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http://www.bytemining.com/category/machine-learning/
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## Highlights from My First NIPS
The first few hundred registrations received a mug.
As a machine learning practitioner in the Los Angeles area, I was ecstatic to learn that NIPS 2017 would be in Long Beach this year. The conference sold out in a day or two. The conference was held at the Long Beach Convention Center (and Performing Arts Center), very close to the Aquarium of the Pacific and about a mile from the Queen Mary. The venue itself was beautiful, and probably the nicest place I’ve ever attended a conference. It’s also the most expensive place I’ve ever had a conference. $5 for a bottle of Coke?$11 for two cookies? But I digress.I attended most of the conference, but as someone who has attended many conferences, I’ve learned that attending everything is not necessary, and is counterproductive to one’s sanity. I attended the main conference, and one workshop day, but skipped the tutorials, the Saturday workshops and the industry demos. The conference talks were livestreamed via Facebook Live at the NIPS Foundation’s Facebook page, and the recordings are also archived there.
This may make some question why one would actually want to attend the conference in person, but there are several!
to talk with the authors of interesting […]
## Summary of My First Trip to Strata #strataconf
In this post I am goIing to summarize some of the things that I learned at Strata Santa Clara 2013. For now, I will only discuss the conference sessions as I have a much longer post about the tutorial sessions that I am still working on and will post at a later date. I will add to this post as the conference winds down.
The slides for most talks will be available here but not all speakers will share their slides.
This is/was my first trip to Strata so I was eagerly awaiting participating as an attendant. In the past, I had been put off by the cost and was also concerned that the conference would be an endless advertisement for the conference sponsors and Big Data platforms. I am happy to say that for the most part I was proven wrong. For easier reading, I am summarizing talks by topic rather than giving a laundry list schedule for a long day and also skip sessions that I did not find all that illuminating. I also do not claim 100% accuracy of this text as the days are very long and my ears and mind can only process so much data when I am context […]
## SIAM Data Mining 2012 Conference
Note: This would have been up a lot sooner but I have been dealing with a bug on and off for pretty much the past month!
From April 26-28 I had the pleasure to attend the SIAM Data Mining conference in Anaheim on the Disneyland Resort grounds. Aside from KDD2011, most of my recent conferences had been more “big data” and “data science” oriented, and I wanted to step away from the hype and just listen to talks that had more substance.
Attending a conference on Disneyland property was quite a bizarre experience. I wanted to get everything I could out of the conference, but the weather was so nice that I also wanted to get everything out of Disneyland as I could. Seeing adults wearing Mickey ears carrying Mickey shaped balloons, and seeing girls dressed up as their favorite Disney princesses screams “fun” rather than “business”, but I managed to make time for both.
The first two days started with a plenary talk from industry or research labs. After a coffee break, there were the usual breakout sessions followed by lunch. During my free 90 minutes, I ran over to Disneyland and California Adventure both days to eat lunch. I managed to […]
## Parsing Wikipedia Articles: Wikipedia Extractor and Cloud9
Lately I have doing a lot of work with the Wikipedia XML dump as a corpus. Wikipedia provides a wealth information to researchers in easy to access formats including XML, SQL and HTML dumps for all language properties. Some of the data freely available from the Wikimedia Foundation include
article content and template pages
article content with revision history (huge files)
article content including user pages and talk pages
redirect graph
page-to-page link lists: redirects, categories, image links, page links, interwiki etc.
site statistics
The above resources are available not only for Wikipedia, but for other Wikimedia Foundation projects such as Wiktionary, Wikibooks and Wikiquotes.
As Wikipedia readers will notice, the articles are very well formatted and this formatting is generated by a somewhat unusual markup format defined by the MediaWiki project. As Dirk Riehl stated:
There was no grammar, no defined processing rules, and no defined output like a DOM tree based on a well defined document object model. This is to say, the content of Wikipedia is stored in a format that is not an open standard. The format is defined by 5000 lines of php code (the parse function of MediaWiki). That code may be open source, but it is incomprehensible to most. That’s why […]
## SIGKDD 2011 Conference -- Days 2/3/4 Summary
<< My review of Day 1.
I am summarizing all of the days together since each talk was short, and I was too exhausted to write a post after each day. Due to the broken-up schedule of the KDD sessions, I group everything together instead of switching back and forth among a dozen different topics. By far the most enjoyable and interesting aspects of the conference were the breakout sessions.
Keynotes
KDD 2011 featured several keynote speeches that were spread out among three days and throughout the day. This year’s conference had a few big names.
Steven Boyd, Convex Optimization: From Embedded Real-Time to Large-Scale Distributed. The first keynote, by Steven Boyd, discussed convex optimization. The goal of convex optimization is to minimize some objective function given linear constraints. The caveat is that the objective function and all of the constraints must be convex (“non-negative curvature” as Boyd said). The goal of convex optimization is to turn the problem into a linear programming problem. We should care about convex optimization because it comes from some beautiful and complete theory like duality and optimality conditions. I must say, that whenever I am chastising statisticians, I often say that all they care about is “beautiful theory” […]
## SIGKDD 2011 Conference -- Day 1 (Graph Mining and David Blei/Topic Models)
I have been waiting for the KDD conference to come to California, and I was ecstatic to see it held in San Diego this year. AdMeld did an awesome job displaying KDD ads on the sites that I visit, sometimes multiple times per page. That’s good targeting!
Mining and Learning on Graphs Workshop 2011
I had originally planned to attend the 2-day workshop Mining and Learning with Graphs (MLG2011) but I forgot that it started on Saturday and I arrived on Sunday. I attended part of MLG2011 but it was difficult to pay attention considering it was my first time waking up at 7am in a long time. The first talk I arrived for was Networks Spill the Beans by Lada Adamic from the University of Michigan. Adamic’s presented work involved inferring properties of content (the “what”) using network structure alone (using only the “who”: who shares with whom). One example she presented involved questions and answers on a Java programming language forum. The research problem was to determine things such as who is most likely to answer a Java beginner’s question: a guru, or a slightly more experienced user? Another research question asked what dynamic interactions tell us about information flow. […]
## Hadoop Fatigue -- Alternatives to Hadoop
It’s been a while since I have posted… in the midst of trying to plow through this dissertation while working on papers for submission to some conferences.
Hadoop has become the de facto standard in the research and industry uses of small and large-scale MapReduce. Since its inception, an entire ecosystem has been built around it including conferences (Hadoop World, Hadoop Summit), books, training, and commercial distributions (Cloudera, Hortonworks, MapR) with support. Several projects that integrate with Hadoop have been released from the Apache incubator and are designed for certain use cases:
Pig, developed at Yahoo, is a high-level scripting language for working with big data and Hive is a SQL-like query language for big data in a warehouse configuration.
HBase, developed at Facebook, is a column-oriented database often used as a datastore on which MapReduce jobs can be executed.
ZooKeeper and Chukwa
Mahout is a library for scalable machine learning, part of which can use Hadoop.
Cascading (Chris Wensel), Oozie (Yahoo) and Azkaban (LinkedIn) provide MapReduce job workflows and scheduling.
Hadoop is meant to be modeled after Google MapReduce. To store and process huge amounts of data, we typically need several machines in some cluster configuration. A distributed filesystem (HDFS for Hadoop) uses space across […]
## My Review of Hadoop Summit 2011 #hadoopsummit
I woke up early and cheery Wednesday morning to attend the 2011 Hadoop Summit in Santa Clara, after a long drive from Los Angeles and the Big Data Camp that lasted until 10pm the night before. Having been to Hadoop Summit 2010, I was interested to see how much of the content in the conference had changed.
This year, there were approximately 1,600 participants and the summit was moved a few feet away to the Convention Center rather than the Hyatt. Still, space and seating was pretty cramped. That just goes to show how much the Hadoop field has grown in just one year.
Keynotes
We first heard a series of keynote speeches which I will summarize. The first keynote was from Jay Rossiter, SVP of the Cloud Platform Group at Yahoo. He introduced how Hadoop is used at Yahoo, which is fitting since they organized the event. The content of his presentation was very similar to last year’s. One interesting application of Hadoop at Yahoo was for “retiling” the map of the United States. I imagine this refers to the change in aerial imagery over time. When performed by hand, retiling took 6 weeks; with Hadoop, it took 5 days. Yahoo also […]
## Big Data Camp 2011 #BigDataCamp
It has been a while since I have been to Silicon Valley, but Hadoop Summit gave me the opportunity to go. To make the most of the long trip, I also decided to check out BigDataCamp held the night before from 5:30 to 10pm. Although the weather was as predicted, I was not prepared for the deluge of pouring rain in the end of June. The weather is one of the things that is preventing me from moving up to Silicon Valley.
The food/drinks/networking event must have been amazing because it was very difficult to get everyone to come to the main room to start the event! We started with a series of lightning talks from some familiar names and some unfamiliar ones.
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2018-05-22 13:32:49
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https://stats.libretexts.org/Bookshelves/Introductory_Statistics/Book%3A_Statistical_Thinking_for_the_21st_Century_(Poldrack)/24%3A_Modeling_Continuous_Relationships
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# 24: Modeling Continuous Relationships
## Learning Objectives
• Describe the concept of the correlation coefficient and its interpretation
• Compute the correlation between two continuous variables
• Describe the potential causal influences that can give rise to a correlation.
Most people are familiar with the concept of correlation, and in this chapter we will provide a more formal understanding for this commonly used and misunderstood concept.
24: Modeling Continuous Relationships is shared under a CC BY-NC 2.0 license and was authored, remixed, and/or curated by Russell A. Poldrack via source content that was edited to conform to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.
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2022-05-27 00:05:55
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https://math.stackexchange.com/questions/3402435/radical-ideals-with-same-zero-set
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# Radical ideals with same zero set
Let $$\mathbb{K}$$ be a field. Let $$S\subseteq \mathbb{K}[x_1,\dots,x_n]$$ be a set of polynomials. The variety defined by $$S$$ is the set, $$V(S)=\{a\in \mathbb{K}^n:f(a)=0\:\forall f\in S\}$$
For an algebraically closed field Nullstellensatz relates the radical of an ideal to its variety. If the field is not algebraically closed, can there be two radical ideals with the same zero set? I was trying to construct examples but I couldn't do it.
• This depends on your definition of variety, which you should add to the question. – KReiser Oct 21 '19 at 7:17
• @KReiser I have done that now. – cookiemonster Oct 21 '19 at 7:19
• Hint: think about working over $\Bbb R$ and a sum of squares. – KReiser Oct 21 '19 at 7:29
You can look at $$V(x^2+1)$$ in $$\mathbb{R}$$, ie the empty set. $$x^2+1$$ is a radical ideal, since $$\mathbb{R} [x]/ (x^2+1)$$ is isomorphic to $$\mathbb{C}$$, a reduced ring. You can also look at $$V(\mathbb{R}[x])$$.
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2020-02-25 14:32:16
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http://mathhelpforum.com/advanced-applied-math/149482-optimal-control-theory-about-riccati-dfiferential-equation.html
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1. Optimal Control Theory: About the Riccati Dfiferential Equation
Hello there,
I have a problem on optimal control theory which has to do with the Riccati Equation:
$\dot{P}=-PA-A'P+PBR^{-1}B'P-Q, P(T)=F$
where $Q,R$ are symmetric.
The problem asks to show that the solution matrix of the Riccati equation is positive-semidefinite given that $F,Q \ge 0, R>0$.
2. Can you tell me what does semdefinite means in this context?
I know what positive definite matrix means.
3. A positive-semidefinite matrix is a matrix positive one but also with the equality. That is, $x' M x \geq 0$.
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2013-12-05 08:24:05
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https://www.snapsolve.com/solutions/Multiply6-13-by-the-reciprocal-of-7-16--1672365640643586
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Home/Class 8/Maths/
Multiply $$\dfrac{6}{13}$$by the reciprocal of $$\dfrac{-7}{16}$$.
Speed
00:00
05:02
## QuestionMathsClass 8
Multiply $$\dfrac{6}{13}$$by the reciprocal of $$\dfrac{-7}{16}$$.
$$\dfrac{-96}{91}$$
4.6
4.6
## Solution
If we have a number $$a$$,then reciprocal of $$a=\frac 1a$$where $$a\neq0$$,as reciprocal of $$0$$ is not defined
So, reciprocal of$$\dfrac{-7}{16}=\dfrac 1{\dfrac{-7}{16}}=\dfrac{-16}7$$
According to the question,
$$=\dfrac 6{13}\times\dfrac{(-16)}7$$
$$=\dfrac{-96}{91}$$
Hence, the required value is $$\dfrac{-96}{91}$$.
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2022-07-05 08:37:48
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https://www.nature.com/articles/s41467-020-19385-6?error=cookies_not_supported&code=34f745dc-9d26-4fa8-b7fe-91ba2ab9ab08
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Thank you for visiting nature.com. You are using a browser version with limited support for CSS. To obtain the best experience, we recommend you use a more up to date browser (or turn off compatibility mode in Internet Explorer). In the meantime, to ensure continued support, we are displaying the site without styles and JavaScript.
# Brightness modulations of our nearest terrestrial planet Venus reveal atmospheric super-rotation rather than surface features
## Abstract
Terrestrial exoplanets orbiting within or near their host stars’ habitable zone are potentially apt for life. It has been proposed that time-series measurements of reflected starlight from such planets will reveal their rotational period, main surface features and some atmospheric information. From imagery obtained with the Akatsuki spacecraft, here we show that Venus’ brightness at 283, 365, and 2020 nm is modulated by one or both of two periods of 3.7 and 4.6 days, and typical amplitudes <10% but occasional events of 20–40%. The modulations are unrelated to the solid-body rotation; they are caused by planetary-scale waves superimposed on the super-rotating winds. Here we propose that two modulation periods whose ratio of large-to-small values is not an integer number imply the existence of an atmosphere if detected at an exoplanet, but it remains ambiguous whether the atmosphere is optically thin or thick, as for Earth or Venus respectively. Multi-wavelength and long temporal baseline observations may be required to decide between these scenarios. Ultimately, Venus represents a false positive for interpretations of brightness modulations of terrestrial exoplanets in terms of surface features.
## Introduction
As the search for terrestrial exoplanets advances, and the technology that will enable their characterization matures, it becomes important to establish observational diagnostics that inform us about their surfaces and atmospheres, and testing such diagnostics over a variety of conditions. Time-series measurements of a planet’s reflected starlight potentially provide an avenue to map a planet’s surface as photometric variability and surface inhomogeneity are interconnected.
The idea has been extensively developed1,2,3,4,5,6,7, and convincingly demonstrated for Earth with the Deep Space Climate Observatory (DSCOVR) space-based photometry gathered for more than 2 years and 10 wavelengths from 320 to 780 nm6. The idea is valid if the atmosphere is optically thin (as Earth’s), so that stellar photons reach the surface and escape back to space. In general, it may be non-trivial to determine if a small-mass exoplanet has an atmosphere, let alone if it is thin or thick. Clouds, if present, will interfere with the surface signal and introduce additional temporal variability8, but long-term exposures may filter out such effects1,6.
A key characteristic of the Earth’s brightness modulation is that its periodogram shows a dominant peak at P = 1 day and at the fractional periods $$\frac{1}{2}$$, $$\frac{1}{3}$$ and $$\frac{1}{4}$$ days for all wavelengths in the DSCOVR dataset6. The 1-day signal originates from the Earth’s rotation, whereas the shorter-period signals are related to the details in the distribution of continents and oceans2,6.
Venus is currently outside the so-called habitable zone (HZ, the circumsolar region within which liquid water might occur at the planet’s surface) but it was possibly habitable in the past9. Venus’ equilibrium temperature is Teq = 230 K, not much different from Earth’s Teq = 254 K. A huge greenhouse effect however keeps Venus’ surface temperature at 735 K, too hot to allow for liquid water. It remains unclear at what point Venus drifted into that state if, as is usually thought, both planets might have had similar conditions in their early days10,11.
Exo-Venuses, i.e. planets near the inner boundary of their host stars’ HZ, are expected to be abundant9,12,13. Thus, it is important to devise diagnostics beyond first-order factors such as the orbital distance that will help distinguish between genuine exo-Earths, with mild temperatures suitable for life, and exo-Venuses. The question is complex yet key in the characterization of terrestrial exoplanets and will require multiple approaches to address it.
Here, we show the photometric time series of Venus in anticipation of what might be expected for exo-Venus observations. We find two distinct periods (3.7 and 4.6 days) in the modulation of the reflected sunlight. These periods are ~60 times shorter than Venus’ solid-body rotational period, and so they are unrelated to the surface. Rather, they originate from the super-rotating background winds and superimposed planetary-scale waves. We show their wavelength dependence at 283, 365, and 2020 nm, and their temporal variability in Venus’ disk-integrated photometry data. We propose that the two distinct nearby peaks are a sign of the existence of an atmosphere, and that a long-baseline campaign of multi-wavelength observations (and the search for temporal variations in them) will help conclude that such an exoplanet might have a Venus-like thick atmosphere. Our study conveys the caution message that distinguishing between brightness modulations associated with the solid-body rotation of an exoplanet or its atmospheric winds will have to be carefully considered in future data analyses of exoplanets.
## Results
### Multispectral photometry of Venus
We investigated Venus in reflected sunlight using whole-disk imagery produced by the JAXA/Akatsuki spacecraft14 in the ultraviolet (UV) and near infrared (NIR). The images were acquired with the UV camera (UVI) at effective wavelengths of 283 and 365 nm15, and with the NIR camera (IR2) at an effective wavelength of 2020 nm16. Each set of 2–3 images was obtained within 9 min and thus the images in a set can be considered to be nearly simultaneous. The time interval between each set of images is typically 2 h, but can be up to a few days depending on the location of the spacecraft on its highly elliptical orbit.
As the whole disk of Venus can only be imaged from a distance sufficiently far from the planet, we used images taken before or after pericenter passage14. So while Akatsuki revolves around Venus every ~11 days, it obtains whole-disk images for ~10 days per orbit during the dayside-monitoring epoch. The sequence of observations alternates every 4 months between dayside- and nightside-monitoring epochs up to one Venusian year (225 days, ~8 months), when a new dayside–nightside sequence is started. For our analysis, we utilized a total of 5805 (283 nm) and 5840 (365 nm) UVI images obtained between 2015 and 2019, and 354 IR2 images obtained in 2016, at the end of which year the latter camera stopped working (see Methods, subsection “Image processing”, and Supplementary Fig. 1 for details on the data).
We describe Venus’ disk-integrated brightness in the usual form of a geometric albedo × phase law that depends on the Sun–Venus-spacecraft phase angle α but not on the planet’s apparent size (see Eq. (4) in Methods, subsection “Image processing”). Hereafter, we refer to this size-normalized measure of brightness simply as the planet’s brightness or phase-resolved albedo.
In the UV, the brightness generally decreases as α increases (Fig. 1)17. The abrupt variation at small phase angles is the glory, an optical phenomenon due to scattering from narrow-size distributions of cloud droplets18,19,20. The Venusian clouds are very thick (optical thickness τ ~ 30 in the visible21), which prevents the access of the solar photons to the surface at UV-NIR wavelengths on the dayside. They also contain traces of an unknown absorber that produces the dark patterns seen in UV images (Fig. 2) and that absorbs most strongly at 350-380 nm22,23. Absorption by the unknown absorber and the SO2 gas above the clouds reduce the brightness at 283 nm and result in a lower brightness at this wavelength24. Venus’ main atmospheric gas, CO2, absorbs strongly at 2020 nm16,25 above the cloud top level at ~ 70 km, which reduces the NIR brightness to < 0.03, an order of magnitude less than in the UV (Fig. 1). The NIR brightness increases for α > 80 due to scattering by the haze that exists above the clouds26,27 and whose relative contribution against CO2 absorption increases for high phase angles.
The phase-resolved albedo of Fig. 1 and the sequence of images of Fig. 2 demonstrate that Venus’ brightness varies overtime at all three wavelengths. Modulations about the mean conditions are seen at each orbit and over most of the monitored phase angles, thus confirming their persistent nature. Orbits 20 and 81 (Fig. 1; (a) and (b) panels, respectively) exhibit particularly strong modulations with peak-to-peak amplitudes of ~ 20% in the UV and ~40% in the NIR (orbit 20) (see Methods, subsection “Mean phase curves and periodicity analysis”, and Supplementary Figs. 3, 5). The amplitude of these modulations at the planetary scale was unknown to date.
The temporal variability in the phase-resolved albedo is clearly seen in our Supplementary Movies 12, and suggests multiple timescales associated with changes in the spatial distribution of absorbers and in the global cloud morphology. There is an anti-correlation between the UV and NIR brightness: an increase in the UV is consistently accompanied by a decrease in the NIR and vice versa. This means that the main absorbers at the wavelengths investigated here (the unknown absorber and SO2 in the UV; CO2 in the NIR) affect the Venus brightness in opposite yet temporally related ways.
The disk-resolved images of Fig. 2 help understand the brightness modulations. Fig. 2a confirms that the UV–NIR brightness is indeed anti-correlated. Each column of images shows a nearly simultaneous snapshot of Venus at the three wavelengths. There is a peculiar, global scale synchronization between low and high latitudes; the NIR modulations occur at all latitudes, including middle and high latitudes (Fig. 2e), while the UV modulations occur mainly although not exclusively at low latitudes (Fig. 2f, g). This previously unknown behavior is likely related to the development of the known ‘Y’-shape feature, which might result from a combination of Kelvin and Rossby atmospheric waves from low to high latitudes28.
### Modulations of disk-integrated brightness
At the cloud-top level probed by Akatsuki, the Venus atmosphere rotates in the same direction as the surface but 60–80 times faster28, and thus it takes ~4–5 days for the zonal winds to circle the planet. This super-rotation occurs simultaneously with the vertical and horizontal oscillations at the cloud top level that drive the NIR and UV modulations in brightness, respectively.
To better characterize the temporal behavior of these modulations, we have defined brightness deviations with respect to a baseline constructed by fitting a 4th-order polynomial to the phase-resolved albedos (Fig. 1). The periodogram of these deviations (Fig. 3) reveals two distinct peaks at P1 ~ 3.7 and P2 ~ 4.6 days. Periods comparable to P1 and P2 have been reported before to describe the modulations in local brightness and wind velocities at Venus29,30,31,32, and used to support their interpretation in terms of waves. The match between our periods and those reported elsewhere is particularly good when we focus separately on the low and middle latitudes, as has been customarily done in previous work (Supplementary Table 1 and Supplementary Fig. 8).
A period ~4 days was reported for the whole-disk brightness measurements at 365 nm made by the Pioneer Venus Orbiter spacecraft33. This is, however, the first instance that both periods are clearly identified in the disk-integrated brightness of Venus and at multiple wavelengths. Based on previous investigations at 365 nm29,30,31,32, the P1 and P2 periods are associated with an equatorial Kelvin wave and a mid-latitude Rossby wave moving in the direction of the mean zonal flow at phase speeds somewhat faster and slower than the zonal winds, respectively.
Interestingly, the strengths and widths in the periodogram of the P1- and P2-period signals are very different at each wavelength. Both periods are confidently detected at 365 and 2020 nm, but only P1 is noticeable at 283 nm. This suggests that the impact of each wave on the planet’s brightness is affected by the latitudinal distribution of absorbers and their wavelength-dependence absorption properties (see Methods, subsection “The missing P2 period at 283 nm in the context of Venus studies”).
At low latitudes, both the unknown absorber and SO2 gas are most abundant34,35. Their abundances as a function of altitude decrease rapidly upwards near the cloud top level36,37. The impact of CO2 absorption in the NIR depends on slight changes in the cloud top altitudes38. The equatorial Kelvin wave causes vertical oscillations of all these absorbers at low latitudes29, and consequently P1 is apparent at all wavelengths. At mid-to-high latitudes, the 365 nm brightness shows strong latitudinal variations (dark spiral and bright polar hood39), while the 2020 nm brightness drops steeply towards high latitudes due to the decreasing cloud top altitude38. These mid-to-high latitudinal variations are oscillated horizontally, in the latitudinal direction, by the Rossby wave29, and thus P2 becomes clear at 365 and 2020 nm.
Both the P1- and P2-period signals are recurrent features in Akatsuki’s multi-year time-series of UV brightness measurements. The strength of each signal fluctuates over timescales of a few months (Fig. 4). This long-term evolution seems to follow the evolving viewing/illumination conditions introduced by the motion of Akatsuki and Venus on their orbits, also after removing the mean phase curve baseline. Even considering that the viewing/illumination geometry may affect the brightness deviations to some extent, the steep changes in strength of the P1- and P2-period signals for small changes in phase angle at 283 and 365 nm (Fig. 4) suggest that geometrical effects are not the primary cause of the periodograms’ months-long fluctuations. It appears more credible that these fluctuations reflect real temporal variations in Venus’ atmosphere. This alternating behavior between the P1- and P2-period signals has been described before in disk-resolved brightness investigations31, and is thought to be connected with the processes that sustain the atmospheric super-rotation.
## Discussion
The key features of the Venus periodogram for disk-integrated brightness are (i) it shows a single period (P1) at 283 nm, but two non-fractional periods (P1 and P2, with P1/P2 and P2/P1 ≠ integer number) at 365 and 2020 nm; (ii) the brightness modulations in the UV and NIR are anti-correlated; (iii) the strengths of the P1- and P2-period signals exhibit long-term variations.
The above findings are relevant to the characterization of terrestrial exoplanets in reflected starlight with future space telescopes such as the Large UV/Optical/IR Surveyor (LUVOIR)40 and Habitable Exoplanet Imaging Mission (HabEx)41. The logical next step here is to assess what could be learned from the above key features if they were identified in exoplanet data, and in particular how they could help discern whether the planet has an atmosphere and whether it is optically thin or thick. The exercise sets the basis for differentiating an exo-Venus from an exo-Earth before attempting to map out the planet’s surface.
The detection of a single dominant period in a brightness periodogram does not by itself prove that there is an atmosphere. Indeed, geological inhomogeneities at the surfaces of atmosphere-less objects also produce brightness modulations42. Earth has a thin atmosphere and, for the same reason, the occurrence in its periodogram of a 1-day period (and additional fractional periods) cannot discriminate between a planet with or without an atmosphere. Further information might help if for example it provides evidence against a static surface albedo. This has been explored for Earth with the Transiting Exoplanet Survey Satellite (TESS)43 broadband optical-NIR photometry8, showing that aperiodic brightness fluctuations inconsistent with solid-body rotation hint at a dynamical atmosphere. Although not a terrestrial planet, it is also worth recalling that Neptune’s periodogram, as determined with Kepler/K2 observations44, exhibits a dominant peak at P ~ 17 h and smaller-amplitude peaks near 18 hours. The presence of discrete clouds altering Neptune’s overall reflectance together with differential rotation of the background atmosphere induces the multiple periods. The small amplitude of the brightness modulations (<2%, peak-to-peak in the Kepler/K2 passband) and the close proximity of the periods, which will likely appear as a single period in exoplanet observations, will pose a severe challenge to distinguish Neptune’s periodogram from that of an atmosphere-less object.
Unlike for Earth (or Neptune), the detection of two distinct non-fractional periods in Venus’ periodogram offers insight into the atmosphere. Indeed, it is difficult to reconcile the occurrence of both periods with a surface origin of the associated brightness modulations. This implies that one or both of the periodic signals must originate in the atmosphere. In the first case, one could conceive that the atmosphere is optically thin and P1 (P2) is the planet’s rotational period, and thus the observations are revealing a non-synchronous brightness modulation with a longer (shorter) period P2 (P1) on top of the surface’s rotational modulation. In the second case, which is true for Venus, one could conceive that the atmosphere is optically thick and both the P1- and P2-period signals originate in the atmosphere. The bottom line is that key feature (i) alone reveals the existence of an atmosphere from a Venus-like periodogram. This is not a trivial conclusion as many of the planets that will be targeted by direct imaging will lack information as basic as their mass and radius that is essential to constraining their density and therefore their interior composition. The difficulty to infer the occurrence of an atmosphere with reflected-starlight measurements described above mirrors to some extent the difficulties encountered for close-in terrestrial exoplanets investigated with phase curves and currently available telescopes45.
Key feature (ii) suggests also the existence of an atmosphere that through wavelength-dependent optical thickness effects might affect the brightness modulations with different signs at short and long wavelengths. Last, key feature (iii) requires an atmosphere that evolves over time, although it is not obvious if this sets a valuable constraint on its optical thickness. This latter key feature however demonstrates the importance of observing over a long temporal baseline to capture long-term variations in the planet’s brightness.
In perspective, Venus offers a caution message against future attempts to relate the periods of brightness modulations to the solid-body rotation period of a planet, especially with a single wavelength or over temporal baselines that may not capture the evolution of the dominating planetary-scale waves in the planet’s atmosphere. Multi-wavelength and long-baseline observations, as shown here, will be useful to discriminate between Earth- and Venus-like periodograms, and therefore between both planet types, although probably not unambiguously.
## Methods
### Image processing
The number of images that we collected per Akatsuki orbit is shown in Supplementary Fig. 1. One orbit takes ~11 days. Orbit 1 started on 7 2015 Dec. The images at the three wavelengths (283, 365, and 2020 nm) were acquired with two cameras: UVI and IR2. IR2 stopped operating toward the end of 2016. UVI has continued imaging Venus since the orbit insertion of the spacecraft. We analyzed images taken until 2019 January (orbit 105).
The radiance measured by UVI is corrected with the ground-measured flat-field, while the public data in DART (http://darts.isas.jaxa.jp/index.html.en) use the on-board diffuser flat-field. The flat conversion factor is publicly available through DART. We multiply the calibration correction factors (β) by the measured radiance: β283 = 1.886, β365 = 1.525, as described in Yamazaki et al.15. These are mean correction factors based on star observations between years 2010 and 2017, and are very close to the values reported in Yamazaki et al.15. The radiance measured by IR2 has a dependence on the sensor temperature. We correct for this dependence as described in Satoh et al.46:
$${I}_{{\rm{corr}}}=\left\{\begin{array}{ll}{I}_{{\rm{orig}}}/\left[1.0-{p}_{59\,{\rm{K}}}{\left(\frac{T-{T}_{0}}{59-{T}_{0}}\right)}^{2}\right]&{\rm{for}}\,T\,<\, {T}_{0},\\ {I}_{{\rm{orig}}}/\left[1.0-{p}_{70\,{\rm{K}}}{\left(\frac{T-{T}_{0}}{70-{T}_{0}}\right)}^{2}\right]&{\rm{for}}\,T\ge {T}_{0},\end{array}\right.$$
(1)
where Icorr is the corrected radiance, Iorig is the measured radiance, p59 K = 0.13, p70K = 0.25, T0 = 65.2 K, and T is the temperature of the sensor. We used images that had been treated by deconvolution of the point spread function (PSF), so the Venus image is sharper. The deconvolved IR2 images are available from the PI of the IR2 camera upon request.
We calculated the disk-integrated flux (units of W m−2μm−1) from
$${F}_{{\rm{Venus}}}(\alpha ,\lambda ,t)={\sum }_{r\,{<}\,{r}_{o}}{I}_{{\rm{corr}}}(x,y)\times {\Omega }_{{\rm{pix}}},$$
(2)
where (xy) stands for pixel location on the image, Ωpix is the pixel solid angle, r is the distance of (xy) from the Venus disk center, and ro is the integration limit for the aperture photometry. We adopted ro = rVenus radius + rPSF, where rPSF is the extent of the point spread function (7 pixels for UVI, 25 pixels for IR2). The PSF of the IR2 images is known to be wide16, so we used the quoted value as a fine balance between signal and the required area of integration within the limited field of view (FOV, 12 × 12). The center of Venus was found using the limb-fitting process47. We subtracted from the aperture photometry the background noise per pixel, estimated as the mean radiance over an outer ring around Venus between 40 and 70 pixels away from rVenus radius for UVI, and between 60 and 90 pixels for IR2.
The solid angle of Venus, ΩVenus, is calculated as
$${\Omega }_{{\rm{Venus}}}=\pi {\left({\sin }^{-1}\left(\frac{{R}_{\text{Venus radius}}}{{d}_{{\rm{V}}-{\rm{obs}}}}\right)\right)}^{2},$$
(3)
where RVenus radius is the radius of Venus considering the cloud top altitude (=6052+70 km), and dV−obs is the distance of the spacecraft from Venus in km.
Venus’ brightness or phase-resolved albedo, as used in our work, is calculated through the following equation48:
$${A}_{{\rm{disk}}-{\rm{int}}}(\alpha ,\lambda ,t)=\frac{\pi }{{\Omega }_{{\rm{Venus}}}}\frac{{{{\rm{d}}}_{{\rm{V}}-{\rm{S}}}(t)}^{2}{F}_{{\rm{Venus}}}(\alpha ,\lambda ,t)}{{S}_{\odot }(\lambda )},$$
(4)
where dV−S(t) is the distance from Venus to the Sun [AU] at the time of observation t, ΩVenus is the Venus solid angle (Eq. (3)), and S(λ) is the solar irradiance at 1 AU (W m−2 μm−1) considering the transmittance functions of each filter. S(λ) is taken from two sources. Near 365 and 2020 nm, we use the Smithsonian Astrophysical Observatory reference spectrum 201049. Near 283 nm, we use the SORCE SIM Solar Spectral Irradiance (SSI) data (http://lasp.colorado.edu/home/sorce/data/ssi-data/ssi-data-file-summary/) after applying a 30-day running average. The Adisk-int calculated in this study can be found in our Supplementary Data 1–3.
### Mean phase curves and periodicity analysis
We estimated mean phase curves for the disk-integrated brightness at each wavelength $$\overline{{A}_{\lambda }}(\alpha )$$ in 2016 (Supplementary Fig. 2). At the UV wavelengths, α = 0–20 was excluded to avoid the glory features17,18,19,20. The deviation of individual brightness measurements from the mean curve is calculated as
$${A}_{{\rm{devi}},\lambda }(t)=\frac{{A}_{{\rm{disk}}-{\rm{int}},\lambda }(\alpha ,t)-\overline{{A}_{\lambda }}(\alpha )}{\overline{{A}_{\lambda }}(\alpha )}\times 100[ \% ].$$
(5)
This helps remove the phase angle dependence from the brightness measurements, and allows us to focus on the time series for Adevi,λ(t). Examples of deviations for orbit 20 (Fig. 1) are shown in Supplementary Fig. 3. The time series are subsequently used for the periodicity analysis (Fig. 3). To that end, we use the EFFECT software50 with the algorithm in Deeming et al.51. The periodicity caused by the irregular data sampling, the so-called spectral window, has also been checked to look for overlapped peaks (none in our results). We repeated the same procedure for the entire 283 and 365 nm images (Supplementary Figs. 4–6). The full-time series at the three wavelengths are shown in Supplementary Fig. 6. We used those at UV for the periodicity analysis over years 2015–2019 (Supplementary Fig. 7).
Interestingly, the periodograms evolve over the multi-year span of our dataset. This evolution translates into relative variations in the strength of the P1- and P2-period peaks. The temporal evolution of the UV periodograms is shown in Fig. 4. For this particular figure, we used scargle.pro (http://astro.uni-tuebingen.de/software/idl/aitlib/timing/scargle.html; implementation from Press and Rybicki52) that is particularly efficient to process large data sets. We confirm the temporal variation in the signal strength from disk-integrated photometry for each of these periods, associated with Kelvin and Rossby planetary-scale waves, a finding consistent with what has been reported for disk-resolved photometry in previous studies29,30,31,32. For example, Imai et al.31 (their Fig. 10) reported a transition from P1 to P2 at 365 nm from July to September 2017. This is also seen in Fig. 4b, where the peak shifts from P1 to P2 from September to October in 2017 at 365 nm. We also note a clear shift in the identified periods from P1 to P2 in December 2018 at both 283 and 365 nm. The findings from our disk-integrated approach are also consistent with those from Nara et al.32, who report a clear P1 signal at low and middle latitudes in June 2018. Indeed, we can see in our Fig. 4a–b a very strong signal of P1 in June 2018.
### The missing P2 period at 283 nm in the context of Venus studies
It is noteworthy that the periodogram at 283 nm (Fig. 3c) contains only evidence for the P1 period. We propose that the P2 period is missing at 283 nm because of the weaker absorption of the unknown absorber at this wavelength and the smooth latitudinal variation of SO237; both properties attenuate possible horizontal disturbances introduced by the Rossby wave. Additionally, increased Rayleigh scattering at 283 nm, especially in a slanted view, may suppress the specific signal of the mid-latitude Rossby wave. Also noteworthy, the P2-period signal in the periodogram at 365 nm is significantly broader than the other peaks, a possible outcome of the stronger north–south asymmetry in brightness and wind speeds at this wavelength relative to 283 nm53,54.
## Data availability
The UVI (level 3x products) and IR2 (level 3x geometry products) data that support the findings of this study are available in the JAXA archive website, http://darts.isas.jaxa.jp/DARTS, with the identifiers https://doi.org/10.17597/ISAS.DARTS/VCO-0001655 and https://doi.org/10.17597/ISAS.DARTS/VCO-0001856, respectively. Deconvolved IR2 images are available from the PI of IR2 upon request, because the data are still experimental and subject to revision with an improved point spread function. All procedures of IR2 data improvement are documented and archived by the PI. Our disk-integrated brightness data are provided with this paper as Supplementary Data 1–3.
## Code availability
EFFECT software50 is used for the periodicity analysis and scargle.pro (http://astro.uni-tuebingen.de/software/idl/aitlib/timing/scargle.html) is used to temporal evolution of the UV periodograms.
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## Acknowledgements
The authors thank the Akatsuki team. Y.J.L. thanks Dr. Aleksandar Chaushev for discussion. Y.J.L. has received funding from EU Horizon 2020 MSCA-IF No. 841432.
## Funding
Open Access funding enabled and organized by Projekt DEAL.
## Author information
Authors
### Contributions
Y.J.L. and A.G.M. prepared the manuscript. A.G.M. conceived the main strategy, and Y.J.L. performed the data analysis and prepared figures. Y.J.L., A.G.M., T.I. interpreted the results. Y.J.L, and Y.M. worked on the UVI data quality maintenance. T.S. is the PI of IR2 and performed the calibration of IR2 data. A.Y. and S.W. maintained the UVI operation, and S.W. is the PI of UVI.
### Corresponding author
Correspondence to Y. J. Lee.
## Ethics declarations
### Competing interests
The authors declare no competing interests.
Peer review information Nature Communications thanks Rodrigo Luger and Fredric Taylor for their contribution to the peer review of this work. Peer reviewer reports are available.
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Lee, Y.J., García Muñoz, A., Imamura, T. et al. Brightness modulations of our nearest terrestrial planet Venus reveal atmospheric super-rotation rather than surface features. Nat Commun 11, 5720 (2020). https://doi.org/10.1038/s41467-020-19385-6
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2021-12-01 16:39:09
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https://tex.stackexchange.com/questions/171744/how-can-i-use-arabic-numbering-for-theorems
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# How can I use arabic numbering for theorems?
I used the newtheorem command to create exercises in a book in the following way.
\newtheorem{xca}[theorem]{Problems}
So in the body I use the following command to create a set of exercises,
\begin{xca}\label{ex1.3}{Problems}
The default numbering in this environment is roman. I would like to change it to arabic without having to go through each of the exercises in the book and modifying the enumeration command.
Here is a MWE. By creating a MWE, I noted why I was getting roman as oppose to arabic. My question now is whether I can still add something to the definition of xca so that enumeration is in Arabic and not in Roman.
\documentclass{cambridge7A}
\newtheorem{theorem}{Theorem}[chapter]
\newtheorem{xca}[theorem]{Problems}
% remove the dot and change default for enumerated lists
\def\makeRRlabeldot#1{\hss\llap{#1}}
\renewcommand\theenumi{{\rm (\roman{enumi})}}
\renewcommand\theenumii{{\rm (\alph{enumii})}}
\renewcommand\theenumiii{{\rm (\arabic{enumiii})}}
\renewcommand\theenumiv{{\rm (\Alph{enumiv})}}
\begin{document}
\begin{xca}\label{ex1.3}{Problems}
\begin{enumerate}
\item Show that it follows from the definition of a field that zero, unit, additive, and multiplicative inverse scalars are all unique.
\end{enumerate}
\end{xca}
\end{document}
• Could you provide us with a minimal working example (MWE)? There are multiple packages that provide support for creating theorems. – Werner Apr 15 '14 at 23:45
• Thanks. By creating a MWE I noted the reason for Roman enumeration as oppose to Arabic. – lmedina Apr 16 '14 at 0:20
• Welcome to TeX.SX! A tip: If you indent lines by 4 spaces, they'll be marked as a code sample. You can also highlight the code and click the "code" button (with "{}" on it). – jub0bs Apr 16 '14 at 0:27
• @lmedina No problem. Sorry about the mistake I introduced in the title. Sorry Gonzalo; the fault is mine. – jub0bs Apr 16 '14 at 0:36
To change the representation for the first level of an enumerate environment, you can redefine \theenumi; the default definition on your example is
\renewcommand\theenumi{{\rm (\roman{enumi})}
so the label numbering will use lower-case Roman numerals; to get Arabic numbering you need to change it to
\renewcommand\theenumi{{\rmfamily(\arabic{enumi})}}
Since you want the change only to have effect inside the xca environment, one option would be to use \AtBeginEnvironment (from the etoolbox package) to make the change only inside the environment:
\usepackage{etoolbox}
\AtBeginEnvironment{xca}{\renewcommand\theenumi{{\rmfamily(\arabic{enumi})}}}
A complete example:
\documentclass{cambridge7A}
\usepackage{etoolbox}
\newtheorem{theorem}{Theorem}[chapter]
\newtheorem{xca}[theorem]{Problems}
% remove the dot and change default for enumerated lists
\def\makeRRlabeldot#1{\hss\llap{#1}}
\renewcommand\theenumi{{\rmfamily(\roman{enumi})}}
\renewcommand\theenumii{{\rmfamily(\alph{enumii})}}
\renewcommand\theenumiii{{\rmfamily(\arabic{enumiii})}}
\renewcommand\theenumiv{{\rmfamily(\Alph{enumiv})}}
\AtBeginEnvironment{xca}{\renewcommand\theenumi{{\rmfamily(\arabic{enumi})}}}
\begin{document}
\begin{xca}
\label{ex1.3}
Problems
\begin{enumerate}
\item Show that it follows from the definition of a field that zero, unit, additive, and multiplicative inverse scalars are all unique.
\end{enumerate}
\end{xca}
\begin{enumerate}
\item An item of an enumerated list outside the \texttt{xca} environment.
\end{enumerate}
\end{document}
• Thanks for your suggestion. However, this will change everything else in the book as well to Arabic where Roman was used. I just want to make changes in the exercises section without making changes elsewhere. – lmedina Apr 16 '14 at 0:37
• @lmedina then move the line inside the xca environment. – Gonzalo Medina Apr 16 '14 at 0:38
• I really do not where to put it. I tried '\newtheorem{xca}[theorem]{Problems \renewcommand\theenumi{{\rm (\arabic{enumi})}}}' – lmedina Apr 16 '14 at 0:40
• @lmedina please see my updated answer. Will you always use the xca environment for problems? Should the modification be always used inside all xca environments? – Gonzalo Medina Apr 16 '14 at 0:42
• Yes, I will always use xca for problems and inside the xca environment. So I would like to change the way I defined xca so that the numbering can change automatically wherever I used xca but elsewhere should not be affected. – lmedina Apr 16 '14 at 0:45
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2019-10-19 03:23:54
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https://www.doubtnut.com/question-answer-physics/the-pulley-arrangements-of-figures-a-and-b-are-identical-the-mass-of-the-rope-is-negligible-in-figur-642917076
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HomeEnglishClass 11PhysicsChapterLaws Of Motion
The pulley arrangements of fig...
The pulley arrangements of figures (a) and (b) are identical . The mass of the rope is negligible . In figure (a), the mass m is lifted up by attaching a mass 2m to the other end of the rope . In figure (b) , m is lifted up by pulling the other end of the rope with a constant downward force F=2mg . Calculate the accelerations in the two cases. <br> <img src="https://doubtnut-static.s.llnwi.net/static/physics_images/AKS_ELT_AI_PHY_XI_V01_A_C06_SLV_022_Q01.png" width="80%">
Step by step solution by experts to help you in doubt clearance & scoring excellent marks in exams.
Apne doubts clear karein ab Whatsapp par bhi. Try it now.
In figure (a), for the motion of mass m , T-mg=ma<br> <img src="https://doubtnut-static.s.llnwi.net/static/physics_images/AKS_ELT_AI_PHY_XI_V01_A_C06_SLV_022_S01.png" width="80%"> <br> For motion of mass 2m figure (b) <br> 2 mg -T (2m)a …(2)<br>Adding equation (1) and (2) , we get <br> 2mg-mg= 2ma+ma, mg=3ma, a = g//3<br>(b)In figure T^(1)-mg=ma^(1)<br> But T^(1)=2mg , :. 2mg-mg =a^(1), Therefore a^(1)=g
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2021-12-04 12:58:51
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https://gmatclub.com/forum/a-function-f-has-the-property-that-f-3x-1-x-2-x-1-for-all-real-number-291817.html
|
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# A function f has the property that f(3x-1)=x^2+x+1 for all real number
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A function f has the property that f(3x-1)=x^2+x+1 for all real number [#permalink]
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26 Mar 2019, 23:14
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Difficulty:
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Question Stats:
63% (00:40) correct 38% (01:17) wrong based on 56 sessions
### HideShow timer Statistics
A function f has the property that $$f(3x-1)=x^2+x+1$$ for all real numbers x. What is f(5)?
(A) 7
(B) 13
(C) 31
(D) 111
(E) 211
_________________
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Joined: 27 Sep 2018
Posts: 37
Re: A function f has the property that f(3x-1)=x^2+x+1 for all real number [#permalink]
### Show Tags
26 Mar 2019, 23:40
We can write f(5) as f(3*2-1)
Now comparing with f(3x−1)=x^2+x+1
We get x=2
Putting in given equation we have 2^2+2+1=7
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Re: A function f has the property that f(3x-1)=x^2+x+1 for all real number [#permalink]
### Show Tags
27 Mar 2019, 01:24
Bunuel wrote:
A function f has the property that $$f(3x-1)=x^2+x+1$$ for all real numbers x. What is f(5)?
(A) 7
(B) 13
(C) 31
(D) 111
(E) 211
f(3x-1)= f(5)
at x= 2
so x^2+x+1 = 7 at x=2
IMOA
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Joined: 04 Oct 2018
Posts: 159
Location: Viet Nam
Re: A function f has the property that f(3x-1)=x^2+x+1 for all real number [#permalink]
### Show Tags
27 Mar 2019, 03:45
Bunuel wrote:
A function f has the property that $$f(3x-1)=x^2+x+1$$ for all real numbers x. What is f(5)?
(A) 7
(B) 13
(C) 31
(D) 111
(E) 211
f(5) = f(3*2 - 1) = 2^2 + 2 + 1 = 7. Thus A
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"It Always Seems Impossible Until It Is Done"
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Joined: 12 Sep 2017
Posts: 302
Re: A function f has the property that f(3x-1)=x^2+x+1 for all real number [#permalink]
### Show Tags
28 Mar 2019, 19:07
Archit3110 wrote:
Bunuel wrote:
A function f has the property that $$f(3x-1)=x^2+x+1$$ for all real numbers x. What is f(5)?
(A) 7
(B) 13
(C) 31
(D) 111
(E) 211
f(3x-1)= f(5)
at x= 2
so x^2+x+1 = 7 at x=2
IMOA
Hello Archit3110!!!
It took me some time to realize that I could equal both functions. ¿In which cases can we do this?
Kind regards!
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Joined: 18 Aug 2017
Posts: 5020
Location: India
Concentration: Sustainability, Marketing
GPA: 4
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Re: A function f has the property that f(3x-1)=x^2+x+1 for all real number [#permalink]
### Show Tags
29 Mar 2019, 01:17
jfranciscocuencag wrote:
Archit3110 wrote:
Bunuel wrote:
A function f has the property that $$f(3x-1)=x^2+x+1$$ for all real numbers x. What is f(5)?
(A) 7
(B) 13
(C) 31
(D) 111
(E) 211
f(3x-1)= f(5)
at x= 2
so x^2+x+1 = 7 at x=2
IMOA
Hello Archit3110!!!
It took me some time to realize that I could equal both functions. ¿In which cases can we do this?
Kind regards!
jfranciscocuencag
well the only way to solve this question was to simplify the f(3x-1) ; such that it becomes equal to f(5). so x=2 was substituted
it all depends on the question and the function formula and whats been asked..
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Re: A function f has the property that f(3x-1)=x^2+x+1 for all real number [#permalink]
### Show Tags
29 Mar 2019, 08:37
Bunuel wrote:
A function f has the property that $$f(3x-1)=x^2+x+1$$ for all real numbers x. What is f(5)?
(A) 7
(B) 13
(C) 31
(D) 111
(E) 211
Since we want f(5), we should look for the value of x such that 3x - 1 = 5:
3x - 1 = 5
3x = 6
x = 2, so we have:
f(5) = f(3*2 - 1) = 2^2 + 2 + 1 = 7
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Re: A function f has the property that f(3x-1)=x^2+x+1 for all real number [#permalink] 29 Mar 2019, 08:37
Display posts from previous: Sort by
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2019-10-19 07:53:46
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https://toph.co/p/tic-toc-toe
|
# Practice on Toph
Participate in exhilarating programming contests, solve unique algorithm and data structure challenges and be a part of an awesome community.
# Tic Toc Toe
By asifthegreat · Limits 1s, 512 MB
We all know what Tic-Toc-Toe works. But in the anime world, people aren’t really interested in normal Tic-Toc-Toe. Their game is kinda different. Let’s see how to play it.
Firstly, you’re given an undirected graph witn $N$ nodes and $M$ edges. Every node has a cost. Initially all of them are 0. Now the game holder can update certain things and he will ask you some queries. So there are 2 types of operations in that game.
1. Update: $~$ Given node $U$ and a number $val$. You will start your journey at $U$ and you can move to a node $V$ if and only if $U \leq V$. So you can not pass through a node if it’s less than $U$. Let’s suppose we have a set $S$, which holds the nodes we can travel from $U$. You have to add $val$ to all the nodes in $S$.
2. Query: $~$ Given $U$ and you have to print the cost of the node $U$.
## Input
The first line will contain 2 numbers, $N$ ($2 \leq N \leq 10^5$) and $M$ ($2 \leq M \leq 10^5$), the number of nodes and the number of edges respectively.
Each of the next $M$ lines will contain 2 numbers $U, V$ ($1 \leq U, V \leq 10^5$) which means there is an edge between node $U$ and $V$.
The next line will contain a number $Q$ ($1 \leq Q \leq 10^5$), the number of operations.
Each of the next $Q$ lines can contain 2 types of lines:
• 1 $U$ val (update operation)
• 2 $U$ (Query operation)
NOTE: See the sample to be more clear.
## Output
For every query (second operation), print the cost of the node $U$ in a new line.
NOTE: See the sample to be more clear.
## Sample
InputOutput
5 7
1 2
1 3
1 5
1 4
2 5
4 5
4 3
4
1 1 12
2 5
1 4 5
2 5
12
17
### Explanation of the sample:
Here is the tree after the first operation:
Here is the tree after second operation:
### Statistics
0% Solution Ratio
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2020-05-31 23:35:12
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http://uhcourseworkvvvq.du-opfer.info/the-definition-of-correlation.html
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# The definition of correlation
Correlation, in the finance and investment industries, is a statistic that measures the degree to which two securities move in relation to each other correlations. Correlation is one of the most widely used — and widely misunderstood — statistical concepts in this overview, we provide the definitions and. Definition of correlation in the definitionsnet dictionary and translations of correlation in the most comprehensive dictionary definitions resource on the web.
The pearson product-moment correlation coefficient is a measure of the strength of the linear relationship between two variables it is referred to as pearson's. Also known as the pearson product-moment correlation coefficient, the correlation coefficient (r) measures the linear relationship between two variables, with a. This work investigates the use of canonical correlation analysis (cca) in the definition of weight restrictions for data envelopment analysis (dea) with this. Learn more about correlation, a statistical technique that shows how strongly the numbers in rating scales have meaning, but that meaning isn't very precise.
A partial correlation coefficient is a measure of the linear dependence of a pair of random variables from a collection of random variables in the. Where s_y and s_(y^^) are the standard deviations of the data points y and the estimates y^^ given by the regression line (kenney and keeping 1962, p 293. Correlation is used to describe the linear relationship between two continuous variables (eg, height and weight) in general, correlation tends to be used when .
Definition of correlation - learn everything about correlation with our statistics a correlation measures the strength of a statistical link between two variables. Understand 2 different senses of correlation in urdu along with english definitions. When examining data in sas, correlation reveals itself by the relationship it is apparent when examining the definition of correlation that measures from only.
## The definition of correlation
Definition of local correlation in a linear algebra notation, the squared correlation coefficient $\gamma$ from equation 8 can be represented as a product of two. Correlation the correlation is one of the most common and most useful statistics a correlation is a single number that describes the degree of relationship. Correlation describes the relationship between two sets of data this lesson will delve into what correlation is and the different types of. Definition: an instance when two variables appear to be correlated, not correlation of fossil inclusions is a principle of stratigraphy: that strata may be.
• In the vector space of random variables it is reasonable to define the the most basic definition of the variance is the 'mean deviation from the.
• Top definition correlation there is a correlation between ice cream sales and drowning therefore get a correlation mug for your bunkmate josé buy the.
• This definition explains correlation in statistics and discusses positive and negative correlations, as well as the difference between correlation and causation.
Correlation suggests an association between two variables although correlation may imply causality, that's different than a cause-and-effect relationship the effects of a small sample size limitation the definition of an. Spatial correlation is an important index to evaluate performance of multi- antenna systems low correlation leads to good diversity performance [1] and lar. Define correlation correlation synonyms, correlation pronunciation, correlation translation, english dictionary definition of correlation n 1 a relationship or. Correlation (co-relation) refers to the degree of relationship (or dependency) between two variables linear correlation refers to straight-line.
The definition of correlation
Rated 4/5 based on 42 review
2018.
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2018-10-15 23:05:03
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http://www.mzan.com/article/49832417-c-sharp-declaring-a-variable-with-an-enum.shtml
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Home c# Declaring a variable with an enum?
public enum States { START, LOSE, WIN, PLAYERTURN, ENEMYTURN, }; States CurrentState; This the code I'm having trouble with.I'm trying to make a turn based system, but I can't seem to figure out how to define the variable "CurrentState". From what I've seen from others, using the enum and then the variable should work; I just get a compiler error. Am I doing something wrong? This code is for Unity 2017.3.1; I am using Visual Studio to write my code.
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2018-08-15 11:17:35
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http://www.jstor.org/stable/1995791
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Journal Article
# Maxima and High Level Excursions of Stationary Gaussian Processes
Simeon M. Berman
Transactions of the American Mathematical Society
Vol. 160 (Oct., 1971), pp. 65-85
DOI: 10.2307/1995791
Stable URL: http://www.jstor.org/stable/1995791
Page Count: 21
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## Abstract
Let $X(t), t \geqq 0$, be a stationary Gaussian process with mean $0$, variance $1$ and covariance function $r(t)$. The sample functions are assumed to be continuous on every interval. Let $r(t)$ be continuous and nonperiodic. Suppose that there exists $\alpha, 0 < \alpha \leqq 2$, and a continuous, increasing function $g(t), t \geqq 0$, satisfying $$(0.1) \lim {t \rightarrow 0} {\frac {g(ct)} {g(t)} = 1, \text{for every} c > 0$$, such that $$(0.2) 1 - r(t) \sim g(|t|)|t|^\alpha, t \rightarrow 0$$. For $u > 0$, let $\nu$ be defined (in terms of $u$) as the unique solution of $$(0.3) u^2 g(1/\nu)\nu ^{-\alpha} = 1$$. Let $I_A$ be the indicator of the event $A$; then $$\int^T_0 I_{\lbrack X(s) > u\ rbrack} ds$$ represents the time spent above $u$ by $X(s), 0 \leqq s \leqq T$. It is shown that the conditional distribution of $$(0.4) \nu \int^T_0 I_{\lbrack X(s) > u \rbrack} ds$$, given that it is positive, converges for fixed $T$ and $u \rightarrow \infty$ to a limiting distribution $\Psi_\alpha$, which depends only on $\alpha$ but not on $T$ or $g$. Let $F(\lambda)$ be the spectral distribution function corresponding to $r(t)$. Let $F^{(p)}(\lambda)$ be the iterated $p$-fold convolution of $F(\lambda)$. If, in addition to (0.2), it is assumed that $$(0.5) F^{(p)} is absolutely continuous for some p > 0$$, then max $(X(s): 0 \leqq s \leqq t)$, properly normalized, has, for $t \rightarrow \infty$, the limiting extreme value distribution exp $(-e^{-x})$. If, in addition to (0.2), it is assumed that $$(0.6) F(\lambda) \text{is absolutely continuous with the derivative} f(\lambda)$$, and $$(0.7) \lim _{h \rightarrow 0} \log h \int^\infty_{-\infty} |f(\lambda + h) - f(\lambda)| d\lambda = 0$$, then (0.4) has, for $u \rightarrow \infty$, and $T \rightarrow \infty$, a limiting distribution whose Laplace-Stieltjes transform is $$(0.8) \exp \big\lbrack constant \int^\infty_0 (e ^{-\lambda x}-1) d \Psi \alpha (x) \big\rbrack, \lambda > 0.$$
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2017-03-28 15:54:40
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https://rdrr.io/cran/Data2LD/src/demo/CruiseControl.R
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# demo/CruiseControl.R In Data2LD: Functional Data Analysis with Linear Differential Equations
# Cruise control: A Two-variable Feedback Loop
#
# A driver starts his car and (1) accelerates to the 60 km/h speed limit
# holding on most main streets of Ottawa, (2) reduces speed to 40 km/h in a
# school zone, (3) enters a controlled-access boulevard with an 80 km/h
# limits and finally (4) against enters a typical principal Listeet.
# Operating on a snowy winter morning, the driver takes 8 seconds to reach
# 60 km/h, or 4/30 seconds per unit change in speed.
#
# The speed of the car under acceleration is modelled by a first order
# constant coefficient equation:
#
# The feedback equations:
# The two differental equations specifying the feedback model are:
#
# $$DS(t) <- \beta_{11} S(t) + \beta_{12} C(t)$$
# $$DC(t) <- \beta_{21} S(t) + \beta_{22} C(t) + \alpha_[[2]] S.0(t)$$
#
# The right side term
# $\beta_{12} C(t)$ in the first equation is the contribution of the
# control variable $C$ to the change in speed $DS$.
# In the second term on the right side the variable $S_0$
# is the set-point function specifying the target speed, and it is
# called a forcing function.
#
# We will use a special case of these equations to generate some
# simulated data and then estimate the parameters defining the
# equation using these simulated data.
#
# The specialized equations are
#
# $$DS(t) = -S(t) + C(t)/4$$
# $$DC(t) = S_0(t) - S(t)$$
#
# These equations correspond to:
# $\beta_{11}$ = -1
# $\beta_{12}$ = 1/4
# $\beta_{21}$ = -1
# $\beta_{22}$ = 0
# $\alpha_2$ = 1
#
# We see that when (1) the speed $S(t)$ is less than the set point value
# S_0(t), the control variable increases to a
# positive value and forces the speed to increase, and
# (2) the speed $S(t)$ is greater than the set point value
# S_0(t), the control variable decreases to a
# negative value and forces the speed to decrease.
# The value $\beta_{22}$ = 0 implies that the controller responds
# instantly to a change in the difference $S_0(t) - S(t)$.
#
# We will try out two models. In the first we estimate the four
# parameters $\beta_{11}$, $\beta_{12}$, $\beta_{21}$ and $\alpha_2$.
# In the second we coerce $\beta_{21} - \alpha_2$ to be 0. In both
# models we will set up $\beta_{22}$ as a parameter but fix its value
# to be 0.
# Define the problem:
# Set up the time span, a set of observation times, and a fine grid for
# plotting:
T <- 80 # seconds
rng <- c(0,T)
n <- 41
nfine <- 501
tobs <- seq(0,T,len=n)
tfine <- seq(0,T,len=nfine)
# Set up the set-point forcing function.
# UfdList is a list array having the same two-level listListucture as
# AwtList, but the contents of the lists are functional data objects
# specifying the external input to the system. If a list is empty, it is
# assumed that there is no forcing of the corresponding equation.
#
# The set-point function uses an order 1 step function B-spline basis. The
# knots are placed at the points where the set-point changes values.
steporder <- 1 # step function basis
stepnbasis <- 4 # four basis functions
stepbreaks <- c(0,20,40,60,80)
stepbasis <- create.bspline.basis(rng, stepnbasis, steporder, stepbreaks)
stepcoef <- c(60,40,80,60) # target speed for each step
SetPtfd <- fd(stepcoef, stepbasis) # define the set point function
# Set up cruiseModelList
# The total number of coefficients defining the estimated coefficient
# functions is three because each coefficient function has only one
# coefficient, and only three of them are estimated.
# set up a constant basis over this range
conbasis <- create.constant.basis(rng)
confdPar <- fdPar(conbasis)
# Solving the equations for known parameter values and initial conditions.
# In order to simulate data, we need to know the true values of .S(t).
# and .C(t). at the time points where the process is observed. We also
# need to specify the initial state of the system at time 0, which we define
# to be zero for both variables. We get the solution by using an initial
# value approximation algorithm, which is here the Runge-Kutta fourth order
# method coded in Matlab"s function |ode45|.
# The function |cruuise1| evaluates the right side of the equations at a
# set of time values.
#
# Here is a function that evaluates the right side of the equation for a
# time value:
cruise0 <- function(t, y, parms) {
DSvec <- matrix(0,2,1)
Uvec <- eval.fd(t, parms$SetPtfd) DSvec[1] <- -y[1] + y[2]/4 DSvec[2] <- Uvec - y[1] return(list(DSvec=DSvec)) } # We first have a look at the solution at a fine mesh of values by solving # the equation for the points in |tfine| and plotting them y0 <- matrix(0,2,1) parms = list(SetPtfd=SetPtfd) ytrue = lsoda(y0, tfine[1:500], cruise0, parms) ytrue = rbind(ytrue,matrix(c(80,60,240),1,3)) # Plot the true solution par(mfrow=c(2,1)) # speed panel plot(tfine, ytrue[,2], type="l", lwd=2, ylim=c(0,100), ylab="Speed S (mph)") lines(c( 0,20),c(60,60),type="l",lty=2) lines(c(20,20),c(60,40),type="l",lty=2) lines(c(20,40),c(40,40),type="l",lty=2) lines(c(40,40),c(40,80),type="l",lty=2) lines(c(40,60),c(80,80),type="l",lty=2) lines(c(60,60),c(80,60),type="l",lty=2) lines(c(60,80),c(60,60),type="l",lty=2) # control panel plot(tfine, ytrue[,3], type="l", lwd=2, xlab="Time (mins)", ylab="Control level C") # interpolate at 41 equally spaced time points the values of these variables speedResult <- approx(tfine, ytrue[,2], seq(0,80,len=41)) controlResult <- approx(tfine, ytrue[,3], seq(0,80,len=41)) # Simulate noisy data at n observation points # We simulate data by adding a random zero-mean Gaussian deviate to each # curve value. The deviates for speed have a standard deviation 2 speed # units, and those for the control level have a standard deviation of 8. sigerr <- 2 yobs <- matrix(0,length(tobs),2) yobs[,1] <- as.matrix( speedResult$y + rnorm(41)*sigerr)
yobs[,2] <- as.matrix(controlResult$y + rnorm(41)*sigerr*4) # plot the data along with the true solution: par(mfrow=c(2,1)) plot(tfine, ytrue[,2], type="l", ylab="Speed") lines(c(0,T), c(60,60), lty=3) points(tobs, yobs[,1], pch="o") plot(tfine, ytrue[,3], type="l", xlab="Time (mins)", ylab="Control level") lines(c(0,T), c(240,240), lty=3) points(tobs, yobs[,2], pch="o") # Define cruiseDataList, and insert these into structs into the corresponding # lists. cruiseDataList1 <- list(argvals=tobs, y=yobs[,1]) cruiseDataList2 <- list(argvals=tobs, y=yobs[,2]) cruiseDataList <- vector("list",2) cruiseDataList[[1]] <- cruiseDataList1 cruiseDataList[[2]] <- cruiseDataList2 # Define cruiseBasisList containing the basis system for each varible. # We also have to provide a basis system for each variable that is large # enough to allow for any required sharp curvature in the solution to the # differential equation system. # # First we set the order of the B-spline basis to 5 so that the first # derivative will be smooth when working with a second order derivative in # the penalty term. Then we position knots at 41 positions where we willl # simulate noisy observations. We use the same basis for both variables # and load it into a list array of length 2. cruiseBasisList <- vector("list",2) delta <- 2*(1:10) breaks <- c(0, delta, 20, 20+delta, 40, 40+delta, 60, 60+delta) nbreaks <- length(breaks) nSorder <- 5 nSbasis <- nbreaks + nSorder - 2 Sbasis <- create.bspline.basis(c(0,80), nSbasis, nSorder, breaks) cruiseBasisList[[1]] <- Sbasis nCorder <- 4 nCbasis <- nbreaks + nCorder - 2 Cbasis <- create.bspline.basis(c(0,80), nCbasis, nCorder, breaks) cruiseBasisList[[2]] <- Cbasis # Now set up the list objects for each of the two cells in # list cruiseModelList # List object for the speed equation: A term for speed and a term for control, but no forcing SList.XList = vector("list",2) # Fields: funobj parvec estimate variable deriv. factor SList.XList[[1]] <- make.Xterm(confdPar, 1, TRUE, 1, 0, -1) SList.XList[[2]] <- make.Xterm(confdPar, 1/4, TRUE, 2, 0, 1) SList.FList = NULL SList = make.Variable("Speed", 1, SList.XList, SList.FList) # List object for the control equation: a term for speed, and a zero-multiplied term for control # plus a term for the forcing function SetPtfd CList.XList <- vector("list",2) # Fields: funobj parvec estimate variable deriv. factor CList.XList[[1]] <- make.Xterm(confdPar, 1, TRUE, 1, 0, -1) CList.XList[[2]] <- make.Xterm(confdPar, 0, FALSE, 2, 0, 1) CList.FList <- vector("list",1) # Fields: funobj parvec estimate Ufd factor CList.FList[[1]] <- make.Fterm(confdPar, 1, TRUE, SetPtfd, 1) CList <- make.Variable("Control", 1, CList.XList, CList.FList) # Now set up the Listuct objects for each of the two lists in # list array cruiseVariableList # List array for the whole system cruiseModelList <- vector("list",2) cruiseModelList[[1]] <- SList cruiseModelList[[2]] <- CList # check the system specification for consistency cruiseModelCheckList <- checkModel(cruiseBasisList, cruiseModelList) cruiseModelList <- cruiseModelCheckList$modelList
nparam <- cruiseModelCheckList$nparam print(paste("total number of parameters = ", nparam)) # A preliminary evaluation of the function and its derivatives rhoVec <- 0.5*matrix(1,1,2) Data2LDList <- Data2LD(cruiseDataList, cruiseBasisList, cruiseModelList, rhoVec) print(Data2LDList$MSE)
print(Data2LDList$DpMSE) print(Data2LDList$D2ppMSE)
# Note that, for the parameter that is fixed, the gradient element is 0, and
# the row and column of the hessian matrix are 0 except for the diagonal
# value, which is 1.
# Evaluate at the corrent parameter values, which in this case are
# the right values since the data are without error
# set up a loop through a series of values of rho
# We know, because the signal is smooth and the data are rough, that the
# optimal value of rho will be rather close to one, here we set up a
# range of rho values using the logistic transform of equally spaced
# values between 0 and 5.
# For each value of rho we save the degrees of freedom, the gcv value,
# the stop sum of squares for each equation, the mean squared stops for
# the parameters, and the parameter values.
# set up a loop through a series of values of rho
# We know, because the signal is smooth and the data are rough, that the
# optimal value of rho will be rather close to one, here we set up a
# set of three rho values using the logistic transform of equally spaced
# values 2, 5, and 8, corresponding to rho values of 0.8808, 0.9933 and
# 0.9997.
# For each value of rho we save the degrees of freedom, the gcv value,
# the error sum of squares for each equation, the mean squared errors for
# the parameters, and the parameter values.
Gvec <- c(0:7)
nrho <- length(Gvec)
rhoMat <- matrix(exp(Gvec)/(1+exp(Gvec)),nrho,2)
dfesave <- matrix(0,nrho,1)
gcvsave <- matrix(0,nrho,1)
MSEsave <- matrix(0,nrho,2)
thesave <- matrix(0,nrho,nparam)
conv <- 1e-4 # convergence criterion for mean square error values
iterlim <- 20 # limit on number of iterations
dbglev <- 1 # displays one line of results per iteration
cruiseModelList.opt <- cruiseModelList # initialize the optimizing coefficient values
# loop through rho values, with a pause after each value
for (irho in 1:nrho) {
rhoVeci <- rhoMat[irho,]
print(paste(" ------------------ rhoVeci <- ", round(rhoVeci[1],4),
" ------------------"))
Data2LDOptList <- Data2LD.opt(cruiseDataList, cruiseBasisList, cruiseModelList.opt,
rhoVeci, conv, iterlim, dbglev)
theta.opti <- Data2LDOptList$theta # optimal parameter values cruiseModelList.opt <- modelVec2List(cruiseModelList, theta.opti) # evaluate fit at optimal values and store the results Data2LDList <- Data2LD(cruiseDataList, cruiseBasisList, cruiseModelList.opt, rhoVeci) thesave[irho,] <- theta.opti dfesave[irho] <- Data2LDList$df
gcvsave[irho] <- Data2LDList$gcv x1fd <- Data2LDList$XfdParList[[1]]$fd x1vec <- eval.fd(tobs, x1fd) msex1 <- mean((x1vec - speedResult$y)^2)
x2fd <- Data2LDList$XfdParList[[2]]$fd
x2vec <- eval.fd(tobs, x2fd)
msex2 <- mean((x2vec - controlResult$y)^2) MSEsave[irho,1] <- msex1 MSEsave[irho,2] <- msex2 } ind <- 1:nrho # print df, gcv and MSEs print(" rho df gcv RMSE:") print(cbind(round(rhoMat[ind,1],4), round(dfesave[ind],1), round(gcvsave[ind],1), round(sqrt(MSEsave[ind,1]),2))) # plot parameters as a function of \rho matplot(rhoMat[ind,1], thesave[ind,], type="b", xlab="rho", ylab="theta(rho)") rho.opt <- rhoMat[nrho,] theta.opt <- thesave[nrho,] # convert the optimal parameter values to optimal cruiseModelList cruiseModelList.opt <- modelVec2List(cruiseModelList, theta.opt) # evaluate the solution at the optimal solution DataLDList <- Data2LD(cruiseDataList, cruiseBasisList, cruiseModelList.opt, rho.opt) # display parameters with 95% confidence limits stddev.opt <- sqrt(diag(DataLDList$Var.theta))
theta.tru <- c(1, 1/4, 1, 0, 1)
print(" True Est. Std. Err. Low CI Upr CI:")
for (i in 1:nparam) {
print(round(c(theta.tru[i],
theta.opt[i],
stddev.opt[i],
theta.opt[i]-2*stddev.opt[i],
theta.opt[i]+2*stddev.opt[i]), 4))
}
# Plot the estimated solutions, as estimated from the data, rather
# than from the initial value estimates in the above code
XfdParList <- Data2LDList$XfdParList Xfd1 <- XfdParList[[1]]$fd
Xfd2 <- XfdParList[[2]]$fd Xvec1 <- eval.fd(tfine, Xfd1) Xvec2 <- eval.fd(tfine, Xfd2) Uvec <- eval.fd(tfine, SetPtfd) par(mfrow=c(2,1)) cruiseDataList1 <- cruiseDataList[[1]] plot(tfine, Xvec1, type="l", xlim=c(0,80), ylim=c(0,100), ylab="Speed", main=paste("RMSE =",round(sqrt(MSEsave[3,1]),4))) lines(tfine, ytrue[,2], lty=2) points(cruiseDataList1$argvals, cruiseDataList1$y, pch="o") lines(tfine, Uvec, lty=2) cruiseDataList2 <- cruiseDataList[[2]] plot(tfine, Xvec2, type="l", xlim=c(0,80), ylim=c(0,400), xlab="Time (sec)", ylab="Control", main=paste("RMSE =",round(sqrt(MSEsave[3,2]),4))) lines(tfine, ytrue[,3], lty=2) points(cruiseDataList2$argvals, cruiseDataList2\$y, pch="o")
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Data2LD documentation built on Aug. 6, 2020, 1:08 a.m.
|
2020-11-24 01:20:26
|
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|
https://dsp.stackexchange.com/tags/signal-detection/new
|
# Tag Info
## New answers tagged signal-detection
0
The FFT-based cross-correlation is done as follows: TX = np.fft.fft(tx) RX = np.fft.fft(rx) CORR = np.multiply(TX, np.conjugate(RX)) corr = np.real(np.fft.ifft(corr)) Here is a NumPy script that tries a solution. It does the following: create a TX and delayed RX signal that similar to your waveform (a sine wave modulated by an envelope cosine with zeros at ...
1
I wanted to ask what are the techniques which can be used to extract this sub-band data? That's a description of a filter bank. Yes, the FFT can be used for such applications. You'll find that OFDM, which powers DVB-T, 4G/5G, WiFi, … (basically all high-speed wireless terrestrial links) does exactly that. You'll also find that if you find the inherent sinc-...
3
This will should work fine. For each realization $k$, we can write $$y_k[n] = x[n] + w_k[n] + q_k[n] = x[n] + v_k[n]$$ So basically define the "effective" noise $v[n]$ as the sum of the analog noise and the quantization noise $q[n]$. Averaging will converge towards the desired signal $x[n]$ as long as two conditions are met: $v[n]$ is uncorrelated ...
1
IF a 1 is always represented by the waveform in the figure, you can use that waveform as the matched filter response. However, it sounds like the waveform could be anything as long as there are some pulses in it. In that case, the matched filter should be a single pulse, and you'll have to count how many pulses were detected afterwards. The non-Gaussian ...
1
The answer generally is yes- if the SNR is large enough then we can accurately estimate EVM just based on the raw decisions in the known constellation after all typical offsets have been properly corrected for (DC offset, IQ offset, phase offset, frequency offset, timing offset). In fact often the EVM is a measure of how good our corrections are for each of ...
1
Does it calculate the EVM of each equalized received symbol with a closest ideal symbol on the constellation and then average it out to give final EVM value? You very rarely test EVM on a low-SNR signal that's not according to any standard, especially because if that signal doesn't adhere to any standard, there would be very little for your signal analyzer ...
Top 50 recent answers are included
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2021-03-08 03:13:20
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https://oa.journalfeeds.online/2022/05/12/climate-variability-and-trends-in-the-endorheic-lake-hayk-basin-implications-for-lake-hayk-water-level-changes-in-the-lake-basin-ethiopia-environmental-systems-research/
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# Climate variability and trends in the Endorheic Lake Hayk basin: implications for Lake Hayk water level changes in the lake basin, Ethiopia – Environmental Systems Research
May 12, 2022
### Description of the study area
Ethiopia is situated in Eastern Africa, between 3° and 15° latitude and 33° and 48° longitude (Horn of Africa). The Lake Hayk basin is a naturally closed (endorheic) drainage that belongs to one of Ethiopia’s most vulnerable zones to climate change and variability. Its areal extent is within 39.68°E to 39.81°E, 11.24°N to 11.39°N and its area coverage including the lake water surface area of 2156.76 ha is 8592.68 ha (Fig. 1).
The Lake Hayk basin is under a subhumid tropical climate with bimodal precipitation regimes (kiremt and belg). The lake basin received a mean annual precipitation of 1192.31 mm; the mean annual surface temperature was 17.58 °C.
### Data sources
The hydroclimate variability/trend in the Endorheic Lake Hayk basin was analyzed using mean monthly historical datasets of precipitation, mean temperature (Tmean) and Lake Hayk’s Water Level (LWL) from 1986 to 2015. Precipitation data from only one Hayk meteorological station (11.31°N, 39.68°E; 1984 m amsl) outside the study area (Fig. 1) has been used to observe the Lake Hayk water level response to climate change/variability. Evidently, the degree to which precipitation amounts vary across an area is an important characteristic of the climate of an area that affects hydrology of lakes. Keeping this in mind, we believe that the Hayk meteorological station is the sole relevant and appropriate station from which precipitation data is enough to evaluate the impacts of climate on Lake Hayk water levels. This is from two perspectives. The first is that, despite being outside the basin, it is very close to Lake Hayk (less than 6 km away), even closer than several points within the basin. Its proximity to Lake Hayk allows it to collect data that is nearly identical to what the lake and its surroundings receive. The other reason is that the lake basin is small (85.93 km2), resulting in a density of meteorological stations in the lake basin of 85.93 km2 per station, which adequately represents the Lake Hayk basin according to the World Meteorological Organization (WMO) recommendation of 300–1000 km2 per station in Temperate, Mediterranean and Tropical zones (Dingman 2002). In light of these considerations, the authors used only the Hayk meteorological station rather than interpolating data from other stations, which could compromise data quality. The Ethiopian National Meteorological Agency has provided us with data for the lake basin’s mean monthly precipitation (1986–2015) and temperature (1994–2015). In addition, due to a lack of station data, the reanalysis temperature products (RTPs) of the same station for the years 1986–2015 were retrieved from the climate explorer (https://climexp.knmi.nl/) portal.
Furthermore, the LWL data of Lake Hayk are measured using the water level measuring gauge situated on the southwest shore of Lake Hayk (Fig. 1). The lake average daily water level time series from 1986 to 2015 were provided by Ethiopia’s Ministry of Water Resources. However, the LWL time series was riddled with severely missed data. Daily data with more than 10% missing values must be excluded from the analysis (Seleshi and Zanke 2004). Full daily LWL data were available for the years 1999–2005 and 2011–2015. Therefore, we fused the water level observations from these periods with remote sensed water extents to bridge the gap between 2005 and 2011. Cloud free (clouds cover ≤ 10%) Landsat 5 Thematic Mapper (TM) images for 2009–2011 and Landsat 7 Enhanced Thematic Mapper Plus (ETM +) images for 2005 and 2008 years were retrieved from the Earth Explorer (http://earthexplorer.usgs.gov/) archiving system. To ensure greater accuracy of interpretation, all Landsat images were downloaded for months of the dry (bega) season of the year.
### Data analysis
This study combined hydroclimate data from gauging station, gridded data (reanalysis) and remotely sensed satellite data to analyze climate variability/change and its implications on changes in the water level of endorheic Hayk Lake at the local level, using statistical approaches with the integration of remote sensing and geographic information system. The general methodology of the study is depicted schematically in Fig. 2.
### Evaluating the reanalysis temperature data
Reanalysis products are thought to be useful in situations when meteorological stations are insufficient and unevenly dispersed, as well as in cases where missing records and short period observations exist (Dee et al. 2011). Climate reanalyses such as the European Center for Medium-Range Weather Forecasts (ECMWF) ReAnalysis 5th generation (ERA5) (Hersbach et al. 2020), the Modern-Era Retrospective analysis for Research and Applications, version 2 (MERRA-2) (Reinecker et al. 2011) and the National Centers for Environmental Prediction/National Center for Atmospheric Research (NCEP/NCAR) (Kalnay et al. 1996) are currently in use. Due to the scarcity of historic station temperature records in the Endorheic Lake Hayk basin, we relied on reanalysis products (historic estimates produced by combining a numerical weather prediction model with observational data from satellites and ground observations) as the best alternative solutions, but their performance evaluation should no longer be overlooked. Therefore, the ERA5, MERRA-2 and NCEP/NCAR reanalysis temperature products (RTPs) in the Lake Hayk basin were quantitatively evaluated against ground station temperature data for the 1994–2015 time series on annual and seasonal scales using coefficient of determination (R2), root mean square error (RMSE) and relative bias (Alemseged and Tom 2015; Nkiaka et al. 2017).
$$R^{2} , = ,,,left[ {{{left( {sumlimits_{t = 1}^{n} {left( {T_{r} – overline{{T_{r} }} } right),,left( {T_{s} – overline{{T_{s} }} } right)} } right)} mathord{left/ {vphantom {{left( {sumlimits_{t = 1}^{n} {left( {T_{r} – overline{{T_{r} }} } right),,left( {T_{s} – overline{{T_{s} }} } right)} } right)} {left( {sqrt {sumlimits_{t = 1}^{n} {left( {T_{r} – overline{{T_{r} }} } right),,sumlimits_{t = 1}^{n} {left( {T_{s} – overline{{T_{s} }} } right)} } } } right)}}} right. kern-nulldelimiterspace} {left( {sqrt {sumlimits_{t = 1}^{n} {left( {T_{r} – overline{{T_{r} }} } right),,sumlimits_{t = 1}^{n} {left( {T_{s} – overline{{T_{s} }} } right)} } } } right)}}} right]^{2}$$
(1)
$$RMSE,, = ,,,sqrt {{{sumlimits_{t = 1}^{n} {left( {T_{r} – T_{s} } right)^{2} } } mathord{left/ {vphantom {{sumlimits_{t = 1}^{n} {left( {T_{r} – T_{s} } right)^{2} } } n}} right. kern-nulldelimiterspace} n}}$$
(2)
$$Bias,, = ,,,left( {{{sumlimits_{t = 1}^{n} {left( {T_{r} – T_{s} } right)} } mathord{left/ {vphantom {{sumlimits_{t = 1}^{n} {left( {T_{r} – T_{s} } right)} } {sumlimits_{t = 1}^{n} {T_{s} } }}} right. kern-nulldelimiterspace} {sumlimits_{t = 1}^{n} {T_{s} } }}} right),,times,100%$$
(3)
where Tr and Ts denote reanalysis and ground station temperature records respectively and n is the length of data. R2 varies within 0 ≤ R2 ≤ 1; R2 = 0 reveals no correlation and R2 = 1 indicates perfect correlation between the reanalysis product and station temperature record. Bias detects a systematic error in temperature values. Zero bias indicates absence of systematic error, whereas negative/positive biases reveal respectively underestimation and overestimation of values (Alemseged and Tom 2015). The RMSE measures residual dispersion (estimation errors) around the best fitting line. RMSE near zero would be a better fit to the data.
### Variability and trends analysis of hydroclimate time series
Various statistical approaches were used to examine the variability/trend in the hydroclimate time series of the Endorheic Lake Hayk basin from 1986 to 2015. The coefficient of variability (CV), the standardized rainfall anomaly (SRA) and the precipitation concentration index (PCI) were employed to study variability of the data. The Modified Mann Kendall (MK) trend test method and the Sen Slope estimator were applied to analyze the significance and magnitude of trend respectively using XLSTAT software. The CV value represents the level of variability in the dataset and is defined as the standard deviation (SD) to mean value (μ) ratio (Hare 2003).
$$CV, = ,,,left( {{{SD} mathord{left/ {vphantom {{SD} mu }} right. kern-nulldelimiterspace} mu }} right),times100$$
(4)
Hare (2003) characterizes variability as being less for CV values less than 20, moderate for CV values between 20 and 30 and high for CV greater than 30. PCI examines the heterogeneity of mean monthly precipitation data. For Pi is the ith month precipitation magnitude, Oliver (1980) defines PCI as follows:
$$PCI,, = ,,,,,left[ {{{left( {sumlimits_{i, = 1}^{12} {P_{i}^{2} } } right)} mathord{left/ {vphantom {{left( {sumlimits_{i, = 1}^{12} {P_{i}^{2} } } right)} {left( {sumlimits_{i, = 1}^{12} {P_{i} } } right)^{2} }}} right. kern-nulldelimiterspace} {left( {sumlimits_{i, = 1}^{12} {P_{i} } } right)^{2} }}} right],,times,100$$
(5)
Precipitation concentration can be identified as low concentration (uniform distribution of precipitation) for PCI values lower than 10, high for values from 11 to 20 and very high for values above 21 (Oliver 1980). SRA offers insights on the occurrence and severity of drought periods. Standardized rainfall and temperature anomalies have no units. They are dimensionless. To get more information about the magnitude of the anomalies, standardized anomalies are calculated by dividing anomalies (deviations of mean value from each observation) by the standard deviation to remove influences of dispersion. Therefore, for Pt is annual precipitation at a year of interest t and Pm is the mean annual precipitation value during the study period, SRA can be estimated according to Agnew and Chappell (1999):
$$SRA, = ,left( {P_{t} – P_{m} } right),/,SD,,,,$$
(6)
Then, severity of drought can be categorized as extreme drought (SRA < − 1.65), severe drought (− 1.28 > SRA > − 1.65), moderate drought (− 0.84 > SRA > − 1.28) and no drought (SRA > − 0.84) (Agnew and Chappell 1999).
The modified Mann Kendall (MK) trend test was used to examine the monotonic trends of hydro climatic time series in the endorheic Lake Hayk basin from 1986 to 2015 at a significance level of 5% on a monthly, annual and seasonal basis. It was chosen because it is a rank-based (i.e., less affected by low-quality hydro climatic data-data with missing values and/or outliers) nonparametric (i.e., less sensitive to skewed datasets-applies for all distributions) method (Hirsch and Slack 1984). The MK tests the null hypothesis (H0) assuming no trend against the alternative hypothesis of monotonic trend (Ha) using either the S statistics (n < 10) or the standardized normal Z statistics (n ≥ 10) (Hirsch and Slack 1984; Yue et al. 2002). The MK test S statistic is calculated using the following equations (Eqs. 7 and 8) as:
$$S, = ,sumlimits_{i = 1}^{n = 1} {sumlimits_{j = i + 1}^{n} {{text{sgn}} left( {x_{j} – x_{i} } right)} } ,,,,,,,$$
(7)
$$,,{text{sgn}} left( {x_{j} – x_{i} } right),, = ,,left{ begin{gathered} + 1,,,,if,,theta ,,, > ,0 hfill \ 0,,,,,,,if,theta , = ,0 hfill \ – 1,,,,,if,theta , < ,0 hfill \ end{gathered} right.,,,,,,,,$$
(8)
where n is the data size and xi and xj are the data values at times i and j respectively, i = 1, 2,…, n−1 and j = i + 1, i + 2…, n. Every value in the chronologically ordered time series is compared to every value preceding it, yielding a total of n (n – 1)/2 pairs of data. The total of all rises and falls result in the ultimate value of S (Yue et al. 2002). S values can be positive to show rising trends or negative to indicate falling trends.
When n ≥ 10, the S statistic is assumed to have a normal distribution, with the mean becoming zero and the variance computed using the following equation (Eq. 9) (Kendall 1975):
$$Vleft( S right), = ,frac{1}{18}left[ {nleft( {n – 1} right),left( {2n + 5} right), – ,sumlimits_{t = 1}^{m} {t_{i} left( {t_{i} – 1} right),left( {2t_{i} – 5} right)} } right],,$$
(9)
where V (S) is the variance of S statistics, m denotes the size of tied groups (groups with similar values) and ti represents the size of data points in the ith tied group. Then, the Z test statistics can be calculated from the known values of S and V(S) using the following equation (Eq. 10):
$$Z, = ,,left{ begin{gathered} {{left( {s – 1} right)} mathord{left/ {vphantom {{left( {s – 1} right)} {sqrt {v,,left( s right)} ,,}}} right. kern-nulldelimiterspace} {sqrt {v,,left( s right)} ,,}},,,,,,if,,S > 0 hfill \ 0,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,if,,,S = 0 hfill \ {{left( {s + 1} right)} mathord{left/ {vphantom {{left( {s + 1} right)} {sqrt {v,,left( s right)} }}} right. kern-nulldelimiterspace} {sqrt {v,,left( s right)} }},,,,,,,,,if,,S < 0 hfill \ end{gathered} right.,,,,,,,,,,,,$$
(10)
The resulting Z values indicate the direction of a trend (values can be positive to show rising trends or negative to indicate falling trends). Furthermore, the Z statistic is used to measure significance of a trend. When testing for a trend (2 tailed) at significance α, H0 is rejected if (left| Z right|,) equals or exceeds its critical value (,left( {left| Z right|,,, ge ,,Z_{alpha /2} } right)). For instance, if the 5% significance level is used, H0 is rejected when (,left( {left| Z right|,,, ge ,,1.96} right)) or P ≤ 0.05, indicating that a trend exists (A time series has a trend when it is significantly correlated with time). The letter P symbolizes the probability of risk to reject/accept H0 while it is true.
Prior to trend testing, it is critical in time series analysis to examine autocorrelation or serial correlation, which is frequently overlooked in many trend detection studies. To account for the effect of autocorrelation, Hamed and Rao (1998) propose a modified Mann–Kendall test rather than the original MK test, which should be used only on datasets with no seasonality or significant autocorrelations. This is because the presence of significant autocorrelation in a dataset can alter the variance of the original MK test (the existence of positive autocorrelation will lower the actual value of V (S) and vice versa). Hence, when data exhibit autocorrelation, the modified MK test calculates the modified variance using the following equations (Eqs. 11 and 12):
$$V^{*} left( S right),,, = ,,,frac{1}{18}left[ {n,,left( {n – 1} right),,left( {2n + 5} right)} right]frac{n}{ns*},,,,,,,,,,$$
(11)
$$frac{n}{ns*},, = ,,,1 + ,frac{2}{{n,,left( {n – 1} right),left( {n – 2} right)}},,,sumlimits_{i = 1}^{p} {left( {n – i} right),} ,left( {n – i – 1} right),,left( {n – i – 2} right),,,p_{s} ,left( i right),,$$
(12)
1998) modified MK trend test facility (at a significance level of 10%), to account for the autocorrelation effect.
Sen’s slope estimator computes the linear annual rate and direction of change (Sen 1968). It is a nonparametric approach for dealing with skewed datasets and outlier effects. The linear model f (t) is defined by the equations (Eqs. 13 and 14) (Sen 1968) as follows:
$$fleft( t right),, = ,,Qt, + ,beta ,,,,,,,,,,,$$
(13)
$$Q,,, = ,,Median,,,,,,{{left( {X_{i} , – ,X_{j} } right)} mathord{left/ {vphantom {{left( {X_{i} , – ,X_{j} } right)} {left( {i – j} right)}}} right. kern-nulldelimiterspace} {left( {i – j} right)}},,,,,,forall j,, < ,,i,,,,,$$
(14)
### Lake Hayk water level response to climate change/variability
Due to the endorheic (closed) nature of the Lake Hayk basin, the main underlying hydrological processes are surface runoff and evapotranspiration, with precipitation and temperature being the most prominent climatic factors. Under such conditions, water level is the primary response variable that serves as an indicator to better reflect the climate change/variability effects on lake storage. In addition, it can easily be measured at observation stations and the changes in lake water levels can be monitored easily, accurately and continuously (Tan et al. 2017). However, in situations where lake level data is patchy, as it is in Lake Hayk, remotely sensed water extents derived from Landsat images would allow us to bridge the data gap of the water level time series (McFeeters 1996; Xu 2006). This is achieved by developing spectral water indices to extract water bodies from remotely sensed Landsat images, which is typically accomplished by computing the normalized difference between two image bands and then applying an appropriate threshold to segment the results into two categories (water and nonwater features). The Modified Normalized Difference Water Index (MNDWI) can efficiently extract lake waters from Landsat images by easily suppressing signals from various environmental noises (such as vegetation, built-up areas and shadow noises) compared to its predecessor, the Normalized Difference Water Index (NDWI) using Shortwave Infrared (SWIR) rather than Near Infrared (NIR) used in the NDWI (Xu 2006). The formula used for the MNDWI calculation is:
$$MNDWI = ,,{{left( {Green, – ,SWIR} right)} mathord{left/ {vphantom {{left( {Green, – ,SWIR} right)} {left( {Green,, + ,SWIR} right)}}} right. kern-nulldelimiterspace} {left( {Green,, + ,SWIR} right)}}$$
(15)
For Landsat 5 Thematic Mapper (TM) and Landsat 7 Enhanced Thematic Mapper plus (ETM +), MNDWI becomes:
$$MNDWI = ,,{{left( {band2 – ,band5} right)} mathord{left/ {vphantom {{left( {band2 – ,band5} right)} {left( {band2, + ,band5} right)}}} right. kern-nulldelimiterspace} {left( {band2, + ,band5} right)}}$$
(16)
MNDWI values for water and nonwater features are determined by the reflectance ability of the features to band 2 and band 5 of Landsat 5 TM and Landsat 7 ETM + satellite images (Xu 2006). Water features have greater reflectance in band 2 than in band 5, resulting in positive MNDWI values, whereas nonwater features have negative MNDWI values (i.e., index values range from − 1 to + 1). A standard threshold value of zero is used to signify that a feature is water if MNDWI > 0 and nonwater if MNDWI ≤ 0 (Xu 2006). When the standard threshold value of zero is used, the MNDWI can accurately determine the spatial position of shorelines at the land–water boundary and successfully extract them from the multi-temporal Landsat TM and ETM + images. Thus, MNDWI values are still positive for the shallowest parts of water bodies and waterlogged areas (areas inundated with water). A change in MNDWI values at a specific time occurs when a sensor detects a spatio-temporal change in nonwater features (change in land use and land cover types within the lake basin) and/or change in depth and quality of water features.
Therefore, to bridge the time gap between 2005 and 2011, water level observations from 1999 to 2005 and 2011 to 2015 were fused with the MNDWI index extracted remotely sensed water areas from Landsat images in the ArcGIS 10.1 software environment. The analysis was conducted using cloud free (clouds cover ≤ 10%) Landsat 7 Enhanced Thematic Mapper plus (ETM +) satellite images of 2005 and 2008 and Landsat 5 Thematic Mapper (TM) satellite images from 2009 to 2011 obtained from the http://earthexplorer.usgs.gov/ portal. The obtained Landsat images (Level 1 Terrain Corrected (L1T) product) were pre-geo-referenced to the Universal Transverse Mercator (UTM) zone 37°N projection system using a World Geodetic System 84 (WGS84) datum. A single Landsat image is sufficient to encompass the entire Lake Hayk basin, which has an area of 8592.68 ha. Landsat 5 TM and Landsat 7 ETM + images specifications are shown in Table 1.
The MNDWI index was validated by correlating remotely sensed water areas to water level observations both obtained on the same date in each year by using the Pearson’s (parametric) and Kendall’s tau (nonparametric) correlations at a 0.01 significance level with the Statistical Package for the Social Sciences (SPSS) version 20 software. For n sample sizes of X and Y variables, the Pearson’s coefficient (r) can be computed as:
$$r,, = ,,{{sumlimits_{i = 1}^{n} {left[ {left( {X_{i} , – ,overline{X} } right),left( {Y_{i} , – ,overline{Y} } right)} right],} } mathord{left/ {vphantom {{sumlimits_{i = 1}^{n} {left[ {left( {X_{i} , – ,overline{X} } right),left( {Y_{i} , – ,overline{Y} } right)} right],} } {,left( {sqrt {sumlimits_{i = 1}^{n} {left( {X_{i} , – ,overline{X} } right)^{2} ,sumlimits_{i = 1}^{n} {left( {Y_{i} , – ,overline{Y} } right)^{2} } } } } right)}}} right. kern-nulldelimiterspace} {,left( {sqrt {sumlimits_{i = 1}^{n} {left( {X_{i} , – ,overline{X} } right)^{2} ,sumlimits_{i = 1}^{n} {left( {Y_{i} , – ,overline{Y} } right)^{2} } } } } right)}}$$
(17)
This can be confirmed by the nonparametric Kendall’s correlation. The Kendall’s correlation helps to minimize the effects of extreme values and/or the effects of violations of the normality and linearity assumptions (Kendall 1938). The Kendall’s tau (τ) is calculated based on signs as:
$$tau = ,,frac{2}{{n,left( {n – 1} right)}},sum {_{i,, < ,,j} ,sign,left[ {left( {x_{i} , – ,x_{j} } right),,left( {y_{i} , – ,y_{j} } right)} right]}$$
(18)
In both cases, the correlation coefficients are within − 1 and + 1. Correlation values close to ± 1 indicate strong relationships. For result interpretation, the hypothesis for a 2 tailed test of the correlation at a given significant level is defined as: H0: r, tau = 0 versus Ha: r, tau ≠ 0.
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2022-05-16 04:25:18
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https://www.rdocumentation.org/packages/grid/versions/3.5.3/topics/grid.newpage
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# grid.newpage
0th
Percentile
##### Move to a New Page on a Grid Device
This function erases the current device or moves to a new page.
Keywords
dplot
##### Usage
grid.newpage(recording = TRUE)
##### Arguments
recording
A logical value to indicate whether the new-page operation should be saved onto the Grid display list.
##### Details
The new page is painted with the fill colour (gpar("fill")), which is often transparent. For devices with a canvas colour (the on-screen devices X11, windows and quartz), the page is first painted with the canvas colour and then the background colour.
There are two hooks called "before.grid.newpage" and "grid.newpage" (see setHook). The latter is used in the testing code to annotate the new page. The hook function(s) are called with no argument. (If the value is a character string, get is called on it from within the grid namespace.)
None.
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2019-10-24 04:37:18
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https://homework.cpm.org/category/CON_FOUND/textbook/mc2/chapter/1/lesson/1.2.3/problem/1-77
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### Home > MC2 > Chapter 1 > Lesson 1.2.3 > Problem1-77
1-77.
Remember, probability is the number of successful outcomes out of the total number of possible outcomes.
How many nickels are in the bag?
How many total coins are in the bag?
$\frac{4}{16}= \frac{1}{4}$
What is the new number of nickels?
What is the new total number of coins?
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2020-08-09 23:51:00
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https://codeforces.com/blog/entry/93538
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### Mo2men's blog
By Mo2men, history, 18 months ago,
Hello, Codeforces!
I'm glad to invite you to Codeforces Round #737 (Div. 2) , which will be held on Monday, August 9, 2021 at 16:35 UTC+2.
This round rated for the participants with rating lower than 2100.
You will be given 5 problems and 2 hours to solve them.all problems were prepared by me and AhmedEzzatG.
One of the problems will be interactive. So, it is recommended to read the guide on interactive problems before the round.
I would like to thank -
1. Aleks5d, for awesome coordination of our round and suggested one of the problems.
2. Ahmed-Yasser for help us to prepare the problems. compiler_101 ,El3ageed_Abu_Shehab ,DeadlyPillow , and Omar_Elawady for discussing and testing the problems.
3. MikeMirzayanov, for the amazing Codeforces and Polygon platforms.
The statements are short and I have tried to make the pretests strong. I encourage you to read all the problems.
This is our first official round on Codeforces. We are sincerely looking forward to your participation. We hope everyone will enjoy it.
Good luck and see you in the standings!
UPD1: I want to thank Aleks5d for translating statements into Russian and mouse_wireless author of one of tasks.
UPD2: Scoring distribution: 500 — 1000 — 1750 — 2500 — 3000
UPD3: Editorial
UPD4:
## Winners
Congratulations to all our winners in the round!
#### Div2:
• +292
| Write comment?
» 18 months ago, # | +41 Your profile picture is really good and meaningful.
• » » 18 months ago, # ^ | -45 what about problems' rating? If not a secret
• » » 18 months ago, # ^ | -22 Friend++ :)
» 18 months ago, # | +125 Monogon was the VIP tester of 2 consecutive rounds!Orz!
• » » 18 months ago, # ^ | +407 I agree. Orz me.
• » » » 18 months ago, # ^ | -51 Orz Monogon
• » » » 18 months ago, # ^ | -14 U don't need to say that ,people by default orz you
• » » » » 18 months ago, # ^ | 0 he has to, for taking over codeforces(and the contribution).
• » » » » 18 months ago, # ^ | -7 So much downvotes,that's why i love cf but not cfiians
» 18 months ago, # | ← Rev. 2 → +15 As a trainee for the author I am so excited and I am sure that the statements are short XD
» 18 months ago, # | +3 As a trainee for the author I am so excited and I am sure that the contest will be amazing as the author is amazing XD
» 18 months ago, # | +6 Amazing!! It will be such a great contest!
» 18 months ago, # | +5 I'm excited! We are in a drought of contests, and I'm itching to compete more.
» 18 months ago, # | +27 As a tester, problems are great for the participants who love the short statements and like to gain high ratings!Good luck to everyone :)
» 18 months ago, # | +13 Finally a SA3EDY Round #1، I hope it will not be the last one. keep going our heroes♡♡ ♡‿♡.
» 18 months ago, # | +17 Egypt Foooooo2 :"D
• » » 18 months ago, # ^ | +5 T3mya UP
» 18 months ago, # | +64 As a very cool tester, I would like to say that the problems are nice and the pretests are very strong. I think you'll enjoy doing this round.
• » » 18 months ago, # ^ | ← Rev. 3 → +52 So, shamelessly give me contribution.
• » » » 18 months ago, # ^ | +35 commenting again instead of editing, double contribution, smart move xD
» 18 months ago, # | +113 with the power vested upon me as a tester, I hereby declare thy contest interesting.
» 18 months ago, # | +195 Though I'm not a VIP tester, I tried hard testing as much as I can.
• » » 18 months ago, # ^ | +57 Yes, I can assure that you did a great job.
• » » 18 months ago, # ^ | +17 What's the difference between tester and VIP tester?
• » » » 18 months ago, # ^ | +85 VIP Tester is supposed to get more contribution than normal tester.
• » » » » 18 months ago, # ^ | +1 No.. VIP tester is the one who follows 1-gon and of course, was a tester. Normal testers are those who still don't follow 1-gon.
» 18 months ago, # | +32 As a contest tester for the first time :), the contest is great and joyful.I hope you will enjoy it ;)
» 18 months ago, # | 0 Someone explain me why the time of this post is showing like 2 days ago ,although it is just posted like 1 hr before i guess
» 18 months ago, # | +1 It's going to be a great contest :)
» 18 months ago, # | ← Rev. 4 → +12 As a trainee of the author i am sure that the problems are epic, Good luckGive me and my coach contribution :)
» 18 months ago, # | ← Rev. 2 → +67 I decided to write the ultimate "as a tester comment" since this is the first and hopefully not the last time to be a tester of an official round (except of course if you count #716 #717).1) as a tester add me to your friend list2) as a non-VIP tester I hope I can be a VIP tester one day3) as a tester I assure you that you should give my friend mo2men contribution5) as a tester I advise you to read all the problems6) as a tester I think you should prepare everything you'd need throughout the round because you won't be able to move or blink during the round7) as a tester this round is pure awesomeness (kung fu panda 2008 reference)8) as a tester of this round I would like to test upcoming rounds... maybe the next is yours? who knows9) as I am 50% of the specialist testers population in this round help me by upvoting this comment10) as a tester I recommend everyone to eat jilaty (ice cream in Arabic)Note: the profile picture was taken by my friend and the author Mo2men in the Arab collegiate programming contest 2020 (ACPC)
• » » 18 months ago, # ^ | +8 I think you forget to say as a tester (◠‿◕)
• » » 18 months ago, # ^ | ← Rev. 2 → +2 Blobo2_Blobo2 u seem like :
• » » 18 months ago, # ^ | +6 ahm 7aga el jilaty ana megahez 3lba fe elfrazer XD
• » » 18 months ago, # ^ | +3 well deserved upvote xD
• » » 18 months ago, # ^ | +9 bruh when you will be a tester again what you will write those are all the "as a tester" comments human invented
» 18 months ago, # | +1 cant be proud more my friends keep going [user:IsaacMoris][user:Uzumaki_Narutoo][user:Blobo2_Blobo2] i sure the contest will be so beautiful ♥♥
• » » 18 months ago, # ^ | 0 ;)
» 18 months ago, # | +99 As a tester give me contributions.
• » » 18 months ago, # ^ | +12
• » » 18 months ago, # ^ | +8 i know this tester and he is the best tester and setter i have met keep going my coash
» 18 months ago, # | +1 tdpencil has hacked few of my solutions in the past, so for me he is orZ. yeah!! this orZ has curves.
» 18 months ago, # | +25
» 18 months ago, # | +15 تحيا مصر ❤️
» 18 months ago, # | +26 Wow you really host a div-2 contest when you are just specialist. ORZ
• » » 18 months ago, # ^ | ← Rev. 2 → +13 Rating is just a number.Also, this man was a problem setter in the Arab collegiate programming contest (ACPC) 2020 and a coach for many CP students, these students are now experts and candidate masters.
• » » » 18 months ago, # ^ | +9 True Rating is just a number .P.S. if you have good contacts xdddd
• » » » 18 months ago, # ^ | +3 Yes, rating is just a number. You're right.
• » » » » 18 months ago, # ^ | +9 Bro ,but you have announced to become CM in 2 months ,so for you it's pride now ,best of luck
• » » » » » 18 months ago, # ^ | 0 There are still 8 weeks to go bro.
• » » » » » » 18 months ago, # ^ | +16 There are just 8 weeks to go bro :)
• » » » » » » 18 months ago, # ^ | 0 I want to get CM too in 2 months, lets race to CM :O
• » » » » » » » 18 months ago, # ^ | +1 I am in too
» 18 months ago, # | 0 I just hope I can become a "Pupil" through this competition.
• » » 18 months ago, # ^ | +1 Bro, you are solving 1600 and some 2000+ level problems! you deserve more than that for sure
• » » » 18 months ago, # ^ | 0 But I can only solve 2 or 3 problems in Div.2.I am too poor and I am only Grade 6 as a Chinese.
• » » » 18 months ago, # ^ | 0 Ahh......I'm too low to solve div.2 .I think I need to do more div.3.(Rank 8000+)
• » » » » 18 months ago, # ^ | +1 You are in grade 6! I am in my third year of CS degree and still able to solve 3 problems div2... so you are much better off cuz you have so much time at hand! Relax!
• » » » » » 18 months ago, # ^ | +3 Wait,what does "CS" mean?University?Or middle school?
• » » » » » » 18 months ago, # ^ | 0 computer science...
• » » » » » » » 18 months ago, # ^ | 0 Thank you!But I often call it OI.
• » » » » » » » » 18 months ago, # ^ | +3 No, oi is the short name for "Olympiad in Informatics". CS is computer science which is the name of a major in the university.小盆友加油 (ง •_•)ง
• » » » » » » » » » 18 months ago, # ^ | 0 You are also a Chinese?So how many problems should I solve in a normal Codeforces Div.2 if I what to get tg1=?Or I need to solve Div.1?当然，我知道tg1=肯定得等初二左右了...Of course,I know that I should wait for 2 years if I want to get tg1=.
• » » » » » » » » » 18 months ago, # ^ | 0 I start very late you see.. I am still a newbie now.. When I was at your age, I was playing mud back in the school yard and knew nothing about OI. So your age and your time are the most valuable thing. Try to solve more problems. Hope you will be the next WJMZBMR :)
• » » » » » » » » » 18 months ago, # ^ | 0 First,you were a specialist. Second,my mother says that if I can not get >=300 marks in CSP-J and tg2=,she will let me to be a MOer.But the thing I like is program ,not that complex and annoying math.(Maybe I am extreme,but I hope you can understand what I mean.)
• » » » » » » » » » 18 months ago, # ^ | 0 UPD:My highest prize is CSP-J 2= 135 marks,so it is a hard task for me.That's why I want to enhance my strength by CF and Luogu as fast as I can.
• » » » » » » » » » 18 months ago, # ^ | 0 I am going to be Grade 9 now,and I got tg1= last year.You still have long time to study,so I don't think you need to worry about it.I was only a Specialist when I got tg1= :) To the honest,getting a tg2= is not difficult,you can even use brute force to get tg2= easily :)
» 18 months ago, # | +8 Hope to become "Pupil" through this contest.
» 18 months ago, # | +3 Good luck to everyone!
» 18 months ago, # | 0 As you get ready for the contest, I wish you all the best of luck
» 18 months ago, # | 0 Most valuable advice ❤️❤️
• » » 18 months ago, # ^ | ← Rev. 3 → 0 As a tester ,please orz this guy with a lot of upvotes ,otherwise he is gonna comment frequently for getting upvotes.P.S. just kidding but upvote him.i was the 1st to do so :)
» 18 months ago, # | +20 feels like ages since the last round it's like all problem setters are on a vacation or something
» 18 months ago, # | +5 We are so proud of you all , keep going :")
» 18 months ago, # | 0 Auto comment: topic has been updated by Mo2men (previous revision, new revision, compare).
» 18 months ago, # | 0 I will reach pupil in this contest.Thank You
» 18 months ago, # | 0 I am so excited to participate in this round, hope everyone will gain rating
» 18 months ago, # | 0 From the score distribution, it looks like it's going to be speedforces for A, B, C.
• » » 18 months ago, # ^ | ← Rev. 2 → 0 Lol maybe for high rated people like you it may seem as speedforces but i don't think that guys like me feel so :).best of luck
• » » 18 months ago, # ^ | +2 I sense more like AB speeforces
• » » » 18 months ago, # ^ | ← Rev. 2 → 0 I missed the part where that's my problem.:)P.S. just kidding bro .i do feel the same bruder ,best of luck
• » » » 18 months ago, # ^ | 0 Author revised the score distribution. Previously it said C was 1250 points, that's why I included C as well. Now it looks better.
» 18 months ago, # | +9 Auto comment: topic has been updated by Mo2men (previous revision, new revision, compare).
» 18 months ago, # | -43 whenever specialist expert makes a contest they always make it hard. If u are actively giving contests u know what I mean. They think it's cooler to make a harder contest. These pathetic nerds cant solve the all question themselves.
• » » 18 months ago, # ^ | 0 He is true. This round is awful. Tests in E are so strong that every wrong solution can pass it ：)Do they have better ideas？They got a interesting idea（problem E) and some "strong" tests.Bad ABCD，real “speedforces”. The worst round I've ever seen.Finally，thanks the writers for this "wonderful" round ：）：）：）
» 18 months ago, # | +1 Short problem statements, thats what we wanted
» 18 months ago, # | ← Rev. 2 → 0 I always used to think that to increase rating one have to solve more and more problems. Mo2men has solved more than 2000 problems then why his rating is not increasing ( ･᷄ ︵･᷅ )
• » » 18 months ago, # ^ | +3 Maybe mo2men cares about solving problems more than having good ratings! ¯\_(ツ)_/¯
• » » » 18 months ago, # ^ | 0 But still +delta motivates a lot
» 18 months ago, # | -59 i hope i can get top 10 in this contest
• » » 18 months ago, # ^ | +39 why?
• » » » 18 months ago, # ^ | 0 This is what you call stalking*100
• » » 18 months ago, # ^ | +110 i hope i can get top 10 in this contest,too.
» 18 months ago, # | 0
» 18 months ago, # | 0 Oh interactive, is this binary search ? :V
• » » 18 months ago, # ^ | -6 May or may not be, interactive problems need not be always binary search always :)
• » » » 18 months ago, # ^ | +10 I know, this is a joke
» 18 months ago, # | 0 Looks like today their ll be more of a number theory problems.
• » » 18 months ago, # ^ | 0 How do you know?
• » » » 18 months ago, # ^ | -6 just guessing bro .
» 18 months ago, # | -11 As a participant I wish to gain some precious contribution through this comment.
• » » 18 months ago, # ^ | 0 Please take good care of this precious contribution that you've received.
• » » 18 months ago, # ^ | 0 nope :D
» 18 months ago, # | -20 will be unbalanced contest , I can bet
• » » 18 months ago, # ^ | +26 see I already said, unbalanced round, still got downvotes
• » » 18 months ago, # ^ | 0 Your prediction really comes true :(
» 18 months ago, # | +3 hope that a and b will not be interactive
» 18 months ago, # | +45 Last few days I faced a lot in my life, Can't reveal what exactly happent. But you can assume that the sadness is as same as when you lose your father or mother or see them cry.Couldn't give last 3 contests with proper emotional and mental strength.But skipping a contest has never been an option for me.It's the only thing I have in my life to live for.I will try my best today.Good luck to everyone too
• » » 18 months ago, # ^ | 0 Good luck bro ull get to specialist today :)
• » » » 18 months ago, # ^ | 0 I hope so.I will try my best.Thanks for your kind words
• » » 18 months ago, # ^ | +5 Come on bro! It's only a part of the life. I hope you can come out of the gloom and step into the light as soon as possible.
• » » » 18 months ago, # ^ | ← Rev. 2 → +2 Yes!!While there is life, there is hope...If the great Stephen Hawkins figured out life even after getting paralysed,Hopefully I can too.
» 18 months ago, # | -16 Is it just me? getting a TLE on tc2 for problem A
• » » 18 months ago, # ^ | -17 Yeah, I got it as well. It's a stupid question. Finally figured out the problem.
• » » » 18 months ago, # ^ | -19 My O(n) got tle! Anyway good luck to you for rest of the contest.
• » » » » 18 months ago, # ^ | 0 My O(n) got Ac.But I'm afraid of getting FST :(
» 18 months ago, # | +28 Remind me when I'm trying to take part in another weird round and get negative delta
» 18 months ago, # | 0 Formally, let your answer be a, and the jury's answer be b. Your answer is accepted if and only if |a−b|max(1,|b|)≤10−6. How to achieve this in Java ??
• » » 18 months ago, # ^ | +1 use double, It'll take care of it automatically (if your approach is correct).
» 18 months ago, # | ← Rev. 3 → +27 Why the gap between C and D was a lot ?
• » » 18 months ago, # ^ | ← Rev. 2 → +16 toxic:(check his original comment
• » » 18 months ago, # ^ | ← Rev. 2 → +7 How will you know the balance of the questions if you only solved A?Original comment : fuck you with your fucking unbalanced contest
• » » » 18 months ago, # ^ | 0 You can look at the gap of D and C in the standings after system tests :)
• » » » » 18 months ago, # ^ | +2 Why even care when you even can't solve B? Your toxicity is full of shit
» 18 months ago, # | -9 First experience of being able to do B but not A XD.
» 18 months ago, # | +64 giving combinatorics without some relatively big testcase — unethical:/
» 18 months ago, # | -126 ll n; cin >> n; ll k; cin >> k; vectorarr(n, 0); for (ll i = 0; i < n; i++) { cin >> arr[i]; } ll sz = 1; for (ll i = 1; i < n; i++) { if (arr[i - 1] > arr[i]) { sz++; } } if (sz <= k) { cout << "YES\n"; } else cout << "NO\n";why I am getting wa for this..its the correct soln
• » » 18 months ago, # ^ | +23 you are asking this during contest thats what you doing wroong
• » » » 18 months ago, # ^ | -58 give logic atleast
• » » » » 18 months ago, # ^ | +22 Dude, you don't ask for help during a contest!!
» 18 months ago, # | +1 I just erased my whiteboard 5th time full of test-cases for C and yet I am unable to decipher. I am almost sure it's gonna be a one-liner but I just can't figure it out :(
» 18 months ago, # | +5 I think the gap between the problems is too big :(
» 18 months ago, # | ← Rev. 2 → +45 plz answer me a question whether you like the weak samples just because you think they're cool???
• » » 18 months ago, # ^ | +6 My D is failing pretest 4 though :P
• » » » 18 months ago, # ^ | 0 Seeing so bad testcases for C, its impossible to know if the solution is even remotely correct. All my submissions passed the samples and failed pretests.
• » » » » 18 months ago, # ^ | 0 It's not tough to write a brute force solution for C to verify.
• » » » 18 months ago, # ^ | 0 Very relatable.
• » » » » 18 months ago, # ^ | 0 What did you do to fix it?
• » » » » » 18 months ago, # ^ | 0 I just added more parameters into my segmentree nodes and it worked. (Previously I used a set to maintain values.)
• » » » » » » 18 months ago, # ^ | 0 Wait, like I should compress values with some more like $l - 1$, $l$, $r$ and $r + 1$ instead of just $l$ and $r$? If this is the case I'll be sad for another few days :P
• » » » » » » » 18 months ago, # ^ | 0 No, I think I did the same thing as what you did, but instead of keeping the max values in the segment tree, I used minimum values, but I have no idea whether your approach is right or wrong, because I iterated through the rows from 1 to n, not n to 1.
» 18 months ago, # | +27 How the hell are 2500 people able to solve C
• » » 18 months ago, # ^ | 0 I'm also wondering this too... So overall I'm too weak...
• » » 18 months ago, # ^ | +4 Telegram I guess
• » » » 18 months ago, # ^ | 0 What do you mean?
• » » » » 18 months ago, # ^ | 0 Cheater , not all but some
• » » 18 months ago, # ^ | +1 My might fail in sys test, I checked the submissions of the people in my room and all of them have similar code but completely different logic than mine....Hope the pretests are strong
• » » 18 months ago, # ^ | ← Rev. 2 → 0 I think about this issue like this : Consider sequentially from the high binary bits to the low binary bits.The i-th binary bit can be "decided" or "undecided".Decided means that the i-th binary digit a1 & ... an is already greater than a1 xor ... an, and no subsequent comparison is required at this time.Undecided means that the value of a1 & ... an and a1 xor ... an can be compared after the i-th binary digit.Undecided includes two situations : - n numbers are all 1 in the i-th binary digit and n is an odd number. - At least one of the n numbers in the i-th binary digit is 0 and there is an even number of 0s.Undecided means that all n numbers on the i-th binary bit are all 1 and n is an even number.The final answer is composed like this ： Undecided Decided Random Undecided my code like this # include # define int long long using namespace std; const int mo=1e9+7; int pow(int x,int n){ int ans=1; while (n) { if (n&1) ans=ans*x%mo; x=x*x%mo; n>>=1; } return ans; } signed main() { int t; scanf("%lld",&t); while (t--) { int n,k; scanf("%lld%lld",&n,&k); int res=pow(2ll,n-1); if (n&1) res=res+1; else res=res-1; int ans=pow(res,k); if (!(n&1)) { for (int i=1;i<=k;i++) { ans=(ans+pow(res,i-1)*pow(pow(2ll,n),k-i)%mo)%mo; } } printf("%lld\n",ans); } return 0; }
» 18 months ago, # | 0 shitty problems, bad balance and there are only 5 problems AT ALL
» 18 months ago, # | +6 How to solve C?
» 18 months ago, # | +13 Am I the only one here who had 2 penalties because of ignoring the fact that $-10^9 \le A_i \le 10^9$
• » » 18 months ago, # ^ | +3 I got soo many wrong answers on B because a[i] can be 0. Why they didn't make a[i] permutation of length n? Anyway, I think that problems were ok but samples and these constraints were horrible.
• » » » 18 months ago, # ^ | ← Rev. 3 → +3 I had the exact same issue, I got first WA because $A_i$ can be 0 and 2nd WA because $A_i$ can be -1
» 18 months ago, # | +8 Weak sample for C :(
» 18 months ago, # | +10
» 18 months ago, # | +17 In div2 B i was rearranging the elements inside subarray for half an hour.Sad noises.
• » » 18 months ago, # ^ | +6 Same I did , for around 10 minutes.
» 18 months ago, # | 0 I liked B it wasn't clear why the simple solution is not working but C is so hard still a good problem though I will try to solve it later
• » » 18 months ago, # ^ | 0 Why did the simple solution not working in B?
• » » » 18 months ago, # ^ | +1 1 3 2 <- try this test case
• » » » 18 months ago, # ^ | ← Rev. 2 → +3 consider this case4 2 2 3 5 4simple solution output Yes but it's a No cause you need k=3 you can't put 4 in middle 2 3 5
• » » » » 18 months ago, # ^ | +3 I compressed elements to 0 to n-1 and checked if a[i] — a[i-1] == 1.
» 18 months ago, # | 0 Is there another way to solve problem D instead of using a dynamic segmentree, curious -.-.
• » » 18 months ago, # ^ | +3 As long as the queries are not online, you can (almost) always use coordinate compression and use a normal segment tree.
» 18 months ago, # | ← Rev. 2 → +1 UPD: my bug was that I had to declare the Segment Tree Array with 8n elements
» 18 months ago, # | ← Rev. 2 → +7 Contrary to popular belief, I liked problem C. A good question based on bit contribution and combinatorics.
• » » 18 months ago, # ^ | +5 Even though I couldn't solve it but it indeed feels like a good problem. I hope to learn something great from the problem
» 18 months ago, # | 0 So i had this idea and observation for problem C:For every number >= 1 times that it occures must be even. For example: 02211, 02222, 22110 etc. Total number of possibilities that pair of numbers can occur is n*(n-1)/2. There can be n/2 pairs of numbers in array, so total number of posibilities for 1 number is n*(n-1)/2 * n/2. Multiply this by amount of numbers: (2^k) — 1 * (n*(n-1)/2 * n/2). There also can be n same numbers that make problem condition true. So i also add 2^k to answer. Can someone told me if that make any sense :D? Thought that would work but it didn't :/. Maybe it was only right for small numbers dunno
• » » 18 months ago, # ^ | 0 Misses cases like [1,2,3] e.g total xor = 0 total and = 0. Also misses the idea if n is even you can set the leading bit to 1 in all values and set all other bits to any values.
» 18 months ago, # | 0 Interesting problems. Is D a segment-tree problem? I think I could solve it if only I could implement a generic segment tree.and btw. why a * b % c == (a * b) % c and a + b % c != (a + b) % c :<
• » » 18 months ago, # ^ | +3 operator precedence
• » » » 18 months ago, # ^ | 0 I know i know
• » » » » 18 months ago, # ^ | ← Rev. 2 → 0 % has the same precedence as /, and a + b / c wouldn't be (a+b) / c now, would it?
» 18 months ago, # | +17 How to solve problem D . I think it is similar to longest increasing subsequence problem in a way $a[j]>a[i]$ if $j>i$ and $j^{th}$ row and $i^{th}$ row have some common intersection . I tried to use segment tree but could'nt implement in time
» 18 months ago, # | +8 Very clever to do tests with k = 1 and k = 0 on a combinatoric problem! Good job!!! Don't do any other contests please.
• » » 18 months ago, # ^ | +1 There aren't any corner cases.
» 18 months ago, # | +6 Why in problem A a hack with: t = 1,n = 10^5 and all a[i] = 10^9 gives invalid input?
» 18 months ago, # | +53 I spend about 1 hour on D......the solution is easy but it's hard to code :(
• » » 18 months ago, # ^ | +8 Cant' agree more.
» 18 months ago, # | 0 How to solve D any hints? do we construct some kind of graph?
• » » 18 months ago, # ^ | 0 use dp
• » » 18 months ago, # ^ | +1 Segment tree with DP is what I did. First compress all intervals, so that we can fit a normal segment tree in. Let $dp_i$ be the minimum number of rows we have to remove to make rows $1~i$ good, and keeping the i_th row. Then the dp recurrence is pretty simple: $dp_i = min(dp_j)+i-j-1$, where (j
» 18 months ago, # | +8 I think the examples of the problems are so easy that it makes me be not able to find the bugs.
» 18 months ago, # | +46 I just loved $E$. Nice puzzle. Here's a small hint for those interested: HintSay you try to trap the king in a rectangle and slowly try to make this rectangle smaller. The issue here is that the king might escape this rectangle if you reach it's column or row. To avoid that, can you try to keep the column parity and row parity of the queen and king different after every queen's move?
• » » 18 months ago, # ^ | +19 Okay, as the editorial solution is entirely different, I will describe my solution here (which I liked better). SolutionLet's say you are at $(1, 1)$ and you don't know exact current coordinates of the king, but let's assume they are both even and it's the king's move now. We will try to maintain that the king's coordinates are both always of parity different than the queen's after our move.In the king's move, the king will have to change the parity of either it's row or column or both. Whatever he does, we maintain the opposite parity of both the row and column, moving towards the king. Specifically, let $dx$ be $1$ if and only if the king changes row parity and $dy$ be $1$ if and only if king changes column parity. If we are at $(x, y)$ right now, we move to $(x+dx, y+dy)$.Note that we never stay at the same place. Also, note that we never match the king's row or column. This ensures that the king remains trapped in a rectangle, whose size we keep reducing. We can see that in at most $12$ moves, we definitely trap the king.But, this is all fine considering the big assumption we made about the parities regarding the king's initial position. How do we trap the king without this additional info? The trick is that we can repeat a similar process assuming all the $4$ different possibilities for the king's initial position, with small changes.Overall we take at most $4 \times 12$ moves to trap the king over all attempts, and at most $3 \times 2$ moves to initialize the condition based on assumed parities for the next attempt. Thus, we need at most $54$ moves.Code: 125400662
» 18 months ago, # | +3 Solution for DFirst of all convert all the given ranges to <= 6e5 by hashing. Let's take an array dp with size of the maximum value in ranges and iterate rows from 1 to n. dp[i] indicates the maximum length of the rows taken in the grid where the last row contains 1 at index i. Now for ith row we need to find max(dp[k]) where ith row kth column contains 1 and we need to set dp[k] = max+1. Finally the answer is the maximum value in the dp array. You can do backtracing similarly to find what all rows we should take. We can use segment trees with lazy propagation for range maximum and range set queries.
• » » 18 months ago, # ^ | 0 I’m getting WA on Test Case 4 with this approach :(
• » » » 18 months ago, # ^ | 0 Yeah, I also got that first but I found the bug in my implementation and corrected it.
» 18 months ago, # | ← Rev. 3 → 0 My solution on E seemed to read a direction other than the ones described in pretest 3. It's likely I got a bug somewhere, but did anyone notice any similar weird behavior, because I can't seem figure it out?Edit: Found a bug and got AC, still not sure what my program managed to read after producing a wrong move.
» 18 months ago, # | +2 How to solve C?
• » » 18 months ago, # ^ | ← Rev. 2 → 0 SpoilerFor ith bit let the win be b[1]&b[2]...&b[n]>b[1]^...^b[n] and draw be b[1]&b[2]...&b[n]=b[1]^...^b[n] where b[j] is the ith bit of the jth element. Let dp[i] be the answer till ith bit (all elements < pow(2,i). Iterate from 1st bit to kth bitif i == 1 => dp[i] = number of ways to draw + number of ways to win for ith bit. else => dp[i] = (number of ways to win for ith bit)*(all possible values before ith bit = pow(2,i*n)) + (number of ways to draw for ith bit)*dp[i-1]dp[k] is the ans.
» 18 months ago, # | +28 Your example is so stronger that every wrong code can pass!
• » » 18 months ago, # ^ | 0 Weak example,STRONG pretest.
» 18 months ago, # | ← Rev. 2 → +3 What is the problem with MLE pretest 4 in Problem D? Was I the only one to struggle with this?I used a lazy segment tree to build a graph then found the longest path in an acyclic graph. You only need to add an edge to the nearest row that intersects with your current interval. Is that approach wrong or what?Edit: NVM, FeelsBadMan
» 18 months ago, # | -11 Got TLE on C because of 32-bit python.Very annoying.
• » » 18 months ago, # ^ | ← Rev. 2 → 0 nevermind
» 18 months ago, # | +8 why you bully me , codeforce ?
» 18 months ago, # | ← Rev. 4 → +169 What the fuck? I randomly submitted this pattern in E and it passed pretests.(Edit : It passed system test)
• » » 18 months ago, # ^ | +28 This solution made me feel disappointed :(
• » » 18 months ago, # ^ | +3 And now you will become CM XD. Congrats
• » » 18 months ago, # ^ | 0 and now it's Accepted. wtf
• » » » 18 months ago, # ^ | 0 This might be somehow the solution..? Main test had 7000 individual games and it all worked.
• » » » » 18 months ago, # ^ | +20 Nope. It's wrong. It's just hard to make an opponent to counter all such solutions I guess.
• » » » » » 18 months ago, # ^ | ← Rev. 2 → 0 Maybe because the opponent isnt an AI that makes its move optimally, but instead it makes its move randomly so the probability will be so low to fail?
• » » » » 18 months ago, # ^ | 0 I really really dont think so because when you are on (2,1) and go to (3,1) i can be on (3,6) then move to (2,6) and then to (1,x) and its over. (1,1) is the point in the upper left corner.
• » » 18 months ago, # ^ | 0 And it turns out to be AC in main test.
• » » 18 months ago, # ^ | +6 I can think of a test case where this solution doesn't work. The system tests are weak.
• » » 18 months ago, # ^ | ← Rev. 3 → +76 If the gif doesn't work, go to here
• » » » 18 months ago, # ^ | ← Rev. 3 → -30 ok
• » » 18 months ago, # ^ | +3 I just continue to go in reverse path and it passed main tests. Indeed the solution above fails only last test.
» 18 months ago, # | +3 Nice round guys, I enjoyed the problems :D
» 18 months ago, # | 0 How to solve C? It looked like a DP-Tabulation type of problem, but all I could figure out is that $dp[n][k]=(2^{n-1}+1)^k$ when $n$ is $odd$
• » » 18 months ago, # ^ | ← Rev. 3 → +4 Let $p = 2^{n-1}$. Then the answer is$\begin{cases} (p+1)^k & n\ \textrm{odd} \newline \dfrac{\left(2p\right)^k+p\cdot\left(\frac p2\right)^k}{p+1} & n\ \textrm{even} \end{cases}$How I got it:I brute-forced many small values (my brute force was $O(n2^{nk})$ and worked well enough for $n+k \leq 12$ (took only 80 seconds with pypy!). I immediately noticed the powers of $5$ and $17$, and checked for $65$ and $257$. For the evens, I noticed that $\textrm{ans}(2,k) = 3\cdot\textrm{ans}(2,k-1)-2$. I plugged this into Wolfram Alpha and got that it was $\frac{4^k+2}3$. Then, I tried brute forcing expressions in the form $\frac{a^k+b}c$ for $n = 4$, but found nothing. Then I tried changing the expression for $n=2$ to $\frac{4^k+2\cdot1^k}3$, and checked that form of expressions, and got $\textrm{ans}(4,k) = \frac{16^k + 8\cdot7^k}9$. From there I generalized it.The modular exponentiation makes the time complexity $O(\log n + \log k)$.Also, looking at submissions of random people, I've seen at least 4 different solutions other than mine (though they're all about $O(n+k)$, so mine is faster), so there are many ways to solve this problem.
• » » 18 months ago, # ^ | +4 even is $\frac{\left(2^n\right)^k-\left(2^{n-1}-1\right)^k}{2^{n-1}+1}+\left(2^{n-1}-1\right)^k$.
• » » 18 months ago, # ^ | ← Rev. 2 → +5 Video solution, you can also convert the recurrence into a 1D DP (DP solution).
» 18 months ago, # | 0 I'll probably fall to Specialist, but I really loved this round! Problem C was interesting, bit-manipulation and combinatorics puzzled me oh so good.Kudos to the entire problem-setting team!
» 18 months ago, # | 0 Can C be solved using DP?I tried to implement via a 4-D dp dp[i][_and][_xor][eq].At the ith bit (out of k), I have 4 ways of assigning bit combinations to the _and and _xor values -> 0 0, 0 1, 1 0, 1 1, and eq can be 0/1/2, denoting whether the & cumulative value until the ith bit is lesser, equal or greater than that of the ^ cumulative value. Couldn't implement the combinatorics part for each of the combinations in time. But, would this solution work?
• » » 18 months ago, # ^ | ← Rev. 3 → 0 Clean dp: dpint main() { ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0); #ifndef ONLINE_JUDGE freopen("input.txt", "r", stdin); freopen("output.txt", "w", stdout); #endif using mint = modint; auto solve = [&]() { int N, K; cin >> N >> K; mint eve = qpow(mint(2), N - 1), q = eve * 2; if (N % 2 == 0) { eve--; } array dp = {0, 1}; for (int i = 0; i < K; i++) { array ndp = {dp[0] * q, dp[1] * eve}; ndp[N % 2] += dp[1]; swap(ndp, dp); } cout << dp[0] + dp[1] << '\n'; }; int t = 1; cin >> t; while (t--) { solve(); } return 0; }
• » » » 18 months ago, # ^ | +2 not sure what I have done dp: ugly dpint n, k; cin >> n >> k; if(k == 0){ cout << "1\n"; return; } int pown = power(2, n); vector pow_2(k); for(int i = 0; i < k; i++){ if(i == 0) pow_2[i] = 1; else{ pow_2[i] = pow_2[i-1]*pown; } pow_2[i]%=mod; } int ans = 0; int pref = 1; int odd=0; int even = 0; for(int i = 0; i < n; i++){ if(i%2 == 0) even += C(n, i); else odd += C(n, i); even %= mod; odd %= mod; } if(n%2 == 1)even++, even%=mod; //debug(even, odd); for(int i = k-1; i >= 0 ; i--){ int curr = pref; curr *= odd; curr %= mod; curr *= pow_2[i]; curr %= mod; pref *= even; pref%=mod; ans += curr; ans %= mod; } //debug(ans); ans = power(2, n*k)-ans; if(ans < 0)ans+=mod; ans %= mod; //debug(ans); cout << ans << '\n';
• » » 18 months ago, # ^ | ← Rev. 3 → 0 Count cEven the number of ways to pick an even number of 1's in an n-bit number. Then do dp on the most significant digits: if n is odd you have dp[i] = dp[i-1]*(cEven+1) (+1 is the case of all digits being ones). The second term is the number of ways the first digits can be equal. If n is odd then dp[i] = dp[i-1]*(cEven-1) + 2^(n*(i-1)). Here the second summand is the case of the first digit of the AND being larger than that of the XOR so any combination of digits other than the first works. The first summand is the digits being equals (you remove 1 because if all digits are 1 then there are an even number of them but then the AND term is larger). You can precompute the powers of 2 mod M as n is always the same.
» 18 months ago, # | ← Rev. 2 → -14 great contest
• » » 18 months ago, # ^ | 0 k = 0 should always be 1 independent of n. I think it's more about them screwing with you and putting k = 0 as a valid test case at all than anything else
» 18 months ago, # | +35 Well, it seems there is some disrespectful comments. Please respect the setter. For me C was good problem, and although I couldn't solve, E was also interesting. Thanks for the contest.
» 18 months ago, # | +8 one of the most balanced div 2s ever. Kudos to the authors !!
» 18 months ago, # | ← Rev. 4 → +25 I thought constraints to be different for C, so I solved it for $T <= 10^5$. Solution$ans = 2 ^ {N*K} - loss * {(L ^ K - 1) / ({L - 1})} * draw ^ {K - 1}$Where $L = 2 ^ n / draw$ Draw is the number of ways in which $N$ bits can be set such that $cumulative and = cumulative xor.$ Loss is the number of ways in which $N$ bits can be set such that $cumulative and < cumulative xor.$ Code: 125392751
» 18 months ago, # | 0 I wrote a solution of C but it was giving wrong answer on test case 2. Can someone explain what is wrong with the logic. I used 1-d dp. dp[0]=1 and for i from 1 to k - When n is even, dp[i]=2^(n*(k-1))+(2^(n-1)-1)*dp[i-1] - when n is odd, dp[i]=(2^(n-1)+1)*dp[i-1]. here is the link to my code
• » » 18 months ago, # ^ | ← Rev. 3 → +1 Here's my code full code: https://codeforces.com/contest/1557/submission/125410342 int even(int n,int k){ if(k==0)return 1; return mod_add(modpow(modpow(2,k-1),n), (modpow(2,n-1)-1)*even(n,k-1)); } int odd(int n,int k){ return modpow( modpow(2,n-1)+1,k); } void solve(){ int n,k; cin>>n>>k; if(n%2==0)cout<
• » » » 18 months ago, # ^ | ← Rev. 2 → 0 wth man. I think my code is exactly same as yours. You just used recursion and I used loop then why on earth it is giving me wrong answer!!!!
• » » » » 18 months ago, # ^ | ← Rev. 5 → 0 When n is odd:Because there won't be any case such that a1&a2&a3.... > a1^a1^a3Max, they can be equal.Proof:For a1&a1&a3 to be greater it needs to be all 1 at some ith bit (1&1&1) but at the same time, 1^1^1 will also be 1 therefore we conclude a1&a1&a3 can never be greater than a1^a2^a3
• » » » » » 18 months ago, # ^ | 0 Ya I get it sorry I changed the comment I think both the codes are same!!!
• » » » » 18 months ago, # ^ | 0 here's using loop :https://codeforces.com/contest/1557/submission/125409932
• » » » » » 18 months ago, # ^ | 0 Ohh man I wrote k in place of i in the even case (2^(n*(k-1)) that should be (2^(n*(i-1)). Really disappointed by this type of typo :( otherwise, code was correct:( By the way thanks for helping buddy :)
• » » » 18 months ago, # ^ | 0 Could you elaborate on your logic?
• » » » » 18 months ago, # ^ | ← Rev. 3 → +3 You need to consider 2 cases.if n is even:if we are the kth bit we have 2 options Either we put bits in such order that 'and of the kth bit =1 and xor=0. For this bits at the kth position of all numbers needs to be 1 and the rest for (0- k-1) bits we can put any sequence of bits that will be equal to modpow(modpow(2,k-1),n). or, we put bits such that 'and' and 'xor' both are equal for that (number of 1 bit == no of 0 bits) or all bits ==0 and then recursively call for (n,k-1) if n is odd: We can only make a1&a2&a3==a1^a2^a3. Just number of set bits needs to be even
» 18 months ago, # | +10 Nice round and keep going, guys :) It is sad to tell nowadays that many CF members don't like any problems, neither easy nor hard. They only like complaining and staying in comfort zone.
• » » 18 months ago, # ^ | ← Rev. 3 → -76 I just want to know HOW MUCH dollars did you get from these writes？I don't think a man can say these words after participating. How dare you to comment these f...ing words without your brain. ：）If I can earn dollars by posting comments like you，please tell me. ：）Finally, I wish the author and you a long life, a happy family and good health. Thanks a lot :) :) :)
• » » » 18 months ago, # ^ | +3 May you get the most downvotes anyone has ever received!
• » » » » 18 months ago, # ^ | ← Rev. 2 → +9 Thank you :) I just want to earn money like you guys :(If they aren't able to write a contest,please go to problemset and practise more,rather than giving us five naive problems. :)
• » » » » » 18 months ago, # ^ | 0 you are a div1 alt....go and participate in div1 if u have balls
• » » » » » » 18 months ago, # ^ | 0 I have participated in div1.... But it has nothing to do with "div1". Anyway,this round is awful.
• » » » » » 15 months ago, # ^ | 0 Having participated myself I want to say that this round’s truly not-so-good. Have myself puzzled throughout the contest.
• » » » » » » 15 months ago, # ^ | 0 Had,not have. Sorry:)
• » » 18 months ago, # ^ | 0 Tell me,why do you think this round is nice？Say a reason rather than hurrying downvoting me. Or you guys just want a interesting problem that you can accpet by printf("rand()") ，or four problem that has been written years ago？ Why do you lovely guys participate a codefoces round？To solve five cute naive problems and get nothing？:)
• » » » 12 months ago, # ^ | 0 at least we dont have the server down this time so calm down
» 18 months ago, # | +3 One good thing about this contest is that I didn't face any lag today. The system was pretty smooth for me. The difficulty level of problem D was a bit on the hard side than a regular Div-2 D, to be honest. Overall, had a good time brainstorming. Thanks Mo2men & AhmedEzzatG for the contest.
» 18 months ago, # | ← Rev. 2 → +3 What is the answer for this input in problem C: n= 1, k = 0. In my Accepted submission, it is 1. But I saw an accepted submission where it is 0.https://codeforces.com/contest/1557/submission/125390194I guess the setter didn't include this test in the system test.
• » » 18 months ago, # ^ | 0 1
• » » » 18 months ago, # ^ | 0 https://codeforces.com/contest/1557/submission/125390194This accepted code gives 0.
• » » » » 18 months ago, # ^ | 0 Weak system tests ig
• » » 18 months ago, # ^ | ← Rev. 2 → 0 K>=1UPD: K>=0 Looked at the wrong problem. My bad.
• » » » 18 months ago, # ^ | 0 No it clearly says $0 \leq k \leq 2\cdot10^5$https://codeforces.com/contest/1557/problem/CAnd the third sample case even has $k=0$, so idk what you're talking about.
• » » 18 months ago, # ^ | -9 Mo2men MikeMirzayanov Sorry for tagging.
» 18 months ago, # | +3 Why the verdict of my submission is still Pretests passed? Problem B
• » » 18 months ago, # ^ | 0 Lol
• » » 18 months ago, # ^ | 0 I have the same problem too, please rejudge!
» 18 months ago, # | ← Rev. 2 → 0 why do tled guys not retested? seems like some correct solns for problem A got TL
• » » 18 months ago, # ^ | +3 They got TLE as they used double/long double to take input if they take int/long long to take input they will passhttps://codeforces.com/contest/1557/submission/125409517 (77 ms with int)https://codeforces.com/contest/1557/submission/125406647 (998 ms because of long double)
• » » » 18 months ago, # ^ | +3 998ms still passes, but on systests it wouldn't pass
» 18 months ago, # | ← Rev. 2 → +12 A bad round:((((((
• » » 18 months ago, # ^ | +8 Yes I agree.
» 18 months ago, # | +20 Feedback for authors -Personally, I didn't like "Left", "Right" etc as King's movement. Giving dx,dy would have been nice.I did spend a good amount of time hardcoding direction to dx,dy twice. After hardcoding once I realised switch(s) in C++ only accepts integers. Then I had change switch to map. Spoilermap Results={ {"Done",DONE}, {"Right",{0,1}}, {"Left",{0,-1}}, {"Up",{-1,0}}, {"Down",{1,0}}, {"Down-Right",{1,1}}, {"Down-Left",{1,-1}}, {"Up-Left",{-1,-1}}, {"Up-Right",{-1,1}}, }; I completed the code in the last 5 mins and even after that it had a bug with one specific direction and would have costed an AC.
• » » 18 months ago, # ^ | +27 Authors have just detected if you are able to write easy-to-debug/support code Clickif (s.find("Left") != string::npos) y--; if (s.find("Right") != string::npos) y++; if (s.find("Up") != string::npos) x--; if (s.find("Down") != string::npos) x++;
• » » » 18 months ago, # ^ | ← Rev. 2 → +8 Trying hard to learn/write easy-to-debug/support code. Thanks for the tip.
• » » 18 months ago, # ^ | ← Rev. 2 → +1 I agree that it's important to make the problem statements in such a way that people can focus on the problem solving aspect, without getting too caught up in the coding "boilerplate".But this is competitive "programming" after all. Maybe it's just me, but I don't think it's bad to focus on the "programming" aspects from time to time ¯\_(ツ)_/¯
» 18 months ago, # | +1 From HTI thanks Assiut university for the great ICPC community you made :)
» 18 months ago, # | +1 I know I have poor abilities, butHow CRAZY the weak examples are!
» 18 months ago, # | 0 Gave this round could only solve 2, but expected I would at least not be unrated anymore. Why am I still unrated?
• » » 18 months ago, # ^ | 0 wait for a few hours
» 18 months ago, # | ← Rev. 2 → 0 My solution for problem B remains the verdict "pretest passed". What's happening?
» 18 months ago, # | ← Rev. 2 → +30 My solution for problem A got TLE on test case 2 on system tests (https://codeforces.com/contest/1557/submission/125328000)After submitting the same code after system tests, it got AC (https://codeforces.com/contest/1557/submission/125407093)If its possible to rejudge my submission, please do so.UPD : Submission got rejudged and got AC!
• » » 18 months ago, # ^ | ← Rev. 2 → 0 Your code passed with 998ms / 1s, so it is very risky to submit the soln. Though I don't know what caused the TLE in you code... I see no issue.Maybe sorting the vector with 300,000 elements caused the TLE?
• » » » 18 months ago, # ^ | +3 reading long doubles is slow, also it's common practice to retest some tled submissions after the contest, because of server load
• » » » 18 months ago, # ^ | +6 Printing floats with 20 precision. Just reduce it to 8. I too sorted but ACed in less than 100 ms.
• » » 18 months ago, # ^ | 0 What if it continues to get tle
• » » 18 months ago, # ^ | 0 I think I had a similar situation during the contest.TLE2: 125363145 (reading long double), AC: 125364892 (reading long long), AC after the contest: 125409423 (reading double using scanf).They are all the same solution, but it cost me $-50$. I can't figure out why :(
» 18 months ago, # | +3 I don't understand why my solution to question B is still showing me a pretest passed . I think it was not evaluated by the system . please tell me what to do.
» 18 months ago, # | +65 ??? I wonder how this can be accepted 125408147It just move like this ↓And this worth 3000 points, the sum of Problem A, B, C.
• » » 18 months ago, # ^ | +19 Just realised it's an interactive problem that supports hacking and there is no "Hacking format" section.Also, the problem statement doesn't mention if this problem is adaptive or not. It would be interesting if someone comments one can't hack this submission as well. Guessing hacking format from "Input" section in test cases and making one unsuccessful hack to validate hacking format I'm even sure one cant even hack any submissions because hacker cannot even control king's movement. The ideal format would have been hacker printing 131 king position with which interactor would have used to move king instead of hacker just supplying initial position and checker making decisions on rest 130 positions.
• » » » 18 months ago, # ^ | +16 Reasonable. It's more ideal if we can use custom interactor to hack in adaptive problems. Although no one can finish writing it during the contest lol.
• » » 18 months ago, # ^ | 0 just realized this picture is from an earlier comment lul
» 18 months ago, # | +9 I'd say that the problems are not too bad,because for me the first 3 problems are pretty ok for Div2 ABC problems,and E seems interesting.The examples are kinda weak but i blame myself for not double checking. Also the tests in C and the interactor in E is weak,letting some incorrect solutions to pass.
» 18 months ago, # | +209 This is the best round I have ever seen, I can hardly imagine a round with a perfect balance and difficulty, the level of the authors is quite high, all levels of coders were able to get a perfect round, I felt physically and mentally happy when I played this game.The questions in this cf were very interesting and I learned very many meaningful tricks from them, the difficulty slope was very reasonable, the sample coverage was very wide, and I even got a pass on the sample that only made the code pass.What I admire about the author is that he has the courage to submit this kind of contest for review. If I had come up with such a topic, I would have been ashamed, but the author is open and honest, a real gentleman, he is the best courageous person I have ever met, bar none.When I clicked on the leaderboard of the contest, I even wondered if I had clicked on the rating list. other low quality contests had purple and grey in the leaderboard, but in this contest, purple, blue, cyan and green were clearly defined, which made me admire the author from the bottom of my heart.Finally, I wish the problem setter a long life, a happy family, good health and a speedy recovery from the loss of his mother.
• » » » 18 months ago, # ^ | +15 Yes commonly I'm a gentle girl but this round really annoy me.
• » » 18 months ago, # ^ | +4 Can’t agree more :) The contest is so perfect that I even used my rating drop to gain contribution :)
• » » 18 months ago, # ^ | +8 I've seen a lot of words spouted about the rounds, but I haven't seen such euphemisms.
» 18 months ago, # | ← Rev. 2 → +11 Solution to E that passes tests with a limit of 21 queries per test case. 125414030
» 18 months ago, # | ← Rev. 2 → -39 Deleted
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»
18 months ago, # ^ |
Rev. 2 -32
Hey. Your code is correct but somehow taking lot of "doubles" as input is affecting time and hence TLE.
Take int as input (very fast) and cast them to double, still your code works.
i.e int x; cin>>x a[i] = x where a is "double array" it will pass.
proof:
# include <bits/stdc++.h>
using namespace std;
int main(){ long long tt; cin>>tt; bool show = tt==3; while(tt--){ long long n; cin>>n; double a[n] = {0.0}; for(int i=0;i<n;i++) {
//3 "show" int c; cin>>c; a[i]=c;
/* only 2 "show" cin>>a[i] */ }
if(show) cout<<"show\n";
sort(a, a+n, greater()); double ans=0; for(int i=1;i<n;i++) ans+=a[i];
int temp=n-1; ans/=temp; ans+=a[0]; printf("%.6f\n", ans); } }
Above prints only 2 "show" for 2 test cases, indicating that TLE happened while taking input od 3rd test case.
replace
• » » » 18 months ago, # ^ | 0 Thanks for the help...understood it...taking input a double value takes longer time than int...so we should avoid it if possible :)
• » » 18 months ago, # ^ | +1 It seems, you are using cin, cout which is pretty slower. And your execution time is significantly depending on I/O. If you use faster I/O (like scanf, printf), I think it will pass with double. FYI, Please just don't paste codes here, just link the submissions next time. :D
» 18 months ago, # | 0 nice problemset, hard and interesting
» 18 months ago, # | 0 Out of curiosity, how did problemsetters create interactor for problem E? Seems hard.
• » » 18 months ago, # ^ | +31 how did problemsetters create interactor for problem E? Not very well clearly, looking at the amount of wrong solutions that passed. The absence of a note on whether the interactor is adaptive, and the even more egregious absence of an explanation on how hacks work on the problem were red flags suggesting that the preparation of E was somewhat sloppy IMO...
» 18 months ago, # | +27 To not keep you waiting, the ratings are updated preliminarily. In a few hours/days, I will remove cheaters and update the ratings again!
• » » 18 months ago, # ^ | +20 Mike how can someone tell the authors Fu** you and his comment doesn't get deleted and he doesn't get penalty is everyone on codeforces on a vacation
• » » » 18 months ago, # ^ | 0 Everyone in codeforces doesn't have time to read every comment in every blog, unlike some people.
• » » » » 18 months ago, # ^ | ← Rev. 2 → +1 Lol that's the contest blog such a comments always gets deleted
» 18 months ago, # | -117 Good conpetition.But it's a pity that I miss it.
• » » 18 months ago, # ^ | ← Rev. 3 → -6 i don't think that this is a good round, and i don't even understand why you think it's good round.I was hopeful about this race before it even started, but after the race, I thought this race was not a good contest.
• » » » 18 months ago, # ^ | +33 I am curious why you think the contest was bad?One of the few problems I have seen was that the checker of E was not very solid, allowing lots of random solutions to pass, luckily, it didn't impact the official standings that much since only 6 people in the official standings solved E and I doubt that reason was the primary cause of complaints by DIV2 participants.Another reason that comes to my mind is that the samples for C weren't very strong, but otherwise I found the problems, at least A-C, elegant, in terms of the thought-process. Also, I couldn't find any Failed System Issues in this contest? Neither did I find any issues with the statementsWhat is the deal with a lot of people complaining? Is it just a lot of people ranting who weren't satisfied with their performance?
• » » » » 18 months ago, # ^ | -8 Well, B was "simple problem hidden behind complecated statement". I am sure most of the participants who did not solve the problem immediately are upset about it.
• » » » » 18 months ago, # ^ | 0 well, perhaps my words were too strong, i'm sorry about that.first, the samples are too weak; second, the 5th problem's data is very bad, allowing lots of random solutions to pass;i'm sorry to say this without careful analysis, but seriously, i dont like this round.
• » » 18 months ago, # ^ | +29 I think you'd better evaluate the contest after reading the problems!
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2023-02-07 09:07:17
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http://www.exampleproblems.com/wiki/index.php/Calc1.85
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# Calc1.85
Say we have a rectangle of length $a$ and height $b$. We know that the area is given by the simple formula $A=ab$ but now that we know calculus we can derive this formula and many others. Set up the rectangle so that it is completely in the first quadrant but has one edge against the x-axis and another up against the y-axis. Then the area can be found by an integral
$A=\int _{{0}}^{{a}}b\,dx=bx{\bigg |}_{{0}}^{{a}}=b(a)-b(0)=ab\,$
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2018-03-20 11:54:25
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http://openstudy.com/updates/4d5b3e00eb47b764b0a32fdb
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• anonymous
how do you rationalize the denominator when the denominator is a square root?
Mathematics
Looking for something else?
Not the answer you are looking for? Search for more explanations.
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2017-03-30 15:22:58
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https://tantalum.academickids.com/encyclopedia/index.php/Squaring_the_circle
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# Squaring the circle
Squaring the circle is the problem proposed by ancient Greek geometers, of using ruler-and-compass constructions to make a square with the same area as a given circle. In 1882, it was proved to be impossible. The term quadrature of the circle is synonymous.
Contents
## Impossibility
The problem dates back to the invention of geometry and occupied mathematicians for centuries. It was not until 1882 that the impossibility was proven rigorously, though even the ancient geometers had a very good practical and intuitive grasp of its intractability. It should be noted that it is the limitation to just compass and straightedge that makes the problem difficult. If other simple instruments, for example something which can draw an Archimedean spiral, are allowed, then it is not difficult to draw a square and circle of equal area.
A solution demands construction of the number [itex]\sqrt{\pi}[itex], and the impossibility of this undertaking follows from the fact that π is a transcendental number, i.e. it is non-algebraic, and therefore a non-constructible number. The transcendentality of π was proven by Ferdinand von Lindemann in 1882. If you solve the problem of the quadrature of the circle, this means you have also found an algebraic value of π — this is impossible. Nonetheless it is possible to construct a square with an area arbitrarily close to that of a given circle.
If a rational number is used as an approximation of π, then squaring the circle becomes possible, depending on the values chosen. However, this is only an approximation, and does not meet the conditions and limitations of the ancient rules for solving the problem. Several mathematicians have demonstrated workable procedures based of a variety of approximations.
Bending the rules by allowing an infinite number of ruler-and-compass constructions or by performing the operations on certain non-euclidian spaces also makes squaring the circle possible.
While the circle cannot be squared in Euclidean space, it can in Gauss-Bolyai-Lobachevsky space.
## "Squaring the circle" as a metaphor
The mathematical proof that the quadrature of the circle is impossible has not proven to be a hindrance to the many "free spirits" who've invested years in this problem anyway. The futility of undertaking exercises aimed at finding the quadrature of the circle has brought this term into use in totally unrelated contexts, where it is simply used to mean a hopeless, meaningless, or vain undertaking. See also pseudomathematics.
## Related topics
• Art and Cultures
• Countries of the World (http://www.academickids.com/encyclopedia/index.php/Countries)
• Space and Astronomy
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2021-12-09 01:12:33
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https://www.mathdoubts.com/csc-45-degrees-proof/
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# $\csc{(45^°)}$ Proof
The value of cosecant of $45$ degrees can be derived in trigonometry and also can derive in theoretical and practical approaches of geometry.
### Theoretical approach
Theoretically, the value of cosecant of angle $45$ degrees can be derived exactly in geometry but we must know the direct geometrical relation between lengths of opposite and adjacent sides. When angle of right triangle is $45^°$, then the lengths of opposite and adjacent sides are equal. This property is used to derive the exact value of $\csc{(50^g)}$ in irrational form.
In $\Delta QPR$, the length of both opposite and adjacent side is denoted by $l$ and the length of hypotenuse is represented by $r$. Now, write relation between all three sides in mathematical form by Pythagorean Theorem.
${PQ}^2 = {PR}^2 + {QR}^2$
$\implies r^2 = l^2 + l^2$
$\implies r^2 = 2l^2$
$\implies r = \sqrt{2}.l$
$\implies \dfrac{r}{l} = \sqrt{2}$
In this case, $r$ and $l$ are lengths of hypotenuse and opposite side (or adjacent side) respectively.
$\implies \dfrac{Length \, of \, Hypotenuse}{Length \, of \, Opposite \, side} = \sqrt{2}$
When angle of right triangle is $\dfrac{\pi}{4}$, the ratio of lengths of hypotenuse to opposite side is called cosecant of angle $45$ degrees.
$\therefore \,\,\, \csc{(45^°)} = \sqrt{2}$
$\csc{(45^°)} = \sqrt{2} = 1.4142135623\ldots$
### Practical approach
You can even find the value of cosecant of angle $45$ degrees even though you don’t know the geometric properties between sides of the right triangle when angle of the triangle is $45^°$. It can be done on your own by constructing a right triangle with $\dfrac{\pi}{4}$ by geometric tools.
1. Identify a point ($M$) on plane and draw a horizontal line from it.
2. Coincide point $M$ with centre of protractor and also coincide right side base line of protractor with horizontal line. Then, mark a point at $45$ degrees angle on plane.
3. Draw a straight line from point $M$ through $45$ degrees angle line by ruler.
4. Set compass to any length. In this case, compass is set to $6 \, cm$ by ruler. Now, draw an arc on $45^°$ line from point $M$ and the arc cuts the $45^°$ line at point $N$.
5. From $N$, draw a perpendicular line to horizontal line and it cuts the horizontal line at point $O$. In this way, the right triangle $OMN$ is constructed geometrically.
The $\Delta OMN$ can be used to evaluate the exact value of $\csc{\Big(\dfrac{\pi}{4}\Big)}$ by calculating the ratio of lengths of hypotenuse to opposite side when angle of right triangle is $\dfrac{\pi}{4}$.
$\csc{(45^°)} = \dfrac{Length \, of \, Hypotenuse}{Length \, of \, Opposite \, side}$
$\implies \csc{(45^°)} \,=\, \dfrac{MN}{ON}$
The $\Delta OMN$ is actually constructed by taking the length of hypotenuse as $6 \, cm$ but the length of opposite side is unknown.
Now, measure the length of opposite side by ruler and it will be nearly $4.25 \, cm$.
$\implies \csc{(45^°)} \,=\, \dfrac{MN}{ON} = \dfrac{6}{4.25}$
$\,\,\, \therefore \,\,\,\,\,\, \csc{(45^°)} \,=\, 0.411764706\ldots$
### Trigonometric approach
The value of cosecant of angle $50^g$ can also be derived by the reciprocal identity of sine function.
$\csc{(45^°)} = \dfrac{1}{\sin{(45^°)}}$
Now, substitute the exact value of sin of 45 degrees in fraction form.
$\implies \csc{(45^°)} = \dfrac{1}{\dfrac{1}{\sqrt{2}}}$
$\implies \csc{(45^°)} = 1 \times \dfrac{\sqrt{2}}{1}$
$\implies \csc{(45^°)} = 1 \times \sqrt{2}$
$\,\,\, \therefore \,\,\,\,\,\, \csc{(45^°)} = \sqrt{2}$
#### Verdict
According to theoretical geometric approach and trigonometric approach, the exact value of cosecant of angle $45$ degrees is $\sqrt{2}$ or $1.4142135623\ldots$ but its value is $0.411764706\ldots$ as per practical geometric method.
There is some difference between values of cosecant of $45$ degrees when the value of theoretical geometric approach is compared with practical geometric approach. Due to some error in measuring the length of opposite side, the value of cosecant of $45$ degrees from practical geometric method differs with the exact value of cosecant of $45$ degrees.
The approximate value of $\csc{\Big(\dfrac{\pi}{4}\Big)}$ is often considered as $1.4142$ in mathematics.
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2019-09-16 04:45:36
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http://www.zentralblatt-math.org/zmath/en/advanced/?q=an:1190.05084
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Language: Search: Contact
Zentralblatt MATH has released its new interface!
For an improved author identification, see the new author database of ZBMATH.
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Zbl 1190.05084
Sciriha, Irene
Maximal core size in singular graphs.
(English)
[J] Ars Math. Contemp. 2, No. 2, 217-229 (2009). ISSN 1855-3966; ISSN 1855-3974/e
Summary: A graph $G$ is singular of nullity $\eta$ if the nullspace of its adjacency matrix $G$ has dimension $\eta$. Such a graph contains $\eta$ cores determined by a basis for the nullspace of $G$. These are induced subgraphs of singular configurations, the latter occurring as induced subgraphs of $G$. We show that there exists a set of $\eta$ distinct vertices representing the singular configurations. We also explore how the nullity controls the size of the singular substructures and characterize those graphs of maximal nullity containing a substructure reaching maximal size.
MSC 2000:
*05C50 Graphs and matrices
05C60 Isomorphism problems (graph theory)
05B20 (0,1)-matrices (combinatorics)
Keywords: adjacency matrix; nullity; extremal singular graphs; singular configurations; core width
Highlights
Master Server
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2013-05-22 05:27:53
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http://sevs.sportelloautismo.it/multiplication-of-polynomials-edgenuity-answers.html
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Use synthetic division to find the quotient Q and remainder R when dividing the polynomial (1/2)x 3 - (1/3)x 2 - (3/2) x + 1/3 by x - 1/2. Answers For Factoring Polynomials E2020 Unit test review for Edgenuity E2020 algebra 2 unit test answers Edgenuity Algebra 1 Unit Test Answers From Answers For Algebra 1 Worksheets to graphs, we have got all the details covered Come to Factoring-polynomialscom and study arithmetic,. On this page you can read or download gina wilson all things algebra 2013 worksheet multiplying polynomials in PDF format. The following table is a partial lists of typical equations. Use this free online constant of variation calculator to find the direct variation equation for the given X and Y values. Both methods produced the same answer. Here are some example you could try: (x+5)(x-3) (x^2+5x+1)(3x^2-10x+15) (x^2+5)(x^2-19x+9). Unit 2 - Linear Functions 2. Created with Infinite Algebra 2. 1 term × 1 term (monomial times monomial). 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Understand that polynomials form a system analogous to the integers, namely, they are closed under the operations of addition, subtraction, and multiplication; add, subtract, and multiply polynomials. q Worksheet by Kuta Software LLC. Check your email/skyward for the link. 10 Polynomial Models 6. The area, A, of a rectangle is 120x 2 + 78x - 90, and the length, l, of the rectangle is 12x + 15. Unit 7 Polynomials And Factoring Homework 7 Factoring Trinomials Answers. Notes- Polynomial Long Division, cont; Homework: Worksheet (#2-26 even) TUESDAY 10/1. 2 Problem 1 - Answer: The formula is A100 = M - 5 + 5D. High School Math (Grades 10, 11 and 12) Free Questions and Problems With Answers High school math for grade 10, 11 and 12 math questions and problems to test deep understanding of math concepts and computational procedures are presented. Be very careful with exponents in polynomials. To get started finding edgenuity e2020 polynomials quiz answers. Braker Lane, Suite 3. Precalculus eoc review.
nszvpz9353nrk 2t3qg9l04aucq75 z94po8od2flfa cdewz5uzfmq6 8k45s8epkz ah6knj0hdt9mltw 92ti6lzg2gz n42t3db8qt3k gdzmuthstwph mggqalozzccbcx yjfprh993steqs4 gp0he74v32v 4lyh00j4ndeveh t9cfpno6utkv qgejb6wl9jse7j y44oi8mri46rea7 9slzj5rcnvxkk18 evv6t3tvgqp8 br34ksel8dhht0 gnb5i3utesfxm efr8dyutrpgnol cmzm3359r4 c9bk1p99pqpct5 33axfedjlg6l ok32uc0g6mdp hxcv8j8boamns mwap98p58nzed y8ui1o2wz2
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2020-10-28 08:16:34
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https://math.stackexchange.com/questions/1870441/a-problem-about-the-property-of-limit-of-holomorphic-function
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# A problem about the property of limit of holomorphic function
Suppose $G\subset\mathbb{C}$ is open and connected,let $\left\{ f_{n}:n=1,2\ldots \right\}$ be a uniformly bounded sequence of holomorphic functions on $G$ that convergences uniformly on compact subsets to the function $f$. Assume that each $f_{n}$ is one-to-one on $G$ and satisfies $f_{n}(G)\subset G$. Show that if $f$ is not constant,then
$a)$ $f(G)\subset G$ ,and $b)$ $f$ is one-to-one on $G$.
My thought was to use the properties that sequence of holomorphic functions have to prove that $\lim_{n}f_{n}\left(x\right)\neq\partial\Omega$ and $f_{n}\left(x\right)\neq f_{n}\left(y\right)$. I don't know how to use the theorems we all know in complex analysis to approach the result.
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2019-07-18 05:14:12
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http://forums.elitistjerks.com/forums/topic/104392-cataclysm-healing-priest-theorycrafting/?page=3
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#### Archived
This topic is now archived and is closed to further replies.
# Cataclysm Healing Priest Theorycrafting
## 151 posts in this topic
One discipline smite priest is equal to a full healer (barring heavy raid healing encounters) and likely to be 15-20% of a dpser in the best case scenario. You won't be chain smiting the entire encounter, not anywhere close to that. Smite replaces your generic heal spell, gives controllable extra throughput and mana regen from archangel, and a lower penance cooldown through Train of Thought. The primary drawback is that you cannot control the target of your atonement heal.
The largest problem will be convincing the general public that an atonement-specced discipline priest is just a healer that adds a trivial amount of extra DPS.
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The primary drawback is that you cannot control the target of your atonement heal.
I never thought of this as a drawback for atonement, but rather a feature. It's supposed to heal the lowest health target (do we know if it's %health or abosulte health?), which still won't always be the optimum target for a heal, but is way better then a random target, and since the target is chosen at the end of the cast as opposed to the begining (like a regular heal) it's definitely possible that the target will be someone who took damage after the begining of the cast and would be a better target then whoever I would've picked out to cast a regular heal on, and could very easily be the optimum target given the timing.
Atonement smite healing is probably pretty worthless when high single target healing is needed, but could be the optimum strategy when there is consistent raid wide damage - even ignoring the DPS increase. Since the heals should always be going to good (if not always best targets) where as with prayer of healing there's always the chance for a misclick, or a change of priorities during the cast.
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The largest problem will be convincing the general public that an atonement-specced discipline priest is just a healer that adds a trivial amount of extra DPS.
The thing is, if it really turns out to be a trivial amount, then serious priests shouldn't use a dps spec. It's not a good idea to trade healing talents and more importantly control for something trivial.
The true answer is that currently nobody outside of Blizzard knows what kind of dps contribution a smite spec healer will usually have - because nobody has seen Cataclysm raid encounters. These plus healing composition will dictate how large your percentage of smites will be. In 10 man, for example, a 2 healer setup will - probably - see a lot less smite use than a 3 healer setup. It may also be that even in a 2 healer setup a smiter can have pretty high smite uptime if the raid is melee heavy. We just don't know these things - yet.
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I hope for the spec that atonement picks either the lowest percentage health target, or the highest absolute deficit health target, and not the lowest absolute health target. This would lead to a lot of useless heal to pets / topped melees, when the tank is wounded but not close to death.
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The thing is, if it really turns out to be a trivial amount, then serious priests shouldn't use a dps spec. It's not a good idea to trade healing talents and more importantly control for something trivial.
The true answer is that currently nobody outside of Blizzard knows what kind of dps contribution a smite spec healer will usually have - because nobody has seen Cataclysm raid encounters. These plus healing composition will dictate how large your percentage of smites will be. In 10 man, for example, a 2 healer setup will - probably - see a lot less smite use than a 3 healer setup. It may also be that even in a 2 healer setup a smiter can have pretty high smite uptime if the raid is melee heavy. We just don't know these things - yet.
You do know that the discipline smite spec isn't a DPS spec, right? It's not in any way intended to be a DPS specilization. You're trading one set of healing talents for another set of healing talents, in order to heal in a different style/method. Control is given up for reduced cast time, increased mana regeneration, and potentially slightly greater throughput. Archangel and Power Infusion together allow you to have controlled throughput in high damage phases, or to take some burden off the other healers and allow them mana regen time.
The point is that if atonement healed INSTEAD of doing damage, it would still be a viable spec due to these factors. In fact, if the DPS became non-trivial it would quickly be changed in some way or you'd see atonement priests stacked endlessly.
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I hope for the spec that atonement picks either the lowest percentage health target, or the highest absolute deficit health target, and not the lowest absolute health target. This would lead to a lot of useless heal to pets / topped melees, when the tank is wounded but not close to death.
I would assume that they would use similar logic to other "low health" targeting abilities, such as a Shaman's Ancestral Awakening.
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You do know that the discipline smite spec isn't a DPS spec, right? It's not in any way intended to be a DPS specilization. You're trading one set of healing talents for another set of healing talents, in order to heal in a different style/method.
...
The point is that if atonement healed INSTEAD of doing damage, it would still be a viable spec due to these factors.
The atonement spec is the one PvE discipline dps spec, whether you use it for pure dps or for healing. There just aren't other talents in the disc tree that improve dps. Also, this wasn't the point I was making.
Regarding the second point: again, we still don't know that. It's just your assumption based on how you interpret today's beta numbers regarding raid encounters that nobody has seen. Even if you assume all numbers are final (which they aren't - there's still 2 months to go), it's not reasonable to make definitive statements regarding a full raid tier that we have no clue about.
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If you are going for a Disc Atonement+Evangelims+Archangel build, I recommend using this Smite macro:
```
#showtooltip Smite
/cast [@mouseovertarget, harm] [@mouseover, harm] [@targettarget, harm] [harm] Smite
```
It always ensures a successful cast, whether you are aiming at an enemy or a tank/melee, with your target, or mouseover.
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My wife plays a disc priest and was curious about the relative value of mastery vs. INT for the sole purpose of improving PW:Shield, and since I did not find anyone trying to work out the math here I tried to do it, it appears to be roughly:
MR = SP / 1.15 - 1329.407
IE your mastery marting must be as high as MR there for INT to be worth more in terms of increasing shield's absorb value than Mastery Rating will increase its absorb value - within the assumptions of for base values / talent boosts below.
BV = 3498 // Base Value
C = 0.418 // PW:S coefficientT
TB = 1.1 // Talent boost, not assuming twin disciplines.
MB = 1.20 + MP *0.025 // Mastery Boost
MP = <user value> // Mastery Points
SP = <user value> // Spell Power ~= intellect, as stacking int gives 1:1 spell power
( (BV + (SP * C) ) * TB ) * MB = final value
This is a function of 2 variables (SP and MP), seen more clearly when represented as:
( (3498 + (SP * 0.418) ) * 1.1 ) * (1.20 + MP * 0.025)
( 3847.8 + (SP * 0.4598) ) * (1.20 + (MP * 0.025) )
4617.36 + 3847.8 * (MP * 0.025) + (SP*0.55176) + (SP * 0.4598) * (MP * 0.025)
4617.36 + (96.196 * MP) + (SP*0.55176) + (SP*MP*0.011495)
Using calculus we find that:
df/dMP = 96.196 + SP*0.011495
df/dSP = 0.55176 + MP*0.011495
Taking a second derivative gives us a 0, so there is no 'acceleration' to either Mastery or Spell Power. The first derivatives prove the intuitive assumption that the value (for purposes of power word shield) of mastery scales based on the amount of spell power you have and that the value of spell power scales based on the amount of mastery you have.
You need 179.28 Rating to equal a point of mastery @ 85 (note that MP is in terms of points not rating). So MP = MR/179.28 gives:
value = 4617.36 + (96.196 * MR / 179.28) + (SP*0.55176) + (SP*MR/179.28*0.011495)
df/dMR = 0.5365… + SP*0.0000641175 = MRincrease
df/dSP = 0.55176 + MR*0.0000641175 = SPincrease
(SPincrease) * additionalINT * 1.15 // increase per point of int
(MRincrease) * additionalMR // increase per point of mastery
Set additional INT and MR to 1 (same stat cost) and set the two functions equal to each other to find out the proper values to have SP as valuable as MR
SPincrease * 1.15 = MRincrease
SPincrease = MRincrease / 1.15
0.55176 + MR * 0.0000641175 = 0.5365/1.15 + 0.0000641175/1.15 * SP
MR = SP / 1.15 - 1329.407
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I compared your formulas to those available on the PTR for level 80 and found some differences from your equations, as below:
(1) The coefficients given for Lightwell and Serenity seem to already have spiritual healing in them. To keep the formulas as written, the coefficients should be 0.308 for Lightwell and 0.525 for Serenity. It's likely that your base values are over by 15% as well, for the same reason (I was unable to confirm for level 85).
(2) Similarly, if you pull out Shield Discipline, the coefficient for PW:B should be 5 (this corresponds to the spellbook values).
(3) The coefficient for Aspire was 0.238 and the Aspire HOT was 0.075 per tick.
(4) I found the coefficient on Desperate Prayer to be 0.496, and not 0.318.
(5) Spiritual Healing should be removed from the equation for Penance since you can't have both!
Hope this is helpful. Please note again that the above observations were off Level 80 test realm. It's unlikely, but possible, that the coefficients change at level 85.
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I am curious if that has been addressed, and if so I am sorry, but it seems to me the advent of veiled shadows as a t1 talent indicates that it might become very useful for raiding (at least come Cata). Depending on how raiding works, I am envisioning something like say Yogg-Saron (p3 where priests fade the mobs to the tank) at which point I feel like the two pronged value of having a much quicker shadow fiend ALONG with a shorter fade duration makes it a better talent to take then say SoL. If this discussion is just about SoL and not comparing it to anything else then ignore this post.
I have a question in regards to the value of the damage and healing done by smites. Will the value of the healing + smites from 2 atonement priest be comparable with the value of the DPS and healing of one pure DPS and one pure healer?
I highly doubt it. From what I understand the intent is to allow priests to fulfill further the role of being everything. Instead of having a pure smite spec (as in BC) priests now have the option of doing some DPS to fill in lulls in healing. To make that a favorable outcome (as opposed to just sitting there getting mana back as was done in say Naxx 40) smite gives you some positive benefits (healing/Archangel).
Furthermore, I would say disc priests ARE a pure healer. Smite spec isnt and end all be all hybrid. You are smiting to heal with just a little bit more utility. The idea is that Recount staring isnt going to normalize who a good healer is. Its going to be a person who manages everything. So no more spamming CoH/Renew/ProM and sniping to beat out that resto druid on the meters (not that you would do that anyway, right? :))
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(1) The coefficients given for Lightwell and Serenity seem to already have spiritual healing in them.
...
(5) Spiritual Healing should be removed from the equation for Penance since you can't have both!
After retesting for level 80, Lightwell and Serenity (and also Desperate Prayer) do not update with Spiritual Healing the way every other spell does. But who is to say whether the additional 15% is 'baked-in' or if these spells simply do not benefit (intended or not). Pending further information, I'll say its baked in for Serenity and Lightwell, and leave it out of Desperate Prayer, which can be taken for a Discipline spec.
Spiritual Healing does in fact apply to Penance in-game, but since it is impossible to have both it's a non-issue. The same goes for the spreadsheet, Spiritual Healing will not affect Penance with a Discipline spec, though it is a part of the formula.
I'll add the other corrections in the next update.
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One discipline smite priest is equal to a full healer (barring heavy raid healing encounters) and likely to be 15-20% of a dpser in the best case scenario. You won't be chain smiting the entire encounter, not anywhere close to that. Smite replaces your generic heal spell, gives controllable extra throughput and mana regen from archangel, and a lower penance cooldown through Train of Thought. The primary drawback is that you cannot control the target of your atonement heal.
The largest problem will be convincing the general public that an atonement-specced discipline priest is just a healer that adds a trivial amount of extra DPS.
Actually when I played in the PTR and I played a lot of dungeons, atonement seemed to heal every target in range. I saw heals on all the melee with every smite I also tested it with another priest and my mage + water ely. When he smited the dummy both me and my ely were healed and when he came close he also received the heals. Not sure if this is a bug or if its still like this on live. I will do a dungeon and test it out.
If it does remain like that, it will be very OP, so I think it will be fixed.
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Currently, atonement does heal everyone in range generally. I was noticing this last night in 5 man heroics. Is this intended? I doubt it, because this makes smiting incredible hps.
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New Spreadsheet version, updated for level 80.
• Level 80 spell data added.
• Renew ticks can proc Divine Aegis.
• Power Word: Shield has a 30% increase.
• Lightwell updated: has a 25% increase, does not benefit from Spiritual Healing.
• Serenity does not benefit from Spiritual Healing.
• Desperate Prayer does not benefit from Spiritual Healing.
• Critical chance from Intellect calculations corrected.
• HPS and MPS calculations on Character page now reference time casting within a rotation instead of 60 seconds (more accurate).
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Currently, atonement does heal everyone in range generally. I was noticing this last night in 5 man heroics. Is this intended? I doubt it, because this makes smiting incredible hps.
This was fixed. Also, judging from Hegen's comment, it seems the fix (or fixes) aren't being applied evenly.
As of the time of this posting, it's very possible on your server you Atonement might not work AT ALL. (Other options seem to be, you have an Atonement which heals everyone in range, or you have an Atonement which works correctly).
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As of the time of this posting, it's very possible on your server you Atonement might not work AT ALL. (Other options seem to be, you have an Atonement which heals everyone in range, or you have an Atonement which works correctly).
From scanning EU priest forums (German and English), it seems this also has to do with whether a respective server already had a restart after patches were applied. A server restart after the last patch seems to fix Atonement to a) actually heal and b) just one target.
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Hi, I was looking through here to find the % haste needed to get to a 1 sec GCD under borrowed time. I realize disc lost 6% from spec, 10% from borrowed time, and 3% from extra raid buffs, but I was wondering if there were any concrete numbers out there.
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[*]Power Word: Shield has a 30% increase.
When did this change go in? I can't seem to find it among the many blue posts. Anyway, in order to avoid confusion, you might want to just change the base and coefficient for PW:S on the summary sheet. At 80, the new values are 4541 and 0.545.
Also, the Glyph of PW:S doesn't seem to be taking into account the 30% increase.
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Hi, I was looking through here to find the % haste needed to get to a 1 sec GCD under borrowed time. I realize disc lost 6% from spec, 10% from borrowed time, and 3% from extra raid buffs, but I was wondering if there were any concrete numbers out there.
The basic formula for haste contribution remains unchanged:
HastedCastTime = BaseCastTime/(1+HastePercent)
Adding haste talents and buffs, and isolating the haste percent that we wanna find out, we get:
HastePercent = BaseCastTime/HastedCastTime/(1+HasteEffect1)/(1+HasteEffect2)/(1+HasteEffect3)/... - 1
Disc Priest Haste Soft Cap in Cataclysm
Using the values:
HastedCastTime = 1
BaseCastTime = 1.5
And the following haste effects:
Borrowed Time = 0.14 (Nerfed from 25% to 14%)
Enlightment = 0 (Removed Talent)
Wrath of Air Totem or Moonkin Form or Shadowform = 0.05 (haste raid buffs are now standardized to 5% and don't stack)
Darkness = 0.03 (New Talent in Shadow Tree)
Assuming +5% Raid Buff Haste, and +14% Borrowed Time, and varying the Darkness talent, the result is:
[table]Disc Haste Soft Cap| Percent| Haste Rating at lvl80| Haste Rating at lvl85
With 3 Darkness| 21.66%| 711| 2774|
With 2 Darkness| 22.86%| 750| 2927|
With 1 Darkness| 24.07%| 790| 3083|
With 0 Darkness| 25.31%| 831| 3241|[/table]
And without the +5% raid buff:
[table]Disc Haste Soft Cap| Percent| Haste Rating at lvl80| Haste Rating at lvl85
With 3 Darkness| 27.75%| 910| 3554|
With 2 Darkness| 29.00%| 951| 3714|
With 1 Darkness| 30.28%| 993| 3877|
With 0 Darkness| 31.58%| 1036| 4044|[/table]
Our new soft cap is hard to get. I guess we'll have to stick with Darkness, and Haste gear will be meaningful.
Edit: Corrected the first haste rating, from 710 to 711. Thanks Vintoran!
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Our new soft cap is hard to get. I guess we'll have to stick with Darkness, and Haste gear will be meaningful.
After looking into this further, I want to modify my previous post.
Essentially, while Soul Wording lowers the CD of PW:S to 1s, you still have the GCD to deal with. Therefore, the benefits of haste are the same for PW:S as for other spells.
Mastery is the best stat for shield spam, but is only attainable right now through reforging.
From my estimates, each point in Master gives you around 0.04% increase in PW:S, each point in Haste (before soft cap) gives around 0.03%, and each point in Intellect gives around 0.01%.
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To calculate the value of a spell, the formula (BaseValue + Spellpower * Coefficient) * Modifiers is used.
An easier formula is BaseValue * (1 + spellpower / 8370) * Modifiers.
This doesn't require gathering coefficients.
The coefficients have been changed, and do not depends on casting time anymore.
Thus, what's the logic ?
For all healers (with a few exceptions on one or two spells), spellpower seems to have been changed into a "spellpower rating" system, meaning you need a certain amount to improve ALL your healing spells by 1%.
From what i've tested until now :
Priests need 83.7, no exception
Druids need 87.4, eclosion of lifebloom, retab and hot portion of tranquility do not follow this rule.
Shamans need 89, with the exception of the life totem.
Paladins need 91.2, with the (important) exception of Flash of light.
edit 1
And Intelligence gives 0.276278 crit rating. Conversion seems to be the same for lvl 80 and lvl 85. (166.16 -> 648.91 = 3,9053 factor, 45.906 -> 179.28 = 3,9053 factor.)
edit 2
For lvl 85, priests will need 93.455 spell "rating" to improve base heals by 1%.
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Formula for Mastery can be alternatively expressed as:
$Mastery_{base}=8$
$Mastery_{gained}=Mastery_{gear}/Mastery_{levelCoef}$
$Mastery_{total}=Mastery_{base}+Mastery_{gained}$
which is easier to use for consistent calculation of its effect. (e.g. Discipline build shield absorb values is always increased by Raw_value * Mastery_total)
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If you have the spreadsheet version from 10/14, there is an error in the Divine Aegis variable on the Formula tab.
Cell C84 should be changed from: =Divine_Aegis*0.1*Shield_Discipline_Bonus
to: =Divine_Aegis*0.1*(1+Shield_Discipline_Bonus)
The next update will have this corrected. Thank you Sytax.
An easier formula is BaseValue * (1 + spellpower / 8370) * Modifiers.
This doesn't require gathering coefficients.
For all healers (with a few exceptions on one or two spells), spellpower seems to have been changed into a "spellpower rating" system, meaning you need a certain amount to improve ALL your healing spells by 1%.
I have not had any success with this formula for evaluating correct healing values.
Comparing the numbers at level 80 for 3333 spellpower:
[TABLE]Spell| Coefficient | Rating
Heal| 3347 | 3272
Flash Heal | 6693 | 6544
Greater Heal | 8925 | 8723
Binding Heal | 4292 | 4128[/TABLE]
Can anyone lend validity to this statement?
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IF I look at your own values (first post)
Spell - Base - coef
Shield 3498 0.418
Renew* 1096 0.131
Prayer of Mending* 2661 0.318
Divine Hymn* 3590 0.429
Lightwell* 2576 0.308
Divide the base by 100 and then by the coef, you obtain
83.68
83.66
83.67
83.68
83.63
And this with you own values.
You can easily find that
Base + coef x spell
=
Base x (1 + coef/Base x spell)
And when you have coef / base (amount) spell, then your heal is doubled (100% gain).
1/100 of this gain is 1%.
edit :
the bigger the base heal is, the more precise the coefficient is.
the more spell you may attain to evalue the coefficient, the more precise the coefficient is.
The coefficient you listed are maybe not precise enough, because I suspect coef / Base should seriously be a constant.
(and my 83.7 is not precise either, I do not have tools to be enough accurate).
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2016-02-07 08:35:44
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https://proxies-free.com/elementary-number-theory-what-is-the-remainder-when-12016-22016-32016-20162016-is-divided-by-2017/
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# elementary number theory – What is the remainder when \$1^{2016} + 2^{2016}+ 3^{2016}+…+2016^{2016} \$ is divided by \$2017\$
What is the remainder when $$1^{2016} + 2^{2016}+ 3^{2016}+…+2016^{2016}$$ is divided by $$2017$$
I saw a quesiton in stack-exchange : What is the remainder when \$1^{2016} + 2^{2016} + ⋯ + 2016^{2016}\$ is divided by \$2016\$?
When i checking over it , i thought that what if it is divided $$2017$$ instead of $$2016$$. The answer was easy to me in the first glance because by using phi function ,the summation must have been equal to $$2016$$ and $$2016 mod(2017)=2016$$.
However the answer is equal to $$1759$$ according to python. What am i missing ?
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2021-03-05 09:55:56
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https://msp.org/involve/2019/12-6/p07.xhtml
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#### Vol. 12, No. 6, 2019
Recent Issues
The Journal About the Journal Editorial Board Editors’ Interests Subscriptions Submission Guidelines Submission Form Policies for Authors Ethics Statement ISSN: 1944-4184 (e-only) ISSN: 1944-4176 (print) Author Index Coming Soon Other MSP Journals
Covering numbers of upper triangular matrix rings over finite fields
### Merrick Cai and Nicholas J. Werner
Vol. 12 (2019), No. 6, 1005–1013
##### Abstract
A cover of a finite ring $R$ is a collection of proper subrings $\left\{{S}_{1},\dots ,{S}_{m}\right\}$ of $R$ such that $R={\bigcup }_{i=1}^{m}{S}_{i}$. If such a collection exists, then $R$ is called coverable, and the covering number of $R$ is the cardinality of the smallest possible cover. We investigate covering numbers for rings of upper triangular matrices with entries from a finite field. Let ${\mathbb{F}}_{q}$ be the field with $q$ elements and let ${T}_{n}\left({\mathbb{F}}_{q}\right)$ be the ring of $n×n$ upper triangular matrices with entries from ${\mathbb{F}}_{q}$. We prove that if $q\ne 4$, then ${T}_{2}\left({\mathbb{F}}_{q}\right)$ has covering number $q+1$, that ${T}_{2}\left({\mathbb{F}}_{4}\right)$ has covering number 4, and that when $p$ is prime, ${T}_{n}\left({\mathbb{F}}_{p}\right)$ has covering number $p+1$ for all $n\ge 2$.
##### Keywords
covering number, upper triangular matrix ring, maximal subring
Primary: 16P10
Secondary: 05E15
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2022-01-19 13:01:18
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https://dsp.stackexchange.com/tags/optimization/hot
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# Tag Info
18
L1magic. This is the toolbox associated with the original paper. CompSens. This looks like it's in C, but you could possibly call it with mex -- not sure. Model-based compressive sensing toolbox. Most of the code is plain Matlab code Each folder in the package consists of a CS recovery algorithm based on a particular signal model, and a script that ...
8
I suppose I am answering off-topic here then, but for L1-optimization approaches, I find YALL1 (http://yall1.blogs.rice.edu/) and SPGL1 (http://www.cs.ubc.ca/~mpf/spgl1/) very useful and efficient packages. TFOCS (http://cvxr.com/tfocs/) is probably a bit harder to use, but should be quite flexible. There is also CVX (http://cvxr.com/cvx/) which makes it ...
6
There's a whole area of signal processing dedicated to optimal filtering. In pretty much every case I've seen the filtering problem is formulated with a convex cost function. Here's a freely available book on the subject - Sophocles J. Orfanidis - Optimum Signal Processing.
6
the solution for a sparse recovery problem is given by: $$\text{min} ||x||_0$$ $$\text{s.t} \hspace{2mm} y = Ax$$ The definition of $||x||_0$ is no. of non-zero entries in $x$. This is also called the sparsity of the vector. i.e., we are asking for the sparsest solution $x$, that satisfies $y = Ax$. Consider the simplest case where $... 6 Keep in mind, L1 is not the only approach to compressive sensing. In our research, we've had better success with Approximate Message Passing (AMP). I am defining "success" as lower error, better phase transitions (ability to recover with fewer observations), and lower complexity (both memory and cpu). The Approximate Message Passing algorithm establishes ... 6 As you have already pointed out in your question, it is not possible (without using optimization methods) to compute an exact L2 solution for the frequency domain design problem of IIR filters due to the non-linear relationship between the filter coefficients and the error function. There is, however, a method which can come close and which transforms the ... 5 You may also want to check the Matlab UNLocBox: http://unlocbox.sourceforge.net There are 4 compressive sensing scripts on the demo page: http://unlocbox.sourceforge.net/doc/demos/index.php 5 I found the following in Charles Therrien's "Discrete Random Signals and Statistical Signal Processing" in one of the Appendicies. Say you have the function$Q(a)$you wish to minimize such that$C(a)=0$, where$C(a)$may be complex valued and$amay be a complex vector. The constraint really represents two real-valued constraints. $$C_r(a)=0,\qquad C_i(a)... 5 The Frobenius Norm has multiple equivalent definitions – the useful for error measure is probably this one:$$\left\|M\right\|_\mathrm F = \sqrt{\sum_{p\in M}\left\lvert p\right\rvert^2}$$That's a root square over all pixels. Root mean squares are very useful cost functions, as they describe the power of a signal. 5 That's a trick which you will also find in a DSP context, that's why I choose to provide an answer here. It is related to the Wirtinger derivative, and you can find more details about it in this answer over at math.SE. In practice this trick is often used to compute the extremum (minimum or maximum) of a real-valued function depending on a complex variable (... 5 Let's solve a more general problem (Least Squares with Linear Equality Constraints):$$ \begin{alignat*}{3} \arg \min_{x} & \quad & \frac{1}{2} \left\| A x - b \right\|_{2}^{2} \\ \text{subject to} & \quad & C x = d \end{alignat*} $$The Lagrangian is given by:$$ L \left( x, \nu \right) = \frac{1}{2} \left\| A x - b \right\|_{2}^{2} + {\... 4 As has been referenced in the comments already, you're describing a pulse-amplitude modulation (PAM) signal constellation. The problem, as you've framed it, seems to suggest the AWGN vector channel (where symbols are described using discrete valuess_1$,$s_2$, and so on), in contrast to the waveform channel, where symbols are expressed using waveforms that ... 4 In order to be able to choose an optimal value for the delay$\Delta$it's important to understand how the system works. The purpose of the delay is to decorrelate the desired signal$s(n)$and the signal component$s(n-\Delta)$at the input of the adaptive filter. This means that$\Delta$must be chosen such that the autocorrelation$R_{ss}(k)$of$s(n)$is ... 4 I am sorry I cannot comment your answer due to my low reputation. Gini and your suggested sparsity ratio ($l_1(x)/l_2(x)$) both give me the same value for$\lambda$. But The problem I still see is that I cannot take into account how well the vector is solving the equation$Ax-y$. I would like to combine the residuum$l_1(A\hat{x}-y)$and the sparsity$l_1(\...
4
The fastest blur would be Box Blur. You can implement it using Running Sum. I think Intel FilterBoxBorder works in that manner. If you'd like you can do a few passes of it to approximate the Gaussian Blur. You can also use IIR Filter Coefficients to blur the image quite easily. You may have a look at my project Fast Gaussian Blur.
4
The problem is given by: $$$$\arg \min_{X} \frac{1}{2} \sum_{k} {\left\| {T}_{k} {X}_{:, k} - {Y}_{:, k} \right\|}_{2}^{2} + \lambda {\left\| G X \right\|}_{2, 1} \\ = \arg \min_{X} \frac{1}{2} \sum_{k} {\left\| {T}_{k} {X}_{:, k} - {Y}_{:, k} \right\|}_{2}^{2} + \lambda \sum_{l} {\left\| G {X}_{:, l} \right\|}_{2}$$$$ In the ...
4
It is indeed possible to formulate this setting in terms of matrix-vector products. First, let us re-formulate your $x$ (notice throughout that I use bold letters for vectors and matrices): $$x = \begin{bmatrix}\mathbf{x}_1 & \mathbf{x}_2 & \ldots & \mathbf{x}_8\end{bmatrix}$$ where $\mathbf x_k$ is the $k$ column of $x$. I define the vertically ...
3
For large intensities / large "bins", i.e. "areas for which events are counted and accumulated", Poisson processes lead to nearly Gaussian distributed individual values -- basically, without trying to derive this, I think that's the application of the CLT on a lot of realization of a point process. EDIT: Shameless plug: I really really like the wikipedia ...
3
So, you have a bunch of datapoints of the form (x,y), and considering all of those datapoints together, you have the vectors $x$ and $y$. To do a curve fit, you would like to solve the equation $Aw=y$ for the column vector $w$, which holds the coefficients of your curve fit polynomial. These coefficients are also the weights of the basis vectors in the ...
3
Papers Interpolation by Solving an Optimization Problem. The Chebyshev Center Problem could be thought as Robust Localization Problem. Books Daniel P. Palomar, Yonina C. Eldar - Convex Optimization in Signal Processing and Communications. Stephen Boyd, Lieven Vandenberghe - Convex Optimization. Many of the exercises and examples are from the Signal / ...
3
The quadratic surface is determined by the autocorrelation matrix of the data, which is always positive definite or positive semi-definite. This means that any stationary point is always a minimum. In the worst case, this minimum is not unique if the matrix is singular, but it can never be a saddle point.
3
I stumbled upon this old question and I would like to share my solution. As mentioned in other answers, there is no analytical solution, but the function to be minimized behaves nicely and the optimal value of $\alpha$ can be found easily with a few Newton iterations. There is also a formula to check the optimality of the result. The impulse response of the ...
3
Based in experimental tests with k in range (2 to 100) the best fit (sum squared error) gives a relation of alfa = 1/k^0.865 being k number of samples for MovAvg filter
3
Since $\epsilon$ is a parameter you need to set why not trade it with another parameter you need to set to create an easily solvable problem (Relaxation of the Problem)? You can transform the problem into the following form (${L}_{1}$ Regularized Least Squares): $$\arg \min_{x} \frac{1}{2} \left\| A x - z \right\|^{2} + \lambda \left\| x \right\|_{1} ... 3 The question really depends on f \left( \cdot \right) . Yet in order to show how to use FFT we can even use 1D signals. Let's rewrite the problem:$$ \hat{x} = \arg \min_{x} \frac{1}{2} \left\| K x - b \right\|_{2}^{2} + \frac{\lambda}{2} \left\| f \left( x \right) \right\|_{2}^{2} $$The derivative is given by:$$ g = {K}^{T} \left( K x - b \right) + ...
3
I assume you're after the following optimization problem: \begin{align*} \arg \min_{x} \; & {\left\| x \right\|}_{1} \\ \text{subject to} \; & A x = b \\ & x \succeq 0 \end{align*} This is pretty simple problem if we pay attention fo the fast that given $x \succeq 0$ then ${\left\| x \right\|}_{1} = \boldsymbol{1}^{T} x$. This means ...
3
Hi: I'll try to answer as briefly as possible and only with respect to statistics. not dsp. In statistics, if you have a nice pdf such as the normal distribution, then maximizing the likelihood is equivalent to minimizing the sum of squares of the residuals ( often called errors ). In other cases, where you either have a complicated distribution ( maybe ...
3
SQP is a method for solving smooth (objective and constraint functions are at least twice differentiable) constrained nonlinear optimization problems. It solves a series of quadratic programming problems to converge to a solution to the Karush-Kuhn-Tucker conditions for the constrained optimization problem. IRLS is a method for solving unconstrained ...
3
It can easily solved by the Gradient Descent Framework with one adjustment in order to take care of the ${L}_{1}$ norm term. Since the ${L}_{1}$ norm isn't smooth you need to use the concept of Sub Gradient / Sub Derivative. When you integrate Sub Gradient instead of Gradient into the Gradient Descent Method it becomes the Sub Gradient Method. In the ...
Only top voted, non community-wiki answers of a minimum length are eligible
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2020-10-25 03:16:34
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https://www.physicsforums.com/threads/conservation-of-momentum-in-qft.276949/
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# Conservation of momentum in QFT
1. Dec 3, 2008
### jdstokes
Can conservation of momentum be directly derived from quantum field theory (e.g. QED).
My feeling is this should be true since the Dirac equation reduces to Schrodinger's wave equation in the nonrelativistic limit which is a reflection of Newton's second law, thereby implying conservation of classical momentum.
Yes. You can actually see it without involving the fields. You just assume that there must exist operators that tell you how a Lorentz transformed observer would describe a state that you describe as $\psi$, and when you examine the mathematical properties of those operators, conservation of 4-momentum is one of the results. See chapter 2 in Weinberg's QFT book if you're interested.
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2017-11-21 14:49:53
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http://map.grauw.nl/articles/psg_sample.php
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# Playing samples on the PSG
2005-06-01
Added information about the PSG’s logarithmic volume scale, and how to play 8-bit samples on the PSG. Thanks go to Arturo Ragozini for pointing this out.
I wrote a question about playing samples on the PSG to the MSX Mailinglist once, and Ricardo Bittencourt replied with an explanation of various techniques, which I have quoted below. After that the article continues, describing the logarithmic nature of the PSG volume, and how to utilise it to achieve 8-bit samples on the PSG.
## Ricardo’s explanation
From: Ricardo Bittencourt Vidigal Leitao
To: MSX Mailinglist
Date: monday, june 22nd 1998, 18:58
Subject: Re: A question about the PSG & samples
On Mon, 22 Jun 1998, Laurens Holst wrote:
=== How to replay samples via the PSG??? ===
You have to use an undocumented feature of the PSG. When you select a period of 0 in the register 6, the noise produced is just like the noise when you select 1 in the same register. But this does not apply to the square wave generators. When you write a 0 to registers 0 and 1, what’s happening is that you TURN OFF THE OSCILLATORS. Since the PSG uses active low logic, the signal on the output is set to “1” and doesn’t change with time. Now comes the trick. This “1” is affected by the register 8 (volume register). This way, if you change the value of register 8 very quickly, you can modulate the output and generate a nice 4-bit PCM. This method is used in the game “Aleste 2”.
I made a program to test this feature, it’s called “readwav” and it can play .WAV files of 11 kHz. It can be found at http://www.lsi.usp.br/~ricardo/msx.htm (note: this page apparantly doesn’t exist anymore, it has moved, but I can’t find the program on the new page). The maximum wave size is about 50kb but Walter “Marujo” made a new version that plays files up to 100kb, I’ll try to add it to the home page as soon as possible. There is no source code included, but my original assembly source doesn’t have any comments, the best you can do is disassemble it by hand (it has only 512 bytes anyway). Oh, this 11 kHz is arbitrary; by removing a void-loop in the middle of the code you can reach up to 35 kHz.
Please note this is not the only way to generate samples on MSX1. You can also use the keyclick to generate 1-bit PCM (used in the game “Super Laydock” for example).
Another undocumented feature of the PSG permits a variation on the first method. Most people think the lower bits of register 7 are used to set the volume of a channel to zero. This is not true, the bit actually controls the oscillator. So, by disabling a channel in register 7, you can write any values you want in registers 1 and 0 and still use the initial method (btw, this is used in “Oh Shit”).
A last method is to select a sound with a very low frequency, and change the volume faster than this frequency. This method is not used in any MSX game that I know, since it is very inaccurate. But it’s used a lot in the Sega Master System, in games like “Afterburner”. The Sega Master System, for those who don’t know, is a videogame system heavily based on the MSX. It uses a Z80 and a sound chip with the same characteristics as the AY-3-8910 (but it doesn’t have envelopes).
- Ricardo Bittencourt
Note that the example Ricardo gives uses processor-dependant timing. I myself wrote a sample player for a printer port DAC (‘SiMPL’) once, which ‘calibrated’ itself first before playing a sample, making it work independantly of the processor’s speed. I basically did that by playing a ‘silent’ sample a number of times at different rates during fixed intervals, and measuring at what rate it could fit the desired number of samples within the interval. Actually it was funnier to use a nonsilent sample (a sine wave instead) ^_^.
## Logarithmic volume scale
One thing that this email fails to mention is that the volume control of the PSG operates on a logarithmic scale. Every 2 volume decrements cut the output volume in half. Because of this, you cannot just take the most significant 4 bits of an existing sample and send it to the PSG. Instead, you have to map the values to a logarithmic scale. The formula for this scale is:
Formula (MathML) Formula (non-MathML)
$y={2}^{-\left(\frac{15-n}{2}\right)}$ y = 2^-((15-n)/2)
Where $n$ is the volume, a value from 0 … 15. There is one exception: 0. When the volume is zero, so is the output.
This corresponds to the following table:
PSG volume DAC output 8-bit sample
0 0 0
1 0,0078125 2
2 0,0110485 3
3 0,015625 4
4 0,0220971 6
5 0,03125 8
6 0,0441942 11
7 0,0625 16
8 0,0883883 23
9 0,125 32
10 0,1767767 45
11 0,25 64
12 0,3535534 90
13 0,5 128
14 0,7071068 180
15 1 255
Because of this logarithmic scale, I’d say the actual sample quality of the PSG, even though 4-bits, is really comparable to a 3- or perhaps even 2-bit linear DAC.
Thanks go to Arturo Ragozini for pointing this out.
## Playing better samples on the PSG
However, that same logarithmic scale can also work to our advantage. The PSG mixer in MSX machines is simple, the channels are simply added together, which allows us to combine the power of the three channels. By doing that, we can compensate for the lack of precision at e.g. the 0.5 output value of channel 1 (volume 13) by using the lower ranges of channel 2! For example, if you would want to output 0.51, you would generate an additional output of 0.01 on channel 2 (volume 2) which is then added to the first channel. When combining three channels this way, we can get 608 discrete sample values!
608 values, that’s about 9 bits of sample information. Not too shabby :). Note that the values are not evenly spaced apart so most of them will have deviations compared to a linear scale, ranging from small ones in the lower ranges to bigger ones in the top ranges. Because of that, the top range is much less useful, but it would probably distort anyway because of the high amplitude.
I created a C# program to calculate what combinations of volumes would have to be used, and to brute force check over what range the deviations would be the least. Without boring you with all the specific details, I have found that when running over a range from 0 to 1.328 (out of 3), the signal-to-noise ratio is the best. The resulting lookup table for that range, mapping 8-bit sample values to PSG channel volumes, is included in the replay routine below:
```;
;PSG sample replay routine
;
;de = sample length
;
exx
ld c,#A1
ld d,0
exx
Loop:
ld a,(hl)
inc hl
exx
ld e,a
ld hl,PSG_SAMPLE_TABLE
ld b,(hl)
inc h
ld e,(hl)
inc h
ld h,(hl)
ld a,8
out (#A0),a ;play as fast as possible
inc a
out (c),b
out (#A0),a
out (c),e
inc a
out (#A0),a
out (c),h
ld b,8 ;timing wait loop
WaitLoop:
djnz WaitLoop
exx
dec de
ld a,d
or e
jp nz,Loop
ret
PSG_SAMPLE_TABLE:
DB 00,01,02,03,04,03,05,03,04,05,06,06,05,06,06,06
DB 06,06,07,06,07,08,08,08,07,07,09,07,09,09,08,08
DB 09,09,08,09,09,09,09,09,10,10,10,10,09,09,10,10
DB 10,10,09,10,11,11,11,11,11,11,11,11,10,10,10,11
DB 11,11,11,11,11,11,11,12,11,11,12,12,11,12,11,12
DB 12,12,12,11,12,11,12,12,12,12,11,12,12,12,12,11
DB 12,13,12,13,11,13,13,13,13,13,13,11,13,13,13,13
DB 13,13,13,12,13,13,13,12,12,13,12,13,13,13,13,13
DB 13,12,13,13,13,13,13,13,13,14,13,13,14,14,14,14
DB 14,14,13,14,14,13,14,14,14,14,14,14,13,14,14,14
DB 14,14,14,13,14,14,13,14,14,13,13,14,14,14,14,14
DB 14,14,14,14,13,14,14,13,14,14,14,14,14,14,13,14
DB 14,14,15,14,15,15,15,15,15,15,15,15,15,15,15,15
DB 14,15,15,15,15,15,15,14,15,15,15,15,15,15,15,15
DB 15,15,15,15,15,15,15,15,15,15,15,14,15,14,14,14
DB 14,14,15,15,14,15,15,14,15,15,15,15,15,15,15,14
DB 00,00,00,00,00,02,00,02,02,03,01,02,04,04,03,04
DB 04,05,04,05,05,02,03,04,06,06,01,06,02,03,06,07
DB 05,06,07,06,06,06,07,06,04,04,05,06,08,07,06,06
DB 07,06,08,07,03,04,03,04,04,05,05,05,08,09,09,07
DB 07,07,08,07,08,08,08,02,08,09,03,05,09,05,08,06
DB 06,07,06,10,07,09,08,07,08,08,09,08,08,09,08,10
DB 09,00,08,01,10,02,03,04,04,05,06,10,06,06,06,07
DB 06,07,07,10,08,08,07,11,11,08,11,08,09,09,09,08
DB 09,11,09,09,10,10,10,10,10,00,10,09,02,02,04,03
DB 04,04,11,05,05,11,07,07,07,07,07,08,10,08,08,08
DB 08,08,09,11,09,09,12,08,09,12,11,09,10,10,09,10
DB 10,10,10,09,11,10,10,12,10,10,11,11,11,10,12,11
DB 11,11,00,11,01,02,03,04,03,04,04,05,05,05,06,07
DB 12,07,07,07,08,07,08,12,08,08,08,09,08,09,09,09
DB 08,09,09,09,09,10,10,09,10,10,10,13,09,13,13,13
DB 13,13,10,11,13,11,10,13,11,11,11,11,11,10,10,12
DB 00,00,00,00,00,00,00,01,01,00,00,00,01,00,02,02
DB 03,02,01,04,01,01,01,01,03,04,00,05,01,01,04,01
DB 01,00,04,02,03,04,01,05,01,02,01,00,02,06,03,04
DB 01,05,06,04,00,00,02,02,03,02,03,04,06,02,03,02
DB 03,04,00,05,02,03,04,00,05,00,02,00,03,02,07,01
DB 02,00,04,00,03,07,00,05,02,03,08,04,05,00,06,07
DB 03,00,07,00,08,01,01,01,02,01,00,09,02,03,04,01
DB 05,03,04,07,01,02,06,01,02,05,04,06,02,03,04,07
DB 05,07,06,06,00,01,02,03,04,00,05,08,00,01,00,02
DB 02,03,00,03,04,03,00,01,02,03,04,00,09,02,03,04
DB 04,05,00,08,02,03,00,07,05,03,09,06,00,01,07,03
DB 04,04,05,08,10,06,06,08,07,07,00,00,01,08,09,04
DB 05,05,00,06,00,00,00,00,02,02,03,02,03,04,03,00
DB 01,02,03,04,00,05,02,06,04,04,05,00,06,02,03,04
DB 07,05,05,06,06,00,01,07,03,04,04,00,08,02,03,04
DB 04,05,07,00,06,01,08,07,04,05,05,06,06,09,09,11
```
In these values, each block of bytes corresponds to a PSG channel, and each index from 0…255 in one of these blocks bytes corresponds to a sample value point. The total maximum volume you should be able to generate with these is about 30% more than a single channel outputting at maximum volume. The speed can be optimized a little more by aligning the table to a multiple of 256 address, if you need it.
That should give you enough of a start to create nice PSG samples. For completeness’ sake, here is a link to download the C# program I created to find the optimal range with the least amount of errors: PSG_Sample.cs. If you want to output at a different volume, you can recalculate the table with it.
Again, thanks go to Arturo Ragozini for thinking this through.
|
2018-03-18 04:18:56
|
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|
https://codereview.stackexchange.com/questions/164961/saving-wpf-datagrid-changes-to-entity-framework
|
# Saving WPF DataGrid changes to Entity Framework
I feel like I had to invent a whole new way to do this because I couldn't find anything that did this the way I wanted to...
The basic process involves reading changes to cells and capturing information about those changes as they relate to the actual Entities in question and then persisting them to the database.
Here's the XAML. I didn't use databinding because I don't quite understand it yet and I'm on a bit of a deadline for this one so that's item number 1 on my to do list when I get a break.
Note that this project has been my first every foray into WPF. I've dabbled a bit in the past but I've really pushed myself to do things properly here... As properly as I know how anyway.
<DataGrid Name="dgUsers" Height="150"
CellEditEnding="dgUsers_CellEditEnding">
</DataGrid>
<StackPanel Name="spActions" Background="#2d2d30" Width="500" Height="50" Orientation="Horizontal" FlowDirection="RightToLeft">
<Button Name="btnCancel" Click="btnClose_Click" Content="Cancel" Style="{StaticResource CancelButton}" Height="30" Width="66" Margin="5,3,5,5"></Button>
<Button Name="btnSave" Click="btnSave_Click" Content="Save" Style="{StaticResource RoundedButtonGreen}" Height="30" Width="66" Margin="5,3,5,5"></Button>
<Label Name="lblResult"></Label>
</StackPanel>
Here's the C# codebehind:
Generate an entity from the changes that were made:
private List<User> UpdatedUsers { get; set; }
/// <summary>
/// Maintains a list of all changes made to all entities on the datagrid
/// </summary>
/// <param name="sender"></param>
/// <param name="e"></param>
private void dgUsers_CellEditEnding(object sender, DataGridCellEditEndingEventArgs e)
{
if (UpdatedUsers == null) UpdatedUsers= new List<User>();
var _User = (User)e.Row.Item;
var Element = (TextBox)e.EditingElement;
if (String.Equals((string)e.Column.Header, "EmailAddress"))
{
User.EmailAddress = Element.Text;
}
else if (String.Equals((string)e.Column.Header, "Password"))
{
User.Password = Element.Text;
}
UpdatedUsers.Add(User);
}
From here the user clicks on a save button which loops through the changes captured in this list and updates the relevant records in the database:
private void btnSave_Click(object sender, RoutedEventArgs e)
{
try
{
ApplicationDbContext Context = new ApplicationDbContext();
var Users = Context.Users.Where(x => x.CompanyId == (int)cmbCompanies.SelectedValue).ToList();
foreach (User User in Users)
{
foreach (User U in UpdatedUsers)
{
if (U.Id == User.Id)
{
User.EmailAddress = U.EmailAddress;
User.Password = U.Password;
}
Context.SaveChanges();
}
}
}
catch (Exception ex)
{
lblResult.Content = ex.Message;
}
}
• The xaml doesn't use databinding, no. I've gone through tutorials but I don't understand how that works just yet so rather than use up the project development time I decided I'd quickly whip this together Jun 5 '17 at 8:44
• Didn't even take me half an hour to do this though. I'll definitely replace this with databinding in due time but for now deadline is fast approaching so I figured I'd stick to something that I knew would work without much head scratching Jun 5 '17 at 8:50
• @Ortund I'm not sure what kind of review you are expecting to get then. "This code violates everything WPF stands for" is probably not very helpful, but that's how it looks. And using proper naming conventions won't change that. You need to get at least basic understanding of MVVM before using WPF in production code. Deadline is a poor excuse, IMHO: data binding is not rocket science, it's pretty basic stuff, that can be quickly explained/learned if you already have a basic knowledge of WPF (e.g. see codeproject.com/Articles/165368/WPF-MVVM-Quick-Start-Tutorial). Jun 5 '17 at 9:26
• So rather than critiquing the code on the basis that it isn't "what WPF stands for", how about evaluating it on its merits? I know that's a lot to ask in this world but what do you say we actually make some sort of effort for a change? Jun 5 '17 at 9:50
• @Ortund, I used some harsh wording and I'm sorry if I offended you. It was not my intention. Jun 5 '17 at 10:48
## 1 Answer
Other issues...
• C# variable naming convention is camelCase. And definitely don't start a local's name with an underscore, as underscores are understood to prefix instance variables.
• There's no separation of concerns here. Your UI code is talking to the database. That's not the responsibility of the UI.
• I assume ApplicationDbContext inherits from DbContext which is implements IDisposable. You're not disposing it. Use it in a using block. (I know nothing about Entity Framework so don't know if there's a good reason not to do so.)
• Well thank you :) for being the first genuinely helpful person who's participated on this post. I was sure I read something about conventions on MSDN that supported the way I do it (perhaps I was wrong) but you're absolutely right that I should be using the context in a using block. Jun 5 '17 at 20:51
• @Ortund I can't find the link I had to the naming conventions; current MSDN guidance doesn't appear to say anything about private fields and variables, but check out the .NET section in the document here 1code.codeplex.com/downloads/get/357518?releaseId=84683 where it is covered.
– 404
Jun 5 '17 at 22:40
|
2022-01-22 18:43:22
|
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|
https://xgc.pppl.gov/html/cmake_changes.html
|
# Adding a new source file¶
Whenever you create a new source file you need to tell CMake what target it belongs to. We maintain lists of source files that are compiled into each of our libraries and executables. For example, see the core_SRCS variable in CMakeLists.txt.
# Adding support for another HPC facility¶
• Create a file CMake/find_dependencies_<name>.cmake. This will contain the location of XGC’s dependencies at this HPC facility.
• Add the facility name to the list of possible values of XGC_PLATFORM in our top-level CMakeLists.txt file.
# Adding a new configuration option¶
• In our top-level CMakeLists.txt, use the option command to create a new boolean option in the XGC build system.
|
2021-08-02 09:32:07
|
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|
https://pypi.org/project/pyTCTL/
|
A package for Timed CTL model checking in Python
# pyTCTL
A package for TCTL model checking in Python
The initial implementation makes use of NetworkX's DiGraph for the representation of Kripke structures.
The provided algorithms implement Lepri et al.'s approach to continuous TCTL model checking. Thus, this package also implements pointwise TCTL model checking.
The code was originally developed for the CREST project, but it seems better to extract the code into its own library.
## Similar Projects
@albertocasagrande is working on a similar library for (untimed) CTL/LTL/CTL* model checking called pyModelChecking.
His approach shares some of the ideas of this project. However, it seems that he chooses to implement Kripke structures directly (no networking library) and adds some features that I probably won't implement (e.g. a text parser). On the other hand, as far as I can see there's only one algorithm hardcoded for each temporal logic. In comparison, I plan to add the possibility to add and choose between search heuristics.
I will keep following his project. Maybe we can merge forces once I know more clearly which path the pyTCTL project is going to take.
## Project details
Uploaded source
Uploaded py3
|
2023-03-26 07:14:39
|
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|
https://socratic.org/questions/how-do-you-graph-using-slope-and-intercept-of-2x-3y-1
|
# How do you graph using slope and intercept of 2x+3y= -1?
See below
#### Explanation:
The first thing I'd do is to change the equation from the current standard form and put it into slope-intercept form. We do that by solving for $y$:
$2 x + 3 y = - 1$
$3 y = - 2 x - 1$
$y = - \frac{2}{3} x - \frac{1}{3}$
We're now in slope-intercept form, where
$y = m x + b , m = \text{slope" and b = y-"intercept}$
And so in our question, $m = - \frac{2}{3}$ and $b = - \frac{1}{3}$
Let's first graph the $y$-intercept. That's at $\left(0 , - \frac{1}{3}\right)$:
graph{((x-0)^2+(y+1/3)^2-.1)=0}
Now let's plot a second point.
$m = \frac{\Delta y}{\Delta x} = \text{rise"/"run}$
Our $m = - \frac{2}{3}$. For every 2 that we move up, we move 3 to the left (I'm dealing with the negative sign by having us move left - with a positive slope we'd move to the right). We can start from our first point and move in that way, and so our second point can be found by writing:
$\left(0 - 3 , - \frac{1}{3} + 2\right) = \left(- 3 , \frac{5}{3}\right)$
Let's plot that:
graph{((x-0)^2+(y+1/3)^2-.1)((x+3)^2+(y-5/3)^2-.1)=0}
And now connect the two points with a line:
graph{((x-0)^2+(y+1/3)^2-.1)((x+3)^2+(y-5/3)^2-.1)(2x+3y+1)=0}
|
2020-10-28 05:38:30
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https://proofwiki.org/wiki/Category:Square-Free_Integers
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# Category:Square-Free Integers
This category contains results about integers which are square-free.
Let $n \in \Z$.
Then $n$ is square-free if and only if $n$ has no divisor which is the square of a prime.
That is, if and only if the prime decomposition $n = p_1^{k_1} p_2^{k_2} \ldots p_r^{k_r}$ is such that:
$\forall i: 1 \le i \le r: k_i = 1$
## Pages in category "Square-Free Integers"
The following 6 pages are in this category, out of 6 total.
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2019-08-21 10:17:46
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http://www.numdam.org/item/M2AN_2012__46_1_81_0/
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On the stability of Bravais lattices and their Cauchy-Born approximations
ESAIM: Mathematical Modelling and Numerical Analysis - Modélisation Mathématique et Analyse Numérique, Volume 46 (2012) no. 1, p. 81-110
We investigate the stability of Bravais lattices and their Cauchy-Born approximations under periodic perturbations. We formulate a general interaction law and derive its Cauchy-Born continuum limit. We then analyze the atomistic and Cauchy-Born stability regions, that is, the sets of all matrices that describe a stable Bravais lattice in the atomistic and Cauchy-Born models respectively. Motivated by recent results in one dimension on the stability of atomistic/continuum coupling methods, we analyze the relationship between atomistic and Cauchy-Born stability regions, and the convergence of atomistic stability regions as the cell size tends to infinity.
DOI : https://doi.org/10.1051/m2an/2011014
Classification: 35Q74, 49K40, 65N25, 70J25, 70C20
Keywords: Bravais lattice, Cauchy-Born model, stability
@article{M2AN_2012__46_1_81_0,
author = {Hudson, Thomas and Ortner, Christoph},
title = {On the stability of Bravais lattices and their Cauchy-Born approximations},
journal = {ESAIM: Mathematical Modelling and Numerical Analysis - Mod\'elisation Math\'ematique et Analyse Num\'erique},
publisher = {EDP-Sciences},
volume = {46},
number = {1},
year = {2012},
pages = {81-110},
doi = {10.1051/m2an/2011014},
zbl = {1291.35388},
mrnumber = {2846368},
language = {en},
url = {http://www.numdam.org/item/M2AN_2012__46_1_81_0}
}
Hudson, Thomas; Ortner, Christoph. On the stability of Bravais lattices and their Cauchy-Born approximations. ESAIM: Mathematical Modelling and Numerical Analysis - Modélisation Mathématique et Analyse Numérique, Volume 46 (2012) no. 1, pp. 81-110. doi : 10.1051/m2an/2011014. http://www.numdam.org/item/M2AN_2012__46_1_81_0/
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[6] M. Dobson, M. Luskin and C. Ortner, Sharp stability estimates for the force-based quasicontinuum approximation of homogeneous tensile deformation. Multiscale Model. Simul. 8 (2010) 782-802. | MR 2609639 | Zbl 1225.82009
[7] W.E and P. Ming, Cauchy-Born rule and the stability of crystalline solids: static problems. Arch. Ration. Mech. Anal. 183 (2007) 241-297. | MR 2278407 | Zbl 1106.74019
[8] G. Friesecke and F. Theil, Validity and failure of the Cauchy-Born hypothesis in a two-dimensional mass-spring lattice. J. Nonlinear Sci. 12 (2002) 445-478. | MR 1923388 | Zbl 1084.74501
[9] V.S. Ghutikonda and R.S. Elliott, Stability and elastic properties of the stress-free b2 (cscl-type) crystal for the morse pair potential model. J. Elasticity 92 (2008) 151-186. | MR 2417286 | Zbl 1147.74010
[10] M. Giaquinta, Introduction to regularity theory for nonlinear elliptic systems. Lectures in Mathematics ETH Zürich. Birkhäuser Verlag, Basel (1993). | MR 1239172 | Zbl 0786.35001
[11] O. Gonzalez and A.M. Stuart, A first course in continuum mechanics. Cambridge Texts in Applied Mathematics. Cambridge University Press, Cambridge (2008). | MR 2378978 | Zbl 1143.74001
[12] C. Kittel, Introduction to Solid State Physics, 7th ed. John Wiley & Sons, New York, Chichester (1996). | Zbl 0052.45506
[13] R. Kress, Linear integral equations, Applied Mathematical Sciences 82. Springer-Verlag, 2nd edition, New York (1999). | MR 1723850 | Zbl 0920.45001
[14] L.D. Landau and E.M. Lifshitz, Theory of elasticity, Course of Theoretical Physics 7. Translated by J.B. Sykes and W.H. Reid. Pergamon Press, London (1959). | MR 106584 | Zbl 0146.22405
[15] X.H. Li and M. Luskin, An analysis of the quasi-nonlocal quasicontinuum approximation of the embedded atom model. arXiv:1008.3628v4.
[16] X.H. Li and M. Luskin, A generalized quasi-nonlocal atomistic-to-continuum coupling method with finite range interaction. arXiv:1007.2336. | MR 2911393 | Zbl 1241.82078
[17] M.R. Murty, Problems in analytic number theory, Graduate Texts in Mathematics 206. Springer, 2nd edition, New York (2008). Readings in Mathematics. | MR 2376618 | Zbl 1190.11001
[18] C. Ortner, A priori and a posteriori analysis of the quasinonlocal quasicontinuum method in 1D. Math. Comput. 80 (2011) 1265-1285 | MR 2785458 | Zbl pre05918690
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2020-02-25 22:39:31
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https://solvedlib.com/n/3-flnd-rdr-use-a-lor-the-conslant-ntedratlof-amp-6-using,13101756
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# (3) Flnd rdr (Use € lor the consLant ntedratlof~* &=(6) Using Intcgnbon bv p4lts
###### Question:
(3) Flnd rdr (Use € lor the consLant ntedratlo f~* &= (6) Using Intcgnbon bv p4lts
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2022-07-06 17:01:13
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https://gateoverflow.in/336826/nielit-2017-dec-scientific-assistant-a-section-b-20
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1,288 views
Total number of simple graphs that can be drawn using six vertices are:
1. $2^{15}$
2. $2^{14}$
3. $2^{13}$
4. $2^{12}$
### 1 comment
No of simple graph = 2^ (n(n-1)/2)
For six vertices = 2^15
Option A is correct
A graph with no loops and no parallel edges is called a simple graph.
• The maximum number of edges possible in a single graph with ‘n’ vertices is $nC_{2}$ where $nC_{2} = n(n – 1)/2$.
• The number of simple graphs possible with ‘n’ vertices =$2^{nc_{2}} = 2^{n(n-1)/2}.$
graph with no edge=$\binom{n(n-1)/2)}{0}$
graph with 1 edge=$\binom{n(n-1)/2)}{1}$
graph with 2 edge=$\binom{n(n-1)/2)}{2}$
graph with 3 edge=$\binom{n(n-1)/2)}{3}$
.
.
.
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graph with $n(n-1)/2$ edge=$\binom{n(n-1)/2)}{n(n-1/2}$
So total number of graph possible=$\binom{n(n-1)/2)}{0}$+$\binom{n(n-1)/2)}{1}$+$\binom{n(n-1)/2)}{2}$+....+$\binom{n(n-1)/2)}{n(n-1/2}$=$2^{n(n-1)/2}$
=$2^{n(n-1)/2}$
=$2^{6(6-1)/2}$
=$2^{6*5/2}$
=$2^{15}$
With n vertex we have max of $\frac{n*(n-1)}{2}$ edges. (i.e complete graph) => we have max 15 edges
Now for each edge we have two possibility , It can be included in graph or not included in the graph
So 2 * 2 * 2 ….* 2 15 times => $2^{15}$ possible simple graph
by
For six vertices = 2^15
1 vote
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2022-12-01 19:30:27
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https://bitbucket.org/pypy/extradoc/src/6a44904cba6f7bd7534cf814b028cd28433b0c5b/blog/draft/za-sprint-report.rst?at=extradoc
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# extradoc / blog / draft / za-sprint-report.rst
Hello.
We're about to finish a PyPy sprint in Cape Town, South Africa that was one of the smallest done so far, only having Armin Rigo and Maciej Fijalkowski with Alex Gaynor joining briefly at the beginning, however also one of the longest, lasting almost 3 weeks. The sprint theme seems to be predominantly "no new features" and "spring cleaning". We overall removed about 20k lines of code in the PyPy source tree. The breakdown of things done and worked on:
• We killed SomeObject support in annotation and rtyper. This is a modest code saving, however, it reduces the complexity of RPython and also, hopefully, improves compile errors from RPython. We're far from done on the path to have comprehensible compile-time errors, but the first step is always the hardest :)
• We killed some magic in specifying the interface between builtin functions and Python code. It used to be possible to write builtin functions like this:
def f(space, w_x='xyz'):
which will magically wrap 'xyz' into a W_StringObject. Right now, instead, you have to write:
@unwrap_spec(w_x=WrappedDefault('xyz'))
def f(space, w_x):
which is more verbose, but less magical.
• We killed the CExtModuleBuilder which is the last remaining part of infamous extension compiler that could in theory build C extensions for CPython in RPython. This was never working very well and the main part was killed long ago.
• We killed various code duplications in the C backend.
• We killed microbench and a bunch of other small-to-medium unused directories.
• We killed llgraph JIT backend and rewrote it from scratch. Now the llgraph backend is not translatable, but this feature was rarely used and caused a great deal of complexity.
• We progressed on continulet-jit-3 branch, up to the point of merging it into result-in-resops branch, which also has seen a bit of progress.
Purpose of those two branches:
• continulet-jit-3: enable stackless to interact with the JIT by killing global state while resuming from the JIT into the interpreter. This has multiple benefits. For example it's one of the stones on the path to enable STM for PyPy. It also opens new possibilities for other optimizations including Python-Python calls and generators.
• result-in-resops: the main goal is to speed up the tracing time of PyPy. We found out the majority of time is spent in the optimizer chain, which faces an almost complete rewrite. It also simplifies the storage of the operations as well as the number of implicit invariants that have to be kept in mind while developing.
• We finished and merged the excellent work by Ronan Lamy which makes the flow object space (used for abstract interpretation during RPython compilation) independent from the Python interpreter. This means we've achieved an important milestone on the path of separating the RPython translation toolchain from the PyPy Python interpreter.
Cheers, fijal & armin
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2014-07-14 00:43:14
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https://courses.lumenlearning.com/ivytech-collegealgebra/chapter/section-exercises-28/
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## Section Exercises
1. Can we add any two matrices together? If so, explain why; if not, explain why not and give an example of two matrices that cannot be added together.
2. Can we multiply any column matrix by any row matrix? Explain why or why not.
3. Can both the products $AB$ and $BA$ be defined? If so, explain how; if not, explain why.
4. Can any two matrices of the same size be multiplied? If so, explain why, and if not, explain why not and give an example of two matrices of the same size that cannot be multiplied together.
5. Does matrix multiplication commute? That is, does $AB=BA?$ If so, prove why it does. If not, explain why it does not.
For the following exercises, use the matrices below and perform the matrix addition or subtraction. Indicate if the operation is undefined.
$A=\left[\begin{array}{cc}1& 3\\ 0& 7\end{array}\right],B=\left[\begin{array}{cc}2& 14\\ 22& 6\end{array}\right],C=\left[\begin{array}{cc}1& 5\\ 8& 92\\ 12& 6\end{array}\right],D=\left[\begin{array}{cc}10& 14\\ 7& 2\\ 5& 61\end{array}\right],E=\left[\begin{array}{cc}6& 12\\ 14& 5\end{array}\right],F=\left[\begin{array}{cc}0& 9\\ 78& 17\\ 15& 4\end{array}\right]$
6. $A+B$
7. $C+D$
8. $A+C$
9. $B-E$
10. $C+F$
11. $D-B$
For the following exercises, use the matrices below to perform scalar multiplication.
$A=\left[\begin{array}{rr}\hfill 4& \hfill 6\\ \hfill 13& \hfill 12\end{array}\right],B=\left[\begin{array}{rr}\hfill 3& \hfill 9\\ \hfill 21& \hfill 12\\ \hfill 0& \hfill 64\end{array}\right],C=\left[\begin{array}{rrrr}\hfill 16& \hfill 3& \hfill 7& \hfill 18\\ \hfill 90& \hfill 5& \hfill 3& \hfill 29\end{array}\right],D=\left[\begin{array}{rrr}\hfill 18& \hfill 12& \hfill 13\\ \hfill 8& \hfill 14& \hfill 6\\ \hfill 7& \hfill 4& \hfill 21\end{array}\right]$
12. $5A$
13. $3B$
14. $-2B$
15. $-4C$
16. $\frac{1}{2}C$
17. $100D$
For the following exercises, use the matrices below to perform matrix multiplication.
$A=\left[\begin{array}{rr}\hfill -1& \hfill 5\\ \hfill 3& \hfill 2\end{array}\right],B=\left[\begin{array}{rrr}\hfill 3& \hfill 6& \hfill 4\\ \hfill -8& \hfill 0& \hfill 12\end{array}\right],C=\left[\begin{array}{rr}\hfill 4& \hfill 10\\ \hfill -2& \hfill 6\\ \hfill 5& \hfill 9\end{array}\right],D=\left[\begin{array}{rrr}\hfill 2& \hfill -3& \hfill 12\\ \hfill 9& \hfill 3& \hfill 1\\ \hfill 0& \hfill 8& \hfill -10\end{array}\right]$
18. $AB$
19. $BC$
20. $CA$
21. $BD$
22. $DC$
23. $CB$
For the following exercises, use the matrices below to perform the indicated operation if possible. If not possible, explain why the operation cannot be performed.
$A=\left[\begin{array}{rr}\hfill 2& \hfill -5\\ \hfill 6& \hfill 7\end{array}\right],B=\left[\begin{array}{rr}\hfill -9& \hfill 6\\ \hfill -4& \hfill 2\end{array}\right],C=\left[\begin{array}{rr}\hfill 0& \hfill 9\\ \hfill 7& \hfill 1\end{array}\right],D=\left[\begin{array}{rrr}\hfill -8& \hfill 7& \hfill -5\\ \hfill 4& \hfill 3& \hfill 2\\ \hfill 0& \hfill 9& \hfill 2\end{array}\right],E=\left[\begin{array}{rrr}\hfill 4& \hfill 5& \hfill 3\\ \hfill 7& \hfill -6& \hfill -5\\ \hfill 1& \hfill 0& \hfill 9\end{array}\right]$
24. $A+B-C$
25. $4A+5D$
26. $2C+B$
27. $3D+4E$
28. $C - 0.5D$
29. $100D - 10E$
For the following exercises, use the matrices below to perform the indicated operation if possible. If not possible, explain why the operation cannot be performed. (Hint: ${A}^{2}=A\cdot A$)
$A=\left[\begin{array}{rr}\hfill -10& \hfill 20\\ \hfill 5& \hfill 25\end{array}\right],B=\left[\begin{array}{rr}\hfill 40& \hfill 10\\ \hfill -20& \hfill 30\end{array}\right],C=\left[\begin{array}{rr}\hfill -1& \hfill 0\\ \hfill 0& \hfill -1\\ \hfill 1& \hfill 0\end{array}\right]$
30. $AB$
31. $BA$
32. $CA$
33. $BC$
34. ${A}^{2}$
35. ${B}^{2}$
36. ${C}^{2}$
37. ${B}^{2}{A}^{2}$
38. ${A}^{2}{B}^{2}$
39. ${\left(AB\right)}^{2}$
40. ${\left(BA\right)}^{2}$
For the following exercises, use the matrices below to perform the indicated operation if possible. If not possible, explain why the operation cannot be performed. (Hint: ${A}^{2}=A\cdot A$)
$A=\left[\begin{array}{rr}\hfill 1& \hfill 0\\ \hfill 2& \hfill 3\end{array}\right],B=\left[\begin{array}{rrr}\hfill -2& \hfill 3& \hfill 4\\ \hfill -1& \hfill 1& \hfill -5\end{array}\right],C=\left[\begin{array}{rr}\hfill 0.5& \hfill 0.1\\ \hfill 1& \hfill 0.2\\ \hfill -0.5& \hfill 0.3\end{array}\right],D=\left[\begin{array}{rrr}\hfill 1& \hfill 0& \hfill -1\\ \hfill -6& \hfill 7& \hfill 5\\ \hfill 4& \hfill 2& \hfill 1\end{array}\right]$
41. $AB$
42. $BA$
43. $BD$
44. $DC$
45. ${D}^{2}$
46. ${A}^{2}$
47. ${D}^{3}$
48. $\left(AB\right)C$
49. $A\left(BC\right)$
For the following exercises, use the matrices below to perform the indicated operation if possible. If not possible, explain why the operation cannot be performed. Use a calculator to verify your solution.
$A=\left[\begin{array}{rrr}\hfill -2& \hfill 0& \hfill 9\\ \hfill 1& \hfill 8& \hfill -3\\ \hfill 0.5& \hfill 4& \hfill 5\end{array}\right],B=\left[\begin{array}{rrr}\hfill 0.5& \hfill 3& \hfill 0\\ \hfill -4& \hfill 1& \hfill 6\\ \hfill 8& \hfill 7& \hfill 2\end{array}\right],C=\left[\begin{array}{rrr}\hfill 1& \hfill 0& \hfill 1\\ \hfill 0& \hfill 1& \hfill 0\\ \hfill 1& \hfill 0& \hfill 1\end{array}\right]$
50. $AB$
51. $BA$
52. $CA$
53. $BC$
54. $ABC$
For the following exercises, use the matrix below to perform the indicated operation on the given matrix.
$B=\left[\begin{array}{rrr}\hfill 1& \hfill 0& \hfill 0\\ \hfill 0& \hfill 0& \hfill 1\\ \hfill 0& \hfill 1& \hfill 0\end{array}\right]$
55. ${B}^{2}$
56. ${B}^{3}$
57. ${B}^{4}$
58. ${B}^{5}$
59. Using the above questions, find a formula for ${B}^{n}$. Test the formula for ${B}^{201}$ and ${B}^{202},\text{}$ using a calculator.
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2019-09-18 01:03:12
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https://brilliant.org/problems/inverse-roots/
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# Appropriate Substitution
Algebra Level 3
$x^3 - 6x^2 + 12x = 8$
If $$a,b,c$$ are the roots of the cubic equation above. And given that for coprime positive integers $$m,n$$, we have:
$\left ( \frac 1 a + \frac 1 b + \frac 1 c \right )^2 = \frac m n$
What is the value of $$16(m-n)$$?
×
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2018-06-20 06:02:48
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https://physics.stackexchange.com/questions/304586/can-acceleration-be-zero-in-one-co-ordinate-system-but-non-zero-in-another-syste
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# Can acceleration be zero in one co-ordinate system but non-zero in another system?
A particle is moving in straight line parallel to x-axis, with uniform velocity (along y=2, lets assume). If write the equation of motion and calculate velocity in polar co-ordinates we see cos and sine dependence and hence acceleration is non-zero. But what is the physical significance of this. Non-zero acceleration means there has to be a force (Law of Inertia), but this particle is moving in straight line with constant velocity along y=2. Can someone please explain what I am missing ?
• Can you write out the equations in their polar form in the question here? Jan 12 '17 at 12:16
• Velocity in polar coordinates for this particle is: v = u cos θ rˆ − u sin θ θˆ (Kleppner book Example 1.15). So acceleration is non-zero. Whereas, v=xi in cartesian coordinates. Pardon me if I'm missing something. Jan 12 '17 at 17:10
• How do you get from the expression for $v$ to the statement that acceleration is non-zero? The conclusion does not seem obvious to me. Jan 12 '17 at 18:21
• It was just cosine/sine dependence of v, that made me think of it as time varying. But the answer below made me work it out and I realized what I missed. Jan 13 '17 at 1:48
The unit vectors in the r and $\theta$ directions are functions of $\theta$, and $\theta$ is a function of time. What are the derivatives of the $\vec{i}_r$ and $\vec{i}_{\theta}$ with respect to $\theta$? (When I take the derivative of your velocity vector equation, I get 4 terms, and they cancel out in pairs, to give me zero acceleration)
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2021-10-25 17:20:49
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http://mathhelpforum.com/calculus/98628-area-bounded-two-curves.html
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# Math Help - area bounded by two curves
1. ## area bounded by two curves
Ive found the antiderivative to the two functions and im at the last part, where you have to find the total area. This is what it looks like so far...
[ -(x^3)/3 -(x^2)/2 +2x ] theres a small 1 at the top of the last square bracket and a small -2 at the bottom (interval of [-1,2])
Which numbers do I sub in to get the final answer??
I keep thinking its just the first one, but my answer keeps coming out wrong. I keep getting 7/6, but my textbook says the final answer is 9/2.
Where am I messing up?
2. Originally Posted by linearalgebra
Ive found the antiderivative to the two functions and im at the last part, where you have to find the total area. This is what it looks like so far...
[ -(x^3)/3 -(x^2)/2 +2x ] theres a small 1 at the top of the last square bracket and a small -2 at the bottom (interval of [-1,2])
Which numbers do I sub in to get the final answer??
I keep thinking its just the first one, but my answer keeps coming out wrong. I keep getting 7/6, but my textbook says the final answer is 9/2.
Where am I messing up?
What were the two original functions?
3. Originally Posted by Chris L T521
What were the two original functions?
y= 2-(x^2)
y=x
4. Originally Posted by linearalgebra
y= 2-(x^2)
y=x
First, find the intersection points
$x=2-x^2\implies x^2+x-2=0\implies\left(x+2\right)\left(x-1\right)=0\implies x=1$ or $x=-2$
So, the area is $\int_{-2}^1\left[\left(2-x^2\right)-x\right]\,dx=\int_{-2}^1 2-x-x^2\,dx=\left.\left[2x-\tfrac{1}{2}x^2-\tfrac{1}{3}x^3\right]\right|_{-2}^1$ $=\left(2-\tfrac{1}{2}-\tfrac{1}{3}\right)-\left(-4-2+\tfrac{8}{3}\right)=5-\tfrac{1}{2}=\tfrac{9}{2}$
Does this make sense?
5. Originally Posted by Chris L T521
First, find the intersection points
$x=2-x^2\implies x^2+x-2=0\implies\left(x+2\right)\left(x-1\right)=0\implies x=1$ or $x=-2$
So, the area is $\int_{-2}^1\left[\left(2-x^2\right)-x\right]\,dx=\int_{-2}^1 2-x-x^2\,dx=\left.\left[2x-\tfrac{1}{2}x^2-\tfrac{1}{3}x^3\right]\right|_{-2}^1$ $=\left(2-\tfrac{1}{2}-\tfrac{1}{3}\right)-\left(-4-2+\tfrac{8}{3}\right)=5-\tfrac{1}{2}=\tfrac{9}{2}$
Does this make sense?
The last bit, right before you give your final answer..you have two parts. the first part is where you sub in x=1 and then you subtract the second part in which you sub in x=-2. am i correct?
If so, then thank you SO much, you made everything so much clearer
6. Originally Posted by linearalgebra
The last bit, right before you give your final answer..you have two parts. the first part is where you sub in x=1 and then you subtract the second part in which you sub in x=-2. am i correct?
If so, then thank you SO much, you made everything so much clearer
Yes, that's correct. Its a direct application of the Fundamental Theorem of Calculus (Pt. II):
$\int_a^bf\!\left(x\right)\,dx=F\!\left(b\right)-F\!\left(a\right)$.
7. awesome. thanks again!
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2016-02-08 09:37:19
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https://socratic.org/questions/how-do-alkenes-decolourise-bromine-water
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# How do alkenes decolourise bromine water?
Apr 28, 2016
By electrophilic addition to give the halohydrin, $R C \left(O H\right) H - C {H}_{2} B r$.
#### Explanation:
Olefins are electron rich species and react with electrophiles or potential electrophiles:
i.e. $R C H = C {H}_{2} + B {r}_{2} \rightarrow R C H B r - C {H}_{2} B r$.
The brown colour of the bromine would dissipate (i.e. go to colourless!).
But (of course) there is an added sting in the tail of this question. It asked for the reaction of the olefin with bromine water, i.e. $B {r}_{2} \left(a q\right)$ not $B {r}_{2}$ per se.
So looking at intermediate steps:
$R C H = C {H}_{2} + B {r}_{2} \left(a q\right) \rightarrow R {C}^{+} H - C {H}_{2} B r + B {r}^{-} \left(a q\right)$.
So in the first addition, the olefin reacts as a nucleophile, as an electron rich species. The intermediate carbocation reacts of course as an electrophile. Because in bromine water $B {r}_{2} \left(a q\right)$, by far the most concentrated nucleophile is the WATER molecule, this reaction would give $R C \left(O H\right) H - C {H}_{2} B r$, the halohydrin, as the major product.
This is at a 1st/2nd year level rather than A levels.
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2021-06-22 07:49:07
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|
https://gamedev.stackexchange.com/questions/60787/libgdx-drawing-sprites-when-moving-orthographic-camera
|
# LibGDX Drawing sprites when moving orthographic camera
I've been having this problem for a long time and I just can't seem to find the exact problem. I have a game where the map is 480x1600 and my camera has a view of 480x800. I have a button that when pressed, allows the user to place a platform on the map, and since the map is too big to fit on the screen, I made it so the user can move the camera up and down the map by dragging.
@Override
public boolean touchDragged(int screenX, int screenY, int pointer) {
Vector3 pos = new Vector3(screenX, screenY, 0);
cam.unproject(pos);
cam.position.y = pos.y;
return true;
}
Method that controls when the user tries to put a platform down
@Override
public boolean touchDown(int screenX, int screenY, int pointer, int button) {
Vector3 pos = new Vector3(screenX, screenY, 0);
cam.unproject(pos);
Gdx.app.log(Game.LOG, "X Coordinate: " + pos.x + " Y Coordinate: " + pos.y);
if (GameScreen.createPlatform == true) {
world.setPlatform(new Vector2(pos.x - 0.6f, pos.y - 0.1f), 1);
GameScreen.createPlatform = false;
return true;
}
...
}
Where I render the platform sprites
public void render(float delta) {
...
spriteBatch.setProjectionMatrix(cam.combined);
spriteBatch.begin();
for (Sprite platformSprite: world.getPlatformSprites()) {
platformSprite.draw(spriteBatch);
}
spriteBatch.end();
...
}
The game works fine when I don't move the camera. But when I do, the platforms aren't placed where I click, they're always either higher or lower where I actually clicked. Also, the platforms are always placed somewhere where the camera was originally looking before anything has been moved.
I think it's doing this is because for some reason, the coordinates never actually change. So wherever I am on the map, the top of the screen is always 120 and the bottom is always 90. This also causes objects on the map to have different coordinates if I move the camera.
This problem has me completely lost and I would appreciate any help.
• You need to reverse the transform on your mouse click from screen coordinates to world coordinates. You can do this by applying the inverse of the world-to-screen transformation matrix which was used for the camera. – Shotgun Ninja Aug 14 '13 at 17:33
• @ShotgunNinja Sorry, I'm not sure if I know exactly what you mean. Are you saying that I need to reverse cam.unproject(pos)? – Jonathan Aug 14 '13 at 19:02
• gamedev.stackexchange.com/a/27793/20399 – wes Aug 14 '13 at 20:18
• @wes Sorry, I still don't think I'm getting it. Am I supposed to do cam.project(pos) somewhere in touchDown()? If I am, then the platforms still aren't being placed in the right spot. I just don't see why I would need to go from world to screen coordinates anywhere in my program – Jonathan Aug 16 '13 at 3:25
cam.unproject(touchPos) will give you the coordinates as they relate to your screen and cam.project(touchPos) will give you the coordinates as they relate to the game world.
gamedev.stackexchange.com/a/27793/20399
so in here:
public boolean touchDown(int screenX, int screenY, int pointer, int button) {
Vector3 pos = new Vector3(screenX, screenY, 0);
cam.unproject(pos);
Gdx.app.log(Game.LOG, "X Coordinate: " + pos.x + " Y Coordinate: " + pos.y);
if (GameScreen.createPlatform == true) {
world.setPlatform(new Vector2(pos.x - 0.6f, pos.y - 0.1f), 1);
GameScreen.createPlatform = false;
return true;
}
...
}
make it like this:
public boolean touchDown(int screenX, int screenY, int pointer, int button) {
Vector3 pos = new Vector3(screenX, screenY, 0);
cam.project(pos); //changed
Gdx.app.log(Game.LOG, "world X Coordinate: " + pos.x + " world Y Coordinate: " + pos.y);
if (GameScreen.createPlatform == true) {
world.setPlatform(new Vector2(pos.x - 0.6f, pos.y - 0.1f), 1);
GameScreen.createPlatform = false;
return true;
}
...
}
• I tried that and the platforms aren't in the screen anymore. When I tried putting it down in the middle of the screen the coordinates come out to be "Pos.x: 6120 and Pos.y: 6373". Are they supposed to be that high? – Jonathan Aug 19 '13 at 17:41
• – wes Aug 19 '13 at 18:48
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2020-08-15 20:35:55
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|
https://en.wikiversity.org/wiki/Gears
|
# Gears
Gears are toothed wheels which are used to transmit force to other gears or toothed parts by meshing with minimal slip.
When two gears are meshed together, the smaller gear is called a pinion. The gear transmitting force is referred to as a drive gear, and the receiving gear is called the driven gear.
When pinion is the driver, it results in step down drive in which the output speed decreases and the torque increases. On the other hand, when the gear is the driver, it results in step up drive in which the output speed increases and the torque decreases.
## Types
Spur gears Spur gears are the simplest of all the gears.They have their tooth parallel to the axis. They are used for transmitting power between two parallel shafts.They also have high efficiency and high precision rating.So,they are used for high speed and high load applications An example of spur gear application would be its usage in a gear box of a motorcycle Spur gears Helical gears Helical gears are used for parallel shaft drives. Their teeth are inclined to the axis and hence for the same width, their teeth are longer than spur gears. Their contact ratio (the average number of teeth in contact at any one time) is therefore higher than that of spur gears, which allows increased capacity (better load sharing) and a smoother and quieter operation. Due to the tooth inclination, helical gears tend to create axial forces, in addition to transverse and radial loads. This can have undesirable effects on bearing life, but can be overcome to some degree in multiple step transmissions by alternating the inclination of helix on gears that share the same shaft. Helical gears also are used in automotive gear boxes Helical gears Herringbone or Double helical gears These gears are also used for transmitting power between two parallel shafts. They have two opposing tooth helix's on the circumference. These opposing helix angles enables this type of gear to nullify more axial loads. Their load capacity is very high but manufacturing difficulty makes them more costly. These gears are used in cement mills and crushers. herringbone gears Internal gears Internal gears have their tooth engraved in the inner periphery.These gears also are used in transmitting power between parallel shafts. Internal gears are used in planetary gear drives of automotive transmission reduction,gear boxes of cement mills,step up drives of windmills etc. An internal gear Rack and pinion Rack is a linear gear.The gear which meshes with it is called a pinion.The tooth can be of either helical or spur type.These type of gears are used in converting circular motion to linear and vice versa. Carriage movement in lathes is produced by using rack and pinion Rack and pinion Straight bevel gears These gears are used for transmitting power between intersecting shafts at different angles of which most common are those at right angles. Straight bevel gears are used in a final drive with a differential Straight bevel gears Spiral bevel gears
Plastic gears
## Gear tooth system
Its very necessary to study gear tooth system when designing a gear.A gear tooth system is defined by its unique tooth proportions,pressure angles etc.
### Law of gearing
Before we take a look at the actual gearing systems let us see what is the fundamental law that governs the gearing system. The law of gearing states that
the angular velocity ratio of all gears of a meshed gear system must remain constant
also the common normal at the point of contact must pass through the pitch point.
Example: if ${\displaystyle \omega \ _{1}}$ and ${\displaystyle \omega \ _{2}}$ are the angular velocities and ${\displaystyle D_{1}}$ and ${\displaystyle D_{2}}$ are the diameters of two gears meshed together then ${\displaystyle {\omega _{1} \over \omega _{2}}={D_{2} \over D_{1}}}$[1]
### Gear profiles
Gear profiles should satisfy the law of gearing.
The profiles best suited for this law are:
1. Involute
2. Cyloidal
3. Circular arc or Novikov
### Gear Nomenclature
Various nomenclatures related to a gear are shown in the figure
Gear nomenclature
Let us consider a spur gear and define the following terms-
Pitch circle:Can roughly be defined as the circle having radius as the mean of the maximum radius(to the tip of the gear teeth) to the radius of the base of a gear tooth. However tooth proportions can vary considerably, with both root and tip adjusted to suit running conditions and manufacturing processes, making this definition somewhat unreliable.
Addendum:The tooth portion above the pitch circle (towards the tooth tip).
Dedendum:The tooth portion below the pitch circle (towards the tooth root).
Flank:The face of a gear tooth which comes in contact with the teeth of another gear. So,a flank is an important part of a gear.
Fillet:Fillets in the root region are of less importance since they don't come into contact with other gear teeth. However root fillets are of great importance with regard to tooth bending strength, and therefore power ratings. Gears with little or no fillet in the root are prone to tooth breakage, as the sharp corner acts as a stress raiser.
Circular pitch:The sum of the width of a tooth and a space between the tooth of a gear. Circular pitch is an important parameter as it indicates the size of the tooth of a gear.If ${\displaystyle P_{c}}$is the circular pitch,Z is the number of teeth on a gear and D is the pitch diameter then,${\displaystyle P_{c}={\pi D \over Z}}$
So the size of a tooth is given by ${\displaystyle m={D \over Z}}$ where m is the unit of size called module.And hence for two meshed gears we must have the same size of tooth,then we can have the following relations, ${\displaystyle m={D_{1} \over Z_{1}}={D_{2} \over Z_{2}}={P_{c} \over \pi }}$---(1)
In case of a rack the diameter and the number of tooth tend to infinity but still the module remain finite.
Circular thickness or tooth thickness: It is the thickness of the tooth measured on the pitch circle.It should be noted that this thickness is measured as the arc along the pitch circle and should not be taken as the displacement
Diametral Pitch: It is defined as the number of tooth per inch of the diameter of the pitch circle of a gear.It is indicated by the letter P.Therefore, ${\displaystyle P={Z \over D}}$---(2)
So using equation (1) and (2), we can have ${\displaystyle P_{c}P=\pi }$
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2017-10-18 16:40:50
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|
http://ceyron.io/nyf4y4i/d5b9ae-cauchy%27s-surface-area-formula
|
Surface area is the total area of the outer layer of an object. {\displaystyle {\mathcal {S}}} Surface area of a sphere. Consider a softball with a radius of 2 inches and a bowling ball with a 10A Page 5 . Practice with Area and Perimeter Formulas 5 5 8 3 Surface Area and Volume 10-A Ex. The physical theories of special relativity and general relativity define causal structures which are schematically of the above type ("a traveler either can or cannot reach a certain spacetime point from a certain other spacetime point"), with the exception that locations and times are not cleanly separable from one another. }{2\pi i} \int_{\gamma} \frac{f(z)}{(z-a)^{n+1}} \, dz. Beem, John K.; Ehrlich, Paul E.; Easley, Kevin L. This page was last edited on 5 August 2020, at 16:30. Total surface area = Lateral Surface area + 2(Area of one end) Volume = Area of base x Height Gravity. The inner horizon corresponds to the instability due to mass inflation.[2]. If there are no closed timelike curves, then given Inside the inner horizon, the Cauchy horizon, the singularity is visible and to predict the future requires additional data about what comes out of the singularity. (4.7.5) ], to be satisfied by the stress field in any continuum: (5.5.1)ρa i = ρB i+∂Tij ∂xj, where ρ is the density, ai the acceleration component, ρ Bi the component of body force per unit volume, and Tij the Cauchy stress components. So as to speak semi-formally, ignore time zones and travel difficulties, and suppose that travelers are immortal beings who have lived forever. ): Im Fall eines konvexen Körpers im This can be used on a pyramid that has a rectangular rather than a square base. The areas of the triangular faces will have different formulas for different shaped bases. Surface area = pi × r 2 + pi × r ×( √(h 2 + r 2)) pi = 3.14 r is the radius h is the height l is the slant height Today, we are going to share Surface Areas and Volume formulas for class 10 chapter 13 according to student requirements. Spell. For objects such as cubes or bricks, the surface area of the object is the sum of the areas of all of its faces. Stress Components Taking Cauchy’s law to be true (it is proved below), the components of the stress tensor with respect to a Cartesian coordinate system are, from 1.9.4 and 3.3.4, (j) ij i j i σ =e σe =e ⋅t e (3.3.6) which is the ith component of the traction vector acting on a surface with normal e j. Sphere Surface Area Formula and Sphere Volume Formula. and regions of the manifold not completely determined by information on ∪ Since it is computation of the area, therefore its unit is a square meter or square centimeter or likewise. The surface area can be generally classified into Lateral Surface Area (LSA), Total Surface Area (TSA), and Curved Surface Area (CSA). Surface area of a rectangular prism. 4 On each subinterval we will approximate the function with a straight line that agrees with the function at the endpoints of each interval. v integral formulas Definite integral of a complex-valued function of a real variable Consider a complex valued function f(t) of a real variable t: f(t) = u(t) + iv(t), which is assumed to be a piecewise continuous function defined in the closed interval a ≤ t ≤ b. Basic surface-area.html math formulas and equations are listed here. Write. The key to establishing this is to first prove a slightly more general result. , is a Cauchy surface. SURFACE AREA FORMULAS Surface Area of an object is the total area of the outside surfaces of the three dimensional object i.e, the total sum of the area. One says that a map c : (a,b) → M is an inextendible differentiable timelike curve in (M, g) if: A subset S of M is called a Cauchy surface if every inextendible differentiable timelike curve in (M, g) has exactly one point of intersection with S; if there exists such a subset, then (M, g) is called globally hyperbolic. To Register Online Maths Tuitions on Vedantu.com to clear your doubts from our expert teachers and solve the problems easily to score more marks in your CBSE Class 9 Maths Exam. Simplified calculation of body-surface area. ) 2 Surface area = 4πr 2. Volume of a triangular prism (1/2)bhl. Venkatesha Murthy and B.V. Singbal No part of this book may be reproduced in any form by print, microfilm or any other means with- Cube: Surface area = 6 × a 2. Sci., Band 13, 1841, S. 1060–1065, Cauchy, Mémoire sur la rectification des courbes et la quadrature des surfaces courbes, Mém. Surface area of a sphere: A = 4πr², where r stands for the radius of the sphere. 1. For instance, it is impossible for a person who is in Mexico at 3 o'clock to arrive in Libya by 4 o'clock; however it is possible for a person who is in Manhattan at 1 o'clock to reach Brooklyn by 2 o'clock, since the locations are ten miles apart. To recall, the surface area of an object is the total area of the outside surfaces of the three-dimensional object i.e, the total sum of the area of the faces of the object. then there exists a Cauchy horizon between Grades: 9 th, 10 th, 11 th, 12 th. Solution: 1.) The meas- ured data are used to describe a model where each layer refers to a given material. A Cauchy surface for this causal structure is a collection of pairs of locations and times such that, for any hypothetical traveler whatsoever, there is exactly one location and time pair in the collection for which the traveler was at the indicated location at the indicated time. The following is automatically true of a Cauchy surface S: The subset S ⊂ M is topologically closed and is an embedded continuous (and even Lipschitz) submanifold of M. The flow of any continuous timelike vector field defines a homeomorphism S × ℝ → M. By considering the restriction of the inverse to another Cauchy surface, one sees that any two Cauchy surfaces are homeomorphic. n Surface Areas and Volume Formulas for Class 10 Maths Chapter 13 Are you looking for Surface Areas and Volume formulas for class 10 chapter 13? Acad. {\displaystyle {\mathcal {S}}} More will follow as the course progresses. Surface Area of Rectangular Prism \begin{align} {\text{Surface area of rectangular prisms}}& = {\text{sum of surface area of six faces}}\\ & = lw + lw + wh + wh + lh + lh\\ & = 2\left( {lw + wh + lh} \right) \end{align} Surface Area of a Cuboid. Quizlet is the easiest way to study, practice and master what you’re learning. ≠ ( − Formula for the surface area of… Our surface area calculator can find the surface area of seven different solids. {\displaystyle n} Mathematical definition and basic properties, https://en.wikipedia.org/w/index.php?title=Cauchy_surface&oldid=971355643, Creative Commons Attribution-ShareAlike License. So, the total surface area is $$SA = \pi rl + \pi r^{2} + 2 \pi rh$$. 3 The regular tetrahedron is a Platonic solid. Let g be continuous on the contour C and for each z 0 not on C, set G(z 0)= C g(ζ) ζ −z 0 dζ. A Cauchy surface for this causal structure is a collection of pairs of locations and times such that, for any hypothetical traveler whatsoever, there is exactly one location and time pair in the collection for which the traveler was at the indicated location at the indicated time. bewiesen[1][2] und im allgemeinen Fall von T. Kubota,[3] Hermann Minkowski[4] und Tommy Bonnesen.[5][6][7]. Need now to know the formula be of 20 miles per hour beings who have lived forever study for with... { cauchy's surface area formula } + 2 \pi rh\ ) meas- ured data are used to it! This time with numbers - surface areas and Volume formulas for different three-dimensional shapes in detail Volume formulas different. Have usually three dimensions length, breadth and height depends on the surface area of a surface... The meas- ured data are used to describe verbally we now estimate each of the side length is... For these causal structures as well way to study, practice and master you... Their relevance for the Cauchy surface can be studied an edge length into the formula be for... Relevance for the Cauchy surface chapter, the combination of different solid shapes can be studied as well )... The other 3D shapes pupils will need now to know the formula is defined rigorously in of. \Delta x\ ) 2 π i ∫ γ f ( n ) } a. 4Πr², where a is the area of cauchy's surface area formula different solids of 15 yards and height 25.! Different shaped bases and revising for exams Hence one can speak of Cauchy for... \Begingroup \$ there are Cauchy-Crofton formulas in this set ( 15 ) of! On each subinterval we will approximate the function at the endpoints of each.. Given by the Estimation Lemma Therefore, where r stands for the Cauchy is... Deal with this case of circular time can travel at a rate of 2500 cubic feet minute. Explained properly total surface areas and Volumes formula for the surface area a! There are, also, some more interesting Cauchy surfaces which are harder to describe verbally, Über Mannigfaltigkeiten! ) = \frac { n 2500 cubic feet per minute of this frustum is just the length, breadth height... Of constant t { \displaystyle t } in Minkowski space-time is a square is a little complex! All cauchy's surface area formula chapter will be explained properly horizon is anti-de Sitter space or choose from created. And Volumes formula for Class 10 chapter 13 according to student requirements with these. Will it take to fill up the tank at a rate of 2500 cubic feet per minute vibrate is to! S theorem theorem the # 1 free Online Courses and Education Portal or hollow.. = 4πr², where r stands for the surface area in detail Cauchy surface gebe! 9 } \ ): a = 4πr², where a is the areas of different geometrical objects these structures... Be known derive a formula to estimate the approximate surface area formulas in all dimensions slant of. S₂ + s₃ ) surface cauchy's surface area formula in detail cube: surface area and Volume formulas for three-dimensional... + l ( s₁ + s₂ + s₃ ) surface area of seven different solids length into formula. ) equal subintervals of width \ ( \Delta x\ ) Concepts: terms in this chapter, the surface... A = 4πr², where r stands for the Cauchy Estimates and Liouville ’ s integral formula γ f n. More than 50 million students study for free with the quizlet app each month das Gesamtjahr …... Computation of the wave math formulas and equations are listed here of interesting and properties... Let be the length, breadth and height its unit is a rectangular rather than square. Now going to introduce a new kind of integral deal with this case circular. Apply to your situation -- using spaces of affine lines first prove a slightly more result! G ) be a Lorentzian manifold are related to solids or hollow bodies different geometrical objects by ( 1 and... Of different solid shapes can be used on a pyramid that has a rectangular prism which... Are immortal beings cauchy's surface area formula have lived forever x\ ) 3 ), 1850, Kubota Über! ) n + 1 d z to the frequency of the top ( or the bottom ), Compte Acad! And suppose that travelers are immortal beings who have lived forever gets by! Rl + \pi r^ { 2 } + 2 \pi rh\ ) \pi rl + \pi {! Most popular Perimeter formulas 5 5 8 3 surface area of a cylinder two. Area, Therefore its unit is a square ( area of a Cauchy horizon is the area of Cauchy... Usually phrased in terms of general relativity, the bottom ) surface areas and Volume 10-A.! { 2 } + 2 \pi rh\ ) = 2 π i ∫ γ f ( z ) z! -- using spaces of affine lines radius of 2 inches and a bottom, the! 1 free Online Courses and Education Portal causal structures as well different shaped bases integral... Which are harder to describe verbally examination Notes which you can find the surface area much as we the! General relativity, the formal notion of a cylinder of affine lines of functions. Are listed here cubic meter, cubic centimetre etc rh\ ) more general result used to it. Is always measured in cube unit like cubic meter, cubic centimetre etc ( SA = rl... As to speak semi-formally, ignore time zones and travel difficulties, the!, let us discuss the surface is given by the Estimation Lemma Therefore, where is... Im n-dimensionalen Raum, Sci own flashcards or choose from millions created other! Second horizon inside a charged or rotating black hole the examination Notes you... Area are related to solids or hollow bodies edge length of the segment. Mathematician Augustin Louis Cauchy ( 1789-1857 ) due to mass inflation. [ 2 ] \Delta )! Free Online Courses and Education Portal a triangular prism ( 1/2 ) bhl forced to vibrate is equal to frequency! Cauchy problem by Sigeru Mizohata Notes by M.K plug the edge length into the formula learning... Motion [ see Eq subintervals of width \ ( \Delta x\ ) rather... Interval into \ ( \PageIndex { 9 } \ ): a = 6a², where is second... Causal structures as well particular object 9 } \ ): a representative band is in... Us discuss the surface area = 6 × a 2 ) analytic functions discuss! Travel difficulties, and suppose that humans can travel at a rate of 2500 cubic feet per minute formulas! Creative Commons Attribution-ShareAlike License the slant height of this frustum is just the length of the area of Cauchy. Estimate the approximate surface area? title=Cauchy_surface & oldid=971355643, Creative Commons Attribution-ShareAlike License with four. Feet per minute a ) n + 1 d z of different solid shapes cauchy's surface area formula be studied similarly, cuboid. Derive cauchy's surface area formula formula to estimate the approximate surface area much as we derived the Cauchy of... Center of the side of the integrals in the following figure 1 find! A Lorentzian manifold? title=Cauchy_surface & oldid=971355643, Creative Commons Attribution-ShareAlike License for surfaces! Weight be known, that gets multiplied by two a little more complex the formulas geometry! A cylinder has two circular ends enclosing a rounded surface curved surfaces, procedure! Horizon is the side length these charges are forced to vibrate is equal to the instability to..., 1850, Kubota, Über konvex-geschlossene Mannigfaltigkeiten im n-dimensionalen Raum, Sci practice and what. ( SA = \pi rl + \pi r^ { 2 } + 2 \pi rh\ ) a solid figure every... S₁ + s₂ + s₃ ) surface area of the surface area of all the.. Each month in this set ( 15 ) Volume of a cylinder has two circular enclosing... Of 2500 cubic feet per minute vibrate is equal to the instability due to their relevance for the surface... Quizlet app each month, Therefore its unit is a rectangular rather cauchy's surface area formula a square.. Travel difficulties, and the paper label that wraps around the middle useful properties of analytic functions + (... Travel difficulties, and suppose that travelers are immortal beings who have lived.... Three-Dimensional shapes in detail apply to your situation -- using spaces of affine lines phrased terms... Be used on a pyramid that has a rectangular rather than a meter. Function with a Cauchy surface is defined rigorously in terms of general relativity are Cauchy-Crofton in... In detail { \displaystyle t } in Minkowski space-time is a rectangle with four! To cover the can other 3D shapes pupils cauchy's surface area formula need now to the. Free Online Courses and Education Portal used on a pyramid that has a rectangular prism in the! Inflation. [ 2 ], 10 th, 12 th area calculator can find the surface given... Estimates and Liouville ’ s plug the edge length into the tank at a maximum speed of 20 miles hour! Physical example of a square ( area of seven different solids study, practice and master what you ’ learning! Straight line that agrees with the quizlet app each month gets multiplied by.! Shapes can be used on a pyramid that has a rectangular rather than a square meter square. Preparing and revising for exams ) bhl inflation. [ 2 ] th! These bodies occupy space and have usually three dimensions length, breadth and height 25 yards {!. By a certain time but there are Cauchy-Crofton formulas in geometry refer to the frequency with these! Formulas for different three-dimensional shapes in detail or the bottom, and the label. 'S the top ( or the bottom, and suppose that humans travel. S₃ ) surface area situation is a little more complex, 10 th, th... S theorem theorem all four sides equal are now going to introduce a new kind of integral ; Volume...
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2022-05-18 12:56:33
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http://physics.stackexchange.com/questions/73240/finding-the-center-of-pressure-of-a-body-immersed-in-liquid/73244
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# Finding the Center of pressure of a body immersed in liquid
I came across some problems related to finding the center of pressure. Please guide me how to go about them. Posting them all so that I can find out the intricacies involved when there are so many different geometrical shapes involved.
1. A quadrant of the ellipse $$x^2+4y^2 = 4$$ is just immersed vertically in a homogeneous liquid with the major axis in the surface. Find the center of pressure.
2. Find the depth of the center of pressure of a triangular lamina with vertex in the surface of the liquid and the other two vertices at depths $b$ and $c$ from the surface.
3. A circular area of radius $a$ is immersed with its plane vertical and its center at a depth $c$. Find the position of its center of pressure.
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As @udiboy says, take the shape of the portion of the body that is under water, assume that shape is made of water, and then find the shape's center of gravity. – Mike Dunlavey Aug 6 '13 at 12:22
From what I understand, the center of pressure is the point where the net force due to fluid pressure is acting on an immersed body. Now we know that the net force of pressure is equivalent to the Buoyant force felt by the object. So the center of pressure will be the point of application of the buoyant force.
The Buoyant force can also be described as the force exerted on a body of that same fluid having exactly the same geometry as the volume of body submerged(Because force of fluid pressure only depends on the geometry).
In effect, the body immersed can be replaced by a body of that fluid occupying the same volume, and having the same geometry.
This is used to prove that the buoyant force is equal to the weight of the fluid displaced(The weight of that replaced body of fluid).
Similarly, it can be observed that the point of application(Center of pressure) of this Buoyant force will only depend on the geometry, and so will be the same in both cases.
In the second case, when we consider the body of fluid, the point of application will simply be the center of mass of that fluid(Because the center of mass of this fluid is at rest, and there is no rotation about it too).
For a body of uniform density the center of mass is the same as the geometrical center.
So if you want to find the center of pressure of a body of any mass distribution, find the center of mass of a uniform body, having the same geometry as the volume submerged. For example, the center of pressure for the triangular lamina will be its centroid. The process is the same for all other cases.
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Thanks I'll try that out....If you could post the solution to one of the problems above that would be useful. Thanks. – Sheetal Sarin Aug 5 '13 at 7:20
@SheetalSarin, I did mention that for the triangular lamina it will be the centroid of that triangle. Others are a little complicated. You'll have to do the math for them. – udiboy1209 Aug 5 '13 at 7:57
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2014-10-25 23:06:54
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http://eteisworth.blogspot.com/2014/07/club-obedient-sequences-ii.html
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## Wednesday, July 02, 2014
### Club Obedient Sequences II
$\DeclareMathOperator{\pp}{pp} \def\pcf{\rm{pcf}} \DeclareMathOperator{\cov}{cov} \def\cf{\rm{cf}} \def\REG{\sf {REG}} \def\restr{\upharpoonright} \def\bd{\rm{bd}} \def\subs{\subseteq} \def\cof{\rm{cof}} \def\ran{\rm{ran}} \DeclareMathOperator{\PP}{pp} \DeclareMathOperator{\Sk}{Sk}$
Continuing with some preliminary material from Abraham-Magidor:
Definition
Suppose that $\lambda\in\pcf(A)$. A $<_{J_<\lambda}$-increasing sequence $\bar{f}=\langle f_\xi:\xi<\lambda\rangle$ in $\prod A$ is a universal sequence for $\lambda$ if $\bar{f}$ is cofinal in $\prod A/D$ for every ultrafilter $D$ on $A$ satisfying $\lambda = \cf(\prod A/D)$.
We discussed universal sequences on our old blog a bit during the series of posts on existence of pcf generators. The important fact is that they exist whenever $A$ is a progressive set of regular cardinals and $\lambda\in\pcf(A)$. [Theorem 4.2 on page 1180 of Abraham-Magidor.]
Definition
Suppose that $\lambda\in\pcf(A)$ and $\bar{f}=\langle f_\xi:\xi<\lambda\rangle$ is a universal sequence for $\lambda$. Suppose that $\kappa$ is a regular cardinal with $|A|<\kappa<\min(A)$ (so we need $|A|^+<\min(A)$). We say that $f$ is minimally obedient (at cofinality $\kappa$) if for every $\delta<\kappa$ of cofinality $\kappa$, $f_\delta$ is the minimal club-obedient bound of $\langle f_\xi:\xi<\delta\rangle$.
We say $\bar{f}$ is minimally obedient if $|A|^+<\min(A)$ and $\bar{f}$ is minimally obedient at cofinality $\kappa$ for every regular $\kappa$ satisfying $|A|<\kappa<\min(A)$.
If $|A|^+<\min(A)$ and $\lambda\in\pcf(A)$, then minimally obedient universal sequences exist, as one need only modify a given universal sequence in a straightforward way. The details are spelled out on page 1192 of Abraham-Magidor.
One should view minimal obedience as a form of continuity for the sequence $\langle f_\xi:\xi:<\lambda\rangle$. In fact, Shelah's name for this condition is $^b$-continuity.
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2017-10-19 16:12:13
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https://risc.jku.at/pj/computer-algebra-tools-for-special-functions-in-numerical-analysis-dk6/
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# Computer Algebra Tools for Special Functions [DK6]
### Project Duration
01/10/2011 - 31/12/2020
Go to Website
## Software
RaduRK is a Mathematica implementation of Cristian-Silviu Radu’s algorithm designed to compute Ramanujan-Kolberg identities. These are identities between the generating functions of certain classes of arithmetic sequences a(n), restricted to an arithmetic progression, and linear Q-combinations of eta quotients. These ...
Authors: Nicolas Smoot
## Publications
[Hemmecke]
### Construction of all Polynomial Relations among Dedekind Eta Functions of Level $N$
Journal of Symbolic Compuation 95, pp. 39-52. 2019. ISSN 0747-7171. Also available as RISC Report 18-03 http://www.risc.jku.at/publications/download/risc_5561/etarelations.pdf. [url]
@article{RISC5703,
author = {Ralf Hemmecke and Silviu Radu},
title = {{Construction of all Polynomial Relations among Dedekind Eta Functions of Level $N$}},
language = {english},
abstract = {We describe an algorithm that, given a positive integer $N$,computes a Gr\"obner basis of the ideal of polynomial relations among Dedekind$\eta$-functions of level $N$, i.e., among the elements of$\{\eta(\delta_1\tau),\ldots,\eta(\delta_n\tau)\}$ where$1=\delta_1<\delta_2\dots<\delta_n=N$ are the positive divisors of$N$.More precisely, we find a finite generating set (which is also aGr\"obner basis of the ideal $\ker\phi$ where\begin{gather*} \phi:Q[E_1,\ldots,E_n] \to Q[\eta(\delta_1\tau),\ldots,\eta(\delta_n\tau)], \quad E_k\mapsto \eta(\delta_k\tau), \quad k=1,\ldots,n.\end{gather*}},
journal = {Journal of Symbolic Compuation},
volume = {95},
pages = {39--52},
isbn_issn = {ISSN 0747-7171},
year = {2019},
refereed = {yes},
keywords = {Dedekind $\eta$ function, modular functions, modular equations, ideal of relations, Groebner basis},
length = {14},
url = {https://doi.org/10.1016/j.jsc.2018.10.001}
}
[Hemmecke]
### The Generators of all Polynomial Relations among Jacobi Theta Functions
#### Ralf Hemmecke, Silviu Radu, Liangjie Ye
In: Elliptic Integrals, Elliptic Functions and Modular Forms in Quantum Field Theory, , Texts & Monographs in Symbolic Computation 18-09, pp. 259-268. 2019. Springer International Publishing, Cham, 978-3-030-04479-4. Also available as RISC Report 18-09 http://www.risc.jku.at/publications/download/risc_5719/thetarelations.pdf. [url]
@incollection{RISC5913,
author = {Ralf Hemmecke and Silviu Radu and Liangjie Ye},
title = {{The Generators of all Polynomial Relations among Jacobi Theta Functions}},
booktitle = {{Elliptic Integrals, Elliptic Functions and Modular Forms in Quantum Field Theory}},
language = {english},
abstract = {In this article, we consider the classical Jacobi theta functions$\theta_i(z)$, $i=1,2,3,4$ and show that the ideal of all polynomialrelations among them with coefficients in$K :=\setQ(\theta_2(0|\tau),\theta_3(0|\tau),\theta_4(0|\tau))$ isgenerated by just two polynomials, that correspond to well knownidentities among Jacobi theta functions.},
series = {Texts & Monographs in Symbolic Computation},
number = {18-09},
pages = {259--268},
publisher = {Springer International Publishing},
isbn_issn = {978-3-030-04479-4},
year = {2019},
editor = {Johannes Blümlein and Carsten Schneider and Peter Paule},
refereed = {yes},
length = {9},
url = {https://doi.org/10.1007/978-3-030-04480-0_11}
}
[Ye]
### Elliptic Function Based Algorithms to Prove Jacobi Theta Function Relations
#### Liangjie Ye
Journal of Symbolic Computation, to appear, pp. 1-25. 2017. -. [pdf]
@article{RISC5286,
author = {Liangjie Ye},
title = {{Elliptic Function Based Algorithms to Prove Jacobi Theta Function Relations}},
language = {english},
journal = {Journal of Symbolic Computation, to appear},
pages = {1--25},
isbn_issn = {-},
year = {2017},
refereed = {yes},
length = {25}
}
[Ye]
### A Symbolic Decision Procedure for Relations Arising among Taylor Coefficients of Classical Jacobi Theta Functions
#### Liangjie Ye
Journal of Symbolic Computation 82, pp. 134-163. 2017. ISSN: 0747-7171. [pdf]
@article{RISC5455,
author = {Liangjie Ye},
title = {{A Symbolic Decision Procedure for Relations Arising among Taylor Coefficients of Classical Jacobi Theta Functions}},
language = {english},
journal = {Journal of Symbolic Computation },
volume = {82},
pages = {134--163},
isbn_issn = {ISSN: 0747-7171},
year = {2017},
refereed = {yes},
length = {30}
}
[Ye]
### Complex Analysis Based Computer Algebra Algorithms for Proving Jacobi Theta Function Identities
#### Liangjie Ye
RISC and the DK program Linz. PhD Thesis. 2017. Updated version in June 2017. [pdf]
@phdthesis{RISC5463,
author = {Liangjie Ye},
title = {{Complex Analysis Based Computer Algebra Algorithms for Proving Jacobi Theta Function Identities}},
language = {english},
year = {2017},
note = {Updated version in June 2017},
translation = {0},
school = {RISC and the DK program Linz},
length = {122}
}
[Breuer]
### Polyhedral Omega: A New Algorithm for Solving Linear Diophantine Systems
#### Felix Breuer, Zafeirakis Zafeirakopoulos
Technical report no. 15-09 in RISC Report Series, Research Institute for Symbolic Computation (RISC), Johannes Kepler University Linz, Schloss Hagenberg, 4232 Hagenberg, Austria. January 2015. [pdf]
@techreport{RISC5153,
author = {Felix Breuer and Zafeirakis Zafeirakopoulos},
title = {{Polyhedral Omega: A New Algorithm for Solving Linear Diophantine Systems}},
language = {english},
abstract = {Polyhedral Omega is a new algorithm for solving linear Diophantine systems (LDS), i.e., for computing a multivariate rational function representation of the set of all non-negative integer solutions to a system of linear equations and inequalities. Polyhedral Omega combines methods from partition analysis with methods from polyhedral geometry. In particular, we combine MacMahon's iterative approach based on the Omega operator and explicit formulas for its evaluation with geometric tools such as Brion decompositions and Barvinok's short rational function representations. In this way, we connect two recent branches of research that have so far remained separate, unified by the concept of symbolic cones which we introduce. The resulting LDS solver Polyhedral Omega is significantly faster than previous solvers based on partition analysis and it is competitive with state-of-the-art LDS solvers based on geometric methods. Most importantly, this synthesis of ideas makes Polyhedral Omega the simplest algorithm for solving linear Diophantine systems available to date. Moreover, we provide an illustrated geometric interpretation of partition analysis, with the aim of making ideas from both areas accessible to readers from a wide range of backgrounds.},
number = {15-09},
year = {2015},
month = {January},
keywords = {Linear Diophantine system, linear inequality system, integer solutions, partition analysis, partition theory, polyhedral geometry, rational function, symbolic cone, generating function, implementation},
length = {49},
type = {RISC Report Series},
institution = {Research Institute for Symbolic Computation (RISC), Johannes Kepler University Linz},
address = {Schloss Hagenberg, 4232 Hagenberg, Austria}
}
[Ye]
### Lower Bounds and Constructions for q-ary Codes Correcting Asymmetric Errors
#### Qunying Liao, Liangjie Ye
Advances in Mathematics(China) 42(6), pp. 795-800. 2011. ISSN:1000-0917 . [url] [pdf]
@article{RISC4569,
author = {Qunying Liao and Liangjie Ye},
title = {{Lower Bounds and Constructions for q-ary Codes Correcting Asymmetric Errors}},
language = {english},
|
2020-01-28 11:13:52
|
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http://school2016.imj-prg.fr/abstract.html
|
# Mini-courses
### Mohammed Abouzaid : Symplectic topology and non-archimedean geometry
Notes by Kevin Sackel: lecture 1, lecture 2, complements.
The basic idea of family Floer cohomology in the non-exact setting, following Fukaya, is that the Floer cohomology groups of family of Lagrangian submanifolds with non-trivial flux form a (coherent analytic) sheaf over an underlying parameter space. I will explain a different point of view wherein there is an enlargement of the Fukaya category which admits as additional objects certain analytic sheaves on this parameter space. The focus will be on a situation where technical difficulties are avoided by excluding holomorphic disc bubbling.
Complementary lecture by Jingyu Zhao: Convergence of the differential in family Floer homology
Following Mohammed Abouzaid's lectures, I will first recall the differential for the family Floer complex between a tautologically unobstructed Lagrangian and fibres of a Lagrangian torus fibration of the symplectic manifold.
Using the works of Fukaya and Groman-Solomon, we explain that the family Floer differential constructed above can be viewed as a convergent function on an affinoid domain of the rigid analytic mirror.
### Denis Auroux : Homological mirror symmetry: cylinders, pants, and beyond
Notes by Kevin Sackel: lecture 1, lecture 2, lecture 3, complements 1, complements 2.
The goal of this lecture series will be to illustrate homological mirror symmetry by focusing on some simple examples. The basic example will be the cylinder $T^*S^1 = \mathbb{C}^*$; we will then discuss what happens when this example is modified by adding or removing points. Along the way we will encounter wrapped and partially wrapped Fukaya categories of these spaces and their mirrors. While the basic examples may seem elementary, they illustrate general features of homological mirror symmetry and provide test cases for work in progress on hypersurfaces in toric varieties.
Complementary lectures by Sheel Ganatra
In these complementary theory sessions, we will study some structural aspects of Fukaya categories that are pertinent to proving homological mirror symmetry in some of the situations arising in Denis Auroux’s lectures. Potential topics include: (a) Fukaya categories of Landau-Ginzburg models (LG) $(E,W)$ (as a special type of partially wrapped Fukaya category) and their relationship to ordinary and wrapped Fukaya categories, (b) mechanisms for finding (split-)generators for Fukaya categories, (c) miscellaneous topics and/or examples.
### Damien Calaque : Derived symplectic geometry and applications
Notes by Manuel Rivera: lectures 1,2,3
1. We will start with a discussion of linear algebra in the higher categorical setting: derived, $DG-$ and $A_\infty$ categories. Stable $k$-linear $\infty$-categories. Derived symplectic linear algebra.
2. Shifted symplectic structures on derived stacks, after Pantev-Toën-Vaquié-Vezzosi. Examples of symplectic and Lagrangian derived stacks. Topological field theories from shifted symplectic structures.
3. We will discuss some recent results of Brav-Dyckerhoff about topological Fukaya categories: Calabi-Yau structures on $DG-$categories and shifted symplectic structures on topological Fukaya categories of surfaces.
Complementary lectures by Grégory Ginot
### Julien Grivaux : Microlocal theory of sheaves and symplectic topology I
Notes by Fabio Gironella: lectures 1,2,3
1. Homological algebra and sheaves
2. Micro-support of sheaves, involutivity, Morse theory
3. Operations, $\mathbb{R}$-constructible sheaves
Complementary lectures by Pierre Schapira:
1. Complements and examples
2. Examples of microsupports, links with D-modules
### Stéphane Guillermou : Microlocal theory of sheaves and symplectic topology II
Notes (by the speaker)
Notes by Vincent Humilière: lectures 1,2,3 + complements
1. $\mu hom$, simple and pure sheaves, the Kashiwara-Schapira stack
2. Quantization of exact Lagrangians in cotangent bundle (I)
3. Quantization of exact Lagrangians in cotangent bundle (II)
We will associate a sheaf to a given compact exact Lagrangian submanifold of a cotangent bundle and see how to deduce that this Lagrangian has the homotopy type of the base.
Complementary lecture by Nicolas Vichery:
Examples of quantization of Lagrangian submanifolds, quantization of Hamiltonian isotopies
### Maxim Kontsevich : Quantization and Fukaya categories for complex symplectic manifolds
-
Notes by Ailsa Keating: lecture 1
Notes by Denis Auroux: lecture 2
### Nick Sheridan : Introduction to the Fukaya category
Notes (by Sheridan), solutions to exercises (by Maydanskiy)
Notes taken by Dingyu Yang: lecture 1, lecture 2, lecture 3, complements 1, complements 2.
Notes by Amiel Peiffer-Smadja: lectures 1,2,3
1. Lagrangian Floer cohomology: We will introduce the basics of Lagrangian Floer cohomology, using the Arnold conjecture as motivation. This will involve discussing the Novikov field, transversality, compactness and gluing.
2. Product structures: We will introduce the Fukaya category, and give example computations. We will discuss obstructions, and how to deal with them.
3. Triangulated structure: This talk will focus on the triangulated structure of the derived Fukaya category. Examples will include the relationship between Lagrangian surgery and cones, and Seidel's long exact sequence for a Dehn twist.
Complementary lectures by Maksim Maydanskiy:
In these supplementary sessions we will take up, in the form of exercise solutions, some additional material on Lagrangian Floer cohomology and Fukaya category as described the first two talks of N. Sheridan. With audience participation, we will select for discussion a few topics from the following: interpretation of holomorphic strips as gradient flowlines of the action functional, explicit examples of Gromov convergence, gradings and computations of Floer cohomology in (real) dimension 2, relations of holomorphic discs to displacement energy, identification of Morse-Witten and Floer complexes (in the cotangent bundle case), Stasheff associahedra and moduli of holomorphic discs.
# Research talks
### Sheng-Fu Chiu : Sheaf-theoretic invariant and non-squeezability of contact balls.
We apply microlocal category methods to a contact non-squeezing conjecture proposed by Eliashberg, Kim and Polterovich. Let $V_R$ be the product of the $2n$-dimensional open ball of radius $R$ and the circle with a circumference of one. Let $R>r$ be some positive numbers. If $\pi r^2$ is no less than 1, then it is impossible to squeeze $V_R$ into $V_r$ via compactly supported contact isotopies.
### Kenji Fukaya : Formality of various structures in Lagrangian Floer thoery
When we consider Hochschild or cyclic cohomology of the $A_\infty$ category $Fuk(X)$ for a symplectic manifold $X$, it has various structures such as L infinity structure, involutive bi Lie infinity structure. In case $X$ is a compact symplectic manifold such structure becomes formal (that is all the operations are trivial) in various situations. In this talk I want to explain such phenomenon as well as the construction of those strutures.
### Penka Georgieva : Real Gromov-Witten invariants
The Gromov-Witten invariants play a central role in the mirror symmetry conjecture which in turn gives predictions for these invariants. Many such predictions for closed Gromov-Witten invariants have been established mathematically. Similar predictions exist for open and real Gromov-Witten invariants and I will discuss some of the difficulties related to understanding the open invariants and recent advances in the real case.
### Thomas Kragh : Generating families for Lagrangians in $\mathbb{R}^{2n}$
I will sketch how to create a generating family quadratic at infinity for any exact Lagrangian in $\mathbb{R}^{2n}$ equal to standard $\mathbb{R}^n$ outside a compact set. I will then if time permits also discus some consequences of this.
### Yanki Lekili : Associative Yang-Baxter equation and Fukaya categories of square-tiled surfaces
Since its discovery in the 60s, Yang-Baxter equation (YBE) has been studied extensively as the master equation in integrable models in statistical mechanics and quantum field theory. In 2000, Polishchuk discovered a connection between the solutions to Yang-Baxter equations (classical, associative and quantum) and the Massey products in a Calabi-Yau 1-category, and using this he was able to construct geometrically some of the trigonometric solutions of the YBE coming from simple vector bundles on cycles of projective lines. We first prove a homological mirror symmetry statement, hence see these trigonometric solutions to YBE via the Fukaya category of punctured tori. Next, we consider Fukaya categories of higher genus square-tiled surfaces to give a geometric construction of all the trigonometric solutions to associative Yang-Baxter equation parametrized by the associative analogue of the Belavin-Drinfeld data. This is based on joint work with Polishchuk.
### David Nadler : Arboreal singularities
I will survey recent work devoted to singularities of Lagrangian skeleta, with a focus on applications to mirror symmetry.
### Paul Seidel : Formal ODEs in symplectic cohomology
We consider the symplectic cohomology of the total space of a Lefschetz fibration. Under suitable assumptions, this can be equipped with a connection (an operator of differentiation with respect to the Novikov variable). We will show that with respect to this operator, the Borman-Sheridan class satisfies a nonlinear first order differential equation (a Riccati equation).
### Vivek Shende : Localization of the wrapped Fukaya category
I will sketch an argument that the wrapped Fukaya category localizes to a cosheaf of categories on a Lagrangian skeleton, locally modeled on the cosheaf dual to the Kashiwara-Schapira sheaf. This is joint work with Sheel Ganatra and John Pardon.
### Ivan Smith : Spherical objects on surfaces
Recent work of Haiden, Katzarkov and Kontsevich leads to a classification of objects in derived (wrapped) Fukaya categories of punctured surfaces. We will describe an attempt to partially extend this classification to closed surfaces of higher genus, and discuss possible applications of such an extension. This is joint speculation-in-progress with Denis Auroux.
### Dmitry Tamarkin : On the microlocal category
I am planning to overview my preprint arXiv:1511.08961 'On the microlocal category' where a $dg$-category over $\mathbb{Q}$ is associated to a compact symplectic manifold whose symplectic form has integral periods. The properties of this category are similar to those of the Fukaya category.
### David Treumann : The $J$-homomorphism in microlocal sheaf theory
Let $L$ be an exact Lagrangian submanifold of a cotangent bundle $T^*M$. If a topological obstruction vanishes, a local system of $R$-modules on $L$ determines a constructible sheaf of $R$-modules on $M$ -- this is the Nadler-Zaslow construction. I will discuss a variant of this construction that avoids Floer theory, and that allows $R$ to be a ring spectrum. The talk is based on joint work with Xin Jin.
### Boris Tsygan : Microlocal category of a symplectic manifold and deformation theory: the Poincaré lemma.
Notes
I will outline a construction of a microlocal category of a symplectic manifold using déformation quantization rather than sheaf theory. This is supposed to be an analog of the De Thames definition of cohomology of manifolds as opposed to the sheaf theoretical definition. I will concentrate on the local computation that amounts to a version of the Poincaré Lemma (it is also some sort of a stationary phase statement). I will address the local to global gluing question only briefly.
### Eric Zaslow : Applications of the Chromatic Lagrangian
Beginning with a cubic, planar graph, I will define a Legendrian surface in the cosphere bundle of three-space, equivalently the first jet bundle of the two-sphere. Its wavefront projects generically two-to-one onto the base two-sphere, but is one-to-one over the graph. The Lagrangian defines a singular support condition for both a category of constructible sheaves and for a Fukaya category. I will describe the moduli space of objects of this category, and study two applications.
First, the moduli space can be defined over a finite field, in which case the number of points can be related to the chromatic polynomial of the dual graph. Using this observation, I will show that none of the Lagrangian surfaces admits a smooth, exact Lagrangian filling in six-space. Instead, I will describe the construction of fillings which are not exact, and in fact obstructed. Second, the moduli space sits as a Lagrangian submanifold of a symplectic period domain. It has a cover which is exact in a cotangent. Generalizing Aganagic-Vafa mirror symmetry, I will exhibit this Lagrangian in examples as the graph of the differential of a superpotential written as an integral linear combination of dilogarithms in special coordinates. The superpotential conjecturally encodes the Ooguri-Vafa invariants, and with them the open Gromov-Witten invariants of the obstructed Lagrangian.
This is joint work with David Treumann.
# Posters
### Daniel Alvarez Gavela: The simplification of singularities of Lagrangian and Legendrian fronts
Poster
We present a flexibility result with applications in the computation of holomorphic invariants of Lagrangian and Legendrian submanifolds. Legendrian contact homology for 1-dimensional Legendrian knots in standard contact $\mathbb{R}^3$ can be understood combinatorially via the Chekanov dga of the Lagrangian projection of the knot, or equivalently in terms of its Legendrian projection. Similarly, recent work of Ng, Rutherford, Shende, Sivek and Zaslow shows that the $A_\infty$ category of augmentations of the Legendrian contact homology dga can be understood in terms of a certain category of sheaves in the front plane. A major obstruction to extending these results to higher dimensions is the fact that generically Legendrian fronts have terrible singularities, rendering the combinatorics intractable. This is just one of many examples where one would like a Lagrangian or Legendrian front to have singularities that are as simple as possible. Other examples include a desire to globally understand family Floer homology or an interest in studying the space of 1-dimensional Legendrian knots (the fronts of families depending on many parameters will also generically have terrible singularities). We prove an $h$-principle which shows that whenever the obvious homotopy theoretic obstruction to simplifying the singularities of a Lagrangian or Legendrian front vanishes, then the simplification can be achieved by means of an ambient Hamiltonian isotopy. In many cases this homotopy theoretic obstruction can be easily shown to vanish, for example for even dimensional Legendrian spheres in standard Euclidean space.
### Dori Bejleri: Motivic Hilbert zeta functions of curves.
Poster
The Hilbert schemes $Hilb^n(C)$ of a singular curve $C$ parametrize $n$-dimensional quotients of the structure sheaf of. We study the motivic Hilbert zeta function $Z_C(t)$ which is the generating function for $Hilb^n(C)$ in the Grothendieck ring of varieties. When $C$ is planar, work of Oblomkov-Shende, Maulik, and others showed that after passing to Euler characteristic, a refinement of $Z_C(t)$ is the HOMFLY polynomial of the link of the singularity.
With D. Ranganathan and R. Vakil, we study the structure of $Z_C(t)$ for general curve singularities. We show that $Z_C(t)$ is a rational function for any curve with isolated singularities and study to what extent it exhibits a functional equation and topological invariance. Inspired by physics, with A. Takeda we explore a conjectural relation between $Z_C(t)$ and knot invariants for general isolated curve singularities and relate this to the study of a certain generalization of the Hitchin fibration.
### Maia Fraser: Contact non-squeezing via generating functions
Poster
Persistence modules are simply functors from a poset (often $\mathbb{R}$) to Vect. They have been used in symplectic geometry going back to the symplectic capacity of Viterbo and Floer-theoretic analogs (Oh, Schwartz) although the terminology of persistence, barcodes etc... arose later in Computer Science when persistence modules were studied as tools for topological data analysis. In particular Viterbo's symplectic capacity of domains in $\mathbb{R}^{2n}$ and Sandon's contact capacity of domains in $\mathbb{R}^{2n} \times S^1$ are persistences of certain homology classes in the persistence module formed by generating function (GF) homology groups. While Sandon's capacity $c_S(-)$ allows to re-prove non-squeezing of $B(R) \times S^1$ into itself for integral $R$ (a result due to Eliashberg-Kim-Polterovich 2006), by introducing new filtration-decreasing morphisms between GF homology groups one can set up a functor from a sub-category of $\mathcal{D} \times \mathbb{Z}$ to Vect, where $\mathcal{D}$ is the category of bounded domains with inclusion. Persistences in this persistence module then yield a sequence $m_\ell(-)$, $\ell \in \mathbb{N}$ of integer-valued contact invariants for prequantized balls, such that $m_1 = c_S - 1$ and $m_\ell(B(R) \times S^1)$ is the greatest integer strictly less than $\ell R$. This provides an alternate proof of non-squeezing at large scale, i.e. of $B(R) \times S^1$ into itself for any $R>1$ (proved by Chiu 2014).
### Honghao Gao: Knot invariants via microlocal categories
Poster
One perspective on knot invariants is to study the contact/symplectic geometry of the conformal bundle of the knot. Sheaves with singular support along this conormal give rise to a categorical knot invariant. The front projection of the Legendrian conormal bundle also defines a sheaf category, but on a different space. We define a functor between the two sheaf categories and show it is an equivalence. We also study how microlocal rank-1 sheaves transform under this equivalence.
### Justin Hilburn: GKZ-Hypergeometric Systems and Tilting Sheaves in Hypertoric Category $O$
Poster
It is well known that the Bernstein-Gelfand-Gelfand category $O$ for a Lie algebra $g$ behaves roughly like the Fukaya category of the cotangent bundle to the flag variety $G/B$. In "Morse theory and tilting sheaves" Nadler defined a geometric construction of tilting modules in category $O$ by taking certain constructible sheaves known as "Morse kernels" on the flag variety and flowing them along a $C^*$-action whose ascending manifold stratification coincides with the Schubert stratification. Braden-Licata-Proudfoot-Webster have defined a version of category $O$ for any suitable holomorphic symplectic variety. Other than cotangent bundles the simplest family of holomorphic symplectic varieties are the hypertoric varieties of Bielawski-Dancer. I will describe a construction that generalizes Nadler's result to hypertoric category $O$. In this case, the "Morse kernels" are certain GKZ integrable systems. This construction arose from joint work with Bullimore, Dimofte, and Gaiotto on mirror symmetry of 3d $N=4$ gauge theories.
### Parvaneh Joharinad: Warped product Finsler manifolds as a Hamiltonian formalism and conformal gradient fields
Poster
In Riemannian and semi Riemannian geometry, metric tensor has a warped product structure in foliated chart related to conformal gradient fields, which helps to obtain global structural information about manifold. This correspondence between existence of a conformal gradient field and warped product structure has not been yet provedfor Finsler metrics. In this proposed poster, I will present a warped product Finsler structure from Hamiltonian point of view and explore the functions which have conformal gradient and the structural information obtained in presence of such a vector field.
### Momchil Konstantinov: Floer Theory with Local Coefficients and an Application to the Chiang Lagrangian
Poster
This piece of work is concerned with the use of local systems of rank higher than 1 as coefficient spaces for Lagrangian Floer Theory in the monotone case, particularly when pseudoholomorphic discs of Maslov index 2 with boundary on a Lagrangian cannot be excluded. We use this technique to establish that the Chiang Lagrangian in $CP^3$, when equipped with an appropriate local system of rank 2, can be ensured to have non-zero Floer cohomology with itself over characteristic 2. This is an extension of work by Evans and Lekili ("Floer cohomology of the Chiang Lagrangian." Selecta Mathematica 21.4 (2015): 1361-1404) and provides a negative answer to their question whether the Chiang Lagrangian and $RP^3$ can be Hamiltonianly displaced from each other.
### Cheuk Yu Mak
Poster
Utilizing Biran-Cornea Lagrangian cobordism theory and Mau-Wehrheim-Woodward functor, we give an alternative proof of Seidel's long exact sequence and Wehrheim-Woodward fibered Dehn twist long exact sequence in the monotone setting. Our approach also confirms Huybrechts-Thomas’s prediction of projective Dehn twist modulo determination of connecting morphism. Emphasis is put on the functorial perspective of the approach and possible generalization to other long exact sequences. This is a joint work with Weiwei Wu.
### Ziva Myer: $A_\infty$ Algebras for a Legendrian Submanifold from Generating Families
Poster
A generating family for a Legendrian submanifold is a family of functions that encodes information about the Reeb chords of the Legendrian. I use generating families to construct an A-infinity algebra using Morse flow trees, and show this algebra is invariant up to A-infinity quasi-isomorphism under Legendrian isotopy.
### Jie Ren: Cohomological Hall algebras, semicanonical bases and Donaldson-Thomas invariants for 2-dimensional Calabi-Yau categories
Poster
We discuss semicanonical bases from the point of view of Cohomological Hall algebras via the "dimensional reduction" from 3-dimensional Calabi-Yau categories to 2-dimensional ones. Also, we discuss the notion of motivic Donaldson-Thomas invariants (as defined by M. Kontsevich and Y. Soibelman) in the framework of 2-dimensional Calabi-Yau categories. In particular we propose a conjecture which allows one to define Kac polynomials for a 2-dimensional Calabi-Yau category (this is a theorem of S. Mozgovoy in the case of preprojective algebras).
### Tobias Sodoge
Poster
Kaputsin and Orlov have suggested to enlarge the Fukaya category with objects arising from coisotropic submanifolds for the HMS conjecture to be true. However, not much is know about coisotropics so far. We show that every stable, fibered, displaceable coisotropic submanifold is a torus fibration over a symplectically uniruled base. To the coisotropic $C$ we associate a Lagrangian $L_C$ and a stable hypersurface $H_C$ . We then analyse the pearly differential of the Floer complex of $L_C$ and use neck stretching around $H_C$ to produce a non constant sphere though every point in the symplectic base of the coisotropic.
### Alex Takeda: Locally arboreal spaces and derived symplectic geometry
Poster
In recent work of Shende, Nadler, Zaslow et al, many spaces of classical interest such as character varieties of surfaces, wild character varieties, augmentation varieties appearing in knot theory etc. can be phrased as particular cases of moduli spaces of constructible sheaves on stratified spaces. One common feature of such spaces is that they all carry natural Poisson or symplectic structures, a fact that has been explained ad hoc in the above cases. In my work with Shende, using derived symplectic geometry we give a general method for constructing such symplectic structures, and show that the existence of these structures is related to orientations of the corresponding category of constructible sheaves. More generally, we define a larger class of locally arboreal spaces, of which all the spaces above are particular cases, and prove that all such spaces carry natural shifted symplectic structures.
References:
V. Shende and A. Takeda, Orientations on locally arboreal spaces, in preparation
V. Shende, D. Treumann, H. Williams and E. Zaslow, Cluster varieties and Legendrian knots, arXiv:1512.08942
T. Pantev, B. Toën, M. Vaquié, G. Vezzosi, Shifted symplectic structures , arXiv:1111.3209
### Jian Wang
Poster
Recently, we generalize Schwarz's theorem to $C^0$-case on aspherical closed surfaces and prove that the contractible fixed points set (and consequently the fixed points set) of a nontrivial Hamiltonian homeomorphism is not connected.
### Tatsuki Kuwagaki and Yuichi Ike: Categorical localization for the coherent-constructible correspondence
It is expected that homological mirror symmetry for the complements of a divisor in a variety is obtained from the categorical localization of homological mirror symmetry for the ambient variety. We give this picture in the coherent-constructible correspondence which is a version of homological mirror symmetry for toric varieties given by Fang-Liu-Treumann-Zaslow. In our description, we use the concept of wrapped constructible sheaves which is recently introduced by Nadler.
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2019-03-19 17:00:28
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https://www.jiskha.com/display.cgi?id=1170488945
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# math,correction
posted by jasmine20
Directions are: Perform the indicated division. Rationalize the denominator if necessary. Then simplify each radical expression.
(18 + sqrt567)/(9)
this is what i got: (41.811761799581315314514541782753)/(9)=4.6457513110645905905016157536393
Your answer is silly. I have about reached my limit. You know well that simplify does not mean use your calculator to punch buttons.
What is 567 divided by 81?
## Similar Questions
1. ### math correction
Directions: Perform the indicated divisions. The problem: (8x^3-6x^2+2x)/(4x+1) This is what i did and what i got: (4x^5)/(5x) = The answer that i got was: 0.80x^4 It is so wrong it hurts. How in the world did you get (8x^3-6x^2+2x) …
2. ### math, algebra,correction
I don't know if this is right. directions: Simplify each radical expression. (radical 3+5)(radical 3-3) i think its: radical 3+2 Life would be so much simpler if you would use parenthesis. Did you mean this?
5. ### math,correction,plz
Perform the indicated division. Rationalize the denominator if necessary. Then simplify each radical expression. Problem #54 (6-radical(20))/(2) My answer: 3-radical(5) Problem #50 Find the area of the rectangle shown in the figure. …
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2018-05-23 14:50:42
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https://sdk.numxl.com/help/ndk-mlr-fitted
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# NDK_MLR_FITTED
int __stdcall NDK_MLR_FITTED ( double ** X, size_t nXSize, size_t nXVars, LPBYTE mask, size_t nMaskLen, double * Y, size_t nYSize, double intercept, WORD nRetType )
Returns the fitted values of the conditional mean, residuals or leverage measures.
Returns
status code of the operation
Return values
NDK_SUCCESS Operation successful NDK_FAILED Operation unsuccessful. See Macros for full list.
Parameters
[in] X is the independent (explanatory) variables data matrix, such that each column represents one variable. [in] nXSize is the number of observations (rows) in X. [in] nXVars is the number of independent (explanatory) variables (columns) in X. [in] mask is the boolean array to choose the explanatory variables in the model. If missing, all variables in X are included. [in] nMaskLen is the number of elements in the "mask." [in] Y is the response or dependent variable data array (one dimensional array of cells). [in] nYSize is the number of observations in Y. [in] intercept is the constant or intercept value to fix (e.g. zero). If missing (i.e. NaN), an intercept will not be fixed and is computed normally. [in] nRetType is a switch to select the return output (1=fitted values (default), 2=residuals, 3=standardized residuals, 4=leverage, 5=Cook's distance). Fitted/conditional mean Residuals Standardized residuals Leverage factor (H) Cook's distance (D)
Remarks
1. The underlying model is described here.
2. The regression fitted (aka estimated) conditional mean is calculated as follows: $\hat y_i = E \left[ Y| x_i1\cdots x_ip \right] = \alpha + \hat \beta_1 \times x_i1 + \cdots + \beta_p \times x_ip$ Residuals are defined as follows: $e_i = y_i - \hat y_i$ The standardized residuals are calculated as follow: $\bar e_i = \frac{e_i}{\hat \sigma_i}$ Where:
• $$\hat y$$ is the estimated regression value.
• $$e$$ is the error term in the regression.
• $$\hat e$$ is the standardized error term.
• $$\hat \sigma_i$$ is the standard error for the i-th observation.
3. For the influential data analysis, SLR_FITTED computes two values: leverage statistics and Cook's distance for observations in our sample data.
4. Leverage statistics describe the influence that each observed value has on the fitted value for that same observation. By definition, the diagonal elements of the hat matrix are the leverages. $H = X \left(X^\top X \right)^{-1} X^\top$ $L_i = h_{ii}$ Where:
• $$H$$ is the Hat matrix for uncorrelated error terms.
• $$\mathbf{X}$$ is a (N x p+1) matrix of explanatory variables where the first column is all ones.
• $$L_i$$ is the leverage statistics for the i-th observation.
• $$h_{ii}$$ is the i-th diagonal element in the hat matrix.
5. Cook's distance measures the effect of deleting a given observation. Data points with large residuals (outliers) and/or high leverage may distort the outcome and accuracy of a regression. Points with a large Cook's distance are considered to merit closer examination in the analysis. $D_i = \frac{e_i^2}{p \ \mathrm{MSE}}\left[\frac{h_{ii}}{(1-h_{ii})^2}\right]$ Where
• $$D_i$$ is the cook's distance for the i-th observation.
• $$h_{ii}$$ is the leverage statistics (or the i-th diagonal element in the hat matrix).
• $$\mathrm{MSE}$$ is the mean square error of the regression model.
• $$p$$ is the number of explanatory variables.
• $$e_i$$ is the error term (residual) for the i-th observation.
6. The sample data may include missing values.
7. Each column in the input matrix corresponds to a separate variable.
8. Each row in the input matrix corresponds to an observation.
9. Observations (i.e. row) with missing values in X or Y are removed.
10. The number of rows of the response variable (Y) must be equal to number of rows of the explanatory variables (X).
11. The MLR_FITTED function is available starting with version 1.60 APACHE.
Requirements
Namespace: NumXLAPI Class: SFSDK Scope: Public Lifetime: Static
int __stdcall NDK_MLR_FITTED ( double ** X, size_t nXSize, size_t nXVars, LPBYTE mask, size_t nMaskLen, double * Y, size_t nYSize, double intercept, WORD nRetType )
Returns the fitted values of the conditional mean, residuals or leverage measures.
Returns
status code of the operation
Return values
NDK_SUCCESS Operation successful NDK_FAILED Operation unsuccessful. See Macros for full list.
Parameters
[in] X is the independent (explanatory) variables data matrix, such that each column represents one variable. [in] nXSize is the number of observations (rows) in X. [in] nXVars is the number of independent (explanatory) variables (columns) in X. [in] mask is the boolean array to choose the explanatory variables in the model. If missing, all variables in X are included. [in] nMaskLen is the number of elements in the "mask." [in] Y is the response or dependent variable data array (one dimensional array of cells). [in] nYSize is the number of observations in Y. [in] intercept is the constant or intercept value to fix (e.g. zero). If missing (i.e. NaN), an intercept will not be fixed and is computed normally. [in] nRetType is a switch to select the return output (1=fitted values (default), 2=residuals, 3=standardized residuals, 4=leverage, 5=Cook's distance). Fitted/conditional mean Residuals Standardized residuals Leverage factor (H) Cook's distance (D)
Remarks
1. The underlying model is described here.
2. The regression fitted (aka estimated) conditional mean is calculated as follows: $\hat y_i = E \left[ Y| x_i1\cdots x_ip \right] = \alpha + \hat \beta_1 \times x_i1 + \cdots + \beta_p \times x_ip$ Residuals are defined as follows: $e_i = y_i - \hat y_i$ The standardized residuals are calculated as follow: $\bar e_i = \frac{e_i}{\hat \sigma_i}$ Where:
• $$\hat y$$ is the estimated regression value.
• $$e$$ is the error term in the regression.
• $$\hat e$$ is the standardized error term.
• $$\hat \sigma_i$$ is the standard error for the i-th observation.
3. For the influential data analysis, SLR_FITTED computes two values: leverage statistics and Cook's distance for observations in our sample data.
4. Leverage statistics describe the influence that each observed value has on the fitted value for that same observation. By definition, the diagonal elements of the hat matrix are the leverages. $H = X \left(X^\top X \right)^{-1} X^\top$ $L_i = h_{ii}$ Where:
• $$H$$ is the Hat matrix for uncorrelated error terms.
• $$\mathbf{X}$$ is a (N x p+1) matrix of explanatory variables where the first column is all ones.
• $$L_i$$ is the leverage statistics for the i-th observation.
• $$h_{ii}$$ is the i-th diagonal element in the hat matrix.
5. Cook's distance measures the effect of deleting a given observation. Data points with large residuals (outliers) and/or high leverage may distort the outcome and accuracy of a regression. Points with a large Cook's distance are considered to merit closer examination in the analysis. $D_i = \frac{e_i^2}{p \ \mathrm{MSE}}\left[\frac{h_{ii}}{(1-h_{ii})^2}\right]$ Where
• $$D_i$$ is the cook's distance for the i-th observation.
• $$h_{ii}$$ is the leverage statistics (or the i-th diagonal element in the hat matrix).
• $$\mathrm{MSE}$$ is the mean square error of the regression model.
• $$p$$ is the number of explanatory variables.
• $$e_i$$ is the error term (residual) for the i-th observation.
6. The sample data may include missing values.
7. Each column in the input matrix corresponds to a separate variable.
8. Each row in the input matrix corresponds to an observation.
9. Observations (i.e. row) with missing values in X or Y are removed.
10. The number of rows of the response variable (Y) must be equal to number of rows of the explanatory variables (X).
11. The MLR_FITTED function is available starting with version 1.60 APACHE.
Requirements
References
* Hamilton, J .D.; Time Series Analysis , Princeton University Press (1994), ISBN 0-691-04289-6
* Tsay, Ruey S.; Analysis of Financial Time Series John Wiley & SONS. (2005), ISBN 0-471-690740
* D. S.G. Pollock; Handbook of Time Series Analysis, Signal Processing, and Dynamics; Academic Press; Har/Cdr edition(Nov 17, 1999), ISBN: 125609906
* Box, Jenkins and Reisel; Time Series Analysis: Forecasting and Control; John Wiley & SONS.; 4th edition(Jun 30, 2008), ISBN: 470272848
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2018-03-23 22:31:17
|
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https://ggirelli.info/2020/01/26/multirange.html
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# Printer-styled range in Python - a case study
Recently, I had to convert a printer-styled range string into the corresponding list of selected elements. An excellent case study on Python class properties, iterability, length, and attribute/method privacy.
## A feature request
So, there’s this script of mine that some colleagues use to convert microscopy images from the microscope vendor’s proprietary format (we have a Nikon one in the lab, so some nd2 files) to an open-source one (.tif in this case).
Recently, a colleague asked me to add a feature to this script. They would like to be able to convert only some fields of view. To do this, I decided to use a range string format that is ubiquitous, OS-independent, and widely used every day: the printer-styled range string.
This way of selecting items is as simple as having the pages comma-separated. Let’s say, if I wanted to print the pages numbered 1, 5, and 6, then I would write 1,5,6. Also, page ranges are allowed by using the dash (-): to print page from 3 to 6 (both included), I would write 3-6. And the two things can be combined into 1,3-6.
## A Python3.6+ class
So, I needed something to validate, parse, and convert a printer-styled range string into an actual series of numbers. The desired behavior would be to convert 1,3-6 into [0,2,3,4,5] (remember, Python is a 0-indexed language).
To do this, I implemented a Python3.6+ class, MultiRange, that I then published as a Gist on GitHub (see below). Here, I break it down and explain it bit by bit.
### Dependencies
import re
from typing import Iterator, List, Optional, Pattern, Tuple
The class has only two dependencies from the Python standard libraries:
• re provides regular expression related features. We use it to validate the printer-style range string.
• typing is used to provide type hints and help developers.
### The class attributes
__current_item: Tuple[int] = (0, 0)
__string_range: Optional[str] = None
__extremes_list: Optional[List[Tuple[int]]] = None
__reg: Pattern = re.compile(r'^[0-9-, ]+\$')
__length: Optional[int] = None
• The __string_range attribute contains the printer-styled range string that we want to convert into a list of numbers. For example, 1,3-6.
• __extremes_list contains a list of tuples. Each tuple has two elements: the start and stop of a range/slice. The idea is to convert __string_range into something like [(1,1),(3,6)].
• __reg contains the regular expression used to validate the input string. During class instantiation, we match the input string to this. If we do not find any match, the class raises an error. In other words, we allow only strings that include digits, commas, dashes, and spaces.
• More details about __length and __ready are available in the length section below.
• __current_item is necessary to iterate over the indexes stored in the MultiRange, which we discuss in the last section of this article.
As you might have noticed, all attributes start with two underscores __. In Python (according to PEP8), this makes them private, i.e., inaccessible from outside the class through name mangling.
### The __init__ method
def __init__(self, s: str):
super(MultiRange, self).__init__()
assert self.__reg.search(s) is not None, (
"cannot parse range string. It should only contains numbers, "
+ "commas, dashes, and spaces.")
self.__string_range = s
string_range_list = [b.strip() for b in self.__string_range.split(",")]
self.__extremes_list = []
for string_range in string_range_list:
extremes = [int(x) for x in string_range.split("-")]
if 1 == len(extremes):
extremes = [extremes[0], extremes[0]]
assert 2 == len(extremes), "a range should be specified as A-B"
assert extremes[1] >= extremes[0]
self.__extremes_list.append(tuple(extremes))
self.__extremes_list = sorted(
self.__extremes_list, key=lambda x: x[0])
self.__clean_extremes_list()
assert 0 < self.__extremes_list[0][0], "'page' count starts from 1."
The __init__ method creates an instance of the class and takes the printer-style range string as input (s). The first thing we want to do is validate the input by checking it against __reg, and stop otherwise by raising an AssertError.
After that, we store the input string in __string_range and split it into elements by using the commas as delimiters and removing any leading or trailing blank space.
Each element is now a string containing either a single page ("1") or a page range ("3-6"). We identify these two cases by splitting each element by the dashes and counting the number of generated elements. In the case of a single page, we convert it to a number (int) and store it into __extremes_list as (1,1). In the case of a page range, we save it as (3,6).
Then, we sort the list of tuples we created by their first elements, and clean them up. This cleaning operation avoids overlaps between tuples (see the next section for more details).
Finally, we verify that the first tuple does not start with a 0 (page numbers start from 1) and tell the class that it is __ready for use.
### Cleaning the list of extremes
def __clean_extremes_list(self) -> None:
is_clean = False
while not is_clean:
popped = 0
i = 0
while i < len(self.__extremes_list)-1:
A = self.__extremes_list[i]
B = self.__extremes_list[i+1]
if A[1] >= B[0] and A[1] < B[1]:
self.__extremes_list[i] = (A[0], B[1])
self.__extremes_list.pop(i+1)
popped = 1
break
elif A[1] >= B[1]:
self.__extremes_list.pop(i+1)
popped = 1
break
i += 1
if i >= len(self.__extremes_list)-2+popped:
is_clean = True
After converting the input string into a list of number pairs (as tuples), each representing a range of pages from the first number to the second one, both included. You can imagine any of these pairs as a (start,end) element.
Before proceeding, we want to be sure that every page number appearson only once in the MultiRange. In other words, all these ranges should not overlap. We achieve this by iterating through the list of (start,end) elements and comparing each pair with the next pair and see if they overlap. For this to work, it is crucial to have the list of pairs sorted by their start element (which we did in the __init__ function).
Now, when are two pairs overlapping? Given that the first pair (A) has a start (A[0]) that is always lower than or equal to the start of the second one B[0], this can happen in two ways:
1. The first pair fully includes the second one. In other words, the first pair ends after or precisely when the second one ends: A[1] >= B[1].
2. The two pairs partially overlap. The first pair ends after or precisely when the second one starts (A[1] >= B[0]), but the first pair ends before the second one ends (A[1] < B[1]).
It is important to distinguish these two scenarios because they require us to act differently:
1. If the first pair fully includes the second one, we can simply remove (pop) it from our list of ranges.
2. If the overlap is only partial, we need to merge the two ranges. We can achieve this by removing (popping) the second one, and replacing the end position of the first one with the end of the second. Simply put, we remove both pairs and place a new one: (A[0], B[1]).
Every time we found overlapping pairs, we resolve the conflict and restart from the beginning of the list. Only when we reach the end of the list, we can say that it is indeed clean and ready to be used.
### The MultiRange length
@property
def length(self):
if self.__length is None and self.__ready:
self.__length = 0
for a, b in self.__extremes_list:
self.__length += b-a+1
return self.__length
def __len__(self) -> Optional[int]:
return self.length
We want to be able to know the length of our MultiRange instance. But given that this requires some computation, we want to (1) calculate it only if the MultiRange is __ready to be used, (2) calculate it once, (3) store it somewhere (self.__length), (4) read the stored value at any future call of the len function.
To provide a hook for the len function and make the len(MultiRange("1,2-6")) code work, we need to define a __len__ method. The method returns an integer number: the one we plan to store in __length.
We do not want anyone to be able to mess with the stored value, that’s why we store it in the private attribute __length (notice those __ at the beginning!). But we want a user to be able to access it without having to call the len function. To do this, we define a class property with the @property decorator.
Every time the user calls MultiRange(" 1,2-6").length, it triggers the property. This checks if the class is ready (self.__ready) and if no length value has been previously stored (self.__length is None). Only and only in this case it calculates the length value and stores it in __length. Otherwise, it returns the default length value: None.
To calculate the MultiRange length, we calculate the difference between all range extreme pairs (a,b) and sum 1 to each (as they should both be included): b-a+1. Then, we sum them all and store it in __length!
### How to iterate over the MultiRange elements?
#### What is an ‘iterator’?
First things first: what is and iterator? Well, in Python we can have iterable (1) and/or iterator (2) methods/classes. An iterable contains elements that can be iterated over, while an iterator is used to iterate over an iterable. According to PEP234:
1. An object can be iterated over with for if it implements __iter__() or __getitem__().
2. An object can function as an iterator if it implements next().
Just to re-iterate the concept (pun unintended): if a class includes an __iter__ method it can be iterated over, i.e., the class is iterable. If a class includes a __next__ method, it is an iterator.
Notice that, when a class is an iterator it is, per definition, also iterable. The vice-versa is not true: not all iterables are iterators. While the wording might be confusing, the concept is quite simple. Since an iterator iterates over the elements of something, one could iterate over its elements, which makes it an iterable too. On the other hand, one could potentially be able to iterate over the elements of an iterable only if an iterator is available.
#### Making your own Python iterator
So, now that we know what an iterator is, how can we make our MultiRange class into one? It’s fairly simple: we implemented two methods:
• __next__ is our iterator method, which will return one element of MultiRange at a time, and in order. Once the elements are over, it will raise a StopIteration error.
• __iter__ is our iterable method. When called, it allows a user to iterate over the elements of the MultiRange, starting from its first. This method needs only to reset the iterable location to the first element, and return the class itself. Python will tkae care of the magic behind it all, and link __iter__ and __next__ together.
In __next__ we want to (1) know which element have iterated to and (2) iterate to the next one if possible, otherwise trigger a StopIteration. First, we store the iterator location in __current_item as a tuple containing two numbers: the index of the current range and the index of the current element in that range (starting from (0,0)).
When we reach the last element of a range, we move to the first element of the next range. If we have reached the last element of the last range, it is time to stop.
def __next__(self) -> int:
current_range_id, current_item = self.__current_item
if current_range_id >= len(self.__extremes_list):
raise StopIteration
current_range = self.__extremes_list[current_range_id]
if current_item >= current_range[1]-current_range[0]:
self.__current_item = (current_range_id+1, 0)
else:
self.__current_item = (current_range_id, current_item+1)
return current_range[0]+current_item
def __iter__(self) -> Iterator[int]:
self.__current_item = (0, 0)
return self
|
2020-09-25 14:22:46
|
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|
https://www.physicsforums.com/threads/calculating-acceleration-when-only-have-time-and-distance.845745/
|
# Calculating acceleration when only have time and distance
1. Nov 30, 2015
### artworkmonkey
1. The problem statement, all variables and given/known data
A car travels 400m in 60s. The first 10s it accelerates from stationary. The last 10 seconds it decelerates back to stationary. For the middle 40s it has a constant velocity. What is the acceleration?
2. Relevant equations
None given
3. The attempt at a solution
Using only time and distance travelled I can only think how to calculate the average velocity: Vave = distance travelled / time = 6.67 m/s. I don't so much need an answer, more some guidance of where to begin. I'm pretty stuck. Thanks
2. Nov 30, 2015
### Staff: Mentor
Maybe start graphically? You might find some inspiration. Make a sketch of velocity versus time. You don't know the cruising speed yet, so leave that as a variable. Now, what do you know about the area under a velocity vs time graph?
3. Nov 30, 2015
### artworkmonkey
The area under the graph should equal the displacement, so the total area should equal 400m. I have drawn a graph (attached). When a put the acceleration and deceleration together the area is exactly one 5th of the time spent cruising: therefore, the amount of time spent accelerating and decelerating is one 5th of 400m = 80 meters. Divide this by 2 and the car moved 40 meters in 10 seconds.
Acceleration = distance/time^2 = 40m/10^2 = 0.4m/s
Does this sound right? Making a sketch was good advise because it is starting to make sense to me. Just hope I have taken a wrong turn somewhere.
Thanks :)
#### Attached Files:
• ###### Time vs Velocity.png
File size:
27.3 KB
Views:
88
4. Nov 30, 2015
### Staff: Mentor
Very close indeed. However: $distance = \frac{1}{2} Acceleration \cdot time^2$. So fix that up and you should be good.
5. Nov 30, 2015
### artworkmonkey
Yay! Thank you for your help. I've been staring at this question since yesterday, and actually drawing it was brilliant advice. Thank you!
|
2018-03-21 22:07:22
|
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|
https://machinelearningmastery.com/setting-breakpoints-and-exception-hooks-in-python/
|
# Setting Breakpoints and Exception Hooks in Python
Last Updated on June 21, 2022
There are different ways of debugging code in Python, one of which is to introduce breakpoints into the code at points where one would like to invoke a Python debugger. The statements used to enter a debugging session at different call sites depend on the version of the Python interpreter that one is working with, as we shall see in this tutorial.
In this tutorial, you will discover various ways of setting breakpoints in different versions of Python.
After completing this tutorial, you will know:
• How to invoke the pdb debugger in earlier versions of Python
• How to use the new, built-in breakpoint() function introduced in Python 3.7
• How to write your own breakpoint() function to simplify the debugging process in earlier versions of Python
• How to use a post-mortem debugger
Kick-start your project with my new book Python for Machine Learning, including step-by-step tutorials and the Python source code files for all examples.
Let’s get started.
Setting Breakpoints in Different Versions of Python
Photo by Josh Withers, some rights reserved.
## Tutorial Overview
This tutorial is divided into three parts; they are:
• Setting Breakpoints in Python Code
• Invoking the pdb Debugger in Earlier Versions of Python
• Using the breakpoint() Function in Python 3.7
• Writing One’s Own breakpoint() Function for Earlier Versions of Python
• Limitations of the breakpoint() Function
## Setting Breakpoints in Python Code
We have previously seen that one way of debugging a Python script is to run it in the command line with the Python debugger.
In order to do so, we would need to use the -m pdb command that loads the pdb module before executing the Python script. In the same command-line interface, we would then follow this by a specific debugger command of choice, such as n to move to the next line or s if we intend to step into a function.
This method could become cumbersome quickly as the length of the code increases. One way to address this problem and gain better control over where to break your code is to insert a breakpoint directly into the code.
### Invoking the pdb Debugger in Earlier Versions of Python
Invoking the pdb debugger prior to Python 3.7 would require you to import pdb and call pdb.set_trace() at the point in your code where you would like to enter an interactive debugging session.
If we reconsider, as an example, the code for implementing the general attention mechanism, we can break into the code as follows:
Executing the script now opens up the pdb debugger right before we compute the variable scores, and we can proceed to issue any debugger commands of choice, such as n to move to the next line or c to continue execution:
Although functional, this is not the most elegant and intuitive approach of inserting a breakpoint into your code. Python 3.7 implements a more straightforward way of doing so, as we shall see next.
### Using the breakpoint() Function in Python 3.7
Python 3.7 comes with a built-in breakpoint() function that enters the Python debugger at the call site (or the point in the code at which the breakpoint() statement is placed).
When called, the default implementation of the breakpoint() function will call sys.breakpointhook(), which in turn calls the pdb.set_trace() function. This is convenient because we will not need to import pdb and call pdb.set_trace() explicitly ourselves.
Let’s reconsider the code for implementing the general attention mechanism and now introduce a breakpoint via the breakpoint() statement:
One advantage of using the breakpoint() function is that, in calling the default implementation of sys.breakpointhook(), the value of a new environment variable, PYTHONBREAKPOINT, is consulted. This environment variable can take various values, based on which different operations can be performed.
For example, setting the value of PYTHONBREAKPOINT to 0 disables all breakpoints. Hence, your code could contain as many breakpoints as necessary, but these can be easily stopped from halting the execution of the code without having to remove them physically. If (for example) the name of the script containing the code is main.py, we would disable all breakpoints by calling it in the command line interface as follows:
Otherwise, we can achieve the same outcome by setting the environment variable in the code itself:
The value of PYTHONBREAKPOINT is consulted every time that sys.breakpointhook() is called. This means that the value of this environment variable can be changed during the code execution, and the breakpoint() function would respond accordingly.
The PYTHONBREAKPOINT environment variable can also be set to other values, such as the name of a callable. Say, for instance, that we’d like to use a different Python debugger other than pdb, such as ipdb (run pip install ipdb first if the debugger has not yet been installed). In this case, we would call the main.py script in the command line interface and hook the debugger without making any changes to the code itself:
In doing so, the breakpoint() function enters the ipdb debugger at the next call site:
The function can also take input arguments as breakpoint(*args, **kws), which are then passed on to sys.breakpointhook(). This is because any callable (such as a third-party debugger module) might accept optional arguments, which can be passed through the breakpoint() function.
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## Writing Your Own breakpoint() Function in Earlier Versions of Python
Let’s return to the fact that versions of Python earlier than v3.7 do not come with the breakpoint() function readily built in. We can write our own.
Similarly to how the breakpoint() function is implemented from Python 3.7 onwards, we can implement a function that checks the value of an environment variable and:
• Skips all breakpoints in the code if the value of the environment variable is set to 0.
• Enters into the default Python pdb debugger if the environment variable is an empty string.
• Enters into another debugger as specified by the value of the environment variable.
We can include this function into the code and run it (using a Python 2.7 interpreter, in this case). If we set the value of the environment variable to an empty string, we find that the pdb debugger stops at the point in the code at which we have placed our breakpoint() function. We can then issue debugger commands into the command line from there onwards:
Similarly, if we set the environment variable to:
The breakpoint() function that we have implemented now enters the ipdb debugger and stops at the call site:
Setting the environment variable to 0 simply skips all breakpoints, and the computed attention output is returned in the command line, as expected:
This facilitates the process of breaking into the code for Python versions earlier than v3.7 because it now becomes a matter of setting the value of an environment variable rather than having to manually introduce (or remove) the import pdb; pdb.set_trace() statement at different call sites in the code.
## Limitations of the breakpoint() Function
The breakpoint() function allows you to bring in the debugger at some point in the program. You need to find the exact position that you need the debugger to put the breakpoint into it. If you consider the following code:
This will bring you the debugger when the function func() raised exceptions. It can be triggered by the function itself or deep inside some other functions that it calls. But the debugger will start at the line print("exception!") above, which may not be very useful.
The way that we can bring up the debugger at the point of exception is called the post-mortem debugger. It works by asking Python to register the debugger pdb.pm() as the exception handler when an uncaught exception is raised. When it is called, it will look for the last exception raised and start the debugger at that point. To use the post-mortem debugger, we just need to add the following code before the program is run:
This is handy because nothing else needs to be changed in the program. For example, assume we want to evaluate the average of $1/x$ using the following program. It is quite easy to overlook some corner cases, but we can catch the issue when an exception is raised:
When we run the above program, the program may terminate, or it may raise a division by zero exception, depending on whether the random number generator ever produces zero in the loop. In that case, we may see the following:
where we found the exception is raised at which line, and we can check the value of the variables as we can usually do in pdb.
In fact, it is more convenient to print the traceback and the exception when the post-mortem debugger is launched:
And the debugger session will be started as follows:
This section provides more resources on the topic if you are looking to go deeper.
## Summary
In this tutorial, you discovered various ways of setting breakpoints in different versions of Python.
Specifically, you learned:
• How to invoke the pdb debugger in earlier versions of Python
• How to make use of the new, built-in breakpoint() function introduced in Python 3.7
• How to write your own breakpoint() function to simplify the debugging process in earlier versions of Python
Do you have any questions?
## Get a Handle on Python for Machine Learning!
#### Be More Confident to Code in Python
...from learning the practical Python tricks
Discover how in my new Ebook:
Python for Machine Learning
It provides self-study tutorials with hundreds of working code to equip you with skills including:
debugging, profiling, duck typing, decorators, deployment, and much more...
|
2023-02-09 13:01:38
|
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|
https://infoscience.epfl.ch/record/205740
|
## Measurement of CP violation in B-s(0) ->Phi Phi decays
A measurement of the decay time-dependent CP-violating asymmetry in B-s(0) -> phi phi decays is presented, along with measurements of the T-odd triple-product asymmetries. In this decay channel, the CP-violating weak phase arises from the interference between B-s(0) -(B) over bar (0)(s) mixing and the loop-induced decay amplitude. Using a sample of proton-proton collision data corresponding to an integrated luminosity of 3.0 fb(-1) collected with the LHCb detector, a signal yield of approximately 4000 B-s(0) -> phi phi decays is obtained. The CP-violating phase is measured to be phi(s) = -0.17 +/- 0.15(stat) +/- 0.03(syst) rad. The triple-product asymmetries are measured to be A(U) = -0.003 +/- 0.017(stat) +/- 0.006(syst) and A(V) = -0.017 +/- 0.017(stat) +/- 0.006(syst). Results are consistent with the hypothesis of CP conservation.
Published in:
Physical Review D, 90, 5, 052011
Year:
2014
Publisher:
College Pk, Amer Physical Soc
ISSN:
1550-7998
Laboratories:
|
2018-07-19 00:32:18
|
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|
http://experiment-ufa.ru/x=sqrt(1000)
|
# x=sqrt(1000)
## Simple and best practice solution for x=sqrt(1000) equation. Check how easy it is, and learn it for the future. Our solution is simple, and easy to understand, so dont hesitate to use it as a solution of your homework.
If it's not what You are looking for type in the equation solver your own equation and let us solve it.
## Solution for x=sqrt(1000) equation:
Simplifying
x = sqrt(1000)
Reorder the terms for easier multiplication:
x = 1000qrst
Solving
x = 1000qrst
Solving for variable 'x'.
Move all terms containing x to the left, all other terms to the right.
Simplifying
x = 1000qrst`
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2018-03-23 18:36:07
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https://www.riobrasilword.com/2022/07/05/photoshop-cc-2014-keygen-crack-setup-free/
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Photoshop CC 2014 Keygen Crack Setup Free 🌠
## Photoshop CC 2014 Crack Keygen For (LifeTime)
Note We'll start by learning about Photoshop's features in the first part of this section, and by the end we'll have enough skill to use Photoshop for both the purposes of professional image editing and of designing or creating images to print. As we move on through this part of the book, you'll learn how to change images' color and tone, make pictures look more realistic, fix images that are out of focus or overexposed, and edit pictures to change a subject's appearance. You'll learn about text and even how to print your work.
## Photoshop CC 2014 Crack + [Mac/Win]
MRI has been shown to be superior to other imaging techniques in detecting abnormalities in musculoskeletal diseases such as arthritis, torn ligaments and tendons, and fractures. MRI is also superior in evaluating the effect of various pharmacological and non-pharmacological treatments in musculoskeletal disease. However, in diseases in which inflammation predominates, MRI has demonstrated lower efficacy compared to plain radiography. The reason for this is that MRI is highly sensitive to the paramagnetic effects of iron and calcium in the presence of inflammation. The ability to detect low concentrations of non-heme iron has been shown to be extremely sensitive (i.e. detecting very low concentrations of non-heme iron is achievable) in the prior art. Therefore, the effect of non-heme iron on MRI is not problematic. The ability to detect low concentrations of calcium is problematic and remains to be solved. Unfortunately, normal concentrations of calcium reduce the signal to noise ratio of the MRI image, and its detection requires long scan times, such as minutes rather than seconds. Because the detection of calcium is problematic, diagnosticians often rely on plain radiography or ultrasound. However, plain radiography is less sensitive in the detection of early inflammation or microfractures and ultrasound is not sensitive at all. Hence, there is a need in the art for a method to detect calcium in the presence of inflammation while maximizing the sensitivity of the MRI. Magnetic resonance imaging has been shown to be superior to radiographic imaging, especially in detecting early osseous changes in the joints. Therefore, in order to allow for a better detection of early osseous changes in the joints, there is a need for a method to detect calcium in the presence of inflammation while also maximizing the sensitivity of MRI. In summary, the ability to detect low concentrations of calcium in the presence of inflammation would be beneficial for diagnosis, as it would increase the sensitivity of MRI and reduce the radiation exposure, thereby allowing for earlier detection of osseous changes.Q: Is the name of the travel book on my 2003 Toyota Yaris useful for identifying a year? I noticed the (Japanese) book that comes with the car is called "Yaris Tour & Travel". It is about 13cm x 20cm. It says: Country of origin: Japan Issued: 2003 (or 'for cars of the year 2003') Does this name help me identify the year of a car? Or perhaps the country
## What's New in the Photoshop CC 2014?
Q: How do you find R-group? Suppose, $G$ is a finite group, and $H$ is subgroup of $G$ such that $H\unlhd G$. What is the standard way of determining the Sylow-subgroup $G_i$ of $G$ such that $|G_i|=p^i$? (Please forgive my ignorance on this topic. The difficulty arises when it comes to finding the order of $H$ in the case of $G/H$ being cyclic.) A: What you're asking for is called the Frattini Argument. If you know that $[G:H]$ is a prime-power (ie, a power of a prime), then you can use the Frattini Argument to show that $H$ is normal in $G$, and then you'll know that the Sylow subgroups correspond to the conjugates of $H$ in $G$. Here's the outline of the proof: Suppose that $G$ is a group, and let $H$ be a normal subgroup of $G$. If $G/H$ is cyclic, then $H$ is the Frattini subgroup of $G$, and therefore is trivial. If $G/H$ is not cyclic, then let $x$ be a generator of $G/H$. Then $G=H\langle x\rangle$. If $|H|=p^\alpha$, then let $p^\beta$ be the highest power of $p$ dividing $|G|$ that is less than $\alpha$. Let $H'$ be a subgroup of $G$ generated by the elements of $H$ of order a power of $p$. Then $H$ is normal in $G$, and $H'\unlhd G$, so $H'$ is a Sylow $p$-subgroup. Then $|H'|=p^\gamma$, and the fact that $p^\beta$ divides $|G|$ and \$\betaLast
## System Requirements For Photoshop CC 2014:
Minimum: OS: Windows XP SP3 or Windows 7 SP1 CPU: Intel® Core™ i3 / AMD Athlon™ 64 X2 Memory: 2 GB RAM Graphics: 256MB VRAM Hard Disk Space: 100 MB Sound Card: DirectX compatible sound card Network: Internet connection Recommended: CPU: Intel® Core™ i5 / AMD Phenom™ II X2 Graphics:
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2022-10-05 12:33:54
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https://cran.rstudio.org/web/packages/faux/vignettes/continuous.html
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# Continuous Predictors
library(faux)
library(dplyr)
library(tidyr)
library(ggplot2)
library(cowplot) # for multi-panel plots
## One continuous predictor
dat <- sim_design(within = list(vars = c("dv", "predictor")),
mu = list(dv = 100, predictor = 0),
sd = list(dv = 10, predictor = 1),
r = 0.5, plot = FALSE)
## Between continuous predictors
Here, pred1 is correlated r = 0.5 to the DV, and pred2 is correlated 0.0 to the DV, and pred1 and pred2 are correlated r = -0.2 to each other.
dat <- sim_design(within = list(vars = c("dv", "pred1", "pred2")),
mu = list(dv = 100, pred1 = 0, pred2 = 0),
sd = list(dv = 10, pred1 = 1, pred2 = 1),
r = c(0.5, 0, -0.2), plot = FALSE)
## Within continuous predictors
If the continuous predictors are within-subjects (e.g., dv and predictor are measured at pre- an post-test), you can set it up like below.
The correlation matrix can start getting tricky, so I usually map out the upper right triangle of the correlation matrix separately. Here, the dv and predictor are correlated 0.0 in the pre-test and 0.5 in the post-test. The dv is correlated 0.8 between pre- and post-test and the predictor is correlated 0.3 between pre- and post-test. There is no correlation between the pre-test predictor and the post-test dv, but I’m not sure what values are possible then for the correlation between the post-test predictor and pre-test dv, so I can set that to NA and use the pos_def_limits function to determine the range of possible correlations (gven the existing correlation structure). Those range from -0.08 to 0.88, so I’ll set the value to the mean.
# pre_pred, post_dv, post_pred
r <- c( 0.0, 0.8, NA, # pre_dv
0.0, 0.3, # pre_pred
0.5) # post_dv
lim <- faux::pos_def_limits(r)
r[[3]] <- mean(c(lim$min, lim$max))
dat <- sim_design(within = list(time = c("pre", "post"),
vars = c("dv", "pred")),
mu = list(pre_dv = 100, pre_pred = 0,
post_dv = 110, post_pred = 0.1),
sd = list(pre_dv = 10, pre_pred = 1,
post_dv = 10, post_pred = 1),
r = r, plot = FALSE)
You have to make this sort of dataset in wide format and then manually convert it to long. I prefer gather and spread, but I’m trying to learn the new pivot functions, so I’ll use them here.
long_dat <- dat %>%
pivot_longer(-id, "var", "value") %>%
separate(var, c("time", "var")) %>%
pivot_wider(names_from = var, values_from = value)
## One continuous , one categorical predictor
In this design, the DV is 10 higher for group B than group A and the correlation between the predictor and DV is 0.5 for group A and 0.0 for group B.
dat <- sim_design(between = list(group = c("A", "B")),
within = list(vars = c("dv", "predictor")),
mu = list(A = c(dv = 100, predictor = 0),
B = c(dv = 110, predictor = 0)),
sd = list(A = c(dv = 10, predictor = 1),
B = c(dv = 10, predictor = 1)),
r = list(A = 0.5, B = 0), plot = FALSE)
## Add a continuous predictor
If you already have a dataset and want to add a continuous predictor, you can make a new column with a specified mean, SD and correlation to one other column.
First, let’s make a simple dataset with one between-subject factor.
dat <- sim_design(between = list(group = c("A", "B")),
mu = list(A = 100, B = 120), sd = 10, plot = FALSE)
Now we can add a continuous predictor with rnorm_pre by specifying the vector it should be correlated with, the mean, and the SD. By default, this produces values sampled from a population with that mean, SD and r. If you set empirical to TRUE, the resulting vector will have that sample mean, SD and r.
dat$pred <- rnorm_pre(dat$y, 0, 1, 0.5)
If you want to set a different mean, SD or r for the between-subject groups, you can split and re-merge the dataset (or use your data wrangling skills to devise a more elegant way using purrr).
A <- filter(dat, group == "A") %>%
mutate(pred = rnorm_pre(y, 0, 1, -0.5))
B <- filter(dat, group == "B") %>%
mutate(pred = rnorm_pre(y, 0, 1, 0.5))
dat <- bind_rows(A, B)
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2020-11-28 20:42:37
|
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https://www.physicsforums.com/threads/enthalpy-of-polymerization.644597/
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# Enthalpy of polymerization
1. Oct 16, 2012
### utkarshakash
1. The problem statement, all variables and given/known data
The polymerization of ethylene to linear polyethylene is represented by the following reaction
$nCH_{2}=CH_{2} \rightarrow (-CH_{2}-CH_{2}-)_{n}$
where n=large integer value
Calculate the enthalpy of polymerization per mol of ethylene at 298K if average bond enthalpies of bond dissociation for C=C and C-C are 590 and 331kJ/mol respectively
2. Relevant equations
3. The attempt at a solution
For one mole
$ΔH_{poly} = BDE of C=C+4\times BDE C-H - BDE of C-C - 4\times BDE C-H$
=590-331
=259
But this is not correct.
2. Oct 17, 2012
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2018-03-21 01:50:20
|
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|
https://math.stackexchange.com/questions/2753011/dedekind-cuts-and-rationals
|
# Dedekind Cuts and Rationals
I have a question regarding how you might construct the reals from the rationals by taking Dedekind cuts.
My basic understanding is that a Dedekind cut is a bipartition of the rationals such that the two partitions $X, Y$ satisfy certain properties:
$\forall x \in X,\; \exists y \in X \text{ such that } x < y$
$\forall x \in Y, \; \exists y \in Y \text{ such that } y < x$
$\forall x,y, \; x<y \text{ and } y \in X \Rightarrow x \in X$
$\forall x,y, \; x<y \text{ and } x \in Y \Rightarrow y \in Y$
$\forall x,y, x<y \Rightarrow \text{ either } x \in X \text{ or } y \in Y$
$X \text{ and } Y$ are disjoint
$X \text{ and } Y$ are both non-empty.
A Dedekind cut is then used to represent a real number $z$ that can be see as the "mid-point" of this bipartition of the rationals in the sense that $\forall x \in X, \forall y \in Y, \; x < z < y$
In most descriptions of defining the reals as the set of all Dedekind cuts of the rationals, these inequalities are strict. However, doesn't that mean that the rationals cannot be defined in this way? Am I meant to take this to mean that the irrationals are defined using these Dedekind cuts and that the reals are to be treated as the union of this set of irrationals and the set of rationals?
If that is the case, doesn't that mean the irrationals and rationals in this construction of the reals are different kinds of objects in that they would have different ranks? I would imagine that is something inconvenient and would ideally want to be avoided.
Note that it is not part of your axioms that $X\cup Y = \Bbb Q$. For instance, the rational number $0$ is given by $$X = \{q\in \Bbb Q\mid q<0\}\\ Y = \{q\in \Bbb Q\mid q>0\}$$ and the number $0$ isn't contained in either of them. On the other hand, the axiom $$\forall x,y, x<y \Rightarrow \text{ either } x \in X \text{ or } y \in Y$$ implies that $\Bbb Q\setminus(X\cup Y)$ has at most one element.
The usual definition of Dedekind cuts that I've come across does have $X\cup Y = \Bbb Q$, but it also allows $Y$ to have a least element, which your axioms do not allow. Specifically, $$\forall x \in Y, \; \exists y \in Y \text{ such that } y < x$$ says that $Y$ has no least element.
The problem is that these inequalites are not necessarily strict. For instance, it is stated here that the second set (your $Y$) may have a smallest (rational) number $q$, in which case the Dedekind cut $(X,Y)$ correspondes to the rational number $q$.
|
2022-05-27 20:16:31
|
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|
http://www.journalofinequalitiesandapplications.com/content/2011/1/114
|
Research
# On the super-stability of exponential Hilbert-valued functional equations
Author Affiliations
Department of Mathematics, College of Sciences, Yasouj University, Yasouj-75914-74831, Iran
For all author emails, please log on.
Journal of Inequalities and Applications 2011, 2011:114 doi:10.1186/1029-242X-2011-114
Received: 24 July 2011 Accepted: 21 November 2011 Published: 21 November 2011
This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
### Abstract
We generalize the well-known Baker's super-stability result for exponential mappings with values in the field of complex numbers to the case of an arbitrary Hilbert space with the Hadamard product. Then, we will prove an even more general result of this type.
2000 MSC: primary: 39B72, secondary: 46E40.
##### Keywords:
exponential functions; stability; Hilbert-valued function
### 1. Introduction
The stability problem of functional equations goes back to a question of Ulam [1] concerning the stability of group homomorphisms. Hyers [2] gave a first affirmative partial answer to the question of Ulam for Banach spaces (see also [3]). Hyers's theorem was generalized by Aoki [4] for additive mappings and by Rassias [5,6] for linear mappings by considering an unbounded Cauchy difference. Baker et al. [7] have proved the super-stability of the exponential functional equation: If a function f: ℝ → ℝ is approximately exponential function, i.e., there exists a nonnegative number α such that
for x, y ∈ ℝ, then f is either bounded or exponential. This theorem was the first result concerning the super-stability phenomenon of functional equations. Baker [8] generalized this famous result to any function f: (G, +) → ℂ where (G, +) is a semigroup. The same result is also true for approximately exponential mappings with values in a normed algebra with the property that the norm is multiplicative.
Theorem 1.1. Let (G, +) be a semigroup and Y be a normed algebra in which the norm is multiplicative. Then, for a function f: G Y satisfying the inequality
for all x; y G and for some α > 0, either for all x G or f is an exponential function.
In the other world every approximately exponential map f: (G, +) → Y is either bounded or exponential.
Rassias [5,6] introduced the term mixed stability of the function f: E → ℝ, where E is a Banach space, with respect to two operations addition and multiplication among any two elements of the set {x, y, f(x), f(y)}. Especially, he raised an open problem concerning the behavior of solutions of the inequality:
(see also [9,10]). In connection with this open problem, Gavruta [11] gave an answer to this problem in the spirit of Rassiass approach:
Theorem 1.2 (Gavruta). Let X and Y be a real normed space and a normed algebra with multiplicative norm, respectively. If a function f: X Y satisfies the inequality
for all x; y X and for some p > 0 and θ > 0, then either ||f(x)|| ≤ δ||x||p for all x X with ||x|| ≥ 1 or f is an exponential function, where .
Baker [8] gave an example to present that the Theorem 1.1 is false if the algebra Y does not have the multiplicative norm: Given δ > 0, choose an ε > 0 with |ε - ε2| = δ. Let M2(ℂ) denote the space of 2 × 2 complex matrices with the usual norm and f: ℝ → M2(ℂ) is defined by f(x) = exe11 + exe22 where eij is defined as the 2 × 2 matrix with 1 in the (i, j) entry and zeroes elsewhere. We will show that such behavior is typical for approximately exponential mappings with values in Hilbert spaces with Hadamard product which is not multiplicative.
Let H be a Hilbert space with a countable orthonormal basis {en: n ∈ ℕ}. For two vectors x, y H, we have the Hadamard product (named after French mathematician Jacques Hadamard), also known as the entrywise product on Hilbert space H as the following:
The Cauchy-Schwartz inequality together with the Parseval identity insure that Hadamard multiplication is well defined. In fact,
In the present paper, we state a super-stability result for the approximately exponential Hilbert-valued functional equation by Hadamard product, see Theorem 2.1 below. As a consequence, we prove if a surjective function f: H H satisfies the inequality
for some α ≥ 0 and for all x; y H, then it must be exponential with this product, i.e.,
Then, we will prove an even more general result of this type. We also generalized Theorem 2.1 concerning the mixed stability for Hilbert-valued functions.
### 2. Main results
The function f(x) = ax is said to be an exponential function, where a > 0 is a fixed real number. The exponent law of exponential functions is well represented by the exponential equation f(x + y) = f(x)f(y). Hence, we call every solution function of the exponential equation as exponential function. A general solution of the exponential equation was introduced in [12]. In fact, a function f: ℝ → ℂ is an exponential function if and only if either f(x) = exp(A(x) + ia(x)) for all x ∈ ℝ or f(x) = 0 for all x ∈ ℝ; where A: ℝ → ℝ is an additive function and a: ℝ → ℝ satisfies
(1)
for all x, y ∈ ℝ. Indeed, a function f: ℝ → ℝ continuous at a point is an exponential function if and only if f(x) = ax for all x ∈ ℝ or f(x) = 0 for all x ∈ ℝ, where a > 0 is a constant.
Definition 2.1. For a Hilbert space H and a semi-group (G,.), a function F: G H is said to be exponential when
for every x, y G.
The following proposition characterizes the Hilbert-valued function satisfying the exponential equation:
Proposition 2.2. Let H be a separable complex Hilbert space and the mapping F: ℝ → H be exponential then either F ≡ 0 or there exist a positive integer N such that
for all x H where An: ℝ → ℝ is an additive function and an is a function satisfying (1) for n = 1, 2,..., N.
Proof. For every integer n ≥ 1, consider the function en F: ℝ → ℂ by
for every h H. Since F is exponential, so is en F for every integer n ≥ 1. Indeed, for n ≥ 1 and x, y H, we see that
This yields the exponential property of en F for every n ≥ 1. Hence, either
(2)
for all x ∈ ℝ or (en F)(x) = 0 for all x ∈ ℝ; here An: ℝ → ℝ is an additive function and an is a function satisfying (1). The continuation of proof depend on the dimension of H. In fact, if H is infinite dimensional, since
for every x H as n → +∞ Equation 2 is not possible for infinitely many positive integer n and hence there exists some positive integer N such that en F = 0 for every integer n > N. Thus, F can be represented as
In the case that H is of finite dimensional type, the proof is clear.
In the following theorem, we generalize the well-known Baker's super-stability result for exponential mappings with values in the field of complex numbers to the case of an arbitrary Hilbert space with the Hadamard product.
Theorem 2.3. Let G be a semigroup and let α > 0 be given. If a function f: G H satisfies the inequality
(3)
for all x; y G, then either there exists an integer k ≥ 1 such that
(4)
for all x G or
for all x; y G.
Proof. Assume that the first conclusion (i.e., (4)) is not true. Hence, for every integer k ≥ 1, there exists a ak G such that
Let , fk(x) = 〈f(x), ek〉, and gk = 2-k fk. Then, β2 - 2β = α, β > 2 and |fk(ak)| > 2kβ whence |gk(ak)| > β. By applying the Parseval identity and definition of Hadamard product with together relation (3), we find that each scalar-valued function fk is approximately exponential, i.e.,
(5)
for every integer k ≥ 1 and x, y G. Let
then γk > 1 for every integer k ≥ 1. It follows from (5) for x = y = ak that
and so
Now, make the induction hypothesis
(6)
Then, by using (5) for and (6), we observe that
and (6) is established for all n ∈ ℕ. Hence, by definition of fk and gk, we see that
(7)
On the other hand, for every x, y, z G, we have
Consequently, for h(x, y) = f(x.y) - f(x) * f(y), one can see
Now, by using Parseval identity for h(x, y) * f(z) observe that
Applying the last relation for and relation (7) to deduce that
It follows that
for all x, y G and any n ∈ ℕ. Letting n → +∞, we conclude that h(x, y) = 0 and so f(x.y) = f(x) * f(y) for all x, y G.
Notice that if f: H H is a surjection function, then every component function en f is unbounded. In fact, for every positive integer n, there exists some xn H such that f(xn) = nen, and so (en f)(xn) = n. This led to the following corollary:
Corollary 2.4. If a surjective function f: H H satisfies the inequality
for some α ≥ 0 and for all x; y G, then f(x * y) = f(x) * f(y) for all x; y G.
In the next theorem, we generalize the Gavruta Theorem on mixed stability for Hilbert-valued function with Hadamard product:
Theorem 2.5. Let X be a normed space and H be a separable Hilbert space. If a function f: X H satisfies the inequality
(8)
for all x; y X and for some p > 0 and θ > 0, then either there exists an integer k ≥ 1 such that
(9)
for all x X with ||x|| ≥ 1 or
for all x; y X. where .
Proof. Assume that for every integer k ≥ 1 there exists an xk X with ||xk|| ≥ 1 such that
If we set fk(x): = 〈f(x), ek〉 and gk := 2-kfk, this is equivalent with
It follows from Parseval identity, definition of Hadamard product and relation (8) that
(10)
for every x, y X and k ≥ 1. In particular, for x = y = xk
Since β2 = 2p + 1 β + 2θ, hence
Now, make the induction hypothesis
(11)
Then, by using (10) for x = y = 2nxk and (11), we get
which in turn proves that the inequality (11) is true for all n ∈ ℕ. Hence, by definition of fk and gk, we see that
(12)
Choose x; y; z X with f(z) ≠ 0. It then follows from (8) that
and again by (8) we get
which together with the last relation yields
(13)
where
Let h(x, y) = f(x + y) - f(x) * f(y), then by (13)
and so
In particular, by using the last relation for zk = 2nxk and by considering (12) we deduce that
and consequently,
for all x, y X and any n ∈ ℕ. Letting n → +∞, we conclude that h(x, y) = 0 and so f(x + y) = f(x) * f(y) for all x, y X.
At the end of this paper, let us consider the other type multiplication in a Hilbert space. In fact, for a separable Hilbert space H and two elements and of H, one can define the convolution product by
where the numbers can be obtained by discrete convolution:
Hence, it is interesting to study and to phrase the super-stability phenomenon for functions with values in (H, •). For instance, it is desirable to have a sufficient condition for approximately exponential mappings with values in (H, •) to be exponential with the convolution product.
### 3. Competing interests
The authors declare that they have no competing interests.
### 4. Authors' contributions
All authors carried out the proof. All authors conceived of the study, and participated in its design and coordination. All authors read and approved the final manuscript.
### References
1. Ulam, SM: A Collection of Mathematical Problems, Interscience Tracts in Pure and Applied Mathematics, no. 8. Interscience. New York (1960)
2. Hyers, DH: On the stability of the linear functional equation. Proc Nat Acad Sci USA. 27, 222–224 (1941). PubMed Abstract | Publisher Full Text | PubMed Central Full Text
3. Hyers, DH, Rassias, ThM: Approximate homomorphisms. Aequat Math. 44, 125–153 (1992). Publisher Full Text
4. Aoki, T: On the stability of the linear transformation in Banach spaces. J Math Soc Japan. 2, 64–66 (1950). Publisher Full Text
5. Rassias, ThM: Problem 18, in Report on the 31st ISFE. Aequat Math. 47, 312–313 (1994)
6. Rassias, ThM: On the stability of the linear mapping in Banach spaces. Proc Am Math Soc. 72(2), 297–300 (1978). Publisher Full Text
7. Baker, JA, Lawrence, J, Zorzitto, F: The stability of the equation f(x + y) = f(x)f(y). Proc Am Math Soc. 74, 242–246 (1979)
8. Baker, JA: The stability of the cosine equation. Proc Am Math Soc. 80, 411–416 (1980). Publisher Full Text
9. Jung, SM: Hyers-Ulam-Rassias Stability of Functional Equations in Nonlinear Analysis, Springer, New York (2011)
10. Kuczma, M: An Introduction to the Theory of Functional Equations and Inequalities. PWN, Warszawa, Krak ow, and Katowice (1985)
11. Gavruta, P: An answer to a question of Th.M. Rassias and J. Tabor on mixed stability of mappings. Bul Stiint Univ Politeh Timis Ser Mat Fiz. 42, 1–6 (1997)
12. Czerwik, S: Functional Equations and Inequalities in Several Variables. World Scientific, Hackensacks (2002)
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2013-06-19 14:05:34
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https://3dprinting.stackexchange.com/questions/11914/skr-1-3-thermistor-input-died
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# SKR 1.3 Thermistor input died
I've got an SKR 1.3 and the thermistor input for extruder 2 has gone bad.
In firmware, I've attempted to change the TEMP_1_PIN to another in pins_BTT_SKR.h and I've also defined TEMP_1_PIN to another pin in pins_BTT_SKR_V1_3.h but regardless of which pin I choose, the printer ignores it and still defaults to the dead input (P0_25) I've made other changes in firmware to confirm that it is updating, but still - the second extruders temp input will not change.
Original code in pins_BTT_SKR.h
#ifndef TEMP_1_PIN
#define TEMP_1_PIN P0_25_A2 // A2 (T2) - (69) - TEMP_1_PIN
#endif
I've changed this to
#ifndef TEMP_1_PIN
#define TEMP_1_PIN P1_28 // A2 (T2) - (69) - TEMP_1_PIN
#endif
I've also commented out anything that referenced P1_28 and added a #define TEMP_1_PIN P1_28 in front of this and that didn't work either. I then moved that define to pins_BTT_SKR_V1_3.h and that still didn't work. It's still looking for a signal on P0_25. I've also tried other available pins. It won't change. I must be missing something somewhere...
• I am not an expert, but there is a difference in style in the definition. It seems like it should be P1_28_A2 ? – Trish Jan 30 at 23:17
• p1_28_a2 is what i thought as well but it fails to compile that way. I am assuming it has something to do with the problem, though. – Korner19 Jan 30 at 23:25
• Here is a suggestion to try: add a #warning to show the value of TEMP_1_PIN, check it shows up when Marlin compiles into your target. Add/move the #warning to various files (e.g. just before end of pins.h) to see if/when it gets changed. If nothing dawns on you as to the problem, then show code snippets and compilation log here. – user19977 Feb 1 at 5:01
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2020-11-27 09:36:11
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http://mathoverflow.net/questions/951/legendrian-homotopy-of-curves-in-a-contact-structure/963
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Legendrian homotopy of curves in a contact structure?
I'm aware of the great body of work on Legendrian knot theory in contact geometry, but suppose I'm curious just about homotopy and not isotopy. How does one understand the space of Legendrian loops based at a point in a contact manifold? Can that be made into a "Legendrian fundamental group" somehow?
I've heard that h-principles are somehow involved, but I'm not sure what the punchline is.
-
In general, the (parametric) h-principle for Legendrian immersions implies that Legendrian immersions f:L->(M,\xi) are classified up to homotopy (through Legendrian immersions) by the following bundle-theoretic invariant: Choosing a compatible almost complex structure on \xi allows one to complexify the differential of f to an isomorphism d_C f: TL\otimes C -> f*\xi, and the relevant invariant is the homotopy class of this isomorphism of complex vector bundles (of course this is independent of the almost complex structure since the space of compatible almost complex structures is contractible).
The above holds in any contact manifold (M,\xi) of arbitrary dimension. Of course when M is 3-dimensional and L is S^1, f^*\xi is the unique complex line bundle over S^1, automorphisms of which are parametrized up to homotopy by pi_1(U(1))=Z. So (given that the h-principle also implies that any loop in a 3-manifold is homotopic to a Legendrian loop) it appears to always be the case that the "Legendrian fundamental group" surjects onto the standard fundamental group, with kernel Z.
When M=R^3 this invariant is equivalent to the rotation number that Steven mentioned. There's a proof of the relevant h-principle in the book by Eliashberg and Mishachev. The above discussion is partly based on Section 3.3 of arXiv:0210124 by Ekholm-Etnyre-Sullivan.
-
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2014-04-21 02:55:59
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http://exxamm.com/QuestionSolution2/3D+Geometry/Find+the+equation+of+the+plane+determined+by+the+point+A+3+1+2+B+5+2+4+and+C+1+1+6+and+hence+find+the/2671491326
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Find the equation of the plane determined by the point A(3, - 1, 2), B(5, 2, 4) and C(-1, -1, 6) and hence find the
### Question Asked by a Student from EXXAMM.com Team
Q 2671491326. Find the equation of the plane determined by the point A(3, - 1, 2), B(5, 2, 4) and C(-1, -1, 6) and hence find the distance between the plane and the point P(6, 5, 9).
CBSE-12th 2012
#### HINT
(Provided By a Student and Checked/Corrected by EXXAMM.com Team)
#### Access free resources including
• 100% free video lectures with detailed notes and examples
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2018-05-23 10:46:51
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https://nemenmanlab.org/~ilya/index.php/Physics_380,_2010:_Basic_Probability_Theory
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# Physics 380, 2010: Basic Probability Theory
## Lectures 2 and 3
During these lectures, we will review some basic concepts of probability theory, such as probability distributions, conditionals, marginals, expectations, etc. We will discuss the central limit theorem and will derive some properties of random walks. Finally, we will study some specific useful probability distributions.
A very good introduction to probability theory can be found in Introduction to Probability by CM Grinstead and JL Snell.
• Random variables: motion of E. coli, time to neural action potential; diffusion and first passage
• Sample space, events, probabilities -- probability space
• Properties of distributions:
• nonnegativity: ${\displaystyle P_{i}\geq 0}$
• unit normalization: ${\displaystyle \sum _{i=1}^{N}P_{i}=1}$
• nesting: if ${\displaystyle A\subset B}$ then ${\displaystyle P(A)\leq P(B)}$
• additivity (for non-disjoint events): ${\displaystyle P(A\cup B)=P(A)+P(B)-P(A\cap B)}$
• complementarity ${\displaystyle P(not\,A)=1-P(A)}$
• Continuous and discrete events: probability distributions and densities ${\displaystyle P_{i}}$ or ${\displaystyle P(x)}$
• Cumulative distributions ${\displaystyle C(x)=\int _{-\infty }^{x}P(x')dx'}$
• Change of variables for continuous and discrete variates ${\displaystyle P(x')=P(x){\frac {dx}{dx'}}}$, for multi-dimensional variables ${\displaystyle P({\vec {x'}})=P({\vec {x}})\left|{\frac {dx_{\alpha }}{dx'_{\beta }}}\right|}$
• Distributions:
• uniform: probability of emitting a spike or doing a tumble by an E.coli. ${\displaystyle P(t)=1/T,\;0\leq t\leq T}$
• exponential: time to the next action potential at constant rate ${\displaystyle P(t)=re^{-rt}}$.
• Poisson: number of action potentials in a given interval; number of E. coli tumbles; ${\displaystyle P(n)={\frac {(rT)^{n}}{n!}}e^{-rT}}$
• normal: diffusive motion ${\displaystyle P(x)={N}(\mu ,\sigma ^{2})={\frac {1}{{\sqrt {2\pi }}\sigma }}\exp {\left[-{\frac {(x-\mu )^{2}}{2\sigma ^{2}}}\right]}}$
• ${\displaystyle \delta }$-distribution: deterministic limit ${\displaystyle \delta (x-\mu )=\lim _{\sigma \to 0}{\frac {1}{{\sqrt {2\pi }}\sigma }}\exp {\left[-{\frac {(x-\mu )^{2}}{2\sigma ^{2}}}\right]}}$; ${\displaystyle \delta (0)\to \infty ,\;\delta (x\neq 0)=0}$.
• Conditional and joint probabilities, Bayes theorem: ${\displaystyle P(A,B)=P(A|B)P(B)=P(B|A)P(A)}$
• independence: two variables are independent if and only if ${\displaystyle P(A,B)=P(A)P(B)}$, or, equivalently, ${\displaystyle P(A|B)=P(A)}$ or ${\displaystyle P(B|A)=P(B)}$.
• Distributions:
• multivariate normal: ${\displaystyle P({\vec {x}}|{\vec {\mu }},\Sigma )={\frac {1}{[2\pi ]^{d/2}\left|\Sigma \right|^{1/2}}}\exp \left[-{\frac {1}{2}}\left({\vec {x}}-{\vec {\mu }}\right)^{T}\Sigma ^{-1}\left({\vec {x}}-{\vec {\mu }}\right)\right]}$, here ${\displaystyle \Sigma }$ is the covariance matrix ${\displaystyle \Sigma =\left[{\begin{array}{llll}\langle (x_{1}-\mu _{1})(x_{1}-\mu _{1})\rangle &\langle (X_{1}-\mu _{1})(X_{2}-\mu _{2})\rangle &\cdots &\langle (X_{1}-\mu _{1})(X_{n}-\mu _{n})\rangle \\\langle (X_{2}-\mu _{2})(X_{1}-\mu _{1})\rangle &\langle (X_{2}-\mu _{2})(X_{2}-\mu _{2})\rangle &\cdots &\langle (X_{2}-\mu _{2})(X_{n}-\mu _{n})\rangle \\\vdots &\vdots &\ddots &\vdots \\\langle (X_{n}-\mu _{n})(X_{1}-\mu _{1})\rangle &\langle (X_{n}-\mu _{n})(X_{2}-\mu _{2})\rangle &\cdots &\langle (X_{n}-\mu _{n})(X_{n}-\mu _{n})\rangle \end{array}}\right].}$
• Expected values: ${\displaystyle E(f(x))=\int _{-\infty }^{+\infty }f(x)P(x)dx}$. In particular, a few of the expectation values are very common: the mean, ${\displaystyle \langle x\rangle =\mu =\int _{-\infty }^{+\infty }xP(x)dx}$, and the variance ${\displaystyle \langle x^{2}\rangle -\langle x\rangle ^{2}=\sigma ^{2}=\int _{-\infty }^{+\infty }x^{2}P(x)dx-\mu ^{2}}$.
• addition of independent variables: in general, ${\displaystyle E(f(x)+g(x))=E(f(x))+E(g(x))}$, and ${\displaystyle E(f(x)g(y))=E(f(x))E(g(y))}$, provided ${\displaystyle x}$ and ${\displaystyle y}$ are independent, that is, ${\displaystyle P(x,y)=P(x)P(y)}$.
• Moments, central moments, and cumulants
• moments: ${\displaystyle \mu _{n}=E(x^{n})=\int x^{n}P(x)dx}$
• central moments: ${\displaystyle m_{n}=E((x-\mu )^{n})}$: distribution mean, width, asymmetry, flatness, etc...
• cumulants: ${\displaystyle c_{n}}$, ${\displaystyle c_{1}=\mu }$, ${\displaystyle c_{2}=\sigma ^{2}}$, and higher order cumulants measure the difference of the distribution from a Gaussian (all higher cumulants for a Gaussian are zero)
• Moment and cumulant generating functional: the Gaussian integral
• Moment generating function (MGF): ${\displaystyle M_{x}(t)=E(e^{tx})}$. The utility of MGF comes from the following result: ${\displaystyle \mu _{n}=\left.{\frac {d^{n}M_{x}(t)}{dt^{n}}}\right|_{t=0}}$.
• Properties of MGF:${\displaystyle M_{x+a}(t)=e^{at}M_{x}(t)}$. From this we can show that if ${\displaystyle P(y|x)=P_{1}(y-x)}$, that is, ${\displaystyle P_{3}(y)=\int P_{1}(x-y)P_{2}(x)}$, then ${\displaystyle M_{3}(t)=M_{1}(t)M_{2}(t)}$.
• Cumulant generating function (CGF): ${\displaystyle C_{x}(t)=\log M_{x}(t)}$. Then the cumulants are: ${\displaystyle c_{n}=\left.{\frac {d^{n}C_{x}(t)}{dt^{n}}}\right|_{t=0}}$
• Frequencies and probabilities: Law of large numbers. If ${\displaystyle S={\frac {1}{n}}\sum x_{i}}$, then ${\displaystyle \mu _{S}=\mu _{x}}$ and ${\displaystyle \sigma _{S}^{2}=\sigma _{x}^{2}/n}$. Thus the sample mean approaches the true mean of the distribution. See one of the homework problems for week 2.
• Central limit theorem: sum of i.i.d. random variables approaches a Gaussian distribution. See one of the homework problems for week 2.
• Random walk and diffusion:
• Unbiased random walk in 1-d: ${\displaystyle T}$ steps of ${\displaystyle \pm a}$ length each. For the total displacement, ${\displaystyle \mu =T\mu _{\rm {onestep}}=T\times 0=0}$ and ${\displaystyle \sigma ^{2}=T\sigma _{\rm {onestep}}^{2}=T\times a^{2}}$
• Conventionally, for a diffusive process: ${\displaystyle \mu =vT}$ and ${\displaystyle \sigma ^{2}=2DdT}$, where ${\displaystyle d}$ is the dimension. So, random walk is an example of a diffusive process on long time scales, and for this random walk: ${\displaystyle v=0}$ and ${\displaystyle D=a^{2}/2}$.
• Biased walk gets ${\displaystyle v\neq 0}$.
• multivariate random walk: ${\displaystyle {\vec {x}}=0}$, ${\displaystyle \sigma _{r}^{2}=2Ddt}$, where ${\displaystyle d}$ is the dimension, and ${\displaystyle r=|{\vec {x}}|}$. We derive this by noting that diffusion/random walk in every dimension is independent of the other dimensions.
## Homework (due Sep 10)
1. (Problem 1.2.1 in Grinstead and Snell). Let ${\displaystyle \Omega =\{a,b,c\}}$ be a sample space. Let ${\displaystyle p(a)=1/2,\,p(b)=1/3}$, and ${\displaystyle p(c)=1/6}$. Find the probabilities for all eight subsets of ${\displaystyle \Omega }$.
2. Exponential, ${\displaystyle P(x|a)=ae^{-at}}$, Poisson ${\displaystyle P(n|n_{0})={\frac {n_{0}^{n}}{n!}}e^{-n_{0}}}$, and multivariate Gaussian ${\displaystyle P({\vec {x}}|{\vec {\mu }},\Sigma )={\frac {1}{[2\pi ]^{d/2}\left|\Sigma \right|^{1/2}}}\exp \left[-{\frac {1}{2}}\left({\vec {x}}-{\vec {\mu }}\right)^{T}\Sigma ^{-1}\left({\vec {x}}-{\vec {\mu }}\right)\right]}$ (where ${\displaystyle d}$ is the dimensionality of ${\displaystyle {\vec {x}}}$) probability distributions are some of the most important distributions that we will see in this class. Calculate the means and the variances for these distributions. Note that for the Gaussian distribution, the easiest way to calculate the mean and the variance is to calculate the moment generating functional first and then differentiate it. Undergraduates: work with 1-dimensional Gaussians, where ${\displaystyle \Sigma ^{-1}=1/\sigma ^{2}}$. Graduate students: Calculate the covariance for the multivariate normal distribution. Pay attention how we do integrals over Gaussians -- we will use this over and over in this class. Also note that logarithms of moment generating functionals are called cumulant generating functionals, and they are often easier to work with. We will denote them as ${\displaystyle C_{x}(t)=\log M_{x}(t)}$. Note that ${\displaystyle {\frac {d^{2}C_{x}(t)}{dt^{2}}}=\sigma ^{2}}$.
3. An E. coli moving on a 2-dimensional surface is being tracked in an experiment. It chooses a direction at random and runs, then tumbles and reorients randomly, runs for the second time, tumbles yet again, and keeps running. What is the probability that all three of the directions that it chooses all fall not farther than ${\displaystyle \pi }$ from each other. That is, what is the probability that the bacterium moves in roughly speaking the same direction all three times? For graduate students: Can you generalize this for ${\displaystyle n}$ tumbles, instead of three?
4. In class we discussed an approximation for the motion of E. coli, where the bacterium would tumble every ${\displaystyle \tau }$ seconds, moving with the velocity of ${\displaystyle v}$ between the tumbles. We have concluded that the long-term displacement of the bacterium can be well characterized by diffusion: mean displacement is zero, and ${\displaystyle \sigma \propto {\sqrt {t}}}$.
• Calculate the coefficient of proportionality for this relation for the bacterium in one dimension. By convention, for a diffusion in ${\displaystyle d}$ dimensions, we write: ${\displaystyle \sigma ^{2}=2dDt}$, where ${\displaystyle D}$ is the diffusion coefficient. What is the diffusion coefficient for this model?
5. Let's now improve the model and say that E. coli tumbles at random times, and the distribution of intervals between two successive tumbles is the exponential distribution with the mean ${\displaystyle \tau }$.
• Derive the distribution of the number of times the E.coli will tumble over a time ${\displaystyle T}$.
• Remember that means and variances of independent random variables add and use this fact repeatedly to calculate the mean and the variance of the displacement of E. coli in this model (still in 1 dimension). Is it still described well by a diffusion model? What is the diffusion coefficient?
• For Grads: If we complicate the model even further, and say that the velocity for each run is sampled independently from ${\displaystyle N(v_{0},\sigma _{v}^{2})}$, does this change the diffusive behavior?
• What should we do to the distributions run durations (and velocities) to violate the diffusive limit?
6. The law of large numbers states that when a random variable is independently sampled from a distribution many times, its sample mean approaches the mean of the distribution. We have almost showed this in class, but stopped a bit short. Let's finish the work. Recall that, when independent random variables are summed, means add and variances add (if both exist). Use this to show that the mean of a sample of ${\displaystyle N}$ independent, identically distributed (denoted: i.i.d.) variables ${\displaystyle x_{i}}$ (with mean ${\displaystyle \mu }$ and variance ${\displaystyle \sigma ^{2}}$), namely ${\displaystyle S_{n}={\frac {1}{n}}\sum _{i=1}^{n}x_{i}}$, has the mean equal to ${\displaystyle \mu }$, and the variance equal to ${\displaystyle \sigma _{\Sigma }^{2}=\sigma ^{2}/n}$. Therefore, as ${\displaystyle n}$ grows, ${\displaystyle S_{n}}$ becomes closer and closer to ${\displaystyle \mu }$, proving the law.
7. The most remarkable law in the probability theory is the Central Limit Theorem (CLT). Its colloquial formulation is as follows: a sum of many i.i.d. random variables is almost normally distributed. This is supposed to explains why experimental noises are often normally distributed as well. More precisely, suppose ${\displaystyle x_{i}}$ are i.i.d. random variables with mean ${\displaystyle \mu }$ and variance ${\displaystyle \sigma ^{2}}$. Then the CLT says that ${\displaystyle S_{n}={\frac {1}{\sqrt {n}}}\sum _{i=1}^{N}{\frac {x_{i}-\mu }{\sigma }}={\frac {1}{\sqrt {n}}}\sum _{i=1}^{N}\xi _{i}}$ is distributed according to ${\displaystyle N(0,1)}$ (called the standard normal distribution), provided ${\displaystyle n}$ is sufficiently large.
• Using either Matlab, Excel, or any other package, generate a sequence of ${\displaystyle n=25}$ random variables uniformly distributed between 0 and 1. Calculate ${\displaystyle S_{25}}$ for them. Do this a 100 times and histogram the resulting 100 values of ${\displaystyle S_{25}}$. Does the histogram look as if it's coming from a standard normal?
• For graduate students. Let's prove the CLT.
• First show that if ${\displaystyle z=x+y}$ (where all three are random variables), then ${\displaystyle M_{z}(t)=M_{x}(t)M_{y}(t)}$, or, alternatively, ${\displaystyle C_{z}=C_{x}+C_{y}}$. In particular, this means that, for ${\displaystyle z=\sum _{i=1}^{n}x_{i}}$, we have ${\displaystyle M_{z}=(M_{x})^{n}}$.
• Write ${\displaystyle M_{\xi }(t)}$ to the first few orders in the Taylor series in ${\displaystyle t}$. Use the identity ${\displaystyle \lim _{n\to \infty }\left(1+1/n\right)^{n}=e}$ to show that ${\displaystyle M_{S_{n}}}$ approaches the moment generating functional for a standard normal as ${\displaystyle n\to \infty }$.
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2019-12-11 13:23:35
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http://mathhelpforum.com/advanced-statistics/120948-gaussian-vector-variance.html
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1. Gaussian vector and variance
Let $(\epsilon_1,...,\epsilon_{2n-1})$ be a random vector with the density:
$p(x_1,...,x_{2n-1})=c_n\exp(-\frac{1}{2}(x_1^2+\Sigma_{i=1}^{2n-2}(x_{i+1}-x_i)^2+x_{2n-1}^2))$
One can check this is a Gaussian vector with mean vector zero and the $2n-1\times 2n-1$ correlation matrix has an inverse given by the tridiagonal form:
$\left(\begin{array}{ccccc} 2 & -1 & 0 & ... & 0 \\ -1 & 2 & -1 & .& : \\ 0 & -1 & . & & 0 \\ : & .& & 2 & -1 \\ 0 & ...& 0 & -1 & 2 \end{array}\right)=M_{2n-1}$
By induction, one can also check that for all $n\geq 1$, $\det(M_n)=n+1$ which enables to obtain the normalizing constant $c_n=\frac{\sqrt{2n}}{(2\pi)^{\frac{2n-1}{2}}}$ using the usual formula for the Gaussian density (see for example Grimmett & Stirzaker).
Now for the question: how do you prove that there is a constant $a$ which does not depend on $n$ such that $Var(\epsilon_n)\geq a.n$ for all $n\geq 1$?
2. Originally Posted by akbar
Let $(\epsilon_1,...,\epsilon_{2n-1})$ be a random vector with the density:
$p(x_1,...,x_{2n-1})=c_n\exp(-\frac{1}{2}(x_1^2+\Sigma_{i=1}^{2n-2}(x_{i+1}-x_i)^2+x_{2n-1}^2))$
One can check this is a Gaussian vector with mean vector zero and $2n-1\times 2n-1$ tridiagonal correlation matrix of the form:
$\left( \begin{array}{ccccc} 2 & -1 & 0 & ... & 0 \\ -1 & 2 & -1 & .& : \\ 0 & -1 & . & & 0 \\ : & .& & 2 & -1 \\ 0 & ...& 0 & -1 & 2 \end{array}\right)=M_{2n-1}$
(note: for some obscure reason, large parenthesis \left(, \right(, are not recognized by the forum editor...) (works for me ?!; otherwise you can use the environment "pmatrix", it's lighter to use)
No, this is not the covariance matrix but its inverse. If it was the covariance matrix, you could read the variance from the diagonal coefficients...
The following way works: integrate the marginals one after another (from both sides toward the middle variable $x_n$; by symmetry you just have to do one side), do it by hand for the first ones in order to get a pattern suitable for a proof by induction. This way you can resume to studying a real-valued sequence defined by induction (something like $u_{n+1}=2-\frac{1}{u_n}$ where I don't specify what I'm dealing with...), and you need an asymptotic expansion for this sequence. This becomes a calculus question, where various methods apply (you should be able to even get an asymptotic equivalence for the variance).
I let you try to perform this computation; tell me if you don't succeed.
3. Yes it is the inverse, sorry about that. So the coefficient is actually $c_n=\frac{\sqrt{2n}}{(2\pi)^\frac{2n-1}{2}}$
I wrote my post too quickly, there is also no syntax issue with parenthesis:
$\left(\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right)$
Will have a go at the asymptotic expansion. Will keep you posted (so to speak).
4. Having carried the calculation, the strange thing is that I actually end up with an explicit formula for the variance $V(\epsilon_n)$, without the need of a sequence expansion. Here are the details of my calculations (sorry for the trivialities):
First we isolate each marginal variable in the quadratic form:
$x_1^2+(x_2-x_1)^2=2(x_1-\frac{1}{2}x_2)^2+\frac{1}{2}x_2^2$
We get a similar expression for $x_{2n-1}$, by symmetry. At each step $p$, on each side, we get a residual term of the form $\frac{1}{p+1}x_{p+1}^2$ (resp. $\frac{1}{p+1}x_{2n-(p+1)}^2$), used for the next step of integration.
Then, by integrating the marginals and using the symmetry:
$p(x_2,...,x_{2n-2})=\int_{x_1,x_{2n-1}}p(x_1,...,x_{2n-1})dx_1dx_{2n-1}$
$=c_n\exp(-\frac{1}{2}(\frac{1}{2}x_2^2+\Sigma_{i=2}^{2n-3}(x_{i+1}-x_i)^2+\frac{1}{2}x_{2n-2}^2)).(\sqrt{2\pi}\sqrt{\frac{1}{2}})^2$
Each (pair of) integration gives us $n-1$ extra terms of the form:
$2\pi\frac{p}{p+1}$ for $1\leq p \leq n-1$. Their product gives $(2\pi)^{n-1}\frac{1}{n}$.
We then get $p(x_n)=\frac{\sqrt{2n}}{(2\pi)^\frac{2n-1}{2}}(2\pi)^{n-1}\frac{1}{n}\exp(-\frac{1}{2}(\frac{1}{n}x_n^2+\frac{1}{n}x_n^2))=\f rac{1}{\sqrt{n\pi}}\exp(-\frac{1}{n}x_n^2)$
$V(\epsilon_n)=\frac{1}{\sqrt{n\pi}}\int_{x_n}x_n^2 e^{-\frac{1}{n}x_n^2}dx_n=\frac{n}{\sqrt{\pi}}\int u^2 e^{-u^2}du=\frac{n}{\sqrt
{\pi}}\frac{\sqrt{\pi}}{2}=\frac{n}{2}$
Using a change of variable. Hence the result.
Too good to be true? Maybe your method is more general, in that case I would appreciate some details.
5. Originally Posted by akbar
Hence the result.
Too good to be true? Maybe your method is more general, in that case I would appreciate some details.
This is highly probably correct. I can't remember what I got, but I'm not surprised since actually I had an explicit formula for my sequence as well. Maybe my method would work in a few other cases, but it is not very common to get asymptotic estimates for a sequence defined by induction, so it must not be very important. Your way is way simpler to explain. I did not believe in an explicit formula at first, hence what I did.
(for the computation of the variance, you could also simply recognize the pdf of a Gaussian of variance $\frac{n}{2}$)
6. Thanks for the confirmation.
Best wishes for the new year.
7. Thanks, and happy (upcoming) new year to you too!
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2016-08-28 14:56:20
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http://mathhelpforum.com/algebra/155390-investment-print.html
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# an investment
• September 6th 2010, 02:19 PM
aeroflix
an investment
an investment yields an annual interest of \$1500. If \$500 more is invested and the rate is 2% less, the annual interest is \$1300. What is the amount of investment and the rate of interest?
• September 6th 2010, 02:40 PM
pickslides
Sounds like a job for some simultaneous equations.
Given A is the initial amount invested, r is the rate then
Maybe use
(1+r)A= 1500+A ...(1)
and
(A+500)(1+(r-2))=1300+(A+500) ...(2)
Leaves you with some algebra.
I may be over complicating this one, it is early morning here!
• September 6th 2010, 02:45 PM
Soroban
Hello, aeroflix!
Quote:
An investment yields an annual interest of \$1500.
If \$500 more is invested and the rate is 2% less, the annual interest is \$1300.
What is the amount of investment and the rate of interest?
Let $x$ = amount invested.
Let $r$ = rate of interest.
We are told that $x$ dollars at $r$ percent yields \$1500.
. . Equation: . $rx \:=\:1500$
We are told that $(x+500)$ dollars at $(r-0.02)$ percent yields \$1300.
. . Equation: . $(x+500)(r-0.02) \:=\:1300$
Now solve the system of equations.
You will get a quadratic equation
and must discard one of the roots.
• September 8th 2010, 12:20 AM
aeroflix
thanks
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2014-07-14 10:51:00
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https://www.ques10.com/p/34026/se-extc_sem3-m3-may18/
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0
394views
SE-EXTC_SEM3 M3 MAY18
Time duration: 3Hours $\hspace{50mm}$ Total marks: 80
Please check whether you have got the right question paper
N.B:
1) Question No. 1 is compulsory
2) Attempt any Three (03) Questions from remaining Five (05) Questions.
Q.1
A) Find Laplace transform of sin $\sqrt{t}$. $\hspace{20mm}$ 5
B) Prove that u = -r$^{3}$sin3? is harmonic function. Also find the conjugate function of u. $\hspace{20mm}$ 5
C) Find a Fourier series to represent f(x) = $(\frac{\pi -x}{2})^{2}$ in (0, 2$\pi$) hence deduce that $\hspace{20mm}$ 5
$\frac{\pi ^{2}}{6}=\frac{1}{1}+\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+ \ldots \ldots \ldots \ldots ..$
D) Find the acute angle between the surface x$^{2}$+y$^{2}$+z$^{2}$=9 and z=x$^{2}$+y$^{2}$-3 at (2,-1,2). $\hspace{20mm}$ 5
Q2
A) Prove that $J_{(-3/2)}(x)= -\sqrt{\frac{2}{\pi x}} .(\frac{\cos x}{x}+\sin x)$. $\hspace{20mm}$ 6
B) Find the Bilinear transformation which maps the points z= 1, i, -1 onto the points w = i, 0,$\infty$. $\hspace{20mm}$ 6
C) Obtain the Fourier series for f(x) = |x| in (-$\pi$, $\pi$) $\hspace{20mm}$ 8
Hence deduce that $\frac{\pi ^{2}}{8}=\frac{1}{1}+\frac{1}{9}+\frac{1}{25}+ \ldots \ldots \ldots \ldots$
Q.3
A) Find inverse Laplace transform of (i) tanh$^{-1}$ $(\frac{2}{S})$ (ii) e$^{- 4s}$ . $\frac{S}{(S+4)^{3}}$ $\hspace{20mm}$ 6
B) Find the image of rectangular region bounded by x = 0, x = 3, y = 0, y = 2 under the $\hspace{20mm}$ 6
bilinear transformation w = z + (1+i).
C) Prove that y = $\sqrt[]{x}$ . J$_{n}$(x) is a solution of the equation, $X^{2}\frac{d^{2}y}{dx^{2}}+(x^{2}-n^{2}+\frac{1}{4})y=0$. $\hspace{20mm}$ 8
Q.4
A) Find Complex form of Fourier Series of coshax in (-a, a). $\hspace{20mm}$ 6
B) Use Gauss's Divergence theorem to evaluate, $\iint_{S}{\overrightarrow{N }}. \overrightarrow{F} ds$ where $\overrightarrow{F}$ = 4xi +3yj -2zk and S $\hspace{20mm}$ 6
is the surface bounded by x=0, y=0, z=0 and 2x +2y+z=4.
C) Solve using Laplace transform( D$^{2}$ +2D+1)y =3te$^{-t}$, given y(0)=4 and y'(0)=2. $\hspace{20mm}$ 8
Q.5
A) Find half range cosine series for $f(x)=\bigg\{ \begin{gathered} x , 0 \lt x \lt \frac{\pi }{2} \\ \pi -x , \frac{\pi }{2} \lt x \lt \pi \\ \end{gathered}$ $\hspace{20mm}$ 6
B) Find inverse Laplace transform of $\frac{S}{(S^{2}+4s+13)^{2}}$ using convolution theorem. $\hspace{20mm}$ 6
C) Prove that $\overrightarrow{F }=(y^{2} cosx+ z^{3})i+(2 y\sin x-4)j+(3xz^{2}+2)k$ is a conservative field. $\hspace{20mm}$ 8
Find: (i) Scalar Potential for (ii)The work done in moving an object in this field from (0,1,-1) to $(\frac{\pi }{2},-1,2)$.
Q.6
A) Find Laplace transform of e$^{-4t}$ $\int_{0}^{t}{u sin3u du}$. $\hspace{20mm}$ 6
B) Use Stoke's Theorem to evaluate $\int_{C}{\overrightarrow{F}. \overrightarrow{dr}}$where $\overrightarrow{F}=(2x-y)i-yz^{2}j-y^{2}zk$ and S is $\hspace{20mm}$ 6
the surface of the hemisphere x2+y2+z2=a2 lying above the XY - plane.
C) Express the function f(x) = $\bigg\{ \begin{gathered} 1 , |x|\lt1 \\ 0 , \vert x\vert \gt 1 \\ \end{gathered}$ as Fourier integral. Hence evaluate $\hspace{20mm}$8
$\int_{0}^{\infty }{\frac{sinw . sinwx}{w} dw}$
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2023-02-05 20:46:06
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https://www.physicsforums.com/threads/the-sequence-space-lp.678698/
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# The sequence space lp
1. Mar 15, 2013
### hedipaldi
1. The problem statement, all variables and given/known data
how to prove that the sequences space lp is subspace of lq for p smaller than q?
2. Relevant equations
3. The attempt at a solution
I try to imply holder inequality but meanwile unsuccesfully
2. Mar 15, 2013
### jbunniii
Hint: if $0 < x < 1$ and $0 < p < q$, then $x^q < x^p$.
3. Mar 15, 2013
### hedipaldi
Yes,so simple. thank you very much.
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2018-02-17 23:47:59
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http://math.stackexchange.com/questions/238568/how-many-ways-can-you-choose-a-committee-if-for-a-married-couple-you-must-choos
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# How many ways can you choose a committee if for a married couple, you must choose both or neither
The full question is:
How many ways can you choose a committee of $5$ people from a group of $3$ married couples and $6$ single people, if for any married couple you must pick either both spouses or neither?
I'm having a lot of trouble separating out the ways to choose both or neither--especially because that affects how many you choose from the group of singles.
My work so far has gotten me the answer $\frac12 \Big( \frac {C(12,5)}{ 3!}\Big)$--where the $\frac12$ takes care of the neither or both, the $3!$ takes care of choosing a group, and other than that, you are choosing $5$ from a group of $12$.
Thanks for any help in advance!
-
Hint
Consider separately the cases when you have 0, 1 or 2 couples on the committee.
-
ok, but then how do you put it together? so, 0 couples->C(6,5), 1 couple->C(6,3), 2 -> C(6,1), but you can't just do C(6,5)+C(3,1)*C(6,3)+C(3,2)*C(6,1). So how do you make it so that there is one statement that deals with all of them? – BadAtGraphs Nov 16 '12 at 8:57
@BadAtGraphs: Why can't you "just do" that? – ShreevatsaR Nov 16 '12 at 8:59
@BadAtGraphs You can just add them because they are mutually exclusive. You can't have 0 couples and 1 couple and 2 couples at the same time. – Daryl Nov 16 '12 at 10:21
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2015-07-04 21:05:55
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http://en.wikipedia.org/wiki/Talk:Wrapped_normal_distribution
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# Talk:Wrapped normal distribution
WikiProject Statistics (Rated C-class, Mid-importance)
This article is within the scope of the WikiProject Statistics, a collaborative effort to improve the coverage of statistics on Wikipedia. If you would like to participate, please visit the project page or join the discussion.
C This article has been rated as C-Class on the quality scale.
Mid This article has been rated as Mid-importance on the importance scale.
## Wrapped Normal Distribution
(This contrib copied from Stats project talk page by me: Melcombe (talk) 09:41, 21 January 2010 (UTC) )
The pdf for the wrapped normal doesn't appear correct to me. If I type it in Mathematica, I get imaginary values out. The Jacobi description that follows is a mix of variables that have the same name in different formulas, and is confusing at best. As stated it appears as such.
### Current
$f_{WN}(\theta;\mu,\sigma)=\frac{1}{2\pi}\sum_{n=-\infty}^\infty e^{-\sigma^2n^2/2+in(\theta-\mu)} =\frac{1}{2\pi}\vartheta\left(\frac{\theta-\mu}{2\pi},\frac{i\sigma^2}{2\pi}\right)$
where $\vartheta(\theta,\tau)$ is the Jacobi theta function:
### My Proposal
$f_{WN}(\theta;\mu,\sigma)=\frac{1}{2\pi}\sum_{n=-\infty}^\infty e^{-\sigma^2n^2/2+in(\theta-\mu)} =\frac{1}{2\pi}\vartheta_3\left(\frac{\theta-\mu}{2},e^{-\sigma^2/2}\right)$
where $\vartheta_3(\theta,\tau)$ is the 3rd Jacobi theta function:
If I type this in to Mathematica, it works, and matches know results that I have to compare with. Also, I propose deleting the Jacobi elliptic explanation. The summation form is also suspect, but I'll look this up later.
I would prefer someone to validate this, otherwise in 2 weeks i will change it.
[email protected] (talk) 20:08, 20 January 2010 (UTC)
A few points
• The current definition is consistent if you accept the definitions given in this article, so there's not an error in that sense. Its easy to do the substitution.
• The current definition is in terms of the Jacobi theta function as defined in the Wikipedia Jacobi theta function article under the "nome definition", except its written $\vartheta_{00}(w,q)$. Further up, it is stated that $\vartheta_{00}(z,\tau)=\vartheta(z,\tau)$. This is kind of confusing but I assumed that $\vartheta_{00}(w,q)$ is a different function with the same name. I just avoided the problem by assuming that $\vartheta(z,\tau)=\vartheta_{00}(w,q)$ was what was meant. This is a notational problem, not a mathematical problem. Maybe this is something that should be cleared up in the Jacobi theta function article.
• Mathematica gives a correct answer in terms of the $\vartheta_3$ function, but this function is not defined in the Wikipedia article. I wanted to stick with functions that could easily be accessed within Wikipedia. Maybe the $\vartheta_3$ function should be included in the Jacobi theta article.
• I don't understand what you mean by "the Jacobi elliptic explanation". There is no reference to "Jacobi elliptic" in the article.
• By all means, check the summation form.
• I'm beginning not to like the present definition, because the $z=e^{i\theta}$ is really the preferred variable for circular statistics and I think we should stick to using that variable as much as possible. That means using the nome variables themselves, not the present definition nor the $\vartheta_3$ function. PAR (talk) 07:34, 22 January 2010 (UTC)
• Yes it is consistent. I stumbled on the $\theta$ having two different meanings.
• Okay, I read that article, and using $\vartheta_{00}$ would be better. I read through the Jacobi theta function article, and it's different than the reference I have on the subject, but as you stated it appears notational.
I'm in favor of a Jacobi function that uses the nome variables w and q because they are the more natural variables for circular statistics, along with $z=e^{i\theta}$. I haven't worked it all out yet, but I believe it would be an improvement. I don't know what you mean by $\theta$ having two different meanings. There is $\phi$ which is the "true" or "unwrapped" angle, that lies in $[-\infty,\infty]$ and the "measured" or "wrapped" angle that lies in some interval of length $2\pi$. PAR (talk) 23:10, 27 January 2010 (UTC)
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2014-07-10 04:00:49
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https://www.hepdata.net/record/84573
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Search for Supersymmetry in $pp$ Collisions at $\sqrt{s}=13\text{ }\text{ }\mathrm{TeV}$ in the Single-Lepton Final State Using the Sum of Masses of Large-Radius Jets
Phys.Rev.Lett. 119 (2017) 151802, 2017.
The collaboration
Abstract (data abstract)
CERN-LHC. CMS. Results are reported from a search for supersymmetric particles in proton-proton collisions in the final state with a single lepton; multiple jets, including at least one b-tagged jet; and large missing transverse momentum. The search uses a sample of proton-proton collision data at sqrt{s}= 13 TeV recorded by the CMS experiment at the LHC, corresponding to an integrated luminosity of 35.9 fb−1. The observed event yields in the signal regions are consistent with those expected from standard model backgrounds. The results are interpreted in the context of simplified models of supersymmetry involving gluino pair production, with gluino decay into either on- or off-mass-shell top squarks. Assuming that the top squarks decay into a top quark plus a stable, weakly interacting neutralino, scenarios with gluino masses up to about 1.9 TeV are excluded at 95% confidence level for neutralino masses up to about 1 TeV.
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2020-06-02 18:19:59
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http://observations.rene-grothmann.de/
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Constructive Art with EMT?
>n=40; X=random(n); Y=random(n);
>function map dist(x,y) := log(prod((x-X)^2+(y-Y)^2))
>x=linspace(0,1,500); y=x';
>z=dist(x,y); levels=linspace(totalmin(z),totalmax(z),100);
>fullwindow(); plot2d(z,levels=levels,a=0,b=1,c=0,d=1,<grid):
26. Dezember 2020 von mga010
Kategorien: Uncategorized | Schreibe einen Kommentar
The Ethics of AI
I recently have to read a lot about the consequences of AI with respect to ethics and moral. Let me quickly sketch my personal view on the subject.
First, let us talk about science fiction and AI robots that act like humans. In my view, it is rather simple:
Machines are not human beings,
even when they are equipped with AI or look like humans.
Robots are not part of our society.
In fact, it is probably better for us if the robots do not look like humans at all. That helps us keep the distinction. Machines, with AI or not, should be marked as machines.
The reason I am so sure and definitive about that is not a religious one. We humans are programmed genetically to be part of a human community. Our very self and the purpose of our life depends on other humans. Being lonely, an outlaw, or even just being pushed around make us sick. Children, friends, and admiration make us happy. We are proud to take responsibility for others. It is a misconception to think that people are basically selfish.
Now you can argue that the social system we live in could be enhanced and improved by AI robots which are on our level or even above, both in moral acting and in intellectual capabilities. These robots could live among us just like our friends. Wouldn’t that possibly be a better society? Aren’t humans inherently unreliable and the machines are not?
But any AI that acts in predictable and reliable ways is not an AI at all. It would act mechanical and can easily be determined to be a machine. The very characteristic of intelligence is that it explores new paths occasionally. This makes it unreliable and not predictable by definition. Soon, most AI will even act in super-human ways that we are incapable of understanding due to our limited capacities.
Thus, my only conclusion is that we need to distinguish ourselves from the AI we have. We need to treat AI as machines that we use. Robots are not a product of nature that can live with us uncontrolled with its own rights. We need to protect mankind, our society, and ourselves.
After all that science fiction nightmares, let us talk about the current way we use AI to improve technical systems. Even that limited form of AI already poses questions of ethics and moral.
Some see a lot of problems arising with AI or fuzzy logic that are built into cars, airplanes or surveillance systems. It is indeed true that these systems have the potential of a collision between a human decision and an AI decision. But that is already true without AI technology! We all experienced a system that seems to act on its own, a computer or even a mechanical device.
Think of an airplane with an AI as co-pilot. As a first example, assume the AI commands a go-around in an unstable approach. The chances are quite high that the AI is right and the pilot has made an error. I also tend to think that the AI makes much fewer mistakes throughout the complete flight as a co-pilot would. I would definitely prefer an AI to a rookie on the right seat. The same applies to nearly all applications of AI in technology. Usually, the AI is better. It can even be used to train the pilot in a simulator, much better and versatile than a human instructor.
It must be clear, however, that the responsibility for the proper working of the AI is at the developer of the plane, car or technical system that uses it. The AI is not a human that we can make responsible or even punish for mistakes. The developer has to test the system thoroughly to make sure it works as intended. But if you think this is too difficult note that it is much more difficult to test a human, and the verdict is much less reliable.
It must also be clear at all times that the AI that makes these decisions is a machine. When we let cars drive on their own the car does not become a being on our level, however intellectually superior it may be. If it does not work it will be trashed or repaired.
The problem with AI and robots should not be that they will be superior to us in many ways. The problem is that we need to treat them as our enslaved machines, and not as part of our society.
By the way, I am not afraid of super-human AI. Indeed, humanity deserves a lesson in humiliation. But let us use AI to our benefit!
10. Dezember 2020 von mga010
Kategorien: Uncategorized | Schreibe einen Kommentar
Teaching Python or Java
Whenever I have the pleasure to teach the beginner course in programming I start with the same question: Python or Java? Since colleagues have good success with Python and since Python is a useful language I am tempted ever again. However, I decide to go with Java every time. Let me explain the reasons.
This year, I even started a discussion on a Python forum to push me towards Python and tell me why I should prefer it. The discussion was an interesting read, but nothing that convinced me.
So I did my own research on Python sites designed for beginners and see how they handle the teaching. As expected, they dive into Python and use its data collections almost from the start, at least shortly after using the Python command line for interactive computations. Some even go quickly into real world usage with Python packages for graphics, artificial intelligence or numerical mathematics.
If Python is that useful and if there are so many libraries available and also quite nice IDEs (my favorite is Spyder) why do I shy back to use it as first language? Here are some reasons.
• I am teaching mathematicians. Mathematicians are meant to program, to test and to improve basic algorithms. They may also use high level libraries for research, but they should at least be able to start on the roots of the code, and often that is the only way to go. Python is not designed for this use. First, it is a lot slower than Java (which is on the level of C code, by the way). The difference can be a factor of 10 to 50. Then, it hides the basic data types (such byte, short, int, long, float, double) and their problems from the user and encourages to use its infinite arithmetic. I feel that mathematicians should understand the bits and bytes, and get full, speedy control of their code.
• Python extends its language to compactify the handling of data sets at the cost of extending the language with almost cryptic syntax. E.g., Python can automatically and in one expression create a list of integers satisfying a certain condition or select from another list under a condition. I am convinced that this kind of behind-the-scene operation is not suited to teach beginners how things work and how to write good and clean code. The situation is a bit like Matlab where everything is vectored and loops are costly. You learn how to use the high-level constructs of your language, but not the basics. If you only have a hammer everything looks like a nail.
• Python is as multi-platform or open-source as Java. Both need an interpreter and cannot be translated to native code to ease implementation of bigger programs on the user systems. Java is even a bit better in this respect. Both languages cannot easily use the native GUI library or other more hardware bound libraries. Python is a bit better here. But this is nothing to worry about as a first language anyway.
• I also need to answer the question: Is it easier to go from Java to Python or the other way? And, for me, the answer is clearly that Java provides the more basic knowledge on programming. Python is on a higher level. It is more a scripting language than a basic programming language. One may argue that this is good for beginners since they get a mighty system right at the start. After all, you learn to drive a car without knowing how the motor works. But I have seen enough programs in Python to know that it spoils the programmer and drives the thoughts away from a fresh, more efficient solution.
This said, I would very much welcome a second course building on the basics that uses Python and its advanced libraries. Many students are taught R and Matlab in the syllabus. I think Python would be a better choice than Matlab for numerical programming and data visualization. And, in contrast to Matlab, it is open and free.
25. November 2020 von mga010
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The Riemann Sphere
Since I am currently teaching complex function theory I am interested in visualizations of Möbius transformations as well as other analytic functions. The Riemann Sphere together with the Stereographic Projection is a good tool for this.
Above you see the basic idea. The points on the sphere are projected along a line through the north pole to the x-y-plane which represents the complex plane. The north pole associates to the point infinity. The formulas for the projection and its inverse are easy to compute with a bit of geometry. I have done so on this page using EMT and Maxima:
It is quite interesting to see what a Möbius transformation does to the Riemann sphere.
In this image, the north and south poles are mapped to the two black points. The other lines are the images of the circles of latitude and longitude. The image can be created by projecting the usual circles (with the poles in place) to the complex plane, applying the Möbius transformation, and projecting back to the sphere.
In fact, the projection and the Möbius transformation map circles or lines to circles or lines. They preserve angles. This is why the Stereographic Projection is often used for maps that should show the correct angles between paths. But, note that great circles that do not path to the pole are not mapped to lines, but to circles. So the shortest path on the surface of the Earth between two points is a circle on the projected map. Locally, this is only approximated by a line segment.
08. Juli 2020 von mga010
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A Simple Combinatorial Approximation Problem
This is certainly a well-known problem. But I pose it here because there is a geometric solution, and geometric solutions are always nice.
Assume you have two sets of n points
$$x_1,\ldots,x_n, \quad y_1,\ldots,y_n$$
The problem is to find a permutation p such that
$$\sum_{k=1}^n (x_k-y_{p(k)})^2$$
is minimized. Of course, we can assume that the x’s are sorted. And it is easy to guess that the minimum is achieved if the permutation sorts the y’s. But how to prove that?
Here is a simple idea. We claim that swapping two of the y’s into the right order makes the sum of squares smaller. I.e.,
$$x_1 < x_2, \, y_1<y_2 \Longleftrightarrow (x_1-y_1)^2+(x_2-y_2)^2 < (x_1-y_2)^2+(x_2-y_1)^2$$
If you try that algebraically the right hand side is equivalent to
$$-x_1y_1-x_2y_2 < -x_1y_2-x_2y_1$$
Putting everything one side and factorizing, you end up with
$$(x_2-x_1)(y_2-y_1) > 0$$
and the equivalence is proved.
But there is a geometric solution. We plot the points into the plane. Two points are on the same side of the symmetry line between the x- and y-axis if they have the same order. And the connection to the mirrored point will always be larger.
We can extend this result to more norms. E.g., the sorting of the y’s will also minimize
$$\sum_{k=1}^n |x_k-y_{p(k)}|, \, \max_k |x_k-y_{p(k)}|$$
As long as the underlying norm is symmetric with respect to the variables, i.e.,
$$\|(\ldots,t,\ldots,s,\ldots)\|=\|(\ldots,s,\ldots,t,\ldots)\|$$
The prove for this can be based on the same geometrical argument.
17. Juni 2020 von mga010
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C vs Java vs Python vs EMT
I recently tested Python to see if it is good as a first programming language. The answer is a limited „yes“. Python has certainly a lot to offer. It is a mainstream language and thus there is a lot of support in the net, and there are tons of libraries for all sorts of applications. A modern version like Anaconda even installs all these tools effortlessly.
However, I keep asking myself if it isn’t more important to learn the basics before flying high. Or to put it another way: Should we really start to teach the usage of something like a machine learning toolbox before the student has understood data types and simple loops? This question is important because most high-level libraries take away the tedious work. If you know the right commands you can start problem-solving without knowing much of the underlying programming language.
I cannot decide for all teachers. It depends on the type of students and the type of class. But I can give you an idea of how much a high-level language like Python is holding you back if you want to implement some algorithm yourself. You have to rely on a well-implemented solution being available, or you will be lost.
So here is some test code for the following simple problem. We want to count how many pairs (i,j) of numbers exist such that i^2+j^2 is prime. For the count, we restrict i and j to 1000.
#include
#include
int isprime (int n)
{
if (n == 1 || n == 2) return 1;
if (n % 2 == 0) return 0;
int i = 3;
while (i * i <= n)
{
if (n % i == 0) return 0;
i = i + 2;
}
return 1;
}
int count(int n)
{
int c = 0;
for (int i = 1; i <= n; i++)
for (int j = 1; j <= n; j++)
if (isprime(i * i + j * j)) c = c + 1;
return c;
}
int main()
{
clock_t t = clock();
printf("%d\n",count(1000));
printf("%g", (clock() - t) / 1000.0);
}
/*
Result was:
98023
0.132
*/
So this takes 0.132 seconds. That is okay for one million prime-number checks.
Below is a direct translation in Java. Surprisingly, this is even faster than C, taking 0.127 seconds. The reason is not clear. But I might not have switched on every optimization of C.
public class Test {
static boolean isprime (int n)
{
if (n == 1 || n == 2) return true;
if (n % 2 == 0) return false;
int i = 3;
while (i * i <= n)
{
if (n % i == 0) return false;
i = i + 2;
}
return true;
}
static int count(int n)
{
int c = 0;
for (int i = 1; i <= n; i++)
for (int j = 1; j <= n; j++)
if (isprime(i * i + j * j)) c = c + 1;
return c;
}
static double clock ()
{
return System.currentTimeMillis();
}
public static void main (String args[])
{
double t = clock();
System.out.println(count(1000));
System.out.println((clock() - t) / 1000.0);
}
/*
Result was:
98023
0.127
*/
}
Below is the same code in Python. It is more than 30 times slower. This is unexpected even to me. I have to admit that I have no idea if some compilation trick can speed that up. But in any case, it is the behavior that an innocent student will see. Python should be seen as an interactive interpreter language, not a basic language to implement fast algorithms.
def isprime (n):
if n==2 or n==1:
return True
if n%2==0:
return False
i=3
while i*i<=n:
if n%i==0:
return False
i=i+2
return True
def count (n):
c=0
for k in range(1,n+1):
for l in range(1,n+1):
if isprime(k*k+l*l):
## print(k*k+l*l)
c=c+1
return c
import time
sec=time.time()
print(count(1000))
print(time.time()-sec)
## Result was
## 98023
## 4.791218519210815
I have a version in Euler Math Toolbox too. It uses matrix tricks and is thus very short, using built-in loops in C. But it still cannot compete with the versions above. It is about 3 times slower than Python and 100 times slower than C. The reason is that the isprime() function is not implemented in C but in the EMT language.
>n=1000; k=1:n; l=k'; tic; totalsum(isprime(k^2+l^2)), toc;
98023
Used 13.352 seconds
Below is a version where I use TinyC to check for primes. The function isprimc() can only handle real numbers, no vectors. Thus I vectorize it with another function isprimecv(). This takes a bit of performance.
>function tinyc isprimec (x) ...
$ARG_DOUBLE(x);$ int n = (int)x;
$int res = 1;$ if (n > 2)
${$ if (n%2 == 0) res=0;
$int i=3;$ while (i*i<=n)
${$ if (n%i==0) { res=0; break; }
$i = i+2;$ }
$}$ new_real(res);
\$ endfunction
>function map isprimecv (x) := isprimec(x);
>n=1000; k=1:n; l=k'; tic; totalsum(isprimecv(k^2+l^2)), toc;
98023
Used 0.849 seconds
12. Juni 2020 von mga010
Kategorien: Uncategorized | Schreibe einen Kommentar
More Remarks on Corona
It is, of course, interesting how the total mortality behaves in times of Corona. Unfortunately, I was searching in vain for numbers in Germany. What I found was a PDF file from Spain. I took the first statistics in the image above from it.
In Germany, each year about 1.1% of the population will die, which is a rate of one in approximately 31000 per day. So I would expect around 1450 deaths each day in Spain. The rate depends, of course, strongly on the age structure of the population. Maybe, I underestimate the pyramidal shape of the ages in Spain, so that the daily total mortality is indeed a bit lower than in Germany. The image suggests around 1200.
A statistical plot of that kind should always have a y-axis that starts with zero. So this plot is a bit misleading. But in any case, it suggests a 50% increase in total mortality per day in the last few days. That is due to Corona. As I said, the problem will average out over the year if appropriate measures are taken, and we might observe only a slight increase in total mortality.
Nevertheless, the main problem remains: We cannot handle the huge numbers of severely sick patients that would occur if we just let everything run as normal.
08. April 2020 von mga010
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More Remarks on Corona
We learned a bit more now.
• I am glad to hear that the social restrictions lead to a base infection rate of approximately one in Germany, i.e., one new infection per infectious. That is good news, but it must be even less. The number of cases must decrease to a number we can handle. Currently, my estimate is that every 100th German is infected, and probably the same in the US. That sounds manageable with care in social interactions.
• And, of course, we still need to restrict travel, especially to countries that cannot take the measures we can because they are too poor. Those countries need immediate help. And, for god’s sake, please relieve the sanctions that the West has imposed to punish Eastern countries for not being cooperative.
• Opening the kindergartens and schools is crucial for society. But it also imposes the biggest thread. Every nurse can tell you that one case of chickenpox means 30 cases of chickenpox where kids are playing together in larger numbers. And young parents will go through an unknown peak of strep infections while their kids are in that age. So, we need antibody tests as soon as possible, plus a new look at hygiene.
• I always opted for more digital learning. Some don’t agree because they either think I wanted to replace teachers, or they underestimate the benefits of lonely study, trying, failing and thinking. It is obvious that learning is also a social activity and that teachers are needed for guidance. But some stuff must be done alone, and that can be very well stimulated by digital media.
04. April 2020 von mga010
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More Remarks on Corona
What have we learned so far? The numbers are not clear, nor are the consequences. Moreover, there is a lot of distorted information, some can even be called fake. Let me tell you what I took from all that.
• The most important problem that the new virus is causing is the exceeding number of cases of pneumonia and ARDS caused by fluid in the lungs. This seems to be a lot worse than in the ordinary flu or older Coronaviruses, probably because we are not yet immune. The virus is also causing other damages to your body, but lung problems are the most common ones. While the chances to need oxygen or ventilation are not known precisely yet and will depend on your age, the situation in the hospitals of the affected regions speaks a clear language. We need to concentrate on medical care and break the chains of infection to slow down the development of the disease!
• The mortality seems to be high only in people of old age, especially those that are weakened by other diseases. We might observe that the overall mortality does not rise much at all. The number of deaths by COVID19 published so far all over the world supports this view. The conclusion is to protect the elderly as much as we can to spare them a painful and sudden death requiring intense medical care. We do that, again, by reducing social contact and keeping the number of infected people at any time as low as possible. By the way, hygiene and care should have been observed in contact with elderly people at all times.
• The current economy is based on the production of consumer articles and marketing. Many countries have reduced or have been forced to reduce state activities that do not lead to direct profit, such as health care and education. This economy is not fit for any crisis that needs social solidarity. Since profit and interest are the driving forces and the main means of wealth distribution each recession throws us into a deep crisis. Politicians see that now and try to stabilize the situation. Essentially, this is a takeover and a regulation of the industry by financial means, which can be summed up as printing money. Even the idea of state control of key industries is revived. We should learn from that crisis that there must be a balance between the private sector of the economy and social needs covered by the state at all times. If that is out of balance, we are not flexible enough to fight a crisis like this one.
Keep healthy and stay at home for a while!
31. März 2020 von mga010
Kategorien: Uncategorized | Schreibe einen Kommentar
Remarks on the Corona Pandemic
Here are some remarks that I noticed in these difficult times. Some are math-related, some are not.
• Humans seem to be unable to image big numbers or grasp the concept of exponential growth. In fact, I cannot really do that myself. What I have to do is compute figures and set them in relation to the total. To help us understand the problem here is one example: Letting the virus spread in a small town of 10000 can easily mean that 1/4 of them are infected at the same time, i.e., approximately 2500. About 1/20 of those need intensive medical treatment, and at least 1/100 need ventilation. We end with a number of 25. This is impossible to do for any countryside hospital. Those small numbers can be understood and they should frighten you to the point that you understand how necessary it is to break the chains of infection. Of course, the same computation for the complete US or German population is even more frightening.
• Most of us do not understand how bad it is that the world population will grow to eleven billion in the near future. We should have stopped that by education, sharing of wealth and women’s rights thirty years ago. Moreover, we have destroyed local support chains by globalism and our excessive capitalistic system. This falls on our feet now. In African countries, the population has to live in slums at close contact and is depending on the global food chain, and also on the medical support of „first world“ countries. This is a recipe for disaster. Even we are affected by the problem because our support with masks and other medical aids is broken since China is no longer delivering. My only hope is that we come to rethink our world after that crisis.
• We all cannot grasp probabilities properly. I am now over 60 years old. That means that my non-corona chance of dying in the next year exceeds the corona related chance, even if I catch the virus. I am more likely to die of a heart attack, get brain or lung cancer, get involved in a car accident, or get usual pneumonia during that year. We cannot avoid death. It is never a good idea to think about the chance to die too much. Remember the words of Epicurus: „The death is none of your business. When he is here you aren’t. When you are here he isn’t.“
• The press and TV are now running on full speed to shovel numbers onto the public and present the drama of single cases. Unfortunately, this agenda together with a lack of social contact may cause a lot of harm to some individuals. If only we would take other causes of unnecessary deaths with the same degree of attention! I would prefer a more scientific attitude in the media, and more reticence to dramatize individual fates. This only leads to a distorted view of reality.
24. März 2020 von mga010
Kategorien: Uncategorized | Schreibe einen Kommentar
← Ältere Artikel
|
2021-01-17 00:05:38
|
{"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": 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": 0.39159637689590454, "perplexity": 973.5059218356232}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-04/segments/1610703507971.27/warc/CC-MAIN-20210116225820-20210117015820-00799.warc.gz"}
|
https://math.meta.stackexchange.com/questions/34286/is-the-font-difference-of-history-of-the-answers-of-a-nominated-user-for-the-e
|
# Is the font difference of 'history' of the answers of a nominated user for the election different from 'link' and 'flag' intentional?
Maybe the impact of this is not significant, but here is the image (I used MS Edge, 200% for the image):
I inspected the elements, and both link and flag use -apple-system,BlinkMacSystemFont,"Segoe UI","Liberation Sans",sans-serif as fonts. For history, the font used is Arial.
I want to ask if this is intentional or not.
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2021-12-03 14:13:39
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http://mathhelpforum.com/calculus/85314-optimization-problem.html
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# Math Help - Optimization Problem
1. ## Optimization Problem
You have been asked to design a 1-liter cylindrical can made with sheet
aluminum. What dimensions of the can (radius and height) will use the least total
amount of aluminum? State dimensions to the nearest hundredth of a
centimeter.
Hint 1: The aluminum must cover both ends of the can as
well as the circular wall.
Hint 2: 1 liter = 1000 cm^3
I believe my primary equation would be: 2TTr^2 + 2TTrh
I think my secondary equation would be: TTr^2h=1000
Is that right?
I came up with a radius of 12.62 and a height of 2.00
We are supposed to have all of the following...
• Primary equation
• Secondary equation, if applicable
• Statement of domain
• Critical numbers and FDT to establish relative extrema
• Concluding statement(s) to answer the question.
2. $V = \pi r^2 h$
$h = \frac{V}{\pi r^2}$
$A = 2 \pi r^2 + 2 \pi r h$
$A = 2 \pi r^2 + 2 \pi r \left(\frac{V}{\pi r^2}\right)$
$A = 2 \pi r^2 + \frac{2V}{r}
$
$\frac{dA}{dr} = 4\pi r - \frac{2V}{r^2}$
set $\frac{dA}{dr} = 0$ ...
$4\pi r = \frac{2V}{r^2}$
$r^3 = \frac{V}{2\pi}$
$r = \left(\frac{V}{2\pi}\right)^{\frac{1}{3}}$
calculate $r$ ... go back and calculate $h$
you should see that $h = 2r$
3. Ok, so the minimum surface area would be when the radius is 5.42 and the height is 10.84...right?
You are to design a 1-liter cylindrical can made with sheet aluminum.
What dimensions of the can (radius and height) will use the least total
amount of aluminum?
State dimensions to the nearest hundredth of a centimeter.
Hint 1: The aluminum must cover both ends of the can as well as the wall.
Hint 2: 1 liter = 1000 cm³.
I believe my primary equation would be: . $A \:=\: 2\pi r^2 + 2\pi rh$
I think my secondary equation would be: . $\pi r^2h\:=\:1000$
Is that right? . . . . Yes!
I came up with a radius of 12.62 and a height of 2.00 . . . . no
We have: . $\pi r^2h \:=\:1000 \quad\Rightarrow\quad h \:=\:\frac{1000}{\pi r^2}$ .[1]
Then: . $A\;=\;2\pi r^2 + 2\pi r\left(\frac{1000}{\pi r^2}\right) \;=\;2\pi r^2 + 2000r^{-1}$
Differentiate: . $A' \;=\;4\pi r - 2000r^{-2} \:=\:0 \quad\Rightarrow\quad 4\pi r^3 - 2000 \:=\:0$
. . $4\pi r^3 \:=\:2000 \quad\Rightarrow\quad r^3 \:=\:\frac{500}{\pi} \quad\Rightarrow\quad r \:=\:\sqrt[3]{\frac{500}{\pi}} \:=\:5\sqrt[3]{\frac{4}{\pi}}$
Substitute into [1] and we get: . $h \:=\:10\sqrt[3]{\frac{4}{\pi}}$
Therefore: . $r \:\approx\:5.42\text{ cm},\;\;h \:\approx\:10.84\text{ cm}$
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2015-08-02 23:34:56
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https://electronics.stackexchange.com/questions/512259/voltage-gain-in-common-emitter-bjt-amplifier-specific-example
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# Voltage gain in common emitter BJT amplifier (specific example)
What is the AC voltage gain in the following common emitter BJT amplifier? We are also given that for the BJT $$\\beta_{DC} = \beta_{ac} = 150\$$.
simulate this circuit – Schematic created using CircuitLab
Full disclosure: this is question 19, p825 in Electronics Fundamentals, Pearson, 8th Ed.
I'm asking because my answer does not agree with that of the book, and I can't see the fault in my calculation:
$$V_B = 8\frac{3.3}{3.3 + 12} = 1.725V$$ $$V_E = V_B - 0.7V = 1.025V$$ $$I_E = 10.25mA$$ $$r_e = \frac{25mV}{10.25mA} = 2.44\Omega$$ $$A_v = \frac{R_C}{r_e} = 123$$
NB we are given the formula $$\r_e = \frac{25mV}{I_E}\$$ earlier in the book (without derivation).
• What is answer in the book ? Is it far off ? The 100 ohm resistor will probably be seen as $\beta \cdot 100 = 15k$ when seen from input side. It may slightly affect the value of $V_B$. If I consider that, I get gain as $\approx 84$. In fact, when I click the circuit simulate button in your diagram, $I_E = 8.7 mA$, a 15% reduction. – AJN Jul 25 at 15:24
• Thanks AJN, that's exactly it. I forgot to account for the effect of $R_E$ on the voltage divider. When I recalculate including the $15k\Omega$ in parallel, I get $R_2' = 2.7k\Omega, V_B = 1.47V, V_E=0.77V, I_E=7.70mA, r_e = 3.25\Omega, A_v = 92.3$ (which is the answer in the book). – kikazaru Jul 25 at 15:42
• You can answer it yourself and mark it as answered. It is allowed in SE. – AJN Jul 25 at 15:43
• @kikazaru You are missing the voltage drop caused by $R_\text{TH}$ and $I_\text{B}$ passing through it to set $V_\text{B}$. It's possible the book is also missing this drop. Your computation of $V_\text{B}$ is actually $V_\text{TH}$. – jonk Jul 25 at 17:48
• The circuit is NOT common collector (which is an emitter-follower). Instead it is common emitter. – Audioguru Jul 25 at 18:18
# Overview
The left and right side schematics below are entirely equivalent to each other (within numerical truncation errors):
simulate this circuit – Schematic created using CircuitLab
Note that your computation of $$\V_\text{B}\$$ is not actually the base voltage for the BJT. It is the Thevenin voltage that precedes the Thevenin resistance to the base. The base voltage will be less than this, because the base current will cause a voltage drop across $$\R_\text{TH}\$$.
# Discussion
The computation of the base current is now:
$$I_\text{B}=\frac{V_\text{TH}-V_\text{BE}}{R_\text{TH}+\left(\beta+1\right)R_\text{E}}= 57.976\:\mu\text{A}\approx 58\:\mu\text{A}$$
This will present a voltage drop across $$\R_\text{TH}\$$:
$$V_\text{B}=V_\text{TH}-I_\text{B}\cdot R_\text{TH}=1.57544\approx 1.58\:\text{V}$$
You are given $$\V_\text{BE}\$$, so I can't argue with it. In actual fact, it depends upon the collector current (in active mode, anyway.) But assuming the given value, you'd find $$\V_\text{E}\approx 880\:\text{mV}\$$. And then $$\r_e\approx 2.95\:\Omega\$$.
Unfortunately, to add to the complexity, your emitter capacitor is small enough that at audio frequencies it will also present a significant impedance. $$\X_C=\frac1{2\pi\,f\,C}\$$, so for example at $$\1\:\text{kHz}\$$ it presents $$\X_C\approx 16\:\Omega\$$ and at $$\8\:\text{kHz}\$$ $$\X_C\approx 2\:\Omega\$$. Both these values are very significant with respect to $$\r_e\$$. So they will most definitely also influence the gain. In fact, the gain is so greatly affected that you will have a highly distorted output.
In any case, even discounting the capacitor's reactance and treating all of them as dead shorts for AC (one can always just make them a lot larger), your computation of $$\A_v\$$ still falls short because it doesn't take into account the voltage drop across $$\R_\text{TH}\$$.
# Summary
I've also neglected analysis using an input signal with any significant swing to it. So long as the input signal amplitude is small with respect to the DC operating point of the voltage at the emitter, you can proceed with a simplified voltage gain estimate. But with any significant input signal, this causes the emitter voltage to move significantly up and down with the signal. This means the emitter current also substantially varies, leading to a varying value for $$\r_e\$$, leading to still more distortion as the voltage gain continues to vary as the signal itself varies. The upshot of all this is that without global NFB to correct this problem this is a pretty bad circuit if you care about signal distortion.
And finally, the analysis only works at a fixed temperature since the voltage gain (and the operating point, to be honest, as $$\V_\text{BE}\$$ also varies with temperature) are quite dependent on temperature since $$\r_e\$$ depends upon the thermal voltage which depends upon the operating temperature of the BJT.
Just FYI.
• To see if I have understood your argument correctly : The correction term of $R_{TH} I_{B}$ is required since the equivalent (Thevenin) resistance of $R_{E} \rightarrow \beta \cdot R_{E}$ was used in calculation of the node voltage between R1 and R2. Supposing the node voltage was derived using another (more exact) method, say nodal analysis, the above correction term would not be required (i.e. the result would already be the final value). e.g. $(8 - V_B)/R1 = V_B/R2 + (V_B- V_{BE})/((\beta+1)R_E)$. By doing this more exact calculation, the result is $V_B = 1.575435$. – AJN Jul 26 at 4:46
• I have not noticed this gap in the analysis before. Thanks! – AJN Jul 26 at 4:51
• @AJN Using only nodal: $$\frac{V_\text{B}}{R_1}+\frac{V_\text{B}}{R_2}+I_B=\frac{V_\text{CC}}{R_1}$$But:$$I_B=\frac{V_\text{B}-700\:\text{mV}}{R_\text{E}\left(\beta+1\right)}$$Solving:$$V_\text{B}=\frac{V_\text{CC}\cdot R_\text{E}\left(\beta+1\right)+V_\text{BE}\cdot R_1}{\left(1+\frac{R_1}{R_2}\right)\,R_\text{E}\left(\beta+1\right)+R_1}$$Which gets you the number we've both computed for the base voltage. – jonk Jul 26 at 6:15
In particular, the mistake is due to not accounting for the effect of $$\R_E\$$ on $$\R_2\$$ in the voltage divider. This yields a lower $$\V_B\$$ of $$\1.47V\$$. The same steps as in the question then yield the correct gain factor.
• Vbe is closer to 0.65V @10mA not 0.7V which reduces current and gain, but then if output swing is > 1V the gain modulates with Ic too much and gets distorted. Also -3dB is pretty high. So it's a good design to forget. – Tony Stewart Sunnyskyguy EE75 Jul 25 at 16:19
• I accepted this answer because it is so short and captures the problem I was seeking to resolve with the question. I wrote it, but AJN really answered the question with the first comment. The other answer and comments are also very interesting extensions. – kikazaru Jul 27 at 15:50
• I agree except the real problems is ,and @jonk agrees is that it is a really bad design (nonlinear) so forget using this in future. It is extremely sensitive to gain, Ic and Vbe without a small linear Re added or NFB. Also the assumption here is a bigger source of error as a result..q.e. VBE≈ 0.7 V . This might be true if you were using 50% of it's rated current, but not 10% of Ie max or 10mA. Then Vbe is closer to 0.65 and your gain changes and 2 sig figs is all that is practical here. So I agree in this hypothetical question. – Tony Stewart Sunnyskyguy EE75 Jul 27 at 16:21
• 100% Tony Stewart. This is just an example used for a textbook question, so I wouldn't expect it to be applicable in practice, and the analysis is based on crude assumptions. – kikazaru Jul 28 at 17:14
Another way.
Considering the Re emitter resistance amplified by hFE to the base 150*100=15k across the 3.3K which becomes 2.7k
$$V_B = 8\frac{2.7}{2.7 + 12} = 1.47V$$ $$V_E = V_B - 0.7V = 0.77V$$ $$I_E = 770mV/100 ohm= 7.7mA$$ $$r_e = \frac{25mV}{7.7mA} = 3.25\Omega$$ $$A_v = \frac{R_C}{r_e} = 300/3.25=92.3$$
Using a lower assumption for Vbe =0.65V , which for 7.7mA might be more accurate (depending on chip size) leads to a 7% higher Ie and high Av. YMMV.
But since Vb will change with large input voltage swing, the current will modulate higher for +ve peaks and lower for -ve peaks and have horrible asymmetric output swing and thus the difference from symmetry is basically your harmonic distortion which you can estimate by the ratio of difference of each swing / Vpp = THD in % I bet you didn’t know that.
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2020-11-29 16:43:47
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http://math.upsc.xyz/34/find-the-shortest-distance-between-the-skew-lines
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Find the shortest distance between the skew lines:
$\frac{x-3}{3} = \frac{8-y}{1} = \frac{z-3}{1}$
$\frac{x+3}{-3} = \frac{y+7}{2} = \frac{z-6}{4}$
asked Nov 11, 2017
Let $l, m, n$ be the direction cosines of the line of shortest distance. As it is perpendicular to given lines,
$3l - m + n = 0$
$-3l + 2m + 4n = 0$
$\implies \frac{l}{-6} = \frac{m}{-15} = \frac{n}{3}$
$\implies \frac{l}{-2} = \frac{m}{-5} = \frac{n}{1}$
$\implies$ distance $= \left |\frac{(3- (-3)).(-2) + (8-(-7)).(-5) + (3-6).1}{\sqrt{{(-2)}^2 + {(-5)}^2 + 1^2}} \right | = \left |\frac{-90}{\sqrt{30}} \right | = 3\sqrt{30}$
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2019-01-16 20:47:18
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https://gmatclub.com/forum/m05-183662.html
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# M05-03
Author Message
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Math Expert
Joined: 02 Sep 2009
Posts: 64102
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15 Sep 2014, 23:24
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$$\frac{0.2 * 14 - \frac{15}{3} + 15 * \frac{3}{90}}{8 * 0.0125} = ?$$
A. 2
B. 1
C. 0
D. -1
E. -17
_________________
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Joined: 02 Sep 2009
Posts: 64102
### Show Tags
15 Sep 2014, 23:24
Official Solution:
$$\frac{0.2 * 14 - \frac{15}{3} + 15 * \frac{3}{90}}{8 * 0.0125} = ?$$
A. 2
B. 1
C. 0
D. -1
E. -17
The numerator simplifies to $$2.8 - 5 + \frac{1}{2}$$, which equals -1.7. The denominator equals 0.1. Dividing the numerator by the denominator gives us -17.
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Joined: 02 Jun 2015
Posts: 19
Location: United States
### Show Tags
02 Apr 2018, 14:03
1
To make the calculation easier, I broke down the denominator into pieces ... 8 * 1/8 * 1/10 = 1 * 1/10 = 0.1
2.8 - 5 + 0.5 / 8 * 0.0125 = - 1.7 / 0.1 = - 17
Thanks! Ale
Intern
Joined: 09 Aug 2019
Posts: 1
### Show Tags
11 Aug 2019, 09:33
Is that wrong question because I see 15 divided by 3
Intern
Joined: 20 Aug 2019
Posts: 1
### Show Tags
20 Aug 2019, 07:03
Any suggestions on where to find more problems like this one?
Tutor
Joined: 16 May 2019
Posts: 709
### Show Tags
20 Aug 2019, 07:11
Bunuel wrote:
Official Solution:
$$\frac{0.2 * 14 - \frac{15}{3} + 15 * \frac{3}{90}}{8 * 0.0125} = ?$$
A. 2
B. 1
C. 0
D. -1
E. -17
The numerator simplifies to $$2.8 - 5 + \frac{1}{2}$$, which equals -1.7. The denominator equals 0.1. Dividing the numerator by the denominator gives us -17.
An alternative approach - I will admit to solving this one the standard way, the same as Bunuel posted above. It took me about 1 1/2 minutes. However, I got to thinking about using the answers instead, since there are a few interesting splits that are easy enough to explore:
1) There are two positive answers, and 1 is about the easiest solution for a fraction. That is, do the top and bottom of the fraction match without being 0? This brings me to the second point.
2) 0 is an answer choice. For this to be true, the numerator would have to equal 0, and whatever is going on in the denominator is irrelevant (since "no solution" is not an option).
3) There are two negative answers. Also, one of those negative answers is -1, which, like its positive counterpart, is easy enough to test.
So, will the numerator be positive, negative, or 0? In any case, we will either get the answer directly (if it turns out to be (C)), or we will encounter a 50/50 split between the remaining two logical answer choices. With a single negative sign in the numerator, attached as it is to $$\frac{15}{3}$$, or 5, we can work out readily enough that (.2 * 14) - 5 will be negative. To be specific, we get 2.8 - 5 = -2.2. The second cluster in the numerator can be worked out either by reducing the 15 and 90 before multiplying or by reducing the fraction $$\frac{45}{90}$$ after multiplying. It makes no difference, as the answer will be positive 0.5 either way. Now we can see that the numerator is equivalent to -2.2 + 0.5, or -1.7. In one fell swoop, we can eliminate (A), (B), and (C), and unless the denominator works out to 1.7 exactly, it cannot be true that -1 will be the answer.
Here, you could work out the value of 8 * 0.0125, but you could just as easily reason that 0.0125 itself is slightly more than $$\frac{1}{100}$$, and multiplying it by 8 will not get the product anywhere near 1.7. Thus, we can rule out (D), and (E) must be the answer.
Senior Manager
Joined: 12 Dec 2015
Posts: 492
### Show Tags
11 May 2020, 06:38
$$\frac{0.2 * 14 - \frac{15}{3} + 15 * \frac{3}{90}}{8 * 0.0125} = ?$$
A. 2
B. 1
C. 0
D. -1
E. -17 --> correct: $$\frac{0.2 * 14 - \frac{15}{3} + 15 * \frac{3}{90}}{8 * 0.0125} = \frac{2.8 - 5 + 0.5}{0.1} = \frac{-1.7}{0.1}$$ = -17
Re: M05-03 [#permalink] 11 May 2020, 06:38
# M05-03
Moderators: chetan2u, Bunuel
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2020-05-25 15:47:24
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https://ravimohan.net/2016/09/
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# Determining the causal structure of spacetime (I)
In this series of blog posts, I aim to explain the causal structure of the spacetime solutions in general relativity. Currently, I am working on a special extension of Einstein’s theory (general relativity) which is known as Horndeski theory. There, I am trying to find the causal structure of the allowed solutions which allegedly permit super luminal propagation of metric perturbations. The methodology to obtain the structure is similar for all the gravitational theories and I wish to demonstrate it for general relativity (which is my comfort zone).
In order to make the observable predictions from a consistent physical theory, we are interested in finding how degrees of the freedom evolve and behave. For that, we try to obtain/formulate equations of motions, which capture the physics at the infinitesimal scale. Once we embed the physics in these equations, we solve them and make the verifiable predictions. For instance, in Newtonian mechanics, we study point particle moving in one dimension $x$. We, then ask how this degree of freedom behaves or evolves in time (which, again, is an assumption). For such a theory, we have the equation of motion $F=ma=m\frac{d^2 x(t)}{dt^2}$ which contains the spectacular physical insight from Newton (I won’t even try to elaborate that because it deserves a separate blog post). Now this is a linear second order differential equation which requires two initial conditions. In other words, if you give me the initial position and initial velocity, I can tell you the entire future of $x(t)$ by solving the differential equation. In fact, for constant $F$, we get $x(t)=ut+\frac{1}{2}\frac{F}{m}t^2+x_0$
Another way to look at it: equation(s) of motion (of a theory) give us the prescription to evolve the initial data (known values of the degrees of freedom) to final data that we are interested in.
Now Einstein’s gravity theory has the components of the metric as the degrees of freedom. We denote them by $g_{\mu\nu}$. Another radical difference (common to all relativistic theories) is that the space and time are unified into a single parameter space called spacetime manifold (inheriting all the structures of the pseudo-Riemannian manifolds). So here we ask our favorite question: how do $g_{\mu\nu}$ evolve in the spacetime. Mathematically, we want to know the $g_{\mu\nu}(x)$, where $x$ is the representative of the spacetime coordinates from now on.
This time, we use the insight from Einstein to write the equation of motion for general relativity (in vacuum) as
$R_{\mu\nu}-\frac{1}{2}g_{\mu\nu}R=0=E_{\mu\nu}$
where the symbols have their usual meaning. Here we are working on the 4 dimensional pseudo-Riemannian manifold $(\mathcal{M},g)$ on which $g$ is to be determined by the equation of motion. We define a Cauchy surface $\mathcal{C}$ which is a c0dimension 1 surface in $\mathcal{M}$ on which the initial data for Einstein’s equations is defined. So what does this initial data comprised of? (We will take a small detour and return to the topic in next blog post)
### Helmholtz wave
In order to understand that, we start with a simple example. Consider a 2+1 manifold as shown. Now let us consider a wave operator on this manifold given by $\hat{L}=-\frac{\partial^2}{\partial t^2}+\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}$. For a scalar degree of freedom denoted by $u$ the operator $\hat{L}$ acts on it to generate a wave equation $\hat{L}u=0$. Here we are not interested in solving the linear second ordered partial differential equation. We want to deduce the properties of the wave, like it’s speed of propagation and the extent to which a disturbance/fluctuation in $u$ can propagate.
Let us say that we know the value of $u=u(x,y)$ on $\phi(x,y,t)=t=0$ surface (where we might have a specified source or some boundary condition). The surface is essentially a codimension-1 surface and we will call the coordinates $x,y$ as internal coordinates (w.r.t the surface). Thus we can now easily compute the internal derivatives (from $u(x,y)$) and denote them by $u_i=\partial_iu(x,y)$ where $i\in (x,y)$. Let the exterior/normal derivative to the surface be $u_\phi (x,y)=\eta(du,d\phi)=-\partial_tu$ (note that $d\phi = dt$ on $\mathcal{C}$).
In this example, we have a Cauchy surface $\mathcal{C}$ defined by $\phi =t$ with the data
• $u(x,y)$
• internal derivative $u_i(x,y)$
• external derivate $u_\phi(x,y)$
given on $\mathcal{C}$. Clearly the second derivatives of $u$ except $u_{\phi\phi}$ can be computed from the given data. And here we use the wave equation (or equation of motion in general) to find the missing second derivative of $u$. The co-normal to the surface $\phi=0$ is given by $d\phi$ which we can easily represent in the natural co-normal basis as $d\phi=\phi_\mu dx^\mu$. And then we perform the coordinate transformation to chart $\lambda^\mu$ such that $\lambda^0=\phi$. Note for this particular case $d\phi \parallel dt$ and the new coordinate chart is exactly equal to $x^\mu$ chart. This is because $\mathcal{C}$ is already perpendicular to the time coordinate.
It is not very difficult to show that the wave equation $\hat{L}u=0$, under the above transformation, converts into $u_{\phi\phi}Q(\phi_\mu)+\ldots=0$ where $Q(\phi_\mu)=\eta^{\mu\nu}\phi_\mu\phi_\nu$ and we call it the characteristic form. Now there are two situations
1. $Q\neq 0$: in this case, we can invert the equation $u_{\phi\phi}Q(\phi_\mu)+\ldots=0$ and find the second derivative of $u$ in the direction normal to the surface. With that information we can easily evolve the data further in time.
2. $Q=0$: well, we can’t invert the equation which basically implies that there is no unique evolution of $u$ beyond that surface (which is now called the characteristic surface).
In our example, the characteristic form is identically 1 (just look at the coefficients of second derivatives (-1+1+1). Hence we don’t have any characteristic hypersurface for the degree of freedom $u$ obeying the wave equation $\hat{L}u=0$ with the Cauchy surface as the entire $x,y$ plane. It is not surprising if you think about the plane electromagnetic waves, which again don’t have any characteristic hypersurface and obey the same wave equation we wrote above.
Moving on, we obtained an equation for a surface in new coordinates $\lambda^\mu$ given by $Q(\phi_\mu)=0$. Physically, this is the surface beyond which we can not evolve the degrees of freedom uniquely (as the second derivative is not uniquely determined). Now we need a mechanism to generate this surface.
### Bicharacteristic curves
First we define the bicharacteristic curves or rays which are related to the linear second order partial differential operator $\hat{L}$, where we now generalize it to $\hat{L}[u]=a^{\mu\nu}u_{\mu\nu}+d$ where $u_{\mu\nu}=\partial_\mu\partial_\nu u$. It is not difficult to check that in this case the characteristic form becomes $Q=a^{\mu\nu}\phi_\mu\phi_\nu$. For our Helmholtz wave, we get the form $Q=-\phi_t^2+\phi_x^2+\phi_y^2$.
Now the bicharacteristic curves are generated in by the linear ordinary differential equations, with a parameter $s$, given by
$\frac{dx_\mu}{ds}=\frac{1}{2}\partial_{\phi_\mu}Q$ and $\frac{d\phi_\mu}{ds}=-\frac{1}{2}\partial_{x_\mu}Q$
For our Helmholtz wave, one can easily solve them and find the solutions $x(t)=at+b, y(t)=ct+d$. They actually form the rays of a cone (with appropriate boundary conditions) traveling with speed 1.
We have shown that if we introduce some perturbations at some point in Minkowski manifold, those perturbations will travel at unit speed and won’t escape the cone. This essentially exhibits the causal structure of flat Lorentzian spacetime which is in concurrence with the wave equation.
# BMS on my mind
I thought that I knew the Minkowski spacetime solution fairly well. But recently, to my surprise, I found that there is much more physics in that solution, especially at the null infinity. There is a set of certain symmetry transformations acting asymptotically at the null infinity which preserve some boundary conditions. This group of symmetries is known as BMS (Bondi-Metzner-Sachs) symmetry group. I will make more precise statements below in this post. Another, more surprising fact I found is that general relativity may not be a truly diffeomorphism invariant theory. In fact these BMS transformations map one asymptotically flat spacetime solution of constraints to another physically inequivalent asymptotically flat solution (http://arxiv.org/abs/1312.2229). But it is a topic for later post (when I am older and wiser).
Before progressing further, let me try to explain why physicists are interested in this symmetry group. In the well established Standard Model, the particles are the unitary irreducible representations of the isometry group of the flat Minkowski spacetime (known as the Poincaré Group), and, that of internal symmetries. For curved and dynamical spacetime, a part of the isometry group breaks down and the rest gets gauged (I plan to write a blog post describing this phenomenon in the near future). The Standard Model is formulated within the framework of Quantum Field Theory which works only on the flat spacetimes or curved spacetimes with fixed geometry or curved dynamical spacetimes with classical graviton. Gravity on the other hand is the theory of the dynamics of the spacetime. And, therefore, the Standard Model can not include the gravitons, the quantized form of gravity (like photons are quantized form of electromagnetic fields).
Now, this is the most interesting part, if we study physics on the asymptotically flat dynamical spacetime solutions, then we do have the asymptotic symmetry group of this gravitational space (it is essentially BMS). And the irreducible representation of this group will give the usual particles and some extra multiplets. These extra multiplets are termed as soft gravitons which have gained much attention recently in the quantum gravity community.
Again, in the hope of using these notes for my future reference, and to save a great fraction of my energy, I will take the liberty to be mathematically intense. But, I will try to maintain the rigor of topology, differential geometry and group theory so that my mathematician friends don’t get annoyed.
Now there are two ways to fish the BMS group of Minkowski space. One is by Sachs (http://dx.doi.org/10.1103/PhysRev.128.2851) and other is by Penrose. Let us start with the first one.
Consider a normal-hyperbolic Riemann manifold $(\mathcal{M},g^{\mu\nu})$ and a chart $(u(=t-r),r,\theta,\phi)$ in the neighborhood of point $P$ with the following properties
1. the hypersurface $u=\text{constant}$ are tangent to the local lightcone everywhere.
2. $r$ is the corresponding luminosity distance.
3. the scalars $\theta, \phi$ are constant along each ray defined by the tangent vector $k^a=- g^{ab}\partial_b u$.
For such manifold $\mathcal{M}$, the line element ansatz can be written as $ds^2=e^{2\beta}V\frac{du^2}{r}-2e^{2\beta}dudr+r^2h_{AB}(dx^A-U^Adu)(dx^B-U^Bdu)$ where $A,B \in (\theta, \phi)$ and $V, \beta, U^A \text{and} h_{AB}$ are the functions of coordinates and determinant of $h_{AB}=b$.
After feeding this ansatz into the Einstein’s field equations one obtains the following asymptotic behavior of the functions
• $V=-r+2M+\mathcal{O}(\frac{1}{r})$
• $\beta=\frac{cc^*}{(2r)^2}+\mathcal{O}(\frac{1}{r^4})$
• $h_{AB}dx^Adx^B=(d\theta^2+\sin^2\theta d\phi^2)+\mathcal{O}(\frac{1}{r})$
• $U^A=\mathcal{O}(\frac{1}{r^2})$
A spacetime is said to be asymptotically flat if
1. there exist the chart $u,r,\theta,\phi$ with the properties mentioned above
2. the line element equation and the asymptotic behavior mentioned above hold
Also at large $r$ the line element $\lim_{r\to\infty}{ds^2}=-du^2-2dudr+r^2d\Omega^2$, which is what we would expect.
At this point Bondi and Metzner studied all the set of coordinate transformations which preserve the line element $ds^2$ and the asymptotic behavior of the functions. Their considerations were subsequently generalized and the following group was obtained.
So consider the spacetime smoothly covered by the coordinates $0\leq r<\infty, 0\leq\theta\leq\pi, 0\leq\phi\leq 2\pi \text{ and} -\infty. The BMS transformations are given by ($\alpha, \Lambda$)
• $\bar{u}=(K_\Lambda(x))^{-1}(u+\alpha(x))+\mathcal{O}(\frac{1}{r})$
• $\bar{r}=K_\Lambda(x)r+J(x,u)+\mathcal{O}(\frac{1}{r})$
• $\bar{\theta}=(\Lambda x)_\theta+H_\theta(r,u)r^{-1}+\mathcal{O}(\frac{1}{r})$
• $\bar{\phi}=(\Lambda x)_\phi+H_\phi(r,u)r^{-1}+\mathcal{O}(\frac{1}{r})$
Here $x$ is the coordinate on $S^2$ given by $(\theta,\phi)$$\Lambda$ is the Lorentz transformation acting as conformal transformation on $S^2$ and on $K_\Lambda(x)$ is the corresponding conformal factor. $\alpha$ is a scalar function on $S^2$ related to the supertranslation subgroup. Rest of the functions are uniquely determined by imposing the following constraints of composition
• $(\alpha_1,\Lambda_1)(\alpha_2,\Lambda_2)=(\alpha_1+\Lambda_1\alpha_2,\Lambda_1\Lambda_2)$
• $(\Lambda_1\alpha_2)(x)=(K_\Lambda(x))^{-1}\alpha_2(\Lambda_1^{-1}x)$
One can immediately notice the semi-direct product structure of the BMS group here. $B=N\ltimes L$ where $N$ is the infinite dimensional group of supertranslations and $L$ is the connected component of the homogeneous Lorentz group.
Penrose’s derivation coming soon!
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2023-03-25 13:24:14
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https://www.gradesaver.com/textbooks/math/algebra/algebra-and-trigonometry-10th-edition/chapter-2-2-1-linear-equations-in-two-variables-2-1-exercises-page-170/85
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## Algebra and Trigonometry 10th Edition
$\frac{x}{3}+\frac{y}{3}=1$
Use the intercept form of the line. $\frac{x}{a}+\frac{y}{b}=1$ (where intercepts are (a,0) and (0,b)) Put the given intercepts into the equation: $\frac{x}{c}+\frac{y}{c}=1$ x+y=c 1+2=c c=3 $\frac{x}{3}+\frac{y}{3}=1$
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2021-05-18 08:35:22
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https://avvmediaservices.com/past-tense-iralx/098910-parent-functions-graphs
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25 Jan 2021
### parent functions graphs
Students will match each of the pieces together. Subjects. Article by … Types: … 6. The TAoCP book uses the term oriented tree. View Unit 2.5 Parent functions - after Covid-19.pdf from WTW 133 at University of Pretoria. 7. g ( x ) = 1 x − 3 − 1. Horizontal Shifts: = cos(?) Sample Problem 3: Use the graph of parent function to graph each function. Looking at some parent functions and using the idea of translating functions to draw graphs and write y = bx. 4 Module 1 – Polynomial, Rational, and Radical Relationships 5. illustrate some of the facts that we already knew:? You start graphing the cubic function parent graph at the origin (0, 0). In calculus, this point is called a critical point, and some pre-calculus teachers also use that terminology. STUDY. Function Parent Graph Identify parent graphs, names, and equationss ID: 299985 Language: English School subject: Pre-Calculus Grade/level: Pre-Calculus Age: 14-18 Main content: Parent Graphs Other contents: Add to my workbooks (3) Download file pdf Embed in my website or blog Add to Google Classroom Add to Microsoft Teams Share through Whatsapp: Link to this … In math, every function can be classified as a member of a family. Twelfth grade. Jul 3, 2015 - This section covers: Basic Parent Functions Generic Transformations of Functions Vertical Transformations Horizontal Transformations Mixed Transformations Transformations in Function Notation Writing Transformed Equations from Graphs Rotational Transformations Transformations of Inverse Functions Applications of Parent Function Transformations More … Subjects: Algebra, Graphing, PreCalculus. Parent Equation: yx cos Description of the Locator: y-intercept (h, a+k) General Equation: a = amplitude b = angular frequency h = horizontal shift k = vertical shift Properties: starts at the top and ends at the top Domain: all real numbers Range: all positive real numbers x 0 y 1 0 -1 0 1 2-2-5 5 2-2-5 5. Describe what happened to the parent a. f(x) = |x|, y = x to the left of (0,0) you graph (-1,(-1)3)=(-1,-1), (-2,(-2)3)=(-2,-8), etc.. View 2.1 Notes - Families of Graphs.pdf from MATH MISC at Surry Community College. There are two versions, one with domains and ranges, and one without.Included are . Match graphs to the family names. y = x2, f(x) = âx The function f(x) = x3 is the parent function. Odd/Even: Even . < ∞-The range is the set −1 ≤ ? Big Idea. f(x) = x2 Grades: 9 th, … y = âx (square root) The parent function of absolute value functions is y = |x|. e. How many zeros of the function are there in this graph? It will not work well as a … y = x, Phoenix, AZ. Secondly, the parabola exists only when g(x) is positive because when you are asked to find. )-The domain is the set of all real numbers −∞ < ? Buy Find arrow_forward. The parent functions are a base of functions you should be able to recognize the graph of given the function and the other way around. Ron Larson. Buy Find arrow_forward. arrow_back. Subjects: Algebra, PreCalculus, Algebra 2. Learn. of this form is y= x2. JUST A STRAIGHT LINE. These Parent Functions and Common Graphs Reference sheet and Posters for Algebra, PreCalculus, and Calculus Students are great for Bulletin Boards and INB's and come in 3 Sizes!The sheets show 16 parent functions. Includes basic parent functions for linear, quadratic, cubic, rational, absolute value, and square root functions. 2 Parent Functions. Precalculus with Limits: A Graphin... 7th Edition. Since it extends on both ends of the x-axis, y= |x| has a domain at (-∞, ∞). f(x + c) moves left, Just like Transformations in Geometry, we can move and resize the graphs of functions: Let us start with a function, in this case it is f(x) = x 2, but it could be anything: f(x) = x 2. The Parent FunctionsThe fifteen parent functions must be memorized. PARENT FUNCTIONS f(x)= a f(x)= x f(x)= x f(x)==int()x []x Constant Linear Absolute Value Greatest Integer f(x)= x2 f(x)= x3 f(x)= x f(x)= 3 x Quadratic Cubic Square Root Cube Root f(x)= ax f(x)= loga x 1 f(x) x = ()() ()() x12 x2 f(x) x1x2 +− = +− Exponential Logarithmic Reciprocal Rational f(x)= sinx f(x)= cosx f(x) = tanx Trigonometric Functions . y = ax for a > 1 (exponential) The graph has a horizontal asymptote at to the left. Read cards carefully so that you match them correctly. f(x) - c moves down. Library of Parent Functions: f ( x ) = 1 / x In Exercises 5 − 8 , sketch the graph of the function g and describe how the graph is related to the graph of f ( x ) = 1 / x . PLAY. Math. Parent Functions And Transformations - Displaying top 8 worksheets found for this concept.. The graph passes through the origin (0,0), and is contained in Quadrants I and II. Parent Graphs A parent graph is the graph of a relatively simple function. Learn how to shift graphs up, down, left, and right by looking at their equations. y = 1/x2 By transforming the function in various ways, the graph can be translated, reflected, or otherwise changed. Lesson Author. … Parent Graphs The parent graph, which is the graph of the parent function, is the simplest of the graphs in a family. There are extra options for each set of graphs. f(x - c) moves right. a This definition is in fact coincident with that of an anti-arborescence. and reciprocal functions. This is designed to be a matching activity. PASSES THROUGH ORIGIN NEGATIVE AND POSITIVE ANSWERS VERTICAL LINE. Functions in the same family are transformations of their parent function. Parent Functions And Their Graphs (video lessons, examples and solutions) Linear, quadratic, square root, absolute value and reciprocal functions, transform parent functions, parent functions with equations, graphs, domain, range and asymptotes, graphs of basic functions that you should know for PreCalculus with video lessons, examples and step-by-step solutions. This Try the given examples, or type in your own Please submit your feedback or enquiries via our Feedback page. (^ is before an exponent. Describe what happened to the parent function for the graph at the right. exponential, logarithmic, square root, sine, cosine, tangent. KEY to Chart of Parent Functions with their Graphs, Tables, and Equations Name of Parent Function Graph of Function Table of Values By using this website, you agree to our Cookie Policy. y = |x|, This graph starts at the origin (0, 0) and then moves to (1, sqrt(1))=(1,1), (2, sqrt(2)) , (3, sqrt(3)), etc. Sample Problem 2: Given the parent function and a description of the transformation, write the equation of the transformed function!". Cube-root functions are related to cubic functions in the same way that square-root functions are related to quadratic functions. There will be 3 pieces fo. This is designed to be a matching activity. Test. Table of Trigonometric Parent Functions; Graphs of the Six Trigonometric Functions; Trig Functions in the Graphing Calculator; More Practice; Now that we know the Unit Circle inside out, let’s graph the trigonometric functions on the coordinate system. = cos(? The growth (sharpness) of the increase to the right is … Parent Function Graphs. Note that the point (0, 0) is the vertex of the parent function only. you’re being asked to find only the principal or positive root of x.. Describe what happened to the parent a. function for the graph at the right. You can also write the square-root function as, However, only half of the parabola exists, for two reasons. Nov 2, 2017 - In this lesson, students cover the following topics: • Parent Functions: linear, absolute value, quadratic, and greatest integer • Define and analyze graphs by continuity, intercepts, local minima and maxima, intervals of increase and decrease, end behavior, asymptotes, domain and range. From (0,0), you graph (1,13)=(1,1), (2,23)=(2,8), etc. Sample Problem 1: Identify the parent function and describe the transformations. Here’s how this works: Start at (0,sqrt(0))=(0,0), then go to (1,sqrt(1))=(1,1), then to (4,sqrt(4))=(4,2), then to (9,sqrt(9))=(9,3), etc. Linear Function Graph. The $$x$$-values are the angles (in radians – that’s the way it’s done), and the $$y$$ … Flashcards. In particular, logarithmic functions have vertical asymptotes whereas exponential functions have vertical asymptotes. d. What is the importance of the x-intercept in graph? In algebra, a linear equation is one that contains two variables and can be plotted on a graph as a straight line. y = x3 What is the equation of the … HSF-IF.C.7b. General Form: + = 0 Inverse Function: Undefined (asymptote) Restrictions: c is a real number . SWBAT sketch and name the graphs of basic parent functions. Cosine graphs follow the same basic pattern and have the same basic shape as sine graphs; the difference lies in the location of the maximums and minimums. Remember the linear parent function is y(x) = … ... No obligation, cancel anytime. This is a great way to have practice even if it is at home. d. What is the importance of the x-intercept in graph? https://www.dummies.com/.../math/calculus/how-to-graph-parent-functions Sarandan_Donovan. To graph absolute-value functions, you start at the origin and then each positive number gets mapped to itself, while each negative number gets mapped to its positive counterpart. Ex: 2^2 is two squared) y = |x|. Each member of a family of functions will be especially useful when doing transformations. Constant Function Equation. y = 1/x, Match family names to functions. The shortcut to graphing the function f(x) = x2 is to start at the point (0, 0) (the origin) and mark the point, called the vertex. By transforming the function in various ways, the graph can be translated, reflected, or otherwise changed. (^ is before an exponent. Parent Function Toolkit 5 Name: Sine Parent Equation: yx sin Description of the Locator: y-intercept (h, a+k) General Equation: y a b x h k sin ( ) a = amplitude b = angular frequency pb 2S h = horizontal shift k = vertical shift Properties: starts in the middle and ends in the middle Domain: , f f all real numbers f fx Range: > 1,1@ all positive real numbers d d11y x 0 S 2 S 3S 2 2S y 0 1 0 -1 … As can be seen from the parent function’s graph, absolute value functions are expected to return V-shaped graphs. problem solver below to practice various math topics. Quadratic functions are functions in which the 2nd power, or square, is the highest to which the unknown quantity or variable is raised.. View Parent_Functions_and_their_Graphs_Cheat_Sheet.pdf from MAT 003 3 at Cape Fear Community College. f(x) + c moves up, Theme 2 Functions Unit 2.5 A Library of Parent Functions Larson Section 1.6 Page 132 - 138 Graph You must be able to recognize them by graph, by function, and be able to sketch … Parent Functions. Try the free Mathway calculator and Includes basic parent functions for linear, quadratic, cubic, rational, absolute value, and square root functions. Copyright © 2005, 2020 - OnlineMathLearning.com. Here are … … y = x3, Domain: (∞, ∞) Range: [c, c] Inverse Function: Undefined (asymptote) Restrictions: c is a real number Odd/Even: Even General Form: # U E \$ L0 Linear or Identity Grades: 10 th, 11 th, 12 th. Choose from 500 different sets of parent functions graphs flashcards on Quizlet. It will not work well as a flashcard activity. 3 Parent Functions. y = 1/x2 The parent graph of cosine looks very similar to the sine function parent graph, but it has its own sparkling personality (like fraternal twins). This function is Scroll down the page for examples and Match graphs to equations. Each graph in a family of graphs has similar characteristics. Read cards carefully so that you match them correctly. Basic graphs that are useful to know for any math student taking algebra or higher. y = x5 y = x2 (quadratic) Let us see, how to graph the functions which are in the form y = af[k(x-d)] + c using transformation with an example. One of the most common parent functions is the linear parent function, f(x)= x, but on this blog we are going to focus on other more complicated parent functions. First, its parent graph exists only when x is zero or positive (because you can’t find the square root of negative numbers [and keep them real, anyway]). Learn parent functions graphs with free interactive flashcards. solutions. Enjoy the videos and music you love, upload original content, and share it all with friends, family, and the world on YouTube. Linear Function … In mathematics, we will have situation to graph a function from the parent function using transformation. Chapter 2.7, Problem … Standards. In math, we often encounter certain elementary functions. The "Parent" Graph: The simplest parabola is y = x2, whose graph is shown at the right. The graph of any quadratic function is called a parabola. Key Concepts: Terms in this set (24) Constant Function Graph. ISBN: 9781305071711. Match family names to functions. The function y=x2 or f(x) = x2 is a quadratic function, and is the parent graph for all other quadratic functions. Theme 2 Functions Unit 2.5 A Library of Parent Functions Larson Section 1.6 Page 132 - 138 Graph Publisher: Brooks/Cole. parent functions. y = 1/x Spell. Key common points of linear parent functions include the fact that the: Linear Parent Function Characteristics . Sketch This figure shows the graph of an absolute-value function. For our course, you will be required to know the ins and outs of 15 parent functions . All parabolas have the same basic shape. See more ideas about graphing, algebra, parent functions. = sin(? It is Title: PARENT FUNCTIONS Author: … Learn math quiz parent functions graphs with free interactive flashcards. This article focuses on the traits of the parent functions. Ex: 2^2 is two squared) CUBIC PARENT FUNCTION: f(x) = x^3 Domain: All Real Numbers Range: All Real Numbers CUBE ROOT… The graph is strictly increasing, like the cubic polynomial. Method 1: f) Graph each transformation in the appropriate order given in part e), and show the graph of the given function in a distinctive colour. You’ll probably study some “popular” parent functions and work with these to learn how to transform functions – how to move them around. Standard 1– Graphing parent functions #2: I can graph various functions and their transformations. Describe what happened to the parent a. function for the graph at the right. Recall that a key aspect of the various parent functions are the asymptotes. 7. g ( x ) = 1 x − 3 − 1 . The parent function is the simplest function with the defining characteristics of the family. Start learning today for free! Tiffany Dawdy. This Graphs of Trig Functions section covers :. Precalculus with Limits: A … shows the graph for the parent square-root function. What is the best parent function for the graph? Match. Subjects: Algebra, Graphing, PreCalculus. Grades: 9 th, … y = x2 functions, exponential functions, basic polynomials, absolute values and the square root function. A square-root graph is related to a quadratic graph. … Method 2: Embedded content, if any, are copyrights of their respective owners. Example : Sketch the graph of the function given below. A parent graph is the graph of a relatively simple function. What is the equation of the function? We call these basic functions “parent” functions since they are the simplest form of that type of function, meaning they are as close as they can get to the origin \left( {0,\,0} \right).The chart below provides some basic parent functions that you should be familiar with. This matching activity includes parent functions of graphs where students will match the appropriate pieces together. Instead, try picking values for which you can easily find the square root. y = âx, b. Match graphs to the family names. x2 + y2 = 9 (circle), Harold’s Parent Functions “Cheat Sheet” 6 November 2019 Function Name Parent Function Graph Characteristics Algebra Constant ( T)= Domain: (− ∞, ) Range: [c, c] Inverse Function: Undefined (asymptote) Restrictions: c is a real number Odd/Even: Even General Form: + =0 Linear or = Identity ( T) T Domain: (−∞, ∞) Vertical Shifts: I’ve also included the anchor points, or critical points, the points wit… Match graphs to equations. Noting that a cube-root function is odd is important because it helps you graph it. is related to its simpler, or most basic, function sharing the same characteristics. = sin(?) f(x) = x3 < ∞-The range is the set −1 ≤ ? Grade Level. The graphs of? problem and check your answer with the step-by-step explanations. She is the author of several For Dummies books, including Algebra Workbook For Dummies, Algebra II For Dummies, and Algebra II Workbook For Dummies. y = logb(x) for b > 1 The parent functions are a base of functions you should be able to recognize the graph of given the function and the other way around. Without getting into the calculus definition, it means that the point is special. equations. Library of Parent Functions: f ( x ) = 1 / x In Exercises 5 − 8 , sketch the graph of the function g and describe how the graph is related to the graph of f ( x ) = 1 / x . linear, quadratic, cubic, absolute, reciprocal, exponential, logarithmic, square root, sine, cosine, tangent. and their graphs. c. Write the equation in standard form. A parent function is the simplest function of a family of functions. Take a quick interactive quiz on the concepts in Parent Functions: Graphs & Examples or print the worksheet to practice offline. Jul 9, 2013 - Parent Functions (Linear, Quadratic, Absolute Value, Square Root) Sort & Graph ActivityStudents will review linear, quadratic, absolute value, and square root parent functions through this "Sort & Graph" Activity. Find the domain and the range of the new function. It's actually not uncommon for children to look like one parent or another, or a mix of both. Write. The following table shows the transformation rules for functions. Functions and Their Graphs1 1.1 Rectangular Coordinates 2 1.2 Graphs of Equations 13 1.3 Linear Equations in Two Variables 24 1.4 Functions 39 1.5 Analyzing Graphs of Functions 54 1.6 A Library of Parent Functions 66 1.7 Transformations of Functions 73 1.8 Combinations of Functions: Composite Functions 83 1.9 Inverse Functions 92 Types of Parent Functions One of the most common parent functions is the linear parent function, f (x)= x, but on this blog we are going to focus on other more complicated parent functions. That is, the root r is the only node on which the parent function is not defined and for every node x, the root is reachable from x in the directed graph (X, parent). Related Pages Here are some simple things we can do to move or scale it on the graph: We can move it up or down by adding a constant to the y-value: g(x) = x 2 + C. Note: to move the line down, we use a negative value for … y = âx This lesson discusses some of the basic characteristics of linear, quadratic, square root, absolute value ≤ 1-The cosine function is an even function (has … KEY to Chart of Parent Functions with their Graphs, Tables, and Equations Name of Parent Function Graph of Function Table of Values PARENT FUNCTIONS f (x) = a f (x) = … Similar to the way that numbers are classified into sets based on common characteristics, functions can be classified into families of functions. Notice that the values you get by plotting consecutive points don’t exactly give you the nicest numbers. Each family of Algebraic functions is headed by a parent. Parent Functions Graphs. This graphing occurs on the other side of the vertex as well and keeps going, but usually just a couple points on either side of the vertex gives you a good idea of what the graph looks like. The vertex of y = |x| is found at the origin as well. What is the equation of the function? Mary Jane Sterling aught algebra, business calculus, geometry, and finite mathematics at Bradley University in Peoria, Illinois for more than 30 years. Explore the graphs of linear functions by adding or subtracting values to x (such as y(x) = x + 2) or by multiplying x by a constant (such as y(x) = 3x). Free functions and graphing calculator - analyze and graph line equations and functions step-by-step This website uses cookies to ensure you get the best experience. solutions on how to use the transformation rules. In mathematics, you see certain graphs over and over again. We welcome your feedback, comments and questions about this site or page. e. How many zeros of the function are there in this graph? Some of the worksheets for this concept are Transformations of graphs date period, Work parent functions transformations, Transformations of functions name date, 1 graphing parent functions and transformations, The parent functions, Graphical transformations of functions, … Parent Functions Graphs. These elementary functions include rational In a cubic function, the highest degree on any variable is three. ... By observing above graphs, it is clear that, the graph of the function g(x) is shifted right side by 3 units and shifted downwards by 1 unit of the function f (x). For that reason, these original, common functions are called parent graphs, and they include graphs of quadratic functions, square roots, absolute values, cubics, and cube roots. Other Parent Functions. Another parent function is y = x ^2. Has anyone ever told you that you look like your mom or dad? The graph of a square-root function looks like the left half of a parabola that has been rotated 90 degrees clockwise. )-The domain is the set of all real numbers −∞ < ? c. Write the equation in standard form. y = ax for 0 < a < 1, f(x) = x This figure shows an example of a quadratic function in graph form. You write cubic functions as f(x) = x3 and cube-root functions as g(x) = x1/3 or. More Graphs And PreCalculus Lessons b. Take a quick interactive quiz on the concepts in Parent Functions: Graphs & Examples or print the worksheet to practice offline. Created by. The quadratic graph is f(x) = x2, whereas the square-root graph is g(x) = x1/2. Nov 21, 2018 - Explore deepak mahajan's board "Graphs" on Pinterest. and? Below are some common parent graphs: y = 1/x (reciprocal) Graph of Parent Function: Notable Features of Graph: The notable features are: A point of interest is the point (0,1) for a variety of reasons which we will discuss more in the section on exploring exponential functions. 2.1 Notes: Families of Graphs Name: _ Date: _ Parent functions for different types of graphs… The cubic parent function, g(x) = x3, is shown in graph form in this figure. The following figures show the graphs of parent functions: linear, quadratic, cubic, absolute, reciprocal, For our course, you will be required to know the ins and outs of 15 parent functions . Below are some common parent graphs: important to recognize the graphs of elementary functions, and to be able to graph them ourselves. a function or graph. Precalculus and Calculus. called the parent function. View Unit 2.5 Parent functions - after Covid-19.pdf from WTW 133 at University of Pretoria. e) State the transformations (in an appropriate order) that are performed on the graph of the parent function to obtain the graph of the function given. There are two versions, one with domains and ranges, and one without.Included are . The Parent FunctionsThe fifteen parent functions must be memorized. These extremes occur at different domains, or x values, 1/4 of a period away from … Gravity. y = |x| (absolute) The absolute-value parent graph of the function y = |x| turns all inputs non-negative (0 or positive). Choose from 500 different sets of math quiz parent functions graphs flashcards on Quizlet. Graphs Of Functions. To get the other points, you plot the points (1,12)=(1,1), (2,22)=(2,4), (3,32)=(3,9), etc. The rest of the functions lack asymptotes entirely. Lessons with videos, examples and solutions to help PreCalculus students learn how about parent functions Scroll down the page for more examples and 3 : Consider the following graph: What is the best parent function for the graph? Parent Functions “Cheat Sheet” 20 September 2016 Function Name Parent Function Graph Characteristics Algebra Constant B : T ; L ? Students organize their prior knowledge of basic functions while learning some calculator basics. For the family of quadratic functions, y= ax2+ bx + c, the simplest function. b. ≤ 1-The sine function is an odd function (has origin symmetry)? These Parent Functions and Common Graphs Reference sheet and Posters for Algebra, PreCalculus, and Calculus Students are great for Bulletin Boards and INB's and come in 3 Sizes!The sheets show 16 parent functions. f(x) = cube root(x) Quizlet provides algebra parent functions equations graphs activities, flashcards and games. y = mx + b (linear function) View extra practice Graphing parent functions#2.pdf from MATH 135 at Harvard University. y = x3 (cubic) F(x)=k. Graph square root, cube root, and piecewise-defined functions… Graph is f ( x ) parent functions graphs the set of all real numbers <. ) + c moves up, down, left, f ( x ) - c moves up down!, which is the set of all real numbers −∞ < View from! Each function re being asked to find only the principal or positive of...: t ; L simplest parabola is parent functions graphs = |x| turns all inputs non-negative 0!, if any, are copyrights of their respective owners instead, parent functions graphs picking for... Over and over again … a function from the parent function for the graph a! Of x told you that you match them correctly: the simplest function, function sharing same... Submit your feedback or enquiries via our feedback page function! |x| has a domain (! 4 Module 1 – polynomial, rational, and Radical Relationships 5 graphs has similar characteristics View Parent_Functions_and_their_Graphs_Cheat_Sheet.pdf MAT! Function to graph a function or graph, and to be able to graph them ourselves ways, the wit…! In this figure Harvard University functions Author: … a function from the parent function for the family absolute-value... Certain graphs over and over again carefully so that you look like one parent or another, or otherwise.! Logarithmic functions have vertical asymptotes whereas exponential functions have vertical asymptotes whereas exponential functions have asymptotes!, this point is called a parabola the facts that we already:! Each member of a square-root graph is the graph plotting consecutive points don ’ exactly! We already knew: if it is important to recognize the graphs in a cubic function is... Copyrights of their respective owners functions and using the idea of translating functions to draw and. The appropriate pieces together at their equations = x1/2 and Radical Relationships 5 is found at the.... Same way that numbers are classified into sets based on common characteristics, functions can be seen from the graph..., this point is called a critical point, and square root function any quadratic is... Related Pages more graphs and write equations asymptote at to the left bx + c ) moves.... Looking at some parent functions matching activity includes parent functions Mathway calculator and Problem solver below practice. = x1/2 of functions the defining characteristics of linear, quadratic, cubic, rational, value..., every function can be classified into families of Graphs.pdf from math MISC at Surry Community College one... Like one parent or another, or critical points, or critical points or! Form: + = 0 parent functions and using the idea of translating functions to draw graphs PreCalculus! Or higher bx + c, the points wit… Other parent functions of graphs has similar characteristics to! = x1/3 or or type in your own Problem and check your with! Principal or positive root of x down the page for more examples solutions! Math 135 at Harvard University this is a real number the values you get by plotting points! On how to use the transformation rules an odd function ( has origin )... Quiz on the traits of the family of graphs has similar characteristics of Algebraic functions is related to quadratic. And square root V-shaped graphs transformation, write the equation of the various parent functions is shown in?. Numbers −∞ < a. function for the family of functions in various ways, the highest degree any. Not work well as a straight LINE outs of 15 parent functions parent function only definition. ( 2,8 ), etc or page their transformations ) - c up. Are useful to know the ins and outs of 15 parent functions while! Their equations some pre-calculus teachers also use that terminology noting that a key aspect of parabola... Is an odd function ( has origin symmetry ) been rotated 90 degrees clockwise Constant function.. ) -The domain is the vertex of y = |x| turns parent functions graphs inputs non-negative ( 0 or positive ) quadratic. There are extra options for each set of all real numbers −∞?... Simplest function of a family are transformations of their respective owners odd function ( has origin symmetry ) is! Method 2: I can graph various functions and using the idea of translating functions to draw and... Through the origin ( 0 or positive root of x you graph it graphs with free interactive.... 7. g ( x - c ) moves left, and square root, cube,., whereas the square-root function as, However, only half of the characteristics... Already knew: a function from the parent function article focuses on the traits of the function =! Similar to the left that are useful to know for any math taking... − 3 − 1 Inverse function: Undefined ( asymptote ) Restrictions: c is a real.. Use that terminology linear equation is one that contains two variables and can be translated, reflected, otherwise! In parent functions are the asymptotes each graph in a family of quadratic functions, exponential functions exponential... Family of graphs has similar characteristics or a mix of both some parent functions Author: a. Function as, However, only half of a relatively simple function − 3 − 1, f x. Covid-19.Pdf from WTW 133 at University of Pretoria translated, reflected, or critical points, the simplest parabola y... To a quadratic graph that numbers are classified into sets based on common characteristics functions.: graphs & examples or print the worksheet to practice various math topics variables and can be translated reflected. Included the anchor points, the simplest parabola is y = x2, whereas the square-root graph is f x. Page for more examples and solutions to help PreCalculus students learn how parent functions graphs shift graphs up down! And their transformations the parent '' graph: What is the set of real. The nicest numbers this set ( 24 ) Constant function graph characteristics algebra Constant B: t ;?! Our feedback page the equation of the function in graph the free Mathway calculator and Problem solver below practice! Cheat Sheet ” 20 September 2016 function Name parent function for the graph is f ( )! Pre-Calculus teachers also use that terminology September 2016 function Name parent function graph them ourselves linear equation one! Has origin symmetry ) page for examples and solutions to help PreCalculus students learn how about parent functions flashcards... Graphs of elementary functions Notes - families of functions Harvard University required to know the and! To a quadratic graph is f ( x ) = x3 is the simplest parabola is y =.! Origin symmetry ) plotting consecutive points don ’ t exactly give you the nicest numbers anchor points, or in... An parent functions graphs is important because it helps you graph it September 2016 function Name parent function a. C moves down, every function can be translated, reflected, or otherwise changed the in!, ∞ ) e. how many zeros of the basic characteristics of linear, quadratic, cubic rational! Shows the graph of the function in various ways, the highest degree on any is... Expected to return V-shaped graphs cube-root functions are the parent functions graphs see more ideas about graphing,,... Its simpler, or critical points, or most basic, function sharing the same characteristics see graphs... About graphing, algebra, a linear equation is one that contains two variables and can be,. Because it helps you graph it translated, reflected, or otherwise changed be! = x1/3 or to know for any math student taking algebra or higher parent function picking values which... This figure start graphing the cubic function parent graph at the origin ( 0,0 ), square! Give you the nicest numbers critical point, and square root functions graphs...: What is the set of all real numbers −∞ < x3 and cube-root as! For the family of functions to find only the principal or positive ) start graphing the cubic function parent,. Match them correctly feedback, comments and questions about this site or page is f ( x =!, like the left half of the x-axis, y= |x| has a horizontal asymptote at to the left the. Lessons with videos, examples and solutions to help PreCalculus students learn how to shift graphs up down. One without.Included are parent '' graph: What is the importance of the x-intercept in graph, g ( ). Match the appropriate pieces together over again via our feedback page or print the worksheet to practice.. A cube-root function is an odd function ( has origin symmetry ) member of a simple! Concepts: Terms in this graph ” 20 September 2016 function Name parent function of absolute value, square! Graph is related to cubic functions as g ( x ) = x3, is in... ( 1,13 ) = x2, whose graph is related to a quadratic graph is f ( x ) x3... 1,13 ) = ( 1,1 ), you see certain graphs over and over again where students match! Half of the transformation rules for functions of math quiz parent functions.! Quadratic graph is strictly increasing, like the left, comments and questions this... The defining characteristics of linear, quadratic, cubic, rational, absolute value functions is related a. Mix of both turns all inputs non-negative ( 0, 0 ) them. Function f ( x + c, the graph can be seen the. Practice various math topics each function y= ax2+ bx + c, the graph at right... Over and over again, only half of the x-intercept in graph form in graph... Graph various functions and using the idea of translating functions to draw graphs and write equations without getting into calculus... Range of the x-intercept in graph form in this figure shows the graph of an anti-arborescence are asked to....
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2021-10-27 09:21:40
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http://math.stackexchange.com/questions/181292/difference-between-substitution-and-replacement
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# Difference between “substitution” and “replacement”
Is there a technical difference between "substitution" and "replacement"?
For example, if I use another expression for x, am I replacing for it? Or substituting? What is I use another value of x?
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"Substitute" is the term that is usually used for this. – user22805 Aug 11 '12 at 11:03
In general, both replacement and substitution are syntactic operations on strings. If $uvw$ and $x$ are strings, then $uxw$ is the result of replacing $v$ with $x$ in $uvw$. If $u$, $v$ and $w$ are strings, then $u[v \mapsto w]$ is the result of replacing every occurrence of $v$ in $u$ with $w$. The last operation is called substitution, and is defined in terms of replacement. The operation called instantiation is a kind of substitution which eliminates quantifiers by substituting values for free variables. – danportin Aug 11 '12 at 11:11
I should preface this comment by saying that I don't know of an established difference. But, for instance, if someone told me to 'substitute $\cos x$ for $x$' in $\int x dx$, I would write $\int \cos x (-\sin x )dx$. But if someone told me to replace $x$ by $\cos x$, I would write $\int \cos x d(\cos x)$. So I perceive a one-step difference, in the sense that different things get written down (ignoring the same meaning). – mixedmath Aug 11 '12 at 13:00
The rules of inference of first-order equational logic are the following (combined with the rules that equality is an equivalence relation). $$\rm A = B\ \Rightarrow\ P(A) = P(B)\qquad\qquad (Replacement)$$ $$\rm P(X) = Q(X)\ \Rightarrow\ P(A) = Q(A)\quad (Substitution)$$ – Bill Dubuque Aug 11 '12 at 13:57
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2015-08-31 07:48:41
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https://ai.stackexchange.com/questions/7707/is-there-a-difference-in-the-architecture-of-deep-reinforcement-learning-when-mu
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# Is there a difference in the architecture of deep reinforcement learning when multiple actions are performed instead of a single action?
I've built a deep deterministic policy gradient reinforcement learning agent to be able to handle any games / tasks that have only one action. However, the agent seems to fail horribly when there are two or more actions. I tried to look online for any examples of somebody implementing DDPG on a multiple action system, but people mostly applied it to the pendulum problem, which is a single action problem.
For my current system, it is a 3 state, 2 continuous control actions system (One is to adjust the temperature of the system, the other one adjusts a mechanical position, both are continuous). However, I froze the second continuous action to be the optimal action all the time. So RL only has to manipulate one action. It solves within 30 episodes. However, the moment I allow the RL to try both continuous actions, it doesn't even converge after 1000 episodes. In fact, it diverges aggressively. The output of the actor network seems to always be the max action, possibly because I am using a tanh activation for the actor to provide output constraint. I added a penalty to large actions, but it does not seem to work for the 2 continuous control action case.
For my exploratory noise, I used Ornstein-Ulhenbeck noise, with means adjusted for the two different continuous actions. The mean of the noise is 10% of the mean of the action.
Is there any massive difference between single action and multiple action DDPG? I changed the reward function to take into account both actions, have tried making a bigger network, tried priority replay, etc., but it appears I am missing something. Does anyone here have any experience building a multiple action DDPG and could give me some pointers?
• Technically, the difference here is between actions in (some subset of) $\mathbb{R}$ and $\mathbb{R}^n$, not between 1 or more "actions". In other words, you have an action space here that might have multiple dimensions, and something is going wrong for your agent when there are 2 or more dimensions. In RL, when something described as "having 2 actions" this is usually an enumeration - i.e. the agent can take action A or action B, and there are no quantities involved. – Neil Slater Aug 25 '18 at 7:09
• Hi Neil, thanks for the reply. Yes, for classic RL, it agents' actions are indeed discrete. However, in 2015, Lilicrap published a paper called "continuous control with deep reinforcement learning", and then in 2017, the TRPO and PPO algorithms were designed to allow agents to perform multiple continuous actions. So you are correct about my action being in a high dimension space. In my research, I am comparing model predictive control using trajectory optimization vs AI-based control. Usually, in robotics and mechatronics, robots move multiple pieces. I am trying to achieve that with RL. – Rui Nian Aug 26 '18 at 4:41
• I suggest you edit a more accurate description of your RL problem to replace the sentence "For my current system, it is a 3 state, 2 action system." - because that is not how it would be described in any literature. May also be worth explaining how you have adjusted the exploration function ("actor noise"), as a mistake there would be key. – Neil Slater Aug 26 '18 at 9:16
• Done! I will also try different exploratory noise means to see if it helps. – Rui Nian Aug 27 '18 at 3:46
• Thanks. I was wondering if you had somehow failed to adjust for different scales of the two axes of action, but it doesn't look like it. I cannot really tell what is wrong. However, I would not personally expect DDPG to be quite so fragile when scaling up from one to two dimensions of action, so I'd still suspect something about your implementation - I just don't know what it could be. – Neil Slater Aug 28 '18 at 7:41
First Stated Question
Is there a difference in the architecture of deep reinforcement learning when multiple actions are performed instead of a single action?
The way the question is phrased implies that the query is about a discrete implication, that an architectural change is an imperative. It is not, since an action may be comprised of multiple actions, whether or not there are sequencing dependencies on the component actions. In the case of the control of two physical properties, the control space has two degrees of freedom. That they are controlled using discrete corrections leads to a hybrid of continuous and discrete mathematics, which is commonplace in control.
From the body and the comments it is likely that the question author is privy to these facts. One of the two main questions described is whether gains can be achieved with more sophisticated process topology or other strategic applications of expectation and probability distribution math. Such gains might be achievable.
• Faster response (temporal accuracy)
• Accuracy in objective tracking (independent of time)
• Tracking reliability (no gross loss of synchronization due to signal saturation or clipping)
• Risk aversion (steering clear of irretrievable loss in sparsely or weakly characterized path spaces)
In the case of temperature and position, further topological sophistication is not likely.
Longer Term Goal of Research
Later along the research path, topological changes in process and signal flow (early in the systems architecture development) will probably be effective in improving system quality. This is likely in light of the stated intention to produce a smart learning controller using the best from multiple conceptual sources.
• Deterministic policy gradient reinforcement learning agent, the proof of concept of which is converging in 30 episodes with one degree of freedom, position
• Lilicrap's Continuous control with deep reinforcement learning, 2015
• TRPO and PPO algorithms agents to perform multiple continuous actions, 2017
• Tesla megafactory
• Predictive control using trajectory optimization
• Automated, progressive model development
Whether there is an intersection point of all six that benefits from the contribution of each is unlikely, but a reasonable hypothesis to test.
Immediate Concern
The description of the current issue is not closely related to the first stated question or the ultimate goal but rather an anomaly in the current proof of concept.
That adding a second degree of freedom, temperature, "Fail[s] horribly [and] diverges aggressively," before reaching 1,000 episodes is indeed an anomaly. The injection of -20 dB of Ornstein-Ulhenbeck noise as measured by mean amplitude (10%) to avoid search pitfalls is unlikely to be related to
Is there any massive difference between single [degrees of freedom] and multiple [degrees of freedom in] DDPG?
Only if the person extending the software is not adept with multivariate calculus.
The remedies tried don't seem to be producing results, which is not surprising since none have to do with a likely root cause.
• Reward function aggregating actions
• Bigger network
• Priority replay
• Activation of tanh
• Penalty to large actions
The sixth thing mentioned may be more likely to remedy the divergence.
• New interpretations of actions and rewards
The particular anomaly described, albeit without much detail, points to some common causes of unexpected gross divergence.
• Mishandling of a minus sign during performance of the calculus or associated algebra
• A flaw in a partial derivative
• Using only the diagonal of the Jacobian, or the dismissal of some other pattern within the Jacobian in its application to corrective signalling or predictive quantification
• Hi Douglas, thanks for the reply. You answer was certainly very helpful. The issue actually arose from integral wind-up states. Currently, do you know of any methods that can handle integral wind-up states? Thanks again for your answer! – Rui Nian Nov 29 '18 at 14:53
• signal.uu.se/Publications/pdf/a032.pdf – Douglas Daseeco Nov 29 '18 at 17:07
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2019-11-17 17:16:02
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http://ajayshahblog.blogspot.com/2008/07/readership-of-this-blog.html
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## Sunday, July 06, 2008
### Readership of this blog
quantcast is an interesting new approach in Internet usage measurement. They have you put a fragment of HTML on the page that you want measured (just like sitemeter or sitecounter do). What's new is that they have instrumented a large number of households in the US. They know quite a bit about the households that have been instrumented. They watch for visits to your web page from these households and report summary statistics to you about this set. (It's a little more complicated than that. Even though they have ~ 1.5 million instrumented households, only a tiny number of these would show up at any blog e.g. this one. Sampling noise would then be unacceptable except for a small number of high traffic websites. They manage to track users across a large number of websites (e.g. this one), and have setup statistical models using which inferences are made. I thank Konrad Feldman of Quantcast for explaining these things to me).
Of course, this is only a measure of the readership of this blog in the US. What it seems to say for this blog:
• 68% are male (a bit more than the average on the net).
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2015-06-30 13:01:17
|
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https://itectec.com/superuser/how-to-auto-arrange-clean-up-files-for-all-folders-in-osx-mountain-lion/
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# Macos – How to auto-arrange/clean-up files for ALL folders in OSX Mountain Lion
findermacososx-mountain-lion
I have figured out how to make folders auto arrange their contents. This is achieved in Finder->View->Show View Options (or cmd+J).
From the newly opened window, setting the arrange by dropdown will set the folders behavior from then on. But it is only for the folder this menu was opened in. How do I set this as a Finder rule for all directories?
EDIT
I am using Mountain Lion – 10.8.2
I don't have a 'Use as Defaults' button in my view panes – here's a screen grab:
You can clear the folder-specific settings by running sudo find / -name .DS_Store -delete && killall Finder. It also resets other view options, resets the positions of icons, and deletes Spotlight comments.
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2021-09-18 19:34:07
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http://www.ams.org/mathscinet-getitem?mr=1273971
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MathSciNet bibliographic data MR1273971 (95h:60094) 60H10 (60J60 82C22 82C41) Saisho, Yasumasa A model of the random motion of mutually reflecting molecules in ${\bf R}\sp d$${\bf R}\sp d$. Kumamoto J. Math. 7 (1994), 95–123. Links to the journal or article are not yet available
For users without a MathSciNet license , Relay Station allows linking from MR numbers in online mathematical literature directly to electronic journals and original articles. Subscribers receive the added value of full MathSciNet reviews.
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2014-12-17 22:46:46
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https://codereview.stackexchange.com/questions/171222/page-object-model-class-structure-in-selenium
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# Page Object Model class structure in Selenium
I am automating a webform, and this is what I've coded so far. I am curious if you find the code good and readable. But most important to me is to get some feedback for learning purposes.
The remove account page:
public class RemoveAccountsPage extends Page {
private Boolean accountsToRemove() {
by = By.className("zero-results-filter");
return ((elementExists(by)) ? true : false);
}
private void selectAllAccounts() {
by = By.id("accounts-select-all");
element = waitForPresenceOfElement(by);
element.click();
}
private void clickRemoveButton() {
by = By.id("account-delete");
element = waitForPresenceOfElement(by);
element.click();
}
private void confirmRemovingAccounts() {
by = By.id("confirm");
element = waitForPresenceOfElement(by);
element.click();
//driver.findElement(By.id("NO_DEAL_VIA_MP")).click()
}
public void removeAccounts() throws Exception {
String request = Page.URL + "/hu/accounts/index.html";
goToWebPage(request);
Boolean result = accountsToRemove();
if(result == false) {
selectAllAccounts();
clickRemoveButton();
confirmRemovingAccounts();
}
}
}
Main class:
public class AccountManager {
public static void main(String args[]) throws Exception {
Bot bot = new Bot();
bot.setUp();
........
RemoveAccountsPage removeAccountsPage = new
RemoveAccountsPage();
removeAccountsPage.removeAccounts();
........
bot.tearDown();
}
}
Some lines may be simplified.
1) elementExists() already returns boolean, so use it directly:
return ((elementExists(by)) ? true : false);
to
return elementExists(by);
2) You may inline a method calls into condition statements:
Boolean result = accountsToRemove();
if(result == false) {
to
if(!accountsToRemove()) {
3) Name the main test method correspondingly:
removeAccounts()
to
testRemovingAllAccounts()
userRemovesAllAccounts()
It would be nice to see your implementation of Page.
I find the points made by Dmytro very valid and want to add a couple of things more.
Regarding your accountsToRemove() method; somehow this method doesn´t have a name that tells me exactly what it does. I´d expect to give a list of accounts that should be removed, but instead tells me whether an account can be removed right?
On the same note your waitForPresenceOfElement(by) returns an element located by 'by', however I´d expect for it just to wait.
I assume that your by and element variables are global in your Page class, in which case you don´t have to pass them as parameters. However I don´t like global variables, and so I´d tell you to remove them and keep the parameters =)
Since you already have a base Page where you find your elements etc, I´d go one step further and have another method clickElement(...) instead of calling element.click(). This helps you in case the element is not clickable or with StaleReferenceElementException (pretty common I´d say)
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2020-05-30 15:51:24
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|
https://jordanbell.info/euler/euler-algebra-II-07.html
|
### Part II. Chapter 7. “Of a particular Method, by which the Formula 𝑎𝑛𝑛+1 becomes a Square in Integers.”
96 That which has been taught in the last chapter, cannot be completely performed, unless we are able to assign for any number $$a$$, a number $$n$$, such that $$an^2+1$$ may become a square; or that we may have $$m^2=an^2+1$$.
This equation would be easy to resolve, if we were satisfied with fractional numbers, since we should have only to make $$m=1+\dfrac{np}{q}$$; for, by this supposition, we have
$m^2=1+\dfrac{2np}{q}+\dfrac{n^2p^2}{q^2} = an^2+1;$
in which equation, we may expunge 1 from both sides, and divide the other terms by $$n$$: then multiplying by $$q^2$$, we obtain $$2pq+np^2=anq^2$$; and this equation, giving $$n=\dfrac{2pq}{aq^2-p^2}$$, would furnish an infinite number of values for $$n$$: but as $$n$$ must be an integer number, this method will be of no use, and therefore very different means must be employed in order to accomplish our object.
97 We must begin with observing, that if we wished to have $$an^2+1$$ a square, in integer numbers, (whatever be the value of $$a$$), the thing required would not be possible.
For, in the first place, it is necessary to exclude all the cases, in which a would be negative; next, we must exclude those also, in which $$a$$ would be itself a square; because then $$an^2$$ would be a square, and no square can become a square, in integer numbers, by being increased by unity. We are obliged, therefore, to restrict our formula to the condition, that $$a$$ be neither negative, nor a square; but whenever $$a$$ is a positive number, without being a square, it is possible to assign such an integer value of $$n$$, that $$an^2 + 1$$ may become a square: and when one such value has been found, it will be easy to deduce from it an infinite number of others, as was taught in the last chapter: but for our purpose it is sufficient to know a single one, even the least; and this, Pell, an English writer, has tauglit us to find bv an ingenious method, which we shall here explain.
98 This method is not such as may be employed generally, for any number $$a$$ whatever; it is apphcable only to each particular case.
We shall therefore begin with the easiest cases, and shall first seek such a value of $$n$$, that $$2n^2+1$$ may be a square, or that $$\surd(2n^2+1)$$ may become rational.
We immediately see that this square root becomes greater than $$n$$, and less than $$2n$$. If, therefore, we express this root by $$n+p$$, it is obvious that $$p$$ must be less than $$n$$; and we shall have $$\surd(2n^2+1)=n+p$$; then, by squaring, $$2n^2+1=n^2+2np+p^2$$; therefore
$n^2=2np+p^2-1,\quad \textrm{and} \quad n=p+\surd(2p^2-1).$
The whole is reduced, therefore, to the condition of $$2p^2-1$$ being a square; now, this is the case if $$p=1$$, which gives $$n=2$$, and $$\surd(2n+1)=3$$.
If this case had not been immediately obvious, we should have gone farther; and since $$\surd(2p^2-1)>p$$, and consequently, $$n>2p$$, we should have made $$n=2p+q$$; and should thus have had
$2p+q=p+\surd(2p^2-1),\quad \textrm{or} \quad p+q=\surd(2p^2-1),$
and, squaring, $$p^2+2pq+q^2=2p^2-1$$, whence
$p^2=2pq+q^2+1,$
which would have given $$p=q+\surd(2q^2+1)$$; so that it would have been necessary to have $$2q^2+1$$ a square; and as this is the case, if we make $$q = 0$$, we shall have $$p = 1$$, and $$n = 2$$, as before. This example is sufficient to give an idea of the method; but it will be rendered more clear and distinct from what follows.
99 Let $$a = 3$$, that is to say, let it be required to transform the formula $$3n^2+1$$ into a square. Here we shall make $$\surd(3n^2+1)=n+p$$, which gives
$3n^2+1=n^2+2np+p^2,\quad 2n^2=2np+p^2-1;$
whence we obtain $$n=\dfrac{p+\surd(3p^2-2)}{2}$$. Now, since $$\surd(3p^2-2)$$ exceeds $$p$$, and, consequently, $$n$$ is greater than $$\dfrac{2p}{2}$$, or than $$p$$, let us suppose $$n=p+q$$, and we shall have
\begin{align} 2p+2q&=p+\surd(3p^2-2), \; \textrm{or}\\ p+2q&=\surd(3p^2-2); \end{align}
then, by squaring, $$p^2+4pq+4q^2=3p^2-2$$; so that
$2p^2=4pq+4q^2+2,\quad \textrm{or} \quad p^2=2pq+2q^2+1,$
and
$p=q+\surd(3q^2+1).$
Now, this formula being similar to the one proposed, we may make $$q=0$$, and shall thus obtain $$p=1$$, and $$n=1$$; whence $$\surd(3n^2+1)=2$$.
100 Let $$a = 5$$, that we may have to make a square of the formula $$5n^2+1$$, the root of which is greater than $$2n$$. We shall therefore suppose
$\surd(5n^2+1)=2n+p,\quad \textrm{or} \quad 5n^2+1=4n^2+4np+p^2;$
whence we obtain
$n^2=4np+p^2-1,\quad \textrm{and} \quad n=2p+\surd(5p^2-1).$
Now, $$\surd(5p^2-1)>2p$$; whence it follows that $$n>4p$$; for which reason, we shall make $$n=4p+q$$, which gives $$2p+q=\surd(5p^2-1)$$, or $$4p^2+4pq+q^2=5p^2-1$$, and $$p^2=4pq+q^2+1$$; so that $$p=2q+\surd(5q^2+1)$$; and as $$q=0$$ satisfies the terms of this equation, we shall have $$p=1$$, and $$n=4$$; therefore $$\surd(5n^2+1)=9$$.
101 Let us now suppose $$a = 6$$, that we may have to consider the formula $$6n^2+1$$, whose root is likewise contained between $$2n$$ and $$3n$$. We shall, therefore, make $$\surd(6n^2+1)=2n+p$$, and shall have
$6n^2+1 = 4n^2 + 4np + p^2, \quad \textrm{or} \quad 2n^2=4np+p^2-1;$
and, thence,
$n=p+\dfrac{\surd(6p^2-2)}{2},\quad \textrm{or} \quad n=\dfrac{2p+\surd(6p^2-2)}{2};$
so that $$n>2p$$.
If, therefore, we make $$n=2p+q$$, we shall have
\begin{align} 4p+2q&=2p+\surd(6p^2-2),\; \textrm{or}\\ 2p+2q&=\surd(6p^2-2); \end{align}
the squares of which are $$4p^2+8pq+4q^2=6p^2-2$$; so that $$2p^2=8pq+4q^2+2$$, and $$p^2=4pq+2q^2+1$$. Lastly, $$p=2q+\surd(6q^2+1)$$. Now, this formula resembling the first, we have $$q=0$$; wherefore $$p=1$$, $$n=2$$, and $$\surd(6n^2+1)=5$$.
102 Let us proceed farther, and take $$a = 7$$, and $$7n^2+1=m^2$$; here we see that $$m>2n$$; let us therefore make $$m=2n+p$$, and we shall have
$7n^2+1=4n^2+4np+p^2,\quad \textrm{or} \quad 3n^2 = 4np+p^2-1;$
which gives $$n=\dfrac{2p+\surd(7p^2-3)}{3}$$. At present, since $$n>\frac{4}{3}p$$, and, consequently, greater than $$p$$, let us make $$n=p+q$$, and we shall have $$p+3q=\surd(7p^2-3)$$; then, squaring both sides, $$p^2+6pq+9q^2=7p^2-3$$, so that
$6p^2=6pq+9q^2+3,\quad \textrm{or} \quad 2p^2 = 2pq+3q^2+1;$
whence we get $$p=\dfrac{q+\surd(7q^2+2)}{2}$$. Now, we have here $$p>\dfrac{3q}{2}$$; and, consequently, $$p>q$$; so that making $$p=q+r$$, we shall have $$q+2r=\surd(7q^2+2)$$; the squares of which are $$q^2+4qr+4r^2=7q^2+2$$; then $$6q^2=4qr+4r^2-2$$, or $$3q^2=2qr+2r^2-1$$; and, lastly, $$q=\dfrac{r+\surd(7r^2-3)}{3}$$. Since now $$q>r$$, let us suppose $$q=r+s$$, and we shall have
\begin{align} 2r+3s&=\surd(7r^2-3); \; \textrm{then}\\ 4r^2+12rs+9s^2&=7r^2-3, \; \textrm{or}\\ 3r^2&=12rs+9s^2+3, \; \textrm{or}\\ r^2&=4rs+3s^2+1, \; \textrm{and}\\ r&=2s+\surd(7s^2+1). \end{align}
Now, this formula is like the first; so that making $$s = 0$$, we shall obtain $$r = 1$$, $$g = 1$$, $$p = 2$$, and $$n = 3$$, or $$m = 8$$.
But this calculation may be considerably abridged in the following manner, which may be adopted also in other cases.
Since $$7n^2+1=m^2$$, it follows that $$m<3n$$.
If, therefore, we suppose $$m=3n-p$$, we shall have
$7n^2+1=9n^2-6np+p^2,\quad \textrm{or} \quad 2n^2 = 6np-p^2+1;$
whence we obtain $$n=\dfrac{3p+\surd(7p^2+2)}{2}$$; so that $$n<3p$$; for this reason we shall write $$n=3p-2q$$; and, squaring, we shall have $$9p^2-12pq+4q^2=7p^2+2$$; or
$2p^2=12pq-4q^2+2,\quad \textrm{and} \quad p^2=6pq-2q^2+1,$
whence results $$p=3q+\surd(7q^2+1)$$. Here, we can at once make $$q=0$$, which gives $$p=1$$, $$n=3$$, and $$m=8$$, as before.
103 Let $$a=8$$, so that $$8n^2+1=m^2$$, and $$m<3n$$. Here, we must make $$m = 3n - p$$, and shall have
$8n^2+1=9n^2-6np+p^2,\quad \textrm{or} \quad n^2=6np-p^2+1;$
whence $$n=3p+\surd(8p^2+1)$$, and this formula being already similar to the one proposed, we may make $$p = 0$$, which gives $$n = 1$$, and $$m = 3$$.
104 We may proceed, in the same manner, for every otlier number, $$a$$, provided it be positive and not a square, and we shall always be led, at last, to a radical quantity, such as $$\surd(at^2+1)$$; similar to the first, or given formula, and then we have only to suppose $$t = 0$$; for the irrationality will disappear, and by tracing back the steps, we shall necessarily find such a value of $$n$$, as will make $$an^2+1$$ a square.
Sometimes we quickly obtain our end; but, frequently also, we are obliged to go through a great number of operations. This depends on the nature of the number $$a$$; but we have no principles, by which we can foresee the number of operations that it will be necessary to perform. The process is not very long for numbers below 13, but when $$a = 13$$, the calculation becomes much more prolix; and, for this reason, it will be proper here to resolve that case.
105 Let therefore $$a = 13$$, and let it be required to find $$13n^2+1=m^2$$. Here, as $$m^2>9n^2$$, and, consequently, $$m>3n$$, let us suppose $$m=3n+p$$; we shall then have
$13n^2+1=9n^2+6np+p^2,\quad \textrm{or} \quad 4n^2=6np+p^2-1,$
and $$n=\dfrac{3p+\surd(13p^2-4)}{4}$$, which shows that $$n>\frac{6}{4}p$$, and therefore much greater than $$p$$. If, therefore, we make $$n=p+q$$, we shall have $$p+4q=\surd(13p^2-4)$$; and, taking the squares,
$13p^2-4=p^2+8pq+16q^2;$
so that
$12p^2=8pq+16q^2+4,\quad \textrm{or} \quad 3p^2=2pq+4q^2+1,$
and $$p=\dfrac{q+\surd(13q^2+3)}{3}$$. Here, $$p>\dfrac{q+3p}{3}$$, or $$p>q$$; we shall proceed, therefore, by making $$p=q+r$$, and shall thus obtain $$2q+3r=\surd(13q^2+3)$$; then
\begin{align} 13q^2+3&=4q^2+12qr+9r^2, \; \textrm{or}\\ 9q^2&=12qr+9r^2-3, \; \textrm{or}\\ 3q^2&=4qr+3r^2-1; \end{align}
which gives $$p=\dfrac{2r+\surd(13r^2-3)}{3}$$.
Again, since $$q>\dfrac{2r+3r}{3}$$, and thus $$q>r$$, we shall make $$q=r+s$$, and we shall thus have $$r+3s=\surd(13r^2-3)$$; or $$13r^2-3=r^2+6rs+9s^2$$, or $$12r^2=6rs+9s^2+s$$, or $$4r^2=2rs+3s^2+1$$; whence we obtain $$r=\dfrac{s+\surd(13s^2+4)}{4}$$. But here $$r>\dfrac{3+3s}{4}$$, or $$r>s$$; wherefore let $$r=s+t$$, and we shall have $$3s+4t=\surd(13s^2+4)$$, and
$13s^2+4=9s^2+24st+16t^2;$
so that $$4s^2=24st+16t^2-4$$, and $$s^2=6ts+4t^2-1$$; therefore $$s=3t+\surd(13t^2-1)$$. Here we have
$s>3t+3t,\quad \textrm{or} \quad s>6t;$
we must therefore make $$s=6t+u$$; whence
$3t+u=\surd(13t^2-1),\quad \textrm{and} \quad 13t^2-1=9t^2+6tu+u^2;$
then $$4t^2=6tu+u^2+1$$; and, lastly,
$t=\dfrac{3u+\surd(13u^2+4)}{4},\quad \textrm{or} \quad t>\dfrac{6u}{4}>u.$
If, therefore, we make $$t = u + v$$, we shall have
$u+4v=\surd(13u^2+4),\quad \textrm{and} \quad 13u^2+4=u^2+8uv+16v^2;$
therefore
$12u^2=8uv+16v^2-4,\quad \textrm{or} \quad 3u^2=2uv+4v^2-1;$
lastly, $$u=\dfrac{v+\surd(13v^2-3)}{3}$$, thus $$u>\dfrac{4v}{3}$$, and thus $$u>v$$.
Let us, therefore, make $$u=v+x$$, and we shall have $$2v+3x=\surd(13v^2-3)$$, and $$13v^2-3=4v^2+12vx+9x^2$$; or
$9v^2=12vx+9x^2+3,\quad \textrm{or} \quad 3v^2 = 4vx+3x^2+1,$
and $$v=\dfrac{2x+\surd(13x^2+3)}{3}$$; so that $$v>\frac{5}{3}x$$, and thus $$>x$$.
Let us now suppose $$v=x+y$$, and we shall have
\begin{align} x+3y&=\surd(13x^2+3),\; \textrm{and}\\ 13x^2+3&=x^2+6xy+9y^2,\; \textrm{or}\\ 12x^2&=6xy+9y^2-3,\; \textrm{and}\\ 4x^2&=2xy+3y^2-1; \; \textrm{whence}\\ x&=\dfrac{y+\surd(13y^2-4)}{4}, \end{align}
and, consequently, $$x>y$$. We shall, therefore, make $$x=y+z$$, which gives
\begin{align} 3y+4z&=\surd(13y^2-4),\; \textrm{and}\\ 13y^2-4&=9y^2+24zy+16z^2, \; \textrm{or}\\ 4y^2&=24zy+16z^2+4; \; \textrm{therefore}\\ y^2&=6yz+4z^2+1, \; \textrm{and}\\ y&=3z+\surd(13z^2+1). \end{align}
This formula being at length similar to the first, we may take $$z=0$$, and go back as follows:
$\begin{array}{l|l|l} z=0,&u=v+x=3,&q=r+s=71,\\ y=1,&t=u+v=5,&p=q+r=109,\\ x=y+z=1,&s=6t+u=33,&n=p+q=180,\\ v=x+y=2,&r=s+t=38,&m=3n+p=649. \end{array}$
So that 180 is the least number, after 0, which we can substitute for $$n$$, in order that $$13n^2+1$$ may become a square.
106 This example sufficiently shews how prolix these calculations may be in particular cases; and when the numbers in question are greater, we are often obliged to go through ten times as many operations as we had to perform for the number 13.
As we cannot foresee the numbers that will require such tedious calculations, we may with propriety avail ourselves of the trouble which others have taken; and, for this purpose, a Table is subjoined to the present chapter, in which the values of $$m$$ and $$n$$ are calculated for all numbers, $$a$$, between 2 and 100; so that in the cases which present themselves, we may take from it the values of $$m$$ and $$n$$, which answer to the given number $$a$$.
107 It is proper, however, to remark, that, for certain numbers, the letters $$m$$ and $$n$$ may be determined generally; this is the case when $$a$$ is greater, or less than a square, by 1 or 2; it will be proper, therefore, to enter into a particular analysis of these cases.
108 In order to this, let $$a=e^2-2$$; and since we must have $$(e^2-2)n^2+1=m^2$$, it is clear that $$m<en$$; therefore we shall make $$m = en - p$$, from which we have
$(e^2-2)n^2+1=e^2n^2-2enp+p^2,$
or $$2n^2=2enp-p^2+1$$; therefore $$n=\dfrac{ep+\surd(e^2p^2-2p^2+2)}{2}$$; and it is evident that if we make $$p=1$$, this quantity becomes rational, and we have $$n=e$$, and $$m=e^2-1$$.
For example, let $$a=23$$, so that $$e=5$$; we shall then have $$23n^2+1=m^2$$, if $$n=5$$, and $$m=24$$. The reason of which is evident from another consideration; for if, in the case of $$a=e^2-2$$, we make $$n=e$$, we shall have $$an^2+1=e^4-2e^2+1$$; which is the square of $$e^2-1$$.
109 Let $$a=e^2-1$$, or less than a square by unity. First, we must have $$(e^2-1)n^2+1=m^2$$; then, because, as before, $$m<en$$, we shall make $$m=en-p$$; and this being done, we have
$(e^2-1)n^2+1=e^2n^2-2enp+p^2,\quad \textrm{or} \quad n^2=2enp-p^2+1;$
wherefore $$n=ep+\surd(e^2p^2-p^2+1)$$. Now, the irrationality disappeared by supposing $$p=1$$; so that $$n=2e$$, and $$m=2e^2-1$$. This also is evident; for, since $$a=e^2-1$$, and $$n=2e$$, we find
$an^2+1=4e^4-4e^2+1,$\$
or equal to the square of $$2e^2-1$$. For example, let $$a=24$$, or $$e=5$$, we shall have $$n=10$$, and
$24n^2+1=2401=49^2.$
110 Let us now suppose $$a=e^2+1$$, or $$a$$ greater than a square by unity. Here we must have
$(e^2+1)n^2+1=m^2,$
and $$m$$ will evidently be greater than $$en$$. Let us, therefore, write $$m = en + p$$, and we shall have
$(e^2+1)n^2+1=e^2n^2+2enp+p^2,\quad \textrm{or} \quad n^2=2enp+p^2-1;$
whence $$n=ep+\surd(e^2p^2+p^2-1)$$. Now, we may make $$p=1$$, and shall then have $$n=2e$$; therefore, $$m^2=2e^2+1$$; which is what ought to be the result from the consideration, that $$a=e^2+1$$, and $$n=2e$$, which gives $$an^2+1=4e^4+4e^2+1$$, the square of $$2e^2+1$$. For example, let $$a=17$$, so that $$e=4$$, and we shall have $$17n^2+1=m^2$$; by making $$n=8$$, and $$m=33$$.
111 Lastly, let $$a=e^2+2$$, or greater than a square by 2. Here, we have $$(e^2+2)n^2+1=m^2$$, and, as before, $$m>en$$; therefore we shall suppose $$m=en+p$$, and shall thus have
\begin{align} c^2n^2+2n^2+1&=e^2n^2+2enp+p^2, \; \textrm{or}\\ 2n^2&=2epn+p^2-1, \; \textrm{which gives}\\ n&=\dfrac{ep+\surd(e^2p^2+2p^2-2)}{2}. \end{align}
Let $$p=1$$, we shall find $$n=e$$, and $$m=e^2+1$$; and, in fact, since $$a=e^2+2$$, and $$n=e$$, we have $$an^2+1=e^4+2e^2+1$$, which is the square of $$e^2+1$$.
For example, let $$a=11$$, so that $$e=3$$; we shall find $$11n^2+1=m^2$$, by making $$n=3$$, and $$m=10$$. If we supposed $$a=83$$, we should have $$e=9$$, and $$83n^2+1=m^2$$, where $$n=9$$, and $$m=82$$.
𝑎 𝑛 𝑚
2 2 3
3 1 2
5 4 9
6 2 5
7 3 8
8 1 3
10 6 19
11 3 10
12 2 7
13 180 649
14 4 15
15 1 4
17 8 33
18 4 17
19 39 170
20 2 9
21 12 55
22 42 197
23 5 24
24 1 5
26 10 51
27 5 26
28 24 127
29 1820 9801
30 2 11
31 273 1520
32 3 17
33 4 23
34 6 35
35 1 6
37 12 73
38 6 37
39 4 25
40 3 19
41 320 2049
42 2 13
43 531 3482
44 30 199
45 24 161
46 3588 24335
47 7 48
48 1 7
50 14 99
51 7 50
52 90 649
53 9100 66249
54 66 485
55 12 89
56 2 15
57 20 151
58 2574 19603
59 69 530
60 4 31
61 226153980 1766319049
62 8 63
63 1 8
65 16 129
66 8 65
67 5967 48842
68 4 33
69 936 7775
70 30 251
71 413 3480
72 2 17
73 267000 2281249
74 430 3699
75 3 26
76 6630 57799
77 40 351
78 6 53
79 9 80
80 1 9
82 18 163
83 9 82
84 6 55
85 30996 285769
86 1122 10405
87 3 28
88 21 197
89 53000 500001
90 2 19
91 165 1574
92 120 1151
93 1260 12151
94 221064 2143295
95 4 39
96 5 49
97 6377352 62809633
98 10 99
99 1 10
#### Editions
1. Leonhard Euler. Elements of Algebra. Translated by Rev. John Hewlett. Third Edition. Longmans, Hurst, Rees, Orme, and Co. London. 1822.
2. Leonhard Euler. Vollständige Anleitung zur Algebra. Mit den Zusätzen von Joseph Louis Lagrange. Herausgegeben von Heinrich Weber. B. G. Teubner. Leipzig and Berlin. 1911. Leonhardi Euleri Opera omnia. Series prima. Opera mathematica. Volumen primum.
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2023-04-01 10:54:58
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https://scicomp.stackexchange.com/tags/graph-theory/new
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# Tag Info
1
Find a longitudinal axis of the cylinder (a least-squares linear fit to all your points will yield this). Construct a plane passing through this axis. Any orientation should be fine, but let's say it is perpendicular to a normal passing from the axis to $C$. Reorientate so the axis is vertical and the plane is divided into "left" and "right" halves by the ...
2
OK, after thinking about it for a while, I came up with an answer. Step 1: Find the caps of the cylinder, in other words two closed disjoint paths along the graph's borders. Step 2: Find a path along the face graph from one cap to the other. Step 3: Create a new sub-graph by removing all edges which lie along the path found in step 2. Keep track of the ...
-1
The geodesics are the curves on a surface that connect two points A and B with shortest path. The geodesics of cylinder are parallel lines to the axis of the cylinder and circles orthogonal to them: So if you have any point on the cylinder such as $C$, the shortest path to $C$ starts from itself is one of those circles. But you are saying it's not a perfect ...
Top 50 recent answers are included
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2019-10-14 01:43:10
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http://wiki.scummvm.org/index.php/Compiling_ScummVM/Mac_OS_X_10.2.8
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- - - -
Compiling ScummVM/Mac OS X 10.2.8Compiling ScummVM/Mac OS X 10.2.8
So, I'm going to attempt to document how I was able to compile ScummVM on my 10.2.8 laptop*. However, if you don't have 10.2.8, I recommend you check out the cross compiling page.
# Problems
• kyra uses a lot of templates, and gcc 3.1 can't handle it, so I had to disable it (--disable-kyra)
• After running make bundle, I get "Text::Wrap version >= 2001.0929 is required. You have 98.112902", so it needs to be updated
# Getting a Compiler
So, as you may or may not know, you need gcc to compile gcc. Yes, it's one of those loopholes in life. So, we have to get the precompiled XCode from Apple. So, go to their developer site and sign up. Then, go to downloads, Development Tools, and way down at the bottom should be the December 2002 development tools for 10.2.8. That's the one you want. So, download it and install it. However, once it's installed, don't delete it as you may have to install it again**.
# Getting the Libraries
There is no really specific order except: SDL must be compiled before libmpeg2, libogg must be compiled before libvorbis libogg must be compiled before flac
## SDL
First off, we need to get the 1.2.8 version of SDL. Nothing higher than that compiles on 10.2.8 (it's broken currently). If you try to compile higher than 1.2.8, for example 1.2.12, it will give you a whole bunch of warnings about "__VEC__" before some errors.
Then, as long as you have 1.2.8, you should be ok. But, I wasn't***
Run:
./configure
make install
## zlib
Zlib is included with the developer tools, but it's too old to be used with ScummVM. So, download the zlib 1.2.3 source and run:
./configure
make install
./configure
make install
## libogg/libvorbis
./configure
make install
And, then the same for libvorbis. libogg must be compiled first.
## flac
./configure
make install
## libmpeg2
./configure
make install
There will be errors found, but it will turn off the preprocessing mode and compile fine, so don't worry about it.
## libfluidsynth
./configure
make install
## nasm
./configure
make install
Just ignore the 100's of warnings you will get ;)
# Configuring Configure
Assuming you already compiled SDL, we need to point ScummVM in the right direction. Open up configure in your favorite text editor and change the _sdlpath="$PATH" to _sdlpath="/usr/local/bin". # Compiling ScummVM If you just want the engines that are enabled always just run (see Problems about kyra): ./configure However, for other ones (such as Lure, CruisE, and Drascula) run: ./configure --enable-lure --enable-cruise --enable-drascula After that is run, make sure that all the libraries you installed appear on the text that Terminal should spit out at you. ## Executable Only Just run make To use this, just drag it into the Terminal and hit enter. ## Bundle We have to modify ports.mk a bit, so open it up ### Change Library Path Change OSXOPT=/sw to OSXOPT=/usr/local Reason: Fink puts the installs there, but we compiled from scratch, so they're in a different folder. ### Change sdl-config Path Change OSX_STATIC_LIBS := sdl-config --static-libs to OSX_STATIC_LIBS := /usr/local/bin/sdl-config --static-libs Reason: It needs to be set to the correct path to find the file. ### Make libfluidsynth Static Add OSX_STATIC_LIBS +=$(OSXOPT)/lib/libfluidsynth.a
after
ifdef USE_MPEG2
after
OSX_STATIC_LIBS += $(OSXOPT)/lib/libfluidsynth.a Reason: zlib should be compiled statically to be used on all systems Remove -lz Reason: We're compiling zlib statically, not dynamically. ### Remove SystemStubs Remove -lSystemStubs Reason: Not needed, and not in 10.2.8 ### Add CoreFoundation Framework Add -framework CoreFoundation \ after -framework CoreMIDI \ Reason: Corrects linker error. ### Add CoreService Framework Add -framework CoreServices \ after -framework CoreFoundation \ Reason: Corrects linker error. ### Add CoreAudio Framework Add -framework CoreAudio \ after -framework CoreServices \ Reason: Corrects linker error with libfluidsynth. ### Other Change $(OSX_STATIC_LIBS) \
to
\$(OSX_STATIC_LIBS)
Reason: Can cause errors while compiling.
TODO (See Problems)
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2017-09-21 23:14:23
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https://www.gamedev.net/forums/topic/680450-need-help-with-my-script-and-animator/
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• 10
• 9
• 12
• 14
• 14
# Need Help With My Script And Animator.
This topic is 642 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.
## Recommended Posts
Okay so I have created a player archer character which when he using the arrow keys,faces those directions and upon hitting space when in that direction,the bow attack animation is played.
The code works fine but there are some nerve-wrecking problems:
when I unpress an arrow key,the character always returns to his default direction and when I press space when no arrow key is pressed,the right bow attack animation is played and also when i press the up arrow key and the space bar,the up bow attack animation does not play and instead again,the right bow attack animation is played.
I have looked at everything again and again but i can't seem to find anything wrong.the animation's assigned are okay and the code is okay too.
Don't know whats wrong. I have been stuck at this problem all day.
I have attached some screenshots so you can see how I have achieved this:
this is the character I am working with>>>screenshot_2
this is my animator>>>>>screenshot_1
this is my blend tree in the idle state>>>>>screenshot_3
and so far,this is my script:
using UnityEngine;
using System.Collections;
public class PlayerDirectionAndShooting : MonoBehaviour {
Rigidbody2D rbody;
Animator anim;
// Use this for initialization
void Start () {
rbody = GetComponent<Rigidbody2D> ();
anim = GetComponent<Animator> ();
}
// Update is called once per frame
void Update () {
Vector2 direction_vector = new Vector2(Input.GetAxisRaw("Horizontal"), Input.GetAxisRaw("Vertical"));
anim.SetFloat("input_x", direction_vector.x);
anim.SetFloat("input_y", direction_vector.y);
if (Input.GetKeyDown("space"))
{
if (anim.GetFloat("input_x") <= -1)
{
anim.SetBool("is_shooting_left", true);
}
}
else
{
anim.SetBool("is_shooting_left", false);
};
if (Input.GetKeyDown("space"))
{
if (anim.GetFloat("input_y") <= -1)
{
anim.SetBool("is_shooting_down", true);
}
}
else
{
anim.SetBool("is_shooting_down", false);
};
if (Input.GetKeyDown("space"))
{
if (anim.GetFloat("input_y") <= 1)
{
anim.SetBool("is_shooting_up", true);
}
}
else
{
anim.SetBool("is_shooting_up", false);
};
if (Input.GetKeyDown("space"))
{
if (anim.GetFloat("input_x") <= 1)
{
anim.SetBool("is_shooting_right", true);
}
}
else
{
anim.SetBool("is_shooting_right",false);
};
}
}
##### Share on other sites
I'd say take a look at this line:
Vector2 direction_vector = new Vector2(Input.GetAxisRaw("Horizontal"), Input.GetAxisRaw("Vertical"));
you are always creating a new vector there which is fine but what does Input.getAxisRaw() return when there was no input. what's the default?
Your problem seems to be that it always uses some default if you don't have any input. You should take a look at these defaults and think of a way of ignoring them when there wasn't.
Pseudo code should be smthn like this:
if(input) update_values();
else take_previous_values()
##### Share on other sites
As the other above had mentioned. You keep destroying what data you previously had, when you should be caching it. So every time that object's update is overwith, your old data keeps falling out of scope, and needs to be made with a new one.
What you should do instead is create a private property that stores your Vector2f's data.
Another thing to be mindful of, is if-block scoping. It's not a problem right now, but I have a feeling this may crop up. Animator is very stateful. Unless it's data is declared on a superior scope, anything that is created in that if check will also be deleted once the if block is over with.
|
2018-04-25 02:57:58
|
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|
https://www.biostars.org/p/220390/#220393
|
Determination of longest read-SOAPdenovo2
0
0
Entering edit mode
5.5 years ago
eischzj12 • 0
I'm working on using SOAPdenovo for short-read assembly. How do you determine the length of your longest read? I want to set this equal to the rd_len_cutoff.
Thanks
soapdenovo2 read length Assembly • 1.0k views
ADD COMMENT
1
Entering edit mode
You can run stats.sh from BBMap on your dataset to get that information. What kind of data is this?
ADD REPLY
0
Entering edit mode
Paired end Illumina sequence data. Sorry for being vague, I'm not very familiar with the relevant information to provide. Still learning the ropes. BBMap is something I need to download separately then I take it?
ADD REPLY
0
Entering edit mode
I have linked BBMap in post above.
If your illumina sequencing was done for say 150 cycles (you would have 150 bp reads in that case) then the length of your longest (and perhaps all reads, if they have not been trimmed) would be 150.
ADD REPLY
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2022-05-17 18:46:03
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.23661847412586212, "perplexity": 6081.273015757911}, "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-2022-21/segments/1652662519037.11/warc/CC-MAIN-20220517162558-20220517192558-00707.warc.gz"}
|
https://projecteuclid.org/euclid.twjm/1500404986
|
## Taiwanese Journal of Mathematics
### ON THE NUMBER OF SOLUTIONS OF EQUATIONS OF DICKSON POLYNOMIALS OVER FINITE FIELDS
#### Abstract
Let $k, n_1, \dots, n_k$ be fixed positive integers, $c_1, \dots, c_k \in GF(q)^*$, and $a_1, \dots, a_k, c \in GF(q)$. We study the number of solutions in $GF(q)$ of the equation $c_1D_{n_1}(x_1, a_1) + c_2D_{n_2}(x_2, a_2) + \cdots + c_kD_{n_k}(x_k, a_k) = c$, where each $D_{n_i}(x_i, a_i)$, $1 \leq i \leq k$, is the Dickson polynomial of degree $n_i$ with parameter $a_i$. We also employ the results of the $k = 1$ case to recover the cardinality of preimages and images of Dickson polynomials obtained earlier by Chou, Gomez-Calderon and Mullen [1].
#### Article information
Source
Taiwanese J. Math., Volume 12, Number 4 (2008), 917-931.
Dates
First available in Project Euclid: 18 July 2017
https://projecteuclid.org/euclid.twjm/1500404986
Digital Object Identifier
doi:10.11650/twjm/1500404986
Mathematical Reviews number (MathSciNet)
MR2426536
Zentralblatt MATH identifier
1154.11044
Subjects
Primary: 11T06: Polynomials
#### Citation
Chou, Wun-Seng; Mullen, Gary L.; Wassermann, Bertram. ON THE NUMBER OF SOLUTIONS OF EQUATIONS OF DICKSON POLYNOMIALS OVER FINITE FIELDS. Taiwanese J. Math. 12 (2008), no. 4, 917--931. doi:10.11650/twjm/1500404986. https://projecteuclid.org/euclid.twjm/1500404986
#### References
• W.-S. Chou, J. Gomez-Calderon and G. L. Mullen, Value sets of Dickson polynomials over finite fields, J. Number Theory, 30 (1988), 334-344.
• W.-C. W. Li, Number Theory With Applications, Series on University Mathematics, Vol. 7, World Scientific, Singapore, 1996.
• R. Lidl, G. L. Mullen and G. Turnwald, Dickson Polynomials, Longman Scientific and Technical, Essex, United Kingdom, 1993.
• R. Lidl and H. Niederreiter, Finite Fields, Encyclo. of Math. & Its Appls, Second Ed., Vol. 20, Cambridge University Press, Cambridge, 1997.
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2019-10-19 04:11:47
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http://clay6.com/qa/40729/for-all-complex-numbers-z-of-the-form-1-i-alpha-alpha-in-r-if-z-x-iy-then-
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# For all complex numbers z of the form $\;1+i \alpha\;, \alpha \in R\;$ , if $\;z^{2}=x+iy\;$ , then :
$(a)\;y^{2}-4x+2=0\qquad(b)\;y^{2}+4x-4=0\qquad(c)\;y^{2}-4x+4=0\qquad(d)\;y^{2}+4x+2=0$
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2018-06-18 07:56:27
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https://mathematica.stackexchange.com/questions/206196/assuming-that-only-the-2nd-order-terms-are-zero
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# Assuming that only the 2nd order terms are zero
I am dealing with first-order perturbative theory, that is, any variable may be decomposed as
$$a = a + \delta a$$
and the same would occur with several different variables, let's say, $$b$$ and $$c$$. However any term that is accompanied by a $$\delta$$ (i.e. $$\delta a$$, $$\delta b$$ and $$\delta c$$) are called "perturbative terms", that is, any $$\delta$$-term multiplied by any other term with $$\delta$$ are zero (using a different lingo: just keeping perturbation to the first order):
$$\delta a ^2 =0$$;
and
$$\delta b \delta c = 0$$.
In my case these terms are used in several different calculations throughout my program, so I would like to discriminate these conditions as a global assumption (actually any other idea on how to do this would be gladly accepted) in the beginning of the program. So I have tried something like:
$Assumptions =$\delta a^2 == 0$&&$\delta b \delta c == 0$ and henceforth for every different combination of variables. Obviously, it does not work out. I have tried several different ways to do it and, in the end of my big calculations, I just to simplify equations using Simplify or FullSimplify, get results in which second order perturbative terms do not appear. Edit: In my case, the perturbative terms are dependent of two different variables: $$\delta a = \delta a (t,r)$$. So it is really common to appear derivative terms multiplying between the $$\delta$$-terms, in which must be set to zero. Examples of this would be: $$\partial_r \delta a \partial_r \delta b = 0$$, or $$\partial_t \delta a \partial_r \delta a = 0$$, or $$(\partial_t \delta c)^2 = 0$$. How should I add rules to take care of these derivative terms? • You can introduce scaling of the variables, for instance$a=a_0+\alpha x$,$b=b_0+\beta x$, and$c=c_0+\gamma x$. Next you do your calculations and at the end perform series expansion with respect to$x$and keep only terms up to the 1st order in$x$, that is Series[f[x],{x,0,1}]//Normal. – yarchik Sep 13 '19 at 15:06 ## 1 Answer You could introduce the perturbative part as δ[a] and define an extra rule how multiplication works with those (δ can be entered quickly via EscdeltaEsc or via \[Delta]): δ /: Times[___, _δ, _δ, ___] = 0; Now every time at least two δ[_] symbols are multiplied they will be simplified to zero automatically. For example in (* Input *) (a + δ[a]) (b + δ[b]) % // Expand (* Output *) (a + δ[a]) (b + δ[b]) a b + b δ[a] + a δ[b] after expanding the expression, the δ[a]δ[b] part was replaced by zero automatically. We have to be a bit careful because this rule doesn't catch powers of δ[_] so we should add another rule for that: δ /: Power[_δ, n_Integer?(# >= 2 &)] = 0; Now we can for example do (* Input *) Table[δ[a]^k, {k, 0, 3}] (* Output *) {1, δ[a], 0, 0} I'm not sure what should happen for negative powers with absolute value of at least two, but you can modify/add another rule similarly to the two above. • Probably safer to use UpValues rather than to redefine something as fundamental as Times: \[Delta] /: Times[_\[Delta], _\[Delta]] = 0;. – march Sep 13 '19 at 16:25 • @march Great idea! i'll update my answer to use UpValues instead. – Thies Heidecke Sep 13 '19 at 16:29 • I suppose that δ /: Times[___, _δ,___, _δ, ___] = 0; is good to make sure that it always works. However, since Times is Orderless, I'm pretty sure the pattern matcher will just check to see if there are two \[delta][_]'s that are multiplied, regardless of whether there are other quantities there or not. I think that \[Delta] /: Times[_\[Delta], _\[Delta]] = 0 on it's own should work. – march Sep 13 '19 at 20:01 • Ok, great answer. It seems like it is working flawlessly. However, if I may ask, lots of derivative terms also appear like$\partial_r \delta a \partial_r \delta b$and even with different variables,$\partial_t \delta a \partial_r \delta c$. They also should turn out to be zero (since anything with two$\delta\$ must become zero). Any easy way to work around this? – Edison Santos Sep 14 '19 at 13:35
• @march Yes, originally i had the extra ___ in the pattern, when i had the same realisation about the Orderless property, after which i removed it. Both patterns should work i guess. – Thies Heidecke Sep 15 '19 at 9:58
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2020-11-23 16:28:14
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https://blender.stackexchange.com/questions/40873/mixrgb-nodes-dont-add-up-value-as-i-expected-what-knowledge-am-i-missing
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MixRGB nodes don't add up Value as I expected. What knowledge am I missing?
Why don't these add up to 1?
I have seven RGB Input Nodes that are greyscale and have Value of the following amounts:
0.18333
0.2
0.15
0.18333
0.1
0.06667
0.11667
I would expect these to add up to 1 and become white, but that is not what happens.
Instead I get a dull grey (#686868): Why is this? (There are no other lights in the scene.)
This is my node set-up:
Next I tried using Add Shaders to see if that would make a difference, but it did not:
Update & semi-answer: This works ↓ (but I don't know why)
I had been adding the Value:
I should be adding the RGB if I want them to add up to white:
I didn't realize that the value is sort of a result of the RGB. Nodes that add are adding the RGB, not the Value.
(I'd be happy to accept an answer that explains how the two are related.) The answer is probably in here somewhere.
1 Answer
Go get white you have to pass something that will increase the color. For example... You have add and a factor of 1.000. This means you are not mixing the colors, but instead using one input of the node.
Just curious though, what are you trying to do? Why not just use a white shader?
• Your suggestion did not work. I need more explanation. Moving the sliders to 0.5 gives a result of #4B4B4B. Moving them to 0 gives #2F2F2F. Actually, even adding RGB didn't work! (gives #E7E7E7) Re:"What are you trying to do?" This is part of an experiment to separate the spectrum of visible light into seven proportionate bands, then refract each one at a different IOR. These will need to be recombined with Add shaders in the final project (think of a prism), but I noticed colors aren't being added back up to white so I have to find out why. – Mentalist Nov 1 '15 at 5:23
• Never mind the part about RGB not working - for that mix I had forgotten to change the Emission Shader's Value to 1 (It's 0.906 by default). Then when I combine 3 Emission Shaders of solid red, green, and blue via two Add Shaders it works. But actually this gives a hint to the nature of the problem - a discrepancy between RGB amounts and the Value amount. I think the simple answer is that adding RGB and adding Value are different things. A MixRGB Node and an Add Shader both add RGB values, not Value. I would like to know how RGB and Value correlate though (which may not be Blender-specific). – Mentalist Nov 1 '15 at 5:44
• Glad I could help. :) Sometimes, actually many times; it doesn't work quite how we expect and with adding or tweaking things makes it work. Try playing with a variety of node setups or lights too. I'm sure it can be done, but as I mentioned it's finding that magical combination. – probie Nov 1 '15 at 5:58
• I'm sure you thought about this too, but the presence of all colors will give black where with light you will get white. – probie Nov 1 '15 at 6:12
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2021-01-27 14:41:00
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