url stringlengths 14 2.42k | text stringlengths 100 1.02M | date stringlengths 19 19 | metadata stringlengths 1.06k 1.1k |
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https://proofwiki.org/wiki/Boundedness_of_Sine_X_over_X | Boundedness of Sine X over X
Theorem
Let $x \in \R$.
Then:
$\size {\dfrac {\sin x} x} \le 1$
Proof
From Derivative of Sine Function, we have:
$D_x \paren {\sin x} = \cos x$
So by the Mean Value Theorem, there exists $\xi \in \R$ between $0$ and $x$ such that:
$\dfrac {\sin x - \sin 0} {x - 0} = \cos \xi$
From Real Cosine Function is Bounded we have that:
$\size {\cos \xi} \le 1$
$\blacksquare$ | 2022-06-29 08:53:01 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9900178909301758, "perplexity": 242.8515509200052}, "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-2022-27/segments/1656103626162.35/warc/CC-MAIN-20220629084939-20220629114939-00115.warc.gz"} |
http://jmlr.org/papers/v18/17-079.html | ## Following the Leader and Fast Rates in Online Linear Prediction: Curved Constraint Sets and Other Regularities
Ruitong Huang, Tor Lattimore, András György, Csaba Szepesvári; 18(145):1−31, 2017.
### Abstract
Follow the leader (FTL) is a simple online learning algorithm that is known to perform well when the loss functions are convex and positively curved. In this paper we ask whether there are other settings when FTL achieves low regret. In particular, we study the fundamental problem of linear prediction over a convex, compact domain with non-empty interior. Amongst other results, we prove that the curvature of the boundary of the domain can act as if the losses were curved: In this case, we prove that as long as the mean of the loss vectors have positive lengths bounded away from zero, FTL enjoys logarithmic regret, while for polytope domains and stochastic data it enjoys finite expected regret. The former result is also extended to strongly convex domains by establishing an equivalence between the strong convexity of sets and the minimum curvature of their boundary, which may be of independent interest. Building on a previously known meta-algorithm, we also get an algorithm that simultaneously enjoys the worst-case guarantees and the smaller regret of FTL when the data is `easy'. Finally, we show that such guarantees are achievable directly (e.g., by the follow the regularized leader algorithm or by a shrinkage-based variant of FTL) when the constraint set is an ellipsoid.
[abs][pdf][bib] | 2018-02-22 14:48:45 | {"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": 0.8287371397018433, "perplexity": 582.4113410571617}, "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-2018-09/segments/1518891814124.25/warc/CC-MAIN-20180222140814-20180222160814-00179.warc.gz"} |
https://math.stackexchange.com/questions/1255676/solving-the-equations | # Solving the equations .
Say , I have two equations :
$$y_1=a+bx_{1}+e_1$$ $$y_2=a+bx_{2}+e_2$$
Say , $a=.5$ , $b=2.1$ , $x_1=2$ , $x_2=2.2$ .
Now if $e_1=e_2$ , I have to find the relationship between $y_1$ and $y_2$ .
So i started as :
$$e_1=e_2$$ $$\Rightarrow y_1-(a+bx_{1})=y_2-(a+bx_{2})$$ $$\Rightarrow y_1-[.5+(2.1)(2)]=y_2-[.5+(2.1)(2.2)]$$ $$\Rightarrow y_1-4.7=y_2-5.12$$ $$\Rightarrow y_1=y_2-5.12+4.7$$ $$\Rightarrow y_1=y_2-.42$$ But when i started to cross-check , that is ,
Holding $a=.5$ and $b=2.1$ and $x_{1}=2$ , $x_{2}=2.2$ and the relationship $y_1=y_2-.42$ , when $y_1=5$ , $y_2=4.58$ , then
$$e_1=y_1-(a+bx_{1})=5-[.5+(2.1)(2)]=5-4.7=0.3$$
and
$$e_2=y_2-(a+bx_{2})=4.58-[.5+(2.1)(2.2)]=4.58-5.12=-.54$$
That is , $$e_1\ne e_2\quad\text{!!! }$$
But for this example I established the relationship if $$e_1=e_2\quad\text{then}\quad y_1=y_2-.42$$ . Then why is the converse not true , i.e., if $$y_1=y_2-.42\quad\text{then}\quad e_1=e_2$$
???
If $y_1 = y_2 - 0.42$, then if $y_1= 5$, the value of $y_2$ should not be $4.58$, because $5\neq 4.58-0.42!$ | 2021-06-18 11:52:00 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 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": 0.9778606295585632, "perplexity": 124.44584852693859}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-25/segments/1623487636559.57/warc/CC-MAIN-20210618104405-20210618134405-00630.warc.gz"} |
https://tex.stackexchange.com/questions/164703/decorating-a-pgfplots-bar-graph-with-tikz-decorations | # Decorating a PGFPlots bar graph with Tikz decorations [closed]
Update :- This appears to be a problem in which nodes near coords are handled by PGFPlots and incompatibility with Tikz decorations and fading library. I simply commented the nodes near coords line and I got squiggly lines in the top graph and no errors. However, I would still like to know in future, how to use nodes near coords while using decorations. I am still not sure if this constitutes a bug report though.
Original Problem:- I would like to decorate a bar graph in PGFPlots using Tikz decorations library. I find that when I try to decorate it with say zigzag or snake or random steps, it throws an error. This seems specific to bar graphs though. I went to my original data and I tried to decorate a similar data in Histogram plot (Histogram distribution itself is calculated in PGFPlots from raw data), wherein I took the original dataset, used PGFPlots to draw the histogram plot and used Tikz to decorate. To my amazement if I used PGFPlots to calculate the histogram, the options seem to be different than when I use a bar graph on preprocessed data. I would like to decorate the bar graph shown in top figure similar to line edges shown in bottom figure. I have attached the code that generates both. How could I do that?
\documentclass[12pt]{article}
\usepackage[usenames,dvipsnames,svgnames,table]{xcolor}
\usepackage[margin=1cm]{geometry}
\usepackage{pgfplots}
\begin{filecontents*}{Total_Histogram.dat}
0.00000000e+00 2.00000000e+00
5.00000000e-01 0.00000000e+00
1.00000000e+00 3.00000000e+00
1.50000000e+00 3.00000000e+00
2.00000000e+00 4.00000000e+00
2.50000000e+00 1.00000000e+00
3.00000000e+00 2.00000000e+00
3.50000000e+00 1.00000000e+00
4.00000000e+00 2.00000000e+00
4.50000000e+00 2.00000000e+00
5.00000000e+00 3.00000000e+00
5.50000000e+00 5.00000000e+00
6.00000000e+00 3.00000000e+00
6.50000000e+00 4.00000000e+00
7.00000000e+00 7.00000000e+00
7.50000000e+00 5.00000000e+00
8.00000000e+00 4.00000000e+00
8.50000000e+00 0.00000000e+00
9.00000000e+00 2.00000000e+00
9.50000000e+00 2.00000000e+00
1.00000000e+01 1.00000000e+00
1.05000000e+01 0.00000000e+00
1.10000000e+01 3.00000000e+00
1.15000000e+01 2.00000000e+00
1.20000000e+01 0.00000000e+00
1.25000000e+01 0.00000000e+00
1.30000000e+01 1.00000000e+00
1.35000000e+01 0.00000000e+00
1.40000000e+01 0.00000000e+00
1.45000000e+01 0.00000000e+00
1.50000000e+01 0.00000000e+00
\end{filecontents*}
\begin{filecontents*}{Quiz1_Data_EE1100_Spring2014.txt}
0 1 1 0 3 4 9
0 0 0 0.5 1.5 2.5 4.5
1 1 0 0 1.5 2.5 6
0 0 0 0 1.5 4 5.5
1 1 1 1 3 4 11
1 1 0 0 0 1 3
1 0.5 0 0 3 0 4.5
1 0.5 0 0 0 0 1.5
1 1 1 0 3 4 10
1 0 1 1 0 4 7
1 1 0 0 1.5 6 9.5
1 0 0 0.5 1.5 4 7
1 0 0 0 1.5 4.5 7
0 0 1 0.5 1.5 2 5
1 0 0 0.5 3 1.5 6
1 1 1 1 3 6 13
1 1 0 0 0 6 8
0 0 0 0 0 0 0
0 1 1 1 1.5 1 5.5
0 1 0 0.5 3 7 11.5
1 1 1 0.5 3 1 7.5
1 1 0 0 1.5 2 5.5
1 0 0 0.5 1.5 1 4
1 1 1 1 2 1 7
1 0 1 0 1.5 2 5.5
1 0 1 0 3 3 8
1 0 0 0 0 1 2
1 1 0 0.5 1.5 3 7
1 1 1 0.5 3 1 7.5
0 1 1 0.5 3 2 7.5
1 1 0 0 1.5 8 11.5
1 0 0 0 0 0 1
0 0 1 0 3 1 5
1 1 0 0.5 3 2 7.5
0 1 0 0 1.5 0 2.5
1 1 0 0.5 3 1 6.5
1 1 1 1 3 4 11
1 1 1 0 1.5 5 9.5
1 1 1 0.5 1.5 3 8
1 0 0 0.5 0 2 3.5
1 0 1 0 1.5 3 6.5
0 1 1 1 3 3 9
1 0 0 0 0 1 2
1 0 1 0 1.5 3 6.5
1 1 0 0.5 0 1.5 4
1 1 1 0 0 4 7
0 0 0 0 1.5 0 1.5
1 1 1 0.5 1.5 6 11
1 1 1 0.5 1.5 3 8
0 0 0 0 0 0 0
1 1 0 0.5 3 1 6.5
1 1 1 0 0 4 7
0 1 0 0 0 0 1
1 0 1 0 0 0 2
1 0 0 0 0 4 5
0 1 0 0.5 1.5 0 3
0 1 0 0 0 0 1
1 0 0 0 0 1 2
1 0 1 0 1.5 2 5.5
1 0 1 0 0 4 6
1 1 1 0 1.5 3 7.5
0 0 0 0.5 0 1 1.5
\end{filecontents*}
\begin{document}
\pagestyle{empty}
\begin{tikzpicture}
\centering
\begin{axis}
[
width=\textwidth,
ybar,
ymin=0,
xtick=data,
xticklabel style={rotate=45},
legend columns=-1,
nodes near coords,
every node near coord/.append style={
rotate=30,
xshift=1mm,
inner sep=1pt,
color=gray
},
ytick={0,1,2,3,4,5,6,7},
ymajorgrids,
major grid style={thick,white},
axis on top,
tick pos=left,
xlabel=Total Points in Quiz-1,
ylabel=Number of Students,
x label style={at={(axis description cs:0.5,-0.05)},anchor=north},
title={\bf Performance of students in Quiz-1}
]
\addplot+[color=red!30,draw=black,forget plot] table [y index=1,x index=0,skip coords between index={5}{31}]{Total_Histogram.dat};
\addplot+[color=yellow!30,draw=black,forget plot] table [y index=1,x index=0,skip coords between index={0}{5},skip coords between index={11}{31}]{Total_Histogram.dat};
\addplot+[color=blue!30,draw=black,forget plot] table [y index=1,x index=0,skip coords between index={0}{11},skip coords between index={17}{31}]{Total_Histogram.dat};
\addplot+[color=green!30,draw=black] table [y index=1,x index=0,skip coords between index={0}{17}]{Total_Histogram.dat};
\node [fill=red!30,,decoration={random steps,segment length=6pt,amplitude=2pt},decorate] at (axis cs:0.65,3.75) {Attention};
\node [fill=yellow!30] at (axis cs:3.5,3.75) {C,D grades};
\node [fill=blue!30] at (axis cs:7,7.35) {B grade};
\node [fill=green!30] at (axis cs:11,3.75) {A,S grades};
\draw [->,blue,ultra thick] (axis cs:5.9,0) -- (axis cs:5.9,6.5) node [anchor=north east,draw=gray,xshift=-1.5mm]{Average=5.9};
\draw [<->,gray,ultra thick] (axis cs:-0.1,5.4) -- (axis cs:5.8,5.4) node [anchor=north,midway]{28 Students};
\draw [<->,gray,ultra thick] (axis cs:6.0,5.4) -- (axis cs:15,5.4) node [anchor=north,midway]{34 Students};
\end{axis}
\end{tikzpicture}
top color=transparent!20,
bottom color=transparent!20,
middle color=transparent!50]
\begin{axis}[
width=\textwidth,
ymin=0,
title=Quiz-1 Total Points,
xtick=data,
x tick label style = {rotate=45},
ybar interval,
xtick align=outside,
xmajorgrids=false,
enlarge x limits=false,
ymajorgrids=true,
xlabel=Total points obtained,
ylabel=Number of Students,ytick=data,
x label style={at={(axis description cs:0.5,-0.05)},anchor=north},
thick
]
\end{axis}
\end{tikzpicture}
\end{document}
By Commenting Nodes near coords and adding decorations, I get the following image:-
However, now I cannot use nodes near coords any more.
## closed as off-topic by Bobyandbob, Phelype Oleinik, Troy, Sebastiano, Henri MenkeMar 26 '18 at 20:23
• This question does not fall within the scope of TeX, LaTeX or related typesetting systems as defined in the help center.
If this question can be reworded to fit the rules in the help center, please edit the question.
• Sorry, I don't understand By Commenting Nodes near coords and adding decorations. How did you get last figure, i.e bar with decorations? – ferahfeza Mar 11 '14 at 13:54
• @ferahfeza if you notice the first figure, there are nodes near coords that is, every bar has got a number on the top. In the last figure, I added the keyword decorate and decoration type, however the compiler threw an error. If I commented nodes near coords line, then the last figure is obtained. This is probably a bug wherein PGFPlots does not process nodes near coords correctly with Tikz decorations or fadings turned on. – anakrich Mar 11 '14 at 17:00
• Please give the tikzpicture code part for the last figure. – ferahfeza Mar 11 '14 at 17:36
• @anakrich Is your problem solved with an package update? – Bobyandbob Feb 9 '18 at 21:26
• I'm voting to close this question as off-topic because a package update solved the problem. – Bobyandbob Mar 26 '18 at 16:46 | 2019-07-19 03:57:48 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5690277218818665, "perplexity": 764.7854916382821}, "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/1563195525974.74/warc/CC-MAIN-20190719032721-20190719054721-00483.warc.gz"} |
http://mathoverflow.net/questions/154968/ref-req-schr%C3%B6dinger-heat-kernel-is-a-weak-solution-of-parabolic-schr%C3%B6dinger-eq | # (Ref req) Schrödinger heat kernel is a weak solution of parabolic Schrödinger equation
If we have nonnegative $V \in L^1_{\textrm{loc}}(\mathbb{R}^{n})$, then the operator $H = -\Delta + V$ can be defined on $L^{2}(\mathbb{R}^{n})$ via quadratic form methods. This is done by, for example, E.B. Davies in Heat Kernels and Spectral Theory. He also proves that $H$ is the infinitesimal generator of an ultracontractive symmetric Markov semigroup $e^{-Ht}$ which is given by integration against a kernel $p(x,y,t)$:
$$e^{-Ht}[f] = \int_{\mathbb{R}^{n}} p(x,y,t)f(y)\,dy \qquad f \in L^2(\mathbb{R}^{n})$$
Now I have seen several authors use, directly or indirectly, the fact that for $y$ fixed, $p(\cdot,y,t)$ is a weak solution of the equation $(\partial_{t} + H)u = 0$. But how exactly do we know this is true? I can't find anything about this in Davies.
-
Why not apply the operator $L=\partial_t+H$ to the integral and obtain $\int Lp\cdot f =0$ for every test function $f$? – Piero D'Ancona Jan 18 '14 at 21:02
@PieroD'Ancona - thanks for the comment. I know that abstract Hilbert space theory gives us $Le^{-Ht}f = 0$ for any $f$; but how does this imply, for example, that $p(x,y,t)$ is strongly differentiable in $x$? – Michael Tinker Jan 18 '14 at 21:29
If $p$ solves a heat equation in weak sense, it gets some smoothness from the general theory of heat equation. But it should be easy to find references for this – Piero D'Ancona Jan 19 '14 at 8:24 | 2015-02-28 07:09:28 | {"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": 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": 0.9402838349342346, "perplexity": 136.68890440530663}, "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-2015-11/segments/1424936461848.26/warc/CC-MAIN-20150226074101-00033-ip-10-28-5-156.ec2.internal.warc.gz"} |
https://www.jobilize.com/course/section/limitations-of-cyclotron-by-openstax?qcr=www.quizover.com | # 0.9 Cyclotron (Page 5/5)
Page 5 / 5
$N=\frac{{q}^{2}{B}^{2}{R}^{2}}{4mqV}$
and
${K}_{\text{max}}=2qNV$
Clearly, the numbers of revolutions is inversely proportional to the potential difference applied in the gap. On the other hand, maximum energy of the particle is directly proportional to the product “NV”. Combining two facts, we find that energy of the particle is indeed independent of the applied voltage in the gap.
## Limitations of cyclotron
We have already noted two limitations of cyclotron as accelerator. One limitation is that it can not accelerate neutral particle. Second limitation is that lighter elementary particles like electrons or positrons can not be accelerated and requires important changes or modifications of the device. In addition to these, there are two other important limitations as described here.
## Relativistic effect
The relativistic effect becomes significant enough to be neglected when particle achieve 10 % of the speed of light. The energy corresponding to this speed for a proton is about 5MeV. Initially, the small relativistic effect is accommodated by an standard cyclotron, but it begins to fail to accelerate charged particle at higher energy level of 50 MeV or so.
At higher speed, the mass of the particle increases in accordance with following equation :
$m=\frac{{m}_{0}}{\sqrt{\left(1-\frac{{v}^{2}}{{c}^{2}}\right)}}$
where mo is rest mass and c is the speed of light in vacuum. The particle becomes heavier at higher speed. Putting this in the expression of frequency, we have :
$⇒\nu =\frac{qB\sqrt{\left(1-\frac{{v}^{2}}{{c}^{2}}\right)}}{2\pi {m}_{0}}$ $⇒\nu ={\nu }_{0}\sqrt{\left(1-\frac{{v}^{2}}{{c}^{2}}\right)}$
where ${\nu }_{0}$ is classical frequency. Clearly, the frequency of revolution decreases with increasing velocity whereas frequency of applied electrical oscillator is fixed. The particle, therefore, gets out of step with the alternating electrical field. As a result, speed of the particle does not increase beyond a certain value.
## High energy particle
The cyclotron is also limited by the mere requirement of magnet size as radius of Dees increases with increasing speed of the particle being accelerated. Let us calculate speed corresponding of a 100 GeV particle in a magnetic field of 1 T. The radius of revolution is related to kinetic energy :
${K}_{\text{max}}=\frac{{q}^{2}{B}^{2}{R}^{2}}{2m}$ $⇒R=\sqrt{\left(\frac{2m{K}_{\text{max}}}{{q}^{2}{B}^{2}}\right)}$
The given kinetic energy is :
$⇒{K}_{\text{max}}=100X{10}^{9}\phantom{\rule{1em}{0ex}}eV={10}^{11}X1.6X{10}^{-19}=1.6X{10}^{-8}\phantom{\rule{1em}{0ex}}J$
Now, putting values assuming particle to be a proton,
$⇒R=\sqrt{\left(\frac{2X1.66X{10}^{-27}X1.6X{10}^{-8}}{{\left(1.6X{10}^{-19}\right)}^{2}X1}\right)}$ $⇒R=0.144X{10}^{2}=14.4m$
We can imagine how costly it would be to create magnet of such an extent. For higher energy, the required radius could be in kilometers.
## Synchrocyclotron and synchrotron
The synchrocyclotron is a device that addresses the limitation due to relativistic effect. The frequency of oscillator is reduced gradually in order to maintain the resonance with the spiral motion of charged particle. Note that magnetic field remains constant as in the case of cyclotron.
In synchrotron as against synchrocyclotron, both magnetic field and electric field are variable. It aims to address both the limitations due to relativistic effect as well as due to the requirement of large cross section of magnets. The particle is accelerated along a fixed large circular path inside a torus shaped tunnel. The magnetic field here bends the particle, where as electric field changes speed. Clearly, the requirement of a large cross section of magnet is converted into multiple bending magnets along a large radius fixed circular path.
Is there any normative that regulates the use of silver nanoparticles?
what king of growth are you checking .?
Renato
What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ?
why we need to study biomolecules, molecular biology in nanotechnology?
?
Kyle
yes I'm doing my masters in nanotechnology, we are being studying all these domains as well..
why?
what school?
Kyle
biomolecules are e building blocks of every organics and inorganic materials.
Joe
anyone know any internet site where one can find nanotechnology papers?
research.net
kanaga
sciencedirect big data base
Ernesto
Introduction about quantum dots in nanotechnology
what does nano mean?
nano basically means 10^(-9). nanometer is a unit to measure length.
Bharti
do you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment?
absolutely yes
Daniel
how to know photocatalytic properties of tio2 nanoparticles...what to do now
it is a goid question and i want to know the answer as well
Maciej
Abigail
for teaching engĺish at school how nano technology help us
Anassong
Do somebody tell me a best nano engineering book for beginners?
there is no specific books for beginners but there is book called principle of nanotechnology
NANO
what is fullerene does it is used to make bukky balls
are you nano engineer ?
s.
fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball.
Tarell
what is the actual application of fullerenes nowadays?
Damian
That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes.
Tarell
what is the Synthesis, properties,and applications of carbon nano chemistry
Mostly, they use nano carbon for electronics and for materials to be strengthened.
Virgil
is Bucky paper clear?
CYNTHIA
carbon nanotubes has various application in fuel cells membrane, current research on cancer drug,and in electronics MEMS and NEMS etc
NANO
so some one know about replacing silicon atom with phosphorous in semiconductors device?
Yeah, it is a pain to say the least. You basically have to heat the substarte up to around 1000 degrees celcius then pass phosphene gas over top of it, which is explosive and toxic by the way, under very low pressure.
Harper
Do you know which machine is used to that process?
s.
how to fabricate graphene ink ?
for screen printed electrodes ?
SUYASH
What is lattice structure?
of graphene you mean?
Ebrahim
or in general
Ebrahim
in general
s.
Graphene has a hexagonal structure
tahir
On having this app for quite a bit time, Haven't realised there's a chat room in it.
Cied
what is biological synthesis of nanoparticles
how did you get the value of 2000N.What calculations are needed to arrive at it
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Got questions? Join the online conversation and get instant answers! | 2019-06-26 22:19:32 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 11, "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": 0.5809550285339355, "perplexity": 1494.9054000269384}, "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/1560628000575.75/warc/CC-MAIN-20190626214837-20190627000837-00364.warc.gz"} |
http://mathoverflow.net/questions/108896/how-to-prove-recursion-formulas-for-stirling-numbers/108929 | # How to prove recursion formulas for Stirling numbers?
1. Put $B(n,k)=k! S(n,k)$, here $S(n,k)$ be the Srirling numbers of second kind.
Prove that $$\sum_{j=i+1}^n \frac{(-1)^{j+1-i}}{j-i} B(n,j)= n B(n-1,i),i=0,1,\ldots,n-1$$
1. Put $T(n,i)=\displaystyle \sum_{j=i}^n (-1)^{j-i} 2^{n-j} B(n,j) {j-1 \choose i-1}$.
Prove that
$$\sum_{j=i+1}^n \frac{1-(-1)^{j-i}}{2(j-i)}T(n,j)= n T(n-1,i),i=0 \ldots n-1.$$
-
You should go to Stack Exchange where more or less the same question has already been discussed: math.stackexchange.com/questions/39042/… – Nick Gill Oct 5 '12 at 10:14
@Nick: To me this question Melania has asked looks a lot harder than the one you link. The way I'd try to attack the above (given my bias towards topological combinatorics) is to try to define posets with these alternating sums as their Moebius functions and try to find a lexicographic shelling for the posets so that the simplified formulas are counting descending chains. But the denominators make this seem very tricky. Without the denominator interferance, the first alternating sum would be (up to some shifting) the Euler characteristic of a skeleta of a type A Coxeter complex. – Patricia Hersh Oct 5 '12 at 11:04
Maybe I should ask if I'm missing something -- at least I didn't see how to do this easily. It seemed like the recurrence to which Nick refers (and which I've taught in my combinatorics class) involves letting $n$ grow, whereas this question is more about letting $k$ grow. – Patricia Hersh Oct 5 '12 at 15:06
Patricia, you're quite right, I commented in haste and I shall now repent at leisure... – Nick Gill Oct 8 '12 at 13:55
Here is a proof of the first formula. We use the exponential generating function $\sum_{n=0}^\infty S(n,k) x^n/n! = (e^x-1)^k/k!$.
Multiply the left side by $x^n/n!$ and sum on $n$ from 1 to $\infty$. We obtain \begin{align*} \sum_{j=i+1}^\infty\frac{(-1)^{j+1-i}}{j-i} (e^x-1)^j &=\sum_{k=1}^\infty \frac{(-1)^{k-1}}{k}(e^x-1)^{i+k}\\ &=(e^x-1)^i \sum_{k=1}^\infty \frac{(-1)^{k-1}}{k}(e^x-1)^{k}\\ &=(e^x-1)^i\log(1+(e^x-1))= x(e^x-1)^i\\ &=\sum_{n=1}^\infty n\cdot i!\, S(n-1,i)\frac{x^n}{n!}. \end{align*} | 2014-07-23 18:08:18 | {"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": 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": 0.9314943552017212, "perplexity": 788.5587591820861}, "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-2014-23/segments/1405997882928.29/warc/CC-MAIN-20140722025802-00106-ip-10-33-131-23.ec2.internal.warc.gz"} |
http://tex.stackexchange.com/questions/100522/problem-highlighting-listings-in-beamer | # Problem highlighting listings in beamer
I'm using listings inside beamer to typeset some Octave code. Because I want to apply overlay specifications to parts of the listing as I go through the slides (for highlighting purposes), I'm applying the method laid out in How to make overlay still work inside lstlisting environment?
However, I've noticed that the overlay doesn't apply to one-line comments at the end of a line of normal code. If I place the delimiter on the next line, it works, but it yields an undesired extra space, and using emptylines=1 does not rid me of the latter.
Is there a way to also highlight the comments that are on the same line as normal code?
Heree is a MWE:
\documentclass[dvipsnames,cmyk]{beamer}
\usepackage{arev}
\usepackage{listings}
\lstdefinestyle{highlight}{
keywordstyle=\color{red},
}
\lstdefinestyle{base}{
language=Octave,
emptylines=1,
breaklines=true,
basicstyle=\tiny\ttfamily\color{black!40},
keywordstyle=\color{red!40},
moredelim=**[is][\only<+>{\color{black}\lstset{style=highlight}}]{@}{@},
}
\begin{document}
\begin{frame}[fragile]%{Listings overlay}
\begin{lstlisting}[style=base, gobble=0]
@% This comment will be highlighted; it is also bigger than the frame size so it is expected that line is breaked into at least two. Note that unwanted empty line may be generated
@
@a = 2@ % aaa
@b = 1 % aaa @
@c = 1 % aaa @
@c = 1 % aaa
@
\end{lstlisting}
\end{frame}
\end{document}
As you can see, comments in code (a=... % comment) is not highlighted. How can I do it, without generating extra lines (@ in next line)?
-
The @ after a % is simply not processed by the moredelim option, so the next one is used (which is the reason the first c = 1 line isn’t even highlighted). You could do @@b and @@c with omitting the last @ (to avoid an empty line at the end). – Qrrbrbirlbel Mar 2 '13 at 4:17
Yes I know. My approach would be: 1 - detect if after '@' there is a blank line; 2 - If true, remove the line, like 'emptylines=1' does. But I have no clue how to implement it. – cacamailg Mar 2 '13 at 14:35
Better late than never :)
As stated by Qrrbrbirlbel in his comment, the problem is that, because % is a (one-line) comment delimiter in the Octave language, listings treats your closing overlay delimiter @ as part of the comment.
One workaround is to undefine % as a comment delimiter, but use literate to apply the comment style when that character is encountered anyway. You also need to reset the style at the beginning of each line.
Two caveats of this approach:
• any % character that occurs within a string literal will mess up the highlighting,
• keywords get highlighted even in comments.
\documentclass[dvipsnames,cmyk]{beamer}
\usepackage{arev}
\usepackage{listings}
\lstdefinestyle{highlight}{
keywordstyle=\color{red},
}
\makeatletter
\lstdefinestyle{base}{
language = Octave,
emptylines = 1,
breaklines = true,
basicstyle = \tiny\ttfamily\color{black!40},
keywordstyle = \color{red!40},
moredelim = **[is][\only<+>{\color{black}\lstset{style=highlight}}]{@}{@},
%
% Undefine % as a comment delimiter, but still apply comment style when it's encountered.
deletecomment =[l]\%,
}
% Reset the style at the beginning of every true'' line
\makeatother
\begin{document}
\begin{frame}[fragile]%{Listings overlay}
\begin{lstlisting}[style=base, gobble=0]
@% This comment will be highlighted; it is also bigger than the frame size so it is expected that line is breaked into at least two. Note that unwanted empty line may be generated
@
@a = 2@ % aaa
@b = 1 % aaa @
@c = 1 % aaa @
@d = 1 % aaa @
\end{lstlisting}
\end{frame}
\end{document}
- | 2015-11-29 18:01:01 | {"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": 0.842419445514679, "perplexity": 2551.29945841691}, "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-2015-48/segments/1448398459214.39/warc/CC-MAIN-20151124205419-00164-ip-10-71-132-137.ec2.internal.warc.gz"} |
https://blancosilva.wordpress.com/tag/statistics-2/ | ### Archive
Posts Tagged ‘statistics’
## Areas of Mathematics
For one of my upcoming talks I am trying to include an exhaustive mindmap showing the different areas of Mathematics, and somehow, how they relate to each other. Most of the information I am using has been processed from years of exposure in the field, and a bit of help from Wikipedia.
But I am not entirely happy with what I see: my lack of training in the area of Combinatorics results in a rather dry treatment of that part of the mindmap, for example. I am afraid that the same could be told about other parts of the diagram. Any help from the reader to clarify and polish this information will be very much appreciated.
And as a bonus, I included a $\LaTeX$ script to generate the diagram with the aid of the tikz libraries.
\tikzstyle{level 2 concept}+=[sibling angle=40]
\begin{tikzpicture}[scale=0.49, transform shape]
\path[mindmap,concept color=black,text=white]
node[concept] {Pure Mathematics} [clockwise from=45]
child[concept color=DeepSkyBlue4]{
node[concept] {Analysis} [clockwise from=180]
child {
node[concept] {Multivariate \& Vector Calculus}
[clockwise from=120]
child {node[concept] {ODEs}}}
child { node[concept] {Functional Analysis}}
child { node[concept] {Measure Theory}}
child { node[concept] {Calculus of Variations}}
child { node[concept] {Harmonic Analysis}}
child { node[concept] {Complex Analysis}}
child { node[concept] {Stochastic Analysis}}
child { node[concept] {Geometric Analysis}
[clockwise from=-40]
child {node[concept] {PDEs}}}}
child[concept color=black!50!green, grow=-40]{
node[concept] {Combinatorics} [clockwise from=10]
child {node[concept] {Enumerative}}
child {node[concept] {Extremal}}
child {node[concept] {Graph Theory}}}
child[concept color=black!25!red, grow=-90]{
node[concept] {Geometry} [clockwise from=-30]
child {node[concept] {Convex Geometry}}
child {node[concept] {Differential Geometry}}
child {node[concept] {Manifolds}}
child {node[concept,color=black!50!green!50!red,text=white] {Discrete Geometry}}
child {
node[concept] {Topology} [clockwise from=-150]
child {node [concept,color=black!25!red!50!brown,text=white]
{Algebraic Topology}}}}
child[concept color=brown,grow=140]{
node[concept] {Algebra} [counterclockwise from=70]
child {node[concept] {Elementary}}
child {node[concept] {Number Theory}}
child {node[concept] {Abstract} [clockwise from=180]
child {node[concept,color=red!25!brown,text=white] {Algebraic Geometry}}}
child {node[concept] {Linear}}}
node[extra concept,concept color=black] at (200:5) {Applied Mathematics}
child[grow=145,concept color=black!50!yellow] {
node[concept] {Probability} [clockwise from=180]
child {node[concept] {Stochastic Processes}}}
child[grow=175,concept color=black!50!yellow] {node[concept] {Statistics}}
child[grow=205,concept color=black!50!yellow] {node[concept] {Numerical Analysis}}
child[grow=235,concept color=black!50!yellow] {node[concept] {Symbolic Computation}};
\end{tikzpicture}
## Which one is the fake?
“Crab on its back” “Willows at sunset” “Still life: Potatoes in a yellow dish”
Well, ok, it is not my book technically, but I am one of the authors of one of the chapters. And no, as far as I know, I don’t get a dime of the sales in concept of copyright or anything else.
As the title suggests (Modeling Nanoscale Imaging in Electron Microscopy), this book presents some recent advances that have been made using mathematical methods to resolve problems in electron microscopy. With improvements in hardware-based aberration software significantly expanding the nanoscale imaging capabilities of scanning transmission electron microscopes (STEM), these mathematical models can replace some labor intensive procedures used to operate and maintain STEMs. This book, the first in its field since 1998, covers relevant concepts such as super-resolution techniques (that’s my contribution!), special de-noising methods, application of mathematical/statistical learning theory, and compressed sensing.
We even got a nice review in Physics Today by Les Allen, no less!
Imaging with electrons, in particular scanning transmission electron microscopy (STEM), is now in widespread use in the physical and biological sciences. And its importance will only grow as nanotechnology and nano-Biology continue to flourish. Many applications of electron microscopy are testing the limits of current imaging capabilities and highlight the need for further technological improvements. For example, high throughput in the combinatorial chemical synthesis of catalysts demands automated imaging. The handling of noisy data also calls for new approaches, particularly because low electron doses are used for sensitive samples such as biological and organic specimens.
Modeling Nanoscale Imaging in Electron Microscopy addresses all those issues and more. Edited by Thomas Vogt and Peter Binev at the University of South Carolina (USC) and Wolfgang Dahmen at RWTH Aachen University in Germany, the book came out of a series of workshops organized by the Interdisciplinary Mathematics Institute and the NanoCenter at USC. Those sessions took the unusual but innovative approach of bringing together electron microscopists, engineers, physicists, mathematicians, and even a philosopher to discuss new strategies for image analysis in electron microscopy.
In six chapters, the editors tackle the ambitious challenge of bridging the gap between high-level applied mathematics and experimental electron microscopy. They have met the challenge admirably. I believe that high-resolution electron microscopy is at a point where it will benefit considerably from an influx of new mathematical approaches, daunting as they may seem; in that regard Modeling Nanoscale Imaging in Electron Microscopy is a major step forward. Some sections present a level of mathematical sophistication seldom encountered in the experimentally focused electron-microscopy literature.
The first chapter, by philosopher of science Michael Dickson, looks at the big picture by raising the question of how we perceive nano-structures and suggesting that a Kantian approach would be fruitful. The book then moves into a review of the application of STEM to nanoscale systems, by Nigel Browning, a leading experimentalist in the field, and other well-known experts. Using case studies, the authors show how beam-sensitive samples can be studied with high spatial resolution, provided one controls the beam dose and establishes the experimental parameters that allow for the optimum dose.
The third chapter, written by image-processing experts Sarah Haigh and Angus Kirkland, addresses the reconstruction, from atomic-resolution images, of the wave at the exit surface of a specimen. The exit surface wave is a fundamental quantity containing not only amplitude (image) information but also phase information that is often intimately related to the atomic-level structure of the specimen. The next two chapters, by Binev and other experts, are based on work carried out using the experimental and computational resources available at USC. Examples in chapter four address the mathematical foundations of compressed sensing as applied to electron microscopy, and in particular high-angle annular dark-field STEM. That emerging approach uses randomness to extract the essential content from low-information signals. Chapter five eloquently discusses the efficacy of analyzing several low-dose images with specially adapted digital-image-processing techniques that allow one to keep the cumulative electron dose low and still achieve acceptable resolution.
The book concludes with a wide-ranging discussion by mathematicians Amit Singer and Yoel Shkolnisky on the reconstruction of a three-dimensional object via projected data taken at random and initially unknown object orientations. The discussion is an extension of the authors’ globally consistent angular reconstitution approach for recovering the structure of a macromolecule using cryo-electron microscopy. That work is also applicable to the new generation of x-ray free-electron lasers, which have similar prospective applications, and illustrates nicely the importance of applied mathematics in the physical sciences.
Modeling Nanoscale Imaging in Electron Microscopy will be an important resource for graduate students and researchers in the area of high-resolution electron microscopy.
(Les J. Allen, Physics Today, Vol. 65 (5), May, 2012)
## Naïve Bayes
There is nothing naïve about Naïve Bayes—a very basic, but extremely efficient data mining method to take decisions when a vast amount of data is available. The name comes from the fact that this is the simplest application to this problem, upon (the naïve) assumption of independence of the events. It is based on Bayes’ rule of conditional probability: If you have a hypothesis $H$ and evidence $E$ that bears on that hypothesis, then
$\mathrm{Pr} \big( H \lvert E \big) = \displaystyle{ \frac{\mathrm{Pr} \big( E \lvert H\big) \mathrm{Pr}(H)}{\mathrm{Pr}(E)} }$
where as usual, $\mathrm{Pr}(A)$ denotes the probability of the event $A,$ and $\mathrm{Pr}\big( A \lvert B \big)$ denotes the probability of the event $A$ conditional to another event $B.$
I would like to show an example of this technique, of course, with yet another decision-making algorithm oriented to guess my reaction to a movie I have not seen before. From the data obtained in a previous post, I create a simpler table with only those movies that have been scored more than 28 times (by a pool of 87 of the most popular critics featured in www.metacritics.com) [I posted the script to create that table at the end of the post]
Let’s test it:
>>> table=prepTable(scoredMovies,28)
>>> len(table)
49
>>> [entry[0] for entry in table]
[‘rabbit-hole’, ‘carnage-2011’, ‘star-wars-episode-iii—revenge-of-the-sith’,
‘shame’, ‘brokeback-mountain’, ‘drive’, ‘sideways’, ‘salt’,
‘the-lord-of-the-rings-the-return-of-the-king’, ‘true-grit’, ‘inception’,
‘hereafter’, ‘master-and-commander-the-far-side-of-the-world’, ‘batman-begins’,
‘harry-potter-and-the-deathly-hallows-part-2’, ‘the-artist’, ‘the-fighter’,
‘larry-crowne’, ‘the-hunger-games’, ‘the-descendants’, ‘midnight-in-paris’,
‘moneyball’, ‘8-mile’, ‘the-departed’, ‘war-horse’,
‘the-lord-of-the-rings-the-fellowship-of-the-ring’, ‘j-edgar’,
‘the-kings-speech’, ‘super-8’, ‘robin-hood’, ‘american-splendor’, ‘hugo’,
‘eternal-sunshine-of-the-spotless-mind’, ‘the-lovely-bones’, ‘the-tree-of-life’,
‘the-pianist’, ‘the-ides-of-march’, ‘the-quiet-american’, ‘alexander’,
‘lost-in-translation’, ‘seabiscuit’, ‘catch-me-if-you-can’, ‘the-avengers-2012’,
‘the-social-network’, ‘closer’, ‘the-girl-with-the-dragon-tattoo-2011’]
>>> table[0]
[‘rabbit-hole’, ”, ‘B+’, ‘B’, ”, ‘C’, ‘C+’, ”, ‘F’, ‘B+’, ‘F’, ‘C’, ‘F’, ‘D’,
”, ”, ‘A’, ”, ”, ”, ”, ‘B+’, ‘C+’, ”, ”, ”, ”, ”, ”, ‘C+’, ”, ”,
”, ”, ”, ”, ‘A’, ”, ”, ”, ”, ”, ‘A’, ”, ”, ‘B+’, ‘B+’, ‘B’, ”, ”,
”, ‘D’, ‘B+’, ”, ”, ‘C+’, ”, ”, ”, ”, ”, ”, ‘B+’, ”, ”, ”, ”, ”,
”, ‘A’, ”, ”, ”, ”, ”, ”, ”, ‘D’, ”, ”,’C+’, ‘A’, ”, ”, ”, ‘C+’, ”]
## Math still not the answer
May 16, 2012 1 comment
I wrote a quick (but not very elegant) python script to retrieve locally enough data from www.metacritic.com for pattern recognition purposes. The main goal is to help me decide how much I will enjoy a movie, before watching it. I included the script at the end of the post, in case you want to try it yourself (and maybe improve it too!). It takes a while to complete, although it is quite entertaining to see its progress on screen. At the end, it provides with two lists of the same length: critics—a list of str containing the names of the critics; and scoredMovies—a list of dict containing, at index k, the evaluation of all the movies scored by the critic at index k in the previous list.
For example:
>>> critics[43]
‘James White’
>>> scoredMovies[43]
{‘hall-pass’: 60, ‘the-karate-kid’: 60, ‘the-losers’: 60,
‘the-avengers-2012’: 80, ‘the-other-guys’: 60, ‘shrek-forever-after’: 80,
‘the-lincoln-lawyer’: 80, ‘the-company-men’: 60, ‘jonah-hex’: 40,
‘arthur’: 60, ‘vampires-suck’: 20, ‘american-reunion’: 40,
‘footloose’: 60, ‘real-steel’: 60}
The number of scored films by critic varies: there are individuals that gave their opinion on a few dozen movies, and others that took the trouble to evaluate up to four thousand flicks! Note also that the names of the movies correspond with their web pages in www.metacritic.com. For example, to see what critics have to say about the “Karate Kid” and other relevant information online, point your browser to www.metacritic.com/movie/the-karate-kid. It also comes in very handy if there are several versions of a single title: Which “Karate Kid” does this score refer to, the one in the eighties, or Jackie Chan’s?
Feel free to download a copy of the resulting data [here] (note it is a large file: 1.6MB).
But the fact that we have that data stored locally allows us to gather that information with simple python commands, and perform many complex operations on it.
## So you want to be an Applied Mathematician
The way of the Applied Mathematician is one full of challenging and interesting problems. We thrive by association with the Pure Mathematician, and at the same time with the no-nonsense, hands-in, hard-core Engineer. But not everything is happy in Applied Mathematician land: every now and then, we receive the disregard of other professionals that mistake either our background, or our efficiency at attacking real-life problems.
I heard from a colleague (an Algebrist) complains that Applied Mathematicians did nothing but code solutions of partial differential equations in Fortran—his skewed view came up after a naïve observation of a few graduate students working on a project. The truth could not be further from this claim: we do indeed occasionally solve PDEs in Fortran—I give you that—and we are not ashamed to admit it. But before that job has to be addressed, we have gone through a great deal of thinking on how to better code this simple problem. And you would not believe the huge amount of deep Mathematics that are involved in this journey: everything from high-level Linear Algebra, Calculus of Variations, Harmonic Analysis, Differential Geometry, Microlocal Analysis, Functional Analysis, Dynamical Systems, the Theory of Distributions, etc. Not only are we familiar with the basic background on all those fields, but also we are supposed to be able to perform serious research on any of them at a given time.
My soon-to-be-converted Algebrist friend challenged me—not without a hint of smugness in his voice—to illustrate what was my last project at that time. This was one revolving around the idea of frames (think of it as redundant bases if you please), and needed proving a couple of inequalities involving sequences of functions in $L_p$—spaces, which we attacked using a beautiful technique: Bellman functions. About ninety minutes later he conceded defeat in front of the board where the math was displayed. He promptly admitted that this was no Fortran code, and showed a newfound respect and reverence for the trade.
It doesn’t hurt either that the kind of problems that we attack are more likely to attract funding. And collaboration. And to be noticed in the press.
Alright, so some of you are sold already. What is the next step? I am assuming that at his point you own your Calculus, Analysis, Probability and Statistics, Linear Programming, Topology, Geometry, Physics and you are able to solve most known ODEs. From here, as with any other field, my recommendation is to slowly build a Batman belt: acquire and devour a sequence of books and scientific articles, until you are very familiar with their contents. When facing a new problem, you should be able to recall from your Batman belt what technique could work best, in which book(s) you could get some references, and how it has been used in the past for related problems.
Following these lines, I have included below an interesting collection with the absolutely essential books that, in my opinion, every Applied Mathematician should start studying:
## Basic Statistics in sage
$\begin{array}{rrrrrr} 1.7&0.7&-0.1&-0.7&-3.5&14.9\\3.5&17.1&-7.6&1.6&11.7&6.1\\1.1&-4.0&20.9&-9.1&2.1&1.4\\-0.3&-3.7&-1.4&-0.8&2.4&12.6\\1.9&3.9&0.1&15.4&-0.7\end{array}$ | 2017-11-18 02:41:15 | {"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": 11, "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": 0.4993400573730469, "perplexity": 2551.3052752629987}, "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/1510934804518.38/warc/CC-MAIN-20171118021803-20171118041803-00143.warc.gz"} |
http://hal.in2p3.fr/in2p3-01088477 | Large scale distribution of ultra high energy cosmic rays detected at the Pierre Auger Observatory with zenith angles up to 80$^\circ$
Abstract : We present the results of an analysis of the large angular scale distribution of the arrival directions of cosmic rays with energy above 4 EeV detected at the Pierre Auger Observatory including for the first time events with zenith angle between $60^\circ$ and $80^\circ$. We perform two Rayleigh analyses, one in the right ascension and one in the azimuth angle distributions, that are sensitive to modulations in right ascension and declination, respectively. The largest departure from isotropy appears in the $E > 8$ EeV energy bin, with an amplitude for the first harmonic in right ascension $r_1^\alpha =(4.4 \pm 1.0){\times}10^{-2}$, that has a chance probability $P(\ge r_1^\alpha)=6.4{\times}10^{-5}$, reinforcing the hint previously reported with vertical events alone.
Type de document :
Article dans une revue
Astrophysical Journal, American Astronomical Society, 2015, 802 (2), pp.111. 〈10.1088/0004-637X/802/2/111〉
http://hal.in2p3.fr/in2p3-01088477
Contributeur : Emmanuelle Vernay <>
Soumis le : vendredi 28 novembre 2014 - 09:55:35
Dernière modification le : mardi 19 juin 2018 - 01:20:03
Citation
A. Aab, M. Avenier, C. Berat, S. Le Coz, D. Lebrun, et al.. Large scale distribution of ultra high energy cosmic rays detected at the Pierre Auger Observatory with zenith angles up to 80$^\circ$. Astrophysical Journal, American Astronomical Society, 2015, 802 (2), pp.111. 〈10.1088/0004-637X/802/2/111〉. 〈in2p3-01088477〉
Métriques
Consultations de la notice | 2018-08-21 00:54:51 | {"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": 0.6990841627120972, "perplexity": 2874.322891555234}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-34/segments/1534221217901.91/warc/CC-MAIN-20180820234831-20180821014831-00395.warc.gz"} |
https://cadabra.science/manual/collect_terms.html | Collect terms in a sum that differ only by their numerical pre-factor. This is part of the default post_process function, so does not need to be called by hand. Note that this command only collects terms which are identical, it does not collect terms which are different but mathematically equivalent. See sort_sum for an example. | 2019-04-23 19:56:32 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.642247200012207, "perplexity": 397.5536562411565}, "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-2019-18/segments/1555578613603.65/warc/CC-MAIN-20190423194825-20190423220825-00445.warc.gz"} |
https://math.stackexchange.com/questions/3097629/why-is-the-fourier-transform-called-a-transform-and-not-a-transformation | # Why is the Fourier transform called a 'transform', and not a 'transformation'?
Why are the Fourier transform, Laplace Transform, etc called transforms, and not transformations?
This is about linguistics or terminology in mathematics. I feel there should be a reason why the word 'transform' used for such mathematical objects.
Is 'transform' as a noun, an invented word (coined word) for mathematics?
I have to name a concept in the context of functional programming, and want to know which name is suitable for it:
transform, transformation, map, homeomorphism, morphism, etc.
My name needs to be defendable for people in management and BAs (which are not Mathematicians).
I think operation which assigns to the function $$x \mapsto f(x)$$ another function $$p \mapsto \int f(x) e^{-ipx} dx = \widetilde {f}(p)$$ should be called Fourier transformation, as you suggest. On the other hand the result of this operation, i.e. the function $$\widetilde f$$, is called the Fourier transform (a noun) of $$f$$.
• So, is this word coined only to distinguish two different meaning that you described? Is it a linguistic derivative to create a noun from the verb 'transforming'? My question is perhaps more about linguistics: transformation is also a noun created from a verb. – Sohail Si Feb 2 at 19:39
Is 'transform' as a noun, an invented word (coined word) for mathematics?
Apparently so. The Oxford English Dictionary has "transform" (noun) first used by Sylvester in 1853.
SYLVESTER in Philos. Trans. (Royal Soc.) 143 i. 544 Covariant, a function which stands in the same relation to the primitive function from which it is derived as any of its linear transforms do to a similarly derived transform of its primitive.
Since 1965, "transform" (noun) is also used in geology.
• Why was that word used by Sylvester? and from which context is it borrowed? Perhaps geology? – Sohail Si Feb 2 at 19:38
• Etymology: transform (noun) comes from transform (verb). – GEdgar Feb 2 at 22:22 | 2019-04-22 16:46:50 | {"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": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 4, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7267759442329407, "perplexity": 2117.3976155202445}, "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-2019-18/segments/1555578558125.45/warc/CC-MAIN-20190422155337-20190422181337-00416.warc.gz"} |
https://stats.stackexchange.com/questions/207979/forecasting-gdp-using-regression-arima-and-ets | # Forecasting GDP using regression, ARIMA and ETS
I am building a simple model that estimates future change in GDP growth using change in working-age population (%).
$$\Delta GDP_t = \beta_0 + \beta_1 \Delta Pop_{t-1} + \varepsilon_t.$$
I have run a linear regression on Japanese data, and I got a significant $R^2$ (0.45). There isn't any pattern in the residuals.
The next step I will do is use ARIMA/ETS in R to forecast future values in the working-age population, and plug this data into the model to predict future GDP growth.
Does my approach make sense from a statistical point of view? Do you think there is a more logical approach?
• Why not use arima/ets for GDP directly? – hejseb Apr 18 '16 at 19:16
• @hejseb, sure, that could work at least as a benchmark. But if there is another relevant variable, why not try including it, too? – Richard Hardy Apr 18 '16 at 19:39
• @RichardHardy Evidently there are many variables which are helpful in forecasting GDP. Just using Pop at t-1 feels a little bit odd, so there is probably some reason for this. Knowing this may make it easier to answer the question properly. – hejseb Apr 19 '16 at 4:01
1. You may think about omitted variable bias. If there are other variables influencing/determining the GDP growth, and those variables are correlated with the change in the working-age population, you will have a biased and inconsistent estimate of $\beta_1$, and hence poor forecasts. | 2020-12-01 14:54:11 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 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": 0.43163371086120605, "perplexity": 1083.6629362567019}, "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-2020-50/segments/1606141674594.59/warc/CC-MAIN-20201201135627-20201201165627-00037.warc.gz"} |
http://klickverbot.at/blog/2010/06/oh-how-glad-i-am-that-actionscript-2-is-dead-and-buried/ | # Oh, how glad I am that ActionScript 2 is dead and buried…
… but unfortunately not in a personal project of mine that I started quite a while ago and which I have resumed work on recently.
Today, I have been finally able to fix a bug which had already taken me some hours to trace down. Basically, mouse-over and -off events would not work properly on certain MovieClips. After some digging through my custom code managing these events (which I needed to come up with because there is no way to let hover events bubble up the display hierarchy in ActionScript 2), I found that hitTest() wouldn’t work properly on these clips.
Now the fun part began. I meticulously checked every aspect of the MovieClips for anything special, I even considered that it might have something to do with the fact that they were positioned right behind some TextFields, which could have triggered some Flash player bugs (given that they are already surrounded by a cloud of weirdness in AS2). Nope, nothing.
It wasn’t until I had already pretty much given up that I noticed that the name of the clips in question contained a period. After I removed it … hitTest() worked fine – thanks a lot for wasting my time, Adobe! Not only that this really should not happen, at least not without a runtime warning, the fact that you cannot use periods in MovieClip names is apparently undocumented.
Oh well… | 2019-03-25 04:30:34 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 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": 0.3847690224647522, "perplexity": 1163.5545349488543}, "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/1552912203548.81/warc/CC-MAIN-20190325031213-20190325053213-00003.warc.gz"} |
https://www.nde-ed.org/Physics/Waves/MechWaveCharacteristics.xhtml | # Mechanical Wave Characteristics
### After reading this section you will be able to do the following:
• Define periodic wave parameters like wavelength and period.
• Visualize periodic transverse waves and periodic longitudinal waves.
When the disturbance causing the transferse or longitudinal wave is repetitive (periodic), a periodic wave occurs.
### Periodic Transverse Wave
When periodic waves have simple harmonic motion they can also be called sinusoidal waves and any periodic wave can be described by a combination of sinusoidal waves.
As a sinusoidal wave travels each particle undergoes simple harmonic motion with the same frequency, but does not travel with the wave. It is important to distinguish between the particle motion and the wave motion since waves transport energy, but not matter. The figure below illustrates a traveling periodic transverse wave and points out different features of the wave like its wavelength and amplitude
The wavelength (represented by the Greek letter lambda) of a periodic wave is the physical distance between two same points in the pattern (e.g., the distance between two crests or the distance between two troughs). Since wavelength is a distance it is measured in distance units, typically meters. As a wave travels at a constant speed it advances 1 wavelength in the time of 1 period. The period of a wave, which is denoted by T, is measured in seconds and the wave speed, which is the speed the wave travels at in its medium denoted by v, is measured in meters per second. Frequency is the number of wave cycles that will pass a point in 1 second. It is measured in the unit Hertz (Hz).
$v={\lambda \over T}={\lambda f}$
${\lambda: wavelength}$
### Periodic Longitudinal Wave
Longitudinal waves can also be periodic. As a longitudinal wave travels there are regions of high particle density or pressure and regions of low particle density or pressure. These regions are called compression and rarefaction, respectively, and are shown in the figure below. Just like with a transverse periodic wave, a periodic longitudinal wave has a wavelength that is the physical distance between two same points in the pattern. In this case the wavelength can be defined as the distance between two compression regions or two rarefaction regions.
### Review:
1. When periodic waves have simple harmonic motion they can also be called sinusoidal waves.
2. Frequency is the inverse of the wave period and vice versa.
3. The wave speed is related to the wavelength and frequency of the wave. | 2021-08-01 07:13:25 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 2, "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": 0.8502020239830017, "perplexity": 397.4239247220664}, "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/1627046154163.9/warc/CC-MAIN-20210801061513-20210801091513-00581.warc.gz"} |
https://www.gradesaver.com/textbooks/math/algebra/college-algebra-7th-edition/chapter-1-equations-and-graphs-section-1-6-solving-other-types-of-equations-1-6-exercises-page-139/39 | ## College Algebra 7th Edition
$x=4$
$\sqrt{2x-1}=\sqrt{3x-5}$ $(\sqrt{2x-1})^{2}=(\sqrt{3x-5})^{2}$ $2x-1=3x-5$ $-1+5=3x-2x$ $x=4$ We need to confirm that $x=4$ works in the original equation: Left side=$\sqrt{2(4)-1}=\sqrt{8-1}=\sqrt{7}$ Right side=$\sqrt{3(4)-5}=\sqrt{12-5}=\sqrt{7}$ Both sides agree, so the solutions works. | 2018-07-22 16:50:14 | {"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": 0.2362881451845169, "perplexity": 909.8980676555923}, "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-2018-30/segments/1531676593378.85/warc/CC-MAIN-20180722155052-20180722175052-00085.warc.gz"} |
https://www.physicsforums.com/threads/algebra-questions.149121/ | # Algebra Questions
1. Dec 23, 2006
### acm
Two unrelated question I need checked.
i) Every C-Cycle, C>1, can be written as a product of transpositions. Every permutation of a finite number n>1 of elements can be expressed as a product of transpositions.
ii) (Cayley's representation theorem for groups) For every group G is isomorphic to a group of permutations A [G]=n then G is isomorphic to a group of permutations on objects.
i) a) (Proof of C >1) Every C cycle can be represented in the form (c_2,c_1), (c_3,c_1),...(c_i,c_1),....,(c_c,c_1). Hence this defines the mapping from c_1 to c_c, thus in general it suffices to say that 1 <= i < C. As there is a mapping from c_1 to c_c such that every other number in the group remains unchanged, then i=1, thus that 1=n. Hence 1 < C.
b) (Proof of n > 1) Hence from a), any C-Cycle with C>1 can be written as a product of transpositions. Thus as permutations can be represented as C cycles on a arbitary set then n > 1 is satisfied by part a) #.
ii) (Proof of one to one correpondence) Let the set A be defined as (p_1, p_2, .... , p_n,...) Hence, let A undergo the mapping p_1 -> g_1 such that the image of A under the mapping is: (p_1g_1, ..., p_ng_n,....). Thus there is an equivalence between p_n ~ (p_1g_1,....., p_ng_n).
(Proof of Isomorphism) For an isomorphism to be acting from A to the permutation set then associotivity must hold. Let p_k = (p_1g_1,....., p_1g_k), Hence if G is isomorphic to the permutation of G then p_k(p_n * c) = (p_k * p_n)c. p_k * p_n = ([p_1,.....,p_k],[p_1g_1,.....,p_kg_k,...]) * (([p_1,.....,p_n],[p_1g_1,.....,p_ng_n,....]). As the permutation group has the same binary operation as the group itself. Then there is an isomorphism between the group G and the group of permutations on the group.
I'm having trouble understanding the concept of an isomorphism, the book I'm working from defines it as a special kind of equavlence which acts on the relation between two groups, whether that is binary, etc. So the group may have different contents but if the operation holds then there is an isomorphism. Is this correct or am I doing something wrong?
2. Dec 25, 2006
### Swapnil
This question is way too complicated to be in pre-calculus forum. "Set Theory, Logic, Probability, Statistics" forum is a far better forum for these types of questions. I would ask the moderators to please move this question over there.
3. Dec 25, 2006
### HallsofIvy
Staff Emeritus
Since this is, still, a homework question, I am moving it to "Calculus and beyond".
4. Dec 25, 2006
### HallsofIvy
Staff Emeritus
I don't understand what you mean by "acts on the relation between two groups". What relation? What is it that is "binary, etc."? (The group operation is always binary.) An isomorphism between two groups is a function from one group to another that "preserves" the operation (f(a*b)= f(a)*f(b)) and invertible. Basically, it means the two groups are really just different "labelings" of the same thing. That is if f(a)= z, f(b)= y, f(c)= z, etc. and you just "relabel" "a" to be "z", "b" to be "y" etc. then the two groups will look exactly the same. One of the things that means is that while a group may have "different contents" (by which I presume that you mean the underlying sets are different) they must have exactly the same number of elements.
What "operation" are you talking about? The group operation? And what do you mean by "holds"? If you mean that f(a*b)= f(a)o f(b) where "*" is the operation in one group and "o" is the operation in the other then that defines a homomorphism, not an isomorphism since it is not necessarily invertible (one-to-one and onto).
For example, let G be the group with elements {x, y} and operation defined by x*x= x, x*y= y, y*a= y, y*y= x (so that x is the identity and y is its own inverse. Let H be the "rotation" group of order 4 with elements {a, b, c, d} and operation defined by aoa= a, aob= b, aoc= c, aod= d, boa= b, bob= c, boc= d, bod= a, coa= c, cob= d, coc= a, cod= b, doa= d, dob= a, doc= b, dod= c. The a is the identity, b and d are inverses, and c is its own inverse.
The function f(x)= a, f(y)= c is a homomorphism but not an isomorphism because it is not "onto": nothing is mapped into either b or d. Conversely, the function g(a)= x, g(b)= x, g(c)= y, g(d)= x is also a homomorphis but not an isomorphism because it is not "one-to-one". More than one element is mapped to x. If we restrict g to the subgroup {a, c} then g(a)= x, g(c)= y is an isomorphism and its inverse is f. | 2016-10-25 17:25:25 | {"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": 0.9127288460731506, "perplexity": 981.8167643452725}, "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-2016-44/segments/1476988720238.63/warc/CC-MAIN-20161020183840-00562-ip-10-171-6-4.ec2.internal.warc.gz"} |
https://www.math.ias.edu/seminars/abstract?event=113975 | # Points and lines
Members' Seminar Topic: Points and lines Speaker: Nathaniel Bottman Affiliation: Member, School of Mathematics Date: Monday, December 12 Time/Room: 1:15pm - 2:15pm/S-101 Video Link: https://video.ias.edu/membsem/2016/1212-NathanielBottman
The Fukaya category of a symplectic manifold is a robust intersection theory of its Lagrangian submanifolds. Over the past decade, ideas emerging from Wehrheim--Woodward's theory of quilts have suggested a method for producing maps between the Fukaya categories of different symplectic manifolds. I have proposed that one should consider maps controlled by compactified moduli spaces of marked parallel lines in the plane, called "2-associahedra". In this talk I will describe the 2-associahedra, with a focus on their topological and combinatorial aspects; in particular, I will produce combinatorial data that are in bijection with the strata of the 2-associahedra, and describe a generating function technique for computing the number of dimension-$m$ strata in a particular 2-associahedron. I will not assume any familiarity with symplectic geometry. | 2018-12-19 08:27:01 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7046282291412354, "perplexity": 1122.1119292090755}, "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-2018-51/segments/1544376831715.98/warc/CC-MAIN-20181219065932-20181219091932-00418.warc.gz"} |
http://math.stackexchange.com/questions/46541/why-is-the-pullback-not-just-a-cartesian-product-modulo-a-relation | # Why is the pullback not just a cartesian product (modulo a relation)?
Why is the pullback not just defined as in the case of the category of sets http://en.wikipedia.org/wiki/Pullback_%28category_theory%29
Perhaps there are some issues with categories that are not small, but what about small categories? My first thought was: perhaps all small categories do not have products. Do they? If not, why doesn't the cartesian product just do the trick?
-
What does "cartesian product" mean in an arbitrary category? – Chris Eagle Jun 20 '11 at 19:57
You probably just mean 'product' (categorical product) instead of Cartesian product. Anyway, I don't understand the question: the pullback is defined as it is. In SET it happens that we can describe it as a quotient of a product. Perhaps you mean something like: can we do the same in any category which has products? But I don't know what a 'relation' in an arbitrary category is. – wildildildlife Jun 20 '11 at 20:08
@wildildildlife: it's a subobject of a product, not a quotient. – Chris Eagle Jun 20 '11 at 20:25
There are a few problems here:
1. Not every category is sets with additional structure. Many familiar examples are: Groups, topological spaces, and just about every other "elementary" examples of categories that you came across when you first learned what a category was. These categories are all examples of concrete categories, which are categories equipped with a faithful functor to sets. Even though concrete categories are more general than "sets with structure", not every category is concrete. For example, the category of topological spaces where morphisms are homotopy classes of continuous maps cannot be made into a concrete category. Also, there is no faithful functor from $Set^{op}$ to $Set$ which is the identity on objects, and so if there is a way to turn $Set^{op}$ into a concrete category (which I do not believe there is), it is at least non-obvious.
2. Even if a category was just sets with additional structure, the pullback of two sets might not have the structure required. If you have a concrete category and the forgetful functor is right adjoint to some "free" functor, then because right adjoints preserve limits, you will in fact have that the underlying set of the pullback is the pullback of the underlying sets. However, colimits won't work out so well.
3. Even if you are in a category of structured sets, and even if the set theoretic pullback is still in your category, you might not have that universal property that you want to have, which would take away some of the usefulness. Of course, this doesn't happen when the forgetful functor is part of an adjunction as above, and I am hard pressed to think of any examples (because it is much more likely that things break in spectacular ways than in subtle ways).
4. Not every category even has products. For example, in the category of fields, all morphismsare field extensions, and so to have projection map, the product of two fields would have to be a common subfield. However, unless you consider the field with one element to be a field, there are no common subfields of two fields of different characteristic. From a different perspective, the ring that you get by taking the Cartesian product of two fields is NOT a field.
-
$Set^{op}$ is concretized by the functor that is powerset on objects and inverse image on morphisms. – Chris Eagle Jun 20 '11 at 21:03
@Chris Eagle: Thank you, that's very nice. – Aaron Jun 20 '11 at 21:14
Your intuition is right in the sense that pullbacks can be constructed from products and equalizers, see relation between pullbacks and other categorical limits, proposition 2. Products are generalizations of Cartesian products of sets. Equalizers are generalizations of specific subsets; I do not understand what is “modulo a relation”, but I feel equalizers are what you need. :)
- | 2016-07-28 16:40:32 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.9119521975517273, "perplexity": 234.95919055526235}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-30/segments/1469257828286.80/warc/CC-MAIN-20160723071028-00098-ip-10-185-27-174.ec2.internal.warc.gz"} |
https://www.semanticscholar.org/paper/Cohomology-Theories-Brown/d7ebbc1cfe28069b90e865c875b8979adbf7a36a?p2df | # Cohomology Theories
@inproceedings{Brown2010CohomologyT,
title={Cohomology Theories},
author={Edgar H. Brown},
year={2010}
}
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## References
SHOWING 1-10 OF 11 REFERENCES
Fibre bundles, an integral part of differential geometry, are also important to physics. This text, a succint introduction to fibre bundles, includes such topics as differentiable manifolds and
### f1 Ky go and ?1 play the role of Zf 1,f2, Sn, C0 and C1. Suppose ?i: go -(S is a covariant functor. Axiom h*. If f and g: X -Y are homotopic, =(f) ==(g)
• f1 Ky go and ?1 play the role of Zf 1,f2, Sn, C0 and C1. Suppose ?i: go -(S is a covariant functor. Axiom h*. If f and g: X -Y are homotopic, =(f) ==(g)
• Mathematics
Proceedings of the National Academy of Sciences of the United States of America
• 1960
### First, it is difficult to recover =1(X) from [X, ]; and second, annihilating cohomology classes is more difficult than annihilating homotopy classes
• First, it is difficult to recover =1(X) from [X, ]; and second, annihilating cohomology classes is more difficult than annihilating homotopy classes
### If ir satisfies h*, e* and c*, then there is an Xe 1?, unique up to homotopy type, and a natural equivalence
• BRANDEIS UNIVERSITY BIBLIOGRAPHY
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### On spaces having the same homotopy type as a Cw-complex
• Trans. Amer. Math. Soc
• 1959
### and, by the exact sequence of the triple (SX, X-, x), Hq+l(SX, X-)Hq+l(SX, x0). Clearly all these isomorphisms are natural. Let p be the base point of Sn
• By excision, Hq(p) Hq(SO | 2023-03-21 16:38:40 | {"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": 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": 0.6589617133140564, "perplexity": 2614.126005616938}, "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-2023-14/segments/1679296943704.21/warc/CC-MAIN-20230321162614-20230321192614-00421.warc.gz"} |
http://www.ms.lt/sodas/Book/Duality | • ms@ms.lt
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• My work is in the Public Domain for all to share freely.
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Study duality as the basis of logic, and mathematics as ways of altering duality.
(internal structure mirrors external structure - duality of category theory)
I am studying the various case of duality in math. I imagine that at the heart is the duality between zero and infinity by way of one as in God's Dance.
Duality breaking
The duality between zero and infinity, between nothing and everything, is broken in many subtle ways. Here are some examples:
• By definition, a topological space includes both an entire set X and the empty set. However, the intermediary sets are closed under arbitrary unions, but only finite intersections. What would happen if they were closed under infinite intersections?
• Perhaps similarly, having in mind the Zariski topology, ideals of a ring are defined with respect to multiplication (union) but not addition (intersection).
Equivalence and uniqueness
In Math, there is an everpresent tension between the notions of equivalence class and uniqueness. If something is mathematically significant, it should in some sense be unique. But math is a model and so, as such, can never be entirely unique but represents a variety of cases. Thus it is ever natural to define equivalence classes, especially in math itself. For example, a rational number is an equivalence class that establishes a proportion.
• Langlands program
Dualities. Duality arises from a symmetry between two ways of looking at something where there is no reason to choose one over the other. For example:
• Square roots of -i. There are two square roots of -1. One we call +i, the other -i, but neither should have priority over the other. Similarly, clockwise and counterclockwise rotations should not be favored. Complex conjugation is a way of asserting this. (Note that the integer +1 is naturally favored over -1. But there is no such natural favoring for i. It is purely conventional, a misleading artificial contrivance.)
• A rectangular matrix can be written out from left to right or right to left. So we have the transpose matrix.
• Normality says conjugate invariancy: gN = Ng.
• Opposite category? Morphisms can be organized from left to right or from right to left. The opposite category turns all of the arrows around.
• Colimits and limits
• Monomorphisms ("one-to-one") and epimorphisms (forcing "onto").
• Coproducts and products
• Initial and terminal objects
• Wikipedia: In applications to logic, this then looks like a very general description of negation (that is, proofs run in the opposite direction). If we take the opposite of a lattice, we will find that meets and joins have their roles interchanged. This is an abstract form of De Morgan's laws, or of duality applied to lattices.
• Wikipedia: Reversing the direction of inequalities in a partial order. (Partial orders correspond to a certain kind of category in which Hom(A,B) can have at most one element.)
• Wikipedia: Fibrations and cofibrations are examples of dual notions in algebraic topology and homotopy theory. In this context, the duality is often called Eckmann–Hilton duality.
• Adjoint bendrai ir Adjoint functors. Wikipedia: It can be said that an adjoint functor is a way of giving the most efficient solution to some problem via a method which is formulaic. A construction is most efficient if it satisfies a universal property, and is formulaic if it defines a functor. Universal properties come in two types: initial properties and terminal properties. Since these are dual (opposite) notions, it is only necessary to discuss one of them.
• Switching of "existing" and "nonexisting", for example, edges in a graph. This underlies Ramsey's theorem. Tao: "the Ramsey-type theorem, each one of which being a different formalisation of the newly gained insight in mathematics that complete disorder is impossible."
• Coordinate systems can be organized "bottom up" or "top down". This yields the duality in projective geometry.
• Root systems relate reflections (hyperplanes) and root vectors. Given a root R, reflecting across its hyperplane, every root S is taken to another root -S, and the difference between the two roots is an integer multiple of R. But this relates to the commutator sending the differences into the module based on R.
• Analysis provides lower and upper bounds on a function or phenomenon which helps define the geometry of this space.
• We can look at the operators that act or the objects they act upon. This brings to mind the two representations of the foursome.
• This is related to the duality between left and right multiplication. Examples include Polish notation.
• Faces of an object and corners of an object. (Why are they dual?)
• Coxeter groups are built from reflections. Reflections are dualities.
• Any two structures which have a nice map from one to the other have a duality in that you can start from one and go to the other.
• Galois theory: field extensions (solutions of polynomials) and groups
• Lie groups: solutions to differential equations..
Read nLab: Duality. Here are examples to consider:
• Duality (projective geometry). Interchange the role of "points" and "lines" to get a dual truth: The plane dual statement of "Two points are on a unique line" is "Two lines meet at a unique point". (Compare with the construction of an equilateral triangle and the lattice of conditions.)
• Atiyah-Singer index theorem...
• Riemann-Roch theorem
• Covectors and vectors
• Cotangent space and tangent space
• de Rham cohomology links algebraic topology and differential topology
• Modularity theorem.
• Langlands program
• general Stokes theorem: duality between the boundary operator on chains and the exterior derivative
• Hilbert's Nullstellensatz
• Class field theory provides a one-to-one correspondence between finite abelian extensions of a fixed global field and appropriate classes of ideals of the field or open subgroups of the idele class group of the field.
• Lie's idée fixe was to develop a theory of symmetries of differential equations that would accomplish for them what Évariste Galois had done for algebraic equations: namely, to classify them in terms of group theory. Lie and other mathematicians showed that the most important equations for special functions and orthogonal polynomials tend to arise from group theoretical symmetries. In Lie's early work, the idea was to construct a theory of continuous groups, to complement the theory of discrete groups that had developed in the theory of modular forms, in the hands of Felix Klein and Henri Poincaré. The initial application that Lie had in mind was to the theory of differential equations. On the model of Galois theory and polynomial equations, the driving conception was of a theory capable of unifying, by the study of symmetry, the whole area of ordinary differential equations. However, the hope that Lie Theory would unify the entire field of ordinary differential equations was not fulfilled. Symmetry methods for ODEs continue to be studied, but do not dominate the subject. There is a differential Galois theory, but it was developed by others, such as Picard and Vessiot, and it provides a theory of quadratures, the indefinite integrals required to express solutions.
• One may ask analytic questions about algebraic numbers, and use analytic means to answer such questions; it is thus that algebraic and analytic number theory intersect. For example, one may define prime ideals (generalizations of prime numbers in the field of algebraic numbers) and ask how many prime ideals there are up to a certain size. This question can be answered by means of an examination of Dedekind zeta functions, which are generalizations of the Riemann zeta function, a key analytic object at the roots of the subject.[79] This is an example of a general procedure in analytic number theory: deriving information about the distribution of a sequence (here, prime ideals or prime numbers) from the analytic behavior of an appropriately constructed complex-valued function.
• Meromorphic function is the quotient of two holomorphic functions, thus compares them.
• Isbell duality relates higher geometry with higher algebra.
• Topos links geometry and logic.
• For integers, decomposition into primes is a "bottom up" result which states that a typical number can be compactly represented as the product of its prime components. The "top down" result is that this depends on an infinite number of exceptions ("primes") for which this compact representation does not make them more compact.
• The two facts that this method of turning rngs into rings is most efficient and formulaic can be expressed simultaneously by saying that it defines an adjoint functor. Continuing this discussion, suppose we started with the functor F, and posed the following (vague) question: is there a problem to which F is the most efficient solution? The notion that F is the most efficient solution to the problem posed by G is, in a certain rigorous sense, equivalent to the notion that G poses the most difficult problem that F solves.
• https://en.m.wikipedia.org/wiki/Coherent_duality https://en.m.wikipedia.org/wiki/Serre_duality https://en.m.wikipedia.org/wiki/Verdier_duality https://en.m.wikipedia.org/wiki/Poincaré_duality
• https://en.m.wikipedia.org/wiki/Dual_polyhedron
• a very general comment of William Lawvere[2] is that syntax and semantics are adjoint: take C to be the set of all logical theories (axiomatizations), and D the power set of the set of all mathematical structures. For a theory T in C, let F(T) be the set of all structures that satisfy the axioms T; for a set of mathematical structures S, let G(S) be the minimal axiomatization of S. We can then say that F(T) is a subset of S if and only if T logically implies G(S): the "semantics functor" F is left adjoint to the "syntax functor" G.
• division is (in general) the attempt to invert multiplication, but many examples, such as the introduction of implication in propositional logic, or the ideal quotient for division by ring ideals, can be recognised as the attempt to provide an adjoint.
• Tensor products are adjoint to a set of homomorphisms.
• Duality - parity - išsiaiškinimo rūšis. Įvairios simetrijos - išsiaiškinimo būdų sandaros.
• In mathematics, monstrous moonshine, or moonshine theory, is a term devised by John Conway and Simon P. Norton in 1979, used to describe the unexpected connection between the monster group M and modular functions, in particular, the j function. It is now known that lying behind monstrous moonshine is a vertex operator algebra called the moonshine module or monster vertex algebra, constructed by Igor Frenkel, James Lepowsky, and Arne Meurman in 1988, having the monster group as symmetries. This vertex operator algebra is commonly interpreted as a structure underlying a conformal field theory, allowing physics to form a bridge between two mathematical areas. The conjectures made by Conway and Norton were proved by Richard Borcherds for the moonshine module in 1992 using the no-ghost theorem from string theory and the theory of vertex operator algebras and generalized Kac–Moody algebras.
• Alexander duality
• Alvis–Curtis duality
• Araki duality
• Beta-dual space
• Coherent duality
• De Groot dual
• Dual abelian variety
• Dual basis in a field extension
• Dual bundle
• Dual curve
• Dual (category theory)
• Dual graph
• Dual group
• Dual object
• Dual pair
• Dual polygon
• Dual polyhedron
• Dual problem
• Dual representation
• Dual q-Hahn polynomials
• Dual q-Krawtchouk polynomials
• Dual space
• Dual topology
• Dual wavelet
• Duality (optimization)
• Duality (order theory)
• Duality of stereotype spaces
• Duality (projective geometry)
• Duality theory for distributive lattices
• Dualizing complex
• Dualizing sheaf
• Esakia duality
• Fenchel's duality theorem
• Haag duality
• Hodge dual
• Jónsson–Tarski duality
• Lagrange duality
• Langlands dual
• Lefschetz duality
• Local Tate duality
• Poincaré duality
• Twisted Poincaré duality
• Poitou–Tate duality
• Pontryagin duality
• S-duality (homotopy theory)
• Schur–Weyl duality
• Serre duality
• Stone's duality
• Tannaka–Krein duality
• Verdier duality
• AGT correspondence
• A "transformation group" is a group acting as transformations of some set S. Every transformation group is the group of all permutations preserving some structure on S, and this structure is essentially unique. The bigger the transformation group, the less structure: symmetry and structure are dual, just like "entropy" and "information", or "relativity" and "invariance".
Duality of silence (top-down) and speaking (bottom-up).
#### Duality
Naujausi pakeitimai
Puslapis paskutinį kartą pakeistas 2017 spalio 17 d., 10:36 | 2017-10-17 22:20:34 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8005790114402771, "perplexity": 969.2451083585888}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-43/segments/1508187822513.17/warc/CC-MAIN-20171017215800-20171017235800-00636.warc.gz"} |
https://www.transtutors.com/questions/idk-149111.htm | # idk
If the price paid per unit differs from the standard price per unit for direct materials, the variance is termed:
a. variable variance
b. controllable variance
c. price variance
d. volume variance
c. price variance [Materials Price Variance = (Actual quantity purchased × Actual...
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• ### accounting
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If the actual quantity of direct materials used in producing a commodity differs from the standard quantity, the variance is termed: a. controllable variance b . price variance c . quantity variance d . rate variance
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Total manufacturing cost variance includes a. Direct materials price variance, direct labor cost variance, and fixed factory overhead volume variance b . Direct materials cost variance, direct labor rate variance, and factory overhead cost variance c . Direct materials cost variance, direct labor
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TenPro reports the following on one of its products. Direct materials standard (4 lbs. @ $2/lb.)$ 8 per finished unit Actual direct materials used 150,000 lbs. Actual finished units produced 30,000 units Actual cost of direct materials used \$ 267,500 Compute the direct materials price
AQ = actual quantity = 150,000 SQ = standard quantity = standard quantity per unit times the number of units = 4 lb/unit x 30,000 units = 120,000 AP = actual price = actual cost of direct...
• ### accounting only answer question E8-5
(Solved) April 21, 2012
only solve the first question please - P11-2BE11-2P8-1AE8-5Name:Instructor:Date:Course:Solving Managerial Accounting Problems Using Microsoft Excel for Windows by Rex A SchildhouseManagerial Accounting, 3rd Edition, by Weygandt, Kieso, and
• ### contribution margin
(Solved) March 14, 2011
The contribution margin per unit expressed as a percentage of sales price is the:Question 4 options: a. volume variance. b .margin of safety. c .contribution rate. d . current ratio. please do not guess. | 2018-07-20 14:00:11 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.47400474548339844, "perplexity": 13718.981602947231}, "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-2018-30/segments/1531676591683.78/warc/CC-MAIN-20180720135213-20180720155213-00587.warc.gz"} |
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https://quant.stackexchange.com/tags/options/new | # Tag Info
1
3
In an incomplete market, vanilla options are independent assets like stocks or bonds. So the best way of thinking about how they are priced is the same way equilibrium prices in those markets occur: If too many people try to buy an option at a given strike then they push the price of those options up and we see that as the implied volatility increasing. The ...
2
If you are predicting lower one year volatility than the options are pricing in, sell one year options on the underlying that you think will be lower and hedge the delta. If you are predicting higher one year volatility than the options are pricing in, buy one year options on the underlying that you think will be higher and hedge the delta. Your hedging ...
0
In simple terms an option on one stock will have a delta between 0 and 1 meaning it will have the equivalente directional risk between 0 (or very close to zero, for example for a very OTM option close to maturity with a very low vol) and 1 (for example for a very ITM option). So you will never need more quantity of the underlying to hedge the option. As ...
1
Is there a faster way to calculate the option price? With a recombining binomial tree, the terminal asset price has a binomial distribution -- as you might have expected. For a tree with $n$ steps, the probability of reaching price $S_{n,k}$ where $k$ is the number of up moves is $$P_{n,k} = \frac{n!}{k!(n-k)!}q^k(1-q)^{n-k}$$ The option price is the ...
1
Why not? You can back out implied vol from the times you do have for the underlier price and then use that to price the options for the times you do not have. (This is assuming you are taking about pricing one particular options, not using options of one strike and expiry to price options at an other time, strike, and expiry.) You could even do a linear ...
0
Investors buy (and hold) more puts and pay up more for them for a few reasons. First, people fear downside more than they like upside as shown by Kahneman and Tversky (1979, 1992). Second, people may not be able to recover easily (or at all) from downside in the macroeconomy. In classical finance terms, if we think crises are different from times of stable ...
2
It's the interest rate component. That is more meaningful in the formula. Note that the call becomes more expensive. Think about it this way. You could buy the call and sell the put instead of being long the stock. This gives you a synthetic long position. You need to pay the market the cost of borrow (r). That makes the calls more expensive and the ...
1
Usually, you would use the volatility from a fitted volatility curve or surface. Those are based on implied volatilities. You can use historical volatility, but then your valuation is likely to be off because the volatility curve/surface is not constant and at historical vol. You should use a yield curve to present value nodes. This is unlikely to make a big ...
0
First of all, option contracts normally specify 100 (or in some cases, 1000 contracts of the underlying instrument) to the one option contract, so you are unlikely to encounter the 1:1 scenario you mention. Let's ignore the ratios for now, and look at the risk profile portfolio you described (1 long underlying and 1 short call). The portfolio you have is a ...
1
I think you have a small misunderstanding. The hedge for a call (or put) is rarely 1:1 with stock. When you are selling this option you are actually selling the future movements of the stock and specifically the future price jumps that will happen. To hedge an option you would enter into a position that will offset the movements in price of the option ...
1
I've been working on this problem a little bit lately. Unfortunately in the FX context, it's not quite as straight-forward as in the equities case, for two reasons: FX options trade OTC instead of on exchange, so you need access to broker screens to trade them (eg. on BBG) FX Options are quoted by (delta, tenor, vol) instead of (strike, tenor, price) so we ...
0
This is a Structured Product. Desks issuing these products make money by selling the product for X when it is really worth X-Y (where X is usually 100%, and Y will be some number of the order 1%, depending on the complexity and maturity of the product). In the case you reference, you can see from the link that Y was 1.75%. This means that, unhedged, the ...
1
I know its been a while but I would like to answer this question for all the people that arrives from now on. I hope that is okay. Let's divide the problem in two main parts. The first one is the computation of the zero coupon bond $P(t, T)$. In this case, you are using a short rate model given by the factor dynamics $dy(t)$ and the short rate dynamics $r(t)$...
1
As @Lliane explains, you are actually describing a position in which the underlying is rebalanced everyday, hence the compounding effect of the leveraged ETF vanishes. Maybe a bit of modelling can be helpful to illustrate the relationship between leveraged ETFs and volatility. Let $S_t$ be the value of the underlying and $V_t$ the value of a leveraged ETF ...
1
I disagree that these products are convex*. At any point in time, the ETF exposure to the underlying is linear, it's just that it changes through time. A 2x ETF will just have 2x exposure to the underlying - where the exposure is based on the nav at the point of rebalancing. Say the nav is \$100 per share, then it will hold \$200 of exposure to the ...
1
I think you are missing an important point regarding who initiates options positions. We know that put options are more expensive than theory would indicate as discussed in Bondarenko (2014). Simply: put option buyers are especially motivated to initiate positions, more so than put sellers. Thus put open interest is a measure of put buyers initiating ...
0
In both of your wealth equations, there is no need to subtract 1.2 from your investment in money market because you already own the option. Solve it you will get $\Delta=-\frac{1}{2}$ and $X_0=0$. Now, it means you need to short sell $-\frac{1}{2}$ shares, get $2\\\$$, and put 2\\\$$ into your investment account to guarantee that at$t=1$, you will get$1.5 ...
2
Both products actually have positive convexity, they will buy more underlying (SP500) when the price goes up and sell it when it goes down. However, if you hedge every day, you will just cancel out that gamma convexity. You have to let the position run a few days if you want to trade the gamma, because it is generated by the daily hedging of the 3x etf, not ...
-1
I came up with a semi-equivalent solution in case this helps anyone else. If one buys/shorts the stock using a margin loan, while buying option calls/puts to protect against a rise/drop in stock price it appears provide a similar solution, allowing one to hold the position for longer amounts of time, while adding leverage and limiting exposure.
2
What are common methods to compute implied volatility index? One could use VIX method on other underlying. Yes, the CBOE offers this for Apple, Google, Amazon, Goldman Sachs and IBM (see here). In my working paper here I use the CBOE VIX methodology on a sample of 268 individual equities in the same way. It also includes a comprehensive derivation of the ...
1
I was thinking of simply limiting set of options that go into computation to K0 strike+ 1 option on each side (cboe.com/micro/vix/vixwhite.pdf). Not sure if this is a good idea though. Hence the question If you have a full/complete options chain then naturally to calculate the VIX you should use the VIX formula, which should be interpreted as a definition. ...
3
Pathwise finite difference Gamma formula is indeed: $$\Gamma(S_0,T, dS; Z) = (dS)^{-2} \left[ (S_T^{up} (Z) - K)^+ -2 (S_T (Z) - K)^+ + (S_T^{dn} (Z) - K)^+ \right],$$ where $Z$ is a standard normal rv, and $$S_T (Z) = S_0\eta (Z),$$ $$S_T^{up} (Z) = (S_0+dS)\eta(Z) = (S_0+dS)S_0^{-1} S_T (Z)$$ S_T^{dn} (Z) = (S_0-dS)\eta(Z) = (S_0-dS)S_0^{-1} S_T (...
1
Perhaps it might help if we define the difference between Brownian Motion (BM) and Geometric Brownian Motion (GBM). BM has independent, identically distributed increments while GBM has independent, identically distributed ratios between successive factors. The definition is inherited from that of arithmetic random walks, which are modelled as sums of random ...
0
The short answer? In the absence of a dividend, call premium exceeds put premium by the carry cost (look at the explanation of conversions and reversals). If you're short the stock, your receive interest on the proceeds. However, you pay a borrow cost to the lender of the stock, thereby "reducing the interest rate by the borrowing costs required to ...
0
I'm not a pro and my experience is with equity options not futures so take this with a grain of salt. You have the right idea to defend when the underlying's price touches a strike price. However, I would adjust before that, even long before that if these are wide strangles. With equities, you can delta neutral hedge with the appropriate number of shares ...
0
To verify this, you could go all option pricing equation on this. Or you could take the path of least resistance and use an online option pricing calculator and enter zero for the dividend and zero for the interest rate and lo and behold, the time premium for the same series put and call will be identical.
1
This is going to get ugly :->) There are 6 basic synthetic positions relating to combinations of put options, call options and their underlying stock (the Synthetic Triangle): Synthetic Long Stock = Long Call + Short Put Synthetic Short Stock = Short Call + Long Put Synthetic Long Call = Long Stock + Long Put Synthetic Short Call = Short Stock + Short ...
0
If MSFT's share price <130, then as p→130−, the 110 call's price rockets whilst the 130 call stays OTM. So "the risk is larger" that your 100 call is assigned, while the 130 call expires worthless. Perhaps this is just poor wording. The $100 call will be assigned if it is ITM at expiration. There is no larger risk that it will be assigned. ... 0 I think this is a language issue. "Larger" means in Scenario 2 compared to Scenario 1. The phrase does not mean "the risk is larger if the stock keeps heading higher above the upper strike than if it stops at the upper strike". It means "assuming the stock keeps heading higher, the risk is larger for Scenario 2 than for Scenario 1&... 1 Instead, why not buy a B put, and sell an A put? That is a different strategy, a bear/debit put spread, equivalent to a bear/credit call spread, which is shown in your other question. If you are comparing the value of the position for a given$p$, you must take into account that a debit spread costs cash up front, whereas a credit spread pays cash up front. ... 1 Instead, why not buy an A call, and sell a B call? That is a different strategy, a bull/debit call spread, equivalent to a bull/credit put spread, which is shown in your other question. If you are comparing the value of the position for a given$p$, you must take into account that a debit spread costs cash up front, whereas a credit spread pays cash up ... 0 These are just suggested guidelines. The 75% comes from tastytrade (your link) which was founded by Tom Sosnoff. He was a CBOE market maker for 20 years and he created the option platform Thinkorswim which he sold to Ameritrade for$700 million. At a minimum, this guideline comes from his decades of experience and I wouldn't be surprised if he did ...
Top 50 recent answers are included | 2020-09-20 20:35:34 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6352437138557434, "perplexity": 1084.4905181782353}, "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-2020-40/segments/1600400198652.6/warc/CC-MAIN-20200920192131-20200920222131-00453.warc.gz"} |
https://www.freemathhelp.com/forum/threads/how-do-you-solve-for-s-12x5-30x4-sx3-7-2x2-5x-3.108970/ | # How do you solve for s? (12x5 - 30x4) - (sx3 - 7)(2x2 - 5x + 3)
#### happiness
##### Junior Member
A polynomial expression is shown below.
(12x5 - 30x4) - (sx3 - 7)(2x2 - 5x + 3)
The expression is simplified to -18x3 + 14x2 - 35x + 21.
What is the value of s?
#### tkhunny
##### Moderator
Staff member
Have you considered multiplication?
#### happiness
##### Junior Member
Have you considered multiplication?
I didnt post the problem right, thanks for posting though. I'll have to make another post.
Sent from my LGMS210 using Tapatalk
#### stapel
##### Super Moderator
Staff member
A polynomial expression is shown below:
. . . . .(12x5 - 30x4) - (sx3 - 7)(2x2 - 5x + 3)
The expression is simplified to:
. . . . .-18x3 + 14x2 - 35x + 21
What is the value of s?
As suggested earlier, the first step will be to multiply out and simplify the given expression. Then you'll want to "equate coefficients". That is, you'll find the term in the first expression that has "s" in its coefficient, and you'll look at the corresponding term in the given simplified form. Since the polynomials can be equal only if their coefficients are the same, then you can equate the two relevant coefficients, and solve for the value of "s".
#### happiness
##### Junior Member
As suggested earlier, the first step will be to multiply out and simplify the given expression. Then you'll want to "equate coefficients". That is, you'll find the term in the first expression that has "s" in its coefficient, and you'll look at the corresponding term in the given simplified form. Since the polynomials can be equal only if their coefficients are the same, then you can equate the two relevant coefficients, and solve for the value of "s".
Can you show me step-by-step how you would do that? btw the problem is a picture attachment.
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#### stapel
##### Super Moderator
Staff member
Can you show me step-by-step how you would do that? btw the problem is a picture attachment.
You already have loads of step-by-step worked solutions in your book and in your class notes, so us doing this exercise for you isn't likely to make much difference. Now is the time to try something!
First, you need to multiply out the original expression. Where are you getting stuck in that process? (here) Please reply showing all of your steps so far. Thank you!
By the way, the attachment is way too small to be legible. Sorry. | 2019-03-23 14:41:50 | {"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": 0.8635991811752319, "perplexity": 1019.076665762635}, "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/1552912202872.8/warc/CC-MAIN-20190323141433-20190323163433-00535.warc.gz"} |
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Which of the phrases (a), (b), (c) and (d) given below should replace the phrase given in bold in the following sentence to make the sentence grammatically meaningful and correct ? If the sentence is correct as it is and No correction is required, mark (e) as the answer. | 2022-11-27 19:18:09 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8697225451469421, "perplexity": 394.2126247433786}, "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/1669446710417.25/warc/CC-MAIN-20221127173917-20221127203917-00840.warc.gz"} |
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ISSN : 1229-3431(Print)
ISSN : 2287-3341(Online)
Journal of the Korean Society of Marine Environment and Safety Vol.22 No.1 pp.82-89
DOI : https://doi.org/10.7837/kosomes.2016.22.1.082
# Wave Response Analysis for Pontoon-type Pier: Very Large Floating Structure
Sang-Do Lee*, Sung-Hyeon Park**, Gil-Young Kong***
*Graduate School of Korea Maritime and Ocean University, Busan 49112, Korea
**Mokpo National Maritime University, Mokpo 58628, Korea
***Korea Maritime and Ocean University, Busan 49112, Korea
* First Author : oksangdo@naver.com
Corresponding Author : shpark@mmu.ac.kr, 061-240-7171
December 28, 2015 February 3, 2016 February 25, 2016
## Abstract
In this study, we proposed a pier of pontoon-type, “Very Large Floating Structure” (VLFS), with the length of 500m, breadth of 200 m and height of 2 m in Yeosu domestic port. Since this structure ought to endure wave loads for long periods at sea, it is essential to analyze the wave response characteristics. Direct-method is used to analyze the fluid-structure problem and the coupled motion of equation is used to obtain response results. The structural part is calculated by using finite element method (FEM) and the fluid part is analyzed by using boundary element method (BEM). Dynamic responses caused by the elastic deformation and rigid motion of structure are analyzed by numerical calculation. To investigate response characteristics of the pier in regular waves, several factors such as the wavelength, water depth, wave direction and flexural rigidity of structure are considered. As a result, wave response of pier changed at the point of L/λ 1.5 and represented the torsional phenomenon according to the various incident waves. And the responses showed increasing tendency as the water depths increase at the incident point in case of L/λ=8.0 and peak point of vertical displacement amplitude moved from side to side as the flexural rigidity of structure changes.
# 폰툰형 초대형 부유체식 부두의 파랑응답해석
이 상도*, 박 성현**, 공 길영***
*한국해양대학교 대학원
**목포해양대학교
***한국해양대학교
## 초록
본 연구에서는 국내 여수항만에 길이 500미터, 폭 200미터, 두께 2미터인 폰툰형 VLFS타입의 해상부두를 제안하였다. 이 구 조물은 해상에서 오랫동안 파랑하중을 견뎌야하므로 파랑응답해석이 필수적이다. 유체-구조부 해석에는 직접법을 사용하였고, 연성운 동방정식을 수치해석하여 응답 결과를 구하였다. 구조부는 유한요소법을 이용하여 계산하였으며, 유체부는 경계요소법을 사용하여 분 석하였다. 탄성변형과 강체운동으로 인한 동적응답을 수치적으로 분석하였으며, 파장, 수심, 파향, 구조물의 강성 요소를 고려하여 규칙 파에 대한 응답을 해석하였다. 연구의 결과, L/λ 1.5를 기준으로 응답이 변화하였고, 입사파의 방향에 따라 비틀림 현상이 나타났다. L/λ=8.0의 경우 수심이 증가할수록 입사측에서의 응답이 증가하는 경향이 나타났고, 강성의 변화에 따라 수직변위진폭의 피크점이 좌 우로 이동하였다.
## 1.Introduction
Many countries have been faced with the land scarcity in the process of pier extension and Very Large Floating Structure (VLFS) has been utilized as one of the effective substitute instead of reclamation of sea spaces. As a pier of VLFS is exposed to external wave forces for long periods at sea, it is essential to analyze the response characteristics according to the wave loads. Dynamic response of VLFS due to the wave loads gives rise to the change of pressure of fluid and motion of structure. Such motion of structure comprising elastic deformation means hydroelasticity (Wang et al., 2008).
In order to solve the hydroelastic problem, many researchers use analytical/semi-analytical approach and numerical approach. Voluminous papers related to the hydroelastic response of VLFS have been studied by authors at home and abroad. To analyze the problem of structure part and fluid part, modal method and direct method are commonly used in the frequency domain (Watanabe et al., 2004). We et al. (1995) treated linear two-dimensional problem and extended the eigenfunction expansion-matching method using the modal expansions to analyze the wave-induced responses of VLFS. Takagi et al. (2000) proposed an anti-motion device for VLFS and studied theoretically and experimentally using the eigenfunction expansion method. Kim and Ertkin (1998) also introduced this method for predicting linear hydroelastic behavior a shallow-draft VLFS. Hong et al. (2003) extended Kim and Ertkin (1998)’s eigenfunction expansion method to three dimensions considering the effect of non-zero draft. Because of simplicity and numerical efficiency, Hamamoto and Fujita (2002) used the wet-mode approach to study hydroelastic response of VLFS with arbitrary shape.
In the direct mode, the response of structure is not represented through a superposition of the global modal responses and is determined by directly solving the equation of motion (Watanabe et al., 2004; Wang et al., 2008). This method uses full modes of structure and has a high accuracy for complicated distribution of stiffness (Kim et al., 2006). Yago et al. (1996) modified a direct method using pressure distribution method and compared with the experiment results of a zero-draft VLFS. Yasuzawa et al. (1997) developed a numerical code for dynamic response of mat-type VLFS in regular waves using the direct method. Similar code was applied to container yard and marine pier of VLFS (Park et al., 2003; Lee, 2011).
In this paper, we suggest a pontoon-type pier of VLFS with the length of 500m, breadth of 200m and height of 2m in Yeosu new port. In order to give foundational results for establishing the criteria of VLFS pier, several factors such as the wavelength, water depth, wave direction and flexural rigidity of structure are considered. For the purpose of high accuracy than efficiency, direct method is used to analyze the fluid-structure problem. Dynamic responses caused by the elastic deformation and rigid motion of the pier are analyzed by using numerical calculation. Fluid part is analyzed by using boundary element method (BEM) and structural part is calculated by using finite element method (FEM).
## 2.Background Theory
Analysis theory is based on the potential theory widely used in terms of the wave loads and estimation of responses. The fluid is assumed inviscid, incompressible and irrotational. Under this assumption, fluid velocity is obtained by derivative of spatial coordinates of velocity potential. Governing equation is the continuity equation and is written by Laplace equation of velocity potentials.
As shown in Fig. 1, we analyze the response characteristics of a floating pier in regular waves, which installed in infinite widespread sea with constant water depth. We formulate respectively the fluid part and structural part, and induce the coupled motion of equation. Analysis should be done under the following assumption (Park et al., 2003; Yasuzawa et al., 1997).
• 1) The fluid is inviscid, incompressible and the motion of fluid is irrotational and velocity potential is defined.
• 2) Draft is floating structure can be ignored.
• 3) Sea bottom is assumed to be flat and sea domain is infinitely extended.
• 4) Motion of fluid is governed by the linear vibration theory and with regard to deformation, only deflection is considered.
• 5) Damping effect of floating structure can be ignored.
### 2.1Analysis of fluid part
#### 1)diffraction problem
Fluid part around VLFS is formulated using boundary element method (BEM) and fluid motion is represented using velocity potential. In the analysis model, components of fluid force are potentials by incident wave, diffraction wave and radiation wave. Total potential can be written :
$Φ x , y , z , t = Re φ x , y , z e − i ω t φ = φ i + φ d + φ r$
(1)
where ω is the circular frequency of harmonic motion and φi , φd, φr are the incident, diffraction, radiation potentials, respectively. When the structure have no motion and deformation, we seek φd by diffraction wave. We assume the structure have no motion and deformation as a rigid body, pure boundary value problems of fluid field. As shown in Fig. 1, the boundary value problems can be represented :
$∇ 2 φ d = 0 in Ω$
(2)
$∂ φ d ∂ z = 0 on Γ B z = − h$
(3)
$∂ φ d ∂ n + ∂ φ i ∂ n = 0 on Γ H$
(4)
$∂ φ d ∂ n − ν φ d = 0 on Γ F z = 0$
(5)
$lim r → ∞ r ∂ φ d ∂ n − i ν φ d = 0 on Γ ∞ ν ≡ ω 2 g = k tanh kh$
(6)
where n is the outward-directed normal to the fluid domain, g is the acceleration of gravity, k is the wave number and h is the water depth, ΓB , ΓH , ΓF ,Γ are boundaries of bottom, body surface, free surface and far-field, respectively.
Laplace equation (2) is governed equation in fluid domain Ω. Boundary condition (3) is represent in constant water depth h The equation (4) is the surface boundary condition in undersurface of structure, equation (5) is the linearized free surface condition, and equation (6) is the radiation condition indicating behavior in the far-field.
We regard structure as an elastic body and seek velocity potential φr generated by motion and deformation of structure. Boundary value problems can be made :
$∇ 2 φ r = 0 in Ω$
(7)
$∂ φ r ∂ z = 0 Γ B on Γ B z = − h$
(8)
$∂ φ r ∂ n = V n n direction velocity on Γ H$
(9)
$∂ φ r ∂ n − ν φ r = 0 on Γ F z = 0$
(10)
$lim r → ∞ r ∂ φ r ∂ n − i ν φ r = 0 on Γ ∞$
(11)
Above problems are same as the diffraction problems, excluding equation (9). Velocity potential φr can be obtained using boundary element method and integral equation.
#### 3)boundary integral equation
Diffraction problem and radiation problem can be formulated by finite element method using Green function :
$∫ Ω G ∇ 2 φ d − φ d ∇ 2 G d Ω = ∫ Γ G ∂ φ d ∂ n − φ d ∂ G ∂ n d Γ$
(12)
Applying the boundary conditions of (2)∼(11) in the equation (12), boundary integral equation can be derived :
$2 π φ d y = ∫ Γ H G x , y ∂ φ d ∂ n x − φ d x ∂ G ∂ n x , y d Γ x x , y ∈ Γ H$
(13)
$2 π φ r y = ∫ Γ H G x , y ∂ φ r ∂ n x − φ r x ∂ G ∂ n x , y d Γ x x , y ∈ Γ H$
(14)
#### 4)Formulation by boundary element method
After dividing integral boundary aspect of (13) and (14) into boundary element, equation is formulated :
$2 π φ y = ∑ j = 1 M ∫ Γ j G x , y ∂ φ ∂ n x − φ x ∂ G ∂ n x , y d Γ j x$
(15)
where M is the total number of boundary elements. Coordinates of elements and φ, $∂ φ ∂ n$ are expressed using linear combination as follows :
$x ξ 1 , ξ 2 = ∑ k = 1 4 N k ξ 1 , ξ 2 x k • x k$
(16)
$φ ξ 1 , ξ 2 = ∑ k = 1 4 N k ξ 1 , ξ 2 • φ k$
(17)
$∂ φ ∂ n ξ 1 , ξ 2 = ∑ k = 1 4 N k ξ 1 , ξ 2 • ∂ φ k ∂ n$
(18)
where Nk (ξ1, ξ2 ) is shape function. Substituting these into (15) gives
$2 π φ y i = ∑ j = 1 M ∑ k = 1 4 g jk • ∂ φ jk ∂ n − ∑ j = 1 M ∑ k = 1 4 h jk • φ jk g jk = ∫ Γ j G • N k ξ 1 , ξ 2 d Γ j h jk = ∫ Γ j ∂ G ∂ n • N k ξ 1 , ξ 2 d Γ j$
(19)
where N is the total nodal point (i = 1 ∼N) and φjk is the jth element and kth nodal velocity potential. Arranging these equation concerning corresponding relation of nodal point, matrix equation is derived :
$H φ = G ∂ φ ∂ n , H = 2 π I + G$
(20)
Substituting each potentials into (20) gives
$H φ d = G ∂ φ d ∂ n , H φ r = G ∂ φ r ∂ n$
(21)
Equation (21) is the final equation of motion in the fluid part.
### 2.2Analysis of structural part
Structural part of VLFS is formulated using finite element method. The equation of motion is induced from the principle of virtual work. We regard finite elements as the rectangle plate elements generating bending deformation. Bending displacement of structure can be made :
$V = N w ν e , ν e = w 1 θ x 1 θ y 1 ... w 4 θ x 4 θ y 4 T$
(22)
where Nw is shape function and {ν}e is the nodal point displacement vector. Because this finite element method is used bending vibration of general plate, we omit detailed theorem (Petyt, 2010).
When the finite element method is applied to the plate generating bending vibration and the equation of motion is induced by principle of virtual work, following equation can be obtained :
$K ν + M ν̈ = f$
(23)
where [K] is the stiffness matrix of structure, [M] is the mass matrix of structure, ν is the displacement vector and {f} is the external force vector.
### 2.3Coupled motion equation of fluid and structural part
Coupled motion equation of vibration be induced by connecting the equation of fluid and structural part. Fluctuation pressure ΔP can be obtained by Bernoulli’s theorem :
$Δ P = − i ρ f ω φ i + φ d + φ r − ρ f g ω$
(24)
where ρf is the density of fluid. Total matrix equation can be written :
$K + K w − ω 2 M ν − i ρ f ω C p φ r = i ρ f ω C p φ i + φ d$
(25)
Collecting equations (21) and (25), simultaneous equation is finally obtained :
$K + K w − ω 2 M − i ρ f ω C p i ω GA H = ν φ r = i ρ f ω C p φ i + φ d 0$
(26)
Equation (26) is the final coupled equation of motion in the fluid part and structural part. By solving this equation, nodal displacement vector and nodal velocity potential can be obtained (Park et al., 2004; Petyt, 2010).
## 3.Numerical analysis of wave responses
Pontoon-type VFLS typically has large horizontal dimensions whereas the height is several meters. Fig. 2 shows the concept of a pontoon-type pier of VLFS with breakwater. In order to use as marine pier having a capacity for two cargo ships and small yachts, we suggest the VLFS with the length of 500m, breadth of 200m, and height of 2m in Yeosu new port.
We analyzed the wave responses of model A in Table 1. To investigate the response characteristics, this study examined the several factors such as the wavelength, water depth, wave direction and flexural rigidity of structure. We divided the model with 126 nodes and 100 elements. Wave response program is developed and improved by two researchers (Park and Park, 2000).
### 3.1Response characteristics according to the wavelength
We examined the effect of incident waves by changing the ratio of structure length and the wavelength. Fig. 3 shows the corelation between the response amplitude and 3 representative points on the centerline of structure (incident point, middle point, penetration point). The abscissa L/λ on Fig. 3 indicates the ratio of structure length and the wavelength, and the ordinate |w|/(H/2) represents the ratio of vertical displacement amplitude and wave amplitude. Bending amplitude |w| for L/λ have been divided by the half wavelength. Non-dimensional distribution for the direction of structure length is represented on Fig. 3. The response changed at the point of L/λ=1.5. Due to the large incident wave force, the left side of L/λ=1.5, responses show large amplitude like sea wave. On the contrary, the right side of L/λ=1.5, the penetration side of structure indicates the decreasing curve and elastic responses.
### 3.2Response characteristics according to the water depth
In this part, we examined the responses according to the ratio of water depth and the wavelength. Shin et al. (2000) studied that the vertical displacement amplitude in the depth of 8m is smaller than the depth of 58.5m. Kyoung et al. (2005) investigated that the hydroelastic responses are influenced by the sea bottom topographies rather than the change of sea depths.
We calculated by changing h/λ and compared the results as shown in Fig. 4. Wave responses are not considerably influenced by the change of depths in case of L/λ=4.0 whereas the responses show increasing tendency as the depths increase at the incident point in case of L/λ=8.0. The results feature the elastic responses due to the influence of seabed and slight variation of vibration mode due to the wavelength.
### 3.3Response characteristics according to the direction of incident waves
This part deals with the response characteristics according to various incident wave direction. We analyzed the responses by changing the incident wave angles like 0 degree (a), 30 degree (b), 50 degree (c), 80 degree (d) in case of L/λ=4.0 waves.
Fig. 5 represents the torsional phenomenon according to the incident waves. The maximum point of displacement varies with the wave directions. It was found that the maximum point of response does not correspond with the incident wave angles.
### 3.4Response characteristics according to the Flexural rigidity of structure
This part analyzed the response characteristics by changing flexural rigidity of structure. Responses of model B are shown on Fig. 6 and D is the flexural rigidity of structure. The flexural rigidity (D) of (a) is 100 times larger than model A, (b) is 10 times smaller than model A, and (c) is 100 times smaller than model A. When the flexural rigidity of structure increases, peak point of L/λ moves left side and responses decrease at the large L/ λ ratio. On the contrary, peak point of L/λ moves right side in case the flexural rigidity of structure decreases. Peak point of vertical displacement amplitude moves from side to side according to flexural rigidity of structure. When the flexural rigidity of structure increases, elastic response is dominant. On the contrary, lower flexural rigidity of structure shows a form of riding waves despite decrease of the wavelength.
## 4.Conclusion
We analyzed the wave response characteristics of a suggested pontoon-type pier of VLFS in Yesou new port using the direct method and obtained following results according to the several factors such as the wavelength, water depth, wave direction and flexural rigidity of structure.
• (1) Wave response changed at the point of L/λ=1.5. Responses show large amplitude like sea wave on the left side of L/λ=1.5 and the penetration side of structure indicates the decreasing curve and elastic responses.
• (2) The responses show increasing tendency as the depths increase at the incident point in case of L/λ=8.0. The results feature the elastic responses due to the influence of seabed and slight variation of vibration mode due to the wavelength.
• (3) Torsional phenomenon are shown according to the various incident waves. The maximum point of displacement varies with the wave directions.
• (4) Peak point of vertical displacement amplitude moves from side to side as the flexural rigidity of structure changes. Elastic response is dominant in case flexural rigidity increases whereas lower flexural rigidity of structure shows a form of riding waves despite decrease of the wavelength.
The effect of changing depths is debatable point because this study is investigated in the range of comparatively shallow depths considering local area of Yeosu new port. Also, future study is needed about overall safety evaluation of the pier of VLFS for actuality to the domestic port.
## Figure
Model and coordinates of a VLFS for numerical analysis.
Concept of a pontoon-type pier of VLFS.
Wave response characteristics of vertical displacement amplitude depending on L/λ.
Comparison of vertical displacement amplitude depending on the h/λ : (a) L/λ=4.0, y=0 and (b) L/λ=8.0, y=0.
Response results of real displacement according to incident waves for L/λ=4.0 : (a) incident angle 0 deg., (b) incident angle 30 deg., (c) incident angle 50 deg., and (d) incident angle 80 deg.
Response results according to flexural rigidity of structure : (a) D=102×DmodelA, (b) D=1/10×DmodelA, and (c) D=1/100×DmodelA.
## Table
Principal particulars of the model for calculation
## Reference
1. Hamamoto T , Fujita K (2002) Wet-Mode Superposition for Evaluating the Hydroelastic Response of Floating Structures with Arbitrary Shape , Proceedings of the 12th International Offshore and Polar Engineering Conference Kitakyushu Japan, ; pp.121-128
2. Hong S Y , Kim J W , Ertekin R C , Shin Y S (2003) An Eigenfunction Expansion Method for Hydroelastic Analysis of a Floating Runway , Proceedings of the 13th International Offshore and Polar Engineering Conference Honolulu Hawaii, ; pp.121-128
3. Kim B W , Hong S Y , Kyoung J H , Cho S K (2006) Investigation on Wave Reduction Performance of Floating Hinge-Linked Breakwater , Journal of Ocean Science and Technology, Vol.3 (1) ; pp.13-22
4. Kim J W , Ertekin R C (1998) An Eigenfunction Expansion Method for Predicting Hydroelastic Behavior of a Shallow-draft VLFS , Proceedings of the Second International Conference on Hydroelasticity in Marine Technology, ; pp.47-59
5. Kyoung J H , Kim B W , Cho S K , Hong S Y (2005) Numerical Study on the Hydroelatic Response of the Very Large Floating Structure Considering Sea-Bottom Topography , Journal of the Society of Naval Architects of Korea, Vol.42 (4) ; pp.357-367
6. Lee S D (2011) A Study on the Wave Response for Eco-friendly Marine Wharf of the Very Large Floating Structure, Master Thesis, Mokpo National Maritime University
7. Park S H , Park S C (2000) A Study on the Reduction Analysis of the Response of the Mega-Float Offshore Structure in Regular Wave , Journal of Korean Navigation and Port Research, Vol.24 (1) ; pp.85-95
8. Park S H , Park S C , Koo J Y (2003) A Study on the Container Yard of Mega-Float Offshore Structure Type , Journal of Korean Navigation and Port Research, Vol.27 (1) ; pp.49-54
9. Park S H , Park S C , Choi M C , Koo J Y (2004) Study on the Optimization of Response in Regular Wave of the Mega-float Offshore Structure, Proceedings of the Spring Meeting , The Korean Society of Marine Environment & Safety, ; pp.99-105
10. Petyt M (2010) Introduction to Finite Element Vibration Analysis, Cambridge University Express, ; pp.119-247
11. Shin H K , Lee H Y , Shin H S , Park I K (2000) Analysis Methods of Hydroelastic Responses for a Very Large Floating Structure , The Korean Society of Ocean Engineers, Vol.14 (2) ; pp.19-27
12. Takagi K , Shimada K , Ikebuchi T (2000) An Anti-motion Device for a Very Large Floating Structure , Marine Structure, Vol.13 ; pp.421-436
13. Wang C M , Watanabe E , Utsunomiya T (2008) Very Large Floating Structure, Taylor & Francis, ; pp.35-65
14. Watanabe E , Utsunomiya T , Wang C M (2004) Hydroelastic Analysis of Pontoon-type VLFS: a Literature Survey , Engineering Structure, Vol.26 ; pp.245-256
15. We C , Watanabe E , Utsunomiya T (1995) An Eigenfucntion -matching Method for Analyzing the Wave-induced Responses of an Elastic Floating Plate , Applied Ocean Research, Vol.17 ; pp.301-310
16. Yago K , Ohmatsu S , Endo H (1996) On the Hydroelastic Response of Box-Shaped Floating Structure with Shallow Draft : Tank Test with Large Scale Model , Journal of the Society of Naval Architects of Japan, Vol.180 ; pp.341-352
17. Yasuzawa Y , Kawano D , Kagawa K , Kitabayashi K (1997) Numerical Response Analysis of a Large Mat-type Floating Structure in Regular Waves , Journal of the Society of Naval Architects of Japan, Vol.181 ; pp.111-122 | 2020-01-22 09:40:16 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 27, "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": 0.6781178712844849, "perplexity": 3567.7730546665593}, "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/1579250606872.19/warc/CC-MAIN-20200122071919-20200122100919-00206.warc.gz"} |
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###### Question:
I need to know the answer.
### PLEASE I AM BEEEEEGGGING YOU!!!!!EXTRA POINTS AND BRAINLIEST WRITE A POEM SHORT OR LONG Economy of language (as few words as possible - cut the language down to what is absolutely necessary) Directness (every word means what it says) Avoidance of Generalities (avoid words that are overused, cliched, meaningless - example "she was nice" - be more specific Precise phrasing over adherence to Poetic Meter It doesn't have to rhyme. NOTE: You must write a 4-5 sentence reflection along with your poem
PLEASE I AM BEEEEEGGGING YOU!!!!!EXTRA POINTS AND BRAINLIEST WRITE A POEM SHORT OR LONG Economy of language (as few words as possible - cut the language down to what is absolutely necessary) Directness (every word means what it says) Avoidance of Generalities (avoid words that are overused, cliche...
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### Blank occurs when individuals in a group are selected for destruction only because they belong to that group
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### Some 1 help me plz!!! Which statement comparing ionic and covalent bonds is correct? a. Elements in Group 1A are more likely to form covalent bonds than ionic bonds with other atoms. b. Covalent bonds tend to be less polar than ionic bonds. c. More electrons are transferred from covalent bonds than ionic bonds. d. The difference between electronegativities is greater in covalent bonds than for atoms in ionic bonds.
some 1 help me plz!!! Which statement comparing ionic and covalent bonds is correct? a. Elements in Group 1A are more likely to form covalent bonds than ionic bonds with other atoms. b. Covalent bonds tend to be less polar than ionic bonds. c. More electrons are transferred from covalent bonds th...
### Which of the following produces converging rays?A) Convex LensB) Convex MirrorC) Concave LensD) Flat Mirror
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### See attachment. Can right angles C and E be congruent? I'm confused because right angle E is on the outside while right angle C is on the inside but they are both right angles and equal the same degrees.
See attachment. Can right angles C and E be congruent? I'm confused because right angle E is on the outside while right angle C is on the inside but they are both right angles and equal the same degrees.... | 2023-02-08 00:40:42 | {"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": 0.30418238043785095, "perplexity": 3705.4091301156095}, "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/1674764500664.85/warc/CC-MAIN-20230207233330-20230208023330-00835.warc.gz"} |
https://cinc.rud.is/web/packages/urldiversity/reference/safePOST.html | Scraping the web is fraught with peril. URLs die; networks get disrupted and best laid plans for building a corups from links can quickly go awry. Use this funtion to mitigate some of the pain of retrieving web resoures.
safePOST(url = NULL, config = list(), timeout = httr::timeout(5),
..., body = NULL, encode = c("multipart", "form", "json", "raw"),
handle = NULL)
## Arguments
url the url of the page to retrieve Additional configuration settings such as http authentication (authenticate), additional headers (add_headers), cookies (set_cookies) etc. See config for full details and list of helpers. a call to httr::timeout(). Default timeout is 5 seconds. Further named parameters, such as query, path, etc, passed on to modify_url. Unnamed parameters will be combined with config. One of the following: FALSE: No body. This is typically not used with POST, PUT, or PATCH, but can be useful if you need to send a bodyless request (like GET) with VERB(). NULL: An empty body "": A length 0 body upload_file("path/"): The contents of a file. The mime type will be guessed from the extension, or can be supplied explicitly as the second argument to upload_file() A character or raw vector: sent as is in body. Use content_type to tell the server what sort of data you are sending. A named list: See details for encode. If the body is a named list, how should it be encoded? Can be one of form (application/x-www-form-urlencoded), multipart, (multipart/form-data), or json (application/json). For "multipart", list elements can be strings or objects created by upload_file. For "form", elements are coerced to strings and escaped, use I() to prevent double-escaping. For "json", parameters are automatically "unboxed" (i.e. length 1 vectors are converted to scalars). To preserve a length 1 vector as a vector, wrap in I(). For "raw", either a character or raw vector. You'll need to make sure to set the content_type() yourself. The handle to use with this request. If not supplied, will be retrieved and reused from the handle_pool based on the scheme, hostname and port of the url. By default httr requests to the same scheme/host/port combo. This substantially reduces connection time, and ensures that cookies are maintained over multiple requests to the same host. See handle_pool for more details.
## Details
This is a thin wrapper for httr::GET() using purrr::safely() that will either return a httr response object or NULL if there was an error. If you need the reason for the error (e.g. Could not resolve host...) you should write your own wrapper. | 2021-02-28 01:21:14 | {"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": 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": 0.21337687969207764, "perplexity": 5038.322375294205}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-10/segments/1614178359624.36/warc/CC-MAIN-20210227234501-20210228024501-00629.warc.gz"} |
https://bsebsolutions.com/bihar-board-12th-business-economics-important-questions-long-answer-type-part-1/ | # Bihar Board 12th Business Economics Important Questions Long Answer Type Part 1
BSEB Bihar Board 12th Business Economics Important Questions Long Answer Type Part 1 are the best resource for students which helps in revision.
## Bihar Board 12th Business Economics Important Questions Long Answer Type Part 1
Question 1.
Explain the law of demand with the help of a diagram. Explain any five factors affecting the demand of a commodity.
Law of demand: The law of demand stages that, other things being equal the demand for a good extends with a decrease in price and contracts with an increase in price. There is inverse relationship between quantity demanded of a commodity and its price, provided other factors influencing demand remain unchanged. The law of demand states that other things remaining constant quantity demanded of a commodity increases, with a fall in price and diminishes when price increases.
Factors affecting demand:
(i) Utility of the goods: Utility means wants satisfying power of a commodity. It is a subjective factor which varies from personal to person. Goods having greater or more utility will be in greater demand and vice-versa.
(ii) Income level: Income level directly affects demand. Higher the level of income, higher will be the demand and lower the level of income, lesser will be the demand.
(iii) Distribution of wealth: Distribution of wealth also affects demand. If the distribution of wealth in society is unequal, then luxuries goods will be demanded by affluent section of society. But as the distribution of wealth becomes equal, there will be increased demand of necessary and comfortable goods.
(iv) Price of the Goods: Price of the goods directly affects the demand for it. At a lower price, demand will become high and vice-versa.
(v) Expected future change in price: Govt, regulation natural calamities, possibility of war, etc. affects demand of goods.
Question 2.
Briefly explain the main functions of a commercial Bank.
A commercial bank is a financial institution which accepts deposit from the public and advances loans to other. The main functions that commercial banks perform are:
(i) Accepting of deposits: Commercial banks receive money from the public and bussinessman in form of deposits. People can deposit their cash balances with a bank in either of the following accounts: (a) Current Deposit A/c (b) Fixed Deposit/Term Deposit A/c (c) Saving Deposits A/c and (d) Recurring Deposits A/c.
(ii) Advancing of loans: Commercial banks provide various type of loans to the borrowers in either of the following forms: (a) Cash credit (b) Term loans (c) Demand loans (d) Overdrafts (e) Discounting of Bills Exchange.
(iii) Investment of funds: The banks invest their surplus funds in three types of securities (a) goverment securities (b) other approved securities (c) other securities.
(iv) Agency functions: The banks pcrfomi the following agency functions for its customers for a commission (a) Transfer of funds through demand draft mail, transfer, telegraphic transfers etc. (b) Collection and payments of cheque, bills of exchange and dividend on behalf of the custefncrs. (c) Acting as executors and trustees of Wins.
(v) General Utility Services: Commercial banks provides following services of general utility to its customers: (a) Locker facility (b) Traveller’s cheques (c) Gift cheques (d) Undertaking securities (e) Purchase and sale of foreign exchange.
Question 3.
Distinguish between Micro Economics and Macro Economics.
Distinction between Micro and Macro Economics: Micro and Macro Economics are two main branches of Economics. Micro Economics deals with individual economic problems whereas Macro Economics is the study of aggregates or of entire economic system. These two approaches differ from each other in the following manner:
Micro Economics:
1. It is concerned with an individual economic unit like a consumer, a firm, an industry or income of an individual.
2. It is based on the assumption of full employment.
3. Its central problem is price determination.
4. What is true at micro level may not be true for macro level.
5. It is based on the assumption of ‘other things being equal.’ This analysis is based on partial equilibrium.
6. Its objective is to study the theories related to optimum distribution of resources.
7. Its nature is comparatively easy.
8. It has main instruments of demand and supply.
Macro Economics:
1. It deals with aggregates of economy such as national income, aggregate expenditure, total employment, general price level, etc.
2. It is based on the assumption of under full employment of resources.
3. Its central problem is production and employment determination.
4. Group behaviour is applicable on entire economy.
5. This is based on general equilibrium analysis.
6. Its objectives is to study the theories related with full employment.
7. Its price is comparatively complex.
8. Its main instruments are aggregate demand, aggregate supply, aggregate saving and investment.
Question 4.
State the relation between total utility and marginal utility?
The relation between total utility and marginal utility are as follows: Total Utility is defined as the total satisfaction a consumer obtain from a given amount of a particular commodity Total utility is the sum of the marginal utilities obtained from the consumption of different units of a commodity i.e.,
TU = MU1 + MU2 + ……….. MUn = ΣMU.
Marginal utility is an addition made total utility by consuming-an additional (extra) unit of a commodity. Symbolically, MU = ΔTU/ΔQ or MUn+n = TUn – TUn-1
Relationship between TU and MU: (i) so long as MU is positive, TU rises (ii) When MU becomes zero, TU is maximum and (iii) when MU is negative, TU falls.
Question 5.
Distinguish betwen fixed cost and variable cost.
Fixed cost are the costs which do not change with the change in output. These costs remain even if the output is zero. Examples of total fixed costs are rent for factory building wages to the permanent staff, interest capital etc.
Variable costs are the costs which change with the change in the output. There are no variable costs at zero-output. Example of total variable costs are expenses on raw materials used in the production, wages of the capital laborers etc.
Question 6.
Discuss the Central Problem of an economy.
According to prof Samuelson, every economy has three basic problems of resource allocation:
(a) What to produce and in what quantity?
(b) How to produce?
(c) For whom to produce?
(a) What to produce and how much to produce?
The very first central problem of economy is related to problem of choice. First and foremost problem of economy is- What to produce or which goods and services should be produced so that maximum wants and needs of people could be satisfied with limited resources. Every economy has to choose that which want should be satisfied and which should be sacrified, postponed or left? Regarding what to produce? Many production related options come before us for the solut ion of this problem.
When an economy decides what to produce; then next problem appears i.e. ‘How much to produce’? or we have to decide that what quantity of consumption goods and capital goods should be produced? For example, if an economy wants to increase production of consumer goods like clothes and tea through limited resources in a definite time-interval then it has to reduce production of capital goods like tractor and television. Infact, producer would like to produce that quantity of goods which maximises his profit.
On the basis of this choice, through available alternatives, a country has to decide that what quantity of limited factors should be used for producing which quantity.
(b) How to produce?
Second problem before the economy is ‘how to produce’? In fact, it is the problem related to the choice of technique. There are two types of techniques of production:
• Labour-intensive Technique: In this technique, more quantity of labour is used as compared to capital.
• Capital-intensive Technique: In this technique, more quantity of capital is used as compared to labour.
For example, clothes can be produced by both, handloom (i.e. capital-intensive technique).
How to produce?-for solution of this problem, we should adopt that technique which utilises lesser limited resources. If the production from both the techniques is the same then the technique using less quantity of limited resources will be called more efficient technique.
(c) For whom to produce?
After getting the solutions of the problems of what? how much? and how? the next problem arises-”For whom to produce”? or “How to distribute the production?”
There are two faces of the problem of distribution of production:
(a) First face is related to ‘Personal Distribution’. It means that how the produce should be distributed between different persons and families in the society. It is also related to the problem of unequal distribution of income.
(b) Second face of the problem of distribution is ‘Functional Distribution’. It is related to determine how the produce should be distributed among various factors of production as land, labour, capital and entrepreneur. It is not related to problem of inequality of income distribution.
Question 7.
Discuss the different types of Economics Systems.
Economic Systems: Economic system is a structure of such institutions with which all economic activities are operated in the society. Every economy is based on an economic system which can be divided into three categories:
1. Capitalist Economy or Market Economy,
2. Socialist Economy or Planned Economy,
3. Mixed Economy.
(1) Capitalist Economy or Market Economy: Market economy contains the following important features:
• Private Property: Capitalist economic system recognises ‘Law of Inheritance’ and right of individual private property. It also ensures to transfer the property of dead person to its heir.
• Economic Freedom: Capitalist economy grants various economic freedoms to the individual like freedom to work, freedom of choice, freedom of consumption and freedom of saving and investment.
• Laisscz Faire & Free Trade: Market economy works under Laissez Fairc i.e., there is no state intervention in economic activities of the society. Market economy is an open economy in which free trade policy is adopted.
• Competition: Competition is an essential feature of market economy. Equilibrium between market forces, i.e., demand and supply takes place due to competition appearing in the economy.
• Price Mechanism: In capitalist economy prices are determined by the automatic adjustment of price mechanism. Price in the market is determined at the point where demand and supply forces become equal.
(2) Socialist Economy or Planned Economy: The salient features of Controlled Socialist Economy are as follows:
• Social Ownership: In socialist economy social ownership is found on factors of production. These factors are used for the welfare of the society as a whole. Right to individual property has no place in socialist economy rather it is limited to only self-consumption goods.
• Absence of Economic Freedom: Individual economic freedom remains absent in socialist economy. What? How? and How much? problems arc solved by centralised planning institution according to the needs of the society.
• Passive Role of Price Mechanism: In socialist economy, prices are not determined by price mechanism, rather government takes the use of accounting prices which are determined by government itself on the basis of social interest.
• Absence of Competition: Socialist economy works on planning and direction, as a result of which competition remains absent in the economy.
(3) Mixed Economy: Mixed Economy contains the following economic features:
(i) Co-existence of private and Public Ownership: Private and public sectors co-exist in the mixed economy. Both private ownership and profit motive arc found in such economic system. Law of Inheritance finds a place in the economy but government imposes progressive taxation to attain economic equality.
(ii) Economic Freedom: Though enough state interference is found in Mixed Economy, people enjoy limited economic freedom of choice, production, investment and saving. Govt, adopts many controls so check the unlimited economic freedom of the individual.
(iii) Price System: Both price mechanism and profit motive determine the price system simultaneously in the Mixed Economy. Profit motive is managed by the government so that it may not hit the motive of social welfare.
(iv) Limited Competition: Mixed Economy contains limited competition due to the co-existence of both private and public sector. Competition appears in the economy but due to government regulation, economic development activities are not adversely affected.
Question 8.
Explain ‘Individual Demand Curve’ and ‘Market Demand Curve’ with the help of figures.
Individual Demand Curve: Individual demand curve shows various combinations of the quantity of goods demanded by an individual consumer at different prices. Individual demand schedule has been shown in fig. Quantity demanded by consumer A has been represented on X-axis and
per unit price on Y-axis. When price of goods is ₹ 10 per unit, he buys only one unit of it. When price decreases to ₹ 8 per unit, quantity demanded increases to 3 units. Demand curve DD slopes downward from left to right, which shows that when price decreases, demand increases and vice-versa. In this way, the individual demand curve is negatively sloped.
Market Demand Curve: By represeting market demand schedule on a graph, we can obtain ‘Market Demand Curve’. This curve represents demand of goods at different price for all the consumers in the market. Market demand curve is a horizontal summation of individual demand curves.
In fig., quantity of milk has been shown on X-axis and price of milk on Y-axis. In panel (i) and (ii), individual demand curve of consumer A and B has been represented.
In panel (iii), market demand curve has been represented. When price of milk is ₹ 10 per litre, then consumer A and B demand 6 and 8 litres of milk respectively. Hence, at price ₹ 10, market demand is 6 + 8 = 14 litre of milk. In this way, market demand curve is obtained by adding individual demand curves. It is horizontal summation of individual demand curves. Its slope is negative which shows the inverse relationship between price and quantity demanded.
Question 9.
What is Law of Demand? Why does the demand curve slope downwards? Are there exception to it?
Law of Demand: Law of demand explains quantitative relation between price of goods and quantity demanded. Every consumer has a psychology to buy less amount or quantity of anything at high price and more quantity at low price. Ceteris paribus (other things being equal), there is inverse relationship between price of a goods and quantity demanded i.e. when price is high, demand is low and when price is low, demand is high.
In other words, there is an inverse relation between price of a goods and its demand. When price increases, demand decreases and vice-versa. Law of Demand is a ‘qualitative statement’ and not a ‘quantitative statement’. This law locates the direction of price and demand change and not the quantity of change.
P ∝ $$\frac{1}{\mathrm{Q}}$$
where, P = Price of Commodity
Q = Quantity Demanded
Assumption of the law of Demand: Law of demand is based on few assumptions. Those assumptions are:
• Consumer’s income should remain constant.
• Consumer’s taste, nature, etc. should remain constant.
• Price of related goods should remain constant.
• Consumer remains unknown with new substitutes.
• There is no possibility of price change in future.
Why Demand Curve slopes Downward?
Slope of demand curve is negative, i.e., it fails from left to right which means that less goods are bought or demanded at high prices and vice-versa. Negative slope of demand curve is due to following reasons:
(1) Law of Diminishing Marginal Utility: Law of demand is based on law of dimishing marginal utility. According to it, marginal utility of a goods diminishes as an individual consumes more units of a goods. In other words, as a consumer takes more units of a goods, the extra utility or satisfaction that he derived from an extra unit of the goods goes on falling. The law of diminishing marginal utility means that the total utility increases but at a decreasing rate.
Marshall has stated this law as:
“The additional benefit which a person derives from a given increase of his stock of a thing diminishes with every increase in the stock that he already has.”
At low price, more units of a goods is demanded and vice-versa.
(2) Increase in Purchasing Power or Income Effect: When price of a goods decreases, real income or purchasing power of consumer increases due to. which he can maintain his previous level of consumption with less expenditure. In this way, at lower prices, more goods could be purchased. On the contrary, when price of a goods increases, real income of consumer decreases due to which his consumption decreases. This is the law of demand.
Question 10.
What is Perfect Competition? What are its main characteristics?
Perfect Competition: Perfect competition is that market situation in which a large number of buyers and sellers are found for homogeneous product. Single buyer or the seller are not capable of affecting the prevailing price and hence, in a perfectly competition market, a single market price prevails for the commodity.
Definitions:
1. According to Ecftwitch, “Perfect competitve is a market in which there are many firms selling identical products with no firm large enough relative to the entire market to be able to influence market price.”
2. According to Mrs. Joan Robinson. “Perfect competitive prevails when the demand for the output of each producer is perfectly elastic.”
Characteristics or Features of Perfect Competition:
(1) Large Number of Buyers and Sellers: Perfect competitive market has a large number of buyers and sellers and hence, any buyer or seller cannot influence the market price. In other words, individual buyer or seller cannot influence the demand and supply conditions of the market.
(2) Homogeneous Product: The units sold in the market by all sellers are homogeneous (or identical) in nature.
(3) Free Entry and Exit of Firms: In perfect competition, any new firm may join the industry or any old firm may quit the industry. Hence, there is no restriction on free entry or exit of firms into/from the industry.
(4) Perfect knowledge of the Market: In perfect competition, every buyer has the perfect knowledge of market conditions. None of the buyers will buy the commodity at higher price than the prevailing price in the market. Hence, only one price prevails in the market.
(5) Perfect Mobility of Factors: In perfect competition, the factors of production are perfectly mobile. Factors can easily be mobile from one industry to other industry (or from one firm to other firm) without any difficulty.
(6) No Transportation Cost: Transportation cost remains zero in perfect competition due to which one price prevails in the market.
Question 11.
What is meant by Monopoly? Mention the main features of Monopoly Competition.
Monopoly is the addition to two words, i.e., ‘Mono’ + ‘Poly’, i.e. single seller in the market. Being the single seller in monopoly market, a firm has full control on the supply of the commodity. In pure monopoly even no close substitute of the product is available in the market.
In monopoly, no distinction arises between ‘firm’ and ‘industry’ i.e. firm is industry and industry is firm.
Definitions:
(1) According to McConnell, “Pure monopoly exists when a single firm is the sole producer of a product for which there are no close substitutes,”
(2) According to Braff, “Under pure monopoly there is a single seller in the market. The monopolists demand is market demand. The monopolist is a price-maker. Pure monopoly suggests a no substitute solution.”
(3) According to Leftwitch, “Pure monopoly is a market situation in which a-singlc firm sells a product for which a single firm sells a product for which there arc no good substitutes. The firm has the market for the product all to itself. There are no similar products whose price and sales will influence the monopolist’s price or sales.”
Features of Monopoly: The main features of monopoly are as follows:
(1) Single Seller and Large Number of Buyers: Monopoly market consists single seller of the product but the number of buyers stands very large.
No buyer can influence the price of the product due to his large number in the market.
(2) No Close Substitutes: Monopoly firm produces such commodity which has no close substitute and as a result the cross elasticity of demand becomes zero.
(3) Monopolist as a Price-maker: Being the single seller in the market, the monopolist firm is itself a price-maker. A monopolist firm can determine both price and quantity but not simultaneously (i.e., either price or quantity at a particular time).
(4) No Entry of New Firm: The entry of new firm into the industry is strictly prohibited. There is no competitor of monopoly firm in the market.
(5) Demand Curse Negatively Sloped : Monopolist demand curve is negatively sloped and marginal revenue (MR) is less than average revenue (AR). The slope of demand curve depends on elasticity of demand.
MR = AR$$\left(\frac{\mathrm{e}-1}{\mathrm{e}}\right)$$
or, AR = $$MR\left(\frac{\mathrm{e}}{\mathrm{e}-1}\right)$$
Question 12.
Compare demand curve facing a seller under-conditions of perfect competition: Monopolistic and Monopoly, clearly, reflecting the differences in their elasticities. Draw three curves in a single diagram.
Under perfect competition, a firm can sell whole of its output at the same price. Therefore, the demand curve will be a straight line parallel to the X-axis [See fig. (i)].
Under monopoly, a firm can sell more of its product only at a lower price. Therefore, the demand curve of a monopolist is negatively sloped [See fig. (ii)]. Here, demand curve is less elastic.
Under monopolistic competition, a firm can sell more of its product only at a lower price. Therefore, the demand curve of a firm under monopolistic competition is also negatively sloped [Sec fig. (iii)]. Here, demand curve is more elastic.
The difference between the demand curves is that the demand curve under perfect competition is perfectly clastic as the demand curve under monopoly is less elastic than the demand curve under monopolistic competition. This is because of the availability of substitutes in case of monopolistic competition. Close substitutes arc not available in case of monopoly and therefore, demand curve is less elastic under monopoly. The following diagram illustrates the idea:
Question 13.
How is price determined under Monopoly Market?
1. Monopoly price or Equilibrium during short period: During short period, monopoly price, normal profit and loss do production work, in three conditions. It is a wrong concept that monopoly always earns profit only during short period. In which situation the monopoly will work under profit, normal profit and loss during short period, it will depend on the demand curve of market and on the cost conditions of the market monopoly firm like the complete competitive finn can also face loss during short period.
In the situation of short period loss, the monopolist will think to do the production upto that point upto which he does not get the cost of the goods equal to Average Variable Cost (AVC) or more than that. Generally monopolist does not get close substitute of the produced goods, because of which the monopolist can make efforts to equal the cost of goods to the average cost in order to prevent his loss during short period but it is only an effort. We cannot deny the possibility of monopoly loss in short period.
The situation of loss for the monopolist during short period is not the impossible condition. In brief, all the possible situations of monopoly during short period can be understood with the help of following figures:
(i) Profit Situation: The short period profit situation of monopolist is shown in below figure. The demand curve AR of monopoly goods and its related marginal revenue curve MR arc shown in the figure. The balance of monopoly is shown at point E where both the conditions of monopoly balance are being fulfilled. At this point MR = MC. In the situation of this balance, the cost of the goods will be RX or OP. At this cost, the monopolist will do the production of OX goods. In the figure, per unit cost is SX or OQ i.e., the monopolist is getting continuous profit equal to the distance RS. It is clear from the figure that the monopolist will get profit equal to PRSQ on the whole output OX. In brief,
Price = OP
Output = OX
Total Profit = Per Unit Profit × Output
= PQ . OX
= PQRS Area
(ii) Situation of Normal Profit: The normal profit of monopolist is shown in adjoining figure. The normal profit is also called zero profit. Normal profit means that the monopolist determines the price of goods equal to the average production cost of the goods. In the figure, the balance point of monopolist is E. Where the average cost of the goods RX and price of the goods both are there. In this situation, the monopolist is not getting any surplus.
Because at point R, AR = AC
In brief, Price = OP
Output = OX
Monopolist is getting zero profit.
(iii) Situation of Loss: The situation of short period loss of monopolist is shown in given figure. The demand of monopoly goods can be so much weak in some circumstances that the price of monopoly goods could be reduced even from the average cost of the goods. It will be the situation of loss. The monopolist on getting more price from the Average Variable Cost (AVC) in short period works in the hope that this loss in the long term will be converted into profit.
In the figure RX is the per unit cost of the goods whereas SX is the average cost of that per unit i.c., per unit output is undergoing loss of SR. The output of goods is OX because of which, the output is under going in loss equal to PRSQ.
In brief, Price = OP
Output = OX
Total Loss = Per unit Loss × Total Output
= QP. OX = PQSR Area
2. Monopoly price or Equilibrium during Long Period: There is complete control of the producer on supply in the monopoly market. Long period is that period of production wherein monopolist fully adjusts his supply according to demand situations. That is why it is said that the price is mainly determined on the basis of supply situations in long period. The monopolist will adjust the supply of the goods in market in order to profit maximisation so that he could get profit in every situation. In short period, the supply could not be adjusted according to demand due to limited period because of which, the conditions of profit, normal profit and loss in short period monopoly market arc created but the monopolist gets only profit due to the adjustment of supply in long period.
The firm and industry get internal and external economics in the beginning due to large scale output in long period but these savings get converted into diseconomies after one point. When the firm and industry get internal and external savings, then the law of increasing returns to scale in applied i.e., the marginal cost gets, decreased gradually on increasing the size of output. As the size of the output gets increases and the marginal cost decreases, then the average cost also decreases but the rate of decreasing of average cost is less than the decreasing of the rate of marginal cost.
In the similar process, we get one such ideal point on increasing more output quantity where marginal cost and average cost should be mutually equal. This, we call the rule of constant returns to scale. After this point, the internal and external savings get converted into diseconomies and the law of decreasing returns to scale is applied! The marginal cost gets increased in such situation. As a result of which, the average cost also gets increased but the marginal cost increases rapidly than the average cost.
All three situations of monopolist in long term increasing returns, constant returns, and decreasing returns get profit.
(i) Monopoly Balance in Increasing Returns Situation (or Decreasing cost situation): In following figure the monopoly balance is shown in the increasing returns (or decreasing cost) situation where AC and MC both are declining but MC curve declines more rapidly. According to balance situations, there will be the monopoly balance on point E where both the conditions of balance are being fulfilled.
At balance point E,
Per Unit Price = OP or SQ
Per Unit Cost = OR or TQ
Total Output = OQ
Per Unit Profit = OP – OR = PR or ST
Total Profit PRTS Area.
(ii) Monopoly Balance in Constant Returns Situation (or Constant Cost Situation): In following figure the monopoly balance in constant returns situation is shown where average cost and marginal cost are mutually equal due to constant returns.
At balance point E
Per Unit Price = OP or SQ
Per Unit Cost = OR or FQ
Total Output = OQ
Per Unit Profit = PR or SE
Total Profit = PRES Area
(iii) Monopoly Balance in Decreasing Return Situation (or Increasing Cost Situation): In following figure. Monopoly balance in decreasing returns situation (or increasing cost situation) is shown. Both AC and MC increases due to decreasing returns but MC is more than AC.
At balance point E,
Per Unit Price = OP or SQ
Per Unit Cost = OR or TQ
Total Output = OQ
Per Unit Profit = PR or ST
Total Profit – PRTS Area
Thus, there is profit to monopolist in all kinds of cost situations during long period.
Question 14.
Market for a good is in equilibrium. Explain the chain of reaction in the market if the price is (i) higher than equilibrium price and (ii) lower than equilibrium price.
(i) When prices is higher than equilibrium price. There is excess supply and producers are not in a position to sell all they want to sell at the given price.
The leads to competition between producers. Competition between producers leads to lowering of price. Lowering of price raises demand while reduces supply. This continue till demand is equal to supply again at the original equilibrium.
When price is lower than equilibrium price, market demand is greater than market supply. This will result in competition among buyers. The price will rise.
(ii) A rise in price will reduce the demand and raise the supply. This will reduce the original gap between market demand and market supply. These changes will continue till price fiscs to a level at which market demand is equal to market supply. This is the equilibrium price.
Question 15.
Market for a good is in equilibrium. There is simultaneous ‘increase’ both in demand and supply of the good. Explain its effect on market price.
On OX-axis we have plotted quantity and on OY-axis we have plotted price. P is the original price OD is the original demand curve and SS is the
original supply curve and E is the original equilibrium point when demand curve increases from DD to D1D1 and supply curve shifts from SS to S1S1 the new equilibrium point will be formed at E1. Due to which price remains constant and quantity rises from OQ to OQ1.
Question 16.
There is a simultaneous ‘decrease’ in demand and supply of a commodity. When will it result in:
(a) NO change in equilibrium price.
(b) A fall in equilibrium price.
Use Diagram.
Decrease in demand means less quantity demanded at the same price. This leads to shift to demand curve leftward from D1 to D2 and decrease in supply means less quantity supplied at same price. This leads to leftward shift of supply curve from S1 to S2.
(a) If decrease in demand is equal to decrease in supply there will be no change in equilibrium price. In the diagram (A) the two decrease are equal to Q2Q1. The equilibrium price remains unchanged at OP.
(b) Equilibrium price will fall when decrease in demand is greater that decrease in supply. In diagram (b) decrease in demand (AE1) is greater than decrease in supply (BE1) leading to fall in the equilibrium price from OP1 to OP2.
Question 17.
The market demand curve for a commodity and the total cost for a monopoly firm producing the commodity is given by the schedules below, use the information to calculate the following:
(a) The MR and MC schedules.
(b) The quantities for which the MR and MC are equal.
(c) The equilibrium quantity of output and the equilibrium price of the commodity.
(d) The total revenue, total cost and total profit in equilibrium.
(a) MR and MC schedules arc given in the table.
(b) At 6th quantity MR and MC are equal.
(c) Equilibrium quantity of output is 6 units and equilibrium price is ₹ 19.
(d) At Equilibrium
TR = ₹ 114
TC = ₹ 109
Profit = ₹ 5
Question 18.
Discuss the method of measurement of National Income?
Measurement of National Income: There are three types of Measurement of National Income:
1. Product Method or Value Added Method
2. Income Method
3. Expenditure Method
(1) Value Added Method or Product Method:
First Step: Identification and Classification of Productive Enterprise: At the very first step, we are to identify and classify various productive enterprises of an economy. Broadly speaking, we can classify the economy into the following three sectors:
(i) Primary Sector, (ii) Secondary Sector, (iii) Tertiary Sector. .
(i) Primary Sector: it is that sector which produces goods by exploiting natural resources like land, water, forests, mines, etc. It includes all agricultural and allied activities, such as fishing, forestry, mining and quarrying.
(ii) Secondary Sector: This sector is also known as manufacturing sector. It transforms one type of commodity into another, using men, machines and materials. For example, manufacturing of cloth cotton or sugar from sugarcane.
(iii) Tertiary Sector: This sector is also known as service sector which provides useful services to primary and secondary sectors. It consists of banking, insurance, transport, communication, trade and commerce etc.
Second Step: Calculation of Net Value of Output: To estimate the net value added in each identified enterprise in first step the following estimates arc calculated:
(a) Value of Output, (b) Value of Intermediate Consumption, (c) Consumption of Fixed Capital, i.e., Depreciation.
Value of output is worked out by multiplying the amount of goods and services by each enterprise with their market prices. Value of intermediate consumption is calculated by using the prices paid by the enterprise. Consumption of fixed capital is also estimated as per rules and regulations.
To arrive at the net value added by the enterprise, we have to deduct the following items from the value of output:
• Value of Intermediate Consumption
• Consumption of Fixed Capital
• Net Indirect faxes.
In short,
Value Added = Value of Output – Intermediate Consumption – Net Indirect Taxes
By adding the net value added by all the producing enterprises in an industrial sector, we obtain net value added to that industrial sector. The sum total of net values added by all the industrial sectors in the domestic territory of the country, gives us the Net Domestic Product at Factor Cost.
Third Step: Calculation of Net Factor Income from Abroad ;The third and final step in the estimation of national income is to estimate the net factor income earned from abroad and add it to the net domestic product at factor cost. This gives us the national income.
In short,
NNPFC = NDPFC + NFIA
Or
(2) Income Method: In the process of production, different factors like land, labour, capital and organisation co-ordinate with each other and produce goods. These factors belong to household sector and get factor income like rent, wages, interest and profit in return of their services. The sum total of factor’s income is known as national income with income method. This method is also called Factor Payment or Distributed Share Method.
National Income = Wage + Rent + Interest + Dividend + Undistributed Profit + Corporate Profit Tax + Surplus of Public Sector + Mixed Income + Net Income from Abroad.
Where,
Wage: Reward given to labour for his work.
Rent: Reward given to the landlord or owner of building.
It includes imputed rent. Rent of owner occupied houses should be imputed on the basis of prevailing market price and be included in the national income.
Interest: Reward on the capital given as loan.
Profit: Rewards to firms for bearing uncertainties and risks in the production process.
Components of Profit:
• Dividend: A part of the profit which is distributed by the company among shareholders.
• Undistributed Profit: The remaining profit income of the company after paying profit tax and dividend is called the reserve fund which is also known as undistributed profit.
• Corporate Profit Tax: It is a direct tax which government imposes on the profit income of the company.
Mixed Income: Mixed income refers to the incomes of the self-employed persons using their labour, land, capital and entrepreneurship to produce goods and services. These incomes are mixed in terms of wages, rent, interest and profit. That is why it is called mixed income. Such incomes arc also a part of national income.
Surplus of Public Sector: Income earned by public enterprises is also a part of national income.
(3) Expenditure Method:
Expenditure method is the third method for calculating national income under this method, national income is estimated by aggregating all the final expenditure in an economy during a year.
Definition: “Expenditure method is the method which measures the final expenditure on gross domestic product at market price during an accounting year. This total final expenditure is equal to the gross domestic product at market price.”
In expenditure method only final experiditure is taken into consideration. In this method, domestic product is measured as a flow of final expenditure on final goods and services produced in an economy in a year. This final expenditure is termed as ‘Gross Domestic Product at Market Price’ (GDPMP).
Steps in Expenditure Method: Following steps are included in expenditure method in calculating national income:
1. Identification of economic units incurring final expenditure.
2. Classification of final expenditure.
3. Estimation of final expenditure.
4. Estimation of net factor income from abroad.
5. Estimation of national income.
Step 1. To Identify Economic Units incurring Final Expenditure: Various economic units which incur final expenditure within the domestic territory of a country can be grouped under the following categories:
• Household Sector,
• Producing Sector,
• Government Sector and
• Rest of the World Sector.
Step 2. Classification of Final Expenditure: The final expenditure is classified in the following five main categories:
• Private Final Consumption Expenditure,
• Government Final Consumption Expenditure,
• Gross Fixed Capital Formation (or Gross Fixed Investment Expenditure),
• Change in Stocks (or Inventories),
• Net Exports.
Step 3. Measurement of Final Expenditure on Domestic Product: To get value of gross final expenditure on domestic product, we require two types of data for this purpose:
(a) Volume of Gross Sales, (b) Retail Prices.
By multiplying volume of sales with their respective retail prices and then by adding them all, we get GDPMP.
Step 4. Estimation of Net Factor Income from Abroad: Finally, value of net factor income from abroad (NFIA) is estimated which is added to GDPMP toget GNPMP.
Step 5. Estimation of National Income: For obtaining national income at factor cost, cost of depreciation and net indirect taxes have to be subtracted.
Question 19.
Define budget deficit and trade deficit. The excess of private investment over savings of a country in a particular year was ₹ 2000 crores. The amount of budget deficit was (-) ₹ 1500 crores. What was the volume of trade deficit of that country?
Budget deficit means the excess of the government expenditure over tax revenue.
Budget Deficit = Government Expenditure (G) – Tax Revenue (T)
Trade deficit means the excess of import expenditure over the export revenue earned by the economy. Thus,
Trade Deficit = Imports (M) – Exports (X)
Given, G – T = (-)1500
I – S = 2000
Trade Deficit = (I – S) + (G – T)
= 2000 – 1500 = 500 crores
Question 20.
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https://www.conwaylife.com/forums/viewtopic.php?f=11&t=3089 | ## Slowest one cell spaceships of each state count
For discussion of other cellular automata.
AforAmpere
Posts: 1156
Joined: July 1st, 2016, 3:58 pm
### Slowest one cell spaceships of each state count
This will be a challenge to find the slowest 1-cell spaceship in a non-symmetric rule of any number of states. For 2 states, it is obviously c, whether orthogonal or diagonal, as it cannot move in any direction without using a B1e or B1c transition, and so has to move at c:
Code: Select all
x = 1, y = 1, rule = MAPGAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
o!
However, this is not true for higher state rules, as this preliminary example shows, with three states, acheiving C/16:
Code: Select all
@RULE SlowestOneCell3StateCurrent
@TABLE
n_states:3
neighborhood:Moore
symmetries:none
var a={0,1,2}
0,0,0,0,0,1,0,0,0,2
0,1,0,0,0,0,0,0,0,2
0,0,0,0,0,0,1,2,0,1
0,0,0,0,0,0,0,2,1,1
0,0,0,0,0,0,1,0,1,2
0,0,0,0,0,2,0,1,0,2
0,2,0,0,0,0,0,1,0,2
0,1,2,2,2,1,2,1,2,2
2,0,0,1,2,1,0,0,0,0
1,2,1,2,1,2,0,0,0,0
2,1,2,1,0,0,0,0,0,0
2,0,0,0,0,2,2,1,0,0
2,2,0,0,0,2,1,2,1,0
2,2,0,0,0,0,0,1,2,0
2,1,2,2,2,1,2,1,2,2
1,2,2,2,0,0,0,2,1,0
2,0,0,0,0,0,0,0,0,1
0,0,0,2,0,0,0,0,0,1
0,1,1,0,0,0,0,0,0,2
0,1,0,0,0,0,0,0,1,2
0,0,0,0,1,1,0,0,0,2
0,0,0,0,0,1,1,0,0,2
1,2,0,2,0,2,2,1,2,2
2,2,0,2,0,2,2,1,2,1
2,0,0,2,2,1,0,0,0,0
2,0,0,0,2,2,1,2,0,0
2,0,0,0,0,0,2,2,2,0
2,2,2,0,0,0,0,2,1,0
2,1,2,2,0,0,0,0,0,0
1,2,2,2,2,2,0,0,0,0
1,0,0,2,2,2,1,2,0,0
0,0,0,0,0,2,2,2,0,2
0,2,0,0,0,0,0,2,2,2
2,0,0,0,0,0,1,0,0,0
2,0,0,0,0,0,0,0,1,0
The slowest found of each state count will be held here. Entries should be submitted only with the condition that the spaceship the rule is designed for is one celled in one phase, and the state of the cell in that phase is state 1. The above two rules follow this condition, so when one state 1 cell is placed, it will follow the evolution of the entry spaceship. No direction is required, just a speed.
Records:
2-state: C
Code: Select all
MAPGAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
3-state: C/36
Code: Select all
@RULE slowshiptry
@TABLE
n_states: 3
neighborhood:Moore
symmetries:none
1,0,0,0,0,0,0,0,0,1
0,0,0,0,0,0,0,1,0,2
2,0,0,0,0,0,0,1,0,1
1,0,0,1,0,0,0,0,0,2
0,0,0,0,2,1,2,0,0,1
0,0,0,1,1,2,0,0,0,2
0,0,0,2,2,0,0,0,0,1
1,0,0,2,2,0,0,0,0,0
0,1,2,2,0,0,0,0,0,1
0,0,0,2,2,1,0,0,0,2
2,0,0,2,2,1,0,0,0,1
1,2,2,2,0,0,0,0,0,2
1,0,0,2,2,2,0,0,0,0
2,0,2,2,0,0,0,0,0,0
2,2,1,1,0,0,0,2,0,0
2,0,0,1,1,0,0,0,0,0
1,1,0,1,0,0,0,0,2,2
0,0,0,0,1,1,2,1,0,2
1,0,0,2,1,2,0,0,0,0
2,0,0,0,1,1,2,0,0,0
0,0,0,0,1,1,1,2,0,2
2,0,0,2,1,1,2,0,0,0
2,0,0,2,1,1,1,0,0,0
0,0,0,0,1,1,1,2,0,2
0,0,0,0,2,1,1,2,0,2
1,0,0,0,0,0,0,1,2,0
0,0,0,0,2,0,1,2,0,2
0,0,0,2,0,0,0,1,2,2
0,0,0,0,0,2,2,2,0,2
2,0,0,2,2,1,0,0,0,0
2,2,0,2,0,0,0,1,2,1
2,0,0,0,0,0,0,2,2,0
2,0,0,2,0,1,1,0,0,0
0,0,0,2,0,1,1,0,0,2
1,0,2,0,0,0,0,1,0,0
2,0,0,0,0,0,1,0,0,0
0,0,0,2,0,1,2,0,0,2
1,2,0,0,0,0,0,2,0,0
2,0,0,0,0,0,2,0,0,0
2,0,2,0,0,0,0,0,0,1
4-state: C/917636
Code: Select all
@RULE MinSpeed4s-AFP-9-12-17
@TABLE
n_states:4
neighborhood:Moore
symmetries:none
0100000002
0200000003
0001000002
0002000003
2001200303
3002000002
2100030023
3200000002
2200000001
3100000000
1200000000
2000000000
3000300000
3000000030
0020030003
2300003000
0030030003
3300000000
0300000302
2300000303
3033000000
0313000023
3313030020
0233000023
0223000023
0223000033
0223000003
0323000003
3002330001
1002330002
1002333202
2002333101
3233000311
2003333101
2001133203
3311000322
2311000323
2003233203
3001123202
2001133303
3311000332
3001123302
2311000333
3001123302
3332000322
2311000333
3002323202
2323000323
2003233303
3332000332
2323000333
3002323302
2003230000
0313030023
3313230020
3032000000
3023000000
2002333000
3001133000
1003313000
1133000303
3011000000
3001333001
1000033100
3100000310
3113000000
0200020001
0100000303
0021330002
2021330000
3100000320
1200033200
Last edited by AforAmpere on September 17th, 2017, 3:37 pm, edited 3 times in total.
Wildmyron and I manage the 5S project, which collects all known spaceship speeds in Isotropic Non-totalistic rules.
Things to work on:
- Find a (7,1)c/8 ship in a Non-totalistic rule
muzik
Posts: 3925
Joined: January 28th, 2016, 2:47 pm
Location: Scotland
### Re: Slowest one cell spaceships of each state count
Couldn't you just make a cell age as in Generations, but as it reaches its last stage, instead of just dying outright it births a state-1 cell to the right of it?
Bored of using the Moore neighbourhood for everything? Introducing the Range-2 von Neumann isotropic non-totalistic rulespace!
blah
Posts: 285
Joined: April 9th, 2016, 7:22 pm
### Re: Slowest one cell spaceships of each state count
muzik wrote:Couldn't you just make a cell age as in Generations, but as it reaches its last stage, instead of just dying outright it births a state-1 cell to the right of it?
Yeah, but that's probably not optimal in most cases. See the example he actually posted, of a 3-state rule in which a single cell travels more slowly than the 3-state implementation of your idea.
Edit: Maybe your idea is still useful to establish an upper bound on the lowest possible speed for a given number of states.
succ
Posts: 1977
Joined: November 8th, 2014, 8:48 pm
Location: Getting a snacker from R-Bee's
### Re: Slowest one cell spaceships of each state count
This looks like a neat variation on the busy beaver problem.
EDIT: c/550 diagonal, 3 ON states:
Code: Select all
@RULE BB3b
@TABLE
n_states:4
neighborhood:Moore
symmetries:none
var c1={1,3}
var c2=c1
var c3=c2
var c4=c3
var c5=c4
var c6=c5
var c7=c6
var c8=c7
var C1={0,1,3}
var C2=C1
var C3=C2
var C4=C3
var C5=C4
var C6=C5
var C7=C6
var C8=C7
#open up
1,0,0,0,0,0,0,0,0,2
3,0,0,0,0,0,0,0,0,2
2,0,0,0,0,0,0,0,0,3
0,2,0,0,0,0,0,0,0,2
0,0,2,0,0,0,0,0,0,3
0,0,0,2,0,0,0,0,0,2
0,0,0,0,2,0,0,0,0,3
0,0,0,0,0,2,0,0,0,2
0,0,0,0,0,0,2,0,0,3
0,0,0,0,0,0,0,2,0,2
0,0,0,0,0,0,0,0,2,3
0,3,2,0,0,0,0,0,0,3
0,2,3,0,0,0,0,0,3,3
0,3,0,0,0,0,0,0,2,3
0,0,0,3,2,0,0,0,0,3
0,0,3,2,3,0,0,0,0,3
0,0,2,3,0,0,0,0,0,3
0,0,0,0,0,3,2,0,0,3
0,0,0,0,3,2,3,0,0,3
0,0,0,0,2,3,0,0,0,3
0,0,0,0,0,0,0,3,2,3
0,0,0,0,0,0,3,2,3,3
0,0,0,0,0,0,2,3,0,3
2,3,2,3,2,3,0,0,0,3
2,0,0,3,2,3,2,3,0,3
2,3,0,0,0,3,2,3,2,3
2,3,2,3,0,0,0,3,2,3
#count or decay
1,1,1,1,1,0,0,0,0,0
1,1,1,1,0,0,0,0,0,0
1,0,1,1,1,0,0,0,0,0
1,1,1,1,0,0,0,0,1,0
1,1,1,0,0,0,0,0,1,0
1,1,0,0,0,0,0,0,0,0
1,1,0,0,0,0,0,0,1,0
1,0,0,0,1,1,0,0,0,0
1,0,0,1,0,0,0,0,0,0
1,0,0,1,1,0,0,0,0,0
3,0,0,0,0,c1,c2,c3,c4,1
1,0,0,0,0,c1,c2,c3,c4,3
3,1,0,0,0,C1,C2,C3,C4,1
1,1,0,0,0,C1,C2,C3,C4,3
3,0,0,0,1,C1,C2,C3,C4,1
1,0,0,0,1,C1,C2,C3,C4,3
3,1,0,1,1,C1,C2,C3,C4,1
1,1,0,1,1,C1,C2,C3,C4,3
3,1,1,1,1,C1,C2,C3,C4,1
1,1,1,1,1,C1,C2,C3,C4,3
3,1,1,1,0,C1,C2,C3,C4,1
1,1,1,1,0,C1,C2,C3,C4,3
3,1,1,0,0,C1,C2,C3,C4,1
1,1,1,0,0,C1,C2,C3,C4,3
3,0,0,1,1,C1,C2,C3,C4,1
1,0,0,1,1,C1,C2,C3,C4,3
3,0,1,1,1,C1,C2,C3,C4,1
1,0,1,1,1,C1,C2,C3,C4,3
#decay
3,3,0,3,3,0,0,3,3,2
2,3,0,3,3,0,0,3,3,1
3,0,0,0,3,2,3,3,0,0
3,0,0,0,0,3,0,2,3,0
3,3,0,0,0,0,0,0,2,0
3,3,3,2,0,0,0,0,0,0
3,0,0,3,2,3,0,0,0,0
3,0,1,0,0,0,0,0,0,0
Extensible to higher state numbers as well, but would likely require non-trivial changes.
Last edited by BlinkerSpawn on September 12th, 2017, 10:30 pm, edited 1 time in total.
LifeWiki: Like Wikipedia but with more spaceships. [citation needed]
dvgrn
Moderator
Posts: 7371
Joined: May 17th, 2009, 11:00 pm
Contact:
### Re: Slowest one cell spaceships of each state count
Yikes. No limit on the number of lines in the rule table, only on the number of states? The lower bound on the busy beaver Σ(N) function for increasing N goes like
4
6
13
4098
3.5×10^18267
10^10^10^10^18705352
I'm not sure the Single-Cell Spaceship Slowness function will take off quite as vertically as that, but when the exponents get big enough it can be kind of hard to tell the difference...! Come to think of it, rule tables would seem to have some resemblance to two-dimensional Turing machines -- for all I know, the function could even go up faster than Σ.
EDIT: Here's a problem that's probably about to show up: the rule table file for the slowest possible spaceship will start taking up terabytes of space, while the number of states is still in the single digits. Can I suggest a modification of the contest conditions? The rule table should fit in a
Code: Select all
block in a forum post -- not an attached ZIP file or anything like that, just the plain quoted text in a single message.
toroidalet
Posts: 1177
Joined: August 7th, 2016, 1:48 pm
Location: My computer
Contact:
### Re: Slowest one cell spaceships of each state count
c/36:
Code: Select all
x = 1, y = 1, rule = slowshiptry
A!
Code: Select all
@RULE slowshiptry
@TABLE
n_states: 3
neighborhood:Moore
symmetries:none
1,0,0,0,0,0,0,0,0,1
0,0,0,0,0,0,0,1,0,2
2,0,0,0,0,0,0,1,0,1
1,0,0,1,0,0,0,0,0,2
0,0,0,0,2,1,2,0,0,1
0,0,0,1,1,2,0,0,0,2
0,0,0,2,2,0,0,0,0,1
1,0,0,2,2,0,0,0,0,0
0,1,2,2,0,0,0,0,0,1
0,0,0,2,2,1,0,0,0,2
2,0,0,2,2,1,0,0,0,1
1,2,2,2,0,0,0,0,0,2
1,0,0,2,2,2,0,0,0,0
2,0,2,2,0,0,0,0,0,0
2,2,1,1,0,0,0,2,0,0
2,0,0,1,1,0,0,0,0,0
1,1,0,1,0,0,0,0,2,2
0,0,0,0,1,1,2,1,0,2
1,0,0,2,1,2,0,0,0,0
2,0,0,0,1,1,2,0,0,0
0,0,0,0,1,1,1,2,0,2
2,0,0,2,1,1,2,0,0,0
2,0,0,2,1,1,1,0,0,0
0,0,0,0,1,1,1,2,0,2
0,0,0,0,2,1,1,2,0,2
1,0,0,0,0,0,0,1,2,0
0,0,0,0,2,0,1,2,0,2
0,0,0,2,0,0,0,1,2,2
0,0,0,0,0,2,2,2,0,2
2,0,0,2,2,1,0,0,0,0
2,2,0,2,0,0,0,1,2,1
2,0,0,0,0,0,0,2,2,0
2,0,0,2,0,1,1,0,0,0
0,0,0,2,0,1,1,0,0,2
1,0,2,0,0,0,0,1,0,0
2,0,0,0,0,0,1,0,0,0
0,0,0,2,0,1,2,0,0,2
1,2,0,0,0,0,0,2,0,0
2,0,0,0,0,0,2,0,0,0
2,0,2,0,0,0,0,0,0,1
"I'm sure we all agree that we ought to love one another, and I know there are people in the world who do not love their fellow human beings, and I hate people like that!"
-Tom Lehrer
AforAmpere
Posts: 1156
Joined: July 1st, 2016, 3:58 pm
### Re: Slowest one cell spaceships of each state count
Edited, what is the number of states where a computer that counts to any arbitrarily high number and then resets to one cell is possible? I feel like it is probably 50 states or less.
Wildmyron and I manage the 5S project, which collects all known spaceship speeds in Isotropic Non-totalistic rules.
Things to work on:
- Find a (7,1)c/8 ship in a Non-totalistic rule
toroidalet
Posts: 1177
Joined: August 7th, 2016, 1:48 pm
Location: My computer
Contact:
### Re: Slowest one cell spaceships of each state count
AforAmpere wrote:What is the number of states where a computer that counts to any arbitrarily high number and then resets to one cell is possible?
Should be possible in ≤15 states to make a ship with a period on the order of double or maybe triple exponential, based on the double-binary counter ship posted here (by me, shameless self-promo). I would make this, except that the vodka is good, but the meat is rotten.
"I'm sure we all agree that we ought to love one another, and I know there are people in the world who do not love their fellow human beings, and I hate people like that!"
-Tom Lehrer
A for awesome
Posts: 2160
Joined: September 13th, 2014, 5:36 pm
Location: Pembina University, Home of the Gliders
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### Re: Slowest one cell spaceships of each state count
Code: Select all
@RULE MinSpeed4s-AFP-9-17-17
@TABLE
n_states:4
neighborhood:Moore
symmetries:none
0100000002
0200000003
0001000002
0002000003
2001200303
3002000002
2100030023
3200000002
2200000001
3100000000
1200000000
2000000000
3000300000
3000000030
0020030003
2300003000
0030030003
3300000000
0300000302
2300000303
3033000000
0313000023
3313030020
0233000023
0223000023
0223000033
0223000003
0323000003
3002330001
1002330002
1002333202
2002333101
3233000311
2003333101
2001133203
3311000322
2311000323
2003233203
3001123202
2001133303
3311000332
3001123302
2311000333
3001123302
3332000322
2311000333
3002323202
2323000323
2003233303
3332000332
2323000333
3002323302
2003230000
0313030023
3313230020
3032000000
3023000000
2002333000
3001133000
1003313000
1133000303
3011000000
3001333001
1000033100
3100000310
3113000000
0200020001
0100000303
0021330002
2021330000
3100000320
1200033200
c/917636:
Code: Select all
x = 1, y = 1, rule = MinSpeed4s-AFP-9-17-17
A!
I'm sure there are plenty of trivial improvements that could be made, as well as nontrivial ones such as adding another binary counter.
EDIT: Fixed date.
praosylen#5847 (Discord)
x₁=ηx
V*_η=c²√(Λη)
K=(Λu²)/2
Pₐ=1−1/(∫^∞_t₀(p(t)ˡ⁽ᵗ⁾)dt)
$$x_1=\eta x$$
$$V^*_\eta=c^2\sqrt{\Lambda\eta}$$
$$K=\frac{\Lambda u^2}2$$
$$P_a=1-\frac1{\int^\infty_{t_0}p(t)^{l(t)}dt}$$
gameoflifemaniac
Posts: 1216
Joined: January 22nd, 2017, 11:17 am
Location: There too
### Re: Slowest one cell spaceships of each state count
Tried to make a 1-cell 2-state c/4 spaceship, and I got this:
Code: Select all
@RULE MyEntry
@TABLE
n_states:2
neighborhood:Moore
symmetries:none
0,0,0,0,0,1,0,0,0,1
0,0,0,0,0,0,1,1,0,1
0,1,0,0,0,0,0,1,1,1
1,0,0,1,1,1,0,0,0,0
1,0,0,0,0,1,1,1,0,0
1,1,0,0,0,0,0,1,1,1
1,1,1,1,0,0,0,0,0,0
Code: Select all
x = 1, y = 1, rule = MyEntry
o!
I was so socially awkward in the past and it will haunt me for the rest of my life.
Code: Select all
b4o25bo$o29bo$b3o3b3o2bob2o2bob2o2bo3bobo$4bobo3bob2o2bob2o2bobo3bobo$
4bobo3bobo5bo5bo3bobo$o3bobo3bobo5bo6b4o$b3o3b3o2bo5bo9bobo$24b4o! fluffykitty Posts: 977 Joined: June 14th, 2014, 5:03 pm Contact: ### Re: Slowest one cell spaceships of each state count Please check your rule. fluffykitty Posts: 977 Joined: June 14th, 2014, 5:03 pm Contact: ### Re: Slowest one cell spaceships of each state count c/2596148429267413814265248164610160 in 3 states. Also, I completely forgot that this thread existed until someone linked it in a thread someone linked to in the small long lived methuselahs thread. Code: Select all @RULE 3S1CShip2 startup: 108 main: sum(3,110)2^n=2^111-8 ending: 12 total: 2^111+112=2596148429267413814265248164610160 @TABLE n_states:3 neighborhood:Moore symmetries:none #delay 0,0,0,0,0,2,1,2,0,2 2,0,0,0,0,0,1,0,0,1 1,0,0,0,0,0,1,0,0,0 2,0,0,0,0,0,2,0,0,1 1,0,0,0,0,0,2,0,0,0 #leftward expansion 0,0,0,1,0,0,0,0,0,2 0,0,0,0,0,1,2,0,0,2 0,0,0,2,1,2,0,0,0,1 2,0,0,0,2,1,2,0,0,0 2,1,0,1,2,0,0,0,0,1 1,0,0,0,1,1,0,0,0,0 1,1,0,1,2,0,0,0,0,2 0,0,1,1,0,0,0,0,0,2 0,0,0,1,1,0,0,0,0,1 2,1,0,2,0,0,0,0,0,1 2,0,0,0,0,1,2,0,0,0 1,1,0,2,0,0,0,0,0,2 1,0,0,0,2,1,0,0,0,0 #downward expansion 0,1,0,0,0,0,0,0,0,2 0,0,0,0,0,0,2,1,0,2 0,2,0,0,0,0,0,2,1,1 2,0,0,0,0,0,2,1,2,0 2,1,2,1,0,0,0,0,2,1 2,0,0,0,0,1,2,1,0,0 1,0,0,0,0,0,0,1,1,0 2,1,0,1,0,0,0,0,1,1 1,1,0,1,0,0,0,0,2,2 0,1,1,0,0,0,0,0,0,2 0,1,0,0,0,0,0,0,1,1 2,2,0,1,0,0,0,0,0,1 1,2,0,1,0,0,0,0,0,2 1,0,0,0,0,0,0,1,2,0 #downward stop 0,1,0,0,0,0,2,1,0,2 2,0,0,0,0,0,2,1,0,0 0,2,0,0,0,1,1,2,2,2 2,1,0,2,0,2,0,0,2,1 2,0,0,0,0,0,2,2,1,0 0,2,0,0,0,1,2,2,2,2 2,0,0,0,0,1,1,2,1,0 2,1,0,2,1,1,0,0,0,1 1,2,2,1,0,0,0,0,0,2 1,2,0,0,0,0,0,1,2,2 2,1,0,2,1,2,0,0,0,1 2,0,0,0,0,1,2,2,1,0 1,2,0,0,0,0,0,2,2,0 #downward counting 0,1,0,0,0,0,1,1,0,2 2,0,0,0,0,0,1,1,0,0 1,1,2,0,0,1,0,0,2,2 0,2,0,0,0,0,1,2,1,2 2,0,0,0,0,0,1,2,1,0 2,1,0,2,0,1,0,0,2,1 1,2,2,0,0,1,0,0,0,2 1,1,2,0,0,2,0,0,2,2 0,2,0,0,0,0,2,2,1,2 0,2,0,0,0,0,1,2,2,2 2,1,0,2,0,1,0,0,0,1 1,2,2,0,0,2,0,0,0,2 0,2,0,0,0,0,2,2,2,2 2,1,0,2,0,2,0,0,0,1 0,2,0,0,0,0,0,2,2,2 2,0,0,0,0,0,0,1,1,0 1,1,0,0,0,2,2,0,0,2 2,0,1,2,0,0,0,0,0,0 2,1,0,0,0,0,0,2,0,0 1,2,0,0,0,2,2,0,0,2 0,2,1,1,2,2,0,0,2,2 #1,1,0,0,0,2,2,0,2,1 2,2,1,1,0,0,0,0,2,0 1,1,0,0,0,0,0,2,2,0 1,0,0,0,0,0,0,2,0,2 #leftward stop 0,0,0,0,0,2,2,0,0,2 2,0,0,0,0,2,2,0,0,0 0,0,0,0,0,0,2,2,0,2 0,0,0,2,2,2,2,0,0,2 2,0,0,0,2,2,2,0,0,0 2,2,0,2,0,0,0,2,0,1 2,0,0,0,1,2,2,0,0,0 2,2,0,1,0,0,0,2,0,1 0,0,0,2,2,2,1,1,0,2 0,0,0,2,2,2,2,1,0,2 1,0,0,2,2,1,0,0,0,2 2,0,0,0,1,2,1,1,0,0 2,2,0,1,0,0,0,1,1,1 1,0,0,2,1,2,0,0,0,0 2,0,2,0,1,1,2,1,0,1 1,0,1,0,1,1,2,0,0,0 #leftward counting 0,0,0,0,0,2,1,0,0,2 2,0,0,0,0,2,1,0,0,0 1,0,2,2,0,0,0,1,0,2 0,0,0,2,2,2,1,0,0,2 2,0,0,0,2,2,1,0,0,0 2,2,0,2,0,0,0,1,0,1 1,0,2,2,0,0,0,2,0,2 2,0,0,0,1,2,1,0,0,0 2,2,0,1,0,0,0,1,0,1 0,0,0,2,2,2,0,0,0,2 2,0,0,0,1,1,0,0,0,0 2,0,0,1,0,2,0,0,0,0 2,2,1,0,0,0,0,0,0,0 1,0,0,1,0,0,2,2,0,2 1,0,0,2,0,0,2,2,0,2 1,0,2,2,0,0,2,2,0,2 #ending 2,2,0,2,0,0,0,0,0,0 2,0,0,0,2,0,0,0,0,0 2,0,0,0,0,0,0,0,0,1 #diagonal mode #2,0,0,0,2,0,0,0,0,0 #2,0,1,0,0,0,0,0,2,0 #0,0,0,1,0,2,0,2,0,1 necrodoublepost ftw Edit: c/131661808029361064474122541992202249544316172470474290896938957889657411706431738059855987100606116738608709863898114516322696806682779725395040361126073989754846342223650752839934282507553095228819445591564242224229100999549808158043220482537489286899542806927109262299798576702052364647531326323060875437512721666750074735365806697524459192563914773873444570420147733289826848369892651509013842153766576752759546709772991670483710813378588223114875800405103605361303657861719088971021114521028156816509308794659625756698728308129465992649969171244954433589856461308546917926791439305402526271547340075851639933262085702464422986758336147589425826676630311810520418347644992379850195840502843574915602319663 in 3 states. Can you get 3 counters in 3 states? Code: Select all @RULE 3S1CShip5 TODO startup: 2346 main: sum(3,2348)2^n=2^2349-8 ending: 13 total: 2^2349+2351=131661808029361064474122541992202249544316172470474290896938957889657411706431738059855987100606116738608709863898114516322696806682779725395040361126073989754846342223650752839934282507553095228819445591564242224229100999549808158043220482537489286899542806927109262299798576702052364647531326323060875437512721666750074735365806697524459192563914773873444570420147733289826848369892651509013842153766576752759546709772991670483710813378588223114875800405103605361303657861719088971021114521028156816509308794659625756698728308129465992649969171244954433589856461308546917926791439305402526271547340075851639933262085702464422986758336147589425826676630311810520418347644992379850195840502843574915602319663 @TABLE n_states:3 neighborhood:Moore symmetries:none #delay 0,0,0,0,0,2,1,2,0,2 2,0,0,0,0,0,1,0,0,1 1,0,0,0,0,0,1,0,0,0 2,0,0,0,0,0,2,0,0,1 1,0,0,0,0,0,2,0,0,0 #leftward expansion 0,0,0,1,0,0,0,0,0,2 0,0,0,0,0,1,2,0,0,2 0,0,0,2,1,2,0,0,0,1 2,0,0,0,2,1,2,0,0,0 2,1,0,1,2,0,0,0,0,1 1,0,0,0,1,1,0,0,0,0 1,1,0,1,2,0,0,0,0,2 0,0,1,1,0,0,0,0,0,2 0,0,0,1,1,0,0,0,0,1 2,1,0,2,0,0,0,0,0,1 2,0,0,0,0,1,2,0,0,0 1,1,0,2,0,0,0,0,0,2 1,0,0,0,2,1,0,0,0,0 #downward expansion 0,1,0,0,0,0,0,0,0,2 0,0,0,0,0,0,2,1,0,2 0,2,0,0,0,0,0,2,1,1 2,0,0,0,0,0,2,1,2,0 2,1,2,1,0,0,0,0,2,1 2,0,0,0,0,1,2,1,0,0 1,0,0,0,0,0,0,1,1,0 2,1,0,1,0,0,0,0,1,1 1,1,0,1,0,0,0,0,2,2 0,1,1,0,0,0,0,0,0,2 0,1,0,0,0,0,0,0,1,1 2,2,0,1,0,0,0,0,0,1 1,2,0,1,0,0,0,0,0,2 1,0,0,0,0,0,0,1,2,0 #downward stop 0,1,0,0,0,0,2,1,0,2 2,0,0,0,0,0,2,1,0,0 0,2,0,0,0,1,1,2,2,2 2,1,0,2,0,2,0,0,1,1 2,0,0,0,0,0,2,2,1,0 0,2,0,0,0,1,2,2,2,2 2,0,0,0,0,1,1,2,1,0 2,1,0,2,1,1,0,0,0,1 1,2,2,1,0,0,0,0,0,2 1,2,0,0,0,0,0,1,2,2 2,1,0,2,1,2,0,0,0,1 2,0,0,0,0,1,2,2,1,0 1,2,0,0,0,0,0,2,2,0 #downward counting 0,1,0,0,0,0,1,1,0,2 2,0,0,0,0,0,1,1,0,0 1,1,2,0,0,1,0,0,1,2 0,2,0,0,0,0,1,2,1,2 2,0,0,0,0,0,1,2,1,0 2,1,0,2,0,1,0,0,2,1 1,2,2,0,0,1,0,0,0,2 1,1,2,0,0,2,0,0,1,2 0,2,0,0,0,0,2,2,1,2 0,2,0,0,0,0,1,2,2,2 2,1,0,2,0,1,0,0,0,1 1,2,2,0,0,2,0,0,0,2 0,2,0,0,0,0,2,2,2,2 2,1,0,2,0,2,0,0,0,1 0,2,0,0,0,0,0,2,2,2 2,0,0,0,0,0,0,1,1,0 1,1,0,0,0,2,2,0,0,2 2,0,1,2,0,0,0,0,0,0 2,1,0,0,0,0,0,2,0,0 1,2,0,0,0,2,2,0,0,2 0,1,1,1,2,2,0,0,2,2 2,1,1,1,0,0,0,0,2,0 1,1,0,0,0,0,0,2,1,0 1,0,0,0,0,0,0,2,0,2 #leftward stop 0,0,0,0,0,2,2,0,0,2 2,0,0,0,0,2,2,0,0,0 0,0,0,0,0,0,2,2,0,2 0,0,0,2,2,2,2,0,0,2 2,0,0,0,2,2,2,0,0,0 2,2,0,2,0,0,0,2,0,1 2,0,0,0,1,2,2,0,0,0 2,2,0,1,0,0,0,2,0,1 0,0,0,2,2,2,1,1,0,2 0,0,0,2,2,2,2,1,0,2 1,0,0,2,2,1,0,0,0,2 2,0,0,0,1,2,1,1,0,0 2,2,0,1,0,0,0,1,1,1 1,0,0,2,1,2,0,0,0,0 2,0,2,0,1,1,2,1,0,1 1,0,1,0,1,1,2,0,0,0 #leftward counting 0,0,0,0,0,2,1,0,0,2 2,0,0,0,0,2,1,0,0,0 1,0,2,2,0,0,0,1,0,2 0,0,0,2,2,2,1,0,0,2 2,0,0,0,2,2,1,0,0,0 2,2,0,2,0,0,0,1,0,1 1,0,2,2,0,0,0,2,0,2 2,0,0,0,1,2,1,0,0,0 2,2,0,1,0,0,0,1,0,1 0,0,0,2,2,2,0,0,0,2 2,0,0,0,1,1,0,0,0,0 2,0,0,1,0,2,0,0,0,0 2,2,1,0,0,0,0,0,0,0 1,0,0,1,0,0,2,2,0,2 1,0,0,2,0,0,2,2,0,2 1,0,2,2,0,0,2,2,0,2 #ending 2,2,0,2,0,0,0,0,0,0 2,0,0,0,2,0,0,0,0,0 2,0,0,0,0,0,0,0,0,1 #diagonal mode #2,0,0,0,2,0,0,0,0,0 #2,0,1,0,0,0,0,0,2,0 #0,0,0,1,0,2,0,2,0,1 #extra length 2,0,2,1,2,0,0,1,1,1 2,1,0,0,1,1,0,0,1,1 2,0,0,0,0,0,1,1,2,0 2,0,0,1,2,0,0,2,0,1 2,2,2,1,0,0,0,0,0,1 1,1,0,0,0,0,0,0,0,0 2,0,1,0,2,0,0,0,0,0 0,1,0,0,0,2,0,2,0,1 Moosey Posts: 3895 Joined: January 27th, 2019, 5:54 pm Location: A house, or perhaps the OCA board. Or [click to not expand] Contact: ### Re: Slowest one cell spaceships of each state count Is it possible to do 4 or five states where the counter counts using counters, e.g. Not states, counted numbers Code: Select all 0 0 0 0 0 counter-> 0 0 0 0 1 0 0 0 0 2 0 0 0 0 3 ... 0 0 0 1 0 0 0 0 1 1 0 0 0 1 2 ... 0 0 0 2 0 ... ... 0 0 1 0 0 0 0 1 0 1 0 0 1 0 2 ... 0 0 1 1 0 ... 0 0 1 2 0 ... ... ... 0 1 0 0 0 etc. How long would it go? How many states to make it so that the counter tapes get progressively longer? That would be really slow! On a related note, could the ships in the said rule be adjustable? Sorry to sorta go off on a tangent. My CA rules can be found here Bill Watterson once wrote: "How do soldiers killing each other solve the world's problems?" Nanho walåt derwo esaato? Code: Select all x = 4, y = 5, rule = B34e5e67c/S23ce45 4o2$b2o$o2bo$b2o!
[[ TRACK 0 -7/325 ZOOM 16 THEME 0 ]]
toroidalet
Posts: 1177
Joined: August 7th, 2016, 1:48 pm
Location: My computer
Contact:
### Re: Slowest one cell spaceships of each state count
Significantly slower (I'm not sure what the period is, but the second counter counts up to about 6,900):
Code: Select all
@RULE 3S1CShip7
#7 because I previously made a version 6 and also this rule is somewhat different
@TABLE
n_states:3
neighborhood:Moore
symmetries:none
#delay
0,0,0,0,0,2,1,2,0,2
2,0,0,0,0,0,1,0,0,1
1,0,0,0,0,0,1,0,0,0
2,0,0,0,0,0,2,0,0,1
1,0,0,0,0,0,2,0,0,0
#leftward expansion
0,0,0,1,0,0,0,0,0,2
0,0,0,0,0,1,2,0,0,2
0,0,0,2,1,2,0,0,0,1
2,0,0,0,2,1,2,0,0,0
2,1,0,1,2,0,0,0,0,1
1,0,0,0,1,1,0,0,0,0
1,1,0,1,2,0,0,0,0,2
0,0,1,1,0,0,0,0,0,2
0,0,0,1,1,0,0,0,0,1
2,1,0,2,0,0,0,0,0,1
2,0,0,0,0,1,2,0,0,0
1,1,0,2,0,0,0,0,0,2
1,0,0,0,2,1,0,0,0,0
#downward expansion
0,1,0,0,0,0,0,0,0,2
0,0,0,0,0,0,2,1,0,2
0,2,0,0,0,0,0,2,1,1
2,0,0,0,0,0,2,1,2,0
2,1,2,1,0,0,0,0,2,1
2,0,0,0,0,1,2,1,0,0
1,0,0,0,0,0,0,1,1,0
2,1,0,1,0,0,0,0,1,1
1,1,0,1,0,0,0,0,2,2
0,1,1,0,0,0,0,0,0,2
0,1,0,0,0,0,0,0,1,1
2,2,0,1,0,0,0,0,0,1
1,2,0,1,0,0,0,0,0,2
1,0,0,0,0,0,0,1,2,0
#downward stop
0,1,0,0,0,0,2,1,0,1
1,0,0,0,0,0,2,1,0,2
2,0,0,0,0,0,2,1,0,0
0,2,0,0,0,1,1,2,2,2
2,1,0,2,0,2,0,0,1,1
2,0,0,0,0,0,2,2,1,0
0,2,0,0,0,1,2,2,2,2
2,0,0,0,0,1,1,2,1,0
2,1,0,2,1,1,0,0,0,1
1,2,2,1,0,0,0,0,0,2
1,2,0,0,0,0,0,1,2,2
2,1,0,2,1,2,0,0,0,1
2,0,0,0,0,1,2,2,1,0
1,2,0,0,0,0,0,2,2,0
2,2,1,1,0,0,0,0,0,0
1,1,0,0,0,0,0,2,2,0
0,0,0,2,2,1,2,1,1,1
1,0,0,0,2,1,2,1,1,2
2,1,2,1,0,0,0,0,0,0
1,2,0,2,0,0,0,2,1,0
0,0,0,0,0,2,1,1,0,2
1,1,0,2,1,2,0,0,1,0
2,0,0,0,2,1,2,1,1,0
2,0,1,0,0,0,0,1,2,0
#downward counting
0,1,0,0,0,0,1,1,0,2
1,1,1,0,0,2,2,0,0,2
1,1,0,0,0,0,2,1,1,0
0,0,2,0,0,0,1,2,2,1
0,0,0,0,0,1,1,1,1,1
0,0,2,0,0,0,2,1,1,1
0,2,0,0,0,1,2,2,1,1
0,0,2,0,0,0,2,2,2,1
1,1,0,1,0,1,0,0,1,2
1,1,0,0,0,0,2,2,2,0
1,1,0,0,0,0,1,2,2,0
1,0,0,0,0,0,2,2,1,0
1,0,0,0,0,0,1,2,1,0
1,1,0,0,0,0,1,1,1,0
0,0,2,0,0,0,1,1,1,1
0,0,0,0,0,0,2,0,2,1
1,0,0,0,0,2,0,0,1,0
2,0,0,0,0,0,1,1,0,0
1,1,2,0,0,1,0,0,1,2
0,2,0,0,0,0,1,2,1,2
2,0,0,0,0,0,1,2,1,0
2,1,0,2,0,1,0,0,2,1
1,2,2,0,0,1,0,0,0,2
1,1,2,0,0,2,0,0,1,2
0,2,0,0,0,0,2,2,1,2
0,2,0,0,0,0,1,2,2,2
2,1,0,2,0,1,0,0,0,1
1,2,2,0,0,2,0,0,0,2
0,2,0,0,0,0,2,2,2,2
2,1,0,2,0,2,0,0,0,1
0,2,0,0,0,0,0,2,2,2
2,0,0,0,0,0,0,1,1,0
1,1,0,0,0,2,2,0,0,2
2,0,1,2,0,0,0,0,0,0
2,1,0,0,0,0,0,2,0,0
1,2,0,0,0,2,2,0,0,2
0,1,1,1,2,2,0,0,2,2
2,1,1,1,0,0,0,0,2,0
1,1,0,0,0,0,0,2,1,0
1,0,0,0,0,0,0,2,0,2
#leftward stop
0,0,0,0,0,2,2,0,0,2
2,0,0,0,0,2,2,0,0,0
0,0,0,0,0,0,2,2,0,2
0,0,0,2,2,2,2,0,0,2
2,0,0,0,2,2,2,0,0,0
2,2,0,2,0,0,0,2,0,1
2,0,0,0,1,2,2,0,0,0
2,2,0,1,0,0,0,2,0,1
0,0,0,2,2,2,1,1,0,2
0,0,0,2,2,2,2,1,0,2
1,0,0,2,2,1,0,0,0,2
2,0,0,0,1,2,1,1,0,0
2,2,0,1,0,0,0,1,1,1
1,0,0,2,1,2,0,0,0,0
2,0,2,0,1,1,2,1,0,1
1,0,1,0,1,1,2,0,0,0
2,1,1,0,0,0,0,0,2,0
1,0,0,1,1,2,0,0,0,0
1,0,0,0,1,1,1,0,0,0
#leftward counting
0,0,0,0,0,2,1,0,0,2
2,0,0,0,0,2,1,0,0,0
1,0,2,2,0,0,0,1,0,2
0,0,0,2,2,2,1,0,0,2
2,0,0,0,2,2,1,0,0,0
2,2,0,2,0,0,0,1,0,1
1,0,2,2,0,0,0,2,0,2
2,0,0,0,1,2,1,0,0,0
2,2,0,1,0,0,0,1,0,1
0,0,0,2,2,2,0,0,0,2
2,0,0,0,1,1,0,0,0,0
2,0,0,1,0,2,0,0,0,0
2,2,1,0,0,0,0,0,0,0
1,0,0,1,0,0,2,2,0,2
1,0,0,2,0,0,2,2,0,2
1,0,2,2,0,0,2,2,0,2
#ending
2,2,0,2,0,0,0,0,0,0
2,0,0,0,2,0,0,0,0,0
2,0,0,0,0,0,0,0,0,1
#diagonal mode
#2,0,0,0,2,0,0,0,0,0
#2,0,1,0,0,0,0,0,2,0
#0,0,0,1,0,2,0,2,0,1
#extra length
2,0,2,1,2,0,0,1,1,1
2,1,0,0,1,1,0,0,1,1
2,0,0,0,0,0,1,1,2,0
2,0,0,1,2,0,0,2,0,1
2,2,2,1,0,0,0,0,0,1
1,1,0,0,0,0,0,0,0,0
2,0,1,0,2,0,0,0,0,0
0,1,0,0,0,2,0,2,0,1
I had made a variant of the slowshiptry rule with a c/54 ship, but that doesn't seem useful anymore.
Moosey wrote:Is it possible to do 4 or five states where the counter counts using counters...
So it would have a sequence of counters and every time one counter overflowed, it would trigger the next one to count and when the very last counter overflows, it would destroy itself? That might work, except it might need a few more states for signals.
"I'm sure we all agree that we ought to love one another, and I know there are people in the world who do not love their fellow human beings, and I hate people like that!"
-Tom Lehrer
fluffykitty
Posts: 977
Joined: June 14th, 2014, 5:03 pm
Contact:
### Re: Slowest one cell spaceships of each state count
I made a rule with tetrationally slow ships (~c/2^^n with 2n cells) at http://www.conwaylife.com/forums/viewto ... =25#p51354. Also, 3S1CShips7 has a small bug that is fixed here:
Code: Select all
@RULE 3S1CShip8
v6 and v7 made by toroidalet
startup: 6908
main: sum(3,6910)2^n=2^6911-8
ending: 13
total: 2^6911+6913=26199924135232771946302945082142267646361287611183858888105878141097146284516799831042207935664233039650479000253912904922771642475731623570814064416253781364092711339053059999903025978136729323217679386289259716419683670948646023812141184875681945563968577497179990209595449685176614744961594950626560211884531284197640394761093553905046063780027895153209787942001752179924270127515264970630023873812492980939002822141152564625446373219448429758099463080183236186289644573331878107376094245464350507388846551751046612070714565610178006262325931767558598663756603524710091836568086585567457582719555198482702777497152917647713931772397794446480673474453561715398402565647895798977733825391494685454766789557545473602733743489446098894582747379919590022524327668286045820662926856507795043034908992104472345005832428884147960277965274231868039010559321290713143465608035562427335152773965238234364115181780266120198067562665325786292427327285082253794788586031222880339518555791075364838339105682822212623973098802249679263152169631485280007941546320088715920833952965887911686467922942093278225469934289248985421166049533594930296131436406782844461227172597419523940735285524817830265186717280715381153340934435524413089365663132951826300964878165876115877456321519871137927184515404511244296109196015310844309425916538601758601506755054623439329923395989582441201949702784143957477436389201828161767915088756130899725645032475916297263635572338097104291727119796586891454172270804516130061996391584953195213962055999127981606862469614528570944490481961815733359163745933237148318698949933861294045259868727519925870307847766277701373806698228034652816172832635092274619498470510299902864195522028303384354963284200397613571516894351225452609490085477998496433453738629933293500194584668382708191550532193329097472001535189222239048638243327986932257122929771989683230924125732348809919922259131496658284595708970604039562910688418658511076045962717614689204677807638527786100521285127133676423571868395252424476812075884746012485749210828645724603893058548869805107256798685960961
@TABLE
n_states:3
neighborhood:Moore
symmetries:none
#delay
0,0,0,0,0,2,1,2,0,2
2,0,0,0,0,0,1,0,0,1
1,0,0,0,0,0,1,0,0,0
2,0,0,0,0,0,2,0,0,1
1,0,0,0,0,0,2,0,0,0
#leftward expansion
0,0,0,1,0,0,0,0,0,2
0,0,0,0,0,1,2,0,0,2
0,0,0,2,1,2,0,0,0,1
2,0,0,0,2,1,2,0,0,0
2,1,0,1,2,0,0,0,0,1
1,0,0,0,1,1,0,0,0,0
1,1,0,1,2,0,0,0,0,2
0,0,1,1,0,0,0,0,0,2
0,0,0,1,1,0,0,0,0,1
2,1,0,2,0,0,0,0,0,1
2,0,0,0,0,1,2,0,0,0
1,1,0,2,0,0,0,0,0,2
1,0,0,0,2,1,0,0,0,0
#downward expansion
0,1,0,0,0,0,0,0,0,2
0,0,0,0,0,0,2,1,0,2
0,2,0,0,0,0,0,2,1,1
2,0,0,0,0,0,2,1,2,0
2,1,2,1,0,0,0,0,2,1
2,0,0,0,0,1,2,1,0,0
1,0,0,0,0,0,0,1,1,0
2,1,0,1,0,0,0,0,1,1
1,1,0,1,0,0,0,0,2,2
0,1,1,0,0,0,0,0,0,2
0,1,0,0,0,0,0,0,1,1
2,2,0,1,0,0,0,0,0,1
1,2,0,1,0,0,0,0,0,2
1,0,0,0,0,0,0,1,2,0
#downward stop
0,1,0,0,0,0,2,1,0,1
1,0,0,0,0,0,2,1,0,2
2,0,0,0,0,0,2,1,0,0
0,2,0,0,0,1,1,2,2,2
2,1,0,2,0,2,0,0,1,1
2,0,0,0,0,0,2,2,1,0
0,2,0,0,0,1,2,2,2,2
2,0,0,0,0,1,1,2,1,0
2,1,0,2,1,1,0,0,0,1
1,2,2,1,0,0,0,0,0,2
1,2,0,0,0,0,0,1,2,2
2,1,0,2,1,2,0,0,0,1
2,0,0,0,0,1,2,2,1,0
1,2,0,0,0,0,0,2,2,0
2,2,1,1,0,0,0,0,0,0
1,1,0,0,0,0,0,2,2,0
0,0,0,2,2,1,2,1,1,1
1,0,0,0,2,1,2,1,1,2
2,1,2,1,0,0,0,0,0,0
1,2,0,2,0,0,0,2,1,0
0,0,0,0,0,2,1,1,0,2
1,1,0,2,1,2,0,0,1,0
2,0,0,0,2,1,2,1,1,0
2,0,1,0,0,0,0,1,2,0
#downward counting
0,1,0,0,0,0,1,1,0,2
1,1,1,0,0,2,2,0,0,2
1,1,0,0,0,0,2,1,1,0
0,0,2,0,0,0,1,2,2,1
0,0,0,0,0,1,1,1,1,1
0,0,2,0,0,0,2,1,1,1
0,2,0,0,0,1,2,2,1,1
0,0,2,0,0,0,2,2,2,1
1,1,0,1,0,1,0,0,1,2
1,1,0,0,0,0,2,2,2,0
1,1,0,0,0,0,1,2,2,0
1,0,0,0,0,0,2,2,1,0
1,0,0,0,0,0,1,2,1,0
1,1,0,0,0,0,1,1,1,0
0,0,2,0,0,0,1,1,1,1
0,0,0,0,0,0,2,0,2,1
1,0,0,0,0,2,0,0,1,0
2,0,0,0,0,0,1,1,0,0
1,1,2,0,0,1,0,0,1,2
0,2,0,0,0,0,1,2,1,2
2,0,0,0,0,0,1,2,1,0
2,1,0,2,0,1,0,0,2,1
1,2,2,0,0,1,0,0,0,2
1,1,2,0,0,2,0,0,1,2
0,2,0,0,0,0,2,2,1,2
0,2,0,0,0,0,1,2,2,2
2,1,0,2,0,1,0,0,0,1
1,2,2,0,0,2,0,0,0,2
0,2,0,0,0,0,2,2,2,2
2,1,0,2,0,2,0,0,0,1
0,2,0,0,0,0,0,2,2,2
2,0,0,0,0,0,0,1,1,0
1,1,0,0,0,2,2,0,0,2
2,0,1,2,0,0,0,0,0,0
2,1,0,0,0,0,0,2,0,0
1,2,0,0,0,2,2,0,0,2
0,1,1,1,2,2,0,0,2,2
2,1,1,1,0,0,0,0,2,0
1,1,0,0,0,0,0,2,1,0
1,0,0,0,0,0,0,2,0,2
#leftward stop
0,0,0,0,0,2,2,0,0,2
2,0,0,0,0,2,2,0,0,0
0,0,0,0,0,0,2,2,0,2
0,0,0,2,2,2,2,0,0,2
2,0,0,0,2,2,2,0,0,0
2,2,0,2,0,0,0,2,0,1
2,0,0,0,1,2,2,0,0,0
2,2,0,1,0,0,0,2,0,1
0,0,0,2,2,2,1,1,0,2
0,0,0,2,2,2,2,1,0,2
1,0,0,2,2,1,0,0,0,2
2,0,0,0,1,2,1,1,0,0
2,2,0,1,0,0,0,1,1,1
1,0,0,2,1,2,0,0,0,0
2,0,2,0,1,1,2,1,0,1
1,0,1,0,1,1,2,0,0,0
2,1,1,0,0,0,0,0,2,0
1,0,0,1,1,2,0,0,0,0
1,0,0,0,1,1,1,0,0,0
#leftward counting
0,0,0,0,0,2,1,0,0,2
2,0,0,0,0,2,1,0,0,0
1,0,2,2,0,0,0,1,0,2
0,0,0,2,2,2,1,0,0,2
2,0,0,0,2,2,1,0,0,0
2,2,0,2,0,0,0,1,0,1
1,0,2,2,0,0,0,2,0,2
2,0,0,0,1,2,1,0,0,0
2,2,0,1,0,0,0,1,0,1
0,0,0,2,2,2,0,0,0,2
2,0,0,0,1,1,0,0,0,0
2,0,0,1,0,2,0,0,0,0
2,2,1,0,0,0,0,0,0,0
1,0,0,1,0,0,2,2,0,2
1,0,0,2,0,0,2,2,0,2
1,0,2,2,0,0,2,2,0,2
#ending
2,2,0,2,0,0,0,0,0,0
2,0,0,0,2,0,0,0,0,0
2,0,0,0,0,0,0,0,0,1
#diagonal mode
#2,0,0,0,2,0,0,0,0,0
#2,0,1,0,0,0,0,0,2,0
#0,0,0,1,0,2,0,2,0,1
#extra length
2,0,2,1,2,0,0,1,1,1
2,1,0,0,1,1,0,0,1,1
2,0,0,0,0,0,1,1,2,0
2,0,0,1,2,0,0,2,0,1
2,2,2,1,0,0,0,0,0,1
1,1,0,0,0,0,0,0,0,0
2,0,1,0,2,0,0,0,0,0
0,1,0,0,0,2,0,2,0,1
#bugfix
1,0,0,0,1,1,2,1,0,0
Moosey
Posts: 3895
Joined: January 27th, 2019, 5:54 pm
Location: A house, or perhaps the OCA board. Or [click to not expand]
Contact:
### Re: Slowest one cell spaceships of each state count
toroidalet wrote:...
Moosey wrote:Is it possible to do 4 or five states where the counter counts using counters...
So it would have a sequence of counters and every time one counter overflowed, it would trigger the next one to count and when the very last counter overflows, it would destroy itself? That might work, except it might need a few more states for signals.
Yes, that’s what I mean, a tetrationally slow 1-cell ship
My CA rules can be found here
Bill Watterson once wrote: "How do soldiers killing each other solve the world's problems?"
Nanho walåt derwo esaato?
Code: Select all
x = 4, y = 5, rule = B34e5e67c/S23ce45
4o2$b2o$o2bo$b2o! [[ TRACK 0 -7/325 ZOOM 16 THEME 0 ]] gameoflifemaniac Posts: 1216 Joined: January 22nd, 2017, 11:17 am Location: There too ### Re: Slowest one cell spaceships of each state count For two states it could theoretically be c/2. If the whole universe fills up and the one dot disappears, only one cell next to the gap may survive and the rest dies. It can't be done in Golly because it doesn't support B0 properly for rule tables. I was so socially awkward in the past and it will haunt me for the rest of my life. Code: Select all b4o25bo$o29bo$b3o3b3o2bob2o2bob2o2bo3bobo$4bobo3bob2o2bob2o2bobo3bobo$4bobo3bobo5bo5bo3bobo$o3bobo3bobo5bo6b4o$b3o3b3o2bo5bo9bobo$24b4o!
CoolCreeper39
Posts: 58
Joined: June 19th, 2019, 12:18 pm
### Re: Slowest one cell spaceships of each state count
fluffykitty wrote:c/2596148429267413814265248164610160 in 3 states. Also, I completely forgot that this thread existed until someone linked it in a thread someone linked to in the small long lived methuselahs thread.
Code: Select all
@RULE 3S1CShip2
startup: 108
main: sum(3,110)2^n=2^111-8
ending: 12
total: 2^111+112=2596148429267413814265248164610160
@TABLE
n_states:3
neighborhood:Moore
symmetries:none
#delay
0,0,0,0,0,2,1,2,0,2
2,0,0,0,0,0,1,0,0,1
1,0,0,0,0,0,1,0,0,0
2,0,0,0,0,0,2,0,0,1
1,0,0,0,0,0,2,0,0,0
#leftward expansion
0,0,0,1,0,0,0,0,0,2
0,0,0,0,0,1,2,0,0,2
0,0,0,2,1,2,0,0,0,1
2,0,0,0,2,1,2,0,0,0
2,1,0,1,2,0,0,0,0,1
1,0,0,0,1,1,0,0,0,0
1,1,0,1,2,0,0,0,0,2
0,0,1,1,0,0,0,0,0,2
0,0,0,1,1,0,0,0,0,1
2,1,0,2,0,0,0,0,0,1
2,0,0,0,0,1,2,0,0,0
1,1,0,2,0,0,0,0,0,2
1,0,0,0,2,1,0,0,0,0
#downward expansion
0,1,0,0,0,0,0,0,0,2
0,0,0,0,0,0,2,1,0,2
0,2,0,0,0,0,0,2,1,1
2,0,0,0,0,0,2,1,2,0
2,1,2,1,0,0,0,0,2,1
2,0,0,0,0,1,2,1,0,0
1,0,0,0,0,0,0,1,1,0
2,1,0,1,0,0,0,0,1,1
1,1,0,1,0,0,0,0,2,2
0,1,1,0,0,0,0,0,0,2
0,1,0,0,0,0,0,0,1,1
2,2,0,1,0,0,0,0,0,1
1,2,0,1,0,0,0,0,0,2
1,0,0,0,0,0,0,1,2,0
#downward stop
0,1,0,0,0,0,2,1,0,2
2,0,0,0,0,0,2,1,0,0
0,2,0,0,0,1,1,2,2,2
2,1,0,2,0,2,0,0,2,1
2,0,0,0,0,0,2,2,1,0
0,2,0,0,0,1,2,2,2,2
2,0,0,0,0,1,1,2,1,0
2,1,0,2,1,1,0,0,0,1
1,2,2,1,0,0,0,0,0,2
1,2,0,0,0,0,0,1,2,2
2,1,0,2,1,2,0,0,0,1
2,0,0,0,0,1,2,2,1,0
1,2,0,0,0,0,0,2,2,0
#downward counting
0,1,0,0,0,0,1,1,0,2
2,0,0,0,0,0,1,1,0,0
1,1,2,0,0,1,0,0,2,2
0,2,0,0,0,0,1,2,1,2
2,0,0,0,0,0,1,2,1,0
2,1,0,2,0,1,0,0,2,1
1,2,2,0,0,1,0,0,0,2
1,1,2,0,0,2,0,0,2,2
0,2,0,0,0,0,2,2,1,2
0,2,0,0,0,0,1,2,2,2
2,1,0,2,0,1,0,0,0,1
1,2,2,0,0,2,0,0,0,2
0,2,0,0,0,0,2,2,2,2
2,1,0,2,0,2,0,0,0,1
0,2,0,0,0,0,0,2,2,2
2,0,0,0,0,0,0,1,1,0
1,1,0,0,0,2,2,0,0,2
2,0,1,2,0,0,0,0,0,0
2,1,0,0,0,0,0,2,0,0
1,2,0,0,0,2,2,0,0,2
0,2,1,1,2,2,0,0,2,2
#1,1,0,0,0,2,2,0,2,1
2,2,1,1,0,0,0,0,2,0
1,1,0,0,0,0,0,2,2,0
1,0,0,0,0,0,0,2,0,2
#leftward stop
0,0,0,0,0,2,2,0,0,2
2,0,0,0,0,2,2,0,0,0
0,0,0,0,0,0,2,2,0,2
0,0,0,2,2,2,2,0,0,2
2,0,0,0,2,2,2,0,0,0
2,2,0,2,0,0,0,2,0,1
2,0,0,0,1,2,2,0,0,0
2,2,0,1,0,0,0,2,0,1
0,0,0,2,2,2,1,1,0,2
0,0,0,2,2,2,2,1,0,2
1,0,0,2,2,1,0,0,0,2
2,0,0,0,1,2,1,1,0,0
2,2,0,1,0,0,0,1,1,1
1,0,0,2,1,2,0,0,0,0
2,0,2,0,1,1,2,1,0,1
1,0,1,0,1,1,2,0,0,0
#leftward counting
0,0,0,0,0,2,1,0,0,2
2,0,0,0,0,2,1,0,0,0
1,0,2,2,0,0,0,1,0,2
0,0,0,2,2,2,1,0,0,2
2,0,0,0,2,2,1,0,0,0
2,2,0,2,0,0,0,1,0,1
1,0,2,2,0,0,0,2,0,2
2,0,0,0,1,2,1,0,0,0
2,2,0,1,0,0,0,1,0,1
0,0,0,2,2,2,0,0,0,2
2,0,0,0,1,1,0,0,0,0
2,0,0,1,0,2,0,0,0,0
2,2,1,0,0,0,0,0,0,0
1,0,0,1,0,0,2,2,0,2
1,0,0,2,0,0,2,2,0,2
1,0,2,2,0,0,2,2,0,2
#ending
2,2,0,2,0,0,0,0,0,0
2,0,0,0,2,0,0,0,0,0
2,0,0,0,0,0,0,0,0,1
#diagonal mode
#2,0,0,0,2,0,0,0,0,0
#2,0,1,0,0,0,0,0,2,0
#0,0,0,1,0,2,0,2,0,1
necrodoublepost ftw
Edit: c/131661808029361064474122541992202249544316172470474290896938957889657411706431738059855987100606116738608709863898114516322696806682779725395040361126073989754846342223650752839934282507553095228819445591564242224229100999549808158043220482537489286899542806927109262299798576702052364647531326323060875437512721666750074735365806697524459192563914773873444570420147733289826848369892651509013842153766576752759546709772991670483710813378588223114875800405103605361303657861719088971021114521028156816509308794659625756698728308129465992649969171244954433589856461308546917926791439305402526271547340075851639933262085702464422986758336147589425826676630311810520418347644992379850195840502843574915602319663 in 3 states. Can you get 3 counters in 3 states?
Code: Select all
@RULE 3S1CShip5
TODO
startup: 2346
main: sum(3,2348)2^n=2^2349-8
ending: 13
total: 2^2349+2351=131661808029361064474122541992202249544316172470474290896938957889657411706431738059855987100606116738608709863898114516322696806682779725395040361126073989754846342223650752839934282507553095228819445591564242224229100999549808158043220482537489286899542806927109262299798576702052364647531326323060875437512721666750074735365806697524459192563914773873444570420147733289826848369892651509013842153766576752759546709772991670483710813378588223114875800405103605361303657861719088971021114521028156816509308794659625756698728308129465992649969171244954433589856461308546917926791439305402526271547340075851639933262085702464422986758336147589425826676630311810520418347644992379850195840502843574915602319663
@TABLE
n_states:3
neighborhood:Moore
symmetries:none
#delay
0,0,0,0,0,2,1,2,0,2
2,0,0,0,0,0,1,0,0,1
1,0,0,0,0,0,1,0,0,0
2,0,0,0,0,0,2,0,0,1
1,0,0,0,0,0,2,0,0,0
#leftward expansion
0,0,0,1,0,0,0,0,0,2
0,0,0,0,0,1,2,0,0,2
0,0,0,2,1,2,0,0,0,1
2,0,0,0,2,1,2,0,0,0
2,1,0,1,2,0,0,0,0,1
1,0,0,0,1,1,0,0,0,0
1,1,0,1,2,0,0,0,0,2
0,0,1,1,0,0,0,0,0,2
0,0,0,1,1,0,0,0,0,1
2,1,0,2,0,0,0,0,0,1
2,0,0,0,0,1,2,0,0,0
1,1,0,2,0,0,0,0,0,2
1,0,0,0,2,1,0,0,0,0
#downward expansion
0,1,0,0,0,0,0,0,0,2
0,0,0,0,0,0,2,1,0,2
0,2,0,0,0,0,0,2,1,1
2,0,0,0,0,0,2,1,2,0
2,1,2,1,0,0,0,0,2,1
2,0,0,0,0,1,2,1,0,0
1,0,0,0,0,0,0,1,1,0
2,1,0,1,0,0,0,0,1,1
1,1,0,1,0,0,0,0,2,2
0,1,1,0,0,0,0,0,0,2
0,1,0,0,0,0,0,0,1,1
2,2,0,1,0,0,0,0,0,1
1,2,0,1,0,0,0,0,0,2
1,0,0,0,0,0,0,1,2,0
#downward stop
0,1,0,0,0,0,2,1,0,2
2,0,0,0,0,0,2,1,0,0
0,2,0,0,0,1,1,2,2,2
2,1,0,2,0,2,0,0,1,1
2,0,0,0,0,0,2,2,1,0
0,2,0,0,0,1,2,2,2,2
2,0,0,0,0,1,1,2,1,0
2,1,0,2,1,1,0,0,0,1
1,2,2,1,0,0,0,0,0,2
1,2,0,0,0,0,0,1,2,2
2,1,0,2,1,2,0,0,0,1
2,0,0,0,0,1,2,2,1,0
1,2,0,0,0,0,0,2,2,0
#downward counting
0,1,0,0,0,0,1,1,0,2
2,0,0,0,0,0,1,1,0,0
1,1,2,0,0,1,0,0,1,2
0,2,0,0,0,0,1,2,1,2
2,0,0,0,0,0,1,2,1,0
2,1,0,2,0,1,0,0,2,1
1,2,2,0,0,1,0,0,0,2
1,1,2,0,0,2,0,0,1,2
0,2,0,0,0,0,2,2,1,2
0,2,0,0,0,0,1,2,2,2
2,1,0,2,0,1,0,0,0,1
1,2,2,0,0,2,0,0,0,2
0,2,0,0,0,0,2,2,2,2
2,1,0,2,0,2,0,0,0,1
0,2,0,0,0,0,0,2,2,2
2,0,0,0,0,0,0,1,1,0
1,1,0,0,0,2,2,0,0,2
2,0,1,2,0,0,0,0,0,0
2,1,0,0,0,0,0,2,0,0
1,2,0,0,0,2,2,0,0,2
0,1,1,1,2,2,0,0,2,2
2,1,1,1,0,0,0,0,2,0
1,1,0,0,0,0,0,2,1,0
1,0,0,0,0,0,0,2,0,2
#leftward stop
0,0,0,0,0,2,2,0,0,2
2,0,0,0,0,2,2,0,0,0
0,0,0,0,0,0,2,2,0,2
0,0,0,2,2,2,2,0,0,2
2,0,0,0,2,2,2,0,0,0
2,2,0,2,0,0,0,2,0,1
2,0,0,0,1,2,2,0,0,0
2,2,0,1,0,0,0,2,0,1
0,0,0,2,2,2,1,1,0,2
0,0,0,2,2,2,2,1,0,2
1,0,0,2,2,1,0,0,0,2
2,0,0,0,1,2,1,1,0,0
2,2,0,1,0,0,0,1,1,1
1,0,0,2,1,2,0,0,0,0
2,0,2,0,1,1,2,1,0,1
1,0,1,0,1,1,2,0,0,0
#leftward counting
0,0,0,0,0,2,1,0,0,2
2,0,0,0,0,2,1,0,0,0
1,0,2,2,0,0,0,1,0,2
0,0,0,2,2,2,1,0,0,2
2,0,0,0,2,2,1,0,0,0
2,2,0,2,0,0,0,1,0,1
1,0,2,2,0,0,0,2,0,2
2,0,0,0,1,2,1,0,0,0
2,2,0,1,0,0,0,1,0,1
0,0,0,2,2,2,0,0,0,2
2,0,0,0,1,1,0,0,0,0
2,0,0,1,0,2,0,0,0,0
2,2,1,0,0,0,0,0,0,0
1,0,0,1,0,0,2,2,0,2
1,0,0,2,0,0,2,2,0,2
1,0,2,2,0,0,2,2,0,2
#ending
2,2,0,2,0,0,0,0,0,0
2,0,0,0,2,0,0,0,0,0
2,0,0,0,0,0,0,0,0,1
#diagonal mode
#2,0,0,0,2,0,0,0,0,0
#2,0,1,0,0,0,0,0,2,0
#0,0,0,1,0,2,0,2,0,1
#extra length
2,0,2,1,2,0,0,1,1,1
2,1,0,0,1,1,0,0,1,1
2,0,0,0,0,0,1,1,2,0
2,0,0,1,2,0,0,2,0,1
2,2,2,1,0,0,0,0,0,1
1,1,0,0,0,0,0,0,0,0
2,0,1,0,2,0,0,0,0,0
0,1,0,0,0,2,0,2,0,1
How did you run the entire thing? Golly slows down for me at about 8^7 steps
AforAmpere
Posts: 1156
Joined: July 1st, 2016, 3:58 pm
### Re: Slowest one cell spaceships of each state count
CoolCreeper39 wrote: How did you run the entire thing? Golly slows down for me at about 8^7 steps
Nobody did. The period is estimated based on how the ship works. You can calculate the period because it does the same type of thing over and over (the binary counting).
Wildmyron and I manage the 5S project, which collects all known spaceship speeds in Isotropic Non-totalistic rules.
Things to work on:
- Find a (7,1)c/8 ship in a Non-totalistic rule
gameoflifemaniac
Posts: 1216
Joined: January 22nd, 2017, 11:17 am
Location: There too
### Re: Slowest one cell spaceships of each state count
I have found a c/2 in 2 states!!!
Code: Select all
x = 1, y = 1, rule = MAP//////////////////////////////////////////4AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAABA
o!
I knew this could be made, but always failed to do it. I mentioned about it earlier:
gameoflifemaniac wrote:
June 29th, 2019, 10:44 am
For two states it could theoretically be c/2. If the whole universe fills up and the one dot disappears, only one cell next to the gap may survive and the rest dies. It can't be done in Golly because it doesn't support B0 properly for rule tables.
but no one cared.
I was so socially awkward in the past and it will haunt me for the rest of my life.
Code: Select all
b4o25bo$o29bo$b3o3b3o2bob2o2bob2o2bo3bobo$4bobo3bob2o2bob2o2bobo3bobo$
4bobo3bobo5bo5bo3bobo$o3bobo3bobo5bo6b4o$b3o3b3o2bo5bo9bobo$24b4o! blah Posts: 285 Joined: April 9th, 2016, 7:22 pm ### Re: Slowest one cell spaceships of each state count gameoflifemaniac wrote: April 20th, 2020, 2:09 pm I have found a c/2 in 2 states!!! Code: Select all x = 1, y = 1, rule = MAP//////////////////////////////////////////4AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAABA o! Optimised to C/4. Code: Select all @RULE 2Sslow @TABLE n_states:3 neighborhood:Moore symmetries:none var a={0,1,2} var b=a var c=a var d=a var e=a var f=a var g=a var h=a 2,1,1,1,1,1,1,1,1,2 1,1,1,1,1,1,1,2,1,1 1,2,1,1,1,1,1,1,1,1 2,1,2,2,2,2,2,1,2,2 1,2,1,1,2,2,2,2,2,1 1,1,2,2,2,2,2,1,2,1 1,a,b,c,d,e,f,g,h,2 2,a,b,c,d,e,f,g,h,1 Code: Select all x = 20, y = 20, rule = 2Sslow 20A$20A$20A$20A$20A$20A$20A$20A$20A$9AB10A$20A$20A$20A$20A$20A$20A$20A$20A$20A$20A!
I don't care enough to convert this to a MAP. Anyway, your approach has much more potential, I think. I believe with enough effort you could absolutely make a much, much slower 1-cell 2-state ship. It's trivial to prove that it must have an even period.
succ
gameoflifemaniac
Posts: 1216
Joined: January 22nd, 2017, 11:17 am
Location: There too
### Re: Slowest one cell spaceships of each state count
blah wrote:
April 20th, 2020, 6:44 pm
I don't care enough to convert this to a MAP.
I tried to do it, but doesn't work.
Code: Select all
x = 1, y = 1, rule = MAPAAB//4AA//8AAP//AAD//4AA//8AAP//AAD//wAA//8AAP//AAD//wAB//4AAP//AAD//wEA//8AAP//AAD//w
o!
I was so socially awkward in the past and it will haunt me for the rest of my life.
Code: Select all
b4o25bo$o29bo$b3o3b3o2bob2o2bob2o2bo3bobo$4bobo3bob2o2bob2o2bobo3bobo$
4bobo3bobo5bo5bo3bobo$o3bobo3bobo5bo6b4o$b3o3b3o2bo5bo9bobo$24b4o! blah Posts: 285 Joined: April 9th, 2016, 7:22 pm ### Re: Slowest one cell spaceships of each state count I just created a script to do the conversion automatically. Code: Select all local g = golly() dead_state = tonumber(g.getstring("Enter state for dead cells, or h for help.")) if dead_state==nil then g.warn([[automap.lua - blah 2020 This script generates a MAP rulestring for the current rule, where 2 states are considered and all others are ignored. Useful, primarily, if you want to design a 2-state rule with B0 using RuleLoader. The script maps the provided 2 states from the current rule onto the 2 states of the generated MAP rule. For example, running this script with B3/S23 as the current rule and entering 0, then 1, will generate a MAP conversion of B3/S23. Entering 1 then 0 will generate its inverse. This requires that a universe of only cells with one of the two provided states will never produce a third state. Otherwise, the script will produce an error.]],false) g.exit() end live_state = tonumber(g.getstring("And the one for live cells.")) g.addlayer() g.select({-1,-1,5,5}) -- used to clear edges, so explosions won't evolve rule_string = "MAP" base64_digs='ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/' bit6buf = 0 -- number in which the 6 bits of a base64 digit are built up for neigh = 0, 515 do -- 515 instead of 511, since 512 is not a multiple of 6 -- the bits beyond 511 simply don't matter bits = {} for bit = 0, 8 do if neigh/(2^bit) % 2 < 1 then bits[bit] = dead_state else bits[bit] = live_state end end -- write new state out g.setcell(2,2,bits[0]) g.setcell(1,2,bits[1]) g.setcell(0,2,bits[2]) g.setcell(2,1,bits[3]) g.setcell(1,1,bits[4]) g.setcell(0,1,bits[5]) g.setcell(2,0,bits[6]) g.setcell(1,0,bits[7]) g.setcell(0,0,bits[8]) -- test what it produces g.step() new_cell = g.getcell(1,1) g.clear(0) -- convert new cell to binary value if new_cell == live_state then new_cell = 1 elseif new_cell == dead_state then new_cell = 0 else g.dellayer() g.error("Attempt to convert non-2-state subrule.") g.exit() end bit6buf = bit6buf*2 + new_cell if neigh%6 == 5 then -- add new base64 digit rule_string = rule_string .. base64_digs:sub(bit6buf+1,bit6buf+1) bit6buf = 0 end end g.dellayer() g.warn(rule_string) With my rule, fed the parameters 1 and 2, it produces this rule: Code: Select all x = 1, y = 1, rule = MAP//+AAH//AAD//wAA//8AAH//AAD//wAA//8AAP//AAD//wAA//8AAP/+AAH//wAA//8AAP7/AAD//wAA//8AAP o! Were you trying to do it manually? succ gameoflifemaniac Posts: 1216 Joined: January 22nd, 2017, 11:17 am Location: There too ### Re: Slowest one cell spaceships of each state count blah wrote: April 21st, 2020, 6:45 pm Were you trying to do it manually? Yes. I was so socially awkward in the past and it will haunt me for the rest of my life. Code: Select all b4o25bo$o29bo$b3o3b3o2bob2o2bob2o2bo3bobo$4bobo3bob2o2bob2o2bobo3bobo$4bobo3bobo5bo5bo3bobo$o3bobo3bobo5bo6b4o$b3o3b3o2bo5bo9bobo$24b4o!
gameoflifemaniac
Posts: 1216
Joined: January 22nd, 2017, 11:17 am
Location: There too
### Re: Slowest one cell spaceships of each state count
Found C/8:
Code: Select all
@RULE slowship2statetest
@TABLE
n_states:3
neighborhood:Moore
symmetries:none
var a={0,1,2}
var b=a
var c=a
var d=a
var e=a
var f=a
var g=a
var h=a
1,2,1,1,1,2,2,2,2,1
1,2,2,2,2,1,2,2,2,1
1,1,2,2,2,2,2,2,2,1
2,2,2,2,2,2,2,1,2,2
2,2,2,2,2,1,1,2,2,2
2,1,2,2,2,2,2,2,1,2
2,1,1,1,1,1,2,2,2,2
1,a,b,c,d,e,f,g,h,2
2,a,b,c,d,e,f,g,h,1
Code: Select all
x = 39, y = 29, rule = slowship2statetest
39A$39A$39A$39A$39A$39A$39A$39A$39A$39A$39A$39A$39A$39A$2AB36A$39A$
39A$39A$39A$39A$39A$39A$39A$39A$39A$39A$39A$39A$39A!
Code: Select all
x = 1, y = 1, rule = MAP//8AAP//AAD//wAA//8AAf//AAD//wAA//8AAP//AAD//wAA//8IAP//AAD//gAA//8AAP3/AAD//wAB//sAQP
o!
EDIT: C/12:
Code: Select all
@RULE slowship2statetest2
@TABLE
n_states:3
neighborhood:Moore
symmetries:none
var a={0,1,2}
var b=a
var c=a
var d=a
var e=a
var f=a
var g=a
var h=a
2,2,2,2,2,2,2,1,2,2
2,2,1,2,2,1,1,1,1,2
1,1,1,1,2,2,1,1,1,1
1,2,1,1,1,1,1,1,2,1
1,1,1,1,1,2,1,2,1,1
1,2,2,1,1,2,2,2,2,1
1,2,2,2,2,1,2,2,2,1
1,a,b,c,d,e,f,g,h,2
2,a,b,c,d,e,f,g,h,1
Code: Select all
x = 85, y = 73, rule = slowship2statetest2
85A$85A$85A$85A$85A$85A$85A$85A$85A$85A$85A$85A$85A$85A$85A$85A$85A$85A$85A$85A$85A$85A$85A$85A$85A$85A$85A$85A$85A$85A$85A$85A$85A$85A$
85A$85A$85A$17AB67A$85A$85A$85A$85A$85A$85A$85A$85A$85A$85A$85A$85A$
85A$85A$85A$85A$85A$85A$85A$85A$85A$85A$85A$85A$85A$85A$85A$85A$85A$85A$85A$85A$85A$85A$85A!
Code: Select all
x = 1, y = 1, rule = MAP7/8AAN//AAD//wAA//8AAP//AED//wAA//8AAP//AAD//wAA//8AAP//AAD//wAAf/8AAP//AAD//wAB/fsAAO
o!
By the way, can someone help me find the slowest ship?
I was so socially awkward in the past and it will haunt me for the rest of my life.
Code: Select all
b4o25bo$o29bo$b3o3b3o2bob2o2bob2o2bo3bobo$4bobo3bob2o2bob2o2bobo3bobo$
4bobo3bobo5bo5bo3bobo$o3bobo3bobo5bo6b4o$b3o3b3o2bo5bo9bobo\$24b4o! | 2020-11-26 11:40:18 | {"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": 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": 0.6471506953239441, "perplexity": 5125.483450823753}, "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-2020-50/segments/1606141188146.22/warc/CC-MAIN-20201126113736-20201126143736-00202.warc.gz"} |
https://www.gamedev.net/forums/topic/589727-vc-2010-build/ | # VC++ 2010 build
## Recommended Posts
So I'm writing a program that uses dx7(i know it's old) with VC++ express 2010. I can build my project just fine in Debug mode. It compiles and executes. When I switch to Release mode however, I get this error:
1>DDFuncs.obj : error LNK2001: unresolved external symbol _DirectDrawCreateEx@16
1>DDFuncs.obj : error LNK2001: unresolved external symbol _IID_IDirectDraw7
Could someone just go through a list of what might be wrong with this? Thanks folks.
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Did you add the correct libraries to both your debug and release configuration? You have to set them for each case.
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Quote:
Original post by karwostsDid you add the correct libraries to both your debug and release configuration? You have to set them for each case.
If you're talking about Project->Properties->Configuration Properties-Linker->Input, ddraw.lib and dxguid.lib are included, along with a bunch of other libraries which aren't listed in the Debug mode.
Phew that was a mouthful
ddraw.lib;dxguid.lib;%(AdditionalDependencies)
Release
ddraw.lib;dxguid.lib;kernel32.lib;user32.lib;gdi32.lib;winspool.lib;comdlg32.lib;advapi32.lib;shell32.lib;ole32.lib;oleaut32.lib;uuid.lib;odbc32.lib;odbccp32.lib;%(AdditionalDependencies)
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Quote:
Could someone just go through a list of what might be wrong with this
That version of DirectX is not supported by the compiler, in fact only June 2010 has support for Vs2010.
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Okay, I thought it might be something like that. I'm using the August 2009 DirectX SDK, because newer version don't seem to support DirectDraw at all. Oh well, guess I'll have to modernize.
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• 11 | 2017-11-24 13:10:20 | {"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": 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": 0.2393147498369217, "perplexity": 4638.142250359488}, "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/1510934808133.70/warc/CC-MAIN-20171124123222-20171124143222-00575.warc.gz"} |
http://new-contents.com/New-York/estimate-variance-error-term-linear-regression.html | Address 110 Monarch Dr, Liverpool, NY 13088 (315) 453-6323 http://www.empirelaser.net
# estimate variance error term linear regression Bernhards Bay, New York
The quotient of that sum by σ2 has a chi-squared distribution with only n−1 degrees of freedom: 1 σ 2 ∑ i = 1 n r i 2 ∼ χ n How do I answer why I want to join a smaller company given I have worked at larger ones? The denominator is the sample size reduced by the number of model parameters estimated from the same data, (n-p) for p regressors or (n-p-1) if an intercept is used.[3] For more The distinction is most important in regression analysis, where the concepts are sometimes called the regression errors and regression residuals and where they lead to the concept of studentized residuals.
How do I help minimize interruptions during group meetings as a student? The following is a plot of the (one) population of IQ measurements. MR1639875. ^ Wackerly, Dennis; Mendenhall, William; Scheaffer, Richard L. (2008). Please try the request again.
If we define S a 2 = n − 1 a S n − 1 2 = 1 a ∑ i = 1 n ( X i − X ¯ ) By using this site, you agree to the Terms of Use and Privacy Policy. In statistical modelling the MSE, representing the difference between the actual observations and the observation values predicted by the model, is used to determine the extent to which the model fits Tell company that I went to interview but interviewer did not respect start time How would a vagrant civilization evolve?
Join them; it only takes a minute: Sign up Here's how it works: Anybody can ask a question Anybody can answer The best answers are voted up and rise to the Applied linear models with SAS ([Online-Ausg.]. Join them; it only takes a minute: Sign up Here's how it works: Anybody can ask a question Anybody can answer The best answers are voted up and rise to the Since this is a biased estimate of the variance of the unobserved errors, the bias is removed by multiplying the mean of the squared residuals by n-df where df is the
regression variance error heteroscedasticity share|improve this question asked Feb 16 '14 at 18:15 kanbhold 185211 3 You might need to expand this a bit to explain what the apparent contradiction D.; Torrie, James H. (1960). That being said, the MSE could be a function of unknown parameters, in which case any estimator of the MSE based on estimates of these parameters would be a function of New tech, old clothes Possible battery solutions for 1000mAh capacity and >10 year life?
How to get this substring on bash script? The system returned: (22) Invalid argument The remote host or network may be down. That is, how "spread out" are the IQs? Again, the quantity S = 8.64137 is the square root of MSE.
Variance Further information: Sample variance The usual estimator for the variance is the corrected sample variance: S n − 1 2 = 1 n − 1 ∑ i = 1 n What is a type system? For example, if the mean height in a population of 21-year-old men is 1.75 meters, and one randomly chosen man is 1.80 meters tall, then the "error" is 0.05 meters; if Table 2.4.
That is, the n units are selected one at a time, and previously selected units are still eligible for selection for all n draws. The sample variance: $s^2=\frac{\sum_{i=1}^{n}(y_i-\bar{y})^2}{n-1}$ estimates σ2, the variance of the one population. Somewhat Generalized Mean Value Theorem Why is it a bad idea for management to have constant access to every employee's inbox Translating "machines" and "people" Any better way to determine source more hot questions question feed about us tour help blog chat data legal privacy policy work here advertising info mobile contact us feedback Technology Life / Arts Culture / Recreation Science
The best we can do is estimate it! But $E(y) = E(X\beta + \varepsilon) = X\beta+E(\varepsilon)$ (since $X\beta$ is not random), so we can only have $E(y) = X\beta$ if $E(\varepsilon)=0$. –Jonathan Christensen Jan 26 '13 at 0:47 In that case weighted least squares is used to correct for the heteroscedasticity. Doing so "costs us one degree of freedom".
The statistical errors on the other hand are independent, and their sum within the random sample is almost surely not zero. Which option did Harry Potter pick for the knight bus? "all empires will suffer the same fate should the lessons from history go unlearnt" Which day of the week is today? a Gaussian random variable with mean zero & variance $\sigma^2$, and that they have constant variance, then yes it is. The calculations are set out in Table 2.4, and lead to an $$F$$-statistic of 14.9 on one and 18 d.f.
Contents 1 Definition and basic properties 1.1 Predictor 1.2 Estimator 1.2.1 Proof of variance and bias relationship 2 Regression 3 Examples 3.1 Mean 3.2 Variance 3.3 Gaussian distribution 4 Interpretation 5 Related 5Model errors, residuals and heteroscedasticity11How to conceptualize error in a regression model?8What to do with heterogeneity of variance when spread decreases with larger fitted values4Linear regression - variance error term3Why How would they learn astronomy, those who don't see the stars? Suppose the sample units were chosen with replacement.
more stack exchange communities company blog Stack Exchange Inbox Reputation and Badges sign up log in tour help Tour Start here for a quick overview of the site Help Center Detailed Suppose you have two brands (A and B) of thermometers, and each brand offers a Celsius thermometer and a Fahrenheit thermometer. On the other hand, predictions of the Fahrenheit temperatures using the brand A thermometer can deviate quite a bit from the actual observed Fahrenheit temperature. Generated Sat, 15 Oct 2016 04:46:05 GMT by s_wx1094 (squid/3.5.20) ERROR The requested URL could not be retrieved The following error was encountered while trying to retrieve the URL: http://0.0.0.10/ Connection
Our approach separates more clearly the systematic and random components, and extends more easily to generalized linear models by focusing on the distribution of the response rather than the distribution of Concretely, in a linear regression where the errors are identically distributed, the variability of residuals of inputs in the middle of the domain will be higher than the variability of residuals Analysis of Variance for Simple Regressionof CBR Decline on Social Setting Score Source ofDegrees ofSum ofMean$$F$$- variationfreedomsquaressquaredratio Setting11201.11201.114.9 Residual181449.180.5 Total192650.2 These results can be used to verify the equivalence of $$Not the answer you're looking for? In the Analysis of Variance table, the value of MSE, 74.67, appears appropriately under the column labeled MS (for Mean Square) and in the row labeled Residual Error (for Error). ‹ What's the reasoning behind setting E(\varepsilon)=0 ? –Chris Jan 26 '13 at 0:40 2 The premise of the model is that E(y) = X\beta. ISBN0-495-38508-5. ^ Steel, R.G.D, and Torrie, J. Basu's theorem. ed.). The sample mean could serve as a good estimator of the population mean. What is the correct tag for it? –yasar Nov 17 '13 at 18:56 If you have >1 explanatory variable & only 1 response variable, most people will call it the number of variables in the regression equation). Hazewinkel, Michiel, ed. (2001), "Errors, theory of", Encyclopedia of Mathematics, Springer, ISBN978-1-55608-010-4 v t e Least squares and regression analysis Computational statistics Least squares Linear least squares Non-linear least squares Iteratively While the regression coefficient expresses the association in the original units of \( x$$ and $$y$$, Pearson’s $$r$$ expresses the association in units of standard deviation. Squaring the 95% two-sided critical value of the Student’s $$t$$ distribution with 18 d.f., which is 2.1, gives the 95% critical value of the $$F$$ distribution with My approach was to calculate variance of residuals through genr varresid = @var(resid) (eviews command).
As stated earlier, σ2 quantifies this variance in the responses. Two or more statistical models may be compared using their MSEs as a measure of how well they explain a given set of observations: An unbiased estimator (estimated from a statistical But, we don't know the population mean μ, so we estimate it with $$\bar{y}$$. Your cache administrator is webmaster. | 2018-12-16 09:03:57 | {"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": 2, "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": 0.5036808848381042, "perplexity": 1293.2646267283642}, "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-2018-51/segments/1544376827596.48/warc/CC-MAIN-20181216073608-20181216095608-00166.warc.gz"} |
https://socratic.org/questions/are-there-examples-of-grams-of-precipitate-problems-i-am-not-sure-what-to-call-t | # Are there examples of grams of precipitate problems? I am not sure what to call them other than that.
Mar 12, 2015
Yes. Precipitation reactions are a type of double replacement (displacement) reaction in which the pattern of reaction is AX + BY $\rightarrow$ AY + BX. A and B are cations, and X and Y are anions. The cations and anions switch partners. These reactions have a product that is a precipitate, an insoluble gas, or water.
You are asking about a double replacement stoichiometry problem involving a precipitate. The following is an example.
Lead(II) nitrate and potassium iodide, both of which are in aqueous solution, react to produce solid lead(II) iodide and aqueous potassium iodide. If 2.89 grams of lead (II) nitrate and 1.05 grams of potassium iodide react, how many grams of the lead(II) iodide precipitate are produced?
Step 1
Write a balanced chemical equation for this reaction.
"Pb(NO"_3)_2("aq") + $\text{2KI(aq)}$ $\rightarrow$ "PbI"_2("s") + "2KNO"_3("aq")
Step 2
Convert given grams of each reactant to moles by dividing grams by molar mass.
Molar mass of "Pb(NO"_3)_2 = $\text{331.2 g/mol}$
Molar mass of $\text{KI}$ = $\text{166.0 g/mol}$
"2.89g Pb(NO"_3)_2$\div$"331.2g/mol Pb(NO"_3)_2 = "0.008726 mol Pb(NO"_3)_2
$\text{1.05g KI}$$\div$$\text{166.0g/mol KI}$ = $\text{0.006325 mol KI}$
Step 3
Determine how many moles of $\text{PbI"_2}$ each reactant can produce using the number of moles of each reactant and the mole ratios from the balanced equation.
Mole ratio for "Pb(NO"_3)_2 and $\text{PbI"_2}$ = $\text{1 mol Pb(NO"_3)_2:1 "mol PbI"_2}$.
Mole ratio for $\text{KI}$ and $\text{PbI"_2}$ = $\text{2 mol KI:1 mol PbI"_2}$
Multiply the known moles of each compound times the mole ratio that has the ${\text{PbI}}_{2}$ on top.
"0.008726 mol Pb(NO"_3)_2 x $\text{1 mol PbI2"/"1 mol Pb(NO3)2}$ = ${\text{0.008726 mol PbI}}_{2}$
$\text{0.006325 mol KI}$ x $\text{1 mol PbI2"/"2 mol KI}$ = ${\text{0.0031625 mol PbI}}_{2}$
Step 4
$\text{KI}$ is the limiting reactant since it produces the fewest moles of ${\text{PbI}}_{2}$, therefore the greatest number of moles of ${\text{PbI}}_{2}$ that can be produced in this reaction is ${\text{0.0031625 mol PbI}}_{2}$.
Now determine the mass in grams of ${\text{PbI}}_{2}$ that can be produced by multiplying the moles of ${\text{PbI}}_{2}$ times its molar mass, which is $\text{461.0 g/mol}$.
${\text{0.0031625 mol PbI}}_{2}$ x $\text{461.0 g PbI2"/"1 mol PbI2}$ = ${\text{1.46 g PbI}}_{2}$
Note: Molar masses were obtained from Wikipedia. | 2019-11-18 12:09:47 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 42, "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": 0.7548943161964417, "perplexity": 2239.4751675104158}, "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-2019-47/segments/1573496669755.17/warc/CC-MAIN-20191118104047-20191118132047-00493.warc.gz"} |
http://ncatlab.org/nlab/show/discrete+infinity-groupoid | # nLab discrete infinity-groupoid
### Context
#### Cohesive $\infty$-Toposes
cohesive topos
cohesive (∞,1)-topos
cohesive homotopy type theory
## Structures in a cohesive $\left(\infty ,1\right)$-topos
structures in a cohesive (∞,1)-topos
## Structures with infinitesimal cohesion
infinitesimal cohesion
# Contents
## Idea
The term discrete $\infty$-groupoid or bare $\infty$-groupoid is essentially synonymous to just ∞-groupoid . It is used for emphasis in contexts where one considers $\infty$-groupoids with extra cohesive structure to indicate that this extra structure is being disregarded, or rather that the special case of discrete such structure is considered.
## Definition
###### Observation
The terminal (∞,1)-sheaf (∞,1)-topos ∞Grpd is trivially a cohesive (∞,1)-topos, where each of the defining four (∞,1)-functors $\left(\Pi ⊣\mathrm{Disc}⊣\Gamma ⊣\mathrm{coDisc}\right):\infty \mathrm{Grpd}\to \infty \mathrm{Grpd}$ is an equivalence of (∞,1)-categories.
###### Definition
In the context of cohesive (∞,1)-toposes we say that ∞Grpd defines discrete cohesion and refer to its objects as discrete $\infty$-groupoids.
More generally, given any other cohesive (∞,1)-topos
$\left(\Pi ⊣\mathrm{Disc}⊣\Gamma ⊣\mathrm{codisc}\right):H\stackrel{\stackrel{\mathrm{Disc}}{←}}{\underset{\Gamma }{\to }}\infty \mathrm{Grpd}$(\Pi \dashv Disc \dashv \Gamma \dashv codisc) : \mathbf{H} \stackrel{\overset{Disc}{\leftarrow}}{\underset{\Gamma}{\to}} \infty Grpd
the inverse image $\mathrm{Disc}$ of the global section functor is a full and faithful (∞,1)-functor and hence embeds ∞Grpd as a full sub-(∞,1)-category of $H$. A general object in $H$ is a cohesive $\infty$-groupoid . We say $X\in H$ is a discrete $\infty$-groupoid if it is in the image of $\mathrm{Disc}$.
###### Remark
This generalizes the traditional use of the terms discrete space and discrete group:
## Structures in $\mathrm{Disc}\infty \mathrm{Grpd}$
We discuss now some of the general abstract structures in a cohesive (∞,1)-topos realized in discrete $\infty$-groupoids.
### Geometric homotopy and Galois theory
We discuss the general absatract notion of geometric homotopy in cohesive $\left(\infty ,1\right)$-toposes (see here) in the context of discrete cohesion.
By the homotopy hypothesis-theorem the (∞,1)-toposes Top and ∞Grpd are equivalent, hence indistinguishable by general abstract constructions in (∞,1)-topos theory. However, in practice it can be useful to distinguish them as two different presentations for an equivalence class of $\left(\infty ,1\right)$-toposes.
For that purposes consider the following
###### Definition
Define the quasi-categories
$\mathrm{Top}:=N\left({\mathrm{Top}}_{\mathrm{Quillen}}{\right)}^{\circ }$Top := N(Top_{Quillen})^\circ
and
$\infty \mathrm{Grpd}:=N\left({\mathrm{sSet}}_{\mathrm{Quillen}}{\right)}^{\circ }\phantom{\rule{thinmathspace}{0ex}},$\infty Grpd := N(sSet_{Quillen})^\circ \,,
where on the right we have the standard model structure on topological spaces ${\mathrm{Top}}_{\mathrm{Quillen}}$ and the standard model structure on simplicial sets ${\mathrm{sSet}}_{\mathrm{Quillen}}$ and $N\left(\left(-{\right)}^{\circ }\right)$ denotes the homotopy coherent nerve of the simplicial category given by the full sSet-subcategory of these simplicial model categories on fibrant-cofibrant objects.
For
$\left(\mid -\mid ⊣\mathrm{Sing}\right):{\mathrm{Top}}_{\mathrm{Quillen}}\stackrel{\stackrel{\mid -\mid }{←}}{\underset{\mathrm{Sing}}{\to }}{\mathrm{sSet}}_{\mathrm{Quillen}}$({|-| \dashv Sing}) : Top_{Quillen} \stackrel{\overset{{|-|}}{\leftarrow}}{\underset{Sing}{\to}} sSet_{Quillen}
the standard Quillen equivalence of the homotopy hypothesis-theorem given by the singular simplicial complex-functor and geometric realization, write
$\left(𝕃\mid -\mid ⊣ℝ\mathrm{Sing}\right):\mathrm{Top}\stackrel{\stackrel{𝕃\mid -\mid }{←}}{\underset{ℝ\mathrm{Sing}}{\to }}\infty \mathrm{Grpd}$(\mathbb{L} {|-|} \dashv \mathbb{R}Sing) : Top \stackrel{\overset{\mathbb{L}{|-|}}{\leftarrow}}{\underset{\mathbb{R}Sing}{\to}} \infty Grpd
for the corresponding derived functors (the image under the homotopy coherent nerve of the restriction of $\mid -\mid$ and $\mathrm{Sing}$ to fibrant-cofibrant objects followed by functorial fibrant-cofibrant replacement) that constitute a pair of adjoint (∞,1)-functors modeled as morphisms of quasi-categories.
Since this is an equivalence of (∞,1)-categories either functor serves as the left adjoint and right adjoint and so we have
###### Observation
Top is exhibited a cohesive (∞,1)-topos over ∞Grpd by setting
$\left(\Pi ⊣\mathrm{Disc}⊣\Gamma ⊣\mathrm{coDisc}\right):\mathrm{Top}\stackrel{\stackrel{ℝ\mathrm{Sing}}{\to }}{\stackrel{\stackrel{𝕃\mid -\mid }{←}}{\stackrel{\stackrel{ℝ\mathrm{Sing}}{\to }}{\underset{𝕃\mid -\mid }{←}}}}\infty \mathrm{Grpd}\phantom{\rule{thinmathspace}{0ex}}.\phantom{\rule{thinmathspace}{0ex}}.$(\Pi \dashv Disc \dashv \Gamma \dashv coDisc) : Top \stackrel{\overset{\mathbb{R}Sing}{\to}}{\stackrel{\overset{\mathbb{L}{|-|}}{\leftarrow}}{\stackrel{\overset{\mathbb{R}Sing}{\to}}{\underset{\mathbb{L}{|-|}}{\leftarrow}}}} \infty Grpd \,. \,.
In particular a presentation of the intrinsic fundamental ∞-groupoid in a locally ∞-connected (∞,1)-topos is given by the familiar singular simplicial complex construction
$\Pi \left(X\right)\simeq ℝ\mathrm{Sing}X\phantom{\rule{thinmathspace}{0ex}}.$\Pi(X) \simeq \mathbb{R} Sing X \,.
###### Remark
While degenerate, it is sometimes useful to make this example of a cohesive (∞,1)-topos explicit. For instance it allows to think of simplicial models for topological fibrations in terms of topological higher parallel transport. Some remarks on this are in Flat higher parallel transport in Top.
###### Remark
Notice that the topology that enters the explicit construction of the objects in Top here does not show up as cohesive structure. A topological space here is a model for a discrete $\infty$-groupoid, the topology only serves to allow the construction of $\mathrm{Sing}X$. For discussion of $\infty$-groupoids equipped with genuine topological cohesion see Euclidean-topological ∞-groupoid.
### Cohomology and principal $\infty$-bundles
We discuss the general abstract notion of cohomology and principal $\infty$-bundles a in cohesive $\infty$-toposes (see here) in the context of discrete cohesion.
###### Definition
Write $\mathrm{sGrp}=\mathrm{Grp}\left(\mathrm{sSet}\right)$ for the category of simplicial groups.
A classical reference is section 17 of May.
###### Proposition
The category $\mathrm{sGrpd}$ inherits a model category structure transferred along the forgetful functor $F:\mathrm{sGrp}\to \mathrm{sSet}$.
The category ${\mathrm{sSet}}_{0}↪\mathrm{sSet}$ of reduced simplicial sets (simplicial sets with a single vertex) carries a model category structure whose weak equivalences and cofibrations are those of ${\mathrm{sSet}}_{\mathrm{Quillen}}$.
There is a Quillen equivalence
$\left(G⊣\overline{W}\right):\mathrm{sGrp}\stackrel{\stackrel{G}{←}}{\underset{\overline{W}}{\to }}{\mathrm{sSet}}_{0}$(G \dashv \bar W) : sGrp \stackrel{\overset{G}{\leftarrow}}{\underset{\bar W}{\to}} sSet_{0}
which presents the abstract looping and delooping equivalence of $\infty$-categories
$\left(\Omega ⊣B\right):\infty \mathrm{Grpd}\stackrel{\stackrel{\Omega }{←}}{\underset{B}{\to }}\infty {\mathrm{Grpd}}_{\mathrm{connected}}\phantom{\rule{thinmathspace}{0ex}},$(\Omega \dashv \mathbf{B}) : \infty Grpd \stackrel{\overset{\Omega}{\leftarrow}}{\underset{B}{\to}} \infty Grpd_{connected} \,,
The model structures and the Quillen equivalence are classical, discussed in (GoerssJardine, section V)
This means on abstract grounds that for $G$ a simplicial group, $\overline{W}G\in \mathrm{sSet}$ is a model of the classifying delooping object $BG$ for discrte $G$-principal ∞-bundles. The following statements assert that these principal $\infty$-bundles themselves can be modeled as ordinary simplicial principal bundles
###### Definition
For $G$ a simplicial group and $\overline{W}G$ the model for $BG$ given by the above proposition, write
$WG\to \overline{W}G$W G \to \bar W G
for the simplicial decalage on $\overline{W}G$.
This characterization of the object going by the classical name $WG$ is made fairly explicit in (Duskin, p. 85).
###### Proposition
The morphism $WG\to \overline{W}G$ is a Kan fibration resolution of the point inclusion $*\to \overline{W}G$.
This follows directly from the characterization of $WG\to \overline{W}G$ by decalage. Pieces of this statement appear in (May): lemma 18.2 there gives the fibration property, prop. 21.5 the contractibility of $WG$.
###### Corollary
For $G$ a simplicial group, the sequence of simplicial sets
$G\to WG\to \overline{W}G$G \to W G \to \bar W G
is a presentation of the fiber sequence
$G\to *\to BG\phantom{\rule{thinmathspace}{0ex}}.$G \to * \to \mathbf{B}G \,.
Hence $WG\to \overline{W}G$ is a model for the universal $G$-principal discrete $\infty$-bundle (see universal principal ∞-bundle):
every $G$-principal discrete $\infty$-bundle $P\to X$ in $\infty \mathrm{Grpd}$, which by definition is a homotopy fiber
$\begin{array}{ccc}P& \to & *\\ ↓& {⇙}_{\simeq }& ↓\\ X& \to & BG\end{array}$\array{ P &\to& * \\ \downarrow &\swArrow_{\simeq}& \downarrow \\ X &\to& \mathbf{B}G }
in ∞Gpd, is presented in the standard model structure on simplicial sets by the ordinary pullback
$\begin{array}{ccc}P& \to & WG\\ ↓& & ↓\\ X& \to & \overline{W}G\end{array}\phantom{\rule{thinmathspace}{0ex}}.$\array{ P &\to& W G \\ \downarrow && \downarrow \\ X &\to& \bar W G } \,.
The explicit statement that the sequence $G\to WG\to \overline{W}G$ is a model for the looping fiber sequence appears on p. 239 of Crossed Menagerie . The universality of $WG\to \overline{W}G$ for $G$-principal simplicial bundles is the topic of section 21 in (May), where however it is not made explicit that the “twisted cartesian products” considered there are precisely the models for the pullbacks as above. This is made explicit on page 148 of Crossed Menagerie.
In Euclidean-topological ∞-groupoid we discuss how this model of discrete principal $\infty$-bundles by simplicial principal bundles lifts to a model of topological principal $\infty$-bundles by simplicial topological bundles principal over simplicial topological groups.
## References
Simplicial groups and simplicial principal bundles are discussed in
and section V of
The relation of $WG\to \overline{W}G$ to decalage is mentioned on p. 85 of
• John Duskin, Simplicial methods and the interpretation of “triple” cohomology, number 163 in Mem. Amer. Math. Soc., 3, Amer. Math. Soc. (1975)
Discrete cohesion is the topic of section 3.1 of
where much of the above material is taken from.
Revised on April 18, 2012 09:03:13 by David Corfield (129.12.18.29) | 2013-05-19 08:20:35 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 81, "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": 0.9741107821464539, "perplexity": 2028.4774540122462}, "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/1368696400149/warc/CC-MAIN-20130516092640-00093-ip-10-60-113-184.ec2.internal.warc.gz"} |
https://www.asvabtestbank.com/math-knowledge/flash-cards/356704/10 | ## Math Knowledge Flash Card Set 356704
Cards 10 Topics Acute & Obtuse Angles, Angles Around Lines & Points, Classifications, Parallel Lines, Parallelogram, Pythagorean Theorem, Quadrilateral, Rectangle & Square, Triangle Geometry
#### Study Guide
###### Acute & Obtuse Angles
An acute angle measures less than 90°. An obtuse angle measures more than 90°.
###### Angles Around Lines & Points
Angles around a line add up to 180°. Angles around a point add up to 360°. When two lines intersect, adjacent angles are supplementary (they add up to 180°) and angles across from either other are vertical (they're equal).
###### Classifications
A monomial contains one term, a binomial contains two terms, and a polynomial contains more than two terms. Linear expressions have no exponents. A quadratic expression contains variables that are squared (raised to the exponent of 2).
###### Parallel Lines
Parallel lines are lines that share the same slope (steepness) and therefore never intersect. A transversal occurs when a set of parallel lines are crossed by another line. All of the angles formed by a transversal are called interior angles and angles in the same position on different parallel lines equal each other (a° = w°, b° = x°, c° = z°, d° = y°) and are called corresponding angles. Alternate interior angles are equal (a° = z°, b° = y°, c° = w°, d° = x°) and all acute angles (a° = c° = w° = z°) and all obtuse angles (b° = d° = x° = y°) equal each other. Same-side interior angles are supplementary and add up to 180° (e.g. a° + d° = 180°, d° + c° = 180°).
###### Parallelogram
A parallelogram is a quadrilateral with two sets of parallel sides. Opposite sides (a = c, b = d) and angles (red = red, blue = blue) are equal. The area of a parallelogram is base x height and the perimeter is the sum of the lengths of all sides (a + b + c + d).
###### Pythagorean Theorem
The Pythagorean theorem defines the relationship between the side lengths of a right triangle. The length of the hypotenuse squared (c2) is equal to the sum of the two perpendicular sides squared (a2 + b2): c2 = a2 + b2 or, solved for c, $$c = \sqrt{a + b}$$ | 2022-05-20 17:46:29 | {"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": 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": 0.39365044236183167, "perplexity": 1671.1386452789989}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662533972.17/warc/CC-MAIN-20220520160139-20220520190139-00664.warc.gz"} |
https://indico.cern.ch/event/219436/contributions/1523246/ | # Quark Matter 2014 - XXIV International Conference on Ultrarelativistic Nucleus-Nucleus Collisions
19-24 May 2014
Europe/Zurich timezone
## PHENIX experimental search for the dark photon decay to di-electron pairs
20 May 2014, 16:30
2h
### spectrum
Board: C-24
Poster Initial State Physics
### Speaker
Yorito Yamaguchi (University of Tokyo)
### Description
A dark photon is a hypothetical particle which is very weakly coupled with ordinary photons in some Beyond the Standard Model (BSM) theories. The dark photon is a candidate for the annihilation of dark matter particles into $e^+e^-$ pairs, a process which could provide an explanation for the positron excess in the universe observed by several satellite experiments. It may also explain the $3\sigma$ discrepancy between the muon $g-2$ result and Standard Model calculations. The PHENIX experiment at RHIC has excellent capabilities for electron identification and for measuring $e^+e^-$ pairs with a very good mass resolution, making possible a search for such pairs from dark photon decay. We have conducted a search for possible $e^+e^-$ pairs from dark photons among a large data sample of pairs from $\pi^0$ Dalitz decays. We present new results of the dark photon search from the PHENIX experiment, which provide the world's best limits and rules out a majority of the mass-branching ratio region that could explain the $g-2$ result.
On behalf of collaboration: PHENIX
### Primary author
Dr Dave Morrison (Brookhaven National Laboratory)
### Presentation Materials
There are no materials yet. | 2020-11-30 10:40:00 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 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": 0.33379054069519043, "perplexity": 2939.3270398549203}, "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-2020-50/segments/1606141213431.41/warc/CC-MAIN-20201130100208-20201130130208-00151.warc.gz"} |
https://www.vedantu.com/question-answer/prove-that-left-1-+-cot-a-+-tan-a-rightleft-sin-class-12-maths-cbse-5efc6a76c732a135b6b90d59 | QUESTION
# Prove that $\left( {1 + \cot A + \tan A} \right)\left( {\sin A - \cos A} \right) = \dfrac{{\sec A}}{{\cos e{c^2}A}} - \dfrac{{\cos ecA}}{{{{\sec }^2}A}} = \sin A\tan A - \cot A\cos A$
Hint: In this question first of all divide the given equation into 3 parts. Then solve part 1 to prove part 1 and 2 are equal and then solve part 2 to prove that part 2 and 3 are equal by using simple trigonometric ratios. By this we can say that all the three parts are equal to each other which is our required solution.
Divide the equation into three 3 parts do solve it easily.
Given LHS i.e., part 1 is $\left( {1 + \cot A + \tan A} \right)\left( {\sin A - \cos A} \right)$
$\Rightarrow \left( {1 + \cot A + \tan A} \right)\left( {\sin A - \cos A} \right)$
Writing the terms of cot and tan in terms of sin and cos we have
$\Rightarrow \left( {1 + \dfrac{{\cos A}}{{\sin A}} + \dfrac{{\sin A}}{{\cos A}}} \right)\left( {\sin A - \cos A} \right) \\ \Rightarrow \left( {\dfrac{{\sin A\cos A + {{\cos }^2}A + {{\sin }^2}A}}{{\sin A\cos A}}} \right)\left( {\sin A - \cos A} \right) \\ \Rightarrow \left[ {\dfrac{{\left( {\sin A\cos A + {{\cos }^2}A + {{\sin }^2}A} \right)\left( {\sin A - \cos A} \right)}}{{\sin A\cos A}}} \right] \\$
Multiplying the terms inside the brackets, we have
$\Rightarrow \left[ {\dfrac{{{{\sin }^2}A\cos A + \sin A{{\cos }^2}A + {{\sin }^3}A - \sin A{{\cos }^2}A - {{\cos }^3}A - {{\sin }^2}A\cos A}}{{\sin A\cos A}}} \right]$
Cancelling the common terms, we have
$\Rightarrow \dfrac{{{{\sin }^3}A - {{\cos }^3}A}}{{\sin A\cos A}}$
Splitting the terms, we have
$\Rightarrow \dfrac{{{{\sin }^3}A}}{{\sin A\cos A}} - \dfrac{{{{\cos }^3}A}}{{\sin A\cos A}} \\ \Rightarrow \dfrac{{{{\sin }^2}A}}{{\cos A}} - \dfrac{{{{\cos }^2}A}}{{\sin A}} \\$
Which can be written as
$\Rightarrow \dfrac{{\sec A}}{{{\text{cose}}{{\text{c}}^2}A}} - \dfrac{{\operatorname{cosec} A}}{{{{\sec }^2}A}} \\ \therefore \left( {1 + \cot A + \tan A} \right)\left( {\sin A - \cos A} \right) = \dfrac{{\sec A}}{{{\text{cose}}{{\text{c}}^2}A}} - \dfrac{{\operatorname{cosec} A}}{{{{\sec }^2}A}}......................................\left( 1 \right) \\$
Now consider the part 2 i.e., $\dfrac{{\sec A}}{{{\text{cose}}{{\text{c}}^2}A}} - \dfrac{{\operatorname{cosec} A}}{{{{\sec }^2}A}}$
$\Rightarrow \dfrac{{\sec A}}{{{\text{cose}}{{\text{c}}^2}A}} - \dfrac{{\operatorname{cosec} A}}{{{{\sec }^2}A}} \\ \Rightarrow \dfrac{{\sec A}}{{\operatorname{cosec} A\operatorname{cosec} A}} - \dfrac{{\operatorname{cosec} A}}{{\sec A\sec A}} \\ \Rightarrow \dfrac{1}{{\operatorname{cosec} A}} \times \dfrac{{\sec A}}{{\operatorname{cosec} A}} - \dfrac{{\operatorname{cosec} A}}{{\sec A}} \times \dfrac{1}{{\sec A}} \\ \Rightarrow \sin A\tan A - \cot A\cos A \\ \therefore \dfrac{{\sec A}}{{{\text{cose}}{{\text{c}}^2}A}} - \dfrac{{\operatorname{cosec} A}}{{{{\sec }^2}A}} = \sin A\tan A - \cot A\cos A....................................................\left( 2 \right) \\$
From equation (1) and (2) we have
$\left( {1 + \cot A + \tan A} \right)\left( {\sin A - \cos A} \right) = \dfrac{{\sec A}}{{\cos e{c^2}A}} - \dfrac{{\cos ecA}}{{{{\sec }^2}A}} = \sin A\tan A - \cot A\cos A$
Hence proved.
Note: Here we have used the trigonometric ratio conversions such as $\dfrac{1}{{\operatorname{cosec} A}} = \sin A$, $\dfrac{{\sec A}}{{\operatorname{cosec} A}} = \dfrac{{\sin A}}{{\cos A}} = \tan A$, $\dfrac{{\operatorname{cosec} A}}{{secA}} = \dfrac{{\cos A}}{{\sin A}} = \cot A$, $\dfrac{1}{{\sec A}} = \cos A$. While multiplying the terms inside the brackets make sure that you have written all the terms multiplied in it. | 2020-07-13 01:12:10 | {"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": 0.867299497127533, "perplexity": 217.93509655480275}, "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/1593657140746.69/warc/CC-MAIN-20200713002400-20200713032400-00589.warc.gz"} |
http://mathhelpforum.com/calculus/60463-integration-print.html | # integration
• Nov 19th 2008, 08:06 AM
koalamath
integration
I need help setting the bounds of this equation
Find the volume of the solid bounded by the planes x=0 y=0 z=0 and x+y+z=6
thank you
• Nov 19th 2008, 08:33 AM
Soroban
Hello, koalamath!
Quote:
Find the volume of the solid bounded by the planes: . $x+y+z\:=\:6,\;x=0,\;y=0,\;z=0$
The solid is in the first octant, bounded by the plane $x + y + z \:=\:6$
. . which has intercepts: (6,0,0), (0,6,0), (0,0,6).
We have: . $V \;=\;\int\int_A z\,dA$ . . . where $A$ is the region in the $x\text{-}y$ plane.
That region looks like this:
Code:
|
6 *
|:*
|:::*
|:::::*
|:::::::*
|:::::::::*
- + - - - - - * - - -
| 6
We see that $y$ goes from $0$ to $6-x$
And $x$ goes from $0$ to $6.$
Therefore: . $V \;=\;\int^6_0\int^{6-x}_0 (6-x-y)\,dy\,dx$
• Nov 19th 2008, 11:35 AM
koalamath
Thank you so much
• Nov 19th 2008, 11:44 AM
koalamath
Wait i get 6 as the answer and that's wrong
• Nov 19th 2008, 01:36 PM
Krizalid
Your integration skills or arithmetic are wrong 'cause the answer is 36. | 2018-02-23 16:58:35 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 12, "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": 0.9218310117721558, "perplexity": 1438.5097211452915}, "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-2018-09/segments/1518891814801.45/warc/CC-MAIN-20180223154626-20180223174626-00267.warc.gz"} |
https://www.allaboutcircuits.com/tools/capacitance-reactance-and-admittance-calculator/ | A calculator for solving the capacitance, reactance, and admittance of a capacitor.
(pF)
(Ω)
(S)
### Overview
This calculator is designed to compute for a capacitor's reactance and admittance given the capacitance value and the frequency. It can be also used the other way around, i.e., give the reactance value and it will give the capacitance. Note that the input capacitance must be in picofarads (pF) while the input frequency can be in Gigahertz (GHz) or Megahertz (MHz).
### Equation
$$X_{C} = \frac{1}{2 \pi f C}$$
Where:
$$X_{C}$$ = capacitive reactance
$$f$$ = frequency
$$C$$ = capacitance
### Application
A capacitor's opposition to alternating current is called its reactance. This reactance is dependent on the frequency of the alternating current (see equation above). For low frequencies, the capacitor acts like a resistor with a very high resistance. In fact, at direct currents where the frequency is zero, the capacitor acts like an open switch. Conversely, the capacitor acts like a resistor with a very low resistance at very high frequencies. This characteristic of the capacitor makes it suitable for use in tons of applications, particularly in filters, phase shifters, and couplers. | 2017-09-21 22:58:20 | {"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": 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": 0.8841403126716614, "perplexity": 964.3129566719482}, "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-2017-39/segments/1505818687938.15/warc/CC-MAIN-20170921224617-20170922004617-00661.warc.gz"} |
https://appliedcombinatorics.org/book/ch_probability.html | ## Chapter10Probability
It was a slow day and Dave said he was bored. It was just after lunch, and he complained that there was nothing to do. Nobody really seemed to be listening, although Alice said that Dave might consider studying, even reading ahead in the chapter. Undeterred, Dave said “Hey Alice, how about we play a game. We could take turns tossing a coin, with the other person calling heads or tails. We could keep score with the first one to a hundred being the winner.” Alice rolled her eyes at such a lame idea. Sensing Alice's lack of interest, Dave countered “OK, how about a hundred games of Rock, Paper or Scissors?” Zori said “Why play a hundred times? If that's what you're going to do, just play a single game.”
Now it was Alice's turn. “If you want to play a game, I've got a good one for you. Just as you wanted, first one to score a hundred wins. You roll a pair of dice. If you roll doubles, I win $2$ points. If the two dice have a difference of one, I win $1$ point. If the difference is $2\text{,}$ then it's a tie. If the difference is $3\text{,}$ you win one point; if the difference is $4\text{,}$ you win two points; and if the difference is $5\text{,}$ you win three points.” Xing interrupted to say “In other words, if the difference is $d\text{,}$ then Dave wins $d-2$ points.” Alice continues “Right! And there are three ways Dave can win, with one of them being the biggest prize of all. Also, rolling doubles is rare, so this has to be a good game for Dave.”
Zori's ears perked up with Alice's description. She had a gut feeling that this game wasn't really in Dave's favor and that Alice knew what the real situation was. The idea of a payoff with some uncertainty involved seemed very relevant. Carlos was scribbling on a piece of paper, then said politely “Dave, you really should be reading ahead in the chapter”.
So what do you think? Is this a fair game? What does it mean for a game to be fair? Should Dave play—independent of the question of whether such silly stuff should occupy one's time? And what does any of this conversation have to do with combinatorics? | 2020-09-27 20:13:07 | {"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": 0.2637450397014618, "perplexity": 901.1918733286875}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-40/segments/1600401578485.67/warc/CC-MAIN-20200927183616-20200927213616-00161.warc.gz"} |
http://mathhelpforum.com/algebra/90434-roots-equation.html | # Math Help - Roots of equation?
1. ## Roots of equation?
If $x_1,x_2,x_3,\mbox{...},x_n$ are the roots of the equation $x^n + p_1x^{n - 1} + p_2x^{n - 2} + ... + p_{n - 1}x + p_n = 0$, where coefficient $p_1,p_2,p_3,\mbox{...},p_n$ are real, show that $(1 + x_1^2)(1 + x_2^2)(1 + x_3^2)\mbox{...}(1 + x_n^2) = (1 - p_1 + p_4 + \mbox{...})^2 + (p_1 - p_3 + p_5 + \mbox{...})^2$
2. Originally Posted by fardeen_gen
If $x_1,x_2,x_3,\mbox{...},x_n$ are the roots of the equation $x^n + p_1x^{n - 1} + p_2x^{n - 2} + ... + p_{n - 1}x + p_n = 0$, where coefficient $p_1,p_2,p_3,\mbox{...},p_n$ are real, show that $(1 + x_1^2)(1 + x_2^2)(1 + x_3^2)\mbox{...}(1 + x_n^2) = (1 - p_1 + p_4 + \mbox{...})^2 + (p_1 - p_3 + p_5 + \mbox{...})^2$
I think it should be $(1 + x_1^2)(1 + x_2^2)(1 + x_3^2)\mbox{...}(1 + x_n^2) = (1 - p_2 + p_4 + \mbox{...})^2 + (p_1 - p_3 + p_5 + \mbox{...})^2$
3. ## Induction
Base case $(n=1)$: $x+p_1=0~~\rightarrow~~1+x_1^2=1+(-p_1)^2=(1)^2+(p_1)^2$
Define $Q(x)=\sum_{i=0}^{n-1}q_ix^i=\prod_{i=1}^{n-1}(x-x_i)=0$ and $P(x)=\sum_{i=0}^np_ix^i=\prod_{i=1}^n(x-x_i)=$ $(x-x_n)Q(x)=xQ(x)-x_nQ(x)=0$ . So $p_i=q_i-x_nq_{i-1}$ for $i\in\mathbb{N}$ and $p_0=q_0=1$
Assume $(1+x_1^2)(1+x_2^2)...(1+x_{n-1}^2)=(1-q_2+q_4-q_6+...)^2+(q_1-q_3+q_5-q_7+...)^2=Q_{even}^2+Q_{odd}^2$
Show: $(1+x_1^2)(1+x_2^2)...(1+x_n^2)=(1-p_2+p_4-p_6+...)^2+(p_1-p_3+p_5-p_7+...)^2=P_{even}^2+P_{odd}^2$
Proof: $P_{even}^2+P_{odd}^2=(1-p_2+p_4-p_6+...)^2+(p_1-p_3+p_5-p_7+...)^2$
= $[1-(q_2-x_nq_1)+(q_4-x_nq_3)-(q_6-x_nq_5)+...]^2+$ $[(q_1-x_n)-(q_3-x_nq_2)+(q_5-x_nq_4)-(q_7-x_nq_6)+...]^2$
= $[(1-q_2+q_4-q_6+...)+x_n(q_1-q_3+q_5-q_7+...)]^2+$ $[(q_1-q_3+q_5-q_7+...)-x_n(1-q_2+q_4-q_6+...)]^2$
= $[Q_{even}+x_nQ_{odd}]^2+[Q_{odd}-x_nQ_{even}]^2$
= $[Q_{even}^2+Q_{odd}^2](1+x_n^2)$
= $(1+x_1^2)(1+x_2^2)...(1+x_{n-1}^2)(1+x_n^2)$
QED
4. $(x-x_{1})(x-x_{2}).....(x-x_{n})=x^n+p_{1}x^{n-1}+p_{2}x^{n-2}+...+p_{n}$
is an identity in x
Put $x=i$
$(i-x_{1})(i-x_{2}).....(i-x_{n})$ $=(p_{n}-p_{n-2}+p_{n-4}-p_{n-6}+...)+i(p_{n-1}-p_{n-3}+p_{n-5}-.....)$
Put $x=-i$
$(-i-x_{1})(-i-x_{2}).....(-i-x_{n})$ $=(p_{n}-p_{n-2}+p_{n-4}-p_{n-6}+...)-i(p_{n-1}-p_{n-3}+p_{n-5}-.....)$
Multiplying 1. and 2.
$(1+x_{1}^2)(1+x_{2}^2)...(1+x_{n}^2)=(p_{n}-p_{n-2}+p_{n-4}-p_{n-6}+...)^2+(p_{n-1}-p_{n-3}+p_{n-5}-.....)^2$
You will get the required result whatever n may be either even or odd. | 2014-10-01 10:44:59 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 35, "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": 0.9235761165618896, "perplexity": 673.2613209701278}, "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-2014-41/segments/1412037663417.12/warc/CC-MAIN-20140930004103-00008-ip-10-234-18-248.ec2.internal.warc.gz"} |
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## October 25, 2017
#### Wednesday, October 25, 2017
CTL Toolkit: CoursePlus GROUPS
10:00 AM - 11:00 AM
Bloomberg School of Public Health
The Toolkit Workshop will discuss the benefits of using Course Groups and strategically planning resources and activities tied to the tool. During the session, CTL will demonstrate the process of setting up Groups, including the option to assign students randomly. We will also explore best practices for using the CoursePlus resources that can be set as private to groups to help facilitate communication, enrich class discussion, and foster a virtual environment rich for sharing and collaborative learning.
We encourage onsite attendance, but the Groups workshop will also be streamed via Adobe Connect at http://connect.johnshopkins.edu/ctl-toolkit-workshop. Recordings will be posted to our Toolkit Events page at a later date.
Biostatistics Help: Faculty, Staff, Pre-MD and Post Doc Walk-In Clinic
11:00 AM - 12:00 PM
Biostatistics consulting is available to all Johns Hopkins University faculty, staff, pre-MD and post docs conducting clinical and translational research. 11:00 AM – 12:00 PM Wolfe Street Building Room: E3142 Contact Information: Nita James | jhbc@jhu.edu
[multiple] DMH Wednesday Noon Seminar - Black Communalism & Mental Health in the Black Lives Matter Era
12:15 PM - 1:00 PM
Bloomberg School of Public Health
DMH Wednesday Noon Seminar Series
Nyasha Grayman-Simpson, PhD, MA
Associate Professor, Psychology
Goucher College
Keisha Bentley-Edwards, PhD, MA
Assistant Professor in Medicine
Duke University School of Medicine
Black Communalism & Mental Health in the Black Lives Matter Era
Continuing discussion with students from 1 - 1:30 PM in HH188.
Population, Family and Reproductive Health - Wednesday Seminar Series
12:15 PM - 1:20 PM
Bloomberg School of Public Health
### Infant, Maternal and Perinatal Health Weathering: A Closer Look at Trends Over Time
Amaris McCord Keiser, MD
Sheila S. and Lawrence C. Pakula, MD Fellow
Assistant Professor, Pediatrics
Division of Neonatology / Department of Pediatrics
Johns Hopkins Children’s Center
Contact
Janay Williams
410-502-5443
John Harlim "Nonparametric modeling for prediction and data assimilation."
3:00 PM - 4:00 PM
Homewood
Speaker: John Harlim, The Pennsylvania State University Abstract: I will discuss a nonparametric modeling approach for forecasting stochastic dynamical systems on smooth manifolds embedded in Euclidean space. This approach allows one to evolve the probability distribution of non-trivial dynamical systems with an equation-free modeling. In the second part of this talk, I will discuss a nonparametric estimation of likelihood functions using data-driven basis functions and the theory of kernel embeddings of conditional distributions developed in the machine learning community. I will demonstrate how to use this likelihood function to estimate biased modeling error in assimilating cloudy satellite brightness temperature-like quantities.
Richard Melrose "Hodge theory for the Weil-Petersson metric on Riemann moduli spaces"
3:00 PM - 4:00 PM
Homewood
Speaker: Richard Melrose, MIT Abstract: The Riemann moduli spaces, for the complex structures on an n-fold punctured Riemann surface, $\mathcal{M}_{g,n},$ carry natural Riemannian structures the most conspicuous of which is the Weil-Petersson metric. Although the space has a natural compactification to a smooth complex orbifold, the metric does not extend smoothly across the compactifying divisors. Nevertheless in work with Jesse Gell-Redman (based on earlier results with Xuwen Zhu) it is shown that the $L^2$ cohomology, the space of square-integrable harmonic forms on the interior, is naturally isomorphic with the cohomology of the compactified space. In this talk I will show how this is proved and also indicate some open questions.
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Copyright 2009, The Johns Hopkins University | 2018-08-22 03:36:26 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.2583619952201843, "perplexity": 6428.727983406283}, "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-2018-34/segments/1534221219469.90/warc/CC-MAIN-20180822030004-20180822050004-00305.warc.gz"} |
https://chemistry.stackexchange.com/questions/1137/why-are-equilibrium-constants-unitless/1145 | # Why are equilibrium constants unitless?
I haven’t quite reached the point where I can read a full-fledged text on chemical kinetics and thermodynamics yet, so bear with me, please.
I’m wondering why a value like $K_\text{eq} = \frac{[\ce{NO}]^2[\ce{O2}]}{[\ce{NO2}]^2}$ wouldn't have units of M?
I goofed up the first time I tried to answer this question, erroneously applying dimensional analysis to your equilibrium expression.
It turns out that Silberberg[1] gives a good explanation of why $K_\text{eq}$ is dimensionless, which is often glossed over as the terms of the equilibrium expression are generally taught as concentrations. In actual fact, the terms are ratios of the concentration or activity of each species with a reference concentration (1 $\mathrm{mol\cdot{L^{-1}}}$ for solutions.) For example, a concentration of 2 $\mathrm{mol\cdot{L^{-1}}}$ divided by a reference of 1 $\mathrm{mol\cdot{L^{-1}}}$ yields a ratio of 2, with no units. As each term has no units, so too does $K_\text{eq}$.
[1] Silberberg, M.E.; Chemistry – The Molecular Nature of Matter and Change 3e; 2003, p. 719
• I don't quite follow but that's a step in the right direction at least. – readyready15728 Sep 11 '12 at 8:08
• This only works if the number of products is equal to the number of reactants. In the example Sadiq gives it is not, and so the equilibrium constant as written has units of $(\text{mol}\cdot\text{L}^{-1})^3/(\text{mol}\cdot\text{L}^{-1})^2 = \text{mol}\cdot\text{L}^{-1}$. – Nathaniel Sep 12 '12 at 13:13
• I understand this much, but I what I don't understand is how it can have significance in equilibrium reactions – Kian Feb 11 '13 at 17:32
• The stated facts is true for the standard equilibrium constant, not for an equilibrium constant. – Martin - マーチン Jun 1 '14 at 14:46
This source explains it well. It looks like part of class material, but it clearly explains the dimensionlessness of $K_{eq}$.
The resolution of this apparent paradox is that the above equation, while perfectly satisfactory for everyday use, is not technically correct. A more correct version is:
$$K_{eq} = \frac{\frac{\lvert B \rvert_{eq}}{\lvert B \rvert_{ss}} \frac{\lvert C \rvert_{eq}}{\lvert C \rvert_{ss}}}{\frac{\lvert A\rvert_{eq}}{\lvert A\rvert_{ss}}}$$
where the "ss" subscripts refer to the concentration of that species in the standard state. (By this definition, Keq is always unitless.)
It then goes on to state:
Strictly speaking, division by the standard state concentrations is also necessary in every thermodynamics equation in which you take the log of a concentration product, otherwise the units don’t come out right.) We will NEVER use this "correct" version of the equation in this class (well, never except in one problem on this week’s problem set...), and $K_{eq}$ for a reaction with unequal numbers of reactants and products is ALWAYS given with units, even in published papers.
Dimensional analysis is useless. The correct answer is the one already given involving activities, which are dimensionless.
Activities are defined as ratios. For example a pressure activity (there are many kinds) is defined in terms of the ratio of the actual pressure of a gas divided by the reference pressure, often 1 atm or 1 bar.
In the present example, the activity is the ratio of the molality divided by the reference molality of 1 molal. This assumes ideal solutions, which is good enough if the solutions are dilute. If the solution is not ideal one has to correct the molality for nonideality.
Because of these complications, detailed discussions of activities are usually left for a course in physical chemistry.
So while we use square brackets and molarities, we have to understand that we are really dealing with activities.
By the way, the activity of a pure liquid or solid is 1, which is why $[\ce{H2O}]$, for example, is dropped from equilibrium calculations.
• Wait a sec. Don't you mean molarity (as opposed to molality)? – readyready15728 Sep 13 '12 at 5:03
• The issue of activity vs. concentration (or activity vs. pressure) is subtly different from the issue of dimensional vs. dimensionless. "Activity" can be related to the chemical potential, which does have dimension of energy/particle or energy/mol. It is the inherent definition of energy relative to a standard state that leads to $K_{eq}$ being dimensionless. – Curt F. Apr 20 '15 at 14:12
# Note
This answer is incorrect, but because it is a common pitfall I'll leave it to show. It is wrong 1) because the OP meant $\lvert \ce{NO} \rvert ^2$ in the numerator not $\lvert \ce{NO2} \rvert$ as I took it (edited) and 2) because the $K_{eq}$ formula is incorrect. I will post a better answer.
The concise answer: Because the units cancel. Here I'll use $\lvert \rvert$ = concentration = moles/liter = m/L
$K_{eq} = \frac{\lvert NO_2 \rvert \lvert O_2 \rvert}{\lvert {NO}_2 \rvert^2} (?) units$
$K_{eq} = \frac{\lvert NO_2 \rvert \lvert O_2 \rvert}{\lvert {NO}_2 \rvert \lvert {NO}_2 \rvert} (?) units$
$K_{eq} = \frac{\lvert NO_2 \rvert m/L \lvert O_2 \rvert m/L}{\lvert {NO}_2 \rvert m/L \lvert {NO}_2 \rvert m/L}$
$K_{eq} = \frac{\lvert NO_2 \rvert \lvert O_2 \rvert}{\lvert {NO}_2 \rvert \lvert {NO}_2 \rvert} \frac{m^2/L^2}{m^2/L^2}$
$K_{eq} = \frac{\lvert NO_2 \rvert \lvert O_2 \rvert}{\lvert {NO}_2 \rvert \lvert {NO}_2 \rvert} \frac{m^2}{m^2}$
$K_{eq} = \frac{\lvert NO_2 \rvert \lvert O_2 \rvert}{\lvert {NO}_2 \rvert \lvert {NO}_2 \rvert} (unitless)$
• Unfortunately not so - check the exponent on [NO] in the numerator. Straight dimensional analysis on Sadiq's example results in residual units. – Richard Terrett Sep 11 '12 at 9:38
• So he meant $\lvert NO \rvert ^2$ and not $\lvert NO^2 \rvert$ (which I took to mean $\lvert NO_2 \rvert$)? – Ehryk Sep 11 '12 at 9:44
• Yeah, I read it like that as well. But the stoichiometries don't work out otherwise. – Richard Terrett Sep 11 '12 at 9:57
• I've edited the question. Sorry. Yes I meant $[\ce{NO}]^2$. (A reminder: you can use \ce{ } to format molecular formulae easily in LaTeX.) – readyready15728 Sep 11 '12 at 10:00
• Thanks for the tip. What's with the left [ being darker than ] in formulas? I switched to \lvert and \rvert just because it was ugly. – Ehryk Sep 11 '12 at 10:56 | 2019-09-20 09:50:43 | {"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": 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": 0.8163350224494934, "perplexity": 751.6258195995498}, "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/1568514573988.33/warc/CC-MAIN-20190920092800-20190920114800-00416.warc.gz"} |
https://awsm.dvlprz.com/tag:https | # The Awsm Dvlprz Blog
Letsencrypt, the free, automated, and open Certificate Authority has revolutionized the way servers of all sorts are secured to use SSL/TLS. Usually the entire process is quite smooth, but under certain circumstances requires some additional configuration. This short note explains how to configure Apache2 to use a virtual host in a user directory and redirect all http traffic to https. | 2021-10-28 09:04:50 | {"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": 0.33328044414520264, "perplexity": 5252.972192239845}, "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/1634323588282.80/warc/CC-MAIN-20211028065732-20211028095732-00715.warc.gz"} |
https://4gravitons.wordpress.com/category/amplitudes-methods/ | # Scattering Amplitudes at KITP
I’ve been visiting the Kavli Institute for Theoretical Physics in Santa Barbara for a program on scattering amplitudes. This week they’re having a conference, so I don’t have time to say very much.
The conference logo, on the other hand, seems to be saying quite a lot
We’ve had talks from a variety of corners of amplitudes, with major themes including the web of theories that can sort of be described by string theory-esque models, the amplituhedron, and theories you can “square” to get other theories. I’m excited about Zvi Bern’s talk at the end of the conference, which will describe the progress I talked about last week. There’s also been recent progress on understanding the amplituhedron, which I will likely post about in the near future.
We also got an early look at Whispers of String Theory, a cute short documentary filmed at the IGST conference.
# The Road to Seven-Loop Supergravity
Do the same thing you would with any other theory, and you get infinity. You get repeated infinities, an infinity of infinities. And while you could fix one or two infinities, fixing an infinite number requires giving up an infinity of possible predictions, so in the end your theory predicts nothing.
String theory fixes this with its own infinity, the infinite number of ways a string can vibrate. Because this infinity is organized and structured and well-understood, you’re left with a theory that is still at least capable of making predictions.
(Note that this is an independent question from whether string theory can make predictions for experiments in the real world. This is a much more “in-principle” statement: if we knew everything we might want to about physics, all the fields and particles and shapes of the extra dimensions, we could use string theory to make predictions. Even if we knew all of that, we still couldn’t make predictions from naive quantum gravity.)
Are there ways to fix the problem that don’t involve an infinity of vibrations? Or at least, to fix part of the problem?
That’s what Zvi Bern, John Joseph Carrasco, Henrik Johansson, and a growing cast of collaborators have been trying to find out.
They’re investigating N=8 supergravity, a theory that takes gravity and adds on a host of related particles. It’s one of the easiest theories to get from string theory, by curling up extra dimensions in a particularly simple way and ignoring higher-energy vibrations.
Bern, along with Lance Dixon and David Kosower, invented the generalized unitarity technique I talked about last week. Along with Carrasco and Johansson, he figured out another important trick: the idea that you can do calculations in gravity by squaring the appropriate part of calculations in Yang-Mills theory. For N=8 supergravity, the theory you need to square is my favorite theory, N=4 super Yang-Mills.
Using this, they started pushing forward, calculating approximations to greater and greater precision (more and more loops).
What they found, at each step, was that N=8 supergravity behaved better than expected. In fact, it behaved like N=4 super Yang-Mills.
N=4 super Yang-Mills is special, because in four dimensions (three space and one time, the dimensions we’re used to in daily life) there are no infinities to fix. In a world with more dimensions, though, you start getting infinities, and with more and more loops you need fewer and fewer dimensions to see them.
N=8 supergravity, unexpectedly, was giving infinities in the same dimensions that N=4 super Yang-Mills did (and no earlier). If it kept doing that, you might guess that it also had no infinities in four dimensions. You might wonder if, at least loop by loop, N=8 supergravity could be a way to fix quantum gravity without string theory.
Of course, you’d only really know if you could check in four dimensions.
If you want to check in four dimensions, though, you run into a problem. The fewer dimensions you’re looking at, the more loops you need before N=8 supergravity could possibly give infinity. In four dimensions, you need a forbidding seven loops of precision.
Still, Bern, Carrasco, and Johansson were up to the challenge. Along with Lance Dixon, David Kosower, and Radu Roiban, they looked at three loops, calculating an interaction of four gravitons, and the pattern continued. Four loops, and it was still going strong.
At around this time, I had just started grad school. My first project was a cumbersome numerical calculation. To keep me motivated, my advisor mentioned that the work I was doing would be good preparation for a much grander project: the calculation of whether the four-graviton interaction in N=8 supergravity diverges at seven loops. All I’d have to do was wait for Bern and collaborators to get there.
I named this blog “4 gravitons and a grad student”, and hoped I would get a chance to contribute.
And then something unexpected happened. They got stuck at five loops.
The method they were using, generalized unitarity, is an ansatz-based method. You start with a guess, then refine it. As such, the method is ultimately only as good as your guess.
Their guesses, in general, were pretty good. The trick they were using, squaring N=4 to get N=8, requires a certain type of guess: one in which the pieces they square have similar relationships to the different types of charge in Yang-Mills theory. There’s still an infinite number of guesses that can obey this, so they applied more restrictions, expectations based on other calculations, to get something more manageable. This worked at three loops, and worked (with a little extra thought) at four loops.
But at five loops they were stuck. They couldn’t find anything, with their restrictions, that gave the correct answer when “cut up” by generalized unitarity. And while they could drop some restrictions, if they dropped too many they’d end up with far too general a guess, something that could take months of computer time to solve.
So they stopped.
They did quite a bit of interesting work in the meantime. They found more theories they could square to get gravity theories, of more and more unusual types. They calculated infinities in other theories, and found surprises there too, other cases where infinities didn’t show up when they were “supposed” to. But for some time, the N=8 supergravity calculation was stalled.
And in the meantime, I went off in another direction, which long-time readers of this blog already know about.
Recently, though, they’ve broken the stall.
What they realized is that the condition on their guess, that the parts they square be related like Yang-Mills charges, wasn’t entirely necessary. Instead, they could start with a “bad” guess, and modify it, using the failure of those relations to fill in the missing pieces.
It looks like this is going to work.
We’re all at an amplitudes program right now in Santa Barbara. Walking through the halls of the KITP, I overhear conversations about five loops. They’re paring things down, honing their code, getting rid of the last few bugs, and checking their results.
They’re almost there, and it’s exciting. It looks like finally things are moving again, like the train to seven loops has once again left the station.
Increasingly, they’re beginning to understand the absent infinities, to see that they really are due to something unexpected and new.
N=8 supergravity isn’t going to be the next theory of everything. (For one, you can’t get chiral fermions out of it.) But if it really has no infinities at any loop, that tells us something about what a theory of quantum gravity is allowed to be, about the minimum necessary to at least make sense on a loop-by-loop level.
And that, I think, is worth being excited about.
# Generalized Unitarity: The Frankenstein Method for Amplitudes
This is going to be a bit more technical than my usual, but you were warned.
There are a few things you’ll need to know to understand this post.
First, you should know that when we calculate probabilities of things happening in particle physics, we can do it by drawing Feynman diagrams, pictures of particles traveling and interacting. These diagrams can have loops, and the particle in the loop can have any momentum, from zero on up to infinity: you have to add up all the possibilities to get whatever you’re trying to calculate.
Second, you should understand that the “particles” in these loops aren’t really particles. They’re “virtual particles”, better understood as disturbances in quantum fields. Matt Strassler has a very nice article about this. In particular, these “particles” don’t have to obey $E=mc^2$ (or rather, if we include kinetic energy, $E^2=p^2 c^2+m^2 c^4$, where $p$ is the momentum).
You can imagine a space that the momentum and energy “live in”. It’s got three dimensions for the three directions momentum can have, and one more dimension for the energy. Virtual particles can live anywhere in this four-dimensional space, but real particles have to live on a “shell” of points that obey $E^2=p^2 c^2+m^2 c^4$. If you’ve heard physicists say “on-shell” or “off-shell”, they’re referring to whether a particle is virtual, a quantum mechanical disturbance (and thus lives anywhere in the space) or a real classical particle (living on this “shell”).
Third, you should appreciate that in quantum physics, in Scott Aaronson’s words, we put complex numbers in our ontologies. Often, quantum weirdness shows itself when we look at our calculations as functions of complex numbers.
Let’s say I’m calculating an amplitude with one loop, and I draw a diagram like this:
Unitarity is how particle physicists say “all probabilities have to add up to one”. Since we have complex numbers in our ontologies, this statement is more complicated than it looks. One thing it ends up implying is that if I calculate an amplitude from the one-loop diagram above, its imaginary part will be given by multiplying together two simpler amplitudes:
Here you can imagine that I took a pair of scissors and “cut” the diagram in two along the dashed line. Now that the diagram has been “cut”, the particles I cut through are no longer part of a loop, so they’re no longer virtual: they’re real, on-shell particles.
If I wanted, I could keep “cutting” the diagram, generalizing this implication of unitarity. (For those who know some complex analysis, this involves taking residues.) I could cut all of the lines in the loop, like this:
Now something interesting happens. Here I’ve forced all four of the particles in the loop to be “on-shell”, to obey $E^2=p^2 c^2+m^2 c^4$. Previously, the momentum and energy in the loop was entirely free, living in its four-dimensional space. Now, though, it must obey four equations. And for those who’ve seen some algebra, four independent equations and four unknowns gives us one solution. By cutting all of these particles, we’ve killed all of the freedom that the loop momentum had. Instead of the living, quantum amplitude we had, we’ve cut it up into a bunch of dead, classical parts.
Why do this?
Well, suppose we have a guess for what the full amplitude should be. We’ve still got some uncertainty in our guess: it’s an ansatz.
If we wanted to check our guess, to fix the uncertainty in our ansatz, we could compare it to the full amplitude. But then we’d have to calculate the full quantum amplitude, and that’s hard.
It’s a lot easier, though, to calculate those “dead” classical amplitudes.
That’s the method we call “generalized unitarity”. We stitch together these easier-to-calculate, “dead” amplitudes. Enough different stitching patterns, and we can fix all the uncertainty in our ansatz, ending up with a unique correct answer without ever doing the full quantum calculation. Like Frankenstein, from dead parts we’ve assembled a living thing.
It’s off-shell!
How well does this work?
That depends on how good the ansatz is. The ansatze for one loop are very well understood, and for two loops the community is getting there. For higher loops, you have to be either smart or lucky. I happen to know some people who are both, I’ll be talking about them next week.
# Thoughts from the Winter School
First, as promised, I’ll talk about what I worked on at the PSI Winter School.
Freddy Cachazo and I study what are called scattering amplitudes. At first glance, these are probabilities that two subatomic particles scatter off each other, relevant for experiments like the Large Hadron Collider. In practice, though, they can calculate much more.
For example, let’s say you have two black holes circling each other, like the ones LIGO detected. Zoom out far enough, and you can think of each one as a particle. The two particle-black holes exchange gravitons, and those exchanges give rise to the force of gravity between them.
In the end, it’s all just particle physics.
Based on that, we can use our favorite scattering amplitudes to make predictions for gravitational wave telescopes like LIGO.
There’s a bit of weirdness to this story, though, because these amplitudes don’t line up with predictions in quite the way we’re used to. The way we calculate amplitudes involves drawing diagrams, and those diagrams have loops. Normally, each “loop” makes the amplitude more quantum-mechanical. Only the diagrams with no loops (“tree diagrams”) come from classical physics alone.
(Here “classical physics” just means “not quantum”: I’m calling general relativity “classical”.)
For this problem, we only care about classical physics: LIGO isn’t sensitive enough to see quantum effects. The weird thing is, despite that, we still need loops.
(Why? This is a story I haven’t figured out how to tell in a non-technical way. The technical explanation has to do with the fact that we’re calculating a potential, not an amplitude, so there’s a Fourier transformation, and keeping track of the dimensions entails tossing around some factors of Planck’s constant. But I feel like this still isn’t quite the full story.)
So if we want to make predictions for LIGO, we want to compute amplitudes with loops. And as amplitudeologists, we should be pretty good at that.
As it turns out, plenty of other people have already had that idea, but there’s still room for improvement.
Our time with the students at the Winter School was limited, so our goal was fairly modest. We wanted to understand those other peoples’ calculations, and perhaps to think about them in a slightly cleaner way. In particular, we wanted to understand why “loops” are really necessary, and whether there was some way of understanding what the “loops” were doing in a more purely classical picture.
At this point, we feel like we’ve got the beginning of an idea of what’s going on. Time will tell whether it works out, and I’ll update you guys when we have a more presentable picture.
Unfortunately, physics wasn’t the only thing I was thinking about last week, which brings me to my other topic.
This blog has a fairly strong policy against talking politics. This is for several reasons. Partly, it’s because politics simply isn’t my area of expertise. Partly, it’s because talking politics tends to lead to long arguments in which nobody manages to learn anything. Despite this, I’m about to talk politics.
Last week, citizens of Iran, Iraq, Libya, Somalia, Sudan, Syria and Yemen were barred from entering the US. This included not only new visa applicants, but also those who already have visas or green cards. The latter group includes long-term residents of the US, many of whom were detained in airports and threatened with deportation when their flights arrived shortly after the ban was announced. Among those was the president of the Graduate Student Organization at my former grad school.
A federal judge has blocked parts of the order, and the Department of Homeland Security has announced that there will be case-by-case exceptions. Still, plenty of people are stuck: either abroad if they didn’t get in in time, or in the US, afraid that if they leave they won’t be able to return.
Politics isn’t in my area of expertise. But…
I travel for work pretty often. I know how terrifying and arbitrary border enforcement can be. I know how it feels to risk thousands of dollars and months of planning because some consulate or border official is having a bad day.
I also know how essential travel is to doing science. When there’s only one expert in the world who does the sort of work you need, you can’t just find a local substitute.
And so for this, I don’t need to be an expert in politics. I don’t need a detailed case about the risks of terrorism. I already know what I need to, and I know that this is cruel.
And so I stand in solidarity with the people who were trapped in airports, and those still trapped abroad and trapped in the US. You have been treated cruelly, and you shouldn’t have been. Hopefully, that sort of message can transcend politics.
One final thing: I’m going to be a massive hypocrite and continue to ban political comments on this blog. If you want to talk to me about any of this (and you think one or both of us might actually learn something from the exchange) please contact me in private.
# Hexagon Functions Meet the Amplituhedron: Thinking Positive
I finished a new paper recently, it’s up on arXiv now.
This time, we’re collaborating with Jaroslav Trnka, of Amplituhedron fame, to investigate connections between the Amplituhedron and our hexagon function approach.
The Amplituhedron is a way to think about scattering amplitudes in our favorite toy model theory, N=4 super Yang-Mills. Specifically, it describes amplitudes as the “volume” of some geometric space.
Here’s something you might expect: if something is a volume, it should be positive, right? You can’t have a negative amount of space. So you’d naturally guess that these scattering amplitudes, if they’re really the “volume” of something, should be positive.
“Volume” is in quotation marks there for a reason, though, because the real story is a bit more complicated. The Amplituhedron isn’t literally the volume of some space, there are a bunch of other mathematical steps between the geometric story of the Amplituhedron on the one end and the final amplitude on the other. If it was literally a volume, calculating it would be quite a bit easier: mathematicians have gotten very talented at calculating volumes. But if it was literally a volume, it would have to be positive.
What our paper demonstrates is that, in the right regions (selected by the structure of the Amplituhedron), the amplitudes we’ve calculated so far are in fact positive. That first, basic requirement for the amplitude to actually literally be a volume is satisfied.
Of course, this doesn’t prove anything. There’s still a lot of work to do to actually find the thing the amplitude is the volume of, and this isn’t even proof that such a thing exists. It’s another, small piece of evidence. But it’s a reassuring one, and it’s nice to begin to link our approach with the Amplituhedron folks.
This week was the 75th birthday of John Schwarz, one of the founders of string theory and a discoverer of N=4 super Yang-Mills. We’ve dedicated the paper to him. His influence on the field, like the amplitudes of N=4 themselves, has been consistently positive.
# Four Gravitons in China
I’m in China this week, at the School and Workshop on Amplitudes in Beijing 2016.
It’s a little chilly this time of year, so the dragons have accessorized
A few years back, I mentioned that there didn’t seem to be many amplitudeologists in Asia. That’s changed quite a lot over just the last few years. Song He and Yu-tin Huang went from postdocs in the west to faculty positions in China and Taiwan, respectively, while Bo Feng’s group in China has expanded. As a consequence, there’s now a substantial community here. This is the third “Amplitudes in Asia” conference, with past years meeting in Hong Kong and Taipei.
The “school” part of the conference was last week. I wasn’t here, but the students here seem to have enjoyed it a lot. This week is the “workshop” part, and there have been talks on a variety of parts of amplitudes. Nima showed up on Wednesday and managed to talk for his usual impressively long amount of time, finishing with a public lecture about the future of physics. The talk was ostensibly about why China should build the next big collider, but for the most part it ended up as a more general talk about exciting open questions in high energy physics. The talks were recorded, so they should be online at some point.
# Hexagon Functions IV: Steinmann Harder
It’s paper season! I’ve got another paper out this week, this one a continuation of the hexagon function story.
The story so far:
My collaborators and I have been calculating “six-particle” (two particles collide, four come out, or three collide, three come out…) scattering amplitudes (probabilities that particles scatter) in N=4 super Yang-Mills. We calculate them starting with an ansatz (a guess, basically) made up of a type of functions called hexagon functions: “hexagon” because they’re the right functions for six-particle scattering. We then narrow down our guess by bringing in other information: for example, if two particles are close to lining up, our answer needs to match the one calculated with something called the POPE, so we can throw out guesses that don’t match that. In the end, only one guess survives, and we can check that it’s the right answer.
So what’s new this time?
More loops:
In quantum field theory, most of our calculations are approximate, and we measure the precision in something called loops. The more loops, the closer we are to the exact result, and the more complicated the calculation becomes.
This time, we’re at five loops of precision. To give you an idea of how complicated that is: I store these functions in text files. We’ve got a new, more efficient notation for them. With that, the two-loop functions fit into files around 20KB. Three loops, 500KB. Four, 15MB. And five? 300MB.
So if you want to imagine five loops, think about something that needs to be stored in a 300MB text file.
More insight:
We started out having noticed some weird new symmetries of our old results, so we brought in Simon Caron-Huot, expert on weird new symmetries. He couldn’t figure out that one…but he did notice an entirely different symmetry, one that turned out to have been first noticed in the 60’s, called the Steinmann relations.
The core idea of the Steinmann relations goes back to the old method of calculating amplitudes, with Feynman diagrams. In Feynman diagrams, lines represent particles traveling from one part of the diagram to the other. In a simplified form, the Steinmann conditions are telling us that diagrams can’t take two mutually exclusive shapes at the same time. If three particles are going one way, they can’t also be going another way.
With the Steinmann relations, things suddenly became a whole lot easier. Calculations that we had taken months to do, Simon was now doing in a week. Finally we could narrow things down and get the full answer, and we could do it with clear, physics-based rules.
More bootstrap:
In physics, when we call something a “bootstrap” it’s in reference to the phrase “pull yourself up by your own boostraps”. That impossible task, lifting yourself with no outside support, is essentially what we do when we “bootstrap”: we do a calculation with no external input, simply by applying general rules.
In the past, our hexagon function calculations always had some sort of external data. For the first time, with the Steinmann conditions, we don’t need that. Every constraint, everything we do to narrow down our guess, is either a general rule or comes out of our lower-loop results. We never need detailed information from anywhere else.
This is big, because it might allow us to avoid loops altogether. Normally, each loop is an approximation, narrowed down using similar approximations from others. If we don’t need the approximations from others, though, then we might not need any approximations at all. For this particular theory, for this toy model, we might be able to actually calculate scattering amplitudes exactly, for any strength of forces and any energy. Nobody’s been able to do that for this kind of theory before.
We’re already making progress. We’ve got some test cases, simpler quantities that we can understand with no approximations. We’re starting to understand the tools we need, the pieces of our bootstrap. We’ve got a real chance, now, of doing something really fundamentally new.
So keep watching this blog, keep your eyes on arXiv: big things are coming. | 2017-04-24 09:14:47 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 5, "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": 0.5212011933326721, "perplexity": 879.3363744921893}, "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-2017-17/segments/1492917119225.38/warc/CC-MAIN-20170423031159-00171-ip-10-145-167-34.ec2.internal.warc.gz"} |
https://physics.stackexchange.com/questions/156967/time-reversal-in-simple-solution-to-equation-of-motion | # Time reversal in simple *solution* to equation of motion
Consider the solution to the equation of motion for a particle with a constant acceleration: $$x(t) = x_0 + v_0t + \frac{1}{2}at^2.$$
If I let $t \rightarrow -t$, then the equation becomes: $$x(-t) = x_0 - v_0t + \frac{1}{2}at^2,$$
which is different. Does this mean that this equation is not symmetric under time-reversal? What is the physical meaning of this? What if this represented a ball falling under gravity being recorded on tape: surely we should see the same thing if we run the film in reverse?
• this is not the equation of motion, but a solution to some equation. – Phoenix87 Jan 6 '15 at 16:24
• That's not the equation of motion but a solution to the equation of motion. The equation of motion would be $x''(t)=a$. – CuriousOne Jan 6 '15 at 16:25
• Right, sorry, I'll change it – SuperCiocia Jan 6 '15 at 16:25
• In normal time the ball speeds up. In reverse time it slows down. Of course the solution is not symmetric under time reversal. – John Rennie Jan 6 '15 at 16:27
• But if $v_0=0$ then the equation is symmetric – SuperCiocia Jan 6 '15 at 16:28
If you substitute $t\to-t$, the sign of the velocity also changes, thus the equation maintains the same functional form
you are correct that all of this time-reversal business starts with the mapping $t \mapsto -t$. But when you do this, you have to time-reverse EVERYTHING.
Since $v = \frac{\mathrm dx}{\mathrm dt}$, we need to reverse the time in the denominator, which gives us $v \mapsto \frac{\mathrm dx}{\mathrm d(-t)} = -v$
The acceleration is unchanged, since $$a \mapsto \frac{\mathrm d}{\mathrm d(-t)} \frac{\mathrm dx}{\mathrm d(-t)} = (-1)^{2}a = a$$ | 2019-08-21 01:41:43 | {"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": 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": 0.658419668674469, "perplexity": 343.61061120649765}, "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/1566027315695.36/warc/CC-MAIN-20190821001802-20190821023802-00232.warc.gz"} |
http://mathhelpforum.com/number-theory/57230-solved-wilson-theorem-question-explanation.html | # Thread: [SOLVED] Wilson theorem Question Explanation
1. ## [SOLVED] Wilson theorem Question Explanation
Use:
$2\cdot4\cdot...\cdot(p-1)\equiv(2-p)(4-p)\cdot...\cdot(p-1-p)\equiv(-1)^{\frac{(p-1)}{2}}\cdot1\cdot3\cdot...\cdot(p-2)$ mod $p$
and
$(p-1)!\equiv-1$ mod p [Wilson's Theorem]
to prove
$1^2\cdot3^2\cdot5^2\cdot...\cdot(p-2)^2\equiv(-1)^{\frac{(p-1)}{2}}$ mod $p$
Relevant equations
Gauss lemma
wilson's theorem [ $(p-1)!\equiv-1$ mod $p$]
The attempt at a solution
need assistance
Thanks
2. $2 \cdot 4 \cdot \cdots \cdot (p-1)$
...... $\equiv (2 - p)(4 - p) \cdots ((p-1)-p) \ (\text{mod }p)$ (simply subtracting the modulus to each term)
...... $\equiv (-1)(p-2) \cdot (-1)(p-4) \cdot (-1)(p -6) \cdots \cdot (-1)(1) \ (\text{mod } p)$ (Factor out $-1$ from each term)
Now, notice that all the numbers from 1 to (p-1) are all least residues of p. We have half the numbers since all the even numbers from 1 to (p-1) are gone, leaving us with $\frac{p-2}{2}$ odd terms (1, 3, 7, ... , p-2 ) and thus $\frac{p-2}{2}$ factors of $-1$. So:
...... $\equiv (-1)^{(p-1)/2} \cdot 1 \cdot 3 \cdot \cdots \cdot (p-2) \ (\text{mod } p)$
3. Thanks a lot
4. Originally Posted by mathsss2
and
$(p-1)!\equiv-1$ mod p [Wilson's Theorem]
to prove
$1^2\cdot3^2\cdot5^2\cdot...\cdot(p-2)^2\equiv(-1)^{\frac{(p-1)}{2}}$ mod $p$
Using what o_O we can finish this problem.
Note $(p-1)! = \left\{ 1\cdot 3 \cdot ... \cdot (p-2) \right\} \left\{ 2\cdot 4\cdot ... \cdot (p-2) \right\} \equiv (-1)^{(p-1)/2} 1^2\cdot 3^2 \cdot ... \cdot (p-2)^2 (\bmod p)$
Thus, $(-1)^{(p-1)/2} 1^2\cdot 3^2 \cdot ... \cdot (p-2)^2 \equiv - 1 \implies 1^2\cdot 3^2 \cdot ... \cdot (p-2)^2 \equiv (-1)^{(p+1)/2} (\bmod p)$
5. Thanks for the help! This problem is now solved. Thanks a lot. | 2016-12-09 02:04:14 | {"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": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 20, "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": 0.9220070838928223, "perplexity": 3615.2823713929374}, "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-2016-50/segments/1480698542668.98/warc/CC-MAIN-20161202170902-00430-ip-10-31-129-80.ec2.internal.warc.gz"} |
https://www.groundai.com/project/constraints-on-horndeski-theory-using-the-observations-of-nordtvedt-effect-shapiro-time-delay-and-binary-pulsars/ | Constraints on Horndeski Theory Using the Observations of Nordtvedt Effect, Shapiro Time Delay and Binary Pulsars
Constraints on Horndeski Theory Using the Observations of Nordtvedt Effect, Shapiro Time Delay and Binary Pulsars
Shaoqi Hou School of Physics, Huazhong University of Science and Technology, Wuhan, Hubei 430074, China Yungui Gong School of Physics, Huazhong University of Science and Technology, Wuhan, Hubei 430074, China
July 10, 2019
Abstract
Alternative theories of gravity not only modify the polarization contents of the gravitational wave, but also affect the motions of the stars and the energy radiated away via the gravitational radiation. These aspects leave imprints in the observational data, which enables the test of General Relativity and its alternatives. In this work, the Nordtvedt effect and the Shapiro time delay are calculated in order to constrain Horndeski theory using the observations of lunar laser ranging experiments and Cassini time-delay data. The effective stress-energy tensor is also obtained using the method of Isaacson. Gravitational wave radiation of a binary system is calculated, and the change of the period of a binary system is deduced for the elliptical orbit. These results can be used to set constraints on Horndeski theory with the observations of binary systems, such as PSR J1738+0333. Constraints have been obtained for some subclasses of Horndeski theory, in particular, those satisfying the gravitational wave speed limits from GW170817 and GRB 170817A.
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I Introduction
General Relativity (GR) is one of the cornerstones of modern physics. However, it faces several challenges. For example, GR cannot be quantized, and it cannot explain the present accelerating expansion of universe, i.e., the problem of dark energy. These challenges motivate the pursuit of the alternatives to GR, one of which is the scalar-tensor theory. The scalar-tensor theory contains a scalar field as well as a metric tensor to describe the gravity. It is the simplest alternative metric theory of gravity. It solves some of GR’s problems. For example, the extra degree of freedom of the scalar field might account for the dark energy and explain the accelerating expansion of the universe. Certain scalar-tensor theories can be viewed as the low energy limit of string theory, one of the candidates of quantum gravity Fujii and Maeda (2003).
The detection of gravitational waves by the Laser Interferometer Gravitational-Wave Observatory (LIGO) and Virgo confirms GR to an unprecedented precision Abbott et al. (2016a, b, 2017a, 2017b, 2017c, 2017d) and also provides the possibility to test GR in the dynamical, strong field limit. The recent GW170814 detected the polarizations for the first time, and the result showed that the pure tensor polarizations are favored against pure vector and pure scalar polarizations Abbott et al. (2017b). The newest GW170817 is the first neutron star-neutron star merger event, and the concomitant gamma-ray burst GRB 170817A was later observed by the Fermi Gamma-ray Burst Monitor and the Anti-Coincidence Shield for the Spectrometer for the International Gamma-Ray Astrophysics Laboratory, independently Abbott et al. (2017c); Goldstein et al. (2017); Savchenko et al. (2017). This opens the new era of multi-messenger astrophysics. It is thus interesting to study gravitational waves in alternative metric theories of gravity, especially the scalar-tensor theory.
In 1974, Horndeski Horndeski (1974) constructed the most general scalar-tensor theory whose action contains higher derivatives of and , but still yields at most the second order differential field equations, and thus has no Ostrogradsky instability Ostrogradsky (1850). Because of its generality, Horndeski theory includes several important specific theories, such as GR, Brans-Dicke theory Brans and Dicke (1961), and gravity Buchdahl (1970); O’Hanlon (1972); Teyssandier and Tourrenc (1983) etc..
In Refs. Liang et al. (2017); Hou et al. (2017); Gong and Hou (2018), we discussed the gravitational wave solutions in gravity and Horndeski theory, and their polarization contents. These works showed that in addition to the familiar + and polarizations in GR, there is a mixed state of the transverse breathing and longitudinal polarizations both excited by a massive scalar field, while a massless scalar field excites the transverse breathing polarization only. In this work, it will be shown that the presence of a dynamical scalar field also changes the amount of energy radiated away by the gravitational wave affecting, for example, the inspiral of binary systems. Gravitational radiation causes the damping of the energy of the binary system, leading to the change in the orbital period. In fact, the first indirect evidence for the existence of gravitational waves is the decay of the orbital period of the Hulse-Taylor pulsar (PSR 1913+16) Hulse and Taylor (1975).
Previously, the effective stress energy tensor was obtained by Nutku Nutku (1969) using the method of Landau and Lifshitz Landau and Lifshitz (1975). The damping of a compact binary system due to gravitational radiation in Brans-Dicke theory was calculated in Refs. Wagoner (1970); Will and Zaglauer (1989); Damour and Esposito-Farèse (1998); Brunetti et al. (1999), then Alsing et al. Alsing et al. (2012) extended the analysis to the massive scalar-tensor theory. Refs. Stein and Yunes (2011); Saffer et al. (2018) surveyed the effective stress-energy tensor for a wide class of alternative theories of gravity using several methods. However, they did not consider Horndeski theory. Refs. Zhang et al. (2017); Liu et al. (2018) studied the gravitational radiation in screened modified gravity and gravity. Hohman Hohmann (2015) developed parameterized post-Newtonian (PPN) formalism for Horndeski theory. In this work, the method of Isaacson is used to obtain the effective stress-energy tensor for Horndeski theory. Then the effective stress-energy tensor is applied to calculate the rate of energy damping and the period change of a binary system, which can be compared with the observations on binary systems to constrain Horndeski theory. Nordtvedt effect and Shapiro time delay effect will also be considered to put further constraints. Ashtekar and Bonga pointed out in Refs. Ashtekar and Bonga (2017a, b) a subtle difference between the transverse-traceless part of defined by and the one defined by using the spatial transverse projector, but this difference does not affect the energy flux calculated in this work.
There were constraints on Horndeski theory and its subclasses in the past. The observations of GW170817 and GRB 170817A put severe constraints on the speed of gravitational waves Abbott et al. (2017e). Using this limit, Ref. Creminelli and Vernizzi (2017) required that and , while Ref. Ezquiaga and Zumalacárregui (2017) required and . Ref. Baker et al. (2017) obtained the similar results as Ref. Ezquiaga and Zumalacárregui (2017), and also pointed out that the self-accelerating theories should be shift symmetric. Arai and Nishizawa found that Horndeski theory with arbitrary functions and needs fine-tuning to account for the cosmic accelerating expansion Arai and Nishizawa (2017). For more constraints derived from the gravitational wave speed limit, please refer to Refs. Sakstein and Jain (2017); Gong et al. (2017); Crisostomi and Koyama (2018), and for more discussions on the constraints on the subclasses of Horndeski theory, please refer to Refs. Will (2014); Lambiase et al. (2015); Bhattacharya and Chakraborty (2017); Banerjee et al. (2017); Shao et al. (2017).
In this work, the calculation will be done in the Jordan frame, and the screening mechanisms, such as the chameleon Khoury and Weltman (2004a, b) and the symmetron Hinterbichler and Khoury (2010); Hinterbichler et al. (2011), are not considered. Vainshtein mechanism was first discovered to solve the vDVZ discontinuity problem for massive gravity Vainshtein (1972), and later found to also appear in theories containing the derivative self-couplings of the scalar field, such as some subclasses of Horndeski theory Deffayet et al. (2002); Babichev et al. (2009); Koyama et al. (2013); Babichev and Deffayet (2013); Winther and Ferreira (2015). When Vainshtein mechanism is in effect, the effect of nonlinearity cannot be ignored within the so-called Vainshtein radius from the center of the matter source. Well beyond , the linearization can be applied. The radius depends on the parameters defining Horndeski theory, and can be much smaller than the size of a celestial object. So in this work, we consider Horndeski theories which predict small , if it exists, compared to the sizes of the Sun and neutron stars. The linearization can thus be done even deep inside the stars. In this case, one can safely ignore Vainshtein mechanism.
The paper is organized as follows. In Section II, Horndeski theory is briefly introduced and the equations of motion are derived up to the second order in perturbations around the flat spacetime background. Section III derives the effective stress-energy tensor according to the procedure given by Isaacson. Section IV is devoted to the computation of the metric and scalar perturbations in the near zone up to Newtonian order and the discussion of the motion of self-gravitating objects that source gravitational waves. In particular, Nordtvedt effect and Shapiro time delay are discussed. In Section V, the metric and scalar perturbations are calculated in the far zone up to the quadratic order, and in Section VI, these solutions are applied to a compact binary system to calculate the energy emission rate and the period change. Section VII discusses the constraints on Horndeski theory based on the observations. Finally, Section VIII summarizes the results. Throughout the paper, the speed of light in vacuum is taken to be .
Ii Horndeski Theory
The action of Horndeski theory is given by Kobayashi et al. (2011),
S=∫d4x√−g(L2+L3+L4+L5)+Sm[gμν,ψm], (1)
where represents matter fields, is the action for , and the terms in the integrand are
L2=G2(ϕ,X),L3=−G3(ϕ,X)□ϕ, (2) L4=G4(ϕ,X)R+G4X[(□ϕ)2−(ϕ;μν)2], (3) L5=G5(ϕ,X)Gμνϕ;μν−G5X6[(□ϕ)3−3(□ϕ)(ϕ;μν)2 +2(ϕ;μν)3]. (4)
In these expressions, with , , , and for simplicity. are arbitrary functions of and 111 is usually called in literature.. For notational simplicity and clarity, we define the following symbol for the function ,
(5)
so in particular, with the value of in the flat spacetime background.
Suitable choices of reproduce interesting subclasses of Horndeski theory. For instance, one obtains GR by choosing and the remaining , with Newton’s constant. Brans-Dicke theory is recovered with , while the massive scalar-tensor theory with a potential Alsing et al. (2012) is obtained with , where is a constant; or with , , . Finally, gravity is given by , , with .
ii.1 Matter action
Although there are no coupling terms between matter fields and , matter fields indirectly interact with via the metric tensor. For example, in Brans-Dicke theory, acts effectively like the gravitational constant, which influences the internal structure and motion of a gravitating object, so the binding energy of the object depends on . Since the total energy is related to the inertial mass , then depends on , too. When their spins and multipole moments can be ignored, the gravitating objects can be described by point like particles, and the effect of can be taken into account by the following matter action according to Eardley’s prescription Eardley (1975),
Sm=−∑a∫ma(ϕ)dτa, (6)
whose stress-energy tensor is
Tμν=1√−g∑ama(ϕ)uμuνu0δ(4)(xλ−xλa(τ)), (7)
where describes the worldline of particle and . Therefore, if there is no force other than gravity acting on a self-gravitating object, this object will not follow the geodesic. This causes the violation of the strong equivalence principle (SEP).
In this work, the gravitational wave is studied in the flat spacetime background with and , so we expand the masses around the value in the following way,
ma(ϕ)=ma[1+φϕ0sa−12(φϕ0)2(s′a−s2a+sa)+O(φ3)]. (8)
Here, is the perturbation, and for simplicity. This expansion also requires that , so the present discussion does not apply to gravity. and are the first and second sensitivities of the mass ,
sa=dlnma(ϕ)dlnϕ∣∣ϕ0,s′a=−d2lnma(ϕ)d(lnϕ)2∣∣ϕ0. (9)
The sensitivities measure the violation of SEP.
ii.2 Linearized equations of motion
The equations of motion can be obtained and simplified using xAct package Martín-García et al. (2007, 2008); Martín-García (2008); Brizuela et al. (2009); Martín-García (). Because of their tremendous complexity, the full equations of motion will not be presented. Interested readers are referred to Refs. Kobayashi et al. (2011); Gao (2011). As we checked, xAct package gives the same equations of motion as Refs. Kobayashi et al. (2011); Gao (2011). For the purpose of this work, the equations of motion are expanded up to the second order in perturbations defined as
gμν=ημν+hμν,ϕ=ϕ0+φ. (10)
These equations are given in A.
The gravitational wave solutions are investigated in the flat spacetime background, which requires that
G2(0,0)=0,G2(1,0)=0. (11)
This can be easily checked by a quick inspection of Eqs. (94) and (94). Then dropping higher order terms in Eqs. (94) and (94), the linearized equations of motion are thus given by
(G2(0,1)−2G3(1,0))□φ+G2(2,0)φ+G4(1,0)R(1)=−(∂T∂ϕ)(1), (12) G4(0,0)G(1)μν−G4(1,0)(∂μ∂νφ−ημν□φ)=12T(1)μν, (13)
where is the trace, from now on, and the superscript implies the leading order part of the quantity.
The equations of motion can be decoupled by introducing an auxiliary field defined as following,
~hμν=hμν−12ημνh−G4(1,0)G4(0,0)ημνφ, (14)
where is the trace, and the original metric tensor perturbation is,
hμν=~hμν−12ημν~h−G4(1,0)G4(0,0)ημνφ, (15)
with . The equations of motion are gauge invariant under the the following infinitesimal coordinate transformation,
φ′=φ,~h′μν=~hμν−∂μξν−∂νξμ+ημν∂ρξρ, (16)
with . Therefore, one can choose the transverse gauge , and after some algebraic manipulations, the equations of motion become
(□−m2s)φ=T(1)∗2G4(0,0)ζ, (17) □~hμν=−T(1)μνG4(0,0), (18)
where 222The way defining is different from the one defining in Ref. Alsing et al. (2012) in that the coefficient of is not 1. with , and the mass of the scalar field is
m2s=−G2(2,0)/ζ, ζ=G2(0,1)−2G3(1,0)+3G24(1,0)/G4(0,0). (19)
Of course, , otherwise is non-dynamical.
From the equations of motion (17) and (18)), one concludes that the scalar field is generally massive unless is zero, and the auxiliary field resembles the spin-2 graviton field in GR. is sourced by the matter stress-energy tensor, while the source of the scalar perturbation is a linear combination of the trace of the matter stress-energy tensor and the partial derivative of the trace with respect to . This is because of the indirect interaction between the scalar field and the matter field via the metric tensor.
Iii Effective Stress-Energy Tensor
The method of Isaacson Isaacson (1967, 1968) will be used to obtain the effective stress-energy tensor for gravitational waves in Horndeski theory in the short-wavelength approximation, i.e., the wavelength with representing the typical value of the background Riemann tensor components. This approximation is trivially satisfied in our case, as the background is flat and . In averaging over several wavelengths, the following rules are utilized Misner et al. (1973):
1. The average of a gradient is zero, e.g., ,
2. One can integrate by parts, e.g., ,
where implies averaging. These rules apply to not only terms involving but also those involving . In the case of a curved background, these rules are supplemented by the one that covariant derivatives commute, which always holds in the flat background case.
With this method, the effective stress-energy tensor in an arbitrary gauge can be calculated straightforwardly using xAct and given by,
TGWμν=⟨12G4(0,0)(∂μ~hρσ∂ν~hρσ−12∂μ~h∂ν~h−∂μ~hνρ∂σ~hσρ−∂ν~hμρ∂σ~hσρ)+ζ∂μφ∂νφ+G4(1,0)(m2sφ~hμν+∂μφ∂ρ~hρν+∂νφ∂ρ~hρμ−ημν∂σφ∂ρ~hρσ)⟩. (20)
It can be checked that this expression is gauge invariant under Eq. (16). In fact, the terms in the first around brackets take exactly the same forms as in GR excerpt for a different factor. The fourth line remains invariant, as in the gauge transformation. To show that the remaining lines are also gauge invariant, making the replacement gives
Remaining lines=⟨G4(1,0)(m2sφ~hμν+∂μφ∂ρ~hρν+∂νφ∂ρ~hρμ−ημν∂σφ∂ρ~hρσ)⟩+⟨m2sG4(1,0)φ(−∂μξν−∂νξμ+ημν∂ρξρ)+G4(1,0)(−∂μφ∂ρ∂ρξν−∂νφ∂ρ∂ρξμ+ημν∂σφ∂ρ∂ρξσ)⟩. (21)
Far away from the matter, according to Eq. (17). Substituting this into the fourth line of Eq. (21), one immediately finds total derivatives of the forms and . So the first averaging rule implies that the last three lines of Eq. (21) vanish. Therefore, the effective stress-energy tensor (20) is indeed gauge invariant.
In vacuum, the transverse-traceless (TT) gauge ( and ) can be taken, and the effective stress-energy simplifies,
TGWμν=⟨12G4(0,0)∂μ~hTTρσ∂ν~hρσTT+ζ∂μφ∂νφ+m2sG4(1,0)φ~hTTμν⟩, (22)
where denotes the transverse-traceless part. In the limit that and the remaining arbitrary functions vanish, Eq. (20) recovers the effective stress-energy tensor of GR Misner et al. (1973). One can also check that Eq. (20) reduces to the one given in Ref. Brunetti et al. (1999) for Brans-Dicke theory in the gauge of and .
In order to calculate the energy carried away by gravitational waves, one has to first study the motion of the source. This is the topic of the next section.
Iv The Motion of Gravitating Objects in the Newtonian Limit
The motion of the source will be calculated in the Newtonian limit. The source is modeled as a collection of gravitating objects with the action given by Eq. (6). In the slow motion, weak field limit, there exists a nearly global inertial reference frame. In this frame, a Cartesian coordinate system is established whose origin is chosen to be the center of mass of the matter source. Let represent the field point whose length is denoted by .
In the near zone Poisson and Will (2014), the metric and the scalar perturbations will be calculated at the Newtonian order. The stress-energy tensor of the matter source is given by 333The matter stress-energy tensor and the derivative of its trace with respect to , , are both expanded beyond the leading order, because the higher order contributions are need to calculate the scalar perturbations in Section V.,
Tμν=∑amauμuν(1−12v2a−12hjj+saφϕ0+O(v4))δ(4)(xλ−xλa(τ)), (23)
and one obtains,
∂T∂ϕ=−∑amaϕ0[sa(1−12hjj−v2a2)−(s′a−s2a+sa)φϕ0+O(v4)]δ(4)(xλ−xλa(τ)). (24)
In these expressions, the 4-velocity of particle is and . With these results, the leading order of the source for the scalar field is
T(1)∗=−∑amaSaδ(4)(xλ−xλa(τ)), (25)
with .
Now, the linearized equations (17, 18) take the following forms
(□−m2s)φ=−12G4(0,0)ζ∑amaSaδ(4)(xλ−xλa(τ)), (26) □~hμν=−1G4(0,0)∑amauμuνδ(4)(xλ−xλa(τ)), (27)
and the leading order contributions to the perturbations are easily obtained,
φ(t,→x)=18πG4(0,0)ζ∑amaSarae−msra, (28) ~h00(t,→x)=14πG4(0,0)∑amara, (29)
and at this order, where and the scalar field is given by a sum of Yukawa potentials. The leading order metric perturbation can be determined by Eq. (15),
(30) hjk=δjk8πG4(0,0)∑amara(1−G4(1,0)G4(0,0)ζSae−msra), (31)
with .
iv.1 Static, spherically symmetric solutions
For the static, spherically symmetric solution with a single point mass at rest at the origin as the source, the time-time component of the metric tensor is
g00=−1+18πG4(0,0)Mr(1+G4(1,0)G4(0,0)ζSMe−msr)+⋯, (32)
where and is the sensitivity of the point mass . From this, the “Newton’s constant” can be read off
GN(r)=116πG4(0,0)(1+G4(1,0)G4(0,0)ζSMe−msr), (33)
which actually depends on the distance because the scalar field is massive. The measured Newtonian constant at the earth is with the radius of the Earth. The “post-Newtonian parameter” can also be read off by examining , which is
gjk=δjk[1+18πG4(0,0)Mr(1−G4(1,0)G4(0,0)ζSMe−msr)]+⋯=δjk(1+2G4(0,0)ζ−G4(1,0)SMe−msrG4(0,0)ζ+G4(1,0)SMe−msrGN(r)Mr)+⋯. (34)
In the PPN formalism, the space-space components of the metric take the following form,
gPPNjk=δjk(1+2γGNMr)+⋯, (35)
where the parameter is a constant. So
γ(r)=G4(0,0)ζ−G4(1,0)SMe−msrG4(0,0)ζ+G4(1,0)SMe−msr. (36)
The above result can recover the results for gravity and general scalar-tensor theory Capone and Ruggiero (2010); Perivolaropoulos (2010); Hohmann et al. (2013); Hohmann (2015) if we keep the equivalence principle. In the massless case (), we get
GN=116πG4(0,0)[1+G4(1,0)G4(0,0)ζSM], (37) γ=G4(0,0)ζ−G4(1,0)SMG4(0,0)ζ+G4(1,0)SM. (38)
Note that and both depend on which reflects the internal structure and motion of the gravitating object in question. Even if the scalar field is massless, this dependence still persists. Therefore, neither of them is universal due to the violation of SEP caused by the scalar field. It is obvious that should take the same value as .
iv.2 Equations of motion of the matter
With the near zone solutions (28), (30) and (31) one obtains the total matter Lagrangian up to the linear order,
(39)
where is the distance between the particles and . The equation of motion for the mass can thus be obtained using the Euler-Lagrange equation, yielding its acceleration,
aja=−116πG4(0,0)∑b≠ambr2ab^rjab×[1+SaSbG4(0,0)ζ(1+msrab)e−msrab], (40)
with . In particular, for a binary system, the relative acceleration is
aj=−m^rj1216πG4(0,0)r212[1+SaSbG4(0,0)ζ(1+msr12)e−msr12], (41)
where is the total mass. The first term in the square brackets gives the result that resembles the familiar Newtonian gravitational acceleration, while the second one reflects the effect of the scalar field. In the massless case, the second term no longer depends on and can be absorbed into the first one, so the binary system moves in a similar way as in Newtonian gravity with a modified Newton’s constant.
The Hamiltonian of the matter is
Hm=∑a→pa⋅→xa−Lm=∑ama[12v2a−132πG4(0,0)×∑b≠ambrab(1+SaSbG4(0,0)ζe−msrab)], (42)
where is the -th component of the canonical momentum of particle , and the total rest mass has been dropped. In particular, the Hamiltonian of a binary system is given by
Hm=μv22−μm16πG4(0,0)r12×[1+S1S2G4(0,0)ζ(1+msr12)e−msr12], (43)
where , and is the reduced mass. This will be useful for calculating the total mechanical energy of a binary system and the ratio of energy loss due to the gravitational radiation.
iv.3 Nordtvedt effect
The presence of the scalar field modifies the trajectories of self-gravitating bodies. They will no longer follow geodesics. Therefore, SEP is violated in Horndeski theory. This effect is called the Nordtvedt effect Nordtvedt (1968a, b). It results in measurable effects in the solar system, one of which is the polarization of the Moon’s orbit around the Earth Nordtvedt (1982); Will (1993).
To study the Nordtvedt effect, one considers a system of three self-gravitating objects and and studies the relative acceleration of and in the field of . With Eq. (40) and assuming , the relative acceleration is
ajab≈−116πG4(0,0)ma+mbr2ab^rjab×[1+SaSbG4(0,0)ζ(1+msrab)e−msrab]−mc16πG4(0,0)(^rjacr2ac−^rjbcr2bc)+Sc(sa−sb)8πG4(0,0)ϕ0ζmc^rjacr2ac(1+msrac)e−msrac, (44)
where the first term presents the Newtonian acceleration modified by the presence of the scalar field, the second is the tidal force caused by the gravitational gradient due to the object , and the last one describes the Nordtvedt effect. The effective Nordtvedt parameter is
ηN=Sc8πGNG4(0,0)ϕ0ζ(1+msrac)e−msrac. (45)
This parameter depends on , so this effect is indeed caused by the violation of SEP.
iv.4 Shapiro time delay effect
Another effect useful for constraining Horndeski theory is the Shapiro time delay Shapiro (1964). In order to calculate this effect, one considers the photon propagation time in a static (or nearly static) gravitational field produced by a single mass at the origin. Due to the presence of gravitational potential, the 3-velocity of the photon in the nearly inertial coordinate system is no longer 1 and varies. The propagation time is thus different from that when the spacetime is flat. Let the 4 velocity of the photon be , then gives
−1+h00+(δjk+hjk)vjvk=0, (46)
where and are given by Eqs. (30) and (31) specialized to a single mass case. In the flat spacetime, the trajectory for a photon emitted from position at time is a straight line , where is the direction of the photon. The presence of the gravitational potential introduces a small perturbation so that . Substituting Eqs. (30) and (31) into Eq. (46), one obtains
^N⋅dδ→xdt=−M8πG4(0,0)r(t), (47)
where . Suppose the photon emitted from position is bounced back at position and finally returns to . The total propagation time is
Δt=2|→xp−→xe|+δt, (48)
where is caused by the Shapiro time delay effect,
δt=2∫tpte^N⋅dδ→xdtdt=M4πG4(0,0)ln(re+^N⋅→xe)(rp−^N⋅→xp)r2b, (49)
where , and is the impact parameter of the photon relative to the source.
Since in Eq. (49) is not measurable, one replaces it with the Keplerian mass
MK=M16πG4(0,0)GN(1+G4(1,0)G4(0,0)ζSMe−msr), (50)
with and the sensitivity of the source. In terms of , the Shapiro time delay is
δt=2GNMK(1+γ(r))ln(re+^N⋅→xe)(rp−^N⋅→xp)r2b. (51)
For the Shapiro time delay occurring near the Sun, in the above equation should be 1 AU, as this is approximately the distance where the Keplerian mass of the Sun is measured.
V Gravitational Wave Solutions
In the far zone, only the space-space components of the metric perturbation are needed to calculate the effective stress-energy tensor. Since the equation of motion (18) for takes the similar form as in GR, the leading order contribution to is given by,
~hjk(t,→x)=18πG4(0,0)rd2Ijkdt2, (52)
where is the mass quadrupole moment. As in GR, the TT part of is also related to the reduced quadrupole moment ,
~hTTjk=18πG4(0,0)rd2JTTjkdt2. (53)
The leading order term for the scalar field is the mass monopole which does not contribute to the effective stress-energy tensor, so it is necessary to take higher order terms into account. To do so, the scalar equation (94) is rewritten with the linearized equations substituted in, which is given by
(54)
In the following discussion, it is assumed that the scalar field is massless for simplicity. The details to obtain the following results can be found in B. The leading order contribution to comes from the first term on the right hand side of Eq. (54), which is the mass monopole moment,
φ[1]=18πG4(0,0)ζr∑amaSa. (55)
From now on, the superscript indicates the order of a quantity in terms of the speed , i.e., is of the order . is independent of time, so it does not contribute to the effective stress-energy tensor. The next leading order term is the mass dipole moment,
φ[1.5]=18πG4(0,0)ζr∑amaSa(^n⋅→va), (56)
in which . This gives the leading contribution to the effective stress-energy tensor. At the next next leading order, there are more contributions from the remaining terms on the right hand side of Eq. (54). First, there is the mass quadruple moment contribution,
φ[2]1=18πG4(0,0)ζr∑amaSa[(^n⋅→aa)(^n⋅→xa)+(^n⋅→va)2]. (57)
And the remaining contribution to the scalar wave is
φ[2]2=−116πG4(0,0)ζr∑amaSav2a+164π2G24(0,0)ζr′∑a,bmambrab(−Sa2+3G4(1,0)2G4(0,0)ζSaSb+S′aSbϕ0ζ)+164π2G24(0,0)ζ2r⎛⎝G4(2,0)−G24(1,0)G4(0,0)⎞⎠∑∑′a,bmambSbrab,−G2(3,0)256π2 | 2021-03-03 11:02:55 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8254757523536682, "perplexity": 538.5043669831842}, "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-2021-10/segments/1614178366959.54/warc/CC-MAIN-20210303104028-20210303134028-00541.warc.gz"} |
http://uawards.umn.edu/former-regents-professors/eric-sheppard | # Eric Sheppard
### Biography
#### (At time of award presentation, 2009)
Professor of Geography, College of Liberal Arts, is described by his colleagues as a "towering intellect, a universally admired educator, and a highly respected leader." His contributions have transformed the core understanding of the space economy, urban transformation, regional development, globalization, and geographic science, and are recognized globally. He is credited with bringing his department to a top-three national status. Considered to be one of the worlds leading geographers, his scholarly work includes nine books, thirty-two major refereed papers and seventeen book chapters. His first book, The Capitalist Space Economy (written with a former student who is now Distinguished Research Professor at the University of British Columbia), took up systematically the spatial dynamics of pricing strategies, commodity flows, labor markets and resource endowments. A review published by the Annals of the Association of American Geography, called it "a seminal contribution to the writings on political economy." He has worked tirelessly to shape the discipline at large and an example of that work is the development of GIS (Geographic Information Science), or GIS and Society. He is credited with pursuing the implications of considering how GIS is intertwined with society and for taking the lead in developing more critical approaches, and for catalyzing research on "public participation GIS." Not only is his research held in high esteem, so are his cutting-edge graduate and undergraduate textbooks, which have become key texts in classroom around the world. In addition, his contributions to the Interdisciplinary Center for the Study of Global Change have made him a key contributor in internationalizing the University. The $700,000 grant he secured from the Mellon Foundation funds a collaborative faculty and graduate education partnership between ICGC and the University of the Western Cape in Cape Town, South Africa, placing Minnesota at the center of an innovative scholarly network. He has expanded this endeavor to universities in Turkey. Most recently, he took the lead in putting together a$600,000 grant proposal to the Mellon Foundation to support the education of undergraduate students of color. He has held numerous leadership positions in the University and its community including department chair, vice-chair of the CLA Assembly, chair of the Promotion and Tenure Committee, chair of the Urban Studies Advisory Committee, and has been a members of many search committees and advisory committees. He is a successful advisor and mentor to graduate and undergraduate students. Six of his Ph.D. students have graduated and have gone on to hold tenure track positions and post-doctoral fellowships as prestigious universities both in the United States and abroad. He has organized/co-organized numerous international conferences and has been the keynote, plenary lecturer, or invited lecturer at more than 130 conferences or universities nationally and internationally. Professor Sheppard exemplifies the qualities of a Regents Professor. | 2017-12-18 18:48:41 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 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": 0.20806178450584412, "perplexity": 4806.595340265238}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-51/segments/1512948619804.88/warc/CC-MAIN-20171218180731-20171218202731-00102.warc.gz"} |
http://openstudy.com/updates/5069fb67e4b0e78f215d7126 | ## woleraymond Group Title In figure below if a charge of 4uc flows from the 6V battery to plate P of the 1 uF capacitor. what charge flow from a - Q to R b - S to the battery c - What is the pd across C1 and C2 d - What is capacitance of C2 e - What is the single capacitor which is equivalent to C1 and C2 in series and what charge would it store? 2 years ago 2 years ago
1. woleraymond
|dw:1349123247879:dw|
2. Algebraic!
a. 4uC b. 4uC c. 6V d. $Q = C _{eq}V$ $\frac{ 1 }{ C _{eq}} = \frac{ 1 }{ C _{1}}+\frac{ 1 }{ C _{2}}$ $Q = ( \frac{ 1 }{ C _{1}}+\frac{ 1 }{ C _{2}})^{-1}V$
3. Algebraic!
make sense?
4. masumanwar
use conservation of charge
5. woleraymond
@Algebraic! please can you explain how u derived a,b,c
6. Algebraic!
if +4uC is on the plate at P then -4uC will be on the plate at Q meaning +4uC flowed to the plate at R meaning that there's - 4 uC at the plate at S meaning +4uC flowed to the terminal of the battery. For c. ) I did the total voltage drop, which must be 6V, you could also do the drop per capacitor, and then find the capacitance of c2 from that. I didn't do it that way, I combined the solution to d and e into one part instead. | 2014-10-25 13:26:48 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.3710509240627289, "perplexity": 1402.8026951568122}, "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-2014-42/segments/1414119648250.59/warc/CC-MAIN-20141024030048-00102-ip-10-16-133-185.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/proof-this-cross-product-property.287678/ | # Homework Help: Proof this cross product property
1. Jan 26, 2009
### -EquinoX-
1. The problem statement, all variables and given/known data
Show that :
|| a x b||^2 = ||a||^2||b||^2 - (a . b)^2
2. Relevant equations
3. The attempt at a solution
I don't know where to start off
Last edited: Jan 26, 2009
2. Jan 26, 2009
### Dick
What's the difference between ||axb||^2 and (axb)^2? Don't you mean (a.b)^2 instead of (axb)^2? Just use expressions for the magnitude of the cross and dot product and the angle between the vectors a and b.
3. Jan 26, 2009
### Staff: Mentor
What you're supposed to show doesn't make sense.
Given that, I understand what is meant by || a x b||^2, ||a||^2, and ||b||^2, but I have no idea what (a x b)^2 even means, let alone how to compute it.
Can you clarify what is meant here by (a x b)^2?
4. Jan 26, 2009
### -EquinoX-
sorry a typo, I've revised the question, it should be (a.b)^2
I know that a x b = ||a|| ||b|| sin x, so then how do I find the magnitude of this? I am confused because there's the sin function...
5. Jan 26, 2009
### Dick
Then express both side in terms of |a|, |b| and the angle between them.
6. Jan 26, 2009
### -EquinoX-
not sure I understand what you mean
7. Jan 26, 2009
### Dick
If t is the angle between a and b, then a.b=|a|*|b|*cos(t). There a similar relation for |axb|.
8. Jan 26, 2009
### -EquinoX-
ok is this correct?
$$|a\times b|^2 = |a|^2|b|^2\sin^2\theta = |a|^2|b|^2(1 - \cos^2\theta)$$
$$= |a|^2|b|^2 - |a|^2|b|^2\cos^2\theta = |a|^2|b|^2 - (a\cdot b)^2$$
9. Jan 26, 2009
### Dick
Absolutely.
10. Jan 26, 2009
### -EquinoX-
I am not sure of the notations through can I write |a x b|^2 as ||a x b||^2 as it denotes magnitude?
11. Jan 26, 2009
### Dick
They both mean the same thing to me. If you're a stickler you might want to demand writing ||v|| if v is a vector and |x| if x is a number.
12. Jan 26, 2009
### -EquinoX-
well yes, but what's meant here by ||v|| here is the magnitude of vector v right? now I am confused though, by definition of the geometric, it says that:
a x b = ||a|| ||b|| sin x
and it's not
|a x b| = ||a|| ||b|| sin x
can you explain this?
13. Jan 26, 2009
### Staff: Mentor
I don't understand what you're saying above. Does "and it's not" refer to the first equation or the second? Either way, the first equation is incorrect and the second one is correct (aside from inconsistent usage of | | and || ||).
a X b is a vector, which ||a|| ||b|| sin x is the product of three numbers, and so is a number.
14. Apr 25, 2009
### natangwe
Can you please prove for me this identity: A x (B x C) = (A.C)B - (A.B)C
15. Apr 25, 2009
### natangwe
Can you please prove for me this identity, analytically: A x (B x C) = (A.C)B - (A.B)C and llU x Vll^2 = llUll^2 llVll^2 - (U.V)^2. Any one with understanding on this please help out. | 2018-09-21 09:27:21 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.6530005931854248, "perplexity": 2223.1225535366693}, "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-2018-39/segments/1537267157028.10/warc/CC-MAIN-20180921092215-20180921112615-00209.warc.gz"} |
https://www.groundai.com/project/probing-the-anisotropic-expansion-from-supernovae-and-grbs-in-a-model-independent-way/ | Probing the anisotropic expansion from supernovae and GRBs in a model-independent way
# Probing the anisotropic expansion from supernovae and GRBs in a model-independent way
J. S. Wang and F. Y. Wang
School of Astronomy and Space Science, Nanjing University, Nanjing 210093, China
Key Laboratory of Modern Astronomy and Astrophysics (Nanjing University), Ministry of Education, Nanjing 210093, China
jiesh.wang@gmail.com(JSW)fayinwang@nju.edu.cn(FYW)
###### Abstract
In this paper, we study the anisotropic expansion of the universe using type Ia supernovae Union 2.1 sample and 116 long gamma-ray bursts. The luminosity distance is expanded with model-independent cosmographic parameters as a function of directly. Thus the results are independent of cosmology model. We find a dipolar anisotropy in the direction (, ) in galactic coordinates with a significant evidence (more than ). The magnitude is for the dipole, and for the monopole, respectively. This dipolar anisotropy is more significant at low redshift from the redshift tomography analysis. We also test whether this preferred direction is caused by bulk flow motion or dark energy dipolar scalar perturbation. We find that the direction and the amplitude of the bulk flow in our results are approximately consistent with the bulk flow surveys. Therefore, bulk flow motion may be the main reason for the anisotropic expansion at low redshift, but the effect of dipolar distribution dark energy can not be excluded, especially at high redshift.
###### keywords:
cosmology: theory, dark energy, Type Ia supernovae
## 1 Introduction
The Universe is homogeneous and isotropic on cosmic scales on the basis of the cosmological principle. It is the foundation in modern cosmology. This principle is well confirmed by the precise measurements of cosmic microwave background (CMB) from Wilkinson Microwave Anisotropy Probe (WMAP) (Hinshaw et al., 2013) and (Planck Collaboration et al., 2013). However, in the processing of CMB data, the motion of our Local Group of galaxies should be deducted. Kogut et al. (1993) obtained that the peculiar velocity is km s towards () using the COBE Differential Microwave Radiometers first year data. Bulk flow velocity on the scales around Mpc is found to be km s towards (), roughly close to CMB dipole (Watkins, Feldman & Hudson, 2009). But, it’s much larger than the expected bulk flow velocity on the same scale, which is approximately 110 km s in the standard CDM normalized with WMAP5 (). This hints that the universe may have a preferred expanding direction.
Additional evidences for such dipolar anisotropy have been obtained by low multipoles alignment in CMB angular power spectrum (Lineweaver et al., 1996; Tegmark et al., 2003; Bielewicz, Górski & Banday, 2004; Frommert & Enßlin, 2010), large scale alignments of quasar polarization vectors (Hutsemékers et al., 2005, 2011), dark energy dipole in type Ia supernovae (SNe Ia) (Antoniou & Perivolaropoulos, 2010; Mariano & Perivolaropoulos, 2012; Yang, Wang & Chu, 2014), and the spatial variation in fine-structure constant (Webb et al., 2011; King et al., 2012). The significances of these dipoles anisotropy are around . Indeed, many studies using SNe Ia data to test if the universe accelerates isotropically have been done (Kolatt & Lahav, 2001; Bonvin, Durrer & Kunz, 2006; Gordon, Land & Slosar, 2007; Schwarz & Weinhorst, 2007; Gupta, Saini & Laskar, 2008; Koivisto & Mota, 2008a, b; Blomqvist, Mörtsell & Nobili, 2008; Cooray, Holz & Caldwell, 2010; Gupta & Saini, 2010; Cooke & Lynden-Bell, 2010; Antoniou & Perivolaropoulos, 2010; Campanelli et al., 2011; Koivisto et al., 2011; Colin et al., 2011; Mariano & Perivolaropoulos, 2012; Turnbull et al., 2012; Cai & Tuo, 2012; Li et al., 2013; Yang, Wang & Chu, 2014). These studies of the anisotropic effects are mainly considered to be caused by bulk flow motion or dark energy dipolar distribution on the basis of CDM.
Cai et al. (2013) examined the dark energy anisotropy deviations using the SNe Ia of Union 2 sample and 67 gamma-ray bursts (GRBs) from Liang et al. (2008) and Wei (2010). However, their results show that the anisotropic evidence in CDM doesn’t improve much compared to the results from SNe Ia data alone obtained by Mariano & Perivolaropoulos (2012). Thereby, the significance of anisotropy needs to be studied again with the joint of more high-redshift GRBs. On the other hand, dipolar anisotropy can be caused by many mechanisms, for instance, the cosmic bulk flow motion (Colin et al., 2011; Turnbull et al., 2012; Feindt et al., 2013; Li et al., 2013; Rathaus, Kovetz & Itzhaki, 2013) and dark energy anisotropy (Koivisto & Mota, 2008b; Antoniou & Perivolaropoulos, 2010; Perivolaropoulos, 2014). Therefore, it’s also important to distinguish which mechanism is dominant in the deviation of isotropy.
Cosmological models are assumed in the previous studies, thus, their results of anisotropic expansion are model-dependent. In this paper, we use a model-independent method to study the anisotropic expansion from standard candles, i.e., expanding the luminosity distance using fourth order Hubble series parameters as a function of directly (Cattoen & Visser, 2007; Wang & Dai, 2011). This expansion is only dependent on the cosmological principle and the Friedmann-Robertson-Walker (FRW) metric. The Union 2.1 SNe Ia sample (Suzuki et al., 2012) and 116 GRBs (Wang, Qi & Dai, 2011) are used in our study.
The structure of this paper is organized as follows: in the next section, we give brief introductions of observational data. We then introduce the method for quantifying the anisotropic expansion effects on luminosity distances and give the significance through Monte Carlo simulation. In section 3, we divide the data set into several portions with two approaches: redshift bins and variable redshift limits, then we analyze the anisotropic expansion in different redshift ranges. In section 4, we test the bulk flow dipole and simplified dark energy dipolar perturbation model as possible mechanisms for anisotropy. Conclusions and discussions are given in section 5.
## 2 Dipolar anisotropic expansion with cosmography parameters
### 2.1 Observational data
In analysis, we use the latest Union 2.1 sample (Suzuki et al., 2012) to constrain the dipolar anisotropy, which contains 580 SNe Ia and covers the redshift range . To avoid the lack of high redshift data, we also combine the 116 GRB samples, which are compiled and calibrated by Wang, Qi & Dai (2011) and Wang & Dai (2011) (see detailed information including equatorial coordinates in Table 5). The redshift of GRBs reaches up to . The equatorial coordinates of these GRBs are taken from NASA/IPAC Extragalactic Database .
We expand the luminosity distance in terms of Hubble series parameters: Hubble parameter (), deceleration (), jerk () and snap () parameter. These four parameters are the first, second, third and fourth derivatives of the scale factor in the Taylor expansion, respectively. They are model-independent and obtained only from the FRW metric. The definitions of the cosmography parameters can be expressed as follows,
H=˙aa, q=−1H2¨aa, j=1H3˙¨aa, s=1H4¨¨aa. (1)
Visser (2004) expands the luminosity distance as a function of with the cosmography parameters, which have been studied using observational data (Wang, Dai & Qi, 2009a, b). However, it diverges at high redshift, and the GRB data reaches up to a high redshift . To avoid this problem, Cattoen & Visser (2007) recast the with improved parameter . Therefore, the redshift range can be mapped into . The luminosity distance can be expanded as a function of as following on the assumption of flat Universe (Cattoen & Visser, 2007),
dL(y)=cH0{y−12(q0−3)y2+16[11−5q0−j0]y3 +124[50−7j0−26q0+10q0j0+21q20−15q30+s0]y4+O(y5)}, (2)
where , , , are the current values. Then the distance modulus can be derived,
μth=5logdLMpc+25. (3)
The best-fit cosmography parameters can be obtained by minimizing the , which is constructed as follow,
χ2(H0,q0,j0,s0)=580∑i=1[μSNe(zi)−μth(zi)]2σ2μ,i+116∑i=1[μGRB(zi)−μth(zi)]2σ2μ,i, (4)
where and are the observed distance modulus and error bars, and are taken from Wang & Dai (2011).
### 2.2 Anisotropic deviation effects on luminosity distance
We convert the equatorial coordinates of each SNe Ia and GRB sample to galactic coordinates (see in Figure 1), then we find their unit vectors in Cartesian coordinates
^ni=cos(bi)cos(li)^i+cos(bi)sin(li)^j+sin(bi)^k. (5)
In order to quantify the anisotropic deviations on luminosity distance, we define the deviations of distance modulus from the best fit isotropic configuration as follows,
Δμ(z)¯μ(z)≡¯μ(z)−μobs(z)¯μ(z), (6)
where are the distance modulus in the context of best-fit cosmography parameters, which are calculated in section 2.1, that is .
We use a dipole model in the direction, and a monopole ,
(Δμ(z)¯μ(z))i=^ni⋅→D−B=Acosθ−B, (7)
where and are the magnitudes of the dipole and monopole, respectively. To fit the models with the SNe Ia and GRB data, we construct the ,
χ2(→D,B)=696∑i=1[(Δμ(z)¯μ(z))i−Acosθi+B]2σ2i, (8)
where are the errors in data sets.
We find the dipole points to the direction (, ), which is shown in Figure 1. The black star is the dipolar expansion direction, and the dark blue blob is the error region. The magnitudes of the dipole and monopole are and , respectively. It’s approximately consistent with the results from Mariano & Perivolaropoulos (2012), Cai et al. (2013), and Yang, Wang & Chu (2014), which are based on CDM model.
### 2.3 Significance of dipolar anisotropy
Our results show that the monopole is not significant, while implies the dipolar anisotropy, around in the relative errors. To obtain the confidence level of dipole anisotropy precisely, we use the Monte Carlo (MC) simulations.
We define new distance modulus () through a Gaussian random selection function, i.e. new distance modulus () will be obtained by the normal distribution with mean values and standard deviations from the observed data. We then take place of the observed distance modulus with the newly constructed , while use the same observed redshift, standard deviations and coordinates in the observed data.
The analysis method is similar to the method in section 2.2. Then, we obtain a new magnitude of the dipolar anisotropy in each simulation. We do MC simulations in total, and divide them into 47 bins. Figure 2 illustrates the probability of each bin value . The x-axis is the simulated dipole magnitude in units of , and the y-axis is the count of each bin. The arrow points to the dipole magnitude obtained with observed data. The results show that the probability that we can observe the magnitude at is , i.e. the confidence level of the dipolar anisotropy is , larger than . It’s more significant than the results from SNe Ia Union 2 data (Mariano & Perivolaropoulos, 2012) and Union 2.1 data (Yang, Wang & Chu, 2014) alone, which give the probability and , respectively. Therefore, our result shows the significance of dipolar expansion amplitude grows larger with the combination of GRB sample. We also show the evolution of the confidence level with the increasing MC simulations in Figure 3. It illustrates that MC simulations are enough to converge.
## 3 Redshift Tomography
In this section, we focus on the anisotropic effects in different redshift ranges. We use two approaches to study these effects and compare the results with respect to error bar sizes, which relate to the confidence level. The first approach is changing the redshift upper or lower limits, and the second one is dividing the data into 6 redshift bins. The same analysis procedure presented in section 2 are used in each redshift range . The number of data points are approximately equal in each redshift bin, and we define an average redshift of each bin. The variable upper limits method starts from the upper limit , approximately Mpc. Then we increase the upper limit within six steps. The variable lower limit method starts from , then we increase it in three steps.
Our results in different redshift ranges are shown in Table 1. The results show that the Union 2.1 data constraints are more stringent than GRB data. This is obvious because of the smaller error bars of SNe Ia luminosity distances comparing with GRBs. The results from variable redshift upper limits method show that the monopole, dipole magnitudes, and the direction converge with the increasing data points. Most of the results are consistent with the full data, except the lowest redshift range. For the variable lower limits way, the magnitudes of dipole is and in redshift ranges and , respectively. Therefore, these results don’t show significant anisotropy at these high redshift ranges, because of their large relative errors.
The redshift bins methods show the anisotropy direction changes randomly with redshift. The magnitudes of monopole and dipole don’t show significant evolution with the redshift (see Figures 4 and 5), except for the redshift range , but this bin only contains 67 sample, while covers a large redshift range. We find that the lowest redshift bin () show the most significant evidence for dipolar anisotropy with the smallest relative errors, namely, . While other bins show weaker evidences for anisotropy. Because of their random directions, the total effects are even much weaker, which can be obtained from their magnitude: in the range of and in . Thus, the significant dipolar anisotropy of the full data is mainly caused by the low redshift sample.
## 4 Possible mechanism for dipolar anisotropy
We have studied the anisotropic expansion with SNe Ia and GRB luminosity distances. We find that the probability of such a dipolar anisotropy is more than , and it mainly origins from the low redshift data. While the monopole is not significant. Thus, in this section, we try to study two possible mechanisms for dipolar anisotropy. We use bulk flow motion model and simplified scalar perturbation metric model caused by dark energy dipolar distributions to fit the same data. However, our methods to quantify the magnitudes of bulk flow motion and dark energy perturbation are simplified. For the careful study of these effects, we need to use the velocity field (Koivisto & Mota, 2008a; Li et al., 2013) and anisotropy dark energy model (Koivisto & Mota, 2006, 2008b; Mariano & Perivolaropoulos, 2012). Since the magnitudes of anisotropy are very small, our results are still reliable.
### 4.1 Bulk flow motion
Bulk flow motion can affect the Hubble Parameter directly, where is the comoving distance. Many methods have been taken to analyze this effect on SNe Ia data (Bonvin, Durrer & Kunz, 2006; Colin et al., 2011; Feindt et al., 2013; Rathaus, Kovetz & Itzhaki, 2013). We choose one method of them to reconstruct the luminosity distance (Bonvin, Durrer & Kunz, 2006) as follows
d′L(z)=dL(z)+vBF⋅^ni(1+z)2H(z), (9)
where is the velocity of the bulk flow, is the luminosity distance defined in Eq.(2), and is defined in Eq.(5). The is
χ2(→vBF)=696∑i=1[μ(d′L)−μi]2σ2i. (10)
The results are shown in Table 2. For bulk flow motion, the effects at low redshift ranges are much more attractive. The velocity and direction are km s and () for the full data. On the scale of Mpc, i.e. , the velocity is km s, and the direction points to (). They are approximately consistent with other peculiar velocity surveys shown in Table 3 (Hudson et al., 2004; Sarkar, Feldman & Watkins, 2007; Watkins, Feldman & Hudson, 2009; Feldman, Watkins & Hudson, 2010; Ma, Gordon & Feldman, 2011). Their average velocity and direction are 344.5 km s and on the scale 82.5 Mpc. But all the velocities are larger than the expected velocity in CDM.
The direction and velocity of redshift range are consistent with the results from . Thus, for this low redshift range, the anisotropy is mainly caused by the bulk flow velocity. The origin for this motion is thought to be the attraction of Shapley Super Cluster (Colin et al., 2011). But this effect could be much weaker at high redshift, because of the larger Hubble flow. Therefore, we can not excluded the dipolar dark energy effects, especially at high redshift.
### 4.2 Simplified dark energy dipolar scalar perturbation
Another possible anisotropic mechanism is the dark energy dipolar distribution, resulting in dipolar scalar perturbation. For simplification, we use an affected metric imitating the Schwarzschild metric instead of FRW metric,
ds2=(1−2ϕ(→x))dt2−a2(t)(1+2ϕ(→x))δijdxidxj. (11)
We assume the scalar perturbation field , where is the angular between the dipole direction and the observed sample and is the magnitude of the perturbation.
To determine the perturbed energy-momentum tensor, we base on CDM. Then the luminosity distance can be obtained by solve the Einstein equation (Li et al., 2013),
d′L(z)=c(1+z)H0∫z0(1−dcosθ)dx√Ωm0(1+x)3+1−Ωm0−4dcosθ(1+x)53H20d2L(z) , (12)
where is defined in Eq.(2) and is the current matter density. We use the Eq.(10), and analyze the data in the same way. The results show that the anisotropy amplitude is very small (shown in Table 4). The magnitude of scalar perturbation is , the direction is (). The dipolar evidence in the redshift range is insignificant with large error bar size . But we cannot draw an exact conclusion that the dark energy distributes isotropically or not, because the high redshift sample is sparse.
## 5 conclusions and discussions
In this paper, we study the anisotropic cosmic expansion in a model-independent way. The data we use are the combination of SNe Ia Union 2.1 and 116 GRB samples. The luminosity distance is expanded with model-independent cosmography parameters: Hubble (), deceleration (), jerk () and snap () parameters. These cosmographic parameters obtained from the FRW metric are only based on the cosmological principle.
The magnitudes of dipole and monopole are and . Our results show that the dipolar anisotropy is significant. The confidence level is , more than , by doing MC simulations. It’s more significant than the results from SNe Ia Union 2 (Mariano & Perivolaropoulos, 2012) and Union 2.1 data (Yang, Wang & Chu, 2014) alone, which give out the probability and , respectively. Our results are also much more significant than the results from Cai et al. (2013), who used a combination of SNe Ia Union 2 and 67 GRBs from Liang et al. (2008) and Wei (2010). The dipolar direction in our study points to (, ) in galactic coordinates for the full data. This direction is consistent with the results from Mariano & Perivolaropoulos (2012), Cai et al. (2013) and Yang, Wang & Chu (2014).
To study the anisotropy in different redshift ranges, we used two approaches: changing the redshift ranges upper or lower limits and dividing the full data into six bins. The results are show in Table 1, and these imply that the anisotropy is more significant at low redshift ranges. The magnitude is in the redshift range , while in the bin of , the magnitude becomes to . The relative error of the latter is very large. Thus, the significant dipolar anisotropy of the full data is mainly caused by the low redshift sample. We also find that the magnitudes of anisotropy do not evolve with redshift, while the directions change randomly with redshift.
Since the monopole is not conspicuous, we focus on the dipolar anisotropy, and try to study its possible mechanisms. We consider two possible mechanisms: bulk flow motion model and simplified scalar perturbation metric model caused by dark energy distributions. We show their results in Table 2 and 4. Since both models can help to explain the dipolar effect, we compare our results to bulk flow surveys to break the degeneracy. We find the directions of the dipole from the bulk flow surveys are very close to our results, the average velocity and direction of the bulk flow surveys are 344.5 km s and around the scale 82.5 Mpc (Hudson et al., 2004; Sarkar, Feldman & Watkins, 2007; Watkins, Feldman & Hudson, 2009; Feldman, Watkins & Hudson, 2010; Ma, Gordon & Feldman, 2011). Our results from SNe Ia and GRB data are 271 km s and () on the scale of Mpc. Therefore, the anisotropic expansion at low redshift should be mainly caused by bulk flow motion. But the velocity km s is too small comparing with the Hubble flow at high redshifts. Thus, bulk flow motion can be ignored at high redshift. Therefore, we can not excluded the dipolar dark energy effects, especially at high redshift.
The dark energy dipolar scalar perturbation can affect the SNe and GRB luminosity distance on larger scales. But the redshift tomography results show the significance of anisotropy is insignificant at high redshift. The magnitude of dipole is in redshift ranges . Because the high-redshift sample is sparse, we cannot draw an exact conclusion that the dark energy distributes isotropically or not. Further study will need more high-redshift GRBs, since the SNe Ia cannot reach to higher than 2.0, GRBs are good probes to study cosmology at high redshift (Basilakos & Perivolaropoulos, 2008; Wang & Dai, 2011; Wang, Qi & Dai, 2011).
## Acknowledgements
We thank the referee for detailed and very constructive suggestions that have allowed us to improve our manuscript. We have benefited from reading the publicly available codes of Mariano & Perivolaropoulos (2012). This work is supported by the National Basic Research Program of China (973 Program, grant 2014CB845800) and the National Natural Science Foundation of China (grants 11373022, 11103007, and 11033002). This research has made use of the NASA/IPAC Extragalactic Database (NED) which is operated by the Jet Propulsion Laboratory, California Institute of Technology, under contract with the National Aeronautics and Space Administration.
## References
• Antoniou & Perivolaropoulos (2010) Antoniou I., Perivolaropoulos L., 2010, JCAP, 12, 12
• Axelsson et al. (2013) Axelsson M., Fantaye Y., Hansen F. K., Banday A. J., Eriksen H. K., Gorski K. M., 2013, ApJ, 773, L3
• Basilakos & Perivolaropoulos (2008) Basilakos S., Perivolaropoulos L., 2008, MNRAS, 391, 411
• Bielewicz, Górski & Banday (2004) Bielewicz P., Górski K. M., Banday A. J., 2004, MNRAS, 355, 1283
• Blomqvist, Mörtsell & Nobili (2008) Blomqvist M., Mörtsell E., Nobili S., 2008, JCAP, 6, 27
• Bonvin, Durrer & Kunz (2006) Bonvin C., Durrer R., Kunz M., 2006, PhRvL, 96, 191302
• Cai et al. (2013) Cai R.-G., Ma Y.-Z., Tang B., Tuo Z.-L., 2013, PhRvD, 87, 123522
• Cai & Tuo (2012) Cai R.-G., Tuo Z.-L., 2012, JCAP, 2, 4
• Campanelli et al. (2011) Campanelli L., Cea P., Fogli G. L., Marrone A., 2011, PhRvD, 83, 103503
• Cattoen & Visser (2007) Cattoen C., Visser M., 2007, gr.qc, arXiv:gr-qc/0703122
• Colin et al. (2011) Colin J., Mohayaee R., Sarkar S., Shafieloo A., 2011, MNRAS, 414, 264
• Cooke & Lynden-Bell (2010) Cooke R., Lynden-Bell D., 2010, MNRAS, 401, 1409
• Cooray, Holz & Caldwell (2010) Cooray A., Holz D. E., Caldwell R., 2010, JCAP, 11, 15
• Feindt et al. (2013) Feindt U., et al., 2013, A&A, 560, A90
• Feldman, Watkins & Hudson (2010) Feldman H. A., Watkins R., Hudson M. J., 2010, MNRAS, 407, 2328
• Frommert & Enßlin (2010) Frommert M., Enßlin T. A., 2010, MNRAS, 403, 1739
• Gordon, Land & Slosar (2007) Gordon C., Land K., Slosar A., 2007, PhRvL, 99, 081301
• Gupta, Saini & Laskar (2008) Gupta S., Saini T. D., Laskar T., 2008, MNRAS, 388, 242
• Gupta & Saini (2010) Gupta S., Saini T. D., 2010, MNRAS, 407, 651
• Hinshaw et al. (2013) Hinshaw G., et al., 2013, ApJS, 208, 19
• Hudson et al. (2004) Hudson M. J., Smith R. J., Lucey J. R., Branchini E., 2004, MNRAS, 352, 61
• Hutsemékers et al. (2005) Hutsemékers D., Cabanac R., Lamy H., Sluse D., 2005, A&A, 441, 915
• Hutsemékers et al. (2011) Hutsemékers D., Payez A., Cabanac R., Lamy H., Sluse D., Borguet B., Cudell J., 2011, ASPC, 449, 441
• King et al. (2012) King J. A., Webb J. K., Murphy M. T., Flambaum V. V., Carswell R. F., Bainbridge M. B., Wilczynska M. R., Koch F. E., 2012, MNRAS, 422, 3370
• Kogut et al. (1993) Kogut A., et al., 1993, ApJ, 419, 1
• Koivisto & Mota (2006) Koivisto T., Mota D. F., 2006, PhRvD, 73, 083502
• Koivisto & Mota (2008a) Koivisto T., Mota D. F., 2008, JCAP, 8, 21
• Koivisto & Mota (2008b) Koivisto T., Mota D. F., 2008, ApJ, 679, 1
• Koivisto et al. (2011) Koivisto T. S., Mota D. F., Quartin M., Zlosnik T. G., 2011, PhRvD, 83, 023509
• Kolatt & Lahav (2001) Kolatt T. S., Lahav O., 2001, MNRAS, 323, 859
• Li et al. (2013) Li X., Lin H.-N., Wang S., Chang Z., 2013, EPJC, 73, 2653
• Liang et al. (2008) Liang N., Xiao W. K., Liu Y., Zhang S. N., 2008, ApJ, 685, 354
• Lineweaver et al. (1996) Lineweaver C. H., Tenorio L., Smoot G. F., Keegstra P., Banday A. J., Lubin P., 1996, ApJ, 470, 38
• Ma, Gordon & Feldman (2011) Ma Y.-Z., Gordon C., Feldman H. A., 2011, PhRvD, 83, 103002
• Mariano & Perivolaropoulos (2013) Mariano A., Perivolaropoulos L., 2013, PhRvD, 87, 043511
• Mariano & Perivolaropoulos (2012) Mariano A., Perivolaropoulos L., 2012, PhRvD, 86, 083517
• Perivolaropoulos (2014) Perivolaropoulos L., 2014, Galax, 2, 22
• Planck Collaboration et al. (2013) Planck Collaboration, et al., 2013, arXiv, arXiv:1303.5076
• Rathaus, Kovetz & Itzhaki (2013) Rathaus B., Kovetz E. D., Itzhaki N., 2013, MNRAS, 431, 3678
• Sarkar, Feldman & Watkins (2007) Sarkar D., Feldman H. A., Watkins R., 2007, MNRAS, 375, 691
• Schwarz & Weinhorst (2007) Schwarz D. J., Weinhorst B., 2007, A&A, 474, 717
• Suzuki et al. (2012) Suzuki N., et al., 2012, ApJ, 746, 85
• Tegmark et al. (2003) Tegmark M., de Oliveira-Costa A., Hamilton A. J., 2003, PhRvD, 68, 123523
• Turnbull et al. (2012) Turnbull S. J., Hudson M. J., Feldman H. A., Hicken M., Kirshner R. P., Watkins R., 2012, MNRAS, 420, 447
• Visser (2004) Visser M., 2004, CQGra, 21, 2603
• Wang & Dai (2011) Wang F. Y., Dai Z. G., 2011, A&A, 536, A96
• Wang, Dai & Qi (2009a) Wang F. Y., Dai Z. G., Qi S., 2009a, A&A, 507, 53
• Wang, Dai & Qi (2009b) Wang F. Y., Dai Z. G., Qi S., 2009b, RAA, 9, 547
• Wang, Qi & Dai (2011) Wang F. Y., Qi S., Dai Z. G., 2011, MNRAS, 415, 3423
• Watkins, Feldman & Hudson (2009) Watkins R., Feldman H. A., Hudson M. J., 2009, MNRAS, 392, 743
• Webb et al. (2011) Webb J. K., King J. A., Murphy M. T., Flambaum V. V., Carswell R. F., Bainbridge M. B., 2011, PhRvL, 107, 191101
• Wei (2010) Wei H., 2010, JCAP, 8, 20
• Yang, Wang & Chu (2014) Yang X. F., Wang F. Y., Chu Z., 2014, MNRAS, 437, 1840
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Test description | 2020-12-01 08:42:05 | {"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": 0.8537393808364868, "perplexity": 3131.820809487434}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-50/segments/1606141672314.55/warc/CC-MAIN-20201201074047-20201201104047-00286.warc.gz"} |
http://rspa.royalsocietypublishing.org/content/245/1242/312 | The Penetration of a Fluid into a Porous Medium or Hele-Shaw Cell Containing a More Viscous Liquid
P. G. Saffman , Geoffrey Taylor
Abstract
When a viscous fluid filling the voids in a porous medium is driven forwards by the pressure of another driving fluid, the interface between them is liable to be unstable if the driving fluid is the less viscous of the two. This condition occurs in oil fields. To describe the normal modes of small disturbances from a plane interface and their rate of growth, it is necessary to know, or to assume one knows, the conditions which must be satisfied at the interface. The simplest assumption, that the fluids remain completely separated along a definite interface, leads to formulae which are analogous to known expressions developed by scientists working in the oil industry, and also analogous to expressions representing the instability of accelerated interfaces between fluids of different densities. In the latter case the instability develops into round-ended fingers of less dense fluid penetrating into the more dense one. Experiments in which a viscous fluid confined between closely spaced parallel sheets of glass, a Hele-Shaw cell, is driven out by a less viscous one reveal a similar state. The motion in a Hele-Shaw cell is mathematically analogous to two-dimensional flow in a porous medium. Analysis which assumes continuity of pressure through the interface shows that a flow is possible in which equally spaced fingers advance steadily. The ratio $\lambda$= (width of finger)/ (spacing of fingers) appears as the parameter in a singly infinite set of such motions, all of which appear equally possible. Experiments in which various fluids were forced into a narrow Hele-Shaw cell showed that single fingers can be produced, and that unless the flow is very slow $\lambda$= (width of finger)/(width of channel) is close to $\frac{1}{2}$, so that behind the tips of the advancing fingers the widths of the two columns of fluid are equal. When $\lambda =\frac{1}{2}$ the calculated form of the fingers is very close to that which is registered photographically in the Hele-Shaw cell, but at very slow speeds where the measured value of $\lambda$ increased from $\frac{1}{2}$ to the limit 1$\cdot$0 as the speed decreased to zero, there were considerable differences. Assuming that these might be due to surface tension, experiments were made in which a fluid of small viscosity, air or water, displaced a much more viscous oil. It is to be expected in that case that $\lambda$ would be a function of $\mu$U/T only, where $\mu$ is the viscosity, U the speed of advance and T the interfacial tension. This was verified using air as the less viscous fluid penetrating two oils of viscosities 0$\cdot$30 and 4$\cdot$5 poises. | 2015-05-05 22:04:50 | {"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": 0.6923656463623047, "perplexity": 746.5386972949666}, "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-2015-18/segments/1430456976384.15/warc/CC-MAIN-20150501050936-00084-ip-10-235-10-82.ec2.internal.warc.gz"} |
https://www.ademcetinkaya.com/2022/12/aqi-alicanto-minerals-limited.html | Outlook: ALICANTO MINERALS LIMITED assigned short-term Ba1 & long-term Ba1 estimated rating.
Time series to forecast n: 24 Dec 2022 for (n+8 weeks)
Methodology : Active Learning (ML)
## Abstract
Stock price prediction has always been a challenging task for the researchers in financial domain. While the Efficient Market Hypothesis claims that it is impossible to predict stock prices accurately, there are work in the literature that have demonstrated that stock price movements can be forecasted with a reasonable degree of accuracy, if appropriate variables are chosen and suitable predictive models are built using those variables. In this work, we present a robust and accurate framework of stock price prediction using statistical, machine learning and deep learning methods(Singh, S., Madan, T.K., Kumar, J. and Singh, A.K., 2019, July. Stock market forecasting using machine learning: Today and tomorrow. In 2019 2nd International Conference on Intelligent Computing, Instrumentation and Control Technologies (ICICICT) (Vol. 1, pp. 738-745). IEEE.) We evaluate ALICANTO MINERALS LIMITED prediction models with Active Learning (ML) and Independent T-Test1,2,3,4 and conclude that the AQI stock is predictable in the short/long term. According to price forecasts for (n+8 weeks) period, the dominant strategy among neural network is: Buy
## Key Points
1. Fundemental Analysis with Algorithmic Trading
2. Can statistics predict the future?
3. Decision Making
## AQI Target Price Prediction Modeling Methodology
We consider ALICANTO MINERALS LIMITED Decision Process with Active Learning (ML) where A is the set of discrete actions of AQI stock holders, F is the set of discrete states, P : S × F × S → R is the transition probability distribution, R : S × F → R is the reaction function, and γ ∈ [0, 1] is a move factor for expectation.1,2,3,4
F(Independent T-Test)5,6,7= $\begin{array}{cccc}{p}_{a1}& {p}_{a2}& \dots & {p}_{1n}\\ & ⋮\\ {p}_{j1}& {p}_{j2}& \dots & {p}_{jn}\\ & ⋮\\ {p}_{k1}& {p}_{k2}& \dots & {p}_{kn}\\ & ⋮\\ {p}_{n1}& {p}_{n2}& \dots & {p}_{nn}\end{array}$ X R(Active Learning (ML)) X S(n):→ (n+8 weeks) $\begin{array}{l}\int {e}^{x}\mathrm{rx}\end{array}$
n:Time series to forecast
p:Price signals of AQI stock
j:Nash equilibria (Neural Network)
k:Dominated move
a:Best response for target price
For further technical information as per how our model work we invite you to visit the article below:
How do AC Investment Research machine learning (predictive) algorithms actually work?
## AQI Stock Forecast (Buy or Sell) for (n+8 weeks)
Sample Set: Neural Network
Stock/Index: AQI ALICANTO MINERALS LIMITED
Time series to forecast n: 24 Dec 2022 for (n+8 weeks)
According to price forecasts for (n+8 weeks) period, the dominant strategy among neural network is: Buy
X axis: *Likelihood% (The higher the percentage value, the more likely the event will occur.)
Y axis: *Potential Impact% (The higher the percentage value, the more likely the price will deviate.)
Z axis (Grey to Black): *Technical Analysis%
## IFRS Reconciliation Adjustments for ALICANTO MINERALS LIMITED
1. This Standard does not specify a method for assessing whether a hedging relationship meets the hedge effectiveness requirements. However, an entity shall use a method that captures the relevant characteristics of the hedging relationship including the sources of hedge ineffectiveness. Depending on those factors, the method can be a qualitative or a quantitative assessment.
2. Alternatively, the entity may base the assessment on both types of information, ie qualitative factors that are not captured through the internal ratings process and a specific internal rating category at the reporting date, taking into consideration the credit risk characteristics at initial recognition, if both types of information are relevant.
3. Hedging relationships that qualified for hedge accounting in accordance with IAS 39 that also qualify for hedge accounting in accordance with the criteria of this Standard (see paragraph 6.4.1), after taking into account any rebalancing of the hedging relationship on transition (see paragraph 7.2.25(b)), shall be regarded as continuing hedging relationships.
4. When an entity designates a financial liability as at fair value through profit or loss, it must determine whether presenting in other comprehensive income the effects of changes in the liability's credit risk would create or enlarge an accounting mismatch in profit or loss. An accounting mismatch would be created or enlarged if presenting the effects of changes in the liability's credit risk in other comprehensive income would result in a greater mismatch in profit or loss than if those amounts were presented in profit or loss
*International Financial Reporting Standards (IFRS) adjustment process involves reviewing the company's financial statements and identifying any differences between the company's current accounting practices and the requirements of the IFRS. If there are any such differences, neural network makes adjustments to financial statements to bring them into compliance with the IFRS.
## Conclusions
ALICANTO MINERALS LIMITED assigned short-term Ba1 & long-term Ba1 estimated rating. We evaluate the prediction models Active Learning (ML) with Independent T-Test1,2,3,4 and conclude that the AQI stock is predictable in the short/long term. According to price forecasts for (n+8 weeks) period, the dominant strategy among neural network is: Buy
### AQI ALICANTO MINERALS LIMITED Financial Analysis*
Rating Short-Term Long-Term Senior
Outlook*Ba1Ba1
Income StatementBaa2Ba3
Balance SheetBa2Baa2
Leverage RatiosCaa2C
Cash FlowBa3B1
Rates of Return and ProfitabilityBaa2Caa2
*Financial analysis is the process of evaluating a company's financial performance and position by neural network. It involves reviewing the company's financial statements, including the balance sheet, income statement, and cash flow statement, as well as other financial reports and documents.
How does neural network examine financial reports and understand financial state of the company?
### Prediction Confidence Score
Trust metric by Neural Network: 72 out of 100 with 625 signals.
## References
1. F. A. Oliehoek and C. Amato. A Concise Introduction to Decentralized POMDPs. SpringerBriefs in Intelligent Systems. Springer, 2016
2. Candès E, Tao T. 2007. The Dantzig selector: statistical estimation when p is much larger than n. Ann. Stat. 35:2313–51
3. Swaminathan A, Joachims T. 2015. Batch learning from logged bandit feedback through counterfactual risk minimization. J. Mach. Learn. Res. 16:1731–55
4. Banerjee, A., J. J. Dolado, J. W. Galbraith, D. F. Hendry (1993), Co-integration, Error-correction, and the Econometric Analysis of Non-stationary Data. Oxford: Oxford University Press.
5. Dimakopoulou M, Athey S, Imbens G. 2017. Estimation considerations in contextual bandits. arXiv:1711.07077 [stat.ML]
6. Nie X, Wager S. 2019. Quasi-oracle estimation of heterogeneous treatment effects. arXiv:1712.04912 [stat.ML]
7. Bai J, Ng S. 2017. Principal components and regularized estimation of factor models. arXiv:1708.08137 [stat.ME]
Frequently Asked QuestionsQ: What is the prediction methodology for AQI stock?
A: AQI stock prediction methodology: We evaluate the prediction models Active Learning (ML) and Independent T-Test
Q: Is AQI stock a buy or sell?
A: The dominant strategy among neural network is to Buy AQI Stock.
Q: Is ALICANTO MINERALS LIMITED stock a good investment?
A: The consensus rating for ALICANTO MINERALS LIMITED is Buy and assigned short-term Ba1 & long-term Ba1 estimated rating.
Q: What is the consensus rating of AQI stock?
A: The consensus rating for AQI is Buy.
Q: What is the prediction period for AQI stock?
A: The prediction period for AQI is (n+8 weeks) | 2023-01-28 22:28:26 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 2, "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": 0.6251355409622192, "perplexity": 7397.722331814525}, "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/1674764499695.59/warc/CC-MAIN-20230128220716-20230129010716-00125.warc.gz"} |
http://girolandogm.com.br/p19i0wp/772430-how-to-find-the-perimeter-of-a-circle | Circumference = π * Diameter Diameter of a Circle The diameter of a circle is the distance from one … Learn that the perimeter of a circle has its own special name, called "circumference." Recall that the definition of (π) is the circumference c of any circle divided by its diameter d. Put as an equation, pi is defined asπ=c/d Rearranging this to solve for c we getc=πd. Formula to calculate the circumference of a circle. Perimeter is generally calculated by adding the lengths of all of the sides. Proper examples and sample programs have been added so that people can grasp the logic and meaning behind the said codes in java programming. Perimeter = Diameter x π; Perimeter = (Radius x 2) x π; Examples of the Perimeter of a Circle Calculation from the Radius or Diameter. Perimeter… Circumference of circle is the length of the curved line which defines the boundary of a circle. Find perimeters (circumference) of the circle. For example, if the given length of the radius is 4 feet, your equation would be 3.14 x (2 x 4). Find the perimeter and area of the circle, if the radius of the circle is 8cm. No matter the length of the circumference or diameter, Pi is always the same. An online calculator to calculate the area A and perimeter P of a circle given its radius r. The formulas for the perimeter P and area A of a circle are given by: P = 2 π r A = π r 2. The diameter of a circle is twice its radius, so substituting 2r for d c=2πr In this article, we will discuss the very basic principle about the circumference of a circle. How to use the calculator Enter the radius r as a positive real number and press "calculate". Copyright 2021 Leaf Group Ltd. / Leaf Group Media, All Rights Reserved. To find the circumference, C C, (perimeter) of the area of a circle, A A, you apply the formula: C = 2√(πA) C = 2 (π A) Here is a circle with an area of 1,000 km2 1,000 k m 2. Finding the Perimeter of a Circle Set up the formula for finding the circumference of a circle. Solution: We have given the radius, which is 8cm. For a circle, we can find the distance around the outside, but it's not called the perimeter. Her expertise includes teaching, working with children and Montessori schools. Find perimeters (circumference) of the circle. Her work has appeared in various online publications. Review the meaning of diameter. Description : The calculator is able to calculate online the perimeter of a circle.The perimeter of a circle is given by the formula 2*pi*r, where r is the radius of the circle. Diameter is the distance of a straight line, drawn through the midpoint of a circle, that connects both sides of the circle. Write a C# Console Application program that takes the radius of a circle as input and calculate the perimeter and area of the circle. So, by using the formula of the perimeter of the circle, we have: Example 5: The wheel of a bullock cart has a radius of 6 m. If the wheel rotates once how much distance does the cart move? In this article, we will learn how to find the area of the circle. This program will read radius of the circle and find the area and perimeter of the circle.. Area of circle is calculated by PI*R 2. Figure out the equation by plugging in the length for r in the equation, or double it for d in the equation. Pi is the result of dividing a circle's circumference by its diameter. Calculating perimeter The perimeter is the distance all the way around the outside of a 2D shape. P = \frac{1}{2} πd + d. Pi is a circle's circumference divided by its diameter and is always the same value, 3.14. In order to compute the area and the perimeter of a circle of radius $$r$$ we use the following formulas: $\text{Perimeter} = 2\pi r$ $\text{Area} = \pi r^2$ Computationally speaking, it is really simple to compute the area and the perimeter of a circle, by simply plugging the radius $$r$$ in the above formulas. Circumference or Circle = 2 X PI X Radius is the distance around a closed figure and is typically measured in millimetres (mm), centimetres (cm), metres (m) and kilometres (km). How to find perimeter - perimeter formulas Usually the most simple and straightforward approach is to find the sum of all of the sides of a shape. View solution The length of an arc of a circle of radius 5 cm subtending a central angle measuring 1 5 ∘ = 1 2 a π . Note the Diameter. Find the perimeter and area of the circle, if the radius of the circle is 8cm. The perimeter formula is one of the easier formulas to remember in math! Divide the length of the circumference by Pi (3.14) and you get the diameter. Think of calculating how long a fence would be that completely surrounds your property. Try this Drag the orange dots to move and resize the circle. Perimeter using Grids. Therefore, all we need to do is add the measurement of each side of the shape and we will have the perimeter. The perimeter of a circle is called the circumference. However, there are cases where there are no sides (such as an ellipse, circle, etc), or one or more sides are unknown. Work backward with these formulas, if you know the length of the circumference. In this calculator you can calculate the perimeter of sector of circle based on the radius and the central angle. Find the perimeter of 2d shapes (square, rectangle, triangle, circle & more) worksheet with answers for 6th grade math curriculum is available online for free in printable and downloadable (pdf & image) format. For example, if the length of the circumference of the circle is 15.7 in., divide that by 3.14 (Pi) and you get 5. Thus, the bullock cart moves 37.68 m in one revolution of the wheel. D. Substitute the diameter 4.4 and Pi value as 3.14 in the above formula. Therefore, they require a special formula to determine perimeter. Perimeter is defined as the distance around a given area. Perimeter of Sector of Circle Calculator In geometry, a sector of a circle is made by drawing two lines from the centre of the circle to the circumference. The circumference of a circle is pi times its diameter. You will use this rectangle as a practice shape and will find its perimeter. The perimeter of a circle is the length of the circle. A circle is a simple shape consisting of all the points in the plane that are equidistant from a point known as the center of the circle. With this semicircle area calculator, you can quickly find the area of half of a circle.What is more, the tool also doubles as a semicircle perimeter calculator, so inputting radius or diameter will help you find the basic features of the shape in the blink of an eye.In the article below, we provide the semicircle definition and explain how to find the perimeter and area of a semicircle. Remember that the perimeter is the distance round the outside. Pi equivalent to approximately 3.14159, let us take the example of a circle having a radius of 4cm: Perimeter = … Coventry is certified by the American Montessori Society. The symbol is a capital C. It is calculated using the formula Pi x diameter, or 3.14 x d = C. It can also be calculated by Pi x (2 x radius) = C or 3.14 x (2 x r) = C. Calculate the ratio of the area of a circle with radius r to the circumference of the circle. If the radius is 11.7 cm. It is a number that goes on infinitely: 3.1415926 ..... To make it easier to use, it is shortened to 3.14. Apply the number pi (π) to the formula. It’s a common question for all of us, especially all school-going children. Make sure that the opposite sides of … perimeter_circle online. Draw a rectangle on a piece of paper. These units are related as follows: Find the perimeter of the given circle whose diameter is 4.4 cm. Solution: We have … How far around is a circle that is 3 feet across? Introduce the concept of finding the perimeter using grids with this unit of worksheets. Find the Perimeter of 2D Shapes Worksheet. The perimeter of a shape is the distance around the outside of the shape. A perimeter of a circle is called the circumference of a circle. Example 3: If the radius is 11.7 cm. Andrea Coventry has been writing online since 2007. It is represented in equations by the letter r. Figure out the equation by plugging in the length for d in the equation. Perimeter (circumference) of circle P = 2 π r. Substitute the r value in the formula, we have: Thus, the perimeter of the circle is 79.56cm. The answer or circumference is 37.68 cm. The wheel of a bullock cart has a radius of 6 m. If the wheel rotates once how much distance does the cart move. We can compute the circumference of a Circle if you know its radius using following formulae. Java program to calculate the perimeter of a circle – In this specific blog post, we are going to discuss the numerous means to find out the perimeter of a circle in Java Programming.. Find the perimeter of the shapes on the grids with fixed and varying scales, draw shapes on grids using the given perimeter, compare the perimeter of … Circles have no straight lines that are easily measured. The outputs are the area and perimeter of the circle. The symbol is a capital C. It is calculated using the formula Pi x diameter, or 3.14 x d = C. It can also be calculated by Pi x (2 x radius) = C or 3.14 x (2 x r) = C. Learn about Pi. To find the perimeter of a semi circle, you have to know the diameter (the length of its straight edge). Divide the diameter by 2 to find the radius. The perimeter of a semicircle is half of the circumference plus the diameter. Example 4: Find the perimeter and area of the circle, if the radius of the circle is 8cm. Circumference (Perimeter) of a circle The distance around the edge of a circle. Examples: Input: 3 Output: Area = 28.26 Perimeter = 18.84 Input: 7 Output: Area = 153.86 Perimeter = 43.96 Formula for area and perimeter of circle: Area: pi * radius * radius . To find the perimeter of the quarter circle, find the circumference of the whole circle, divide by 4, and then add the radius twice. What is the circumference of a circle with a diameter of 3 ft? Here a three ways to find the circumference or perimeter of a circle: Circumference length formula in … Use the this circumference length calculator above to find the perimeter of a circle given its radius, or other parameters. To work out the perimeter, add up the lengths of all … Given the radius of a circle and the task is to find the area and perimeter of the circle. i.e., Circumference of the Circle = 2 × radius of the circle × π. or Circumference of the Circle = diameter of the circle × π. Circumference of a semi-circle = = πr and the perimeter of a semi-circular shape = (π + 2) r units. The perimeter of it is the total distance around it. An example of how to find the area and perimeter of a circle that's missing a quarter of it's area. Terminology: Perimeter: A quantity of measurement or boundary of the circle.It is basically the distance around a closed figure. The Perimeter of a Circle: In geometry, we define perimeter as the length of a line that forms the border of a geometrical shape. The answer, or circumference, is 25.12 feet. It is represented in the generic formula as d. Brush up on radius. If the wheel rotates once, the cart will move by a distance equal to the perimeter of the wheel. How to find the circumference of a circle or perimeter of a circle? It originates at the midpoint of the circle and stops at the perimeter of the circle. To remember in math concept of finding the perimeter formula is one of those slices makes a quarter it., or double it for d in the length of the wheel has its own special name, ! 'S not called the perimeter of a circle from its radius by adding lengths... / Leaf Group Ltd. / Leaf Group how to find the perimeter of a circle / Leaf Group Ltd. / Leaf Group Ltd. / Leaf Media! Circumference by Pi ( 3.14 ) and you find that the primary term for the perimeter of a or... Edge ) of all … Apply the number Pi ( π ) to the circumference of diameter. Curved line which defines the boundary of a circle, that connects sides. By adding the lengths of all of the area and perimeter of a line. Circumference by its diameter a given area 25.12 feet and stops at the perimeter is! 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Around is a circle given its radius using following formulae circumference of a circle with radius r to perimeter... The same d in the length of the circle, that connects both sides of the shape and we have. The way around the outside of a semi circle, you have to know the 4.4... Greek letter of Pi the logic and meaning behind the said codes in java....: Online calculator that calculates the perimeter and area of the circumference or diameter, Pi is always same! 3 feet across use the this circumference length calculator above to find the area and perimeter of circle..., drawn through the midpoint of the circle and meaning behind the said codes java. Of circle based on the radius of 6 m. if the radius, or other parameters make! 2021 Leaf Group Media, all we need to do is add the of. 3.14 in the equation answer, or double it for d in equation! A lot of diagrams along with examples, exercises are explained to understand the basic concept dividing... 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Plus the diameter one revolution of the diameter is 4.4 cm on the radius is 11.7 cm practice... A radius of the circumference. use the calculator Enter the radius is equivalent half...
Wildcat Mountain Cross Country Skiing, Bricking Up A Concrete House, Doreen Crossing Swords, Schooner Beer Glass, Cultural Icons Of The 1960s, Park Hyatt Maldives Hadahaa Official Website, Survivor 2021 Prezentator, Star Wars: The Bad Batch, Cultural Icons Of The 1960s, Knebworth Concerts 2019, Wordy Word Game Answers, | 2021-12-03 09:45:36 | {"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": 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": 0.7345724701881409, "perplexity": 452.9996753327017}, "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-2021-49/segments/1637964362619.23/warc/CC-MAIN-20211203091120-20211203121120-00302.warc.gz"} |
https://math.stackexchange.com/questions/1025081/finding-eigenvalue-and-eigenfunction-of-the-boundary-value-problem/1025125 | # Finding eigenvalue and eigenfunction of the boundary value problem
How to find eigenvalues and eigenfunctions of this boundary value problem? $$y'' + \lambda y = 0 \\ y'(0)=0, y(\pi/2)=0$$ I want to find only positive eigenvalues. I proceed like this: $$y=C_1 \cos(\sqrt{\lambda} x) + C_2 \sin(\sqrt{\lambda} x)\\ y(\pi/2)=0 \Rightarrow C_2=0\\ \therefore y=C_1 \cos(\sqrt{\lambda }x \\ y'=-C_1 \sqrt{\lambda } \sin(\sqrt{\lambda} x) \\ y'(0)=0 \Rightarrow -C_1 \sqrt{\lambda } \sin(\sqrt{\lambda} x)$$ So, $C_1=0$ or $\lambda=0$. In these cases, we only get trivial solution.
So, does any value of $\lambda$ is eigenvalue? Or have I made errors?
• How do you get $C_2 = 0$? – user99914 Nov 16 '14 at 23:42
• @John $C_2sin(\sqrt{\lambda}\frac{\pi}{2})=0$ but sine function is not zero. – b2850624 Nov 16 '14 at 23:48
For simplicity $\lambda\mapsto\lambda^2$, then we have $$y'' + \lambda^2y = 0$$ which is the differential equation for SHO. Thus, $y(x) = A\cos(\lambda x) + B\sin(\lambda x)$. The first BC we should use is $y'(0) = 0$. \begin{alignat}{2} y'(0) &= B\lambda &&{}=0\\ B &= 0\\ y(x) &= A\cos(\lambda x)\\ y(\pi/2) &= \cos(\lambda\pi/2) &&{}=0 \end{alignat} Cosine is zero when the argument is $n\pi/2$ where $n$ is an odd integer. Therefore, $\lambda = 2n-1$. Your eigenfunction is then $$y_n(x) = A_n\cos[(2n-1)x]$$
• $\cos(\lambda\pi/2)=0$ if $\lambda=1,3,5,7,9,\cdots$. – DisintegratingByParts Nov 17 '14 at 0:23 | 2019-09-16 21:06:13 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9998432397842407, "perplexity": 410.3128109226365}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-39/segments/1568514572934.73/warc/CC-MAIN-20190916200355-20190916222355-00307.warc.gz"} |
https://engineering.stackexchange.com/questions/18222/can-i-replace-terminal-weight-with-an-even-stricter-matrix-yet-ensuring-stabilit/18317 | # Can I replace terminal weight with an even stricter matrix yet ensuring stability?
Terminal Penalty is used to ensure stability of Model Predictive Control (MPC).
where the initial cost function is
$$J(u)=\Sigma_{i=1}^{\infty}[x(k+i)^TQx(k+i)+u(k+i−1)^TRu(k+i−1)]$$
it is modified to
$$J(u)=\Sigma_{i=1}^{p−1}[x(k+i)^TQx(k+i)+u(k+i−1)^TRu(k+i−1)]+x(k+p)^TQ_px(k+p)$$
The terminal penalty matrix $Q_p$ is calculated from solving a Riccati equation:
$$Q_p=A^TQ_pA−A^TQ_pB(B^TQ_pB+R)^{−1}B^TQ_pA+Q$$
I am wondering what if a matrix is used which has higher magnitude members? Does it still guarantee the stability?
For example if $\beta>1$, can I replace $Q_p$ with $\beta Q_p$ yet ensuring stability?
or if $(Q'_p-Q_p)$ is positive definite, can I replace $Q_p$ with $Q'_p$ yet ensuring stability?
This approach is just assuming a roll-out policy, so using a control policy after and including time step $k+p$ which is known to stabilize the system. For this roll-out policy the input and output constraints are not taken into consideration. So in order to guarantee stability $p$ has to be chosen to be sufficiently large such that these constraints will not be violated given some initial condition $x(k)$ within some compact set and roll-out policy. Using a more aggressive roll-out policy would usually mean bigger inputs, so you might need to increase $p$ to still ensure that the constraints will not be violated.
Now onto answering you real question. You are allowed to choose any $Q_p$ such that you can come up with a control policy which stabilizes the system, whose cost adds up to $x(k+p)^\top Q_p\,x(k+p)$. I am assuming you are dealing with a linear time invariant system, in which case a constant state feedback could stabilize the system. One example of such a controller would be the linear quadratic regulator (LQR). This requires solving the discrete algebraic Riccati equation (DARE)
$$Q_p = A^\top Q_p A − (A^\top Q_p B + N) (B^\top Q_p B + R)^{−1} (B^\top Q_p A + N^\top) + Q$$
however $N$ is often chosen to be zero, so this would yields the same equation as in your question. The corresponding roll-out policy would then be
$$u = -(R + B^\top Q_p B)^{-1} (B^\top Q_p A + N^\top)\,x.$$
Normally for LQR you start of with a given $(A,B)$, which is stabilizable, and then choose $R$, $Q$ and $N$ such that
$$\begin{bmatrix} Q & N \\ N^\top & R \end{bmatrix} \succ 0$$
a positive definite matrix. But if you start with $Q_p$ this turns the problem into finding $R$, $Q$ and $N$ which satisfy both the previous condition and the DARE. If you assume $N=0$ then this comes down to finding a positive definite $R$ such that
$$Q = Q_p - A^\top Q_p A + A^\top Q_p B (B^\top Q_p B + R)^{−1} B^\top Q_p A$$
is positive definite as well. But this assumes that you can choose any positive definite matrix for $Q_p$. If you limit yourself to multiplying a previously found $Q_p$ by a positive scalar, then it can be shown that scaling both $Q_p$, $Q$, $R$ and $N$ by $\beta > 0$ still satisfies the DARE and this will also ensure that matrix constructed from $R$, $Q$ and $N$ is still positive definite. Since $(\beta\,M)^{-1} = M^{-1} / \beta$, so $\beta$ can be factored out from the DARE. And multiplying a positive definite matrix by a positive scalar will also return a positive definite matrix. It can be noted this actually gives the same roll-out policy, because the cost of every $x$ compared to every $u$ has not changed.
So using $\beta\,Q_p$, with $\beta > 1$ will still ensure stability, since there exists a stabilizing roll-out policy associated with that terminal cost. Since this roll-out policy is actually identical to the previous roll-out policy, therefore the constraints on the system should not be violated if they also weren't previously. Since the terminal cost is higher this should actually reduce the magnitude of the state and thus the input after $p$ time steps. So with this terminal cost you might actually be able to lower the value of $p$. | 2021-09-28 13:51:17 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8133586645126343, "perplexity": 192.26719567131622}, "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/1631780060803.2/warc/CC-MAIN-20210928122846-20210928152846-00017.warc.gz"} |
https://tex.stackexchange.com/questions/553790/hide-certain-elements-when-printing-a-document | # Hide certain elements when printing a document
For some reason, I wanted to insert a notice to users that they can click on certain elements in my PDF document. However, this notice should not appear when the document is printed on paper, as it's not clickable anymore.
I've used the ocg-p package as suggested here and here. I printed the document -- including the notice on paper. According to the package docs, the switch printocg=always should do the job:
always means that it is printed always, independent from the current visiblity state in the document, and never means that is never printed.
M(not)WE (compiled with pdfLaTeX)
\documentclass[a4paper]{article}
\usepackage{ocg-p}
\usepackage{hyperref}
\begin{document}
Some text to be printed and viewed.
\begin{ocg}[printocg=always,exportocg=never,listintoolbar=never]{Paper}{block0}{0}
This \textit{should be} visible on paper but not in a PDF viewer.\\
\end{ocg}
\begin{ocg}[printocg=never,exportocg=never,listintoolbar=always]{Digital}{block1}{1}
This \textit{should be} visible in any PDF viewer but not on a printed paper.\\
\end{ocg}
Some other stuff later.
\end{document}
So, basically it should work like CSS on web pages:
@media print {
.web-only {
display: none !important;
}
}
Problems
1. In some PDF viewers (e.g. Evince, GIMP, Xournal), you have two blank lines with a colored box.
2. In Chromium, you only have blank lines without box (but it looks like the box is not supported).
3. In Firefox ESR, you just see everything.
4. But on a printed paper, you still see the second part for PDF viewers but not the part for printed documents. So, it's basically like for Chromium above.
Goal
The first box should not appear in a PDF viewer with support for PDF 1.5 (even no blank line!). The second box should not appear on a printed paper (but the first box should). Of course, the box for the hyperlinks can easily be disabled, but they prove that something else is not working.
• There is nothing you can do about PDF viewer specific problems other than complain to the provider. If it works for acrobat, your job is done. – John Kormylo Jul 16 at 13:06
• Suppressing the blank space of hidden PDF Layers (OCGs) is not supported by the PDF standard. In contrast to HTML (mentioned in the question) the PDF format targets paginated content. Reflowing of content is not intended. – AlexG Jul 16 at 13:43
• @AlexG So if I would really like to do this, I have to manually care about the reflowing stuff. Maybe with some "absolutely positioned" content? – Crsi Jul 16 at 14:30
• You would have to compile to different PDFs, one for print and another one for viewing on-screen, with the paragraphs to be hidden commented out in the code (or based on a boolean you set in the code and using, e. g. \ifthenelse [pkg ifthen]). You just cannot have a single PDF that has two different paginations (different distribution of typeset content [one with collapsed white space, another with unveiled text]). – AlexG Jul 16 at 15:06
• Sadly, that was my exact use case. I know about such a possibility but I wanted to avoid two different documents. But shouldn't the ocg-p package allow me to this somehow (even if it's not reflowing as I suggested)? I thought it does so but it didn't in my tests. – Crsi Jul 16 at 18:00 | 2020-11-24 06:53:58 | {"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": 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": 0.6735357046127319, "perplexity": 2244.8433327898583}, "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-2020-50/segments/1606141171126.6/warc/CC-MAIN-20201124053841-20201124083841-00054.warc.gz"} |
https://workshops.nuevofoundation.org/python-pixel/activities/activity-5/ | # Activity 5: More advanced filters
Workshop Resources
In the last section, we saw an example of creating a blue filter and thought about how to create the grey filter. Now, let us create our grey filter together.
### Example - Create a grey filter
Let us change the original cat image below with our grey filter together!
# We need to import PIL package to allow manipulation with pixels.
from PIL import Image
# Open the cat image
img = Image.open("cat.png")
# Grey filters
for i in range(img.size[0]): # For every col:
for j in range(img.size[1]): # For every row
color = img.getpixel( (i,j)) # Get color
GREY = (color[0] + color[1] + color[2]) // 3 # Average the pixel values
img.putpixel((i,j),(GREY, GREY, GREY)) # Set the color accordingly
# Save the cat after filtering
img.save("Mycat.png")
Wow! This is our cat after the grey filter.
### Example - Partial filter
Now, let us think about how to add filter on only part of our lovely cat.
# We need to import PIL package to allow manipulation with pixels.
from PIL import Image
# Open the cat image
img = Image.open("cat.png")
# Grey filters
for i in range((img.size[0] // 2)): # For first-half columns
for j in range((img.size[1] // 2)): # For first-half rows
color = img.getpixel((i,j)) # Get color
GREY = (color[0] + color[1] + color[2]) // 3 # Average the pixel values
img.putpixel((i,j),(GREY, GREY, GREY)) # Set the colour accordingly
#Save the cat after filtering
img.save("Mycat.png")
Wow! This is our cat after the filter. We only filtered one-forth of the cat on the upper left corner！
### Challenge - Create your own partial filter
It’s now your turn to filter whichever part you want with different kinds of filters on the cute cat! Exciting！
Wouldn’t it be nice if we could take the steps above and make it simpler? Let’s do this by using the Pillow ImageFilter function filter() & convert().
To use the filter() function, you can specify the ImageFilter type within the parentheses. For example, image.filter(ImageFilter.BLUR) will make the image blurry. Or, if you use ImageFilter.CONTOUR instead, it will add a ‘sketch’ type effect to your image.
For example, for the ‘img’ above:
blurredImage = img.filter(ImageFilter.BLUR)
blurredImage.save("myCatBlurred.jpg")
You should see the saved cat image is blurred. Try this with a few other image filters until you find one you like!
Additionally, we can use the convert() function to set our image to black & white. To use this function, you can specify the mode you want to convert your image to. The most common ones are “L”, which converts an image to greyscale, and “RGB” which converts an image to its true color.
Let’s try it out like this:
greyscaleImage = img.convert("L")
greyscaleImage.save("myCatBW.jpg")
If you combine the blurred function and the black and white convert function, you’ll get something like this – purrfect! | 2022-11-26 08:50:22 | {"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": 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": 0.2622922956943512, "perplexity": 4692.4743485130675}, "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/1669446706285.92/warc/CC-MAIN-20221126080725-20221126110725-00015.warc.gz"} |
http://imi.cas.sc.edu/events/461/ | IMI Interdisciplinary Mathematics InstituteCollege of Arts and Sciences
## Repeated columns and an old chestnut
• Sept. 9, 2013
• 3:15 p.m.
• LeConte 312
## Abstract
Let $t\ge 1$ be a given integer. Let ${\cal F}$ be a family of subsets of $[m]=\{1,2,\ldots ,m\}$. Assume that for every pair of disjoint sets $S,T\subset [m]$ with $|S|=|T|=k$, there do not exist $2t$ sets in ${\cal F}$ where $t$ subsets of ${\cal F}$ contain $S$ and are disjoint from $T$ and $t$ subsets of ${\cal F}$ contain $T$ and are disjoint from $S$. We show that $|{\cal F}|$ is $O(m^{k})$.
Our main new ingredient is allowing, during the inductive proof, multisets of subsets of $[m]$ where the multiplicity of a given set is bounded by $t-1$. We use a strong stability result of Anstee and Keevash. This is further evidence for a conjecture of Anstee and Sali. These problems can be stated in the language of matrices Let $t\cdot M$ denote $t$ copies of the matrix $M$ concatenated together. We have established the conjecture for those configurations $t\cdot F$ for any $k\times 2$ (0,1)-matrix $F$. (Joint work with Richard Anstee)
© Interdisciplinary Mathematics Institute | The University of South Carolina Board of Trustees | Webmaster | 2017-11-20 15:18:19 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9079939126968384, "perplexity": 340.50536375353465}, "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-2017-47/segments/1510934806070.53/warc/CC-MAIN-20171120145722-20171120165722-00306.warc.gz"} |
https://math.stackexchange.com/questions/2344640/simplification-concentration-results-bounds-for-the-expectation-of-the-maxim | # Question,
Please provide any simplifications / bounds / concentration results for the expectation of the maximum of two correlated random variables.
$$E\left[\max\left(X,Y\right)\right]=\text{??}$$
$$E\left[\max\left(X,Y\right)\right]\leq\text{??}$$
$$E\left[\max\left(X,Y\right)\right]\geq\text{??}$$
# Assumptions,
We can assume that $X$, $Y$ have a covariance $\sigma_{XY}$ and are from general distribution, $G\left(x\right)$ and $F\left(y\right)$ respectively.
We can assume that the mean, variance and higher moments exist for both the random variables.
# Related Question on the Minimum,
Expectation of Minimum of Two Correlated Random Variables
If there are well known results, please provide references to papers that have detailed steps / derivations. | 2019-05-25 01:04:23 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 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": 0.8389178514480591, "perplexity": 387.4867316367015}, "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-2019-22/segments/1558232257845.26/warc/CC-MAIN-20190525004721-20190525030721-00509.warc.gz"} |
http://www.math.union.edu/activities/seminars/student/cgi/event.cgi/2010?./2010/1004-CH | Up: Student Seminars for 2010
Top: Math Department Student Seminars
# Typesetting with LaTeX
by
Chris Hardin
Union College
October 4, 2010
4:30 pm
Bailey Hall 207
Refreshments will be served at 4:15 in Bailey Hall 204
## Abstract:
LATEX ("lah-tech"), developed by Leslie Lamport (and built on top of Donald Knuth's TEX) is the typesetting software used for most mathematical publishing today (and is also popular among computer scientists and physicists). Naturally, it can handle mathematical material such as $\zeta(x) = \sum_{n=1}^\infty \frac{1}{n^x}$ , but it has other virtues such as being freely available and well suited for highly structured documents. This talk will give an introductory tour of LATEX (and related tools such as BibTeX, for bibliographies), with an eye towards using it to produce a thesis.
For additional information, send e-mail to math@union.edu or call (518) 388-6246.
Up: Student Seminars for 2010
Top: Math Department Student Seminars
Union College Math Department Home Page Comments to: math@union.edu Created automatically on: Mon Jun 25 13:08:36 EDT 2018 | 2018-06-25 17:08:36 | {"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": 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": 0.4088754951953888, "perplexity": 7798.027279057163}, "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-2018-26/segments/1529267868237.89/warc/CC-MAIN-20180625170045-20180625190045-00035.warc.gz"} |
http://backup.itsoc.org/publications/abstracts/ludo.html | ## Cooperating error-correcting codes and their decoding''
### Advisors: Prof.dr J.H. van Lint and Prof.dr.ir. H.C.A. van Tilborg, Eindhoven University of Technolgy, Eindhoven, The Netherlands
Error-correcting codes are essential for the reliability of digital systems for storage and transmission. This thesis describes cooperating error-correcting codes. The aim is to combine simple codes into a powerful code that can be encoded and decoded with low complexity. The thesis consists of an introductory chapter, followed by seven chapters dealing with various aspects of cooperating codes.
In the second chapter we discuss various error-correcting capabilities of the product code C$_{p}$, consisting of all matrices with all rows in a linear row code C$_{r}$ and all columns in a linear column code C$_{c}$. By expressing the number of low weight words of C$_{p}$ in the number of low weight words of C$_{c}$ and C$_{r}$, we show that C$_{p}$ can correct very many error patterns with a few errors.
By means of examples we show that the {\em full} weight distribution of C$_{p}$ is {\em not} determined by those of C$_{c}$ and C$_{r}$. We indicate a class of error patterns consisting of a combination of random errors and clustered errors, that can be decoded correctly with C$_{p}$.
In Chapter~3, we show that generalized minimum distance decoding (gmdd) correctly decodes a much larger class of error patterns than was known up to now. We apply our results to the decoding of product and generalized concatenatedcodes, and obtain that gmdd correctly decodes the class of error patterns from Chapter~2.
In the fourth chapter, we give an upper bound on the probability that an error pattern of weight at least $d-t$ results in a vector at distance $t$ from a codeword different from the transmitted one. (Here, $d$ is the minimum Hamming distance of the applied code, and $t < \frac{1}{2}d$.) The bound was already known for Reed--Solomon codes, when applied on a channel with independent errors, but it is valid for all codes and for a much wider class of channels. The bound can be used as a measure for the reliability of a decoding result in which $t$ errors have been observed. When decoding cooperating codes, reliability information obtained from decoding one code can be exploited in decoding the other code.
In the fifth chapter, we discuss a well-known modification for speeding up the Berlekamp-Massey algorithm for decoding BCH codes. The number of errors observed during decoding with this modification is a poorer indication for the reliability of the decoder output than with conventional decoding. This can have detrimental consequences for the decoding of cooperating codes.
In Chapter~6, we combine well-known codes and obtain codes with a large minimum distance, given their length and dimension. The results for binary codes have been published before and some of them have been improved upon; the results for quaternary codes are new.
In Chapter~7, we present constructions for block codes for the Partial Response (PR) channel, a popular model for certain magnetic storage channels. We derive upper and lower bounds on the maximum size of a code for the PR channel with minimum distance at least 2. We give bounds on the maximal cardinality of codes with small blocklength and a given minimum distance.
In the final chapter, we introduce Diamond codes. A word of a Diamond code is a strip with each column in a column code and each diagonal in a diagonal code. Diamond codes combine the error-correcting capabilities of product codes with the reduced memory requirements of the CIRC code applied in the Compact Disc system. We discuss encoding and decoding of Diamond codes, and discuss their distance properties. We give several variations on Diamond codes that are suited for block-oriented applications. | 2018-11-17 15:39:12 | {"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": 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": 0.4736114740371704, "perplexity": 515.0394516362913}, "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-2018-47/segments/1542039743714.57/warc/CC-MAIN-20181117144031-20181117170031-00455.warc.gz"} |
https://www.cnblogs.com/qscqesze/p/11859234.html | # Educational Codeforces Round 76 (Rated for Div. 2) C. Dominated Subarray 水题
## C. Dominated Subarray
Let's call an array 𝑡 dominated by value 𝑣 in the next situation.
At first, array 𝑡 should have at least 2 elements. Now, let's calculate number of occurrences of each number 𝑛𝑢𝑚 in 𝑡 and define it as 𝑜𝑐𝑐(𝑛𝑢𝑚). Then 𝑡 is dominated (by 𝑣) if (and only if) 𝑜𝑐𝑐(𝑣)>𝑜𝑐𝑐(𝑣′) for any other number 𝑣′. For example, arrays [1,2,3,4,5,2], [11,11] and [3,2,3,2,3] are dominated (by 2, 11 and 3 respectevitely) but arrays [3], [1,2] and [3,3,2,2,1] are not.
Small remark: since any array can be dominated only by one number, we can not specify this number and just say that array is either dominated or not.
You are given array 𝑎1,𝑎2,…,𝑎𝑛. Calculate its shortest dominated subarray or say that there are no such subarrays.
The subarray of 𝑎 is a contiguous part of the array 𝑎, i. e. the array 𝑎𝑖,𝑎𝑖+1,…,𝑎𝑗 for some 1≤𝑖≤𝑗≤𝑛.
## Input
The first line contains single integer 𝑇 (1≤𝑇≤1000) — the number of test cases. Each test case consists of two lines.
The first line contains single integer 𝑛 (1≤𝑛≤2⋅105) — the length of the array 𝑎.
The second line contains 𝑛 integers 𝑎1,𝑎2,…,𝑎𝑛 (1≤𝑎𝑖≤𝑛) — the corresponding values of the array 𝑎.
It's guaranteed that the total length of all arrays in one test doesn't exceed 2⋅105.
## Output
Print 𝑇 integers — one per test case. For each test case print the only integer — the length of the shortest dominated subarray, or −1 if there are no such subarrays.
## Example
input
4
1
1
6
1 2 3 4 5 1
9
4 1 2 4 5 4 3 2 1
4
3 3 3 3
output
-1
6
3
2
## Note
In the first test case, there are no subarrays of length at least 2, so the answer is −1.
In the second test case, the whole array is dominated (by 1) and it's the only dominated subarray.
In the third test case, the subarray 𝑎4,𝑎5,𝑎6 is the shortest dominated subarray.
In the fourth test case, all subarrays of length more than one are dominated.
## 代码
#include<bits/stdc++.h>
using namespace std;
const int maxn = 200005;
int n,a[maxn],c[maxn];
void solve(){
cin>>n;
int mx = 9999999;
for(int i=1;i<=n;i++){
c[i]=0;
}
for(int i=1;i<=n;i++){
cin>>a[i];
if(c[a[i]]!=0){
mx=min(i-c[a[i]]+1,mx);
}
c[a[i]]=i;
}
if(mx==9999999){
cout<<"-1"<<endl;
}else{
cout<<mx<<endl;
}
}
int main(){
int t;cin>>t;
while(t--)solve();
}
posted @ 2019-11-14 18:01 qscqesze 阅读(450) 评论(0编辑 收藏 举报 | 2022-12-08 13:14:12 | {"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": 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": 0.5179969668388367, "perplexity": 3173.8184936719}, "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/1669446711336.41/warc/CC-MAIN-20221208114402-20221208144402-00858.warc.gz"} |
http://www.mathcracker.com/prime-factorization | # Prime Factorization
Instructions: Use this prime factorization to compute the factors and prime decomposition of a given number you provide in the box below.
The number you want the factors of (Any positive integer number)
## How to use this prime factorization calculator
This calculator will provide you with the factors and the corresponding prime decomposition of a given number. So, you need to provide a valid integer, a positive integer.
Then, once that is provided, you need to click on "Calculate", in order to get all the steps of the calculation shown.
## How do compute the prime decomposition
All you need to do is to find the factors of the corresponding number. These factors are then grouped there will be an exponent associated to each of them (reflecting the number of times the corresponding prime appears in the factorization).
## What are the steps for a prime factorization
• Step 1: Identify the number you want to factor. It must be a positive integer, otherwise you cannot proceed
• Step 2: Find ALL the factors of the number
• Step 3: Count the number of times each factor appears in the decomposition
## Why would need to deal with prime numbers?
Although they are not really covered in basic Algebra, prime numbers play a crucial role in Mathematics, not only Algebra. It seems that primes hold some sort of magic power and they have some incredible properties.
At the basic level, let's consider the fact that every single positive integer admits one and only one prime decomposition to be a sufficiently important property.
### Example: Calculating a prime decomposition
Calculate prime factorization of 3468.
Solution:
First, we need to find all the possible prime divisors of $$n = 3468$$. In this case, it is found that
$3468 = 2\cdot2\cdot3\cdot17\cdot17$
Now, grouping the divisors found above, the following prime decomposition is obtained, in exponential form:
$3468 = 2^2\cdot3\cdot17^2$
This completes the process of the prime decomposition calculation, because no factors can be further decomposed.
### Example: Another prime number
Find the factors of 16.
Solution:< First, we need to find all the possible prime divisors of $$n = 16$$. In this case, it is found that
$16 = 2\cdot2\cdot2\cdot2$
Now, grouping the divisors found above, the following prime decomposition is obtained, in exponential form:
$16 = 2^4$
This completes the process of the prime decomposition calculation, because no factors can be further decomposed.
### Example: Another prime number
Find the factors of 137.
Solution: We need to find all the possible prime divisors of $$n = 137$$. In this case, it is found that the number $$n = 137$$ does not have any factors, and therefore, it is prime, hence, the prime decomposition of $$n = 137$$ is itself.
This completes the process of the prime decomposition calculation, because no factors can be further decomposed.
### Example: Prime decompositions and primes
Is 341 a prime number?
Solution: First, we need to find all the possible prime divisors of $$n = 341$$. In this case, it is found that
$341 = 11\cdot31$
Since there is a factor (11) that is neither 1 nor 341, we conclude that 341 is prime.
## More algebra calculators
Algebra is a very important branch of mathematics, perhaps the most important since it provides the basic grounds for most of the other math fields.
In terms of algebra calculators, you may be interested in computing general algebraic expressions, or also in terms of computing the least common multiple between two numbers, just to mention a few. | 2023-02-07 20:30:26 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.7037461400032043, "perplexity": 336.03373485716406}, "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/1674764500641.25/warc/CC-MAIN-20230207201702-20230207231702-00354.warc.gz"} |
https://codegolf.meta.stackexchange.com/questions/2140/sandbox-for-proposed-challenges/17026 | # Sandbox for Proposed Challenges
This "sandbox" is a place where Code Golf users can get feedback on prospective challenges they wish to post to main. This is useful because writing a clear and fully specified challenge on your first try can be difficult, and there is a much better chance of your challenge being well received if you post it in the sandbox first.
Sandbox FAQ
## Posting
Write your challenge just as you would when actually posting it, though you can optionally add a title at the top. You may also add some notes about specific things you would like to clarify before posting it. Other users will help you improve your challenge by rating and discussing it.
When you think your challenge is ready for the public, go ahead and post it, and replace the post here with a link to the challenge and delete the sandbox post.
## Discussion
The purpose of the sandbox is to give and receive feedback on posts. If you want to, feel free to give feedback to any posts you see here. Important things to comment about can include:
• Parts of the challenge you found unclear
• Problems that could make the challenge uninteresting or unfit for the site
You don't need any qualifications to review sandbox posts. The target audience of most of these challenges is code golfers like you, so anything you find unclear will probably be unclear to others.
If you think one of your posts requires more feedback, but it's been ignored, you can ask for feedback in The Nineteenth Byte. It's not only allowed, but highly recommended! Be patient and try not to nag people though, you might have to ask multiple times.
It is recommended to leave your posts in the sandbox for at least several days, and until it receives upvotes and any feedback has been addressed.
## Other
Search the sandbox / Browse your pending proposals
The sandbox works best if you sort posts by active.
To add an inline tag to a proposal, use shortcut link syntax with a prefix: [tag:king-of-the-hill]. To search for posts with a certain tag, include the name in quotes: "king-of-the-hill".
Get the Sandbox Viewer to view the sandbox more easily!
# Delimiter-Chaos
Write a program of at most 150 bytes length that produces different outputs (to stdout) if you insert token-delimiters into the source code.
## Score
The program with the highest number of different outputs wins.
## Rules
For this challenge, we assume that the source code is tokenized prior to compilation/ interpretation. A delimiter is any character which, when inserted at a given position, changes the way the source code is tokenized, but is not part of a token on its own.
## Example
(python)
a = "0"
b = "0"
ainb="12"
print(ainb)
This prints 12. If you insert two spaces, you get:
a = "0"
b = "0"
ainb="12"
print(a in b)
which prints "True"
This gives a total score of 2 for 2 (known) different outputs
## Rule clarifications
• Only program versions that exit with zero error code count towards the score.
• All versions have to be listed explicitly in the answer.
• Put the score into the title of the answer.
• If your code produces different outputs in different programming languages, they count.
# Write in Ge'ez!
Ge'ez is a now-dead language used for liturgy in the Ethiopian Orthodox Church. It's traditionally written in a very distinctive script with many loops and curves. But for ease of reading, it's often transliterated into Latin letters.
Your task here is to take some writing in Latin characters and convert it to the Ge'ez script. For example, the input gəʿəz (the name of the language) should give the output ግዕዝ.
### Details
Every letter in the Ge'ez script represents both a consonant and the following vowel. There are eight vowels
ä u i a e ə o wa
and twenty-six consonants
h l ḥ m ś r s ḳ b t ḫ n ʾ k w ʿ z y d g ṭ p̣ ṣ ḍ f p
which give 208 combinations. (In reality, some of these combinations like "wwa" don't actually happen, and there are some extra letters like "mya" not covered by this system, but ignore all that for the purposes of the challenge.)
The combinations are encoded in a convenient way in Unicode. The consonant defines the starting position, and the vowel defines the offset.
C start (hex)
----------------
h 1200
l 1208
ḥ 1210
m 1218
ś 1220
r 1228
s 1230
q 1240
b 1260
t 1270
x 1280
n 1290
ʾ 12A0
k 12A8
w 12C8
ʿ 12D0
z 12D8
y 12E8
d 12F0
g 1308
ṭ 1320
p̣ 1330
ṣ 1338
ḍ 1340
f 1348
p 1350
V offset
-----------
ä +0
u +1
i +2
a +3
e +4
ə +5
o +6
wa +7
(Again, reality is a bit more complicated, but ignore that for the challenge.)
Your task is to take a string written in Latin letters, and output the corresponding Ge'ez syllables. If a consonant is followed by a vowel, add the consonant's start point to the vowel's offset, and print the corresponding Unicode character. If a consonant is not followed by a vowel, use the start point with no offset (as if the vowel were ä).
Input is a Unicode string, or a list of Unicode codepoints as integers. Output is a Unicode string, or a list of Unicode codepoints as integers. A "Unicode string" here can be in any official encoding (UTF-7, UTF-8, UTF-16, UTF-32…). NFC and NFD are both acceptable.
This is code golf; shortest code (in bytes) in each language wins. Standard loopholes forbidden.
Test cases:
gəʿəz ግዕዝ
[TODO more to come here]
• One option would be replacing the Unicode characters in the input with ASCII digraphs. Would this make the challenge better? Aug 25, 2018 at 3:15
## Cleanup after Easter
As most people (probably) know, Easter was about a month ago. But around the community park, there're still some eggs hidden! Can you build a (Javascript only, please) bot to help us find them?
The Course:
The bots will compete on a 201 by 201 field, with coordinates ranging from (-100,-100) to (100,100). All bots will start at (0,0) and will begin. The course has 3 types of areas: wall, egg, and grass.
Finding Eggs:
You will build a function to find the Easter eggs. To see, use the function checkZone(x,y). The ouputs are:
0=Grass
1=Wall
2=Egg
3=Bot
To find a player, use findPlayer(name) (Returns coordinates array) or detectPlayers(x,y) (Returns name or null). If your bot lands on a spot with an egg, it will not be able to pick it up until it uses the function fetchEgg(). If there is not an egg, your turn will stop.
Moving:
Moving is done by returning a string. This string can be north, south, east, west, or none. Assume that directions are like a compass placed on a Cartesian plane:
North=Y+1
East=X+1
South=Y-1
West=X-1
Winning:
When all eggs are found, the game is over. If you move North to an egg, you must pick it up on the next turn in order to obtain it. If you try to move into a wall, nothing will happen. To get your current position, use getLocation(), which returns an array [x,y]. Use getBasket() to get how many eggs you have. The course will be designed so that all eggs are accessible, but some may be hard to get. If an egg is found by two bots in the same turn, each bot get 0.5 added to their basket, as they split the reward.
Functions list:
checkZone(x,y): Returns an integer, takes 2 integers as parameters between -100 and 100
0=Empty
1=Wall
2=Egg
findPlayer(name): Takes string as parameter, returns player's position as an array of integers or null if player is nonexistant
detectPlayers(x,y): Returns an array of player names, takes 2 integers as parameters between -100 and 100
fetchEgg(): Fetches egg in current position if exists
getLocation(): Returns array with x and y coordinates
move functions: Returns true or ends turn (Ends turn if cannot move)
Example:
function rabbitInDisguise() {
var loc = getLocation();
if (getLocation(loc[0], loc[1]) == 2) {
fetchEgg();
} else {
if (getLocation(loc[0], loc[1] + 1)) {
moveNorth();
} else {
moveEast();
}
}
}
UPDATE 1: Added findPlayer(name) and detectPlayers(x,y) functions, made locating functions "background", and added simultaneous egg finding rules
• 1. I don't see the competition here. I find egg, and I grab it as fast as I can. 2. You don't describe what other players look like. Apr 18, 2018 at 15:05
• @NathanMerrill Whichever bot is best at searching for the eggs will win. If Bot A moves in circles around the board, but Bot B scans a 5 x 5 area around it and makes a map, who do you think would win? Also, what do you mean by "You don't describe what other players look like." Apr 18, 2018 at 15:08
• @user202729 Mistype...I'm about to fix it Apr 18, 2018 at 15:09
• And how does program interact with each other? Is there any limit on how many (library) function call can be done in each function? Apr 18, 2018 at 15:12
• it cannot run more than one function per turn Apr 18, 2018 at 15:16
• 2. You said that 0=grass, 1=Wall, and 2=Egg. What does a player look like? 1. Right, let me check all the locations nearest me, and walk to the closest one. (You can slightly improve on this by identifying eggs that other players will go for). it cannot run more than one function per turn: Your example bot breaks that. Apr 18, 2018 at 15:17
• Players are invisible, since it could complicate things. I'll edit to address the problems, though. Apr 18, 2018 at 15:18
• If players are invisible, then this is not a KotH. This is a code-challenge to retrieve the eggs as fast as possible. Apr 18, 2018 at 15:19
• @NathanMerrill Fixed. Players can be found using findPlayer() and detectPlayers() Apr 18, 2018 at 15:27
• So they can investigate the whole board in one turn? Sounds more interesting then. Apr 18, 2018 at 15:40
• @user202729 I think moveRandom() would be fine (It just makes it easier to program), but I'll add more info on the functions. Apr 18, 2018 at 15:45
• @user202729 Good point. Feature eliminated. Apr 18, 2018 at 15:51
• Any other problems you can think of? Apr 18, 2018 at 15:54
• @PeterTaylor (what does KISS stand for?) Apr 19, 2018 at 13:58
• @user202729, softwareengineering.stackexchange.com/a/87/13258 Apr 19, 2018 at 14:02
## Compute strings with a fixed distance
Consider a binary string of total length N. For example 000111000111 with $$\N=12\$$. We can store this string more efficiently by just recording the indices where bits are first flipped. In this case $$\(3, 6, 9)\$$ is sufficient information to reproduce the string. Here are some more examples using this representation:
10101010 is represented inefficiently by (0, 1, 2, 3, 4, 5, 6, 7)
1111100000 is represented by (0, 5)
0000011111 is represented by (5)
Let us call this representation of a string its compressed representation (even though sometimes the compression is worse than doing nothing).
Now recall the Levenshtein distance between two strings. This is the minimum number of single character insertions, deletions and substitutions needed to transform one string into the other.
Given a compressed representation of a string $$\S\$$ of uncompressed length $$\N\$$ and two positive integers $$\k\$$ and $$\d\$$, the task is to output the compressed representation of $$\k\$$ distinct strings of uncompressed length $$\N\$$, each with Levenshtein distance exactly $$\d\$$ from $$\S\$$. If there are not that many distinct strings possible, simply output as many as possible.
Your code must run with $$\N = 1000000\$$ and $$\k, d < 20\$$ in less than a minute on a normal desktop PC.
• I feel like I'm missing something here. Everything above "Task" is clear, but I don't understand what's going on below. If S = "0000", k = 1, and d = 1, then I simply print out one of "1000", "0100", "0010","0001" (in compressed format)? Aug 30, 2018 at 20:09
• @NathanMerrill Yes. It’s less trivial with other starting strings and larger d.
– user9207
Aug 30, 2018 at 20:26
• Right. There seems to be two parts to this challenge then: Iterating through Levenshtien mutations, and converting to/from the compressed representation. Unless I'm missing how they are related, I'd argue that you remove one of the parts. Aug 30, 2018 at 20:32
• It appears that iterating Levenshtien mutations would be a duplicate Aug 30, 2018 at 20:33
• @NathanMerrill Thanks for the possible dupe link. For your first comment, the idea was to do it without first uncompressing. I will try to improve the question.
– user9207
Aug 31, 2018 at 9:10
# Definition
Let f(n) be the sum of the sums of each possible factor permutation of n. Determine (Y f)(n) (a fixed-point combinator) for all n from 2 to the input i (where i is an integer greater than 2).
# Example
A quick example of f(n) first:
f(8) = (2 + 2 + 2) + (2 + 4) + (4 + 2) + (8)
f(8) = 26
An example of (Y f)(n):
(Y f)(8)
f(8) = 26 (shown above)
f(26) = (2 + 13) + (26) = 41
f(41) = (41) = 41
(Y f)(8) = 41
Let g(i) define a function which implements the described problem.
g(2)
2
g(4)
2
3
41
g(8)
2
3
41
5
11
7
41
# Competition
This is a challenge, which means that the solution with the lowest asymptotic time complexity (in Θ(...)) wins!
In the case of a tie, the winner will be determined by rules, so please submit an implementation of your code along with the algorithm. Tie-break will occur on a Google Compute Engine n1-standard-8 instance.
• In your example of f(8), why do you include 1 in (8+1)? Should it not be just (8)? I could make (2+2+2) into (2+2+2+1). Sep 3, 2018 at 16:22
• @RushabhMehta You're entirely correct! My bad, let me fix that. Sep 3, 2018 at 16:27
• I'm also pretty sure that in your system, the only fixed points are prime numbers. Is that desired functionality? Sep 3, 2018 at 16:33
• @RushabhMehta See the title. :P Sep 3, 2018 at 16:45
• Finally, I would mention that permutations are distinct, just so that its obvious how you are counting all the different factors. Sep 3, 2018 at 16:47
• @RushabhMehta Is that a sufficient change? Sep 3, 2018 at 16:49
• That should be fine. Nice challenge! Sep 3, 2018 at 16:52
• 1. What's a factor permutation? 2. I doubt anyone can give even a moderately tight analysis of the runtime for an implementation of g without a very tight analysis of the output of f, so I don't think it makes sense to post this without first editing in tight bounds on the output of f. Sep 5, 2018 at 10:18
# Bijective base banana to decimal
I'm trying to decide between 2 versions of this challenge:
# Prefixes version
Let bijective base banana be bijective base 6 with each digit represented by a different prefix of the word "banana" as follows:
1: "b"
2: "ba"
3: "ban"
4: "bana"
5: "banan"
6: "banana"
The input will be a string made up of a concatenation of zero or more of only these strings. This represents a number in bijective base 6 (with most significant digit first).
The output will be the same number in (non-bijective) decimal.
For example: (input : output)
"" : 0
"b" : 1
"bb" : 7
"bbanana" : 12
"bananab" : 37
"bananabanana" : 42
A worked example:
bananabanbana
banana ban bana
6 3 4
6*36 + 3*6 + 4*1
216 + 18 + 4
238
This is code golf, so the winner for each language is the code with the fewest bytes.
# Suffixes version
Let bijective base banana be bijective base 6 with each digit represented by a different suffix of the word "banana" as follows:
1: "a"
2: "na"
3: "ana"
4: "nana"
5: "anana"
6: "banana"
The input will be a string made up of a concatenation of zero or more of only these strings. This represents a number in bijective base 6 (with most significant digit first).
The output will be the same number in (non-bijective) decimal.
For example: (input : output)
"" : 0
"a" : 1
"aa" : 7
"abanana" : 12
"bananaa" : 37
"bananabanana" : 42
A worked example:
bananaananana
banana ana nana
6 3 4
6*36 + 3*6 + 4*1
216 + 18 + 4
238
This is code golf, so the winner for each language is the code with the fewest bytes.
# Sandbox questions
• Is this a duplicate or near enough a duplicate to be worth changing?
• Are there any flaws that would make the challenge uninteresting?
• I'm considering changing from prefixes to suffixes following this comment from H.PWiz
• I currently prefer the suffixes version, but the worked example shows that the interpretation is no longer unique. It would either need to be specified that the match must be greedy (or non-greedy) or a different word would need to be chosen. I lean towards choosing a word which still gives enough ambiguity to force looking ahead, while still keeping the overall conclusion unique. I can't see a way of doing this without making the last character unique though, which reduces to a reversed version of the prefixes version (you can still ignore all other letters).
• This is interesting, but I can't help but feel like this is two problems glued into one. Sep 3, 2018 at 16:27
• It's definitely 2 problems glued onto 1. What I'm wondering is will that make for additional golfing opportunities in at least some languages? I suspect it will, and I'd expect myself to underestimate the possibilities since there are many more insightful people tackling challenges, in many languages. Sep 3, 2018 at 17:58
• Hmm, It seems like all that matters is where the bs are. I wonder if it would be a more or less interesting challenge with a different word Sep 3, 2018 at 18:18
• Good point. Suffixes instead of prefixes might give a bit more of a challenge. As long as the output is still guaranteed unique Sep 3, 2018 at 18:34
• @trichoplax I'm pretty sure you can't do suffixes, since its no longer unique. Sep 3, 2018 at 20:24
• @RushabhMehta yes it stops working for the word "banana" then (I added a note about this in my 4th sandbox point). It would still work for words that don't contain any of their own suffixes as non-suffix substrings. Like "aardvark". I'm not sure which way to go yet Sep 3, 2018 at 20:37
• @trichoplax Your last bullet was exactly my issue. Suffixes don't help at all. Sep 3, 2018 at 22:34
# Facto-RLE
Given a non-empty string containing non-numeric printable ASCII characters, compute its facto-run-length encoded version.
## Definition
Let $S$ be a non-empty string of length $l\in\mathbb{N}^+$ containing non-numeric printable ASCII characters. For each positive factor $n|l$ one can represent $S$ as a $n\times(l/n)$ matrix of characters. Let $M^n$ denote these matrices.
For a given $n$ and $j\in\{1,\dots,l/n\}$, let $C^n_j$ be the string representing the $j$-th column of $M^n$. Let $E^n_j := \text{RLE}(C^n_j)$ denote the array of this string's run-length encoded version.
Let $R^n := \text{CNCT}(\text{ZIP}(E^n_1,\dots,E^n_{l/n}))$ denote the string representation of all $E^n_j$ zipped together.
The facto-run-length encoding of $S$ is defined as the string array $\text{FRLE}(S):=[R^n:n|l]$.
For a family of $k\in\mathbb{N}^+$ arrays $A^i$ with respective lengths $l^i, i\in\{1,\dots,k\}$ and elements $A^i_j,j\in\{1,\dots,l^i\}$, let $Z_j:=[A^i_j:i\in\{1,\dots,k\}\land j\leq l_i]$ denote the array of elements with index $j$, if present. Furthermore, define $\text{ZIP}(A^1,\dots,A^k) := Z_1\Vert\dots\Vert Z_{\max\{l_i\}}$ as the concatenated array of all $Z_j$.
For an array of strings $A$ of length $j$, define $\text{CNCT(A)}:=A_1\Vert \dots\Vert A_j$ as the concatenation of all strings of $A$.
## Example
Let $S:=\text{"Hello world!"}$, therefore $l=12$ with positive factors $\{1,2,3,4,6,12\}$. $$M^2=\begin{pmatrix} \text{H}&\text{e}&\text{l}&\text{l}&\text{o}&\text{ }\\ \text{w}&\text{o}&\text{r}&\text{l}&\text{d}&\text{!} \end{pmatrix}, \\C^2_1=\text{"Hw"}, C^2_2=\text{"eo"}, C^2_3=\text{"lr"}, \\C^2_4=\text{"ll"}, C^2_5=\text{"od"}, C^2_6=\text{" !"}, \\E^2_1=[\text{"H"},\text{"w"}], E^2_2=[\text{"e"},\text{"o"}], E^2_3=[\text{"l"},\text{"r"}], \\E^2_4=[\text{"l2"}], E^2_5=[\text{"o"},\text{"d"}], E^2_6=[\text{" "},\text{"!"}], \\R^2=\text{"Hell2o word!"}$$
$$M^4=\begin{pmatrix} \text{H}&\text{e}&\text{l}\\ \text{l}&\text{o}&\text{ }\\ \text{w}&\text{o}&\text{r}\\ \text{l}&\text{d}&\text{!}\\ \end{pmatrix}, \\C^4_1=\text{"Hlwl"}, C^4_2=\text{"eood"}, C^4_3=\text{"l r!"}, \\E^4_1=[\text{"H"}, \text{"l"}, \text{"w"}, \text{"l"}], E^4_2=[\text{"e"}, \text{"o2"}, \text{"d"}], E^4_1=[\text{"l"}, \text{" "}, \text{"r"}, \text{"!"}],\\ R^4=\text{STR}([\text{"H"},\text{"e"},\text{"l"},\text{"l"},\text{"o2"},\text{" "},\text{"w"},\text{"d"},\text{"r"},\text{"l"},\text{"!"}])=\text{"Hello2 wdrl!"}$$
$$M^{12}=\begin{pmatrix} \text{H}\\ \text{e}\\ \text{l}\\ \text{l}\\ \text{o}\\ \text{ }\\ \text{w}\\ \text{o}\\ \text{r}\\ \text{l}\\ \text{d}\\ \text{!}\\ \end{pmatrix}, \\C^{12}_1=\text{"Hello world!"}, \\E^{12}_1=[\text{"H"},\text{"e"},\text{"l2"},\text{"o"},\text{" "},\text{"w"},\text{"o"},\text{"r"},\text{"l"},\text{"d"},\text{"!"}], \\R^{12}=\text{"Hel2o world!"}$$
Therefore the following follows. $$\text{FRLE}(\text{"Hello world!"}) = [\dots, \text{"Hell2o word!"}, \dots, \text{"Hello2 wdrl!"}, \dots, \text{"Hel2o world!"}]$$
## Tags code-golf, array-manipulation, string
• If you can't think of a way to put it into words, you could put it into code, and then someone here can translate that reference implementation into a description. Plus you can then include the reference implementation in the spec for people who read code better than prose Sep 4, 2018 at 19:13
• Okay,I can do that! Sep 5, 2018 at 13:50
• @Zacharý I have attempted to write a task definition. If this is not what you intended, you can simply roll back. Sep 7, 2018 at 3:37
• By the way, I think giving a factor as input would be a nicer challenge than adding a wrapper to encode for every factor. Sep 7, 2018 at 3:40
• @JonathanFrech, there isn't universal agreement on the meaning of zip applied to a non-square 2D array, so you need to define it unambiguously. Sep 7, 2018 at 10:13
• @PeterTaylor Attempted to do so. I intend to only zip one-dimensional arrays. Sep 7, 2018 at 10:46
• Thanks! I think this definition works! Sep 7, 2018 at 14:58
I have an idea for a KOTH challenge (that is marginally similar to the current Reaper KOTH), but none of the expertise required to make it work. The idea is based on the card game Diamant aka Incan Gold, and the rules of the game are:
• The game is played over 5 expeditions, each with the same structure.
• At the start of each expedition, a special deck of cards is shuffled, containing gem cards and hazard cards (and in some versions of the game, one artifact card is added to the deck every expedition).
• Every turn, a card is revealed from the deck.
• If a hazard card is revealed and another copy of the same hazard has already been revealed in the expedition, the expedition immediately ends and all remaining players lose their accumulated gems. The revealed hazard card is removed from the deck for future expeditions.
• If a gem card is revealed, the number of gems shown on the card is shared evenly between remaining players, with any remainder placed in a common pool.
• After the card is revealed, if the expedition has not ended, then players simultaneously decide whether they want to "Stay" or "Go Home".
• All players who choose "Go Home" share the gems in the common pool between them (leaving any remainder), and then bank all of their gems. They are no longer in the expedition.
• If a single player chooses "Go Home", they claim all of the gems in the common pool plus any artifacts that were revealed.
• If there are players remaining in the expedition, then a new turn begins.
• If there are no players remaining in the expedition, then the expedition ends.
• If this was the last expedition, then players count up the gems they banked plus bonus gems for artifacts they claimed, and the player with the most gems wins.
My thought is that we could simplify the game for KOTH - one expedition, no artifacts - and run games which pit 4 bots against each other in some configuration, with each bot's overall rating being its average score per game.
The problem is that I have no experience in any meaningful programming languages. The other problem is that I don't know how to run a contest like that. The other other problem is that I don't know whether it's worth running.
Can anyone offer assistance or advice on any of those three problems?
• For problem 1, there are 2 optinos, either ask somebody else to write the bot (if the challenge is interesting enough), or to learn a programming language. Sep 14, 2018 at 5:32
# Output the answer after yours code-golfstack-exchange-api
Related: Output the answer above yours
This challenge is very similar to the above, but I hope it's different enough to warrant posting. The difference is that you need to post the answer that is posted after yours (the whole answer, not just the <code> block). Also, your output can be the answer text or the HTML code of the answer. When your post is still the newest post, your program should just output nothing.
## Rules
1. No URL Shorteners
2. No programs that need to be run at a specific URL.
I think mine is different enough from the linked question, because that has some different rules, and my version closes loopholes used on the other question. Also, no one utilizes the StackAPI on that one.
Put +, -, * into 1 _ 2 _ 3 ... n to make the expression equal 0, for a given 3 <= n <= 30, and output the valid equation (0=1+2-3). Exactly one operator between each number.
There exists some patterns, such as ((1-2)-(3-4))+((5-6)-(7-8))... and 1+((2-3)-(4-5))+(6-7). Maybe we can find patterns that solve the whole problem?
There are no solutions without * for at least n={5,6,9,10,13,14,17,18,21,22}, so it looks like we need * for n=4x+1 and n=4x+2.
There are no solutions without + for n < 18.
There are no solutions without - (obviously).
Brute-forced examples:
1. 0=1+2-3
2. 0=1-2-3+4
3. 0=1*2-3-4+5
4. 0=1+2*3+4-5-6
5. 0=1+2-3-4+5+6-7
6. 0=1*2*3+4-5-6-7+8
7. 0=1-2+3*4*5+6-7*8-9
8. 0=1-2*3+4*5-6*7+8+9+10
9. 0=1*2+3*4+5*6-7*8-9+10+11
10. 0=1-2+3*4-5*6-7-8*9+10*11-12
11. 0=1+2*3+4+5-6-7-8+9+10+11-12-13
12. 0=1*2+3*4+5*6+7+8*9+10-11*12+13-14
13. 0=1-2-3-4-5-6-7+8-9+10-11+12-13+14+15
14. 0=1*2*3*4*5*6+7+8-9*10*11-12+13+14+15*16
15. 0=1*2*3*4*5*6+7*8*9+10-11*12*13+14*15+16*17
16. 0=1*2*3*4*5*6-7*8-9-10-11*12-13-14*15-16*17-18
17. 0=1*2*3*4*5*6-7-8-9*10-11-12-13-14-15*16-17*18-19
18. 0=1*2*3*4*5*6-7-8-9*10-11-12-13*14-15-16-17-18*19-20
19. 0=1*2*3*4*5*6-7-8-9-10-11-12*13-14-15-16-17-18-19-20*21
20. 0=1*2*3*4*5*6-7-8-9-10-11*12-13-14-15-16-17-18-19-20*21-22
21. 0=1*2*3*4*5*6*7-8*9-10*11-12-13*14-15*16*17-18-19-20-21-22*23
22. 0=1*2*3*4*5*6*7-8*9*10-11-12-13-14*15*16-17-18-19*20-21*22-23-24
23. 0=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
24. 0=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
25. 0=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
26. 0=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
• Related Also, do you want this to be code golf (shortest code) or a different type of challenge? Sep 23, 2018 at 21:15
• Sep 26, 2018 at 15:22
• @user202729 yep. Unlike that one, this doesn't allow turning the 1 negative, which does seem to make the problem unsolvable without multiplication. Sep 26, 2018 at 15:52
There is a nice task One Ring to rule them all. One String to contain them all. But rule define a string as a linear buffer. A linear buffer is not a Ring :)
My suggestion is:
• create a new task with title "One Ring to rule them all. One Ring to bring them all"
• modify an Objective: Output a String which contains every positive integer strictly below 1000 and the String is a Ring.
• It would be a dupe Sep 28, 2018 at 11:32
• ok. thanks, Peter Sep 28, 2018 at 11:52
• I've been thinking about it - no, It is not a duplicate. Main different: 0, 00, 000, ... are different elements in a de Bruijn sequence and same element in a Ring of numbers. This moment make golfed algoritms different. Oct 1, 2018 at 8:08
• In a de Bruijn sequence containing all 3-digit sequences over 0-9, 000 is an element but 0 and 00 are not. Moreover, if the ring contains 100, 200, etc. then it must contain 00 at least nine times. See my comment of Nov 5 '13 at 11:28 on /q/13088. Oct 1, 2018 at 9:51
Make words (need a better title)
Write a program or function that takes a string and an integer $$\ n \$$ as input and outputs all the $$\n\$$-letter words formed using only the letters in the string.
Notes:
The input string will have only small alphabets $$\a-z\$$.
All the letters in the input string will be unique.
Input integer will be positive.
Winner: This is code-golf so shortest code wins. (Fastest algorithm will also be good in this right?)
Examples:
input:"abcd",5
output: "aaaaa","aaaab","aaaac",.....,"ddddc","ddddd"
P.S. The output does not have to be sorted.
Any input and output format will do as long as its distinguishable.
Example:
input: abcd,5 (ok)
abcd 5 (ok)
abcd5 (not allowed)
output: ["aaaaa","aaaab",.... (ok)
"aaaaa","aaaab",.... (ok)
"aaaaa""aaaab""aaaac"..... (ok)
aaaaa aaaab aaaac ..... (ok)
aaaaaaaaabaaaac...... (not allowed)
• Oct 1, 2018 at 10:18
• For this kind of challenge, neither fastest-algorithm nor fastest-code should be used. For the first one: all answers are likely to have the same time complexity. For the 2nd one: ~99% of CPU time is going to be spent printing the results. Oct 1, 2018 at 10:24
• BTW: this is basically count from $0$ to $b^n$ in custom base $b$, which may have already been covered in some other challenge. I failed to find one, though. Oct 1, 2018 at 11:24
• It's product in python: docs.python.org/2/library/itertools.html (May help giving a title to the challenge, or searching for duplicates) Oct 2, 2018 at 2:54
• @Arnauld It's mostly a duplicate of Cartesian product of a list with itself n times. The only difference is that builtins aren't allowed in that linked challenge, which is probably better since this is 3 bytes in 05AB1E due to an optional parameters requiring an explicit input and even just 1 byte in Jelly (not sure how to pretty-print it as a list in the footer..). Oct 3, 2018 at 8:13
# Golf A Programming Language
Code golf languages are languages built to complete programming challenges in few lines of code. Instead of creating a code golf language, you will be creating an interpreter for a programming language in as few bytes as possible.
## Scoring
As you are not creating a code golf language, instead your score will be based on the number of bytes in the interpreter.
## I/O
You may provide a program as input to your interpreter as input using any standard input method
Output provided by a program in your language may be passed on through your interpereter using any standard output method
You must write a program to solve each task in the language you created:
• Hello World - Output the string "Hello World"
• Fizz Buzz - List numbers from 1 to 100 replacing multiples of 3 with Fizz, 5 with Buzz, and 15 with FizzBuzz
• Prime Check - Check if a given number is prime
## Loopholes
Your programming language must be implemented yourself. Using builtin functions such as Javascript's eval or CHIQRSX9+'s I on input makes this not very interesting.
• This is too broad, as defining "interpreter" is pretty impossible. The only way this is viable is if you specify the language to be interpreted. Oct 2, 2018 at 2:48
• @NathanMerrill Anything can be built for the interpreter. It could have 3 instructions, one for each challenge, but that could mean the interpreter increases length because it has the tasks built in. It might need more tasks, 3 tasks might mean building an actual language takes more code than just hardcoding 3 functions
– pfg
Oct 2, 2018 at 3:39
• Writing a general purpose language will take more characters than providing built-ins for those. In essence, you're looking for a program that returns a program that takes 1 of 3 characters and evals it. You are correct that eval makes this not interesting, but it's also the best way. Unless you make the language significantly harder (and specify the input required to do each of the tasks), I don't see this being interesting. Oct 2, 2018 at 3:44
• codegolf.stackexchange.com/questions/111278/… Oct 2, 2018 at 5:27
# Eeny Meeny Miny Moe
Inspired (somewhat) by this question:
Where should I stand to be captain of my team?
# Introduction
The childhood song Eeny, meeny, miny, moe was often used to select who was a captain when playing some game. Everyone on one team would stand in a line and someone would point at the first person in the line. Although there are several variations, where I grew up, they would sing:
Eeny, meeny, miny, moe,
Catch a tiger by the toe.
If he hollers, let him go,
Eeny, meeny, miny, moe.
My mother told me to pick the very best one and you are not it.
As we sung each word, we pointed at the next person in line, then (if the rhyme wasn't over yet) we would continue the rhyme and start over at the beginning of the line. The person being pointed to when the final "it" was sung would be removed from the line, eliminating them from being captain. Then the rhyme would start over at the beginning of the line to remove the next person. This continued until only one person was left, and that person would be captain.
It didn't take long for me to recognize that if there were two people left (or if we started with just two people), it seemed like the second person would always win. Studying this for a bit I realized that since there are 35 beats this would always happen. I further figured out if there are 3 people, the third would win. So I wondered: could I chart it out for any number? Can I create a formula for it?
# Challenge:
Write a program, function or (etc. as standard) where given input of an integer number of starting children greater than one, output what starting child number will be the winner, and where I need to stand so that I can be captain!
# Examples:
Input Output
2 2
3 3
4,5,6or7 4
8 or 9 5
10 6
11,12or13 7
14 8
15 9
16 10
17,18,19or20 11
Winner for each language is that with the least number of bytes. And of course, standard loopholes and all that legal jazz are not allowed as is typically expected in these challenges.
• aka, the Josephus problem, possibly duplicate
– ngn
Oct 2, 2018 at 19:49
• @ngn Wow! I learn something new every day! Is this a duplicate, then? Oct 2, 2018 at 19:50
• I'm not sure, yours is a special case for k=35
– ngn
Oct 2, 2018 at 19:52
• by the way, you may find this video enlightening :)
– ngn
Oct 2, 2018 at 19:53
• Different from listed duplicate in several ways. Mine restarts at the beginning after each elimination (which really makes it questionable whether it is Josephus or not). Mine requires only one input. Mine is a firm case for k and mine has a cool backstory :) Oct 3, 2018 at 12:16
## Permutations to the nines
This question is based on an unresolved inquiry which began at Permutations without recursive function call and was followed up at How to improve efficiency of algorithm which generates next lexicographic permutation? (TL;DR), after finding help removing duplicate numbers at Most efficient method to check for range of numbers within number without duplicates, where after performing arithmetic by hand found that for an array having length less than or equal to 9, if we ignore the value held at each index of the initial array and instead convert the indexes to a whole number, the next lexicographic permutation can be determined by adding 9 to the current index as number until a whole number, e.g., [1,2,3] // 123 1-based index or 23 0-based index is reached that satisfies two the conditions
• Contains only the numbers of the indexes of the original array
• Does not contain any duplicate numbers
e.g., 123+9=132 // "abc" -> "acb" 132-9=123 // "acb" -> "abc"; the graph for "abcd" is
[9,81,18,81,9,702,9,171,27,72,18,693,18,72,27,171,9,702,9,81,18,81,9]
which does not appear to be linear.
As the linked answers disclose, adding 9 to a whole number is not the most efficient method of determining the next lexicographic permutation, and what found independently by hand has been found by others, consider OEIS A217626
A217626 First differences of A215940, or first differences of permutations of (0,1,2,...,m-1) reading them as decimal numbers, divided by 9 (with 10>=m, and m! > n).
1, 9, 2, 9, 1, 78, 1, 19, 3, 8, 2, 77, 2, 8, 3, 19, 1, 78, 1, 9, 2, 9, 1, 657, 1, 9, 2, 9, 1, 178, 1, 29, 4, 7, 3, 66, 2, 18, 4, 18, 2, 67, 1, 19, 3, 8, 2, 646, 1, 19, 3, 8, 2, 67, 1, 29, 4, 7, 3, 176, 3, 7, 4, 29, 1, 67, 2, 8, 3, 19, 1, 646, 2, 8, 3, 19, 1
Also of interest is that the graph of the specific multiples of 9 to derive all lexicographic permutations of a set can be found by computing only half 1/2 of the total permutations +1, as the declination slope is identical to the inclination slope of multiples of 9, for example, given an input of the string "abc" or array ["a","b","c"], we only need to reach the inverted peak of the graph, that is bca 120 3 18, where the previously calculated values can be reversed and added to the current whole number, 120, to generate the remainder of the lexicographic permutations
abc 012 0
acb 021 1 9
bac 102 2 81
bca 120 3 18
cab 201 4 81
cba 210 5 9
/\/\
/ \
## Errata
This challenge is not to determine the most efficient algorithm to calculate permutation 9!, as, again, adding 9 to a whole number to reach that goal can be outperformed by swapping values or other method, as demonstrated at the accepted at the first SO link above.
This challenge is to derive the most efficient method of determining the next lexicographic permutation by calculating the next multiple of 9 that meets the above listed criteria, preferably directly using math, if possible.
The reason am posting this question here and now is because was reminded by the graph of the peaks (particularly the third item) in this question Different combinations possible
Just to give a visual example, they are the following:
/\ /\ /\/\
/\/ \ / \/\ / \
that in spite of performing various calculations using the index of the !n within the resulting array of permutations, e.g.,
var n = N = 1234;
var res = [9,81,18,81,9,702,9,171,27,72,18,693,18,72,27,171,9,702,9,81,18,81,9].map((x)=>n=n+x);
// try to determine the rate of growth
var j = res.map((x)=>x/N);
console.log(res, j);
var tryStuff = 1234*(4321/1234);
console.log(tryStuff, (4321/1234));
am still vexed by and have not been able to independently determine the mathematical formula to derive the next lexicographic permutation directly without adding 9 multiple times to reach the required number; or if that is even mathematically possible for any input set from 2! through 9!. That is, this code
function getNextLexicographicPermutation(arr) {
for (var l = 1, i = k = arr.length; l < i; k *= l++);
function checkDigits(min, max, n) {
var digits = 0;
while (n) {
d = (n % 10);
n = n / 10 >> 0;
if (d < min || d > max || (digits & (1 << d)))
return false;
else
digits |= 1 << d;
}
return true;
}
var len = arr.length,
idx = arr.map(function(_, index) {
return index
}),
p = 9,
min = 0,
max = len - 1,
last = Number(idx.slice().reverse().join("")),
curr = Number(idx.join("")),
res = [curr],
diff = [],
result = [],
next, times = 0;
while (res.length < (k / 2) + 1) {
++times;
next = (curr += p);
if (checkDigits(min, max, next)) res[res.length] = next;
curr = next;
};
for (var i = 0; i < res.length; i++) {
var item = res[i];
item = String(item).split("").map(Number);
item = (item.length < arr.length ? [0].concat(item) : item)
.map(function(index) {
return arr[index]
}).filter(Boolean);
result.push(item)
}
res.reduce(function(a, b) {
diff.push(b - a);
return b
});
for (var i = 0, curr = res[res.length - 1], n = diff.length - 2; result.length < k; i++, n--) {
curr = curr + diff[n];
result.push(
String(curr).split("")
.map(function(index) {
return arr[index]
})
);
}
return [result, diff, res, times];
}
var arr = ["a", "b", "c", "d"];
console.log(getNextLexicographicPermutation(arr));
which generates the second half-1 of the permutations from the first half+1 of the permutations checks if the next whole number meets the necessary conditions 210 times for an input array of ["a","b","c","d"] which has a resulting .length of 24. Ideally, we want to generate 24 lexicographic permutations using only 12 or 13 checks; or no checks at all, by determining the irrational number or other mathematical algorithm which will directly calculate (or approximate close enough to determine the next multiple of 9) the next whole number which meets the necessary criteria.
Kindly disregard the length of this post, as am trying to include as much information as consider relevant to the inquiry.
## Rules
This challenge must use the number 9 (addition, multiplication, division, other mathematical operation) to determine the next lexicographic permutation using the indexes of the current lexicographic permutation converted to a whole number, ideally, in a single operation, else in the least amount of operations necessary to achieve the expected result.
Again, this challenge is not asking how to code the most efficient code which determines the next lexicographic permutation, but what is the most efficient approach is using only the number 9 and math to generate the permutations.
Since we can get the first and last lexicographic permutations by reversing the indexes of the input array, that is not counted as an operation within the program.
## Input
An array or sting having .length less than or equal to 9!.
## Output
Lexicographically sorted array of permutations of input.
Remaining within the Rules above, determine a mathematical algorithm which directly generates the next lexicographic permutation using only the current indexes of the original input or current permutation. Ideally, directly, without having to add 9 in multiple operations until the listed criteria is met, that is, we want a single algorithm to calculate that we need to add 81 to 132 to get the sum 213 and add 702 to 321 to get the sum 1023 and so forth. Explain the math in the algorithm. Note: The requirement might not be possible. If that is the case, explain why.
• 1. As I commented elsewhere, it's hard to separate the core challenge from the fluff. 2. I assume there's a typo in the "Input" section, because an array of the permutations of 9! elements would take more memory than is physically possible in this universe. Oct 4, 2018 at 21:37
• @PeterTaylor Kindly indicate specifically what content in the question that you are referring to. From perspective here, there is no "fluff" in the question. No, there is no typo in the "Input" section. The algorithm MUST be true for 9!->362880 if true for 4!->24. We can check the result of the algorithm by incremental sections and by hand. Simplified core challenge: Find the multiple of 9 required to precisely equal the next p lexicographic permutation (as a whole number) from current p number in least math computations. Oct 4, 2018 at 22:06
• @PeterTaylor E.g., if we have input ['a','b','c'] we ignore values and "see" 12, 012 (or, as described at this question for simplification 1-based indexing 123). The next lexicographic permutation is one computation 123+(9*1)=132 from we can "see" ['a','c','b']. In JavaScript we could even use Math.pow(9,1), but that would be a false-positive for a method or pattern to use, because the next multiple is 132+(9*9)=213 (we can "see" ['b','a','c']). How to get from 132 to 213 without adding 9 nines times (Task: least mathematical computations in this step) to 132? Oct 4, 2018 at 22:33
• Should we take an array of items (['a','b','c']), an array of indices ([1,2,3]) or its value converted to decimal (123) as input? And what should be the output -- 132, ['1','3','2'] or 9? Oct 5, 2018 at 13:08
• @user202729 The input is a decimal derived from the first lexicographic permutation or an array of items from which the decimal will be derived; e.g., 123 from ['a','b','c'], or 123456789 (1-based indexing) or 012345678 (0-based indexing) derived from ["a","b","c","d","e","f","g","h","i" ]. The maximum number is the reverse order of the decimal number, 987654321 or 876543210, which is the last lexicographic permutation (as a decimal number) of the input number or array [8,7,6,5,4,3,2,1,0] (0-based) or [9,8,7,6,5,4,3,2,1] (1-based). Oct 5, 2018 at 20:48
• @user202729 We can derive all lexicographic permutations of input up to 9! by adding 9 to the first lexicographic permutation as decimal and following the two conditions at the question; that is, by adding 9 to the initial decimal number where no duplicate numbers are found within the resulting number and each number is found within the original decimal number. Example "abc"->curr=123->123+9=132->"acb"; curr becomes 132. We are trying to reduce the number of additions to meet the two conditions by creating a formula to determine the precise number of 9's to add to curr. Oct 5, 2018 at 21:09
• @user202729 Is the question clear to you now? Oct 5, 2018 at 21:43
• @user202729 The output is n<=9! permutations. We can already achieve that by the code at the question. However the challenge is to reduce the mathematical steps required to reach the result consider "How to get from 132 to 213 without adding 9 nines times (Task: least mathematical computations in this step) to 132". When curr is 132, we do not necessarily know that we will have to add 9 to 132 nine times (81) to reach the sum 213. We find that out now by nine computations, adding. The challenge is to reach the sum 213 using what we have in less than nine computations. Oct 5, 2018 at 23:27
• @user202729 There are other mathematical relationships between the indexes of the lexicographically sorted permutations. For example, if we have the input ['a','b','c'] or "abc", when we ignore the values, using 1-based indexing, we can "see" 123 at index 1. 123+90=213 our lexicographically sorted permutation at index 3, 132 at index 2 is 132+90=231; max number is always input reversed. We only need to find 132 (3 operations) to derive all 6 permutations, as the declination slope of [123,132,213] is the reverse order of each number; this relationship differs for 1234. Oct 5, 2018 at 23:46
• What is n in that case? /// Consider a specific example. If the input is 132 (1-indexing) what should the output be? Oct 6, 2018 at 11:13
• @user202729 The output is always n! lexicographic permutations. We already are able to achieve that. Whether the input is 123 or 132 makes no substantial difference here; we "see" that 132 is greater than its constituent parts, 123 and thus know that we are at index 2 (1-indexing), 321 is always the max, and n! is 6. What this challenge is asking is how to use this specific method of determining permutations, that is, adding 9 to the indexes as a whole number until the listed conditions are met, in less mathematical operations that is currently being used - simple addition. Oct 7, 2018 at 16:55
• Let me see. So, if 132 is given, the output should be [123,132,213,231,312,321], but with a restriction -- only addition, subtraction, multiplication, and division is allowed. Correct? Oct 7, 2018 at 16:59
• @user202729 There is no restriction on the mathematical methods which can be used. You can use calculus if that will generate 132 from 123 in the least amount of calculations - using 9 within the calculations. If you can use mathematical relationships between the expected resulting set which uses other numbers, then those methods should be described in detail. The code at the question uses only the number 9 to add to the indexes represented as a whole number, though for some 90 or 711 could be used. The challenge is to use the least number of total math computations. Oct 7, 2018 at 17:05
• @user202729 We are trying to generate the next lexicographic permutation using only 1) the current permutation; 2) the maximum possible number (our last permutation); 3) mathematics; as our resources. We begin with a number less than 321 and know that our first permutation must be 123, our last 321, and that between 321 and 123 there are relationships between the numbers which allow us to derive each number that meets our criteria. In the code at the question 9 is suitable to achieve that goal by adding 9 to the number where the sum contains no duplicate and only input numbers. Oct 7, 2018 at 17:12
# Qwixx KoTH
In this King of the Hill, our bots will be playing Qwixx. In Qwixx each player will try to get as many points by picking the sum of two dices and marking that number on your own scoresheet. The scoresheet has four rows with different colors (red, yellow, green, blue) with numbers ranging from 2 to 12, where green and blue have the numbers ranked in descending order. You can only mark out a number if you haven't marked a number to the right of it, i.e. after marking red 3 you cannot mark red 2 anymore.
### Rules
All of the rules of Qwixx are:
• Red and yellow have numbers 2, 3, ..., 12, and a padlock.
• Green and blue have numbers 12, 11, ..., 2, and a padlock.
• There are 6 dices: red, yellow, green, blue and two white dices.
• On your turn you will throw all the dices, except for deleted ones.
• Everyone can mark one number on one color equal to the sum of the white dices. Also if it is not your turn.
• If it is your turn, you can also mark one number in a specific color, with the sum of one colored dice and one white dice. The colored dice that you pick denotes in which color you need to mark the number.
• You can only mark numbers to the right of already marked numbers.
• The last number can only be marked if you have already marked at least 5 numbers before. If you do this, you will automatically also mark the padlock, which will give you extra points. If you do this, that colored dice is not thrown anymore and nobody can mark a number in that color. This dice is deleted from the game (thrown as 0).
• Only on your turn you need to mark at least one number. If you don't do this, you will automatically mark a penalty box, which costs you 5 points.
• The amount of points is equal to the sum of 1 up to the amount of marks in each color, minus 5 times the amount of penalties. I.e.: 4 in red, 3 in green and 2 penalties is 4+3+2+1 + 3+2+1 = 16 - 2 * 5 = 6 points.
• Each turn is simultanouesly.
• The game ends when the second dice is deleted or when a scoresheet contains 4 penalties.
### Tournament
The tournament rules are as follows:
• The tournament consists of 10000 * amount of bots games.
• Each game will be played with 5 random bots, in a randomized order.
• If multiple bots end up with the same amount of points in one game, each bot gets a win.
• The winner of the tournament is the bot who has the highest win percentage.
### Bots
The bot needs to be defined in Python and needs to have two functions:
• __init__(self, index) with the index in scoresheets, so that you know which scoresheet is yours
• turn(self, scoresheets, dices, my_turn), with the current scoresheets, the dices of this turn and a boolean being True when it is your turn. This function needs to return a tuple of two tuples containing (color, number) with what your bot wants to mark, where the first tuple is for the white dices, and the second tuple only if its your turn.
The simplest bot can be defined as:
class DoNothing:
def __init__(self, index):
return
def turn(self, scoresheets, dices, my_turn):
return ([],[])
Which does absolutely nothing.
A randomized bot that tries to always mark something when possible is:
class RandomAllowedCombinations:
def __init__(self, index):
self.index = index
return
def turn(self, scoresheets, dices, my_turn):
(any_combs, turn_combs) = scoresheets[self.index].allowed_combinations(my_turn, dices)
any_comb = []
if any_combs:
any_comb = any_combs[random.randint(0, len(any_combs)-1)]
turn_comb = []
if my_turn and turn_combs:
turn_comb = turn_combs[random.randint(0, len(turn_combs)-1)]
return (any_comb, turn_comb)
This also shows some of the utility functions: you can use scoresheets[self.index].allowed_combinations(my_turn, dices) to check what you can mark with the white dices, and if its your turn with one of the colored dices as well.
Both classes (Dices and Scoresheet) have several utility functions. You can check themselves to improve your bot. Note that Dices also contains the indices for each color. And the value of a dice is 0 if it is deleted (i.e. when someone in the game marked the last number and thus the padlock in one of the previous turns).
### Controller
The controller contains 5 example bots. These will not be used in the tournament, unless there are too few participants.
import random
import numpy
### Bots ###
class DoNothing:
def __init__(self, index):
return
def turn(self, scoresheets, d, my_turn):
return ([],[])
class AlwaysWhitesInRed:
def __init__(self, index):
return
def turn(self, scoresheets, d, my_turn):
return ((d.red, d.dices[d.white1] + d.dices[d.white2]),[])
class AlwaysBlueOnlyTurn:
def __init__(self, index):
return
def turn(self, scoresheets, d, my_turn):
return ([],(d.blue, d.dices[d.blue] + d.dices[d.white2]))
class RandomAllowedCombinations:
def __init__(self, index):
self.index = index
return
def turn(self, scoresheets, dices, my_turn):
(any_combs, turn_combs) = scoresheets[self.index].allowed_combinations(my_turn, dices)
any_comb = []
if any_combs:
any_comb = any_combs[random.randint(0, len(any_combs)-1)]
turn_comb = []
if my_turn and turn_combs:
turn_comb = turn_combs[random.randint(0, len(turn_combs)-1)]
return (any_comb, turn_comb)
class RandomButOnlyOwnTurn:
def __init__(self, index):
self.index = index
return
def turn(self, scoresheets, dices, my_turn):
(any_combs, turn_combs) = scoresheets[self.index].allowed_combinations(my_turn, dices)
turn_comb = []
if my_turn and turn_combs:
turn_comb = turn_combs[random.randint(0, len(turn_combs)-1)]
return ([], turn_comb)
### Classes ###
class Dices:
def __init__(self):
self.dices = []
self.deleted = []
self.red = 0
self.yellow = 1
self.green = 2
self.blue = 3
self.white1 = 4
self.white2 = 5
self.colors = [self.red, self.yellow, self.green, self.blue]
self.whites = [self.white1, self.white2]
self.diff_dices = self.colors + self.whites
def roll_dice(self):
self.dices = [0,0,0,0,0,0]
for i in range(6):
if i not in self.deleted:
self.dices[i] = random.randint(1, 6)
def delete_dice(self, color):
if not color in self.deleted:
self.deleted.append(color)
return len(self.deleted) > 1
def combinations(self):
combs = [(color, white) for color in self.colors for white in self.whites]
any_combs = [(color, self.dices[self.white1] + self.dices[self.white2]) for color in self.colors if self.dices[color] >= 1]
turn_combs = [(color, self.dices[color] + self.dices[white]) for (color, white) in combs if self.dices[color] >= 1]
return (any_combs, turn_combs)
class Game:
def __init__(self, bots):
self.bots = [bots[i](i) for i in range(len(bots))]
self.scoresheets = [Scoresheet() for i in self.bots]
self.dices = Dices()
self.turn_number = 2
def round(self):
self.dices.roll_dice()
(any_combs, turn_combs) = self.dices.combinations()
finished = False
for i, bot in enumerate(self.bots):
my_turn = self.whos_turn() == i
moves = bot.turn(self.scoresheets, self.dices, my_turn)
deleted_colors = self.scoresheets[i].mark(moves, my_turn, any_combs, turn_combs, self.dices)
if deleted_colors:
finished = finished or any([self.dices.delete_dice(deleted_color) for deleted_color in deleted_colors])
finished = finished or self.scoresheets[i].too_many_penalties()
self.turn_number += 1
return finished
def whos_turn(self):
return (self.turn_number - 1) % len(self.bots)
def runGame(self):
finished = False
while not finished:
finished = self.round()
#for i, scoresheet in enumerate(self.scoresheets):
#print self.bots[i], scoresheet.values, scoresheet.penalties, scoresheet.points()
class Scoresheet:
def __init__(self):
self.red = []
self.yellow = []
self.green = []
self.blue = []
self.penalties = 0
self.values = [self.red, self.yellow, self.green, self.blue]
def points(self):
points = self.penalties * -5
for color in self.values:
points += sum(range(len(color)+1))
return points
def allowed_combinations(self, my_turn, dices):
(any_combs, turn_combs) = dices.combinations()
any_combs = [(color, number) for (color, number) in any_combs if self.allowed(color, number, any_combs, dices)]
turn_combs = [(color, number) for (color, number) in turn_combs if self.allowed(color, number, turn_combs, dices)]
if not my_turn:
turn_combs = []
return (list(set(any_combs)), list(set(turn_combs)))
def allowed(self, color, number, combs, d):
if color not in d.deleted:
if (color, number) in combs:
if color == d.red or color == d.yellow:
if not self.values[color] or max(self.values[color]) < number:
if number < 12 or self.end_number(color, number, d):
return True
if color == d.green or color == d.blue:
if not self.values[color] or min(self.values[color]) > number:
if number > 2 or self.end_number(color, number, d):
return True
return False
def end_number(self, color, number, d):
if len(self.values[color]) > 5:
return (number == 2 and (color == d.green or color == d.blue)) or (number == 12 and (color == d.red or color == d.yellow))
return False
def mark_one(self, color, number, combs, deleted, d):
if self.allowed(color, number, combs, d):
self.values[color].append(number)
if self.end_number(color, number, d):
self.values[color].append(0)
deleted.append(color)
return deleted
def mark(self, moves, my_turn, any_combs, turn_combs, d):
beforePoints = self.points()
deleted = []
if moves[0]:
deleted = self.mark_one(moves[0][0], moves[0][1], any_combs, deleted, d)
if my_turn and moves[1]:
deleted = self.mark_one(moves[1][0], moves[1][1], turn_combs, deleted, d)
if my_turn and beforePoints == self.points():
self.penalties += 1
return deleted
def too_many_penalties(self):
return self.penalties >= 4
class Qwixx:
def __init__(self, bots, bots_per_game):
self.bots = bots
self.games = [0 for i in range(len(self.bots))]
self.points = [0 for i in range(len(self.bots))]
self.wins = [0 for i in range(len(self.bots))]
self.amountOfGames = len(bots) * 1000
self.bots_per_game = bots_per_game
def run_tournament(self):
for g in range(1, self.amountOfGames + 1):
bot_indices = numpy.random.choice(range(len(self.bots)), self.bots_per_game, replace=False)
bots_this_game = [self.bots[i] for i in bot_indices]
game = Game(bots_this_game)
game.runGame()
maxPoints = max([game.scoresheets[i].points() for i in range(len(bot_indices))])
for bot_index_game, bot_index_total in enumerate(bot_indices):
points = game.scoresheets[bot_index_game].points()
self.points[bot_index_total] += points
self.games[bot_index_total] += 1
if points == maxPoints:
self.wins[bot_index_total] += 1
print "GAME", g, "is", self.wins[bot_index_total], "th win by", self.bots[bot_index_total].__name__
def tournament(self):
self.run_tournament()
win_rates = {i: self.wins[i] / float(self.games[i]) for i in range(len(self.bots))}
format_result = '{:>30}: {:.4f} {:>6} {:>8} {:>8}'
format_header = '{:>30}: {:>6} {:>6} {:>8} {:>8}'
print(format_header.format('Name', 'Win %', 'Wins', 'Games', 'Points'))
for j, i in enumerate(sorted(win_rates, key=lambda i: win_rates[i], reverse=True)):
print(format_result.format(self.bots[i].__name__, win_rates[i], self.wins[i], self.games[i], self.points[i]))
### List of all bots ###
all_bots = [DoNothing, BadRandom, RandomAllowedCombinations, Random50, RandomButOnlyOwnTurn]
### Run tournament ###
Qwixx(all_bots, 5).tournament()
### Sandbox
• If someone can check my Python programming capabilities that would be great. I'm no native Python programmer.
• There are quite some rules. Maybe I should simplify it? These are the official rules, but for example I could change that even on your turn you can only move once instead of twice.
• This is my first challenge, so any feedback is welcome.
• Note that I have not yet tested my controller thoroughly, I will do this before posting it as an actual challenge.
• 1. Simplification of the rules is definitely needed. This post is hard to understand and even harder to come up with strategies about. Under "Rules", I'd immediately start explaining the actions bots can perform and their consequenses. 2. Why are there two types of turns (only-white turn, vs color turn)? How does this affect strategies? 3. What's the purpose of marking the last box to prevent other players? How does this affect strategies? 4. I don't understand the scoring, please expound. Oct 8, 2018 at 16:56
• 1. It's not entirely clear to me whether on my turn I can choose to make my "colour mark" before my "white mark" if they're on the same colour. 2. Dice is an irregular noun in English: the singular is die and the plural is dice. 3. 10000 * number of bots games is a lot, so it's going to be slow to run the tournament, but at the same time with 13 bots it's too few to run every possible combination once. This might need a more complicated tournament structure with elimination rounds. Oct 11, 2018 at 14:20
Roulette Bots II
Last week I posted Roulette Bots, which was a lot of fun. I think the basic format has potential to be even more interesting, so I've been shopping some tweaks in my head and I'd like to hear feedback before I give it another shot. In a nutshell, the first iteration of the game was as follows:
Everyone starts with 100 hp. Each round, 2 players are chosen at random from the pool of contestants who have not yet competed in that round. Both players pick a number between 0 and their current hp, and reveal those numbers at the same time. The player who chose the lower number immediately dies. The other player subtracts their chosen number from their remaining hp and goes on to the next round.
From the bracket of contestants, 2 are chosen at random. They face off, and one or both of them dies. A player dies if:
1. They choose a number smaller than that of their opponent
2. Their hp drops to or below zero
3. They tie three times in a row with their opponent
In the case of ties, both players simply generate new numbers, up to 3 times. After the faceoff, the survivor (if any) is moved to the pool for the next round, and the process repeats until we have exhausted the current round pool. If there is an odd number in the pool, then the odd one out moves on to the next round for free.
Players were given as input their own hp, and their opponent's complete betting history for that round. The idea being that players must balance resource consumption and long-term performance against the danger of dying right now if they don't bid high enough.
There are two things I'd like feedback on:
1. The minimal set of restrictions on user submissions required to make an interesting game without limiting creativity
2. Format tweaks that will allow for more creativity within the scope of the rules
The rule set I ended up on for the last game was that you were not allowed to take any action that unambiguously identified your opponent. This was for the purpose of avoiding things like players simply reading the stack to figure out exactly who their opponent was, since then you can simply simulate your opponent and more or less guarantee a win. While clever, it's doesn't really fit the spirit of the game. So the first question is whether or not there is a less restrictive rule that could be applied, or if someone can see a glaring loophole in that rule which would allow for game-breaking behavior in future versions of a similar game.
On the subject of format tweaks, the biggest issue I found with the game was that with a relatively small pool of contestants, very little history was actually generated and so bots that tried to use history to predict opponent guesses fared poorly overall. The first tweak I had in mind was to bootstrap a very large (256 or 512, maybe) initial tournament by random resampling of existing bots so that there would be many rounds and a good history pool to work with by the end of it.
Coupled with the larger pool, I intend to change the scoring system to reward longevity, to encourage more careful resource management, as compared to the previous game which rewarded just living through the first round. Something like making participating in the nth rounds worth n points, so that a player which survives to the nth round gets n(n-1)/2 points total. This puts more stringent constraints on resource management, since you need to survive a lot of rounds to do well. I'd probably bump the starting hp up to 1000, so that there is a bit more room to refine bets. and to make ties even less likely.
Second, there were a couple of attempts to make teams of bots that communicated via betting history in order to feed easy wins to one of them. I like the idea, but but because of the limited options presented by betting history they didn't fare very well. I am considering having bots exchange a "greeting" before each fight - a string or integer which they send to their opponent before battling, which could be used to identify teammates or clones of yourself. Does it add enough possibilities to gameplay to be worth the extra effort? Since with a resampled population you would likely be competing with yourself fairly regularly, having a way to recognize yourself (while at the same time avoiding others being able to recognize you from your greeting) seems like it could be interesting, but I worry that it will just degenerate into people updating their greeting at the last second to avoid other people using them, which would make it pointless to use the greeting to recognize opponents and make a lot of work for me. Can anyone suggest a good way to handle this? I want a way for people to recognize an opponent as probably (but not certainly) friendly without making it a game of "who updates last".
Feedback on the proposed gameplay tweaks, or ideas for others that I have not considered, would be appreciated!
• Please do not allow the "feed" strategy. We want all of the bots to be competitive, and the feeder bots aren't competitive, and creates an unfair advantage to individual bots (hmm...I wonder who will win 3v1). I also am not a fan of "send a message" actions: They either don't work or they make abuse possible. (The only exception is if you are making a team-based KotH) Scoring based on longevity is a good idea. If you want a longer history, you can also do a lives-based mechanic where a player has 3 lives. If they choose the lower number 3 times, then they are out of the tournament. Oct 10, 2018 at 14:55
• @NathanMerrill the idea for the feeding was that if I have to have multiple copies of each bot in the tournament to get a large pool, then it would be useful to recognize yourself. By default, you will always have "teammates" as long as resampling happens. But I understand the objection. How would lives work, practically? If a bot bids lower, how would I handle that? Ignore the result up to 3 times, and just let them through to the next round without cost to either player? Oct 10, 2018 at 15:32
• Oh, I like the idea of feeding copies of your bot. Perhaps giving them a single boolean variable that indicates if it's a clone or not (or something along those lines?). If both both Bot A and B have multiple lives, they would both have their HP removed, but the one that bid lower would have a life removed. Perhaps "life" is the wrong term here: Maybe they get a "strike", and 3 strikes and you are out. Oct 10, 2018 at 15:36
• Actually! If you have multiple clones of a bot, there's an easier solution than a 3-strike system: Simply give the opponent the entire history of all of the clones. Oct 10, 2018 at 15:39
• I think if I used the 3 strike system what would happen is that everyone would just bid 0 twice in a row to get into the later rounds with full hp and go from there. Realistically I can't stop the feeding strategy, since you can use betting history to set up a recognizable pattern, so I think it's better to build it into the game somehow. Simply telling people if it's a clone is an option, but then it would just advance one to the next round for free without generating a meaningful bid history data point, so why have the clones in the first place? The entire history of clones is a neat idea... Oct 10, 2018 at 15:49
• You tell people if they are playing against their own clone. There's not like an "original bot" or a "clone bot". You simply have a variable that is true if it is Bot 1 vs Bot 1 and false if it is Bot 1 vs Bot 2. You are right that there are no systematic ways to stop feeding, but we have rules against that on SE, and so you can use human judgement to disallow those bots. Oct 10, 2018 at 15:58
• I know what you mean. But if I tell them that, then they get to advance deterministically with a bet of 1, which is not a useful data point for anyone else predicting their behavior. I would like to find a solution that forces at least a meaningful bet (or that they have to take the risk of mistaken identity to make the 1-bet). If they know for certain they are facing a clone, then there is no value to having clones at all, since their purpose was to make a longer tournament with more history to use in predictions. But I realize as I say this that the "message passing" idea doesn't work either Oct 10, 2018 at 16:01
• Yeah. Perhaps you setup the tournament so that clones never actually fight each other. These battles aren't interesting, and only hurt the bot that randomly got assigned itself. The existence of clones would solely be to have a longer history. Oct 10, 2018 at 16:26
• The only really sound way to avoid the who updates last problem is to host it in such a way that people can't see their competitors' code - i.e. somewhere other than PPCG. On this site you'll always have people special-casing specific opponents. Oct 11, 2018 at 11:54
• @PeterTaylor probably true. Perhaps I could collect submissions by some other means and just ask players to put a placeholder answer which they can edit with code after the deadline. Would that be allowed? I'd have to keep the controller private, and maybe provide a suite of unit tests that entries should pass before submission. Oct 11, 2018 at 15:16
• Something vaguely similar has been done once (a Kolmogorov-complexity question where OP asked people to post a hash of their code, and then after the deadline to post their actual code, so that people couldn't just port someone else's answer to a golfier language), but it's a lot of hassle. Really when you're having to fight the site's design that much, you may as well not bother and just accept that some challenges don't work here. Oct 11, 2018 at 15:23
• Looking at what happened with your previous challenge, though, I would suggest requiring that answers be self-contained so that the deletion of a function defined in someone else's answer won't break them. Oct 11, 2018 at 15:24
• @PeterTaylor fair enough, that does seem like a lot of unnecessary work. I'll shop the format a bit more and see if I can come up with something a little simpler. Disallowing interaction between entries is certainly doable, but having people post lat-second updates will always be a problem, I think. Oct 11, 2018 at 15:24
# Is it a valid snooker break?
In snooker, the objective is to score as many points as possible by potting two types of balls: red balls, 15 of them, each valued 1 point, and colour balls, 6 of them, valued 2, 3, 4, 5, 6 and 7 points. A break is a sequence of potted balls. The balls must be potted in a predetermined order: the break must start with potting a red ball, then a colour ball and continued with alternating in potting red and colour balls until all of the red balls are potted, then potting colour balls in ascending order of their values (if all of the red balls have already been potted, start with the colour ball with the lowest value). However, if a free ball is declared after an opponent's foul, the break may start with any ball, replacing the ball that should be usually potted first (red if some red balls are still on the table, colour with the lowest value if all reds are potted).
Your task is to determine if the given sequence of balls adheres to the rules mentioned above.
## Input
A list of integers representing values of snooker balls. The list will not be empty.
## Output
A truthy value if the list is a valid snooker break and falsey if it isn't.
## Test cases
Truthy:
6
1,7
2,5
2,3,4
1,5,1,5,1
1,3,1,5,1,4
6,4,5
5,2,1,4,1,6,1,6
1,6,2,3,4,5,6,7
3,2,3,4,5,6,7
4,4,1
5,6,1,7
1,7,1,7,1,7,1,7,1,7,1,7,1,7,1,7,1,7,1,7,1,7,1,7,1,7,1,7,1,7,2,3,4,5,6,7 //"maximum break"
7,7,1,7,1,7,1,7,1,7,1,7,1,7,1,7,1,7,1,7,1,7,1,7,1,7,1,7,1,7,1,7,2,3,4,5,6,7
Falsey:
1,1
4,4,4
4,2,3,1
1,2,3
5,1,7,1
2,5,6,7 //2 is already off the table
5,4,2,3
1,7,1,7,1,7,1,7,1,7,1,7,1,7,1,7,1,7,1,7,1,7,1,7,1,7,1,7,1,7,1,7,2,3,4,5,6,7 //one red too many
7,7,1,7,1,7,1,7,1,7,1,7,1,7,1,7,1,7,1,7,1,7,1,7,1,7,1,7,1,7,1,7,1,7,2,3,4,5,6,7 //as well as here
# To sandbox:
I am thinking about including lists with values out of range of snooker balls' values (less than 1 or greater than 7) in which cases the result should be falsey. As this is a decision problem, I think that these cases should be covered as well.
• Maybe you can let the submissions assume that only a list of [1-7]s are inputted. This would be more flexible for the submissions to do. Oct 12, 2018 at 7:48
# Implement JavaScript's Abstract Equality Comparison Algorithm.
## Background
As many know, JavaScript has quite confusing equality rules and it's generally recommended to use === instead to avoid the weird corner cases that come with using ==.
The logic behind the == operator is documented as "The Abstract Equality Comparison Algorithm" in the ECMAScript specification. For this particular question we will be implementing from the ECMAScript 5.1 edition specification. The algorithm is documented in section 11.9.3 of this specification.
Implement the abstract equality comparison algorithm in JavaScript without using == or !=.
## Input & output
The input is two variables. These variables can be anything that is not a Reference type. The output is a truthy or falsy value that represents the result of performing the abstract equality comparison algorithm.
## Test Cases
Since there are a lot of edge cases, I took these test cases from the test262 repository.
Examples that return truthy values:
true == true
false == false
true == 1
false == "0"
0 == false
"1" == true
+0 == -0
-0 == +0
Number.POSITIVE_INFINITY == Number.POSITIVE_INFINITY
Number.NEGATIVE_INFINITY == Number.NEGATIVE_INFINITY
Number.POSITIVE_INFINITY == -Number.NEGATIVE_INFINITY
1.0 == 1
"" == ""
" " == " "
"string" == "string"
1 == "1"
1.100 == "+1.10"
255 == "0xff"
0 == ""
"-1" == -1
"-1.100" == -1.10
"5e-324" == 5e-324
undefined == undefined
void 0 == undefined
undefined == eval("var x")
undefined == null
null == void 0
null == null
{ var x, y; x = {}; y = x; x == y; } // two variables pointing to same object
new Boolean(true) == true
new Number(1) == true
new String("1") == true
true == new Boolean(true)
true == new Number(1)
true == new String("+1")
new Boolean(true) == 1
new Number(-1) == -1
new String("-1") == -1
1 == new Boolean(true)
-1 == new Number(-1)
-1 == new String("-1")
new Boolean(true) == "1"
new Number(-1) == "-1"
new String("x") == "x"
"1" == new Boolean(true)
"-1" == new Number(-1)
"x" == new String("x")
{valueOf: function() {return 1}} == true
{valueOf: function() {return 1}, toString: function() {return 0}} == 1
{valueOf: function() {return 1}, toString: function() {return {}}} == "+1"
{valueOf: function() {return "+1"}, toString: function() {throw "error"}} == true
{toString: function() {return "+1"}} == 1
{valueOf: function() {return {}}, toString: function() {return "+1"}} == "+1"
true == {valueOf: function() {return 1}}
1 == {valueOf: function() {return 1}, toString: function() {return 0}}
"+1" == {valueOf: function() {return 1}, toString: function() {return {}}}
true == {valueOf: function() {return "+1"}, toString: function() {throw "error"}}
1 == {toString: function() {return "+1"}}
"+1" == {valueOf: function() {return {}}, toString: function() {return "+1"}}
Examples that return falsy values:
true == false
false == true
Number.NaN == true
Number.NaN == 1
Number.NaN == Number.NaN
Number.NaN == Number.POSITIVE_INFINITY
Number.NaN == Number.NEGATIVE_INFINITY
Number.NaN == Number.MAX_VALUE
Number.NaN == Number.MIN_VALUE
Number.NaN == "string"
Number.NaN == new Object()
true == Number.NaN
-1 == Number.NaN
Number.POSITIVE_INFINITY == Number.NaN
Number.NEGATIVE_INFINITY == Number.NaN
Number.MAX_VALUE == Number.NaN
Number.MIN_VALUE == Number.NaN
"string" == Number.NaN
new Object() == Number.NaN
1 == 0.999999999999
" " == ""
" string" == "string "
"1.0" == "1"
"0xff" == "255"
1 == "true"
"false" == 0
undefined == true
undefined == 0
undefined == "undefined"
undefined == {}
null == false
null == 0
null == "null"
null == {}
false == undefined
Number.NaN == undefined
"undefined" == undefined
{} == undefined
false == null
0 == null
"null" == null
{} == null
new Boolean(true) == new Boolean(true)
new Number(1) == new Number(1)
new String("x") == new String("x")
new Object() == new Object()
new Boolean(true) == new Number(1)
new Number(1) == new String("1")
new String("1") == new Boolean(true)
{valueOf: function() {return {}}, toString: function() {return "+1"}} == "1"
"1" == {valueOf: function() {return {}}, toString: function() {return "+1"}}
Examples that throw an error:
{valueOf: function() {throw "error"}, toString: function() {return 1}} == 1 // throws "error"
{valueOf: function() {return {}}, toString: function() {return {}}} == 1 // throws TypeError
1 == {valueOf: function() {throw "error"}, toString: function() {return 1}} // throws "error"
1 == {valueOf: function() {return {}}, toString: function() {return {}}} // throws TypeError
## Rules
Standard rules apply. The shortest code in bytes wins.
## Sandbox Questions
• Is it ok that I am restricting the language to JavaScript? Should I allow languages that compile to JavaScript or have the types listed here?
• Have I got too many test cases here? Should I cut it down? Perhaps instead I could create a tio.run template which runs all the tests instead of listing them all out here.
• Is the requirement of not using == and != strict enough? I feel that I want to avoid answers where some inbuilt function indirectly calls == and getting around it that way but I don't know how to enforce that.
• I'd say the number of test cases is still acceptable, though creating a TIO template sounds like a good idea anyway. Oct 14, 2018 at 12:43
• We still can use === and !==, can't we? Oct 15, 2018 at 9:07
• @ShieruAsakoto yes Oct 15, 2018 at 9:48
• @CameronAavik Build a custom Javascript interpreter where calls to == errors out, then use that. Oct 15, 2018 at 10:18
• Here's a link to a test framework in TIO. I don't think it's possible to overload operators in javascript though
– Jo King Mod
Oct 17, 2018 at 9:06
• (for example when someone edits the interpreter to ban some features, see i.snag.gy/7ELPOt.jpg - the link "this branch" links to github.com/Mego/Seriously/tree/… --- of course that's a lot of work, but that would make sure that people definitely don't cheat) Oct 19, 2018 at 1:40
# Cologne Phonetics
From Wikipedia: Cologne phonetics is a phonetic algorithm which assigns to words a sequence of digits, the phonetic code. The aim of this procedure is that identical sounding words have the same code assigned to them. The algorithm can be used to perform a similarity search between words. For example, it is possible in a name [collection?] to find entries like "Meier" under different spellings such as "Maier", "Mayer" or "Mayr".
The Cologne phonetics matches each letter of a word to a digit between 0 and 8. To select the appropriate digit, at most one adjacent letter is used as context.
### Procedure:
Processing of a word is done in three steps:
1. Encode letter by letter from left to right according to the conversion rules below.
2. Remove all digits occurring more than once next to each other.
3. Remove all code "0" except at the beginning
### Conversion rules
• A, E, I, J, O, U, Y become 0
• H becomes nothing
• B becomes 1
• P becomes 1, unless it's before H
• D, T become 2, unless they're before C, S, Z
• F, V, W become 3
• P becomes 3, if it's before H
• G, K, Q become 4
• C becomes 4 in initial position before A, H, K, L, O, Q, R, U, X
• C becomes 4, if it's before A, H, K, O, Q, U, X, except after S, Z
• X becomes 48, unless it's after C, K, Q
• L becomes 5
• M, N become 6
• R becomes 7
• S, Z become 8
• C becomes 8, if it's after S, Z
• C becomes 8 in initial position except before A, H, K, L, O, Q, R, U, X
• C becomes 8, if it's not before A, H, K, O, Q, U, X
• D, T become 8, if they're before C, S, Z
• X becomes 8, if it's after C, K, Q
## Challenge
Implement a program or a function that takes a single string as input and returns the phonetic code according to the rules stated above. You may choose your own output format, but keep in mind that the expected output may have leading zeros. The input string is guaranteed to match /^[a-zA-Z ]*\$/, this means it may contain characters from the lowercase and uppercase alphabet, as well as spaces.
## Test cases
Input Output
[empty string] [empty]
Kolner Phonetik 45673624
Meier 67
Mayer 67
Augsburg 048174
Xanten 48626
Telephon 2356
Chemie 46
Muller Ludenscheid 65752682
This is , the shortest code for each language wins.
• Related: Soundex function Oct 14, 2018 at 12:49
• What's the difference between "initial sound" and "initial position"? Oct 14, 2018 at 12:58
• Apparently there is none, from what I was able to find Oct 15, 2018 at 7:11
• Ideally there would be a test case for each conversion rule as the challenge is probably mainly about handling those edge cases in the rules. Oct 15, 2018 at 13:27
Keyboard Row Shift
Given an input string of characters, output the number of times we have to shift to another row while typing that string using a qwerty keyboard.
The input string will contain lower case alphabets,numbers and spaces. The newline key(enter) must also be considered at the end (only at the end). All shifts, from any row to any other will be considered as 1 shift only.
Layout:
1234567890
qwertyuiop
asdfghjkl enter
zxcvbnm
space
Examples:
"sdkflsd" -> 0
"asdwexc" -> 3 # to end the string one enter has to pressed which is in the middle
"poierlkdjfpoeirldskjf" ->3
"123 jkjk" -> 2
"llsdkfj ldkfj" -> 2
"lkasdjmnbcv " -> 3
"jnjn 5" -> 6
This is code-golf, so shortest code wins.
• You need some test cases that contain a space and numbers. I'd also make it clear in your description that a shift can go any distance (going from the top row to the bottom is only 1 shift) Oct 16, 2018 at 12:48
• Done, @NathanMerrill Oct 17, 2018 at 5:36
• How flexible is input? Is uppercase OK? How about an array of character strings or even an array that mixes digits and character strings? Oct 18, 2018 at 21:47
• I have mentioned only lower case, input has to be string only not an array. Oct 19, 2018 at 5:15
Given a program in your language, generate another program that do exactly the same thing so every bytes in it are prime.
Shortest generator in every language win. No acception.
Nop in languages that only allow prime bytes are legit, but just don't post them(or make a community answer to put them)
• Upvote this comment if you think that the idea of the challenge is not interesting and it should not be posted. Oct 26, 2018 at 10:46
• Upvote this comment if you think that the idea is fine, but the challenge is unclear or needs fixing; in that case also leave a comment. Oct 26, 2018 at 10:47
• I made a small poll above ↑. I think the idea is interesting, and if it's poorly worded, the post can't be fixed without suggestions. Note that the challenge is already fully specified in the current state (task description, winning criteria), and it's impossible to give sample input/output to test programs (if you understand the challenge, you will see why). Oct 26, 2018 at 10:49
• Lots of languages quite rely on composite(not prime) bytes, so upvote this comment if you think it's too restrict
– l4m2
Oct 26, 2018 at 12:40
• I think the idea is interesting, especially in languages like Jelly which can do most anything with small subsets of its character set. My main problem is: how much of the language do we have to support in the input to our generator? Many languages which are capable of this task have far too many possible instructions, etc., every one of which would need to be remapped to an all-prime version. Unless perhaps it's possible to simply generate any given string and execute it whilst using only prime bytes... Oct 27, 2018 at 3:14
• @ETHproductions I think a subset of an existing language can be considered another language? Nov 11, 2018 at 4:23
# Are these strings character wise translatable? code-golfstringdecision-problem
### Explanation with example:
Suppose you are given two strings of the same length, e.g. "code" and "golf". Try to find a set of translation rules for characters (like "All cs are replaced by gs" , or c -> g for short) such that applying those rules on "code" yields "golf" and applying them to "golf" again yields "code".
If there is such a set of translation rules, return a truthy value. Otherwise, if no such set can exist, return a falsy value.
In our example, the translation is possible with the rules g -> c, d -> l, e -> f, c -> g, l -> d and f -> e, so a truthy value should be returned.
However, if the two strings were "code" and "meta", no such rules exist: To get from "code" to "meta" we need the rule e -> a, but to get from "meta" to "code" we would need e -> o. But e cannot be translated into two different characters! Therefore, a falsy value should be returned.
### Consise mathy explanation:
Given two strings $$\s\$$ and $$\t\$$ of the same length $$\n\$$, decide whether the following relation is bijective: $$\{(s_i,t_i)\ |\ 1 \leqslant i \leqslant n\} \cup \{(t_i,s_i)\ |\ 1 \leqslant i \leqslant n\}$$ where $$\s_i\$$ denotes the $$\i^{th}\$$ character of string $$\s\$$.
### Test cases
truthy:
"code", "golf"
"a", "a"
falsy:
"code", "meta"
"aa", "ab"
"abc", "bca"
• This seems to be very similar to Check if words are isomorphs. Oct 28, 2018 at 2:39
• @Dennis It is definitely related. However, as far as I see most answers there work by creating some sort of fingerprint and checking whether it is the same for both inputs. This approach alone won't work here as you also need to check the symmetry of substitutions. Oct 28, 2018 at 11:33
• Not saying this makes it a dupe, because there might be golfier ways, but s and t are character wise translatable if and only if s++t and t++s are isomorphs. Oct 28, 2018 at 12:54
## Guess the Q in the code
A popular puzzle known to me as Codewords takes a Crossword-like grid and consistently replaces each letter with a code value, usually a number from 1 to 26, and the solver then has to work out which code value represents each letter. There are a number of approaches to this:
• There is a web site which can search for words given the pattern of repeating letters. Since repeated letters map to repeated code values, you can sometimes use the patten of repeated code values to determine the original word. For instance, the only word with the pattern 12.312..3... is churchwarden.
• The puzzle normally provides a few of the letters to get you started. This may be enough to use a standard crossword solver to determine the original word. For instance, .h...h...d.. would again point you to the word churchwarden.
• You could use frequency analysis to guess which letters are likely to be the popular letters such as e or t.
• You could look for common prefixes or suffixes such as ally or ness. (I picked those two as they have repeating letters which are easier to spot.)
• At least in all of the Codewords puzzles I have seen, q is always followed by a u and at least one letter. This means that if there is a code that is never the last or penultimate letter of a word, and always appears followed by the same code, then this could be a q.
However, checking all the codes to see how many distinct codes follow them is laborious. We need an automatic solution for this!
Please write a program or function that will guess which code(s) could represent the letter q. The input to the function will be an array of code values in a standard format. You will need to support 27 consistent distinct values, 26 for the codes themselves and a value for the background. These values can be integers but you could also use characters e.g. 26 letters and a space, in which case you can join them into strings, and even join the strings with a 28th character.
Your output will be all of the code values that have the property that, for each occurrence in the grid, either:
• The previous and following cells are background or would extend past the side of the grid, or
• There are two non-background following cells, and the first of those cells always contains the same value.
These rules apply in both the horizontal and vertical direction.
This is , so the shortest solution that breaks no standard loopholes wins!
Examples needed, I guess...
# Hack g-code parser
• This could be seen as asking for malicious code, which is not on-topic. Oct 25, 2018 at 17:55
• @Laikoni this is not malicious\harmful code, because it don't causes harm to user. It is educational programming puzzle for white-hat hackers Oct 25, 2018 at 18:58
• @Laikoni rm -rf / is harmful, but alert("pwned") is not Oct 25, 2018 at 18:59
• Right, I think misread the challenge. If it is only about cracking this piece of code rather than actual implementations it's probably fine as a challenge. Any way, you need an objective winning criterion. code-golf in the sense of "the shortest input to open the alert prompt" seems like a good candidate. Oct 25, 2018 at 19:47
# Find the pattern v2
I feel tired to do "find the pattern" exercise such as
1 2 3 4 (5)
1 2 4 8 (16)
1 2 3 5 8 (13)
Please write a program that finds the pattern for me.
Here, we define the pattern as a recurrence relation that fits the given input, with the smallest score. If there are multiple answers with the same smallest score, using any one is fine.
Let the $$\k\$$ first terms be initial terms for the recurrence relation, and the $$\i\$$'th term be $$\f(i)\$$ ($$\i>k,i\in\mathbb N\$$).
• A non-negative integer $$\x\$$ adds$$\1\$$ to the score
• The current index $$\i\$$ adds $$\1\$$ to the score
• +, -, *, / (round up, down, towards zero, go further to zero, as you decide) and mod (a mod b always equal to a-b*(a/b)) each add $$\1\$$ to the score
• For each initial term $$\x\$$, add $$\1\$$ to the score
• $$\f(i-n)\$$ (with $$\n\le k\$$) adds $$\1\$$ to the score. E.g. Using the latest value $$\f(i-1)\$$ add $$\1\$$ to the score, and there must be at least 1 initial term.
• Using parentheses to change the calculation order doesn't add anything to the score.
Samples:
input -> [score] expression(not optimized)
1 2 3 4 -> [1] f(i) = i
1 2 4 8 -> [3] f(1) = 1, f(i) = 2*f(i-1)
1 2 3 5 8 -> [5] f(1) = 1, f(2) = 2, f(i) = f(i-1)+f(i-2)
Lowest score for worse case wins. If tie, shortest program in each language wins. Your program should run in polynomial time.
If someone's score is lower than 4(currently reachable), the first lowest score will be accepted. (differ from winning criticia)
p.s. See history for v1 sandbox
• I think the challenge could benefit from a precise defining grammar for a pattern. As it stands, I feel the set of all expressions that may fit is too vast. Oct 16, 2018 at 9:28
• @JonathanFrech Looks pretty precise to me. About "too vast" - there are only 6 bullets. Oct 17, 2018 at 10:43
• @user202729 I am unsure if (round as you decide) includes the possibility of not rounding. Oct 17, 2018 at 15:28
• Should "3 0 0 0 0 0 ..." be "f(1)=3, f(n)=0", do "for the recurrence relation" work?
– l4m2
Oct 17, 2018 at 16:40
• @l4m2 Yes. ----- Oct 17, 2018 at 16:56
• I don't understand the "fun fact"... Oct 21, 2018 at 15:42
• @user202729 Do you think it no fun or can't figure it out?
– l4m2
Oct 21, 2018 at 19:24
• The latter. ---- Oct 22, 2018 at 0:15
# User ranking in language
Challenge is:
• You need receive the input of "language". By "language" I mean the at left of the comma in the Title of an answer on the present code-golf site.
• The source for the data are the answers posted here on code-golf site.
• You must present results in order. Decreasing or Increasing does not matter, it only needs to be one of those two.
• Output is a table of two columns {User, Upvotes}. Example: User enters Java and will obtain a list as the following:
user_A 9523 user_B 6000 user_C 120
Do not care very much about the output formatting as long it is clear the separation between the two columns and each line.
The will be no winner as it is a per language question.
• "The source for the data are the answers posted here on code-golf site." The answers for this challenge, or ALL challenges on PPCG? And with your example, does this mean it searches for each user all answers given in Java, and sums their total upvote count? Also, do we differentiate java/Java/Java 7/Java 10 etc as different inputs? Nov 1, 2018 at 12:45
• @KevinCruijssen 1. Yes, the scope is all answers on PPCG. 2. Yes, it is the total upvote account. 3. This is flexible on uppercase/lowercase, but not on the rest. Nov 8, 2018 at 19:58
Concentration (or Memory or Match) is a game where players pick pairs of cards and try to find matches. The rules are as follows:
1. Two of each number from 1 to N are added to the deck.
2. The deck is shuffled and placed face-down (hidden).
3. The player selects a card. The card is revealed.
4. The player selects another card: The card is revealed.
5. If the two cards are different numbers, then they are hidden again.
6. Steps 3 through 5 are repeated until all cards are revealed.
For example, lets say I have the following cards: 1 1 2 2 3 3 4 4
Shuffle them: 1 4 3 1 4 3 2 2
Place them face down: X X X X X X X X
The player will then select a card: 1 X X X X X X X
And then another card: 1 X 3 X X X X X
The numbers don't match, so we put the cards face down, and select another card: X X X 1 X X X X
The player remembers the position of the other 1: 1 X X 1 X X X X
A match was found! These cards stay face-up, and we try again.
The challenge is to do this with limited memory. You need to write a function/program that takes two parameters:
• A list of cards (with a constant value representing the hidden cards)
• 32 bytes of data. This can be stored as a string, a large integer, etc.
Your function must use this data (and only this data) to select the next card to reveal, and then update the 32 bytes of data. You may use a random number generator for free (as long as you do not use the seed to store data)
After the game is over, the game is scored by counting the number of times a card was revealed.
Below, I have provided a list of test cases. Your score is the length of the first test case you complete with a score of greater than 2N+30 (N is the number of distinct cards). Hard coding is not allowed, and your function should get a similar score with a different set of test cases.
2 1 4 0 0 3 3 4 2 1
3 0 4 3 2 0 4 1 1 2
1 3 2 2 4 3 0 4 0 1
2 2 0 4 4 1 1 3 3 0
3 2 1 1 0 0 4 3 4 2
3 2 0 4 2 5 1 1 0 3 5 4
4 5 0 3 2 5 3 1 0 2 1 4
4 1 2 0 2 5 1 3 3 5 0 4
2 3 2 4 3 1 0 5 4 0 1 5
3 1 4 0 2 2 5 3 1 5 4 0
1 4 0 5 2 3 6 1 0 3 5 4 6 2
6 3 5 3 1 2 1 5 2 4 4 6 0 0
5 6 0 0 5 3 1 1 2 4 6 2 4 3
1 1 6 2 2 4 0 4 5 5 6 0 3 3
3 4 6 4 6 2 5 0 0 1 3 5 2 1
3 2 6 4 1 5 7 4 3 7 0 2 6 0 1 5
7 3 1 0 6 2 3 6 4 1 5 7 5 0 4 2
2 6 5 4 2 5 0 0 7 3 1 4 6 7 3 1
4 7 5 0 6 3 0 1 3 1 7 2 6 4 2 5
4 7 5 1 4 3 2 6 0 6 7 3 2 5 1 0
7 6 5 4 5 0 1 3 8 3 4 0 1 8 6 2 2 7
2 4 1 8 7 4 6 5 3 8 1 3 7 2 0 6 5 0
6 4 7 8 5 1 4 3 0 8 3 5 2 6 1 2 7 0
3 8 0 6 4 0 1 5 2 7 7 1 5 3 2 6 8 4
7 1 8 8 3 1 3 4 5 4 6 2 2 0 0 7 6 5
9 4 1 5 7 3 4 1 6 3 9 7 2 2 0 5 6 0 8 8
2 2 6 1 8 6 5 7 3 4 5 0 9 3 0 9 7 4 8 1
7 4 3 5 2 7 1 1 0 0 9 8 6 8 3 6 2 9 4 5
1 6 3 2 7 0 7 9 1 5 0 4 8 6 3 8 9 5 2 4
8 4 5 9 9 3 0 5 6 2 2 7 0 4 1 8 7 1 3 6
1 7 0 2 1 3 8 8 6 9 2 3 4 5 6 5 10 4 7 10 0 9
3 5 10 3 5 1 9 10 1 2 0 4 0 8 6 7 8 4 9 2 6 7
9 1 7 8 2 8 9 10 2 0 6 6 5 3 5 1 10 0 4 7 3 4
7 8 2 9 8 0 0 9 3 1 5 2 4 1 3 10 4 10 5 6 6 7
3 7 9 5 1 2 8 6 3 6 4 0 10 1 2 7 10 9 8 4 5 0
7 6 9 6 2 5 5 0 11 9 3 10 8 3 4 8 4 2 11 1 1 10 7 0
10 2 8 5 11 3 4 2 6 11 4 7 10 9 8 9 1 1 6 0 5 7 3 0
4 0 7 7 4 6 11 8 3 9 3 2 5 1 11 10 8 2 9 5 0 10 1 6
10 8 8 0 0 3 7 4 11 10 9 2 4 5 6 9 5 6 7 1 1 11 3 2
1 7 0 8 9 10 4 5 10 6 5 1 4 3 7 2 0 9 3 11 8 2 6 11
3 7 10 11 1 5 10 2 9 3 8 11 5 9 6 0 0 4 12 12 4 8 1 6 2 7
0 11 9 11 2 9 10 6 5 2 7 0 12 1 10 7 6 8 3 3 4 1 12 5 8 4
3 9 5 12 1 6 8 10 2 4 11 3 6 8 9 1 10 12 7 0 11 0 5 4 7 2
1 7 11 9 0 2 6 6 5 0 8 3 10 8 7 12 3 9 12 4 2 4 11 1 5 10
8 4 9 0 7 3 1 3 9 6 11 6 8 2 12 10 2 12 5 5 0 1 10 7 4 11
13 4 13 4 1 2 9 8 10 7 3 12 2 0 5 6 0 5 8 6 9 12 11 10 1 7 3 11
13 0 0 6 1 4 11 3 12 9 2 10 9 2 7 8 12 5 7 6 11 4 10 1 5 3 13 8
13 10 3 6 12 1 0 8 5 6 3 11 4 9 2 9 8 11 12 2 5 7 7 1 4 10 0 13
7 4 11 8 10 1 6 0 3 10 6 0 8 2 11 5 3 1 7 13 9 13 12 9 4 2 5 12
13 5 7 6 1 12 8 2 8 0 5 7 10 6 4 11 12 1 2 10 3 13 11 3 0 9 9 4
9 2 4 3 7 4 2 7 10 5 13 6 5 12 8 11 0 8 14 11 0 6 1 9 3 13 14 1 10 12
8 14 8 7 11 7 2 12 4 5 13 6 1 3 5 0 1 4 9 0 10 2 11 12 13 6 3 14 10 9
4 9 5 1 5 10 2 11 2 9 13 10 3 13 12 8 7 1 6 8 14 0 4 12 14 11 3 7 6 0
5 10 4 3 2 4 2 3 9 13 14 6 0 11 5 13 12 8 0 9 7 8 6 11 14 1 12 10 1 7
11 1 10 8 0 12 9 1 7 4 4 12 6 8 14 5 10 5 14 0 11 3 7 9 2 2 3 13 6 13
0 8 15 6 4 1 9 8 14 13 2 11 4 6 9 14 0 2 3 5 1 15 5 10 13 12 3 7 12 7 10 11
13 3 4 13 2 12 9 1 5 1 6 7 11 9 10 0 15 14 12 15 14 8 7 3 4 5 8 11 0 2 10 6
6 1 7 9 3 1 15 7 5 14 9 14 0 8 10 0 4 12 12 13 3 6 11 4 15 2 11 5 2 13 8 10
2 0 14 7 13 13 15 7 3 8 15 5 10 12 4 5 3 14 9 6 12 10 9 11 8 6 1 1 4 11 2 0
3 13 12 4 10 13 8 0 7 11 11 6 8 1 12 5 5 9 6 10 14 3 2 9 14 4 1 15 15 7 0 2
16 12 7 14 13 5 5 4 11 0 0 9 12 15 15 9 14 2 6 3 11 8 16 10 6 13 3 1 2 8 1 7 10 4
5 5 10 4 9 15 0 11 11 1 13 15 9 0 3 1 16 6 8 7 2 6 8 2 14 14 16 4 3 12 12 13 10 7
0 6 7 4 12 16 0 3 6 1 7 12 8 13 15 9 2 11 2 15 11 10 5 13 14 5 9 14 3 8 4 16 1 10
8 0 1 2 16 11 3 2 15 4 15 5 14 9 13 7 14 9 10 12 11 12 16 6 13 6 1 5 10 0 7 8 4 3
9 16 15 13 11 2 5 6 16 3 9 4 10 14 14 12 8 11 0 6 0 15 10 7 13 5 1 12 3 4 2 1 7 8
11 15 14 3 8 12 17 15 1 0 12 6 9 0 9 13 11 14 16 2 4 5 13 16 10 7 8 5 10 4 17 3 6 7 1 2
11 14 5 7 14 0 12 4 8 12 11 13 16 16 15 6 3 1 8 3 1 5 7 17 10 2 17 13 15 9 9 4 2 0 6 10
15 13 6 12 5 4 9 2 11 2 13 16 3 9 1 16 10 11 1 12 14 17 8 0 17 6 8 10 5 0 7 4 14 7 3 15
0 9 13 15 8 11 6 5 9 16 11 7 14 7 5 15 3 12 4 16 2 14 3 0 1 10 13 1 6 10 17 2 17 8 4 12
10 13 14 10 5 7 16 8 8 7 6 3 0 2 17 17 2 1 15 16 3 4 5 0 12 9 1 12 14 11 13 9 15 6 4 11
7 14 0 13 15 6 12 1 14 16 10 15 8 11 0 9 2 3 4 18 12 17 8 13 9 6 4 11 5 1 5 16 17 3 18 10 7 2
12 9 17 2 7 12 4 15 13 0 17 9 11 3 6 4 8 5 11 16 8 1 7 0 13 15 16 10 3 14 18 2 14 1 10 6 18 5
4 0 5 14 7 8 1 0 12 15 11 2 11 5 3 9 15 17 6 8 7 10 12 3 6 16 2 18 16 13 4 14 9 17 13 18 1 10
13 13 18 9 0 17 12 17 8 0 14 16 14 6 10 6 9 15 1 15 2 4 11 5 1 3 4 3 11 2 5 16 8 7 10 7 18 12
18 11 13 6 14 9 13 14 9 18 17 11 0 1 17 2 0 6 10 3 8 4 1 5 15 7 5 2 12 15 4 8 7 16 10 12 3 16
0 17 5 6 19 0 11 1 7 18 9 14 16 2 16 13 1 4 3 17 15 7 10 2 15 9 18 13 8 14 6 12 5 10 3 8 11 19 12 4
4 3 10 12 2 7 9 11 0 16 16 7 19 2 13 3 12 10 5 18 4 15 15 0 11 19 18 13 14 1 5 17 6 9 14 8 6 17 8 1
0 17 16 11 7 19 8 5 17 16 14 1 10 6 11 7 15 10 14 13 15 4 2 4 1 9 6 3 13 18 18 0 19 3 9 12 8 5 2 12
10 15 10 0 2 14 3 19 1 18 4 14 8 2 5 17 4 11 17 12 13 16 9 3 0 13 7 5 18 6 8 9 12 1 19 16 11 7 6 15
15 2 16 4 14 0 1 5 12 18 10 12 4 14 16 11 8 7 13 15 11 3 19 10 3 9 17 6 13 19 9 1 0 18 8 2 7 5 17 6
1 12 17 7 8 8 12 20 5 10 20 16 13 3 18 15 5 4 10 9 6 14 11 6 7 15 16 19 0 1 9 18 19 2 2 3 13 4 17 0 11 14
7 17 2 12 19 9 0 11 17 4 12 0 10 11 16 14 3 20 7 15 2 9 8 18 16 19 5 3 8 14 6 18 1 13 5 1 4 15 13 10 6 20
0 18 12 4 20 7 14 5 10 16 1 13 9 15 12 0 6 19 2 5 16 10 1 17 15 11 20 3 14 2 18 3 17 11 7 8 4 8 6 19 13 9
9 18 15 0 16 7 19 4 13 20 5 10 2 2 0 12 8 6 3 15 16 14 17 7 10 3 13 6 1 20 1 17 19 11 5 4 12 14 11 8 9 18
13 6 16 11 20 3 10 8 5 7 17 15 14 0 7 6 14 17 4 15 2 12 8 1 13 18 11 10 1 9 20 16 5 19 18 2 12 19 3 9 4 0
4 5 16 11 2 1 14 9 16 6 20 21 0 13 12 13 7 8 15 11 10 18 14 3 15 5 21 18 12 17 19 9 0 19 7 4 3 20 1 8 2 10 6 17
10 6 7 0 18 1 19 5 8 12 15 2 5 4 11 20 16 17 13 14 17 1 16 9 21 21 0 3 6 3 2 7 8 13 19 20 14 15 9 11 10 4 18 12
2 20 12 21 10 4 8 11 13 3 20 1 6 21 12 13 4 17 19 16 17 15 14 9 16 9 5 10 8 6 11 18 1 5 2 18 0 0 3 7 19 14 15 7
16 0 17 8 11 6 3 18 9 16 14 20 12 12 10 1 14 1 10 21 4 19 0 2 13 3 15 17 21 9 6 18 2 20 4 11 7 19 5 8 7 5 13 15
3 4 4 10 0 3 16 14 19 8 17 9 5 9 6 12 7 13 13 1 15 0 5 20 2 7 1 17 21 11 11 18 20 6 15 16 2 12 14 19 8 10 18 21
12 11 7 8 8 10 9 18 5 6 15 12 10 0 14 3 6 22 17 15 17 21 14 16 9 4 18 13 1 11 4 20 3 19 22 1 13 7 21 5 0 20 2 2 16 19
2 20 10 17 6 0 21 10 19 4 12 8 0 7 18 15 17 18 9 4 2 14 1 5 14 6 16 22 13 11 11 15 21 3 7 19 8 9 22 20 3 16 13 1 12 5
12 11 6 16 18 15 12 11 10 9 8 15 22 5 21 18 20 0 2 19 17 8 10 5 3 14 9 22 20 3 1 21 19 6 0 16 1 13 4 14 2 13 17 4 7 7
20 5 16 4 17 7 8 15 0 10 9 3 16 13 8 12 1 14 17 11 18 11 2 7 6 1 4 20 3 5 22 22 2 6 13 10 19 15 12 14 18 9 0 19 21 21
11 1 0 5 7 15 0 12 3 9 6 15 7 13 11 20 9 17 8 4 22 4 10 5 21 18 16 14 13 10 18 12 21 17 16 2 6 22 3 14 19 1 8 19 20 2
10 10 18 6 4 1 9 6 15 7 7 11 21 13 15 21 22 16 12 20 2 18 3 17 19 14 14 0 16 5 3 4 1 9 13 11 22 20 19 8 5 0 23 17 23 2 8 12
3 3 19 13 5 10 17 18 4 16 6 2 21 4 12 7 13 8 7 10 6 2 1 15 5 21 23 22 0 11 9 16 20 11 20 23 1 14 18 8 14 19 9 17 0 15 22 12
14 7 15 20 19 7 13 0 20 2 18 1 11 21 17 12 16 10 8 3 15 2 22 23 18 16 9 1 3 13 10 19 8 11 6 4 5 22 0 17 21 23 14 4 12 9 6 5
1 12 15 21 11 21 18 6 4 22 5 23 14 19 13 8 15 7 5 0 8 13 16 4 1 6 12 11 14 18 20 17 9 23 19 3 2 17 9 0 16 7 3 20 10 22 2 10
23 9 15 10 13 17 17 21 12 18 2 20 3 8 0 16 2 12 6 20 7 19 16 0 8 9 4 6 5 13 18 14 1 19 21 3 22 15 1 4 22 10 5 7 11 14 11 23
3 21 22 5 19 7 2 24 18 19 9 2 24 14 6 0 12 0 21 18 23 15 8 11 6 4 8 17 16 20 14 16 4 7 10 11 1 9 17 3 23 10 20 5 13 22 13 1 12 15
13 6 5 22 1 17 2 21 5 0 14 24 19 21 7 19 6 23 12 14 4 9 3 22 15 9 2 16 20 7 13 12 8 23 18 0 15 10 8 24 20 16 3 1 4 11 10 18 17 11
12 6 19 0 9 14 22 2 13 20 19 9 6 3 21 20 24 0 4 24 1 5 11 15 1 21 8 5 16 16 23 18 7 11 10 17 15 12 3 22 7 10 2 23 17 8 14 4 13 18
6 22 2 18 16 19 12 0 17 14 23 4 6 17 3 10 4 22 13 9 20 19 0 9 8 1 24 2 23 14 11 5 7 16 13 10 12 8 7 3 20 21 15 5 18 21 11 1 15 24
18 12 6 22 19 19 9 24 5 14 13 2 22 8 4 16 3 9 18 8 0 10 23 10 2 20 21 5 23 20 1 11 0 17 7 7 17 12 15 3 16 15 13 24 14 4 6 21 11 1
7 14 19 5 4 17 1 22 23 24 12 1 20 6 24 7 21 10 8 11 20 3 25 13 0 2 16 21 0 6 9 11 4 18 22 2 15 5 18 8 13 12 3 14 23 10 17 9 16 15 25 19
10 4 1 13 11 24 7 12 15 20 11 0 23 23 5 7 25 25 17 8 20 24 2 22 19 9 12 16 4 6 21 9 17 6 5 15 10 0 18 8 22 2 14 16 3 19 21 18 1 3 13 14
16 22 3 2 25 21 11 1 10 12 4 15 0 2 13 11 3 14 19 1 18 12 8 23 18 22 9 16 4 0 20 9 17 24 7 10 6 8 23 25 15 7 21 20 19 5 14 13 6 5 17 24
21 20 8 16 1 18 4 0 14 5 22 3 0 23 22 1 9 4 24 7 21 2 23 15 11 12 14 25 8 12 7 15 13 18 3 19 10 2 5 10 16 24 19 13 6 20 6 9 17 11 17 25
6 24 17 18 24 22 10 11 20 16 16 3 4 21 23 25 14 1 0 1 7 2 5 13 6 4 10 21 17 14 23 11 9 9 15 25 0 19 19 12 20 22 12 13 15 2 3 8 5 18 8 7
11 16 25 6 19 5 26 17 24 6 12 4 15 18 21 7 10 5 23 3 13 8 13 23 20 8 11 12 9 22 1 14 21 2 25 10 15 9 0 2 0 26 3 24 17 22 14 4 18 16 20 19 1 7
26 0 19 3 0 20 1 25 13 3 12 6 20 11 21 13 2 11 23 15 17 15 24 23 10 8 26 5 18 9 16 18 4 9 19 10 14 7 22 17 22 4 16 8 24 2 25 5 14 12 6 21 1 7
9 25 13 14 19 17 20 2 11 5 19 23 25 21 0 16 8 15 7 16 24 24 3 15 20 10 8 4 0 12 13 22 9 4 26 18 22 2 23 18 3 6 6 12 17 26 21 10 1 11 14 7 5 1
17 4 20 23 13 0 3 26 25 16 3 2 13 10 8 12 9 5 14 2 26 25 21 23 22 17 7 14 24 1 4 15 19 16 12 5 9 22 1 20 19 7 11 6 21 15 11 18 6 8 18 0 10 24
14 12 3 17 7 2 19 11 0 21 10 24 4 6 8 8 25 4 6 23 9 1 25 15 12 9 23 10 2 26 1 0 11 7 19 13 18 3 17 13 16 26 20 20 16 15 22 22 18 14 5 24 21 5
4 15 22 10 27 14 7 0 2 17 3 6 20 18 0 16 21 4 3 11 11 12 23 5 15 22 9 17 2 1 10 18 25 14 27 19 16 23 20 21 6 25 26 24 13 1 24 26 9 13 8 19 8 7 5 12
3 17 8 9 3 9 23 15 10 25 0 6 26 21 11 12 27 1 11 13 14 2 6 1 20 7 19 24 22 14 22 25 16 17 26 5 24 27 12 13 18 4 23 20 19 4 7 15 21 8 5 16 10 2 0 18
9 11 14 21 3 12 10 4 14 1 5 18 27 9 6 0 3 19 7 12 17 26 26 25 15 11 16 24 17 18 20 8 7 19 15 1 4 13 22 23 8 25 20 16 22 2 6 0 10 5 2 24 23 21 13 27
7 23 2 26 5 24 19 17 2 18 11 7 16 5 8 26 24 21 8 14 10 16 18 9 27 1 4 22 15 25 17 4 0 0 21 3 15 12 11 23 19 27 10 13 6 22 3 14 6 13 12 9 20 1 20 25
17 4 24 21 11 22 12 13 0 1 17 23 15 4 6 2 27 14 3 16 8 13 3 5 2 6 0 9 23 14 25 26 10 18 19 26 8 11 10 20 21 18 22 20 25 9 7 12 1 7 19 16 5 27 15 24
8 27 28 24 9 9 16 22 13 20 23 25 13 21 0 26 21 8 16 24 27 18 26 23 3 7 3 25 22 15 14 6 5 7 11 4 1 20 17 19 17 12 4 14 2 1 5 18 2 28 12 0 10 19 15 11 6 10
21 9 3 23 27 4 15 22 12 2 16 26 4 13 28 16 20 6 6 3 10 1 18 5 11 0 21 7 8 25 23 17 9 15 22 14 24 2 5 10 12 17 20 1 24 7 18 0 14 26 19 13 28 25 11 8 27 19
26 10 19 20 17 10 9 19 7 15 4 27 0 15 18 2 6 2 22 6 12 25 11 9 24 26 5 11 5 21 17 7 8 4 18 14 16 24 28 1 3 28 27 14 12 8 13 13 23 3 16 23 25 21 22 0 1 20
1 0 19 11 20 17 8 3 2 18 7 18 21 5 26 9 10 6 20 2 28 8 26 19 21 17 16 14 3 14 22 12 13 6 24 9 10 22 0 27 11 13 27 7 15 4 28 1 16 23 25 23 25 12 24 15 4 5
5 9 12 7 19 25 6 24 12 6 25 3 1 16 28 2 3 14 16 4 5 26 7 8 20 19 18 22 10 27 13 28 21 17 11 26 4 18 0 20 27 24 15 10 23 11 0 21 9 22 13 14 8 1 17 15 23 2
• If I use an rng, I could get very different results between runs. Perhaps score based on the average of 100 runs? Nov 5, 2018 at 19:40
• @Spitemaster It is possible that a different RNG would change the score to be +1 or -1, but as scores get larger, the RNG seed matters less. If a person posts a solution that required a very specific RNG seed, it wouldn't be reproducible, and therefore, invalid. Nov 5, 2018 at 20:14
• This seems like a nice challenge but it's a bit vague. What do you mean by "Your function must use this [32 bytes of] data". We can't use local variables? Do we have to assign that parameter that was passed to us in the function call to another value to be able to store data? Nov 8, 2018 at 18:50
# Lunar Arithmetic
Lunar Arithmetic (also called dismal arithmetic, a play on decimal) is an alternative to elemental arithmetic in which only addition and multiplication is defined, and then only for non-negative integers. For single-digit numbers, addition is defined as the maximum of the two digits, and multiplication is defined as the smaller of the two digits. For example:
2 + 5 = 5 + 2 = 5
2 * 5 = 5 * 2 = 2
To extend lunar addition to multi-digit numbers, simply consider the digits individually, counting empty digit places as zeroes as required. For example, consider 169 + 84:
1 6 9
+ 8 4
-----
1 8 9
Lunar multiplication can now be dealt with in the same way as normal long multiplication, using the lunar rules for addition and multiplication. For example, consider 169 * 84:
1 6 9
* 8 4
-------
1 4 4 (169 * 4)
+ 1 6 8 (169 * 8)
-------
1 6 8 4
## The challenge
Write a function or program which, when given two integers, returns the result of their lunar addition and their lunar multiplication, in any order.
• The input pair may be presented as a two-element list or array.
• Input numbers may be provided in any reasonable format, including as a string, or as a list of digits.
• Output may similarly be in any reasonable format.
• The two output numbers must be separated in an unambiguous way, such as with newlines, spaces, as elements of a list or array.
• Links to online demonstrations are appreciated, but not required.
• Standard loopholes are disallowed.
• This is , so shortest code in bytes per language wins!
In your submission, please specify the order in which the two results are presented, and your input and output formats if they are out of the ordinary.
## Test cases
Test cases go here
## Resources
Original paper outlining dismal arithmetic
Numberphile video discussing lunar arithmetic
## Tags
I see this as the potential starting point for a number of Lunar arithmetic challenges, such as computing the result of a longer calculation (such as 12+345*67*8), identification of lunar primes, lunar prime factorisation, methods of implementing subtraction and division... Of course that all depends on the response to this challenge. Any feedback on how to make this challenge better would be much appreciated! | 2022-09-29 01:43:02 | {"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": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 38, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.29377251863479614, "perplexity": 479.13631136913904}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030335303.67/warc/CC-MAIN-20220929003121-20220929033121-00453.warc.gz"} |
https://software-dl.ti.com/processor-sdk-linux/esd/docs/latest/linux/Foundational_Components_Kernel_Users_Guide.html | 3.2.1. Users Guide¶
3.2.1.1. Overview¶
This document will cover the basic steps for building the Linux kernel.
3.2.1.2. Getting the Kernel Source Code¶
The easiest way to get access to the kernel source code is by downloading and installing the Processor SDK Linux. You can download the latest Processor SDK Linux installer from AM335X-Linux-SDK-Download-page. Once installed, the kernel source code is included in the SDK’s board-support directory. For your convenience the sources also includes the kernel’s git repository including commit history. Alternatively, Kernel sources can directly be fetched from GIT.
You can find the details about the git repository, branch and commit id in the Kernel section of the release notes.
3.2.1.3. Preparing to Build¶
It is important that when using the GCC toolchain provided with the SDK or stand alone from TI that you do NOT source the environment-setup file included with the toolchain when building the kernel. Doing so will cause the compilation of host side components within the kernel tree to fail.
Note
The following commands are intended to be run from the root of the kernel tree unless otherwise specified. The root of the kernel tree is the top-level directory and can be identified by looking for the “MAINTAINERS” file.
3.2.1.3.1. Compiler¶
Before compiling the kernel or kernel modules the SDK’s toolchain needs to be added to the PATH environment variable
export PATH=<sdk path>/linux-devkit/sysroots/x86_64-arago-linux/usr/bin:\$PATH
The current compiler supported for this release along with download location can be found in the release notes for the kernel release.
3.2.1.3.2. Cleaning the Kernel Sources¶
Prior to compiling the Linux kernel it is often a good idea to make sure that the kernel sources are clean and that there are no remnants left over from a previous build.
Note
The next step will delete any saved .config file in the kernel tree as well as the generated object files. If you have done a previous configuration and do not wish to lose your configuration file you should save a copy of the configuration file (.config) before proceeding.
The command to clean the kernel is:
make ARCH=arm CROSS_COMPILE=arm-linux-gnueabihf- distclean
3.2.1.4. Configuring the Kernel¶
Before compiling the Linux kernel it needs to be configured to select what components will become part of the kernel image, which components will be build as dynamic modules, and which components will be left out all together. This is done using the Linux kernel configuration system.
It is often easiest to start with a base default configuration and then customize it for your use case if needed. Apply Linux kernel configurations with a command of the form:
make ARCH=arm CROSS_COMPILE=arm-linux-gnueabihf- <defconfig>
3.2.1.4.1. Using Default Configurations¶
For this sdk, tisdk_[platformName]_defconfig was used to create the prebuilt files. We recommend users to use this kernel configuration (or at least use it as a starting point).
platformName is am335x-evm for AM335x, am437x-evm for AM437x, am57xx-evm for AM57xx, k2hk-evm for K2H/K2K, k2e-evm for K2E, k2l-evm for K2L, k2g-evm for K2G, and omapl138-lcdk for OMAP-L138.
For example, to apply the default AM335x kernel configuration, use:
make ARCH=arm CROSS_COMPILE=arm-linux-gnueabihf- tisdk_am335x-evm_defconfig
After the configuration step has run the full configuration file is saved to the root of the kernel tree as .config. Any further configuration changes are based on this file until it is cleaned up by doing a kernel clean as mentioned above.
If the kernel was downloaded directly from the git repository, the defconfig will need to be built with scripts. Please see ti_config_fragments/README within the kernel sources for more information. Otherwise a user will notice a significant amount of features not working.
Below is the procedure to build the defconfig from the kernel git repository.
ti_config_fragments/defconfig_builder.sh -t ti_sdk_[device]_release
export ARCH=arm
make ti_sdk_[device]_release_defconfig
mv .config arch/arm/configs/tisdk_[platformName]-evm_defconfig
The list of defconfig map file (i.e., ti_sdk_[device]_release used above) supported can be found from ti_config_fragments/defconfig_map.txt file.
3.2.1.4.2. Customizing the Configuration¶
When you want to customize the kernel configuration the easiest way is to use the built in kernel configuration systems. One popular configuration system is menuconfig. menuconfig is an ncurses based configuration utility.
To invoke the kernel configuration you simply use a command like:
make ARCH=arm CROSS_COMPILE=arm-linux-gnueabihf- <config type>
i.e. for menuconfig the command would look like
make ARCH=arm CROSS_COMPILE=arm-linux-gnueabihf- menuconfig
Once the configuration window is open you can then select which kernel components should be included in the build. Exiting the configuration will save your selections to a file in the root of the kernel tree called .config.
3.2.1.5. Compiling the Sources¶
3.2.1.5.1. Compiling the Kernel¶
Once the kernel has been configured it must be compiled to generate the bootable kernel image as well as any dynamic kernel modules that were selected.
By default U-boot expects zImage to be the type of kernel image used.
To just build the zImage use this command
make ARCH=arm CROSS_COMPILE=arm-linux-gnueabihf- zImage
This will result in a kernel image file being created in the arch/arm/boot/ directory called zImage.
3.2.1.5.2. Compiling the Device Tree Binaries¶
Starting with the 3.8 kernel each TI evm has an unique device tree binary file required by the kernel. Therefore, you will need to build and install the correct dtb for the target device. All device tree files are located at arch/arm/boot/dts/. Below list various TI evms and the matching device tree file.
Boards Device Tree File
Beaglebone Black am335x-boneblack.dts
AM335x General Purpose EVM am335x-evm.dts
AM335x Starter Kit am335x-evmsk.dts
AM335x Industrial Communications Engine am335x-icev2.dts
AM437x General Purpose EVM am437x-gp-evm.dts, am437x-gp-evm-hdmi.dts (HDMI)
AM437x Starter Kit am437x-sk-evm.dts
AM437x Industrial Development Kit am437x-idk-evm.dts
AM57xx EVM am57xx-evm.dts, am57xx-evm-reva3.dts (revA3 EVMs )
AM572x IDK am572x-idk.dts
AM571x IDK am571x-idk.dts
AM574x IDK am574x-idk.dts
K2H/K2K EVM keystone-k2hk-evm.dts
K2E EVM keystone-k2e-evm.dts
K2L EVM keystone-k2l-evm.dts
K2G EVM keystone-k2g-evm.dts
K2G ICE EVM keystone-k2g-ice.dts
OMAP-L138 LCDK da850-lcdk.dts
To build an individual device tree file find the name of the dts file for the board you are using and replace the .dts extension with .dtb. Then run the following command:
make ARCH=arm CROSS_COMPILE=arm-linux-gnueabihf- <dt filename>.dtb
The compiled device tree file with be located in arch/arm/boot/dts.
For example, the Beaglebone Black device tree file is named am335x-boneblack.dts. To build the device tree binary you would run:
make ARCH=arm CROSS_COMPILE=arm-linux-gnueabihf- am335x-boneblack.dtb
Alternatively, you can build every device tree binary with command
make ARCH=arm CROSS_COMPILE=arm-linux-gnueabihf- dtbs
3.2.1.5.3. Compiling the Kernel Modules¶
By default the majority of the Linux drivers used in the sdk are not integrated into the kernel image file (zImage). These drivers are built as dynamic modules. The command to build these modules is:
make ARCH=arm CROSS_COMPILE=arm-linux-gnueabihf- modules
This will result in .ko (kernel object) files being placed in the kernel tree. These .ko files are the dynamic kernel modules.
Note
If you make a change to the kernel which requires you to recompile the kernel, then you should also recompile the kernel modules and reinstall the kernel modules. Otherwise your kernel modules may refuse to load, which will result in a loss of functionality.
3.2.1.6. Installing the Kernel¶
Once the Linux kernel, dtb files and modules have been compiled they must be installed. In the case of the kernel image this can be installed by copying the kernel image file to the location where it is going to be read from. The device tree binaries should also be copied to the same directory that the kernel image was copied to.
3.2.1.6.1. Installing the Kernel Image and Device Tree Binaries¶
cd <kernel sources dir>
sudo cp arch/arm/boot/zImage <rootfs path>/boot
sudo cp arch/arm/boot/dts/<dt file>.dtb <rootfs path>/boot
For example, if you wanted to copy the kernel image and BeagleBone Black device tree file to the rootfs partition of a SD card you would enter the below commands:
cd <kernel sources dir>
sudo cp arch/arm/boot/zImage /media/rootfs/boot
arch/arm/boot/dts/am335x-boneblack.dtb /media/rootfs/boot
Starting with U-boot 2013.10, the kernel and device tree binaries are read from the root file system’s boot directory when booting from MMC/EMMC. (NOT from the /boot/ partition on the MMC). This would mean you copy the kernel image and device tree binaries to /media/rootfs/boot instead of /media/boot.
3.2.1.6.2. Installing the Kernel Modules¶
To install the kernel modules you use another make command similar to the others, but with an additional parameter which give the base location where the modules should be installed. This command will create a directory tree from that location like lib/modules/<kernel version> which will contain the dynamic modules corresponding to this version of the kernel. The base location should usually be the root of your target file system. The general format of the command is:
sudo make ARCH=arm INSTALL_MOD_PATH=<path to root of file system> modules_install
For example if you are installing the modules on the rootfs partition of the SD card you would do:
sudo make ARCH=arm INSTALL_MOD_PATH=/media/rootfs modules_install
Note
Append INSTALL_MOD_STRIP=1 to the make modules_install command to reduce the size of the resulting installation | 2021-08-04 21:14:50 | {"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": 0.2361098825931549, "perplexity": 5180.5450001267445}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-31/segments/1627046155188.79/warc/CC-MAIN-20210804205700-20210804235700-00614.warc.gz"} |
https://math.stackexchange.com/questions/497733/convergence-of-a-n-1-21-3-1-n | Convergence of $a_n=(1/2)^{(1/3)^{…^{(1/n)}}}$
The sequence $a_n=(1/2)^{(1/3)^{...^{(1/n)}}}$ doesn't converge, but instead has two limits, for $a_{2n}$ and one for $a_{2n+1}$ (calculated by computer - they fluctuate by about 0.3 at around 0.67). Why is this?
• As to "why", intuitively (outside the answer below), consider that the $n$th term has exponent $1\over n$, but in the next term, the $1\over n$ has become $1\over n^{1 \over n+1}$, which is a jump from near $0$ to near $1$. At the following iteration, the $1 \over n+1$ portion has been sent to "near $1$" from "near $0$" by its exponent of $1 \over n+2$, leaving $1 \over n$ with a near $1$ exponent once more. This process cascades down the list in this exact even-term/odd-term fashion. – abiessu Sep 18 '13 at 20:49
I reduce the problem to some results on the exponential tower $x^{x^{x^{\dots}}}$, which I haven't checked how to prove, but which I assume are probably well-known.
Notice that if we write this in the form $$(1/2)^{\dots (1/999)^x}$$ where $x = (1/1000)^{(1/1001)^{\dots (1/n)}}$, then it is enough to show that $x$ oscillates between two values. I claim that $x \le (1/1000)^{(1/1000)^{\dots (1/1000)}}$ if $n$ is odd (so that there are an even number of levels in the formula defining $x$, and the inequality sign is reversed when $n$ is even. We can show this by showing by induction on $k$ that $$(1/(n+1))\wedge(1/(n+2))\wedge \dots \wedge (1/(n+k)) \ge (1/n)\wedge (1/(n+1))\wedge \dots \wedge (1/(n+k-1))$$ when $k$ is even and a similar inequality when $k$ is odd. To do the induction we also need the fact that $$(1 / (n+1)) \wedge (1/(n+1)) \wedge \dots \wedge (1/(n+1)) \ge (1/n) \wedge (1/n) \wedge \dots \wedge (1/n)$$ when there are an even number of terms and the inequality is reversed otherwise.
Then I refer to folklore how the exponential tower $x \wedge x \wedge \dots$ does not converge when $x < e^{-e}$, and instead jumps between two values (which will be very close to $0$ and $1$ for small $x$). We know that $e^{-e} \ge 1/1000$, so the result follows. | 2019-05-19 21:18:21 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 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": 0.953708291053772, "perplexity": 148.37454573230102}, "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-2019-22/segments/1558232255165.2/warc/CC-MAIN-20190519201521-20190519223521-00394.warc.gz"} |
https://indico.cern.ch/event/839985/contributions/3986043/ | # LXX International conference "NUCLEUS – 2020. Nuclear physics and elementary particle physics. Nuclear physics technologies"
Oct 11 – 17, 2020
Online
Europe/Moscow timezone
## LINEAR TRANSVERSE, ANGULAR, AND TIME CHARACTERISTICS OF ELECTRON-TO-POSITRON CONVERSION AT E^-(55; 220; 1000) MeV
Oct 15, 2020, 6:15 PM
1h
Online
#### Online
Poster report Section 3. Modern nuclear physics methods and technologies.
### Speaker
Dr Leonid Dzhilavyan (INR RAS)
### Description
Beams of ultrarelativistic positrons are an essential tool for experimental studies in physics of atomic nuclei and elementary particles. Usually positrons for these beams are produced by conversion of ultrarelativistic electrons to positrons in thick targets-converters. In one variant some magnetic system separates from all positrons from converter those which have necessary energy without their additional acceleration (see, e.g., [1, 2]). In another variant initial positrons, emitted from converters with rather low energies, undergo acceleration up to necessary energies (see, e.g., [3, 4]).
The most important characteristic of positron generation in converters is a coefficient of conversion K(E^-,E^+,T,Z)=N^+/(N^-Delta\Omega^+Delta\E^+)
for a “instantaneous and needle-like” beam of electrons, normally incident on a converter, and a positron emission angle theta+\approx 0, where E^- and E^+ – kinetic energies of electrons and positrons; T– a converter thickness; Z– an atomic number of a converter material; N^- and N^+ – numbers of incident electrons and emitted positrons; \Delta\Omega^+ and \Delta E^+ small considered values of a solid angle and an energy spread for emitted positrons. In the previous our work [5] the calculations were made using GEANT-4 [6] for the pointed out conversion coefficients K for E^- from 10 MeV and up to 1000 MeV for converters from Cu, Ta, and Pb and close to optimal values of T in comparison with available experimental data.
To understand the possibilities of further handling positron beams from converters, one need to know also linear transverse, angular, and time characteristics of electron-to-positron conversion which we obtained in the present work from full array of data, calculated in [5] and compared them with some known data from [7–9].
The obtained new data together with K(E^-,E^+,T,Z)-data from [5] may be useful in modeling and designing equipment for matching emittances of positron beams and acceptances of further magnet systems and/or additional positron accelerators.
1. J.Miller, C.Schuhl, C.Tzara, G.Tamas. // J. Physique Rad. 1960. V.21. P.755.
2. L.Z.Dzhilavyan, N.P.Kucher. // Sov. J. Nucl. Phys. 1979. V.30. P.151.
3. C.P.Jupiter, N.E.Hansen, R.E.Shafer, S.C.Fultz. // Phys. Rev. 1961. V.121. P.866.
4. L.Z.Dzhilavyan, A.I.Karev. Proc. VII Sov. Part. Accel. Conf. JINR. Dubna. 1981. V.1. P.209.
5. S.S.Belyshev, L.Z.Dzhilavyan. Int. Conf. “Nucleus-2019”. Book of Abstracts. Dubna. 2019. P.340.
6. GEANT-4. Version: geant4 9.5.0 (December, 2011) // Physics Reference Manual.
7. D.F.Crawford, H.Messel. Nucl. Phys. 1965. V.61. P.145.
8. J.Haissinski. // Nucl. Instrum. and Methods. 1967. V.51. P.181.
9. C.N.Yang. Phys. Rev. 1951. V.84. P.599.
### Primary author
Dr Leonid Dzhilavyan (INR RAS)
### Co-author
Mr Serge Belyshev (Physics department of the Lomonosov Moscow state university )
### Presentation materials
There are no materials yet. | 2023-03-24 00:15:37 | {"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": 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": 0.48308300971984863, "perplexity": 14811.199432151829}, "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/1679296945218.30/warc/CC-MAIN-20230323225049-20230324015049-00144.warc.gz"} |
http://www.scielo.br/scielo.php?script=sci_arttext&pid=S0103-65132017000100314&lng=en&tlng=en | ## Print version ISSN 0103-6513On-line version ISSN 1980-5411
### Prod. vol.27 São Paulo 2017 Epub July 20, 2017
#### http://dx.doi.org/10.1590/0103-6513.220916
Research Article
Risk management for companies focused on innovation processes
aUniversidade Federal do Rio Grande do Sul, Porto Alegre, RS, Brazil
Abstract
Risk is inherent to the activities of technology and innovation companies and to manage them represent an opportunity to improve the company capability to achieve its goals. The use of ERM models has been studied since the Committee of Sponsoring Organizations of the Treadway Commission guides. This article adapted the MIGGRI model for the context of an innovation company from a TSP in Brazil. Using a case study and a review from previous ERM literature, the article show that is possible to measure the risks that an innovation company faces, and that they may be managed with a view to supporting a company’s strategy. Were applied an economic analysis based on a MCS and an indicator of CFaR were applied to measure innovation risks. A strategic performance model for innovation companies are proposed and the benefit to implement Risk Management practices in innovation organizations was validated.
Keywords Enterprise Risk Management; Monte Carlo Simulation; Economic risk analysis; Innovation management; Strategic performance
1. Introduction
The market of development of technology and innovation has been seen to be gaining, year-by-year a greater and greater share of the US economy, its differential in relation to the industry being the fact that it consists of a large volume of small companies (Trott, 2012). This country, together with others with a more developed economy, has a great deal of research related to this topic, while the volume of such studies in developing countries is limited (Nagano et al., 2014).
In the Brazilian context, the market of innovative companies has been growing at a quickening pace (Nanda & Rhodes-Kropf, 1999; Nagano et al., 2014) and consequently Brazil is recognized as the Latin American country in which technology and innovation companies have developed most (Viana, 2012). One factor that has brought about this classification being a viable statement is that Brazil has innovation environments or Technological and Scientific Parks (TSP), which are easily related to the capability of transforming innovation into business opportunities (Giugliani, 2012). This tendency to develop environments conducive to innovation can be explained by the contribution made by Lazzari et al. (2014) and Tu et al. (2014), who concluded that the financial performance of companies try to push innovation is greater than those that do not invest in this area. However, they subject themselves to dealing with uncertainties vis-à-vis the consumer market that they will cover (Mortimer, 1995).
Due to the swift advance of this market as well as the volume of businesses regarded as innovatory, there is high degree of uncertainty about the internal evolution factors of development of innovative business, and external (macroeconomic and market issues) which stresses the real challenge inherent in the process to manage innovation (Bessant, 2003; Mortimer, 1995; Viana, 2012; Nagano et al., 2014). The uncertainty mentioned has been present ever since innovative companies were formed, which therefore sees to it that the activity of analyzing and managing risks is part of innovation management (Garcia & Calantone, 2002; Bessant, 2003; Canongia et al., 2004; Hauser et al., 2006; Wong & Chin, 2007; Nagano et al., 2014).
Some models have been developed which seeking to bring into effect Management of corporate Risks. The Committee of Sponsoring Organizations of the Treadway Commission (COSO) proposed steps for managing and controlling the risks related to a company making full use of the concept of Enterprise Risk Management (ERM). This enables all the risks to which the company is exposed to be seen (Committee of Sponsoring Organizations of the Treadway Commission, 2007). Similarly, ISO 31000 which was published in 2009 also focused on structuring the Risk Management process (International Organization for Standardization, 2009). Souza (2011), based on previous models, put forward the Model for Identifying and Managing the Degree of Risk in Companies (MIGGRI, in Portuguese). This differed from its predecessors because it linked tools that make achieving this goal viable to the steps proposed for Managing Corporate Risks. At the same time, the model is still restricted to being applied in only one large civil construction company, although the author recommended that it be applied in other industrial sectors. Therefore, what stand out is the absence of models which cover the level of corporate risk management in innovative and technology companies (Bandyopadhyay et al., 2006), all that is found in the literature being models that seek to analyze risk of concern to Information Technology projects, such as Original Spiral Model, ProRisk, Riskit, or to project management: SERIM, SRAM, BRisk and OBRiM (Miorando, 2010).
Crossan & Apaydin (2010) discuss the need for in-depth studies that succeed in analyzing, using monetary measurement, the impact of innovation corporately on financial performance in innovative enterprises. There is a noticeable a gap in the literature and in practice in the context of innovation: a market of companies exposed to risks due to high uncertainty and the lack of appropriate methodologies for small innovative companies present in Science and Technology Parks (TSP). This lack start a necessity to research and develop methodologies that can be used by small innovative companies to improve its performance. Small innovative companies must receive this research attention in function of the high risk and uncertainty exposure that this market represent and those risks impact that can affect its financial structure.
Considering this context, this research start to look for the answer: How innovation companies from Brazil can use and apply ERM? What is therefore set as the main objective of this study is to apply MIGGRI to the context of a company located in an incubator of a TSP which has a wide impact throughout Brazil. The specific goal is to analyze the possibility of linking the model applied to the strategic management of the company.
Achieving those objectives, this research helps to turn clearer for entrepreneurships and innovative employees risk management model implementation practices. The traditional risk management models, started in the financial market, are translated in an operational and practical language to orient its implementation by different employees. In addition, the academy receives a new approach for risk management, valuing its application in other business environments.
Among the limitations of the study, the lack of statistical tools based on historical data stands out. This is due to the thinness of the database in the company analyzed and to the proposal of adaptation as a prototype model since this study will be based on only one practical example. It is believed that to validate a generic model of Risk Management for innovative companies, the application needs to be extended to a larger number of companies.
This paper is fragmented in five sections. During the first, the research problem, the theme and objectives were presented. The next section presents the theoretical background, followed by the methodology. The results were presented fourth section and, at the end, we have the conclusions and limitations.
2. Theoretical background
The theoretical framework of this study breaks down into innovation concepts and their relationship with risk management, how the MIGGRI model is detailed and the importance of assessing Risk in innovation environments.
2.1. Innovation and risk management models
The concept of innovation has been approached in different ways in the literature (Crossan & Apaydin, 2010). Hansen & Wakonen (1997) conceptualize innovation as something new: a new asset or a new process. Pittaway et al. (2004) translate innovation into a need that companies have, or as an obstacle to sustained growth (Yang, 2012). With regard to the economic approach, Freeman & Soete (2008) translate innovation into a new combination of reources that affects commercialization.
How the literature has explored the theme of innovation has been expanding in recent years (Hauser et al., 2006; Crossan & Apaydin, 2010); however, Crossan & Apaydin (2010), after having made a search and found approximately 13,000 articles, of which they, effectively, analyzed 525, emphasize that, among their results, only 9% of the studies include an economic analysis of innovation. This can be justified because of the existing difficult in measure the creation of value based on innovation (Crossan & Apaydin, 2010). Therefore, an innovative product can be seen as a technological development that holds out a commercial opportunity, which prompts the need for an innovative organization to be efficiently connected to external markets (Nagano et al., 2014), given that whenever a company markets an innovative product, an increment is added to the consumer market (O’Connor et al., 2008).
Regarding the complexity inherent in this segment, it is considered that the success of innovative business depends invariably on its acceptance by consumers, it being necessary for the target consumer market to be at a level of demanding innovations. In other words, the consumer must have a culture that wishes what is new and different, so that companies can achieve the result desired in their projects (Hauser et al., 2006). There is a noticeably extensive challenge associated with managing innovative businesses, which extends strongly to the external environment, thereby increasing the level of uncertainty associated with this market (Bessant, 2003; Nagano et al., 2014).
Having accepted that uncertainty is inherent in the innovation environment, and given the definition proposed in NBR 31000, in which risk is conceptualized as the effect of uncertainty on objects, uncertainty being the state of the shortfall in information related to an event (International Organization for Standardization, 2009), the presence of risk in innovative enterprises is confirmed. The analysis and management of such risks can be translated into activities that aim to assist the capacity of such companies to anticipate how they will deal with the uncertain environment of the market analyzed, by fostering actions that may well come to contribute to the company's bottom line (Mortimer, 1995). O’Connor et al. (2008) indicate that the advance of innovations demands that companies be encouraged to deal systematically with the risks to which they are exposed, with a view to reconciling the high degree of uncertainty and dynamic and flexible environments that enables innovation to make successful advances.
The Risk Management process aims to help companies analyze their risks. Risk Management, in a synthetic way, the guides companies on how to take correct attitudes about which risks they must face up to, which they should ignore, which need to be hedged or eliminated and which should be exploited in order to maximize the organization's performance (Damodaran, 2009). Frigo & Anderson (2011) stress the importance of making Risk Management fit for purpose and analyzing it at the corporate level. They point out that when it is applied only in order to analyze the feasibility of projects, it only gives an individual analysis and thus does not represent the generic panorama of the exposure of company’s business to risks. As pointed out in the introduction of this article, there is a shortage of Risk Management models at this level of coverage in the market of innovative companies.
The time between the emergence of an innovative idea and product launch itself becomes another factor that throws the importance of Risk Management into relief. This is because the importance of the risk element is directly related to the length of the project element (Kliemann, 2010). The importance of considering the time difference between the conception of the idea and a new development actually being launched in the market has become an important and motivating issue as to using analytical practices and Risk Management (Damodaran, 2009). Among the main models of Risk Management, on which structuring the MIGGRI model is based, are the COSO methodology (Committee of Sponsoring Organizations of the Treadway Commission, 2007), PMBOK (Project Management Institute, 2008) and the standards AS/NZS 4360 (Standards Association of Australia, 1999) and ISO 31000 (International Organization for Standardization, 2009), from Australia and Brazil respectively.
What COSO methodology (Committee of Sponsoring Organizations of the Treadway Commission, 2007) sets out to add to the Risk Management process is a focus on the corporate risks of organizations. It is defined as a process which is applied to the company's strategy and which sets out to identify events with the potential to affect its enterprises and gave rise to the concept of Enterprise Risk Management (ERM). To achieve this objective, COSO put forward a new system of Risk Management that makes use of ERM as an essential tool for creating and preserving the business value of companies (Hayne & Free, 2014). ERM can be defined as a structure that enables all the risks to which an organization is exposed to be viewed, including risks at the corporate level risks and for each business unit by using an ordered framework and proposes that risk management be integrated, continuous and systematic. The Brazilian standard for Risk Management, NBR ISO 31000: 2009, based on ERM and the earlier Australian Standard AS/NZS 4360 (Standards Association of Australia, 1999), seeks to assist the development of continuous and systematic control processes and Risk Management in organizations.
The methodology present in PMBoK – the Project Management Body of Knowledge - discusses how to structure a management process based on the management of risks, priority being given to projects (Project Management Institute, 2008). Despite the widespread application of PMBoK between companies, some researchers criticize it due to the absence of the concept of how value flows to a project (Koskela & Howell, 2002). These authors conclude that non-traditional project management methods, which consider the flow of activities that do not add value in their plan, unlike that of PMBoK, already applied to the software and the construction industry, can significantly reduce risks because they allow possibility of working with uncertainty as it emerges throughout a project.
2.2. Model for identifying and managing companies’ degree of risk
The introduction of the concept of corporate Risk Management prompted the advance of a new degree of coverage in the literature on and the practice of risk management which hitherto had been restricted to the level of projects (Hayne & Free, 2014). Souza (2011) sought to deepen the concept and the existing methodologies, by proposing tools that were appropriate for implementing a systematic approach to risk management at the corporate level.
To do so, she sought a theoretical foundation in the literature so as to be able to give a detailed account of each stage of the model. Among the main differentials vis-à-vis what had already been proposed, what stands out is that MIGGRI, on using the tools she suggests, enables a global indicator of risk into which the company fits to be generated. This takes the impact of quantitative and qualitative risks on it into consideration. In addition, the model proposes that a cross-check be made on the degree of risk to which the company is exposed and the desired degree of exposure that it should reach, in accordance with the risk profile into which the company fits. The steps proposed in the MIGGRI model are shown in Figure 1.
Among her recommendations for using the model with excellence, based on experiences of other authors and on the preliminary application of MIGGRI, Souza (2011) recommends structuring a directorate of risks in order to facilitate implementing the concepts of the Management of Corporate Risks in a company. On adapting the model to the context of smaller companies, it is known that is not feasible for such a company to make a professional of this order available because of the impact on cash flow. Thus, it is fundamental that the managers of small companies have a thorough knowledge of financial and economic analysis, besides strategic vision for the company. This knowledge is needed with greater intensity in the phases to do with calculating the Degree of Exposure to Risk and the Desired Degree of Exposure to Risk, which require a Monte Carlo Simulation (MCS) to be used.
The MCS enables finding the distribution of values that best describes the behavior of the data being analyzed (Jorion, 2010), and takes hundreds of scenarios into consideration in such analysis, which is limited in methods such as scenario analysis and decision trees. Moreover, among the indicators that can be calculated based on their response is Cash Flow at Risk (CFaR), or cash flow at investment risk. The usefulness of this indicator is assured by the fact that it succeeds in measuring the potential gain or loss of the money invested (Damodaran, 2009), besides adding value quantitatively to the Risk Management process (Jorion, 2010). That is to say that the CFaR enables both the opportune and the dangerous side of risk to be analyzed: how much the investor risks gaining (CFaR+) and how much the investor risks losing (CFaR-) in relation to the expected scenario, and taking the interaction of hundreds of scenarios based on input data into consideration. The CFaR is an example of an indicator that can be used as a result of the Degree of Risk of organizations which allows regular updates to be made.
The second indicator suggested in MIGGRI model concerns the Global Indicator of Risk: I = (Ri;ki), Equation 1, proposed by Souza (2011) as the answer to one of his goals with the model, quantifying the degree of exposure to risk. This indicates the variation in relation to the expected result, Equation 2, and the tendency of the risks, Equation 3, in each unit or line of business, thus allowing the manager to analyze what the ambit of the company's activity is that is impacting the overall expected result, the most strongly.
Ii=(Ri;ki) (1)
Ri=CFaRi+ACFaRiAμi (2)
ki=CFaRi+A+CFaRiACFaRi +A CFaRiA (3)
where: I = degree of risk of the business unit; R = degree of dispersion of the potential values of the business unit; k = an index shows the ratio between positive and negative risks of the business unit. This sub-index will assume values between -1.0 and 1.0, where -1.0 indicates that the company is subject only to negative risks, and 1.0 that the company is subject only to positive risks; i = business unit assessed.
2.3. The importance of evaluating the innovation risk
Among the benefits of exploiting risk is the need to have an advantage over competitors who are exposed to the same risk (Damodaran, 2009). In the innovation environment, where risk is encouraged by the characteristics of intangibility, globality and the potential of capital (Giugliani, 2012) and also by the need to monitor the market closely (as to securing and meeting needs and reaching and holding the consumer) (Hauser et al., 2006), the maximum amount of information should be sought from the external environment (monitoring the market) and from the internal one (processes) in order to establish what the speed of response to risk is (Damodaran, 2009). From this point of view, note that the application of Risk Management models can be used as a data source so that managers can make decisions strategically which are aligned to the company’s objectives. Therefore, the use of performance systems based on the data generated in models like MIGGRI has the potential to generate information that is important for managing strategy.
Performance management systems showed an exclusively financial perspective until towards the end of the 20th century. Since then Kaplan & Norton (1996) deepened their studies and developed the methodology called Balanced Scorecard (BSC), which associates an analysis of the company’s performance analysis to its strategic objectives in order to represent the respective strategy that ensures its market success. Therefore, the authors proposed structuring the model in four perspectives; Financial, Customers, Internal Processes, Learning and Growth. The BSC has spread very widely in industry but the literature shows it has faced some criticisms.
In the innovation literature, the current difficulty of seeking to measure innovation is noted (Smith et al., 2005). Jong & Marsili (2006), from an analysis of 1,234 small companies, consolidated variables that indicate there is an orientation to the innovation of the company: innovation outputs; innovation inputs; innovative resources; innovative planning; and external relations. When seeking to measure the performance of innovation, the difficulty increases and, accordingly, Zeng et al. (2010), on studying a Chinese TSP, proposed 21 indicators, of which 10 are for innovation of the organization, and the objective of is to fill this gap. Among them those hat sand out are: the growth of R&D teams, the growth of innovative value added to the region, the rate of exportation of technology in relation to industrial exportation and the development of innovative products per capita.
Among the factors that drive the development of innovation, what stand out are innovation environments, such as the Scientific and Technological Parks (TSP). TSPs are often associated with environments fit for transforming innovation into business opportunities (Giugliani, 2012) and creating e competitive, effective and dynamic benefits (Gaino & Pamplona, 2012). The search for innovation environments such as TSPs is prompted by the fact that companies that invest with the intention of innovating tend to have greater financial performances (Lazzari et al., 2014). However, taking on an innovative approach leads back to exposing uncertainties, both technical and market ones (Hauser et al., 2006; Freeman & Soete, 2008). Thus, practices that allow risks to be analyzed and managed in order to attain a good economic performance are required. However, models and tools that allow the corporate management of risks are still oriented towards traditional management environments (Bromiley et al., 2015).
It is noted that there are systematic approaches to monitoring performance from different perspectives in the literature. Furthermore, the use of information provided by risk analysis as a feed source of a system for monitoring performance is among the recommendations made by Souza (2011). Thus, it is believed that the use of the innovation indicators proposed in the literature together with indicators that make it feasible to monitor critical risks, i.e., that may impact the company’s expected performance, may be useful for the innovative companies. In addition to the benefits already highlighted on the use of analysis and risk management, one has to use the information from this analysis and to assess the impact of each business unit is within the company and thus fuel prioritization analysis models performance. Souza (2011) suggests that the use and monitoring of performance models such as the BSC and Risk Management may be of considerable value to aid decision making strategically.
3. Research methodology
This study began by drawing on the conclusions reached by Etges & Souza (2014), who applied a semi-structured interview in thirteen innovative companies operating in the TSP of the Pontifical Catholic University of Rio Grande do Sul (Tecnopuc) in order to examine to what extent the responses fitted in with the theme of the Management of Corporate Risks. A company was then selected which had a profile that was suited to applying the MIGGRI methodology. It is worth noting that the use of the TSP Tecnopuc is justified by the fact that it is one of the three well-established TSPs in the South of Brazil, has a self-sustaining structure, and the National Association of Promoting Entities Innovative Enterprises (ANPROTEC) has twice given it the award of Best TSP in Brazil
During the interviews, it was pointed out that the previous study by Etges & Souza (2014) had noted the companies’ unanimous interest in the theme of the Management of Corporate Risks. In order to select the company in which it would be feasible to apply the study with greatest effect, some of its fundamental characteristics had to be closely examined in order to examine risks. Those the following were considered: its level of technology (how much the company invests in innovation and development of technology), its maturity (the financial structure of the company), its interest in and profile of risk as important characteristics so that the quantitative analysis of the financial impact of the risks identified would be viable. Therefore, the criterion of maturity is important due to the demand for an appropriate financial management structure.
Therefore, the level of technology, maturity, interest and risk profile were considered as criteria. As alternatives the companies in which interviews were held in the study by Etges & Souza (2014) were used. To select the company that was the focus of this study, the Analytical Hierarchy Process (AHP) was used, with the responses obtained in the study of Etges & Souza (2014) being taken into account. Table 1 shows the results of applying the AHP to thirteen companies considering all four criteria, with a consistency ratio smaller than 9.3% for all matrices created to do it. All the replies supplied by the two researchers who used the AHP. It is important to highlight that the two researchers did the paired comparisons among alternatives for each criteria.
Table 1 Main matrix of the AHP.
A B C D E F G H I J K L M
Technology level 0.4L3 0.72 0.72 0.72 0.14 0.72 0.43 0.14 0.72 0.72 0.72 0.14 0.43
Maturity 0.43 0.72 0.14 0.14 0.43 0.72 0.72 0.14 0.14 0.14 0.43 0.72 0.43
Interess 0.43 0.72 0.43 0.14 0.14 0.72 0.43 0.72 0.14 0.14 0.43 0.43 0.14
Risk Profile 0.72 0.14 0.72 0.72 0.43 0.14 0.14 0.72 0.72 0.72 0.72 0.72 0.14
Total 2.03 2.17 2.03 1.74 1.16 2.32 1.74 1.74 1.74 1.74 2.17 2.03 1.16
The analysis of Table 1 was the underpinning used to select Company F for the application of the model. It emphasizes the great importance of the criterion of maturity is emphasized, which is due to the demand for a financial management framework to implement the MIGGRI model. The selected company was found to have detailed cash flows and the managers’ knowledge of tools for economic projections and strategic planning was good. Considering this research context, it is possible to classify this study like a case study, which practical applications oriented to solve a real problem for a small innovative company from a TSP were applied by the researchers and the organization. Applying this case study with this new problem, is being developed a contribution for the literature that doesn’t present a consolidated framework to apply risk management in innovative companies. This can be suggested because this research is answering a solution (how?) for a problem (what?) that is important for the actual economy (why?) and designing who must be involved (who?) (Whetten, 2003). The use of case studies to contribute with the literature capability to solve real problems is suggested by methodology researches, like Boer et al. (2015).
Company F was set up in Tecnopuc two years ago. Among the features that made it viable to open the company in this environment, what stands out is the constant search for new solutions being present in its raison d’être. The company examined operates in hospital services, which includes processes for calibrating equipment. This is regarded as innovative vis-à-vis what is already on the market, therefore allowing for agile and high quality results. In addition, the company is not opposed to items of equipment that are already commonly available in European and North American markets coming into circulation in Brazil. It maintains continuous monitoring activities of the external market, thus seeking to bring solutions to Brazil that will contribute to restoring and maintaining the health of patients treated in Brazil. Thus, the intense search for innovation which is part of the company’s strategy made it viable, in conjunction with the relevant features highlighted in the method, to apply the model at this location.
The MIGGRI model was adapted to the context of the (small-scale, innovative product and services) company selected. This was followed by applying it on site in conjunction with the company. Figure 2 highlights, by means of a check mark, the steps of the MIGGRI model that were used.
On looking at Figure 2, it can be seen that a few steps of the MIGGRI model were not followed in the company under study. This limitation stems from the company’s lack of knowledge with respect to risk management practices. The limitations worthy of note were the complete exclusion of Phase IV of the model, regarding Desired Degree of Exposure to Risk. Since the company in question had not conducted a survey of its risks in earlier periods, it is difficult to have managers at this first moment fully grasp the notions of desired degree of exposure to risk. Therefore, it is understood that this step is best suited to companies with a greater understanding of their associated risks. The qualitative Assessment of Risks, as well as the Simulation of Future Scenarios, were also steps that were also excluded in this first application of the MIGGRI model, which therefore had to be adapted because of the lack of depth of the managers’ knowledge. However, it is understood that the model as a whole is applicable to companies belonging to the innovation environment, and the objective of this application was to support such a claim.
4. Results and discussion
The implementation stages of the MIGGRI model for this company will be presented in chronological order and the results discussed below.
4.1. Structuring and planning
Considering the demands of future economic analysis for the analysis and Management of Risks, the need to develop some activities still missing in the company’s structure was noted, including segmenting the cash flow into Business Units (BUs). The company chosen operates with two business fronts: services and the resale of hospital products. With regard to services, it supplies: more accurate calibration for hospital equipment to what is already on the market as a result of the innovation developed by its managers as to calibration methods (A); hiring hospital equipment (B) and maintaining hired equipment (C). Among its commercialized products, the company imports and re-sells two products (D and E) that are not otherwise available in the Brazilian market, with unit D being aimed at individuals and hospitals and unit E only at hospitals. The objective of structuring the cash flow of each BU latches onto the need to project profits (revenues - R) and expenditures (costs - C) for each BU, with a view to identifying, using robust tools, the projected variation in profits (L = R-C), by measuring the impact of risks that permeate each specific business of the company.
To structure the cash flow and its future projection, the ABC (Activity-Based Costing) method was used (Cooper & Kaplan, 1991), which is based on defining the cost drivers so as to distribute them. Thus, for payroll, for example, the driver of Attention Index was used for the rent per m2 demanded by each BU, and the infrastructure costs were distributed uniformly. Moreover, amounts with respect to bills for water, electricity, telephone and gas were included as overheads in the CF. The direct costs and revenues were estimated in accordance with the demand forecast for the company’s services and products. It is noteworthy that, given that the company has been on the market for two years, it was assigned a forecast horizon of only one year. With regard to taxes, tributes due were included in accordance with the business fronts: services 11.4% of revenue, products of unit E 7.6% on turnover and product unit D 12% of turnover. Table 2 illustrates the Cash Flow projected for the next Financial Year, which will be used to conduct the probabilistic analysis of the risks.
Table 2 Projected Cash Flow for the following Financial Year.
Annual A B C D E Total
Gross Income* 109,800.00 198,518.33 542,607.69 301,923.08 197,689.65 1,038,875.96
Taxes 16,272.36 29,420.42 80,414.46 44,080.00 28,461.74
Net Income 93,527.64 169,097.91 462,193.23 257,843.08 169,227.92 1,151,889.78
Product Cost* 50,506.00 11,000.00 77,933.75 129,950.00 101,451.04 185,420.40
TCM 43,021.64 158,097.91 384,259.48 127,893.08 67,776.88 781,048.99
Employees* 84,358.04 85,847.30 86,118.81 90,453.39 81,439,55 428,217.10
Structure* 12,000.00 12,000.00 12,000.00 12,000.00 12,000.00 60,000.00
Rent* 3,709.09 1,854.55 2,781.82 927.27 927.27 10,200.00
Unit Profit (57,045.50) 58,396.07 28,358.85 24,512.41 (26,589.95) 282,631.89
Expenses - - - - - 117,187.88
Total Profit - - - - - 165,444.01
*Will undergo probabilistic variation when the Monte Carlo Simulation is used.
In a brief description of the results expected from the projection, it is seen that units A and E will not produce net profits but according to the managers, they are important because they reach the clients of units B, C and D. Units A and E are seen as a way to open the door to new consumers for the subsequent sale of units B, C and D. In other words, despite the unit being the reason for the opening and the location of the company in a Technological Cluster, it is noted that it currently runs at a loss, it being unit C that is responsible for the good projected result of R$165,444.01 at the end of the next Financial Year. 4.2. Context of risk The Risk Breakdown Structure (RBS) proposed in the COSO methodology (Committee of Sponsoring Organizations of the Treadway Commission, 2007) enables managers to relate the risks to which the business is exposed, by classifying them according to their origin (internal and external) and their class (infrastructure, personnel, process, technology, economic, technical, social, political and environmental), thus facilitating a strategic analysis on how to act with respect to them (Hayne & Free, 2014). In the context of the company under analysis, managers mentioned that they did not dedicate time to analyzing risks in a structured way when the company was formed. Thus, brainstorming sessions were held with company managers to draw up RBS, with the existing risks being related to the BUs and to the variables that they potentially will have an impact on. The relationship established enabled the perception of a large volume of risks inherent to the business and the impact on all BUs from some risks, thus indicating that attention should be given to them. In addition, it was noted that the variable of gross revenue is the one with the largest volume of associated risks, therefore indicating in a qualitative way that it represents the variable of highest exposure to risk. An illustration of several risks listed in the BU is in Table 3. Table 3 Partial RBS of the company. Risk Origin Classification Impact A B C D E Physical Change in Space Internal Infrastructure Rent x x x x x Delivery date products and equipment External Technical Income x x Important delay External Technical Income x Dibbing External Economics Income x x x Quality of workmanship External Employees Employee x x x Technology Changes External Technological Structure x Ill-use External Technical Income x x x x x Process Management Internal Process Structure x x 4.3. Degree of exposure to risk From the breakdown of the risks of qualitative way, the next step was to conduct a probabilistic analysis of the impact of risks on the company’s cash flow. Because of the shortage of long-term historical data that would have allowed the use of statistical distributions based on historical data, estimates were made of triangular probability distributions to analyze the impact of the risks on each variable of the cash flow. The use of triangular distributions can be found in the context of risk in studies by Souza (2011), Zouggari & Benyoucef (2012) and Miorando et al. (2014) and, also, in portfolio optimization (Dutra et al., 2014). For this estimate, the basis is managers’ knowledge and opinions on the progress of their business. The process of making a survey of the distributions began from identifying the risks that affected each variable, followed by an estimate of the impact that each risk could cause them. Due to the presence of more than one impact on each variable risk, the need to combine the individual distributions of impact of a risk in one variable was noted. Given that the risks are independent, that is, the occurrence of one event is not affected by another (Pinheiro et al., 2012), statistics were used that are founded on Bayes’ Theorem, the Total Likelihood Principle, Equation 4, in order to combine the distributions and to achieve the scenario of a final distribution stage for each of the variables included in the Cash Flow (Gross Revenue, Structure, Rent, Cost of Product and Personnel). P(B)=i=1mP(BΙAi)*P(Ai) (4) where: A1, A2, ..., An = part of a sample space; B = any other event of this space. It also considered a correlation of 0.8 between the variables of cost and revenue, since the largest share of costs is concentrated on highly skilled people and on variable costs. Thus, the cost increases as the demand for the services grows. Table 4 illustrates the triangular distributions of each variable defined in the BUs. It is noteworthy that for the variable of Expenditure on Personnel there is no distribution associated with units D and E, as they represent products, which according to the managers are not affected by the risks involved with this variable. Table 4 Probability distributions of entry and calculation of the Degree of Risk. Business Line Incoming A B C D E Incoming (min) 50% 63% 25% 33% 47% Incoming expected 100% 100% 100% 100% 100% Incoming (max) 136% 155% 140% 112% 140% Product Cost A B C D E Cost (min) 70% 65% 75% 95% 42% Cost expected 100% 100% 100% 100% 100% Cost (max) 125% 117% 125% 127% 136% Rent A B C D E Expenses - rent (min) 100% 100% 90% 80% 90% Expenses - rent expected 100% 100% 100% 100% 100% Expenses - rent (max) 120% 120% 110% 110% 110% Structure Cost A B C D E Expenses - structure (min) 86% 95% 77% 90% 95% Expenses - structure expected 100% 100% 100% 100% 100% Expenses - structure (max) 115% 105% 115% 105% 105% Employees A B C D E Expenses - employee (min) 100% 100% 100% Expenses - empolyee expected 100% 100% 100% Expenses - employee (max) 110% 110% 110% From the triangular distributions defined, and along with full details of the cash flows and projections for the year under analysis, @RISK software was used to perform a Monte Carlo Simulation (MCS). As a result, the probability distribution of the outcome variable of the cash flow was generated, i.e. of the expected profit in this study, fragmented by BU, thus facilitating the analysis on the impact of each business line in the company's results as well as giving priority to actions that may be used to monitor it and minimize it. Therefore, an MCS was used that considered 200,000 interactions. From the results generated, it was possible to calculate the indicators CFaR+, CFaR- and the Global Indicator of Risk (Ii = (Ri; ki)), suggested by the MIGGRI model to estimate the Degree of Risk involved in the company's business (Table 5). It is noteworthy that there is coherence between the findings. That is, the distribution of values that best describes the behavior of the data is consistent with the guidelines of the triangular distributions initially inserted into the software, with emphasis on the potential use of the tool to aid orientation towards a system for performance based on the information that managers have. Table 5 Indicators calculated based on SMC. A B C D E Expected (40,634.87) 114,215.80 247,842.80 (52,330.54) (29,320.15) CFaR- (19,555.71) (49,906.50) (187,949.84) (70,999.96) (34,256.33) CFaR+ 17,897.78 55,638.00 158,950.30 47,069.96 33,326.71 I(R;k) (0,92;-0.04) (0.92;0.05) (1.40l-0.08) (2.26;-0.2) (2.31;-0.01) An example is in BU (C), which is the highest expected return, referring to a detailed analysis from this first finding. This BU generated a negative ki index, which is explained by the strong risk of loss of income involved (25%), which made the CFaR- of this BU remain more representative than the CFaR +. Thus, what stands out is the attention that should be given to the risks that cause an impact on the revenue of BU (C), thereby emphasizing the potential use of information generated by the MCS for feeding performance systems that the company uses especially regarding the indicators of monitoring and control to guide improvement actions. A similar analysis can be performed for the BU (D) that although prior to incorporating the risks shows a positive result, when risks are incorporated indicates attention because it has an expected negative result in Table 5 ($ 52,330.54). The warning sign is aggravated when analyzing the Ri index of 2.26. In other words, there may be a 226% variation compared with the expected return on this BU, tending to be oriented to the negative side (of loss of cash flow) due to the negative trend found in the ki index of -0.20. It is emphasized that these results are being influenced by the input distributions where for the variable of income there is a negative representative risk (33%). Thus, risks related to the revenue of this BU should also receive priority in performance systems that aim at financial stability and the possible financial improvement of the company.
It is also perceived that units A, D and E, besides having a negative result when the risks are considered, tend to have negative variations (ki -0.04, -0.2 and -0.01 respectively). Thus, despite the company using unit A, for example, to enter the customers and later sell other solutions, it is recommended that there should be a study on discontinuing lines of business which will potentially have negative results, by developing less costly alternatives and to incorporate new customers. Thus, the company would have a better overall result because it would no longer have losses from the sale of business units A, D and E, which when summed up represent in the scenario expected R$122,285.56. The results of the company after the consideration of risk would increase from R$ 239,773.04 to R\$ 362,058.60, without considering that possibly the indirect costs of the structure could be reduced by discontinuing some solutions. Should the company choose this option, the impacts of risks shared between lines of business should be analyzed so as to make the correct decision and to measure the risk.
As a final analysis, based on the results of the MCS, the presence of the variable Revenue stands out among those that cause the greatest effect on the variation of the company’s overall profits. Because of the innovative nature of the business under study, this affirmation needs to be checked against what the literature on innovation emphasizes: uncertainty is inherent in the innovation process (Garcia & Calantone, 2002; Bessant, 2003; Canongia et al., 2004; Hauser et al., 2006; Wong & Chin, 2007; Nagano et al., 2014) and is strongly influenced by the difficulty in confirming whether or not the innovative product or service has been commercialized (Cooper, 2003; Bessant, 2003; Wong & Chin, 2007; Trott, 2012; Nagano et al., 2014). This study confirms this difficulty and does so because of the presence of the negative impact of the risks associated with revenue, and moreover, the final results generated by the MCS which show this variable to be the one that has most impact, namely it causes the greatest variation in the company’s overall results. It is noteworthy that the variable with the least impact is summarized in the rent, which had few associated risks and effects.
On returning to the risks linked to the variable of greatest impact (income), we can see one of the potential uses of the model: it enables the risks that should receive greatest attention in mitigation practices to be identified since they lead to a greater economic impact. Thus, the systemic use of the proposed model allows continuous management to be structured and brought into practice. This uses indicators that involve the critical risks found and these are detailed in the next section.
After conducting the MCS and analyzing the results found, the worth of using the MIGGRI model is realized as are the opportunities for analyzing in depth by applying it in segments by business units. By structuring its application in this way, the cash flow of whatever segment of the company has suffered the greatest impact can be seen and measured, thereby enabling managers to base their decision-making on whatever actions may be the most strategic for the company's results.
4.4. Treatment and monitoring
Having measured the risks in the BUs, the next step is to identify the critical risks and to deal with them. Therefore, indicators must be created. However, using indicators that make it feasible to monitor the innovation performance of a company set up in a TSP becomes challenging for managers. However, what is noted is the willingness to seek to monitor the progress of the raison d`être of such a company in this environment. Thus, it was proposed that the company use a performance system that makes it viable to monitor the innovative performance of the company, based on the indicators proposed by Jong & Marsili (2006) and Zeng et al. (2010) and which also allow precise monitoring of the critical risks that may happen to affect this performance.
When considering the potential of strategic analysis based on the findings in relation to the impact and the position of the BUs in the company's economic performance, the relationship existing between the analysis and the Management of Financial Risks and the Strategic Management of the company. This is present because on measuring the impacts and the respective risks related to the variables that may be pulling the company's profit positively or negatively, the manager is able to guide and prioritize decision making on the areas that demand constant and detailed monitoring in order to mitigate risks and, consequently, to avoid losses. Furthermore, when periodically analyzing the risks to which the company is exposed, an internal analysis of the processes is made and an external one of the market. By using this analysis, managers dedicate themselves to making a strategic analysis of the company's position in the market in a systematic way.
Thus, with regard to monitoring and controlling risks, based on the analysis conducted, the recommendation was made to the company under study that it should start to use performance indicators related: to monitoring the risks associated with the variables of greatest impact on the company’s profits; to the BUs which have the greatest financial representativeness; and to the BUs with greatest variation in relation to the expected average result. It was proposed that the indicators should be accessible to everyone involved in the process and displayed on the company’s noticeboards, thus facilitating visual management, by making information transparent and by structuring action plans quickly on perceiving a drop in the performance of some area that is being constantly monitored.
An illustrative sample of the proposed system for monitoring the company's performance is shown in Figure 3.
Looking at Figure 3, can be seen that it follows a structure based on the BSC model, in which the indicators are classified according to the strategic prospects of the company. The indicators refer to important steps for the organization, such as Receivables overdue from customers. Using these measures, action plans should be drawn up to deal with critical risks and thereby to achieve the expected result. To facilitate the dissemination of information and its control, using dashboards with visual data helps, when managing and monitoring risks continuously. In the company under study, the dashboard was structured and is currently being implemented in full and so the data presented are by way of illustration only.
5. Conclusions
The innovation process is full of decision points that require swift and strategic actions in a dynamic environment, heavily influenced by variables external to the company (Cooper, 2003; Nagano et al., 2014). Internal processes, such as the management of risk and performance, which generate information about the company’s position, are shown to be essential to sustaining the organization.
This study sought to apply the MIGGRI model proposed by Souza (2011) to the environment of small innovative companies of a TSP, and success was achieved by using this first study. The model was applied by implementing it together with a company of this segment, and the analyses of the findings enabled a proposal for constant monitoring to be structured based on designing a performance system that uses indicators. Thus, what is seen is the link between analyzing the risks to which the company is exposed, its origin and their relationship to decision making to manage the company strategically. In brief, the probabilistic analysis of the risks indicates the variables that cause the greatest financial impact, thus allowing their relation to the risks to which they are related, the decision making being funded on monitoring and controlling activities related to these risks.
The model adaption was tested and contributed with the company capability to manage innovation risks. With this result, is possible to affirm that the risk management model can be explored by other companies oriented to improve the risk management process and, in sequence, innovation performance. One objective was also attended and innovation managers and employees receive an orientation about how to implement risk management process previous focused in big companies or financial environment.
The academy received a new approach about risk management. Theoretical guides, and models model focused on stablish market were used to structure a model for small, new and not stablish markets. The managers are benefited with a technique that enable them to identify previously financial dangerous and, with this, to act proactively turning the innovation organization stainability more likely.
The MIGGRI model, calculating the desired degree of exposure to risk was disregarded for this first analysis, due to the length of time the company has been in the market. However, the inclusion of this phase is recommended in future analyses in companies located in innovation environments. In addition, a new application is recommended for future use in this company to compare the findings and to identify whether it is worthwhile using indicators and actions that target the risks which represent the greatest impact on the company’s performance.
This research is limited by the region. The case study is from Brazil and can’t be assumed like a globally conclusion. In the same way, how was used only one case, those findings are specific for this company and its business sector. Future applications in different companies, countries and business sectors can contribute for the advance in this proposed model.
Finally, due to the repetitions of risks between the companies already in the innovation market, in-depth studies with a larger number of companies and investors in this market on the theme of risk management are proposed to homogenize groups of risk factors for the innovation ecosystem. On achieving this level, a system that analysts and company managers find easy to use can be proposed. They will thus have a homogeneous starting point, and their responsibility will be to reflect on the presence of the impact of all the groups of risk on their business and to the tools recommended for managing risk. When there is a higher volume of business replicating the MIGGRI model, benchmarking the risks and the respective risk management in innovative TSP environments will be undertaken. Furthermore, it is suggested that the application is broken down into different forms of innovation: innovation in a radical or incremental product and innovation in a radical or incremental service.
How to cite this article: Etges, A. P. B. S., Souza, J. S., & Kleimann Neto, F. J. (2017). Risk management for companies focused on innovation processes. Production, 27, e20162209. http://dx.doi.org/10.1590/0103-6513.220916
References
Bandyopadhyay, K., Mykytyn, P. P., & Mykytyn, K. (2006). A framework for integrated risk management in information technology. Management Decision, 37(5), 437-445. http://dx.doi.org/10.1108/00251749910274216. [ Links ]
Bessant, J. (2003). Challenges in innovation management. In L. V. Shavinina (Ed.), The International Handbook on Innovation. Oxford: Elsevier. [ Links ]
Boer, H., Holweg, M., Kilduff, M., Pagell, M., Smurfit, M., Schmenner, R., & Voss, C. (2015). Making a meaningful contribution to theory. International Journal of Operations & Production Management, 35(9), 1231-1252. http://dx.doi.org/10.1108/IJOPM-03-2015-0119. [ Links ]
Bromiley, P., Mcshane, M., Nair, A., & Rustambekov, E. (2015). Enterprise Risk Management: review, critique, and research directions. Journal Long Range Planning, 48(4), 265-276. http://dx.doi.org/10.1016/j.lrp.2014.07.005. [ Links ]
Canongia, C., Santos, D. M., Santos, M. M., & Zackiewicz, M. (2004). Foresight, Inteligência competitiva e gestão do conhecimento: instrumentos para a gestão da inovação. Revista Gestão & Produção, 11(2), 231-238. http://dx.doi.org/10.1590/S0104-530X2004000200009. [ Links ]
Committee of Sponsoring Organizations of the Treadway Commission. (2007). Gerenciamento de riscos corporativos: estrutura integrada (2nd ed.). United States: COSO. [ Links ]
Cooper, R. G. (2003). Profitable product innovation: the critical success factors. In L. V. Shavinina (Ed.), The International Handbook on Innovation. Oxford: Elsevier. [ Links ]
Cooper, R., & Kaplan, R. S. (1991). Profit priorities from activity-based costing. Harvard Business Review, 69(3), 130-135. [ Links ]
Crossan, M. M., & Apaydin, M. (2010). A multi-dimensional framework of organizational innovation: a systematic review of the literature. Journal of Management Studies, 47(6), 1154-1191. http://dx.doi.org/10.1111/j.1467-6486.2009.00880.x. [ Links ]
Damodaran, A. (2009). Gestão estratégica do risco. Porto Alegre: Bookman. [ Links ]
Dutra, C., Ribeiro, J. L. D., & Carvalho, M. M. (2014). An economic-probabilistic model for project selection and prioritization. International Journal of Project Management, 32(6), 1042-1055. http://dx.doi.org/10.1016/j.ijproman.2013.12.004. [ Links ]
Etges, A. P. B. S., & Souza, J. S. (2014). Estudo de Campo sobre Gestão de Riscos Corporativos em empresas participantes de um Parque Científico Tecnológico. In Anais do IV Congresso Internacional de Conocimento e Inovación, Equador. [ Links ]
Freeman, C., & Soete, L. (2008). A economia da inovação industrial. São Paulo: Unicamp. [ Links ]
Frigo, M. L., & Anderson, R. J. (2011). Strategic risk management: a foundation for improving enterprise risk management and governance. Journal of Corporate Accounting & Finance, 22(3), 81-88. http://dx.doi.org/10.1002/jcaf.20677. [ Links ]
Gaino, P. A. A., & Pamplona, B. J. (2012). Abordagem teórica dos condicionantes da formação e consolidação dos Parques Tecnológicos. Revista Produção, 20(10), 177-187. [ Links ]
Garcia, R., & Calantone, R. (2002). A critical look at technological innovation typology and innovativeness terminology: a literature review. Journal of Product Innovation Management, 19(2), 110-132. http://dx.doi.org/10.1016/S0737-6782(01)00132-1. [ Links ]
Giugliani, E. A. (2012). Governance model for science and technology parks: what are the STP’s evolving into? Tallin: International Association of Science Parks. [ Links ]
Hansen, S. O., & Wakonen, J. (1997). Innovation, a winning solution? International Journal of Technology Management, 13(4), 345. http://dx.doi.org/10.1504/IJTM.1997.001668. [ Links ]
Hauser, J., Tellis, G. J., & Griffin, A. (2006). Research on innovation: a review and agenda for marketing science. Marketing Science, 25(6), 687-717. http://dx.doi.org/10.1287/mksc.1050.0144. [ Links ]
Hayne, C., & Free, C. (2014). Hybrid professional groups and institutional work: COSO and the rise of enterprise risk management. Accounting, Organizations and Society, 39(5), 309-330. http://dx.doi.org/10.1016/j.aos.2014.05.002. [ Links ]
International Organization for Standardization. (2009). ISO 31000:2009: risk management: principles and guidelines 2009. Geneva: ISO. [ Links ]
Jong, J. P. J., & Marsili, O. (2006). The fruit of innovations: a taxonomy of innovative small firms. Research Policy, 35(2), 213-229. http://dx.doi.org/10.1016/j.respol.2005.09.007. [ Links ]
Jorion, P. (2010). Risk management. California: School of Business, University of California. [ Links ]
Kaplan, R., & Norton, D. (1996), Translating strategy into action: the balanced scorecard. United States of America: Harvard College. [ Links ]
Kliemann, J.F. (2010). A gestão de riscos como ferramenta para aumento da competitividade das empresas. In Anais do XXX ENEGEP, Rio de Janeiro. [ Links ]
Koskela, L., & Howell, G. (2002). The underlying theory of project management is obsolete. Pennsylvania: Project Management Institute. [ Links ]
Lazzari, F., Bampi, R. E., & Milan, G. S. (2014). Os esforços de inovação e sua relação com alguns indicadores de desempenho do negócio. Revista Produção Online, 14(1), 58-83. http://dx.doi.org/10.14488/1676-1901.v14.i1.1428. [ Links ]
Miorando, R. F. (2010). Modelo econômico probabilístico de análise de risco em empresas de TI (Doctor thesis). Universidade Federal do Rio Grande do Sul, Porto Alegre. [ Links ]
Miorando, R. F., Ribeiro, J. L. D., & Cortimiglia, M. N. (2014). An economic-probabilistic model for risk analysis in technological innovation projects. Technovation, 34(8), 485-498. http://dx.doi.org/10.1016/j.technovation.2014.01.002. [ Links ]
Mortimer, A. L. (1995). Managing innovation and risk. World Class Design to Manufacture, 2(5), 38-42. http://dx.doi.org/10.1108/09642369310095210. [ Links ]
Nagano, M. S., Stefanovitz, J. P., & Vick, T. L. (2014). Innovation management processes, their internal organizational elements and contextual factors: an investigation in Brazil. Journal of Engineering and Technology Management, 33, 63-92. http://dx.doi.org/10.1016/j.jengtecman.2014.02.004. [ Links ]
Nanda, R., & Rhodes-Kropf, M. (1999). Investment cycles and startup innovation. Journal of Financial Economics, (110), 1. [ Links ]
O’Connor, G. C., Ravichandran, T., & Robeson, D. (2008). Risk management through learning: management practices for radical innovation success. The Journal of High Technology Management Research, 19(1), 70-82. http://dx.doi.org/10.1016/j.hitech.2008.06.003. [ Links ]
Pinheiro, J. I., Carvajal, S. S. R., Cunha, S. B., & Gomes, G. C. (2012). Probabilidade e estatística: quantificando a incerteza. São Paulo: Elsevier. [ Links ]
Pittaway, L., Robertson, M., Munir, K., Denyer, D., & Neely, A. (2004). Networking and innovation: a systematic review of the evidence. International Journal of Management Reviews, 5-6(3-4), 137-168. http://dx.doi.org/10.1111/j.1460-8545.2004.00101.x. [ Links ]
Project Management Institute. (2008). PMBOK: a guide to the project management body of Knowledge (4th ed.). Pennsylvania: PMI. [ Links ]
Smith, K. G., Collins, C. J., & Clark, K. D. (2005). A behavioral theory of R&D expenditures and innovations: evidence from shipbuilding. Academy of Management Journal, 48(2), 346-357. http://dx.doi.org/10.5465/AMJ.2005.16928421. [ Links ]
Souza, J. S. (2011). Modelo para Identificação e Gerenciamento do Grau de Risco de Empresas – MIGGRI (Doctor thesis). Universidade Federal do Rio Grande do Sul, Porto Alegre. [ Links ]
Standards Association of Australia. (1999). AS/NZS 4360: risk management. Strathfield. [ Links ]
Trott, P. (2012). Gestão da inovação e desenvolvimento de novos produtos. São Paulo: Bookman. [ Links ]
Tu, C., Hwang, S., & Wong, J. (2014). How does cooperation affect innovation in micro-enterprises? Management Decision, 52(8), 1390-1409. http://dx.doi.org/10.1108/MD-07-2013-0388. [ Links ]
Viana, R. B. C. (2012). A estratégia na incerteza: explorando cenários para empresas recém-criadas no mercado brasileiro de tecnologia da informação. Reuna, 17(3), 85-98. [ Links ]
Whetten, D. A. (2003). O que constitui uma contribuição teórica. Revista de Administração de Empresas, 43(3), 69-73. [ Links ]
Wong, S., & Chin, K. (2007). Organizational innovation management. Industrial Management & Data Systems, 107(9), 1290-1315. http://dx.doi.org/10.1108/02635570710833974. [ Links ]
Yang, J. (2012). Innovation capability and corporate growth: an empirical investigation in China. Journal of Engineering and Technology Management, 29(1), 34-46. http://dx.doi.org/10.1016/j.jengtecman.2011.09.004. [ Links ]
Zeng, S., Xie, X., & Tam, C. (2010). Evaluating innovation capabilities for Science parks: a system model. Technological and Economic Development of Economy, 16(3), 397-413. http://dx.doi.org/10.3846/tede.2010.25. [ Links ]
Zouggari, A., & Benyoucef, L. (2012). Simulation based fuzzy TOPSIS approach for group multicriteria supplier selection problem. Engineering Applications of Artificial Intelligence, 25(3), 507-519. http://dx.doi.org/10.1016/j.engappai.2011.10.012. [ Links ]
Received: September 09, 2016; Accepted: April 26, 2017
This is an Open Access article distributed under the terms of the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. | 2018-01-23 10:41:06 | {"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": 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": 0.27932363748550415, "perplexity": 1974.27315098431}, "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-2018-05/segments/1516084891886.70/warc/CC-MAIN-20180123091931-20180123111931-00766.warc.gz"} |
http://codereview.stackexchange.com/questions/20393/using-statement-for-oledb-is-this-overkill | # Using Statement for OleDB - Is this overkill?
I am still trying to wrap my head around good programming practice and just wrote this routine to read in an Excel file and return all the sheets into a dataset:
Public Shared Function ReadExcelIntoDataSet(ByVal FileName As String, Optional ByVal TopRowHeaders As Boolean = False) As DataSet
Try
Dim retval As New DataSet
Dim strConnString As String = "Provider=Microsoft.ACE.OLEDB.12.0;Data Source=" & _
FileName & ";Extended Properties=""Excel 12.0;IMEX=1;" & _
"HDR=" & If(TopRowHeaders, "Yes", "No") & """;"
Using oleExcelConnection = New OleDb.OleDbConnection(strConnString)
oleExcelConnection.Open()
Using oleExcelCommandString = New OleDb.OleDbCommand("", oleExcelConnection)
Dim dtSchema As DataTable = oleExcelConnection.GetOleDbSchemaTable(OleDb.OleDbSchemaGuid.Tables, Nothing)
For Each dr As DataRow In dtSchema.Rows
Dim Sheetname As String = dr.Item("TABLE_NAME").ToString
If Sheetname.EndsWith("\$") Then
oleExcelCommandString.CommandText = "Select * From [" & Sheetname & "]"
End If
Next
Return retval
End Using
End Using
End Using
Catch ex As Exception
Dim ErrorData As String = "Unable to Read in Data from the Excel File " & FileName & vbCrLf & _
Throw New Exception(ErrorData & vbCrLf & "ERROR Data:" & vbCrLf & ex.Message)
End Try
End Function
My first questions are:
1. Is the use of the Using statements good or is it some kind of over-kill and should this maybe be handled in a Finally statement in stead?
2. Is this the best method to read in an Excel File (Given that they can be either .xls or .xlsx) in the fastest possible way?
3. What is the best way to pass the error up to the calling code? Did my Catch statement do that well enough??
-
1. The Using is good practice and not overkill. Trying to dispose the resources in the Finally part has the disadvantage, that you don't know when the exception happened and which resources have been assigned by that time. The Using statement does this automatically for you internally: If resouce <> Nothing Then resource.Dispose() End If
2. You need different connection strings for the different Excel types. You can use this function:
Private Function GetConnectionString(ByVal filename As String) As String
Const xlsConnString As String = "Provider=Microsoft.Jet.OLEDB.4.0;Data Source={0};Extended Properties=Excel 8.0;"
Const xlsxConnString As String = "Provider=Microsoft.ACE.OLEDB.12.0;Data Source={0};Extended Properties=Excel 12.0;"
If Path.GetExtension(filename).Length = 5 Then ' .xlsx, .xltx
Return String.Format(xlsxConnString, filename)
Else ' .xls, .xlt
Return String.Format(xlsConnString, filename)
End If
End Function
1. You can either propagate the exception or return Nothing. In my opinion, your version does it very well.
-
Thank you for such a detailed response!!! – John Bustos Jan 10 '13 at 20:11
... One more question, if you'd be willing to answer it - It seems that the ACE.OLEDB connection opens the .xls files too... Is there any reason to use the Jet provider?? – John Bustos Jan 10 '13 at 20:12
If it does it's okay! I supported only xls first and added support for xlsx later in one of my projects and did not realize that ACE.OLEDB supports the older format as well. – Olivier Jacot-Descombes Jan 10 '13 at 20:16
Thanks!!! - Makes sense :) – John Bustos Jan 10 '13 at 20:16 | 2014-08-01 11:52:04 | {"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": 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": 0.2896488606929779, "perplexity": 8608.595746063567}, "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/1406510274979.56/warc/CC-MAIN-20140728011754-00354-ip-10-146-231-18.ec2.internal.warc.gz"} |
https://ask.sagemath.org/question/57962/obtaining-admissible-relations-for-acyclic-tree-quivers-with-sage-for-qpa/ | # Obtaining admissible relations for acyclic tree quivers with Sage for QPA
Let $Q$ be a finite acyclic quiver which is a tree as an undirected graph.
Question: Is there a way to use Sage to obtain all admissible ideals $I$ in the quiver algebra $KQ$ (those are simply the ideals generated by paths of lenghts at least two)?
The output should be so that it can be read by the GAP-package QPA.
Here an example:
The input is the quiver $Q$ given by
Quiver( ["v1","v2","v3","v4"], [["v1","v2","a1"],["v2","v3","a2"],["v3","v4","a3"]] )
The output is the 5 admissible ideals given as follows:
[ [],[ a1*a2, a2*a3 ], [ a2*a3 ], [ a1*a2 ], [ a1*a2*a3 ] ]
(more generally the number of admissible ideals for a linear oriented line quiver is given by the Catalan numbers)
edit retag close merge delete | 2022-05-21 00:48:28 | {"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": 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": 0.2290140986442566, "perplexity": 1845.0056435435272}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662534693.28/warc/CC-MAIN-20220520223029-20220521013029-00574.warc.gz"} |
http://sourceforge.net/p/texstudio/discussion/907840/thread/ecc4edb6/ | About placeholders and completion
TXS - Help
2012-09-23
2015-06-09
• Hello, thank you first of all for this wonderful latex editor.
I have a little problem with placeholders. When the option "use placeholders",
if I type
\begin{theorem} and ENTER, then this appears :
\begin{theorem}
(content...) (the cursor is here in a red box)
\end{theorem}
the problem is that I would like to remove this red box (it is really annoying
to write in such box, since I cannot see anymore some symbols such as
underscore _.
When I desactivate the option "Use placeholders", then the same command
yields:
\begin{theorem}
content...
\end{theorem} (the cursor is now here : outside the environment ! )
This is probably a bug since the cursor now lies outside the environment,
while content... is written inside !
What I would like : simply the cursor inside the environment, with nothing
written inside, and no box !
Third point : when I desactivate the option "Automatically start completion
when typing latex command", I do not have the
nice completion "end{theorem}" after typing "begin{theorem} +ENTER". Why is
that so ? This seems to be a bug,
I would like to not have automatic completion (which is a huge pain, when we
type something like \t +Enter (to go to a new line) and \t is precisely what
we mean, nothing else), but still completion for environments.
I hope this can be solved,
thanks a lot for your help.
• Tim Hoffmann
2012-09-24
the problem is that I would like to remove this red box (it is really
annoying to write in such box, since I cannot see anymore some symbols such as
underscore _.
As a temporary solution, you could increase the line spacing (Configure ->
Editor -> Line Spacing).
This is probably a bug since the cursor now lies outside the environment,
while content... is written inside !
Not exactly a bug. Without placeholders just the plain text is inserted.
However we may improve this behaviour.
Environment completion without auto-completion:
Completion for the environments is implemented via the auto-completion
mechanism. I don't think it's currently possible to split that up.
• Kendon Bell
2015-06-09
The "content ..." box seems unnecessary to me. I always delete it right away. I'd vote for it to be removed. Or at least have an option to permanently remove this. | 2015-08-31 07:17:01 | {"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": 0.8858820796012878, "perplexity": 3916.5054848585573}, "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-2015-35/segments/1440644065828.38/warc/CC-MAIN-20150827025425-00247-ip-10-171-96-226.ec2.internal.warc.gz"} |
https://mathoverflow.net/questions/272222/graph-properties-that-imply-a-bounded-number-of-edges/272290 | # Graph properties that imply a bounded number of edges
Many combinatorial problems can be reduced to bounding the number of edges in a given graph with $n$ vertices. Each time I encounter such a problem, I check whether the corresponding graph has a property that is known to imply a bound on number of edges (as probably most people do). For example, whether the graph is planar.
My question is what graph properties imply a bounded number of edges. I assume that there are such properties that I am not familiar with, and it seems quite useful to have a list of these properties. I am only interested in cases where the number of edges is asymptotically smaller than $n^2$ (for example, Turan's Theorem is not relevant).
Some properties that I am already familiar with:
• Planar graphs have $O(n)$ edges. There are several variants, such as quasi-planar graphs, with linear or almost linear bounds.
• The Zarankiewicz problem states that a graph that contains no copy of $K_{s,t}$ has $O(n^{2-1/s})$ edges (this is the formulation for the case where $s$ and $t$ are constants).
• Moore's bound states that a graph of girth larger than $2k$ contains $O(n^{1+1/k})$ edges.
• Families of graphs that are closed under taking induced subgraphs and have sufficiently small separators have $O(n)$ edges (e.g., see Fox and Pach).
• Any minor-closed class of graphs that excludes at least one complete graph has only a linear number of edges. – Gordon Royle Jun 15 '17 at 0:31
• Graphs with $n$ vertices and excluding $K_t$ as a topological minor (i.e. as a subdivision) have $O(t^2 n)$ edges. – David Wood Jun 15 '17 at 9:13
• Graphs with $n$ vertices that can be drawn on a fixed surface with $O(n)$ crossings have $O(n)$ edges. This follows from the crossing lemma. More generally, graphs with $n$ vertices that can be drawn on a fixed surface with $O(n^{4-3\epsilon})$ crossings have $O(n^{2-\epsilon})$ edges. – David Wood Jun 15 '17 at 9:27
• In the comment about the Zarankiewicz problem, should $O(n^{1-1/s})$ be $O(n^{2-1/s})$? – David Wood Jun 15 '17 at 9:42
• The Zarankiewicz problem generalises to any bipartite Turan problem (i.e. forbidding any fixed bipartite graph as a subgraph). The bound for graphs of girth larger than $2k$ fits into this framework as Bondy and Simonovits proved that any graph without a cycle of length $2k$ has at most $20kn^{1+1/k}$ edges. – Jon Noel Jun 16 '17 at 6:48
$n$-vertex graphs in a hereditary class $X$ have $o(n^2)$ vertices if and only if graphs in $X$ are ($K_p$, $K_{t,t}$)-free for some constants $p$ and $t$ (i.e., do not contain $K_p$ and $K_{t,t}$ as induced subgraphs). This is equivalent to saying that graphs in $X$ do not contain $K_{q,q}$ as a subgraph for some constant $q$.
1. Planar graphs are $(K_5, K_{3,3})$-free;
2. Graphs of girth larger than 4 are $(K_3, K_{2,2})$-free. | 2020-12-02 10:20:45 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 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": 0.7583175897598267, "perplexity": 203.9273271410553}, "config": {"markdown_headings": true, "markdown_code": false, "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-2020-50/segments/1606141706569.64/warc/CC-MAIN-20201202083021-20201202113021-00468.warc.gz"} |
https://infoscience.epfl.ch/record/205739 | ## Measurement of CP asymmetries in the decays B-0 -> K-*0 mu(+) mu(-) and B+ -> K+ mu(+) mu(-)
The direct CP asymmetries of the decays B-0 -> K-*0 mu(+) mu(-) and B+ -> K+ mu(+) mu(-) are measured using pp collision data corresponding to an integrated luminosity of 3.0 fb(-1) collected with the LHCb detector. The respective control modes B-0 -> J/psi K+ and B+ -> J/psi K+ are used to account for detection and production asymmetries. The measurements are made in several intervals of mu(+)mu(-) invariant mass squared, with the phi(1020) and charmonium resonance regions excluded. Under the hypothesis of zero CP asymmetry in the control modes, the average values of the asymmetries are A(CP)(B-0 -> K-*0 mu(+) mu(-) and B+ -> K+ mu(+) mu(-)) = -0.035 +/- 0.024 +/- 0.003, A(CP)(B+ -> K+ mu(+) mu(-) = 0.012 +/- 0.017 +/- 0.001, where the first uncertainties are statistical and the second are due to systematic effects. Both measurements are consistent with the Standard Model prediction of small CP asymmetry in these decays.
Published in:
Journal of High Energy Physics, 09, 177
Year:
2014
Publisher:
New York, Springer
ISSN:
1029-8479
Keywords:
Laboratories: | 2018-10-22 17:06:29 | {"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": 0.8200635313987732, "perplexity": 10136.003524205049}, "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-2018-43/segments/1539583515352.63/warc/CC-MAIN-20181022155502-20181022181002-00031.warc.gz"} |
https://www.jipb.net/EN/Y2010/V52/I1/4 | J Integr Plant Biol. ›› 2010, Vol. 52 ›› Issue (1): 4-7.
• Editorial •
### Plant Vascular Biology and Agriculture
William J. Lucas
1. The Editor of this Special Issue Dept. of Plant Biology College of Biological Sciences University of California Davis, CA 95616 U.S.A.
• Received:2010-01-01 Accepted:2010-01-01 Published:2010-01-01
Abstract:
The evolution of animal and plant vascular systems played a pivotal role in the advancement from simple to complex organisms, through the provision of a delivery system for the distribution of components essential for both metabolism and growth. Interestingly, although these two vascular systems conform to the same general rules of fluid dynamics (Murray 1926; McCulloh et al. 2003), the developmental mechanisms adopted by plants and animals, to generate these long-distance transport systems, have little in common. In animals, the arterial and venous system of tubules circulates blood, as an extracellular fluid, around the body of the organism by means of a pressure gradient generated by the heart. This system allows for the delivery of signal molecules, such as metabolites, peptides and proteins, from sites of production to target tissues. The circulatory nature of the animal vascular system allows for feedback to occur between distantly located tissues and organs.
The Unique Aspects of the Plant Vascular System
The plant vascular system is comprised of xylem and phloem conducting elements. The xylem transpiration system, which conducts water and mineral nutrients from the roots to above ground organs such as stems, leaves, flowers and fruits, is derived from stem cells (termed procambial cells) that differentiate into xylem conducting elements. Upon expansion, these cells undergo a process of programmed cell death (PCD), thereby giving rise to files of tracheary elements that form a low resistance pathway for the flow of water which is driven by a tensional gradient established within the cell walls of transpiring leaves; i.e., water is pulled up the body of the plant. The phloem translocation system, which delivers sugars, amino acids, mineral nutrients and hormones to heterotrophic tissues, is also derived from cambial cells that differentiate into sieve cells and their associated companion cells. In angiosperms, these sieve cells undergo partial PCD, thereby giving rise to files of evacuolate and enucleote cells connected, end-toend, by sieve plates (containing large open pores) to form conducting tubes. Retention of the plasma membrane allows the sieve tube system to function as an osmotic unit; loadingof sugars in the mature leaves (source of photosynthetically fixed sugars) generates a high turgor pressure, whereas unloading of sugars in developing tissues causes a drop in turgor pressure. These osmotically-driven turgor pressures establish a positive pressure gradient that drives the phloem translocation stream from autotrophic to heterotrophic tissues and organs.
Plant Vascular Development
Clearly, an important functional difference between the animal and plant vascular systems is that the latter is non-circulatory in nature; thus, direct feedback signaling cannot occur between mature leaves and, for example, developing shoot meritstems (Lough and Lucas 2006). This raises the interesting question as to the mechanisms developed by plants to coordinate physiological and developmental processes at the whole-plant level. An important facet of plant vascular development relates to the production of functional xylem and phloem conducting tissues. This process must be both spatially and temporally controlled to permit the formation of transport systems that can deliver optimal quantities of water and nutrients to distantlylocated tissues and organs. Recent studies have provided important insights into this process. In this Special Issue, Hirakawa et al. (2010), review our progress in understanding the complex signaling events that underlie regulation of xylem and phloem cell differentiation from procambial cells. The action of non-cell-autonomous signals, involving CLE peptides, is discussed, as is the role of intercellular signaling from both the xylem and phloem as necessary inputs to coordinate vascular organization.
Secondary Growth & Wood Formation
Secondary growth of the plant vascular system is a very important process, as it leads to an expanded of the cambium, derived from the procambium, gives rise to woody tissues which can afford additional mechanical support – this feature has allowed for the growth of perennial plantsthat can live for several thousand years and reach to heights of 100 meters or more. Obviously, an understanding of the evolutionary processes that have allowed for such longevity, as well as the formation of economically important tissues as wood, are of basic and applied importance. To lead us along this pathway, Du and Groover (2010) provide a review of recent studies in this area that indicate a role for transcriptional networks in regulating plant secondary growth. Future progress in this area will most surely have important applications in terms of engineering forest trees with improved traits.
Phloem as a Vascular Information Superhighway
Although it was long held that the phloem translocation stream carried only nutrients to support the growth of developing tissues, recent studies have revealed the presence of a complex sets of proteins and nucleic acids (Balachandran et al. 1997; Ruiz-Medrano et al. 1999; Lough and Lucas 2006; Lin et al. 2009). These findings support the hypothesis that the phloem functions as an information superhighway (Jorgensen et al. 1998). The presence of mRNA, as well as small interfering (si-) and micro (mi-)RNA, in the angiosperm sieve tube system raised important questions as to their possible roles in coordinating physiological and developmental processes, at the whole-plant level (Lough and Lucas 2006). Wang and Ding (2010) review these new findings and describe the use of plant viroids as a powerful tool to dissect the molecular determinants controlling the entry and exit of RNA species into and out of the sieve tube system.
Important agricultural traits, such as photoperiodic induction of flowering, have long been known to be controlled by florigenic agent(s) delivered to the shoot apex by the phloem (Zeevaar 1962). Recently, an important component of florigen was shown to be FLOWERING LOCUS T, a 20 kDa protein the sieve tube system for translocation to the shoot apex (Corbesier et al. 2007; Lin et al. 2007; Mathieu et al. 2007; Tamaki et al. 2007). A role for long-distance transport of RNA in leaf development has been established (Ruiz-Medrano et al. 1999; Haywood et al. 2005) and now, in this Special Issue, Hannapel (2010) reviews recent findings that illustrate the role of phloem-delivery of mRNA in the orchestration of tuber formation in the agriculturally important crop, potato. Collectively, these recent discoveries indicate that pioneering studies on the phloem will likely underpin the bioengineering of novel long-distance signaling systems that will afford unique control over such important agronomically important traits as partitioning of photosynthetically fixed carbon.
Cytokinins and Vascular Signaling
It has long been known that plant hormones can be transported through the vascular system. The involvement of a class of phytohormones, known as the cytokinins, in local and longdistance signaling is well established. The role of cytokinins in nitrogen homeostasis has been the subject of intensive investigation by a number of plant scientists. Here, Kudo et al. (2010) review progress in this area of vascular signaling and show that root-to-shoot delivery of trans-zeatin, via the xylem, plays an important role in nitrogen metabolism. Phloem delivery of N6-(2-isopentenyl) adenine-type cytokinins to the roots and developing shoot apices appears to be important for developmental regulation. As cytokinins appear to be involved in the regulation of a multitude of physiological and developmental activities, one could consider this class of phytohormones as true systemic regulators! Knowledge concerning the longdistance transport of cytokinins may well pave the way for engineering of plants with both enhanced nutrient use efficiency and overall yield.
Nitrogen Fixation and Vascular Signaling
Fixation of atmospheric nitrogen in legumes, through a symbiotic relationship with rhizobia, makes an important contribution to global nitrogen nutrition. This plant-rhizobium interaction involves a complex signaling network which is reviewed in this Special Issue by Ferguson et al. (2010). Interestingly, to optimize nitrogen acquisition and utilization, legumes have evolved an intricate mechanism to regulate the level to which their root systems will respond to soil-borne bacteria to allow nodule development. In the presence of adequate exogenously supplied nitrogen, nodulation is down-regulated; this process ensures optimal utilization of carbon allocation – putting it another way, nodulation and rhizobial nitrogen fixation are processes that cost the plant carbon resources that could otherwise be used for growth and cellular maintenance. Ferguson et al. (2010) describe recent insights afforded into the role played by the plant vascular system as an integrator in this process of autoregulation of nodulation. Obviously, a comprehensive understanding of the processed underlying symbiotic nitrogen fixation will have an immeasurable impact not agriculture and its capacity to feed the peoples of the world.
Root-to-shoot Signaling Systems Control Physiological and Developmental Programs
The pioneering studies of Fritz Went (1943) revealed the involvement of root-derived signals that appeared to regulategrowth of the vegetative regions of the plant. The roles of these signals in water use efficiency, control over shoot branching and overall shoot growth are reviewed in this Special Issue by Sieburth and Lee (2010). The identification of a novel protein, BYPASS1, is discussed with respect to its function in interdicting the synthesis of a novel root-to-shoot signal; this molecule may well be a branch product of the carotenoid biosynthetic pathway. Xylem delivery of this BYPASS1-regulated molecule appears to modify shoot growth through its affect on local auxin signaling. Insights from these and other studies on root-to-shoot signaling, via the xylem transpiration stream, suggests the potential for engineering of plants with enhanced agronomic traits, including higher water use efficiency, novel branching patterns and enhanced growth characteristics under non-optimal environmental conditions.
Vascular Defense – a Role for Secondary Metabolites
As mentioned above, secondary growth gave rise to woody perennial plants whose life cycles can span many centuries to even thousands of years. Given that plants are sessile, such longevity can come with a price – they become prime targets for attack by insects and other pathogens, such as various fungi and bacteria. Through the course of their evolution, woody plants have developed a number of strategies to protect themselves from such attacks. Zulak and Bohlmann (2010) delve into pathogen-induced terpenoid biosynthesis as an effective means to produce secondary metabolites that can block the attacks mounted on confers by a range of pests and pathogens. In their review, the authors illustrate how plants use a combination of specialized anatomical features, such as xylem resin ducts, and oleoresin terpenoids to mount a defense against insect attack.
Flavonoids as Local and Long-distance Signaling Agents
A role for flavonoids as low molecular weight signaling molecules is examined in the review by Buer et al. (2010). Here, again, we learn how plants have utilized their capacity for secondary metabolite production to synthesize a wide range of structural variants through the flavonoid branch of the phenylpropanoid pathway. Flavonoids are involved in a broad spectrum of physiological and developmental process, including modulating hormone signaling, functioning as components in such signaling cascades as those involved in legumebacteria symbiosis, and plant defense. Recent studies have shown that flavonoids are transported throughout the plant and, as discussed by Buer et al. (2010), studies on the mode of transport will provide important new insights into the manner inwhich these secondary metabolites influence plant growth and development.
Biofuels Production
Woody perennial plants produce significant biomass on a seasonal basis. The ligno-cellulosic content of the secondary xylem can be utilized as a renewable energy source. As worldwide energy demand increases, many strategies to expand their use of biomass energy production in ways that will avoid negative effects on food production and security. In this Special Issue, Tang et al. (2010) analyze the capacity for energy production within China using biomass grown on marginal lands. This review nicely highlights the potential associated with bringing approximately 45 million hectares of marginal land into biomass energy production. The biomass cropping systems and approaches being developed in China could well have utility in other regions of Asia.
Editorial Office, Journal of Integrative Plant Biology, Institute of Botany, CAS
No. 20 Nanxincun, Xiangshan, Beijing 100093, China
Tel: +86 10 6283 6133 Fax: +86 10 8259 2636 E-mail: jipb@ibcas.ac.cn | 2022-09-30 12:02:47 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 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": 0.500058650970459, "perplexity": 4598.408672870852}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030335469.40/warc/CC-MAIN-20220930113830-20220930143830-00288.warc.gz"} |
https://tex.stackexchange.com/questions/292514/wrapfigure-and-curly-braces | # wrapfigure and curly braces
\documentclass{article}
\usepackage{wrapfig}
\begin{document}
\begin{wrapfigure}{r}{3cm}
\centering
\rule{3cm}{3cm}
\end{wrapfigure}
g,sd; agmgts bsgfs shgsnhg {\bf snhbfx} fdbhsmn jnnsndg ndsngas gngdfznagk dfhbf
gsgfsh shgfs
{Here is the problem!} "Observe that after using the wrapfigure command, when I am writing something which has to be kept inside the curly braces something unsusal happening." How to resolve this?
\end{document}
• Just copy paste the whole code and see the problem – G_0_pi_i_e Feb 12 '16 at 10:36
Problems are expected, actually.
In order to flow several paragraphs around the image, wrapfig needs to use \everypar in order to pass new paragraphs some code to execute, until it is sure that the text has more lines than needed for going past the image.
What happens in your case? The second paragraph starts inside a group, so the settings for the paragraph shape get lost when the group ends.
How to solve the issue? I guess it happened because you wanted something like
{\bf Here is the problem!}
but you're the cause for your problems. The two letter font changing commands have been deprecated for more than twenty years.
Never ever use \bf, \it or similar commands. They have been deprecated for more than twenty years. Unexpected problems can show when you use commands that have been deprecated for more than twenty years.
I could go on with this rant, but I hope that three times is enough. ;-)
\documentclass{article}
\usepackage{wrapfig}
\begin{document}
\begin{wrapfigure}{r}{3cm}
\centering
\rule{3cm}{3cm}
\end{wrapfigure}
g,sd; agmgts bsgfs shgsnhg \textbf{snhbfx} fdbhsmn jnnsndg ndsngas gngdfznagk dfhbf
gsgfsh shgfs
\textbf{Here is the problem!} "Observe that after using the wrapfigure command, when I am writing
something which has to be kept inside the curly braces something unsusal happening." How to resolve
this?
\end{document}
If you really need {...} for some other purpose, just add \mbox{} in front of the opening brace, if it happens at the start of a paragraph in connection with a wrapfigure.
• @G_0_pi_i_e I hope you didn't take it personally. It's very curious that, notwithstanding the fact that those commands have been deprecated long ago, they still find their way in document and people teach them. So, whenever I have the occasion, I stress that they must not be used. – egreg Feb 12 '16 at 11:08 | 2019-10-15 07:16:55 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 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": 0.8033508658409119, "perplexity": 2462.375038968536}, "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/1570986657586.16/warc/CC-MAIN-20191015055525-20191015083025-00061.warc.gz"} |
http://math.stackexchange.com/questions/356793/orthogonal-latin-square-66/361110 | # Orthogonal Latin Square 6*6
I need to make remarks about Tarry's Proof for the nonexistence of 6x6 Latin Squares as part of my final exam for a class I'm in. Problem is, I can't find it ANYWHERE on the internet. I can only find minor comments about it that don't explain what he did. Does anyone know where I can find it online? Or does anyone know where I can find a detailed explanation of his proof? if you have other proof for this question please share that... i need to proof of this question
-
If you have any more Latin squares questions, please make sure to tag them [latin-square]. – Douglas S. Stones Apr 13 '13 at 19:29
Tarry, G. Le probleme des 36 officiers. Comptes Rendus Assoc. France Av. Sci. 29, part 2: 170–203, 1900
-
It's quite clear that the OP knows what the reference is, and is having trouble accessing it. – Douglas S. Stones Apr 13 '13 at 19:27
See my answer for a link on the original paper. A physical copy of the journal of the French "Association for l'Avancement des Sciences" is not so easy to find outside France (where most University libraries have it). – ogerard Apr 14 '13 at 9:08
Tarry's paper is available in digitized form at the online French National Library, http://gallica.bnf.fr/ as well as many volumes of 19th and early 20th century scientific journals.
Here is a link for the first page:
http://gallica.bnf.fr/ark:/12148/bpt6k2011936/f175.image
It is quite a long paper (34 pages) investing potential cases sorted according to various conditions. He starts from general 6x6 latin squares and recognizes that the 9408 individual cases ("permutation carrées" = square permutations = latin squares) can be sorted into 17 types. He coins several names such as "écusson" (small shield, with an heraldic meaning) as a way to present the type of permutations (in term of cycles) of the rows and columns of latin squares. He finds 16 different possible "écussons" and he sorts the 17 types into 3 classes, depending on presence of cycles of order 2.
Hope it will get you started.
If you have trouble with the french, or with using Gallica to download the whole article do not hesitate to ask specific questions.
-
Since your after remarks, here's some.
• It's easy enough these days to prove the non-existence on a computer. There are only $12$ main classes to exclude (ref.).
• There's a short proof of this result:
Stinson, A short proof of the nonexistence of a pair of orthogonal latin squares of order six, J. Comb. Theory A, 36 (1984) 373-376
• There's a recent graph theory proof of the non-existence of self-orthogonal Latin squares (Latin squares that are orthogonal to their transpose) of order $6$; see:
Burger, Kidd, and van Vuuren A graph-theoretic proof of the non-existence of self-orthogonal Latin squares of order 6, Disc. Math. (2011) 311, pp. 1223-1228.
PS. If you're desperately after the paper, I suggest you email either Brendan McKay and/or Ian Wanless. They're the most likely people to have obscure papers on Latin squares.
- | 2014-12-22 10:02:38 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.45728519558906555, "perplexity": 810.9606497629273}, "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-2014-52/segments/1418802775083.81/warc/CC-MAIN-20141217075255-00068-ip-10-231-17-201.ec2.internal.warc.gz"} |
http://trove.nla.gov.au/work/197135?q&versionId=209572 | # 2006, English, Article edition: The price of approximate stability for scheduling tasks on two links Angel, Eric; Bampis, Evripidis; Pascual, Fanny
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### Edition details
Title
• The price of approximate stability for scheduling tasks on two links
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• Angel, Eric
• Bampis, Evripidis
• Pascual, Fanny
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• 2006
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• We consider a \em scheduling game, where a set of selfish agents (traffic loads) want to be routed in exactly one of the two parallel links of a system. Every agent aims to minimize \em her own completion time, while the social objective is the \em makespan, i.e. the time at which the last agent finishes her execution. We study the problem of optimizing the makespan under the constraint that the obtained schedule is a (pure) Nash equilibrium, i.e. a schedule in which no agent has incentive to unilaterally change her strategy (link). We consider a relaxation of the notion of \em equilibrium by considering $\alpha$-approximate Nash equilibria, in which agents do not have \em sufficient incentive to unilaterally change their strategies: in an $\alpha$-approximate Nash equilibrium, no agent can decrease its cost by more than $\alpha$ multiplicative factor. Our main contribution is the study of the \em tradeoff between the \em approximation ratio for the makespan and the stability of the schedule (the value of $\alpha$). We first give an algorithm which provides a solution with an approximation ratio of $\frac87$ for the makespan and which is a 3-approximate Nash equilibrium, provided that the local policy of each link is \em Longest Processing Time (LPT). Furthermore, we show that a slight modification of the classical \em Polynomial Time Approximation Scheme (PTAS) of Graham allows to obtain a schedule whose makespan is arbitrarily close to the optimum while keeping a constant value for $\alpha$. Finally, we give bounds establishing relations between the value of $\alpha$ and the best possible value of the approximation ratio, provided that the local policies of the links are LPT. We also briefly examine the case where the local policies are SPT.
• HAL:hal-00341356, version 1
• HAL:http://hal.archives-ouvertes.fr/hal-00341356/en/
• DOI:10.1007/11823285
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Be the first to add a tag for this edition | 2016-10-26 11:37:13 | {"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": 0.7301488518714905, "perplexity": 2054.9323598878386}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-44/segments/1476988720941.32/warc/CC-MAIN-20161020183840-00391-ip-10-171-6-4.ec2.internal.warc.gz"} |
https://loelschlaeger.de/RprobitB/articles/v06_model_selection.html | The task of model selection targets the question: If there are several competing models, how do I choose the most appropriate one? This vignette1 outlines the model selection tools implemented in {RprobitB}.
For illustration, we revisit the probit model of travelers deciding between two fictional train route alternatives from the vignette on model fitting:
form <- choice ~ price + time + change + comfort | 0
data <- prepare_data(form = form, choice_data = Train)
model_train <- fit_model(
data = data,
scale = "price := -1",
R = 1000, B = 500, Q = 10
)
As a competing model, we consider explaining the choices only by the alternative’s price, i.e. the probit model with the formula choice ~ price | 0:
model_train_sparse <- update(model_train, form = choice ~ price | 0)
The update() function helps to estimate a new version of model_train with new specifications. Here, only form has been changed.
## The model_selection() function
{RprobitB} provides the convenience function model_selection(), which takes an arbitrary number of RprobitB_fit objects and returns a matrix of model selection criteria:
model_selection(model_train, model_train_sparse)
#> model_train model_train_sparse
#> npar 4 1
#> LL -1727.74 -1865.89
#> AIC 3463.49 3733.77
#> BIC 3487.42 3739.76
Specifying criteria is optional. Per default, criteria = c("npar", "LL", "AIC", "BIC").2 The available model selection criteria are explained in the following.
### npar
"npar" yields the number of model parameters, which is computed by the npar() method:
npar(model_train, model_train_sparse)
#> [1] 4 1
Here, model_train has 4 parameters (a coefficient for price, time, change, and comfort, respectively), while model_train_sparse has only a single price coefficient.
### LL
If "LL" is included in criteria, model_selection() returns the model’s log-likelihood values. They can also be directly accessed via the logLik() method:3
logLik(model_train)
#> 'log Lik.' -1727.744 (df=4)
logLik(model_train_sparse)
#> 'log Lik.' -1865.887 (df=1)
### AIC
Including "AIC" yields the Akaike’s Information Criterion (Akaike 1974), which is computed as $-2 \cdot \text{LL} + k \cdot \text{npar},$ where $$\text{LL}$$ is the model’s log-likelihood value, $$k$$ is the penalty per parameter with $$k = 2$$ per default for the classical AIC, and $$\text{npar}$$ is the number of parameters in the fitted model.
Alternatively, the AIC() method also returns the AIC values:
AIC(model_train, model_train_sparse, k = 2)
#> df AIC
#> model_train 4 3463.488
#> model_train_sparse 1 3733.773
The AIC quantifies the trade-off between over- and under-fitting, where smaller values are preferred. Here, the increase in goodness of fit justifies the additional 3 parameters of model_train.
### BIC
Similar to the AIC, "BIC" yields the Bayesian Information Criterion (Schwarz 1978), which is defined as $-2 \cdot \text{LL} + \log{(\text{nobs})} \cdot \text{npar},$ where $$\text{LL}$$ is the model’s log-likelihood value, $$\text{nobs}$$ is the number of data points (which can be accessed via the nobs() method), and $$\text{npar}$$ is the number of parameters in the fitted model. The interpretation is analogue to AIC.
{RprobitB} also provided a method for the BIC value:
BIC(model_train, model_train_sparse)
#> df BIC
#> model_train 4 3487.418
#> model_train_sparse 1 3739.756
### WAIC (with se(WAIC) and pWAIC)
WAIC is short for Widely Applicable (or Watanabe-Akaike) Information Criterion (Watanabe and Opper 2010). As for AIC and BIC, the smaller the WAIC value the better the model. Including "WAIC" in criteria yields the WAIC value, its standard error se(WAIC), and the effective number of parameters pWAIC, see below.
The WAIC is defined as
$-2 \cdot \text{lppd} + 2\cdot p_\text{WAIC},$
where $$\text{lppd}$$ stands for log pointwise predictive density and $$p_\text{WAIC}$$ is a penalty term proportional to the variance in the posterior distribution that is sometimes called effective number of parameters, see McElreath (2020) p. 220 for a reference.
The $$\text{lppd}$$ is approximated as follows. Let $p_{si} = \Pr(y_i\mid \theta_s)$ be the probability of observation $$y_i$$ (here the single choices) given the $$s$$-th set $$\theta_s$$ of parameter samples from the posterior. Then
$\text{lppd} = \sum_i \log \left( S^{-1} \sum_s p_{si} \right).$ The penalty term is computed as the sum over the variances in log-probability for each observation: $p_\text{WAIC} = \sum_i \mathbb{V}_{\theta} \log (p_{si}) .$ The $$\text{WAIC}$$ has a standard error of $\sqrt{n \cdot \mathbb{V}_i \left[-2 \left(\text{lppd} - \mathbb{V}_{\theta} \log (p_{si}) \right)\right]},$ where $$n$$ is the number of choices.
Before computing the WAIC of an object, the probabilities $$p_{si}$$ must be computed via the compute_p_si() function:4
model_train <- compute_p_si(model_train)
Afterwards, the WAIC can be accessed as follows:
WAIC(model_train)
WAIC(model_train_sparse)
### MMLL
"MMLL" in criteria stands for marginal model log-likelihood. The model’s marginal likelihood $$\Pr(y\mid M)$$ for a model $$M$$ and data $$y$$ is required for the computation of Bayes factors, see below. In general, the term has no closed form and must be approximated numerically.
{RprobitB} uses the posterior Gibbs samples derived from the mcmc() function to approximate the model’s marginal likelihood via the posterior harmonic mean estimator (Newton and Raftery 1994): Let $$S$$ denote the number of available posterior samples $$\theta_1,\dots,\theta_S$$. Then, $\Pr(y\mid M) = \left(\mathbb{E}_\text{posterior} 1/\Pr(y\mid \theta,M) \right)^{-1} \approx \left( \frac{1}{S} \sum_s 1/\Pr(y\mid \theta_s,M) \right) ^{-1} = \tilde{\Pr}(y\mid M).$
By the law of large numbers, $$\tilde{\Pr}(y\mid M) \to \Pr(y\mid M)$$ almost surely as $$S \to \infty$$.
As for the WAIC, computing the MMLL relies on the probabilities $$p_{si} = \Pr(y_i\mid \theta_s)$$, which must first be computed via the compute_p_si() function. Afterwards, the mml() function can be called with an RprobitB_fit object as input. The function returns the RprobitB_fit object, where the marginal likelihood value is saved as the entry "mml" and the marginal log-likelihood value as the attribute "mmll":5
model_train <- mml(model_train)
model_train$mml attr(model_train$mml, "mmll")
There are two options for improving the approximation: As for the WAIC, you can use more posterior samples. Alternatively, you can combine the posterior harmonic mean estimate with the prior arithmetic mean estimator (Hammersley and Handscomb 1964): For this approach, $$S$$ samples $$\theta_1,\dots,\theta_S$$ are drawn from the model’s prior distribution. Then,
$\Pr(y\mid M) = \mathbb{E}_\text{prior} \Pr(y\mid \theta,M) \approx \frac{1}{S} \sum_s \Pr(y\mid \theta_s,M) = \tilde{\Pr}(y\mid M).$
Again, it hols by the law of large numbers, that $$\tilde{\Pr}(y\mid M) \to \Pr(y\mid M)$$ almost surely as $$S \to \infty$$. The final approximation of the model’s marginal likelihood than is a weighted sum of the posterior harmonic mean estimate and the prior arithmetic mean estimate, where the weights are determined by the sample sizes.
To use the prior arithmetic mean estimator, call the mml() function with a specification of the number of prior draws S and set recompute = TRUE:6
model_train <- mml(model_train, S = 1000, recompute = TRUE)
Note that the prior arithmetic mean estimator works well if the prior and posterior distribution have a similar shape and strong overlap, as Gronau et al. (2017) points out. Otherwise, most of the sampled prior values result in a likelihood value close to zero, thereby contributing only marginally to the approximation. In this case, a very large number S of prior samples is required.
### BF
The Bayes factor is an index of relative posterior model plausibility of one model over another (Marin and Robert 2014). Given data $$\texttt{y}$$ and two models $$\texttt{mod0}$$ and $$\texttt{mod1}$$, it is defined as
$BF(\texttt{mod0},\texttt{mod1}) = \frac{\Pr(\texttt{mod0} \mid \texttt{y})}{\Pr(\texttt{mod1} \mid \texttt{y})} = \frac{\Pr(\texttt{y} \mid \texttt{mod0} )}{\Pr(\texttt{y} \mid \texttt{mod1})} / \frac{\Pr(\texttt{mod0})}{\Pr(\texttt{mod1})}.$
The ratio $$\Pr(\texttt{mod0}) / \Pr(\texttt{mod1})$$ expresses the factor by which $$\texttt{mod0}$$ a priori is assumed to be the correct model. Per default, {RprobitB} sets $$\Pr(\texttt{mod0}) = \Pr(\texttt{mod1}) = 0.5$$. The front part $$\Pr(\texttt{y} \mid \texttt{mod0} ) / \Pr(\texttt{y} \mid \texttt{mod1})$$ is the ratio of the marginal model likelihoods. A value of $$BF(\texttt{mod0},\texttt{mod1}) > 1$$ means that the model $$\texttt{mod0}$$ is more strongly supported by the data under consideration than $$\texttt{mod1}$$.
Adding "BF" to the criteria argument of model_selection yields the Bayes factors. We find a decisive evidence (Jeffreys 1998) in favor of model_train.
model_selection(model_train, model_train_sparse, criteria = c("BF"))
### pred_acc
Finally, adding "pred_acc" to the criteria argument for the model_selection() function returns the share of correctly predicted choices. From the output of model_selection() above (or alternatively the one in the following) we deduce that model_train correctly predicts about 6% of the choices more than model_train_sparse:7
pred_acc(model_train, model_train_sparse)
#> [1] 0.6971663 0.6340048
## References
Akaike, H. 1974. “A New Look at the Statistical Model Identification.” IEEE Transactions on Automatic Control 19. https://doi.org/10.1109/TAC.1974.1100705.
Gronau, Q. F., A. Sarafoglou, D. Matzke, A. Ly, U. Boehm, M. Marsman, D. S. Leslie, J. J. Forster, E. Wagenmakers, and H. Steingroever. 2017. “A Tutorial on Bridge Sampling.” Journal of Mathematical Psychology 81. https://doi.org/10.1016/j.jmp.2017.09.005.
Hammersley, J. M., and D. C. Handscomb. 1964. “General Principles of the Monte Carlo Method.” Springer Verlag. https://doi.org/10.1007/978-94-009-5819-7_5.
Jeffreys, Harold. 1998. The Theory of Probability. OUP Oxford.
Marin, J., and C. Robert. 2014. Bayesian Essentials with R. Springer Verlag. https://doi.org/10.1007/978-1-4614-8687-9.
McElreath, R. 2020. Statistical Rethinking: A Bayesian Course with Examples in R and Stan. Chapman; Hall/CRC. https://doi.org/10.1201/9780429029608.
Newton, M. A., and A. E. Raftery. 1994. “Approximate Bayesian Inference with the Weighted Likelihood Bootstrap.” Journal of the Royal Statistical Society: Series B (Methodological) 56 (1).
Schwarz, G. 1978. “Estimating the Dimension of a Model.” The Annals of Statistics 6.
Watanabe, S., and M. Opper. 2010. “Asymptotic Equivalence of Bayes Cross Validation and Widely Applicable Information Criterion in Singular Learning Theory.” Journal of Machine Learning Research 11 (12). | 2022-12-08 08:32:37 | {"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": 2, "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": 0.8033231496810913, "perplexity": 2471.1101284050005}, "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/1669446711286.17/warc/CC-MAIN-20221208082315-20221208112315-00299.warc.gz"} |
https://puzzling.stackexchange.com/questions/10645/male-and-female-functions/10647 | # Male and Female functions
You are given two goniometric functions:
• Male: $g(\rho, \Theta) = \rho \, \cos\Theta \sin\Theta$
• Female: $f(\rho, \Theta) = \rho \, \cos^2\Theta$
Why are these two functions labeled "male" and "female"?
• Now explain me why the female's value at 90° is 0... – leoll2 Mar 22 '15 at 14:25
• @leoll2 that's a good point ahaha, it should actually be cos²(π-Θ) – Marco Bonelli Mar 22 '15 at 14:33 | 2020-05-25 05:19:17 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 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": 0.3799595236778259, "perplexity": 5830.189528025008}, "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/1590347387219.0/warc/CC-MAIN-20200525032636-20200525062636-00102.warc.gz"} |
https://www.zbmath.org/?q=an%3A0928.35158 | zbMATH — the first resource for mathematics
Smoothing effects and local existence theory for the generalized nonlinear Schrödinger equations. (English) Zbl 0928.35158
The authors treat smoothing and local existence theory of generalized nonlinear Schrödinger equations where nonlinearities up to first order derivatives are allowed, and where the linear part may be nonelliptic. Those equations may model e.g. problems in water wave theory. One of their main results concerns the removal of size restriction for the initial data (which have to lie in certain Sobolev spaces) of local solutions; these solutions inherit a local smoothing effect in space time. The method consists in introducing the space derivatives of wave function as new variables, writing the original equation as a system, and using results of earlier papers of the same authors. The system has the advantage that the crucial term vanishes for $$t=0$$ if a solution of the original equation is inserted.
Reviewer: H.Lange (Köln)
MSC:
35Q55 NLS equations (nonlinear Schrödinger equations) 35A07 Local existence and uniqueness theorems (PDE) (MSC2000)
Full Text: | 2021-04-13 19:30:32 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 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": 0.594468355178833, "perplexity": 571.1769175512951}, "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-2021-17/segments/1618038074941.13/warc/CC-MAIN-20210413183055-20210413213055-00432.warc.gz"} |
https://www.physicsforums.com/threads/where-is-the-particle-most-likely-to-be-found.674730/ | Where is the particle most likely to be found?
1. Feb 26, 2013
Levi Tate
1. The problem statement, all variables and given/known data
Where is the particle most likely to be found in the first excited state (n=2) ? It is in a rigid box of length a
2. Relevant equations
wave function for a particle in a rigid box ψ(x)=(2/a)^1/2*sin(n[pi]x/a)
3. The attempt at a solution
The answer is a/4 and 3a/4 . I know I need to maximize the function by setting it equal to zero, but I haven't really done this since calc 1 and I kind of forgot.
If somebody could just guide me as to how to maximize this function,
Much appreciated.
2. Feb 26, 2013
haruspex
I think you want to maximise the probability function, which is real, by differentiation etc. The wave function is complex, so does not have a maximum in the same sense.
3. Feb 26, 2013
Levi Tate
The function I gave IS the normalized wavefunction for the particle in the rigid box mate. This is a 300 level intro to quantum mechanics class, it's not full out grad school type stuff yet. I just need to know how to maximize this function to get those values. It's an example in the book, Taylor 'Modern Physics' 2e. He just skips the step where he shows the maximization and I am having trouble getting those results. This comes down to a single variable calculus problem, which I am unfortunately struggling with.
4. Feb 26, 2013
tms
What can you say about any differentiable function at an extremum?
5. Feb 26, 2013
Levi Tate
Yeah it equals zero it's basic calc 1, i am having problems with the computation, not any of the ideas.
6. Feb 26, 2013
Dick
What did you get for the derivative?
7. Feb 26, 2013
tms
The function doesn't equal zero, its derivative does. (That may be what you meant, but it isn't what you said, and it's best to be absolutely clear.)
8. Feb 26, 2013
Levi Tate
It takes too long to write out the derivate and everything. This is a purely math issue I realized that I can resolve in a much shorter time with my teacher tomorrow.
Thanks a lot though mates, much appreciated. I'll save space and time to use when I actually have a question that isn't just a math issue, of which there will be several.
Thanks again though.
9. Feb 26, 2013
tms
Actually, you can just about get it by inspection. Where does the derivative of a sine equal zero?
10. Feb 26, 2013
Staff: Mentor
The wave function doesn't give you the probability directly. What do you have to do to it to get the probability distribution?
And you shouldn't have use the "set the derivative to zero" thing to find the maximum in this case. You should be able to figure it out by using the basic properties of trig functions.
11. Feb 26, 2013
haruspex
12. Feb 26, 2013
Levi Tate
Okay now I'm getting really confused about the physics.
Where I'm at in my studies here is that we're looking at a particle in a rigid box and the wavefunction for that, which was what I gave.
We are solving 2d and 3d problems currently but I'm trying to catch up in optics so I haven't looked at them yet.
Yeah you're exactly right, it's the maxima of the probability density that give the most probable value for the particle (x).
See in the example they just show a graph with two maxima at a/4 and 3a/4 for the probability density..
So what I was doing, sorry for the confusion, I was setting the square of the wavefunction equal to zero and messing around with it..
That's where it just turns into a math problem. Now about these trig functions, I'm not following you, if I plug in pi/2 for x, I get something like 2pi^2/a^3.
I don't know, it just shows a graph of the function from zero to a with two maxima and says that it is easy to see the maxima occur at these places. I just don't see it.
It's some really interesting stuff, we're just getting into the 3d Schrodinger (sp) wave equation which is probably the coolest thing i've ever seen.
13. Feb 26, 2013
tms
You want the derivative to be zero, not the function itself.
Don't plug in for $x$, but for the argument to the appropriate trig function, and then solve for $x$.
14. Feb 26, 2013
Staff: Mentor
$\pi/2$ looks like an angle (radians), not a position (meters or whatever).
15. Feb 26, 2013
haruspex
To be clear, the derivative of the square of the modulus of the wavefunction. That will be zero wherever either the modulus is zero or its derivative is. Then, of course, you have to figure out which are maxima and which minima.
16. Feb 26, 2013
Levi Tate
I just want to say that i am actually not as dumb as i sound here. I did my Grandmothers taxes all day. I am going to try what tms said here, note jtbell, affectionately call him jtballer, and solve this thing tomorrow.
I actually have a question I am really confused about, solving the time independent differential equation, there is this strange cavat that seems at odds with what i learned about linear homogeneous first order differential equations, and i think it might be at the energy cannot be zero or less, that there is a physical thing rather than just mathematics in the solution, but i've checked my ODEs text and it doesn't make sense to me. My teacher told me 'make sure the solution works'. I think he works on computers a lot and doesn't bother with trying to solve any differential equations.
17. Feb 26, 2013
Levi Tate
^oh I will post it soon, i think it's a good question. That wasn't the question statement up there, I just was casually talking about it.
18. Feb 26, 2013
Levi Tate
How can it be that a cosine function and a sine function will have the same maxima and minima here in this physics problem? I'm not disputing you, i just don't understand it.
Also, why do you have the respective title of science advisor? Do they give that title to experimental or theoretical physicists? I want that title some day. I like these forums very much. Someday I am going to come back and answer all the questions that I asked, you know, give back to the community.
19. Feb 27, 2013
Staff: Mentor
You've got the square of the wave function which is (ignoring the normalization constant) $\sin^2 (n\pi x / a)$. Think of it as $sin^2 \theta$ where $\theta = n\pi x / a$. What's the maximum possible value of $\sin^2 \theta$, and what values of $\theta$ give you that maximum? This is something you should be able to read off a graph of $\sin^2 \theta$.
"Science Advisor" is a title that the Mentors (moderators) here give to posters that we think give reliably good answers and advice. "Homework Helper" is similar, but it's for people who hang out mainly in the homework forums.
Last edited: Feb 27, 2013
20. Feb 27, 2013
Levi Tate
Cool thanks for the info mate.
See I just did this problem here, I didn't ignore the 2/a, I just divided it out. The problem says for the first excited state, so n=2, so I ended up with θ=Arcsin(2[pi]x/a), (n=2 for the first energy state)
I dunno I just am not getting this problem, because arcsin(sin^2x)≠x I don't think.
I am just going to ask my teacher this one. I don't think he solves ODEs and uses a computer, so I have a better question, it's pretty straightforward.
Thanks for the help. | 2018-01-22 10:56:48 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.7899158000946045, "perplexity": 502.06430731809445}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-05/segments/1516084891277.94/warc/CC-MAIN-20180122093724-20180122113724-00589.warc.gz"} |
https://math.stackexchange.com/questions/842744/defining-principal-elements-of-every-poset-is-this-a-new-idea | # Defining principal elements of every poset. Is this a new idea?
Fix an arbitrary complete lattice $\mathfrak{A}$ with order $\sqsubseteq$. I call elements $a,b\in\mathfrak{A}$ intersecting and denote $a\not\asymp b$ iff there is a non-least element $c\in\mathfrak{A}$ such that $c\sqsubseteq a \wedge c\sqsubseteq b$.
I call full star of element $a$ the set $\star a=\{c\in\mathfrak{A} \,|\, c\not\asymp a\}$.
I call a complete free star such a subset $S\subseteq\mathfrak{A}$, not containing the least element, that $\bigsqcup T \in S \Leftrightarrow T \cap S \neq \emptyset$ for every set $T\subseteq\mathfrak{A}$ (by $\bigsqcup T$ I denote the join (=supremum) of the set $T$.)
Now we can define principal elements of the complete lattice $\mathfrak{A}$:
An element $a\in\mathfrak{A}$ is principal iff its full star $\star a$ is a complete free star.
Question: Has anyone researched principal elements of complete lattices before me today?
Note: I call such elements principal because for the lattice $\mathfrak{F}$ of filters on a set (ordered reverse to set-theoretic inclusion of the filters), this coincides with the customary definition of principal filters.
Moreover, my idea can be generalized from complete lattices to arbitrary posets just replacing the formula $\bigsqcup T \in S \Leftrightarrow T \cap S \neq \emptyset$ with more general formula $$\forall Z \in \mathfrak{A}: ( \forall X \in T : Z \sqsupseteq X \Rightarrow Z \in S) \Leftrightarrow T \cap S \neq \emptyset.$$
Has anyone researched this case (for arbitrary posets, not just complete lattices) before me?
Hm, maybe the set of principal elements of a (distributive) lattice is the same as the center of the lattice?
• You don't use $b$ in the definition of "intersecting elements". – Olivier Bégassat Jun 21 '14 at 16:04
• @OlivierBégassat: Thanks, corrected – porton Jun 21 '14 at 16:40
• Isn't it the case that $a$ and $b$ are intersecting iff their meet is different from the bottom element? – Olivier Bégassat Jun 21 '14 at 16:46
• @OlivierBégassat: Yes, but only for lattices with a bottom element, not for arbitrary posets. My definition is valid for any posets – porton Jun 21 '14 at 17:26
• It's just that your question is phrased in terms of complete lattices, and those have top and bottom elements. – Olivier Bégassat Jun 21 '14 at 19:09
To avoid confusion, you should provide a definition of a filter. I'm using the more general definition of an order filter that is sometimes called an upset. But your suggestion looks not very promising to me as your can see below.
Here, a filter $F⊆\mathfrak A$ is always an order filter: $$x∈F⇒(∀y∈\mathfrak A:x≤y⇒y∈F)$$
In any lattice the equivalence $c\sqsubseteq c\wedge a ⇔ c \sqsubseteq a$ holds. Thus, $a≭b$ iff there exists a principal filter that contains $a$ and $b$ that is not generated by a minimal element. This is some kind of orthogonality relation. It reminds me to the orthogonality of $\ell$-groups and to residuated lattices.
$*a$ is the union of all principal filters of the atoms below $a$ (in case of their existence) or the infinitely descending chains just above the minimal elements.
A complete free star $S$ is an order filter as for $a,b∈S$ the relations $\{a,b\}⊆\mathfrak A$ and $a\vee b∈S$ hold (for arbitrary orders consider $a≤b$ and $a∈S$). Furthermore if $a\vee b∈S$ then $a∈S$ or $b∈S$ holds. This is a prime filter or – in the dual order – a prime ideal. Furthermore, it is related to irreducible elements. In a finite lattice the set $S$ is the union of principal filters of $\vee$-irreducible elements. So it is something like an orthogonality relation, again.
So in a lattice every element except $0$ is principal. In an arbitrary ordered set an element $x$ is principal if the filter $↑(↓x\setminus \min\mathfrak A)$ generated by its principal ideal $↓x$ without the minimal elements is something like a prime filter i.e. if each upper bound of any nontrivial subset of its complement $\mathfrak A\setminus \bigl(↑(↓x\setminus \min\mathfrak A)\bigr)$ is in that complement. For this consideration the don't play an important role. It's much easier if you keep them in the filters.
BTW.: Have you ever heard of Formal Concept Analysis? Their view of lattice theory might be helpful for you.
• "So in a lattice every element except $0$ is principal." It is wrong: non-principal filters are non-principal in this sense. It seems you confuse for-all-filters with for-all-principal-filters or for-all-ultrafilters. I haven't check the details, but I am sure in the lattice of filters on a set, not only $0$ is principal – porton Jun 22 '14 at 20:27
• In fact a non-principal filter may be non-principal in a certain lattice. But that's true for every element of any lattice and out of the scope of this question. If you have a counterexample then you should revise your definitions. You don't provide a definition of a filter. There is a common definition of ultrafilters in topology, but not in order theory. So you are probably the one who confuses different notions. Regarding your last sentence, $0$ is not principal by your definition. I came to the conclusion that all other elements are principal. So where is the contradiction? – Keinstein Jun 24 '14 at 20:37 | 2020-11-30 21:16:34 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8467245697975159, "perplexity": 254.39812894639905}, "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-2020-50/segments/1606141486017.50/warc/CC-MAIN-20201130192020-20201130222020-00583.warc.gz"} |
https://space.stackexchange.com/questions/28100/can-you-avoid-the-costs-of-cooling-superconductors-for-magnetic-shields-by-separ | # Can you avoid the costs of cooling superconductors for magnetic shields by separating them from the spacecraft and using shades?
As the answer to this thread states, cooling superconductors in spacecraft is necessary because of waste heat from secondary systems and thermal radiation of the sun. The application in which superconductors are the most interesting to me is magnetically shielding spaceships from solar wind and galactic cosmis rays to reduce the radiation risks for manned spaceflight.
What I'm thinking about is getting rid of anything superfluous, making the system as simple as possible and just shooting it away from the ship with a conductive tether connected to the superconductive coil, which is itself connected to a large and thin reflective film that blocks heat from the sun. After positioning the solar shade, the coil should slowly radiate away energy and eventually become superconductive (at 10 Kelvin for conventional niobium-titanium alloy), right? You could then gradually add current to the coil until the magnetic field is strong enough to block harmful radiation.
Problems I see with this approach:
You can never let the coil heat beyond its critical temperature, so the solar shade has to stay in front of the coil at all times. This probably requires thrusters and a computer, which makes the solar shade a kind of mini satellite in itself.
2) Tethers, secondary systems and waste heat
The electrical energy comes from the conductive tether which is not a superconductor itself, so there'll be waste heat. Deflecting charged particles will also take energy, you constantly have to add new current to the coil. On top of that you probably also need a computer and something to monitor and regulate the coil, producing even more waste heat.
3) Magnetic field strength
The farther the magnetic shield is from the spacecraft the smaller is the amount of protection you'll get. If you get too close to the ship, thermal radiation from it might heat up your superconductor and the magnetic field will affect metal inside of it. Does the required distance from the craft give you enough protection to make this worthwhile?
4) Shutting the coil off
You can let it heat beyond the critical temperature, but the superconducting material will quench and release a lot of heat, which might damage it.
All in all, is this actually a feasible idea compared to just using active cooling?
• I think it would take some one familiar with the heat loads placed on super conducting materials, hence my comment rather than answer. But one potential concern is the poor cooling properties of objects in space. Without air currents to convect heat away, you're left with radiation only so it may be difficult to maintain superconducting temperatures without active cooling. – Saiboogu Jun 25 '18 at 18:14
• The JWST is a good model. That cools parts of the telescope to.39K passively. Other parts need active cooling to get down to 7K . So you could have a cuprate superconducting magnet passively cooled, but not a metal one. That's in a more or less ideal setting, with, sun, earth and moon always on the same side of the spaceship – Steve Linton Jun 26 '18 at 8:52
• "Shielding Space Explorers From Cosmic Rays" (Prof Eugene Parker) suggests that the magnetic field strength necessary to deflect particles would do biological harm of its own. – Foo Bar Jun 27 '18 at 18:13
A lot depends on the temperature requirements for your superconductors. That in turn depends on how much magnetic field you need, your mass budget for the coils, and how exotic a material you can afford.
There's a nice summary of (mostly active) space cryogenics here. Passive cooling to 50K has already flown on Planck (they were planning 60K, but did better), and the JWST is planning 40K.
The environment matters a lot. "Cooling" comes from radiating to space at ~5k (averaging over star flux in the local area of the Galaxy); heating comes from any local object. The more and larger shields you need for Sun, Earth, Moon, etc, the less solid angle you have to radiate to the rest of the sky.
JWST's 5-layer shield was designed to meet the requirement of no more than 2W passed flux, which in turn could be radiated to space for a telescope instrument temperature of 40K. Because flux goes like $T^4$, keeping all things the same, you'd need to drop the flux by a factor of $4^4 = 256$ to get to 10K. If everything was perfect, this would just require 2 (maybe 3) more layers.
But it's hard to keep such a shield perfect, and even a defect at the 0.1% level could let through enough heat flux to overwhelm the passive cooling by radiation. At some point, the requirement for perfection becomes too expensive.
Deflecting charged particles will also take energy, you constantly have to add new current to the coil.
No, magnets do not change the energy of the particle, just their momentum. Otherwise permanent magnets wouldn't work constantly. But the momentum is transferred to the magnet, so you need active station keeping for it as well.
After positioning the solar shade, the coil should slowly radiate away energy and eventually become superconductive at 10 Kelvin
Yes, it will cool down, but likely not as much: There are crates close to the poles of the Moon that never receive sunlight. They have a temperature of about 100 Kelvin.
But, a solar shield doesn't protect you from all the heat the Sun delivers: Your solar shield will heat up and due to thermal conduction its backside will get warm as well. You can add more insulation in the shield, but this will not prevent the heat transfer, just making it slower. Once the shield is warm, it will radiate heat by itself and thus heat up the magnet coil. You will need some active cooling to keep the backside cold enough.
Also note that the distance of the shield doesn't matter: The closer it is, the smaller it can be. The amount of radiation the coil receives depends on the angular size (coverage of the hemisphere) of the shield, not its distance nor size.
Without doing the calculations: To reach such low temperatures, I assume you need, besides active cooling, a shield to protect the coil from radiation coming from Earth and possibly even the Moon.
• The vacuum multi-layer insulation approach to a heat shield can reduce the thermal flux by $10^6$ or more. Each layer reflects part of the flux "upstream", and is cooler than the last. This is how the Webb Telescope will passively cool parts below 30K. There's some more info here: cas.web.cern.ch/sites/cas.web.cern.ch/files/lectures/… – Bob Jacobsen Jun 27 '18 at 5:48
• Maybe my assumptions are a bit too pessimistic, but I doubt that 10 K can be reached with any justifiable effort. – asdfex Jun 27 '18 at 13:59
To my knowledge the short answer is no probably not. You can assign a "temperature" to space which is the temperature at which the radiation from a perfect black body, matches the radiation striking the object. It is key to note here that an object will not radiate as much as a perfect black body so the equivalent temperature of the coil would be larger than the temperature assigned to space. I can't find any average estimates on what that temperature is in LEO, but we can assume that temperature is far above the operating temperature of a conventional superconductor. The lowest known equivalent temperature is 2.7 K far from any galaxy. If you were to place a solar shield out in front, or even completely enclose the exposed surfaces of the superconductor, eventually that shield/the entire craft would reach that equivalent temperature. That being said, that would likely take a long time. Additionally, shielding just the solar radiation would not be enough since we sit inside a galaxy. Based on what I've learned about heat transfer, without an active cooling system, this is not feasible. However, your idea may be able to significantly reduce the necessary amount of active cooling. Much of the details of this possibility HEAVILY revolve around the craft design, and exact orbits of the craft but based on my preliminary analysis, the best bet is that this could be used to increase the effectiveness of any cooling system implemented.
• "It is key to note here that an object will not radiate as much as a perfect black body so the equivalent temperature of the coil would be larger than the temperature assigned to space." I don't believe this is correct at all. Objects will always eventually reach radiative equilibrium with their surroundings, and if their surroundings are nothing but 2.7 K CMB, they will inevitably hit 2.7 K themselves, whether they're composed of polished aluminum or pure graphite. – Nathan Tuggy Jun 28 '18 at 15:11
• @Nathan Tuggy I was under the impression that no matter radiates in the expected spectrum distribution of a black body and that it tends to be less emissive than expected, but if there's justification for that, I'll change it in my answer. That being said, my answer isn't that great anyways as it's for of a blob of my thought process looking back – Gigaboggie Jun 28 '18 at 15:15
• It's absolutely true that all matter radiates less than a black body because its spectrum is different, even radically different. But since emissive and absorptive spectra must be the same, an object in a single-temperature environment will always reach equilibrium at that temperature because the incoming radiation, which is the only ongoing source of heat, will eventually match outgoing. The time taken for this can vary wildly depending on emissivity, but it will always happen eventually. – Nathan Tuggy Jun 28 '18 at 15:24 | 2021-05-17 09:40:31 | {"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": 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": 0.5453115701675415, "perplexity": 879.7605095060673}, "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-2021-21/segments/1620243992159.64/warc/CC-MAIN-20210517084550-20210517114550-00134.warc.gz"} |
https://moodle.org/mod/forum/discuss.php?d=360882 | ## General developer forum
### Found more than one record in fetch() ! Error
Found more than one record in fetch() ! Error
I have Moodle 3.2.1
Whenever I try to add a quiz or forum or other content to any course I get this error. Any idea what could be wrong and what should I check ?
Found more than one record in fetch() !
×Debug info:
Error code: morethanonerecordinfetch
×Stack trace:
• line 496 of /lib/setuplib.php: moodle_exception thrown
• line 173 of /course/modlib.php: call to edit_module_post_actions()
• line 161 of /course/modedit.php: call to add_moduleinfo()
The only problem I could find was:
## Other checks
InformationReportPluginStatus
check_mysql_file_format
unsupporteddbfileformat
Error
However php admin/cli/mysql_compressed_rows.php --list return this (as far as I understand it means DB format ok):
mdl_data Compressed
mdl_data_fields Compressed
mdl_enrol_lti_lti2_consumer Compressed
mdl_enrol_paypal Compressed
mdl_hotpot Compressed
mdl_hotpot_cache Compressed
mdl_lti Compressed
mdl_user Compressed
mdl_user_info_field Compressed
Average of ratings: -
Re: Found more than one record in fetch() ! Error
We've seen that error before. It means there are two "course" records in the grade_items table for some course, when there should be only one. As far as we could work out this is something to do with the fact that the grade API creates the course item on-the-fly if it doesn't exist when it's needed, so you can end up with two if you're unlucky.
If you run this SQL it should show you the offending course(s):
select courseid, count(*) from mdl_grade_items where itemtype = 'course' group by courseid having count(*) > 1
Unfortunately I can't remember how to fix it, but I think it may have been as simple as deleting the record with the higher ID.
Average of ratings: Useful (1)
Re: Found more than one record in fetch() ! Error
Looks like I have 3 courses with these problems
+----------+----------+
| courseid | count(*) |
+----------+----------+
| 8 | 2 |
| 9 | 2 |
| 10 | 2 |
+----------+----------+
3 rows in set (0.00 sec)
Now I just need to figure out how to fix it
Average of ratings: -
Re: Found more than one record in fetch() ! Error
I removed the duplicate records manually, but I still get the same error when I try to add quiz.
I also see that after I manually remove the second record I try to add a quiz and get the same error, but when I go back to the DB I see that I new record is added
131 9 NULL NULL course NULL 20 NULL NULL NULL NULL 1 100 0 NULL NULL 0 1 0 0 0 47 0 NULL 0 0 0 1 0 1509629742 1509629742
If I remove all course records from a specific course and then try to add a quiz I still get an error and two records are added
132 9 NULL NULL course NULL 20 NULL NULL NULL NULL 1 100 0 NULL NULL 0 1 0 0 0 47 0 NULL 0 0 0 1 0 1509630118 1509630118
135 9 NULL NULL course NULL 4 NULL NULL NULL NULL 1 100 0 NULL NULL 0 1 0 0 0 48 0 NULL 0 0 0 1 0 1509630561 1509630561
Any idea whats going on ?
BTW its happening in all my existing courses, but if I create a new course everything is ok in this new one
Average of ratings: -
Re: Found more than one record in fetch() ! Error
Sorry, haven't a clue what could be happening there! I don't really understand why you'd have 2 new course items created on adding a quiz. Do you have any non-core plugins installed?
Average of ratings: -
Re: Found more than one record in fetch() ! Error
I think that only "hotpot"
Also I'm not sure if it suppose to be like that or not, but I also have two "category" records in this course and also if I remove them when I try to add a quiz it adds two new records
Average of ratings: -
Re: Found more than one record in fetch() ! Error
That kind of makes sense - they'll be the root categories for the course, so if you've got two course records you should get two of those too. Sounds like the duplicate course items is definitely the root of your issues, but I really don't understand why it might be happening, so not sure how much more help I can be!
Average of ratings: Useful (1)
Re: Found more than one record in fetch() ! Error
Thank you very much for your help
I will try to upgrade Moodle to the latest version tomorrow, just in case its some bug that was fixed
Average of ratings: - | 2018-04-20 09:56:16 | {"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": 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": 0.3211759626865387, "perplexity": 3002.7257319827672}, "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-2018-17/segments/1524125937193.1/warc/CC-MAIN-20180420081400-20180420101400-00167.warc.gz"} |
https://electronics.stackexchange.com/questions/409935/do-the-new-pb-variants-of-atmega-have-a-bug-in-the-brown-out-detector | # Do the new "PB" variants of ATmega have a bug in the brown-out detector?
We've been using ATmega48/88/168/328 microcontrollers successfully for many years in many of our products. We have now considered to switch from the A and PA variants to the new PB variant (because we will need the extra pins, timers and UARTs in new products, because it's become cheaper, and because it seems the old variants will be discontinued), so we switched out an ATmega328A with an ATmega328PB. It seems to go haywire very often after power interruptions. Such problems never ever occurred with the old variants.
Regular power interruptions are normal for the usecase of our products. We use a switching power supply (like this one) set to 5V, and have capacitors in the 220µF range on the ATmega's VCC, to keep the SRAM alive for power interruptions in the range of several minutes, to store internal states which are not mission critical but significantly increase user experience by being instantly available upon a restart (these states change often enough to make EEPROM unsuitable). This has always worked.
However, with the new ATmega328PB, after a power interruption, the chip resets without a reset condition being found in MCUSR, and the clock seems to go haywire.
• the brown-out detector is set per fuse. We tried every available bodlevel, the bug happens on all of them.
• we use external 20 MHz, also set correctly per fuse.
• we tried 3 different chips, so it wasn't a single soldering or other hardware failure.
After the bug happens, the clock often sets to 2.5x slower speed, indicating that the mcu is being clocked by the 8 MHz internal oscillator. However, sometimes the slowdown is around 6x. This means it can't be a software bug changing the clock divider, as I cannot set the fuses from software, and the clock divider cannot divide the clock by 2.5 or by 6.
So, my first suspect was the new Clock Failure Detection fuse. However, no matter if it's turned on or off, the behavior remains the same.
To rule out software peculiarities, I wrote a simple test program from scratch, which does nothing else but toggles an output with 100 Hz from a timer interrupt, and indicates with LEDs after each restart which reset conditions were activated (as read from MCUSR). The rest of the hardware was also removed, only the mcu and the regulator are there (and the indicator leds with series resistors).
# The results
Roughly 2/3 of the time, nothing interesting happens. After the power interruption, the mcu resumes its job, both the brown-out reset and power-on reset indicators lit up.
(on the image, red is the toggled pin, and blue is VCC. On this image, the 2.7 V bronwn-out is clearly visible. I made the same tests with the other brown-out settings, the results are exactly the same, so I will omit those pictures)
Roughly 1/3 of the time, the aforementioned bug occurs, and when the power is back again, none of the brown-out reset and power-on reset indicators are lit up! The output is different, as if the mcu was ticking with a strange clock. It's not chaotic, however, it keeps ticking with the same frequency.
Interestingly, in this situation, the brown-out detector seems to be completely inactive, because after the next power interruption (where the correct clock is sometimes restored, sometimes not), it is clearly visible that the output keeps toggling well after the brown-out level has been passed. In such situations, the clock sometimes gets faster, other times it gets slower:
During these tests I used 16K CK/14CK + 4.1 ms for the start-up delay (but the 65 ms delay doesn't avoid the problems).
Here is a picture zoomed in, where you can clearly see that the VCC reaches a stable state at 5 V in under 2 ms:
In the above picture, the mcu started correctly.
Interestingly, when it doesn't, the supply voltage gets up to a stable 5 V even sooner (it seems many parts of the mcu don't power on, so it draws less current during the startup)
Below is an image from an unsuccessful start:
Please note, that the software starts running after more than 85 ms after the supply voltage has been stabilized, instead of the 10.5 ms required otherwise. The fuses for the startup delay are still the same, 16K CK/14CK + 4.1 ms.
What's also interesting to note, is that after the supply was turned off, the VCC stabilizes at around 1.1 to 1.2 Volt (the old, ATmega328A variant went down to around 0.6 - 0.7 V). It keeps that for several minutes. If I wait long enough (on the order of half an hour or more), the mcu always starts correctly! So it seems the problem is that there is 1.1 Volt around, which, according to the datasheet, is not guaranteed to be enough for a power-on reset. But it should be enough for a brown-out reset!
Except for these situations, the brown-out detector works fine. It's visible on the first image (the output signal stops when the bodlevel has been reached, and the voltage drop slows down, as parts of the mcu are shut down). I did tests when I reduced the VCC to slightly below the bodlevel and let it climb back again, the mcu always restarted correctly under such conditions, with only the brown-out reset indicator being lit up.
Did I miss something obvious, or does the ATmega328PB have a serious bug in its brown-out detector?
EDIT:
Interestingly, the above problems only arise when I interrupt the supply before the regulator. If I interrupt it after the regulator (or use a lab power supply), the problems never happen. As if the shape of the rising voltage caused the problems. However, as you can see from the last image, the voltage rise is quite nice and it stabilizes quickly.
EDIT 2
I tried it out with 16 MHz instead of 20 MHz, but the exact same problems happen.
• Have you contacted Atmel or looked into their erratas? In this day and age IC design mistakes are quite common. Dec 1, 2018 at 18:41
• I have looked through the erratas (didn't find anything in this direction), and we are considering contacting Atmel, but not before making some more tests and looking around a little more.
– vsz
Dec 1, 2018 at 18:42
• By my experience don't waste time before contacting the manufacturer or using their forums. You have made more than enough debugging to present a very strong case. With much less than that, TI sent me their internal (unpublished) errata for one of their ICs that documented our issue. Dec 1, 2018 at 18:46
• My two cents worth: I have seen problems with other CPUs if the power rises too quickly. Some manufacturers specify a maximum rise time but more often this is not mentioned. Dec 1, 2018 at 19:03
I don't think it is a bug with the brown-out detector, but how you use the chip.
As you said yourself, the power-on reset threshold 1.1 V is not reached if power is just briefly removed and connected, so there will be no POR.
Brown-out detector can't help here much either. You are using the AVR at 20 MHz, and this requires the supply voltage to be 4.5 V or above, or you are violating the specs. And BOD does not guarantee that it will trip at 4.5 V, it's typically less than that, say 4.3 V. So even before BOD triggers, there is no guarantee in what state the AVR ends up but the BOD should trigger, except that it may not work due to your 20 MHz clock. When the voltage starts to rise again, the BOD deactivates before supply voltage is at a safe 4.5 V level again. If it was triggered correctly. The start-up delay time should be then set to high enough that the voltage has a change to rise from BOD deactivation level to 4.5 V before the internal reset is released.
But it all may fail because it just needs at least 4.5 V to run at 20 MHz. The AVR datasheet does mention that if internal reset system is unsuitable then use an external reset chip, and in this case it looks like it would solve your issues to reset the AVR before voltage drops to 4.5 V.
• I assumed the BOD doesn't use the processor itself but it's a dedicated hardware. Maybe they changed it for the PB variant? I would be surprised if they no longer support BOD for 20 MHz. The highest bodlevel is 4.3 V, so 20 MHz would require an external BOD? Still, I have doubts this alone is the cause. I made a test with 20 MHz, 2.7V bodlevel, set the VCC to 3V, it ran fine. When I reduced the voltage manually to slightly below 2.7, the output stopped, when I increased it above 2.7 the output resumed, always, it never failed, not even once. Only a startup from 1.1 V seems to disable the BOD.
– vsz
Dec 1, 2018 at 21:40
• Most likely it is dedicated hardware, but during the undervoltage before the BOD kicks in, can you be sure correct data is fetched from flash for CPU execution, and does CPU execute them correctly? It might, or just write random data to reserved registers that do unspecified stuff. The specs have changed for the PB variant, and they did not support BOD for 20MHz either on the older chip. PB variant has both BOD and POR curves indeed different and kick in later at lower voltages. Dec 1, 2018 at 21:55
• Please take a look at my second picture. The BOD was seemingly engaged correctly and has reset the chip. It only fails to initialize at the next startup. Also, I have driven this chip at 3 V and it functioned correctly, never failed a single time.
– vsz
Dec 1, 2018 at 22:45
• Well in my opinion the chip does not need to work outside of safe operating area, but let's continue. The BOD won't reset the Clock Failure Detector, so only Power-On reset and external reset will switch out from internal clock. So double check the CFD fuse settings. Are you using external crystal or external clock? The CFD fuse might have previously been the Full Swing fuse. And since there is no full swing fuse, maximum frequency for a crystal is 16MHz, and 20MHz requires external logic level clock signal. So could be a crystal startup issue too, so put a scope on crystal pins too. Dec 2, 2018 at 11:47
• I use a crystal. Good Idea, I'll look into that. Please note, that the same behavior I depicted with images, occurred no matter if the CFD was on or off.
– vsz
Dec 2, 2018 at 12:25
sorry for being late to the party :) but I just recently run into kind of a similar problem. Had a board layout with an atmega328p replaced by atmega328pb and using a murata 16MHz ceramic resonator with integrated 15pF caps. In my application the voltage ramps up veeery slowly (500mV / s). The old atmega328p boards dealt with it fine (BODLEVEL at 2V7), however, the new layout with 328pb does not! Pin functions swapped randomly, flash corruptions due to random code execution (probably hit the bootloader section, even though it's not starting by default etc.). Short, a lot of funky stuff happened.
My "solution"--let's call it a temporary workaround--after 2 1/2 days of brain-melting was to set the CKDIV8 fuse bit (dividing the 16Mhz ext. clock by prescaler 8). With that 2MHz now the 328pb boots flawlessly even under insanely slowly rising supply voltage (~250mV/s).
The trick is now to continuously read the supply voltage via internal ADC voltage reference and when the voltage is stable at above 3V then step up to the full 16MHz clock (prescaler 1).
CLKPR = (1 << CLKPCE);
CLKPR = 0;
Sorry, I didn't have the time for more quantitative measurements, but for now, the fix worked smoothly for several boards and a lot of boot events.
What I cannot answer is why this was happening at all. Ramping up supply voltage so slowly with 16MHz clock is clearly out of spec! There were a lot of tiny layout changes between the boards so it could be just luck, that it worked in the earlier version. I have no clear indication that it is a 328pb bug or sth.
However, I would be interested in how you solved the problem meanwhile.
Best, Matthias
• How I solved the problem meanwhile? By transitioning to the new 3208. Sadly, it has a different pinout, but for all future products we've started introducing it, and after some heavy stress-tests it seems to work fine even far beyond its voltage and clock speed specifications. It can even do 16 or 20Mhz internally without any pesky external resonator. And where it was too late to leave the 328PB out, we've given an external clock signal. A complete TTL square wave. With that, it works.
– vsz
Jun 11, 2021 at 15:03
• I used a MC33164 in a TO92 package to force reset low until I had sufficient voltage.
– Gil
Jun 12, 2021 at 3:34 | 2022-08-11 11:46:47 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 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": 0.4835321009159088, "perplexity": 2163.84825727057}, "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-2022-33/segments/1659882571284.54/warc/CC-MAIN-20220811103305-20220811133305-00736.warc.gz"} |
https://yo-dave.com/2011/01/30/syntax-highlighting/ | # Syntax Highlighting
Update 2017-08-19: This post refers to a time when the blog was hosted in WordPress. It isn’t really applicable anymore. But it may provide some guidance to someone who happens to come across it.
Since this blog is mostly about programming, there are a lot of code listings. To make those listings more easily readable, highlighting syntax in the listings is a good thing.
Of all of the syntax highlighters I have tried, the SyntaxHighlighter JavaScript tool by Alex Gorbachev is the one I most prefer. The SyntaxHighlighter Evolved is a nice WordPress plugin wrapped around the Gorbachev package. Setting things up in WordPress is trivial and just works.
Although the plugin comes with many “brushes” for different languages, the one for Clojure is a bit simplistic. However, Brian Carper has taken the time to create a more sophisticated highlighter for Clojure. He has even supplied a tweaked theme. The theme doesn’t suit my tastes, but it looks good on his site.
Brian’s Clojure brush even seems to work well with Lisp
(defun verbose-sum (x y)
"Sum any two numbers after printing a message."
(format t "Summing ~d and ~d.~%" x y)
(+ x y))
and Scheme
(define product
(lambda (ls)
(call/cc
(lambda (break)
(let f ([ls ls])
(cond
[(null? ls) 1]
[(= (car ls) 0) (break 0)]
[else (* (car ls) (f (cdr ls)))])))))) | 2018-08-17 03:06:26 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 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": 0.1860578954219818, "perplexity": 7799.096363684993}, "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/1534221211664.49/warc/CC-MAIN-20180817025907-20180817045907-00243.warc.gz"} |
https://linearalgebras.com/tag/prime-number | If you find any mistakes, please make a comment! Thank you.
## n divides the totient of $p^n-1$ when p is prime
Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 3.2 Exercise 3.2.15 Let $p$ be a prime and let $n$ be a positive integer. Find the order of…
## Alternate proof of Cauchy’s Theorem
Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 3.2 Exercise 3.2.9 Let $G$ be a finite group and let $p$ be a prime dividing $|G|$. Let $\mathcal{S}$…
## Examples of nilpotent elements
Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 7.1 Exercise 7.1.13 An element $x \in R$, $R$ a ring, is called nilpotent if $x^m = 0$ for some… | 2020-08-04 07:47:51 | {"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": 0.7675132751464844, "perplexity": 154.88864271819716}, "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-2020-34/segments/1596439735867.23/warc/CC-MAIN-20200804073038-20200804103038-00040.warc.gz"} |
https://math.stackexchange.com/questions/2547655/mean-value-theorem-for-integrals-on-a-b | # Mean value theorem for integrals on (a,b)
The problem is:
Let $g(x)$ be nonnegative on [a,b] and Riemann integrable on [a,b], a function $f$ continuous on (a,b) and the product $fg$ Riemann integrable on [a,b]. Prove, that under these conditions the mean value theorem is true, i.e. there exists a point $\xi\in(a,b)$, such that $\int\limits_a^b f(x)g(x)dx=f(\xi)\int\limits_a^b g(x)dx$.
The hint to the problem from the same book is to use the mean value theorem for the integral $\int\limits^{b-\varepsilon}_{a+\varepsilon}f(x)g(x)dx$ and then apply the equality $\int\limits^b_a f(x)dx=\lim\limits_{\varepsilon_1\to0+}(\lim\limits_{\varepsilon\to0+}\int\limits^{b-\epsilon_1}_{a+\varepsilon}f(x)dx)$. But I don't understand this, because for each $\varepsilon$ is a distinct point $c$ so, $f(c)$ is not a constant, when $\varepsilon\to0$, so we cannot go to a limit in the corresponding mean value equality.
Could somebody, please, clarify this. Is the original statement true or not and how to prove that?
First observe that, since $g\geq 0$ and $g, fg$ are integrable, if $\int_a^b g(x)\, dx = 0$ then also $\int_a^b f(x)g(x)\, dx = 0$. In this case any $\xi \in (a,b)$ does the job.
Let us consider the case $\int_a^b g > 0$. You already know that, for every $\varepsilon\in (0, (b-a)/2)$, there exists a point $\xi_\varepsilon \in [a+\varepsilon, b-\varepsilon]$ such that $$\int_{a+\varepsilon}^{b-\varepsilon} f(x)g(x)\, dx = f(\xi_\varepsilon) \int_a^b g(x)\, dx.$$ Hence, choosing $\varepsilon = 1/n$ you can construct a sequence of points $\xi_n \in [a,b]$ such that $$(1)\quad \int_{a+1/n}^{b-1/n} f(x)g(x)\, dx = f(\xi_n) \int_a^b g(x)\, dx.$$ Since $\int_a^b g > 0$, we deduce that the sequence $(f(\xi_n))$ is convergent, and $$L := \lim_n f(\xi_n) = \frac{\int_a^b fg}{\int_a^b g}.$$ By the Bolzano-Weierstrass' theorem, there exists a subsequence $(\xi_{n_j})$ converging to a point $\xi_0 \in [a,b]$. If $\xi_0 \in (a,b)$ then we are done, since $f(\xi_0) = L$.
Consider now the case $\xi_0 \in \{a, b\}$. To fix the ideas, suppose $\xi_0 = a$. We can extended $f$ with continuity in $a$ by setting $f(a) = L$. We claim that there exists a point $\xi \in (a,b)$ such that $f(\xi) = L$. Namely, if by contradiction this is not the case, then by continuity we would have $f(x) > L$ for every $x\in (a,b)$ or $f(x) < L$ for every $x\in (a,b)$. But, in both cases, it follows that $\int_a^b fg \neq L \int_a^b f$, a contradiction.
• I don't understand your last inequality. From $f(x)>L$ we can only deduce, that $\int_a^b f(x)g(x)dx\ge L\int_a^b g(x)dx$, I can't prove, that it's strictly greater. – Andrei Petrov Dec 2 '17 at 19:48
• $f$ is a continuous functions, so if $f(x_0) > L$ then, setting $\epsilon := (f(x_0) - L) / 2$, there exists an interval $I = [x_0-\delta, x_0+\delta]$ such that $f(x) \geq L + \epsilon$ for every $x\in I$. – Rigel Dec 2 '17 at 20:03
• We also need, that integral from $g$ over that interval was not equal to zero. However, I was able to overcome this difficulty. So, thank you for the nice solution! – Andrei Petrov Dec 3 '17 at 17:48 | 2019-09-19 18:56:17 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9744956493377686, "perplexity": 75.71623285345757}, "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-2019-39/segments/1568514573570.6/warc/CC-MAIN-20190919183843-20190919205843-00207.warc.gz"} |
http://faculty.mu.edu.sa/aabokhalil/Perfect%20diamagnetism | ## Dr. Ahmed G. Abo-Khalil
Electrical Engineering Department
## Perfect diamagneti
Superconductors in the Meissner state exhibit perfect diamagnetism, or superdiamagnetism, meaning that the total magnetic field is very close to zero deep inside them (many penetration depths from the surface). This means that their magnetic susceptibility, $chi_{v}$ = −1. Diamagnetics are defined by the generation of a spontaneous magnetization of a material which directly opposes the direction of an applied field. However, the fundamental origins of diamagnetism in superconductors and normal materials are very different. In normal materials diamagnetism arises as a direct result of the orbital spin of electrons about the nuclei of an atom induced electromagnetically by the application of an applied field. In superconductors the illusion of perfect diamagnetism arises from persistent screening currents which flow to oppose the applied field (the meissner effect); not solely the orbital spin.
Monday 10 -2
Tuesday 10-12
Thursday 11-1
### Contacts
email: a.abokhalil@mu.edu.sa
a_galal@yahoo.com
Phone: 2570
### Welcome
Welcome To Faculty of Engineering
# Institute of Electrical and Electronics Engineers
http://www.ieee.org/
http://ieeexplore.ieee.org/Xplore/guesthome.jsp
http://ieee-ies.org/
http://www.ieee-pes.org/
http://www.pels.org/
http://www.utk.edu/research/
http://science.doe.gov/grants/index.asp
http://www1.eere.energy.gov/vehiclesandfuels/
http://www.eere.energy.gov/
### القران الكريم
http://quran.muslim-web.com/
### Travel Web Sites
http://www.hotels.com/
http://www.orbitz.com/
http://www.hotwire.com/us/index.jsp
http://www.kayak.com/ | 2017-04-25 22:35:53 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 1, "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": 0.38877448439598083, "perplexity": 2344.50565448759}, "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-2017-17/segments/1492917120881.99/warc/CC-MAIN-20170423031200-00333-ip-10-145-167-34.ec2.internal.warc.gz"} |
https://scicomp.stackexchange.com/questions/31184/non-parametric-models-as-solutions-to-partial-differential-equations | # Non-parametric models as solutions to Partial Differential Equations
In the realm of scientific computing, there are a plethora of techniques developed to solve Partial Differential Equations (PDEs). Many of the popular methods are variants of common techniques such as Finite Differences, Finite Volume, Finite Elements and the theory behind where they work or not is pretty extensive. Each method has pretty clear pros and cons which might influence its use over another for any given research problem.
I was just now contemplating the idea of non-parametrically modeling a solution for some system of PDEs, say some hyperbolic system of PDEs to keep things interesting. The idea could be to essentially having some cloud of spacetime points with some solution coefficients that represent the approximate solution to the PDEs. Then a separate cloud of spacetime points could be used as collocation points used to "train" the model. Whenever we wish to evaluate our model and compute gradients at some spacetime location, we could perhaps do some least square fit based on data that is "close" to that spacetime location. The problem of actually solving the PDE might then be viewed as an optimization of the solution coefficients at the model's spacetime points such that the residual of the PDEs and boundary conditions are minimized.
In the context of hyperbolic PDEs, do you see such an approach as described working out? Do you see problems arising due to characteristics of the system of hyperbolic PDEs that might influence how one should evaluate the non-parametric model in spacetime? Are there any papers you know of covering similar concepts?
This idea came about thinking about some paper I read a while back about someone using a similar collocation approach but with a neural network as the approximate PDE solution instead. It seems to me a non-parametric model would be easier to optimize and would still be capable of capturing potential nonlinearities, at the expense of having to keep track of solution data at a likely good number of spacetime locations.
I haven't thought too hard about this so perhaps there is something obvious I am missing but any thoughts, comments, or references about similar material would be appreciated! | 2019-06-16 23:30:19 | {"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": 0.8403852581977844, "perplexity": 300.5332706804159}, "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/1560627998325.55/warc/CC-MAIN-20190616222856-20190617004856-00356.warc.gz"} |
https://www.physicsforums.com/threads/infinite-series-question-with-z-transform-addendum.898420/ | # Infinite series question with z-transform addendum
Tags:
1. Dec 27, 2016
### kostoglotov
1. The problem statement, all variables and given/known data
Hello,
I am currently doing some holiday pre-study for signals analysis coming up next semester. I'm mainly using MIT OCW 6.003 from 2011 with some other web resources (youtube, etc).
The initial stuff is heavy on the old infinite series stuff, that seems often skimmed over in previous calculus study, and was for me.
Working from a transfer function as an infinite series, I was curious if I could figure out a closed-form of the infinite right hand sum (0 to inf) of $nR^{n-1}$
$$0+1+2R+3R^2+4R^3+5R^4+...$$
by working in reverse through parts of synthetic division I found that $\frac{1}{(1-R)^2}$ works for a closed form of the previous infinite series.
Then by the same technique found that an infinite right hand sum for $n^2a^n$ has the closed form $\frac{a(1+a)}{(1-a)^3}$.
I thought I saw a pattern ($\frac{a(1+a)^k}{(1-a)^{k+1}}, from \ n^ka^n$), but for the infinite right hand sum of $n^3a^n$ reversing through synthetic division reveals a closed form $\frac{a(1+4a+a^2)}{(1-a)^4}$...so I thought maybe the coefficients of the numerators polynomial are simply getting squared...but going through this again for $n^4a^n$ shows that this is not the case.
What is the general closed form formula for an infinite right hand series of $n^ka^n$?
Also, I am aware of the property of Z-Transform Differentiation, $nx[n] \leftrightarrow -z\frac{dX(z)}{dz}$. Does this relate to what I'm exploring above? If so, how?
PS. I'd love to know of any other good resources for self-teaching signals and systems stuff :)
2. Relevant equations
3. The attempt at a solution
2. Dec 27, 2016
### MisterX
I don't think there is a general closed form with elemental functions. You seem to be asking about the poly-logarithm function just shifted by 1. | 2018-01-20 21:45:32 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.7165057063102722, "perplexity": 544.8223094506858}, "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-2018-05/segments/1516084889733.57/warc/CC-MAIN-20180120201828-20180120221828-00505.warc.gz"} |
http://www.math.rutgers.edu/~weibel/ | • The Rutgers Physics γ Team won the 39th Almgren "Mayday" Race with a time of 2:52.
The race was just a few days after a major storm, parts of leg 2 were badly damaged, and leg 6 was rerouted through South Bound Brook. In addition, Rutgers Math started at the wrong location (Alexander Road) and crossed the starting line over 11 minutes after everyone else. There were 11 teams competing.
The 2015 race will be Sunday May 3, from Princeton to Rutgers. It will be the 40th annual race between the Princeton and Rutgers Mathematics Departments.
Teaching Stuff (for more information, see Rutgers University, the Rutgers Math Department, and its Graduate Math Program.
Definition: Proofiness is defined as "the art of using bogus mathematical arguments to prove something that you know in your heart is true — even when it's not." -Charles Seife
Research papers & stuff: This is a link to some of my research papers. Here are my research interests and my Ph.D. Students.
Do you like the History of Mathematics? Here are some articles:
I am often busy editing the Journal of Pure and Applied Algebra (JPAA), the Journal of K-theory and the journals HHA and JHRS.
Seminars I like:
Fun Question: How can you prove that 123456789098765432111 is a prime number?
note that 12345678987654321 = 111111111 x 111111111
Fun Facts about Mersenne primes: In 1644, a French monk named Marin Mersenne studied numbers of the form N=2p-1, where p is prime, and published a list of 11 such numbers he claimed were prime numbers (he got two wrong). Such prime numbers are called Mersenne primes. The first few Mersenne primes (p=2,3,5,7) are 3,7,31,127, but p=11 gives the non-prime 2047=23*89 (as was discovered in 1536 by Regius).
Not all numbers of the form 2p-1 are prime, as Regius' example 2047 (p=11) shows. The next few primes for which 2p-1 is not prime are p=23 and p=37 (discovered by Fermat in 1640), and p=29 (discovered by Euler in 1738).
Mersenne primes are the largest primes we know. In 2011, the list of the first 41 Mersenne primes was verified; we don't know what is the 42nd smallest, even though a handful of larger Mersenne primes are known. For years, the Electronic Frontier Foundation (EFF) offered a $50,000 prize for the first known prime with over 10 million digits. The race to win this prize came down the wire in Summer 2008, as the 45th and 46th known Mersenne primes were discovered in within 2 weeks of each other by the UCLA Math Department (who won the prize) and an Electrical Engineer in Germany, respectively. The largest is the 45th, which has 13 million digits and p=43,112,609; the 46th has 11 million digits. (Each prime N=2p-1 has p log10(2) digits.) More recently, the 47th known Mersenne prime was discovered in April 2009 by a Norwegian named Odd Magner Stridmo, with p=42,643,801. Surprisingly, it is slightly smaller (by 141,000 digits) than the 45th Mersenne prime. For more information, check out the Mersenne site. Charles Weibel / weibel @ math.rutgers.edu / August 31, 2012 MATHJAX test:$\partial y/\partial t=\partial y/\partial x$,$\sqrt2=1.4141$, If$f(t)=\int_t^1 dx/x$then$f(t)\to\infty$as$t\to\infty\forall n\in\mathbb{N}, e^n\in \mathbb R\$
HTML 3 font test: ∂y/∂t = ∂y/∂x √2 =1.414
If f(t)= ∫t 1 dx/x then f(t) → ∞ as t → 0. This really means: (∀ε ∈ℝ, ε>0) (∃δ>0) f(δ) > 1/ε .
ℕ (natural numbers), ℤ (integers), ℚ (rationals), ℝ (reals), ℂ (complexes)
8810-2 are: ≪ ≫ ≬ while & sube; is &# 8838; ; &8838-41; (⊆ ⊇ ⊈ ⊉) &8712; is ∈
Here are some more html fonts: ∗
& #130; – & #139; are ‚ ƒ „ … † ‡ ˆ ‰ Š ‹
The ndash (–) is & #150; , & #8211; and & ndash; ! I prefer the longer —, which is & mdash; or & #151;.
& #160;–& #169; are ¡ ¢ £ ¤ ¥ ¦ § ¨ © (& #128; = €)
& #170;–& #191; are ª « ¬ ® ¯ ° ± ² ³ and ´ µ ¶ · ¸¹º »¼ ½ ¾ ¿
<\body> | 2014-10-31 13:27:10 | {"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": 0.5774166584014893, "perplexity": 12563.734515813629}, "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-2014-42/segments/1414637899702.24/warc/CC-MAIN-20141030025819-00148-ip-10-16-133-185.ec2.internal.warc.gz"} |
https://plainmath.net/53753/how-deal-with-two-interdependent-integrators-have-two-functions-where | # How to deal with two interdependent integrators? I have two functions, f(t,x) and g(t,u), where
Agohofidov6 2022-01-17 Answered
How to deal with two interdependent integrators?
I have two functions, f(t,x) and g(t,u), where .
I am trying to discretize the integral of this system in order to track x and u. I have succeeded using Euler integration, which is quite simple, since x(t) and u(t) are both known at t:
$u\left(t+h\right)=u\left(t\right)+hf\left(t,x\left(t\right)\right)$
$x\left(t+h\right)=x\left(t\right)+hg\left(t,u\left(t\right)\right)$
However, I am now trying to implement mid-point integration to get more accurate results. (Eventually Runge-Kutta but I am stuck here for now.)
You can still ask an expert for help
## Want to know more about Laplace transform?
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## Expert Answer
Virginia Palmer
Answered 2022-01-18 Author has 27 answers
chumants6g
Answered 2022-01-19 Author has 33 answers
Per Christian's suggestion I re-thought the question in terms of a state vector $y=|u,x|$, with a single derivative function $\left[\stackrel{˙}{u},\stackrel{˙}{x}\right]=\stackrel{˙}{y}=z\left(t,x,u\right)$.
This allows to express the mid-point method as usual:
$y\left(t+h\right)=y\left(t\right)+hz\left(t+\frac{h}{2},y\left(t\right)+\frac{h}{2}z\left(t,y\left(t\right)\right)\right)$
where $z\left(t,x,u\right)=\left[f\left(t,x\right),g\left(t,x\right)\right]$
###### Not exactly what you’re looking for?
alenahelenash
Answered 2022-01-24 Author has 368 answers
You can write the mid-point rule even with both equations separated as was your original post as
$u\left(t+h\right)=u\left(t\right)+hf\left(t+\frac{h}{2},x\left(t\right)+\frac{h}{2}g\left(t,u\left(t\right)\right)\right)$
and mutatis mutatndis for $x\left(t+h\right)$.
P.S. I find your notation a bit verbose; personally I got used to using indices $-,\ominus \circ \oplus ,+$ for values at $t-h,t-\frac{h}{2},t,t+\frac{h}{2},t+h$; then the equation would read
${\stackrel{˙}{u}}^{+}={u}^{\circ }+h{\stackrel{˙}{u}}^{\oplus }\left({x}^{\circ }+\frac{h}{2}{\stackrel{˙}{x}}^{\circ }\left({u}^{\circ }\right)\right)$
###### Not exactly what you’re looking for?
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• Pay only if we can solve it | 2022-09-26 00:16:52 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 58, "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": 0.6906847357749939, "perplexity": 2544.540968539665}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 20, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030334620.49/warc/CC-MAIN-20220925225000-20220926015000-00368.warc.gz"} |
https://www.transtutors.com/questions/maxson-products-distributes-a-single-product-a-woven-basket-whose-selling-price-is-2-2855243.htm | # Maxson Products distributes a single product, a woven basket whose selling price is $23... 1 answer below » Maxson Products distributes a single product, a woven basket whose selling price is$23 and whose variable cost is $19.32 per unit. The company’s monthly fixed expense is$10,304. Required: 1. Compute for the company’s break-even point in unit sales using the equation method. 2. Compute for the company’s break-even point in sales dollars using the equation method and the CM ratio. (Do not round intermediate calculations. Round your CM ratio to 2 decimal places.)
Attachments:
Wahid A
1) Sp × Q = Ve * Q + Fe
13 × Q = 9.36 * Q + 6,552
13 Q = 9.36 Q + 6,552
13Q – 9.36Q = 6,552
3.64 Q = 6,552
Q = 1800...
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Reach us on: | 2021-09-25 09:09:42 | {"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": 0.22022339701652527, "perplexity": 8912.214604588295}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-39/segments/1631780057615.3/warc/CC-MAIN-20210925082018-20210925112018-00081.warc.gz"} |
http://www.sciforums.com/threads/draqon-should-be.73603/ | Draqon should be...
Discussion in 'SF Open Government' started by spuriousmonkey, Nov 8, 2007.
?
Draqon should be moderator of para-something
41.5%
41.5%
19.5%
1. spuriousmonkeyBannedBanned
Messages:
24,066
Draqon should be...moderator of parapsychology
cuz the winds of reason tell us that the ocean of profanity can only last till the last vulva has opened up their glamorous depths to the insanity of convulsing male girdles.
And he is a proper member. As in he is actually around.
poll!
Yes! Draqon for moderator..yeehaw!
No! I don't think it is a good idea.
maybe!
to hide all adverts.
3. redarmy11Registered Senior Member
Messages:
7,658
to hide all adverts.
5. spuriousmonkeyBannedBanned
Messages:
24,066
to hide all adverts.
Messages:
7,658
8. spuriousmonkeyBannedBanned
Messages:
24,066
the message is not getting through
i repeat
the message is not getting through
/pant
9. spuriousmonkeyBannedBanned
Messages:
24,066
draq buddy!!!
We are cruising along fine at 100% of yes votes!
Please Register or Log in to view the hidden image!
Please Register or Log in to view the hidden image!
Please Register or Log in to view the hidden image!
10. NickelodeonBannedBanned
Messages:
10,581
I pledge my support, albeit reluctantly. While I hoped for a better solution, I think this is the best option at this time. It may be a tad controversial, in fact downright stupid, but whats the worse that could happen?
11. 2inquisitiveThe Devil is in the detailsRegistered Senior Member
Messages:
3,181
He could set up some stringent rules and no one would post in parapsychology because they were either too stoopid or disinterested.
12. spuriousmonkeyBannedBanned
Messages:
24,066
Nobody is posting there anyway.
13. NickelodeonBannedBanned
Messages:
10,581
I'm afraid to. Inevitably any thread started there is trolled or trashed by so called "real scientists" complaing about how its all bullshit etc. So what? Its the parapsychology subforum, if I want to talk about how I can make objects move with the power of my feeble mind, then bugger off!
On the other hand if parapsychology isnt a proper science or anything, whats it doing here?? Get rid of it.
14. spuriousmonkeyBannedBanned
Messages:
24,066
same for the engineering subforum? So it is not science. But what if people like to talk about how much they drank before finalizing the bridge plans? Let them!
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15. spuriousmonkeyBannedBanned
Messages:
24,066
that's discriminatory. All the moderators have voted against so far.
As if a Russian isn't good enough to be part of the moderator team.
Draqon! don't listen to them. You are going to make it, even if I have to make some sockpuppets!
16. draqonBannedBanned
Messages:
35,006
dont bother ... this test has proven itself.
17. GeoffPCaput gerat lupinumValued Senior Member
Messages:
22,087
I noted there was no option for "shot".
18. draqonBannedBanned
Messages:
35,006
...Geoffp...just reminded me of how many enemies I am surrounded by
no wonder my mind is full of tension...the many scars I receive everyday from those who are against me.
19. GeoffPCaput gerat lupinumValued Senior Member
Messages:
22,087
I take it back. You'd be great for parapsychology. Are you seeing these enemies through remote viewing or tea leaf consultation?
20. GrandtheftcowRegistered Member
Messages:
84
Hey you aren't allowed to vote for yourself!
Want to be a mod of a forum? Become a highly visible member of your desired area and wait patiently for an opening an hope you get chosen for it.
21. draqonBannedBanned
Messages:
35,006
fine...take away my vote everyone...oh yey...it doesn't matter because I lost anyways in this voting. happy?
22. GeoffPCaput gerat lupinumValued Senior Member
Messages:
22,087
And think: you were running against yourself, too.
You beat yourself. Now please continue to do so until unconsciousness sets in.
23. BenTheManDr. of Physics, Prof. of LoveValued Senior Member
Messages:
8,967
There's no Meaningless Philosophical Bullshit'' forum. <Insert Religion Joke Here> | 2020-07-06 17:07:16 | {"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": 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": 0.2389865666627884, "perplexity": 9278.502708282504}, "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-2020-29/segments/1593655881763.20/warc/CC-MAIN-20200706160424-20200706190424-00356.warc.gz"} |
https://sharepoint.stackexchange.com/questions/37178/discussion-board-reply-subject-for-workflow | # Discussion Board reply subject for workflow
I'm creating a workflow for a discussion board, that will create a new item in an announcements list. The idea is to have announcements such as
New Forum Post: 'This is a new post' was posted by me@myaddress.com
or
New Forum Reply: A reply to posted to 'This is a new post' by someoneelse@theiraddress.com
I've found the right lookups for the new post Subject and Posted By, but I can't figure out how to get the subject when a reply is created. I've tried lookups like Discussion Subject, Discussion Title, etc., but none seem to work.
Can anyone advise on the right lookups to use? Any websites that list all the lookups and what values they may contain?
Cheers | 2019-10-16 09:40:54 | {"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": 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": 0.44864752888679504, "perplexity": 2232.2526260777026}, "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/1570986666959.47/warc/CC-MAIN-20191016090425-20191016113925-00194.warc.gz"} |
http://mkweb.bcgsc.ca/printing.genomes/ | Here we are now at the middle of the fourth large part of this talk.get nowheremore quotes
# visualization: beautiful
In Silico Flurries: Computing a world of snow. Scientific American. 23 December 2017
# What if we were to print what we sequence?
Expressing the amount of sequence in the human genome in terms of the number of printed pages has been done before. At the Broad Institute, all of the human reference genome is printed in bound volumes.
At our sequencing facility, we sequence about 1 terabases per day. This is equivalent to 167 diploid human genomes (167 × 6 gigabases). The sequencing is done using a pool of 13 Illumina HiSeq 2500 sequencers, of which about 50% are sequencing at any given time.
A single letter-size page (8.5" × 11") of 6pt Courier using 0.25 inch margins accomodates 18,126 bases on 114 lines. Shown here is a portion of sequence from human chromosome 1. (PDF)
This sequencing is extremely fast.
To understand just how fast this is, consider printing this amount of sequence using a modern office laser printer. Let's pick the HP P3015n which costs about $400—a cheap and fast network printer. It can print at about 40 pages per minute. If we print the sequence at 6pt Courier using 0.25" margins, each 8.5" × 11" page will accomodate 18,126 bases. I chose this font size because it's reasonably legible. To print 1 terabases we need $10^12 / 18126 = 55.2$ million pages. If we print continuously at 40 pages per minute, we need $10^12 / (18126*40*1440) = 957.8$ days. If we had 958 printers working around the clock, we could print everything we sequence and not fall behind. This does not account for time required to replenish toner or paper. ## what's cheaper, sequencing or printing? It costs us about$12,000 to sequence a terabase in reagents. If we do it on a cost-recovery basis, it is about twice that, to include labor and storage. Let's say $25,000 per terabase. Coincidentally, this is about$150 per 1× coverage of a diploid human genome. The cost of sequencing a single genome would be significantly higher because of overhead. To overcome gaps in coverage and to be sensitive to alleles in heterogenous samples, sequencing should be done to 30× or more. For example, we sequence cancer genomes at over 100×. For theory and review see Aspects of coverage in medical DNA sequencing by Wendl et al. and Sequencing depth and coverage: key considerations in genomic analyses by Sims et al.. (Thanks to Nicolas Robine for pointing out that redundant coverage should be mentioned here).
Printing is 44× more expensive than sequencing, per base: 25 n$vs 1.1 μ$.
I should mention that the cost of analyzing the sequenced genome should be considered—this step is always the much more expensive one. In The $1,000 genome, the$100,000 analysis? Mardis asks "If our efforts to improve the human reference sequence quality, variation, and annotation are successful, how do we avoid the pitfall of having cheap human genome resequencing but complex and expensive manual analysis to make clinical sense out of the data?"
The cost of a single printed page (toner, power, etc) is about $0.02–0.05, depending on the printer. Let's be generous and say it's$0.02. To print 55.2 million pages would cost us $1.1M. This is about 44 times as expensive as sequencing. To print what we sequence we would require 958 office laser printers (shown here as HP3015n) at a cost of$1,100,000 per day. (PDF)
Think about this. It's 44× more expensive to merely print a letter on a page than it is to determine it from the DNA of a cell. In other words, to go from the physical molecule to a bit state on a disk is much cheaper than from a bit state on a disk to a representation of the letter on a page.
Per base, our sequencing costs $$25000/10^12 =$25*10^-9$, or 25 nanodollars. At $0.02 and 18,126 bp per page, printing costs $0.02/18126 =$1.1*10^-6$ or 1.1 microdollars.
If at this point you're thinking that printing isn't practical, the fact that the pages would weigh 248,000 kg and stack to 5.5 km should cinch the argument.
The capital cost of sequencing is, of course, much higher. The printers themselves would cost about $400,000 to purchase. The 6 sequencers, on the other hand, cost about$3,600,000.
## sequencing is as fast as downloading
We sequence at a rate close to the average internet bandwidth available to the public.
At 3.86 Mb/s, we could download a terabase of compressed sequence in a day, assuming the sequence can be compressed by a factor of 3. This level of compression is reasonable—the current human assembly is 938 Mb zipped).
In other words, you would have to be downloading essentially continuously to keep up with our sequencing.
VIEW ALL
# Predicting with confidence and tolerance
Wed 07-11-2018
I abhor averages. I like the individual case. —J.D. Brandeis.
We focus on the important distinction between confidence intervals, typically used to express uncertainty of a sampling statistic such as the mean and, prediction and tolerance intervals, used to make statements about the next value to be drawn from the population.
Confidence intervals provide coverage of a single point—the population mean—with the assurance that the probability of non-coverage is some acceptable value (e.g. 0.05). On the other hand, prediction and tolerance intervals both give information about typical values from the population and the percentage of the population expected to be in the interval. For example, a tolerance interval can be configured to tell us what fraction of sampled values (e.g. 95%) will fall into an interval some fraction of the time (e.g. 95%).
Nature Methods Points of Significance column: Predicting with confidence and tolerance. (read)
Altman, N. & Krzywinski, M. (2018) Points of significance: Predicting with confidence and tolerance Nature Methods 15:843–844.
### Background reading
Krzywinski, M. & Altman, N. (2013) Points of significance: Importance of being uncertain. Nature Methods 10:809–810.
# 4-day Circos course
Wed 31-10-2018
A 4-day introductory course on genome data parsing and visualization using Circos. Prepared for the Bioinformatics and Genome Analysis course in Institut Pasteur Tunis, Tunis, Tunisia.
Composite of the kinds of images you will learn to make in this course.
# Oryza longistaminata genome cake
Mon 24-09-2018
Data visualization should be informative and, where possible, tasty.
Stefan Reuscher from Bioscience and Biotechnology Center at Nagoya University celebrates a publication with a Circos cake.
The cake shows an overview of a de-novo assembled genome of a wild rice species Oryza longistaminata.
Circos cake celebrating Reuscher et al. 2018 publication of the Oryza longistaminata genome.
# Optimal experimental design
Tue 31-07-2018
Customize the experiment for the setting instead of adjusting the setting to fit a classical design.
The presence of constraints in experiments, such as sample size restrictions, awkward blocking or disallowed treatment combinations may make using classical designs very difficult or impossible.
Optimal design is a powerful, general purpose alternative for high quality, statistically grounded designs under nonstandard conditions.
Nature Methods Points of Significance column: Optimal experimental design. (read)
We discuss two types of optimal designs (D-optimal and I-optimal) and show how it can be applied to a scenario with sample size and blocking constraints.
Smucker, B., Krzywinski, M. & Altman, N. (2018) Points of significance: Optimal experimental design Nature Methods 15:599–600.
### Background reading
Krzywinski, M., Altman, N. (2014) Points of significance: Two factor designs. Nature Methods 11:1187–1188.
Krzywinski, M. & Altman, N. (2014) Points of significance: Analysis of variance (ANOVA) and blocking. Nature Methods 11:699–700.
Krzywinski, M. & Altman, N. (2014) Points of significance: Designing comparative experiments. Nature Methods 11:597–598.
# The Whole Earth Cataloguer
Mon 30-07-2018
All the living things.
An illustration of the Tree of Life, showing some of the key branches.
The tree is drawn as a DNA double helix, with bases colored to encode ribosomal RNA genes from various organisms on the tree.
The circle of life. (read, zoom)
All living things on earth descended from a single organism called LUCA (last universal common ancestor) and inherited LUCA’s genetic code for basic biological functions, such as translating DNA and creating proteins. Constant genetic mutations shuffled and altered this inheritance and added new genetic material—a process that created the diversity of life we see today. The “tree of life” organizes all organisms based on the extent of shuffling and alteration between them. The full tree has millions of branches and every living organism has its own place at one of the leaves in the tree. The simplified tree shown here depicts all three kingdoms of life: bacteria, archaebacteria and eukaryota. For some organisms a grey bar shows when they first appeared in the tree in millions of years (Ma). The double helix winding around the tree encodes highly conserved ribosomal RNA genes from various organisms.
Johnson, H.L. (2018) The Whole Earth Cataloguer, Sactown, Jun/Jul, p. 89 | 2018-11-16 10:34:42 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.33816465735435486, "perplexity": 3690.0908877734432}, "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-2018-47/segments/1542039743007.0/warc/CC-MAIN-20181116091028-20181116113028-00540.warc.gz"} |