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https://math.stackexchange.com/questions/2741157/shortest-distance-of-0-c-from-a-parabola?noredirect=1 | # Shortest distance of $(0,c)$ from a parabola
Find the shortest distance of the point $(0, c)$ from the parabola $y=x^2$ where $0 \leq c\leq 5$.
The point on the parabola $(x,x^2)$.
$\mathscr{l}(x)=\sqrt{x^2+(y-c)^2}=\sqrt{x^2+(x^2-c)^2}\\\mathscr{l}(y)=\sqrt{y+(y-c)^2}$
If I follow the expression $\mathscr{l}(y)=0$ then its fine and i'll get $\mathscr{l}(y)$ is minimum at $y=\frac{2c-1}{2}$ and $\mathscr{l}_{min}=\frac{\sqrt{4c-1}}{2}$. $$\frac{d\mathscr{l}(y)}{dy}=\frac{1+2(y-c)}{2\sqrt{y+(y-c)^2}}\\ \frac{d\mathscr{l}(y)}{dy}=0\implies y=\frac{2c-1}{2}\\ \mathscr{l}_{min}=\mathscr{l}(\frac{2c-1}{2})=\frac{\sqrt{4c-1}}{2}$$
My Attempt
But, if I consider $\mathscr{l}(x)$
$$\frac{d\mathscr{l}(x)}{dx}=\frac{2x+2(x^2-c).2x}{2\sqrt{x^2+(x^2-c)^2}}=\frac{x+2x(x^2-c)}{\sqrt{x^2+(x^2-c)^2}}=\frac{x\big[1+2(x^2-c)\big]}{\sqrt{x^2+(x^2-c)^2}}$$ $$\mathscr{l}'(x)=0\implies x\big[1+2(x^2-c)\big]=0\\ \implies x=0 \text{ or } x^2=\frac{2c-1}{2}\\ \implies x=0 \text{ or } x=\sqrt{\frac{2c-1}{2}}\text{ as }x\geq 0\\$$ How do I deal with the critical point $x=0$, and why do i not get this case if I use $\mathscr{l}(y)$ ?
Note: I have checked a similar problem link which does not completely address my post here.
To avoid confusion, I use $\ell$ to denote the distance. So, $\ell=\sqrt{x^2+(x^2-c)^2}=\sqrt{y+(y-c)^2}$.
$$\frac{d\ell}{dx}=\frac{d\ell}{dy}\cdot\frac{dy}{dx}=2x\frac{d\ell}{dy}$$
$\displaystyle \frac{d\ell}{dx}=0$ if and only if $\displaystyle \frac{d\ell}{dy}=0$ or $x=0$.
That's why we have one more critical point when differentiating w.r.t. $x$.
$$\frac{d\ell}{dx}=\frac{x\big[2x^2-(2c-1))\big]}{\sqrt{x^2+(x^2-c)^2}}$$
If $c>0.5$, $\displaystyle x=\pm\sqrt{\frac{2c-1}{2}}$ are local minimum points and $x=0$ is a local maximum point by the first derivative test.
If $0\le c\le 0.5$, $x=0$ is the only critical point which is a local minimum point by the first derivative test.
Note: When $0\le c<0.5$, $\displaystyle \frac{d\ell}{dy}>0$ for all $y\ge0$ and hence $\ell$ attains its least value when $y=0$.
Alternatively, we can use
$$\ell=\sqrt{x^4-(2c-1)x^2+c^2}=\sqrt{\left(x^2-\frac{2c-1}{2}\right)^2+\frac{4c-1}{4}}$$
If $c>0.5$, $\ell$ is the least if $\displaystyle x^2=\frac{2c-1}{2}$.
If $0\le c\le 0.5$, $\displaystyle \frac{2c-1}{2}\le0$ and $\ell$ is the least if $\displaystyle x=0$.
• Thanx. Could u pls explain a bit more on how you reached the conclusion that "if $c>0.5$, $x=\pm\frac{2c-1}{2}$ are local minimum and $x=0$ is a local maximum and if $0\leq c\leq 0.5$, $x=0$ is the only critical point which is local minimum" – ss1729 Apr 17 '18 at 16:53
• If $c<0.5$, it is impossible that $2x^2-(2c-1)$ as $-(2c-1)$ would be positive. So the derivative is zero only when $x=0$. If $x$ is slightly larger than $0$, then $\frac{2x^2-(2c-1)}{\sqrt{x^2+(x^2-c)^2}}$ is positive and hence $\frac{x[2x^2-(2c-1)]}{\sqrt{x^2+(x^2-c)^2}}$ is positive. If $x$ is slightly smaller than $0$, then $\frac{2x^2-(2c-1)}{\sqrt{x^2+(x^2-c)^2}}$ is positive and hence $\frac{x[2x^2-(2c-1)]}{\sqrt{x^2+(x^2-c)^2}}$ is negative. By first derivative test, $x=0$ is a minimum point. – CY Aries Apr 17 '18 at 16:59
• If $c>0.5$, $2x^2-(2c-1)=0$ if $x=\pm\sqrt{\frac{2c-1}{2}}$. When $x$ is slightly larger than $\sqrt{\frac{2c-1}{2}}$, $2x^2-(2c-1)>0$ $x>0$ and $\sqrt{x^2+(x^2-c)^2}>0$. So, the derivative is positive. When $x$ is slightly smaller than $\sqrt{\frac{2c-1}{2}}$, $2x^2-(2c-1)<0$ $x>0$ and $\sqrt{x^2+(x^2-c)^2}>0$. So, the derivative is negative. By the first derivative test, $x=\sqrt{\frac{2c-1}{2}}$ is a minimum point. The case for $x=-\sqrt{\frac{2c-1}{2}}$ can be similarly argued. – CY Aries Apr 17 '18 at 17:04
• When $c>0.5$, the derivative is also $0$ when $x=0$. When $x$ is slightly larger than $0$, $x>0$, $2x^2-(2c-1)<0$ and $\sqrt{x^2+(x^2-c)^2}>0$. So the derivative is negative. When $x$ is slightly smaller than $0$, $x<0$, $2x^2-(2c-1)<0$ and $\sqrt{x^2+(x^2-c)^2}>0$. So the derivative is positive. By the first derivative test, $x=0$ is a maximum point. – CY Aries Apr 17 '18 at 17:08
• $c>\frac{1}{2}\implies 2c>1\implies 2c-1>0$ and $x>0\implies 2x^2>0$. From there how do you conclude $2x^2-(2c-1)<0$ ? – ss1729 Apr 17 '18 at 17:18
A projection of $P(0,c)$ on the parabola is the point $Q(x,x^2)$ with $x\geq 0$ such that $x^2+(x^2-c)^2$ is minimized. This happens only if $\frac{d}{dx}\left[x^2+(x^2-c)^2\right]=0$. In particular, if $c\leq \frac{1}{2}$ the closest point is the vertex of the parabola$^{(*)}$ and the wanted distance is $c$. If $c>\frac{1}{2}$ the closest point in the first quadrant is $\left(\sqrt{c-\frac{1}{2}},c-\frac{1}{2}\right)$ and the wanted distance equals $\sqrt{c-\frac{1}{4}}$.
(*) It makes sense: the osculating circle at the vertex of the parabola has its center at $\left(0,\frac{1}{2}\right)$.
Remarkably, to project a generic point on a parabola is not task which can be solved by straightedge and compass only. Indeed, by assuming $c>0$, the projection of $P(c,0)$ on the parabola is the point $Q(x,x^2)$ with $x>0$ such that $(x-c)^2+x^4$ is minimized. This happens only if $$\frac{d}{dx}\left[(x-c)^2+x^4\right]=0\quad \Longleftrightarrow\quad 2x^3+x=c \quad \stackrel{x=\sqrt{\frac{2}{3}}z}{\Longleftrightarrow}\quad 4z^3+3z=3c\sqrt{\frac{3}{2}}$$ and since $\sinh(3\theta)=4\sinh^3(\theta)+3\sinh(\theta)$, the solution is given by $$x = \sqrt{\frac{2}{3}}\sinh\left(\frac{1}{3}\text{arcsinh}\left(3c\sqrt{\frac{3}{2}}\right)\right)$$ and the shortest distance is $$\sqrt{c^2+\frac{1}{3}\sinh^2\left(\frac{1}{3}\text{arcsinh}\left(3c\sqrt{\frac{3}{2}}\right)\right)-\sqrt{\frac{3}{2}}\sinh\left(\frac{1}{3}\text{arcsinh}\left(3c\sqrt{\frac{3}{2}}\right)\right)}$$ which over the interval $\left(0,\frac{1}{2}\right)$ is well-approximated by the simpler expression $\frac{c^2}{1+2c^2}$, over the interval $\left[\frac{1}{2},5\right)$ is well-approximated by the simpler expression $\sqrt{c^2+\frac{4}{3}}-1$. | {"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": 0.9424198865890503, "perplexity": 86.42359926512408}, "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-2021-04/segments/1610703536556.58/warc/CC-MAIN-20210123063713-20210123093713-00352.warc.gz"} |
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1.7
## Uppsala University
Skip to 0 minutes and 8 secondsSo what was the Paris Agreement all about? Well, there are annual negotiations by world leaders on issues of climate change. They occur every year in different cities around the globe. And this year was a particularly important year. They had them in Paris. And some people may remember that we've previously had a big event in Copenhagen in 2009, and a long time ago, in 1997, we had one in Kyoto in Japan. So this is the latest of the big annual events that we had.
Skip to 0 minutes and 37 secondsAnd prior to Paris, virtually all countries in the world gave what was called a pledge, an INDC, an Intended Nationally Determined Contribution, an awful acronym, but basically a pledge, a promise of what emission reductions that country would make between 2020 and 2030. So that all went in prior to the Paris negotiations and from all countries around the globe. And when you add up all of those pledges, what you find is that the temperature that we would get across the globe is about three to four degrees C temperature rise as an average. And, I think it's worth bearing in mind that that's a global average. And what's really important as well, what are regional repercussion of that?
Skip to 1 minute and 23 secondsAnd they could be as high as eight to twelve degrees C, repercussions in different parts of the world. So very severe heat waves and severe droughts and so forth as well. So what's the global average? We must remember that actually, that plays out regionally very differently to that. And of course, we have weather, which is what we live with day to day, not the climate. So the global averages can be a little bit misleading. And that a three to four degree temperature rise is a huge change in the planet. The difference between now and an ice age is only about five degrees. So we're talking about a significant move of that sort of order of magnitude.
Skip to 1 minute and 59 secondsIt's a commonly held view that we need to hold to less than two degrees C, that was the discussion prior to Paris and has been for quite a few years. A two degree C rise. But it's also worth bearing in mind, many of the poorer parts of the world have been arguing for much lower than that, 1 to and 1.5 degrees C, because even at 2 degrees C, many people will die. They'll be poor. They'll be generally in the southern hemisphere. They'll be low emitters. And they'll typically be nonwhite. So there are many people who will suffer, even at two degrees centigrade. And yet that's been the principle focus of international discussions up until Paris.
Skip to 2 minutes and 35 secondsSo there's a big gap between what was going into Paris, the pledges, and what people have been talking about internationally, these two degrees centigrade. So we're really aiming for three or four, even higher degree temperature rise, but really what we have is goal of no more than two. And Paris is really about discussing how do we reconcile those two. Now the Paris agreement was a 32-page document, the usual sort of legal boring document that comes out of these negotiations. And I think it was a triumph. There are lots of small triumphs, but there are two very big triumphs to me.
Skip to 3 minutes and 13 secondsThe first of these is that in having world leaders come together, virtually all world leaders, and say that they all agree climate change is a hugely important issue, they are basically saying they all accept the science. So there's been a long history of very sceptical or even scientific denial about climate change, by many people, often getting very well funded by the fossil fuel companies and so forth. So there has been a very strong opposition to the science of climate change, but often not by scientists, very seldom by scientists. And I think having the world leaders come together, and they actually said, we believe in the science, not the sceptics. And I think that is really very important.
Skip to 3 minutes and 56 secondsSo that is a major triumph and significant credit goes to the French in pulling all of this together and their diplomacy and getting negotiations to work so well. The second triumph relates to a particular paragraph within the agreement, within the 32-page agreement. And this paragraph says that the global community is going to aim to hold the temperature well below two degrees centigrade and pursue efforts to keep the temperature rise to only one and a half degree centigrade, which is what the poor part of the world in particular have been asking for. They're also going to do that on the basis of the best science.
Skip to 4 minutes and 30 secondsThat's the IPCC, is probably the best way of thinking that, but obviously that evolves over time as well. And in line with equity, in other words, the poor parts of the world will take longer to reduce their emissions than the wealthy parts of the world. So 2 degrees C, 1.5, based on the best science, and equity. These are really important. These are enshrined within the Paris agreement. And we can all hold our leaders, whether that's our government leaders, our organisational leaders. So these people can be held to account, this is what we have internationally signed up for. There are some problems with the Paris agreement.
Skip to 5 minutes and 5 secondsFirstly, it's about climate change and you would think there'd be some reference to fossil fuels. Nowhere in the 32-page document is there any reference to fossil fuels or de-carbonization. Aviation and shipping emissions are exempt. So no country takes responsibility for those. They're simply just held separate to the agreement. The emissions from aviation and shipping are the same as adding the emissions from UK and Germany together. And they're growing much more rapidly than our national emissions. So aviation and shipping are hugely important. The voluntary pledges do not add up to two degrees C, much nearer three to four degrees centigrade. And we're not going to review those pledges until 2023 or possibly even a little later.
Skip to 5 minutes and 44 secondsBy that time, we will have put another 300 billion tonnes of carbon dioxide in the atmosphere. By the time we get to review the pledges that we all know are too weak for the commitments in the Paris agreement, by the time we get to review them, we would have put another 300 billion tonnes of carbon in the atmosphere. The only way that all of this can be squared, we can make any sense at all of this, is if we assume certain types of technology, which do not exist at the moment, that will suck carbon dioxide out of the atmosphere in many decades to come.
Skip to 6 minutes and 17 secondsSo I sometimes refer to them as a Doctor Strangelove technology, based on quite a famous film about a mad scientist. Now, perhaps we should research these things, but I think to assume that they work is a real fundamental mistake. And yet that was embedded in the Paris agreement, but without any words to say so. You have to look behind the text to find this. The two other, well, one other major point I think, is that there is some money put aside in the Paris agreement for the poorer parts of the world to mitigate their emissions and to respond to the climate change that we have imposed upon them. It's about $100 billion every single year. Skip to 6 minutes and 53 secondsNow there's is no agreement who's going to pay that or where it's going to come from, but I think it's worth bearing in mind that the subsidy in 2015 to fossil fuels, the direct and indirect subsidy, the indirect in the health impacts mostly from fossil fuels, was about$5.3 trillion, according to the IMF, the International Monetary Fund. So about 53 times more in 2015, a single year, than we're going to give the poorer parts of the world to deal with climate change. Or indeed, it's about 1/30th of the size of the UK economy.
Skip to 7 minutes and 21 secondsSo I think it is a derisory sum, that we've given such a small amount of money to the poor parts of the world to deal with climate change.
# The 2° target in the Paris Agreement
In this video Kevin Anderson, visiting professor in climate change leadership, Uppsala University, discusses the 2°C target in the Paris Agreement in more depth.
He notes that the target to limit global warming to +2°C above pre-industrial temperatures is a dangerous target:
It’s a commonly held view that we need to hold to less than 2°C. […] But it is also worth bearing in mind that poorer parts of the world have been arguing for much lower than that, 1.5°C, because even at 2°C many people will die. They will be poor, they will be generally in the Southern hemisphere, they will be low-emitters, and they will typically be non-white.
There were two very big triumphs in the Paris Agreement, according to Anderson. Virtually all world leaders came together and agreed that climate change is a serious issue - so science won over the climate change deniers. Furthermore, world leaders were able to agree to limit global warming to 2°C or even 1.5°C, and to do this on the basis of the best science and equity.
However, much like Stabinsky said in the previous step, Anderson also notes that in sum the voluntary pledges of intended nationally determined contributions (INDCs) of the signatories are currently nowhere close of meeting the 2°C target, and that the treaty relies on speculative negative emissions technologies, which imply that we would find a way to capture carbon from the atmosphere and sequester it. Furthermore, the impact of aviation, international shipping or the need for decarbonization are not taken into account in the Agreement.
It is about climate change and you think that there would be some reference to fossil fuels. [But] nowhere in the 32 page document is there any reference to fossil fuels or decarbonization.
The Paris Agreement nevertheless provides a good starting point for climate action and leadership, he argues:
[W]e can all hold our leaders, that’s our government leaders, organisational leaders […] to account: this is what we have internationally signed up for.
### Reflection
What are your experiences of climate change in your life time? Are the changes sudden or slow and harder to detect? How are human and natural communities affected?
Add your thoughts in the comments section below (the pink box with a plus) and add your stories, photos and videos on Twitter hashtag #FLccleadership | {"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": 0.257629930973053, "perplexity": 975.6998886559791}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-34/segments/1534221215487.79/warc/CC-MAIN-20180820003554-20180820023554-00590.warc.gz"} |
https://www.physicsforums.com/threads/product-of-total-differentials.746932/ | # Product of total differentials
1. Apr 4, 2014
### taimoortalpur
Dear All,
I am unable to understand a simple mathematics relation. I spent 2-3 hours to Google multi-variable mathematics and have studied some concepts, still i am missing/confusing some basics. The problem I have at hand is following.
Vector p can be written as
p = (p1, p2, p3) = n(sin θ3 cos φ, sin θ3 sin φ, cos θ3)
As defined earlier, p1, p2, and p3 are the x1, x2, and x3 components of vector p and therefore dp1dp2 can be written as shown in attached figure.
Kindly help me understand the multiplication of these two total differentials dp1*dp2. I cannot find any easy reference to product of total differentials. I will appreciate reference to any material.
Many thanks for your help and support.
Regards,
Taimoor.
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2. Apr 4, 2014
### mathman
Similar Discussions: Product of total differentials | {"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": 0.940414547920227, "perplexity": 2296.084081437435}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-51/segments/1512948597585.99/warc/CC-MAIN-20171217210620-20171217232620-00489.warc.gz"} |
http://math.stackexchange.com/questions/220134/absorbing-time-in-0-of-a-simple-left-drifted-markov-chain-on-non-negative-inte | # Absorbing time in $0$ of a simple left-drifted Markov chain on non-negative integers
Let $M$ denote the Markov chain on states $\{0, 1, 2, ...\}$ with absorbing state $0$. For $i \geq 1$, let the transition probabilities be $p$ for $(i, i-1)$ and $1-p$ for $(i, i+1)$. Further, assume $p > 1/2$, thus, there is a drift towards $0$. I am interested in the mean absorbing/hitting time in state $0$, when started at state $n > 0$, denoted as $\mathbb{E}\{h(n)\}$. Due to the drift towards $0$, this should easily be determinable by using the linearity of the expectation on each single step?
More precise, the expected change ('increment') $\Delta$ on each single step of a walk that did not yet reach $0$, is $\mathbb{E(\Delta)} = (-1) \cdot p + 1 \cdot (1-p) = 1 - 2p < 0$. Thus, 'on average' we make progress $|1 - 2p|$ towards $0$ on each step, so we should hit $0$ after $h := \frac{n}{|1-2p|}$ many time steps, since $n + \mathbb{E}\{h \cdot \Delta\} = n + h \mathbb{E}\{\Delta\} = n - n = 0$. Thus, it should hold that $\mathbb{E}\{h(n)\} = h$.
I suppose that there is something wrong with this easy argument, since I fail to find a formal proof for it; I cannot adopt the hitting time definition to this. So my two questions are:
(1) Is this approach right or wrong, and how can $\mathbb{E}\{h(n)\}$ be determined correctly?
(2) If this argument is wrong, shouldn't it at least provide an upper bound on the hitting time?
-
(1) Show that $\mathbb E(h(n))=n\mathbb E(h(1))$.
(2) Show that $\mathbb E(h(1))=1+(1-p)\mathbb E(h(2))$.
(3) Conclude that, indeed, $\mathbb E(h(n))=n/(2p-1)$ for every $n\geqslant1$.
Further, this approach can even easily be extended to the case of a chain with alternating probabilities $p_1$, $p_2$, started at even $n$. (But for pattern periods $k$ greater than 2, the pendant of (2) runs into trouble then). The fundamental different way of thinking is, that this approach groups the random walks by their first visit of some nearer node (independently of the individual walk lengths so far), while my approach tried to deal with the expected progress of each random walk after each batch of exactly $k$ steps (here, $k=1$). Thanks a lot for your help! – cubic lettuce Oct 25 '12 at 9:30 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9843376874923706, "perplexity": 161.4828821000478}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-22/segments/1464049276131.97/warc/CC-MAIN-20160524002116-00211-ip-10-185-217-139.ec2.internal.warc.gz"} |
https://indico.desy.de/indico/event/3241/other-view?view=standard | # Workshop on Multi Parton Interaction
chaired by , ,
from to (Europe/Berlin)
Description Multiple parton interactions are one of the most common, yet poorly understood, phenomena at the LHC. The results of the first LHC measurements differ substantially from predictions provided by Monte Carlo simulations. The modelling implemented in Monte Carlo generators aims to describe multiple scattering throughout the whole range of scales, from soft to hard. At the same time, much work is ongoing to improve theoretical understanding of multiple scatterings at both low and high pT, and study the related phenomenology at the LHC. Therefore it is extremely important to combine theoretical efforts in order to achieve a better description of multiple interactions, and, in general, the underlying event. Topics to be discussed will include: - Improving theoretical understanding of the multiple interaction (MPI) - Implementation in general purpose MC programs. - Current development on multi parton PDF and use of them in MC programs - Multi parton interaction contribution to signal and background processes - Recent experimental results for LHC (and Tevatron?) and plans for future measurements of MPI at the LHC - Continued development of better MC event generators UPLOADING slides: - Hit the 'Manage' button - the modification code is mpi10
Go to day
• Monday, September 13, 2010
• 14:00 - 15:30 Afternoon session I. Location: Seminar Room 4
• 14:00 Measurement of the Underlying Event activity at 900 GeV and 7 TeV with CMS detector 45' Speaker: Livio Fano' Material:
• 14:45 Multiparton interactions as seen by an ATLAS Higgs physicist 45' Speaker: Bill Murray Material:
• 15:30 - 16:00 Coffee Break
• 16:00 - 17:30 Afternoon session II. Location: Seminar Room 4
• 16:00 Minimum Bias Physics at LHCb 45' Speaker: Michael Schmelling Material:
• 16:45 ATLAS leading track UE analysis 45' Speaker: Deepak Kar Material:
• Tuesday, September 14, 2010
• 09:00 - 10:30 Morning session I. Location: Seminar Room 4
• 09:00 MPI in Pythia 8 45' Speaker: Richard Corke Material:
• 09:45 MPI model in Herwig++ 45' Speaker: Stefan Gieseke Material:
• 10:30 - 11:00 Coffee break
• 11:00 - 12:30 Morning session II. Location: Seminar Room 4
• 11:00 Minimum Bias and the Underlying Event in Sherpa 45' Speaker: Korinna Zapp Material:
• 11:45 Multiple interactions, diffraction and the BFKL pomeron 45' Speaker: Gösta Gustafson Material:
• 12:30 - 14:00 Lunch break
• 14:00 - 15:30 Afternoon session I. Location: Seminar Room 4
• 14:00 MPI: general features and consistency requirements 45' Speaker: Daniele Treleani Material:
• 14:45 Double Parton Distributions incorporating pQCD Evolution and Sum Rule Constraints 45' Speaker: Jonathan Gaunt Material:
• 15:30 - 16:00 Coffee break
• 16:00 - 18:00 Afternoon session II. Location: Sem. 4 & Hörsaal
• 16:00 Multiparton interactions: some theoretical considerations 45' Speaker: Markus Diehl Material:
• 16:45 Break 15'
• 17:00 Parton Distribution Functions 1h0' ( Hörsaal ) Speaker: James Stirling Material:
• 19:30 - 22:00 Workshop dinner
• Wednesday, September 15, 2010
• 09:00 - 11:15 Morning session I. Location: Seminar Room 4
• 09:00 Probing double parton scattering with same-sign W pairs at the LHC 45' Speaker: Steve Kom Material:
• 09:45 MPI in Vector Boson plus jet(s) production at the LHC 45' Speaker: Ezio Maina Material:
• 10:30 Dynamical Characteristics of Double Parton Scattering 45' Speaker: Ed Berger Material:
• 11:15 - 11:30 Coffee break
• 11:30 - 13:00 Morning session II. Location: Seminar Room 4
• 11:30 4-jet correlations at Tevatron and LHC 45' Speaker: Boris Blok Material:
• 12:15 Recombination within multi-chain contributions in pp scattering 45' Speaker: Jochen Bartels Material:
• 13:00 - 14:00 Lunch break | {"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": 0.6931478977203369, "perplexity": 20813.375657611694}, "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/1531676591837.34/warc/CC-MAIN-20180720213434-20180720233434-00012.warc.gz"} |
https://pytorch.org/docs/stable/torchvision/utils.html | # torchvision.utils¶
torchvision.utils.make_grid(tensor, nrow=8, padding=2, normalize=False, range=None, scale_each=False, pad_value=0)[source]
Make a grid of images.
Parameters
• tensor (Tensor or list) – 4D mini-batch Tensor of shape (B x C x H x W) or a list of images all of the same size.
• nrow (python:int, optional) – Number of images displayed in each row of the grid. The final grid size is (B / nrow, nrow). Default: 8.
• padding (python:int, optional) – amount of padding. Default: 2.
• normalize (bool, optional) – If True, shift the image to the range (0, 1), by the min and max values specified by range. Default: False.
• range (tuple, optional) – tuple (min, max) where min and max are numbers, then these numbers are used to normalize the image. By default, min and max are computed from the tensor.
• scale_each (bool, optional) – If True, scale each image in the batch of images separately rather than the (min, max) over all images. Default: False.
• pad_value (python:float, optional) – Value for the padded pixels. Default: 0.
Example
See this notebook here
torchvision.utils.save_image(tensor, fp, nrow=8, padding=2, normalize=False, range=None, scale_each=False, pad_value=0, format=None)[source]
Save a given Tensor into an image file.
Parameters
• tensor (Tensor or list) – Image to be saved. If given a mini-batch tensor, saves the tensor as a grid of images by calling make_grid.
• - A filename (fp) –
• format (Optional) – If omitted, the format to use is determined from the filename extension. If a file object was used instead of a filename, this parameter should always be used.
• **kwargs – Other arguments are documented in make_grid. | {"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": 0.33001384139060974, "perplexity": 7670.56125865402}, "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-2020-16/segments/1585371660550.75/warc/CC-MAIN-20200406200320-20200406230820-00409.warc.gz"} |
https://crypto.stackexchange.com/questions/43486/reducing-the-output-of-aes-to-64-bit | # Reducing the output of AES to 64 bit
Is it possible to reduce the output of AES after encryption into 64 bit with the possibility of a reversible before decryption Without any error
• I would investigate Karnaugh Maps, and the Quine-McCluskey algorithm if you're interested in reducing Boolean expressions. en.wikipedia.org/wiki/Karnaugh_map en.wikipedia.org/wiki/Quine%E2%80%93McCluskey_algorithm Feb 2 '17 at 1:59
• If you want to encrypt something non-block sized (vs get a 64 bit PRP) then consider using a mode such as CTR. Feb 2 '17 at 1:59
• FFX mode. But it's slow and complicated. Feb 2 '17 at 3:05
• What does "the possibility of a reversible before decryption Without any error" mean? Your question starts out reasonably, but seems to trail off into ungrammatical nonsense. Feb 2 '17 at 11:12
There is no way to reduce the output of a single AES encryption operation to 64 bits with the possibility of decryption without error.
There are however ways to use AES to build a 64-bit cryptogram that can be decrypted without error. The simplest is: encipher the constant zero with AES, keep the first 64 bits, XOR with the plaintext to get the cryptogram. Problem is, that's insecure if the AES key is used more than once.
There are ways to allow several uses with the same key:
• If an initial IV setup is tolerable (e.g. the IV overhead negligible compared to many later 64-bit cryptograms delivered in order), CTR mode is fine, as noted by Thomas M. DuBuisson.
• Construct a 64-bit symmetric Feistel cipher using AES encryption for the round function (AES-encipher the concatenation of the round number and 32-bit input); with 10 rounds for good measure, and perhaps using addition modulo $2^{32}$ rather than XOR for combination of the output of the round function (so that the cipher is not always an even permutation, which would be noticeable after $2^{64}-2$ plaintext/ciphertext pairs), that's a secure 64-bit block cipher. Beware however that with this method (or any 64-bit block cipher) alone, identical 64-bit plaintexts will be recognizable as identical 64-bit ciphertexts.
• FFX mode is another possibility, as noted by CodesInChaos.
The AES itself is an standard with a block size of 128 bits and 3 different key sizes (128, 192 and 256). The Rijndael, the winner of the AES contest, was proposed with this block size but with 5 different key sizes (add 160 and 224 to the previous 3).
This suggest a bigger flexibility in the Rijndael. In fact, playing with the parameters and the maths behind the Rijndael one can produce almost any arbitrary block and key size combination. Don't forget that this is not AES, but Rijndael.
Apart from this generalization, there are other algorithms designed designed to be 64 bit block like mCyrpton, present, katan/ktantan or Prince to mention some that I know. The cryptoanalysis made by the community is proportional to its impact. The the Rijndael has been more tested (on its original setup).
Apart from the CTR block cipher mode mention on a comment of a previous answer, you have other possibilities to think using an already tested implementation of the AES: Format preserving encryption.
Any solution you apply depends on where you apply, what are your needs. But focus, don't implement your crypto unless you are doing a research or it is going to be publicly review before and during production phase. | {"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": 0.45049238204956055, "perplexity": 1782.3255466890114}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-49/segments/1637964358570.48/warc/CC-MAIN-20211128164634-20211128194634-00078.warc.gz"} |
https://www.bas.ac.uk/data/our-data/publication/effect-of-plasma-density-on-diffusion-rates-due-to-wave/ | # Effect of plasma density on diffusion rates due to wave particle interactions with chorus and plasmaspheric hiss: extreme event analysis
Wave particle interactions play an important role in controlling the dynamics of the radiation belts. The purpose of this study is to estimate how variations in the plasma density can affect diffusion rates resulting from interactions between chorus waves and plasmaspheric hiss with energetic particles and the resulting evolution of the energetic electron population. We perform a statistical analysis of the electron density derived from the plasma wave experiment on the CRRES satellite for two magnetic local time sectors corresponding to near midnight and near noon. We present the cumulative probability distribution of the electron plasma density for three levels of magnetic activity as measured by Kp. The largest densities are seen near L* = 2.5 while the smallest occur near L* = 6. The broadest distribution, corresponding to the greatest variability, occurs near L* = 4. We calculate diffusion coefficients for plasmaspheric hiss and whistler mode chorus for extreme values of the electron density and estimate the effects on the radiation belts using the Salammbô model. At L* = 4 and L* = 6, in the low density case, using the density from the 5th percentile of the cumulative distribution function, electron energy diffusion by chorus waves is strongest at 2 MeV and increases the flux by up to 3 orders of magnitude over a period of 24 h. In contrast, in the high density case, using the density from the 95th percentile, there is little acceleration at energies above 800 keV at L* = 6, and virtually no acceleration at L* = 4. In this case the strongest energy diffusion occurs at lower energies around 400 keV where the flux at L* = 6 increases 3 orders of magnitude.
### Details
Publication status:
Published
Author(s):
Authors: Sicard-Piet, A., Boscher, D., Horne, R. B. ORCID record for R. B. Horne, Meredith, N. P. ORCID record for N. P. Meredith, Maget, V.
On this site: Nigel Meredith, Richard Horne
Date:
1 August, 2014
Journal/Source:
Annales Geophysicae / 32
Page(s):
1059-1071
Digital Object Identifier (DOI):
https://doi.org/10.5194/angeo-32-1059-2014 | {"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": 0.9610900282859802, "perplexity": 1568.4618070188083}, "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/1579250619323.41/warc/CC-MAIN-20200124100832-20200124125832-00151.warc.gz"} |
https://www.aimsciences.org/article/doi/10.3934/nhm.2017007 | # American Institute of Mathematical Sciences
June 2017, 12(2): 173-189. doi: 10.3934/nhm.2017007
## The Riemann solver for traffic flow at an intersection with buffer of vanishing size
1 Department of Mathematics, Penn State University, University Park, Pa. 16802, USA 2 Department of Mathematical Sciences, Norwegian University of Science and Technology, NO-7491 Trondheim, Norway
Received December 2015 Revised March 2016 Published May 2017
Fund Project: The first author was partially supported by NSF, with grant DMS-1411786: "Hyperbolic Conservation Laws and Applications".
The paper examines the model of traffic flow at an intersection introduced in [2], containing a buffer with limited size. As the size of the buffer approaches zero, it is proved that the solution of the Riemann problem with buffer converges to a self-similar solution described by a specific Limit Riemann Solver (LRS). Remarkably, this new Riemann Solver depends Lipschitz continuously on all parameters.
Citation: Alberto Bressan, Anders Nordli. The Riemann solver for traffic flow at an intersection with buffer of vanishing size. Networks & Heterogeneous Media, 2017, 12 (2) : 173-189. doi: 10.3934/nhm.2017007
##### References:
show all references
##### References:
The flux $f_k$ as a function of the density $\rho$, along the $k$-th road
Constructing the solution of the the Riemann problem, according to the limit Riemann solver (LRS), with two incoming and two outgoing roads. The vector $\mathbf{f}=(\bar f_1,\bar f_2)$ of incoming fluxes is the largest point on the curve $\gamma$ that satisfies the two constraints $\sum_{i\in\mathcal{I}} \gamma_i(s) \theta_{ij}\leq \omega_j$, $j\in\mathcal{O}$
Left: an incoming road which is initially free. For $t_1<t<t_2$ part of the road is congested (shaded area). Right: an outgoing road which is initially congested. For $0<t<t_3$ part of the road is free (shaded area). In both cases, a shock marks the boundary between the free and the congested region
The two cases in the proof of Theorem 2.3. Left: none of the outgoing roads provides a restriction on the fluxes of the incoming roads. The queues are zero. Right: one of the outgoing roads is congested and restricts the maximum flux through the node
A case with three incoming roads. For large times, the first two roads become free, while the third road remains congested
[1] Constantine M. Dafermos. A variational approach to the Riemann problem for hyperbolic conservation laws. Discrete & Continuous Dynamical Systems - A, 2009, 23 (1&2) : 185-195. doi: 10.3934/dcds.2009.23.185 [2] Yanbo Hu, Wancheng Sheng. The Riemann problem of conservation laws in magnetogasdynamics. Communications on Pure & Applied Analysis, 2013, 12 (2) : 755-769. doi: 10.3934/cpaa.2013.12.755 [3] Yu Zhang, Yanyan Zhang. Riemann problems for a class of coupled hyperbolic systems of conservation laws with a source term. Communications on Pure & Applied Analysis, 2019, 18 (3) : 1523-1545. doi: 10.3934/cpaa.2019073 [4] João-Paulo Dias, Mário Figueira. On the Riemann problem for some discontinuous systems of conservation laws describing phase transitions. Communications on Pure & Applied Analysis, 2004, 3 (1) : 53-58. doi: 10.3934/cpaa.2004.3.53 [5] Zhi-Qiang Shao. Lifespan of classical discontinuous solutions to the generalized nonlinear initial-boundary Riemann problem for hyperbolic conservation laws with small BV data: shocks and contact discontinuities. Communications on Pure & Applied Analysis, 2015, 14 (3) : 759-792. doi: 10.3934/cpaa.2015.14.759 [6] Weishi Liu. Multiple viscous wave fan profiles for Riemann solutions of hyperbolic systems of conservation laws. Discrete & Continuous Dynamical Systems - A, 2004, 10 (4) : 871-884. doi: 10.3934/dcds.2004.10.871 [7] Anupam Sen, T. Raja Sekhar. Structural stability of the Riemann solution for a strictly hyperbolic system of conservation laws with flux approximation. Communications on Pure & Applied Analysis, 2019, 18 (2) : 931-942. doi: 10.3934/cpaa.2019045 [8] Mauro Garavello, Francesca Marcellini. The Riemann Problem at a Junction for a Phase Transition Traffic Model. Discrete & Continuous Dynamical Systems - A, 2017, 37 (10) : 5191-5209. doi: 10.3934/dcds.2017225 [9] Alberto Bressan, Fang Yu. Continuous Riemann solvers for traffic flow at a junction. Discrete & Continuous Dynamical Systems - A, 2015, 35 (9) : 4149-4171. doi: 10.3934/dcds.2015.35.4149 [10] Boris Andreianov, Mohamed Karimou Gazibo. Explicit formulation for the Dirichlet problem for parabolic-hyperbolic conservation laws. Networks & Heterogeneous Media, 2016, 11 (2) : 203-222. doi: 10.3934/nhm.2016.11.203 [11] Tatsien Li, Wancheng Sheng. The general multi-dimensional Riemann problem for hyperbolic systems with real constant coefficients. Discrete & Continuous Dynamical Systems - A, 2002, 8 (3) : 737-744. doi: 10.3934/dcds.2002.8.737 [12] Tai-Ping Liu, Shih-Hsien Yu. Hyperbolic conservation laws and dynamic systems. Discrete & Continuous Dynamical Systems - A, 2000, 6 (1) : 143-145. doi: 10.3934/dcds.2000.6.143 [13] Rinaldo M. Colombo, Mauro Garavello. A Well Posed Riemann Problem for the $p$--System at a Junction. Networks & Heterogeneous Media, 2006, 1 (3) : 495-511. doi: 10.3934/nhm.2006.1.495 [14] Meixiang Huang, Zhi-Qiang Shao. Riemann problem for the relativistic generalized Chaplygin Euler equations. Communications on Pure & Applied Analysis, 2016, 15 (1) : 127-138. doi: 10.3934/cpaa.2016.15.127 [15] Vladimir S. Gerdjikov, Rossen I. Ivanov, Aleksander A. Stefanov. Riemann-Hilbert problem, integrability and reductions. Journal of Geometric Mechanics, 2019, 11 (2) : 167-185. doi: 10.3934/jgm.2019009 [16] Alberto Bressan, Marta Lewicka. A uniqueness condition for hyperbolic systems of conservation laws. Discrete & Continuous Dynamical Systems - A, 2000, 6 (3) : 673-682. doi: 10.3934/dcds.2000.6.673 [17] Gui-Qiang Chen, Monica Torres. On the structure of solutions of nonlinear hyperbolic systems of conservation laws. Communications on Pure & Applied Analysis, 2011, 10 (4) : 1011-1036. doi: 10.3934/cpaa.2011.10.1011 [18] Stefano Bianchini. A note on singular limits to hyperbolic systems of conservation laws. Communications on Pure & Applied Analysis, 2003, 2 (1) : 51-64. doi: 10.3934/cpaa.2003.2.51 [19] Xavier Litrico, Vincent Fromion, Gérard Scorletti. Robust feedforward boundary control of hyperbolic conservation laws. Networks & Heterogeneous Media, 2007, 2 (4) : 717-731. doi: 10.3934/nhm.2007.2.717 [20] Fumioki Asakura, Andrea Corli. The path decomposition technique for systems of hyperbolic conservation laws. Discrete & Continuous Dynamical Systems - S, 2016, 9 (1) : 15-32. doi: 10.3934/dcdss.2016.9.15
2018 Impact Factor: 0.871 | {"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": 0.7475422620773315, "perplexity": 2302.1723645703387}, "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-2020-16/segments/1585371830894.88/warc/CC-MAIN-20200409055849-20200409090349-00249.warc.gz"} |
https://qiskit.org/documentation/apidoc/verification.html | # Verification (qiskit.ignis.verification)¶
## Quantum Volume¶
qv_circuits([qubit_lists, ntrials, qr, cr]) Return a list of square quantum volume circuits (depth=width) QVFitter([backend_result, …]) Class for fitters for quantum volume.
## Randomized Benchmarking¶
Randomization benchmarking (RB) is a well-known technique to measure average gate performance by running sequences of random Clifford gates that should return the qubits to the initial state. Qiskit Ignis has tools to generate one- and two-qubit gate Clifford RB sequences simultaneously, as well as performing interleaved RB, purity RB and RB on the non-Clifford CNOT-Dihedral group.
randomized_benchmarking_seq([nseeds, …]) Generate generic randomized benchmarking (RB) sequences. RBFitter(backend_result, cliff_lengths[, …]) Class for fitters for randomized benchmarking. InterleavedRBFitter(original_result, …[, …]) Class for fitters for interleaved RB, derived from RBFitterBase class. PurityRBFitter(purity_result, npurity, …) Class for fitter for purity RB. CNOTDihedralRBFitter(cnotdihedral_Z_result, …) Class for fitters for non-Clifford CNOT-Dihedral RB. Abstract base class (ABS) for utils for various groups and sets of gates for randomized benchmarking. Clifford([num_qubits, table, phases]) Clifford Operator Class. CliffordUtils([num_qubits, group_tables, …]) Class for util functions for the Clifford group. CNOTDihedral(n_qubits) CNOT-dihedral Object Class. DihedralUtils([num_qubits, group_tables, …]) Class for util functions for the CNOT-dihedral group. count_gates(qobj, basis, qubits) Take a compiled qobj and output the number of gates in each circuit. gates_per_clifford(transpiled_circuits_list, …) Take a list of transpiled QuantumCircuit and use these to calculate the number of gates per Clifford. calculate_1q_epg(gate_per_cliff, epc_1q, qubit) Convert error per Clifford (EPC) into error per gates (EPGs) of single qubit basis gates. calculate_2q_epg(gate_per_cliff, epc_2q, …) Convert error per Clifford (EPC) into error per gate (EPG) of two qubit cx gates. calculate_1q_epc(gate_per_cliff, epg_1q, qubit) Convert error per gate (EPG) into error per Clifford (EPC) of single qubit basis gates. calculate_2q_epc(gate_per_cliff, epg_2q, …) Convert error per gate (EPG) into error per Clifford (EPC) of two qubit cx gates. coherence_limit([nQ, T1_list, T2_list, gatelen]) The error per gate (1-average_gate_fidelity) given by the T1,T2 limit. twoQ_clifford_error(ngates, gate_qubit, gate_err) The two qubit Clifford gate error given measured errors in the primitive gates used to construct the Clifford (see arxiv:1712.06550).
## Tomography¶
state_tomography_circuits(circuit, …[, …]) Return a list of quantum state tomography circuits. process_tomography_circuits(circuit, …[, …]) Return a list of quantum process tomography circuits. basis Quantum tomography basis StateTomographyFitter(result, circuits[, …]) Maximum-Likelihood estimation state tomography fitter. ProcessTomographyFitter(result, circuits[, …]) Maximum-Likelihood estimation process tomography fitter. TomographyFitter(result, circuits[, …]) Base maximum-likelihood estimate tomography fitter class marginal_counts(counts[, meas_qubits, pad_zeros]) Compute marginal counts from a counts dictionary. combine_counts(counts1, counts2) Combine two counts dictionaries. expectation_counts(counts) Converts count dict to an expectation counts dict. count_keys(num_qubits) Return ordered count keys.
## Topological Codes¶
RepetitionCode(d[, T]) Implementation of a distance d repetition code, implemented over T syndrome measurement rounds. GraphDecoder(code[, S]) Class to construct the graph corresponding to the possible syndromes of a quantum error correction code, and then run suitable decoders. lookuptable_decoding(training_results, …) Calculates the logical error probability using postselection decoding. postselection_decoding(results) Calculates the logical error probability using postselection decoding.
## Accreditation¶
AccreditationCircuits(target_circ[, …]) This class generates accreditation circuits from a target. Class for fitters for accreditation QOTP(circ, num[, two_qubit_gate, …]) Performs a QOTP (or random compilation) on a generic circuit. QOTPCorrectCounts(qotp_counts, qotp_postp) Corrects a dictionary of results, shifting the qotp | {"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": 0.5024316906929016, "perplexity": 6628.909247786975}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-34/segments/1596439737233.51/warc/CC-MAIN-20200807231820-20200808021820-00552.warc.gz"} |
https://codereview.stackexchange.com/questions/128641/project-euler-8-maximum-product-of-13-adjacent-digits/128668 | # Project Euler 8: Maximum product of 13 adjacent digits
Problem 8 on Project Euler which asks
Find the thirteen adjacent digits in the 1000-digit number that have the greatest product. What is the product?
This is my solution in Visual C#.
class ProblemEight
{
static byte[] input = new byte[] { 7, 3, 1, 6, 7, 1, 7, 6, 5, 3, 1, 3, 3, 0, 6, 2, 4, 9, 1, 9, 2, 2, 5, 1, 1, 9, 6, 7, 4, 4, 2, 6, 5, 7, 4, 7, 4, 2, 3, 5, 5, 3, 4, 9, 1, 9, 4, 9, 3, 4, 9, 6, 9, 8, 3, 5, 2, 0, 3, 1, 2, 7, 7, 4, 5, 0, 6, 3, 2, 6, 2, 3, 9, 5, 7, 8, 3, 1, 8, 0, 1, 6, 9, 8, 4, 8, 0, 1, 8, 6, 9, 4, 7, 8, 8, 5, 1, 8, 4, 3, 8, 5, 8, 6, 1, 5, 6, 0, 7, 8, 9, 1, 1, 2, 9, 4, 9, 4, 9, 5, 4, 5, 9, 5, 0, 1, 7, 3, 7, 9, 5, 8, 3, 3, 1, 9, 5, 2, 8, 5, 3, 2, 0, 8, 8, 0, 5, 5, 1, 1, 1, 2, 5, 4, 0, 6, 9, 8, 7, 4, 7, 1, 5, 8, 5, 2, 3, 8, 6, 3, 0, 5, 0, 7, 1, 5, 6, 9, 3, 2, 9, 0, 9, 6, 3, 2, 9, 5, 2, 2, 7, 4, 4, 3, 0, 4, 3, 5, 5, 7, 6, 6, 8, 9, 6, 6, 4, 8, 9, 5, 0, 4, 4, 5, 2, 4, 4, 5, 2, 3, 1, 6, 1, 7, 3, 1, 8, 5, 6, 4, 0, 3, 0, 9, 8, 7, 1, 1, 1, 2, 1, 7, 2, 2, 3, 8, 3, 1, 1, 3, 6, 2, 2, 2, 9, 8, 9, 3, 4, 2, 3, 3, 8, 0, 3, 0, 8, 1, 3, 5, 3, 3, 6, 2, 7, 6, 6, 1, 4, 2, 8, 2, 8, 0, 6, 4, 4, 4, 4, 8, 6, 6, 4, 5, 2, 3, 8, 7, 4, 9, 3, 0, 3, 5, 8, 9, 0, 7, 2, 9, 6, 2, 9, 0, 4, 9, 1, 5, 6, 0, 4, 4, 0, 7, 7, 2, 3, 9, 0, 7, 1, 3, 8, 1, 0, 5, 1, 5, 8, 5, 9, 3, 0, 7, 9, 6, 0, 8, 6, 6, 7, 0, 1, 7, 2, 4, 2, 7, 1, 2, 1, 8, 8, 3, 9, 9, 8, 7, 9, 7, 9, 0, 8, 7, 9, 2, 2, 7, 4, 9, 2, 1, 9, 0, 1, 6, 9, 9, 7, 2, 0, 8, 8, 8, 0, 9, 3, 7, 7, 6, 6, 5, 7, 2, 7, 3, 3, 3, 0, 0, 1, 0, 5, 3, 3, 6, 7, 8, 8, 1, 2, 2, 0, 2, 3, 5, 4, 2, 1, 8, 0, 9, 7, 5, 1, 2, 5, 4, 5, 4, 0, 5, 9, 4, 7, 5, 2, 2, 4, 3, 5, 2, 5, 8, 4, 9, 0, 7, 7, 1, 1, 6, 7, 0, 5, 5, 6, 0, 1, 3, 6, 0, 4, 8, 3, 9, 5, 8, 6, 4, 4, 6, 7, 0, 6, 3, 2, 4, 4, 1, 5, 7, 2, 2, 1, 5, 5, 3, 9, 7, 5, 3, 6, 9, 7, 8, 1, 7, 9, 7, 7, 8, 4, 6, 1, 7, 4, 0, 6, 4, 9, 5, 5, 1, 4, 9, 2, 9, 0, 8, 6, 2, 5, 6, 9, 3, 2, 1, 9, 7, 8, 4, 6, 8, 6, 2, 2, 4, 8, 2, 8, 3, 9, 7, 2, 2, 4, 1, 3, 7, 5, 6, 5, 7, 0, 5, 6, 0, 5, 7, 4, 9, 0, 2, 6, 1, 4, 0, 7, 9, 7, 2, 9, 6, 8, 6, 5, 2, 4, 1, 4, 5, 3, 5, 1, 0, 0, 4, 7, 4, 8, 2, 1, 6, 6, 3, 7, 0, 4, 8, 4, 4, 0, 3, 1, 9, 9, 8, 9, 0, 0, 0, 8, 8, 9, 5, 2, 4, 3, 4, 5, 0, 6, 5, 8, 5, 4, 1, 2, 2, 7, 5, 8, 8, 6, 6, 6, 8, 8, 1, 1, 6, 4, 2, 7, 1, 7, 1, 4, 7, 9, 9, 2, 4, 4, 4, 2, 9, 2, 8, 2, 3, 0, 8, 6, 3, 4, 6, 5, 6, 7, 4, 8, 1, 3, 9, 1, 9, 1, 2, 3, 1, 6, 2, 8, 2, 4, 5, 8, 6, 1, 7, 8, 6, 6, 4, 5, 8, 3, 5, 9, 1, 2, 4, 5, 6, 6, 5, 2, 9, 4, 7, 6, 5, 4, 5, 6, 8, 2, 8, 4, 8, 9, 1, 2, 8, 8, 3, 1, 4, 2, 6, 0, 7, 6, 9, 0, 0, 4, 2, 2, 4, 2, 1, 9, 0, 2, 2, 6, 7, 1, 0, 5, 5, 6, 2, 6, 3, 2, 1, 1, 1, 1, 1, 0, 9, 3, 7, 0, 5, 4, 4, 2, 1, 7, 5, 0, 6, 9, 4, 1, 6, 5, 8, 9, 6, 0, 4, 0, 8, 0, 7, 1, 9, 8, 4, 0, 3, 8, 5, 0, 9, 6, 2, 4, 5, 5, 4, 4, 4, 3, 6, 2, 9, 8, 1, 2, 3, 0, 9, 8, 7, 8, 7, 9, 9, 2, 7, 2, 4, 4, 2, 8, 4, 9, 0, 9, 1, 8, 8, 8, 4, 5, 8, 0, 1, 5, 6, 1, 6, 6, 0, 9, 7, 9, 1, 9, 1, 3, 3, 8, 7, 5, 4, 9, 9, 2, 0, 0, 5, 2, 4, 0, 6, 3, 6, 8, 9, 9, 1, 2, 5, 6, 0, 7, 1, 7, 6, 0, 6, 0, 5, 8, 8, 6, 1, 1, 6, 4, 6, 7, 1, 0, 9, 4, 0, 5, 0, 7, 7, 5, 4, 1, 0, 0, 2, 2, 5, 6, 9, 8, 3, 1, 5, 5, 2, 0, 0, 0, 5, 5, 9, 3, 5, 7, 2, 9, 7, 2, 5, 7, 1, 6, 3, 6, 2, 6, 9, 5, 6, 1, 8, 8, 2, 6, 7, 0, 4, 2, 8, 2, 5, 2, 4, 8, 3, 6, 0, 0, 8, 2, 3, 2, 5, 7, 5, 3, 0, 4, 2, 0, 7, 5, 2, 9, 6, 3, 4, 5, 0 };
static long product, max = 0;
public static void SolutionEight()
{
for (short i = 0; i < 987; i++)
{
product = (long)input[i]
* input[i + 1]
* input[i + 2]
* input[i + 3]
* input[i + 4]
* input[i + 5]
* input[i + 6]
* input[i + 7]
* input[i + 8]
* input[i + 9]
* input[i + 10]
* input[i + 11]
* input[i + 12];
max = product > max ? product : max;
}
Console.WriteLine(max);
}
}
It gives the correct answer and it's best run time out of 20,000 runs in the release build has been 0.0956389122543305 milliseconds on an Intel Core i5-5200U @2.2Ghz processor.
How can I speed it up further?
[BENCHMARK]
Here are the benchmarks of all the awesome solutions provided in the answers. The implementations were run on an Intel Core i5-5200U @2.2Ghz processor with 8GB RAM. The fastest time was calculated out of 20,000 runs and Console.WriteLine(...) was not called in any implementation.
NOTE: I've tried my best to run all implementation on same standards without introducing my own optimizations (string to int conversions for the input were removed wherever necessary)
JNS' Bitshift optimization
• x64 (Debug) - 0.241196671392629 milliseconds
• x64 (Release) - 0.131561820759616 milliseconds
• x64 (Debug) - 0.0797768487584903 milliseconds
• x64 (Release) - 0.070446223172702 milliseconds
Risky Martin's reciprocal multiplication (implemented by brian_o)
• x64 (Debug) - 0.0121298132615248 milliseconds
• x64 (Release) - 0.0181947198922873 milliseconds
Dennis_E's Queue optimization
• x64 (Debug) - 0.033590252108838 milliseconds
• x64 (Release) - 0.0284584080366544 milliseconds
Falco's Dividend-Factor multiplication
• x64 (Debug) - 0.0149290009372613 milliseconds
• x64 (Release) - 0.00793103174792009 milliseconds
Domi1819's Casting optimization
• x64 (Debug) - 0.0102636881443672 milliseconds
• x64 (Release) - 0.005598375351473 milliseconds
Brian_o's Meticulous Zero-skip
• x64 (Debug) - 0.00933062558578834 milliseconds
• x64 (Release) - 0.00419878151360475 milliseconds
Zonker.in.Geneva and Mathreadler's Logarithmic approach (implemented by brian_o)
• x64 (Debug) - 0.0121298132615248 milliseconds
• x64 (Release) - 0.00466531279289417 milliseconds
[UPDATE]
If you are looking for a larger dataset to test your algorithm, I've generated 1 million random numbers using George Marsaglia's CMWC (Complementary Multiply With Carry) Generator (source code here).
Here is the File
• Are you allowed to use parallel processing? – 410_Gone May 18 '16 at 9:20
• Isn't the point of project Euler that you figure this out for yourself? projecteuler.net/about – Dominic Cronin May 18 '16 at 11:59
• I've already completed the Euler challenge. Correct solution in under a second (I reached 0.9ms) But interested to see how far it can be optimized – Paras May 18 '16 at 12:09
• Remember to compile and benchmark as both x86 and x64. You may get surprising results. – brian_o May 18 '16 at 21:26
• This is not the first Project Euler question on Code Review. Actually if you check this tag project-euler you can see more of them. Additionally, the purpose of naming the question with Project Euler is so people are aware what they are looking and not accidently spoil themselves, because honestly if one wants to look for the solution they are already here on the internet. The "project euler: x" helps those who don't want to spoil themselves of the answer. You can read more about it here – Paras May 19 '16 at 12:04
Inspired by Zonker.in.Geneva and mathreadler's ideas.
This doesn't do any multiplying or dividing to find the correct index. Only at the very end, when it has found the best run, it calculates that runs product.
private static readonly byte[] logs = { 0, 0, 69, 109, 138, 160, 179, 194, 207, 219 };
public static long SolutionEightAlt19()
{
int bestScore = 0;
uint bestIndex = 0;
int runningScore = 0;
uint prevUsable = 0;
for (uint i = 0; i < 1000; ++i)
{
if (input[i] == 0)
{
prevUsable = 0;
runningScore = 0;
}
else
{
++prevUsable;
runningScore += logs[input[i]];
if (prevUsable > 13) runningScore -= logs[input[i - 13]];
if (prevUsable >= 13 && runningScore > bestScore)
{
bestScore = runningScore;
bestIndex = i - 12;
}
}
}
return (long)(input[bestIndex] * input[bestIndex + 1] * input[bestIndex + 2] * input[bestIndex + 3] * input[bestIndex + 4] * input[bestIndex + 5] * input[bestIndex + 6]) * (input[bestIndex + 7] * input[bestIndex + 8] * input[bestIndex + 9] * input[bestIndex + 10] * input[bestIndex + 11] * input[bestIndex + 12]);
}
• +1: This was my idea also, though you might be able to further improve it with if (input[i] == 0) { .. i = i+12; } – RBarryYoung May 19 '16 at 18:43
• Hmm, my i+12 doesn't work at all because it still has to load the pipline. duh. My bad, you got it right. – RBarryYoung May 20 '16 at 18:27
• This one clocks out at 0.0713792857312808 milliseconds on my machine! Interesting solution – Paras May 21 '16 at 1:37
• All of the answers are great but I'm picking this one because it's very inspiring and possibly scalable with minor effort! Thanks everyone – Paras May 26 '16 at 22:00
• How did you pick the scale for the logarithm approximations? According to the back of my envelope, 255 for log(9) is barely accurate enough for 13 digits, 219 is dangerously inaccurate, and 130 happens to show the lowest relative error of all 8-bit scale values. – greybeard Apr 6 at 17:42
Using your base algorithm with a timing unit I get an average of 130 ticks (0.013 milliseconds) per iteration. (Note: running on Debug)
class ProblemEight
{
static byte[] input = { 7, 3, 1, 6, 7, 1, 7, 6, 5, 3, 1, 3, 3, 0, 6, 2, 4, 9, 1, 9, 2, 2, 5, 1, 1, 9, 6, 7, 4, 4, 2, 6, 5, 7, 4, 7, 4, 2, 3, 5, 5, 3, 4, 9, 1, 9, 4, 9, 3, 4, 9, 6, 9, 8, 3, 5, 2, 0, 3, 1, 2, 7, 7, 4, 5, 0, 6, 3, 2, 6, 2, 3, 9, 5, 7, 8, 3, 1, 8, 0, 1, 6, 9, 8, 4, 8, 0, 1, 8, 6, 9, 4, 7, 8, 8, 5, 1, 8, 4, 3, 8, 5, 8, 6, 1, 5, 6, 0, 7, 8, 9, 1, 1, 2, 9, 4, 9, 4, 9, 5, 4, 5, 9, 5, 0, 1, 7, 3, 7, 9, 5, 8, 3, 3, 1, 9, 5, 2, 8, 5, 3, 2, 0, 8, 8, 0, 5, 5, 1, 1, 1, 2, 5, 4, 0, 6, 9, 8, 7, 4, 7, 1, 5, 8, 5, 2, 3, 8, 6, 3, 0, 5, 0, 7, 1, 5, 6, 9, 3, 2, 9, 0, 9, 6, 3, 2, 9, 5, 2, 2, 7, 4, 4, 3, 0, 4, 3, 5, 5, 7, 6, 6, 8, 9, 6, 6, 4, 8, 9, 5, 0, 4, 4, 5, 2, 4, 4, 5, 2, 3, 1, 6, 1, 7, 3, 1, 8, 5, 6, 4, 0, 3, 0, 9, 8, 7, 1, 1, 1, 2, 1, 7, 2, 2, 3, 8, 3, 1, 1, 3, 6, 2, 2, 2, 9, 8, 9, 3, 4, 2, 3, 3, 8, 0, 3, 0, 8, 1, 3, 5, 3, 3, 6, 2, 7, 6, 6, 1, 4, 2, 8, 2, 8, 0, 6, 4, 4, 4, 4, 8, 6, 6, 4, 5, 2, 3, 8, 7, 4, 9, 3, 0, 3, 5, 8, 9, 0, 7, 2, 9, 6, 2, 9, 0, 4, 9, 1, 5, 6, 0, 4, 4, 0, 7, 7, 2, 3, 9, 0, 7, 1, 3, 8, 1, 0, 5, 1, 5, 8, 5, 9, 3, 0, 7, 9, 6, 0, 8, 6, 6, 7, 0, 1, 7, 2, 4, 2, 7, 1, 2, 1, 8, 8, 3, 9, 9, 8, 7, 9, 7, 9, 0, 8, 7, 9, 2, 2, 7, 4, 9, 2, 1, 9, 0, 1, 6, 9, 9, 7, 2, 0, 8, 8, 8, 0, 9, 3, 7, 7, 6, 6, 5, 7, 2, 7, 3, 3, 3, 0, 0, 1, 0, 5, 3, 3, 6, 7, 8, 8, 1, 2, 2, 0, 2, 3, 5, 4, 2, 1, 8, 0, 9, 7, 5, 1, 2, 5, 4, 5, 4, 0, 5, 9, 4, 7, 5, 2, 2, 4, 3, 5, 2, 5, 8, 4, 9, 0, 7, 7, 1, 1, 6, 7, 0, 5, 5, 6, 0, 1, 3, 6, 0, 4, 8, 3, 9, 5, 8, 6, 4, 4, 6, 7, 0, 6, 3, 2, 4, 4, 1, 5, 7, 2, 2, 1, 5, 5, 3, 9, 7, 5, 3, 6, 9, 7, 8, 1, 7, 9, 7, 7, 8, 4, 6, 1, 7, 4, 0, 6, 4, 9, 5, 5, 1, 4, 9, 2, 9, 0, 8, 6, 2, 5, 6, 9, 3, 2, 1, 9, 7, 8, 4, 6, 8, 6, 2, 2, 4, 8, 2, 8, 3, 9, 7, 2, 2, 4, 1, 3, 7, 5, 6, 5, 7, 0, 5, 6, 0, 5, 7, 4, 9, 0, 2, 6, 1, 4, 0, 7, 9, 7, 2, 9, 6, 8, 6, 5, 2, 4, 1, 4, 5, 3, 5, 1, 0, 0, 4, 7, 4, 8, 2, 1, 6, 6, 3, 7, 0, 4, 8, 4, 4, 0, 3, 1, 9, 9, 8, 9, 0, 0, 0, 8, 8, 9, 5, 2, 4, 3, 4, 5, 0, 6, 5, 8, 5, 4, 1, 2, 2, 7, 5, 8, 8, 6, 6, 6, 8, 8, 1, 1, 6, 4, 2, 7, 1, 7, 1, 4, 7, 9, 9, 2, 4, 4, 4, 2, 9, 2, 8, 2, 3, 0, 8, 6, 3, 4, 6, 5, 6, 7, 4, 8, 1, 3, 9, 1, 9, 1, 2, 3, 1, 6, 2, 8, 2, 4, 5, 8, 6, 1, 7, 8, 6, 6, 4, 5, 8, 3, 5, 9, 1, 2, 4, 5, 6, 6, 5, 2, 9, 4, 7, 6, 5, 4, 5, 6, 8, 2, 8, 4, 8, 9, 1, 2, 8, 8, 3, 1, 4, 2, 6, 0, 7, 6, 9, 0, 0, 4, 2, 2, 4, 2, 1, 9, 0, 2, 2, 6, 7, 1, 0, 5, 5, 6, 2, 6, 3, 2, 1, 1, 1, 1, 1, 0, 9, 3, 7, 0, 5, 4, 4, 2, 1, 7, 5, 0, 6, 9, 4, 1, 6, 5, 8, 9, 6, 0, 4, 0, 8, 0, 7, 1, 9, 8, 4, 0, 3, 8, 5, 0, 9, 6, 2, 4, 5, 5, 4, 4, 4, 3, 6, 2, 9, 8, 1, 2, 3, 0, 9, 8, 7, 8, 7, 9, 9, 2, 7, 2, 4, 4, 2, 8, 4, 9, 0, 9, 1, 8, 8, 8, 4, 5, 8, 0, 1, 5, 6, 1, 6, 6, 0, 9, 7, 9, 1, 9, 1, 3, 3, 8, 7, 5, 4, 9, 9, 2, 0, 0, 5, 2, 4, 0, 6, 3, 6, 8, 9, 9, 1, 2, 5, 6, 0, 7, 1, 7, 6, 0, 6, 0, 5, 8, 8, 6, 1, 1, 6, 4, 6, 7, 1, 0, 9, 4, 0, 5, 0, 7, 7, 5, 4, 1, 0, 0, 2, 2, 5, 6, 9, 8, 3, 1, 5, 5, 2, 0, 0, 0, 5, 5, 9, 3, 5, 7, 2, 9, 7, 2, 5, 7, 1, 6, 3, 6, 2, 6, 9, 5, 6, 1, 8, 8, 2, 6, 7, 0, 4, 2, 8, 2, 5, 2, 4, 8, 3, 6, 0, 0, 8, 2, 3, 2, 5, 7, 5, 3, 0, 4, 2, 0, 7, 5, 2, 9, 6, 3, 4, 5, 0 };
static long product, max = 0;
static long[] timings = new long[1000];
public static void SolutionEight()
{
Stopwatch watch = new Stopwatch();
for (int iterations = 0; iterations < timings.Length; iterations++)
{
watch.Restart();
for (short i = 0; i < 987; i++)
{
product = (long)input[i]
* input[i + 1]
* input[i + 2]
* input[i + 3]
* input[i + 4]
* input[i + 5]
* input[i + 6]
* input[i + 7]
* input[i + 8]
* input[i + 9]
* input[i + 10]
* input[i + 11]
* input[i + 12];
max = product > max ? product : max;
}
watch.Stop();
timings[iterations] = watch.ElapsedTicks;
}
Console.WriteLine(timings.Average());
}
}
Skipping zeros
if (input[i + 12] == 0)
i += 13;
Skipping zeros after setting the new max value gets it down to ~60 ticks (0.006 milliseconds).
This is covered in most of the other answers, but it's a very significant performance improvement that shouldn't be omitted. Check the other answers for better explanations.
Optimizing multiplication
product = (long)(input[i] * input[i + 1] * input[i + 2] * input[i + 3] * input[i + 4] * input[i + 5] * input[i + 6])
* (input[i + 7] * input[i + 8] * input[i + 9] * input[i + 10] * input[i + 11] * input[i + 12]);
By simply adding two pairs of braces, you can get it down to ~30 ticks (0.003 milliseconds) per operation. Magic, huh?
In .NET, multiplying two bytes will not return a byte. In fact, all integer multiplications in .NET will result in an int, as long as there are no longs involved. (If there is a long involved, both sides will be casted to long and a "long multiplication" is performed)
In your code, long multiplications are done (since 9^13 can overflow an int the idea is correct) however long multiplication is less performant than int multiplication. In your sample 12 long multiplications are performed.
The idea behind this optimization is to split the multiplication into smaller parts, specifically into blocks of 6 and 5 int multiplications (since 9^7 can not overflow ints) and then doing a final long multiplication to gain performance.
Other optimizations that I could think of did not provide any significant improvement.
• Wow, that was very astute pointer! * 0 * – Paras May 18 '16 at 12:55
• I've reached 0.0643813165419395 milliseconds just by applying these two simple refactors! Thanks – Paras May 18 '16 at 13:00
Implementing a shifting frame approach where you just divide by the leading digit and multiply by the last instead of multiplying all 13 digits over and over again is unfortunately not a good approach for 13 digits. Division is usually much slower on arithmetic units than multiplication (I remember a factor of 10-15 for some CPUs) so a single division might be slower than 13 multiplications. So I tried a different approach: I just multiply the dividends and factors and wait unti all the multiplied factors are bigger than the dividends and only then do a division and multiplication.
And I skip zeroes.
You could implement some shifting frame logic like this:
Let a product p_i be the product of all digits d_i...d_(i+1)
A product p_(i+1) can only be bigger than p_i if the digit d_(i+13) is bigger than d_i
Because: p_(i+1) = p_i / d_i * d_(i+13)
So you can skip the multiplication if the next digit is smaller or equal. This could make the algorithm twice as fast...
You can expand this, by multiplying the divisor_digits divs = d_i * d_(i+1)... and multiplying the new digits prods = d_(i+13) * d_(i+14) and only calculate the new maximum if prods > divs, then the new maximum is p_i / divs * prods
Results are the same as the original implementation and runtime is about 3-4 times faster in a first benchmark.
Java-Code:
private static long shiftingFrame( final int[] input )
{
final long cutoff = Long.MAX_VALUE / 10;
final int totalDigits = input.length;
int i = skipZeroes( input, 0, 13 );
long maxprod = i < totalDigits ? multiplyDigits( input, i ) : 0;
long lastprod = maxprod;
i = i + 12;
long diffactor = 1;
long mults = 1;
for ( ++i; i < totalDigits; ++i )
{
final int digit = input[ i ];
if ( digit == 0 )
{
i = skipZeroes( input, i, 13 );
if ( i >= totalDigits )
{
break;
}
lastprod = multiplyDigits( input, i );
i = i + 12;
mults = 1;
if ( lastprod >= maxprod )
{
maxprod = lastprod;
diffactor = 1;
}
else
{
diffactor = maxprod / lastprod;
}
}
else
{
final int spareDigit = input[ i - 13 ];
mults *= digit;
diffactor *= spareDigit;
if ( mults > diffactor || diffactor > cutoff )
{
lastprod = multiplyDigits( input, i - 12 );
mults = 1;
if ( lastprod > maxprod )
{
maxprod = lastprod;
diffactor = 1;
}
else
{
diffactor = maxprod / lastprod;
}
}
}
}
return maxprod;
}
private static long multiplyDigits( final int[] input, final int start )
{
return (long) ( input[ start ] * input[ start + 1 ] * input[ start + 2 ] * input[ start + 3 ]
* input[ start + 4 ] * input[ start + 5 ] * input[ start + 6 ] )
* ( input[ start + 7 ] * input[ start + 8 ] * input[ start + 9 ] * input[ start + 10 ]
* input[ start + 11 ] * input[ start + 12 ] );
}
private static int skipZeroes( final int[] input, int start, final int digits )
{
final int length = input.length;
int end = start + digits;
for ( int i = start; i < end; ++i )
{
if ( i >= length )
{
return length;
}
if ( input[ i ] == 0 )
{
start = i + 1;
end = start + digits;
}
}
return start;
}
• Over the thumb Benchmark timings in Java: Orginal: 14,5 ms, ZerroSkipping: 11,5 ms and ShiftFrame: 4,3 ms (Average runtime over a loop with 100.000, and warmup: 10.000 rounds) - Benchmarking code: pastebin.com/R4uvgRNP – Falco May 18 '16 at 11:27
• I converted the code to C# (had to remove all the finals) and the algorithm's fastest clocked out at 0.0779107236413326 milliseconds! It's awesome! – Paras May 18 '16 at 12:35
• I have made some more benchmarks and robustness-tests, I have changed the code a little and added a check for long-overflow, which could occur with high numbers. It is now safer and still as fast, if you want to use the code, please update these changes :-) - I am also trying around with other data-types, sometimes floating point arithmetics are faster than long calculations... – Falco May 18 '16 at 14:09
• Great! Keep us updated. Try if you can simplify the nested structures, maybe use a more lightweight data type (memory access is faster) – Paras May 18 '16 at 14:37
One obvious optimisation would be to skip sections that have a 0 in them. In the code below, I'm checking if the 13th digit is a 0 and if so, skipping the next 13 numbers, their product is always going to be 0...
int i = 0;
do
{
product = (long)input[i]
* input[i + 1]
* input[i + 2]
* input[i + 3]
* input[i + 4]
* input[i + 5]
* input[i + 6]
* input[i + 7]
* input[i + 8]
* input[i + 9]
* input[i + 10]
* input[i + 11]
* input[i + 12];
max = product > max ? product : max;
if (0 != input[i + 12]) i++;
else i += 13;
}
while (i < 987);
On a modern cpu, you can gain some benefit from parallel execution. Since the number of executions required is quite small, you need to avoid starting/ending threads however as this can have quite an overhead so you want to use threads from the thread pool. You also want to minimise contention between the threads, so you don't want them all updating the same Max variable. A basic strategy is to split the processing into a number of sections (where the number of sections is the number of CPUs).
Each window, calculates the max for that window, then compares it with the max for the other windows. The calculation stage can then but opitimised using the various other techniques suggested.
This results in code something like the following (I haven't really validated the upper bounds checking however it does result in the correct answer so I've assumed it is close enough).
public static void SolutionEight()
{
int windowSize = 1000 / threads;
for (int i = 0; i < threads; i++)
{
int windowStart = i;
tasks[windowStart] = Task<long>.Run(() => { return GetMax(windowStart * windowSize, (windowStart + 1) * windowSize - ((windowStart == threads - 1) ? 12 : 0)); });
}
long max = 0;
{
max = max > result ? max : result;
}
Console.WriteLine(max);
}
static public long GetMax(int start, int end)
{
long max = 0;
for(int i = start; i < end - 12; i++)
{
if (HasZero(i)) continue;
// Optimised as suggested by @Falco
if (i > 0 && input[i - 1] > input[i + 12]) continue;
// Optimised as suggestd by @domi1819
long product = (long)(input[i]
* input[i + 1]
* input[i + 2]
* input[i + 3]
* input[i + 4]
* input[i + 5]
* input[i + 6])
*( input[i + 7]
* input[i + 8]
* input[i + 9]
* input[i + 10]
* input[i + 11]
* input[i + 12]);
max = product > max ? product : max;
}
return max;
}
static bool HasZero(int i)
{
return !(input[i] != 0 &&
input[i + 1] != 0 &&
input[i + 2] != 0 &&
input[i + 3] != 0 &&
input[i + 4] != 0 &&
input[i + 5] != 0 &&
input[i + 6] != 0 &&
input[i + 7] != 0 &&
input[i + 8] != 0 &&
input[i + 9] != 0 &&
input[i + 10] != 0 &&
input[i + 11] != 0 &&
input[i + 12] != 0);
}
• What if the number at index i+13 is also 0? Is that a risk we must take or is there a way around that as well? – Paras May 18 '16 at 8:23
• @ParasDPain When the 13th digit is 0, you could go into an if that scans until it gets 13 digits that don't have 0. Whether or not it's worth while would depend on the number of 0s within 13 digits of each other. There's the same risk at the start of the set of numbers, although obviously you can visually check that this doesn't exist for the specific problem. – forsvarir May 18 '16 at 8:34
• That's a very extreme solution! I love it! Was waiting for a thread based implementation. – Paras May 18 '16 at 13:13
• The fastest run of your implementation clocked at 0.159553697516981 milliseconds. I tried removing the for and linearize the code but only got upto 0.14695... milliseconds – Paras May 18 '16 at 14:14
• @ParasDPain interesting, I get about a 50% boost over your original solution, but I'm running on a fairly old i7 laptop. It may be that your CPU is running through the code to quickly, so the cost of threading is higher than the cost of execution. – forsvarir May 18 '16 at 14:25
I implemented Risky Martin's proposal of using reciprocal multiplication, and it seems to be my best performer so far. Who knew?
I was kind of surprised because I thought that performing 1.0 / input[i - 13] (a division) would be just as expensive as dividing, but maybe there's an optimization because one is a long and the other is a byte? Voodoo? Or maybe I'm just wrong?
When generalizing, be careful for precision errors, but it seems fine for this particular input.
public static long SolutionEightAlt8()
{
double best = 0;
double prevProduct = 1;
uint prevUsable = 0;
for (uint i = 0; i < 1000; ++i)
{
if (input[i] == 0)
{
prevUsable = 0;
prevProduct = 1;
}
else
{
++prevUsable;
prevProduct *= input[i];
if (prevUsable > 13)
{
double recip = 1.0 / input[i - 13];
prevProduct *= recip;
}
best = prevProduct > best ? prevProduct : best;
}
}
return (long)best;
}
• Reciprocal is not just a division. There are faster ways to compute them than dividing the number 1.0 by them and we need to cast the "input[i]" to double although the compiler probably does that for you if you have not switched on your pedantic flags. But good contribution anyway. Compile with "-Ofast" should do such numerical short cuts automatically but can also destroy precision. – mathreadler May 18 '16 at 17:29
• Really interesting idea, however this is still (a tiny tiny bit) slower than my implementation on my current hardware (not the same I used for the benchmarks of my answer) Benchmark Code. – Domi May 18 '16 at 17:40
• I was thinking you could store the reciprocals in an array, so there would be no division in the loop. So you could write something like prevProduct *= reciprocals[input[i - 13]]; It's interesting that calculating the reciprocal in the loop is faster than just dividing in the loop. – Risky Martin May 19 '16 at 0:40
• @RiskyMartin I actually tried that in a subsequent experiment and was VERY surprised to learn that it didn't make a difference! Since there are only 10 possible digits, I baked a precomputed "here's the reciprocal for any digit" right into the class as a static readonly array. But no change! Maybe it gets cached in some way? – brian_o May 19 '16 at 1:07
• Since division is exact (dividing two integers to produce an integer), you should use 2adic division rather than real division. I assume long does arithmetic modulo 2^64? The presence of even divisors makes things a bit more complicated than usual; what you can do is make your table satisfy inverse[y] * y == 8 (mod 2^64). Then, to divide x by y, you can do your arithmetic with long and compute (x * inverse[y]) >> 3. – user14393 May 19 '16 at 17:32
Skipping 0s is good, but really you want to make sure none of your 13 consecutive digits is 0.
In your example data, starting at index 157, you have the following string of digits (874715852386305071569329096329522744304355766896648) which I've split up into runs of 0s and non-0s:
874 715 852 386 3 (13 digits)
0
5
0
715 693 29
0
963 295 227 443 (12 digits)
0
435 576 689 664 8 (13 digits)
By only skipping the first 0 (because digits[i+13] == 0), you're performing 24 unnecessary calculations, all of which equal zero.
Also, once you confirm that none of the 13 consecutive digits is 0, then if digits[i+13] == 1, you can skip over it, because either the product is the same (if digits[i] == 1) or less (if digits[i] > 1).
• I'm expanding on the Skipping Zeroes section listed above, trying to optimize it further....Sorry if i wasn't clear. But, i did see (after I posted) that brian_o is already on that path – Zonker.in.Geneva May 18 '16 at 17:05
Here's one that does the unrolled multiplication and also does 0 sequence skipping. I made a similar function that did division and a single multiplication, but this one is faster.
public static long SolutionEightAlt3()
{
long best = 0;
uint prevUsable = 0;
for (uint i = 0; i < 1000; ++i)
{
if (input[i] == 0)
{
prevUsable = 0;
}
else
{
++prevUsable;
if (prevUsable >= 13)
{
long prevProduct = (long)(input[i - 12] *
input[i - 11] *
input[i - 10] *
input[i - 9] *
input[i - 8] *
input[i - 7]) *
(long)(input[i - 6] *
input[i - 5] *
input[i - 4] *
input[i - 3] *
input[i - 2] *
input[i - 1] *
input[i]);
best = prevProduct > best ? prevProduct : best;
}
}
}
return best;
}
EDIT (More reasoning):
One way to think about solution is to say that there are 988 substrings, and you need to find the best one.
Calculating the score for a substring at index i has a certain cost. If you can optimize that cost, great. (see domi1819's answer, or my original (late to the party!).
It's worth it to think about whether you can shortcut the cost of a given index i's cost by using i-1 to bootstrap the cost caluclation: divide by the non-shared "old" digit, multiply by the "new" digit. In my experiments, with my hardware, I found that calculating the score outright (non-bootstrap, 13 optimized number multiplication) outperformed the divide-and-multiply methodology. So I'm sticking with the 13-num-multiply method. But the important part is, I checked! If we were looking for 26-digit strings, the answer would almost certainly be different.
Your first method did 987 scorings. But as others have talked about, the presence of zeroes makes many of those calculations unnecessary.
So figuring out which scorings can be obviated is very important, BUT there's a tradeoff: zero-skipping is not free!
• no zero skipping: 987 scorings
• domi1819: (CHEAP zero skipping methodology!) 430 scorings for this input
• my solution: (more expensive zero skipping methodology) 263 scorings for this input
On my hardware, my method seems to perform better than domi1819's, and I believe it's because I'm doing fewer scorings. I've been playing around with the algorithm and trying to do cheaper zero-skipping. My best implementation so far is 330 scorings with even faster overall performance than my currently posted answer. I feel there's still some room for improvement.
You mentioned parallelization. Might be a good strategy, but remember, just like zero skipping it's not free! I have yet to run experiments, but sounds like fun. My gut tells me it wouldn't be worth it with input this small, but who knows! Measure!
Final Edit:
I think I've gone as far as I'm willing to go. This is my final, best-performing function (for this input, using my hardware, compiling to x86). It does some zero skipping (moderate cost), and ends up computing 330 scores. My other more sophisticated attempts to compute fewer scores haven't outperformed it, maybe because they spend too much time figuring out what not to compute, or maybe because they mess up cache lines. Who knows.
public static long SolutionEightAlt15()
{
long best = 0;
for (uint i = 0; i < 988; ++i)
{
long product = (long)(input[i ] * input[i + 1] * input[i + 2] * input[i + 3 ] * input[i + 4 ] * input[i + 5 ] * input[i + 6]) *
(input[i + 7] * input[i + 8] * input[i + 9] * input[i + 10] * input[i + 11] * input[i + 12]);
best = product > best ? product : best;
if (product == 0)
{
// skip forward. you might still be inside an invalid region, but
// you'll just end up skipping forward again. decent balance of cost
// and efficacy for the tested input
uint j;
for (j = 12; j >= 0; --j)
{
if (input[i + j] == 0) break;
}
i += j;
}
}
return best;
}
• Thanks and I like the iterative checker for consecutive 0s . Also, is it necessary to explicitly cast the long twice, wouldn't once suffice? – Paras May 18 '16 at 13:17
• @ParasDPain I'm pretty sure it's unnecessary, but I don't think it costs anything at runtime. I also think it helps readability by making intention clear. (If I learned it actually DOES cost something at runtime, I'd remove it.) – brian_o May 18 '16 at 14:02
• Actually, the IL code of (long)int1 * int2 casts both ints to longs. When there's a long on one side, the compiler decides to cast other side to long as well to perform a long multiplication. – Domi May 18 '16 at 16:10
• Yes, and you can confirm it yourself by disassembling long l = (long)int1 * int2;, which results in ldloc.X (load variable X onto stack), conv.i8 (cast stack top to long), ldloc.Y (load variable Y onto stack), conv.i8 (cast stack top to long), mul (multiply two top stack elements and store result on stack) and stloc.Z (store stack top into variable Z and remove stack top). As you can see it's doing two casts although you only told it to cast the left side. – Domi May 18 '16 at 16:41
One optimization is to use bitshift for numbers that are multiple of 2. Not sure if that is significant, but in therory it should be faster ;). The code example below has also a few structual optimizations:
public class ProblemEight
{
static readonly long[] input = new long[] { 7, 3, 1, 6, 7, 1, 7, 6, 5, 3, 1, 3, 3, 0, 6, 2, 4, 9, 1, 9, 2, 2, 5, 1, 1, 9, 6, 7, 4, 4, 2, 6, 5, 7, 4, 7, 4, 2, 3, 5, 5, 3, 4, 9, 1, 9, 4, 9, 3, 4, 9, 6, 9, 8, 3, 5, 2, 0, 3, 1, 2, 7, 7, 4, 5, 0, 6, 3, 2, 6, 2, 3, 9, 5, 7, 8, 3, 1, 8, 0, 1, 6, 9, 8, 4, 8, 0, 1, 8, 6, 9, 4, 7, 8, 8, 5, 1, 8, 4, 3, 8, 5, 8, 6, 1, 5, 6, 0, 7, 8, 9, 1, 1, 2, 9, 4, 9, 4, 9, 5, 4, 5, 9, 5, 0, 1, 7, 3, 7, 9, 5, 8, 3, 3, 1, 9, 5, 2, 8, 5, 3, 2, 0, 8, 8, 0, 5, 5, 1, 1, 1, 2, 5, 4, 0, 6, 9, 8, 7, 4, 7, 1, 5, 8, 5, 2, 3, 8, 6, 3, 0, 5, 0, 7, 1, 5, 6, 9, 3, 2, 9, 0, 9, 6, 3, 2, 9, 5, 2, 2, 7, 4, 4, 3, 0, 4, 3, 5, 5, 7, 6, 6, 8, 9, 6, 6, 4, 8, 9, 5, 0, 4, 4, 5, 2, 4, 4, 5, 2, 3, 1, 6, 1, 7, 3, 1, 8, 5, 6, 4, 0, 3, 0, 9, 8, 7, 1, 1, 1, 2, 1, 7, 2, 2, 3, 8, 3, 1, 1, 3, 6, 2, 2, 2, 9, 8, 9, 3, 4, 2, 3, 3, 8, 0, 3, 0, 8, 1, 3, 5, 3, 3, 6, 2, 7, 6, 6, 1, 4, 2, 8, 2, 8, 0, 6, 4, 4, 4, 4, 8, 6, 6, 4, 5, 2, 3, 8, 7, 4, 9, 3, 0, 3, 5, 8, 9, 0, 7, 2, 9, 6, 2, 9, 0, 4, 9, 1, 5, 6, 0, 4, 4, 0, 7, 7, 2, 3, 9, 0, 7, 1, 3, 8, 1, 0, 5, 1, 5, 8, 5, 9, 3, 0, 7, 9, 6, 0, 8, 6, 6, 7, 0, 1, 7, 2, 4, 2, 7, 1, 2, 1, 8, 8, 3, 9, 9, 8, 7, 9, 7, 9, 0, 8, 7, 9, 2, 2, 7, 4, 9, 2, 1, 9, 0, 1, 6, 9, 9, 7, 2, 0, 8, 8, 8, 0, 9, 3, 7, 7, 6, 6, 5, 7, 2, 7, 3, 3, 3, 0, 0, 1, 0, 5, 3, 3, 6, 7, 8, 8, 1, 2, 2, 0, 2, 3, 5, 4, 2, 1, 8, 0, 9, 7, 5, 1, 2, 5, 4, 5, 4, 0, 5, 9, 4, 7, 5, 2, 2, 4, 3, 5, 2, 5, 8, 4, 9, 0, 7, 7, 1, 1, 6, 7, 0, 5, 5, 6, 0, 1, 3, 6, 0, 4, 8, 3, 9, 5, 8, 6, 4, 4, 6, 7, 0, 6, 3, 2, 4, 4, 1, 5, 7, 2, 2, 1, 5, 5, 3, 9, 7, 5, 3, 6, 9, 7, 8, 1, 7, 9, 7, 7, 8, 4, 6, 1, 7, 4, 0, 6, 4, 9, 5, 5, 1, 4, 9, 2, 9, 0, 8, 6, 2, 5, 6, 9, 3, 2, 1, 9, 7, 8, 4, 6, 8, 6, 2, 2, 4, 8, 2, 8, 3, 9, 7, 2, 2, 4, 1, 3, 7, 5, 6, 5, 7, 0, 5, 6, 0, 5, 7, 4, 9, 0, 2, 6, 1, 4, 0, 7, 9, 7, 2, 9, 6, 8, 6, 5, 2, 4, 1, 4, 5, 3, 5, 1, 0, 0, 4, 7, 4, 8, 2, 1, 6, 6, 3, 7, 0, 4, 8, 4, 4, 0, 3, 1, 9, 9, 8, 9, 0, 0, 0, 8, 8, 9, 5, 2, 4, 3, 4, 5, 0, 6, 5, 8, 5, 4, 1, 2, 2, 7, 5, 8, 8, 6, 6, 6, 8, 8, 1, 1, 6, 4, 2, 7, 1, 7, 1, 4, 7, 9, 9, 2, 4, 4, 4, 2, 9, 2, 8, 2, 3, 0, 8, 6, 3, 4, 6, 5, 6, 7, 4, 8, 1, 3, 9, 1, 9, 1, 2, 3, 1, 6, 2, 8, 2, 4, 5, 8, 6, 1, 7, 8, 6, 6, 4, 5, 8, 3, 5, 9, 1, 2, 4, 5, 6, 6, 5, 2, 9, 4, 7, 6, 5, 4, 5, 6, 8, 2, 8, 4, 8, 9, 1, 2, 8, 8, 3, 1, 4, 2, 6, 0, 7, 6, 9, 0, 0, 4, 2, 2, 4, 2, 1, 9, 0, 2, 2, 6, 7, 1, 0, 5, 5, 6, 2, 6, 3, 2, 1, 1, 1, 1, 1, 0, 9, 3, 7, 0, 5, 4, 4, 2, 1, 7, 5, 0, 6, 9, 4, 1, 6, 5, 8, 9, 6, 0, 4, 0, 8, 0, 7, 1, 9, 8, 4, 0, 3, 8, 5, 0, 9, 6, 2, 4, 5, 5, 4, 4, 4, 3, 6, 2, 9, 8, 1, 2, 3, 0, 9, 8, 7, 8, 7, 9, 9, 2, 7, 2, 4, 4, 2, 8, 4, 9, 0, 9, 1, 8, 8, 8, 4, 5, 8, 0, 1, 5, 6, 1, 6, 6, 0, 9, 7, 9, 1, 9, 1, 3, 3, 8, 7, 5, 4, 9, 9, 2, 0, 0, 5, 2, 4, 0, 6, 3, 6, 8, 9, 9, 1, 2, 5, 6, 0, 7, 1, 7, 6, 0, 6, 0, 5, 8, 8, 6, 1, 1, 6, 4, 6, 7, 1, 0, 9, 4, 0, 5, 0, 7, 7, 5, 4, 1, 0, 0, 2, 2, 5, 6, 9, 8, 3, 1, 5, 5, 2, 0, 0, 0, 5, 5, 9, 3, 5, 7, 2, 9, 7, 2, 5, 7, 1, 6, 3, 6, 2, 6, 9, 5, 6, 1, 8, 8, 2, 6, 7, 0, 4, 2, 8, 2, 5, 2, 4, 8, 3, 6, 0, 0, 8, 2, 3, 2, 5, 7, 5, 3, 0, 4, 2, 0, 7, 5, 2, 9, 6, 3, 4, 5, 0 };
public static void SolutionEight()
{
var max = 0L;
for (short i = 0; i < input.Length - ADJACENT_DIGITS_LENGTH; i++)
{
var product = 1L;
for (int j = 0; j < ADJACENT_DIGITS_LENGTH; j++)
{
var val = input[i + j];
switch (val)
{
case 1:
break;
case 2:
product <<= 1;
break;
case 4:
product <<= 2;
break;
case 8:
product <<= 3;
break;
default:
product *= val;
break;
}
}
max = product > max ? product : max;
}
Console.WriteLine(max);
}
}
Edit: Added the case 1 in code above.
Maybe it is also possible to let the compiler optimize the code. That requires the compile to know the concrete number at compile time:
switch(val)
{
case 2:
product *= 2;
break;
case 3:
product *= 3;
break;
//...
}
• I ran the solution with the same clocking mechanism, and out of 20,000 runs the fastest took 0.181480667643583 milliseconds . The bit shifts must've made it faster but the 2nd for loop slows it down as well. Can you elaborate on if var is faster and the use of Math.Max(...), etc. – Paras May 18 '16 at 6:47
• var doesn't have any effect of performance because it is actually a long (the compiler knows that it must be a long). I think Math.Max(...) has also no effect on performance... but i didn't checked it. – JanDotNet May 18 '16 at 6:52
• @ParasDPain: You are right... the nested for loop slows it down (I am fixed to optimize code for readablity that I didn't see it ^^). Therefore another optimization would be to remove the first for loop. That could be done by using T4 templates. – JanDotNet May 18 '16 at 6:55
• There is no difference between using var or the explicity type because it results in the same compilation. – JanDotNet May 18 '16 at 7:10
• @ParasDPain That's a different question (and one which has many answers round the network). – Philip Kendall May 18 '16 at 10:17
One optimization is to skip sequences that contain a 0. Another is if the previous sequence doesn't start with a 0, calculate the current sequence by dividing the previous product by first digit in the previous sequence and multiplying by the last digit in the current sequence.
I hope you had code to convert the 1000 digits from the string into an array of digits. I know you're going for speed, but a lot of times readability and maintainability is more important, and in those cases you'd want to calculate a product using loops or LINQ instead of coding an increment to the index 13 times. You should also calculate the currently hardcoded value of 987 using something like int count = input.Count() - sequenceLength;.
• Agreed, I'm currently trying to implement the skip for 0 and the frame-shifting to reduce multiplication operations. Agreed readability is important but I'm competing with friends so ready to sacrifice it. And yes of course I wrote a function to convert the string to a byte array – Paras May 18 '16 at 8:36
• I'm not sure if a shifting frame will be a benefit, in many Arithmetic Units division is about 10-15 times slower than multiplication, which would make the single division maybe more expensive than 13 multiplications... but you can test it... – Falco May 18 '16 at 9:08
• I'll test and post results – Paras May 18 '16 at 9:30
• @ParasDPain Instead of dividing, you could also try storing the reciprocals of the digits in an array and multiplying them or using a corresponding multiply shift operation. – Risky Martin May 18 '16 at 10:39
You are basically doing all multiplications 13 times.
When you have the product of 13 numbers, a1 * a2 * ... * a13, all you have to do to find the next product is divide by a1 and multiply by a14. You don't have to multiply a2 through a13 all over again.
The obvious way to handle this is by using a Queue.
The 2nd optimization you can make is to start a new sequence of 13 when you encounter a 0. (If you don't check for 0, you'll get a DivideByZeroException)
First, let's make an extension method to get all the digits from a number-string:
public static IEnumerable<int> Digits(this string s) {
foreach (char c in s) yield return c - '0';
}
Then solve problem 8:
public long SolveProblem008() {
private const string input = "731..."; //I'm not going to display the entire string here...
return GreatestConsecutiveProduct(13, input.Digits());
}
static long GreatestConsecutiveProduct(int length, IEnumerable<int> digits) {
var buffer = new Queue<int>(length);
long product = 1L, max = long.MinValue;
foreach (int input in digits) {
if (buffer.Count < length) { //We don't have 13 digits yet
if (input == 0) { //Encountered a 0: start from scratch
product = 1L;
buffer.Clear();
} else {
buffer.Enqueue(input);
product *= input;
}
} else {
if (input == 0) { //Encountered a 0: start from scratch
product = 1L;
buffer.Clear();
} else {
product /= buffer.Dequeue();
buffer.Enqueue(input);
product *= input;
if (product > max) max = product;
}
}
}
return max;
}
• I ran it and am getting the wrong answer - 2090188800 (possibly an overflow, the max product exceeds int.MaxValue). Could you please confirm that – Paras May 18 '16 at 10:18
• And it was, I changed the type of the method, product and Queue to long and got the correct answer but still the fastest run of 20,000 runs only reached 0.122231195173827, tells me the implementation can be improved – Paras May 18 '16 at 10:21
• @ParasDPain Yes, I forgot. The problem used to be to find a sequence of 5 consecutive numbers (if I remember correctly) which fits well within an int. They changed it to 13 later. I edited it to return a long – Dennis_E May 18 '16 at 10:24
• This optimization will help for very large n (>13), but on most arithmetic unit division is much slower (factor 10-15) than multiplication. So a single division will probably be slower than 13 multiplications. And this is without taking pipelining, branch prediction and L1/L2 caches into account... – Falco May 18 '16 at 10:47
• Have you actually tested this and found that it improves performance? This seems.. most unlikely really. As Falco says the optimisation itself is more than questionable for the number of digits we're talking about, but the real problem is the horribly unpredictable branch you have in there. And that's just the two major things - the overhead from the queue probably won't matter too much, but is far from free anyhow. – Voo May 18 '16 at 13:24
Seems someone had already tried my first instinct to make a "moving frame" that divides an old number in the tail and multiplies a new in the front (1mul and 1 div instead of 12 mul). However divs are more slow than muls and also screw up the pipelines so maybe that is why it did not work very well.
Another popular approach in engineering is to do dyadic subdivision, calculate pairs of products, then pairs of those and so on.
So first level would be the original numbers n1,n2,n3,.... Second level would be n1*n2,n3*n4,... third level n1*n2*n3*n4,.... and so on. We can stop at 8 since that will be the power of two which is just below 13.
So the first product will be "first eight * third four * 13th number". In best case two multiplications instead of 12. I can leave the implementation as an exercise, after all the project is aimed to help people get better at programming.
I just realized I have no way to test this. Even current state of my unoptimized code with optimization flags takes only 7-17 µs to run, which is far too little to do any reliable testing. Maybe if we can come up with some mockup data which is a lot longer so it forces every algorithm to take longer.
• I'll post 1 million random number sequence in the question to benchmark algorithm's efficiency with data size – Paras May 21 '16 at 3:13
• Sounds great. Maybe post a link to a file instead of all the number in the question text. – mathreadler May 21 '16 at 3:29
• I've updated the question with a link to the file (sorry about the delay) – Paras May 21 '16 at 14:40
I had another thought which is a variation on the shifting frame...
The problem with the shifting frame idea is that division is much slower than multiplication.
But...the 13 consecutive digits that determine the maximum product are also the same consecutive digits that determine the maximum SUM....
So, if we employ the shifting frame method but subtract the first element and add the next element, we avoid divisions. We only keep track of the index which will give us the greatest sum. At the end of our loop, we do one single multiplication based on 13 consecutive digits starting at the GreatestSumIndex.
Naturally, we still want to avoid any sequences that have any zeros.
As it's approaching 02:00am, I'll work on the code in the morning. Unless someone else wants a stab at it....
• "the 13 consecutive digits that determine the maximum product are also the same consecutive digits that determine the maximum SUM" Unfortunately, nope. 5555555555555 has a smaller digit sum than 9191919191916, but its digit product is way bigger. – brian_o May 19 '16 at 2:29
• It will work with a log law though $\log(ab) = log(a)+log(b)$, but sadly most ways of calculating logs is rather slow. – mathreadler May 19 '16 at 15:56
• mathreadler You only need to calculate the log of each digit once. – user1008646 May 19 '16 at 17:03
• Yes I agree, but that calculation can still be slow. I read in other stackexchange and overflow sites that c implementations of log can take 35 cycles on modern intel cpus where a multiplication would only take a few or less than one in average if there is a good pipeline. But maybe it's worth it if it's only once per digit. – mathreadler May 21 '16 at 4:19
• @mathreadler You can make this work without doing any log calculations at runtime. Throw the log representation of digits 1-9 into a static array. – brian_o May 22 '16 at 13:31 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.18072983622550964, "perplexity": 219.79853997575853}, "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/1573496668772.53/warc/CC-MAIN-20191116231644-20191117015644-00003.warc.gz"} |
https://www.physicsforums.com/threads/find-the-temperature-coefficient-of-metal-x.320653/ | # Find the temperature coefficient of metal X
1. Jun 18, 2009
### icystrike
1. The problem statement, all variables and given/known data
Metal X
1) since the graph does not show that V is linearly related with I , is it right for me to apply the formula
R(50*C)=R(20*C) (1+ a lΔTl )
to find the temperature coefficient of metal X ?
2) R/Ω is linearly related with T/*C ?
If yes, will R=(dR/dT)T+c ?
Whereby dR/dT and c are constants that a fixed for the similar metal
2. Relevant equations
3. The attempt at a solution
Last edited: Jun 18, 2009
2. Jun 18, 2009
### icystrike
Re: Resistance
please can someone answer to my question..
3. Jun 18, 2009
### LowlyPion
4. Jun 18, 2009
### icystrike
Re: Resistance
the graph is not part of the measure lab assignment.
I ask this question simply because of curiosity.
Thank you for answering my question , your help is greatly appreciated.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook
Similar Discussions: Find the temperature coefficient of metal X | {"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": 0.8736603856086731, "perplexity": 4552.2706693596265}, "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-2017-43/segments/1508187823997.21/warc/CC-MAIN-20171020082720-20171020102720-00179.warc.gz"} |
https://mathematica.stackexchange.com/questions/103989/resolve-does-not-resolve-this-statement-with-quantifier | # Resolve does not resolve this statement with quantifier
I'm trying to use Resolve to get a condition on a parameter for which the statement is true, but it doesn't look like it's working:
Resolve[Exists[a, a > 0, Element[Sqrt[100 + a^4]/a, Rationals]], Rationals]
The statement does in fact have parameter values that make it true. For example, 3/2 or 20/3 are some instances of a that work. How can I have Mathematica solve this problem?
I do not know how to get Resolve to find your answers; however, Reduce can find solutions. Rewrite your equation with $a=u/v$, where $u$ and $v$ are coprime integers greater than zero. Hence, $\sqrt{u^4+100*v^4}/(u*v)=c/d$, where $c$ and $d$ are coprime integers greater than zero. If the square root returns an irrational, then there is no solution. If the square root returns an integer $m$, then we have a solution.
Block[{r, u, v, m},
Flatten[
ParallelTable[
r = {ToRules[
Reduce[{m^2 == u^4 + 100*v^4, u>0, m>0, GCD[u,v]==1}, Integers]
]};
If[r =!= {}, {u, v, m} /. r, {}],
{v, 1, 1000}],
1]]
The solutions come in pairs. If $u/v$ is a solution, then so is $10*v/u$, both giving the same result $c/d$. Up to $v=5000$ there are 4 solutions $\{u,v\}$ of {3,2}, {20,3}, {1519,492}, and {4920,1519}.
If using Reduce is a cheat, then consider writing $m^2-x^2=100*y^2$, where $x=u^2$ and $y=v^2$. Assume $y=1,4,9,...$ is given, find the divisors pairs $p*q=100*y^2$, and $\{m,x\}$ via $p=m-x$ and $q=m+x$. Save solutions with square $x$.
Short answer is: I don't believe you can make Mathematica solve this because math does not know how to solve this.
The longer version might not be 100% correct, since I am not a mathematician, but it goes somewhere along this:
The problem you're trying to solve here boils down to the question of "Is a specific rational number a square of another rational number?". And there is no constructive solution to this problem. The only known way to list all numbers that are squares of other numbers is to actually calculate them one by one from a list of all rationals.
So that's what you can do with your problem. You can write an algorithm that will get you first 10, 100, 1000 or however much of rationals that solve it, but there still will be infinitely more of those to go, and there is no simpler way to get them all.
• Thinks for your advices. – yode Jan 16 '16 at 12:27 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 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": 0.6603620648384094, "perplexity": 283.9268292651532}, "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-2019-47/segments/1573496670821.55/warc/CC-MAIN-20191121125509-20191121153509-00338.warc.gz"} |
http://the-biologist-is-in.blogspot.com/2014_02_01_archive.html | // Twitter Cards // Prexisting Head The Biologist Is In: February 2014
## Wednesday, February 26, 2014
### The Trouble with Seeds (1/3)
For someone interested in breeding plants, that some plants often don't produce seeds can be a major barrier. A breeder would be limited to looking for selectable mutations, rather than using the more general mixing and segregation of genetics to increase crop diversity.
Many of our common vegetables, fruits, and landscape plants are traditionally propagated by clonal divisions. In some cases (apples, pears, etc.) the complex genetic diversity of the crop means that every seedling would produce a distinct plant. The market demands for consistent production then encourage growers to clone the plants. In other cases, the clonal tradition is enforced by the plants themselves due to their inability to produce seeds. (I'll later discuss a third category of plants which have such long life-cycles that breeding projects become difficult to undertake.)
Researchers have figured out tricks to get seed and allow breeding to be done with plants that might otherwise prefer not to.
The Trouble with Seeds (1/3): Garlic, Horseradish, Potato Onion, Walking Onion, and Banana.
The Trouble with Seeds (2/3): Pineapple, Lily of the Valley, Potato, and Sweet Potato.
The Trouble with Seeds (3/3): Babington's Leek, Crosnes, and Bur Oak.
Most of this post is an accumulation of information from other sources (in colored quotes below) about how to get seeds from these crops. I've also included my thoughts and experiences where I felt something needed to be clarified or extended.
How to get garlic (Allium sativum) to set seed; notes on growing.
[1] "Most of our seed production is done with severed scapes kept in water rather than by leaving the whole plant in the ground."
"Garlic cultivars having flowers with purple anthers are much more likely to be male fertile and to produce seed than those with yellow anthers."
"Garlic umbels have both bulbils for asexual reproduction and flowers for sexual reproduction. Bulbils and flowers compete for the plant's resources. With certain exceptions, if the plant is left to its own the bulbils win and the flowers wither and die before they can produce seed. The bulbils must be removed from the umbel in order to tilt the balance toward seed production."
"Interestingly, in subsequent generations of seed-produced plants the bulbils are often far fewer and may not require removal for successful seed production."
"When the scape becomes nearly straight, the spathe (bract or leaf covering the umbel) should be slit open to examine the umbel's development. As soon as the bulbils have developed they should be removed. Bulbil removal is rather tedious and time consuming, but not particularly difficult after a bit of practice."
"Bulbil removal is a combination of plucking them out with tweezers and rocking them out to dislodge them."
"Because bulbils as well as flowers are usually still developing at the time bulbil removal begins, it is usually necessary to return a week or so after the initial procedure to remove any bulbils that subsequently developed."
"Initial seed yields are typically quite low, but subsequent generations of seed-produced plants yield significantly more seeds, sometimes more than 600 per umbel. Early efforts may be difficult, but later efforts can be exceptionally rewarding."
"Seed from garlic plants that have previously been propagated only by asexual means (cloves, bulbils) have a low germination rate ranging from 10% to an optimistic 35% at best."
"However, studies have shown that subsequent generations of seed-produced plants typically have a much higher germination rate, sometimes as high as 100%."
"First generation seedlings often exhibit a high frequency of unfavorable characteristics, such as stunted growth, deformed leaves, limited root development, and chlorophyll deficiencies. Subsequent generations of seed-produced garlic exhibit increasing vigor and a lower frequency of genetically deficient plants."
"Garlic seed has a period of dormancy and should not be planted immediately after harvest."
"Garlic seeds should be given a bleach soak prior to planting to help protect them from contamination, followed by a cold treatment to shorten dormancy. Soak the garlic seeds in a 1% solution of household bleach (1 teaspoon bleach in 2 cups water) for 20 minutes, rinse the seeds, distribute the seeds on moist paper towels, place the seeds in a plastic sack, and store in a refrigerator for approximately four weeks."
"Like the seed, newly harvested garlic bulbs, including rounds, have a natural period of dormancy. Various things affect dormancy but temperature is a major factor. Warm temperatures lengthen dormancy and cold temperatures shorten dormancy. We do not have a definitive recommendation, but as a starting point we suggest waiting at least a month after harvesting the rounds before replanting them in the fall. We have not found this to be an issue for us, but if necessary temporary storage in the refrigerator could be employed to help break dormancy sooner."
"Once the rounds are planted in the fall the seed-produced garlic is on a normal cycle and can be grown out just as one would with the rest of the garlic crop. Large rounds from vigorous seedlings should yield fully-developed plants and divided bulbs at harvest the following year. You can use these fully-developed plants to produce second generation seeds, though doing so severely diminishes the bulb and thus inhibits or sacrifices replication of what is essentially a new cultivar. You can also opt not to produce second generation seed from these plants and instead harvest the bulbs and cloves at maximum size for planting in the fall. This delays the next generation of seed-produced garlic until the following year, when you can use some plants for producing seed, and some for continued replication of the new cultivar via cloves. There isn’t a right or wrong way in this regard, and you may want to try some both ways. If you have a particularly promising cultivar, however, you may want to asexually replicate and preserve it before using some of its plants for a second generation of seed production."
Garlic can be grown from grocer-bought bulbs, though it can be difficult to break their hormone-induced dormancy. If you want to grow your own garlic, look for bulbs grown by local organic farmers. They are less likely to have been prepared for long-storage and more likely to grow when given sunlight and moisture.
I haven't successfully grown garlic (my last attempt was overgrown by zucchini), but I really like the idea of growing them from seed and exploring the genetic diversity hidden within them.
How to get horseradish (Armoracia rusticana) to set seed.
[1] "Horseradish may be an interspecific hybrid and is generally reported to be sterile. However, viable seed has been produced (14.1-7)"
[2] "I found an article about getting seed from horseradish, but misplaced it. The trick was to girdle the step just above the storage portion of the root, then replant. By cutting off the flow of carbohydrates to the roots, more carbs are reserved to nourish the developing seeds. Similar tricks have been used to get self-seeds from self-incompatible Easter Lilies, Narcissus and Amaryllis (Hippeastrum)."
Although there is limited information available about seeds of this plant, I expect the first generations of seeds would (like garlic) have very low viability. Across successive generations of reproduction by seed, viability should increase as chromosomal abnormalities are filtered out by selection. After several generations, it might be possible to develop a reliably seed-propagated horseradish line.
I'm mostly interested in the potential to improve horseradish as a perennial crop for additional uses from its root. The leaves [3,4] and flowers [5] are edible, having a bitter-pungency when steamed. Shorter or softer leaves might be useful. Larger flower heads, like broccoli, would be interesting. There are lots of potential ways to take this plant that have not been explored.
There is likely to be a lot of interesting genetic diversity hidden in horseradish, as recessive mutations have been accumulating for as long as we have been propagating it clonally. The first few generations of plants produced from seed will likely have some fun surprises.
How to get potato onions (Allium cepa var. aggregatum) to set seed; notes on growing.
[1] "One fall, I left a few Potato Onions in the ground to see if they would survive the winter. They survived just fine. However, these over-wintered Potato Onions all produced flowers the following summer. I had never seen any of my Potato Onions flower before. The over-wintering had triggered their flowering response."
"I planted one row of seeds, and as a control, I planted another row of Potato Onion bulbs. The Potato Onion seeds sprouted and grew, but were not like the control row which had been planted with the original clones."
"I found that these larger onions from seed did exhibit the trait of a long storage life. They easily stored the whole winter and well into the following summer."
"The following spring, I again planted two rows of Potato Onions: one row was from the same bulbs I had grown out for a decade, and the experimental row was planted with some of the very best large bulbs which came from the seeds. Again, the original bulbs were completely uniform. But, the experimental row showed a multiplicity of different traits. Some bulbs grew enormous top growth. Others went immediately to seed. Some developed a nest of small bulbs, others a nest of large bulbs. Some nests had many bulbs, other nests had only three or four bulbs."
During the winter of 2012/2013 I purchased some red onions. One was left alone all winter and when I was planting my garden in the spring, I decided to plant it. Typical onions are considered biennial plants, where the plant spends the first year storing energy in the onion and then uses that energy to flower in the second year. I was looking forward to seeing the flowers and to collecting seeds that I could later grow.
The onion grew nice bluish-green foliage and made a nice presence in my garden… but it never flowered. When the garden season ended, I found the single onion had divided into three. The three onions were smaller than I could get at the local grocer, but they were larger than any others I had grown in Minnesota.
I set aside the new onions for storage during the winter of 2013/2014. Again, they held up to storage very well. Now that spring is beginning to beckon in the near future, the onions have started to grow small roots. I'll probably grow them in a planter this year, so I can more easily keep an eye on them and note the details of their growth.
This onion seems to be behaving like a potato onion (they multiply and store very well), even though nobody seems to claim they have dark-red potato onions, so I've decided to continue growing them and increase their numbers. Maybe I've accidentally found an onion that has the characteristics needed to be a nice dark-red potato onion.
I'm still hoping to get some seed from this onion. I might have to try leaving a few out in the cold next winter, or store them in a temperature-controlled fridge, to trick them into going to flower. Ideally, I would have several different types of onions in bloom at once so I could get some hybridization going on to work towards an onion well-suited to my garden conditions.
How to get walking onions (Allium cepa x fistulosum) to set seed.
[1] "In A. x proliferum, the parental chromosomes derived from A. fistulosum and A. cepa were unequivocally identified by GISH, proving the hybrid status of the crop."
The walking onion was formerly considered to be A. cepa var. proliferum, but is now known to be an interspecific hybrid between A. cepa (common onion) and A. fistulosum (Welsh onion).
[2] "Walking onions do not form true seed, even though they will get a few blossoms. They grow 'top sets' instead, a cluster of bulbils at the top of the stalk where the seeds would normally be."
This plant is propagated by planting the bulbils produced on the flower stalk. It gets the name "walking onion" because the plant will normally arch over the flower stalk and place the new formed bulbils onto the ground some few feet away, walking across a garden over several years.
Nobody seems to be talking about saving true seed for walking onions. It does produce flowers, but they are starved of nutrients by the adjacent bulbils and wither. There is the potential to generate viable seed by removing the bulbils early like can be done with garlic. Because nobody seems to be talking about doing this, it looks like something I will have to experiment with.
One useful aspect of walking onions is that they are very cold hardy. They have survived several winters in the garden of my parents' Minnesota house without any effort given to protect them. Transferring some of this cold-hardiness to an onion with larger bulbs would be a worthwhile project.
[3] "If I remember right, two growers on this forum have reported obtaining true seeds from Egyptian onions, and successfully growing offspring from them. I think that in both cases A. cepa was flowering nearby."
This suggests that this onion will set seed when a genetically compatible onion is nearby. If the onions are not self-compatible, the population consisting of clones would normally prevent any fertilization. As this type of onion is derived from an inter-species cross, it may accept pollen from more than one species.
How to get bananas (Musa spp.) to set seed.
Normal bananas already set seed without our intervention, but unless you live in the tropics you're unlikely to encounter a normal banana. The bananas common in the stores outside the tropics are the result of a peculiar quirk of chromosomes.
Most banana species can readily hybridize. Some species have two copies of each chromosome (diploid), while others have four copies of each chromosome (tetraploid). If a diploid and tetraploid banana happen to cross, the resulting progeny will have three copies of each chromosome (triploid). Such a triploid banana will grow and mature normally, but will encounter a severe difficulty when it tries to make gametes.
It turns out that our common biology has a very hard time making gametes when the number of each chromosome is not divisible by two. In the case of the banana, a triploid plant will succeed in making a fruit, but it will fail to make any seeds.
Plant breeders get around this quirk by determining how many copies of each chromosome are found in different species/varieties of banana [1,2], then using this knowledge to intentionally create new triploid types that can be screened for useful traits. Our beloved store-bought-banana is a genetic dead end, but this doesn't prevent the generation of new varieties.
I like the idea of a banana that can grow in Minnesota, though I know it is a somewhat unrealistic idea. There are some banana species which can survive in more northern climates [3] and these provide the potential for breeding one that can survive in my Minnesota yard.
Most of the lovely plant diagrams in this post were derived from public-domain images hosted at botanicalillustrations.org. Some other diagrams are public-domain images from the same era that I found via google. I chose the original images which depicted the plants under discussion in the way I appreciated and then subtracted out the yellowed background of the page using my favorite image editor (GIMP)
## Thursday, February 20, 2014
### Making My Own Carrots
In 2013 I grew a batch of mixed bed (4ft x 6ft) of several carrot varieties, including a range of different colored forms. I had never grown carrots in my current garden, so I didn't know which varieties would work well and which would utterly fail. My main hope was that several types would prosper and I would get lots of mixed carrots to eat over the fall and winter seasons.
I harvested all the carrots during the second snowfall, digging and cleaning every single plant. There were plenty of largish carrots along with lots of small carrots. The brand of carrot culture that I applied (sow thinly and let the plants fight it out, without intervention) was particularly difficult for some varieties. A short, dwarf-rooted type ("Paris Market") almost disappeared in the resulting jungle. The red ("Atomic Red") variety from the Burpee Kaleidoscope mix tasted great, but performed poorly overall. The yellow ("Solar Yellow") and white ("Lunar White") ones from the same mix, on the other hand, did very well. The orange ("Bambino") and purple-skinned orange ("Cosmic Purple") ones from the same mix, came somewhere in between.
I ended up with several pounds of carrots and have been eating them all winter.
A secondary hope was that I would be able to save some of the largest carrots, those that agreed most with my local conditions and gardening style, for seed growing. Over several years, this would let me develop a locally-adapted carrot variety of my very own.
Carrots are a biennial plant, producing flowers/seeds during their second year. Since I wasn't confident in my carrots ability to survive in the ground all through the harsh Minnesota winter, I set some aside in the fridge for growing in the spring. For these, I trimmed the greens short, but left the growth point intact. The roots were then trimmed to fit into a quart-sized ziplock bag in the back of my fridge.
At the beginning of February, I checked in on the stored roots. Several had started to grow new shoots, while several others had rotted to mush. The rot took most of the orange ("Bambino") carrots, so it appears that color will be under-represented in the genetics of my developing population.
I decided that it was time to force the remaining roots, in fear of losing more to rot and so they could get an early start on the growing season. (Admittedly, the desire to see some green growth this deep in winter was a major factor in this decision.) I trimmed all the roots to about three inches long and placed them upright in a wide-bottom glass cylinder. I had enough roots such that they would cross-brace each other and remain upright in this container.
I happened to have a deep purple carrot ("Purple Haze", "Purple Dragon", or something) in the fridge, among leftovers from a farmers' market foray before my own carrots were ready… and I like purple, so I trimmed it like the others and added it to the forcing container.
If I am lucky, the roots will bloom early enough to let me grow the resulting seed this year, having been tricked into living their two year life-cycle in one year. If they don't bloom early, I will have had a nice windowsill plant for most of the winter and will have the seeds for next year.
In either scenario, the resulting next generation plants will contain a mix of F1s from the saved varieties. There are very few wild carrots ("Queen Anne's Lace") around where I live, so I shouldn't have any problems with weedy genetics getting mixed into my carrot population. The next generation, a few years from now, should then show a riotous mix of traits as the various alleles segregate in the F2 progeny.
Part 2: Carrot flowers.
Part 3: Generation 2.
## Sunday, February 9, 2014
### Genetics of Squash Shape (2/2)
In my first blog post, I examined a small biology puzzle: the classical model of squash shape genetics (at left) didn't seem to match what I was seeing in my own garden.
The classical model involves two genes ('A' and 'B', respectively). The dominant allele of both genes must be present (A_B_) to produce a disk-shaped squash. If one gene is present in the dominant form and the other is in the recessive form (A_bb or aaB_), spherical-shaped squash will result. If only the recessive alleles of both genes are present (aabb), then elongated squash will result. After some digging, I found this model comes from a 1927 paper by Edmund W. Sinnott in the research journal, The American Naturalist.
This model predicts that crossing a (AABB) PattyPan to a (aabb) Zucchini should result in (AaBb) progeny plants with disc-shaped squash. The progeny plants I grew, instead each produced intermediate/elongated-fat squash. I assumed this meant my original PattyPan squash was a hybrid that contained both recessive alleles (AaBb) and so I then calculated the probabilities of producing (aabb) progeny from crossing the potential male parents (PattyPan and Zucchini) to the female parent (PattyPan). The best probability I calculated was $$p = \frac{1}{16}$$. I wasn't pleased with this result and decided I needed further data.
The magic of the internet then made its presence known: Ottawa Gardener was forwarded to my original post in a discussion of their recent post. They had grown Zucchini, PattyPan, and some Pumpkins, then tossed some of the PattyPan squash to their chickens. The next year when they moved the chicken run, up came a batch of hybrid squash. Most appeared to be intermediate between the Zucchini and the PattyPan, with a few looking intermediate between the PattyPan and the Pumpkin. (I've rearranged their photo to make the diagram at left.)
Because there are three potential male parents (and two offspring types), the calculations of probability get somewhat more intricate and I won't go into them here. The detail I found most interesting was the recreation of the intermediate/elongated-fat shaped squash at the bottom-left.
Upon digging into the research literature further, I found a 1910 paper by R.A. Emerson, again in The American Naturalist. In this paper, the author describes the result of crossing "White Scallop" (disc) and "Yellow Crookneck" (elongated) squash as being intermediate in shape. This is contrary to the Sinnott model of shape genetics and is the result that both I and Ottawa Gardener observed in our gardens.
It seems like the Emerson result was never followed up on and the Sinnott result erroneously became the standard model of squash shape genetics that has been used in textbooks ever since.
I'm really happy to know that my garden results are consistent with results from others, but I still need more data.
The simplest model I've come up with requires there to be a third gene (C) that allows the first two genes (A & B) to produce disc-shaped squash. Sorting out the genetics of a cross involving three genes is much harder than a cross involving one or two genes. Fortunately, I've got several hundred F2 seeds from the F1 plants I grew.
I plan to grow several F2s this coming year and I've managed to find homes for a few more in others' gardens. It may take several years of this to collect sufficient data to get an idea of what is going on.
Would you be willing to grow some of my experimental squash seeds, then send me photos/measurements of the fruit that grow? Get a message to me and we'll work out how to get seeds to you.
Citations and notes :
1. Emerson RA. The Inheritance of Sizes and Shapes in Plants. 1910, The American Naturalist 44: 739-746. (https://archive.org/details/jstor-2455667)
• "Yellow Crookneck" x "White Scallop" -> F1 "intermediate".
• This matches the results I found in my garden
2. Sinnott EW. Inheritance of Fruit Shape in Curcurbita pepo. 1922, Botanical Gazette 74: 95-103. (https://archive.org/details/jstor-2470204)
• Sphere x Scallop -> F1 disc -> F2 (3 disc):(1 sphere).
3. Sinnott EW. A Factorial Analysis of Certain Shape Characters in Squash Fruits. 1927, The American Naturalist 61: 333-344. (http://www.jstor.org/discover/10.2307/2456386)
• Sphere(#103) x disc -> F1 disc -> F2 (3 disc):(1 sphere).
• Sphere(#22) x disc -> F1 disc -> F2 (3 disc):(1 spheroid).
• Sphere(#103) x sphere(#22) -> F1 disc -> F2 (9 disc):(6 sphere):(1 elongate).
Part 1 | {"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": 0.38377004861831665, "perplexity": 4002.564829279384}, "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-2018-26/segments/1529267860570.57/warc/CC-MAIN-20180618144750-20180618164750-00016.warc.gz"} |
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ݒȂ | {"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": 0.9843529462814331, "perplexity": 14754.560999988009}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662527626.15/warc/CC-MAIN-20220519105247-20220519135247-00577.warc.gz"} |
https://usa.cheenta.com/rectangle-problem-geometry-prmo-2017-question-13/ | Categories
# Rectangle Problem | Geometry | PRMO-2017 | Question 13
Try this beautiful Rectangle Problem from Geometry from PRMO 2017, Question 13. You may use sequential hints to solve the problem.
Try this beautiful Rectangle Problem from Geometry, from PRMO 2017.
## Rectangle Problem – Geometry – PRMO 2017, Question 13
In a rectangle $A B C D, E$ is the midpoint of $A B ; F$ is a point on $A C$ such that $B F$ is perpendicular to $A C$; and FE perpendicular to BD. Suppose $\mathrm{BC}=8 \sqrt{3}$. Find AB.
• $9$
• $24$
• $11$
### Key Concepts
Geometry
Triangle
Trigonometry
Answer:$24$
PRMO-2017, Problem 13
Pre College Mathematics
## Try with Hints
We have to find out the value of $AB$. Join $BD$. $BF$ is perpendicular on $AC$.
Let $\angle \mathrm{BAC}=\theta$
since $\mathrm{E}$ is mid point of hypotenous $\mathrm{AB}$ of right $\Delta \mathrm{AFB}$, therefore $A E=F E=B E$
Can you now finish the problem ……….
Therefore
Therefore$\angle E F A=\angle F A E=\theta$
and $\angle \mathrm{FEB}=\angle \mathrm{EAF}+\angle \mathrm{EFA}=2 \theta$
$\Rightarrow \angle E B D=90^{\circ}-\angle B E F=90^{\circ}-2 \theta$
But $\angle \mathrm{FAE}=\angle \mathrm{CAB}=\angle \mathrm{DBA}$
Therefore $\theta=90^{\circ}-2 \theta \Rightarrow \theta=30^{\circ}$
Therefore in $\Delta \mathrm{ABC}, \tan \theta=\frac{\mathrm{BC}}{\mathrm{AB}}$
Can you finish the problem……..
Therefore $A B=B C \cot \theta=8 \sqrt{3} \cot 30^{\circ}=24$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.8509973883628845, "perplexity": 1699.7767416521438}, "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/1623487653461.74/warc/CC-MAIN-20210619233720-20210620023720-00628.warc.gz"} |
http://wikimechanics.org/muon | The Muon
## Seed-Aggregate Model
We describe the mechanics of muons by weaving together various chains of events that link absorption and emission processes. The archetypal history that these interactions modify is
$\Psi \left( \,\sf{\mu}^{-} \right) = \left( \sf{\Omega}_{1}, \sf{\Omega}_{2}, \sf{\Omega}_{3} \ \ldots \ \right)$
where each orbital cycle $\sf{\Omega}$ is a bundle of 40 Anaxagorean sensations; 8 right-side and 12 left-side somatic sensations; 2 hot, 2 freezing, 2 steamy and 2 cool thermal sensations; 4 red, 4 green and 4 white visual sensations. Here is a visual representation of the muon made by combining the icons of its components
Each Anaxagorean sensation defines a seed. So a muon can be modeled as the seed-aggregate
$\sf{\Omega} \left(\, \sf{\mu}^{-} \right) = \mathrm{4}\sf{U} + \mathrm{2}\sf{B} + \mathrm{2}\sf{T} + \mathrm{2}\sf{S } + \mathrm{2}\sf{C} + \mathrm{4}\sf{M} + \mathrm{4}\sf{A} + \mathrm{8}\sf{O} + \mathrm{12}\bar{\sf{O}}$
## A Quark Model of the Muon
The quark hypothesis proposes that pairs of seeds are always bound together as quarks. So to make a more detailed model of the muon, seeds are associated in these pairs
+ →
+ →
+ →
+ →
+ →
+ →
+ →
so that a muon consists of twenty quarks. And here is a visual representation of the muon that combines their icons. The back row of quarks is the same as the front row.
\begin{align} \sf{\Omega} \left(\, \sf{\mu}^{-} \right) = \mathrm{4}\bar{\sf{u}} &+ \mathrm{2}\bar{\sf{b}} + \mathrm{2}\sf{t} + \mathrm{2}\bar{\sf{s}} + \mathrm{2}\sf{c} + \mathrm{4}\bar{\sf{m}}+ \mathrm{4}\sf{a} \end{align}
Related WikiMechanics articles.
page revision: 173, last edited: 29 Jun 2018 16:18 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9112300872802734, "perplexity": 4325.849692519406}, "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/1542039742322.51/warc/CC-MAIN-20181114232605-20181115014605-00434.warc.gz"} |
https://soar.wichita.edu/handle/10057/5407/browse?rpp=20&sort_by=1&type=title&etal=-1&starts_with=M&order=ASC | Now showing items 1360-1379 of 2432
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(Wichita State University, 2011-07)
As mobile hand-held devices such as cell phones and personal digital assistants are battery operated, it is important to minimize their energy consumption. The limited lifetime of the battery is always a problem for all ... | {"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": 0.252810537815094, "perplexity": 5623.37272349283}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-33/segments/1659882570692.22/warc/CC-MAIN-20220807181008-20220807211008-00177.warc.gz"} |
http://www.khanacademy.org/math/multivariable-calculus/partial_derivatives_topic/divergence | If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.
# Divergence
3 videos
Is a vector field "coming together" or "drawing apart" at a given point in space. The divergence is a vector operator that gives us a scalar value at any point in a vector field. If it is positive, then we are diverging. Otherwise, we are converging!
### Divergence 1
VIDEO 10:20 minutes
Introduction to the divergence of a vector field.
### Divergence 2
VIDEO 10:52 minutes
The intuition of what the divergence of a vector field is.
### Divergence 3
VIDEO 10:48 minutes
Analyzing a vector field using its divergence. | {"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": 0.870096743106842, "perplexity": 1915.379885219781}, "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-2015-06/segments/1422115862432.8/warc/CC-MAIN-20150124161102-00201-ip-10-180-212-252.ec2.internal.warc.gz"} |
http://mathquill.com/ | # MathQuill by Han and Jeanine
WYSIWYG math with only HTML, CSS and JS
Try typing x^2:
Backspace and try 3/4, try \sqrt x, try \sin\theta .
Move around with the arrow keys, click, drag, cut and paste like any other textbox.
See our thorough demo for more features, textbox variants and the public API.
MathQuill renders LaTeX math as semantically rich HTML and presents it as human-readable math by styling with CSS. Showcasing the power of the HTML DOM, this math is WYSIWYG editable.
Current development is proudly supported by Desmos, whose awesome graphing calculator makes extensive use of MathQuill.
### Usage
Want to use it on your website? See the project README on GitHub.
### Project Contact
Han (laughinghan@gmail.com) | {"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": 0.18943071365356445, "perplexity": 21000.48290070916}, "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-2016-07/segments/1454701160822.87/warc/CC-MAIN-20160205193920-00136-ip-10-236-182-209.ec2.internal.warc.gz"} |
https://socratic.org/questions/what-are-the-x-intercept-s-of-the-graph-of-y-12-x-2-x | Algebra
Topics
# What are the x-intercept(s) of the graph of y + 12 = x^2 + x?
Jun 11, 2017
See a solution process below:
#### Explanation:
To find the $x$-intercepts we need to set $y$ to $0$ and solve for $x$:
$y + 12 = {x}^{2} + x$ becomes:
$0 + 12 = {x}^{2} + x$
$12 - \textcolor{red}{12} = {x}^{2} + x - \textcolor{red}{12}$
$0 = {x}^{2} + x - 12$
$0 = \left(x + 4\right) \left(x - 3\right)$
Solution 1)
$x + 4 = 0$
$x + 4 - \textcolor{red}{4} = 0 - \textcolor{red}{4}$
$x + 0 = - 4$
$x = - 4$
Solution 2)
$x - 3 = 0$
$x - 3 + \textcolor{red}{3} = 0 + \textcolor{red}{3}$
$x - 0 = 3$
$x = 3$
The $x$-intercepts are: $- 4$ and $3$
Or
$\left(- 4 , 0\right)$ and $\left(3 , 0\right)$
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https://www.physicsforums.com/threads/magnetic-photons.194225/ | # Magnetic photons?
1. Oct 27, 2007
### Urvabara
Here is the Abdus Salam's http://www.cc.jyu.fi/~hetahein/tiede/salam.pdf [Broken] "MAGNETIC MONOPOLE AND TWO PHOTON THEORIES OF C-VIOLATION".
Questions:
1. What are those $$\chi$$-fields and $$\chi$$-particles?
2. What does non-minimal mean?
3. What are those weird symbols: ig over m and e over m? There are no fraction bars.
4. B-particles = magnetic photons?
5. Does all mass-less vector mesons emitted by hadrons interact with atomic electric fields (or decay into leptons)?
6. Current status of C-violation in physics?
Last edited by a moderator: May 3, 2017
2. Oct 30, 2007
### blechman
They are proposed fermions that have magnetic charge. This is what the quantum magnetic current would look like in QFT.
Minimal operators are the lowest-order operator out of an infinite collection of operators. Minimal coupling is a condition for the standard model of particle physics. When considering higher-order corrections in a QFT, you often must include non-minimal operators. But it is usually sufficient to start with a minimal set, and build on it later. Notice that the non-minimal operators are suppressed by the mass (see below).
A typesetting error. They are ratios.
Salam was proposing the existence of a second photon with "magnetic" couplings rather than the ordinary QED photon with "electric" couplings. Such a new field would have different transformation laws.
I'm confused by this point. I can't put it into context, since the link you sent does not have bibliographical information, so I'm not sure when it was that Salam wrote this. Maybe someone else can answer this (or if you can clarify the question for me, I'll give it another try). At a guess: yes, they do. There's been no substantiated evidence of a magnetic photon, or a magnetic monopole.
C is maximally violated in weak nuclear interactions involving W-boson decays (there are no L-handed anti-neutrinos). Again, I would need to see *when* Salam wrote this paper to have an idea of what he was thinking about: he was, after all, the guy that figured out how the weak nuclear force worked, but was that before or after this paper?
3. Oct 30, 2007
### Urvabara
Thanks for the reply!
Ok. I have updated the file.
A. Salam (1966). "Magnetic monopole and two photon theories of C-violation". Physics Letters 22: 683-684.
4. Nov 5, 2007
### Urvabara
Anything new about this?
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https://www.mathreference.org/example/page/id/95/lg/en | # Exercise ID95
Algebra → Roots → Simplifying
[Level: ] [Number of helps: 0] [Number of pictures: 0] [Number of steps: 9] [Number of characters: 0]
Evaluate the following expression:
$\sqrt{6-4\sqrt{2}}$
$\sqrt{6-4\sqrt{2}}$ $=\sqrt{6-\sqrt{{{4}}^{2}2}}$ $=\sqrt{6-\sqrt{32}}$ $=\sqrt{\frac{6+\sqrt{{6}^{2}-32}}{2}}-\sqrt{\frac{6-\sqrt{{6}^{2}-32}}{2}}$ $=\sqrt{\frac{6+\sqrt{36-32}}{2}}-\sqrt{\frac{6-\sqrt{36-32}}{2}}$ $=\sqrt{\frac{6+\sqrt{4}}{2}}-\sqrt{\frac{6-\sqrt{4}}{2}}$ $=\sqrt{\frac{6+2}{2}}-\sqrt{\frac{6-2}{2}}$ $=\sqrt{4}-\sqrt{2}$ $=2-\sqrt{2}$
$\sqrt{6-4\sqrt{2}}=2-\sqrt{2}$
$\sqrt{A±B}=\sqrt{\frac{A+\sqrt{{A}^{2}-B}}{2}}±\sqrt{\frac{A-\sqrt{{A}^{2}-B}}{2}}$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 12, "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": 0.8050463795661926, "perplexity": 9188.62010632788}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-09/segments/1550247504790.66/warc/CC-MAIN-20190221132217-20190221154217-00032.warc.gz"} |
http://www.whxb.pku.edu.cn/EN/10.3866/PKU.WHXB201912068 | Acta Physico-Chimica Sinica ›› 2020, Vol. 36 ›› Issue (11): 1912068.
• Article •
### Natural Polyphenol Tannic Acid as an Efficient Electrolyte Additive for High Performance Lithium Metal Anode
Qin Ran1, Tianyang Sun1, Chongyu Han1, Haonan Zhang2, Jian Yan2, Jinglun Wang1,*()
1. 1 Key Laboratory of Theoretical Organic Chemistry and Functional Molecule, Ministry of Education, School of Chemistry and Chemical Engineering, Hunan University of Science and Technology, Xiangtan 411201, Hunan Province, P. R. China
2 Soundon New Energy Technology Co. Ltd., Xiangtan 411201, Hunan Province, P. R. China
• Received:2019-12-27 Accepted:2020-03-06 Published:2020-03-17
• Contact: Jinglun Wang E-mail:jlwang@hnust.edu.cn
• Supported by:
the Doctoral Foundation of Hunan University of Science and Technology, China(E518B1);2019 Undergraduate Student Scientific Research Innovation Plan "Challenge Cup Project" of Hunan University of Science and Technology, China(TZ9003)
Abstract:
As the application of lithium-ion batteries in advanced consumer electronics, energy storage systems, plug-in hybrid electric vehicles, and electric vehicles increases, there has emerged an urgent need for increasing the energy density of such batteries. Lithium metal anode is considered as the "Holy Grail" for high-energy-density electrochemical energy storage systems because of its low reduction potential (-3.04 V vs standard hydrogen electrode) and high theoretical specific capacity (3860 mAh·g-1). However, the practical application of lithium metal anode in rechargeable batteries is severely limited by irregular lithium dendrite growth and high reactivity with the electrolytes, leading to poor safety performance and low coulombic efficiency. Recent research progress has been well documented to suppress dendrite growth for achieving long-term stability of lithium anode, such as building artificial protection layers, developing novel electrolyte additives, constructing solid electrolytes, using functional separator, designing composite electrode or three-dimensional lithium-hosted material. Among them, the use of electrolyte additives is regarded as one of the most effective and economical methods to improve the performance of lithium-ion batteries. As a natural polyphenol compound, tannic acid (TA) is significantly cheaper and more abundant compared with dopamine, which is widely used for the material preparation and modification in the field of lithium-ion batteries. Herein, TA is first reported as an efficient electrolyte film-forming additive for lithium metal anode. By adding 0.15% (mass fraction, wt.) TA into the base electrolyte of 1 mol·L-1 LiPF6-EC/DMC/EMC (1 : 1 : 1, by wt.), the symmetric Li|Li cell exhibited a more stable cyclability of 270 h than that of only 170 h observed for the Li|Li cell without TA under the same current density of 1 mA·cm-2 and capacity of 1 mAh·cm-2 (with a cutoff voltage of 0.1 V). Electrochemical impedance spectroscopy (EIS), scanning electron microscopy (SEM), Fourier-transform infrared (FTIR) spectroscopy, cyclic voltammetry (CV), and energy-dispersive X-ray spectroscopy (EDS) analyses demonstrated that TA participated in the formation of a dense solid electrolyte interface (SEI) layer on the surface of the lithium metal. A possible reaction mechanism is proposed here, wherein the small amount of added polyphenol compound could have facilitated the formation of LiF through the hydrolysis of LiPF6, following which the resulting phenoxide could react with dimethyl carbonate (DMC) through transesterification to form a cross-linked polymer, thereby forming a unique organic/inorganic composite SEI film that significantly improved the electrochemical performance of the lithium metal anode. These results demonstrate that TA can be used as a promising film-forming additive for the lithium metal anode.
MSC2000:
• O646 | {"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": 0.21257475018501282, "perplexity": 7981.387191989743}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-45/segments/1603107879537.28/warc/CC-MAIN-20201022111909-20201022141909-00264.warc.gz"} |
http://www.soundandvision.com/content/aperion-audio-intimus-633-t-ii-surround-speaker-system-measurements | # TOWER SPEAKER REVIEWS
## Aperion Audio Intimus 633-T (II) Surround Speaker System Measurements
Measurements
Measurement Conditions
From the midbass up the results shown are pseudo anechoic, taken in a real room at one meter using the LMS measurement system from LinearX Systems, gated to minimize the effects of the room. This pseudo-anechoic response is then combined with the nearfield measurements of the woofers and port.
The measurements shown were taken with the Aperions' grilles in place. The nominal impedance and sensitivity are estimates derived from the test graphs. The effective bass limit is, by our convention, is taken as the minus 10dB point in the response (relative to the level at 1kHz, unless stated otherwise). In a real room this falling response will be enhanced by room gain. The gain, and therefore the actual response, however, will vary with the room.
Measurement Summary
Both the Aperion Intimus 633-T (II) and Intimus 634-VAC should be very easy loads to drive. Any modern, properly functioning amplifier or receiver should have no problem with them.
Their measured responses are respectable at their prices. Their off-axis measurements are also good. In particular, the three-way, 634-VAC center channel speaker is relatively (but not completely) free of the off-axis response suckouts (caused by comb filtering interference patterns between drivers) that are present in nearly all two-way, woofer-tweeter-woofer center channel designs we have measured.
Details: Intimus 633-T (II)
Ported cabinet tuning: 34Hz
Minimum impedance: 4.9Ω at 118Hz
Estimated nominal impedance: 7Ω
Sensitivity: 88dB/2.83V/m
Effective bass limit (-10dB): 28Hz
Estimated difficulty to drive: Easy
Discussion: Aperion Intimus 633-T (II)
The horizontal response (Fig.1) is quite smooth, with a one notable exception: the peaks at 700Hz and 1.7kHz, with a small valley in between. This appears too low in frequency to be a culprit in the brightness heard in the listening tests. There is no other obvious evidence here of serious high frequency response anomalies. The response with the grille removed, however (not shown) is about 2dB hotter above 4kHz, though it does not have the modest bump visible around 10kHz in the graph. The speaker was clearly balanced to provide the flattest response with its grille in place.
Fig.1: Aperion Intimus 633-T (II) pseudo-anechoic response off the horizontal axis at 45 degrees (red) and 60 degrees (blue).
The horizontal off-axis response falls off relatively smoothly at higher frequencies. Note the off-axis suckout at 2kHz in the vertical response, however (Fig.2). It's most serious above the tweeter (red curve). The optimum ear height for listening to the 633-T (II) would appear to be either on or, counter intuitively, just slightly above the tweeter axis (just a few degrees above will reduce the 1.7kHz peak without risking the dip that develops at a higher, 15-degrees above.)
Fig.2: Aperion Intimus 633-T (II) pseudo-anechoic response at 15 degrees above (red) and 15 degrees below (blue) the tweeter.
Details: Aperion Intimus 634-VAC Center
Sealed cabinet tuning: 63Hz
Minimum impedance: 4.9Ω at 111Hz
Estimated nominal impedance: 8Ω
Sensitivity: 87dB/2.83V/m
Effective bass limit (-10dB): 44Hz
Estimated difficulty to drive: Easy
Discussion: Aperion Intimus 634-VAC Center
The front averaged response of the 634-VAC (Fig.3) is more uneven than the response of the 633-T (II). I was surprised, in fact, to see the double peak, with a valley in the middle, in the response from 500Hz to 2.5kHz. This caused no clearly audible coloration in my room. The peak at about 12kHz does smooth out somewhat when the grille is removed, and while this does increase the output from 7kHz-15kHz by about 2dB, the overall top end response of the 634-VAC is smoother with the grille off. This may explain why I preferred an odd configuration for serious listening: grilles on for the 633-T (II) and off for the 634-VAC.
Fig.3: Aperion Intimus 634-VAC pseudo-anechoic response off the horizontal axis at 45 degrees (red) and 60 degrees (blue).
The wide off axis curves, taken at 45 and 60 degrees, aren't as smooth as we have measured from some other three-way center channel speakers, including the smaller Aperion 533-VAC. But the latter measured much leaner through the midbass. The suckout in the 634-VAC's off-axis response at 400Hz-500Hz is likely due to the interaction between the two laterally displaced woofers.
Fig.4: Aperion Intimus 634-VAC pseudo-anechoic response at 15 degrees above (red) and 15 degrees below (blue) tweeter.
The vertical response of the 634-VAC center (Fig.4) indicates that the best vertical listening height is on or slightly below the tweeter axis.
This is a respectable set of measurements for such relatively inexpensive speakers. I would, however, like to see a smoother response in both the 633-T (II) and 634-VAC through the upper midrange, and the elimination of the off-axis dip at 500Hz in the 634-VAC.
All figures: Violet curve: pseudo-anechoic response on tweeter axis, averaged across a 30-degree horizontal window, combined with nearfield woofer and port responses. All measurements taken at 1-meter.
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(OP)
New on the form. Glad I found this site.
Story: shown washer received from supplier, measured by QE department found non-conforming, disposition rejected.
Supplier yelling the parts meet the print.
Inspectors using X min and X max methods to determine the worst case.
Question: what would be the theoretical values (absolute minimum and absolute maximum)?
Supplier states:
x min.: .7285
X Max.: .7745
QE/Inspectors :
X min.: .7335
X Max.: .7665.
Parts measured on CMM are closer to the supplier values (and obviously they are defending their calculated numbers)
Algorithms used: maximum inscribed and minimum circumscribed circles (not LSQ)
Which theoretical values you agree with (if any)? Supplier or QE/Inspector?
Since you are using the max inscribed and min circumscribed methods, then it doesn't equate with measuring X, because there could be form error on the ID or OD. (Think of a sudden bubble in the shape of one of those circles.)
Kedu,
You failed to state the drawing interpretation standard. Assuming ASME Y14.5-2009, I believe the supplier has correctly calculated the wall thickness extremes allowed by the tolerances.
Similar questions have been discussed here quite a few times. One recent example is thread1103-433578: Stack-up concept, and that should lead you to more.
pylfrm
Kedu,
As per ASME Y14.5, I get xmin=0.7335. This occurs when you have the largest ID, the smallest OD, and maximum positional error. The value xmax occurs under a different set of conditions, and I am not sure I would care about it.
--
JHG
(OP)
Belanger,
That is exactly the gist of my question: to correlate theoretical value with CMM measurements (specified algorithms) and open setup inspection measurements (x min and X max)
Then first step has been (and still is an open question): what would be the theoretical values?
Pylfrm,
A master document (used by company accords the board) specifies ASME Y14.5-2009. So, you are correct.
Thank you for related discussion.
May I ask you how you came up with the conclusion that the supplier is correct? The examples shown have no LMC combined with RFS. Should I understand it is the same as LMC on ID combined with MMB on OD?
Drawoh,
Our QE got the same x min. values as yours, but not the supplier. Looks like pylfrm, agrees more with folks from manufacturing hence a state of confusion persists.
Such problems with using 2-point measurements as stand-ins for functional gages combined with non-functional datum references. At the least, I am unable to envision a case where there is some mechanism that centers the washer on the OD and then is concerned with the LMC of the hole. It makes me wonder what problem the originator of the requirement was trying to solve.
The distance from the center to the maximum OD is 3.005/2 (this is the center of the RFS datum feature)
The distance from the maximum OD to the far side is 2.995 (if the part is oddly shaped, there could be a dent or flat)
The distance from the center to the LMB is (1.508+.02)/2
so 3.005/2 - 2.995 + (1.508+.2)/2 = -.7285.
No one else seems to believe in showing their work. Any reason for withholding that info?
(edit to fix typo in calculation)
Full transparency: here is my work:
VC for ID: 1.508 + .020 =1.528
RC for ID: 1.492 -.016 -.020 = 1.456
X min: [2.985 (RMB) for datum feature A – 1.528 (VC)] /2 = .7285
X max.: [3.005 (perfect form at MMC required) – 1.456 (RC)] / 2 = .7745
I will feel “safe” if one of the experts around this forum ratify (or rebuke for that matter) my calculated values.
3DDave,
=(3-0.005)/2-1.508/2-0.02/2
...and I got 0.7335.
Remember, it is LMC/B of the OD, not MMC/B.
--
JHG
Agree with greenimi approach.
------------------
min x = { LMC of OD - outer boundary of ID (aka VC) } /2
min x = (2.995-1.528)/2
min x = .7335
----------------
max x = { MMC of OD - inner boundary of ID (aka RC) } /2
max x = (3.005-1.456)/2
max x = .7745
----------------
VC = 1.508 + .020 = 1.528
RC = 1.492 - .020 - .016 = 1.456
AndrewTT,
Looks like you agree with my approach, but not with my results
x min is different versus what I got.
check your OD LMC value (3.000-.005 = 2.995)
Drawoh, it's the center of the RFS condition, which means that the basically max diameter part can have a notch or a flat. If the parts are guaranteed to be perfect form at every section at every size, then it wouldn't matter. Which is why using 2-point measurements for this task is not useful.
For what it is worth, I agree with greenimi's numbers (I would not just call 2.985 "RMB", but rather "LMB" of datum feature A.)
When it comes to the value associated with the OD in the stack for Xmin, it should not just be 2.995. As 3DDave said, the OD can be oddly shaped (there could be a dent or flat) and that will affect the Xmin value. We went through this kind of exercise at least a few times in last 2-3 months on this forum.
I would, however, like to ask the OP additional question. What is the purpose of this exercise? I hope the inspectors do realize that even if they check that the value of "wall thickness" falls within calculated <Xmin;Xmax> range, there can still be a non-conformance against one the three characteristics used in the calculations.
So are we saying that the OD can be made as small as 2.995 and then a local form error of .010 (from rule #1) can get you down to 2.985 locally? Trying to figure out how the 2.985 is showing up.
AndrewTT,
The local size is still 2.995, but only in a strategic spot. The actual mating envelope is at 3.005, thus the local bite/dent/flat can penetrate the AME with the full local variation allowed.
The more important question is whether the subject parts will work in your assembly.
Mike Halloran
Pembroke Pines, FL, USA
(OP)
All,
Thank you for the enlightenment. Look like supplier is correct. So be it.
Pmarc,
We have these parts in inspection and are measured for size (ID, OD, -caliper, ring gage and cylindrical pins), but when comes for position at LMC no gage is available . So, consequently (should I say naturally) inspectors asked for a min/max values that can be measured with a caliper. (Another thought floating around has been to put these parts on the CMM, on a regular basis, but that “brain storming” - suggestion has been quickly scrapped and considered ridiculous for a long term)
Therefore, calipers are good enough (wrongfully or not) – since supplier is also checking them with the calipers.
The dilemma has been:
What are the values inspectors should abide to? and from here things have gotten out of control hence my original questions.
May I ask, what are the risks you foresee using this inspection method-calipers—correlated with correct x min/x max values? Accept bad parts? Reject good ones? Either one? Both?
Caliper inspection cannot perfectly represent a comparison against boundaries.
It's not clear what the goal of this dimensioning scheme was to begin with; there should be an engineering analysis showing the reasons for all the tolerances, but I expect there isn't one because the requirements conflict from what would be expected.
For example - the smaller the hole, the more restrictive it is in relation to whatever passes through it, but the greater the distance it will be allowed to be off-center, so it would not be able to be centered wrt to the datum feature. But the datum feature reference suggests the being exactly on center is the desired goal.
One could do this inspection optically - make a transparency that has the LMB of the hole completely blacked out and set it behind a mechanism with an iris that contracts around the OD. If any light is seen at the border of the LMB simulator then the hole is too far off center. A separate check would see that the hole ID is not out of spec.
Kedu,
Inspecting position tolerance at LMC with calipers is quite a simplification, but I understand why the inspectors want to take this approach (not that I agree with it, though).
If sizes of OD and ID are known it should be quite easy to roughly estimate actual positional error of ID with respect to OD based on measurement of "wall thickness" on at least one side of ID.
For example, if measured values of OD and ID along x axis are 3.000 and 1.500 respectively and measured "wall thickness" on one of the sides of ID is .730, this means (again, very roughly speaking) that the center of ID is .020 (.020 = 3.000/2 - 1.5000/2 - .730) off of its true position, resulting in total .040 of actual position tolerance error.
Notice that for this set of made up numbers OD and ID values are within specification limits, the measured "wall thickness" is within <.7285;.7745> range, yet the position of the ID is out of spec. (.040 vs. allowable .028).
Quote (pmarc)
(I would not just call 2.985 "RMB", but rather "LMB" of datum feature A.)
Thank you for correcting my terminology.
Kedu,
Assume the outside diameter size tolerance is satisfied. If the minimum wall thickness is at least (3.005 - (1.508 + .020))/2 = 0.7385, the position tolerance is satisfied. If the minimum wall thickness is less than 0.7285, the position tolerance is not satisfied. If the minimum wall thickness is between 0.7285 and 0.7385, more information is required to determine whether the position tolerance is satisfied.
By the way, how thick are these washers? Do you know anything about their intended function?
pylfrm
(OP)
pylfrm,
Thickness: gage 0000 (13/32)
Support for a sealing surface O-Ring, washer centered on OD, but O-Ring is not allowed to be extruded from ID, hence need of minimum material between ID and OD.
I would remove "x max" entirely. You can already reliably measure OD and ID. "x min" is a bit of a stab but doable. After that, there's no point, and additional measurement can only cause conflict.
WAIT!
Subject part is a spacer?
Or an anti-extrusion washer?
What material?
What do all of the mating parts look like?
Mike Halloran
Pembroke Pines, FL, USA
(OP)
pmarc,
Took me a couple of weeks to go over all the simple stacks recommended and found here on eng-tips.
May I ask, why do you think datum displacement has no influence over min-max calculations? Otherwise stated, if datum feature A is modified at MMB X min/ x max stays exactly the same. Why is that?
Since our washer is assembled with clearance, has been some meeting "brainstorming sessions" to change the print to show A(M) in order to allow for functional gaging.
The title of this thread is perfect
I kind of brought all those stackup discussions together (I was involved in some) and here is what I have concluded. Please feel free to agree or disagree with my statements:
Regardless of which modifier is used (MMC, LMC, RFS) same min-max. values (same wall thickness) are obtained as a result of the stackups calculation if datum feature is modified at RMB or MMB.
(Simple OD-ID association)
Example 1: Feature at MMC: same min-max values regardless if datum feature is RMB or MMB.
Example 2: Feature at LMC: same min-max values regardless if datum feature is RMB or MMB
Example 3: Feature at RFS: same min-max values regardless if datum feature is RMB or MMB
Example 4: Datum feature at LMB: same min-max values regardless if feature is RFS or MMC.
Wall thickness calculations (min-max) a.k.a. mutual diameter relationship offers much similarity between RFS, MMC (and LMC to a certain extent).
Probably a good question for the experts (pmarc I am sure knows to answer it) is:
What would be ASME theoretical explanation why datum displacement/ datum shift does not have an effect in the above stackup calculations?
Is the default Rule#1 the main culprit?
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Register now while it's still free! | {"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": 0.39864131808280945, "perplexity": 3691.555327900188}, "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-2018-26/segments/1529267863886.72/warc/CC-MAIN-20180620202232-20180620222232-00214.warc.gz"} |
https://www.jrevell.com/microsoft-dynamics-nav-either-the-caller-does-not-have-the-required-permission-or-the-specified-path-is-read-only-error/ | # Microsoft Dynamics Nav – Either the caller does not have the required permission or the specified path is read-only – Error
Today on Microsoft Dynamics Nav 2013R2, I kept getting the following error, which stumped me for a bit…
`Either the caller does not have the required permission or the specified path is read-only.`
After much investigation, I found a solution to the problem, delete the following folder and restart the service.
`C:\ProgramData\Microsoft\Microsoft Dynamics NAV\71\Server\{Nav Instance Name}\users\default`
Upon restarting the service, I was able to connect back to Nav without receiving the error. | {"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": 0.8624874353408813, "perplexity": 2408.847399699905}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-04/segments/1547584336901.97/warc/CC-MAIN-20190123172047-20190123194047-00193.warc.gz"} |
http://mathhelpforum.com/statistics/22569-just-wondering.html | 1. ## just wondering
A student moving into a new flat buys a box of 20 light bulbs. Unknown to the student, exactly one of the bulbs is defective. Find:
b) The conditional probability that the second bulb used is not defective, given that the first bulb used is not defective;
If the first bulb is not defective then you have 19 bulbs from which you can choose, and one of these is defective so the probability is $\frac{18}{19}$.
This is the easy way to work it out but I am having trouble using the actual formula which may come in handy for other examples:
Let A be the event that the 2nd bulb is not defective.
Let B be the event that the 1st bulb is not defective.
Then $P(A|B)=\frac{P(A\cap B)}{P(B)}$
$P(B) = \frac{19}{20}$ but how can you work out $P(A\cap B))$?
$P(A\cap B) = P(A|B)\cdot P(B)$
$P(A\cap B) = \frac{18}{19}\cdot \frac{19}{20}=\frac{18}{20}$
So how can we work out that $P(A\cap B) = \frac{18}{20}$ from what we are given?
I apologise for any offence my pedantry has caused
2. $\begin{array}{l}
P(X \cap Y) = P(X|Y)P(Y) \\
P(A \cap B) = P(B)P(A|B) = \left( {\frac{{19}}{{20}}} \right)\frac{{18}}{{19}} = \frac{{18}}{{20}} \\
\end{array}$
3. Let me rephrase my question:
A student moving into a new flat buys a box of 20 light bulbs. Unknown to the student, exactly one of the bulbs is defective. Let A be the event the 2nd bulb is defective, and let B be the event the 1st bulb is defective. What is $P(A\cap B)$?
4. Originally Posted by slevvio
Let me rephrase my question:
A student moving into a new flat buys a box of 20 light bulbs. Unknown to the student, exactly one of the bulbs is defective. Let A be the event the 2nd bulb is defective, and let B be the event the 1st bulb is defective. What is $P(A\cap B)$?
$P(A \cap B) = \left( {\frac{2}{{20}}} \right)\left( {\frac{1}{{19}}} \right)$
5. Thank you but once again I have asked the wrong thing.
A student moving into a new flat buys a box of 20 light bulbs. He takes a number of bulbs from the box without replacement. Unknown to the student, exactly one of the bulbs is defective. Find:
Let A be the event the 2nd bulb is NOT DEFECTIVE and B the event that the 1st bulb is NOT DEFECTIVE. Find $P(A\cap B)$. | {"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": 0.8557793498039246, "perplexity": 426.1286728354519}, "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/1508187820930.11/warc/CC-MAIN-20171017072323-20171017092323-00347.warc.gz"} |
https://www.shaalaa.com/question-bank-solutions/to-construct-a-triangle-similar-to-a-given-abc-with-its-sides-8-5-of-the-corresponding-sides-of-abc-draw-a-ray-bx-such-that-cbx-is-an-acute-angle-division-of-a-line-segment_258668 | # To construct a triangle similar to a given ΔABC with its sides 8/5 of the corresponding sides of ΔABC draw a ray BX such that ∠CBX is an acute angle - Mathematics
MCQ
Fill in the Blanks
To construct a triangle similar to a given ΔABC with its sides 8/5 of the corresponding sides of ΔABC draw a ray BX such that ∠CBX is an acute angle and X is on the opposite side of A with respect to BC. Then minimum number of points to be located at equal distances on ray BX is ______.
• 5
• 8
• 13
• 3
#### SolutionShow Solution
To construct a triangle similar to a given ΔABC with its sides 8/5 of the corresponding sides of ΔABC draw a ray BX such that ∠CBX is an acute angle and X is on the opposite side of A with respect to BC. Then minimum number of points to be located at equal distances on ray BX is 8.
Explanation:-
To construct a triangle similar to a given triangle with its sides m/n of the corresponding sides of given triangle, the minimum number of points to be located at equal distance is equal to the greater of m and n in m/n.
Here, m/n=8/5
So the minimum number of points to be located at equal distance on ray BX is 8.
Concept: Division of a Line Segment
Is there an error in this question or solution?
#### APPEARS IN
NCERT Mathematics Exemplar Class 10
Chapter 10 Construction
Exercise 10.1 | Q 5 | Page 114
Share | {"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": 0.6543681025505066, "perplexity": 821.9164843663262}, "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/1679296945323.37/warc/CC-MAIN-20230325095252-20230325125252-00352.warc.gz"} |
https://socratic.org/questions/a-triangle-has-sides-a-b-and-c-if-the-angle-between-sides-a-and-b-is-pi-12-the-a-6 | Trigonometry
Topics
# A triangle has sides A, B, and C. If the angle between sides A and B is (pi)/12, the angle between sides B and C is (5pi)/12, and the length of B is 1, what is the area of the triangle?
Mar 11, 2016
$a r e a = 0 , 25$
#### Explanation:
$a r e a = 0 , 25$
##### Impact of this question
561 views around the world | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 2, "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": 0.5731030702590942, "perplexity": 602.3342438696709}, "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/1631780058450.44/warc/CC-MAIN-20210927120736-20210927150736-00459.warc.gz"} |
https://codedump.io/share/cpp0DXqtR5H8/1/execute-node-with-php | pinug - 1 year ago 208
Javascript Question
# execute node with php
i want to execute a javascript file with node(node.js) without using windows terminal, so i used php's exec() function to execute the file, but it didn't work.
What is the problem ? Did i write the wrong windows command ? Did i write wrong full path of node? (i wrote it but maybe it's a backslash problem)? Windows don't know the "node" command ?
``console.log('Welcome to Node.js !');``
``````<?php
echo exec('C:\Program Files (x86)\nodejs\node.exe welcometonode.js');//not working
echo exec('node welcometonode.js');//not working
echo exec('C:\\Program Files (x86)\\nodejs\\node.exe welcometonode.js');//not working
echo exec('C:\Program Files (x86)\nodejs\node welcometonode.js');//not working
?>``````
``````exec("\"C:/Program Files (x86)/nodejs/node.exe\" \"C:/path/to/script/welcometonode.js\""); | {"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": 0.9604382514953613, "perplexity": 27600.048954131587}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-13/segments/1521257647475.66/warc/CC-MAIN-20180320130952-20180320150952-00522.warc.gz"} |
https://squ.pure.elsevier.com/en/publications/investigation-of-physico-chemical-properties-for-the-1-butyl-3-me | # Investigation of Physico-chemical Properties for the 1-Butyl-3-methylimidazolium Tetrafluoroborate ([Bmim][BF4])–Diethylenetriamine (DETA) System for CO2 Capture
*Corresponding author for this work
Research output: Contribution to journalArticlepeer-review
1 Citation (Scopus)
## Abstract
This paper presents the experimental values of densities, viscosities and refractive indices for 1-butyl-3-methylimidazolium tetrafluoroborate ([Bmim][BF4])–diethylenetriamine (DETA) mixtures over the entire composition range and temperature range 298.15–333.15 K. The physico-chemical properties such as excess molar volume, partial molar volume, partial molar volume at infinite dilution, apparent molar volume, apparent molar volume at infinite dilution, isobaric thermal expansion coefficient and excess isobaric thermal expansion coefficient were calculated from the experimental density data. The excess Gibbs energy and excess entropy of flow were determined from the viscosity data using Eyring’s theory of rate process. The specific interactions between molecules in mixtures were explained by analyzing the behavior of the thermodynamic excess/deviation properties. To determine the correlating accurately with the utilized models including the Redlich–Kister, modified Graber, Lorentz–Lorenz and McAllister equations, the adjustable parameters and standard deviations were determined. Moreover, the qualitative analysis of these properties was further legitimized with Fourier-transform infrared (FTIR) spectroscopic studies.
Original language English 578-610 33 Journal of Solution Chemistry 48 5 https://doi.org/10.1007/s10953-019-00868-0 Published - May 1 2019
## Keywords
• Correlation accuracy
• Density
• Excess/deviation properties
• Refractive index
• Specific molecular interactions
• Viscosity
## ASJC Scopus subject areas
• Biophysics
• Biochemistry
• Molecular Biology
• Physical and Theoretical Chemistry | {"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": 0.8147364854812622, "perplexity": 10355.203967453017}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-17/segments/1618038879305.68/warc/CC-MAIN-20210419080654-20210419110654-00229.warc.gz"} |
https://www.gradesaver.com/textbooks/math/statistics-probability/elementary-statistics-12th-edition/chapter-9-inferences-from-two-samples-9-4-two-dependent-samples-matched-pairs-basic-skills-and-concepts-page-474/6 | ## Elementary Statistics (12th Edition)
$\mu_d$ is between -2.7280538 and 2.8457008.
Null hypothesis:$\mu_d=0$, alternative hypothesis:$\mu_d$ is more than 0. The degree of freedom: 34-1=33. The corresponding critical value using the table: $t_{\alpha/2}=t_{0.05}=1.692.$ The margin of error: $E=t_{\alpha/2}\frac{s_d}{\sqrt{n}}=1.692\frac{9.604106}{\sqrt{34}}=2.7868773.$ Hence the confidence interval $\mu_d$ is between $\overline{d}-E$=0.0588235-2.7868773=-2.7280538 and $\overline{d}+E$=0.0588235+2.7868773=2.8457008. | {"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": 0.954786479473114, "perplexity": 3959.920148468634}, "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/1544376823183.3/warc/CC-MAIN-20181209210843-20181209232843-00378.warc.gz"} |
https://www.gradesaver.com/textbooks/math/algebra/elementary-algebra/chapter-2-real-numbers-2-2-addition-and-subtraction-of-rational-numbers-concept-quiz-2-2-page-57/7 | ## Elementary Algebra
When finding a common denominator, we may use any number that is a multiple of the denominators of all of the fractional addends. Since $20$, $40$, and $60$ are all multiples of both $4$ and $5$, any of them can be used as a common denominator. Since $20$ is the $\text{least}$ common multiple of $4$ and $5$, it is the $\text{least}$ common denominator. | {"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": 0.9757952690124512, "perplexity": 70.13248264670855}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-04/segments/1547584415432.83/warc/CC-MAIN-20190123213748-20190123235748-00193.warc.gz"} |
http://events.berkeley.edu/index.php/calendar/sn/?event_ID=120228&date=2018-10-01&tab=all_events | ## Differential Geometry Seminar: Existence of infinitely many minimal hypersurfaces in closed manifolds
Seminar | October 1 | 3:10-4 p.m. | 939 Evans Hall
Antoine Song, Princeton
Department of Mathematics
In the early 80's, Yau conjectured that in any closed $3$-manifold there should be infinitely many minimal surfaces. I will review previous contributions to the question and present a proof of the conjecture, which builds on min-max methods developed by F. C. Marques and A. Neves. A key step is the construction by min-max theory of a sequence of closed minimal surfaces in a manifold N with non-empty stable boundary, and I will explain how to achieve this via the construction of a non-compact cylindrical manifold.
lott@math.berkeley.edu | {"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": 0.7090538740158081, "perplexity": 709.1293267525017}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-39/segments/1568514573735.34/warc/CC-MAIN-20190919204548-20190919230548-00316.warc.gz"} |
http://bmjopen.bmj.com/content/7/7/e015580 | Article Text
Robot-assisted surgery in a broader healthcare perspective: a difference-in-difference-based cost analysis of a national prostatectomy cohort
1. Vibe Bolvig Hyldgård1,2,
2. Karin Rosenkilde Laursen2,
3. Johan Poulsen3,4,
4. Rikke Søgaard2,5
1. 1 Health Economics, DEFACTUM, Central Denmark Region, Aarhus, Denmark
2. 2 Department of Public Health, Aarhus University, Aarhus, Denmark
3. 3 Department of Urology, Aalborg University Hospital, Aalborg, Denmark
4. 4 Department of Urology, King’s College Hospital, London, UK
5. 5 Department of Clinical Medicine, Aarhus University, Aarhus, Denmark
1. Correspondence to Vibe Bolvig Hyldgård; vibe.bolvig{at}rm.dk
## Abstract
Objective To estimate costs attributable to robot-assisted laparoscopic prostatectomy (RALP) as compared with open prostatectomy (OP) and laparoscopic prostatectomies (LP) in a National Health Service perspective.
Patients and methods Register-based cohort study of 4309 consecutive patients who underwent prostatectomy from 2006 to 2013 (2241 RALP, 1818 OP and 250 LP). Patients were followed from 12 months before to 12 months after prostatectomy with respect to service use in primary care (general practitioners, therapists, specialists etc) and hospitals (inpatient and outpatient activity related to prostatectomy and comorbidity). Tariffs of the activity-based remuneration system for primary care and the Diagnosis-Related Grouping case-mix system for hospital-based care were used to value service use. Costs attributable to RALP were estimated using a difference-in-difference analytical approach and adjusted for patient-level and hospital-level risk selection using multilevel regression.
Results No significant effect of RALP on resource-use was observed except for a marginally lower use of primary care and fewer bed days as compared with OP (not LP). The overall cost consequence of RALP was estimated at an additional €2459 (95% CI 1377 to 3540, p=0.003) as compared with OP and an additional €3860 (95% CI 559 to 7160, p=0.031) as compared with LP, mainly due to higher cost intensity during the index admissions.
Conclusions In this study from the Danish context, the use of RALP generates a factor 1.3 additional cost when compared with OP and a factor 1.6 additional cost when compared with LP, on average, based on 12 months follow-up. The policy interpretation is that the use of robots for prostatectomy should be driven by clinical superiority and that formal effectiveness analysis is required to determine whether the current and eventual new purchasing of robot capacity is best used for prostatectomy.
• cost analysis
• economics
• prostate cancer
• prostatectomy
• robot-assisted surgery
• robotics and laparoscopy
This is an Open Access article distributed in accordance with the Creative Commons Attribution Non Commercial (CC BY-NC 4.0) license, which permits others to distribute, remix, adapt, build upon this work non-commercially, and license their derivative works on different terms, provided the original work is properly cited and the use is non-commercial. See: http://creativecommons.org/licenses/by-nc/4.0/
## Statistics from Altmetric.com
### Strengths and limitations of this study
• A broad healthcare sector perspective with 12 months follow-up of a national cohort.
• A strong analytical approach including a quasi-experimental difference-in-difference design in combination with the use of regression-based adjustment for selection.
• Adjustment for body mass index could not be undertaken due to this information not being available in national register data.
• A proportion of patients had missing values regarding cancer stage but these patients did not seem to be different from patients with complete data.
## Introduction
The most common cancer among men older than 50 years is prostate cancer.1 The incidence has increased notably since the diagnostic prostate-specific antigen test was introduced and, in accordance, the incidence of prostatectomy has increased rapidly.1–3 Internationally, the transition from open prostatectomy (OP) to laparoscopic prostatectomy (LP) was much slower than the on-going transition from LP to robot-assisted laparoscopic prostatectomy (RALP), which is today the most frequently used technique in North America and in some parts of Europe.4 As a consequence of the rapid dissemination of RALP, the literature comparing RALP to LP is scarce.
The minimally invasive methods LP and RALP have been found to hold some perioperational advantages over OP such as less bleeding and fewer complications of, for example, urinary incontinence.1 5–8 The literature is, however, not definite in terms of whether these benefits of the minimally invasive approaches can be achieved equally with or without robot support.2 4 9 It has been argued that robot technology has a particular advantage in obese patients but, again, this has been questioned by a recent study demonstrating similar oncological and pathological outcomes when comparing RALP to LP and OP in obese patients.10
In comparison with not using robot support, the use of robot support leads to significantly higher costs due to the capital binding in the robot, maintenance costs and surgical supplies.4 11 12 However, there could be cost savings in the longer-term and in a broader healthcare sector perspective that outweighs the additional cost of the surgical procedure. These could flow from the better process outcomes such as less bleeding and fewer bed days. Despite the obvious relevance of a broader perspective, the literature is characterised by focussing solely on admission costs or just operating costs. The overall consequences of the dissemination of the robot technology to healthcare costs are therefore to a large extent uncertain. The objective of this study is to estimate the costs attributable to RALP as compared with OP and LP in a broad healthcare sector perspective and using a time horizon that allows for clinical manifestation of the postoperative advantages of robot support.
## Patients and methods
### Design
A national-scale cohort was followed from 1 year before to 1 year after prostatectomy. A quasi-experimental difference-in-difference (DID) design13 was combined with regression to adjust for pretreatment covariates (risk selection into surgical technique).14 Data were collected in connection with a Danish health technology assessment (HTA) of robot-assisted surgery, which this study is a further development of.15
### Study population
Consecutive men who underwent prostatectomy in Denmark in the period 1 January 2006 to 1 august 2013 were identified from the National Patient Registry,16 using the procedural codes KKEC00, KKEC00A, KKEC00B, KKEC00C, KKEC01, KKEC01A, KKEC01B, KKEC01C, KZXX00 and ZPW00002. To enhance comparability of the patients an inclusion criterion was that the robot-assisted technique should be available at the given hospital at the time of the prostatectomy.
### Data sources
Individual-level register data were extracted from national administrative registries including The Danish National Patient Register,16 The Danish Civil Registration System,17 and The Danish National Health Service Register.18 Costs were drawn from the registries for the diagnosis related grouping system (DRG) and the Danish outpatient grouping system (DAGS).19
### Costs
A healthcare sector perspective was applied in this study. Thus, the study included service use within the primary sector (general practitioners, medical specialists, therapists and other privately practicing specialists) and within the hospital sector (inpatient and outpatient hospital-based activity). Primary care service was valuated via the activity-based fees and hospital-based care via the DRG/DAGS-tariffs that were used at the time of service provision. The DRG tariffs for prostatectomy cover the activity from the day of admission to the day of discharge (preparation, surgery, remobilisation and discharge) whereas follow-up visits and other events after discharge,for example, caused by complications, are therefore separately reimbursed. The specific tariffs for prostatectomy are shown in online supplementary table 1. The higher tariff of the robot-assisted surgery (on average €4525) thus refers to the rather expensive instrument kit required for each surgery, robot maintenance costs and longer operating time. The theoretical interpretation of the DRG tariff is an average long-term cost. The influence of the lack of person-individual variation in the DRG tariff as a cost estimate for the admission for prostatectomy was informed by conducting sensitivity analysis where the number of bed days was added as a proxy for cost intensity. Other sensitivity analyses included adjustment for experience with robot and patient volume, as well as restrictions to the two most recent years and exclusion of the tariffs from the costs. Costs are reported in Euros (2014 price year).
### Identification of relevant aspects of risk selection
Characteristics that affect the choice of surgical method were identified in a literature review. Patient-level characteristics included age, cancer stage and comorbidity and hospital-level characteristics included organisational structure around the technology such as specialisation of staff. The identified characteristics were defined for the study population based on information from national registries: age (years), tumour size and nodal involvement based on the TNM-classification,20 comorbidity as defined by the Charlson Comorbidity Index,21 geographical region of the treating centre, level of experience by time of surgery (to-date volume of prostatectomies using the particular technology), and organisational structure of the surgical department, referring to whether the robot is used within a single department, used across several departments or used in a robotic centre. Finally, dummies for year of surgery were specified in order to be able to adjust for changes in DRG tariffs over the years.
### Statistical analysis
Summary statistics including Pearson's χ2 tests for categorical variables and ANOVA for continuous variables were used to describe patient's characteristics. All analysis followed a DID design where the costs attributable to prostatectomy were estimated as the differences between comparators (OP, LP and RALP) of differences in resource use and costs between 12 month periods before and after prostatectomy.13 To further handle risk selection (as described in the previous section) regression models were used to adjust the DID-estimates for covariates identified to affect selection into surgical technique.14 Regressions were specified as multilevel regressions due to the patient-level being nested in the hospital-level (centres treating more than one patient) in order not to underestimate SE. The validity of regression models was visually inspected based on conventional regression diagnostic plots and found to be robust.
Results are reported as arithmetic means with 95% CI based on bootstrapping with 5000 replicates due to the skewed nature of the data. All tests were two-sided with a 5% significance level. The statistical analyses were performed in Stata SE V.13.1.
### Ethics
The study was conducted in accordance with the Person Data Act and hence was approved by relevant authorities (The Danish Data Protection Agency) (Journal number 2007-58-0010). Consent is not required for register-based studies according the Danish Ethical Committee system.
## Results
Of the 4309 patients included in this study 52% underwent RALP, 42% underwent OP and 6% underwent LP (cf. online supplementary table 2 for procedure volume overtime). There were 22 conversions from either RALP or LP to OP, which were categorised according to the intended technique. The characteristics of the cohort are shown in table 1. The treatment groups were clinically similar in age, though the RALP group was younger than the OP and LP group (median age 64 vs 65 years) (p<0.001). The choice of surgical technique differed geographically and with regard to the organisation of the robot technology (p<0.001). Cancer severity was routinely registered for a proportion of patients only, which could be due to the fact that nodal involvement and metastases are rarely an issue for prostatectomy candidates. However, in case of no nodal involvement patients were less likely to have received a minimally invasive technique (p≤0.001).
Table 1
Comparison of descriptive characteristics (n (% of treatment group))
Service use per patient, including length of stay, and the unadjusted mean costs of the patients’ healthcare are depicted in table 2,3 respectively. All treatment groups had statistically significant higher service use in the year following the surgery. No differences were found when comparing RALP to LP but OP was associated with 2.6 extra bed days and slightly higher primary care service use (0.5 more contacts) compared with RALP. This was, however, not reflected in the costs, as RALP was associated with the highest costs primarily caused by differences in inpatient care (table 3).
Table 2
Healthcare service use in relation to prostatectomy
Table 3
Healthcare costs in relation to prostatectomy
Figure 1 illustrates the cost patterns overtime. The process of getting referred by the general practitioner to the hospital for diagnosis and later treatment seems to be reflected as a rise of costs in the primary care sector, is followed by a rise in outpatient care and later in inpatient care at the time of the prostatectomy. Outpatient follow-up is clearly evident but is not set at a fixed time. No clear differences stood out except for higher inpatient costs of RALP at the time of the index prostatectomy. Both in the year prior to and after the prostatectomies included in this study the patterns are rather similar especially for OP and RALP while LP fluctuates more due to fewer patients having received this surgical technique.
Figure 1
Time series graphics for the unadjusted mean costs (€). Month zero marks the time of prostatectomy, price year 2014. LP, laparoscopic prostatectomy; OP, open prostatectomy; RALP, robot-assisted laparoscopic prostatectomy.
Table 4 illustrates DID-estimates similar to those of table 3 except that multivariate modelling is used to adjust for eventual residual risk selection not handled by the DID-analytical strategy. Results support the unadjusted results as significant differences are revealed when RALP is held against OP and LP respectively. The adjusted costs attributable to RALP varied as RALP was associated with an extra €3860 (95% CI €559 to €7160) when held against LP and €2459 (95% CI €1377 to €3540) when compared with OP.
Table 4
Adjusted estimates of the costs attributable to robot-assisted laparoscopic prostatectomy (RALP): main model compared with extended model, which includes adjustment for tumour size and nodal involvement
Costs were significantly higher when patients were operated in Region of Southern Denmark or North Denmark Region (p<0.05), and when they were operated in hospitals with a robotic centre (p<0.05).
An extended model was applied to assess the role of informative missings on cancer severity. Adding cancer severity to the model did not substantially affect the cost attributable to RALP. Tumours categorised as T3–T4 were associated with significant additional costs for all surgical techniques and having missing data with respect to nodal involvement was associated with decreased costs, but there was no significant interaction between either tumour size or nodal involvement and surgical technique.
Restricting the main model to activity during the two most recent years (2012 and 2013) does not significantly alter the findings (the average attributable costs increases from €2459 to €3889 compared with OP and reduces from €3860 to €3359 compared with LP).
In order to directly analyse the contribution of the index admission versus the after-period for the costs attributable to RALP, sensitivity analyses restricting the costs to the after-period alone show comparable after-periods for LP and RALP whereas the after-period for OP is characterised by significantly more activity (€2332 (95% CI €1287 to €2777)).
## Discussion
Practically all prostatectomies performed in Danish hospitals over a period of 8 years were included in this analysis, which focused on the broad healthcare sector consequences of using robot technology. The cost of RALP was found to be higher than the costs of both OP and LP due to the difference in DRG tariffs across these surgical techniques. No evidence was found of RALP impacting service use when compared with LP, however, some reduction in bed days in the after-period was found when compared with OP. Hence, the main contribution of this study is an important piece of evidence that, when considering a broad healthcare sector perspective and a longer time horizon than the index admission, the use of RALP does not seem to generate cost consequences that can outweigh the additional cost associated with the index surgery.
A recent study by Hughes et al estimated the resource use in the postoperative phase after prostatectomy in a hospital perspective and found that RALP led to costs savings, when the cost of the index surgery was excluded from the equation.22 This study is in many ways similar to the present in that it is based on a large sample and considers extra-index-surgery consequences of using robot technology. It has however a couple of weaknesses, that is, circumvented in the present study. First, it includes patients who were referred to centres not offering robot technology and who could have different profiles than those referred to centres offering robot technology. Second, the investigators did not analytically handle the fact that patients were selected into surgical technique. It thus remains unclear whether the difference between the present results of no cost saving and Hughes et al's finding of a cost saving is due to these weaknesses or whether they are simply do to differences between the British and the Danish context.
Previous studies have assessed the costs of robot technology in an analytical perspective restricted to hospital costs of the index surgery. Kim et al found that RALP, despite shorter hospital stays, was associated with higher operation costs than OP by an average that more or less corresponds to the difference in Danish DRG tariffs between surgical techniques (mean $11 932 vs$9390; p<0.001).23 Similarly, Bolenz et al found hospital costs to be higher for RALP compared with LP and OP, which was a bit lower but still within the level of the difference in the Danish DRG tariffs (median $6752 for RALP,$5687 LP and \$4437 for OP; p<0.001).12 These studies were conducted in the USA, that is, not normally considered to be comparable as a setting due to different system structures and price levels.
The strengths of this study relates to the design where a cohort of consecutive patients are observed and where appropriate analytical effort is made into handling selection for surgical techniques. The hybrid DID-design in combination with regression-based adjustment for pretreatment covariates serves to minimise the effect of selection bias, which can be an important issue in observational designs that may have been chosen as the only option or in priority of external validity. This design has the ability to cleanse out exogenous factors, such as time and to isolate the costs related to the prostatectomy from the costs related to, for example, chronic comorbidities or other time invariant patient characteristics.24 The design is particularly powerful when combined with extra means for handling selection and multilevel multivariate regression was here used to adjust for hospital-level characteristics as well as patient-level characteristics that could have caused confounding. It should also be mentioned that we were able to validate the consecutiveness of data and the coding of surgical techniques by comparing register data to the independent clinical database UroLap, which supported that data were truly representing consecutive patients and which gave no reason to suspect misclassification.25
In the early stages of this work, we suspected that the cost implications of robot technology would be affected by centre volume and experience with the technology. We thus included variables in the regression model for these organisational-level covariates but they appeared to be insignificant contributors and were thus excluded from the reported main model. Also, we sought to assess whether there was any effect modification from point at the learning curve by including interaction terms between the dummies of year of surgery and the cost consequences of robot technology but again, these turned out to be insignificant and were thus left out in the main model. The geographical variations found could reflect patient heterogeneity caused by both cultural and structural variations such as different waiting times and referral practice.
The main weakness of this study lies in the premises of basing it on registry data, where severity and other clinical details are not routinely recorded. One variable of relevance to choice of surgical technique would be body mass index (BMI).26 Another weakness concerns the missing values on cancer stage, as it appeared that doctors are not routinely registering TNM status in relation to prostatectomy. Tumour size was registered for about 50% of patients while nodal involvement and metastasis were registered for around 25% of patients only. Whether this reflects irrelevance of registration in relation to the choice of surgical technique and expected outcome or other reasons is unclear, but conducting parallel analysis with and without TNM status did not substantially affect results. And more importantly, patients with missing values on the TNM status did not seem to be different from patients with complete data. A number of sensitivity analyses were undertaken to address limitations of the study. First, the use of national tariffs as an expression for the patient-level cost of hospital service ignores patient-level and hospital-level variation. For example, differences in coefficient of usage are not reflected in the tariffs. A sensitivity analysis where the number of bed days was included in the model was therefore undertaken and confirmed that variation captured in bed days had no influence on the main result. This analysis is, however, no full compensation for the lack of patient-level variation and this limits the interpretation of the analysis to the broad-sector consequences of using robot technology as opposed to the technical efficiency or productivity that characterises the operation of the robot technology. Also, it should be noted that time dummies were included in the base-case model in order to take out variation that was due to changes in the DRG tariffs overtime. If centres in the future administer the robot technology (and other surgical techniques for that sake) in a more of less efficient way, for example, by operating more patients per robot this will affect the cost of index surgery (and should lead to an adjustment of the DRG tariff) whereas the main focus of this analysis, the broader-sector cost consequences, should be unaffected if the quality level is kept.
Further research seems warranted as RALP is here found to be overall more costly than its alternatives while there appears to be limited evidence for a clinical benefit to the patients. At best, a randomised controlled trial comparing RALP to both LP and OP should be conducted and followed by a cost-effectiveness evaluation. LP is a relatively rare choice of surgical approach in Denmark although it has been found to create health outcomes and functional outcomes comparable to those of RALP.3 9 27 However, there is evidence that RALP is a superior choice with regards to the risk of erectile dysfunction.28 If this was also the case in the present cohort, it was not reflected in the number of visits to neither hospitals nor the primary healthcare sector.
## Conclusions
In this study from the Danish context, the use of RALP generates a factor 1.3 additional cost when compared with OP and a factor 1.6 additional cost when compared with LP, on average, based on 12 months follow-up. The policy interpretation is that the use of robots for prostatectomy should be driven by clinical superiority and that formal effectiveness analysis is required to determine whether the current and eventual new purchasing of robot capacity is best used for prostatectomy.
## Acknowledgments
The authors would like to thank Line Stjernholm Tipsmark (LST) for assistance in requisitioning data and for performing preliminary analyses.
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View Abstract
## Footnotes
• Contributors Drafting the manuscript: VBH. Analysis and interpretation: VBH, RS. Statistical analysis: VBH, KRL. Concept and design: JP, RS. Acquisition of data: RS, KRL, LST. Critical revision of manuscript: JP, RS, KRL, VBH. Supervision: RS.
• Funding This research received no specific grant from any funding agency in the public, commercial or not-for-profit sectors.
• Competing interests None declared.
• Provenance and peer review Not commissioned; externally peer reviewed.
• Data sharing statement Unfortunately no additional data can be made available due to legal restrictions.
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https://artvinyl.de/matplotlib/sonth/mazda/38172993a18066908e-sector-area-definition-geometry | ### sector area definition geometry
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mikebledsoe
Welcome to Monday morning with Mike and max and today we're going to be talking about freedom because you know what it is the fourth of July and we are americans and so of course not only are we totally into the conversation today but we're into it. Just make use of radians instead of degrees. Worksheets have become an integral part of the education system They represent the wishes and interests of everyday citizens like you and me as well as large multi national corporations These will be used to access Drafting Board The parent volunteers exchange groups at 15-minute intervals, eventually working with all four Solve the following problems. Solution: Step 1: Find the area of the entire circle using the area formula A = r 2. Figure 2 Using the arc length and the radius to find the measure of the associated central angle. Search: Second Prison Bunker Access Code. OADB is also a sector. Solution. The sector area is recalculated as you drag. Area of a Sector and Arc Length Geometry BowerPower.net . Area. Then, the area of a sector of circle formula is calculated using the unitary method. For example, 0.28 x 78.5 = 21.89. The simplicity of the central angle formula originates from the definition of a radian. Find m AOB. Then, the area of a sector of circle formula is calculated using the unitary method. Download 200 SAT/ACT math practice questions (and support the mathplane sites!) Let's learn about how to calculate the area of a sector. The sector is $$\frac{1}{6}$$ of the full area. Lets work out a couple of example problems involving the area of a sector. Calculations of geometric shapes and solids: the Semicircle. Search: Desmos Double Integral Calculator. The area of a circle segment is slightly more difficult to calculate. Sector ( ) is part of the disk, which is bounded by two radii and an arc between the radii. So, the area of the sector is about 923.2 cm2. Sector can also be referred as Wedge. What is a sector? Round decimal answers to the nearest tenth. Students should also know that a circle has 360 degrees. a portion of a military front or area of operation.
Reduce 8/32 to . Webinars. Area = 1/3 22/7 42 42 . The Quadrant and Semicircle are two special types of Sector: Quarter of a circle is called a Quadrant. Then, the area of a sector of circle formula is calculated using the unitary method. The same method may be used to find arc length - all you need to remember is the formula for a circle's circumference. Area of a Semicircle. A = R ( 2 h + a) Volume, V. The volume of spherical sector, either open spherical sector or spherical cone, is equal to one-third of the product of the area of the zone and the radius of the sphere. Again, you will be multiplying the percent by the area of the whole circle. Equation of a Circle. Reduce 8/32 to . A radian is a unit of angular size, where 1 radian is defined as a central angle () whose arc length is equal to the radius Now, if you are still hungry, take a look at the sector area calculator to calculate the area of each pizza slice! So, m AOB = 90. More About Sector. A sector is a wedge of a circle made from two radii. To play a "Power Chord " for rock songs, we play the first note of a chord , and then another note 5 tones higher. Plug the sector's central angle measurement into the formula. Lets work out a couple of example problems involving the area of a sector. of the circle. The area, A, of a circle with radius r is A = r 2. Calculations at a semicircle. is the arc of sector OADB. The formula to find the area of a sector is A =. Next Generation Science Standards. A sector of a circle is a region bounded by two radii and an arc of the circle. = (130/360) x 3.14 x 28 x 28. Common Core State Standards. NGSS. CCSS Math. Area of the sector = /360 r 2. circle, a radius, a diameter, a central angle, and an arc. Looking to improve test scores? Solution: The sector A O B has angle 60 . Therefore A = 1 6 3.1416 ( 10) 2 = 52.36 square. The formula for the area of a sector is (angle / 360) x x radius2. Sector area. Radii, the plural of radius, are line segments that start on the outside and end at the center of the circle. The sector of a circle formula in radians is: A =. Search: Knowledge Matters Diversification Math Quiz Answers. Thus, when the angle is , area of sector, OPAQ =. Discuss what a sector is and how to find the area of a sector. The region of the circle cut off from the rest of the circle by this chord is called as segment of a circle. Helps your students by becoming CK-12 Certified! Introduce the term radian, and briefly explain its connection to the work in Example 1. Definition: The part of a circle enclosed by two radii of a circle and their intercepted arc . In a circle with radius r and centre at O, let POQ = (in degrees) be the angle of the sector. All it needs is; the radius and angle to find the area of sector. For example, if the central angle is 100 degrees, _____ Math Comic #172 - "Pink Floyd" - (1-8-15) Visit the Weekly webcomic gallery _____ Remember the area of a circle = $$\pi r^2$$ The sector area is: $$\frac{1}{6} \times \pi \times 4^2 = 8.4~\text{cm}^2$$ The area between an arc and two radii of a circle is called Sector. This lesson assumes that people already know how to calculate the Circumference and the Area of a Circle. is the arc of sector OACB. | {"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": 0.8781455755233765, "perplexity": 563.4344678944242}, "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/1664030337480.10/warc/CC-MAIN-20221004054641-20221004084641-00003.warc.gz"} |
https://www.johndcook.com/blog/2020/11/18/rotating-symbols-in-latex/ | Rotating symbols in LaTeX
Linear logic uses an unusual symbol, an ampersand rotated 180 degrees, for multiplicative disjunction.
The symbol is U+214B in Unicode.
I was looking into how to produce this character in LaTeX when I found that the package `cmll` has two commands that produce this character, one semantic and one descriptive: `\parr` and `\invamp` [1].
This got me to wondering how you might create a symbol like the one above if there wasn’t one built into a package. You can do that by using the `graphicx` package and the `\rotatebox` command. Here’s how you could roll your own par operator:
` \rotatebox[origin=c]{180}{\&}`
There’s a backslash in front of the & because it’s a special character in LaTeX. If you wanted to rotate a K, for example, there would be no need for a backslash.
The `\rotatebox` command can rotate any number of degrees, and so you could rotate an ampersand 30° with
` \rotatebox[origin=c]{30}{\&}`
to produce a tilted ampersand.
Related posts
[1] The name `\parr` comes from the fact that the operator is sometimes pronounced “par” in linear logic. (It’s not simply `\par` because LaTeX already has a command `\par` for inserting a paragraph break.)
The name `\invamp` is short for “inverse ampersand.” Note however that the symbol is not an inverted ampersand in the sense of being a reflection; it is an ampersand rotated 180°.
One thought on “Rotating symbols in LaTeX”
1. Eric Rasmusen
Thank you for this good post. You and readers of this post will like the handout I made based on it for my 7th graders :
https://www.rasmusen.org/special/Cedars_School/03.03_Latex_rotating_symbols.pdf
They had great fun with this. I started by showing them the output from a command with 45 and 90 and asked them what they thought would happen if I replaced it with 180, 270, 0, 360, 720. Then we tried replacing center=c with l and r. | {"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": 0.9194025993347168, "perplexity": 1061.5446859011113}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030335058.80/warc/CC-MAIN-20220927194248-20220927224248-00223.warc.gz"} |
https://www.physicsforums.com/threads/what-are-everyday-nonlinear-examples.394750/ | # What are everyday nonlinear examples?
1. Apr 12, 2010
### UseAsDirected
What are everyday nonlinear" examples?
Hello!
Is there a simple way to identify a nonlinear equation or physical system by looking at it?
I have sifted through material about unpredictability, chaos, fractals, and the other buzzwords encompassing nonlinear systems", and have glossed over mathematical explanations covered in Wiki articles, but do not seem to understand how to identify an algebraic nonlinear example other than variable cannot be separated", superimposed," is "non-homogenous". I am seeking a basic explanation for rather young kids in a gifted physics program.
For example, is an exponential, logarithm, root, or quadratic nonlinear as they do not conform to lines"? (Notwithstanding that I can just plot it against two logs to get a line.) And also, the above examples are predictable," no? Are they still nonlinear?
Does nonlinearity imply that despite knowing the initial conditions the outcome cannot be predicted?
(I solved the intersection of a quadratic and a linear equation, found two points, and am concluding the system is nonlinear" because the nonlinear" shape of x^2 (parabolic) causes the equation to be a system" of solutions (more than one point satisfies the bounds).
I am seeking an elementary school explanation and basic examples.
Thanks,
-E
2. Apr 12, 2010
### CRGreathouse
Re: What are everyday nonlinear" examples?
In my mind most things are nonlinear. Treating things (within a fixed range) as though they were linear is a mathematical trick that lets us work with complicated things as if they were simple.
Your examples are all simple, predictable, and nonlinear.
No, that's "chaotic", not "nonlinear". But multivariate nonlinear systems are often chaotic.
3. Apr 12, 2010
### UseAsDirected
Re: What are everyday nonlinear" examples?
Thank you for the response. I get it now that nonlinear systems can be simple and deterministic.
And, does sensitivity" to initial conditions imply chaos? I read about these buzzwords and that due to such and such sensitivity hither tither system is chaotic" or not-deterministic. Do I conflate the two, chaos and non-determinism? I am trying to build up a catalog of understanding. Thanks again.
-E
4. Apr 13, 2010
### IttyBittyBit
Re: What are everyday nonlinear" examples?
No, sensitivity to initial conditions is not enough for chaos. To have chaos, you also need to visit every point in the phase space arbitrarily closely an infinite number of times, at any large time frame (this is just a rough explanation; to really define chaos you need to get nitty gritty with higher mathematics). What does this mean? Consider the function x^n as n goes to infinity. This function is sensitive to initial conditions, since starting with 1 it will just give 1, but start at any number greater than 1, say 1.000000000001, and it will give infinity. So it's really sensitive to initial conditions. But it's not chaotic; it's simple and predictable.
The logistic map is a really simple (and good) introduction to chaos, you might want to read about it if you haven't already: http://en.wikipedia.org/wiki/Logistic_map
And non-determinism doesn't imply chaos either. A random (not pseudorandom!) sequence of numbers is non-deterministic, but non-chaotic.
5. Apr 13, 2010
### Staff: Mentor
Re: What are everyday nonlinear" examples?
The Mandelbrot set is determined by a very simple nonlinear equation: z = z2 + c, where z and c are complex numbers. I might be wrong, but I think that the Mandelbrot set is deterministic in the sense that given a complex number you can determine whether it is in the set or not, but very small changes in input values lead to very different outcomes, so the set is very sensitive to changes in initial conditions, hence chaotic. There's a nice animation about halfway down the page at http://en.wikipedia.org/wiki/Mandelbrot_set, in the section titled Zoom animation.
Another example of a simple, non-linear equation with chaotic behavior is in the Bifurcation topic here http://mathworld.wolfram.com/Bifurcation.html. The graph is generated by various values of r in the equation xn = rxn - 1(1 - xn - 1). If you look at the graph, the two left-most red lines are at r = 3.44 and x = .44, x = .85. Substituting .44 for x0 in the equation above gives x1 = 3.44*.44*(1 - .44) ~ .85. Substituting this value in the equation gives x1 = 3.44*.85(1 - .85) ~ .44. Varying r by a little bit causes a small variation in the output values, but varying r by a little more causes bifurcations at around r = 3.45, and more at around 3.545, but the system really goes bonkers at r = 3.57 or so.
6. Apr 13, 2010
### zhentil
Re: What are everyday nonlinear" examples?
Here's a way to think about it: suppose you're measuring the population of bunnies. You know that right now there are 100 bunnies and that the population is growing at the rate of four bunnies per week. If this data allows you to determine the population at all future times, the system is linear. Otherwise, it's nonlinear.
More generally, a system is linear if and only if knowledge of the function and its derivative at any point allows you to completely determine the function.
7. Apr 13, 2010
### Staff: Mentor
Re: What are everyday nonlinear" examples?
I think you are confusing linear with deterministic. The population growth of rabbits is NOT linear.
8. Apr 13, 2010
### UseAsDirected
Re: What are everyday nonlinear" examples?
Thanks for the responses.
To Ittybitty; hi, you wrote:
>This function is sensitive to initial conditions, since starting with 1 it will just give 1, but start >at any number greater than 1, say 1.000000000001, and it will give infinity. So it's really >sensitive to initial conditions. But it's not chaotic; it's simple and predictable.
What is sensitive" about it? I raised a larger number, 1.00001 to 50 and is 1.0005, 1.00001^500 = 1.005, and 1.00001^5,000 = 1.051.
It doesn't seem sensitive at all.
I have another, perhaps more pragmatic physics education questions.
Q.) Are their key *science* concepts to non-linearity? Or, is it just mathematics?
Thanks,
-E
9. Apr 13, 2010
### CRGreathouse
Re: What are everyday nonlinear" examples?
Here's an example of a reasonably chaotic (though deterministic) system I worked with a few years back.
Take a starting value, say 1. Repeatedly apply the tangent function until it is in a given range, say [319, 320).
10. Apr 13, 2010
### UseAsDirected
Re: What are everyday nonlinear" examples?
I just did it in Excel and it's quite cool!
So, it is chaotic yet ordered?
11. Apr 13, 2010
### Staff: Mentor
Re: What are everyday nonlinear" examples?
Ittybitty said x^n as n goes to infinity. 50, 500, and 5000 are insignificantly small in comparison to infinity.
12. Apr 13, 2010
### IttyBittyBit
Re: What are everyday nonlinear" examples?
Ok, first of all, every single one of you (Mark44, zhentil, UseAsDirected) are making the same, very grave, mistake: assuming that determinism and chaos are incompatible.
In fact, one of the defining marks of chaos is that it has to be deterministic.
let me repeat:
chaos is always deterministic.
In fact, that's why we call it deterministic chaos.
Non-deterministic 'chaos' is just randomness. And randomness is not chaos, it's just randomness. Like the throw of a dice or what age you will die at.
This is actually a common misconception.
Now another misconception is about linear vs. nonlinear systems. The problem arises from the fact that 'linear' is typically taken to loosely mean 'easy' in engineering courses. The word 'linear' does not have a mathematical definition (perhaps the closest thing to mathematical definition would be that a linear system is something that satisfies the superposition principle), but things like 'linear transforms on vector spaces' or 'systems of linear equations' do. Thus it is wrong, in my opinion, to make sweeping assertions about something being 'linear' or not; we have to study the concept in the context it is meant to be studied.
Yup.
13. Apr 13, 2010
### UseAsDirected
Re: What are everyday nonlinear" examples?
In this example, throwing of die is random? I think the fall of die is calculable, provided all the minutest details are known.
And, what is the relationship between determinism" and predictability"?
Thanks,
-E
14. Apr 14, 2010
### IttyBittyBit
Re: What are everyday nonlinear" examples?
Well it is possible for something to be deterministic, yet not possible for us to be able to predict it. In chaos, usually you have this process where, as you progress forwards in time, the exact details of the initial conditions become more and more important. Take the weather for example (a chaotic process). We have the equations to model it, it's just that we can never know all the initial conditions with 100% accuracy. Thus our prediction ability is limited to just a few days in the future. We can never hope to track every single child across the world blowing bubbles into the wind, for example.
You know about the butterfly effect right? That a single butterfly, flapping it's wings in, say, china, can lead to the difference between a hurricane striking or not striking a city on the coast of the US. The interesting thing about chaos is that this is guaranteed to happen; a flap of a butterfly's wings will, without a doubt, be translated into a storm being created or not.
But with randomness, the picture is different. If something is truly random, we can't even predict it on the shortest time-scale.
You are right, throwing a dice is not in fact random. It's actually very hard to construct a perfectly random sequence. Some people have done this with quantum devices, but even those are based upon the assumption that the quantum world is truly random, something that has not been proven.
15. Apr 14, 2010
### UseAsDirected
Re: What are everyday nonlinear" examples?
The butterfly's act of flapping is guaranteed to create a storm or not? Is or not" part a typo? Has it been shown experimentally that a butterfly's flapping guarantees the initiation of a storm? If we cannot micro-analyse the initial conditions of a system, how can we possibly demonstrate this?
Q.) What is the key scientific concept in non-linearity[, if there is one]? Or, does it belong to mathematics?
Thanks,
-E
16. Apr 14, 2010
### IttyBittyBit
Re: What are everyday nonlinear" examples?
No it's not a typo, I just explained it really badly.
What I meant to say is: To model the state of a system as it evolves in time, we eventually need to know finer and finer details about the initial conditions, with no limit to how far we have to go.
This has been proven with a mathematical analysis of the subject of chaos, btw. In fact, it is one of the defining characteristics of chaos. A system that 'forgives' perturbations in the initial conditions smaller than a certain threshold is not deemed chaotic.
About linear/nonlinear systems: you might want to start with researching the superposition principle.
17. Apr 14, 2010
### epenguin
Re: What are everyday nonlinear" examples?
The questions really are several, I will divide my answers.
Although to CRGreathouse's mind most things are nonlinear to my mind enough things are linear enough for linearity to be useful and essential. That is if they are a bit nonlinear, e.g. if the restoring force in an oscillation curves a bit you can still treat it acceptably depending on accuracy required as linear and get useful results. The concepts are still useful. The qualitative behaviour carries over. For instance even the simple pendulum never has a linear restoring force, it depends on sine of displacement not displacement. Nevertheless even when not quite right the period is still independent, exactly or approximately I don't remember, of maximum or initial displacement i.e. also amplitude. Therefore this is not essentially nonlinear. It is linear for small displacements still.
'Nonlinear science' - think its practitioners think this way - is when you have essentially qualitatively different behaviour from the linear. Thus the nonlinear version of the simple pendulum would be the grandfather clock. Its final period and trajectory or amplitude is independent of initial displacement - called a 'limit cycle'. That is a qualitative difference between the two dynamics.
Last edited: Apr 14, 2010
18. Apr 14, 2010
### epenguin
Re: What are everyday nonlinear" examples?
For students I would have recommended the system mentioned by Mark44. If you have just a hand graphics scientific calculator, and a very simple programme, they can have fun and surprises with that system.
A best short account and introduction to it in a few pages is probably still "Simple Mathematical Models with very complicated dynamics" by Robert May, Nature, 1976
The kind of diag. on the cover of the book below is key to throwing light on the strange behaviours. It is a look behind the iteration - I think you can see what it is. Here r is above 1 and below 3 I think and you see x homes in on a single stationary point. (For r above 2 it homes in in the oscillatory manner shown.)
One of the concepts to come out of chaos studies with this simple system was universality. That is Mark's illustration is with the simplest formula but it didn't much matter what the formula is as long as it has an extremum basically - it can even be a 'tent' - a straight line up and then down like ^. Even more surprising, chaos set on for the same ('universal') value of the controlling parameter (height of maximum) whatever the function chosen, and qualitatively the approach to it via period doublings was the same for different functions! Students can have fun with these too.
http://img576.imageshack.us/i/chaosholmgren311.png/ [Broken]
For books the above one by Holmgren is at least short. I found it disappointing - it told me everything I had already worked out and nothing I wanted to know - e.g. proper explanation of universality. The combination of things it expected you to know (topological terminology) and the elementary things it thought needed lengthy explanation were to me a bit disconcerting, but basically the math is elementary.
For the teacher a longer and wider and better book - but as I say longer - is 'Chaos and Fractals' by Pietgen Jurgens and Saupe. High school math (in Europe) is enough for it - there is more but it is explained.
Those 3 refs should keep you busy quite some time!
Last edited by a moderator: May 4, 2017
19. Apr 14, 2010
### epenguin
Re: What are everyday nonlinear" examples?
You ask for 'everyday examples' and can you identify chaos? I believe it is not easy to identify from observation whether a dynamic is really chaotic, but I will leave this to the experts. In the past engineering sought to avoid it, but these days some work on how to exploit it.
The examples often given e.g. above are so simple to enable tractability and not meant to represent reality but rather principle they are called 'toy models'.
But at least chaos and non-linearity are fairly new paradigms of dynamics. Before the seventies if they came up they tended to be swept under the carpet.
The irregular behaviour of a dripping tap in some conditions of flow is supposed to be chaos. You can at least suspect the erratic and intermittent (equally annoying) behaviour of an old fluorescent lighting tube is chaos. That behind some physical and mental pathologies where the patient is quite unpredictably well and not, or behind some economics phenomena can look like a chaotic dynamic. I would like to hear of better examples.
Last edited: Apr 14, 2010
20. Apr 14, 2010
### zhentil
Re: What are everyday nonlinear" examples?
Note the use of the conditional "if".
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http://umj.imath.kiev.ua/article/?lang=en&article=9624 | 2019
Том 71
№ 7
# Spectral exp nsion of self-conjoint operators on generalized elements of a Hilbert space
Kats G. I.
Abstract
The author discusses spectral expansions of generalized eigen-elements belonging to systems derived in the article (Ukrainian Mathematical Journal, vol. 12, N 1, 1960). The necessary and sufficient condition of the possibility of such expansions was obtained. Some generalizations are discussed, as well as a supplement to the estimate of the growth of eigen-functions and the interpretation of «eigen-subspaces» of self-adjoint operators.
Citation Example: Kats G. I. Spectral exp nsion of self-conjoint operators on generalized elements of a Hilbert space // Ukr. Mat. Zh. - 1961. - 13, № 4. - pp. 13-33.
Full text | {"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": 0.953271746635437, "perplexity": 1926.4852179694062}, "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/1566027321140.82/warc/CC-MAIN-20190824130424-20190824152424-00124.warc.gz"} |
http://math.stackexchange.com/questions/245557/maximize-a-sum | # Maximize a sum.
Let $m$ be a fixed positive integer. Consider two sequences of non-negative integers $a_1, a_2, ..., a_m$ and $b_1, b_2, ..., b_m$ which both taken together add up to $n$, that is, $\sum a_i + \sum b_i = n$, where $n$ is fixed positive integer.
Then what is the maximum value that the sum $\sum a_ib_i$ can attain?
Intuitively it seems that the answer should be $[\frac{n^2}{4}]$. I'm wondering how to prove it rigorously.
-
oh yes, thanks, I will remove it. – dineshdileep Nov 27 '12 at 9:42
## 2 Answers
Since $(a_i+a_j)(b_i+b_j)\ge a_ib_i+a_jb_j$, the sum never decreases when we replace $a_i$ by $a_i+a_j$, $b_i$ by $b_i+b_j$ and $a_j$ and $b_j$ by $0$. Thus we can assume $a_i=b_i=0$ for $i\ne1$, and then it's a simple one-dimensional optimization to show that $a_1b_1$ is maximal for $a_1=\lfloor n/2\rfloor$ and $b_1=\lceil n/2\rceil$, with maximal value $\lfloor n/2\rfloor\lceil n/2\rceil=\lfloor n^2/4\rfloor$.
-
a really good one – dineshdileep Nov 27 '12 at 10:03
Edit: The following answer does not take into account that the $a_i$, $b_i$ should be integers.
Your conjecture is correct. For a proof you can do without calculus:
Given the condition $a+b=C$ the product $$ab={1\over 4}\bigl((a+b)^2-(a-b)^2\bigr)={1\over 4}\bigl(C^2-(a-b)^2\bigr)$$ is largest when $a=b$. Therefore we should choose $a_i=b_i$ $\ (1\leq i\leq m)$.
The problem now is to maximize $Q:=\sum_{i=1}^m a_i^2$ under the condition $\sum_{i=1}^m a_i={\displaystyle{n\over2}}$.
When $a>0$, $a'>0$ then $$a^2+a'^2 <(a+a')^2 + 0\ .$$ It follows that replacing any pair $a_i>0$, $a_{i'}>0$ by the pair $a_i+a_{i'}$, $0$ will increase $Q$, so that $Q$ becomes maximal when $a_1={n\over2}$ and all other $a_i=0$. Therefore $Q_{\max}={n^2\over4}$ indeed.
-
What if $n$ is odd? – N N Nov 27 '12 at 9:56 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 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": 0.9937332272529602, "perplexity": 97.9814829889682}, "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-2016-26/segments/1466783403826.29/warc/CC-MAIN-20160624155003-00061-ip-10-164-35-72.ec2.internal.warc.gz"} |
http://blog.slcg.com/2012/02/mutual-fund-expense-analyzer-tool-for.html | ## Monday, February 27, 2012
### Mutual Fund Expense Analyzer: A Tool for Calculating Mutual Fund Fees and Expenses
By Geng Deng, PhD, FRM, Tim Husson, PhD and Olivia Wang, PhD
Every mutual fund investor should know how important fees and expenses are in determining the net return of his investment. Compared with other factors affecting a mutual fund’s or an Exchange Traded Fund (ETF)’s return, such as market returns, fees and expenses are more stable over time and it is therefore easier to predict their effect on a fund’s future performance. However, comparing fees and expenses across funds can be tedious and confusing, as different funds can use different fee structures.
Luckily, FINRA provides a free online tool which allows investors to compare the fees and expenses charged by over 18,000 mutual funds, ETFs and Exchange Traded Notes (ETNs). For example, we could choose three of the largest mutual funds, PIMCO Total Return Fund Institutional Class with ticker PTTRX, Fidelity Cash Reserves Fund with ticker FDRXX, and SPDR S&P 500 ETF with ticker SPY to compare their cost structures. The output is shown in Table 1:
Table 1
PIMCO Total Return Fund Institutional Class Fidelity Cash Reserves Fund SPDR S&P 500 ETF Data as of Data as of Data as of 2/14/2012 2/22/2012 n/a Ticker Symbol PTTRX FDRXX SPY Investment Amount $10,000.00$10,000.00 $10,000.00 Estimated Return You Selected 3.00% 3.00% 3.00% Holding Period 10 10 10 Fund Value After 10 Year(s)$12,834.99 $12,951.02$13,318.76 Profit/Loss $2,834.99$2,951.02 $3,318.76 Total Fees & Sales Charges$522.50 $422.25$104.22 Total Fees $522.50$422.25 $104.22 Total Sales Charges$0.00 $0.00$0.00
Several inputs are required when generating this table. Here, we assume the initial investment amount is \$10,000 and the investment horizon is ten years for all three funds. Also, we assume the annual return is 3% for all three funds. This assumption of uniform returns across the funds is not realistic, but it serves the purpose of making the cost structure more transparent.
This table makes clear the advantage of ETFs in terms of fees and expenses over conventional mutual funds: the total fees and sales changes for PIMCO total return fund is about five time of that of the SPDR S&P 500 ETF. Indeed, another piece of information provided by this tool (not shown in the table) is the annual operating expenses for each fund: while the expense is 0.09% for SPDR S&P 500 ETF, it is 0.46% for PIMCO Total Return Fund Institutional Class. This tool also outputs a wide variety of other data, including Morningstar ratings, investment objectives, and average annual returns.
We are glad that FINRA is making an effort to help investors make well-informed decisions. Although fee structures are not the only factor an investor should consider when choosing a mutual fund or ETF, this tool at least provides an objective comparison of known costs. | {"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": 0.19259393215179443, "perplexity": 5914.054238598711}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-04/segments/1547583763149.45/warc/CC-MAIN-20190121050026-20190121072026-00579.warc.gz"} |
https://math.stackexchange.com/questions/2050705/the-formula-for-the-volumn-of-three-vectors-in-mathbbr4 | # The formula for the volumn of three vectors in $\mathbb{R}^4$
For two vectors $v_1,v_2 \in \mathbb{R}^3$,there is a formula for the area of parallelogram generated by $v_1,v_2$. That is given by the cross product $|v_1 \times v_2|$ or $|det(v_1,v_2,\frac{v_1\times v_2}{|v_1\times v_2|})|$.
I am wondering if such formula exists in $\mathbb{R}^4$.
More precisely, let $w_1,w_2,w_3\in \mathbb{R}^4$ be independent vectors in $\mathbb{R}^4$. Then what is the volumn of parallelepiped generated by $w_1,w_2,w_3$?
I suppose that if there is a vector $u_1\in \mathbb{R}^4$ such that $u_1 \perp w_i$ for $1\le i \le 3$ and $|u_1|=1$, then the volumn should be $|det(w_1,w_2,w_3,u_1)|$. But I couldn't find a formula for such $u_1$ using $w_1,w_2,w_3$. (Namely, I am trying to find some concept similar to the cross product in $\mathbb{R}^3$.)
Could you shed me a light on this?
Any comment would be highly appreciated.
• Also see for example this thread, or this. Dec 9 '16 at 5:37
The following works for $k$ vectors $u_1,u_2,\ldots,u_k\in\Bbb{R}^n$ ($k<n$). Let $M$ be the $k\times n$ matrix having those $k$ vectors as rows. Let $V$ be the (hyper)volume of the $k$-dimensional parallelotope with edges $u_1,u_2,\ldots$. Then $$V^2=\det MM^T.\qquad(*)$$
A way to see this is as follows. Any transformation of the type $u_i\mapsto u_i+cu_j$ leaves $V$ unchanged (think: a shearing map transforming a parallelopiped to another with same base and height). It also amount to premultiplying $M$ (and postmultiplying $M^T$) by an elementary matrix with determinant $1$ (leaving $\det MM^T$ unchanged). But, a sequence of such transformations performs Gram-Schmidt orthogonalization on the set $u_1,\ldots,u_k$. So if the formula $(*)$ holds for an orthogonal set of vectors, it holds for all vectors.
But, the entries of the matrix $MM^T$ are the inner products $(u_i,u_j)$. So when $u_1,u_2,\ldots,u_k$ are orthogonal, that matrix is diagonal with elements $||u_i||^2$. The claim follows.
• Thank you very much! May I ask one more? What is the precise definition of volumn of k-dimensional parallelotope gen'd by $\mu_1,\cdots,\mu_k$? I am wondering whether the following is right. Let $V$ be the subspace of $\mathbb{R}^n$ generated by $\mu_1,\cdots,\mu_k$. And choose an orthonormal basis ${w_{k+1},\cdots ,w_{n}}$ of $V^{\perp}$. Then I suppose the volumn of k-dimensional parallelotope gen'd by $\mu_1,\cdots,\mu_k$ is $det(\mu_1,\cdots,\mu_k,w_{k+1},\cdots ,w_{n})$. Right? Dec 9 '16 at 6:25 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.9790319800376892, "perplexity": 170.2550680289161}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-39/segments/1631780057733.53/warc/CC-MAIN-20210925172649-20210925202649-00018.warc.gz"} |
https://arxiv.org/list/nucl-th/1906?100 | # Nuclear Theory
## Authors and titles for Jun 2019
[ total of 166 entries: 1-25 | 26-50 | 51-75 | 76-100 | ... | 151-166 ]
[ showing 25 entries per page: fewer | more | all ]
[1]
Title: The structure of cold neutron star with a quark core within the MIT and NJL models
Comments: 18 pages, 4 figures, 4 tables Iranian J. Sci. Tech. A (2019) accepted for publication
Journal-ref: Iran J Sci Technol Trans Sci 43 (2019) 2691
Subjects: Nuclear Theory (nucl-th); High Energy Astrophysical Phenomena (astro-ph.HE); High Energy Physics - Phenomenology (hep-ph)
[2]
Title: Magnetic field in nuclear collisions at ultra high energies
Authors: V. A. Okorokov (National Research Nuclear University MEPhI)
Comments: 9 pages, 3 figures, 1 table
Journal-ref: Phys. 1, 183, 2019
Subjects: Nuclear Theory (nucl-th)
[3]
Title: The possibility of $^{14}$C cluster as a building block of medium mass nuclei
Journal-ref: Phys. Rev. C 101, 034304 (2020)
Subjects: Nuclear Theory (nucl-th); Nuclear Experiment (nucl-ex)
[4]
Title: An update on fine-tunings in the triple-alpha process
Comments: 9 pages, 2 figures, contribution to the EPJA topical issue on "The tower of effective (field) theories and the emergence of nuclear phenomena"
Journal-ref: Eur. Phys. J. A 56 (2020) 89
Subjects: Nuclear Theory (nucl-th); Solar and Stellar Astrophysics (astro-ph.SR); High Energy Physics - Phenomenology (hep-ph); High Energy Physics - Theory (hep-th)
[5]
Title: EoS from terrestrial experiments: static and dynamic polarizations of nuclear density
Comments: 13 pages, 6figures, To appear in the AIP Conference Proceedings of the Xiamen-CUSTIPEN Workshop on the EOS of Dense Neutron-Rich Matter in the Era of Gravitational Wave Astronomy (January 3 - 7, 2019, Xiamen, China
Subjects: Nuclear Theory (nucl-th)
[6]
Title: Shell model results for $^{47-58}$Ca isotopes in the $fp$, $fpg_{9/2}$ and $fpg_{9/2}d_{5/2}$ model spaces
Comments: Accepted in Journal of Physics G: Nuclear and Particle Physics (2020)
Journal-ref: J.Phys. G: Nucl. Part. Phys. 47 (2020) 065105 (13pp)
Subjects: Nuclear Theory (nucl-th); Nuclear Experiment (nucl-ex)
[7]
Title: Sound velocity in dense stellar matter with strangeness and compact stars
Comments: 10 pages, 5 figures, 1 table, review article to appear in Chinese Physics C
Journal-ref: Chinese Physics C 45, 055104 (2021)
Subjects: Nuclear Theory (nucl-th); High Energy Astrophysical Phenomena (astro-ph.HE); Solar and Stellar Astrophysics (astro-ph.SR); High Energy Physics - Phenomenology (hep-ph)
[8]
Title: Turning the nuclear energy density functional method into a proper effective field theory: reflections
Authors: R.J. Furnstahl
Comments: 14 pages, 2 figures. Contribution to the EPJA topical issue: "The tower of effective (field) theories and the emergence of nuclear phenomena". v2: Corrections and added references based on referee reports
Subjects: Nuclear Theory (nucl-th)
[9]
Title: Mapping the Phases of Quantum Chromodynamics with Beam Energy Scan
Comments: Review article on the physics of beam energy scan in relativistic nuclear collisions, updated version-2 matching the published paper [Physics Reports 853 (2020) pp. 1-87]
Journal-ref: Physics Reports 853 (2020) pp. 1-87
Subjects: Nuclear Theory (nucl-th); High Energy Physics - Lattice (hep-lat); High Energy Physics - Phenomenology (hep-ph); Nuclear Experiment (nucl-ex)
[10]
Title: Limiting fragmentation as an initial state probe in heavy ion collisions
Journal-ref: Phys. Rev. C 100, 054901 (2019)
Subjects: Nuclear Theory (nucl-th); High Energy Physics - Phenomenology (hep-ph)
[11]
Title: Adiabatic projection method with Euclidean time subspace projection
Comments: 13 pages, 5 figures, version accepted for publication in EJPA
Journal-ref: Eur. Phys. J. A (2019) 55: 144
Subjects: Nuclear Theory (nucl-th); High Energy Physics - Lattice (hep-lat)
[12]
Title: Analytical forms of wave function and form factors of deuteron
Authors: V.I. Zhaba
Comments: 12 pages, in Ukrainian, 3 figures, 4 tables
Journal-ref: Visnyk Lviv University. Series Physics. 56 (2019) 43-55
Subjects: Nuclear Theory (nucl-th)
[13]
Title: "Splitting" magnetic catalysis effect prevents vacuum superconductivity in strong magnetic fields
Authors: Gaoqing Cao
Journal-ref: Phys. Rev. D 100, 074024 (2019)
Subjects: Nuclear Theory (nucl-th); High Energy Physics - Phenomenology (hep-ph)
[14]
Title: Influence of initial-state momentum anisotropy on the final-state collectivity in small collision systems
Journal-ref: Phys. Rev. C 100, 064905 (2019)
Subjects: Nuclear Theory (nucl-th); High Energy Physics - Phenomenology (hep-ph); Nuclear Experiment (nucl-ex)
[15]
Title: Diffusion of charm quarks in jets in high-energy heavy-ion collisions
Journal-ref: Eur.Phys.J. C79 (2019) no.9, 789
Subjects: Nuclear Theory (nucl-th); High Energy Physics - Theory (hep-th)
[16]
Title: Neutron matter at the interface(s): static response and effective mass
Comments: 12 pages, 8 figures; v2 corresponds to published version
Journal-ref: Eur. Phys. J. A 56, 112 (2020)
Subjects: Nuclear Theory (nucl-th); High Energy Astrophysical Phenomena (astro-ph.HE); Quantum Gases (cond-mat.quant-gas)
[17]
Title: Superfluid Phase Transitions and Effects of Thermal Pairing Fluctuations in Asymmetric Nuclear Matter
Journal-ref: Sci Rep 9, 18477 (2019)
Subjects: Nuclear Theory (nucl-th); High Energy Astrophysical Phenomena (astro-ph.HE); Quantum Gases (cond-mat.quant-gas); Superconductivity (cond-mat.supr-con)
[18]
Title: Renormalizing random-phase approximation by using exact pairing
Comments: 26 pages and 10 figures. Accepted in Physical Review C
Subjects: Nuclear Theory (nucl-th)
[19]
Title: Near-threshold $η^\prime$ meson production in ${π^-}A$ reactions
Authors: E.Ya.Paryev
Journal-ref: Nucl.Phys.A 988, 24 (2019)
Subjects: Nuclear Theory (nucl-th); Nuclear Experiment (nucl-ex)
[20]
Title: Absence of hindrance in microscopic $^{12}$C+$^{12}$C fusion study
Journal-ref: Phys. Rev. C 100, 024619 (2019)
Subjects: Nuclear Theory (nucl-th)
[21]
Title: Light (anti)nuclei production in Cu+Cu collisions at $\sqrt{s_{\rm{NN}}}=200$GeV
Journal-ref: Eur. Phys. J. A (2019) 55: 160
Subjects: Nuclear Theory (nucl-th); Nuclear Experiment (nucl-ex)
[22]
Title: Finite size effects on cumulants of the critical mode
Authors: Moussa Agah Nouhou (SUBATECH, Nantes), Marcus Bluhm (Wroclaw U. & SUBATECH, Nantes), Anna Borer, Marlene Nahrgang, Taklit Sami, Nathan Touroux (SUBATECH, Nantes)
Comments: Proceedings for the 12th International Conference on The Critical Point and Onset of Deconfinement (CPOD 2018), September 24 - 28, 2018, Mon-Repos, Corfu Island, Greece
Subjects: Nuclear Theory (nucl-th); High Energy Physics - Phenomenology (hep-ph)
[23]
Title: New method of analytic continuation of elastic-scattering data to the negative-energy region and asymptotic normalization coefficients for $^{17}$O and $^{13}$C
Journal-ref: Phys. Rev. C 100, 024627 (2019)
Subjects: Nuclear Theory (nucl-th)
[24]
Title: Properties of isospin asymmetric matter derived from chiral effective field theory
Title: $p_T$-Dependent Particle Number Fluctuations From Principal Component Analyses in Hydrodynamic Simulations of Heavy-Ion Collisions | {"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": 0.8997360467910767, "perplexity": 14585.15375761807}, "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-2021-49/segments/1637964362589.37/warc/CC-MAIN-20211203030522-20211203060522-00346.warc.gz"} |
https://mathoverflow.net/users/12039/thomas-connor?tab=topactivity | Thomas Connor
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7 A rank 3 geometry for the sporadic simple group of Suzuki 5 Geometric interpretation of $BN$-pairs 5 Proof of an 'easy' exercise in a book of Tits 4 Lists of small groups 4 What is the automorphism group of this geometry?
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7 A rank 3 geometry for the sporadic simple group of Suzuki 2 Ask for recommendations for textbook on mathematical logic 1 Which mathematical ideas have done most to change history? 1 Find elements $\rho_i$ such that $H < B : [\langle H, \rho_i \rangle : H ] = 2$
### Tags (11)
8 gr.group-theory × 8 1 geometry × 3 7 incidence-geometry × 4 1 involutions × 2 7 sporadic-groups × 2 0 buildings × 2 2 books 0 intuition 2 lo.logic 0 soft-question
### Bookmarks (5)
43 Generating finite simple groups with $2$ elements 16 How to find more (finite almost simple) groups with a given Sylow subgroup 8 Maximal number of directed edges in suitable simple graphs on $n$ vertices without directed triangles. 3 cosets question 0 Find elements $\rho_i$ such that $H < B : [\langle H, \rho_i \rangle : H ] = 2$ | {"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": 0.480582058429718, "perplexity": 1541.5242099431057}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-25/segments/1623487612537.23/warc/CC-MAIN-20210614135913-20210614165913-00302.warc.gz"} |
https://mathematics21.org/2016/06/23/about-each-regular-paratopological-group-is-completely-regular-article/ | # About “Each regular paratopological group is completely regular” article
In this blog post I consider my attempt to rewrite the article “Each regular paratopological group is completely regular” by Taras Banakh, Alex Ravsky in a more abstract way using my theory of reloids and funcoids.
The following is a general comment about reloids and funcoids as defined in my book. If you don’t understand them, restrict your mind to the special case ${f}$ to be a quasi-uniform space and ${(\mathsf{FCD}) f}$ is the corresponding quasi-proximity.
${\langle f \rangle^{\ast}}$ is the closure operator corresponding to a funcoid ${f}$. I also denote the image ${\mathscr{P} X \rightarrow \mathscr{P} Y}$ of a function ${f : X \rightarrow Y}$ as ${\langle f \rangle^{\ast}}$.
I will also denote ${f^{\circ}}$ the interior funcoid for a co-complete funcoid ${f}$ (for the special case if ${f}$ is a topological space ${f^{\circ}}$ is the interior operator of this space). It is defined in the file addons.pdf (not yet in my book).
By definition (slightly generalizing the special case if ${f}$ is a quasi-uniform space) an endo-reloid ${f}$ on a set ${U}$ is normal when ${\langle (\mathsf{FCD}) f \rangle^{\ast} A \sqsubseteq \langle ((\mathsf{FCD}) f)^{\circ} \rangle^{\ast} \langle (\mathsf{FCD}) f \rangle^{\ast} \langle F \rangle^{\ast} A}$ for every entourage ${F \in \mathrm{up}\, f}$ of ${f}$ and every set ${A \subseteq U}$.
Then it appear “obvious” that this definition of normality is equivalent to the formula:
$\displaystyle (\mathsf{FCD}) f \sqsubseteq ((\mathsf{FCD}) f)^{\circ} \circ (\mathsf{FCD}) f \circ (\mathsf{FCD}) f.$
However, I have failed to prove it. Here is my attempt
$\langle ((\mathsf{FCD}) f)^{\circ} \rangle^{\ast} \langle (\mathsf{FCD}) f \rangle^{\ast} \langle (\mathsf{FCD}) f \rangle^{\ast} A = \\ \langle ((\mathsf{FCD}) f)^{\circ} \rangle^{\ast} \langle (\mathsf{FCD}) f \rangle^{\ast} \bigsqcap_{F \in \mathrm{up} f} \langle F \rangle^{\ast} A = \\ \langle ((\mathsf{FCD}) f)^{\circ} \rangle^{\ast} \bigsqcap_{F \in \mathrm{up} f} \langle (\mathsf{FCD}) f \rangle^{\ast} \langle F \rangle^{\ast} A = ? ?$
The further step fails because in general ${\langle ((\mathsf{FCD}) f)^{\circ} \rangle^{\ast} \bigsqcap_{F \in \mathrm{up} f} S \neq \bigsqcap_{F \in \mathrm{up} f} \langle \langle ((\mathsf{FCD}) f)^{\circ} \rangle^{\ast} \rangle^{\ast} S}$.
So as now my attempt has failed. Please give me advice how to overcome this shortcoming of my theory. | {"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": 42, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9535009264945984, "perplexity": 395.22533676759065}, "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/1600400245109.69/warc/CC-MAIN-20200926200523-20200926230523-00026.warc.gz"} |
https://sources.gentoo.org/proj/qemu-kvm.git/tree/pc-bios/openbios-ppc?h=qemu-kvm-0.12.4-gentoo&id=b0293e54af32bcf08f4b4e5b4b472534de343972 | summaryrefslogtreecommitdiff log msg author committer range
blob: 3941c0bf5e40cf6ffc29225e2b5b10044000ed16 (plain)
ofshex dumpascii
0000 7f 45 4c 46 01 02 01 00 00 00 00 00 00 00 00 00 00 02 00 14 00 00 00 01 ff f0 80 00 00 00 00 34 .ELF...........................4
0020 00 04 81 44 00 00 00 00 00 34 00 20 00 03 00 28 00 0a 00 09 00 00 00 01 00 00 00 94 ff f0 00 00 ...D.....4.....(................
0040 ff f0 00 00 00 04 80 48 00 05 2c 88 00 00 00 07 00 00 00 04 00 00 00 01 00 04 80 dc ff ff ff fc .......H..,.....................
0060 ff ff ff fc 00 00 00 04 00 00 00 04 00 00 00 05 00 00 00 01 64 74 e5 51 00 00 00 00 00 00 00 00 ....................dt.Q........
0080 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 07 00 00 00 04 60 00 00 00 60 00 00 00 4b ff ff fc ....................`...`...K...
00a0 38 21 00 10 80 01 00 34 7c 08 03 a6 80 01 00 38 7c 0f f1 20 80 01 00 3c 7c 09 03 a6 80 01 00 40 8!.....4|......8|......<|......@
00c0 7c 01 03 a6 80 01 00 00 80 41 00 08 80 61 00 0c 80 81 00 10 80 a1 00 14 80 c1 00 18 80 e1 00 1c |........A...a..................
00e0 81 01 00 20 81 21 00 24 81 41 00 28 81 61 00 2c 81 81 00 30 80 21 00 04 4c 00 00 64 7c 68 02 a6 .....!.\$.A.(.a.,...0.!..L..d|h..
0100 48 00 86 dc 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 H...............................
0120 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
0140 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
0160 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
0180 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 48 00 22 78 00 00 00 00 00 00 00 00 ....................H."x........
01a0 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
01c0 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
01e0 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
0200 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
0220 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
0240 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
0260 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
0280 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 4b ff fe 69 00 00 00 00 00 00 00 00 ....................K..i........
02a0 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
02c0 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
02e0 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
0300 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
0320 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
0340 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
0360 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
0380 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 7c 31 43 a6 7c 20 00 a6 54 21 00 3e ....................|1C.|...T!.>
03a0 7c 20 01 24 7c 30 42 a6 38 21 ff b0 90 01 00 00 7c 11 42 a6 90 01 00 04 90 41 00 08 90 61 00 0c |..\$|0B.8!......|.B......A...a..
03c0 90 81 00 10 90 a1 00 14 90 c1 00 18 90 e1 00 1c 91 01 00 20 91 21 00 24 91 41 00 28 91 61 00 2c .....................!.\$.A.(.a.,
03e0 91 81 00 30 7c 08 02 a6 90 01 00 34 7c 00 00 26 90 01 00 38 7c 09 02 a6 90 01 00 3c 7c 01 02 a6 ...0|......4|..&...8|......<|...
0400 90 01 00 40 38 21 ff f0 3c 60 ff f1 38 63 84 f8 7c 69 03 a6 4e 80 04 21 4b f0 00 0e 00 00 00 00 ...@8!..<`..8c..|i..N..!K.......
0420 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
0440 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
0460 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
0480 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 7c 31 43 a6 7c 20 00 a6 54 21 00 3e ....................|1C.|...T!.>
04a0 7c 20 01 24 7c 30 42 a6 38 21 ff b0 90 01 00 00 7c 11 42 a6 90 01 00 04 90 41 00 08 90 61 00 0c |..\$|0B.8!......|.B......A...a..
04c0 90 81 00 10 90 a1 00 14 90 c1 00 18 90 e1 00 1c 91 01 00 20 91 21 00 24 91 41 00 28 91 61 00 2c .....................!.\$.A.(.a.,
04e0 91 81 00 30 7c 08 02 a6 90 01 00 34 7c 00 00 26 90 01 00 38 7c 09 02 a6 90 01 00 3c 7c 01 02 a6 ...0|......4|..&...8|......<|...
0500 90 01 00 40 38 21 ff f0 3c 60 ff f1 38 63 84 64 7c 69 03 a6 4e 80 04 21 4b f0 00 0e 00 00 00 00 ...@8!..<`..8c.d|i..N..!K.......
0520 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
0540 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
0560 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
0580 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 4b ff fb 69 00 00 00 00 00 00 00 00 ....................K..i........
05a0 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
05c0 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
05e0 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
0600 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
0620 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
0640 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
0660 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
0680 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 4b ff fa 69 00 00 00 00 00 00 00 00 ....................K..i........
06a0 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
06c0 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
06e0 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
0700 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
0720 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
0740 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
0760 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
0780 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 4b ff f9 69 00 00 00 00 00 00 00 00 ....................K..i........
07a0 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
07c0 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
07e0 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
0800 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
0820 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
0840 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
0860 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
0880 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 7c 71 43 a6 7c 7b 02 a6 60 63 20 00 ....................|qC.|{..`c..
08a0 7c 7b 03 a6 7c 71 42 a6 4c 00 00 64 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 |{..|qB.L..d....................
08c0 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
08e0 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
0900 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
0920 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
0940 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
0960 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
0980 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 4b ff f7 69 00 00 00 00 00 00 00 00 ....................K..i........
09a0 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
09c0 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
09e0 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
0a00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
0a20 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
0a40 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
0a60 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
0a80 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 4b ff f6 69 00 00 00 00 00 00 00 00 ....................K..i........
0aa0 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
0ac0 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
0ae0 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
0b00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
0b20 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
0b40 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
0b60 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
0b80 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 4b ff f5 69 00 00 00 00 00 00 00 00 ....................K..i........
0ba0 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
0bc0 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
0be0 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
0c00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
0c20 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
0c40 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
0c60 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
0c80 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 4b ff f4 69 00 00 00 00 00 00 00 00 ....................K..i........
0ca0 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
0cc0 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
0ce0 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
0d00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
0d20 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
0d40 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
0d60 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
0d80 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 4b ff f3 69 00 00 00 00 00 00 00 00 ....................K..i........
0da0 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
0dc0 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
0de0 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
0e00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
0e20 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
0e40 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
0e60 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
0e80 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 4b ff f2 69 00 00 00 00 00 00 00 00 ....................K..i........
0ea0 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
0ec0 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
0ee0 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
0f00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
0f20 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
0f40 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
0f60 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
0f80 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 4b ff f1 69 00 00 00 00 00 00 00 00 ....................K..i........
0fa0 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 4b ff f1 49 00 00 00 00 00 00 00 00 ....................K..I........
0fc0 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
0fe0 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
1000 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
1020 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
1040 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
1060 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
1080 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 4b ff f0 69 00 00 00 00 00 00 00 00 ....................K..i........
10a0 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
10c0 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
10e0 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
1100 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
1120 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
1140 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
1160 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
1180 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 4b ff ef 69 00 00 00 00 00 00 00 00 ....................K..i........
11a0 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
11c0 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
11e0 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
1200 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
1220 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
1240 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
1260 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
1280 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 4b ff ee 69 00 00 00 00 00 00 00 00 ....................K..i........
12a0 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
12c0 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
12e0 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
1300 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
1320 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
1340 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
1360 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
1380 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 4b ff ed 69 00 00 00 00 00 00 00 00 ....................K..i........
13a0 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
13c0 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
13e0 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
1400 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
1420 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
1440 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
1460 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
1480 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 4b ff ec 69 00 00 00 00 00 00 00 00 ....................K..i........
14a0 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
14c0 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
14e0 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
1500 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
1520 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
1540 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
1560 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
1580 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 4b ff eb 69 00 00 00 00 00 00 00 00 ....................K..i........
15a0 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
15c0 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
15e0 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
1600 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
1620 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
1640 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
1660 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
1680 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 4b ff ea 69 00 00 00 00 00 00 00 00 ....................K..i........
16a0 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
16c0 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
16e0 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
1700 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
1720 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
1740 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
1760 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
1780 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 4b ff e9 69 00 00 00 00 00 00 00 00 ....................K..i........
17a0 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
17c0 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
17e0 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
1800 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
1820 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
1840 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
1860 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
1880 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
18a0 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
18c0 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
18e0 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
1900 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
1920 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
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17720 4e 80 00 20 3d 60 ff f5 38 00 ff ff 81 2b 90 18 39 29 00 01 55 2a 10 3a 91 2b 90 18 3d 20 ff f5 N...=`..8....+..9)..U*.:.+..=...
17740 39 29 9c 88 7c 09 51 2e 4e 80 00 20 4e 80 00 20 3d 60 ff f5 38 00 ff ff 81 2b 90 18 39 29 00 01 9)..|.Q.N...N...=`..8....+..9)..
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177e0 3d 20 ff f4 3b 49 61 18 7f 43 d3 78 4b ff 5d ad 7c 7d 1b 78 7f 63 db 78 48 00 2b 99 3c 80 ff f4 =...;Ia..C.xK.].|}.x.c.xH.+.<...
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17860 3b 81 00 48 3c a0 ff f4 7f 66 db 78 38 a5 70 28 38 80 00 40 7f 83 e3 78 48 00 3b 0d 38 e0 00 01 ;..H<....f.x8.p(8..@...xH.;.8...
17880 38 de 00 28 38 80 00 04 7f 85 e3 78 38 60 00 01 3f a0 ff f4 4b ff 65 31 7f 83 e3 78 3b bd 70 30 8..(8......x8`..?...K.e1...x;.p0
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178c0 38 84 51 08 7f a5 eb 78 38 c0 00 04 7f 63 db 78 4b ff 5e fd 7f 43 d3 78 4b ff 5c c1 7c 7b 1b 78 8.Q....x8....c.xK.^..C.xK.\.|{.x
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17900 93 e9 9b 8c 80 01 00 a4 7f e3 fb 78 bb 41 00 88 38 21 00 a0 7c 08 03 a6 4e 80 00 20 94 21 ff f0 ...........x.A..8!..|...N....!..
17920 3c 80 ff f5 7c 08 02 a6 3c 60 ff f4 38 84 9b 8c 38 63 70 34 38 84 00 04 90 01 00 14 4b ff 55 89 <...|...<`..8...8cp48.......K.U.
17940 80 01 00 14 38 21 00 10 7c 08 03 a6 4e 80 00 20 94 21 ff f0 3d 40 ff f5 7c 08 02 a6 3d 20 ff f5 ....8!..|...N....!..=@..|...=...
17960 3c 80 ff f5 90 01 00 14 39 29 9c 88 38 00 00 00 3c 60 ff f4 38 84 9b 8c 38 63 5f 6c 38 84 00 08 <.......9)..8...<`..8...8c_l8...
17980 81 6a 90 18 39 6b 00 01 55 68 10 3a 91 6a 90 18 7c 09 41 2e 4b ff 55 31 80 01 00 14 38 21 00 10 .j..9k..Uh.:.j..|.A.K.U1....8!..
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179c0 3c 60 ff f4 38 63 70 1c 4b ff 5b 5d 3c 80 ff f5 3c 60 ff f4 38 84 9b 8c 38 63 4f 10 38 84 00 0c <`..8cp.K.[]<...<`..8...8cO.8...
179e0 4b ff 54 e5 3c 80 ff f4 7f a3 eb 78 38 84 60 c4 38 a0 00 01 4b ff 5e 81 3c 80 ff f4 7f a3 eb 78 K.T.<......x8.`.8...K.^.<......x
17a00 38 84 60 d4 38 a0 00 00 4b ff 5e 6d 3c 80 ff f4 3c a0 ff f4 38 84 51 08 38 a5 70 20 7f a3 eb 78 8.`.8...K.^m<...<...8.Q.8.p....x
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17a60 38 a0 00 12 4b ff 5e 11 4b ff 0d 45 2f 83 00 00 41 9e 00 18 3c 80 ff f4 7f a3 eb 78 38 84 60 18 8...K.^.K..E/...A...<......x8.`.
17a80 38 a0 00 12 4b ff 5d f1 3c 80 ff f1 3c 60 ff f4 38 84 76 10 38 63 70 44 4b ff 61 4d 3c 60 ff f4 8...K.].<...<`..8.v.8cpDK.aM<`..
17aa0 3f a0 ff f5 38 63 70 54 3b bd 9b 8c 4b ff 5a 79 3c 60 ff f4 38 9d 00 10 38 63 70 74 4b ff 54 09 ?...8cpT;...K.Zy<`..8...8cptK.T.
17ac0 3d 00 ff f5 3d 20 ff f5 39 29 9c 88 38 00 00 00 3c 60 ff f4 38 9d 00 14 81 68 90 18 38 63 59 74 =...=...9)..8...<`..8....h..8cYt
17ae0 39 6b 00 01 55 6a 10 3a 91 68 90 18 7c 09 51 2e 4b ff 53 d5 3c 60 ff f4 3c 80 ff f1 38 63 70 7c 9k..Uj.:.h..|.Q.K.S.<`..<...8cp|
17b00 38 84 76 50 4b ff 60 e1 80 01 00 34 bb a1 00 24 38 21 00 30 7c 08 03 a6 4e 80 00 20 94 21 ff e0 8.vPK.`....4...\$8!.0|...N....!..
17b20 7c 08 02 a6 90 01 00 24 7c a0 2b 78 7c 85 23 78 38 80 00 00 bf c1 00 18 38 e1 00 08 7c de 33 78 |......\$|.+x|.#x8.......8...|.3x
17b40 7c 06 03 78 4b ff f9 f1 2f 83 00 01 40 9d 00 18 88 01 00 08 3b e3 ff fe 38 81 00 0a 2f 80 00 00 |..xK.../...@.......;...8.../...
17b60 41 9e 00 0c 38 81 00 09 3b e0 ff ff 7f c3 f3 78 7f e5 fb 78 48 00 2a 75 80 01 00 24 7f e3 fb 78 A...8...;......x...xH.*u...\$...x
17b80 bb c1 00 18 38 21 00 20 7c 08 03 a6 4e 80 00 20 3d 60 ff f5 81 2b 90 00 98 69 03 c4 7c 00 06 ac ....8!..|...N...=`...+...i..|...
17ba0 81 2b 90 00 39 29 03 c5 88 69 00 00 7c 00 06 ac 54 63 06 3e 4e 80 00 20 3d 60 ff f5 81 2b 90 00 .+..9)...i..|...Tc.>N...=`...+..
17bc0 98 69 03 ce 7c 00 06 ac 81 2b 90 00 39 29 03 cf 88 69 00 00 7c 00 06 ac 54 63 06 3e 4e 80 00 20 .i..|....+..9)...i..|...Tc.>N...
17be0 3d 60 ff f5 81 2b 90 00 98 69 03 d4 7c 00 06 ac 81 2b 90 00 39 29 03 d5 88 69 00 00 7c 00 06 ac =`...+...i..|....+..9)...i..|...
17c00 54 63 06 3e 4e 80 00 20 3d 40 ff f5 81 6a 90 00 39 2b 03 da 88 09 00 00 7c 00 06 ac 39 2b 00 80 Tc.>N...=@...j..9+......|...9+..
17c20 88 09 00 00 7c 00 06 ac 98 6b 03 c0 7c 00 06 ac 81 2a 90 00 39 29 03 c1 88 69 00 00 7c 00 06 ac ....|....k..|....*..9)...i..|...
17c40 54 63 06 3e 4e 80 00 20 94 21 ff 90 38 80 00 4f 38 60 00 1f 38 00 00 5f 3d 20 ff f4 39 40 00 00 Tc.>N....!..8..O8`..8.._=...9@..
17c60 98 81 00 19 98 81 00 1a 39 29 35 0c 98 61 00 1f 99 41 00 20 98 81 00 21 99 41 00 22 98 61 00 2c ........9)5..a...A.....!.A.".a.,
17c80 98 01 00 18 38 00 ff 82 98 01 00 1b 99 41 00 24 38 00 00 52 98 01 00 1c 99 41 00 25 38 00 ff be ....8........A.\$8..R.....A.%8...
17ca0 98 01 00 1d 99 41 00 26 38 00 00 0f 98 01 00 23 99 41 00 27 38 00 ff 97 39 40 00 00 98 01 00 28 .....A.&8......#.A.'8...9@.....(
17cc0 38 00 00 09 98 01 00 29 38 00 00 28 98 01 00 2b 38 00 ff a3 bf 81 00 60 98 01 00 2f 3b 80 ff bf 8......)8..(...+8......`.../;...
17ce0 3b a0 ff 8f 9b 81 00 1e 9b a1 00 2a 38 00 ff ff 39 61 00 08 7c a9 84 aa 7c ab 85 aa 9b a1 00 2d ;..........*8...9a..|...|......-
17d00 9b 81 00 2e 98 01 00 30 38 00 00 10 39 01 00 18 7c 09 03 a6 7d 21 52 14 7d 68 52 14 39 4a 00 01 .......08...9...|...}!R.}hR.9J..
17d20 88 09 00 08 98 0b 00 19 42 00 ff ec 38 00 00 0c 3d 60 ff f5 99 41 00 4b 98 01 00 41 38 00 00 0f ........B...8...=`...A.K...A8...
17d40 39 20 00 00 98 01 00 43 99 21 00 42 38 00 00 0e 38 cb 90 00 98 01 00 4c 99 21 00 4d 38 00 ff ff 9......C.!.B8...8......L.!.M8...
17d60 98 01 00 4e 99 21 00 44 38 00 00 01 98 01 00 50 99 21 00 45 38 00 00 03 98 01 00 51 99 21 00 52 ...N.!.D8......P.!.E8......Q.!.R
17d80 38 00 00 02 99 21 00 46 99 21 00 47 99 21 00 48 99 21 00 49 99 21 00 4a 98 01 00 53 38 00 00 67 8....!.F.!.G.!.H.!.I.!.J...S8..g
17da0 81 2b 90 00 98 09 03 c2 7c 00 06 ac 39 40 00 00 81 2b 90 00 99 49 03 c4 7c 00 06 ac 81 2b 90 00 .+......|...9@...+...I..|....+..
17dc0 99 49 03 c5 7c 00 06 ac 38 00 00 01 81 2b 90 00 98 09 03 c4 7c 00 06 ac 88 a1 00 50 81 2b 90 00 .I..|...8....+......|......P.+..
17de0 60 a0 00 20 98 09 03 c5 7c 00 06 ac 38 e0 00 02 81 2b 90 00 98 e9 03 c4 7c 00 06 ac 81 2b 90 00 `.......|...8....+......|....+..
17e00 88 01 00 51 98 09 03 c5 7c 00 06 ac 39 00 00 03 81 2b 90 00 99 09 03 c4 7c 00 06 ac 81 2b 90 00 ...Q....|...9....+......|....+..
17e20 88 01 00 52 98 09 03 c5 7c 00 06 ac 38 00 00 04 81 2b 90 00 98 09 03 c4 7c 00 06 ac 81 2b 90 00 ...R....|...8....+......|....+..
17e40 98 e9 03 c5 7c 00 06 ac 81 2b 90 00 99 49 03 c4 7c 00 06 ac 81 2b 90 00 99 09 03 c5 7c 00 06 ac ....|....+...I..|....+......|...
17e60 38 00 00 11 81 2b 90 00 98 09 03 d4 7c 00 06 ac 81 2b 90 00 88 01 00 29 98 09 03 d5 7c 00 06 ac 8....+......|....+.....)....|...
17e80 38 00 00 19 39 60 00 00 7c 09 03 a6 39 41 00 18 81 26 00 00 99 69 03 d4 7c 00 06 ac 81 26 00 00 8...9`..|...9A...&...i..|....&..
17ea0 7c 0a 58 ae 98 09 03 d5 7c 00 06 ac 39 6b 00 01 42 00 ff e0 38 00 00 09 3d 20 ff f5 7c 09 03 a6 |.X.....|...9k..B...8...=...|...
17ec0 39 09 90 00 39 40 00 00 38 e1 00 18 81 28 00 00 99 49 03 ce 7c 00 06 ac 7d 27 52 14 81 68 00 00 9...9@..8....(...I..|...}'R..h..
17ee0 88 09 00 2e 98 0b 03 cf 7c 00 06 ac 39 4a 00 01 42 00 ff dc 38 00 00 15 3d 20 ff f5 7c 09 03 a6 ........|...9J..B...8...=...|...
17f00 39 09 90 00 39 40 00 00 38 e1 00 18 81 68 00 00 39 2b 03 da 88 09 00 00 7c 00 06 ac 39 2b 00 80 9...9@..8....h..9+......|...9+..
17f20 88 09 00 00 7c 00 06 ac 99 4b 03 c0 7c 00 06 ac 81 68 00 00 39 2b 00 80 88 09 00 00 7c 00 06 ac ....|....K..|....h..9+......|...
17f40 7d 27 52 14 88 09 00 19 98 0b 03 c0 7c 00 06 ac 81 68 00 00 39 2b 00 80 88 09 00 00 7c 00 06 ac }'R.........|....h..9+......|...
17f60 39 4a 00 01 42 00 ff a8 38 00 00 00 98 0b 03 c8 7c 00 06 ac 3d 20 ff f4 38 00 00 40 3d 00 ff f5 9J..B...8.......|...=...8..@=...
17f80 39 29 35 0c 7c 09 03 a6 39 69 00 10 39 48 90 00 81 2a 00 00 88 0b 00 00 98 09 03 c9 7c 00 06 ac 9)5.|...9i..9H...*..........|...
17fa0 81 2a 00 00 88 0b 00 01 98 09 03 c9 7c 00 06 ac 81 2a 00 00 88 0b 00 02 98 09 03 c9 7c 00 06 ac .*..........|....*..........|...
17fc0 39 6b 00 03 42 00 ff cc 81 28 90 00 38 00 00 ff 98 09 03 c6 7c 00 06 ac 81 68 90 00 39 2b 03 da 9k..B....(..8.......|....h..9+..
17fe0 88 09 00 00 7c 00 06 ac 39 2b 00 80 88 09 00 00 7c 00 06 ac 39 40 00 20 99 4b 03 c0 7c 00 06 ac ....|...9+......|...9@...K..|...
18000 38 00 00 01 81 28 90 00 98 09 03 c4 7c 00 06 ac 81 28 90 00 98 a9 03 c5 7c 00 06 ac 81 68 90 00 8....(......|....(......|....h..
18020 39 2b 03 da 88 09 00 00 7c 00 06 ac 99 4b 03 c0 7c 00 06 ac bb 81 00 60 38 21 00 70 4e 80 00 20 9+......|....K..|......`8!.pN...
18040 94 21 ff e0 7c 08 02 a6 90 01 00 24 bf 41 00 08 3f a0 ff f5 7c 7a 1b 78 38 60 00 02 7c 9b 23 78 .!..|......\$.A..?...|z.x8`..|.#x
18060 7c bf 2b 78 7c de 33 78 4b ff fb 29 3b 80 00 02 81 3d 90 00 9b 89 03 c4 7c 00 06 ac 54 63 06 36 |.+x|.3xK..);....=......|...Tc.6
18080 81 3d 90 00 60 63 00 04 98 69 03 c5 7c 00 06 ac 38 60 00 04 4b ff fa fd 81 3d 90 00 38 00 00 04 .=..`c...i..|...8`..K....=..8...
180a0 98 09 03 c4 7c 00 06 ac 60 63 00 04 81 3d 90 00 54 63 06 3e 98 69 03 c5 7c 00 06 ac 81 3d 90 00 ....|...`c...=..Tc.>.i..|....=..
180c0 98 09 03 ce 7c 00 06 ac 81 3d 90 00 9b 89 03 cf 7c 00 06 ac 38 60 00 05 4b ff fa e1 81 3d 90 00 ....|....=......|...8`..K....=..
180e0 38 00 00 05 98 09 03 ce 7c 00 06 ac 70 63 00 ef 81 3d 90 00 98 69 03 cf 7c 00 06 ac 38 00 00 06 8.......|...pc...=...i..|...8...
18100 81 3d 90 00 98 09 03 ce 7c 00 06 ac 38 00 00 00 81 3d 90 00 98 09 03 cf 7c 00 06 ac 39 60 00 00 .=......|...8....=......|...9`..
18120 39 40 00 00 48 00 00 20 7c 1b 48 ae 7c 1a 49 ae 39 29 00 01 42 00 ff f4 39 6b 00 01 3b 7b 00 10 9@..H...|.H.|.I.9)..B...9k..;{..
18140 3b 5a 00 20 7f 8b f0 00 40 9c 00 30 38 1f 00 01 7f 8a f8 00 39 20 00 00 7c 09 03 a6 41 9d 00 10 ;Z......@..08.......9...|...A...
18160 3c 00 80 00 7f 9f 00 00 40 9e ff cc 38 00 00 01 7c 09 03 a6 4b ff ff c0 3f a0 ff f5 38 60 00 02 <.......@...8...|...K...?...8`..
18180 4b ff fa 11 38 00 00 02 81 3d 90 00 98 09 03 c4 7c 00 06 ac 54 63 06 36 81 3d 90 00 60 63 00 03 K...8....=......|...Tc.6.=..`c..
181a0 98 69 03 c5 7c 00 06 ac 38 60 00 04 4b ff f9 e5 81 3d 90 00 38 00 00 04 98 09 03 c4 7c 00 06 ac .i..|...8`..K....=..8.......|...
181c0 70 63 00 fb 81 3d 90 00 98 69 03 c5 7c 00 06 ac 81 3d 90 00 98 09 03 ce 7c 00 06 ac 38 00 00 00 pc...=...i..|....=......|...8...
181e0 81 3d 90 00 98 09 03 cf 7c 00 06 ac 38 60 00 05 4b ff f9 c9 81 3d 90 00 38 00 00 05 98 09 03 ce .=......|...8`..K....=..8.......
18200 7c 00 06 ac 60 63 00 10 81 3d 90 00 54 63 06 3e 98 69 03 cf 7c 00 06 ac 38 00 00 06 81 3d 90 00 |...`c...=..Tc.>.i..|...8....=..
18220 98 09 03 ce 7c 00 06 ac 38 00 00 0e 81 3d 90 00 98 09 03 cf 7c 00 06 ac 80 01 00 24 bb 41 00 08 ....|...8....=......|......\$.A..
18240 38 21 00 20 7c 08 03 a6 4e 80 00 20 94 21 ff e0 7c 08 02 a6 90 01 00 24 bf 81 00 10 3f a0 ff f5 8!..|...N....!..|......\$....?...
18260 81 7d 90 00 39 2b 03 da 88 09 00 00 7c 00 06 ac 39 2b 00 80 88 09 00 00 7c 00 06 ac 38 00 00 10 .}..9+......|...9+......|...8...
18280 98 0b 03 c0 7c 00 06 ac 81 7d 90 00 39 2b 00 80 88 09 00 00 7c 00 06 ac 38 00 00 0c 98 0b 03 c0 ....|....}..9+......|...8.......
182a0 7c 00 06 ac 81 5d 90 00 7d 49 53 78 8c 09 00 80 7c 00 06 ac 39 6a 03 da 88 0b 00 00 7c 00 06 ac |....]..}ISx....|...9j......|...
182c0 88 09 00 00 7c 00 06 ac 38 00 00 14 98 0a 03 c0 7c 00 06 ac 81 7d 90 00 39 2b 00 80 88 09 00 00 ....|...8.......|....}..9+......
182e0 7c 00 06 ac 3b 80 00 00 9b 8b 03 c0 7c 00 06 ac 81 7d 90 00 39 2b 00 80 88 09 00 00 7c 00 06 ac |...;.......|....}..9+......|...
18300 38 00 00 02 98 0b 03 c4 7c 00 06 ac 38 00 00 03 81 3d 90 00 98 09 03 c5 7c 00 06 ac 38 60 00 04 8.......|...8....=......|...8`..
18320 4b ff f8 71 81 3d 90 00 38 00 00 04 98 09 03 c4 7c 00 06 ac 70 63 00 fb 81 3d 90 00 98 69 03 c5 K..q.=..8.......|...pc...=...i..
18340 7c 00 06 ac 38 60 00 05 4b ff f8 71 81 3d 90 00 38 00 00 05 98 09 03 ce 7c 00 06 ac 70 63 00 8f |...8`..K..q.=..8.......|...pc..
18360 81 3d 90 00 60 63 00 10 98 69 03 cf 7c 00 06 ac 38 00 00 06 81 3d 90 00 98 09 03 ce 7c 00 06 ac .=..`c...i..|...8....=......|...
18380 38 00 00 0e 81 3d 90 00 98 09 03 cf 7c 00 06 ac 38 00 00 0a 81 3d 90 00 98 09 03 d4 7c 00 06 ac 8....=......|...8....=......|...
183a0 81 3d 90 00 9b 89 03 d5 7c 00 06 ac 38 00 00 0b 81 3d 90 00 98 09 03 d4 7c 00 06 ac 38 00 00 0f .=......|...8....=......|...8...
183c0 81 3d 90 00 98 09 03 d5 7c 00 06 ac 38 60 00 17 4b ff f8 11 81 3d 90 00 38 00 00 17 98 09 03 d4 .=......|...8`..K....=..8.......
183e0 7c 00 06 ac 54 63 06 fe 81 3d 90 00 60 63 00 a0 98 69 03 d5 7c 00 06 ac 38 60 00 09 4b ff f7 e5 |...Tc...=..`c...i..|...8`..K...
18400 81 3d 90 00 38 00 00 09 98 09 03 d4 7c 00 06 ac 54 63 06 34 81 3d 90 00 60 63 00 0f 98 69 03 d5 .=..8.......|...Tc.4.=..`c...i..
18420 7c 00 06 ac 81 7d 90 00 39 2b 03 da 88 09 00 00 7c 00 06 ac 38 00 00 20 98 0b 03 c0 7c 00 06 ac |....}..9+......|...8.......|...
18440 80 01 00 24 bb 81 00 10 38 21 00 20 7c 08 03 a6 4e 80 00 20 94 21 ff e0 38 60 00 10 7c 08 02 a6 ...\$....8!..|...N....!..8`..|...
18460 90 01 00 24 bf 81 00 10 4b ff f7 a1 3f a0 ff f5 81 7d 90 00 39 2b 03 da 88 09 00 00 7c 00 06 ac ...\$....K...?....}..9+......|...
18480 39 2b 00 80 88 09 00 00 7c 00 06 ac 38 00 00 10 98 0b 03 c0 7c 00 06 ac 81 7d 90 00 39 2b 00 80 9+......|...8.......|....}..9+..
184a0 88 09 00 00 7c 00 06 ac 54 63 06 36 60 63 00 01 98 6b 03 c0 7c 00 06 ac 81 3d 90 00 39 29 00 80 ....|...Tc.6`c...k..|....=..9)..
184c0 88 09 00 00 7c 00 06 ac 38 60 00 02 4b ff f6 c5 81 3d 90 00 38 00 00 02 98 09 03 c4 7c 00 06 ac ....|...8`..K....=..8.......|...
184e0 60 63 00 0f 81 3d 90 00 54 63 06 3e 98 69 03 c5 7c 00 06 ac 38 60 00 04 4b ff f6 99 81 3d 90 00 `c...=..Tc.>.i..|...8`..K....=..
18500 38 00 00 04 98 09 03 c4 7c 00 06 ac 60 63 00 04 81 3d 90 00 54 63 06 3e 98 69 03 c5 7c 00 06 ac 8.......|...`c...=..Tc.>.i..|...
18520 38 60 00 05 4b ff f6 95 81 3d 90 00 39 60 00 05 99 69 03 ce 7c 00 06 ac 70 63 00 ef 81 3d 90 00 8`..K....=..9`...i..|...pc...=..
18540 98 69 03 cf 7c 00 06 ac 38 00 00 06 81 3d 90 00 98 09 03 ce 7c 00 06 ac 81 3d 90 00 99 69 03 cf .i..|...8....=......|....=...i..
18560 7c 00 06 ac 38 00 00 0a 81 3d 90 00 98 09 03 d4 7c 00 06 ac 3b 80 00 20 81 3d 90 00 9b 89 03 d5 |...8....=......|...;....=......
18580 7c 00 06 ac 38 00 00 0b 81 3d 90 00 98 09 03 d4 7c 00 06 ac 38 00 00 00 81 3d 90 00 98 09 03 d5 |...8....=......|...8....=......
185a0 7c 00 06 ac 38 60 00 17 4b ff f6 39 81 3d 90 00 38 00 00 17 98 09 03 d4 7c 00 06 ac 54 63 06 fe |...8`..K..9.=..8.......|...Tc..
185c0 81 3d 90 00 60 63 00 e0 98 69 03 d5 7c 00 06 ac 38 60 00 09 4b ff f6 0d 81 3d 90 00 38 00 00 09 .=..`c...i..|...8`..K....=..8...
185e0 98 09 03 d4 7c 00 06 ac 54 63 06 34 81 3d 90 00 98 69 03 d5 7c 00 06 ac 81 7d 90 00 39 2b 03 cc ....|...Tc.4.=...i..|....}..9+..
18600 88 09 00 00 7c 00 06 ac 70 00 00 f3 98 0b 03 c2 7c 00 06 ac 81 7d 90 00 39 2b 03 da 88 09 00 00 ....|...p.......|....}..9+......
18620 7c 00 06 ac 9b 8b 03 c0 7c 00 06 ac 80 01 00 24 bb 81 00 10 38 21 00 20 7c 08 03 a6 4e 80 00 20 |.......|......\$....8!..|...N...
18640 3d 60 ff f5 81 2b 90 00 98 69 03 c8 7c 00 06 ac 54 84 f6 be 81 2b 90 00 98 89 03 c9 7c 00 06 ac =`...+...i..|...T....+......|...
18660 54 a5 f6 be 81 2b 90 00 98 a9 03 c9 7c 00 06 ac 81 2b 90 00 54 c6 f6 be 98 c9 03 c9 7c 00 06 ac T....+......|....+..T.......|...
18680 4e 80 00 20 94 21 ff e0 3c e0 ff f5 39 00 00 00 bf a1 00 14 81 27 90 00 3b e7 90 00 99 09 03 c0 N....!..<...9........'..;.......
186a0 7c 00 06 ac 81 27 90 00 3b a0 00 04 39 49 01 ce 7f a0 57 2c 7c 00 06 ac 39 69 01 d0 7d 00 5f 2c |....'..;...9I....W,|...9i..}._,
186c0 7c 00 06 ac 38 00 00 08 7c 00 57 2c 7c 00 06 ac 7d 00 5f 2c 7c 00 06 ac 38 00 00 09 7c 00 57 2c |...8...|.W,|...}._,|...8...|.W,
186e0 7c 00 06 ac 7d 00 5f 2c 7c 00 06 ac 38 c0 00 01 7c c0 57 2c 7c 00 06 ac 7c 60 5f 2c 7c 00 06 ac |...}._,|...8...|.W,|...|`_,|...
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24220 4b ff 60 bd 2f 83 00 00 40 9e 00 44 38 bd 00 68 81 3d 00 68 80 05 00 14 7f 89 00 00 40 9e 00 10 K.`./...@..D8..h.=.h........@...
24240 80 a5 00 0c 2f 85 00 00 40 9e 00 08 7d 25 4b 78 38 9d 02 1c 38 61 00 08 4b ff e9 b9 3b ff 00 02 ..../...@...}%Kx8...8a..K...;...
24260 2f 83 00 00 41 9e 00 14 48 00 00 5c 38 9d 02 1c 38 61 00 08 4b ff ec 29 3b c1 04 28 3c a0 ff f2 /...A...H..\8...8a..K..);..(<...
24280 7f e6 fb 78 38 a5 42 48 38 61 00 08 38 80 00 00 7f c7 f3 78 4b ff fb 35 7c 7f 1b 79 40 82 00 28 ...x8.BH8a..8......xK..5|..y@..(
242a0 38 60 04 28 4b fe 43 bd 7f c4 f3 78 38 a0 04 28 7c 7d 1b 78 4b ff 63 35 7f a3 eb 78 93 fd 04 24 8`.(K.C....x8..(|}.xK.c5...x...\$
242c0 48 00 00 08 38 60 00 00 80 01 08 74 bb a1 08 64 38 21 08 70 7c 08 03 a6 4e 80 00 20 94 21 fe d0 H...8`.....t...d8!.p|...N....!..
242e0 7c 08 02 a6 90 01 01 34 bf 41 01 18 7c 7f 1b 78 7c 3d 0b 78 7c a3 2b 78 7c dc 33 78 4b ff 64 e5 |......4.A..|..x|=.x|.+x|.3xK.d.
24300 3d 20 ff f4 3b 69 79 d8 94 7d 00 08 7c 7a 1b 78 7f a3 eb 78 7f 64 db 78 4b ff 62 15 7c 7e 1b 79 =...;iy..}..|z.x...x.d.xK.b.|~.y
24320 41 82 00 14 88 1e 00 00 2f 80 00 00 41 9e ff e4 48 00 01 24 7f 43 d3 78 3b a0 00 01 4b fe 43 05 A......./...A...H..\$.C.x;...K.C.
24340 48 00 00 fc 4b ff 60 4d 2f 83 00 05 40 9e 00 5c a0 1f 02 10 2f 80 00 02 40 9e 00 d0 80 01 00 08 H...K.`M/...@..\..../...@.......
24360 2f 80 00 00 40 9e 00 c4 38 00 00 05 39 60 00 01 39 20 00 00 7c 09 03 a6 48 00 00 08 7d 49 03 78 /...@...8...9`..9...|...H...}I.x
24380 7c 1e 58 ae 55 2a 40 2e 39 6b 00 01 2f 80 00 00 41 9e 00 08 42 00 ff e8 80 1f 02 48 7f 80 48 00 |.X.U*@.9k../...A...B......H..H.
243a0 40 9e 00 88 48 00 00 44 3b a1 00 0c 38 a0 01 00 38 9f 00 10 7f a3 eb 78 4b ff ed c1 7f c3 f3 78 @...H..D;...8...8......xK......x
243c0 7f a4 eb 78 4b ff 63 bd 2f 83 00 00 40 9e 00 5c a0 1f 02 10 2f 80 00 02 40 9e 00 2c 80 01 00 08 ...xK.c./...@..\..../...@..,....
243e0 2f 80 00 00 40 9e 00 20 2f 9c 00 00 41 9e 00 44 7f 83 e3 78 7f e4 fb 78 38 a0 04 20 4b ff 61 ed /...@.../...A..D...x...x8...K.a.
24400 48 00 00 30 80 c1 00 08 3c a0 ff f2 7f e3 fb 78 38 a5 42 48 7f 87 e3 78 38 80 00 00 4b ff f9 ad H..0....<......x8.BH...x8...K...
24420 7c 7d 1b 78 48 00 00 10 3b a0 00 01 48 00 00 08 3b a0 00 00 7f 43 d3 78 4b fe 42 09 80 01 01 34 |}.xH...;...H...;....C.xK.B....4
24440 7f a3 eb 78 bb 41 01 18 38 21 01 30 7c 08 03 a6 4e 80 00 20 2f 80 00 3a 40 9e ff 50 4b ff fe e8 ...x.A..8!.0|...N.../..:@..PK...
24460 94 21 ff e0 7c 08 02 a6 90 01 00 24 bf a1 00 14 83 a3 00 00 7f a3 eb 78 4b ff ef 35 7f a3 eb 78 .!..|......\$...........xK..5...x
24480 4b fe 41 c1 80 01 00 24 bb a1 00 14 38 21 00 20 7c 08 03 a6 4e 80 00 20 2c 05 00 00 4d 82 00 20 K.A....\$....8!..|...N...,...M...
244a0 20 05 00 20 2f 80 00 00 41 9d 00 14 7c 00 00 d0 7c 83 00 30 38 80 00 00 4e 80 00 20 7c 89 04 30 ..../...A...|...|..08...N...|..0
244c0 7c 60 28 30 7c 84 28 30 7d 23 03 78 4e 80 00 20 2c 05 00 00 4d 82 00 20 20 05 00 20 2f 80 00 00 |`(0|.(0}#.xN...,...M......./...
244e0 41 9d 00 14 7c 00 00 d0 7c 64 06 30 7c 63 fe 70 4e 80 00 20 7c 69 00 30 7c 80 2c 30 7c 63 2e 30 A...|...|d.0|c.pN...|i.0|.,0|c.0
24500 7d 24 03 78 4e 80 00 20 2c 05 00 00 4d 82 00 20 20 05 00 20 2f 80 00 00 41 9d 00 14 7c 00 00 d0 }\$.xN...,...M......./...A...|...
24520 7c 64 04 30 38 60 00 00 4e 80 00 20 7c 69 00 30 7c 80 2c 30 7c 63 2c 30 7d 24 03 78 4e 80 00 20 |d.08`..N...|i.0|.,0|c,0}\$.xN...
24540 94 21 ff f0 2f 83 00 00 7c 08 02 a6 93 e1 00 0c 90 01 00 14 3b e0 00 00 40 9c 00 10 20 84 00 00 .!../...|...........;...@.......
24560 7c 63 01 90 3b e0 00 01 2f 85 00 00 40 9c 00 10 20 c6 00 00 7c a5 01 90 6b ff 00 01 38 e0 00 00 |c..;.../...@.......|...k...8...
24580 48 00 00 5d 2f 9f 00 00 7c 69 1b 78 7c 8a 23 78 41 9e 00 0c 21 4a 00 00 7d 29 01 90 80 01 00 14 H..]/...|i.x|.#xA...!J..})......
245a0 83 e1 00 0c 7d 44 53 78 7d 23 4b 78 38 21 00 10 7c 08 03 a6 4e 80 00 20 94 21 ff f0 38 e0 00 00 ....}DSx}#Kx8!..|...N....!..8...
245c0 7c 08 02 a6 90 01 00 14 48 00 00 15 80 01 00 14 38 21 00 10 7c 08 03 a6 4e 80 00 20 94 21 ff d0 |.......H.......8!..|...N....!..
245e0 7c ec 3b 78 7c 08 02 a6 90 01 00 34 7c a0 33 79 bf 01 00 10 7c df 33 78 7c be 2b 78 7c 99 23 78 |.;x|......4|.3y....|.3x|.+x|.#x
24600 7c 78 1b 78 3b 80 00 00 3b a0 00 01 40 82 00 14 4b fd ba ed 3b 40 00 00 3b 60 00 00 48 00 00 cc |x.x;...;...@...K...;@..;`..H...
24620 57 ca 08 3c 57 80 08 3c 57 e9 0f fe 57 ab 0f fe 2f 9e 00 00 7d 29 53 78 7d 6b 03 78 57 aa 08 3c W..<W..<W...W.../...})Sx}k.xW..<
24640 57 e0 08 3c 41 9c 00 18 7c 1f 03 78 7d 3e 4b 78 7d 5d 53 78 7d 7c 5b 78 4b ff ff c8 3b 40 00 00 W..<A...|..x}>Kx}]Sx}|[xK...;@..
24660 3b 60 00 00 48 00 00 34 41 9d 00 20 40 9a 00 0c 7f 9f c8 40 41 9d 00 14 7f 3f c8 10 7f 1e c1 10 ;`..H..4A...@......@A....?......
24680 7f 7b e8 14 7f 5a e1 14 7c df 33 78 7c be 2b 78 7c 9d 23 78 7c 7c 1b 78 57 a0 f8 7e 57 cb f8 00 .{...Z..|.3x|.+x|.#x||.xW..~W...
246a0 57 ea f8 7e 57 89 f8 00 7d 29 03 78 57 c8 f8 7e 57 87 f8 7e 7d 6b 53 78 7f 80 eb 79 7d 05 43 78 W..~W...}).xW..~W..~}kSx...y}.Cx
246c0 7c e3 3b 78 7d 66 5b 78 7d 24 4b 78 7f 9e c0 40 7f 1e c0 00 40 82 ff 94 2f 8c 00 00 41 9e 00 0c |.;x}f[x}\$Kx...@....@.../...A...
246e0 93 0c 00 00 93 2c 00 04 80 01 00 34 7f 64 db 78 7f 43 d3 78 bb 01 00 10 38 21 00 30 7c 08 03 a6 .....,.....4.d.x.C.x....8!.0|...
24700 4e 80 00 20 94 21 ff e0 7c 08 02 a6 90 01 00 24 38 e1 00 08 4b ff fe c9 80 01 00 24 80 61 00 08 N....!..|......\$8...K......\$.a..
24720 80 81 00 0c 38 21 00 20 7c 08 03 a6 4e 80 00 20 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ....8!..|...N...................
24740 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
24760 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
24780 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
247a0 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
247c0 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
247e0 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
24800 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
24820 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
24840 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
24860 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
24880 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
248a0 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
248c0 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
248e0 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
24900 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
24920 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
24940 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
24960 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
24980 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
249a0 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
249c0 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
249e0 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
24a00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
24a20 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
24a40 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
24a60 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
24a80 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
24aa0 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
24ac0 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
24ae0 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
24b00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
24b20 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
24b40 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
24b60 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
24b80 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
24ba0 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
24bc0 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
24be0 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
24c00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
24c20 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
24c40 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
24c60 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
24c80 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
24ca0 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
24cc0 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
24ce0 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
24d00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
24d20 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
24d40 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
24d60 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
24d80 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
24da0 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
24dc0 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
24de0 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
24e00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
24e20 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
24e40 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
24e60 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
24e80 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
24ea0 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
24ec0 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
24ee0 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
24f00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
24f20 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
24f40 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
24f60 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
24f80 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
24fa0 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
24fc0 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
24fe0 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
25000 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
25020 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
25040 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
25060 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................................
25080 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 04 00 00 ff f4 52 f8 00 00 40 00 ..........................R...@.
250a0 00 00 40 00 00 00 00 80 00 00 00 80 00 00 00 20 00 00 00 20 07 de 29 00 ff f0 8b fc 00 09 00 00 ..@...................).........
250c0 ff f4 53 04 00 00 40 00 00 00 40 00 00 00 00 80 00 00 00 80 00 00 00 20 00 00 00 20 07 de 29 00 ..S...@...@...................).
250e0 ff f0 8b fc 00 0a 00 00 ff f4 53 14 00 00 40 00 00 00 40 00 00 00 00 80 00 00 00 80 00 00 00 20 ..........S...@...@.............
25100 00 00 00 20 07 de 29 00 ff f0 8b fc 80 04 00 00 ff f4 53 24 00 00 80 00 00 00 80 00 00 00 00 80 ......)...........S\$............
25120 00 00 00 80 00 00 00 20 00 00 00 20 14 dc 93 80 ff f0 8b 6c 00 08 00 00 ff f4 53 34 00 00 80 00 ...................l......S4....
25140 00 00 80 00 00 00 00 80 00 00 00 80 00 00 00 20 00 00 00 20 14 dc 93 80 ff f0 8b 6c 10 08 00 00 ...........................l....
25160 ff f4 53 34 00 00 80 00 00 00 80 00 00 00 00 80 00 00 00 80 00 00 00 20 00 00 00 20 14 dc 93 80 ..S4............................
25180 ff f0 8b 6c 70 00 00 00 ff f4 53 34 00 00 80 00 00 00 80 00 00 00 00 80 00 00 00 80 00 00 00 20 ...lp.....S4....................
251a0 00 00 00 20 14 dc 93 80 ff f0 8b 6c 70 02 00 00 ff f4 53 34 00 00 80 00 00 00 80 00 00 00 00 80 ...........lp.....S4............
251c0 00 00 00 80 00 00 00 20 00 00 00 20 14 dc 93 80 ff f0 8b 6c 80 0c 00 00 ff f4 53 40 00 00 80 00 ...................l......S@....
251e0 00 00 80 00 00 00 00 80 00 00 00 80 00 00 00 20 00 00 00 20 14 dc 93 80 ff f0 8b 6c 00 0c 00 00 ...........................l....
25200 ff f4 53 50 00 00 80 00 00 00 80 00 00 00 00 80 00 00 00 80 00 00 00 20 00 00 00 20 1d cd 65 00 ..SP..........................e.
25220 ff f0 8b 30 00 3c 00 00 ff f4 53 5c 00 00 80 00 00 00 80 00 00 00 00 80 00 00 00 80 00 00 00 80 ...0.<....S\....................
25240 00 00 00 80 1d cd 65 00 ff f0 89 ec ff f4 53 6c 10 57 48 01 80 80 00 00 80 0c 00 00 80 00 00 00 ......e.......Sl.WH.............
25260 00 10 00 00 f0 00 00 00 10 00 00 00 80 00 00 00 00 01 00 00 00 00 00 00 00 40 00 00 09 0b 09 0b .........................@......
25280 ff f4 53 74 10 6b 00 1f f2 80 00 00 f2 c0 00 00 f2 00 00 00 02 00 00 00 80 00 00 00 10 00 00 00 ..St.k..........................
252a0 f2 00 00 00 00 80 00 00 00 00 00 00 01 00 00 00 08 09 0a 0b ff f4 53 7c 10 57 00 02 fe c0 00 00 ......................S|.W......
252c0 fe e0 00 00 80 00 00 00 7f 00 00 00 80 00 00 00 01 00 00 00 fe 00 00 00 00 80 00 00 fd 00 00 00 ................................
252e0 01 00 00 00 15 16 17 18 ff f4 56 14 ff f4 56 30 ff f0 a1 b8 ff f4 56 38 ff f0 a1 bc ff f4 51 90 ..........V...V0......V8......Q.
25300 ff f4 56 48 ff f0 a6 fc ff f4 56 50 ff f0 a2 38 ff f4 56 48 ff f0 a6 68 ff f4 56 50 ff f0 a2 4c ..VH......VP...8..VH...h..VP...L
25320 ff f4 56 58 ff f0 a5 e0 ff f4 56 5c ff f0 a2 60 ff f4 56 64 ff f0 a5 3c ff f4 56 14 ff f4 56 48 ..VX......V\...`..Vd...<..V...VH
25340 ff f0 a4 c8 ff f4 56 50 ff f0 a2 74 ff f4 56 70 ff f4 56 8c ff f0 a4 34 ff f4 56 94 ff f0 a3 a8 ......VP...t..Vp..V....4..V.....
25360 4f 70 65 6e 42 49 4f 53 02 04 ff 00 ff 00 00 00 a3 e7 1b 18 00 01 52 ac 00 00 05 84 28 73 65 6d OpenBIOS..............R.....(sem
25380 69 73 29 87 00 00 00 80 00 00 00 00 00 00 00 00 28 6c 69 74 29 85 00 80 00 00 00 0c 00 00 00 02 is).............(lit)...........
253a0 28 64 6f 29 84 00 00 80 00 00 00 1c 00 00 00 07 28 3f 64 6f 29 85 00 80 00 00 00 2c 00 00 00 08 (do)............(?do)......,....
253c0 28 6c 6f 6f 70 29 86 80 00 00 00 3c 00 00 00 09 28 2b 6c 6f 6f 70 29 87 00 00 00 80 00 00 00 4c (loop).....<....(+loop)........L
253e0 00 00 00 0a 64 75 70 83 00 00 00 80 00 00 00 60 00 00 00 0e 32 64 75 70 84 00 00 80 00 00 00 70 ....dup........`....2dup.......p
25400 00 00 00 0f 3f 64 75 70 84 00 00 80 00 00 00 80 00 00 00 10 6f 76 65 72 84 00 00 80 00 00 00 90 ....?dup............over........
25420 00 00 00 11 32 6f 76 65 72 85 00 80 00 00 00 a0 00 00 00 12 70 69 63 6b 84 00 00 80 00 00 00 b0 ....2over...........pick........
25440 00 00 00 13 64 72 6f 70 84 00 00 80 00 00 00 c0 00 00 00 14 32 64 72 6f 70 85 00 80 00 00 00 d0 ....drop............2drop.......
25460 00 00 00 15 6e 69 70 83 00 00 00 80 00 00 00 e0 00 00 00 16 72 6f 6c 6c 84 00 00 80 00 00 00 f0 ....nip.............roll........
25480 00 00 00 17 72 6f 74 83 00 00 00 80 00 00 01 00 00 00 00 18 2d 72 6f 74 84 00 00 80 00 00 01 10 ....rot.............-rot........
254a0 00 00 00 19 73 77 61 70 84 00 00 80 00 00 01 20 00 00 00 1a 32 73 77 61 70 85 00 80 00 00 01 30 ....swap............2swap......0
254c0 00 00 00 1b 3e 72 82 80 00 00 01 40 00 00 00 1c 72 3e 82 80 00 00 01 4c 00 00 00 1d 72 40 82 80 ....>r.....@....r>.....L....r@..
254e0 00 00 01 58 00 00 00 1e 64 65 70 74 68 85 00 80 00 00 01 64 00 00 00 1f 64 65 70 74 68 21 86 80 ...X....depth......d....depth!..
25500 00 00 01 74 00 00 00 20 72 64 65 70 74 68 86 80 00 00 01 84 00 00 00 21 72 64 65 70 74 68 21 87 ...t....rdepth.........!rdepth!.
25520 00 00 00 80 00 00 01 94 00 00 00 22 2b 81 00 80 00 00 01 a8 00 00 00 23 2d 81 00 80 00 00 01 b4 ..........."+..........#-.......
25540 00 00 00 24 2a 81 00 80 00 00 01 c0 00 00 00 25 75 2a 82 80 00 00 01 cc 00 00 00 26 6d 75 2f 6d ...\$*..........%u*.........&mu/m
25560 6f 64 86 80 00 00 01 d8 00 00 00 27 61 62 73 83 00 00 00 80 00 00 01 e8 00 00 00 28 6e 65 67 61 od.........'abs............(nega
25580 74 65 86 80 00 00 01 f8 00 00 00 29 6d 61 78 83 00 00 00 80 00 00 02 08 00 00 00 2a 6d 69 6e 83 te.........)max............*min.
255a0 00 00 00 80 00 00 02 18 00 00 00 2b 6c 73 68 69 66 74 86 80 00 00 02 28 00 00 00 2c 72 73 68 69 ...........+lshift.....(...,rshi
255c0 66 74 86 80 00 00 02 38 00 00 00 2d 3e 3e 61 83 00 00 00 80 00 00 02 48 00 00 00 2e 61 6e 64 83 ft.....8...->>a........H....and.
255e0 00 00 00 80 00 00 02 58 00 00 00 2f 6f 72 82 80 00 00 02 68 00 00 00 30 78 6f 72 83 00 00 00 80 .......X.../or.....h...0xor.....
25600 00 00 02 74 00 00 00 31 69 6e 76 65 72 74 86 80 00 00 02 84 00 00 00 32 64 2b 82 80 00 00 02 94 ...t...1invert.........2d+......
25620 00 00 00 33 64 2d 82 80 00 00 02 a0 00 00 00 34 6d 2a 82 80 00 00 02 ac 00 00 00 35 75 6d 2a 83 ...3d-.........4m*.........5um*.
25640 00 00 00 80 00 00 02 b8 00 00 00 36 40 81 00 80 00 00 02 c8 00 00 00 37 63 40 82 80 00 00 02 d4 ...........6@..........7c@......
25660 00 00 00 38 77 40 82 80 00 00 02 e0 00 00 00 39 6c 40 82 80 00 00 02 ec 00 00 00 3a 21 81 00 80 ...8w@.........9l@.........:!...
25680 00 00 02 f8 00 00 00 3b 2b 21 82 80 00 00 03 04 00 00 00 3c 63 21 82 80 00 00 03 10 00 00 00 3d .......;+!.........<c!.........=
256a0 77 21 82 80 00 00 03 1c 00 00 00 3e 6c 21 82 80 00 00 03 28 00 00 00 3f 3d 81 00 80 00 00 03 34 w!.........>l!.....(...?=......4
256c0 00 00 00 40 3e 81 00 80 00 00 03 40 00 00 00 41 3c 81 00 80 00 00 03 4c 00 00 00 42 75 3e 82 80 ...@>......@...A<......L...Bu>..
256e0 00 00 03 58 00 00 00 43 75 3c 82 80 00 00 03 64 00 00 00 44 73 70 40 83 00 00 00 80 00 00 03 70 ...X...Cu<.....d...Dsp@........p
25700 00 00 00 45 6d 6f 76 65 84 00 00 80 00 00 03 80 00 00 00 46 66 69 6c 6c 84 00 00 80 00 00 03 90 ...Emove...........Ffill........
25720 00 00 00 47 28 65 6d 69 74 29 86 80 00 00 03 a0 00 00 00 48 28 6b 65 79 3f 29 86 80 00 00 03 b0 ...G(emit).........H(key?)......
25740 00 00 00 49 28 6b 65 79 29 85 00 80 00 00 03 c0 00 00 00 4a 65 78 65 63 75 74 65 87 00 00 00 80 ...I(key)..........Jexecute.....
25760 00 00 03 d0 00 00 00 4b 68 65 72 65 84 00 00 80 00 00 03 e4 00 00 00 4c 68 65 72 65 21 85 00 80 .......Khere...........Lhere!...
25780 00 00 03 f4 00 00 00 4d 64 6f 62 72 61 6e 63 68 88 00 00 80 00 00 04 04 00 00 00 4e 64 6f 3f 62 .......Mdobranch...........Ndo?b
257a0 72 61 6e 63 68 89 00 80 00 00 04 18 00 00 00 4f 75 6e 61 6c 69 67 6e 65 64 2d 77 40 8c 00 00 80 ranch..........Ounaligned-w@....
257c0 00 00 04 2c 00 00 00 50 75 6e 61 6c 69 67 6e 65 64 2d 77 21 8c 00 00 80 00 00 04 44 00 00 00 51 ...,...Punaligned-w!.......D...Q
257e0 75 6e 61 6c 69 67 6e 65 64 2d 6c 40 8c 00 00 80 00 00 04 5c 00 00 00 52 75 6e 61 6c 69 67 6e 65 unaligned-l@.......\...Runaligne
25800 64 2d 6c 21 8c 00 00 80 00 00 04 74 00 00 00 53 69 6f 63 40 84 00 00 80 00 00 04 8c 00 00 00 54 d-l!.......t...Sioc@...........T
25820 69 6f 77 40 84 00 00 80 00 00 04 9c 00 00 00 55 69 6f 6c 40 84 00 00 80 00 00 04 ac 00 00 00 56 iow@...........Uiol@...........V
25840 69 6f 63 21 84 00 00 80 00 00 04 bc 00 00 00 57 69 6f 77 21 84 00 00 80 00 00 04 cc 00 00 00 58 ioc!...........Wiow!...........X
25860 69 6f 6c 21 84 00 00 80 00 00 04 dc 00 00 00 59 69 81 00 80 00 00 04 ec 00 00 00 5a 6a 81 00 80 iol!...........Yi..........Zj...
25880 00 00 04 f8 00 00 00 5b 63 61 6c 6c 84 00 00 80 00 00 05 04 00 00 00 5c 73 79 73 2d 64 65 62 75 .......[call...........\sys-debu
258a0 67 89 00 80 00 00 05 14 00 00 00 5d 24 69 6e 63 6c 75 64 65 88 00 00 80 00 00 05 28 00 00 00 5e g..........]\$include.......(...^
258c0 24 65 6e 63 6f 64 65 2d 66 69 6c 65 8c 00 00 80 00 00 05 3c 00 00 00 5f 73 74 61 74 65 85 00 80 \$encode-file.......<..._state...
258e0 00 00 05 54 00 00 00 04 00 00 00 00 66 6f 72 74 68 2d 6c 61 73 74 8a 80 00 00 05 64 00 00 00 04 ...T........forth-last.....d....
25900 00 01 52 18 6c 61 74 65 73 74 86 80 00 00 05 7c 00 00 00 04 00 01 52 18 62 61 73 65 84 00 00 80 ..R.latest.....|......R.base....
25920 00 00 05 90 00 00 00 04 00 00 00 10 2f 63 82 80 00 00 05 a4 00 00 00 03 00 00 00 01 2f 77 82 80 ............/c............../w..
25940 00 00 05 b4 00 00 00 03 00 00 00 02 2f 6c 82 80 00 00 05 c4 00 00 00 03 00 00 00 04 2f 6e 82 80 ............/l............../n..
25960 00 00 05 d4 00 00 00 03 00 00 00 04 2f 78 82 80 00 00 05 e4 00 00 00 03 00 00 00 08 31 81 00 80 ............/x..............1...
25980 00 00 05 f4 00 00 00 03 00 00 00 01 32 81 00 80 00 00 06 04 00 00 00 03 00 00 00 02 33 81 00 80 ............2...............3...
259a0 00 00 06 14 00 00 00 03 00 00 00 03 2d 31 82 80 00 00 06 24 00 00 00 03 ff ff ff ff 30 81 00 80 ............-1.....\$........0...
259c0 00 00 06 34 00 00 00 03 00 00 00 00 6d 79 2d 73 65 6c 66 87 00 00 00 80 00 00 06 44 00 00 00 03 ...4........my-self........D....
259e0 00 00 00 00 64 65 63 69 6d 61 6c 87 00 00 00 80 00 00 06 5c 00 00 00 01 00 00 00 20 00 00 00 0a ....decimal........\............
25a00 00 00 05 a8 00 00 03 08 00 00 00 10 68 65 78 83 00 00 00 80 00 00 06 74 00 00 00 01 00 00 00 20 ............hex........t........
25a20 00 00 00 10 00 00 05 a8 00 00 03 08 00 00 00 10 6f 63 74 61 6c 85 00 80 00 00 06 98 00 00 00 01 ................octal...........
25a40 00 00 00 20 00 00 00 08 00 00 05 a8 00 00 03 08 00 00 00 10 63 75 72 72 65 6e 74 87 00 00 00 80 ....................current.....
25a60 00 00 06 bc 00 00 00 04 00 00 05 84 6c 61 73 74 84 00 00 80 00 00 06 e4 00 00 00 01 00 00 06 e8 ............last................
25a80 00 00 02 d8 00 00 00 10 23 6f 72 64 65 72 86 80 00 00 06 f8 00 00 00 04 00 00 00 01 63 6f 6e 74 ........#order..............cont
25aa0 65 78 74 87 00 00 00 80 00 00 07 14 00 00 00 05 00 00 55 24 00 00 00 10 76 6f 63 61 62 75 6c 61 ext...............U\$....vocabula
25ac0 72 69 65 73 3f 8d 00 80 00 00 07 2c 00 00 00 03 ff ff ff ff 74 72 75 65 84 00 00 80 00 00 07 4c ries?......,........true.......L
25ae0 00 00 00 03 ff ff ff ff 66 61 6c 73 65 85 00 80 00 00 07 60 00 00 00 03 00 00 00 00 28 69 6d 6d ........false......`........(imm
25b00 65 64 69 61 74 65 29 8b 00 00 00 80 00 00 07 74 00 00 00 01 00 00 06 08 00 00 01 c4 00 00 00 74 ediate)........t...............t
25b20 00 00 02 e4 00 00 06 08 00 00 02 78 00 00 01 34 00 00 03 20 00 00 00 10 28 63 6f 6d 70 69 6c 65 ...........x...4........(compile
25b40 2d 6f 6e 6c 79 29 8e 80 00 00 07 90 00 00 00 01 00 00 06 08 00 00 01 c4 00 00 00 74 00 00 02 e4 -only).....................t....
25b60 00 00 06 18 00 00 02 78 00 00 01 34 00 00 03 20 00 00 00 10 69 6d 6d 65 64 69 61 74 65 89 00 80 .......x...4........immediate...
25b80 00 00 07 cc 00 00 00 01 00 00 06 fc 00 00 02 d8 00 00 07 94 00 00 00 10 63 6f 6d 70 69 6c 65 2d ........................compile-
25ba0 6f 6e 6c 79 8c 00 00 80 00 00 08 04 00 00 00 01 00 00 06 fc 00 00 02 d8 00 00 07 d0 00 00 00 10 only............................
25bc0 66 6c 61 67 73 3f 86 80 00 00 08 2c 00 00 00 01 00 00 05 e8 00 00 05 b8 00 00 01 b8 00 00 01 c4 flags?.....,....................
25be0 00 00 02 e4 00 00 00 20 00 00 00 7f 00 00 02 6c 00 00 00 10 69 6d 6d 65 64 69 61 74 65 3f 8a 80 ...............l....immediate?..
25c00 00 00 08 4c 00 00 00 01 00 00 08 50 00 00 06 08 00 00 02 6c 00 00 06 08 00 00 03 44 00 00 00 10 ...L.......P.......l.......D....
25c20 63 6f 6d 70 69 6c 65 2d 6f 6e 6c 79 3f 8d 00 80 00 00 08 84 00 00 00 01 00 00 08 50 00 00 06 18 compile-only?..............P....
25c40 00 00 02 6c 00 00 06 18 00 00 03 44 00 00 00 10 5b 81 00 82 00 00 08 b4 00 00 00 01 00 00 06 48 ...l.......D....[..............H
25c60 00 00 05 68 00 00 03 08 00 00 00 10 5d 81 00 80 00 00 08 d8 00 00 00 01 00 00 06 38 00 00 05 68 ...h........]..............8...h
25c80 00 00 03 08 00 00 00 10 61 6c 6c 6f 74 85 00 80 00 00 08 f4 00 00 00 01 00 00 03 f8 00 00 01 b8 ........allot...................
25ca0 00 00 04 08 00 00 00 10 2c 81 00 80 00 00 09 14 00 00 00 01 00 00 03 f8 00 00 05 e8 00 00 09 18 ........,.......................
25cc0 00 00 03 08 00 00 00 10 63 2c 82 80 00 00 09 30 00 00 00 01 00 00 03 f8 00 00 05 b8 00 00 09 18 ........c,.....0................
25ce0 00 00 03 20 00 00 00 10 61 6c 69 67 6e 85 00 80 00 00 09 50 00 00 00 01 00 00 05 e8 00 00 03 f8 ........align......P............
25d00 00 00 05 e8 00 00 06 08 00 00 01 c4 00 00 02 6c 00 00 01 c4 00 00 05 e8 00 00 06 08 00 00 01 c4 ...............l................
25d20 00 00 02 6c 00 00 09 18 00 00 00 10 6e 75 6c 6c 2d 61 6c 69 67 6e 8a 80 00 00 09 74 00 00 00 01 ...l........null-align.....t....
25d40 00 00 03 f8 00 00 00 74 00 00 09 78 00 00 03 f8 00 00 01 34 00 00 01 c4 00 00 06 48 00 00 03 a4 .......t...x.......4.......H....
25d60 00 00 00 10 77 2c 82 80 00 00 09 bc 00 00 00 01 00 00 03 f8 00 00 06 08 00 00 02 6c 00 00 09 18 ....w,.....................l....
25d80 00 00 03 f8 00 00 05 c8 00 00 09 18 00 00 03 2c 00 00 00 10 6c 2c 82 80 00 00 09 ec 00 00 00 01 ...............,....l,..........
25da0 00 00 05 d8 00 00 03 f8 00 00 05 d8 00 00 06 08 00 00 01 c4 00 00 02 6c 00 00 01 c4 00 00 05 d8 .......................l........
25dc0 00 00 06 08 00 00 01 c4 00 00 02 6c 00 00 09 18 00 00 03 f8 00 00 05 d8 00 00 09 18 00 00 03 38 ...........l...................8
25de0 00 00 00 10 3c 3e 82 80 00 00 0a 1c 00 00 00 01 00 00 03 44 00 00 02 98 00 00 00 10 28 74 6f 29 ....<>.............D........(to)
25e00 84 00 00 80 00 00 0a 6c 00 00 00 01 00 00 05 e8 00 00 01 b8 00 00 03 08 00 00 00 10 3e 62 6f 64 .......l....................>bod
25e20 79 85 00 80 00 00 0a 88 00 00 00 01 00 00 05 e8 00 00 06 08 00 00 02 3c 00 00 01 b8 00 00 00 10 y......................<........
25e40 62 6f 64 79 3e 85 00 80 00 00 0a a8 00 00 00 01 00 00 05 e8 00 00 06 08 00 00 02 3c 00 00 01 c4 body>......................<....
25e60 00 00 00 10 72 65 76 65 61 6c 86 80 00 00 0a cc 00 00 00 01 00 00 05 94 00 00 02 d8 00 00 06 fc ....reveal......................
25e80 00 00 03 08 00 00 00 10 72 65 63 75 72 73 69 76 65 89 00 81 00 00 0a f0 00 00 00 01 00 00 0a f4 ........recursive...............
25ea0 00 00 00 10 72 65 63 75 72 73 65 87 00 00 00 81 00 00 0b 18 00 00 00 01 00 00 05 94 00 00 02 d8 ....recurse.....................
25ec0 00 00 05 e8 00 00 01 b8 00 00 09 34 00 00 00 10 6e 6f 6f 70 84 00 00 80 00 00 0b 34 00 00 00 01 ...........4....noop.......4....
25ee0 00 00 00 10 65 6e 76 69 72 6f 6e 6d 65 6e 74 3f 8c 00 00 80 00 00 0b 5c 00 00 00 05 00 00 0b a0 ....environment?.......\........
25f00 00 00 00 10 6e 6f 2d 65 6e 76 69 72 6f 6e 6d 65 6e 74 3f 8f 00 00 00 80 00 00 0b 78 00 00 00 01 ....no-environment?........x....
25f20 00 00 00 e4 00 00 07 78 00 00 00 10 72 65 73 6f 6c 76 65 2d 6f 72 69 67 8c 00 00 80 00 00 0b 9c .......x....resolve-orig........
25f40 00 00 00 01 00 00 03 f8 00 00 00 a4 00 00 05 e8 00 00 01 b8 00 00 01 c4 00 00 01 34 00 00 03 08 ...........................4....
25f60 00 00 00 10 28 69 66 29 84 00 00 82 00 00 0b c0 00 00 00 01 00 00 00 20 00 00 04 30 00 00 09 34 ....(if)...................0...4
25f80 00 00 03 f8 00 00 06 48 00 00 09 34 00 00 00 10 28 74 68 65 6e 29 86 82 00 00 0b f0 00 00 00 01 .......H...4....(then)..........
25fa0 00 00 0b c4 00 00 00 10 74 6d 70 2d 63 6f 6d 70 2d 64 65 70 74 68 8e 80 00 00 0c 1c 00 00 00 04 ........tmp-comp-depth..........
25fc0 ff ff ff ff 74 6d 70 2d 63 6f 6d 70 2d 62 75 66 8c 00 00 80 00 00 0c 3c 00 00 00 04 48 03 14 14 ....tmp-comp-buf.......<....H...
25fe0 73 65 74 75 70 2d 74 6d 70 2d 63 6f 6d 70 8e 80 00 00 0c 58 00 00 00 01 00 00 05 68 00 00 02 d8 setup-tmp-comp.....X.......h....
26000 00 00 06 48 00 00 03 44 00 00 04 30 00 00 00 2c 00 00 03 f8 00 00 0c 5c 00 00 02 d8 00 00 04 08 ...H...D...0...,.......\........
26020 00 00 09 34 00 00 06 08 00 00 09 34 00 00 01 78 00 00 0c 40 00 00 03 08 00 00 08 f8 00 00 00 10 ...4.......4...x...@............
26040 65 78 65 63 75 74 65 2d 74 6d 70 2d 63 6f 6d 70 90 00 00 80 00 00 0c 74 00 00 00 01 00 00 01 78 execute-tmp-comp.......t.......x
26060 00 00 0c 40 00 00 02 d8 00 00 03 44 00 00 04 30 00 00 00 44 00 00 06 38 00 00 0c 40 00 00 03 08 ...@.......D...0...D...8...@....
26080 00 00 00 20 00 00 00 10 00 00 09 34 00 00 0c 5c 00 00 02 d8 00 00 00 74 00 00 02 d8 00 00 04 08 ...........4...\.......t........
260a0 00 00 06 48 00 00 05 68 00 00 03 08 00 00 05 e8 00 00 01 b8 00 00 03 e8 00 00 00 10 69 66 82 81 ...H...h....................if..
260c0 00 00 0c d8 00 00 00 01 00 00 0c 78 00 00 00 20 00 00 04 30 00 00 09 34 00 00 03 f8 00 00 06 48 ...........x.......0...4.......H
260e0 00 00 09 34 00 00 00 10 74 68 65 6e 84 00 00 82 00 00 0d 44 00 00 00 01 00 00 0b c4 00 00 0c dc ...4....then.......D............
26100 00 00 00 10 65 6c 73 65 84 00 00 82 00 00 0d 74 00 00 00 01 00 00 00 20 00 00 04 1c 00 00 09 34 ....else.......t...............4
26120 00 00 03 f8 00 00 06 48 00 00 09 34 00 00 01 34 00 00 0b c4 00 00 00 10 28 62 65 67 69 6e 29 87 .......H...4...4........(begin).
26140 00 00 00 80 00 00 0d 90 00 00 00 01 00 00 00 10 28 61 67 61 69 6e 29 87 00 00 00 80 00 00 0d c8 ................(again).........
26160 00 00 00 01 00 00 00 10 28 75 6e 74 69 6c 29 87 00 00 00 80 00 00 0d e0 00 00 00 01 00 00 00 10 ........(until).................
26180 28 77 68 69 6c 65 29 87 00 00 00 80 00 00 0d f8 00 00 00 01 00 00 00 10 28 72 65 70 65 61 74 29 (while).................(repeat)
261a0 88 00 00 80 00 00 0e 10 00 00 00 01 00 00 00 10 72 65 73 6f 6c 76 65 2d 64 65 73 74 8c 00 00 80 ................resolve-dest....
261c0 00 00 0e 28 00 00 00 01 00 00 03 f8 00 00 05 e8 00 00 01 b8 00 00 01 c4 00 00 09 34 00 00 00 10 ...(.......................4....
261e0 62 65 67 69 6e 85 00 81 00 00 0e 44 00 00 00 01 00 00 0c 78 00 00 00 20 00 00 0d cc 00 00 09 34 begin......D.......x...........4
26200 00 00 03 f8 00 00 00 10 61 67 61 69 6e 85 00 82 00 00 0e 6c 00 00 00 01 00 00 00 20 00 00 0d e4 ........again......l............
26220 00 00 09 34 00 00 00 20 00 00 04 1c 00 00 09 34 00 00 0e 48 00 00 0c dc 00 00 00 10 75 6e 74 69 ...4...........4...H........unti
26240 6c 85 00 82 00 00 0e 94 00 00 00 01 00 00 00 20 00 00 0d fc 00 00 09 34 00 00 00 20 00 00 04 30 l......................4.......0
26260 00 00 09 34 00 00 0e 48 00 00 0c dc 00 00 00 10 77 68 69 6c 65 85 00 81 00 00 0e c8 00 00 00 01 ...4...H........while...........
26280 00 00 0c 78 00 00 00 20 00 00 0e 14 00 00 09 34 00 00 00 20 00 00 04 30 00 00 09 34 00 00 03 f8 ...x...........4.......0...4....
262a0 00 00 06 48 00 00 09 34 00 00 01 34 00 00 00 10 72 65 70 65 61 74 86 82 00 00 0e fc 00 00 00 01 ...H...4...4....repeat..........
262c0 00 00 00 20 00 00 0e 2c 00 00 09 34 00 00 00 20 00 00 04 1c 00 00 09 34 00 00 0e 48 00 00 0b c4 .......,...4...........4...H....
262e0 00 00 0c dc 00 00 00 10 6c 65 61 76 65 73 86 80 00 00 0f 3c 00 00 00 04 00 00 11 50 72 65 73 6f ........leaves.....<.......Preso
26300 6c 76 65 2d 6c 6f 6f 70 8c 00 00 80 00 00 0f 74 00 00 00 01 00 00 0f 78 00 00 02 d8 00 00 0d cc lve-loop.......t.......x........
26320 00 00 00 94 00 00 0e 14 00 00 04 30 00 00 00 2c 00 00 00 74 00 00 02 d8 00 00 01 34 00 00 03 f8 ...........0...,...t.......4....
26340 00 00 00 a4 00 00 01 c4 00 00 01 34 00 00 03 08 00 00 0e 2c 00 00 04 1c ff ff ff c4 00 00 03 f8 ...........4.......,............
26360 00 00 01 c4 00 00 09 34 00 00 0f 78 00 00 03 08 00 00 00 10 64 6f 82 81 00 00 0f 90 00 00 00 01 .......4...x........do..........
26380 00 00 0c 78 00 00 0f 78 00 00 02 d8 00 00 03 f8 00 00 00 20 00 00 00 30 00 00 09 34 00 00 06 48 ...x...x...............0...4...H
263a0 00 00 0f 78 00 00 03 08 00 00 00 10 3f 64 6f 83 00 00 00 81 00 00 0f fc 00 00 00 01 00 00 0c 78 ...x........?do................x
263c0 00 00 0f 78 00 00 02 d8 00 00 00 20 00 00 00 40 00 00 09 34 00 00 03 f8 00 00 03 f8 00 00 0f 78 ...x...........@...4...........x
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https://arxiv.org/abs/1907.10118v2 | # Title:Young star cluster populations in the E-MOSAICS simulations
Abstract: We present an analysis of young star clusters (YSCs) that form in the E-MOSAICS cosmological, hydrodynamical simulations of galaxies and their star cluster populations. Through comparisons with observed YSC populations, this work aims to test models for YSC formation and obtain an insight into the formation processes at work in part of the local galaxy population. We find that the models used in E-MOSAICS for the cluster formation efficiency and high-mass truncation of the initial cluster mass function ($M_\mathrm{c,\ast}$) both quantitatively reproduce the observed values of cluster populations in nearby galaxies. At higher redshifts ($z \geq 2$, near the peak of globular cluster formation) we find that, at a constant star formation rate (SFR) surface density, $M_\mathrm{c,\ast}$ is larger than at $z=0$ by a factor of four due to the higher gas fractions in the simulated high-redshift galaxies. Similar processes should be at work in local galaxies, offering a new way to test the models. We find that cluster age distributions may be sensitive to variations in the cluster formation rate (but not SFR) with time, which may significantly affect their use in tests of cluster mass loss. By comparing simulations with different implementations of cluster formation physics, we find that (even partially) environmentally-independent cluster formation is inconsistent with the brightest cluster-SFR and specific luminosity-$\Sigma_\mathrm{SFR}$ relations, whereas these observables are reproduced by the fiducial, environmentally-varying model. This shows that models in which a constant fraction of stars form in clusters are inconsistent with observations.
Comments: 20 pages, 17 figures, accepted for publication in MNRAS Subjects: Astrophysics of Galaxies (astro-ph.GA) DOI: 10.1093/mnras/stz2721 Cite as: arXiv:1907.10118 [astro-ph.GA] (or arXiv:1907.10118v2 [astro-ph.GA] for this version)
## Submission history
From: Joel Pfeffer [view email]
[v1] Tue, 23 Jul 2019 20:09:59 UTC (5,723 KB)
[v2] Wed, 25 Sep 2019 10:34:16 UTC (6,093 KB) | {"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": 0.7821996212005615, "perplexity": 2593.906182726919}, "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/1579251694176.67/warc/CC-MAIN-20200127020458-20200127050458-00120.warc.gz"} |
https://mathematica.stackexchange.com/questions/57760/how-to-understand-the-usage-of-inner-and-outer-figuratively/57777 | # How to understand the usage of Inner and Outer figuratively?
## Description:
In Mathematica the functions like Thread, Inner, Outer etc. are very important and are used frequently.
For the function Thread:
Thread[f[{a, b, c}]]
{f[a], f[b], f[c]}
Thread[f[{a, b, c}, x]]
{f[a, x], f[b, x], f[c, x]}
Thread[f[{a, b, c}, {x, y, z}]]
{f[a, x], f[b, y], f[c, z]}
And I understand the Usage1, Usage2, Usage3 easily as well as I use them masterly.
However I always cannot master the usage of Inner and Outer so that I must refer to the Mathematica Documentation every time when I feel I need using them.
I find that I cannot master them owing to that I cannot understand the results of Inner and Outer clearly. Namely, I always forget what construct they generate when executed.
The typical usage cases of Inner and Outer shown as below:
Inner Usage:
Inner[f, {a, b}, {x, y}, g]
g[f[a, x], f[b, y]]
Inner[f, {{a, b}, {c, d}}, {x, y}, g]
{g[f[a, x], f[b, y]], g[f[c, x], f[d, y]]}
Inner[f, {{a, b}, {c, d}}, {{x, y}, {u, v}}, g]
{{g[f[a, x], f[b, u]], g[f[a, y], f[b, v]]},
{g[f[c, x], f[d, u]], g[f[c, y], f[d, v]]}}
Outer Usage:
Outer[f, {a, b}, {x, y, z}]
{{f[a, x], f[a, y], f[a, z]}, {f[b, x], f[b, y], f[b, z]}}
Outer[f, {{1, 2}, {3, 4}}, {{a, b}, {c, d}}]
{{{{f[1, a], f[1, b]}, {f[1, c], f[1, d]}},
{{f[2, a], f[2, b]}, {f[2, c], f[2, d]}}},
{{{f[3, a], f[3, b]}, {f[3, c], f[3, d]}},
{{f[4, a], f[4, b]}, {f[4, c], f[4, d]}}}}
## Questions:
1. How to master the usage Inner and Outer? Namely, how can I use them without referring to the Mathematica Documentation?
2. How to understand the result of Out[3],Out[4],Out[5] figuratively? Namely, by using graphics or other way.
• I recommend that you download and work through Leonid Shifrin's Mathematica programming: an advanced introduction. It's free and answers a lot of question you ask. – m_goldberg Aug 20 '14 at 10:12
• I don't know why you are leaving. All I can say is that I hope you are not taking SE too seriously. It's just a website, a tool to get help and learn from. When you find yourself spending too much time on it, it's good to take a break. I do that from time to time. But don't let it affect you emotionally. – Szabolcs Nov 9 '16 at 7:43
• @Szabolcs In fact, I made a mistake and Moderator R.M pointed it out some time ago. And I did affected by M.SE emotionly but I don't know why. – xyz Nov 9 '16 at 7:59
I think of Outer just like nikie showed.
Inner is a generalization of matrix multiplication. I like the picture from the Wikipedia page.
To calculate an entry of matrix multiplication, you first pair list entries (a11,b12) and (a12,b22). You "times/multiply" those pairs (a11*b12) and (a12*b22), and then you "plus/add" all the results (a11*b12)+(a12*b22). Note that you "times" before you "plus" in matrix multiplication which helps me remember the order of arguments for Inner.
listL={{a11,a12},{a21,a22},{a31,a32},{a41,a42}};
listR={{b11,b12,b13},{b21,b22,b23}};
Inner[times,listL,listR,plus]
Animated Mathematica Functions contains cool animated illustrations of the way a number of built-in functions work. Among them are
Outer
• @kguler...I am learning so much this week...nice – ubpdqn Aug 20 '14 at 12:52
• Loved them when they first came out; still love them today. – J. M.'s discontentment May 4 '15 at 10:02
• @J. M., same here -- especially the sound effects:) – kglr May 4 '15 at 10:10
• @Guesswhoitis., I know you are J.M :) Welcome back! – xyz May 5 '15 at 5:24
• @Leandro I am not sure why you are addressing me here. This is kglr's answer, not mine; I only edited it. Also the answer starts with a link to a collection of these animations: reference.wolfram.com/legacy/flash – Mr.Wizard Jul 26 '16 at 18:21
Not sure if that's what you're looking for: This is the image I always have in mind for Outer[f,{a,b,c},{x,y,z}]:
args = {{a, b, c}, {x, y, z}};
TableForm[Outer[f, args[[1]], args[[2]]], TableHeadings -> args]
(i = Inner[List, Range@3, Range@3, List]) // MatrixForm;
(o = Outer[List, Range@3, Range@3]) // MatrixForm
p1 = ListLinePlot[i, Mesh -> All, PlotStyle -> Red, PlotTheme -> "Detailed"];
p2 = ListLinePlot[o, Mesh -> All, PlotStyle -> Blue, PlotTheme -> "Detailed"];
Legended[Show[p2, p1, PlotRange -> All], LineLegend[{Red, Blue}, {"Inner", "Outer"}]]
• like you answer+1 – ubpdqn Aug 20 '14 at 12:04
• +1 for the compactness. It might have been even more immediate (specially in the first example) with: (i = Inner[List, {a, b, c}, Range@3, List]) // MatrixForm. – Trad Dog Apr 8 '16 at 8:34
I think of Outer like nikie's answer shows. Here's a similar view of Inner. Think of the arguments in columns. Apply f to each row and g to the result.
args = {{a, b, c}, {x, y, z}};
Format[g[e__]] := Column[{g, e},
Dividers -> {None, {False, True, False}}, Alignment -> Center];
Inner[f, Sequence @@ args, g]
• Might I suggest f@@{a,x} etc.? – Timothy Wofford Aug 20 '14 at 13:49
• Thanks. I wanted a divider, but I hate dealing with tables/grids in Mma. I'd thought about f[a, x], too (i.e., no Format-ting). I was trying to emphasize the columns. – Michael E2 Aug 20 '14 at 13:52 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.22817695140838623, "perplexity": 7327.25267881869}, "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/1600402101163.62/warc/CC-MAIN-20200930013009-20200930043009-00573.warc.gz"} |
https://www.computer.org/csdl/trans/tc/1973/04/01672314-abs.html | Issue No. 04 - April (1973 vol. 22)
ISSN: 0018-9340
pp: 347-358
A. Maruoka , Department of Electrical Communications, Faculty of Engineering, Tohoku University
ABSTRACT
The flexible network is a network composed of flexible logical multivalued cells that can be adjusted at will to perform any functions. The network is assumed to have a fixed-interconnection pattern and to have a fixed-input variable assignment. The following question is answered. What is the necessary and sufficient range of the flexibility of the cells' functions to obtain all the output functions the circuit could realize? Furthermore, the relation between the cells functions that realize the same output functions is derived and the number of different realizable output functions is counted.
INDEX TERMS
Cellular logic, flexible logic, integrated circuit, Maitra cascade, tree network.
CITATION
N. Honda, A. Maruoka, "Logical Networks of Flexible Cells", IEEE Transactions on Computers, vol. 22, no. , pp. 347-358, April 1973, doi:10.1109/T-C.1973.223721 | {"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": 0.856941282749176, "perplexity": 2998.841183598818}, "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/1492917122726.55/warc/CC-MAIN-20170423031202-00121-ip-10-145-167-34.ec2.internal.warc.gz"} |
http://mathhelpforum.com/calculus/36418-integral.html | 1. ## Integral
I have a portion of ad admittance exam coming up tomorrow and I was wondering if someone could give me some integrals I could solve and then you could check my work...I am getting fairly good at integrals but practice never hurt! Now I dont want ridculously difficult integrals...but I don't want too easy of ones either xD
If someoen could give me some I would be very greatful
Mathstud
2. Hello,
Try this one :
$\int_n^{n+1} \arctan \left(\frac{1}{\sqrt{t}} \right) dt$
3. A couple I found from the MIT integration bee from a while back:
$\int \left(\frac{1}{\ln x} + \ln \left(\ln x\right)\right)dx$
$\int \sin(101x) \cdot \sin^{99} (x) dx$
$\int e^{\arccos (x)} dx$
Have fun?
4. Originally Posted by Moo
Hello,
Try this one :
$\int_n^{n+1} \arctan \left(\frac{1}{\sqrt{t}} \right) dt$
Thanks a lot Moo! I appreciate it...this is a fairly easy one...or at least I hope xD
$\int{arctan\bigg(\frac{1}{\sqrt{x}}\bigg)dx}=\int{ arcot(\sqrt{x})dx}$
Then you would go
Let $u=\sqrt{x}\Rightarrow{u^2=x}$
Then $dx=2udu$
So we have $\int{2u\cdot{arcot(u)du}}$
Using parts we have $\int{2u\cdot{arcot(u)}}du=u^2\cdot{arcot(u)}+\int\ frac{u^2}{1+u^2}$
THis is the same as
$u^2\cdot{arcot(u)}+\bigg[\int\frac{u^2+1}{u^2+1}du-\int\frac{1}{u^2+1}du\bigg]$
Which then obviously goes to
$u^2\cdot{arcot(u)}+u-arctan(u)$
Back subbing we get $x\cdot{arcot(\sqrt{x})}+\sqrt{x}-arctan(\sqrt{x})$
Whew I dont know why that took so long to write ...I showed all my steps just so you could critique me. Thanks
5. Actually, I've never used cot function...
Yes, the answer looks like yours
Try this one (I don't have the answer yet) :
$\int_0^1 \sqrt[3]{1-x^7}-\sqrt[7]{1-x^3} dx$
6. Originally Posted by o_O
A couple I found from the MIT integration bee from a while back:
$\int \left(\frac{1}{\ln x} + \ln \left(\ln x\right)\right)dx$
$\int \sin(101x) \cdot \sin^{99} (x) dx$
$\int e^{\arccos (x)} dx$
Have fun?
Ok well lets do these one at a time...starting with number three
$\int{e^{arccos(x)}dx}$
Letting $u=arcos(x)\Rightarrow{cos(u)=x}$
Then $dx=-sin(u)$
So we have $\int-\sin(u)e^udu$
then by parts we haev $-sin(u)e^{u}+\int\cos(u)e^{u}du$
which then goes to $-sin(u)e^{u}+\cos(u)e^{u}-\int{-\sin(u)e^{u}du}$
Adding the like integrals and dividing by 2 we get $\int-\sin(u)e^{u}du=\frac{e^{u}}{2}(\cos(u)-\sin(u))$
back subbing we get $\int{e^{arccos(x)}dx}=\frac{e^{arccos(x)}}{2}(x-\sqrt{1-x^2})$
7. Originally Posted by Moo
Actually, I've never used cot function...
Yes, the answer looks like yours
Try this one (I don't have the answer yet) :
$\int_0^1 \sqrt[3]{1-x^7}-\sqrt[7]{1-x^3} dx$
Aww moo...Do I really have to do that one...you would have to use the binomial coefficient power series
8. Originally Posted by Mathstud28
Aww moo...Do I really have to do that one...you would have to use the binomial coefficient power series
You said you wanted to train
9. Originally Posted by Moo
You said you wanted to train
lasjflsdfkjasldfkjsadf;ljk......ok I will do it then
10. Originally Posted by o_O
A couple I found from the MIT integration bee from a while back:
$\int \left(\frac{1}{\ln x} + \ln \left(\ln x\right)\right)dx$
$\int \sin(101x) \cdot \sin^{99} (x) dx$
$\int e^{\arccos (x)} dx$
Have fun?
For the third one we have $\int\frac{1}{\ln(x)}-\ln(\ln(x))dx$
Letting $u=\ln(x)\Rightarrow{e^u=x}$
then $dx=e^{u}$
so we have $\int{\frac{e^{u}}{u}+e^{u}\ln(u)du}$
which by splitting the integrals gives $\int{\frac{e^{u}}{u}du}+\int{e^{u}\ln(u)}dy$
leaving the first as it is we do integration by parts on the second to reveal
$\int{\frac{e^{u}}{u}+e^{u}\ln(u)}du=\int\frac{e^{u }}{u}du+\bigg[e^{u}\ln(u)-\int\frac{e^{u}}{u}du\bigg]$
cancelling like terms we get $\int{\frac{e^u}{u}+e^{u}\ln(u)}du=e^{u}\ln(u)$
Back subbing we get $\int\bigg[\frac{1}{\ln(x)}+\ln(\ln(x))dx\bigg]=e^{\ln(x)}\cdot{\ln(\ln(x))}=x\ln(\ln(x))+C$
11. Originally Posted by Moo
Actually, I've never used cot function...
Yes, the answer looks like yours
Try this one (I don't have the answer yet) :
$\int_0^1 \sqrt[3]{1-x^7}-\sqrt[7]{1-x^3} dx$
$e\int_0^{1}\bigg[\ln\bigg(\frac{1}{x}\bigg)\bigg]^{\frac{1}{3}}dx-e\int_0^{1}\bigg[\ln\bigg(\frac{1}{x}\bigg)\bigg]^{\frac{1}{7}}dx$?
12. Originally Posted by Mathstud28
Ok well lets do these one at a time...starting with number three
$\int{e^{arccos(x)}dx}$
Letting $u=arcos(x)\Rightarrow{cos(u)=x}$
Then $dx=-sin(u)$
So we have $\int-\sin(u)e^udu$
then by parts we haev $-sin(u)e^{u}+\int\cos(u)e^{u}du$
which then goes to $-sin(u)e^{u}+\cos(u)e^{u}-\int{-\sin(u)e^{u}du}$
Adding the like integrals and dividing by 2 we get $\int-\sin(u)e^{u}du=\frac{e^{u}}{2}(\cos(u)-\sin(u))$
back subbing we get $\int{e^{arccos(x)}dx}=\frac{e^{arccos(x)}}{2}(x-\sqrt{1-x^2})$
Originally Posted by Mathstud28
For the third one we have $\int\frac{1}{\ln(x)}-\ln(\ln(x))dx$
Letting $u=\ln(x)\Rightarrow{e^u=x}$
then $dx=e^{u}$
so we have $\int{\frac{e^{u}}{u}+e^{u}\ln(u)du}$
which by splitting the integrals gives $\int{\frac{e^{u}}{u}du}+\int{e^{u}\ln(u)}dy$
leaving the first as it is we do integration by parts on the second to reveal
$\int{\frac{e^{u}}{u}+e^{u}\ln(u)}du=\int\frac{e^{u }}{u}du+\bigg[e^{u}\ln(u)-\int\frac{e^{u}}{u}du\bigg]$
cancelling like terms we get $\int{\frac{e^u}{u}+e^{u}\ln(u)}du=e^{u}\ln(u)$
Back subbing we get $\int\bigg[\frac{1}{\ln(x)}+\ln(\ln(x))dx\bigg]=e^{\ln(x)}\cdot{\ln(\ln(x))}=x\ln(\ln(x))+C$
Are these right o_O?
13. Looks good to me. You made some typos but that's probably due to messing up while typing up your work.
$\int_{0}^{\infty} xe^{-x^{3}}dx$
14. Originally Posted by o_O
Looks good to me. You made some typos but that's probably due to messing up while typing up your work.
$\int_{0}^{\infty} xe^{-x^{3}}dx$
$\sum_{n=0}^{\infty}\frac{(-1)^{n}x^{3n+2}}{(3n+2)n!}\bigg|_{0}^{\infty}$ | {"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": 58, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9307116270065308, "perplexity": 1387.3690238638685}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-04/segments/1484560280133.2/warc/CC-MAIN-20170116095120-00140-ip-10-171-10-70.ec2.internal.warc.gz"} |
https://www.deepdyve.com/lp/springer_journal/the-structure-of-the-wake-generated-by-a-submarine-model-in-yaw-EyOmFjGdBL | # The structure of the wake generated by a submarine model in yaw
The structure of the wake generated by a submarine model in yaw The turbulent wake of a submarine model in yaw was investigated using stereoscopic particle image velocimetry at $$Re_{L} = 2.4 \times 10^{6}.$$ R e L = 2.4 × 10 6 . The model (DARPA SUBOFF idealized submarine geometry) is mounted in a low-speed wind tunnel using a support that mimics the sail, and it is yawed so that the body moves in the plane normal to the support. The measurements reveal the formation of a pair of streamwise vortices that are asymmetric in strength. The weaker vortex quickly diffuses, and in the absence of further diffusion, the stronger vortex maintains its strength even at the furthest downstream location. It is suggested that the flow fields obtained here using a semi-infinite sail as a support will be similar to those obtained using a finite length sail since its tip vortex would not interact significantly with the body vortices present in the wake, at least for a considerable distance downstream of the stern $$(x/D > 24).$$ ( x / D > 24 ) . Hence, a submarine in yaw is expected to generate wakes which are inherently more persistent than one in pitch, and the strong asymmetries in yaw are expected to produce a net rolling moment on the body. http://www.deepdyve.com/assets/images/DeepDyve-Logo-lg.png Experiments in Fluids Springer Journals
# The structure of the wake generated by a submarine model in yaw
, Volume 56 (6) – Jun 3, 2015
9 pages
/lp/springer_journal/the-structure-of-the-wake-generated-by-a-submarine-model-in-yaw-EyOmFjGdBL
Publisher
Springer Journals
Subject
Engineering; Engineering Fluid Dynamics; Fluid- and Aerodynamics; Engineering Thermodynamics, Heat and Mass Transfer
ISSN
0723-4864
eISSN
1432-1114
D.O.I.
10.1007/s00348-015-1997-4
Publisher site
See Article on Publisher Site
### Abstract
The turbulent wake of a submarine model in yaw was investigated using stereoscopic particle image velocimetry at $$Re_{L} = 2.4 \times 10^{6}.$$ R e L = 2.4 × 10 6 . The model (DARPA SUBOFF idealized submarine geometry) is mounted in a low-speed wind tunnel using a support that mimics the sail, and it is yawed so that the body moves in the plane normal to the support. The measurements reveal the formation of a pair of streamwise vortices that are asymmetric in strength. The weaker vortex quickly diffuses, and in the absence of further diffusion, the stronger vortex maintains its strength even at the furthest downstream location. It is suggested that the flow fields obtained here using a semi-infinite sail as a support will be similar to those obtained using a finite length sail since its tip vortex would not interact significantly with the body vortices present in the wake, at least for a considerable distance downstream of the stern $$(x/D > 24).$$ ( x / D > 24 ) . Hence, a submarine in yaw is expected to generate wakes which are inherently more persistent than one in pitch, and the strong asymmetries in yaw are expected to produce a net rolling moment on the body.
### Journal
Experiments in FluidsSpringer Journals
Published: Jun 3, 2015
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https://www.keyword-suggest-tool.com/search/position+velocity+and+acceleration+equations/ | # Position velocity and acceleration equations
Position velocity and acceleration equations keyword after analyzing the system lists the list of keywords related and the list of websites with related content, in addition you can see which keywords most interested customers on the this website
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### Rectilinear Motion - Math24
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From the instantaneous position r = r(t), instantaneous meaning at an instant value of time t, the instantaneous velocity v = v(t) and acceleration a = a(t) have the general, coordinate-independent definitions; =, = = Notice that velocity always points in the direction of motion, in other words for a curved path it is the tangent vector.Loosely speaking, first order derivatives are related to ...
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DA: 13 PA: 50 MOZ Rank: 82 | {"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": 0.778839111328125, "perplexity": 1553.3547698980435}, "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/1590347410535.45/warc/CC-MAIN-20200530231809-20200531021809-00052.warc.gz"} |
http://math.stackexchange.com/questions/203686/intuitively-concretely-actually-what-happens-when-you-multiply-a-matrix-by-its?answertab=active | # Intuitively (concretely actually) what happens when you multiply a matrix by its transpose?
The construct $A^TA$ for $A$ any $m \times n$ matrix seems to appear often in formulae and results. For example I was reading that square root of eigenvalues of $A^TA$ (an $n \times n$ matrix) are singular values of $A$. If I do it on paper it seems like the diagonal contains squares and the off diagonals contain all possible quadratic combinations (and of course symmetric). What is a good way to think of this construct or how is it used other than in that singular value statement?
-
Let $\langle v, w \rangle$ be the usual dot product on $\mathbb{R}^n$. Then any symmetric bilinear form on $\mathbb{R}^n$ can be uniquely represented in the form $\langle v, Bw \rangle$ where $B$ is some symmetric matrix. (It is a good idea to see explicitly how the coefficients of $B$ determine the coefficients of the corresponding quadratic form $\langle v, Bv \rangle$, which you can think of as determining an ellipse $\langle v, Bv \rangle = 1$.)
In particular, when you change coordinates so that your old vectors $v$ are replaced by new vectors $Av$ for some matrix $A$, then the dot product behaves on new vectors like
$$\langle Av, Aw \rangle = \langle v, A^T A w \rangle$$
so $A^T A$ is a matrix describing a bilinear form related to the dot product by change of coordinates. More concretely, you can think of $A^T A$ as describing the coefficients of an equation for an ellipse $\langle Av, Av \rangle = 1$ related to the unit sphere $\langle v, v \rangle = 1$ by a change of coordinates.
-
IMO, many of those behaviors resemble "squared norms", making $A^T A$ a better candidate for a quadratic function of $A$ than $A^2$ is. This is especially so when you consider complex matrices and use conjugate transpose instead of transpose. Or if you view $A$ as a collection of column vectors.
$A^TA$ is sort of projection operator onto the row space or coimage space (without normalization), while $A^+A$ is the 'real' projector, where $A^+$ is the pseudoinverse of $A$. Actually, when the left inverse $A_L^{-1}$ exists (full column rank), $A^+=A_L^{-1}=(A^TA)^{-1}A^T$. | {"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": 0.9594332575798035, "perplexity": 130.76763888981074}, "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-2014-10/segments/1394011173115/warc/CC-MAIN-20140305091933-00010-ip-10-183-142-35.ec2.internal.warc.gz"} |
https://studyadda.com/question-bank/force-and-laws-of-motion_q45/2404/210935 | • # question_answer A bullet of mass $20\,\,g$ is fired from a rifle of $8\,\,kg$ with an initial velocity of$100\,\,m\,\,{{s}^{-1}}$. Find the final velocity of the recoil of the rifle. A) $-0.25\,\,m\,\,{{s}^{-1}}$ B) $2.5\,\,m\,\,{{s}^{-1}}$ C) $25\,\,m\,\,{{s}^{-1}}$ D) $250\,\,m\,\,{{s}^{-1}}$
Initial momentum$=0$ Final momentum$=(0.02\times 100)+8\times v$ According to the law of conservation of momentum, $(0.02\times 100)+8v=0$, $8v=-2$ $v=-\frac{1}{4}m{{s}^{-1}}=-0.25\,\,m\,\,{{s}^{-1}}$ | {"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": 0.9950562715530396, "perplexity": 1977.0229221130553}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-34/segments/1596439735881.90/warc/CC-MAIN-20200804161521-20200804191521-00126.warc.gz"} |
https://bitbucket.org/asksol/python-stomp/ | # python-stomp
Bitbucket is a code hosting site with unlimited public and private repositories. We're also free for small teams!
Close
# stompy - Python STOMP client library
This is useful for connecting to and communicating with Apache ActiveMQ (an open source Java Message Service (JMS) message broker) or other brokers with support for the STOMP protocol.
The majority of the methods available take a single argument; a dictionary. This dictionary should contain the necessary bits you need to pass to the STOMP server. It is outlined in each method exactly what it needs to work.
For specifics on the protocol, see the STOMP protocol specification.
This library is basically a Python implementation of Perl's Net::Stomp.
To enable the ActiveMQ Broker for STOMP add the following to the activemq.xml configuration:
<connector>
<serverTransport uri="stomp://localhost:61613"/>
</connector>
See http://bitbucket.org/asksol/python-stomp/ for latest code.
# Recent activity
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Tip: You can also navigate files with Alt+j (next) and Alt+k (previous) and view the file with Alt+o. | {"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": 0.15772423148155212, "perplexity": 12489.635109208873}, "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-2014-10/segments/1394010693428/warc/CC-MAIN-20140305091133-00001-ip-10-183-142-35.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/does-a-uniformly-accelerating-charge-radiate-what-about-radiation-reaction-force.174708/ | 1. Jun 21, 2007
### user299792458
It is often stated that ANY accelerating charge radiates, so this includes uniformly accelerating charges. But the radiation reaction force is proportional to the THIRD derivative of x, so it vanishes when acceleration is constant. What's the deal here?
Here's a graph which supposedly shows that the energy carried away by radiation (which is proportional to acceleration squared) is different than the work done by the radiation reaction force (which is proportional to velocity times the THIRD derivative of x). What's the deal with all of this?
And here's the full paper where this graph came from:
It seems the paper states that this is some kind of an unsolved problem or something. Is any of this true?
2. Jun 21, 2007
### Meir Achuz
"It seems the paper states that this is some kind of an unsolved problem or something. Is any of this true?"
I'm not going to read the paper, but it is still an unsolved problem.
The x triple dot result is only for some special cases, not including uniform acceleration. | {"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": 0.9656160473823547, "perplexity": 343.0978990672631}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-04/segments/1484560281162.88/warc/CC-MAIN-20170116095121-00259-ip-10-171-10-70.ec2.internal.warc.gz"} |
https://brilliant.org/problems/a-problem-by-dominick-hing-5/ | # A number theory problem by Dominick Hing
Number Theory Level 2
How many positive integral divisors does the number 2000 have?
× | {"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": 0.9512204527854919, "perplexity": 4589.708733369142}, "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/1476988720468.71/warc/CC-MAIN-20161020183840-00451-ip-10-171-6-4.ec2.internal.warc.gz"} |
https://ch00ftech.com/2013/05/02/vendor-bender/?replytocom=196032 | # Vendor Bender
I found out this week that sometimes goods and services purchased in China can be of low quality.
Disclaimer If you purchased a QR clock and are reading this, don't worry. I have every intention of shipping you exactly what you paid for. If your clock is found to have any issues or if you are in any way dissatisfied, don't hesitate to contact me, and I will do whatever I can to make it right.
I just spent last weekend installing the LED modules on the QR clock PCBs that I discussed a few posts ago. At the time of writing, I was really impressed with the overall quality of the soldering job on the PCBs I received from Myro, and although I wasn't exactly looking forward to the prospect of soldering for 20 to 25 hours, I was excited to finish my QR clocks and ship them to my customers.
Unfortunately, I ran into a few hiccoughs.
# Displays
The QR clocks require nine 8x8 LED matrices to work, and they work best with displays with square pixels. Square pixel LED matrices are apparently a rarity, so I had to get these special ordered from a manufacturer that I found on Alibaba. You can read all about it here.
I got samples from two different vendors and was very impressed by the one I ended up choosing. Their displays were bright, evenly lit, and didn't show any defects like I found with the first vendor I picked out:
Soon after my fundraiser ended, I ordered 500 LED modules from the better manufacturer. I only needed 405 for the 45 preorders I received, but I figured it'd be a good idea to order a bunch extra to install into my extra PCBs and sell to any interested customers in the future.
I was informed shortly after placing my order that there was going to be an unexpectedly long lead time on these displays of 35 days. This wasn't a problem since I was waiting for the PCBs anyway, and sure enough, almost exactly 35 days later, I got the shipment.
They were for the most part very well packaged, and showed very little if any cosmetic damage.
I wasn't planning on trusting them entirely though. Part of the reason I decided to install the LED modules myself was that it would give me the opportunity to test all of the modules before installing them. Removing the modules isn't impossible with a hot air gun, but I'd rather just choose not to install them in the first place.
Well, about 10 modules in, I found a defective module:
No bother. Set it aside and move on...
Huh...
Well, I mean, I bought some extras right?
Alright, I'll save you some scrolling:
38 of the 500 displays I ordered showed at least one pixel missing. That's a 7.6% failure rate. More importantly, these things cost me $2.67 a piece, so that's around$100 worth of product I'll be throwing in the trash.
The good news is that I still had more than enough displays to fulfill all of my preorders with around 5 complete clocks to spare (I'm still in the process of assembling the extras as of time of writing).
The bad news is that some of the modules that passed my initial test on the breadboard showed problems after they were soldered in place. It seems that some have intermittent internal connections that may have been broken through the soldering process. I'm really hoping that none of the working ones I've already installed and shipped have this problem.
I'm not certain if this level of drop out is to be expected with parts like this. I contacted the vendor two days ago and still haven't gotten a reply. If it comes down to it, I might try to file a complaint with Alibaba, but I'm not naive enough to expect this issue to be resolved in my favor.
Update
Not two hours after I posted this, I got this email:
I very much regret the difficulties you have been with our company.
I checked with our production Dept again, we would like to remake 30pcs goods for you. (This was in response to an email I wrote before I found the last 8)
After finished, I will text them one by one. Please don't worry. And the shipping cost, we will pay it.
Apologize for the inconvenience cause by our error. Kindly contact me if any problem!
Looks like everything's coming up Milhouse!
This is a major inconvenience for me, but at least they're taking the effort to make the situation right.
In addition to this issue, the new LEDs are not quite as bright as the samples I was sent for the prototype:
This is not as big of a deal however because at maximum brightness, the original QR clock prototype will often force a cellphone camera to increase the shutter speed to the point where you start to see refresh artifacts:
These interfere with the ability to scan the display. I tried mitigating this effect in the firmware by increasing the refresh rate, but even after maxing out around 240Hz, the problem still persists.
Basically, in any environment where the maximum brightness of the sample LEDs is required for contrast, the clock couldn't be scanned anyway because of refresh artifacts. Besides, the new LEDs are plenty bright for indoor use anyway.
Still would have been nice if the samples matched the product though.
# PCBs
In my previous article where I discussed ordering these QR clock PCBs, I raved about how good they looked. After soldering on the displays and installing firmware however, I was less excited:
Something like 50% of the assembled clocks had at least one row of LEDs blacked out.
Because all of these LEDs were on the same row of the same display module, I had reason to believe that the problem existed outside the displays themselves. The LEDs in the display are addressed by row and column, so an entirely dead row of LEDs tells me that there's something wrong with the components addressing them.
A quick visual inspection of the parts on the backside showed no immediate issues. I ran some more solder over the IC to clear up any potential cold solder joints, but it didn't fix the problem. Only after adding some solder to the current limiting resistors did I see something I've never seen before.
The resistor split in half.
Seriously, right down the center.
Apparently, a large portion of these resistors had very very small hairline fractures through them which are invisible to the naked eye.
Let's run through one such case. A resistance measurement across this 62 resistor shows a resistance of:
Infinity.
A (very very) close visual inspection shows a very faint crack:
And any attempt to rework this resistor renders it in twain:
This is one of the stranger things I've ever seen happen to a PCB. The good news is that it's very easy to fix by replacing the broken resistors with fresh new ones. The bad news is that it can be difficult to find. Some of the resistors were held together so well that the display appeared to be working flawlessly until it was jostled and the halves of the resistor very slightly separated.
As a precaution, I manually flexed the PCB of every clock I shipped out to locate broken resistors. I believe I found all of them (over 30), but there's still a possibility that more will pop up after shipment.
Though the problem was easy to resolve, it still doesn't answer the question of what the heck happened here. I contacted Myro to ask such a question. I went through the typical round of back and forth and finger pointing:
Do you think somewhere is shorting caused it or resistors problem? Or current is big?
Until I finally got a fairly straight answer:
I just checked your order. There was a information about V-cut. Dring separated boards , they felt V-cut too thin, which v-cut was not enoug for separated boards. Maybe they did press too much and broke the resistors. We promise do not happen again. We will control the V-cut for a panel. Or Use machine to cut to single board.
Translated to English, when PCBs are manufactured, they are often made in panels. These panels are large and easier to deal with than multiple smaller PCBs. The components are soldered on in this panel state, and then the panels are separated. One of the methods of separation or "depanelization" is "V-scoring" where channels are cut in the PCB to make it weaker at the separating edges. Once panelization is complete, the boards need to simply be flexed, and they should break along the v-scored edges.
In this case, the v-score channel wasn't deep enough, so the PCBs had to be bent exceptionally far in order to separate the boards. I imagine it went down something like this:
I suspect the edges were bent downwards away from the soldered on components. Because the resistors were on the outside of this curve, they were literally pulled end from end and split in two.
As of writing, I am very displeased with my service from Myro. They have not yet offered any kind of compensation for the time that I've wasted cleaning up their mistakes, and I will not be using them or promoting them again in my future projects.
Besides, in addition to these broken resistors, I also saw a few of these:
and some cases where resistors were missing entirely.
Really shoddy work.
Update
After some more negotiation, I got this:
Thanks. We refunded $30 to your account. You can use it for your future orders. The margins on their assembly services are very tight, so they couldn't afford to pay me for my time, but it's still something. I would have rather gotten a cash refund, but since I've had so few problems with Myro produced PCBs, I think I will continue to use their PCB fab services and avoid their assembly especially when I have pre-orders to worry about. # Conclusion To be clear, I'm not disparaging outsourcing manufacturing in general. The system works, and a lot of really high quality products get manufactured in China. I just ran into some bad luck with the vendors I chose who sent me poor quality work after I had already verified the quality of some of their work earlier. Despite these setbacks, the QR clocks were all extensively tested and shown to work and have already begun to ship. If any problems do crop up in the field, I'm ready to pounce on customer complaints and resolve them professionally. Besides, I never planned to turn a profit with this project but instead to learn as much as I could about the whole process. I'll elaborate on this with a later article, but if you were to work out the math, I will probably be making less than a dollar an hour on this whole arrangement. The lessons learned, however are invaluable. Still, I would have liked to have learned these lessons when the quantities were smaller and stakes were lower. ## 21 thoughts on “Vendor Bender” 1. The information you share on the process of the clocks creation is truly invaluable! It really sheds light on what does it mean to work with manufacturers from China. 2. Great article, but you need a title that is less apt to induce forehead slapping. Of course goods from China are low quality - that's what we all get when we shop to the lowest bidder! It shouldn't take a project like this to realize that quality control is not something you see in products sourced overseas. It's just a shame you couldn't find a source outside the PRC for those display modules. • Not all goods from China are low quality. For example, nearly every Apple product is made in China, and they are famous for the quality of their products. I've done product test manufacturing work in some of these facilities, and as long as you have a tight enough product spec and thorough testing protocols, you can produce some really awesome stuff. Besides, I didn't order from the lowest bidder. I immediately tossed out the first manufacturer I talked to (which were half the price) because of the poor quality of their products. The subtitle was meant to poke fun at the typical China stereotype. Some stuff in China is very cheap, but some of it can be very high quality if you take the necessary precautions. I went through several steps to vet my vendors and only ordered from those who I had a preexisting relationship with, but still got screwed by what I'm going to attribute to bad luck. The point of this post wasn't to disparage China as a place to get things manufactured, it was to document the particular problems I came across, and how I dealt with them. • I agree with Ch00f about Chinese quality. I have experienced many of the same things that he's gone through outsourcing to China. My Company has a factory in Shenzhen that produces outstanding quality work, so in fact not all Chinese vendors produce poorly made products and it is very stereotypical indeed that everything or most is inferior from China. With PCB design and manufacturing your "one off" for mass production and bringing it to market is a dynamic nightmare in itself. Copy and pasting your single eagle drawing to make a large board to ease cost with V-Grooving along with several other methods and tricks takes a lot of trial and error to gain the experience to make perfect production runs on the first try. Knowing the other dynamics involved with parts, like components breaking by scoring the v-groove, plus waste while manufacturing and so many things I could mention can and will save you huge amounts of lead time, time to produce on the machine and a whole lot of money. I would like to share a link to an awesome video in 2 parts that I guarantee will educate anyone from a very experienced point of view about larger scale PCB design for manufacture. I thought I knew mostly what I needed to know from experience, but this video I caught a couple of years ago has really given me the edge to do it perfectly the first time. It is very informative and entertaining I promise. Now many will already know several things mentioned already, but I promise there's a ton of knowledge you'll find in there that will enlighten anyone on this subject experienced or not. I hope this helps you Ch00f as well as others as much as it has helped me. Daves 2 part video can be found here: http://www.youtube.com/watch?v=VXE_dh38HjU • I have no personal experience with Chinese manufacturing, but both anecdotes sound very familiar- samples from third party vendors are good to fantastic, but when it comes to the production run the quality seems to inevitably drop a fair bit. It is also well known that quality out of China can be fantastic, but when you scratch under the surface these top quality products are produced in China in a factory either entirely owned and controlled by the manufacturer or it is more of a 'first party' arrangement where the manufacturer and vendor have very close ties (read Apple and Foxconn etc). 3. Thanks for the interesting read! I'm glad you took the time to explain the 'V-cut', that one had me scratching my head hahaha 4. Interesting article. You are actually very lucky that the vendor is willing to work with you at all- in many cases, the overseas supplier will just cut their losses and move on. Best of luck! 5. Great post. This is the biggest problem with outsourcing to China and buying from Alibaba. Outside of the ones you already mentioned, do you recommend any other PCB manufacturers and sellers on Alibaba? Thanks! 6. You can get high-quality products out of China, but (and I'm sure you know this) companies like Apple and Amazon have staff over there, watching over the manufacturing process, to make sure that proper procedure is followed, and no corners are cut. For readers of this blog who would like more info on the subject of outsourcing to China, there's a great book: "Poorly Made in China: An Insider's Account of the China Production Game" by Paul Midler. It's about non-electronic products but will give insight on how the economy works as a whole. 7. What a nightmare - you have unreliable LED modules you may or may not be able to catch in test, and you have unreliable assembly that you may or may not be able to catch in test. Prepare for a high return rate and watch your profits dwindle. Sorry - I feel your pain. 8. Yeah, i ordered a console for fireworks off alibaba a while back. The problems seem to be soldering, i.e. they solder in chain really really fast. add shipping, vibrations and whatnot and i guess one or two can come loose. Its not that big of a deal, as you can repair it yourself....but the fact that they dont check the product at all or very little once it is done is of some concern. How fucking hard can it be to power it and check all buttons and functions? Apparently it seems checking product for quality is more expensive than refunding/replacing/etc. 9. "Something like 50% of the assembled clocks had at least one row of LEDs blacked out." 50% of the assembled boards had manufacturing defects, and the best Myro could do was offer you a$30 credit for a future order?! That's crazy. Is there no guarantee on their assembly work? What's their incentive to assemble things properly, if they can have a 50% failure rate but still charge you for the entire production run?
10. It's great.
Thank you so much! | {"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": 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": 0.38350093364715576, "perplexity": 1792.7988507913196}, "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/1631780058415.93/warc/CC-MAIN-20210927090448-20210927120448-00532.warc.gz"} |
http://mathhelpforum.com/math-philosophy/205512-infinity-plus-1-a-print.html | # Infinity plus 1
• October 17th 2012, 03:55 AM
astartleddeer
Infinity plus 1
I have no idea what I'm attempting to get at with this one.
Anyway, the following must give a real number of some sort.
$\infty - 1$
$\infty + 1$
$(-\infty) - 1$
$(-\infty) + 1$
Thus, that real number will always be $\infty + 1$ for an example.
But this now leads me to the conclusion, that any number can be made greater than infinity.
$\infty + 2$
$\infty + 3$
$\infty + \infty = (2)(\infty)$
What is going on at this end of the spectrum for the real number line?
• October 17th 2012, 04:22 AM
astartleddeer
Re: Infinity plus 1
If I assume the distance to the edge of space is infinity and I travel a distance $\infty + 1$ and I turn around - What would I see?
Do I exist at a distance $\infty + 1?$
Ok, let's assume I don't. So now I travel a distance of $\infty - 1$ and I remain looking forward - What would I see now?
I must continue to exist at this distance if I have not surpassed $\infty$
• October 17th 2012, 07:57 AM
Plato
Re: Infinity plus 1
Quote:
Originally Posted by astartleddeer
I have no idea what I'm attempting to get at with this one.
Anyway, the following must give a real number of some sort.
$\infty - 1$
$\infty + 1$
$(-\infty) - 1$
$(-\infty) + 1$
Why in the world you think that?
$\infty$ is not a real number.
Or as my favorite philosophy professor would say "infinity is where mathematicians hid their ignorance".
• October 17th 2012, 02:59 PM
johnsomeone
Re: Infinity plus 1
What you're speculating about is either:
1) whether you actually won your grade school argument when, after that brat said "Well I call 'first' infinity times!", you replied with the devastating "Fine - then I call 'first' infinity plus one times!" (The answer is you did win!)
or
2) the ordinal numbers.
I won't even try to explain it. Just google "ordinal number".
The gist is this: Instead of thinking of the natural numbers as being about "counting", so like "number of things in a set", it's thinking of the natural numbers as being about "what's the next one greater than this one?". The former are cardinals ("the cardinality of a set"), and the later are called ordinals. These two notions are the same for finite numbers, but once things become infinite, they're very different.
• October 27th 2012, 05:56 AM
HallsofIvy
Re: Infinity plus 1
Quote:
Originally Posted by astartleddeer
I have no idea what I'm attempting to get at with this one.
Anyway, the following must give a real number of some sort.
Since $\infty$ is not a real number, this basic assumption is not true.
Quote:
$\infty - 1$
$\infty + 1$
$(-\infty) - 1$
$(-\infty) + 1$
Thus, that real number will always be $\infty + 1$ for an example.
But this now leads me to the conclusion, that any number can be made greater than infinity.
$\infty + 2$
$\infty + 3$
$\infty + \infty = (2)(\infty)$
What is going on at this end of the spectrum for the real number line?
• October 27th 2012, 06:41 AM
Deveno
Re: Infinity plus 1
Quote:
Originally Posted by astartleddeer
I have no idea what I'm attempting to get at with this one.
Anyway, the following must give a real number of some sort.
why do you assert this?
Quote:
$\infty - 1$
$\infty + 1$
$(-\infty) - 1$
$(-\infty) + 1$
Thus, that real number will always be $\infty + 1$ for an example.
But this now leads me to the conclusion, that any number can be made greater than infinity.
$\infty + 2$
$\infty + 3$
$\infty + \infty = (2)(\infty)$
What is going on at this end of the spectrum for the real number line?
the general rule for binary operations (like "+") is:
a thing + another same kind of thing = still the same kind of thing (one apple + one orange equals how many bananas? hmm?)
real numbers are FINITE: that is, every real number is less than some integer (we can think of unit lengths "covering" a line segment whose length is a given real number).
infinity is...erm, not finite (hence the name). there's no point on the real line where we can say "the real numbers end here".
but let's pretend that the far-off horizon we can never reach is "out there, somewhere".
the only thing that makes sense is:
∞+1 = ∞. or, in general:
∞+r = ∞-r = ∞, for any (non-infinite) real number r.
one might suspect (and perhaps might be able to prove) that if r > 0, r*∞ = ∞, as well. opinions differ as to whether we should distinguish between -∞ and ∞ (the reasons are complicated).
but now we have a curious problem, what should 0*∞ be? the temptation is to say 0*∞ = 1, but this leads to more problems that you can imagine.
another problem comes when we try to imagine what ∞-∞ might be.
what happens is: as soon as you throw ∞ into the mix, it breaks the ALGEBRA of the real numbers. so ask yourself:
which is more useful: the power of algebra for modelling and solving problems, or the ability to say, "we can use infinity now!" (just how many things have you ever encountered that ARE infinite, anyway?).
• December 10th 2012, 04:05 PM
FelipeAbraham
Re: Infinity plus 1
Hi.
Well, infact there is a bunch of number theories that encompasses infinite and infinitezimal numbers.
Try Ordinal numbers, Hiperreal numbers and Surreal numbers, for example.
With my best regards.
• December 18th 2012, 03:55 AM
astartleddeer
Re: Infinity plus 1
Hi, | {"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": 34, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7142674326896667, "perplexity": 1098.2735532792165}, "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/1469257825365.1/warc/CC-MAIN-20160723071025-00054-ip-10-185-27-174.ec2.internal.warc.gz"} |
http://mathhelpforum.com/differential-equations/148900-undetermined-coefficients.html | # Math Help - Undetermined Coefficients
1. ## Undetermined Coefficients
Find a particular solution to y+25y=10sin(5t)
I chose y_p to be Acos(5t)+Bsin(5t).
After simplifying, my equation becomes 0=0, which leads to nothing.
Would I multiply my particular solution by t or a higher power of t? In general, it is still unclear to me when to modify particular solutions.
2. You have to modify the particular solution when the homogeneous solution contains what you would normally have tried for the particular solution. For this equation, what is the homogeneous solution?
3. Originally Posted by krtica
Find a particular solution to y+25y=10sin(5t)
I chose y_p to be Acos(5t)+Bsin(5t).
After simplifying, my equation becomes 0=0, which leads to nothing.
Would I multiply my particular solution by t or a higher power of t? In general, it is still unclear to me when to modify particular solutions.
Well...
I assume you meant $y'+25y=10\sin(5t)$
Ok then,
If $y_p = A\cos(5t) + B\sin(5t)$ then
$y_p' = -5A\sin(5t) + 5B\cos(5t)$
So we have...
$y'+25y= -5A\sin(5t) + 5B\cos(5t) + 25A\cos(5t) + 25B\sin(5t)$
$\sin(5t)(-5A + 25B) + \cos(5t)(5B + 25A)$
So you get, $25B - 5A = 10$ and $5B + 25A = 0$
=> $B = -5A$
=> $-125A - 5A = 10$
=> $A = -\frac{1}{13}$
$B = \frac{5}{13}$
4. I apologize, I meant to type y"+25y=10sin(5t).
5. You need to find the homogeneous solution. That is, solve the following DE: $\ddot{y}(t)+25 y(t)=0$.
I think you'll find, if you do so, that the solution includes the particular solution you were thinking of using as an ansatz. Therefore, you need to modify your ansatz. (Ansatz, in case you haven't heard the term, is a German word that essentially means "your guess" or "your working hypothesis".) | {"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": 0.9465299844741821, "perplexity": 972.2743444127689}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-52/segments/1419447554115.86/warc/CC-MAIN-20141224185914-00028-ip-10-231-17-201.ec2.internal.warc.gz"} |
https://sriasat.wordpress.com/page/2/ | ## A combinatorial identity
$(*)\qquad\qquad\qquad\displaystyle\sum_{n\ge i\ge j\ge k\ge 0}\binom ni\binom ij\binom jk=4^n$.
One way to prove it is to count the number of possible triples $(A,B,C)$ of sets with $C\subseteq B\subseteq A\subseteq S=\{1,\dots,n\}$. This can be done in two ways. First, if $A, B, C$ contain $i,j,k$ elements respectively then the number of such triples clearly equals the left-hand side of $(*)$. On the other hand, since each element of $S$ must belong to one of the four disjoint regions in the picture below
the number of such triples equals precisely $4^n$.
An algebraic proof can be obtained by repeated use of the binomial theorem:
$\displaystyle 4^n=\sum_{n\ge i\ge 0}\binom ni3^i=\sum_{n\ge i\ge j\ge 0}\binom ni\binom ij2^j=\sum_{n\ge i\ge j\ge k\ge 0}\binom ni\binom ij\binom jk$.
Similarly, one obtains the more general identity
$\displaystyle\sum_{n\ge i_m\ge\cdots\ge i_1\ge 0}\binom{n}{i_1}\binom{i_1}{i_2}\cdots\binom{i_{m-1}}{i_m}=(m+1)^n$.
This also follows from the multinomial theorem since the left-hand side equals
$\displaystyle\sum_{n\ge i_m\ge\cdots\ge i_1\ge 0}\frac{n!}{(n-i_1)!(i_1-i_2)!\cdots (i_{m-1}-i_m)!i_m!}=(m+1)^n$.
Filed under Combinatorics
## Jordan-Hölder theorem
I am cramming for my algebra comprehensive exam so I will probably be posting stuff like this for a while.
A chain of subgroups $\{e\}=H_0\triangleleft H_1\triangleleft\cdots\triangleleft H_n=G$ of a finite group $G$ with $H_{i+1}/H_i$ simple for all $i$ is called a composition series of $G$. (The $H_{i+1}/H_i$ are called composition factors.) The theorem in question states that every group has a composition series, and the composition factors in any two such series are unique up to reordering.
We can easily prove the first part of the theorem, that any finite group has a composition series, using the extreme principle. Let $n\ge 0$ be the largest integer for which there is a chain $\{e\}=H_0\triangleleft H_1\triangleleft\cdots\triangleleft H_n=G$ of subgroups. (Such an $n$ exists because there is trivially a chain with $n=1$, and our group is finite.) We claim that the composition factors in this case are simple. If not, WLOG $H_{i+1}/H_i$ has a non-trivial normal subgroup $\widetilde N$. By correspondence, $\widetilde N=N/H_i$ for some $H_i\triangleleft N\triangleleft H_{i+1}$. But then
$\{e\}=H_0\triangleleft H_1\triangleleft\cdots\triangleleft H_i\triangleleft N\triangleleft H_{i+1}\cdots\triangleleft H_n=G$
is a longer chain, a contradiction.
While the proof of the general theorem requires some non-trivial group theory (e.g. see here), we can easily prove the theorem for finite abelian groups using elementary number theory. For such a group, the orders of the composition factors must exactly be the prime factors—listed with multiplicity—of the order of the group. So the result follows immediately from the fundamental theorem of arithmetic.
Filed under Algebra, Number theory
## A new AM-GM inequality
I recently found the following inequalities useful:
$\displaystyle \left(x+\frac{a_1+\cdots+a_n}{n}\right)^n\ge (x+a_1)\cdots(x+a_n)\ge \left(x+\sqrt[n]{a_1\cdots a_n}\right)^n$,
for any $x,a_1,\dots,a_n\ge 0$.
To prove the left half simply apply AM-GM to the product
$\displaystyle (x+a_1)\cdots (x+a_n)\le \left(\frac{(x+a_1)+\cdots+(x+a_n)}{n}\right)^n$.
The right half follows directly from Hölder’s inequality.
More generally, one has
$\displaystyle\left(\frac{a_1+\cdots+a_n}{n}+\frac{b_1+\cdots+b_n}{n}\right)^n\ge (a_1+b_1)\cdots(a_n+b_n)\\\ge \left(\sqrt[n]{a_1\cdots a_n}+\sqrt[n]{b_1\cdots b_n}\right)^n$
for $a_1,\dots,a_n,b_1,\dots,b_n\ge 0$. The proof goes similarly.
One obtains similar inequalities for any number of vectors $a,b,c,\dots$.
As an example, for $a_i=i$ we get the inequalities
$\displaystyle \left(x+\frac{n+1}{2}\right)^n\ge (x+1)\cdots (x+n)>\left(x+\frac ne\right)^n$
using the fact that $\sqrt[n]{n!}/n\to 1/e$ from above.
Filed under Algebra
## Hilbert’s nullstellensatz
Leaving it here because I want to study this at some point.
I had occasion recently to look up the proof of Hilbert’s nullstellensatz, which I haven’t studied since cramming for my algebra qualifying exam as a graduate student. I was a little unsatisfied with the proofs I was able to locate – they were fairly abstract and used a certain amount of algebraic machinery, which I was terribly rusty on – so, as an exercise, I tried to find a more computational proof that avoided as much abstract machinery as possible. I found a proof which used only the extended Euclidean algorithm and high school algebra, together with an induction on dimension and the obvious observation that any non-zero polynomial of one variable on an algebraically closed field has at least one non-root. It probably isn’t new (in particular, it might be related to the standard model-theoretic proof of the nullstellensatz, with the Euclidean algorithm and high school algebra taking…
View original post 3,546 more words
Filed under Algebra
## Midy’s theorem
Let $b$ have order $d$ modulo $p$. Then the base $b$ expansion of $1/p$ has period $L=(b^d-1)/p$ of length $d$. To see this, note that if $b^d-1=mp$, then
$\displaystyle\frac 1p=\frac{m}{b^d-1}=\frac{m}{b^d}+\frac{m}{b^{2d}}+\frac{m}{b^{3d}}+\cdots$.
Note also that $aL for any $1\le a\le p-1$. Hence $a/p$ has period $aL$. Now suppose that $d$ is even. Since $b$ has order $d$ modulo $p$, it follows that $b^{d/2}\equiv -1\pmod p$. Hence $p-a\equiv b^{d/2}a\pmod p$. This means that at their midpoints the two numbers $aL$ and $(p-a)L$ are mirror images of one another. This means that splitting $aL$ midway into two equal parts and adding them gives $b^{d/2}-1$, i.e., a string of $(b-1)$‘s in base $b$. This is known as Midy’s theorem.
For example, with $p=7$ and $b=10$ we get $d=6$, and $L=142857$. Split $L$ into two equal parts $142$ and $857$, adding which gives
$142+857=999$.
In general, if $m$ is any divisor of $d$, then
\begin{aligned} b^d-1&=(b^m-1)(1+b^{d/m}+b^{2d/m}+\cdots+b^{(m-1)d/m})\\&\equiv 0\pmod p\end{aligned}
so
$b^{d/m}+b^{2d/m}+\cdots+b^{(m-1)d/m}\equiv -1\pmod p$
and so splitting $L$ into $m$ equal parts and adding them will always give a multiple of $b^{d/m}-1$. | {"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": 75, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9374959468841553, "perplexity": 356.97279348979544}, "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/1524125944479.27/warc/CC-MAIN-20180420155332-20180420175332-00011.warc.gz"} |
http://mathoverflow.net/users/10280/user02138?tab=questions&sort=votes | # user02138
less info
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bio website user02138.myopenid.com location Cambridge, MA 02138 age member for 4 years seen Jun 17 at 18:26 profile views 386
My mathematical interests include Topological Quantum Field Theory, Algebraic Topology, Number Theory and Combinatorics. Litterarum radices amarae, fructus dulces. (Bitter are the roots of study, but how sweet their fruit.) — Cato
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### Counting Lattice Points in Real Polytopes
apr 19 '11 at 9:13 Gil Kalai 8,936
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### What are the consequences of allowing the ABC-conjecture $\kappa_{\epsilon}$ to also vary with $\omega(abc)$?
nov 3 '10 at 19:27 user02138 275
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views | {"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": 0.29835519194602966, "perplexity": 7177.622495181267}, "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-2014-42/segments/1414119648891.34/warc/CC-MAIN-20141024030048-00000-ip-10-16-133-185.ec2.internal.warc.gz"} |
https://wiki-en.twistly.xyz/wiki/Factor_theorem | # Factor theorem
In algebra, the factor theorem is a theorem linking factors and zeros of a polynomial. It is a special case of the polynomial remainder theorem.[1]
The factor theorem states that a polynomial ${\displaystyle f(x)}$ has a factor ${\displaystyle (x-k)}$ if and only if ${\displaystyle f(k)=0}$ (i.e. ${\displaystyle k}$ is a root).[2]
## Factorization of polynomials
Two problems where the factor theorem is commonly applied are those of factoring a polynomial and finding the roots of a polynomial equation; it is a direct consequence of the theorem that these problems are essentially equivalent.
The factor theorem is also used to remove known zeros from a polynomial while leaving all unknown zeros intact, thus producing a lower degree polynomial whose zeros may be easier to find. Abstractly, the method is as follows:[3]
1. "Guess" a zero ${\displaystyle a}$ of the polynomial ${\displaystyle f}$. (In general, this can be very hard, but math textbook problems that involve solving a polynomial equation are often designed so that some roots are easy to discover.)
2. Use the factor theorem to conclude that ${\displaystyle (x-a)}$ is a factor of ${\displaystyle f(x)}$.
3. Compute the polynomial ${\textstyle g(x)={\frac {f(x)}{(x-a)}}}$, for example using polynomial long division or synthetic division.
4. Conclude that any root ${\displaystyle x\neq a}$ of ${\displaystyle f(x)=0}$ is a root of ${\displaystyle g(x)=0}$. Since the polynomial degree of ${\displaystyle g}$ is one less than that of ${\displaystyle f}$, it is "simpler" to find the remaining zeros by studying ${\displaystyle g}$.
### Example
Find the factors of
${\displaystyle x^{3}+7x^{2}+8x+2.}$
To do this one would use trial and error (or the rational root theorem) to find the first x value that causes the expression to equal zero. To find out if ${\displaystyle (x-1)}$ is a factor, substitute ${\displaystyle x=1}$ into the polynomial above:
{\displaystyle {\begin{aligned}x^{3}+7x^{2}+8x+2&=(1)^{3}+7(1)^{2}+8(1)+2\\&=1+7+8+2\\&=18\end{aligned}}}
As this is equal to 18 and not 0. This means ${\displaystyle (x-1)}$ is not a factor of ${\displaystyle x^{3}+7x^{2}+8x+2}$. So, we next try ${\displaystyle (x+1)}$ (substituting ${\displaystyle x=-1}$ into the polynomial):
${\displaystyle (-1)^{3}+7(-1)^{2}+8(-1)+2.}$
This is equal to ${\displaystyle 0}$. Therefore ${\displaystyle x-(-1)}$, which is to say ${\displaystyle x+1}$, is a factor, and ${\displaystyle -1}$ is a root of ${\displaystyle x^{3}+7x^{2}+8x+2.}$
The next two roots can be found by algebraically dividing ${\displaystyle x^{3}+7x^{2}+8x+2}$ by ${\displaystyle (x+1)}$ to get a quadratic:
${\displaystyle {x^{3}+7x^{2}+8x+2 \over x+1}=x^{2}+6x+2,}$
and therefore ${\displaystyle (x+1)}$ and ${\displaystyle x^{2}+6x+2}$ are factors of ${\displaystyle x^{3}+7x^{2}+8x+2.}$ Of these, the quadratic factor can be further factored using the quadratic formula, which gives as roots of the quadratic ${\displaystyle -3\pm {\sqrt {7}}.}$ Thus the three irreducible factors of the original polynomial are ${\displaystyle x+1,}$ ${\displaystyle x-(-3+{\sqrt {7}}),}$ and ${\displaystyle x-(-3-{\sqrt {7}}).}$
## References
1. ^ Sullivan, Michael (1996), Algebra and Trigonometry, Prentice Hall, p. 381, ISBN 0-13-370149-2.
2. ^ Sehgal, V K; Gupta, Sonal, Longman ICSE Mathematics Class 10, Dorling Kindersley (India), p. 119, ISBN 978-81-317-2816-1.
3. ^ Bansal, R. K., Comprehensive Mathematics IX, Laxmi Publications, p. 142, ISBN 81-7008-629-9. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 39, "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": 0.8793827295303345, "perplexity": 300.1147544980567}, "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/1627046152129.33/warc/CC-MAIN-20210726120442-20210726150442-00659.warc.gz"} |
http://compgroups.net/comp.text.tex/xcolor-rowcolors-and-array-environments/182691 | COMPGROUPS.NET | Post | Groups | Users | Stream | Browse | About | |
### xcolor, \rowcolors and array environments
• Email
• Follow
Is it a bug or a feature? (Or how do I disable alternating row colors
in math mode, while still using them outside?)
\documentclass{article}
\usepackage[table]{xcolor}
\rowcolors{2}{red!50}{blue!50}
\begin{document}
$a = \left(\begin{array}{c} 1\\ 2\\ 3 \end{array}\right)$
\end{document}
Thanks,
Dominik
0
See related articles to this posting
Am 11.09.2010 02:42, schrieb Dominik Menke:
> Is it a bug or a feature? (Or how do I disable alternating row colors
> in math mode, while still using them outside?)
>
> \documentclass{article}
> \usepackage[table]{xcolor}
> \rowcolors{2}{red!50}{blue!50}
> \begin{document}
> $a = \left(\begin{array}{c} \hiderowcolors 1\\ 2\\ 3 \end{array}\right)$
> \end{document}
Herbert
0
On Sep 11, 5:28=A0am, Herbert Voss <Herbert.V...@FU-Berlin.de> wrote:
> Am 11.09.2010 02:42, schrieb Dominik Menke:> Is it a bug or a feature? (O=
r how do I disable alternating row colors
> > in math mode, while still using them outside?)
>
> > \documentclass{article}
> > \usepackage[table]{xcolor}
> > \rowcolors{2}{red!50}{blue!50}
> > \begin{document}
> > $a =3D \left(\begin{array}{c} > > \hiderowcolors > =A0 1\\ 2\\ 3 \end{array}\right)$
>
> > \end{document}
>
> Herbert
That is a little inconvenient when there are a lot of
math alignments. It would be better if xcolor.sty
defined a switch to just turn table colors off for the
rest of the document or current group (and one
to turn them back on, of course). I had some success
with the following. However, I didn't test it very much.
\makeatletter
\def\stoprowcolors{\@rowcolorsfalse}
\def\startrowcolors{\@rowcolorstrue}
\makeatother
\everymath{\stoprowcolors}
The problem with \hiderowcolors is that it must occur within
the relevant alignment (its definition starts with \noalign).
A command like \stoprowcolors would be usable elsewhere.
It would also be nice if xcolor.sty could provide a "nomath"
option for row colors.
Dan
0
> It would also be nice if xcolor.sty could provide a "nomath"
> option for row colors.
I'd prefer that way too. I also tried to rewrite the \rowcolors
command, by I already failed on the extraction from xolors.sty...
Dominik
0
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https://math.stackexchange.com/questions/3218924/prove-that-a-covering-projection-of-graphs-is-a-local-isomorphism | # Prove that a covering projection of graphs is a local isomorphism
This question comes from section 1.3 of Topological Graph Theory by Tucker and Gross.
Let $$B$$ be a group acting freely on a graph $$G$$ meaning that the only automorphism defined from the group action with a fixed point (on both vertices and edges) is the identity automorphism. The quotient $$G/B$$ is the graph with vertex orbits as vertices and edge orbits as edges. The endpoints of an edge orbit $$[e]$$ are the vertex orbits containing vertices incident to any edges in $$[e]$$. We define the quotient map $$q_B \colon G \rightarrow G/B$$ to be the map that maps vertices and edges to their orbits.
A covering projection is defined to be a map between graphs $$p \colon \tilde{G} \rightarrow G$$ such that there exists an isomorphism $$i \colon \tilde{G}/B \colon \rightarrow G$$ with $$p = i \circ q_B$$.
The star of a vertex $$v$$, $$\text{star}(v)$$, is the subgraph consisting of $$v$$, the neighbors of $$v$$, and each edge connecting $$v$$ to a neighbor. A map between graphs $$f$$ is a local isomorphism if for each vertex $$v$$ the image $$f(\text{star}(v))$$ is isomorphic to $$\text{star}(v)$$.
The exercise asks to show that if $$p \colon \tilde{G} \rightarrow G$$ is a covering projection and $$G$$ is a simple graph then $$p$$ is a local isomorphism.
Here's my approach so far. It should be sufficient to show that $$q_B$$ is a local isomorphism. This means that no two vertices or edges in $$\text{star}(v)$$ should lie in the same orbit under the action of $$B$$. I have been trying to show that two vertices from $$\text{star}(v)$$ lie in the same orbit, then the action defined by $$B$$ must have a fixed point coming from an automorphism other than the identity. If anyone could help me fill in the details for this argument, or has a completely different approach, it would be much appreciated. | {"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": 26, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9037062525749207, "perplexity": 61.513601502128246}, "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/1566027330968.54/warc/CC-MAIN-20190826042816-20190826064816-00171.warc.gz"} |
https://plato.stanford.edu/entries/physics-experiment/app2.html | ## Appendix 2: The Discovery of CP Violation: A Persuasive Experiment
A group at Princeton University, led by Cronin and Fitch, decided to test CP conservation. The experimenters were quite aware of the relevance of their experiment to the question of CP violation, but they did not expect to observe it. As Val Fitch, one of the group leaders remarked, “Not many of our colleagues would have given credit for studying CP invariance, but we did so anyway” (Fitch 1981, p. 991). A preliminary estimate indicated that the CP phase of the experiment would detect about 7500 $$\ce{K2^0}$$ decays and thus reduce the limit on CP violation from the then current limit of 1/300 (0.3%) to 1/7500 (For details of this episode see Franklin (1986, Ch. 3)).
The experimental beam contained only $$\ce{K2^0}$$ mesons. (The $$\ce{K1^0}$$ meson has a much shorter lifetime than the $$\ce{K2^0}$$ meson, so that if we start with a beam containing both types of particles, after a time only the $$\ce{K2^0}$$ mesons will remain). The experimental apparatus detected two charged particles from the decay of the $$\ce{K2^0}$$ meson. The vector momentum of each of the two decay products from the $$\ce{K2^0}$$ beam and the invariant mass $$m^*$$ were computed assuming that each product had the mass of a pion:
$m^* = [(E_1 + E_2)^2 - (\bp_1 + \bp_2)^2]^{\bfrac{1}{2}},$
where E and $$\bp$$ are the energy and vector momenta of the pions, respectively. If both particles were indeed pions from $$\ce{K2^0}$$ decay, $$m^*$$ would equal the $$\ce{K2^0}$$ mass. The experimenters also computed the vector sum of the two momenta and the angle between this sum and the direction of the $$\ce{K2^0}$$ beam. This angle should be zero for two-body decays, but not, in general, for three-body decays.
Figure 3. Angular distributions in three mass ranges for events with $$\cos(\theta) \gt 0.9995$$. From Christenson et al. (1964).
This was exactly what the Princeton group observed (Christenson et al. 1964). As seen clearly in Figure 3, there is a peak at the $$\ce{K^0}$$ mass, 498 MeV/c2, for events with $$\cos(\theta)$$ greater than 0.9999 ($$\cos(\theta)$$ approximately equal to 1 means $$\theta$$ is approximately equal to 0). No such peak is seen in the mass regions just above or just below the $$\ce{K^0}$$ mass. The experimenters reported a total of $$45\pm 9$$ two-pion $$\ce{K2^0}$$ decays out of a total of 22,700 $$\ce{K2^0}$$ decays. This was a branching ratio of $$(1.95\pm 0.2) \times 10^{-3}$$, or approximately 0.2 percent.
The most obvious interpretation of the Princeton result was that CP symmetry was violated. This was the view taken in three out of four theoretical papers written during the period immediately following the report of that result. The Princeton result had persuaded most of the physics community that CP symmetry was violated. The remaining theoretical papers offered alternative explanations.[1] These alternatives relied on one or more of three arguments: (1) the Princeton results are caused by a CP asymmetry (the local preponderance of matter over antimatter) in the environment of the experiment, (2) $$\ce{K2^0}$$ decay into two pions does not necessarily imply CP violation, and ( 3) the Princeton observations did not arise from two-pion $$\ce{K2^0}$$ decay. This last argument can divided into the assertions that (3a) the decaying particle was not a $$\ce{K2^0}$$ meson, (3b) the decay products were not pions, and (3c) another unobserved particle was emitted in the decay. Included in these alternatives were three suggestions that cast doubt on well-supported fundamental assumptions of modern physics. These were: (1) pions are not bosons, (2) the principle of superposition in quantum mechanics is violated, and (3) the exponential decay law fails. Although by the end of 1967 all of these alternatives had been experimentally tested and found wanting, the majority of the physics community had accepted CP violation by the end of 1965, even though all the tests had not yet been completed. As Prentki, a theoretical particle physicist, remarked, this was because in some cases “the price one has to pay in order to save CP becomes extremely high,” and because other alternatives were “even more unpleasant” (Prentki 1965).
This is an example of what one might call a pragmatic solution to the Duhem-Quine problem.[2] The alternative explanations and the auxiliary hypotheses were refuted, leaving CP violation unprotected. One might worry that other plausible alternatives were never suggested or considered. This is not a serious problem in the actual practice of physics. No fewer than ten alternative explanations of the Princeton result were offered, and not all of them were very plausible. Had others been suggested they, too, would have been considered by the physics community. Consider the model of Nishijima and Saffouri (1965). They explained two-pion $$\ce{K2^0}$$ decay by the existence of a “shadow” universe in touch with our “real” universe only through the weak interactions. They attributed to the two pion decay observed to the decay of the $$\ce{K^0}$$ from the shadow universe. This implausible model was not merely considered, it was also experimentally tested. Everett (1965) noted that if the $$\ce{K^0}$$, the shadow $$\ce{K^0}$$ postulated by Nishijima and Saffouri existed, then a shadow pion should also exist, and the decays of the $$\ce{K+}$$ into a positive pion and a neutral pion and of the $$\ce{K+}$$ into a positive pion and a neutral shadow pion and should occur with equal rates. The presence of the shadow pion could be detected by measuring the ordinary $$\ce{K+}$$ branching ratio in two different experiments, one in which the neutral pion was detected and one in which it was not. If the shadow pion existed the two measurements would differ. They didn’t. There was no shadow pion and thus, no $$\KOp$$
What was the difference between the episodes of parity nonconservation and CP violation. In the former parity nonconservation was immediately accepted. No alternative explanations were offered. There was a convincing and decisive set of experiments. In the latter at least ten alternatives were proposed, and although CP violation was accepted rather quickly, the alternatives were tested. In both cases there are only two classes of theories, those that conserve parity or CP, and those that do not. The difference lies in the length and complexity of the derivation linking the hypothesis to the experimental result, or to the number of auxiliary hypotheses required for the derivation. In the case of parity nonconservation the experiment could be seen by inspection to violate mirror symmetry (See Figure 1). In the CP episode what was observed was $$\ce{K2^0}$$ decay into two pions. In order to connect this observation to CP conservation one had to assume (1) the principle of superposition, (2) that the exponential decay law held to 300 lifetimes, (3) that the decay particles were both “real” pions and that pions were bosons, (4) that no other particle was emitted in the decay, (5) that no other similar particle was produced, and (6) that there were no external conditions present that might regenerate $$\ce{K1^0}$$ mesons. It was these auxiliary assumptions that were tested and eliminated as alternative explanations by subsequent experiments.
The discovery of CP violation called for a theoretical explanation, a call that is still unanswered. | {"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": 0.9640079736709595, "perplexity": 748.1079079318991}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-39/segments/1631780057083.64/warc/CC-MAIN-20210920161518-20210920191518-00059.warc.gz"} |
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# Effect of a partial coverage of quasar broad-line regions by intervening H$$_2$$-bearing clouds
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### Abstract
We consider the effect of a partial coverage of quasar broad-line regions (QSO BLRs) by intervening H$$_2$$-bearing clouds when a part of quasar (QSO) radiation passes by a cloud not taking part in formation of an absorption-line system in the QSO spectrum. That leads to modification of observable absorption line profiles and consequently to a bias in physical parameters derived from standard absorption line analysis. In application to the H$$_2$$ {absorption} systems the effect has been revealed in the analysis of H$$_2$$ absorption system in the spectrum of Q~1232+082 (Ivanchik et al. 2010, Balashev et al. 2011). We estimate a probability of the effect to be detected in QSO spectra. To do this we derive distribution of BLR sizes of high-z QSOs from Sloan Digital Sky Survey (SDSS) Data Release 9 (DR9) catalogue and assume different distributions of cloud sizes. We conclude that the low limit of the probability is about $$11\%$$. The latest researches shows that about a fifth of observed H$$_2$$ absorption systems can be partially covered. Accounting of the effect may allow to revise significantly physical parameters of interstellar clouds obtained by the spectral analysis.
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###### Journal
1508.04750
10.1007/s10509-015-2459-4
Galaxy astrophysics | {"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": 0.9187095761299133, "perplexity": 3707.6622452104707}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-26/segments/1560628000044.37/warc/CC-MAIN-20190626013357-20190626035357-00052.warc.gz"} |
http://mathhelpforum.com/calculus/213998-fourier-transform.html | # Math Help - fourier transform
1. ## fourier transform
Hi members,
I have some problems with fourier tranfrom.
See the two attached Pdf files.Questions are on the pdf files. | {"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": 0.9974910616874695, "perplexity": 16151.938910788946}, "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-2015-40/segments/1443738095178.99/warc/CC-MAIN-20151001222135-00207-ip-10-137-6-227.ec2.internal.warc.gz"} |
https://plainmath.net/6804/find-exponential-function-that-fits-experimental-data-collected-over | Find an exponential function that fits the experimental data collected over time
Find an exponential function that fits the experimental data collected over time t.
$\begin{array}{|cccccc|}\hline t& 0& 1& 2& 3& 4\\ y& 600.00& 630.00& 661.50& 694.58& 729.30\\ \hline\end{array}$
You can still ask an expert for help
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Macsen Nixon
Let the exponential function be, $y=b\left(a{\right)}^{t}$
$600=b\left(a{\right)}^{0}$
$b=600$ And at $t=1$
$630=600\left(a{\right)}^{1}$
Therefore, the exponential function is $y=600\left(\frac{21}{20}{\right)}^{t}$. | {"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": 0.8533432483673096, "perplexity": 1197.2858097974467}, "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/1656103355949.26/warc/CC-MAIN-20220628050721-20220628080721-00607.warc.gz"} |
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# If y is 50 percent more than 50 percent of 200, what percentage
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If y is 50 percent more than 50 percent of 200, what percentage [#permalink]
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15 Sep 2018, 08:18
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If y is 50 percent more than 50 percent of 200, what percentage of 200 is y?
(A) 25%
(B) 100/3%
(C) 50%
(D) 200/3%
(E) 75%
New question!!!..
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Re: If y is 50 percent more than 50 percent of 200, what percentage [#permalink]
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15 Sep 2018, 08:35
chetan2u wrote:
If y is 50 percent more than 50 percent of 200, what percentage of 200 is y?
(A) 25%
(B) 100/3%
(C) 50%
(D) 200/3%
(E) 75%
New question!!!..
y = 50% of 200 of 150% of 200 is $$\frac{50}{100}*200*\frac{150}{100}$$ = $$150$$
Now, a% of 200 = 150
Hence, $$a = 75$$, Answer must be (E)
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If y is 50 percent more than 50 percent of 200, what percentage [#permalink]
### Show Tags
15 Sep 2018, 11:44
chetan2u wrote:
If y is 50 percent more than 50 percent of 200, what percentage of 200 is y?
(A) 25%
(B) 100/3%
(C) 50%
(D) 200/3%
(E) 75%
New question!!!..
We don't need 200 - multiply the multipliers
If y is 50 percent more than 50 percent of 200, what percentage of 200 is y?
The base of comparison is the same number. We could put any number in for 200. (Try 60, e.g.)
y as a percentage of the base number (200, 60, x) does not change
Let $$200 = x$$
Multiply the multipliers:
$$y = (1.5*.5x)=.75x$$
$$\frac{y}{x}=.75=75$$ %
Answer E
Algebraically
(1) "y is 50 percent more than"
(2) "50 percent of 200"
Solve (2) first. It has a "counting" number
=>
(2) 50 percent of 200 = 100
Then
(1) y is 50 percent more than 100: y = 150
(3) What percentage of 200 is y? Reverse:
$$y$$ (150) IS what percentage OF 200?
$$y=\frac{x}{100}*200$$
$$150=\frac{x}{100}*200$$
$$(\frac{150}{200}*100)=(.75*100)=75$$ %
Answer E
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If y is 50 percent more than 50 percent of 200, what percentage [#permalink] 15 Sep 2018, 11:44
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Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | {"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": 0.4009193480014801, "perplexity": 8372.967454056965}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-22/segments/1558232257939.82/warc/CC-MAIN-20190525084658-20190525110658-00323.warc.gz"} |
https://www.sparrho.com/item/deterministic-quantum-network-for-distributed-entanglement-and-quantum-computation/1b57be8/ | Deterministic Quantum Network for Distributed Entanglement and Quantum Computation
Research paper by I. Cohen, K. Mølmer
Indexed on: 22 Feb '18Published on: 22 Feb '18Published in: arXiv - Quantum Physics
Abstract
We propose a simple architecture for a scalable quantum network, in which the quantum nodes consist of qubit systems confined in cavities. The nodes are deterministically coupled by transmission and reflection of photons, which are disentangled from the qubits at the end of each operation. A single photon can generate an entangling controlled phase (C-PHASE) gate between any selected number of qubits in the network and forms the basis for universal quantum computing, distributed over multiple processor units. We analyze this network and we show that the gate fidelity is high, also when more qubits are involved, as requested, e.g., in an efficient Grover search. In our derivation we consider qubit transition energies in the optical regime, however, we stress that it can be readily generalized to other architectures where the nodes may be coupled, e.g., by microwave photons. | {"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": 0.9106332063674927, "perplexity": 1184.0257092763713}, "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-2018-13/segments/1521257651780.99/warc/CC-MAIN-20180325025050-20180325045050-00006.warc.gz"} |
https://yutsumura.com/welcome-to-problems-in-mathematics/ | Welcome to Problems in Mathematics
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https://codegolf.stackexchange.com/questions/141315/what-an-odd-function/141424 | # What an Odd Function
Your task here will be to implement a function1 that forms a permutation on the positive integers (A bijection from the positive integers onto themselves). This means that each positive integer should appear exactly once in the permutation. The catch is your function should have a larger probability of outputting an odd number than an even number.
Now this may seem strange or impossible. Surely there are just as many odd numbers as even numbers? And while this intuition is correct for finite sets it actually does not hold for infinite sets. For example take the following permutation:
1 3 2 5 7 4 9 11 6 13 15 8 17 19 10 21 23 12 25 27 14 29 31 16 33 35 18 37 39 20 41 43 22 45 47 24 49 51 26 53 55 ...
If you take any subsection of the sequence with size greater than $$\1\$$ you will have at least as many odd numbers as even numbers, thus it seems that the probability of any random term being odd is greater than that of being even. You will also note every number odd or even number will eventually appear in the sequence and can only appear once. Thus the sequence is a true permutation.
### Definition of Probability
To avoid confusion or ambiguity I am going to clearly lay out what is meant by probability in this question.
Let us say we have a function $$\f\$$. The probability of a number being odd will be defined as the limit of ratio odd members of the set to the size of the set $$\f\{1\dots n\}\$$ as $$\n\$$ tends towards infinity.
$$\lim_{n\rightarrow \infty}\dfrac{\left|\{x : x \in \{1\dots n\}, \mathrm{odd}(f(x))\}\right|}{n}$$
For example the aforementioned function would have a probability of being odd of $$\2/3\$$.
This is so answers will be scored in bytes with less bytes being better.
### Extra Challenges
Here are some fun ideas to play around with and perhaps try to implement. These are just for fun and do not affect scoring in any way. Some of these are not even valid solutions to this challenge, and an answer which only includes solutions to challenges 2 or 3 is not a valid answer, and is liable to be deleted.
• Write a permutation with an odd probability of $$\1\$$. (this is possible)
• Write a permutation that has more odd numbers than even numbers in $$\f\{1\dots n\}\$$ for any $$\n\$$ but has a odd probability of $$\1/2\$$.
• Write a permutation that has no defined probability (that is there is no limit).
1: Here function will mean program or function. It is just a piece of code that takes input and produces output.
# Jelly, 7 bytes
Æf^<¥4P
Swaps 2s and 3s in the input's prime factorization. The probability of odds is 2/3.
Try it online!
### How it works
Æf^<¥4P Main link. Argument: n
Æf Compute all prime factors of n, with duplicates.
¥4 Combine the two links to the left into a dyadic chain and call it with
right argument 4.
< Compare each prime factor with 4. Yields 1 for 2 and 3, 0 otherwise.
^ Bitwise XOR the results with the corresponding prime factors.
This maps 2 to 3, 3 to 2, and all other primes to themselves.
P Take the product of the resulting primes.
• This answer is quite clever. I believe I understand why it works but you might want to include a proof that it works because I found it unintuitive at first. – Wheat Wizard Sep 2 '17 at 16:32
• Proof that this is a permutation: the function is its own inverse. Proof of ratio: chance that an output is odd is the chance that the original had no factors 3, which is exactly when it's not divisible by three. That chance is 2/3. – tomsmeding Sep 3 '17 at 8:40
# Husk, 11 10 bytes
-1 byte thanks to Leo, and a slightly different function
This has an odd probability of 1
!uΣz:NCNİ1
Try it online!
It indexes the sequence:
[1,2,3,5,7,9,11,4,13,15,17,19,21,23,25,27,29,6,31,33]
1 odd, 1 even, 5 odd, 1 even, 9 odd, 1 even, 13 odd...
### Explanation
! Index the following sequence (1-indexed)
u remove duplicates [1,2,3,5,7,9,11,4,13,15...]
Σ Concatenate [1,1,2,3,5,3,7,9,11,4,13..]
z: Zipwith append [[1,1],[2,3,5],[3,7,9,11]..
N Natural numbers
CNİ1 Odd numbers cut into lists of lengths [[1],[3,5],[7,9,11]...]
corresponding to the Natural numbers
f n=n:2*n+1:2*n+3:f(n+2)
(f 0!!)
Implements the example sequence [1,3,2,5,7,4,9,11,6,13,15,8,17,19,10,21,...].
Try it online!
For reference: old version, 34 bytes (-1 byte thanks to @xnor):
(!!)$do e<-[0,2..];[e,2*e+1,2*e+3] Try it online! • Save a paren: (!!)$do ... – xnor Sep 2 '17 at 1:24
# Husk, 8 bytes
!uΣzeİ1N
Try it online!
This implements the example sequence (1,3,2,5,7,4...).
### Explanation
!uΣzeİ1N
ze zip together
İ1 the odd numbers
N with the natural (positive) numbers
Σ flatten the resulting list
u remove duplicates
! index into the obtained sequence with the input
Everybody does Challenge 1, so let's do the other two.
{($_==1)+$_-(-1)**($_%%2)} Try it online! It's just 1 3 2 5 4 7 6... In an even number of terms, there are always 2 more odd numbers than even. In an odd number, 1 more. However this has clearly limit of (n+2)/(2n+2) -> ½. # Perl 6, 70 bytes — Challenge 3 {((1,),(2,4),->@a,@ {@(map(@a[*-1]+2*(*+1),^(4*@a)))}...*).flat[$_-1]}
Try it online!
Admittedly, this is horribly golfed. It indexes a sequence that contains 2⁰ odd numbers, then 2¹ even, then 2² odd, then 2³ even, and so on.
The probability after n such "blocks", if n is odd, is (2⁰+2²+2⁴+...+2ⁿ⁻¹)/(2ⁿ-1). The sum in the numerator is equal to ⅓(4½(n+1) - 1) = ⅓(2n+1 - 1). So the probability after odd number of blocks is ⅔ (in the limit).
If we add one more block (and strike an even count of them n+1), however, we didn't add any odd numbers (numerator stays the same) but there is now (2n+1 - 1) numbers in total. The parentheses cancel and we get the probability of ⅓ (in the limit).
This is apparently supposed to have 2 different cluster points, ⅓ and ⅔, to make sure that the limit doesn't exist, but this doesn't really prove it. My attempt to do a solid, rigorous proof can be found in this Math.SE answer: https://math.stackexchange.com/a/2416990/174637. Bashing mistakes is welcome.
# Perl 6, 39 bytes — The core challenge.
{my$l=$_ div 3;$_%3??2*($_-$l)-1!!2*$l}
Try it online!
Though I posted this answer because of the challenges 2 and 3 which offered a pleasant little mathy puzzle, there is a strict requirement for all answers to contain a solution to the core challenge. Here it is then.
This is the example sequence.
• These are extra challenges. For this to be a valid answer you must supply a solution to the core challenge. A solution to challenge 1 is also a solution to the core challenge, but a solution to challenges 2 or 3 is not. – Peter Taylor Sep 2 '17 at 7:23
• Well, the extra challenges are what's interesting on this question for me. The core challenge is not. But I added some solution anyway. – Ramillies Sep 2 '17 at 11:34
• I asked for a proof that your response to Challenge 3 has no limit in this Math.SE question: math.stackexchange.com/questions/2416053/… – Kevin Sep 4 '17 at 5:05
• @Kevin, thanks for asking. I think I may have confused you. I was quite sure it was OK. The only thing is that I often prove things quite rigorously for myself, just for the peace of mind (because your feet may slip quite easily, esp. when handling infinite objects like this) — and I haven't done it here. That's all what I wanted to say. – Ramillies Sep 4 '17 at 17:33
• @Kevin — so after all, I overcame my laziness (a heroic deed!) and made the proof. I posted it as an answer to your Math.SE question. Hopefully it's OK (doing this kind of work at the night isn't really a good idea :—)). It turned out that it isn't nearly as horrible as I initially thought. – Ramillies Sep 4 '17 at 23:06
# Brain-Flak, 120 bytes
(({})<{{({}[()]<({}(()[{}]))>)}{}({}[({})]<({}<>{}<({}())>)><>)}>)<>({}[()]){{}((<>{}<>[{}]){}[()])<>}{}{(({}){})<>{}}<>
Try it online!
Performs the following function:
This function generates the sequence
2 4 1 6 3 5 7 8 9 11 13 15 17 19 21 10 23 25 27 29...
The function has an odd probability of 1
# R, 82 bytes (Extra challenge 1)
f<-function(n){if(sqrt(n)==floor(sqrt(n))){2*sqrt(n)}else{2*(n-floor(sqrt(n)))-1}}
Try it Online!
If the input is a perfect square, gives an even number. Otherwise, gives an odd number. The perfect squares have natural density 0, which means that this sequence gives odd numbers with probability 1.
• 58 bytes – Giuseppe Sep 6 '17 at 17:28
• 56 bytes – Giuseppe Sep 26 '17 at 15:37
• 53 bytes – Giuseppe Jan 24 '18 at 22:42
# C (gcc), 29 bytes
f(n){return n&3?n+n/2|1:n/2;}
Try it online!
Every fourth number is even:
1 3 5 7 9 11 13 15 17 19 21 23 25 27 29
2 4 6 8 10
# Extra challenge 1, 52 bytes
f(n,i){for(i=0;n>>i/2;i+=2);return n&n-1?2*n-i-1:i;}
Try it online!
Returns 2*(x+1) if n equals 2x and consecutive odd numbers otherwise:
1 3 5 7 9 11 13 15 17 19 21 23 25
2 4 6 8 10
• return n&n-1?2*n-i-1:i; can be return~-n&n?2*n+~i:i for -2. In addition, but return can be n= to save some more bytes. – Kevin Cruijssen Mar 12 '20 at 16:11
# Brain-Flak, 140138 136 bytes
({}<(((()())()()))((<>[()])[()()])>){({}[()]<(({}(({}({}))[({}[{}])]))[({}[{}])]<>(({}(({}({}))[({}[{}])]))[({}[{}])])<>)>)}{}({}<{}{}>)
Try it online!
## Explanation
This performs a similar function to the one suggested in the question.
2 3 1 4 7 5 6 11 9 8 15 13 10 17 15 ...
It works mostly based on a snippet I made to roll the stack for size 3 stacks.
(({}(({}({}))[({}[{}])]))[({}[{}])])
We set up two stacks one with accumulator values (two odd one even) and one with the numbers 4 4 2. Each iteration we roll both stacks and add the top of the left stack to the top of the right stack.
(({}(({}({}))[({}[{}])]))[({}[{}])]<>(({}(({}({}))[({}[{}])]))[({}[{}])])<>)
This will increment each odd number by 4 and the one even number by 2. As we loop through we get a pattern of 2 odd 1 even, with every positive integer being hit. Thus we just loop n times with n being the input. This has an asymptotic probability of 2/3.
# Jelly, 10 bytes
ÆE;0ṭ2/FÆẸ
The probability of odds is 2/3.
Try it online!
### How it works
ÆE;0ṭ2/FÆẸ Main link. Argument: n
ÆE Compute the exponents of n's prime factorization.
;0 Append a 0.
2/ Reduce all pairs by...
ṭ tack, appending the left argument to the right one.
This inverts all non-overlapping pairs of exponents.
F Flatten the result.
ÆẸ Consider the result a prime factorization and compute the corresponding
integer.
# JavaScript, 23 bytes
n=>n/2+n/2%2+(n%4&&n-1)
Output: 1, 3, 5, 2, 7, 9, 11, 4, 13, 15, 17, 6, 19, 21, 23, 8...
• For all n = 4k:
• f(n) = n/2 = 2k
• For all n = 4k + b
• f(n) = n/2 + b/2 + n - 1 = 3/2 * (4k + b) + 1/2 * b - 1 = 6k + 2b - 1
Challenge 2:
n=>n^(n>1)
Output: 1, 3, 2, 5, 4, 7, 6, 9, 8, 11, 10, 13, 12, 15, 14
• n=>n%4?1.5*n|1:n/2 is 5 bytes shorter. – nwellnhof Sep 2 '17 at 16:01
# C, 80 bytes
#define R if(k++==n)return
i,j,k;f(n){for(i=k=1,j=2;;i+=4,j+=2){R i;R i+2;R j;}}
Implementation of the example permutation from the question.
Try it online!
## Batch, 36 bytes
@cmd/cset/an=%1*2,(-~n*!!(n%%3)+n)/3
Implements the sequence given in the question.
## CJam (21 bytes)
{2b_(&!\_,2*\1$~+2b?} Online demo showing the first 32 outputs. This is an anonymous block (function). This is also a solution to challenge 1: the numbers mapped to even numbers are the powers of 2, so the density of even numbers in the first n outputs is lg(n)/n which tends to zero. ### Dissection { e# Declare a block; let's call the input x 2b e# Convert to base 2 _(& e# Copy, pull out first digit (1) and set intersection with rest ! e# Boolean not, so empty set (i.e. power of 2) maps to 1 and non-empty e# to 0 \_,2* e# Take another copy, find its length, and double it \1$~+ e# Take the original base 2 array and append ~(2L) = -2L-1
2b e# Convert from base 2, to get 2x-2L-1
? e# Take the 2L if it was a power of 2, and the 2x-2L-1 otherwise
}
# Perl 40 bytes
$,=$";$i=4;{say$i-3,$i/2,($i+=4)-5;redo}
# Brain-Flueue, 88 bytes
({}<(((<>)[()])[()()])>)<>(((()())()()))<>{({})({})({})({}[()]<({}<>({})<>)>)}{}{}({}){}
Try it online!
## Explanation
This implements the same function as my last answer but uses Brain-Flueue's FIFO model to cut some corners. Here are the first couple terms it generates.
2 3 1 4 7 5 6 11 9 8 15 13 10 17 15 ...
The first part of the code is just a bit of setup, we put 0,-1,-3 on the first stack and 2,4,4 on the second stack. The 2,4,4 will be used to cycle through even and odd numbers just as I did in my Brain-Flak answer.
We then loop n times, each time adding the top of the left stack to the right stack. Since Brain-Flueue uses queues as opposed to stacks the values naturally roll as we touch them preventing the need for extra code.
• What is the difference between Flueue and Flak? – RaviRavioli Jan 24 '18 at 22:16
• @tfbninja Flueue uses a Queue instead of a stack. – Wheat Wizard Jan 24 '18 at 22:36
• but...you're using the bflk interpreter... how do you make it different – RaviRavioli Jan 25 '18 at 0:22
• @tfbninja The -lflueue argument. – Wheat Wizard Jan 25 '18 at 3:15
# Python 2, 46104 55 bytes
lambda n:2*((n-int(n**.5))+.5,n**.5-1)[n!=1>0==n**.5%1]
Try it online!
Misread the question, now properly implemented a function that can be used to generate a sequence instead of one that generates a sequence. Also excluded 0 from the set of possible outputs.
Probability of finding an odd positive integer now converges to 1.
• This should return an number, not a set / list as far as I understood – Mr. Xcoder Sep 1 '17 at 22:59
• Also, this is not a correct permutation, since it contains 0. – Mr. Xcoder Sep 1 '17 at 23:05
• @Mr.Xcoder Thanks for noticing. – Jonathan Frech Sep 2 '17 at 17:52
# Jelly, 9 bytes
Ḥ€’ĖUẎQ⁸ị
Try it online!
Implements 1, 3, 2, 5, 7, 4, 9, 11, 6, ... (probability 2/3).
# 05AB1E, 11 bytes
·ÅÉā‚ø˜Ù¹<è
Try it online!
# Pyth, 9 bytes
*Fmxd<d4P
You can use this code to verify the ratio of odd numbers up to a certain point. Substitute 10000 with your desired limit (don't set it much higher, because it memory errors).
Km*Fmxk<k4PdS10000clf%T2KlK
The above gives roughly 0.667. The true probability of odd occurrences is 2/3. This approach is an equivalent implementation of Dennis' answer.
# Explanation
*Fmxd<d4P Full program.
P Prime factors.
m Map over ^.
x Bitwise XOR between:
d The current prime factor.
<d4 The integer corresponding to the boolean value of current factor < 4.
*F Product of the list.
# Java 8, 20 bytes
n->n%4>0?n+n/2|1:n/2
Some things I tried myself ended up being a few bytes longer or slightly incorrect..
Implements: 1,3,5,2,7,9,11,4,13,15,17,6,19,21,23,8,25,27,29,10,31,33,35,12,37,...
with a probability of 3/4.
Try it here.
## Lua, 67 53 bytes
Explanation coming when I'm done golfing this :)
This program takes an integer via command-line arguments as input and prints out the nth element of the exemple sequence to STDOUT
n=...print(n%3<1 and n/3*2or n+math.floor(n/3)+n%3-1)
### Explanations
n=... -- shorthand for the argument
print( -- prints out the result of the following ternary
n%3<1 -- if n is divisible by 3
and n/3*2 -- prints out the corresponding even number
or n+math.floor(n/3)+n%3-1) -- else prints out the odd number
The even numbers of this sequence are both the nth even number and the n multiple of 3, hence the formula n%3*2 is sufficient to generate them.
For odd numbers, it's a little bit more tough. Based on the fact that we can find them depending on the current n, we have the following table :
n | 1 2 4 5 7 8 10 11
target | 1 3 5 7 9 11 13 15
target-n| +0 +1 +1 +2 +2 +3 +3 +4
Let's call the value target-n i, we can see that each time n%3==2, i is incremented. There goes our formula :
n+math.floor(n/3)+n%3-1
The odd numbers are based on n on which we add i.
The value of i increments at the same rate as the euclidian division by 3, with an offset. math.floor(n/3) gives us the rate of increment and n%3-1 gives us the offset, making it happens on n%3==2 instead of n%3==0.
• One byte could easily be saved by removing an unnecessary space (...and (n/...). – Jonathan Frech Sep 2 '17 at 17:55
• @JonathanFrech was able to save 2 at this spot by totally removing the parenthesis as and n/3*2or works just as fine – Katenkyo Sep 6 '17 at 17:23
# 05AB1E, 6 bytes
ÒD4‹^P
Port of @Dennis' Jelly answer, so make sure to upvote him!
The probability of odds is therefore also $$\\frac{2}{3}\$$.
Explanation:
Ò # Get the prime factors of the (implicit) input-integer, counting multiplicities
D # Duplicate this list
4‹ # Check for each value whether it's smaller than 4 (1 for 2 and 3; 0 otherwise)
^ # Bitwise-XOR the two lists at the same indices
# (2 becomes 3; 3 becomes 2; everything else remains unchanged)
P # And take the product of this
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http://mathhelpforum.com/differential-geometry/110478-prove-sin-x-uniformly-continuous-print.html | # prove sin(x) is uniformly continuous.
Show 40 post(s) from this thread on one page
Page 1 of 2 12 Last
• Oct 25th 2009, 06:36 PM
cgiulz
prove sin(x) is uniformly continuous.
Prove directly from the definition of uniform continuity that $\sin(x)$ is uniformly continuous on $(-\infty,\infty).$
(Hint: a unit circle may help to make an estimation.)
• Oct 25th 2009, 07:01 PM
Krizalid
prove that $|\sin x-\sin y|\le|x-y|$ holds for $x,y\in\mathbb R$ and you're done.
• Oct 25th 2009, 08:45 PM
cgiulz
Anything wrong with this argument?
given $\epsilon,\delta > 0$
$|\sin(x) - \sin(y)| \leq |x - y| < \epsilon = \delta$
• Oct 25th 2009, 08:49 PM
Bruno J.
Double post.
• Oct 25th 2009, 08:51 PM
Bruno J.
$|m-n|\leq |m|-|n|$ is wrong. Take m=0, n=1 for example.
• Oct 25th 2009, 08:53 PM
cgiulz
Quote:
Originally Posted by Bruno J.
$|m-n|\leq |m|-|n|$ is wrong. Take m=0, n=1 for example.
Typo; I was trying to apply the triangle inequality. It is not even necessary anyway. Any thoughts on the edited post?
• Oct 25th 2009, 08:55 PM
Jose27
Since the derivative is bounded, using the mean value theorem you get that the function is uniformly continous.
• Oct 25th 2009, 09:04 PM
redsoxfan325
Quote:
Originally Posted by cgiulz
Prove directly from the definition of uniform continuity that $\sin(x)$ is uniformly continuous on $(-\infty,\infty).$
(Hint: a unit circle may help to make an estimation.)
Here are some tools that will help you.
$|\sin x-\sin y|=\left|2\sin\left(\frac{x-y}{2}\right)\cos\left(\frac{x+y}{2}\right)\right|$
$|\cos\theta|\leq1$ and you can use the unit circle to show that $|\sin\theta|\leq|\theta|$.
Let $\delta=\epsilon$ and don't look back.
• Oct 25th 2009, 09:06 PM
cgiulz
Quote:
Originally Posted by Jose27
Since the derivative is bounded, using the mean value theorem you get that the function is uniformly continous.
Yes, but how can it be done directly from the definition?
• Oct 25th 2009, 09:07 PM
redsoxfan325
Quote:
Originally Posted by Jose27
Since the derivative is bounded, using the mean value theorem you get that the function is uniformly continous.
The only thing is that this requires him to assume $\sin x$ is differentiable and furthermore that it's derivative is $\cos x$. (If they haven't proved that $\sin x$ is continuous yet, it's unlikely they've proved its derivative is $\cos x$.)
• Oct 25th 2009, 09:11 PM
cgiulz
Quote:
Originally Posted by redsoxfan325
The only thing is that this requires him to assume $\sin x$ is differentiable and furthermore that it's derivative is $\cos x$. (If they haven't proved that $\sin x$ is continuous yet, it's unlikely they've proved its derivative is $\cos x$.)
This is true.
Go yanks (Wink)
• Oct 25th 2009, 09:16 PM
redsoxfan325
Quote:
Originally Posted by cgiulz
Go yanks (Wink)
(Angry)
I hope the Phillies can pull it off two years in a row.
• Oct 25th 2009, 09:22 PM
Jose27
You could say that $\sin x$ is unif. cont. in $[0,2 \pi]$ and since it's periodic it's unif. cont. in $\mathbb{R}$
Edit. Apparently I keep avoiding the definition of uniform cont.
• Oct 25th 2009, 09:32 PM
cgiulz
Quote:
Originally Posted by cgiulz
Anything wrong with this argument?
given $\epsilon,\delta > 0$
$|\sin(x) - \sin(y)| \leq |x - y| < \epsilon = \delta$
I think this is correct, is it not?
• Oct 25th 2009, 09:34 PM
redsoxfan325
That is true, yes, but I think you need to show that $|\sin x-\sin y|\leq|x-y|$, not just state it.
Show 40 post(s) from this thread on one page
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https://mail.python.org/pipermail/python-list/2003-December/233428.html | # How to delete this file ???
Peter Hansen peter at engcorp.com
Mon Dec 1 22:57:34 CET 2003
DCK wrote:
>
> Hello again
> Really sorry for long time between me letters. All problem gone, after i've
> used raw string (r"\\path\to\any\file") Really thanks for all answers :)
Note that it's almost always better to use forward slashes instead.
In this case, "//path/to/any/file" would work wonderfully, and is much
Yes, it does work with Windows. The only time it doesn't is when using
command line programs which interpret the forward slash as an option,
and insist on backslashes.
(An unfortunate additional area is that os.path for Windows normalizes
things to use backslashes, and therefore path comparisons can get a little
tricky if you use forward slashes but aren't careful about normalizing
with os.path.normpath() all the time.)
-Peter | {"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": 0.9451194405555725, "perplexity": 9575.024293163917}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-50/segments/1480698540928.63/warc/CC-MAIN-20161202170900-00009-ip-10-31-129-80.ec2.internal.warc.gz"} |
http://mathhelpforum.com/differential-geometry/146114-image-t.html | # Math Help - Image of T
1. ## Image of T
If $H$ is a Hilbert Space and $T:H\rightarrow H$ a bounded operator. Given that $T$ is bounded below $\|Tx\|\geq\delta\|x\|$ for some $\delta>0$ and that $ker(T^*)=0$ show that the image of $im(T)$ of $T$ is closed.
This is a revision question to prove that if an operator is bounded below and the kernel of the adjoint is zero then it is invertible. The question requires that you prove $T$ is bijective. I have proved injective and done most of the surjection proof, I just now need to show that the image of T is equal to the closure of the image of T i.e. $im(T)$ is closed.
I am assuiming that you need to show that the compliment is open somehow, it seems clear that the compliment is the empty set so open and closed. Any help would be great.
2. Originally Posted by ejgmath
If $H$ is a Hilbert Space and $T:H\rightarrow H$ a bounded operator. Given that $T$ is bounded below $\|Tx\|\geq\delta\|x\|$ for some $\delta>0$ and that $ker(T^*)=0$ show that the image of $im(T)$ of $T$ is closed.
(You don't need the information about $\ker(T^*)$ for this part of the question.) If $Tx_n\to y$ then $(Tx_n)$ is a Cauchy sequence. The condition $\|Tx\|\geq\delta\|x\|$ (applied to $x=x_m-x_n$) shows that $(x_n)$ is also Cauchy and therefore converges to a limit z say. Then Tz = y, which shows that y is in im(T). Therefore im(T) is closed. | {"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": 24, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9757036566734314, "perplexity": 97.96837805376404}, "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-2014-35/segments/1408500815756.79/warc/CC-MAIN-20140820021335-00171-ip-10-180-136-8.ec2.internal.warc.gz"} |
http://metar.gr/forum/ni2%2B-electron-configuration-a4ad73 | Learn with Videos. What is the mass of 4.50 L of sunflower oil? Thus, it can either have a tetrahedral geometry or square planar geometry. of protons = 28. The splitting pattern and electron configuration for both isotropic and octahedral ligand fields are compared below. An electron configuration can quickly and simply tell a reader how many electron orbitals an atom has as well as the number of electrons populating each of its orbitals. A good place to start when trying to figure out the electron configuration of an ion is the electron configuration of the neutral parent atom.. Homework Equations Ni : [Ar]4s 2 3d 8 The Attempt at a Solution Ni +: [Ar]4s 1 3d 8 Apparently this solution is wrong and should be Ni +: [Ar]3d 9 but I do not know why. Answer to: 1. … => No. No. 8 min. B: What would be the predicted electron configuration for a Ni2+ ion? Expert Answer 100% (11 ratings) Fe3+ Ni. Electron orbitals are differently-shaped regions around an atom's nucleus where electrons are mathematically likely to be located. 0 2. Need an editable periodic table to edit? The electron configurations diagrams for d1 through d10 with large and small $$\delta$$ are illustrated in the figures below. This trend can be explained based on the properties of the ligands. Materials: Electron Configuration of Nickel (Ni) [Complete, Abbreviated, Uses ... Electron Configuration of Nickel (Ni) [Complete, Abbreviated, Uses ... https://1.bp.blogspot.com/-oTqJKC4GssU/XUXvqhn8GSI/AAAAAAAAf2U/TwlISA_krXso5JRLysBdZBeKI5RLGfJCwCLcBGAs/s320/electron-configuration-of-nickel.webp, https://1.bp.blogspot.com/-oTqJKC4GssU/XUXvqhn8GSI/AAAAAAAAf2U/TwlISA_krXso5JRLysBdZBeKI5RLGfJCwCLcBGAs/s72-c/electron-configuration-of-nickel.webp, https://materials.gelsonluz.com/2019/08/electron-configuration-of-nickel-ni.html. As far as I know 4s has less energy than 3d so, why does one electron promote from 4s to 3d after one ionization? 2 min. Since the configuration of Fe 3+ has five d electrons, we would expect to see five unpaired spins in complexes with Fe. How to find unpaired electrons in electron configuration . It states that the subshells are arranged in increasing order of their n + l value where n is the number of main shell and l is the number of subshell. Doing a question from TBR, I came across an answer that says "Nickel dication (Ni2+) has the electronic configuration ls22s22p63s23p63d8." P5.Q. Nevertheless, check the complete configuration and other interesting facts about Nickel that most people don't know. This is true for [FeF 6] 3-; however, [Fe(CN) 6] 3-only has one unpaired electron, making it a weaker magnet. 1s22s22p63s23p6 1s22s22p63s23p63d6 1s22s22p63s23p64s23d10 1s22s22p63s23p63d8 1s22s22p63s23p64s23d8 what is the ground-state electron configuration of a neutral atom of nickel?-0. *Response times vary by subject and question complexity. A) Write the electron configuration for Ni2+. Groups 13 - 18 are the P block. To write the configuration for the Titanium ions, first we need to write the electron configuration for just Titanium (Ti). Some are hard to memorise (or predict), so what is the electron configuration of an atom of Ni? Electron Configuration Chart for All Elements in the Periodic Table. As the outermost shell is of n=4 so electron will eject out from S orbital so it's configuration … What is the electron configuration for Ni2? dont know what the electron configuration if for: a) N2. C r = 2 4, M n = 2 5, F e = 2 6, N i = 2 8) Pauli's Exclusion Principle. This problem has been solved! => Electronic configuration = (2,8,14,4) But, Ni2+ implies it lost two electrons, => Electronic configuration of Ni2+ = (2,8,14,2). Source(s): https://shorte.im/baARo. . The 2 electrons are taken from the p orbital as it requires less energy to remove 2 electrons. the atomic number of N is 7 so, 7 electrons. All but (II) What is the ground-state electron configuration of the ion Cu2+? Previous question Next question Get more help from Chegg. It comes down to stability of the sub-shell and the orbital. 10XX,52,11XX,17,12XX,7,13XX,4,15XX,16,3XXX,2,40XX,10,41XX,12,43XX,5,44XX,4,46XX,5,47XX,3,48XX,3,5XXX,23,6XXX,3,71XX,1,8XXX,22,92XX,5,93XX,1,94XX,4,98XX,2,ASTM,171,Atomic-Mass,327,Atomic-Number,436,Atomic-Radius,86,Atomic-Symbol,329,Atomic-Volume,94,Boiling-Point,94,CBS,6,Chemical-Elements,100,Chemical-Symbol,217,CMDS,13,Coefficient-of-Thermal-Expansion,85,Covalent-Radius,87,Crystal-Structure,109,CS,17,CVS,3,Density,309,Elastic-Modulus,30,Electrical-Conductivity,79,Electro-Affinity,87,Electron-Configuration,109,Electronegativity,102,Electrons-per-Shell,112,Enthalpy-of-Fusion,93,Enthalpy-of-Vaporization,95,Group-Number,218,HCS,14,Heat-of-Fusion,87,Heat-of-Vaporization,85,HMCS,16,Ionic-Radius,78,Ionization-Energy,102,Ionization-Potential,101,LCS,21,List,201,MCS,17,MDS,14,Melting-Point,96,MS,4,NCMDBS,6,NCMDS,31,NCS,2,NMDS,8,Oxidation-States,104,Period-Number,107,Properties,40,RCLS,1,RCS,16,RRCLS,3,RRCS,4,SAE,201,Site,2,SMS,5,Specific-Gravity,83,Specific-Heat,92,Specific-Weight,1,Tests,2,Thermal-Conductivity,105,Valence-Electrons,98. What is the ground-state electron configuration of Se2-? Note that when writing the electron configuration for an atom like Fe, the 3d is usually written before the 4s. Or [Ar] 3d8 It has a role as a cofactor. The electronic configuration of nickel in its ground state is. 1 s 2 2 s 2 2 p 6 3 s 2 3 p 6 3 d 10 One point is earned for the correct configuration. .) Thus, the electron configuration for an Ni2+ion is 1s22s22p63s23p63d8. Nevertheless, check the complete configuration and other interesting facts about Nickel that most people don't know. Which of the following three sets consist of atoms or ions with the same electron configuration in the ground state? Gelson Luz is a Mechanical Engineer, expert in welding and passionate about materials. Median response time is 34 minutes and may be longer for new subjects. The third ionization energy of bromine is the energy required for which of the following processes? Ni2+: 1s2.2s2 2p6. Fe. D Ni2+ F V Hg2+ A [Kr]4d3 B [Ar]3d5 C [Ar]452 D [Ar]3d8 E [Ar]3d2 F [Xe]4f145d10 G [Ar]3d10 DZn2+ A MO3+ Ev Ti2+ E V3+ This problem has been solved! 3s2 3p6 3d8. Concepts and reason The concept of this question is that the filling of electrons in the shells and subshells of an atom takes place on the basis of Aufbau rule. Ni2+ 6. Correct Electron Configuration for Chromium (Cr) Half-filled and fully filled subshell have got extra stability. brainly.in/question/5057016. Atomic no. C: How many unpaired electrons are there in the observed ground state configuration of a neutral ruthenium (Ru) atom? Thus, the electron configuration for a P3-ion is 1s22s22p63s23p6. Since CN− ion is a strong field ligand, it causes the pairing of unpaired 3d electrons. (At. [Ar] 3d9 [Ar] 4s2 3d7 [Ar] 4s2 3d10 4p1 [Ar] 4s1 3d8 [Ar] 3d9. *a. V5+ *b. Cr3+ *c. Ni2+ *d. Fe3+ 2. B) Write The Electron Configuration For Br−. Copper and silver both have exceptional electron configurations because they both have full d orbitals at the expense of an s orbital. Favourite answer. That means that the d is filled, and the s is partially filled. Correct Electron Configuration for Copper (Cu) There are 118 elements in the periodic table. D: Which group (column) of elements in the d-block has a different ground state distribution of electrons in the outermost s and d-orbitals for neutral atoms of each of its three elements? Ni2+: From Table 2.2, the electron configuration for at atom of nickel is 1s22s22p63s23p63d84s2. The relative energy of the repulsion energy and splitting energy defines the high-spin and low-spin states . This means that a neutral titanium atom will contain #22# protons in its nucleus and #22# electrons surrounding its nucleus. Write the electronic configuration of (i) mn4+, (ii) fe3+ (iii) cr2+ and zn2+ mention the number of unpaired electrons in each case. Show transcribed image text. Write the noble gas configurations of. Each element has a unique atomic structure that is influenced by its electronic configuration, which is the distribution of electrons across different orbitals of an atom. NCERT Solutions for Class 12 Chemistry Chapter 8 the D and F Block Elements is an important study material which has answers to textbook and important questions from the previous year and sample papers and chemistry Class 12 chapter 8 exercise solutions.. Is nickel special? The electron configurations highlighted in red (d 3, low spin d 6, d 8, and d 10) do not exhibit Jahn-Teller distortions. Thank you for watching all the articles on the topic Electron Configuration for Ni, Ni2+, and Ni3+ (Nickel and Nickel Ions).All shares of bluevelvetrestaurant.com are very good. This problem has been solved! Favourite answer. Hund's rule of maximum multiplicity. 1s22s22p63s23p6 1s22s22p63s23p63d6 1s22s22p63s23p64s23d10 1s22s22p63s23p63d8 1s22s22p63s23p64s23d8 Sunflower oil has a density of 0.920 g/mL. 4 Answers. V5+,Cr3+,Ni2+,Fe3+ Determine If The Ion Is Diamagnetic Or Paramagnetic. Papildinformāciju par to, kā mēs izmantojam jūsu informāciju, varat iegūt mūsu Privātuma politikā un Sīkfailu politikā. To write the configuration for the Cobalt ions, first we need to write the electron configuration for just Cobalt (Co). Answer to Give the electron configurations for the following ions: P5+, P3+, Sn4+, Se2+, I+, and Ni2+.. Possible oxidation states are +2,3. (a) Write the complete electron configuration (e.g., 1s2 2s2. Anonymous. See the answer. See the answer. So, its 1s2, 2s2, 2px1, 2py1, 2pz1. This give us the (correct) configuration of: 1s2 2s2 2p6 3s2 3p6 3d5 4s1. The noble gas electron configuration is a type of shortcut to writing out the full electron configuration of an element. CN – will cause pairing of electrons. Therefore we have (still incorrect) 1s 2 2s 2 2p 6 3s 2 3p 6 3d 9 4s 2. A) Write the electron configuration for Ni2+. The +1/2 spin electron shown has a half arrow pointing up where as the -1/2 spin electron is shown as a half arrow pointing down. Ag Happens to be one of the 11 exceptions in the periodic table to the common trends, this happens because Ag it is found more stable at the configuration with 4d^10 5s^1 rather than a 5s^2 4d^9 which is what would be expected. In this video we will write the electron configuration for B 3+, the Boron ion. Ni2+: From Table 2.2, the electron configuration for at atom of nickel is 1s22s22p63s23p63d84s2. Since it has a +2 charge its configuration is the same as Fe (iron). Learn more about electronic configuration and unpaired electrons. See the answer. Mēs un mūsu partneri saglabāsim jūsu ierīcē informāciju un/vai piekļūsim jūsu ierīces informācijai ar sīkfailu un līdzīgu tehnoloģiju starpniecību, lai nodrošinātu jums personalizētas reklāmas un saturu, iegūtu ar reklāmu un saturu saistītus datus, gūtu ieskatu par auditoriju un veiktu produktu izstrādi. Electron configuration of Nickel is [Ar] 3d8 4s2. An atom's electron configuration is a numeric representation of its electron orbitals. Groups 3 - 12 are D block. Ag. In this example, the electron configuration for Ni2+ still kept its 3d8, but lost the 4s2 (became 4s0… Electron configuration was first conceived under the Bohr model of the atom, and it is still common to speak of shells and subshells despite the advances in understanding of the quantum-mechanical nature of electrons.. An electron shell is the set of allowed states that share the same principal quantum number, n (the number before the letter in the orbital label), that electrons may occupy. Electron Configuration For Ag 1. Lv 4. The relative energy of the repulsion energy and splitting energy defines the high-spin and low-spin states . On the other hand d 1, d 2, low spin d 4, low spin d 5, low spin d 7, and d 9, would be expected to exhibit Jhan-Teller distortion. To write the configuration for the Manganese ions, first we need to write the electron configuration for just Manganese (Mn). and the separate Lanthanides, Actinides series being the F block. Electron Configuration and Oxidation States of Nickel. These electronic configurations correspond to a variety of transition metals. The electron configuration of Si is: [Ne] 3s2 3p2. Stability of completely or half filled and exchange energy related to spin of electrons. Considering both weak and strong ligand fields, a Tanabe–Sugano diagram shows the energy splitting of the spectral terms with the increase of the ligand field strength. of Ni is 28. Nickel with an Oxidation State of +2: Ni2+: [Ar] 4s03d8. How is the electron configuration of . Aufbau principle of filling electrons. for Zn2+. ... 4s2 3d3 . Expert Answer 100% (11 ratings) Si2+ is Si with 2 less electrons so the electron configuration is: [Ne] 3s2. Nickel(2+) is a nickel cation in which the nickel carries a double positive charge. Notice that the electron configurations for d 1, d 2, d 3, d 8, d 9, and d 10 are the same no matter what the magnitude of $$\Delta$$. P5.Q. 24 min. Write orbital diagrams for each of these ions. Thus, the electron configuration for an Ni2+ ion is 1s22s22p63s23p63d8. Example Definitions Formulaes. [Ar] 4s^2 3d^6 . Search for "Gelson Luz" in your favorite browser to learn more. Therefore, one of the 4s2 electrons jumps to the 3d5 so that it is half-filled (see video below). Include complete electron configuration, abbreviated electron configuration (nobl gas notation), and orbital notation electron configuration for each ion. This give us the (correct) configuration of: 1s2 2s2 2p6 3s2 3p6 3d5 4s1. n atomic physics and quantum chemistry, the electron configuration is the distribution of electrons of an atom or molecule (or other physical structure) in atomic or molecular orbitals. that is only the electron configuration of nickel, a nickel (II) cation would lose the 2 electrons in the 4s and be 1s2 2s2 2p6 3s2 3p6 3d8 Answer Save. In the case of Nickel the abbreviated electron configuration is [Ar] 3d8 4s2. V5+,Cr3+,Ni2+,Fe3+ This problem has been solved! Maybe add your school logo, work team or anything else to maker your paper look cool? To write the configuration for the Manganese ions, first we need to write the electron configuration for just Manganese (Mn). Thank you for watching all the articles on the topic Electron Configuration for Ni, Ni2+, and Ni3+ (Nickel and Nickel Ions).All shares of bluevelvetrestaurant.com are very good. Special Cases and Exceptions Notating cations: When you’re dealing with cations, it’s very similar to … Correct Electron Configuration for Chromium (Cr) Half-filled and fully filled subshell have got extra stability. of electrons = 28. Electron Configuration. brainly.in/question/4711738 These solutions are easy to understand and are given in a simple way. Relevance. 2.9 (12 pts). Understand the noble gas electron configuration. Ni would be [Ar] 3d8 4s2 and since it has lost 2 electrons to become Ni^2+, the electron config would be [Ar]3d8. Check all that apply: Gd3+ P3+ Ag3+ Pt2+ Cr6+ Ni2+ 1 Answer Chemistry5 years ago. => No. So the full electron configuration for Palladium would start with: 1s 2 2s 2 2p 6 3s 2 3p 6 3d 10 4s 2 4p 6 Then you get to the critical point of deciding whether to fill it in as (4d 8, 5s 2) or (4d 10). Ni is in the +2 oxidation state i.e., in d 8 configuration. what is the ground-state electron configuration of a neutral atom of nickel?-0. => 2 electrons in the first orbit , 8 electrons in the second orbit , 14 electrons in the third orbit and 2 electrons in the fourth orbit. In order to become an ion with a plus two charge, it must lose two electrons—in this case the 4selectrons. Question: A) Write The Electron Configuration For Ni2+. Therefore, one of the 4s2 electrons jumps to the 3d5 so that it is half-filled (see video below). This leads to many more electron configuration states than is the case for the free ion. Which of the following is the correct complete electron configuration notation for Ni2+ ? 1 decade ago. It is diamagnetic in nature due to the unpaired electron. Or simply Ni2+: [Ar] 3d8. Hence V5+ions have the same electron configuration as argon: [V5+] = [Ar] = 1s2 2s2 2p6 view the full answer. Jūs jebkurā brīdī varat mainīt savas izvēles, atverot jūsu privātuma vadīklas. Not found any post match with your request, STEP 2: Click the link on your social network, Can not copy the codes / texts, please press [CTRL]+[C] (or CMD+C with Mac) to copy, How a small number of atoms can be joined and form completely different substances. Get 1:1 help now from expert Chemistry tutors It is a divalent metal cation, a metal cation allergen, a nickel cation and a monoatomic dication. The d electron configuration of C r 2 +, M n 2 +, F e 2 + a n d N i 2 + a r e 3 d 4, 3 d 5, 3 d 6 a n d 3 d 8 respectively. indiavision. Chemistry Q&A Library Which of the following is the correct complete electron configuration notation for Ni2+ ? Write nobel gas configuration for . Nickel Overview Nickel Complete Electron Configuration 1s2 2s2 2p6 3s2 3p6 4 s2 3 d8 Abbreviated Electron Configuration [Ar] 3d8 4s2 Sources Question: Match Each Transition Metal Ion With Its Condensed Ground-state Electron Configuration. The best way to understand this concept is to break the periodic table according to their configuration. Groups 1 and 2 are the S block. Yahoo ir daļa no Verizon Media. 6 years ago. Cu. b) N2^2+ c) N2-d) N2^2-need help thank you. In order to become an ion with a plus two charge, it must lose two electrons—in this case the 4s electrons. Question: A) Write The Electron Configuration For Ni2+. For example, the electron configuration of the neon atom is 1s2 2s2 2p6, using the notation explained below. Lai atļautu Verizon Media un mūsu partneriem veikt savu personas datu apstrādi, atlasiet 'Piekrītu' vai atlasiet 'Pārvaldīt iestatījumus', lai iegūtu papildinformāciju un pārvaldītu savas izvēles. There are 4 CN− ions. B) Write The Electron Configuration For Br−. 1s-2,2s-2,2p-6,3s-2,3p-6,3d-8,4s-2. Give the electron configurations for the following ions: P5+, P3+, Sn4+, Se2+, 1+, and Ni2+. B) Write the electron configuration for Br−. So the electron configuration of Ni2+ is,. 7. Give the electron configurations for the following ions: P5+, P3+, Sn4+, Se2+, 1+, and Ni2+. The periodic table is a tabular display of the chemical elements organized on the basis of their atomic numbers, electron configurations, and chemical properties. Homework Statement Which is the electron configuration of the Ni + ion? The energy of the isotropic field is the same as calculated for the high spin configuration in Example 1: $E_{\text{isotropic field}}= 7 \times 0 + 2P = 2P \nonumber$ 2.9 (12 pts). 5 min. Therefore the Iron electron configuration will be 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 6. Finding electronic configration of elements. Which one of the following aqua complex will exhibit the minimum paramagnetic behaviour? Electron Configuration . Informācija par jūsu ierīci un interneta savienojumu, tostarp jūsu IP adrese, Pārlūkošanas un meklēšanas darbības Verizon Media tīmekļa vietņu un lietotņu lietošanas laikā. In the case of Nickel the abbreviated electron configuration is [Ar] 3d8 4s2. In this case, titanium, #"Ti"#, is located in period 4, group 4 of the periodic table and has an atomic number of #22#. Include complete electron configuration, abbreviated electron configuration (nobl gas notation), and orbital notation electron configuration for each ion.
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## Tainted Group Title Medal will be given @nitz Let f(x) = x – 2 and g(x) = 3x + 4. What is f(g(x)) one year ago one year ago Edit Question Delete Cancel Submit
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Technically, we're doing\[f(3x + 4) \]Notice that \(f(\color{#C00}{x}) = \color{#C00}{x} - 2\). And so \(f(\color{#C00}{3x + 4}) = \color{#C00}{3x + 4}- 2\)
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https://byjus.com/neet/chemical-equilibrium-formulas-for-neet/ | # Chemical Equilibrium Formulas For Neet
When the system consists of more than one substance and the composition of the system does not vary with time, the system is said to be in chemical equilibrium. The chemical composition of a system at equilibrium must be uniform and there should be no net chemical reaction taking place.
## Types of Reaction
1. Reversible reaction
2. Irreversible reaction
## State of Equilibrium
Most of the chemical reactions do not go to completion in a closed system and attain a state of equilibrium.
Another approach comes from kinetics as developed by Guldberg and Waage (1863). Equilibrium is said to have reached in a physical or chemical system when the rate of forward and reverse processes are equal.
At equilibrium,
Rate of forward reaction = Rate of backward reaction
## Law of Chemical Equilibrium or Mass Action
At equilibrium,
rf = rb
Where Kc = kf / kb
Where [ ] is used to indicate “molar concentration”.
## Equilibrium Constant in terms of Partial Pressure
In a gaseous system, we can use partial pressure terms in places of active masses. Thus, for the above gaseous equilibrium,
Where Kp is an equilibrium constant in terms of partial pressure.
## Equilibrium Constant in terms of Mole Fraction
Where Kx is the equilibrium constant in terms of mole fraction.
## Relation between Kp and Kc
Kp = Kc (RT)Δn
where R is gas constant 0.0821 liter.atm/degree/mole and Δn is the total number of molecules of the product – total number of molecules of the reactants.
## Relation between Equilibrium Constant and Standard Free Energy
ΔGo = – RT ln Keq
(or)
ΔGo = – 2.303RT log Keq
## Reaction Quotient (Q)
The reaction quotient has the same algebraic form as Keq but the current concentrations nor specifically the equilibrium concentrations are used in calculations.
The expression for Q is
## Degree of Dissociation
Degree of dissociation (ɑ) = No. of molecules dissociated/Total no.of molecules taken.
Dissociation constant, Kdiss
## Factors affecting Equilibrium
### Le Chatelier’s Principle
If an equilibrium system is subjected to a change of any one of the factors such as temperature, pressure or concentration the equilibrium shifts in the direction that ends the annual effect of that change.
Important conclusions of the principle are
### Pressure change
The increase of pressure of an equilibrium system shifts the equilibrium in the direction in which there is a decrease in volume.
### Temperature change
The increase of temperature favours the reaction which takes place with the absorption of heat. The effect of a decrease in temperature is just the reverse of it.
### Concentration
When a substance is added its active mass increases. Hence the reaction will proceed in the direction in which the increase in active mass is used up.
### Effect of adding an inert gas
At constant volume equilibrium is unaffected. At constant pressure, the volume of the system increases.
## Thermodynamics of Equilibrium
Van’t Hoff Equation =
(or)
Chemical equilibrium is the stage of a reversible reaction at which the active masses of the reactants and the products become constant in the mixture and do not change with time. It is the state of a reversible reaction at which measurable properties like colour, density, pressure and concentration are nearly unchangeable.
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http://tex.stackexchange.com/questions/88237/aligned-numerator-and-denominator-in-frac-or-any-other-variant | # Aligned numerator and denominator in \frac or any other variant
Is there a way to align the numerator and the denominator of a fraction? Consider two fractions (from a SIAM paper)
The denominator (k!) is nicely centred in the first fraction. However in the second fraction the brackets in the numerator and the denominator are not aligned. As the exponent of the numerator gets bigger, the misalignment gets worse. In general, the strategy of centring numerator and denominator is correct. However, it would be good to have the option of aligning brackets in specific instances like this.
MWE:
\documentclass[preview]{standalone}
\usepackage{amsmath}
\begin{document}
$\dfrac{(x_i - x)^{N+1}}{(N+1)!}$
\end{document}
-
The numerator and denominator are horizontally centered with respect to one another; this is by far the most common typographic treatment of fractions. – egreg Dec 26 '12 at 16:43
With the mathtools package providing the \mathrlap macro you could do (in the denominator) (N+1)\mathrlap{!}\hphantom{^{N+1}} or, in this simple case, (N+1)^{N+1}\hfill. For the latter already exist a then-abused macro: \cfrac[l]{(x_i - x)^{N+1}}{(N+1)^{N+1}}. – Qrrbrbirlbel Dec 26 '12 at 16:54
@devendra Although I would recommend to use just the default LaTeX style in your case, here are some related questions: How to improve the looks of a fraction?, \frac{1-z^{n+1}}{1-z} doesn't look very good … – Qrrbrbirlbel Dec 26 '12 at 17:01
Related Question: Align denominator of fraction to left. – Peter Grill Dec 26 '12 at 17:07
@devendra: I would recommend not deleting the question (but perhaps others disagree). Am sure others will come up with a similar question in the future (see Related Question I linked to) and if the answer is that is a bad idea to do this that may save them the time to post another question. Knowing what not to do is usually a good thing to know. – Peter Grill Dec 26 '12 at 17:13
As illustrated in the highest-voted answer to Align denominator of fraction to left, \hfill will do the trick.
Here's a macro, \myfrac, that puts the \hfill either in the numerator or the denominator as necessary
\newcommand{\myfrac}[2]{%
\setbox0\hbox{$#1$} % put the numerator in box0
\dimen0=\wd0 % measure box0
\setbox1\hbox{$#2$} % put the denominator in box1
\dimen1=\wd1 % measure box1
\ifdim\wd0<\wd1 % if box0 is narrower than box1
\dfrac{#1\hfill}{#2} % put \hfill in the numerator
\else
\dfrac{#1}{#2\hfill} % otherwise put \hfill in the denominator
\fi
}
Complete MWE
\documentclass{article}
\usepackage{amsmath}
\newcommand{\myfrac}[2]{%
\setbox0\hbox{$#1$} % put the numerator in box0
\dimen0=\wd0 % measure box0
\setbox1\hbox{$#2$} % put the denominator in box1
\dimen1=\wd1 % measure box1
\ifdim\wd0<\wd1 % if box0 is narrower than box1
\dfrac{#1\hfill}{#2} % put \hfill in the numerator
\else
\dfrac{#1}{#2\hfill} % otherwise put \hfill in the denominator
\fi
}
\begin{document}
\begin{itemize}
\item[Original] $\dfrac{(x_i - x)^{N+1}}{(N+1)!}$
\item[Test 1] $\myfrac{(x_i - x)^{N+1}}{(N+1)!}$
\item[Test 2] $\myfrac{(N+1)!}{(x_i - x)^{N+1}}$
\end{itemize}
\end{document}
-
Wouldn't \newcommand{\myfrac}[2]{\dfrac{#1\hfill}{#2\hfill}} produce the same (horrible) output? – egreg Dec 26 '12 at 23:55
@egreg ah, yes, of course :) My code is unnecessarily complicated- thanks for catching :) – cmhughes Dec 27 '12 at 1:54 | {"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": 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": 0.9973961710929871, "perplexity": 1931.218978925029}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-49/segments/1416931010938.65/warc/CC-MAIN-20141125155650-00162-ip-10-235-23-156.ec2.internal.warc.gz"} |
https://brilliant.org/problems/sqrt-s-can-ruin-your-lives-2/ | # $$\sqrt{ }$$s Can Ruin Your Lives!
Calculus Level 3
If $$f(1)=1$$ and $$f'(1)=2$$, find the closed form of $$\displaystyle \lim_{x \to 1} \frac{\sqrt{f(x)} - 1}{\sqrt{x} - 1}$$.
Bonus: Evaluate this limit without applying L'Hôpital's Rule. | {"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": 0.994512677192688, "perplexity": 657.2155412681632}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-30/segments/1500549427749.61/warc/CC-MAIN-20170727062229-20170727082229-00259.warc.gz"} |
https://www.gradesaver.com/textbooks/math/algebra/intermediate-algebra-for-college-students-7th-edition/chapter-9-section-9-5-exponential-and-logarithmic-equations-exercise-set-page-726/48 | ## Intermediate Algebra for College Students (7th Edition)
$x=1000$
The given expression involves common logarithm. Common logarithm has a base of $10$. RECALL: $\log_b{x}= y \longrightarrow b^y=x$ Use the rule above with $b=10$ to obtain: $10^3=x \\1000=x$ | {"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": 0.5948444604873657, "perplexity": 1423.157213722738}, "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/1537267156724.6/warc/CC-MAIN-20180921013907-20180921034307-00170.warc.gz"} |
https://arizona.pure.elsevier.com/en/publications/fundamental-limit-of-resolving-two-point-sources-limited-by-an-ar-2 | # Fundamental limit of resolving two point sources limited by an arbitrary point spread function
Ronan Kerviche, Saikat Guha, Amit Ashok
Research output: Contribution to journalArticlepeer-review
## Abstract
Estimating the angular separation between two incoherently radiating monochromatic point sources is a canonical toy problem to quantify spatial resolution in imaging. In recent work, Tsang et al. showed, using a Fisher Information analysis, that Rayleigh’s resolution limit is just an artifact of the conventional wisdom of intensity measurement in the image plane. They showed that the optimal sensitivity of estimating the angle is only a function of the total photons collected during the camera’s integration time but entirely independent of the angular separation itself no matter how small it is, and found the information-optimal mode basis, intensity detection in which achieves the aforesaid performance. We extend the above analysis, which was done for a Gaussian point spread function (PSF) to a hard-aperture pupil proving the information optimality of image-plane sinc-Bessel modes, and generalize the result further to an arbitrary PSF. We obtain new counterintuitive insights on energy vs. information content in spatial modes, and extend the Fisher Information analysis to exact calculations of minimum mean squared error, both for Gaussian and hard aperture pupils.
Original language English (US) Unknown Journal Published - Jan 17 2017
• General | {"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": 0.9716386198997498, "perplexity": 1423.5477756093583}, "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-2021-10/segments/1614178360853.31/warc/CC-MAIN-20210228115201-20210228145201-00057.warc.gz"} |
http://www.cs.utah.edu/~suresh/web/teaching/accountability-in-data-mining/ | http://www.cs.utah.edu/~suresh
suresh at cs utah edu
Ph: 801 581 8233
Room 3404, School of Computing
50 S. Central Campus Drive,
Salt Lake City, UT 84112.
Wednesdays 1:25-2:45
MEB 3105
## Overview
The fruits of data mining pervade every aspect of our life. We are recommended books and movies; given differential pricing for insurance; screened for potential terror threats; diagnosed with various diseases; and targeted for political advertising. The ability to sift through massive data sets with sophisticated algorithms has resulted in applications with impressive predictive power.
Since the internals of a learning process can be complex and opaque, it is important to verify that the results satisfy the properties claimed by the algorithm. The importance of this goes beyond merely checking the algorithm itself. Validation mechanisms also provide a way for users to demand accountability from authorities who might use the results of mining to affect users in some way (by changing their insurance rates, or even putting them on a terrorism watchlist). As the results of data mining affect more and more of our lives, the more crucial it is that the user be able to validate decisions made on their behalf and that affect them.
This is a broad (and fluid!) topic: we will investigate technical challenges in building a framework for validating data mining outcomes, and we will also make forays into the problem of “explanation”: how does one even “explain” the results of data mining (let alone validate it). Some of the topics we will encounter will include interactive proofs, privacy mechanisms, theories of explanations in science, and homomorphic encryption.
## Class Mechanics
Participants in the seminar will be expected to
• present at least one paper
• read all the assigned readings and participate in class discussions
In addition, each student will prepare notes for one lecture (scribing format will be provided).
## Schedule of Readings
#### Jan 9: Seminar overview.
We will focus our study of accountability in data mining on three aspects: checking if a computation was performed accurately, protecting data from unwarranted access/computations, and attempting to explain the results of data mining. In all of this, our goal will be to understand what the theory offers, how effective it is, and what questions remain to be asked.
• Kevin Slavin’s talk on how algorithms are taking over the world (it also has a transcript).
• Cory Doctorow’s story Human Readable is an engaging tale of what happens when algorithms control all aspects of our world, and how difficult it is to demand accountability for the behavior of machines (note that the story is available as a podcast, or as part of the anthology With a Little Help).
• Databuse: Digital Privacy and the Mosaic, by Benjamin Wittes. On why privacy as a holistic term is not sufficient for our modern digital data needs.
• Your next mayor: a computer, by Will Doig @ Salon.com. A short speculation on cities of the future.
• Six Provocations for big data, by danah boyd and Kate Crawford. Implications of the data revolution for how we think about data, science, knowledge, and ethics.
#### Scribe: Suresh Venkat.
Jan 16: Checking computations: some background.
An overview of the theory of interactive proofs: in brief, how a weak verifier, with some well chosen queries, can verify the results of a much harder computation presented by an all-powerful prover.
Presenter: Amirali Abdullah.
Scribe: John Moeller (notes)
Jan 23: Checking computations: the big-data problem.
In the world of big data, your ability to check a computation is limited even more so by lack of access to all the data. We will look at new research that does verification of a computation even if you can only watch as data streams by.
Presenter: Chenxu Ding
Scribe: Mina Ghashami
Jan 30: Checking computations: a matter of trust.
Resource limits are one problem with verifying computations. Trust is another. Suppose you are trying to get answers from an untrusted server, and also have limited access to a trusted entity. Can you ensure that computations are correct ? We will start with a study of Merkle trees: a key data structure that shows up in this setting, and go onto an exploration of authenticated data structures.
Also look at Section 2 of Charalampos Papamanthou’s Ph.D thesis for a good overview of authenticated data structures and the basic model.
Presenter: Samira Daruki
Feb 6: Checking computations: Outsourcing computation.
In the cloud, no one knows you’re a machine. Or something like that anyway. As we outsource more computations to third parties, how can we check that the computations are being done correctly ?
Presenter: Swetha Machanavajhala
Scribe: Amirali Abdullah
Feb 13: NO class: (Suresh in San Diego)
Feb 20: Protecting your data: Asking hidden queries.
Search engines mine the queries you ask to learn things about you. Can you hide your queries and still get the answers you need ?
Scribe: Samira Daruki
Feb 27: Protecting your data: Computing without revealing your data
In the Millionaire’s Problem, two millionaires want to figure out who has more money without revealing their actual worth to each other. In general, you might have $n$ players that each possess a variable $x_i$, and they wish to compute a function $f(x_1, \ldots, x_n)$ without anyone learning anyone else’s input.
Readings: Chapter 1 from this book.
Presenter: Yan Zheng
Scribe:
Mar 6: NO class: (Suresh in Germany)
Mar 20: Protecting your data: Anonymizing data
Instead of performing computations on hidden data, you can release a modified version of your data. How should you do that, and what guarantees can you enforce on the data ? We look at some traditional statistical notions of privacy preserving data publishing.
Presenter: Parasaran Raman
Scribe: Swetha Machanavajhala
Mar 27: Protecting your data: Differential Privacy
Differential privacy brings cryptographic principles to the problem of data privacy. How does it work ?
Readings: Differential Privacy (Cynthia Dwork), Differential Privacy Survey
Presenter: John Moeller
Scribe: Samira Daruki
Apr 3: Questions on differential privacy
Some of the papers that we came across in our discussions:
Papers on private/secure near neighbor search:
• In the MPC setting, there’s a (not too) large line of work here, most of which bases on the Yao protocols to compute distances in a distributed manner, add up sums, do comparisons, etc.
• The next is the setting where the server is blind to both the data and the user query. http://www.cs.utah.edu/~lifeifei/papers/snnicde.pdf
• And there’s also a setting where the server has to be blind to the query, but is allowed to know the database. (I.e, a map/GPS based system where the user wants to be guaranteed his locational privacy, but the data is publically available.) Here is one of the most cited papers. Usually, it’s based on a suitable spatial transform and Hilbert / space filling curve. http://dl.acm.org/citation.cfm?id=1376631
Graph Privacy:
Privacy and Clustering:
• Multi-Party computation of Clustering: (Analyst gets to look at scrambled data)
• Data owner publishes clustering results Anonymously (Model 3)
Apr 10: Understanding results: Concise explanations.
Occam’s razor expresses the idea that a model used to explain data should be concise. We will briefly review different approaches for achieving conciseness in models (VC-dimension, MDL, sparsity and so on)
Presenter: Scribe:
Apr 17: Understanding results: Explaining recommendations.
Recommendation engines like Netflix and Amazon will often generate “explanations” along with the recommendation. For example, “We think you’ll like this movie because you watched that one”, or “You should buy this because your friends bought things similar to it” and so on. How does one generate good explanations ? what makes an explanation more effective ? We will look at this in the context of recommendation systems.
Presenter: Parasaran Raman
Scribe:
Apr 24: Understanding results: Causality versus correlation.
One of the easiest mistakes in data mining is confusing correlation with causality. But what is causality anyway ? We’ll take a brief look at Judea Pearl’s theory of counterfactuals and how that might lead to better ways of generating hypotheses from a mining process. | {"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": 0.18793563544750214, "perplexity": 2718.325899957894}, "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-2014-41/segments/1410657133485.50/warc/CC-MAIN-20140914011213-00045-ip-10-196-40-205.us-west-1.compute.internal.warc.gz"} |
https://bibbase.org/network/publication/lande-arnold-themeasurementofselectiononcorrelatedcharacters-1983 | The Measurement of Selection on Correlated Characters. Lande, R. & Arnold, S. J. Evolution, 37(6):1210--1226, 1983. ArticleType: research-article / Full publication date: Nov., 1983 / Copyright © 1983 Society for the Study of Evolution
Multivariate statistical methods are derived for measuring selection solely from observed changes in the distribution of phenotypic characters in a population within a generation. Selective effects are readily detectable in characters that do not change with age, such as meristic traits or adult characters in species with determinate growth. Ontogenetic characters, including allometric growth rates, can be analyzed in longitudinal studies where individuals are followed through time. Following an approach pioneered by Pearson (1903), this analysis helps to reveal the target(s) of selection, and to quantify its intensity, without identifying the selective agent(s). By accounting for indirect selection through correlated characters, separate forces of directional and stabilizing (or disruptive) selection acting directly on each character can be measured. These directional and stabilizing selection coefficients are respectively the parameters that describe the best linear and quadratic approximations to the selective surface of individual fitness as a function of the phenotypic characters. The theory is illustrated by estimating selective forces on morphological characters influencing survival in pentatomid bugs and in house sparrows during severe weather conditions.
@article{lande_measurement_1983,
title = {The {Measurement} of {Selection} on {Correlated} {Characters}},
volume = {37},
issn = {00143820},
url = {http://www.jstor.org/stable/2408842},
abstract = {Multivariate statistical methods are derived for measuring selection solely from observed changes in the distribution of phenotypic characters in a population within a generation. Selective effects are readily detectable in characters that do not change with age, such as meristic traits or adult characters in species with determinate growth. Ontogenetic characters, including allometric growth rates, can be analyzed in longitudinal studies where individuals are followed through time. Following an approach pioneered by Pearson (1903), this analysis helps to reveal the target(s) of selection, and to quantify its intensity, without identifying the selective agent(s). By accounting for indirect selection through correlated characters, separate forces of directional and stabilizing (or disruptive) selection acting directly on each character can be measured. These directional and stabilizing selection coefficients are respectively the parameters that describe the best linear and quadratic approximations to the selective surface of individual fitness as a function of the phenotypic characters. The theory is illustrated by estimating selective forces on morphological characters influencing survival in pentatomid bugs and in house sparrows during severe weather conditions.},
number = {6},
urldate = {2010-09-23TZ},
journal = {Evolution},
author = {Lande, Russell and Arnold, Stevan J.},
year = {1983},
note = {ArticleType: research-article / Full publication date: Nov., 1983 / Copyright © 1983 Society for the Study of Evolution},
pages = {1210--1226}
} | {"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": 0.41773703694343567, "perplexity": 4474.376598362107}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-31/segments/1627046154127.53/warc/CC-MAIN-20210731234924-20210801024924-00343.warc.gz"} |
http://mathhelpforum.com/advanced-algebra/158997-proving-intersection-subspace.html | # Math Help - Proving intersection is a subspace
1. ## Proving intersection is a subspace
Let $W_1$ and $W_2$ be two subspaces of a vector space $V$ over the field $\mathbb{F}$. Prove that the intersection of $W_1$ and $W_2$ is a subspace of $V$.
Can someone check my working for this question
Let $x\in W_1, y\in W_1$
Since $W_1$ is a subspace, it is closed under addition and so $x+y\in W_1$
Similarly, let $x\in W_2, y\in W_2$
Since $W_2$ is a subspace, it is closed under addition and so $x+y\in W_2$
$\Rightarrow x+y\in W_1\cap W_2\Rightarrow$ Closure under addition
Let $\lambda\in\mathbb{R}, x\in W_1, W_2$
$\lambda x\in W_1$ since $W_1$ is a subspace
$\lambda x\in W_2$ since $W_2$ is a subspace
Clearly, $\lambda x\in W_1\cap W_2\Rightarrow$ Closure under scalar multiplication
$\Therefore W_1 \cap W_2$ is a subspace of $V$
What about the condition for non-empty? This condition always confuses me. Why is it that the zero vector has to be included for it to be non-empty?
2. Originally Posted by acevipa
Let $W_1$ and $W_2$ be two subspaces of a vector space $V$ over the field $\mathbb{F}$. Prove that the intersection of $W_1$ and $W_2$ is a subspace of $V$.
Can someone check my working for this question
Let $x\in W_1, y\in W_1$
Since $W_1$ is a subspace, it is closed under addition and so $x+y\in W_1$
Similarly, let $x\in W_2, y\in W_2$
Since $W_2$ is a subspace, it is closed under addition and so $x+y\in W_2$
$\Rightarrow x+y\in W_1\cap W_2\Rightarrow$ Closure under addition
Let $\lambda\in\mathbb{R}, x\in W_1, W_2$
$\lambda x\in W_1$ since $W_1$ is a subspace
$\lambda x\in W_2$ since $W_2$ is a subspace
Clearly, $\lambda x\in W_1\cap W_2\Rightarrow$ Closure under scalar multiplication
$\Therefore W_1 \cap W_2$ is a subspace of $V$
What about the condition for non-empty? This condition always confuses me. Why is it that the zero vector has to be included for it to be non-empty?
well, since $\displaystyle W_1$ and $\displaystyle W_2$ are subspaces, they each contain 0. so 0 will be in their intersection, and hence, the intersection is non-empty.
containing zero is important for vector spaces and are hence important for subspaces. so the subspaces must have 0 in them. it is not to say that they have to have 0 to be non-empty per sae, but it is usually easy to verify that the zero vector is in there. so you always check that. and if 0 is not in there, then you're in trouble. you wouldn't have a vector space at all. just some other kind of set.
3. Thanks a lot. I understand that well now. | {"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": 46, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.966009259223938, "perplexity": 114.23683313047148}, "config": {"markdown_headings": true, "markdown_code": false, "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-2014-35/segments/1408500816424.18/warc/CC-MAIN-20140820021336-00242-ip-10-180-136-8.ec2.internal.warc.gz"} |
https://math.stackexchange.com/questions/534659/selection-of-dirichlet-parameters | Selection of Dirichlet parameters
Suppose that I have to draw from a Dirichlet distribution a multinomial distribution over three events: $$\theta \sim Dir(\alpha_1, \alpha_2, \alpha_3)$$
If $$\alpha_1 = \alpha_2 = \alpha_3 \land \alpha_1 + \alpha_2 + \alpha_3 >> 3$$ $\theta$ distribution is likely to be an uniform distributioin over the three events.
On the contrary, if: $$\alpha_1 = \alpha_2 = \alpha_3 \land \alpha_1 + \alpha_2 + \alpha_3 < 3$$ $\theta$ is likely to be a multinomial distribution that puts the majority of its mass on one of those three events.
My question is: how should I choose the $\alpha_1, \alpha_2, \alpha_3$ if I want $\theta$ to be a distribution that assigns high probability to two events and low probability to the third one?
In other words, I would like that the three most probable values for $\theta$ are:
• $\theta_1 = 0.5, \theta_2=0.5, \theta_3=0$
• $\theta_1 = 0.5, \theta_2=0, \theta_3=0.5$
• $\theta_1 = 0, \theta_2=0.5, \theta_3=0.5$
Furthermore, is that easy to generalize this to the case in which we have $N>3$ events? | {"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": 0.9408092498779297, "perplexity": 215.85478460550945}, "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-2019-26/segments/1560627999817.30/warc/CC-MAIN-20190625092324-20190625114324-00352.warc.gz"} |
https://www.vedantu.com/physics/periodic-function | # Periodic Function
View Notes
## What is a Periodic Function?
One can understand periodic function meaning as the motion that occurs repetitively over the course of fixed time intervals. Periodic function examples include rocking a chair, which is a circular motion. In other words, one can also define a periodic function as the motion that returns to its initial position after a fixed duration of time.
### Understanding the Difference Between Periodic and Oscillatory Motion
After going through the periodic function definition, one can easily get confused with oscillatory motion at first glance. But, not all periodic functions are oscillatory at the same time. One of the biggest differences between the two is that, while periodic motions can be repetitive at times, oscillatory motion is only constrained at an equilibrium point.
For better understanding, one can take the example of a bob of a pendulum. It oscillates along its equilibrium position in a periodic manner. During its movement, the displacement takes place from zero to positive to negative passing through its initial position. Such a motion is periodic and oscillatory at the same time. Another aspect of the oscillating motion is Simple Harmonic Motion (SHM), where the restoring force of the periodic motion is directly proportional to that of its displacement.
The Formula for Periodic Function
One can define the periodic function f, along with a non-zero constant in the same case:
f (x+P) = f (x)
The function is applicable for all the values of x in the same domain. While the constant P is termed as the period of a function.
Derivation of Periodic Function Equation
For an oscillating object, its periodic function can be defined as:
f(t) = Acosωt
With the cosine part repeating itself after a certain point of time, it can defined as:
cosθ = cos(θ+2π)
cos(ωt) = cos(ωt+2π) ——(1)
Considering the time period to be T:
f(t) = f(t+T)
Acosωt = Acosω(t+T)
Acosωt = Acos(ωt + ωT) ——(2)
So from equation 1 and 2, we can derive:
ωT = 2π
Thus, T = 2πω
### Time Period of Periodic Function
The time period of the periodic function is given by;
T = 2πω, ω is the angular frequency of the oscillating object.
### Frequency of Periodic Function
We all know that the frequency is given by the total number of oscillations per unit time. For periodic motion, frequency is given by;
F = 1/T;
F = 1/ (2πω);
### Solved Questions
Example 1
State whether a motion can be periodic but not oscillatory or not?
a) True
b) False
Solution
The answer is option A. You can often find motions that are periodic but not oscillatory. For example, a uniform circular motion is a periodic motion, but there is no restoring force being applied on it. So, it is not an oscillatory motion.
Example 2
For a given pendulum, if l is the length of the bob, while its mass is m, and it is moving along the circular arc with angle θ. So if a spherical mass M is placed at the end of the circle, what is the momentum of the sphere gained by the moving bob?
a) Infinity
b) Zero
c) Constant
d) Unity
Solution
The answer is B. The sphere will not attain any momentum through the bob at the end of the circle. This is because, at the end of the circle, the velocity of the bob becomes zero.
Example 3
Let us assume that a 2 kg body is suspended from a stretchable spring. So, if someone pulls down the spring, it is released with an oscillating motion vertically. What is the name of the force that is applied to the body, when the spring passes through its mean position?
a) Force equal to the pull
b) Force equal to the weight of the body
c) Force equal to gravity
d) Conservative force
Solution
The answer is B. It is imperative to understand that at the mean position, the total acceleration of the body is zero. So, the resultant force that is applied by the spring is the same as that of the weight of the body.
FAQ (Frequently Asked Questions)
1. Define SHM (Simple Harmonic Motion) with reference to motion.
Motion can be defined as the type of phenomenon in which a body can change its position within a given time frame. Based on the characteristic of the motion, they can be further divided into different types. Some of the most prominent types of motion include:
• Reciprocating motion
• Linear motion
• Rotary motion
• Oscillating motion
Simple Harmonic Motion or SHM, on the other hand, is a type of oscillating motion. In this type of motion, the total force exerted on the object is actually restoring force. In other words, SHM can be portrayed as periodic motion. In this motion, the object tends to move to and fro along a fixed line. The pendulum of a clock is one of the best examples of simple harmonic motion.
2. Define angular frequency with context to periodic motion.
Angular frequency, mainly represented using the Greek alphabet, ω (Omega), is used for periodic motion. Angular frequency can be defined as the angular displacement that occurs per unit time. In other words, the rate of change of sinusoidal waveform (waveform describing a smooth periodic oscillation) can be defined as angular velocity too. It is mainly expressed in the form of an argument of a sine function.
For example,
y(t) = sin(θ(t)) = sin(ωt) = sin(2πft)
or, ω=2πf
The SI unit of angular frequency is radians per second. | {"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": 0.8309762477874756, "perplexity": 477.9881621801939}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-04/segments/1610703513062.16/warc/CC-MAIN-20210117143625-20210117173625-00418.warc.gz"} |
https://math.stackexchange.com/questions/913893/is-this-statement-meaningful-if-one-of-the-elements-is-undefined | # Is this statement meaningful if one of the elements is undefined?
Am I allowed to say a statement like $\max\left\lbrace a,b\right\rbrace$ if it turns out that the element $b$ is undefined, or simply does not exist? Would the result be $a$, or is the whole statement invalid?
edit: based on the answers I've seen, I think I will have to break my proof into separate cases. I always want to avoid that when possible, but I would rather make sense in what I write.
• The "painting" $\max\{a,b\}$ is not a statement, but an expression. The value of this expression is undefined if one of $a$ or $b$ (supposedly in ${\mathbb R}$) is undefined. Aug 30, 2014 at 14:49
Hint: $$\max\left\lbrace a,b\right\rbrace=\dfrac{a+b+|a-b|}{2}$$
If you don't know $b$ then you can't determine $|a-b|$ and the rest of the formula ;)
• Wonderful answer. Welcome to the community! Aug 30, 2014 at 13:38
• You're very kind, many thanks for the welcome! Aug 30, 2014 at 13:50
The $\text{max}$ operator requires comparison among arguments, or alternatively, requires that the arguments have values which can be used to compute $\operatorname{max}\{a, b\} = \dfrac{a+b+|a-b|}2$. Since we cannot compare an incomparable (since it is undefined) number $b$ with any other number(s), operating on them is rendered meaningless.
It is analogous to asking what $f(x) = x^2$ when evaluated at the undefined value of $b$.
You're "allowed" to say what you like, as long as you can define what it means.
If it's convenient for your proof to define some quantity that's equal to the maximum of $a$ and $b$ except in the cases where $b$ is undefined, in which case it's equal to $a$, then go right ahead.
The only remaining issue is whether you refer to this quantity as $\max\{a,b\}$, or give it some other name. If you're going to use properties of $\max$ that aren't clearly true of your quantity, then it would be wise to give it a different name. For example using $\max\{a,b\} \ge a$ won't cause any problems. Using $\max\{a,b\} \ge b$ might lead you to make errors, since it's not true (or meaningful even) in the cases for which $b$ is undefined.
If one of the elements is $\infty$ or $-\infty$ (as a limit), then you can extend the definition of $max\{a, b\}$ to be meaningful. I can't think of any other instance where it would make sense to use an argument that you would call undefined. | {"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": 0.6817096471786499, "perplexity": 160.09445881230914}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-27/segments/1656103036077.8/warc/CC-MAIN-20220625160220-20220625190220-00212.warc.gz"} |
http://www.cheatcodes.com/guide/complete-walkthrough-and-hacking-guide-enter-the-matrix-ps2-40531/ | # Complete Walkthrough and Hacking Guide - Guide for Enter the Matrix
Scroll down to read our guide named "Complete Walkthrough and Hacking Guide" for Enter the Matrix on PlayStation 2 (PS2), or click the above links for more cheats.
============================================================
Contents
============================================================
1) Data Log
2) Introduction and Basic Terms
2.2) Special Thanks
3) Drive Content
3.1) A:
3.2) B:
3.3) V:
3.4) Ram:
3.5) Logos:
4) Guide to weapon drops
5) Sparks Training Construct
6) Special Code Information // Secrets
7) Speed guide to open Multiplayer Mode
8) Ghost / Niobe Game Walkthrough
8.1) Post Office
8.2) City Driving
8.3) City Rooftops
8.4) The Airport
8.5) The Aqueduct
8.6) The Sewers
8.7) The Chateau
8.8) Under The City
8.9) The Zen Garden
8.10) The Freeway
8.11) The Power plant
8.12) Chinatown
8.13) The Skyscraper
8.14) Onboard The Logos
9) The Obligatory Legal Bit
(If you want a copy just ask!)
============================================================
1) Data Log
============================================================
16.05.03 Inception Date, This is pretty much final
though. Addition of new codes when available.
20.05.03 Addition of decodes, more secrets etc.
21.05.03 Addition of a complete walkthrough, call me crazy !
Niobe only just now, Ghost update VERY shortly..
??.??.?? Will update when have more
============================================================
2) Introduction and Basic terms
============================================================
Welcome to yet another quick-faq, It started quick as hacking only, I since gave in
to the call and created a complete walkthrough. This FAQ specialises on the
Hacking side of the Matrix, As such does touch on the game itself. Yet now does
hold a full guide for the in-game itself.
This guide is based on the XBOX version, however no doubt it is cross platform.
Enjoy with an open mind.
There are NO details of patches / bugs / fixes regarding the PC version.
No technical support here im afraid..
GAME
------------------------------------------------------------
appear
after a time limit to give you a hand.
* Experiment with different 'states' ie door opening, normal action and they turn
the
knob, run and they elbow barge the door through, use focus and they kick it in.
Try different directions when you fight also ie press throw to throw or use back
and throw to sweep their legs and punch them away (smiths special move..)
This works for Niobe, yet is different for Ghost, characters also have different
moves.
* Try to fight as much as possible to build up focus faster, if you are in a hurry
disarm them as you pass, this will often stop the threat.
* In a firefight make good use of strafe.
* Don't be afraid to waste time, or wait for your character to heal and recooperate
focus. Especially on harder levels, you may need to.
HACKING
------------------------------------------------------------
* BEGINNERS - type HELP then enter, this will then tell you what this
command will do, it will also give an example, if you are not sure just
try. use HELP DIR, this will tell you about DIR and what it does. Now
type DIR to open it in your command prompt bar on the left. Use this
to make life easier and select DIR then what you want to look at
using the command prompt bar also.
* All uppercase wording are commands you can use / write.
If you have ever used DOS or the like this is probably quite straight forward.
Hacking is all about experimentation (and a couple of tricks up your sleeve!).
* Once you have typed a command, it will then be in the left column to use.
this will save typing.
* Always use DIR to explore new drive/ folders when you open them.
Please mail me if you find this guide hard to understand, I have attempted
to make it as simple as possible.
------------------------------------------------------------
2.2) Special Thanks
------------------------------------------------------------
* Larry and Andy Wachoski, for good reason !
* Again thanks to neoseeker.com for inspiration
love you guys !
* My sister for making it possible to see reloaded at the cinema a
day before anyone else in the country !
* All those who have helped me in the past, and those who
have yet to help..
* Thanks to anklecutter for Persephone's phone number !
Thanks to leoking13 for decode information..
An inpressive matrix hack from grandthefteggo, thanks.
More website secrets from Jae, Nick P.and Sinuouser
* I could write anything here,so, if you were not afraid what would you do ?
============================================================
3) Drive Content
============================================================
3.1) A: Drive
------------------------------------------------------------
Folder - dir a:\system
REBOOT
This will reset all that has been achieved. (hacking)
(I don't advise ever using this, as it wipes all your progress, some people
will try to use it to clear cheats / disks etc. however if you have more
than one save with something activated - this will not work. The best
thing, and fastest is to reset the machine itself,)
HELP
This can be used before a command, it will explain its use
CLS
This will clear the screen
This opens up the B: drive, Use password 'GUEST'
Alternatively use FREEMIND or COWBOYCURTIS and receive a
message from Morpheus, use 1-5 to play fragments.
Caution - You can only login once (unless you reboot)
DIR
This will show all folders / files in an area, select using the
command prompt bar on the left.
EXIT
This followed by 'Y' will quit
Use this to open a text file, once they have been found use
the command prompt to select the text you want.
ECHO
Use echo to write a string of words. ie. ECHO [ENTER TEXT]
Folder - dir a:\tools
VIBRATION
Use to make a vibration [1-100] for fun
BEEP
To make notes sequence [1-8] for fun
CHEAT
Enter a cheat code using this command, codes are stored later
in this faq.
Folder - dir\text
NOTEZ.TXT
MANUAL1
Information of portkey technology (use READ)
MANUAL2
crack the codes at the websites : (use READ)
www.thematrix.com - (23631BE6) hidden on this site is
www.danielinstitute.com - leads to an alternate enterthematrix
site, register with Atari to get codes emailed to you.
see secrets section for more..
------------------------------------------------------------
3.2) B: Drive
------------------------------------------------------------
Folder dir\id
list of in game characters that have been seen in game - unseen are locked
Also opened by progression. Use VIEW command :
swata.img Armoured swat
neo.img Neo
abel.img Abel
ajack.img Agent Jackson
ajohn.img Agent Johnson
asmith.img Agent Smith
athomp.img Agent Thompson
axel.img Axel
ballard.img Ballard
bane.img Bane
cain.img Cain
cujo.img Cujo
dman.img Doberman
kmaker.img Key Maker
mero.img Merovingian
morph.img Morpheus
police.img Police
guard.img Guard
seraph.img Seraph
soren.img Soren
sparks.img Sparks
swatst.img Swat standard
swatse.img Sewer swat
swatsn.img Swat sniper
trinity.img Trinity
twins.img Twins
Folder - dir\guns
list of in-game weapons that have been picked up - unused are locked
also unlocked by progression (use VIEW to open)
mp5.img MP-5N
m95snip.img M95 Sniper Rifle
9000s.img 9000 S
92fs.img Beretta 92Fs
380.img .380 Mustang
com.img RO635
infinity.img 1911 .45 cal
m16.img M16 Commando
cross.img Antique Double Crossbow
50ae.img .50AE
18c.img 18C Auto pistol
c4.img Demolition charge
33sg1.img 33SG1 Sniper Rifle
g36.img G36 w/AG36
ump.img UMP
gp25.img GP-25 Chloro-bromo
m240.img M240 Machine gun
mac11.img MAC-11
590shot.img 590 Entry shotgun
sg552.img SG-552 Commando
p229s.img P229s
ssshot.img Street sweeper shotgun
Folder - dir\fmv
list of cutscenes that have been viewed in game - unseen are locked
Alternate character fmv stays locked (use PLAY to view)
0.fmv Opening sequence - 'red or blue'
1.fmv The last transmission - 'reality is a bitch'
2.fmv Axle's dust off - '72 hours'
3.fmv Meeting of captains - 'anything for an entrance'
4.fmv The keymaker - 'bring it to him'
5.fmv GHOST: Persephone - 'back off'
6.fmv NIOBE: Persephone - 'taste of tears'
7.fmv Zion - 'its good to be home'
8.fmv NIOBE: Deadbolt - 'something to eat that requires teeth'
9.fmv GHOST: Trinity joins in - 'sneaky'
10.fmv GHOST: Trinity loses - 'celibacy is hands on job'
11.fmv Seraph and Choices - 'I must apologise'
12.fmv Operator - 'i cant watch this'
13.fmv Hell on the highway - 'go kick his ass'
14.fmv A plan in motion - 'time is always against us'
15.fmv GHOST: Sniper run - 'we should be okay!'
16.fmv Meltdown - 'kaboom?'
17.fmv Broken lines - 'it didnt work'
18.fmv GHOST: Oracle - 'only as a friend'
19.fmv NIOBE : Oracle - 'do i know you?'
20.fmv Smart sentinels - 'when will it end'
21.fmv Closing and teaser trailer
Folder - dir\maps
list of levels that have been played - others locked
Also opened via progression, (use VIEW to check details)
Full details of map use and drops later in faq
B2SW.IMG - Drainage Canal - JXTRR10
B5APN.IMG - North Concourse PNSRZ10
B5APT.IMG - Airport Tunnels - RKHMS10
B3CW.IMG - 2nd Floor West - JDZMT10
B3CS.IMG - Courtyard - ZKHBD10
B6SK.IMG - Skyscraper - RHFTQ10
B6CT.IMG - Warehouse - ZSZQH10
B5PP.IMG - Transformer Field - JDHQL10
B1PO.IMG - 2nd Floor Po Boxes - B1AXXF2
Folder - dir\cars
list of cars in game, unseen are locked
Also opened via progression, (use VIEW to check details)
police.img Chevy Cop Car 2001
fbird.img Pontiac Firebird HO Custom 1967
agents.img Chevrolet Caprice Classic 2001
coupe.img Cadillac Coupe de Ville 1960
swat.img GMC Yukon XL 2001
Folder - dir\tools
tools is locked - enter password
to crack this password enter a 5 digit binary number (zeros / ones)
each attempt will tell you how many are correct and haw many are
in the right place, this is more trial and error than anything.
once you have the correct amount of 0/1's just rearrange them.
VIEW
Use to look at img files on b: drive
VIRTUAL
Enter password FROZENFISH to access, then complete puzzle before
timer runs out if you complete the puzzle as blue, neoblue.bat will run
warning you,(careful.wav) telling you to take the red backdoor. you must
type TRACEKILL before the timer runs out.
complete the red puzzle however and get neored.bat (welcome.wav) you
can now access V: drive
__]__
___/___
/] ]
the correct symbol looks a little like the one above..
TRACEKILL
Use to kill a trace program (if you fail VIRTUAL)..
DROP
This drops weapon in the matrix - use VIEW map to get codes
more details later in faq
PLAY
Use to play unlocked fmv footage
DECODE
Use to get passwords from codes off internet sites / emails etc
------------------------------------------------------------
3.3) V: Drive
------------------------------------------------------------
Folder - dir\tools
ROOTSEARCH
Also opens v:\root folder
DIAL
Use to call a hardline number, secret numbers noted later in faq
Call Trinity on number below for PORTKEY code
Folder - dir\root
MAIL
001 949 555 0112
001 714 555 0187
001 213 555 0142 - Morpheus
001 310 555 0111 - Trinity with code 942
PORTKEY
To use a communications port (enter key number)
use code 942 to gain the CRACKer file on you v: drive
CRACK
Use this and password that trinity gave you - 8RAM
(aghr22m5) to open RAM: Drive
------------------------------------------------------------
3.4) RAM: Drive
------------------------------------------------------------
Folder dir\trinity
TRAINING
To force a training program into the matrix, Use this to
access (*.dsk) files in program folder
HANDSHAKE
This will allow you to talk to sparks, use EMP.exe when he says
Shows your stats for the saved game.. Use this to read (*.bio) files
Folder dir\prog
*.dsk files are kept here for training, more info later in faq
sword.dsk
Folder dir\bio
*.bio files are kept here use READBIO to check information
------------------------------------------------------------
3.5) Logos: Drive
------------------------------------------------------------
Folder dir\sparks
EMP
Sparks has disabled this command
Folder dir\art
This holds preproduction art file (x31 *.img files)
Use VIEW to check them out
============================================================
4) Guide to weapon drops
============================================================
By using the Hacking system it is possible to drop weapons from the construct
directly into the matrix.
It is fairly straight forward -
Firstly use VIEW to check each map, on the bottom right is a code
Now use DROP [enter code here] to activate the change in the matrix
Now quit and save, when you replay a level the bonus weapons will be there.
Note : a DROP will not survive a reset, you will need to hack again.
Map B2SW.IMG - Drainage Canal
Code JXTRR10,
Level Waterway
Weapons Mac 11, SS Shotgun, Grenades
Actual Location Found near the ambush near the end of the area
Map B5APN.IMG - North Concourse
Code PNSRZ10
Level Ghost Only, North point (Airport)
Weapon M240
Actual Location Found on the desk to the right when you see the chopper outside
Map B5APT.IMG - Airport Tunnels
Code RKHMS10
Level Niobe only, Jackson in steam (Airport)
Actual Location Take a right at the junction, then next right then left down the
small opening
Map B3CW.IMG - 2nd Floor West
Code JDZMT10
Level Niobe Only, Persephone room (The Chateau)
Weapons 1911 pistol, SS shotgun, Ump, Entry Shotgun
Actual Location Found in the last left room after Vlad, near end of area
Map B3CS.IMG - Courtyard
Code ZKHBD10
Level Ghost only - Secret Passage 2
Weapon Crossbow, UMP, Entry Shotgun
Actual Location Found at the start of the area
Map B6SK.IMG - Skyscraper
Code RHFTQ10
Level Vertigo (skyscraper)
Weapon .50AE Pistol
Actual Location Found at start straight ahead
Map B6CT.IMG - Warehouse
Code ZSZQH10
Level Hard Line Pursuit (Chinatown)
Actual Location Found at start of area
Map B5PP.IMG - Transformer Field
Code JDHQL10
Level Transformer field
Weapon 18C Auto Pistol, SG-552, G36, Grenades
Actual Location Found Straight ahead at start of area
Map B1PO.IMG - 2nd Floor Po Boxes
Code B1AXXF2
Level Got It! (Post Office)
Weapons Entry Shotgun, MP5
Actual Location Found on the shelf near the Osiris package
============================================================
5) Sparks Training Construct
============================================================
Sparks Training (door one)
------------------------------------------------------------
* Once you complete the game (either character) wait til the end of the credits
a code will be shown for a short period (see below). When you enter this
it will open sparks training construct. From the load screen tap right to enter
this.
* Only one door is open initially. go to the wall and press the left button to start
(red one on the far wall) the first training.
* Pick up weapons from the wall rack and kill swat before they reach the line.
Once you have completed a high score of 20 plus on the training the 2nd door
will be open.
* You can also now take part in the second training. Press the 2nd
button on the right (green).
* Use the sniper rifle to shoot the platform under the feet of the guards, so they
fall, do this for a score of 20 plus. This is the 1st room clear.
Sparks Training (door two)
------------------------------------------------------------
* Now make your way through the second door in the hallway and take out the
swat on the way. Go through the door at the end.
* You will now be in a room with a cage in the centre, Straight ahead you will see
a large metal door - this is to the 2nd training room. (only open once you
complete training two)
* If you take the low door opening to the right. Take out the 2x swat high up on
the level (climb up if you wish)
Continue through to the next room on the right hand side back wall.
In the room of boxes are more swat, like a survival mode they will come from
the 2x side rooms with increasing numbers. When you are bored leave the room.
* In the 2nd training room, through the double metal doors are 2x buttons again.
Press the right red one to start training three.
* Within the time limit kill 20 plus swat guys, this will unlock training four.
This is the easiest just use sweep and keep them at bay.
* Now press the left green button. (training four)
* Training four requires you to knock off as many swats as you can within
the time limit.. in 3 minutes try to sweep then kick swat off the side. Or jump
kicks work well, dont fall off yourself or training is over.
* Complete the final training (20 plus swat) and unlock...?
============================================================
6) Special Code Information / secrets
============================================================
Secret Phone numbers:
------------------------------------------------------------
Persephone's number - 001-949-555-0101 (anklecutter)
Cheat codes (only working ones noted here):
------------------------------------------------------------
Use CHEAT [code here] to activate, repeat to disable cheat.
I don't condone the use of cheats, yet as part of the guide here they are
They are really for replay value i.e Infinite focus, Dont spoil the game for
yourself..
1 3 d 2 c 7 7 f Opens sparks training construct (loading section)
d 5 c 5 5 d 1 e Opens multiplayer mode (loading section)
1 d d f 2 5 5 6 Infinite ammo
0 0 3 4 a f f f All weapons
7 f 4 d f 4 5 1 Infinite health
f f f f f f f 1 Invisibility
6 9 e 5 d 9 e 4 Infinite Focus
7 8 6 7 f 4 4 3 Increased speed for logos
Decode codes:
------------------------------------------------------------
Use DECODE [enter code here] to activate
All these codes DO work, however depending on your progression they may
not be recognised.
H A B X G W T 7 Z R Z C 7 C I R S 7 K E Z I R N L - take your game further
D C A Y O E 7 V D S Z Z C K O - red solution
E C H U M X B I A O E F - swordplay
E A B X M S X A M A 7 E U N G D - virtual drive
I A D Y S F N Y C - sparks
Training files *.dsk:
------------------------------------------------------------
Complete the game with either character to get sword.dsk
once activated this will take the place of throw, caution you cannot kill some
Merovingian as it can also take place of the stake. reset the game itself to
remove the training program.
Websites :
------------------------------------------------------------
atari to get codes emailed to you.
Try entering the site pressing the yellow button in the top right. Click on the
message once the page has loaded and move your cursor to get to the picture
underneath. Select the monitors to get access to pictures / messages from shiny.
Either move the cursor over the squares or click on the boxes to clear them.
www.thematrix.com - another route to whatisthematrix site, if your quick a link
to http//:hackthematrix.warnbros.com can be found, here is a keyboard with
buttons allocated to certain words.
If you select one of these it will ask for a code, using the same code that is on
the
site in the code box will bring up video clips and such like. Dated but fun none
the
les.
Similar to www.whatisthematrix.com near mainframe on the top right is a 'code'
button, just another register ?
High Bandwidth only - Look to the top right of the page, you should see a
yellow round button - this states 'options'. Press it, this will open a panel. To
the right of 'low bandwidth' is a metal square, press this to reveal a green
button. Press this for 'binary access' A large panel will drop with ZION83N6
printed on it, now click on access panel 3 to open it. Use the switches to
enter code 01101111 press enter to get to the Hexidecimal key.
Thanks to grandthefteggo for this one..
Only one code so far:
1010-0011-1011-0001-1010-0100-0010-1000 (A3B1A428)
============================================================
7) Speed Guide to get Multiplayer Mode
============================================================
For any deeper information check the above FAQ..
Type any uppercase
DIR B:\TOOLS, this is locked enter password, see above to decipher
VIRTUAL, password FROZENFISH - complete the red puzzle shape to get
Then CRACK 8RAM, you can know access RAM:
HANDSHAKE, then EMP at Spark's request will shutdown the system with
multiplayer available.
Use the load screen to pick level, keep tapping left to get to multiplayer mode
The Subway Morpheus vs Agent Smith
The Dojo Niobe Vs Trinity
The Pit Seraph Vs Ballard
The Ballroom Niobe Vs Ghost
The Atrium Sewing Woman Vs Janitor
The Alley Firebird man Vs PoliceCar Man ?!
Is there is more to multiplayer - watch the trailer to see niobe vs swat in the
dojo ?!
Just not open yet ? More training ?
MUTLTIPLAYER IS NOT AVAILABLE ON PC VERSION
============================================================
8) Ghost / Niobe Game Walkthrough
============================================================
This guide is based on the level rather than the character as some are repeated
They are in chronological order. If one character only, it will be noted.
In this guide Niobe is the primary character.
------------------------------------------------------------
8.1) The Post Office
------------------------------------------------------------
Weapons :
SMG 9MM
9000s Pistol
92Fs Pistol
18c Auto Pistol
.380 Pistol
Entry Shotgun
MP5
Enemies:
Guards
Police
8.1.1) Closing Time
Fairly straight forward, head straight ahead and take out the guard, keep going and
hit the button for the elevator [cut scene] now turn left to the newly open door.
Take
out the guard here also. Go through the door.
8.1.2) Behind The Scenes
Head straight forward past the lockers and take your right near the end (taking care
of the guard on the way) then turn left, go to the back wall, then right again
towards
the door. Through the door - follow the corridor. watch out for a guard coming up
from behind. Go through the door on the right hand side. Head to your right and
take out the guards here, Go down the bottom end and take a left. All the way
down then left again (clear this area too) then your first right. Now head left all
the
way to the back wall (with a locker on) more guards are here. Now head left to the
locker on the far wall. Go through the opening (another two guards) and dive/kick
through the window on the left side of the corridor. Press the RED button in here
on the right hand side [cut scene]. Go back through the window and head back the
way you came, get to the locker near the green and white brick wall and turn right,
then left through the new opening you made. take out the guard and go to the door
on the far left. Turn to the right after the door and take out the guards. Head
through
the left area past all the computers. On the far left back desk is a MEDIKIT. Go
through the opening to the right and head left down the corridor. Take the last door
on the right. Taking out the guards on the way, follow the wall to the right, then
turn left, head towards the last office cubicle in front of you. Turn left again
before
you enter it (just after the wall) head forwards and take to door on your right
after
the office cubicle. Follow the corridor to the right. Go through the last door on
the
right.
8.1.3) Epicenter
Jump onto the fence cage in front of you and climb over it. Keeping an eye out for
guards, go straight ahead. When you get to the wall at the end turn right, at the
corner turn left and head straight up again. In the next sorting area go to the top
right hand corner, through the gap, go straight ahead to the shuttered door at the
end of the area. Take out the guards on the way. [cut scene].
8.1.4) Unexpected Arrival
The elevator opens to trouble, deal with the police first in this area. Head up and
turn left at the end, take out the police behind the fork lift, this will crash
into the
wall starting a fire (and all the sprinklers). Keep following the corridor to the
left
and left again. Taking care of all the police. On top of a box at the next corner
leading to the right is a MEDIKIT. Follow the right turn and right again through
the
gap. Head all the way forward into the caged area. [cut scene]. You are now high
up, the easiest way is to ignore the police below and focus straight ahead to the
opposite side of the cage entrance (if you do find yourself down below you can
use the caged area wall to climb up) turn to the left and jump along the vent
pipe you are stood on or it will break (if it does break hold focus and grab the
edge)
Keep heading away from the caged area along the right hand side. Get on top of
the room to the right and jump through the skylight.
8.1.5) Redirected
Head through the door in front of you once you take out the police. Make your way
through the caged storage area to the opening on the left hand side. Turn right
and pick up the pistols from the table in front of you. Use your focus and shoot the
police on the other side of the cage. Head round to the left through the next cage
area and turn right after the yellow barriers. Head straight forward. [cut scene].
Head towards the new area straight ahead. Either climb the cage or break through
the office glass to get to the other side. Zig zag your way past all the boxes.
Keep your eye out for police on your way. Run under the low shutter at the end.
NIOBE - Turn left and take out all the police in the next area. Head up the stairs
at
the end and onto the belt. [cut scene]
GHOST - Go through the door on the left, turn right and gp all the way to the end.
Turn left and enter the cubicle. Turn left again and go through the door. To the
left
is a window (another way into this room from the other offices) kill the police in
here. Head right and when you can get over to the left area of the floor. Go
through
the door here, near the yellow barrier. Go back through the caged area and through
the door. There is a MEDIKIT in here, press the red button on the right side then
turn left so your facing back on yourself and take the door at the end of the area.
Take the door to the right.
8.1.6) Backtracking (GHOST ONLY)
Head all the way down the slopes and take out the police, open the door here. Turn
left, you need to head back through the sorting area. Head forward and right at the
first machine, go all the way to the end, as far as you can , then turn right. Now
head to the left hand opening and go through. Turn left again and head through the
area ahead. Get to the back right corner of the room, then climb the fence to the
left. At the bottom of the slopes infront is a door to the right, take it.
8.1.7.) Got It!
The pick up you need is straight ahead in the left most PO BOX. But first if you
have activated the DROP hack pick up all your weapons from the shelf on the
right hand side. A MEDIKIT is here also. When you get to the PO BOX a [cut scene]
will trigger. Head straight ahead past the next column and turn right and run
towards the lift. Take out all the police to open the lift doors. Stay away from
tear
gas for obvious reason. When you emerge from the lift head towards the front
doors of the building. Try to open either one to trigger the shutters. Now enjoy a
lobby type fight. Clear the area of police. If you need a breather hide in the
toilets
on the left hand side of the lobby. When the area is clear head back towards the
lift where a shutter is in the way. Take the door to your right. Go through the next
door ahead then through the opening to the right in the next room. Head up the
stairs and through the door. Then up more stairs and another door. Yes more
stairs and a door at the top. Take them.
8.1.8) Big Distraction
[cut scene] Either make your way down the stairs to ground level, Or if you are
impatient focus jump off the edge. Go straight ahead and turn right. This level is
very linear, just keep going straight anyway you can, taking out all the police on
the way. Finally after getting back through the demolished sorting area, you will
get to a door on your right. Go in and take the stairs. Go through the door at the
top.
8.1.9) Breakout!
Open the door to the left, go through and take out the police that will emerge
from behind you. Straight ahead are a couple of MEDIKITS on the desk. Keep
moving as police will continue to come from behind. Take the other door in this
room. Head through the storage room to the door at the other end. You need
to head around to the left to another door when you enter this area. Use your
focus and RUN as the police will keep re spawning and alot will be shooting at you
,open the door to trigger the [cut scene].
------------------------------------------------------------
8.2) City Driving
------------------------------------------------------------
As Niobe when you drive remember to change your view, although it is more
cinematic the default view is restrictive (good fun though). The chase view
will make life easier for you
8.2.1) Get To The Hardline (NIOBE)
Go straight ahead and through the speed trap ! Stay on the right side of the road
and up the embankment. When the road joins up again get on the left hand side
watch for an overpass above the road, head left of the column holding it up to cut
out for a police car coming out of a right junction, this is the right you need to
take. Keep following the road ahead, keep on the left and turn down the street to
the left after the orange building on the corner. Again follow the road. You will
reach
right and head back the way you came, at the orange building circle round it to
the right through the alley. Now head back towards the roadblock (the secret to
survive is to keep moving) repeat this if you need to until it says roadblock
cleared
8.2.1) Get To The Hardline (GHOST)
As gunner you have two roles, clear the path infront of civilians and take out any
police cars that show up. Apart from that you are along for the ride. Luckily
as Ghost there is no survival section.
------------------------------------------------------------
8.3) City Rooftops
------------------------------------------------------------
KEEP MOVING !
Weapons:
.50AE
Enemies:
Police
Agent
Run straight ahead towards the ladders in the distance, jumping from rooftop to
rooftop. Climb the ladders and continue forward to the next set. Climb them to the
top. [cut scene] Turn left but keep on the right side of the roof, jump to the
on the next building in front of you. Climb up and follow the roof round to the
right.
Get onto the next rooftop (step) and make your way to the top right corner to jump
to the next building, Head straight forward, then right when you get to the wall.
Jump to the next roof and turn left. Take out the two police infront to build up you
focus. Head right over the corrugated, sloped roof. Jump off the end of the roof
to the damaged wooden floor of the next building. You will fall through, from the
level you land on jump across the building to the ledge in front. Make your way
right and jump to the ladder in front. (If you do fall to the lowest level of the
building
go through the opening to the right and follow it round to the opening opposite the
building.
Head to the top left edge of the building you are on and again jump diagonally to
the damaged wood section, Drop to the next level and take out the police. Ignore
the MEDIKIT to the right of the stairs if you can - he is coming. Head up the stairs
and right out the door. Head straight towards the ladder jumping the gap on the way
familiar ?) Climb the ladder to the right (yes the first one) [cut scene].
------------------------------------------------------------
8.4) The Airport
------------------------------------------------------------
New Weapons:
M16
33SG Sniper Rifle
M95 Sniper Rifle
SG - 552
M240 - Get this for Ghost using the DROP hack - think of it as a helicopter buster.
As there are two choppers for you to knock out I suggest conserving ammo for
these encounters.
New Enemies:
Standard Swat
8.4.1) Check In (NIOBE ONLY)
[cut scene] Focus jump straight ahead over the balcony, turn 180 degrees and
head behind the check out to the left hand side.
8.4.2) The Belts (NIOBE ONLY)
[cut scene] Head left and press the GREEN button on the wall to open the shutter,
head through, continue straight to the next button and do the same again. Go
straight forward to the stairs and up to the control room, take out the swat on the
way. Press the RED button in the control room to start the belts. Make sure you
clear the room or the swat will turn the button off when you leave the room. Head
up the stairs on the right side of the room to get some STUN GRENADES. Follow
the walkway and go through the door at the end. Open the next door and take out
the swat guys. Focus run across the walkway, kick the end door through. Go
through the next door, take out the swat ahead and head a little way up the
walkway jump off it onto the first conveyor belt on the right, run with it to the
orange
hole in the wall..
8.4.3) Jackson In Steam (NIOBE ONLY)
Open the door on the right side of the room, Head forward past the pipes, Take out
all the swat, beware of the steam. If you have activated the DROP hack the head
right at the junction, take out the swat on the stairs but carry on. Take the next
right, then the small opening on the left near the end (kill the swat guard) collect
your goodies. Go back to the stairs where you fought the swat, head down them
and go straight ahead, keep going straight ahead at the junction to the centre
core (MEDIKIT down below in the centre). Take the right exit out the area and head
straight forward up the stairs (you should see 3L) turn left at the top of the
stairs
- still taking out all the swat teams that keep coming. Take the 2nd right that
should read 1R - 7R, this is the corridor you need. Follow the corridor left then
right.
8.4.4) Jackson In Steam 2 (NIOBE ONLY)
[cut scene] RUN ! Use focus to get a bit of distance between you and Agent
Jackson, Keep running and follow the tunnel ahead, take out any swat as quick as
possible. Dont take the lower floor passage as it takes time to climb back up, you
also still take damage from above. There is a MEDIKIT at the end of the passage
however. The high walkway seems to have a similar problem. Eventually you will
reach a ladder at the end, climb up it.
8.4.5) Hangars (NIOBE ONLY)
Turn left and run to the far wall, take out the four swat when they come round the
corner Turn right and follow the hangar to the other side. Turn to the right and
look
at the plane. Head towards the front of the plane and climb the scaffolding. At the
top silently strangle the sniper. Jump from the scaffold to the top of the plane and
run towards the tail end. Jump onto the scaffold at this end (MEDIKIT here) and
climb to the top again. Climb up the boxes and jump onto the walkway. Take a
right and open the door on your left. Make your way through the corridor and take
the left door at the end. On the walkway head further left to the yellow ladder.
Climb down and use your sniper rifle (if you have one) to take out the swat, then
activate the lift button (on the right front of the lift) as it lowers, jump to the
wing of
the plane, run along and open the plane door (it is possible to just skip the next
bit
towards the front of the plane. On the left side is another door to exit the plane.
On
the scaffold take out the swat high up, than make your way down to ground level.
Head towards the back of the hanger to make your way to the other side as before,
as before four swat will come round the corner for you. Run along the back of the
hangar to trigger the [cut scene]
8.4.6) The Bowels (NIOBE ONLY)
On this level you can use the Sniper rifle (night vision sight) to make
your way through, Follow the small tunnel and turn right after the door to take
out the swat sniper. Pinch his gun and make your way up the corridor. The next
sniper is just after the second narrow section of corridor, where the corridor
starts
to turn left. The next is just after the next narrow part of corridor. Keep going
until
the next narrow part as 3x swat guys will run towards you. 3x more are behind
them. Keep your distance and stand your ground, or run in and disarm them all.
As you get closer to the stairs 3x more will run down at you, be careful. Remember
to use the walls. At the top of the stairs are 2x guys that will through grenades.
Make you way up. At the top use the left opening (tarmac access) [cut scene]
8.4.7) Catching A Plane (NIOBE ONLY)
[cut scene] Clear the room of swat, use the cover as much as you can. You are
heading to the back right of the room, to a yellow ladder. At the top turn right to
climb up another ladder [cut scene] head through the door on the left. Run over
the walkway then through the next door, shoot from the walkway at the swat below
dont go past the steam. Head left to the ladder and slide down it. Then down the
next. Head towards the plane. [cut scene].
8.4.8) Agent On Board (NIOBE ONLY)
When you start note the button on the right of Axel, head towards the front of the
plane, defeat the swat and make your way round the cargo netting in the middle
of the plane , on the left is the rack with green parachutes on, pick them up.
Head to the back of the plane, on your way are some boxes, on the top is a
MEDIKIT, if you find you need it later. when you get to Axel [cut scene]
Before you do anything press the green button on the left side of the back of the
plane, this will open the rear, now turn around and face the Agent. Make sure
he is never armed, now make you way to the edge of the plane. Use focus to knock
him down, then drop kick him out of the plane. [cut scene]
Note : You will receive a message stating a focus increase bonus, You get this
regardless throughout the game, yet this is the only place it will advise you.
Some people are concerned that if you fall out the plane when you get this
message - it wont show next time. Don't worry you will get it message or not.
8.4.1) Concourse (GHOST ONLY)
Head round to the right to the security desk. clear the area anyway you wish. Now
carry on . Clear the area around gate E1, Go behind the desk for a MEDIKIT. Take
the first left to head through the toilets, take out the swat at the other end.
Follow
the back corridor and go through the male toilets. More swat are waiting at the
other end. Clear the area around gate E2 and again take the first left, this time
through a customs area. Open the door at the end.
8.4.2) North Point (GHOST ONLY)
Head forward through the open door, This is an ideal place for a flashbang if you
have one, either way clear the room. Pick up the stun grenades from the back right
corner. Now head down the small opening on the right hand wall. Throw a grenade
next to the windowed area on the right side. You can now get through. Jump past
the window and clear are E3 of guards. Head all the way to the end and take out
the guards. Take the left opening to lead you to area E4. If you activated the DROP
hack command then your weapon cache is behind the desk in this area. There is
also a MEDIKIT here. Swat will come out of the surrounding areas, take care of
them. Head all the way forward to the next area where the chopper is waiting for
you. Move towards the window.
8.4.3) North Point 2 (GHOST ONLY)
As you face the window there is a monument structure in front of you and two walls
coming inwards on the left and right. Get close to the right hand wall, this will
protect you from the helicopter fire. This boss like others follows a pattern.
First it will strafe left past the window shooting, stay next to the right wall to
avoid
being hit.
At the far left of its run it will turn 180 degrees and start heading back right,
this is
your chance. Use focus to get as much time as possible and get to the window,
open fire with preferable the M240 (DROP hack) if you have it if not then the MP5.
The M240 Will take the chopper down in seconds! If however you are stuck with
the MP5 then you have some work to do. Try to get at least 2 full clips into the
chopper as it heads to the right. Now turn around, If you have any grenades wait
1 second then through them towards the Pentium 4 sign on the wall.
This is the third part, 4x swat guys will fast rope down through the ceiling.
Take them out as best you can, try to keep your focus up by fighting, the grenade
will help. When you have head back to the right hand wall where you started.
The pattern will start again. Repeat this about 3 times to finish it off.
Once the chopper is down, some swat will open the back right side of the room,
take them out and go through. Past gate W4 and through the shop. Take out the
swat at the far end asap while they are in a group. At the far end head up the
stairs
8.4.4) Monorail (GHOST ONLY)
Head up the stairs to the first right. If you look down to the left walkway you will
see a firefight there. Help by taking out the swat and the gate infront of you will
open. Head all the way up the walkway taking out the swat as you go. Head down
the stairs and to the end.
8.4.5) Revolving Restaurant (GHOST ONLY)
[cut scene] Turn around and head up the stairs as you need to initially clear the
area of swat. Make your way to the top, with the swat gone face the bar, then turn
left. If you focus jump straight ahead you will get to the Piano. Stand on top and
turn around. You must fight off the swat from here until the ladders come around
again. [cut scene]
8.4.6) Terminal (GHOST ONLY)
Take out the swat straight ahead, focus jump over the barrier to get to the lower
area. Once you have cleared the area the shutter on the first floor will open. Kill
the
swat that have just entered and go through the new opening. Open the right hand
shutter and make your way to the right. Taking out the swat as you go. Stay on the
right hand side and take the last opening to the left at the end for the escalators.
Turn right and head up them taking out more swat. At the top make your way round
to the right. Head all the way forward taking out the swat and head towards the
open door at the far end of the upper area.
8.4.7) Control Tower (GHOST ONLY)
[cut scene] Head straight forward and silent kill the sniper. Use the sniper rifle
to
clear the stairway below of swat (near the door on the 3rd level), ignore those on
the roof. It will now tell you to make a move downstairs. go. Turn around and take
the yellow stairs. Switch your weapon to take out the swat as you descend. Go
through the open door ahead when you see it. Silent kill the swat on the left of the
balcony. Get back up to the top of the tower again.
8.4.8) Control Tower 2 (GHOST ONLY)
[cut scene] As soon as you get control look left at the plane, zoom right in with
your sniper rifle and use focus to give you time. Now start shooting at the front
wheels of the plane, you are successful when you get a [cut scene] now for a
tough fight. Switch to the best automatic weapon you have. If you still have ammo
for the M240 you are laughing. Get yourself into a corner with the best view of the
control tower you can and keep an eye out for the agent in the chopper. As soon
as he appears slam on focus and keep moving and shooting him (cartwheel etc.)
Repeat this about 3 or 4 times to defeat him (only once though with the M240)
Head down the stairs when you can [cut scene]
------------------------------------------------------------
8.5) The Aqueduct
------------------------------------------------------------
8.5.1) Retrieving Axel (GHOST ONLY)
Ghost gets to drive as well ! This mission is in three parts, due to the driving
physics this can be a little annoying at times though. Other parts are fun.
Go straight ahead down the Aqueduct. If you hit a bump at a bad angle, let of the
acceleration for the landing until you right yourself. Avoid the boxes if you can.
If
the [lane gets too far ahead you will fail the mission, so keep moving.
Dont worry about passing the plane and getting ahead, just keep going forward.
Once Axel has jumped out the plane you can relax for this bit, you can even take
your time on this bit. Stay out of the water and keep making your way forward.
This is where other swat vans will chase you, not too much of a threat. Just keep
your speed up. Keep an eye out for arrows telling you which route to take. ie the
first set of three small yellow arrows will lead you through the gate to the left.
If you
do miss them, just back up and take the route. There are some sharp 90 degree
turns here, try not to break, just release the accelerator and slide round them.
You will reach the final part, chasing after the car. Stay as close as you can.
Just watch them and follow the route they take. Dont be afraid to break and stay
behind them. (Instead of overtaking.) Get to the part where there are two tunnels
splitting the aqueduct and get in one for the [cut scene].
------------------------------------------------------------
8.6) The Sewers
------------------------------------------------------------
New Weapons:
SS Shotgun
Demolition Charges
2x MP5
2x Mac 11
New Enemies:
Sewer Swat
8.6.1) Abyss
Head straight forward along the pipe and follow it round to the right. Jump over the
wall and carry on straight. Keep following it round to the right and climb up the
ladder at the end, again follow the path round to the right and up the slope. At
the
end turn around and climb higher up the slope. Then use the stairs to double-back
again. Climb up onto the generator at the top, then climb the ladder above it. Run
along the ledge then make your way right around the pipe to climb the ladder that
is attached to it. At the top take out the swat nearby, also the ones on the
opposite walkway to help Ballard. Keep making your way along the walkway to the
ladder at the far end. Get your Sniper rifle ready, When you get to the top Ballard
will be under fire again, turn around and use focus to buy yourself some time.
Two snipers are on the highest level on the left walkway, near the part which is
connected to the ceiling. There is also a sniper on the same level as Ballard to
the left again, just round the far corner of the shutter, If your not sure look for
the
bullets to give them away. Now head towards the area where you took out the two
and to the right, past the ladder you will see Ballard, walk towards him.
8.6.2) Abyss 2
[cut scene] Take out your sniper rifle to clear the area, First kill the swat just
near
the phone, then as you face the centre structure a swat is on the far right walkway
and one on the far left. Now looking at the centre aim high to see a sniper on the
top level aiming at you, take him out. Head back and use the phone. [cut scene]
With your new grenades, take out the fan at the end of the walkway and go through
8.6.3) Pumped
Head left down the pipe, then take the last right. Drop down into the area and take
out the swat in the room to the right, go through an make your way round right
again and clear the next room. Head through this one and take the left corridor.
Follow the corridor round and clear out more swat. Stay next to the left wall and
right staying on the top level. Jump the gap straight ahead and keep moving to
drop down the other side, as a few swat are now far behind you. Follow the arrow
to the ladder on your left, climb it and go near the steam to change the objective.
Now turn to your left and make your way back towards the way you came,
towards the swat, jump back over the gap and go straight forward, dropping down
the gap slightly to your left, the caged door infront is now open (under the
generator), head through and down the hole.
Head straight forward and keep to the left wall, past the water. Get to the last
opening on the right, here climb up onto the shelving on the right, and up onto the
ledge, take out the swat nearby. Drop down to the left , near the pipe. Head
forwards through the opening. Take out the swat on the way and turn right and
right again to find a small ledge between walls, climb on this then climb up to
the left hand side. Turn left and look at the RED pipe on the ceiling, that is what
you are following, jump and make you way to the far end of the room. Drop down
to the machine that the pipe is connected to and press the RED button. This will
cut the steam. Make your way back to the beginning of the room, yet take the
right before the end (the other end of the RED pipe leads here, in this room follow
the ladder up. Now go straight forward following the right hand wall round to the
left, get to the hole in the floor.
8.6.4) Ice And Corrupt
Ice and Corrupt are under fire, take out the swat nearby, now follow the ground
level as far as you can to find another swat guy. Ice will tell you to climb up. Now
make your way towards the start, near the first swat guys is a ledge, now on your
right hand side. climb it. If your having trouble put your gun away. Now turn right
and climb onto the small ledge next to the large generator, next climb onto the
generator itself. Turn right and walk along to trigger a small [cut scene]. When
you get up take out any nearby swat and take the left opening next to the
damaged spinning mechanism. head through here taking out any swat in the way
in
for it again. Head left then left again, follow the area round to a small ledge on
the
right, climb this then climb up to the higher level. Right infront of you is a large
spinning mechanism, Take a look at the base which isn't moving, There are 3
pieces of it that stick out, all the same colour orange as the mechanism. Change
to 1st person view and shoot at the first one directly next to your feet. They are
waist height and have little white lights on. When you hit them they will disappear
make you way round and get all three. Now take out any remaining swat while
the mechanism destroys itself. Now facing the mess you have just made turn to
your left, jump into the pipe you have opened.
8.6.5) Waterway
[cut scene] Drop down into the pipe and head left, run up to the swat and silent
kill him, run and follow the pipe further and take out swat number two. Now take
a right and head up the ladder. Go left and over the bridge, take out the swat on
the way, Run and focus jump off the ledge in front to get to the wall on the other
side (this cuts out some of the stairs) now make your way down casually or focus
jump back across to get to the bottom. Follow the path and kill the swat on the
walkway. Slide down the ladder on the left side of the walkway. Now turn 180
degrees and head away from the ladder to the far wall, pick up the flashlight.
(it looks like a gun, but is an add-on, Also a word of warning for those of you who
cheat, you may have difficulty picking it up) now when you have your gun out, it
will
by the ladder back off until it explodes. Take out any swat that come down, go to
1st person and try to take out the ones high up also. Now climb back up the
ladder and carry on. Follow the corridor the pipe on the left, go through that and
follow the swat guy to the right, stay on the upper level. As soon as you see them
throw a grenade to take them all out in one go, now carry on forward to the grate
at the end (drop to the low level) Wurm will wave, wait for him to leave then shoot
the two swat that appear behind him. Now turn around (this can be tough) get back
on the high level using the ladder on your left. If you have used the DROP hack
then check behind the pipe above the grating to the left for your weapons cache.
Now stay on the left side and make your way towards the other far end of the
tunnel, due to the restrictive view of the tunnel, you are either attacking on a
wide
angle (open) or on a tight corner. Take your time, and dont feel bad for hanging
back and recharging (the getaway anyone !?) after each team of swat. When you
get to the far end try to grenade the large group of swat. The tunnel splits in two
at
this end, take the right opening to continue. Focus jump across the broken gap,
then drop down the other side of the pipe.
8.6.6) Waterway 2
If you like take a left and a left, this will take you down to the swat you just
jumped
past and a MEDIKIT. Again make you way along the high left side of the tunnel,
take you last opening on the left. Then right down the stairs, Now take a right,
then
a right and follow the narrow tunnel. Drop into the larger area and take out the
swat
Now head right around the edge of the hole to the far side and slide down the
ladder, more swat for you at the bottom. Go through the only exit to the left. Take
out the lone swat above the plank to the left, then climb up to where he was.
Emerge from the broken pipe and stay next to the right hand wall once you have
cleared away the swat. Keep making you way around the room, once you have
crossed the white pipe, jump and grab hold of the red pipe above, shimmy along
over the water, they will drop automatically so keep moving. When you get to the
ledge drop down into the broken pipe below. Follow it along to a left hand pipe. Go
down this and kill the swat on the left at the bottom. Now travel along the opposite
way. Climb up the ladder. Follow the path all the way to an enclosed room. Take
out the surrounding swat - then go down the hole in the centre.
8.6.7) Breathing Room
Follow the tunnel all the way to its end then take the right hand opening, head
forward until there is an explosion. Now turn right and head forward to take out the
sewer swat, turn right again to head towards the newly opened room, drop down
the hole in the centre. Follow the tunnel again, it will open to absolute chaos.
Grenade happy sewer swat are everywhere, head left up the tunnel on the right
hand side is a grate (up some stairs) with a MEDIKIT behind. Nearby are stairs to
the left - take them, then right up more stairs. Make your way along the next fan
area taking out the swat on the way. When you get to the far side, take the small
tunnel on the right hand side. You will get to another fan area similar to the one
before, make you way to the damaged area up ahead. Either make your way
carefully across the left edge or focus jump the gaps in the middle to get to the
back wall (watch for two swat guys hiding behind the built up walls). Take the
opening on the back left. Make your way through the next room round to the right
again taking out the swat on the way. Take the tunnel at the end and run through,
when it opens up, drop down to the left and take out more swat during your
descent. At the bottom of the area drop down the hole. Then down the next one.
8.6.8) Malachi And Bane
pipe, get onto the ledge and make you way right, after taking out the swat jump
onto the RED pipe above and shimmy along to drop onto the red boxes. Take out
the swat nearby, now head left towards the ladder at the end. Dont climb down
Turn right and focus jump to the platform just left of the pipe. You should land
near
the barrels, now jump diagonally into the pipe itself (through the break in the
railings.) Drop down from the pipe to the next area. Head right through it taking
out
the swat, then right again, Then left and left again to exit the area. Head all the
way
through the pipe.
8.6.9) Malachi And Bane 2
[cut scene] Turn left to see the ladder, go round the far side of it to climb up.
Make sure you have your sniper rifle out. Kill the swat at the top, then shoot the
one opposite. To keep everyone alive use your focus the best you can, swat guys
come from far left, far right and from the top (centre walkway) keep taking them
down asap. I find the M95 Sniper rifle the best as has a faster rate of fire, easier
target. From the left, swat can come in groups of 2x. Kill about each swat about
2-3 times to finish this area [cut scene].
------------------------------------------------------------
8.7) The Chateau
------------------------------------------------------------
New Weapons:
Wooden Stake.
1911 Pistol
UMP
Crossbow
New Enemies:
Cujo
Doberman - to take out the 'vampire' programs here keep attacking until they
are stunned, your character will automatically stake them at this stage.
Experiment with different types of stake moves.
8.7.1) Great Hall (NIOBE AND GHOST)
[cut scene]
Follow the path of the halls left, left then right. [cut scene]
8.7.3) The Attic (NIOBE ONLY)
[cut scene] Head straight forward through the opening, take out the three vamps
opening, take out the vamps on the way but keep going straight until you reach the
far back wall, now turn left, keep going and take your first right, A secret door
will
be open here. Walk through it.
8.7.4) Persephone Bedroom (NIOBE ONLY)
Turn to your right and walk up to the red chest of drawers, Persephone's number is
on here. Now make your way towards the bed and take the white door on the right
nearby. Head all the way through to the room at the end, Pick up the weapons and
make your way back to the bedroom. Take out the two vamps first, Now work your
magic on Vlad. The second white door in the room is now open, take it. Head
through and open the next door infront. Now follow the corridor round to the left,
If you have activated the DROP hack take the last white door on your left. in here
are your goodies and a MEDIKIT. Go back into the hall and take the double doors
at the end. Press the button for the lift (right hand side) and walk in.
8.7.2) West Wing (GHOST ONLY)
Head straight ahead through the room (over the table if you wish) and through the
double doors at the far end. Take out the vamps in the kitchen. Head over to the
left hand opening. Then left again. Take out the vamp that appears in here. Now go
through the door he has opened. Take out the vamps in this room on the right.
head into the elevator to the right of the pool tables.
8.7.3) Atrium (GHOST ONLY)
Go through the door to the right, follow the corridor to the atrium and take out the
vamps here. Head all the way through and take out the next vamp in the next room
Now head through the passage that has opened behind the bookshelf on the left
wall.
8.7.4) Secret Passage (GHOST ONLY)
Head forward and fall through the floor, Go straight ahead and climb the right hand
wall at the end. At the top take out the vamp who comes through the wall. Go
through the opening and climb to the top of the wall infront. Climb up higher, then
drop down into the attic space. Keep fighting vamps on here until the window is
blown open. Keep away from the edges of the room as they catch fire.
Jump through the new opening.
8.7.5) Secret Passage 2 (GHOST ONLY)
If you have activated the DROP hack all the weapons will be by your feet to the
next balcony. Here is a MEDIKIT and two vamps for you. After they are sorted
jump to the next balcon to the right. One more balcony then enter the french
window to the left. Take out the vamp here and use the back right opening. Follow
the corridor and down the stairs to another museum type area. Kill the vamps here
too. Head all the way through and down the next corridor to a third area. Kill
another couple of vamps and take the large opening at the back right side of the
room. Again another area with more vamps, exit using the double doors they came
in from. [cut scene]
8.7.6) Merovingian's Office (NIOBE AND GHOST)
Head down the hall and through the open white door to the right, take out the vamp,
go through the door he came in, turn right down the corridor and open the double
white doors to your right. Run in [cut scene] Take out the vamp that comes in. Now
head back out into the corridor and take the newly opened left white door. Another
vamp will burst in, sort him out then go through the door he opened in front of you.
Take a right up the stairs then turn left, open the door in the corner on your
right.
Run into the cinema and head towards the projector, past all the chairs. Now turn
around and fight the vamps. Head back under the screen to see the two doors you
passed are now open on your left. Go in and run to the door at the end.
8.7.7) Garage Hallway (NIOBE AND GHOST)
Take your first right and silent kill the vamp looking the other way (do you
recognise this place from the film - running keymaker..) take out his mate while
your at it. Make your way to the far end and turn right, get to the door.
Run through the door that opens infront of you
8.7.9) The Dungeon (NIOBE AND GHOST)
[cut scene] Make your way down the spiral staircase (anti clockwise) then turn
right, head through the opening in front. Down the stairs and into the hall. Take
out
the two vamps here, in the far right and left corners are MEDIKITS. Get to the back
of the room to find more stairs leading down. Get into the circle arena to fight
CUJO
When he gets to about half damage he will send a few of vamps at you and shoot
from afar, take these vamps out asap and CUJO will come back for a beating.
Directly opposite where you came in is the exit, climb out of the arena and climb
up into the opening. Make your way up the stairs.
8.7.10) Cain And Abel (NIOBE AND GHOST)
Follow the route of the dungeon, through the gates and onwards [cut scene] open
the door and you will pick up the alternate character, now head out the door and
back the way you came. [cut scene]. As Cain and Abel are werewolf programs
you cannot kill them. Instead attack them to do so damage and force them near
the prisoners in the cells, they will hold onto Cain and Abel. Now pick up your
partner again and make a run back to the start of the level. [cut scene]
------------------------------------------------------------
8.8) Under The City
------------------------------------------------------------
8.8.1) Twins In Pursuit (NIOBE)
Fairly straightforward, keep moving as fast as possible, keep your eye on the
arrow, a sharp left takes you to the exit. [cut scene]
8.8.1) Twins In Pursuit (GHOST)
Only one job on this level, Keep looking ahead and keep the way clear infront.
Shoot any civilian cars ahead to keep the speed up, ignore the twins.
------------------------------------------------------------
8.9) The Zen Gardens
------------------------------------------------------------
8.9.1) Trinity And Ghost (GHOST ONLY!)
Enjoy yourself and fight Trinity.
------------------------------------------------------------
8.10) The Freeway
------------------------------------------------------------
8.10.1) Chase Morpheus (NIOBE)
Same again, keep moving, use ghost whenever agents get close. [cut scene]
more of it, use ghost to clear police out the way.
8.10.2) The Truck (NIOBE)
[cut scene] More driving more police, when you get to the semi, keep it in sight
it will slow down to meet you eventually. [cut scene - best one no less!]
8.10.1) Chase Morpheus (GHOST)
Take out the police ahead until [cut scene] now face backwards and just keep
shooting at the agents to keep them at bay. Take out any police on the way.
You will finally get a [cut scene] and the agents will leave your trail. Now
concentrate your fire forwards at any police that get near. [cut scene] Continue
taking out the police.
8.10.2) The Truck (GHOST)
[cut scene] Keep taking out police again, stay looking ahead.
When you get near the truck, try not to hit civilian cars as debris slows you down
and Niobe will drive round them. Keep going you will eventually reach the truck.
[cut scene]
------------------------------------------------------------
8.11) The Power plant
------------------------------------------------------------
New Weapons:
G36
GP-25 Launcher
New Enemies:
Armoured swat
8.11.1) Reactor Foundation (NIOBE ONLY)
Make your way along the large tunnel, as you reach the door take out the guard.
Clear the next area of guards and make your way to the top right end of the room
Go through the opening and follow it round to the end. Make your way down and
take out all the guards. Head to the far end and take both the ladders up. Turn to
the right and climb up the cage. Head forwards and take out the other guard, at the
end is more cage to climb. At the top wait for ghost to take out the lights. Now go
to the opening and stay close to the left wall, go along the edge then climb up
higher on more cage. At the top, turn right and focus run jump over the gap and
off the edge [cut scene]. Ghost will help, but you must take out the ground swat.
Turn left and start there. Keep going straight and then go right through the opening
Take out the three swat here, hiding behind concrete pillars. Follow it round to the
right, the right again. The ladder you need is on your left hand side. Turn left
and
run across the bridges in front of you. At the end turn left and go to the edge (
there
is a MEDIKIT on the far right of the ledge you stand on). Look at the door on the
ground floor and slightly to the right, Jump down there. Now go through the door /
opening and follow the slope down.
8.11.2) Reactor Foundation 2 (NIOBE ONLY)
Carry on down the slope and turn right, in the next area take a right after
killing the
swat, follow the tunnel area till it opens out to a place with pillars, take out
the swat
behind them. Make your way left through this area, then take the opening on the
right at the bottom. Again follow the passage round until you get to a fluorescent
lit
room. take the left opening out of here and climb up the ladder. Take the right hand
exit out of this area, then immediately turn left, Due to the snipers here get
behind
cover quickly. Now focus run forward along this area between bits of cover and the
snipers on your right wont hit you. Run round to the right then into the left
opening.
Follow the passage again to another sniper area, yep, Focus run again and get
behind cover, take out any swat on your side. Make your way up then left ( behind
a pillar straight ahead is a MEDIKIT if you need it) take the right opening. Follow
this passage way to the exit. The pit area you enter at the other end is full of
swat
for you to take care of. take a left then your first right. There are many ways
through the pit, once you get to high ground you can see them. If you decide to get
to high ground before clearing the area make use of focus run to keep you alive.
There is either a set of ladders take your first right, then right again to see
them.
Or from where you are looking up the pit, there is a rectangle block of concrete on
your left, make your way round it to find a lower part, climb onto that, then onto
the
block itself. You can now either jump left to the ladder area, or right to a
concrete
path that will lead to the end of the pit. Jump from the path to the wooden plank
(slope) when you see it, keep running along the high area and along another plank
to get to the other end of the pit. Head towards the left hand door.
8.11.1) Reactor Construction (GHOST ONLY)
Like Niobe head along the open tunnel roound to the right, as you reach the door
take out the guard. Clear the next area of guards and make your way to the top
right end of the room. Go through the opening and follow it round to the end. Now
round and down, take out the guard here. Cut through the small door opening and
turn left. then right, head over to the far right edge near the caging. The left
part of
the caging is missing so focus jump from here over to the ledge ahead. Head left
past the ladders and focus jump to the left to get to the generator. Press the green
button to get [cut scene]. Now focus jump back to the ledge you came from, to the
far left side of it and grab hold of the ladder, climb up and then up the ladder you
passed earlier. At the top head right, follow the corridor round. [cut scene]
You know need to sniper the snipers ! The first is dead ahead infront just zoom in,
the second zoom out and he is on the walkway just slightly to the right, the third
is
high left on a yellow ledge. The last one is is even higher up and slightly right
above
the third. Niobe now needs covering fire from swat on the ground. Look straight
ahead and down for the first. He runs from left to right, the second is behind the
corner of the wall on the right. They respawn so keep at them.
When you get the all clear head along the ledge to the left. Change your weapon
if you wish and take out the swat here. Climb up the caging to the right. Clear the
swat here at the top and cover Niobe again. Get you and your rifle to the ledge.
Here she makes a run round the far end of the room (right to left) take out the
swat on the far left of the pit, they are on the same level as her. Get to the left
of
the ledge and climb the caging that sticks out when you get the all clear. At the
top head straight along the ledge and climb higher up more metal bars. Run round
to the right and climb up. Make your way over more structure then, make your
way down and to the right. Keep Heading down and take out the swat here. Now
drop down the hole in the floor to the left. Head through the opening behind you
8.11.2) Reactor Construction 2 (GHOST ONLY)
Follow the tunnel through until it open out, if you were cautious you would see the
heads of swat guys over the edge of the inner wall, either focus run round the edge
of the room or take them out as you go. When you get to the other side, follow
more tunnels. Climb up the crates to the top and turn left. Run and focus jump to
the platform ahead. Focus jump again over the gap, now drop down to the right.
Turn around and head through the opening to the right. Follow the tunnels up the
stairs and focus jump right to the crates, focus jump again to the concrete
structure ahead. Jump from structure to structure, forward then right at the end,
stay on the right side and jump into the opening on the right where the ladder
and stop behind the last pillar in the centre, use this for cover, pop out with
focus
to take out the four swat high up. Climb up the crates and to where they were.
Head forward and drop down to the lower level, take out the swat there. Follow the
area round to the right.
8.11.3) Nuclear Shipping
[cut scene] NIOBE from inside the control room - take the right exit, then turn
left
and make your way to the shutters, the top left (left of shutter one is the door)
[cut scene] GHOST from outside the control room - head to the end of the building
to the left and look left to the container in the middle of the area, head towards
the
rear of it and climb up onto the small container, now jump up onto the large
container itself. Make your way towards the front and silent kill the sniper on top.
Now head towards the exit door - the far left shutter one is just right of the door.
Go through the door, Now focus run right across the room, disarm any swat on the
ground level and stay near cover. Kick the door at the far end open.
8.11.4) Transformer Field
If you have activated the DROP hack the weapons cache will be directly infront of
you. When you get through the door, the next area has a large structure in the
centre, head towards it and use the stairs that circle the outside to get to the
top.
Take out the few swat on the way. At the top press the green button to get a [cut
scene] and the Chloro-bromo launcher (be careful with this one!) Now dont be
tempted to just jump over the barrier, its fun but you die. Take the long way, at
least at first and go down the walkway. Run down halfway then dive off at the
corner - the swat will be straight ahead. Use your new toy, at this distance you
will
be fine. Now turn to your right to the open doorway and make your way over there.
Head round the structure and straight ahead to a couple of swat, take them out
then turn right and run up the corridor. Ignore the room on the right, just keep
going
8.11.5) Transformer Field 2
I wont bore you with this level, it is lay'ed out quite linear, just make your way
through, taking out all the armoured swat (who refuse to die) on the way.
At the first junction you get to the arrow will tell you to go left, There is
nothing to
the right. Be careful in the area full of pillars, they hide everywhere. Shortly
after
is a left turn although the arrow states to go straight, there are a couple of swat
here and a MEDIKIT. The next Junction is a reverse, the arrow says left, but you
can carry straight on, straight is a swat and another MEDIKIT.
When you get to the end, climb up the ladder. [cut scene]
8.11.6) Generator Turbine (NIOBE ONLY)
Stay on the high beam an follow it round to the left, then right, Focus jump to the
roof of the control tower, now climb the ladder infront of you up the side of the
generator. There is another ladder infront of you, slide down that one. Run
backwards down the stairs directly behind you. (This is the fastest way, there are
many ways to do this level however. ie drop down and silent kill the sniper, use his
rifle to clear the control room, then jump left and take the walkway round to the
control room) [cut scene] from facing the controls turn left and take that route
down lower, take out the swat here. Pick up any weapons / ammo from this
floor that you need then take the right hand door. Turn left and run to the door
straight ahead on the other side of the room. Make your way through both areas
and come out into a corridor, turn right and run up that way. Take the last door on
the left. Follow it round to the right , just keep going.
8.11.7) Generator Turbine 2 (NIOBE ONLY)
Go through the door nearby [cut scene] now head round to the right, (in the office
here is a MEDIKIT) Keep going all the way down, when you get past reactor 2 (the
door number on the left) the next office has a MEDIKIT. Carry on and enter door 1
Keep going through all the doors until you get to a ladder, climb it. Follow the
gantry round and climb the next ladder. One final short ladder for you then plant
the bomb. Take out the swat that drop in and slide down the ladder. Make your
way over the other side of the walkway and down the slopes. Jump over to door 2
when you can. Go through the doors and take out the swat, then climb the ladder
over on the left. Follow the walkway to door 3, carry on the walkways to get to the
highest area. Focus jump to the central area when you have taken out the swat.
Slide down the ladder infront, turn around and go through door 4. Some choice
now, take the middle walkway and run for the door at the far end. After door 5 make
your way to the top area and into the middle again, clear out the swat and climb
down the ladder infront of you. Make you way down the final walkway and through
the doors, keep going for a [cut scene]
8.11.8) Core Control (NIOBE ONLY)
[cut scene]
8.11.9) The Core (NIOBE ONLY)
Make your way right along the walkway, Keep going all the way round until you get
to the stairs leading down, keep taking out the swat as you go. Keep going til more
stairs, then you will find the walkway leads to the centre core, start making your
way back up the inner walkway, climb the ladders when you reach them. Follow
the gantry to the right, and up the stairs. Keep moving and taking out swat (a lot
I know) more stairs will lead you higher. Take the next set of stairs when you can.
Run along the last bit of walkway..well done.
8.11.6) Nuclear Waste Sector (GHOST ONLY)
[cut scene] Head straight forward into the first room and take out the swat, watch
out for the flammable tanks, use them to take out swat, but make sure you are no
where near. Continue to the wide doors on the other side of the room. Go through
the small area ahead, then through another set of wide doors. The next area again
is full of tanks and swat, take out the tanks on your way round to clear the swat
with the fire as they hide round objects. In the glass area on the high right is a
MEDIKIT. Another set of wide doors at the end, take these through another small
area, take out the swat here then move on. Another large room, use the tanks
again and make your way around the right side, climb up onto the black walkway
on the right and over to the wide doors to exit. Head through the last small area
8.11.7) Nuclear Waste Sector (GHOST ONLY)
Go through the doors in front. Go through the next large area, similar to one before
use the tanks again, work your way through and take out more swat, again through
a wide door to a smaller area, Get through this one and open the door. In the next
area climb up onto the computer units straight ahead and drop over them to the
walkway. Again use the tanks to make life easier. Clear the swat and get through
the far door. Yep another smaller area here for cleaning, take care of the swat and
walkway, again use the tanks to take down the swat. At the end of the walkway
walkway all the way left for a MEDIKIT. Take the door ahead to move on. Another
smaller area, then after the door a big firefight for you. Take the ladder to the
left,
at the top shoot the tanks in the back wall. The run round to the right, up the
small
stairs taking out the swat as you go. Shoot the far tanks on the right to clear the
room. [cut scene]
8.11.8) Core Control Room (GHOST ONLY)
[cut scene] Another sniper section for you, First sniper is just right of the
central
core, look for a piece of walkway that is just and end. due to the colours you may
need to zoom a little to see them. Second one is directly ahead and down, look
at the piece of walkway that is left of the stairs. Finally is a guy over on the
left, he
is on the walkway just near the door on the left hand side. Now turn around and
head towards the opening behind you, throw in a grenade as soon as you see a
swat guy and use the wall for cover. This should take care of them, if not pop in
the
room and finish the remainder. Niobe will need help again so run back. Niobe is
making a move from the very right, upwards on a walkway. Swat are heading from
the top right of the screen (watch for their bullets). One is on the same level as
Niobe, a second is where you shot the first guy, on a small part of gantry below,
a third is below him on another similar piece of gantry. Again turn around and take
on the swat team. If you need focus go in and fight them / waste time. Afterwards
again Niobe needs cover. This time she is on the left, Use bullets and focus to find
them, they are on the right walkway across from the door. For the last time swat
will come in the control room. Clear the room. [cut scene]
8.11.9) Core Control Room 2 (GHOST ONLY)
Take out more swat from the room and stay away from the tear gas. Make sure
you damage the computer servers in the centre of the room. Head down the
stairs (left or right side of the room) and clear the swat away. MEDIKIT here in the
centre. An agent will appear. Head back upstairs. Use focus and hit the agent
into a damaged server. [cut scene].
8.11.10) Agent Escape (NIOBE ONLY)
[cut scene] Go straight forward and into the vents ahead. drop down the shaft [cut
scene] take out the swat, turn around and run straight forward. Head towards the
shutter / elevator at the far end. Turn around again and run the opposite way when
the elevator stops. Focus run to avoid explosions and dive at the window at the end
------------------------------------------------------------
8.12) Chinatown
------------------------------------------------------------
New weapons:
2x .50AE
New Enemies :
Agent Smith
Seraph ?!
Information :
There is more than one path through many of these levels, some are similar, some
more diverse. If you find a better / faster way through - mail me.
8.12.1) Seraph's Tea House
[cut scene] Fight against Seraph, at about a quarter energy left you will win
You will go to the skyscraper next before returning to Chinatown.
8.12.2) Smiths Trap
Turn around and open the door, turn right and right again, head down the stairs.
Keep turning right and going down stairs until you see the red door infront of you.
Open this to get to the street. Run right to the bottom of the alley, then turn
left.
Keep going straight ahead, there is a slight bend in the next alley. Get to the end
then take a left, then right then a left, run all the way down here. turn left and
run
some more. At the end a quick right then left. You will see a white door ahead.
Go through another like it then turn right, keep running straight, you will see an
agent appear ahead and a red door open on your right, take the red door. Open the
next one that follows it, now run to the base of the stairs and get up them. Climb
to the top of the building. Now turn left and drop off the other side.
8.12.3) On Foot
Drop down to the street and run left, go round the obstacles but keep going
straight, at the end turn right down the alley, turn right again and go down the
stairs, follow the tunnel. When you get up the stairs at the other end run to the
back of the police car and get the grenades, kill the police on the way. This will
change the objective. Run to the right, then left at the end and through the red
door
, then the 2nd red door. Run right and take out the cops on the way. Go through
another red door at the end of the passageway. Run right through the tire shop and
take the white door on the right. Now run left up the street until you can take a
right
run all the way. Past the agent and take your next left. Get to the fire escape on
the left of the street and climb up to the very top. Open the door at the top, then
the white door infront. Turn to your right and open the door you see there.
8.12.4) Hard Line Pursuit
If you have activated the DROP hack command, your weapons will be by your feet,
Run ahead and right down the stairs, turn right and drop down the hole under the
barrel, take the red door out of the room. Run right along the street then left,
take
out the police here. Keep running until you get to the phonebox. You will see a
swat setting a demo charge here - so keep moving. Run towards the swat van,
dodging bullets, it will explode so take the ladder nearby on the left hand side. At
the top you can get a grenade launcher if you don't have one already. Run to the
right, use the grenade launcher to get the agent out the way if you wish, but jump
off the edge to ground level - yet keep running in the same direction. Turn left at
the
end. Run all the way down and to the right hand side, keep on the right and follow
the wall when you get to the waterfront. Stay on the right and jump / climb the
boxes to get past the fence. Cut through the alley up ahead. Turn right then left
then climb up and jump over the fence.
Go through the door to the left to enter the slaughter house. Head right up the
stairs, then left up more stairs. Turn right and run to the back red door, past the
agent. Outside turn right and run along the waterfront again. Focus jump forward
off the edge, keep moving forward, jump the wall and run right. Get to the end and
run left down the alley, then left again when you can. At the barrels go through the
opening to the left, head right through the fish house, take the right opening to
get
out of there. Turn left and run straight ahead, after the small bridge run right
towards the building. Kick the green door in on the left.
8.12.6) The Church
Run up the stairs ahead, go straight forward on this floor to the end, then turn
right
to the stairs, right again up another set, then another. Now turn right and go
straight to the end, then left through the narrow passage, then left again. Finally
right up some more stairs. At the top turn right and focus run left towards the next
building, focus jump the gap and stay on the left side of the building, run all the
way round the building to the otherside. Where there is another plank, jump to the
next building (i find focus jumping quicker and more fun than running along the
planks) again make your way around it, then drop down the other side, now focus
through the damaged building in front. Come out the other side and jump more
planks.Take a small left then right and go through another run down building. Jump
across two more gaps to get to the red door at the end. Go in another small left
then right will take you a corridor where you can see an opening. Take the left
passage and run up the stairs (although for a bit of flair you can get to the
opening
and climb up. Now focus jump up to the next building. Using the surrounding
buildings focus jump back to a point higher up on the stairs!) At the top make your
run to the right towards the wooden steps at the end. [cut scene]
------------------------------------------------------------
8.13) The Skyscraper
------------------------------------------------------------
8.13.1) Vertigo
Move ! Turn right and make your right again (if you activated the DROP hack then
all the weapons you gained will be here), following the line of the corridor, leap
over the crate in front and head towards the opening ahead. Turn left slightly to
get
through the next opening, then jump and climb up the crates to get right. Jump off
the top of the crates to get left round the barriers, go past the barrel and make
your
way right round the edge of the office. Turn right and cut through the room, now
turn left at the end to continue. When the agent appears infront, dive left through
the window and open the door. Run right up the stairs, turn left and jump out the
window onto the scaffold. Run alongside the building, past the agent shooting out
the windows, keep running until you see the caution sign, focus jump through this.
Keep running along the wooden planks round the building. An agent Smith will fall
and break the one you stand on. Run through the open window to your right and
then jump back out the window on the opposite wall. Run right, then left then right
along the planks, focus jump here off the edge, straight forward ( dont take the
slope) turn around and run into the open window. then up the slope and out the
other side. Run straight ahead and down the slope. When you get to the bottom
either keep moving round the edge of the building / or use the insides as a shortcut
to avoid being shot. Keep moving til you get to an up lope (near an agent) Make
your way all the way up to the top and focus jump forward off the edge, towards
the side of the building [cut scene] run right and open the door. In the office head
towards the back left of the room, an agent will break the wall, nip past him and
through the opening. Run to the lift.
You will now go back to Chinatown..
------------------------------------------------------------
8.14) Onboard The Logos
------------------------------------------------------------
8.14.1) Tunnels Of The Real (NIOBE)
What can I say, keep moving, keep firing, you can't go wrong on this level !
8.14.2) The Rabbit Hole (NIOBE)
Same again yet on this one, you barely have to move, never mind fire.
8.14.1) Tunnels Of The Real (GHOST)
You have to do a little more work as GHOST, primarily face to the rear of the ship
and shoot away at the sentinels. If the radar above shows activity to the left side
of the ship (the front) change view and take care of them (although this is rare and
not often many) generally keep firing backwards until [cut scene]
8.14.2) The Rabbit Hole (GHOST)
More of the same Im afraid, keep at it until [cut scene] Finally the last stretch.
Use your guns to keep the bomb from the ship, just keep shooting to the rear
until [cut scene].
============================================================
9) Legal Bit
============================================================
Please do not use without my permission, Use can be granted
under the agreement that nothing shall be changed. The above
and all contents herein are copyright to myself. I being
known as Dead Rabbit. Misuse and/or alterations of this
body will be perceived as violatory. With this in mind
further proceedings on my part; if I so wish; may be applied
for.
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https://pure.mpg.de/pubman/faces/ViewItemFullPage.jsp?itemId=item_2465555 | English
# Item
ITEM ACTIONSEXPORT
Statistical Gravitational Waveform Models: What to Simulate Next?
Doctor, Z., Farr, B., Holz, D. E., & Pürrer, M. (2017). Statistical Gravitational Waveform Models: What to Simulate Next? Physical Review D, 96: 123011. doi:10.1103/PhysRevD.96.123011.
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Doctor, Zoheyr, Author
Farr, Ben, Author
Holz, Daniel E., Author
Pürrer, Michael1, Author
Affiliations:
1Astrophysical and Cosmological Relativity, AEI-Golm, MPI for Gravitational Physics, Max Planck Society, ou_1933290
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Free keywords: Astrophysics, High Energy Astrophysical Phenomena, astro-ph.HE,General Relativity and Quantum Cosmology, gr-qc
Abstract: Models of gravitational waveforms play a critical role in detecting and characterizing the gravitational waves (GWs) from compact binary coalescences. Waveforms from numerical relativity (NR), while highly accurate, are too computationally expensive to produce to be directly used with Bayesian parameter estimation tools like Markov-chain-Monte-Carlo and nested sampling. We propose a Gaussian process regression (GPR) method to generate accurate reduced-order-model waveforms based only on existing accurate (e.g. NR) simulations. Using a training set of simulated waveforms, our GPR approach produces interpolated waveforms along with uncertainties across the parameter space. As a proof of concept, we use a training set of IMRPhenomD waveforms to build a GPR model in the 2-d parameter space of mass ratio $q$ and equal-and-aligned spin $\chi_1=\chi_2$. Using a regular, equally-spaced grid of 120 IMRPhenomD training waveforms in $q\in[1,3]$ and $\chi_1 \in [-0.5,0.5]$, the GPR mean approximates IMRPhenomD in this space to mismatches below $4.3\times 10^{-5}$. Our approach can alternatively use training waveforms directly from numerical relativity. Beyond interpolation of waveforms, we also present a greedy algorithm that utilizes the errors provided by our GPR model to optimize the placement of future simulations. In a fiducial test case we find that using the greedy algorithm to iteratively add simulations achieves GPR errors that are $\sim 1$ order of magnitude lower than the errors from using Latin-hypercube or square training grids.
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Dates: 2017-06-162017
Publication Status: Published in print
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Rev. Method: -
Identifiers: arXiv: 1706.05408
URI: http://arxiv.org/abs/1706.05408
DOI: 10.1103/PhysRevD.96.123011
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Title: Physical Review D
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Publ. Info: Lancaster, Pa. : American Physical Society
Pages: - Volume / Issue: 96 Sequence Number: 123011 Start / End Page: - Identifier: ISSN: 0556-2821
CoNE: /journals/resource/111088197762258 | {"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": 0.43540528416633606, "perplexity": 11065.16472132716}, "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/1566027313536.31/warc/CC-MAIN-20190818002820-20190818024820-00100.warc.gz"} |
http://www.ck12.org/book/CK-12-Physical-Science-Concepts-For-Middle-School/r7/section/2.10/ | <img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# 2.10: Conservation of Mass
Difficulty Level: At Grade Created by: CK-12
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Practice Conservation of Mass
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If you build a campfire like this one, you start with a big pile of logs. As the fire burns, the pile of logs slowly shrinks. By the end of the evening, all that’s left is a small pile of ashes. What happened to the matter that you started with? Was it destroyed by the fire?
### Where’s the Matter?
It may seem as though burning destroys matter, but the same amount, or mass, of matter still exists after a campfire as before. Look at the sketch in Figure below. It shows that when wood burns, it combines with oxygen and changes not only to ashes but also to carbon dioxide and water vapor. The gases float off into the air, leaving behind just the ashes. Suppose you had measured the mass of the wood before it burned and the mass of the ashes after it burned. Also suppose you had been able to measure the oxygen used by the fire and the gases produced by the fire. What would you find? The total mass of matter after the fire would be the same as the total mass of matter before the fire.
Q: What can you infer from this example?
A: You can infer that burning does not destroy matter. It just changes matter into different substances.
### Law of Conservation of Mass
This burning campfire example illustrates a very important law in science: the law of conservation of mass. This law states that matter cannot be created or destroyed. Even when matter goes through a physical or chemical change, the total mass of matter always remains the same.
Q: How could you show that the mass of matter remains the same when matter changes state?
A: You could find the mass of a quantity of liquid water. Then you could freeze the water and find the mass of the ice. The mass before and after freezing would be the same, showing that mass is conserved when matter changes state.
### Summary
• Burning and other changes in matter do not destroy matter. The mass of matter is always the same before and after the changes occur.
• The law of conservation of mass states that matter cannot be created or destroyed.
### Vocabulary
• law of conservation of mass: Law stating that matter cannot be created or destroyed in chemical reactions.
### Practice
At the following URL, apply the law of conservation of mass to a scene from a Harry Potter film. Then answer the questions below.
1. What is the mass of the professor in kilograms? What is the mass of the cat in kilograms? (Hint: 1 pound = 0.45 kilograms)
2. The scene must be magic because it defies the law of conservation of mass. Explain why.
### Review
1. What is the law of conservation of mass?
2. Describe an example of the law of conservation of mass.
### Vocabulary Language: English
law of conservation of mass
law of conservation of mass
Law stating that matter cannot be created or destroyed in chemical reactions.
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https://forums.torque3d.org/viewtopic.php?t=1000 | ### how i can get rid of the texture spots in terrain?
There are no stupid questions, just stupid answers.
#### how i can get rid of the texture spots in terrain?
subabrain
Posts: 190
Joined: Sat Aug 20, 2016 4:58 pm
Hello again,
okay - a question again
and this is:
how can iget rid of the texture spots on a terrain?
i hope you know what i mean 8)
example:
Greetz
Robert
#### Re: how i can get rid of the texture spots in terrain?
Duion
Posts: 962
Joined: Sun Feb 08, 2015 1:51 am
Do the terrain textures correctly.
Explaining everything would take too much time but I give you some basics:
Here is an example of one of my materials:
new TerrainMaterial()
{
diffuseMap = "art/terrains/grass/ter_grass_short_01_B";
diffuseSize = "200";
detailMap = "art/terrains/grass/ter_grass_short_01_D";
detailDistance = "100";
macroSize = "30";
internalName = "ter_grass_short_01";
macroStrength = "0.5";
detailSize = "4";
macroMap = "art/terrains/grass/ter_grass_short_01_M";
parallaxScale = "0";
normalMap = "art/terrains/grass/ter_grass_short_01_N";
};
The diffuse map gives the color of the terrain and it should be really big and the detail map gives the details and it should be without or almost without color, since it will blend to the diffuse map. Macro map is like detail map just bigger, for medium range and normal map for shadowing details and parallax.
You can also look at the textures how I made them: https://github.com/Duion/Uebergame/tree ... ains/grass
Or just play the game and see how its done in realtime.
#### Re: how i can get rid of the texture spots in terrain?
LukasPJ
Posts: 379
Joined: Tue Feb 03, 2015 7:25 pm
So, I actually didn't find any official tutorials on this topic, so here's a little rough explanation by me.
In order to avoid the repetitiveness you have a selection of different textures you can apply to the terrain.
The diffuse texture provides color for the terrain, you should use it for the over-all color and not for finer details. If you only have a diffuse texture, then it's okay if it looks blurry, it just shouldn't look repetitive and it should reflect the color of the terrain at this area. For example, green for grass or white for snow.
The detail texture is used for the up-close texture of the terrain. This is usually a gray-scale image, or even better a high-pass filtered image. It applies a detail to the underlying diffuse color, for example pebbles for rocks or straws for grass. It can contain colour, but often the best results come with less color in the detail texture, as you can then have different nuances to your detailed texture based on the diffuse texture.
The macro texture, afaik, has the same role as the detail-texture but is used for medium-distance detail instead of the close-up.
Lastly, the normal texture contains normal and height-map information for even cooler detail.
What I often do, is to set the diffuse texture to a huge texture which is as big as the terrain-heightmap (e.g. 2048px * 2048px). This allows me to paint the overall terrain-color anyway I want and with no repetition. E.g.
Then I can control the micro-detail with the detail and macro maps.
#### Re: how i can get rid of the texture spots in terrain?
subabrain
Posts: 190
Joined: Sat Aug 20, 2016 4:58 pm
hey,
ah ok now i understand!
Thank you very much!
Greetz
Robert
#### Re: how i can get rid of the texture spots in terrain?
Jason Campbell
Posts: 253
Joined: Fri Feb 13, 2015 2:51 am
Thanks also, I've been using torque for a long time and never bothered to look into the macro layer.
#### Re: how i can get rid of the texture spots in terrain?
Duion
Posts: 962
Joined: Sun Feb 08, 2015 1:51 am
The macro layer helps most to break up the repetitiveness of the texture, if setup well it is almost impossible to spot where the texture repeats.
This is an advantage you have in Torque3D over other engines, since I have not seen macro maps yet in other engines.
But the macro map gives you also problems, since each texture adds to the ground it can get very dark or bright, so all your textures have to be very low in contrast.
So if you make terrain textures try to set them to medium brighntess, like the brightness value goes from 0-255, make the overall texture around 128 brightness level, that way it will combine well.
#### Re: how i can get rid of the texture spots in terrain?
LukasPJ
Posts: 379
Joined: Tue Feb 03, 2015 7:25 pm
The macro layer helps most to break up the repetitiveness of the texture, if setup well it is almost impossible to spot where the texture repeats.
This is an advantage you have in Torque3D over other engines, since I have not seen macro maps yet in other engines.
But the macro map gives you also problems, since each texture adds to the ground it can get very dark or bright, so all your textures have to be very low in contrast.
So if you make terrain textures try to set them to medium brighntess, like the brightness value goes from 0-255, make the overall texture around 128 brightness level, that way it will combine well.
That's odd, it shouldn't work that way. I fixed an issue like this a long time ago do you still see this behaviour?
It's should not add them up, but rather interpolate between them iirc
#### Re: how i can get rid of the texture spots in terrain?
Duion
Posts: 962
Joined: Sun Feb 08, 2015 1:51 am
As far as I know the textures use the multiplication method, not add, maybe I expressed myself wrong.
But the bright and dark spots still add up, since they get multiplied, thats why I said to make the textures mostly neutral brightness, since if they are bright they brighten everything up and if they are darker than average they darken everything.
#### Who is online
Users browsing this forum: No registered users and 1 guest | {"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": 0.4154563248157501, "perplexity": 2198.482927335752}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-22/segments/1526794863277.18/warc/CC-MAIN-20180520092830-20180520112830-00005.warc.gz"} |
http://www.sawaal.com/physics-questions-and-answers/a-bucket-of-water-is-hung-from-a-spring-balance-a-piece-of-iron-is-suspended-in-the-water-without-to_8388 | 11
Q:
# A bucket of water is hung from a spring balance. A piece of iron is suspended in the water without touching the sides or touching the bottom of the bucket. The reading of the spring balance
A) will increase B) will decrease C) does not change D) varies with the increase of depth of the immersion of the iron piece
Explanation:
When the piece of iron is lowered into the water, the water level in the bucket increases. And so the pressure at the bottom of bucket also increases. Thus the total thrust at the botto also increases. Therefore the spring balance reading increases.
Subject: Physics
Q:
A disk and a sphere of same radius but different masses roll on two inclined planes of the same altitude and length. Which one of the two objects gets to the bottom of the plane first ?
A) Depends on their masses B) Sphere C) Disk D) Both reach at the same time
Explanation:
Time does not depend on mass, else
$\inline \fn_jvn T \prec \sqrt{\left ( 1 +\frac{k^{2}}{R^{2}} \right )}$
$\inline \fn_jvn \frac{k^{2}}{R^{2}}$ => is least for 'sphere' and
hence least time is taken by sphere.
Filed Under: Physics
Exam Prep: AIEEE
1 58
Q:
A potentiometer wire is 100 cm long and a constant potential difference is maintained across it. Two cells are connected in series first to support one another and then in opposite direction. The balance points are obtained at 50 cm and 10 cm from the positive end of the wire in the two cases. The ratio of the emf's is :
A) 3:4 B) 5:4 C) 5:1 D) 3:2
Explanation:
In series : E1 + E2 = K(50)
In parallel : E1 - E2 = K(10)
E1 + E2 / E1 - E2 = 5/1
=> E1/E2 = 3/2
The emf's are in the ratio of 3:2.
Filed Under: Physics
Exam Prep: AIEEE
2 72
Q:
What is the maximum velocity with which a body of mass 'm' must enter a vertical loop of radius 'R'so that it can complete the loop ?
A) $\inline \fn_jvn \sqrt{2gR}$ B) gR C) $\inline \fn_jvn \sqrt{5gR}$ D) $\inline \fn_jvn \sqrt{2g/R}$
A) A B) B C) C D) D
Explanation:
To complete the vertical loop, the minimum speed required at the lowest point = $\inline \fn_jvn \sqrt{5gR}$ .
Filed Under: Physics
Exam Prep: AIEEE
1 17
Q:
Which of the following group of elements are diamagnetic ?
A) Argon, copper, silver B) Hydrogen, argon, copper C) Oxygen, copper, silver D) Hydrogen, oxygen, Argon
Explanation:
Filed Under: Physics
Exam Prep: AIEEE
1 26
Q:
At the Curie temperature, the ferromagnetic materials get converted into
A) Diamagnetic materials B) Non-magnetic materials C) Paramagnetic materials D) None of the others | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 6, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8310799598693848, "perplexity": 1785.2038773569532}, "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/1492917120338.97/warc/CC-MAIN-20170423031200-00540-ip-10-145-167-34.ec2.internal.warc.gz"} |
http://mathematica.stackexchange.com/questions/54925/controlling-quality-of-discretized-region-meshes | # Controlling quality of discretized region meshes
I want to discretize 3D constructive solids, supported by Mathematica 10 regions functionality. Quality of resulting meshes must be controllable, and meshes must be useful for further computation.
This should be easy, but the trivial example below stuns me.
(* Simple derived region in 3D with a sharp edge: *)
r = RegionDifference[Ball[{0, 0, 0}], Ball[{0, 0, 1}]];
(* RegionPlot3D with PlotPoints handles this reasonably well: *)
RegionPlot3D[r, PlotPoints -> 50]
(* This should create a discretized mesh, and it does, but with poor quality: *)
DiscretizeRegion[r]
Edges are noticeably jaggy. This is understandable, and usually option like PlotPoints fixes these kind of issues. DiscretizeRegion has numerous options which should affect quality of its' output. I think I've played with all of them.
Biggest difference I've seen has been is through use of MaxCellMeasure, but interestingly enough, same ugly jaggies stay there even if complexity of the mesh itself increases. I would need an option like PlotPoints which apparently affects grid resolution of a marching cubes style algorithm in RegionPlot3D, but there isn't one for DiscretizeRegion. Any suggestions?
-
In principal you should be able to do
r = RegionDifference[Ball[{0, 0, 0}], Ball[{0, 0, 1}]];
rp = RegionPlot3D[r, PlotPoints -> 50];
DiscretizeGraphics[rp]
Unfortunately, this does not work and is hopefully improved in a future version.
One thing you can do, however, is use the finite element mesh generator for this:
Needs["NDSolveFEM"]
m = ToElementMesh[r,
"BoundaryMeshGenerator" -> {"RegionPlot", "SamplePoints" -> 50},
"MeshOrder" -> 1]
You can then convert the ElementMesh to a MeshRegion simply by using
MeshRegion[m]
Note that in this case MeshRegion does not show it's output since it's beyond a threshold (too big). It has some 300000 tetrahedron elements.
You can force it to display with
Show[MeshRegion[m]]
The "MeshOrder" is set to one (= linear elements) and the "SamplePoints" correspond to the PlotPoints in RegionPlot. There is more information about boundary mesh generation in ToBoundaryMesh, ToElementMesh, ElementMesh and the related mesh generation and element mesh visualization tutorial.
One last thing I should add, is that the 3D stuff does have some rough edges.
-
Whoa, thanks! I'm a bit surprised FEM exposes this, instead of DiscretizeRegion, where I'd expect it to live now. Hoping to see those rough edges (pun intended) smoothed out in the future! – kirma Jul 15 '14 at 16:00
@kirma, DiscretizeRegion should give proper results via AccuraryGoal and PrecidionGoal but that's not implemented fully for 3D. The FEM stuff exposes this to allow maximum 'hackability' and in this case that allows for a workaround. If you do find stuff that does not work, please, by all means send it to support such that it can be looked at and get fixed. Thanks. – user21 Jul 15 '14 at 16:09
good to see you back alive, user21 :-). Is it possible to refine the grid only near the edge, in one way or another? – Albert Retey Jul 15 '14 at 18:47
@AlbertRetey, well yes back to get the geometry and FEM stuff started ;-). You could use "MaxBoundaryCellMeasure" to refine the boundary; this works well in 2D. However, in 3D this does not yet influence the sampling of the boundary. Also, since this region has some thin parts to it it would generate quite a lot more elements. – user21 Jul 16 '14 at 6:59
Here is a way to do what you want using DiscretizeGraphics as suggested by @user21. No need to call NDSolveFEM
r = RegionDifference[Ball[{0, 0, 0}], Ball[{0, 0, 1}]]
rp = RegionPlot3D[r, PlotPoints -> 50]
Now we discretize the Graphics object with the following clever replacements that somehow works:
DiscretizeGraphics[Normal[rp /. {(PlotRange -> _) :>
PlotRange -> All, (Lighting -> _) :> Lighting -> Automatic}]]
To get even finer mesh, increase PlotPoints in RegionPlot (this will increase the time it takes to render the RegionPlot), then proceed as before. I used PlotPoints -> 100 for the image below:
-
@user21, I can't seem to get the mesh as fine as that produced from NDSolveFEM in your answer. I've tried various options: MeshQualityGoal, MaxCellMeasure etc. They don't seem to have any effect. Unless I increase PlotPoints in RegionPlot which takes a while to render. Any suggestions? – RunnyKine Jul 22 '14 at 21:16
but using NDSolveFEM is the best part of it.... ;-) I think that MaxCellMeasure is not working is a bug. MeshQualityGoal has no much to do with the number of tets generated. It tries to generate tets that have certain quality see here – user21 Jul 23 '14 at 7:11 | {"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": 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": 0.4942346215248108, "perplexity": 2931.186189708247}, "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-2015-14/segments/1427131298692.48/warc/CC-MAIN-20150323172138-00173-ip-10-168-14-71.ec2.internal.warc.gz"} |
http://www.purplemath.com/learning/viewtopic.php?f=17&t=1399 | Could really understand the question
Standard deviation, mean, variance, z-scores, t-tests, etc.
Could really understand the question
Question :
A random sample of 500 components was taken from a large consignment and 60 were found to be defective . Obtain the 98% confidence limits for the percentage of defective components in the consignment
(Note that significant value of Z at 2% level of significance is 2.33)
please just point me to the right direction
zorro
Posts: 28
Joined: Sat Jun 12, 2010 9:26 am
Re: Could really understand the question
zorro wrote:Question :
A random sample of 500 components was taken from a large consignment and 60 were found to be defective . Obtain the 98% confidence limits for the percentage of defective components in the consignment
(Note that significant value of Z at 2% level of significance is 2.33)
please just point me to the right direction
Where are you stuck?
Martingale
Posts: 350
Joined: Mon Mar 30, 2009 1:30 pm
Location: USA
Re: Could really understand the question
How do i know what is the standard deviation $\sigma$ in this problem
from the question i can figure out that
n = 500
$\bar{X}$ = 60
zorro
Posts: 28
Joined: Sat Jun 12, 2010 9:26 am
Re: Could really understand the question
zorro wrote:How do i know what is the standard deviation $\sigma$ in this problem
from the question i can figure out that
n = 500
$\bar{X}$ = 60
you need to use
$\left[\hat{p}-Z_{\alpha/2}\sqrt{\frac{\hat{p}(1-\hat{p})}{n}},\hat{p}+Z_{\alpha/2}\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\right]$
Where $\hat{p}=\frac{X}{n}$
Martingale
Posts: 350
Joined: Mon Mar 30, 2009 1:30 pm
Location: USA
Re: Could really understand the question
Martingale wrote:you need to use
$\left[\hat{p}-Z_{\alpha/2}\sqrt{\frac{\hat{p}(1-\hat{p})}{n}},\hat{p}+Z_{\alpha/2}\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\right]$
Where $\hat{p}=\frac{X}{n}$
Could u please tell what is the name of the formula ??
zorro
Posts: 28
Joined: Sat Jun 12, 2010 9:26 am
Re: Could really understand the question
zorro wrote:Could u please tell what is the name of the formula ??
It's the $(1-\alpha)100\%$ confidence interval for population proportions (assuming normality)
Martingale
Posts: 350
Joined: Mon Mar 30, 2009 1:30 pm
Location: USA
Re: Could really understand the question
Thanks All
zorro
Posts: 28
Joined: Sat Jun 12, 2010 9:26 am | {"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": 9, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8726205825805664, "perplexity": 2181.2390223909206}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-52/segments/1418802770400.105/warc/CC-MAIN-20141217075250-00008-ip-10-231-17-201.ec2.internal.warc.gz"} |
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19 Oct 2010, 19:39
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KUDOS
I am completely walking on air right now. After months of studying (mostly quant), I finally took the test today for the first time, and I did better than I ever dreamed. Here's my story.
I'm 31/F, liberal arts degree from fairly well-respected state university, 8 years of journalism experience. I had never considered business school until April, when I took a "finance for non-finance professionals" seminar offered by one of my trade association clients. That was my "aha!" moment: I'd been told all through school that I was no good at math, so I'd avoided anything that had to do with it ever since. Long story short, that seminar showed me otherwise. It made me realize that business school might be a good option for me.
When I first started looking into it, I thought, "Maybe I can get into the part-time program at [low-ranked local university]." My original plan was to go part time, so that was where I looked first. I bought several GMAT books (mainly Kaplan Premier and OGMAT). It quickly became apparent that I needed a LOT of work on quant (reality check: I had not done long division by hand in so long that I actually had to Google "how to do long division"), while my verbal skills, with a little brushing up, were in good shape. My job for the past 8 years has basically entailed reading comprehension, sentence correction and critical reasoning, so I didn't focus much on that until the end.
Quant was a long, hard slog. I spent at least an hour a day reading study materials and doing practice questions; the results of the practice quizzes made me want to cry more than once, but I kept at it, day after day. Finally, in August, I signed up for two related Foundations of Math online seminars through Manhattan GMAT. They came with MGMAT's Foundations of Math book, which was a GODSEND. I cannot recommend it highly enough. While the OG and Kaplan books often assumed I knew things I didn't know, MGMAT's Foundations of Math walked me through the basics in ways I could almost always understand easily. It really made the difference for me. After I mastered that book, I was able to circle back to the Kaplan and OG books and tackle the material that had previously been beyond me. The Web seminars themselves were somewhat helpful, but the book was invaluable.
I took four of the six Manhattan GMAT practice tests that came with the seminars and scored 690, 700, 690 and 730. I spent a lot of time studying the explanations for the problems I got wrong (and the ones I got right!), and I liked the assessment tools that allowed me to compare all the tests at once and see at a glance which areas of math were my greatest weaknesses. This tool also helped me shore up my weaknesses on the verbal section.
The day before the test, I got a mani-pedi and went to see "Secretariat" (fluffy, no strenuous thinking required), then went home to wait for bedtime. I thought I was going to jump out of my skin, I was so nervous. I barely slept. I got to the testing center so early that I had to wait downstairs for half an hour before I could go up. I thought I was going to be sick!
OK, so, the test itself. I'm pretty sure I messed up the Analysis of an Issue essay (my outline wasn't very good), but I feel good about the Analysis of an Argument essay. We'll see in a few weeks. The math was ... oh, man. I thought for sure I was bombing it. I encountered several questions I hadn't seen in all my months of prep and no questions on a few topics at which I'd worked very, very hard, but that's the luck of the draw. (Lesson: Study as many question types as you can! Don't get hung up on any one topic!) By the time I got to the verbal section, my head was spinning. As the final question approached, my apprehension mounted -- I was incredibly nervous about my score. I clicked through all the demographic questions and finally got to the score page.
740. SERIOUSLY? WHAT?? Um... I may have done some silent fist-pumps. Wink
I had hoped to get at least 700. 740 (44Q/47V) was beyond my wildest expectations. I hit the 68th percentile in math -- considering I probably started at the 10th percentile and relied entirely on self-study, I'm thrilled with that number. 99th percentile on verbal; I made some adjustments after a few practice tests that put me in the 93rd percentile, and apparently it worked. Again, Manhattan GMAT's extensive practice test analysis was very helpful in preparing for the verbal section (to be fair, I did not take any other company's practice tests).
Overall: 97th percentile. I will be on Cloud Nine for at least a week.
I have a lot of work to do in the application process, but this score opens the door to a lot of possibilities. Needless to say, I'm now planning to go full time next fall (a conclusion I came to over several months of research -- as a career changer, it seems like the wisest choice). I can't wait to see what the future holds! In the meantime, I'll be bolstering my quant skills with an accounting or statistics course this spring through the USDA grad school. My message to anyone who thinks they can't handle the math is: TRY! I thought I couldn't and I was wrong. It took a LOT of studying, but I got there. You can too!
Best of luck to everyone who has yet to take the test!
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19 Oct 2010, 21:08
I will TAKE and RUN with the 47V
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19 Oct 2010, 21:14
seriously ! I will call anyone who takes V44 - 47 a magician !
mainhoon wrote:
I will TAKE and RUN with the 47V
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19 Oct 2010, 21:14
mainhoon wrote:
I will TAKE and RUN with the 47V
I REALLY hope high verbal scores are valuable to B-schools! It seems like the majority of posters on this and other GMAT forums have higher quant scores. I keep telling myself that business schools want a good mixture of applicants -- some with better quant scores and some with better verbal scores, right?
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19 Oct 2010, 21:17
Hey erdehoff
Congratulations !
Regarding your Manhattan. Can you please breakdown your Verbal scores for us. specially during the last 10 days.
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19 Oct 2010, 22:18
nusmavrik wrote:
Hey erdehoff
Congratulations !
Regarding your Manhattan. Can you please breakdown your Verbal scores for us. specially during the last 10 days.
I am happy to try! However, I'm not sure how much good it will do. For me, verbal comes naturally; my job for the past 8 years has been to do all the things the GMAT tests for in the verbal section (sentence correction, reading comprehension, critical reasoning).
Sentence correction is something I believe many people can learn if they study the rules of English grammar hard enough; basically, there are rules, and once you learn them, they make sense. Algebra works the same way, in my experience -- the difference is that I know the rules of English grammar far better than I know the rules of algebra, hence the score differential!
I will tell you that the results of my practice test (all through Manhattan GMAT) told me a lot about where I needed to focus my attention when it came to the verbal section. I took four practice tests in the past three weeks with scores of 690, 700, 690 and 730; my verbal scores were (respectively) 99th percentile, 93rd, 93rd and 99th. While I focused my studies over the past few months on the quantitative section, I did spend a fair amount of time in the past 2 weeks analyzing my verbal results and studying the questions I missed and the ones I got right. I liked the way Manhattan GMAT allowed me to analyze all my verbal scores at once and zero in on my weaknesses; because I did not take any other company's practice tests, I cannot say how it compares to Kaplan, Veritas, etc. I liked the detailed explanations of each answer, which I imagine most test prep companies will provide with their practice tests.
I wish I had a magic formula to recommend to students who want to improve their verbal score. I do think that much of the verbal score stems from analytical skill, and that is something that transcends English ability.
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19 Oct 2010, 22:44
WOW V 47 !!
This verbal score is dream score for any gmat taker - especially if he is Non-Native.
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20 Oct 2010, 05:15
gr88 score!! gud luck 4 ur app!!!
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20 Oct 2010, 12:22
Congratulations, its a great score. Can you provide some further details about the test itself - tougher word problems youve encountered, combinatorics and probabilities etc, the topics of the RC passages (whatever you can remember)... Thank you!
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20 Oct 2010, 12:46
Great Score....Congratulations.
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20 Oct 2010, 13:19
Awesome score! Do not worry about your Quants score, give you best effort in application.
Good luck!
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23 Oct 2010, 10:48
reg123456 wrote:
Congratulations, its a great score. Can you provide some further details about the test itself - tougher word problems youve encountered, combinatorics and probabilities etc, the topics of the RC passages (whatever you can remember)... Thank you!
Thanks! I'm not sure how much I can say about the specifics (they make you sign a nondisclosure agreement) and I barely remember any of the verbal section at all -- by that point I was just trying to buckle down and get through it. I can't remember any of the RC topics. I remember very little of the math; I did get one question that was similar to a question I'd already done, so all I had to do was look at the notes I'd already taken for the answer.
The main thing I remember is that I had a really hard time figuring out work-rate problems during my studies; I must have spent weeks studying those damned things -- and I didn't get a single one on the actual test! Luck of the draw, I suppose.
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Display posts from previous: Sort by | {"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": 0.3259255290031433, "perplexity": 3525.952041073119}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-36/segments/1471982295966.49/warc/CC-MAIN-20160823195815-00070-ip-10-153-172-175.ec2.internal.warc.gz"} |
https://export.arxiv.org/abs/2001.02006 | cond-mat.str-el
(what is this?)
# Title: Correlations and incipient antiferromagnetic order within the linear Mn chains of metallic Ti$_4$MnBi$_2$
Abstract: We report measurements on Ti$_4$MnBi$_2$, where a crystal structure involving linear chains of Mn ions suggests one-dimensional magnetic character. The electrical resistivity is metallic, consistent with the results of electronic structure calculations that find a robust Fermi surface albeit with moderate electronic correlations. Curie-Weiss fit to the magnetic susceptibility finds that the Mn moments are in the low-spin $S = 1/2$ configuration. Neutron diffraction measurements detect weak antiferromagnetic order within the Mn chains, with further evidence for the small staggered moment coming from the entropy associated with the ordering peak in the specific heat as well as from the results of spin-polarized electronic structure calculations. The antiferromagnetic moments are apparently associated with the $d_{x^{2}-y^{2}}$ and $d_{xy}$ orbitals of Mn while the remaining Mn orbitals are delocalized. Strong quantum fluctuations, possibly related to an electronic instability that forms the Mn moment or to the one-dimensional character of Ti$_4$MnBi$_2$, nearly overcome magnetic order.
Comments: 13 pages and 7 figures Subjects: Strongly Correlated Electrons (cond-mat.str-el) Journal reference: Phys. Rev. B 102, 014406 (2020) DOI: 10.1103/PhysRevB.102.014406 Cite as: arXiv:2001.02006 [cond-mat.str-el] (or arXiv:2001.02006v1 [cond-mat.str-el] for this version)
## Submission history
From: Abhishek Pandey [view email]
[v1] Tue, 7 Jan 2020 12:52:27 GMT (3566kb)
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https://hal-centralesupelec.archives-ouvertes.fr/hal-01572142 | # Robust Calibration of Radio Interferometers in Non-Gaussian Environment
Abstract : The development of new phased array systems in radio astronomy, as the low frequency array (LOFAR) and the square kilometre array (SKA), formed of a large number of small and flexible elementary antennas, has led to significant challenges. Among them, model calibration is a crucial step in order to provide accurate and thus meaningful images and requires the estimation of all the perturbation effects introduced along the signal propagation path, for a specific source direction and antenna position. Usually, it is common to perform model calibration using the a priori knowledge regarding a small number of known strong calibrator sources but under the assumption of Gaussianity of the noise. Nevertheless, observations in the context of radio astronomy are known to be affected by the presence of outliers which are due to several causes, e.g., weak non-calibrator sources or man made radio frequency interferences. Consequently, the classical Gaussian noise assumption is violated leading to severe degradation in performances. In order to take into account the outlier effects, we assume that the noise follows a spherically invariant random distribution. Based on this modeling, a robust calibration algorithm is presented in this paper. More precisely, this new scheme is based on the design of an iterative relaxed concentrated maximum likelihood estimation procedure which allows to obtain closed-form expressions for the unknown parameters with a reasonable computational cost. This is of importance as the number of estimated parameters depends on the number of antenna elements, which is large for the new generation of radio interferometers. Numerical simulations reveal that the proposed algorithm outperforms the state-of-the-art calibration techniques.
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Journal articles
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Cited literature [69 references]
https://hal-centralesupelec.archives-ouvertes.fr/hal-01572142
Contributor : Remy Boyer <>
Submitted on : Friday, August 4, 2017 - 5:39:15 PM
Last modification on : Saturday, May 1, 2021 - 3:47:27 AM
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### Citation
Virginie Ollier, Mohammed Nabil El Korso, Rémy Boyer, Pascal Larzabal, Marius Pesavento. Robust Calibration of Radio Interferometers in Non-Gaussian Environment. IEEE Transactions on Signal Processing, Institute of Electrical and Electronics Engineers, 2017, 65 (21), pp.5649-5660. ⟨10.1109/TSP.2017.2733496⟩. ⟨hal-01572142⟩
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