Lied 2020, Rolls-royce Silver Shadow Neupreis, Promi Namen Witze, Nuno Bettencourt 2020, Griechische Nachnamen Mit S, Wie Sehen Blitzer An Ampeln Aus, Seiler Und Speer I Hob A Pech Chords, | 2022-11-30 17:33:22 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7533661127090454, "perplexity": 2000.5544409892}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446710765.76/warc/CC-MAIN-20221130160457-20221130190457-00878.warc.gz"} |
https://homework.cpm.org/category/CON_FOUND/textbook/gc/chapter/1/lesson/1.2.1/problem/1-57 | Home > GC > Chapter 1 > Lesson 1.2.1 > Problem1-57
1-57.
For each equation below, solve for $x$. Show all work. The answers are provided so that you can check them. If you are having trouble with any solutions and cannot find your errors, you may need to see your teacher for extra help (you can ask your team as well).
1. $5x-2x+x=15$
$\begin{array}{r} 5x-2x+x=15\\ 5x-x=15\\ 4x=15 \end{array}$
$x=3.75$
1. $3x-2-x=7-x$
$x=3$
1. $3\left(x-1\right)=2x-3+3x$
$$$3(x-1)=2x-3+3x\\ \; \; \; 3x-3=2x−3+3x\\ \; \; \; 3x-3=-3+5x\\ \qquad -3=2x-3\\ \qquad \quad \; 0=2x$$$
$x=0$
1. $3(2-x)=5(2x-7)+2$
$x=3$
1. $\frac { 26 } { 57 } = \frac { 849 } { 5 x }$
$\frac{26}{57}=\frac{849}{5x}$
$130x=48393$
$x\approx372.25$
1. $\frac{4x+1}{3}=\frac{x-5}{2}$
$x=-3.4$ | 2021-01-23 02:28:39 | {"extraction_info": {"found_math": true, "script_math_tex": 17, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 2, "x-ck12": 0, "texerror": 0, "math_score": 0.7147796750068665, "perplexity": 1241.2474706670732}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-04/segments/1610703531702.36/warc/CC-MAIN-20210123001629-20210123031629-00174.warc.gz"} |
https://demo7.dspace.org/items/efc785df-f3b4-44ee-842a-470181c0bf73 | ## Meson-Photon Transition Form Factors in the Charmonium Energy Range
##### Authors
Rosner, Jonathan L.
##### Description
The study of electron-positron collisions at center-of-mass energies corresponding to charmonium production has reached new levels of sensitivity thanks to experiments by the BES and CLEO Collaborations. Final states $\gamma P$, where $P$ is a pseudoscalar meson such as $\pi^0$, $\eta$, and $\eta'$ can arise either from charmonium decays or in the continuum through a virtual photon: $e^+ e^- \to \gamma^* \to \gamma P$. Estimates of this latter process are given at center-of-mass (c.m.) energies corresponding to the $J/\psi(1S)$, $\psi(2S)$, and $\psi(3770)$ resonances.
Comment: 7 pages, to be submitted to Phys. Rev. D. Version referring to new CLEO data
##### Keywords
High Energy Physics - Phenomenology | 2022-12-06 08:30:50 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9429165124893188, "perplexity": 3081.9495610201484}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446711074.68/warc/CC-MAIN-20221206060908-20221206090908-00688.warc.gz"} |
https://www.gamedev.net/forums/topic/652431-directional-light-shadow-mapping-and-matrices/ | • ### Announcements
GameDev.net and CRC Press have teamed up to bring a free ebook of content curated from top titles published by CRC Press. The freebook, Practices of Game Design & Indie Game Marketing, includes chapters from The Art of Game Design: A Book of Lenses, A Practical Guide to Indie Game Marketing, and An Architectural Approach to Level Design. The GameDev.net FreeBook is relevant to game designers, developers, and those interested in learning more about the challenges in game development. We know game development can be a tough discipline and business, so we picked several chapters from CRC Press titles that we thought would be of interest to you, the GameDev.net audience, in your journey to design, develop, and market your next game. The free ebook is available through CRC Press by clicking here. The Curated Books The Art of Game Design: A Book of Lenses, Second Edition, by Jesse Schell Presents 100+ sets of questions, or different lenses, for viewing a game’s design, encompassing diverse fields such as psychology, architecture, music, film, software engineering, theme park design, mathematics, anthropology, and more. Written by one of the world's top game designers, this book describes the deepest and most fundamental principles of game design, demonstrating how tactics used in board, card, and athletic games also work in video games. It provides practical instruction on creating world-class games that will be played again and again. View it here. A Practical Guide to Indie Game Marketing, by Joel Dreskin Marketing is an essential but too frequently overlooked or minimized component of the release plan for indie games. A Practical Guide to Indie Game Marketing provides you with the tools needed to build visibility and sell your indie games. With special focus on those developers with small budgets and limited staff and resources, this book is packed with tangible recommendations and techniques that you can put to use immediately. As a seasoned professional of the indie game arena, author Joel Dreskin gives you insight into practical, real-world experiences of marketing numerous successful games and also provides stories of the failures. View it here. An Architectural Approach to Level Design This is one of the first books to integrate architectural and spatial design theory with the field of level design. The book presents architectural techniques and theories for level designers to use in their own work. It connects architecture and level design in different ways that address the practical elements of how designers construct space and the experiential elements of how and why humans interact with this space. Throughout the text, readers learn skills for spatial layout, evoking emotion through gamespaces, and creating better levels through architectural theory. View it here. Learn more and download the ebook by clicking here. Did you know? GameDev.net and CRC Press also recently teamed up to bring GDNet+ Members up to a 20% discount on all CRC Press books. Learn more about this and other benefits here.
Followers 0
# OpenGL Directional light, shadow mapping and matrices..
## 14 posts in this topic
I'm using OpenGL and I've managed to get spotlight shadowmapping to work. I have the position and direction of the spotlight. The view matrix is constructed using glm::lookAt(), and the projection matrix is a perspective matrix.
For directional lights however, all I have is the light direction.
1) What is the view matrix and how to construct it, given only light direction?
2) What is the projection matrix and how is it constructed?
3) Do you still do a shadow map pass with directional lights before the actual lighting pass?
Thanks
0
##### Share on other sites
With a directional light, some things are simpler than a spotlight, like setting up the transforms. A directional light uses a very simple orthographic projection, as opposed to a perspective projection.
However, most things are more complex because a directional light in theory affects everything out to infinity, but in practice your shadow texture is only a finite size and you don't want to stretch it over too large of an area or the resulting shadows will look very pixely. Shadowmapping with a directional light is all about optimizing the shadow texture usage to make the projection of shadow map texels onscreen as small as possible.
In addition to the light direction, you'll need to define a frustum for your directional shadow camera. An orthographic frustum is just a box oriented so that the z-axis is aligned with the projection direction. The box should include everything that casts or receives a shadow, but at the same time you want the box to be as small as possible because your shadow texture will be stretched over the x and y extents of the box. Minimizing the z-extents of the box isn't quite as critical, but still important, as this range is mapped into your shadow texture bit depth.
You'll want to calculate the position and size of this box based on your main camera's frustum. If your main camera can see very far, like say 1000 meters, you won't be able to cover the whole area seen by your main camera with just one shadow texture. A single shadow texture will only adequately cover a few tens of meters from the camera (depending on shadow texture resolution). If you want to cover more area then that, you may want to look into cascaded shadow mapping and similar techniques.
Once your shadow camera frustum is defined, the matrices are quite simple to set up.
1. The view matrix is just a transform with only position and rotation, centered in the center of the box and rotated to align with the box.
2. The projection matrix is just a pure scale matrix that maps the extents of the box into clip space (so each axis is mapped into a -1 to 1 range).
2
##### Share on other sites
The view matrix - why is it in the center of the box, and not infront of it? Otherwise wont half of the geometry be outside the view? And aligned, as in looking down the -Z axis?
Projection matrix - how would you calculate such a scale matrix? Does glm::ortho, which creates an orthographic matrix, do the same thing?
Edited by KaiserJohan
0
##### Share on other sites
The view matrix, I chose to put the origin of the view transform in the center of the box because it simplifies constructing the projection matrix. You can put the view transform at one end of the box or the other if you prefer, but then you'll need to introduce a translation component into the projection matrix to compensate. It's just easier to put it at the center.
And yes, the -Z axis.
The projection matrix would just be:
1/he.x 0 0 0
0 1/he.y 0 0
0 0 1/he.z 0
0 0 0 1
(where "he" is the half-extents of your box).
Also you could use glOrtho( -he.x, he.x, -he.y, he.y, -he.z, he.z ) to define the projection matrix if you like. (I may have flipped the signs on z there)
It is probably safer to use glOrtho, as I'm not completely sure about the range of clipping coordinates in z.
Edited by lunkhound
1
##### Share on other sites
I've got shadow mapping working, using the following matrices:
Mat4 viewMatrix = LookAt(cameraPosition + (directionalLight.mLightDirection * Z_FAR/2.0f), -directionalLight.mLightDirection, Vec3(0.0f, 1.0f, 0.0f));
Mat4 lightVP = CreateOrthographicMatrix(-Z_FAR, Z_FAR, -Z_FAR, Z_FAR, Z_NEAR, Z_FAR) * viewMatrix;
The problem is when moving around the camera, so does the shadows move around, which is unrealistic. What is the proper way to build the view matrix?
Edited by KaiserJohan
0
##### Share on other sites
What you have there doesn't look right at all. It isn't that easy.
I would do it something like this:
1. Find a vector orthogonal to the light direction.
2. Use the cross-product to find another vector orthogonal to the other two. Now you have 3 orthogonal unit vectors.
3. Form a rotation matrix from the 3 vectors. The light direction should be the z-axis, and the other two are x and y. Make sure the handedness is correct. The x-axis cross y-axis should equal the z-axis, not the negative z-axis. This rotation matrix transforms directions from world space into the light space.
4. Generate a list of the 8 corner vertices of the main camera's frustum. You'll need the main camera's position, z-near, z-far, horizontal field of view, and vertical field of view.
5. Loop over the 8 corner vertices, and apply the rotation matrix to each in turn and record the minimum and maximum on the x, y, and z axes. This gives the extents of the box I mentioned in an earlier post.
6. Find the coordinates of the center of the box in light space i.e. ((xmin + xmax)*0.5, (ymin + ymax)*0.5, (zmin + zmax)*0.5)
Now it is straightforward to assemble the shadowmap view matrix. The 3x3 rotation matrix goes into the rotation part of the 4x4 matrix. For the translation part of the matrix, you want to the box-center coordinates NEGATED. And the last row of the matrix is just (0 0 0 1).
1
##### Share on other sites
Thanks for the reply. I've tried it, but I do not seem to get any shadows. Here is an image:
I tried it with the following code, as per your suggestions:
for (const RenderableLighting::DirectionalLight& directionalLight : lighting.mDirectionalLights)
{
Vec3 lightDir = directionalLight.mLightDirection; // = Vec3(0.0f, 1.0f, 1.0f)
Vec3 perpVec1(1.0f, 0.0f, 0.0f);
Vec3 perpVec2(0.0f, 1.0f, -1.0f);
Mat3 rotationMatrix; // using glm, so matrices are column-major
rotationMatrix[0] = perpVec1;
rotationMatrix[1] = perpVec2;
rotationMatrix[2] = lightDir;
assert(glm::cross(perpVec1, perpVec2) == lightDir);
Vec3 corners[8] = {Vec3(-25.0f, -25.0f, 25.0f), Vec3(-25.0f, -25.0f, -25.0f), Vec3(-25.0f, 25.0f, 25.0f), Vec3(-25.0f, 25.0f, -25.0f),
Vec3(25.0f, -25.0f, 25.0f), Vec3(25.0f, -25.0f, -25.0f), Vec3(25.0f, 25.0f, 25.0f), Vec3(25.0f, 25.0f, -25.0f) };
for (uint32_t i = 0; i < 8; i++)
corners[i] = corners[i] * rotationMatrix;
float minX = corners[0].x, minY = corners[0].y, minZ = corners[0].z, maxX = corners[0].x, maxY = corners[0].y, maxZ = corners[0].z;
for (uint32_t i = 0; i < 8; i++)
{
if (corners[i].x < minX)
minX = corners[i].x;
if (corners[i].x > maxX)
maxX = corners[i].x;
if (corners[i].y < minY)
minY = corners[i].y;
if (corners[i].y > maxY)
maxY = corners[i].y;
if (corners[i].z < minZ)
minZ = corners[i].z;
if (corners[i].z > maxZ)
maxZ = corners[i].z;
}
Mat4 viewMatrix(rotationMatrix);
viewMatrix[3][0] = -(minX + maxX) * 0.5f;
viewMatrix[3][1] = -(minY + maxY) * 0.5f;
viewMatrix[3][2] = -(minZ + maxZ) * 0.5f;
viewMatrix[0][3] = 0.0f;
viewMatrix[1][3] = 0.0f;
viewMatrix[2][3] = 0.0f;
viewMatrix[3][3] = 1.0f;
Mat4 lightVP = glm::ortho(minX, minX, minY, maxY, minZ, maxZ) * viewMatrix;
GLCALL(glDrawBuffer(GL_NONE));
GLCALL(glBindFramebuffer(GL_DRAW_FRAMEBUFFER, mGBuffer.mFramebuffer));
GLCALL(glDrawBuffer(GBuffer::GBUFFER_COLOR_ATTACHMENT_FINAL));
DirLightLightingPass(directionalLight, gNDCBiasMatrix * lightVP, lighting.mGamma, lighting.mScreenSize);
}
Do you spot anything that stands out? Besides the view/projection calculations, the rest should be fine
Edited by KaiserJohan
0
##### Share on other sites
Well, the rotation matrix has some problems for starters. The rows of the rotation matrix all need to be unit magnitude (some of yours aren't), and they need to be perpendicular to each other (also not true for your matrix). So what you've got there is a matrix that scales, skews, and rotates.
Another problem that jumps out is that you aren't calculating the corners of the main camera's frustum. Instead you've got the corners of a 50 x 50 x 50 box centered on the origin. Your main camera has a perspective projection matrix, so the shape of it isn't a box, it is more of a truncated pyramid.
If you take your main camera's view-projection matrix, invert it (note, NOT a simple transpose), and then run the corners of a 2 x 2 x 2 box centered on the origin through it, that should give the corners of your camera's frustum.
Also your orthographic matrix is also wrong. You should calculate the half-extents of the box from your minX, maxX, etc, and express the ortho in terms of the half-extents.
1
##### Share on other sites
I'm using the cross product to find out the perpendicular vectors; I forgot to normalize lightDirection - the following should give me the proper rotation matrix no?
Vec3 lightDir = glm::normalize(directionalLight.mLightDirection); // = Vec3(0.0f, 1.0f, 1.0f)
Vec3 perpVec1 = glm::cross(lightDir, Vec3(0.0f, 0.0f, 1.0f));
Vec3 perpVec2 = glm::normalize(glm::cross(lightDir, perpVec1));
Mat3 rotationMatrix; // using glm, so matrices are column-major
rotationMatrix[0] = perpVec1;
rotationMatrix[1] = perpVec2;
rotationMatrix[2] = lightDir;
assert(glm::dot(lightDir, perpVec1) == 0);
assert(glm::dot(lightDir, perpVec2) == 0);
As for the corners of the camera frustrum, I'm thinking one step at a time, the simplest would be to just use static boundaries like the box I posted, it would work just as well for the moment, right?
When you say half-extents, you mean the min/max divided by two, like this?
Mat4 lightVP = glm::ortho(minX/2.0f, minX/2.0f, minY/2.0f, maxY/2.0f, minZ/2.0f, maxZ/2.0f) * viewMatrix;
0
##### Share on other sites
You also need to normalize the first cross product there (perpVec1).
If the 50x50x50 box encloses your scene, then yes that should work fine as a temporary hack.
The extents would be:
Vec3 extents( maxX - minX, maxY - minY, maxZ - minZ );
The half extents would just be 0.5 * extents:
Vec3 he = extents * 0.5;
Then you want something like ortho( -he.x, he.x, -he.y, he.y, -he.z, he.z );
Otherwise, you'll have unwanted translation in the projection matrix. Translation that already exists in the view matrix so you'd be applying it twice.
1
##### Share on other sites
Thanks, I've got shadows working with the static bounding box, I'll be trying what you wrote in an earlier post about inverting the camera view/projection matrix. One thing though, what do you mean by
run the corners of a 2 x 2 x 2 box centered on the origin through it
?
0
##### Share on other sites
The corners of a 2x2x2 box at the origin are 8 vertices where each component is either a 1 or a -1. Vec3(1,1,1), Vec3(-1,1,1), .. and so on. This corresponds to the view volume in clip-space. The view-projection matrix maps from world space to clip space (it is the concatenation of the view matrix and the projection matrix). So if you invert the view-projection matrix, it will map from clip space to world space. If you apply the inverse-view-projection matrix to each of the 8 corners, you should end up with the 8 corners of the view volume in world-space. The inverse-view-projection matrix will be a 4x4 matrix, so each of your vertices will need to be "promoted" to a Vec4 with a w-component of 1, and the result will be in homogeneous coordinates so you will need to divide by w after the transform.
1
##### Share on other sites
If I understod correctly;
Mat4 rotationMatrix(Vec4(perpVec1, 0.0f), Vec4(perpVec2, 0.0f), Vec4(lightDir, 0.0f), Vec4(0.0f, 0.0f, 0.0f, 1.0f)); // again, column major
Vec4 corners[8] = { Vec4(-1.0f, -1.0f, 1.0f, 1.0f), Vec4(-1.0f, -1.0f, -1.0f, 1.0f), Vec4(-1.0f, 1.0f, 1.0f, 1.0f), Vec4(-1.0f, 1.0f, -1.0f, 1.0f),
Vec4(1.0f, -1.0f, 1.0f, 1.0f), Vec4(1.0f, -1.0f, -1.0f, 1.0f), Vec4(1.0f, 1.0f, 1.0f, 1.0f), Vec4(1.0f, 1.0f, -1.0f, 1.0f) };
const Mat4 inverseVPMatrix = glm::inverse(viewProjectionMatrix); // the cameras view-projection matrix
for (uint32_t i = 0; i < 8; i++)
{
corners[i] = corners[i] * inverseVPMatrix;
corners[i] /= corners[i].w;
}
for (uint32_t i = 0; i < 8; i++)
corners[i] = corners[i] * rotationMatrix;
Looking at the corners after multiplying with the inverse VPMatrix and dividing by w, the values all seem to be in the range of [-2.0, 2.0]. which can't be right?
Edited by KaiserJohan
0
##### Share on other sites
That doesn't sound right. I don't see anything wrong with the code there. There must be something wrong with the viewProjection matrix, make sure view and projection are concatenated in the correct order. For column major matrices, it should be projection * view, not the other way around.
1
##### Share on other sites
Turned out the issue was with wrong order of multiplication of the corners and inverse VP matrix. This is the correct one:
corners[i] = inverseVPMatrix * corners[i];
Perhaps due to glm being column major?
I consider this topic solved; big thanks for all the help! I'm gonna investigate cascaded shadow mapping next,
0
## Create an account
Register a new account
Followers 0
• ### Similar Content
• So it's been a while since I took a break from my whole creating a planet in DX11. Last time around I got stuck on fixing a nice LOD.
A week back or so I got help to find this:
https://github.com/sp4cerat/Planet-LOD
In general this is what I'm trying to recreate in DX11, he that made that planet LOD uses OpenGL but that is a minor issue and something I can solve. But I have a question regarding the code
He gets the position using this row
vec4d pos = b.var.vec4d["position"]; Which is then used further down when he sends the variable "center" into the drawing function:
if (pos.len() < 1) pos.norm(); world::draw(vec3d(pos.x, pos.y, pos.z));
Inside the draw function this happens:
draw_recursive(p3[0], p3[1], p3[2], center); Basically the 3 vertices of the triangle and the center of details that he sent as a parameter earlier: vec3d(pos.x, pos.y, pos.z)
Now onto my real question, he does vec3d edge_center[3] = { (p1 + p2) / 2, (p2 + p3) / 2, (p3 + p1) / 2 }; to get the edge center of each edge, nothing weird there.
But this is used later on with:
vec3d d = center + edge_center[i]; edge_test[i] = d.len() > ratio_size; edge_test is then used to evaluate if there should be a triangle drawn or if it should be split up into 3 new triangles instead. Why is it working for him? shouldn't it be like center - edge_center or something like that? Why adding them togheter? I asume here that the center is the center of details for the LOD. the position of the camera if stood on the ground of the planet and not up int he air like it is now.
Full code can be seen here:
https://github.com/sp4cerat/Planet-LOD/blob/master/src.simple/Main.cpp
If anyone would like to take a look and try to help me understand this code I would love this person. I'm running out of ideas on how to solve this in my own head, most likely twisted it one time to many up in my head
Toastmastern
• I googled around but are unable to find source code or details of implementation.
What keywords should I search for this topic?
Things I would like to know:
A. How to ensure that partially covered pixels are rasterized?
Apparently by expanding each triangle by 1 pixel or so, rasterization problem is almost solved.
But it will result in an unindexable triangle list without tons of overlaps. Will it incur a large performance penalty?
How to ensure proper synchronizations in GLSL?
GLSL seems to only allow int32 atomics on image.
C. Is there some simple ways to estimate coverage on-the-fly?
In case I am to draw 2D shapes onto an exisitng target:
1. A multi-pass whatever-buffer seems overkill.
2. Multisampling could cost a lot memory though all I need is better coverage.
Besides, I have to blit twice, if draw target is not multisampled.
• By mapra99
Hello
I am working on a recent project and I have been learning how to code in C# using OpenGL libraries for some graphics. I have achieved some quite interesting things using TAO Framework writing in Console Applications, creating a GLUT Window. But my problem now is that I need to incorporate the Graphics in a Windows Form so I can relate the objects that I render with some .NET Controls.
To deal with this problem, I have seen in some forums that it's better to use OpenTK instead of TAO Framework, so I can use the glControl that OpenTK libraries offer. However, I haven't found complete articles, tutorials or source codes that help using the glControl or that may insert me into de OpenTK functions. Would somebody please share in this forum some links or files where I can find good documentation about this topic? Or may I use another library different of OpenTK?
Thanks!
• Hello, I have been working on SH Irradiance map rendering, and I have been using a GLSL pixel shader to render SH irradiance to 2D irradiance maps for my static objects. I already have it working with 9 3D textures so far for the first 9 SH functions.
In my GLSL shader, I have to send in 9 SH Coefficient 3D Texures that use RGBA8 as a pixel format. RGB being used for the coefficients for red, green, and blue, and the A for checking if the voxel is in use (for the 3D texture solidification shader to prevent bleeding).
My problem is, I want to knock this number of textures down to something like 4 or 5. Getting even lower would be a godsend. This is because I eventually plan on adding more SH Coefficient 3D Textures for other parts of the game map (such as inside rooms, as opposed to the outside), to circumvent irradiance probe bleeding between rooms separated by walls. I don't want to reach the 32 texture limit too soon. Also, I figure that it would be a LOT faster.
Is there a way I could, say, store 2 sets of SH Coefficients for 2 SH functions inside a texture with RGBA16 pixels? If so, how would I extract them from inside GLSL? Let me know if you have any suggestions ^^.
• By KarimIO
EDIT: I thought this was restricted to Attribute-Created GL contexts, but it isn't, so I rewrote the post.
Hey guys, whenever I call SwapBuffers(hDC), I get a crash, and I get a "Too many posts were made to a semaphore." from Windows as I call SwapBuffers. What could be the cause of this?
Update: No crash occurs if I don't draw, just clear and swap.
static PIXELFORMATDESCRIPTOR pfd = // pfd Tells Windows How We Want Things To Be { sizeof(PIXELFORMATDESCRIPTOR), // Size Of This Pixel Format Descriptor 1, // Version Number PFD_DRAW_TO_WINDOW | // Format Must Support Window PFD_SUPPORT_OPENGL | // Format Must Support OpenGL PFD_DOUBLEBUFFER, // Must Support Double Buffering PFD_TYPE_RGBA, // Request An RGBA Format 32, // Select Our Color Depth 0, 0, 0, 0, 0, 0, // Color Bits Ignored 0, // No Alpha Buffer 0, // Shift Bit Ignored 0, // No Accumulation Buffer 0, 0, 0, 0, // Accumulation Bits Ignored 24, // 24Bit Z-Buffer (Depth Buffer) 0, // No Stencil Buffer 0, // No Auxiliary Buffer PFD_MAIN_PLANE, // Main Drawing Layer 0, // Reserved 0, 0, 0 // Layer Masks Ignored }; if (!(hDC = GetDC(windowHandle))) return false; unsigned int PixelFormat; if (!(PixelFormat = ChoosePixelFormat(hDC, &pfd))) return false; if (!SetPixelFormat(hDC, PixelFormat, &pfd)) return false; hRC = wglCreateContext(hDC); if (!hRC) { std::cout << "wglCreateContext Failed!\n"; return false; } if (wglMakeCurrent(hDC, hRC) == NULL) { std::cout << "Make Context Current Second Failed!\n"; return false; } ... // OGL Buffer Initialization glClear(GL_DEPTH_BUFFER_BIT | GL_COLOR_BUFFER_BIT); glBindVertexArray(vao); glUseProgram(myprogram); glDrawElements(GL_TRIANGLES, indexCount, GL_UNSIGNED_SHORT, (void *)indexStart); SwapBuffers(GetDC(window_handle));
• 11
• 19
• 14
• 23
• 11 | 2017-07-27 22:56:15 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.25764045119285583, "perplexity": 3341.547171963903}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-30/segments/1500549429548.55/warc/CC-MAIN-20170727222533-20170728002533-00258.warc.gz"} |
https://web2.0calc.com/questions/solve-for-x_26 | +0
# Solve for X
0
282
1
b(5px - 3c) = a(qx - 4)
Aug 20, 2017
#1
+178
0
Input: b(5px-3c) = a(qx-4)
Intepretation: Solve for $$x$$ in $$b(5px+3c) = a(qx-4)$$
Expand:
$$5bpx-3bc=aqx-4a$$
Move the $$x$$ terms to the right, integer ones to the left:
$$5bpx-aqx=3bc-4a$$
Pull the x out from the left term:
$$(5bp-aq)x=3bc-4a$$
Divide both sides by $$5bp-aq$$:
$$x=\frac{3bc-4a}{5bp-aq}$$
Done :D
Aug 20, 2017 | 2019-01-23 19:11:33 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.961338460445404, "perplexity": 13011.376855374625}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-04/segments/1547584336901.97/warc/CC-MAIN-20190123172047-20190123194047-00278.warc.gz"} |
http://support.gams.com/gams:get_the_lagrangian_multiplier_for_the_constraints | # GAMS Support Wiki
### Site Tools
gams:get_the_lagrangian_multiplier_for_the_constraints
# Lagrangian Multiplier for Constraints
Q: During nonlinear programming, we need the Lagrangian multiplier for the constraints. Is is possible for me to get the value from my solution?
The Lagrange multipliers, also known as marginals, can be accessed after a solve as follows.
Say the constraint of interest is called `CON(j,k)` and you want to set `lamda(j,k)` to the value of its marginal after the solve. Then anywhere after the solve statement, just write:
` lamda(j,k) = CON.M(j,k) ;` | 2017-11-20 07:46:21 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8721042275428772, "perplexity": 1332.7470643201102}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-47/segments/1510934805923.26/warc/CC-MAIN-20171120071401-20171120091401-00242.warc.gz"} |
https://tex.stackexchange.com/questions/140278/plot-3d-graphs-of-functions-defined-implicitly | # plot 3d graphs of functions defined implicitly
How can I plot 3d graphs of functions defined implicitly (quadratic forms, for Linear Algebra course notes -- I'd like to include lots of examples)? As far as I can see it is not possible with pgfplots and not through gnuplot either. Is there any package that will help with that?
For example, parabolic and hyperbolic cylinders, hyperboloids, ellipsoids, etc. Concrete examples:
2xy + 2xz = 1 (hyperbolic cylinder)
(x^2)/(2^2) + (y^2)/(3^2) + (z^2)/(2^2) = 1 (ellipsoid)
I also see that Maxima can do this:
(%11) hc:2*x*y+2*x*z=2;
(%i2) draw3d(enhanced3d=true,implicit(hc,x,-5,5,y,-5,5,z,-5,5));
This will work fine (the hyperbolic cylinder is correctly plotted on the screen), but I don't know what backend Maxima uses for this, and I'd like to use a plain LaTeX method, or something that could be called from LaTeX, as I may have to send the document for others to compile themselves on different environments.
• please give an example with the function that doesn't work. – percusse Oct 23 '13 at 14:54
• Well, it's not that there is a function that doesn't work, but rather that there doesn't seem to be any package that will plot functions defined implicitly. I'll add one example function to the description. – Jay Oct 23 '13 at 14:57
• If available, Maxima uses gnuplot to plot graphics. To plot implicit functions with gnuplot see this FAQ, this not-so-FAQ and also the related questions Plotting an implicit function using pgfplots – giordano Oct 23 '13 at 16:07
• Yes -- I have seen the question about implicit functions with pgfplots, but that's for 2d only. Implicit functions of two variables (that is, defined using three variables) are a bit trickier. So far I have been parameterizing them (I tell Maxima to solve the equation for z, then use the result), but that's not a perfect solution...) – Jay Oct 23 '13 at 16:11
• I see. With gnuplot I fear it's impossible to directly plot such functions because you can't plot them in 4d and project onto 3d. I think you can only solve the equation numerically, and plot the result stored in a text file. This is what Maxima does, all in all (in your home directory Maxima should leave the gnuplot script used to create the last plot). – giordano Oct 23 '13 at 16:22
Asymptote contour3 package draws 3D surfaces described as the null space of real-valued functions of (x, y, z). Note that the images here are rendered into raster format (png).
% impsurf.tex :
%
\documentclass[10pt,a4paper]{article}
\usepackage{lmodern}
\usepackage{subcaption}
\usepackage[inline]{asymptote}
\usepackage[left=2cm,right=2cm,top=2cm,bottom=2cm]{geometry}
\begin{asydef}
settings.outformat="png";
settings.render=8;
import graph3;
import contour3;
currentlight=light(gray(0.8),ambient=gray(0.1),specular=gray(0.7),
specularfactor=3,viewport=true,dir(42,48));
pen bpen=rgb(0.75, 0.7, 0.1);
material m=material(diffusepen=0.7bpen
,ambientpen=bpen,emissivepen=0.3*bpen,specularpen=0.999white,shininess=1.0);
\end{asydef}
%
\begin{document}
%
\begin{figure}
\captionsetup[subfigure]{justification=centering}
\centering
\begin{subfigure}{0.49\textwidth}
\begin{asy}
size(200,0);
currentprojection=orthographic(camera=(9,10,4),up=Z,target=O,zoom=1);
// ellipsoid
real f(real x, real y, real z) {return (x^2)/(2^2) + (y^2)/(3^2) + (z^2)/(2^2)-1;}
draw(surface(contour3(f,(-3,-3,-3),(3,3,3),32)),m
,render(compression=Low,merge=true));
xaxis3(Label("$x$",1),-4,4,red);
yaxis3(Label("$y$",1),-4,4,red);
zaxis3(Label("$z$",1),-4,4,red);
\end{asy}
%
\caption{$(\frac{x}{2})^2+(\frac{y}{3})^2+(\frac{z}{2})^2= 1$ (ellipsoid)}
\label{fig:1a}
\end{subfigure}
%
\begin{subfigure}{0.49\textwidth}
\begin{asy}
size(200,0);
currentprojection=orthographic(camera=(9,4,4),up=Z,target=O,zoom=1);
// hyperbolic cylinder
real f(real x, real y, real z) {return 2*x*y + 2*x*z-1;}
draw(surface(contour3(f,(-3,-3,-3),(3,3,3),32)),m
,render(compression=Low,merge=true));
xaxis3(Label("$x$",1),-4,4,red);
yaxis3(Label("$y$",1),-4,4,red);
zaxis3(Label("$z$",1),-4,4,red);
\end{asy}
%
\caption{$2xy + 2xz = 1$ (hyperbolic cylinder)}
\label{fig:1b}
\end{subfigure}
\caption{}
\label{fig:1}
\end{figure}
%
\end{document}
%
% Process:
%
% pdflatex impsurf.tex
% asy impsurf-*.asy
% pdflatex impsurf.tex
An example for the hyperbolic cylinder:
\documentclass[pstricks]{standalone}
\usepackage{pst-solides3d,pst-math}
\begin{document}
\begin{pspicture}(-4,-4)(4,4)% the main 2D area
\psset{lightsrc=viewpoint,viewpoint=50 -200 20 rtp2xyz,Decran=30}
\pstVerb{/constA 1 def /constB 1 def }
\defFunction[algebraic]{hcyl0}(u,v)
{ constA*SINH(u) }% x=f(u)
{ constB*COSH(u) }% y=f(u)
{ v } % z=f(v)
\defFunction[algebraic]{hcyl1}(u,v)
{ constA*SINH(u) }% x=f(u)
{ -constB*COSH(u) }% y=f(u)
{ v } % z=f(v)
\psSolid[object=surfaceparametree,base=-2 2 -3 3,
fillcolor=red!40,function=hcyl0,linewidth=0.1\pslinewidth,ngrid=25]
\psSolid[object=surfaceparametree,base=-2 2 -3 3,
fillcolor=red!40,function=hcyl1,linewidth=0.1\pslinewidth,ngrid=25]
\gridIIID[Zmin=-3,Zmax=3](-4,4)(-4,4)
\end{pspicture}
\end{document}
• +1 but you've parametrized the surfaces... the OP asked about implicit curves – cmhughes Oct 23 '13 at 20:26
• implicit curves make only sense when they cannot be converted into another type. – user2478 Oct 23 '13 at 20:28
• Plotting implicit curves and surfaces is not so efficient in PSTricks as the current implementation does not use the most optimal algorithm. – kiss my armpit Oct 24 '13 at 4:14
• Herbert: Implicit surfaces can also make sense when it is important to be able to change the (3d) bounding box easily. Additionally, it's easier to compute the intersection of a parametrized surface with an implicit surface than with another parametrized surface. – Charles Staats Oct 24 '13 at 23:15
This answer is based on g.kov's excellent answer, but uses the relatively new smoothcontour3 module to produce a nicer-looking surface. The smoothcontour3 module has been incorporated into Asymptote version 2.33 (released 11 May 2015, just a little too late for the Tex Live 2015 cutoff), so if you have that version or later, you should not need to download the module. However, if you have an earlier version of Asymptote (for instance, the version incorporated into TeX Live 2015), you need to download the module from the line above or from the asymptote base code and put it in the same directory as the file you want to compile.
The code (again, heavily based on g.kov's code):
% impsurfsmooth.tex :
%
\documentclass[10pt,a4paper]{article}
\usepackage{lmodern}
\usepackage{subcaption}
\usepackage{asypictureB}
\usepackage[left=2cm,right=2cm,top=2cm,bottom=2cm]{geometry}
settings.outformat="png";
settings.render=8;
import graph3;
import smoothcontour3;
pen bpen = rgb(0.75, 0.7, 0.1);
material m = material(diffusepen=0.7bpen, ambientpen=bpen, emissivepen=0.3*bpen,
specularpen=0.999white, shininess=1.0);
%
\begin{document}
%
\begin{figure}
\captionsetup[subfigure]{justification=centering}
\centering
\begin{subfigure}[b]{0.49\textwidth}
\begin{asypicture}{name=ellipsoid}
size(200,0);
currentprojection = orthographic(9,10,4);
// ellipsoid
real f(real x, real y, real z) {
return (x^2)/(2^2) + (y^2)/(3^2) + (z^2)/(2^2) - 1;
}
draw(implicitsurface(f, (-3,-3,-3), (3,3,3), overlapedges=true), surfacepen=m);
xaxis3(Label("$x$",1),-4,4,red);
yaxis3(Label("$y$",1),-4,4,red);
zaxis3(Label("$z$",1),-4,4,red);
\end{asypicture}
%
\caption{$(\frac{x}{2})^2+(\frac{y}{3})^2+(\frac{z}{2})^2= 1$ (ellipsoid)}
\label{fig:1a}
\end{subfigure}
%
\begin{subfigure}[b]{0.49\textwidth}
\begin{asypicture}{name=hyperboliccylinder}
size(200,0);
currentprojection=orthographic(9,4,4);
// hyperbolic cylinder
real f(real x, real y, real z) {
return 2*x*y + 2*x*z - 1;
}
draw(implicitsurface(f, (-3,-3,-3), (3,3,3), overlapedges=true), surfacepen=m);
xaxis3(Label("$x$",1),-4,4,red);
yaxis3(Label("$y$",1),-4,4,red);
zaxis3(Label("$z$",1),-4,4,red);
\end{asypicture}
%
\caption{$2xy + 2xz = 1$ (hyperbolic cylinder)}
\label{fig:1b}
\end{subfigure}
\caption{}
\label{fig:1}
\end{figure}
%
\end{document}
%
% Process:
%
% pdflatex --shell-escape impsurfsmooth.tex | 2020-01-25 09:09:18 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7756137847900391, "perplexity": 3111.3244021544247}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-05/segments/1579251671078.88/warc/CC-MAIN-20200125071430-20200125100430-00085.warc.gz"} |
https://nosco.ch/ai/ml/nn_backward.php | (EN) A-AA+ Sitemap Search
# Neural Networks: Backward Propagation
Suppose we have M training examples $x^{(m)}, m = 1, .. M$ with known outcomes ("labels") $y^{(m)}, m = 1, .. M$. Each label y is a K-vector with all zeros except one 1 at the index k when the example belongs to the class k. Using the forward propagation we compute the predictions $h_\Theta(x^{(m)}), m = 1, .. M$. The measure for the deviation for one example $m$ is based on the logistic regression cost that is summed over all K classes:
$$C^{(m)}(\Theta) = - \sum_{k=1}^K \left[y_k^{(m)} \log (h_\Theta (x^{(m)})_k) + (1 - y_k^{(m)})\log (1 - h_\Theta(x^{(m)})_k)\right] \tag{1}$$
The total cost is the average cost of all examples plus a regularization term $R(\Theta)$:
\begin{align} C(\Theta)& = \frac{1}{M} \sum_{m=1}^M C^{(m)}(\Theta) + R(\Theta) \tag{2} \\ R(\Theta)& = \frac{\lambda}{2M} \sum_{i,j,l;j \ne 0} (\Theta_{ij}^{(l)})^2 \tag{3} \end{align}
Our goal is to find the parameters $\Theta$ that minimize the cost. For this purpose we need to compute the partial derivatives of $C(\Theta)$ with respect to all parameters $\Theta_{ij}^{(l)}$:
$$D_{ij}^{(l)} = \dfrac{\partial}{\partial \Theta_{ij}^{(l)}}C(\Theta) \tag{4}$$
First we will neglect the regularization term. Since the total cost is a linear function (sum) of the costs of the individual examples we need only to compute the partial derivatives for a specific example $x^{(m)}$. In the following we will drop the superscript $(m)$ to simplify the notation.
Notice that a change of $\Theta_{ij}^{(l)}$ propagates to the next level only to $z_i^{(l+1)}$. So, the chain rule for derivation gives us:
\begin{align} \dfrac{\partial C}{\partial \Theta_{ij}^{(l)}} & = \dfrac{\partial C}{\partial z_i^{(l+1)}} \dfrac{\partial z_i^{(l+1)}}{\partial \Theta_{ij}^{(l)}} \\ & = \delta^{(l+1)}_i \dfrac{\partial z_i^{(l+1)}}{\partial \Theta_{ij}^{(l)}} \tag{5} \\ & = \delta^{(l+1)}_i a_j^{(l)} ; \text{ where: } i \ge 1, j \ge 0, l \lt L \end{align}
where we introduced the auxiliary terms:
$$\delta^{(l)}_i = \dfrac{\partial C}{\partial z_i^{(l)}} \tag{6}$$
These terms can be computed backward starting with the output layer L. The cost for one example can be written as:
$$C = - \sum_{k=1}^K \left[y_k \log (a_k^{(L)}) + (1 - y_k)\log (1 - a_k^{(L)})\right] \tag{7}$$
We also need the derivative of the sigmoid function $g(z)$:
$$g'(z) = \dfrac {d g(z)} {d z} = g(z) (1 - g(z)) \tag{8}$$
Now we can compute $\delta^{(L)}_i$ for the output layer as follows:
\begin{align} \delta^{(L)}_i & = \dfrac{\partial C}{\partial z_i^{(L)}} \\ & = \dfrac{\partial C}{\partial a_i^{(L)}} \dfrac{\partial a_i^{(L)}}{\partial z_i^{(L)}} \\ & = \dfrac{\partial C}{\partial a_i^{(L)}} g'(z_i^{(L)}) \\ & = \left[ - \dfrac{y_i}{a_i^{(L)}} + \dfrac{1 - y_i}{1 - a_i^{(L)}} \right] a_i^{(L)} (1 - a_i^{(L)}) \\ & = a_i^{(L)} - y_i \tag{9} \end{align}
The equation (9) can be written in vector form:
$$\delta^{(L)} = a^{(L)} - y \tag{10}$$
For inner layers $l, l \lt L$ we use the multivariate chain rule ($i \ge 1$):
\begin{align} \delta^{(l)}_i & = \dfrac{\partial C}{\partial z_i^{(l)}} \\ & = \sum_{j \ge 1} \dfrac{\partial C}{\partial z_j^{(l+1)}} \dfrac{\partial z_j^{(l+1)}}{\partial z_i^{(l)}} \\ & = \sum_{j \ge 1} \delta_j^{(l+1)} \dfrac{\partial z_j^{(l+1)}}{\partial z_i^{(l)}} \tag{11} \end{align}
Because
\begin{align} z_j^{(l+1)} & = \sum_{k \ge 0} \Theta_{jk}^{(l)} a_k^{(l)} \\ & = \Theta_{j0} + \sum_{k \ge 1} \Theta_{jk}^{(l)} g(z_k^{(l)}) \tag{12} \end{align}
we get (i ≥ 1, j ≥ 1):
$$\dfrac{\partial z_j^{(l+1)}}{\partial z_i^{(l)}} = \Theta_{ji}^{(l)} g'(z_i^{(l)}) \tag{13}$$
Substituing the expression (13) into (11) we get (i ≥ 1):
$$\delta^{(l)}_i = \sum_{j \ge 1} \Theta_{ji}^{(l)} \delta_j^{(l+1)} g'(z_i^{(l)}) \tag{14}$$
The formula (14) can be written in vector form as:
$$[0; \delta^{(l)}] = (\Theta^{(l)})^T \delta^{(l+1)} \ .*\ [0; g'(z^{(l)})] \tag{15}$$
Here we used the Hadamard product ".*" of two vectors that is defined as the vector whose elements are the products of the elements of the respective vectors ("elementwise product"). We also had to add the first zero elements to $\delta^{(l)}$ and $g'(z^{(l)})$ to cope with the fact that the summation in (14) starts with $j = 1$ whereas the matrix $\Theta^{(l)}$ has a first column $\Theta_{i0}^{(l)}$.
We have derived all ingredients to compute the partial derivatives $\partial C^{(m)} / \partial \Theta_{ij}^{(l)}$. The computation of the derivatives of the regularization term $R(\Theta)$ is trivial:
$$\dfrac{\partial R(\Theta)}{\partial \Theta_{ij}^{(l)}} = \begin{cases} \dfrac{\lambda}{M} \Theta_{ij}^{(l)} , & \text {if } j \ne 0 \tag{16} \\ 0 , & \text {if j = 0} \end{cases}$$
### Algorithm
1. Set up the matrices $D^{(l)}, l = 1, 2, .., L-1$ that correspond to the partial derivatives (Eq. 4). The dimension of the matrix $D^{(l)}$ is $N_l \times (N_{l-1}+1)$. Initialize all these matrices with 0.
2. For the training examples $x^{(m)}, m = 1, 2, .., M$ do:
1. Set $a^{(1)} = x^{(1)}$. This is an N-vector.
2. Perform forward propagation to compute $a^{(l)}, l = 2, 3, ..,L$. The vector $a^{(l)}$ has the dimension $N_l$.
3. Compute the difference between the predicted values and the labeled values:
$\delta^{(L)} = a^{(L)} - y$.
All these vectors have the dimension $N_L = K$, where K is the number of classes.
4. Compute $\delta^{(L-1)}, \delta^{(L-2)},\dots,\delta^{(1)}$ using (Eq. 16):
$[0; \delta^{(l)}] = (\Theta^{(l)})^T \delta^{(l+1)} \ .*\ [0; g'(z^{(l)})]$
The dimensions are:
$\dim(\delta^{(l)}) = N_l$
$\dim((\Theta^{(l)})^T) = (N_l + 1) \times N_{l+1}$
$\dim(\delta^{(l+1)}) = N_{l+1}$
$\dim(g'(z^{(l)})) = N_l$
5. Add the partial derivatives of the example $m$ to the accummulated partial derivatives according to (Eq. 5) written in matrix form:
$D^{(l)} := D^{(l)} + \delta^{(l+1)} (a^{(l)})^T, l = 1, \dots L-1$
3. End the loop over the M examples.
4. Compute the average of the accumated values and add the regularization term:
$D^{(l)} := \dfrac{1}{M} D^{(l)} + \dfrac{\lambda}{M} \hat{\Theta}^{(l)} , l = 1, \dots L-1$
The term $\hat{\Theta}^{(l)}$ denotes the matrix $\Theta^{(l)}$ with the first column set to 0. | 2020-07-08 07:54:24 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 5, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9992313385009766, "perplexity": 773.641024768242}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-29/segments/1593655896905.46/warc/CC-MAIN-20200708062424-20200708092424-00233.warc.gz"} |
https://physics.stackexchange.com/questions/438222/in-quantum-mechanics-why-is-langle-t-rangle-frac-langle-p2-rangle2m-r | # In quantum mechanics, why is $\langle T\rangle=\frac{\langle p^2 \rangle}{2m}$ rather than $\langle T\rangle=\frac{\langle p \rangle^2}{2m}$?
I'm a newbie reading quantum mechanics from "Inroduction to Quantum Meachanics" by Griffiths and in the early pages of the book the author defines:
$$\langle x\rangle =\int_{-\infty}^{\infty} x|\Psi(x,t)|\,dx = \int_{-\infty}^{\infty} \Psi^* (x)\Psi \,dx,$$
$$\langle v\rangle = \frac{d}{dt}\left(\langle x\rangle\right)= -\frac{i\hbar}{m}\int_{-\infty}^{\infty} \Psi^*\frac{\partial\Psi}{\partial x} \,dx,$$
$$\langle p\rangle = m\langle v\rangle= -i\hbar\int_{-\infty}^{\infty} \Psi^*\frac{\partial\Psi}{\partial x} \,dx,$$
so to me the author seems to be working out with expectations, which made perfect sense to me. I then googled the expression for kinetic energy and I was expecting to find out that:
$$\langle T\rangle=\frac{\langle p \rangle^2}{2m},$$
$$\langle T\rangle=\frac{\langle p^2 \rangle}{2m}.$$
Why is this? I don't understand what happened in the case of kinetic energy. Why isn't the author now working with expected momentum in the case of expected kinetic energy? Can you perhaps show me a derivation of $$\langle T\rangle$$ and more importantly, explanation on why it is done like that? In the book, the author says that generally:
$$\langle Q(x, p)\rangle = \int \Psi^*Q(x, \frac{\hbar}{i}\frac{\partial}{\partial x})\Psi\,dx,$$
with advising that every $$p$$ should be replaced with $$\frac{\hbar}{i}\frac{\partial}{\partial x}$$ when calculating the expectation of interest. The why-part for this was however a bit non-existing.
• The first answer here might help you: physics.stackexchange.com/questions/424800/… – Martin C. Nov 1 '18 at 13:11
• In the mean time, I will explain with intuition rather than math. Momentum has a direction, kinetic energy does not. You can have a mean momentum of $0$ but a non-zero mean kinetic energy. Performing the average of the square of momentum fixes this. – Aaron Stevens Nov 1 '18 at 13:13
• @MartinC. Thank you, I actually checked that answer before but it didn't really open up to me unfortunately :/ – jjepsuomi Nov 1 '18 at 13:16
• @AaronStevens thank you, the intuition already helped :) Of course the details are still in the mist for me. Any other book recommendation where this might be explicitly derived? – jjepsuomi Nov 1 '18 at 13:19
• If $A = B$, then $\langle A \rangle = \langle B \rangle$, because we can do the same thing to both sides of an equation. That's absolutely all there is to it. If you think $H = p^2/2m$, then $\langle H \rangle = \langle p^2 / 2m \rangle$. – knzhou Nov 1 '18 at 14:08
Why is this?
For concreteness, let's look at a specific example for which $$\langle T \rangle \ne \frac{\langle P \rangle^2}{2m}$$
Consider the case that we have a particle with state vector (working in 1D for simplicity)
$$|\psi\rangle = \frac{1}{\sqrt{2}}\left(|+p\rangle + |-p\rangle\right)$$
where $$p \ne 0$$ and $$P\,|\pm p\rangle = \pm p\,|\pm p\rangle$$ (these are eigenkets of the momentum operator).
Clearly, the expectation value of momentum is
$$\langle P\rangle = \langle\psi|P|\psi\rangle = \frac{1}{2}\left(+p -p\right) = 0$$
This is because the momentum measurement has equal chance of yielding $$+p$$ and $$-p$$.
However, a kinetic energy measurement measurement can only yield
$$T = \frac{(\pm p)^2}{2m} = \frac{p^2}{2m}$$
and so
$$\langle T \rangle = \frac{p^2}{2m} \ne \frac{\langle P \rangle^2}{2m} = 0$$
If you think about it, this really doesn't come down to QM and just depends on how you take averages. QM only comes into play if you actually want to calculate those averages given the state vector of the system.
We know that $$T=\frac{p^2}{2m}$$, so the average of this is then $$\langle T\rangle=\left\langle\frac{p^2}{2m}\right\rangle=\frac{\langle p^2\rangle}{2m}$$
Since, in general, $$\langle p^2\rangle\neq\langle p \rangle^2$$, this is where we end up.
If you want to find this value using the position basis, then we invoke QM: $$\langle T\rangle=-\frac{\hbar^2}{2m}\int\Psi^*\frac{\partial^2}{\partial x^2}\Psi\ dx$$
This is because in the position basis, the $$P^2$$ operator is $$-\hbar^2\frac{\partial^2}{\partial x^2}$$.
• Thank you, it helped me a bit, but I still 100% don't get it. The reason why I think it confuses me, is because in the case of velocity, we used expected position. In the case of momentum, we used expected velocity. But in the case of kinetic energy we did NOT use expected momentum. Do you happen to know any books perhaps where this is explained in detail? :) – jjepsuomi Nov 1 '18 at 13:57
• @jjepsuomi Which part? How averages work with functions of other variables, or how you calculate those averages? Keep in mind that the heart of your question (why $\langle p^2\rangle$ instead of $\langle p \rangle^2$) isn't about QM – Aaron Stevens Nov 1 '18 at 13:59
• @jjepsuomi The kinetic energy uses the expectation value of the momentum squared. The answer explicitly shows how this works: $\langle T\rangle=\left\langle\frac{p^2}{2m}\right\rangle=\frac{\langle p^2\rangle}{2m}$. – Martin C. Nov 1 '18 at 14:11
• Thank you, I think I now got it, huh it was that simple then x) – jjepsuomi Nov 1 '18 at 14:14
• @jjepsuomi Sorry, I didn't realize you just wanted an example worked through showing that $\langle p\rangle^2\neq\langle p^2\rangle$ for a specific case. – Aaron Stevens Nov 1 '18 at 14:25 | 2019-06-25 03:58:29 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 24, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7750585675239563, "perplexity": 316.4945317964932}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-26/segments/1560627999787.0/warc/CC-MAIN-20190625031825-20190625053825-00374.warc.gz"} |
https://rubandpaint.pl/railway/2021_5_21_4662/ | +86 0371 8654 9132
## size distribution d80 significance - mayukhportfolio.co
size distribution d80 significance. What is Particle Size Distribution D50, D50 Particle Size ... - AimSizer. Particle size distribution D50 is fully explained and regularly updated on this webpage by senior professionals of Aimsizer Scientific. Last Update: 2014-12-22. Read more.
## particle size distribution d80 - BINQ Mining
Jan 24, 2013 · size distribution d80 significance – Basalt Crusher. about size distribution d80 significance. The particle-size distribution (PSD) of a powder, or granular material, or particles dispersed in fluid, »More detailed
## How to Interpret Particle Size Distribution Data
Jul 29, 2013 · How to Interpret Particle Size Distribution Data. Post navigation. Previous. Next. For a great presentation on How to Interpret Particle Size Distribution Data D50
## particle size d10 d50 d90 p80 - BINQ Mining
Mar 02, 2013 · explain particle size d80, d50 – Crusher South Africa. size distribution d80 significance Typical Particle Size Plot Summary statistics for particle size distribution d50, d10, Can someone explain what P80 and D50 mean? »More detailed
The D50 is the size in microns that splits the distribution with half above and half below this diameter. The Dv50 (or Dv0.5) is the median for a volume distribution, Dn50 is used for number distributions, and Ds50 is used for surface distributions. Since the primary result from laser diffraction is a volume distribution, the default D50 cited ...
## What is Particle Size Distribution D50, D50 Particle Size ...
Aug 02, 2021 · Particle Size Distribution D50 is also known as median diameter or medium value of particle size distribution, it is the value of the particle diameter at 50% in the cumulative distribution. Particle Size Distribution D50 is one of an important parameter characterizing particle size. For example, if D50=5.8 um, then 50% of the particles in the ...
The particle-size distribution (PSD) of a powder, or granular material, or particles dispersed in fluid, is a list of values or a mathematical function that defines the relative amount, typically by mass, of particles present according to size. Significant energy is usually required to disintegrate soil, etc. particles into the PSD that is then called a grain size distribution.
## Particle Size Distribution Effects that Should be ...
a steeper size distribution curve is produced. This was demonstrated by Armstrong (1960), who compared open and closed circuit ball milling and rod milling in the laboratory. He concluded that the size distribution from a laboratory rod mill gave a similar-shaped size distribution to that of a closed circuit laboratory ball mill.
## difference between k80 and p80 grind size
Difference between k80 and p80 grind size particle size distribution d80 binq mining, significance read particle size p80 definition hoteleldoradobenin tests of significance - yale university tests of significance, for a sample of size n, the t distribution will have n-1 degrees
## A basic guide to particle characterization
Figure 5: Illustration of the D[4,3] and D[3,2] on a particle size distribution where a significant proportion of fines are present. Percentiles For volume weighted particle size distributions, such as those measured by laser diffraction, it is often convenient to report parameters based upon the maximum
## Understanding & Interpreting Particle Size Distribution ...
The D50 is the size in microns that splits the distribution with half above and half below this diameter. The Dv50 (or Dv0.5) is the median for a volume distribution, Dn50 is used for number distributions, and Ds50 is used for surface distributions. Since the primary result from laser diffraction is a volume distribution, the default D50 cited ...
## D10, D50, D90 - for DLS? Yes not just diffraction - with ...
Aug 25, 2016 · The parameter D90 should more correctly be labeled as Dv (90). It signifies the point in the size distribution, up to and including which, 90% of the total volume of material in the sample is ‘contained’. For example, if the D90 is 844nm, this means that 90% of the sample has a size of 844nm or smaller. The definition for D50 or Dv (50 ...
## A basic guide to particle characterization
Figure 5: Illustration of the D[4,3] and D[3,2] on a particle size distribution where a significant proportion of fines are present. Percentiles For volume weighted particle size distributions, such as those measured by laser diffraction, it is often convenient to report parameters based upon the maximum
## ACTTR Inc. - The Meaning of D10 D50 D90 in Particle Size ...
Jul 21, 2020 · The particle size distribution of powers is typically expressed by these three values. They are often described in academic research reports or quality control reports. "D" means Distribution particle size distribution. The length unit, D10 represents the 10% of particles in the powders are smaller than this size. Typically, the unit is μm.
## What is particle size distribution test?
May 11, 2020 · A Particle Size Distribution Analysis (PSD) determines and reports information about the size and range of particles representative of a given material. This analysis can be performed using a variety of techniques; the most suitable will be determined based on
## crushing d80 product size iron ore invest benefit in egypt
d80 crusher scresizes wikipedia Hitlers Holly 2020619 iron Sand crusher sizes of screen d80 crusher screen sizes wikipedia d80 crusher screen size size distribution d80 significance sand crushersand d80 crusher screen size d80 d80 product size iron ore d80 crusher screen size Impact Crusher Which Gives 4 To 40 Mesh Partical...
## Sieve Analysis, Particle Size Analysis - AboutCivil.Org
Mar 21, 2017 · It is determined from the grain size distribution curve at the point where the curve crosses a horizontal line through the 10% passing value on the y axis. Other D sizes are found in a similar manner. The D50 size, called the median grain size, is the grain diameter for which half the sample (by weight) is smaller and half is larger. Two ...
## Method of Particle-size Evaluation of Ground Material ...
May 31, 2018 · After wet grinding of a sample using NP-100, the particle size distribution is measured by the laser diffraction method. Laser diffraction method. In this method, the sample is irradiated by laser light and the particle-size distribution of the sample is determined from the intensity pattern of diffracted and scattered light.
## Mechanical Analysis of Soil - Memphis
Particle-Size Distribution Curve End of Part 1 Particle-Size Distribution Curve The results of mechanical analysis (sieve and hydrometer analyses) are generally presented by semi-logarithmic plots known as particle-size distribution curves. The particle diameters are plotted in log scale, and the corresponding percent finer in arithmetic scale.
## Investigating of the effect of ore work index and particle ...
Apr 01, 2016 · These experiments were conducted for the same 3 sample with providing d80 of 1000 μm to show the effect of increasing feed size on the modeling process.The 1000 μm feed size of same copper samples were obtained by grinding ores in laboratory jaw and roll crusher until the d80 of product reached 1000 μm.Therefore, six ore samples were treated for grinding experiments that provided at
## Grinding Size Estimation and Beneficiation Studies Based ...
chemical structures. These differences lead to significant variations of size reduction, size distribution, and in consequence the selection of enrichment process steps. The chemical properties of these ores are presented in Table 1. Table 1. Chemical analysis of chromite ores Assay (%) ore A ore B ore C Cr 2 O 3 12.07 16.30 23.83 FeO 10.11 8 ...
## What do D60, D30 and D10 mean in soil? What do Cu & Cc ...
Answer (1 of 3): First, let us take a look at this grain size distribution curve. In this curve, x axis - Particle size(dia) in mm in log scale y axis - Percentage finer. From this curve, we can find the D60, D30 and D10 of a particular soil. D60 - 60 % of the soil particles are finer than t...
## Sample size estimation and power analysis for clinical ...
In unequal sample size of 1: 2 (r = 0.5) with 90% statistical power of 90% at 5% level significance, the total sample size required for the study is 48. Sample size estimation with two proportions . In study based on outcome in proportions of event in two populations (groups), such as percentage of complications, mortality improvement ...
## particle size distribution d80 - tytansystem.pl
particle size distribution d80 - Mining. 2013-1-24 Particle size d80: d80 is the average particle size as determined by a screen on which 20% by weight of the particles will remain and 80% Distribution of fines, » More detailed. Get Price. particle size d10 d50 d90 p80 - Mining. More
## (PDF) Correlation of the Blaine value and the d80 size of ...
Correlation of the Blaine value and the d80 size of the cement particle size distribution [Korrelation Zwischen dem Blaine-Wert und der d80 Korngrößenverteilung im Zement] January 2008 ZKG ...
## (PDF) Correlation of the Blaine value and the d(80) size ...
E = 10 1.74 x 10 –4 x F Bl. + 0.035 x w i + 0.4714 (1) where, E is the specific grinding energy (kWh/tonne), F Bl is the cement fineness (Blaine) in cm 2 /g and. w i is the clinker work index ...
## 7.2.2.2. Sample sizes required - NIST
For a one-sided test at significance level $$\alpha$$, look under the value of 2$$\alpha$$ in column 1. Note that this table is based on the normal approximation (i.e., the standard deviation is known). Sample Size Table for Two-Sided Tests
## Particle Size Distribution Dependent on Principle of ...
The fact that particle size distribution is dependent on principle of measurement is an extremely substantive problem arising from the very concept of "particle size distribution." "Particle size distribution" is an index (means of expression) indicating what sizes (particle size) of particles are present in what proportions (relative particle ...
## Particle Size Essentials Guidebook - Horiba
Learn why particle size is important, how to interpret particle size distribution calculations, result interpretation, setting specifications and more. HORIBA's full line of particle characterization instruments are explained in detail as well as how to select the right particle size analyzer for your application.
## Sieve analysis - Wikipedia
A sieve analysis (or gradation test) is a practice or procedure used in civil engineering and chemical engineering to assess the particle size distribution (also called gradation) of a granular material by allowing the material to pass through a series of sieves of progressively smaller mesh size and weighing the amount of material that is stopped by each sieve as a fraction of the whole mass.
## Solved An important factor in solid missile fuel is the ...
Transcribed image text: An important factor in solid missile fuel is the particle size distribution. Significant problems occur if the particle sizes are too large. From production data in the past, it has been determined that the particle size (in micrometers) distribution is characterized by the following function. f(x) 3x4, x>1 0, elsewhere (a) Verify that this is a valid density function.
## Why is D10 called effective diameter (geotechnical ...
Answer (1 of 4): Nearly 120 years ago a guy named Allen Hazen set out on a goal to identify, from a soil mass, the particular size/diameter of soil grains that can be used to speculate various properties of the whole soil mass. ALLEN HAZEN (1869-1930) He co...
## 7.1 The Sampling Distribution of the Sample Mean (σ Un ...
7.1 The Sampling Distribution of the Sample Mean (σ Un-known) Learning Objectives. By the end of this chapter, the student should be able to: Construct and interpret confidence intervals for means when the population standard deviation is unknown. Carry out hypothesis tests for means when the population standard deviation is unknown.
## Recent Studies on the Application of Fluidised-Bed ...
Figure 2 Particle size distribution (a) and mass and zinc distribution by particle size (b) for the sphalerite ore used in the study. The p80 of the particle size distribution is 750 Pm. 0 20 40 60 80 100 100 1000 Mass passing, % Particle size, um 0 10 20 30 40 50 1180 - 850-250 425 - 250 850 - 425 Distribution, % Particle size range, um mass Zn
## Central Limit Theorem (CLT) Definition
Aug 04, 2021 · The central limit theorem (CLT) states that the distribution of sample means approximates a normal distribution as the sample size gets larger, regardless of the population's distribution. Sample ...
## Example of Sieve Analysis Data Calculations, Graphs, and Data
size of 0.72, sorting of 1.10, skewness of 0.06, and kurtosis of 1.23. The sample we have just calculated since its values are so close to these we could conclude that most likely these samples are from the same depositional environment. Return to Sieve Analysis Lab Exercise Return to | 2022-01-18 02:33:27 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7405782341957092, "perplexity": 2688.8447156289485}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320300658.84/warc/CC-MAIN-20220118002226-20220118032226-00448.warc.gz"} |
http://atcoder.noip.space/contest/abc188/a | # Home
Score : $100$ points
### Problem Statement
A basketball game is being played, and the score is now $X$-$Y$. Here, it is guaranteed that $X \neq Y$.
Can the team which is behind turn the tables with a successful three-point goal?
In other words, if the team which is behind earns three points, will its score become strictly greater than that of the other team?
### Constraints
• $0 \le X \le 100$
• $0 \le Y \le 100$
• $X \neq Y$
• $X$ and $Y$ are integers.
### Input
Input is given from Standard Input in the following format:
$X$ $Y$
### Output
If the team which is behind can turn the tables with a successful three-point goal, print Yes; otherwise, print No.
3 5
### Sample Output 1
Yes
The team with $3$ points is behind.
After a successful $3$-point goal, it will have $6$ points, which is greater than that of the other team - $5$.
Thus, we should print Yes.
16 2
### Sample Output 2
No
The gap is too much. The team which is behind cannot overtake the other by getting $3$ points.
12 15
### Sample Output 3
No
A $3$-point goal will tie the score but not turn the tables. | 2021-11-27 19:51:29 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.19350989162921906, "perplexity": 2484.043484054716}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-49/segments/1637964358233.7/warc/CC-MAIN-20211127193525-20211127223525-00382.warc.gz"} |
http://tex.stackexchange.com/tags/equations/new | # Tag Info
3
you should really have posted these as two separate questions, instead of adding on to the first. but you could try \documentclass{article} \usepackage{amsmath} \begin{document} $$\mathbf{q}=(\mu_1, \mu_2, \ldots ,\mu_{n-2}, \mu_n )^{\text{T}} \label{2}$$ and L = \log(l) = ...
5
3
I do not see the benefit of using such complicated equation numbers. Numbering equations within sections is more than enough IMHO. Nevertheless, the additional level can be suppressed, if the subsection counter is zero. Also \numberwithin does not define a transitive closure. Thus the equation number needs to be reset for each \section, too. ...
2
Using \numberwithin{equation}{subsection} changes the counter output format and shifts the resetting of the equation counter from section to subsection level (i.e. each time \stepcounter{subsection} is used. However, this means that an orphane equation after a \section, but before \subsection will just use the old counter value, from a previous equation. ...
2
Use the package chngcntr: In your preamble: \usepackage{chngcntr} In the Introduction section: \counterwithout{equation}{chapter} After the Introduction section: \counterwithin{equation}{chapter}
3
1
With fancyvrb package you can get it \documentclass{article} \usepackage{lipsum} \usepackage{fancyvrb} \newsavebox{\FVerbBox} \newenvironment{FVerbatim} {\VerbatimEnvironment \begin{center} \begin{lrbox}{\FVerbBox} \begin{BVerbatim}} {\end{BVerbatim} \end{lrbox} \mbox{\usebox{\FVerbBox}} \end{center}} \begin{document} \lipsum*[2] ...
3
The best, i.e., "most LaTeX-y" way to typeset this expression is to set up a macro named, say, \norm, that takes one argument -- the term(s) to be encased in double vertical bars. With the method shown in the example below, it's easy to change the size of the bars, if needed, by providing an optional argument to the \norm macro. \documentclass{article} ...
1
The amsmath package already provides the needed tools: Bmatrix for the multiline matrix with curly braces and bmatrix for the matrix with square brackets. \documentclass[a4paper,12pt,numbers=noenddot]{scrartcl} \usepackage{amsmath} \begin{document} Here is an equation \nabla_Q N_I = \begin{Bmatrix} N_{I,1} \\ N_{I,2} \\ N_{I,3} ...
1
Here is a way. I define an lrcases and a dlrcases environment, analogous to the (d)cases and (d)rcases environments from mathtools. The code is borrowed from @Gonzalo Medina: \documentclass{article} \usepackage[utf8]{inputenc} \usepackage{mathtools, bm} \makeatletter \newcases{dlrcases}{\quad}{% ...
2
Here you go. I could not replicate the font other than using a Sans Serif family (it appears bold in mine). Output Code \documentclass[margin=10pt]{standalone} \usepackage{tikz} \usetikzlibrary{arrows.meta,calc,positioning} \tikzset{ every node/.style={draw=gray, rounded corners, text centered,text width=3cm}, } \begin{document} ...
3
I'm not sure why the flalign environment seems to have become very fashionable. Here are two solutions, based on the alignedat inner environment. Inside the equation* or flalign* environment (if you really want left alignment) I define a local abbreviation for the labels. If you have several of these constructions, move the command in the preamble, so ...
6
I don't think there's a ready-made package for Lagrangean equations. Applying a bit more formatting -- and not using in a LaTeX document -- may be desirable. I suggest using a gather* environment, with a nested alignedat environment. \documentclass{article} \usepackage{mathtools} \begin{document} \begin{gather*} \max U(x,y)\\ \shortintertext{subject ...
2
This gets (like some other answers) the proper math spacing and (unlike the other answers), gets the proper (right) alignment on the numbers, using tabular stacks. \documentclass{article} \usepackage{tabstackengine} \stackMath \setstacktabulargap{0pt} \begin{document} \tabularCenterstack{rrcr}{ \textit{Angle}: & -45^{\circ} \le& \theta ... 2 the equation number was a syntax error extra { (don't ignore error messages!) You can get bold math italic using \bm from the package of the same name. Also: don't use \emph in math mode, and you do not need \limits here. \documentclass[a4paper,12pt,numbers=noenddot]{scrartcl} \usepackage{mathtools} \usepackage{bm} \begin{document} ... 11 I would use alignat for multiple alignments. \documentclass{article} \usepackage{amsmath} \begin{document} \begin{alignat*}{2} \textit{Angle}: &\ & -45^{\circ} &\le \theta \le +45^{\circ} \\ \textit{Bins}: &\ & -135^{\circ} &< \theta < -45^{\circ} \\ \textit{Slant}: &\ & +45^{\circ} ... 4 You could use tabular or array instead of flalign*: \documentclass{article} \usepackage{array,amsmath} \begin{document} \begin{tabular}{r>{}r<{}*{4}{>{}l<{}}} \textit{Angle}: & -45^{\circ} & \le & \theta & \le & +45^{\circ} \\ \textit{Bins}: & -135^{\circ} & < & \theta & < & ... 4 Try: \begin{flalign*} \textit{Angle}: && -45^{\circ} &\le \theta \le +45^{\circ} \\ \textit{Bins}: && -135^{\circ} &< \theta < -45^{\circ} \\ \textit{Slant}: && +45^{\circ} &< \theta < +135^{\circ} \\ \textit{Tilt}: && -180^{\circ} &\le \theta \le -135^{\circ} \ or\ ... 2 Here you are: it is enough to use \min. I also replaced the pair of || with a \norm command, defined with the help of the mathtools package. The star version adds a pair of implicit \left … \right on both sides of the \Vert delimiters. If you want to fine-tune the size of the delimiters, use an optional argument instead, such as \norm[\big]{…} instead. ... 6 I would write \min instead of \textnormal {min} \limits, and I would use \lVert and \rVert instead of ||. $$\lVert \mathbf{x}^{1} - \overline{\mathbf{x}}^{2} \rVert = % \min_{ \mathbf{x}^{2} \subseteq \Gamma^{2} } % \lVert \mathbf{x}^{1} - \mathbf{x}^{2} ({\pmb \xi}) \rVert$$ 1 Here's a solution that uses an array environment to align the various terms. \documentclass{article} \usepackage{amsmath,array} \begin{document} \[ \setlength\arraycolsep{0pt} \begin{array}{c @{} >{{}}c<{{}} c >{{}}c<{{}} l @{{}={}} c @{} >{{}}c<{{}} l} \mathrm{B} & + & \mathrm{R} & \cdot & \mathrm{t}_1 & ... 2 One way to align the content as desired is to use a \makebox to center the text in the desired amount of space: Notes: I would highly recommend you use the siunitx package as well for typesetting anything with units. Code: \documentclass{article} \usepackage{mathtools} \usepackage{siunitx} \sisetup{output-decimal-marker={,}} ... 1 Some variations of Kormylo's answer: The cells in the left column uses an inner tabular. The second line is indented to emphasize the term in the first line. No fixed column widths. The equation numbers can be referenced, using amsmath macros for the formatting of the equation number. Full example: \documentclass{report} \usepackage[utf8]{inputenc} ... 1 You can use m-type columns and flalign for left aligning the numbered equation. Some vertical space corrections are needed, but they're hidden in a macro, so you don't need to specify them each time. \documentclass{report} \usepackage[T1]{fontenc} \usepackage[utf8]{inputenc} \usepackage[ngerman]{babel} \usepackage[intlimits]{amsmath} ... 1 The p{6cm} is equivalent to \parbox[t]{6cm}, but \parbox[c]{6cm} would do better for this. Nor do you really need a display math environment when \displaystyle ... will do. Especially, if you put the equation number into its own column. \documentclass{report} \usepackage[intlimits]{amsmath} \usepackage{here} \usepackage{floatflt} ... 0 Here is a way. I used threeparttable in order to have the caption left-aligned w.r.t. the table. Off topic: you shouldn't load here since float defines an H option. \documentclass{report} \usepackage[utf8]{inputenc} \usepackage{lmodern} \usepackage[german]{babel} \usepackage[showframe, nomarginpar]{geometry} \usepackage[intlimits]{mathtools} ... 6 Square roots are very sensitive to ascenders and descenders. Just look at the two of them and you'll realize that the culprit is the j in the second one: \documentclass{article} \begin{document} x_{ij} = \frac{ \sum_{t}{e_i(t)e_j(t)} } { \sqrt{\sum_t{e_{i\vphantom{j}}^2(t)}} \sqrt{\sum_t{e_j^2(t)}} ... 7 \documentclass{article} \begin{document} $$x_{ij} = \frac{ \sum_{t}{e_i(t)e_j(t)} } { \sqrt{\strut\sum_t{e_i^2(t)}} \sqrt{\strut\sum_t{e_j^2(t)}} }$$ \end{document} 1 More of a workaround than a regular answer to your question: using \limits. \documentclass{article} \begin{document} $$x_{ij} = \frac{ \sum\limits_{t}{e_i(t)e_j(t)} } { \sqrt{\sum\limits_t{e_i^2(t)}} \sqrt{\sum\limits_t{e_j^2(t)}} }$$ ... 3 Perhaps you can use an extensible harpoon that is has a width proportionate to the other content: \documentclass{article} \usepackage{mathtools,graphicx,accents} \newcommand{\vect}[1]{\accentset{\xrightharpoonup{\hphantom{\scalebox{.4}{#1}}}}{#1}} \begin{document} \[ \vect{L^2((0,T);W^{1,2}(\Omega)')} \end{document} The content is scaled to 40% ...
2
For line numbering to be done correctly the math environments has to be wrapped using the \begin{linenomath*} and \end{linenomath*} code: \documentclass[12pt]{article} \usepackage{amsmath} \usepackage{lineno} \linenumbers \begin{document} For line numbering to be done correctly the math environments has to be wrapped using the ''linenomath code as ...
2
\documentclass[journal]{IEEEtran} \usepackage{amsmath} \usepackage{lipsum} \usepackage{stfloats} \begin{document} \title{\huge equation position control in two column paper} \maketitle \enlargethispage{-2cm} \section{First} \label{first} \begin{picture}(0,0) \put(0,-600){\hspace{-\parindent}\parbox{\textwidth}{% \hrulefill \vspace*{4pt} \normalsize ...
0
3
Here, I provide \undereq{} to be used in the \underbrace subscript. \documentclass{article} \usepackage{amsmath,stackengine,graphicx} \stackMath \newcommand\undereq[1]{% \stackunder[2pt]{\mkern1mu\rotatebox{90}% {$\scriptstyle=\mkern-3mu$}}{\scriptstyle \mathstrut#1}% } \begin{document} \[ = ...
1
This should get you going. I've introduced a few commands to make it easier to typeset some parts of the equations in a more automated way, or if you wish to change something all over the document at once. *They're not strictly necessary for the obtained result. \documentclass{article} \usepackage{amsmath,mathtools} \newcommand\partt[1][]{% \ifmmode ...
0
The equation environment (including equation*) already centres its contents, so there's no need for center. Additionally, center adds vertical space, as discussed in [When should we use \begin{center} instead of \centering?]When should we use \begin{center} instead of \centering?) Only use \begin{equation*} <stuff> \end{equation*} If you need to ...
5
The large space is due to the center environment, which adds vertical unneeded space and does nothing useful, since equation center its contents by default. So, simply remove the center environment. MWE \documentclass{article} \usepackage{amsmath,amssymb} \begin{document} \textbf{Solution} \begin{equation*} (1+x)(1+y)\geqslant{4} \end{equation*} ...
1
\begin{tabular}{ccc} 1 & 2 & 3 \ \ 6 & 7 & 8 \end{tabular} You will get 1 2 3 6 7 8 so you can write down B,R and R_max in the first line as you said and want and others in the second.
1
You can place the equations in an array, see http://en.wikibooks.org/wiki/LaTeX/Mathematics#Matrices_and_arrays For instance \begin{matrix} -1 & 3 \\ 2 & -4 \end{matrix} = \begin{matrix*}[r] -1 & 3 \\ 2 & -4 \end{matrix*} will produce
3
Trying to center the quote while placing the number on the right was too difficult using the list environment, so I used a \parbox instead. The main difference is that a \parbox will not break at the end of a page. \documentclass[a4paper,11pt,oldfontcommands]{memoir} \usepackage{pdfpages} \usepackage{tikz} \usetikzlibrary{positioning} ...
3
Here is a colorful suggestion using tcolorbox. \documentclass[a4paper,11pt,oldfontcommands]{memoir} \usepackage[most]{tcolorbox} \usepackage{showframe} \usepackage[colorlinks=true]{hyperref} \newtcolorbox[auto counter,number within=section]{myquote}[2][]{% colback=red!5!white,colframe=red!75!black,fonttitle=\bfseries, enlarge left by=1cm,enlarge right by ...
5
This will get you started. You can modify it for different numbering (e.g by section as 1.1, 1.2 etc change the within=none to within=section). This example has a table of quotes with page numbers. You can easily modify it so you have no quote caption (or only a caption within the table of quotes), and have a different layout with the number on the right ...
3
You could use an aligned environment inside an equation environment. \documentclass{article} \usepackage{amsmath} % for 'aligned' environment \begin{document} \begin{aligned} \dot{R}&=\frac{66}{364}\, P_R+ \frac{\partial V}{\partial P_R} &\qquad \dot{\theta}&=2P_{\theta}\Bigl(\frac{13}{168r^2}-\frac{33}{364R^2}\Bigr) ...
2
Instead of all those math ($…$) and center environments, and since you do not seem to wish any particular alignment, I would recommand an equation environment combined with a gathered environment (with a bit of additional vertical space here between the two lines, but it's not mandatory) from the amsmath package. It does the same job, but much more shortly ...
4
If you want to have some more alignment than you have right now I would recommend the following approach: % arara: pdflatex \documentclass{article} \usepackage{mathtools} \begin{document} \begin{alignedat}{2} \dot{R}&=\frac{66}{364}\,P_R+ \frac{\partial V}{\partial P_R} \qquad ...
1
Here is one way, using a stack. The top set are your originals; the bottom set are my replacement, "without changing their format," as requested. \documentclass{article} \usepackage{stackengine} \stackMath \begin{document} \begin{center} \$\dot{R}=\frac{66}{364}\ P_R+ \frac{\partial V}{\partial P_R}\qquad ...
0
The reason for seeing references only on a per-chapter basis is the following: you have not set a master document. In order to do so, open the main .tex file in Texmaker (the one with all the input/includes of all your chapters). Then select Options / Define Current Document as "Master Document".
2
You also can play with arraycolsep: \documentclass[11pt]{beamer} \usetheme{Warsaw} \usepackage[utf8]{inputenc} \usepackage{amsmath,adjustbox,mathtools} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{graphicx} \begin{document} \begin{frame} \frametitle{LU Factorization of A} \setlength\arraycolsep{3.25pt} \mathbf{LU} = ...
1
You can use \resizebox to scale the equation. Remember to re-enter math mode after it. \documentclass[11pt]{beamer} \usetheme{Warsaw} \usepackage[utf8]{inputenc} \usepackage{amsmath,adjustbox,mathtools} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{graphicx} \begin{document} \begin{frame} \frametitle{LU Factorization of A} ...
3
For example, as noted in my comment, one could replace all \left with \biggl and all \right with \biggr. There are different sizes, as well: \big, \Big, etc. that you can search for on the site. Your question will likely be marked duplicative, because the question has been asked many times, for example, here: Error while equation splitting with align ...
Top 50 recent answers are included | 2015-05-23 03:09:39 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 5, "x-ck12": 0, "texerror": 0, "math_score": 0.9978853464126587, "perplexity": 5841.743216443612}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-22/segments/1432207927104.48/warc/CC-MAIN-20150521113207-00159-ip-10-180-206-219.ec2.internal.warc.gz"} |
https://nforum.ncatlab.org/discussion/7518/clone/ | # Start a new discussion
## Not signed in
Want to take part in these discussions? Sign in if you have an account, or apply for one below
## Discussion Tag Cloud
Vanilla 1.1.10 is a product of Lussumo. More Information: Documentation, Community Support.
• CommentRowNumber1.
• CommentAuthorzskoda
• CommentTimeNov 29th 2016
New stub for the notion of clone in universal algebra, including a reference relating it to operads.
• CommentRowNumber2.
• CommentAuthorTim_Porter
• CommentTimeNov 29th 2016
Zoran: the categorical form of the full clone on a functor was one dual form of categorical shape theory if I remember it rightly.
• CommentRowNumber3.
• CommentAuthorMike Shulman
• CommentTimeJan 25th 2017
• (edited Jan 25th 2017)
Where on earth did the word “clone” come from?
And if a clone is really the same as a Lawvere theory, why not merge it with Lawvere theory and add a redirect?
• CommentRowNumber4.
• CommentAuthorTodd_Trimble
• CommentTimeJan 25th 2017
I don’t know where ’clone’ came from, but it’s been around a rather long time, before Lawvere’s thesis I believe.
Clones and Lawvere theories and cartesian operads are virtually the same; you could say that a clone is to a Lawvere theory as a multicategory is to a monoidal category (so a clone is something like a cartesian multicategory with one object). Thus I believe clones and cartesian operads are synonyms, with maybe a slight difference in how they are usually packaged. I have notes on this here.
• CommentRowNumber5.
• CommentAuthorTim_Porter
• CommentTimeJan 25th 2017
• (edited Jan 25th 2017)
The ‘full clone’ on a functor was a term that occurred in describing categorical shape theory, but I would have to check up how! I seem to remember the use involved Fred Linton’s work on theories. (Edit: It seems I have repeated myself … At least I said the same thing both times!)
• CommentRowNumber6.
• CommentAuthorKarol Szumiło
• CommentTimeJan 25th 2017
Todd, maybe you could add some of these remarks as comments on MO. What I wrote in my answer is literally everything that I know about clones (I’ve learned about them solely because I wanted to understand the example of affine spaces), so I’m not familiar with all these subtleties.
• CommentRowNumber7.
• CommentAuthorMike Shulman
• CommentTimeJan 25th 2017
Well, we also have cartesian multicategory, so if that would be more appropriate we could redirect “clone” there. But it doesn’t make sense to me to have two pages with different names about essentially the same object.
• CommentRowNumber8.
• CommentAuthorTodd_Trimble
• CommentTimeJan 25th 2017
Before we do that, let’s pause and consider again the analogy, written as the proportion clone: cartesian multicategory :: Lawvere theory: cartesian monoidal category, where redirecting clone to cartesian multicategory would be analogous to redirecting Lawvere theory to cartesian monoidal category. I wouldn’t think we’d want to do the latter.
• CommentRowNumber9.
• CommentAuthorMike Shulman
• CommentTimeJan 26th 2017
An even better analogy is clone : cartesian multicategory :: operad : symmetric multicategory, and the latter two are separate pages. So, I guess. But I hate to have so much duplication; I notice that we also have Lawvere theory and algebraic theory as separate pages. At least the pages should link to each other; I added some.
• CommentRowNumber10.
• CommentAuthorTodd_Trimble
• CommentTimeJan 26th 2017
• (edited Jan 26th 2017)
As for Lawvere theory and algebraic theory: the latter was written after the former as a separate article because the full generality of algebraic theory (of unbounded rank) could not of course be written under Lawvere theory which is traditionally refers just to the finitary single-sorted case. The connection between the two is implicitly acknowledged by adopting the phrase “large Lawvere theory” (red herring-like). Personally, I think it’s okay on occasion to admit some redundancy; the articles are slightly different in style, with the Lawvere theory article giving some extra intuition and motivation – appropriate for readers encountering the categorical concept first in its baby form – whereas the algebraic theory article is somewhat more hard-core in style.
There’s yet another article, infinitary Lawvere theory. I believe Andrew Stacey started that (and I see you were the last editor). I had wanted to pursue a different approach but didn’t want to write all over what I regarded as Andrew’s (couldn’t think of how to do it gracefully), so I started algebraic theory also for that reason.
Edit: Apologies; it wasn’t started by Andrew; it was started by Toby.
• CommentRowNumber11.
• CommentAuthorMike Shulman
• CommentTimeJan 26th 2017
You make good points. I would like it, though, if we could have some more obvious way of pointing out at the beginning of an article that some other page(s) are about “almost the same subject”, as opposed to merely a related notion, so that someone finding the page from elsewhere won’t miss out on something that they would probably be interested in just because it’s on a page with a different name.
• CommentRowNumber12.
• CommentAuthorMike Shulman
• CommentTimeDec 1st 2019
The Idea and Definition sections of the page clone are a bit contradictory. Is there a difference between an “abstract” clone (= one-object cartesian multicategory) and a “concrete” clone (consisting of operations on a particular set)?
• CommentRowNumber13.
• CommentAuthorTodd_Trimble
• CommentTimeDec 2nd 2019
The definition section reads weirdly to me. It might be possible to embed any abstract clone into an endomorphism clone (even that’s not so clear to me), but if that’s so, I’d think there should then be many ways of doing it. Do we then make a distinction between the resulting subclone supported on an set $S$ and the resulting one supported on a different set $S'$? (There could even be abstractly isomorphic but different such subclones supported on the same set.)
I’m thinking the definition section should be rewritten. Miles Gould’s thesis seems as good a source as any.
• CommentRowNumber14.
• CommentAuthorMike Shulman
• CommentTimeDec 5th 2019
Clarified the difference between concrete and abstract clones (following wikipedia for terminology). I don’t have time to add any more, but now at least the idea and definition sections aren’t contradictory.
• CommentRowNumber15.
• CommentAuthorvarkor
• CommentTimeSep 3rd 2020
Clarify that, though an abstract clone is equivalent to a cartesian operad, it is presented differently. Therefore, it is somewhat unreasonable to define a clone to be a cartesian operad.
• CommentRowNumber16.
• CommentAuthorMike Shulman
• CommentTimeSep 3rd 2020
Hm, I’m not sure I agree. A group is a group, regardless of whether we define it in terms of multiplication and inversion or in terms of some other operation like $(g,h)\mapsto g h^{-1}$.
• CommentRowNumber17.
• CommentAuthorMike Shulman
• CommentTimeSep 3rd 2020
Or, closer to home, an operad is an operad, whether we present it with May-style composition $g \circ (f_1,\dots, f_n)$ or Markl-style composition $g \circ_i f$.
• CommentRowNumber18.
• CommentAuthorvarkor
• CommentTimeSep 3rd 2020
That’s true. Perhaps it would fairer to say that an abstract clone is an alternate presentation of a cartesian operad (cf. the one on [cartesian multicategory]). I do think “abstract clone” should be taken to mean the specific presentation: otherwise there is no distinction between abstract clones, cartesian operads, and Lawvere theories — whereas, in practice, the different perspectives are helpful. In this case, perhaps the page for abstract clone and for cartesian multicategory should be merged, with abstract clone given its own subsection on the cartesian multicategory page? | 2020-10-29 22:05:20 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 5, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7618255019187927, "perplexity": 2176.0906108487716}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-45/segments/1603107905965.68/warc/CC-MAIN-20201029214439-20201030004439-00340.warc.gz"} |