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https://www.physicsforums.com/threads/expressing-equivalent-impedance.691989/	# Expressing equivalent impedance 1. May 16, 2013 ### Color_of_Cyan 1. The problem statement, all variables and given/known data Determine the equivalent impedance of the following network: [Broken] (My bad, the element at the far right where the j2ohm is was supposed to be an inductor, fixed now. The numbers are correct too.) 2. Relevant equations R(series) = R1 + R2? R(parallel) = [ (1/R1) + (1/R2) ]-1 except with impedance now 3. The attempt at a solution Not sure if I can do algebra with the imaginary components... but the right branch would simplify to (4Ω - j4Ω/3) Then this is in parallel with j4Ω, simplifying to (12Ω + j4Ω + j12Ω)/(j4Ω(12Ω - j4Ω)). All I want to know, is if this is really the right track? Can I really algebraically manipulate the imaginary components like this (maybe I Just need to review algebra now)? Any other tips would be helpful, thanks. Last edited by a moderator: May 6, 2017 2. May 16, 2013 ### Simon Bridge ... just treat then as you would a variable - except that you have extra simplifications: j.j=-1 and 1/j = -j. At the end you want to express it as z=a+jb - which may mean rationalizing the denominator. But you are basically on the right track. 3. May 16, 2013 ### SammyS Staff Emeritus Yes, this should work fine. Is that element at the far right a resistor or an inductor ? I see that you have now revised the image to show it as an inductor. #### Attached Files: • ###### eetest2prob3.jpg File size: 10.1 KB Views: 67 Last edited: May 16, 2013 4. May 16, 2013 ### Color_of_Cyan Thank you. Although, what happens when you end up having to multiply (Ω * jΩ)? Or you shouldn't have to do this? Because acting on what you said, I ended up getting (continued from the work in my last post): (4-j3Ω)/(12Ω-j4Ω) all to the -1 power then 12Ω - j4Ω/4 - j3Ω and rationalizing the denominator I get (12Ω - j4Ω)(4 - j3Ω) which is where I ask the above question. Zeq would then be [ (12Ω - j4Ω)((4 - j3Ω) ] + 2Ω 5. May 16, 2013 ### Simon Bridge $\omega \cdot j\omega = j\omega^2$ ... pretty much as you'd expect. ... but I think you need to distinguish between the frequency symbol $\omega$ and the symbol for the unit $\Omega$ - or it will get confusing. i.e. A capacitor with impedance "$^-j4\Omega$" - makes sense if the cap-omega on the diagram is intended to be the unit - but your last post gives me pause to wonder if it's a mistake - like the inductor on the far right given a resistor symbol. You can use w or ω for frequency if you don't want to use LaTeX. 6. May 16, 2013 ### Color_of_Cyan No, not a mistake, the "-j4Ω" is correctly there. The only incorrect thing about the problem statement was the resistor in place of an inductor Sammy pointed out there. This might sound dumb but can I ask, where you are bringing in frequency / the lower-case omega in to this from though? I meant this was going to mean getting "Ohms squared" in the equation, unless I put down the complex numbers / units wrong somehow for simplifying the impedances. And doing what you say again and actually continuing trying to simplify & foiling from my last post would give: Zeq = (48Ω -j36Ω2 -j16Ω + 12Ω2) + 2Ω so Zeq = 50Ω -j36Ω2 -j16Ω + 12Ω2 Seems all algebra now, but is there any way to ideally simplify this more? 7. May 16, 2013 ### Staff: Mentor There should be no squared Ohms in the results; everything should shake out to Ohms. Perhaps something has gone awry during your simplification steps (perhaps the rationalization operations?). Frequency doesn't enter into the problem as given since all of the impedances are specified as constants. 8. May 16, 2013 ### Simon Bridge For a reactive component, the impedance depends on the frequency. Inductive impedance is directly proportional to the frequency and capacitative impedance is inversely proportional to frequency. See the contradiction? From your last reply - the $\Omega$ are units and should not appear as a variable in your equations. Please spell out how you are using the $\Omega$ symbol when you next reply. 9. May 18, 2013 ### Color_of_Cyan Okay, I made some mistakes in the math before here so I'm going to start over again here: The right-most elements are in parallel so [ -1/4Ωj + 1/2Ωj ]-1 = ( 1/4Ωj )2 = 4Ωj , which would then be in series with 4Ω so combined would be 4Ω + 4Ωj so at this point would look like....: [Broken] (and I'll leave the "resistor" there this time, just for reference) So trying to combine the right two parallels again: = [ 1/(4Ω+4Ωj) + 1/4Ωj ]-1 = [ (4Ωj+4Ωj+4Ω)/4Ωj(4Ω+4Ωj) ]-1 = [ (8Ωj+4Ω)/4Ωj(4Ω+4Ωj) ]-1 = 4Ωj(4Ω+4Ωj)/(8Ωj+4Ω) But how do I simplify this now? I would want to factor 4Ωj from the top, so I think that would be 4Ωj(2 - j)/4Ωj(4Ω +4Ωj) but not sure. The final answer not simplified from that point would be (or would have been) =[ 4Ωj(4Ω+4Ωj)/(8Ωj+4Ω) ] + 2Ω Last edited by a moderator: May 6, 2017 10. May 18, 2013 ### Staff: Mentor The idea is to clear the imaginary term from the denominator, yielding a canonical "a + jb" form of result. Look up: rationalization of complex numbers. 11. May 18, 2013 ### Color_of_Cyan Ugh, so you multiply by the OPPOSITE conjugate.... thanks! That would mean simplify: [ 4Ωj(4Ω+4Ωj)/(8Ωj+4Ω) ] (4Ω - 8Ωj)/(4Ω - 8Ωj) but would that mean I get Ω2 again if I want to really foil say ie the denominator (8Ωj + 4Ω)(-8Ωj + 4Ω) when I multiply the 4Ω by 4Ω? Or was I just supposed to forget about even mentioning the "Ω" here and just worry about the "j"? 12. May 18, 2013 ### Staff: Mentor If you carry out the multiplications and divisions diligently, the units will reduce to simple Ω's. Consider the units of the factors involved: $$\frac{Ω \cdot Ω}{Ω} \cdot \frac{Ω}{Ω} = \frac{Ω^2}{Ω} \cdot \frac{Ω}{Ω}$$ How does that reduce? You never ignore units when you multiply or divide. The units are associated with the values and the mathematical operations apply to all parts of the expressions. 13. May 19, 2013 ### Color_of_Cyan I see now, and I almost got it, I think. So if you get stuff like j3 it "cancels" down to 1 power of j because j*j = -1? If that happens you have to switch the signs too for each coefficient, right? After simplifying the two parallel resistors above with your help I got: (-64Ω3/80Ω2) + (192Ω3/80Ω2)j =(-4/5)Ω + (12/5)Ωj So by adding the series 2Ω in, my final answer would just be Zeq = (6/5)Ω +(12/5)Ωj But still not sure if it's right. 14. May 19, 2013 ### Staff: Mentor The imaginary term looks okay, but the real term does not. 15. May 19, 2013 ### Color_of_Cyan Ok, I looked over it again and got: (+64Ω3/80Ω2) + (192Ω3/80Ω2)j =(+4/5)Ω + (12/5)Ωj Zeq = (14/5)Ω +(12/5)Ωj 16. May 20, 2013 ### Staff: Mentor Yup, that looks good.	{"extraction_info": {"found_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": 0.859269380569458, "perplexity": 2921.6806282735333}, "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/1508187825464.60/warc/CC-MAIN-20171022203758-20171022223758-00363.warc.gz"}
https://www.cnn.com/2015/11/06/world/two-degrees-question-ice-ages/index.html	Ice ages have come and gone many times in Earth's history. What accounts for the Earth's current rapid warming trend? CNN — As part of CNN’s Two Degrees series on climate change, Jonas Martin of Seattle asked the following question, which travels a long way back in time: “There have been hundreds, if not thousands of ice ages in the past, before humans walked the Earth. How was this extreme climate change possible without humans, and isn’t climate change natural?” This is a popular question that we have received more than a dozen times in various forms. And it’s a fair question: If the Earth has cycled naturally between extremes larger than where we are now and even where we could be heading, why should someone believe that humans are causing the recent warming? The other part of this view point is if the Earth reaches these extremes periodically on its own, why should people worry if it does so because of human actions? The quick answer to the first part of the question, which is likely already known by most who pose the question in the first place, is yes: The Earth’s climate has changed, many times, going back millions of years. These large-scale climate shifts consist of colder ice ages followed by much warmer interglacial periods, characterized by melting ice sheets and higher sea levels. The causes of these shifts were indeed natural. After all, humans didn’t begin to significantly interfere with the climate system through the burning of fossil fuels until the Industrial Revolution. ### Our distant past helps us see our future So how can we say that the recent warming we have seen is caused by humans and not just this natural cycle? By knowing about these natural cycles and studying them, it makes us even more confident that the recent warming we have seen over the past 150 or so years is caused by humans and our emission of greenhouse gasses into the atmosphere. Scientists understand the natural processes behind the previous warm and cold periods that lead to ice ages. They occur in regular patterns called Milankovitch cycles. These cycles occur because the Earth’s orbit around the sun is not constant. The shape of the orbit changes, the tilt of the Earth on its axis changes and the even the direction of the axis changes over time. All of these changes result in varying amounts of energy (i.e. heat) that the planet receives from the sun. That, of course, determines how warm or cold the planet becomes. While these cycles do impact Earth in much the same way as we are seeing now, they happen very, very slowly. These cycles take place on 100,000 year time frames, and the amount of warming we have seen, even though it is “only” about 1.5°F (0.85° C) since 1880, would take many thousands of years to occur if the process were occurring purely naturally. Also, when you plot these orbital cycles out, we should be in a “cooling” phase of the cycle – not warming. The other important fact we learn when looking at these long-term cycles is that greenhouse gases, namely carbon dioxide and methane, move up and down with the global temperature. When greenhouse gases are high, the Earth is warm, when they are low, the Earth is cool. But again, when these changes are occurring naturally, they take thousands to tens of thousands of years to occur, whereas humans have caused this to occur in just over 100 years. In fact, humans have pushed the level of carbon dioxide in our atmosphere to levels not seen in millions of years. The last time levels were this high, sea levels were several meters higher and temperatures were several degrees warmer. So yes, the climate changes naturally, in much the same way it is changing now, but it happens much, much slower. That in turn gives the Earth, and its various life forms, time to adapt. When these changes occur rapidly, you can have mass extinctions.	{"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": 0.8180264234542847, "perplexity": 838.7429809724892}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-49/segments/1637964362969.51/warc/CC-MAIN-20211204094103-20211204124103-00036.warc.gz"}
https://intellipaat.com/community/2363/a-heuristic-overestimation-underestimation	2 views I am confused about the terms overestimation/underestimation. I perfectly get how A* algorithm works, but I am unsure of the effects of having a heuristic that overestimates or underestimate. Is overestimation when you take the square of the direct birdview-line? And why would it make the algorithm incorrect? The same heuristic is used for all nodes. Is underestimation when you take the square root of the direct birdview-line? And why is the algorithm still correct? I can't find an article which explains it nice and clear so I hope someone here has a good description. by (108k points) You are overestimating when the heuristic's estimate is higher than the actual final path cost. You're underestimating when it's lower (you don't have to underestimate, you just have to not overestimate; correct estimates are fine). If your graph's edge costs are all 1, then the examples you give would provide overestimates and underestimates, though the plain coordinate distance also works peachy in Cartesian space. Assuming F(N) = distance from node N to a goal node, G(N) = known the smallest distance from the start to node N, H(N) = estimate of distance from N to the goal node. If H(N) is an overestimate then if the algorithm reaches the goal node by some other path with distance < F(N) then it will never look at node N, because H(N) is then seemingly worse than an already established path. And so even if the _actual_ distance via N is optimal this path will then be overlooked. Which is incorrect. If, on the other hand, H(N) is an underestimate then if the algorithm reaches the goal node by some path with distance D, and the actual distance via N is less than D, then it is guaranteed that F(N) will also, be less than D, so the algorithm will continue to look at N (and perhaps find a path shorter than D). Overestimating doesn't exactly make the algorithm "incorrect"; what it means is that you no longer have an admissible heuristic, which is a condition for A* to be guaranteed to produce optimal behavior. With an inadmissible heuristic, the algorithm can wind up doing tons of superfluous work examining paths that it should be ignoring, and possibly finding suboptimal paths because of exploring those. Whether that actually occurs depends on your problem space. It happens because the path cost is 'out of joint' with the estimated cost, which essentially gives the algorithm messed up ideas about which paths are better than others.	{"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": 0.9243650436401367, "perplexity": 809.2474739218387}, "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/1659882570921.9/warc/CC-MAIN-20220809094531-20220809124531-00698.warc.gz"}
https://www.statistics-lab.com/category/%E8%B4%9D%E5%8F%B6%E6%96%AF%E5%88%86%E6%9E%90%E4%BB%A3%E5%86%99/	## 统计代写\|贝叶斯分析代写Bayesian Analysis代考\|STAT4102 statistics-lab™ 为您的留学生涯保驾护航在代写贝叶斯分析Bayesian Analysis方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写贝叶斯分析Bayesian Analysis代写方面经验极为丰富，各种代写贝叶斯分析Bayesian Analysis相关的作业也就用不着说。 • Statistical Inference 统计推断 • Statistical Computing 统计计算 • (Generalized) Linear Models 广义线性模型 • Statistical Machine Learning 统计机器学习 • Longitudinal Data Analysis 纵向数据分析 • Foundations of Data Science 数据科学基础 ## 统计代写\|贝叶斯分析代写Bayesian Analysis代考\| Do Not Forget the Importance of the Variance in the TNormal Distribution The variance captures our uncertainty about the weighted function. Because the TNormal for ranked nodes is always in the range $[0,1]$ any variance above $0.5$ would be considered very high (you should try it out on a simple weighted mean example). You may need to experiment with the variance to get it just right. In each of the previous examples the variance was a constant, but in many situations the variance will be dependent on the parents. For example, consider the $\mathrm{BN}$ in Figure $9.40$ that is clearly based on a definitional idiom. In this case system quality is defined in terms of the quality of two subsystems $S 1$ and $S 2$. It seems reasonable to assume all nodes are ranked and that the NPT for System quality should be a TNormal whose mean is a weighted mean of the parents. Assuming that the weights of $S 1$ and $S 2$ are equal we therefore define the mean of the TNormal as wmean $(S 1, S 2)$. However, it also seems reasonable to assume that the variance depends on the difference between the two subsystem qualities. Consider, for example these two scenarios for subsystems $S 1$ and $S 2$ : 1. Both $S 1$ and $S 2$ have “medium” quality. 2. $S 1$ quality is “very high,” while $S 2$ quality is “very low.” If the variance in the TNormal expression is fixed at, say $0.1$, then the System Quality in both scenarios 1 and 2 will be the same-as is shown in Figure 9.41(a) and (b). Specifically, the system quality in both cases is medium but with a lot of uncertainty. However, it seems logical to assume that there should be less uncertainty in scenario 1 (when both subsystems have the same, medium, quality) than in scenario 2 (when both subsystems have very different levels of quality). To achieve the required result we therefore have to ensure that the variance in the TNormal expression is a function of the difference in subsystem qualities. Setting the variance as abs(S1-S2)/5 produces the required result as shown in Figure 9.41(c) and (d). The use of a variable variance also enables us to easily implement the measurement idiom in the case where all the nodes of the idiom are ranked. This is explained in Box 9.12. The special case of indicator nodes is shown in Box 9.13. ## 统计代写\|贝叶斯分析代写Bayesian Analysis代考\|Elicitation Protocols and Cognitive Biases We are aiming to build a scientific model, so open, factual, and honest discussion of the risks, our beliefs (i.e., theories) about how they interrelate, and what the probabilities are is of the utmost importance. The elicitor (the modeler/risk analyst) and the elicitee (the subject matter expert) must be mutually respectful of each other’s professionalism, skills, and objectives. Attributes of a good elicitation protocol involve elicitors making an effort to understand subject matter sufficiently to probe and challenge discussion in order to allow experts to sharpen and refine thinking. Similarly, more accurate probabilities are elicited when people are asked for reasons for them, but the BN structure supplies some or all of this, thus making this easier than when asking for probabilities alone. Without these prerequisites the elicitation exercise will be futile. Some practical advice on how to elicit numbers from experts is provided in O’Hagan et al (2006). Box $9.14$ provides some examples of what has been used, based primarily on Spetzler and von Holstein 1975 (also known as the Stanford Elicitation Prototcol). There is plenty of advice on how not to perform elicitation from the field of cognitive psychology as pioneered by Kahneman and colleagues (1982). A summary (by no means exhaustive) of the well-known biases is listed next and we recommend that these be presented and discussed with experts as part of any pre-elicitation training: • Ambiguity effect-Avoiding options for which missing information makes the probability seem unknown. • Attentional bias-Neglecting relevant data when making judgments of a correlation or association. • Availability heuristic-Estimating what is more likely by what is more available in memory, which is biased toward vivid, unusual, or emotionally charged examples. • Base rate neglect-Failing to take account of the prior probability. This was at the heart of the common fallacious reasoning in the Harvard medical study described in Chapter 2 . It is the most common reason for people to feel that the results of Bayesian inference are nonintuitive. • Bandwagon effect – Believing things because many other people do (or believe) the same. Related to groupthink and herd behavior. • Confirmation bias-Searching for or interpreting information in a way that confirms one’s preconceptions. • Déformation professionnelle-Ignoring any broader point of view and seeing the situation through the lens of one’s own professional norms. ## 统计代写\|贝叶斯分析代写贝叶斯分析代考\|不要忘记方差在t正态分布中的重要性 1. $S 1$和$S 2$都是中等质量。 2. $S 1$质量“非常高”，而$S 2$质量“非常低”。如果TNormal表达式中的方差固定在，比如$0.1$，那么在场景1和场景2中的系统质量将是相同的，如图9.41(a)和(b)所示。具体地说，在这两种情况下，系统质量是中等的，但有很大的不确定性然而，假设场景1(当两个子系统具有相同的中等质量时)的不确定性应该比场景2(当两个子系统具有非常不同的质量水平时)的不确定性更低似乎是合乎逻辑的。因此，为了达到所需的结果，我们必须确保TNormal表达式中的方差是子系统质量差异的函数。将方差设为abs(S1-S2)/5会产生如图9.41(c)和(d)所示的结果变量方差的使用还使我们能够轻松地实现度量习惯用法，在这种情况下，习惯用法的所有节点都是排序的。这将在框9.12中解释。指示节点的特殊情况在框9.13中显示统计代写\|贝叶斯分析代写贝叶斯分析代考\|启发式协议和认知偏差我们的目标是建立一个科学的模型，所以公开、实事求是和诚实地讨论风险，我们的信念(即理论)是如何相互联系的，以及概率是什么是最重要的。激发者(建模师/风险分析师)和被激发者(主题专家)必须相互尊重对方的专业知识、技能和目标。一个好的诱导协议的属性包括诱导者努力充分理解主题，以探索和挑战讨论，以便让专家们提高和精炼思维。类似地，当人们被问及其原因时，会引出更准确的概率，但BN结构提供了部分或全部这些，因此比单独询问概率更容易。没有这些先决条件，启发练习将是徒劳的O’Hagan等人(2006)就如何从专家那里引出数字提供了一些实用的建议。Box $9.14$提供了一些已经使用的例子，主要基于Spetzler和von Holstein 1975(也称为斯坦福启发协议)。Kahneman和他的同事(1982)在认知心理学领域率先提出了很多关于如何不进行诱导的建议。下面是对众所周知的偏见的总结(并非详尽无遗)，我们建议将这些偏见作为任何预诱导培训的一部分与专家讨论: • 歧义效应—避免信息缺失使概率看起来未知的选项。注意偏差-在对相关或关联做出判断时忽略相关数据。 • 可用性启发式-通过记忆中更多的可用性来估计什么更有可能发生，这偏向于生动的、不寻常的或情绪化的例子。 • 基准率忽略-未考虑先验概率。这就是第二章中描述的哈佛医学研究中常见谬误推理的核心。人们觉得贝叶斯推断的结果是非直观的，这是最常见的原因。 • 从众效应-相信一些事情，因为许多其他人也这么做(或相信)。与群体思维和从众行为有关。 • 确认偏误——以一种证实某人先入为主的方式搜索或解释信息。 • Déformation professionnelle-忽略任何更广泛的观点，通过自己的专业规范来看待情况 ## 有限元方法代写 tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。 ## MATLAB代写 MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。 ## 统计代写\|贝叶斯分析代写Bayesian Analysis代考\|STATS3023 statistics-lab™ 为您的留学生涯保驾护航在代写贝叶斯分析Bayesian Analysis方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写贝叶斯分析Bayesian Analysis代写方面经验极为丰富，各种代写贝叶斯分析Bayesian Analysis相关的作业也就用不着说。 • Statistical Inference 统计推断 • Statistical Computing 统计计算 • (Generalized) Linear Models 广义线性模型 • Statistical Machine Learning 统计机器学习 • Longitudinal Data Analysis 纵向数据分析 • Foundations of Data Science 数据科学基础 ## 统计代写\|贝叶斯分析代写Bayesian Analysis代考\|Hints and Tips When Working with Ranked Nodes and NPTs We have found that the set of weighted functions (i.e., WMEAN, WMIN, WMAX, and MIXMINMAX) is sufficient to generate almost any ranked node NPT in practice where the ranked node’s parents are all ranked. In cases where the weighted function does not exactly capture the requirements for the node’s $\mathrm{NPL}^{\prime}$ it is usually possible to get to what you want by manually tweaking the NPT that is generated by a weighted function. For example, Figure $9.37$ shows a part of the table that is automatically generated for the node $Y$ as specified in Figure 9.31. You will note that the probability of $Y$ being “very high” when both parents are “very low” is very close to 0 but not equal to 0 . If you really want this probability to be 0 then you can simply enter 0 manually into that cell. ## 统计代写\|贝叶斯分析代写Bayesian Analysis代考\|Exploit the Fact That a Ranked Node Parent Has an Underlying Numerical Scale In many real-world models you will find that nodes that are not ranked nodes will have one or more parents that are ranked. In such situations you can exploit the underlying numerical property of the ranked node parent to define the NPT of the child node. For example, it makes sense to extend the model of Figure $9.35$ by adding a Boolean node called Release Product? which is true when the product has been sufficiently well tested to be released and false otherwise. The extended model is shown in Figure 9.38. We could as usual define the NPT for the new Boolean node manually (it has 10 entries). But it makes much more sense and is far simpler to exploit the fact that the node $Y$ has an underlying numerical value between 0 and 1. Since we have a 5-point scale we know that if $Y$ is above $0.5$ then the quality is at least “medium.” If the value is $0.7$ then the quality is in the middle of the “high” range. So, suppose that previous experience suggests that testing effectiveness needs to be “high” in order for the product to be released without too many problems. Then we can simply define the NPT of the node Release product? by the expression: if $(\mathrm{Y}>0.7$, “True”, “False”). The effect of running the resulting model with some observations is shown in Figure 9.39. . ## 统计代写\|贝叶斯分析代写贝叶斯分析代考\|利用分级节点父节点具有底层数值尺度的事实 if $(\mathrm{Y}>0.7$， “True”， “False”)。运行结果模型和一些观察结果的效果如图9.39所示 ## 有限元方法代写 tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。 ## MATLAB代写 MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。 ## 统计代写\|贝叶斯分析代写Bayesian Analysis代考\|MAST90125 statistics-lab™ 为您的留学生涯保驾护航在代写贝叶斯分析Bayesian Analysis方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写贝叶斯分析Bayesian Analysis代写方面经验极为丰富，各种代写贝叶斯分析Bayesian Analysis相关的作业也就用不着说。 • Statistical Inference 统计推断 • Statistical Computing 统计计算 • (Generalized) Linear Models 广义线性模型 • Statistical Machine Learning 统计机器学习 • Longitudinal Data Analysis 纵向数据分析 • Foundations of Data Science 数据科学基础 ## 统计代写\|贝叶斯分析代写Bayesian Analysis代考\|Weighted Averages A common simple approach to quantitative risk assessment is to use a weighted average score to combine risks and produce an overall “risk score” as shown in Table 9.7. This is purely arithmetical and is easily implemented in a spreadsheet, such as Excel. Here we have identified three risks to a project: Risk $\mathrm{A}$, Risk $\mathrm{B}$ and Risk $\mathrm{C}$ with respective probabilities $10 \%, 20 \%$ and $80 \%$ and “weights” 3,2 , and 1 . This produces an overall weighted average risk score of $25 \%$. As we saw in Chapter 3, this is the “risk register” approach that can be viewed as the extension of the simple approach to risk-assessment in which we define risk as probability times impact. Specifically, the impacts are viewed as relative “weights.” For all of the reasons discussed in Chapter 3 we do not recommend this approach to risk assessment, but there may be many reasons why we would want to incorporate weighted averages into a BN. For example, we might wish to use a weighted average as a score to determine which new car to buy based on criteria such as price, quality, and delivery time. Although the weighted average is deterministic (and therefore can be computed in Excel) the values for the criteria could be based on a range of uncertain factors and relationships that require a BN model in which the weighted average is just a component. Fortunately, it is possible to replicate weighted averages (using the same example probabilities and weights as Table 9.7) in a BN as shown in Figure 9.23. Each of the risk factors is represented by a Boolean node whose “probability” is simply specified as the “True” value in the NPTso, for example, since Risk A has probability $10 \%$ we set its NPT as “True” $=10 \%$. The Risk Score node is also Boolean but it makes sense to replace the labels “False” and “True” with “Low” and “High,” respectively. The key to ensure we can replicate the weighted average calculation is to introduce the labelled node Weights whose states correspond to the three risk node weights. The normalised weights are used in the NPT for this node. ## 统计代写\|贝叶斯分析代写Bayesian Analysis代考\|Alternative Weighted Functions The weighted mean is not the only natural function that could be used as the mean of the TNormal ranked node NPTs. Suppose, for example, that in Figure $9.26$ we replace the node Quality of Testing Process with the node Testing Effort as shown in Figure 9.35. In this case we elicit the following information: • When $X_1$ and $X_2$ are both “very high” the distribution of $Y$ is heavily skewed toward “very high.” • When $X_1$ and $X_2$ are both “very low” the distribution of $Y$ is heavily skewed toward “very low.” • When $X_1$ is very low and $X_2$ is “very high” the distribution of $Y$ is centered toward “very low.” • When $X_1$ is very high and $X_2$ is “very low” the distribution of $Y$ is centered toward “low.” Intuitively, the expert is saying here that, for testing to be effective, you need not just to have good people but also to put in the effort. If either the people or the effort is insufficient, then the result will be poor. However, really good people can compensate to a small extent for lack of effort. A simple weighted mean for $Y$ will not produce an NPT to satisfy these elicited requirements (you can try it out by putting in different weights; you will never be able to satisfy both of the last two elicited constraints). Informally, $Y$ ‘s mean is something like the minimum of the parent values, but with a small weighting in favor of $X_1$. The necessary function, which we call the weighted min function (WMIN), is what is needed in this case. The general form of this function (together with analogous WMAX and the mixture function MIXMINMAX) is shown in Box 9.11. You need not know the details because the function is built into AgenaRisk, so it is sufficient to know what the effect of the function is with different values. ## 统计代写\|贝叶斯分析代写贝叶斯分析代考\|备选加权函数 • 当$X_1$和$X_2$都是“非常高”时，$Y$的分布严重偏向于“非常高”。当$X_1$和$X_2$都是“非常低”时，$Y$的分布严重偏向于“非常低”。 • 当$X_1$非常低，$X_2$非常高时，$Y$的分布以“非常低”为中心。 • 当$X_1$非常高，$X_2$是“非常低”时，$Y$的分布以“低”为中心。 ## 有限元方法代写 tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。 ## MATLAB代写 MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。 ## 统计代写\|贝叶斯分析代写Bayesian Analysis代考\|STAT4102 statistics-lab™ 为您的留学生涯保驾护航在代写贝叶斯分析Bayesian Analysis方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写贝叶斯分析Bayesian Analysis代写方面经验极为丰富，各种代写贝叶斯分析Bayesian Analysis相关的作业也就用不着说。 • Statistical Inference 统计推断 • Statistical Computing 统计计算 • (Generalized) Linear Models 广义线性模型 • Statistical Machine Learning 统计机器学习 • Longitudinal Data Analysis 纵向数据分析 • Foundations of Data Science 数据科学基础 ## 统计代写\|贝叶斯分析代写Bayesian Analysis代考\|The Crucial Independence Assumptions Take a look again at the BN model of Figure $7.3$ and the subsequent calculations we used. Using the terminology of Chapter 5 what we have actually done is use some crucial simplifying assumptions in order to avoid having to work out the full joint probability distribution of: (Norman late, Martin late, Martin oversleeps, Train strike) We will write this simply as $(N, M, O, T)$ For example, in calculating the marginal probability of $\operatorname{Martin}$ late $(M)$ we assumed that $M$ was dependent only on Martin oversleeps $(O)$ and Train strike $(T)$. The variable Norman late $(N)$ simply did not appear in the equation because we assume that none of these variables are directly dependent on $N$. Similarly, although $M$ depends on both $O$ and $T$, the variables $O$ and $T$ are independent of each other. These kind of assumptions are called conditional independence assumptions (we will provide a more formal definition of this later). If we were unable to make any such assumptions then the full joint probability distribution of $(N, M, O, T)$ is (by the chain rule of Chapter 5) $$P(N, M, O, T)=P(N \mid M, O, T) P(M \mid O, T) P(O \mid T) P(T)$$ However, because $N$ directly depends only on $T$ the expression $P(N \mid M, O, T)$ is equal to $P(N \mid T)$, and because $O$ is independent of $T$ the expression $P(O \mid T)$ is equal to $P(O)$. Hence, the full joint probability distribution can be simplified as: $$P(N, M, O, T)=P(N \mid T) P(M \mid O, T) P(O) P(T)$$ and this is exactly what we used in the computations. ## 统计代写\|贝叶斯分析代写Bayesian Analysis代考\|Structural Properties of BNs In $\mathrm{BNs}$ the process of determining what evidence will update which node is determined by the conditional dependency structure. The main formal area of guidance for building sensible BN structures therefore requires some understanding of different types of relationships between variables and the different ways these relationships are structured. Generally we are interested in the following problem. Suppose that variable $A$ is linked to both variables $B$ and $C$. There are three different ways the links can be directed as shown in Figure 7.8. Although $B$ and $C$ are not directly linked, under what conditions in each case are $B$ and $C$ independent of $A$ ? Knowing the answer to this question enables us to determine how to construct appropriate links, and it also enables us to formalize the different notions of conditional independence that we introduced informally in Chapter $6 .$ The three cases in Figure $7.8$ are called, respectively, serial, diverging, and converging connections. We next discuss each in turn. Consider the example of a serial connection as shown in Figure 7.9. Suppose we have some evidence that a signal failure has occurred $(B)$. Then clearly this knowledge increases our belief that the train is delayed $(A)$, which in turn increases our belief that Norman is late $(C)$. Thus, evidence about $B$ is transmitted through $A$ to $C$ as is shown in Figure 7.10. However, now suppose that we know the true status of $A$; for example, suppose we know that the train is delayed. Then this means we have hard evidence for A (see Box $7.5$ for an explanation of what hard and uncertain evidence are and how they differ). ## 统计代写\|贝叶斯分析代写Bayesian Analysis代考\|The Crucial Independence Assumptions $$P(N, M, O, T)=P(N \mid M, O, T) P(M \mid O, T) P(O \mid T) P(T)$$ $$P(N, M, O, T)=P(N \mid T) P(M \mid O, T) P(O) P(T)$$ ## 有限元方法代写 tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。 ## MATLAB代写 MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。 ## 统计代写\|贝叶斯分析代写Bayesian Analysis代考\|MAST90125 statistics-lab™ 为您的留学生涯保驾护航在代写贝叶斯分析Bayesian Analysis方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写贝叶斯分析Bayesian Analysis代写方面经验极为丰富，各种代写贝叶斯分析Bayesian Analysis相关的作业也就用不着说。 • Statistical Inference 统计推断 • Statistical Computing 统计计算 • (Generalized) Linear Models 广义线性模型 • Statistical Machine Learning 统计机器学习 • Longitudinal Data Analysis 纵向数据分析 • Foundations of Data Science 数据科学基础 ## 统计代写\|贝叶斯分析代写Bayesian Analysis代考\|Accounting for Multiple Causes Norman is not the only person whose chances of being late increase when there is a train strike. Martin is also more likely to be late, but Martin depends less on trains than Norman and he is often late simply as a result of oversleeping. These additional factors can be modeled as shown in Figure 7.3. You should add the new nodes and edges using AgenaRisk. We also need the probability tables for each of the nodes Martin oversleeps (Table 7.3) and Martin late (Table 7.4). The table for node Martin late is more complicated than the table for Norman late because Martin late is conditioned on two nodes rather than one. Since each of the parent nodes has two states, true and false (we are still keeping the example as simple as possible), the number of combinations of parent states is four rather than two. If you now run the model and display the probability graphs you should get the marginal probability values shown Figure 7.4(a). In particular, note that the marginal probability that Martin is late is equal to $0.446$ (i.e. $44.6 \%$ ). Box $7.1$ explains the underlying calculations involved in this. But if we know that Norman is late, then the probability that Martin is late increases from the prior $0.446$ to $0.542$ as shown in Figure 7.4(b). Box $7.1$ explains the underlying calculations involved. ## 统计代写\|贝叶斯分析代写Bayesian Analysis代考\|Using Propagation to Make Special When we enter evidence and use it to update the probabilities in the way we have seen so far we call it propagation. In principle we can enter any number of observations anywhere in the BN model and use propagation to update the marginal probabilities of all the unobserved variables. This can yield some exceptionally powerful types of analysis. For example, without showing the computational steps involved, if we first enter the observation that Martin is late we get the revised probabilities shown in Figure 7.5(a). What the model is telling us here is that the most likely explanation for Martin’s lateness is Martin oversleeping; the revised probability of a train strike is still low. However, if we now discover that Norman is also late (Figure 7.5(b)) then Train strike (rather than Martin oversleeps) becomes the most likely explanation for Martin being late. This particular type of (backward) inference is called explaining away (or sometimes called nonmonotonic reasoning). Classical statistical tools alone do not enable this type of reasoning and what-if analysis. In fact, as even the earlier simple example shows, BNs offer the following benefits: • Explicitly model causal factors – It is important to understand that this key benefit is in stark contrast to classical statistics whereby prediction models are normally developed by purely data-driven approaches. For example, the regression models introduced in Chapter 2 use historical data alone to produce equations relating dependent and independent variables. Such approaches not only fail to incorporate expert judgment in scenarios where there is insufficient data, but also fail to accommodate causal explanations. We will explore this further in Chapter $9 .$ • Reason from effect to cause and vice versa-A BN will update the probability distributions for every unknown variable whenever an observation is entered into any node. So entering an observation in an “effect” node will result in back propagation, that is, revised probability distributions for the “cause” nodes and vice versa. Such backward reasoning of uncertainty is not possible in other approaches. • Reduce the burden of parameter acquisition-A BN will require fewer probability values and parameters than a full joint probability model. This modularity and compactness means that elicitation of probabilities is easier and explaining model results is made simpler. ## 统计代写\|贝叶斯分析代写Bayesian Analysis代考\|Using Propagation to Make Special • 显式建模因果因素——重要的是要了解，这一关键优势与经典统计形成鲜明对比，经典统计通常通过纯粹的数据驱动方法开发预测模型。例如，第 2 章介绍的回归模型仅使用历史数据来生成与因变量和自变量相关的方程。这种方法不仅无法在数据不足的情况下纳入专家判断，而且无法适应因果解释。我们将在本章中进一步探讨9. • 从结果到原因的原因，反之亦然 – 每当将观察输入任何节点时，BN 都会更新每个未知变量的概率分布。因此，在“影响”节点中输入观察结果将导致反向传播，即修改“原因”节点的概率分布，反之亦然。这种对不确定性的反向推理在其他方法中是不可能的。 • 减少参数获取的负担——与完整的联合概率模型相比，BN 将需要更少的概率值和参数。这种模块化和紧凑性意味着概率的引出更容易，模型结果的解释也变得更简单。 ## 有限元方法代写 tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。 ## MATLAB代写 MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。 ## 统计代写\|贝叶斯分析代写Bayesian Analysis代考\|STATS3023 statistics-lab™ 为您的留学生涯保驾护航在代写贝叶斯分析Bayesian Analysis方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写贝叶斯分析Bayesian Analysis代写方面经验极为丰富，各种代写贝叶斯分析Bayesian Analysis相关的作业也就用不着说。 • Statistical Inference 统计推断 • Statistical Computing 统计计算 • (Generalized) Linear Models 广义线性模型 • Statistical Machine Learning 统计机器学习 • Longitudinal Data Analysis 纵向数据分析 • Foundations of Data Science 数据科学基础 ## 统计代写\|贝叶斯分析代写Bayesian Analysis代考\|Second-Order Probability Recall the Honest Joe’s and Shady Sam’s example we encountered in Chapter 4 (Box 4.7). In that example we expressed a belief in the chance of the die being fair or not, as a probability, while being aware that the fairness is also expressed as a probability. At first glance this looks very odd indeed since it suggests we are measuring a “probability about a probability.” We call this a second-order probability. If we think of the fairness of the die, and its chance of landing face up on a 6 , as a property of the die and how it is thrown, then we are expressing a degree of belief in the chance of the die having a given value. This is no different from expressing a degree of belief in a child reaching a given height when they mature in the sense that height and chance are both unknown properties of a thing that is of interest to us. Example $6.10$ shows how we might model such second-order probabilities in practice. Let us assume someone has smuggled a die out of either Shady Sam’s or Honest Joe’s, but we do not know which casino it has come from. We wish to determine the source of the die from (a) a prior belief about where the die is from and (b) data gained from rolling the die a number of times. We have two alternative hypotheses we wish to test: Joe (“die comes from Honest Joe’s”) and Sam (“die comes from Shady Sam’s”). The respective prior probabilities for these hypotheses are: $$P(\text { Joe })=0.7 \quad P(\text { Sam })=0.3$$ This is justified by the suspicion that our smuggler may be deterred by the extra personal risk in smuggling a die from Shady Sam’s compared with Honest Joe’s. The data consists of 20 rolls of the die, observing there was one “6” and nineteen “not 6 ” results. So, we need to compute the likelihoods $\mathrm{P}($ Joe I data) and $\mathrm{P}($ Sam I data) and combine these (by Bayes theorem) with our prior beliefs about the hypotheses to get our posterior beliefs. To compute the likelihoods, recall that we believed that a die from Honest Joe’s was fair, with a chance of a 6, $p=1 / 6$, and from Shady Sam’s it was unfair, say, $p=1 / 12$. We can use these assumptions with the Binomial distribution to generate the likelihoods we need for the data, $X$ successes in 20 trials, given each of our hypotheses: $$P(X=x \mid p, 20)=\left(\begin{array}{l} 20 \ x \end{array}\right) p^x(1-p)^{20-x}$$ The results are shown in Table 6.4. Notice that we are expressing beliefs in hypotheses that are equivalent to beliefs about probabilities, in this case $$P(\text { Joe })=P(p=1 / 6)$$ and $$P(\text { Sam })=P(p=1 / 12)$$ From Table $6.4$ we can see that we now should favor the hypothesis that the die was sourced from Shady Sam’s rather than Honest Joe’s. This conclusion reverses our prior assumption (which had favored Honest Joe’s). ## 统计代写\|贝叶斯分析代写Bayesian Analysis代考\|A Very Simple Risk Assessment Problem Since it is important for Norman to arrive on time for work, a number of people (including Norman himself) are interested in the probability that he will be late. Since Norman usually travels to work by train, one of the possible causes for Norman being late is a train strike. Because it is quite natural to reason from cause to effect we examine the relationship between a possible train strike and Norman being late. This relationship is represented by the causal model shown in Figure $7.1$ where the edge connects two nodes representing the variables “Train strike” $(T)$ to “Norman late” $(N)$. It is obviously important that there is an edge between these variables since $T$ and $N$ are not independent (using the language of Chapter 5); common sense dictates that if there is a train strike then, assuming we know Norman travels by train, this will affect his ability to arrive on time. Common sense also determines the direction of the link since train strike causes Norman’s lateness rather than vice versa. To ensure the example is as simple as possible we assume that both variables are discrete, having just two possible states: true and false. Let us assume the following prior probability information: 1. The probability of a train strike is $0.1$ (and therefore the probability of no train strike is 0.9). This information might be based on some subjective judgment given the most recent news or it might be based on the recent frequency of train strikes (i.e. one occurring about every 10 days). So the prior probability distribution for the variable “Train strike” is as shown in Table 7.1. 2. The probability Norman is late given that there is a train strike is $0.8$ (and therefore the probability Norman is not late given that there is a train strike is $0.2$ ). The probability Norman is late given that there is not a train strike is $0.1$ (and therefore the probability Norman is not late given that there is not a train strike is 0.9). So, the (conditional) probability distribution for “Norman late” given “Train strike” is as shown in Table $7.2$. ## 统计代写\|贝叶斯分析代写Bayesian Analysis代考\|Second-Order Probability $$P(\text { Joe })=0.7 \quad P(\text { Sam })=0.3$$ $$P(X=x \mid p, 20)=(20 x) p^x(1-p)^{20-x}$$ $$P(\text { Joe })=P(p=1 / 6)$$ $$P(\mathrm{Sam})=P(p=1 / 12)$$ ## 统计代写\|贝叶斯分析代写Bayesian Analysis代考\|A Very Simple Risk Assessment Problem 1. 火车撞车的概率是0.1（因此没有火车撞击的概率是 0.9）。该信息可能基于给定最新消息的一些主观判断，也可能基于最近的火车罢工频率（即大约每 10 天发生一次）。因此，变量“火车罢工”的先验概率分布如表 7.1 所示。 2. 鉴于火车罢工，诺曼迟到的概率是0.8（因此，鉴于火车罢工，诺曼不迟到的概率是0.2）。鉴于没有火车罢工，诺曼迟到的概率是0.1（因此，鉴于没有火车罢工，诺曼不迟到的概率是 0.9）。因此，给定“火车罢工”的“诺曼晚点”的（条件）概率分布如表所示7.2. ## 有限元方法代写 tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。 ## MATLAB代写 MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。 ## 统计代写\|贝叶斯分析代写Bayesian Analysis代考\|STAT4102 statistics-lab™ 为您的留学生涯保驾护航在代写贝叶斯分析Bayesian Analysis方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写贝叶斯分析Bayesian Analysis代写方面经验极为丰富，各种代写贝叶斯分析Bayesian Analysis相关的作业也就用不着说。 • Statistical Inference 统计推断 • Statistical Computing 统计计算 • (Generalized) Linear Models 广义线性模型 • Statistical Machine Learning 统计机器学习 • Longitudinal Data Analysis 纵向数据分析 • Foundations of Data Science 数据科学基础 ## 统计代写\|贝叶斯分析代写Bayesian Analysis代考\|Probability Notation Where There Are Different Consider the experiment of rolling two fair dice. There are many different outcomes of interest for this experiment including the following: • The sum of the two dice rolled (let’s call this outcome $X$ ). The highest number die rolled (let’s call this outcome $Y$ ). These two different outcomes of interest have different sets of elementary events. Outcome $X$ has eleven elementary events: 2,3,4,5,6,7,8,9,10, 11, 12 . • Outcome $Y$ has six elementary events: 1, 2, 3, 4, 5, 6 . If we are not careful about specifying the particular outcome of interest for the experiment, then there is the potential to introduce genuine ambiguity when calculating probabilities. For example, consider the elementary event ” 2 .” What is the probability of observing this event for this experiment? In other words what is $P(2)$ ? The answer depends on whether we are considering outcome $X$ or outcome $Y$ : • For outcome $X$, the probability $P(2)$ is $1 / 36$ because there are 36 different ways to roll two dice and only one of these, the roll $(1,1)$, results in the sum of the dice being 2 . • For outcome $Y$, the probability $P(2)$ is $1 / 12$ because of the 36 different ways to roll two dice there are three ways, the rolls $(1,2),(2,1)$ and $(2,2)$, that result in the highest number rolled being 2 . Because of this ambiguity it is common practice, when there are different outcomes of interest for the same experiment to include some notation that identifies the particular outcome of interest when writing down probabilities. Typically, we would write $P(X=2)$ or $P(Y=2)$ instead of just $P(2)$. The notation extends to events that comprise more than one elementary event. For example, consider the event $E$ defined as “greater than 3”: • For outcome $X$, the event is $E$ is equal to ${4,5,6,7,8,9,10,11,12}$. • For outcome $Y$, the event is $E$ is equal to ${4,5,6}$. We calculate the probabilities as • For event $X, P(E)=11 / 12$. • For event $Y, P(E)=3 / 4$. Typically we would write $P(X=E)$ or $P(X \geq 3)$ for the former and $P(Y=E)$ or $P(Y \geq 3)$ for the latter. In this example the outcomes $X$ and $Y$ can be considered as variables whose possible values are their respective set of elementary events. In general, if there is not an obviously unique outcome of interest for an experiment, then we need to specify each outcome of interest as a named variable and include this name in any relevant probability statement. ## 统计代写\|贝叶斯分析代写Bayesian Analysis代考\|Probability Distributions Consider the experiment of selecting a contractor to complete a piece of work for you. We are interested in the outcome “quality of the contractor.” Since, as discussed in Box 5.5, this is just one of many possible outcomes of interest for this experiment (others might be price of contractor, experience of contractor, etc.) it is safest to associate a variable name, say $Q$, with the outcome “quality of the contractor.” Let us assume that the set of elementary events for $Q$ is {very poôr, poōr, averāge, good, very good}. On the basis of our previous experience with contractors, or purely based on subjective judgment, we might assign the probabilities to these elementary events for $Q$ as shown in the table of Figure 5.2(a). Since the numbers are all between 0 and 1 , and since they sum to 1 , this assignment is a valid probability measure for $Q$ (i.e., for the experiment with outcome $Q$ ) because it satisfies the axioms. A table like the one in Figure 5.2(a), or equivalent graphical representations like the ones in Figure 5.2(b) and Figure 5.2(c), is called a probability distribution. In general, for experiments with a discrete set of elementary events:There is a very common but somewhat unfortunate notation for probability distributions. The probability distribution for an outcome such as $Q$ of an experiment is often written in shorthand as simply: $P(Q)$. If there was an event referred to as $Q$ then the expression $P(Q)$ is ambiguous since it refers to two very different concepts. Generally it will be clear from the context whether $P(Q)$ refers to the probability distribution of an outcome $Q$ or whether it refers to the probability of an event $Q$. ## 统计代写\|贝叶斯分析代写Bayesian Analysis代考\|Probability Notation Where There Are Different • 掷出的两个骰子的总和（我们称这个结果为X）。掷出的最高点数（我们称这个结果为是的）。这两种不同的兴趣结果具有不同的基本事件集。结果X有十一个基本事件：2,3,4,5,6,7,8,9,10,11,12。 • 结果是的有六个基本事件：1、2、3、4、5、6。如果我们在指定实验感兴趣的特定结果时不小心，那么在计算概率时就有可能引入真正的歧义。 • 对于结果X, 概率磷(2)是1/36因为掷两个骰子有 36 种不同的方法，而其中只有一种，掷骰子(1,1)，结果骰子的总和为 2 。 • 对于结果是的, 概率磷(2)是1/12因为掷两个骰子有 36 种不同的方式，所以有三种方式，掷骰子(1,2),(2,1)和(2,2)，这导致滚动的最高数字为 2 。由于这种模糊性，通常的做法是，当同一实验有不同的感兴趣结果时，在写下概率时包含一些标识感兴趣的特定结果的符号。通常，我们会写磷(X=2)或者磷(是的=2)而不仅仅是磷(2). • 对于结果X, 事件是和等于4,5,6,7,8,9,10,11,12. • 对于结果是的, 事件是和等于4,5,6. 我们计算概率为 • 活动X,磷(和)=11/12. • 活动是的,磷(和)=3/4. 通常我们会写磷(X=和)或者磷(X≥3)对于前者和磷(是的=和)或者磷(是的≥3)对于后者。在这个例子中，结果X和是的可以被认为是变量，其可能值是它们各自的基本事件集。一般来说，如果一个实验没有明显独特的感兴趣的结果，那么我们需要将每个感兴趣的结果指定为一个命名变量，并将这个名称包含在任何相关的概率陈述中。 ## 有限元方法代写 tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。 ## MATLAB代写 MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。 ## 统计代写\|贝叶斯分析代写Bayesian Analysis代考\|STATS3023 statistics-lab™ 为您的留学生涯保驾护航在代写贝叶斯分析Bayesian Analysis方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写贝叶斯分析Bayesian Analysis代写方面经验极为丰富，各种代写贝叶斯分析Bayesian Analysis相关的作业也就用不着说。 • Statistical Inference 统计推断 • Statistical Computing 统计计算 • (Generalized) Linear Models 广义线性模型 • Statistical Machine Learning 统计机器学习 • Longitudinal Data Analysis 纵向数据分析 • Foundations of Data Science 数据科学基础 ## 统计代写\|贝叶斯分析代写Bayesian Analysis代考\|When Frequentist and Subjective Approaches Merge Consider the following two statements: 1. There is a $50.9 \%$ chance that a new born baby in the United Kingdom is a girl. 2. There is a $5 \%$ chance of the Spurs winning the FA Cup next year. On the surface there seems to be no doubt that statement 1 is explained by a frequentist argument: Over the last 100 years $50.9 \%$ of all births recorded in the United Kingdom have been girls. There is also no doubt that statement 2 has no such frequentist explanation (and hence must be subjective) since there is only one FA Cup next year, and we cannot somehow play the tournament many times in the same year and count the number of occasions on which the Spurs win. But if we dig a little deeper here, things get rather murky. The $50.9 \%$ figure in statement 1 is actually based on many years of data that may disguise crucial trend information. Suppose we discover that the percentage of girls born is increasing; say a hundred years ago $48.5 \%$ of babies were girls compared with $51.2 \%$ last year. Then surely the probability of a randomly selected newborn being a girl now is higher than $50.9 \%$ (and higher than $51.2 \%$ if the figures have been steadily increasing). And what exactly do we mean by a “randomly” selected baby. Surely, what we are interested in are specific babies such as “the next baby born to Mrs. Roberts of 213 White Hart Lane, London N17.” In that case the frequency data may need to be adjusted to take account of specific factors relevant to Mrs. Roberts. Both the general trend adjustments and the case specific adjustments here clearly require the subjective judgment of relevant experts. But that means, according to the frequentists, that their own approach is no longer valid since, as we saw earlier: • The measure cannot be validated • Different experts will give different subjective measures ## 统计代写\|贝叶斯分析代写Bayesian Analysis代考\|The Basics of Probability In discussing the difference between the frequentist and subjective approaches to measuring uncertainty, we were careful in Chapter 4 not to mention the word probability. That is because we want to define probability in such a way that it makes sense for whatever reasonable approach to measuring uncertainty we choose, be it frequentist, subjective, or even an approach that nobody has yet thought of. To do this in Section $5.2$ we describe some properties (called axioms) that any reasonable measure of uncertainty should satisfy; then we define probability as any measure that satisfies those properties. The nice thing about this way of defining probability is that not only does it avoid the problem of vagueness, but it also means that we can have more than one measure of probability. In particular, we will see that both the frequentist and subjective approaches satisfy the axioms, and hence both are valid ways of defining probability. In Section $5.3$ we introduce the crucial notion of probability distributions. In Section $5.4$ we use the axioms to define the crucial issue of independence of events. An especially important probability distribution-the Binomial distribution-which is based on the idea of independent events, is described in Section 5.5. Finally in Section $5.6$ we will apply the lessons learned in the chapter to solve some of the problems we set in Chapter 2 and debunk a number of other probability fallacies. ## 统计代写\|贝叶斯分析代写Bayesian Analysis代考\|When Frequentist and Subjective Approaches Merge 1. 有一个50.9%在英国刚出生的婴儿是女孩的可能性。 2. 有一个5%马刺明年有机会赢得足总杯。从表面上看，似乎毫无疑问，陈述 1 可以通过一个常客论点来解释：过去 100 年50.9%在英国记录的所有新生儿中都是女孩。 • 无法验证该措施 • 不同的专家会给出不同的主观衡量标准 ## 有限元方法代写 tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。 ## MATLAB代写 MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。 ## 统计代写\|贝叶斯分析代写Bayesian Analysis代考\|MAST90125 statistics-lab™ 为您的留学生涯保驾护航在代写贝叶斯分析Bayesian Analysis方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写贝叶斯分析Bayesian Analysis代写方面经验极为丰富，各种代写贝叶斯分析Bayesian Analysis相关的作业也就用不着说。 • Statistical Inference 统计推断 • Statistical Computing 统计计算 • (Generalized) Linear Models 广义线性模型 • Statistical Machine Learning 统计机器学习 • Longitudinal Data Analysis 纵向数据分析 • Foundations of Data Science 数据科学基础 ## 统计代写\|贝叶斯分析代写Bayesian Analysis代考\|Frequentist versus Subjective View of Uncertainty When we consider statements about uncertain events like The next toss on a coin will be a head. • A hurricane will destroy the White House within the next 5 years. what we really want to do is measure the uncertainty of such events. In other words we want to be able to make statements like • There is a 1 in 2 (or equivalently $50 \%$ ) chance that the next toss on a coin will be a head. • There is a 1 in 10 million (or equivalently $0.000001 \%$ ) chance that a hurricane will destroy the White House within the next 5 years. Although these statements are superficially similar, there are fundamental differences between them, which come down to the nature of the experiments that give rise to these outcomes. Specifically, whether the following assumptions are reasonable: • Assumption 1 (repeatability of experiment)-The experiment is repeatable many times under identical conditions. • Assumption 2 (independence of experiments)-Assuming the experiment is repeatable then the outcome of one experiment does not influence the result of any subsequent experiment. ## 统计代写\|贝叶斯分析代写Bayesian Analysis代考\|What If You Toss 100 Consecutive Heads Your job (as a risk expert) is not to calculate the chance of tossing a head. Rather, your job is to calculate the chance that the next toss of the coin is a head (just look back at the original problems posed at the start of this chapter). So, if you observe what is known to be a fair coin being tossed 100 times and each time the result is heads, what do you believe are the chances of the next coin being heads? A frequentist, given the fair coin assumption, would insist the answer is still $50 \%$. This is because the frequentist, with these assumptions, does not actually require any coin tosses to take place in practice. To the frequentist, the fair coin assumption means that the chance is always $50 \%$ on each throw. In other words, in making a prediction the frequentist must ignore the actual data that has been seen. The 100 consecutive heads would simply be considered a freak coincidence, that is, no more or less likely than any other random sequence of heads and tails. But then, the frequentist must ignore, for example, 1. The possibility that a fair coin can be tossed in such a way that makes heads more likely 2. That the coin tossed was not actually the fair coin assumed In fact, we will see that such assumptions are irrational given the type of actual data observed. Only the subjective approach coupled with Bayesian reasoning will work effectively in such cases. ## 统计代写\|贝叶斯分析代写Bayesian Analysis代考\|Frequentist versus Subjective View of Uncertainty • 飓风将在未来 5 年内摧毁白宫。我们真正想做的是衡量此类事件的不确定性。换句话说，我们希望能够做出如下陈述 • 有 2 个中的 1 个（或等效的50%) 下一次掷硬币的机会是正面。 • 1000 万分之一（或同等0.000001%) 飓风将在未来 5 年内摧毁白宫的可能性。尽管这些陈述表面上相似，但它们之间存在根本差异，这归结为产生这些结果的实验的性质。具体来说，以下假设是否合理： • 假设 1（实验的可重复性）——实验在相同条件下可重复多次。 • 假设 2（实验的独立性）——假设实验是可重复的，那么一个实验的结果不会影响任何后续实验的结果。 ## 统计代写\|贝叶斯分析代写Bayesian Analysis代考\|What If You Toss 100 Consecutive Heads 1. 可以以更容易出现正面的方式投掷公平硬币的可能性 2. 投掷的硬币实际上并不是假设的公平硬币事实上，我们将看到，鉴于观察到的实际数据类型，这种假设是不合理的。在这种情况下，只有结合贝叶斯推理的主观方法才能有效地发挥作用。 ## 有限元方法代写 tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。 ## MATLAB代写 MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。 ## 统计代写\|贝叶斯分析代写Bayesian Analysis代考\|STAT4102 statistics-lab™ 为您的留学生涯保驾护航在代写贝叶斯分析Bayesian Analysis方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写贝叶斯分析Bayesian Analysis代写方面经验极为丰富，各种代写贝叶斯分析Bayesian Analysis相关的作业也就用不着说。 • Statistical Inference 统计推断 • Statistical Computing 统计计算 • (Generalized) Linear Models 广义线性模型 • Statistical Machine Learning 统计机器学习 • Longitudinal Data Analysis 纵向数据分析 • Foundations of Data Science 数据科学基础 ## 统计代写\|贝叶斯分析代写Bayesian Analysis代考\|Correlation Coefficient and p-Values The correlation coefficient is a number between $-1$ and 1 that determines whether two paired sets of data (such as those for height and intelligence of a group of people) are related. The closer to 1 the more “confident” we are of a positive linear correlation and the closer to-1 the more confident we are of a negative linear correlation (which happens when, for example, one set of numbers tends to decrease when the other set increases as you might expect if you plotted a person’s age against the number of toys they possess). When the correlation coefficient is close to zero there is little evidence of any relationship. Confidence in a relationship is formally determined not just by the correlation coefficient but also by the number of pairs in your data. If there are very few pairs then the coefficient needs to be very close to 1 or $-1$ for it to be deemed “statistically significant,” but if there are many pairs then a coefficient closer to 0 can still be considered “highly significant.” The standard method that statisticians use to measure the “significance” of their empirical analyses is the $p$-value. Suppose we are trying to determine if the relationship between height and intelligence of people is significant and have data consisting of various pairs of values (height, intelligence) for a set of people; then we start with the “null hypothesis,” which, in this case is the statement “height and intelligence of people are unrelated.” The $p$-value is a number between 0 and 1 representing the probability that the data we have arisen if the null hypothesis were true. In medical trials the null hypothesis is typically of the form that “the use of drug X to treat disease $\mathrm{Y}$ is no better than not using the drug.” The calculation of the $p$-value is based on a number of assumptions that are beyond the scope of this discussion, but people who need $p$-values can simply look them up in standard statistical tables (they are also computed automatically in Excel when you run Excel’s regression tool). The tables (or Excel) will tell you, for example, that if there are 100 pairs of data whose correlation coefficient is $0.254$, then the $p$-value is $0.01$. This means that there is a 1 in 100 chance that we would have seen these observations if the variables were unrelated. A low $p$-value (such as $0.01$ ) is taken as evidence that the null hypothesis can be “rejected.” Statisticians say that a $p$-value of $0.01$ is “highly significant” or say that “the data is significant at the $0.01$ level.” A competent researcher investigating a hypothesized relationship will set a $p$-value in advance of the empirical study. Typically, values of either $0.01$ or $0.05$ are used. If the data from the study results in a $p$-value of less than that specified in advance, the researchers will claim that their study is significant and it enables them to reject the null hypothesis and conclude that a relationship really exists. ## 统计代写\|贝叶斯分析代写Bayesian Analysis代考\|Spurious Correlations Although the preceding examples illustrate the danger of reading too much into dubious correlations between variables, the relationships we saw there did not arise purely by chance. In each case some additional common factors helped explain the relationship. But many studies, including unfortunately many taken seriously, result in claims of causal relationships that are almost certainly due to nothing other than pure chance. Although nobody would seriously take measures to stop Americans drinking beer in order to reduce Japanese child mortality, barely a day goes by when some decision maker or another somewhere in the world takes just as irrational a decision based on correlations that turn out to be just as spurious. For example, on the day we first happened to be drafting this section (16 March 2009) the media was buzzing with the story that working night shifts resulted in an increased risk of breast cancer. This followed a World Health Organization study and it triggered the Danish government to make compensation awards to breast cancer sufferers who had worked night shifts. It is impossible to state categorically whether this result really is an example of a purely spurious correlation. But it is actually very simple to demonstrate why and how you will inevitably find a completely spurious correlation in such a study-which you might then wrongly claim is a causal relationship-if you measure enough things. ## 有限元方法代写 tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。 ## MATLAB代写 MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。	{"extraction_info": {"found_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": 0.8717528581619263, "perplexity": 1179.69587335848}, "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-20221004054641-20221004084641-00795.warc.gz"}
https://www.physicsforums.com/threads/integration-velocity-to-displacement-or-position.172410/	# Integration (Velocity to Displacement or Position) 23 File size: 16.5 KB Views: 55 2. ### G01 2,698 EDIT: I realize you may have work in your document, but I can't yet see it. So, if you have work in the document, ignore my lecture below:) According to the forum rules, you must show some work to get help. What have you tried? Where are you stuck? Etc.? 3. ### dt19 47 This is the integration in LaTeX, if anyone else can't see it: $v = \frac{e^{\frac{t - 1205.525}{-100}}-142000}{30}$ $s = \int v dt$ $s = \int \frac{e^{\frac{t - 1205.525}{-100}}-142000}{30} dt$ Do you know how to integrate exponential functions? EDIT: I don't know what is up with the LaTeX, but there should be only one expression on each line. So ignore the bit after the second equals sign on the first line. And there shouldn't be an 's' after the fraction on the first line. No idea why it is doing this, there's nothing wrong with the code as I put it in. Last edited: Jun 1, 2007 4. ### G01 2,698 First, I would sugest splitting the integral up into as many smaller integrals as possible. HINT:$$e^{a+b}=e^ae^b$$ Using this relationship, plus splitting the integral up, you should end up with two smaller, easier integrals of known forms. Last edited: Jun 1, 2007 5. ### kieran1black2 23 i have already tried that... it didnt seem to work... could you show it in latex? so i can see what im doin wrong? 6. ### jackiefrost 137 The 1/30 comes can "come out" of the integrand, right? And the subtraction (e to the blah minus 140000) just yields two integrals: Integral[e to the blah dt] minus Integral[142000 dt]. So that leaves the tricky part: Integral[e to the blah]. That exponent can be broken into two fairly simple expressions by going ahead and doing the division by -100... get it? 7. ### kieran1black2 23 what i ended up with is (e^-t/100)/(-30/100) x e^(-1205.525/-100)/t - 142000/30t that is after integration... is that what you meant?	{"extraction_info": {"found_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": 0.9656022787094116, "perplexity": 1537.6650197878837}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-22/segments/1432207929411.0/warc/CC-MAIN-20150521113209-00286-ip-10-180-206-219.ec2.internal.warc.gz"}
https://papers.nips.cc/paper/2009/hash/0f96613235062963ccde717b18f97592-Abstract.html	Authors Anne Hsu, Thomas Griffiths Abstract <p>A classic debate in cognitive science revolves around understanding how children learn complex linguistic rules, such as those governing restrictions on verb alternations, without negative evidence. Traditionally, formal learnability arguments have been used to claim that such learning is impossible without the aid of innate language-specific knowledge. However, recently, researchers have shown that statistical models are capable of learning complex rules from only positive evidence. These two kinds of learnability analyses differ in their assumptions about the role of the distribution from which linguistic input is generated. The former analyses assume that learners seek to identify grammatical sentences in a way that is robust to the distribution from which the sentences are generated, analogous to discriminative approaches in machine learning. The latter assume that learners are trying to estimate a generative model, with sentences being sampled from that model. We show that these two learning approaches differ in their use of implicit negative evidence -- the absence of a sentence -- when learning verb alternations, and demonstrate that human learners can produce results consistent with the predictions of both approaches, depending on the context in which the learning problem is presented.</p>	{"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": 0.8890227675437927, "perplexity": 861.1473522348729}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-10/segments/1614178349708.2/warc/CC-MAIN-20210224223004-20210225013004-00140.warc.gz"}
http://math.stackexchange.com/questions/60199/finding-the-limit-of-a-quotient/60201	# Finding the limit of a quotient I am trying to find the limit of $(x^2-6x+5)/(x-5)$ as it approaches $5$. I assume that I just plug in $5$ for $x$ and for that I get $0/0$ but my book says $4$. I try and factor and I end up with $(25-30+5)/(5-5)$ which doesnt seem quite right to me but I know that if I factor out $5$ and get rid of the $5-5$ (although that would make it $1-1$ wouldn't it?) that leaves me with $5-6+5$ which is $4$. What do I need to do in this problem? - Hint: x=5 is a root of x^2-6x+5 hence x^2-6x+5 is x-5 times... something which you might want to compute. – Did Aug 27 '11 at 22:46 The limit of $(x^2-6x+5)/(x-5)$ as it approaches $5$ is $5$. Presumably you're trying to find the limit of $(x^2-6x+5)/(x-5)$ as $x$ approaches $5$? – joriki Aug 27 '11 at 22:47 Perform Polynomial Long Division. – Bill Dubuque Aug 27 '11 at 22:53 @Jordan: Really, if you're interested in a tutor and you don't know of a local resource, I've tutored over Skype in the past. If you're interested, shoot me an email. My email is very easy to find on my blog (but not something I post on this forum). – mixedmath Jun 13 '12 at 12:23 ## 3 Answers Let $P(x)=x^{2}-6x+5$ and $Q(x)=x-5$. Since $P(x)$ and $Q(x)$ are continuous and $P(5)=Q(5)=0$, $\frac{P(5)}{Q(5)}$ is undetermined. You have two alternatives: 1. manipulate algebraically $\frac{P(x)}{Q(x)}=\frac{x^{2}-6x+5}{x-5}$. 2. use L'Hôpital's rule $$\lim_{x\rightarrow 5}\frac{P(x)}{Q(x)}=\lim_{x\rightarrow 5}\frac{P^{\prime }(x)}{Q^{\prime }(x)}=\lim_{x\rightarrow 5}\frac{2x-6}{1}=2\cdot 5-6=4.$$ In option 1, since $P(5)=0$, you know that you can factor $P(x)$ as $$P(x)=x^{2}-6x+5=(x-5)(x-c).$$ You can compute $c=1$, by solving the equation $$x^{2}-6x+5=0.$$ Instead you can perform a long division, as suggested by Bill Dubuque, to evaluate $P(x)/Q(x)=x-1$. So, $$P(x)=x^{2}-6x+5=(x-5)(x-1)$$ and $$\lim_{x\rightarrow 5}% \frac{P(x)}{Q(x)}=\lim_{x\rightarrow 5}\frac{(x-5)(x-1)}{x-5}% =\lim_{x\rightarrow 5}(x-1)=5-1=4.$$ You are allowed to divide $P(x)$ and $Q(x)$ by $x-5$, because you perform a limiting process, and you actually never make $x=5$, which means $x-5$ is never equal to $0$. - $(x^2-6x+5)$ = $(x-1)(x-5)$ cancel out the $x-5$ and you don't have to worry about dividing by zero. $x-1$ is the end result. Plug in $5$ for $x$: $5-1 = 4$ - How do I know to factor it to that instead of something else? – user138246 Aug 27 '11 at 22:46 It's called factoring a quadratic, and the result is unique, up to constants. – The Chaz 2.0 Aug 27 '11 at 22:47 Oh is that some stuff I need to memorize? – user138246 Aug 27 '11 at 22:49 @Jordan: It is called factoring. In this case quadratic factoring: purplemath.com/modules/factquad.htm – Dair Aug 27 '11 at 22:54 Oh okay, I was just wondering if that was the quadratic formula that was used. – user138246 Aug 27 '11 at 22:55 Anyway, you can also do it from first principles without any factoring tricks: Let $x = 5 + \epsilon$. Then, when $x \ne 5$, and thus $\epsilon \ne 0$, \begin{aligned} \frac{x^2 - 6x + 5}{x-5} &= \frac{(5 + \epsilon)^2 - 6(5 + \epsilon) + 5}{5 + \epsilon - 5} \\ &= \frac{(25 + 10\epsilon + \epsilon^2) - (30 + 6\epsilon) + 5}{\epsilon} \\ &= \frac{4\epsilon + \epsilon^2}{\epsilon} = 4 + \epsilon. \end{aligned} We thus see that $$\lim_{x \to 5} \frac{x^2 - 6x + 5}{x-5} = \lim_{\epsilon \to 0}\, 4 + \epsilon = 4.$$ (Addendum: The reason for choosing that particular substitution is simple: we want to know what happens when $x$ gets close to $5$; $\epsilon = x - 5$ tells how close $x$ is to $5$. In particular, if the limit as $x \to 5$ is well defined, then after the substitution and simplification we should end up with the limit plus some terms that vanish as $\epsilon \to 0$, as we indeed do. If, instead, we ended up with some terms like $1/\epsilon$ that diverge as $\epsilon \to 0$, then we'd know that the limit was not well defined.) - Why can x = 5+ e? – user138246 Aug 27 '11 at 23:23 @Jordan Carlyon: Why can't it? More seriously, for any real number $x$, we can write it in the form $x = 5 + \epsilon$, where $\epsilon = x - 5$. It's also then easy to show that $\epsilon = 0$ if and only if $x = 5$. – Ilmari Karonen Aug 27 '11 at 23:27 What is e? Or is that just a variable like x or y? – user138246 Aug 27 '11 at 23:29 @Jordan Carlyon: Yes, it's just a variable. (Mathematicians often like to use the Greek letter epsilon ($\epsilon$) for "very small" quantities — especially ones which tend to zero in some limit we're interested in. But that's just a matter of tradition; any other symbol would work just as well.) – Ilmari Karonen Aug 27 '11 at 23:34 But I don't understand why or how that is being used and what the purpose of it is. I have never seen anything like that before, why would I want to do that? Why can I just change a variable to a limit plus a different variable? What allows me to do that and how do I know that it if beneficial to solving the problem without guessing? – user138246 Aug 27 '11 at 23:38	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.9297829866409302, "perplexity": 289.6663713527486}, "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-2016-22/segments/1464049278244.7/warc/CC-MAIN-20160524002118-00063-ip-10-185-217-139.ec2.internal.warc.gz"}
https://pub.uni-bielefeld.de/publication/1779329	# Two modes of thinking - also relevant for the learning of mathematics? Wachsmuth I (1981) For the learning of mathematics 2(2): 38-45. Journal Article \| Published \| English Author Publishing Year ISSN PUB-ID ### Cite this Wachsmuth I. Two modes of thinking - also relevant for the learning of mathematics? For the learning of mathematics. 1981;2(2):38-45. Wachsmuth, I. (1981). Two modes of thinking - also relevant for the learning of mathematics? For the learning of mathematics, 2(2), 38-45. Wachsmuth, I. (1981). Two modes of thinking - also relevant for the learning of mathematics? For the learning of mathematics 2, 38-45. Wachsmuth, I., 1981. Two modes of thinking - also relevant for the learning of mathematics? For the learning of mathematics, 2(2), p 38-45. I. Wachsmuth, “Two modes of thinking - also relevant for the learning of mathematics?”, For the learning of mathematics, vol. 2, 1981, pp. 38-45. Wachsmuth, I.: Two modes of thinking - also relevant for the learning of mathematics? For the learning of mathematics. 2, 38-45 (1981). Wachsmuth, Ipke. “Two modes of thinking - also relevant for the learning of mathematics?”. For the learning of mathematics 2.2 (1981): 38-45. All files available under the following license(s): This Item is protected by copyright and/or related rights. [...] Main File(s) File Name Access Level Open Access	{"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": 0.849557638168335, "perplexity": 2706.94959983815}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-26/segments/1529267863650.42/warc/CC-MAIN-20180620143814-20180620163814-00159.warc.gz"}
http://www.ck12.org/statistics/Central-Limit-Theorem/lecture/Distribution-of-Sample-Mean/r1/	<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" /> # Central Limit Theorem ## The central limit theorem states that the sampling distribution of any statistic will be normal or nearly normal, if the sample size is large enough. 0% Progress Practice Central Limit Theorem Progress 0% Distribution of Sample Mean Explores how to find the sample mean and standard deviation using the Central Limit Theorem when given information about the population mean and standard deviation.	{"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": 0.994452178478241, "perplexity": 734.4890780749087}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-48/segments/1448398471441.74/warc/CC-MAIN-20151124205431-00077-ip-10-71-132-137.ec2.internal.warc.gz"}
https://tex.stackexchange.com/questions/276675/chapter-header-spacing-spacing-after-sections	# Chapter header spacing, spacing after sections I am trying to create double spacing throughout this document - after "Chapter 1" and "after Dimension Reduction in Symmetric Spaces", etc. I also want a top margin of an 1 inch. Can anyone tell me how this might be done? Thanks! I have included the tex code and the pdf output - test.pdf I also added a pdf below created in Word that has the formatting that I'd to implement in tex - Chapter 1.pdf \documentclass[12pt,fleqn, letterpaper]{report} \usepackage{sectsty} \usepackage{lipsum} \chapterfont{\centering} \sectionfont{\fontsize{12}{15}\selectfont \centering \mdseries} \chapterfont{\fontsize{12}{15}\selectfont \centering \mdseries} \subsectionfont{\fontsize{12}{15}\selectfont \centering \mdseries} \usepackage{mathptmx} \usepackage[nodisplayskipstretch]{setspace} \usepackage[left=.88in,right=.88in,top=.88in,bottom=.88in,includefoot] {geometry} \doublespacing \begin{document} \chapter{Dimension Reduction in Symmetric Spaces} \label{chpt:Dim_Red} \vspace{-15pt} In this chapter PGA procedures and their computations for data in three types of manifolds, the space of positive definite matrices, the special orthogonal group and the unit spheres, are specified and analyzed. \section{The Space of Positive Definite Matrices, $P(n)$} \label{sec:pos_def} \subsection{Geometry of $P(n)$} \label{ssec:posdef_geo} A real symmetric matrix . . . \end{document} I wouldn't do it at any cost. Here is a way to jump in to the Bay of Bengal. \titleformat{\chapter}[display] {\filcenter\fontsize{12}{15}\selectfont \mdseries}{\chaptertitlename\ \thechapter}{0pt}{} \titleformat{\section}{\fontsize{12}{15}\selectfont \centering \mdseries} \titleformat{\subsection}{\fontsize{12}{15}\selectfont \centering \mdseries} \titlespacing{\chapter} {0pt}{0pt}{0.7\baselineskip} \titlespacing{\section} {0pt}{0pt}{0pt} \titlespacing{\subsection} {0pt}{0pt}{0pt} Full code: \documentclass[12pt,fleqn, letterpaper]{report} \usepackage{titlesec} \usepackage{lipsum} \titleformat{\chapter}[display] {\filcenter\fontsize{12}{15}\selectfont \mdseries}{\chaptertitlename\ \thechapter}{0pt}{} \titleformat{\section}{\fontsize{12}{15}\selectfont \centering \mdseries} \titleformat{\subsection}{\fontsize{12}{15}\selectfont \centering \mdseries} \titlespacing{\chapter} {0pt}{0pt}{0.7\baselineskip} \titlespacing{\section} {0pt}{0pt}{0pt} \titlespacing{\subsection} {0pt}{0pt}{0pt} \usepackage{mathptmx} \usepackage[nodisplayskipstretch]{setspace} \usepackage[left=.88in,right=.88in,top=.88in,bottom=.88in,includefoot] {geometry} \doublespacing \begin{document} \chapter{Dimension Reduction in Symmetric Spaces} \label{chpt:Dim_Red} \vspace{-15pt} In this chapter PGA procedures and their computations for data in three types of manifolds, the space of positive definite matrices, the special orthogonal group and the unit spheres, are specified and analyzed. definite matrices, the special orthogonal group and the unit spheres, are % \section{The Space of Positive Definite Matrices, $P(n)$} \label{sec:pos_def} %definite matrices, the special orthogonal group and the unit spheres, are % \subsection{Geometry of $P(n)$} \label{ssec:posdef_geo} A real symmetric matrix . . . definite matrices, the special orthogonal group and the unit spheres, are \end{document}	{"extraction_info": {"found_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": 0.9644038081169128, "perplexity": 3106.65070019365}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-30/segments/1563195528208.76/warc/CC-MAIN-20190722180254-20190722202254-00321.warc.gz"}
http://math.stackexchange.com/questions/282241/why-doesnt-dirichlet-function-have-a-derivative-in-x-0?answertab=votes	# Why doesn't Dirichlet function have a derivative in X=0 $\newcommand{\dirichlet}{\mathop{\rm dirichlet}\nolimits}$ I'm trying to find two examples for the following criterias: 1. A method that is continuous in exactly one point but doesn't have a derivative in that point 2. A method that is continuous in exactly one point and does have a derivative in that point After looking deeper at some examples, I found out that $f(x) = x\cdot\dirichlet(x)$ doesn't have a deriviate at $x = 0$ however $f(x) = x^2\cdot \dirichlet(x)$ does have. I can't understand the difference between the two. Any helps could help. - First, $\Delta(x)$ is nowhere continuous, e.g. $\lim_{x\to y}\Delta(x)$ does not exist for any $y$. The definition of the derivative is given by a limit, so if you are looking for the derivative of $x^p\Delta(x)$ for $p\geq 1$ at zero, by the definition you have $$\lim_{x\to 0}\frac{x^p\Delta(x) - 0^p\Delta(0)}{x-0} = \lim_{x\to0}x^{p-1}\Delta(x)$$ and the question now is, when the latter limit does exist. As we discussed already, it does not exist for $p-1=0$ i.e. $p=1$ hence $x\Delta(x)$ does not have a derivative at zero. However, since $$\lim_{x\to 0}x^{p-1} = 0$$ for $p>1$ and $\Delta(x)$ is bounded, the limit $\lim_{x\to0}x^{p-1}\Delta(x)$ does exist and equals zero for all $p>1$. @vondip: it's a power. Like you used $x$ or $x^2$ - I just gave an answer for a more general case. What I meant is that when the power is $p=1$ then the derivative does not exist since we have just to take a limit of the Dirichlet function alone. However, whenever $p>1$ (in particular, $p=2$) the limit also involves some continuous function (which is $x^{p-1}$) which converges to zero, and that's why the limit does exist. – Ilya Jan 19 '13 at 21:10	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 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": 0.9788733124732971, "perplexity": 82.51787811828046}, "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/1408500833525.81/warc/CC-MAIN-20140820021353-00411-ip-10-180-136-8.ec2.internal.warc.gz"}
https://infoscience.epfl.ch/record/150036	Infoscience Conference paper # Drift stabilization of ballooning modes in an inward-Shifted LHD configuration A drift-magnetohydrodynamic theory is applied to a background anisotropic pressure equilibrium state to generate a drift corrected ballooning mode equation. The ratio of the mode frequency to the hot particle drift frequency constitutes the critical expansion parameter. The fast particles thus contribute weakly to the instability driving mechanism and also to the diamagnetic drift stabilisation. This equation is used to model the inward-shifted Large Helical Device (LHD) configuration. In the single-fluid limit, a weakly ballooning unstable band that encompasses a third of the plasma volume develops in the core of the plasma at low $left<\eta_{dia} ight>$ that becomes displaced towards the edge of the plasma at the experimentally achieved $left<\eta_{dia} ight>simeq 5%$. Finite diamagnetic drifts (mainly due to the thermal ions) effectively stabilise these ballooning structures at all values of $left<\eta_{dia} ight>$. The validity of the large hot particle drift approximation is verified for hot to thermal ion density ratios that remain smaller than 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": 0.9315932393074036, "perplexity": 2151.2522319027016}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-39/segments/1505818695113.88/warc/CC-MAIN-20170926070351-20170926090351-00320.warc.gz"}
https://graz.pure.elsevier.com/en/publications/convolution-quadrature-based-bem-in-acoustics-for-absorbing-bound-2	Convolution Quadrature based BEM in acoustics for absorbing boundary conditions Research output: Contribution to journalArticleResearchpeer-review Abstract In many fields of engineering the acoustic behavior has to be determined, e.g. the sound distribution in a room or the sound radiation into the surrounding. Often, the goal is to obtain a sound pressure field such that disturbing noise is reduced to an acceptable level. In room acoustics, sound absorbing materials are often used to obtain this goal. The mathematical description is done with the wave equation and absorbing boundary conditions. The numerical treatment can be done with Boundary Element methods, where the absorbing boundary results in a Robin boundary condition. This boundary condition connects the Neumann trace with the Dirichlet trace of the time derivative. Here, an indirect formulation, which uses the single layer potential, is used as basic boundary integral equation. The convolution quadrature method is applied for time discretisation, which allows a simple formulation of the Robin boundary condition in the Laplace domain. Convergence studies with a refinement in space and time show the expected rates. A realistic example for indoor acoustics, the computation of the sound pressure level in a staircase of the University of Zurich, show the suitability of this approach in determining the indoor acoustics. The absorbing boundary condition shows the expected behavior. (© 2016 Wiley-VCH Verlag GmbH & Co. KGaA, Weinheim) Original language English 23-26 4 Proceedings in Applied Mathematics and Mechanics 16 1 https://doi.org/10.1002/pamm.201610007 Published - 1 Oct 2016 Cite this In: Proceedings in Applied Mathematics and Mechanics , Vol. 16, No. 1, 01.10.2016, p. 23-26. Research output: Contribution to journalArticleResearchpeer-review @article{0a4024e56b7c4b8799cc3a94b8da8695, title = "Convolution Quadrature based BEM in acoustics for absorbing boundary conditions", abstract = "In many fields of engineering the acoustic behavior has to be determined, e.g. the sound distribution in a room or the sound radiation into the surrounding. Often, the goal is to obtain a sound pressure field such that disturbing noise is reduced to an acceptable level. In room acoustics, sound absorbing materials are often used to obtain this goal. The mathematical description is done with the wave equation and absorbing boundary conditions. The numerical treatment can be done with Boundary Element methods, where the absorbing boundary results in a Robin boundary condition. This boundary condition connects the Neumann trace with the Dirichlet trace of the time derivative. Here, an indirect formulation, which uses the single layer potential, is used as basic boundary integral equation. The convolution quadrature method is applied for time discretisation, which allows a simple formulation of the Robin boundary condition in the Laplace domain. Convergence studies with a refinement in space and time show the expected rates. A realistic example for indoor acoustics, the computation of the sound pressure level in a staircase of the University of Zurich, show the suitability of this approach in determining the indoor acoustics. The absorbing boundary condition shows the expected behavior. ({\circledC} 2016 Wiley-VCH Verlag GmbH & Co. KGaA, Weinheim)", keywords = "CQM, Absorbing boundary conditions, integral equations", author = "Dominik P{\"o}lz and Stefan Sauter and Martin Schanz", year = "2016", month = "10", day = "1", doi = "10.1002/pamm.201610007", language = "English", volume = "16", pages = "23--26", journal = "Proceedings in Applied Mathematics and Mechanics", issn = "1617-7061", publisher = "Wiley-Blackwell", number = "1", } TY - JOUR T1 - Convolution Quadrature based BEM in acoustics for absorbing boundary conditions AU - Pölz, Dominik AU - Sauter, Stefan AU - Schanz, Martin PY - 2016/10/1 Y1 - 2016/10/1 N2 - In many fields of engineering the acoustic behavior has to be determined, e.g. the sound distribution in a room or the sound radiation into the surrounding. Often, the goal is to obtain a sound pressure field such that disturbing noise is reduced to an acceptable level. In room acoustics, sound absorbing materials are often used to obtain this goal. The mathematical description is done with the wave equation and absorbing boundary conditions. The numerical treatment can be done with Boundary Element methods, where the absorbing boundary results in a Robin boundary condition. This boundary condition connects the Neumann trace with the Dirichlet trace of the time derivative. Here, an indirect formulation, which uses the single layer potential, is used as basic boundary integral equation. The convolution quadrature method is applied for time discretisation, which allows a simple formulation of the Robin boundary condition in the Laplace domain. Convergence studies with a refinement in space and time show the expected rates. A realistic example for indoor acoustics, the computation of the sound pressure level in a staircase of the University of Zurich, show the suitability of this approach in determining the indoor acoustics. The absorbing boundary condition shows the expected behavior. (© 2016 Wiley-VCH Verlag GmbH & Co. KGaA, Weinheim) AB - In many fields of engineering the acoustic behavior has to be determined, e.g. the sound distribution in a room or the sound radiation into the surrounding. Often, the goal is to obtain a sound pressure field such that disturbing noise is reduced to an acceptable level. In room acoustics, sound absorbing materials are often used to obtain this goal. The mathematical description is done with the wave equation and absorbing boundary conditions. The numerical treatment can be done with Boundary Element methods, where the absorbing boundary results in a Robin boundary condition. This boundary condition connects the Neumann trace with the Dirichlet trace of the time derivative. Here, an indirect formulation, which uses the single layer potential, is used as basic boundary integral equation. The convolution quadrature method is applied for time discretisation, which allows a simple formulation of the Robin boundary condition in the Laplace domain. Convergence studies with a refinement in space and time show the expected rates. A realistic example for indoor acoustics, the computation of the sound pressure level in a staircase of the University of Zurich, show the suitability of this approach in determining the indoor acoustics. The absorbing boundary condition shows the expected behavior. (© 2016 Wiley-VCH Verlag GmbH & Co. KGaA, Weinheim) KW - CQM KW - Absorbing boundary conditions KW - integral equations U2 - 10.1002/pamm.201610007 DO - 10.1002/pamm.201610007 M3 - Article VL - 16 SP - 23 EP - 26 JO - Proceedings in Applied Mathematics and Mechanics JF - Proceedings in Applied Mathematics and Mechanics SN - 1617-7061 IS - 1 ER -	{"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": 0.91289222240448, "perplexity": 786.7357717118142}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-43/segments/1570987823061.83/warc/CC-MAIN-20191022182744-20191022210244-00500.warc.gz"}
http://en.wikipedia.org/wiki/Polynomial_expansion	# Polynomial expansion For locally approximating a function with a polynomial, see Taylor series. In mathematics, an expansion of a product of sums expresses it as a sum of products by using the fact that multiplication distributes over addition. Expansion of a polynomial expression can be obtained by repeatedly replacing subexpressions that multiply two other subexpressions, at least one of which is an addition, by the equivalent sum of products, continuing until the expression becomes a sum of (repeated) products. During the expansion, simplifications such as grouping of like terms or cancellations of terms may also be applied. Instead of multiplications, the expansion steps could also involve replacing powers of a sum of terms by the equivalent expression obtained from the binomial formula; this is a shortened form of what would happen if the power were treated as a repeated multiplication, and expanded repeatedly. It is customary to reintroduce powers in the final result when terms involve products of identical symbols. Simple examples of polynomial expansions are the well known rules $(x+y)^2=x^2+2xy+y^2$ $(x+y)(x-y)=x^2-y^2$ when used from left to right. A more general single-step expansion will introduce all products of a term of one of the sums being multiplied with a term of the other: $(a+b+c+d)(x+y+z)=ax+ay+az+bx+by+bz+cx+cy+cz+dx+dy+dz$ An expansion which involves multiple nested rewrite steps is that of working out a Horner scheme to the (expanded) polynomial it defines, for instance $1+x(-3+x(4+x(0+x(-12+x\cdot 2))))=1-3x+4x^2-12x^4+2x^5$. The opposite process of trying to write an expanded polynomial as a product is called polynomial factorization. ## Expansion of a polynomial written in factored form Two expressions can be multiplied by using the commutative law, associative law and distributive law. (To multiply more than 2 expressions, just multiply 2 at a time) To multiply two factors, each term of the first factor must be multiplied by each term of the other factor. If both factors are binomials, the FOIL rule can be used, which stands for "First Outer Inner Last," referring to the terms that are multiplied together. For example, expanding $(x+2)(2x-5)\,$ yields $2x^2-5x+4x-10 = 2x^2-x-10.$ ## Expansion of (x+y)n Main article: Binomial theorem When expanding $(x+y)^n$, a special relationship exists between the coefficients of the terms when written in order of descending powers of x and ascending powers of y. The coefficients will be the numbers in the (n + 1)th row of Pascal's triangle. For example, when expanding $(x+y)^6$, the following is obtained: ${\color{red}1}x^6+{\color{red}6}x^5y+{\color{red}{15}}x^4y^2+{\color{red}{20}}x^3y^3+{\color{red}{15}}x^2y^4+{\color{red}{6}}xy^5+{\color{red}1}y^6 \,$	{"extraction_info": {"found_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": 9, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8425243496894836, "perplexity": 411.75282688741015}, "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/1418802771909.45/warc/CC-MAIN-20141217075251-00109-ip-10-231-17-201.ec2.internal.warc.gz"}
https://math.stackexchange.com/questions/2246853/can-all-linear-operators-on-functions-be-represented-as-a-convolution-of-the-inp	# Can all linear operators on functions be represented as a convolution of the input function with the operator's impulse response? Where $f, g$ are functions, $\delta$ is the Dirac Delta distribution ('aka Delta function'), and L is a linear (LTI) operator, if $g = L(f)$, then can we say in all cases $g = Li * f$ , where $Li$ is defined as $L(\delta)$, i.e. the impulse response? My questions: Is this true? Any references would be appreciated. If $L$ is well-defined linear operator $L^1(\mathbb{R}) \to L^\infty(\mathbb{R})$ then $L[f(y)](x) = h \ast f(x)$ for some $h \in L^\infty(\mathbb{R})$ if and only if $$\forall a, \qquad L[f(y-a)](x) = L[f(y)](x-a)$$ That's why we say "linear and time-invariant (LTI) operators". The same theoy of LTI operators works in $L^\infty(\mathbb{R}),L^2(\mathbb{R}), C^0(\mathbb{R}), D(\mathbb{R})$ and $D'(\mathbb{R})$. The Dirac delta $\delta$ belongs to this last space : it is a distribution. • @user45664 Yes $L$ is LTI and if it converges then $h(x) = L[\delta](x) = \lim_{\epsilon \to 0} L[\frac{1_{\|y\| <\epsilon}}{2 \epsilon}](x)$ – reuns Apr 22 '17 at 18:22 • @user45664 Assume that $f \ast h = 0$ for every $f$ so that $h(x) = 0$ "everywhere". Then $h(0) = 0$ or $h(0) = 1$ doesn't change anything under convolution. So it is unique up to isolated values (sets of measure $0$). – reuns Apr 22 '17 at 18:35 • So you need to state it like this : if $L$ is LTI and $h(x) = \lim_{\epsilon \to 0} L[\frac{1_{\|y\| <\epsilon}}{2 \epsilon}](x)$ is well-defined, then $L[f] = f \ast h$ @user45664 – reuns Apr 22 '17 at 18:37 • @user45664 Yes, this is a rectangle approaching $\delta$ in the limit, and a rigorous definition of $\delta$. – reuns Apr 22 '17 at 18:51	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9593177437782288, "perplexity": 332.2264384978903}, "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-2019-26/segments/1560627999000.76/warc/CC-MAIN-20190619143832-20190619165832-00380.warc.gz"}
https://istopdeath.com/factor-k2-k-2/	# Factor k^2-k-2 Consider the form . Find a pair of integers whose product is and whose sum is . In this case, whose product is and whose sum is . Write the factored form using these integers. Factor k^2-k-2	{"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": 0.923883855342865, "perplexity": 940.4084047908802}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764499816.79/warc/CC-MAIN-20230130101912-20230130131912-00778.warc.gz"}
http://weblib.cern.ch/collection/Theses?ln=ka&as=1	# Theses უკანასკნელი დამატებები: 2021-09-25 15:09 Thermal efficiency study of the ALPHA-g cryostat - an analysis and heat simulation / Kuhn, Philipp Benjamin This thesis presents a thermal study of the ALPHA-g cryostat at CERN [...] CERN-THESIS-2021-144 - 77 p. 2021-09-24 19:28 High energy physics: from quantum field theory to Higgs coupling measurement / Ranken, Evan Altair This document summarizes work in theoretical and experimental high-energy physics completed as part of the author’s doctoral studies, while providing context and motivation for these projects [...] CERN-THESIS-2020-366 - 210 p. 2021-09-23 14:19 / Alhawiti, Fawaz Mutlaq S There is a need to answer the questions that the Standard Model theory SM fails to explain [...] CERN-THESIS-2021-143 - 2021-09-21 21:50 Design Testing of the Small Monitored Drift Tubes for the Phase II Upgrade of the ATLAS Muon Spectrometer / Minnella, Joseph Dominick The new high luminosity environment at the Large Hadron Collider necessitates an upgrade of detector technology [...] CERN-THESIS-2021-142 - 35 p. 2021-09-20 18:19 Search for Higgs boson pair production in the single-lepton $WWb\bar{b}$ channel with the ATLAS detector / D'Amico, Valerio A search for the Standard Model (SM) Higgs boson pair production in the $WWb\bar{b}$ single-lepton channel is presented [...] CERN-THESIS-2020-365 - 168 p. 2021-09-20 17:50 Search for exotic decays of the Higgs boson using photons with the Compact Muon Solenoid experiment / Wamorkar, Tanvi The search for answers to the fundamental questions of nature are being explored by the experiments at the Large Hadron Collider at CERN [...] CERN-THESIS-2021-141 - 197 p. 2021-09-20 14:19 Mechanical characterisation of Nb3Sn Rutherford cable stacks / Wolf, Felix Josef Nb3Sn Rutherford cables are used in CERN’s superconducting 11 T dipole and MQXF quadrupole magnets, which are proposed for the instantaneous luminosity (rate of particle collisions) upgrade of the Large Hadron Collider (LHC) by a factor of ﬁve to a High Luminosity-Large Hadron Collider (HL-LHC).. [...] CERN-THESIS-2021-140 - 159 p. 2021-09-19 23:34 Searches for new physics with leptons and invisible particles at the ATLAS experiment at the LHC / Gallardo, Gabriel The Standard Model of Particle Physics (SM), while successful at describing almost all subatomic phenomena observed to date, has some glaring open questions: the Higgs boson mass is unstable at high energies, there is no dark matter candidate, and the prediction and measurement of $(g − 2)_\mu$ ar [...] CERN-THESIS-2021-139 - 194 p. 2021-09-19 19:42 Collider Physics Measurements in High Jet Multiplicity Final States / Mikuni, Vinicius Massami The CMS Experiment has collected an unprecedented amount of data during the 2016-2018 data taking period [...] CERN-THESIS-2021-138 - 159 p. 2021-09-19 11:00 Analysis of Data from a Network of Pixel Detectors / Fiedler, Petr This thesis deals with design and implementation of a bundle of scripts automating the upload of specific files using Rucio middleware from server atlastpx2.utef.cvut.cz at Institute of Experimental and Applied Physics at Czech Technical University to Worldwide LHC Computing Grid coordinated by Euro [...] CERN-THESIS-2018-490 - 49 p.	{"extraction_info": {"found_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": 0.9493507146835327, "perplexity": 3483.354800779124}, "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/1631780058467.95/warc/CC-MAIN-20210927181724-20210927211724-00415.warc.gz"}
https://www.arxiv-vanity.com/papers/1403.3772/	# Ludics Characterization of Multiplicative-Additive Linear Behaviours Christophe Fouqueré Université Paris 13, Sorbonne Paris Cité, LIPN, F–93430, Villetaneuse, France CNRS, UMR 7030, F–93430, Villetaneuse, France. Myriam Quatrini IML –FRE 35-29, Université Aix-Marseille & CNRS, F–13284 Marseille Cedex 07, France. ###### Abstract Ludics is a logical theory that J.-Y. Girard developed around . At first glance, it may be considered as a Brouwer-Heyting-Kolmogorov interpretation of Logic as a formula is denoted by the set of its proofs. More primitively, Ludics is a theory of interaction that models (a variant of) second-order multiplicative-additive Linear Logic. A formula is denoted by a set of objects called a behaviour, a proof by an object that satisfies some criteria. Our aim is to analyze the structure of behaviours in order to better understand and refine the usual notion of formulas or types. More precisely, we study properties that guarantee a behaviour to be recursively decomposable by means of multiplicative-additive linear connectives and linear constants.111This work has been partially funded by the French ANR projet blanc “Locativity and Geometry of Interaction” LOGOI ANR-10-BLAN-0213 02. ## 1 Introduction Ludics is a logical theory that J.-Y. Girard developed around . At first glance, it may be considered as a Brouwer-Heyting-Kolmogorov interpretation of Logic as a formula is denoted by the set of its proofs. More primitively, Ludics is a theory of interaction, where interaction is the fact that a meeting happens between two objects together with the dynamical process that this meeting creates. This notion is primitive in the sense that the main objects of Ludics, called designs, are defined with respect to interaction: they are objects between which meetings may happen and on which rewriting processes may be described. Hence, among the computational theoretical and methodological frameworks, Ludics is ontologically closer to Game Semantics than to Proof Theory or Type Theory. Indeed, if interaction corresponds to cut and cut-elimination in Proof Theory, and to application rule and normalization in Type Theory, it presupposes more primitive notions fixed: formulas and formal proofs in Proof Theory, types and terms in Type Theory. On the opposite, the concept of play in Game Theory serves as a means for interaction. However we should notice that, in Game Semantics, the definition of a game is external to interaction, whereas in Ludics the corresponding notion of behaviour is internal: it is a closure of a set of designs with respect to interaction. In other words, Ludics may be considered as an untyped computational theoretical framework, but with types subsequently recovered. In [12], Terui showed that such a notion of interactive type may be applied with success to the study of formal grammars. Our aim is to analyze the structure of interactive types in order to better understand and refine the usual notion of formulas or types. More precisely, we characterize in this paper behaviours that correspond to (linear) logical formulas. We give properties that guarantee a behaviour to be recursively decomposable by means of multiplicative-additive linear connectives and linear constants. First of all, essential finiteness (or uniform boundedness when infinite sums are accepted) ensures that such a decomposition indeed terminates on constants. Additive decomposition is immediate as it is already present in Ludics. Multiplicative decomposition is a more complex problem to tackle. For that purpose, two notions turn out to be fundamental: (1) incarnation, a specific concept of Ludics, enables to characterize which part of a behaviour is used during interaction, (2) the presentation of a design as a set of paths instead of a set of chronicles as originally introduced by Girard. These notions help also study the relation between Ludics and Game semantics [7, 8, 1]. In a previous work [9], the presentation of a design as a set of paths was the key point that allows the authors to compute the incarnation of a set of designs without computing the behaviour. To be able to fully characterize behaviours that are linearly decomposable, we investigate two new notions: visitability and regularity. A visitable path is a path that may be travelled by interaction. A regular behaviour is such that its set of visitable paths is exactly the set of positive-ended chronicles of its incarnation, stable by shuffle and dual operations, where a shuffle of two paths is an interleaving of actions that respects polarity. With that given, we prove the following result: iff is a regular uniformly bounded behaviour. where , defined inductively in the following way: The paper is organized as follows. In section 2, we recall the main facts concerning Ludics. In particular, we make explicit the equivalence between the two presentations of a design, as set of paths versus set of chronicles. We define what is a shuffle of paths. To our knowledge, the shuffle operation, largely used in combinatorics and to study parallelism (see for example [6, 11]), appears in Logics only to study non-commutativity [2, 3]. We give a few properties concerning orthogonality in terms of path travelling, introducing visitable paths, i.e., paths that are visited by orthogonality. In section 3 additive and multiplicative operations on behaviours are studied with respect to visitability. The main result is a characterization of a tensor of behaviours mainly as a shuffle operation on paths. Section 4 is devoted to prove our main theorem already stated above. ## 2 Ludics from Paths, Incarnation from Visitable Paths ### 2.1 Chronicles and Paths The main objects of Ludics, the designs, are defined by Girard [10] in order to be the support of interaction and of its dynamics. The interaction between two designs occurs when their respective bases share a same address in dual positions. The dynamic part of interaction is decomposed in elementary steps called actions (moves in Game Semantics). The (dynamics of the) interaction consists in following two dual alternate sequences of actions, one in each design. ###### Definition 2.1 (Base, Action, Sequence) • A base is a non-empty finite set of sequents of pairwise disjoint addresses: , …, such each is a finite set of addresses, at most one may be empty and the other contain each exactly one address. An address is noted as a sequence of integers. • An action is • either a triple: a polarity that may be positive () or negative (); an address that is the focus of ; a finite set of integers called a ramification. When used in an interaction, the action creates the finite set of new addresses on which the interaction may continue. An action with focus is said justified by . • or the positive action daimon denoted by . • A sequence of actions is based on if each action of is either justified by a previous action in , or has its focus in , or is a daimon which should be, in this case, the last action of . An action of is initial when the focus of is in . For ease of reading, one may put as superscript of an action its polarity: is a positive action whereas is a negative action. Two kinds of sequences of actions, paths and chronicles, may equivalently be used to define designs. Roughly speaking, a path is an alternate sequence that allows to recover the justification relation between actions, a chronicle is a path with an additional constraints on the justification relation: a negative action should be justified by the immediate previous action in the sequence. ###### Definition 2.2 (Path, Chronicle) • A path based on is a finite sequence of actions based on such that • Alternation: The polarity of actions alternate between positive and negative. • Justification: A proper action is either justified, i.e., its focus is built by one of the previous actions in the sequence, or it is called initial with a focus in one of (resp. ) if the action is negative (resp. positive). • Negative jump: (There is no jump on positive action) Let be a subsequence of , - If is a positive proper action justified by a negative action then there is a sub-sequence of such that , and for all , is immediately preceded by and is justified by in . - If is an initial positive proper action then its focus belongs to one and either is the first action of and is empty, or is immediately preceded in by a negative action with a focus hereditarily justified by an element of . • Linearity: Actions have distinct focuses. • Daimon: If present, a daimon ends the path. If it is the first action in then one of is empty. • Totality: If there exists an empty , then is non-empty and begins either with or with a positive action with a focus in . • A chronicle based on is a non-empty path such that each non-initial negative action is justified by the immediate previous (positive) action. The polarity of a path is given by the polarity of its first action if the path is not empty: a negative path is a path with a first action that is negative. A positive-ended path is a path whose last action is positive. Abusively, we may say that an empty path is also negative or positive, a positive-ended path may be empty.	{"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": 0.9330598711967468, "perplexity": 844.9539839226426}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320305242.48/warc/CC-MAIN-20220127072916-20220127102916-00543.warc.gz"}
http://physics.stackexchange.com/questions/103820/what-is-the-role-of-the-speed-of-light-in-mass-energy-equivalency	# What is the role of the speed of light in mass-energy equivalency? [duplicate] Where does $c$ squared come into play in the equation $E=mc^2$. Multiplication obviously but how does energy equal mass times the speed of light? - ## marked as duplicate by DumpsterDoofus, Brandon Enright, Kyle Kanos, joshphysics, Qmechanic♦Mar 17 '14 at 8:36 I'm not sure this is actually a duplicate, since neither of the linked questions actually address the role of the constant $c$ in the equation. I thought we already had a question about that particular point, but I can't find it. – David Z Mar 17 '14 at 2:01 This mass–energy relation states that the universal proportionality factor between equivalent amounts of energy and mass is equal to the speed of light squared. This also serves to convert units of mass to units of energy, no matter what system of measurement units is used. To find out how much energy an object has, multiply the mass of the object by the square of the speed of light. Now we multiphy by $c^2$ because when mass is converted into energy, this energy is by definition moving at $c$. Pure energy is understood in terms of electromagnetic radiation which always moves at the speed of light	{"extraction_info": {"found_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": 0.961922287940979, "perplexity": 288.59692155616176}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-18/segments/1429246655962.81/warc/CC-MAIN-20150417045735-00209-ip-10-235-10-82.ec2.internal.warc.gz"}
https://stats.stackexchange.com/questions/267024/what-if-do-not-use-any-activation-function-in-the-neural-network	# What if do not use any activation function in the neural network? [duplicate] or, for example, is it good to use activation function only for a last layer? As I know, if there are no activation functions in neural network, feedforward will be like simple matrix multiplication, but I don't understand why this is bad. ## marked as duplicate by Franck Dernoncourt, kjetil b halvorsen, Carl, John, gung♦Mar 13 '17 at 12:22 Consider a two layer neural network. Let $x \in \mathbb{R}^n$ be your input vector, and consider a single layer without an activation function and weight matrix $A$ and bias $b$, it would computer $$Ax + b$$ A second layer (without activation and weight matrix $C$ and bias $d$) would then compute $$C(Ax + b)+d$$ This is equal to $$CAx + Cb + d$$ This is equivalent to a single layer neural network with weight matrix $CA$ and bias vector $Cb+d$. It is well known that single layer neural networks cannot solve some "simple" problems, for example, they cannot solve the XOR problem. Suppose we have a single layer neural network $Ax + b$ that can solve the XOR problem. The matrix is of the form $A = (w_{1}, w_{2})$ since it takes two inputs and outputs a single value. Then $$0w_1 + 0w_2 + b \leq 0 \iff b \leq 0 \\ 0w_1 + 1w_2 + b > 0 \iff b > -w_2 \\ 1w_1 + 0w_2 + b > 0 \iff b > -w_1 \\ 1w_1 + 1w_2 + b \leq 0 \iff b \leq -w_1 - w_2$$ Suppose all of the left hand sides are true (which is required to solve the XOR problem) then we know $b \leq 0$. The second and third line gives that $w_1$ and $w_2$ are also negative (since they are less than $b$ which is negative). The fourth line gives $b \leq -w_1 - w_2$ then since $b$ is negative $2b \leq b \leq -w_1 - w_2$ but then $b > -w_2$ and $b > -w_1$ gives $2b > -w_1 - w_2$ which is a contradiction. Hence the single layer neural network cannot solve the XOR problem. Less formally, $Ax+b$ defines a line which separates the plane such that all points are classified according to the side of the line they lay on. Try drawing a straight line such that $(0,0)$ and $(1,1)$ are on one side and $(0,1)$ and $(1,0)$ is on the other, you will not be able to. Introducing non-linear activation functions between the layers allows for the network to solve a larger variety of problems. To be more precise the Universal approximation theorem states that a feed-forward network with a single hidden layer containing a finite number of neurons, can approximate continuous functions on compact subsets of $\mathbb{R}^n$, under mild assumptions on the activation function. • this is nice explanation, but I still don't understand why 1 layer can not approximate xor function. I can not imagine why it is so. – Dmitry Nalyvaiko Mar 12 '17 at 17:45 • I've added a proof @DmitryNalyvaiko – HBeel Mar 12 '17 at 18:21 • thanks, but when I wrote "1 layer", I meant layer "CAx + Cb + d" about which you wrote at the beggining . you considered network with 2 input layers and 1 output layer. what proof will be in case of network, for example, with size [2, 3, 1] where 2 is input size, 3 - hidden layer size, 1 - output layer size? – Dmitry Nalyvaiko Mar 12 '17 at 20:20 • > Less formally, Ax+bAx+b defines a line which separates the plane such that all points are classified according to the side of the line they lay on. Try drawing a straight line such that (0,0)(0,0) and (1,1)(1,1) are on one side and (0,1)(0,1) and (1,0)(1,0) is on the other, you will not be able to. how to do that with curve? – Dmitry Nalyvaiko Mar 12 '17 at 20:39 • @DmitryNalyvaiko so if you have [2, 3, 1] you would have $A \in \mathbb{R}^{3,2}$ and $C \in \mathbb{R}^{1,3}$ so $CA \in \mathbb{R}^{1,2}$, so the proof is still valid by calling elements $CA = (w_1, w_2)$. Similarly for the bias – HBeel Mar 13 '17 at 11:05	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 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": 2, "x-ck12": 0, "texerror": 0, "math_score": 0.8236714601516724, "perplexity": 253.6622535177109}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-43/segments/1570987836368.96/warc/CC-MAIN-20191023225038-20191024012538-00088.warc.gz"}
http://www.in-nomine.org/tag/mathjax/	# The rendering equation The rendering equation, the keystone of computer graphics rendering: $$L_o(x, \vec w) = L_e(x, \vec w) + \int_\Omega f_r(x, \vec w’, \vec w) L_i(x, \vec w’) (\vec w’ \cdot \vec n) \mathrm{d}\vec w’$$ (This is at the same time also a test for MathJax support on my weblog in order to provide correct mathematical typesetting using proper typography instead of some lame pixelated image.)	{"extraction_info": {"found_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": 0.9507491588592529, "perplexity": 3734.6664430574397}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-35/segments/1408500959239.73/warc/CC-MAIN-20140820021559-00324-ip-10-180-136-8.ec2.internal.warc.gz"}
http://www.zora.uzh.ch/41491/	Quick Search: Browse by: Zurich Open Repository and Archive Permanent URL to this publication: http://dx.doi.org/10.5167/uzh-41491 # Pato, M; Agertz, O; Bertone, G; Moore, B; Teyssier, R (2010). Systematic uncertainties in the determination of the local dark matter density. Physical Review D, 82(2):023531-7pp. Preview Accepted Version PDF (Accepted manuscript, Version 2) 456kB View at publisher Preview Accepted Version PDF (Accepted manuscript, Version 1) 455kB ## Abstract A precise determination of the local dark matter density and an accurate control over the corresponding uncertainties are of paramount importance for dark matter (DM) searches. Using very recent high-resolution numerical simulations of a Milky Way like object, we study the systematic uncertainties that affect the determination of the local dark matter density based on dynamical measurements in the Galaxy. In particular, extracting from the simulation with baryons the orientation of the Galactic stellar disk with respect to the DM distribution, we study the DM density for an observer located at ˜8kpc from the Galactic center on the stellar disk, ρ0. This quantity is found to be always larger than the average density in a spherical shell of the same radius ρ¯0, which is the quantity inferred from dynamical measurements in the Galaxy, and to vary in the range ρ0/ρ¯0=1.01-1.41. This suggests that the actual dark matter density in the solar neighborhood is on average 21% larger than the value inferred from most dynamical measurements, and that the associated systematic errors are larger than the statistical errors recently discussed in the literature. ## Citations 67 citations in Web of Science® 69 citations in Scopus®	{"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": 0.9679673910140991, "perplexity": 1636.5043762915448}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-18/segments/1429246639191.8/warc/CC-MAIN-20150417045719-00177-ip-10-235-10-82.ec2.internal.warc.gz"}
http://mathhelpforum.com/calculus/20387-proving-function-injective.html	# Math Help - Proving a function to be injective 1. ## Proving a function to be injective Hi all, I'll get right to the question: Suppose you are given functions f:A->B and g:B->C such that the composite function g(f(x)) is injective, prove that f is injective. I'm pretty sure I have it, but the last step seems a bit of a jump, and although i can't seem to find counter example to refute my proof, i'd like to make sure! here's what i think: Suppose x1 not= x2 for x1 and x2 in a set A =>g(f(x1)) not=g(f(x2)) Therefore f(x1) not= f(x2) because if they WERE equal to each other, it would mean the composite function g(f(x1)) would =g(f(x2)) and we know this isn't true because it was explicitly stated that x1 not= x2 and that for the composite function to remain injective (another thing that is true at start of question) g(f(x1)) must not=g(f(x2)) with x1 not= x2. So by contradiction f(x1) not= f(x2) and so f is injective. Erm, it all just melts my brain reading over it, it's like it loops itself so many times...but here's hoping someone can help if i am wrong! Edit: How do I use proper math notation here, i've seen it on a few posts, but haven't figured it out yet ;/ 2. Originally Posted by scorpio1 Suppose you are given functions f:A->B and g:B->C such that the composite function g(f(x)) is injective, prove that f is injective. Suppose x1 not= x2 for x1 and x2 in a set A =>g(f(x1)) not=g(f(x2)) herefore f(x1) not= f(x2) because if they WERE equal to each other, it would mean the composite function g(f(x1)) would =g(f(x2))	{"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": 0.825650155544281, "perplexity": 1448.8122818633321}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-22/segments/1432207931085.38/warc/CC-MAIN-20150521113211-00284-ip-10-180-206-219.ec2.internal.warc.gz"}
http://www.cs.mcgill.ca/~rwest/wikispeedia/wpcd/wp/w/Wave.htm	# Wave A wave is a disturbance that propagates through space or spacetime, often transferring energy. While a mechanical wave exists in a medium (which on deformation is capable of producing elastic restoring forces), waves of electromagnetic radiation (and probably gravitational radiation) can travel through vacuum, that is, without a medium. Waves travel and transfer energy from one point to another, with little or no permanent displacement of the particles of the medium (there is little or no associated mass transport); instead there are oscillations around fixed positions. For many years, scientists have been trying to work out the problem of energy transfer from one place to another - especially sound and light energy. ## Characteristics Surface waves in water Periodic waves are characterized by crests (highs) and troughs (lows), and may usually be categorized as either longitudinal or transverse. Transverse waves are those with vibrations perpendicular to the direction of the propagation of the wave; examples include waves on a string and electromagnetic waves. Longitudinal waves are those with vibrations parallel to the direction of the propagation of the wave; examples include most sound waves. When an object bobs up and down on a ripple in a pond, it experiences an orbital trajectory because ripples are not simple transverse sinusoidal waves. A = At deep water. B = At shallow water. The circular movement of a surface particle becomes elliptical with decreasing depth. 1 = Progression of wave 2 = Crest 3 = Trough Ripples on the surface of a pond are actually a combination of transverse and longitudinal waves; therefore, the points on the surface follow orbital paths. All waves have common behaviour under a number of standard situations. All waves can experience the following: • Reflection – the change of direction of waves, due to hitting a reflective surface. • Refraction – the change of direction of waves due to them entering a new medium. • Diffraction – the circular spreading of waves that happens when the distance between waves move through an opening of equal distance. • Interference – the superposition of two waves that come into contact with each other. • Dispersion – the splitting up of waves by frequency. • Rectilinear propagation – the movement of waves in straight lines. ### Polarization A wave is polarized if it can only oscillate in one direction. The polarization of a transverse wave describes the direction of oscillation, in the plane perpendicular to the direction of travel. Longitudinal waves such as sound waves do not exhibit polarization, because for these waves the direction of oscillation is along the direction of travel. A wave can be polarized by using a polarizing filter. ### Examples Examples of waves include: • Ocean surface waves, which are perturbations that propagate through water. • Radio waves, microwaves, infrared rays, visible light, ultraviolet rays, x-rays, and gamma rays make up electromagnetic radiation. In this case, propagation is possible without a medium, through vacuum. These electromagnetic waves travel at 299,792,458 m/s in a vacuum. • Sound - a mechanical wave that propagates through air, liquid or solids. • Seismic waves in earthquakes, of which there are three types, called S, P, and L. • Gravitational waves, which are fluctuations in the gravitational field predicted by general Relativity. These waves are nonlinear, and have yet to be observed empirically. • Inertial waves, which occur in rotating fluids and are restored by the Coriolis force. ## Mathematical description Waves can be described mathematically using a series of parameters. The amplitude of a wave (commonly notated as A, or another letter) is a measure of the maximum disturbance in the medium during one wave cycle. In the illustration to the right, this is the maximum vertical distance between the baseline and the wave. The units of the amplitude depend on the type of wave — waves on a string have an amplitude expressed as a distance (meters), sound waves as pressure (pascals) and electromagnetic waves as the amplitude of the electric field (volts/meter). The amplitude may be constant (in which case the wave is a c.w. or continuous wave), or may vary with time and/or position. The form of the variation of amplitude is called the envelope of the wave. The wavelength (denoted as λ) is the distance between two sequential crests (or troughs). This generally has the unit of metres; it is also commonly measured in nanometres for the optical part of the electromagnetic spectrum. A wavenumber k can be associated with the wavelength by the relation $k = \frac{2 \pi}{\lambda}$. Waves can be represented by simple harmonic motion. The period T is the time for one complete cycle for an oscillation of a wave. The frequency f (also frequently denoted as ν) is how many periods per unit time (for example one second) and is measured in hertz. These are related by: $f=\frac{1}{T}$. In other words, the frequency and period of a wave are reciprocals of each other. The angular frequency ω represents the frequency in terms of radians per second. It is related to the frequency by: $\omega = 2 \pi f = \frac{2 \pi}{T}$. There are two velocities that are associated with waves. The first is the phase velocity, which gives the rate at which the wave propagates, is given by $v_p = \frac{\omega}{k}$. The second is the group velocity, which gives the velocity at which variations in the shape of the wave's amplitude propagate through space. This is the rate at which information can be transmitted by the wave. It is given by $v_g = \frac{\partial \omega}{\partial k}$ ### The wave equation The wave equation is a differential equation that describes the evolution of a harmonic wave over time. The equation has slightly different forms depending on how the wave is transmitted, and the medium it is traveling through. Considering a one-dimensional wave that is travelling down a rope along the x-axis with velocity v and amplitude u (which generally depends on both x and t), the wave equation is $\frac{1}{v^2}\frac{\partial^2 u}{\partial t^2}=\frac{\partial^2 u}{\partial x^2}. \$ In three dimensions, this becomes $\frac{1}{v^2}\frac{\partial^2 u}{\partial t^2} = \nabla^2 u$, where $\nabla^2$ is the Laplacian. The velocity v will depend on both the type of wave and the medium through which it is being transmitted. A general solution for the wave equation in one dimension was given by d'Alembert. It is $u(x,t)=F(x-vt)+G(x+vt). \$ This can be viewed as two pulses travelling down the rope in opposite directions; F in the +x direction, and G in the -x direction. If we substitute for x above, replacing it with directions x, y, z, we then can describe a wave propagating in three dimensions. The Schrödinger equation describes the wave-like behaviour of particles in quantum mechanics. Solutions of this equation are wave functions which can be used to describe the probability density of a particle. Quantum mechanics also describes particle properties that other waves, such as light and sound, have on the atomic scale and below. ### Traveling waves Waves that remain in one place are called standing waves - e.g. vibrations on a violin string. Waves that are moving are called traveling waves, and have a disturbance that varies both with time t and distance z. This can be expressed mathematically as: $u = A(z,t) \cos (\omega t - kz + \phi)\,$ where A(z,t) is the amplitude envelope of the wave, k is the wave number and φ is the phase. The phase velocity vp of this wave is given by: $v_p = \frac{\omega}{k}= \lambda f,$ where λ is the wavelength of the wave. ### Propagation through strings The speed of a wave traveling along a string (v) is directly proportional to the square root of the tension (T) over the linear density (μ): $v=\sqrt{\frac{T}{\mu}}.$ ## Transmission medium The medium that carries a wave is called a transmission medium. It can be classified into one or more of the following categories: • A linear medium if the amplitudes of different waves at any particular point in the medium can be added. • A bounded medium if it is finite in extent, otherwise an unbounded medium. • A uniform medium if its physical properties are unchanged at different locations in space. • An isotropic medium if its physical properties are the same in different directions.	{"extraction_info": {"found_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": 12, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9504244923591614, "perplexity": 450.06374528243435}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-43/segments/1508187824675.67/warc/CC-MAIN-20171021081004-20171021101004-00640.warc.gz"}
http://mathhelpforum.com/differential-equations/246285-spring-mass-help.html	1. ## Spring-Mass help Need help solving this spring mass system. Any help is appreciated. Attached Thumbnails 2. ## Re: Spring-Mass help Looks like an interesting problem. But to "help" we really need to know what kind of help you need. Do you know what a "k value of 8 lbs/ft" means? Do you know what an "equilibrium position" is? The basic physics law here is "force equals mass times acceleration". 3. ## Re: Spring-Mass help I'm not sure how to go about the problem at all. What formula to use, what to plug into where...it's all confusing to me. A step-by-step lay out of what to do would be very helpful. 4. ## Re: Spring-Mass help First fix units (unless you choose to carry units along with all your variables): Distances are feet. Forces are pounds (lbs). Time is seconds. Remember that a person's "weight" is not the same as a person's mass. Weight = mass times g, where g is the acceleration due to gravity at Earth's surface. Second, understand the coordinate system (which isn't the one I would've chosen, but the problem specifies it, so you should go with it). The bridge is at x = -100. The non-bridge end of the free-hanging (no person attached) chord is at x = 0. The positive x direction is down. The water is at x = 120. The force of gravity is in the positive x-direction. The spring force from the chord is in the negative x-direction. The forces acting on a jumper are: gravity alone when -100 <= x <= 0, and both gravity and the chord when x >= 0. That "k value" means the spring constant for use in Hook's Law. For Question #2, I'm not entirely sure of the meaning, but I my best guess is that it means this: By equilibrium position it means the value of x where, eventually, the bungee jumper would hang after the oscillation died down... it would be some positive x-value. It's the point where there's no net force on the jumper. By "velocity at the equilibrium position" it means, I think, the instantaneous velocity at the moment when the jumper first reaches that equilibrium x value on their downward fall (and since it will be positive, the jumper will have a lot of downward velocity at that moment, and so is continuing down post it towards the water). 5. ## Re: Spring-Mass help equilibrium position is where $F_{net} = 0 \implies kx = W$ 6. ## Re: Spring-Mass help Appreciate the help but still lost on what equation I should plug all of this information in to. 7. ## Re: Spring-Mass help solving this problem involves more than one equation ... in addition to the equation I provided for finding the position of equilibrium you will need, at a minimum, one formula from kinematics with uniform acceleration; formulas for mechanical energy (kinetic energy, elastic potential energy, and gravitational potential energy) in order to use conservation of energy principles; and equations for motion of a spring oscillator. 8. ## Re: Spring-Mass help What equation to find how long she will be in free fall? 9. ## Re: Spring-Mass help Originally Posted by afeldler What equation to find how long she will be in free fall? one of these ... $\displaystyle v_f = v_0 - gt$ $\displaystyle \Delta y = v_0 t - \frac{1}{2}gt^2$ $\displaystyle \Delta y = \frac{1}{2}(v_0 + v_f)t$	{"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": 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": 0.8337050676345825, "perplexity": 971.5803834352981}, "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/1566027314959.58/warc/CC-MAIN-20190819201207-20190819223207-00210.warc.gz"}
http://physics.stackexchange.com/questions/114958/why-is-the-space-time-interval-squared	# Why is the space-time interval squared? The space-time interval equation is this: $$\Delta s^2=\Delta x^2+\Delta y^2+\Delta z^2-(c\Delta t)^2$$ Where, $\Delta x, \Delta y, \Delta z$ and $\Delta t$ represent the distances along various coordinates according to an observer, and $\Delta s$ is the space-time interval. All observers agree on the space-time interval, it is constant. My question is why is it squared? If we had in equation like this: $$\Delta s'=\Delta x^2+\Delta y^2+\Delta z^2-(c\Delta t)^2$$ $\Delta s'$ would be constant as well. It would also never be imaginary. It would have units of $[length]^2$ instead of $[length]$ though. Is there a theoretical or practical reason that we define the space-time interval based on squaring, or is it just to make it look similar to Pythagoras' theorem/give it simpler units or something else entirely? - Well it's a generalized Pythagorean theorem, it's natural to define it like that, as in any (pseudo)metric space: it's a "distance" in the correct units. However, if you want to define the square as another quantity, you are welcome to use that notation. – orion May 30 '14 at 20:41 You are correct when you point out that any function of $\Delta x^2 + \Delta y^2 + \Delta z^2 - \Delta t^2$ will be constant and agreed on by all observers. So we could define $\Delta s$ to be its cosine...if all we were interested in was getting an invariant. You are also right when you point out the dimensional issue. Measure time in light-centimeters, and distance along the x,y,z axes in centimetres. Then length is measured in centimetres, and so is time.... Then the right hand side has units cm$^2$, and hence, so does the left hand side. Using cosine or other, similar functions like the identity function you suggest, would produce a quantity that did not even have the units of length (and so, could not be proper time). Now, definitions are arbitrary, so you could define Ps to be equal to $\Delta x^2 + \Delta y^2 + \Delta z^2 - \Delta t^2$ if you want, and you could give it any name you want. But would you be able to express the fundamental laws of Physics in terms of that quantity? It is a requirement of the principle of relativity that it be an invariant, and either Ps or cos(Ps) would satisfy that, but it is desirable that it make life easy for us in our formulas, since doing Physics is already hard enough. There are important reasons why we want to use the square root function instead of cosine or instead of the identity function which one of the other answers insists on. There is more to it than just to make it look like Pythagoras' Theorem or make it look like pre-relativistic physics. These reasons do not become apparent until you get to General Relativity, or at least to Differential Geometry. This is your question, rephrased: Why do we want to study an invariant quantity with dimensions of length? (Which is the same as time). The answer is that we want to be able to define $s$, the proper time, or, as I am expressing it, "the length of a path". It will be given by a line integral $s = \int ds$ along the path, and will be invariant for all observers. To an observer who is moving along that path, it will appear to be the elapsed time. Now it is quite basic that if first 2 cm of time elapse, and then another 3, the total elapsed time is 5 cm. So we need an additive quantity. Neither cosine nor Ps are additive, as simple examples show, but if we define $ds^2 = dx^2+ dz^2+dy^2-dt^2$, then it will be additive, by the higher-dimensional non-Euclidean analogue to Pythagoras' Theorem. that is why the squaring occurs, and it is indeed squaring a quantity $ds$, and when finite intervals are involved along straight lines, it is indeed the square of a quantity $\Delta s$ defined as $$\Delta s = \sqrt {\Delta x^2 + \Delta y^2 + \Delta z^2 - \Delta t^2}.$$ SHORT ANSWER We square $\Delta s$ so that we get an additive quantity along world-lines. - the advanced answer is, you want it to be a tensor, so it has to be linear, so you have to take the square root of those squares for it to be linear. – joseph f. johnson May 30 '14 at 3:40 I thought it would have to do with general relativity. – PyRulez May 30 '14 at 10:55 Did Einstein know all this when he was defining the interval, that it would be necessary? Or did he have hunch that happened to work out? – PyRulez May 31 '14 at 23:40 I suppose he wanted a space-time interval to be measured in dimensions of length: that is simple physical intuition. Later, special relativity was taught with differential geometry concepts because, in fact, Einstein was thinking about General Relativity from the very start of relativity, even during special relativity. And a tensor has to be linear, so it has to have dimensions of length, not length squared. – joseph f. johnson Jun 1 '14 at 2:25 Well, length squared could of turned out to be additive, but it didn't luckily. – PyRulez Jun 1 '14 at 11:47 and Δs is the space-time interval. Actually, many (most?) will say that the spacetime interval is $\Delta s^2$. In other words, $\Delta s^2$ is not the squared interval; it is the symbol for the interval. Since this has been questioned in a comment, I provide some references below: Here is the definition of the spacetime-interval. Suppose, as measured by a certain experimenter, two events are separated by a time $t$ and a spatial distance $x$. Then in terms of these numbers the spacetime-interval between the two events is the quantity $$s^2=x^2-c^2 t^2.\tag{17.1}$$ Notice that this is written as the square of a number $s$. The pacetime-interval is the quantity $s^2$, not $s$. In fact, we will not often deal with $s$ itself. The reason is that $s^2$ is not always positive, unlike distance in space. If $ct$ is larger than $x$ in Equation 17.1 then $s^2$ will be negative. In order to avoid taking the square-root of a negative number, physicists usually just calculate $s^2$ and leave it at that. You should just regard $s^2$ as a single symbol, rather than as the square of something. What immediate information does the spacetime interval gives us? If the spacetime interval between events A and B is negative, then either $t_1$ or $t_2$ is negative. It follows that events A and B are timelike related, as illustrated in figure 12$a$. In this case it is possible for an inertial observer to be present at both events A and B. The elapsed time such an observer would measure between A and B is simply the square root of minus the spacetime interval, $\Delta t=\sqrt{-\text(interval)}$. Also, from Space-time intervals: The interval is defined by $$\Delta s^2 = \Delta x^2+\Delta y^2+\Delta z^2-(c\Delta t)^2$$ Note that the symbol $\Delta s^2$ is usually taken as a fundamental quantity and not the square of some other quantity $\Delta s$. And Sean Carroll writes in "Lecture Notes on General Relativity": The interval is defined to be $s^2$, not the square root of this quantity. Is than an theoretical or practical reason that we define the space-time interval based on squaring Theoretically, the interval is the Minkowski dot (inner) product of a displacement four-vector with itself $$\Delta s^2 = x^{\mu}x_{\mu}$$ which is invariant under the Lorentz transformation. This is analogous to the length squared of the 3-D displacement vector $$l^2 = \mathbf x \cdot \mathbf x$$ However, the Minkowski inner product is not positive definite; the inner product can be positive or negative. Practically, the sign of the interval determines whether the four-displacement is time-like or space-like (the interval is light-like if the interval is zero). If the interval is time-like then the proper time is $$\tau = \sqrt{\frac{\|\Delta s^2\|}{c^2}}$$ If the interval is space-like, the proper distance is $$\sigma = \sqrt{\|\Delta s^2\|}$$ - So it is to look like Pythagoras' theorem, and they get to avoid imaginary numbers. Tricky. – PyRulez May 30 '14 at 0:11 @PyRulez, who's they? – Alfred Centauri May 30 '14 at 0:46 One can easily quote a mixed-up textbook to support your point, but you are quite wrong. In relativity, dimensions of length and dimensions of time are the same. Therefore, if the lengths are squared on one side, they have to be length squared on the other, too. And, it is so (delta x)^2, as you can see when you try integrating along a geodesic to get s instead of ds. – joseph f. johnson May 30 '14 at 1:18 @AlfredCentauri I suppose Einstein and the text book companies. CONSPIRACY jk. – PyRulez May 30 '14 at 10:58 @josephf.johnson, I've added additional references that show that you are quite wrong; my statement that "many will say that the spacetime interval is $\Delta s^2$" is a fact. Whether you agree with those authors or consider them "mixed up" not does not make my opening statement incorrect. Indeed, it appears likely that you don't quite understand what the authors are saying given your 2nd sentence above. – Alfred Centauri May 30 '14 at 12:00 or is it just to make it look similar to Pythagoras' theorem ...? If you take a look at Einstein's "Relativity: The Special and General Theory", you will see in the Appendix I (just before equation (10)) that in order to derive the interval equation, Einstein actually did begin with the Pythagorean Theorem in 3D, which he put like this: $$r = \sqrt{x^2 + y^2 + z^2} = ct$$ This way he showed the vector of light traveling in a three dimensional space. He then transformed the equation in a number of ways, but the Pythagorean Theorem was the source of the whole equation. (And I got the downvote because ... I reminded the history? Well, I guess it does not pay to study the sources ...) EDIT: PyRulez commented below that: "This doesn't really explain why the space-time interval was squared (only the distances need to be)". Well, $x$ ($\Delta x$) is a distance, $y$ is a distance, $z$ is a distance and $ct$ - as I showed above (or what simply follows from the fact that it is velocity multiplied by time) - is also a distance. Now, what do you call the result of adding and subtracting distances (squared)? Einstein called it a "line element" or "linear element" and he wrote in "The Foundation of the General Theory of Relativity" (p. 119): "The magnitude of the linear element pertaining to points of the four-dimensional continuum in infinite proximity, we call ds". If we have a continuum and we add/subtract what we call distances in this continuum, then we must obtain a distance as a result. - The shortest euclidean distance between three points namely 1,2,3 follows that $dist(1,3)=dist(1,2)+dist(2,3)$. where $dist(x,y)$ is the vector between points x and y. Now we from daily experience know that space in itself without time is euclidean. Now this linearity relation should be carried over to space-time. Why ? Because if there are three simultaneous events for an observer , then their space-time distance must equal the euclidean distance and thus will follow the linearity condition. So we expect that The space-time interval between any events a,b and c must also follow the relation $dist^* (a,c)=dist^* (a,b)+dist^* (b,c)$ where $dist^*$ is the space-time interval distance vector. which will be followed only if the unit of space interval is length and not length$^2$. This linearity relation also makes the math easier and lets us do things in special relativity similar to pre-relativity days like defining velocity,kinetic energy,momentum in a similar way newton did and they follow the same kind of vector/scalar addition respectively the way they did in pre-relativity days. Now if you still did everything the same way like defining momentum to be $p=$ $m$ x (new metric) $/$ proper time and energy to be having units $p^2/2m$. Where new metric denotes the metric to be $\Delta s'$ in the question. You won't be having the relations like energy conservation,momentum conservation to hold true in the same mathematical form they used to do earlier in pre-relativity mechanics. So either treat minkowski metric to be of the dimensions $length$ or completely change the way you defined momentum,energy and everything before relativity so that your theory remains consistent with the universe. The latter seems a very daunting task to do than the former. To summarise : . Our equations retain their old pre-relativity mathematical form is the core reason why we take space-time interval to be of the units of length. Also our distance still is a vector quantity. - @PyRulez I have edited the answer. – Iota May 29 '14 at 21:32 -1; The premise of the answer is false: the correct statement is the triangle inequality, the rest does not seem to answer the question. – Robin Ekman May 30 '14 at 0:38 The answer is poorly organised, but basically right on target. To have units of length, it has to be s^2, not s. To be additive along geodesics, it also has to be s^2, not s. – joseph f. johnson May 30 '14 at 1:15 Triangle inequality is one more condition that is there along with linearity. – Iota Jun 28 '14 at 8:13	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9252429008483887, "perplexity": 384.02695147230645}, "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/1464049278385.58/warc/CC-MAIN-20160524002118-00087-ip-10-185-217-139.ec2.internal.warc.gz"}
http://physics.stackexchange.com/questions/15467/snells-law-of-refraction	# Snell's Law of Refraction I was told that "Snell's law of refraction implies that a light ray in an isotropic medium travels from point a to point b in stationary time." Why is this true? Thanks. - Let's answer this in reverse. We'll find a path of stationary time and show it satisfies Snell's law. The classic analogy and derivation for this problem are in the Feynman Lectures. I'll basically be reproducing that argument. Feynman imagines the material with a lower index of refraction (faster speed) as a beach, and high index of refraction as the ocean. Point A is on the beach and point B is in the ocean. You're trying to go from one to the other in the fastest possible time. The fastest possible time is also a stationary time, simply because minimums are stationary points. There are many paths you can take from point A to point B. Here are three examples: We want to find the place where nearby paths take the same amount of time. If we have two nearby paths, one spends more time on the beach and the other spends more time in the ocean. Let's draw two nearby paths, then draw circles with their centers at A and B. The circles indicate points that are the same distance away, so the little extra bits beyond the circles indicate the extra distance traveled on the beach and in the ocean. The purple segments are the extra distances traveled on the beach and in the ocean. We don't want them to be the same distance. Instead, we want them to take the same amount of time to travel. That way the extra time spent on the beach cancels the extra time spent in the ocean. Notice that we can solve this by zooming in closely on the region with the purple lines. When we zoom in very close and make the purple lines right next to each other, the green circles can be approximated by their tangent lines. The purple line on top is then perpendicular to the top green circle's tangent and the purple line on bottom is perpendicular to the bottom circle's tangent. (This follows because radii of circles are perpendicular to tangents.) So we have two small right triangles, one on the beach and one in the water. These two triangles have the same hypotenuse, that little horizontal segment between them. Call it $h$. Next we draw a line perpendicular to the beach/ocean interface. The angle that the path on land make with this line is $\theta_1$ and the angle that the path in the water makes is $\theta_2$. (It doesn't matter which of the two paths you use to measure this angle; in the limit where the two paths are very close, the angles are the same.) Then, by trigonometry, we see that the extra distance traveled on the beach is $h\sin\theta_1$ and the extra distance traveled in the water is $h\sin\theta_2$. The time is takes to travel a distance is proportional to the distance multiplied by the index of refraction, so setting these times equal to each other and canceling out the common $h$, we get $$n_1\sin\theta_1 = n_2 \sin\theta_2$$ - Thanks, Mark. Love the illustrations! :-) – Light Oct 7 '11 at 9:27 What is "stationary" in this context? – Georg Oct 7 '11 at 12:14 Really excellent answer. I've never seen stationary time explained in quite that way. – Colin K Oct 7 '11 at 16:55 @Georg It means that nearby paths take the same amount of time. en.wikipedia.org/wiki/Fermat's_principle – Mark Eichenlaub Oct 7 '11 at 16:56 @Colin Thanks - although credit goes to Feynman (or whoever he got the argument from)! – Mark Eichenlaub Oct 7 '11 at 16:57	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 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": 0.8016082644462585, "perplexity": 346.42730881288145}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-10/segments/1394011250185/warc/CC-MAIN-20140305092050-00089-ip-10-183-142-35.ec2.internal.warc.gz"}
http://clay6.com/qa/4552/evaluate-begin-cos-15-sin-15-sin-75-cos-75-end-	Browse Questions # Evaluate : $\begin{vmatrix} cos 15^{\circ} & sin 15^{\circ} \\ sin 75^{\circ} & cos 75^{\circ} \end{vmatrix}$ Toolbox: • The determinant of a square matrix of order $2 \times 2$ is $\|A\|=a_{11} \times a_{22} - a_{12} \times a_{21}$ • $\cos (A+B) =\cos A \cos B- \sin A \sin B$ • $\cos 90^{\circ}=0$ Let $\Delta=\begin{vmatrix} cos 15^{\circ} & sin 15^{\circ} \\ sin 75^{\circ} & cos 75^{\circ} \end{vmatrix}$ on expanding we get, $\Delta =\cos 15^{\circ} \cos 75^{\circ}-\sin 15^{\circ}\sin 75^{\circ}$ This is of the form $\cos (A+B)=\cos A \cos B-\sin A\sin B$ Therefore $\cos 15^{\circ} \cos 75^{\circ}-\sin 15^{\circ}\sin 75^{\circ}=\cos (15+75)$ $=\cos 90^{\circ}$ But $\cos 90^{\circ}=0$ Therefore $\Delta=0$	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 3, "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": 0.9984795451164246, "perplexity": 974.2778082651412}, "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/1492917122619.71/warc/CC-MAIN-20170423031202-00564-ip-10-145-167-34.ec2.internal.warc.gz"}
https://irmar.univ-rennes1.fr/seminaire/cryptographie/leonardo-colo	# Supersingular isogeny Diffie-Hellman Supersingular isogeny graphs have been used in the Charles–Goren–Lauter cryptographic hash function and the supersingular isogeny Diffie–Hellman (SIDH) protocole of De\,Feo and Jao. A recently proposed alternative to SIDH is the commutative supersingular isogeny Diffie–Hellman (CSIDH) protocole, in which the isogeny graph is first restricted to $\FF_p$-rational curves $E$ and $\FF_p$-rational isogenies then oriented by the quadratic subring $\ZZ[\pi] \subset \End(E)$ generated by the Frobenius endomorphism $\pi: E \rightarrow E$. We introduce a general notion of orienting supersingular elliptic curves and their isogenies, and use this as the basis to construct a general oriented supersingular isogeny Diffie-Hellman (OSIDH) protocole. By imposing the data of an orientation by an imaginary quadratic ring $\OO$, we obtain an augmented category of supersingular curves on which the class group $\Cl(\OO)$ acts faithfully and transitively. This idea is already implicit in the CSIDH protocol, in which supersingular curves over $\FF_p$ are oriented by the Frobenius subring $\ZZ[\pi] \simeq \ZZ[\sqrt{-p}]$. In contrast we consider an elliptic curve $E_0$ oriented by a CM order $\OO_K$ of class number one. To obtain a nontrivial group action, we consider $\ell$-isogeny chains, on which the class group of an order $\OO$ of large index $\ell^n$ in $\OO_K$ acts, a structure we call a whirlpool. The map from $\ell$-isogeny chains to its terminus forgets the structure of the orientation, and the original base curve $E_0$, giving rise to a generic supersingular elliptic curve. Within this general framework we define a new oriented supersingular isogeny Diffie-Hellman (OSIDH) protocol, which has fewer restrictions on the proportion of supersingular curves covered and on the torsion group structure of the underlying curves. Moreover, the group action can be carried out effectively solely on the sequences of moduli points (such as $j$-invariants) on a modular curve, thereby avoiding expensive isogeny computations, and is further amenable to speedup by precomputations of endomorphisms on the base curve $E_0$.	{"extraction_info": {"found_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": 0.9326203465461731, "perplexity": 729.4008016914509}, "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-2021-43/segments/1634323588246.79/warc/CC-MAIN-20211028003812-20211028033812-00615.warc.gz"}
http://mathhelpforum.com/calculus/159397-sketching-surfaces-quick-easy-question.html	# Thread: Sketching Surfaces (Quick and Easy Question) 1. ## Sketching Surfaces (Quick and Easy Question) Why is that, $x= \sqrt{y^2 + z^2}$ Is a single cone while, $x^2 = 9y^2 + z^2$ consists of 2 cones, tip to tip? I know that the second surface consists of elliptical cones, but I don't see how that could make any difference. How come for, $x= \sqrt{y^2 + z^2}$ my solutions manual only shows one cone? Couldn't I rewrite this as, $x^{2} = y^{2} + z^{2}$ Certainly there are negative values of x that will satisfy this equation as well, right? 2. right (0,0,0) is the vertex of the cone. And {x<0} is another branch of the second cone. While for the first one only the {x>0} branch is allowed. 3. Originally Posted by xxp9 right (0,0,0) is the vertex of the cone. And {x<0} is another branch of the second cone. While for the first one only the {x>0} branch is allowed. I can see that for the first one only x>0 is allowed, but why? 4. Originally Posted by jegues I can see that for the first one only x>0 is allowed, but why? Becaues for real numbers, all functions are single valued. $\sqrt{a}$ is defined as the non-negative number x such that $x^2= a$. $\sqrt{4}= 2$, not -2. Square roots are always non-negative. 5. Originally Posted by HallsofIvy Becaues for real numbers, all functions are single valued. $\sqrt{a}$ is defined as the non-negative number x such that $x^2= a$. $\sqrt{4}= 2$, not -2. Square roots are always non-negative. In which case or scenario do we end up seeing the plus or minus case then?	{"extraction_info": {"found_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": 10, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8919681906700134, "perplexity": 1091.391005380429}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-50/segments/1480698542939.6/warc/CC-MAIN-20161202170902-00343-ip-10-31-129-80.ec2.internal.warc.gz"}
https://goldbook.iupac.org/terms/view/C01197	## compensation effect https://doi.org/10.1351/goldbook.C01197 In a considerable number of cases plots of $$T\ \Delta ^{\ddagger }S$$ vs. $$\Delta ^{\ddagger}H$$, for a series of reactions, e.g. for a reaction in a range of different solvents, are straight lines of approximately unit slope. Therefore, the terms $$\Delta ^{\ddagger}H$$ and $$T\ \Delta ^{\ddagger }S$$ in the expression partially compensate, and $\Delta ^{\ddagger}G = \Delta ^{\ddagger}H - T\ \Delta ^{\ddagger}S$ often is a much simpler function of solvent (or other) variation than $$\Delta ^{\ddagger}H$$ or $$T\ \Delta ^{\ddagger }S$$ separately.	{"extraction_info": {"found_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": 0.8204837441444397, "perplexity": 565.2094637479569}, "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/1679296948951.4/warc/CC-MAIN-20230329054547-20230329084547-00389.warc.gz"}
http://mathhelpforum.com/algebra/199567-rational-exponents-solve-please.html	# Math Help - rational exponents? Solve please 1. ## rational exponents? Solve please (-3a1/4)(9a)-3/2 2. ## Re: rational exponents? Solve please Originally Posted by zbest1966 (-3a1/4)(9a)-3/2 Here are some properties you should know: $a^ma^n = a^{m+n}$ $(ab)^n = a^nb^n$ $a^{m/n} = \sqrt[n]{a^m}$ 3. ## Re: rational exponents? Solve please I understand the properties but how do you solve it. 4. ## Re: rational exponents? Solve please Originally Posted by zbest1966 I understand the properties but how do you solve it. If you understood the properties, I suspect you would not be having difficulties with the problem. By the way, we have nothing to "solve" here, this is not an equation. Rather, we are "simplifying" the expression. $\left(-3a^{1/4}\right)(9a)^{-3/2}$ $=-3a^{1/4}\cdot9^{-3/2}a^{-3/2}$ $=-\frac{3a^{1/4}}{9^{3/2}a^{3/2}}$ $=-\frac{3a^{1/4}}{27a^{3/2}}$ Now, use the properties $\frac{a^m}{a^n} = a^{m-n}$ and $a^{-n} = \frac1{a^n}$ to finish it: $=-\frac{a^{-5/4}}{9}=-\frac1{9a^{5/4}}$ As a final step, you may want to convert $a^{5/4}$ to radical form, depending on the problem's instructions. 5. ## Re: rational exponents? Solve please Originally Posted by zbest1966 I understand the properties but how do you solve it. You could actually apply the properties you say you understand or you could ask someone else to solve it for you. I see you chose the second solution method.	{"extraction_info": {"found_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": 11, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9295567274093628, "perplexity": 1373.6791951447674}, "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-2016-07/segments/1454701151880.99/warc/CC-MAIN-20160205193911-00273-ip-10-236-182-209.ec2.internal.warc.gz"}
https://chemicalstatistician.wordpress.com/tag/energy/	## Physical Chemistry Lesson of the Day – Intensive vs. Extensive Properties An extensive property is a property that depends on the size of the system. Examples include An intensive property is a property that does not depend on the size of the system. Examples include As you can see, some intensive properties can be derived from extensive properties by dividing an extensive property by the mass, volume, or number of moles of the system. ## Physical Chemistry Lesson of the Day – The First Law of Thermodynamics The change in internal energy of a system is defined to be the internal energy of a system in its final state subtracted by the internal energy of the system in its initial state. $\Delta U = U_{final} - U_{initial}$. However, since we cannot measure the internal energy of a system directly at any point in time, how can we calculate the change in internal energy? The First Law of Thermodynamics states that any change in the internal energy of a system is equal to the heat absorbed the system plus any work done on the system. Mathematically, $\Delta U = q + w$. Recall that I am using the sign convention in chemistry. The value of $q$ and $w$ can be positive or negative. • A negative $q$ denotes heat released by the system. • A negative $w$ denotes work done by the system.	{"extraction_info": {"found_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": 6, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8799349665641785, "perplexity": 188.14890344685003}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-30/segments/1563195529007.88/warc/CC-MAIN-20190723064353-20190723090353-00377.warc.gz"}
http://math.stackexchange.com/questions/86886/chernoff-hoeffding-like-bound-for-kl-divergence-of-sampled-distribution-and-true	# Chernoff-Hoeffding-like bound for KL divergence of sampled distribution and true distribution Let $X_1, X_2, \ldots$ be i.i.d. Bernoulli random variables with mean $\mu$, and $\hat{\mu}_n = \frac{1}{n}\sum_i X_i$. Then the Chernoff-Hoeffding bound says that for all $c > 0$, $$P\left(\hat{\mu}_n - \mu > c\right) \leq e^{-2 n c^2}.$$ Now, is there any similar bound for $KL(\hat{\mu}_n\|\|\mu)$ instead of $(\hat{\mu}_n - \mu)$? [NOTE: $KL(\hat{\mu}_n\|\|\mu)$ means the KL-divergence between Bernoulli($\hat{\mu}_n$) and Bernoulli($\mu$).] One thing I know is that for Bernoulli, $KL(p\|\|q) \geq 2(p-q)^2$, and hence, $$P\left(\hat{\mu}_n - \mu > c\right) \leq P\left(KL(\hat{\mu}_n \|\| \mu) > 2c^2\right),$$ but it doesn't help. Thanks. - What do you mean by $KL(\hat \mu \\| \mu)$? Note that $\mu$ is a number. – Srivatsan Nov 30 '11 at 2:47 I edited the question to clarify it: $KL(\hat{\mu}_n\|\|\mu)$ means the KL-divergence between Bernoulli($\hat{\mu}_n$) and Bernoulli($\mu$) – cresmoon Nov 30 '11 at 11:33 I don't follow the question completely; in particular, the expression $\newcommand{\KL}{\mathrm{KL}}$ $\KL(\hat \mu \\| \mu)$ does not make sense to me because $\mu$ is a real number rather than a distribution. (Edit: The question is now edited.) Nevertheless I will mention that Sanov's theorem seems to be related to what the OP has in mind. Suppose that the true distribution is $q$ and the empirical distribution is $\hat p$. Sanov's theorem allows us to measure the KL divergence between $\hat p$ and $q$; in fact, it does a bit more, as we will see. (I am quoting the theorem from the wikipedia page.) Sanov's theorem. Let $A$ be a set of probability distributions over an alphabet $X$, and let $q$ be an arbitrary distribution over $X$ (where $q$ may or may not be in $A$). Suppose $x_1, x_2, \ldots, x_n$ are $n$ i.i.d. samples from $q$. Then the probability that the empirical distribution of the sample lies inside $A$ is at most $(n+1)^{\|X\|} 2^{-n \delta}$, where $$\delta := \inf_{p \in A} \ \KL(p \\| q) .$$ The OP's question then follows as a Corollary. If $\hat p$ is the empirical distribution, then $$\Pr[\KL(\hat p \\| q) \geqslant \delta] \leqslant (n+1)^{\|X\|} 2^{- n \delta}.$$ Proof. Apply Sanov's theorem with $A$ defined as the set of distributions $p$ such that $\KL(p \\| q) \geqslant \delta$. - Thanks. It seems to be what I'm looking for. – cresmoon Nov 30 '11 at 11:37	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 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": 0.9937741756439209, "perplexity": 145.42265725209597}, "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/1393999671637/warc/CC-MAIN-20140305060751-00066-ip-10-183-142-35.ec2.internal.warc.gz"}
http://www.komal.hu/verseny/feladat.cgi?a=feladat&f=P4752&l=en	Mathematical and Physical Journal for High Schools Issued by the MATFUND Foundation # Problem P. 4752. (September 2015) P. 4752. The string of a bow is pulled by a force of 200 N, such that the arrow is moved backwards by a distance of 50 cm. The applied force is proportional to the displacement of the midpoint of the string. With this bow an arrow of mass 40 g is shot vertically upwards. To what maximum height can the arrow fly if 40% of the elastic energy is used? (4 pont) Deadline expired on October 12, 2015. ### Statistics: 188 students sent a solution. 4 points: 155 students. 3 points: 18 students. 2 points: 10 students. 1 point: 3 students. 0 point: 1 student. Unfair, not evaluated: 1 solution. Problems in Physics of KöMaL, September 2015	{"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": 0.9184837341308594, "perplexity": 1209.7874474271262}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-09/segments/1518891812938.85/warc/CC-MAIN-20180220110011-20180220130011-00410.warc.gz"}
https://mathoverflow.net/questions/314648/if-pi-and-sigma-agree-at-almost-all-places-then-the-central-character-of	# If $\Pi$ and $\Sigma$ agree at almost all places, then the central character of $\Pi$ corresponds to $\operatorname{Det} \Sigma$ Let $$\Sigma$$ be an $$n$$-dimensional representation of the global Weil group $$W_F$$ for a number field $$F$$, and $$\Pi$$ an automorphic representation of $$\operatorname{GL}_n(\mathbb A_F)$$. Suppose that at every place $$v$$ of $$F$$ where $$\Sigma_v$$ and $$\Pi_v$$ are both unramified, we have $$\Pi_v \leftrightarrow \Sigma_v$$ under the local Langlands correspondence. In Guy Henniart's proof on the local Langlands correspondence, he claims that $$\operatorname{Det} \Sigma$$ then corresponds to the central character of $$\Pi$$ under the isomorphism $$W_F^{\operatorname{ab}} \cong \mathbb A_F^{\ast}/F^{\ast}$$ of global class field theory. Why should this be the case? It is true that at almost all places $$v$$, $$\operatorname{Det} \Sigma_v$$ identifies with the central character of $$\Pi_v$$ under the isomorphism $$F_v^{\ast} \cong W_{F_v}^{\operatorname{ab}}$$. We have for $$a = (a_v) \in \mathbb A_F^{\ast}$$, $$\varpi_{\Pi}(a) = \prod\limits_v \varpi_{\Pi_v}(a_v)$$ so it seems like something could go wrong with $$\varpi_{\Pi}$$ at the ramified places. Now that I have thought about it a little more, I think this just comes down to the following fact: If $$\chi = \otimes \chi_v$$ is a character of $$\mathbb A_F^{\ast}/F^{\ast}$$, and $$S$$ is a finite set of places of $$F$$, then $$\chi$$ is completely determined by the characters $$\chi_v : v \not\in S$$. This is because if $$\eta = \otimes_v \eta_v$$ is another character of $$\mathbb A_F^{\ast}/F^{\ast}$$ with $$\eta_v = \chi_v$$ for all $$v \not\in S$$, then $$\chi \eta^{-1}$$ defines a character of $$\prod\limits_{v \in S} k_v^{\ast}$$ which is trivial on the diagonal embedding $$\Delta$$ of $$k^{\ast}$$. But $$\Delta$$ is dense in this product by weak approximation, so in fact $$\chi = \eta$$ on $$\mathbb A_F^{\ast}$$. Now $$\operatorname{Det} \circ \Sigma$$ corresponds to a character of $$\mathbb A_F^{\ast}/F^{\ast}$$ which agrees at almost all places with the central character of $$\Pi$$ (also trivial on $$F^{\ast}$$), so they must be the same character.	{"extraction_info": {"found_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": 42, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9774358868598938, "perplexity": 51.167241527688624}, "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/1579250591763.20/warc/CC-MAIN-20200118023429-20200118051429-00437.warc.gz"}
http://mathonline.wikidot.com/the-ring-of-q-2	The Ring of Q(√2) # The Ring of Q(√2) Recall from the Rings page that if $+$ and $$ are binary operations on the set $R$, then $R$ is called a ring under $+$ and $$ denoted $(R, +, )$ when the following are satisfied: • 1. For all $a, b \in R$ we have that $(a + b \in R)$ (Closure under $+$). • 2. For all $a, b, c \in R$, $a + (b + c) = (a + b) + c$ (Associativity of elements in $R$ under $+$). • 3. There exists an $0 \in R$ such that for all $a \in R$ we have that $a + 0 = a$ and $0 + a = a$ (The existence of an identity element $0$ of $R$ under $+$). • 4. For all $a \in R$ there exists a $-a \in R$ such that $a + (-a) = 0$ and $(-a) + a = 0$ (The existence of inverses for each element in $R$ under $+$). • 5. For all $a, b \in R$ we have that $a + b = b + a$ (Commutativity of elements in $R$ under $+$). • 6. For all $a, b \in R$ we have that $a b = b * a$ (Closure under $$). • 7. For all $a, b, c \in R$, $a (b * c) = (a * b) * c$ (Associativity of elements in $R$ under $$). • 8. There exists a $1 \in R$ such that for all $a \in R$ we have that $a 1 = a$ and $1 * a = a$ (The existence of an identity element $1$ of $R$ under $$). • 9. For all $a, b, c \in R$ we have that $a (b + c) = (a * b) + (b * c)$ and $(a + b) * c = (a * c) + (b * c)$ (Distributivity of $$ over $+$). We will now look at the ring of $\mathbb{Q}(\sqrt{2})$. Let $\mathbb{Q}(\sqrt{2})$ be the set of real numbers of the form $a + b\sqrt{2}$ where $a, b \in \mathbb{Q}$. Let $+$ denote standard addition and $$ denote standard multiplication. We will show that then $\mathbb{Q}(\sqrt{2})$ forms a ring with these operations by verifying each of the axioms above. Let $x, y, z \in \mathbb{Q} (\sqrt{2})$ where $x = a + b\sqrt{2}$, $y = c + d \sqrt{2}$ and $z = e + f \sqrt{2}$ for $a, b, c, d, e, f \in \mathbb{Q}$. Clearly the $\mathbb{Q}(\sqrt{2})$ is closed under $+$ since: (1) \begin{align} \quad x + y = (a + b \sqrt{2}) + (c + d \sqrt{2}) = (a + c) + (b + d)\sqrt{2} \in \mathbb{Q} (\sqrt{2}) \end{align} The operation $+$ is also associative since: (2) \begin{align} \quad x + (y + z) = (a + b \sqrt{2}) + [(c + d\sqrt{2}) + (e + f \sqrt{2})] = (a + b\sqrt{2}) + [(c + e) + (d + f)\sqrt{2}] \\ = [(a + c) + (b + d)\sqrt{2}] + (e + f\sqrt{2}) = [(a + b\sqrt{2}) + (c + d\sqrt{2})] + (e + f\sqrt{2}) = (x + y) + z \end{align} The identity element of $+$ is $0 = 0 + 0 \sqrt{2}$ since: (3) \begin{align} \quad x + 0 = (a + b\sqrt{2}) + (0 + 0\sqrt{2}) = (a + 0) + (b + 0)\sqrt{2} = a + b\sqrt{2} = x \end{align} (4) \begin{align} \quad 0 + x = (0 + 0\sqrt{2}) + (a + b\sqrt{2}) = (0 + a) + (0 + b)\sqrt{2} = a + b\sqrt{2} = x \end{align} For each element $x$, the additive inverse of $x$ is $-x = -a -b\sqrt{2}$ since: (5) \begin{align} \quad x + (-x) = (a + b\sqrt{2}) + (-a - b\sqrt{2}) = (a - a) + (b - b)\sqrt{2} = 0 + 0\sqrt{2} = 0 \end{align} (6) \begin{align} \quad (-x) + x = (-a - b \sqrt{2}) + (a + b\sqrt{2}) = (-a + a) + (-b + b)\sqrt{2} = 0 + 0\sqrt{2} = 0 \end{align} The operation $+$ is commutative since: (7) \begin{align} \quad x + y = (a + b\sqrt{2}) + (c + d\sqrt{2}) = (a + c) + (b + d)\sqrt{2} = (c + a) + (d + b)\sqrt{2} = (c + d\sqrt{2}) + (a + b\sqrt{2}) = y + x \end{align} The set $\mathbb{Q}(\sqrt{2})$ is closed under $$ since: (8) \begin{align} \quad x y = (a + b\sqrt{2}) * (c + d\sqrt{2}) = ac + ad\sqrt{2} + bc\sqrt{2} + 2bd = (ac + 2bd) + (ad + bc)\sqrt{2} \in \mathbb{Q}(\sqrt{2}) \end{align} The operation $$ is associative since: (9) \begin{align} \quad x (y * z) = (a + b\sqrt{2}) * [(c + d\sqrt{2}) * (e + f\sqrt{2})] = (a + b\sqrt{2}) * [(ce + 2df) + (cf + de)\sqrt{2}] \\ = (ace + 2adf) + (acf + ade)\sqrt{2} + (bce + 2bdf)\sqrt{2} + 2(bcf +bde) \\ \end{align} And: (10) \begin{align} \quad (x * y) * z = [(a + b\sqrt{2}) * (c + d\sqrt{2})] * (e + f\sqrt{2}) = [(ac + 2bd) + (ad + bc)\sqrt{2}] * (e + f\sqrt{2}) \\ = (ace + 2bde) + (acf + 2bdf)\sqrt{2} + (ade + bce)\sqrt{2} + 2(adf + bcf) \end{align} Comparing the two equations above we see that they are indeed equal. The identity element for $$ is $1 + 0\sqrt{2} \in \mathbb{Q} (\sqrt{2})$. Left distributivity also holds for $$ since: (11) \begin{align} \quad x * (y + z) = (a + b\sqrt{2}) * [(c + d\sqrt{2}) + (e + f\sqrt{2})] \\ = (a + b\sqrt{2}) * [(c + e) + (d + f)\sqrt{2}] = (ac + ae) + (ad + af)\sqrt{2} + (bc + be)\sqrt{2} + 2(bd + bf) \end{align} And: (12) \begin{align} \quad x * y + x * z = (a + b \sqrt{2}) * (c + d \sqrt{2}) + (a + b\sqrt{2}) * (e + f\sqrt{2}) = [(ac +2bd) + (ad + bc)\sqrt{2}] + [(ae + 2bf) + (af + be)\sqrt{2}] \end{align} Comparing the two equations above and we see that they are equal. We can similarly show that right distributivity holds for $$, which is left to the reader. Therefore the set $\mathbb{Q}(\sqrt{2})$ satisfies all of the ring axioms with respect to the operations of $+$ and $$ and so $(\mathbb{Q}(\sqrt{2}), +, *)$ is a ring.	{"extraction_info": {"found_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": 12, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.996463418006897, "perplexity": 659.9563748261689}, "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/1590347419593.76/warc/CC-MAIN-20200601180335-20200601210335-00100.warc.gz"}
http://mathhelpforum.com/trigonometry/9151-correct-me-if-im-wrong-but.html	Thread: Correct me if im wrong.. but 1. Correct me if im wrong.. but The function of $cos \pi/6 =$ the cofunction of $sin \pi/6$ or something similar... (forgot how it was worded) and would something like -2sin(4x)sin(-2x) be the same as 2sin(4x)sin(2x) 2. Originally Posted by 3deltat The function of $cos \pi/6 =$ the cofunction of $sin \pi/6$ or something similar... (forgot how it was worded) and would something like -2sin(4x)sin(-2x) be the same as 2sin(4x)sin(2x) Sine and Cosine are Co-functions. Meaning, $\sin x=\cos \left( \frac{\pi}{2} -x \right)$ In the same way, Tangent and Cotangent are cofunctions. And, Secant and Cosecant are cofunctions. You can use this mnemonic to remember the properties of trigonometric cofunctions. 3. Sine and Cosine are Co-functions. Meaning, In the same way, Tangent and Cotangent are cofunctions. And, Secant and Cosecant are cofunctions. Ah. So the statement "The function of $cos \pi/6$ Would = the cofunction of $sin \pi/6$ ? =) 4. Originally Posted by 3deltat Ah. So the statement "The function of $cos \pi/6$ Would = the cofunction of $sin \pi/6$ ? =) Thus, the by the cofunction identities, $\cos \frac{\pi}{6}=\sin \left( \frac{\pi}{2} - \frac{\pi}{6} \right)=\sin \frac{\pi}{3}$ 5. Originally Posted by 3deltat would something like -2sin(4x)sin(-2x) be the same as 2sin(4x)sin(2x) As sine is an odd function $sin(-x) = -sin(x)$. Thus $-2sin(4x)sin(-2x) = -2sin(4x) \cdot -sin(2x) = 2sin(4x)sin(2x)$.	{"extraction_info": {"found_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": 0.9032608270645142, "perplexity": 3612.7263318652567}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-13/segments/1490218188914.50/warc/CC-MAIN-20170322212948-00331-ip-10-233-31-227.ec2.internal.warc.gz"}
https://socratic.org/questions/5766b0c87c01495abb84a1fc	Chemistry Topics # Question #4a1fc Jun 19, 2016 Celsius scale sets the zero to the temperature of the melting ice and the 100 to the temperature of boiling ice, then it divides the mechanical dilatation of some fluid (a gas or mercury) in 100 parts and each step is one degree. The Kelvin scale uses the same approach to divide the scale but it set up the zero to the minimum temperature that can be reached by a body (called the absolute zero). This was selected under the idea that a minimum zero was possible when all the atoms of an object stop to move. In reality with quantum mechanics we discovered that we cannot have an absolute zero because the particles will move under the influence of the uncertainty principle that does not allow the observer to be sure that a particle has a precise velocity (in our case zero). Because of this, the Kelvin scale was redefined saying that the triple point of water (when it is gas, liquid and solid together) is at 273.16 degrees. Then, because in the Celsius scale the triple point of water is at 0.01 degrees, the absolute zero of the Kelvin scale is at -273.15 Celsius. To simplify: the zero of the Kelvin scale is more or less when the atoms are not moving and it correspond to -273.15 Celsius. The zero of the Celsius is when the ice is melting and it is at 273.15 Kelvin. Except for this shift in the scale, the everything else is the same. ##### Impact of this question 4604 views around the world You can reuse this answer	{"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": 0.8852442502975464, "perplexity": 489.98334857349937}, "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-2019-39/segments/1568514573832.23/warc/CC-MAIN-20190920050858-20190920072858-00143.warc.gz"}
https://www.physicsforums.com/threads/acceleration-of-the-center-of-mass-of-this-cylinder.82126/	Acceleration of the center of mass of this cylinder 1. Jul 13, 2005 sterlinghubbard A 2.81 kg hollow cylinder with inner radius 0.29 m and outer radius 0.5 m rolls without slipping when it is pulled by a horizontal string with a force of 47.7 N, as shown in the diagram below. Its moment of inertia about the center of mass is .5m(r(out)^2 + r(in)^2). What is the accelereation of the cylinder's center of mass? Answer in units of m/s^2. What am I doing wrong? I found the Torque of the hollow cylinder by T = F(r). Then I found the angular acceleration by Torque = Interia * Alpha. Inertia was found using the supplied forumula. After finding the angular acceleration I found the Tangential Acceleration by TangentialAcceleration = radius * AngularAcceleration. What am I doing wrong? 2. Jul 14, 2005 siddharth Isn't the moment of inertia of a hollow cylinder $$\frac{1}{2} M (R_1^2 + R_2^2)$$ So, your value of M/2 is not 0.5 but 2.81/2 3. Jul 14, 2005 sterlinghubbard Hence, .5M which is the same as 2.81/2. 4. Jul 14, 2005 siddharth Oh, you mean 0.5 * 2.81 . Didn't see that, sorry. There wil be a torque due to friction, the value of which is not known. So, I don't think you can use the above equations alone to get the answer. Have you applied Newton's second law in the horizontal direction? (ie, F-f = ma). Then eliminate f using all the equations and solve for a. That should give you the correct answer. Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook Similar Discussions: Acceleration of the center of mass of this cylinder	{"extraction_info": {"found_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": 0.9426729679107666, "perplexity": 707.1451971171626}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-51/segments/1512948523222.39/warc/CC-MAIN-20171213123757-20171213143757-00589.warc.gz"}
https://www.esaral.com/tag/biot-savarts-law/	Magnetic Effect of Electric Current Class 12 Notes \| Introduction Class 9-10, JEE & NEET The electricity and magnetism are linked to each other and it is proved when the electric current passes through the copper wire, it produces a magnetic effect. The electromagnetic effects first time noticed by Hans Christian Oersted. Oersted discovered a magnetic field around a conductor carrying an electric current. The magnetic field is a quantity, which has both magnitude and direction. The direction of a magnetic field is usually taken to be the direction in which, a north pole of the compass needle moves inside it. So here you will get Magnetic Effect of Electric Current Class 12 complete Notes to prepare for Boards as well as for JEE & NEET Exams. Oersted discovered a magnetic field around a conductor carrying an electric current. Other related facts are as follows: (a) A magnet at rest produces a magnetic field around it while an electric charge at rest produces an electric field around it. (b) A current-carrying conductor has a magnetic field and not an electric field around it. On the other hand, a charge moving with a uniform velocity has an electric as well as a magnetic field around it. (c) An electric field cannot be produced without a charge whereas a magnetic field can be produced without a magnet. (d) No poles are produced in a coil carrying current but such a coil shows north and south polarities. (e) All oscillating or an accelerated charge produces E.M. waves also in addition to electric and magnetic fields. ### Unit of Magnetic field UNIT OF $\overrightarrow{\mathrm{B}}:$ MKS weber/metre $^{2},$ SI tesla, CGS maxwell cm’ or gauss. One Tesla $=$ one (weber/m’) $=10^{4}$ (maxwell/cm’) $=10^{4}$ gauss ## Biot-Savart’s Law With the help of experimental results, Biot and Savart arrived at a mathematical expression that gives the magnetic field at some point in space in terms of the current that produces the field. That expression is based on the following experimental observations for the magnetic field $\overrightarrow{\mathrm{d} B}$ at a point $P$ associated with a length element $\overrightarrow{\mathrm{d} \ell}$ of a wire carrying a steady current I. $\mu_{0}$ is called permeability of free space $\frac{\mu_{0}}{4 \pi}=10^{-7}$ henry/meter. $1(\mathrm{H} / \mathrm{m})=1 \frac{\mathrm{T} \mathrm{m}}{\mathrm{A}}=1 \frac{\mathrm{Wb}}{\mathrm{Am}}=1 \frac{\mathrm{N}}{\mathrm{A}^{2}}=1 \frac{\mathrm{Ns}^{2}}{\mathrm{c}^{2}}$ DIMENSIONS of $\mu_{0}=\left[\mathrm{M}^{\prime} \mathrm{L}^{\prime} \mathrm{T}^{-2} \mathrm{A}^{-2}\right]$ For vaccum $: \sqrt{\frac{1}{\mu_{0} \varepsilon_{0}}}=\mathrm{c}=3 \times 10^{8} \mathrm{m} / \mathrm{s}$ ### Biot-Savart law in Vector form [Note: Static charge is a source of electric field but not of magnetic field, whereas the moving charge is a source of electric field as well as magnetic field.] the direction of $\mathrm{d} \mathrm{B}$ is perpendicular to the plane determined by $\overrightarrow{\mathrm{d} \ell}$ and $\overrightarrow{\mathrm{r}}$ (i.e. if $\overrightarrow{\mathrm{d} \ell}$ and $\overrightarrow{\mathrm{r}}$ lie in the plane of the paper then $\overrightarrow{\mathrm{dB}}$ is $\perp$ to plane of the paper). In the figure, direction of $\overrightarrow{\mathrm{dB}}$ is into the page. (Use right hand screw 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": 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": 0.8637047410011292, "perplexity": 324.6875910293497}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-21/segments/1620243990929.24/warc/CC-MAIN-20210512131604-20210512161604-00323.warc.gz"}
https://www.physicsforums.com/threads/belief-network-problem.198639/	# Belief network problem 1. Nov 16, 2007 ### sensitive I am having problem solving this exercise. The problem actually comes with a diagram but I do not know and I do not think i can draw it in the forum. The exercise is based on car starting(Heckerman 1995) Since I can't draw the network diagram here but values of probability are given but first let me define all the variables B - Battery G - Gauge F - Fuel T - Turnover S - Start N - No Y - Yes P(B = N) = 0.02 p(F = N) = 0.05 P(G = N\|B = Y, F = Y) = 0.04 P(G = N\|B = Y, F = N) = 0.97 P(G = N\|B = N, F = Y) = 0.10 P(G = N\|B = N, F = N) = 0.99 P(T = N\|B = Y) = 0.03 P(T = N\|B = N) = 0.98 P(S = N\|T = Y, F = Y) = 0.01 P(S = N\|T = Y, F = N) = 0.92 P(S = N\|T = N, F = Y) = 1.0 P(S = N\|T = N, F = N) = 1.0 It was asked to calculate p(F = N\|S = N) Im thinking of Bayesian but I got stuck somewhere so I think it is the wrong approach since S depend on F and T NOT F alone. Im thinking of the other approach and came up with an expression P(F = N\|S = N) = P(S = N\|F = N)/P(F) = P(S = N, B, G, T\|F = N)/P(F) But I am not sure how to compute and put the figures together. Any input/help is appreciated. Thank you 2. Nov 16, 2007 ### EnumaElish How do you get P(F = N\|S = N) = P(S = N\|F = N)/P(F) ? P(F = N\|S = N) = P(S = N & F = N)/P(S = N) and P(S = N\|F = N) = P(S = N & F = N)/P(F = N) so P(S = N & F = N) = P(F = N\|S = N)P(S = N) = P(S = N\|F = N)P(F = N). 3. Nov 16, 2007 ### sensitive So the Bayesian approach was right. I taught I was wrong at the first place because using Bayesian ended up with the following P(S = N\|F = N) P(F = N)/ P(S = N) but from the diagram I have and as you can see from the probabilities, S depend on both F and T and in the expression above we want to know the probability of S = N given that F = N (in other words the probability that the engine will not start given that the fuel tank was empty). 4. Nov 17, 2007 ### sensitive I am still having trouble solving P(S = N\|F = N). plz help.... 5. Nov 19, 2007 ### EnumaElish Below, I assume that the notation (A\|B,C) means (A\|B)\|C = A\|(B\|C), and neither A\|(B & C) nor (A\|B) & C. (If anyone disagrees, please post your opinion.) P(S = N\|F = N) = P(S = N\|T = Y, F = N) P(T = Y) + P(S = N\|T = N, F = N) P(T = N) so you should first derive P(T = Y) and P(T = N). You can derive P(T=N) from: P(B = N) = 0.02 P(T = N\|B = Y) = 0.03 P(T = N\|B = N) = 0.98 using a formula similar to the one in the previous paragraph of this post. Then, P(T=Y) = 1 - P(T=N). Last edited: Nov 19, 2007	{"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": 0.8999514579772949, "perplexity": 2230.6111427207643}, "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-2017-09/segments/1487501170186.50/warc/CC-MAIN-20170219104610-00061-ip-10-171-10-108.ec2.internal.warc.gz"}
https://brilliant.org/problems/shenanigans-2/	# Constant Shenanigans! Geometry Level 4 $\large \sum_{k=1}^{1000} \sqrt{\sqrt[\phi]{(k^{1 + \sqrt{5}})}\left(\dfrac{\left[(1+\sin(k\tau))(1-\sin(k\tau)\right]^{\left \lfloor \pi \right \rfloor}}{\left[\frac{1}{2}(e^{ik\tau} + e^{-ik\tau})\right]^{2\left \lfloor e \right \rfloor}}\right)} = \ ?$ Details and Assumptions: • $$\tau$$ is the almighty circle constant. • $$\pi$$ is the lesser circle constant. • $$\phi$$ is the golden ratio. • $$e$$ is the base of the natural logarithm. • $$i$$ is the imaginary unit. • $$\lfloor x \rfloor$$ is the greatest integer function. ×	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9356080889701843, "perplexity": 3423.7404335686374}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-22/segments/1558232255165.2/warc/CC-MAIN-20190519201521-20190519223521-00070.warc.gz"}
http://math.stackexchange.com/questions/251494/kernel-and-image-of-projection-matrix	# Kernel and Image of projection Matrix This is my HW (Linear algebra 2) and I need to find projection matrix, kernel and image of the projection. V=$R^2$ So I have: subspace of V $sp{(2,-3)}$ Than I found the projection matrix is: $$\begin{matrix} 4/13 & -6/13\\ -6/13 & 9/13 \\ \end{matrix}$$ but now I need to find kernel and image... I don't remember how to do that and I searched and google and I know I saw I need to find Ax=0 So is my kernel is 0?! I don't totally understand this - What does "$sp(2,-3)$" mean? – wj32 Dec 5 '12 at 11:58 I would guess the space is $\mathbb{R}^2$ and $sp(2,-3)$ is the span of the vector $(2,-3)$. But the OP should answer that. – Julian Kuelshammer Dec 5 '12 at 12:01 so is it always depend on the determinant?? if it was non zero? and what about the image? – Mary Dec 5 '12 at 12:19 You can find the kernel by solving the linear equation $Ax=0$. What the image is, is obvious if you recall what the projection map does, i.e. that it projects onto the given subspace. You should include some more information about what you already know. – Julian Kuelshammer Dec 5 '12 at 12:21 What is your definition of projection matrix? How did you come up with this projection matrix? – Julian Kuelshammer Dec 5 '12 at 12:23 The projection you constructed is precisely the orthogonal projection onto the span of $(2,-3)$. So the span of $(2,-3)$ is the image. Being an orthogonal projection, its kernel is the orthogonal of its image, so the kernel is $V^\perp$, which you can write as the span of $(3,2)$. My bad, I didn't read the question carefully when typing. The image can never be all of $\mathbb R^2$, unless your projection is the identity matrix. I've edited the answer. – Martin Argerami Dec 5 '12 at 14:48 Yes.$\ \ \ \ \ \ \$ – Martin Argerami Dec 5 '12 at 15:30	{"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": 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": 0.9314219355583191, "perplexity": 562.3612775534513}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-35/segments/1408500832155.37/warc/CC-MAIN-20140820021352-00147-ip-10-180-136-8.ec2.internal.warc.gz"}
https://www.physicsforums.com/threads/photoelectric-question-3.203984/	Photoelectric question#3 1. Dec 11, 2007 neelakash 1. The problem statement, all variables and given/known data Why is it that even for incident radiation that is monochromatic,photoelectrons are emitted with a spread of velocities? 2. Relevant equations 3. The attempt at a solution Possibly, the incident radiation is partly used up in increasing the thermal energy of the crystal...Now,we claim that all the energy of the absorbed photon is given to an emitted electron.Here we are neglecting the energy taken up by the lattice. I think,for photoelectric effect it is possible to have KE(max)=h(nu)-(phi) when we neglect the energy taken up by the lattice.If we neglect the energy exciting the lattice,we must replace KE(max) by KE. This also explains why the photoelectric measurements are very sensitive to the nature of emitting surface.If it is a conductor,standard photoelectric current is obtained---here the energy is totally used up emitting electrons(I do not know if for suitable conductors,the velocity spread is observed);for non-conductors,current falls to drastically low values.Here,the energy is converted to heat... Last edited: Dec 11, 2007 Can you offer guidance or do you also need help? Similar Discussions: Photoelectric question#3	{"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": 0.9583346843719482, "perplexity": 2014.2595190438908}, "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-2017-43/segments/1508187824225.41/warc/CC-MAIN-20171020135519-20171020155519-00191.warc.gz"}
https://www.physicsforums.com/threads/electron-in-g-orbit.71992/	# Electron in G orbit 1. Apr 18, 2005 ### yogi Does an electron in circular earth orbit emit radiation? 2. Apr 18, 2005 ### Garth A good question, similar to the question: "Does an electron 'sitting' on a laboratory bench emit radiation as by the equivalence principle it is accelerating upwards at an acceleration g?" If the answer to the orbiting scenario is 'yes', then the electron might be expected to spiral inwards to conserve energy; however where does the energy of radiation of an emitting supported electron on a lab bench come from? Garth Last edited: Apr 18, 2005 3. Apr 18, 2005 ### Meir Achuz yogi: Yes. An e in Earth orbit behaves classically because its radial quantum number is so high. It would emit what is called Larmor radiation at the rate: P=2e^2 R^2 \omega^2/3c^2. (This is in Gaussian units.) You can put numbers in to see that this radiated power is tiny. Garth: The EP does not hold for radiation by electric charges. There is another post about that today. 4. Apr 18, 2005 ### Garth Why not? It would seem logical to expect that all physical processes within this universe's space-time would be subject to the the geometric effects and principles of GR. There has been dicsussion on this subject on these Forums some time ago. Garth 5. Apr 19, 2005 ### yogi Seems the references cited in the other thread support the notion that 1) an electron in freefall will radiate, and 2) an observer comoving in the freefalling frame does not see the radiation? So unless there is some identifiable local physics that tell the electron in free fall when and how to radiate - the radiation must be due to a global property. If we place a charged spherical capacitor in orbit we should expect to detect radio frequence signals? - what about a charged parallel plate capacitor? Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook Similar Discussions: Electron in G orbit	{"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": 0.8471104502677917, "perplexity": 1916.1781418245257}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "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/1508187826642.70/warc/CC-MAIN-20171023202120-20171023222120-00854.warc.gz"}
http://tex.stackexchange.com/questions/144490/how-do-i-typeset-a-tenfold-powering-a-tower-of-powers-with-latex	# How do I typeset a tenfold powering (a tower-of-powers) with LaTeX? How do I write x^x^x^x^x^x^x^x^x^x? My LaTeX distribution doesn't wanna eat it, it stops swallowing stuff at the double power x^x^x. What I'm aiming at is essentially building a telescope out of X's stacked on top of each other. Don't ask, that's my sense of humour. - $x^{\scriptscriptstyle x^{x^{x^{x^{x^{x^{x^{x^{x}}}}}}}}}$ – egreg Nov 11 '13 at 21:30 I'm surprised noone has said $$x\uparrow\uparrow10$$ just because. – You Nov 11 '13 at 23:53 @You It would in fact only be $x \uparrow 10$; two \uparrows and you would have x \uparrow 10 exponentiation steps. (But this is almost entirely beside the point. :)) – Sean Allred Nov 12 '13 at 0:59 According to mathworld.wolfram.com/PowerTower.html, you would need two arrows. – Cephalopod Nov 12 '13 at 8:44 LaTeX vs. Viagra. – Nicholas Hamilton Nov 12 '13 at 13:27 \documentclass{article} \def\powertower#1#2{#1\ifnum#2>1 ^{\powertower{#1}{\numexpr#2-1\relax}}\fi} \begin{document} $\powertower{x}{100}$ \end{document} - This is the TeX version of Stairway to heaven. :) Amazing! – Paulo Cereda Nov 11 '13 at 23:45 @PauloCereda: maybe so, but that x at the bottom left is obviously under a lot of stress, and ready to fall over at any moment. – wasteofspace Nov 11 '13 at 23:50 @wasteofspace: I think David is inspired because we were singing The London bridge is falling down in the chatroom yesterday. :) – Paulo Cereda Nov 11 '13 at 23:53 @PauloCereda a minor sin compared to your earlier one of referencing a picture of tower bridge to accompany the lyrics – David Carlisle Nov 12 '13 at 0:11 You could have called this macro \towerofpower in honor of a certain soul/funk band. :-) – Mico Nov 12 '13 at 14:23 \documentclass[12pt]{article} \begin{document} $x^{x^{x^{x^{x^{x^{x^{x^{x^x}}}}}}}}$ \end{document} - I'm curious—how hard would it be to generalize this as a \PowerTower{x}{10} (power tower reference) – Sean Allred Nov 11 '13 at 21:54 @Sean: done. :) – Paulo Cereda Nov 11 '13 at 23:25 @PauloCereda Yay!! – Sean Allred Nov 12 '13 at 1:00 I dedicate this code to Sean. :) Long live expl3! :) \documentclass{article} \usepackage{expl3} \usepackage{xparse} \ExplSyntaxOn \cs_new:Npn \paulo_epicrecursion:nn #1 #2 { #1^{ \int_compare:nTF { #2 > 1 } { \paulo_epicrecursion:nn { #1 } { \int_eval:n { #2 - 1 } } } { #1 } } } \NewDocumentCommand{ \powertower } { m m } { \paulo_epicrecursion:nn { #1 } { #2 } } \ExplSyntaxOff \begin{document} $\powertower{x}{10}$ \end{document} The output: Update: egreg brilliantly noted that the above code generates a long list of nested \int_eval:n calls. The optimization is presented as follows: \documentclass{article} \usepackage{xparse} \ExplSyntaxOn \cs_new:Npn \paulo_epicrecursion:nn #1 #2 { #1^{ \int_compare:nTF { #2 > 1 } { \paulo_epicrecursion:nx { #1 } { \int_eval:n { #2 - 1 } } } { #1 } } } \NewDocumentCommand{ \powertower } { m m } { \paulo_epicrecursion:nn { #1 } { #2 } } \cs_generate_variant:Nn \paulo_epicrecursion:nn { nx } \ExplSyntaxOff \begin{document} $\powertower{x}{10}$ \end{document} The trick here is the generated variant. As egreg explained, the first call doesn't need the full expansion, but the recursive ones do. - +1 for \powertower and epicrecursion. (Edit: Ah, I saw Sean's comment only after writing this one.) – Torbjørn T. Nov 11 '13 at 23:26 @TorbjørnT.: :) – Paulo Cereda Nov 11 '13 at 23:39 superb, absolutely superb :) – cmhughes Nov 11 '13 at 23:40 The command name is wrong: it should be \paulo_epicrecursion:nn because it has two arguments. – egreg Nov 11 '13 at 23:49 @egreg: oopsie, I forgot to update it when I added the second argument. Thanks, fixed. :) – Paulo Cereda Nov 11 '13 at 23:50 Here's a shorter version with LaTeX3 functions: \documentclass{article} \usepackage{xparse} \ExplSyntaxOn \NewDocumentCommand{ \powertower } { m m } { #1 \prg_replicate:nn { #2 - 1 } { ^\c_group_begin_token #1 } \prg_replicate:nn { #2 - 1 } { \c_group_end_token } } \ExplSyntaxOff \begin{document} $\powertower{x}{10}$ \end{document} Limitation: the tower maximum size is 254 because of the maximum nested group level is 255. This limitation on the nested groups is also in the other solutions, of course. Here's a picture of $\powertower{x}{253}$ (I was trying with standalone that adds a level of grouping and forgot to change). With LuaLaTeX the limit is pushed to 499; at 500 the semantic nest size is overflown. - As always, brilliant! :) – Paulo Cereda Nov 12 '13 at 0:20 Very elegant; it's the simpler solutions that seem to be harder to think of, sometimes. :) – Sean Allred Nov 12 '13 at 0:56 What? afraid of going beyond 255? not me! This was obtained as $\epictower{x}{600}$. This reproduces exactly what TeX would have done (were it not for the limitation to 255 group levels) inclusive of a quite odd feature of TeX regarding extra horizontal space on the right. update 2015. I forgot to say in 2013 what I was referring too: the horizontal whitespace extending to the the right of the power towers, on the side of the character glyphs. The image with the 600 x's does not show the resulting (big) extra horizontal space because the figure was clipped to the contained ink. The images below with shorter staircases display using fboxes the extra whitespace, look on right of the rightmost characters the distance to the fbox frame. You can also experiment with David Carlisle's macro inside an fbox (with \fboxsep=0pt or rather -0.4pt). In the next picture, on top left what is produced by \epictower, on the right and bottom the original thing, to demonstrate that horizontal and vertical placement is correct (the font size is set to 60pt and then standard size). And here is the code: \documentclass{article} % use this for 600 x's! % \usepackage [paperheight=65cm,paperwidth=100cm]{geometry} \usepackage {geometry} \pagestyle{empty} \makeatletter \newdimen \epic@x \newdimen \epic@y \newdimen \epic@extrax \newdimen \epic@extray \newbox \epic@one \newbox \epic@two \newcommand{\epictower}[2]{\ifcase #2\relax 1\or #1\or {#1}^{#1}\or {#1}^{{#1}^{#1}}\else \epic@tower {#1}{#2}\fi } \def\epic@tower #1#2{% \sbox\epic@one{\m@th $\scriptscriptstyle #1$}% \sbox\epic@two{\m@th $\scriptscriptstyle {#1}^{#1}$}% \epic@x=\wd\epic@one \epic@extrax=\wd\epic@two \epic@y=\ht\epic@two \epic@extray=\ht\epic@one \toks@ \expandafter{\the\numexpr #2-3}% {#1}^{{#1}^{% \setlength{\unitlength}{1sp}% \begin{picture}({\numexpr (#2-2)\epic@x+(#2-3)\epic@extrax\relax},% {\numexpr (#2-2)\epic@y+\epic@extray\relax}) \count@\z@ \loop \put (\numexpr \count@ \epic@x\relax,% \numexpr \count@ * \epic@y\relax){\copy\epic@one}% \ifnum\the\toks@>\count@ \repeat \end{picture}}}} \makeatother \begin{document}\thispagestyle{empty} \fboxsep-.4pt \fontsize{60}{60} \newcommand{\testtexpowers}[1]{% \setbox 1 \hbox{$\scriptscriptstyle #1$}% \setbox 2 \hbox{$\scriptscriptstyle #1^{\fbox{\copy1}}$}% \setbox 3 \hbox{$\scriptscriptstyle #1^{#1^{\fbox{\copy1}}}$}% \setbox 4 \hbox{$\scriptscriptstyle #1^{#1^{#1^{\fbox{\copy1}}}}$}% \setbox 5 \hbox{$\scriptscriptstyle #1^{#1^{#1^{#1^{\fbox{\copy1}}}}}$}% \setbox 6 \hbox{$\scriptscriptstyle #1^{#1^{#1^{#1^{#1^{\fbox{\copy1}}}}}}$}% \setbox 7 \hbox{$\scriptscriptstyle #1^{#1^{#1^{#1^{#1^{#1^{\fbox{\copy1}}}}}}}$}% \setbox 8 \hbox{$\scriptscriptstyle #1^{#1^{#1^{#1^{#1^{#1^{#1^{\fbox{\copy1}}}}}}}}$}% \setbox 9 \hbox{$\scriptscriptstyle #1^{#1^{#1^{#1^{#1^{#1^{#1^{#1^{\fbox{\copy1}}}}}}}}}$}% \leavevmode\rlap{\rlap{\rlap{\rlap{\rlap{\rlap{\rlap{\rlap{\fbox{\box1}}\fbox{\box2}}\fbox{\box3}}\fbox{\box4}}\fbox{\box5}}\fbox{\box6}}\fbox{\box7}}\fbox{\box8}}\fbox{\box9}} \testtexpowers{x} \testtexpowers{X} \testtexpowers{xX} \clearpage \fbox{$\epictower {x}{1}$}\fbox{$x$}\hrule\fbox{$x$} \hrule \fbox{$\epictower {x}{2}$}\fbox{$x^x$}\hrule\fbox{$x^x$} \hrule \fbox{$\epictower {x}{3}$}\fbox{$x^{x^x}$}\hrule\fbox{$x^{x^x}$} \hrule \fbox{$\epictower {x}{4}$}\fbox{$x^{x^{x^x}}$}\hrule\fbox{$x^{x^{x^x}}$} \hrule \fbox{$\epictower {x}{5}$}\fbox{$x^{x^{x^{x^x}}}$}\hrule\fbox{$x^{x^{x^{x^x}}}$} \hrule \fbox{$\epictower {x}{6}$}\fbox{$x^{x^{x^{x^{x^x}}}}$}\hrule\fbox{$x^{x^{x^{x^{x^x}}}}$} \hrule \clearpage \fbox{$\epictower {X}{1}$}\fbox{$X$}\hrule\fbox{$X$} \hrule \fbox{$\epictower {X}{2}$}\fbox{$X^X$}\hrule\fbox{$X^X$} \hrule \fbox{$\epictower {X}{3}$}\fbox{$X^{X^X}$}\hrule\fbox{$X^{X^X}$} \hrule \fbox{$\epictower {X}{4}$}\fbox{$X^{X^{X^X}}$}\hrule\fbox{$X^{X^{X^X}}$} \hrule \fbox{$\epictower {X}{5}$}\fbox{$X^{X^{X^{X^X}}}$}\hrule\fbox{$X^{X^{X^{X^X}}}$} \hrule \fbox{$\epictower {X}{6}$}\fbox{$X^{X^{X^{X^{X^X}}}}$}\hrule\fbox{$X^{X^{X^{X^{X^X}}}}$} \hrule \end{document} % comment-out the preceding, sets page dimensions in geometry % only limited by TeX's \maxdimen ! $\epictower{x}{600}$ \end{document} - it is easy to modify \epictower to print the exponents along an arbitrary algebraically parametrized curve, for nice special effects. Size of the bounding box will need extra coding if the curve zig-zags; or one may just set the bounding box to the space taken by x^{x^x} for simplicity. – jfbu Nov 12 '13 at 14:23 As more than three powers get very unpleasant to read, I would really recommend using another notation. As already proposed in the comments (hence a community wiki answer), you can represent power towers by the arrow notation: (x\uparrow\uparrow10) You might want to watch Numberphiles YouTube video about Grahams number to understand this notation. This notation is much easier to read. In fact, you can read it at all in contrast to the big tower of powers notation. - The recursion: this notation was introduced by Someone that is already a legend...=). – Andrea L. Nov 14 '13 at 12:50 This is a version of David's answer, but without fancy etex for those being stuck with Knuth TeX. \newcount\powercount \def\powertower#1#2{ #1\powercount=#2 $\powertower{x}{100}$	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9385421872138977, "perplexity": 4171.763115261339}, "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/1464049276415.60/warc/CC-MAIN-20160524002116-00065-ip-10-185-217-139.ec2.internal.warc.gz"}
https://tlu.tarilabs.com/cryptography/building-on-bulletproofs/link.html	# Building on Bulletproofs Cathie Yun Cryptographer ## Summary In this post Cathie explains the basics of the Bulletproofs zero knowledge proof protocol. She then goes further to explain specific applications built on top of Bulletproofs, which can be used for range proofs or constraint system proofs. The constraint system proof protocol was used to build a confidential assets scheme called Cloak and a confidential smart contract language called ZkVM.	{"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": 0.8710153698921204, "perplexity": 3612.9082721872437}, "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-2021-31/segments/1627046150264.90/warc/CC-MAIN-20210724094631-20210724124631-00008.warc.gz"}
https://fearlessmath.net/mod/page/view.php?id=60	## Video and Examples Area is the number unit squares that fit inside a figure. A unit square is a square that is one unit long by one unit wide where the unit of length can be millimeters, inches, feet, miles, etc. To find the area of a rectangle, you need to imagine it being cut into its unit squares, and then count the number of unit squares inside that rectangle. In this 4-by-6 rectangle it is pretty easy to see that the rectangle can be cut into 24 unit squares. If each unit square is 1 cm-by-1 cm, then we say the area of the rectangle is 24 square centimeters. This is often written as 24 cm2. Rather than counting all the unit squares in the 4-by-6 rectangle, a quicker method for finding the number of unit squares in a rectangle is to multiply the length of the base by the height. 6 cm • 4 cm = 24 cm2 What is the area of a rectangle with a base of 5 units and a height of 8 units? As we have already learned, you need to imagine the rectangle cut into its unit square and then counting the unit squares. A shortcut is to multiply the base by height. Area = base • height Area = bh Area =(5 units)(8 units) = 40 2 Area of other shapes Finding the area of other shapes is essentially the same as finding the area of a rectangle. You need to count the number of unit squares inside the figure. This is the case whether the figure is a rectangle, triangle, trapezoid, or circle. For each figure, there is a formula for finding the number of unit squares rather than having to count the unit squares one-by-one. These formulas will be explain in other parts of this website. Directions: Use the "height" and "base" sliders to create a rectangle. The "Show/Hide Grid" determines whether the grid is showing. Use the slider to choose whether to display the perimeter or the area of the rectangle. • Find a formula for calculating the perimeter of the rectangle. • Find a formula for calculating the area of the rectangle. # Self-Check Question 1 Find the area of this rectangle. [show answer] A = bh A = (3)(4) = $12 \,u^2$ Question 2 Find the area of this rectangle. [show answer] A = bh A = (6)(5) = $30 \,u^2$ Question 3 A rectangle with a base of 6 centimeters has an area of $78 \,cm^2$. What is the height of this rectangle? [show answer] A = bh 78 = 6h $h=\frac{78}{6}=13\mbox{ cm}$	{"extraction_info": {"found_math": true, "script_math_tex": 4, "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": 0.8232834339141846, "perplexity": 367.0217228876436}, "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/1656104209449.64/warc/CC-MAIN-20220703013155-20220703043155-00512.warc.gz"}
https://www.simonsfoundation.org/event/cca-seminar-series-anatoly-spitkovsky/	# CCA Seminar Series: Anatoly Spitkovsky • Speaker • Anatoly Spitovsky, Ph.D.Professor, Department of Astrophysical Sciences, Princeton University Date & Time Title: Particle acceleration in astrophysical shocks: lessons from kinetic simulations. Abstract: Shocks in low density plasmas (so-called “collisionless shocks”) are ubiquitous throughout the Universe; and are thought to produce nonthermal particles that extend over decades in energy. I will describe the progress in modeling collisionless shock structure and particle acceleration using ab-initio kinetic simulations, focusing on the current understanding of magnetic field amplification mechanisms, the conditions necessary for particle injection into the acceleration process, and the physics behind the electron-to-ion ratio in shock acceleration. These results will be applied to understanding morphologies of shocks in supernova remnants and galaxy clusters, and to explaining the relative abundances of different species in galactic cosmic rays.	{"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": 0.8510206937789917, "perplexity": 3969.1856988864033}, "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/1573496670821.55/warc/CC-MAIN-20191121125509-20191121153509-00112.warc.gz"}
https://byjus.com/ncert-solutions-class-2-maths-chapter-15-how-many-ponytails/	# NCERT Solutions For Class 2 Maths Chapter 15 How Many Ponytails? Addition of numbers plays an important role not only in Mathematics but also in daily life. NCERT Solutions for Class 2 Maths Chapter 15 How Many Ponytails?, will educate kids about the addition of numbers. Here we provide you with the most detailed NCERT Solutions for Class 2 Maths in PDF format. The solutions have been written in a simple language, which is easy to understand for kids. The NCERT Solutions for Class 2 Maths will help your kid to practice the questions of the textbook in advance. You can download the solutions in PDF format from the link given here. ## Download the PDF of NCERT Solutions For Class 2 Maths Chapter 15 How Many Ponytails? ### Access answers to Maths NCERT Solutions for Class 2 Maths Chapter 15 How Many Ponytails? Fruit Seller The fruit seller has many fruits for you. Question: 1 Look at them and find out what the different fruits are. The different fruits are bananas, apples, oranges and mangoes. Question: 2 Count and write	{"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": 0.8867965936660767, "perplexity": 1677.5387147635533}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-27/segments/1656104669950.91/warc/CC-MAIN-20220706090857-20220706120857-00505.warc.gz"}
http://tex.stackexchange.com/questions/44271/multline-inside-an-item/44272	# multline inside an item Is there a way to use a multline inside an \item element of enumerate WITHOUT a new line between the index of list and the formula? \begin{enumerate}[i.] \item $checkorder( (\lambda,\lambda,\lambda) ) \triangleq true$ \item \begin{multline}$$checkorder( (a,(u,U_1,U_2),(v,V_1,V_2)) ) \triangleq \\ (u = \lambda \vee u \leq_A a) \wedge (v = \lambda \vee a \leq_A v) \wedge \\ checkorder(U_1) \, \wedge checkorder(U_2) \wedge checkorder(V_1) \wedge checkorder(V_2)$$ \end{multline} \item $checkorder(T) \triangleq false$ negli altri casi \end{enumerate} this is what i have: and this is what i would want: - The amsmath alignment environments are only for display math. But mathtools augments the available environments: \documentclass[a4paper]{article} \usepackage{mathtools,enumerate,amssymb} \begin{document} \begin{enumerate}[i.] \item $\mathit{checkorder}( (\lambda,\lambda,\lambda) ) \triangleq \mathit{true}$ \item $\begin{multlined}[t] \mathit{checkorder}( (a,(u,U_1,U_2),(v,V_1,V_2)) ) \triangleq \\ (u = \lambda \vee u \leq_A a) \wedge (v = \lambda \vee a \leq_A v) \wedge {} \\ \mathit{checkorder}(U_1) \wedge \mathit{checkorder}(U_2) \wedge \mathit{checkorder}(V_1) \wedge checkorder(V_2) \end{multlined}$ \item $\mathit{checkorder}(T) \triangleq \mathit{false}$ negli altri casi \end{enumerate} \end{document} Notice a couple of refinements: long function names should be inside \mathit, to get a better looking font in this case; the \wedge before \\ should be followed by {} to get correct spacing. Never use $$ in LaTeX. - thanks! why never use$$ ? – tyranitar Feb 12 '12 at 11:52 See Why is $...$ preferable to ?. On this site appreciation for an answer is expressed by clicking on the upper arrow besides the text; usually, after a delay to see if other more useful answers come along, one accepts the better one. – egreg Feb 12 '12 at 12:08 +1 By the same rational, is it not more preferable to use $$...$$ instead of $...$ I reference Joseph Wright's comments in Are ( and ) preferable to \$? – cmhughes Feb 12 '12 at 17:14 I had a similar issue with \multline environment. So I came up with a forced solution. Here is what I did: \item \abovedisplayskip-15pt \belowdisplayskip\abovedisplayskip \begin{setlength}{\multlinegap}{0pt} \begin{multline} 2(x^2y-2xy^3+6xy)-3(-5xy+3x^2y\\-7xy^3) +5(xy^3-xy+x^2y) \end{multline} \end{setlength} Of course, egreg's solution is by far the best and mine is was just out of desperation in finding an immediate solutions. I know that my code above is not the best but hey it solved my problem. I have decided to implement egreg's answer into my code as it seems to better and more concise. -	{"extraction_info": {"found_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": 2, "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": 0.9292135834693909, "perplexity": 1975.7523733253704}, "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/1418802771091.111/warc/CC-MAIN-20141217075251-00043-ip-10-231-17-201.ec2.internal.warc.gz"}
https://www.physicsforums.com/threads/force-needed-to-make-a-disk-climb-a-step.230010/	# Force needed to make a disk climb a step • Start date • #1 5 0 [SOLVED] Force needed to make a disk climb a step Howdy ho. First off, let me say that I'm a newbie here, and I'm not sure if this is the correct place to put this doubt. Here's the deal: ## Homework Statement We have a disk, and we need to know the force F needed to have it climb the step, with 50mm. The force is applied at the top of the disk. ............F <---- ..... ....................\|........\| ..................\|............\| ___________\|............\| ........50mm \| \|........\| .................\|___....._______ Sorry for that draw, it's the best I could do. The only known data is: the step has 50mm; the disk is uniform and weights 40N; no friction is to be considered. - ## The Attempt at a Solution I've tried solving it in several ways. I've considered a normal vertical reaction 1 between the ground and the disk, and considered a normal diagonal reaction 2 between the disk and the tip of the step. I also considered that the angle this reaction 2 makes is 45º. So, I concluded that, in the vertical axis, we have: 40 = N1 + N2.sin45º. After this, we would have in the horizontal axis the force F and N2.cos45º. I thought that one condition for the disk to climb the step would be: F > N2.cos45º. However, this wouldn't be enough. So I used the moment. The moment the force F causes in the center of the disk is F.r . We do not know r, so this would lead to another variable in another equation. So, unless there is missing some data here, I don't know how to solve it. I have the solution, and it says the force should be F=17,88N . Sorry for any language error, english is not my native language, and is not the language I'm using in classes, so I may have mistaken some technical terms. Related Introductory Physics Homework Help News on Phys.org • #2 Doc Al Mentor 44,871 1,119 Taking moments is the right approach. Hint: You can choose any pivot point you like, so choose wisely. • #3 5 0 I can't seem to get a solution. I tried using the tip of the step as the pivot. But the distance between this pivot and the 3 forces (weight, reaction 1 and force F) all depend on R. Besides, the sum of all moments is depending on reaction 1, which is unknown. Thanks for any help • #4 alphysicist Homework Helper 2,238 1 Hi Towelie, Was this a book problem? This answer does depend on R so they would have to give it to you somehow. (The answer you have of F=17.88 N corresponds to a radius of 15 cm.) • #5 5 0 No, it's in a work sheet, made by my professor. It is indeed possible some data is missing, it wouldn't be the first time. But thanks for all your help! • #6 Doc Al Mentor 44,871 1,119 I tried using the tip of the step as the pivot. But the distance between this pivot and the 3 forces (weight, reaction 1 and force F) all depend on R. You need to be given R. Besides, the sum of all moments is depending on reaction 1, which is unknown. At the instant the disk begins to rise up, what is that reaction force? • #7 5 0 That would be 0, am I correct? • #8 Doc Al Mentor 44,871 1,119 That would be 0, am I correct? You are correct. • #9 5 0 Ok. So I assumed the instant the disk is rising up, that is, N1=0. I also assumed R=0,15 , like alphysicist said. In this scenario, all we need is the moment caused by force F to be bigger than the moment caused by weight: F . 0,25 > 40 . 0,1118 (=) F > 17,888N The same result from my solutions. So I think it is correct! Thank you very much for your help, Doc Al and alphysicist. • Last Post Replies 5 Views 5K • Last Post Replies 8 Views 21K • Last Post Replies 11 Views 1K • Last Post Replies 8 Views 5K • Last Post Replies 5 Views 3K • Last Post Replies 3 Views 1K • Last Post Replies 10 Views 6K • Last Post Replies 0 Views 4K	{"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": 0.9367918968200684, "perplexity": 886.050974108554}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-05/segments/1579251681412.74/warc/CC-MAIN-20200125191854-20200125221854-00386.warc.gz"}
http://hea-www.harvard.edu/astrostat/slog/groundtruth.info/AstroStat/slog/2008/parametric-bootstrap-vs-nonparametric-bootstrap/index.html	#### Parametric Bootstrap vs. Nonparametric Bootstrap The following footnotes are from one of Prof. Babu’s slides but I do not recall which occasion he presented the content. – In the XSPEC packages, the parametric bootstrap is command FAKEIT, which makes Monte Carlo simulation of specified spectral model. – XSPEC does not provide a nonparametric bootstrap capability. Parametric Bootstrap: $$X_1^,…,X_n^ \sim F(\cdot;\theta_n)$$ Both $$\sqrt{n} \sup_x \|F_n(x)-F(x;\theta_n)\|$$ and $$\sqrt{n} \sup_x \|F_n^(x)-F(x;\theta_n^)\|$$ have the same limiting distribution.[1] Nonparametric Bootstrap:$$X_1^,…,X_n^ \sim F_n.$$ A bias correction $$B_n(x)=F_n(x)-F(x;\theta_n)$$ is needed. $$\sqrt{n} \sup_x \|F_n(x)-F(x;\theta_n)\|$$ and $$\sqrt{n} \sup_x \|F_n^(x)-F(x;\theta_n^)-B_n(x)\|$$ have the same limiting distribution.[2] 1. In the XSPEC packages, the parametric bootstrap is command FAKEIT, which makes Monte Carlo simulation of specified spectral model.[] 2. XSPEC does not provide a nonparametric bootstrap capability[]	{"extraction_info": {"found_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": 0.9533606171607971, "perplexity": 4154.281332799808}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-47/segments/1510934806736.55/warc/CC-MAIN-20171123050243-20171123070243-00074.warc.gz"}
http://www.purplemath.com/learning/viewtopic.php?p=4980	## Solving an equation with logarithms Complex numbers, rational functions, logarithms, sequences and series, matrix operations, etc. phiv Posts: 2 Joined: Fri Nov 05, 2010 1:13 am Contact: ### Solving an equation with logarithms Hello, I am having trouble isolating W in the following equation, because it is outside and inside the logarithm: $R=W \cdot log_2(1+\frac{P}{W})$ I tried a number of things, but I always end up with a W as an exponent, and a W that is not an exponent, and I am unable to isolate W. Thank you, Martingale Posts: 350 Joined: Mon Mar 30, 2009 1:30 pm Location: USA Contact: ### Re: Solving an equation with logarithms phiv wrote:Hello, I am having trouble isolating W in the following equation, because it is outside and inside the logarithm: $R=W \cdot log_2(1+\frac{P}{W})$ I tried a number of things, but I always end up with a W as an exponent, and a W that is not an exponent, and I am unable to isolate W. Thank you, you can't solve for w explicitly. If you want a solution you will have to do it numerically. phiv Posts: 2 Joined: Fri Nov 05, 2010 1:13 am Contact: ### Re: Solving an equation with logarithms Turns out you are correct. The professor solved using trial and error values for W (ie. numerically).	{"extraction_info": {"found_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": 2, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8839014768600464, "perplexity": 829.5517396969653}, "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-2016-07/segments/1454701146196.88/warc/CC-MAIN-20160205193906-00336-ip-10-236-182-209.ec2.internal.warc.gz"}
https://arxiv-export-lb.library.cornell.edu/abs/2208.06306?context=cs.CC	cs.CC (what is this?) # Title: Wasserstein Complexity of Quantum Circuits Abstract: Given a unitary transformation, what is the size of the smallest quantum circuit that implements it? This quantity, known as the quantum circuit complexity, is a fundamental property of quantum evolutions that has widespread applications in many fields, including quantum computation, quantum field theory, and black hole physics. In this letter, we obtain a new lower bound for the quantum circuit complexity in terms of a novel complexity measure that we propose for quantum circuits, which we call the quantum Wasserstein complexity. Our proposed measure is based on the quantum Wasserstein distance of order one (also called the quantum earth mover's distance), a metric on the space of quantum states. We also prove several fundamental and important properties of our new complexity measure, which stand to be of independent interest. Finally, we show that our new measure also provides a lower bound for the experimental cost of implementing quantum circuits, which implies a quantum limit on converting quantum resources to computational resources. Our results provide novel applications of the quantum Wasserstein distance and pave the way for a deeper understanding of the resources needed to implement a quantum computation. Comments: 7+7 pages Subjects: Quantum Physics (quant-ph); Computational Complexity (cs.CC); High Energy Physics - Theory (hep-th); Mathematical Physics (math-ph) Cite as: arXiv:2208.06306 [quant-ph] (or arXiv:2208.06306v1 [quant-ph] for this version) ## Submission history From: Kaifeng Bu [view email] [v1] Fri, 12 Aug 2022 14:44:13 GMT (58kb,D) Link back to: arXiv, form interface, contact.	{"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": 0.8260981440544128, "perplexity": 715.6793133940315}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446711069.79/warc/CC-MAIN-20221206024911-20221206054911-00384.warc.gz"}
http://math.stackexchange.com/questions/426932/why-are-vandermonde-matrices-invertible	# Why are Vandermonde matrices invertible? A Vandermonde-matrix is a matrix of this form: $$\begin{pmatrix} x_0^0 & \cdots & x_0^n \\ \vdots & \ddots & \vdots \\ x_n^0 & \cdots & x_n^n \end{pmatrix} \in \mathbb{R}^{(n+1) \times (n+1)}$$. condition ☀ : $\forall i, j\in \{0, \dots, n\}: i\neq j \Rightarrow x_i \neq x_j$ Why are Vandermonde-matrices with ☀ always invertible? I have tried to find a short argument for that. I know some ways to show that in principle: • rank is equal to dimension • all lines / rows are linear independence • determinant is not zero • find inverse According to proofwiki, the determinant is $$\displaystyle V_n = \prod_{1 \le i < j \le n} \left({x_j - x_i}\right)$$ There are two proofs for this determinant, but I've wondered if there is a simpler way to show that such matrices are invertible. - Simpler than knowing its determinant and checking very easily it can't be zero by the given data? I doubt it... – DonAntonio Jun 22 '13 at 14:48 You have to prove the determinant first. I'm with you that checking it with the determinant is easy, but I guess there are easier ways than the two proves from proofwiki to show that it is invertible. – moose Jun 22 '13 at 14:59 The Vandermonde determinant is non-zero $\implies$ the vandermonde matrix is invertible – Ale Jan 23 at 8:25 This is not entirely dissimilar to the answer already posted by Chris Godsil, but I'll post this anyway, maybe it can provide slightly different angle for someone trying to understand this. We want to show that the matrix $$\begin{pmatrix} x_0^0 & \cdots & x_0^n \\ \vdots & \ddots & \vdots \\ x_n^0 & \cdots & x_n^n \end{pmatrix}$$ is invertible. It suffices to show that the rows of this matrix are linearly independent. So let us assume that $c_0v_0+c_1v_1+\dots+c_nv_n=\vec 0=(0,0,\dots,0)$, where $v_k=(1,x_k^1,\dots,x_k^n)$ is the $k$-the row written as a vector and $c_0,\dots,c_n\in\mathbb R$. Then we get on the $k$-th coordinate $$c_0+c_1x_k+c_2x_k^2+\dots+c_kx_k^n=0,$$ which means that $x_k$ is a root of the polynomial $p(x)=c_0+c_1x+c_2x^2+\dots+c_kx^n$. Now if the polynomial $p(x)$ of degree at most $n$ has $(n+1)$ different roots $x_0,x_1,\dots,x_n$, it must be the zero polynomial and we get that $c_0=c_1=\dots=c_k=0$. This proves that the vectors $v_0,v_1,\dots,v_n$ are linearly independent. (And, in turn, we get that the given matrix is invertible.) - Let $V$ be your Vandermonde matrix. If $p(t)=a_0+a_1t+\cdots+a_nt^n$ and $\alpha$ is the vector of coefficients of $p$, then the entries of $V\alpha$ are the values of $p$ on the points $x_0,\ldots,x_n$. Now for $r=0,\ldots,n$ choose polynomials $p_r$ of degree $n$ so that $p(x_r)=1$ and $p(x_s)=0$ if $s\ne r$. Then the matrix with the coefficients of the polynomials $p_r$ as its columns is the inverse of $V$. This is, of course, just a way of viewing Lagrange interpolation. - Why is the matrix with the coefficients of the polynomials $p_r$ as its columns the inverse of $V$? – moose Jun 22 '13 at 17:58 Because when you multiply it by $V$ you get the identity. The point is that $V$ applied to the coefficients of $p_r$ returns the $r$-th standard basis vector. – Chris Godsil Jun 22 '13 at 21:15 Ok, but when you write "now [...] choose polynomials [...] so that $p(x_r)=1$ and $p(x_s) = 0$ if $s \neq r$" don't you already use that $V$ is invertible? Why can you choose polynomials like this? – moose Jun 23 '13 at 6:30 Well, to get $p_r$ first take the product $q_r(x)=\prod_{i\ne r}(x-x_r)$ which is zero except at $x_r$ and not zero at $x_r$. Now set $p_r(x)=q_r(x)/q_r(x_r)$. – Chris Godsil Jun 23 '13 at 13:24 For any $n+1$ distinct numbers $x_0, \ldots, x_n \in \mathbb{R}$, let $V(x_0,\ldots,x_n)$ and $D(x_0,\ldots,x_n)$ be a Vandermonde matrix and its determinant: $$V(x_0,\ldots,x_n) = \begin{pmatrix} x_0^0 & \cdots & x_0^n \\ \vdots & \ddots & \vdots \\ x_n^0 & \cdots & x_n^n \end{pmatrix}\quad\text{ and }\quad D(x_0,\ldots,x_n) = \det V(x_0,\ldots,x_n)$$ It is clear $D(x_0) = 1 \ne 0$. Let us assume $D(x_0,\ldots,x_{m}) \ne 0$ for some $m < n$ and consider what happens to $D(x_0,\ldots,x_{m+1})$. Viewed as a function of its last argument $x_{m+1}$, $D(x_0,\ldots,x_m,x)$ is a polynomial in $x$ with degree at most $m+1$. Since this polynomial vanishes at $m+1$ different values $x_0, \ldots, x_{m}$ already, it cannot vanish on any other $x$ (in particular at $x_{m+1}$). Otherwise, $D(x_0,\ldots,x_m,x)$ will be identically zero. We know that $D(x_0,\ldots,x_m,x)$ isn't the zero polynomial. The leading coefficient of $x^{m+1}$ in $D(x_0,\ldots,x_m,x)$ is proportional to $D(x_0,\ldots,x_m)$ which is non-zero by induction assumption. By induction, we can conclude $D(x_0,\ldots,x_n) \ne 0$ and hence $V(x_0,\ldots,x_n)$ is invertible. -	{"extraction_info": {"found_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": 0.9722539186477661, "perplexity": 188.7976424128931}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-41/segments/1410657137190.70/warc/CC-MAIN-20140914011217-00073-ip-10-234-18-248.ec2.internal.warc.gz"}
https://www.nag.com/numeric/cl/nagdoc_latest/html/s/s10acc.html	# NAG C Library Function Document ## 1Purpose nag_cosh (s10acc) returns the value of the hyperbolic cosine, $\mathrm{cosh}x$. ## 2Specification #include #include double nag_cosh (double x, NagError fail) ## 3Description nag_cosh (s10acc) calculates an approximate value for the hyperbolic cosine, $\mathrm{cosh}x$. For $\left\|x\right\|\le {E}_{1}\text{, }\mathrm{cosh}x=\frac{1}{2}\left({e}^{x}+{e}^{-x}\right)$. For $\left\|x\right\|>{E}_{1}$, the function fails owing to danger of setting overflow in calculating ${e}^{x}$. The result returned for such calls is $\mathrm{cosh}{E}_{1}$, i.e., it returns the result for the nearest valid argument. The value of machine-dependent constant ${E}_{1}$ may be given in the Users' Note for your implementation. ## 4References NIST Digital Library of Mathematical Functions ## 5Arguments 1: $\mathbf{x}$doubleInput On entry: the argument $x$ of the function. 2: $\mathbf{fail}$NagError Input/Output The NAG error argument (see Section 3.7 in How to Use the NAG Library and its Documentation). ## 6Error Indicators and Warnings NE_ALLOC_FAIL Dynamic memory allocation failed. See Section 2.3.1.2 in How to Use the NAG Library and its Documentation for further information. NE_INTERNAL_ERROR An internal error has occurred in this function. Check the function call and any array sizes. If the call is correct then please contact NAG for assistance. See Section 2.7.6 in How to Use the NAG Library and its Documentation for further information. NE_NO_LICENCE Your licence key may have expired or may not have been installed correctly. See Section 2.7.5 in How to Use the NAG Library and its Documentation for further information. NE_REAL_ARG_GT On entry, ${\mathbf{x}}=〈\mathit{\text{value}}〉$. Constraint: $\left\|{\mathbf{x}}\right\|\le {E}_{1}$. The function has been called with an argument too large in absolute magnitude. There is a danger of overflow. The result returned is the value of $\mathrm{cosh}x$ at the nearest valid argument. ## 7Accuracy If $\delta$ and $\epsilon$ are the relative errors in the argument and result, respectively, then in principle $ε≃xtanh⁡x×δ.$ That is, the relative error in the argument, $x$, is amplified by a factor, at least $x\mathrm{tanh}x$. The equality should hold if $\delta$ is greater than the machine precision ($\delta$ is due to data errors etc.) but if $\delta$ is simply a result of round-off in the machine representation of $x$ then it is possible that an extra figure may be lost in internal calculation round-off. The behaviour of the error amplification factor is shown by the following graph: Figure 1 It should be noted that near $x=0$ where this amplification factor tends to zero the accuracy will be limited eventually by the machine precision. Also, for $\left\|x\right\|\ge 2$ $ε∼xδ=Δ$ where $\Delta$ is the absolute error in the argument $x$. ## 8Parallelism and Performance nag_cosh (s10acc) is not threaded in any implementation. None. ## 10Example This example reads values of the argument $x$ from a file, evaluates the function at each value of $x$ and prints the results. ### 10.1Program Text Program Text (s10acce.c) ### 10.2Program Data Program Data (s10acce.d) ### 10.3Program Results Program Results (s10acce.r)	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 29, "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": 0.9534562826156616, "perplexity": 1387.1360132820087}, "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/1631780060882.17/warc/CC-MAIN-20210928184203-20210928214203-00556.warc.gz"}
https://www.physicsforums.com/threads/heat-capacity-of-air-at-constant-volume.188252/	# Heat Capacity of Air at Constant Volume 1. ### s.p.q.r 25 Hi I have an ongoing dispute with my mate on this one, please help to clarify this before I open up a can of whoop ass on that sorry mo-fo. 300 litres of air are compressed into a 3 litre tank. What is the heat capacity of this air? 2. ### Hootenanny 9,677 Staff Emeritus What do you think it is? 3. ### s.p.q.r 25 The Cp J mol is 29.19. But because I ask for constant volume, it is definately lower then this. This is what I think. I can find no references to constant volume anywhere and unfortunately I have no teacher to ask as I study archaeology, not physics. 4. ### Andrew Mason 6,876 Air is almost entirely a diatomic gas, $\gamma = C_p/C_v = 1.4$ (7/5) AM 5. ### s.p.q.r 25 Hi, Thanks for the reply. Is 1.4 per gram or mol? Also, How can you measure a gram of gas and how much is 1 mol? Cheers. 8,952 7. ### Loren Booda Is heat capacity independent of volume for an ideal gas? Stupid question - gas performs work while being compressed. Last edited: Oct 2, 2007 8. ### mgb_phys 8,952 For an ideal gas heat capcity just depends on the amount (number of moles) present and the number of vibration states of the molecular. For a real gas it also depends on the pressure because the molecules close to each other change the vibration state/bond energy. 9. ### Loren Booda In a modification of the "ideal gas" law, I seem to recall an equation with correction terms for the volume and pressure, respectively. Has anyone run across this? 10. ### s.p.q.r 25 Hi, This ratio of 1.4, does this just mean that you divide the constant pressure capacity (1.020J/g) by 1.4? 11. ### Andrew Mason 6,876 $\gamma = 1.4$ is the ratio of the specific heat (heat flow per gram or per mole per degree K change in temperature) at constant pressure to the specific heat at constant volume. $\gamma = C_p/C_v$. What you want to find is Cv. You also have to find the number of moles of air in this container to find its heat capacity (heat flow per degree K change in Temp.). AM Last edited: Oct 8, 2007	{"extraction_info": {"found_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": 0.8265289664268494, "perplexity": 1493.2365298673146}, "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-2015-32/segments/1438042987628.47/warc/CC-MAIN-20150728002307-00058-ip-10-236-191-2.ec2.internal.warc.gz"}
https://brilliant.org/problems/inspired-by-none/	# One two three four five Algebra Level pending $\large \sum_{n=0}^\infty \frac1{(n+1)(n+2)(n+3)(n+4)(n+5)}$ If the series above can be written as $$\frac 1A$$, find the value of $$A$$. ×	{"extraction_info": {"found_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": 0.9374438524246216, "perplexity": 1501.8289902866159}, "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-20161020183840-00221-ip-10-171-6-4.ec2.internal.warc.gz"}
http://dtubbenhauer.com/research.html	Research subjects 5. Higher category theory (especially for $n=2$) and categorification. In short: My main research interest is $2$-categorical representation theory (of e.g. categorified Coxeter groups), categorification (of e.g. quantum groups) and its applications in representation theory, low dimensional topology and algebraic geometry. In particular, I am interested in algebraic, combinatorial and diagrammatic aspects of categorification. I am also interested in related topics as for example representation theoretical questions about Hecke/Brauer algebras or Lie groups and modular representation theory.	{"extraction_info": {"found_math": true, "script_math_tex": 1, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.8137586116790771, "perplexity": 905.5456917120295}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-17/segments/1524125937193.1/warc/CC-MAIN-20180420081400-20180420101400-00078.warc.gz"}
http://jmlr.org/papers/v21/18-101.html	## Regularized Gaussian Belief Propagation with Nodes of Arbitrary Size Francois Kamper, Sarel J. Steel, Johan A. du Preez; 21(101):1−42, 2020. ### Abstract Gaussian belief propagation (GaBP) is a message-passing algorithm that can be used to perform approximate inference on a pairwise Markov graph (MG) constructed from a multivariate Gaussian distribution in canonical parameterization. The output of GaBP is a set of approximate univariate marginals for each variable in the pairwise MG. An extension of GaBP (labeled GaBP-m), allowing for the approximation of higher-dimensional marginal distributions, was explored by Kamper et al. (2019). The idea is to create an MG in which each node is allowed to receive more than one variable. As in the univariate case, the multivariate extension does not necessarily converge in loopy graphs and, even if convergence occurs, is not guaranteed to provide exact inference. To address the problem of convergence, we consider a multivariate extension of the principle of node regularization proposed by Kamper et al. (2018). We label this algorithm slow GaBP-m (sGaBP-m), where the term 'slow' relates to the damping effect of the regularization on the message passing. We prove that, given sufficient regularization, this algorithm will converge and provide the exact marginal means at convergence, regardless of the way variables are assigned to nodes. The selection of the degree of regularization is addressed through the use of a heuristic, which is based on a tree representation of sGaBP-m. As a further contribution, we extend other GaBP variants in the literature to allow for higher-dimensional marginalization. We show that our algorithm compares favorably with these variants, both in terms of convergence speed and inference quality. [abs][pdf][bib]	{"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": 0.936896800994873, "perplexity": 692.9141174950462}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-29/segments/1593655887046.62/warc/CC-MAIN-20200705055259-20200705085259-00154.warc.gz"}
http://math.stackexchange.com/questions/213739/special-differential-equation-continued/213751	# Special Differential Equation (Continued) As a second part of my problem I end up with the differential equation looking like: $$\frac{d^2 y}{dx^2} + \frac{1}{x}\frac{dy}{dx} - \frac{a}{x^2}y - \frac{c}{x}y + b x e^{-x^2/p^2}y - d e^{-x^2/p^2}y = 0.$$ It is more complex that my previous question. Can someone suggestion a solution method for this? - As before, what's the domain of $x$? – Pragabhava Oct 14 '12 at 17:48 It has to be >0. and all coefficients a-d are non-zero. – nagendra Oct 14 '12 at 17:51 The first suggestion would be to write it in a easy to look form, i.e. $$\frac{d^2 y}{d x^2} + \frac{1}{x} \frac{d y}{d x} + \left(-\frac{a}{x^2} - \frac{c}{x} + bxe^{-x^2/p^2}-de^{-x^2/p^2}\right) y = 0$$ – Pragabhava Oct 14 '12 at 17:54 i was wondering, if the nature of the equation is to blow up due to a singularity at x=0, would be incorrect to change x to x + \delta where \delta<<1 and then proceed with the solution? (sorry, I couldn't get latex to do a greek delta for me ) – drN Oct 14 '12 at 18:10 After transfering the ODE of the form $p(x)\dfrac{d^2y}{dx^2}+q(x)\dfrac{dy}{dx}+r(x)y=0$ to the ODE of the form $\dfrac{d^2z}{dx^2}+f(x)z=0$ by considering the method in mathworld.wolfram.com/…;, eqworld.ipmnet.ru/en/methods/methods-ode/Khorasani2003.pdf claims that $\dfrac{d^2z}{dx^2}+f(x)z=0$ have method to solve generally for general $f(x)$ . But how is the reliability of eqworld.ipmnet.ru/en/methods/methods-ode/Khorasani2003.pdf? – doraemonpaul Oct 14 '12 at 22:19 The added complication makes closed form solutions even less likely, but you still have $x=0$ as a regular singular point with indicial roots $\pm \sqrt{a}$, and corresponding series solutions.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8911192417144775, "perplexity": 668.8672892901128}, "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-2014-52/segments/1418802767828.26/warc/CC-MAIN-20141217075247-00162-ip-10-231-17-201.ec2.internal.warc.gz"}
https://usm.uni-muenchen.de/CAST/cms/index.php/16-research/137-active-galactic-nuclei-the-galactic-center	### The life cycle of starbursting circumnuclear discs High-resolution observations from the submm to the optical wavelength regime resolve the central few 100 pc region of nearby galaxies in great detail. They reveal a large diversity of features: thick gas and stellar discs, nuclear starbursts, inflows and outflows, central activity, jet interaction, etc. Concentrating on the role circumnuclear discs play in the life cycles of galactic nuclei, we employ 3D adaptive mesh refinement hydrodynamical simulations with the RAMSES code to self-consistently trace the evolution from a quasi-stable gas disc, undergoing gravitational (Toomre) instability, the formation of clumps and stars and the disc’s subsequent, partial dispersal via stellar feedback. Our approach builds upon the observational finding that many nearby Seyfert galaxies have undergone intense nuclear starbursts in their recent past and in many nearby sources star formation is concentrated in a handful of clumps on a few 100 pc distant from the galactic centre. We show that such observations can be understood as the result of gravitational instabilities in dense circumnuclear discs. By comparing these simulations to available integral field unit observations of a sample of nearby galactic nuclei, we find consistent gas and stellar masses, kinematics, star formation and outflow properties. Important ingredients in the simulations are the self-consistent treatment of star formation and the dynamical evolution of the stellar distribution as well as the modelling of a delay time distribution for the supernova feedback. The knowledge of the resulting simulated density structure and kinematics on pc scale is vital for understanding inflow and feedback processes towards galactic scales. ## Radiative transfer modelling of AGN tori Active Galactic Nuclei (AGN) are thought to be enshrouded by geometrically and optically thick toroidal dust distributions, the so-called molecular tori. We derive observable properties for a self-consistent model of such toroidal gas and dust distributions, where the geometrical thickness is achieved and maintained with the help of X-ray heating and radiation pressure due to the central engine (Wada et al., 2012). We simulate spectral energy distributions (SEDs) and images with the help of dust continuum radiative transfer calculations with RADMC-3D (Dullemond et al., 2012). This results - for the first time - in time-resolved SEDs and images for a physical model of the central obscurer. We find that temporal changes are mostly visible at shorter wavelengths, close to the combined peak of the dust opacity -- which follows a loal dust extinction law -- as well as the central source spectrum and are caused by variations in the column densities of the generated outflow. Because of the three-component morphology of the hydrodynamical models -- a thin disc with high-density filaments (the remainder of the initial condition and determined by the inflow from larger scales), a surrounding fluffy component (the obscurer) and a low-density outflow along the rotation axis - we find dramatic differences depending on the observed wavelength: whereas the mid-infrared images are dominated by the elongated appearance of the outflow cone, the long wavelength emission is mainly given by the cold and dense disc component. Overall, we find good agreement with observed characteristics, especially for those models, which show clear outflow cones in combination with a geometrically thick distribution of gas and dust, as well as a geometrically thin, but high column density disc in the equatorial plane. ### by Andreas Burkert We investigate the origin, structure and evolution of the small gas cloud, G2, which is on an orbit almost straight into the Galactic central supermassive black hole (Gillessen et al., 2012). The movie shows the evolution of the cloud in an idealised hot atmosphere, using the hydrodynamics code PLUTO. With the help of such simulations and analytical calculations, we find that G2 is a sensitive probe of the hot accretion zone of Sgr A* and we can show that the observed structure and evolution of G2 can be well reproduced if it formed in pressure equilibrium with the surrounding in 1995 at a distance from the SMBH of 7.6e16 cm. Another possibility is that G2 is the head of a larger, shell-like structure that formed at apocenter. Our numerical simulations show that this scenario explains not only G2's observed kinematical and geometrical properties but also the observations of a low surface brightness gas tail that trails the cloud. In 2013, while passing the SMBH G2 will break up into a string of droplets that within the next 30 years mix with the surrounding hot gas and trigger cycles of AGN activity. Click on the image to see the ESO press release movie (Credit: ESO/MPE/M.Schartmann/L.Calçada)! Highlights Pair of galaxy clusters in a wide, warm web of gas Ausgerechnet! Unser Universum Gas perturbations reveal protoplanets The central parsecs of the low-luminosity active galaxy NGC 1052: evidence for a truncated accretion disc Clusters of Small Clumps in High-Redshift Disk Galaxies Magneticum Pathfinder: The evolution of the universe in an unmatched extend The complex Interplay between Spin, Mass, and Morphology in Galaxies Magneticum sheds new light on recently discovered Fast Radio Bursts (FRBs) A Disk-Disk Major Merger Event in a Cosmological Hydrodynamical Zoom-Simulation The formation of filamentary bundles in turbulent molecular clouds G2 modelled as a mass-losing source of gas Supernova-driven galactic winds The Dark Halo-Spheroid Conspiracy and the Origin of Elliptical Galaxies Phd Award 2012 of the Astronomische Gesellschaft CAST group outing 2012 Evolution of the Galactic Centre Cloud G2 Universe Cluster PhD Thesis Award 2011 Evolution of Molecular Clouds in Spiral Galaxies by Clare Dobbs Globular Clusters Black Hole Correlation by Andreas Burkert Cosmological Resimulations by Ludwig Oser Star Formation in the Galactic Centre by Christian Alig A New Model for the Antennae Galaxies by Simon Karl Simulating the Bullet Cluster by Chiara Mastropietro Triggered Star Formation by Matthias Gritschneder The Mystery of Sedna by Hagen Schulte in den Bäumen The Formation of Fossil Galaxy Groups in the hierarchical Universe by Elena D'Onghia Molecular Cloud Formation in Colliding Flows by Fabian Heitsch Comparison of hydro codes on planet-disk interaction problem by Pawel Ciecielag Orbital Structure of Galaxies in N-Body Simulations by Roland Jesseit	{"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": 0.8222268223762512, "perplexity": 2563.6622293166997}, "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-2021-21/segments/1620243991269.57/warc/CC-MAIN-20210516105746-20210516135746-00161.warc.gz"}
https://www.lmfdb.org/L/2/75/15.14	## Results (1-50 of at least 1000) Next Label $\alpha$ $A$ $d$ $N$ $\chi$ $\mu$ $\nu$ $w$ prim arith $\mathbb{Q}$ self-dual $\operatorname{Arg}(\epsilon)$ $r$ First zero Origin 2-75-25.6-c1-0-2 $0.773$ $0.598$ $2$ $3 \cdot 5^{2}$ 25.6 $$1.0 1 -0.239 0 2.65124 Modular form 75.2.g.c.31.3 2-75-25.6-c1-0-0 0.773 0.598 2 3 \cdot 5^{2} 25.6$$ $1.0$ $1$ $-0.318$ $0$ $1.81778$ Modular form 75.2.g.b.31.2 2-75-25.4-c1-0-2 $0.773$ $0.598$ $2$ $3 \cdot 5^{2}$ 25.4 $$1.0 1 0.136 0 2.93222 Modular form 75.2.i.a.4.2 2-75-25.19-c1-0-3 0.773 0.598 2 3 \cdot 5^{2} 25.19$$ $1.0$ $1$ $0.0951$ $0$ $3.38793$ Modular form 75.2.i.a.19.3 2-75-15.8-c1-0-0 $0.773$ $0.598$ $2$ $3 \cdot 5^{2}$ 15.8 $$1.0 1 -0.213 0 2.17607 Modular form 75.2.e.a.68.2 2-75-25.21-c1-0-4 0.773 0.598 2 3 \cdot 5^{2} 25.21$$ $1.0$ $1$ $0.239$ $0$ $3.72896$ Modular form 75.2.g.c.46.3 2-75-75.62-c1-0-4 $0.773$ $0.598$ $2$ $3 \cdot 5^{2}$ 75.62 $$1.0 1 -0.0263 0 2.94763 Modular form 75.2.l.a.62.6 2-75-15.2-c1-0-1 0.773 0.598 2 3 \cdot 5^{2} 15.2$$ $1.0$ $1$ $-0.286$ $0$ $1.64009$ Modular form 75.2.e.a.32.1 2-75-25.11-c1-0-5 $0.773$ $0.598$ $2$ $3 \cdot 5^{2}$ 25.11 $$1.0 1 0.131 0 3.77890 Modular form 75.2.g.b.61.2 2-75-75.38-c1-0-2 0.773 0.598 2 3 \cdot 5^{2} 75.38$$ $1.0$ $1$ $0.0303$ $0$ $2.31203$ Modular form 75.2.l.a.38.4 2-75-75.8-c1-0-1 $0.773$ $0.598$ $2$ $3 \cdot 5^{2}$ 75.8 $$1.0 1 -0.176 0 1.57733 Modular form 75.2.l.a.8.5 2-75-75.8-c1-0-3 0.773 0.598 2 3 \cdot 5^{2} 75.8$$ $1.0$ $1$ $-0.181$ $0$ $2.10229$ Modular form 75.2.l.a.8.2 2-75-75.8-c1-0-0 $0.773$ $0.598$ $2$ $3 \cdot 5^{2}$ 75.8 $$1.0 1 0.360 0 0.219153 Modular form 75.2.l.a.8.1 2-75-75.2-c1-0-0 0.773 0.598 2 3 \cdot 5^{2} 75.2$$ $1.0$ $1$ $-0.451$ $0$ $1.49292$ Modular form 75.2.l.a.2.1 2-75-1.1-c1-0-1 $0.773$ $0.598$ $2$ $3 \cdot 5^{2}$ 1.1 $$1.0 1 0 0 3.01549 Elliptic curve 75.b Modular form 75.2.a.b Modular form 75.2.a.b.1.1 2-75-15.8-c1-0-3 0.773 0.598 2 3 \cdot 5^{2} 15.8$$ $1.0$ $1$ $0.197$ $0$ $3.36245$ Modular form 75.2.e.b.68.1 2-75-25.19-c1-0-0 $0.773$ $0.598$ $2$ $3 \cdot 5^{2}$ 25.19 $$1.0 1 -0.215 0 1.10927 Modular form 75.2.i.a.19.1 2-75-25.6-c1-0-1 0.773 0.598 2 3 \cdot 5^{2} 25.6$$ $1.0$ $1$ $-0.160$ $0$ $2.14907$ Modular form 75.2.g.a.31.1 2-75-75.17-c1-0-3 $0.773$ $0.598$ $2$ $3 \cdot 5^{2}$ 75.17 $$1.0 1 -0.0682 0 2.78804 Modular form 75.2.l.a.17.6 2-75-25.16-c1-0-5 0.773 0.598 2 3 \cdot 5^{2} 25.16$$ $1.0$ $1$ $0.454$ $0$ $3.56709$ Modular form 75.2.g.c.16.1 2-75-75.47-c1-0-7 $0.773$ $0.598$ $2$ $3 \cdot 5^{2}$ 75.47 $$1.0 1 -0.360 0 4.83712 Modular form 75.2.l.a.47.1 2-75-75.38-c1-0-3 0.773 0.598 2 3 \cdot 5^{2} 75.38$$ $1.0$ $1$ $-0.279$ $0$ $2.42220$ Modular form 75.2.l.a.38.8 2-75-75.2-c1-0-2 $0.773$ $0.598$ $2$ $3 \cdot 5^{2}$ 75.2 $$1.0 1 -0.0303 0 2.29496 Modular form 75.2.l.a.2.4 2-75-75.8-c1-0-7 0.773 0.598 2 3 \cdot 5^{2} 75.8$$ $1.0$ $1$ $0.307$ $0$ $4.28069$ Modular form 75.2.l.a.8.8 2-75-25.4-c1-0-3 $0.773$ $0.598$ $2$ $3 \cdot 5^{2}$ 25.4 $$1.0 1 -0.0167 0 3.35318 Modular form 75.2.i.a.4.4 2-75-15.8-c1-0-1 0.773 0.598 2 3 \cdot 5^{2} 15.8$$ $1.0$ $1$ $-0.0208$ $0$ $2.65308$ Modular form 75.2.e.b.68.2 2-75-5.4-c1-0-3 $0.773$ $0.598$ $2$ $3 \cdot 5^{2}$ 5.4 $$1.0 1 0.323 0 3.95467 Modular form 75.2.b.a.49.1 2-75-75.23-c1-0-0 0.773 0.598 2 3 \cdot 5^{2} 75.23$$ $1.0$ $1$ $-0.344$ $0$ $0.655154$ Modular form 75.2.l.a.23.2 2-75-25.6-c1-0-3 $0.773$ $0.598$ $2$ $3 \cdot 5^{2}$ 25.6 $$1.0 1 0.198 0 2.93910 Modular form 75.2.g.b.31.1 2-75-75.53-c1-0-1 0.773 0.598 2 3 \cdot 5^{2} 75.53$$ $1.0$ $1$ $-0.233$ $0$ $1.52137$ Modular form 75.2.l.a.53.4 2-75-25.14-c1-0-1 $0.773$ $0.598$ $2$ $3 \cdot 5^{2}$ 25.14 $$1.0 1 0.122 0 2.71360 Modular form 75.2.i.a.64.2 2-75-1.1-c1-0-2 0.773 0.598 2 3 \cdot 5^{2} 1.1$$ $1.0$ $1$ $0$ $0$ $3.24466$ Elliptic curve 75.c Modular form 75.2.a.c Modular form 75.2.a.c.1.1 2-75-75.62-c1-0-5 $0.773$ $0.598$ $2$ $3 \cdot 5^{2}$ 75.62 $$1.0 1 0.344 0 3.21972 Modular form 75.2.l.a.62.2 2-75-25.11-c1-0-2 0.773 0.598 2 3 \cdot 5^{2} 25.11$$ $1.0$ $1$ $-0.109$ $0$ $1.95996$ Modular form 75.2.g.c.61.2 2-75-75.62-c1-0-7 $0.773$ $0.598$ $2$ $3 \cdot 5^{2}$ 75.62 $$1.0 1 0.0880 0 4.05244 Modular form 75.2.l.a.62.8 2-75-75.8-c1-0-6 0.773 0.598 2 3 \cdot 5^{2} 75.8$$ $1.0$ $1$ $0.207$ $0$ $3.82721$ Modular form 75.2.l.a.8.6 2-75-75.47-c1-0-5 $0.773$ $0.598$ $2$ $3 \cdot 5^{2}$ 75.47 $$1.0 1 0.176 0 3.30404 Modular form 75.2.l.a.47.5 2-75-75.2-c1-0-3 0.773 0.598 2 3 \cdot 5^{2} 75.2$$ $1.0$ $1$ $0.0219$ $0$ $2.32786$ Modular form 75.2.l.a.2.6 2-75-75.17-c1-0-7 $0.773$ $0.598$ $2$ $3 \cdot 5^{2}$ 75.17 $$1.0 1 0.233 0 3.59345 Modular form 75.2.l.a.17.4 2-75-75.38-c1-0-7 0.773 0.598 2 3 \cdot 5^{2} 75.38$$ $1.0$ $1$ $0.451$ $0$ $4.71810$ Modular form 75.2.l.a.38.1 2-75-25.9-c1-0-3 $0.773$ $0.598$ $2$ $3 \cdot 5^{2}$ 25.9 $$1.0 1 0.237 0 4.10334 Modular form 75.2.i.a.34.4 2-75-25.21-c1-0-1 0.773 0.598 2 3 \cdot 5^{2} 25.21$$ $1.0$ $1$ $-0.198$ $0$ $1.73835$ Modular form 75.2.g.b.46.1 2-75-75.23-c1-0-6 $0.773$ $0.598$ $2$ $3 \cdot 5^{2}$ 75.23 $$1.0 1 0.0263 0 3.45848 Modular form 75.2.l.a.23.6 2-75-25.6-c1-0-5 0.773 0.598 2 3 \cdot 5^{2} 25.6$$ $1.0$ $1$ $0.433$ $0$ $4.12523$ Modular form 75.2.g.c.31.1 2-75-25.19-c1-0-2 $0.773$ $0.598$ $2$ $3 \cdot 5^{2}$ 25.19 $$1.0 1 0.0167 0 3.13315 Modular form 75.2.i.a.19.4 2-75-25.21-c1-0-0 0.773 0.598 2 3 \cdot 5^{2} 25.21$$ $1.0$ $1$ $-0.433$ $0$ $1.41239$ Modular form 75.2.g.c.46.1 2-75-25.16-c1-0-0 $0.773$ $0.598$ $2$ $3 \cdot 5^{2}$ 25.16 $$1.0 1 -0.0280 0 1.30193 Modular form 75.2.g.b.16.1 2-75-75.47-c1-0-1 0.773 0.598 2 3 \cdot 5^{2} 75.47$$ $1.0$ $1$ $-0.0767$ $0$ $2.05633$ Modular form 75.2.l.a.47.4 2-75-25.21-c1-0-2 $0.773$ $0.598$ $2$ $3 \cdot 5^{2}$ 25.21 $$1.0 1 -0.0258 0 2.48103 Modular form 75.2.g.c.46.2 2-75-75.2-c1-0-4 0.773 0.598 2 3 \cdot 5^{2} 75.2$$ $1.0$ $1$ $-0.169$ $0$ $2.53702$ Modular form 75.2.l.a.2.3	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, 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https://www.precisionmicrodrives.com/content/ab-025-using-spice-to-model-dc-motors/	# AB-025 : Using SPICE to Model DC Motors ### Introduction SPICE (Simulation Program with Integrated Circuit Emphasis) is an open source program for simulating electrical circuits. It enables engineers to model the behaviour of their circuits in software, which reduces prototyping costs and time. As DC motors have mechanical properties – i.e. they cannot be simulated accurately only considering electrical properties – it can be difficult to use them in SPICE. This Application Bulletin guides you through the principle of operation and extends the DC motor model for vibration motors and gear motors. ### Principles of Operation The equivalent circuit for a DC motor consists of an inductor, a resistor and a voltage source in series. These represent the coil inductance, coil resistance, and back EMF respectively. DC Motor Equivalent Circuit, Including Back EMF Voltage Source The back EMF voltage source is dependant on the speed of the motor and the torque constant Kτ. W, therefore, need a mechanical system model in order to calculate the speed of the motor. The SPICE engine doesn’t explicitly support mechanical models, however, it is facilitated by the use of electrical equivalent circuits. Basic Electrical and Mechanical Equivalent of a DC Motor Example parameters for the circuit: .PARAM K_t = 900u .PARAM K_EMF = 900u .PARAM L_Motor = 50u .PARAM R_motor = 5.5 .PARAM R_loss = 300n .PARAM L_inertia = 15n Current-dependent voltage sources are used to communicate information between the electrical and mechanical equivalent circuits. Note the two 0V voltage sources, V_sense_1 and V_sense_2. These do not alter the behaviour of the circuit and are simply used to provide convenient current measurements. The back EMF voltage source depends on the current sensed by V_Sense_2. ### Justification for the Mechanical Equivalent Circuit The table below outlines the variables in the mechanical circuit. Electrical Variable description, symbol, unit Mechanical Equivalent description, symbol, unit Voltage, V, [V] ≡ [kg⋅m2⋅A^-1⋅s^-3 ] Torque, τ, [N⋅m] ≡ [kg⋅m2⋅s^-2] Current, I, [A] Angular velocity, ω, [rad⋅s^-1] Resistance, R, [Ω] ≡ [kg⋅m2⋅A^-2⋅s^-3] Coefficient of viscous friction, μ, [kg⋅m2⋅s^-1] Inductance, L, [H] ≡ [kg⋅m2⋅A^-2⋅s^-2] Moment of Inertia, I, [kg⋅m2] Important Note: We can see from the above that Current and the Moment of Inertia share the same symbol, I. Pay careful attention to wither you’re working with the electrical or the mechanical parameters to avoid mistakes. The rotor torque in a DC motor is determined by the current through the coils and the torque constant, Kτ. Torque is represented as the voltage V_torque in the mechanical equivalent circuit. The resulting speed of the rotor, ω, is represented by the current in the mechanical circuit. So what limits this speed of the rotor? Anything in the mechanical circuit which limits the current, such as the resistor R_loss. The voltage dropped over R_loss depends on the current, as per Ohm’s law: $$V = I \times R$$ This acts to reduce the rotor torque, and it a velocity dependent loss. In other words, it represents a velocity dependent friction. The equivalent equation is: $$F = \mu \times \omega$$ Where μ is the coefficient of friction. This is a simple model of viscous friction (see more in the conclusion), and does not fully convey the complexities of friction in DC motors. It does, however, clearly illustrate the role of friction as an energy sink in the mechanical system which acts to reduce the speed of the motor. What does the inductor represent? L_inertia limits the current in the mechanical circuit when there is a change in voltage V_torque. The voltage dropped across the motor is given (remember, here I is current): $$V = – L \frac {dI}{dt}$$ The mechanical equivalent for which is (I now represents the moment of inertia): $$\tau = I \times \alpha = I \times \frac{d \omega}{dt}$$ Where $$\alpha$$ is angular acceleration. It is therefore clear that in our model the inductance L_inertia is the electrical equivalent variable of the moment of inertia of the rotor. ### Modelling the Transition to Steady State Behaviour Assume initially that the motor is unpowered (V_drive = 0V). After some time, a DC voltage is applied (V_drive = 3V). This change in voltage causes a change of current in the circuit. The coil inductance opposes this change and generates a voltage proportional to dI/dt which is known as back EMF. This is observed as the gradient of the initial current inrush (the greater the inductance, the lower the rate of change, the shallower the gradient). After the voltage is applied the motor begins to turn. The faster the motor turns, the greater the back EMF in the coils. This back EMF not only reduces the height of the initial current inrush but in the steady state means that the current draw is significantly lower than the motor terminal resistance alone would permit. This can be proven with a quick calculation from DC motor datasheet, such as the 106-002, by taking the Rated Voltage and dividing it by the Typical Terminal Resistance. For example, using the 106-002 we calculate $$\frac{3V}{16 \Omega} = 187.5 mA$$, which is much higher than the Typical N/L current of 17mA – caused by the back EMF limiting the current flow. The peak inrush current is identified as the first stationary point in the plot below (marker B). It is specified in our datasheets as “Max. Start Current”. Note that the peak inrush current is limited by back EMF, which is a function of the motor speed. Therefore a finite drive signal rise time, say 10μs instead of 0μs, affects this value. Remember that an instant transition between voltages is not physical. In a real application, it can be useful to limit the rise time of the driving signal for just this reason. Simulated DC motor performance A – finite $$\frac{dI}{dt}$$ due to coil inductance B – height of peak limited by back EMF C – current draw reaches steady state value as back emf and friction balance D – driving signal (green) terminated, large back emf spike Note: that a large negative voltage spike at D as the motor coil inductance resists the changing current. These spikes are why flyback diodes are recommended (such as the Schottky flyback diode recommended here). It can be seen that the coil inductance acts to limit the initial current inrush, whereas the back EMF acts to reduce the steady state current. We can model a motor in a stall condition as a sudden large increase in the mechanical equivalent circuit resistance. ### Determining Rotor Position The voltage at C1 provides the angular position of the rotor in radians. ### Modelling a Vibration Motor We can use this simple DC motor to plot the periodic acceleration of a test sled caused by a vibration motor. We assume: • a test sled of mass 100g • an eccentric point mass of 5g at radius 2mm from the rotor axis • the moment of inertia of the rotor to be negligible relative to the moment of inertia of the eccentric mass • the mass of the motor to be negligible relative to the mass of the test sled Considering the centripetal force of a mass rotating about a fixed point: $$F = m_{eccentric} \times r \times \omega ^{2}$$ so along an axis, say x, this force is given by: $$F_{x} = m_{eccentric} \times r \times \omega ^{2} cos(\theta)$$ Where theta ($$\theta$$) is the angular position of the rotor and r is the radius of the eccentric point mass. By Newton’s second law, we set equation 4 equal to the magnitude force upon the test sled, mass m_sled: $$m_{eccentric} \times r \times \omega ^{2} = m_{sled} a_{sled}$$ or using equation 6 and equating forces along a given axis x: $$m_{eccentric} \times r \times \omega ^{2} cos( \theta )= m_{sled} a _{sled,x}$$ Rearranging for $$a _{sled,x}$$: $$a_{sled,x} = \frac{m_{eccentric}\cdot r \cdot \omega ^{2}\cdot cos(\theta)}{m_{sled,x}} [m \cdot s^{-2}]$$ or in units of gravitational acceleration (where $$g= 9.81[ms^{-2}]$$ ): $$a_{sled,x} = \frac{m_{eccentric}\cdot r \cdot \omega ^{2}\cdot cos(\theta)}{ 9.81 \cdot m_{sled,x}} [g]$$ Inserting the constants assumed earlier, the following trace is added to the LTSpice plot window: .00001I(V_sense_2)I(V_sense_2)sin(V(n003)57.329) where V(n003) is the voltage measured at the capacitor in the integrator circuit (the particular node value may well be different to this in your model), and I(V_sense_2) is the current measured in the mechanical equivalent circuit. By default, LTSpice computes trigonometric functions with arguments in degrees. Thus, the argument in Trace Equation 1 is converted from radians by the multiplication factor of 57.329 ( = 360/(2*pi) ) . An example of this trace output can be seen in the stall condition model extension (here). ### Dimensional Analysis of Equations 9 and 11 Using the table in figure 3, we can perform a dimensional analysis of the final equations in order to check their validity. Analysing the units of the right hand side of equation 9: $$\frac{[m_{eccentric}]\cdot [r] \cdot [\omega ^{2}]\cdot cos(\theta)}{[m_{sled,x}]}$$ $$= \frac{ kg \cdot m \cdot s^{-2} }{kg}$$ $$= m \cdot s^{-2}$$ Which gives the expected result – the units are those of acceleration. What units would we expect to be displayed in the SPICE trace? Performing the same analysis on equation 11: $$.00001 \cdot I(V_{sense_2}) \cdot I(V_{sense_2}) \cdot sin(V(n_{003}) \cdot 57.329) = [A] \cdot [A] = [A^{2}]$$ As we have already included the gravitational constant and mass of the test sled in the multiplier “0.00001”, 1A2 in the trace window represents an acceleration of 1G, as normalised for a 100g test sled. ### Behavioural Extensions to the Model Extensions to the basic DC motor model are presented, with the intention of emulating the behaviour of real-world motors. #### 1) Separation of Rotor/Eccentric Mass Inertia, Inclusion of Gravity The model below improves upon the earlier model by separating the inertia of the rotor and eccentric mass. Additional voltage sources are also included in the mechanical equivalent circuit, including a source to account for the effect of gravity on the eccentric mass. An optional load torque voltage source is also included. Separation of Motor and Eccentric Mass Inertia in the Mechanical Circuit #### 2) Non-Zero Turn-On Voltage Gradually increase the driving voltage from $$V = 0$$ to $$V = V_{rated}$$ on a real-life DC motor and it will exhibit a finite turn-on voltage. At this voltage, the current through the motor coils results in a sufficient torque to overcome the inertia and static friction of the system and the rotor begins to turn. This behaviour is exhibited in the “typical performance characteristics” plots on our vibration motor datasheets as an apparent discontinuity in motor speed. A circuit Which Models a Finite Turn-on Voltage We define the forward voltage drop of a diode (D1) as 0.6V. Once V_drive exceeds this 0.6V limit, the diode conducts, and the voltage at A exceeds the threshold voltage of the voltage controlled switch. If the input V_drive drops below 0.7V at some later time, the voltage at A is clamped by C1, and thus this mechanical system remains unaffected by the static friction limit (which is correct, as the motor is moving). Example model parameters; PULSE(0 3 100m 100m 100m 100m) .tran 0 500m 0 1m .model turnoff SW(Vt = 1u Ron = 1n) .model turn_on D( Vfwd = 0.6 Rrev = 1n Ron = 1n) #### 3) Modelling a Reduction Gearbox A gearbox is modelled by the inclusion of an additional mechanical equivalent circuit. The current from the “internal” mechanical circuit is measured and divided by the reduction gear ratio. This determines the speed at the shaft of the motor. Additional terms for friction and inertia are included, as well as a voltage source into which a load torque profile can be set. A model of a gearmotor. Two integrator circuits represent the rotor and the (post-gearbox) shaft A “shaft feedback” voltage source is included in the internal mechanical circuit, the magnitude of which depends on the loading at the gearbox output shaft. #### 4) Stall Condition To model a transient stall, simply use a voltage controlled switch in the mechanical path. Triggering the opening of this switch at a given time t effectively introduces infinite mechanical resistance. A DC Motor Equivalent with a Voltage Dependant ‘Stall’ Switch in the Mechanical Section The “stall” switch is controlled by the voltage source V_stall. Example model parameters; .model SW SW(Vt = 0.0 Ron = 1p) .MODEL Stall SW(Vt = 0.01 Ron = 1n) Make sure that Ron is set, as it defaults (in LTSpice) to 1 Ω, which is much greater than the mechanical circuit resistance. This is a general principle to be adhered to – when placing parts in the model, ensure you are aware of the component default values, don’t assume them to be zero. Vibration Motor Characteristics, Stalled at 150ms and Turned off at 250ms Vibration Motor Characteristics, Stalled at 150ms and Released at 210ms Note that the 40ms between release and turn-off is insufficient for the back-EMF (red) to reach its maximum value. Note that it is also possible to model stall condition by using .STEP to perform parameter sweeps. ### A Note on the .STEP Directive The .STEP directive can be used to run sequential simulations with different component values. For example, to model a change in the rotor moment of inertia, simply vary the inertia of the mechanical equivalent circuit between two simulations as below: .STEP param L_inertia .00000002 1.00000002 1 The STEP parameter above is used to run two simulations. The inductance of L_inertia is varied from .00000002 to 1.00000002 in steps of 1. This .STEP directive will, therefore, run two consecutive simulations. The first simulation is a simple driving pulse of duration 100ms. The second run simulates a stall condition. This is modelled (using the .STEP directive) as a massive increase in L_inertia. ### Conclusion A SPICE model for a DC motor is presented. The model utilises a mechanical equivalent circuit to calculate the speed of the motor, taking into account the inertia and losses of the physical system. This is also used in determining the back EMF of the motor coils. Behavioural extensions to the model are presented, including a modification to model a stalled motor, inclusion of a gearbox, and finite turn-on voltage. The complexity of the model required by a designer is application dependent. Once satisfied with the behaviour of the model, the designer can define a subcircuit (.SUBCKT) and create a simple two-wire symbol for the device. For high precision position systems, it will be necessary to improve the handling of friction in the model. Techniques such as the generalized Maxwell slip model (below) are well covered in the literature, and considered outside the scope of this bulletin.	{"extraction_info": {"found_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": 2, "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": 0.8518326282501221, "perplexity": 1389.9581643851798}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-50/segments/1606141180636.17/warc/CC-MAIN-20201125012933-20201125042933-00215.warc.gz"}
http://math.stackexchange.com/users/21231/jalaj?tab=activity	Jalaj Reputation 361 Top tag Next privilege 500 Rep. Access review queues Sep24 awarded Autobiographer Dec14 awarded Yearling Mar20 accepted Is it mathematically correct to write $a \mod n \equiv b$? Mar18 comment Why are all the interesting constants so small? Waow. What a number! Mar18 answered Limit of $\arctan(x)/x$ as $x$ approaches $0$? Mar16 comment Is it mathematically correct to write $a \mod n \equiv b$? Thanks everyone! Mar16 comment Is it mathematically correct to write $a \mod n \equiv b$? What's the difference in the two notations? Should we not have $\equiv$ sign when relating an integer with an equivalence class? Mar16 asked Is it mathematically correct to write $a \mod n \equiv b$? Mar5 comment Trying to calculate RSA decryption key Is it not supposed to be $d \equiv e^{-1} \mod \phi(n)$ where $\phi(n)=(p-1)(q-1)$? Feb13 suggested rejected edit on Convergence/Divergence of infinite series $\sum_{n=1}^{\infty} \frac{(\sin(n)+2)^n}{n3^n}$ Feb11 revised Rank of matrices Made it more readable with latex entries instead of text style entry. Feb11 suggested approved edit on Rank of matrices Feb11 answered Formula to calculate when you will have a certain amount of money in your bank account Feb3 comment Equilateral triangle whose vertices are lattice points? Ohh yes! Thanks for pointing it out. Feb3 answered Equilateral triangle whose vertices are lattice points? Jan31 comment Perfect Squares ending in 576 I know, that's why I mentioned that modular arithmetic helps you out in these scenarios. Jan31 revised Perfect Squares ending in 576 added 44 characters in body Jan31 answered Perfect Squares ending in 576 Jan31 answered Trigonometric limit Jan31 comment Balanced Incomplete Block Design for magazine editor It is correct. I thought you are actually interested in constructing a BIBD and not just get the parameters.	{"extraction_info": {"found_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": 0.8696572780609131, "perplexity": 1480.6625999149387}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-27/segments/1435375095632.21/warc/CC-MAIN-20150627031815-00303-ip-10-179-60-89.ec2.internal.warc.gz"}
https://pbelmans.ncag.info/blog/2018/01/28/fortnightly-links-54/	• Alice Rizzardo and Michel Van den Bergh: A $k$-linear triangulated category without a model gives an example of (drumroll...) a $k$-linear triangulated category without a model, i.e. without an enhancement (as a dg category, or equivalently an $\mathrm{A}_\infty$-category), where $k$ is a field. It nicely complements the earlier (and much easier) preprint A note on non-unique enhancements by the same authors, where an example of $k$-linear triangulated category is given with two enhancements. Are there more possible pathologies of triangulated categories, linear over a field, that aren't resolved yet? • Remy van Dobben de Bruyn's preprints and drafts are available on his personal webpage. If you like your algebraic geometry to be done over fields of positive characteristic, then you're in for a treat, with some really nice results. • Tutorial: visualizing root systems is a tutorial for (drumroll...) visualising root systems in Sage. It is really useful that you can draw these things to get a feel for what the computations in a root system look like, and the results look stunning.	{"extraction_info": {"found_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": 0.820469081401825, "perplexity": 659.5781885457126}, "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/1573496668525.62/warc/CC-MAIN-20191114131434-20191114155434-00375.warc.gz"}
http://www.aot-math.org/&url=http:/www.aot-math.org/article_53654.html	# Linear preservers of two-sided right matrix majorization on $\mathbb{R}_{n}$ Document Type: Original Article Authors Rafsanjan University of Vali Asr Abstract A nonnegative real matrix $R\in \textbf{M}_{n,m}$ with the property that all its row sums are one is said to be row stochastic. For $x, y \in \mathbb{R}_{n}$, we say $x$ is right matrix majorized by $y$ (denoted by $x\prec_{r} y$) if there exists an $n$-by-$n$ row stochastic matrix $R$ such that $x=yR$. The relation $\sim_{r}$ on $\mathbb{R}_{n}$ is defined as follows. $x\sim_{r}y$ if and only if $x\prec_{r} y\prec_{r} x$. In the present paper, we characterize the linear preservers of $\sim_{r}$ on $\mathbb{R}_{n}$, and answer the question raised by F. Khalooei [Wavelet Linear Algebra \textbf{1} (2014), no. 1, 43--50]. Keywords ### History • Receive Date: 06 September 2017 • Revise Date: 03 December 2017 • Accept Date: 03 December 2017	{"extraction_info": {"found_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": 0.8693584203720093, "perplexity": 591.0467425753662}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-39/segments/1568514573331.86/warc/CC-MAIN-20190918193432-20190918215432-00431.warc.gz"}
https://byjus.com/question-answer/the-number-of-solutions-of-the-equation-x-y-z-10-where-x-y-and/	Question # The number of solutions of the equation $$x+y+z=10$$ where $$x,y$$ and $$z$$ are positive integers A 36 B 55 C 72 D 45 Solution ## The correct option is A $$36$$Given equation is $$x+y+z=10$$where $$x,y$$ and $$z$$ are positive integersFor any pair of positive integers n and k, the number of k-tuples of positive integers whose sum is n is equal to the number of $$(k − 1)$$-element subsets of a set with $$(n − 1)$$ elements.Both of these numbers are given by the binomial coefficient $${\displaystyle \textstyle {n-1 \choose k-1}}.$$$$\therefore$$ Required number of solutions $$={ _{ }^{ (10-1) }{ C } }_{ (3-1) }={ _{ }^{ 9 }{ C } }_{ 2 }=\cfrac { 9\times 8 }{ 2 } =36$$Maths Suggest Corrections 0 Similar questions View More People also searched for View More	{"extraction_info": {"found_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": 0.9137437343597412, "perplexity": 162.06787021745458}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320303884.44/warc/CC-MAIN-20220122194730-20220122224730-00564.warc.gz"}
https://fearlessmath.net/mod/page/view.php?id=112	## Find the Missing Angle on a Line Find the missing angle on a line. Sample: Take a look at these angles. Notice that they form a straight angle. Find the missing angle. Duane Habecker, 03/14/2010, Created with GeoGebra # Self-check Q1: What is the measure of the missing angle? Q2: What is the measure of the missing angle? Q3: What is the measure of the missing angle?	{"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": 0.9864176511764526, "perplexity": 1107.6851509065216}, "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/1631780057524.58/warc/CC-MAIN-20210924110455-20210924140455-00572.warc.gz"}
https://mathoverflow.net/questions/30536/t-structures-and-higher-categories	# t-structures and higher categories? I'm curious to find out where the viewpoint of higher categories may be useful so here is a somewhat vague question (which may or may not have a reasonable answer). Given a triangulated category, one can consider the set of all possible t-structures on it. Simple examples where one can compute things by hand indicate that this is something complicated but not hopelessly so. See for example the paper http://arxiv.org/abs/0909.0552 by Jon Woolf which describes a three parametric family of t-structures on the constructible bounded derived category of $\mathbf{P}^1(\mathbf{C})$ stratified by a point and its complement. Some of these t-structures are more interesting than others and there is one that is the most interesting of them all since by taking the bounded derived category of its heart one gets back the triangulated category one started with. (For that t-structure the heart is the category of perverse sheaves on $\mathbf{P}^1(\mathbf{C})$.) On the other hand, the set of t-structures on a triangulated category is interesting since there lurks somewhere the conjectural motivic t-structure whose existence implies Grothendieck's standard conjectures. See the recent paper http://arxiv.org/abs/1006.1116 by Beilinson. On the triangulated categories page of the n-category lab website it says "Therefore, all the structure and properties of a triangulated category is best understood as a 1-categorical shadow of the corresponding properties of stable (infinity,1)-categories". See http://ncatlab.org/nlab/show/triangulated+category. Note that this is quite a strong statement, since it is referring to all, and not just some, properties and structure of a triangulated category. So I'd like to ask: is there a higher categorical analog of a t-structure? More generally, how does the higher categorical viewpoint help one understand the set of all (or maybe all "nice" in an appropriate sense) t-structures on a given trangulated category, provided it is the homotopy category of a stable $(\infty,1)$ category? upd: as Mike points out in the comments, the answer to the first question is yes and it is given by proposition 6.15 of Lurie's Stable Infinity Categories. The second, more "philosophical" question remains. • I'm far from an expert, but I thought that stable infinity categories were the higher categorical analog of t-structures. My understanding is that it's actually a strength of the higher categorical theory that being stable is a property rather than an extra structure (as in the case of a t-structure on a triangulated category). You're probably already aware of arxiv.org/abs/math/0608228. – Mike Skirvin Jul 4 '10 at 18:51 • Mike -- thanks! The first question seems to be answered by proposition 6.15 there. But even for stable $(\infty,1)$ categories a t-structure is still an extra structure, not property. – algori Jul 4 '10 at 19:21 • I guess you are implicitly assuming your triangulated categories are small? It is certainly not the case otherwise that there is necessarily a set of t-structures and things do seem a little hopeless in that case. – Greg Stevenson Jul 4 '10 at 21:14 • Greg -- yes, to be completely precise one should fix a Grothendieck universe and require that set of objects belong to it; then so will the set of t-structures. But somehow it seems to me that these set theoretical subtleties are not very important here, which is why I'm willing to sweep them under the carpet. – algori Jul 4 '10 at 21:52 • algori: I think I misread the point of your original question: actually you want to formulate things in a purely infinity-categorical way? If so, of course it's reasonable, as you wrote, to define a t-structure as a certain kind of localization as in DAG I Prop. 6.15. I think it's worth keeping in mind, though, that this structure is completely encoded at the homotopy level. Admittedly, I have no idea whether all the structure of the collection of all t-structures still lives there. – Thomas Nevins Jul 6 '10 at 20:56 I wanted to contribute something because nobody's really explained why a t-structure on a stable $\infty$-category is the same thing as a t-structure on its homotopy category. It's not just because someone defined it as such in Definition 1.2.1.4 of Higher Algebra; it's because the most natural generalization is no generalization at all. Roughly speaking, a t-structure on a triangulated category should have the following three properties: • h1. It determines a full subcategory $\mathcal D_{\geq 0}$ closed under the shift functor $\Sigma$. • h2. $hom(X,Y) = 0$ if $X \in \mathcal D_{\geq 0}$ and $Y \in \mathcal D_{\leq -1}$. • h3. Any object fits into an exact triangle, sandwiched between an object in $\mathcal D_{\geq 0}$ and $\mathcal D_{\leq -1}$. If you think about the algebraic manipulations for which one utilizes t-structures, the natural generalization of a t-structure to a stable $\infty$-category ought to be as follows: • 1$.$ It determines a full subcategory $\mathcal C_{\geq 0}$ closed under the shift functor $\Sigma$. • 2$.$ $hom(X,Y)$ is contractible if $X \in \mathcal C_{\geq 0}$ and $Y \in \mathcal C_{\leq -1}$. • 3$.$ Any object fits into an fiber sequence, sandwiched between an object in $\mathcal C_{\geq 0}$ and $\mathcal C_{\leq -1}$. If $\mathcal C$ is stable, one defines a triangulated structure on its homotopy category $\mathcal D = ho \mathcal C$ by demanding that its exact triangles are precisely those arising from fiber sequences. Then we see immediately that (1)-(3) imply (h1)-(h3). That is, we have a map from "oo-categorical" t-structures on $\mathcal C$ to t-structures on $ho \mathcal C$. What might be surprising is that the other direction holds, but it's actually an easy exercise to show that any collection of data satisfying (h1)-(h3) uniquely determines data satisfying the properties (1)-(3). In a nutshell: (h1) implies (1) obviously. (h1) and (h2) together imply (2) because you can write a fiber sequence for $X[i]$ and $X[i+1]$, then apply the hom functor to get a fiber sequence of mapping spaces. The long exact sequence of homotopy groups will prove that every homotopy group of hom(X,Y) vanishes. Finally, by definition of exact triangles for $ho\mathcal C$, (h3) also implies (3). So we have an interesting phenomenon where the set of t-structures can't distinguish between two oo-categories with equivalent homotopy categories. This is one of the main reasons that Bridgeland stability conditions don't have an obvious generalization to see higher-homotopical structures beyond the homotopy category. • If I think about the algebraic manipulations for which one utilizes t-structures, what comes to my mind are the torsion and torsionfree classes of a torsion theory. And when I think to them, what comes to my mind are reflective factorization systems. I appreciate your explanation, but I have to disagree that as you say "the natural generalization [here] is no generalization at all": the ambient where axioms (1,2,3) live says $\Sigma$ is a genuine (homotopy) colimit, in opposition to the classical setting; this makes (has made, working out our paper) a incredible difference [continue] – Fosco Aug 25 '14 at 7:12 • Since for example you don't have to ask that $\mathcal C_\ge$ / $\mathcal C_\le$ is closed under $\Sigma$ / $\Omega$ any more; it's automatic from its being coreflective / reflective. Another point, maybe a minor point: what is a t-structure? This could be a personal opinion (undoubtedly it is) but definitions should be self-explanatory; when they are not, something is missing. And, this is absolutely not an arrogant statement, but in our setting the definition of t-structure is self-explanatory [continue] – Fosco Aug 25 '14 at 7:18 • t-structures are built up from those data (a pair of subcategories with properties blah blah) since they come from factorization systems: orthogonality of the objects come from orthogonality between arrows, closure under loop/suspension comes from (co)reflectivity, as well as the presence of suitable fiber sequences; t-structures abound since in a stable $\infty$-category it is extremely simple to be a simple factorizations system (Remark 3.12 of our work). Not to mention all the other results: the heart of a t-structure is abelian since you do the only possible thing [continue] – Fosco Aug 25 '14 at 7:22 • like in the most self-explanatory Algebra proofs: you have some data, you use it in the only meaningful way. And "Harder-Narashiman filtrations"... These are blatantly Postnikov towers, which are a well known byproduct of factorization systems in $\omega$ stages. From our point of view it's rather obvious that these decompositions exist, induced from the given factorization system / t-structure: every object admits a "tower" $X\to A_1\to \dots A_n\to \dots \to Y$ where each $A_i$ has fiber $K_i[n_i]$ and $K_i$ lives in the heart. – Fosco Aug 25 '14 at 7:27 • here is Domenico's opinion (he has to post from his phone but didn't succeed): I don't think Hiro is meaning that the $\infty$-point of view is saying nothing here. On the contrary, I think he is pointing out that by speaking of the $\infty$-categorical version one is really speaking of the very same thing as the classical version on the homotopy category. Not something more general nor more special. It is precisely this that tells that one can use $\infty$-eyes to look at the old thing. And that with $\infty$-eyes one can see better than with 1-eyes, no doubt. [continue] – Fosco Aug 25 '14 at 10:22 So I'd like to ask: is there a higher categorical analog of a t-structure? As Mike Skirvin pointed out in a comment, higher categorical analog of t-structures have been introduced by Lurie. A more up-to-date reference might be Higher Algebra ($\S$ 1.2.1). More generally, how does the higher categorical viewpoint help one understand the set of all (or maybe all "nice" in an appropriate sense) t-structures on a given trangulated category, provided it is the homotopy category of a stable (∞,1) category? I guess the answer can be found at the same place. There, Lurie says that "there is a bijective correspondence between $t$-localizations of $\mathcal C$ (a stable $\infty$-category) and $t$-structures on the triangulated category $h\mathcal C$. The higher categorical point-of-view also seems to be useful to understand the yoga of derived functors in a more conceptual way. In Section 1.3 of the same reference (Higher Algebra) it is explained that if $\mathcal A$ is an abelian category with enough injectives, then its derived $\infty$-category $\mathcal D^-(\mathcal A)$ is stable, admits a $t$-structure, has homotopy category the standard derived category, and satisfies the following universal property: there is a canonical equivalence of abelian categories $\mathcal A\to \mathcal D^-(\mathcal A)^\heartsuit$, and if $\mathcal C$ is a stable $\infty$-category with a left-complete $t$-structure then any right exact functor $\mathcal A\to\mathcal C^\heartsuit$ extends (in an essentially unique way) to an exact functor $\mathcal D^-(\mathcal A)\to \mathcal C$. I feel the urge to answer you even if I'm not an expert (!), since now I can say that yes, one can do extremely better than referring a t-structure to its homotopy category. In this work D. Fiorenza and I prove that t-structures on a stable $\infty$-category $\cal C$ are uniquely determined by a reflective, normal factorization system on $\cal C$. In a few words, a factorization system $(\cal E,M)$ is reflective when both classes satisfy the 3-for-2 property (the same of weak equivalences in a model category), and normal when the class $\cal E$/$\cal M$ satisfy a rather peculiar pullback/pushout stability property, which can be packaged in a single self-dual request: denoting $R,S$ respectively the reflection and coreflection functors associated to the reflective factorization $(\cal E,M)$ (see nLab page for an explanation: the correspondence whose $\infty$-categorical analogue we exploit was first presented in the Cassidy-Hebert-Kelly paper "Reflective subcategories, localizations and factorization systems"), this factorization is normal in a stable category $\cal C$ iff the square $$\begin{array}{ccc} SA & \to &A \\ \downarrow && \downarrow \\ 0 &\to & RA \end{array}$$ is a pullback/pushout. Read between the lines and notice that $R,S$ are nothing more than the truncation functors $\tau_{\ge 0}, \tau_{<0}$ of the t-structure, and the universality request for a normal, reflective factorization system is telling you that "every object lies in a distinguished triangle blah blah". And now, fortified by $\infty$-category theory, re-read the definition of t-structure with a new eye :) Theorem 3.13 in the paper is the gist of the discussion: notice that the bijection of the correspondence established there holds for simple, well-motivated categorical reasons. Proofs are incredibly easy-to-follow, so easy that an entire section contains exercises for the reader! And finally, there's no need to refer to the homotopy category even in the definition of the heart of the t-structure, which can be characterized in terms of the factorization system alone: abelianity of the heart is now automatic (see Exercises 4.1, 4.2) as a consequence of "adjacency" (Remark 4.2) of the torsion and torsion-free classes of the t-structure. A more precise discussion of the points we leave as exercises will appear in a subsequent paper (denoted as [FLa] in the bibliography), together with examples from algebraic topology (Bousfield "localization" is a t-structure on spectra), algebraic geometry (I guess I have to thank the OP for giving the references above) and other realms: hope you'll be one of our readers! • Are you saying that a semiorthogonal decomposition is not a t-structure where the reflection and coreflection $\tau_\ge, \tau_\le$ are exact? – Fosco Aug 24 '14 at 17:08 • Oh, I am sorry for the misunderstanding. I don't know the answer yet, but to the best of my knowledge a semiorthogonal deocmposition is simply a t-structure where the reflection and coreflection (the truncation in the two directions) are exact functors; this entails that the two "adjacent" subcategories of the t-structure are triangulated (or $\infty$-stable) subcategories. I warmly invite anybody to contradict me if I'm wrong, I'm still learning! – Fosco Aug 24 '14 at 17:26	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8742501735687256, "perplexity": 498.5709047296563}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-39/segments/1568514573173.68/warc/CC-MAIN-20190918003832-20190918025832-00172.warc.gz"}
https://www.seykota.com/rm/Bernoulli_approach/index.htm	The Bernoulli Approach Using The Bernoulli Principle to Explain Lift "It would be better for the true physics if there were no mathematicians on earth." Daniel Bernoulli (Feb 8, 1700 - Mar 17, 1782) Daniel Bernoulli was the son and nephew of the respected mathematicians Johann Bernoulli and Jakob Bernoulli. He was influenced by Sir Issac Newton (1642-1727) and by Leonhard Euler (1707-83). From 1725-1749 he won prizes for work in astronomy, gravity, tides, magnetism, ocean currents, and the behavior of ships at sea. Around 1738 he wrote Hydrodynamica, established the basis for the kinetic theory of gases, discovered how to measure blood pressure, considered the basic properties of fluid flow, pressure, density, and velocity, and presented the fundamental relationship now known as the Bernoulli Principle. He also proposed the Saint Petersburg paradox, concerning a strategy for geometrical betting. Accounting for lift with the Bernoulli Principle Thousands of Internet sites, many by educational institutions and by students, account for lift with the Bernoulli Principle. Typically, they interpret Bernoulli's Principle as: high velocity fluid causes low pressure. They use this relationship to account for a variety of lift phenomena, such as airplanes, curve balls in baseball, ping-pong balls sticking in funnels, perfume atomizers, levitators and other such devices. A standard Bernoulli Principle diagram, such as this, appears in numerous textbooks and on countless web sites. Origin of the Bernoulli Principle The Bernoulli Principle derives from the physics of a closed hydraulic loop of constant density, with varying cross sections and with no internal flow motivation. From these constraints, since the total energy in such a system must be everywhere equal, a decrease in pressure (thermal) energy must balance an increase in kinetic energy (given constant gravitational energy). Since fluid flowing through a narrow cross section flows faster, it has higher kinetic energy and must have lower pressure. From this, the popular interpretation of Bernoulli's Principle: fast flow means low pressure. Standard Bernoulli Principle Diagram A standard pipe diagram, such as this, appears in textbooks and on web sites. It assumes steady, incompressible (constant density), friction-free, unmotivated flow. It seems unclear from this diagram (1) how to motivate the fluid flow and (2) that the fluid flow is actually part of the same constant-density closed loop. The Fundamental Bernoulli Principle Assumptions The Bernoulli Principle assumes (1) a closed-loop system of varying cross section, (2) constant density of the fluid and (3) no flow motivation. While this situation might obtain in a theory, in practice there is simply no such thing as a closed hydraulic loop in which fluid circulates with no flow motivation. The velocity in such a loop would have to be zero, so the equation would only be valid at v = 0, at which point lift would also be zero. Scientists interested in understanding lift have had no alternative simple scientific rule to relate lift to fluid flow. So despite the fundamental assumptions, the Bernoulli Principle came to fill the conceptual vacuum and became the standard explanation for lift. The Mathematics of the Bernoulli Principle The Bernoulli Principle is a statement of energy balance. It correctly states that for certain conditions (steady flow, no friction and incompressible fluid), the sum of the thermal, kinetic and gravitational energies is constant at all locations. Thus, for fluid moving through a system of pipes of varying cross section, the fluid that moves through narrow pipes, moves fast and has low pressure while the fluid that moves through wide pipes moves slowly and has high pressure. The formula for this principle is: ET + EK + EG = k For systems that do not involve gravity change, this becomes ET + EK = k Thermal energy + kinetic energy is everywhere constant. Now since ET = PV and EK = Mv²/2 PV + Mv²/2 = k The product of Pressure and Volume plus 1/2 the product of Mass and the square of velocity is everywhere constant. So for flow from a large diameter pipe to a small diameter pipe within a closed system, velocity (v) increases so pressure (P) falls. Students often interpret Bernoulli's Principle to say that that fast flow "causes" low pressure. In practice, the opposite may be true since it takes a pressure gradient to "cause" fluids to flow. Continuing, Dividing both sides by volume, P + dv2/2 = k and, rearranging, P = k - dv2/2 So, an increase in velocity indicates a decrease in pressure. Bernoulli's Law, by Jakob Bernoulli is an altogether different concept; it states that a large number of items chosen at random from a population will, on the average, have the characteristics of the population. Bernoulli's Principle, by Daniel Bernoulli has to do with fluid flow. Text Book Examples The examples below, from standard textbooks, show examples of using the Bernoulli Principle to explain lift. (Click on the problem number, or the tab at the top of the page, to view the problem and solution.) 1. Halliday and Resnick Problem # HR-74-P 2. Halliday and Resnick Problem # HR-75-P No diagram. 3. Potter and Foss Problem # 2.32 Conclusions and Observations The Bernoulli Principle appears widely in explanations for lift, even in those which seem to compromise the qualifying assumptions of incompressibility (constant density), no flow motivation, and a closed fluid loop. Proponents of the application of Bernoulli's Principle claim that the problem with the assumptions are mostly minor artifacts and point out that the density assumption becomes moot if the velocity is low enough to assure incompressible flow. However, at low velocity, approaching zero, as the mathematical problems disappear, so does the lift. Other factors seem to favor choosing the Bernoulli Principle to address lift problems. There has been no easily comprehensible theory of lift which acknowledges the importance of density effects. A closed-form derivation of the Navier-Stokes equation for compressible flow presents a formidable and daunting mathematical challenge. Numerical solutions seem to require substantial computational power and re-formulations of basic fluid dynamics equations into unfamiliar forms compatible with integral calculus. Above all, the Bernoulli Principle to explain lift has stood the test of time and has gained wide popularity and acceptance.	{"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": 0.9288181662559509, "perplexity": 1201.0540488498375}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-27/segments/1656104496688.78/warc/CC-MAIN-20220704202455-20220704232455-00674.warc.gz"}
https://forum.bebac.at/forum_entry.php?id=17713&category=22&order=time	## Initial sample size guess for the Potvin methods [Two-Stage / GS Designs] Hi Hötzi, are you sure we are on the same page? I am talking about the initial guess for the sample size iteration, not the sample size itself. Pass or fail! ElMaestro	{"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": 0.8652024269104004, "perplexity": 2672.804780394018}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-21/segments/1620243988961.17/warc/CC-MAIN-20210509062621-20210509092621-00229.warc.gz"}
https://www.physicsforums.com/threads/work-negative-sign.574657/	# Work negative sign 1. Feb 5, 2012 ### xzibition8612 1. The problem statement, all variables and given/known data See attachment 2. Relevant equations 3. The attempt at a solution Why is it -umg(50)? Why negative? Also I think this has something to do with conservative force and friction, and I'm very confused on this. Any help would be appreciated thanks. File size: 31.2 KB Views: 72 File size: 29.1 KB Views: 75 2. Feb 5, 2012 ### Curious3141 When the force is constant, Work = Force*Displacement. In this case, it's negative because energy is being taken away from the cars by the frictional force, while they're being decelerated to a stop. The energy dissipated by friction is equal to the drop in kinetic energy (as each car comes to rest). This is conservation of energy. μ is the coefficient of kinetic friction. mg is the weight (equal to the normal force on the car). 50ft is the distance through which car 1 moves. BTW, there's an error in the second part of the solution. The expression for car 2 should be $\sqrt{(2)(20)(0.5)(32.2)}$ (i.e. 20, not 50), since 20ft is the distance car 2 takes to be brought to rest. The final stopping distance was calculated correctly, though. Similar Discussions: Work negative sign	{"extraction_info": {"found_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": 0.8480625152587891, "perplexity": 862.9576037281661}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-47/segments/1510934806225.78/warc/CC-MAIN-20171120203833-20171120223833-00228.warc.gz"}
https://okayama.pure.elsevier.com/en/publications/measurement-of-the-muon-neutrino-charged-current-single-%CF%80-product	# Measurement of the muon neutrino charged-current single π+ production on hydrocarbon using the T2K off-axis near detector ND280 The T2K Collaboration Research output: Contribution to journalArticlepeer-review 8 Citations (Scopus) ## Abstract We report the measurements of the single and double differential cross section of muon neutrino charged-current interactions on carbon with a single positively charged pion in the final state at the T2K off-axis near detector using 5.56×1020 protons on target. The analysis uses data control samples for the background subtraction and the cross section signal, defined as a single negatively charged muon and a single positively charged pion exiting from the target nucleus, is extracted using an unfolding method. The model-dependent cross section, integrated over the T2K off-axis neutrino beam spectrum peaking at 0.6 GeV, is measured to be σ=(11.76±0.44(stat)±2.39(syst))×10-40 cm2 nucleon-1. Various differential cross sections are measured, including the first measurement of the Adler angles for single charged pion production in neutrino interactions with heavy nuclei target. Original language English 012007 Physical Review D 101 1 https://doi.org/10.1103/PhysRevD.101.012007 Published - Jan 21 2020 ## ASJC Scopus subject areas • Physics and Astronomy (miscellaneous) ## Fingerprint Dive into the research topics of 'Measurement of the muon neutrino charged-current single π+ production on hydrocarbon using the T2K off-axis near detector ND280'. Together they form a unique fingerprint.	{"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": 0.966696560382843, "perplexity": 4523.161504959418}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-33/segments/1659882571472.69/warc/CC-MAIN-20220811133823-20220811163823-00125.warc.gz"}
https://edurev.in/course/quiz/attempt/18484_Test-Magnetostatics-2/393c2c59-24ba-4705-8823-d13fc43acf6a	Test: Magnetostatics- 2 20 Questions MCQ Test Topicwise Question Bank for Electrical Engineering \| Test: Magnetostatics- 2 Description Attempt Test: Magnetostatics- 2 \| 20 questions in 60 minutes \| Mock test for Electrical Engineering (EE) preparation \| Free important questions MCQ to study Topicwise Question Bank for Electrical Engineering for Electrical Engineering (EE) Exam \| Download free PDF with solutions QUESTION: 1 A circular conductor of 1 cm radius has an internal magnetic field where , r0 is the radius of the conductor and is the unit vector. The total current in the conductor is given by Solution: By Ampere's law, Hence, current enclosed Given, r0 = 1 cm = 0.01 m QUESTION: 2 What is the value of mutual inductance between an infinitely long straight conductor along the y-axis and a rectangular single turn coil situated in x-y plane with its corners located at point (a, 0), (a + d, 0), (a, h) and (a + d, h) as shown in figure below? Solution: The mutual inductance is given by Flux density at any point P in the x-y plane is and QUESTION: 3 Assertion (A): The force acting between two parallel wires carrying currents is directly proportional to the individual currents and inversely proportional to the square of the distance between them, Reason (R): The force is repulsive if the two currents I1 and I2 are in the same directions and attractive if in opposite directions. Solution: The force acting between the two parallel wires carrying currents I2 and I2 respectively is given by Thus, Hence, assertion is a correct statement. Reason is a false statement because F will be attractive if the two currents I2 and I2 will be in the same directions and repulsive if in opposite directions. QUESTION: 4 The energy stored in a magnetic field is given by Solution: QUESTION: 5 Lorentz force law is Solution: QUESTION: 6 The equation is the generalization of Solution: QUESTION: 7 The unit of relative permeability is Solution: Relative permeability, μr is a dimensionless quantity. QUESTION: 8 Consider the volume current density distribution in cylindrical co-ordinates as Where a and b are inner and outer radii of the cylinder as shown in figure below Now, consider the value of magnetic field intensity in various regions I, II and III respectively obtained as: Which of the above found values of H are correct for the respective regions? Solution: • For region-I (0 < r < a), J (r, ϕ, z) = 0 Using Ampere’s law, we have or , H·2πr = 0 or H = 0 • For region - II Again, or, • For region - III (b < r < ∞), J = 0 Again, QUESTION: 9 Assertion (A): It is not possible to have isolated magnetic charges. Reason (R): The magnetic flux lines always close upon themselves. Solution: Since, magnetic flux lines always close upon themselves therefore it is not possible to have isolated magnetic charges (or poles). QUESTION: 10 Match List-l with List-lI and select the correct answer using the codes given below the fists: List-l A. Gauss’s law for magnetostatic fields B. Gauss’s law for electrostatic fields C. Conservativeness of electrostatic fields D. Ampere's law List-lI Codes: A B C D (a) 5 4 1 2 (b) 3 1 4 5 (c) 3 4 1 5 (d) 5 1 4 2 Solution: QUESTION: 11 The magnetic vector potential is given by The total magnetic flux crossing the surface ϕ = π/2 , 1 ≤ ρ ≤ 2 m , 0 ≤ z ≤ 5 m is Solution: The total magnetic flux crossing the surface ϕ = π/2 , 1 ≤ ρ ≤ 2 m , 0 ≤ z ≤ 5 m is given as: = 3.75 Wb QUESTION: 12 Which of the following is not a source of magnetostatic fields? Solution: An accelerated charge will produce both time varying electric and magnetic field called electromagnetic field. QUESTION: 13 Consider the following statements associated with the characteristics of static magnetic field: 1. It is solenoidal. 2. Magnetic flux lines are not closed in certain cases. 3. The total number of flux lines entering a given region is equal to the total number of flux lines leaving the region. 4. It is conservative. Which of the above statements is/are not correct? Solution: • Static magnetic field is non-conservative unlike static electric field. Hence, statement-4 is not correct. • Magnetic flux lines are always closed. Hence, statement-2 is not correct. • Statement 1 and 3 are correct. QUESTION: 14 What are the units for the product of two values whose respective units are Henrys and Amperes? Solution: The Henry, the unit for mutual inductance is equivalent to V-sec/Ampere. So, H x Ampere = Volt-sec = V-s QUESTION: 15 What are the possible dimensions for a rectangular coil that has a magnetic flux of 9.5 webers when in a magnetic field of strength 19 Tesla at an angle of 60° from the perpendicular to the plane of the coil? Solution: Magnetic flux φB = BA cosθ or, Hence, the area of the rectangular coil should be 1 m2. Only option (d) is matching the answer i.e. 250 cm x 40 cm = 10000 cm2 = 1 m2. QUESTION: 16 What is the change in magnetic fiux in a coil of area 5 m2 as its orientation relative to the perpendicular of a uniform 3.0 T magnetic field changes from 45° to 90°? Solution: Flux ϕ = B A cosθ At θ = 45°, ϕ1= 3 x 5 x cos 45° = 11 Wb At θ = 90°, ϕ2 = 3 x 5 x cos 90° = 0 Wb ∴ Change in flux = ϕ2 - ϕ1 = -11 Wb QUESTION: 17 If a vector field is solenoidal, which of these is true? Solution: (Non-existence of monopole) QUESTION: 18 If , the value of around the closed circular quadrant shown in the given figure is Solution: QUESTION: 19 The Maxwell’s equation, is based on Solution: QUESTION: 20 Given a vector field in cylindrical coordinates. For the contour as shown below, is Solution: Use Code STAYHOME200 and get INR 200 additional OFF Use Coupon Code	{"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": 0.9342736005783081, "perplexity": 1529.0390303357258}, "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-2022-21/segments/1652662545548.56/warc/CC-MAIN-20220522125835-20220522155835-00324.warc.gz"}
https://eprints.lancs.ac.uk/id/eprint/63712/	# Search for the associated production of a b quark and a neutral supersymmetric Higgs boson which decays to tau pairs Collaboration, D0 and Bertram, Iain and Borissov, Guennadi and Fox, Harald and Williams, Mark and Ratoff, Peter and Love, Peter and Sopczak, Andre (2010) Search for the associated production of a b quark and a neutral supersymmetric Higgs boson which decays to tau pairs. Physical review letters, 104 (15). ISSN 1079-7114 Preview PDF PhysRevLett.104.151801.pdf - Published Version ## Abstract We report results from a search for production of a neutral Higgs boson in association with a $b$ quark. We search for Higgs decays to $\tau$ pairs with one $\tau$ subsequently decaying to a muon and the other to hadrons. The data correspond to 2.7fb$^{-1}$ of $\ppbar$ collisions recorded by the D0 detector at $\sqrt{s} = 1.96$TeV. The data are found to be consistent with background predictions. The result allows us to exclude a significant region of parameter space of the minimal supersymmetric model. Item Type: Journal Article Journal or Publication Title: Physical review letters © 2010 The American Physical Society Submitted to Phys. Rev. Letters Uncontrolled Keywords: /dk/atira/pure/subjectarea/asjc/3100 Subjects: Departments: ID Code: 63712 Deposited By: Deposited On: 26 Apr 2013 08:53 Refereed?: Yes Published?: Published	{"extraction_info": {"found_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": 0.9464766979217529, "perplexity": 3154.662915167715}, "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/1659882572021.17/warc/CC-MAIN-20220814083156-20220814113156-00780.warc.gz"}
https://brilliant.org/discussions/thread/sophie-germains-identitydivisibilitynumber-theory/	# Sophie Germain’s identity(divisibility,number theory) The most useful formula in competitions is the fact that $a-b \| a^n-b^n$ for all n, and $a+b \| a^n+b^n$ for odd n.We have $a^2-b^2=(a-b)(a+b)$. But a sum of two squares such as $x^2 + y^2$ can only be factored if 2xy is also a square. Here you must add and subtract 2xy. The simplest example is the identity of Sophie Germain: $a^4 + 4b^4 = a^4 + 4a^2.b^2 + 4b4 - 4a^2.b^2 = (a^2 + 2b^2)2 - (2ab)^2 = (a^2 + 2b^2 + 2ab)(a^2 + 2b^2 - 2ab)$ Some difficult Olympiad problems are based on this identity. For instance, in the 1978 Kurschak Competition, we find the following problem which few students solved. example:1 $n > 1 ⇒ n^4 + 4^n$ is never a prime. If n is even, then $n^4 +4^n$ is even and larger than 2. Thus it is not a prime. So we need to show the assertion only for odd n. But for odd $n = 2k + 1$, we can make the following transformation, getting Sophie Germain’s identity: $n^4 + 4^n = n^4 + 4·4^{2k} = n^4 + 4 · (2k)^4$ which has the form $a^4 + 4b^4$. This problem first appeared in the Mathematics Magazine 1950. It was proposed by A. Makowski, a leader of the Polish IMO-team. Quite recently, the following problem was posed in a Russian Olympiad for 8th graders: Note by Chakravarthy B 7 months, 2 weeks ago This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science. When posting on Brilliant: • Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused . • Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone. • Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge. MarkdownAppears as italics or _italics_ italics bold or __bold__ bold - bulleted- list • bulleted • list 1. numbered2. list 1. numbered 2. list Note: you must add a full line of space before and after lists for them to show up correctly paragraph 1paragraph 2 paragraph 1 paragraph 2 [example link](https://brilliant.org)example link > This is a quote This is a quote # I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world" # I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world" MathAppears as Remember to wrap math in $$ ... $$ or $ ... $ to ensure proper formatting. 2 \times 3 $2 \times 3$ 2^{34} $2^{34}$ a_{i-1} $a_{i-1}$ \frac{2}{3} $\frac{2}{3}$ \sqrt{2} $\sqrt{2}$ \sum_{i=1}^3 $\sum_{i=1}^3$ \sin \theta $\sin \theta$ \boxed{123} $\boxed{123}$ Sort by: Did anyone understand this? - 7 months, 2 weeks ago Yeah I did........ although, there is a typo.....The question should be $n^4+4^n$ instead of $n^4+4n$ - 7 months, 1 week ago Ok. I changed it. - 7 months, 1 week ago No you didn't. It is still the same....... - 7 months, 1 week ago Once check - 7 months, 1 week ago Yup, now it is fine....!! - 7 months, 1 week ago	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 20, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9672008752822876, "perplexity": 2109.754587027088}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-43/segments/1570986668569.22/warc/CC-MAIN-20191016113040-20191016140540-00132.warc.gz"}
https://socratic.org/questions/how-do-you-simplify-3a-2-2a-3-2	Algebra Topics # How do you simplify (-3a^2)(2a^3)^2? Sep 23, 2015 The answer is $- 12 {a}^{8}$. #### Explanation: $\left(- 3 {a}^{2}\right) {\left(2 {a}^{3}\right)}^{2}$ Apply exponent rule ${\left({a}^{m}\right)}^{n} = \left({a}^{m \cdot n}\right)$. Simplify ${\left(2 {a}^{3}\right)}^{2}$ to $\left(4 {a}^{6}\right)$ . Because in $\left(- 3 {a}^{2}\right) {\left(2 {a}^{3}\right)}^{2}$ the square is being applied to both terms i.e ${\left(2\right)}^{2}$ and ${\left({a}^{3}\right)}^{2}$ Now, $\left(- 3 {a}^{2}\right) \left(4 {a}^{6}\right)$ Apply exponent rule $\left({a}^{m} \cdot {a}^{n}\right) = {a}^{m + n}$. Simplify. $- 3 \times 4 \times {a}^{2 + 6} =$ $- 12 {a}^{8}$ ##### Impact of this question 183 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": 12, "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": 0.9050043821334839, "perplexity": 4747.103582191817}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-39/segments/1568514574588.96/warc/CC-MAIN-20190921170434-20190921192434-00384.warc.gz"}
https://www.physicsforums.com/threads/finite-differencing-dynamic-boundary.826108/	# Finite Differencing Dynamic Boundary 1. Aug 3, 2015 ### joshmccraney Hi PF! I'm using a finite differencing scheme to solve the following $$h_t = h h_{zz} + 2h_z^2$$ where the subscripts denote partial derivatives. The difficulty I'm facing is the boundary conditions are dynamic, and move with time $t$. This makes choosing a $\Delta z$ very difficult and unintuitive for me. I have already developed a MatLab code that works pretty well but I feel my $\Delta z$ should change in time. If anyone knows any references or has advice on this issue of dealing with the $\Delta z$ please let me know. Thanks so much! Josh 2. Aug 4, 2015 ### soarce Try to change the time variable in order to achieve time independent boundary conditions. 3. Aug 4, 2015 ### joshmccraney I've already performed a similarity transform, which can actually be solved analytically under certain physical situations! But I'm trying to solve before the transform, so moving boundary conditions is all I can work with. 4. Aug 4, 2015 5. Aug 4, 2015 Thanks! 6. Aug 4, 2015 ### Staff: Mentor On a problem like this, what I would do is change variables, so that the z variable is as follows: $$z^=\frac{(z-z_{left}(t))}{(z_{right}(t)-z_{left}(t))}$$ You would have to transform the differential equation so that the partial derivative with respect to t is at constant z versus z. But the advantage would be that the boundary conditions could always be applied at z* = 0 and z* = 1. Basically what you are doing is taking the complexity out of the boundary conditions and putting it into the differential equation where it can be handled more easily. Chet 7. Aug 5, 2015 ### joshmccraney Chet, this is exactly what the department chair recommended. We then have $$\frac{\partial}{\partial z} = \frac{1}{z_r(t)-z_l(t)}\frac{\partial}{\partial z^}$$Unfortunately at the moment I need to solve this problem in the $z$ space, not the $z^$ space. This being said, later I'll look into this method. Would you mind if I asked you for advice then if I run into any trouble? I do have a well-prepared strategy to proceed, but I'm wondering how to use finite differencing techniques to solve for the last spacial node in my domain? Currently I'm using a forward time centered space scheme. During each time loop, I use FD to recalculate each $h$ value, which gives me something like this $$\frac{h_i^{j+1}-h_i^j}{\Delta t} \approx h_i^j \left( \frac{h_{i+1}^j-2h_i^j+h_{i-1}^j}{\Delta z^2} \right) + 2\left( \frac{h_{i+1}^j-h_{i-1}^j}{2 \Delta z} \right)^2$$ where the subscript denotes space and the superscript denotes time. To find $h^{j+1}_{right}$ should I allow $h_{right}^j=0$ since that is what the right end was at time $j$ and then use a backward space technique for the first derivative term? Last edited: Aug 5, 2015 8. Aug 5, 2015 ### Staff: Mentor You need actual boundary conditions at the two ends. What are the boundary conditions? chet 9. Aug 6, 2015 ### Strum I am not quite sure I understand this method. What is $z_{left}$ and $z_{right}$? What happens if the moving boundary is part of the solution to the problem, as in a Stefan problem? 10. Aug 6, 2015 ### Staff: Mentor Josh, Do you feel like you can explain this to Strum? Chet 11. Aug 12, 2015 ### Strum So Chet. It doesn't look like Josh wants to chip in. Would you mind explaining how this would generalize to moving boundary problems? 12. Aug 12, 2015 ### Staff: Mentor zleft(t) and zright(t) are the time-dependent locations of the left- and right boundaries of the region being analyzed. Both boundaries are assumed to be moving. To complete the problem statement, Josh would have to provide the boundary conditions on h at these locations. We would like to convert the problem statement from h = h (z,t) to h = h(z,t) so that, in the z - t framework, the boundaries are not moving. The transformation that I gave in post #6 will do this. But, one has to transform the differential equation properly. We start out by writing: $$dh=\left(\frac{\partial h}{\partial t}\right)_{z^}dt+\left(\frac{\partial h}{\partial z^}\right)_{t}dz^$$ Then we evaluate the partial derivatives in the differential equation as follows: $$\left(\frac{\partial h}{\partial z}\right)_t=\left(\frac{\partial h}{\partial z^}\right)_{t}\left(\frac{\partial z^}{\partial z}\right)_t=\left(\frac{\partial h}{\partial z^}\right)_{t}\frac{1}{(z_{right}(t)-z_{left}(t))}$$ $$\left(\frac{\partial h}{\partial t}\right)_z=\left(\frac{\partial h}{\partial t}\right)_{z^} +\left(\frac{\partial h}{\partial z^}\right)_{t} \left(\frac{\partial z}{\partial t}\right)_z= \left(\frac{\partial h}{\partial t}\right)_{z^} -\frac{z^}{(z_{right}(t)-z_{left}(t))}\left(\frac{\partial h}{\partial z^}\right)_{t} \left(\frac{dz_{right}}{dt}-\frac{dz_{left}}{dt}\right)$$ This is then substituted into the differential equation. Hope this makes sense. Chet 13. Aug 19, 2015 ### joshmccraney Thanks Chet! Sorry it has taken so long for me to thank you, but thanks so much! 14. Aug 21, 2015 ### joshmccraney And sorry Strum, I recently changed my email address, and that coupled with a summer vacation meant I was scarce; I'm sorry and didn't intend to be rude. At the risk of sounding redundant to what Chet said, $z_{left}$ is the left boundary position of $z$ in the problem. If you imagine $z$ as the length of a candle, then perhaps $z_{left}$ is the height, and since the candle is presumable burning, $z_{left}$ would be getting smaller in time. Obviously the differential equation above, while it can be re-cast as the heat equation, is definitely not a candle; in fact, it's capillary corner flow! The same logic extends to $z_{right}$. Chet's proposed change of variables implies $z^* \in [0,1]$. This is a common substitution in non-dimentionalizing pressure terms in navier-stokes equations, if you've studies those at all. Again, I'm terribly sorry for the late delays, and the only reason I was able to respond to Chet was via my phone, which took some work. And Chet, $h(z_{left}) = 1$ and $h(z_{right}) = 0$. I'll work with this some and let you know what I get, although it seems in your previous post you already did the heavy lifting. Similar Discussions: Finite Differencing Dynamic Boundary	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9112725257873535, "perplexity": 667.2298581519331}, "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/1512948615810.90/warc/CC-MAIN-20171218102808-20171218124808-00353.warc.gz"}
https://www.physicsforums.com/threads/find-the-magnitude-of-the-velocity-vector-on-impact.269403/	# Find the magnitude of the velocity vector on impact 1. Nov 4, 2008 ### ilovedeathcab (resolved) find the magnitude of the velocity vector on impact Thelma and Louise drive their '66 T-Bird convertible off a cliff. They are going 25.9 m/s and the cliff is 97 meters high. Find the magnitude of their velocity vector the moment they impact the ground. Make sure you do the following in your answer: DO NOT include units. We'll deal with that in the next question. Give precision to tenths (#.#). so given y is -97, and vi is 25.9 m/s equations for horizontal projectile motion for vf are vyf=-g(t) vf=sqrt of (vxf^2+vyf^2) so far for t i had 4.447seconds ____ \/-2-97 ----------= 4.447 9.81 Last edited: Nov 5, 2008 2. Nov 4, 2008 ### mgb_phys You don't need the time. You can use vf2 = vi2 + 2as for the vertical velocity Since you ignore air resistance there is no change in the horizontal velocity. 3. Nov 4, 2008 ### merryjman And think of vxf and vyf as being two legs of a right triangle. The answer you're looking for is the hypotenuse of this triangle. 4. Nov 4, 2008 ### ilovedeathcab what is 2as???? 5. Nov 4, 2008 ### mgb_phys 2 acceleration * distance It's one of the standard motion equations - although some books use different letters. Similar Discussions: Find the magnitude of the velocity vector on impact	{"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": 0.9401460289955139, "perplexity": 2505.2846636894506}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-09/segments/1518891808539.63/warc/CC-MAIN-20180217224905-20180218004905-00036.warc.gz"}
http://mathhelpforum.com/calculus/142453-rotated-about.html	# Math Help - Rotated about the... Okay so I'm having trouble formulating this integral. The region is bounded by: ${x}^{2}-{y}^{2}=7$ and $x=4$ rotated about: $y=5$ I tried using shells.. It didn't really work for me Thanks, CC 2. Originally Posted by CalculusCrazed Okay so I'm having trouble formulating this integral. The region is bounded by: ${x}^{2}-{y}^{2}=7$ and $x=4$ rotated about: $y=5$ I tried using shells.. It didn't really work for me Thanks, CC $V=2\pi \int_{\sqrt{7}}^4 x \cdot {\color{red}2}\sqrt{x^2-7}\, dx=\ldots =2\cdot 18\pi=36\pi$? (use substitution $u := x^2-7$ to solve this integral) 3. Originally Posted by CalculusCrazed Okay so I'm having trouble formulating this integral. The region is bounded by: ${x}^{2}-{y}^{2}=7$ and $x=4$ rotated about: $y=5$ I tried using shells.. It didn't really work for me Thanks, CC washers w/r to x ... $R = 5 + \sqrt{x^2-7} $ $r = 5 - \sqrt{x^2-7}$ $R^2 - r^2 = 25 - (x^2 - 7) = 32 - x^2$ $V = \pi \int_{\sqrt{7}}^4 32 - x^2 \, dx$ 4. Okay so you did R meaning the radius of the big circle and r for the little. How did you find the bounds? 5. Originally Posted by CalculusCrazed Okay so you did R meaning the radius of the big circle and r for the little. How did you find the bounds? I sketched a graph ... upper branch is $y = \sqrt{x^2-7}$ lower branch is $y = -\sqrt{x^2-7}$ R = 5 - lower branch r = 5 - upper branch 6. Okay, so why did you do R^2 - r^2? 7. Originally Posted by CalculusCrazed Okay, so why did you do R^2 - r^2? volume by the method of washers ... a washer is a disk with a hole in it. $V = \pi \int_a^b [R(x)]^2 - [r(x)]^2 \, dx$ The Washer Method for Solids of Revolution 8. The little circle minus the big circle. Okay I guess I just don't get how you got root 7 to 4 9. Originally Posted by CalculusCrazed The little circle minus the big circle. Okay I guess I just don't get how you got root 7 to 4 $x = \sqrt{7}$ is the vertex of the hyperbola at the left end of the defined region to be rotated. $x = 4$ is the right boundary of the region to be rotated. I recommend you meet with your instructor ... this is all very basic material which should have been covered in class. 10. ...plug in zero for y in other words.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 20, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9150923490524292, "perplexity": 824.9551746574956}, "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-2014-41/segments/1410657137190.70/warc/CC-MAIN-20140914011217-00281-ip-10-234-18-248.ec2.internal.warc.gz"}
https://www.physicsforums.com/threads/a-problem-about-logarithmic-inequality.874512/	# A problem about logarithmic inequality Tags: 1. Jun 5, 2016 ### Ashes Panigrahi 1. The problem statement, all variables and given/known data If a,b,c are positive real numbers such that ${loga}/(b-c) = {logb}/(c-a)={logc}/(a-b)$ then prove that (a) $a^{b+c} + b^{c+a} + c^{a+b} >= 3$ (b) $a^a + b^b + c^c >=3$ 2. Relevant equations A.M $>=$ G.M 3. The attempt at a solution Using the above inequation, I am able to get (b) but I have no idea to do (a) keeping in mind the exponents as a sum whereas in the question the denominators are as a difference. 2. Jun 6, 2016 It doesn't look like a one or two step problem, but I did get the result from the set of logarithmic equations (by combining a couple of them) that abc=1. It does appear when a=b=c=1 that the equality holds. For any other case, you then need to show that the "greater than" sign applies. 3. Jun 8, 2016 ### Ray Vickson It is not allowed to have $a = b = 1$ (or even $a = b = c \neq 1$) because that would involve division by zero in the constraint $$\log(a)/(b-c) = \log(b)/(c-a) = \log(c)/(a-b)$$ Even trying to take limits as $a \to 1$, $b \to 1$ and $c \to 1$ will not really work, because depending on exactly how those limits are taken, the quantities involved in the constraint can still fail to exist, or can exist but have more-or-less arbitrary values. 4. Jun 8, 2016 The use of the "logs" to state the problem appears to be largely algebraic. After processing these equations, the result emerges that abc=1 and the logs wind up being removed. The problem could actually begin with, given a,b, c positive real, such that abc=1, prove the two inequalities. I think you will find that the equality holds only for the case a=b=c=1. I did not do the proof in its entirety, partly because on a homework problem we are not supposed to present the whole solution, but I think the only case where the equal sign would apply is for a=b=c=1. (I agree, this puts a zero in the denominators of the original statement of the problem.) 5. Jun 8, 2016 ### Staff: Mentor We may assume that either $0 < a < b < c$ or $0 < a < c < b$. As @Ray Vickson pointed out they cannot be equal. Neither can one of them equal one for then all of them would be equal one, especially they would be equal. The rest is simply about cases. Assume now $0<a<b<c$. If $c < 1$ then $\frac{\log a}{b-c} > 0$ and $\frac{\log b}{c-a} < 0$, a contradiction. In the same way $b < 1$ leads to $\log a < \log b < 0$ and $\frac{\log a}{b-c} > 0$ and $\frac{\log b}{c-a} < 0$, a contradiction. If $a < 1$ then $\frac{\log a}{b-c} > 0$ and $\frac{\log c}{a-b} < 0$, a contradiction. This means $a, b, c > 1$ and both assertions are true. (The second case $0 < a < c < b$ should be analogue.) The principle behind it is the following: The assumption is symmetric in $(a,b,c),$ i.e. with positive signum. On the other side there has to be a change in signs if not all of them are positive factors which means greater than one for the nominators. 6. Jun 8, 2016 ### Ray Vickson I suspect that as long as $a,b,c$ are valid (no division by 0) the inequalities to be shown will be strict. 7. Jun 8, 2016 I will show the algebra that gives the result $abc=1$. Removing the log factors on both sides, $a^{c-a}=b^{b-c}$ and $a^{a-b}=c^{b-c}$. (also $b^{a-b}=c^{c-a}$). Taking the first two equations and multiplying left sides and right sides, $a^{c-b}=(bc)^{b-c}$ so that $a^{c-b}=1/(bc)^{c-b}$. Thereby $(abc)^{c-b}=1$. I think it can be concluded that $abc=1$. (You can write similar equations by combining the other pairs of equations also with the result that $abc=1$).The way the problem is written with the factors in the denominators places additional restrictions, but otherwise, it would be possible to have $a=b=c=1$. To proceed from here, I would let $c=1/(ab)$ and write the inequalities as functions of $a$ and $b$. There are basically two cases that need to be tested, $ab>1$ and $ab<1$. The way the problem is written yes, if one of the three letters is equal to 1, then all 3 must equal one (log 1=0, etc), but that makes the denominators all zero. Instead of using logs, they could simply have presented the condition $abc=1$. (with a,b, c positive real). In that case, the problem would be less restrictive, but the same inequalities would apply. Last edited: Jun 8, 2016 8. Jun 8, 2016 If $a<b<c$ then $c>1$. But $b$ could be greater than one or less than one, and $a<1$. Please double-check your logic. 9. Jun 8, 2016 ### Staff: Mentor The logic is ok, but you are right: there is no solution at all. Three positive real numbers can be ordered. They cannot be pairwise equal for this meant a division by zero. Therefore there has to be at least one change of sign in the denominators: $b-c, c-a, a-b$. To be equal this means there is a change in the sign of the nominators. However, the nominators are somehow linearly ordered, e.g. $0 < a < b < c,$ and the denominators are cyclic. Since the logarithm is strictly monotone, the two orderings cannot be true at the same time. In your example with $0 < a < b < c$ we get $\frac{\log a}{b-c} = \frac{\log b}{c-a}=\frac{\log c}{a-b}=\frac{\log a}{-}=\frac{\log b}{+}=\frac{\log c}{-}$. On the other hand $\log a < \log b < \log c$ and the nominator in the middle cannot have a different sign than the two left and right. The assumption is always wrong. 10. Jun 8, 2016 A very good observation. Yes, the problem statement is clearly inconsistent. (unless we are missing the obvious). I agree with your assessment. 11. Jun 9, 2016 ### Staff: Mentor Well that rules out a<b<c, but what about a<c<b? $$\frac{\log a}{b-c} = \frac{\log b}{c-a}=\frac{\log c}{a-b}=\frac{\log a}{+}=\frac{\log b}{+}=\frac{\log c}{-}$$ Clearly log(c) has a different sign than the others, so 0<a<b<1<c. The equalities are cyclic, but swapping the parity makes a difference. Edit: no, doesn't work either. c-a is the largest denominator, so log(b) has to be the largest numerator in magnitude, but log(a) is larger. Both cases are excluded, but the second one is a bit more subtle. 12. Jun 9, 2016 ### Ray Vickson There is no need to check other cases. The three numbers must all be different. Let's call the smallest of the three $a$, the largest of the three $c$ and the one in the middle $b$. In the original problem there were no characteristics distinguishing $a, b$ or $c$, so there were no a priori ways to distinguish between them. Basically, it is just saying "without loss of generality, assume $a < b < c$", and we do that type of thing all the time. 13. Jun 9, 2016 @fresh_42 made a very good observation. I've looked hard to try to find an exception to this that would make it work, but it seems the problem in the way it is formulated is inconsistent. Writing the expressions without the logs: $a^{1/(b-c)}=b^{1/(c-a)}=c^{1/(a-b)}$ apparently it is impossible for all three terms to be greater than 1 or all three terms to be less than one. If two of them are greater than one, the third must be less than one, etc... Last edited: Jun 9, 2016 14. Jun 9, 2016 ### Staff: Mentor Okay, checking that in more detail, swapping a and b doesn't change the equalities, but that is not as obvious as an a->b->c->a rotation. Draft saved Draft deleted Similar Discussions: A problem about logarithmic inequality	{"extraction_info": {"found_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": 2, "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": 0.8799536228179932, "perplexity": 373.12281806162287}, "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-2017-43/segments/1508187824357.3/warc/CC-MAIN-20171020211313-20171020231313-00563.warc.gz"}
http://math.stackexchange.com/questions/103786/direct-proof-that-pi-is-not-constructible	# Direct proof that $\pi$ is not constructible Is there a direct proof that $\pi$ is not constructible, that is, that squaring the circle cannot be done by rule and compass? Of course, $\pi$ is not constructible because it is transcendental and so is not a root of any polynomial with rational coefficients. But is there a simple direct proof that $\pi$ is not a root of polynomial of degree $2^n$ with rational coefficients? The kind of proof I seek is one by induction on the height of a tower of quadratic extensions, one that ultimately relies on a proof that $\pi$ is not rational. Does any one know of a proof along these lines or any other direct proof? I just want a direct proof that $\pi$ is not constructible without appealing to transcendence. - So far as I know, there is no proof that $\pi$ doesn't have degree $2^n$ that doesn't also prove $\pi$ is transcendental. I leave this as a comment, rather than an answer, because I don't know how to substantiate it. – Gerry Myerson Jan 30 '12 at 0:20 @lhf: I don't see how knowing that $\pi$ is irrational helps. There are lots of irrational algebraic numbers. – Zhen Lin Jan 30 '12 at 0:30 @Zhen Lin: What lhf means is, I think, consider a tower of quadratic extensions, starting from Q. Note that $\pi\not\in \mathbb Q$ as an induction basis and then, by some magic induction argument, if $\pi\not\in F\implies \pi\not\in F(u)$, where $u^2\in F$. – Myself Jan 30 '12 at 0:49 @Marc: that's still not the question being asked in the first sentence. Most irreducible polynomials of degree $2^n$ do not have constructible numbers as their roots; in fact, generically they have Galois group $S_{2^n}$. – Qiaochu Yuan Jan 30 '12 at 22:52 Ha. Dave Renfro's sci.math post was a response to one of mine. I guess when the conversation turns to $\pi$, things naturally keep coming 'round.... – Gerry Myerson Jul 4 '12 at 4:00 I've just read in the book The Number $\pi$ by Eymard and Lafon that no such proof is known. “The proof that it is impossible to square the circle does not involve direct demonstration of the non-constructibility of the number $\pi$. As far as we are aware, it is not known how to do this! One proves that it is not algebraic, which is much more restrictive, then one uses the fact that a constructible number is algebraic.” [§4.2, p. 134] I see that the book was written $8$ years ago, maybe the information is not up to date, though I doubt that it's not [+1] – Belgi Oct 9 '12 at 11:09	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.9015530943870544, "perplexity": 226.6359560239649}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-18/segments/1461860117405.91/warc/CC-MAIN-20160428161517-00057-ip-10-239-7-51.ec2.internal.warc.gz"}
http://math.stackexchange.com/questions/307309/show-that-a-map-fx-to-y-is-onto-iff-ff-1c-c-for-all-subsets-c-subse	# Show that a map $f:X\to Y$ is onto iff $f(f^{-1}(C))=C$ for all subsets $C\subseteq Y$. Show that a map $f:X\to Y$ is onto iff $f(f^{-1}(C))=C$ for all subsets $C\subseteq Y$. My workings so far: Because this is an if and only if proof we need to show it both ways. First let's assume $f$ is onto, that is, $\forall y\in Y\ \ \exists x\in X$ such that $f(x)=y$. Now $$f^{-1}(C)=\{ x\in X\| f(x)\in C \}$$ Let $f^{-1}(C)=D \subseteq X$ for simplicity. This means $D$ is the subset of $X$ containing all $x$ that get mapped to an element in $C$ under the map $f$. Now $$f(f^{-1}(C))=f(D)=\{ y\in Y \| \exists x \in D \| f(x)=y \}$$ We can be sure this set is well defined by the fact that $f$ is onto. By the previous definition of $D$ it seems rather trivial that $f(D)=C$. Is this enough to prove the statement in this direction. Maybe using the more elaborate notation $$f(f^{-1}(C))=\{ y\in Y \| (\exists x \in \{ x\in X\| f(x)\in C \}) \| f(x)=y \}$$ Makes it a little bit more obvious (if possibly more confusing). Now for the other direction we assume that $f(f^{-1}(C))=C$ for all $C\subseteq Y$. Here I really have no idea how to get started. If anyone could point me in the right direction and possible check if argument for the first part is rigorous enough that would be great! - Regarding your workings: The set $f(f^{-1}(C)) = f(D)$ is always defined, even if $f$ isn’t onto. And yes, it can be seen quite easily by looking long enough at the definition of the set that it is indeed $C$ if $f$ is onto. But exercises like this often try to provoke a more low-level approach, as if asked: How low-level can you go? Use the definition of set-equality. Hint: Show more generally $f(f^{-1}(C)) \subseteq C$ for all $C \subseteq Y$ without assuming the surjectivity of $f$. Do this by following an arbitrary element of the left hand side. Then, assuming surjectivity of $f$, show for every $c ∈ C$ there is an $b ∈ f^{-1}(C)$ such that $f(b) = c$, so $c ∈ f(f^{-1}(C))$, meaning $C \subseteq f(f^{-1}(C))$. For the other direction, consider singletons $C = \{y\}$ for $y ∈ Y$.	{"extraction_info": {"found_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": 0.9873498678207397, "perplexity": 58.91413908564528}, "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-2015-18/segments/1429246656168.61/warc/CC-MAIN-20150417045736-00154-ip-10-235-10-82.ec2.internal.warc.gz"}
http://mathhelpforum.com/pre-calculus/44496-find-asymptotes-rational-function.html	# Math Help - Find Asymptotes of Rational Function 1. ## Find Asymptotes of Rational Function Find the vertical, horizontal and obligue asymptotes, if any, of the rational function below. F(x) = (x -1)/(x - x^3) 2. (x-1)/(x (1-x^2) = (x-1)/(x (1-x)(1+x) ) = -1/(x (1+x) ) Take note that since we canceled 1-x, then there is a hole at x = 1. That is, f(x) is not defined at x = 1. There is no oblique asymptotes. The vertical asymptote is at x = 0 and x = -1 (These are the points that makes the function undefined). The horizontal asymptote is zero since the limit as x goes to ± infinity equals 0. In Algebra, you would probably know it by this way: - Degree of Numerator > Degree of Denominator, then no horizontal asymptote exist. - Degree of Numerator = Degree of Denominator, then horizontal asymptotes is the y = a/b, where a and b is the coefficients for the leading terms in numerator and denominator, respectively. - Degree of Numerator < Degree of Denominator, then horizontal asymptote is y = 0 3. Originally Posted by Chop Suey (x-1)/(x (1-x^2) = (x-1)/(x (1-x)(1+x) ) = -1/(x (1+x) ) Take note that since we canceled 1-x, then there is a hole at x = 1. That is, f(x) is not defined at x = 1. There is no oblique asymptotes. The vertical asymptote is at x = 0 and x = -1 (These are the points that makes the function undefined). The horizontal asymptote is zero since the limit as x goes to ± infinity equals 0. In Algebra, you would probably know it by this way: - Degree of Numerator > Degree of Denominator, then no horizontal asymptote exist. - Degree of Numerator = Degree of Denominator, then horizontal asymptotes is the y = a/b, where a and b is the coefficients for the leading terms in numerator and denominator, respectively. - Degree of Numerator < Degree of Denominator, then horizontal asymptote is y = 0 The graph verifies what you have done! (Good work! ) --Chris 4. ## It's clear.... Both replies are clear. Thanks a million!	{"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": 0.9940406680107117, "perplexity": 838.904635505047}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-40/segments/1443736672441.2/warc/CC-MAIN-20151001215752-00152-ip-10-137-6-227.ec2.internal.warc.gz"}
https://patrickmilano.com/2019/08/07/sums-of-two-cubes/	# Sums of Two Cubes This post was inspired by a joke from Futurama in which two robots laugh at the fact that their serial numbers (3370318 and 2716057) are both expressible as the sum of two cubes. I’ll explain how to express these numbers as sums of two cubes and discuss some of the relevant mathematics. ## Modular Arithmetic Modular arithmetic is a basic yet powerful technique in number theory. Whereas usual arithmetic involves the integers, modular arithmetic involves the remainders of integers after division by an integer . Integers that have the same remainder after division by are equivalent in the following sense: we say that and are congruent modulo if is a multiple of . This is written . The set (that is, the set of all integers that are congruent to modulo ) is called the congruence class of . Everyone who can tell time has used modular arithmetic. Hours are measured modulo 12 (or modulo 24 in military time). 5 hours past 10:00 is 3:00 rather than 15:00. We can write to express this as a congruence relation. It turns out to be very helpful to consider cubes modulo 9. This is because every cube is congruent to either 0, 1, or -1 modulo 9. You can check this by cubing 0, 1, 2, 3, 4, 5, 6, 7, and 8 (the only possible remainders after division by 9). For example, . Because cubes can only be congruent to 1, 0, or -1 modulo 9, the sum of two cubes can only be -2, -1, 0, 1, or 2 modulo 9. Therefore any integer that is congruent to 3, 4, 5, or 6 modulo 9 cannot be expressed as the sum of two cubes. A similar argument shows that any integer that is congruent to 3 or 4 modulo 7 is not expressible as the sum of two cubes. You can check this by cubing 0, 1, 2, 3, 4, 5, and 6. ## Finding two cubes If an integer is in the right congruence classes modulo 9 and modulo 7, we would like to know if it is possible to express as the sum of two cubes. To do this, we will use the following factorization. This reduces the problem of finding two cubes that add up to to finding a factorization of the form . Note that when is outside the ellipse . Because of this, we may assume that is the smaller divisor whenever is large enough. I leave it to the reader to verify that is sufficient. We can simplify our computations by changing coordinates. The equation becomes . Solving for in terms of and , we obtain If and are integers, then so are and . It is also true that if and are integers then and are integers, but this is trickier to show. We know that is an integer, and so must be divisible by 4. This is only possible if and are either both even or both odd. (This is yet another helpful application of modular arithmetic: even integers are exactly those that are congruent to 0 modulo 2; odd integers are those which are congruent to 1 modulo 2). From the equations and we see that and are integers. The upshot of all this is that we can check whether a given integer is a sum of two cubes by (1) finding all factorizations where and (2) checking whether is an integer. Here’s an implementation in Python. In [1]: import math In [2]: def print_cube_sum(n,a,b): # Prints n = a^3 + b^3 with appropriate parentheses if a or b is negative. # This only supports nonnegative n. # Make sure a is the smaller of a and b so that b > 0. if a > b: a, b = b, a s = str(n) + ' = ' + str(b) + '^3' if a < 0: s += ' + (' + str(a) + ')^3' elif a > 0: s += ' + ' + str(a) + '^3' print(s) return In [3]: def find_two_cubes(n): if n % 9 not in {0, 1, 2, 7, 8}: return if n % 7 not in {0, 1, 2, 5, 6}: return # Find divisors c such that n = cd and c <= d. for c in range(1, math.floor(math.sqrt(n))+1): if n % c == 0 and 4n - c3 >= 0: d = math.sqrt((4n - c*3)/(3c)) if d.is_integer(): # If d is an integer, then a and b are integers and n = a^3 + b^3. a = int((c - d) // 2) b = int((c + d) // 2) print_cube_sum(n,a,b) return Let’s test this for the numbers 1 through 100, and also for Flexo and Bender’s serial numbers. In [4]: for n in range(1,101): find_two_cubes(n) 1 = 1^3 7 = 2^3 + (-1)^3 8 = 2^3 9 = 2^3 + 1^3 16 = 2^3 + 2^3 19 = 3^3 + (-2)^3 26 = 3^3 + (-1)^3 27 = 3^3 28 = 3^3 + 1^3 35 = 3^3 + 2^3 37 = 4^3 + (-3)^3 54 = 3^3 + 3^3 56 = 4^3 + (-2)^3 61 = 5^3 + (-4)^3 63 = 4^3 + (-1)^3 64 = 4^3 65 = 4^3 + 1^3 72 = 4^3 + 2^3 91 = 6^3 + (-5)^3 91 = 4^3 + 3^3 98 = 5^3 + (-3)^3 In [5]: # Express Flexo's serial number as a sum of two cubes. find_two_cubes(3370318) 3370318 = 119^3 + 119^3 In [6]: # Express Bender's serial number as a sum of two cubes. find_two_cubes(2716057) 2716057 = 952^3 + (-951)^3 Notice that the algorithm failed to find the expression . This is because of our earlier assumption that . When , the inequality is false. This is the only special case that our algorithm missed. It’s interesting to note that some integers that are in the right congruence classes modulo 9 and modulo 7 are not expressible as the sum of two cubes. 20, for example, is congruent to 2 modulo 9 and -1 modulo 7, yet it is not expressible as a sum of two cubes. ## Broughan’s Theorem A very nice paper by Kevin A. Broughan explains why some numbers (like 20) aren’t expressible as a sum of two cubes. Broughan’s main theorem is this: The equation has a solution in positive integers and if and only if the following three conditions are satisfied: 1. There exists a divisor of with such that 2. for some positive integer , and such that 3. is a perfect square. has a solution in positive integers and if and only if the following three conditions are satisfied: 1. There exists a divisor of with such that 2. for some positive integer , and such that 3. is a perfect square. An interesting corollary of Broughan’s theorem is that no prime number (except for 2) is expressible as a sum of two positive cubes. ## One thought on “Sums of Two Cubes” • ##### Russell Schwartzsays: Doesn’t it follow immediately from the factorization $a^3 + b^3 = (a+b)(a^2 – ab + b^2)$ that no odd prime is expressible as the sum of two positive cubes?	{"extraction_info": {"found_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": 0.8159223198890686, "perplexity": 322.7630843953666}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-27/segments/1656103626162.35/warc/CC-MAIN-20220629084939-20220629114939-00469.warc.gz"}
https://hsm.stackexchange.com/questions/6721/what-are-philolaos-even-odd-numbers	# What are Philolaos' “even-odd” numbers? Number, indeed, has two proper kinds (ιδια ειδη), odd and even, and a third mixed together from both, the even-odd(αρτιοπέριττον). Of each of the two kinds there are many shapes, of which each thing itself gives signs. (Philolaus F5 = Stobaeus, Eclogae 1.21.7c; tr. Huffman 1993) The quote is from Horky's Plato and Pythagoreanism, 2013, (p.141), it is commented twice further in text. On p185 the author writes " the so-called even-odd (αρτιοπέριττον), which appears to refer to the “one.” Aristotle, too, knew this fragment, and he is explicit in associating the “one” with the “even-odd” class and stating that it is derived from the “even” and the “odd.” On p190, disappointingly, he says "however It is beyond the scope of this study to examine in more detail the significance of the concept of “mixture” of Forms or classes". The “one” is not the number preceding two which Ancient greeks apparently did not accept as a number. Philolaos quote might suggest that it is set apart as it has no shape but I am tempted to read it as suggesting that the mixed numbers have multiple shapes, while the numbers from the proper kinds have each just one. So what are the 3 kinds of numbers? The odd numbers are perhaps the ones that we also call odd, (starting with 3), while the even numbers are seen as, so to say, 'strictly even', the powers of 2, that is 2,4,8,16 and the rest, that is 6,10,12 etc. as even-odd. References and comments (or should this be moved to Philosophy SE)? • According to Liddell & Scott, perseus.tufts.edu/hopper/…, an even-odd is two times an odd number. – Michael E2 Nov 19 '17 at 0:35 • Thanks but LS sends us back to F5 and Aristotle's "one" (Arist.Fr.199); to Plutarch (?= Plu.2.1139f ) and what Ph.1.3?. – sand1 Nov 19 '17 at 9:25 • First, let me say that while I am familiar with a couple of relevant texts, I don't have a comprehensive knowledge of the area, and thus my opinions should not be thoroughly relied upon. Perhaps you're suggesting LS on even-odd has been superseded (plausible). I thought you were asking just about the definitions of three terms. I'm not sure of the context, the Pythagoreans or broadly "Ancient greeks." The Ancient greeks were a heterogeneous bunch, and perhaps the Pythagoreans were, too. For instance, Euclid gives a different classification of numbers (even-times odd etc.)..... – Michael E2 Nov 19 '17 at 15:13 • ...The LS reference to an Aristotle fragment, which can be tracked down, seems at odds with the definition in LS. The fragment raises more questions than it answers, which is the way with fragments, I suppose. It seems to me that Aristotle might be criticizing the Pythagorean concept of even-odd, as opposed to stating or explaining it. – Michael E2 Nov 19 '17 at 15:13 • I'd guess he means the classification scheme given by Euclid, book 7, definitions 8-10. Euclid classifies numbers according to this scheme in book 9, propositions 32-34. – Marius Kempe Nov 19 '17 at 19:00 Aristotle comments on the Pythagorean theory in Met, Book I (A), 986a14-986a22: these thinkers also consider that number is the principle both as matter for things and as forming their modifications and states, and hold that the elements of number are the even and the odd, and of these the former is unlimited, and the latter limited; and the $1$ proceeds from both of these (for it is both even and odd), and number from the $1$; and the whole heaven, as has been said, is numbers. But see also Thomas Heath, A History of Greek Mathematics. Volume I (1921), page 71: The explanation of this strange view might apparently be that the unit, being the principle of all number, even as well as odd, cannott itself be odd and must therefore be called even-odd. There is, however, another explanation, attributed by Theon of Smyrne to Aristotle, to the effect that the unit when added to an even number makes an odd number, but when added to an odd number makes an even number: which could not be the case if it did not partake of both species. Philolaus' Fr.5 is discussed at lenght in: Carl Huffman, Philolaus of Croton: Pythagorean and Presocratic, Cambridge UP (1993), page 178-on. There are various interpretations, including the possibility of an interpolation. See page 190: In summary, I believe that the even-odd is a derived class of numbers whose first member is, as the ancient tradition indicates, the one, but which also includes numbers that consist of even and odd numbers combined in ratios (e.g. $2 : 1 , 4 : 3$, and $3 : 2$ ) . This class of numbers corresponds to the third class of things in Fr.2, which consists of members that are harmonized from both limiting and unlimited constituents. The even-odd numbers are the numbers by which these harmonized things are known. This connection of course remains conjectural, but I believe that it is a plausible way to make sense of both Fr.2 and Fr.5 of Philolaus and Aristotle's testimony that there was a connection between the even - odd dichotomy and the unlimited - limited dichotomy, although my suggestion does not identify the two as Aristotle does. • Using today terms,one could say that for the ancient Greeks, before the discovery of irrationals, numbers were integers>1, even or odd, and rationals. So, including 1 among the even-odd makes sense but not the explanation given by Theon and others, as adding any odd number changes parity. – sand1 Nov 21 '17 at 16:48 • @sand1 - Strictly speaking, for ancient Greeks numbers where only naturals. Then they have ratios between magnitudes, and magnitudes can be "measured" by numbers. You can see the post: irrationality-of-the-square-root-of-2. – Mauro ALLEGRANZA Nov 21 '17 at 16:50 • Notice also that for Euclid 'irrational' (ἄλογος – a better translation is 'ratioless') means something different from the modern 'irrational' – Euclid's term for his equivalent of the modern concept is 'incommensurable'. (See the definitions of book 10 of the Elements.) – Marius Kempe Nov 21 '17 at 21:15 • Also, in Euclid a number doesn't measure a magnitude – only objects that have a ratio to each other can measure each other (strictly, if the measuring one is equal to or less than the measured one). – Marius Kempe Nov 22 '17 at 14:06	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8027545213699341, "perplexity": 1407.6660636266129}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-39/segments/1568514573071.65/warc/CC-MAIN-20190917121048-20190917143048-00075.warc.gz"}
http://sagemath.org/doc/reference/arithgroup/sage/modular/arithgroup/congroup_gamma.html	Congruence Subgroup $$\Gamma(N)$$¶ class sage.modular.arithgroup.congroup_gamma.Gamma_class(args, *kwds) The principal congruence subgroup $$\Gamma(N)$$. are_equivalent(x, y, trans=False) Check if the cusps $$x$$ and $$y$$ are equivalent under the action of this group. ALGORITHM: The cusps $$u_1 / v_1$$ and $$u_2 / v_2$$ are equivalent modulo $$\Gamma(N)$$ if and only if $$(u_1, v_1) = \pm (u_2, v_2) \bmod N$$. EXAMPLE: sage: Gamma(7).are_equivalent(Cusp(2/3), Cusp(5/4)) True image_mod_n() Return the image of this group modulo $$N$$, as a subgroup of $$SL(2, \ZZ / N\ZZ)$$. This is just the trivial subgroup. EXAMPLE: sage: Gamma(3).image_mod_n() Matrix group over Ring of integers modulo 3 with 1 generators ( [1 0] [0 1] ) index() Return the index of self in the full modular group. This is given by $\begin{split}\prod_{\substack{p \mid N \\ \text{p prime}}}\left(p^{3e}-p^{3e-2}\right).\end{split}$ EXAMPLE:: sage: [Gamma(n).index() for n in [1..19]] [1, 6, 24, 48, 120, 144, 336, 384, 648, 720, 1320, 1152, 2184, 2016, 2880, 3072, 4896, 3888, 6840] sage: Gamma(32041).index() 32893086819240 ncusps() Return the number of cusps of this subgroup $$\Gamma(N)$$. EXAMPLES: sage: [Gamma(n).ncusps() for n in [1..19]] [1, 3, 4, 6, 12, 12, 24, 24, 36, 36, 60, 48, 84, 72, 96, 96, 144, 108, 180] sage: Gamma(30030).ncusps() 278691840 sage: Gamma(2^30).ncusps() 432345564227567616 nirregcusps() Return the number of irregular cusps of self. For principal congruence subgroups this is always 0. EXAMPLE: sage: Gamma(17).nirregcusps() 0 nu3() Return the number of elliptic points of order 3 for this arithmetic subgroup. Since this subgroup is $$\Gamma(N)$$ for $$N \ge 2$$, there are no such points, so we return 0. EXAMPLE: sage: Gamma(89).nu3() 0 reduce_cusp(c) Calculate the unique reduced representative of the equivalence of the cusp $$c$$ modulo this group. The reduced representative of an equivalence class is the unique cusp in the class of the form $$u/v$$ with $$u, v \ge 0$$ coprime, $$v$$ minimal, and $$u$$ minimal for that $$v$$. EXAMPLES: sage: Gamma(5).reduce_cusp(1/5) Infinity sage: Gamma(5).reduce_cusp(7/8) 3/2 sage: Gamma(6).reduce_cusp(4/3) 2/3 TESTS: sage: G = Gamma(50); all([c == G.reduce_cusp(c) for c in G.cusps()]) True sage.modular.arithgroup.congroup_gamma.Gamma_constructor(N) Return the congruence subgroup $$\Gamma(N)$$. EXAMPLES: sage: Gamma(5) # indirect doctest Congruence Subgroup Gamma(5) sage: G = Gamma(23) sage: G is Gamma(23) True sage: TestSuite(G).run() Test global uniqueness: sage: G = Gamma(17) True sage: G2 = sage.modular.arithgroup.congroup_gamma.Gamma_class(17) sage: G == G2 True sage: G is G2 False sage.modular.arithgroup.congroup_gamma.is_Gamma(x) Return True if x is a congruence subgroup of type Gamma. EXAMPLES: sage: from sage.modular.arithgroup.all import is_Gamma sage: is_Gamma(Gamma0(13)) False sage: is_Gamma(Gamma(4)) True Previous topic Congruence Subgroup $$\Gamma_0(N)$$ Next topic The modular group $${\rm SL}_2(\ZZ)$$	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.821165144443512, "perplexity": 1142.2371759675002}, "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-2015-18/segments/1429246652114.13/warc/CC-MAIN-20150417045732-00229-ip-10-235-10-82.ec2.internal.warc.gz"}
https://www.lmfdb.org/EllipticCurve/Q/7056/bg/	# Properties Label 7056.bg Number of curves $4$ Conductor $7056$ CM $$\Q(\sqrt{-7})$$ Rank $0$ Graph # Related objects Show commands: SageMath sage: E = EllipticCurve("bg1") sage: E.isogeny_class() ## Elliptic curves in class 7056.bg sage: E.isogeny_class().curves LMFDB label Cremona label Weierstrass coefficients j-invariant Discriminant Torsion structure Modular degree Faltings height Optimality CM discriminant 7056.bg1 7056bq4 $$[0, 0, 0, -262395, 51731946]$$ $$16581375$$ $$120495224844288$$ $$[2]$$ $$28672$$ $$1.7628$$ $$-28$$ 7056.bg2 7056bq3 $$[0, 0, 0, -15435, 907578]$$ $$-3375$$ $$-120495224844288$$ $$[2]$$ $$14336$$ $$1.4163$$ $$-7$$ 7056.bg3 7056bq2 $$[0, 0, 0, -5355, -150822]$$ $$16581375$$ $$1024192512$$ $$[2]$$ $$4096$$ $$0.78989$$ $$-28$$ 7056.bg4 7056bq1 $$[0, 0, 0, -315, -2646]$$ $$-3375$$ $$-1024192512$$ $$[2]$$ $$2048$$ $$0.44332$$ $$\Gamma_0(N)$$-optimal $$-7$$ ## Rank sage: E.rank() The elliptic curves in class 7056.bg have rank $$0$$. ## Complex multiplication Each elliptic curve in class 7056.bg has complex multiplication by an order in the imaginary quadratic field $$\Q(\sqrt{-7})$$. ## Modular form7056.2.a.bg sage: E.q_eigenform(10) $$q + 4 q^{11} + O(q^{20})$$ ## Isogeny matrix sage: E.isogeny_class().matrix() The $$i,j$$ entry is the smallest degree of a cyclic isogeny between the $$i$$-th and $$j$$-th curve in the isogeny class, in the LMFDB numbering. $$\left(\begin{array}{rrrr} 1 & 2 & 7 & 14 \\ 2 & 1 & 14 & 7 \\ 7 & 14 & 1 & 2 \\ 14 & 7 & 2 & 1 \end{array}\right)$$ ## Isogeny graph sage: E.isogeny_graph().plot(edge_labels=True) The vertices are labelled with LMFDB labels.	{"extraction_info": {"found_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": 0.9777349233627319, "perplexity": 3266.1993231719493}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446710902.80/warc/CC-MAIN-20221202114800-20221202144800-00770.warc.gz"}
https://www.nag.com/numeric/nl/nagdoc_26.2/nagdoc_cl26.2/html/f01/f01jhc.html	# NAG C Library Function Document ## 1Purpose nag_matop_real_gen_matrix_frcht_exp (f01jhc) computes the Fréchet derivative $L\left(A,E\right)$ of the matrix exponential of a real $n$ by $n$ matrix $A$ applied to the real $n$ by $n$ matrix $E$. The matrix exponential ${e}^{A}$ is also returned. ## 2Specification #include #include void nag_matop_real_gen_matrix_frcht_exp (Integer n, double a[], Integer pda, double e[], Integer pde, NagError fail) ## 3Description The Fréchet derivative of the matrix exponential of $A$ is the unique linear mapping $E⟼L\left(A,E\right)$ such that for any matrix $E$ $eA+E - e A - LA,E = oE .$ The derivative describes the first-order effect of perturbations in $A$ on the exponential ${e}^{A}$. nag_matop_real_gen_matrix_frcht_exp (f01jhc) uses the algorithms of Al–Mohy and Higham (2009a) and Al–Mohy and Higham (2009b) to compute ${e}^{A}$ and $L\left(A,E\right)$. The matrix exponential ${e}^{A}$ is computed using a Padé approximant and the scaling and squaring method. The Padé approximant is then differentiated in order to obtain the Fréchet derivative $L\left(A,E\right)$. ## 4References Al–Mohy A H and Higham N J (2009a) A new scaling and squaring algorithm for the matrix exponential SIAM J. Matrix Anal. 31(3) 970–989 Al–Mohy A H and Higham N J (2009b) Computing the Fréchet derivative of the matrix exponential, with an application to condition number estimation SIAM J. Matrix Anal. Appl. 30(4) 1639–1657 Higham N J (2008) Functions of Matrices: Theory and Computation SIAM, Philadelphia, PA, USA Moler C B and Van Loan C F (2003) Nineteen dubious ways to compute the exponential of a matrix, twenty-five years later SIAM Rev. 45 3–49 ## 5Arguments 1: $\mathbf{n}$IntegerInput On entry: $n$, the order of the matrix $A$. Constraint: ${\mathbf{n}}\ge 0$. 2: $\mathbf{a}\left[\mathit{dim}\right]$doubleInput/Output Note: the dimension, dim, of the array a must be at least ${\mathbf{pda}}×{\mathbf{n}}$. The $\left(i,j\right)$th element of the matrix $A$ is stored in ${\mathbf{a}}\left[\left(j-1\right)×{\mathbf{pda}}+i-1\right]$. On entry: the $n$ by $n$ matrix $A$. On exit: the $n$ by $n$ matrix exponential ${e}^{A}$. 3: $\mathbf{pda}$IntegerInput On entry: the stride separating matrix row elements in the array a. Constraint: ${\mathbf{pda}}\ge {\mathbf{n}}$. 4: $\mathbf{e}\left[\mathit{dim}\right]$doubleInput/Output Note: the dimension, dim, of the array e must be at least ${\mathbf{pde}}×{\mathbf{n}}$. The $\left(i,j\right)$th element of the matrix $E$ is stored in ${\mathbf{e}}\left[\left(j-1\right)×{\mathbf{pde}}+i-1\right]$. On entry: the $n$ by $n$ matrix $E$ On exit: the Fréchet derivative $L\left(A,E\right)$ 5: $\mathbf{pde}$IntegerInput On entry: the stride separating matrix row elements in the array e. Constraint: ${\mathbf{pde}}\ge {\mathbf{n}}$. 6: $\mathbf{fail}$NagError Input/Output The NAG error argument (see Section 3.7 in How to Use the NAG Library and its Documentation). ## 6Error Indicators and Warnings NE_ALLOC_FAIL Dynamic memory allocation failed. See Section 2.3.1.2 in How to Use the NAG Library and its Documentation for further information. On entry, argument $〈\mathit{\text{value}}〉$ had an illegal value. NE_INT On entry, ${\mathbf{n}}=〈\mathit{\text{value}}〉$. Constraint: ${\mathbf{n}}\ge 0$. NE_INT_2 On entry, ${\mathbf{pda}}=〈\mathit{\text{value}}〉$ and ${\mathbf{n}}=〈\mathit{\text{value}}〉$. Constraint: ${\mathbf{pda}}\ge {\mathbf{n}}$. On entry, ${\mathbf{pde}}=〈\mathit{\text{value}}〉$ and ${\mathbf{n}}=〈\mathit{\text{value}}〉$. Constraint: ${\mathbf{pde}}\ge {\mathbf{n}}$. NE_INTERNAL_ERROR An internal error has occurred in this function. Check the function call and any array sizes. If the call is correct then please contact NAG for assistance. See Section 2.7.6 in How to Use the NAG Library and its Documentation for further information. NE_NO_LICENCE Your licence key may have expired or may not have been installed correctly. See Section 2.7.5 in How to Use the NAG Library and its Documentation for further information. NE_SINGULAR The linear equations to be solved for the Padé approximant are singular; it is likely that this function has been called incorrectly. NW_SOME_PRECISION_LOSS ${e}^{A}$ has been computed using an IEEE double precision Padé approximant, although the arithmetic precision is higher than IEEE double precision. ## 7Accuracy For a normal matrix $A$ (for which ${A}^{\mathrm{T}}A=A{A}^{\mathrm{T}}$) the computed matrix, ${e}^{A}$, is guaranteed to be close to the exact matrix, that is, the method is forward stable. No such guarantee can be given for non-normal matrices. See Section 10.3 of Higham (2008), Al–Mohy and Higham (2009a) and Al–Mohy and Higham (2009b) for details and further discussion. ## 8Parallelism and Performance nag_matop_real_gen_matrix_frcht_exp (f01jhc) is threaded by NAG for parallel execution in multithreaded implementations of the NAG Library. nag_matop_real_gen_matrix_frcht_exp (f01jhc) makes calls to BLAS and/or LAPACK routines, which may be threaded within the vendor library used by this implementation. Consult the documentation for the vendor library for further information. Please consult the x06 Chapter Introduction for information on how to control and interrogate the OpenMP environment used within this function. Please also consult the Users' Note for your implementation for any additional implementation-specific information. The cost of the algorithm is $O\left({n}^{3}\right)$ and the real allocatable memory required is approximately $9{n}^{2}$; see Al–Mohy and Higham (2009a) and Al–Mohy and Higham (2009b). If the matrix exponential alone is required, without the Fréchet derivative, then nag_real_gen_matrix_exp (f01ecc) should be used. If the condition number of the matrix exponential is required then nag_matop_real_gen_matrix_cond_exp (f01jgc) should be used. As well as the excellent book Higham (2008), the classic reference for the computation of the matrix exponential is Moler and Van Loan (2003). ## 10Example This example finds the matrix exponential ${e}^{A}$ and the Fréchet derivative $L\left(A,E\right)$, where $A = 1 2 2 2 3 1 1 2 3 2 1 2 3 3 3 1 and E = 1 0 1 2 0 0 0 1 4 2 1 2 0 3 2 1 .$ ### 10.1Program Text Program Text (f01jhce.c) ### 10.2Program Data Program Data (f01jhce.d) ### 10.3Program Results Program Results (f01jhce.r)	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 65, "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": 0.9598303437232971, "perplexity": 1952.3428450425567}, "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-2021-31/segments/1627046151641.83/warc/CC-MAIN-20210725080735-20210725110735-00462.warc.gz"}
https://web.math.princeton.edu/~jonfick/ETSM_AY11-12.html	Ergodic Theory and Statistical Mechanics Seminar AY2011-2012 (See current year) Thursdays 2:00-3:30pm, Room 601, Fine Hall, Princeton University. Contact Information: Date: May 18th 2012 Room: Fine 322 Speaker: Marian Gidea (Institute for Advanced Study) Title: Perturbations of geodesic flows producing unbounded growth of energy Abstract: We consider a geodesic flow on a manifold endowed with some generic Riemannian metric. We couple the geodesic flow with a time-dependent potential driven by an external dynamical system, which is assumed to satisfy some recurrence condition. We prove that there exist orbits whose energy grows unboundedly at a linear rate with respect to time; this growth rate is optimal. In particular, we obtain unbounded growth of energy in the case when the external dynamical system is quasi-periodic, of rationally independent frequency vector (not necessarily Diophantine). Our result generalizes Mather's acceleration theorem and is related to Arnold's diffusion problem. It also extends some earlier results by Delshams, de la Llave and Seara. Date: May 10th 2012 Speaker: Han Li (Yale University) Title: Effective discreteness of the 3-dimensional Markov spectrum Abstract: Let the set O={non-degenerate, indefinite, real quadratic forms in 3-variables with determinant 1}. We define for every form Q in the set O, the Markov minimum m(Q)=min{\|Q(v)\|: v is a non-zero integral vector in $R^3$}. The set M={m(Q): Q is in O} is called the 3-dimensional Markov spectrum. An early result of Cassels-Swinnerton-Dyer combined with Margulis' proof of the Oppenheim conjecture asserts that, for every a>0 {M \intersect (a, \infty)} is a finite set. In this lecture we will show that #{M \intersect (a, \infty)}<< a^{-26}. This is a joint work with Prof. Margulis, and our method is based on dynamics on homogeneous spaces. Date: May 3rd 2012 Speaker: L. H. Eliasson (Université Paris Diderot) Title: Reducibility for the quasi-periodic linear Schrödinger and wave equations. Abstract: We shall discuss reducibility of these equations on the torus with a small potential that depends quasi-periodically on time. Reducibility amounts to "reduce" the equation to a time-independent linear equation with pure point spectrum in which case all solutions will be of Floquet type. For the Schrödinger equation, this has been proven in a joint work with S. Kuksin, and for the wave equation we shall report on a work in progress with B. Grebert and S. Kuksin. Date: April 26th 2012 Speaker: Akhtam Dzhalilov (Stony Brook University) Title: Invariant Measures, Conjugations and Renormalizations of Circle Maps with Break points Abstract: An important question in circle dynamics is regarding the absolute continuity of an invariant measure. We will consider orientation preserving circle homeomorphisms with break points, that is, maps that are smooth everywhere except for several singular points at which the first derivative has a jump. It is well known that the invariant measures of sufficiently smooth circle diffeomorphisms are absolutely continuous w.r.t. Lebesgue measure. But in the case of homeomorphisms with break points the results are quite different. We will discuss conjugacies between two circle homeomorphisms with break points. Consider the class of circle homeomorphisms with one break point $b$ and satisfying the Katznelson-Ornsteins smoothness condition i.e. $Df$ is absolutely continuous on $[b, b + 1]$ and $D^2f \in L^p(S^1, dl)$, $p > 1$. We will formulate some results concerning the renormalization behavior of such circle maps. Date: April 19th 2012 Speaker: Jon Chaika (University of Chicago) Title: Badly approximable directions on flat surfaces and bounded geodesics in Teichmueller space. Abstract: In this talk we show that directions on flat surfaces which are poorly approximated by saddle connection directions are a winning set for Schmidt's game. This extends a result of Schmidt for the torus and strengthens a result of Kleinbock and Weiss. It is equivalent to saying that the Teichmueller geodesic flow is bounded. We go on to show that the set of bounded geodesics is winning as a subset of projective measured laminations, answering a question of McMullen. This is joint work with Yitwah Cheung and Howard Masur. Date: April 12th 2012 Speaker: Percy Wong (Princeton University) Title: Local semicircle law in the bulk for Gaussian $\beta$-ensemble (paper available on arXiv) Date: April 5th 2012 Speaker: Xiuyuan Cheng (Princeton University) Title: The Spectrum of Random Kernel Matrices Abstract: We consider n-by-n matrices whose (i, j)-th entry is f(X_i^T X_j), where X_1, ...,X_n are i.i.d. standard Gaussian random vectors in R^p, and the kenrel function f is a real-valued function. We study the weak limit of the spectral density when p and n go to infinity and p/n = \gamma which is a constant. The limiting spectral density is dictated by a cubic equation involving its Stieltjes transform, and the parameters of the cubic equation are decided by the Hermite expansion of the rescaled kernel function. While the case of kernel functions that are differentiable at the origin has been previously resolved by El-Karoui (2010), our result is applicable to non-smooth kernel functions, e.g. the sign function. For this larger class of kernel functions, we obtain a new family of limiting densities, which includes the Marcenko-Pastur distribution and Wigner’s semi-circle as special cases. Date: March 29th 2012 Speaker: Jon Fickenscher (Princeton University) Title: Minimal Interval Exchange Transformations with Many Ergodic Measures Abstract: We will give a brief survey of interval exchange transformations, especially results on minimality and unique ergodicity. In particular, we shall prove that a known upper bound on the number distinct ergodic probability measures for a minimal IETs is sharp. This is a generalization of known examples for IETs with order-reversing permutations. This talk will contatin information from the following paper, available on arXiv. Date: March 15th 2012 Speaker: Jinxin Xue (Institute for Advanced Study) Title: Noncollision singularities in planar two-center-two-body problem Abstract: In this work we study a model called planar 2-center-2-body problem. In the plane, we have two fixed centers Q_1=(-\chi,0), Q_2=(0,0) of masses 1, and two moving bodies Q_3 and Q_4 of masses \mu. They interact via Newtonian potential. Q_3 is captured by Q_2, and Q_4 travels back and forth between two centers. Based on a construction of Gerver, we prove that there is a Cantor set of initial conditions which lead to solutions of the Hamiltonian system whose velocities are accelerated to infinity within finite time avoiding all early collisions. We consider this model as a simplified model for the planar four-body problem case of the Painleve conjecture. This is a joint work with Dmitry Dolgopyat. Date: March 8th 2012 Speaker: Ilya Vinogradov (Princeton University) Title: Effective bisector estimate for PSL(2,C) with applications to circle packings Abstract: Let Gamma be a non-elementary discrete geometrically finite subgroup of PSL(2,C). Under the assumption that the critical exponent of Gamma is greater than 1 we prove an effective bisector counting theorem for Gamma. We then apply this Theorem to the Apollonian circle packing problem to get power savings and to compute the overall constant. The proof relies on spectral theory of Gamma\PSL(2,C). Date: Wednesday, February 29th 2012 at 3:00pm, Fine 110 *note special date, time and place* Speaker: Sergei Suslov (Arizona State University) Title: On a Hidden Symmetry of Simple Harmonic Oscillators Abstract: Since the original 1926 Schroedinger's paper, there was a misconception that the “simple” harmonic oscillator can be solved only by the separation of variables, which results in a traditional “static” electron density distribution. It is not entirely accurate and a nontrivial oscillator hidden symmetry group, found by Niederer in 1973, provides "dynamic solutions". The phase space oscillations of the electron position and linear momentum probability distributions are computer animated and some possible applications to quantum optics are briefly discussed. A visualization of the Heisenberg Uncertainty Principle is presented. In addition, these "dynamic harmonic states" possess the nontrivial Berry phase, which may be use for their identification. The corresponding phase is evaluated in terms of elementary functions. In view of importance of the simple harmonic oscillators in numerous applications, these results will be interesting to everybody who is going to study and/or teach quantum mechanics --- It may help better understand a general concept of quantum motion. Date: February 23rd 2012 Speaker: Itzhak Fouxon (Tel Aviv University) Title: Effect of emergent distinguishability of particles in a non-equilibrium chaotic system Abstract: We consider the behavior of classical particles which evolution consists of free motion interrupted by binary collisions. The fluid of hard balls and the dilute gas with arbitrary short-range interactions are treated, where the total number of particles is moderate (say, five particles). We assume that the decay of correlations, characteristic for chaotic systems, holds (it can be considered proved for hard balls). We show that the numbers of collisions of a given particle with other particles grow effectively as a biased random walk. This is used to prove that over indefinitely long periods of time each particle has preferences: it systematically collides more with certain particles and less with others. Thus certain particles are effectively attracted and certain others are repelled, making the particles effectively distinguishable. The effect is of statistical origin and it reminds of entropic forces. Date: February 16th 2012 Speaker: Yakov Sinai (Princeton University) Title: Binary correlations of the Moebius function Last update: 2012-09-19	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.928814172744751, "perplexity": 1021.7521482906583}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-43/segments/1539583514355.90/warc/CC-MAIN-20181021203102-20181021224602-00441.warc.gz"}
http://physics.stackexchange.com/questions/61954/why-doesnt-this-metric-cover-all-of-de-sitter-space	Why doesn't this metric cover all of de Sitter space? This represents a confused attempt to work through a problem in Carroll's Spacetime and Geometry. Supposedly I should be able to use the geodesic equation, $$\frac{d^2x^\mu}{d\lambda^2}+\Gamma^\mu_{\rho\sigma}\frac{dx^\sigma}{d\lambda}\frac{dx^\rho}{d\lambda}=0$$ to show why this metric $$ds^2=-dt^2+e^{2Ht}[dx^2 + dy^2 + dz^2]$$ fails to cover the entire manifold of de Sitter space. I am urged to combine these two equations to solve for the affine parameter $\lambda$ as a function of $t$, and then show that the geodesics of the space reach $t=-\infty$ in a finite value of the affine parameter, demonstrating what I sought to prove. I begin by parametrizing the metric in terms of the proper time, $\tau$ $(\lambda = \tau)$, to yield: $$-1 =-\frac{dt^2}{d\lambda^2}+e^{2Ht}[\frac{dx^2}{d\lambda^2} + \frac{dy^2}{d\lambda^2} + \frac{dz^2}{d\lambda^2}]$$ And from the geodesic equation, I derive: $$\frac{d^2 x^i}{d\lambda^2} = -2H\left(\frac{dt}{d\lambda}\right)\left(\frac{dx^i}{d\lambda}\right)$$ To solve any one of these three equations, say, for example, the first one, I make the subsitution: $$\nu = \frac{dx}{d\lambda}, t'=\frac{dt}{d\lambda}$$ then $$\nu'=-2H\nu t'\\\nu'/t'=\frac{d\nu}{d\lambda}\frac{d\lambda}{dt}=-2H\nu=\frac{d\nu}{dt}\\\frac{d\nu}{dt}=-2H\nu\Rightarrow\nu = C_1e^{-2Ht}$$ Substituting back into the metric, I'm left with: $$1 = \left(\frac{dt}{d\lambda}\right)^2 - e^{2Ht}\left(\sum_i C_i^2 e^{-4Ht}\right)$$ Define $\sum_i C_i^2 = \alpha$, yielding: $$\frac{dt}{d\lambda} = \sqrt{1 + \alpha e^{-2Ht}}$$ I can't seem to find a nice analytical solution for $\lambda(t)$ that demonstrates that $t \rightarrow -\infty$ in a finite value of the affine parameter without relying on a computer, and this doesn't leave me with any nice physical (or mathematical) intuition for what's going on in the problem. Have I made an error somewhere, or is this, as Walter Cronkite (and my GR professor) would say, "the way it is"? Can anyone point me in the right direction, or show me how to proceed? Thanks in advance. - Just looking at your last equation (haven't check through the derivation): in the $t\to -\infty$ limit the exponential dominates and you can simplify the square root then separate variables. – Michael Brown Apr 23 '13 at 3:19	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 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": 0.925923764705658, "perplexity": 204.33271761011963}, "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-2014-23/segments/1405997885796.93/warc/CC-MAIN-20140722025805-00046-ip-10-33-131-23.ec2.internal.warc.gz"}
http://www.jgaa.info/getPaper?id=118	Distributing Unit Size Workload Packages in Heterogeneous Networks Vol. 10, no. 1, pp. 51-68, 2006. Regular paper. Abstract The task of balancing dynamically generated work load occurs in a wide range of parallel and distributed applications. Diffusion based schemes, which belong to the class of nearest neighbor load balancing algorithms, are a popular way to address this problem. Originally created to equalize the amount of arbitrarily divisible load among the nodes of a static and homogeneous network, they have been generalized to heterogeneous topologies. Additionally, some simple diffusion algorithms have been adapted to work in dynamic networks as well. However, if the load is not divisible arbitrarily but consists of indivisible unit size tokens, diffusion schemes are not able to balance the load properly. In this paper we consider the problem of balancing indivisible unit size tokens on heterogeneous systems. By modifying a randomized strategy invented for homogeneous systems, we can achieve an asymptotically minimal expected overload in l1, l2 and l∞ norm while only slightly increasing the run-time by a logarithmic factor. Our experiments show that this additional factor is usually not required in applications. Submitted: September 2004. Revised: April 2005. Communicated by Tomasz Radzik article (PDF) BibTeX	{"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": 0.8347136378288269, "perplexity": 641.7628605280606}, "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-2017-51/segments/1512948514051.18/warc/CC-MAIN-20171211203107-20171211223107-00551.warc.gz"}
http://mathhelpforum.com/statistics/18807-sum-integers.html	1. ## sum of integers Find the sum of the first 100 positive even integers. Explain how this sum can be found quickly. Sn = (n/2)(a1 + an) is that the correct formula? 2. Originally Posted by aikenfan Find the sum of the first 100 positive even integers. Explain how this sum can be found quickly. Sn = (n/2)(a1 + an) is that the correct formula? that is the correct formula. note that we can consider an ordered list of even integers as an arithmetic sequence. we use that formula to find the sum of the terms of an arithmetic sequence. you know that $a_1 = 2$, and we want $n = 100$, so now the only thing left for you to find is $a_100$ 3. Hello, aikenfan! Find the sum of the first 100 positive even integers. Explain how this sum can be found quickly. Sn = (n/2)(a1 + an) is that the correct formula? That formula will work, but I prefer the general Sum Formula: . . $S_n \;=\;\frac{n}{2}\left[2a_1 + (n-1)d\right]$ It is more complicated, but we don't need to calculate the 100th term. We have: . $a_1 = 2,\;d = 2,\;n = 100$ Therefore: . $S_{100} \;=\;\frac{100}{2}\left[2\!\cdot\!2 + (100-1)2\right] \;=\; 10,100$ 4. I am not sure how to calculate the 100th term? an = dn + c? 5. Originally Posted by aikenfan I am not sure how to calculate the 100th term? an = dn + c? you would use the formula $a_n = a_1 + (n - 1)d$ (which is the general formula for the nth term of an arithmetic sequence) where $a_n$ is the nth term, $a_1$ is the first term, $n$ is the current number of the term, and $d$ is the common difference. you would use the values Soroban gave thus to find $a_{100}$ you would compute $a_{100} = 2 + (100 - 1)(2)$ 6. an = dn + c Isn't this the same thing? I get 200 either way... 7. Originally Posted by aikenfan an = dn + c Isn't this the same thing? I get 200 either way... well, if you define c as a_1 - d, then yes. but i've never seen anyone use "c" in these formulas 8. Sn = (100/2)(2+200) = 10100	{"extraction_info": {"found_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": 13, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9727883338928223, "perplexity": 476.6847560171458}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-26/segments/1498128321426.45/warc/CC-MAIN-20170627134151-20170627154151-00326.warc.gz"}
https://homework.cpm.org/category/CC/textbook/cca2/chapter/12/lesson/12.2.3/problem/12-136	### Home > CCA2 > Chapter 12 > Lesson 12.2.3 > Problem12-136 12-136. Without using a calculator, solve these two problems from a college entrance exam. 1. If $x^{−4/3} + 1 = 17$, find $x^{−1/3}$. Solve the first equation for $x$, then use that value to find the answer. 2. If $\log(A)=a$, find $\log\left(\frac{A}{100}\right)$. $a - 2$	{"extraction_info": {"found_math": true, "script_math_tex": 6, "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": 0.896737277507782, "perplexity": 2008.6875796896525}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662545875.39/warc/CC-MAIN-20220522160113-20220522190113-00008.warc.gz"}
https://www.physicsforums.com/threads/rotating-wheel-question.48379/	# Rotating wheel question 1. Oct 17, 2004 ### mickeychief Can anyone help me in solving this question? At t=0, a wheel is rotating about a fixed axis at a constant angular acceleration has an angular velocity of 2 rad/s. Two seconds later it has turned 5 complete revolutions what is the angular acceleration of this wheel?? 2. Oct 17, 2004 ### cepheid Staff Emeritus The question says that the angular acceleration is constant, so why would it have changed two seconds later? Are you sure you were not asked to solve for the angular velocity? 3. Oct 17, 2004 ### mickeychief I typed the question just as it was written. I am having trouble finding the formula for the solution, maybe it is written wrong?? 4. Oct 17, 2004 ### pervect Staff Emeritus I would say that the first step is to write the angular velocity of the wheel as a function of time. Have you tried to do this at all? 5. Oct 18, 2004 ### HallsofIvy Staff Emeritus The problem didn't say the angular acceleration had changed. It gave the initial angular velocity and the total angle moved in 2 seconds and asked you to calculate the constant angular acceleration. mickeychief: this is just like straight line acceleration except that you use angles instead of length: If the accelaration is a, then the velocity after t is at+ v0,where v0 is the initial velocity, and the "distance" moved is (1/2)at2+ v0t. Since you are told that v0= 2 rad/sec and that, with t= 0, the angle ("distance") moved was "5 complete revolutions"= 5(2pi)= 10pi radians, you have (1/2)a(4)+ 2(2)= 10pi. Solve for a.	{"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": 0.9727357625961304, "perplexity": 1049.499174942977}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-50/segments/1480698541910.39/warc/CC-MAIN-20161202170901-00303-ip-10-31-129-80.ec2.internal.warc.gz"}
https://www.questionsolutions.com/cylindrical-plate-subjected-three-cable-forces/	# The cylindrical plate is subjected to the three 2 The cylindrical plate is subjected to the three cable forces which are concurrent at point D. Express each force which the cables exert on the plate as a Cartesian vector, and determine the magnitude and coordinate direction angles of the resultant force. Image from: Hibbeler, R. C., S. C. Fan, Kai Beng. Yap, and Peter Schiavone. Statics: Mechanics for Engineers. Singapore: Pearson, 2013. #### Solution: Show me the final answers↓ We will first write the locations of points A, B, C, and D in Cartesian vector form. Using the diagram, the points are at the following locations: $A:(0.75i+0j+0k)$ m $B:(-0.75\sin30^0i+0.75\cos30^0j+0k)\,=\,(-0.375i+0.65j+0k)$ m $C:(-0.75\cos45^0i-0.75\sin45^0j+0k)\,=\,(-0.53i-0.53j+0k)$ m $D:(0i+0j+3k)$ m (Note that the radius of the circle is 0.75 m. Using our sine and cosine functions, we can then figure out the x and y coordinates of each point) We will now write position vectors for points from A to D, B to D, and C to D. $r_{AD}\,=\,\left\{(0-0.75)i+(0-0)j+(3-0)k\right\}$ $r_{AD}\,=\,\left\{-0.75i+0j+3k\right\}$ m $r_{BD}\,=\,\left\{(0-(-0.375))i+(0-0.65)j+(3-0)k\right\}$ $r_{BD}\,=\,\left\{0.375i-0.65j+3k\right\}$ m $r_{CD}\,=\,\left\{(0-(-0.53))i+(0-(-0.53))j+(3-0)k\right\}$ $r_{CD}\,=\,\left\{0.53i+0.53j+3k\right\}$ m A position vector, denoted $\mathbf{r}$ is a vector beginning from one point and extending to another point. It is calculated by subtracting the corresponding vector coordinates of one point from the other. If the coordinates of point A was $(x_A,y_A,z_A)$ and the coordinates of point B was$(x_B,y_B,z_B)$, then $r_{AB}\,=\,(x_B-x_A)i+(y_B-y_A)j+(z_B-z_A)k$ We will now find the magnitude of each position vector. magnitude of $r_{AD}\,=\,\sqrt{(-0.75)^2+(0)^2+(3)^2}\,=\,3.09$ m magnitude of $r_{BD}\,=\,\sqrt{(0.375)^2+(-0.65)^2+(3)^2}\,=\,3.09$ m magnitude of $r_{CD}\,=\,\sqrt{(0.53)^2+(0.53)^2+(3)^2}\,=\,3.09$ m The magnitude is equal to the square root of the sum of the squares of the vector. If the position vector was $r\,=\,ai+bj+ck$, then the magnitude would be, $r_{magnitude}\,=\,\sqrt{(a^2)+(b^2)+(c^2)}$. In the simplest sense, you take each term of a vector, square it, add it together, and then take the square root of that value. Let us now write the unit vector for each position vector. $u_{AD}\,=\,\left(-\dfrac{0.75}{3.09}i+0j+\dfrac{3}{3.09}k\right)$ $u_{BD}\,=\,\left(\dfrac{0.375}{3.09}i-\dfrac{0.65}{3.09}j+\dfrac{3}{3.09}k\right)$ $u_{CD}\,=\,\left(\dfrac{0.53}{3.09}i+\dfrac{0.53}{3.09}j+\dfrac{3}{3.09}k\right)$ The unit vector is each corresponding unit of the position vector divided by the magnitude of the position vector. If the position vector was $r\,=\,ai+bj+ck$, then unit vector, $u\,=\,\dfrac{a}{\sqrt{(a^2)+(b^2)+(c^2)}}+\dfrac{b}{\sqrt{(a^2)+(b^2)+(c^2)}}+\dfrac{c}{\sqrt{(a^2)+(b^2)+(c^2)}}$ We can now write each force in Cartesian form by multiplying the given magnitudes of each force by the unit vector. $F_A\,=\,6\left(-\dfrac{0.75}{3.09}i+0j+\dfrac{3}{3.09}k\right)$ $F_A\,=\,\left\{-1.46i+0j+5.82k\right\}$ kN $F_B\,=\,8\left(\dfrac{0.375}{3.09}i-\dfrac{0.65}{3.09}j+\dfrac{3}{3.09}k\right)$ $F_B\,=\,\left\{0.97i-1.68j+7.77k\right\}$ kN $F_C\,=\,5\left(\dfrac{0.53}{3.09}i+\dfrac{0.53}{3.09}j+\dfrac{3}{3.09}k\right)$ $F_C\,=\,\left\{0.86i+0.86j+4.85k\right\}$ kN Next, we will find the resultant force. $F_R\,=\,F_A+F_B+F_C$ $F_R\,=\,(-1.46+0.97+0.86)i+(0-1.68+0.86)j+(5.82+7.77+4.85)k$ $F_R\,=\,\left\{0.37i-0.82j+18.44k\right\}$ kN To figure out the coordinate direction angles, we need to find the magnitude of the resultant force. magnitude of $F_R\,=\,\sqrt{(0.37)^2+(-0.82)^2+(18.44)^2}\,=\,18.46$ kN We can now write the coordinate direction angles by taking the cosine inverse of each component of the resultant force divided by it’s magnitude. $\alpha\,=\,\cos^{-1}\left(\dfrac{0.37}{18.46}\right)\,=\,88.8^0$ $\beta\,=\,\cos^{-1}\left(\dfrac{-0.82}{18.46}\right)\,=\,92.5^0$ $\gamma\,=\,\cos^{-1}\left(\dfrac{18.44}{18.46}\right)\,=\,2.66^0$ #### Final Answers: $F_A\,=\,\left\{-1.46i+0j+5.82k\right\}$ kN $F_B\,=\,\left\{0.97i-1.68j+7.77k\right\}$ kN $F_C\,=\,\left\{0.86i+0.86j+4.85k\right\}$ kN $F_R\,=\,\left\{0.37i-0.82j+18.44k\right\}$ kN magnitude of $F_R\,=\,18.46$ kN $\alpha\,=\,88.8^0$ $\beta\,=\,92.5^0$ $\gamma\,=\,2.66^0$ ## 2 thoughts on “The cylindrical plate is subjected to the three” • questionsolutions Post author Sorry for the late reply. We use sin30 and cos30 to figure out the exact point where B lies with respective to the x and y axes. Using sin30 we can figure out the x length, and using cos30 we can figure out the y length. Remember that sin is opposite over hypotenuse while cos is adjacent over hypotenuse.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 45, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9601085186004639, "perplexity": 701.1428610325576}, "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-2022-49/segments/1669446710698.62/warc/CC-MAIN-20221129132340-20221129162340-00065.warc.gz"}
http://math.stackexchange.com/questions/606559/evaluate-limit-of-a-sequence-nbhm-2013	# Evaluate limit of a sequence… NBHM $2013$ Question is to Evaluate $$\lim_{n\rightarrow \infty} \sin((2n\pi + \frac{1}{2n\pi}) \sin(2n\pi + \frac{1}{2n\pi}))$$ All I could do was to see that $$\sin(2n\pi + \frac{1}{2n\pi}))=\sin( \frac{1}{2n\pi})$$ Just because $\sin(2n\pi+\theta)=\sin(\theta)$.. So, we now have $$\lim_{n\rightarrow \infty} \sin((2n\pi + \frac{1}{2n\pi}) \sin( \frac{1}{2n\pi}))$$ Now, as $\lim _{x\rightarrow \infty}x\sin(\frac{1}{x})=1$ we would have $$\lim_{n\rightarrow \infty}2n\pi \sin( \frac{1}{2n\pi})=1$$ So, we now have $$\lim_{n\rightarrow \infty} \sin(1 + \frac{1}{2n\pi} \sin( \frac{1}{2n\pi}))$$ Now, as $\sin(x)$ is bounded and $\frac{1}{2n\pi} \rightarrow 0$ we would have $$\lim_{n\rightarrow \infty}\frac{1}{2n\pi} \sin( \frac{1}{2n\pi})=0$$ So, we would now left with : $$\lim_{n\rightarrow \infty} \sin(1)=\sin(1)$$ After all i would like to say that as $\sin (x)$ is continuous I can take limits inside. So, we have $$\lim_{n\rightarrow \infty} \sin((2n\pi + \frac{1}{2n\pi}) \sin(2n\pi + \frac{1}{2n\pi}))=\sin 1$$ I would like somebody to check if I have done correctly and I would be thankful if any body can let me know if there is anything more to be specified to do so. Thank you... :) - I am not seeing here any mistake – Dutta Dec 14 '13 at 14:20 This is in NBHM 2013 Ph.D exam question paper... – Praphulla Koushik Dec 14 '13 at 14:21 I did not see your heading first – Dutta Dec 14 '13 at 14:22 It is alright :) I have kept it just for my quick reference.... :) – Praphulla Koushik Dec 14 '13 at 14:23 @PraphullaKoushik: really, this is a Ph.D. exam question? Seems pretty basic for that. – Eckhard Dec 14 '13 at 14:23 Question is to Evaluate $$\lim_{n\rightarrow \infty} \sin((2n\pi + \frac{1}{2n\pi}) \sin(2n\pi + \frac{1}{2n\pi}))$$ As $\sin(2n\pi+\theta)=\sin(\theta)$ we would have : $$\sin(2n\pi + \frac{1}{2n\pi}))=\sin( \frac{1}{2n\pi})$$ So, we now have $$\lim_{n\rightarrow \infty} \sin((2n\pi + \frac{1}{2n\pi}) \sin( \frac{1}{2n\pi}))$$ Now, as $\lim _{x\rightarrow \infty}x\sin(\frac{1}{x})=1$ we would have $$\lim_{n\rightarrow \infty}2n\pi \sin( \frac{1}{2n\pi})=1$$ So, we now have $$\lim_{n\rightarrow \infty} \sin(1 + \frac{1}{2n\pi} \sin( \frac{1}{2n\pi}))$$ Now, as $\sin(x)$ is bounded and $\frac{1}{2n\pi} \rightarrow 0$ we would have $$\lim_{n\rightarrow \infty}\frac{1}{2n\pi} \sin( \frac{1}{2n\pi})=0$$ So, we would now left with : $$\lim_{n\rightarrow \infty} \sin(1)=\sin(1)$$ After all i would like to say that as $\sin (x)$ is continuous I can take limits inside. So, we have $$\lim_{n\rightarrow \infty} \sin((2n\pi + \frac{1}{2n\pi}) \sin(2n\pi + \frac{1}{2n\pi}))=\sin 1$$ -	{"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": 0.9665473103523254, "perplexity": 488.8312430917567}, "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-2015-48/segments/1448398471436.90/warc/CC-MAIN-20151124205431-00033-ip-10-71-132-137.ec2.internal.warc.gz"}

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