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http://mathcs.chapman.edu/~jipsen/structures/doku.php/wajsberg_hoops	Wajsberg hoops Definition A Wajsberg hoop is a hoop $\mathbf{A}=\langle A, \cdot, \rightarrow, 1\rangle$ such that $(x\rightarrow y)\rightarrow y = (y\rightarrow x)\rightarrow x$ Remark: Lattice operations are term-definable by $x\wedge y=x\cdot(x\rightarrow y)$ and $x\vee y=(x\rightarrow y)\rightarrow y$. Morphisms Let $\mathbf{A}$ and $\mathbf{B}$ be Wajsberg hoops. A morphism from $\mathbf{A}$ to $\mathbf{B}$ is a function $h:A\rightarrow B$ that is a homomorphism: $h(x\cdot y)=h(x)\cdot h(y)$, $h(x\rightarrow y)=h(x)\rightarrow h(y)$, $h(1)=1$ Example 1: Properties Classtype variety decidable no yes yes Congruence regular & yes Radim Belohlovek, On the regularity of MV-algebras and Wajsberg hoops, Algebra Universalis, 44, 2000, 375–377MRreview\\\hline Finite members $\begin{array}{lr} f(1)= &1\\ f(2)= &1\\ f(3)= &\\ f(4)= &\\ f(5)= &\\ f(6)= &\\ f(7)= &\\ \end{array}$	2017-02-20 20:07:38	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8521950840950012, "perplexity": 5354.8414886757855}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-09/segments/1487501170609.0/warc/CC-MAIN-20170219104610-00400-ip-10-171-10-108.ec2.internal.warc.gz"}
https://robotics.stackexchange.com/tags/self-driving/hot	# Tag Info 8 Yes, a state matrix with zero rows and/or columns makes sense and is viable. It typically signify pure integrators in the system. In the example you give, $$\dot{v} = -\frac{b}{m} v +\frac{1}{m} u$$ where $v$ is the speed, $u$ is the externally applied force, and $bv$ is some viscous damping force. Now if the viscous damping coefficient is zero (no ... 3 How do self-driving bots usually deal with transient objects, e.g., parked cars on the side of roads when they can come and go? No. In most of the open-source slams, dynamic objects are ignored which means they are just mapped as a stationary object. But there are few papers that deal with this in the way you think. These aren't moving objects at the time ... 3 You can use both! ROS nodes can be written in python. I'm doing a similar project and the reusability of ros code is a nice thing to have in mind. ROS also have many things already implemented (even that you want to build everything from scratch). The message types of ROS will also provide an easier to integrate application, for future projects and ... 2 What you are describing seems a lot like reinforcement learning, and yes, it works well for this sort of scenario. It is a popular enough approach that you should be able to find starter code in whatever language you prefer. This paper looks like it could be interesting for you. Basic set-up Typically the way the problem is set up is that we want to find a ... 2 1 Axial (or thrust) force refers to the force applied along the axis of rotation, not the load that is applied to the motor. (This load would be a torque, anyways... not a force). Probably fine for a tractor, not fine for a F1 racecar. This is a question you need to answer based on your application. There might be. You can adapt drills with anti-backdrive pins ... 1 High level path planning is present in any navigation system. You can select a goal and a start position (or your current position) and a (drivable) path will be planned by any navigation system, incl. google maps. Tesla cars use a neural network based control system. Video and radar feeds are combined and a semantic segmented map of the environment is ... 1 Tom Sheridan from MIT used to say (I am paraphrasing from 25 years ago so I am sure to have some of this wrong) that robots which operate only in free space should update around 10 Hz; if they interact with environmental impedances then they should update around 100 Hz; if they have to go from free space to hard contact with a stiff environment, then at ... 1 First Simulators available like Gazebo, Yes there are many simulators, You can check v-rep (free) or whebots or matlab (non-free) for example. What is the difference between a pure software simulation and a real world (say RC) scale model simulation? The pure simulation don't count noise and non-accurate parameters you use For example when you specify ... Only top voted, non community-wiki answers of a minimum length are eligible	2021-12-06 14:01:48	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4636234939098358, "perplexity": 835.6161161660698}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-49/segments/1637964363301.3/warc/CC-MAIN-20211206133552-20211206163552-00122.warc.gz"}
http://winterland.me/2019/03/05/aeson's-mysterious-lazy-parsing/	# Aeson's mysterious lazy parsing Today I’m hacking on a new JSON parser for my stdio package, of course i reused quite a lot of code from aeson. Kudos, Bryan O’Sullivan! While this time i really hacked into the core, I learned something i never really though about. Aeson provide two flavor of parsing entrances: decode and decode' (and similarly eitherDecode and eitherDecode'). The document on decode says that This function parses immediately, but defers conversion.. I have never actually even though about it carefully because this looks quite obvious to a Haskell veteran like me. defers conversion? It must be the costs of convert numbers, strings, arrays, etc. are too high, so that during parsing we suspend the conversion and return a result containing thunks. So if our JSON instances only use part of the keys, we can save some time. OK, that’s my interpretation of those docs for a long time, and i always prefer decode to decode' since thunking those conversions makes sense to me. But when i started to hack a JSON value parser from ground, suddenly I realized this is far more complex than my first look, a question is begging me to answer: How could be this conversion deferring possible, when parsing should tell us if the JSON is valid? There must be something wrong here. After parsing we should have something like Either String a, if we defer the conversions, how could we know it’s Right? If the deferred conversion somehow met problems, and we’ve already given a Right result, a bottom undefined will be produced? No this’s just unacceptable, we must do all the conversions before we can be sure the result is Right. But what’s the point of this lazy vs strict arrangement then? Reading source quickly leads us to the difference before json and json' parser: i.e. it’s the conversions between JSON bytes and Value, aeson’s intermediate JSON representation, makes this differences. In json case, thunks are produced during parsing. But where’re those thunks? What makes things more mysterious is that Value use strict field all the way: Look at those strict fields! It means that if we force a Value to WHNF, e.g. to know if it’s a Number or a String, we have to force the Scientific or Text payload. The only possible places for thunks are the payload of the payload of Objects and Arrays, which are payloads of a HashMap or a Vector respectively. So during building these HashMaps and Vectors, we write thunks instead of WHNF data structures. But why the laziness doesn’t bring bottom problem? If we have parsed the whole JSON bytes without pack results into HashMap or Vector, so where’re they? Reading the source of internal lazy parsers reveal the answer: Parser is classic CPS parser from attoparsec package, In CPS parser’s world, a parser will parse its own part and call next parser if itself runs successfully. To get a successful parsing result, we have to run though all the parsers in a line, convert everything into a proper result and pass on. Look at the loop above, it will parse all the elements of a JSON array into a list before call next continuation. Now the important part comes: Since aeson uses accumulator style to build the list, a reverse is neccessary, but this monadic return is not strict! so we build a thunk to defer the vector building process, but the list of elements is already there, in our memory, built as the parsing loop goes. So yes all parsing work is done, and we don’t have to worry about bottoms coming from deferred conversions: the only deferred conversion here is just a conversion between list of elements to vector of elements, which should never fail. Similarly in HashMap case, we have already parsed all the key-values into a list of key-value pairs. The loop body of the Object parser is like this: Now things are clear, we never miss a thing during parsing, and we only deferred two conversions when using aeson’s lazy paring. Now the question becomes how does this laziness helped things? Well if we modify the benchmarks in aeson’s repo, change the json parser to a strict json', suddenly the performance dropped by ~40%. This big performance gap lies on Vector and HashMap building, some further benchmarks indicate it’s the HashMap building which is costing: calculating hashes and build a HAMT trees is not a easy job. If everything stops at this stage, and we never access fields of the HashMap, then we can say laziness have improved our performance. But this is just a false positive like many others, as long as we want to access a single field from the HashMap, the thunk will be entered and building cost will be paid, even if you don’t want every field of the HashMap, of course the Value nested inside the field still may contain thunks, but we’re actually not saving time in common case, where the final step of JSON parsing is to covert a JSON Value to a Haskell record. From the memory perspective, lazy parsing is more problematic: we are hold lists instead of packed Vectors or HashMap, which means many indirection pointers and boxes can not be GCed, all the nodes of the JSON document is parsed anyway, and they are holded within Haskell lists, which are holded by conversion thunks. That’s just awful! My conclusion: avoid use lazy parsing in aeson if possible, it’s not worked like what I expected, and brings no benefits in most of the case.	2019-05-23 08:04:19	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.435698926448822, "perplexity": 2602.119447370937}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-22/segments/1558232257156.50/warc/CC-MAIN-20190523063645-20190523085645-00073.warc.gz"}
https://pyabc.readthedocs.io/en/latest/code.html	# Contributing Code¶ ## Testing¶ We’re commited to testing our code. Tests are run on Travis CI. We encourage to test whatever possible. However, it might not always be easy to test code which is based on random sampling. We still encourage to provide general sanity and intergation tests. We highly encourage a test-driven development (TDD) style. ### Python (PEP8)¶ We try to respect the PEP8 standard. We run flake8 as part of the test suite. The tests won’t pass if flake8 complains. ## Writing tests¶ Test can be written with pytest or the unittest module. ## Versioning scheme¶ For version numbers, we use A.B.C, where • C is increased for bug fixes • B is increased for new features • A for API breaking, backwards incompatible changes. That is, we follow the versioning scheme suggested by the Python packaging guide. First, a so called “wheel” is created via python setup.py bdist_wheel A wheel is essentially a zip archive which contains the source code and the binaries (if any). This archive is uploaded using twine twine upload dist/pyabc-x.y.z-py3-non-any.wheel replacing x.y.z by the appropriate version number. For a more in depth discussion see also the section on distributing packages of the Python packaging guide	2018-09-20 07:45:16	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.32087433338165283, "perplexity": 7164.80680311081}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-39/segments/1537267156418.7/warc/CC-MAIN-20180920062055-20180920082055-00206.warc.gz"}
http://tug.org/pipermail/texhax/2004-September/002701.html	# [texhax] Importing .eps files in LaTeX M Senthil Kumar senthil at www.cdfd.org.in Thu Sep 2 13:53:25 CEST 2004 > \begin{figure}[p] > \begin{center} > \includegraphics{figuur1-uit-mayura.eps} > \end{center} > \end{figure} > These commands are placed in the middle of my paper. I want the file to > appear in this place. But that doesn't happen, it appears at the end of my > paper. I've tried almost everything: scaling it, setting the box size, etc. Hi, Try removing the \begin{figure} and \end{figure} lines and putting the \includegraphics{figure.eps} line wherever you want the figure to appear. HTH, Senthil	2017-12-17 19:47:20	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9232143759727478, "perplexity": 9205.002689960083}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-51/segments/1512948597485.94/warc/CC-MAIN-20171217191117-20171217213117-00553.warc.gz"}
https://www.nature.com/articles/nmat2981?error=cookies_not_supported&code=cd5f0671-18ff-4493-92cc-fa0712efbb0b	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. # Nodeless superconducting gap in AxFe2Se2 (A=K,Cs) revealed by angle-resolved photoemission spectroscopy ## Abstract Pairing symmetry is a fundamental property that characterizes a superconductor. For the iron-based high-temperature superconductors1,2, an s±-wave pairing symmetry has received increasing experimental3,4,5,6,7,8,9,10,11 and theoretical12,13,14,15,16,17,18,19,20,21 support. More specifically, the superconducting order parameter is an isotropic s-wave type around a particular Fermi surface, but it has opposite signs between the hole Fermi surfaces at the zone centre and the electron Fermi surfaces at the zone corners. Here we report the low-energy electronic structure of the newly discovered superconductors, AxFe2Se2 (A=K,Cs) with a superconducting transition temperature (Tc) of about 30 K. We found AxFe2Se2 (A=K,Cs) is the most heavily electron-doped among all iron-based superconductors. Large electron Fermi surfaces are observed around the zone corners, with an almost isotropic superconducting gap of ~10.3 meV, whereas there is no hole Fermi surface near the zone centre, which demonstrates that interband scattering or Fermi surface nesting is not a necessary ingredient for the unconventional superconductivity in iron-based superconductors. Thus, the sign change in the s± pairing symmetry driven by the interband scattering as suggested in many weak coupling theories12 becomes conceptually irrelevant in describing the superconducting state here. A more conventional s-wave pairing is probably a better description. This is a preview of subscription content, access via your institution ## Relevant articles • ### Spectroscopic evidence of superconductivity pairing at 83 K in single-layer FeSe/SrTiO3 films Nature Communications Open Access 14 May 2021 • ### Visualization of the electronic phase separation in superconducting KxFe2−ySe2 Nano Research Open Access 19 October 2020 • ### Two-gap superconductivity in CaFe0.88Co0.12AsF revealed by temperature dependence of the lower critical field Hc1c (T) npj Quantum Materials Open Access 02 July 2019 ## Access options Get just this article for as long as you need it \$39.95 Prices may be subject to local taxes which are calculated during checkout ## References 1. 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Preprint at http://arxiv.org/abs/1012.6017 (2010). ## Acknowledgements This work is supported in part by the National Science Foundation of China, Ministry of Education of China, Science and Technology Committee of Shanghai Municipal, and National Basic Research Program of China (973 Program) under grant Nos. 2011CB921802 and 2011CBA00102. ## Author information Authors ### Contributions ARPES measurements were done by Y.Z., L.X.Y., M.X., Z.R.Y., F.C., C.H., H.C.X., J.J. and B.P.X. M.M. and S.K. helped with the experiment at UVSOR. J.J.Y., X.F.W. and X.H.C. provided the samples. Y.Z., L.X.Y. and D.L.F. analysed the ARPES data. J.P.H. and D.L.F wrote the paper. D.L.F. was responsible for project direction, planning and infrastructure. ### Corresponding author Correspondence to D. L. Feng. ## Ethics declarations ### Competing interests The authors declare no competing financial interests. ## Supplementary information ### Supplementary Information Supplementary Information (PDF 678 kb) ## Rights and permissions Reprints and Permissions Zhang, Y., Yang, L., Xu, M. et al. Nodeless superconducting gap in AxFe2Se2 (A=K,Cs) revealed by angle-resolved photoemission spectroscopy. Nature Mater 10, 273–277 (2011). https://doi.org/10.1038/nmat2981 • Accepted: • Published: • Issue Date: • DOI: https://doi.org/10.1038/nmat2981 • ### Research Progress of FeSe-based Superconductors Containing Ammonia/Organic Molecules Intercalation • Han-Shu Xu • Shusheng Wu • Kaibin Tang Topics in Current Chemistry (2022) • ### Spectroscopic evidence of superconductivity pairing at 83 K in single-layer FeSe/SrTiO3 films • Yu Xu • Hongtao Rong • X. J. Zhou Nature Communications (2021) • ### Visualization of the electronic phase separation in superconducting KxFe2−ySe2 • Yujie Chen • Juan Jiang • Yulin Chen Nano Research (2021) • ### Superconductivity and Jahn-Teller Distortion in s±-Wave Iron-Based Superconductors • P. K. Parida • S. Sahoo Brazilian Journal of Physics (2021) • ### Iron-Based Chalcogenide Spin Ladder BaFe2X3 (X = Se,S) • Shan Wu • Benjamin A. Frandsen • Robert Birgeneau Journal of Superconductivity and Novel Magnetism (2020)	2023-04-01 11:54:16	{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8058741688728333, "perplexity": 10498.61835256814}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296949958.54/warc/CC-MAIN-20230401094611-20230401124611-00608.warc.gz"}
https://howlingpixel.com/i-en/Stellar_dynamics	# Stellar dynamics Stellar dynamics is the branch of astrophysics which describes in a statistical way the collective motions of stars subject to their mutual gravity. The essential difference from celestial mechanics is that each star contributes more or less equally to the total gravitational field, whereas in celestial mechanics the pull of a massive body dominates any satellite orbits.[1] Historically, the methods utilized in stellar dynamics originated from the fields of both classical mechanics and statistical mechanics. In essence, the fundamental problem of stellar dynamics is the N-body problem, where the N members refer to the members of a given stellar system. Given the large number of objects in a stellar system, stellar dynamics is usually concerned with the more global, statistical properties of several orbits rather than with the specific data on the positions and velocities of individual orbits.[1] The motions of stars in a galaxy or in a globular cluster are principally determined by the average distribution of the other, distant stars. Stellar encounters involve processes such as relaxation, mass segregation, tidal forces, and dynamical friction that influence the trajectories of the system's members. Stellar dynamics also has connections to the field of plasma physics. The two fields underwent significant development during a similar time period in the early 20th century, and both borrow mathematical formalism originally developed in the field of fluid mechanics. ## Key Concepts Stellar dynamics involves determining the gravitational potential of a substantial number of stars. The stars can be modeled as point masses whose orbits are determined by the combined interactions with each other. Typically, these point masses represent stars in a variety of clusters or galaxies, such as a Galaxy cluster, or a Globular cluster. From Newton's second law an equation describing the interactions of an isolated stellar system can be written down as, ${\displaystyle m_{i}{\frac {d\mathbf {r_{i}} }{dt}}=\sum _{i=1 \atop i\neq j}^{N}{\frac {Gm_{i}m_{j}\left(\mathbf {r} _{i}-\mathbf {r} _{j}\right)}{\left\\|\mathbf {r} _{i}-\mathbf {r} _{j}\right\\|^{3}}}}$ which is simply a formulation of the N-body problem. For an N-body system, any individual member, ${\displaystyle m_{i}}$ is influenced by the gravitational potentials of the remaining ${\displaystyle m_{j}}$ members. In practice, it is not feasible to calculate the system's gravitational potential by adding all of the point-mass potentials in the system, so stellar dynamicists develop potential models that can accurately model the system while remaining computationally inexpensive.[2] The gravitational potential, ${\displaystyle \Phi }$, of a system is related to the gravitational field, ${\displaystyle \mathbf {\vec {g}} }$ by: ${\displaystyle \mathbf {\vec {g}} =-\nabla \Phi }$ whereas the mass density, ${\displaystyle \rho }$, is related to the potential via Poisson's equation: ${\displaystyle \nabla ^{2}\Phi =4\pi G\rho }$ ### Gravitational Encounters and Relaxation Stars in a stellar system will influence each other's trajectories due to strong and weak gravitational encounters. An encounter between two stars is defined to be strong if the change in potential energy between the two is greater than or equal to their initial kinetic energy. Strong encounters are rare, and they are typically only considered important in dense stellar systems, such as the cores of globular clusters.[3] Weak encounters have a more profound effect on the evolution of a stellar system over the course of many orbits. The effects of gravitational encounters can be studied with the concept of relaxation time. A simple example illustrating relaxation is two-body relaxation, where a star's orbit is altered due to the gravitational interaction with another star. Initially, the subject star travels along an orbit with initial velocity, ${\displaystyle \mathbf {v} }$, that is perpendicular to the impact parameter, the distance of closest approach, to the field star whose gravitational field will affect the original orbit. Using Newton's laws, the change in the subject star's velocity, ${\displaystyle \delta \mathbf {v} }$, is approximately equal to the acceleration at the impact parameter, multiplied by the time duration of the acceleration. The relaxation time can be thought as the time it takes for ${\displaystyle \delta \mathbf {v} }$ to equal ${\displaystyle \mathbf {v} }$, or the time it takes for the small deviations in velocity to equal the star's initial velocity. The relaxation time for a stellar system of ${\displaystyle N}$ objects is approximately equal to: ${\displaystyle t_{\text{relax}}\backsimeq {\frac {0.1N}{\ln N}}t_{\text{cross}}}$ where ${\displaystyle t_{\text{cross}}}$ is known as the crossing time, the time it takes for a star to travel across the galaxy once. The relaxation time identifies collisionless vs. collisional stellar systems. Dynamics on timescales less than the relaxation time are defined to be collisionless. They are also identified as systems where subject stars interact with a smooth gravitational potential as opposed to the sum of point-mass potentials.[2] The accumulated effects of two-body relaxation in a galaxy can lead to what is known as mass segregation, where more massive stars gather near the center of clusters, while the less massive ones are pushed towards the outer parts of the cluster.[3] ## Connections to statistical mechanics and plasma physics The statistical nature of stellar dynamics originates from the application of the kinetic theory of gases to stellar systems by physicists such as James Jeans in the early 20th century. The Jeans equations, which describe the time evolution of a system of stars in a gravitational field, are analogous to Euler's equations for an ideal fluid, and were derived from the collisionless Boltzmann equation. This was originally developed by Ludwig Boltzmann to describe the non-equilibrium behavior of a thermodynamic system. Similarly to statistical mechanics, stellar dynamics make use of distribution functions that encapsulate the information of a stellar system in a probabilistic manner. The single particle phase-space distribution function, ${\displaystyle f(\mathbf {x} ,\mathbf {v} ,t)}$, is defined in a way such that ${\displaystyle f(\mathbf {x} ,\mathbf {v} ,t)\,{\text{d}}\mathbf {x} \,{\text{d}}\mathbf {v} }$ represents the probability of finding a given star with position ${\displaystyle \mathbf {x} }$ around a differential volume ${\displaystyle {\text{d}}\mathbf {x} }$ and velocity ${\displaystyle {\text{v}}}$ around a differential volume ${\displaystyle {\text{d}}\mathbf {v} }$. The distribution is function is normalized such that integrating it over all positions and velocities will equal unity. For collisional systems, Liouville's theorem is applied to study the microstate of a stellar system, and is also commonly used to study the different statistical ensembles of statistical mechanics. In plasma physics, the collisionless Boltzmann equation is referred to as the Vlasov equation, which is used to study the time evolution of a plasma's distribution function. Whereas Jeans applied the collisionless Boltzmann equation, along with Poisson's equation, to a system of stars interacting via the long range force of gravity, Anatoly Vlasov applied Boltzmann's equation with Maxwell's equations to a system of particles interacting via the Coulomb Force.[4] Both approaches separate themselves from the kinetic theory of gases by introducing long-range forces to study the long term evolution of a many particle system. In addition to the Vlasov equation, the concept of Landau damping in plasmas was applied to gravitational systems by Donald Lynden-Bell to describe the effects of damping in spherical stellar systems.[5] ## Applications Stellar dynamics is primarily used to study the mass distributions within stellar systems and galaxies. Early examples of applying stellar dynamics to clusters include Albert Einstein's 1921 paper applying the virial theorem to spherical star clusters and Fritz Zwicky's 1933 paper applying the virial theorem specifically to the Coma Cluster, which was one of the original harbingers of the idea of dark matter in the universe.[6][7] The Jeans equations have been used to understand different observational data of stellar motions in the Milky Way galaxy. For example, Jan Oort utilized the Jeans equations to determine the average matter density in the vicinity of the solar neighborhood, whereas the concept of asymmetric drift came from studying the Jeans equations in cylindrical coordinates.[8] Stellar dynamics also provides insight into the structure of galaxy formation and evolution. Dynamical models and observations are used to study the triaxial structure of elliptical galaxies and suggest that prominent spiral galaxies are created from galaxy mergers.[1] Stellar dynamical models are also used to study the evolution of active galactic nuclei and their black holes, as well as to estimate the mass distribution of dark matter in galaxies. ## References 1. ^ a b c Murdin, Paul (2001). "Stellar Dynamics". Encyclopedia of Astronomy and Astrophysics. Nature Publishing Group. p. 1. ISBN 978-0750304405. 2. ^ a b Binney, James; Tremaine, Scott (2008). Galactic Dynamics. Princeton: Princeton University Press. pp. 35, 63, 65, 698. ISBN 978-0-691-13027-9. 3. ^ a b Sparke, Linda; Gallagher, John (2007). Galaxies in the Universe. New York: Cambridge. p. 131. ISBN 978-0521855938. 4. ^ Henon, M (June 21, 1982). "Vlasov Equation?". Astronomy and Astrophysics. 114: 211–212. 5. ^ Lynden-Bell, Donald (1962). "The stability and vibrations of a gas of stars". Monthly Notices of the Royal Astronomical Society. 124: 279–296. 6. ^ Einstin, Albert (2002). "A Simple Application of the Newtonian Law of Gravitation to Star Clusters" (PDF). The Collected Papers of Albert Einstein. 7: 230–233 – via Princeton University Press. 7. ^ Zwicky, Fritz (2009). "Republication of: The redshift of extragalactic nebulae". General Relativity and Gravitation. 41: 207–224. 8. ^ Choudhuri, Arnab Rai (2010). Astrophysics for Physicists. New York: Cambridge University Press. pp. 213–214. ISBN 978-0-521-81553-6. Astrophysics Astrophysics is the branch of astronomy that employs the principles of physics and chemistry "to ascertain the nature of the astronomical objects, rather than their positions or motions in space". Among the objects studied are the Sun, other stars, galaxies, extrasolar planets, the interstellar medium and the cosmic microwave background. Emissions from these objects are examined across all parts of the electromagnetic spectrum, and the properties examined include luminosity, density, temperature, and chemical composition. Because astrophysics is a very broad subject, astrophysicists apply concepts and methods from many disciplines of physics, including classical mechanics, electromagnetism, statistical mechanics, thermodynamics, quantum mechanics, relativity, nuclear and particle physics, and atomic and molecular physics. In practice, modern astronomical research often involves a substantial amount of work in the realms of theoretical and observational physics. Some areas of study for astrophysicists include their attempts to determine the properties of dark matter, dark energy, black holes, and other celestial bodies; whether or not time travel is possible, wormholes can form, or the multiverse exists; and the origin and ultimate fate of the universe. Topics also studied by theoretical astrophysicists include Solar System formation and evolution; stellar dynamics and evolution; galaxy formation and evolution; magnetohydrodynamics; large-scale structure of matter in the universe; origin of cosmic rays; general relativity and physical cosmology, including string cosmology and astroparticle physics. Box orbit In stellar dynamics, a box orbit refers to a particular type of orbit that can be seen in triaxial systems, i.e. systems that do not possess a symmetry around any of its axes. They contrast with the loop orbits that are observed in spherically symmetric or axisymmetric systems. In a box orbit, a star oscillates independently along the three different axes as it moves through the system. As a result of this motion, it fills in a (roughly) box-shaped region of space. Unlike loop orbits, the stars on box orbits can come arbitrarily close to the center of the system. As a special case, if the frequencies of oscillation in different directions are commensurate, the orbit will lie on a one- or two-dimensional manifold and can avoid the center. Such orbits are sometimes called "boxlets". Cartwheel Galaxy The Cartwheel Galaxy (also known as ESO 350-40 or PGC 2248) is a lenticular galaxy and ring galaxy about 500 million light-years away in the constellation Sculptor. It is an estimated 150,000 light-years diameter, and has a mass of about 2.9–4.8 × 109 solar masses; its outer ring has a circular velocity of 217 km/s.It was discovered by Fritz Zwicky in 1941. Zwicky considered his discovery to be "one of the most complicated structures awaiting its explanation on the basis of stellar dynamics."An estimation of the galaxy's span resulted in a conclusion of 150,000 light years, which is slightly larger than the Milky Way. Chandrasekhar's variational principle In astrophysics, Chandrasekhar's variational principle provides the stability criterion for a static barotropic star, subjected to radial perturbation, named after the Indian American astrophysicist Subrahmanyan Chandrasekhar. Chandrasekhar's white dwarf equation In astrophysics, Chandrasekhar's white dwarf equation is an initial value ordinary differential equation introduced by the Indian American astrophysicist Subrahmanyan Chandrasekhar, in his study of the gravitational potential of completely degenerate white dwarf stars. The equation reads as ${\displaystyle {\frac {1}{\eta ^{2}}}{\frac {d}{d\eta }}\left(\eta ^{2}{\frac {d\varphi }{d\eta }}\right)+(\varphi ^{2}-C)^{3/2}=0}$ with initial conditions ${\displaystyle \varphi (0)=1,\quad \varphi '(0)=0}$ where ${\displaystyle \varphi }$ measures the density of white dwarf, ${\displaystyle \eta }$ is the non-dimensional radial distance from the center and ${\displaystyle C}$ is a constant which is related to the density of the white dwarf at the center. When ${\displaystyle C=0}$, this equation reduces to Lane–Emden equation with polytropic index ${\displaystyle 3}$. In the Lane–Emden equation, the density at the centre can be scaled out of the equation, but for white-dwarfs, the central density is directly tied to the equation. Chandrasekhar limit The Chandrasekhar limit () is the maximum mass of a stable white dwarf star. The currently accepted value of the Chandrasekhar limit is about 1.4 M☉ (2.765×1030 kg).White dwarfs resist gravitational collapse primarily through electron degeneracy pressure (compare main sequence stars, which resist collapse through thermal pressure). The Chandrasekhar limit is the mass above which electron degeneracy pressure in the star's core is insufficient to balance the star's own gravitational self-attraction. Consequently, a white dwarf with a mass greater than the limit is subject to further gravitational collapse, evolving into a different type of stellar remnant, such as a neutron star or black hole. Those with masses under the limit remain stable as white dwarfs.The limit was named after Subrahmanyan Chandrasekhar, an Indian astrophysicist who improved upon the accuracy of the calculation in 1930, at the age of 20, in India by calculating the limit for a polytrope model of a star in hydrostatic equilibrium, and comparing his limit to the earlier limit found by E. C. Stoner for a uniform density star. Importantly, the existence of a limit, based on the conceptual breakthrough of combining relativity with Fermi degeneracy, was indeed first established in separate papers published by Wilhelm Anderson and E. C. Stoner in 1929. The limit was initially ignored by the community of scientists because such a limit would logically require the existence of black holes, which were considered a scientific impossibility at the time. That the roles of Stoner and Anderson are often forgotten in the astronomy community has been noted. Chandrasekhar potential energy tensor In astrophysics, Chandrasekhar potential energy tensor provides the gravitational potential of a body due to its own gravity created by the distribution of matter across the body, named after the Indian American astrophysicist Subrahmanyan Chandrasekhar. The Chandrasekhar tensor is a generalization of potential energy in other words, the trace of the Chandrasekhar tensor provides the potential energy of the body. Division on Dynamical Astronomy The Division on Dynamical Astronomy (DDA) is a branch of the American Astronomical Society that focuses on the advancement of all aspects of dynamical astronomy, including celestial mechanics, solar system dynamics, stellar dynamics, as well as the dynamics of the interstellar medium and galactic dynamics, and coordination of such research with other branches of science. It awards the Brouwer Award every year, which was established to recognize outstanding contributions to the field of Dynamical Astronomy, including celestial mechanics, astrometry, geophysics, stellar systems, galactic and extra galactic dynamics. The Division also awards the Vera Rubin Early Career Prize for promise of continued excellence for an astronomer no more than 10 years beyond receipt of their doctorate. Dynamical friction In astrophysics, dynamical friction or Chandrasekhar friction, sometimes called gravitational drag, is loss of momentum and kinetic energy of moving bodies through gravitational interactions with surrounding matter in space. It was first discussed in detail by Subrahmanyan Chandrasekhar in 1943. Emden–Chandrasekhar equation In astrophysics, the Emden–Chandrasekhar equation is a dimensionless form of the Poisson equation for the density distribution of a spherically symmetric isothermal gas sphere subjected to its own gravitational force, named after Robert Emden and Subrahmanyan Chandrasekhar. The equation was first introduced by Robert Emden in 1907. The equation reads ${\displaystyle {\frac {1}{\xi ^{2}}}{\frac {d}{d\xi }}\left(\xi ^{2}{\frac {d\psi }{d\xi }}\right)=e^{-\psi }}$ where ${\displaystyle \xi }$ is the dimensionless radius and ${\displaystyle \psi }$ is the related to the density of the gas sphere as ${\displaystyle \rho =\rho _{c}e^{-\psi }}$, where ${\displaystyle \rho _{c}}$ is the density of the gas at the centre. The equation has no known explicit solution. If a polytropic fluid is used instead of an isothermal fluid, one obtains the Lane–Emden equation. The isothermal assumption is usually modeled to describe the core of a star. The equation is solved with the initial conditions, ${\displaystyle \psi =0,\quad {\frac {d\psi }{d\xi }}=0\quad {\text{at}}\quad \xi =0.}$ The equation appears in other branches of physics as well, for example the same equation appears in the Frank-Kamenetskii explosion theory for a spherical vessel. The relativistic version of this spherically symmetric isothermal model was studied by Subrahmanyan Chandrasekhar in 1972. Fesenkov Astrophysical Institute Fesenkov Astrophysical Institute (Астрофизический институт имени В. Г. Фесенкова, АФИФ), or FAPHI, is a research institute in Almaty, Kazakhstan. The institute was founded in 1941 as the Institute for Astronomy and Physics of the Kazakh branch of the USSR Academy of Sciences, when a group of Soviet astronomers was evacuated during World War II from the European parts of the USSR to Almaty. In 1948 G.A. Tikhov had organized an independent sector of astrobotany, and in 1950 astronomers established the Astrophysical Institute of the Kazakh SSR. In 1989 the Institute was renamed after Vasily Fesenkov, one of its founders. FAPHI conducts both observational and theoretical research. The prime objects of observations are the Sun, outer planets, comets, Herbig Ae/Be stars, and active galaxies. The topics of theoretical research include stellar dynamics and computational astrophysics, active galactic nuclei, cosmology, physics of comets and interstellar medium. The institute runs three observational bases in mountains near Almaty: Kamenskoe Plateau Observatory, Assy-Turgen Observatory and Tien Shan Astronomical Observatory. FAPHI is a member of the International Astronomical Union. Jack G. Hills Jack G. Hills is a theorist of stellar dynamics. He worked on the Oort cloud; the inner part of it, the Hills cloud, was named after him. He spent much of his professional career at Los Alamos National Laboratory, which named him a Laboratory Fellow in 1998. Jeans equations The Jeans equations describe the motion of a collection of stars in a gravitational field. If n = n(x,t) is the density of stars in space, as a function of position x = (x1x2x3) and time t, v = (v1v2v3) is the velocity, and Φ = Φ(x,t) is the gravitational potential, the Jeans equations may be written as ${\displaystyle {\frac {\partial n}{\partial t}}+\sum _{i}{\frac {\partial (n\langle {v_{i}}\rangle )}{\partial x_{i}}}=0,}$ ${\displaystyle {\frac {\partial (n\langle {v_{j}}\rangle )}{\partial t}}+n{\frac {\partial \Phi }{\partial x_{j}}}+\sum _{i}{\frac {\partial (n\langle {v_{i}v_{j}}\rangle )}{\partial x_{i}}}=0\qquad (j=1,2,3.)}$ Here, the <…> notation means an average at a given point and time (x,t), so that, for example, ${\displaystyle \langle {v_{1}}\rangle }$ is the average of component 1 of the velocity of the stars at a given point and time. The second set of equations may alternately be written as ${\displaystyle n{\frac {\partial \langle {v_{j}}\rangle }{\partial t}}+\sum _{i}n\langle {v_{i}}\rangle {\frac {\partial {\langle {v_{j}}\rangle }}{\partial x_{i}}}=-n{\frac {\partial \Phi }{\partial x_{j}}}-\sum _{i}{\frac {\partial (n\sigma _{ij}^{2})}{\partial x_{i}}}\qquad (j=1,2,3.)}$ where ${\displaystyle \sigma _{ij}^{2}=\langle {v_{i}v_{j}}\rangle -\langle {v_{i}}\rangle \langle {v_{j}}\rangle }$ measures the velocity dispersion in components i and j at a given point. The Jeans equations are analogous to the Euler equations for fluid flow and may be derived from the collisionless Boltzmann equation. They were originally derived by James Clerk Maxwell but were first applied to stellar dynamics by James Jeans. A Lindblad resonance, named for the Swedish galactic astronomer Bertil Lindblad, is an orbital resonance in which an object's epicyclic frequency (the rate at which one periapse follows another) is a simple multiple of some forcing frequency. Resonances of this kind tend to increase the object's orbital eccentricity and to cause its longitude of periapse to line up in phase with the forcing. Lindblad resonances drive spiral density waves both in galaxies (where stars are subject to forcing by the spiral arms themselves) and in Saturn's rings (where ring particles are subject to forcing by Saturn's moons). Lindblad resonances affect stars at such distances from a disc galaxy's centre where the natural frequency of the radial component of a star's orbital velocity is close to the frequency of the gravitational potential maxima encountered during its course through the spiral arms. If a star's orbital speed around the galactic centre is greater than that of the part of the spiral arm through which it is passing, then an inner Lindblad resonance occurs - if smaller, then an outer Lindblad resonance. At an inner resonance, a star's orbital speed is increased, moving the star outwards, and decreased for an outer resonance causing inward movement. Michel Hénon Michel Hénon (French: [enɔ̃]; 1931 in Paris – 7 April 2013 in Nice) was a French mathematician and astronomer. He worked for a long time at the Nice Observatory. In astronomy, Hénon is well known for his contributions to stellar dynamics. In the late 1960s and early 1970s he made important contributions on the dynamical evolution of star clusters, in particular globular clusters. He developed a numerical technique using Monte Carlo methods to follow the dynamical evolution of a spherical star cluster much faster than the so-called n-body methods. In mathematics, he is well known for the Hénon map, a simple discrete dynamical system that exhibits chaotic behavior. He published a two-volume work on the restricted three-body problem. N-body units N-body units are a completely self-contained system of units used for N-body simulations of self-gravitating systems in astrophysics. In this system, the base physical units are chosen so that the total mass, M, the gravitational constant, G, and the virial radius, R, are normalized. The underlying assumption is that the system of N objects (stars) satisfies the virial theorem. The consequence of standard N-body units is that the velocity dispersion of the system, v, is ${\displaystyle \scriptstyle {\frac {1}{2}}{\sqrt {2}}}$ and that the dynamical or crossing time, t, is ${\displaystyle \scriptstyle 2{\sqrt {2}}}$. The use of standard N-body units was advocated by Michel Hénon in 1971. Early adopters of this system of units included H. Cohn in 1979 and D. Heggie and R. Mathieu in 1986. At the conference MODEST14 in 2014, D. Heggie proposed that the community abandon the name "N-body units" and replace it with the name "Hénon units" to commemorate the originator. NEMO (Stellar Dynamics Toolbox) NEMO (Not Everybody Must Observe) is a toolkit for stellar dynamics. At its core it manipulates an n-body system (snapshot), but can also derive or compute orbits, derive images and extract tables to take to other analysis systems. Schönberg–Chandrasekhar limit In stellar astrophysics, the Schönberg–Chandrasekhar limit is the maximum mass of a non-fusing, isothermal core that can support an enclosing envelope. It is expressed as the ratio of the core mass to the total mass of the core and envelope. Estimates of the limit depend on the models used and the assumed chemical compositions of the core and envelope; typical values given are from 0.10 to 0.15 (10% to 15% of the total stellar mass). This is the maximum to which a helium-filled core can grow, and if this limit is exceeded, as can only happen in massive stars, the core collapses, releasing energy that causes the outer layers of the star to expand to become a red giant. It is named after the astrophysicists Subrahmanyan Chandrasekhar and Mario Schönberg, who estimated its value in a 1942 paper. They estimated it to be ${\displaystyle \operatorname {({\frac {\operatorname {M} _{ic}}{M}})} _{SC}=0.37({\frac {\operatorname {\mu } _{e}}{\operatorname {\mu } _{ic}}})^{2}}$ The Schönberg–Chandrasekhar limit comes into play when fusion in a main-sequence star exhausts the hydrogen at the center of the star. The star then contracts until hydrogen fuses in a shell surrounding a helium-rich core, both of which are surrounded by an envelope consisting primarily of hydrogen. The core increases in mass as the shell burns its way outwards through the star. If the star's mass is less than approximately 1.5 solar masses, the core will become degenerate before the Schönberg–Chandrasekhar limit is reached, and, on the other hand, if the mass is greater than approximately 6 solar masses, the star leaves the main sequence with a core mass already greater than the Schönberg–Chandrasekhar limit so its core is never isothermal before helium fusion. In the remaining case, where the mass is between 1.5 and 6 solar masses, the core will grow until the limit is reached, at which point it will contract rapidly until helium starts to fuse in the core. Subrahmanyan Chandrasekhar Subrahmanyan Chandrasekhar ; 19 October 1910 – 21 August 1995) was an Indian American astrophysicist who spent his professional life in the United States. He was awarded the 1983 Nobel Prize for Physics with William A. Fowler for "...theoretical studies of the physical processes of importance to the structure and evolution of the stars". His mathematical treatment of stellar evolution yielded many of the current theoretical models of the later evolutionary stages of massive stars and black holes. The Chandrasekhar limit is named after him. Chandrasekhar worked on a wide variety of physical problems in his lifetime, contributing to the contemporary understanding of stellar structure, white dwarfs, stellar dynamics, stochastic process, radiative transfer, the quantum theory of the hydrogen anion, hydrodynamic and hydromagnetic stability, turbulence, equilibrium and the stability of ellipsoidal figures of equilibrium, general relativity, mathematical theory of black holes and theory of colliding gravitational waves. At the University of Cambridge, he developed a theoretical model explaining the structure of white dwarf stars that took into account the relativistic variation of mass with the velocities of electrons that comprise their degenerate matter. He showed that the mass of a white dwarf could not exceed 1.44 times that of the Sun – the Chandrasekhar limit. Chandrasekhar revised the models of stellar dynamics first outlined by Jan Oort and others by considering the effects of fluctuating gravitational fields within the Milky Way on stars rotating about the galactic centre. His solution to this complex dynamical problem involved a set of twenty partial differential equations, describing a new quantity he termed 'dynamical friction', which has the dual effects of decelerating the star and helping to stabilize clusters of stars. Chandrasekhar extended this analysis to the interstellar medium, showing that clouds of galactic gas and dust are distributed very unevenly. Chandrasekhar studied at Presidency College, Madras (now Chennai) and the University of Cambridge. A long-time professor at the University of Chicago, he did some of his studies at the Yerkes Observatory, and served as editor of The Astrophysical Journal from 1952 to 1971. He was on the faculty at Chicago from 1937 until his death in 1995 at the age of 84, and was the Morton D. Hull Distinguished Service Professor of Theoretical Astrophysics. Formation Evolution Spectral classification Remnants Hypothetical Nucleosynthesis Structure Properties Star systems Earth-centric observations Lists Related articles Major subfields of astronomy and astrophysics This page is based on a Wikipedia article written by authors (here). Text is available under the CC BY-SA 3.0 license; additional terms may apply. Images, videos and audio are available under their respective licenses.	2019-06-27 03:57:29	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 42, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6835880279541016, "perplexity": 776.2223136557807}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-26/segments/1560628000613.45/warc/CC-MAIN-20190627035307-20190627061307-00404.warc.gz"}
https://zbmath.org/?q=ut%3Aanalytic+map	## Found 1,471 Documents (Results 1–100) 100 MathJax ### A splitting result for real submanifolds of a Kähler manifold. (English)Zbl 07566852 MSC: 22E45 53D20 Full Text: ### A matryoshka of Brocard porisms. (English)Zbl 07561828 MSC: 51M04 51N20 53A04 Full Text: ### A non-Borel special alpha-limit set in the square. (English)Zbl 07554511 MSC: 03E15 37E99 37B10 Full Text: Full Text: ### Invariant measures for random expanding on average Saussol maps. (English)Zbl 07544522 MSC: 37H15 37C30 Full Text: Full Text: Full Text: ### Continuous linear images of spaces $$C_p(X)$$ with the weak topology. (English)Zbl 07541289 MSC: 46A03 46E10 54C35 Full Text: ### The fibers of the ramified Prym map. (English)Zbl 07538709 MSC: 14H40 14H15 14H45 Full Text: Full Text: Full Text: Full Text: ### Shortest paths in arbitrary plane domains. (English)Zbl 07525916 MSC: 54F50 51M25 Full Text: Full Text: ### Bifurcation analysis for a class of cubic switching systems. 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(English)Zbl 07523752 MSC: 51N15 51M15 51M09 Full Text: Full Text: Full Text: Full Text: ### Conformally symmetric triangular lattices and discrete $$\vartheta$$-conformal maps. (English)Zbl 1485.30003 MSC: 30C35 30G25 Full Text: ### Open and surjective mapping theorems for differentiable maps with critical points. (English)Zbl 1483.26012 MSC: 26B10 26B05 30G30 Full Text: ### Liftable vector fields, unfoldings and augmentations. (English)Zbl 07430028 MSC: 58K40 58K20 32S05 Full Text: Full Text: Full Text: ### Stability of the Denjoy-Wolff theorem. (English)Zbl 1480.30006 Reviewer: Weiwei Cui (Lund) MSC: 30D05 30F45 Full Text: Full Text: Full Text: Full Text: ### On the distribution of periods of holomorphic cusp forms and zeroes of period polynomials. (English)Zbl 1477.11094 MSC: 11F67 30C15 Full Text: ### A moment map interpretation of the Ricci form, Kähler-Einstein structures, and Teichmüller spaces. (English)Zbl 1472.53092 Novikov, Sergey (ed.) et al., Integrability, quantization, and geometry II. Quantum theories and algebraic geometry. Dedicated to the memory of Boris Dubrovin 1950–2019. Providence, RI: American Mathematical Society (AMS). Proc. Symp. Pure Math. 103, Part 2, 223-255 (2021). Full Text: Full Text: Full Text: ### Perverse sheaves on semi-abelian varieties. (English)Zbl 1470.32089 MSC: 32S60 14F17 55N25 Full Text: Full Text: Full Text: Full Text: ### A higher-order tangent map and a conjecture on the higher Nash blowup of curves. (English)Zbl 1471.14009 MSC: 14B05 14M25 32S45 Full Text: Full Text: Full Text: Full Text: Full Text: Full Text: ### Tubular neighborhoods of orbits of power-logarithmic germs. (English)Zbl 1473.37025 MSC: 37C15 37C05 37C35 Full Text: Full Text: Full Text: ### On the distribution of Julia sets of holomorphic maps. (English)Zbl 07552992 MSC: 37F10 30D05 Full Text: Full Text: Full Text: ### Estimate of the rate of global convergence of circle packings. (Chinese. English summary)Zbl 07494910 MSC: 52C26 52C15 30G25 Full Text: ### A glimpse into Thurston’s work. (English)Zbl 1479.57040 Ohshika, Ken’ichi (ed.) et al., In the tradition of Thurston. Geometry and topology. Cham: Springer. 1-58 (2020). Full Text: Full Text: Full Text: Full Text: Full Text: ### Topological classification and finite determinacy of knotted maps. (English)Zbl 1457.32015 MSC: 32C05 32S55 Full Text: ### Korovkin-type results on convergence of sequences of positive linear maps on function spaces. (English)Zbl 1459.41012 MSC: 41A36 46E15 Full Text: Full Text: ### Characterization of the response maps of alternating-current networks. (English)Zbl 1453.94167 MSC: 94C05 05C82 05C90 Full Text: ### Hölder regularity and exponential decay of correlations for a class of piecewise partially hyperbolic maps. (English)Zbl 1455.37031 MSC: 37D30 37D20 37C30 Full Text: Full Text: Full Text: Full Text: Full Text: ### Locally analytic representations in the étale coverings of the Lubin-Tate moduli space. (English)Zbl 1469.11456 MSC: 11S31 14G22 Full Text: ### On the integral Hodge conjecture for real varieties. I. (English)Zbl 1465.14055 MSC: 14P99 57N65 14E99 Full Text: ### On the compactness of Grunsky differential operators. (English)Zbl 1461.30053 MSC: 30C62 32G15 Full Text: Full Text: ### Karnaugh maps of logical systems and applications in digital circuit design. (English)Zbl 1478.94152 MSC: 94C05 94C15 94C30 Full Text: ### Fixed points, symmetries, and bounds for basins of attraction of complex trigonometric functions. (English)Zbl 1486.30078 MSC: 30D05 37F10 28A80 Full Text: ### The Abel map for surface singularities. II: Generic analytic structure. (English)Zbl 1444.32030 MSC: 32S05 32S25 Full Text: ### Dynamically affine maps in positive characteristic. (English)Zbl 1444.14082 Moree, Pieter (ed.) et al., Dynamics: topology and numbers. Conference, Max Planck Institute for Mathematics, Bonn, Germany, July 2–6, 2018. Providence, RI: American Mathematical Society (AMS). Contemp. Math. 744, 125-156 (2020). MSC: 14K15 37P55 37C30 Full Text: Full Text: ### Generalized Mandelbrot and Julia sets in a family of planar angle-doubling maps. (English)Zbl 1453.37040 Bohner, Martin (ed.) et al., Difference equations and discrete dynamical systems with applications. ICDEA 24, Dresden, Germany, May 21–25, 2018. Proceedings of the 24th international conference on difference equations and applications. Cham: Springer. Springer Proc. Math. Stat. 312, 21-54 (2020). Full Text: Full Text: ### One-side continuity of meromorphic mappings between real analytic hypersurfaces. (English)Zbl 1446.32010 MSC: 32H04 32V99 Full Text: Full Text: ### On the limit cycles of a class of discontinuous piecewise linear differential systems. (English)Zbl 1441.34028 MSC: 34A36 34C05 34C07 Full Text: Full Text: Full Text: Full Text: ### Transcendental holomorphic maps between real algebraic manifolds in a complex space. (English)Zbl 1442.14179 Reviewer: Rong Du (Shanghai) Full Text: Full Text: Full Text: Full Text: Full Text: Full Text: all top 5 all top 5 all top 5 all top 3 all top 3	2022-08-10 20:31:49	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7336525321006775, "perplexity": 12316.209934655653}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-33/segments/1659882571210.98/warc/CC-MAIN-20220810191850-20220810221850-00256.warc.gz"}
http://cs.brown.edu/courses/cs019/2015/Assignments/tour-guide.html	Tour Guide No doubt you remember the campus tour—the walking backwards, the explanation of meal plans, the intramural sports speech.... It’s a fine tradition, but the admissions department is running out of willing volunteers. That’s where you come in: you’ve been asked to write a program to calculate tours for campus visitors. 1Dijkstra’s Algorithm The map data you receive will contain many sites that are not tour stops, such as intersections and intermediate landmarks. All map sites, together with the paths between them, form a graph. To go from one tour stop to another, your program should be able to find the shortest path from any point in this graph to every other point in the graph. In this assignment you will implement Dijkstra’s algorithm using the Loc type to represent points in the map of Brown’s campus. A Loc includes its name and location, as well as its neighbors and the distances in meters to those neighbors. Every location will have a unique, unambiguous name. A DLoc represents the result of a shortest-path search for a point in the graph. It contains the location of the point, the length of the shortest path to that point, and the name of the previous location in that path. The Loc and DLoc types are defined as follows: data Loc: \| loc(name :: String, latitude :: Number, longitude :: Number, neighbors :: List, distances :: List) end data DLoc: \| root(site :: Loc) \| dloc(site :: Loc, dist :: Number, prev :: String) end All instances of dloc and root are relative to the starting point of a shortest-path search (see below). Your program should take in a Loc representing the starting point for your path, and get-loc :: (String -> Loc), which is a function to fetch a Loc by name. In other words, get-loc is your way of accessing information from the graph about locations, as is further explained in Importing Data. Your program should return a List<DLoc> indicating the shortest total distance from the starting point to each location on the map of Brown’s Campus, as well as the previous stop in the shortest path to each location. Use the root case of DLoc to represent the starting point of your shortest-path-search. Your implementation of Dijkstra’s algorithm should have the following contract: fun dijkstra(start :: Loc, get-loc :: (String -> Loc)) -> List: Your implementation should run in $$O([s \rightarrow s \cdot \log s])$$ time, where $$s$$ is the number of vertices in the graph. Note that a complete graph (with an edge between every pair of vertices) would have $$(s \cdot (s-1))/2$$ edges, making this impossible. However, graph representations of real maps are not so densely connected, and you may assume that each vertex has only some constant number of neighbors. 2Campus Tour (Extra Credit) In addition to finding the shortest path between points, you may extend your program for extra credit to be able to find a path through lists of predefined “tours”, such as a “Campus Art Tour” or a “Freshman Dorm Tour”. These tours will be supplied by the Admissions Office. Even if a stop appears multiple times in the list, it only needs to be visited once. These tours have no fixed order. Your program should simply move toward the closest unvisited stop in the provided list of tours, even if this results in tours being interleaved. Tours will be represented by the Tour type. A Tour stores the name of a tour and a list of its stops’ names: data Tour: \| tour(name :: String, stops :: List) end The Admissions Office wants a program that provides the following tour-calculating function: fun campus-tour(tours-list :: List, start-lat :: Number, start-lon :: Number, get-loc :: (String -> Loc)) -> List: where the output is a list of strings indicating the names of each stop on the calculated tour. campus-tour produces a list of location names to walk through, ordered from first to last (including intermediate sites like street corners), to complete the given tours according to the specifications above. Since the user might not begin their tour standing at a site on the map, your campus-tour should first find the site closest to the starting position, and then continue to the nearest selected tour stop. Remember, you will be computing a single path that visits all stops in all the selected tours. Keep in mind that if you aren’t given any tours or the tours you are given contain no locations, you don’t have to go anywhere, so make sure your output reflects this. 3Importing Data The support code for this assignment contains the map data for Brown’s campus, as well as additional support code you need to write Tour Guide. Links to shared copies of the support code have been provided for you. In addition to the Loc and Tour data types described above, the file mapdata.arr contains the following definitions and helper functions: fun get-dist(latA :: Number, lonA :: Number, latB :: Number, lonB :: Number) -> Number takes the latitude and longitude of two locations, and returns the approximate distance in meters between them. fun get-loc-maker(loc-list :: List<Loc>) -> (String -> Loc) takes a List<Loc> and returns a function (which we call get-loc) that can be supplied to dijkstra or campus-tour. We guarantee that this generated function runs in $$O([s \rightarrow 1])$$ where $$s$$ is the number of vertices in the graph.To create a get-loc with this performance, the support code uses a library named string-dict. This uses hashing, which is described in the textbook. Note, however, that you are not permitted to use string-dict. See an example in the template file for proper usage. Below are actual map data for Brown’s campus. You are welcome to play around with these data in the REPL and use them in your tests. However, you should not use these data in your implementation because we might use different graphs when grading. brown-loc-list :: List<Loc> represents all of the locations on the Brown campus map that your program will be processing. You could supply this list to get-loc-maker to obtain a get-loc function, which may be used as an argument for dijkstra or campus-tour. brown-tours :: List<Tour> supplied by the Admissions Office. 4Analysis You should also hand in an analysis document addressing the following questions: 1. Analyze the worst-case running time of your dijkstra function. Show your work, as always! 2. [If you do the extra credit portion] Analyze the worst-case running time of your campus-tour function. Show your work, as always! 3. [If you do the extra credit portion] As is, your campus-tour always visits nearest unvisited selected stop. Is this optimal? That is, does this algorithm create the shortest path visiting all the selected tour stops? If so, explain why this is true. If not, provide a counter-example. 5Built-Ins In addition to the allowed functions in the general guidelines, you can also use sets. 6Template Files Implementation Code Final Tests 7Handing In To submit, return to the Captain Teach assignments page: https://www.captain-teach.org/brown-cs019/assignments/	2017-10-20 12:23:43	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.24995586276054382, "perplexity": 1598.3389621580804}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-43/segments/1508187824104.30/warc/CC-MAIN-20171020120608-20171020140608-00524.warc.gz"}
https://plainmath.net/14070/steps-platform-above-ground-hekeeps-straight-moment-feettouch-ground	# A 75.0-kg man steps off a platform 3.10 m above the ground. Hekeeps his legs straight as he falls, but at the moment his feettouch the ground his knee A 75.0-kg man steps off a platform 3.10 m above the ground. Hekeeps his legs straight as he falls, but at the moment his feettouch the ground his knees begin to bend, and, treated as aparticle, he moves an additional 0.60 m before coming torest. a) what is the speed at the instant his feet touch theground? b) treating him as a particle, what is his acceleration(magnitude and direction) as he slows down, if the acceleration isassumed to be constant? c) draw his free-body diagram (see section 4.6). in termsof forces on the diagram, what is the net force on him? usenewton's laws and the results of part (b) to calculate the averageforce his feet exert on the ground while he slows down. expressthis force in newtons and also as a multiple of his weight. • Questions are typically answered in as fast as 30 minutes ### Solve your problem for the price of one coffee • Math expert for every subject • Pay only if we can solve it Leonard Stokes Here are the detailed explanations: ###### Not exactly what you’re looking for? content_user a) The speed when his feet touch the ground is, $$v=\sqrt{2gh}$$ $$=\sqrt{2(9.80\ m/s^2)(3.10m)}$$ $$=7.8\ m/s$$ b) Use kinematic equation to find the acceleration when he showdown is, $$v^2=u^2+2as$$ since initial speed (u=0) is zero $$v^2=2as$$ $$a=\frac{v^2}{2s}$$ $$=\frac{(7.80\ m/s^2)}{2(0.60\ m)}$$ $$=50.6\ m/s^2$$ c) Since the net force that his feet exert on the ground is same as the force that the ground exerts on his feet. Apply newton second law for net force is, $$F_{m}=ma$$ $$ma=F-mg$$ $$F=m(a+g)$$ $$=75.0\ kg)(50.6\ m/s^2+9.80\ m/s^2)$$ $$=4537.5\ N$$ Fraction of his weigh is, $$\frac{F}{mg}=\frac{m(a+g)}{mg}$$ $$=\frac{a}{g}+1$$ $$=\frac{50.6\ m/s^2}{9.80\ m/s^2}+1$$ $$F=6.16mg$$	2022-01-22 03:35:44	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.664188027381897, "perplexity": 1911.69730517962}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320303729.69/warc/CC-MAIN-20220122012907-20220122042907-00290.warc.gz"}
https://math.stackexchange.com/questions/2030323/is-there-a-more-efficient-method-to-determine-the-last	# Is there a more efficient method to determine the last three digits of $2^{8064}$ ? Here's my method I use two powerful hints : -The first one consists to take the number $\pmod{1000}$ because we want the last three digits. -The second one is a powerful theorem from Euler-Gauss which shows that if we have an integer of the form $p_1^{k_1}...p_l^{k_l}$ with $p_1,...,p_l$ prime numbers then for every integer $a$ we have $a^{({p_1}^{k_1}-{p_1}^{k_1-1})...{(p_l^{k_l}-p_l^{k_l-1})}}\equiv 1 \pmod{p_1^{k_1}...p_l^{k_l}}$ and $\gcd(p_1^{k_1}...p_l^{k_l},a)=1$. Applying these two results I get : $1000=2^3\times5^3$ which means that $2^{4\times 100}\equiv 1 \pmod{1000}\Leftrightarrow 2^{400}\equiv 1 \pmod{1000}$. We find a period for the power of $2$. But the problem is that $\gcd(2,1000)\ne 1$. So it won't work with this theorem and its simple hypothesis. Now using @Joffan's argument on Carmichael, we have the fact that the order of $2$ is $100$. So $2^{100}\equiv 1 \pmod{1000}$. So $2^{100\times 80+64} \equiv 2^{64}\equiv (((2^8)^2)^2)^2 \equiv 616 \pmod{1000}$ However, excepting computing it is there a faster method ? • $2^{64}=2^{8^2}$, but $2^{8^2}\not=(2^8)^2$. – Barry Cipra Nov 25 '16 at 16:10 • @BarryCipra Indeed you're right I re-edit – Maman Nov 25 '16 at 16:13 • Be careful using Euler's theorem here! $2$ and $1000$ aren't coprime, so Euler's theorem is not applicable at all. You need to use the Chinese reminder theorem first. More to the point, $2^4 \not \equiv 1 \pmod{10}$. – Arthur Nov 25 '16 at 16:19 • Further to @BarryCipra 's point, $2^{64} = (2^{32})^2 = ((2^{16})^2)^2 = (((2^8)^2)^2)^2$ – Joffan Nov 25 '16 at 16:19 • @Arthur It means that $2$ is not invertible in $(\mathbb{Z}/1000\mathbb{Z})^{\times}$ ? – Maman Nov 25 '16 at 16:22 You could also use Exponentiation by Squaring, in which you basically notice that $a^{2n}=(a^2)^n$ and $a^{2n+1}=aa^{2n}$. With modulo arithmetic, you can also use the identity $ab\mod n=(a\mod n)(b\mod n)$ You might also get a bonus in speed by using Montgomery Reductions, which just converts the multiplication to an easier form. How about finding reminder of $2^{8061}$ divided by $125$ and then multiply by 8. $2^{8061}=2(5-1)^{4030}=2(125m+{4030\choose 2}25-40305+1)=125k-40300+2 =125n+702\equiv 77 \pmod{125}$ So reminder you're looking for is $778=616$ • OK then we conclude by CRT it's faster indeed... – Maman Nov 25 '16 at 16:48 • Nice. Note that we straightforwardly have $2^{4061}\equiv 2^{61} \bmod 125$ by Euler for smaller numbers. – Joffan Nov 25 '16 at 19:31 • @Joffan true, thank you! – Djura Marinkov Nov 25 '16 at 20:45 The result for the order of $2$ modulo $1000$ is valid, but with conditions - since $2^3 \mid 1000$ but $2^4$ does not, we have that $2^{k+400} \equiv 2^k \bmod 1000$ for $k\ge 3$. The $400$ is Euler's totient $\phi(1000)$; we can actually do better with Carmichael's reduced totient $\lambda(1000)=100$ and get $2^{k+100} \equiv 2^k \bmod 1000$ for $k\ge 3$, although that is not needed on this occasion. Anyway this validates $2^{8064}\equiv 2^{64} \bmod 1000$, since $64>3$, and then we can used exponentiation-by-squaring on $2^8=256$ to get: \begin{align} 2^8 &\equiv 256 \bmod 1000\\ 2^{16} &\equiv 256^2\equiv 65536 \equiv 536 \bmod 1000\\ 2^{32} &\equiv 536^2\equiv 287296 \equiv 296 \bmod 1000\\ 2^{64} &\equiv 296^2\equiv 87616 \equiv 616 \bmod 1000\\ \end{align} So $616$ are your required final digits of $2^{8064}$ • Thank you, so we can use the Euler-Fermat theorem but with very restrictive conditions ! – Maman Nov 25 '16 at 17:14 • So what could be the theorem with this hypothesis ? I tried $6$ and $60$ it fails for instance. – Maman Nov 25 '16 at 19:53 • $6$ is not prime, of course, so you need to look at the component prime powers there. So we need to get to a power of $2$ to saturate factors of $2$. $\phi(60) =16$ , $\lambda(60) = 4$ so we should find that $6^{k+4} \equiv 6^{k} \bmod 60$ for all $\fbox{$k\ge 2$}$. Successive powers of $6, \bmod 60$: $6,36, 36, 36, 36, 36, \ldots$ so the assertion is true. A perhaps-confusing factor is that although the order of $6$ divides $\lambda(60)$, in this case it is: $1$. – Joffan Nov 25 '16 at 20:20	2019-10-19 10:46:42	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.892081081867218, "perplexity": 437.5368270262202}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-43/segments/1570986692723.54/warc/CC-MAIN-20191019090937-20191019114437-00077.warc.gz"}
http://www.blkmage.net/tag/number-theory/	# Suugaku Girl supplementary handout for chapter 2: Prime numbers She doesn't seem that excited So last time, Tetra was being enlightened by MC-kun about definitions. This actually arises from MC-kun using prime numbers as a motivating example. Primes are megas important in mathematics and even more important today. The entire branch of mathematics called number theory is all about studying the properties of prime numbers. They’re so useful that we’ve done stuff like extend the notion of prime elements to algebraic structures called rings or apply analytic techniques to learn more about them, but we’ll stick with elementary number theory for now. Now, for hundreds of years, we’d been studying number theory only because it’s cool and mathematicians love prime numbers. Last time, I mentioned some examples of math preceding useful applications. Well, number theory is a really good example of that, because in the 70s, we found a use for it, which is its main use today, in cryptography. There have been some new techniques using some algebra as well, but for the most part, modern cryptography relies on the hardness of factoring primes. Neat! Okay, so we’re back to the original question that MC-kun tries to get Tetra to answer, which is, what is a prime number? Definition. An integer $p$ is prime if and only if $p\geq 2$ and the only positive divisors of $p$ are 1 and itself. MC-kun explains that the motivation for excluding 1 from the definition of a prime number is because we want to be able to say that we can write every number as a unique product of prime numbers. This is very useful, because now we know we can break down every number like this and we can tell them apart because they’re guaranteed to have a unique representation. This is called unique prime factorization. Theorem. Let $a > 0$ be an integer. Then we can write $a = p_1p_2\cdots p_k$ for some primes $p_1,\dots,p_k$. This representation is unique up to changing the order of terms. We can show this by induction on $a$. We’ve got $a=2$ so that’s pretty obvious. So let’s say that every integer $k\lt a$ can be decomposed like this and suppose we can’t decompose $a$ into prime numbers, assuming $a$ itself isn’t already a prime since it would just be its own prime decomposition. Then we can factor $a=cd$ for some integers $c$ and $d$. But both $c$ and $d$ are less than $a$, which means they can be written as a product of primes, so we just split them up into their primes and multiply them all together to get $a$. Tada. As a sort of side note, I mentioned before that primes are so useful that we wanted to be able to extend the idea of prime elements into rings. Well, it turns out for certain rings, it isn’t necessarily true that numbers will always have a unique representation when decomposed into primes. This is something that comes up in algebraic number theory, which is named so because it involves algebraic structures and techniques. This was invented while we were trying to figure out if Fermat’s Last Theorem was actually true (which needed this and other fun mathematical inventions from the last century that implies that Fermat was full of shit when he said he had a proof). So at the end of the chapter, after Tetra gets her chair kicked over by the megane math girl, we’re treated to a note that acts as a sort of coda to the chapter that mentions that there are infinitely many primes. How do we know this? Suppose that there are only finitely many primes. Then we can just list all of the prime numbers, like on Wikipedia or something. So we’ve got our list of primes $p_1,p_2,\dots,p_k$. So let’s make a number like $N=1+p_1\cdots p_k$. Well, that number is just a regular old number, so we can break it down into its prime factors. We already know all the primes, so it has to be divisible by one of them, let’s say $p_i$. Now we want to consider the greatest common divisor of the two numbers, which is just the largest number that divides both of them. We’ll denote this by $\gcd(a,b)$. So since $p_i$ is a factor of $N$, we’ve got $\gcd(N,p_i)=p_i$. But then that gives us $p_i=\gcd(N,p_i)=\gcd(p_i,1)=1$ by a lemma that says that for $a=qb+r$, we have $\gcd(a,b)=\gcd(b,r)$. This means that we have $p_i=1$, which is a contradiction, since 1 isn’t a prime number, and so I guess there are actually infinitely many primes. So the nice thing is that we won’t run out of prime numbers anytime soon, which is very useful because as we get more and more computing power, we’ll have to increase the size of the keys we use in our cryptosystems. Luckily, because factoring is so hard, we don’t need to increase that size very much before we’re safe for a while. Or at least until we develop practical quantum computers. # The (1st annual?) π day anime and mathematics lecture Happy $\pi$ day. We’ll begin with an obligatory video. One of the reasons I enjoyed Summer Wars so much is because the main character’s superpower is math. Well, okay, you say, he’s really good at math, but so what? A lot of people complain about the implausibility of OZ, but those of us with a basic understanding of cryptography and number theory will have been drawn to Kenji’s quick problem solving work with an eyebrow raised. So let’s talk about why Kenji is a wizard. Kenji doing some friggin mathematics We’ll start with modular arithmetic, which Kenji mentions to Natsuki on the train ride to Ueda. When we divide numbers, we often end up with remainders. Suppose we divide some integer $k$ by $N$ and we get a remainder of $r$. Then we say that $k$ and $r$ are equivalent $\bmod{N}$ and we denote that by $k = r \bmod{N}$. Because it’s how division works, for any integer $k, r$ will be some number from $0$ to $N-1$. It turns out a lot of arithmetic operations work the same way in modular arithmetic: adding, subtracting, and multiplying numbers and then taking the modulus of the result will give you the same number as adding, subtracting, and multiplying the moduli of the numbers you started out with. However, division doesn’t work as we would expect it to. So we have to think about division (or the equivalent operation) differently. Instead of thinking of division as splitting a group of stuff into smaller groups, we’ll think of it as multiplying by an inverse. What’s an inverse? Well, we can try thinking of it in terms of addition. It’s pretty intuitive that subtraction is the opposite of addition. If we have some integer $k$, then the additive inverse of $k$ is $-k$. When we add $k$ and $-k$, we get $0$, the additive identity. The identity is just the special number that we can add to anything and get that same thing back unchanged ($n+0 = n$). In the same way, if we multiply $k$ by its inverse, $k^{-1}$, then we’ll get $1$, since $k \times 1$ is just $k$ again. What this means is that the inverse of $k \bmod{N}$ is just some other number $j$ from $0$ to $N-1$ such that $j\cdot k = 1 \bmod{N}$ and it’s just multiplication again. Now, the problem with this is that it’s not guaranteed that there’s always an inverse hanging around in $\bmod{N}$. In particular, if $k$ and $N$ share any divisors, then $k$ won’t have an inverse $\bmod{N}$. This is interesting because it also tells us that if we consider integers mod a prime number $P$, then every integer $\bmod{P}$ has an inverse, since $P$ doesn’t share any divisors with any integers from $0$ to $P-1$. We call these things that have inverses units. So if we have a unit $k$, then $k^m$ is also a unit, for any integer $m$. We even have a funny function $\phi$ defined such that $\phi(n)$ is the number of units in $\bmod{n}$. Love Machine solicits help So taking everything we’ve learned, we can set up a cryptosystem! The one we’ll be looking at is called RSA, after the guys who invented it. We have Bob who wants to securely send a message $M$ to Alice. Alice chooses two prime numbers $p$ and $q$ and figures out $m = pq$. She also goes and figures out $\phi(m)$, which happens to be $(p-1)(q-1)$. Finally, she picks some integer $k$, a unit in $\bmod{\phi(m)}$. She lets everyone know $m$ and $k$, but she keeps $p$, $q$, and $\phi(m)$ secret. So Bob wants to send $M$, which is just his message conveniently in number form. He makes $M$ into a number between $0$ and $m$, and if $M$ is too big, he can just break it up into chunks. Bob figures out the smallest $b$ such that $b = a^k \bmod{m}$ and sends $b$ over to Alice. Now, since Alice has $k$ and $\phi(m)$, she can also find $k^{-1}$ pretty easily. Once she has that, she can get the original message by figuring out $b^{k^{-1}} = (M^k)^{k{-1}} = M \bmod{m}$, since $kk^{-1} = 1 \bmod \phi(m)$. The interesting thing here is that all of the information is out there for someone to encrypt a message to send to Alice, but no one is able to decrypt it. Well, they’re able to decrypt it if they know what $p$ and $q$ are, since once they’ve got that, they can get $\phi(m)$. But it turns out getting $p$ and $q$ from $m$ (which Alice just throws up on the interwebs) is really hard. And it really works for reals, because RSA is pretty widely deployed for things like keeping your credit card information safe while you send it through the tubes to Amazon. A conveniently displayed ciphertext Let’s go back and think about units some more. Of course, there are only $N$ numbers in the integers $\bmod{N}$, so there’s a point at which $k^m$ is just $1$ again and starts over. If $k^m = 1 \bmod{N}$, we say that $m$ is the order of $k$. But why do we care about finding the order of $k$? It turns out finding the order of elements is very, very similar to factoring an integer into primes and other related problems, like discrete logarithms. If we can find orders of elements, it won’t be too hard to figure out how to factor a number. In this case, the eavesdropper wants to figure out what $p$ and $q$ are, so they’ll want to factor $m$. And it turns out a lot of other public-key cryptosystems (like elliptic curves) are based on the difficulty of factoring. How hard could it be? Well, we could just check every possibility, which doesn’t seem that bad for a number like 48, but once we get into numbers that are hundreds of digits long, that might start to suck. It turns out the fastest known algorithms for order finding take approximately $e^{O(\log N \log \log N)^{\frac{1}{2}}}$ steps. Current key lengths for RSA are at least 1024 bits, which would give us about 4.4 x 1029 operations. Assuming three trillion operations per second (3 GHz), it’d take a PC about 4.7 billion years. Sure, you could just throw more powerful computers at it, but they’d just double the key size and suddenly, you’d need to do 1044 operations. It's a lot easier to write math than typeset it Well, that’s not entirely true. One of the breakthroughs in quantum computing was coming up with a fast algorithm for factoring. It turns out quantum order finding takes $O((\log N)^2 \log \log N \log \log \log N)$ steps, which, for a 1024-bit key is just over 60 operations. Doubling the key-size to 2048 bits only increases the number of operations by just over 20. Unfortunately (or fortunately, because we’d be pretty screwed if someone could easily break RSA right now), we haven’t built any quantum computers that large yet, nor are we capable of doing so anytime soon. tl;dr – Kenji is a quantum computer.	2018-06-23 02:12:03	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7994477152824402, "perplexity": 207.40378009069317}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-26/segments/1529267864919.43/warc/CC-MAIN-20180623015758-20180623035758-00144.warc.gz"}
https://adventures.michaelfbryan.com/posts/rust-best-practices/bad-habits/	When you are coming to Rust from another language you bring all your previous experiences with you. Often this is awesome because it means you aren’t learning programming from scratch! However, you can also bring along bad habits which can lead you down the wrong rabbit hole or make you write bad code. This one is a pet peeve of mine. In most C-based languages (C, C#, Java, etc.), the way you indicate whether something failed or couldn’t be found is by returning a “special” value. For example, C#’s String.IndexOf() method will scan an array for a particular element and return its index. Returning -1 if nothing is found. That leads to code like this: string sentence = "The fox jumps over the dog"; int index = sentence.IndexOf("fox"); if (index != -1) { string wordsAfterFox = sentence.SubString(index); Console.WriteLine(wordsAfterFox); } You see this sort of “use a sentinel value to indicate something special” practice all the time. Other sentinel values you might find in the wild are "", or null (someone once referred to this as their “billion-dollar mistake”). The general reason why this is a bad idea is that there is absolutely nothing to stop you from forgetting that check. That means you can accidentally crash your application with one misplaced assumption or when the code generating the sentinel is far away from the code using it. We can do a lot better in Rust, though. Just use Option! By design, there is no way to get the underlying value without dealing with the possibility that your Option may be None. This is enforced by the compiler at compile time, meaning code that forgets to check won’t even compile. let sentence = "The fox jumps over the dog"; let index = sentence.find("fox"); // let words_after_fox = &sentence[index..]; // Error: Can't index str with Option<usize> if let Some(fox) = index { let words_after_fox = &sentence[fox..]; println!("{}", words_after_fox); } (playground) Back in the 70’s, a naming convention called Hungarian Notation was developed by programmers writing in languages where variables are untyped or dynamically typed. It works by adding a mnemonic to the start of a name to indicate what it represents, for example the boolean visited variable might be called bVisited or the string name might be called strName. You can still see this naming convention in languages Delphi where classes (types) start with T, fields start with F, arguments start with A, and so on. type TKeyValue = class private FKey: integer; FValue: TObject; public property Key: integer read FKey write FKey; property Value: TObject read FValue write FValue; function Frobnicate(ASomeArg: string): string; end; C# also has a convention that all interfaces should start with I, meaning programmers coming to Rust from C# will sometimes prefix their traits with I as well. trait IClone { fn clone(&self) -> Self; } In this case, just drop the leading I. Rust’s syntax guarantees that it just isn’t possible to confuse a trait for a normal type, so it isn’t helping anyone. This is in contrast with C# where interfaces and classes are largely interchangeable. This is also seen inside functions where people will conjure up new names for something as they convert it from one form to another. Often these names are silly or contrived, providing negligible additional information to the reader. let account_bytes: Vec<u8> = read_some_input(); let account_str = String::from_utf8(account_bytes)?; let account: Account = account_str.parse()?; I mean, if we’re calling String::from_utf8() we already know account_str will be a String so why add the _str suffix? Unlike a lot of other languages, Rust encourages shadowing variables when you are transforming them from one form to another, especially when the previous variable is no longer accessible (e.g. because it’s been moved). let account: Vec<u8> = read_some_input(); let account = String::from_utf8(account)?; let account: Account = account.parse()?; This is arguably superior because we can use the same name for the same concept. Other languages frown on shadowing because it can be easy to lose track of what type a variable contains (e.g. in a dynamically typed language like JavaScript) or you can introduce bugs where the programmer thinks a variable has one type but it actually contains something separate. Neither of these is particularly relevant to a strongly typed language with move semantics like Rust, so you can use shadowing freely without worrying about shooting yourself in the foot. ## An Abundance of Rc<RefCell<T>> Link to heading A common pattern in Object Oriented languages is to accept a reference to some object so you can call its methods later on. On its own there is nothing wrong with this, Dependency Injection is a very good thing to do, but unlike most OO languages Rust doesn’t have a garbage collector and has strong feelings on shared mutability. Perhaps this will be easier to understand with an example. Say we are implementing a game where the player needs to beat up a bunch of monsters until they have inflicted a certain amount of damage (I dunno, maybe it’s for a quest or something). We create a Monster class which has a health property and a takeDamage() method, and so we can keep track of how much damage has been inflicted we’ll let people provide callbacks that get called whenever the monster receives damage. type OnReceivedDamage = (damageReceived: number) => void; class Monster { health: number = 50; takeDamage(amount: number) { amount = Math.min(this.health, amount); this.health -= amount; } on(event: "damaged", callback: OnReceivedDamage): void { } } Let’s also create a DamageCounter class which tracks how much damage we’ve inflicted and lets us know when that goal is reached. class DamageCounter { damageInflicted: number = 0; reachedTargetDamage(): boolean { return this.damageInflicted > 100; } onDamageInflicted(amount: number) { this.damageInflicted += amount; } } Now we’ll create some monsters and keep inflicting a random amount of damage until the DamageCounter is happy. const counter = new DamageCounter(); const monsters = [new Monster(), new Monster(), new Monster(), new Monster(), new Monster()]; monsters.forEach(m => m.on("damaged", amount => counter.onDamageInflicted(amount))); while (!counter.reachedTargetDamage()) { // pick a random monster const index = Math.floor(Math.random()monsters.length); const target = monsters[index]; // then damage it a bit const damage = Math.round(Math.random() 50); target.takeDamage(damage); console.log(Monster ${index} received${damage} damage); } (TypeScript Playground) Now let’s port this code to Rust. Our Monster struct is fairly similar, although we need to use Box<dyn Fn(u32)> for a closure which accepts a single u32 argument (all closures in JavaScript are heap allocated by default). type OnReceivedDamage = Box<dyn Fn(u32)>; struct Monster { health: u32, } impl Monster { fn take_damage(&mut self, amount: u32) { for callback in &mut self.received_damage { } } } } impl Default for Monster { fn default() -> Self { Monster { health: 100, received_damage: Vec::new() } } } Next comes our DamageCounter, nothing interesting here. #[derive(Default)] struct DamageCounter { damage_inflicted: u32, } impl DamageCounter { fn reached_target_damage(&self) -> bool { self.damage_inflicted > 100 } fn on_damage_received(&mut self, damage: u32) { self.damage_inflicted += damage; } } And finally our code that inflicts damage. fn main() { let mut counter = DamageCounter::default(); let mut monsters: Vec<_> = (0..5).map(\|_\| Monster::default()).collect(); for monster in &mut monsters { } while !counter.reached_target_damage() { let index = rng.gen_range(0..monsters.len()); let target = &mut monsters[index]; let damage = rng.gen_range(0..50); target.take_damage(damage); println!("Monster {} received {} damage", index, damage); } } (playground) But herein lies our first problem, when we try to compile the code rustc gives us not one, but four compile errors for the monster.add_listener() line 🤣 error[E0596]: cannot borrow counter as mutable, as it is a captured variable in a Fn closure --> src/main.rs:47:48 \| \| ^^^^^^^ cannot borrow as mutable error[E0499]: cannot borrow counter as mutable more than once at a time --> src/main.rs:47:39 \| \| ---------^^^^^^^^------------------------------------ \| \| \| \| \| \| \| borrows occur due to use of counter in closure \| \| counter was mutably borrowed here in the previous iteration of the loop \| cast requires that counter is borrowed for 'static error[E0597]: counter does not live long enough --> src/main.rs:47:48 \| \| ------------------^^^^^^^---------------------------- \| \| \| \| \| \| \| borrowed value does not live long enough \| \| value captured here \| cast requires that counter is borrowed for 'static ... 60 \| } \| - counter dropped here while still borrowed error[E0502]: cannot borrow counter as immutable because it is also borrowed as mutable --> src/main.rs:50:12 \| \| ----------------------------------------------------- \| \| \| \| \| \| \| first borrow occurs due to use of counter in closure \| \| mutable borrow occurs here \| cast requires that counter is borrowed for 'static ... 50 \| while !counter.reached_target_damage() { \| ^^^^^^^ immutable borrow occurs here There are a number of things wrong with this line, but it can be boiled down to: • The closure captures a reference to counter • The counter.on_damage_received() method takes &mut self so our closure needs a &mut reference. We add the closures in a loop so we end up taking multiple &mut references to the same object at the same time • Our listener is a boxed closure without any lifetime annotations, meaning it needs to own any variables it closes over. We would need to move the counter into the closure, but because we do this in a loop we’ll have a “use of moved value” error • After passing the counter to add_listener() we try to use it in our loop condition Overall it’s just a bad situation. The canonical answer to this is to wrap the DamageCounter in a reference-counted pointer so we can have multiple handles to it at the same time, then because we need to call a &mut self method we also need a RefCell to “move” the borrow checking from compile time to run time. fn main() { - let mut counter = DamageCounter::default(); + let mut counter = Rc::new(RefCell::new(DamageCounter::default())); let mut monsters: Vec<_> = (0..5).map(\|_\| Monster::default()).collect(); for monster in &mut monsters { + let counter = Rc::clone(&counter); + })); } - while !counter.reached_target_damage() { + while !counter.borrow().reached_target_damage() { let index = rng.gen_range(0..monsters.len()); let target = &mut monsters[index]; ... } } (playground) Well… it works. But this approach tends to get messy, especially when you are storing non-trivial things like a Rc<RefCell<Vec<Foo>>>> (or its multi-threaded cousin Arc<Mutex<Vec<Foo>>>>) inside structs 1. It also opens you up to situations where the RefCell might be borrowed mutably multiple times because your code is complex and something higher up in the call stack is already using the RefCell. With a Mutex this will cause a deadlock while the RefCell will panic, neither of which is conducive to a reliable program. A much better approach is to change your API to not hold long-lived references to other objects. Depending on the situation, it might make sense to take a callback argument in the Monster::take_damage() method. struct Monster { health: u32, } impl Monster { fn take_damage(&mut self, amount: u32, on_damage_received: impl FnOnce(u32)) { } } impl Default for Monster { fn default() -> Self { Monster { health: 100 } } } ... fn main() { let mut counter = DamageCounter::default(); let mut monsters: Vec<_> = (0..5).map(\|_\| Monster::default()).collect(); while !counter.reached_target_damage() { let index = rng.gen_range(0..monsters.len()); let target = &mut monsters[index]; let damage = rng.gen_range(0..50); println!("Monster {} received {} damage", index, damage); } } (playground) A nice side-effect of this is that we get rid of all the callback management boilerplate, meaning this version is only 47 lines long instead of the Rc<RefCell<_>> version’s 62. Other times it may not be acceptable to give take_damage() a callback parameter, in which case you could return a “summary” of what happened so the caller can decide what to do next. impl Monster { fn take_damage(&mut self, amount: u32) -> AttackSummary { } } struct AttackSummary { } ... fn main() { let mut counter = DamageCounter::default(); let mut monsters: Vec<_> = (0..5).map(\|_\| Monster::default()).collect(); while !counter.reached_target_damage() { let index = rng.gen_range(0..monsters.len()); let target = &mut monsters[index]; let damage = rng.gen_range(0..50); let AttackSummary { damage_received } = target.take_damage(damage); println!("Monster {} received {} damage", index, damage); } } (playground) This is my preferred solution; from experience, it tends to work well for larger codebases or when the code is more complex. Another hang-over from writing a lot of C is using the wrong integer type and getting frustrated because you need to cast to/from usize all the time. I’ve seen people run into this so many times in the wild, especially when indexing. The underlying problem is that C programmers are all taught to use int for indexing and for-loops, so when they come to Rust and they need to store a list of indices, the programmer will immediately reach for a Vec<i32>. They then get frustrated because Rust is quite strict when it comes to indexing and standard types like arrays, slices, and Vec can only be indexed using usize (the equivalent of size_t), meaning their code is cluttered with casts from i32 to usize and back again. There are a number of perfectly legitimate reasons for why Rust only allows indexing by usize: • It doesn’t make sense to have a negative index (accessing items before the start of a slice is UB), so we can avoid an entire class of bugs by indexing with an unsigned integer • A usize is defined to be an integer with the same size as a normal pointer, meaning the pointer arithmetic won’t have any hidden casts • The std::mem::size_of() and std::mem::align_of() functions return usize Of course, when stated this way the solution is clear. Choose the right integer type for your application but when you are doing things that eventually be used for indexing, that “right integer type” is probably usize. <rant> There’s an old Rust koan on the User Forums by Daniel Keep that comes to mind every time I see a grizzled C programmer reach for raw pointers or std::mem::transmute() because the borrow checker keeps rejecting their code: Obstacles. You should go read it. It’s okay, I’ll wait. Too often you see people wanting to hack around privacy, create self-referencing structs, or create global mutable variables using unsafe. Frequently this will be accompanied by comments like “but I know this program will only use a single thread so accessing the static mut is fine” or “but this works perfectly fine in C”. The reality is that unsafe code is nuanced and you need to have a good intuition for Rust’s borrow checking rules and memory model. I hate to be a gate keeper and say “you must be this tall to write multi-threaded unsafe code” 2, but there’s a good chance that if you are new to the language you won’t have this intuition and are opening yourself and your colleagues up to a lot of pain. It’s fine to play around with unsafe if you are trying to learn more about Rust or you know what you are doing and are using it legitimately, but unsafe is not a magical escape hatch which will make the compiler stop complaining and let you write C with Rust syntax. </rant> A common practice in C is to prefix functions with the name of the library or module to help readers understand where it comes from and avoid duplicate symbol errors (e.g. rune_wasmer_runtime_load()). However, Rust has real namespaces and lets you attach methods to types (e.g. rune::wasmer::Runtime::load()). Just use them - it’s what they are there for. The for-loop and indexing is the bread and butter for most C-based languages. let points: Vec<Coordinate> = ...; let differences = Vec::new(); for i in 1..points.len() [ let current = points[i]; let previous = points[i-1]; differences.push(current - previous); ] (playground) However, it’s easy to accidentally introduce an off-by-one error when using indexing (e.g. I needed to remember to start looping from 1 and subtract 1 to get the previous point) and even seasoned programmers aren’t immune from crashing due to an index-out-of-bounds error. In situations like these, Rust encourages you to reach for iterators instead. The slice type even comes with high-level tools like the windows() and array_windows() methods to let you iterate over adjacent pairs of elements. let points: Vec<Coordinate> = ...; let mut differences = Vec::new(); for [previous, current] in points.array_windows().copied() { differences.push(current - previous); } (playground) You could even remove the for-loop and mutation of differences altogether. let differences: Vec<_> = points .array_windows() .copied() .map(\|[previous, current]\| current - previous) .collect(); (playground) Some would argue the version with map() and collect() is cleaner or more “functional”, but I’ll let you be the judge there. As a bonus, iterators can often allow better performance because checks can be done as part of the looping condition instead of being separate3 (Alice has a good explanation here). Once you start drinking the Kool-Aid that is Rust’s iterators you can run into the opposite problem - when all you have is a hammer everything looks like a nail. Long chains of map(), filter(), and and_then() calls can get quite hard to read and keep track of what is actually going on, especially when type inference lets you omit a closure argument’s type. Other times your iterator-based solution is just unnecessarily complicated. As an example, have a look at this snippet of code and see if you can figure out what it is trying to do. pub fn functional_blur(input: &Matrix) -> Matrix { assert!(input.width >= 3); assert!(input.height >= 3); // Stash away the top and bottom rows so they can be // directly copied across later let mut rows = input.rows(); let first_row = rows.next().unwrap(); let last_row = rows.next_back().unwrap(); let top_row = input.rows(); let middle_row = input.rows().skip(1); let bottom_row = input.rows().skip(2); let blurred_elements = top_row .zip(middle_row) .zip(bottom_row) .flat_map(\|((top, middle), bottom)\| blur_rows(top, middle, bottom)); let elements: Vec<f32> = first_row .iter() .copied() .chain(blurred_elements) .chain(last_row.iter().copied()) .collect(); Matrix::new_row_major(elements, input.width, input.height) } fn blur_rows<'a>( top_row: &'a [f32], middle_row: &'a [f32], bottom_row: &'a [f32], ) -> impl Iterator<Item = f32> + 'a { // stash away the left-most and right-most elements so they can be copied across directly. let &first = middle_row.first().unwrap(); let &last = middle_row.last().unwrap(); // Get the top, middle, and bottom row of our 3x3 sub-matrix so they can be // averaged. let top_window = top_row.windows(3); let middle_window = middle_row.windows(3); let bottom_window = bottom_row.windows(3); // slide the 3x3 window across our middle row so we can get the average // of everything except the left-most and right-most elements. let averages = top_window .zip(middle_window) .zip(bottom_window) .map(\|((top, middle), bottom)\| top.iter().chain(middle).chain(bottom).sum::<f32>() / 9.0); std::iter::once(first) .chain(averages) .chain(std::iter::once(last)) } (playground) Believe it or not, but that’s one of the more readable versions I’ve seen… Now let’s look at the imperative implementation. pub fn imperative_blur(input: &Matrix) -> Matrix { assert!(input.width >= 3); assert!(input.height >= 3); // allocate our output matrix, copying from the input so // we don't need to worry about the edge cases. let mut output = input.clone(); for y in 1..(input.height - 1) { for x in 1..(input.width - 1) { let mut pixel_value = 0.0; pixel_value += input[[x - 1, y - 1]]; pixel_value += input[[x, y - 1]]; pixel_value += input[[x + 1, y - 1]]; pixel_value += input[[x - 1, y]]; pixel_value += input[[x, y]]; pixel_value += input[[x + 1, y]]; pixel_value += input[[x - 1, y + 1]]; pixel_value += input[[x, y + 1]]; pixel_value += input[[x + 1, y + 1]]; output[[x, y]] = pixel_value / 9.0; } } output } (playground) I know which version I prefer. In most other mainstream languages it is quite common to see the programmer write a check before they do an operation which may throw an exception. Our C# IndexOf() snippet from earlier is a good example of this: int index = sentence.IndexOf("fox"); if (index != -1) { string wordsAfterFox = sentence.SubString(index); Console.WriteLine(wordsAfterFox); } Closer to home, you might see code like this: let opt: Option<_> = ...; if opt.is_some() { let value = opt.unwrap(); ... } or this: let list: &[f32] = ...; if !list.is_empty() { let first = list[0]; ... } Now both snippets are perfectly valid pieces of code and will never fail, but similar to sentinel values you are making it easy for future refactoring to introduce a bug. Using things like pattern matching and Option help you avoid this situation by making sure the only way you can access a value is if it is valid. if let Some(value) = opt { ... } if let [first, ..] = list { ... } Depending on where it is used and how smart LLVM or your CPU’s branch predictor are, this may also generate slower code because the fallible operation (opt.unwrap() or list[index] in that example) needs to do unnecessary checks 3. In many languages, it is normal to call an object’s constructor and initialize its fields afterward (either manually or by calling some init() method). However, this goes against Rust’s general convention of “make invalid states unrepresentable”. Say you are writing an NLP application and have a dictionary containing all the possible words you can handle. This is one way you could create the dictionary: let mut dict = Dictionary::new(); // read the file and populate some internal HashMap or Vec However, writing Dictionary this way means it now has two (hidden) states - empty and populated. All downstream code that uses the Dictionary will assume it’s been populated already and write code accordingly. This may include doing things like indexing into the dictionary with dict["word"] which may panic if "word" isn’t there. Now you’ve opened yourself up to a situation where passing an empty dictionary to code that expects a populated dictionary may trigger a panic. But that’s completely unnecessary. Just make sure the Dictionary is usable immediately after constructing it instead of populating it after the fact. let dict = Dictionary::from_file("./words.txt")?; impl Dictionary { fn from_file(filename: impl AsRef<Path>) -> Result<Self, Error> { let mut words = Vec::new(); for line in text.lines() { words.push(line); } Ok(Dictionary { words }) } } Internally the Dictionary::from_file() might create an empty Vec and populate it incrementally, but it won’t be stored in the Dictionary’s words field yet so there is no assumption that it is populated and useful. How frequently you fall into this anti-pattern depends a lot on your background and coding style. Functional languages are often completely immutable so you’ll fall into the idiomatic pattern naturally. After all, it’s kinda hard to create a half-initialized thing and populate it later when you aren’t allowed to mutate anything. On the other hand, OO languages are much happier to let you initialize an object after it has been constructed, especially because object references can be null by default and they have no qualms about mutability… You could argue this contributes to why OO languages have a propensity for crashing due to an unexpected NullPointerException. To point out the obvious, a really nice property of immutable objects is that you can rely on them to never change. However, in languages like Python and Java, immutability isn’t transitive - i.e. if x is an immutable object, x.y isn’t guaranteed to be immutable unless it was explicitly defined that way. This means it’s possible to write code like this… class ImmutablePerson: def __init__(self, name: str, age: int, addresses: List[str]): self._name = name self._age = age @property def name(self): return self._name @property def age(self): return self._age @property Then someone else comes along and accidentally messes up the address list as part of their normal code. def send_letters(message: str, addresses: List[str]): # Note: the post office's API only works with with uppercase letters so we # need to pre-process the address list client = PostOfficeClient() person = ImmutablePerson("Joe Bloggs", 42, ["123 Fake Street"]) send_letters( ) While I admit the example is a bit contrived, it’s not uncommon for functions to modify the arguments they are given. Normally this is fine, but when your ImmutablePerson assumes its addresses field will never change, it’s annoying for some random piece of code on the other side of the project to modify it without you knowing. The typical solution to this is to preemptively copy the list so even if the caller tries to mutate its contents, they’ll be mutating a copy and not the original addresses field. class ImmutablePerson: ... @property In general, you’ll see defensive copies being used anywhere code wants to be sure that another piece of code won’t modify some shared object at an inopportune time. Considering this is an article about Rust, you’ve probably guessed what the root cause of this is - a combination of aliasing and mutation. You’ve also probably guessed why defensive copies aren’t really necessary when writing Rust code - lifetimes and the “shared immutable XOR single mutable” rule for references means it just isn’t possible for code to modify something without first asking its original owner for mutable access or explicitly opting into shared mutation by using a type like std::sync::Mutex<T>. 1. Out of curiosity, how many people noticed there are 4 >’s in Rc<RefCell<Vec<Foo>>>> but only 3 <’s? ↩︎	2023-02-08 20:00:42	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 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": 0.2040858268737793, "perplexity": 4970.241109102977}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764500904.44/warc/CC-MAIN-20230208191211-20230208221211-00293.warc.gz"}
https://openstax.org/books/elementary-algebra/pages/1-4-multiply-and-divide-integers	Elementary Algebra 1.4Multiply and Divide Integers Elementary Algebra1.4 Multiply and Divide Integers Learning Objectives By the end of this section, you will be able to: • Multiply integers • Divide integers • Simplify expressions with integers • Evaluate variable expressions with integers • Translate English phrases to algebraic expressions • Use integers in applications Be Prepared 1.4 A more thorough introduction to the topics covered in this section can be found in the Prealgebra chapter, Integers. Multiply Integers Since multiplication is mathematical shorthand for repeated addition, our model can easily be applied to show multiplication of integers. Let’s look at this concrete model to see what patterns we notice. We will use the same examples that we used for addition and subtraction. Here, we will use the model just to help us discover the pattern. We remember that $a·ba·b$ means add a, b times. Here, we are using the model just to help us discover the pattern. The next two examples are more interesting. What does it mean to multiply 5 by $−3?−3?$ It means subtract 5, 3 times. Looking at subtraction as “taking away,” it means to take away 5, 3 times. But there is nothing to take away, so we start by adding neutral pairs on the workspace. Then we take away 5 three times. In summary: $5·3=15−5(3)=−155(−3)=−15(−5)(−3)=155·3=15−5(3)=−155(−3)=−15(−5)(−3)=15$ Notice that for multiplication of two signed numbers, when the: • signs are the same, the product is positive. • signs are different, the product is negative. We’ll put this all together in the chart below. Multiplication of Signed Numbers For multiplication of two signed numbers: Same signs Product Example Two positives Two negatives Positive Positive $7·4=28−8(−6)=487·4=28−8(−6)=48$ Different signs Product Example Positive · negative Negative · positive Negative Negative $7(−9)=−63−5·10=−507(−9)=−63−5·10=−50$ Example 1.46 Multiply: $−9·3−9·3$ $−2(−5)−2(−5)$ $4(−8)4(−8)$ $7·6.7·6.$ Try It 1.91 Multiply: $−6·8−6·8$ $−4(−7)−4(−7)$ $9(−7)9(−7)$ $5·12.5·12.$ Try It 1.92 Multiply: $−8·7−8·7$ $−6(−9)−6(−9)$ $7(−4)7(−4)$ $3·13.3·13.$ When we multiply a number by 1, the result is the same number. What happens when we multiply a number by $−1?−1?$ Let’s multiply a positive number and then a negative number by $−1−1$ to see what we get. $−1·4−1(−3)Multiply.−43−4is the opposite of4.3is the opposite of−3.−1·4−1(−3)Multiply.−43−4is the opposite of4.3is the opposite of−3.$ Each time we multiply a number by $−1,−1,$ we get its opposite! Multiplication by $−1−1$ $−1a=−a−1a=−a$ Multiplying a number by $−1−1$ gives its opposite. Example 1.47 Multiply: $−1·7−1·7$ $−1(−11).−1(−11).$ Try It 1.93 Multiply: $−1·9−1·9$ $−1·(−17).−1·(−17).$ Try It 1.94 Multiply: $−1·8−1·8$ $−1·(−16).−1·(−16).$ Divide Integers What about division? Division is the inverse operation of multiplication. So, $15÷3=515÷3=5$ because $15·3=5.15·3=5.$ In words, this expression says that 15 can be divided into three groups of five each because adding five three times gives 15. Look at some examples of multiplying integers, to figure out the rules for dividing integers. $5·3=15so15÷3=5−5(3)=−15so−15÷3=−5(−5)(−3)=15so15÷(−3)=−55(−3)=−15so−15÷(−3)=55·3=15so15÷3=5−5(3)=−15so−15÷3=−5(−5)(−3)=15so15÷(−3)=−55(−3)=−15so−15÷(−3)=5$ Division follows the same rules as multiplication! For division of two signed numbers, when the: • signs are the same, the quotient is positive. • signs are different, the quotient is negative. And remember that we can always check the answer of a division problem by multiplying. Multiplication and Division of Signed Numbers For multiplication and division of two signed numbers: • If the signs are the same, the result is positive. • If the signs are different, the result is negative. Same signs Result Two positives Two negatives Positive Positive If the signs are the same, the result is positive. Different signs Result Positive and negative Negative and positive Negative Negative If the signs are different, the result is negative. Example 1.48 Divide: $−27÷3−27÷3$ $−100÷(−4).−100÷(−4).$ Try It 1.95 Divide: $−42÷6−42÷6$ $−117÷(−3).−117÷(−3).$ Try It 1.96 Divide: $−63÷7−63÷7$ $−115÷(−5).−115÷(−5).$ Simplify Expressions with Integers What happens when there are more than two numbers in an expression? The order of operations still applies when negatives are included. Remember My Dear Aunt Sally? Let’s try some examples. We’ll simplify expressions that use all four operations with integers—addition, subtraction, multiplication, and division. Remember to follow the order of operations. Example 1.49 Simplify: $7(−2)+4(−7)−6.7(−2)+4(−7)−6.$ Try It 1.97 Simplify: $8(−3)+5(−7)−4.8(−3)+5(−7)−4.$ Try It 1.98 Simplify: $9(−3)+7(−8)−1.9(−3)+7(−8)−1.$ Example 1.50 Simplify: $(−2)4(−2)4$ $−24.−24.$ Try It 1.99 Simplify: $(−3)4(−3)4$ $−34.−34.$ Try It 1.100 Simplify: $(−7)2(−7)2$ $−72.−72.$ The next example reminds us to simplify inside parentheses first. Example 1.51 Simplify: $12−3(9−12).12−3(9−12).$ Try It 1.101 Simplify: $17−4(8−11).17−4(8−11).$ Try It 1.102 Simplify: $16−6(7−13).16−6(7−13).$ Example 1.52 Simplify: $8(−9)÷(−2)3.8(−9)÷(−2)3.$ Try It 1.103 Simplify: $12(−9)÷(−3)3.12(−9)÷(−3)3.$ Try It 1.104 Simplify: $18(−4)÷(−2)3.18(−4)÷(−2)3.$ Example 1.53 Simplify: $−30÷2+(−3)(−7).−30÷2+(−3)(−7).$ Try It 1.105 Simplify: $−27÷3+(−5)(−6).−27÷3+(−5)(−6).$ Try It 1.106 Simplify: $−32÷4+(−2)(−7).−32÷4+(−2)(−7).$ Evaluate Variable Expressions with Integers Remember that to evaluate an expression means to substitute a number for the variable in the expression. Now we can use negative numbers as well as positive numbers. Example 1.54 When $n=−5,n=−5,$ evaluate: $n+1n+1$ $−n+1.−n+1.$ Try It 1.107 When $n=−8,n=−8,$ evaluate $n+2n+2$ $−n+2.−n+2.$ Try It 1.108 When $y=−9,y=−9,$ evaluate $y+8y+8$ $−y+8.−y+8.$ Example 1.55 Evaluate $(x+y)2(x+y)2$ when $x=−18x=−18$ and $y=24.y=24.$ Try It 1.109 Evaluate $(x+y)2(x+y)2$ when $x=−15x=−15$ and $y=29.y=29.$ Try It 1.110 Evaluate $(x+y)3(x+y)3$ when $x=−8x=−8$ and $y=10.y=10.$ Example 1.56 Evaluate $20−z20−z$ when $z=12z=12$ and $z=−12.z=−12.$ Try It 1.111 Evaluate: $17−k17−k$ when $k=19k=19$ and $k=−19.k=−19.$ Try It 1.112 Evaluate: $−5−b−5−b$ when $b=14b=14$ and $b=−14.b=−14.$ Example 1.57 Evaluate: $2x2+3x+82x2+3x+8$ when $x=4.x=4.$ Try It 1.113 Evaluate: $3x2−2x+63x2−2x+6$ when $x=−3.x=−3.$ Try It 1.114 Evaluate: $4x2−x−54x2−x−5$ when $x=−2.x=−2.$ Translate Phrases to Expressions with Integers Our earlier work translating English to algebra also applies to phrases that include both positive and negative numbers. Example 1.58 Translate and simplify: the sum of 8 and $−12,−12,$ increased by 3. Try It 1.115 Translate and simplify the sum of 9 and $−16,−16,$ increased by 4. Try It 1.116 Translate and simplify the sum of $−8−8$ and $−12,−12,$ increased by 7. When we first introduced the operation symbols, we saw that the expression may be read in several ways. They are listed in the chart below. $a−ba−b$ $aa$ minus $bb$ the difference of $aa$ and $bb$ $bb$ subtracted from $aa$ $bb$ less than $aa$ Be careful to get a and b in the right order! Example 1.59 Translate and then simplify the difference of 13 and $−21−21$ subtract 24 from $−19.−19.$ Try It 1.117 Translate and simplify the difference of 14 and $−23−23$ subtract 21 from $−17.−17.$ Try It 1.118 Translate and simplify the difference of 11 and $−19−19$ subtract 18 from $−11.−11.$ Once again, our prior work translating English to algebra transfers to phrases that include both multiplying and dividing integers. Remember that the key word for multiplication is “product” and for division is “quotient.” Example 1.60 Translate to an algebraic expression and simplify if possible: the product of $−2−2$ and 14. Try It 1.119 Translate to an algebraic expression and simplify if possible: the product of $−5−5$ and 12. Try It 1.120 Translate to an algebraic expression and simplify if possible: the product of 8 and $−13.−13.$ Example 1.61 Translate to an algebraic expression and simplify if possible: the quotient of $−56−56$ and $−7.−7.$ Try It 1.121 Translate to an algebraic expression and simplify if possible: the quotient of $−63−63$ and $−9.−9.$ Try It 1.122 Translate to an algebraic expression and simplify if possible: the quotient of $−72−72$ and $−9.−9.$ Use Integers in Applications We’ll outline a plan to solve applications. It’s hard to find something if we don’t know what we’re looking for or what to call it! So when we solve an application, we first need to determine what the problem is asking us to find. Then we’ll write a phrase that gives the information to find it. We’ll translate the phrase into an expression and then simplify the expression to get the answer. Finally, we summarize the answer in a sentence to make sure it makes sense. Example 1.62 How to Apply a Strategy to Solve Applications with Integers The temperature in Urbana, Illinois one morning was 11 degrees. By mid-afternoon, the temperature had dropped to $−9−9$ degrees. What was the difference of the morning and afternoon temperatures? Try It 1.123 The temperature in Anchorage, Alaska one morning was 15 degrees. By mid-afternoon the temperature had dropped to 30 degrees below zero. What was the difference in the morning and afternoon temperatures? Try It 1.124 The temperature in Denver was $−6−6$ degrees at lunchtime. By sunset the temperature had dropped to $−15−15$ degrees. What was the difference in the lunchtime and sunset temperatures? How To Apply a Strategy to Solve Applications with Integers. 1. Step 1. Read the problem. Make sure all the words and ideas are understood 2. Step 2. Identify what we are asked to find. 3. Step 3. Write a phrase that gives the information to find it. 4. Step 4. Translate the phrase to an expression. 5. Step 5. Simplify the expression. 6. Step 6. Answer the question with a complete sentence. Example 1.63 The Mustangs football team received three penalties in the third quarter. Each penalty gave them a loss of fifteen yards. What is the number of yards lost? Try It 1.125 The Bears played poorly and had seven penalties in the game. Each penalty resulted in a loss of 15 yards. What is the number of yards lost due to penalties? Try It 1.126 336. Checking Account Reymonte has a balance of $−49−49$ in his checking account. He deposits $281 to the account. What is the new balance? Everyday Math 337. Stock market Javier owns 300 shares of stock in one company. On Tuesday, the stock price dropped$12 per share. What was the total effect on Javier’s portfolio? 338. Weight loss In the first week of a diet program, eight women lost an average of 3 pounds each. What was the total weight change for the eight women? Writing Exercises 339. In your own words, state the rules for multiplying integers. 340. In your own words, state the rules for dividing integers. 341. Why is $−24≠(−2)4?−24≠(−2)4?$ 342. Why is $−43=(−4)3?−43=(−4)3?$ Self Check After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section. On a scale of 1–10, how would you rate your mastery of this section in light of your responses on the checklist? How can you improve this?	2020-12-01 21:06:29	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 278, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7186406254768372, "perplexity": 1423.95018772528}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-50/segments/1606141681524.75/warc/CC-MAIN-20201201200611-20201201230611-00406.warc.gz"}
http://lifelongprogrammer.blogspot.com/2008/11/	### Notes on Cygwin Open “System Variables “, add the path to Cygwin's bin directory to the end of “Path” variable, then we can use unix commands such as ls, ps, in windows command-prompt. Set up Environment Variables The CYGWIN variable is used to configure many global settings for the Cygwin runtime system. The PATH environment variable is used by Cygwin applications as a list of directories to search for executable files to run. The HOME environment variable is used by many programs to determine the location of your home directory The TERM environment variable specifies your terminal type. It is automatically set to cygwin if you have not set it to something else. The LD_LIBRARY_PATH environment variable is used by the Cygwin function dlopen () as a list of directories to search for .dll files to load. Customizing bash In home directory, you can see the following files: .bash_profile(or .profile),.bashrc, .inputrc .bash_profile is executed when bash is started as login shell, e.g. from the command bash --login. This is a useful place to define and export environment variables and bash functions that will be used by bash and the programs invoked by bash. It is a good place to redefine PATH if needed. such as adding a ":." to the end of PATH to also search the current working directory, PATH=$PATH:. unsetting MAILCHECK or deﬁne MAILPATH to point to your existing mail inbox in order to avoid delays you should. .bashrc is similar to .profile but is executed each time an interactive bash shell is launched. It serves to deﬁne elements such as aliases. bashrc is not called automatically for login shells. You need source it from .bash_profile. .inputrc controls how programs using the readline library (including bash) behave. It is loaded automatically. Consider the following settings: # Ignore case while completing set completion-ignore-case on # Make Bash 8bit clean set meta-flag on set convert-meta off set output-meta on Tools that do not use readline for display, such as less and ls, require additional settings, which could be put in your .bashrc: alias less=’/bin/less -r’ alias ls=’/bin/ls -F --color=tty --show-control-chars’ Using Cygwin Mapping path names Cygwin supports both Win32- and POSIX-style paths, where directory delimiters may be either forward or back slashes. UNC pathnames (starting with two slashes and a network name) are also supported. Cygwin maintains a special internal POSIX view of the Win32 ﬁle system that allows these programs to successfully run under Windows. Cygwin uses this mapping to translate between Win32 and POSIX paths as necessary. The Cygwin Mount Table The mount utility program is used to to map Win32 drives and network shares into Cygwin’s internal POSIX directory tree.Whenever Cygwin generates a POSIX path from a Win32 one, it uses the longest matching preﬁx in the mount table. Invoking mount without any arguments displays Cygwin’s current set of mount points. Whenever Cygwin cannot use any of the existing mounts to convert from a particular Win32 path to a POSIX one, Cygwin will automatically default to an imaginary mount point under the default POSIX path /cygdrive. Z:\ would be automatically converted to /cygdrive/Z. mount; mount c: /c Additional Path-related Information Symbolic links can also be used to map Win32 pathnames to POSIX, but symbolic links cannot set the default ﬁle access mode. ln -s d:/yuanyun/ws /ws The .exe extension Executable program ﬁlenames end with .exe but the .exe need not be included in the command, so that traditional UNIX names can be used. but, for programs that end in .bat and .com, you cannot omit the extension. If a shell script myprog and a programmyprog.exe coexist in a directory, the program has precedence and is selected for execution of myprog. The gcc compiler produces an executable named filename.exe when asked to produce filename. This allows many makeﬁles written for UNIX systems to work well under Cygwin. Unfortunately, the install and strip commands do distinguish between filename and filename.exe. They fail when working on a non-existing filename even if filename.exe exists, thus breaking some makeﬁles. This problem can be solved by writing install and strip shell scripts to provide the extension ".exe" when needed. The /proc ﬁlesystem Cygwin supports the /proc virtual ﬁlesystem. The ﬁles in this directory are representations of various aspects of your system. cat /proc/cpuinfo One unique aspect of the Cygwin /proc ﬁlesystem is /proc/registry, which displays the Windows registry with each KEY as a directory and each VALUE as a ﬁle. As anytime you deal with the Windows registry, use caution since changes may result in an unstable or broken system. cd /proc/registry;ls The @pathnames To circumvent the limitations on shell line length in the native Windows command shells, Cygwin programs expand their arguments starting with "@" in a special way. If a ﬁle pathname exists, the argument @pathname expands recursively to the content of pathname. Double quotes can be used inside the ﬁle to delimit strings containing blank space. Embedded double quotes must be repeated. Cygwin Utilities cygcheck cygpath The cygpath program is a utility that converts Windows native ﬁlenames to Cygwin POSIX-style pathnames and vice versa. It can be usedwhen a Cygwin programneeds to pass a ﬁle name to a native Windows program, or expects to get a ﬁle name from a nativeWindows program. dumper The dumper utility can be used to create a core dump of running Windows process. This core dump can be later loaded to gdb and analyzed. One common way to use dumper is to plug it into cygwin’s Just-In-Time debugging facility by adding error_start=x:\path\to\dumper.exe to the CYGWIN environment variable. If error_start is set this way, then dumper will be started whenever some program encounters a fatal error. dumper can be also be started from the command line to create a core dump of any running process. Unfortunately, because of a Windows API limitation, when a core dump is created and dumper exits, the target process is terminated too. kill -f, --force force, using win32 interface if necessary The kill program allows you to send arbitrary signals to other Cygwin programs. The usual purpose is to end a running program from some other window when ^C won’t work, but you can also send program-speciﬁed signals such as SIGUSR1 to trigger actions within the program, like enabling debugging or re-opening log ﬁles. Each program deﬁnes the signals they understand. You may need to specify the full path to use kill from within some shells, including bash, the default Cygwin shell. This is because bash deﬁnes a kill builtin function; To make sure you are using the Cygwin version, use: /bin/kill --version mount Cygdrive mount points Whenever Cygwin cannot use any of the existingmounts to convert froma particular Win32 path to a POSIX one, Cygwin will, instead, convert to a POSIX path using a default mount point: /cygdrive. The mount utility can be used to change this default automount preﬁx through the use of the "--change-cygdrive-preﬁx" option. We can set the automount preﬁx to /: mount --change-cygdrive-prefix / ps -a, --all show processes of all users -e, --everyone show processes of all users -W, --windows show windows as well as cygwin processes regtool: View or edit the Win32 registry umount -A, --remove-all-mounts remove all mounts -c, --remove-cygdrive-prefix remove cygdrive prefix -s, --system remove system mount (default) -S, --remove-system-mounts remove all system mounts -u, --user remove user mount -U, --remove-user-mounts remove all user mounts Using Cygwin effectively with Windows Many Windows utilities provide a good way to interact with Cygwin’s predominately command-line environment In cygwin, you can also call windows commands, such as ipconfig, net.exe, notepad.exe. Most of these tools support the /? switch to display usage information. Pathnames Windows programs do not understand POSIX pathnames, so any arguments that reference the ﬁlesystem must be in Windows (or DOS) format or translated. Cygwin provides the cygpath utility for converting between Windows and POSIX paths. notepad.exe "$(cygpath -aw "Desktop/Phone Numbers.txt")" A few programs require a Windows-style, semicolon-delimited path list, which cygpath can translate from a POSIX path with the -p option. javac -cp "$(cygpath -pw "$CLASSPATH")" hello.java The cygutils package Unix tools such as tr can convert between CRLF and LF endings, but cygutils provides several dedicated programs: conv, d2u, dos2unix, u2d, and unix2dos. Creating shortcuts with cygutils The cygutils package includes a mkshortcut utility for creating standard Microsoft .lnk ﬁles. Printing with cygutils There are several options for printing from Cygwin, including the lpr found in cygutils. The easiest way to use cygutils’ lpr is to specify a default device name in the PRINTER environment variable. You may also specify a device on the command line with the -d or -P options, which will override the environment variable setting. A device name may be a UNC path (\\server_name\printer_name), a reserved DOS device name (prn, lpt1), or a local port name that is mapped to a printer share. lpr sends raw data to the printer; no formatting is done. Many, but not all, printers accept plain text as input. If your printer supports PostScript, packages such as a2ps and enscript can prepare text ﬁles for printing. The ghostscript package also provides some translation from PostScript to various native printer languages. Additionally, a native Windows application for printing PostScript, gsprint, is available from the Ghostscript website. Programming with Cygwin Using GCC with Cygwin Console Mode Applications gcc –help,gcc hello.c -o hello.exe,g++ Welcome.cpp -o Welcome.exe The g++ command signifies that the C++ complier should be used instead of the C compiler. Compiling Programs with Multiple Source Files Compiling a program, which has two or more source files, can be accomplished two ways. The first method requires listing all the files on the command line. The second method takes advantage of Cygwin’s wild-card character(). g++ .cpp -o Fig06_05 The STLPort Library 2. The library must be installed from the Cygwin prompt. 3. Enter the command make -f gcc-cygwin.mak to start the creation of the install files 4. When completed, enter make -f gcc-cygwin.mak install to begin the installation of the new library The new library should now be installed and working. To test it, compile a program using g++ -I /usr/Local/Include/STLPort File.cpp -L / usr/Local/lib -lSTLPort_Cygwin -o ExecutableFile. Using the GDB Debugger Before you can debug your program, you need to prepare your program for debugging. What you need to do is add -g to all the other ﬂags you use when compiling your sources to objects. gcc -g myapp.c -o myapp, g++ Debug.cpp -o Debug -g What this does is add extra information to the objects (they get much bigger too) that tell the debugger about line numbers, variable names, and other useful things. These extra symbols and debugging information give your program enough information about the original sources so that the debugger can make debugging much easier for you. To invoke GDB, simply type gdb Debug.exe, then (gdb) will appear to prompt you to enter commands, like run or help. If your program crashes and you’re trying to ﬁgure out why it crashed, the best thing to do is type run and let your program run. After it crashes, you can type where to ﬁnd out where it crashed, or info locals to see the values of all the local variables. There’s also a print that lets you look at individual variables or what pointers point to. If your program is doing something unexpected, you can use the break command to tell gdb to stop your program when it gets to a speciﬁc function or line number: break 47, break my_function when you type run your programwill stop at that "breakpoint" and you can use the other gdb commands to look at the state of your program at that point, modify variables, and step through your program’s statements one at a time. You may specify additional arguments to the run command to provide command-line arguments to your program. Debugging with command line arguments myprog -t foo --queue 47 gdb myprog (gdb) run -t foo --queue 47 Use Cygwin in Eclipse.org CDT Use Linux's telnet on cygwin In order to use Linux's telnet on cygwin instead of windows version, we just need to install inetutils package (in category Net) . Some Little Trick: Problem: When I connect to some Lunix/Unix machines, and use vi, it reports 'Unknown terminal: cygwin'. This is because the "cygwin" terminal type is cygwin's emulation of a UNIX terminal in a windows "dos box". Since it's not a "standard" terminal type, it's not included in a lot of terminfo databases. Solution: Add 'export TERM=xterm' to .bashrc or .kshrc etc. Trick2: Use backslash(\) directly as path delimiter on cygwin On cygwin, we can not directly call 'cd D:\dira\dirb' to change directory, we can type 'cd D:/dira/dirb' or 'cd D:\\dira\\dirb'.But this is somewhat inconvenient. We can use one little trick, add sinle quote to the path. cd 'D:\dira\dirb' So now, we can just copy the dir path from windows explore, and paste in cygwin's command line, Trick3: Display Chinese in Cygwin Set home directory if have not done it, and change to home directory. 1. edit ~/.inputrc, and add the following lines: set meta-flag on set input-meta on set convert-meta off set output-meta on 2. edit ~/.bash_profile, and add the following line alias ls='ls –show-control-chars' Resources Cygwin User's Guide DiveIntoCygwinGCC.pdf http://ras52-tech.blogspot.com/2007/01/telnet-on-cygwin.html http://www.nabble.com/Could-I-use-backslash-directly-as-path-delimiter-on-cygwin--to21273102.html#a21273528 http://pinglunliao.blogspot.com/2006/05/chinese-in-cygwin.html ### Take Notes from Mastering Regular Expressions Introduction to Regular Expression Searching Text Files: Egrep egrep is freely available for many systems, including DOS, MacOS, Windows, Unix, and so on. Egrep Metacharacters ^ (caret) and $(dollar) represents the start and end of the line of text. Character Classes: Matching any one of several characters [], usually called a character class, list the characters you want to allow at that point in the match. Within a character class, the character-class metacharacter '-' (dash) indicates a range of characters: <H[1-6]>,[0-9] and [a-z], [0-9A-Z_!.?]. A dash is a metacharacter only when it is within a character class and is not the first character listed in the class; otherwise it matches the normal dash character. The question mark and period at the end of the class are usually regular-expression metacharacters, but only when not within a class. The only special characters within the class in [0-9A-ZR!.?] are the two dashes. ^cat$,^$:an empty line (with nothing in it, not even spaces). ^: Since every line has a beginning, every line will match even lines that are empty! Negated character classes [^] The leading ^ in the class "negates" the list. [^1-6] matches a character that's not 1 through 6. ^ is a line anchor outside a class, but a class metacharacter inside a class (but, only when it is immediately after the class's opening bracket; otherwise, it's not special inside a class). A negated character class means "match a character that's not listed" and not "don't match what is listed." A convenient way to view a negated class is that it is simply a shorthand for a normal class that includes all possible characters except those that are listed. Matching Any Character with Dot The metacharacter . is a shorthand for a character class that matches any character. In 03[-./]19[-./]76, the dots are not metacharacters because they are within a character class. The dashes are also not class metacharacters in this case because each is the first thing after [ or [^. The list of metacharacters and their meanings are different inside and outside of character classes. Knowing the target text well is an important part of wielding regular expressions effectively. Alternation: Matching any one of several subexpressions \| means "or." Bob\|Robert In gr[a\|e]y, the '\|' character is just a normal character, like a and e. Alternation reaches far, but not beyond parentheses. A character class can match just a single character in the target text. With alternation, since each alternative can be a full-fledged regular expression in and of itself, each alternative can match an arbitrary amount of text. Character classes are almost like their own special mini-language (with their own ideas about metacharacters, for example), while alternation is part of the "main" regular expression language. Also, take care when using caret or dollar in an expression that has alternation. Compare ^From\|Subject\|Date:• with ^(From\|Subject\|Date):•. The first is composed of three alternatives, so it matches "^From or Subject or Date: •," which is not particularly useful. We want the leading caret and trailing : • to apply to each alternative. We can accomplish this by using parentheses to "constrain" the alternation: ^(From\|Subject\|Date):• Ignoring Differences in Capitalizatio: -i egrep's command-line option "-i" tells it to do a case-insensitive match. Word Boundaries: \< and \> \<cat\>, \<cat or cat\> The "start of a word" is simply the position where a sequence of alphanumeric characters begins; "end of word" is where such a sequence ends. "start and end of word" is better phrased as "start and end of an alphanumeric sequence," Summary of Metacharacters Seen So Far. Metacharacter Name Matches . dot any one character [] character class any character listed [^] negated character class any character not listed ^ caret the position at the start of the line$ dollar the position at the end of the line \< backslash less-than the position at the start of a word \> backslash greater-than the position at the end of a word not supported by all versions of egrep \| or; bar matches either expression it separates () parentheses used to limit scope of \| The rules about which characters are and aren't metacharacters (and exactly what they mean) are different inside a character class. Dot is a metacharacter outside of a class, but not within one. Conversely, a dash is a metacharacter within a class (usually), but not outside. Moreover, a caret has one meaning outside, another if specified inside a class immediately after the opening [, and a third if given elsewhere in the class. Don't confuse alternation with a character class. The class [abc] and the alternation (a\|b\|c) effectively mean the same thing, but the similarity in this example does not extend to the general case. A character class can match exactly one character, and that's true no matter how long or short the specified list of acceptable characters might be. Alternation, on the other hand, can have arbitrarily long alternatives, A negated character class is simply a notational convenience for a normal character class that matches everything not listed. Thus, [^x] doesn't mean "match unless there is an x," but rather "match if there is something that is not x." The difference is subtle, but important. The first concept matches a blank line, for example, while [^x] does not. The useful -i option discounts capitalization during a match Optional Items: ? The question mark attaches only to the immediately-preceding item. Other Quantifiers: Repetition: + (plus) and * (star) <HR•+SIZE •* = •* [0-9]+ •> Defined range of matches: intervals Some versions of egrep support a metasequence for providing your own minimum and maximum: {min,max}. This is called the interval quantifier. Parentheses and Backreferences Uses for parentheses: to limit the scope of alternation \|, and to group multiple characters into larger units to which you can apply quantifiers like question mark and star. In many regular-expression flavors, parentheses can "remember" text matched by the subexpression they enclose. Backreferencing allows matching new text that is the same as some text matched earlier in the expression. egrep -i '\<([a-z]+) +\1\>' files With tools that support backreferencing, parentheses "remember" the text that the subexpression inside them matches, and the special metasequence \1 represents that text later in the regular expression, whatever it happens to be at the time. Use \1, \2, \3, etc., to refer to the first, second, third, etc. sets. Pairs of parentheses are numbered by counting opening parentheses from the left, ([a-z])([0-9])\1\2. The Great Escape: \ ega\.att\.com, $$[a-zA-Z]+$$ matchs a word within parentheses. A backslash used in this way is called an "escape" when a metacharacter is escaped, it loses its special meaning and becomes a literal character. When used before a non-metacharacter, a backslash can have different meanings depending upon the version of the program. For example \<, \>, \1, etc. as metasequences. Variable names: [a-zA-Z_][a-zA-Z_0-9] A string within double quotes: "[^"]" Dollar amount (with optional cents): \$[0-9]+(\.[0-9][0-9])? An HTTP/HTML URL <http://[-a-z0-9_.:]+/[-a-z0-9_:@&?=+,.!/~%$]\.html?\> egrep -i '\<http://[^ ]\.html?\>' files... Time of day, such as "9:17 am" or "12:30 pm" [0-9]?[0-9]:[0-9][0-9]•(am\|pm) (1[012]?[1-9]):[0-5][0-9]•(am\|pm) [01]?[0-9]\|2[0-3]:[0-5][0-9] Egrep Metacharacter Summary Items to Match a Single Character Metacharacter Matches .dot Matches any one character []character class Matches any one character listed [^]negated character class Matches any one character not listed \chares(caped character) Items Appended to Provide "Counting": The Quantifiers ?question One allowed, but it is optional star Any number allowed, but all are optional +plus At least one required; additional are optional {min,max}specified range Min required, max allowed Items That Match a Position ^caret Matches the position at the start of the line $dollar Matches the position at the end of the line \<word boundary Matches the position at the start of a word \>word boundary Matches the position at the end of a word Other \|alternation Matches either expression it separates ()parentheses Limits scope of alternation, provides grouping for the quantifiers, and "captures" for backreferences \1, \2, ...backreference Matches text previously matched within first, second, etc., set of parentheses. Three reasons for using parentheses are constraining alternation, grouping, and capturing. Character classes are special, and have their own set of metacharacters totally distinct from the "main" regex language Alternation and character classes are fundamentally different, providing unrelated services that appear, in only one limited situation, to overlap A negated character class is still a "positive assertion"even negated, a character class must match a character to be successful. Because the listing of characters to match is negated, the matched character must be one of those not listed in the class. The useful -i option discounts capitalization during a match. There are three types of escaped items: 1. The pairing of \ and a metacharacter is a metasequence to match the literal character 2. The pairing of \ and selected non-metacharacters becomes a metasequence with an implementation-defined meaning (for example, \< often means "start of word"). 3. The pairing of \ and any other character defaults to simply matching the character (that is, the backslash is ignored). Items governed by a question mark or star don't need to actually match any characters to "match successfully." They are always successful, even if they don't match anything. Extended Introductory Examples Ensure that each file contained 'ResetSize' exactly as many times as 'SetSize'. perl -0ne 'print "$ARGV\n" if s/ResetSize//ig != s/SetSize//ig' perl -w programFile: -w tells Perl to check program more carefully and issue warnings about items it thinks to be dubious. Matching Text with Regular Expressions if ( $celsius =~ m/^[0-9]+$/) The m means to attempt a regular expression match, while the slashes delimit the regex itself. =~ links a regex search with the target string to be searched. The operator == tests whether two numbers are the same. (The operator eq is used to test whether two strings are the same.) Side Effects of a Successful Match Use the metacharacter \1 within the regular expression to refer to some text matched earlier during the same match attempt, and use the variable $1 in subsequent code to refer to that same text after the match has been successfully completed. Non-Capturing Parentheses: (?:) The benefits of this are twofold. One is that by avoiding the unnecessary capturing, the match process is more efficient. Another is that, overall, using exactly the type of parentheses needed for each situation may be less confusing later to someone reading the code who might otherwise be left wondering about the exact nature of each set of parentheses. \b normally matches a word boundary, but within a character class, it matches a backspace. Temperature-conversion program final listing if ($input =~ m/^([-+]?[0-9]+(\.[0-9]+)?)\s([CF])$/i) {$InputNum = $1; # Save to named variables to make the ...$type = $3; # ... rest of the program easier to read. if ($type =~ m/c/i) { } else { } } else { # The initial regex did not match, so issue a warning. } 2. Perl can check a string in a variable against a regex using the construct $variable =~ m/regex/. The m indicates that a match is requested, while the slashes delimit (and are not part of) the regular expression. The whole test, as a unit, is either true or false. 4. Among the more useful shorthands that Perl and many other flavors of regex provide are: \t, a tab character \b, backspace \n, a newline character \r, a carriage-return character \s matches any "whitespace" character (space, tab, newline, formfeed, and such) \S anything not \s \w [a-zA-Z0-9_] (useful as in \w+, ostensibly to match a word) \W anything not \w, i.e., [^a-zA-Z0-9_] \d [0-9], i.e., a digit \D anything not \d, i.e., [^0-9] 5. The /i modifier makes the test case-insensitive. Also /g ("global match") and /x ("free-form expressions"). 6. (?:) non-capturing parentheses can be used for grouping without capturing. 7. After a successful match, Perl provides the variables$1, $2,$3, etc., which hold the text matched by their respective () parenthesized subexpressions in the regex. Subexpressions are numbered by counting open parentheses from the left, starting with one. Subexpressions can be nested, as in (Washington(•DC)?) Raw () parentheses can be intended for grouping only, but as a byproduct, they still capture into one of the special variables. Modifying Text with Regular Expressions: $var =~ s/regex/replacement/ The regex is the same as with m//, but the replacement is actually a Perl string in its own right,. You can include references to variables, including$1, $2, and so on to refer to parts of what was just matched. Perl provides the catch-all \b, which matches start-of-word or end-of-word metacharacters:$var =~ s/Jeff/Jeffrey/; $var =~ s/\bJeff\b/Jeffrey/;$var =~ s/\bJeff\b/Jeff/i Just what does $var =~ s/\bJeff\b/Jeff/i do? Example: Form Letter$letter =~ s/=FAMILY=/$family/g; The /g "global replacement" modifier instructs the s/// to replace all occurrences. Example: Prettifying a Stock Price Always take the first two digits after the decimal point, and take the third digit only if it is not zero. Then, remove any other digits:$price =~ s/(\.\d\d[1-9]?)\d/$1/ Automated Editing perl -p -i -e 's/sysread/read/g' file This runs the Perl program s/sysread/read/g. (the -e flag indicates that the entire program follows right there on the command line.) The -p flag results in the substitution being done for every line of the named file, and the -i flag causes any changes to be written back to the file when done. There is no explicit target string for the substitute command to work on (that is, no$var =~ ) because conveniently, the -p flag implicitly applies the program, in turn, to each line of the file. Also, because I used the /g modifier, I'm sure to replace multiple occurrences that might be in a line. A Small Mail Utility Perl’s magic "<>" operator gives the next line of input when you assign from it to a normal $variable, as with "$variable = <>". It is just Perl's funny way to express a kind of a getline (). while ($line = < >){ if ($line =~ m/^\s$/ ) { # If we have an empty line... last; # this immediately ends the 'while' loop.} if ($line =~ m/^Subject: (.)/i) { $subject =$1;} if ($line =~ m/^From: (\s+) $$([^()])$$/i) {$reply_address = $1;$from_name = $2; } }$line =~ s/^/\|> /; The substitute searches for ^, which of course immediately matches at the beginning of the string. It doesn't actually match any characters, though, so the substitute "replaces" the "nothingness" at the beginning of the string with '\|>•'. In effect, it inserts '\|>•' at the beginning of the string. Perl's defined function indicates whether the variable has a value, while the die function issues an error message and exits the program. Adding Commas to a Number with Lookaround Regular expressions generally work left-to-right. Lookaround matchs positions within the text and doesn't "consume" text. Positive lookahead (?=), such as (?=\d). Lookbehind (?<=) looks back (toward the left), such as (?<=\d) Lookaround doesn't "consume" text (?=Jeffrey), matches only the marked location. Lookahead uses its subexpression to check the text, but only to find a location in the text at which it can be matched, not the actual text it matches. The combined expression, (?=Jeffrey)Jeff effectively matches "Jeff" only if it is part of "Jeffrey." It is effectively the same as Jeff(?=rey). Approaches to the "Jeffs" Problem s/\bJeffs\b/Jeff's/g s/\b(Jeff)(s)\b/$1'$2/g s/\bJeff(?=s\b)/Jeff'/g s/(?<=\bJeff)(?=s\b)/'/g This regex doesn't actually "consume" any text. It uses both lookahead and lookbehind to match positions of interest, at which an apostrophe is inserted. Very useful to illustrate lookaround. s/(?=s\b)(?<=\bJeff)/'/g This is exactly the same as the one above, but the two lookaround tests are reversed. Because the tests don't consume text, the order makes no difference to whether there's a match. Which "Jeffs" solutions would preserve case when applied with /i? s/\b(Jeff)(s)\b/$1'$2/g and s/(?<=\bJeff)(?=s\b)/'/g To preserve case, you've got to either replace the exact characters consumed (rather than just always inserting 'Jeff's'), or not consume any letters. Back to the comma example ... "Locations having digits on the right in exact sets of three, and at least some digits on the left." The second requirement is simple enough with lookbehind. $pop =~ s/(?<=\d)(?=(\d\d\d)+$)/,/g; print "The US population is $pop\n"; Also (?<=\d)(?=(?:\d\d\d)+$) $text =~ s/(?<=\d)(?=(\d\d\d)+(?!\d))/,/g; Four Types of Lookaround Type Regex Successful if the enclosed subexpression ... Positive Lookbehind (?<=......) successful if can match to the left Negative Lookbehind (?<!......) successful if can not match to the left Positive Lookahead (?=......) successful if can match to the right Negative Lookahead (?!......) successful if can not match to the right Commafication without lookbehind Lookbehind is not as widely supported (nor as widely used) as lookahead.$text =~ s/(\d)(?=(\d\d\d)+(?!\d))/$1,/g; When one iteration ends, the next picks up the inspection of the text at the point where the previous match ended. Does$text =~ s/(\d)((\d\d\d)+\b)/$1,$2/g "commafy" a number? This won't work the way we want. It leaves results such as "281,421906." This is because the digits matched by (\d\d\d)+ are now actually part of the final match, and so are not left "unmatched" and available to the next iteration of the regex via the /g. When one iteration ends, the next picks up the inspection of the text at the point where the previous match ended. The whole point of using lookahead was to get the positional check without actually having the inspected text check count toward the final "string that matched." Actually, this expression can still be used to solve this problem. With each such application, one more comma is added (to each number in the target string, due to the /g modifier). while ( $text =~ s/(\d)((\d\d\d)+\b)/$1,$2/g ) { # Nothing to do inside the body of the while -- we merely want to reapply the regex until it fails } Text-to-HTML Conversion Separating paragraphs ^ and$ normally refer not to logical line positions, but to the absolute start- and end-of-string positions. Most regex-endowed languages give us an easy solution, an enhanced line anchor match mode in which the meaning of ^ and $to change from string related to the logical-line related meaning. With Perl, this mode is specified with the /m modifier:$text =~ s/^\s$/<p>/mg; \s can match a newline. This means that if we have several blank lines in a row, ^\s$ is able to match them all in one shot. The fortunate result is that the replacement leaves just one <p> instead of the several in a row we would otherwise end up with. $text =~ s/\b(username regex\@hostname regex)\b/<a href="mailto:$1">$1<\/a>/g; Perl allows picking our own delimiters. s!regex!string!modifiers or s{regex}{string}modifiers. Matching the username and hostname The /x modifier does two simple but powerful things for the regular expression. First, it causes most whitespace to be ignored, so you can "free-format" the expression for readability. Secondly, it allows comments with a leading #. Specifically, /x turns most whitespace into an "ignore me" metacharacter, and # into an "ignore me, and everything else up to the next newline" metacharacter . They aren't taken as metacharacters within a character class (which means that classes are not free-format, even with /x), and as with other metacharacters, you can escape whitespace and # that you want to be taken literally. Of course, you can always use \s to match whitespace, as in m/<a \s+ href=>/x. /x applies only to the regular expression, and not to the replacement string. Putting it together undef$/; # Enter "file-slurp" mode $text = <>; # Slurp up the first file given on the command line.$text =~ s/&/&/g; # Make the basic HTML ... $text =~ s/</</g; # ... characters &, <, and > ...$text =~ s/>/>/g; # ... HTML safe. $text =~ s/^\s$/<p>/mg; # Separate paragraphs. $text =~ s{ \b # Capture the address to$1 ... ( \@ [-a-z0-9]+(\.[-a-z0-9]+)\.(com;edu;info) # hostname ) \b }{<a href="mailto:$1">$1</a>}gix; # Turn HTTP URLs into links ... $text =~ s{ \b # Capture the URL to$1 ... ( http:// [-a-z0-9]+(\.[-a-z0-9]+)\.(com\|edu\|info) \b # hostname ( / [-a-z0-9_:\@&?=+,.!/~'%\$] # Optional path (?<![.,?!]) # Not allowed to end with [.,?!] )? ) }{<a href="$1">$1</a>}/gix; Building a regex library The same expression is used for each of the two hostnames, which means that if we ever update one, we have to be sure to update the other. Rather than keeping that potential source of confusion, consider the three instances of$HostnameRegex in this modified snippet from our program: $HostnameRegex = qr/[-a-z0-9]+(\.[-a-z0-9]+)\.(com\|edu\|info)/i; # Turn email addresses into links ...$text =~ s{ \b # Capture the address to $1 ... ( \w[-.\w] # username \@$HostnameRegex # hostname ) \b }{<a href="mailto:$1">$1</a>}gix; # Turn HTTP URLs into links ... $text =~ s{ \b # Capture the URL to$1 ... ( http:// $HostnameRegex \b # hostname ( / [-a-z0-9_:\@&?=+,.!/~'%\$] # Optional path (?<![.,?!]) #not allowed to end with [.,?!] )? ) }{<a href="$1">$1</a>}gix; qr operator converts the regex provided into a regex object. Later you can use that object in place of a regular expression, or even as a subexpression of some other regex. Why '$' and '@' sometimes need to be escaped You'll notice that the same '$' is used as both the end-of-string metacharacter, and to request interpolation (inclusion) of a variable. Normally, there's no ambiguity to what '$' means, but within a character class it gets a bit tricky. Since it can't possibly mean end-of-string within a class, in that situation Perl considers it a request to interpolate (include from) a variable, unless it's escaped. If escaped, the '$' is just included as a member of the class. That's what we want this time, so that's why we have to escape the dollar sign in the path part of the URL-matching regex. Perl uses @ at the beginning of array names, and Perl string or regex literals allow arrays to be interpolated. If we wish a literal @ to be part of a regex, we must escape it so that it's not taken as an array interpolation. That Doubled-Word Thing $/ = ".\n"; while (<>) { next if !s/\b([a-z]+)((?:\s<<[^>]+>)+)(\1\b)/\e[7m$1\e[m$2\e[7m$3\e[m/ig; s/^(?:[^\e]\n)+//mg; # Remove any unmarked lines. s/^/$ARGV: /mg; # Ensure lines begin with filename. print; } Double-word example in Perl$/ = ".\n"; # Sets a special "chunk-mode"; chunks end with a period-newline combination while (< >) { next unless s{# (regex starts here) ### Need to match one word: \b # Start of word ... . ( [a-z]+ ) # Grab word, filling $1 (and \1). ### Now need to allow any number of spaces and/or <TAGS> ( #Save what intervenes to$2. (?: # (Non-capturing parens for grouping the alternation) \s # Whitespace (includes newline, which is good). \| # -or- <[^>]+> # Item like <TAG>. )+ # Need at least one of the above, but allow more. ) ### Now match the first word again: (\1\b) # \b ensures not embedded. This copy saved to $3. #(regex ends here) } # Above is the regex. The replacement string is below, followed by the modifiers, /i, /g, and /x {\e[7m$1\e[m$2\e[7m$3\e[m}igx; s/^(?:[^\e]+\n)+//mg; # Remove any unmarked lines. s/^/$ARGV: /mg; # Ensure lines begin with filename. print; } ❶ Because the doubled-word problem must work even when the doubled words are split across lines, I can't use the normal line-by-line processing I used with the mail utility example. Setting the special variable$/ (yes, that's a variable) as shown puts the subsequent <> into a magic mode such that it returns not single lines, but more-or-less paragraph-sized chunks. The value returned is just one string, but a string that could potentially contain many of what we would consider to be logical lines. ❷ Did you notice that I don't assign the value from <> to anything? When used as the conditional of a while like this, <> magically assigns the string to a special default variable.[] That same variable holds the default string that s/// works on, and that print displays. Using these defaults makes the program less cluttered, but also less understandable to someone new to the language, so I recommend using explicit operands until you're comfortable. [] The default variable is $_ (yes, that's a variable too). It's used as the default operand for many functions and operators. ❻ The variable$ARGV magically provides the name of the input file. Combined with /m and /g, this substitution tacks the input filename to the beginning of each logical line remaining in the string. Cool! Moving bits around: operators, functions, and objects You'll also notice that the regular expressions are located not in the main text-processing part of the program, but at the start, in the initialization section. The Pattern.compile function merely analyzes the string as a regular expression, and builds an internal "compiled version" that is assigned to a Pattern variable (regex1, etc.). Then, in the main text-processing part of the program, that compiled version is applied to text with regex1.matcher(text), the result of which is used to do the replacement. public class TwoWord { public static void main(String [] args) { Pattern regex1 = Pattern.compile("\\b([a-z]+)((?:\\s<\\<[^>]+\\>)+)(\\1\\b)", Pattern.CASE_INSENSITIVE); String replace1 = "\033[7m$1\033[m$2\033[7m$3\033[m"; Pattern regex2 = Pattern.compile("^(?:[^\\e]\\n)+", Pattern.MULTILINE); Pattern regex3 = Pattern.compile("^([^\\n]+)", Pattern.MULTILINE); // For each command-line argument.... for (int i = 0; i < args.length; i++) { try { BufferedReader in = new BufferedReader(new FileReader(args[i])); String text; // For each paragraph of each file..... while ((text = getPara(in)) != null) { // Apply the three substitutions text = regex1.matcher(text).replaceAll(replace1); text = regex2.matcher(text).replaceAll(""); text = regex3.matcher(text).replaceAll(args[i] + ":$1"); System.out.print(text); } } catch (IOException e) { System.err.println("can't read ["+args[i]+"]: " + e.getMessage());} } } // Routine to read next "paragraph" and return as a string static String getPara(BufferedReader in) throws java.io.IOException { StringBuffer buf = new StringBuffer(); String line; while ((line = in.readLine()) != null && (buf.length() == 0 ;; line.length() != 0)) {buf.append(line + "\n");} return buf.length() == 0 ? null : buf.toString(); } } Overview of Regular Expression Features and Flavors Care and Handling of Regular Expressions Regex handling in Java 1. Inspect the regular expression and compile it into an internal form that matches in a case-insensitive manner, yielding a "Pattern" object. 2. Associate it with some text to be inspected, yielding a "Matcher" object. 3. Actually apply the regex to see if there is a match in the previously-associated text, and let us know the result. 4. If there is a match, make available the text matched within the first set of capturing parentheses. Regex handling in Python import re; R = re.compile("^Subject:(.)", re.IGNORECASE); M = R.search(line) if M: subject = M.group(1) A Search-and-Replace Example $text =~ s{ \b # Capture the address to$1 ... ( @ [-\w]+(\.[-\w]+)+\.(com;edu;info) # hostname ) \b }{<a href="mailto:$1">$1</a>}gix; Search and replace in Java import java.util.regex.; // Make regex classes easily available Pattern r = Pattern.compile( "\\b \n"+ "# Capture the address to $1 ... \n"+ "( \n"+ " \\w[-.\\w]* # username \n"+ " @ \n"+ " [-\\w]+(\\.[-\\w]+)\\.(com\|edu\|info) # hostname \n"+ ") \n"+ "\\b \n", Pattern.CASE_INSENSITIVE\|Pattern.COMMENTS); Matcher m = r.matcher(text); text = m.replaceAll("<a href=\"mailto:$1\">$1</a>"); Note that each '\' wanted in a string's value requires '\\' in the string literal, so if you're providing regular expressions via string literals as we are here, \w requires '\\w'. For debugging, System.out.println(r.pattern()) can be useful to display the regular expression as the regex function actually received it. One reason that I include newlines in the regex is so that it displays nicely when printed this way. Another reason is that each '#' introduces a comment that goes until the next newline; so, at least some of the newlines are required to restrain the comments. Perl uses notations like /g, /i, and /x to signify special conditions (these are the modifiers for replace all, case-insensitivity, and free formatting modes 135), but java.util.regex uses either different functions (replaceAll versus replace) or flag arguments passed to the function (e.g., Pattern.CASE_INSENSITIVE and Pattern.COMMENTS). Search and Replace in Other Languages Awk uses an integrated approach, /regex/, to perform a match on the current input line, and uses "var ~ " to perform a match on other data. Sub function is used for substitution. sub(/mizpel/, "misspell") To replace all matches within the line, awk does not use any kind of /g modifier, but a different operator altogether: gsub(/mizpel/, "misspell"). Strings, Character Encodings, and Modes Strings as Regular Expressions Strings in Java Java string literals are like those presented in the introduction, in that they are delimited by double quotes, and backslash is a metacharacter. Common combinations such as '\t' (tab), '\n' (newline), '\\' (literal backslash), etc. are supported. Using a backslash in a sequence not explicitly supported by literal strings results in an error. Strings in Python Python uses either single quotes or double quotes to create strings, Python also offers "triple-quoted" strings of the form '''''' and """""", which are different in that they may contain unescaped newlines. All four types offer the common backslash sequences such as \n, but have the same twist that PHP has in that unrecognized sequences are left in the string verbatim. Contrast this with Java and C# strings, for which unrecognized sequences cause an error. Like PHP and C#, Python offers a more literal type of string, its "raw string." Similar to C#'s @"" notation, Python uses an 'r' before the opening quote of any of the four quote types. For example, r"\t\x2A" yields \t\x2A. Unlike the other languages, though, with Python's raw strings, all backslashes are kept in the string, including those that escape a double quote (so that the double quote can be included within the string): r"he said \"hi\"\." results in he said \"hi\"\.. This isn't really a problem when using strings for regular expressions, since Python's regex flavor treats \" as ", but if you like, you can bypass the issue by using one of the other types of raw quoting: r'he said "hi"\.' Regex literals in Perl$str =~ m/(\w+)/; can also be written as: $regex = '(\w+)';$str =~ $regex; or perhaps:$regex = "(\\w+)"; $str =~$regex; When a regex is provided as a literal, Perl provides extra features that the regular-expression engine itself does not, including: interpolation of variables, support for a literal-text mode via \Q\E (113). Optional support for a \N{name} construct, which allows you to specify characters via their official Unicode names. For example, you can match '¡Hola!' with \N{INVERTED EXCLAMATION MARK}Hola!. In Perl, a regex literal is parsed like a very special kind of string. In fact, these features are also available with Perl double-quoted strings. The point to be aware of is that these features are not provided by the regular-expression engine. Since the vast majority of regular expressions used within Perl are as regex literals, most think that \Q\E is part of Perl's regex language, but if you ever use regular expressions read from a configuration file (or from the command line, etc.), it's important to know exactly what features are provided by which aspect of the language. Richness of encoding-related support Sometimes things are not as simple as they might seem. For example, the \b of Sun's java.util.regex package properly understands all the word-related characters of Unicode, but its \w does not (it understands only basic ASCII). Regular expressions for programs that work with Unicode often support a \unum metasequence that can be used to match a specific Unicode character (117). It's important to realize that \uC0B5 is saying "match the Unicode character U+C0B5," and says nothing about what actual bytes are to be compared, which is dependent on the particular encoding used internally to represent Unicode code points. If the program happens to use UTF-8 internally, that character happens to be represented with three bytes. But you, as someone using the Unicode-enabled program, don't normally need to care. Characters versus combining-character sequences Perl and PCRE (and by extension, PHP's preg suite) support the \X metasequence, which fulfills what many might expect from dot ("match one character") in that it matches a base character followed by any number of combining characters. Regex Modes and Match Modes Case-insensitive match mode Common Metacharacters and Features Character Representations 1. Character Shorthands: \n, \t, \a, \b, \e, \f, \r, \v, ... 2. Octal Escapes: \num 3. Hex/Unicode Escapes: \xnum, \x{num}, \unum, \Unum, ... 4. Control Characters: \cchar Character Classes and Class-Like Constructs 1. Normal classes: [a-z] and [^a-z] 2. Almost any character: dot 3. Exactly one byte: \C 4. Unicode Combining Character Sequence: \X 5. Class shorthands: \w, \d, \s, \W, \D, \S 6. Unicode properties, blocks, and categories: \p{Prop}, \P{Prop} 7. Class set operations: [[a-z]&&[^aeiou]] 8. POSIX bracket-expression "character class": [[:alpha:]] 9. POSIX bracket-expression "collating sequences": [[.span-ll.]] 10. POSIX bracket-expression "character equivalents": [[=n=]] 11. Emacs syntax classes Anchors and Other "Zero-Width Assertions" 1. Start of line/string: ^, \A 2. End of line/string: $, \Z, \z 3. Start of match (or end of previous match): \G 4. Word boundaries: \b, \B, \<, \>, ... 5. Lookahead (?=), (?!); Lookbehind, (?<=), (?<!) Comments and Mode Modifiers 1. Mode modifier: (?modifier), such as (?i) or (?-i) 2. Mode-modified span: (?modifier:), such as (?i:) 3. Comments: (?#) and # 4. Literal-text span: \Q\E Grouping, Capturing, Conditionals, and Control 1. Capturing/grouping parentheses: (), \1, \2, ... 2. Grouping-only parentheses: (?:) 3. Named capture: (?<Name>) 4. Atomic grouping: (?>) 5. Alternation: \|\| 6. Conditional: (?if then\|else) 7. Greedy quantifiers: , +, ?, {num,num} 8. Lazy quantifiers: ?, +?, ??, {num,num}? 9. Possessive quantifiers: +, ++, ?+, {num,num}+ Character Representations This group of metacharacters provides visually pleasing ways to match specific characters that are otherwise difficult to represent. Character shorthands Many utilities provide metacharacters to represent certain control characters that are sometimes machine-dependent, and which would otherwise be difficult to input or to visualize: \a Alert (e.g., to sound the bell when "printed") Usually maps to the ASCII <BEL> character, 007 octal. \b Backspace Usually maps to the ASCII <BS> character, 010 octal. (With many flavors, \b is a shorthand only within a character class, a word-boundary metacharacter outside) \e Escape character Usually maps to the ASCII <ESC> character, 033 octal. \f Form feed Usually maps to the ASCII <FF> character, 014 octal. \n Newline On most platforms (including Unix and DOS/Windows), usually maps to the ASCII <LF> character, 012 octal. On MacOS systems, usually maps to the ASCII <CR> character, 015 octal. With Java or any .NET language, always the ASCII <LF> character regardless of platform. \r Carriage return Usually maps to the ASCII <CR> character. On MacOS systems, usually maps to the ASCII <LF> character. With Java or any .NET language, always the ASCII <CR> character regardless of platform. \t,\v Octal escape \num Implementations supporting octal (base 8) escapes generally allow two- and three digit octal escapes to be used to indicate a byte or character with a particular value. For example, \015\012 matches an ASCII CR/LF sequence. Octal escapes can be convenient for inserting hard-to-type characters into an expression. In Perl, for instance, you can use \e for the ASCII escape character, but you can't in awk. Since awk does support octal escapes, you can use the ASCII code for the escape character directly: \033. Some implementations, as a special case, allow \0 to match a NUL byte. Some allow all one-digit octal escapes, but usually don't if backreferences such as \1 are supported. When there's a conflict, backreferences generally take precedence over octal escapes. Some allow four-digit octal escapes, usually to support a requirement that any octal escape begin with a zero (such as with java.util.regex). You might wonder what happens with out-of-range values like \565 (8-bit octal values range from \000 until only \377). It seems that half the implementations leave it as a larger-than-byte value (which may match a Unicode character if Unicode is supported), while the other half strip it to a byte. In general, it's best tolimit octal escapes to \377 and below. Hex and Unicode escapes: \xnum, \x{num}, \unum, \Unum, ... Similar to octal escapes, many utilities allow a hexadecimal (base 16) value to be entered using \x, \u, or sometimes \U. If allowed with \x, for example, \x0D\x0A matches the CR/LF sequence. Control characters: \cchar Many flavors offer the \cchar sequence to match control characters with encoding values less than 32 (some allow a wider range). For example, \cH matches a Control-H, which represents a backspace in ASCII, while \cJ matches an ASCII linefeed (which is often also matched by \n, but sometimes by \r, depending on the platform. Details aren't uniform among systems that offer this construct. You'll always be safe using uppercase English letters as in the examples. With most implementations, you can use lowercase letters as well, but Sun's Java regex package, for example, does not support them. And what exactly happens with non-alphabetics is very flavor-dependent, so I recommend using only uppercase letters with \c. Related Note: GNU Emacs supports this functionality, but with the rather ungainly metasequence ?\^char (e.g., ?\^H to match an ASCII backspace). .............Table 3-7. A Few Utilities and the Octal and Hex Regex Escapes Their Regexes Support Character Classes and Class-Like Constructs Normal classes: [a-z] and [^a-z] [] is never a metacharacter within a class, while [-] usually is. Some metasequences, such as \b, sometimes have a different meaning within a class than outside of one . Almost any character: dot In some tools, dot is a shorthand for a character class that can match any character, while in most others, it is a shorthand to match any character except a newline. It's a subtle difference that is important when working with tools that allow target text to contain multiple logical lines (or to span logical lines, such as in a text editor). Concerns about dot include: In some Unicode-enabled systems, such as Sun's Java regex package, dot normally does not match a Unicode line terminator (109). A match mode (111) can change the meaning of what dot matches. The POSIX standard dictates that dot not match a NUL (a character with the value zero), although all the major scripting languages allow NULLs in their text (and dot matches them). Dot versus a negated character class When working with tools that allow multiline text to be searched, take care to note that dot usually does not match a newline, while a negated class like [^"] usually does. This could yield surprises when changing from something such as "." to "[^"]". The matching qualities of dot can often be changed by a match modesee. Exactly one byte Perl and PCRE (and hence PHP) support \C, which matches one byte, even if that byte is one of several that might encode a single character (on the other hand, everything else works on a per-character basis). This is dangerousits misuse can cause internal errors, so it shouldn't be used unless you really know what you're doing. Perl and PHP support \X as a shorthand for \P{M}\p{M}, which is like an extended . (dot). It matches a base character (anything not \p{M}), possibly followed by any number of combining characters (anything that is \p{M}). Unicode uses a system of base and combining characters which, in combination, create what look like single, accented characters like à ('a' U+0061 combined with the grave accent '`' U+0300). You can use more than one combining character if that's what you need to create the final result. For example, if for some reason you need 'Ç̆', that would be 'c' followed by a combining cedilla '¸' and a combining breve '' (U+0063 followed by U+0327 and U+0306). Besides the fact that \X matches trailing combining characters, there are two differences between it and dot. One is that \X always matches a newline and other Unicode line terminators (109), while dot is subject to dot-matches-all match-mode (111), and perhaps other match modes depending on the tool. Another difference is that a dot-matches-all dot is guaranteed to match all characters at all times, while \X doesn't match a leading combining character. Class shorthands: \w, \d, \s, \W, \D, \S \d Digit Generally the same as [0-9] or, in some Unicode-enabled tools, all Unicode digits. \D Non-digit Generally the same as [^\d] \w Part-of-word character Often the same as [a-zA-Z0-9_]. Some tools omit the underscore, while others include all alphanumerics in the current locale. If Unicode is supported, \w usually refers to all alphanumerics; notable exceptions include java.util.regex and PCRE (and by extension, PHP), whose \w are exactly [a-zA-Z0-9_]. \W Non-word character Generally the same as [^\w]. \s Whitespace character On ASCII-only systems, this is often the same as [• \f\n\r\t\v]. Unicode-enabled systems sometimes also include the Unicode "next line" control character U+0085, and sometimes the "white space" property \p{Z} (described in the next section). \S Non-whitespace character Generally the same as [^\s]. Simple class subtraction: [[a-z]-[aeiou]] .NET offers a simple class "subtraction" nomenclature, which allows you to remove from what a class can match those characters matchable by another class. For example, the characters matched by [[a-z]-[aeiou]] are those matched by [a-z] minus those matched by [aeiou], i.e. that are non-vowel lower-case ASCII. As another example, [\p{P}-[\p{Ps}\p{Pe}]] is a class that matches characters in \p{P} except those matchable by [\p{Ps}\p{Pe}], which is to say that it matches all punctuation except opening and closing punctuation such as and (. Full class set operations: [[a-z] && [^aeiou]] Sun's Java regex package supports a full range of set operations (union, subtraction, intersection) within character classes. OR allows you to add characters to the class by including what looks like an embedded class within the class. AND does a conceptual "bitwise AND" of two sets, keeping only those characters found in both sets. It is achieved by inserting the special class metasequence && between two sets of characters. AND is less confusing in that [\p{InThai}&&\P{Cn}] is normally read as "match only characters matchable by \p{InThai} and \P{Cn}," although it is sometimes read as "the list of allowed characters is the intersection of \p{InThai} and \P{Cn}." POSIX bracket-expression "character class": [[:alpha:]] A POSIX character class is one of several special metasequences for use within a POSIX bracket expression. An example is [:lower:], which represents any lowercase letter within the current locale. For English text, [:lower:] is comparable to a-z. Since this entire sequence is valid only within a bracket expression, the full class comparable to [a-z] is [[:lower:]]. Yes, it's that ugly. But, it has the advantage over [a-z] of including other characters, such as ö, ñ, and the like if the locale actually indicates that they are "lowercase letters." The exact list of POSIX character classes is locale dependent, but the following are usually supported: [:alnum:] alphabetic characters and numeric character [:alpha:] alphabetic characters [:blank:] space and tab [:cntrl:] control characters [:digit:] digits [:graph:] non-blank characters (not spaces, control characters, or the like) [:lower:] lowercase alphabetics [:print:] like [:graph:], but includes the space character [:punct:] punctuation characters [:space:] all whitespace characters ([:blank:], newline, carriage return, and the like) [:upper:] uppercase alphabetics [:xdigit:] digits allowed in a hexadecimal number (i.e., 0-9a-fA-F). POSIX bracket-expression "collating sequences" : [[.span-ll.]] A locale can have collating sequences to describe how certain characters or sets of characters should be ordered. For example, in Spanish, the two characters ll (as in tortilla) traditionally sort as if they were one logical character between l and m, and the German ß is a character that falls between s and t, but sorts as if it were the two characters ss. These rules might be manifested in collating sequences named, for example, span-ll and eszet. A collating sequence that maps multiple physical characters to a single logical character, such as the span-ll example, is considered "one character" to a fully compliant POSIX regex engine. This means that [^abc] matches a 'll' sequence. A collating sequence element is included within a bracket expression using a[..] notation: torti[[.span-ll.]]a matches tortilla. A collating sequence allows you to match against those characters that are made up of combinations of other characters. It also creates a situation where a bracket expression can match more than one physical character. POSIX bracket-expression "character equivalents" : [[=n=]] Some locales define character equivalents to indicate that certain characters should be considered identical for sorting and such. For example, a locale might define an equivalence class 'n' as containing n and ñ, or perhaps one named 'a' as containing a, à, and ´. Using a notation similar to [::], but with '=' instead of a colon, you can reference these equivalence classes within a bracket expression: [[=n=][=a=]] matches any of the characters just mentioned. If a character equivalence with a single-letter name is used but not defined in the locale, it defaults to the collating sequence of the same name. Locales normally include normal characters as collating sequences [.a.], [.b.], [.c.], and so onso in the absence of special equivalents, [[=n=][=a=]] defaults to [na]. Emacs syntax classes GNU Emacs doesn't support the traditional \w, \s, etc.; rather, it uses special sequences to reference "syntax classes" : \schar: matches characters in the Emacs syntax class as described by char \Schar: matches characters not in the Emacs syntax class \sw matches a "word constituent" character, and \s- matches a "whitespace character." These would be written as \w and \s in many other systems. Emacs is special because the choice of which characters fall into these classes can be modified on the fly, so, for example, the concept of which characters are word constituents can be changed depending upon the kind of text being edited. Anchors and Other "Zero-Width Assertions" Anchors and other "zero-width assertions" don't match actual text, but rather positions in the text. Start of line/string: ^, \A Caret ^ matches at the beginning of the text being searched, and, if in an enhanced line-anchor match mode, after any newline. In some systems, an enhanced-mode ^ can match after Unicode line terminators as well. When supported, \A always matches only at the start of the text being searched, regardless of any match mode. End of line/string:$, \Z, \z $has a variety of meanings among different tools, but the most common meaning is that it matches at the end of the target string, and before a string-ending newline, as well. The latter is common, to allow an expression like s$ (ostensibly, to match "a line ending with s") to match '...s', a line ending with s that's capped with an ending newline. Two other common meanings for $are to match only at the end of the target text, and to match before any newline. In some Unicode systems, the special meaning of newline in these rules are replaced by Unicode line terminators(Java, for example, offers particularly complex semantics for$ with respect to Unicode line terminators). A match mode can change the meaning of $to match before any embedded newline (or Unicode line terminator as well). When supported, \Z usually matches what the "unmoded"$ matches, which often means to match at the end of the string, or before a string-ending newline. To complement these, \Z matches only at the end of the string, period, without regard to any newline. Start of match (or end of previous match): \G Word boundaries: \b, \B(Not word-boundary), \<, \>, ... Lookahead (?=⋯), (?!⋯); Lookbehind, (?<=⋯), (?<!⋯) The most restrictive rule exists in Perl and Python, where the lookbehind can match only fixed-length strings. For example, (?<!;\w) and (?<!this\|that) are allowed, but (?<!books?) and (?<^\w+:) are not, as they can match a variable amount of text. The next level of support allows alternatives of different lengths within the look behind, so (?<!books?) can be written as (?<!book\|books). PCRE (and as such the preg suite in PHP) allows this. The next level allows for regular expressions that match a variable amount of text, but only if it's of a finite length. This allows (?<!books?) directly, but still disallows(?<!^\w+:) since the \w+ is open-ended. Sun's Java regex package supports this level. The fourth level, however, allows the subexpression within lookbehind to match any amount of text, including the (?<!^\w+:) example. This level, supported by Microsoft's .NET languages, is truly superior to the others, but does carry a potentially huge efficiency penalty if used unwisely. (When faced with lookbehind that can match any amount of text, the engine is forced to check the lookbehind subexpression from the start of the string, which may mean a lot of wasted effort when requested from near the end of a long string.) Mode modifier: (?modifier), such as (?i) or (?-i) Many flavors now allow some of the regex and match modes to be set within the regular expression itself. A common example is the special notation (?i), which turns on case-insensitive matching, and (?-i), which turns it off. For example, <B>(?i)very(?-i)</B> has the very part match with case insensitivity, while still keeping the tag names case-sensitive. This example works with most systems that support (?i), including Perl, PHP, java.util.regex, Ruby,[] and the .NET languages. It doesn't work with Python or Tcl, neither of which support (?-i). With most implementations except Python, the effects of (?i) within any type of parentheses are limited by the parentheses (that is, turn off at the closing parentheses). So, the (?-i) can be eliminated by wrapping the case-insensitive part in parentheses and putting (?i) as the first thing inside: <B>(?:(?i)very)</B>. Common Mode Modifiers Letter Mode I case-insensitivity match mode x free-spacing and comments regex mode s dot-matches-all match mode m enhanced line-anchor match mode Mode-modified span: (?modifier :⋯), such as (?i:⋯) Using a syntax like (?i:⋯), a mode-modified span turns on the mode only for what's matched within the parentheses. The <B>(?:(?i)very)</B> example is simplified to <B>(?i:very)</B>. When supported, this form generally works for all mode-modifier letters the system supports. Tcl and Python are two examples that support the (?i) form, but not the mode-modified span (?i:⋯) form. Literal-text span: \Q⋯\E The special sequence \Q⋯\E turns off all regex metacharacters between them, except for \E itself. (If the \E is omitted, they are turned off until the end of the regex.) It allows what would otherwise be taken as normal metacharacters to be treated as literal text. This is especially useful when including the contents of a variable while building a regular expression. With the Perl code m/\Q$query\E/i, a$query of 'C:\WINDOWS\' becomes C\:\\WINDOWS\\, resulting in a search that finds the original 'C:\WINDOWS\' as the user expects. This feature is less useful in systems with procedural and object-oriented handling, as they accept normal strings. While building the string to be used as a regular expression, it's fairly easy to call a function to make the value from the variable "safe" for use in a regular expression. PHP has the preg_quote function; Java has a quote method. The only regex engines that I know of that support \Q⋯\E are java.util.regex and PCRE. Perl supports \Q⋯\E within regex literals, but not within the contents of variables that might be interpolated into them. Grouping, Capturing, Conditionals, and Control Capturing/Grouping Parentheses: (⋯) and \1, \2, ... Common, unadorned parentheses generally perform two functions, grouping and capturing. One of the most common uses of parentheses is to pluck data from a string. The text matched by a parenthesized subexpression (also called "the text matched by the parentheses") is made available after the match in different ways by different programs, such as Perl's $1,$2, etc. (A common mistake is to try to use the \1 syntax outside the regular expression; something allowed only with sed and vi.) Grouping-only parentheses: (?:⋯) Entire match First set of parentheses Perl $&$1 PHP $matches[0]$matches[1] Python MatchObj.group(0) MatchObj.group(1) Ruby $&$1 Java MatcherObj.group() MatcherObj.group(1) vi & \1 Named capture: (?<Name>⋯) Python, PHP's preg engine, and .NET languages support captures to named locations. Python and PHP use the syntax (?P<name>⋯), while the .NET languages use (?<name>⋯). for .Net: \b(?<Area>\d\d\d\)-(?<Exch>\d\d\d)-(?<Num>\d\d\d\d)\b for Python/PHP: \b(?P<Area>\d\d\d\)-(?P<Exch>\d\d\d)-(?P<Num>\d\d\d\d)\b Program can then refer to each matched substring through its name. RegexObj.Groups["Area"] in C#, RegexObj.group("Area") in Python, and \$matches["Area"] in PHP. Within the regular expression itself, the captured text is available via \k<Area> with .NET, and (?P=Area) in Python and PHP. With Python and .NET (but not with PHP), you can use the same name more than once within the same expression. For example, to match the area code part of a US phone number, which look like '(###)' or '###-', you might use (shown in .NET syntax): ⋯(?:$$(?<Area>\d\d\d)$$\|(?<Area>\d\d\d)-)⋯. When either set matches, the three-digit code is saved to the name Area. Atomic grouping: (?>⋯) Atomic grouping, (?>⋯), will be very easy to explain once the important details of how the regex engine carries out its work is understood (☞169). Here, I'll just say that once the parenthesized subexpression matches, what it matches is fixed (becomes atomic, unchangeable) for the rest of the match, unless it turns out that the whole set of atomic parentheses needs to be abandoned and subsequently revisited. A simple example helps to illustrate this indivisible, "atomic" nature of text matched by these parentheses. The string '¡Hola!' is matched by ¡.!, but is not matched if . is wrapped with atomic grouping, ¡(?>.)!. In either case, . first internally matches as much as it can ('¡'), but the inability of the subsequent ! to match wants to force the .* to give up some of what it had matched (the final '!'). That can't happen in the second case because .* is inside atomic grouping, which never "gives up" anything once the matching leaves them. Although this example doesn't hint at it, atomic grouping has important uses. In particular, it can help make matching more efficient (☞171), and can be used to finely control what can and can't be matched (☞269). Alternation: ⋯\| ⋯ \| ⋯ Alternation has very low precedence, so this and\|or that matches the same as (this and)\|(or that), and not this (and\|or) that, even though visually, the and\|or looks like a unit. Most flavors allow an empty alternative, as in (this\|that\|). The empty subexpression can always match, so this example is comparable to (this\|that)?. Conditional: (?if then \|else) This construct allows you to express an if/then/else within a regex. The if part is a special kind of conditional expression discussed in a moment. Both the then and else parts are normal regex subexpressions. If the if part tests true, the then expression is attempted. Otherwise, the else part is attempted. (The else part may be omitted, and if so, the '\|' before it may be omitted as well.) The kinds of if tests available are flavor-dependent, but most implementations allow at least special references to capturing subexpressions and lookaround. Greedy quantifiers: , +, ?, {num,num} Intervals {min,max}or \{min,max \} Lazy quantifiers: ?, +?, ??, {num,num}? Quantifiers are normally "greedy," and try to match as much as possible. Conversely, these non-greedy versions match as little as possible, just the bare minimum needed to satisfy the match. Possessive quantifiers: *+, ++, ?+, {num,num}+ Currently supported only by java.util.regex and PCRE (and hence PHP), possessive quantifiers are like normally greedy quantifiers, but once they match something, they never "give it up." ## Labels Java (159) Lucene-Solr (110) Interview (59) All (58) J2SE (53) Algorithm (45) Soft Skills (36) Eclipse (34) Code Example (31) Linux (24) JavaScript (23) Spring (22) Windows (22) Web Development (20) Nutch2 (18) Tools (18) Bugs (17) Debug (15) Defects (14) Text Mining (14) J2EE (13) Network (13) PowerShell (11) Chrome (9) Design (9) How to (9) Performance (9) UIMA (9) html (9) Http Client (8) Maven (8) Security (8) bat (8) blogger (8) Big Data (7) Google (7) Guava (7) JSON (7) ANT (6) Database (6) Scala (6) Shell (6) css (6) Cache (5) IDE (5) Tips (5) adsense (5) xml (5) AIX (4) Code Quality (4) GAE (4) Git (4) Jackson (4) Memory Usage (4) Miscs (4) OpenNLP (4) Spark (4) Testing (4) ads (4) Android (3) Apache Spark (3) Concurrency (3) Eclipse RCP (3) English (3) IBM (3) JAX-RS (3) Jetty (3) Script (3) regex (3) seo (3) .Net (2) Apache (2) Architecture (2) Batch (2) Build (2) C# (2) C/C++ (2) CSV (2) Career (2) Cassandra (2) Distributed (2) Fiddler (2) Firefox (2) Google Drive (2) Gson (2) Html Parser (2) Http (2) Image Tools (2) JQuery (2) Jersey (2) LDAP (2) Life (2) Logging (2) Python (2) Storage (2) Text Search (2) xml parser (2) AOP (1) AspectJ (1) Cloud (1) Codility (1) Data Mining (1) Exif (1) FindBugs (1) Greasemonkey (1) HTML5 (1) Httpd (1) I18N (1) JDK8 (1) JMX (1) Mac (1) Mobile (1) Netbeans (1) Notes (1) Perl (1) Problems (1) Quality (1) Redhat (1) Redis (1) Review (1) RxJava (1) boilerpipe (1) htm (1) ongoing (1) procrun (1) rss (1)	2017-08-16 13:22:42	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4618697166442871, "perplexity": 6988.8378026228975}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-34/segments/1502886101966.48/warc/CC-MAIN-20170816125013-20170816145013-00223.warc.gz"}
https://deborahhindi.com/clarenza/integration-by-parts-fraction-example.php	Integrating algebraic fractions 1 mathcentre.ac.uk Lecture 13. Integration By Partial Fractions, Partial Fractions Deposition Fall 2010, Integration By Parts: Prof X-Squared... 00:00. Example 3 (Kristakingmath ## Integration using partial fractions University of Auckland 7.4 Integration by Partial Fractions www.math.uci.edu. Integration by Parts: Example 4. The integration by parts formula gives (which comes up doing the same ideas used in partial fraction decomposition), Integration by Parts: Indefinite Integrals; Integration by Partial Fractions Exercises. BACK; NEXT ; Example 1. Example 2. Find. Show Answer. Integration by Parts: Integration by Partial Fractions Examples. BACK; NEXT ; Example 1. Decompose . If we take x = -5, for example, Integration using partial fractions and is very useful for preparing such functions for integration The sum of partial fractions includes (see examples below math worksheet a maths polynomial partial fraction integration questions integration by parts partial fractions worksheet with solutions Integration by Parts: Indefinite Integrals; Integration by Partial Fractions Exercises. BACK; NEXT ; Example 1. Example 2. Find. Show Answer Again in this example it is easier to integrate the derivative than the thing itself: Partial fractions; вЂњIntegration by parts.вЂќ From Math Insight. Integration using partial fractions and is very useful for preparing such functions for integration The sum of partial fractions includes (see examples below gives the formula for integration by parts. Two other well-known examples are when integration by parts is applied to a function expressed as a product of 1 and Integration: Integration by Partial Fractions Step 1 If you are integrating a rational function p(x) q(x) where degree of p(x) is greater than degree of Integration By Parts: A "Loopy" Example Integration by Prof. , Coefficients of a Partial Fraction Decomposition; Integration of Rational functions. for example. Note that integration by parts will not be enough to help integrate a (in short Partial Fraction In this section we will be looking at Integration by Parts integration techniques in this example and do integration by parts. In your later math Integration by Parts; Integration by Partial Fraction Decomposition is a procedure where we can вЂњdecomposeвЂќ a proper rational Partial Fraction Term. Example Math explained in easy language, Example: what is the integral of sin(x) ? Integration by Parts. See Integration by Parts. THE METHOD OF INTEGRATION BY PARTS For example, if , then the differential of is . kouba@math.ucdavis.edu . Integration: By Parts & By Partial Fractions This simple example illustrates this simple concept of adding WORSHEET ON INTEGRATION BY PARTS AND PARTIAL FRACTIONS Integration using partial fractions and is very useful for preparing such functions for integration The sum of partial fractions includes (see examples below Integration using partial fractions and is very useful for preparing such functions for integration The sum of partial fractions includes (see examples below Practice finding indefinite integrals using the method of integration by parts. If you're seeing this message, Integrating using linear partial fractions. a Quotient Rule Integration by Parts formula, apply the resulting integration formula to an example, Integration: Integration by Partial Fractions Step 1 If you are integrating a rational function p(x) q(x) where degree of p(x) is greater than degree of ### 7. Integration by Parts Learn math while you play with it! Lecture 22 Integration By Parts A "Loopy" Example. Free Pre-Algebra, Algebra, Trigonometry, Calculus, Geometry, Statistics and Chemistry calculators step-by-step, It shows you how the concept of Integration by Parts can be applied to solve problems using the Cymath solver. Cymath is an online math equation solver Examples. ### Calculus II Integration by Parts - Pauls Online Math Notes Integration by Partial Fractions Exercises Shmoop. math worksheet a maths polynomial partial fraction integration questions integration by parts partial fractions worksheet with solutions https://en.wikipedia.org/wiki/Integral math worksheet a maths polynomial partial fraction integration questions integration by parts partial fractions worksheet with solutions. • Integration by Partial Fractions Exercises Shmoop • Integration by Substitution Parts & Partial Fractions • Chapter 6 Integration partial fractions and improper • Integration by Parts Example 4 S.O.S. Mathematics • Lecture 13. Integration By Partial Fractions, Partial Fractions Deposition Fall 2010, Integration By Parts: Prof X-Squared... 00:00. Example 3 (Kristakingmath Integration by parts . A special rule, integration by parts, can often be used to integrate the product of two functions. It is appropriate when one of the functions Integration using partial fractions and is very useful for preparing such functions for integration The sum of partial fractions includes (see examples below INTEGRATION-BY-PARTS & INTEGRATION OF RATIONAL FUNCTIONS "Integration-by-Parts" is a technique of limited applicability that exchanges an Partial Fraction Practice finding indefinite integrals using the method of integration by parts. If you're seeing this message, Integrating using linear partial fractions. This brings us to an integration technique known as integration by parts, used to solve such integrals. For example, write the rational fraction as the sum of Free Pre-Algebra, Algebra, Trigonometry, Calculus, Geometry, Statistics and Chemistry calculators step-by-step Integration by Parts; Integration by Partial Fraction Decomposition is a procedure where we can вЂњdecomposeвЂќ a proper rational Partial Fraction Term. Example Advanced Math Solutions Integration by parts is essentially the reverse of the product rule. this is one of the more intuitive examples. Integrating algebraic fractions 1 integration by substitution, and that is the case for the п¬Ѓrst three of the examples. We call fractions Integrating algebraic fractions (1) The integral of an algebraic fraction can often be found by first expressing the fraction as the sum of its partial fractions. Another example is the fraction (x (x+ 1) = (x 3) 4 (x 1) The technique of integration by partial fractions is based on a deep all of the cases of integration Another example is the fraction (x (x+ 1) = (x 3) 4 (x 1) The technique of integration by partial fractions is based on a deep all of the cases of integration Integration by using this formula is called integration by parts. Example 1. Compute \(\int {x Two such methods - Integration by Parts, and Reduction to Partial Fractions are discussed here. Example вЂ“ To find the partial fractions corresponding to \ How to Integrate by Partial Fractions. partial fractions can be used to simplify integration. We have now split up the fraction into its constituent parts. How to Integrate by Partial Fractions. partial fractions can be used to simplify integration. We have now split up the fraction into its constituent parts. Integration by Parts: Indefinite Integrals; Integration by Partial Fractions Exercises. BACK; NEXT ; Example 1. Example 2. Find. Show Answer MIXED INTEGRATION PRACTICE For example, any function in a INTEGRATION BY PARTS. Integration by parts is the integration counterpart to the product rule for Integration: By Parts & By Partial Fractions This simple example illustrates this simple concept of adding WORSHEET ON INTEGRATION BY PARTS AND PARTIAL FRACTIONS Advanced Math Solutions Integration by parts is essentially the reverse of the product rule. this is one of the more intuitive examples. Example 1 в€«x 4 ln(x) dx. The logical two halves of this are x 4 and ln(x). In the ILATE table, x 4 is "A" for algebraic, and ln(x) is "L" for logarithmic. ## Integration by Parts S.O.S. Math Integration by Substitution Parts & Partial Fractions. Another example is the fraction (x (x+ 1) = (x 3) 4 (x 1) The technique of integration by partial fractions is based on a deep all of the cases of integration, Integration by parts . A special rule, integration by parts, can often be used to integrate the product of two functions. It is appropriate when one of the functions. ### Integration By Partial Fractions Worksheet integration 7.4 Integration by Partial Fractions www.math.uci.edu. Integration of Logarithmic Functions. The following are some examples of integrating logarithms via U-substitution: Using integration by parts once again,, In this section we will be looking at Integration by Parts integration techniques in this example and do integration by parts. In your later math. Integration of Logarithmic Functions. The following are some examples of integrating logarithms via U-substitution: Using integration by parts once again, Chapter 6: Integration: partial fractions and improper integrals Course 1S3, 2006вЂ“07 April 5, 2007 These are just summaries of the lecture notes, and few details math worksheet a maths polynomial partial fraction integration questions integration by parts partial fractions worksheet with solutions Integration by Parts: Example 4. The integration by parts formula gives (which comes up doing the same ideas used in partial fraction decomposition) Chapter 6: Integration: partial fractions and improper integrals Course 1S3, 2006вЂ“07 April 5, 2007 These are just summaries of the lecture notes, and few details Integration: Integration by Partial Fractions Step 1 If you are integrating a rational function p(x) q(x) where degree of p(x) is greater than degree of In this section we will use partial fractions to rewrite integrands into a form that will allow Integration Techniques. Integration by Parts; For example Integration by reduction formulae integration by parts, integration by trigonometric substitution, for example I n-1 or I n-2. 7.4 Integration by Partial Fractions Example Here we write the integrand as a polynomial plus a rational function 7 x+2 whose denom- Example 1 в€«x 4 ln(x) dx. The logical two halves of this are x 4 and ln(x). In the ILATE table, x 4 is "A" for algebraic, and ln(x) is "L" for logarithmic. 25/01/2011В В· Examples showing how various functions can be Irrational Algebraic Fraction Of The Following The Integration By Parts Twice To Regain The Original Integration By Parts: A "Loopy" Example Integration by Prof. , Coefficients of a Partial Fraction Decomposition; Integration by Parts A-Level Mathematics revision section of Revision Maths looking at Integration by Parts which is very useful in integration: Example. Find Practice finding indefinite integrals using the method of integration by parts. If you're seeing this message, Integrating using linear partial fractions. In this section we will use partial fractions to rewrite integrands into a form that will allow Integration Techniques. Integration by Parts; For example In this section we will use partial fractions to rewrite integrands into a form that will allow Integration Techniques. Integration by Parts; For example Example 1 в€«x 4 ln(x) dx. The logical two halves of this are x 4 and ln(x). In the ILATE table, x 4 is "A" for algebraic, and ln(x) is "L" for logarithmic. Lecture 13. Integration By Partial Fractions, Partial Fractions Deposition Fall 2010, Integration By Parts: Prof X-Squared... 00:00. Example 3 (Kristakingmath Example: What is в€« x cos(x) dx ? OK, we have x multiplied by cos(x), so integration by parts is a good choice. First choose which functions for u and v: Lecture 13. Integration By Partial Fractions, Partial Fractions Deposition Fall 2010, Integration By Parts: Prof X-Squared... 00:00. Example 3 (Kristakingmath Integration using partial fractions University of Auckland. This section contains lecture video excerpts, lecture notes, problem solving videos, and a worked example on integration by parts., math worksheet a maths polynomial partial fraction integration *questions integration by parts partial fractions worksheet with solutions. ### Integration by Parts Maths Tutor INTEGRATION BY PARTIAL FRACTIONS. Again in this example it is easier to integrate the derivative than the thing itself: Partial fractions; вЂњIntegration by parts.вЂќ From Math Insight., Integration: By Parts & By Partial Fractions This simple example illustrates this simple concept of adding WORSHEET ON INTEGRATION BY PARTS AND PARTIAL FRACTIONS. ### Calculus II Integration by Parts - Pauls Online Math Notes Integration by Partial Fractions Examples shmoop.com. Integration by parts . A special rule, integration by parts, can often be used to integrate the product of two functions. It is appropriate when one of the functions https://en.wikipedia.org/wiki/Partial_fractions_in_integration Chapter 6: Integration: partial fractions and improper integrals Course 1S3, 2006вЂ“07 April 5, 2007 These are just summaries of the lecture notes, and few details. Integration by parts is based on the derivative of a product of 2 functions. Integration by Partial Fractions; So for this example, Integration by Parts: Integration by Partial Fractions Examples. BACK; NEXT ; Example 1. Decompose . If we take x = -5, for example, Integration of Rational functions. for example. Note that integration by parts will not be enough to help integrate a (in short Partial Fraction a Quotient Rule Integration by Parts formula, apply the resulting integration formula to an example, Integration by parts includes integration of two functions which are in multiples. The famous ILATE rule is followed in integration by parts to ease the calculation. Example: What is в€« x cos(x) dx ? OK, we have x multiplied by cos(x), so integration by parts is a good choice. First choose which functions for u and v: Integration by Parts; Integration by Partial Fraction Decomposition is a procedure where we can вЂњdecomposeвЂќ a proper rational Partial Fraction Term. Example ... (Fraction) Notation; Integration by Substitution: Definite Integrals; Integration by Parts: Indefinite Integrals; Some Tricks; Integration by Parts: Example 2 Lecture 13. Integration By Partial Fractions, Partial Fractions Deposition Fall 2010, Integration By Parts: Prof X-Squared... 00:00. Example 3 (Kristakingmath Integrating algebraic fractions (1) The integral of an algebraic fraction can often be found by first expressing the fraction as the sum of its partial fractions. Practice finding indefinite integrals using the method of integration by parts. If you're seeing this message, Integrating using linear partial fractions. Integration: By Parts & By Partial Fractions This simple example illustrates this simple concept of adding WORSHEET ON INTEGRATION BY PARTS AND PARTIAL FRACTIONS Integration: By Parts & By Partial Fractions This simple example illustrates this simple concept of adding WORSHEET ON INTEGRATION BY PARTS AND PARTIAL FRACTIONS In this section we will use partial fractions to rewrite integrands into a form that will allow Integration Techniques. Integration by Parts; For example a Quotient Rule Integration by Parts formula, apply the resulting integration formula to an example, gives the formula for integration by parts. Two other well-known examples are when integration by parts is applied to a function expressed as a product of 1 and Integration by Parts: Integration by Partial Fractions Examples. BACK; NEXT ; Example 1. Decompose . If we take x = -5, for example, The purpose of integration by parts is to replace a difficult integral with one that Excel in math and let's find the answer to the example above This brings us to an integration technique known as integration by parts, used to solve such integrals. For example, write the rational fraction as the sum of Integration by Parts: Integration by Partial Fractions Examples. BACK; NEXT ; Example 1. Decompose . If we take x = -5, for example, It shows you how the concept of Integration by Parts can be applied to solve problems using the Cymath solver. Cymath is an online math equation solver Examples Another example is the fraction (x (x+ 1) = (x 3) 4 (x 1) The technique of integration by partial fractions is based on a deep all of the cases of integration Discussion of discounted cash flow analysis for real estate or the gross rent last week if he knew of a way to value an income property using Rental method of valuation example Rugby An income approach is a real estate appraisal When using the income approach for purchasing a rental The cost approach valuation method	2021-10-25 22:24:12	{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9281934499740601, "perplexity": 728.3433043223356}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-43/segments/1634323587770.37/warc/CC-MAIN-20211025220214-20211026010214-00131.warc.gz"}
http://mathoverflow.net/questions/31025/intuitive-explanation-for-the-use-of-matrix-factorizations-in-knot-theory	# Intuitive explanation for the use of matrix factorizations in knot theory Hello! I read through parts of Khovanov/Rozansky's paper on the categorification of the HOMFLY polynomial using Matrix Factorizations. Technically, I can follow (though it seems to me that quite a lot of details are missing and tedious to fill in) - intuitively, however, I have no idea why one is lead to consider matrix factorizations when studying knot theory, in particular the RT invariants obtained from interpreting colored tangles as morphisms between modules over the quantum group. Until now, it feels quite mysterious to me why Khovanov and Rozansky choose particular potentials like $x_1^n+x_2^n-x_3^n-x_4^n$ in their construction, and why one should expect that in the end we get something invariant under the Reidemeister moves. Can somebody explain to me the motivation behind this construction? What is the relation between the morphism of modules over the quantum group a wide edge represents and the matrix factorization associated to it? Thank you! - Many knot homologies are expected to have Floer-theoretic interpretations. However, in Floer theory often the chain "complex" $CF(L_0,L_1)$ is not a complex but rather an a-infty bimodule over a pair of curved a-infty algebras; matrix factorizations are more or less a special case of these, where the curvature of the ainfty algebras is a multiple of the identity. Under some nice (monotone or exact) assumptions $CF(L,L)$ has differential that squares to zero, since the curvatures of the a-infty algebras on both sides occur with opposite sign and so cancel if the Lagrangians are the same (or related by a symplectomorphism). But even in the monotone or exact situations $CF(L_0,L_1)$ can have a differential which gives a matrix factorization. Now Manolescu has proposed a symplectic interpretation of Khovanov-Rozansky for links, which looks like $CF(L,\beta(L))$ where \beta is a braid presentation of the link, but so far there doesn't seem to be a proposal for graphs. Probably for graphs the invariant would be the homotopy type of $CF(L_0,L_1)$ (or some more general quilted Floer chain group) which under suitable monotonicity assumptions will be a matrix factorization. Note that Kamnitzer has a proposal for a Floer-theoretic version for arbitrary G; it would be interesting to see if one could extend this proposal to the graph case, which one would probably need for an exact triangle.	2014-03-11 14:08:56	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8662356734275818, "perplexity": 381.7195762825097}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-10/segments/1394011207526/warc/CC-MAIN-20140305092007-00042-ip-10-183-142-35.ec2.internal.warc.gz"}
https://assignmentstutors.com/physical-assessment-technique-practice/	# Physical Assessment Technique Practice Physical Assessment Techniques Practice: Conduct a focused assessment utilizing either inspection, palpation, or auscultation as the physical assessment technique. Write a summary of the assessment and the skill utilized. Answer the following 3 questions in the summary. What skill was utilized during the focused assessment? Describe how you felt during the assessment? What can you do to improve your technique the next time? 0 replies	2023-03-20 12:54:49	{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9238009452819824, "perplexity": 10016.175011167052}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296943483.86/warc/CC-MAIN-20230320114206-20230320144206-00441.warc.gz"}
https://www.gradesaver.com/textbooks/math/other-math/CLONE-547b8018-14a8-4d02-afd6-6bc35a0864ed/chapter-7-measurement-7-4-problems-solving-with-metric-measurement-7-4-exercises-page-507/12	## Basic College Mathematics (10th Edition) Convert the length of the fabric to m. $10m30cm=10m+0.30m=10.3m$ Divide by 3 to find the amount of fabric for each window. $10.3\div3=3.4\overline3$	2019-10-18 13:25:26	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4587976634502411, "perplexity": 898.5584075717923}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-43/segments/1570986682998.59/warc/CC-MAIN-20191018131050-20191018154550-00114.warc.gz"}
http://mathoverflow.net/revisions/113824/list	MathOverflow will be down for maintenance for approximately 3 hours, starting Monday evening (06/24/2013) at approximately 9:00 PM Eastern time (UTC-4). 3 added 2 characters in body The maximal subgroups of $A_n$ are given by the O'Nan-Scott Theorem. They lie in one of the following classes: 1) $A_n \cap (S_{n-k} \times S_k)$, that is the stabiliser of a $k$-set. 2) $A_n \cap (S_a wr S_b)$ where $n=ab$, that is the stabiliser of a partition. 3) $A_n\cap AGL(d,p)$ where $n=p^d$ for some prime $p$. 4) $A_n \cap (S_m wr S_k)$ where $n=m^k$, that is the stabiliser of a cartesian power. 5) $A_n \cap (T^{k+1}.(Out(T) \times S_{k+1}))$ where $n=\|T\|^k$ and $T$ is a finite nonabelian simple group 6) an almost simple group acting primitively on $n$ points. For classical groups the main result is Aschbacher's Theorem in the paper pointed out by Rivin. More details of the structure of the subgroups is given in the book by Kleidman and Liebeck. For exceptional groups of Lie type there are papers by Liebeck and Seitz as noted by Barnea. Both these results and Aschbacher's Theorem have the same philosophy as the O'Nan-Scott Theorem, namely that a maximal subgroup is either one of a small number of natural families that are usually stabilisers of some geometric structure, or is almost simple. For sporadic simple groups, all the information in in the online Atlas. They are all known except for the monster. In this case there are a couple of possibilities for maximal subgroups but where it is not known if they are actually subgroups. 2 added 18 characters in body The maximal subgroups of $A_n$ are given by the O'Nan-Scott Theorem. They lie in one of the following classes: 1) $A_n \cap (S_{n-k} \times S_k)$, that is the stabiliser of a $k$-set. 2) $A_n \cap (S_a wr S_b)$ where $n=ab$, that is the stabiliser of a partition. 3) $A_n\cap AGL(d,p)$ where $n=p^d$ for some prime $p$. 4) $A_n \cap (S_m wr S_k)$ where $n=m^k$, that is the stabiliser of a cartesian power. 5) $A_n \cap (T^{k+1}.(Out(T) \times S_{k+1}))$ where $n=\|T\|^k$ and $T$ is a finite nonabelian simple group 6) an almost simple group acting primitively on $n$ points. For classical groups the main result is Aschbacher's Theorem in the paper pointed out by Rivin. More details of the structure of the subgroups is given in the book by Kleidman and Liebeck. For exceptional groups of Lie type there are papers by Liebeck and Seitz as noted by Barnea. Both these results and Aschbacher's Theorem have the same philosophy as the O'Nan-Scott Theorem, namely that a maximal subgroup is either one of a small number of natural families that are usually stabilisers of some geometric structure, or is almost simple. For sporadic simple groups, all the information in in the online Atlas. They are all known except for the monster. In this case there are a couple of possibilities for maximal subgroups but it is not known if they are actually subgroups. 1 The maximal subgroups of $A_n$ are given by the O'Nan-Scott Theorem. They lie in one of the following classes: 1) $A_n \cap (S_{n-k} \times S_k)$, that is the stabiliser of a $k$-set. 2) $A_n \cap (S_a wr S_b)$ where $n=ab$, that is the stabiliser of a partition. 3) $A_n\cap AGL(d,p)$ where $n=p^d$ for some prime $p$. 4) $A_n \cap (S_m wr S_k)$ where $n=m^k$, that is the stabiliser of a cartesian power. 5) $A_n \cap (T^{k+1}.(Out(T) \times S_{k+1}))$ where $n=\|T\|^k$ and $T$ is a finite nonabelian simple group 6) an almost simple group acting primitively on $n$ points. For classical groups the main result is Aschbacher's Theorem in the paper pointed out by Rivin. More details of the structure of the subgroups is given in the book by Kleidman and Liebeck. For exceptional groups of Lie type there are papers by Liebeck and Seitz. Both these results and Aschbacher's Theorem have the same philosophy as the O'Nan-Scott Theorem, namely that a maximal subgroup is either one of a small number of natural families that are usually stabilisers of some geometric structure, or is almost simple. For sporadic simple groups, all the information in in the online Atlas. They are all known except for the monster. In this case there are a couple of possibilities for maximal subgroups but it is not known if they are actually subgroups.	2013-06-20 02:18:09	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8138908743858337, "perplexity": 234.87498747775018}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2013-20/segments/1368710006682/warc/CC-MAIN-20130516131326-00094-ip-10-60-113-184.ec2.internal.warc.gz"}
https://dsp.stackexchange.com/questions/55226/frequency-resolution-problem/55231	# Frequency Resolution Problem Im trying to tackle the following problem while still not having a firm idea on what "frequency resolution" means : Suppose we sample a continuous time signal with sampling period Ts = 1/2000, and then use a window of length 1000 on the resulting discrete time signal. If we transform it using a 2000 point DFT what would its frequency resolution be ? Can anyone help me figure this out ? • Do you want potential plot resolution with interpolation, peak location estimation resolution given an S/N, result bin separation, or peak separation resolution with a separation criteria? All these produce different frequency resolutions for the same length DFT. – hotpaw2 Feb 1 at 21:33 Depends a bit on what you are trying to achieve. If you do an FFT of length $$N$$ of a signal sampled at sampled at a rate of $$F_s$$, then many people would say your frequency resolution is $$\frac{F_s}{N}$$. Whether that's correct or not, really depends on how exactly you define frequency resolution and what you are planning to do with it. What's really happening is that you sample a frequency domain function with a sampling interval of $$\frac{F_s}{N}$$. As soon as you pick an FFT size, you are sampling in both domains with the sampling intervals being $$\frac{1}{F_s}$$ in time and $$\frac{F_s}{N}$$ in frequency. Frequency domain sampling has all the same properties, requirements and problems as time domain sampling, you can get aliasing, you can interpolate, there is assumed periodicity in the other domain, etc. By simply applying the sampling theorem we could argue that the frequency resolution required to fully characterize a signal is simply the inverse of the length in the time domain. This works well for signals that are inherently time bound, such as the impulse response of an LTI system. However it's not practical for long continuous signals. In this case you need to pick a frequency resolution that's "good enough" for your application, and that really depends on the requirements and goal of your specific application. The Sampling is given by $${T}_{s} = \frac{1}{2000}$$ [Sec]. The Window length is 1000 Samples. Since the Window length must be equal to the data length we infer the data length is 1000 samples which means the sampling time is $$0.5$$ [Sec]. The Bin resolution in DFT is the ration between the sampling interval to the number of DFT Samples, which in this case is 2000. Hence the bin resolution is $$\frac{1}{4000}$$ [Hz].	2019-11-15 09:44:14	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 9, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6916500926017761, "perplexity": 510.85644830720844}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-47/segments/1573496668618.8/warc/CC-MAIN-20191115093159-20191115121159-00553.warc.gz"}
http://mathematica.stackexchange.com/questions/25993/how-does-mathematicas-findsequencefunction-work	How does Mathematica's FindSequenceFunction work? Mathematica's built-in function FindSequenceFunction can find out a closed form of the general terms of a given integer sequence. Anyone knows how it works? Is there any related reference papers? - You already posted this question here stackoverflow.com/q/16772823/353410. You may want to delete one of them – belisarius May 28 '13 at 12:10 Irrespective of what is implemented in Mathematica, the math of it is explained in lecture notes (pdf‌) by Manuel Kauers. – Sasha May 28 '13 at 14:51 Running TracePrint[] on an instance of FindSequenceFunction[], with the option TraceInternal set to True, reveals that Mathematica does a number of things (lookups, sequence transforms, etc.) to analyze a sequence given to FindSequenceFunction[]; a lot of the functions called are in the hidden contexts DiscreteFindSequenceFunctionDump and DiscreteFindSequenceFunctionDataDump . – Ｊ. Ｍ. May 28 '13 at 15:03 @Sasha Thanks a lot! – user1123187 May 29 '13 at 7:30	2014-04-17 01:39:28	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.21268844604492188, "perplexity": 2487.3775538343702}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-15/segments/1397609526102.3/warc/CC-MAIN-20140416005206-00440-ip-10-147-4-33.ec2.internal.warc.gz"}
https://par.nsf.gov/biblio/10374690-antenna-beam-characterization-global-cm-experiment-leda-its-impact-signal-model-parameter-reconstruction	Antenna beam characterization for the global 21-cm experiment LEDA and its impact on signal model parameter reconstruction ABSTRACT Cosmic dawn, the onset of star formation in the early universe, can in principle be studied via the 21-cm transition of neutral hydrogen, for which a sky-averaged absorption signal, redshifted to MHz frequencies, is predicted to be O(10–100) mK. Detection requires separation of the 21-cm signal from bright chromatic foreground emission due to Galactic structure, and the characterization of how it couples to instrumental response. In this work, we present characterization of antenna gain patterns for the Large-aperture Experiment to detect the Dark Ages (LEDA) via simulations, assessing the effects of the antenna ground-plane geometries used, and measured soil properties. We then investigate the impact of beam pattern uncertainties on the reconstruction of a Gaussian absorption feature. Assuming the pattern is known and correcting for the chromaticity of the instrument, the foregrounds can be modelled with a log-polynomial, and the 21-cm signal identified with high accuracy. However, uncertainties on the soil properties lead to percentage changes in the chromaticity that can bias the signal recovery. The bias can be up to a factor of two in amplitude and up to few per cent in the frequency location. These effects do not appear to be mitigated by larger ground planes, conversely gain more » Authors: ; ; ; ; ; ; Publication Date: NSF-PAR ID: 10374690 Journal Name: Monthly Notices of the Royal Astronomical Society Volume: 515 Issue: 2 Page Range or eLocation-ID: p. 1580-1597 ISSN: 0035-8711 Publisher: Oxford University Press 2. Abstract The detection of the Epoch of Reionization (EoR) delay power spectrum using a ”foreground avoidance method” highly depends on the instrument chromaticity. The systematic effects induced by the radio-telescope spread the foreground signal in the delay domain, which contaminates the EoR window theoretically observable. Applied to the Hydrogen Epoch of Reionization Array (HERA), this paper combines detailed electromagnetic and electrical simulations in order to model the chromatic effects of the instrument, and quantify its frequency and time responses. In particular, the effects of the analogue receiver, transmission cables, and mutual coupling are included. These simulations are able to accurately predict the intensity of the reflections occurring in the 150-m cable which links the antenna to the back-end. They also show that electromagnetic waves can propagate from one dish to another one through large sections of the array due to mutual coupling. The simulated system time response is attenuated by a factor 104 after a characteristic delay which depends on the size of the array and on the antenna position. Ultimately, the system response is attenuated by a factor 105 after 1400 ns because of the reflections in the cable, which corresponds to characterizable k∥-modes above 0.7 $h\,\,\rm {Mpc}^{-1}$ at 150 MHz.more » 5. ABSTRACT Future generations of radio interferometers targeting the 21 cm signal at cosmological distances with N ≫ 1000 antennas could face a significant computational challenge in building correlators with the traditional architecture, whose computational resource requirement scales as $\mathcal {O}(N^2)$ with array size. The fundamental output of such correlators is the cross-correlation products of all antenna pairs in the array. The FFT-correlator architecture reduces the computational resources scaling to $\mathcal {O}(N\log {N})$ by computing cross-correlation products through a spatial Fourier transform. However, the output of the FFT-correlator is meaningful only when the input antenna voltages are gain- and phase-calibrated. Traditionally, interferometric calibration has used the $\mathcal {O}(N^2)$ cross-correlations produced by a standard correlator. This paper proposes two real-time calibration schemes that could work in parallel with an FFT-correlator as a self-contained $\mathcal {O}(N\log {N})$ correlator system that can be scaled to large-N redundant arrays. We compare the performance and scalability of these two calibration schemes and find that they result in antenna gains whose variance decreases as 1/log N with increase in the size of the array.	2023-01-31 04:32:40	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6581001877784729, "perplexity": 1194.3458010834597}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764499842.81/warc/CC-MAIN-20230131023947-20230131053947-00497.warc.gz"}
https://tex.stackexchange.com/questions/178594/widecheck-over-two-hat?noredirect=1	Widecheck over two \hat [duplicate] I need to write a formula with a \widecheck over two \hat. The equation is the following: \documentclass[a4paper,11pt]{article} \usepackage{amsmath} \usepackage{amsfonts,amssymb,mathabx} \begin{document} $$\tilde{A}^\mu_\Lambda = \widehat{\widehat{G}\cdot\widehat{j^\mu_\Lambda}}.$$ \end{document} If I write this code I obtain a lot of errors (undefined control sequence, missing number treated as zero, illegal unit of measure (pt inserted), etc...). If I use \widehat instead of \widecheck, the result is the same. I have problems also writing \widecheck{\hat{G} \cdot j} because the result is not correct (G is translated to the right and overwrites j). How can I write the \widecheck? marked as duplicate by egreg, Guido, Werner, Benedikt Bauer, Martin SchröderMay 16 '14 at 21:21 • Welcome to TeX.SX! This is another example of the bug described in Why do arguments to nested tilde or breve commands reappear when amsmath is used?; the solutions shown there seem to apply to your case as well. – egreg May 16 '14 at 19:32	2019-10-19 05:11:54	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 1, "x-ck12": 0, "texerror": 0, "math_score": 0.9587044715881348, "perplexity": 3384.641282009578}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-43/segments/1570986688826.38/warc/CC-MAIN-20191019040458-20191019063958-00096.warc.gz"}
https://www.aimsciences.org/article/doi/10.3934/cpaa.2007.6.1131	Article Contents Article Contents # An asymptotic convergence result for a system of partial differential equations with hysteresis • A partial differential equation motivated by electromagnetic field equations in ferromagnetic media is considered with a relaxed rate dependent constitutive relation. It is shown that the solutions converge to the unique solution of the limit parabolic problem with a rate independent Preisach hysteresis constitutive operator as the relaxation parameter tends to zero. Mathematics Subject Classification: 35K55, 47J40, 35B40. Citation:	2023-04-01 03:37:33	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 1, "x-ck12": 0, "texerror": 0, "math_score": 0.8261480927467346, "perplexity": 657.3275111681273}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296949701.0/warc/CC-MAIN-20230401032604-20230401062604-00589.warc.gz"}
https://zbmath.org/?q=an:0876.34017	# zbMATH — the first resource for mathematics Solvability of singular nonlinear two-point boundary value problems. (English) Zbl 0876.34017 This paper discusses singular nonlinear boundary value problems. In particular the differential equation $\frac{1}{h(t)} (h(t)y'(t))'= g(t)F(t,y(t))=0, \qquad 0<t<1$ with Dirichlet or mixed boundary data is examined. Under suitable assumptions (see Theorem 1 and Theorem 2) the authors establish the existence of a solution. Although the results of the paper are new, the technique involved however is standard. The paper is well written and some nice examples are given at the end to illustrate the theory involved. ##### MSC: 34B15 Nonlinear boundary value problems for ordinary differential equations ##### Keywords: singular nonlinear boundary value problems Full Text: ##### References: [1] Bobisud, L.E., Existence of solutions for nonlinear singular boundary value problems, Applic. analysis, 35, 43-57, (1990) · Zbl 0666.34017 [2] Bobisud, L.E.; O’regan, D.; Royalty, W.D., Singular boundary value problems, Applic. analysis, 23, 233-243, (1986) · Zbl 0584.34012 [3] Chan, C.Y.; Hon, C., A constructive solution for a generalized Thomas-Fermi theory of ionized atoms, Q. appl. math., 45, 591-599, (1987) · Zbl 0639.34021 [4] Dunninger, D.R.; Kurtz, J.C., A priori bounds and existence of positive solutions for singular nonlinear boundary value problems, SIAM J. math. analysis, 17, 595-609, (1986) · Zbl 0595.34016 [5] Dunninger, D.R.; Kurtz, J.C., Existence of solutions for some nonlinear singular boundary value problems, J. math. analysis applic., 115, 369-405, (1986) · Zbl 0616.34012 [6] Gatica, J.A.; Oliker, V.; Waltman, P., Singular nonlinear boundary value problems for second order ordinary differential equations, J. diff. eqns, 79, 62-78, (1989) · Zbl 0685.34017 [7] Mawhin, T.; Omana, W., Two point boundary value problems for nonlinear perturbations of some singular linear differential equations at resonance, Commentat. math. univ. carolinae, 30, 3, 537-550, (1989) · Zbl 0686.34014 [8] Bobisud, L.E.; O’regan, D.; Royalty, W.D., Solvability of some nonlinear boundary value problems, Nonlinear analysis, 12, 855-869, (1988) · Zbl 0653.34015 [9] Callegari, A.; Nachmann, A., A nonlinear singular boundary value problem in the theory of pseudoplastic fluids, SIAM J. appl. math., 38, 275-281, (1980) · Zbl 0453.76002 [10] Luning, C.D.; Perry, W.L., Positive solutions of negative exponent generalized emder-Fowler boundary value problems, SIAM J. math. analysis, 12, 874-879, (1981) · Zbl 0478.34021 [11] O’regan, D., Positive solutions to singular and nonsingular second order boundary value problems, J. math. analysis applic., 142, 40-52, (1989) · Zbl 0689.34015 [12] O’regan, D., Existence of positive solutions to some singular and nonsingular second order boundary value problems, J. diff. eqns, 84, 228-251, (1990) · Zbl 0706.34030 [13] O’regan, D., Some existence principles and some general results for singular nonlinear two point boundary value problems, J. math. analysis applic., 166, 24-40, (1992) · Zbl 0756.34030 [14] Taliaferro, S., A nonlinear singular boundary value problem, Nonlinear analysis, 3, 897-904, (1979) · Zbl 0421.34021 [15] Wang, Junyu, Singular two-point boundary value problems, Northeastern math. J., 3, 218-291, (1987) · Zbl 0669.34023 This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching.	2021-07-31 09:19:17	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7005234360694885, "perplexity": 2873.1690942111863}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-31/segments/1627046154085.58/warc/CC-MAIN-20210731074335-20210731104335-00137.warc.gz"}
http://iprc.soest.hawaii.edu/users/zuojun/papers/mean_necc/abs_NECC.html	Yu-et-al 2000 Abstract # Influence of equatorial dynamics on the Pacific North Equatorial Countercurrent ### Z. Yu , J.P. McCreary, W.S. Kessler, and K.A. Kelly J. of Phys. Oceanogr., 30, 3179-3190, 2000. The Pacific North Equatorial Countercurrent (NECC) is generally not well simulated in numerical models. In this study, the causes of this problem are investigated by comparing model solutions to observed NECC estimates. The ocean model is a general circulation model of intermediate complexity. Solutions are forced by climatological and interannual wind stresses, ${\bf \tau }=(\tau ^{x},\tau ^{y})$, from Florida State University and the European Centre for Medium-Range Weather Forecasts. Estimates of the observed NECC structure and transport are prepared from expendable bathythermograph data and from the ocean analysis product of NOAA/National Center for Environmental Prediction. In solutions forced by climatological winds, the NECC develops a discontinuity (right panels) in the central Pacific that is not present in the observations. The character of the error suggests that it arises from the near-equatorial (5S--5N) zonal wind stress, $\tau ^{x}$, being relatively too strong compared to the y-derivative of the wind stress curl term, (curl~${\bf \tau })_{y}$, associated with the Intertropical Convergence Zone. This is confirmed in solutions forced by interannual winds, which exhibit a wide range of responses from being very similar to the observed NECC to being extremely poor, the latter occurring when near-equatorial $\tau ^{x}$ is relatively too strong. Our results show further that the model NECC {\it transport} is determined mainly by the strength of (curl~${\bf \tau })_{y}$, but that its {\it structure} depends on near-equatorial $\tau ^{x}$; thus, NECC physics involves equatorial as well as Sverdrup dynamics. Only when the two forcing features are properly prescribed do solutions develop a NECC with both realistic spatial structure and transport.	2013-05-24 10:00:16	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4643688201904297, "perplexity": 4034.4621582401955}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2013-20/segments/1368704517601/warc/CC-MAIN-20130516114157-00053-ip-10-60-113-184.ec2.internal.warc.gz"}
https://dmtcs.episciences.org/3009	## Florian Heigl ; Clemens Heuberger - Analysis of Digital Expansions of Minimal Weight dmtcs:3009 - Discrete Mathematics & Theoretical Computer Science, January 1, 2012, DMTCS Proceedings vol. AQ, 23rd Intern. Meeting on Probabilistic, Combinatorial, and Asymptotic Methods for the Analysis of Algorithms (AofA'12) - https://doi.org/10.46298/dmtcs.3009 Analysis of Digital Expansions of Minimal Weight Authors: Florian Heigl ; Clemens Heuberger Extending an idea of Suppakitpaisarn, Edahiro and Imai, a dynamic programming approach for computing digital expansions of minimal weight is transformed into an asymptotic analysis of minimal weight expansions. The minimal weight of an optimal expansion of a random input of length $\ell$ is shown to be asymptotically normally distributed under suitable conditions. After discussing the general framework, we focus on expansions to the base of $\tau$, where $\tau$ is a root of the polynomial $X^2- \mu X + 2$ for $\mu \in \{ \pm 1\}$. As the Frobenius endomorphism on a binary Koblitz curve fulfils the same equation, digit expansions to the base of this $\tau$ can be used for scalar multiplication and linear combination in elliptic curve cryptosystems over these curves. Volume: DMTCS Proceedings vol. AQ, 23rd Intern. Meeting on Probabilistic, Combinatorial, and Asymptotic Methods for the Analysis of Algorithms (AofA'12) Section: Proceedings Published on: January 1, 2012 Imported on: January 31, 2017 Keywords: dynamic programming,limit distribution,digital expansions,Hamming weight,elliptic curve cryptography,Frobenius endomorphism,minimal expansions,[INFO.INFO-DS] Computer Science [cs]/Data Structures and Algorithms [cs.DS],[INFO.INFO-DM] Computer Science [cs]/Discrete Mathematics [cs.DM],[MATH.MATH-CO] Mathematics [math]/Combinatorics [math.CO],[INFO.INFO-CG] Computer Science [cs]/Computational Geometry [cs.CG]	2023-02-07 08:42:08	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5933665037155151, "perplexity": 3238.3201426119194}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764500392.45/warc/CC-MAIN-20230207071302-20230207101302-00101.warc.gz"}
http://fortranwiki.org/fortran/show/Source+code+editors	# Fortran Wiki Source code editors There are many editors that support Fortran syntax highlighting. (TODO expand this list) • Code::Blocks IDE for Fortran An IDE for Fortran 2003 and 2008. • Emacs – V 23.x and later come with f90-mode supporting Fortran 2003. • gedit – free-format Fortran 95 • Vim – Supports Fortran syntax up to Fortran 95, but see below. • NEdit – GUI editor with some Fortran support. It used to be a good choice, but is rather obsolete compared to gedit. • Eclipse with Photran – For the IDE users. • Geany: GUI editor with support for many languages as well as Fortran; see below. • Sublime Text 2 – A beautiful text editor (paid but no enforced time limit for free use) that supports Fortran syntax via Textmate bundle. • Kate: – Has syntax highlighting and indentation for Fortran. An enhanced Vim syntax file that includes Fortran 2003 keywords is available here: The following can be added to \$HOME/.vimrc. It maps <shift>-F to toggle between fixed and free format Fortran source. In case it is hit by mistake, <ctrl>-F is mapped to re-detect the syntax. nmap <S-F> :set syntax=fortran<CR>:let b:fortran_fixed_source=!b:fortran_fixed_source<CR>:set syntax=text<CR>:set syntax=fortran<CR> nmap <C-F> :filetype detect<CR> When editing Fortran on Linux I use gedit which is adequate. When editing Fortran on Microsoft Windows (TM) there are lots of editors available, none of them (in my experience) really good enough. Here are my thoughts on a few that I’ve used more than a few times: • Code::Blocks IDE for Fortran (http://darmar.vgtu.lt/): Recommended! Originally developed for C++, this advanced IDE has many features of eclipse and supports most features of Fortran 2003 and 2008. The environment is quite user friendly and easy to use. It is one of the few true IDEs for Fortran. I think the support of the Fortran community will help this IDE to become a very powerful general purpose and fast Fortran programming tool. • Jedit - written in Java - when opening more than one file this has no tabs but a drop-down list at the top, which I find less convenient. But it has good Fortran90 aware syntax highlighting (I had to make a small change to the configuration file to prevent every line starting with C being treated as a comment, but that was not difficult). But the print system is a Java-special, not using the regular Windows print menu, so you can’t alter your printer defaults, e.g. to print 2 pages per sheet of paper. This is an annoyance. • Gedit - there is a windows binary to download which worked with no trouble, but there is no manual and no help (missing file). The syntax highlighting did work for Fortran90, but it did not understand the .f90 or .f95 file extension, and had to be set each time, which was very tedious. I could not find the relevant file on the website, nor when I untarred a regular gedit distribution. Looking at the help on a Linux box there did not seem to be a way of associating file extensions with language highlighting. • Notepad++ - has tabbed browsing, and uses regular Windows printing, but the syntax highlighting was set for Fortran77, and was really not useful. After being annoyed by it for some time, I found it best to disable the highlighting entirely. In principle one could define one’s own Fortran90 syntax file, but I couldn’t be bothered to do this. Apart from this it’s a fairly good editor. • Notetab light - the free version of a commercial editor notetab. This is another adequate editor: regular Windows printing, no tabbed file opening, no syntax highlighting. It seems to have no particular features that at least one of the others has. • Geany(http://www.geany.org/): This is a free open source editor with many capabilities which also supports programming in Fortran. There are several plugins to deal with project management, debugging etc. The editor lets you for extensive customization for a programming language and even adding new one! It is available for both Windows and Linux. • Simply Fortran(http://simplyfortran.com/): lightweight IDE for MS Windows and compatible operating systems. Made with GNU Fortran in mind, but configurable with most compilers, it comes with an intergrated development environment, a graphical debugger, and a collection of other development accessories. In early stages of development as of now, has some bugs, but the author’s working on them. • Programmer’s Notepad - A general source code editor for MS Windows. It can be configured to build Fortran files with arbitrary command lines, or to invoke make in a particular directory. Like many editors, it is C-centric, but the Fortran support is adequate.	2013-05-21 22:15:49	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.46569570899009705, "perplexity": 3761.1902813928236}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2013-20/segments/1368700795821/warc/CC-MAIN-20130516103955-00027-ip-10-60-113-184.ec2.internal.warc.gz"}
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Spectroscopy/Magnetic_Resonance_Spectroscopies/Nuclear_Magnetic_Resonance/NMR%3A_Experimental/Diffusion_Ordered_Spectroscopy_(DOSY)	# Diffusion Ordered Spectroscopy (DOSY) Diffusion Ordered SpectroscopY (DOSY) utilizes magnetic field gradients to investigate diffusion processes occurring in solid and liquid samples. ## Theory ### Spin Diffusion In the classic formation of a spin-echo (i.e. 90o-$$\tau$$-180o-2$$\tau$$) the intensity of the echo will damp according to T2 relaxation processes. However, in the presence of inhomogeneities in the magnetic field, if a spin drifts to a location in which the magnetic field is different, the Larmor frequency changes and the spin will not refocus with the remaining spins. We can then re-write the bloch equation to account for diffusion processes $\frac{dM}{dt}=\gamma MB-\frac{M_x+M_y}{T_2}-\frac{M_z-M_0}{T_1}+D\nabla ^2M$ Typically, the changes in the Larmor frequency have a negligible effect. However, application of a Gradient, G, will induce a large change in larmor frequencies experienced at places in the sample. The principle behind the DOSY experiment is the formation of a magnetic field gradient echo. This is a spin-echo in which a magnetic field gradient is applied during the spin evolution. Thus spins that diffuse during this time will not refocus in the spin echo, as the application of the gradient will impart their own magnetic field. ## Pulse Program The sequence consists of a conventional Hahn echo (spin echo) sequence in which 2 gradient pulses are applied at equal timings after the 90 and 180 degree pulses. The gradients are opposite in magnitude. The basic principle relies on the diffusion of spins in the sample. Initially the 90 degree pulse tips the magnetization into the x-y plane, where spins begin to precess with their characteristic Larmor frequencies. The application of a gradient sometime after the 90 degree pulse, encodes a spatial component to the spin. That is the gradient is not uniform over the sample and therefore, the processional frequencies will change. Next the application of the 180 flips the magnetization to refocus the spins. However, the spins, due to the application of the gradient will not refocus. The application of a gradient at the same time (but with opposite direction) will refocus the spins at a total time of 2 $$\tau$$. If a spin diffuses to a different place in the sample, the refocusing will not occur, leading the dampening of the echo intensity. Below is the basic pulse sequence for a pulsed field gradient experiment. ## Processing Processing the data is fairly straight forward. Apply a Fourier Transformation along the F2 dimension. The F1 dimension, in which either the time or the gradient was incremented remains in the time domain. The echo intensity of a given peak is then described by: $\ln \mu (g_a, t_c)-\ln \mu(0, t_c)=-C_n \gamma ^2 D \delta ^2 g_a ^2 t_D$ where $$\mu((g_a, t_c)$$ is the echo amplitude after the gradient application, $$ln m(0, t_c)$$ is the echo amplitude with no gradient applied, Cn is a constant that depends on the particular pulse sequence used, $$\gamma$$ is the gyromagnetic ratio, D is the diffusion coefficient, $$\delta$$ is the width of the applied gradient, and tis the diffusion time. ## Analysis Use this feature to ...	2019-04-24 15:46:27	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6631754636764526, "perplexity": 931.3692875900101}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-18/segments/1555578650225.76/warc/CC-MAIN-20190424154437-20190424180437-00327.warc.gz"}
http://cs.stackexchange.com/questions/10020/modeling-timing-characterists-of-an-architecture	# Modeling timing characterists of an architecture I am building a system and I have a couple of architectures in mind. I want to have an idea of which architecture is likely to be most performant (quickest). I can make different decisions like 1) Do everything in one thread 2) Separate this part into one thread, that part into another thread since it can run in parallel and so on Instead of actually building the system in two different ways and seeing how it performs, is there a way to quickly model/prototype the system where I can express the above concepts and the times it takes for inter thread I/O etc (of which I have a very good estimate) and use the model/prototype to estimate which architecture is better? - Maybe you need a process calculus focused on timing? I cannot help further, I don't know about PC, but it could be an interesting trail to follow. – didierc Feb 21 '13 at 21:50 Essentially, what you want to do is called Synthesis (constructing a system from specifications). In discrete time, this problem is $2-EXP$-complete. In continuous time I think it's even harder. Try this( google.co.il/…) for reference. – Shaull Feb 22 '13 at 10:29	2016-07-01 18:49:12	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.629098653793335, "perplexity": 705.3673538748878}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-26/segments/1466783403823.74/warc/CC-MAIN-20160624155003-00063-ip-10-164-35-72.ec2.internal.warc.gz"}
https://www.eng-tips.com/viewthread.cfm?qid=479057	× INTELLIGENT WORK FORUMS FOR ENGINEERING PROFESSIONALS Are you an Engineering professional? Join Eng-Tips Forums! • Talk With Other Members • Be Notified Of Responses • Keyword Search Favorite Forums • Automated Signatures • Best Of All, It's Free! Eng-Tips's functionality depends on members receiving e-mail. By joining you are opting in to receive e-mail. #### Posting Guidelines Promoting, selling, recruiting, coursework and thesis posting is forbidden. # Near field antenna to tell if radiated emissions are better or worse after modification to circuit? ## Near field antenna to tell if radiated emissions are better or worse after modification to circuit? (OP) Hi, We have an SMPS in an earthed metal enclosure A, which is wired to another SMPS (12W) in another earthed enclosure B. B contains 4 SMPS chargers of 3W each. A is totally enclosed, with only a few holes of dimension much smaller than the wavelength of the problem frequencies of 50MHz to 200MHz. B is half open (so people can take out the chargers). We want to adjust the Y caps in the SMPS in A. We then want to see if this makes the radiated emissions better or worse. Can we do this simple analysis with a near field probe connected to a EMC Analyser eg SA1002A by Laplace instruments? SA1002A Should we use a B field near field probe, or an E field near field probe? ..................................*.. Here is a better description of the system... Please help our kiddies post-lockdown pager system game kit pass radiated EMC? I am doing this with a company in South Africa for charity. Its as attached. It is an earthed metal enclosure A, which contains a 20W offline SMPS and a 12V battery. This enclosure feeds a 12V cable to enclosure B which contains a 10W , 12V to 5V SMPS. This 5V is then fed to each of four removable pagers, (so they get charged up) which the kiddies can remove and talk to each other. (they will be in a “hospitals and doctors” game) It fails EMC at the moment. (Radiated emissions failure at frequencies from 80MHz to 190MHz) We wish to sell it into the EU, where they need some post-Covid relief as much as everyone else. Should the cable between A and B be earth screened and the earth screen connected to both enclosure A and B? The enclosure B is open at the front, so is it therefore a waste of time to earth enclosure B? As such, is it also a waste of time to take earthed 12V cable from A to B? The wires running from the 12v_to_5v SMPS to each pager, should they be loomed as close as possible to the walls of enclosure B? The enclosure B has to be open at the front so that kiddies can grab the pagers. We are thinking we may need a common mode choke at the point in enclosure A where the 12V cable leaves A to go to B? Also, we could put a common mode choke at the input and output of the 12v_to_5v SMPS. (NiZn torroid based common mode chokes with just a few turns) We could also put a flux band round the offline flyback transformer in A. We could also increase the Y capacitor across the flyback transformer in A. At the moment its only 100pF. We don’t want PCB respins if poss as the customers don’t want to pay much. https://files.engineering.com/getfile.aspx?folder=... ### RE: Near field antenna to tell if radiated emissions are better or worse after modification to circuit? (OP) It is well known that passing radiated emissions to EU standard with an SMPS that is not surrounded in a metal enclosure is pretty well impossible. Even if it is in a metal enclosure, it will still fail radiated emissions if a cable leaves the enclosure to go somewhere else. (as in the attached). In that case, a common mode choke and Y caps will be needed in order to stop the cable from radiating. Would you agree? ### RE: Near field antenna to tell if radiated emissions are better or worse after modification to circuit? "It fails EMC at the moment. " - that's good, you have the first step, as long as you know how badly you are failing (X dB). I often use a near field probe to do before and after comparisons before I take a unit back to the test lab (\$). If your mod shows an X + 3 dB or better improvement you are probably okay, more margin is better. Yes, chokes on cables are almost always required for EMC. Z #### Red Flag This Post Please let us know here why this post is inappropriate. Reasons such as off-topic, duplicates, flames, illegal, vulgar, or students posting their homework. #### Red Flag Submitted Thank you for helping keep Eng-Tips Forums free from inappropriate posts. 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Register now while it's still free!	2022-09-29 05:19:22	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.19195322692394257, "perplexity": 5134.908063954979}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030335304.71/warc/CC-MAIN-20220929034214-20220929064214-00275.warc.gz"}
https://pos.sissa.it/363/235/	Volume 363 - 37th International Symposium on Lattice Field Theory (LATTICE2019) - Main session Study of thermal SU(3) supersymmetric Yang-Mills theory and near-conformal theories from the gradient flow G. Bergner, C. Lopez,* S. Piemonte *corresponding author Full text: pdf Pre-published on: 2020 January 04 Published on: Abstract We compute the renormalization group flow of the mass anomalous dimension in adjoint QCD with $N_{f}=1$, $3/2$, and 2 Dirac fermions, using the gradient flow. Preliminary results are in agreement with at least a near-conformal scenario in all cases. At the largest flavor numbers we obtain the strongest indication for an IR conformal fixed point scenario. Moreover, we provide results for the thermal phase transitions in SU(3) supersymmetric Yang-Mills theory. We find hints for a connection between chiral and center symmetries in terms of a single first order phase transition where chiral symmetry is restored and center symmetry gets broken. Open Access	2020-01-24 21:11:56	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.48607945442199707, "perplexity": 1488.3171523670683}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-05/segments/1579250625097.75/warc/CC-MAIN-20200124191133-20200124220133-00497.warc.gz"}
http://slideplayer.com/slide/753623/	# Improper Integrals I. ## Presentation on theme: "Improper Integrals I."— Presentation transcript: Improper Integrals I Improper Integrals Definition An integral is improper if either: the interval of integration is infinitely long or if the function has singularities in the interval of integration (or both). Improper Integrals Improper integrals cannot be defined as limits of Riemann sums. Neither can one approximate them numerically using methods based on evaluating Riemann sums. IMPROPER INTEGRALS Examples 1 The integral is improper because the interval of integration is infinitely long. IMPROPER INTEGRALS Examples 2 is improper because the integrand has a singularity. IMPROPER INTEGRALS Examples 3 is improper because the integrand has a singularity and the interval of integration is infinitely long. Improper Integrals Definition Assume that the function f takes finite values on the interval [a, ∞). If the limit exists and is finite, the improper integral converges, and Improper Integrals Example Hence the integral converges. Improper Integrals Definition Assume that the function f takes finite values on the interval [a, ∞). If the limit does not exists or is not finite, the improper integral diverges Improper Integrals Example Hence the integral diverges. Improper Integrals Definition Assume that the function f has a singularity at x = a. If the limit exists and is finite, the improper integral converges, and Improper Integrals Example Hence the integral converges. Improper Integrals Definition Assume that the function f has a singularity at x = a. If the limit does not exist or is not finite, the improper integral diverges. Improper Integrals Example Hence the integral diverges. Improper Integrals Definition If the function f has a singularity at a point c, a < c < b, then the improper integral converges if and only if both improper integrals and converge. In this case Improper Integrals Example Hence the integral converges. Improper Integrals Definition If the function f has a singularity at a point c, a < c < b, then the improper integral diverges if either or diverges. Improper Integrals Example Neither limits exists. The integral diverges. Improper Integrals Warning The integral diverges. Trying to compute that integral by the Fundamental Theorem of Calculus, one gets This is an incorrect computation. Summary An integral is improper if either: the interval of integration is infinitely long or if the function has singularities in the interval of integration (or both). Such integrals cannot be defined as limits of Riemann sums. They must be defined as limits of integrals over finite intervals where the function takes only finite values.	2021-11-30 12:47:19	{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.996390163898468, "perplexity": 547.6327747671368}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-49/segments/1637964358973.70/warc/CC-MAIN-20211130110936-20211130140936-00468.warc.gz"}
https://threesixty360.wordpress.com/2008/10/19/	## Archive for October 19th, 2008 ### Using calculus to generate the quadratic formula October 19, 2008 I get (unreasonable considerable) joy in finding fancy new proofs of elementary results, proofs that might come under the heading of mathematics made difficult or obscure. (I’ve never seen Linderholm’s book, but suspect it would appeal to a twisted part of my psyche.) I don’t quite understand the psychology behind this liking, but there it is nonetheless. One of my favorite examples follows: solving quadratic equations using integration to complete the square. Suppose $f(x)=ax^2+bx+c$, and we want to find the solutions to $f(x)=0$. Note that $f'(x) = 2ax+b$ and $f(0) = c$, and thus it must be that $f(x) = c+ \int_0^x 2at+b \; dt$. We compute this antiderivative using the change of variables $w=2at+b, \quad dw = 2a\;dt$, which leads to $c + \int_b^{2x+b} \frac1{2a} \; w \; dw$. This last expression is equal to $c + \left( \frac1{4a} (2ax+b)^2 - \frac{b^2}{4a} \right)$. Thus the roots of $f(x) = 0$ are found by solving the equation $c + \frac{ (2ax+b)^2 - b^2 }{4a} = 0$. This leads to $(2ax+b)^2 -b^2 = -4ac$, and thus $2ax+b = \pm \sqrt{b^2 - 4ac}$, and so finally we arrive at the roots $x = \frac{ -b \pm \sqrt{b^2 - 4ac}}{2a}$. Forthcoming: Using Fubini’s Theorem to prove Integration by Parts	2019-06-27 02:10:39	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 13, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.963051438331604, "perplexity": 338.8159718112735}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-26/segments/1560628000610.35/warc/CC-MAIN-20190627015143-20190627041143-00435.warc.gz"}
https://math.stackexchange.com/questions/1266463/evaluating-area-using-an-integral-in-polar-coordinates	# Evaluating area using an integral in polar coordinates I am trying to find the area of a circle which is given by the polar parameterization $$r(\phi) = \cos\phi + \sin\phi.$$ I can evaluate it in 2 ways and don't know why I get different answers. First way. Using geometry. This circle circumscribes a square with a side of length 1. So the diagonal of the square, due to Pifagor's theorem , equals to $\sqrt{2}$. Which means that the radius of the circle is $\frac{\text{diagonal}}{2}=\frac{\sqrt{2}}{2}$. So the area of the circle should be equal to $\pi/2$. Second way. I use definite integral to find the area. As I was taught the area equals to $$\frac{1}{2}\int _0 ^{2\pi} r(\phi)^2 d\phi.$$ Using this formula I get that area equals $\pi$. Please tell me where am I wrong. I am sure that the integral is solved correctly (checked it in wolfram). P.S. I also attached the graph from Wolfram: The issue is that your parameterization traces around the circle twice as $\phi$ varies over the interval $[0, 2 \pi]$ of integration. To see that it does so more than once, just observe that your integrand $$r(\phi)^2 = (\cos \phi + \sin \phi)^2$$ satisfies $$r(\phi + \pi) = r(\phi).$$ By that symmetry, (1) to compute the the area it's enough to integrate over any interval of length $\pi$, and (2) using your computation gives $$A = \int_0^{\pi} r(\phi)^2 d\phi = \frac{1}{2}\int_0^{\pi} r(\phi)^2 d\phi = \frac{\pi}{2},$$ which coincides with the result from your geometric argument.	2020-06-02 15:47:58	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9424876570701599, "perplexity": 119.06476935003562}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-24/segments/1590347425148.64/warc/CC-MAIN-20200602130925-20200602160925-00119.warc.gz"}
https://www.esaral.com/q/the-number-of-words-86162	# The number of words Question: The number of words (with or without meaning) that can be formed from all the letters of the word "LETTER" in which vowels never come together is__________. Solution: For vowels not together Number of ways to arrange $\mathrm{L}, \mathrm{T}, \mathrm{T}, \mathrm{R}=\frac{4 !}{2 !}$ Then put both $\mathrm{E}$ in 5 gaps formed in ${ }^{5} C_{2}$ ways. $\therefore$ No. of ways $=\frac{4 !}{2 !} \cdot{ }^{5} C_{2}=120$	2023-03-22 07:20:51	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6417974829673767, "perplexity": 710.7607501305964}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296943750.71/warc/CC-MAIN-20230322051607-20230322081607-00116.warc.gz"}
https://mathoverflow.net/questions/373642/what-is-the-image-of-the-diagonal-map-on-the-cohomology-of-lie-groups	# What is the image of the diagonal map on the cohomology of Lie groups Consider a simple Lie group $$G$$ and its mod $$p$$ cohomology $$H^(G, \mathbb{Z}_p)$$. A good reference is the book Mimura, Mamoru; Toda, Hirosi, Topology of Lie groups, I and II. Transl. from the Jap. by Mamoru Mimura and Hirosi Toda, Translations of Mathematical Monographs. 91. Providence, RI: American Mathematical Society (AMS). iv, 451 p. (1991). ZBL0757.57001. Consider the diagonal map $$\Delta: H^(G, \mathbb{Z}_p) \rightarrow H^(G, \mathbb{Z}_p) \otimes H^(G, \mathbb{Z}_p)$$. Let $$x_i$$ for $$i=1,...,n$$ denote the generators of $$H^*(G, \mathbb{Z}_p)$$. Assume there is an element $$x^a_i\otimes x^b_j$$ contained in $$\Delta(x_m)$$, with $$a,b \neq 0$$. Question 1: Is it true that at least one of the numbers $$a, b$$ is equal to $$1$$? Question 2: Is it true that an element $$x_ix_j \otimes x_l$$ or $$x_i \otimes x_jx_l$$ can never be contained in $$\Delta(x_m)$$? I assume that both answers are "yes", after checking many examples calculated in the mentioned book and also a paper on this topic released by Mimura and Kono. The second question basically asks, wether the diagonal map can "mix" generators. In the book the diagonal map is named $$\phi$$ and $$\phi'(x):= \phi(x) - 1\otimes x - x\otimes 1$$. The results are presented in terms of $$\phi'$$, which shouldnt change anything. • Is it "tmage" or "image"? – efs Oct 9, 2020 at 0:38 • What do you call the "diagonal map"? The coproduct induced by the multiplication $G \times G \to G$? Oct 9, 2020 at 8:30 • Could you be more precise about the meaning of the word "contain" here? Oct 9, 2020 at 13:19 • By "generators", do you mean an additive basis (as $\mathbb{F} _p$ vector space) or something else? Oct 9, 2020 at 15:46 • Yes $\Delta$ is also known the coproduct map. Usually $\Delta(x)$ is a sum of elements $a \otimes b$. We say that $\Delta(x)$ contains a certain $a \otimes b$ if it ocurrs in this sum. The generators generate $H(G, \mathbb{Z}_p)$ additively and multiplicatively. We consider the ring structure of $H$. – nxir Oct 9, 2020 at 16:22	2022-05-21 19:27:38	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 18, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8241018652915955, "perplexity": 365.4811528496378}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662540268.46/warc/CC-MAIN-20220521174536-20220521204536-00085.warc.gz"}
https://plus.google.com/communities/100886145872320383869	Defeating the Data Tsunami, one algorithm at a time. See all Members (1,471) ## Stream Join this community to post or comment ### Igor Carronowner Conferences / Meetings / Meetups - In the Sunday Morning Insight entry entitled The Hardest Challenges We Should be Unwilling to Postponed, I mentioned a challenge set up by Isabelle Guyon entitled the AutoML challenge ( http://codalab.org/AutoML, her presen... 1 ### Igor Carronowner Job announcements - Jerome Bobin sent me the following opportunity the other day: POST-DOC : LEARNING REPRESENTATIONS FOR LARGE-SCALE MULTIVARIATE DATA. The concept of sparsity and sparse signal representations has led to the development... 1 ### Igor Carronowner Blogs (Nuit Blanche, ....) - Streaming is bound to take a centerstage role if we are talking about Big Data...or Data as we call it in Texas. Today, we have an algorithm for matrix completion, paper on verifying the properties of certain streams and some... 1 ### Igor Carronowner Blogs (Nuit Blanche, ....) - 1 ### Igor Carronowner Blogs (Nuit Blanche, ....) - Dick just let me know of the following: Dear Igor, You might want to post: Landau, E. (2015). Glitter Cloud May Serve as Space Mirror. http://www.jpl.nasa.gov/news/news.php?feature=4553&utm_<wbr></wbr>source=iContact&utm_... 1 ### Igor Carronowner Blogs (Nuit Blanche, ....) - Hat tip to Thomas Arildsen for pointing out this preprint: Compressed Sensing Recovery via Nonconvex Shrinkage Penalties by Joseph Woodworth, Rick Chartrand The ℓ0 minimization of compressed sensing is often relaxed to ℓ1... 4 1 ### Igor Carronowner Conferences / Meetings / Meetups - This is the where the meetup will be streamed starting at around 7:15PM Paris time. The theme of today's meetup will about (mostly) Deep Learning. It will be held in conjunction with the Deep Learning Paris meetup . W... 1 ### Igor Carronowner Blogs (Nuit Blanche, ....) - Unless I am mistaken, this is the first time I see RandNLA operation performed using a GPU. It is not the last. GPU Accelerated Randomized Singular Value Decomposition and Its Application in Image Compression by Hao... 1 ### Igor Carronowner Blogs (Nuit Blanche, ....) - Here are some theoretical developments for sparse coding, an item discovered experimentally twenty years ago. Dictionary Learning with Few Samples and Matrix Concentration by Kyle Luh, Van Vu Let A be an n×n matrix, X ... 1 ### Igor Carronowner Videos - 1 This is a community of people interested in Compressive Sensing, Advanced Matrix Factorization, Randomized Linear Algebra and any other means of defeating the curse of dimensionality. Theoreticians all the way to hardware hackers are welcome. ### Igor Carronowner Blogs (Nuit Blanche, ....) - George sent me the following a few days ago: Dear Igor, My name is Huaijin (George) Chen, a PhD student at Rice working with Ashok Veeraraghavan. We recently got our paper "FPA-CS: Focal Plane Array-based Compressiv... 1 ### Igor Carronowner Blogs (Nuit Blanche, ....) - Off of the press, here is the thesis of Jeremy Vila one of the Map Makers: Empirical-Bayes Approaches to Recovery of Structured Sparse Signals via Approximate Message Passing. In recent years, there have been mass... 2 1 ### Igor Carronowner Blogs (Nuit Blanche, ....) - SLOPE is Adaptive to Unknown Sparsity and Asymptotically Minimax by Weijie Su, Emmanuel Candes We consider high-dimensional sparse regression problems in which we observe $y = X \beta + z$, where $X$ is an $n \times p$ desig... 1 ### Igor Carronowner Blogs (Nuit Blanche, ....) - Here is the video: Towards a Learning Theory of Causation by David Lopez-Paz, Krikamol Muandet, Bernhard Schölkopf and Iliya Tolstikhin We pose causal inference as the problem of learning to classify probability distribut... 1 ### Igor Carronowner Blogs (Nuit Blanche, ....) - Here is a senior thesis of interest: Turning Big Data into Small Data: Hardware Aware Approximate Clustering with Randomized SVD and Coresets, Tarik Adnan Moon Organizing data into groups using unsupervised learning algori... 1 ### Igor Carronowner Job announcements - Mark Plumbley just sent me the following: Dear Igor, We think that these PhD opportunities might be of interest to people on Nuit Blanche. They are part of a new "MacSeNet" EU-funded Marie Sklodowska-Curie Innovative Trainin... 1 ### Thomas Arildsen Discussion - Looks interesting and well-written - I will need to take a closer look. By Joseph Woodworth and . Abstract: The $\ell^0$ minimization of compressed sensing is often relaxed to $\ell^1$, which yields easy computation using the shrinkage mapping known as soft thresholding, and can be shown to recover the original solution under certain hypotheses. Recent work has derived a general class of ... 1 comment on original post 2 ### Igor Carronowner Implementations - Laurent just let me know of the release of an implementation for blind deconvolution: Dear Igor You have been kind enough to publicize the following paper (and you contributed to the build-up as you are indeed in the acknow... 1 ### Igor Carronowner Blogs (Nuit Blanche, ....) - Optimizing random features for tensors: Tensor machines for learning target-specific polynomial features by Jiyan Yang, Alex Gittens Recent years have demonstrated that using random feature maps can significantly decrease... 1	2015-04-26 21:36:14	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6459318995475769, "perplexity": 7342.659080314485}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-18/segments/1429246656168.61/warc/CC-MAIN-20150417045736-00220-ip-10-235-10-82.ec2.internal.warc.gz"}
http://images.planetmath.org/stationaryincrement	# stationary increment A stochastic process $\{X(t)\mid t\in T\}$ of real-valued random variables $X(t)$, where $T$ is a subset of $\mathbb{R}$, is said have stationary increments if the probability distribution function for $X(s+t)-X(s)$ is fixed (the same) for all $s\in T$ such that $s+t\in T$. In other words, the distribution for $X(s+t)-X(s)$ is a function of “how long” or $t$, not “when” or $s$. A stochastic process that possesses both stationary increments and independent increments is said to have stationary independent increments. Title stationary increment StationaryIncrement 2013-03-22 15:01:25 2013-03-22 15:01:25 CWoo (3771) CWoo (3771) 9 CWoo (3771) Definition msc 60G51 stationary independent increment	2018-06-18 05:40:19	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 10, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9684715867042542, "perplexity": 912.241259001851}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-26/segments/1529267860089.11/warc/CC-MAIN-20180618051104-20180618071104-00423.warc.gz"}
http://www.cgl.uwaterloo.ca/smann/GSInfo/latex.html	Stephen Mann Introduction University of Waterloo, Waterloo, Ontario, Canada This subpage contains information about preparing your thesis in LaTeX that should be useful to my graduate students. Others may find it useful, too, although getting access to the scripts won't be as easy (but see my software page for some of them). See this IST page to get a current latex template (it was at the bottom when I looked). Although given the University's tendancy to change links without providing proper forwarding, if this page is dead then you may have to search do a web search on the universty's webpage to hopefully find the latest and greatest. chap1.tex - an example chap1.tex file (used in an older thesis.tex) with some examples of how to do some common things. ## Useful scripts /u/smann/bin/Latex - Latex requires that you run it multiple times, with an invokation to bibenter at some point. This script will do the proper sequence of invocations of latex/bibtex for you, calling latex the minimum number of times needed. /u/smann/bin/Pdflatex - pdflatex requires that you run it multiple times, with an invokation to bibenter at some point. This script will do the proper sequence of invocations of pdflatex/bibtex for you, calling pdflatex the minimum number of times needed. /u/smann/bin/bibenter - This is a Tcl/Tk script to ease the process of entering bibliographies in bibtex format. /u/smann/bin/pstex2pdf - If you make your figures using xfig, you can set the "special" flag on text (Edit the text from xfig to set this flag), export the figure as Combined PS/latex, and use a special macro package from within LaTeX to get LaTeX fonts in your figure. The last part is a bit of a pain. The pstex2pdf script will take the two files (.pstex and pstex_t) and combine them, creating a PDF file (.pdf) and a PostScript file (.ps). You can then include the figure using \includegraphics like any other figure.	2018-01-24 11:48:53	{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9139701724052429, "perplexity": 2128.018334530681}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-05/segments/1516084894125.99/warc/CC-MAIN-20180124105939-20180124125939-00757.warc.gz"}
https://tex.stackexchange.com/questions/657848/drawing-a-curved-plot-in-tikz	# Drawing a curved plot in tikz I need some help in tikz. I don't know how to draw a curved plot like this red one on the picture, I have already prepared the axis, dashed line and these arrows. There is my code: \begin{tikzpicture} \begin{axis}[ axis lines = left, xmin=0, xmax=10, ymin=0, ymax=10, xtick = \empty, ytick = {0}, clip = false, ] \node [below] at (current axis.right of origin) {$t$}; \node [left] at (current axis.above origin) {$E$}; \addplot[color = grey!60!white, dashed, thick] coordinates {(0,4) (8,4)}; \draw[-{Triangle}, grey!80!white, opacity = 0.8] (1.5,0.1) to (1.5, 3.9); \draw[-{Triangle}, grey!80!white, opacity = 0.8] (7,0.1) to (7, 1.9); \draw[-{Triangle}, grey!80!white, opacity = 0.8] (4.5,4.1) to (4.5, 6.9); \node [right] at (1.5,2) {$E_s$}; \node [right] at (7,1) {$E_p$}; \node [right] at (4.5,5.5) {$E_a$}; \end{axis} \end{tikzpicture} Assuming the red curve is just representative and not a specific function, this is probably easier with just straightforward TikZ commands. Use \coordinate to name specific points, then \draw to draw the axes, grid and red curve, as well as the arrows. Use nodes to place labels on the arrows. The only complicated part is the brace with 𝛥H. You need the decorations.pathreplacing library for that. If you want more control over the curve, you can read about Bézier curves in the TikZ manual (sections 2.4 or 14.3). \documentclass{article} \usepackage{tikz} \usetikzlibrary{arrows.meta, decorations.pathreplacing} \begin{document} \begin{tikzpicture}[>=Straight Barb] \draw[gray!30] (-.5,-.2) grid (8.8,6.7); \draw[very thick, <->] (0,6.5)node[below left=2pt]{$E$} -- (0,0) -- (8.5,0); \coordinate (A) at (0,2.5); \coordinate (B) at (1.5,2.5); \coordinate (B0) at (1.5,0); \coordinate (C) at (3.5, 4.5); \coordinate (C0) at (3.5, 2.5); \coordinate (D) at (5.8,1.2); \coordinate (D0) at (5.8,0); \draw[thick, red, looseness=.7] (A) -- (B) to[out=0, in=180] (C) to[out=0, in=180] (D) --++ (2,0); \draw[dashed] (B) --++ (6.5,0); \draw[thick, gray, <->] (B) --node[right]{$E_s$} (B0); \draw[thick, gray, <->] (C) --node[right]{$E_a$} (C0); \draw[thick, gray, <->] (D) --node[right]{$E_p$} (D0); \draw[decorate, decoration=brace, thick] (8.1,2.5)--node[right]{$\Delta H$}(8.1,1.2); \end{tikzpicture} \end{document}	2022-09-27 20:45:43	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 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": 0.9352438449859619, "perplexity": 4872.985482627769}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030335058.80/warc/CC-MAIN-20220927194248-20220927224248-00508.warc.gz"}
https://www.physicsforums.com/threads/inverse-laplace-transform-bromwitch-integral.783007/	# Inverse Laplace transform. Bromwitch integral Inverse Laplace transform $$\mathcal{L}^{-1}[F(p)]=\frac{1}{2\pi i}\int^{c+i\infty}_{c-i\infty}F(s)e^{st}dp=f(t)$$ Question if we integrate along a straight line in complex plane where axis are $$Re(p)$$, $$Im(p)$$, why we integrate from $$c-i \ínfty$$ to $$c+\infty$$? So my question is, because $$Im(p)$$ are also real numbers why we integrate from $$c-i\infty$$ to $$c+i\infty$$. Last edited: ## Answers and Replies Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post? jasonRF Gold Member I'm guessing that you meant that the integral is with respect to $s$, not $p$. Anyway, for the traditional Laplace transform, F(s) = \int^{\infty}_{0} dt e^{-s t} f(t) $F(s)$ is only analytic for some right half plane (that is $\Re(s)>s_0$). So for the inverse transform you pick a $c>s_0$ for your path. Does that help? jason Yes. Mistake. I understand that. I am not very good in complex analysis. My question is if [tex[Im(s)[/tex] are real numbers why I integrate from $$c-i\infty$$ to $$c+i\infty$$? Why I have this $$i$$? Thanks for the answer.	2020-09-24 10:32:14	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9759367108345032, "perplexity": 460.85386322126345}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-40/segments/1600400217623.41/warc/CC-MAIN-20200924100829-20200924130829-00485.warc.gz"}
http://pziq.umood.it/how-to-plot-hyperboloid-in-matlab.html	# How To Plot Hyperboloid In Matlab Ezsurf tager to indgange: håndtaget til den funktion, du vil plotte og en vektor af maksimale og minimale x og y-koordinaterne for din plot. Vary the Parameter b to Change the Cylinder. Instructions for Using WinPlot. With the command gca we get the handle to the current axes with which it is possible to set axis bounds. Verify that what you are looking at is the hyperboloid of one sheet X^2 + Y. When I run it, it prompts me to enter in CSV file. A parametric vector equation for the line is therefore. Before learning how to graph a hyperbola from its equation, get familiar with the vocabulary words and diagrams below. At least one of the ranges must have arguments evaluating to real constants; the other range may have. x and y should now both be column vectors with the sa. This is a very useful tool in all types of scientific and math based research allowing the user. For the vertices, I got the lengths of the semimajor axes and the center of the ellipse. [Note - Do not mark o for bad reasoning regarding boundary points, or the lack. The usual Cartesian coordinate system can be quite difficult to use in certain situations. Relevant equations I found that there are matrix related to it, where the matrix is given by h=fspecial('gaussian',256); and it is all zeros. See the second figure. created via numpy. Did you know that the orbit of a spacecraft can sometimes be a hyperbola? A spacecraft can use the gravity of a planet to alter its path and propel it at high speed away from the planet and back out into space using a technique called "gravitational slingshot". e position of the samples has been normalized within each unit using a lin ear correlation and all results have been plotted using dierent colors for each F. Contribute to RInterested/PLOTS development by creating an account on GitHub. Find here the Matlab function to plot the $$\epsilon$$ function : epsilon_oliver_pharr. Open MATLAB on your. Try a scatter plot Select the category "XY (Scatter)" and go with the defaults. while differential impedance formed hyperboloid with the mean. Loading Unsubscribe from Mathispower4u? Cancel Unsubscribe. As long as you have the mathematical equation describing that hyperboloid, you should be able to generate some data and then draw it. the forces applied to the corner nodes act only on ½ of the element edge), however we applied a constant magnitude along the plates edge. Since each pair (x,y) in the domain determines a unique value of z, the graph of a function must satisfy the "vertical line test" already familiar from single-variable calculus. write a program in matlab to Plot the following standard surfaces: (a) ellipsoid, (b) hyperboloid of one sheet, (c) hyperboloid of two sheets, (d) cone, (e) elliptic paraboloid, (f) hyperbolic paraboloid, (g) torus, (h) right helicoid. What are the traces in planes z=k for kc< k=c kc>. Doubly-Curved and Warped Surfaces: Sphere, torus, oblate ellipsoid, conoid, serpentine, paraboloid, hyperboloid (Definition) 5. Example 1 Example 2 (yz= 1) Sketch the surface yz= 1. Platform: Matlab, Scripts; Publisher. Berechnungen mit Hypozykloiden. The designer can shift these points to alter the shape of the curve. At least for starters, include the following commands x=-3. B) Slice for x = 1. Get the free "Surface Plot in R3" widget for your website, blog, Wordpress, Blogger, or iGoogle. Since the surface of a sphere is two dimensional, parametric equations usually have two variables (in this case #theta# and #phi#). 2 Surfaces in R3 [M] 12:Use Maple/Gnuplot to plot the following func-tions f: R 2!R. I've talked about the various procedures for fitting different types of curves on this blog before, but today I want to show you a technique for hyperbolic curve fitting in Excel. The calculator can be used to calculate. Figure 21 Plots of beta Hay : a) as a function of the half-angle of the conical indenter (for a Poisson’s ratio of 0. Loading Unsubscribe from Mathispower4u? Cancel Unsubscribe. A hyperbola is a conic section defined as the locus of all points in the plane such as the difference of whose distances from two fixed points , (foci) is a given positive constant and. MATLAB EXAMPLES 185 %% plot hyperboloid of one sheet t=linspace(-10,10,25);% parameterizesz-value [x,y,z]=cylinder(sqrt(0. Right from mathematics formula chart to algebra course, we have every part included. Creating a plot using commands. My very first introduction to programming was in middle school during the first week of class. Visualize the radiation pattern from an antenna in polar coordinates. Learn more about example, 3d, plot, plotting, multivariate, ezplot, ezsurf, mesh, surf, xy, yz, xz, plane MATLAB. 14 - Find du if u = ln(1 + se2t). The final. That is, it is either the x-axis, the y-axis, or the z-axis. 2 Surfaces in R3 [M] 12:Use Maple/Gnuplot to plot the following func-tions f: R 2!R. Plotting points in polar isn't technically supported yet, but you can use the work-around I described above: Define these. Merrill algebra 2, maths tests ks2, free math percentage application, plot differential equations on matlab 2nd order, Prentice hall physics solution, how to find the inscribed and circumscribed rectangles in a quadratic. Submit your pictures and your code. A true brotherhood: the locker room, bus trips, flights, dorms, Court St. Numerical simulations confirm the theoretical model. while differential impedance formed hyperboloid with the mean. Brug MATLAB bygget s - in " ezsurf "-funktionen til hurtigt at afbilde din hyperboloid. In the applet, you'll see two cones joined at their apexes. i) f(x;y) = cos(x+ y). Basic concepts of applying adaptive ﬁlters in practical applications are also highlighted. Matlab worksheet for Section 12. Platform: Matlab, Scripts; Publisher. Professional Interests: Human machine interface, integration of software and hardware, image processing, microscopy, convex optimization. In this work, we derive analytically from first principles a generalized FN equation that (i) is applicable to an arbitrary potential (or surface) and (ii) contains as a correction a new function of only. This example shows Chebfun2 being used to represent parameterized surfaces. List of Practical (Using any software/MATLAB) Practical/Lab work to be performed on a Computer. 14 - Find the linear approximation of the function Ch. Tcheir1, and M. The article introduces a novel compliant-body pump, with a high volumetric efficiency: A flexible-matrix-composite hyperboloid structure. We will also discuss finding the area between two polar curves. Introduction to the Hyperbolic Functions in Mathematica. A better solution is to have MATLAB read the file without destroying the labels. I am a novice at Matlab and highly interested in learning more about Matlab. There are no x’s in the equation yz = 1. Find out how to use it and use it to plot a hyperboloid of one sheet and a hyperboloid of two sheets. How to draw a Circle in Matlab This also works in GNU-Octave, FreeMat, Scilab and Scicoslab 2. "Three-dimensional" redirects here. For other uses, see Moscow (disambiguation). u² = y² + z². Please help me in this. • A shell is a thin structure composed of curved sheets of material, so that the curvature plays an important role in the structural behavior, realizing a spatial form • Motivation: • A shell is the most efficient way of using the material, and can be very useful in case o storage of fluids and solids (uniform loads). The coarea of this map will be the surface area of the hyperboloid. At least one of the ranges must have arguments evaluating to real constants; the other range may have. The following implicit function plots an ellipse with semi-major axis radius or 5 and semi-minor axis radius of 4 and rotated -45 degrees: 20. — A Matlab Companion for Multivariable Calculus: Hyperboloid 102 if 277 plot 16 24 35 plot3 35 48 Polar. the forces applied to the corner nodes act only on ½ of the element edge), however we applied a constant magnitude along the plates edge. Single-download, text-based, Directory of web links to essential Physics resources. ⇤ I can name the 6 quadric surfaces, write their equation, and sketch their graph. You have just find two parameters so you need only one point more, e. I would like to plot a hyperboloid of the form. Hyperbolic manifold in the hyperboloid model: Matlab's sylvester solver without input used to plot a cost function restricted to a 1D or 2D subspace of a. Conversely, an equation for a hyperbola can be found. The hyperboloid model of conductance was employed to determine ΔG since this model provided the best fit to the conductance of TEM nanopores as discussed before. We graph this by using the MATLAB command for a cylinder. In addition, we are able to exhibit the beautiful structure of the spectrum and the close links between the eigenfunctions, the rays of geometrical optics, and the geometry of the damping region. But it’s easier when you have personalized tools to give each and every student what he or she needs to succeed. Fungrim currently consists of 263 symbols (named mathematical objects), 1198 entries (definitions, formulas, tables, plots), and 42 topics (listings of entries). It is also known as secondary hyperboloid reflector or sub-reflector. Contribute to RInterested/PLOTS development by creating an account on GitHub. Unlike other graphing software, we have an enormous library of mathematical functions at our ﬁngertips. Merrill algebra 2, maths tests ks2, free math percentage application, plot differential equations on matlab 2nd order, Prentice hall physics solution, how to find the inscribed and circumscribed rectangles in a quadratic. The Plot Function. Its purpose was described % above. In the applet, you'll see two cones joined at their apexes. Distances on S 2, d S, are measured along great circle arcs, while hyperbolic distances. That is, it is either the x-axis, the y-axis, or the z-axis. Imagine these cones are of infinite height (but shown with a particular height here for practical reasons) so we can see the extended. Before learning how to graph a hyperbola from its equation, get familiar with the vocabulary words and diagrams below. Learn more about 3d plots, plot, plotting. Afterwards the inertans, resistance and compliance are kept constant in pairs. Is there any trick for that? I'm afraid I. MATHEMATICA. % % The 'figure' statement tells Matlab to open a new figure window. A hyperbola is a type of conic section that looks somewhat like a letter x. This banner text can have markup. The 2-dim DEq dy/dt module is designed to generate plots in x,. How is it I could do it? Then you can use surf() to plot it. Provide details and share your research! But avoid … Asking for help, clarification, or responding to other answers. Numbers and Vector. B) Slice for x = 1. 3), and b) as a function of the Poisson’s ratio for a Berkovich indenter. meshgrid function, which builds two-dimensional grids from. Drawing conics in Matplotlib apr 21, 2016 geometry algebraic-geometry python numpy matplotlib. The two plots below extend the range of our acoust radiation detection contours for various energy values. Search MathWorks. With the command gca we get the handle to the current axes with which it is possible to set axis bounds. This example shows Chebfun2 being used to represent parameterized surfaces. On the perpendicular through S, to the x-axis, mark the line segment SP of length MR to get the point P of the hyperbola. Plotting X Y data Input XYRange referencing the X and Y. 3D Function Plots in Origin. MTH243 (Calculus for Functions of Several Variables) MATLAB. RevolutionPlot3D[{fx, fz}, {t, tmin, tmax}] generates a plot of the surface obtained by rotating the parametric curve with x, z. Therefore, before you start this activity, you want to be familiar with the concept of cylindical coordinates. 359 (3/24/08) 2 x z 2 x2 −z2 = −1 The intersection of The level surface x2 +y2 −z2 = −1 x2 +y2 −z2 = −1 with the xz-plane FIGURE 15 FIGURE 16 Example 10 Describe all the level surfaces of k(x,y,z) = x2 +y2 − z2. One common form of parametric equation of a sphere is: #(x, y, z) = (rho cos theta sin phi, rho sin theta sin phi, rho cos phi)# where #rho# is the constant radius, #theta in [0, 2pi)# is the longitude and #phi in [0, pi]# is the colatitude. (b) Use Green’s theorem to find the area of the region R (Hint: The vector field ~ F ( x, y ) = x ~ j may be helpful) You've reached the end of your free preview. MATLAB Central contributions by Chaowei Chen. I am working on hyperbolic navigation system, I am not able to plot directly hyperbola in MATLAB. figure(2) mesh(x,y,z) %or. Several data sets of sample points sharing the same x-coordinates can be fitted at once by passing in a 2D-array that contains one dataset per column. Conic Sections - interactive 3-D graph. But it’s easier when you have personalized tools to give each and every student what he or she needs to succeed. Running sphere alone can plot it too. It supports line plots, scatter plots, piecewise constant plots, bar plots, area plots, mesh{ and surface plots, patch plots, contour plots, quiver plots, histogram plots, box plots, polar axes,. 14 - If u = x2y3 + z4, where x = p + 3p2, y = pep, and Ch. This is a hyperboloid between the two focii i and j. Plot: A) Slice for x = 2. Parametric equation of the hyperbola In the construction of the hyperbola, shown in the below figure, circles of radii a and b are intersected by an arbitrary line through the origin at points M and N. mat, which contains the variables theta and rho. A unied plot composing the re sults of the magnetic susceptibility and SIRM from the 3 proles is presented in order to better distinguish the dierent units by means of magnetic properties. Find a parametrization of the hyperboloid of one sheet given by (use cylindrical coordinates). Chapter 12: Functions of Several Variables Vladimir A. modeled as a rotational hyperboloid and the dielectric liquid was cyclohexane. Plotting of graphs of the functions and to illustrate the effect of on the graph. MATHEMATICA ACHSENBESCHRIFTUNG PLOT - Settings for the option PlotRange. Find more Mathematics widgets in Wolfram\|Alpha. Math 265 Matlab Projects – Roettger, Spring 09 Project Contour Situation: Matlab’s gradient command for a function f uses difference quotients to approximate the gradient. I am a novice at Matlab and highly interested in learning more about Matlab. See also Linear Explorer, Quadratic Explorer and General Function Explorer. hyperbolic_space. Read "Application of a three-dimensional hyperbolic location system to the false killer whale (Pseudorca crassidens) depredation issue, International Journal of Global Environmental Issues" on DeepDyve, the largest online rental service for scholarly research with thousands of academic publications available at your fingertips. Vary the Parameter b to Change the Cylinder. It does read the file properly and return the misspelled words, however they are all on line 1. Want to read both pages?. then the space becomes a normed vector space. Simulation and experiment design Our software and algorithms are developed in C/C++ using Player/Gazebo [11] and interfaced with MATLAB for high-level conﬁguration speciﬁcation for both. hyperbolic_interface. Is there a way to plot a point in polar? Comment actions Permalink. Graphing Matlab - Free download as PDF File (. Parametric Surfaces. The usual Cartesian coordinate system can be quite difficult to use in certain situations. In this example, we will draw. MATLAB: Elementární vlastnosti jazyka. y 2 b 2 x 2 a 2 FIG. A partial derivative can also be performed in Matlab. 5 shows that first differential of Z lead to conversion of bi-conical plot to single hyperboloid plot which expands and contracts with respect to swinging frequency resembling a hyperboloid. Surface and Mesh Plots Representing gridded data as surface and mesh plots; Volume Visualization Representing gridded volume data as iso, slice, and stream plots; Polygons Shaded polygons × MATLAB Command. Discover what MATLAB. Zisserman • Bayesian Decision Theory • Bayes decision rule • Loss functions • Likelihood ratio test • Classifiers and Decision Surfaces • Discriminant function • Normal distributions • Linear Classifiers • The Perceptron • Logistic Regression Decision Theory. (b) Find the surface area of this torus Problem 3 Plot the vector field ~ F (x, y) = x x 2 + y 2, 1 x 2 + y 2 on the rectangle-10 ≤ x ≤ 10,-5 ≤ y ≤ 5. MATLAB can be used as a drawing board for pictures and diagrams. Sketching ellipsoid, hyperboloid, elliptic cone, paraboloid using Cartesian coordinate. The applet was created with LiveGraphics3D. Torus Made by Revolving a Circle of Radius 1 Centered at (0,2) About the x Axis. Plot hyperboloid of one sheet The equation of the shape we’ll graph is x2 + y2 = 0:003z2 + 0:05: In this case, you can see that at each value of z we have x2 + y2 = #. Mathematica plots such a line by means of its 3-dimensional parametric plotting com-mand, ParametricPlot3D. il Moodle Course Page Please visit this page regularly for all administrative information: homework assignments, course adminstrator, books and course materials, exams and grading policy, faculty and teaching assistants lists, office hours, study units, etc. 72\) should be most applicable for a Berkovich indenter, which is more like a cone than a paraboloid of revolution. edu is a platform for academics to share research papers. Call the nexttile function to create an axes object and return the object as ax1. x = y2 z2 vs. NPTEL provides E-learning through online Web and Video courses various streams. Best Graphing Calculator Online We have the most sophisticated and comprehensive TI 84 type graphing calculator online. All data in Fungrim is represented in symbolic, semantic form designed to be usable by computer algebra software. When the stress state lies on the surface the material is said to have reached its yield point and the material is said to have become plastic. Assignment! Bonus Problem: Change this last piece of code to graph the hyperboloid of one sheet determined by: z 2+ 4 = x + y2: Print it out from a few di erent angles. Identify the axis The axis of a hyperbolic paraboloid is one of the three coordinate axes. Then let's plot the sphere and -plane together in the same plot. For example, you might have noticed in figure 16. Parameters: x: array_like, shape (M,). Example: Let's define "sEqn" to be the equation of the sphere whose center is and radius is 2. List of Practical (Using any software/MATLAB) Practical/Lab work to be performed on a Computer. Wolfram Universal Deployment System Instant deployment across cloud, desktop, mobile, and more. If not given, they are assumed to be integer indices, i. Right from mathematics formula chart to algebra course, we have every part included. how to plot 3d hyperbola ?. At least one of the ranges must have arguments evaluating to real constants; the other range may have. Find more Mathematics widgets in Wolfram\|Alpha. , home/away games, championships. Before learning how to graph a hyperbola from its equation, get familiar with the vocabulary words and diagrams below. Polar Chart（极坐标图）： Snip20180214_15. Using your favorite plotting software (i. We graph this by using the MATLAB command for a cylinder. First, note that a vector in the direction of L is PQ= (–2, –1, 1). Just as with ellipses, writing the equation for a hyperbola in standard form allows us to calculate the key features: its center, vertices, co-vertices, foci, asymptotes, and the lengths and positions of the transverse and conjugate axes. The program package SOM Toolbox contains extra MATLAB functions developed by the SOM Programming Team of our laboratory. This example shows Chebfun2 being used to represent parameterized surfaces. But, the mathematical description of circles can get quite confusing, since there is a set equation for a circle, including symbols for the radius, and center of the circle. In 3D space, the graph of a second degree equation in x, y and z is one of six quadric surfaces. Parameters: x: array_like, shape (M,). ContourPlot3D treats the variables x, y, and z as local, effectively using Block. Workings On A Hyperbola. Come to Rational-equations. MATHEMATICA ACHSENBESCHRIFTUNG PLOT - Settings for the option PlotRange. Operator notation must be used, that is, the procedure name is given without parameters specified, and the ranges must be given simply in the form a. The hyperboloid of one sheet is an example of the doubly ruled surface. The calculator can be used to calculate. Create Polar Line Plot. Specify the plotting interval by specifying the second argument to fimplicit3. How to draw a Circle in Matlab This also works in GNU-Octave, FreeMat, Scilab and Scicoslab 2. I was able to generate the hyperboloid surface using parametric equations. You can plot the previous inputed equation against the time function, this would verfiy for you if the inputed formula is working and no bug issues will occur later. Random data points that we have just generated in 3D-space. It was assumed that. My very first introduction to programming was in middle school during the first week of class. Dismiss Join GitHub today. GitHub is home to over 40 million developers working together to host and review code, manage projects, and build software together. MATLAB in particular hides many of the details of data structures and data manipulation from the user. The variable rho is a measure of how intensely the antenna radiates for each value of theta. Here we demonstrate how MATLAB could be used to design and model objects, inspired by [1, Section 2. js: Define ordinal axis from JSON data; Put percent labels next to legend instead of in the slice; Plotting A Hyperboloid; Is an if-let or a normal if. u² = y² + z². The main purpose of this paper is constructing and plotting the surfaces that are generated from the motion of inextensible curves in. y 2 b 2 x 2 a 2 FIG. y: array_like, shape (M,) or (M, K). ContourPlot3D treats the variables x, y, and z as local, effectively using Block. Two-element vector of form [min max] — Use the same plotting interval of [min max] for x, y, and z. (a) Plot this surface using ezmesh (use axis equal to get a nicer picture). Provide details and share your research! But avoid … Asking for help, clarification, or responding to other answers. Several data sets of sample points sharing the same x-coordinates can be fitted at once by passing in a 2D-array that contains one dataset per column. Try the following commands in Matlab to generate a helix and examine it from various points of view. 3d Surface Generator software free downloads and reviews at WinSite. Conic Sections - interactive 3-D graph. Mathematica plots such a line by means of its 3-dimensional parametric plotting com-mand, ParametricPlot3D. For example, to generate the surface of the elliptic hyperboloid of two sheets described by the equation, Plotting the direction field of a vector fields in 2D. In this course we will use Mathematica computer algebra system (CAS), which is available in computer labs at URI. How to Graph a Hyperbola Think of a hyperbola as a mix of two parabolas — each one a perfect mirror image of the other, each opening away from one another. The arguments to a MATLAB [non-ez] plotting function, such as surf, plot, plot3, mesh, or contour, are two or three identically shaped arrays. Hence, this seemingly old problem - discovering. – Hyperboloid of one sheet – Hyperboloid of two sheets – Elliptic paraboloid – Hyperbolic paraboloid – Elliptic cone (degenerate) (traces) 2 2 2 Ax By Cz Dx Ey F + + + + + = 0 Quadric Surfaces. Unlike Matlab, Maple provides a plotting procedure which allows you to an ellipsoid can look like the top part of a hyperboloid of two sheets or like a paraboloid. * Large array of primitives for model creation (e. Come to Rational-equations. I would like to graph y=211. Want to read both pages?. Cubic Function Explorer. Since is undefined, our plot has an asymptote there. Siddique2 1Mathematics Department, King Saud University, Riyadh, Saudi Arabia 2Mathematics Faculty, ECPI University, USA Abstract-The role of technology and using specialized software in the educational process is growing in recent times. How to draw a Circle in Matlab This also works in GNU-Octave, FreeMat, Scilab and Scicoslab 2. Right from mathematics formula chart to algebra course, we have every part included. y = z2 x2 This are both hyperbolic paraboloids, one rst with the x-axis perpendicular to the 'saddle' and the other with the. The arguments to a MATLAB [non-ez] plotting function, such as surf, plot, plot3, mesh, or contour, are two or three identically shaped arrays. 3 c mathcentre January 9, 2006. MATLAB can be used as a drawing board for pictures and diagrams. An optimization approach is then proposed to fit the model to real input SRP data and estimate the position of the acoustic source. This article is meant to inform new MATLAB users how to plot an anonymous function. This is the bi-weekly visible open thread. Look through the examples and example code in the "specialized plotting". How to plot an elliptical surface in Matlab? - MATLAB Answers - MATLAB Central. Spirals by Polar Equations top Archimedean Spiral top You can make a spiral by two motions of a point: There is a uniform motion in a fixed direction and a motion in a circle with constant speed. Perhaps the most straightforward way to prepare such data is to use the np. Note that in all quadratic cases below, the z-axis is the axis of symmetry. 1 SYLLABUS FOR B. This command basically generates a volume. The surface is given by equation z = y^2 - x^2, and I must have some thickness - It will be printed on 3D printer. Initially, stenciling is not used so if you look (by holding down the left mouse button and moving) at the dinosaur from "below" the floor, you'll see a bogus dinosaur and appreciate how the basic technique works. (a) Parametrize the hyperboloid and plot it. Free 3d Surface Generator Shareware and Freeware. X = range(M), Y = range(N). with operating impedance between the cones. Use MathJax to format equations. MATLAB in particular hides many of the details of data structures and data manipulation from the user. hyperbolic_interface. We find that the mathematical notation and the multidimensional capabilities of Matlab greatly simplify code development and maintenance compared to other programming languages. Use your calculator or MatLab to plot a few, for k = 1,2,3,4 and −2 ≤ x ≤ 2, on the same set of axes. New sidebar ad fo…. Only use this command if you want to lose everything. Surfaces, Volumes, and Polygons. ); however, there are occasions when a design is dependent on mathematical functions or equations to describe its geometry/topology. (vii) Matrix operation (addition, multiplication, inverse, transpose). The input parameters are a matrix of magnitudes, Zp, and a list of property,value pairs that modify the default plot behavior. Evolution of epsilon as a function of the power law exponent m of the unloading curve. Open an example of the pgfplots package in Overleaf. Learn more about 3d plots, plot, plotting. If r is a list, np. \] The following figures summarizes the most important ones. Find here the Matlab function to plot the $$\epsilon$$ function : epsilon_oliver_pharr. Numerical simulations confirm the theoretical model. Then, we use the command legend to add an annotation to the figure. Area of 3D plot. Stewart 14. The arguments to a MATLAB [non-ez] plotting function, such as surf, plot, plot3, mesh, or contour, are two or three identically shaped arrays. EDIT: Yes, this is a clue to the cause -- the other nearest neighbour surface plot in my collection, produced with MATLAB's TriScatteredInterp , produces exactly. The volume of an ellipsoid is given by the following formula:. Learn more about area of ploygon, numerical integration. All data in Fungrim is represented in symbolic, semantic form designed to be usable by computer algebra software. Each column of Zp contains information along a single half-meridian and each row gives height values along a circular arc. Particularly, students will • learn about parametrization of quadric surfaces. Hyperboloid of two sheets − −+= If you draw all traces on one coordinate axis, can you guess how the surface should look like? • See the Matlab illustration in class for more understanding of different traces & views • download Matlab file eg_hyper2. In this section we will take a look at the basics of representing a surface with parametric equations. Uživatelský přístup. Saddle (hypar) Surfaces. A, B, C, E, F-five distinct points 'directrix'=dir-dir is the line which is the directrix of the hyperbola 'focus'=fou-fou is a point which is the focus of the hyperbola 'eccentricity'=ecc-ecc is a constant bigger than one denoting the eccentricity of the hyperbola 'vertices'=ver-where ver is a list of two points which is the vertices of the hyperbola. Hyperbolic manifold in the hyperboloid model: Matlab's sylvester solver without input used to plot a cost function restricted to a 1D or 2D subspace of a. The hyperbolic identities Introduction The hyperbolic functions satisfy a number of identities. Random data points that we have just generated in 3D-space. = plot3( ) enables MatLab to change the properties of this particular plot using the set command. I am a novice at Matlab and highly interested in learning more about Matlab. It can be used to construct a Poincaré disk model as a projection viewed from (t=-1,x 1 =0,x 2 =0), projecting the upper half hyperboloid onto the unit disk at t=0. ^ instead of * , / , ^ when you want to evaluate expressions in x , y , z element by element. The six listed are: elliptic cone elliptic paraboloid hyperbolic paraboloid ellipsoid hyperboloid of one sheet hyperboloid of two sheet I need help with putting an equation to its standard form and identifying the quadric surface given an equation. RGB Triplet Hexadecimal Color Code. You can use the following code to get the desired results. You have just find two parameters so you need only one point more, e. Whether you’re teaching a group of five or 500, reaching every student can be challenging. Contribute to RInterested/PLOTS development by creating an account on GitHub. Plotting angled hyperboloid in MATLAB. Zisserman • Bayesian Decision Theory • Bayes decision rule • Loss functions • Likelihood ratio test • Classifiers and Decision Surfaces • Discriminant function • Normal distributions • Linear Classifiers • The Perceptron • Logistic Regression Decision Theory. At least for starters, include the following commands x=-3. Mathematica tutorial on plotting a hyperbolic paraboloid. Create the top plot by passing ax1 to the plot function. Find out how to use it and use it to plot a hyperboloid of one sheet and a hyperboloid of two sheets. The final. This paper presents a novel approach for indoor acoustic source localization using sensor arrays. ) * Selection, zooming, 3-D mouse-only based handling, etc. Submit your pictures and your code. The applet was created with LiveGraphics3D. 4 shows that fault caused permanent shift in the operating impedance. How to draw a Circle in Matlab This also works in GNU-Octave, FreeMat, Scilab and Scicoslab 2. Post about anything you want, ask random questions, whatever. For other uses, see 3D (dis. Inmatning allmänna parabel i MATLAB med "p = 1" ger "y = / (4 1) (x ^ 2. Examples of this might be the curvature of a lens, the design of a custom spring, wind-foil. RevolutionPlot3D[fz, {t, tmin, tmax}] generates a plot of the surface of revolution with height fz at radius t. Includes all the functions and options you might need. Wolfram Universal Deployment System Instant deployment across cloud, desktop, mobile, and more. Examples: Parabolic Cylinder. Books Recommended. Andrew James Marston. = plot3( ) enables MatLab to change the properties of this particular plot using the set command. EDIT: Yes, this is a clue to the cause -- the other nearest neighbour surface plot in my collection, produced with MATLAB's TriScatteredInterp , produces exactly. Random data points that we have just generated in 3D-space. The stereographic projection (Fig. Examples of evaluating Mathematica functions applied to various numeric and exact expressions that involve the hyperbolic functions or return them are shown. Someone who doesn't know the distinction might be tempted to search for help on TeX or. Plotting the graphs of the polynomial of degree 4 and 5. Unlike Matlab, Maple provides a plotting procedure which allows you to an ellipsoid can look like the top part of a hyperboloid of two sheets or like a paraboloid. Page 1 of 2 10. For eksempel:. This is a very useful tool in all types of scientific and math based research allowing the user. Even though MATLAB’s plotting procedure looks like a really simplistic approach, it’s actually quite useful for any data you want to plot quickly. Example 5 (12. The graph of a function z = f(x,y) is also the graph of an equation in three variables and is therefore a surface. See also Linear Explorer, Quadratic Explorer and General Function Explorer. There is no built-in MATLAB command to read this data, so we have to write an m-file to do the job. Introduction to the Hyperbolic Functions in Mathematica. I am working on hyperbolic navigation system, I am not able to plot directly hyperbola in MATLAB. legend A key that identifies the color, gradient, picture, texture, or pattern assigned to each data series in a chart. Hyperboloid Pen Holder This hyperboloid pen is a lovely pen holder with enough space to hold 16 pens with ease. :return: if r is a scalar, np. We assume that you already know how to use MATLAB’s plotting commands to graph plane curves of the form y = f (x). Plotting X Y data Input XYRange referencing the X and Y. Anton Betten1. This doesn't. Then you can use surf () to plot it. Ellipsoid dimensions. Using the region function to plot a 3d graph to within a certain limit. x-coordinates of the M sample points (x[i], y[i]). The designer can shift these points to alter the shape of the curve. arXiv physics papers Terms Topic Distributions Per Term. HW #1: DUE MONDAY, FEBRUARY 4, 2013 1. Programování v kombinaci MATLAB a Maple. Lectures 5 & 6: Classifiers Hilary Term 2007 A. Download Plotting of polar equations can be facilitated by the fol- lowing steps: 1. We start by plotting two simple quadratic surfaces that are commonly taught in multivariate calculus: a cone and a Hyperboloid of one sheet. The variable rho is a measure of how intensely the antenna radiates for each value of theta. Tingnan ang profile ni Aldrich Zeno sa LinkedIn, ang pinakamalaking komunidad ng propesyunal sa buong mundo. x and y should now both be column vectors with the sa. dic This class can parse, analyze words and interprets sentences. Introduction to adaptive ﬁlters This chapter introduces the fundamental principles of adaptive ﬁltering and commonly used adaptive ﬁlter structures and algorithms. write a program in matlab to Plot the following standard surfaces: (a) ellipsoid, (b) hyperboloid of one sheet, (c) hyperboloid of two sheets, (d) cone, (e) elliptic paraboloid, (f) hyperbolic paraboloid, (g) torus, (h) right helicoid. The energy difference after the electric ﬁeld redistributes, was calculated in the ﬁnite element analysis software Comsol. Conversely, an equation for a hyperbola can be found. Plotting in Scilab www. Aldrich ay may 2 mga trabaho na nakalista sa kanilang profile. Torus Made by An Elliptic Hyperboloid with Axis Along the y Axis:. The plot should appear as shown in this EquationExplorer plot:. 3B, I Na traces of one representative cell are shown during an activation step far above activation threshold (left plot) and near the threshold (right plot). Tvorba a vlastnosti M-souborů. Right from mathematics formula chart to algebra course, we have every part included. Ellipsoids. Mahesh says: 13 Nov 2015 at 11:55 pm [Comment permalink] Isn't there a dt at the end for the curve length integral?. New sidebar ad fo…. ; Die reelle Funktion ist streng monoton steigend und besitzt in = ihren einzigen Wendepunkt. To plot the graph of a function, you need to take the following steps − Define x, by specifying the range of values for the variable x, for which the function is to be plotted Define the function, y = f (x) Call the plot command, as plot (x, y). Below is an example of creating and plotting the values of the X squared graph from -10 to +10. Find a parametrization for S and plot this surface using ezmesh. Ellipsoid dimensions. ContourPlot3D has attribute HoldAll and evaluates f and g after assigning numerical values to x, y and z. Geometry homework help radicals. Find here the Matlab function to plot the $$\epsilon$$ function : epsilon_oliver_pharr. Their connection on the one hand with integrable systems and, on the other, with Lie–Poisson systems motivates the research for optimal numerical schemes to solve them. Jayaram says: 13 Mar 2015 at 9:54 am [Comment permalink] A very useful and interesting article for practical applications. The graph of a function z = f(x,y) is also the graph of an equation in three variables and is therefore a surface. " Håndtaget er derefter " @ HYP " (uden anførselstegn). This gives the axis along which the hyperboloid opens. Just as with ellipses, writing the equation for a hyperbola in standard form allows us to calculate the key features: its center, vertices, co-vertices, foci, asymptotes, and the lengths and positions of the transverse and conjugate axes. Quiver3 Matlab Quiver3 Matlab. Visualize this radiation pattern by plotting the data in polar coordinates using the polarplot function. Surfaces, Volumes, and Polygons. The hyperboloid model of conductance was employed to determine ΔG since this model provided the best fit to the conductance of TEM nanopores as discussed before. how to plot 3d hyperbola ?. 359 (3/24/08) 2 x z 2 x2 −z2 = −1 The intersection of The level surface x2 +y2 −z2 = −1 x2 +y2 −z2 = −1 with the xz-plane FIGURE 15 FIGURE 16 Example 10 Describe all the level surfaces of k(x,y,z) = x2 +y2 − z2. Graph a Circle - powered by WebMath. Intermediate: PSIM, MI Power, Matlab/coding, RSCAD/RTDS, Pspice, Proteus Advanced: Matlab/Simulink, PSCAD, Multisim, CERTIFICATIONS The investigation was carried out along with three dimensional plots to visualize the locus of impedance in resistance, reactance and time axis. Geometric Optimization of Solar Concentrating Collectors using Quasi-Monte Carlo Simulation. There is a limit, however, which you can set, to how finely the Wolfram Language will ever sample a function. ^ instead of * , / , ^ when you want to evaluate expressions in x , y , z element by element. a quick temporary workaround for the negative theta range problem for those who need it is to replace theta with "theta + a2pi" or "theta + 360a" where "a" is a scaling factor. For example, the following routine plots the line L through the points P(1, 2, 3) and Q(-1, 1, 4). stl files (more on this later too). Central Authentication Service. Matlab ignores any line that begins with % Matlab has many built in matrix functions and operators. Casse grain is another type of feed given to the reflector antenna. legend A key that identifies the color, gradient, picture, texture, or pattern assigned to each data series in a chart. Ezsurf tager to indgange: håndtaget til den funktion, du vil plotte og en vektor af maksimale og minimale x og y-koordinaterne for din plot. Create Polar Line Plot. Drawing conics in Matplotlib apr 21, 2016 geometry algebraic-geometry python numpy matplotlib. hyperbolic_polygon) HyperbolicRegularPolygon (class in sage. Want to read both pages?. dic This class can parse, analyze words and interprets sentences. The plot function usually takes two arguments (but can take one). Class Activity: Parametrization and Visualization This class activity is intended for two 50-minute class periods. Specify the plotting interval by specifying the second argument to fimplicit3. A member of the set of numbers that locates a point in a Cartesian coordinate system. This paper presents a novel approach for indoor acoustic source localization using sensor arrays. This is a conscious and deliberate decision. The arguments to a MATLAB [non-ez] plotting function, such as surf, plot, plot3, mesh, or contour, are two or three identically shaped arrays. A hyperbola is a type of conic section that looks somewhat like a letter x. how to plot 3d hyperbola ?. i need help im down to my last submission out of 10 please help! Use polar coordinates to find the volume of the given solid. MATLAB Programming 2:0:0 ; 2. Surfaces, Volumes, and Polygons. Thanks a lot in advance. Find out how to use it and use it to plot a hyperboloid of one sheet and a hyperboloid of two sheets. Is their any way to this using Matlab or R ? I am not able to figure it out. The results given in this paper demonstrate the capability of such a drive system in washing machine applications where simplicity, reliability and stability are more important issues. It can be used to construct a Poincaré disk model as a projection viewed from (t=-1,x 1 =0,x 2 =0), projecting the upper half hyperboloid onto the unit disk at t=0. The following table shows the number of hours five car salespeople worked and the number of cars they sold. Quadric Surfaces - Traces Traces are cross sections parallel to a plane. Plot: A) Slice for x = 2. edu is a platform for academics to share research papers. I need it to read into the paragraph and return the different lines and I have no idea how to do. The Java applet did not load, and the above is only a static image representing one view of the applet. Try the following commands in Matlab to generate a helix and examine it from various points of view. A hyperbola is a type of conic section that looks somewhat like a letter x. plex quadric such as an ellipsoid, paraboloid or hyperboloid and the state of the art requires 9 [51,49,9] to 12 [4] points to perform a generic quadric ﬁt. ContourPlot3D is also known as a level surface or iso surface. Creating a plot using commands. Introduction. Problem 1: What is wrong with the following argument (from Mathematical Fallacies, Flaws, and Flimﬂam - by Edward Barbeau): There is no point on the parabola 16y = x2 closest to (0,5). Their connection on the one hand with integrable systems and, on the other, with Lie–Poisson systems motivates the research for optimal numerical schemes to solve them. Note that the equations on this page are true only for ellipses that are aligned with the coordinate plane, that is, where the major and minor axes are parallel to the coordinate system. http://mathispower4u. hyperbolic_interface. Mathematica plots such a line by means of its 3-dimensional parametric plotting com-mand, ParametricPlot3D. Ketika kami (tim penyusun) membaca kembali kamus tersebut, ternyata terdapat gelaran musik yg disajikan dng menggu- nakan alat musik bambu yg tersusun dl ra. In each plot, three traces represent I Na in the detached cell, the cell approached to the obstacle, and the cell subsequently moved away, respectively. Using Excel, plot each point on the same graph where the first coordinate is the number of hours and the second coordinate is the number of cars sold (hours, cars). where F is the distance from the center to the foci along the transverse axis, the same axis that the vertices are on. In this type, the feed is located at the vertex of the paraboloid, unlike in the parabolic reflector. Verify that what you are looking at is the hyperboloid of one sheet X^2 + Y. Create the top plot by passing ax1 to the plot function. A script normally contains following kinds of. These involve numeric and symbolic calculations and plots. In fact it is one of the three doubly ruled surfaces (besides the plane and the hyperboloid), having two distinct independent families of lines generating the surface. Moscow Mосква (Russian) — Federal city — Cloc. Another notable work is the hyperboloid by Zuber et al. Mathematica plots such a line by means of its 3-dimensional parametric plotting com-mand, ParametricPlot3D. Cubic Function Explorer. We begin with a brief discussion of how MATLAB does its plotting. A hyperbola is a type of conic section that looks somewhat like a letter x. b, rather than as an equation. Conversely, an equation for a hyperbola can be found. The 3D function plot gallery is a collection of 3D Parametric Function Plots which can be plotted in Origin. Therefore, before you start this activity, you want to be familiar with the concept of cylindical coordinates. SOLUTION The y2-term is positive, so the transverse axis is vertical. In 3D space, the graph of a second degree equation in x, y and z is one of six quadric surfaces. The regions we look at in this section tend (although not always) to be shaped vaguely like a piece of pie or pizza and we are looking for the area of the region from the outer boundary (defined by the polar equation) and the origin/pole. $(4,0,9)$ from the graph. Answer to: Consider the level surface given by x^2 - y^2 + z^2 = 2. A hyperbola is a type of conic section that looks somewhat like a letter x. 2 Surfaces in R3 [M] 12:Use Maple/Gnuplot to plot the following func-tions f: R 2!R. This is a hyperboloid between the two focii i and j. MATLAB The polynomial for displacement has been extracted using the elliptic parabolid, elliptic hyperboloid and general 3rd order polynomial functions. hyperbolic_regular_polygon) HyperbolicSpace() (in module sage. Using pgfplots, I'm trying to create a quiver plot on a hyperboloid surface something like this. Please pick the appropriate calculator from below to begin. But, the mathematical description of circles can get quite confusing, since there is a set equation for a circle, including symbols for the radius, and center of the circle. You mean, 3d Hyperboloid? Hyperboloid (Wikipedia) -> Link. com page 4/17 Step 2: Multiple plot and axis setting In this example we plot two functions on the same figure using the command plot twice. This banner text can have markup. This command basically generates a volume. 359 (3/24/08) 2 x z 2 x2 −z2 = −1 The intersection of The level surface x2 +y2 −z2 = −1 x2 +y2 −z2 = −1 with the xz-plane FIGURE 15 FIGURE 16 Example 10 Describe all the level surfaces of k(x,y,z) = x2 +y2 − z2. The turtle module is an extended reimplementation of the same-named module from the Python standard distribution up to version Python 2. 1:2pi; subplot(2,2,1); plot(x,sin(x)); subplot(2,2,2);. hyperboloid of two sheets determined by: z 2= x + y + 4 : Print it out from a few di erent angles. it's basically drawing a bunch of different graphs of different ranges that. The applet is not loading because it looks like you do not have Java installed. Maple, Matlab, Wolframalpha, etc. Dobrushkin,Lippitt Hall 202C, 874-5095,[email protected] Identify the axis The axis of a hyperbolic paraboloid is one of the three coordinate axes. Load the file antennaData. legend A key that identifies the color, gradient, picture, texture, or pattern assigned to each data series in a chart. The last line plots the vector field. edu is a platform for academics to share research papers. You can plot the previous inputed equation against the time function, this would verfiy for you if the inputed formula is working and no bug issues will occur later. Homework question 3: The command to plot a surface that cannot be written as a function of x and y is called implicitplot. Quadric Surfaces in Matlab In this activity we will use cylindrical coordinate transformations to assist in the plotting of quadric surfaces , the three-dimensional analogue of the conic sections. Anton Betten1. In 3D space, the graph of a second degree equation in x, y and z is one of six quadric surfaces. This is analogous to the spherical stereographic projection, and can be better understood by analogy with the more familiar sphere, S 2. This is a very useful tool in all types of scientific and math based research allowing the user. Is there a way to plot a point in polar? Comment actions Permalink. then the space becomes a normed vector space. Direct link to this comment: Functions of two Variables. Graphing Matlab - Free download as PDF File (. Principal curvature directions along with the surface normal, define a 3D orientation frame at a surface point. Problem 12). Using your favorite plotting software (i. For example, the following routine plots the line L through the points P(1, 2, 3) and Q(–1, 1, 4). Dismiss Join GitHub today. 7, Functions of three variables p. Occasionally we get sloppy and just refer to it simply as a paraboloid; that wouldn't be a problem, except that it leads to confusion with the hyperbolic paraboloid. with operating impedance between the cones. The other type is the hyperboloid of two sheets, and it is illustrated by the graph of x 2 - y 2 - z 2 = 1, shown below. The xz trace is found by setting y = 0. MATLAB Central contributions by Yury. Because it's such a neat surface, with a fairly. EDIT: Yes, this is a clue to the cause -- the other nearest neighbour surface plot in my collection, produced with MATLAB's TriScatteredInterp , produces exactly. Example: Let's define "sEqn" to be the equation of the sphere whose center is and radius is 2. The high volumetric efficiency is attributed to the geometry of the pump’s structure (hyperboloid) as well as the high negative effective Poisson’s ratio of the 3-layer [θ/β/θ] flexible-matrix-composite (carbon/polyurethane) laminate adopted for the body of the pump. ) Selection, zooming, 3-D mouse-only based handling, etc. Use MathJax to format equations. png 例：蜗牛线（蚌线、蚶线）例：心脏线例：伯努利双纽线例：阿基米德螺线例：对数螺线例：双曲. When a voltage was applied, a thin channel was ionized and made conducting by a strong electric ﬁeld. There is a limit, however, which you can set, to how finely the Wolfram Language will ever sample a function. array of shape (size, dim). Whether to project points always on the upper or lower branch of the hyperboloid, or on both based on the sign of the last coordinate. Graphs of Hyperbolic Functions The graphs and properties such as domain, range and asymptotes of the 6 hyperbolic functions: sinh(x), cosh(x), tanh(x), coth(x), sech(x) and csch(x) are explored using an applet. surf(X,Y,Z) creates a three-dimensional surface plot, which is a three-dimensional surface that has solid edge colors and solid face colors. \] The following figures summarizes the most important ones. The plot should appear as shown in this EquationExplorer plot:. Here the cylindrical axis must be the x-axis rather than the usual z-axis. The two plots below extend the range of our acoust radiation detection contours for various energy values. NPTEL provides E-learning through online Web and Video courses various streams. MATLAB allows you to display your plots however you choose. Quadric Surfaces The graph in 2D space of a second degree equation in x and y is an ellipse, parabola or hyperbola. As with the ellipse the focus is at the point and the directrix is the line. Chapter 12: Functions of Several Variables. MATLAB has the sphere() function. In the applet, you'll see two cones joined at their apexes. C++, rather than in MATLAB or JAVA. Tingnan ang profile ni Aldrich Zeno sa LinkedIn, ang pinakamalaking komunidad ng propesyunal sa buong mundo. Find more Mathematics widgets in Wolfram\|Alpha. A convex shaped reflector, which acts as a hyperboloid is placed opposite to the feed of the antenna. Active 5 years, 10 months ago. MATLAB in particular hides many of the details of data structures and data manipulation from the user. 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https://terrytao.wordpress.com/page/3/	Over on the polymath blog, I’ve posted (on behalf of Dinesh Thakur) a new polymath proposal, which is to explain some numerically observed identities involving the irreducible polynomials $P$ in the polynomial ring ${\bf F}_2[t]$ over the finite field of characteristic two, the simplest of which is $\displaystyle \sum_P \frac{1}{1+P} = 0$ (expanded in terms of Taylor series in $u = 1/t$). Comments on the problem should be placed in the polymath blog post; if there is enough interest, we can start a formal polymath project on it. In this blog post, I would like to specialise the arguments of Bourgain, Demeter, and Guth from the previous post to the two-dimensional case of the Vinogradov main conjecture, namely Theorem 1 (Two-dimensional Vinogradov main conjecture) One has $\displaystyle \int_{[0,1]^2} \|\sum_{j=0}^N e( j x + j^2 y)\|^6\ dx dy \ll N^{3+o(1)}$ as ${N \rightarrow \infty}$. This particular case of the main conjecture has a classical proof using some elementary number theory. Indeed, the left-hand side can be viewed as the number of solutions to the system of equations $\displaystyle j_1 + j_2 + j_3 = k_1 + k_2 + k_3$ $\displaystyle j_1^2 + j_2^2 + j_3^2 = k_1^2 + k_2^2 + k_3^2$ with ${j_1,j_2,j_3,k_1,k_2,k_3 \in \{0,\dots,N\}}$. These two equations can combine (using the algebraic identity ${(a+b-c)^2 - (a^2+b^2-c^2) = 2 (a-c)(b-c)}$ applied to ${(a,b,c) = (j_1,j_2,k_3), (k_1,k_2,j_3)}$) to imply the further equation $\displaystyle (j_1 - k_3) (j_2 - k_3) = (k_1 - j_3) (k_2 - j_3)$ which, when combined with the divisor bound, shows that each ${k_1,k_2,j_3}$ is associated to ${O(N^{o(1)})}$ choices of ${j_1,j_2,k_3}$ excluding diagonal cases when two of the ${j_1,j_2,j_3,k_1,k_2,k_3}$ collide, and this easily yields Theorem 1. However, the Bourgain-Demeter-Guth argument (which, in the two dimensional case, is essentially contained in a previous paper of Bourgain and Demeter) does not require the divisor bound, and extends for instance to the the more general case where ${j}$ ranges in a ${1}$-separated set of reals between ${0}$ to ${N}$. In this special case, the Bourgain-Demeter argument simplifies, as the lower dimensional inductive hypothesis becomes a simple ${L^2}$ almost orthogonality claim, and the multilinear Kakeya estimate needed is also easy (collapsing to just Fubini’s theorem). Also one can work entirely in the context of the Vinogradov main conjecture, and not turn to the increased generality of decoupling inequalities (though this additional generality is convenient in higher dimensions). As such, I am presenting this special case as an introduction to the Bourgain-Demeter-Guth machinery. We now give the specialisation of the Bourgain-Demeter argument to Theorem 1. It will suffice to establish the bound $\displaystyle \int_{[0,1]^2} \|\sum_{j=0}^N e( j x + j^2 y)\|^p\ dx dy \ll N^{p/2+o(1)}$ for all ${4, (where we keep ${p}$ fixed and send ${N}$ to infinity), as the ${L^6}$ bound then follows by combining the above bound with the trivial bound ${\|\sum_{j=0}^N e( j x + j^2 x^2)\| \ll N}$. Accordingly, for any ${\eta > 0}$ and ${4, we let ${P(p,\eta)}$ denote the claim that $\displaystyle \int_{[0,1]^2} \|\sum_{j=0}^N e( j x + j^2 y)\|^p\ dx dy \ll N^{p/2+\eta+o(1)}$ as ${N \rightarrow \infty}$. Clearly, for any fixed ${p}$, ${P(p,\eta)}$ holds for some large ${\eta}$, and it will suffice to establish Proposition 2 Let ${4, and let ${\eta>0}$ be such that ${P(p,\eta)}$ holds. Then there exists ${0 < \eta' < \eta}$ such that ${P(p,\eta')}$ holds. Indeed, this proposition shows that for ${4, the infimum of the ${\eta}$ for which ${P(p,\eta)}$ holds is zero. We prove the proposition below the fold, using a simplified form of the methods discussed in the previous blog post. To simplify the exposition we will be a bit cavalier with the uncertainty principle, for instance by essentially ignoring the tails of rapidly decreasing functions. Given any finite collection of elements ${(f_i)_{i \in I}}$ in some Banach space ${X}$, the triangle inequality tells us that $\displaystyle \\| \sum_{i \in I} f_i \\|_X \leq \sum_{i \in I} \\|f_i\\|_X.$ However, when the ${f_i}$ all “oscillate in different ways”, one expects to improve substantially upon the triangle inequality. For instance, if ${X}$ is a Hilbert space and the ${f_i}$ are mutually orthogonal, we have the Pythagorean theorem $\displaystyle \\| \sum_{i \in I} f_i \\|_X = (\sum_{i \in I} \\|f_i\\|_X^2)^{1/2}.$ For sake of comparison, from the triangle inequality and Cauchy-Schwarz one has the general inequality $\displaystyle \\| \sum_{i \in I} f_i \\|_X \leq (\# I)^{1/2} (\sum_{i \in I} \\|f_i\\|_X^2)^{1/2} \ \ \ \ \ (1)$ for any finite collection ${(f_i)_{i \in I}}$ in any Banach space ${X}$, where ${\# I}$ denotes the cardinality of ${I}$. Thus orthogonality in a Hilbert space yields “square root cancellation”, saving a factor of ${(\# I)^{1/2}}$ or so over the trivial bound coming from the triangle inequality. More generally, let us somewhat informally say that a collection ${(f_i)_{i \in I}}$ exhibits decoupling in ${X}$ if one has the Pythagorean-like inequality $\displaystyle \\| \sum_{i \in I} f_i \\|_X \ll_\varepsilon (\# I)^\varepsilon (\sum_{i \in I} \\|f_i\\|_X^2)^{1/2}$ for any ${\varepsilon>0}$, thus one obtains almost the full square root cancellation in the ${X}$ norm. The theory of almost orthogonality can then be viewed as the theory of decoupling in Hilbert spaces such as ${L^2({\bf R}^n)}$. In ${L^p}$ spaces for ${p < 2}$ one usually does not expect this sort of decoupling; for instance, if the ${f_i}$ are disjointly supported one has $\displaystyle \\| \sum_{i \in I} f_i \\|_{L^p} = (\sum_{i \in I} \\|f_i\\|_{L^p}^p)^{1/p}$ and the right-hand side can be much larger than ${(\sum_{i \in I} \\|f_i\\|_{L^p}^2)^{1/2}}$ when ${p < 2}$. At the opposite extreme, one usually does not expect to get decoupling in ${L^\infty}$, since one could conceivably align the ${f_i}$ to all attain a maximum magnitude at the same location with the same phase, at which point the triangle inequality in ${L^\infty}$ becomes sharp. However, in some cases one can get decoupling for certain ${2 < p < \infty}$. For instance, suppose we are in ${L^4}$, and that ${f_1,\dots,f_N}$ are bi-orthogonal in the sense that the products ${f_i f_j}$ for ${1 \leq i < j \leq N}$ are pairwise orthogonal in ${L^2}$. Then we have $\displaystyle \\| \sum_{i = 1}^N f_i \\|_{L^4}^2 = \\| (\sum_{i=1}^N f_i)^2 \\|_{L^2}$ $\displaystyle = \\| \sum_{1 \leq i,j \leq N} f_i f_j \\|_{L^2}$ $\displaystyle \ll (\sum_{1 \leq i,j \leq N} \\|f_i f_j \\|_{L^2}^2)^{1/2}$ $\displaystyle = \\| (\sum_{1 \leq i,j \leq N} \|f_i f_j\|^2)^{1/2} \\|_{L^2}$ $\displaystyle = \\| \sum_{i=1}^N \|f_i\|^2 \\|_{L^2}$ $\displaystyle \leq \sum_{i=1}^N \\| \|f_i\|^2 \\|_{L^2}$ $\displaystyle = \sum_{i=1}^N \\|f_i\\|_{L^4}^2$ giving decoupling in ${L^4}$. (Similarly if each of the ${f_i f_j}$ is orthogonal to all but ${O_\varepsilon( N^\varepsilon )}$ of the other ${f_{i'} f_{j'}}$.) A similar argument also gives ${L^6}$ decoupling when one has tri-orthogonality (with the ${f_i f_j f_k}$ mostly orthogonal to each other), and so forth. As a slight variant, Khintchine’s inequality also indicates that decoupling should occur for any fixed ${2 < p < \infty}$ if one multiplies each of the ${f_i}$ by an independent random sign ${\epsilon_i \in \{-1,+1\}}$. In recent years, Bourgain and Demeter have been establishing decoupling theorems in ${L^p({\bf R}^n)}$ spaces for various key exponents of ${2 < p < \infty}$, in the “restriction theory” setting in which the ${f_i}$ are Fourier transforms of measures supported on different portions of a given surface or curve; this builds upon the earlier decoupling theorems of Wolff. In a recent paper with Guth, they established the following decoupling theorem for the curve ${\gamma({\bf R}) \subset {\bf R}^n}$ parameterised by the polynomial curve $\displaystyle \gamma: t \mapsto (t, t^2, \dots, t^n).$ For any ball ${B = B(x_0,r)}$ in ${{\bf R}^n}$, let ${w_B: {\bf R}^n \rightarrow {\bf R}^+}$ denote the weight $\displaystyle w_B(x) := \frac{1}{(1 + \frac{\|x-x_0\|}{r})^{100n}},$ which should be viewed as a smoothed out version of the indicator function ${1_B}$ of ${B}$. In particular, the space ${L^p(w_B) = L^p({\bf R}^n, w_B(x)\ dx)}$ can be viewed as a smoothed out version of the space ${L^p(B)}$. For future reference we observe a fundamental self-similarity of the curve ${\gamma({\bf R})}$: any arc ${\gamma(I)}$ in this curve, with ${I}$ a compact interval, is affinely equivalent to the standard arc ${\gamma([0,1])}$. Theorem 1 (Decoupling theorem) Let ${n \geq 1}$. Subdivide the unit interval ${[0,1]}$ into ${N}$ equal subintervals ${I_i}$ of length ${1/N}$, and for each such ${I_i}$, let ${f_i: {\bf R}^n \rightarrow {\bf R}}$ be the Fourier transform $\displaystyle f_i(x) = \int_{\gamma(I_i)} e(x \cdot \xi)\ d\mu_i(\xi)$ of a finite Borel measure ${\mu_i}$ on the arc ${\gamma(I_i)}$, where ${e(\theta) := e^{2\pi i \theta}}$. Then the ${f_i}$ exhibit decoupling in ${L^{n(n+1)}(w_B)}$ for any ball ${B}$ of radius ${N^n}$. Orthogonality gives the ${n=1}$ case of this theorem. The bi-orthogonality type arguments sketched earlier only give decoupling in ${L^p}$ up to the range ${2 \leq p \leq 2n}$; the point here is that we can now get a much larger value of ${n}$. The ${n=2}$ case of this theorem was previously established by Bourgain and Demeter (who obtained in fact an analogous theorem for any curved hypersurface). The exponent ${n(n+1)}$ (and the radius ${N^n}$) is best possible, as can be seen by the following basic example. If $\displaystyle f_i(x) := \int_{I_i} e(x \cdot \gamma(\xi)) g_i(\xi)\ d\xi$ where ${g_i}$ is a bump function adapted to ${I_i}$, then standard Fourier-analytic computations show that ${f_i}$ will be comparable to ${1/N}$ on a rectangular box of dimensions ${N \times N^2 \times \dots \times N^n}$ (and thus volume ${N^{n(n+1)/2}}$) centred at the origin, and exhibit decay away from this box, with ${\\|f_i\\|_{L^{n(n+1)}(w_B)}}$ comparable to $\displaystyle 1/N \times (N^{n(n+1)/2})^{1/(n(n+1))} = 1/\sqrt{N}.$ On the other hand, ${\sum_{i=1}^N f_i}$ is comparable to ${1}$ on a ball of radius comparable to ${1}$ centred at the origin, so ${\\|\sum_{i=1}^N f_i\\|_{L^{n(n+1)}(w_B)}}$ is ${\gg 1}$, which is just barely consistent with decoupling. This calculation shows that decoupling will fail if ${n(n+1)}$ is replaced by any larger exponent, and also if the radius of the ball ${B}$ is reduced to be significantly smaller than ${N^n}$. This theorem has the following consequence of importance in analytic number theory: Corollary 2 (Vinogradov main conjecture) Let ${s, n, N \geq 1}$ be integers, and let ${\varepsilon > 0}$. Then $\displaystyle \int_{[0,1]^n} \|\sum_{j=1}^N e( j x_1 + j^2 x_2 + \dots + j^n x_n)\|^{2s}\ dx_1 \dots dx_n$ $\displaystyle \ll_{\varepsilon,s,n} N^{s+\varepsilon} + N^{2s - \frac{n(n+1)}{2}+\varepsilon}.$ Proof: By the Hölder inequality (and the trivial bound of ${N}$ for the exponential sum), it suffices to treat the critical case ${s = n(n+1)/2}$, that is to say to show that $\displaystyle \int_{[0,1]^n} \|\sum_{j=1}^N e( j x_1 + j^2 x_2 + \dots + j^n x_n)\|^{n(n+1)}\ dx_1 \dots dx_n \ll_{\varepsilon,n} N^{\frac{n(n+1)}{2}+\varepsilon}.$ We can rescale this as $\displaystyle \int_{[0,N] \times [0,N^2] \times \dots \times [0,N^n]} \|\sum_{j=1}^N e( x \cdot \gamma(j/N) )\|^{n(n+1)}\ dx \ll_{\varepsilon,n} N^{3\frac{n(n+1)}{2}+\varepsilon}.$ As the integrand is periodic along the lattice ${N{\bf Z} \times N^2 {\bf Z} \times \dots \times N^n {\bf Z}}$, this is equivalent to $\displaystyle \int_{[0,N^n]^n} \|\sum_{j=1}^N e( x \cdot \gamma(j/N) )\|^{n(n+1)}\ dx \ll_{\varepsilon,n} N^{\frac{n(n+1)}{2}+n^2+\varepsilon}.$ The left-hand side may be bounded by ${\ll \\| \sum_{j=1}^N f_j \\|_{L^{n(n+1)}(w_B)}^{n(n+1)}}$, where ${B := B(0,N^n)}$ and ${f_j(x) := e(x \cdot \gamma(j/N))}$. Since $\displaystyle \\| f_j \\|_{L^{n(n+1)}(w_B)} \ll (N^{n^2})^{\frac{1}{n(n+1)}},$ the claim now follows from the decoupling theorem and a brief calculation. $\Box$ Using the Plancherel formula, one may equivalently (when ${s}$ is an integer) write the Vinogradov main conjecture in terms of solutions ${j_1,\dots,j_s,k_1,\dots,k_s \in \{1,\dots,N\}}$ to the system of equations $\displaystyle j_1^i + \dots + j_s^i = k_1^i + \dots + k_s^i \forall i=1,\dots,n,$ but we will not use this formulation here. A history of the Vinogradov main conjecture may be found in this survey of Wooley; prior to the Bourgain-Demeter-Guth theorem, the conjecture was solved completely for ${n \leq 3}$, or for ${n > 3}$ and ${s}$ either below ${n(n+1)/2 - n/3 + O(n^{2/3})}$ or above ${n(n-1)}$, with the bulk of recent progress coming from the efficient congruencing technique of Wooley. It has numerous applications to exponential sums, Waring’s problem, and the zeta function; to give just one application, the main conjecture implies the predicted asymptotic for the number of ways to express a large number as the sum of ${23}$ fifth powers (the previous best result required ${28}$ fifth powers). The Bourgain-Demeter-Guth approach to the Vinogradov main conjecture, based on decoupling, is ostensibly very different from the efficient congruencing technique, which relies heavily on the arithmetic structure of the program, but it appears (as I have been told from second-hand sources) that the two methods are actually closely related, with the former being a sort of “Archimedean” version of the latter (with the intervals ${I_i}$ in the decoupling theorem being analogous to congruence classes in the efficient congruencing method); hopefully there will be some future work making this connection more precise. One advantage of the decoupling approach is that it generalises to non-arithmetic settings in which the set ${\{1,\dots,N\}}$ that ${j}$ is drawn from is replaced by some other similarly separated set of real numbers. (A random thought – could this allow the Vinogradov-Korobov bounds on the zeta function to extend to Beurling zeta functions?) Below the fold we sketch the Bourgain-Demeter-Guth argument proving Theorem 1. I thank Jean Bourgain and Andrew Granville for helpful discussions. Let ${\lambda}$ denote the Liouville function. The prime number theorem is equivalent to the estimate $\displaystyle \sum_{n \leq x} \lambda(n) = o(x)$ as ${x \rightarrow \infty}$, that is to say that ${\lambda}$ exhibits cancellation on large intervals such as ${[1,x]}$. This result can be improved to give cancellation on shorter intervals. For instance, using the known zero density estimates for the Riemann zeta function, one can establish that $\displaystyle \int_X^{2X} \|\sum_{x \leq n \leq x+H} \lambda(n)\|\ dx = o( HX ) \ \ \ \ \ (1)$ as ${X \rightarrow \infty}$ if ${X^{1/6+\varepsilon} \leq H \leq X}$ for some fixed ${\varepsilon>0}$; I believe this result is due to Ramachandra (see also Exercise 21 of this previous blog post), and in fact one could obtain a better error term on the right-hand side that for instance gained an arbitrary power of ${\log X}$. On the Riemann hypothesis (or the weaker density hypothesis), it was known that the ${X^{1/6+\varepsilon}}$ could be lowered to ${X^\varepsilon}$. Early this year, there was a major breakthrough by Matomaki and Radziwill, who (among other things) showed that the asymptotic (1) was in fact valid for any ${H = H(X)}$ with ${H \leq X}$ that went to infinity as ${X \rightarrow \infty}$, thus yielding cancellation on extremely short intervals. This has many further applications; for instance, this estimate, or more precisely its extension to other “non-pretentious” bounded multiplicative functions, was a key ingredient in my recent solution of the Erdös discrepancy problem, as well as in obtaining logarithmically averaged cases of Chowla’s conjecture, such as $\displaystyle \sum_{n \leq x} \frac{\lambda(n) \lambda(n+1)}{n} = o(\log x). \ \ \ \ \ (2)$ It is of interest to twist the above estimates by phases such as the linear phase ${n \mapsto e(\alpha n) := e^{2\pi i \alpha n}}$. In 1937, Davenport showed that $\displaystyle \sup_\alpha \|\sum_{n \leq x} \lambda(n) e(\alpha n)\| \ll_A x \log^{-A} x$ which of course improves the prime number theorem. Recently with Matomaki and Radziwill, we obtained a common generalisation of this estimate with (1), showing that $\displaystyle \sup_\alpha \int_X^{2X} \|\sum_{x \leq n \leq x+H} \lambda(n) e(\alpha n)\|\ dx = o(HX) \ \ \ \ \ (3)$ as ${X \rightarrow \infty}$, for any ${H = H(X) \leq X}$ that went to infinity as ${X \rightarrow \infty}$. We were able to use this estimate to obtain an averaged form of Chowla’s conjecture. In that paper, we asked whether one could improve this estimate further by moving the supremum inside the integral, that is to say to establish the bound $\displaystyle \int_X^{2X} \sup_\alpha \|\sum_{x \leq n \leq x+H} \lambda(n) e(\alpha n)\|\ dx = o(HX) \ \ \ \ \ (4)$ as ${X \rightarrow \infty}$, for any ${H = H(X) \leq X}$ that went to infinity as ${X \rightarrow \infty}$. This bound is asserting that ${\lambda}$ is locally Fourier-uniform on most short intervals; it can be written equivalently in terms of the “local Gowers ${U^2}$ norm” as $\displaystyle \int_X^{2X} \sum_{1 \leq a \leq H} \|\sum_{x \leq n \leq x+H} \lambda(n) \lambda(n+a)\|^2\ dx = o( H^3 X )$ from which one can see that this is another averaged form of Chowla’s conjecture (stronger than the one I was able to prove with Matomaki and Radziwill, but a consequence of the unaveraged Chowla conjecture). If one inserted such a bound into the machinery I used to solve the Erdös discrepancy problem, it should lead to further averaged cases of Chowla’s conjecture, such as $\displaystyle \sum_{n \leq x} \frac{\lambda(n) \lambda(n+1) \lambda(n+2)}{n} = o(\log x), \ \ \ \ \ (5)$ though I have not fully checked the details of this implication. It should also have a number of new implications for sign patterns of the Liouville function, though we have not explored these in detail yet. One can write (4) equivalently in the form $\displaystyle \int_X^{2X} \sum_{x \leq n \leq x+H} \lambda(n) e( \alpha(x) n + \beta(x) )\ dx = o(HX) \ \ \ \ \ (6)$ uniformly for all ${x}$-dependent phases ${\alpha(x), \beta(x)}$. In contrast, (3) is equivalent to the subcase of (6) when the linear phase coefficient ${\alpha(x)}$ is independent of ${x}$. This dependency of ${\alpha(x)}$ on ${x}$ seems to necessitate some highly nontrivial additive combinatorial analysis of the function ${x \mapsto \alpha(x)}$ in order to establish (4) when ${H}$ is small. To date, this analysis has proven to be elusive, but I would like to record what one can do with more classical methods like Vaughan’s identity, namely: Proposition 1 The estimate (4) (or equivalently (6)) holds in the range ${X^{2/3+\varepsilon} \leq H \leq X}$ for any fixed ${\varepsilon>0}$. (In fact one can improve the right-hand side by an arbitrary power of ${\log X}$ in this case.) The values of ${H}$ in this range are far too large to yield implications such as new cases of the Chowla conjecture, but it appears that the ${2/3}$ exponent is the limit of “classical” methods (at least as far as I was able to apply them), in the sense that one does not do any combinatorial analysis on the function ${x \mapsto \alpha(x)}$, nor does one use modern equidistribution results on “Type III sums” that require deep estimates on Kloosterman-type sums. The latter may shave a little bit off of the ${2/3}$ exponent, but I don’t see how one would ever hope to go below ${1/2}$ without doing some non-trivial combinatorics on the function ${x \mapsto \alpha(x)}$. UPDATE: I have come across this paper of Zhan which uses mean-value theorems for L-functions to lower the ${2/3}$ exponent to ${5/8}$. Let me now sketch the proof of the proposition, omitting many of the technical details. We first remark that known estimates on sums of the Liouville function (or similar functions such as the von Mangoldt function) in short arithmetic progressions, based on zero-density estimates for Dirichlet ${L}$-functions, can handle the “major arc” case of (4) (or (6)) where ${\alpha}$ is restricted to be of the form ${\alpha = \frac{a}{q} + O( X^{-1/6-\varepsilon} )}$ for ${q = O(\log^{O(1)} X)}$ (the exponent here being of the same numerology as the ${X^{1/6+\varepsilon}}$ exponent in the classical result of Ramachandra, tied to the best zero density estimates currently available); for instance a modification of the arguments in this recent paper of Koukoulopoulos would suffice. Thus we can restrict attention to “minor arc” values of ${\alpha}$ (or ${\alpha(x)}$, using the interpretation of (6)). Next, one breaks up ${\lambda}$ (or the closely related Möbius function) into Dirichlet convolutions using one of the standard identities (e.g. Vaughan’s identity or Heath-Brown’s identity), as discussed for instance in this previous post (which is focused more on the von Mangoldt function, but analogous identities exist for the Liouville and Möbius functions). The exact choice of identity is not terribly important, but the upshot is that ${\lambda(n)}$ can be decomposed into ${\log^{O(1)} X}$ terms, each of which is either of the “Type I” form $\displaystyle \sum_{d \sim D; m \sim M: dm=n} a_d$ for some coefficients ${a_d}$ that are roughly of logarithmic size on the average, and scales ${D, M}$ with ${D \ll X^{2/3}}$ and ${DM \sim X}$, or else of the “Type II” form $\displaystyle \sum_{d \sim D; m \sim M: dm=n} a_d b_m$ for some coefficients ${a_d, b_m}$ that are roughly of logarithmic size on the average, and scales ${D,M}$ with ${X^{1/3} \ll D,M \ll X^{2/3}}$ and ${DM \sim X}$. As discussed in the previous post, the ${2/3}$ exponent is a natural barrier in these identities if one is unwilling to also consider “Type III” type terms which are roughly of the shape of the third divisor function ${\tau_3(n) := \sum_{d_1d_2d_3=1} 1}$. A Type I sum makes a contribution to ${ \sum_{x \leq n \leq x+H} \lambda(n) e( \alpha(x) n + \beta(x) )}$ that can be bounded (via Cauchy-Schwarz) in terms of an expression such as $\displaystyle \sum_{d \sim D} \| \sum_{x/d \leq m \leq x/d+H/d} e(\alpha(x) dm )\|^2.$ The inner sum exhibits a lot of cancellation unless ${\alpha(x) d}$ is within ${O(D/H)}$ of an integer. (Here, “a lot” should be loosely interpreted as “gaining many powers of ${\log X}$ over the trivial bound”.) Since ${H}$ is significantly larger than ${D}$, standard Vinogradov-type manipulations (see e.g. Lemma 13 of these previous notes) show that this bad case occurs for many ${d}$ only when ${\alpha}$ is “major arc”, which is the case we have specifically excluded. This lets us dispose of the Type I contributions. A Type II sum makes a contribution to ${ \sum_{x \leq n \leq x+H} \lambda(n) e( \alpha(x) n + \beta(x) )}$ roughly of the form $\displaystyle \sum_{d \sim D} \| \sum_{x/d \leq m \leq x/d+H/d} b_m e(\alpha(x) dm)\|.$ We can break this up into a number of sums roughly of the form $\displaystyle \sum_{d = d_0 + O( H / M )} \| \sum_{x/d_0 \leq m \leq x/d_0 + H/D} b_m e(\alpha(x) dm)\|$ for ${d_0 \sim D}$; note that the ${d}$ range is non-trivial because ${H}$ is much larger than ${M}$. Applying the usual bilinear sum Cauchy-Schwarz methods (e.g. Theorem 14 of these notes) we conclude that there is a lot of cancellation unless one has ${\alpha(x) = a/q + O( \frac{X \log^{O(1)} X}{H^2} )}$ for some ${q = O(\log^{O(1)} X)}$. But with ${H \geq X^{2/3+\varepsilon}}$, ${X \log^{O(1)} X/H^2}$ is well below the threshold ${X^{-1/6-\varepsilon}}$ for the definition of major arc, so we can exclude this case and obtain the required cancellation. A basic estimate in multiplicative number theory (particularly if one is using the Granville-Soundararajan “pretentious” approach to this subject) is the following inequality of Halasz (formulated here in a quantitative form introduced by Montgomery and Tenenbaum). Theorem 1 (Halasz inequality) Let ${f: {\bf N} \rightarrow {\bf C}}$ be a multiplicative function bounded in magnitude by ${1}$, and suppose that ${x \geq 3}$, ${T \geq 1}$, and ${ M \geq 0}$ are such that $\displaystyle \sum_{p \leq x} \frac{1 - \hbox{Re}(f(p) p^{-it})}{p} \geq M \ \ \ \ \ (1)$ for all real numbers ${t}$ with ${\|t\| \leq T}$. Then $\displaystyle \frac{1}{x} \sum_{n \leq x} f(n) \ll (1+M) e^{-M} + \frac{1}{\sqrt{T}}.$ As a qualitative corollary, we conclude (by standard compactness arguments) that if $\displaystyle \sum_{p} \frac{1 - \hbox{Re}(f(p) p^{-it})}{p} = +\infty$ for all real ${t}$, then $\displaystyle \frac{1}{x} \sum_{n \leq x} f(n) = o(1) \ \ \ \ \ (2)$ as ${x \rightarrow \infty}$. In the more recent work of this paper of Granville and Soundararajan, the sharper bound $\displaystyle \frac{1}{x} \sum_{n \leq x} f(n) \ll (1+M) e^{-M} + \frac{1}{T} + \frac{\log\log x}{\log x}$ is obtained (with a more precise description of the ${(1+M) e^{-M}}$ term). The usual proofs of Halasz’s theorem are somewhat lengthy (though there has been a recent simplification, in forthcoming work of Granville, Harper, and Soundarajan). Below the fold I would like to give a relatively short proof of the following “cheap” version of the inequality, which has slightly weaker quantitative bounds, but still suffices to give qualitative conclusions such as (2). Theorem 2 (Cheap Halasz inequality) Let ${f: {\bf N} \rightarrow {\bf C}}$ be a multiplicative function bounded in magnitude by ${1}$. Let ${T \geq 1}$ and ${M \geq 0}$, and suppose that ${x}$ is sufficiently large depending on ${T,M}$. If (1) holds for all ${\|t\| \leq T}$, then $\displaystyle \frac{1}{x} \sum_{n \leq x} f(n) \ll (1+M) e^{-M/2} + \frac{1}{T}.$ The non-optimal exponent ${1/2}$ can probably be improved a bit by being more careful with the exponents, but I did not try to optimise it here. A similar bound appears in the first paper of Halasz on this topic. The idea of the argument is to split ${f}$ as a Dirichlet convolution ${f = f_1 * f_2 * f_3}$ where ${f_1,f_2,f_3}$ is the portion of ${f}$ coming from “small”, “medium”, and “large” primes respectively (with the dividing line between the three types of primes being given by various powers of ${x}$). Using a Perron-type formula, one can express this convolution in terms of the product of the Dirichlet series of ${f_1,f_2,f_3}$ respectively at various complex numbers ${1+it}$ with ${\|t\| \leq T}$. One can use ${L^2}$ based estimates to control the Dirichlet series of ${f_2,f_3}$, while using the hypothesis (1) one can get ${L^\infty}$ estimates on the Dirichlet series of ${f_1}$. (This is similar to the Fourier-analytic approach to ternary additive problems, such as Vinogradov’s theorem on representing large odd numbers as the sum of three primes.) This idea was inspired by a similar device used in the work of Granville, Harper, and Soundarajan. A variant of this argument also appears in unpublished work of Adam Harper. I thank Andrew Granville for helpful comments which led to significant simplifications of the argument. In the previous set of notes we established the central limit theorem, which we formulate here as follows: Theorem 1 (Central limit theorem) Let ${X_1,X_2,X_3,\dots}$ be iid copies of a real random variable ${X}$ of mean ${\mu}$ and variance ${0 < \sigma^2 < \infty}$, and write ${S_n := X_1 + \dots + X_n}$. Then, for any fixed ${a < b}$, we have $\displaystyle {\bf P}( a \leq \frac{S_n - n \mu}{\sqrt{n} \sigma} \leq b ) \rightarrow \frac{1}{\sqrt{2\pi}} \int_a^b e^{-t^2/2}\ dt \ \ \ \ \ (1)$ as ${n \rightarrow \infty}$. This is however not the end of the matter; there are many variants, refinements, and generalisations of the central limit theorem, and the purpose of this set of notes is to present a small sample of these variants. First of all, the above theorem does not quantify the rate of convergence in (1). We have already addressed this issue to some extent with the Berry-Esséen theorem, which roughly speaking gives a convergence rate of ${O(1/\sqrt{n})}$ uniformly in ${a,b}$ if we assume that ${X}$ has finite third moment. However there are still some quantitative versions of (1) which are not addressed by the Berry-Esséen theorem. For instance one may be interested in bounding the large deviation probabilities $\displaystyle {\bf P}( \|\frac{S_n - n \mu}{\sqrt{n} \sigma}\| \geq \lambda ) \ \ \ \ \ (2)$ in the setting where ${\lambda}$ grows with ${n}$. Chebyshev’s inequality gives an upper bound of ${1/\lambda^2}$ for this quantity, but one can often do much better than this in practice. For instance, the central limit theorem (1) suggests that this probability should be bounded by something like ${O( e^{-\lambda^2/2})}$; however, this theorem only kicks in when ${n}$ is very large compared with ${\lambda}$. For instance, if one uses the Berry-Esséen theorem, one would need ${n}$ as large as ${e^{\lambda^2}}$ or so to reach the desired bound of ${O( e^{-\lambda^2/2})}$, even under the assumption of finite third moment. Basically, the issue is that convergence-in-distribution results, such as the central limit theorem, only really control the typical behaviour of statistics in ${\frac{S_n-n \mu}{\sqrt{n} \sigma}}$; they are much less effective at controlling the very rare outlier events in which the statistic strays far from its typical behaviour. Fortunately, there are large deviation inequalities (or concentration of measure inequalities) that do provide exponential type bounds for quantities such as (2), which are valid for both small and large values of ${n}$. A basic example of this is the Chernoff bound that made an appearance in Exercise 47 of Notes 4; here we give some further basic inequalities of this type, including versions of the Bennett and Hoeffding inequalities. In the other direction, we can also look at the fine scale behaviour of the sums ${S_n}$ by trying to control probabilities such as $\displaystyle {\bf P}( a \leq S_n \leq a+h ) \ \ \ \ \ (3)$ where ${h}$ is now bounded (but ${a}$ can grow with ${n}$). The central limit theorem predicts that this quantity should be roughly ${\frac{h}{\sqrt{2\pi n} \sigma} e^{-(a-n\mu)^2 / 2n \sigma^2}}$, but even if one is able to invoke the Berry-Esséen theorem, one cannot quite see this main term because it is dominated by the error term ${O(1/n^{1/2})}$ in Berry-Esséen. There is good reason for this: if for instance ${X}$ takes integer values, then ${S_n}$ also takes integer values, and ${{\bf P}( a \leq S_n \leq a+h )}$ can vanish when ${h}$ is less than ${1}$ and ${a}$ is slightly larger than an integer. However, this turns out to essentially be the only obstruction; if ${X}$ does not lie in a lattice such as ${{\bf Z}}$, then we can establish a local limit theorem controlling (3), and when ${X}$ does take values in a lattice like ${{\bf Z}}$, there is a discrete local limit theorem that controls probabilities such as ${{\bf P}(S_n = m)}$. Both of these limit theorems will be proven by the Fourier-analytic method used in the previous set of notes. We also discuss other limit theorems in which the limiting distribution is something other than the normal distribution. Perhaps the most common example of these theorems is the Poisson limit theorems, in which one sums a large number of indicator variables (or approximate indicator variables), each of which is rarely non-zero, but which collectively add up to a random variable of medium-sized mean. In this case, it turns out that the limiting distribution should be a Poisson random variable; this again is an easy application of the Fourier method. Finally, we briefly discuss limit theorems for other stable laws than the normal distribution, which are suitable for summing random variables of infinite variance, such as the Cauchy distribution. Finally, we mention a very important class of generalisations to the CLT (and to the variants of the CLT discussed in this post), in which the hypothesis of joint independence between the variables ${X_1,\dots,X_n}$ is relaxed, for instance one could assume only that the ${X_1,\dots,X_n}$ form a martingale. Many (though not all) of the proofs of the CLT extend to these more general settings, and this turns out to be important for many applications in which one does not expect joint independence. However, we will not discuss these generalisations in this course, as they are better suited for subsequent courses in this series when the theory of martingales, conditional expectation, and related tools are developed. Kevin Ford, James Maynard, and I have uploaded to the arXiv our preprint “Chains of large gaps between primes“. This paper was announced in our previous paper with Konyagin and Green, which was concerned with the largest gap $\displaystyle G_1(X) := \max_{p_n, p_{n+1} \leq X} (p_{n+1} - p_n)$ between consecutive primes up to ${X}$, in which we improved the Rankin bound of $\displaystyle G_1(X) \gg \log X \frac{\log_2 X \log_4 X}{(\log_3 X)^2}$ to $\displaystyle G_1(X) \gg \log X \frac{\log_2 X \log_4 X}{\log_3 X}$ for large ${X}$ (where we use the abbreviations ${\log_2 X := \log\log X}$, ${\log_3 X := \log\log\log X}$, and ${\log_4 X := \log\log\log\log X}$). Here, we obtain an analogous result for the quantity $\displaystyle G_k(X) := \max_{p_n, \dots, p_{n+k} \leq X} \min( p_{n+1} - p_n, p_{n+2}-p_{n+1}, \dots, p_{n+k} - p_{n+k-1} )$ which measures how far apart the gaps between chains of ${k}$ consecutive primes can be. Our main result is $\displaystyle G_k(X) \gg \frac{1}{k^2} \log X \frac{\log_2 X \log_4 X}{\log_3 X}$ whenever ${X}$ is sufficiently large depending on ${k}$, with the implied constant here absolute (and effective). The factor of ${1/k^2}$ is inherent to the method, and related to the basic probabilistic fact that if one selects ${k}$ numbers at random from the unit interval ${[0,1]}$, then one expects the minimum gap between adjacent numbers to be about ${1/k^2}$ (i.e. smaller than the mean spacing of ${1/k}$ by an additional factor of ${1/k}$). Our arguments combine those from the previous paper with the matrix method of Maier, who (in our notation) showed that $\displaystyle G_k(X) \gg_k \log X \frac{\log_2 X \log_4 X}{(\log_3 X)^2}$ for an infinite sequence of ${X}$ going to infinity. (Maier needed to restrict to an infinite sequence to avoid Siegel zeroes, but we are able to resolve this issue by the now standard technique of simply eliminating a prime factor of an exceptional conductor from the sieve-theoretic portion of the argument. As a byproduct, this also makes all of the estimates in our paper effective.) As its name suggests, the Maier matrix method is usually presented by imagining a matrix of numbers, and using information about the distribution of primes in the columns of this matrix to deduce information about the primes in at least one of the rows of the matrix. We found it convenient to interpret this method in an equivalent probabilistic form as follows. Suppose one wants to find an interval ${n+1,\dots,n+y}$ which contained a block of at least ${k}$ primes, each separated from each other by at least ${g}$ (ultimately, ${y}$ will be something like ${\log X \frac{\log_2 X \log_4 X}{\log_3 X}}$ and ${g}$ something like ${y/k^2}$). One can do this by the probabilistic method: pick ${n}$ to be a random large natural number ${{\mathbf n}}$ (with the precise distribution to be chosen later), and try to lower bound the probability that the interval ${{\mathbf n}+1,\dots,{\mathbf n}+y}$ contains at least ${k}$ primes, no two of which are within ${g}$ of each other. By carefully choosing the residue class of ${{\mathbf n}}$ with respect to small primes, one can eliminate several of the ${{\mathbf n}+j}$ from consideration of being prime immediately. For instance, if ${{\mathbf n}}$ is chosen to be large and even, then the ${{\mathbf n}+j}$ with ${j}$ even have no chance of being prime and can thus be eliminated; similarly if ${{\mathbf n}}$ is large and odd, then ${{\mathbf n}+j}$ cannot be prime for any odd ${j}$. Using the methods of our previous paper, we can find a residue class ${m \hbox{ mod } P}$ (where ${P}$ is a product of a large number of primes) such that, if one chooses ${{\mathbf n}}$ to be a large random element of ${m \hbox{ mod } P}$ (that is, ${{\mathbf n} = {\mathbf z} P + m}$ for some large random integer ${{\mathbf z}}$), then the set ${{\mathcal T}}$ of shifts ${j \in \{1,\dots,y\}}$ for which ${{\mathbf n}+j}$ still has a chance of being prime has size comparable to something like ${k \log X / \log_2 X}$; furthermore this set ${{\mathcal T}}$ is fairly well distributed in ${\{1,\dots,y\}}$ in the sense that it does not concentrate too strongly in any short subinterval of ${\{1,\dots,y\}}$. The main new difficulty, not present in the previous paper, is to get lower bounds on the size of ${{\mathcal T}}$ in addition to upper bounds, but this turns out to be achievable by a suitable modification of the arguments. Using a version of the prime number theorem in arithmetic progressions due to Gallagher, one can show that for each remaining shift ${j \in {\mathcal T}}$, ${{\mathbf n}+j}$ is going to be prime with probability comparable to ${\log_2 X / \log X}$, so one expects about ${k}$ primes in the set ${\{{\mathbf n} + j: j \in {\mathcal T}\}}$. An upper bound sieve (e.g. the Selberg sieve) also shows that for any distinct ${j,j' \in {\mathcal T}}$, the probability that ${{\mathbf n}+j}$ and ${{\mathbf n}+j'}$ are both prime is ${O( (\log_2 X / \log X)^2 )}$. Using this and some routine second moment calculations, one can then show that with large probability, the set ${\{{\mathbf n} + j: j \in {\mathcal T}\}}$ will indeed contain about ${k}$ primes, no two of which are closer than ${g}$ to each other; with no other numbers in this interval being prime, this gives a lower bound on ${G_k(X)}$. Klaus Roth, who made fundamental contributions to analytic number theory, died this Tuesday, aged 90. I never met or communicated with Roth personally, but was certainly influenced by his work; he wrote relatively few papers, but they tended to have outsized impact. For instance, he was one of the key people (together with Bombieri) to work on simplifying and generalising the large sieve, taking it from the technically formidable original formulation of Linnik and Rényi to the clean and general almost orthogonality principle that we have today (discussed for instance in these lecture notes of mine). The paper of Roth that had the most impact on my own personal work was his three-page paper proving what is now known as Roth’s theorem on arithmetic progressions: Theorem 1 (Roth’s theorem on arithmetic progressions) Let ${A}$ be a set of natural numbers of positive upper density (thus ${\limsup_{N \rightarrow\infty} \|A \cap \{1,\dots,N\}\|/N > 0}$). Then ${A}$ contains infinitely many arithmetic progressions ${a,a+r,a+2r}$ of length three (with ${r}$ non-zero of course). At the heart of Roth’s elegant argument was the following (surprising at the time) dichotomy: if ${A}$ had some moderately large density within some arithmetic progression ${P}$, either one could use Fourier-analytic methods to detect the presence of an arithmetic progression of length three inside ${A \cap P}$, or else one could locate a long subprogression ${P'}$ of ${P}$ on which ${A}$ had increased density. Iterating this dichotomy by an argument now known as the density increment argument, one eventually obtains Roth’s theorem, no matter which side of the dichotomy actually holds. This argument (and the many descendants of it), based on various “dichotomies between structure and randomness”, became essential in many other results of this type, most famously perhaps in Szemerédi’s proof of his celebrated theorem on arithmetic progressions that generalised Roth’s theorem to progressions of arbitrary length. More recently, my recent work on the Chowla and Elliott conjectures that was a crucial component of the solution of the Erdös discrepancy problem, relies on an entropy decrement argument which was directly inspired by the density increment argument of Roth. The Erdös discrepancy problem also is connected with another well known theorem of Roth: Theorem 2 (Roth’s discrepancy theorem for arithmetic progressions) Let ${f(1),\dots,f(n)}$ be a sequence in ${\{-1,+1\}}$. Then there exists an arithmetic progression ${a+r, a+2r, \dots, a+kr}$ in ${\{1,\dots,n\}}$ with ${r}$ positive such that $\displaystyle \|\sum_{j=1}^k f(a+jr)\| \geq c n^{1/4}$ for an absolute constant ${c>0}$. In fact, Roth proved a stronger estimate regarding mean square discrepancy, which I am not writing down here; as with the Roth theorem in arithmetic progressions, his proof was short and Fourier-analytic in nature (although non-Fourier-analytic proofs have since been found, for instance the semidefinite programming proof of Lovasz). The exponent ${1/4}$ is known to be sharp (a result of Matousek and Spencer). As a particular corollary of the above theorem, for an infinite sequence ${f(1), f(2), \dots}$ of signs, the sums ${\|\sum_{j=1}^k f(a+jr)\|}$ are unbounded in ${a,r,k}$. The Erdös discrepancy problem asks whether the same statement holds when ${a}$ is restricted to be zero. (Roth also established discrepancy theorems for other sets, such as rectangles, which will not be discussed here.) Finally, one has to mention Roth’s most famous result, cited for instance in his Fields medal citation: Theorem 3 (Roth’s theorem on Diophantine approximation) Let ${\alpha}$ be an irrational algebraic number. Then for any ${\varepsilon > 0}$ there is a quantity ${c_{\alpha,\varepsilon}}$ such that $\displaystyle \|\alpha - \frac{a}{q}\| > \frac{c_{\alpha,\varepsilon}}{q^{2+\varepsilon}}.$ From the Dirichlet approximation theorem (or from the theory of continued fractions) we know that the exponent ${2+\varepsilon}$ in the denominator cannot be reduced to ${2}$ or below. A classical and easy theorem of Liouville gives the claim with the exponent ${2+\varepsilon}$ replaced by the degree of the algebraic number ${\alpha}$; work of Thue and Siegel reduced this exponent, but Roth was the one who obtained the near-optimal result. An important point is that the constant ${c_{\alpha,\varepsilon}}$ is ineffective – it is a major open problem in Diophantine approximation to produce any bound significantly stronger than Liouville’s theorem with effective constants. This is because the proof of Roth’s theorem does not exclude any single rational ${a/q}$ from being close to ${\alpha}$, but instead very ingeniously shows that one cannot have two different rationals ${a/q}$, ${a'/q'}$ that are unusually close to ${\alpha}$, even when the denominators ${q,q'}$ are very different in size. (I refer to this sort of argument as a “dueling conspiracies” argument; they are strangely prevalent throughout analytic number theory.) Chantal David, Andrew Granville, Emmanuel Kowalski, Phillipe Michel, Kannan Soundararajan, and I are running a program at MSRI in the Spring of 2017 (more precisely, from Jan 17, 2017 to May 26, 2017) in the area of analytic number theory, with the intention to bringing together many of the leading experts in all aspects of the subject and to present recent work on the many active areas of the subject (e.g. the distribution of the prime numbers, refinements of the circle method, a deeper understanding of the asymptotics of bounded multiplicative functions (and applications to Erdos discrepancy type problems!) and of the “pretentious” approach to analytic number theory, more “analysis-friendly” formulations of the theorems of Deligne and others involving trace functions over fields, and new subconvexity theorems for automorphic forms, to name a few). Like any other semester MSRI program, there will be a number of workshops, seminars, and similar activities taking place while the members are in residence. I’m personally looking forward to the program, which should be occurring in the midst of a particularly productive time for the subject. Needless to say, I (and the rest of the organising committee) plan to be present for most of the program. Applications for Postdoctoral Fellowships and Research Memberships for this program (and for other MSRI programs in this time period, namely the companion program in Harmonic Analysis and the Fall program in Geometric Group Theory, as well as the complementary program in all other areas of mathematics) remain open until Dec 1. Applications are open to everyone, but require supporting documentation, such as a CV, statement of purpose, and letters of recommendation from other mathematicians; see the application page for more details. Let ${X_1,X_2,\dots}$ be iid copies of an absolutely integrable real scalar random variable ${X}$, and form the partial sums ${S_n := X_1 + \dots + X_n}$. As we saw in the last set of notes, the law of large numbers ensures that the empirical averages ${S_n/n}$ converge (both in probability and almost surely) to a deterministic limit, namely the mean ${\mu= {\bf E} X}$ of the reference variable ${X}$. Furthermore, under some additional moment hypotheses on the underlying variable ${X}$, we can obtain square root cancellation for the fluctuation ${\frac{S_n}{n} - \mu}$ of the empirical average from the mean. To simplify the calculations, let us first restrict to the case ${\mu=0, \sigma^2=1}$ of mean zero and variance one, thus $\displaystyle {\bf E} X = 0$ and $\displaystyle {\bf Var}(X) = {\bf E} X^2 = 1.$ Then, as computed in previous notes, the normalised fluctuation ${S_n/\sqrt{n}}$ also has mean zero and variance one: $\displaystyle {\bf E} \frac{S_n}{\sqrt{n}} = 0$ $\displaystyle {\bf Var}(\frac{S_n}{\sqrt{n}}) = {\bf E} (\frac{S_n}{\sqrt{n}})^2 = 1.$ This and Chebyshev’s inequality already indicates that the “typical” size of ${S_n}$ is ${O(\sqrt{n})}$, thus for instance ${\frac{S_n}{\sqrt{n} \omega(n)}}$ goes to zero in probability for any ${\omega(n)}$ that goes to infinity as ${n \rightarrow \infty}$. If we also have a finite fourth moment ${{\bf E} \|X\|^4 < \infty}$, then the calculations of the previous notes also give a fourth moment estimate $\displaystyle {\bf E} (\frac{S_n}{\sqrt{n}})^4 = 3 + O( \frac{{\bf E} \|X\|^4}{n} ).$ From this and the Paley-Zygmund inequality (Exercise 42 of Notes 1) we also get some lower bound for ${\frac{S_n}{\sqrt{n}}}$ of the form $\displaystyle {\bf P}( \|\frac{S_n}{\sqrt{n}}\| \geq \varepsilon ) \geq \varepsilon$ for some absolute constant ${\varepsilon>0}$ and for ${n}$ sufficiently large; this indicates in particular that ${\frac{S_n \omega(n)}{\sqrt{n}}}$ does not converge in any reasonable sense to something finite for any ${\omega(n)}$ that goes to infinity. The question remains as to what happens to the ratio ${S_n/\sqrt{n}}$ itself, without multiplying or dividing by any factor ${\omega(n)}$. A first guess would be that these ratios converge in probability or almost surely, but this is unfortunately not the case: Proposition 1 Let ${X_1,X_2,\dots}$ be iid copies of an absolutely integrable real scalar random variable ${X}$ with mean zero, variance one, and finite fourth moment, and write ${S_n := X_1 + \dots + X_n}$. Then the random variables ${S_n/\sqrt{n}}$ do not converge in probability or almost surely to any limit, and neither does any subsequence of these random variables. Proof: Suppose for contradiction that some sequence ${S_{n_j}/\sqrt{n_j}}$ converged in probability or almost surely to a limit ${Y}$. By passing to a further subsequence we may assume that the convergence is in the almost sure sense. Since all of the ${S_{n_j}/\sqrt{n_j}}$ have mean zero, variance one, and bounded fourth moment, Theorem 24 of Notes 1 implies that the limit ${Y}$ also has mean zero and variance one. On the other hand, ${Y}$ is a tail random variable and is thus almost surely constant by the Kolmogorov zero-one law from Notes 3. Since constants have variance zero, we obtain the required contradiction. $\Box$ Nevertheless there is an important limit for the ratio ${S_n/\sqrt{n}}$, which requires one to replace the notions of convergence in probability or almost sure convergence by the weaker concept of convergence in distribution. Definition 2 (Vague convergence and convergence in distribution) Let ${R}$ be a locally compact Hausdorff topological space with the Borel ${\sigma}$-algebra. A sequence of finite measures ${\mu_n}$ on ${R}$ is said to converge vaguely to another finite measure ${\mu}$ if one has $\displaystyle \int_R G(x)\ d\mu_n(x) \rightarrow \int_R G(x)\ d\mu(x)$ as ${n \rightarrow \infty}$ for all continuous compactly supported functions ${G: R \rightarrow {\bf R}}$. (Vague convergence is also known as weak convergence, although strictly speaking the terminology weak-* convergence would be more accurate.) A sequence of random variables ${X_n}$ taking values in ${R}$ is said to converge in distribution (or converge weakly or converge in law) to another random variable ${X}$ if the distributions ${\mu_{X_n}}$ converge vaguely to the distribution ${\mu_X}$, or equivalently if $\displaystyle {\bf E}G(X_n) \rightarrow {\bf E} G(X)$ as ${n \rightarrow \infty}$ for all continuous compactly supported functions ${G: R \rightarrow {\bf R}}$. One could in principle try to extend this definition beyond the locally compact Hausdorff setting, but certain pathologies can occur when doing so (e.g. failure of the Riesz representation theorem), and we will never need to consider vague convergence in spaces that are not locally compact Hausdorff, so we restrict to this setting for simplicity. Note that the notion of convergence in distribution depends only on the distribution of the random variables involved. One consequence of this is that convergence in distribution does not produce unique limits: if ${X_n}$ converges in distribution to ${X}$, and ${Y}$ has the same distribution as ${X}$, then ${X_n}$ also converges in distribution to ${Y}$. However, limits are unique up to equivalence in distribution (this is a consequence of the Riesz representation theorem, discussed for instance in this blog post). As a consequence of the insensitivity of convergence in distribution to equivalence in distribution, we may also legitimately talk about convergence of distribution of a sequence of random variables ${X_n}$ to another random variable ${X}$ even when all the random variables ${X_1,X_2,\dots}$ and ${X}$ involved are being modeled by different probability spaces (e.g. each ${X_n}$ is modeled by ${\Omega_n}$, and ${X}$ is modeled by ${\Omega}$, with no coupling presumed between these spaces). This is in contrast to the stronger notions of convergence in probability or almost sure convergence, which require all the random variables to be modeled by a common probability space. Also, by an abuse of notation, we can say that a sequence ${X_n}$ of random variables converges in distribution to a probability measure ${\mu}$, when ${\mu_{X_n}}$ converges vaguely to ${\mu}$. Thus we can talk about a sequence of random variables converging in distribution to a uniform distribution, a gaussian distribution, etc.. From the dominated convergence theorem (available for both convergence in probability and almost sure convergence) we see that convergence in probability or almost sure convergence implies convergence in distribution. The converse is not true, due to the insensitivity of convergence in distribution to equivalence in distribution; for instance, if ${X_1,X_2,\dots}$ are iid copies of a non-deterministic scalar random variable ${X}$, then the ${X_n}$ trivially converge in distribution to ${X}$, but will not converge in probability or almost surely (as one can see from the zero-one law). However, there are some partial converses that relate convergence in distribution to convergence in probability; see Exercise 10 below. Remark 3 The notion of convergence in distribution is somewhat similar to the notion of convergence in the sense of distributions that arises in distribution theory (discussed for instance in this previous blog post), however strictly speaking the two notions of convergence are distinct and should not be confused with each other, despite the very similar names. The notion of convergence in distribution simplifies in the case of real scalar random variables: Proposition 4 Let ${X_1,X_2,\dots}$ be a sequence of scalar random variables, and let ${X}$ be another scalar random variable. Then the following are equivalent: • (i) ${X_n}$ converges in distribution to ${X}$. • (ii) ${F_{X_n}(t)}$ converges to ${F_X(t)}$ for each continuity point ${t}$ of ${F_X}$ (i.e. for all real numbers ${t \in {\bf R}}$ at which ${F_X}$ is continuous). Here ${F_X(t) := {\bf P}(X \leq t)}$ is the cumulative distribution function of ${X}$. Proof: First suppose that ${X_n}$ converges in distribution to ${X}$, and ${F_X}$ is continuous at ${t}$. For any ${\varepsilon > 0}$, one can find a ${\delta}$ such that $\displaystyle F_X(t) - \varepsilon \leq F_X(t') \leq F_X(t) + \varepsilon$ for every ${t' \in [t-\delta,t+\delta]}$. One can also find an ${N}$ larger than ${\|t\|+\delta}$ such that ${F_X(-N) \leq \varepsilon}$ and ${F_X(N) \geq 1-\varepsilon}$. Thus $\displaystyle {\bf P} (\|X\| \geq N ) = O(\varepsilon)$ and $\displaystyle {\bf P} (\|X - t\| \leq \delta ) = O(\varepsilon).$ Let ${G: {\bf R} \rightarrow [0,1]}$ be a continuous function supported on ${[-2N, t]}$ that equals ${1}$ on ${[-N, t-\delta]}$. Then by the above discussion we have $\displaystyle {\bf E} G(X) = F_X(t) + O(\varepsilon)$ and hence $\displaystyle {\bf E} G(X_n) = F_X(t) + O(\varepsilon)$ for large enough ${n}$. In particular $\displaystyle {\bf P}( X_n \leq t ) \geq F_X(t) - O(\varepsilon).$ A similar argument, replacing ${G}$ with a continuous function supported on ${[t,2N]}$ that equals ${1}$ on ${[t+\delta,N]}$ gives $\displaystyle {\bf P}( X_n > t ) \geq 1 - F_X(t) - O(\varepsilon)$ for ${n}$ large enough. Putting the two estimates together gives $\displaystyle F_{X_n}(t) = F_X(t) + O(\varepsilon)$ for ${n}$ large enough; sending ${\varepsilon \rightarrow 0}$, we obtain the claim. Conversely, suppose that ${F_{X_n}(t)}$ converges to ${F_X(t)}$ at every continuity point ${t}$ of ${F_X}$. Let ${G: {\bf R} \rightarrow {\bf R}}$ be a continuous compactly supported function, then it is uniformly continuous. As ${F_X}$ is monotone increasing, it can only have countably many points of discontinuity. From these two facts one can find, for any ${\varepsilon>0}$, a simple function ${G_\varepsilon(t) = \sum_{i=1}^n c_i 1_{(t_i,t_{i+1}]}}$ for some ${t_1 < \dots < t_n}$ that are points of continuity of ${F_X}$, and real numbers ${c_i}$, such that ${\|G(t) - G_\varepsilon(t)\| \leq \varepsilon}$ for all ${t}$. Thus $\displaystyle {\bf P} G(X_n) = {\bf P} G_\varepsilon(X_n) + O(\varepsilon)$ $\displaystyle = \sum_{i=1}^n c_i(F_{X_n}(t_{i+1}) - F_{X_n}(t)) + O(\varepsilon).$ Similarly for ${X_n}$ replaced by ${X}$. Subtracting and taking limit superior, we conclude that $\displaystyle \limsup_{n \rightarrow \infty} \|{\bf P} G(X_n) - {\bf P} G(X)\| = O(\varepsilon),$ and on sending ${\varepsilon \rightarrow 0}$, we obtain that ${X_n}$ converges in distribution to ${X}$ as claimed. $\Box$ The restriction to continuity points of ${t}$ is necessary. Consider for instance the deterministic random variables ${X_n = 1/n}$, then ${X_n}$ converges almost surely (and hence in distribution) to ${0}$, but ${F_{X_n}(0) = 0}$ does not converge to ${F_X(0)=1}$. Example 5 For any natural number ${n}$, let ${X_n}$ be a discrete random variable drawn uniformly from the finite set ${\{0/n, 1/n, \dots, (n-1)/n\}}$, and let ${X}$ be the continuous random variable drawn uniformly from ${[0,1]}$. Then ${X_n}$ converges in distribution to ${X}$. Thus we see that a continuous random variable can emerge as the limit of discrete random variables. Example 6 For any natural number ${n}$, let ${X_n}$ be a continuous random variable drawn uniformly from ${[0,1/n]}$, then ${X_n}$ converges in distribution to the deterministic real number ${0}$. Thus we see that discrete (or even deterministic) random variables can emerge as the limit of continuous random variables. Exercise 7 (Portmanteau theorem) Show that the properties (i) and (ii) in Proposition 4 are also equivalent to the following three statements: • (iii) One has ${\limsup_{n \rightarrow \infty} {\bf P}( X_n \in K ) \leq {\bf P}(X \in K)}$ for all closed sets ${K \subset {\bf R}}$. • (iv) One has ${\liminf_{n \rightarrow \infty} {\bf P}( X_n \in U ) \geq {\bf P}(X \in U)}$ for all open sets ${U \subset {\bf R}}$. • (v) For any Borel set ${E \subset {\bf R}}$ whose topological boundary ${\partial E}$ is such that ${{\bf P}(X \in \partial E) = 0}$, one has ${\lim_{n \rightarrow \infty} {\bf P}(X_n \in E) = {\bf P}(X \in E)}$. (Note: to prove this theorem, you may wish to invoke Urysohn’s lemma. To deduce (iii) from (i), you may wish to start with the case of compact ${K}$.) We can now state the famous central limit theorem: Theorem 8 (Central limit theorem) Let ${X_1,X_2,\dots}$ be iid copies of a scalar random variable ${X}$ of finite mean ${\mu := {\bf E} X}$ and finite non-zero variance ${\sigma^2 := {\bf Var}(X)}$. Let ${S_n := X_1 + \dots + X_n}$. Then the random variables ${\frac{\sqrt{n}}{\sigma} (\frac{S_n}{n} - \mu)}$ converges in distribution to a random variable with the standard normal distribution ${N(0,1)}$ (that is to say, a random variable with probability density function ${x \mapsto \frac{1}{\sqrt{2\pi}} e^{-x^2/2}}$). Thus, by abuse of notation $\displaystyle \frac{\sqrt{n}}{\sigma} (\frac{S_n}{n} - \mu) \rightarrow N(0,1).$ In the normalised case ${\mu=0, \sigma^2=1}$ when ${X}$ has mean zero and unit variance, this simplifies to $\displaystyle \frac{S_n}{\sqrt{n}} \rightarrow N(0,1).$ Using Proposition 4 (and the fact that the cumulative distribution function associated to ${N(0,1)}$ is continuous, the central limit theorem is equivalent to asserting that $\displaystyle {\bf P}( \frac{\sqrt{n}}{\sigma} (\frac{S_n}{n} - \mu) \leq t ) \rightarrow \frac{1}{\sqrt{2\pi}} \int_{-\infty}^t e^{-x^2/2}\ dx$ as ${n \rightarrow \infty}$ for any ${t \in {\bf R}}$, or equivalently that $\displaystyle {\bf P}( a \leq \frac{\sqrt{n}}{\sigma} (\frac{S_n}{n} - \mu) \leq b ) \rightarrow \frac{1}{\sqrt{2\pi}} \int_{a}^b e^{-x^2/2}\ dx.$ Informally, one can think of the central limit theorem as asserting that ${S_n}$ approximately behaves like it has distribution ${N( n \mu, n \sigma^2 )}$ for large ${n}$, where ${N(\mu,\sigma^2)}$ is the normal distribution with mean ${\mu}$ and variance ${\sigma^2}$, that is to say the distribution with probability density function ${x \mapsto \frac{1}{\sqrt{2\pi} \sigma} e^{-(x-\mu)^2/2\sigma^2}}$. The integrals ${\frac{1}{\sqrt{2\pi}} \int_{-\infty}^t e^{-x^2/2}\ dx}$ can be written in terms of the error function ${\hbox{erf}}$ as ${\frac{1}{2} + \frac{1}{2} \hbox{erf}(t/\sqrt{2})}$. The central limit theorem is a basic example of the universality phenomenon in probability – many statistics involving a large system of many independent (or weakly dependent) variables (such as the normalised sums ${\frac{\sqrt{n}}{\sigma}(\frac{S_n}{n}-\mu)}$) end up having a universal asymptotic limit (in this case, the normal distribution), regardless of the precise makeup of the underlying random variable ${X}$ that comprised that system. Indeed, the universality of the normal distribution is such that it arises in many other contexts than the fluctuation of iid random variables; the central limit theorem is merely the first place in probability theory where it makes a prominent appearance. We will give several proofs of the central limit theorem in these notes; each of these proofs has their advantages and disadvantages, and can each extend to prove many further results beyond the central limit theorem. We first give Lindeberg’s proof of the central limit theorem, based on exchanging (or swapping) each component ${X_1,\dots,X_n}$ of the sum ${S_n}$ in turn. This proof gives an accessible explanation as to why there should be a universal limit for the central limit theorem; one then computes directly with gaussians to verify that it is the normal distribution which is the universal limit. Our second proof is the most popular one taught in probability texts, namely the Fourier-analytic proof based around the concept of the characteristic function ${t \mapsto {\bf E} e^{itX}}$ of a real random variable ${X}$. Thanks to the powerful identities and other results of Fourier analysis, this gives a quite short and direct proof of the central limit theorem, although the arguments may seem rather magical to readers who are not already familiar with Fourier methods. Finally, we give a proof based on the moment method, in the spirit of the arguments in the previous notes; this argument is more combinatorial, but is straightforward and is particularly robust, in particular being well equipped to handle some dependencies between components; we will illustrate this by proving the Erdos-Kac law in number theory by this method. Some further discussion of the central limit theorem (including some further proofs, such as one based on Stein’s method) can be found in this blog post. Some further variants of the central limit theorem, such as local limit theorems, stable laws, and large deviation inequalities, will be discussed in the next (and final) set of notes. The following exercise illustrates the power of the central limit theorem, by establishing combinatorial estimates which would otherwise require the use of Stirling’s formula to establish. Exercise 9 (De Moivre-Laplace theorem) Let ${X}$ be a Bernoulli random variable, taking values in ${\{0,1\}}$ with ${{\bf P}(X=0)={\bf P}(X=1)=1/2}$, thus ${X}$ has mean ${1/2}$ and variance ${1/4}$. Let ${X_1,X_2,\dots}$ be iid copies of ${X}$, and write ${S_n := X_1+\dots+X_n}$. • (i) Show that ${S_n}$ takes values in ${\{0,\dots,n\}}$ with ${{\bf P}(S_n=i) = \frac{1}{2^n} \binom{n}{i}}$. (This is an example of a binomial distribution.) • (ii) Assume Stirling’s formula $\displaystyle n! = (1+o(1)) \sqrt{2\pi n} n^n e^{-n} \ \ \ \ \ (1)$ where ${o(1)}$ is a function of ${n}$ that goes to zero as ${n \rightarrow \infty}$. (A proof of this formula may be found in this previous blog post.) Using this formula, and without using the central limit theorem, show that $\displaystyle {\bf P}( a \leq 2\sqrt{n} (\frac{S_n}{n} - \frac{1}{2}) \leq b ) \rightarrow \frac{1}{\sqrt{2\pi}} \int_{a}^b e^{-x^2/2}\ dx$ as ${n \rightarrow \infty}$ for any fixed real numbers ${a. The above special case of the central limit theorem was first established by de Moivre and Laplace. We close this section with some basic facts about convergence of distribution that will be useful in the sequel. Exercise 10 Let ${X_1,X_2,\dots}$, ${Y_1,Y_2,\dots}$ be sequences of real random variables, and let ${X,Y}$ be further real random variables. • (i) If ${X}$ is deterministic, show that ${X_n}$ converges in distribution to ${X}$ if and only if ${X_n}$ converges in probability to ${X}$. • (ii) Suppose that ${X_n}$ is independent of ${Y_n}$ for each ${n}$, and ${X}$ independent of ${Y}$. Show that ${X_n+iY_n}$ converges in distribution to ${X+iY}$ if and only if ${X_n}$ converges in distribution to ${X}$ and ${Y_n}$ converges in distribution to ${Y}$. (The shortest way to prove this is by invoking the Stone-Weierstrass theorem, but one can also proceed by proving some version of Proposition 4.) What happens if the independence hypothesis is dropped? • (iii) If ${X_n}$ converges in distribution to ${X}$, show that for every ${\varepsilon>0}$ there exists ${K>0}$ such that ${{\bf P}( \|X_n\| \geq K ) < \varepsilon}$ for all sufficiently large ${n}$. (That is to say, ${X_n}$ is a tight sequence of random variables.) • (iv) Show that ${X_n}$ converges in distribution to ${X}$ if and only if, after extending the probability space model if necessary, one can find copies ${Z_1,Z_2,\dots}$ and ${Z}$ of ${X_1,X_2,\dots}$ and ${X}$ respectively such that ${Z_n}$ converges almost surely to ${Z}$. (Hint: use the Skorohod representation, Exercise 29 of Notes 0.) • (v) If ${X_1,X_2,\dots}$ converges in distribution to ${X}$, and ${F: {\bf R} \rightarrow {\bf R}}$ is continuous, show that ${F(X_1),F(X_2),\dots}$ converges in distribution to ${F(X)}$. Generalise this claim to the case when ${X}$ takes values in an arbitrary locally compact Hausdorff space. • (vi) (Slutsky’s theorem) If ${X_n}$ converges in distribution to ${X}$, and ${Y_n}$ converges in probability to a deterministic limit ${Y}$, show that ${X_n+Y_n}$ converges in distribution to ${X+Y}$, and ${X_n Y_n}$ converges in distribution to ${XY}$. (Hint: either use (iv), or else use (iii) to control some error terms.) This statement combines particularly well with (i). What happens if ${Y}$ is not assumed to be deterministic? • (vii) (Fatou lemma) If ${G: {\bf R} \rightarrow [0,+\infty)}$ is continuous, and ${X_n}$ converges in distribution to ${X}$, show that ${\liminf_{n \rightarrow \infty} {\bf E} G(X_n) \geq {\bf E} G(X)}$. • (viii) (Bounded convergence) If ${G: {\bf R} \rightarrow {\bf R}}$ is continuous and bounded, and ${X_n}$ converges in distribution to ${X}$, show that ${\lim_{n \rightarrow \infty} {\bf E} G(X_n) = {\bf E} G(X)}$. • (ix) (Dominated convergence) If ${X_n}$ converges in distribution to ${X}$, and there is an absolutely integrable ${Y}$ such that ${\|X_n\| \leq Y}$ almost surely for all ${n}$, show that ${\lim_{n \rightarrow \infty} {\bf E} X_n = {\bf E} X}$. For future reference we also mention (but will not prove) Prokhorov’s theorem that gives a partial converse to part (iii) of the above exercise: Theorem 11 (Prokhorov’s theorem) Let ${X_1,X_2,\dots}$ be a sequence of real random variables which is tight (that is, for every ${\varepsilon>0}$ there exists ${K>0}$ such that ${{\bf P}(\|X_n\| \geq K) < \varepsilon}$ for all sufficiently large ${n}$). Then there exists a subsequence ${X_{n_j}}$ which converges in distribution to some random variable ${X}$ (which may possibly be modeled by a different probability space model than the ${X_1,X_2,\dots}$.) The proof of this theorem relies on the Riesz representation theorem, and is beyond the scope of this course; but see for instance Exercise 29 of this previous blog post. (See also the closely related Helly selection theorem, covered in Exercise 30 of the same post.)	2016-07-28 18:16:18	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 748, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.946721613407135, "perplexity": 185.66886913653488}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-30/segments/1469257828313.74/warc/CC-MAIN-20160723071028-00273-ip-10-185-27-174.ec2.internal.warc.gz"}
https://math.stackexchange.com/questions/580006/how-can-i-prove-that-these-integrals-do-not-converge	# How can I prove that these integrals do not converge? • the first : When $x \to \pi$, we have $\sin x = \sin (\pi -x) \sim \pi - x$ and $\int_1^{\pi} \frac{\pi}{\pi -x}$ diverges.	2019-04-21 06:44:47	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9531650543212891, "perplexity": 143.48142296495462}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-18/segments/1555578530253.25/warc/CC-MAIN-20190421060341-20190421082341-00177.warc.gz"}
https://www.codingame.com/playgrounds/29924/computing-with-data/parseq---part-i	# Computing with Data elgeish 106.3K views ## ParSeq - Part I Before delving into the details of those features, let's take a look at a familiar example, rewritten using ParSeq this time:	2019-06-20 19:46:03	{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8955725431442261, "perplexity": 4717.350680385726}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-26/segments/1560627999273.24/warc/CC-MAIN-20190620190041-20190620212041-00113.warc.gz"}
http://math.stackexchange.com/questions/305989/sum-of-two-random-variables-is-random-variable	# Sum of two random variables is random variable How to show: If $X$ and $Y$ are random variables on a probability space $(\Omega, F, \mathbb P)$, then so is $X+Y$. The definition of a random variable is a function $X: \Omega \to \mathbb R$ with the property that $\{\omega\in\Omega: X(\omega)\leq x\}\in F$ for each $x\in\mathbb R$. and further more, how to approach $X+Y$ and $\min\{X,Y\}$? - This was asked (and answered) pretty recently on the site. – Did Feb 17 '13 at 7:34 There are a number of ways to do it. A standard trick for proving things like this is by noticing that $\{X + Y < x\} = \displaystyle \bigcup_{r \in \mathbb{Q}} \{X < r\} \cap \{Y < x - r\}$, and that showing that this is in $F$ is enough to show that $X + Y$ is measurable. Then use the properties of $X$, $Y$, and $\sigma-$algebras to deduce that this set is measurable. Obvioulsy whenever $X(\omega) < r$ and $Y(\omega) < x-r$, $X(\omega)+Y(\omega) < x$. On the other hand, if $X(\omega)+Y(\omega) = z < x$, take some rational $r$ with $X(\omega) < r < X(\omega) + x - z$, and you have $Y(\omega) = z - X(\omega) < x - r$. – Robert Israel Feb 17 '13 at 6:28 In fact, a random variable is a measurable function from $\Omega$ to $\mathbb{R}$. $$\{X+Y>x\}=\{X>x-Y\}=\bigcup_{q\in \mathbb{Q}}\{X>q>x-Y\}=\bigcup_{q\in \mathbb{Q}}(\{X>q\ \}\bigcap\{Y>x-q\})=\bigcup_{q\in \mathbb{Q}}(\{X\le q\ \}^c\bigcap\{Y\le x-q\}^c)$$ Since $\{X\le q\ \}^c\in \mathcal{F}\ ,\ \{Y\le x-q\}^c\in \mathcal{F}$, $\mathbb{Q}$ is countable and $\mathcal{F}$ is a $\sigma-field$, we obtain $\{X+Y>x\}=\{X+Y\le x\}^c\in \mathcal{F}$. Then $$\{\omega:X(\omega)+Y(\omega)\le x\}\in \mathcal{F}$$ $$\{\omega:min\{X(\omega),Y(\omega)\}\le x\}=\{\omega:X(\omega)\le x\}\bigcup\{\omega:Y(\omega)\le x\}\in \mathcal{F}$$ By definition, $X+Y,and\ min\{X,Y\}$ are both random variables.	2016-07-26 16:05:47	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9636544585227966, "perplexity": 78.05777808886829}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-30/segments/1469257824995.51/warc/CC-MAIN-20160723071024-00008-ip-10-185-27-174.ec2.internal.warc.gz"}
http://www2.macaulay2.com/Macaulay2/doc/Macaulay2-1.19/share/doc/Macaulay2/Macaulay2Doc/html/_printing__Precision.html	# printingPrecision -- current precision for printing numbers ## Synopsis • Usage: printingPrecision = n • Inputs: • Consequences: • Henceforth, inexact numbers are printed with at most n digits of precision. Meaningless digits will not be displayed. The special case where n=0 is interpreted as meaning n=infinity, and this case is used when a number appears alone on an output line to display all the meaningful digits. ## Description i1 : 1/3p100 o1 = .333333333333333333333333333333 o1 : RR (of precision 100) i2 : {1/3p100} o2 = {.333333} o2 : List i3 : printingPrecision o3 = 6 i4 : printingPrecision = 16 o4 = 16 i5 : {1/3p100} o5 = {.3333333333333333} o5 : List i6 : printingPrecision = 0 o6 = 0 i7 : {1/3p100} o7 = {.333333333333333333333333333333} o7 : List For complex numbers, if printingAccuracy is set to its default value of -1, the two parts of the number are treated together (although a digit further further to the right of the point may sometimes be displayed in the smaller part). i8 : printingAccuracy o8 = -1 i9 : printingPrecision = 16 o9 = 16 i10 : {1p100e12/3+1p100/3ii} o10 = {333333333333.3333+.33333ii} o10 : List i11 : printingAccuracy = 10 o11 = 10 i12 : {1p100e12/3+1p100/3ii} o12 = {333333333333.3333+.3333333333ii} o12 : List	2023-02-03 19:29:48	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.307500958442688, "perplexity": 5394.705007412726}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764500074.73/warc/CC-MAIN-20230203185547-20230203215547-00146.warc.gz"}
https://zbmath.org/?q=an:0852.44004	# zbMATH — the first resource for mathematics On a new integral transformation of ramifying analytic functions. (English) Zbl 0852.44004 Gliklikh, Yu. E. (ed.), Methods and applications of global analysis. Voronezh: Voronezh University Press. Nov. Global’nom Anal. 101-145 (1993). This paper is an elementary introduction to the theory of an integral transformation of complex analytic functions. It consists of two parts. The first part is aimed at the consideration of the two-dimensional case. Here we consider first the most simple version of the transformation – the so-called $$\partial/ \partial \xi$$-transformation. From the general point of view it is the technical pattern (which is, however of interest by itself) for constructing of the general (two-dimensional) transformation. The second part contains the consideration of the case of arbitrary dimensions. Except for the necessary definitions, we formulate here the main theorems on commutation of the introduced transformation with differentiation and multiplication by the independent variable. These theorems are the main ones in applications. Later on, the particular cases of the general transformation are derived from the general formula for the transformation. These particular cases are the $${\mathcal R}$$-transformation and the $$\partial/ \partial \xi$$-transformation which was our starting point in the first part. For the entire collection see [Zbl 0843.00021]. ##### MSC: 44A15 Special integral transforms (Legendre, Hilbert, etc.) 35G10 Initial value problems for linear higher-order PDEs 35K25 Higher-order parabolic equations	2021-05-08 20:54:54	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7095495462417603, "perplexity": 633.8704998085981}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-21/segments/1620243988923.22/warc/CC-MAIN-20210508181551-20210508211551-00262.warc.gz"}
https://dsp.stackexchange.com/questions/16211/analyzing-spectrum-of-a-signal?noredirect=1	# Analyzing Spectrum of a signal? my_signal = 8192cos(2pi6(0:7)/8) + 1i* 8192sin(2pi6(0:7)/8); my_signal_fft = fft(my_signal); plot(abs(my_signal_fft)); This is matlab code. What do you notice about the spectrum? Is it symmetrical? Obviously not, because I think it has an imaginary part. Would you enlighten me for the analysis of this signal's spectrum and why having an imaginary part doesn't make it symmetrical? How is the other half gone? • There are a lot of questions here. Would it be correct to summarize by asking "Why does an imaginary component result in an asymmetrical spectrum?" May 13 '14 at 12:35 $$x(n)=A[\cos(2\pi fn)+i\sin(2\pi fn)]=Ae^{i2\pi fn}$$ This is a complex exponential with normalized frequency $f$ (normalized by the sampling frequency). The bins of the signal's FFT of length $N$ correspond to normalized frequencies $$f_i=\frac{i-1}{N},\quad i=1,2,\ldots, N$$ In your case $N=8$ and $f=6/8=f_7$ (using the Matlab indexation, which starts with index $1$). So the frequency of your signal $x(n)$ lies exactly on the FFT grid. This is why there is one peak at index $7$ and all other values are $0$ (apart from numerical inaccuracies). If $x(n)$ were real-valued, its FFT would satisfy the following symmetry condition: $$X_k=X^_{N-k}$$ However, for complex-valued signals, such as yours, there is in general no symmetry. • I'm sorry I'm a noob in dsp yet, can you please elaborate "The bins of the signal" and "So the frequency of your signal x(n) lies exactly on the FFT grid." where else it can lie, what is meant with FFT grid? May 13 '14 at 12:53 • @Anarkie: The bins of the FFT are simply the values of the FFT. You have 8 bins and each bin corresponds to a certain frequency given by the formula in my answer. The frequency of your complex exponential could be anywhere between two bins, but it so happens that it exactly corresponds to one of the bin frequencies. So you see exactly one nice peak at this particular FFT bin. May 13 '14 at 12:55 • – jojek May 13 '14 at 12:56 • @MattL. Sorry for the late response, I understand fi calculation but just don't understand why the peak is at 6/8 index 7, why not 2/8, why not 5/8? May 25 '14 at 19:43 • @Anarkie: Simply because in the argument of your sine and cosine functions you have 2pi(0:7)6/8, so 6/8 is the relative frequency. And because Matlab's indices start with 1, not with 0, the corresponding FFT index is not 6 but 7. May 25 '14 at 19:50	2021-10-19 22:21:44	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 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": 0.7673286199569702, "perplexity": 647.7543367787584}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-43/segments/1634323585281.35/warc/CC-MAIN-20211019202148-20211019232148-00460.warc.gz"}
https://en.wikisource.org/wiki/Translation:Measurements_of_Becquerel_rays	# Translation:Measurements of Becquerel rays Measurements of Becquerel rays. The Experimental Confirmation of the Lorentz-Einstein Theory (1908) by Alfred Bucherer, translated from German by Wikisource Measurements of Becquerel rays. The Experimental Confirmation of the Lorentz-Einstein Theory. By A. H. Bucherer (Bonn) To the extent by which the theory of electricity occupies other areas of the phenomenal world, and in this way is becoming the foundation of the whole of physics, the desire for a consistent theory of the actual electrodynamic phenomena increases. It appeared for a long time, as if the Faraday-Maxwell-concept of the aether as the mediator of the electromagnetic processes, would serve as a sufficient and definite foundational hypothesis. Now, by further developing the aether concept in an epistemological way, one suddenly arrived at a dualistic view of aether and matter. The aether was interpreted as something that exists separately from matter, and thus one was faced with the questions, as to whether the aether is moving along with matter, or whether it is at rest. It was recognized soon, that the hypothesis of the stationary aether is the simpler one, and Lorentz made it to the foundation of his older theory of electrons. In this theory, the aether occurs as "quasi-matter". Thus the aether must be included in a dynamic system, in order to maintain the validity of the third axiom of Newton. A uniformly moving charge ${\displaystyle A}$ per se exerts another force upon a resting ${\displaystyle B}$, as ${\displaystyle B}$ upon ${\displaystyle A}$. The great inner improbability of this assumption caused difficulties for the older theory; even more the inner contradiction consisting in the fact, that a reference system shall be defined by the aether, although it represents an infinitely extended homogeneous medium. However, the demonstration was most disastrous, that no influence of annual motion of Earth through the aether could be found in the optical phenomena, which was contrary to the requirement of the theory. Although this theory was occasionally further elaborated mathematically, it was definitely clear for the physicists, that further progression can only be made on the basis of the principle of relativity. The requirement was to be made, that at uniform motion of two bodies ${\displaystyle A}$ and ${\displaystyle B}$ relative to each other, their interaction shall be independent from whether ${\displaystyle A}$ or ${\displaystyle B}$ are to be assumed as being at rest or in motion. The path to such a theory was paved by Lorentz in the year 1904. He showed, that by a suitable transformation of time and coordinates, the influence of uniform motion upon the optics of moving systems in Maxwell's equations, is vanishing, and that all observations known up to then, are in agreement with the additional consequences of this new theory. The deformation which the body's are undergoing by their motion, is characteristic for this theory. All dimensions, which coincide with the direction of motion, are contracted in the ratio ${\displaystyle {\sqrt {1-\beta ^{2}}}}$, where ${\displaystyle \beta }$ means the ratio of the body's velocity to the speed of light. Then it was shown by Einstein, that one arrives at the exact same experimental consequences, when "local time" as introduced by Lorentz, is defined as time per se, and at the same time the space coordinates in Maxwell's equations are so transformed, that they are in agreement with this definition of time. The relativity principle is clearly emphasized in Einstein's version. While in Lorentz's version, the deformation and the kinetic energy are still definitely localized, this localization becomes relative according to Einstein. Einstein and Planck alluded to the important consequences following from the relativity principle. I recall, that the equations of motion assume the classical from of the Lagrangian equations after some transformations; and that one, starting from the principle of least action, arrives at important conclusions concerning the entropy and temperature of moving bodies. Also the extension of the concept of mechanical mass (which is dependent on velocity and energy content, and which also has relative character) is highly remarkable, namely the law of the constancy of mass is now logically connected with the law of the conservation of energy. The law of the conservation of the center of gravity becomes widened, by extending it to radiating systems as well. This is because electromagnetic radiation is connected with an emanation of mass. That this theory will also be fundamentally important for astronomy, and that it is destined to achieve a particular agreement with the astronomical observation by extension of Newton's law, shall also be mentioned. The concept of the aether experiences a remarkable transformation. Because when a pure translatory motion of the system doesn't influence the phenomena arising in it, then properties must ascribed to the aether as the mediator of these processes, which are incompatible with the previous concept of the aether. The previous dualistic view of aether and matter must be replaced by a monistic one. Thus, the relativity principle presents itself as a far-reaching and surprisingly unifying principle. This principle peremptorily requires a direct experimental test. – It was clear from the outset, that only such phenomena can be used as confirmations of the validity of the competing theories, where the bodies are moving with high velocities. For that, measurements of Becqerel rays are suitable, and W. Kaufmann underwent the difficult task, to conduct experiments in this direction. Kaufmann's method is known to all of us, and also that Kaufmann drew the conclusion with certainty, that the relativity theory shall be considered as disproved by his experiments. A situation of unique difficulty was created by this result. Some physicists now further developed the relativity principle, with the expectation that more precise measurements nevertheless would eventually bring a decision, while others (including me) interpreted Kaufmann's result as decisive. Since all other observations alluded to the existence of some kind of relativity principle which was still unknown, I thus developed a new relativity principle, which, however, only had the character of a calculation rule. Kaufmann's measurements were compatible with this principle, and (as I was thinking at that time) it was only about the investigation of the deviation of electrons flying obliquely towards the magnetic field. Here, differences with respect to Maxwell's theory should occur. A clarification of the state of facts could only be achieved by new experiments, conducted with essentially increased precision. For that purpose, I created a new experimental arrangement, which I already described in the Physikalische Zeitschrift.[1] The method chosen, allowed for testing my relativity principle, i.e. an investigation of the deflection of electrons flying obliquely towards the field direction, as well as testing the Lorentz-Einstein relativity principle and the initial theory of Maxwell, thus the same question which formed the subject of Kaufmann's investigation. Becquerel rays shall fly through a condenser field, and the electric force acting upon the electrons shall be compensated by superposition of a uniform magnetic field, which is parallel to the plates of the condenser. After leaving the condenser, the magnetic field alone acts upon the rays. The deflected electrons fall upon a photomicrograph film, so that the deflection can be measured. Since the force stemming from the magnetic field is proportional to the velocity of the electrons, then the compensation can only exist for a quite definite velocity, and only electrons of this velocity can traverse the condenser field undeflected, and therefore they can leave. The details of the experimental arrangement are as follows: The condenser consists of two circular plates lying horizontally, whose diameter is ca. 8 cm and whose mutual distance amounts to ca. ¼ mm. As radiation source, a granule of radium salt in the form of a sphere (namely fluoride instead of the previously used bromide) is brought between the plates in the center of the condenser. Since the specific concentration of radium in fluoride is more than double as in bromide, the time of exposition becomes quite essentially diminished by using fluoride, which is of great importance in these experiments. The condenser is located in a cylindric tin consisting of brass, namely at half the height from the ground, so that its surfaces are located exactly perpendicular to the cylinder axis, which passes through the center of the condenser. The cylindric tin (which is very exactly formed) has an inner diameter of ca. 16 cm and an inner height of 8 cm. This tin can be airtightly sealed by sanded glass covers, so that it can be evacuated. The air pump employed was a Gaede pump, which worked excellently. By suitable drilling, the cables (being isolated from an accumulator battery) were inserted into the brass tube. The photographic plate is pressed by two springs against the interior wall of the tin. The latter can be inserted into the interior of the solenoid, whose rectangular cross-section is adapted to the dimensions of the tin. The solenoid is 103 cm long and has two windings of 103 turns each. The field strength achievable with the solenoid, was ca. 140 Gauss. The purpose of my arrangement can be easily recognized. Since the directions of the rays are namely forming all possible angles ${\displaystyle \alpha }$ with the direction of the magnetic force, then the force (occurring according to Maxwell's theory) can assume all possible values. If the electrodynamic force and the electric one ${\displaystyle \epsilon F}$ is compensated, it is ${\displaystyle {\frac {u}{v}}=\beta ={\frac {F}{v\ H\ \sin \alpha }}.}$ By means of this arrangement, rays of a certain velocity find quite automatically the angle, at which the compensation occurs at a given field strength, and which allows them to leave the condenser. Thus electrons of all velocities and corresponding masses will hit upon the film, and thus produce a curve which allows to determine the mass as a function of velocity. Consequently, a single exposition suffices to test the various theories of the electron. The equations of motion of the electrons assume the following form: I ${\displaystyle {\begin{cases}{\frac {d}{dt}}(m{\dot {x}})=0\\{\frac {d}{dt}}(m{\dot {y}})=\epsilon H{\dot {z}}\\{\frac {d}{dt}}(m{\dot {z}})=-\epsilon H{\dot {y}}\end{cases}}}$ Here, ${\displaystyle \epsilon ,\ m}$ mean the specific charge and mass of the electron. The direction of increasing ${\displaystyle x}$ is the direction of the magnetic field ${\displaystyle {\mathfrak {H}}}$. While the ${\displaystyle X}$- and ${\displaystyle Y}$-axes coincide with the plane of the condenser plates, ${\displaystyle z}$ is perpendicular to this plane. Within the condenser, the direction of motion forms the angle ${\displaystyle \alpha }$ with ${\displaystyle {\mathfrak {H}}}$. Due to the spiral motion in the pure magnetic field, however, the direction deviates from the initial one, so that the impact point ${\displaystyle P}$ upon the film, lies in the direction ${\displaystyle \theta }$ which somewhat deviates from ${\displaystyle \alpha }$. If one integrates the preceding equations and if one sets ${\displaystyle \varphi ={\tfrac {\epsilon }{m}}Ht}$, then it follows: II ${\displaystyle {\begin{cases}{\dot {x}}=u\ \cos \alpha \\{\dot {y}}=u\ \sin \alpha \cos \varphi \\{\dot {z}}=-u\ \sin \alpha \sin \varphi .\end{cases}}}$ If one furthermore sets ${\displaystyle OD=DP=\alpha }$, and ${\displaystyle u={\frac {F}{H\ \sin \alpha }},}$ then one obtains by repeated integration: III ${\displaystyle {\begin{cases}x=a\ \cos \alpha +{\frac {m}{\epsilon }}{\frac {F}{H^{2}}}\varphi \cot \alpha \\y=a\ \sin \alpha +{\frac {m}{\epsilon }}{\frac {F}{H^{2}}}\sin \varphi \\z={\frac {m}{\epsilon }}{\frac {F}{H^{2}}}(1-\cos \varphi ).\end{cases}}}$ If one denotes the values of ${\displaystyle x,y,z}$ and ${\displaystyle \varphi }$ (present at the impact point of the electron) by index ${\displaystyle \Theta }$, and if one furthermore sets ${\displaystyle {\tfrac {F}{2\alpha H^{2}}}=K}$, then one finds: IV ${\displaystyle {\begin{cases}(1)&{\frac {x_{0}}{2a}}=\cos \Theta ={\frac {1}{2}}\cos \alpha +K{\frac {m}{\epsilon }}\varphi _{0}\cot \alpha \\(2)&{\frac {y_{0}}{2a}}=\sin \Theta ={\frac {1}{2}}\sin \alpha +K{\frac {m}{\epsilon }}\varphi _{0}\sin \varphi _{0}\\(3)&{\frac {z_{0}}{2a}}=K{\frac {m}{\epsilon }}(1-\cos \varphi _{0}).\end{cases}}}$ By insertion of (3) into (1) and (2) it follows: (1a) ${\displaystyle \sin \Theta ={\frac {1}{2}}\sin \alpha +{\frac {z_{0}\sin \ \varphi _{0}}{2a(1-\cos \varphi _{0})}}}$ (2a) ${\displaystyle \cos \Theta ={\frac {1}{2}}\cos \alpha +{\frac {z_{0}\varphi _{0}\cot \alpha }{2a(1-\cos \varphi _{0})}}}$ ${\displaystyle \alpha }$ is given from these equations, and from that the velocity of the ray hitting at ${\displaystyle P}$. Since ${\displaystyle \theta }$ slightly differs from ${\displaystyle \alpha }$, one proceeds just so, as if the electron (which is hitting at ${\displaystyle P}$) would have moved with the previously calculated velocity at a circular path in a vertical plane, which passes through the radium granule and ${\displaystyle P}$. Now, the force acting in the magnetic field is: (4) ${\displaystyle {\frac {m\ u^{2}}{r}}=\epsilon Hu\ \sin \alpha .}$ However, as it is shown by a simple calculation, it is: (5) ${\displaystyle {\frac {1}{r}}={\frac {a^{2}}{2z}}\left(1+{\frac {z^{2}}{\alpha ^{2}}}\right)}$ Now if one considers, that according to Lorentz: (6) ${\displaystyle {\frac {\epsilon }{m}}={\frac {\epsilon }{m_{0}}}(1-\beta ^{2})^{\frac {1}{2}},}$ then the relations (4), (5) and (6) give (7) ${\displaystyle {\frac {\epsilon }{m_{0}}}={\frac {2zv}{\alpha ^{2}\left(1+{\frac {z^{2}}{\alpha ^{2}}}\right)H\ \sin \alpha }}\tan \arcsin \beta }$ While according to Maxwell under employment of Abraham's formula, and when we set ${\displaystyle \tanh \delta =\beta }$: (8) ${\displaystyle {\frac {\epsilon }{m_{0}}}={\frac {2zv}{a^{2}\left(1+{\frac {z^{2}}{\alpha ^{2}}}\right)H\ \sin \alpha }}\left\{{\frac {3}{4\beta }}{\frac {2\delta -\tan h\ 2\delta }{\tan h\ 2\delta }}\right\}}$ Obviously only that theory is valid, for which ${\displaystyle {\tfrac {\epsilon }{m_{0}}}}$ is a real constant for any value of ${\displaystyle \beta }$ within the observational errors. The experiments. I omit at this place the experiments, which I have undertaken to test my relativity principle. The report shall suffice, that I refuted this theory by my experiments. Therefore, it is still only about the question: whether the Lorentz-Einstein theory or Maxwell's theory. In the investigation of this question, I was led by the following viewpoints: 1. It was desirable, to investigate a velocity range (being as large as possible) of rays, because only in this way, the sought velocity function can be determined with certainty. Rays that are too fast had to be excluded, because the percentage errors become too great due to their slight deflectability. 2. Since the specific charge assumes (at very slight velocities) the same value for all relevant theories, also rays being as slow as possible had to be investigated in order to determine ${\displaystyle {\tfrac {\epsilon }{m_{0}}}}$. It was indeed achieved by me, that rays of ⅓ of the speed of light are deflected and radiographically fixated. This is also of especial importance, because the investigation remains confined to a single area. It was avoided to resort to comparisons with cathode-rays values. In my point of view, the previous measurements of cathode rays were namely made (with exception of Bestelmeyer's) under hardly controllable circumstances, by using an energy equation for the calculation of the velocity, which hardly accounts for the complicated energy changes in the vicinity of the cathode. The processes taking place at the cathode, are too little investigated, to serve as a foundation of the calculations. That also the Zeeman effect (at the current state of research) can give no information about the specific charge of the electron, can be seen from the deviating values, which have been given from the investigation of the spectra of various metals at strong and weak fields. 3. A main requirement was the precision of the measurement of the apparatus constants. I believe that I have gone so far in this respect, as it was allowed by the current physical technology. The apparatuses were made with great skill and understanding by the renowned firm M. Wolz in Bonn. In the following, I give a short overview concerning the most important auxiliary measurements. I. The electric field. The electric field of the condenser was given from the measurement of the potential difference of an accumulator battery of 320 elements, and the thickness of the quartz plates which determined the distance of the condenser plates. After every measurement, the potential difference was measured by the compensation method. The thickness measurement is based on the following arrangement: The arm balance of a fine balance was used as a lever, whose rotation axis was the knife edge; the other end of the lever rested upon an optical plane plate. On the arm balance, a vertically located mirror was mounted, in which a fine platinum wire was mirrored. A fine cathetometer of Wolz was adjusted to the image, and if one moved the quartz plate (which was to be measured) between lever and optical plate, then the mirror image was displaced. The cathetometer was again adjusted and read. A simple calculation then gave the thickness of the plate of 0,25075 mm. II. The magnetic field. The solenoid field was so measured, that its magnetic effect upon a magnetic needle suspended in the interior at a quartz string, was compensated by a coil (being exactly measurable and winded on marble) which was moved over the solenoid. ${\displaystyle H=23,24\partial }$ was given as the average field strength, while ${\displaystyle H=23,19\partial }$ in the center of the solenoid, where ${\displaystyle \partial }$ was measured in Ampere. The current was provided by the urban center, and was regulated by means of a Siemens precision-Ampere-meter and a constant resistance. The constancy was so good in general, that one could be assured, that the solenoid current and thus the magnetic field remained constant up to one-thousandth. The results. Every single of the obtained curves allowed to determine the specific charge of the electron as a function of velocity, and thus to decide the question concerning the sought natural law. For the purpose of this report, however, I preferred to calculate (from a series of curves ) the maxima ${\displaystyle Z_{m}}$ of deflection, which were read by means of a cathetometer. Thus one obtains results, which were achieved under manifold experimental conditions. One avoids the already given and somewhat complicated calculations, whose discussion would lead too far at this place. In the following table, I have put together the results. Regarding the first series, it is still to be noticed, that the small deviation of the value of ${\displaystyle {\tfrac {\epsilon }{m_{0}}}}$ is probably explained by the difficulty of current regulation, which is required at so small rays. A look upon the both last columns shows, that the decision turns out in favor of the Lorentz-Einstein theory. Number of epxeriment βm H in Gauss Zm in mm ${\displaystyle {\frac {\epsilon }{m_{0}}}\times 10^{-7}}$according to Lorentz ${\displaystyle {\frac {\epsilon }{m_{0}}}\times 10^{-7}}$according to Maxwell 10 u. 1187133 0,31780,37920,42860,51600,6879 104,54115,76127,35127,54127,54 16,3714,4513,510,186,23 1,6951,7061,7061,7041,705 1,6761,6781,6701,6481,578 It remains, to come back to the possibility of the inclusion of a certain correction at the values of deflection. My calculations are related to deflections, which the rays (being compensated within the condenser) experience in the pure magnetic field. However, besides those normal rays, also anomalous ones occur, namely rays that didn't experience a complete compensation of the acting forces withing the condenser, and therefore traverse the condenser in a curved path. They are making the radiographic curve somewhat less clear, and are possibly displacing somewhat the center-of-gravity line of the curve. I have calculated these extreme rays, and convinced myself that they didn't noticeably influence the results. Finally I still want to shortly remark about the obvious constancy of the values of ${\displaystyle {\tfrac {\epsilon }{m_{0}}}}$; it is an advantage of the method used by me, that (as the calculation shows) small errors in the determination of the electric field strength are only slightly influencing this constancy, provided, that these slight errors are made evenly at all experiments. Thus when the value of ${\displaystyle {\tfrac {\epsilon }{m_{0}}}=1,705\times 10^{7}}$ contains a possible error, then the essential result of my experiment is not altered by that. This result is the confirmation of the relativity principle. Supplement. As a calculation shows, which was conducted only afterwards and which is based on a successful experiment, the action of the protection ring around the condenser is not without influence as I believed and also expressed at the beginning. Theoretically, this marginal action is just so, as if the radius of the condenser would be increased by a small amount – here 0,31 mm. If one applies this correction, than ${\displaystyle {\tfrac {\epsilon }{m_{0}}}\times 10^{-7}}$ becomes 1,730; 1,730; 1,729; 1,730 according to Lorentz for the experiments No. 8, 7, 13, 3 respectively. Discussion. Bestelmeyer: I would like to ask the reader to provide the following dimensions. How large was the distance between the plates (reader: ¼ mm). How large was the diameter of the interior condenser (reader: 40 mm). How was the length of the path between the plates (4 cm) and outside of the plates? (4 cm), and how large was the deflection of the rays? (16 mm down to 6,23 mm). There, I would like to allude to a difficulty at the measurements. Electrons are moving rectilinear between the condenser plates, for which the electric and magnetic force is equal. However, besides the rays which are passing through rectilinearly in the condenser, also such ones are traversing which have a substantially greater and a substantially smaller velocity. For the dimensions that you have given now, those are (in my estimation) ca. ± 10 perc. which are passing through with different velocity, and when one determines now from the deflection of the value of ${\displaystyle e/m}$, then it is not sure as to whether it belongs to the rays of average velocity, or to the ones having a velocity which is 10 perc. higher or lower. Thus one must know the initial radiation distribution. If they are altogether of the same amount, then one can assume that the deflection also corresponds to the mean velocity. Bucherer: I have taken pains to investigate this error strictly mathematically. For that purpose, I have solved a transcendental equation from that I measured the velocities and the radius of curvature. It is given, that these rays don't matter. This is different with Bestelmeyer's method. There, the slit is wider. When you look at the image, then you will see that the extra rays play no role at all. The image can only stem from the central rays, which are normally passing through, that is, which have no curvature. I want to write down here some numbers, which take this point into account; namely, ${\displaystyle z}$ (i.e. the deflection experienced by the extra rays) corresponds to the following velocities: βm = 0,516;β = 0,47;β = 0,588; ${\displaystyle z_{m}}$ = 10,18 mm${\displaystyle z}$ = 0,971 cm${\displaystyle z}$ = 0,954 cm For rays that are passing through rectilinearly, i.e. the normal rays, ${\displaystyle z_{m}=10,18}$ mm, and eventually for faster rays, the deflection is 0,954 cm. You see, that the deflection of these rays lies in the vicinity of the normal rays, so that (at most) a small widening can occur by that. The distance of the condenser plates is ½ mm, this at most gives a width of up to ¾ mm; actually, the curve is not wider. This clearly proves, that the extra rays play no role. Wien: I would like to ask the reader, if he maybe has evidence, why the experiment of Kaufmann has led to results which are different from his ones. Bucherer: I don't want to start a criticism of Kaufmann's experiments, without expressly confirming, that I highly esteem the pioneering work of Kaufmann. When I now pass to a criticism of Kaufmann's experiments, then I want to allude at first to the difficulty in the measurement of such a small curve. The velocities, which come into account, are 0,8 to 0,56 of the speed of light. I haven't taken into account the lower values of the deflection, since the percentage errors are too high at this place. In the area, which come into question for Kaufmann, the curve is already very small as well. I precisely looked at the curve and discovered an asymmetry of 5 perc. at one portion, which is as much as the difference of the theories in this range of velocity. I alerted Kaufmann to this fact, so he has measured again and has actually found the deviation of 5 perc. Another point is as follows: I have measured the resistance of the condenser; Kaufmann assumes it to be infinitely great. When the resistance of the condenser doesn't vanish, then an error is inserted by that, which I estimate to 1 perc. in Kaufmann's experiment. I myself have found the following relations (a drawing is following, in which condenser, resistances, and the battery are indicated schematically). When the switched resistance doesn't vanish against that of the condenser, then the potential difference of the battery is evidently not relevant. This error alone amounts to 0,06 perc. at my experiments. (Numbers are given again.) You see, already there you get noticeable resistances. At Kaufmann's experiments, the resistances already amounted several megohm. Also the voltage measurements of Kaufmann are invalid in my view. Another error is possibly that one: At such exact measurements as they come in question here, the plate may not be so compressed, that the pressure occurs in the middle, but the pressure may only exerted upon the quartz plates, otherwise the pales will be easily bend. When one studies interference phenomena, one sees that when it is pressed from the middle, surely a dozen rings are leaving. Then, the difficulty of the magnetic field has to be mentioned. Kaufmann uses a permanent magnet of 145 Gauss; he had an armature, which was removed and then the magnetism possibly changes with time. Also at this place, a source of error possibly enters. Bestelmeyer: I don't believe, that the question concerning the sources of error mentioned by me, can be decided here. Thus I only want to say, that I'm not convinced that the mentioned sources of error actually play no role here. Here, it is about very precise measurements. When one considers the smaller dimensions of the apparatus, then the curves approximately show the same sharpness, as in my experiments. For the faster rays, the velocity function according to Lorentz's theory, is ca. 1,24, according to Abraham it is 1,19. The larger velocities correspond the smaller values of deflection. The smaller velocities correspond to the larger deflections. Even at great velocities, the difference between the theory of Lorentz and that of Abraham only amounts to a few tenths of a millimeter at the measurement; I believe, that the measurements can possibly be imprecise by that amount. Though I don't want to definitely assert this here, without a more detailed examination of the numbers. Bucherer: I'm responding to this: Here, the values of ${\displaystyle z}$ are of such kind, that one can assume without further ado, that the extra rays have no effect. Take an optical analogue. Imagine, that the condenser plates are sooty, then the extra rays will only come to deflection in small number through the condenser (drawing). They cannot exert an influence, this is totally excluded. Here, this influence lies only at rays having velocities between the speed of light and 0,95. This bundle was overlooked by Bestelmeyer at all. I have conducted measurements also with respect to this. Therefore, I have taken (at the results represented by me here) only the maxima of the curves, to avoid exactly this point, and the influence is vanishing at those maximi. Minkowski (Göttingen): I want to express my delight, that the experimental results speak in favor of Lorentz's theory when compared with the one of the rigid electron. That one day this will be the case, could not be doubted at all from the theoretical standpoint. The rigid electron is in my view a monster when put together with Maxwell's equations, whose innermost harmony is the relativity principle. When one approaches Maxwell's equations on the basis of the idea of the rigid electron, then it just appears to me, as if one goes into a concert and one has plugged ones ears with cotton. One has to admire in the highest the courage and the force of the school of the rigid electron, which jumps over the widest mathematical hurdles with a fabulous approach, in the hope to fall upon experimental ground at the other side. Yet the rigid electron is not a working hypothesis, it is a working hindrance. Bucherer: The situation is mostly represented in a wrong way. It is actually not about the decision between Maxwell's and Lorentz's theory; indeed, Maxwell's theory is already superseded for a long time by the experiments of Michelson and Morley, as well as Trouton and Noble. Minkowski: Not Maxwell's and Lorentz's theories are the actual oppositions, but the rigid and the non-rigid electron, i.e. the Zeppelin- and the Parseval electron. Historically, I want to add that the transformation, which play the main role in the relativity principle, were first mathematically discussed by Voigt in the year 1887. Already then, Voigt drew some consequences with their aid, in respect to the principle of Doppler. Voigt: Minkowski is reminiscent of an old paper of mine. There, it is about the applications of Doppler's principle, which occur in special parts, though not on the basis of the electromagnetic theory, but on the basis of the elastic theory of light. However, already then some of the same consequences were given, which were later gained from the electromagnetic theory. 1. 8, 430, 1907. This work is a translation and has a separate copyright status to the applicable copyright protections of the original content. Original: This work is in the public domain in the United States because it was published before January 1, 1926. The author died in 1927, so this work is also in the public domain in countries and areas where the copyright term is the author's life plus 80 years or less. This work may also be in the public domain in countries and areas with longer native copyright terms that apply the rule of the shorter term to foreign works. This work is released under the Creative Commons Attribution-ShareAlike 3.0 Unported license, which allows free use, distribution, and creation of derivatives, so long as the license is unchanged and clearly noted, and the original author is attributed.	2021-05-15 03:15:55	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 69, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8072918653488159, "perplexity": 674.659751526644}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-21/segments/1620243991812.46/warc/CC-MAIN-20210515004936-20210515034936-00607.warc.gz"}
https://tex.stackexchange.com/questions/250714/trees-and-arrows	# Trees and arrows I need to draw an arrow, using \draw[semithick,->] (t)..controls +(south west:5) and +(south:5) .. (wh);. I want the arrow to connect e3 at the bottom of the following tree to the single e3 occurring higher up in it: \documentclass[12pt,a4paper]{article} \usepackage{tikz-qtree} \usepackage{tikz-qtree-compat} \usepackage{ulem} \begin{document} \begin{tikzpicture} \Tree [.{$<$t$_{1}$,\hspace{0.1cm} \\$\lbrack$ \hspace{0.3cm} $\rbrack$, $\lbrack$ \hspace{0.3cm} $\rbrack$ $>$} [.{$<$(e$_{3}$ $\rightarrow$ t$_{1}$) $\rightarrow$ t$_{1}$),\hspace{0.1cm} \\$\lbrack$ e$_{3}$ $\rbrack$ $>$} \node(wh){something} ;] [.{$<$(e$_{3}$ $\rightarrow$ t$_{1}$), \hspace{0.1cm} \\$\lbrack$ e$_{2}$ $\rbrack$, \hspace{0.1cm} \\$\lbrack$ \hcancel{e$_{3}$} \thinspace $\rbrack$ $>$ } [.{e$_{3}$} ] [.\node[draw]{{$<$t$_{1}$, \hspace{0.1cm} \\$\lbrack$ e$_{2}$ $\rbrack$, \hspace{0.1cm} \\$\lbrack$ e$_{3}$ $\rbrack$ $>$ }}; [.\node[draw]{e$_{2}$ }; ] [.{$<$(e$_{2}$ $\rightarrow$ t$_{1}$), \hspace{0.1cm} \\$\lbrack$ e$_{3}$ $\rbrack$ $>$} [.{$<$(e$_{3}$ $\rightarrow$ e$_{2}$ $\rightarrow$ t$_{1}$),\hspace{0.1cm} \\$\lbrack$ $\emptyset$ $\rbrack$ $>$} {$<$(e$_{3}$ $\rightarrow$ e$_{2}$ $\rightarrow$ t$_{1}$),\hspace{0.1cm} \\$\lbrack$ $\emptyset$ $\rbrack$ $>$} ] [.\node[draw]{e$_{3}$ }; [. e$_{3}$; ] ] ] ] ] ] ] ] \end{tikzpicture} \end{document} How can I do this? • Did you forget a } somewhere? I can't compile this code. – Alenanno Jun 17 '15 at 9:17 • It works for me... strange – user65526 Jun 17 '15 at 9:50 • I also can't get the MWE to compile (an undefined control sequence error). It looks like there are several places whether _ is used outside math mode. I don't see a reason to keep going in and out of math mode within your type specifications. I would recommend posting a much simpler MWE (or at least a screenshot of the output you get iwth the one you posted) so we can see what's going on with the arrow. The code snippet for the arrow looks fine -- you should just be able to put that after the last bracket in the tree and before \end{tikzpicture}. – Jason Zentz Jun 17 '15 at 12:17 Simply use \nodes with names for those elements and then use the names to connect them: \documentclass[12pt,a4paper]{article} \usepackage{tikz-qtree} \usepackage{tikz-qtree-compat} \usepackage{ulem} \def\hcancel#1{}% provissional definition; delete this line in your actual code \begin{document} \begin{tikzpicture} \Tree [.{$<$t$_{1}$,\hspace{0.1cm} \\$\lbrack$ \hspace{0.3cm} $\rbrack$, $\lbrack$ \hspace{0.3cm} $\rbrack$ $>$} [.{$<$(e$_{3}$ $\rightarrow$ t$_{1}$) $\rightarrow$ t$_{1}$),\hspace{0.1cm} \\$\lbrack$ e$_{3}$ $\rbrack$ $>$} \node(wh){something} ; ] [.{$<$(e$_{3}$ $\rightarrow$ t$_{1}$), \hspace{0.1cm} \\$\lbrack$ e$_{2}$ $\rbrack$, \hspace{0.1cm} \\$\lbrack$ \hcancel{e$_{3}$} \thinspace $\rbrack$ $>$ } [. \node (ue) {e$_{3}$}; ] [.\node[draw]{{$<$t$_{1}$, \hspace{0.1cm} \\$\lbrack$ e$_{2}$ $\rbrack$, \hspace{0.1cm} \\$\lbrack$ e$_{3}$ $\rbrack$ $>$ }}; [.\node[draw]{e$_{2}$ }; ] [.{$<$(e$_{2}$ $\rightarrow$ t$_{1}$), \hspace{0.1cm} \\$\lbrack$ e$_{3}$ $\rbrack$ $>$} [.{$<$(e$_{3}$ $\rightarrow$ e$_{2}$ $\rightarrow$ t$_{1}$),\hspace{0.1cm} \\$\lbrack$ $\emptyset$ $\rbrack$ $>$} {$<$(e$_{3}$ $\rightarrow$ e$_{2}$ $\rightarrow$ t$_{1}$),\hspace{0.1cm} \\$\lbrack$ $\emptyset$ $\rbrack$ $>$} ] [.\node[draw]{e$_{3}$ }; [. \node (le) {e$_{3}$}; ] ] ] ] ] ] \draw[semithick,->] (le)..controls +(south west:5) and +(south:5) .. (ue); \end{tikzpicture} \end{document} Since I didn't know where the \hcancel command came from, I provided a simple definition \def\hcancel#1{} to make the example code compilable. Remove that definition in your actual code.	2020-08-05 10:54:21	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 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": 0.7821471095085144, "perplexity": 10348.268348696842}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-34/segments/1596439735939.26/warc/CC-MAIN-20200805094821-20200805124821-00144.warc.gz"}
https://computergraphics.stackexchange.com/questions/2342/how-to-blend-world-space-normals/2353	How to blend World Space Normals I am trying to blend two world space normals inside a shader. One comes from a tangent space normal map converted into world space using a classic TBN matrix and the other one is a mesh normal map in world space. I found some interesting resources here : But those blending technics seem to be only available for tangent space normals, especially the Reoriented Normal Mapping (RNM). I tried to apply the RNM technic with unpack already done. n1 += vec3(0, 0, 1); n2 = vec3(-1, -1, 1); return n1 dot(n1, n2) / n1.z - n2; But this doesn't give expected results and I don't get why. Is there a way to apply the RNM blending on world space normals ? Thanks a lot. Edit: Here are some output and the result I am having with RNM. The A normal is not disturbed by the B normal, it gives some strange massively reoriented results. float3 nv = dot(normal, viewDir); color.rgb = nv; • Do you have any physical interpretation for the two normal maps? e.g. one is high-frequency and one is low-frequency and applied "on top of it" (which is the Reoriented Normal Mapping case)? I would expect that you would need to feed the mesh's normal/tangent/bitangent into the equation somehow. You might consider simply transforming the world space normal into tangent space, applying RNM, then transforming the result to world space—that's only one extra 3x3 mul. Apr 19 '16 at 14:17 • Thanks for your answer @JohnCalsbeek but using baked tangent space normal doesn't provide good enough results that's why I am using a world space normal, especially a world space bent normal map, and I try to apply on top of if the regular normal map. – MaT Apr 19 '16 at 19:24 • I've added some screenshot to illustrate the issue I am having. – MaT Apr 19 '16 at 20:01 Huge thanks to @MJP who answered this. The aim is to avoid the simplification made when using tangent space normals. Here is the paper : Blending in detail But only implement equation (4) which gives you this. float3 ReorientNormal(in float3 u, in float3 t, in float3 s) { // Build the shortest-arc quaternion float4 q = float4(cross(s, t), dot(s, t) + 1) / sqrt(2 * (dot(s, t) + 1)); // Rotate the normal return u * (q.w * q.w - dot(q.xyz, q.xyz)) + 2 * q.xyz * dot(q.xyz, u) + 2 * q.w * cross(q.xyz, u); }	2021-10-19 19:16:17	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.27837038040161133, "perplexity": 3533.6259538089876}, "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-2021-43/segments/1634323585280.84/warc/CC-MAIN-20211019171139-20211019201139-00560.warc.gz"}
https://math.stackexchange.com/questions/3262955/intersection-of-subspaces-and-linear-independence	# Intersection of Subspaces and Linear independence Suppose $$U_1$$ and $$U_2$$ are subspaces of a finite-dimensional vector space. Let $$u_1,...,u_m$$ be a basis of $$U_1\cap U_2$$, thus dimension of the intersection is $$m$$. $$\textbf{The part I don't understand is:}$$ Because $$u_1,...,u_m$$ is a basis of $$U_1\cap U_2$$, it is linearly independent in $$U_1$$. Why is this true? Reference: Axler, Sheldon J. $$\textit{Linear Algebra Done Right}$$, New York: Springer, 2015. • A basis for $U_1 \cap U_2$ is by definition as subset $\beta \subset U_1 \cap U_2$ which is linearly independent and one which spans $U_1 \cap U_2$. Is there something else which is bothering you? Also, the concept of linear independence is one which depends on the underlying field of scalars, not on the space $V$ of consideration – peek-a-boo Jun 15 '19 at 5:42 Since $$u_1, ..., u_m$$ is a basis for $$U_1 \cap U_2$$, it is linearly independent in $$U_1 \cap U_2$$ (by the definition of a basis). Therefore, for scalars $$c_1,..., c_m$$ in the vector space, we have $$c_1u_1 +... + c_mu_m = 0 \implies c_1=c_2=...=c_m = 0$$ (by the definition of linear independence). This implication holds true for the subspace $$U_1$$ as well, since each of $$u_1,...,u_m$$ are in $$U_1$$ (since $$u_1,...,u_m$$ are in $$U_1$$ and $$U_2$$). Therefore, $$u_1,...,u_m$$ is linearly independent in $$U_1$$. Note: $$U_1\cap U_2\subset U_1$$. • Chris, thanks for the response. I was thinking about this in 2D and 3D. Start with two planes, $p1$ and $p2$. If we have non-intersecting and non-parallel lines in this section of two planes, it will be LI. This set of lines would be still LI in each of the plane, $p1$ and $p2$. Is this correct? – Frank Swanton Jun 15 '19 at 6:19	2021-06-23 19:11:24	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 24, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9583065509796143, "perplexity": 128.90816905215317}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-25/segments/1623488539764.83/warc/CC-MAIN-20210623165014-20210623195014-00034.warc.gz"}
https://www.zbmath.org/?q=an%3A1206.11069	# zbMATH — the first resource for mathematics Arboreal Galois representations. (English) Zbl 1206.11069 Summary: Let $$G_{\mathbb{Q}}$$ be the absolute Galois group of $$\mathbb{Q}$$, and let $$T$$ be the complete rooted $$d$$-ary tree, where $$d \geq 2$$. In this article, we study “arboreal” representations of $$G_{\mathbb{Q}}$$ into the automorphism group of $$T$$, particularly in the case $$d = 2$$. In doing so, we propose a parallel to the well-developed and powerful theory of linear $$p$$-adic representations of $$G_\mathbb{Q}$$. We first give some methods of constructing arboreal representations and discuss a few results of other authors concerning their size in certain special cases. We then discuss the analogy between arboreal and linear representations of $$G_{\mathbb{Q}}$$. Finally, we present some new examples and conjectures, particularly relating to the question of which subgroups of $$\operatorname{Aut}(T)$$ can occur as the image of an arboreal representation of $$G_{\mathbb{Q}}$$. ##### MSC: 11F80 Galois representations 11R32 Galois theory 20E08 Groups acting on trees 20E18 Limits, profinite groups Full Text: ##### References: [1] Aitken W., Hajir F., Maire C., (2005) Finitely ramified iterated extensions. Int. Math. Res. Not. 14, 855–880 · Zbl 1160.11356 · doi:10.1155/IMRN.2005.855 [2] Boston, N.: Galois groups of tamely ramified p-extensions. In: Proceedings of Journées Arithmetiques 2005, special issue of Journal de Théorie des Nombres de Bordeaux. To appear (2006) · Zbl 1123.11038 [3] Boston N., Jones R. Densely settled groups and arboreal Galois representations. preprint (2006) [4] Breuil, C., Conrad, B., Diamond, F., Taylor, R.: On the modularity of elliptic curves over Q: wild 3-adic exercises. J. Amer. Math. Soc. 14(4), 843–939 (2001) (electronic) · Zbl 0982.11033 [5] Fontaine, J.-M., Mazur, B.: Geometric Galois representations. In: Coates, J., Elliptic curves, modular forms, & Fermat’s last theorem (Hong Kong, 1993), Ser. Number Theory, I Internat. Press, Cambridge, MA, pp. 41–78. [6] Jones, R.: The density of prime divisors in the arithmetic dynamics of quadratic polynomials. Available at http://www.arxiv.org . · Zbl 1193.37144 [7] Labute J. Mild pro p-groups and Galois groups of p-extensions of Q. J. Reine Angew. Math. Yau, S.-T.(eds.) 596, (2006), 155–182. · Zbl 1122.11076 [8] Markšaitis G.N., (1963) On p-extensions with one critical number.Izv. Akad. Nauk SSSR Ser. Mat. 27, 463–466 [9] Nekrashevych V. (2005) Self-similar groups, Vol. 117 of Mathematical Surveys and Monographs. American Mathematical Society, Providence, RI · Zbl 1087.20032 [10] Odoni R.W.K. The Galois theory of iterates and composites of polynomials. Proc. London Math. Soc. (3) 51(3), (1985), 385–414. · Zbl 0622.12011 [11] Serre J.-P. (1972), Propriétés galoisiennes des points d’ordre fini des courbes elliptiques. Invent. Math. 15(4): 259–331 · Zbl 0235.14012 · doi:10.1007/BF01405086 [12] Serre J.-P. (1987), Sur les représentations modulaires de degré 2 de $${\mathrm Gal}(\overline{Q}/Q)$$ . Duke Math. J. 54(1): 179–230 · Zbl 0641.10026 · doi:10.1215/S0012-7094-87-05413-5 [13] Shimura G., (1966) A reciprocity law in non-solvable extensions. J. Reine Angew. Math. 221, 209–220 · Zbl 0144.04204 · doi:10.1515/crll.1966.221.209 [14] Stoll M., (1992) Galois groups over Q of some iterated polynomials. Arch. Math. (Basel) 59(3): 239–244 · Zbl 0758.11045 [15] Wiles A. (1995), Modular elliptic curves and Fermat’s last theorem’. Ann. of Math. (2) 141(3): 443–551 · Zbl 0823.11029 · doi:10.2307/2118559 This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching.	2021-04-20 23:20:34	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5968572497367859, "perplexity": 2347.789533290018}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-17/segments/1618039491784.79/warc/CC-MAIN-20210420214346-20210421004346-00617.warc.gz"}
https://www.biostars.org/p/166312/	GenomeStudio genotype format to VCF? 3 1 Entering edit mode 6.9 years ago always_learning ★ 1.1k Dear All, We are working on some project to check concordance between Genotype Sample and NGS sample. We are using MEGA Genotype Array. We have processed genotype sample using GenomeStudio from Illumina and following procedure given at Is there a tool to transform GenomeStudio genotype format to VCF? . Generated (Genotype) VCF has some very weird entries with Chromosome as 0 and Position as 0 as well. I am not sure why these entries are coming with Genotype VCF. As we are using same genotype file for concordance with Sequencing VCF file and we are getting concordance around 80-85% only which ideally should be more then 90%. Could some one help me with some additional input about any improvements specially why We are getting Chromosome as 0 and Position as 0 . Apart from this we are not getting QUAL score as well for any of VCF entries . Thanks Genotype • 6.2k views 0 Entering edit mode 6.9 years ago vassialk ▴ 200 NextGene, DNALasergene, GeneousPro and CLCGenomics can help you to play with data. 0 Entering edit mode 6.9 years ago Sam ★ 4.5k If it is not a must for you to work on the vcf data, you may try and convert the genome studio data into plink format. Then you can use the --vcf function from plink to convert the vcf format into plink format. That will allow you to easily check the concordance. However, for those SNPs with chromosome 0 and position 0, something might be wrong. How do you call the VCF file from the NGS samples? side note: vassialk, it is not really helpful to keep posting the commercial software here without actually answering the question. 0 Entering edit mode We have used plink format and then converted in vcf format only. 0 Entering edit mode So then you don't need to transform the genomestudio format into vcf? If you already got the vcf files, then you should follow Robert's suggestion to use bcftools for the concordance estimation. 0 Entering edit mode 6.9 years ago Robert Sicko ▴ 630 Chr0 probes in the genome studio project are problem probes (mapping to multiple locations, etc.) and should be omitted prior to calculating concordance. You can also generate a VCF from genome studio using: https://github.com/jaredo/chiamante/wiki After that, you can use bcftools and intersect your array.vcf with sequencing.vcf to create union.vcf. Then compare concordance between array.vcf & sequencing.vcf for only locations in union.vcf	2022-09-26 03:40:50	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.17112915217876434, "perplexity": 6393.048442551945}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030334644.42/warc/CC-MAIN-20220926020051-20220926050051-00276.warc.gz"}
https://paperswithcode.com/paper/delayed-acceptance-abc-smc	# Delayed acceptance ABC-SMC 7 Aug 2017Richard G. EverittPaulina A. Rowińska Approximate Bayesian computation (ABC) is now an established technique for statistical inference used in cases where the likelihood function is computationally expensive or not available. It relies on the use of a~model that is specified in the form of a~simulator, and approximates the likelihood at a~parameter value $\theta$ by simulating auxiliary data sets $x$ and evaluating the distance of $x$ from the true data $y$... (read more) PDF Abstract	2020-06-04 01:43:09	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8339635729789734, "perplexity": 1201.603153014802}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-24/segments/1590347436828.65/warc/CC-MAIN-20200604001115-20200604031115-00192.warc.gz"}
https://tex.stackexchange.com/questions/579934/how-increase-the-font-size-of-an-equation-with-beginequation	# How increase the font size of an equation with begin{equation}? I have the following code in RMarkdown that prints an equation like: # in a LaTeX environment I would write: \usepackage{amsmath} \begin{document} $$\label{eq: eq1} \psi_{j} = \frac{\gamma_{i,j,k}^s}{\sqrt{w_{i,j,n}}}$$ \end{document} Yet, I would like to increase the font size of the equation. All I found online doesn't work. Can anyone help me? Thanks! • What do you mean? Having a bigger font size in equations? Jan 21, 2021 at 20:55 • @Bernard exactly Jan 21, 2021 at 20:57 • put \large before the \begin or \Large or \huge Jan 21, 2021 at 20:58 • @DavidCarlisle it makes all that comes after the equation also bigger, is there a way to limit it only to the equation? Jan 21, 2021 at 21:01 • put it in a group, or put \normalsize after Jan 21, 2021 at 21:04	2022-05-27 00:33:52	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 2, "x-ck12": 0, "texerror": 0, "math_score": 0.6176277995109558, "perplexity": 1614.5453605739367}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662627464.60/warc/CC-MAIN-20220526224902-20220527014902-00079.warc.gz"}
http://paperity.org/papers/100	#### Species diversity of the genus Riccia L. (Marchantiales, Ricciaceae) in Maranhão state, Brazil Ricciaceae is a little-known liverwort family in northeastern Brazil. Fieldwork in 4 localities in Maranhão state yielded 4 species of Riccia, with 2 taxa, R. mauryana and R. weinionis, representing new state records. This paper describes the species diversity of the genus Riccia in Maranhão state, and provides descriptions, ecological notes, and illustrations for each species. #### In-hospital care, complications, and 4-month mortality following a hip or proximal femur fracture: the Spanish registry of osteoporotic femur fractures prospective cohort study SummaryWe have characterised 997 hip fracture patients from a representative 45 Spanish hospitals, and followed them up prospectively for up to 4 months. Despite suboptimal surgical delays (average 59.1 hours), in-hospital mortality was lower than in Northern European cohorts. The secondary fracture prevention gap is unacceptably high at 85%.PurposeTo characterise inpatient care... #### Narcolepsy: Pathophysiology and Neuropsychological Changes Narcolepsy is now recognized as a distinctive disorder with specific pathophysiology and neurochemical abnormalities. Findings on the role of the neuropeptide hypocretin are opening new avenues of research and new strategies for therapy. Recently, neuropsychological and electrophysiological studies have provided evidence for reduced memory performance on standard memory tests in... #### 4D gauge theories with conformal matter Abstract One of the hallmarks of 6D superconformal field theories (SCFTs) is that on a partial tensor branch, all known theories resemble quiver gauge theories with links comprised of 6D conformal matter, a generalization of weakly coupled hypermultiplets. In this paper we construct 4D quiverlike gauge theories in which the links are obtained from compactifications of 6D... #### A Modified Lindstedt–Poincaré Method for a Strongly Nonlinear System with Quadratic and Cubic Nonlinearities A modified Lindstedt–Poincaré method is presented for extending the range of the validity of perturbation expansion to strongly nonlinear oscillations of a system with quadratic and cubic nonlinearities. Different parameter transformations are introduced to deal with equations with different nonlinear characteristics. All examples show that the efficiency and accuracy of the... #### Comparison of Trigeminal and Postherpetic Neuralgia Although postherpetic neuralgia and trigeminal neuralgia (tic douloureux) are common causes of facial pain, they have very little in common aside from lancinating pain (other qualities of pain in each disorder are different). Each disorder affects different areas of the face and the treatment of each is quite dissimilar. The pathogenesis of these two disorders quite likely... #### Dual upconversion nanophotoswitch for security encoding Light-harvesting lanthanide ions (Ln3+) doped NaYF4 inks could provide polychromatic patterns for opposing counterfeiting commodity infestation because of their distinctive upconversion photoluminescence (UPL) properties. Herein, three kinds of core-triple-shell Ln3+ ions doped NaYF4 upconversion nanocrystals (UCNCs) are synthesized through modified high-temperature... #### The Extreme Climate Event Database (EXCEED): Development of a picture database composed of drought and flood stimuli The present study introduces the Extreme Climate Event Database (EXCEED), a picture database intended to induce emotionally salient stimuli reactions in the context of natural hazards associated with global climate change and related extreme events. The creation of the database was motivated by the need to better understand the impact that the increase in natural disasters... #### Dennis Ross poetry submission By Dennis Ross, Published on 09/15/18 #### The Female of Bertrana Hieroglyphica Petrunkevitch (Araneae, Argiopidae) Four species of this very interesting genus are known at the present time. Keyserling (1884) established the genus and described the first known species from a group of females from Peras, Brazil. Two other species have been described from South America by Simon (1893); one of these was from Peru and the other from Para, Brazil. Dr. Petrunkevitch was the first to describe a male... #### Influenza Infection Screening Tools Fail to Accurately Predict Influenza Status for Hospitalized Patients During Pandemic H1N1 Influenza Season BACKGROUND: Following the severe acute respiratory syndrome outbreak in 2003, hospitals have been mandated to use infection screening questionnaires to determine which patients have infectious respiratory illness and, therefore, require isolation precautions. Despite widespread use of symptom-based screening tools in Ontario, there are no data supporting the accuracy of these... #### Are Navy Weight Management Programs Ensuring Sailor Physical Readiness? An Analysis at Naval Medical Center San Diego The obesity epidemic in the USA includes active duty service members in the military and effects physical readiness. At the Naval Medical Center San Diego command, the Health & Wellness Department is charged with administering the Weight Management Programs (WMP) for sailors in the San Diego area to ensure military physical readiness requirements. The optimal allocation of... #### Glowbal IP: An Adaptive and Transparent IPv6 Integration in the Internet of Things The Internet of Things (IoT) requires scalability, extensibility and a transparent integration of multi-technology in order to reach an efficient support for global communications, discovery and look-up, as well as access to services and information. To achieve these goals, it is necessary to enable a homogenous and seamless machine-to-machine (M2M) communication mechanism... #### Application of decomposition to hyperbolic, parabolic, and elliptic partial differential equations The decomposition method is applied to examples of hyperbolic, parabolic, and elliptic partial differential equations without use of linearizatlon techniques. We consider first a nonlinear dissipative wave equation; second, a nonlinear equation modeling convectlon-diffusion processes; and finally, an elliptic partial differential equation. #### Clinical applications of proton MR spectroscopy in the diagnosis of brain tumours There are few but important problems in magnetic resonance (MR) diagnosis of the brain tumours such as predicting the grade, exact definition of the tumour borders, differentiation of the cystic tumours from abscess, the tumoral core from peritumoral oedema, and the tumour recurrence from radiation necrosis. MR spectroscopy (MRS) can add more information to MR imaging (MRI) in... #### Microwave Assisted Synthesis of Some Biologically Active Benzothiazolotriazine Derivatives Synthesis of some biologically active benzothiazolotriazine derivatives by microwave irradiation is reported. 2-Amino-6-substituted benzothiazoles 1 on treatment with benzaldehyde in anhydrous ethanol afforded 2-benzylidenoimino-6-substitutedbenzothiazoles 2 which underwent cyclisation with ammoniumthiocyanate in dioxane to give 2-phenyl benzothiazolo [3,2-α]-s-triazine-4-[3H... #### Directly grown nanostructured electrodes for high-power and high-stability alkaline nickel/bismuth batteries Bismuth oxide (Bi2O3) has received great attention as an anode material for alkaline nickel/bismuth (Ni/Bi) batteries due to its high theoretical capacity and easy preparation. However, the generally poor conductivity of metal oxides and the instability of Bi2O3 during cycling severely limit the device performance. Herein, we present the use of directly grown Bi2O3 nanoflake film... #### Computing the Number of Induced Copies of a Fixed Graph in a Bounded Degree Graph In this paper we show that for any graph H of order m and any graph G of order n and maximum degree $$\Delta$$ one can compute the number of subsets S of V(G) that induces a graph isomorphic to H in time $$O(c^m \cdot n )$$ for some constant $$c = c(\Delta ) >0$$. This is essentially best possible (in the sense that there is no $$c^{o(m)}poly(n)$$-time algorithm under the... #### Evaluating the psychometric quality of school connectedness measures: A systematic review Introduction There is a need to comprehensively examine and evaluate the quality of the psychometric properties of school connectedness measures to inform school based assessment and intervention planning. Objective To systematically review the literature on the psychometric properties of self-report measures of school connectedness for students aged six to 14 years. Methods A...	2019-02-19 09:25:40	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.3382816016674042, "perplexity": 6183.241891911191}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-09/segments/1550247489729.11/warc/CC-MAIN-20190219081639-20190219103639-00143.warc.gz"}
https://stats.stackexchange.com/questions/434130/independence-of-a-gaussian-random-variable-and-the-product-of-another-gaussian-r	# Independence of a Gaussian random variable and the product of another Gaussian random variable and a Bernoulli random variable Let $$X$$ and $$Y$$ be two independent Gaussian random variables with mean $$0$$ and variance $$σ^2_X$$ and $$σ^2_Y$$ respectively. Let $$Z$$ be a random variable measurable with respect to $$σ(Y)$$ and suppose that $$Z$$ assumes only value $$1$$ or $$−1$$. Show that $$(ZX, Y)$$ is a 2-dimensional Gaussian random vector and determine its variance and covariance matrix. Say also if $$ZX$$ and $$Z$$ are independent. In an attempt to solve this, I first decided to find the distribution of $$ZX$$ by using the law of Total Probability. I assumed independence for $$Z$$ and $$X$$ since they come from different sigma algebras. As I seek the distribution of $$ZX$$, I asked my about the chance that $$ZX \le t$$ for some arbitrary real value $$t$$. To handle the discreteness of $$Z$$, consider enumerating its possible values: $$\Pr[ZX \le t] = \Pr[ X \le t \text{ and }Z=1] + \Pr[X \le t \text{ and } Z=-1].$$ Now, can you please show me how to continue from here? First, $$Z$$ and $$X$$ are independent since they come from independent sigma-algebras. Next, let $$\Pr[Z=1]=p$$. You missed minus sign in equality: $$\Pr[ZX \le t] = \Pr[ X \le t \text{ and }Z=1] + \Pr[-X \le t \text{ and } Z=-1]$$ continue using independence $$\Pr[ZX \le t] =p\Pr[X\le t]+(1-p)\Pr[-X\le t].$$ Note that $$X$$ is centered Gaussian random variable and then $$-X$$ has the same distribution as $$X$$, so both probabilities above coincide: $$\Pr[-X\le t]=\Pr[X\le t]$$ for any $$t$$. Then $$\Pr[ZX \le t] =\Pr[X\le t]\cdot(p+(1-p))=\Pr[X\le t].$$ It follows that $$ZX$$ is Gaussian random variable with mean $$0$$ and variance $$\sigma_X^2$$.	2023-03-30 06:06:17	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 35, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9414543509483337, "perplexity": 66.3044340689466}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296949097.61/warc/CC-MAIN-20230330035241-20230330065241-00588.warc.gz"}
https://coolgyan.org/calculators/simple-pendulum-calculator/	Simple Pendulum Calculator Formula : T = 2 Π Lg . Enter the unknown value as ‘ x ‘ Length(L) = m Acceleration of Gravity (g) = m/s2 Period (T) = s Simple Pendulum Calculator is a free online tool that displays the time period of a given simple pendulum. CoolGyan’S online simple pendulum calculator tool makes the calculation faster, and it shows the time period value in a fraction of seconds. How to Use the Simple Pendulum Calculator? The procedure to use the simple pendulum calculator is as follows: Step 1: Enter the length (L), acceleration of gravity (g) and x for the unknown in the respective input fields Step 2: Now click the button “Calculate the Unknown” to get the time period Step 3: Finally, the value of x will be displayed in the output field What is Meant by the Simple Pendulum? A simple pendulum can be defined as a point mass attached to a light inextensible string and suspended from fixed support. The vertical line passing through the fixed support is the mean position of the simple pendulum. Also, the vertical distance between the point of suspension and the center of mass of the suspended body (when it is in mean position) is called the length of the simple pendulum and is denoted by L. The time period of a simple pendulum is given by the formula: T = 2π/ω0 = 2π × √(L/g)	2021-07-30 20:42:12	{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.802362859249115, "perplexity": 604.0280198866434}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-31/segments/1627046153980.55/warc/CC-MAIN-20210730185206-20210730215206-00230.warc.gz"}
https://codegolf.meta.stackexchange.com/questions/2140/sandbox-for-proposed-challenges?page=3&tab=votes	# Sandbox for Proposed Challenges This "sandbox" is a place where Code Golf users can get feedback on prospective challenges they wish to post to main. This is useful because writing a clear and fully specified challenge on your first try can be difficult, and there is a much better chance of your challenge being well received if you post it in the sandbox first. Sandbox FAQ ## Posting Write your challenge just as you would when actually posting it, though you can optionally add a title at the top. You may also add some notes about specific things you would like to clarify before posting it. Other users will help you improve your challenge by rating and discussing it. When you think your challenge is ready for the public, go ahead and post it, and replace the post here with a link to the challenge and delete the sandbox post. ## Discussion The purpose of the sandbox is to give and receive feedback on posts. If you want to, feel free to give feedback to any posts you see here. Important things to comment about can include: • Parts of the challenge you found unclear • Problems that could make the challenge uninteresting or unfit for the site You don't need any qualifications to review sandbox posts. The target audience of most of these challenges is code golfers like you, so anything you find unclear will probably be unclear to others. If you think one of your posts needs more feedback, but it's been ignored, you can ask for feedback in The Nineteenth Byte. It's not only allowed, but highly recommended! It is recommended to leave your posts in the sandbox for at least several days, and until it receives upvotes and any feedback has been addressed. ## Other Search the sandbox / Browse your pending proposals The sandbox works best if you sort posts by active. To add an inline tag to a proposal use shortcut link syntax with a prefix: [tag:king-of-the-hill]. To search for posts with a certain tag, include the name in quotes: "king-of-the-hill". Get the Sandbox Viewer to view the sandbox more easily! # Capsa, a card game KotH! Capsa, known by many times, including the name Big Two in English, is a popular card game in East Asia and South East Asia, especially throughout China, Hong Kong, Indonesia, Macau, Malaysia, Singapore and Taiwan. There are many variations and house rules. The rules of this particular variation are thus: • This game will be played between exactly four bots. The cards are dealt between everyone, so that everyone has 13 cards. • Rank is ordered with 2s before As as follows: 2, A, K, Q, J, 10, 9, 8, 7, 6, 5, 4, 3, hence, the name Big Two. • Suits are ordered as follows: Spades, Hearts, Clubs, Diamonds • Ordering runs rank first then suit. Thus, 7 of Hearts > 7 of Diamonds > 6 of Hearts, the lowest card is the 3 of Diamonds, and the highest, the 2 of Spades You can play cards in sets of one, two or five cards, (singles, pairs, or five-card poker hands). Each set must always be bigger than the one before. With singles, play may proceed in this way: 1 3H, 2 4H, 3 5D, 4 5C, 1 10S, 2 Pass, 3 JD, 4 Pass, 1 2S, 3 Pass Pairs are ordered by the higher suit in the pair. 6C6H is of a lower rank than 6D6S. Play may proceed in this way: 3 3D3C, 4 6D6H, 1 6D6S, 2 JHJS, 3 Pass, 4 ADAC, 1 Pass, 2 Pass Poker hands are ordered in the following way, from lowest in rank to highest in rank: • Straight: Five cards that are consecutive in rank, e.g. 6H 7S 8D 9H 10C or JS QH KD AS 2C. Rank is determined by the highest card, with suit used as a tie-breaker. • Flush: Five cards with the same suit, e.g. 5H 7H 10H QH AH Rank is determined first by suit, then by highest card. • Full house: A three of a kind with a two of a kind, e.g. JD JS JH 3D 3S. Rank is determined by the triple, without regard for the pair. • Straight flush: Five cards that are consecutive in rank and are all of the same suit, e.g. 6H 7H 8H 9H 10H. Ranked the same as straights, with suit as a tie-breaker • Four-of-a-kind: Four cards of the same rank, with any 5th card, e.g. 9D 9C 9H 9S 4S This hand is known as the bomb. Wins any round of poker hands it is played in, unless someone else plays their own bomb. Any five-card hand that is higher than the previous five-card hand played is eligible. For example, you can play a full house on a straight. Rules of play: • At the beginning of the game, the cards are dealt between everyone, so that everyone has 13 cards. The first player is the one that holds the 3 of Diamonds and they must play this card first, whether singly, with another 3 in a pair, or in a poker hand. • A round begins with the first player playing a single card, a pair or a poker hand. Every other player either respond with the same number of cards (you must play singles on singles, never a pair or a poker hand), or that player passes for that round. • A player may pass even when they have a playable card, but they must pass if none of their cards are high enough in rank, that is, if they have no playable cards. For example, in a round of poker hands, if you only have a flush as a five-card set, but another player has already played a full house, you must pass, as your flush is too low. • The round ends when all but one player passes. That last player wins the round and starts the next round. • The game ends when one player has played all of their cards. Possible scoring systems: • Bots are judged by the number of games they win over (TBD) games that they play. So if we have seven bots to test, we'll play them until every bot has played more than (TBD) games. • Bots are judged by the number of cards they have left at the end of the game. (The winner will obviously have 0 cards at the end of the game). • "The cards are dealt between everyone, so that everyone has 13 cards" belongs in the "rules of play" section, even if that means having to repeat it. Dec 30 '15 at 11:09 • @PeterTaylor Done. If you have any comments on the controller linked at the bottom of this post, I'd love to hear them. Thanks Dec 30 '15 at 11:27 # Let's Play Unikong In honor of April Fool's day, we shall have an epic battle to see who can play Unikong best. Or, rather, whose program can play it best. ## Goal Write a program in any language to play the game Unikong. It should seek to try and score as high as possible. ## Scoring Whomever's program has the highest average score, over 10 games wins. ## Rules Standard rules apply. Additionally, your program must actually play the game, not change the score variable by some other means, and you can read the variables from the game to avoid trolls and downvotes, but not modify any variables to make it easier. I will run the tests myself and will use the first 10 runs to calculate the score. Please include any specific instructions needed to run your program. Notes: This would be my first question, so hopefully I got the format right. Are the rules clear enough? Do I need to clarify anything? • Do not require a video, run the programs yourself and see how far they get without your input. No pressing continue. Apr 1 '16 at 17:09 • Uh, will that link exist beyond today? If not then I think you'd need to find a way to replicate it (not sure about copyright here...). Anyway, I'm not really sure what counts as cheating. I think it will be hard to ban everything that's bad, so it might be better to write a controller that only allows looking at certain variables and only allows the basic player input. Apr 1 '16 at 17:09 • @FryAmTheEggman I've changed the link to something that will remain after April Fool's. Apr 1 '16 at 22:02 ## Specification In this challenge, your task is to reverse the lengths of an array-of-arrays, while keeping its concatenation intact. More explicitly, your input is an array of arrays of nonnegative integers, which you may assume to fit in the native int type of your language. The input may be an empty array or an array of empty arrays, or it may contain arrays of different lengths. You can take the input in any reasonable format. Your output shall be another array of arrays, again in any reasonable format. The concatenation of the output shall be equal to the concatenation of the input, so it contains the same integers in the same order. However, the sequence of lengths in the output shall be the reverse of that of the input. ## Example Consider the input array A = [[4,10],[0],[],[3,3,2],[1]] The concatenation of A is B = [4,10,0,3,3,2,1] and its length sequence is C = [2,1,0,3,1] The correct output is [[4],[10,0,3],[],[3],[2,1]] since it's the unique array with concatenation B and length sequence reverse(C). ## Rules and scoring You con write a full program or a function. The lowest byte count wins, and standard loopholes are disallowed. ## Test cases TODO: make more [] -> [] [[]] -> [[]] [[],[],[1]] -> [[1],[],[]] [[1,2],[4,5,6]] -> [[1,2,4],[5,6]] [[4,10],[0],[],[3,3,2],[1]] -> [[4],[10,0,3],[],[3],[2,1]] I'm debating whether I should guarantee that the input is non-empty, and/or only contains non-empty arrays. In some languages (like J), empty arrays make the challenge significantly harder, but on the other hand, I don't want it to be too easy either. • Just an FYI "reshape" operations pretty heavily trivialise this. I don't think it would be crazy to ban them, but it's probably fine with them also. Jun 7 '16 at 17:12 • I'd prefer the more general challenge of splitting an array the same way as another array of arrays with equally many elements. – xnor Jun 8 '16 at 0:28 • @FryAmTheEggman I'd prefer not to explicitly ban any builtins. If that leads to 3-byte Jelly answers, so be it. :P I also suspect Jelly would win anyway... Jun 8 '16 at 14:35 • @xnor That would be essentially be this challenge, but with only arrays-of-arrays. Do you think they would be different enough not to be duplicates? Jun 8 '16 at 14:39 • Even if the depth is fixed at 2, the shortest approach would most likely still be the be the same in many languages. It's a borderline dupe, and since it takes only one gold badge user to close a duplicate, it will probably end up closed. Jun 8 '16 at 17:23 • @Zgarb I'm not sure if I'd count it as a dupe, but I think reversed input vs general array wouldn't make much difference there. I expect most solutions would just work with the reversed input as if it were a general array. – xnor Jun 8 '16 at 20:50 • @xnor Reversed input has the added complexity that the same array must be used for content and shape. That's not a big problem for some languages, but I'd expect another approach to outgolf the pop/map approach with the added overhead of creating a second reference for the reversed array in, e.g., Python. I admit I haven't tried it yet though. Jun 9 '16 at 0:24 • I think what you call "concatenation" of the array is called "flattening". Oct 13 '16 at 18:24 # The Secret Handshake This is based off of this sandbox comment. This is a challenge of secrecy. The goal of this KOTH is to write a program that is capable of identifying itself amongst a crowd of other programs. In order to do this, you must develop a secret handshake which will be recognized only by other copies of the same program. A single game involves every program competing at once. There will be five instances of each program in the arena, and the winner will be the first program to correctly identify the four other copies. After a large number of games, the submission with the most victories will be the overall winner. ## The Gameflow: 1. Each program receives a number which tells the number of bots in the arena, which is five times the number of submissions. The bots are arranged in a circle, and they are each considered ID #0 from their own perspective, with N-1 being the maximum ID number. 2. Then the game cycle starts. At this point, each bot will be awaiting input. 1. Your bot will receive input consisting of an ID number and an optional message. • For example, 7 hi means that bot #7 said hi to you. • If the ID number is 0, then there will not be a message. This would occur if it is your turn but there is no message to receive. 2. Now, your bot is allowed to output a guess consisting of four ID numbers. • If those four numbers are the IDs of your teammates, then your team will win that game. • You will receive no confirmation of an incorrect guess. 3. Next, you must output a message to send. The message will be a destination ID number followed by up to 3* characters. Example messages: 7 4 w 12 #?Q. This number is subject to change. Larger messages make it harder to fake a secret handshake. I hope that a very short message forces people to use multi-step handshakes. It might also be interesting to limit it to 1-character messages. # Sandbox Notes Something that I haven't quite figured out is how the controller program will determine which bot gets to move each turn. I suppose it would be simplest go in order: Each bot has an "unread message queue" and receives/sends one message each turn. The bot immediately after you (ID #1) then moves next. Alternatively, there could be a set turn order which is unrelated to the ID order, simply to make it impossible for one bot to tell which bot moves next. • 1. I don't think it needs a turn order: the only problem with adding simultaneous moves would be the need to account for ties. (You are going to run it more than once to pick a winner, right?) 2. I would keep the guess per turn without penalty for guessing wrong. A good strategy will allow a team to win before anyone wins by blind guessing. Too harsh a penalty for bad guesses will bias the game too much in favour of the last person to update their bot so that it fakes the responses of other teams. Nov 11 '14 at 10:43 • The idea is quite amusing, but something tells me there must exist a not too hard to find optimal strategy to maximize the chances of winning. The fight would likely occur around sending fake messages to the competitors. Second problem I see is, the first contestants will be at a huge disadvantage since the new players will see exactly how their code works. So much for secrecy. I think the challenge would be more interesting if the code was kept secret, but this is hardly compatible with the spirit of this site. – user16991 Jan 24 '15 at 6:48 • @PhiNotPi are you still interested in this challenge? Would you be willing to allow me to implement this? Jul 29 '16 at 20:23 • @RohanJhunjhunwala Sure, you can implement this. I don't think I ever started on a controller for it, mainly because I don't think this challenge will be as fun in practice as in theory. Jul 30 '16 at 13:17 # Solve the Nonogram! It is time to embark on a perilous quest to defeat the British Intelligence. The aim of this challenge is to write the shortest code that will solve a Nonogram. # What is a Nonogram? The rules are simple. You have a grid of squares, which must be either filled in black or left blank. Beside each row of the grid are listed the lengths of the runs of black squares on that row. Above each column are listed the lengths of the runs of black squares in that column. Your aim is to find all black squares. In this puzzle type, the numbers are a form of discrete tomography that measures how many unbroken lines of filled-in squares there are in any given row or column. For example, a clue of "4 8 3" would mean there are sets of four, eight, and three filled squares, in that order, with at least one blank square between successive groups. [1][2] So the solution to the above Nonogram would be: # Implementation Details You can chose to represent the Nonogram however you would like and take it as an input in whatever way you deem fit for your language. Same goes for output. The aim of this challenge is to literally just get the job done; if you can solve the monogram with whatever output your program gives, that is valid. One caveat is you can't use an online solver :) You are, of course, free to use any language you want and since this is code golf, the entries will be sorted in the order: accuracy -> length of code -> speed. This problem is very algorithmically challenging in that there is no completely efficient solution to it and as such, you won't be penalized for not being able to solve larger ones, although your answer will be heavily rewarded if it is able to handle big cases (see bonus). As a benchmark, my solution works for up to roughly 50x50 within 5-10 mins. # Bonus I actually learnt about Nonograms from a cryptographic Christmas card released by the British Intelligence here. The first part was basically a massive 25x25 Nonogram. If your solution is able to solve this, you will get kudos :) To make your life easier in terms of data entry, I have provided how I represented the data for this specific puzzle for your free use. The first 25 lines are the row clues, followed by a '-' separator line, followed by 25 lines of the col clues, followed by a '#' separator line, and then a representation of the grid with the square clues filled in. 7 3 1 1 7 1 1 2 2 1 1 1 3 1 3 1 1 3 1 1 3 1 1 6 1 3 1 1 3 1 5 2 1 3 1 1 1 2 1 1 7 1 1 1 1 1 7 3 3 1 2 3 1 1 3 1 1 2 1 1 3 2 1 1 4 1 4 2 1 2 1 1 1 1 1 4 1 3 2 1 1 1 2 5 3 2 2 6 3 1 1 9 1 1 2 1 2 1 2 2 3 1 3 1 1 1 1 5 1 1 2 2 5 7 1 2 1 1 1 3 1 1 2 1 2 2 1 1 3 1 4 5 1 1 3 1 3 10 2 1 3 1 1 6 6 1 1 2 1 1 2 7 2 1 2 5 - 7 2 1 1 7 1 1 2 2 1 1 1 3 1 3 1 3 1 3 1 1 3 1 1 5 1 3 1 1 3 1 1 4 1 3 1 1 1 1 2 1 1 7 1 1 1 1 1 7 1 1 3 2 1 2 1 8 2 1 2 2 1 2 1 1 1 2 1 7 3 2 1 1 2 3 1 1 1 1 1 4 1 1 2 6 3 3 1 1 1 3 1 1 2 5 2 2 2 2 1 1 1 1 1 2 1 1 3 3 2 1 8 1 6 2 1 7 1 4 1 1 3 1 1 1 1 4 1 3 1 3 7 1 1 3 1 1 1 2 1 1 4 1 3 1 4 3 3 1 1 2 2 2 6 1 7 1 3 2 1 1 # 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 0 0 0 1 1 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 1 1 0 0 0 0 1 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 # There can be only 1! Your task is, given a positive integer n, to generate an expression that equals to the number n. The catch is: you're only allowed the number 1 in the output. The operators at your disposal are: • +, -, and / • sqrt (as s) • ceil and floor (as c and f respectively) • ! (factorial) • The factorial, in this case, only works for positive integers. You are also allowed to stack 1's together, so something like 11 is acceptable in the output. However, they count as the same amount of 1's that's in the number (so 11 counts as 2 1's). You must also include brackets in the output, so that the expression in the output, when executed through the order of operations, will result in the input. ## Examples: • Input = 24, one possible output = (1+1+1+1)! • Input = 11, one possible output = 11 • Input = 5, one possible output = c(s((1+1+1+1)!)) • The ceiling of the square root of 24 is 5. ## Rules: • You are guaranteed that the input is a positive integer from 1 to 2^31-1. • Your program must work for any positive integer up to 2^31-1, even if they are not tested. • Your program must finish processing all outputs for all numbers in the set in 1 hour. • The results for every run of the program must be exactly the same - also, no seeds. • You are not allowed to have imaginary numbers anywhere in the output (so no s(some negative number)). • You are also not allowed to have numbers larger than 2^31-1 anywhere in the output, even when they are sqrted or /ed (so no (((1+1+1)!)!)! or ((1+1+1+1)!)!). ## Set of Numbers: 945536, 16878234, 32608778, 42017515, 48950830, 51483452, 52970263, 54278649, 63636656, 78817406, 89918907, 90757642, 95364861, 102706605, 113965374, 122448605, 126594161, 148064959, 150735075, 154382918, 172057472, 192280850, 194713795, 207721209, 220946392, 225230299, 227043979, 241011012, 248906099, 249796314, 250546528, 258452706, 276862988, 277140688, 280158490, 286074562, 308946627, 310972897, 322612091, 324445400, 336060042, 346729632, 349428326, 352769482, 363039453, 363851029, 392168304, 401975104, 407890409, 407971913, 425780757, 459441559, 465592122, 475898732, 482826596, 484263150, 506235403, 548951531, 554295842, 580536366, 587051904, 588265985, 588298051, 590968352, 601194306, 607771869, 618578932, 626776380, 667919873, 681786366, 689854904, 692055400, 697665495, 711608194, 734027104, 750869335, 757710567, 759967747, 777616154, 830071127, 833809927, 835873060, 836438554, 836945593, 863728236, 864158514, 871273503, 881615667, 891619600, 897181691, 918159061, 920521050, 924502226, 929983535, 943162304, 950210939, 950214176, 962610357, 974842859, 988572832 (These are 100 random numbers from 1 to 1 billion.) ## Scoring System: Your score is determined like so: • Your program will be tested against the random numbers in the set. • You must provide the output generated using the numbers random numbers in the set (either inside your answer or as a pastebin link). • Your then have two "scores": A primary score and a secondary score. • Your primary score is (no. of 1's in output)(no. of operators in output). If your primary score is the lowest, you win. • Your secondary score is your byte-count, and is only used in the case of a tie-breaker - the person with the lowest byte-count wins. ## Meta: • Anything that I need to clear up? • Is this challenge a dupe? • Is implicit multiplication allowed? Sep 11 '16 at 1:23 • @LegionMammal978 No, it is not allowed. You must use . Sep 11 '16 at 1:43 • For the tiebreak, do parentheses contribute to the operator count? Sep 11 '16 at 1:49 • Does the code need to be deterministic (that is, do random algorithms need to set the PRNG seed to ensure the same results each time)? Sep 11 '16 at 1:51 • Surely at least one output must be hardcoded? Sep 11 '16 at 4:45 • Related, though I don't believe it's a dupe. Sep 11 '16 at 8:47 • code-challenge can't be used along with other scoring tags. Sep 11 '16 at 11:18 # Telegraphy Golf: Decode Baudot Code ## Background In 1870 Émile Baudot invented Baudot Code, a fixed-length character encoding for telegraphy. He designed the code to be entered from a manual keyboard with just five keys; two operated with the left hand and three with the right: The right index, middle and ring fingers operate the I, II, and III keys, respectively, and the left index and middle fingers operate IV and . (Henceforth I'll use their Western Arabic numerals, i.e. 1, through 5.) Characters are entered as chords. To enter the letter "C," for example, the operator presses the 1, 3, and 4 keys simultaneously, whereupon a rotating brush arm reads each key in sequence and transmits a current or, for keys not depressed, no current. The result is, in modern terms, a 5-bit least-significant-bit-first binary encoding, in which our example, "C," is encoded as 10110. ### 5 bits?? You might be thinking that 5 bits, which can express at most 32 unique symbols, isn't enough for even all of the English letters and numerals, to say nothing of punctuation. Baudot had a trick up his sleeve, though: His character set is actually two distinct sets: Letters and Figures, and he defined two special codes to switch between them. Letter Shift, which switches to Letters mode, is activated by pressing the 5 key alone (00001), and Figure Shift is activated with the 4 key (00010). ## Challenge Your challenge is to write a program or function that decodes Baudot Code transmissions. A real transmission would begin with some initialization bits, plus a start and stop bit before and after each character, but we're going to skip those and only worry about the 5 unique bits for each character. Input and output formats are discussed below. ### Baudot's Code There are two different versions of Baudot Code: Continental and U.K. We're going use the U.K. version, which doesn't include characters like "É" from Baudot's native French. We're also going to leave out all of the symbols in the U.K. version that aren't among the printable ASCII characters. You will only have to decode the characters in the table below, all of which are printable ASCII characters except the final three control characters that are explained below the table. The "Ltr" column shows the characters in Letter mode and "Fig" shows the Figure mode characters: Encoding Encoding Ltr Fig 12345 Ltr Fig 12345 --- --- -------- --- --- -------- A 1 10000 P + 11111 B 8 00110 Q / 10111 C 9 10110 R - 00111 D 0 11110 S 00101 E 2 01000 T 10101 F 01110 U 4 10100 G 7 01010 V ' 11101 H 11010 W ? 01101 I 01100 X 01001 J 6 10010 Y 3 00100 K ( 10011 Z : 11001 L = 11011 - . 10001 M ) 01011 ER ER 00011 N 01111 SP FS 00010 O 5 11100 LS SP 00001 / 11000 The last three rows in the right column are control characters: • ER is Erasure. Baudot's telegraphy machines would print an asterisk-like symbol for this character to tell the reader that the preceding character should be ignored, but we're going to be even nicer to the reader and actually omit (do not print) the preceding character. It acts the same in both Letter and Figure mode. • FS is Figure Shift. This switches the character set from Letters to Figures. If the decoder is already in Figure mode, FS is treated as a Space (ergo SP in the "Ltr" column). When the decoder is in Figure mode it stays in Figure mode until an LS character is received. • LS is Letter Shift. It switches the character set from Figures to Letters. If the decoder is already in Letter mode, LS is treated as a Space. When in Letter mode the decoder stays in Letter mode until an FS character is received. The decoder always starts in Letter mode. Here's an example with Figure Shift, Letter Shift, and Space: 01011 10000 00100 00001 00010 10000 11100 00001 10101 11010 M A Y LS/SP FS/SP 1 5 LS/SP T H This yields the message MAY 15TH. As you can see, the first 00001 (Letter Shift/Space) character acts as a space, because the decoder is already in Letter mode. The next character, 00010 (Figure Shift/Space) switches the decoder to Figure mode to print 15. Then 00001 appears again, but this time it acts as Letter Shift to put the decoder back in Letter mode. For your convenience, here are the characters in a format that's perhaps easier to digest in an editor, sorted by code: A,1,10000\|E,2,01000\|/,,11000\|Y,3,00100\|U,4,10100\|I,,01100\|O,5,11100\|SP,FS,00010\|J,6,10010\|G,7,01010\|H,,11010\|B,8,00110\|C,9,10110\|F,,01110\|D,0,11110\|LS,SP,00001\|-,.,10001\|X,,01001\|Z,:,11001\|S,,00101\|T,,10101\|W,?,01101\|V,',11101\|ER,ER,00011\|K,(,10011\|M,),01011\|L,=,11011\|R,-,00111\|Q,/,10111\|N,,01111\|P,+,11111 ### Input Input will be a string, array, or list of bits in least-significant-bit-first order. Each character will be represented by a quintet of 5 bits. Bits may be in any reasonable format, e.g. a binary string, an array of 0s and 1s, a string of "0" and "1" characters, a single very large number, etc., as long as it maps directly to the bits of the transmission. Every transmission will have at least one printable quintet and at most 255 quintets (printable or otherwise), i.e. 5–1,275 bits inclusive. The input can contain only the bits of the transmission, with two allowed exceptions: Any number of leading or trailing 0 bits and/or, for string input, a single trailing newline may be added to the transmission. Leading or trailing bits or characters cannot be added before or after each quintet, i.e. you cannot pad each quintet to 8 bits or separate quintets with any additional bits, e.g. "01111\n11100". Notes & edge cases 1. The transmission will contain only the characters in the "Ltr" and "Fig" columns in the table above. You will never receive e.g. 01110 in Figure mode, because it is absent from the "Fig" column. 2. It is assumed that the decoder will always be in Letter mode at the beginning of a transmission. However, the first character may be an FS character to switch to Figure mode immediately. 3. When the decoder is in Letter mode, it may receive an LS character, and when it is in Figure mode it may receive an FS character. In either event a Space character must be printed (see Output). 4. The ER character will never be the first character in a transmission, nor will it ever immediately follow an LS, FS, or another ER. 5. An FS character may immediately follow an LS character and vice versa. 6. Neither the LS nor FS character will be the last character in any transmission. 7. The / and - characters may be received in either Letter mode (codes 11000 and 10001, respectively) or Figure mode (10111 and 00111). ### Output Output may be in any reasonable format, the most reasonable being ASCII (or UTF-8, for which all of the represented characters are the same as ASCII). Please indicate in your answer if your output is in another encoding or format. Notes • The space character (see 3. above) should be an ASCII space (0x20) or your encoding's equivalent, i.e. what you get when you press the space bar. ## Winning This is code golf. The shortest code in bytes wins. ## Restrictions • Standard loopholes are forbidden. • Trailing spaces and/or a single trailing newline are allowed. Leading spaces or other characters (that are not part of the transmission) are disallowed. • You may not use any built-in or library functions that decode Baudot Code (or any of its descendants, e.g. Murray Code, ITA-1, etc.). ## Test Cases Input: 001101000010100111101110010101 Output: BAUDOT Input: 11010010001001100011110111101111100 Output: HELLO Input: 01011100000010000001000101000011100000011010111010 Output: MAY 15TH Input: 00010001000001000001011101110011100101010010110101010001111100101 Output: 32 FOOTSTEPS Input: 10110000110101011100111100001111011010000001101110 Output: GOLF Input: 000100011000001111100000100010110111001100010110010000111111 Output: 8D =( :P Input: 0000100001000010000100010001111011111011000011100010001 • This is quite similar to the Morse decoding question. The main thing it adds is the three control characters. It might be worth adding a cross-reference in a comment after posting the question. Sep 19 '16 at 21:56 • Can the transmission start with erasure? Sep 20 '16 at 1:00 • @Sp3000 No. Good catch. I had a note to that effect but apparently accidentally deleted it. Sep 20 '16 at 1:07 • Can an erasure follow a shift that is acting as a space? Sep 20 '16 at 14:16 • @Phlarx Nope. ER will never follow LS or FS. Sep 20 '16 at 14:17 • Thank you for your feedback, all. I've posted the challenge: codegolf.stackexchange.com/questions/94056/… Sep 21 '16 at 13:23 • Don't forget to take of this answer in the Sandbox. " When you think your challenge is ready for the public, go ahead and post it, replace the post here with a link to the challenge and delete it. " Sep 22 '16 at 19:29 my first post on here, be gentle ;) # Find all anagrams within a text Somehow I stumbled upon an implementation of a school assignment from about a year ago, and after having seen many amazing and mindblowing code-golf solutions on here, I thought it's time I bring my own challenge and see how much you guys can blow my mind again ;) ## The assignment Given a text, find all words that have at least one other word in the text as an anagram (case insensitive). Multiple occurrences of the same word are not counted. The output shall be grouped by words that are an anagram of each other. ## Rules • How you handle input/output is up to you. Function-parameters, file-io, standard in/out, whatever works the best for you. • You must be able handle any non-empty input as long as you don't run into language or memory limitations. • The output does not have any fixed formatting. That means you may put them each group at a line, or put them all at one line but use different delimiters, a 2d array, some other exotic data-structure your language of choice happens to have, as long as it makes reasonable sense, it is considered correct. (This means that for example if you are just writing a function, that function does not need to display the output, it could just provide it as a return-value.) Just keep in mind the requirement that the words that are an anagram of each other should be grouped together. • The order in which the output appears does not matter. That applies to the order of the groups as well as the order of the words within the group. • A group of only one word is invalid, since that fails the "have at least one other word in the text as an anagram" requirement. (just omit them from your output ^^) • Each word should only appear once in the output • All interpunction characters are stripped away from the word before checking for anagrams. That means that "it's" and "its" are the same word (and thus both are an anagram of "sit"). My sample program at the bottom uses http://en.cppreference.com/w/cpp/string/byte/ispunct as check if a character is an interpuntion character. If your language has such a method, you may use it. Otherwise take the characters from the default C locale as specified on there: !"#$%&'()+,-./:;<=>?@[\]^_{\|}~ • All other characters are part of the word and treated as is, so "a" and "á" are not the same. ## Example Given the following input text (the actual text I was given as example by school :P, no idea where this text is coming from...) Parts of the world have sunlight for close to 24 hours during summer. Dan went to the north pole to lead an expedition during summer. He had a strap on his head to identify himself as the leader. Dan had to deal with the sun never going down for 42 consecutive days and his leadership strap soon became a blindfold. He wondered what kind of traps lay ahead of him. the following output would be correct: • 24, 42 • deal, lead • and, dan • parts, strap, traps Or this would also be correct: 24, 42 \| deal, lead \| and, dan \| parts, strap, traps This one would not 24, 42 , deal, lead , and, dan , parts, strap, traps (since the groups are not obvious) ## My own (non golfed) version to check The is the exact program I submitted to school back then. You may use it to check your own results. Added bonus: If it happens to be that this program has a bug (I haven't found them yet) your submission is allowed to have it as well, since it is used to check the result. (In that case you are of course not required to have said bugs) #include <string> #include <iostream> #include <fstream> #include <map> #include <set> #include <algorithm> #include <cstring> void stringRemoveInterpunction(std::string& string); void stringToLower(std::string& string); std::string stringToAnagramIdentifier(std::string word); /* * due to use of std::ispunct and std::tolower it may not work for text with non-ascii characters??! / int main(int argc, char argv[]) { if (argc < 2) { std::cerr << "usage: " << argv[0] << " <filename>" << std::endl; return 1; } std::ifstream fileStream(argv[1]); if (!fileStream) { std::cerr << "Could not open file " << argv[1] << std::endl; return 2; } // map to store the anagrams, key is so called "anagram identifier", value is a list of the words. std::map<std::string, std::set<std::string>> anagrams; // read words separated by whitespace from the file for (std::string word; fileStream >> word;) { // remove interpunction & convert to lowercase, since casing should be ignored stringRemoveInterpunction(word); stringToLower(word); // add to anagrams-store anagrams[stringToAnagramIdentifier(word)].insert(word); } // display all the anagrams for (auto anagram : anagrams) { // skip entries which contains only one item, no anagrams found if (anagram.second.size() <= 1) { continue; } // output a comma-separated list of the anagrams auto anagramIterator = anagram.second.begin(); std::cout << anagramIterator++; while (anagramIterator != anagram.second.end()) { std::cout << ", " << anagramIterator++; } std::cout << std::endl; } return 0; } void stringRemoveInterpunction(std::string& string) { string.erase(std::remove_if(string.begin(), string.end(), std::ptr_fun<int, int>(&ispunct)), string.end()); } void stringToLower(std::string& string) { std::transform(string.begin(), string.end(), string.begin(), std::ptr_fun<int, int>(&std::tolower)); } std::string stringToAnagramIdentifier(std::string word) { // sort the characters std::sort(word.begin(), word.end()); return word; } # Sandbox Questions • Do i need to add other tags, or is just code-golf enough? • I'm not completely sure about the upper-limit of the input text. My idea was that the code should be able to handle any size input as long as its within the memory-limits of the language. Like you don't have to write "memory optimal code" or something, but also shouldn't asume it is smaller than X. I could also just pick an upper limit of "1 kilobyte" or something to avoid any uncertainty about the requirements I think this is fine now as it is. • Someone in the comments below asked how to handle special characters like$?() so I took a look at how my "check program" handled that and it strips them away before doing the anagram check. So I added a rule for that, but while writing that I felt it makes it needlessly complicated and I'm considering ditching that rule and altering my check-program to reflect that (but then I cant claim its the exact same anymore :( ) • Any other parts that are not clear? • What happens if there are 3 that are anagrams, such as eat ate tea? Are they all printed in one line / group? Dec 19 '16 at 17:27 • @Flp.Tkc yes, like "parts, strap, traps" in the example Dec 19 '16 at 17:29 • I suggest you remove the grouping rule, since it's not really the interesting part of the challenge. It just adds code and limits the solutions. It's apparent which of the words that are together anyway. I do suggest they have to be grouped though, but without the need for delimiters. Dec 20 '16 at 23:24 • I think the memory rule is fine. I'm quite sure people will write a script that in theory would work for any input length if it wasn't for language or memory limitations. I also suggest you guarantee at least one character in the input. Otherwise people would need to add code just to handle empty input and that's not the interesting part of the challenge. Dec 20 '16 at 23:29 • You should add rules regarding special characters. Are it's, sit and its anagrams? What about hyphens? Can there be any special characters such as $?() etc? How are they treated? Dec 20 '16 at 23:31 • @StewieGriffin The grouping thing is a natural result of my solution to the problem back then, and imo its a fun part of the challenge, so I'm not totally sure about removing it. About the memory limit: I'm totally fine with the guarantee of at least one character if you think empty input needs special handling (i didnt think it would, but I dont really care :P) And yes, you are correct about special characters. I'd say "it's" is two words: "it" and "s". All special characters are ignored, so essentially regarded as whitespace between words. I'll add that Dec 20 '16 at 23:35 • welp, turns out I handled special characters differently, I strip them away from the word, so "it's" becomes "its", guess that'll be the rule then since I want to keep as close as possible to the original program Dec 20 '16 at 23:48 • @StewieGriffin I added a rule about special character handling, but I'm not completely sure about it, thoughts? (see my added "sandbox question") Dec 21 '16 at 0:05 • parsing and anagrams are relevant tags, maybe strings too, but I'm not sure about that one. Dec 21 '16 at 7:43 • !"#$%&'()+,-./:;<=>?@[]^_{\|}~ are all the non-alphanumeric ASCII-characters, except \ . I suggest you include the missing two symbols in the list to ignore, and say: "All non-alphanumeric characters (except spaces and newlines) must be trimmed away. So, it's and its are the same word." (You might want to rephrase that since my English isn't perfect, but something along those lines. If this is the rule then R2D2 and dr.22 will be anagrams, I'm not sure if that the desired behavior..? Dec 21 '16 at 10:17 • PS! I don't mean to be difficult, it's just that in my experience you'll get these questions sooner or later. So it's a good thing to sort it out while it's still in the Sandbox :) Dec 21 '16 at 10:20 • @StewieGriffin those two characters should have been included, look at the list from en.cppreference.com/w/cpp/string/byte/ispunct where I got it from, somehow I messed up the copy-paste :( And about "what is desired behaviour?" Im honestly not sure, I never really thought about it before you mentioned it and then I looked at how my sample program handled it. I remember putting in the ispunct-trim for stuff like commas after a word, im not sure about characters within the word... Dec 21 '16 at 10:26 • You can say that the input will not have any special characters except ' and -. Those two must be trimmed away. I think you must include those two, since these can be found in many texts. Dec 21 '16 at 10:34 # PPCG Handwriting OCR code-challengetest-battery (insert logo here once I make it) Given an image consisting of handwritten text, output the text that is written. The image of the handwritten text will be generated by taking characters from one or more handwriting samples given by PPCG users. ## Rules • You may not hardcode your program to only recognize the samples in the corpus. • There will be sufficient spacing between characters to avoid ambiguity. • Only ASCII alphanumeric characters (those matching the regex [A-Za-z0-9], i.e. uppercase and lowercase English letters and digits) will be present in the input. • Inputs will be formed by concatenating individual characters from the handwriting samples. • The test cases used for scoring may be modified at any point if I feel it is necessary to do so. Reasons may include but are not limited to: needing more test cases to have a single winner, removing problematic test cases, and fixing errors in test cases. Aside from the above considerations, there are no guarantees on the appearance of the handwriting, as these are actual handwriting samples and thus can have drastic variances. ## Score Your score will be the number of test cases that are correctly recognized, divided by the total number of test cases. The highest score wins. In the event of a tie, the first submission to reach the high score wins. Additional test cases may be added to break ties. ## Handwriting Samples This Imgur album contains the handwriting samples, as well as the names of the users who contributed them. I've made a chat room for submitting handwriting samples. The more samples I get, the better this challenge will be, so please take a few minutes and submit a sample! • You may not hardcode your program to only recognize the samples in the corpus.: Can we tailor our code to be better for the samples than other inputs though? Aug 2 '17 at 16:08 • The test cases used for scoring may be modified at any point: Will current answers be modified? Aug 2 '17 at 16:09 • @TheLethalCoder 1) No. 2) Answers will be run on the test cases any time the answer or the test cases change, and the scores will be updated accordingly. – user45941 Aug 2 '17 at 16:11 • Hadn't seen that loophole before :) Aug 2 '17 at 16:13 • Can I nitpick again and say handwriting OCR is called ICR? :P Aug 3 '17 at 9:06 • Can you add an example input as well? Aug 3 '17 at 9:07 • @TheLethalCoder 1) ICR is when a program tries to learn what handwriting looks like via machine learning. OCR is just parsing written/text input into data. This is OCR, not ICR. 2) I'll add some examples once I get more samples and finish writing my generator. – user45941 Aug 3 '17 at 9:08 • Can you assume a minimum height for the characters? Aug 3 '17 at 9:28 • @TheLethalCoder Aside from the above considerations, there are no guarantees on the appearance of the handwriting, as these are actual handwriting samples and thus can have drastic variances. – user45941 Aug 3 '17 at 9:29 • To be honest with no restrictions this is going to be very hard, Do you lose points for returning extra information? Assume the input Hello if I return H.e.l.l.o. is that 100% for that test case or do I lose and get something like 50%? Aug 3 '17 at 9:31 • And does the output have to be in the correct order? Aug 3 '17 at 9:58 • @TheLethalCoder It's all or nothing. Getting 100% on a test battery challenge should be hard. – user45941 Aug 3 '17 at 9:58 • Please update imgur album Sep 18 '17 at 4:28 • @Pavel I will once I finish cutting up the images. My free time has been limited these past few weeks. – user45941 Sep 18 '17 at 4:29 ## 2 Spooky 4 Me In terms of halloween, some things are just too spooky for me... Feel like we need some serious doots from skeletons to fuel our hallowed weens. So, in the spirit of that end, print the following, exactly as it is shown, if and only if the input does not equal "DOOT" (in all caps ONLY): _.---._ .' '. :) (: \ (@) (@) / \ A / ) ( \"""""/ '._.' .=. .---._.-.=.-._.---. / ':-(_.-: :-._)-:' \ / /' (__.-: :-.__) '\ \ / / (___.-' '-.___) \ \ / / (___.-'^'-.___) \ \ / / (___.-'='-.___) \ \ / / (____.'='.____) \ \ / / (___.'='.___) \ \ (_.: '---'.=.'---' :._) :\|\| __ _.=._ __ \|\|: :\|\| ( '.-.=.-.' ) \|\|: :\|\| \ '.=.' / \|\|: :\|\| \ .=. / \|\|: :\|\| .-'.'-._.-'.'-. \|\|: .:::\ ( ,): O O :(, ) /:::. \|\|\|\| ' / /''--'--''\ \ ' \|\|\|\| '''' / / \ \ '''' / / \ \ / / \ \ / / \ \ / / \ \ / / \ \ /.' '.\ (_)' '(_) \\. .// \\. .// \\. .// \\. .// \\. .// \\. .// \\. .// ///) (\\\ ,///' '\\\, ///' '\\\ ""' '"" However, if the input DOES equal "DOOT", in all caps only, print this instead: _.---._ .' '. :) (: \ (@) (@) / \ A / ) ( \"""""/ '._.' ' '''' _''\| .=. @=====*===::_ \| .---._.-.=.-._.---. (( \-@\|_) )) .\| / ':-(_.-: :-._)-:' \ ]--------'" / /' (__.-: :-.__) '\ \ \|\|: / / (___.-' '-.___) \ \ \|\|: / / (___.-'^'-.___) \ \ \|\|: / / (___.-'='-.___) \ \\|\|: / / (____.'='.____) \ \|\|: / / (___.'='.___) \\|\|: (_.: '---'.=.'---' :._) :\|\| __ _.=._ __ :\|\| ( '.-.=.-.' ) :\|\| \ '.=.' / :\|\| \ .=. / :\|\| .-'.'-._.-'.'-. .:::\ ( ,): O O :(, ) \|\|\|\| ' / /''--'--''\ \ '''' / / \ \ / / \ \ / / \ \ / / \ \ / / \ \ / / \ \ /.' '.\ (_)' '(_) \\. .// \\. .// \\. .// \\. .// \\. .// \\. .// \\. .// ///) (\\\ ,///' '\\\, ///' '\\\ ""' '"" # Rules • Trailing newlines and spaces are allowed. • The design is horizontally symmetric, if you find inconsistencies let me know. Doot it up, and enjoy! (Yes, I'm going to make it more official when posting on the actual SE) • What about this challenge ensures that the same old techniques won't be the best ones (i.e. that it adds value to the site)? Oct 13 '17 at 23:01 • @PeterTaylor worse? Better? 10x worse? 10x better? I don't really know what makes the challenge unique beyond a formal proof that it is, but if the users enjoy it; why not allow it... current event challenges attract new users. Oct 16 '17 at 23:03 # Check equation proofs in a ring The recent to prove that (-a)×(-a) = a×a attracted a number of faulty submissions, because there wasn't an easy way to verify the proofs. So, let's write some proof checkers. ### Input Your program should take a sequence of strings representing expressions in a ring. Valid expressions consist of: • Single-lowercase-letter variables (a to z) • Two constants: the additive identity 0 and multiplicative identity 1 • Compound expressions: (X+Y), (XY) and (-X), where X and Y stand for subexpressions. (The parentheses must always be present, and there must be no whitespace.) • All strings except the first and last represent valid expressions. • Each expression (after the first) can be obtained from the preceding expression, by substituting one of the ring axioms in the expression exactly once. Output truthy if these conditions are met. Otherwise, output falsey. You may assume that the first and last strings in the input are valid expressions. But your program must check the intermediate strings. ### The ring axioms For this challenge, use the following substitution rules (and do not use any others). Substitutions can go both left-to-right and right-to-left. 1. (X+(Y+Z)) = ((X+Y)+Z) 2. (X+0) = X 3. (X+(-X)) = 0 4. (X+Y) = (Y+X) 5. (X(YZ)) = ((XY)Z) 6. (X1) = X 7. (1X) = X 8. (X(Y+Z)) = ((XY)+(XZ)) 9. ((X+Y)Z) = ((XZ)+(YZ)) ## Scoring Proof checkers are traditionally small, so that people can review them easily. Therefore, your program should be written in as few bytes as possible. Is the input format fair for most languages and approaches? Usually, code-golf problems should not require input validation. However, I thought it would be appropriate behaviour for a proof checker. I think the current spec still accommodates regex-based solutions. The format for the original challenge also listed which axiom was used for each step. I could include this but I doubt that it improves the challenge much. ## Test cases ### Valid proofs (0+a) (a+0) a (a(-1)) ((a(-1))+0) ((a(-1))+(a+(-a))) (((a(-1))+a)+(-a)) (((a(-1))+(a1))+(-a)) ((a((-1)+1))+(-a)) ((a(1+(-1)))+(-a)) ((a0)+(-a)) ### Invalid proofs These proofs are missing intermediate steps. (0+a) a (-0) ((-0)+0) 0 ((a0)+(-a)) (0+(-a)) This is simply untrue, so no proof should ever be accepted. (ab) (ba) ### Invalid expressions Your proof checker should reject if these appear partway through a proof. a+b (a+-b) (a + b) 1+ 42 • You should include the ring axioms to make the post self-contained. Oct 18 '17 at 16:23 At the ASCII Stock Exchange, each character has a price. If a character is used more often, its price rises, otherwise the price decreases over time. Initially, each character has price 10. [Meta: Is this too low/high?] After each answer, the prices change as follows: • Each character that is not used in the answer's code has its price decreased by one, except when the price is already one, in which case it stays one. • If a character is used n-times, then its price increases by n. We define the score of a piece of code as the sum of the prices of its characters. ### Example For the sake of simplicity, we only consider characters A, B and C for this example. The challenge itself works with all bytes from \0 to \255. Initially, we have the following prices: A -> 10, B -> 10, C -> 10 If the code of the first answer is BAAA, then is has a score of 40 (computed by taking the previous character prices into account) and the prices change to A -> 13, B -> 11, C -> 9 If the next answer is CC, it has a score of 18 and the prices are updated to A -> 12, B -> 10, C -> 11 Your objective is to write a program or function which calculates the score of a given piece of code in dependence of a list of previous answers which all influence the initial prices in the way described above. The goal is to do so while minimizing the submission's score itself in the context of this challenge, that is your submission should be able to calculate its own score by taking a list of all previous submissions and its own source code as input. The answer with the lowest score in each language wins. ## Rules • You may take a list of strings and a string as input, or require that the string to be scored is the first/last element of the list, or any other reasonable input format. • You may not answer twice in a row. • If an answer in language X has already been posted, you may only post another answer in language X if your submission achieves a lower score than the previous answer and the code is not identical. • For this challenge only major releases of languages are considered their own language (e.g. Java 7 vs. Java 8). If there already is an answer in version A of a language and you have an answer in version B of the language and are in doubt whether version B is different enough from version A to be treated as different language, make sure that your code is not valid in version A. To avoid having to copy all previous answers in order to calculate your submission's score, the chain will maintain a score calculator on TIO. Click on the link to the calculator in the previous answer and enter your code into the input field to calculate its score. Then add your code as a command-line argument, generate a new link and include it in your answer for the next submission. If you wrote answer number 42 in Haskell with a score of 100, please format it as <code> ## Test Cases These test cases are in the format list of strings, string to score -> result. [], "BAAA" -> 40 ["BAAA"], "CC" -> 18 ["abc"], "abc" -> 33 ["ab12", "aa22", "31a"], "ac23" -> 42 ["a","b","c","d","e","f","g","h","i","j"], "123" -> 3 ## Meta • Any idea what could be a good initial price? I picked the number 10 rather arbitrarily. • Letting the cost of unused characters decrease until 0 might lead to score 0 answers. Do you think this is a problem and the minimum cost should be 1? Minimum changed to 1. • I'm unsure what range of characters is a sensible choice. Limiting answers to printable ASCII plus white space would make things easier but also exclude a lot of languages. Another possibility would be to allow the whole byte range from \0 to \255. Then also golfing languages could participate, albeit to score them they would need to be converted to their byte form which usually contains a lot of unprintable characters. The score calculator is able to handle unprintables, but I don't know how to insert them into the text fields on TIO. All bytes from \0 to '\255' are allowed. • Is the winning criterion suitable for answer chaining? • A leader board snippet would be nice, but I don't know how to modify the existing ones. If someone could provide such a snippet, I would be very grateful. • Relevant tags? • I think decreasing the scores to 1 is more sensible, because otherwise I could write a submission using all (or even just one of) the zero-score characters a billion times (say, in a comment), keeping a "minimum" score, but then subsequent submissions would have completely absurd scores, and it'd be easy to outgolf later, leading to a boring answer chain. Oct 2 '17 at 18:08 • You may want to correct "enter your code into the input field to calculate its score. Then add your code as a command-line argument," Oct 2 '17 at 19:55 • I think allowing all 256 bytes is a good idea because languages like Jelly, 05AB1E, etc. will more than likely use more than the printable ASCII chars. Oct 2 '17 at 20:30 • "While in principal Python 2.7.14 and Python 2.7.13 are different languages according to the site rules, I recommend to avoid using the fact that it is technically allowed as an excuse to post boring answers." Recommendations do not work, especially when they're as imprecise as this. (Are you "recommending" not treating Python 2 and Python 3 as separate languages? I have no idea). If you want to ban boring exploitation of the convention on different interpreters, ban it outright, but give a clear definition of how lines should be drawn. Oct 2 '17 at 21:21 Note: If you are seeing this first, you might want to sort by active. ## Unhappy numbers ascii art Draw a square (or a rectangle as close to a square as possible) that represents the cycle of an unhappy number. [ short description of unhappy numbers here + example ] [ square formating rules ] Unhappy integer. ASCII art. ### Example Input: 4 Output: 4 - 16 - 37 20 58 42 - 145 - 89 • The cycle of an unhappy number is a constant. – J B Mar 7 '11 at 23:42 • @JB: thanks, I will rephrase the question. I didn't mean the 4 cycle. (Why did I chose 4 as an example? :/ ) Mar 8 '11 at 4:56 • @Eelvex This is an nice challenge. Are you still interested in finishing it? Feb 7 '16 at 19:06 • This challenge proposal has been inactive for over a month. I would like to take ownership of the challenge and make it ready for posting. Please let me know within the next 2 weeks if you have any objections and would still like to finish and post this challenge yourself. Feb 10 '16 at 0:19 • @JAtkin, no objections. Feb 12 '16 at 11:20 • If this was posted, can you please delete it? Aug 17 '17 at 16:58 # Tower Builder king-of-the-hill The king is coming to visit! Your city is competing with another neighboring city to attract his attention. Each player owns their own tower within a city full of other players. Each turn you place 1 blocks on your tower or another tower in your city. After 100 turns, the king will only visit the city with the highest tower. If he visits your city, then you will gain points equal to the number of blocks in your personal tower. In the case of a tie, neither city is visited. Games include all players, and each game will randomly arrange players into cities. Your score is the total score across all games. • Player identifiers are randomly generated at the start of the tournament, but are consistent from game to game. • In addition, contrary to past KoTHs: I allow saving state from game to game (but not between tournaments) • You will always have complete information, including: • The size of everybody's towers (including the other city) • The actions players have taken • The current score of all players • How is always playing on the tallest tower in your city a bad strategy? Maybe the other city will do the same, but you won't lose. Jul 22 '17 at 22:47 • @aschepler because the number of points you get is height of your personal tower. If you only build on another's tower, and never your own, you'll never get any points. Jul 23 '17 at 1:55 • @NathanMerrill it took me a few minutes to understand what you even meant by that. you probably need to make this more clear Aug 7 '17 at 0:24 • If the points are only awarded to the player who is visited by the king, it would seem that putting 99 blocks on the city tower and 1 block on your personal tower guarantees victory, no? Jan 30 '18 at 17:22 • @AdmBorkBork there's no singular "city tower". Furthermore, you are still competing against the players in your city. Your score is your total score across many games. Jan 30 '18 at 17:40 • Oh, the part that I missed was that multiple players share a city. I think that should be made more clear because that will drastically change strategy. I read it as you have a personal tower and everyone also gets their own city with a separate tower, and it's only the city's tower that matters for the king. Jan 30 '18 at 17:45 • This needs some clarification about arrangement. The game will randomly arrange players into one of () cities, or a city with () others in it – pfg Jan 30 '18 at 19:10 • @pfg It'll definitely be one of () cities, but I'm not sure how many. My intuition says "2", but I can't give a solid reason. Any suggestions? Jan 30 '18 at 19:21 • 2 or 3 per city or total cities? 3 seems good per city because with just two the strategy is much easier @Nathan Merrill – pfg Jan 30 '18 at 19:23 • Hmmm...I'm thinking 100 players in 2 cities, and allow for duplicate entries. Jan 30 '18 at 19:28 • It seems to me that if you allow saving state you can simplify the last point to "You will be notified of the actions of every player" and let those who want to analyse it calculate the sizes of the towers, scores, etc. Normally I would favour supplying calculated information to save the players from re-implementing the game logic, but here the logic is simpler than the serialisation would be. Jan 31 '18 at 8:43 • @PeterTaylor I already provide complete information about the current game (including its history). The difference here is that you are allowed to persist information from game to game, meaning you can remember players who are antagonistic. Jan 31 '18 at 15:04 • My point is that the communication between the server and the bots would be simpler if you didn't provide that information, and you don't need to provide it because it can be calculated easily. Jan 31 '18 at 15:07 # Chess ASCII Art, Knight In honor of the world chess championship, in the shortest possible program, output the following ASCII art piece ,...., ,::::::< ,::/^\". ,::/, e. ,::; \| '. ,::\| \___,-. c) ;::\| \ '-' ;::\| \ ;::\| _.=\ ;:\|.= _.=\ '\|_.= __\ \_..== / .'.___.-'. / \ ('--......--') /'--......--'\ "--......--" This is a code-golf challenge • You might want to make sure all the lines are aligned properly (they could be fine, since I'm on mobile and I know it can display differently, but it looks bent to me). Nov 23 '18 at 19:59 • you're right, it was a little bent, I've reformatted it Nov 23 '18 at 20:17 • Seems straightforward enough Nov 24 '18 at 1:23 • lol, akin to image compression of pixel art in a very specific case :) I like the idea. Nov 24 '18 at 22:22 • The very worst is approx 145 bytes + "verbatim output this". Be fun to see much better ones :) Nov 24 '18 at 22:33 • Isn't the World Chess Championship already over? According to google it ended nov. 28th. ;) Did you forgot to post it? Dec 3 '18 at 9:08 • Yeah, I made this post on November 23rd, cross-posting from sandbox eventually slipped my mind Dec 3 '18 at 14:42 • @Thaufeki You could still post it, or are you going to wait a year? ;) Dec 4 '18 at 10:22 • I would have to wait two, next one isn't until 2020! I'll post it now Dec 4 '18 at 14:43 # Is this checkmate? Input A chess position in FEN format. You can assume the input is a valid chess position. Output Two distinct consistent outputs for checkmate or not. Examples 8/8/8/8/8/5BKN/8/7k b - - 93 47 Mate • I suggest wording similar to “two distinct consistent outputs for checkmate or not” Mar 12 '19 at 11:52 • @Quintec Thanks. – user9207 Mar 12 '19 at 11:58 • I'd recommend keeping it all self-contained and having a description for the FFN format, as well as a few more test cases – Jo King Mod Mar 12 '19 at 21:41 • Is restricted to the FFN format a part of the challenge? Why not allow it in any reasonable format? – tsh Mar 13 '19 at 5:47 • @Quintec why not just say truthy or falsey? – qwr Jun 20 '19 at 14:48 • I suggest removing the restriction to FEN format, as it doesn't really add anything to the challenge, and specifying that the output be a truthy or falsey value as that is the usual spec for these types of challenges. Aug 26 '19 at 13:28 ## Lexicographically earliest valid UTF-8 byte sequence permutation There are currently 1,114,112 possible Unicode characters (code points). Each character has a unique valid byte sequence in the UTF-8 encoding. Different characters have different length encodings: • ASCII characters have a 1-byte encoding 00-7F. • The next 1920 characters have a 2-byte encoding C2 80-DF BF. • The rest of the BMP has a 3-byte encoding E0 A0 80-ED 9F BF and EE 80 80-EF BF BF. • The other planes have a 4-byte encoding F0 90 80 80-F4 8F BF BF. It's possible for two strings (specific non-normalised sequences of Unicode code points) of Unicode to have byte sequences that are permutations of each other in a number of ways: • One string could simply be a permutation of the other at the Unicode level, e.g. ab (61 62) and ba (62 61). • UTF-8 continuation bytes could be switched between two characters, e.g. ¡â (C2 A1 C3 A2) and ¢á (C2 A2 C3 A1). • UTF-8 continuation bytes could be switched within a character, e.g. ᴵ (E1 B4 B5) and ᵴ (E1 B5 B4). For this challenge I would like you to write a program or function that finds the string whose UTF-8 byte sequence is lexicographically earliest of all such sequences that are permutations of the UTF-8 byte sequence of a given Unicode string. For example, if your input is ᵴ¢ába (E1 B5 B4 C2 A2 C3 A1 62 61) your output would be ab¡âᴵ (61 62 C2 A1 C3 A2 E1 B4 B5). Note however that some byte sequences are not valid UTF-8 (e.g. E0 80 A0 which is an overlong encoding for a space) so you need to take care to avoid these. It would be helpful if your "Try It Online" or similar link includes a footer that helps demonstrate the correctness of your output, where this is not obvious from the I/O format or code. This is , so the shortest program or function that breaks no standard loopholes wins! • For test cases, it will probably be a good idea to provide both strings and hex since I'd guess many languages will have to try both. Also this probably needs at least a link to an explanation of UTF8 continuation bytes. Jun 29 '19 at 16:57 • "permutation of its canonical" should be "permutation of the input's canonical". Jun 29 '19 at 17:00 • @FryAmTheEggman ... but I provided the hex? I'm not sure what I'm missing... – Neil Jun 29 '19 at 23:24 • @EriktheOutgolfer How about "I would like you to write a program or function that returns the Unicode string whose canonical UTF-8 byte sequence is the lexicographically earliest of all such sequences that are permutations of the canonical UTF-8 byte sequence of a given Unicode string"? – Neil Jun 29 '19 at 23:27 • I meant in the example that you actually had, and presumably some number of test cases. I only mentioned it because I thought it was odd that you did it in the explanation but not the example. Jun 29 '19 at 23:54 • @Neil Looks good. :) Jun 30 '19 at 16:34 • I'm wary of the use of the word "canonical" in this question, because it raises issues in my mind about normalisation of Unicode strings. I think that the intended challenge is really about byte arrays with constraints on the most significant bits, and I think it would be better to make that explicit (and to make the constraints explicit). Jul 2 '19 at 11:16 • @PeterTaylor There is that, but I wanted to exclude sequences such as E0 80 A0. – Neil Jul 2 '19 at 13:23 • @Neil, I think there's a miscommunication here. I'm saying that instead of talking about Unicode strings the question should explicitly state the FSS-UTF constraints on byte sequences, and maybe rule out encoding UTF-16 surrogate codepoints and codepoints greater than 0x10FFFF. Jul 2 '19 at 13:48 • @PeterTaylor OK but I really wanted this to be a string question rather than a byte sequence question... – Neil Jul 2 '19 at 16:20 • The problem then is dealing with the lexicographically first rearrangement of C3 A9 (é in normal form C) being 65 CC 81 (é in normal form D). Jul 2 '19 at 16:29 • @PeterTaylor Is that possible to do just by permuting the byte sequence? – Neil Jul 2 '19 at 16:34 • No, and that's why I'm arguing that the question should be phrased in terms of byte sequences rather than strings. Jul 2 '19 at 17:40 • There’s a mistake in the example: á is C3A1 and ¡ is C2A1. Good challenge. From the sound of it I/O will be flexible; this seems sensible since it keeps it open to more languages. Jul 2 '19 at 17:54 • @PeterTaylor I wanted it to be clear that these byte sequences must be a valid UTF-8 encoding of a Unicode string. I've tried rewriting the question a bit... – Neil Jul 2 '19 at 23:55 • Can you add some test cases? Jul 27 '19 at 14:35 • @MilkyWay90 Ok I added some – Wheat Witch Mod Jul 27 '19 at 14:47 • Okay, I give /support (also, you may want to add disallowing standard loopholes and using any default io method to finish it up) Jul 27 '19 at 14:48 • @MilkyWay90 Those are already standard I am not going to be making my post any more cluttered with stuff that adds nothing. – Wheat Witch Mod Jul 27 '19 at 14:53 # Move arrows along a contour Posted here • I've edited in a question since it needs 2D for clarity. Jul 26 '19 at 8:48 • @Adám Only the arrows move. "+-\|" always stay in place, or are "hidden" behind an arrow. So, you second example is correct (I deleted the first one) Jul 26 '19 at 9:02 • Continuing @Adám's question: Are shapes always separated by at least one space, or can they be next to each other like ++++\n++++, and we have to determine if it's a ++++\n++++ or +--+\n+--+ based on the directions the arrows are facing? I.e. is this a possible/valid input, and are those outputs correct? Jul 26 '19 at 11:26 • @GalenIvanov I didn't have anything but arrows move, and not it doesn't follow, because you couldn't tell what was behind the arrows, -s or +s which is what would make the two possible answers. Jul 26 '19 at 11:28 • @KevinCruijssen Example 2 has two adjacent shapes with no (vertical) spacing. Jul 26 '19 at 11:29 • @Adám Ah, you're right. As for your question however, you'd still know +<<+\n+>>+ is +--+\n+--+ due to the directions of the arrows in combination with the rule "when an arrow is on a corner, it keeps its current direction and changes it only after the turn is taken". See the pastebin in my previous comment for some test cases where you do know it's ++++\n++++ instead, because of the arrow directions. +^<+\n+>v+ will be two ++++\n++++ boxes, but +<<+\n+>>+ will be one +--+\n+--+ box. Jul 26 '19 at 11:37 • @KevinCruijssen Good point, but OP actually never says that the input is an obtainable state, though it does make sense. I still have a feeling that there could be ambiguous cases. Jul 26 '19 at 11:41 • @KevinCruijssen The shapes will always be separated by at least one space, I'll add it to the description. Jul 26 '19 at 11:44 • @GalenIvanov Probably better indeed to not have to deal with confusing ambiguous cases. In that case I would also have at least a newline separation, so test case 2 should be slightly modified. :) Jul 26 '19 at 11:45 • @Adám Oh, I see - indeed I need to correct the second case to have a vertical space between the shapes. Jul 26 '19 at 11:51 • Even single tracks can be incredibly difficult: [" ++ ","++++","++^+"," ++ "] only has one possible output: [" ++ ","++++","+++>"," ++ "] Jul 26 '19 at 11:55 • @AdámYes, this is the only output. Do you think a condition that sharp turns are forbidden will help? (this means no two + can be adjacent) Jul 26 '19 at 12:21 • @GalenIvanov Maybe that's indeed better to reduce confusion and make the challenge someone more manageable. Although you can still deduct the solution in Adam's comment above, having no spaces inside the space makes it rather difficult to parse correct. Always having at least one \|/- between two + will always give the shapes always spaces, making it easier to parse individual shapes. In which case Adam's one would become [" +-+ "," \| \| ","+-+ +-+","\| \|","+-+ ^-+"," \| \| "," +-+ "] -> [" +-+ "," \| \| ","+-+ +-+","\| \|","+-+ +>+"," \| \| "," +-+ "] Jul 26 '19 at 13:55 • @KevinCruijssen Thanks, I added clarification. Jul 26 '19 at 14:09 • @GalenIvanov Maybe also change the one Adam edited in, since it's still with ++ below one-another. :) Jul 26 '19 at 15:08 • I found a lot of similar questions, but not quite the same. Would it be considered a duplicate? Aug 1 '19 at 8:17 • we've had double factorial closed before as a dupe of the vanilla factorial question Aug 1 '19 at 12:03 • @Giuseppe Thanks! I found that post, but I thought the variable factorial range might make it considerably different from the original challenge. Aug 1 '19 at 12:12 • Can we take input as two integers? Aug 1 '19 at 12:49 • Can you give a definition for the multi-factorials above 2? Aug 2 '19 at 4:55 • @Adám That makes the challenge somewhat easier, but I suppose it is fair. I added the other input option. Aug 2 '19 at 7:04 • @MilkyWay90 Yes! Totally skipped that part, oops. Is it clear enough like this? Aug 2 '19 at 7:05 • @KevinCruijssen Input numbers can be in any format, of course. I thought this was included in the default I/O rules, buy I will specify this in the challenge. Aug 2 '19 at 12:39 • @KevinCruijssen In the input, at least one factorial sign is expected, i.e. a positive integer. I suppose the correct return value for a factorial number of zero would just be the base, but as the challenge is to calculate a factorial, this seems like an unreasonable restriction. The same goes for a factorial number lager than the base. If the factorial number is equal to the base, the base should be returned as per the generalized formula. I will add additional examples to clear this up. Thanks for the feedback! Aug 2 '19 at 12:39 • @KevinCruijssen The range(start,stop,step) function in Python 3 returns a generator with an initial value of start (or 0 if not specified) and a final value of stop-1. For example, the list representation of range(1,4) is [1,2,3]. If this list is sliced as [1,2,3][::-1], the returned list is [3,2,1], which can also be achieved with range(3,0,-1). When the slicing operator is applied to the generator directly, a new generator is returned which generates the sliced list instead. Does that answer your question? Also, range(420,0,-30) would be a cleaner approach in this case. Aug 2 '19 at 13:08 • @Jitse Ah ok, now it makes more sense. I indeed knew range(1,4) is [1,2,3]. And I also knew [::-1] reverses the order, since I see it used in answers every now and then. Seeing the range(1,4)[::-1] == range(3,0,-1) now I'm not sure why I didn't see it myself when I asked you the question. And yes, range(value,0,-n) would indeed be clearer than range(1,value+1)[::-n]. ;) Aug 2 '19 at 13:24 • Seems good except maybe you should specify that $n!^{(k)} = n\underbrace{!\ldots!}{k}$ (n!^{(k)} = n\underbrace{!\ldots!}{k}) Aug 2 '19 at 16:47 # Cliquish Program Challenge: Write a program that accepts a character (or byte, see additional information) as input. Then: 1. If the character is contained within the source code, output a different character also in your source code. 2. If the character is not contained within the source code, output a different character also not in your source code. This is a , so the shortest program (in bytes) wins. • Your program must consist of at least 2 distinct characters. • Your program must have at least 2 possible outputs. • Your program does not have to be deterministic; it may output a character at random (with any distribution), so long as it conforms to the above criteria. • Your program may optionally take a byte as input instead of characters. If you do, the code page the input is written in must contain at least each distinct byte in the source code, as well as at least two different bytes not in the source code. • You may take input and give output in any reasonable way. For example, you may take input as a function parameter, a command-line argument, a line from STDIN, a triple-nested array containing a single character, etc. You could output via return value, STDOUT, exit code (if applicable), fax output, etc. The input and output formats must be consistent, however. • Your output can only consist of the required character, optionally followed by one trailing newline. Prompt information (such as ans =) is exempt from this rule; such unpreventable output is acceptable. • Slightly more interesting than most generalised quines. Have you checked quine for dupes? Aug 2 '19 at 7:32 • @PeterTaylor related related related. I couldn't find any exact dupes. Aug 2 '19 at 19:39 • What if your program contains all but one possible character? What encoding are the characters in? UTF-8? latin-1? ASCII? Aug 2 '19 at 20:18 • @Beefster Hopefully I've addressed your first question. As per your second, I believe we have a standard consensus on what a character is Aug 11 '19 at 18:13 • By induction the program must have 4 possible outputs. Inside/outside of the source × two included in each (if choose one, output the other). Btw I don't understand the title Aug 22 '19 at 7:05 # ASCII Maze Unrendering 3000 Posted • Part of the wall for the current only test case seems a bit too narrow. Aug 28 '19 at 2:37 • I think this could be considered a special case of this challenge, with slight differences in the voxel style. Aug 29 '19 at 6:56 • @flawr You're right. As written, this is a duplicate. Do you know if the reverse has been done, taking the 3d version and returning the original? Aug 29 '19 at 17:52 • I don't think the reverse has been done, but I'm not sure this would make a good challenge. I'm still thinking about how one could make this. I mean the 2d version you propose would make it a bit simpler. Aug 29 '19 at 19:10 • Yes. The full 3d version would almost certainly be impossible, because most voxels would be completely blocked. Here, every block is visible, and a human can figure out where the walls are, so a computer should be able to do it too. Aug 29 '19 at 19:17 • @Night2 Indeed, didn't notice that point, I fixed the samples in the rules to include AA-000-AA Sep 20 '19 at 6:25 • Are the letters mandatory to be uppercased, or could we output in lowercase as well? Sep 20 '19 at 9:51 • @KevinCruijssen Yes, letters are mandatory to be uppercase, also I feel like you ask that because of a builtin somewhere :P Sep 20 '19 at 12:36 • Fine by me. I was mainly asking because it isn't a core-part of the challenge, and in some challenges lowercase/uppercase/mixed case is all allowed. But you're indeed right that lowercase would save a byte in my language of choice 05AB1E, where A is the lowercase alphabet builtin, and I'd need an additional u to uppercase it. ;p But since number plates are always uppercase, I can understand to keep the uppercase mandatory here as well. I had prepared a 24-byter, but will throw it away for now since I feel this can be done shorter.. Will try again later when it's posted to main. :) Sep 20 '19 at 12:41 • Another question, or more recommendation: allowed both 0-based and 1-based indexing for answers. I see your test cases are 0-based, but some languages use 1-based indexing instead. Sep 20 '19 at 12:44 • @KevinCruijssen Sure, no problem with 1-based indexing, going to fix the rules to add that. As I've never used such languages, I often forget about them Sep 20 '19 at 13:19 • I would say if the format irl has to be in uppercase then following that would be better. I've got myself an 85-byte JS answer that conforms to the irl format though. Sep 21 '19 at 1:06 • @ShieruAsakoto That's what I also advocate for, minus all the dumb real life rules that would just make it unfun to do. Anyway, going to post this Sep 23 '19 at 18:41 # Almost Illegal Strings Posted. • Does the challenge need to have the robber code output "Well done!"? What if they could just submit any code with the substring that runs? – xnor Oct 24 '20 at 10:24 • @xnor That's fair enough - I just struggle to define 'program that runs'. Would you consider zero exit code and no stderr output reasonable? Or are there some languages that output to stderr even in a valid program? Oct 24 '20 at 12:00 • I think this is too similar to the original "illegal strings" question which basically turned into a cops-and-robbers anyway Oct 24 '20 at 18:26 • @Sisyphus I was thinking you could just copy whatever condition Find an Illegal String uses. But it doesn't seem to be that rigorous, saying "The compiler/interpreter/runtime must give an error when given any source code that contains your string as a substring." I think requiring no output to STDERR is probably fine. Maybe though some languages give warnings to STDERR that golfers typically ignore? Note that defaults allow programs that print then crash, so just requiring output doesn't preclude a fatal error after. – xnor Oct 24 '20 at 21:12 • Rule suggestion: commenting out the almost illegal string is not allowed Oct 28 '20 at 15:33 • I think it's too thorny to define what comments are in a way that works across languages. One objective way to handle it would be to let cops specify a set of characters that are not in their code, where they might include their language's comment character(s) or quote literals. – xnor Oct 29 '20 at 0:40 • @xnor Ok, I've rewritten the challenge to just require that the program does not error, and feel the rules are fairly straightforward and watertight. Nov 1 '20 at 5:50 • @Beefster I considered this, and something like xnor's suggestion of banning characters. However, I think it ruins the purity of the challenge a bit, and for most languages it's very easy to avoid being in a comment (newline + end of comment block will do it). There are some languages which have 'inescapable comments', such as raw strings with specific delimiters, or something like Perl's __END__, but they're sacrifices I'm willing to make. Nov 1 '20 at 5:52 • The problem with not banning comments (or including things within strings, for that matter) is that it's quite easy to cop out and write a hello world with the almost illegal string appended in a comment. That takes out all the challenge. xnor's suggestion solves that mostly, but that does end up limiting cops a little bit to make the robbers' job an actual challenge. Nov 2 '20 at 16:25 • @Beefster Apologies, I have not been clear about my reasoning. My logic is that for the vast majority of languages, a cop can 'comment proof' their string trivially by adding a newline (to escape a single line comment) and an end of comment block (to end a multiline comment). For example, the string \nx"""x''' (with a literial newline) escapes all Python comments. You can do this in most languages - or am I missing something? Nov 3 '20 at 3:12 • @Beefster It looks like you were right - it is too hard without allowing Cops to ban characters. I'd like to apologise for not taking your feedback more seriously while the question was in the sandbox. Nov 6 '20 at 1:10 # Print random integers until 0 • Presumably, if the first number is zero, we just output zero and exit? Jan 16 at 18:57 • "integers may be separated by any non-digit, non-empty separator" can we output in zero-padded form (00 01 .. 98 99) so that a separator isn't necessary? What about codepoints/byte values (NULL - c)? Can it be a list-like object rather than a separated string? I think this would be better with more sequence-like IO rules Jan 16 at 18:59 • @pxeger Yes, if the first number is zero, you just output zero and exit. I don't think the standard is to output with zero-padding, so I'll say no on that. Byte values and list objects are both standard forms of outputting a bunch of numbers, so of course, those are allowed Jan 16 at 19:07 • Also, I assume it is not necessary that the program follows the indicated procedure (which would be unobservable anyway), as long as the output has the same statistical properties. So if a different procedure is used, perhaps the answer writer should justify it. (For example: generate a number K geometrically distributed with parameter 1/99, then output K nonzero numbers, then output a 0) Jan 16 at 19:07 • @LuisMendo Yeah, I'm pretty sure that "you don't have to follow the letter of the challenge so long as the behaviour is the same" is a standard rule, but in case not, I've edited in a sentence along with your example Jan 16 at 19:10 • I didn't fully define my example, sorry. You may want to change it to "then output K independent numbers with a uniform distribution on the set {1, 2, ..., 99}". That corresponds to the procedure you specify with a uniform distribution on {0, 1, ..., 99}. Jan 16 at 19:21 • @cairdCoinheringaahing are built-in functions for generating pseudo-random numbers forbidden? If not, is the challenge a very trivial one or am I missing something? Jan 27 at 23:17 • @Davide Trivial challenges are not necessarily bad, since it encourages participation in relatively hard-to-use or minimal esolangs. And I think this one is actually good because I smell many unexpected approaches to golf the problem. (Also, banning built-ins are considered bad.) Jan 28 at 0:26 • @Bubbler thank you so much for this train of information Jan 28 at 0:35 • @Davide Nope, all builtins are allowed. As Bubbler said, trivial challenges are only really bad if there’s no room for interesting solutions, which I don’t think is the case here Jan 28 at 10:06 # Remove Nth occurrences • There is this but it got closed, Everything is clear, but can you replace, "the n" with "the integer n" in the last point of your assumptions list. For some reason that made me read the sentence multiple times. Feb 23 at 12:08 • Is there a reason for limiting the array values to 1-9? Feb 23 at 19:49 • @pxeger Because the values don't matter, so there's no reason to complicate it more. Plus, it can allow for some interesting string based approaches by taking $A$ as a single string Feb 23 at 19:52 • @pxeger It certainly helps that 0 cannot be part of the array, because then I can use it to erase values in APL. I'm sure other languages can take advantage of that too – user Feb 24 at 1:48 # To raise $$\ e \$$ to the power of a matrix code-golfmathmatrix Posted ## Meta • Is this clear enough? • Is this a duplicate? • Any other feedback? • why not require exact calculation, it makes for a harder challenge but a more interesting one Apr 1 at 19:54 • @rak1507 it's irrational so an exact value can't be calculated Apr 1 at 19:55 • looked like there were methods on wikipedia but I could be wrong Apr 1 at 21:45 • Note that the exponential of a 9x9 matrix of 100's exceeds what floats can represent. You might want to lower the 100 bound or make allowances for that. – xnor Apr 2 at 6:30 • I think the precision convergence rule is too restrictive and too tied to that specific power series method, and some loose accuracy bound would allow more varied methods. For instance, one can approximate $e^M \approx (I+M/n)^n$ for large $n$. – xnor Apr 2 at 6:38 • @xnor I didn't read much into the matrix exponential (because I couldn't find the article on Wikipedia and was too busy watching the rest of the 3b1b video :P), so I hadn't not really realised there were other ways to compute it. What would you recommend? I still like the idea of requiring it to be observed to converge within floating point limits, because it adds an extra layer of challenge rather than just "repeat this step 100 times", but maybe that just isn't practical Apr 2 at 6:50 • I think convergence within floating point limits would unfortunately be hard and finicky here because of the nature of exponentials. The same way that $e^{100}$ and $e^{100.001}$ differ a lot, small errors in the computation can accumulate into huge ones. Also, the values in the output might be extremely small and become represented as zero. I'd have to think more about bounds, but maybe something like every entry within either 1% or 1e-4 of the true one should work. – xnor Apr 2 at 7:03 • My solution to floating-point errors is "the result should be within [insert error bound here] relative error for the given test cases". (The bolded part is VERY important. FP computation methods often have errors dependent on the magnitude of the input, so it is very hard to judge if an implementation is valid, even if the possible input range is specified. Explicitly giving the test cases makes it much easier to test submissions. Also, you need to craft the test cases carefully so that you don't accidentally ban a valid method or allow invalid methods.) Apr 4 at 23:28 # Non-quining infinite printer Seems like the title could be better but I'm not sure what to do instead I have heard that a monkey typing random keys on a typewriter, given infinite time, will eventually type out the entire works of Shakespeare, and in fact type out every possible string of characters of any length. This sounds to me like the basis for a profitable business venture in publishing. Unfortunately, however, as a result of previous failed business ventures, I am legally barred from possessing either monkeys or typewriters, so I'll instead need a program. I want this program to provably generate every possible string of characters, assuming infinite time and memory. Repetition is fine, as is overlap, as long as every possible string appears somewhere in the output. There's a catch, though. I imagine once my business gets off the ground and people realize the potential profits, they might want to get their hands on my program. The trouble is that since I'm outputting every possible string of text, in theory I'll eventually end up outputting the program itself, leaving it open to be stolen. To prevent this, I want the program never to output its own source code. It should still output every other possible string, just not itself (or, obviously, any strings which it is a substring of). Because my funds are currently very tight, I can't afford to pay for more bytes than are necessary, so I'm seeking the shortest possible program that does the job (This is a code-golf challenge, shortest answer in bytes wins). Is it communicated well enough what this challenge is asking for? Should I add a TL;DR and/or a more technical explanation of what's being looked for? • I've had enough run-ins with the law in previous ventures, I'd like this one to go smoothly. So no abusing loopholes in the program, please. • Given that this program is my financial plan for the next infinity years, I'd like some reassurance that it actually does what it's supposed to. Please provide at least a brief explanation of why your code works, since it can't exactly be tested. • Any character encoding is fine, please specify though. The exception is that your source code must be printable in the encoding you use. Is this a reasonable way to handle this that is fair to all languages? • No reading your own source, because that makes quine-related holes uncool, and uncoolness does not fit with my businesses' brand persona. • How are the strings in the output separated? Some previous challenges about "print all possible strings" were closed as unclear because of this. Jun 7 at 1:06 • @Bubbler The strings don't have to be separated. What I mean is that every string should be a substring of the "main" output if that makes sense. Put another way, running a regex match on the output for any string of characters (other than the source) should return at least one result. Would adding something to that effect clear up that confusion? Jun 7 at 1:13 • It should still output every other possible string, just not itself (or, obviously, any strings which it is a substring of) Okay, the task makes sense now. I think you'll need to add that information somewhere before the sentence I quoted. Now I wonder if the task is actually possible... Jun 7 at 1:39 • @Bubbler Sure, it's possible. One approach is to store the program's source in a string, and then run some sort of code that generates every possible string, checking after each generation whether the generated string contains the source string, and only printing if it doesn't. There might be some trouble in making sure the source isn't present in a combination of two consecutively generated strings (i.e. if the source code is AB, then outputting XA and BX` consecutively is an issue), but a simple solution to that is to insert a character not present in the source between each pair. Jun 7 at 8:02 • Here's an implementation in Python tio.run/… Jun 7 at 8:03 • @jeje An easily thought solution is just use an unused char as split, but that likely be longer – l4m2 Jul 1 at 16:40 • What do mean by "all possible strings"? Is this restricted to all strings consisting of printable ASCII? The entire Unicode table? Jul 8 at 1:14	2021-10-17 02:23:45	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 1, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.35602959990501404, "perplexity": 985.0438743956939}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-43/segments/1634323585120.89/warc/CC-MAIN-20211017021554-20211017051554-00217.warc.gz"}
https://brilliant.org/problems/how-many-solutions-10/	# How many solutions Level pending Given $$a$$ and $$b$$ are integers, how many distinct solutions for $$(a,b)$$ does the following equation have? $ab=a+b$ ×	2017-03-27 04:49:20	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5385374426841736, "perplexity": 553.3309381129199}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-13/segments/1490218189403.13/warc/CC-MAIN-20170322212949-00184-ip-10-233-31-227.ec2.internal.warc.gz"}
http://www.physicsforums.com/showthread.php?t=156859	# tensor product by Ragnar Tags: product, tensor P: 24 Could someone tell me what the tensor product is and give an example? P: 2,955 Quote by Ragnar Could someone tell me what the tensor product is and give an example? The tensor product is a way of formulating a new tensor from other tensors. If you are given the tensors A, B, C, D, ... then the tensor product TP is also a tensor and is represented by the relation TP = A@B@C@D@... The "@" is being used for the product operator which is a symbol which actually looks like an x surrounded by a zero. Suppose A, B, C, D are vectors. We feed in the 1-forms m, n, o, p as follows TP(m,n,o,p) = A(m)@B(n)@C(o)@D(p) The value of the tensor TP on the one forms is defined in this way. An example is really trivial and you can call the above an example. The tensors don't need to be vectors on the right. They just need to be tensors. Notice that TP is a tensor of rank 4 since it takes in 4 1-forms. Pete Related Discussions Linear & Abstract Algebra 11 Calculus & Beyond Homework 0 Differential Geometry 3 Differential Geometry 17 Differential Geometry 5	2014-04-17 21:38:43	{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8533309102058411, "perplexity": 438.1870575114641}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-15/segments/1397609532128.44/warc/CC-MAIN-20140416005212-00302-ip-10-147-4-33.ec2.internal.warc.gz"}
https://deepai.org/publication/simultaneous-transmission-of-classical-and-quantum-information-under-channel-uncertainty-and-jamming-attacks	# Simultaneous transmission of classical and quantum information under channel uncertainty and jamming attacks We derive universal codes for simultaneous transmission of classical messages and entanglement through quantum channels, possibly under attack of a malignant third party. These codes are robust to different kinds of channel uncertainty. To construct such universal codes, we invoke and generalize properties of random codes for classical and quantum message transmission through quantum channels. We show these codes to be optimal by giving a multi-letter characterization of regions corresponding to the capacity of compound quantum channels for simultaneously transmitting and generating entanglement with classical messages. Also, we give dichotomy statements in which we characterize the capacity of arbitrarily varying quantum channels for simultaneous transmission of classical messages and entanglement. These include cases where the malignant jammer present in the arbitrarily varying channel model is classical (chooses channel states of product form) and fully quantum (is capable of general attacks not necessarily of product form). • 53 publications • 3 publications • 3 publications 11/18/2019 ### Universal superposition codes: capacity regions of compound quantum broadcast channel with confidential messages We derive universal codes for transmission of broadcast and confidential... 01/11/2018 ### Universal random codes: Capacity regions of the compound quantum multiple-access channel with one classical and one quantum sender We consider the compound memoryless quantum multiple-access channel (QMA... 02/04/2022 ### Beware of Greeks bearing entanglement? Quantum covert channels, information flow and non-local games Can quantum entanglement increase the capacity of (classical) covert cha... 09/19/2018 ### One-shot Capacity bounds on the Simultaneous Transmission of Classical and Quantum Information We study the communication capabilities of a quantum channel under the m... 01/03/2018 ### Secure communication over fully quantum Gel'fand-Pinsker wiretap channel In this work we study the problem of secure communication over a fully q... 12/08/2021 ### Reliable Simulation of Quantum Channels The Quantum Reverse Shannon Theorem has been a milestone in quantum info... 09/05/2019 ### Playing Games with Multiple Access Channels Communication networks have multiple users, each sending and receiving m... ## 1 Abstract We derive universal codes for simultaneous transmission of classical messages and entanglement through quantum channels, possibly under attack of a malignant third party. These codes are robust to different kinds of channel uncertainty. To construct such universal codes, we invoke and generalize properties of random codes for classical and quantum message transmission through quantum channels. We show these codes to be optimal by giving a multi-letter characterization of regions corresponding to capacity of compound quantum channels for simultaneously transmitting and generating entanglement with classical messages. Also, we give dichotomy statements in which we characterize the capacity of arbitrarily varying quantum channels for simultaneous transmission of classical messages and entanglement. These include cases where the malignant jammer present in the arbitrarily varying channel model is classical (chooses channel states of product form) and fully quantum (is capable of general attacks not necessarily of product form). ## 2 Introduction In real world communication using quantum or classical systems, the parameter determining the channel in use may belong to an uncertainty set, rendering the protocols that assume the channel to be perfectly known practically obsolete. Given such uncertainty, when using the channel many times, as done in Shannon theoretic information processing tasks, assuming the channel to be memoryless or fully stationary is not realistic. In this paper, we consider three models that include channel uncertainty without attempting to reduce it via techniques such as channel identification or tomography. We refer to these models as the compound, arbitrarily varying and fully quantum arbitrarily varying channel models. Each of these models are considered here for transmission of entanglement and classical messages simultaneously between a sender and receiver. Informally, the first two channel models consist of a set of quantum channels known to the communicating parties. In the compound model, communication is done under the assumption that asymptotically, one of the channels from this set (unknown to the parties) is used in a memoryless fashion. The codes used in this model therefore have to be reliable for the whole family of memoryless channels for large enough values of . In the arbitrarily varying model, given a number of channel uses , an adversarial party chooses the sequence unknown to the communication parties, to yield the channel . The adversary may choose this sequence knowing the encoding procedure used by the sender. The code in use therefore has to be reliable for the whole family of memoryless channels. Finally, in the third channel model, namely that of the fully quantum arbitrarily varying, the assumption of memoryless communication is dropped. Here, the adversary may choose channel states that are not necessarily of the product form mentioned in the previous model. The quantum channel has different capacities for information transmission. One may consider the capacity of the channel for public ([18, 28]) or private ([15, 13]) classical message transmission, entanglement transmission or entanglement generation ([15]) to name a few. These communication scenarios have been considered subsequently under channel uncertainty ([7, 24, 6, 11, 1, 3]). Simultaneous transmission of classical and quantum messages, the subject of this work, has also been of interest([16]). This includes scenarios where the communication parties would like to enhance their classical message transmission by sharing quantum information primarily at their disposal or vice versa([5, 19, 20]). The body of research in this area is clearly interesting, when regions beyond those achieved by simple time-sharing between established classical message and quantum information transmission codes are reached. Simultaneous transmission of classical messages and entanglement is a nontrivial problem even if capacity achieving codes for the corresponding univariate transmission goals are at hand. It was already observed in [16] for perfectly known quantum channels that the naive time sharing strategy is generally insufficient to achieve the full capacity region. Examples of channels where coding beyond time-sharing is indispensable does not depend on constructing pathologies. They are readily found even within the standard arsenal of qubit quantum channels, e.g. the dephasing qubit channels [16]. We derive codes for simultaneous transmission of classical messages and entanglement that are robust to the three types of uncertainty mentioned above. The codes used here for the compound model, are different from those used for the point to point communication in [16] when considering the special case of . Given that the input state approximation techniques used therein prove insufficient in presence of channel state uncertainty, in the present work we use the decoupling approach first established in [23]. We combine robust random codes for classical message transmission from [24] and a generalization of (decoupling based) entanglement transmission codes from [7] to construct appropriate simultaneous codes for compound quantum channels under the maximal error criterion. We show that these codes are optimal by giving a multi-letter characterization of the capacity of compound quantum channels with no assumption on, the size of the underlying uncertainty set. We use the asymptotic equivalence of the two tasks of entanglement transmission and entanglement generation to include the capacity region corresponding to simultaneous transmission of classical messages and generation of entanglement between the two parties. Next, we convert the codes derived for the compound channel, using Ahlswede’s robustification and elimination techniques ([1]) to derive suitable codes for arbitrarily varying quantum channels. This is possible given that the error functions associated with codes corresponding to the compound model decay to zero exponentially. We derive a dichotomy statement ([1]), for the simultaneous classical message and entanglement transmission through AVQCs under the average error criterion. This dichotomy is observed when considering two scenarios where the communicating parties do and do not have access to unlimited common randomness, yielding the common-randomness and deterministic capacity regions of the channel model respectively. Therefore, we show that firstly, the common-randomness capacity region of the arbitrarily varying channel is equal to that of the compound channel , namely the compound channel generated by the convex hull of the uncertainty set of channels . Secondly, if the deterministic capacity of the arbitrarily varying channel is not the point , it is equal to the common-randomness capacity of the channel. We give a necessary and sufficient condition for the deterministic capacity region to be be the point . This condition is known as symmetrizablity of the channel (see [3] and [9]). Finally, we show that the codes derived here, can be used for fully quantum AVCs where the jammer is not restricted to product states, but can use general quantum states to parametrize the channel used many times. This model has been introduced in Section 8 along with the main result and related work for fully quantum AVCs and hence here, we avoid further explanation of the techniques used there. The task of simultaneous transmission of classical messages and entanglement was first considered by Devetak and Shor in [16] in case of a memoryless quantum channel under assumption that the channels state is perfectly known to its users. The authors derived a multi-letter characterization of the capacity region in this setting which also classified the naïve time-sharing approach as being suboptimal for simultaneous transmission. A code construction sufficient to achieve also the rate pairs lying outside the time-sharing region was derived using a ”piggy-backing” technique. A specialized construction introduced in [15] allows to encode the identity of the classical message into the coding states of an underlying entanglement transmission code. The mentioned strategy to optimally combine different communication tasks in quantum channel coding was afterwards used and further developed in different directions. We explicitly mention subsequent research activity by Hsieh and Wilde [19, 20] where the idea of ”piggy backing” classical messages onto quantum codes was extended to include entanglement assistance. The resulting code construction being sufficient to achieve each point in the three-dimensional rate region for entanglement-assisted classical/quantum simultaneous transmission leads to a full (multi-letter) characterization of the ”Quantum dynamic capacity” of a (perfectly known) quantum channel [21] (see the textbook [29] for an up-to-date pedagocial presentation of the mentioned results). In order to derive classically enhanced quantum codes being robust against channel uncertainty, we refine the construction entanglement transmission codes for compound quantum channels from [7, 8] instead of elaborating on the usual approach building up on codes from [15]. In fact, it was noticed earlier that deriving entanglement generation codes from secure classical message transmission codes (the strategy which the arguments in [15] follow) seems to be not suitable when the channel is a compound quantum channel. In the first section following this introduction, we introduce the notation used in this work. Precise definitions of the channel models, codes used in different scenarios along with capacity regions and finally the main results in form of Theorem 5 and Theorem 12, are given in Section 4. In Section 5, we present preliminary coding results for entanglement transmission (Section 5.1) and classical message transmission (Section 5.2). The entanglement transmission codes introduced in this section are a generalization of the random codes in [7] and [8] to accommodate conditional typicality of the input on words from many copies of an alphabet. The classical message transmission codes are those from [24] that prove sufficient for our simultaneous coding purposes. Equipped with these results, we move on to Section 6, to prove the coding results for the compound channel model. In this section, after proving a converse for the capacity region in Theorem 5, we prove the direct part in two steps. In the first step, we show that capacity regions that correspond to the case where the sender is restricted to inputting maximally entangled pure states are achieved. In the second step, we prove achievablity of capacity regions corresponding to general inputs, using elementary methods that are less involved that the usual BSST type results used for this generalization in [7] and [8]. In Section 7, after proving a converse for the capacity region under the arbitrarily varying channel model, we prove coding results in this model by converting the compound channel model codes using Ahlswede’s robustification method. This, assumes unlimited common randomness available to the legal parties. We then use an instance of elimination to show that if the deterministic capacity region is not the point , negligible amount of common randomness per use of the channel is sufficient to achieve the same capacity region. Also in this section, we prove necessity and sufficiency of symmetrizablity condition for the case where the deterministic capacity region is the point . Finally, in Section 8, we generalize these results to the case of quantum jammer by proving Theorem 31. ## 3 Notations and conventions All Hilbert spaces are assumed to have finite dimensions and are over the field . All alphabets are also assumed to have finite dimensions. We denote the set of states by . Pure states are given by projections onto one-dimensional subspaces. To each subspace , we can associate unique projection whose range is the subspace and we write for the maximally mixed state on , i.e. πF:=qFtr(qF). The set of completely positive trace preserving (CPTP) maps between the operator spaces and is denoted by . Thus , plays the role of the input Hilbert space to the channel (traditionally owned by Alice) and is channel’s output Hilbert space (usually in Bob’s possession). stands for the set of completely positive trace decreasing maps between and . will denote in what follows, the group of unitary operators acting on . For a Hilbert space , we will always identify with a subgroup of . For any projection we set . Each projection defines a completely positive trace decreasing map given by for all . In a similar fashion, any defines a by for . We use the base two logarithm which is denoted by . The von Neumann entropy of a state is given by S(ρ):=−tr(ρlogρ). The coherent information for and is defined by Ic(ρ,N):=S(N(ρ))−S((idHA⊗N)(\|ψ⟩⟨ψ\|)) where is an arbitrary purification of the state . We also use to denote entropy exchange. A useful equivalent definition of is given in terms of and any complementary channel where denotes the Hilbert space of the environment. Due to Stinespring’s dilation theorem, can be represented as N(ρ)=trHe(vρv∗) for where is a linear isometry. The complementary channel of is given by ^N(ρ):=trHB(vρv∗). The coherent information can then be written as Ic(ρ,N)=S(N(ρ))−S(^N(ρ)). (1) This quantity can also be defined in terms of the bipartite state with σ:=idHA⊗N(\|ψ⟩⟨ψ\|) as I(A⟩B,σ):=S(σB)−S(σ) where is the marginal state given by and we have the identity Ic(ρ,N)=I(A⟩B,σ). As a measure of closeness between two states , we may use the fidelity . The fidelity is symmetric in the input and for a pure state , we have . A closely related quantity is the entanglement fidelity, which for and , is given by with an arbitrary purification of the state . Another quantity that will be significant in the present work is the quantum mutual information (see e.g [29]). For a state , the quantum mutual information is defined as I(X;B,ρ):=S(ρX)+S(ρB)−S(ρ) where and are marginal states of . For the approximation of arbitrary compound channels (introduced in the next section) by finite ones we use the diamond norm , given for any by ∥N∥⋄:=supn∈Nmaxa∈L(Cn⊗H),∥a∥1=1∥(idn⊗N)(a)∥1, where is the identity channel. We state the following facts about (see e.g [31]). First, for all . Thus, , where denotes the unit sphere of the normed space . Moreover, for arbitrary linear maps . Throughout this work we have made use of the idea of nets to approximate arbitrary compound quantum channels using ones with finite uncertainty sets. This idea is presented in Appendix A by Definition 38 and proceeding two lemmas. We use exponentially as or we say approaches (goes to) zero exponentially, if is a strictly positive constant. For and both approaching zero exponentially, we use if . We use to denote the closure of set and finally, we use to denote the group of permutations on elements such that for each and . ## 4 Basic definitions and main results We consider two channel models of compound and arbitrarily varying quantum channels. They are both generated by an uncertainty set of CPTP maps. For the purposes of the present work, when considering the arbitrarily varying channel model, we assume finiteness of the generating uncertainty set. This assumption is absent in the case of the compound channel model. ### 4.1 The compound quantum channel Here, we consider quantum compound channels. Let be a set of CPTP maps. The compound quantum channel generated by is given by family . In other words, using instances of the compound channel is equivalent to using instances of one of the channels from the uncertainty set. The users of this channel may or may not have access to the Channel State Information (CSI). We will often use the set to index members of . A compound channel is used times by the sender Alice, to convey classical messages from a set to a receiver Bob. At the same time, the parties would like to communicate quantum information. Here, we consider two scenarios in which quantum information can be communicated between the parties. Classically Enhanced Entanglement Transmission (CET): While transmitting classical messages using instances of the compound channel, the sender wishes to transmit the maximally entangled state in her control to the receiver. The subspace with and , quantifies the amount of quantum information transmitted. More precisely: ###### Definition 1. An CET code for , is a family with • , • with and • . ###### Remark 2. We remark that as defined above, for each we have a entanglement transmission code for . For every and , we define the following performance function for this communication scenario when instances of the channel have been used, P(CCET,N⊗ns,m):=F(\|m⟩⟨m\|⊗ΦAB,idFA,n⊗R∘N⊗ns∘Pm(ΦAA)), where is a maximally entangled state on and R:=∑m∈[M1,n]\|m⟩⟨m\|⊗Rm. Classically Enhanced Entanglement Generation (CEG): In this scenario, while transmitting classical messages, Alice wishes to establish a pure state shared between her and Bob. As the maximally entangled pure state shared between the parties is an instance of such a pure state, it can be proven that the previous task achieved in CET, achieves the task laid out by this one, but the opposite is not necessarily true. More precisely: ###### Definition 3. An CEG code for , is a family , where is a pure state on and • with and • . The relevant performance functions for this task, for every and , are P(CCEG,N⊗ns,m):=F(\|m⟩⟨m\|⊗Φ,idFA,n⊗R∘N⊗ns(Ψm)), (2) with maximally entangled on . Averaging over the message set , will give us the corresponding average performance functions for each , ¯¯¯¯P(CX,N⊗ns):=1M1,n∑m∈[M1,n]P(CX,N⊗ns,m), for . For each scenario, we define the achievable rates. ###### Definition 4. Let A pair of non-negative numbers is called an achievable X rate for the compound channel , if for each exists a number , such that for each we find and X code such that 1. for , are simultaneously fulfilled. We also define X ”average-error-rates” by averaging the performance functions in the last condition over . We define the X capacity region of by CX(J):={(R1,R2)∈R+0×R+0:(R1,R2) is achievable X rate for J}. (3) Also the capacity region corresponding to average error criteria is defined as ¯¯¯¯CX(J):={(R1,R2)∈R+0×R+0:(R1,R2) is achievable X average-error-rate for % J}. (4) Moreover, let be an alphabet, , and be a pure state for all . Given the state ω(M,p,Ψ):=∑x∈Xp(x)\|x⟩⟨x\|⊗idHA⊗M(Ψx), (5) we introduce the following set, ^C(Ns,p,Ψ):={(R1,R2)∈R+0×R+0:R1≤I(X;B,ω(Ns,p,Ψ))∧R2≤I(A⟩BX,ω(Ns,p,Ψ))} with denoting collectively. We will also use 1lA:={(1lx1,1lx2):(x1,x2)∈A}. The following statement is the first main result of this paper. ###### Theorem 5. Let be any compound quantum channel. Then CCET(J)=¯¯¯¯CCET(J)=CCEG(J)=¯¯¯¯CCEG(J)=cl(∞⋃l=11l⋃p,Ψ⋂s∈S^C(N⊗ls,p,Ψ)) holds. This theorem is proven in the following steps. In Section 6.1, we prove that is a subset of the set on the rightmost set in the above equalities. In Section 6.2, we prove that the rightmost set is a subset of . Together with the operational inclusions CCET(J)⊂CCEG(J) and CX(J)⊂¯¯¯¯CX(J) for , we conclude the equalities in the statement of the theorem. ### 4.2 The arbitrarily varying quantum channel The arbitrarily varying quantum channel generated by a set of CPTP maps with input Hilbert space and output Hilbert space , is given by family of CPTP maps , where Nsl:=Ns1⊗…Nsl (sl∈Sl). We use to denote the AVQC generated by . To avoid further technicalities, we always assume for the AVQC generating sets appearing in this paper. Most of the results in this paper may be generalized to the case of general sets by clever use of approximation techniques from convex analysis together with continuity properties of the entropic quantities which appear in the capacity characterizations (see [3]). ###### Definition 6. An random CET code for is a probability measure on , where • , • ,. • The sigma-algebra is chosen such that the function gsl(P(m),R(m)):=F(\|m⟩⟨m\|⊗ΦAB,idH⊗lA⊗R∘Nsl∘P(m)(ΦAA)) (6) is measurable with respect to , for all . In (6), is a maximally entangled state on and . • We further require that contains all the singleton sets. The case where is deterministic, namely is equal to unity on a singleton set and zero otherwise, gives us a deterministic CET codes for . Abusing the terminology, we also refer to the singleton sets as deterministic codes. ###### Definition 7. A non-negative pair of real numbers is called an achievable CET rate pair for with random codes and average error criterion, if there exists a random CET code for with members of singleton sets notified by such that 1. , The random CET capacity region with average error criterion of is defined by ¯¯¯¯¯Ar,CET (J):={(R1,R2):(R1,R2) is achievable CET rate pair for J with random codes and average error criterion}. ###### Definition 8. A non-negative pair of real numbers is called an achievable deterministic CET rate for with average error criterion, if there exists a deterministic CET code for with 1. , Correspondingly we define the following capacity region, ¯¯¯¯¯Ad,CET (J):={(R1,R2):(R1,R2) is achievable deterministic CET rate pair for J with average error criterion}. The deterministic CET codes defined here, are entanglement transmission codes for each . More precisely we have the following definition. ###### Definition 9. An , , entanglement transmission code for AVQC is a pair with with and . The corresponding performance function for this task is F(ΦAB,idH⊗nA⊗R∘Nsn∘P(ΦAA)), sn∈Sn. Essential to the statement of our results is the concept of symmetrizablity defined in the following. ###### Definition 10. Let with be an AVQC. 1. is called -symmetrizable for , if for each finite set with , there is a map such that for all ∑sl∈Slp(ρi)(sl)Nsl(ρj)=∑sl∈Slp(ρj)(sl)Nsl(ρi). (7) 2. We call symmetrizable if it is -symmetrizable for all . ###### Remark 11. The above definition for symmetrizablity was first established in [3], generalizing the concept of symmetrization for classical AVQCs from [17]. This definition for symmetrizablity was meaningfully simplified in [9], to require checking of the condition (7) for two input states only (K=2). We prove the following result to be the second main result of this paper. ###### Theorem 12. Let with be an AVQC. The following hold. 1. implies where is the CET capacity of compound channel with average error criterion defined in the previous section and conv(J):={Nq:Nq:=∑s∈Sq(s)Ns,q∈P(S)}. 2. if and only if is symmetrizable. ## 5 Universal random codes for quantum channels In this section we prove universal random coding results for entanglement transmission and classical message transmission over quantum channels. Most of the statements below, are implicitly contained in the literature. We state some properties of these codes that stem from their random nature and prove useful when deriving CET codes stated in Section 6. Before proceeding with the following two sections in which we introduce appropriate entanglement transmission and classical message transmission coding results and for the reader’s convenience, we present briefly the concept of types used in the remainder of this section. For more information on the concept of types, see e.g. [29]. For , the word that is a string of letters and the state with spectral decomposition , we define the -typical (frequency typical) projection qδ,l(ρ):=∑xl∈Tlp,δ\|xl⟩⟨xl\|, where is the set of -typical sequences in , defined by Tlp,δ:={xl:∀x∈X,\|1lN(x\|xl)−p(x)\|≤δ ∧ p(x)=0⟺N(x\|xl)=0} (9) where is the number of occurrences of letter in word . For each , we consider the set of types over alphabet , defined as T(X,l):={λ:Tlλ≠∅}, where (). ### 5.1 Entanglement transmission codes In this section, we prove universal entanglement transmission coding results that are to be combined with suitable classical message transmission codes introduced in the next section. The following lemma is a generalization of random entanglement transmission codes obtained in [7] and [8], where a in turn generalization of the decoupling lemma from [23] has been obtained. As stated in the following lemma, there are two points to be remarked about these codes. First, the random nature of these codes gives us an encoding state (outcome of the random encoding operation) with a tensor product structure, that is of interest for the present work. Therefore at this stage, we skip the de-randomization step that seemed natural in the original work. Secondly, the integration over unitary groups with respect to the normalized Haar measure done in the random encoding operation therein, is replaced here by an average over the elements of discrete and finite subsets of representations of the unitary group known as unitary designs (see e.g. [26]). The product structure of the encoding state can be used for an instance of channel coding stated later on. This becomes clear when the tensor product structure of the average state is used to accommodate typicality. For where is some finite alphabet, and , we introduce the following notation. For the tuple where for , we define Gxl:=Gx1⊗⋯⊗Gxl, where and clearly, . Then denotes the maximally mixed state on (correspondingly denotes the maximally mixed state on for ), a purification of (correspondingly denotes a purification of ) and is a unitary design (see Theorem 16) for . The following lemma reduces to Theorem 5 of [7] when . ###### Lemma 13. Let be any compound quantum channel and alphabet be given. For subspaces with and , there exists , such that for all , we find for each , a subspace and a family of entanglement transmission codes with and 1. , with defined in (5) for , 2. with exponentially as , 3. . The ingredients to prove this lemma are presented here in form of two lemmas prior to the main proof. The following two lemmas reduce to Lemma 5 and 6 from [7]111see Lemmas 44 and 45 for the statements. when . Following these lemmas, we state Theorem 16 based on which we replace the integration with respect to Haar measure, with an average over a subset of the unitary groups called unitary designs. In short, the entanglement transmission codes in [7] were derived given a number and subspace . Here, we derive codes for a subspace , with a tensor product structure determined by word (see the description above Lemma 13). ###### Lemma 14. Let be a probability distribution with on an alphabet . For and , there exist a real number , functions , with and and an orthogonal projection satisfying 1. . The last inequality implies ∥qδ,lρxlqδ,l∥22≤2−(S(ρxl)−lϕ(δ)). ###### Proof. Let for each , be the frequency typical projection associated with state in terms of Lemma 44. We show that the projection operator has the properties listed in the statement above. We have tr(ρxlqδ,l) =tr(⨂x∈Aρ⊗Nxxq(x)δ,Nx)=∏x∈Atr(	2022-08-12 21:40:34	{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8808583617210388, "perplexity": 579.7202289087636}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-33/segments/1659882571758.42/warc/CC-MAIN-20220812200804-20220812230804-00245.warc.gz"}
http://math.stackexchange.com/questions/99401/heat-invariants-on-a-one-dimensional-riemannian-manifold	Heat Invariants on a one - dimensional Riemannian manifold I am trying to understand the asymptotic heat trace expansion $$\text{Tr}(e^{-t\triangle_g}) \backsim \sum_{k \geq 0} t^{k - \frac{n}{2}}c_{2k} \quad (t \to 0^+)$$ that is associated with the Laplace - Beltrami operator $$\triangle_g = \frac{1}{\sqrt{\text{det}(g)}} \sum^n_{i,j = 1} \frac{\partial}{\partial x_i} g^{ij} \sqrt{\text{det}(g)} \frac{\partial}{\partial x_j}$$ where $g$ denotes the metric on the n - dimensional Riemannian Manifold $(M,g)$ that we consider. Now, I understand that the first invariant $c_0$ is equal to the Volume of $M$ and the next coefficient is associated with the Scalar curvature (is that correct?). If I take $M$ to be 1-dimensional compact and boundaryless, then this means the second coefficient should evaluate to zero, is that correct ? Many thanks for your help! - 1 Answer There is only one closed (i.e. compact and boundaryless) 1-manifold, the sphere $S^1$. Furthermore, the circumference is in that case the only metric invariant, i.e. two 1-dimensional spheres $(S^1, g)$ and $(S^1, h)$ are isometric if and only if they have the same circumference. So restricting to the case of $M = S^1$ with circumference $l$, the asymptotic of the trace of the heat operator is given by $$\operatorname{Tr} e^{-t\Delta} \sim \frac{l}{\sqrt{4\pi}} t^{-\frac{1}{2}}.$$ So you are right, in this case all higher heat invariants vanish and the only information stored in the asymptotic of the trace of the heat operator are the dimension and the circumference, i.e. the heat invariants give in this case a complete set of isometry invariants. All this can be found, e.g., in Chapter 1.1 of Rosenberg: The Laplacian on a Riemannian manifold, London Mathematical Society Student Texts 31, Cambridge University Press, 1997. - 1 dimensional spheres are often called circles:) – Thomas Rot Jan 16 '12 at 11:26 thanks a lot! that helped a great deal! – harlekin Jan 16 '12 at 11:28	2014-09-02 01:53:40	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 2, "x-ck12": 0, "texerror": 0, "math_score": 0.962844967842102, "perplexity": 217.94568378586743}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-35/segments/1409535921318.10/warc/CC-MAIN-20140909054110-00407-ip-10-180-136-8.ec2.internal.warc.gz"}
https://tex.stackexchange.com/questions/281500/referencing-with-svjour3-springer	# Referencing with svjour3 (Springer) I am unable to do referencing with natbib using springer 's svjour3 template. If I add natbib in the following preamble, it gives me error. I wanna use author year style for citation and alphabetical references with my bib file. what am I doing wrong? \RequirePackage{fix-cm} \documentclass[smallextended]{svjour3} \smartqed \bibliographystyle{spbasic} \usepackage{natbib} \usepackage{graphicx} \usepackage{mathptmx} \usepackage{amsmath} \usepackage{cite} \usepackage{latexsym} \clearpage \begin{document} • You can use natbib as a global option and kick ouy both packages, natbib and cite, from the preamble. Harder to say more, since 8gives me error* isn't specific. please provide a [minimal working example](dickimaw-books.com/latex/minexample/index.html0. – Johannes_B Dec 4 '15 at 8:11 • For sure, do not load both cite and natbib. – Mico Dec 4 '15 at 8:18 • @Johannes_B Thank you for response. kicking out both cite and natbib does not solve my issue. If I add only \usepackage{natbib}, it does not solve my problem. Excuse me I am very beginner, I dont know how to explain my problem in technological language as minimal working example. – Amber Dec 4 '15 at 9:22 • It is no technological language, it is just a code example. A compilable one. Just take half an hour to read the linked material and we will be able to help you within minutes. Without understanding a problem, a solution will achieved by guessing and looking into a crystal ball for days. And that is very unsatisfying. – Johannes_B Dec 4 '15 at 9:36 • Maybe you just need to delete the auxiliary files. not sure. – Johannes_B Dec 4 '15 at 9:36 As explained above, keeping only natbib leads to something that works \RequirePackage{fix-cm} \documentclass[smallextended]{svjour3} \smartqed \usepackage{natbib} \usepackage{graphicx} \usepackage{mathptmx} \usepackage{amsmath} \usepackage{latexsym} \begin{filecontents}{mybib.bib} @BOOK{foo, AUTHOR = {J. Doe and F. Foo},	2020-03-29 10:09:20	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.710055410861969, "perplexity": 3100.820847089874}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-16/segments/1585370494064.21/warc/CC-MAIN-20200329074745-20200329104745-00048.warc.gz"}
https://www.nature.com/articles/s41467-021-25780-4?error=cookies_not_supported&code=5d124e3a-60f4-4f15-9444-d44325943344	## Introduction Imposing a new periodicity on a crystal leads to the reorganization of the electronic bands of the parent compound through their back-folding on the new Brillouin zone. New periodicities may be engineered in designer materials, for instance in artificial heterostructures with Moiré minigaps, or emerge due to a structural or electronic phase transition. The charge density wave state is an electronic order where the charge density develops a spatial modulation concomitantly to a periodic distortion of the atomic lattice and the opening of a gap in the quasi-particle spectrum. By reducing the electronic band energy, this gap compensates for the elastic and Coulomb energy costs associated with the formation of the CDW. It also lowers the degeneracy of the electronic states at the crossings of the folded bands. These are the points in the band structure of the parent compound that are connected by the wavevector of the new periodicity. Although a gap should open at all the crossings of the folded bands, previous studies only focused on the primary CDW gap around the Fermi-level, which leads to the highest energy gain of the reconstructed system. The existence of secondary gaps and associated charge modulations (CMs) remains largely unexplored. In many cases, only a tiny fraction of the electrons are involved in the CDW formation. Therefore, the CDW gap manifests only as a slight reduction of the density of states (DOS)—which can depend on momentum—rather than a full depletion of the DOS. This makes it challenging to measure the CDW gap using spectroscopic probes such as angle-resolved photoemission spectroscopy (ARPES) and scanning tunnelling spectroscopy. This is particularly true for 2H-NbSe21,2,3,4,5 (hereinafter simply NbSe2). However, the effect of the redistributed electrons can be readily detected in topographic STM images, even for minute changes brought upon by the opening of the CDW gap as demonstrated in the following. NbSe2 is an iconic material of correlated electron physics. It hosts a nearly commensurate charge density wave below TCDW = 33.5 K and a superconducting (SC) order below TSC = 7.2 K6,7,8,9,10. NbSe2 is a layered material with a three-fold symmetric crystal structure around the direction perpendicular to the layers (Fig. 1a). Each unit cell is composed of two slabs of Se–Nb–Se trilayers, where the Se lattices are 60° rotated, while the Nb atoms are aligned on top of each other in a trigonal prismatic coordination with the Se atoms. The Fermi surface (FS) of NbSe2 is mainly determined by the bonding and antibonding combinations of the Nb-4d orbitals11,12 leading to double-walled barrel-shaped pockets around the K and Γ points of the hexagonal Brillouin-zone2,3,13,14 (Fig. 1b). The charge-ordered state consists of three CDWs, which form along the three equivalent ΓM directions with wavevectors $$(1-\delta )\frac{2}{3}\| {{\Gamma }}M\|$$, where δ ≈ 0.02 and depends on temperature10. In real space, this yields a locally commensurate 3a0 × 3a0 superstructure delimited by discommensurations15,16, where a0 is the atomic periodicity. The 3a0 × 3a0 reconstruction is readily accessible to topographic STM imaging. Its bias-dependent contrast has been the focus of previous studies, with particular emphasis on the contrast inversion expected in a classic Peierls scenario between images acquired above and below the CDW gap17,18, and on the role of defects in stabilizing the CDW5. Sacks et al.18 calculate the bias dependence of the CDW phase in a perturbative approach, considering a single band normal state description of NbSe2. They find that the phase-shift of the CDW component of the local DOS can be very different from the 180° expected in a one-dimensional (1D) case (Supplementary Note VII) when changing the imaging bias across the Fermi level (EF). However, their model does not reproduce the full bias dependence of the CDW amplitude and phase that we find. ## Results and discussion ### Bias dependent STM topography, CDW phase and amplitude In Fig. 2a–c, we present a selection from numerous topographic STM images of the same region on a cleaved NbSe2 surface at different sample biases (Vb) between −0.5 V and 0.5 V. They show a triangular atomic lattice with a superimposed 3a0 × 3a0 CDW modulation (see also Supplementary Fig. 1a), consistent with previous STM studies of unstrained bulk NbSe24,5,16,17,19,20,21,22. A defect-free region with a well resolved CDW outlined in red is magnified in Fig. 2d, f, h for each Vb. In order to identify the origin of the bias dependence of the topographic contrast in these images, we separate the atomic lattice and CDW contributions using Fourier filtering (Supplementary Note I). This analysis shows that the bias-dependent STM contrast is due to the changing CDW signal (Fig. 2e, g, i) since the corresponding atomic lattice contrast remains unchanged (see Supplementary Fig. 1). The observed CDW pattern can be modelled as the sum of three plane waves as described in ref 16. While each plane wave has its own phase φi(r), which depends on a selected reference point, the dephasing parameter Θ(r) = φ1(r) + φ2(r) + φ3(r) mod 360 is uniquely defined for each particular CDW pattern, independent of any reference point. Θ(r) represents the internal CDW structure, quantifying the local relative position of the wavefronts of the three CDWs. In Fig. 2j–l, we show Θ(r) corresponding to the STM images in Fig. 2a–c, respectively. They were obtained by fitting the CDW contrast following the method described in ref. 16. Each bias voltage is characterized by a dominant dephasing parameter (Fig. 2j–l), except in the vicinity of defects discussed later. This visual assessment is confirmed by the peaked histograms of Θ(r) (Supplementary Fig. 2). Fitting a Gaussian to these histograms allows to extract a well-defined dephasing parameter Θ0(Vb) for each imaging bias (Supplementary Fig. 3). For a quantitative analysis of the bias dependence of Θ0, we note that a given local CDW structure is represented by any arbitrary combination of φi(r) summing up to the same dephasing parameter, in particular the one where all three phases are equal. Moreover, the threefold symmetry of the system implies there is no privileged plane wave among the three used to describe the CDW. These observations allow us to map the problem to a one-dimensional (1D) description with an apparent CDW phase φ0(Vb) = Θ0(Vb)/3 (Supplementary Note III), and model φ0(Vb) to understand the bias dependent CDW pattern. Plotting φ0(Vb) in Fig. 3a reveals a striking non-monotonic bias dependence, with an inflexion point around −0.15 V and a minimum slightly above the Fermi-level (EF = 0 V). This dependence is robust as long as φ0(Vb) is extracted from topographic STM images away from defects (Supplementary Fig. 4). Close to defects, the dephasing parameter Θ0(Vb) is different and tends to depend much less on imaging bias (Supplementary Fig. 5). This is consistent with earlier findings that defects (and impurities) can act as strong pinning centres23,24 locking the local phase of the CDW or driving the formation of CDW domains25,26. The CDW amplitude can be extracted in a similar way to the phase, by fitting the histogram of the local amplitudes ai(r) of each plane wave measured over the entire field of view with a Gaussian, and extracting the peak value ai(Vb). The bias dependence and magnitude of ai(Vb) is nearly the same for all three CDWs (Supplementary Fig. 6b). For the analysis, we consider the average of these three amplitudes at each bias A0(Vb) = (a1(Vb) + a2(Vb) + a3(Vb))/3 plotted in Fig. 3b. ### Modelling and calculations To understand the bias dependence of the CDW amplitude and phase in Fig. 3, we simulate topographic STM traces using a 1D model system (Supplementary Note VIII). In the simplest configuration corresponding to the Peierls reconstruction, we consider the contribution to the tunneling current of a single CM and its associated gap centred on EF (Fig. 4a–c). In this case, traces at the same polarity are always in phase, whereas traces at opposite polarities always show contrast inversion (or a 180° phase shift in the present harmonic model). The latter, often considered as an identifying hallmark of the CDW state27, clearly does not reproduce the data in Fig. 3a. A single CM can only produce two sets of STM traces differing by contrast inversion in the vicinity of the gap. To generate a more complex bias dependence of the phase, we consider the possibility of a second CM whose associated gap opens in another band and away from EF (Fig. 4d–f). If these two harmonic CMs are in phase (Fig. 4d), the resulting STM traces are either in-phase or 180° out of phase (Fig. 4e), unable to reproduce the data in Fig. 3a. To generate more structures in the bias dependence of the phase, we need to introduce a phase shift between the two CMs (Figs. 4g–i). This leads to a phase that is no longer bi-modal, limited to two values differing by 180° as in Fig. 4c, f. It takes many different values (Fig. 4i), where the precise bias dependence is defined by the magnitude of the two gaps, their position relative to EF and by the relative phase shift between the two CMs. The simulated STM topographic traces in Figs. 4b, e and h also reveal a pronounced bias-dependent imaging amplitude with distinct line-shapes in the three model cases discussed above (Supplementary Fig. 7). The broad parameter space of our 1D model makes it challenging to run a self-converging fit to the data. Visually optimizing the size and position of the two gaps in Fig. 4i, we find a range of parameters (Supplementary Note X) simultaneously reproducing the experimental bias dependent CDW phase and amplitude data remarkably well (Fig. 3). As for the relative phase between the two CMs, it is chosen to minimize the Coulomb interaction of the CMs and to conform with the strong commensuration energy that locally locks them to the lattice. Reducing the Coulomb energy is obtained by introducing a phase shift between the two CMs, which can only be ±120° (Fig. 4g) to satisfy the lock-in criterion with the lattice given the 3a0 periodicity of the CDW. In the following, we turn to theoretically modelling multiple CDW gaps on different bands in NbSe2. We deploy self-consistent calculations to include the CDW gap within the random phase approximation on the two-dimensional two-band tight-binding fit to the NbSe2 band structure constrained by ARPES3. The corresponding FS shown in Fig. 1b consists of inner (red) and outer (blue) bands originating from symmetric and antisymmetric combinations of the Nb d$${}_{3{z}^{2}-{r}^{2}}$$ orbitals. The model (see “Methods”) was previously shown to accurately reproduce the full range of experimental measurements on the charge-ordered state12,28. The resulting DOS for the gapped and ungapped cases in each band are shown in Fig. 5a. To emphasize the DOS reduction accompanying the CDW phase transition, we plot the difference between the gapped and ungapped DOS for each band in Fig. 5b. Our theoretical modelling shows a clear gap on the inner band at EF, consistent with the gap measured by ARPES around the K-point3. Interestingly, Fig. 5b reveals further DOS reductions, for example near 100 meV on the inner band and −50 meV on the outer band. These features are indicative of CDW gaps opening away from EF in addition to the (primary) gap at EF, supporting the simple model we propose to understand the bias dependence of the CDW appearance in STM images of NbSe2. According to Fig. 5b, there could even be more than two gaps. Consequently, we have included up to n = 8 gaps to our 1D model. However, the agreement with the data is similar for n = 2 and n = 3 (Supplementary Note XI), and we see no improvements adding more gaps. In summary, the remarkable match between the bias dependence of the CDW contrast in STM topography and the simple 1D model proposed here provides compelling evidence that the CDW in NbSe2 is composed of at least two out-of-phase CMs on the inner and outer bands. While a 180° phase shift between these two CMs would minimize the Coulomb energy, the complex bias dependence of the CDW amplitude and phase observed by STM can only be reproduced when considering also the commensuration energy. This highlights the importance of the coupling of charge order to the lattice, which manifests in the formation of discommensurations15,16 and ultimately enables the observation of the multiband CMs uncovered here. The present study further highlights the power of topographic imaging to gain unique insight into detailed features of the CDW too faint to be detected accurately by tunneling spectroscopy. The formation of multiple modulations in response to new periodicities of a primary transition directly observed here, is extremely general and should in principle be present in all charge (and spin) density wave materials, and suggests new directions to explore in the physics of spatially modulated electronic orders. ## Methods ### Crystal growth and STM measurements Single crystals of NbSe2 were grown via iodine-assisted chemical vapour transport and cleaved in-situ at room temperature. STM experiments were done in UHV (base pressure below 2 10−10 mbar) using tips mechanically cut from a PtIr wire and conditioned in-situ on a clean Ag(111) single crystal. The bias voltage was applied to the sample. STM images were recorded in constant current mode with at least 64 pixel/nm resolution. Details of the CDW amplitude and phase fitting procedure can be found in ref 16. ### Self-consistent calculations We carried out a 22-orbital Slater–Koster tight-binding fit29 to the NbSe2 band structure, constrained by ARPES measurements2,3 and local density approximation numerical calculations11. This provided not only the band structure, but the orbital composition of the bands. We found that the two bands crossing the Fermi level are predominantly composed of the niobium $${d}_{3{z}^{2}-{r}^{2}}$$ orbitals (>60% across the Brillouin zone). A third, small, pancake-shaped pocket centred on Γ derives primarily from the selenium p-orbitals and is therefore not expected to mix significantly with the other bands, in-keeping with experimental observations that it plays no role in the charge ordering. We then re-fit the two bands of interest using only the two relevant orbitals; the fit was indistinguishable from the phenomenological fit to ARPES data provided in ref. 3. The Coulomb interaction can be neglected in NbSe2, as the large DOSs at the Fermi level leads to strong screening. This remains true down to the lowest temperatures (above the SC transition at 7.2 K) since the CDW gap only opens on small regions of the FS. The relevant interaction is the electron-phonon coupling, for which we constructed an analytic expression following Ref. 30. It has long been suggested that the CDW in NbSe2 originates not from FS nesting, but from the dependence of the electron–phonon coupling on the momentum transfer in the phonon-mediated electron–electron scattering11. Our calculation includes the dependence of the coupling on the ingoing and outgoing electron momenta, as well as the orbital composition of the electronic bands scattered between. Only by taking all of these factors into account were we able to find a consistent explanation of the full range of experimental observations, including ARPES2,3, scanning tunneling spectroscopy/microscopy4, and inelastic X-ray scattering31. Our model has only one free parameter, the overall magnitude of the electron–phonon coupling, which we fixed using TCDW = 33.5 K. We modelled the effect of the CDW on the electronic band structure using the random phase approximation. We employed the Nambu–Gor’kov method to work within the gapped phase; this method consists of promoting the electronic Green’s function to a 9 × 9 matrix, representing the tripling of the real-space unit cell in both lattice directions induced by the CDW formation. The CDW gap appears in off-diagonal elements, and diagonalisation then results in a gapped electronic band structure. We solved for the CDW gap self-consistently at high-symmetry points in the Brillouin zone, and used the results to constrain a tight-binding fit for the gap structure. We assumed the gap to be independent of energy. The CDW gap serves as an order parameter, and so our model naturally accounts for long-range CDW correlations. Further details of the method are given in refs. 12,28. We found that a CDW gap opens at the Fermi level on the outer pocket centred on the K-point, along the MK line, in agreement with ARPES2,3. However, since the order parameter is non-zero at all points in the Brillouin zone, and at all energies, gaps also open wherever band crossings are introduced. This is the origin of the multiband CDW, evidenced by the suppression of DOS at energies below EF seen in Fig. 5. We calculated the DOS at different energies, with and without the CDW gap, by summing over the Brillouin zone the spectral function found from the electronic Green’s function.	2023-02-05 06:01:54	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6911817193031311, "perplexity": 1249.1086278895605}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764500215.91/warc/CC-MAIN-20230205032040-20230205062040-00120.warc.gz"}
http://nbviewer.jupyter.org/url/norvig.com/ipython/Cheryl-and-Eve.ipynb	# When Cheryl Met Eve: A Birthday Story¶ Peter Norvig, May 2015 The "Cheryl's Birthday" logic puzzle made the rounds, and I wrote code that solves it. There I said that one reason for solving the problem with code rather than pencil and paper is that you can do more with code. Gabe Gaster proved me right when he tweeted that he had extended my code to generate a new list of dates that satisfies the constraints of the puzzle: January 15, January 4, July 13, July 24, July 30, March 13, March 24, May 11, May 17, May 30 In this notebook, I verify Gabe's result, and find some other variations on the puzzle. # Code for Original Cheryl's Birthday Puzzle¶ Let's recap the puzzle and the code to solve it. I've refactored my previous code, in a few ways: • DATES is now a global variable, not a constant. • I switched to sets of possible dates, not lists. • I define a new function, satisfy, which is similar to the built-in function filter, except that satisfy returns a set, not a list, and it allows you to specify more than one function that the candidate elements must satisfy. • I changed the name statement3 to albert1, and so on. In [1]: # Albert and Bernard just became friends with Cheryl, and they want to know when her birthday is. # Cheryl gave them a set of 10 possible dates: from __future__ import division, print_function CHERYL_DATES = { 'May 15', 'May 16', 'May 19', 'June 17', 'June 18', 'July 14', 'July 16', 'August 14', 'August 15', 'August 17'} # Cheryl then tells Albert and Bernard separately the month and the day of the birthday respectively. def tell(part): "Cheryl tells a part of her birthdate to someone; return a new set of possible dates that match the part." return {date for date in DATES if part in date} def Month(date): return date.split()[0] def Day(date): return date.split()[1] def know(possible_dates): "A person knows the birthdate if they know there is exactly one possible date." return len(possible_dates) == 1 # Albert and Bernard make three statements: def albert1(date): "Albert: I don't know when Cheryl's birthday is, but I know that Bernard does not know too." told = tell(Month(date)) return (not know(told) and all(not know(tell(Day(d))) for d in told)) def bernard1(date): "Bernard: At first I don't know when Cheryl's birthday is, but I know now." at_first = tell(Day(date)) return (not know(at_first) and know(satisfy(at_first, albert1))) def albert2(date): "Albert: Then I also know when Cheryl's birthday is." return know(satisfy(tell(Month(date)), bernard1)) # So when is Cheryl's birthday? def cheryls_birthday(dates): "Return a list of the possible dates for which Albert and Bernard's statements are true." global DATES DATES = dates # Assign to global DATES here; used by tell return satisfy(dates, albert1, bernard1, albert2) def satisfy(items, *predicates): "Return the subset of items that satisfy all the predicates." return {item for item in items if all(p(item) for p in predicates)} In [2]: cheryls_birthday(CHERYL_DATES) Out[2]: {'July 16'} In [3]: know(cheryls_birthday(CHERYL_DATES)) Out[3]: True # Verifying Gabe's Version¶ Gabe tweeted these ten dates: In [4]: GABE_DATES = [ 'January 15', 'January 4', 'July 13', 'July 24', 'July 30', 'March 13', 'March 24', 'May 11', 'May 17', 'May 30'] We can verify that they do indeed make the puzzle work, giving a single known birthdate: In [5]: cheryls_birthday(GABE_DATES) Out[5]: {'July 30'} # Creating Our Own Versions¶ If Gabe can do it, we can do it! Our strategy will be to repeatedly pick a random sample of dates, and check if they solve the puzzle. We'll limit ourselves to a subset of dates (not all 366) to make it more likely a random selection will have multiple dates with the same month and day (otherwise Albert and Bernard would know right away): In [6]: some_dates = {mo + ' ' + d1 + d2 for mo in ('March', 'April', 'May', 'June', 'July') for d1 in '12' for d2 in '56789'} Here's how to pick a random sample of the dates: In [7]: import random def sample(dates=some_dates, n=10): "Return a random sample of dates." return set(random.sample(dates, n)) In [8]: sample() Out[8]: {'April 26', 'July 16', 'July 26', 'June 18', 'June 25', 'June 27', 'March 18', 'March 27', 'May 17', 'May 28'} Now we need to cycle through samples until we hit one that works. I anticipate wanting to solve other puzzles besides the original cheryls_birthday, so I'll make the puzzle be a parameter of the function pick_dates. Note that pick_dates returns two things: the ten dates that form the puzzle, and the one date that is the solution. In [9]: def pick_dates(puzzle=cheryls_birthday): "Pick a set of dates for which the puzzle has a unique solution." while True: dates = sample() solutions = puzzle(dates) if know(solutions): return dates, solutions.pop() In [10]: pick_dates() Out[10]: ({'April 18', 'April 19', 'July 15', 'July 26', 'June 17', 'March 18', 'March 27', 'May 18', 'May 19', 'May 26'}, 'May 26') Great! We can make a new puzzle, just like Gabe. But how often do we get a unique solution to the puzzle (that is, the puzzle returns a set of size 1)? How often do we get a solution where Albert and Bernard know, but we the puzzle solver doesn't (that is, a set of size greater than 1)? How often is there no solution (size 0)? Let's make a Counter of the number of times each length-of-solution occurs: In [11]: from collections import Counter def counter(puzzle=cheryls_birthday, N=10000): "Try N random samples and count how often each possible length-of-solution appears." return dict(Counter(len(puzzle(sample())) for _ in range(N))) In [12]: counter(cheryls_birthday) Out[12]: {0: 8765, 1: 589, 2: 645, 3: 1} This says that about 6% of the time we get a unique solution (a set of len 1). A little more often than that we get an ambiguous solution (with 2 or more possible birth dates). And most of the time, the sample of dates leads to no solution. # A New Puzzle: Enter Eve¶ Now let's see if we can create a more complicated puzzle. We'll introduce a new character, Eve, give her a statement, and alter the rest of the puzzle slightly: Albert and Bernard just became friends with Cheryl, and they want to know when her birthday is. Cheryl wrote down a list of 10 possible dates for all to see. Cheryl then writes down the month and shows it just to Albert, and also writes down the day and shows it just to Bernard. Albert: I don't know when Cheryl's birthday is, but I know that Bernard does not know either. Bernard: At first I didn't know when Cheryl's birthday is, but I know now. Albert: Then I also know when Cheryl's birthday is. Eve: Hi, my name is Eve and I'm an evesdropper. It's what I do! I peeked and saw the first letter of the month and the first digit of the day. When I peeked, I didn't know Cheryl's birthday, but after listening to Albert and Bernard I do. And it's a good thing I peeked, because I couldn't have figured it out without peeking. So when is Cheryl's birthday? We can easily code this up: In [13]: def cheryls_birthday_with_eve(dates): "Return a list of the possible dates for which Albert, Bernard, and Eve's statements are true." global DATES DATES = dates return satisfy(dates, albert1, bernard1, albert2, eve1) def eve1(date): """Eve: I peeked and saw the first letter of the month and the first digit of the day. When I peeked, I didn't know Cheryl's birthday, but after listening to Albert and Bernard I do. And it's a good thing I peeked, because I couldn't have figured it out without peeking.""" at_first = tell(first(Day(date))) & tell(first(Month(date))) return (not know(at_first) and know(satisfy(at_first, albert1, bernard1, albert2)) and not know(satisfy(tell(''), albert1, bernard1, albert2))) def first(seq): return seq[0] Note: I admit I "cheated" a bit here. Remember that the function tell tests for (part in date). For that to work for Eve, we have to make sure that the first letter is distinct from any other character in the date (it is—because only the first letter is uppercase) and that the first digit is distinct from any other character (it is—because in some_dates I carefully made sure that the first digit is always 1 or 2, and the second digit is never 1 or 2). Also note that tell('') denotes the hypothetical situation where Cheryl told Eve nothing, and thus is equal to DATES. I have no idea if it is possible to find a set of dates that works for this puzzle. But I can try: In [14]: pick_dates(puzzle=cheryls_birthday_with_eve) Out[14]: ({'April 17', 'July 26', 'July 28', 'June 15', 'June 16', 'June 26', 'March 28', 'March 29', 'May 17', 'May 28'}, 'July 26') That was easy. How often is a random sample of dates a solution to this puzzle? In [15]: counter(cheryls_birthday_with_eve) Out[15]: {0: 9391, 1: 311, 2: 297, 3: 1} About half as often as for the original puzzle. # An Even More Complex Puzzle¶ Let's make the puzzle even more complicated by making Albert wait one more time before he finally knows: Albert and Bernard just became friends with Cheryl, and they want to know when her birtxhday is. Cheryl wrote down a list of 10 possible dates for all to see. Cheryl then writes down the month and shows it just to Albert, and also writes down the day and shows it just to Bernard. Albert: I don't know when Cheryl's birthday is, but I know that Bernard does not know either. Bernard: At first I didn't know when Cheryl's birthday is, but I know now. Albert: I still don't know. Eve: Hi, my name is Eve and I'm an evesdropper. It's what I do! I peeked and saw the first letter of the month and the first digit of the day. When I peeked, I didn't know Cheryl's birthday, but after listening to Albert and Bernard I do. And it's a good thing I peeked, because I couldn't have figured it out without peeking. Albert: OK, now I know. So when is Cheryl's birthday? Let's be careful in coding this up; Albert's second statement is different; he has a new third statement; and Eve's statement uses the same words, but it now implicitly refers to a different statement by Albert. We'll use the names albert2c, eve1c, and albert3c (c for "complex") to represent the new statements: In [16]: def cheryls_birthday_complex(dates): "Return a list of the possible dates for which Albert, Bernard, and Eve's statements are true." global DATES DATES = dates return satisfy(dates, albert1, bernard1, albert2c, eve1c, albert3c) def albert2c(date): "Albert: I still don't know." return not know(satisfy(tell(Month(date)), bernard1)) def eve1c(date): """Eve: I peeked and saw the first letter of the month and the first digit of the day. When I peeked, I didn't know Cheryl's birthday, but after listening to Albert and Bernard I do. And it's a good thing I peeked, because I couldn't have figured it out without peeking.""" at_first = tell(first(Day(date))) & tell(first(Month(date))) return (not know(at_first) and know(satisfy(at_first, albert1, bernard1, albert2c)) and not know(satisfy(tell(''), albert1, bernard1, albert2c))) def albert3c(date): "Albert: OK, now I know." return know(satisfy(tell(Month(date)), eve1c)) Again, I don't know if it is possible to find dates that works with this story, but I can try: In [17]: pick_dates(puzzle=cheryls_birthday_complex) Out[17]: ({'July 25', 'July 27', 'July 28', 'June 18', 'June 26', 'March 17', 'March 18', 'May 16', 'May 19', 'May 26'}, 'June 26') It worked! Were we just lucky, or are there many sets of dates that work? In [18]: counter(cheryls_birthday_complex) Out[18]: {0: 9091, 1: 908, 2: 1} Interesting. It was actually easier to find dates that work for this story than for either of the other stories. ## Analyzing a Solution to the Complex Puzzle¶ Now we will go through a solution step-by-step. We'll use a set of dates selected in a previous run: In [19]: DATES = { 'April 28', 'July 27', 'June 19', 'June 16', 'July 15', 'April 15', 'June 29', 'July 16', 'May 24', 'May 27'} Let's find the solution: In [20]: cheryls_birthday_complex(DATES) Out[20]: {'July 27'} Now the first step is that Albert was told "July": In [21]: tell('July') Out[21]: {'July 15', 'July 16', 'July 27'} And no matter which of these three dates is the actual birthday, Albert knows that Bernard would not know the birthday, because each of the days (15, 16, 27) appears twice in the list of possible dates. In [22]: all(not know(tell(Day(d))) for d in tell('July')) Out[22]: True Next, Bernard is told the day: In [23]: tell('27') Out[23]: {'July 27', 'May 27'} There are two dates with a 27, so Bernard did not know then. But only one of these dates satisfies Albert's statement: In [24]: satisfy(tell('27'), albert1) Out[24]: {'July 27'} So now Bernard knows. Poor Albert still doesn't know: In [25]: satisfy(tell('July'), bernard1) Out[25]: {'July 15', 'July 16', 'July 27'} Then along comes Eve. She evesdrops the "J" and the "2": In [26]: tell('J') & tell('2') Out[26]: {'July 27', 'June 29'} Two dates, so Eve doesn't know yet. But only one of the dates satisfies the statements: In [27]: satisfy(tell('J') & tell('2'), albert1, bernard1, albert2c) Out[27]: {'July 27'} But she wouldn't have known if she had been told nothing: In [28]: satisfy(tell(''), albert1, bernard1, albert2c) Out[28]: {'July 15', 'July 16', 'July 27'} What about Albert? After hearing Eve's statement he finally knows: In [29]: satisfy(tell('July'), eve1c) Out[29]: {'July 27'} # What Next?¶ If you like, there are many other directions you could take this: • Could you create a puzzle that goes one or two rounds more before everyone knows? • Could you add new characters: Faith, and then George, and maybe even a new Hope? • Would it be more interesting with a different number of possible dates (not 10)? • Should we include the year or the day of the week, as well as the month and day? • Perhaps a puzzle that starts with Richard Smullyan announcing that one of the characters is a liar. • Or you could make a puzzle harder than the hardest logic puzzle ever. • It's up to you ...	2018-10-22 02:29:33	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 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": 0.4930668771266937, "perplexity": 3099.3222511295676}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-43/segments/1539583514443.85/warc/CC-MAIN-20181022005000-20181022030500-00110.warc.gz"}
http://playzona.net/North-Carolina/how-to-use-the-trial-and-error-method.html	Providing great personal service to your computer needs at a very reasonable price compared to others! Address Mooresville, NC 28115 (661) 369-2804 # how to use the trial and error method Mount Mourne, North Carolina If t = 2, then tÂ³ + t = 2Â³ + 2 = 10 . This method can be applied to solving the quadratic equation p(x) = 0. x2 = 2 + 1/(2.333 333) = 2.428 571 Repeat this until you get an answer to a suitable degree of accuracy. We can all take turns equaling 3.Sample ProblemFactor the polynomial x2 + 4x â€“ 5.We can factor this quadratic polynomial into two binomials of the form:(x + m)(x + n)We need Bitte versuche es spÃ¤ter erneut. All Rights Reserved. Quadratic equations are equations of the form ax2 + bx + c = 0, where a, b, and c are constants. You can use the free Mathway calculator and problem solver below to practice Algebra or other math topics. Logging outâ€¦ Logging out... Neither m nor n make an appearance alongside the first term in the final polynomial, which is probably just as well, since that x appears to be busy squaring itself.Let's see While trial and error is an extremely powerful tool that can be used to solve problems, it also has some weaknesses. Create a wire coil Compute the kangaroo sequence Java String/Char charAt() Comparison QQ Plot Reference Line not 45° Why doesn't a single engine airplane rotate along the longitudinal axis? So a and c could be -2 and 1, or 2 and -1.And b and d could be -3 and 1, or 3 and -1.We'll try all the possible factorizations and This is shown in the last video on this page. SchlieÃŸen Weitere Informationen View this message in English Du siehst YouTube auf Deutsch. Trial and error is typically good for problems where you have multiple chances to get the correct solution. Use it to check your answers. Is the measure of the sum equal to the sum of the measures? You observe that for $x<1$ the logarithm is negative and that the two functions are both monotonically increasing, but the logarithm grows slower. in a court-room, or laboratory). Assume m and n are integers. Home Math By Grades Pre-K Kindergarten Grade 1 Grade 2 Grade 3 Grade 4 Grade 5 Grade 6 Grades 7 & 8 Grades 9 & 10 Grades 11 & 12 Basic There are a number of important factors that makes trial and error a good tool to use for solving problems. Wird verarbeitet... Therefore, we can factor our original polynomial like this:x2 + 4x + 3 = (x + 1)(x + 3)If we let m = 3 and n = 1 we'll have the It involves rearranging the equation you are trying to solve to give an iteration formula. Melde dich bei YouTube an, damit dein Feedback gezÃ¤hlt wird. The possible factors are ±1 and ±6 or ±2 and ±3. Upon computing the square root of both sides, we see that either x - 3 = 1 or x - 3 = -1. Once in a while, though, trinomials go through mood swings and stop cooperating, and then we have a bit more begging and pleading to do. Solution: Note that by adding 1 to both sides, we obtain a square polynomial on the left, namely x2 - 6x + 9 = 1, which by identity (P.7.2b) becomes (x An example of situations where you wouldn't want to use trial and error are diffusing a bomb or performing an operation on a patient. Anmelden Statistik 19.640 Aufrufe 65 Dieses Video gefÃ¤llt dir? If the method works, the person using it has acquired the correct solution to a problem. Linear Equations >> P.11. after testing $n=48$, $n=44$, $n=42$, $n=43$) that the upper limit for $n$ implied by $(0)$ is $n\le n_2=43$. If none of this trial-and-erroring can get a quadratic polynomial out of its bad mood, about all there is left to do is take it for ice cream and then put Let's try our other option. (3x + 1)(x â€“ 1) = 3x2 â€“ 2x â€“ 1Ah, that's more like it. It is primarily used to solve the problem. In summary: $$8n^2<64n\log_2n\iff 2\le n\le 43.$$ share\|cite\|improve this answer answered Mar 9 '13 at 11:14 Hagen von Eitzen 208k16206379 By what principle do the derivatives imply that $n_1\le Thus, the solutions of the quadratic equation are x = 2 and x = 4. Four such systems are identified: Natural selection which "educates" the DNA of the species, The brain of the individual (just discussed); The "brain" of society-as-such (including the publicly held body of Chen, S. We didn't actually need to check all the possible factorizations, but it's easier to check them all than it is to figure out which ones we could safely ignore. Wird geladen... However, typical simple examples of bogosort do not track which orders of the list have been tried and may try the same order any number of times, which violates one of Diese Funktion ist zurzeit nicht verfÃ¼gbar. Trial and error is primarily used to find a single solution to a single problem. Wird geladen... JSTOR4535886. Trial and error is used best when it is applied to situations that give your large amounts of time and safety to come up with a solution. Is turning off engines before landing "Normal"? Wird geladen... Your cache administrator is webmaster. More precisely, the derivative of the left hand side in$(1)$is$1$, that of the right hand side is$ \frac{8}{n\ln 2}\$. Anmelden 19 Wird geladen... Hutchinson, London & Praeger, New York. The more you practice factoring, the less error you'll run into, because you'll learn to see which trials will work without having to write down all the steps. Solution: By trial and error, we find x2 - 7x + 10 = (x - 2)(x - 5). Wiedergabeliste Warteschlange __count__/__total__ Solve Simple Equations By Trial And Error Method - Maths Algebra We Teach Academy Maths AbonnierenAbonniertAbo beenden4.8584Â Tsd. But not randomly! However, this is not a good technique for problems that don't give you multiple chances to find a solution. 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All Rights Reserved.IT Training and Consulting current community blog chat Mathematics Mathematics Meta your communities Sign up or log in to customize your list.	2018-10-23 12:58:03	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.28681299090385437, "perplexity": 1037.973563766521}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-43/segments/1539583516135.92/warc/CC-MAIN-20181023111223-20181023132723-00488.warc.gz"}
https://psychology.stackexchange.com/tags/sensory-perception/hot	# Tag Info Accepted ### How can someone asleep recognize a very brief sound? Short answer The auditory system remains active during sleep. Background Filtering of sensory input during sleep is a recognized phenomenon and indeed the senses are typically lulled during sleep. ... • 19.6k Accepted ### Does synesthesia lack symmetry? Generally spoken, synesthesia is unidirectional. For example, grapheme–color synesthesia (i.e., letter–color and digit–color synesthesia) is the most prevalent type of synesthesia. The presentation of ... • 19.6k ### Is there a term in psychology for when a tool is perceived as an extension of your body? Short answer Possible interesting terms are: distal attribution (externalization) body transfer illusion (rubber hand illusion) embodiment Background This is a very interesting, yet difficult ... • 19.6k Accepted ### Is there a difference between physiological stimulations and psychological stimulations? Psychology and physiology are at different levels of explanation or levels of analysis. The answer depends entirely on how you view the relationship between such levels and in particular (mental) ... Accepted ### Difference between thoughts and sensations Short answer Sensations are different from thoughts and are separated in the spatial and temporal domain. The distinction between thoughts and perceptions, however, is less well defined, but can still ... • 19.6k ### In psychophysics, why are log luminance rather than absolute luminance values reported? In general, subjective sensation increases linearly with the the log of physical intensity, which is described by Fechner's law. We are sensitive to small variations when light is dim, but we need ... • 19.6k ### If light travels at c, and the human nervous system's speed/perception speed<c, why aren't we not seeing or blind at some times? I think you are confusing speed with rate. The speed of light is how long it take to for light to get from one point to another, not how often light "events" arrive at the eye. To give an analogy, ... • 407 ### Difference between active and passive touch? The difference is in whether the animal has voluntary control over the touch. If the animal touches another object by moving its body to initiate the touch, then the it is active touch. If the animal ... • 5,824 Accepted ### Is inhibitory brain circuitry involved in cross-modal sensory perception? As far as I know, there is just one article that explicitly mentions the generation of cross-modal qualia, i.e., visual qualia in response to tactile stimulation (Ortiz et al., 2011). Before ... • 19.6k Accepted ### Is there knowledge of the receptive field patterns of cortical columns in associative brain regions? Short answer Associative brain areas are not retinotopically organized. Only lower visual areas more upstream from these areas are organized in such a predictable, low-level way. Higher up, things get ... • 19.6k Accepted ### Can binaural beats be generated with carrier tones outside the audible frequency range? Short answer No, infrasonic or ultrasonic sound cannot generate binaural beats. Background Binaural beats are generated in the brain and are associated with the frequency bands of the EEG. Binaural ... • 19.6k Accepted ### for a persistent perceptual experience, why is video able to have a lower frame rate than audio? Sound is pressure waves; young humans can hear (aka, detect pressure waves) up to about 20 kHz. To produce these high frequency waves with a speaker with a time-domain signal, it is necessary to have ... • 5,817 Accepted ### How quick a flash of light is invisible It depends on ambient light, but in darkness, humans can detect as few as several photons, perhaps even down to a single photon (though all detection at these low levels is probabilistic - see Tinsley ... • 5,817 ### What causes the sensation of taste when Dimethyl Sulfoxide (DMSO) is absorbed through the skin? Short answer It may not be so much the direct action of DMSO on the olfactory sensory system, but the smell of one of its metabolites that is excreted via the pulmonary system and the skin after ... • 19.6k ### Does the retina contribute in distinguishing lines and borders? Short answer Yes, retinal circuitry enhances the perception of contrast. Background In the retina, the neurons that guide the visual signal to the brain, the retinal ganglion cells (RGCs), process ... • 19.6k ### for a persistent perceptual experience, why is video able to have a lower frame rate than audio? I don't have a full answer, but it might get things started... You are mixing up two concepts frame rate and sampling rate. In a video presented at 24 fps each frame, potentially, has a wide range of ... • 2,617 ### How are tactile pleasure and pain differentiated in the somatosensory cortex? Obviously, pain and touch receptors are represented by entirely different receptors in the skin, i.e. nociceptors and mechanoreceptors, respectively. Each has physically different afferents that carry ... • 19.6k ### Synchronization of perception of sensory information The synchronization of sensory information is called multisensory integration: Multisensory integration, also known as multimodal integration, is the study of how information from the different ... • 18k ### Synchronization of perception of sensory information Short answer The brain actively integrates and synchronizes sensory inputs, up to the point that it actually delays one modality to match it with another. Background Your question is all about ... • 19.6k Accepted ### Is thermoception part of the sense of touch in the 5 human senses or is it a 6th separate sense? Short answer Heat receptors are often grouped under the 'skin receptors', and hence are bundled along with touch (pressure, vibration, stretch), cold and pain receptors. However, ciliopathy is a ... • 19.6k ### Multistable perception with three possibilities Here’s an image I found that triggers tristable perception (as opposed to just bistable perception): The three possible interpretations are A big cube with a smaller cube in front of it A big cube ... 1 vote Accepted ### Has there been any neuroscientific study of polydactyly? The extra finger seen in polydactyly is often connected only with a bit of skin, but it may contain bone and even joints (Fig. 1). According to the Boston's children's hospital, the extra finger (or ... • 19.6k 1 vote ### The differences between sensory distortions and hallucinations The differences between illusion and hallucinations are quite clear and have been known for over 100 years. For example, hallucinations are when you see things in total darkness or hear things when ... • 169 1 vote Accepted ### What scientific evidence is there for the definable real world quality of redness independent our perception? Short answer None. Background First off, I am not familiar with the principles as laid out in your question posed by Goethe and Feigenbaum (I'll look into these people, thanks for the pointer!). ... • 19.6k 1 vote ### What is the conceptual difference between causal inference and 'prediction' Correct me if I'm wrong here, but after some more reading and thinking this is what I took away: both the term 'prediction' and 'cause' here refer to the maximum likelihood of a distribution ($Y$), ... • 161 1 vote ### What are the definitions of 'multi-channel coding' and 'opponent channel coding'? Short answer Multi-channel coding in color vision refers to the different photoreceptors in the retina. Opponent-channel coding refers to the opposing color pairs: the red-green and yellow-blue ... • 19.6k 1 vote ### Can binaural beats be generated with carrier tones outside the audible frequency range? No, the carrier frequency for a binaural beat needs to be less than approximately 1500 Hz. The binaural beat arises because when two tones with slightly different frequencies are added, the ... • 2,617 1 vote ### How are positions and counts of higher concepts encoded in sparse representations? You are basically asking how to bind different concepts together based off their representation in neurons. The one way I know how to do this is using the Semantic Pointer Architecture (SPA). To ... • 8,783 1 vote Accepted ### How are positions and counts of higher concepts encoded in sparse representations? Two ideas on this so far: I think we have neurons representing multiple occurrences of a given feature, for example one neuron for "one face", one for "two faces", etc. At some number it doesn't ... • 751 Only top scored, non community-wiki answers of a minimum length are eligible	2022-07-02 08:22:33	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4310303032398224, "perplexity": 3453.2140783116697}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-27/segments/1656103989282.58/warc/CC-MAIN-20220702071223-20220702101223-00065.warc.gz"}
https://discuss.codechef.com/questions/144193/adaroks-editorial	× Setter- Alei Reyes Tester- Pranjal Jain Editorialist- Abhishek Pandey Medium-Hard # PROBLEM: Given a chessboard with kings placed on it, you need to find minimum number of rooks to place such that kings cannot reach each other. Rooks cannot be attacked by the kings for simplicity. # QUICK-EXPLANATION: Key to AC- Realize that since $NM \leq 1024$, at least one of the dimensions must be less than $32$. Now if we remove rows where kings are, and merge groups of empty rows into a single row, there can be at most $16$ rows where we can put rooks. Bitmasking + Clever Brute Force is the way to go. Without loss of generality, assume $N \leq M$, otherwise we can simply transpose the matrix and obtain the same result. This caps $N$ to $N \leq 32$. Now, we need to see how many rows we can put rooks on. Out of these $32$ rows, remove rows where kings are placed as rook cannot be placed there without attacking the king. Once its done, see that you might get some consecutive rows which don't have king on them. Placing a rook on any of the row gives same effect, as long as we place it in the same column. Hence, merge them all into single row. We see that this limits maximum number of rows we can place rooks on to $N \leq 16$. Try all possible subsets of rows on which we can place rooks. With rows decided, we now only need to see which columns we have to place rook. This problem can be shown equivalent to finding minimum number of lines to cut given set of intervals (where every interval must be cut by a line). The solution to this is, sort the intervals and greedily pick an interval to cut. All intervals intersecting with this interval are also cut by line and hence are removed. Repeat until no interval is left. # EXPLANATION: The editorial will be modularized into parts. You can feel free to jump at and start from the part unclear to you, provided you are clear with the parts before it :) 1. How to arrive at $N \leq 16$ result- First convince yourself that the board's area is $\leq 1024$. Now, this means, that at least one dimension must be $\leq \sqrt{1024}$, i.e. $\leq 32$. Without loss of generality, assume that $N \leq M$, as if thats not the case, we can rotate the matrix by $90^\circ$ and get same result. Now, put kings alternatively on these $32$ rows. That is, keep one king at row $1$, next at row $3$...and so on. We see that we have exactly $16$ rows with kings, and $16$ free rows. We are to see that we can, at no instance, have more than $16$ free rows. Now, addition of any more kings cannot increase number of free rows, as a rook cannot be placed in same row as of king. Also, if we remove any of the king, then the group of free rows will merge, again decreasing number of free rows. Also, we can see that the configuration we chose is a maximal one, no other configuration exists which has more free rows than this. Hence, number of rows where rooks can be put (or free rows as I called earlier) are $\leq 16$ 2. Getting bit-masking (without dp :p) into the picture- The above observation of $N \leq16$ makes life veryy convenient. The reason being, now we have $2^16$ possible choices of keeping rooks on the rows. This is, $=65536$ combinations to try, at max. Hence, we can simply try all the combinations! Once we are decided on which rows we have to put the rooks, I think you all can agree that the problem becomes somewhat simpler, and something which can be done with much less effort than what we imagined on reading the statement. Next part deals with how to calculate the final answer. In case you wish to do an independent try without any more spoilers from editorial, this is your checkpoint :p We store the column and row numbers of positions where kings are placed, so as to not to place rooks on them. Now, lets say we are placing rooks at row $i$ and row $j$, as per our current combination. Obviously, there should not be two kings on same column, else they can reach each other (recall that rooks are being placed ONLY at rows specified by our mask). Let me call the part of board bounded by row $i$ and row $j$ as $region$. Now, you can model the problem as, "Given possible intervals of free columns, where we have to place rooks (recall we need to place rooks between $2$ kings in same region!), what are minimum number of rooks to place such that each interval has at least $1$ rook assigned to it." The solution for this is to, sort all such intervals and greedily place rooks at larger interval, and eliminate intervals where rooks are already set. "I cannot understand how to make these intervals." Now, for all columns from $[1,M]$, start from row $i+1$ and iterate up to row $j-1$. Make a vector $v$ to store column numbers of any kings up see. This ensures columns are put in sorted order in $v$. Now, lets say we are looking at king present on index $idx$ on $v$. Our interval would be, all free columns after column of king at $v[idx-1]$ and all columns before those of king at $v[idx]$. A rough implementation is below- View Content # SOLUTION View Content View Content $Time$ $Complexity=O(2^{min(N,M)}MN)$ $Space$ $Complexity=O(NM)$ # CHEF VIJJU'S CORNER :D 1. Given $NM$ dimensional chess board, how many knights can you put at it such that no knight attacks each other? Solution 2. Given a $NN$ dimensional chess board, what if we ask to place bishops instead of knights in above problem. Can you draw a recurrence for it? Solution 3. Setter's Solution- View Content 4. Tester's Solution- View Content 5. Related Problems- This question is marked "community wiki". 15.4k12066 accept rate: 18% 19.8k350498541 toggle preview community wiki: Preview By Email: Markdown Basics • italic or _italic_ • bold or __bold__ • image?![alt text](/path/img.jpg "title") • numbered list: 1. Foo 2. Bar • to add a line break simply add two spaces to where you would like the new line to be. • basic HTML tags are also supported • mathemetical formulas in Latex between \$ symbol Question tags: ×1,406 ×160 ×145 ×18	2019-02-23 10:46:51	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6540793776512146, "perplexity": 906.7101722464514}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-09/segments/1550249500704.80/warc/CC-MAIN-20190223102155-20190223124155-00309.warc.gz"}
http://www.ams.org/mathscinet-getitem?mr=2869073	MathSciNet bibliographic data MR2869073 (2012k:35199) 35J92 (35A01 35A23 35B07 35J20 46E35) Su, Jiabao; Tian, Rushun Weighted Sobolev type embeddings and coercive quasilinear elliptic equations on \$\Bbb{R}\sp {N}\$$\Bbb{R}\sp {N}$. Proc. Amer. Math. Soc. 140 (2012), no. 3, 891–903. Article 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.	2014-07-31 12:23:42	{"extraction_info": {"found_math": true, "script_math_tex": 1, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9941163063049316, "perplexity": 9087.678056821838}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-23/segments/1406510273012.22/warc/CC-MAIN-20140728011753-00452-ip-10-146-231-18.ec2.internal.warc.gz"}
https://math.stackexchange.com/questions/2371897/center-of-a-triangle-free-connected-graph-atmost-two-vertex	Center of a triangle free connected graph atmost two vertex • Any graph $G$, $d(u,v)$ denote the length of the shortest path from $u$ to $v$. • $\epsilon (u) =$ max$_{v \in V(G)}d(u,v)$ is called eccentricity of $u$ and $r$ = min$\epsilon (u)$ denote the radius of a graph $G.$ • Center of a graph $G$ is the set of all vertex $u$ such that $\epsilon (u) = r$. Let $G$ be a triangle free graph. Then center of $G$ has atmost two vertices. I want to show by contradiction. Suppose that the center of $G$ has atleast three vertex says $u,v,w$. Then there exist $x,y,z \in V(G)$ such that $\epsilon (u) = \epsilon (v) = \epsilon (w) = d(u,x) = d(v,y) = d(w,z)$. If $u$ is not adjacent with $v$, then $d(u,v) > d(u,x )$, which is a contradiction. Thus $u,v,w$ form a triangle which is a contradiction. • Not sure how you determined $d(u, v)>d(u, x)$. – Dark Logician Jul 26 '17 at 4:56 • I feel that it should be right , but i do not know how to prove. – user120386 Jul 26 '17 at 5:17 • If $G$ is a regular $n$-gon, isn't every vertex in the center by symmetry? – stewbasic Jul 26 '17 at 6:11 • maybe there's something missing in the question, but the picture in the Wikipedia definition of the center of a graph gives a counterexample to that statement: en.wikipedia.org/wiki/Graph_center#/media/File:Graphcenter.svg – user96233 Jul 29 '17 at 20:07 • This statement is used in some theorem.Thats why I had asked this question here. Thanks to all – user120386 Jul 31 '17 at 7:32	2019-07-19 20:59:38	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8593131303787231, "perplexity": 152.5757466419507}, "config": {"markdown_headings": false, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-30/segments/1563195526359.16/warc/CC-MAIN-20190719202605-20190719224605-00422.warc.gz"}
https://physics.stackexchange.com/questions/575882/deducing-fusion-rules-of-non-abelian-fluxons	# Deducing fusion rules of non-abelian fluxons I have been reading about non-abelian fluxons in John Preskill's lectures notes on topological quantum computing and I do not understand how he deduced the fusion rules for fluxons in the example he gave. ## $$S_3$$ fluxons example Preskill's example is non-abelian fluxons of the group $$G = S_3$$, the permutation group on three elements. As shown on page 31, the total list of particles in the theory is given by $$\begin{array}{\|c\|c\|c\|c\|} \hline \text{Type}& \text{Flux} & \text{Charge} & \text{Dim} \\ \hline A & e& [+]& 1\\ \hline B & e & [-] &1\\ \hline C & e & [2]&2\\ \hline D & (12) & [+] &3\\ \hline E & (12) &[-] &3\\ \hline F & (123) &[1] &2\\ \hline G & (123) &[\omega] &2\\ \hline H & (123) & [\omega] &2\\ \hline \end{array}$$ where the flux labels are the possible conjugacy classes $$\alpha$$ of $$S_3$$ and the charge labels are the irreps of normaliser $$N(\alpha)$$ of the corresponding conjugacy class. It is stated that when we fuse two particles together we apply the following rule: The flux of the composite can belong to any of the conjugacy classes that can be obtained as a product of representatives of the classes that label the two constituents. Finding the charge of the composite is especially tricky, as we must decompose a tensor product of representations of two different normalizer groups as a sum of representations of the normalizer of the product flux. He provides the example that if we fuse two $$D$$ particles, we have in equation (9.47) $$D \times D = A + C + F + G + H$$ ## My question I do not undestand how this fusion rule was deduced. Preskill states that the possible charges of the composite must be formed by taking tensor products of the charges of the constituent particles. For the case of $$D$$ particles, the charge of these particles is the trivial representation $$[+]$$. If we take a tensor product of two trivial representations, it should surely form another trivial representation that will not decompose into a direct sum of irreps, i.e. $$[ + ] \otimes [+] = [+]$$, so how do we get the sum above? Preskill does not give much detail on exactly what the kinematics of these particles are. I would expect for a composite fluxon with flux labelled by a conjugacy class $$\alpha$$ and charge given by a representation of the corresponding normaliser $$R_{(\alpha)}^i$$, the Hilbert space that describes this system will be something like $$H_{\alpha}^i = \mathcal{H}_\alpha \otimes R_{(\alpha)}^i$$ where $$\mathcal{H}_\alpha$$ describes the flux degrees of freedom and $$R_{(\alpha)}^i$$ describes the charge degrees of freedom. So for example for a particle of type $$D$$ I would have $$H_D = \mathcal{H}_{(12)} \otimes R_{(12)}^{[+]}$$ so for fusion of two $$D$$ fluxons, I would write $$H_D \otimes H_D = (\mathcal{H}_{(12)} \otimes R_{(12)}^{[+]}) \otimes (\mathcal{H}_{(12)} \otimes R_{(12)}^{[+]}) \stackrel{?}{=} H_A \oplus H_C \oplus H_F \oplus H_G \oplus H_H$$ which I expect will decompose into irreps much like (9.47) above? Is this what John Preskill did to deduce the fusion rule?	2021-07-25 02:19:45	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 20, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7808429002761841, "perplexity": 296.59611074215223}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-31/segments/1627046151563.91/warc/CC-MAIN-20210725014052-20210725044052-00335.warc.gz"}
http://opencyto.org/gatingsets.html	View on GitHub # OpenCyto ## A Robust BioConductor Framework for Automated Flow Data Analysis OpenCyto Analysis of HVTN080 # OpenCyto Analysis of HVTN080 The HVTN080 study flow cytometry data has been preprocessed (loaded in R from the raw FCS files and the FlowJo workspaces). The gating set is available here, and has been gated using OpenCyto. We have gated the data using this OpenCyto template. The manually gated data is available here, and contains the manual gates from the FlowJo workspaces. The raw FCS files are available from flowrepository.org/id/FR-FCM-ZZ7U. The data in the gating sets has already been compensated and transformed using the compensation and transformation information stored in the FlowJo workspaces. The FlowJo workspaces for this data will be posted soon. The code to perform the gating and extract the features is here. ### Visualization of the gating hierarchy We load the gated data and visualize the results. suppressPackageStartupMessages( {require(flowWorkspace) require(flowViz) require(knitr) require(data.table) require(MIMOSA) require(GFMisc) require(scales) require(lme4) require(contrast) require(multcomp) require(COMPASS) }) opts_chunk\$set(list(message=FALSE,warning=FALSE)) WORKDIR<-"/Users/gfinak/Dropbox/GoTeam/Projects/Paper-OpenCytoPipeline/HVTN080/reproducible/" AUTO<-"./gs_auto_clean" MANUAL<-"./gs_manual_clean" setwd(WORKDIR) ## loading R object... ## Done manual<-load_gs(MANUAL) ## loading R object... ## Done Here is a view of the gating hierarchy for the manually gated data. plot(manual[[1]]) And here is the same for the OpenCyto gated data. plot(auto[[1]]) ### Visualize gating layouts for manual and automated gating We can view dotplots of each gate from selected samples for each of the OpenCyto and manually gated data. First, the OpenCyto gated data: plotGate(auto[[1]],scales=list(cex=c(1.2,1.2)),par.strip.text=list(cex=1.4),xlab=list(fontsize=12),ylab=list(fontsize=12),par.settings=list(gate.text=list(cex=1)),path=2) Then the manually gated data: plotGate(manual[[1]],scales=list(cex=c(1.2,1.2)),par.strip.text=list(cex=1.4),xlab=list(fontsize=12),ylab=list(fontsize=12),par.settings=list(gate.text=list(cex=1)),path=2)	2017-04-23 07:50:27	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 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": 0.429565966129303, "perplexity": 11216.219417048136}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-17/segments/1492917118310.2/warc/CC-MAIN-20170423031158-00554-ip-10-145-167-34.ec2.internal.warc.gz"}
http://math.stackexchange.com/tags/pi/new	# Tag Info 2 Consider the primes, written in order: $p_1, p_2, p_3,\ldots$ That is, $p_1=2, p_2=3, p_3=5, p_4=7, p_5=11, \ldots$. Now consider the sequence $$\frac{p_1-1}{p_1}, \frac{p_2-1}{p_2}, \frac{p_3-1}{p_3},\ldots$$ Much like in the OP, this is a sequence of fractions, all irreducible, with an increasingly larger prime dividing the denominator. Yet the limit ... 6 You proof idea breaks because it could prove that $0\notin\mathbb Q$ by considering the sequence $$a_n = 2^{-n}$$ We have $$\lim_{n\to\infty} a_n = 0$$ but the denominator is exactly $2^n$, wich diverges. Generally, you assume $$\lim_{n\to\infty} \frac{a_n}{b_n} = \frac ab \Rightarrow \lim_{n\to\infty} a_n = a \wedge \lim_{n\to\infty} b_n = b$$ wich is ... 0 The correct answer to the question what the circumference of a circle with diameter $d$ would be $\pi \cdot d$. Of course this is not a satisyfing answer. But since this ridiculous number $\pi$ cannot even be described by the root of a polynomial with coefficients in $\mathbb Q$ we can only approximate $\pi$. This is not a bad thing though. For most, if not ... 16 As I mentioned in the comments, the sizes of the squares are a little more complicated than what you're hoping for. I don't know of a simple expression for the size of each square, but you can get each one by solving a quadratic equation. I wrote a program to do so; it draws all the squares whose sizes are above a small threshold. Here's the result. Hope it ... 1 Your way of filling the corners will only (as far as I can see) a triangular part of them. It's going to be hard to fill them with squares. Chris says the same in his comment except that he actually tries to tell you what you might do (and he caught you miscounting on the number of smaller squares in each step). 2 The integral of any function over a set with measure $0$ is equal to $0$. 3 What will happen if $$(C+4)\pi^2+B\pi+(A-48)=0?$$ 0 Ah, 42 and The Hitchhiker's Guide to the Galaxy. Would you like to know how this is connected to the 24-dimensional Leech lattice? :) Given the Ramanujan-type formula, $$\sum_{n=0}^\infty \left(\frac{(2n)!}{(n!)^2}\right)^3\cdot \frac{An+B}{(C)^{n+1/2}}=\frac1\pi$$ there are relatively simple expressions for $A,C$. Define, $$A(\tau) = ... 2 The formula,$$\sum_{k=0}^\infty\frac{2^{-5k}(6k+1)((2k-1)!!)^3}{4(k!)^3} = \frac{1}{\pi}\tag1$$or its equivalent form,$$\sum_{k=0}^\infty \frac{(2k)!^3}{k!^6}\frac{6k+1}{(2^8)^{k+1/2}} = \frac{4}{\pi}\tag2$$and the similar,$$\sum_{k=0}^\infty (-1)^k \frac{(2k)!^3}{k!^6}\frac{6k+1}{(2^9)^{k+1/2}} = \frac{8}{\pi}\tag3$$belong to the Ramanujan-type ... 1 After you get the angle at or below 2\pi the reference angle is just the angle 0\le x\le\frac{\pi}{2} that is the distance between your angle and the x axis. So in quadrant I it is just x. Quadrant II it is \pi -x Quadrant III it is x-\pi Quadrant IV it is 2\pi - x Some example reference angles in degrees... 30 is 30. 90 is 90. 95 is 85. ... 0 In quadrant III, subtract \pi from the value. In quadrant II, subtract the value from \pi. You don't need to subtract anything if it's in quadrant I! Note: I may have misunderstood the question - I'm not sure if this is right. 7 2\sqrt2 and \sqrt2 are two distinct irrational numbers s.t. the statement: '2\sqrt2+\sqrt2 is rational or 2\sqrt2-\sqrt2 is rational' is not true. \sqrt2 and 2-\sqrt2 are two distinct irrational numbers s.t. the statement: '\sqrt2+(2-\sqrt2) is rational or \sqrt2-(2-\sqrt2) is rational' is true. This illustrates that the statement cannot be ... 12 This seems to be an open problem. It is a conjecture that the statement is false, i.e. that \pi + e and \pi - e are irrational. According to Wikipedia this remains unproven. (Just imagine the impact of the discovery of an equation such as \pi=e+\frac{4233108252.........3123782}{31238295213.......0591231} ... unbelievable!) Remark that at least one of ... 1 \tau\equiv\dfrac{C}{R}=2\pi. See tauday.com It is a symbol used by \tauists that define \tau to be the true circle constant. 1 This is probably not the standard way at all but one possible way to derive the value for \pi is to consider the following diagram. We know the circumference of the circle is 2\pi r so we set \|OA\|= 1 Now we get a poor approximation to by saying it is: 2 \cdot \|AB\| = 2 \cdot \sqrt{2} \approx 2.82 If we bisect AB at E and draw a segment from O ... 10 Yes, of course there is a 0 in the decimal expansion of \pi=3.1415926535897932384626433832795\underline{0}2884197.... 0 A number of different ways of showing this exist. First notice that the area has to be (\text{constant}\cdot r^2) because the area of a region of any shape in a plane must be proportional to the square of the distances. E.g. if you multiply all distances by 3, then the area is multiplied by 9. And "constant" in this case means it's the same number ... 2 In around 250 BC, Archimedes expressed \pi as a limit. He constructed sequences of inscribed and circumscribed polygons whose perimeters were lower and upper bounds of the value of \pi respectively, such that the perimeters converged to the same value. I do not know how rigorously the ancient Greeks were capable of proving that the perimeters truly were ... 3 Say you're trying to approximate a number by rational numbers p/q. Usually, the bigger q is, the better your chance of approximating the number closely. On the other hand, the smaller q is, the simpler the approximation. In the case of \pi, if you want to have a better approximation than 22/7, you have to go all the way up to q = 57. (See this.) ... 2 3.14 is two decimal places. That's the only justification I can give. When doing hand calculations (or sliderule calculations), carrying out more than three significant digits is cumbersome. Remember this: with N digits in a multiplication, you have N^2 digits in the final answer before you can round again. In the world of engineering, often you can't ... 4 This is the same as proving that$$\frac{1}{16}(133-37\sqrt{5})>\pi$$and it follows from the fact that the continued fraction of the LHS is:$$ [3;7,15,1,660,\ldots] $$while the continued fraction of \pi is:$$ [3;7,15,1,292,\ldots]. 2 If you were actually generating random real numbers, the probability of any of those falling exactly on the circle line (a set of measure zero) would be zero and thus by the law of large numbers not alter the outcome of the Monte Carlo simulation as the number of points goes to infinity, independent of whether you count those points as "inside", "outside", ... Top 50 recent answers are included	2014-12-21 21:35:46	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8621282577514648, "perplexity": 236.80358807000798}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-52/segments/1418802772416.132/warc/CC-MAIN-20141217075252-00122-ip-10-231-17-201.ec2.internal.warc.gz"}
https://www.math.ntnu.no/conservation/2003/042.html	### On the Convergence Rate of Vanishing Viscosity Approximations Alberto Bressan and Tong Yang Abstract: Given a strictly hyperbolic, genuinely nonlinear system of conservation laws, we prove the a priori bound $\big\\|u(t,\cdot)-u^\ve(t,\cdot)\big\\|_{\L^1}= \O(1)(1+t)\cdot \sqrt\ve\|\ln\ve\|$ on the distance between an exact BV solution $u$ and a viscous approximation $u^\ve$, letting the viscosity coefficient $\ve\to 0$. In the proof, starting from $u$ we construct an approximation of the viscous solution $u^\ve$ by taking a mollification $u*\varphi_{\strut \sqrt\ve}$ and inserting viscous shock profiles at the locations of finitely many large shocks, for each fixed $\ve$. Error estimates are then obtained by introducing new Lyapunov functionals which control shock interactions, interactions between waves of different families and by using sharp decay estimates for positive nonlinear waves. Paper: Available as PostScript (512 Kbytes) or gzipped PostScript (200 Kbytes; uncompress using gunzip). Author(s): Alberto Bressan, <bressan@sissa.it> Tong Yang, <matyang@cityu.edu.hk> Publishing information: To appear in Comm. Pure Appl. Math. Comments: Submitted by: <matyang@cityu.edu.hk> June 17 2003. [ 1996 \| 1997 \| 1998 \| 1999 \| 2000 \| 2001 \| 2002 \| 2003 \| All Preprints \| Preprint Server Homepage ] © The copyright for the following documents lies with the authors. Copies of these documents made by electronic or mechanical means including information storage and retrieval systems, may only be employed for personal use. Conservation Laws Preprint Server <conservation@math.ntnu.no> 2003-10-24 14:20:03 UTC	2018-12-19 11:24:11	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8069725036621094, "perplexity": 1315.2197107842028}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-51/segments/1544376832259.90/warc/CC-MAIN-20181219110427-20181219132427-00307.warc.gz"}
http://openturns.github.io/openturns/latest/user_manual/_generated/openturns.HMatrix.html	# HMatrix¶ class HMatrix(args) Hierarchical matrices. Hierarchical matrices (or HMatrix) are a compressed representation of dense matrices. In many applications, matrix coefficients represent an interaction between two degrees of freedom; when these interactions are smooth, it is possible to approximate sub-blocks by a local low-rank approximation B =~ UV^T where B has dimension (m,n), U (m,k), and V (n,k). Of course, this is interesting only if k is much lower than m and n. In order to obtain this compressed representation, several different steps must be performed: 1. Clustering: creation of rows and columns cluster trees Vertices where interactions are computed are reordered to improve locality. A binary space partition algorithm is used to recursively divide vertex set. Root cell contains all vertices. At each recursion step, a cell is divided into two new cells until it contains less than a given number of vertices. Space partition is performed orthogonally to original axis, by cutting its longest dimension. • The ‘median’ clustering algorithm divides a cell into two cells containing the same number of degrees of freedom. • The ‘geometric’ clustering algorithm divides a cell into two cells of the same geometric size • The ‘hybrid’ clustering algorithm is a mix. It first performs a ‘median’ bisection; if volumes of these new cells are very different, a ‘geometric’ clustering is performed instead. 2. Admissibility: creation of an empty HMatrix structure The first step created a full binary tree for rows and columns degrees of freedom. We will now create a hierarchical representation of our matrix by checking whether some blocks can be replaced by low-rank approximations. The whole matrix represents the interactions of all rows degrees of freedom against all columns degrees of freedom. It can not be approximated by a low-rank approximation, and thus it is replaced by 4 blocks obtained by considering interactions between rows and columns children nodes. This operation is performed recursively. At each step, we compute axis aligned bounding boxes rows_bbox and cols_bbox: if min(diameter(rows_bbox), diameter(cols_bbox)) <= etadistance(rows_bbox, cols_bbox) then we consider that interaction between rows and columns degrees of freedom can have a local low-rank approximation, and recursion is stopped. Otherwise, we recurse until bottom cluster tree is reached. The whole matrix is thus represented by a 4-tree where leaves will contain either low-rank approximation or full blocks. The eta parameter is called the admissibility factor, and it can be modified. 3. Assembly: coefficients computations The hierarchical structure of the matrix has been computed during step 2. To compute coefficients, we call the assemble method and provide a callable to compute interaction between two nodes. Full blocks are computed by calling this callable for the whole block. If compression method is ‘SVD’, low-rank approximation is computed by first computing the whole block, then finding its singular value decomposition, and rank is truncated so that error does not exceed assemblyEpsilon. This method is precise, but very costly. If compression method is a variant of ACA, only few rows and columns are computed. This is much more efficient, but error may be larger than expected on some problems. 4. Matrix computations Once an HMatrix is computed, usual linear algebra operations can be performed. Matrix can be factorized in-place, in order to solve systems. Or we can compute its product by a matrix or vector. But keep in mind that rows and columns are reordered internally, and thus results may differ sensibly from standard dense representation (for instance when computing a Cholesky or LU decomposition). Methods assemble(args) Assemble matrix. assembleReal(callable, symmetry) Assemble matrix. assembleTensor(callable, outputDimension, …) Assemble matrix by block. compressionRatio() Compression ratio accessor. copy() Copy matrix. dump(name) Save matrix to a file. factorize(method) Factorize matrix. fullrkRatio() Block ratio accessor. gemm(transA, transB, alpha, a, b, beta) Multiply matrix in-place self=alphaop(A)op(B)+betaself. gemv(trans, alpha, x, beta, y) Multiply vector in-place y=alphaop(A)x+betay. getClassName() Accessor to the object’s name. getDiagonal() Diagonal values accessor. getId() Accessor to the object’s id. getImplementation(args) Accessor to the underlying implementation. getName() Accessor to the object’s name. getNbColumns() Accessor to the number of columns. getNbRows() Accessor to the number of rows. norm() Compute norm value. scale(alpha) Scale matrix in-place A=alphaA. setName(name) Accessor to the object’s name. solve(args) Solve linear system op(A)x=b, after A has been factorized. solveLower(args) Solve lower linear system op(L)x=b, after A has been factorized. transpose() Transpose matrix in-place. __init__(args) Initialize self. See help(type(self)) for accurate signature. assemble(args) Assemble matrix. Parameters: f : HMatrixRealAssemblyFunction or HMatrixTensorRealAssemblyFunction Assembly function. symmetry : str Symmetry flag, either N or L assembleReal(callable, symmetry) Assemble matrix. Parameters: f : assembly function Callable that takes i,j int parameters and returns a float symmetry : str Symmetry flag, either N or L assembleTensor(callable, outputDimension, symmetry) Assemble matrix by block. Parameters: f : assembly function Callable that takes i,j int parameters and returns a Matrix outputDimension : int Block dimension symmetry : str Symmetry flag, either N or L compressionRatio() Compression ratio accessor. Returns: ratio : 2-tuple of int Numbers of elements in the compressed and uncompressed forms. copy() Copy matrix. As factorization overwrites matrix contents, this method is useful to get a copy of assembled matrix before it is factorized. Returns: matrix : HMatrix Matrix copy. dump(name) Save matrix to a file. Parameters: fileName : str File name to save to. factorize(method) Factorize matrix. Parameters: method : str Factorization method, either one of: LDLt, LLt or LU fullrkRatio() Block ratio accessor. Returns: ratio : 2-tuple of int Numbers of elements in full blocks and low rank blocks. gemm(transA, transB, alpha, a, b, beta) Multiply matrix in-place self=alphaop(A)op(B)+betaself. Parameters: transA : str Whether to use A or A^t: either N or T. transB : str Whether to use B or B^t: either N or T. alpha : float Coefficient a : HMatrix Multiplied matrix A. b : HMatrix Multiplied matrix B. beta : float Coefficient. gemv(trans, alpha, x, beta, y) Multiply vector in-place y=alphaop(A)x+betay. Parameters: trans : str Whether to use A or A^t: either N or T. alpha : float Coefficient x : sequence of float Vector to multiply. beta : float Coefficient. y : Point Vector multiplied in-place. getClassName() Accessor to the object’s name. Returns: class_name : str The object class name (object.__class__.__name__). getDiagonal() Diagonal values accessor. Returns: diag : Point Diagonal values. getId() Accessor to the object’s id. Returns: id : int Internal unique identifier. getImplementation(args) Accessor to the underlying implementation. Returns: impl : Implementation The implementation class. getName() Accessor to the object’s name. Returns: name : str The name of the object. getNbColumns() Accessor to the number of columns. Returns: nbColumns : int Number of columns. getNbRows() Accessor to the number of rows. Returns: nbRows : int Number of rows. norm() Compute norm value. Returns: norm : float Frobenius norm. scale(alpha) Scale matrix in-place A=alphaA. Parameters: alpha : float Coefficient. setName(name) Accessor to the object’s name. Parameters: name : str The name of the object. solve(args) Solve linear system op(A)x=b, after A has been factorized. Parameters: b : sequence of float or Matrix Second term of the equation, vector or matrix. trans : bool Whether to solve the equation with A (False) or A^t (True). Defaults to False. x : Equation solution, vector or matrix. solveLower(args) Solve lower linear system op(L)*x=b, after A has been factorized. Parameters: b : sequence of float or Matrix Second term of the equation, vector or matrix. trans : bool Whether to solve the equation with L (False) or L^t (True). Defaults to False. x : Equation solution, vector or matrix. transpose() Transpose matrix in-place.	2018-12-12 17:19:04	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.39907556772232056, "perplexity": 3697.0119478705255}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-51/segments/1544376824059.7/warc/CC-MAIN-20181212155747-20181212181247-00616.warc.gz"}
http://www.pygsti.info/tutorials/13_GST_on_2_qubits.html	# An example of how to run GST on a 2-qubit system¶ This tutorial gives an overview of the typical steps used to perform an end-to-end (i.e. experimental-data-to-report) Gate Set Tomography analysis on a 2-qubit system. The steps are very similar to the single-qubit case; the main differences are: • the use of more general syntax when constructing 2-qubit gate sets • the increased number of fiducial and germ gate sequences • the increased run time required to compute GST estimates • a different report style In [1]: import pygsti ### Step 1: Construct the desired 2-qubit gateset¶ There are several ways to do this, as outlined by the comments in the cell below. In [2]: # via build_gateset: # [4] = a 4-dimensional Hilbert (state) space # [('Q0','Q1')] = interpret this 4-d space as that of two qubits 'Q0', and 'Q1' (note these labels must begin with 'Q'!) # "Gix" = gate label; can be anything that begins with 'G' and is followed by lowercase letters # "X(pi/2,Q1)" = pi/2 single-qubit x-rotation gate on the qubit labeled Q1 # "CX(pi,Q0,Q1)" = controlled pi x-rotation using qubits Q0 (control) and Q1 (target) # "rho0" = prep label; can be anything that begins with "rho" # "E1" = effect label; can be anything that begins with "E" # "2" = a prep or effect expression indicating a projection/preparation of the 3rd (b/c 0-based) computational basis element # 'dnup': ('rho0','E2') = designate the SPAM label "dnup" to mean preparation using "rho0" (a prep label) and measuring the outcome "E2" (an effect label) # "pp" = create all of these gate & SPAM operators in the Pauli-product basis. gs_target = pygsti.construction.build_gateset( [4], [('Q0','Q1')],['Gix','Giy','Gxi','Gyi','Gcnot'], [ "X(pi/2,Q1)", "Y(pi/2,Q1)", "X(pi/2,Q0)", "Y(pi/2,Q0)", "CX(pi,Q0,Q1)" ], prepLabels=['rho0'], prepExpressions=["0"], effectLabels=['E0','E1','E2'], effectExpressions=["0","1","2"], spamdefs={'upup': ('rho0','E0'), 'updn': ('rho0','E1'), 'dnup': ('rho0','E2'), 'dndn': ('rho0','remainder') }, basis="pp") # Note that you can also explicity add identity operations, e.g. "I(Q0)" to get the same gate set, # and that this same syntax could be used for non-entangling 2-qubit gates, e.g. "X(pi/2,Q0):X(pi/2,Q1)". gs_targetB = pygsti.construction.build_gateset( [4], [('Q0','Q1')],['Gix','Giy','Gxi','Gyi','Gcnot'], [ "I(Q0):X(pi/2,Q1)", "I(Q0):Y(pi/2,Q1)", "X(pi/2,Q0):I(Q1)", "Y(pi/2,Q0):I(Q1)", "CX(pi,Q0,Q1)" ], prepLabels=['rho0'], prepExpressions=["0"], effectLabels=['E0','E1','E2'], effectExpressions=["0","1","2"], spamdefs={'upup': ('rho0','E0'), 'updn': ('rho0','E1'), 'dnup': ('rho0','E2'), 'dndn': ('rho0','remainder') }, basis="pp") # If you're lucky and your gateset is one of pyGSTi's "standard" gate sets, you can just import it. from pygsti.construction import std2Q_XYCNOT gs_targetC = std2Q_XYCNOT.gs_target #check that these are all the same assert(abs(gs_target.frobeniusdist(gs_targetB)) < 1e-6) assert(abs(gs_target.frobeniusdist(gs_targetC)) < 1e-6) ### Step 2: Obtain lists of fiducial and germ gate sequences¶ These are the building blocks of the gate sequences performed in the experiment. Typically, these lists are either given to you by the folks at Sandia National Labs (email pygsti@sandia.gov), provided by pyGSTi because you're using a "standard" gate set, or computed using "fiducial selection" and "germ selection" algorithms (which are a part of pyGSTi, but not covered in this tutorial). In [3]: #If you know the fiducial strings you can create a list manually. Note # that in general there can be different "preparation" and "measurement" # (or "effect") fiducials. prep_fiducials = pygsti.construction.gatestring_list( \ [ (), ('Gix',), ('Giy',), ('Gix','Gix'), ('Gxi',), ('Gxi','Gix'), ('Gxi','Giy'), ('Gxi','Gix','Gix'), ('Gyi',), ('Gyi','Gix'), ('Gyi','Giy'), ('Gyi','Gix','Gix'), ('Gxi','Gxi'), ('Gxi','Gxi','Gix'), ('Gxi','Gxi','Giy'), ('Gxi','Gxi','Gix','Gix') ] ) effect_fiducials = pygsti.construction.gatestring_list( \ [ (), ('Gix',), ('Giy',), ('Gxi',), ('Gyi',), ('Gix','Gxi'), ('Gxi','Giy'), ('Gyi','Gix'), ('Gyi','Giy'), ('Gxi','Gxi') ] ) #Or, if you're lucky, you can just import them prep_fiducialsB = std2Q_XYCNOT.prepStrs effect_fiducialsB = std2Q_XYCNOT.effectStrs #check that these are the same assert(prep_fiducials == prep_fiducialsB) assert(effect_fiducials == effect_fiducialsB) #Use fiducial sequences to create a "spam specifiers" object, telling # GST which preparation and measurement fiducials to follow and precede which # state preparation and effect operators, respectively. specs = pygsti.construction.build_spam_specs( prepStrs=prep_fiducials, effectStrs=effect_fiducials, prep_labels=gs_target.get_prep_labels(), effect_labels=gs_target.get_effect_labels() ) #Alternatively, if you're lucky, you can grab the specs directly: specsB = std2Q_XYCNOT.specs assert(specs[0] == specsB[0]) In [4]: #germ lists can be specified in the same way. In this case, there are # 71 germs required to do honest GST. Since this would crowd this tutorial # notebook, we create some smaller lists of germs manually and import the # full 71-germ list from std2Q_XYCNOT germs4 = pygsti.construction.gatestring_list( [ ('Gix',), ('Giy',), ('Gxi',), ('Gyi',) ] ) germs11 = pygsti.construction.gatestring_list( [ ('Gix',), ('Giy',), ('Gxi',), ('Gyi',), ('Gcnot',), ('Gxi','Gyi'), ('Gix','Giy'), ('Gix','Gcnot'), ('Gxi','Gcnot'), ('Giy','Gcnot'), ('Gyi','Gcnot') ] ) germs71 = std2Q_XYCNOT.germs ### Step 3: Data generation¶ Now that fiducial and germ strings have been found, we can generate the list of experiments needed to run GST, just like in the 1-qubit case. As an additional input we'll need a list of lengths indicating the maximum length strings to use on each successive GST iteration. In [5]: #A list of maximum lengths for each GST iteration maxLengths = [1,2,4] #Create a list of GST experiments for this gateset, with #the specified fiducials, germs, and maximum lengths. We use #"germs4" here so that the tutorial runs quickly; really, you'd #want to use germs71! listOfExperiments = pygsti.construction.make_lsgst_experiment_list(gs_target.gates.keys(), prep_fiducials, effect_fiducials, germs4, maxLengths) #Create an empty dataset file, which stores the list of experiments # and zerod-out columns where data should be inserted. Note the use of the SPAM # labels in the "Columns" header line. pygsti.io.write_empty_dataset("tutorial_files/My2QDataTemplate.txt", listOfExperiments, "## Columns = upup count, updn count, dnup count, dndn count") In [6]: #Generate some "fake" (simulated) data based on a depolarized version of the target gateset gs_datagen = gs_target.depolarize(gate_noise=0.1, spam_noise=0.001) ds = pygsti.construction.generate_fake_data(gs_datagen, listOfExperiments, nSamples=1000, sampleError="multinomial", seed=2016) #if you have a dataset file with real data in it, load it using something like: ### Step 4: Run GST using do_long_sequence_gst¶ Just like for 1-qubit GST, we call the driver routine do_long_sequence_gst to compute the GST estimates. Usually for two qubits this could take a long time (hours) based on the number of gate sequences used. However, since we chose an incomplete set of only 4 germs and set our maximum max-length to 4, this will run fairly quickly (~20min). Some notes about the options/arguments to do_long_sequence_gst that are particularly relevant to 2-qubit GST: • mxBasis indicates which basis the target gateset matrices are in -- this should be the same as that used to create the gateset. Here 'pp' denotes the Pauli-product basis. • advancedOptions expects a dictionary of with option names as the keys. Relevant options are: • memoryLimitInBytes gives an estimate of how much memory is available to use on your system (in bytes). This is currently not a hard limit, and pyGSTi may require 50-100% more memory than this "limit". So you'll need to be conservative in the value you place here: if your machine has 10GB of RAM, set this to 3 or 5 GB initially and increase it as you see how much memory is actually used using a separate OS performance monitor tool. • depolarizeLGST gives an amount (in [0,1]) to depolarize the initial LGST estimate that is used as the initial guess for long-sequence GST. In practice, we find that in the larger 2-qubit Hilbert space, the LGST estimate may be so poor as to adversely affect the subsequent long-sequence GST (e.g. very slow convergence). Depolarizing the LGST estimate remedies this. If you're unsure what to put here, either don't specify depolarizeLGST at all (the same as using 0.0), or just use 0.1. • verbosity tells the routine how much detail to print to stdout. If you don't mind waiting a while without getting any output, you can leave this at its default value (2). If you can't standing wondering whether GST is still running or has locked up, set this to 3. In [7]: import time start = time.time() results = pygsti.do_long_sequence_gst(ds, gs_target, prep_fiducials, effect_fiducials, germs4, maxLengths, gaugeOptRatio=0.1, 'depolarizeLGST' : 0.1, 'verbosity' : 3} ) end = time.time() print "Total time=%f hours" % ((end - start)/3600.0) #If you wanted to, you could pickle the results for later analysis: #pickle.dump(results, open("MySavedResults.pkl", "w")) LGST: Singular values of I_tilde (truncating to first 16 of 16) = [ 6.43752847 2.13984198 2.10172622 1.26244223 1.20551035 1.03646394 0.84525607 0.80289546 0.53731306 0.51858772 0.3602247 0.34371294 0.32000311 0.21656255 0.20857989 0.17889827] --- LGST --- --- Gauge Optimization to TP (L-BFGS-B) --- 9s 0.0000029910 10s 0.0000029320 10s 0.0000029152 11s 0.0000029119 The resulting TP penalty is: 2.91191e-06 The gauge matrix found (B^-1) is: [[ 1.00002822e+00 8.39858500e-05 1.57419351e-05 -1.19777920e-04 1.11635776e-04 -1.23539072e-04 1.55511014e-05 1.60194489e-04 7.10550394e-05 -7.48321849e-05 1.51163644e-05 -7.44678475e-06 -1.47453707e-04 1.20325507e-04 4.50074480e-05 5.06034200e-05] [ 9.57099317e-08 9.99999891e-01 3.38597359e-08 -3.33879578e-08 -1.20002564e-07 8.68053526e-08 -4.87320470e-08 1.06466050e-07 5.82125274e-08 -1.23279168e-07 -2.30488576e-08 -4.99390795e-08 6.37590181e-09 -1.70250024e-09 3.85781928e-08 -2.63039550e-08] [ -1.15954038e-07 3.40968125e-08 9.99999886e-01 5.60386072e-08 1.75338235e-07 -6.05421109e-08 1.60770014e-07 -8.32288560e-08 -5.59783787e-08 -8.46655340e-08 -8.55066818e-08 6.14220469e-08 4.16334591e-08 -1.33552443e-08 9.25059311e-08 -2.84811484e-08] [ 5.30762317e-08 -2.55120591e-08 5.86324169e-08 9.99999940e-01 -8.72020801e-08 5.48248943e-08 -1.24927330e-07 5.84452449e-08 2.31331476e-08 1.85611424e-08 6.02073500e-08 -3.80871765e-08 -1.58868120e-08 -7.78051935e-09 -4.21297202e-08 -1.80981295e-08] [ 2.24974244e-07 -1.23079246e-07 1.74092273e-07 -9.46780679e-08 9.99999669e-01 1.09788466e-07 -2.61275186e-07 1.92026636e-07 1.09918741e-07 1.81611291e-08 8.88348708e-08 -1.16547563e-07 -3.84043140e-08 1.27416517e-08 -9.40887003e-08 1.22757129e-08] [ -8.19648861e-08 9.48236286e-08 -5.77669813e-08 5.68033401e-08 1.16739460e-07 9.99999771e-01 6.44337895e-08 -6.95462728e-08 -6.68246562e-08 7.99054977e-09 -4.72844780e-08 5.24959284e-08 4.76577934e-08 -5.47939801e-09 1.57644666e-08 -2.13655663e-08] [ 1.32496206e-07 -4.97334177e-08 1.60390884e-07 -1.25827206e-07 -2.61901594e-07 6.37329675e-08 9.99999631e-01 1.43151453e-07 2.36079566e-08 8.85507695e-08 1.33625048e-07 -8.97300164e-08 -1.07270235e-08 -3.04798899e-08 -1.30568369e-07 -3.88010163e-08] [ -1.42864606e-07 9.56406275e-08 -8.68652824e-08 5.91301055e-08 1.81324157e-07 -6.59000505e-08 1.41724304e-07 9.99999837e-01 -6.91441516e-08 8.30384372e-08 9.63017387e-09 6.11775356e-08 1.87768967e-08 -1.88856015e-08 2.86472269e-09 3.52135731e-08] [ -9.95528764e-08 5.43359991e-08 -5.73393951e-08 2.09582838e-08 1.07067584e-07 -6.81194738e-08 2.34486159e-08 -6.56950158e-08 9.99999917e-01 1.34094945e-08 -2.20307664e-08 4.34461268e-08 3.75641527e-08 -2.51169494e-08 1.67742737e-08 -3.71469154e-08] [ 1.38905119e-08 -1.16711451e-07 -8.25050476e-08 1.59305824e-08 2.55599206e-08 3.46241804e-09 8.97273295e-08 8.60906518e-08 1.66230714e-08 9.99999675e-01 -1.64715314e-07 -1.60419603e-10 7.28000514e-08 -2.06213845e-08 1.83043816e-07 -1.00247390e-07] [ -3.73807529e-08 -2.52745098e-08 -8.61908971e-08 6.24644517e-08 8.58004713e-08 -4.43326792e-08 1.33051622e-07 6.67563243e-09 -2.37933778e-08 -1.63536446e-07 9.99999839e-01 4.55700538e-08 4.16986251e-08 7.38695141e-09 1.25015175e-07 -3.95174775e-08] [ 8.08316678e-08 -4.98558828e-08 6.14134984e-08 -3.83978072e-08 -1.16385192e-07 5.21376190e-08 -8.96869723e-08 6.17063006e-08 4.36038630e-08 -4.85188507e-10 4.56713777e-08 9.99999949e-01 -1.27826919e-08 -2.84311139e-09 -3.19485048e-08 -4.77104388e-10] [ 3.94569766e-08 1.18595428e-08 4.36894622e-08 -8.45370030e-09 -3.60706196e-08 5.39199919e-08 -1.10509873e-08 7.91704459e-09 3.54211127e-08 8.09728531e-08 3.79727527e-08 -1.25042519e-08 9.99999943e-01 3.28528078e-08 -4.81974882e-08 4.07443257e-08] [ -2.62468968e-08 -8.14364539e-09 -1.56203917e-08 -1.09385775e-08 7.78636889e-09 -7.06368755e-09 -3.08404170e-08 -1.38455559e-08 -2.53792451e-08 -2.55286383e-08 1.01832175e-08 -3.14778344e-09 3.67858533e-08 9.99999962e-01 1.10908289e-08 -2.53014329e-08] [ 4.55023290e-08 3.55660903e-08 9.15062553e-08 -4.17888200e-08 -9.71116816e-08 1.69285311e-08 -1.30979639e-07 2.63354100e-09 1.57478414e-08 1.82341209e-07 1.25782532e-07 -3.20897556e-08 -4.50810030e-08 9.56636383e-09 9.99999870e-01 5.14374636e-08] [ -2.34361981e-08 -3.22552208e-08 -3.01549612e-08 -1.13179896e-08 4.09751604e-09 -1.28762843e-08 -4.03250186e-08 2.60443792e-08 -4.20858312e-08 -9.62602428e-08 -3.98623492e-08 -6.46475012e-10 5.10314278e-08 -3.33087848e-08 4.90161255e-08 9.99999925e-01]] The gauge-corrected gates are: rho0 = 0.5001 0 0 0.5000 0 0 0 0 0 0 0 0 0.5000 0 0 0.5000 E0 = 0.5810 -0.0380 0.0670 0.4619 -0.0469 -0.0101 0.0010 -0.0659 0.0740 -0.0710 0.1000 0.0650 0.4529 -0.0479 0.0710 0.4980 E1 = 0.5170 0.0460 -0.0610 -0.4531 -0.0659 0.0169 0.0300 0.0701 0.0660 0.0090 -0.0380 -0.0420 0.3559 0.0511 -0.0620 -0.4180 E2 = 0.4915 -0.0255 0.0695 0.3494 0.0476 -0.0496 -0.0565 0.0676 -0.0715 0.0915 0.0125 -0.0895 -0.4426 0.0166 -0.0745 -0.3965 Gix = 0.9997 0.0003 -0.0001 0.0002 0.0004 -0.0004 0.0003 -0.0002 -0.0002 0.0001 -0.0002 0.0002 0 0 0 0 -0.0096 0.9236 -0.0096 0.0096 0.0057 0.0136 0.0057 -0.0057 -0.0019 0.0402 -0.0019 0.0019 0.0073 -0.0118 0.0073 -0.0073 -0.0847 0.1121 -0.0847 -0.9153 -0.0279 0.0152 -0.0279 0.0279 0.0396 -0.0455 0.0396 -0.0396 0.0156 -0.0296 0.0156 -0.0156 -0.0905 0.0663 0.9095 0.0905 0.0086 0.0717 0.0086 -0.0086 -0.0088 -0.0141 -0.0088 0.0088 -0.0167 0.0207 -0.0167 0.0167 0.0073 -0.0376 0.0073 -0.0073 0.8880 0.0633 -0.1120 0.1120 0.0328 -0.0573 0.0328 -0.0328 0.0010 -0.0249 0.0010 -0.0010 -0.0069 -0.0144 -0.0069 0.0069 0.0349 0.8330 0.0349 -0.0349 0.0045 0.0465 0.0045 -0.0045 -0.0241 0.0221 -0.0241 0.0241 0.0134 -0.0366 0.0134 -0.0134 -0.1173 0.2052 -0.1174 -0.8826 0.0832 -0.1196 0.0832 -0.0832 0.0252 -0.0304 0.0252 -0.0252 -0.0339 0.0608 -0.0339 0.0339 0.0309 -0.0237 1.0309 -0.0309 0.0229 0.0485 0.0229 -0.0229 -0.0020 0.0846 -0.0020 0.0020 0.0031 -0.0023 0.0031 -0.0031 0.0042 0.0215 0.0042 -0.0042 0.9040 -0.0273 -0.0960 0.0960 0.0114 -0.0179 0.0114 -0.0114 -0.0128 0.0221 -0.0128 0.0128 -0.0776 0.0353 -0.0776 0.0776 -0.0379 0.8649 -0.0379 0.0379 -0.0221 0.0283 -0.0221 0.0221 -0.0199 -0.0019 -0.0199 0.0199 0.0058 0.1160 0.0058 -0.0058 -0.1566 0.0590 -0.1566 -0.8434 0.0248 -0.0308 0.0248 -0.0248 0.0152 -0.0539 0.0152 -0.0152 -0.0082 0.0242 -0.0082 0.0082 -0.0518 0.0841 0.9482 0.0518 0.0077 -0.0076 0.0077 -0.0077 -0.0236 -0.0055 -0.0236 0.0236 0.0011 -0.0043 0.0011 -0.0011 -0.0484 0.0685 -0.0484 0.0484 0.9038 -0.0013 -0.0962 0.0962 0.0157 -0.0017 0.0157 -0.0157 0.0015 -0.1531 0.0015 -0.0015 -0.0558 0.1375 -0.0558 0.0558 0.0136 0.8740 0.0136 -0.0136 -0.0317 0.0003 -0.0317 0.0317 0.0497 0.0281 0.0497 -0.0497 -0.0479 0.1197 -0.0479 0.0479 -0.1037 0.1154 -0.1037 -0.8963 0.0038 0.0325 0.0038 -0.0038 0.0092 -0.0939 0.0092 -0.0092 -0.0087 0.0261 -0.0087 0.0087 -0.0718 0.1017 0.9282 0.0718 Giy = 0.9998 0 -0.0002 0 0.0004 0.0002 0.0004 -0.0001 0 -0.0002 -0.0002 0.0001 0 0 0.0002 0 0.0905 -0.1291 0.0635 0.9095 0.0130 0.0729 0.0366 -0.0130 -0.0198 0.0394 -0.0254 0.0198 -0.0129 0.0141 0.0027 0.0129 0.0136 -0.0046 0.9403 -0.0136 0.0099 -0.0152 -0.0301 -0.0099 0.0098 -0.0290 0.1028 -0.0098 0.0078 0.0105 0.0433 -0.0078 -0.0793 -0.9316 -0.1088 0.0793 -0.0276 0.0613 -0.0202 0.0276 0.0101 -0.0190 -0.0384 -0.0101 -0.0117 0.0182 0.0071 0.0117 0.0182 0.0263 0.0069 -0.0182 0.9106 0.0413 0.0157 0.0894 0.0169 0.0179 0.0266 -0.0169 0.0056 -0.0411 0.0361 -0.0056 0.0011 -0.0583 -0.0503 -0.0011 0.1213 -0.1151 0.2036 0.8787 -0.0289 0.0800 -0.2570 0.0289 -0.0477 0.0368 -0.0489 0.0477 0.0472 -0.0097 0.1059 -0.0472 -0.0444 -0.0318 0.7954 0.0444 0.0181 0.1481 -0.0141 -0.0181 0.0143 0.0407 -0.0267 -0.0143 -0.0384 -0.0092 -0.0423 0.0384 -0.0880 -0.9282 -0.0077 0.0880 0.0298 0.0534 -0.0321 -0.0298 -0.0008 0.0326 0.0004 0.0008 -0.0328 0.0297 0.0009 0.0328 0.0024 -0.0027 -0.0592 -0.0024 0.8954 0.1361 0.0460 0.1046 0.0189 -0.0261 0.0369 -0.0189 0.0051 -0.0569 0.0516 -0.0051 -0.0665 0.0376 -0.0971 0.0665 0.1467 -0.2685 0.2310 0.8533 -0.0207 -0.0097 -0.0274 0.0207 -0.0275 0.0287 -0.0969 0.0276 0.0443 -0.0563 0.0911 -0.0444 -0.0186 -0.0176 0.6869 0.0186 0.0210 -0.0128 0.0300 -0.0210 0.0032 0.0132 0.0118 -0.0032 0.0218 0.0566 -0.0804 -0.0218 -0.1102 -0.8269 -0.1283 0.1102 0.0123 -0.0366 0.0775 -0.0123 -0.0061 -0.0163 0.0075 0.0061 -0.0044 -0.0067 -0.0503 0.0044 -0.0083 -0.0396 -0.0119 0.0083 0.8917 0.0910 -0.0126 0.1083 -0.0073 0.0458 -0.0014 0.0073 0.0121 -0.0827 -0.0405 -0.0121 -0.0439 0.0446 -0.0998 0.0439 0.1244 -0.1036 0.0858 0.8756 -0.0004 -0.0584 -0.0241 0.0004 0 -0.0396 0.0496 0 -0.0367 0.0247 -0.1258 0.0367 -0.0087 -0.0709 0.8316 0.0087 -0.0234 0.0363 -0.0054 0.0234 -0.0201 -0.0389 0.0050 0.0201 -0.0422 0.0367 -0.0213 0.0422 -0.0735 -0.9080 -0.1131 0.0735 Gxi = 0.9998 0 -0.0002 0 0.0003 -0.0002 0.0004 -0.0002 0 -0.0003 0 0 0 0 0.0002 0 0.0047 0.8900 -0.0175 0.0079 -0.0191 0.0727 -0.0288 -0.0336 0.0097 -0.1638 0.0527 0.0029 -0.0202 0.0909 0.0141 0.0076 0.0003 0.0139 0.8864 0.0352 0.0431 -0.0996 0.0687 -0.0543 0.0226 0.0869 -0.1024 0.0129 0.0103 0.0086 0.0400 -0.0458 -0.0023 -0.0213 -0.0295 0.9072 -0.0030 0.0554 0.0412 0.0099 -0.0124 -0.0012 -0.0026 -0.0827 0.0178 0.0001 0.0290 0.0772 0.0161 -0.0236 0.0098 -0.0228 0.9069 0.0134 0.0458 -0.0024 -0.0163 0.0184 0.0250 0.0096 -0.0032 0.0479 -0.0139 0.0098 -0.0407 0.0396 -0.0698 0.0412 0.0752 0.8277 0.0931 -0.0189 -0.0311 0.0459 -0.0645 0.0316 -0.0034 -0.0272 0.0913 0.0029 0.0099 -0.0485 0.0486 0.0093 -0.0513 0.1281 0.7737 -0.0244 -0.0079 0.0570 0.0384 0.0272 -0.0521 -0.0220 -0.0949 0.0329 -0.0748 0.0869 -0.0967 0.0410 0.0696 -0.0917 0.1698 0.9294 -0.0241 0.0680 -0.0752 -0.0097 0.0076 -0.0494 0.0490 0.0263 -0.1161 0.0140 -0.0308 0.0088 0.0508 0.0578 -0.0369 -0.0770 -0.0857 -0.0004 -0.0284 -0.0215 -0.9031 0.0196 -0.0299 0.0103 0.0391 -0.1246 0.0342 0.0623 -0.0758 0.1302 -0.0379 -0.0784 0.0504 -0.1748 -0.0104 0.0511 -0.0081 -0.7991 0.0160 -0.0934 -0.0306 0.0558 -0.1444 -0.0077 0.0035 -0.0831 0.1200 -0.0200 -0.0024 -0.0004 -0.0085 -0.0358 0.0356 0.0061 -0.7437 0.0027 -0.0166 0.0298 -0.0269 -0.0925 0.0375 -0.0138 0.0740 0.0048 -0.0580 0.0067 -0.0712 -0.0511 0.0076 -0.0202 0.0429 -0.8985 -0.1049 -0.0287 0.0027 -0.0031 0.1204 -0.0214 -0.0186 0.0303 0.8818 0.0300 -0.0314 0.0102 0.1005 0.0238 0.0236 0.0075 -0.0347 -0.0828 -0.0385 -0.0785 0.0874 -0.0388 0.1013 0.1295 -0.0091 0.9498 -0.0711 -0.1041 0.0486 0.0351 0.0231 0.0647 0.0238 -0.0534 -0.0747 0.0083 -0.0382 0.0712 0.0883 0.0123 -0.0178 -0.0374 0.9377 0.0499 -0.0317 -0.0769 0.1331 -0.0004 -0.0116 0.0422 -0.0038 -0.0870 -0.0022 -0.0110 0.0171 0.1202 0.0137 -0.0587 -0.0274 0.8876 0.0048 -0.0475 -0.0290 0.0939 Gyi = 0.9999 0 0 0 0 -0.0001 -0.0002 0.0002 -0.0002 -0.0004 -0.0002 0 0.0001 0 0.0002 -0.0003 -0.0045 0.9592 -0.0417 0.0082 0.0269 0.0345 0.0025 0.0165 0.0024 0.0700 -0.0040 0.0169 0.0383 0.0681 0.0392 -0.0419 0.0265 -0.0612 0.9153 0.0050 -0.0451 -0.0112 0.0614 0.0653 0.0083 -0.0745 0.0431 0.0132 0.0023 -0.0437 0.0508 -0.0338 -0.0188 0.0105 -0.0318 0.9196 0.0499 -0.0727 0.0297 0.0633 -0.0199 0.0381 -0.0379 0.0142 0.0157 0.0329 0.0375 0.0835 0.0944 -0.0096 -0.0065 -0.0078 -0.0830 -0.0987 -0.0049 -0.0337 0.0657 -0.0730 -0.0577 -0.0163 0.9070 0.0125 0.0297 0.0065 -0.0489 0.1514 -0.0854 0.0255 0.0932 -0.1829 0.1620 -0.0181 -0.0304 0.1502 -0.1137 0.0130 0.0169 0.9211 0.0419 0.0065 -0.0303 0.0277 0.1267 0.0120 0.0041 -0.0537 -0.0564 -0.0089 0.0113 0.0562 0.1981 -0.0314 0.0277 0.0169 0.8783 -0.0094 -0.0649 0.0518 -0.0567 0.1108 -0.0123 0.1464 0.0093 -0.0863 0.0363 0.0715 0.0466 0.0613 0.0403 -0.0323 0.0775 0.9138 -0.0137 0.0417 0.0103 0.0183 0.0106 0.0495 0.0492 -0.0232 0.8959 -0.0105 0.0439 0.0092 -0.0054 0.0307 0.0110 0.0008 0.0757 -0.0114 0.0616 0.0581 -0.1002 0.1057 -0.0459 -0.0074 0.0787 0.7633 0.0277 0.0913 -0.0977 0.1478 -0.1416 -0.0362 0.0238 -0.0623 0.0129 0.0051 -0.0427 0.1298 -0.1416 0.0569 -0.0080 -0.0156 0.7881 0.0087 0.0047 -0.0204 0.0212 -0.0336 0.0450 -0.0097 0.0659 -0.0140 -0.0454 -0.0466 -0.1087 -0.0041 0.0572 0.0335 0.1136 0.9146 0.0090 0.0210 -0.0190 -0.0401 -0.1093 0.0042 -0.0377 -0.0073 -0.8962 -0.0586 0.0436 -0.0378 -0.1176 0.0087 -0.0896 0.0085 0.1028 0.0146 0.0028 0.0139 -0.0282 -0.1129 -0.0535 -0.0187 0.0035 -0.8299 0.1027 -0.0027 -0.0829 -0.0206 -0.1480 -0.0652 -0.0244 0.0456 -0.0450 0.0714 -0.0149 0.0491 -0.0946 0.0148 0.0081 0.0365 -0.8601 -0.0857 0.0692 0.0950 -0.0219 -0.0093 0.0028 -0.0435 0.1704 -0.0027 -0.0101 0.0252 0.0010 -0.0971 -0.0199 0.0608 0.0047 -0.8601 0.0139 -0.0005 0.0020 -0.1379 0.0024 -0.0227 -0.0197 0.1049 Gcnot = 0.9998 0.0002 0 -0.0001 0.0002 0 -0.0002 -0.0002 -0.0001 0.0003 0.0003 0 0 0 -0.0002 0 -0.0101 0.9337 0.0171 -0.0238 0.0002 -0.0161 -0.0007 -0.0296 -0.0150 0.0252 -0.0060 0.0295 -0.0278 0.0427 -0.0480 0.0035 0.0171 -0.0372 0.0503 -0.0246 -0.0458 0.0751 -0.1735 0.0317 -0.0157 0.0208 0.1745 -0.0830 0.0185 -0.0130 0.9192 0.0319 -0.0980 0.0898 0.0129 0.0058 -0.0116 -0.0057 -0.0395 -0.1539 0.0535 0.0092 0.0128 0.1485 0.0709 -0.0876 -0.0376 0.9061 0.0065 0.0109 0.0104 0.0068 0.0283 0.0034 -0.0850 -0.0490 0.0090 0.9338 -0.0363 0.0003 -0.0248 0.0575 -0.0240 -0.0119 0.0091 0.0086 -0.0052 -0.0129 -0.1390 0.0382 -0.0408 -0.0382 0.9097 0.0077 0.0429 0.1283 -0.0254 -0.0124 -0.0245 0.0077 -0.0482 0.0479 -0.0873 0.0198 -0.0175 -0.0773 -0.0334 -0.8585 0.0386 0.0023 -0.1245 -0.0806 0.0174 -0.0357 0.0808 -0.0242 -0.0043 0.0162 0.0470 -0.0427 -0.0683 0.0227 0.8300 0.0740 0.0758 -0.1093 -0.0430 0.0264 0.0349 0.0038 0.0129 -0.0109 -0.0183 0.0625 -0.0087 -0.0112 -0.0289 -0.9237 -0.0019 0.0021 -0.0252 0.0807 -0.1811 -0.0234 0.0299 -0.0576 0.0039 0.0197 0.0049 0.0170 -0.0197 -0.0208 -0.8493 -0.0150 -0.0962 0.0458 0.0372 -0.0463 0.1115 0.0104 0.0531 -0.0798 0.1156 0.0760 -0.0120 0.0464 0.0568 0.0126 -0.0466 0.0821 -0.0117 -0.0137 -0.0384 0.0752 0.0382 -0.7338 -0.0128 -0.0086 -0.0450 0.0034 0.0007 0.0143 0.0181 0.0470 -0.1100 0.0713 -0.0854 0.0328 -0.0322 0.0347 0.9257 0.0794 -0.0140 -0.0251 0.0281 0.0185 -0.0318 0.0045 -0.0053 -0.0546 -0.0929 0.0807 0.0104 0.0307 0.1217 0.0967 0.0020 0.0893 0.9129 -0.0007 -0.0036 0.0761 -0.0266 -0.0250 -0.0885 -0.0124 0.0939 -0.0560 0.1213 0.0835 0.0554 -0.0201 -0.0811 -0.0449 -0.0132 0.9309 -0.0334 -0.0039 -0.0230 0.0204 0.8297 0.0443 -0.0731 0.0076 -0.0323 -0.1205 0.0210 -0.0987 -0.0290 0.0159 0.0254 -0.0725 -0.0057 0.0089 0.0966 -0.0901 -0.0234 0.8849 -0.0333 -0.0480 0.1533 0.0394 -0.0254 -0.0274 -0.1438 0.0208 -0.0864 0.1124 0.0119 -0.0165 --- Iterative MLGST: Beginning iter 1 of 3 : 666 gate strings --- --- Minimum Chi^2 GST --- Memory estimates: (4 spam labels,666 gate strings, 1263 gateset params, 16 gate dim) Peristent: 0.0251528 GB Intermediate: 1.60566 GB Limit: 3 GB 28861.3: p in (0.000951297,0.996945), weights in (31.6712,1025.28), gs in (-0.825202,0.873666), maxLen = 7, nClipped=0 28861.3: p in (0.000951297,0.996945), weights in (31.6712,1025.28), gs in (-0.825202,0.873666), maxLen = 7, nClipped=0 28861.3: p in (0.000951297,0.996945), weights in (31.6712,1025.28), gs in (-0.825202,0.873666), maxLen = 7, nClipped=0 1.32908e+22: p in (-1e+06,1e+06), weights in (31.6244,3162.28), gs in (-578.981,527.413), maxLen = 7, nClipped=2664 1.24539e+22: p in (-1e+06,1e+06), weights in (31.6244,3162.28), gs in (-57.9337,52.8221), maxLen = 7, nClipped=2664 1.0055e+18: p in (-145757,137378), weights in (31.6244,3162.28), gs in (-5.86512,5.54628), maxLen = 7, nClipped=2655 755700: p in (-0.0319906,0.992101), weights in (31.7484,3162.28), gs in (-1.42235,1.36452), maxLen = 7, nClipped=9 15737.9: p in (-0.00187681,1.0009), weights in (31.6244,3162.28), gs in (-0.848526,0.895988), maxLen = 7, nClipped=2 3044.49: p in (-0.000462267,1.00011), weights in (31.6244,3162.28), gs in (-0.897282,0.938826), maxLen = 7, nClipped=2 1101.61: p in (0.000375036,0.998276), weights in (31.6501,1632.91), gs in (-0.932097,0.953545), maxLen = 7, nClipped=0 2775.12: p in (-0.00376492,0.999225), weights in (31.635,3162.28), gs in (-1.05549,1.01607), maxLen = 7, nClipped=2 1036.16: p in (0.000703814,0.997577), weights in (31.6612,1191.99), gs in (-0.9385,0.956302), maxLen = 7, nClipped=0 1031.62: p in (0.000989783,0.996885), weights in (31.6721,1005.15), gs in (-0.953545,1.00667), maxLen = 7, nClipped=0 1068.63: p in (0.00107062,0.996694), weights in (31.6752,966.458), gs in (-1.01181,1.03151), maxLen = 7, nClipped=0 1029.78: p in (0.00106071,0.996675), weights in (31.6755,970.961), gs in (-0.959929,1.0099), maxLen = 7, nClipped=0 1029.82: p in (0.00104157,0.996653), weights in (31.6758,979.84), gs in (-0.964455,1.01763), maxLen = 7, nClipped=0 1029.76: p in (0.00105947,0.996638), weights in (31.6761,971.528), gs in (-0.961223,1.01176), maxLen = 7, nClipped=0 1029.76: p in (0.00105721,0.996632), weights in (31.6762,972.567), gs in (-0.961333,1.00557), maxLen = 7, nClipped=0 1029.76: p in (0.00106047,0.996631), weights in (31.6762,971.069), gs in (-0.961269,1.01046), maxLen = 7, nClipped=0 1029.76: p in (0.00106059,0.99663), weights in (31.6762,971.014), gs in (-0.961747,1.01037), maxLen = 7, nClipped=0 1029.76: p in (0.00106059,0.99663), weights in (31.6762,971.014), gs in (-0.961747,1.01037), maxLen = 7, nClipped=0 Sum of Chi^2 = 1029.76 (1998 data params - 1023 model params = expected mean of 975; p-value = 0.108871) 2Delta(log(L)) = 1031.6 --- Iterative MLGST: Beginning iter 2 of 3 : 1041 gate strings --- --- Minimum Chi^2 GST --- Memory estimates: (4 spam labels,1041 gate strings, 1263 gateset params, 16 gate dim) Peristent: 0.0393154 GB Intermediate: 2.50974 GB Limit: 3 GB 3868.67: p in (0.00106059,0.99663), weights in (31.6762,971.014), gs in (-0.961747,1.01037), maxLen = 8, nClipped=0 3868.67: p in (0.00106059,0.99663), weights in (31.6762,971.014), gs in (-0.961747,1.01037), maxLen = 8, nClipped=0 3868.67: p in (0.00106059,0.99663), weights in (31.6762,971.014), gs in (-0.961747,1.01037), maxLen = 8, nClipped=0 2.09317e+22: p in (-1e+06,1e+06), weights in (31.6244,3162.28), gs in (-482.821,529.093), maxLen = 8, nClipped=4164 1.983e+22: p in (-1e+06,1e+06), weights in (31.6244,3162.28), gs in (-48.2442,52.8539), maxLen = 8, nClipped=4164 1.98886e+16: p in (-14492.3,23042.7), weights in (31.6244,3162.28), gs in (-4.78646,5.23007), maxLen = 8, nClipped=4130 14246.7: p in (-0.00289212,0.996332), weights in (31.6809,3162.28), gs in (-1.02231,1.15982), maxLen = 8, nClipped=1 2471.37: p in (0.00101164,0.996763), weights in (31.6741,994.231), gs in (-0.954342,0.994824), maxLen = 8, nClipped=0 2177.06: p in (0.00120096,0.996249), weights in (31.6822,912.507), gs in (-0.972775,0.958881), maxLen = 8, nClipped=0 2672.99: p in (-0.000231468,0.996694), weights in (31.6752,3162.28), gs in (-1.04897,0.992802), maxLen = 8, nClipped=1 2157.54: p in (0.00120422,0.995963), weights in (31.6868,911.272), gs in (-0.979694,0.960486), maxLen = 8, nClipped=0 2158.07: p in (0.0012273,0.995797), weights in (31.6894,902.661), gs in (-0.964203,0.959713), maxLen = 8, nClipped=0 2157.3: p in (0.00120935,0.99578), weights in (31.6897,909.336), gs in (-0.976654,0.960488), maxLen = 8, nClipped=0 2157.32: p in (0.00119948,0.995695), weights in (31.6911,913.07), gs in (-0.968298,0.958575), maxLen = 8, nClipped=0 2157.28: p in (0.00120715,0.99569), weights in (31.6912,910.165), gs in (-0.974894,0.960096), maxLen = 8, nClipped=0 2157.28: p in (0.00120678,0.995648), weights in (31.6918,910.304), gs in (-0.976964,0.960305), maxLen = 8, nClipped=0 2157.28: p in (0.00120715,0.99569), weights in (31.6912,910.165), gs in (-0.974894,0.960096), maxLen = 8, nClipped=0 Sum of Chi^2 = 2157.28 (3123 data params - 1023 model params = expected mean of 2100; p-value = 0.187744) 2Delta(log(L)) = 2163.58 --- Iterative MLGST: Beginning iter 3 of 3 : 1582 gate strings --- --- Minimum Chi^2 GST --- Memory estimates: (4 spam labels,1582 gate strings, 1263 gateset params, 16 gate dim) Peristent: 0.0597474 GB Intermediate: 3.81404 GB Limit: 3 GB Maximum eval sub-tree size = 1244 Memory limit and/or MPI has imposed a division of the evaluation tree: Size of original tree = 1582 Size of original gatestring_list = 1582 Tree is split into 2 sub-trees Sub-tree lengths = [1242, 433] (Sum = 1675) >> sub-tree 0: Size of evalTree = 1242 Size of gatestring_list = 1242 Max in use at once = (smallest tree size for mem) = 1242 >> sub-tree 1: Size of evalTree = 433 Size of gatestring_list = 340 Max in use at once = (smallest tree size for mem) = 341 4580.63: p in (0.00120715,0.99569), weights in (31.6912,910.165), gs in (-0.974894,0.960096), maxLen = 10, nClipped=0 4580.63: p in (0.00120715,0.99569), weights in (31.6912,910.165), gs in (-0.974894,0.960096), maxLen = 10, nClipped=0 4580.63: p in (0.00120715,0.99569), weights in (31.6912,910.165), gs in (-0.974894,0.960096), maxLen = 10, nClipped=0 3.16155e+22: p in (-1e+06,1e+06), weights in (31.6244,3162.28), gs in (-536.519,392.614), maxLen = 10, nClipped=6328 2.97799e+22: p in (-1e+06,1e+06), weights in (31.6244,3162.28), gs in (-53.681,39.3079), maxLen = 10, nClipped=6328 8.9276e+18: p in (-399627,575710), weights in (31.6244,3162.28), gs in (-5.3972,4.35141), maxLen = 10, nClipped=6285 20207.2: p in (0.00131702,0.9938), weights in (31.7213,871.372), gs in (-1.13362,1.26681), maxLen = 10, nClipped=0 3795.61: p in (0.00120813,0.995879), weights in (31.6881,909.795), gs in (-0.975194,0.963297), maxLen = 10, nClipped=0 3789.87: p in (0.00115156,0.995449), weights in (31.695,931.873), gs in (-0.979597,0.957151), maxLen = 10, nClipped=0 3766.28: p in (0.00133187,0.995169), weights in (31.6994,866.5), gs in (-0.967818,0.955476), maxLen = 10, nClipped=0 3790.36: p in (0.00125192,0.995228), weights in (31.6985,893.742), gs in (-0.974918,0.974374), maxLen = 10, nClipped=0 3765.31: p in (0.00140506,0.995109), weights in (31.7004,843.632), gs in (-0.968218,0.955654), maxLen = 10, nClipped=0 3765.33: p in (0.0014075,0.995093), weights in (31.7007,842.899), gs in (-0.965199,0.954485), maxLen = 10, nClipped=0 3765.31: p in (0.00140337,0.995091), weights in (31.7007,844.14), gs in (-0.967142,0.955345), maxLen = 10, nClipped=0 3765.31: p in (0.00140419,0.995083), weights in (31.7008,843.892), gs in (-0.966037,0.955355), maxLen = 10, nClipped=0 3765.31: p in (0.00140337,0.995091), weights in (31.7007,844.14), gs in (-0.967142,0.955345), maxLen = 10, nClipped=0 Sum of Chi^2 = 3765.31 (4746 data params - 1023 model params = expected mean of 3723; p-value = 0.309897) 2Delta(log(L)) = 3774.19 --- Last Iteration: switching to ML objective --- --- MLGST --- Memory estimates: (4 spam labels,1582 gate strings, 1263 gateset params, 16 gate dim) Peristent: 0.0596413 GB Intermediate: 3.81404 GB Limit: 3 GB Maximum eval sub-tree size = 1244 Memory limit and/or MPI has imposed a division of the evaluation tree: Size of original tree = 1582 Size of original gatestring_list = 1582 Tree is split into 2 sub-trees Sub-tree lengths = [1242, 433] (Sum = 1675) >> sub-tree 0: Size of evalTree = 1242 Size of gatestring_list = 1242 Max in use at once = (smallest tree size for mem) = 1242 >> sub-tree 1: Size of evalTree = 433 Size of gatestring_list = 340 Max in use at once = (smallest tree size for mem) = 341 Least squares msg = Both actual and predicted relative reductions in the sum of squares are at most 0.000001 ; flag = 1 Maximum log(L) = 1885.77 below upper bound of -3.59017e+06 2Delta(log(L)) = 3771.55 (4746 data params - 1023 model params = expected mean of 3723; p-value = 0.285059) 2Delta(log(L)) = 3771.55 Total time=0.780860 hours ### Step 5: Create report(s) using the returned Results object¶ The Results object returned from do_long_sequence_gst is able to generate several different types of reports. Most of these are designed to display single-qubit results (for historical reasons). The "general"-type report was designed with 2-qubit data presentation in mind, so creating reports with create_general_report will be the most useful. In [8]: results.gatesets['final estimate'].set_basis("pp", [4]) print results.gatesets['final estimate'].get_basis_name() print results.gatesets['final estimate'].get_basis_dimension() pp [4] In [9]: results.create_general_report_pdf(filename="tutorial_files/easy_2q_general.pdf",verbosity=2) Generating tables * Generating table: targetSpamBriefTable Generating table: bestGatesetSpamBriefTable Generating table: bestGatesetSpamParametersTable Generating table: bestGatesetVsTargetTable Generating table: bestGatesetSpamVsTargetTable Generating table: bestGatesetGaugeOptParamsTable Generating table: bestGatesetChoiEvalTable Generating table: datasetOverviewTable Generating table: bestGatesetEvalTable Generating table: bestGatesetRelEvalTable Generating table: targetGatesBoxTable Generating table: bestGatesetErrGenBoxTable Generating table: fiducialListTable Generating table: prepStrListTable Generating table: effectStrListTable Generating table: germList2ColTable Generating table: progressTable * Generating plots * -- LogL plots (4): 1 Generating figure: colorBoxPlotKeyPlot 2 Generating figure: bestEstimateSummedColorBoxPlot 3 Generating special: bestEstimateColorBoxPlotPages * Merging into template file * Latex file(s) successfully generated. Attempting to compile with pdflatex... Initial output PDF tutorial_files/easy_2q_general.pdf successfully generated. Final output PDF tutorial_files/easy_2q_general.pdf successfully generated. Cleaning up .aux and .log files. Now open tutorial_files/easy_2q_general.pdf to see the results. You've run 2-qubit GST! In [ ]:	2017-09-23 09:15:29	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5017713308334351, "perplexity": 2280.4139739750653}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-39/segments/1505818689615.28/warc/CC-MAIN-20170923085617-20170923105617-00039.warc.gz"}
https://www.rdocumentation.org/packages/VGAM/versions/1.0-4/topics/slash	VGAM (version 1.0-4) # slash: Slash Distribution Family Function ## Description Estimates the two parameters of the slash distribution by maximum likelihood estimation. ## Usage slash(lmu = "identitylink", lsigma = "loge", imu = NULL, isigma = NULL, gprobs.y = ppoints(8), nsimEIM = 250, zero = NULL, smallno = .Machine\$double.eps*1000) ## Arguments lmu, lsigma Parameter link functions applied to the $$\mu$$ and $$\sigma$$ parameters, respectively. See Links for more choices. imu, isigma Initial values. A NULL means an initial value is chosen internally. See CommonVGAMffArguments for more information. gprobs.y Used to compute the initial values for mu. This argument is fed into the probs argument of quantile to construct a grid, which is used to evaluate the log-likelihood. This must have values between 0 and 1. nsimEIM, zero See CommonVGAMffArguments for information. smallno Small positive number, used to test for the singularity. ## Value An object of class "vglmff" (see vglmff-class). The object is used by modelling functions such as vglm, and vgam. ## Details The standard slash distribution is the distribution of the ratio of a standard normal variable to an independent standard uniform(0,1) variable. It is mainly of use in simulation studies. One of its properties is that it has heavy tails, similar to those of the Cauchy. The general slash distribution can be obtained by replacing the univariate normal variable by a general normal $$N(\mu,\sigma)$$ random variable. It has a density that can be written as $$f(y) = \left\{ \begin{array}{cl} 1/(2 \sigma \sqrt(2 \pi)) & if y=\mu, \\ 1-\exp(-(((y-\mu)/\sigma)^2)/2))/(\sqrt(2 pi) \sigma ((y-\mu)/\sigma)^2) & if y \ne \mu. \end{array} \right .$$ where $$\mu$$ and $$\sigma$$ are the mean and standard deviation of the univariate normal distribution respectively. ## References Johnson, N. L. and Kotz, S. and Balakrishnan, N. (1994) Continuous Univariate Distributions, 2nd edition, Volume 1, New York: Wiley. Kafadar, K. (1982) A Biweight Approach to the One-Sample Problem Journal of the American Statistical Association, 77, 416--424. rslash, simulate.vlm. ## Examples Run this code # NOT RUN { sdata <- data.frame(y = rslash(n = 1000, mu = 4, sigma = exp(2))) fit <- vglm(y ~ 1, slash, data = sdata, trace = TRUE) coef(fit, matrix = TRUE) Coef(fit) summary(fit) # } Run the code above in your browser using DataCamp Workspace	2022-07-07 16:43:01	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 2, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6911505460739136, "perplexity": 2736.261849894063}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-27/segments/1656104495692.77/warc/CC-MAIN-20220707154329-20220707184329-00225.warc.gz"}
https://electronics.stackexchange.com/questions/570426/why-put-a-resistor-in-series-with-the-positive-terminal-in-a-non-inverting-op-am	# Why put a resistor in series with the positive terminal in a non inverting op-amp circuit? I've tried to find the answer, but failed. Why do we add R1 in series with the positive terminal for a non-inverting amplifier? What is the best value we can assume or choose for it? It's to compensate the effect of bias current in the non-inverting amplifier. The compensating resistor value equals the parallel combination of R2 and R3. The input current creates a voltage drop across R1 that offsets the voltage across the combination of R2 and R3. • Input currents have higher temperature coefficients than input voltages, so the thermal DC errors are dominated by input resistance if those are not balanced. For a '741, resistive imbalance should be under 1k ohm for best performance. Jun 10, 2021 at 17:17 • I think it's worth to mention that in order for this "trick" to work, the op amp's input offset current should be less than an order of magnitude of bias current. :) Jun 12, 2021 at 10:46 • I wouldn't call it a trick so much as a design technique. I first came across it in some 1968/1969 op amp handbooks out of the old National Semiconductor, authored by some of the early pioneers in integrated analog circuit design like Bob Widlar and Robert Dobkins. Jun 12, 2021 at 13:38 • Do you know the "circuit" who's name is Widlar (current source). :) Used intensively in OPamps. – user288518 Jun 12, 2021 at 15:20 • somewhat related: electronics.stackexchange.com/a/45723/7036 Jun 13, 2021 at 7:38 In the "old times", it was for "impedance reasons" and "symmetry". The simplest value is $$\R_2 \parallel R_3\$$, so it should be: $$\20 \times 100/(100+20) \implies \$$ nearest $$\16.7 \text{ k}\Omega\$$. The exact calculation is much more complicated, because you have to take into account variation of all offset characteristics versus temperature, power supplies and so on ... And sometimes, it can be used for minimizing noise. If the "compensation resistor R1" is no longer really needed, the resulting offset voltage at the output can still be compensated by simple techniques. And if it is really necessary, it is also interresting to cite a technique of global compensation of the offset quantities very particular in instrumentation amplifiers operating at very low voltage level. In these "OPamps", the overall output offset is compensated by a "long-term integration method, sampling and holding "a correction voltage reinjected into the input circuit. A less "restrictive" current circuit is, for example, an amplifier for ECG electrocardiography where the main problem is above all "common mode voltages". . Anyway, be caution with simulators ! Here is a simulation I made with microcap8 or 12 of SpectrumSoftware. Do you see the problem ? It is not the software ! I have also noted that the internal resistance of the generator is not shown. Some OPAs also need a resistance (50 ohm) to ground ... for RF impedance matching. That is not the case here. I forgot an important element in the offset compensation method of yesteryear. To do this, we used a potentiometer, between the two emittors of the two differential transistors of the input circuit, whose cursor was connected either to + Vs or to -Vs. This element has now disappeared, since it is now possible to "adjust" in the factory (by laser trimming) the symmetry of the assembly during the manufacture of the operational amplifier. The main objective is to "place" the operation point of the "transistors" in the "linear" part of the characteristics. This method (laser adjustment of the built-in resistors) is also used on some "differential current measuring amplifiers" even with a high applied voltage (~ 400V). Image to explain how microelectonics is ... see my comments lower. https://slideplayer.fr/slide/3251080/ Example of opamp schematic and layout, GLADE VLSI Design https://peardrop.co.uk/ • What do you mean by "OPAs"? Op amps (operational amplifiers)? Or a particular series of (high-bandwidth) op amps whose part names start with "OPA"? Or something else? Jun 12, 2021 at 0:58 • Symmetry of what? The influence of bias currents? Or something else? Jun 12, 2021 at 1:00 • OPAs is what you say. Operational Amplifiers generally speaking. For symmetry, you must go into the amplifier and how is designed its internal topology, essentially based on differential amplifier, for some reason of temperature stability. Must see the schematic for understanding. – user288518 Jun 12, 2021 at 8:38 • However, the topology described is not sufficient in terms of the diagram. It must also be at the level of the final achievement on the chip. Since this is microelectronics, you have to look at the most basic level of realization, and it's not that simple. Let us add the fact that a differential stage is not made up of two transistors but at least four placed in a crossed way! And placed in the 4 corners of a square so that the temperature (outside or inside released by the components themselves) has the least possible influence on the characteristics of the assembly! – user288518 Jun 12, 2021 at 8:51 • Microelectronic technology is an architectural "marvel of ingenuity" in the true sense of the word ... not many people know the difficulty. Remember that a certain current component is like drawing plans for a city to the nearest mm with a square area of 100 km side! Sorry for this long comment ... – user288518 Jun 12, 2021 at 8:51 Besides matching the input impedance to null the offset voltage caused by the input bias currents, another reason is to limit currents in the case of an input overvoltage condition. In the case of overvoltage (input beyond power rails) most op amps can tolerate a few mA of input current through their internal rail clamping diodes without damage. A resistor of, e.g., 10 kΩ allows the input voltage to exceed the rails by some 10s of volts. Although some op amps do contain some serial input resistance, this is usually kept rather small for noise reasons. As a result, external current limiting is almost always necessary when inputs can leave the range between the supply rails. Why put a resistor in series with the positive terminal in a non inverting op-amp circuit? A compensation resistor is sometimes added in series to the non-inverting terminal of an op-amp when the non-inverting input signal comes from a low impedance source. Although there are a variety of possible benefits that accrue from the presence of a compensation resistor, they all are related to the idea of balancing the impedance from each of the op-amp inputs to ground. To achieve such balance, the value of $$\R1\$$ should be $$R1 = (R2 \|\| R3) - Rsource$$ Where $$\Rsource\$$ is the output impedance of the source driving the non-inverting input. The input stage of any op-amp is some sort of differential pair. (It may not be a "pair", there is sometimes 4 or more transistors, but the idea is the same.) The output of a differential pair is a highly non-linear function of the inputs. Op-amps rely upon negative feedback to attain linearity. The gain of a differential pair varies according to how evenly the transistors conduct. If the pair is evenly balanced the gain is highest. When one transistor conducts much more than the other, the gain can be very low. While the gain of an op-amp based amplifier is set by the feedback network, the bandwidth of the amplifier will suffer if the open-loop gain of the op-amp is below its optimal. One reason for using the compensation resistor, then, is to ensure the differential pair is operating at its maximum gain point, rather than off to the side where its gain is highly reduced, and by so doing, ensuring the maximum bandwidth is available for the op-amp. Another reason for using the compensation resistor is to reduce input offset voltage. There are yet other reasons that have been offered, and I cannot evaluate their significance apart from actual circuits with actual op-amps. But they all based upon balancing the differential pair. What Mike said is true. I am providing more information for those who want to dig a little deeper. Write down the equation which provides the value of Vo, the output of the op-amp, as a function of Vin. Start with Vin = 1 Volt to make it easy. In a perfect world, the input impedances of an op-amp are infinite. The output impedance of the op-amp is zero. Same is true of the voltage source Vin. Also, the differential voltage between + and - inputs to the op-amp is zero. Furthermore, the open loop gain of an op-amp is infinite. In the real world, none of this is true. The bias currents, which are necessary to activate the differential input circuits of the op-amp, are exceedingly small which means that what is said at the beginning are fair assumptions to make to address the present problem. The bias current flowing in the parallel combination of R2 and R3 produces a voltage at the inverting input of the op-amp which can be cancelled at the non-inverting input of the op-amp with the addition of R1 with the value of R2\|\|R3. In other words, the error produced at the output of the op-amp caused by the input offset voltage, because of the bias currents, can be eliminated. # Simple but difficult to understand Nice answers... but let me point out that it is still difficult for a beginner to navigate in the sense of all this. I know this from personal experience as a student in the 70's and I have described it in my Wikibooks story. I had the misfortune of a teacher who knew but did not understand op-amp circuits to teach me. And when she recited the definitions of input bias current, input offset current, input offset voltage, CMRR, etc... to me, I did not understand anything… # Input bias currents ## Examples 741 op-amp. For example, I could not understand why there were input currents at all and why they passed through the input sources especially in the case when they exited the op-amp inputs (as in 741 with its PNP input transistors). According to my naive ideas at the time, I thought that the input source produced current and passed it through the op-amp inputs (ie, the op-amp was a load)... but here the hell, it was just the opposite - the op-amp produced current and passed it through the input source. TTL gates. I had the same problem with TTL circuits. Here the teacher (very well) understood these circuits and now maybe this was a "problem"-:) In the lab, he made me investigate 7400 in a very strange and incomprehensible way to me - by including a variable resistor between one of the inputs and ground and changing its resistance as an input quantity. I even had to draw the transfer characteristic Vout = f(Rin) in my lab report. What kind of nonsense was that? All this would have been clear to me if they had drawn the current paths (loops) in the internal structure of these circuits… ## The phenomenon So, the phenomenon here is that the circuit (op-amp here) produces the input bias currents and passes them via the input circuits to ground. And if we insert some resistance in these paths, voltage drop will appear across them… and it will serve as another input voltage that is connected in series with the genuine input voltage - Fig. 1. Fig. 1. Conceptual circuit diagram for the input circuit of an op-amp with input p-n-p transistors Undesired voltage drop. In some cases, this voltage drop can be undesired. An example is the voltage drop across the equivalent resistance Re = R2\|\|R3 in the OP's non-inverting amplifier. Desired voltage drop. In other cases, it is desired and we intentionally create it by inserting a resistor (R1 in the OPs circuit; RB1, RB2 in Fig. 2 below). We use the fact that the differential amplifier (in the op-amp input stage) has two opposite inputs (inverting and non-inverting) so their voltages are subtracted and if they are equal, the output voltage will be zero. Fig. 2. Differential pair with RB1 and RB2 Note that both input bias currents are created by the negative power supply V- (not by the emitter current "source"). Voltage compensation. This is a simple but widely used circuit trick. We can name it "passive voltage compensation". There is also a more sophisticated "active voltage compensation" (see my Wikibooks story) based on the negative feedback principle. # Input offset current In the real op-amp, there is some difference between the input bias currents. So, if we want a perfect compensation, the two equivalent resistances will not be equal. If there is such a possibility, we can adjust one of them (R1 in the OPs circuit) to equalize the voltage drops across them. # Input offset voltage I had a problem understanding this parameter as well. The misleading word here was "voltage"... and I was looking to see where that voltage appeared (I imagined the op-amp was producing it-:) Much later I realized that this is not a voltage but a difference (asymmetry) between the two halves of the differential amplifier… and to compensate for it, we could apply the required voltage (until zeroing the op-amp output voltage). So, in the definition, this is an indirect assessment of something non-electrical (construction) by means of the electrical quantity voltage; in circuits, this is a compensating voltage needed to make the circuit symmetric. # Generalizing It is not difficult to notice that all these imperfections (asymmetries) can be compensated in the same way - by adding a compensating voltage "produced" by inserting a resistor in the current path. So, we can adjust the one or the other equivalent resistance (connected to the inverting or non-inverting input) to obtain a full compensation of all imperfections at once.	2022-05-23 05:49:19	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 6, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.49340954422950745, "perplexity": 1115.324846386706}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662555558.23/warc/CC-MAIN-20220523041156-20220523071156-00047.warc.gz"}
http://mathhelpforum.com/calculus/85623-conv-div-comparison.html	# Math Help - conv. and div. by comparison 1. ## conv. and div. by comparison just a question: while using comparisons to decide whether the integral converges or diverges we know that 1/x div 1/x^2 conv 2. Originally Posted by alex83 just a question: while using comparisons to decide whether the integral converges or diverges we know that 1/x div 1/x^2 conv ask yourself this question for $x>1$ $\frac{1}{x^2}$ and $\frac{1}{x^3}=\frac{1}{x^2}\frac{1}{x}$	2014-08-22 07:17:56	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 3, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9547852873802185, "perplexity": 6493.86955285397}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-35/segments/1408500823169.67/warc/CC-MAIN-20140820021343-00332-ip-10-180-136-8.ec2.internal.warc.gz"}
https://www.computer.org/csdl/trans/tc/1990/07/t0962-abs.html	The Community for Technology Leaders Issue No. 07 - July (1990 vol. 39) ISSN: 0018-9340 pp: 962-965 ABSTRACT <p>Properties of some shuffle/exchange type permutation sets are studied from operational points of view. The permutation sets studied are Omega , Omega /sup -1/, Psi , L, and U; Omega and Omega /sup -1/ are, respectively, the omega and inverse omega permutation sets, Omega identical to ( Omega intersection Omega /sup -1/); and L and U are, respectively, the admissible lower and upper triangular permutation sets. Several intuitive operations are introduced. Based on these operations, important known results relating to these sets are readdressed. The recursive nature of the sets is also discussed.</p> INDEX TERMS shuffle/exchange type permutation sets; intuitive operations; multiprocessor interconnection networks. CITATION S.-T. Huang, "Notes on Shuffle/Exchange Type Permutation Sets", IEEE Transactions on Computers, vol. 39, no. , pp. 962-965, July 1990, doi:10.1109/12.55699	2017-09-19 21:00:03	{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8845300078392029, "perplexity": 8079.404957089097}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-39/segments/1505818686034.31/warc/CC-MAIN-20170919202211-20170919222211-00716.warc.gz"}
https://math24.net/geometric-series.html	Geometric Series A sequence of numbers {an} is called a geometric sequence if the quotient of successive terms is a constant, called the common ratio. Thus an+1/an = q or an+1 = qan for all terms of the sequence. It's supposed that q ≠ 0 and q ≠ 1. For any geometric sequence: ${a_n} = {a_1}{q^{n - 1}}.$ A geometric series is the indicated sum of the terms of a geometric sequence. For a geometric series with q ≠ 1, ${S_n} = {a_1} + {a_2} + \ldots + {a_n} = {a_1}\frac{{1 - {q^n}}}{{1 - q}},\;\; q \ne 1.$ We say that the geometric series converges if the limit $$\lim\limits_{n \to \infty } {S_n}$$ exists and is finite. Otherwise the series is said to diverge. Let $S = \sum\limits_{n = 0}^\infty {{a_n}} = {a_1}\sum\limits_{n = 0}^\infty {{q^n}}$ be a geometric series. Then the series converges to $$\frac{{{a_1}}}{{1 - q}}$$ if $$\left\| q \right\| \lt 1,$$ and the series diverges if $$\left\| q \right\| \gt 1.$$ Solved Problems Click or tap a problem to see the solution. Example 1 Find the sum of the first $$8$$ terms of the geometric sequence $3,6,12, \ldots$ Example 2 Find the sum of the series $1 - 0,37 + 0,37^2 - 0,37^3 + \ldots$ Example 1. Find the sum of the first $$8$$ terms of the geometric sequence $3,6,12, \ldots$ Solution. Here $${a_1} = 3$$ and $$q = 2.$$ For $$n = 8$$ we have ${S_8} = {a_1}\frac{{1 - {q^8}}}{{1 - q}} = 3 \cdot \frac{{1 - {2^8}}}{{1 - 2}} = 3 \cdot \frac{{1 - 256}}{{\left( { - 1} \right)}} = 765.$ Example 2. Find the sum of the series $1 - 0,37 + 0,37^2 - 0,37^3 + \ldots$ Solution. This is an infinite geometric series with ratio $$q = -0,37.$$ Hence, the series converges to $S = \sum\limits_{n = 0}^\infty {{q^n}} = \frac{1}{{1 - \left( { - 0,37} \right)}} = \frac{1}{{1 + 0,37}} = \frac{1}{{1,37}} = \frac{{100}}{{137}}.$ See more problems on Page 2.	2022-05-19 05:00:55	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9339092373847961, "perplexity": 292.5118467587692}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662525507.54/warc/CC-MAIN-20220519042059-20220519072059-00752.warc.gz"}
http://mathstat.ysu.edu/mathjaxexample.html	# MATHEMATICS 1571 ## Final Examination Review Probems 1. For the function $$f$$ defined by $$f(x) = 2x^2 - 5x$$ find the following: 1. $$f(a + b)$$ 2. $$f(2x) - 2f(x)$$ 2. Find the domain of $$g$$ if 1. $$g(x) = \sqrt{x^2-3x-3}$$ 2. $$g(x) = \dfrac{x + 2}{x^3 - x}$$ 3. The graph of $$f(x) = \sqrt{x + 3} -4$$ is the same as the graph of $$f(x) = \sqrt{x}$$ except that it is moved how? 4. For the functions $$f$$ and $$g$$, find $$f \circ g$$, $$g \circ f$$, and the domains of each. 1. $$f(x) = \dfrac{1}{x^2 - 1}$$, $$g(x) = \sqrt{x}$$ 2. $$f(x) = x^2 + 1$$, $$g(x) = \dfrac{x}{x-1}$$ 5. Find the intercepts of the following equations. Also determine whether the equations are symmetric with respect to the $$y$$-axis or the origin. 1. $$y = x^4 + x^3 + x^2$$ 2. $$y = \dfrac{1}{x^3 - 3x}$$ 3. $$y = 2 - \|x\|$$	2022-01-18 15:54:23	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7960128784179688, "perplexity": 380.905101375191}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320300934.87/warc/CC-MAIN-20220118152809-20220118182809-00155.warc.gz"}
http://www.koreascience.or.kr/article/ArticleFullRecord.jsp?cn=HNSHCY_2015_v37n4_469	SEMI-CYCLOTOMIC POLYNOMIALS • Journal title : Honam Mathematical Journal • Volume 37, Issue 4, 2015, pp.469-472 • Publisher : The Honam Mathematical Society • DOI : 10.5831/HMJ.2015.37.4.469 Title & Authors SEMI-CYCLOTOMIC POLYNOMIALS LEE, KI-SUK; LEE, JI-EUN; Kim, JI-HYE; Abstract The n-th cyclotomic polynomial $\small{{\Phi}_n(x)}$ is irreducible over $\small{\mathbb{Q}}$ and has integer coefficients. The degree of $\small{{\Phi}_n(x)}$ is $\small{{\varphi}(n)}$, where $\small{{\varphi}(n)}$ is the Euler Phi-function. In this paper, we define Semi-Cyclotomic Polynomial $\small{J_n(x)}$. $\small{J_n(x)}$ is also irreducible over $\small{\mathbb{Q}}$ and has integer coefficients. But the degree of $\small{J_n(x)}$ is $\small{\frac{{\varphi}(n)}{2}}$. Galois Theory will be used to prove the above properties of $\small{J_n(x)}$. Keywords n-th cyclotomic polynomial;semi-cyclotomic polynomial;irreducible polynomial; Language English Cited by References 1. J. R. Bastida and R. Lyndon, Field Extensions and Galois Thory, Encyclopedia of Mathmatics and Its Application, Addison-Wesley Publishing Company (1984). 2. T. W. Hungerford, Abstract Algebra An Introduction, Brooks/Cole, Cengage Learning (2014). 3. S. Lang, Algebra, Addison-Wesley Publishing Company (1984). 4. P. Ribenboim, Algebraic Numbers, John Wiley and Sons Inc (1972).	2018-11-17 10:54:24	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 11, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6038737893104553, "perplexity": 2785.9475650830577}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-47/segments/1542039743353.54/warc/CC-MAIN-20181117102757-20181117124757-00505.warc.gz"}
https://jmanton.wordpress.com/2010/11/21/introduction-to-the-legendre-transform/	Home > Informal Classroom Notes > Introduction to the Legendre Transform Introduction to the Legendre Transform I have not come across an introduction to the Legendre transform which was entirely intuitive and satisfying. The article Making Sense of the Legendre Transform is one such attempt and has a number of attractive features. However, like other attempts, it immediately links the Legendre transform to re-parametrising a function by its derivative, yet this link is not made crystal clear. Below, I present an alternative introduction to the Legendre transform which takes as its starting point the fact that a convex set is uniquely defined by the collection of its supporting hyperplanes. My purpose is not to be rigorous, but instead, to present just enough facts for the reader to be comfortable with the basic ideas surrounding the Legendre transform and to tune the reader’s receptiveness to other pedagogic material on Legendre transforms. The Legendre Transform Readers are invited to draw their favourite convex function $f: \mathbb{R} \rightarrow \mathbb{R}$ on a piece of paper. Treating the graph of $f$ as the boundary of a cup, we can fill the inside of it. The resulting shape (the boundary plus the inside) is called the epigraph of the convex function. (Precisely, the epigraph of $f$ is $\{(x,z) \mid z \geq f(x)\}$.) Observe that the epigraph of a convex function is a convex set. A line (or more generally, a hyperplane, if we were working in a higher dimension) is called a supporting hyperplane of a convex set in $\mathbb{R}^2$ if it intersects the convex set and the convex set is contained on just one side of it. (This implies that the points of intersection are boundary points.) For example, the horizontal line $y=0$ is a supporting hyperplane for the epigraph of $y=x^2$. Going further, given a gradient $m$, we can draw the line $y=mx+c$ on the same graph as $y = x^2$, and we observe that if $c$ is a large negative number then the line does not intersect $y=x^2$ (or its epigraph) at all, and as $c$ is increased, there comes a time when $y=mx+c$ first touches $y=x^2$. This value of $c$ makes $y=mx+c$ a supporting hyperplane for the epigraph of $y=x^2$. As $c$ increases further, there will be points of the epigraph of $y=x^2$ on either side of the line $y=mx+c$. For the function $y=e^x$, it is seen that if $m$ is negative, there are no corresponding supporting hyperplanes. In general, given a convex function $f: \mathbb{R} \rightarrow \mathbb{R}$, for each slope $m$ there is at most one value of $c$ making $y=mx+c$ a supporting hyperplane of the epigraph of $f$. Furthermore, the set of values of $m$ for which a supporting hyperplane exists forms a convex set. Recall that in convex analysis, it is a very useful fact that a convex set is defined by its supporting hyperplanes. Therefore, if we know all the supporting hyperplanes of (the epigraph of) $f$ then we should be able to reconstruct $f$. Shortly, we will endeavour to do this from first principles. Perhaps the simplest way of “storing” what the supporting hyperplanes of $f$ are, is to graph $c$ versus $m$. Indeed, for each value of $m$, there is at most one value of $c$ for which $y=mx+c$ is a supporting hyperplane, hence plotting $m$ on the horizontal axis and $c$ on the vertical axis produces the graph of a function. In fact, it can be proved that this graph will always be concave if $f$ is convex. To visualise this, pick a point $x$ and draw a tangent line to the graph $y=x^2$ at the point $(x,x^2)$. This is a supporting hyperplane with $m$ equal to $\frac{df}{dx}=2x$ and $c$ equal to the $y$-intercept of the tangent line, namely $-x^2$. Now, as $x$ goes from $-\infty$ to $\infty$, it can be seen that $m$ monotonically increases and $c$ initially increases and then decreases. Furthermore, observe that plotting the points $(m,c)$ corresponding to the supporting hyperplanes is the same as plotting the points $\{(2x,-x^2) \mid x \in \mathbb{R}\}$ which is the same as plotting the graph $\{(m,c) \mid c = -\frac{m^2}{4}\}$ of the function $c = -\frac{m^2}{4}$. Plotting $-c$ versus $m$ will produce a convex function if $f$ is convex, and this is the convention which has proved expedient. Clearly, there is no material difference between plotting $-c$ or $c$ as a function of $m$. Formally, we have just seen that to any convex function $f$ we can associate a function $h$ such that $y=mx+c$ is a supporting hyperplane of (the epigraph of) $f$ if and only if $c = -h(m)$ (where the negative sign is just a matter of convention). The function $h$ thus defined is called the Legendre transform of $f$. As mentioned earlier, $h$ is not necessarily defined on the whole of $\mathbb{R}$ but only on a convex subset of $\mathbb{R}$. To overcome this minor notational inconvenience, it is customary to set $h(m)$ equal to infinity for values of $m$ for which it would otherwise be undefined. This is a standard trick in convex analysis for avoiding the need to keep track explicitly of the set on which a convex function is defined. Can we think of another way of storing the set of supporting hyperplanes of $f$? Each hyperplane $y=mx+c$ is represented by the pair of coefficients $(m,c)$. If we had chosen to fix $c$ instead, we would have found that depending on the value of $c$, there could be multiple values of $m$ for which $y=mx+c$ is a supporting hyperplane. It seems sensible then to represent the set of all points $(m,c)$ corresponding to supporting hyperplanes by using the aforementioned function $h$. Therefore, I choose to interpret the Legendre transform as what one would arrive at if charged with the task of writing down in a nice and simple way what the supporting hyperplanes are of the epigraph of a convex function $f$. As promised earlier, let’s see how a function $f$ can be recovered from its Legendre transform $h$. Recall that for each $m$ (for which $h(m)$ is finite), the line $y=mx-h(m)$ is a supporting hyperplane. In particular, we know that at least one point of the graph of $f$ lies on this line, and we know that every single point of the graph of $f$ lies on or above this line. By drawing all the supporting hyperplanes, we can see intuitively that they therefore trace out $f$. In fact, readers familiar with envelopes of curves will see that this is the same idea here, and for those not familiar, clicking on the link will bring up the Wikipedia entry with a nice figure showing how the supporting hyperplanes trace out the function. For simplicity, consider the special case when $h$ is differentiable. (A convex function is differentiable at all but at most a countable number of points, and even at such points, left and right derivatives still exist.) Choose an $m$. How can we find an $x$ such that the point $(x,mx-h(m))$ on the supporting hyperplane $y = mx-h(m)$ belongs to the graph of $f$? If we perturb $m$, we get another supporting hyperplane $y = (m+\epsilon)x-h(m+\epsilon)$. This perturbed hyperplane, as we will call it, intersects with the original hyperplane $y = mx-h(m)$ at the point with $x$ coordinate given by $x = \frac{1}{\epsilon}\left[h(m+\epsilon)-h(m)\right]$ which approaches $h'(m)$, the derivative of $h$, as $\epsilon \rightarrow 0$. For positive $\epsilon$, the perturbed hyperplane $y = (m+\epsilon)x-h(m+\epsilon)$ will lie above the original hyperplane whenever $x$ is larger than the point of intersection. Similarly, if $\epsilon$ is negative, the perturbed hyperplane will lie above the original hyperplane whenever $x$ is smaller than the point of intersection. Therefore, the point of intersection in the limit $\epsilon \rightarrow 0$ is precisely the point which also lies on the graph of $f$. (If this is not clear, re-read the third sentence of the preceding paragraph.) Thus, the point $(h'(m),mh'(m)-h(m))$ lies on the graph of $f$ for all $m$ for which $h$ is defined and differentiable. (If $h$ is not differentiable at $m$, we would have a range of valid $x$ values, namely, those values lying between the left derivative and the right derivative of $h$ at $m$. We would therefore obtain a line segment belonging to the graph of $f$. For the moment though, this level of detail is a distraction.) For completeness — and because something unexpected will reveal itself — let’s give a formula for the graph of $h$ if we are given only $f$, under the simplifying assumption that $f$ is differentiable. As hinted at earlier, it can be shown that the supporting hyperplanes are the same as the tangent lines of $f$. The tangent line of $f$ at the point $(x_0,f(x_0))$ is simply $y - y_0 = m(x-x_0)$ where $m=f'(x_0)$ and $y_0 = f(x_0)$. The $y$-intercept is thus $-mx_0+y_0$, and hence the point $(m,mx_0-y_0) = (f'(x_0),x_0f'(x_0)-f(x_0))$ lies on the graph of $h$. Comparing this with the previous paragraph, we see excitedly that going from $f$ to the graph of $h$ has exactly the same form as going from $h$ to the graph of $f$, and indeed, it can be established rigorously that if $h$ is the Legendre transform of a convex function $f$, then $f$ is the Legendre transform of $h$. The Legendre transform is its own inverse. We close this section by deriving the standard formula for the Legendre transform. Let $f$ be an arbitrary function. (It need not even be convex.) Recall the idea given earlier about computing the Legendre transform, namely, we draw the line $y=mx+c$ for $c$ a very small number and gradually increase $c$ until the line first intersects the graph of $f$. That is to say, we want the smallest value (or, if the smallest value does not exist, the infimum) of $c$ for which $y=mx+c$ intersects the graph of $f$. Therefore, it is equally valid to look for the set of all values of $c$ for which $y=mx+c$ intersects the graph of $f$, then choose the infimum of this set. This set is easily found by fixing $m$ and looking in turn at each point $(x_0,f(x_0))$ on the graph of $f$ and seeing for what value of $c$ the line $y=mx+c$ passes through this point. The line passing through $(x_0,f(x_0))$ with gradient $m$ is simply $y-f(x_0)=m(x-x_0)$ and its $y$-intercept is thus $-mx_0+f(x_0)$. The infimum of $c$ for which $y=mx+c$ intersects the graph of $f$ is therefore $\inf_{x_0} -mx_0+f(x_0)$. The Legendre transform is defined to be the negative of this, by convention, therefore, we have arrived at the mathematical definition of the Legendre transform: $h(m) = \sup_x mx-f(x)$. The Legendre Transform in Physics Often the Legendre transform is introduced by saying that it allows a function $f(x)$ to be re-parametrised by its derivative, meaning precisely: Given a slope $m$, first find the value of $x$ such that $f'(x)=m$ then return the value of $f$ at this point. This is written concisely as $f(x(m))$, where $x$ is now considered a function of $m$. This is useful in physics; see for example Making Sense of the Legendre Transform. Starting from first principles, assume we are given a differentiable and strictly convex function $f: \mathbb{R} \rightarrow \mathbb{R}$ and we are given a slope $m$, and we wish to write down an explicit expression for the function implicitly defined by the requirement that $m$ is mapped to $f(x)$ where $x$ satisfies $f'(x)=m$. With ideas from the previous section fresh in our minds, we know that we can find the point $x$ for which $f'(x)=m$ by finding the supporting hyperplane $y=mx+c$ with slope $m$ for the epigraph of $f$, and seeing where the supporting hyperplane intersects the graph of $f$. If $h$ is the Legendre transform of $f$, then we know the hyperplane is $y = mx-h(m)$. Therefore, finding $\{x \mid f'(x)=m\}$ is equivalent to finding $\{x \mid mx-h(m)=f(x)\}$. The (possibly) good news is that the latter expression involves the function $f$ and not its derivative, but the bad news is that this expression is still an implicit one for $x$. The most important observation — and one of the reasons for including the derivations in the previous section — is that provided the derivative exists, we have that $x = h'(m)$ satisfies $f'(x)=m$. (Recall the discussion about how the graph of $f$ can be recovered by thinking about what perturbed hyperplanes tell us.) Therefore, it is not the Legendre transform but its derivative which allows us to write down an explicit expression for the value of $x$ for which $f'(x)=m$. Furthermore, note from the previous section that $f(h'(m))$ can be written in terms of $h$ alone, as $mh'(m)-h(m)$. Although this section opened by considering how to write down an explicit expression for $f(x(m))$, and the Legendre transform was found to be a valuable stepping-stone, the fact of the matter is that $f(x(m))$ is not necessarily well-defined unless the derivative of $f$ exists and is injective, whereas the Legendre transform is always well-defined and has useful properties. Therefore, although we may have originally thought we wanted an expression for $f(x(m))$ as an intermediate step in achieving an ultimate objective (e.g., reformulating a physical law in a more convenient way — see Making Sense of the Legendre Transform), it is mathematically advantageous to work with the Legendre transform instead. In the “worst case” we will need to require that the Legendre transform $h$ of $f$ is differentiable and write $f(x(m))$ as $f(h'(m))$ (or $mh'(m)-h(m)$), but it may be possible that an alternative manipulation can be found for achieving the ultimate objective which uses $h$ directly and does not require $h'(m)$ to exist. That is to say, there is nothing to be lost and a chance of something to be gained by working with $h$ rather than $f(x(m))$ simply because the latter can be obtained from the former in a simple way. Summary • Within convex analysis, a convex set is uniquely defined by its supporting hyperplanes, and some properties of a convex set are better elucidated by examining its supporting hyperplanes. • A function is convex if and only if its epigraph is a convex set. • The Legendre transform is what one would come up with if charged with the task of describing the set of all supporting hyperplanes of the epigraph of a convex function. • The Legendre transform is its own inverse (when applied to a convex function). • The Legendre transform is related to the concept of the envelope of a curve. • If $h$ is the Legendre transform of $f$ then, if $h'(m)$ exists, the point $x=h'(m)$ has the property that $f'(x)=m$. • The Legendre transform (precisely, its derivative) therefore allows a function $f$ to be re-parametrised by its derivative. This is often the motivation given for the usefulness of the Legendre transform. • In trying to derive from first principles a re-parametrisation for $f$ in terms of its derivative, the Legendre transform arises naturally, as an intermediate step. 1. November 23, 2010 at 7:49 am I found McCallaugh’s expanation in “Tensor Methods in Statistics” very intuitive. Dual relationship between function and it’s legendre transform is clear from the figure on page 156 http://yaroslavvb.com/upload/legendre-transform.png 2. January 11, 2011 at 4:54 pm Thanks for the clear explanation! I think a lot of people, even researchers who use the Legendre transform routinely, don’t understand this geometric perspective. 3. January 26, 2011 at 11:44 am I liked your explanation so much I created a bunch of pictures illustrating it graphically (also including the multidimensional version of the transform), and put it on my website here: http://maze5.net/?page_id=733 • January 26, 2011 at 6:25 pm Thanks, Nick. Your pictures are very informative and visually appealing. 4. May 19, 2013 at 9:41 am I was writing my own explanation of the Legendre Transform when I came across this page. Finally, someone gives a simple explanation where the sup operator comes from. Thank you. Btw, here is the article I wrote: http://www.onmyphd.com/?p=legendre.fenchel.transform. 5. March 18, 2014 at 12:38 am Reblogged this on simplicity. 6. October 2, 2018 at 5:00 pm It’s beautiful! Thank you! 7. January 28, 2020 at 1:48 pm For some reason I can’t quite explicate, this makes me want to search Mamikon’s work (EG with Apostol) for places where some of this might be echoed…. 1. May 16, 2011 at 1:25 pm 2. May 12, 2017 at 4:04 am 3. July 1, 2017 at 1:49 am 4. December 22, 2018 at 11:14 am	2021-07-25 03:08:20	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 226, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8562749028205872, "perplexity": 132.99618731988505}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-31/segments/1627046151563.91/warc/CC-MAIN-20210725014052-20210725044052-00320.warc.gz"}
https://www.physicsforums.com/threads/questions-about-representation-theory-of-lie-algebra.984205/	# Questions about representation theory of Lie algebra • A Gold Member ## Summary: SL(2, C), SU(2), SU(3) ## Main Question or Discussion Point I have confusions about representation theory. In the following questions, I will try to express it as best as possible. For this thread say representation is given as ρ: L → GL(V) where L is the Lie group(or symmetry group for a physicist) GL(V) is the general linear transformations of V(of some dimension) Questions: 1. Often the irreducible representations are labelled along the columns of Reps, dimensions; see page 20 in this. What do they exactly mean? As I see it reps are the representation space V of the given symmetry group(SU(2) in this case) and dimension is referring to the dimension of V. Am I correct? 2. Each entry in the columns is a new irreducible representation(is this correct?), how does the increase in weight relates to the increase in dimensions? What happens here? 3. Take a look at page 35 of this book(attached): Introduction to quarks and Partons by F. E. Close(1979). There he labels the multiplet by different quark names. By doing this he is constructing a representation space V namely (u, d, s) and the SU(3) group acts on this space, correct? So given a higher dimensional representation space(say the octuplet (1, 1)) how does the SU(3) act on it? #### Attachments • 58.5 KB Views: 20 Related Topology and Analysis News on Phys.org fresh_42 Mentor Maybe you should have posted this in a physics forum, since at least your reference on slide 9 seems to be physics. Summary:: SL(2, C), SU(2), SU(3) I have confusions about representation theory. In the following questions, I will try to express it as best as possible. For this thread say representation is given as ρ: L → GL(V) where L is the Lie group(or symmetry group for a physicist) GL(V) is the general linear transformations of V(of some dimension) Questions: 1. Often the irreducible representations are labelled along the columns of Reps, dimensions; see page 20 in this. What do they exactly mean? I can only assume that this refers to the weight spaces of a representation: how many ladder up and ladder down operators we have. As I see it reps are the representation space V of the given symmetry group(SU(2) in this case) and dimension is referring to the dimension of V. Am I correct? Looks like. In general you must be careful whether the dimension of the Lie group, or Lie algebra, or the dimension of the representation is meant. 2. Each entry in the columns is a new irreducible representation(is this correct?), how does the increase in weight relates to the increase in dimensions? What happens here? See the theorem under section 13 in here: https://www.physicsforums.com/insights/lie-algebras-a-walkthrough-the-representations/ It is the other way around. The increase of $\dim V$ allows, or better enforces more weight spaces. $V$ is written as a direct sum of invariant subspaces, and the invariance is that of the operators given by the Cartan subalgebra. Ladder up and ladder down is impossible at the end of the ladder, and its length is given by $\dim V$. 3. Take a look at page 35 of this book(attached): Introduction to quarks and Partons by F. E. Close(1979). There he labels the multiplet by different quark names. By doing this he is constructing a representation space V namely (u, d, s) and the SU(3) group acts on this space, correct? So given a higher dimensional representation space(say the octuplet (1, 1)) how does the SU(3) act on it? $V_0$ is a the two dimensional subspace, which is invariant under the two operators of the Cartan subalgebra. If $\dim V=8$, then there is one way up and one way down: $V=V_0\oplus V_{-1} \oplus V_{1}$. This should correspond to $\mathfrak{su}(3)\cong \operatorname{ad} \mathfrak{su}(3)$ itself: diagonal, upper triangular and lower triangular matrices. Gold Member I understand now. The confusion was that I thinking that the representation space V always has the same dimensions as n in SU(n). I didn't think that there can be higher-dimensional V or even lower ones. I have one more question: The dimension of V, it has to be either equal or more than the n in SU(n), is this correct? In other words, is it possible to construct a doublet for SU(3)? Or a triplet for SU(4)? In general you must be careful whether the dimension of the Lie group, or Lie algebra, or the dimension of the representation is meant. Yes, I see that now. fresh_42 Mentor I have one more question: The dimension of V, it has to be either equal or more than the n in SU(n), is this correct? No. The question is, which kind of representation you are looking for. E.g. you can always define $\mathfrak{g}.V\equiv 0$, and $V$ can have any dimension. Sure, the trivial representation is not very illuminating, but allowed. The dimension must at least be the dimension of the Lie algebra if we require a faithful representation, i.e. an embedding $\mathfrak{g}\hookrightarrow \operatorname{GL}(V)$. I suppose there are cases in between, too, i.e. with $0< \dim V < \dim \mathfrak{g}$, although not faithful. In other words, is it possible to construct a doublet for SU(3)? Or a triplet for SU(4)? I'm not sure about the representations of $SU(n)$ with $n>2$ in general. The case $n=2$ is rather simple and easy to calculate, see the theorem in the link, or this one here: https://arxiv.org/pdf/math-ph/0005032.pdf, chapter 5. I had to do it first, or search the internet. The plan goes as follows: 1. Write $\mathfrak{su}(n)= \mathfrak{h} +\sum_{\alpha\in \phi} \mathfrak{su}(n)_\alpha$ with a Cartan subalgebra $\mathfrak{h}$ and an according root system $\phi$. 2. In case of $\mathfrak{g}=\mathfrak{sl}(n)$ this decomposition is diagonal, upper and lower triangular matrices. The basis change towards $\mathfrak{su}(n)$ makes this is a bit more inconvenient, as the matrices look different. 3. Then choose a representation space $V$ and calculate the subspace $V_0$ defined by $\mathfrak{h}.v=0$. 4. Look what repeated applications $Y^k.V_0$ do (ladder down), where $Y$ is a basis vector of $\mathfrak{su}(n)_\alpha$ and determine the weights. 5. Do the same for $X^k.V_0$ (ladder up). 6. This should give you a decomposition of $V$ along the weights and how the various operators act among them. So or similar is the task. It's a bit nasty to actually calculate it, so I would probably rather search for the solution on the internet. I'm sure the irreducible, finite dimensional representations for $\mathfrak{su}(n)$ are fully cataloged. Let's see whether your question has an easy answer. Assume we have $V=V_0 \oplus V_+ \oplus V_-$ all one dimensional, say spanned by $v_0,v_+,v_-$, and $\mathfrak{h}.V_0=0$ with $\mathfrak{h}\subseteq \mathfrak{su}(n)$ CSA. Let us further assume we have at least three positive roots $\gamma \in \{\,\alpha, \beta, \alpha+\beta \,\}$ and $$[H,X_\gamma]=\gamma(H)X_\gamma\, \; , \; \,[H,Y_\gamma]=-\gamma(H)Y\; , \;[X_\gamma,Y_\gamma]=H_\gamma\; , \;[X_\alpha,X_\beta]=X_{\alpha+\beta}\; , \;[Y_\alpha,Y_\beta]=-Y_{\alpha+\beta}$$ This means we have for $v_i \in \{\,v_0,v_+,v_-\,\}$ \begin{align} H_\gamma(v_0) &=0\, , \,H_\gamma(v_+)=h_1(\gamma)v_+\, , \,H_\gamma(v_-)=h_{-1}(\gamma)v_-\\ X_\gamma(v_0) &=x_0(\gamma)v_+\, , \,X_\gamma(v_+)=0\, , \,X_\gamma(v_-)=x_{-1}(\gamma)v_0\\ Y_\gamma(v_0) &=y_0(\gamma)v_-\, , \,Y_\gamma(v_+)=y_1(\gamma)v_0\, , \,Y_\gamma(v_-)=0 \end{align} Now we have to check whether the Jacobi identity holds or if there need to be adjustments on the coefficients. But beside some signs the multiplication table should be ok. Finally we have to check, whether there are values $h_{\pm 1}(\gamma), x_{0}(\gamma), x_{-1}(\gamma), y_{0}(\gamma), y_{1}(\gamma)$ such that the above satisfy the conditions of a representation, i.e. $[A,B](v)=A(B(v))-B(A(v))$. As this is basically the same as in the case of $\mathfrak{su}(2)$, it will be sufficient to concentrate on what is new, and that are the composite roots $\gamma=\pm(\alpha+\beta)$. Now you can either try yourself whether there is a solution for the coefficients $h_{\pm 1}(\gamma), x_{0}(\gamma), x_{-1}(\gamma), y_{0}(\gamma), y_{1}(\gamma)$ which are linear in $\gamma$, or search for "representations + SU(3)". Playing a bit with $[X_{\alpha+\beta},Y_{\alpha+\beta}]=H_{\alpha+\beta}$ on $\{\,v_0,v_+,v_-\,\}$ are only three equations, nine if we add $[X_\alpha,X_\beta]=X_{\alpha+\beta},[Y_\alpha,Y_\beta]=Y_{\alpha+\beta}$, i.e. doable. Last edited: Gold Member No. The question is, which kind of representation you are looking for. E.g. you can always define g.V≡0g.V≡0\mathfrak{g}.V\equiv 0, and VVV can have any dimension. Sure, the trivial representation is not very illuminating, but allowed. The dimension must at least be the dimension of the Lie algebra if we require a faithful representation, i.e. an embedding g↪GL(V)g↪GL⁡(V)\mathfrak{g}\hookrightarrow \operatorname{GL}(V). I suppose there are cases in between, too, i.e. with 0<dimV<dimg0<dim⁡V<dim⁡g0< \dim V < \dim \mathfrak{g}, although not faithful. Actually this was sufficient(and I understood the rest as well), thank you very much for your time. fresh_42 Mentor Actually this was sufficient(and I understood the rest as well), thank you very much for your time. I may have to correct my statement a bit. If $\dim V = \dim \mathfrak{g}$ then we always have $\operatorname{ad}\, : \,\mathfrak{g} \hookrightarrow \mathfrak{gl(g)}$ in case of semisimple Lie algebras. I wrote $\operatorname{GL}(V)$ which is a mistake. That would have been a group representation, but we are in the world of Lie algebras here. The condition on the dimensions was the lazy one, will say if $\mathfrak{g}$ and $V$ are of equal dimension, then we can get a faithful representation. But(!), if e.g. $\mathfrak{g}=\mathfrak{su}(3)$ then we have an eight dimensional Lie algebra. If we choose $\dim V= 3$ then $\dim \mathfrak{gl}(V)=9$ and we can have a faithful representation at least in principle, i.e. $\mathfrak{su}(3)$ fits in $\mathfrak{gl}(V)$. And since $su(3)$ are $3\times 3$ matrices, they act on $\mathbb{C}^3$ naturally, and the commutator delivers a Lie algebra representation. Hence there is a three dimensional and faithful representation. So your question had indeed an easy answer: Yes there is a three dimensional, faithful and irreducible (triplet) representation for $\mathfrak{su}(3)$, defined by: $A.v=A(v)$ and $[A,B]=AB-BA$ where $A,B \in \mathfrak{su}(3)$ and $v\in \mathbb{C}^3$. Sorry, that I haven't seen it earlier. Here is another important fact: A representation is a Lie algebra hiomomorphism $\varphi\, : \,\mathfrak{g}\longrightarrow \mathfrak{gl}(V)$. This implies that its kernel is an ideal of $\mathfrak{g}$. The Lie algebras $\mathfrak{su}(n)$, however, are simple, i.e. they have no ideals. Thus $\varphi$ is either trivial or faithful in this case. Hence there are no non-trivial representations of $\mathfrak{su}(n)$ with a too small representation space. Say $\dim V=m$ and $\dim \mathfrak{su}(n)=n^2-1$. Then we must have $n^2-1 \leq m^2 = \dim \mathfrak{gl}(V)$. This means $m \geq n$ for non trivial representations. Last edited:	2020-02-26 23:57:00	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.912712574005127, "perplexity": 406.58576735945263}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-10/segments/1581875146562.94/warc/CC-MAIN-20200226211749-20200227001749-00116.warc.gz"}
https://mathematics.huji.ac.il/event/nt-ag-lunch-tba-8	# NT & AG Lunch: Yakov Varshavsky Title: "GL(n,F)\GL(n,A)/GL(n,O) over function fields and vector bundles on curves" Abstract: The starting point of the geometric approach to the theory of automorphic forms over function fields is a beautiful observation of Weil asserting that there is a natural bijection between the two-sided quotient GL(n,F)\GL(n,A)/GL(n,O) and the set of isomorphism classes rank n vector bundles on a curve. The goal of my talk will be to explain this result and to give some applications. Key words: adeles and ideles in the function field case, algebraic curves, line and vector bundles on curves, Picard group, Riemann-Roch theorem. ## Date: Mon, 03/12/2018 - 13:00 to 14:00 ## Location: Faculty lounge, Math building	2020-02-27 08:03:01	{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.841502845287323, "perplexity": 973.7582690502638}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-10/segments/1581875146665.7/warc/CC-MAIN-20200227063824-20200227093824-00076.warc.gz"}
https://docs.wavesenterprise.com/en/1.1.2/how-the-platform-works/anchoring.html	# Anchoring¶ One of the main ideas behind private blockchain is that transactions are processed by a certain number of participants known in advance. In a private blockchain there is a threat of information spoofing, because the number of participants is quite small if we compare it to the public blockchain where there are no restrictions to join the network. When using PoS consensus algorithm, the threat of overwriting a blockchain becomes real. To increase the confidence of the private blockchain participants in data placed there, the anchoring mechanism was developed. Anchoring allows checking the data for invariability. The guarantee of invariability is achieved by publishing data from a private blockchain to a larger network, where data spoofing is unlikely due to a larger number of participants and blocks. Published data represent a signature and a height of blocks in a private network. Mutual connectivity of two or more networks increases their resistance, because as a result of a long-range attack as forgery or alteration of data resulting from a long-range attack would require attacking all connected networks. ## How does anchoring work in the Waves Enterprise blockchain¶ Targetnet anchoring scheme Anchoring process is shown below: 1. Anchoring configurations are set in the configuration file of the private blockchain node. You should use recommended values for the configurations to avoid anchoring malfunctioning. 2. Each height-range an anchoring transaction, that contains data of the block at current-height - threshold, is broadcasted to the Targetnet by the anchoring node. The Data Transaction with a key-value list is used as an anchoring transaction. The node then requests height of the broadcasted transaction. 3. The node then checks the Targetnet height each 30 seconds until its height reaches the height of the created transaction + height-above. 4. When required Targetnet height is reached and the presence of previously created data transaction is confirmed, another anchoring data transaction is created in the private blockchain. ## Transaction structure for anchoring¶ Targetnet transaction consists of the following fields: • height - the height of the chosen block from the private blockchain. • signature - the signature of the chosen block from the private blockchain. The private blockchain transaction consists of the following fields: • height - the height of the chosen block from the private blockchain. • signature - the signature of the chosen block from the private blockchain. • targetnet-tx-id - the Targetnet anchoring transaction ID. • targetnet-tx-timestamp - the timestamp of the Targetnet anchoring transaction. ## Errors during the anchoring¶ Errors during the anchoring can occur at any step. In case of any error in the private blockchain the Data Transaction containing the error code and the description is always published. The error transaction includes the following data: • height - the height of the chosen block from the private blockchain. • signature - the signature of the chosen block from the private blockchain. • error-code - the error code. • error-message - the error message. Error types Code Message Possible cause 0 Unknown error An unknown error occurred during the send of the transaction to the Targetnet 1 Fail to create data transaction for Targetnet Creating of the transaction to be sent to the Targetnet failed 2 Fail send transaction to Targetnet The transaction publication to the Targetnet failed (it could be a JSON request error) 3 Invalid http status of response from Targetnet transaction broadcast The Targetnet has returned an HTTP code other than 200 after the transaction publication 4 Fail to parse http body of response from Targetnet transaction broadcast The Targetnet has returned an unknown JSON after the transaction publication 5 Targetnet return transaction with id='$mainnetTxId' but it differ from transaction that we sent id='$sentTxId The Targetnet has returned mismatched ID after the transaction publication 6 Targetnet didn't respond on transaction info request The Targetnet has not responded to the request about the transaction info 7 Fail to get current height in Targetnet Failed to get current Targetnet height 8 Anchoring transaction in Targetnet disappeared after height rise enough The anchoring transaction has disappeared from the Targetnet after its height evened height-above value 9 Fail to create sidechain anchoring transaction Fail to public the anchoring transaction in the private blockchain 10 Anchored transaction in sidechain was changed during Targetnet height arise await, looks like a rollback has happened Anchored transaction in sidechain was changed during Targetnet height arise await, looks like a rollback has happened	2020-09-19 09:14:08	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 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": 0.17605814337730408, "perplexity": 4534.200900937508}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-40/segments/1600400191160.14/warc/CC-MAIN-20200919075646-20200919105646-00417.warc.gz"}
http://dict.cnki.net/h_51766738066.html	全文文献工具书数字学术定义翻译助手学术趋势更多 gamma 在外科学分类中的翻译结果: 查询用时：0.195秒历史查询 gamma 很抱歉，暂未找到该词条在当前类别下的译词。您可以查看在所有学科下的译词。 “gamma”译为未确定词的双语例句 Gamma Nail in the Treatment of the Intertrochanteric Fractures of Femur (Primary Report of 20 Cases) Gamma钉治疗股骨粗隆间骨折20例初步报告短句来源 Modified Gamma Nailing for the Treatment ofPeritrochanteric Fracture of the Femur 改良Gamma钉手术治疗股骨转子周围部骨折短句来源 Gamma Nail for the Treatment of 32 Cases with Intertrochanteric Fracture Gamma钉治疗股骨转子间粉碎性骨折——附32例分析短句来源 Methods Totally 174 type A2 (AO/OTA) intertrochanteric fractures were treated with Gamma nails (type Asia-Pacific,Stryker) from February 2002 to December 2005.Their average age was 72.6 years old. 方法2002年2月～2005年12月采用Gamma钉内固定治疗174例AO/OTA分型为A2型股骨转子间骨折患者,其中A2．2型107例,A2．3型67例; 闭合复位135例,开放复位39例。短句来源 They were respectively treated with five kinds of operations:48 patients with dynamic hip screw(DHS),27 patients with proximal femoral anatomical plate,23 patients with Gamma nail,21 patients with proximal femoral nail(PFN),8 patients with tow-way cannulated compressive screw. 其中动力髋螺钉(DHS)固定48例,股骨近端解剖型钢板固定27例,Gamma钉固定23例,股骨近端钉(PFN)固定21例,双头空心加压螺纹钉固定8例。短句来源更多我想查看译文中含有：的双语例句 gamma It is also shown that on the nilmanifold $\Gamma\backslash (H^3\times H^3)$ the balanced condition is not stable under small deformations. For $\gamma\in\mathbb{R}$ let $C(\gamma)$ be the set of all $f\in{\mathcal S}^\prime$ for which$\sum_{n=0}^{\infty}\,\|(f,h_n)\|\,(n+1)^{\gamma}>amp;lt;\infty$, where $(h_n)_{n=0,1,\dots}$ is the orthonormal base of Hermite functions. We show that $C(\gamma)\subset S_0$ if and only if $\gamma\geq\frac{1}{4}$ and that $S_0\subset C(\gamma)$ if and only if $\gamma\leq{-}\frac{1}{4}$. Using these results we give an explicit solution of the problem of optimal reconstruction of functions from Sobolev's classes $W^{\gamma}_{p}(M^{d})$ in $L_{q}(M^{d}), 1 \leq q \leq p \leq \infty$. This result generalizes the characterization of Fourier series of distributions with a distributional point value given in [5] by $\lim_{x\rightarrow\infty}\sum_{-x\leq n\leq ax}a_{n}e^{inx_{0}}=\gamma\ (\mathrm{C},k)\,$. 更多	2019-09-15 22:28:49	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6091127395629883, "perplexity": 4731.427907388012}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-39/segments/1568514572436.52/warc/CC-MAIN-20190915215643-20190916001643-00209.warc.gz"}
https://zbmath.org/?q=an:1423.13126	# zbMATH — the first resource for mathematics On the approximate periodicity of sequences attached to non-crystallographic root systems. (English) Zbl 1423.13126 Summary: We study Fomin-Zelevinsky’s mutation rule in the context of non-crystallographic root systems. In particular, we construct approximately periodic sequences of real numbers for the non-crystallographic root systems of rank 2 by adjusting the exchange relation for cluster algebras. Moreover, we describe matrix mutation classes for types $$H_3$$ and $$H_4$$. ##### MSC: 13F60 Cluster algebras Full Text: ##### References: [1] Armstrong, [Drew Armstrong] Drew, Braid groups, clusters, and free probability: An Outline from the AIM workshop · Zbl 1191.05095 [2] Felikson, [Felikson 12] A.; Shapiro, M.; Tumarkin, P., Cluster algebras of finite mutation type via unfoldings., Int. Math. Res. Notices, 8, 1768-1804, (2012) · Zbl 1283.13020 [3] Fomin, [Fomin and Zelevinsky 02] S.; Zelevinsky, A., Cluster algebras I: foundations, J. Am. Math. Soc., 15, 2, 497-529, (2002) · Zbl 1021.16017 [4] Fomin, [Fomin and Zelevinsky 03] S.; Zelevinsky, A., Cluster algebras II: finite type classification, Invent. Math., 154, 1, 63-121, (2003) · Zbl 1054.17024 [5] Gu, [Gu 11] W., A decomposition algorithm for the oriented adjacency graph of the triangulations of a bordered surface with marked points, Electr. J. Combin., 18, 1, (2011) · Zbl 1217.05213 [6] Gu, [Gu 12] W., The decomposition algorithm for skew-symmetrizable exchange matrices, Elect. J. Comb., 19, 2, (2012) · Zbl 1256.65020 [7] Lawson, [Lawson] J., Minimal mutation-infinite quivers · Zbl 1401.13067 [8] Warkentin, [Warkentin 14] M., Exchange graphs via quiver mutation This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching.	2021-12-04 13:38:12	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6235173344612122, "perplexity": 4973.632667487334}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-49/segments/1637964362992.98/warc/CC-MAIN-20211204124328-20211204154328-00100.warc.gz"}
https://bioinformatics.mdanderson.org/public-software/ngchm/navigating-clustered-heat-maps/	### Department of Bioinformatics and Computational Biology Home > Public Software > NG-CHM > Using the Viewer ## Using the NG-CHM Viewer We have developed a series of tutorials and videos describing key features of the NG-CHM Viewer: Topic Video Quick Tour (ppt) N/A Display Components and Interactivity Features (video) Panning and Zooming (video) Selections (video) Color schemes (video) Covariate Bars (video) ### Introduction to Next-Generation (Clustered) Heat Maps Next-Generation (Clustered) Heat Maps (NG-CHM) are highly interactive, dynamic (clustered) heat maps that enable the user to see an overview of the entire heat map, and via navigation controls, to zoom and pan across the heatmap to see details of the heatmap at many levels of resolution. Other controls enable searching for specific heatmap entries, generating production quality PDFs, and linking out to information related to rows, columns, and individual heatmap entries. The user interface is mostly intuitive, although a few less commonly used features can only be accessed with the use of specific keys. This page describes how to navigate a NG-CHM using the NG-CHM Viewer. It presumes that you are interacting with a NG-CHM that has already been loaded on a NG-CHM web server. In addition to these pages, we have also produced a video describing the capabilities of the NG-CHM viewer. ### Launching the NG-CHM Viewer The NG-CHM Viewer is a web application. You can use most web browsers including Firefox, Safari and Chrome. Although many features will work on IE, some will not. (We are striving to eliminate such issues, but new feature development is our first priority at this time.) You do not need any special plugins or other browser software installed, but you will need to have Javascript enabled. If you know the name of the map, you can point a browser directly to it using the following URL format: http://bioinformatics.mdanderson.org/chm/chm.html?name=YOUR-NGCHM-NAME You can also see a searchable list of available maps by not specifying a specific NG-CHM name. For instance: http://bioinformatics.mdanderson.org/chm/ These examples use our NG-CHM server at MD Anderson. If the NG-CHM you wish to access is on a different server, you will need to substitute that server’s URL. ### Help and Feedback If you have any comments or questions about the CHM tool, contact Bradley Broom .	2019-06-20 10:01:41	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.18664462864398956, "perplexity": 4318.728102590272}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-26/segments/1560627999200.89/warc/CC-MAIN-20190620085246-20190620111246-00255.warc.gz"}
https://quant.stackexchange.com/tags/cvar/hot	Tag Info Accepted Rockafellar-Uryasev mean-CVaR optimiztion $VaR_\alpha$ is a scalar choice variable in the minimization problem. In the Rockafeller-Uryasev paper, it is simply called $\alpha\in R$. (C.f., the program described in Theorem 2 of that paper, or ... • 393 Accepted Question on Rockafellar's Paper for optimisation of CVaR On 1, I suspect that is a typo and that the second formula should sum to r. On 2, that is applying well-known techniques in how to handle piece-wise linear functions in an optimizer. For instance, ... • 5,291 Accepted CVAR alternatives for optimization Following the comments and the edits to the question, I'll try to show how conditional Value-at-Risk (aka Expected Tail Loss) can be minimised for a portfolio. We start with the implementation ... • 2,856 Accepted Confidence Interval on Monte-Carlo-CVaR One: Your VaR CI relies on normal approximation and might be (very) bad depending on the number of samples and the target function (P&L). Often it is better to use the exact approach based on the ... • 1,933 CVaR is concave risk measure or convex? CVaR is a convex function in the underlying portfolio (measured as for instance absolute value or profit). I won't get into proving anything so instead I am going to link the first result from Google ... • 1,585 Accepted CVaR formulation If $Y=-\pi(\mu,D)$ then the first formula is $$\mathrm{CVaR}_\eta(-Y)=\max_{\nu\in R}\left\{\nu+\frac1\eta E((-Y-\nu)^-)\right\}$$ where $X^-=\min (X,0)$ and $X^+=\max(X,0)$. Note that $(-X)^-=-(X^+)$.... Accepted How to minimize $CVaR_{\alpha}(\min(X,d))$, where $X$ is a random variable and d is the decision variable? The minimum value is always attained at $d=0$. In this proof, I will assume that the distribution of the random variable $X$ is absolutely continuous and monotonically increasing, and thus the CDF ... • 136 What are the advantages of $EVaR$ over $CVaR$? The entropic value at risk (EVaR) is a coherent risk measure, developed to tackle some computational inefficiencies of the CVaR. It is the tightest possible upper bound for traditional VaR and CVaR, ... • 2,815 Accepted Questions about VaR and CVaR. Is there any relation between $VaR_{\alpha}(X)$ and $VaR_{\alpha}(-X)$, or $CVaR_{\alpha}(X)$ and $CVaR_{\alpha}(-X)$? We consider the case where the distribution function $F$ of $X$ is strictly increasing. Then \begin{align} VaR_{\alpha}(X) &= \inf\{x: P(X >x) \le \alpha \}\\ &=\inf\{x: F(x)\ge 1-\alpha \... • 20.4k Accepted Calculate CVaR for a portfolio This sounds correct, however step 2 is a little vague, so I will try to restate the steps here for you. The assets in your portfolio must be priced with respect to a set of risk factors (e.g. ... • 1,335 Accepted How to calculate the distortion function for CVaR? I have solved it myself. The key was to realize that for $X \geq 0$ and $S_X(t) = \mathbb{P}(X>t)$ $$\int_0^\infty S(t) dt = \int_0^1 F_X^{-1}(u) du = \mathbb{E}\left[X \right].$$ This is ... • 81 Accepted How to prove the following relation of Conditional Value-at-Risk and Value-at-Risk? A slightly different take here: How to prove the following relation of Conditional Value-at-Risk and Value-at-Risk? Let $F$ be the cumulative distribution function of $X$. We assume that $F$ is continuous. Then, for $x\ge 0$, \begin{align} F^{-1}(x) = \inf\{s: F(s) \ge x \}. \end{align} Moreover, \begin{align} \... • 20.4k CVaR is concave risk measure or convex? It does not even matter if it’s concave or convex wrt global optimisation, both concave and convex functions have global optimal points albeit the only difference is maximum vs minimum which is easily ... Accepted When is the VAR equal to the CVAR That is incredibly unlikely for a continuous distribution -- though possible for a distribution with a part that is not absolutely continuous, i.e. is atomic. The way to see this is to remember that ... • 2,740 Accepted 1 vote Accepted Elicitability of risk measures From Ziegel (2013) : The risk of a financial position is usually summarized by a risk measure. As this risk measure has to be estimated from historical data, it is important to be able to verify and ... • 82 Only top scored, non community-wiki answers of a minimum length are eligible	2022-05-23 21:47:40	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8757399916648865, "perplexity": 684.6598388241467}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662561747.42/warc/CC-MAIN-20220523194013-20220523224013-00140.warc.gz"}
https://math.eretrandre.org/tetrationforum/showthread.php?tid=1622	Real Multivalued/Parametrized Iteration Groups bo198214 Administrator Posts: 1,615 Threads: 101 Joined: Aug 2007 08/15/2022, 08:02 AM (This post was last modified: 08/15/2022, 08:07 AM by bo198214.) I was shock puzzled when Leo.M mentioned the possibility that a superfunction might not result from or not have an iteration group. This question arose from actual real valued superfunctions of $$b^z$$ with b around eta minor. There the fixed point derivative is negative and so would imaginably not allow for e.g. a half iterate. However the given superfunctions are valid superfunctions S, satisfying $$S(x+1) = f(S(x))$$ however they show oscillatory behaviour are hence not injective as we are used to from positive fixed point multipliers/derivatives. In those cases you can just define $$f^{\circ t}(x) = S(t+S^{-1}(x))$$ and you have an iteration (semi) group particularly satisfying $$f^{\circ s+t }(x) = f^{\circ s}(f^{\circ t}(x)),\quad f^{\circ 1} (x)=f(x)$$ So how can this be done in case S in not invertible?! Using Multivalued functions! As a research object I started with the edge case $$f(x)=-x$$ (multiplier -1). This case is so edge that it often is not even called parabolic fixed point, because $$f(f(x))=x$$. However it serves our research purpose. One can easily give the regular iteration group: $$f^{\mathfrak{R} t}(z) = (-1)^t z = e^{\pi i t}z$$ which is complex valued, the superfunction would be $$S^{\mathfrak{R}}(z)=e^{\pi i z}$$. Now we know from the real valued Fibonacci extension that this strange "realifyer" is used as a real replacement for $$(-c)^t$$ $$\rho(t) = \cos(\pi t) c^t$$ It satisfies $$\rho(t+1)=(-c)\rho(t)$$ and $$\rho(1)=(-c)^1$$ but is completely real valued. And typically it can be used in conjuction with the inverse Schröder function, or the regular Superfunction to get a real valued solution, when we replace $$(-c)^t$$ with it: So in our case we can define a superfunction for $$f(x)=-x$$ by: $$S(x)=\cos(\pi x)$$ I started with this edge case c=1 because I can easily invert this function, while it much more difficult when $$c\neq 1$$. So our iteration semigroup can be defined as $$f^{\circ t}(x)= \cos\left(\pi \left(t\pm\frac{\arccos(x)}{\pi}\right)\right) = \cos(\pi t \pm \arccos(x))$$ we need to choose the right branch of $$\arccos$$. Visually this gives an Aha! We have a circle as half iterate and and an ellipse as 1/3 iterate. But we can check (pink line) that it indeed are proper iterates, as long as we choose the right branch, though it does not even have the fixed point a 0 - very edge case. And these are ellipses in the geometrical sense, the real thing - not such something that looks like it. You can convince yourself when expanding $$\cos(\pi t \pm \arccos(x))$$. So this gives already a rich result for such a simple function, but it becomes even better for $$c\neq 1$$. Here instead of cumbersomely trying to compute inverse we just use a parametrized version, which is quite simple to obtain an iteration group from a given Superfunction S: $$(x,y) = (S(s),S(s+t))$$ that's it already. This time the iterates have the same fixed point, as the original function, but - as they are not regular at the fixed point - they behave wildly around the fixed point (not too wild though). So, conclusion: Oscillating superfunctions are still associated with continuous iteration (semi)groups, only that they are multivalued now and one has to be careful with choosing the right branch (which is that way with most complex functions anyways). Later I want to transfer that to our Fibonacci LFT - so stay tuned tommy1729 Ultimate Fellow Posts: 1,703 Threads: 374 Joined: Feb 2009 08/15/2022, 11:49 PM but those iterations probably fail hard on the complex plane ?? that is for complex x and/or s and/or t. regards tommy1729 JmsNxn Ultimate Fellow Posts: 985 Threads: 117 Joined: Dec 2010 08/17/2022, 02:58 AM This was the original comments by myself and MphLee when Leo first came on the scene. This is all Riemann surface territory. Riemann surface stuff gets wild when you talk about iterations. These are well known constructions; but it classifies itself under "dynamics on a Riemann surface". It behaves very similar to the normal dynamics we have; which is why I keep mentioning Milnor. We either have $$\mathbb{C}$$ (Euclidean) or $$\mathbb{D}$$ (A simply connected domain) or $$\widehat{\mathbb{C}}$$ (The Riemann Surface). These are just three Riemann surfaces that we use as our base, especially with most of the work done here. Choosing an arbitrary Riemann surface $$S$$--which can be as wild as possible, and taking iterates of $$f^{\circ n} : S \to S$$--we get everything Leo is talking about. The trouble is, talking about it as multivalued functions won't give you all the juice that Riemann surfaces will give you. It's like drinking orange juice compared to eating oranges. Whether we process our oranges first (Project the Riemann surface into the space of multivalued functions), or we just eat oranges (Prove everything with Riemann surfaces, that Milnor and most of complex dynamics set up). If you want to go down the Riemann surface route though. We should refer to this more broadly. Where the preimage of the $$\beta$$ tetration of $$\eta^-$$ is Riemann surface, and constructing an action from this riemann surface to $$\mathbb{C}$$ creates a multivalued semi group at $$\eta^-$$. This becomes: $$f^{\circ t}(z) = F\left(t+\mu(\mathcal{F}(z))\right)\\$$ Where $$F$$ is the beta iteration with period $$2 \pi i/\lambda$$, and $$\mu : S \to \mathbb{C}$$ and $$\mathcal{F}(z) = \{y \in \mathbb{C}\,\|\, F(z) = y\}$$. Then we just note that $$\{\forall z \mathcal{F}(z)\} = S$$; and we are just choosing a projection schema (a multivalued function). Let's go down the rabbit hole, bo! I loved Leo's original thesis. And I loved the idea. The trouble I had, was that it was the image, and not the Riemann surface preimage. tommy1729 Ultimate Fellow Posts: 1,703 Threads: 374 Joined: Feb 2009 08/17/2022, 12:25 PM (08/17/2022, 02:58 AM)JmsNxn Wrote: This was the original comments by myself and MphLee when Leo first came on the scene. This is all Riemann surface territory. Riemann surface stuff gets wild when you talk about iterations. These are well known constructions; but it classifies itself under "dynamics on a Riemann surface". It behaves very similar to the normal dynamics we have; which is why I keep mentioning Milnor. We either have $$\mathbb{C}$$ (Euclidean) or $$\mathbb{D}$$ (A simply connected domain) or $$\widehat{\mathbb{C}}$$ (The Riemann Surface). These are just three Riemann surfaces that we use as our base, especially with most of the work done here. Choosing an arbitrary Riemann surface $$S$$--which can be as wild as possible, and taking iterates of $$f^{\circ n} : S \to S$$--we get everything Leo is talking about. The trouble is, talking about it as multivalued functions won't give you all the juice that Riemann surfaces will give you. It's like drinking orange juice compared to eating oranges. Whether we process our oranges first (Project the Riemann surface into the space of multivalued functions), or we just eat oranges (Prove everything with Riemann surfaces, that Milnor and most of complex dynamics set up). If you want to go down the Riemann surface route though. We should refer to this more broadly. Where the preimage of the $$\beta$$ tetration of $$\eta^-$$ is Riemann surface, and constructing an action from this riemann surface to $$\mathbb{C}$$ creates a multivalued semi group at $$\eta^-$$. This becomes: $$f^{\circ t}(z) = F\left(t+\mu(\mathcal{F}(z))\right)\\$$ Where $$F$$ is the beta iteration with period $$2 \pi i/\lambda$$, and $$\mu : S \to \mathbb{C}$$ and $$\mathcal{F}(z) = \{y \in \mathbb{C}\,\|\, F(z) = y\}$$. Then we just note that $$\{\forall z \mathcal{F}(z)\} = S$$; and we are just choosing a projection schema (a multivalued function). Let's go down the rabbit hole, bo! I loved Leo's original thesis. And I loved the idea. The trouble I had, was that it was the image, and not the Riemann surface preimage. let me get this straith $\mathcal{F}$ send you to the right complex value and mu sends you to the right branch ? regards tommy1729 tommy1729 Ultimate Fellow Posts: 1,703 Threads: 374 Joined: Feb 2009 08/17/2022, 12:26 PM (08/15/2022, 11:49 PM)tommy1729 Wrote: but those iterations probably fail hard on the complex plane ?? that is for complex x and/or s and/or t. regards tommy1729 right ? JmsNxn Ultimate Fellow Posts: 985 Threads: 117 Joined: Dec 2010 08/18/2022, 04:31 AM Honestly, no idea Tommy. My knowledge of Riemann surfaces is largely cursory, and more so I learnt pieces of it by way of the link between Elliptic curves and Algebraic curves. In theory, there should be a lift (I believe that's the word), such that we can write: $$f^{\circ t} (z) : S \to S\\$$ Where we are sort of sending the preimages to themselves. Then we can only care about the Riemann surface itself; and in proving things about the Riemann surface we prove things about the iteration on the complex plane--and figure out how and where it is holomorphic if it could be. For example if $$S = \{ y \| y = F(z)\}$$ then we define a coordinate $$s_0 \in S$$ such that $$f(s_0)$$ is the value of $$y$$ for $$F(z+1)$$. And now we are acting on the preimage. Then we perform iterations as such. I haven't done much Riemann surface stuff in a while, but this is kind of what you are doing. This isn't very different from how you define the Riemann surface of $$\log$$. It's just $$\{y \| y = e^z\}$$. Now we are talking about Dynamics on a Riemann surface; and from there, yes, you choose a projection of this riemann surface into the complex plane (you choose your branch of $$\log$$). bo198214 Administrator Posts: 1,615 Threads: 101 Joined: Aug 2007 08/18/2022, 06:56 AM (08/17/2022, 02:58 AM)JmsNxn Wrote: If you want to go down the Riemann surface route though. We should refer to this more broadly. (08/17/2022, 02:58 AM)JmsNxn Wrote: Let's go down the rabbit hole, bo! I loved Leo's original thesis. And I loved the idea. The trouble I had, was that it was the image, and not the Riemann surface preimage. No, I totally don't want to go that rabbit hole! This would be a lot of theory to digest, with possibly no effect on the questions we have - rather it would pose a lot of new questions! Remember, I am beyond my limit already ... And I found already what I wanted. Its not just that you can say: well we doing everything multivalued (like when you draw a picture of arctan with stacked copies along y) and somehow some choice of branch will work - but it is rather the idea of (analytic) continuation that solves the puzzle - the graph is one continuous line, one does not need to care which branch cut gets glued to which other branch cut. But what exactly you mean by Leo's original thesis? JmsNxn Ultimate Fellow Posts: 985 Threads: 117 Joined: Dec 2010 08/18/2022, 07:30 AM (08/18/2022, 06:56 AM)bo198214 Wrote: But what exactly you mean by Leo's original thesis? Oh, I apologize here. Leo made a bunch of introductory posts about a year or so ago. And it's very similar to this construction. And Mphlee and I were talking to him about it, and asking him questions. So I'm a big fan of the idea when he first provided it. I don't mean thesis, thesis. I Just mean the constructions he's developed when referring to multivalued iterations. And I'm just referring to it as a thesis, as though this is an idea he's originated and "defending like a thesis". Again, Bo. I apologize, it's the casual nature of my language. I don't mean thesis, I mean his introduction and discussion from a year or so ago. bo198214 Administrator Posts: 1,615 Threads: 101 Joined: Aug 2007 08/18/2022, 07:39 AM (08/18/2022, 07:30 AM)JmsNxn Wrote: (08/18/2022, 06:56 AM)bo198214 Wrote: But what exactly you mean by Leo's original thesis? Oh, I apologize here. Leo made a bunch of introductory posts about a year or so ago. And it's very similar to this construction. And Mphlee and I were talking to him about it, and asking him questions. So I'm a big fan of the idea when he first provided it. I don't mean thesis, thesis. I Just mean the constructions he's developed when referring to multivalued iterations. And I'm just referring to it as a thesis, as though this is an idea he's originated and "defending like a thesis". Again, Bo. I apologize, it's the casual nature of my language. I don't mean thesis, I mean his introduction and discussion from a year or so ago. There is totally no need to apologize for that I asked a question! (And in a lot of other posts you made there was also no reason at all to apologize ... there is no fault on you) Also I didn't read it in the sense of master thesis or phd thesis. Just wanted to know what it is! JmsNxn Ultimate Fellow Posts: 985 Threads: 117 Joined: Dec 2010 08/18/2022, 07:53 AM Here you go! https://math.eretrandre.org/tetrationfor...p?tid=1318 Where Leo. W as a user makes their introduction and draws out a lot of cool shit, and we bicker a bit about Riemann surfaces, lol. Then, https://math.eretrandre.org/tetrationfor...p?tid=1539 Where we enter a lot of observations by Mphlee, which are relevant when we concern ourselves with the group property $$f \circ g = g \circ f$$ and the admission of an exponential $$f \circ g(t) = g(t+1)$$. This is where Leo makes his introduction, and I mean that as an "introduction to a good paper". « Next Oldest \| Next Newest » Possibly Related Threads… Thread Author Replies Views Last Post Range of complex tetration as real Daniel 2 183 10/22/2022, 08:08 PM Last Post: Shanghai46 From complex to real tetration Daniel 3 176 10/21/2022, 07:55 PM Last Post: Daniel Cost of real tetration Daniel 1 204 09/30/2022, 04:41 PM Last Post: bo198214 Constructive real tetration Daniel 1 224 09/01/2022, 05:41 AM Last Post: JmsNxn Real and complex tetration Daniel 1 288 08/17/2022, 01:56 AM Last Post: JmsNxn Complex to real tetration Daniel 1 219 08/14/2022, 04:18 AM Last Post: JmsNxn Constructing a real valued Fibonacci iteration--its relation to $$1/1+z$$ JmsNxn 7 441 08/13/2022, 12:05 AM Last Post: JmsNxn Challenging the concept of real tetration Daniel 2 371 07/04/2022, 03:09 PM Last Post: Daniel Complex to real tetration via Kneser Daniel 3 517 07/02/2022, 02:22 AM Last Post: Daniel Real and complex tetration Daniel 3 472 06/22/2022, 11:16 PM Last Post: tommy1729 Users browsing this thread: 1 Guest(s)	2022-11-26 10:03:00	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8017851114273071, "perplexity": 1269.0230785257927}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446706285.92/warc/CC-MAIN-20221126080725-20221126110725-00092.warc.gz"}
https://dsp.stackexchange.com/questions/47454/digital-filter-simulating-hydroacoustic-signal-distortion	# Digital filter simulating hydroacoustic signal distortion Preambule: I'm designing a sound model for my small submarine game. Model is running on the server, and I want to present the client with a mono-channel wav-stream from his hydrophone (20kHz discretization should suffice, I target 20Hz-10kHz band). I want that signal to be relatively-realistic. Specifically, I need to account for the fact that water dissipates high frequencies much faster. Sound model: I plan to pre-generate (with help of some RNG and IFFT) raw propeller time-domain sound sample from it's intensity spectrum. Then I plan to amplitude-modulate it with shifted shaft rpm x propeller blade count sine wave to get that propeller beat. At this point i need to materialize that signal on the player's sensor. It will probably involve two filters: • Some filter (wich is the filter in question) wich will account for water dissipating signal energy. • Bandpass filter, wich corresponds to this particular sensor's sensitivity band. • Simple amplitude multiplication to account for range, sensor directivity etc. According to R.J. Urik "Principles of Underwater Sound", on the distance of r meters from the target, band level (dB) in frequency bin f of such sound can be approximated using following formula: [1]: ResultBL = SourceBL - 10 * log10(r * r) - r * F(f) or without wave front expansion term (wich is trivial): [2]: ResultBL = SourceBL - r * F(f) where F(f) is some smooth monotonic function (big radical with multiple squared frequencies and constants) of frequency. 1. Is it possible to design a time-domain digital filter for given range r, that implements\approximates the signal distortion [2]? 2. What type of filter would you recommend? Problem is not realtime, but calculations should be fast. I'm not limited to causal filters, so I'm sensing something like non-causal IIR? 3. What algorithms are applicable to this problem. 4. If such algorithm involves transfer function, how should I transform\approximate F(f) term in it? 5. Will it be fast? What factors affect it's speed? 6. Will such filter work good in the whole band (20Hz-10kHz)? 7. Will such filter work well on large intensity level variances, e.g. on both very weak and very strong signals? 8. Is such synthesis computationally-expensive? Thank you. I wouldn’t bother much with the precise shape of the filter because any real passive sonar will boost the higher frequencies at the receiver and any realistic source levels are going to around 70 years old and most likely a double screw, and aspect dependent. Cavitation is depth dependent. A single pole low pass filter with a corner around 10 Hz is probably ok. Fidelity is one of those things that can become an obsession. There is ambient noise and self noise associated with hydro acoustic flow. Higher frequencies exhibit hull shading. No one band covers all the signal types of interest. If you want to be obsessive, look at Mike Porter's Ocean Acoustics web site. http://oalib.hlsresearch.com/ I'm partial to RAM PE and there at least was a broad band (decomposition of narrow band) demo for the MATLAB version. There was a video game that came out around 20 years ago that collaborated with a company named SONOLYST. They actually got an Oscar for sound effects for the Hunt for Red October movie. I like Urick. I actaully meet him a few times and took a week long class from him but you might find Ross a bit more helpful. Mechanics of Underwater Noise, Ross, D. isbn={9781483160467}, url={https://books.google.com/books?id=sdwgBQAAQBAJ, 2013, Elsevier Science • I have no illusion about actual source levels, game is fiction. Physical effects during transmission are however important: I cannot pass the same signal to high-frequency and low-frequency hydrophones, the latter one will have an unfair detection range. The same screw must be heard much better on 1kHz than on 2k-10kHz bow hydrophone. Screw count and aspect dependency, as well as all the other stuff, like flow noise are no problem: they are easily approximated. My problem is with diffirential frequency distortion by water, wich I don't know how to perform efficiently. – Boris-Barboris Feb 27 '18 at 16:09 • Themore common term dispersion, see my edits – user28715 Feb 27 '18 at 16:40 • Good links, thank you. Some of the internal links are broken, but I will contact the author. – Boris-Barboris Feb 27 '18 at 16:46 • Urick passed away in 1996, so that's a dead link as well. There is a legend about Urick. He was very interested in deep sound channel (DSC) propagation and a technique to probe the DSC is to use implosive sound sources, typically an explosive. Urick went around the TV repair shops in Silver Springs and purchased burnt out TV picture tubes and used these for impulsive sources. He developed calibration curves for brand, size, and crush depth. – user28715 Feb 27 '18 at 18:17 • @StanleyPawlukiewicz Interesting, others have considered bulbs as excellent implosive sources too. CRTs must implode much more deeply though which makes sense for DSC...not to mention the environmental impact... (Can you perhaps edit the book reference so that it either becomes a link or renders as plain text (code) please?) – A_A Feb 28 '18 at 10:12 In addition to the very good points raised by Stanley Pawlukiewicz, it is indeed a question of level of detail and you are probably looking for a "good enough" approximation. You can however set up a generative model for the type of sound you are after, with a set of parameters that you could vary to simulate different conditions. In general, engine sound is integrated impulses. If you obtain a high sampling frequency (undistorted) recording of the exhaust of a motorbike (or car) at idle speed, you will observe that you can actually count the Revolutions Per Minute (RPM) from the sound recording. In fact, you can still get a rather accurate RPM count (for a motorbike) even if you throttle the engine up to moderate RPM where, you can still observe the pulsating output of a piston engine. If you keep throttling the bike to high RPM, the pulses will start merging and the determination of RPM will be very sensitive. If you are recording a car, this limit will be at lower RPM because it has more cylinders and generates more pulses per unit of time. The sound that reaches your ears from a motorbike or a car is the impulsive noise of the engine filtered by the exhaust. Jet engines suck in air, ignite it and get an "impulse" of boost by the sudden expansion of the gas. This happens very quickly but essentially, what you hear is filtered by the "flute pipe" around the turbine. This is very obvious in Pulsejets, less obvious in early jets that still had ignition chambers and even less obvious in axial jet's that are essentially large "flutes". Underwater sound, the sound that reaches a hydrophone would be the sum of the engine noise coupled to the water through the hull of the boat PLUS the sound the screw makes in the water. A propeller in the water is different than a propeller in the air, because water is (almost) incompressible. Therefore, the propeller doesn't "sound", until it starts cavitating. Therefore, for general "shipping", what you hear in the water at some range $r$, is mostly engine noise, reduced in amplitude with adapted inverse square law and filtered. The filter is the combined effect of the way the engine is coupled to the water (through the hull) and the frequency response of sea water for the "channel" the sound travels in. So, a wooden fishing boat with an inboard engine on rubber buffers sounds differently than a recreational outboard engine craft. The inboard engine "knocks" against larger surface that is coupled to the water. Its low frequency content will be much stronger than high frequencies. Now, sound propagation in sea water depends on a number of factors, but there are models that can return a "frequency response" based on these factors (for more information please see this link). So, bringing all this together: 1. Set up a generator of impulses. One impulse for each piston chamber of your engine. The amplitude of the impulse is proportional to the size of the engine, the frequency of the impulses is proportional to the RPM of the engine. 2. Apply the "engine-hull" impulse response. This is what a single impulse would sound like in the water. You might be able to isolate one from existing recordings. 3. Apply the "hull-sea" impulse response. This is what a single impulse would sound like as it travels through the "water channel" from the source to the hydrophone. In shallow water you get a huge amount of reflections and depending on the surrounding environment, you might even hear reflections BEFORE the direct sound. In deep water, sound can be trapped in "channels", which means that at certain positions between source and hydrophone you might not even be able to "listen" to the source. It vanishes. 4. Reduce the amplitude taking into account $r$. You can work with Finite Impulse Response filters or more complex Digital Waveguide models. It is probably better to do this offline and blend between files (depending on distance and bathymetry for example) rather than operate the generative model in realtime. Hope this helps. • 1). All valid points for simulation, but I just need a game, so I can throw away a lot of complex real effects. It does not to have to be realistic, but it has to be intuitive - non-contradictory with expectations, like propeller beat is there, and high frequencies fade faster than low on the range. 2). Why FIR? I was adwised elsewhere to look at Parks-McClellan, but looking at pictures from mathworks.com/help/signal/ref/firpm.htm, it looks too imprecise. I don't need a causal filter. Are there alternatives? – Boris-Barboris Feb 28 '18 at 15:14 • @Boris-Barboris 1) The effects mentioned here are not complex. 2) The Parks-McClellan algorithm leads to the design of FIR filters (?). – A_A Feb 28 '18 at 15:18 • 2) Yes. I find a lot of info on FIR design, becaues they are popular. However, I am not in need of causal FIR filter per se. I also hear stuff like better IIR performance, as well as better quality. Wich actually goes well with my experience with numerical modelling and ODE solving, where implicit non-causal schemes are generally much better then explicit ones. That makes me seek alternatives to popular FIR methods preemptively. – Boris-Barboris Feb 28 '18 at 15:24 • @Boris-Barboris The criterion for choosing a filter type is not its popularity. To receive answers closer to what you are asking, you need to provide some specifications in terms of cut off frequencies, pass pand width, pass pand ripple, etc. If all you want to do is playback a loop recording that gets low pass filtered at some variable cut off depending on $r$ then you can use something as simple as a modulated sinc filter (to change the cut off in realtime). – A_A Feb 28 '18 at 15:48 • The original question contains all the info. I need a filter for some smooth monotonic frequency response function, key word is "some", filter design method should be able to work with any such smooth monotonic function. No ripple\passband\stopband\cutoff, i need a direct (wich takes a table of frequency:desired-response-magnitude pairs) method, something like matlab's yulewalk, wich I will try as well. – Boris-Barboris Feb 28 '18 at 15:57	2020-02-24 03:13:20	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.39757147431373596, "perplexity": 1728.5151751470055}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-10/segments/1581875145869.83/warc/CC-MAIN-20200224010150-20200224040150-00509.warc.gz"}
https://www.edaboard.com/threads/interrupt-service-routine-for-pic16f877.215399/	# Interrupt Service Routine for PIC16f877 Status Not open for further replies. #### eefinni ##### Newbie level 3 Hi all, I had tried all means to exit the ISR but it seems impossible.. I'd read about RETFIE but it is for ASM. I had also read about instructions set for RETFIE but I don't know how to implement it. Is there anyway to exit from the ISR? Here is my code: PHP: //Define compiler error message. #ifndef __CPU_16F877__ #error "This program is tailored for PIC16F877 controller" #endif //Include required header file. #include "io16f877.h" //the hardware register defination file. #include "USART.h" #include "DELAYS.h" //List out all inputs and outputs #define LED_1 RA0 #define LED_2 RA1 #define LED_3 RA2 #define IR RC1 #define RED RB5 #define Yellow RB6 volatile unsigned int DataCount; volatile unsigned char DATA[3]; /******************************************************************** This function initializes the IO ports. *******************************************************************/ void initialize_IO_ports(void) { //set the digital IO ports as per requirement. ADCON1 = 0x07; //set PortA as digital I/O TRISA = 0x00; //PortA as output. TRISB = 0x00; TRISC = 0x8F; //clear the output ports at the beginning. PORTA = 0x00; PORTC=0x00; PORTB=0x00; } #pragma vector=0x04 __interrupt void isr(void) { //if(DataCount==0) // PCL=PCLATH; //if(RCIF) //{ RCIE=0; DATA[DataCount]=RCREG; if(DataCount!=2) DataCount++; CREN=0; RCIF=0; //} if(DATA[0]=='0') LED_1=1; if(DATA[1]=='1') LED_2=1; if(DATA[2]=='2') LED_3=1; //if(DataCount==2) // PCLATH=(PCL+1); //if(DataCount==2) //if(DataCount==2) //GIE=0; //RCIF=1; //STATUS=0x009; //DelayUs(1); //DelayUs(1); Yellow=1; DelayMs(100); Yellow=0; RCIE=1; //INDF=0x009; //DelayMs(100); //GIE=1; return; //break; } // define the main function void main() { int i,k=0; char send[3]; RCIF=0; GIE=1; PEIE=1; send[0]='0'; send[1]='2'; send[2]='2'; // initialization of the peripherals to be used by // calling the respective initialization functions. initialize_IO_ports(); init_uart(); while(1) { if(IR==1) { k++; RED=1; } if(k==1) { GIE=1; for(i=0;i<3;i++) { CREN=1; while(TXIF==0); TXREG=send[i]; DelayMs(10); } } if(k==2) { GIE=1; send[0]='0'; send[1]='1'; send[2]='2'; for(i=0;i<3;i++) { CREN=1; while(TXIF==0); TXREG=send[i]; DelayMs(10); } } if(k==3) { GIE=1; k=0; Yellow=1; send[0]='1'; send[1]='1'; send[2]='2'; for(i=0;i<3;i++) { CREN=1; while(TXIF==0); TXREG=send[i]; DelayMs(10); } } / if(DATA[0]=='0') LED_1=1; if(DATA[1]=='1') LED_2=1; if(DATA[2]=='2') LED_3=1; */ DelayMs(100); } //endWhile. }//endmain #### bigdogguru ##### Administrator What compiler are you using? #### bigdogguru ##### Administrator I just figured that out. Unfortunately, it looks like IAR no longer supports their PIC compilers. I couldn't find a user manual for the PIC16 version of their compiler on their website. ---------- Post added at 11:08 ---------- Previous post was at 10:46 ---------- Just briefly looking over your code and without studying it thoroughly. It looks like you're not testing to see what exactly triggered the interrupt, therefore you're not resetting or disabling that particular interrupt. So when you exit the ISR the global interrupt flag is reenable and BINGO you're right back in the ISR. So you are basically stuck in the interrupt service routine. You should be able to run the program in debug mode and figure out what exactly is triggering the interrupt. My suggestion would be test for the actual interrupt and clear it. If I sound like I'm babbling it is due to the fact it's 4AM here. #### eefinni ##### Newbie level 3 I just figured that out. Unfortunately, it looks like IAR no longer supports their PIC compilers. I couldn't find a user manual for the PIC16 version of their compiler on their website. ---------- Post added at 11:08 ---------- Previous post was at 10:46 ---------- Just briefly looking over your code and without studying it thoroughly. It looks like you're not testing to see what exactly triggered the interrupt, therefore you're not resetting or disabling that particular interrupt. So when you exit the ISR the global interrupt flag is reenable and BINGO you're right back in the ISR. So you are basically stuck in the interrupt service routine. You should be able to run the program in debug mode and figure out what exactly is triggering the interrupt. My suggestion would be test for the actual interrupt and clear it. If I sound like I'm babbling it is due to the fact it's 4AM here. Wow.. You should seriously get some sleep! Really grateful for your replies! I'm actually making a small program to test its interrupt. Basically I wired the chip to send and receive to itself. When triggered, the interrupt that caused by RCIF, will start the ISR. I did try to clear & set RCIF, GIE and even PEIE! Non did work.. I'm trying to use the instruction set of the PIC. The 14bit word which specifies the instruction type or operands.. Do you know which register I should write the instruction set to perform the RETFIE function? #### bigdogguru ##### Administrator Now that I've had three hours of sleep, let me see if I can convey my thoughts a little clearer. In regards to the RETFIE instruction, all it does is GIE=1 and set PC to its prior value. So embedding the RETFIE instruction is not really going to help matters. What I believe is happening, I take a more thorough look at your code shortly, is that an interrupt you were not expecting is triggered, calling your ISR which is not prepared to handle this unexpected interrupt. Because it is not prepared for this particular interrupt, it doesn't reset its interrupt flag. Therefore, once the ISR is exited and the GIE is again set, BINGO you're right back in the ISR again and the vicious cycle starts all over again. If you can debug your program, set a breakpoint inside the ISR. Then run the program and when you hit the breakpoint, check all the interrupt flags to determine which interrupt is calling the ISR. Also, you could and should disable any maskable interrupts you do not intend to service. I'll check your program shortly and let you know what else I find. eefinni ### eefinni Points: 2 Helpful Answer Positive Rating #### eefinni ##### Newbie level 3 Really grateful to you! I've tried to run it on the debugger. I couldn't set my cursor to the main. It kept looping to the initialization part of the code! Not even to the ISR!! But most importantly, You are right! I saw in asm that there is a RETFIE in my ISR! Do you know of any free debugger software where I could run my code to check which interrupt triggers the ISR and did not manage to clear when it ends? Status Not open for further replies.	2022-09-28 23:00:55	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.30022960901260376, "perplexity": 4498.329876247553}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "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-2022-40/segments/1664030335286.15/warc/CC-MAIN-20220928212030-20220929002030-00063.warc.gz"}
http://www.mzan.com/article/4764611-java-security-invalidalgorithmparameterexception-the-trustanchors-parameter-mus.shtml	Home java.security.InvalidAlgorithmParameterException: the trustAnchors parameter must be non-empty on Linux, or why is the default truststore empty # java.security.InvalidAlgorithmParameterException: the trustAnchors parameter must be non-empty on Linux, or why is the default truststore empty Bozho 1# Bozho Published in 2011-01-21 22:44:08Z This question already has an answer here: Error - trustAnchors parameter must be non-empty 28 answers When you google for this exception: java.security.InvalidAlgorithmParameterException: the trustAnchors parameter must be non-empty, multiple results appear. However there is no definitive solution, only guesses. The problem arises (in my case at least) when I try to use open a connection over SSL. It works fine on my windows machine, but when I deploy it to the linux machine (with sun's jre installed) it fails with the above exception. The problem is that the default truststore of the JRE is empty for some reason (size of only 32 bytes, whereas it is 80kb on windows). When I copied my jre/lib/security/cacerts file from windows to linux, it worked fine. The question is - why is the linux jre having an empty trust store? Note that this happens on an Amazon EC2 instance, with the AMI linux, so it might be due to some amazon policies (I think java was pre-installed, but I'm not sure) bestsss 2# The standard Sun JDK for linux has an absolutely ok cacerts and overall all files in the specified directory. The problem is the installation you use. Attila Szegedi 3# Attila Szegedi Reply to 2012-07-24 01:41:56Z If this happens to you with an OpenJDK install on Mac OS X (as opposed to Linux), and you do have the official Mac OS X Java (i.e. latest Java 6) installed through Software Update, you can just do this: cd $OPENJDK_HOME/Contents/Home/jre/lib/security ln -s /System/Library/Java/Support/CoreDeploy.bundle/Contents/Home/lib/security/cacerts ln -s /System/Library/Java/Support/Deploy.bundle/Contents/Home/lib/security/blacklist ln -s /System/Library/Java/Support/Deploy.bundle/Contents/Home/lib/security/trusted.libraries where $OPENJDK_HOME is the root directory of your OpenJDK install, typically OPENJDK_HOME=/Library/Java/JavaVirtualMachines/1.7.0u.jdk. This is identical to how official Java installs on Mac OS X acquire these files - they also just symlink them from those system bundles. Works for Lion, not sure for earlier versions of the OS. Andrew 4# I have avoided this error (Java 1.6.0 on OSX 10.5.8) by putting a dummy cert in the keystore, such as keytool -genkey -alias foo -keystore cacerts -dname cn=test -storepass changeit -keypass changeit Surely the question should be "Why can't java handle an empty trustStore?" Paulo Henrique Lellis Gonalves 5# Paulo Henrique Lellis Gonalves Reply to 2013-03-21 03:59:05Z Make sure that you have valid cacerts in the JRE/security, otherwise you will not bypass the invalid empty trustAnchors error. In my Amazon EC2 Opensuse12 installation, the problem was that the file pointed by the cacerts in the JRE security directory was invalid: $java -version java version "1.7.0_09" OpenJDK Runtime Environment (IcedTea7 2.3.4) (suse-3.20.1-x86_64) OpenJDK 64-Bit Server VM (build 23.2-b09, mixed mode)$ ls -l /var/lib/ca-certificates/ -rw-r--r-- 1 root 363 Feb 28 14:17 ca-bundle.pem $ls -l /usr/lib64/jvm/jre/lib/security/ lrwxrwxrwx 1 root 37 Mar 21 00:16 cacerts -> /var/lib/ca-certificates/java-cacerts -rw-r--r-- 1 root 2254 Jan 18 16:50 java.policy -rw-r--r-- 1 root 15374 Jan 18 16:50 java.security -rw-r--r-- 1 root 88 Jan 18 17:34 nss.cfg So I solved installing an old Opensuse 11 valid certificates. (sorry about that!!) $ ll total 616 -rw-r--r-- 1 root 220065 Jan 31 15:48 ca-bundle.pem -rw-r--r-- 1 root 363 Feb 28 14:17 ca-bundle.pem.old -rw-r--r-- 1 root 161555 Jan 31 15:48 java-cacerts I understood that you could use the keytool to generate a new one (http://mail.openjdk.java.net/pipermail/distro-pkg-dev/2010-April/008961.html). I'll probably have to that soon. regards lellis user2322889 6# Have the same issue. Resolved it by installing ca-certificate bundle from Mozilla: $zypper in ca-certificates-mozilla The following NEW package is going to be installed: ca-certificates-mozilla 1 new package to install. Retrieving package ca-certificates-mozilla-1.85-8.8.1.noarch (1/1), 143.7 KiB (239.1 KiB unpacked) Retrieving: ca-certificates-mozilla-1.85-8.8.1.noarch.rpm.....................[done] Installing: ca-certificates-mozilla-1.85-8.8.1 ...............................[done] Additional rpm output: Updating certificates in /etc/ssl/certs... 144 added, 0 removed. creating /var/lib/ca-certificates/ca-bundle.pem ... creating /var/lib/ca-certificates/java-cacerts ... 144 added, 0 removed.$ ll /var/lib/ca-certificates/ total 392 drwxr-xr-x 2 root root 4096 Apr 26 07:25 ./ drwxr-xr-x 30 root root 4096 Apr 25 15:00 ../ -rw-r--r-- 1 root root 220196 Apr 26 07:25 ca-bundle.pem -rw-r--r-- 1 root root 161555 Apr 26 07:25 java-cacerts P.S. $cat /etc/SuSE-release openSUSE 12.2 (x86_64) VERSION = 12.2 CODENAME = Mantis$ java -version java version "1.7.0_09" OpenJDK Runtime Environment (IcedTea7 2.3.4) (suse-3.20.1-x86_64) OpenJDK 64-Bit Server VM (build 23.2-b09, mixed mode) Aaron Digulla 7# Aaron Digulla Reply to 2016-04-21 16:00:06Z My solution on Windows was to either run console window as Administrator or change the environment variable MAVEN_OPTS to use a hardcoded path to trust.jks (e.g. 'C:\Users\oddros') instead of '%USERPROFILE%'. My MAVEN_OPTS now looks like this: -Djavax.net.ssl.trustStore=C:\Users\oddros\trust.jks -Djavax.net.ssl.trustStorePassword=changeit Ryan Shillington 8# Ryan Shillington Reply to 2013-07-26 20:29:15Z My cacerts file was totally empty. I solved this by copying the cacerts file off my windows machine (that's using Oracle Java 7) and scp'd it to my Linux box (OpenJDK). cd %JAVA_HOME%/jre/lib/security/ scp cacerts mylinuxmachin:/tmp and then on the linux machine cp /tmp/cacerts /etc/ssl/certs/java/cacerts It's worked great so far. Manuel Darveau 9# Manuel Darveau Reply to 2013-11-18 01:47:10Z Not the answer to the original question but when trying to resolve a similar issue, I found that the Mac OS X update to Maverics screwed up the java install (the cacert actually). Remove sudo rm -rf /Library/Java/JavaVirtualMachines/*.jdk and reinstall from http://www.oracle.com/technetwork/java/javase/downloads/index.html user987339 10# This happens because Access Privilege varies from OS to OS. Windows access hierarchy is different from Unix. However, this could be overcome by following these simple steps: Increase accessibility with AccessController.doPrivileged(java.security.PrivilegedAction subclass) Set your own java.security.Provider subclass as security property. a. Security.insertProviderAt(new , 2); Set your Algorythm with Security.setProperty("ssl.TrustManagerFactory.algorithm" , “XTrust509”); The Camster 11# The Camster Reply to 2014-06-25 18:36:52Z I can generate this error by setting system property trustStore to a missing jks file. For example System.setProperty("javax.net.ssl.keyStore", "C:/keystoreFile.jks"); System.setProperty("javax.net.ssl.keyStorePassword", "mypassword"); System.setProperty("javax.net.ssl.trustStore", "C:/missing-keystore.jks"); System.setProperty("javax.net.ssl.trustStorePassword", "mypassword"); This code does not generate a FileNotFound exception for some reason, but exactly the InvalidAlgorithmParameter exception listed above. Kind of a dumb answer, but I can reproduce. fuzzyanalysis 12# Had the same issue on Ubuntu 14.10 with java-8-oracle installed. Solved installing ca-certificates-java package: sudo apt-get install ca-certificates-java I got this error in Ubuntu. I saw that /usr/lib/jvm/java-8-openjdk-amd64/jre/lib/security/cacerts was a broken link to /etc/ssl/certs/java/cacerts. That lead me to this bug: https://bugs.launchpad.net/ubuntu/+source/ca-certificates-java/+bug/983302 The README for ca-certificates-java eventually showed the actual fix: run update-ca-certificates -f apt-get install ca-certificates-java didn't work for me. It just marked it as manually installed.	2018-02-23 00:21:39	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 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": 0.3352128267288208, "perplexity": 10204.70987318948}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-09/segments/1518891814300.52/warc/CC-MAIN-20180222235935-20180223015935-00526.warc.gz"}