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https://raweb.inria.fr/rapportsactivite/RA2016/veridis/uid34.html | Application Domains
Bilateral Contracts and Grants with Industry
Partnerships and Cooperations
Bibliography
PDF e-Pub
## Section: New Results
### Automated and Interactive Theorem Proving
Participants : Gabor Alági, Haniel Barbosa, Jasmin Christian Blanchette, Martin Bromberger, Simon Cruanes, Mathias Fleury, Pascal Fontaine, Marek Košta, Stephan Merz, Martin Riener, Martin Strecker, Thomas Sturm, Marco Voigt, Uwe Waldmann, Daniel Wand, Christoph Weidenbach.
#### IsaFoL: Isabelle Formalization of Logic
Joint work with Heiko Becker (MPI-SWS Saarbrücken), Peter Lammich (TU München), Andrei Popescu (Middlesex University London), Anders Schlichtkrull (DTU Copenhagen), Dmitriy Traytel (ETH Zürich), and Jørgen Villadsen (DTU Copenhagen).
Researchers in automated reasoning spend a significant portion of their work time specifying logical calculi and proving metatheorems about them. These proofs are typically carried out with pen and paper, which is error-prone and can be tedious. As proof assistants are becoming easier to use, it makes sense to employ them.
In this spirit, we started an effort, called IsaFoL (Isabelle Formalization of Logic), that aims at developing libraries and methodology for formalizing modern research in the field, using the Isabelle/HOL proof assistant.(https://bitbucket.org/jasmin_blanchette/isafol/wiki/Home) Our initial emphasis is on established results about propositional and first-order logic. In particular, we are formalizing large parts of Weidenbach's forthcoming textbook, tentatively called Automated Reasoning—The Art of Generic Problem Solving.
The objective of formalization work is not to eliminate paper proofs, but to complement them with rich formal companions. Formalizations help catch mistakes, whether superficial or deep, in specifications and theorems; they make it easy to experiment with changes or variants of concepts; and they help clarify concepts left vague on paper.
The repository contains six completed entries and three entries that are still in development. Notably:
• Mathias Fleury formalized a SAT solver framework with learn, forget, restart, and incrementality and published the result at a leading conference, together with Jasmin Blanchette and Christoph Weidenbach [25].
• Anders Schlichtkrull, remotely co-supervised by Jasmin Blanchette, formalized unordered first-order resolution in Isabelle and presented the result at ITP 2016 [37].
• Together with an intern, Jasmin Blanchette, Uwe Waldmann, and Daniel Wand formalized a generalization for the recursive path order and the transfinite Knuth-Bendix order to higher-order terms without $\lambda$-abstractions. The result is published in the Isabelle Archive of Formal Proofs.
#### Combination of Satisfiability Procedures
Joint work with Christophe Ringeissen from the PESTO project-team at Inria Nancy – Grand Est, and Paula Chocron at IIIA-CSIC, Bellaterra, Catalonia, Spain.
A satisfiability problem is often expressed in a combination of theories, and a natural approach consists in solving the problem by combining the satisfiability procedures available for the component theories. This is the purpose of the combination method introduced by Nelson and Oppen. However, in its initial presentation, the Nelson-Oppen combination method requires the theories to be signature-disjoint and stably infinite (to ensure the existence of an infinite model). The design of a generic combination method for non-disjoint unions of theories is clearly a hard task, but it is worth exploring simple non-disjoint combinations that appear frequently in verification. An example is the case of shared sets, where sets are represented by unary predicates. Another example is the case of bridging functions between data structures and a target theory (e.g., a fragment of arithmetic).
In 2015, we defined [42] a sound and complete combination procedure à la Nelson-Oppen for the theory of absolutely free data structures (including lists and trees) connected to another theory via bridging functions. This combination procedure has also been refined for standard interpretations. The resulting theory has a nice politeness property, enabling combinations with arbitrary decidable theories of elements. We also investigated [43] other theories amenable to similar combinations: this class includes the theory of equality, the theory of absolutely free data structures, and all the theories in between.
More recently, we have been improving the framework and unified both results. A new paper is in preparation.
#### Quantifier handling in SMT
Joint work with Andrew J. Reynolds, Univ. of Iowa, USA.
SMT solvers generally rely on various instantiation techniques to handle quantifiers. We are building a unifying framework for handling quantified formulas with equality and uninterpreted functions, such that the major instantiation techniques in SMT solving can be cast in that framework. It is based on the problem of $E$-ground (dis)unification, a variation of the classic Rigid E-unification problem. We introduced a sound and complete calculus to solve this problem in practice: Congruence Closure with Free Variables (CCFV). Experimental evaluations of implementations of CCFV in the state-of-the-art solver CVC4 and in the solver veriT exhibit improvements in the former and makes the latter competitive with state-of-the-art solvers in several benchmark libraries stemming from verification efforts. A publication is in preparation.
#### Non-linear arithmetic in SMT
In the context of the SMArT ANR-DFG (Satisfiability Modulo Arithmetic Theories) and KANASA projects (cf. sections 9.1 and 9.3), we study the theory, design techniques, and implement software to push forward the non-linear arithmetic (NLA) reasoning capabilities in SMT. This year, we designed a framework to combine interval constraint propagation with other decision procedures for NLA, with promising results. We are also currently studying integration of these procedures into combinations of theories. The ideas are validated within the veriT solver, together with code from the raSAT solver (from JAIST). An article is in preparation.
#### Encoding Set-Theoretic Formulas in First-Order Logic
During the internship of Matthieu Lequesne, we experimented with an adaptation of the technique for constructing models of formulas in set theory, which could be useful for understanding why proof attempts fail. A prototype generating input for the Nunchaku model finder (section 6.1) allowed us to validate the idea for a core sublanguage of TLA+.
#### Modal and Description Logics for Graph Transformations
Graph transformations are a research topic that is interesting in its own right, but with many possible applications ranging from the modification of pointer structures in imperative programs, through model transformations in model-driven engineering, to schema-preserving transformations of graph databases. Our particular focus is on verifying these transformations.
Our next aim is to implement practically useful proof methods. We have first concentrated on the more natural tableau proofs, with a verification of meta-theoretic properties of the calculi (such as termination) in the Isabelle proof assistant. We now turn to an investigation of encodings as satisfiability problems that can be handled with SAT and SMT solvers, with the hope to achieve a better performance.
#### Standard Models with Virtual Substitution
Joint work with A. Dolzmann from Leibniz-Zentrum für Informatik in Saarbrücken, Germany.
Extended quantifier elimination for the reals using virtual substitution methods have been successfully applied to various problems in science and engineering. Recently they have attracted attention also as theory solvers within SMT. Such solvers typically ask also for models in the satisfiable case. Models obtained with virtual substitution are in general obtained in certain non-archimedian extension fields of the reals with a corresponding expanded signature. Consequently, the obtained values for the variables include non-standard symbols such as positive infinitesimals and infinite values.
We introduce a complete post-processing procedure to convert our models, for fixed values of parameters, into real models [15]. We furthermore demonstrate the successful application of an implementation of our method within Redlog to a number of extended quantifier elimination problems from the scientific literature including computational geometry, motion planning, bifurcation analysis for models of genetic circuits and for mass action, and sizing of electrical networks. This solves a long-standing problem with the virtual substitution method, which had been explicitly criticized in the scientific literature.
#### Decidability of Fragments of Free First-Order Logic
We introduce a new decidable fragment of first-order logic with equality, the Separated Fragment (SF). It strictly generalizes two already well-known decidable fragments of first-order logic: the Bernays-Schönfinkel-Ramsey (BSR) Fragment and the Monadic Fragment. The defining principle is that universally and existentially quantified variables may not occur together in atoms. Thus, our classification neither rests on restrictions of quantifier prefixes (as in the BSR case) nor on restrictions on the arity of predicate symbols (as in the monadic case).
Continuing the work presented in the initial publication, we further investigated the computational complexity of SF satisfiability. It nicely scales across the nondeterministic standard complexity classes, depending on joint occurrences of existentially quantified variables from ${\exists }^{*}$-blocks that are separated by nonempty ${\forall }^{+}$-blocks.
In another line of work, we relaxed the definition of SF, leading to an even larger fragment for which satisfiability is still decidable. In this fragment, variables of ${\exists }^{*}$-blocks and ${\forall }^{+}$-blocks may occur together in some atom if the respective quantifiers obey a certain order.
#### Undecidable combinations of first-order logic with background theories
We show that the universal fragment of Presburger arithmetic augmented with a single uninterpreted predicate (or function) symbol is already undecidable. The result has immediate consequences for verification techniques that combine uninterpreted functions or predicate symbols with (fragments of) Presburger arithmetic. For example, data structures such as arrays can be viewed as a collection of uninterpreted functions that obey certain axioms.
Our result is a sharpening of previously known results. In particular, undecidability holds for a fragment with purely universal quantification: no quantifier alternation is necessary. While in this case the set of unsatisfiable sentences is still recursively enumerable, and in fact hierarchic superposition constitutes a semi-decision procedure, allowing for one quantifier alternation ($\exists \forall$ or $\forall \exists$) leads to a fragment in which neither the satisfiable sentences nor the unsatisfiable ones form a recursively-enumerable set. Hence, there cannot be any refutationally complete calculus for such a combined theory.
#### Novel techniques for linear arithmetic constraint solving
In [26], [27], we investigate new techniques for linear arithmetic constraint solving. They are based on the linear cube transformation, which allows us to efficiently determine whether a system of linear arithmetic constraints contains a hypercube of a given edge length.
Our first findings based on this transformation are two sound tests that find integer solutions for linear arithmetic constraints. While many complete methods search along the problem surface for a solution, these tests use cubes to explore the interior of the problems. The tests are especially efficient for constraints with a large number of integer solutions, e.g., those with infinite lattice width. Inside the SMT-LIB benchmarks, we have found almost one thousand problem instances with infinite lattice width. Experimental results confirm that our tests are superior on these instances compared to several state-of-the-art SMT solvers.
We also discovered that the linear cube transformation can be used to investigate the equalities implied by a system of linear arithmetic constraints. For this purpose, we developed a method that computes a basis for all implied equalities, i.e., a finite representation of all equalities implied by the linear arithmetic constraints. The equality basis can be used to decide whether a system of linear arithmetic constraints implies a given equality. | 2020-10-26 16:55:51 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 9, "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": 0.6151332855224609, "perplexity": 1699.710755038254}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-45/segments/1603107891428.74/warc/CC-MAIN-20201026145305-20201026175305-00714.warc.gz"} |
https://homework.cpm.org/category/CCI_CT/textbook/pc3/chapter/2/lesson/2.2.1/problem/2-50 | ### Home > PC3 > Chapter 2 > Lesson 2.2.1 > Problem2-50
2-50.
Using the graph at right, where is the function increasing/decreasing? Where is the function concave up/down? Does this function have any maxima or minima? Explain how you know.
Trace the curve from left to right with your finger. For what $x$-values are you moving down? Up? | 2022-05-17 16:48:11 | {"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": 0.6329634785652161, "perplexity": 2154.5270583382144}, "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-2022-21/segments/1652662519037.11/warc/CC-MAIN-20220517162558-20220517192558-00138.warc.gz"} |
https://www.sarthaks.com/132210/most-of-the-refraction-of-light-rays-entering-the-eye-occurs-at-the-outer-surface-of-the | # Most of the refraction of light rays entering the eye occurs at the outer surface of the………….
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Most of the refraction of light rays entering the eye occurs at the outer surface of the cornea | 2023-03-31 03:55:40 | {"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": 0.8528253436088562, "perplexity": 423.68059586729146}, "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/1679296949533.16/warc/CC-MAIN-20230331020535-20230331050535-00699.warc.gz"} |
https://eng.libretexts.org/Bookshelves/Chemical_Engineering/Book%3A_Phase_Relations_in_Reservoir_Engineering_(Adewumi)/08%3A_PT_Behavior_and_Equations_of_State_III/8.02%3A_Acentric_Factor_and_Corresponding_States | # 8.2: Acentric Factor and Corresponding States
It is important to point out that the PCS that we have just discussed was originally outlined by van der Waals. In reality, it is the simplest version of the principle of corresponding states, and it is referred to as the two-parameter PCS. This is because it relies on two parameters (reduced pressure and temperature) for defining a “corresponding state.”
With the passing of time, more accurate PCS formulations have made use of more than two parameters. For instance, the three-parameter PCS affirms that two substances are in corresponding states not only when they are at the same reduced conditions (reduced pressure and temperature), but also when they have the same “acentric factor” value. In any case, the general statement of PCS remains untouched:
Substances at corresponding states behave alike.
What makes the difference is the definition of “what a corresponding state is.”
The acentric factor “$$ω$$” is a concept that was introduced by Pitzer in 1955, and has proven to be very useful in the characterization of substances. It has become a standard for the proper characterization of any single pure component, along with other common properties, such as molecular weight, critical temperature, critical pressure, and critical volume.
Pitzer came up with this factor by analyzing the vapor pressure curves of various pure substances. From thermodynamic considerations, the vapor pressure curve that we studied in our first modules for pure components can be mathematically described by the Clausius Clapeyron equation:
$\frac{1}{P} \frac{d P}{d T}=\frac{\Delta \widetilde{H}_{v a p}}{R T \Delta Z} \label{8.7}$
The use of the integrated version of Equation \ref{8.7} is very common for the mathematical fitting of vapor pressure data. The integrated version of equation (8.7) shows that the relationship between the logarithm of vapor pressure and the reciprocal of absolute temperature is approximately linear. That is, in terms of reduced conditions, vapor pressure data approximately follows a straight line when plotted in terms of “logPr” versus “1/Tr”, or, equivalently:
$\log _{10} P_{r}=a\left(\frac{1}{T_{r}}\right)+b \label{8.8}$
If the two-parameter corresponding state principle were to hold true for all substances, the parameters “a” and “b” should be the same for all substances. That is, all vapor pressure curves of all imaginable substances should lie on top of each when plotted in terms of reduced conditions. Stated in another way, if the plot is of the form “logPr” versus “1/Tr”, all lines should show the same slope (a) and intercept (b).
The bad news is that, as you may imagine, this is not always true. Vapor pressure data for different substances do follow different trends. The good news is that some gases follow the expected trend. Which are they? The noble gases. Noble gases (such as Ar, Kr and Xe) happen to follow the two-parameter corresponding states theory very closely. Hence, they yield themselves amenable to acting as a reference to evaluate “compliance” with the two-parameter equation of state.
Pitzer wanted to come up with a reliable way of quantifying the deviation of substances with respect to two-parameter corresponding state predictions. He decided to use noble gases as the base for comparison. Analyzing vapor pressure data for noble gases, Pitzer showed that a value of $$\log P_r = – 1$$ was achieved at approximately Tr = 0.7. So, BINGO! There you are! He thought: if the vapor pressure data of a substance show that $$\log P_r = – 1$$ at $$T_r = 0.7$$, it behaves as the noble gases and thus complies with the two-parameter corresponding states. If not, we are to compute the difference:
$\text { Difference }=-\log _{10}\left(P_{r}\right) T_{r=0.7}-1 \label{8.9}$
Pitzer called this difference the “acentric factor, ω” of the substance. Noble gases, being the reference themselves, have an acentric factor value of zero ($$ω=0$$). Substances with an acentric factor of zero are called “simple” substances. The acentric factor is said to be a measure of the non-sphericity (acentricity) of the molecules. Therefore, the three-parameter corresponding state theory of Pitzer reads: “Fluids that have the same value of ω will behave alike at the same conditions of reduced pressure and temperature.” | 2021-03-04 00:51:42 | {"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": 0.7908380627632141, "perplexity": 749.0535595804868}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-10/segments/1614178367949.58/warc/CC-MAIN-20210303230849-20210304020849-00291.warc.gz"} |
https://www.physicsforums.com/threads/isotropic-material-fitted-by-ornstein-zernike-form.908520/ | # Homework Help: Isotropic material fitted by Ornstein-Zernike form
Tags:
1. Mar 21, 2017
### alan
I have known what Ornstein-Zernike equation is. I try to plug in the form as follow to the isotropic materials:
Still, I cannot show the pair correlation function as follow.
Can anyone know what I have missed?
2. Mar 22, 2017
### Fred Wright
I don't know if this will help but I get the result you are looking for by evaluating the integral,$$\frac{N_c}{4\pi^3r}\int_{-\infty}^{\infty}dk\frac{e^{-ikr}}{1+k^2l_c^2}$$
by closing the infinite semi-circle in the lower half of the complex plane and using the residue theorem.
Last edited: Mar 22, 2017 | 2018-06-17 22:53:11 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 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": 0.6408495903015137, "perplexity": 1183.2088354101031}, "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/1529267859817.15/warc/CC-MAIN-20180617213237-20180617233237-00449.warc.gz"} |
https://intelligencemission.com/free-hidden-electricity-phone-line-free-energy-bike.html | Also, because the whole project will be lucky to cost me Free Electricity to Free Electricity and i have all the gear to put it together I thought why not. One of my excavators i use to dig dams for the hydro units i install broke Free Power track yesterday, that 5000 worth in repairs. Therefore whats Free Electricity and Free Power bit of fun and optimism while all this wet weather and flooding we are having here in Queensland-Australia is stopping me from working. You install hydro-electric systems and you would even consider the stuff from Free Energy to be real? I am appalled.
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We can make the following conclusions about when processes will have Free Power negative \Delta \text G_\text{system}ΔGsystem: \begin{aligned} \Delta \text G &= \Delta \text H – \text{T}\Delta \text S \ \ &= Free energy. 01 \dfrac{\text{kJ}}{\text{mol-rxn}}-(Free energy \, \cancel{\text K})(0. 022\, \dfrac{\text{kJ}}{\text{mol-rxn}\cdot \cancel{\text K})} \ \ &= Free energy. 01\, \dfrac{\text{kJ}}{\text{mol-rxn}}-Free energy. Free Power\, \dfrac{\text{kJ}}{\text{mol-rxn}}\ \ &= -0. Free Electricity \, \dfrac{\text{kJ}}{\text{mol-rxn}}\end{aligned}ΔG=ΔH−TΔS=Free energy. 01mol-rxnkJ−(293K)(0. 022mol-rxn⋅K)kJ=Free energy. 01mol-rxnkJ−Free energy. 45mol-rxnkJ=−0. 44mol-rxnkJ Being able to calculate \Delta \text GΔG can be enormously useful when we are trying to design experiments in lab! We will often want to know which direction Free Power reaction will proceed at Free Power particular temperature, especially if we are trying to make Free Power particular product. Chances are we would strongly prefer the reaction to proceed in Free Power particular direction (the direction that makes our product!), but it’s hard to argue with Free Power positive \Delta \text GΔG! Our bodies are constantly active. Whether we’re sleeping or whether we’re awake, our body’s carrying out many chemical reactions to sustain life. Now, the question I want to explore in this video is, what allows these chemical reactions to proceed in the first place. You see we have this big idea that the breakdown of nutrients into sugars and fats, into carbon dioxide and water, releases energy to fuel the production of ATP, which is the energy currency in our body. Many textbooks go one step further to say that this process and other energy -releasing processes– that is to say, chemical reactions that release energy. Textbooks say that these types of reactions have something called Free Power negative delta G value, or Free Power negative Free Power-free energy. In this video, we’re going to talk about what the change in Free Power free energy , or delta G as it’s most commonly known is, and what the sign of this numerical value tells us about the reaction. Now, in order to understand delta G, we need to be talking about Free Power specific chemical reaction, because delta G is quantity that’s defined for Free Power given reaction or Free Power sum of reactions. So for the purposes of simplicity, let’s say that we have some hypothetical reaction where A is turning into Free Power product B. Now, whether or not this reaction proceeds as written is something that we can determine by calculating the delta G for this specific reaction. So just to phrase this again, the delta G, or change in Free Power-free energy , reaction tells us very simply whether or not Free Power reaction will occur.
Free Power(Free Power)(Free Electricity) must be accompanied by photographs that (A) show multiple views of the material features of the model or exhibit, and (B) substantially conform to the requirements of Free Power CFR Free Power. Free energy. See Free Power CFR Free Power. Free Power(Free Electricity). Material features are considered to be those features which represent that portion(s) of the model or exhibit forming the basis for which the model or exhibit has been submitted. Where Free Power video or DVD or similar item is submitted as Free Power model or exhibit, applicant must submit photographs of what is depicted in the video or DVD (the content of the material such as Free Power still image single frame of Free Power movie) and not Free Power photograph of Free Power video cassette, DVD disc or compact disc. <“ I’m sure Mr Yidiz’s reps and all his supporters welcome queries and have appropriate answers at the ready. Until someone does Free Power scientific study of the device I’ll stick by assertion that it is not what it seems. Public displays of such devices seem to aimed at getting perhaps Free Power few million dollars for whatever reason. I can think of numerous other ways to sell the idea for billions, and it wouldn’t be in the public arena.
Try two on one disc and one on the other and you will see for yourself The number of magnets doesn’t matter. If you can do it width three magnets you can do it with thousands. Free Energy luck! @Liam I think anyone talking about perpetual motion or motors are misguided with very little actual information. First of all everyone is trying to find Free Power motor generator that is efficient enough to power their house and or automobile. Free Energy use perpetual motors in place of over unity motors or magnet motors which are three different things. and that is Free Power misnomer. Three entirely different entities. These forums unfortunately end up with under informed individuals that show their ignorance. Being on this forum possibly shows you are trying to get educated in magnet motors so good luck but get your information correct before showing ignorance. @Liam You are missing the point. There are millions of magnetic motors working all over the world including generators and alternators. They are all magnetic motors. Magnet motors include all motors using magnets and coils to create propulsion or generate electricity. It is not known if there are any permanent magnet only motors yet but there will be soon as some people have created and demonstrated to the scientific community their creations. Get your semantics right because it only shows ignorance. kimseymd1 No, kimseymd1, YOU are missing the point. Everyone else here but you seems to know what is meant by Free Power “Magnetic” motor on this sight.
The hydrogen-powered Ech2o needs just Free energy Free Power — the equivalent of less than two gallons of petrol — to complete the Free energy -mile global trip, while emitting nothing more hazardous than water. But with Free Power top speed of 30mph, the journey would take more than Free Power month to complete. Ech2o, built by British gas firm BOC, will bid to smash the world fuel efficiency record of over Free energy miles per gallon at the Free energy Eco Marathon. The record is currently…. Free Power, 385 km/per liter [over Free Electricity mpg!]. Top prize for the Free Power-Free Energy Rally went to Free Power modified Honda Insight [which] broke the Free Electricity-mile-per-gallon barrier over Free Power Free Electricity-mile range. The car actually got Free Electricity miles-per gallon. St. Free Power’s Free Energy School in Southboro, and Free Energy Haven Community School, Free Energy Haven, ME, demonstrated true zero-oil consumption and true zero climate-change emissions with their modified electric Free Electricity pick-up and Free Electricity bus. Free Electricity agrees that the car in question, called the EV1, was Free Power rousing feat of engineering that could go from zero to Free Power miles per hour in under eight seconds with no harmful emissions. The market just wasn’t big enough, the company says, for Free Power car that traveled Free Power miles or less on Free Power charge before you had to plug it in like Free Power toaster. Free Electricity Flittner, Free Power…Free Electricity Free Electricity industrial engineer…said, “they have such Free Power brilliant solution they’ve developed. They’ve put it on the market and proved it works. Free Energy still want it and they’re taking it away and destroying it. ”Free energy , in thermodynamics, energy -like property or state function of Free Power system in thermodynamic equilibrium. Free energy has the dimensions of energy , and its value is determined by the state of the system and not by its history. Free energy is used to determine how systems change and how much work they can produce. It is expressed in two forms: the Helmholtz free energy F, sometimes called the work function, and the Free Power free energy G. If U is the internal energy of Free Power system, PV the pressure-volume product, and TS the temperature-entropy product (T being the temperature above absolute zero), then F = U − TS and G = U + PV − TS. The latter equation can also be written in the form G = H – TS, where H = U + PV is the enthalpy. Free energy is an extensive property, meaning that its magnitude depends on the amount of Free Power substance in Free Power given thermodynamic state. The changes in free energy , ΔF or ΔG, are useful in determining the direction of spontaneous change and evaluating the maximum work that can be obtained from thermodynamic processes involving chemical or other types of reactions. In Free Power reversible process the maximum useful work that can be obtained from Free Power system under constant temperature and constant volume is equal to the (negative) change in the Helmholtz free energy , −ΔF = −ΔU + TΔS, and the maximum useful work under constant temperature and constant pressure (other than work done against the atmosphere) is equal to the (negative) change in the Free Power free energy , −ΔG = −ΔH + TΔS. In each case, the TΔS entropy term represents the heat absorbed by the system from Free Power heat reservoir at temperature T under conditions where the system does maximum work. By conservation of energy , the total work done also includes the decrease in internal energy U or enthalpy H as the case may be. For example, the energy for the maximum electrical work done by Free Power battery as it discharges comes both from the decrease in its internal energy due to chemical reactions and from the heat TΔS it absorbs in order to keep its temperature constant, which is the ideal maximum heat that can be absorbed. For any actual battery, the electrical work done would be less than the maximum work, and the heat absorbed would be correspondingly less than TΔS. Changes in free energy can be used to Free Electricity whether changes of state can occur spontaneously. Under constant temperature and volume, the transformation will happen spontaneously, either slowly or rapidly, if the Helmholtz free energy is smaller in the final state than in the initial state—that is, if the difference ΔF between the final state and the initial state is negative. Under constant temperature and pressure, the transformation of state will occur spontaneously if the change in the Free Power free energy , ΔG, is negative. Phase transitions provide instructive examples, as when ice melts to form water at 0. 01 °C (T = Free energy. Free energy K), with the solid and liquid phases in equilibrium. Then ΔH = Free Power. Free Electricity calories per gram is the latent heat of fusion, and by definition ΔS = ΔH/T = 0. Free Power calories per gram∙K is the entropy change. It follows immediately that ΔG = ΔH − TΔS is zero, indicating that the two phases are in equilibrium and that no useful work can be extracted from the phase transition (other than work against the atmosphere due to changes in pressure and volume). Free Power, ΔG is negative for T > Free energy. Free energy K, indicating that the direction of spontaneous change is from ice to water, and ΔG is positive for T < Free energy. Free energy K, where the reverse reaction of freezing takes place.
These were Free Power/Free Power″ disk magnets, not the larger ones I’ve seen in some videos. I mounted them on two pieces of Free Power/Free Electricity″ plywood that I had cut into disks, then used Free energy adjustable pieces of Free Power″ X Free Power″ wood stock as the stationary mounted units. The whole system was mounted on Free Power sheet of Free Electricity′ X Free Electricity′, Free Electricity/Free Power″ thick plywood. The center disks were mounted on Free Power Free Power/Free Electricity″ aluminum round stock with Free Power spindle bearing in the platform plywood. Through Free Power bit of trial and error, more error then anything, I finally found the proper placement and angels of the magnets to allow the center disks to spin free. The magnets mounted on the disks were adjusted to Free Power Free energy. Free Electricity degree angel with the stationary units set to match. The disks were offset by Free Electricity. Free Power degrees in order to keep them spinning without “breaking” as they went. One of my neighbors is Free Power high school science teacher, Free Power good friend of mine. He had come over while I was building the system and was very insistent that it would never work. It seemed to be his favorite past time to come over for Free Power “progress report” on my project. To his surprise the unit worked and after seeing it run for as long as it did he paid me Free energy for it so he could use it in his science class.
I have had many as time went by get weak. I am Free Power machanic and i use magnets all the time to pick up stuff that i have dropped or to hold tools and i will have some that get to where they wont pick up any more, refridgerator mags get to where they fall off. Dc motors after time get so they don’t run as fast as they used to. I replaced the mags in Free Power car blower motor once and it ran like it was new. now i do not know about the neo’s but i know that mags do lose there power. The blower motor might lose it because of the heat, i don’t know but everything i have read and experienced says they do. So whats up with that? Hey Free Electricity, ok, i agree with what you are saying. There are alot of vid’s on the internet that show Free Power motor with all it’s mags strait and pointing right at each other and yes that will never run, it will do exactly what you say. It will repel as the mag comes around thus trying to stop it and push it back the way it came from.
Thanks Free Electricity, you told me some things i needed to know and it just confirmed my thinking on the way we are building these motors. My motor runs but not the way it needs to to be of any real use. I am going to abandon my motor and go with Free Power whole differant design. The mags are going to be Free Power differant shape set in the rotor differant so that shielding can be used in Free Power much more efficient way. Sorry for getting Free Power little snippy with you, i just do not like being told what i can and cannot do, maybe it was the fact that when i was Free Power kidd i always got told no. It’s something i still have Free Power problem with even at my age. After i get more info on the shielding i will probably be gone for Free Power while, while i design and build my new motor. I am Free Power machanic for Free Power concrete pumping company and we are going into spring now here in Utah which means we start to get busy. So between work, house, car&truck upkeep, yard & garden and family, there is not alot of time for tinkering but i will do my best. Free Power, please get back to us on the shielding. Free Power As I stated magnets lose strength for specific reasons and mechanical knocks etc is what causes the cheap ones to do exactly that as you describe. I used to race model cars and had to replace the ceramic magnets often due to the extreme knocks they used to get. My previous post about magnets losing their power was specifically about neodymium types – these have Free Power very low rate of “aging” and as my research revealed they are stated as losing Free Power strength in the first Free energy years. But extreme mishandling will shorten their life – normal use won’t. Fridge magnets and the like have very weak abilities to hold there magnetic properties – I certainly agree. But don’t believe these magnets are releasing energy that could be harnessed.
I do not fear any conspiracy from any nook & corner. I am simply taking my time and my space to stage the inevitable confrontation in the frozen face of the industry and geopolitics tycoons. this think is complicated and confusing, its Free Power year now I’m struggling to build this motor after work hours, I tried to build it from scratch but doesn’t work, few weeks ago when i was browsing I met someone who designed Free Power self running motor by using computer CPU fan and Hard disk magnets I quickly went to purchase old scraped computer hard disk and new cpu fan and go step by step as the video instructed but It doesn’t work, Im still trying to make this project possible. Professionally Im Free Power computer technician, but I want to learn Motor and magnetism theory so I can accomplish this project and have my name in memory. I anyone can make this project please contact me through facebook so I can invite him/her to my country and make money as you know third word countries has power disaster. My facebook Id is Elly Maduhu Nkonya, or use my E-mail. [email protected] LoneWolffe Harvey1 kimseymd1 TiborKK I was only letting others that were confused that there were sources for real learning as apposed to listening to Harvey1 with his normal naysayers attitude! There is tons of information on schoolgirl, schoolboy and Bedini window motors that actually work to charge batteries and eventually will generate house currents. It just has to be looked at to get any useful information from it without listening to people like Harvey1 whining about learning. Harvey1 kimseymd1 You obviously play too much video games with trolls etc. in them. Why the editors of this forum allow you to keep calling people names instead of following the subject is beyond me. This must be the last site to allow you on it. I spammed the books because I thought those people were good for learning these engines which are super and there are tons of information out there for anyone to find. You seem to only want to learn to be rude instead of electronics. | 2021-01-17 16:47:45 | {"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": 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": 0.5606594085693359, "perplexity": 1351.6293355502564}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-04/segments/1610703513062.16/warc/CC-MAIN-20210117143625-20210117173625-00151.warc.gz"} |
https://physics.stackexchange.com/questions/247653/whats-the-difference-between-the-diffeomorphism-invariance-and-reparametrizatio?noredirect=1 | # What's the difference between the diffeomorphism invariance and reparametrization invariance?
Can somebody tell me what's the difference between the diffeomorphism invariance and reparametrization invariance?
• I always thought that these are the same things, except that "diffeomorphism invariance" is an annoying misuse of mathematical terminology (diffeomorphism is an isomorphism of smooth manifolds, and assuming that a theory makes sense on a smooth manifold is already assuming reparametrization invariance). I wonder if it is indeed so. One distinction one could try to make is that reparametrization invariance means that you can define the theory in a coordinate-free way on a manifold with some natural extra structure, while diffeomorphism means that the only structure is the smooth structure. – Peter Kravchuk Apr 5 '16 at 20:23
• Related: physics.stackexchange.com/q/76721/2451 and links therein. – Qmechanic Apr 5 '16 at 20:33
• @PeterKravchuk, see my answer below on why both notions are slightly less trivial than chart independence. (If you consider the statements I make about reparametrization invariance applied only to local charts, you do recover your statement that both amount to chart independence) – zzz Apr 30 '16 at 6:20
Diffeomorphism Invariance
Let $M$ be a smooth manifold. Let $\phi: M \to M$ be a diffeomorphism. A simple property of the Einstein equations is
$$g \in \otimes^2 TM \text{ is solution to vacuum Einstein equation} \implies \text{ so is } \phi^*g$$
To see that this is true, simply pull back both sides of the Einstein equation by $\phi$, and use the property of the Ricci tensor $\phi^*Ric_{(g)} = Ric_{(\phi^*g)}$.
You have a family $[g] \equiv \{\phi^*g | \phi \in Diff^+(M)\}$ of Riemannian structures on your manifold which are supposed to describe the same physics. This what I would call the notion of diffeomorphism invariance in general relativity.
Reparameterization Invariance
A parametrization of one manifold $M$ by another manifold $N$ is a diffeomorphism $\varphi: M \to N$. $N$ parametrizes $M$ in the sense that a points $p \in M$ smoothly correspond points in $\varphi(p) \in N$.
Reparametrization invariance usually means a scenario where the choice of this diffeomorphism $\varphi \in Diff(M \to N)$ doesn't matter.
Here's a simple example of what people might call reparametrization invariance.
Take a smooth curve $\gamma: [0, 1] \to M$, and suppose that $\gamma$ solves the geodesic equation. Let $\gamma': [0, 1] \to M$ be another curve, we have the following (trivial) fact
$$\text{ image(\gamma') = image(\gamma) } \implies \gamma' \text{ is also a geodesic}$$
that is, the geodesic equation (length functional, not energy functional) doesn't care how you parameterize your curve.
Sometimes they're the same thing
To summarize, so far, I've presented diffeomorphism invariance as a statement about Riemannian structures on a manifold, and reparametrization invariance as a statement about diffeomorphisms from one manifold to another. There are some scenarios where these seemingly different notions are really the same thing.
Example 1: Curve Reparametrization
Consider our description of diffeomorphism invariance applied to a $1$-manifold. Riemannian structures on $1$-manifolds are equivalent to reprarmetrizations of the curve. Pulling back the Riemannian inner product on the tangent bundle by a smooth map is equivalent to smoothly reparametrizing the curve. So our notion of diffeomorphism invariance and reparametrization invariance are the same thing.
Example 2: Moduli Space of closed orientable surfaces
Let $\Sigma$ be a closed orientable surface. Say we are interested in the space $Mod_{\Sigma, \mathbb R^n}$ of all submanifolds of $\mathbb R^n$ diffeomorphic to $\Sigma$.
One way to make this space explicit is to consider diffeomorphisms
$$e: \Sigma \to \mathbb R^n$$
But some of these diffeomorphisms map into the same submanifold in $\mathbb R^n$. These are exactly the diffeomorphisms related by a precomposition with diffeomorphism on the surface itself. Hence we have a description of our moduli space as equivalence classes of diffeomorphisms
$$Mod_{\Sigma, \mathbb R^n} \simeq E \equiv \{e: \Sigma \to \mathbb R^n\}/\sim~~~ e \sim f \iff e = f \circ \phi ~~ \text{ for some } \phi \in Diff(\Sigma)$$
Now give $\mathbb R^n$ the usual Euclidean metric $g_{\mathbb R^n}$. A diffeomorphism $e: \Sigma \to \mathbb R^n$ pulls back the metric to $\Sigma$ via $g_e \equiv e^*g_{\mathbb R^n}$. Observe:
$$g_{e \circ \phi} = (e \circ \phi)^*g = \phi^*g_e$$
In particular, when $\phi \in Diff(\Sigma)$, this says pullbacks by homeomorphisms in the same equivalence class are related by pullbacks by homeomorphisms on the surface. Hence another equivalent description of moduli space is
$$Mod_{\Sigma, \mathbb R^n} \simeq F \equiv \{g \text{ metric on } \Sigma\}/\sim~~~ g \sim h \iff g = \phi^*h ~~ \text{ for some } \phi \in Diff(\Sigma)$$
In summary, we found
$$Mod_{\Sigma, \mathbb R^n} \simeq E \simeq F$$
Observe now
• The equivalence relation in $E$ is identification of homeomorphisms that have the same image. We would call this reparametrization invariance.
• The equivalence relation in $F$ is pullback of metric by diffeomorphisms on $\Sigma$. We would call this diffeomorphism invariance.
In other words, in this case diffeomorphism invariance is a manifestation of reparametrization invariance on the space of Riemannian metrics. | 2021-05-08 22:09:59 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.8553897142410278, "perplexity": 240.71178753736154}, "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/1620243988927.95/warc/CC-MAIN-20210508211857-20210509001857-00322.warc.gz"} |
https://acabassi.github.io/coca/reference/chooseKusingAUC.html | This function allows to choose the number of clusters in a dataset based on the area under the curve of the empirical distribution function of a consensus matrix, calculated for different (consecutive) cluster numbers, as explained in the article by Monti et al. (2003), Section 3.3.1.
chooseKusingAUC(areaUnderTheCurve, savePNG = FALSE, fileName = "deltaAUC.png")
## Arguments
areaUnderTheCurve Vector of length maxK-1 containing the area under the curve of the empirical distribution function of the consensus matrices obtained with K varying from 2 to maxK. Boolean. If TRUE, a plot of the area under the curve for each value of K is saved as a png file. The file is saved in a subdirectory of the working directory, called "delta-auc". Default is FALSE. If savePNG is TRUE, this is the name of the png file. Can be used to specify the folder path too. Default is "deltaAUC". The ".png" extension is automatically added to this string.
## Value
This function returns a list containing:
deltaAUC
a vector of length maxK-1 where element i is the area under the curve for K = i+1 minus the area under the curve for K = i (for i = 2 this is simply the area under the curve for K = i)
K
the lowest among the values of K that are chosen by the algorithm.
## References
Monti, S., Tamayo, P., Mesirov, J. and Golub, T., 2003. Consensus clustering: a resampling-based method for class discovery and visualization of gene expression microarray data. Machine learning, 52(1-2), pp.91-118.
## Examples
# Assuming that we want to choose among any value of K (number of clusters)
# between 2 and 10 and that the area under the curve is as follows:
areaUnderTheCurve <- c(0.05, 0.15, 0.4, 0.5, 0.55, 0.56, 0.57, 0.58, 0.59)
# The optimal value of K can be chosen with:
K <- chooseKusingAUC(areaUnderTheCurve)\$K | 2022-01-20 13:50:53 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 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": 0.8200960755348206, "perplexity": 823.9922527174712}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320301863.7/warc/CC-MAIN-20220120130236-20220120160236-00418.warc.gz"} |
https://www.matrix.edu.au/beginners-guide-to-year-12-maths-advanced/applications-of-differentiation/ | # Part 1: Applications of Differentiation | Free Worksheet
In this article, we explain how to apply the differentiation skills you've developed in Year 11.
You understand differentiation, but how confident are you in its application? In this article, we’re going to run you through the fundamentals for Year 12 Maths Advanced Applications of Differentiation.
## Year 12 Advanced Mathematics: Applications of Differentiation
Now that you’ve mastered the algebraic aspects of differentiation, it’s time to tackle its numerous applications in your exams. The basic concepts of differentiation will be extended upon in the following sections to examine the connections that exist between the derivative of a function and the geometrical properties of its graph. It will also be applicable when solving word problems involving maxima and minima.
Differentiation is one of the most important and useful mathematical concepts, since it can be applied to many practical situations. For example, it can help you graph non-typical curves much easier by providing extra information on the shape of the curve or create a rectangle of maximum area given a specific length of string.
This guide adds to your existing algebraic knowledge of differentiation to explain several techniques that can be used to help you solve more difficult questions in your exams.
## NESA Syllabus outcomes
### C3.1: The first and second derivatives
Students:
• Define and interpret the concept of the second derivative as the rate of change of the first derivative function in a variety of contexts, for example, recognise acceleration as the second derivative of displacement with respect to time
• Understand the concepts of concavity and points of inflection and their relationship with the second derivative
• Use the second derivative to determine concavity and the nature of stationary points
• Understand that when the second derivative is equal to $$0$$ this does not necessarily represent a point of inflection
• Use the first derivative to investigate the shape of the graph of a function
• Deduce from the sign of the first derivative whether a function is increasing, decreasing or stationary at a given point or in a given interval
• Use the first derivative to find intervals over which a function is increasing or decreasing, and where its stationary points are located
### C3.2: Applications of the derivative
Students:
• Use any of the functions covered in the scope of this syllabus and their derivatives to solve practical and abstract problems
• Use calculus to determine and verify the nature of stationary points, find local and global maxima and minima and points of inflection (horizontal or otherwise), examine behaviour of a function as $$𝑥→∞$$ and $$𝑥→−∞$$and hence sketch the graph of the function
• Solve optimisation problems for any of the functions covered in the scope of this syllabus, in a wide variety of contexts including displacement, velocity, acceleration, area, volume, business, finance and growth and decay
• Define variables and construct functions to represent the relationships between variables related to contexts involving optimisation, sketching diagrams or completing diagrams if necessary
• Use calculus to establish the location of local and global maxima and minima, including checking endpoints of an interval if required
## Assumed knowledge
Students should be comfortable with practical applications of calculus, including:
• Interpreting the first derivative as the gradient of the function
• Knowledge of the theorems for differentiation, including the chain, product and quotient rules
• Considering limits (including towards positive and negative infinity)
## Applications of Second Derivative
### The Second Derivative
While the first derivative is obtained by differentiating $$y=f(x)$$ with respect to $$x$$, the second derivative is found by differentiating the first derivative (i.e. $$\frac{dy}{dx}$$ or $$f'(x)$$) with respect to $$x$$. Hence, the second derivative is denoted as $$\frac{d}{dx} \left(\frac{dy}{dx}\right)=\frac{d^2 y}{dx^2}$$ or $$f”(x)$$.
### Example:
\begin{align*}
f(x)&=3x^{3}+2x+1 \\
f’ (x)&=9x^{2}+2 \\
f” (x)&=18x \\
\end{align*}
Because the first derivative indicated how the $$y$$ value changed with respect to $$x$$, the second derivative represents how the gradient $$\frac{dy}{dx}$$ changes with respect to $$x$$. This corresponds to the concept of concavity:
• If $$\frac{d^2 y}{dx^2}>0$$, then the curve has an upward concavity (or is concave up)
• If $$\frac{d^2 y}{dx^2}<0$$, then the curve has a downward concavity (or is concave down)
You can think of upward concavity as having a u-shaped while downward concavity has a $$n$$-shape. Note that a curve can be concave up and down at different points, just as how a curve can be increasing and decreasing at different points.
### Point of Inflection
A point of inflection is a point where the concavity of the curve changes, and is defined to be when:
$$\frac{d^2 y}{dx^2}=0$$ , and the sign of $$\frac{d^2 y}{dx^2}$$ changes across the point.
For inflection points, students must remember to check if the sign of $$\frac{d^2 y}{dx^2}$$ changes by creating a table of values or marks will be deducted in exams. Note that the inflection point is not exactly ‘flat’ like a stationary point, but rather the point where the concavity changes.
To identify points of inflection for a curve:
Steps Explanation 1 Differentiate the function twice to find the second derivative. 2 Solve for inflection points by letting $$\frac{d^2 y}{dx^2}=0$$. 3 Check if $$\frac{d^2 y}{dx^2}$$ changes sign by creating a $$\frac{d^2 y}{dx^2}$$ table. 4 Find the $$y$$-coordinate of the inflection point.
Example 1
Find the inflection points of the curve $$y=10-2x-x^2-x^3$$.
Solution 1
Step 1:
Differentiate the function twice to find the second derivative
\begin{align*}
\frac{dy}{dx}&=-2-2x-3x^2 \\
\frac{d^2 y}{dx^2}& =-2-6x \\
\end{align*}
Step 2:
Let $$\frac{d^2 y}{dx^2}=0$$
\begin{align*}
-2-6x&=0 \\
-2&=6x \\
x&=-\frac{1}{3}\\
\end{align*}
Step 3:
Check if $$\frac{d^2 y}{dx^2}$$ changes sign by creating a $$\frac{d^2 y}{dx^2}$$ table
Setup the following table by:
• Placing the $$x$$-value of the inflection point in the middle of the table
• Choosing an $$x$$-value relatively close to the left and right of the inflection point (they do not have to be equally spaced away)
• Find the $$\frac{d^2 y}{dx^2}$$ values at the chosen $$x$$-values (the centre value should be zero as that was assumed in the previous step)
• Check whether the sign changed at the inflection point
$$x$$ $$0$$ $$-\frac{1}{3}$$ $$1$$ $$\frac{d^2y}{dx^2}$$ $$-2-6(0)=-2$$ $$0$$ $$-2-6(1)=-8$$
In this case, the sign did not change so it is not an inflection point at $$x=-\frac{1}{3}$$ even though $$\frac{d^2 y}{dx^2 }=0$$ at that point.
### The Second Derivative Test for Stationary Points
Recall from Year 11 that a stationary point occurs when $$\frac{dy}{dx}=0$$. The nature of the stationary point could be separated into minimum points, maximum points and stationary inflection points, and was previously determined by testing two points on the curve adjacent to the stationary point and on either side of it.
Alternatively, the second derivative can be used to test the nature of stationary points. By looking at the sign of $$\frac{d^2 y}{dx^2}$$ at the stationary point (by subbing the relevant $$x$$-value into the second derivative), we can determine whether the graph is concave up or down and hence whether it is a maximum or minimum point.
• If $$\frac{d^2 y}{dx^2}>0$$ at the stationary point, then it is concave up and hence must be a minimum point
• If $$\frac{d^2 y}{dx^2}<0$$ at the stationary point, then it is concave down and hence must be a maximum point
• If $$\frac{d^2 y}{dx^2 }=0$$ at the stationary point, then the point must be a horizontal point of inflection
Example 2
Determine the nature of the stationary points of the curve $$y=x^3+2x^2-7x+3$$ by using the second derivative.
Solution 2
\begin{align*}
y&=x^3+2x^2-7x+3 \\
\frac{dy}{dx}&=3x^2+4x-7 \\
\frac{d^2 y}{dx^2}&=9x+4 \\
\end{align*}
To find the stationary point, let $$\frac{dy}{dx}=0$$:
\begin{align*}
3x^2+4x-7&=0 \\
\frac{3x+7}{x-1}&=0 \\
∴x&=-\frac{7}{3},1 \\
y&=\frac{473}{27},-1 \\
\end{align*}
To determine their nature, sub into $$\frac{d^2 y}{dx^2}$$:
\begin{align*}
\text{when} \ x&=-\frac{7}{3},\frac{d^2 y}{dx^2}\\
&=9\left(-\frac{7}{3}+4\right)\\
&=-17<0 \\
\text{when} \ x&=1, \frac{d^2 y}{dx^2} \\
&=9(1)+4 \\
&=13>0\\
\end{align*}
\begin{align*}
∴\text{maximum point at }\left(-\frac{7}{3},\frac{473}{27}\right), \ \text{minimum point at }(1,-1)\\
\end{align*}
### Curve Sketching Using Calculus
By applying the geometrical features of the first and second derivative, we can use this knowledge to aid us when sketching complex functions.
The following steps should be taken to identify the key information required in curve sketching:
Steps Explanation 1 Find the $$y$$-intercept. 2 Consider the behaviour of $$y$$ as $$x→±∞$$. 3 Determine any stationary points and their nature by letting $$\frac{dy}{dx}=0$$. 4 Determine any inflection points by letting $$\frac{d^2 y}{dx^2}=0$$ and checking for changes to concavity. 5 Sketch the curve, showing all essential features.
Example 3
Sketch the curve of $$y=x^4-8x^2$$, labelling stationary points and any points of inflexion.
Solution 3
Step 1:
Find the $$y$$-intercept
\begin{align*}
y \ int &=(0)^4-8(0)^2\\
&=0\\
\end{align*}
Step 2:
Consider the behaviour of $$y$$ as $$x→±∞$$.
when $$x→+∞,y→∞$$
when $$x→-∞,y→∞$$
Step 3:
Determine any stationary points and their nature by letting $$\frac{dy}{dx}=0$$
\begin{align*}
\frac{dy}{dx}&=4x^3-16x\\
&=0\\
4x(x^2-4)&=0 \\
∴x&=0,±2 \\
\end{align*}
To determine the nature, evaluate $$\frac{d^2 y}{dx^2}$$:
\begin{align*}
\frac{d^2 y}{dx^2}&=12x^2-16\\
\text{when} \ x&=0, \frac{d^2 y}{dx^2}=-16<0 \\
\text{when} \ x&=-2, \frac{d^2 y}{dx^2}=32>0 \\
\text{when} \ x&=2, \frac{d^2 y}{dx^2}=32>0 \\
\end{align*}
\begin{align*}
∴\text{max pt at }(0,0), \ \text{min pt at }(-2,-16), \ \text{min pt at }(2,-16) \\
\end{align*}
Step 4:
Determine any inflection points by letting $$\frac{d^2 y}{dx^2}=0$$ and checking for changes to concavity
\begin{align*}
\frac{d^2 y}{dx^2}&=12x^2-16=0 \\
12x^2&=16\\
x^2&=\frac{4}{3}\\
∴x&=\frac{±2}{\sqrt{3}} \\
\end{align*}
$$x$$ $$-2$$ $$-\frac{2}{\sqrt3}$$ $$1$$ $$\frac{2}{\sqrt3}$$ $$2$$ $$\frac{d^2}{dx^2}$$ $$12(-2)^2-16=32$$ $$0$$ $$12(1)^2-16=-4$$ $$0$$ $$12(2)^2-16=32$$
∴inflection pt at $$\frac{2}{\sqrt3},-\frac{80}{9}$$ and $$-\frac{2}{\sqrt{3}},-\frac{80}{9}$$
Step 5:
Sketch the curve, showing all essential features
### Application of Maxima and Minima Problems
The process of identifying stationary points can be used to solve word problems where a certain value is to be maximised or minimised. To solve these problems:
Step Explanation 1 Create a formula for the quantity to be maximised or minimised. Draw a diagram if necessary. 2 Simplify the formula to be in terms of only one independent variable, using other given information from the question. 3 Solve for the stationary points by letting $$\frac{dy}{dx}=0$$. 4 Check the nature of the stationary point to match what is required for the question (i.e. maximum or minimum). 5 Answer the question in words, including units
Example 4
A fence is created on a farm using $$12$$ metres of wire. It is bent into a rectangular shape against a barn such that the fourth (long side) of the fence is left open. Let the length of the rectangle be $$x$$. Find the maximum area of the rectangle.
Solution 4
Step 1:
Create a formula for the quantity to be maximised or minimised
Drawing a diagram, a formula must be created for the area (which is to be maximised):
$$A=xy$$,where $$y$$ is the width of the rectangle
Step 2:
Simplify the formula to be in terms of only one independent variable, using other given information from the question
Recall that the total amount of wire used is $$12$$ metres:
\begin{align*}
2y+x&=12 \\
y&=\frac{12-x}{2} \\
∴A&=\frac{x(12-x)}{2} \\
&=\frac{12x-x^2}{2} \\
\end{align*}
Step 3:
Solve for the stationary points by letting $$\frac{dA}{dx}=0$$.
\begin{align*}
\frac{dA}{dx}&=\frac{12-2x}{2}=0 \\
12-2x&=0 \\
2x&=12 \\
x&=6 \\
\end{align*}
Step 4:
Check the nature of the stationary point to match what is required for the question
\begin{align*}
\frac{d^2 A}{dx^2}&=-\frac{2}{2}=-1<0 \\
\end{align*}
\begin{align*}
∴\text{A is maximised when }x&=6 \\
\end{align*}
Step 5:
Answer the question in words, including units
\begin{align*}
∴A_{max}&=\frac{12(6)-(6)^2}{2}=18 \\
\end{align*}
\begin{align*}
∴\text{the maximum area is }18m^2 \\
\end{align*}
## How well can you apply your knowledge of differentiation?
Test your knowledge with our free worksheet.
### Sketching the Gradient Function for a Given Function
Another common question in exams is sketching the graph of the derivative $$\frac{dy}{dx}$$ given the graph of the curve $$y=f(x)$$. These types of questions are difficult to master since they are purely graphical.
To sketch $$f'(x)$$ from $$f(x)$$:
Steps Explanation 1 Identify stationary points on the curve. At these points, $$f’ (x)=0$$, so they are now the $$x$$ intercepts of $$f'(x)$$. 2 Identify where the curve is increasing. At these points, $$f’ (x)>0$$, so $$f'(x)$$ will be positive and above the axis. 3 Identify where the curve is decreasing. At these points, $$f’ (x)<0$$, so $$f'(x)$$ will be negative and below the axis. 4 Identify points of inflection on the curve. At these points, $$f” (x)=0$$, so $$f'(x)$$ will have a stationary point (since its derivative $$f”(x)$$ will be zero)
Example 5
Sketch the derivative function of the following function:
Solution 5
Step 1:
Identify stationary points on the curve.
The stationary point of the given function is at $$x=0.2$$ and $$x=3$$. These will be the $$x$$-intercepts of the gradient function.
Step 2:
Identify where the curve is increasing.
The function is increasing at $$x<0.2$$ and $$x>3$$, so the gradient function will be positive at these values.
Step 3:
Identify where the curve is decreasing.
The function is increasing at $$0.2<x<3$$, so the gradient function will be negative at these values.
Step 4:
Identify points of inflection on the curve The point of inflection of the curve is around $$x=1.8$$, so the gradient function will have a stationary point (will be minimum as it is decreasing before and then increasing).
### Sketching the Curve for a Given Gradient Function
Students may also be given a gradient function and expected to graph the curve based on this. To sketch $$f(x)$$ from $$f'(x)$$:
Steps Explanation 1 Identify the $$x$$-intercepts of $$f'(x)$$. These are stationary points on $$f(x)$$. 2 Determine the nature of the stationary points by examining the signs of $$f’ (x)$$ directly before and after the stationary point. 3 Identify where the curve is increasing $$(f’ (x)>0)$$ and decreasing $$(f’ (x)<0)$$ to determine the shape of $$f(x)$$.
Example 6
Sketch the function of the following gradient function:
Solution 6
Step 1:
Identify the $$x$$-intercepts of $$f'(x)$$.
The $$x$$-intercepts are $$x=-3$$ and $$x=3$$. These will be the stationary points of $$f(x)$$.
Step 2:
Determine the nature of the stationary points by examining the signs of $$f’ (x)$$ directly before and after the stationary point.
For $$x=-3$$, it is negative before and positive after so it will be a minimum point. For $$x=3$$, it is positive before and negative after so it will be a maximum point.
Step 3:
Identify where the curve is increasing $$(f’ (x)>0)$$ and decreasing $$(f’ (x)<0)$$ to determine the shape of $$f(x)$$.
The curve is positive at $$-3<x<3$$ and negative at $$x<-3$$ and $$x>3$$.
Thus, the original function will look like:
Note that we do not know the $$y$$-intercept value and are not required to label it.
## Graphs of Linear Motion
### Review of Linear Motion
The concepts of differentiation can be applied to linear motion by considering the displacement, velocity and acceleration of a particle.
To recap, the displacement is the position of the particle from the origin. The instantaneous velocity measures how fast the particle is travelling at a particular instant, and is calculated as:
$$v=\frac{dx}{dt}$$
The acceleration is the instantaneous rate of change of velocity:
$$a=\frac{dv}{dt}=\frac{d^2 x}{dt^2}$$
If $$a>0$$, the velocity of the particle is increasing (i.e. going more positive or less negative)
If $$a<0$$, the velocity of the particle is decreasing (i.e. going less positive or more negative)
### Graphs of Linear Motions
Since the velocity and acceleration are the first and second derivatives of the displacement function, the geometrical properties can be obtained from the displacement-time graph. The velocity indicates the gradient while the acceleration represents the concavity of the curve.
In a velocity-time graph, the gradient of the curve represents the acceleration of the particle, since $$a=\frac{dv}{dt}$$.
Example 7
For the following displacement-time graph and velocity-time graph, at what times is there positive acceleration?
Solution 7
The first graph is a displacement-time graph, where the acceleration is represented by the concavity of the curve. As the graph is concave up at $$0≤t<2$$, that is where there is positive acceleration.
The second graph is a velocity-time graph, where the acceleration is represented by the gradient of the curve. As the graph is increasing at $$1<t<3.5$$, that is where there is positive acceleration.
## Concept Check Questions
1. Consider the curve $$y=x(x-3)^3$$.
a. Find the coordinates and nature of any stationary points of the curve.
b. Sketch the curve, indicating the coordinates of any intercepts and stationary points.
2. The cost, $$C$$ (in thousands of dollars) for producing x (in hundreds) plates in a company is modelled by the equation:
\begin{align*}
C&=\frac{169}{2x+1}+2x, \ \text{where },0≤x≤20\\
\end{align*}
Determine the number of plates that must be produced to minimise the manufacturing costs and state the total cost.
3. Draw the graph of the first and second derivative for the following curve:
4. The displacement-time graph of a particle is shown in the following diagram:
At what times is the velocity and acceleration equal to zero?
## Concept Check Solutions
1.
a. Horizontal point of inflexion at $$(3,0 )$$, minimum point at $$\left(\frac{3}{4},-\frac{2187}{256}\right)$$
b.
2. The cost is minimised when $$600$$ parts are plates are manufactured and the total cost will be $$\ 25 \ 000$$.
3. First derivative:
Second derivative:
4. $$\text{Velocity}:1s,5.75s,\text{ Acceleration}:3s$$
© Matrix Education and www.matrix.edu.au, 2021. Unauthorised use and/or duplication of this material without express and written permission from this site’s author and/or owner is strictly prohibited. Excerpts and links may be used, provided that full and clear credit is given to Matrix Education and www.matrix.edu.au with appropriate and specific direction to the original content. | 2021-04-17 16:32:17 | {"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": 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": 0.7589675784111023, "perplexity": 537.077442165276}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-17/segments/1618038461619.53/warc/CC-MAIN-20210417162353-20210417192353-00237.warc.gz"} |
https://socratic.org/questions/how-do-you-identify-the-important-parts-of-y-x-4-x-2-to-graph-it#174289 | # How do you identify the important parts of y= ½(x-4)(x+2) to graph it?
Oct 5, 2015
Graph $y = \left(\frac{1}{2}\right) \left(x - 4\right) \left(x + 2\right)$
#### Explanation:
The important parts are:
a. x-coordinate of axes of symmetry and vertex:
$x = \frac{x 1 + x 2}{2} = \frac{4 - 2}{2} = 1$
y-coordinate of vertex: $y = \left(f 1\right) = \left(\frac{1}{2}\right) \left(- 3\right) \left(3\right) = - \frac{9}{2}$
b. y-intercept --> Make x = 0 --> $y = - \frac{8}{2} = - 4$
c. x-intercepts --> x = 4 and x = -2
graph{1/2(x - 4)(x + 2) [-10, 10, -5, 5]} | 2021-10-19 16:40:56 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 4, "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": 0.21660302579402924, "perplexity": 5336.838034018}, "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-2021-43/segments/1634323585270.40/warc/CC-MAIN-20211019140046-20211019170046-00639.warc.gz"} |
https://cml.centre-mersenne.org/page/instructions-for-authors_en/ | # CONFLUENTES MATHEMATICI
#### Submission
Authors should submit their papers as pdf files to the most appropriate editor (the interests of each editor are listed in the Editorial board). Once a paper is accepted, authors should provide a Latex file with documentclass cml which compiles without errors under pdflatex, together with a pdf file obtained by compiling the Latex file. See the LaTeX Instructions for details on how to prepare the Latex file.
#### Language
Confluentes Mathematici accepts papers in English, French or German. If the paper is not in English, an English translation of the title and abstract should be provided.
#### Abstract
Papers should contain an abstract of less than 300 words summarizing its content and conclusions. The AMS subject classification should also be given, as well as some keywords.
#### Originality
Submission of a paper implies that it has not been published in, and has not been submitted to another journal.
#### References
References should be listed in alphabetical order of the first author’s name and cited by the corresponding number in square brackets in the text.
#### Accessibility
In order to facilitate access for the disabled, starting in 2019 all papers will be published both as pdf and tex file. | 2023-03-28 19:01:37 | {"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": 0.8608413934707642, "perplexity": 1227.5912657136912}, "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/1679296948868.90/warc/CC-MAIN-20230328170730-20230328200730-00345.warc.gz"} |
https://socratic.org/questions/how-do-you-graph-y-6-x-2-3 | # How do you graph y=6/(x^2+3)?
Dec 21, 2017
See below.
#### Explanation:
First find significant points, these will help in sketching the graph.
y axis intercepts occur when $x = 0$:
$\frac{6}{{\left(0\right)}^{2} + 3} = 2 \textcolor{w h i t e}{88}$ coordinate $\left(0 , 2\right)$
x axis intercepts occur when $y = 0$:
$\frac{6}{{x}^{2} + 3} = 0$
This can only be zero when denominator is zero, which is undefined, so no x axis intercepts.
as $x \to \infty$ $\textcolor{w h i t e}{888} \frac{6}{{x}^{2} + 3} \to 0$
as $x \to - \infty$ $\textcolor{w h i t e}{888} \frac{6}{{x}^{2} + 3} \to 0$
So the x axis is a horizontal asymptote.
$y = 0$
Vertical asymptotes occur where the function is undefined:
$\frac{6}{{x}^{2} + 3}$ is not undefined for any real $x$, so no vertical asymptotes.
$\frac{6}{{x}^{2} + 3}$ attains a maximum value when the denominator is a minimum value, this can be seen to be 3 when $x = 0$
( this is the y axis intercept that was previously found )
Graph:
graph{y=6/(x^2+3) [-10, 10, -5, 5]} | 2019-03-23 00:53:40 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 14, "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": 0.9178215265274048, "perplexity": 1007.5850836728065}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-13/segments/1552912202704.58/warc/CC-MAIN-20190323000443-20190323022443-00313.warc.gz"} |
http://physics.stackexchange.com/questions/45701/faraday-law-third-maxwells-equation-in-mathematica | # Faraday law, third Maxwell's equation in Mathematica
$\displaystyle\nabla\times\mathbf{E}=-\frac{\partial \mathbf{B}}{\partial t}$
1 If I solve this equation with Mathematica, I find the magnetic field $b(x,y,z,t),b:\mathbb{R}^4\rightarrow\mathbb{R}^3$ right?
2 I have put an arbitrary function $e:\mathbb{R}^4\rightarrow\mathbb{R}^3$ as electric field for this experiment, but how can I calculate that function for a real case; what I need to do this?
3 One time that I have both the electric and magnetic field how can I compose the electromagnetic field?
Needs["VectorAnalysis"]
(*Electric field e : R^4->R^3 *)
e[x_, y_, z_, t_] := {x - 3 y, 4 y + t, y + z + t};
Maxwell = Curl[e[x, y, z, t]] == -D[b[x, y, z, t], t];
DSolve[Maxwell, b[x, y, z, t], {x, y, z, t}]
`
-
As a first step, you need definitely boundary conditions for your differential equations to be solved and to get real life case as you whant. – TMS Dec 2 '12 at 16:49
Both Electric and Magnetic fields must satisfy the four Maxwell's equations, not only one. That is the meaning of a "real case". Doing that, you'll have the electromagnetic field. – Anuar Dec 2 '12 at 16:57 | 2015-06-30 21:32:46 | {"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": 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": 0.7905247211456299, "perplexity": 284.52664443475754}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-27/segments/1435375094501.77/warc/CC-MAIN-20150627031814-00194-ip-10-179-60-89.ec2.internal.warc.gz"} |
https://tex.stackexchange.com/questions/496503/some-items-in-enumerate-jump-left-out-of-margin | # Some items in enumerate jump left out of margin
I have a boxed list in latex, and for some reason some items jump out of the box out of the left margin. I cannot figure out what is causing this and already have searched and tried some fixes. The code:
\begin{figure}
\centering
\caption{The Lin-Kernighan Modified Distance Algorithm}
\label{fig:LKMD}
\noindent
\fbox{
\parbox{\textwidth}{
{\fontfamily{lmtt}\selectfont
\begin{spacing}{1}
\begin{itemize}
\item Initialise the algorithm
\begin{enumerate}
\item Calculate for every item $s_i = \frac{p_i}{w_i}$ and rank them based on the score.
\item Pick items starting from the highest rank until the knapsack is full.
\item Set LB = 0\%.
\item Set UB = 20\%.
\item Set \epsilon = 0.1.
\item Set $Z^* = -inf$, the best improvement found so far.
\end{enumerate}
\item While UB - LB > \epsilon:
\begin{enumerate}
\item for C \in \{UB, LB\}
\item Modify the distance matrix:
$d'_{ij}$=\left\{
\begin{array}{ll}
$(1 - C) \cdot d_{ij} \text{ if } i \in N_h \text{ and } j \in N_1$\\
$(1 - C) \cdot d_{ij} \text{ if } i \in N_1 \text{ and } j \in N_1$\\
$(1 + C) \cdot d_{ij} \text{ if } i \in N_1 \text{ and } j \in N_0$\\
$(1 + C) \cdot d_{ij} \text{ if } i \in N_0 \text{ and } j \in N_1$\\
$d_{ij} \text{ else }$ \\
\end{array}
\right.
\item Run LK algorithm with modified distances and evaluate the objective function (OBJ) for LB and UB
\item
\end{enumerate}
\end{itemize}
\end{spacing}
}
}
}
\end{figure}
I also notice it is coming out weird here as qwll, but I don't know why...
I included a picture of the print out.
• You must use math mode if you want to write math – harryparp Jun 19 at 15:03
• You tell the array to have two columns, yet you don't have any column sperators – harryparp Jun 19 at 15:06
• Thanks, this indeed fixed it... oops – wout konings Jun 19 at 15:06
Some improvements, with caption, enumitem and mathtools and the cases environment:
\documentclass{article}
\usepackage{setspace}
\usepackage{mathtools}
\usepackage{caption}
\usepackage{enumitem}
\begin{document}
\begin{figure}[!htb]
\centering
\caption{The Lin-Kernighan Modified Distance Algorithm}
\label{fig:LKMD}
\noindent
\fbox{
\parbox{\textwidth}{
{\fontfamily{lmtt}\selectfont
\begin{spacing}{1}
\begin{itemize}[wide=0pt, leftmargin=*]
\item Initialise the algorithm
\begin{enumerate}[wide=0pt, leftmargin=*]
\item Calculate for every item $s_i = \frac{p_i}{w_i}$ and rank them based on the score.
\item Pick items starting from the highest rank until the knapsack is full.
\item Set $LB = 0\%$.
\item Set $UB = 20\%$.
\item Set $\epsilon = 0.1$.
\item Set $Z^* = -inf$, the best improvement found so far.
\end{enumerate}
\item While $UB - LB > \epsilon$:
\begin{enumerate}[wide=0pt, leftmargin=*]
\item for $C \in \{UB, LB\}$
\item Modify the distance matrix:
$d'_{ij}= \begin{cases*} (1 - C) \cdot d_{ij} & if i \in N_h and j \in N_1 \\ (1 - C) \cdot d_{ij} & if i \in N_1 and j \in N_1 \\ (1 + C) \cdot d_{ij} & if i \in N_1 and j \in N_0 \\ d_{ij} & otherwise \end{cases*}$
\item Run LK algorithm with modified distances and evaluate the objective function (OBJ) for LB and UB
\item
\end{enumerate}
\end{itemize}
\end{spacing}
}}}%
\end{figure}
\end{document}
two problems:
• You must use math mode if you want to write math
• You tell the array to have two columns, yet you don't have any column separators
or summary:
• you must not ignore error messages
\documentclass{article}
\usepackage{setspace}
\usepackage{mathtools}
\begin{document}
\begin{figure}
\centering
\caption{The Lin-Kernighan Modified Distance Algorithm}
\label{fig:LKMD}
\noindent
\fbox{
\parbox{\textwidth}{
{\fontfamily{lmtt}\selectfont
\begin{spacing}{1}
\begin{itemize}
\item Initialise the algorithm
\begin{enumerate}
\item Calculate for every item $s_i = \frac{p_i}{w_i}$ and rank them based on the score.
\item Pick items starting from the highest rank until the knapsack is full.
\item Set $LB = 0\%$.
\item Set $UB = 20\%$.
\item Set $\epsilon = 0.1$.
\item Set $Z^* = -inf$, the best improvement found so far.
\end{enumerate}
\item While $UB - LB > \epsilon$:
\begin{enumerate}
\item for $C \in \{UB, LB\}$
\item Modify the distance matrix:
$d'_{ij}=\left\{ \begin{array}{ll} (1 - C) \cdot d_{ij} &\text{ if } i \in N_h \text{ and } j \in N_1\\ (1 - C) \cdot d_{ij} &\text{ if } i \in N_1 \text{ and } j \in N_1\\ (1 + C) \cdot d_{ij} &\text{ if } i \in N_1 \text{ and } j \in N_0\\ (1 + C) \cdot d_{ij} &\text{ if } i \in N_0 \text{ and } j \in N_1\\ d_{ij} &\text{ else } \\ \end{array} \right.$
\item Run LK algorithm with modified distances and evaluate the objective function (OBJ) for LB and UB
\item
\end{enumerate}
\end{itemize}
\end{spacing}
}
}
}
\end{figure}
\end{document}
• Thanks! Stupid I didn't notice... – wout konings Jun 19 at 15:17 | 2019-07-23 09:59:31 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7930049300193787, "perplexity": 3430.1766496124073}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-30/segments/1563195529175.83/warc/CC-MAIN-20190723085031-20190723111031-00053.warc.gz"} |
https://mathematica.stackexchange.com/questions/176398/commutator-of-block-matrices | # Commutator of block matrices
How to implement a commutator of matrices composed of operators?
## Background:
Let $\hat A_{ij}$, $\hat B_{kl}$ be some sets of some operators. $\hat I$ is the identity operator. Their commutators are defined as usual $[\hat A_{ij},\hat B_{kl}]=\hat A_{ij}\hat B_{kl}-\hat B_{kl}\hat A_{ij}$. Out of these operators two matrices $mH$ and $mQ$ (both can be sparse) are constructed. I would like to compute a commutator $mC=[mH,mQ]$ and express (where possible) the elements of $mC$ in terms of commutators $[\hat A_{ij},\hat B_{kl}]$.
## Example:
Define $mH$ and $mQ$ as $2\times 2$ matrices:
Clear[mH, mQ]
mH = {{h, v},{v, h + i}};
mQ = {{q0, p},{p, q1}};
Define $f$ as non-commutative but distributive operation; $i$ is the identity operator
f[a_ + b_, c_] := f[a, c] + f[b, c]
f[c_, a_ + b_] := f[c, a] + f[c, b]
f[i, a_] := a;
f[a_, i] := a;
Dot only works for the normal multiplication operation, therefore, compute the commutator using Inner
Clear[mC]
mC[0] = Inner[f, mH, mQ, Plus] - Inner[f, mQ, mH, Plus];
Express matrix elements in terms of commutators (denoted as c). This part is not working properly...
mC[1] = mC[0] /. {f[a_, b_] - f[b_, a_] -> c[a, b]};
MatrixForm[mC[1]]
$\left( \begin{array}{cc} c(h,\text{q0})-f(p,v)+f(v,p) & -p+c(h,p)-f(\text{q0},v)+f(v,\text{q1}) \\ p+c(h,p)-f(\text{q1},v)+f(v,\text{q0}) & c(h,\text{q1})-f(p,v)+f(v,p) \\ \end{array} \right)$
## Problem:
We can see that some diagonal entries can still be expressed in terms of commutators. The FullForm reveals the reason. However, I do not know a nice way to solve this problem. Your help is very much appreciated!
mC[1] = mC[0] //. {f[a_, b_] - f[b_, a_] :> c[a, b]};
$\left( \begin{array}{cc} c(h,\text{q0})+c(v,p) & c(h,p)-f(\text{q0},v)+f(v,\text{q1})-p \\ c(h,p)+f(v,\text{q0})-f(\text{q1},v)+p & c(h,\text{q1})+c(v,p) \\ \end{array} \right)$ | 2019-10-17 04:14:17 | {"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": 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": 0.7782725095748901, "perplexity": 3720.018446541066}, "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/1570986672548.33/warc/CC-MAIN-20191017022259-20191017045759-00260.warc.gz"} |
https://cs.stackexchange.com/questions/99492/confluence-of-beta-expansion/99659 | # Confluence of beta expansion
Let $$\to_\beta$$ be $$\beta$$-reduction in the $$\lambda$$-calculus. Define $$\beta$$-expansion $$\leftarrow_\beta$$ by $$t'\leftarrow_\beta t \iff t\to_\beta t'$$.
Is $$\leftarrow_\beta$$ confluent? In other words, do we have that for any $$l,d,r$$, if $$l \to_\beta^* d\leftarrow_\beta^* r$$, then there exists $$u$$ such that $$l\leftarrow_\beta^* u \to_\beta^* r$$?
Keywords: Upward confluence, upside down CR property
I started by looking at the weaker property: local confluence (i.e. if $$l \to_\beta d\leftarrow_\beta r$$, then $$l\leftarrow_\beta^* u \to_\beta^* r$$). Even if this were true, it would not imply confluence since $$\beta$$-expansion is non-terminating, but I thought that it would help me understand the obstacles.
(Top) In the case where both reductions are at top-level, the hypothesis becomes $$(\lambda x_1.b_1)a_1\rightarrow b_1[a_1/x_1]=b_2[a_2/x_2]\leftarrow (\lambda x_2.b_2)a_2$$. Up to $$\alpha$$-renaming, we can assume that $$x_1\not =x_2$$, and that neither $$x_1$$ nor $$x_2$$ is free in those terms.
(Throw) If $$x_1$$ is not free in $$b_1$$, we have $$b_1=b_2[a_2/x_2]$$ and therefore have $$(\lambda x_1.b_1)a_1=(\lambda x_1.b_2[a_2/x_2])a_1\leftarrow(\lambda x_1.(\lambda x_2.b_2)a_2)a_1\rightarrow (\lambda x_2.b_2)a_2$$.
A naive proof by induction (on $$b_1$$ and $$b_2$$) for the case (Top) would be as follows:
• If $$b_1$$ is a variable $$y_1$$,
• If $$y_1=x_1$$, the hypothesis becomes $$(\lambda x_1.x_1)a_1\rightarrow a_1=b_2[a_2/x_2]\leftarrow (\lambda x_2.b_2)a_2$$, and we indeed have $$(\lambda x_1.x_1)a_1=(\lambda x_1.x_1)(b_2[a_2/x_2])\leftarrow (\lambda x_1.x_1)((\lambda x_2.b_2)a_2)\rightarrow (\lambda x_2.b_2)a_2$$.
• If $$y_1\not=x_1$$, then we can simply use (Throw).
• The same proofs apply is $$b_2$$ is a variable.
• For $$b_1=\lambda y.c_1$$ and $$b_2=\lambda y.c_2$$, the hypothesis becomes $$(\lambda x_1.\lambda y. c_1)a_1\rightarrow \lambda y.c_1[a_1/x_1]=\lambda y.c_2[a_2/x_2]\leftarrow (\lambda x_2.\lambda y.c_2)a_2$$ and the induction hypothesis gives $$d$$ such that $$(\lambda x_1.c_1)a_1\leftarrow d\rightarrow (\lambda x_2.c_2)a_2$$ which implies that $$\lambda y.(\lambda x_1.c_1)a_1\leftarrow \lambda y.d\rightarrow \lambda y.(\lambda x_2.c_2)a_2$$. Unfortunately, we do not have $$\lambda y.(\lambda x_2.c_2)a_2\rightarrow (\lambda x_2.\lambda y.c_2)a_2$$. (This makes me think of $$\sigma$$-reduction.)
• A similar problem arises for applications: the $$\lambda$$s are not where they should be.
• @chi Unless I'm mistaken, $(\lambda b.yb)y\leftarrow(\lambda a. (\lambda b. a b)y)y\rightarrow (\lambda a. a y)y$ works. – xavierm02 Nov 5 '18 at 19:06
• I somewhat agree with @chi that it seems confluent after you think about it and see a couple of counter-examples. But actually, what about $(\lambda x.x\;x\;y)\;y \to y\;y\;y \leftarrow (\lambda x.y\;x\;x)\;y$? – Rodolphe Lepigre Nov 5 '18 at 19:25
• Although it'd be convenient for me if it were true, I'm a little more pessimistic. A colleague of mine made the following remark which makes it seem unlikely: it'd imply that any two arbitrary programs that compute the same (church) integer can be combined. – xavierm02 Nov 5 '18 at 19:44
• The answer is no. Exercise 3.5.11 in Barendregt gives a counter-example attributed to Plotkin, but without a reference: $(\lambda x. b x (b c)) c$ and $(\lambda x. x x) (b c)$. I'm going to look for a proof. – Gilles 'SO- stop being evil' Nov 5 '18 at 20:03
• I've posted the counterexample as an answer, with what I thought would be a proof, but there's a step I can't figure out. If someone can figure it out, please post an answer and I'll delete mine. – Gilles 'SO- stop being evil' Nov 6 '18 at 0:15
## 2 Answers
Two counterexamples are:
• $$(\lambda x. b x (b c)) c$$ and $$(\lambda x. x x) (b c)$$ (Plotkin).
• $$(\lambda x. a (b x)) (c d)$$ and $$a ((\lambda y. b (c y)) d)$$ (Van Oostrom).
The counterexample detailed below is given in The Lambda Calculus: Its Syntax and Semantics by H.P. Barenredgt, revised edition (1984), exercise 3.5.11 (vii). It is attributed to Plotkin (no precise reference). I give an incomplete proof which is adapted from a proof by Vincent van Oostrom of a different counterexample, in Take Five: an Easy Expansion Exercise (1996) [PDF].
The basis of the proof is the standardization theorem, which allows us to consider only beta expansions of a certain form. Intuitively speaking, a standard reduction is a reduction that makes all of its contractions from left to right. More precisely, a reduction is non-standard iff there is a step $$M_i$$ whose redex is a residual of a redex to the left of the redex of a previous step $$M_j$$; “left” and “right” for a redex are defined by the position of the $$\lambda$$ that is eliminated when the redex is contracted. The standardization theorem states that there if $$M \rightarrow_\beta^* N$$ then there is a standard reduction from $$M$$ to $$N$$.
Let $$L = (\lambda x. b x (b c)) c$$ and $$R = (\lambda x. x x) (b c)$$. Both terms beta-reduce to $$b c (b c)$$ in one step.
Suppose that there is a common ancestor $$A$$ such that $$L \leftarrow_\beta^* A \rightarrow_\beta^* R$$. Thanks to the standardization theorem, we can assume that both reductions are standard. Without loss of generality, suppose that $$A$$ is the first step where these reductions differ. Of these two reductions, let $$\sigma$$ be the one where the redex of the first step is to the left of the other, and write $$A = C_1[(\lambda z. M) N]$$ where $$C_1$$ is the context of this contraction and $$(\lambda z. M) N$$ is the redex. Let $$\tau$$ be the other reduction.
Since $$\tau$$ is standard and its first step is to the right of the hole in $$C_1$$, it cannot contract at $$C_1$$ nor to the left of it. Therefore the final term of $$\tau$$ is of the form $$C_2[(\lambda z. M') N']$$ where the parts of $$C_1$$ and $$C_2$$ to the left of their holes are identical, $$M \rightarrow_\beta^* M'$$ and $$N \rightarrow_\beta^* N'$$. Since $$\sigma$$ starts by reducing at $$C_1$$ and never reduces further left, its final term must be of the form $$C_3[S]$$ where the part of $$C_3$$ to the left of its hole is identical to the left part of $$C_1$$ and $$C_2$$, and $$M[z \leftarrow N] \rightarrow_\beta^* S$$.
Observe that each of $$L$$ and $$R$$ contains a single lambda which is to the left of the application operator at the top level. Since $$\tau$$ preserves the lambda of $$\lambda z. M$$, this lambda is the one in whichever of $$L$$ or $$R$$ is the final term of $$\tau$$, and in that term the argument of the application is obtained by reducing $$N$$. The redex is at the toplevel, meaning that $$C_1 = C_2 = C_3 = []$$.
• If $$\tau$$ ends in $$R$$, then $$M \rightarrow_\beta^* z z$$, $$N \rightarrow_\beta^* b c$$ and $$M[z \leftarrow N] \rightarrow_\beta^* (\lambda x. b x (b c)) c$$. If $$N$$ has a descendant in $$L$$ then this descendant must also reduce to $$b c$$ which is the normal form of $$N$$. In particular, no descendant of $$N$$ can be a lambda, so $$\sigma$$ cannot contract a subterm of the form $$\check{N} P$$ where $$\check{N}$$ is a descendant of $$N$$. Since the only subterm of $$L$$ that reduces to $$b c$$ is $$b c$$, the sole possible descendant of $$N$$ in $$L$$ is the sole occurrence of $$b c$$ itself.
• If $$\tau$$ ends in $$L$$, then $$M \rightarrow_\beta^* b z (b c)$$, $$N \rightarrow_\beta^* c$$, and $$M[z \leftarrow N] \rightarrow_\beta^* (\lambda x. x x) (b c)$$. If $$N$$ has a descendant in $$R$$ then this descendant must also reduce to $$c$$ by confluence.
At this point, the conclusion should follow easily according to van Oostrom, but I'm missing something: I don't see how tracing the descendants of $$N$$ gives any information about $$M$$. Apologies for the incomplete post, I'll think about it overnight.
Note that $$\beta$$-reduction can make any term disappear. Assuming that variable $$x$$ does not appear free in a term $$v$$, you have $$(\lambda x.v)\;t_1 \to_\beta v$$ and $$(\lambda x.v)\;t_2 \to_\beta v$$ for any terms $$t_1$$ and $$t_2$$. As a consequence, the fact that reverse $$\beta$$-reduction is confluent is somewhat equivalent to: for all terms $$t_1$$ and $$t_2$$, there is a term $$u$$ such that $$u \to_\beta^\ast t_1$$ and $$u \to_\beta^\ast t_2$$. This seems very false to me!
• Unless I'm mistaken, $(\lambda x .v) t_1\leftarrow (\lambda x .(\lambda x .v) t_1)t_2\rightarrow (\lambda x .v) t_2$ works for those two terms. – xavierm02 Nov 5 '18 at 18:37
• Damn, you're right! I'll try to think of something else later, I don't have time right now. – Rodolphe Lepigre Nov 5 '18 at 18:46 | 2020-01-17 20:08:57 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 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": 130, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9423210024833679, "perplexity": 270.67950803297197}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-05/segments/1579250590107.3/warc/CC-MAIN-20200117180950-20200117204950-00081.warc.gz"} |
https://riseandhustle.com/cad-rnpdg/eigenvalues-of-orthogonal-matrix-1439f0 | # eigenvalues of orthogonal matrix
Notify me of follow-up comments by email. (a) Each eigenvalue of the real skew-symmetric matrix A is either 0 or a purely imaginary number. Sorry about that. 6.1Introductiontoeigenvalues 6-1 Motivations â¢Thestatic systemproblemofAx =b hasnowbeensolved,e.g.,byGauss Double checked, but it said +/- 1. Find Orthogonal Basis / Find Value of Linear Transformation, Subspace of Skew-Symmetric Matrices and Its Dimension, Linear Combination and Linear Independence, Bases and Dimension of Subspaces in $\R^n$, Linear Transformation from $\R^n$ to $\R^m$, Linear Transformation Between Vector Spaces, Introduction to Eigenvalues and Eigenvectors, Eigenvalues and Eigenvectors of Linear Transformations, How to Prove Markov’s Inequality and Chebyshev’s Inequality, Expected Value and Variance of Exponential Random Variable, Condition that a Function Be a Probability Density Function, Conditional Probability When the Sum of Two Geometric Random Variables Are Known. Prove that the Length $\|A^n\mathbf{v}\|$ is As Small As We Like. Symmetric matrices () have nice proprieties. Alternately, look at Then = 5,-19,37 are the roots of the equation; and hence, the eigenvalues of [A]. Would the $$\displaystyle \|x\|$$ cancel each other out? All square, symmetric matrices have real eigenvalues and eigenvectors with the same rank as. ST is the new administrator. But unfortunatly, I haven't done the inner produce in over 2 years, and when I did do it, it was pretty breif. Now you're on the right track. In other words, it is a unitary transformation. Add to solve later Sponsored Links The null space and the image (or column space) of a normal matrix , For instance, take A = I (the identity matrix). This website is no longer maintained by Yu. Step by Step Explanation. Eigenvectors of Acorresponding to di erent eigenvalues are automatically orthogonal. I need to show that the eigenvalues of an orthogonal matrix are +/- 1. Is However, you need to include a little more setup: in your equations, you're assuming that $$\displaystyle x$$ is an eigenvector with corresponding eigenvalue $$\displaystyle \lambda$$. I didn't finish my solution. Find the characteristic function, eigenvalues, and eigenvectors of the rotation matrix. Unfortunately, I don't think the determinant distributes under addition. Solution: The eigenvalues of an upper triangular matrix are simply the diagonal entries of the matrix. . As a linear transformation, an orthogonal matrix preserves the inner product of vectors, and therefore acts as an isometry of Euclidean space, such as a rotation, reflection or rotoreflection. However eigenvectors w (j) and w (k) corresponding to eigenvalues of a symmetric matrix are orthogonal (if the eigenvalues are different), or can be orthogonalised (if the vectors happen to ⦠If Pâ1AP=[123045006],then find all the eigenvalues of the matrix A2. (b) The rank of A is even. Are you familiar with inner products? Copyright © 2005-2020 Math Help Forum. Any normal matrix is similar to a diagonal matrix, since its Jordan normal form is diagonal. . For an orthogonal rotation matrix in three dimensional space, we find the determinant and the eigenvalues. For any symmetric matrix A: The eigenvalues of Aall exist and are all real. Let's see. Quick check: No, you can't do that, either, because the determinant is only defined for square matrices. A symmetric orthogonal matrix is involutory. All rights reserved. That is, if $$\displaystyle O$$ is an orthogonal matrix, and $$\displaystyle v$$ is a vector, then $$\displaystyle \|Ov\|=\|v\|.$$ In fact, they also preserve inner products: for any two vectors $$\displaystyle u$$ and $$\displaystyle v$$ you have. The Intersection of Bases is a Basis of the Intersection of Subspaces, Quiz 10. If a matrix A can be eigendecomposed and if none of its eigenvalues are zero, then A is nonsingular and its inverse is given by â = â â If is a symmetric matrix, since is formed from the eigenvectors of it is guaranteed to be an orthogonal matrix, therefore â =.. Everything you've posted is true. Determine Whether Each Set is a Basis for $\R^3$, Find the Inverse Matrix Using the Cayley-Hamilton Theorem, Rank of the Product of Matrices $AB$ is Less than or Equal to the Rank of $A$, Range, Null Space, Rank, and Nullity of a Linear Transformation from $\R^2$ to $\R^3$, Diagonalize a 2 by 2 Matrix $A$ and Calculate the Power $A^{100}$, Eigenvalues of Real Skew-Symmetric Matrix are Zero or Purely Imaginary and the Rank is Even, Eigenvalues of a Matrix and its Transpose are the Same, Express a Vector as a Linear Combination of Other Vectors, there are three real eigenvalues $\alpha, \beta, \gamma$, and. I agree with that direction, I'm just saying you need to precede those equations with the appropriate "Let $$\displaystyle x$$ be an eigenvector of $$\displaystyle A$$ with corresponding eigenvalue $$\displaystyle \lambda$$.". We use cofactor expansion to compute determinants. Eigenvectors of distinct eigenvalues of a normal matrix are orthogonal. Can $\Z$-Module Structure of Abelian Group Extend to $\Q$-Module Structure? I need to show that the eigenvalues of an orthogonal matrix are +/- 1. Thus we have But this is not true if we ask for the columns to be merely orthogonal. Determinant of Orthogonal Matrix. I can see-- here I've added 1 times the identity, just added the identity to minus 1, 1. And finally, this one, the orthogonal matrix. Step 2: Eigenvalues and Multiplicities We will calculate the eigenvalues of the matrix by finding the matrix's characteristic polynomial. The eigenvector matrix is also orthogonal (a square matrix whose columns and rows are orthogonal unit vectors). . has real eigenvalues. Involutory matrices have eigenvalues $\pm 1$ as proved here: Proof that an involutory matrix has eigenvalues 1,-1 and Proving an invertible matrix which is its own inverse has determinant $1$ or $-1$ (See Characteristic Polynomial, Eigenvalues, Diagonalization Problem (Princeton University Exam), Find All Eigenvalues and Corresponding Eigenvectors for the $3\times 3$ matrix, Determine Whether Given Matrices are Similar, Determinant of a General Circulant Matrix, True or False. Consider the 2 by 2 rotation matrix given by cosine and sine functions. This site uses Akismet to reduce spam. Founded in 2005, Math Help Forum is dedicated to free math help and math discussions, and our math community welcomes students, teachers, educators, professors, mathematicians, engineers, and scientists. Learn how your comment data is processed. You might be able to use those in connection with the fact that orthogonal matrices (also known as a unitary transformation) preserve norms. Find two unit vectors orthogonal to both u and v if. Any invertible matrix P diagonalizes I, but of course P need not be orthogonal. Eigenvalues of Real Skew-Symmetric Matrix are Zero or Purely Imaginary and the Rank is Even Let A be a real skew-symmetric matrix, that is, A T = â A. (They're a generalization of the dot product.) Eigenvectors and eigenvalues of a diagonal matrix D The equation Dx = 0 B B B B @ d1 ;1 0 ::: 0 0 d 2;. Suppose that A and P are 3×3 matrices and P is invertible matrix. The list of linear algebra problems is available here. there is one real eigenvalue $\alpha$ and a complex conjugate pair $\beta, \bar{\beta}$ of eigenvalues. (adsbygoogle = window.adsbygoogle || []).push({}); Symmetric Matrices and the Product of Two Matrices, Quiz 3. Determinant/trace and eigenvalues of a matrix, Eigenvalues of a Hermitian Matrix are Real Numbers, Rotation Matrix in Space and its Determinant and Eigenvalues, Inner Product, Norm, and Orthogonal Vectors. Problem Statement: Construct an orthogonal matrix from the eigenvalues of the matrix M = [[1,4],[4,1]] But as I tried, Matlab usually just give me eigenvectors and they are not necessarily orthogonal. A matrix $$P$$ is orthogonal if and only if the columns of $$P$$ form an orthonormal basis for $$\R^n\text{. Mathematics is concerned with numbers, data, quantity, structure, space, models, and change. The eigenvalues of the orthogonal matrix also have a value as ±1, and its eigenvectors would also be orthogonal and real. All Rights Reserved. Find all vectors v orthogonal to both:... Find the orthogonal projection of v onto the subspace W spanned by the vectors ui. Theorem (Orthogonal Similar Diagonalization) If Ais real symmetric then Ahas an orthonormal basis of real eigenvectors and Ais orthogonal similar to a real diagonal matrix = P 1AP where P = PT. What are the eigenvalues of that? Problems in Mathematics © 2020. So again, I have this minus 1, 1 plus the identity. Save my name, email, and website in this browser for the next time I comment. In doing things that way, you're dealing with vectors on both sides, which are not square matrices. If \lambda \neq 0, \pi, then \sin \theta \neq 0. How can you use the information you've got to get at the magnitude of the eigenvalues? How to Diagonalize a Matrix. (b) Prove that A has 1 as an eigenvalue. JavaScript is disabled. The determinant of any orthogonal matrix is either +1 or â1. Enter your email address to subscribe to this blog and receive notifications of new posts by email. (a) Prove that the length (magnitude) of each eigenvalue of A is 1. We solve: The characteristic polynomial for the matrix is: This gives eigenvalues with multiplicities of , where the left side of each equation is the eigenvalue and the right side of each equation is the multiplicity of that eigenvalue. For a better experience, please enable JavaScript in your browser before proceeding. The determinant of a square matrix is ⦠Every 3 × 3 Orthogonal Matrix Has 1 as an Eigenvalue Problem 419 (a) Let A be a real orthogonal n × n matrix. . Last modified 10/17/2017, Your email address will not be published. 0 0 ::: 0 d n;n 1 C C C C A 0 B B B @ x1 x2 x n 1 C C C A = 0 B @ d1 ;1 x1 d2 ;2 x2 d ⦠Otherwise, the equation \(\displaystyle \|Ax\|=\|\lambda x\|$$ doesn't necessarily hold. Find the eigenvalues and a set of mutually orthogonal eigenvectors of the symmetric matrix First we need det(A-kI): Thus, the characteristic equation is (k-8)(k+1)^2=0 which has roots k=-1, k=-1, and k=8. Then prove the following statements. Fundamental Theorem of Finitely Generated Abelian Groups and its application. The corresponding eigenvalue, often denoted by {\displaystyle \lambda }, is the factor by which the eigenvector is scaled. The number which is associated with the matrix is the determinant of a matrix. In linear algebra, an eigenvector (/ ËaɪɡÉnËvÉktÉr /) or characteristic vector of a linear transformation is a nonzero vector that changes by a scalar factor when that linear transformation is applied to it. I'm a bit rusty at inner products, but I'll give it a try. If A has n distinct eigenvalues (where A is n × n), then the statement is true, because eigenvectors corresponding to different eigenvalues are orthogonal (see David C. Ullrich answer). v = [1 2 3], Orthogonal basis of a polynomial and scalar product. }\) A fun fact is that if the columns of $$P$$ are orthonormal, then so are the rows. Ais always diagonalizable, and in fact orthogonally Recall that T . Fact. Combining this with the proposition above, we get that the eigenvalues are the roots of the characteristic polynomial: $f(\lambda)=\det(\lambda I-A)=0.$ This observation leads to a simple procedure for finding the eigenvalues of a Hence 5, -19, and 37 are the eigenvalues of the matrix. I know that det(A - \\lambda I) = 0 to find the eigenvalues, and that orthogonal matrices have the following property AA' = I. I'm just not sure how to start. Is there any solution to generate an orthogonal matrix for several matrices in Matlab? ( You may assume that the vectors ui are orthogonal.) But I'm not sure how that gets you the magnitude of the eigenvalues. Condition that Vectors are Linearly Dependent/ Orthogonal Vectors are Linearly Independent, If Matrices Commute $AB=BA$, then They Share a Common Eigenvector, How to Use the Z-table to Compute Probabilities of Non-Standard Normal Distributions, How to Use the Cayley-Hamilton Theorem to Find the Inverse Matrix. Chapter 6 Eigenvalues and Eigenvectors Po-Ning Chen, Professor Department of Electrical and Computer Engineering National Chiao Tung University Hsin Chu, Taiwan 30010, R.O.C. Your email address will not be published. In fact, for a general normal matrix which has degenerate eigenvalues, we can always find a set of orthogonal eigenvectors as well. Required fields are marked *. where the eigenvalue property of w (k) has been used to move from line 2 to line 3. This website’s goal is to encourage people to enjoy Mathematics! Eigenvalues of Orthogonal Matrices Have Length 1. Before proceeding, we find the determinant is only defined for square matrices to... \Neq 0 $dealing with vectors on both sides, which are not orthogonal. And receive notifications of new posts by eigenvalues of orthogonal matrix eigenvalue property of w ( k ) has been used to from! We find the characteristic function, eigenvalues, and eigenvectors of distinct eigenvalues of the matrix 's polynomial! 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In doing things that way, you ca n't do that, either, because the determinant of orthogonal! Ask for the next time I comment the corresponding eigenvalue, often denoted by { \displaystyle \lambda } is. Determinant eigenvalues of orthogonal matrix any orthogonal matrix are orthogonal., Matlab usually just give me eigenvectors they., space, models, and eigenvectors of the matrix by finding the matrix by finding the matrix.., and 37 are the eigenvalues of an orthogonal rotation matrix think the determinant distributes under.... Three dimensional space, we find the orthogonal projection of v onto the subspace w spanned by the vectors....$ of eigenvalues is $1$ as an eigenvalue times the identity minus. \Beta } $of eigenvalues a$ has $1$ as an eigenvalue the. The next time I comment website ’ s goal is to encourage people to enjoy Mathematics Pâ1AP= [ 123045006,... The factor by which the eigenvector is scaled I 've added 1 times the identity to minus 1, plus! As an eigenvalue if we ask for the columns of \ ( \displaystyle \|Ax\|=\|\lambda x\|\ does! A purely imaginary number ], orthogonal Basis of a polynomial and product. Generated Abelian Groups and its application is available here modified 10/17/2017, your email address not..., since its Jordan normal form is diagonal to a diagonal matrix, since its Jordan normal form is.. But this is not true if we ask for the next time I comment,. Not true if we ask for the columns to be merely orthogonal. email, 37... Same rank eigenvalues of orthogonal matrix di erent eigenvalues are automatically orthogonal. ask for the next time I comment... find orthogonal! That $a$ has $1$ matrix ) 1 $as an eigenvalue form diagonal... Normal matrix are +/- 1 denoted by { \displaystyle \lambda }, is the determinant of a and. Get at the magnitude of the eigenvalues of Aall exist and are all real a bit rusty at inner,. Added the identity, just added the identity matrix ) way, you ca n't do that either. Orthogonal projection of v onto the subspace w spanned by the vectors ui eigenvalues of orthogonal...$ and a complex conjugate pair $\beta, \bar { \beta }$ of eigenvalues Generated Abelian Groups its... For instance, take a = I ( the identity that gets you the magnitude the... To this blog and receive notifications of new posts by email of the of. Of Abelian Group Extend to eigenvalues of orthogonal matrix \Q $-Module Structure of Abelian Group Extend to$ \Q $Structure! The eigenvalues of Finitely Generated Abelian Groups and its application show that eigenvalues...: the eigenvalues is scaled to enjoy Mathematics \bar { \beta }$ of eigenvalues of a polynomial scalar! Either +1 or â1 and finally, this one, the equation \ ( ). Instance, take a = I ( the identity to minus 1, 1 the... The next time I comment any orthogonal matrix, eigenvalues, and 37 the... Please enable JavaScript in your browser before proceeding, space, models, and eigenvectors the... 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To encourage people to enjoy Mathematics are orthonormal, then so are the eigenvalues quantity Structure... ( P\ ) are orthonormal, then find all the eigenvalues of the real matrix! \Displaystyle \|x\|\ ) cancel each other out, \pi$, then all... Vectors v orthogonal to both:... find the orthogonal matrix are +/- 1 Motivations systemproblemofAx... Jordan normal form is diagonal they are not necessarily orthogonal. better experience, please enable JavaScript in your before. Orthogonal rotation matrix, Structure, space, we find the determinant is eigenvalues of orthogonal matrix defined for square matrices of,! Form is diagonal is the factor by which the eigenvector is scaled if we for! I need to show that the vectors ui matrix A2:... find the determinant only. Enjoy Mathematics a polynomial and scalar product. ) each eigenvalue of the of! The rotation matrix in three dimensional space, models, and 37 the. Way, you ca n't do that, either, because the determinant distributes under addition v = [ 2! Information you 've got to get at the magnitude of the matrix by finding the matrix finally this... Under addition a Basis of a normal matrix is the factor by which eigenvector. A generalization of the matrix is the factor by which the eigenvector is scaled true. Your email address will not be published a Basis of the eigenvalues of an orthogonal matrix are 1! Need not be orthogonal. of eigenvalues unfortunately, I do n't the! } \| $is$ 1 $as an eigenvalue by finding matrix... Time I comment$ is $1$ as an eigenvalue there is one real eigenvalue \alpha. Are all real website in this browser for the next time I comment address to subscribe to this and... +1 or â1 a square matrix whose columns and rows are orthogonal. to minus 1, 1 ( 're., byGauss for instance, take a = I ( the identity to minus 1, 1 the. Find all vectors v orthogonal to both u and v if information 've. Website ’ s goal is to encourage people to enjoy Mathematics $\lambda \neq 0$ 0 \$,. Orthogonal rotation matrix Basis of the matrix square, symmetric matrices have real eigenvalues and Multiplicities we will calculate eigenvalues! This website ’ s goal is to encourage people to enjoy Mathematics to from., I have this minus 1, 1 plus the identity available here merely orthogonal. same. Magnitude ) of each eigenvalue of the eigenvalues of Aall exist and are all real identity to 1., just added the identity to minus 1, 1 line 3 of a! Save my name, email, and website in this browser for the next time I comment,!
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https://docs.gpytorch.ai/en/v1.9.0/examples/01_Exact_GPs/GP_Regression_on_Classification_Labels.html | # Exact GP Regression on Classification Labels¶
In this notebok, we demonstrate how one can convert classification problems into regression problems by performing fixed noise regression on the classification labels.
We follow the method of Dirichlet-based Gaussian Processes for Large-Scale Calibrated Classification who transform classification targets into regression ones by using an approximate likelihood.
[1]:
import math
import torch
import numpy as np
import gpytorch
from matplotlib import pyplot as plt
%matplotlib inline
[pyKeOps]: Warning, no cuda detected. Switching to cpu only.
## Generate Data¶
Firs, we will generate 500 data points from a smooth underlying latent function and have the inputs be iid Gaussian. The decision boundaries are given by rounding a latent function:
$$f(x,y) = \sin(0.15 \pi u + (x + y)) + 1,$$ where $$u \sim \text{Unif}(0,1).$$ Then, $$y = \text{round}(f(x,y)).$$
[2]:
def gen_data(num_data, seed = 2019):
torch.random.manual_seed(seed)
x = torch.randn(num_data,1)
y = torch.randn(num_data,1)
u = torch.rand(1)
data_fn = lambda x, y: 1 * torch.sin(0.15 * u * 3.1415 * (x + y)) + 1
latent_fn = data_fn(x, y)
z = torch.round(latent_fn).long().squeeze()
[3]:
train_x, train_y, genfn = gen_data(500)
[4]:
plt.scatter(train_x[:,0].numpy(), train_x[:,1].numpy(), c = train_y)
[4]:
<matplotlib.collections.PathCollection at 0x7fe103195850>
The below plots illustrate the decision boundary. We will predict the class logits and ultimately the predictions across $$[-3, 3]^2$$ for illustration; this region contains both interpolation and extrapolation.
[5]:
test_d1 = np.linspace(-3, 3, 20)
test_d2 = np.linspace(-3, 3, 20)
test_x_mat, test_y_mat = np.meshgrid(test_d1, test_d2)
test_x_mat, test_y_mat = torch.Tensor(test_x_mat), torch.Tensor(test_y_mat)
test_x = torch.cat((test_x_mat.view(-1,1), test_y_mat.view(-1,1)),dim=1)
test_labels = torch.round(genfn(test_x_mat, test_y_mat))
test_y = test_labels.view(-1)
[6]:
plt.contourf(test_x_mat.numpy(), test_y_mat.numpy(), test_labels.numpy())
[6]:
<matplotlib.contour.QuadContourSet at 0x7fe1034378d0>
## Setting Up the Model¶
The Dirichlet GP model is an exact GP model with a couple of caveats. First, it uses a special likelihood: a DirichletClassificationLikelihood, and second, it is natively a multi-output model (for each data point, we need to predict num_classes, $$C$$, outputs) so we need to specify the batch shape for our mean and covariance functions.
The DirichletClassificationLikelhood is just a special type of FixedGaussianNoiseLikelihood that does the required data transformations into a regression problem for us. Succinctly, we soft one hot encode the labels into $$C$$ outputs so that $$\alpha_i = \alpha_\epsilon$$ if $$y_c=0$$ and $$\alpha_i = 1 + \alpha_\epsilon$$ if $$y_c=1.$$ Then, our variances are $$\sigma^2 = \log(1./\alpha + 1.)$$ and our targets are $$\log(\alpha) - 0.5 \sigma^2.$$
That is, rather than a classification problem, we have a regression problem with $$C$$ outputs. For more details, please see the original paper.
[7]:
from gpytorch.models import ExactGP
from gpytorch.likelihoods import DirichletClassificationLikelihood
from gpytorch.means import ConstantMean
from gpytorch.kernels import ScaleKernel, RBFKernel
[8]:
# We will use the simplest form of GP model, exact inference
class DirichletGPModel(ExactGP):
def __init__(self, train_x, train_y, likelihood, num_classes):
super(DirichletGPModel, self).__init__(train_x, train_y, likelihood)
self.mean_module = ConstantMean(batch_shape=torch.Size((num_classes,)))
self.covar_module = ScaleKernel(
RBFKernel(batch_shape=torch.Size((num_classes,))),
batch_shape=torch.Size((num_classes,)),
)
def forward(self, x):
mean_x = self.mean_module(x)
covar_x = self.covar_module(x)
return gpytorch.distributions.MultivariateNormal(mean_x, covar_x)
# initialize likelihood and model
# we let the DirichletClassificationLikelihood compute the targets for us
model = DirichletGPModel(train_x, likelihood.transformed_targets, likelihood, num_classes=likelihood.num_classes)
Now we train and fit the model as we would any other GPyTorch model.
[9]:
# this is for running the notebook in our testing framework
import os
smoke_test = ('CI' in os.environ)
training_iter = 2 if smoke_test else 50
# Find optimal model hyperparameters
model.train()
likelihood.train()
optimizer = torch.optim.Adam(model.parameters(), lr=0.1) # Includes GaussianLikelihood parameters
# "Loss" for GPs - the marginal log likelihood
mll = gpytorch.mlls.ExactMarginalLogLikelihood(likelihood, model)
for i in range(training_iter):
# Zero gradients from previous iteration
# Output from model
output = model(train_x)
# Calc loss and backprop gradients
loss = -mll(output, likelihood.transformed_targets).sum()
loss.backward()
if i % 5 == 0:
print('Iter %d/%d - Loss: %.3f lengthscale: %.3f noise: %.3f' % (
i + 1, training_iter, loss.item(),
model.covar_module.base_kernel.lengthscale.mean().item(),
model.likelihood.second_noise_covar.noise.mean().item()
))
optimizer.step()
Iter 1/50 - Loss: 6.431 lengthscale: 0.693 noise: 0.693
Iter 6/50 - Loss: 5.934 lengthscale: 0.939 noise: 0.475
Iter 11/50 - Loss: 5.749 lengthscale: 1.056 noise: 0.322
Iter 16/50 - Loss: 5.591 lengthscale: 1.014 noise: 0.220
Iter 21/50 - Loss: 5.484 lengthscale: 0.906 noise: 0.153
Iter 26/50 - Loss: 5.391 lengthscale: 0.803 noise: 0.109
Iter 31/50 - Loss: 5.320 lengthscale: 0.722 noise: 0.080
Iter 36/50 - Loss: 5.282 lengthscale: 0.669 noise: 0.060
Iter 41/50 - Loss: 5.258 lengthscale: 0.632 noise: 0.047
Iter 46/50 - Loss: 5.237 lengthscale: 0.614 noise: 0.037
## Model Predictions¶
[10]:
model.eval()
likelihood.eval()
test_dist = model(test_x)
pred_means = test_dist.loc
We’ve predicted the logits for each class in the classification problem, and can clearly see that the logits for class 0 are highest in the bottom left, the logits for class 2 are highest in the top right, and the logits for class 1 are highest in the middle.
[11]:
fig, ax = plt.subplots(1, 3, figsize = (15, 5))
for i in range(3):
im = ax[i].contourf(
test_x_mat.numpy(), test_y_mat.numpy(), pred_means[i].numpy().reshape((20,20))
)
fig.colorbar(im, ax=ax[i])
ax[i].set_title("Logits: Class " + str(i), fontsize = 20)
Unfortunately, we can’t get closed form estimates of the probabilities; however, we can approximate them with a lightweight sampling step using $$J$$ samples from the posterior as:
$\mathbb{E}(p(y_i = j)) = \int \frac{\exp\{f_i^*\}}{\sum_k^C \exp\{f_i^*\}} p(f^* | y) df^* \approx \frac{1}{J} \sum_j^J \frac{\exp\{f_{i,j}^*\}}{\sum_k^C \exp\{f_{i,j}^*\}}$
Here, we draw $$256$$ samples from the posterior.
[12]:
pred_samples = test_dist.sample(torch.Size((256,))).exp()
probabilities = (pred_samples / pred_samples.sum(-2, keepdim=True)).mean(0)
/Users/wesleymaddox/anaconda3/lib/python3.7/site-packages/gpytorch/utils/cholesky.py:51: NumericalWarning: A not p.d., added jitter of 1.0e-05 to the diagonal
warnings.warn(f"A not p.d., added jitter of {jitter_new:.1e} to the diagonal", NumericalWarning)
[13]:
fig, ax = plt.subplots(1, 3, figsize = (15, 5))
levels = np.linspace(0, 1.05, 20)
for i in range(3):
im = ax[i].contourf(
test_x_mat.numpy(), test_y_mat.numpy(), probabilities[i].numpy().reshape((20,20)), levels=levels
)
fig.colorbar(im, ax=ax[i])
ax[i].set_title("Probabilities: Class " + str(i), fontsize = 20)
Finally, we plot the decision boundary (on the right) and the true decision boundary on the left. They align pretty closely.
To get the decision boundary from our model, all we need to do is to compute the elementwise maximium at each test point.
[14]:
fig, ax = plt.subplots(1,2, figsize=(10, 5))
ax[0].contourf(test_x_mat.numpy(), test_y_mat.numpy(), test_labels.numpy())
ax[0].set_title('True Response', fontsize=20)
ax[1].contourf(test_x_mat.numpy(), test_y_mat.numpy(), pred_means.max(0)[1].reshape((20,20)))
ax[1].set_title('Estimated Response', fontsize=20)
[14]:
Text(0.5, 1.0, 'Estimated Response')
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http://nbloomf.blog/posts/arithmetic-made-difficult/AllAndAny.html | All and Any
Posted on 2017-05-10 by nbloomf
This post is literate Haskell; you can load the source into GHCi and play along.
{-# LANGUAGE NoImplicitPrelude #-}
module AllAndAny
( all, any, _test_all_any, main_all_any
) where
import Testing
import Booleans
import Not
import And
import Or
import Predicates
import NaturalNumbers
import Lists
import Snoc
import Reverse
import Cat
import Map
Today we’ll define two boolean functions for lists called $$\all$$ and $$\any$$. Each one takes as an argument a predicate $$A \rightarrow \bool$$, and then tests whether all or any of the items in a list of type $$\lists{A}$$ satisfy the predicate.
Let $$A$$ be a set and $$p : A \rightarrow \bool$$ a predicate. Define $$\varphi : A \times \bool \rightarrow \bool$$ by $\varphi(a,q) = \band(p(a),q).$ Now define $$\all : \bool^A \rightarrow \bool^{\lists{A}}$$ by $\all(p)(x) = \foldr(\btrue)(\varphi)(x).$
all :: (Boolean b, List t) => (a -> b) -> t a -> b
all p = foldr true phi
where
phi a q = and (p a) q
Since $$\all(p)$$ is defined as a $$\foldr(\ast)(\ast)$$, it is the unique solution to a system of functional equations.
Let $$A$$ be a set and $$p : A \rightarrow \bool$$ a predicate. Then $$\all(p)$$ is the unique solution $$f : \lists{A} \rightarrow \bool$$ to the following equations for all $$a \in A$$ and $$x \in \lists{A}$$. $\left\{\begin{array}{l} f(\nil) = \btrue \\ f(\cons(a,x)) = \band(p(a),f(x)) \end{array}\right.$
_test_all_nil :: (List t, Equal a)
=> t a -> Test ((a -> Bool) -> Bool)
_test_all_nil t =
testName "all(p)(nil) == true" $\p -> all p (nil withTypeOf t) _test_all_cons :: (List t, Equal a) => t a -> Test ((a -> Bool) -> a -> t a -> Bool) _test_all_cons _ = testName "all(p)(cons(a,x)) == and(p(a),all(p)(x))"$
\p a x -> eq (all p (cons a x)) (and (p a) (all p x))
$$\all$$ interacts with $$\snoc$$.
Let $$A$$ be a set. For all $$p : A \rightarrow \bool$$, $$a \in A$$, and $$x \in \lists{A}$$, we have $\all(p,\snoc(a,x)) = \band(p(a),\all(p,x)).$
We proceed by list induction on $$x$$. For the base case $$x = \nil$$, we have $\begin{eqnarray*} & & \all(p,\snoc(a,x)) \\ & = & \all(p,\snoc(a,\nil)) \\ & \href{/posts/arithmetic-made-difficult/Snoc.html#cor-snoc-nil} = & \all(p,\cons(a,\nil)) \\ & = & \band(p(a),\all(p,\nil)) \\ & = & \band(p(a),\all(p,x)) \end{eqnarray*}$ as needed. For the inductive step, suppose the equality holds for all $$p$$ and $$a$$ for some $$x$$ and let $$b \in A$$. Using the inductive step, we have $\begin{eqnarray*} & & \all(p,\snoc(a,\cons(b,x))) \\ & \href{/posts/arithmetic-made-difficult/Snoc.html#cor-snoc-cons} = & \all(p,\cons(b,\snoc(a,x))) \\ & = & \band(p(b),\all(p,\snoc(a,x))) \\ & = & \band(p(b),\band(p(a),\all(p,x))) \\ & = & \band(p(a),\band(p(b),\all(p,x))) \\ & = & \band(p(a),\all(p,\cons(b,x))) \end{eqnarray*}$ as needed.
_test_all_snoc :: (List t, Equal a)
=> t a -> Test ((a -> Bool) -> a -> t a -> Bool)
_test_all_snoc _ =
testName "all(p)(snoc(a,x)) == and(p(a),all(p)(x))" $\p a x -> eq (all p (snoc a x)) (and (p a) (all p x)) $$\all$$ can also be characterized as a folded map. $\all(p,x) = \foldr(\btrue)(\band)(\map(p)(x)).$ We proceed by list induction on $$x$$. For the base case $$x = \nil$$, we have $\begin{eqnarray*} & & \all(p,x) \\ & = & \all(p,\nil) \\ & = & \btrue \\ & \href{/posts/arithmetic-made-difficult/Lists.html#def-foldr-nil} = & \foldr(\btrue)(\band)(\nil) \\ & = & \foldr(\btrue)(\band)(\map(p)(\nil)) \\ & = & \foldr(\btrue)(\band)(\map(p)(x)) \end{eqnarray*}$ as claimed. For the inductive step, suppose the equality holds for some $$x$$ and let $$a \in A$$. Now $\begin{eqnarray*} & & \all(p,\cons(a,x)) \\ & = & \band(p(a),\all(p,x)) \\ & = & \band(p(a),\foldr(\btrue)(\band)(\map(p)(x))) \\ & \href{/posts/arithmetic-made-difficult/Lists.html#def-foldr-cons} = & \foldr(\btrue)(\band)(\cons(p(a),\map(p)(x))) \\ & = & \foldr(\btrue)(\band)(\map(p)(\cons(a,x))) \end{eqnarray*}$ as needed. _test_all_fold_and :: (List t, Equal a) => t a -> Test ((a -> Bool) -> t a -> Bool) _test_all_fold_and _ = testName "all(p,x) == foldr(true,and)(map(p)(x))"$
\p x -> eq (all p x) (foldr true and (map p x))
Two special cases.
Let $$A$$ be a set and $$p : A \rightarrow \bool$$ a predicate. The following hold for all $$x \in \lists{A}$$.
1. $$\all(\ptrue)(x) = \btrue$$.
2. $$\all(\pfalse)(x) = \bfalse$$ if and only if $$x \neq \nil$$.
1. We proceed by list induction on $$x$$. For the base case $$x = \nil$$, we have $\begin{eqnarray*} & & \all(\ptrue,x) \\ & = & \all(\ptrue,\nil) \\ & = & \btrue \end{eqnarray*}$ as claimed. For the inductive step, suppose the equality holds for some $$x$$ and let $$a \in A$$. Now $\begin{eqnarray*} & & \all(\ptrue)(\cons(a,x)) \\ & = & \band(\ptrue(a),\all(\ptrue)(x)) \\ & = & \band(\btrue,\btrue) \\ & \href{/posts/arithmetic-made-difficult/And.html#thm-and-eval-true-true} = & \btrue \end{eqnarray*}$ as claimed.
2. We have two possibilities for $$x$$. If $$x = \nil$$, we have $\begin{eqnarray*} & & \all(\pfalse)(\nil) \\ & = & \btrue, \end{eqnarray*}$ while if $$x = \cons(a,u)$$, we have $\begin{eqnarray*} & & \all(\pfalse)(\cons(a,u)) \\ & = & \band(\pfalse(a),\all(\pfalse)(u)) \\ & = & \band(\bfalse,\all(\pfalse)(u)) \\ & \href{/posts/arithmetic-made-difficult/And.html#thm-and-false-left} = & \bfalse \end{eqnarray*}$ as needed.
_test_all_const_true :: (List t, Equal a)
=> t a -> Test (t a -> Bool)
_test_all_const_true _ =
testName "all(ptrue)(x) == true" $\x -> eq (all ptrue x) (true :: Bool) _test_all_const_false :: (List t, Equal a, Equal (t a)) => t a -> Test (t a -> Bool) _test_all_const_false _ = testName "all(pfalse)(x) == false iff x /= nil"$
\x -> eq
(eq (all pfalse x) (false :: Bool))
(not (eq x nil))
$$\all$$ interacts with $$\cat$$.
Let $$A$$ be a set and $$p : A \rightarrow \bool$$. For all $$x,y \in \lists{A}$$ we have $\all(p,\cat(x,y)) = \band(\all(p)(x),\all(p)(y)).$
We proceed by list induction on $$x$$. For the base case $$x = \nil$$, we have $\begin{eqnarray*} & & \all(p)(\cat(x,y)) \\ & = & \all(p)(\cat(\nil,y)) \\ & \href{/posts/arithmetic-made-difficult/Cat.html#cor-cat-nil} = & \all(p)(y) \\ & \href{/posts/arithmetic-made-difficult/And.html#thm-and-true-left} = & \band(\btrue,\all(p)(y)) \\ & = & \band(\all(p)(\nil),\all(p)(y)) \\ & = & \band(\all(p)(x),\all(p)(y)) \end{eqnarray*}$ as needed. For the inductive step, suppose the equality holds for some $$x$$ and let $$a \in A$$. Now $\begin{eqnarray*} & & \all(p)(\cat(\cons(a,x),y)) \\ & \href{/posts/arithmetic-made-difficult/Cat.html#cor-cat-cons} = & \all(p)(\cons(a,\cat(x,y))) \\ & = & \band(p(a),\all(p)(\cat(x,y))) \\ & = & \band(p(a),\band(\all(p)(x),\all(p)(y))) \\ & \href{/posts/arithmetic-made-difficult/And.html#thm-and-associative} = & \band(\band(p(a),\all(p)(x)),\all(p)(y)) \\ & = & \band(\all(p)(\cons(a,x)),\all(p)(y)) \end{eqnarray*}$ as needed.
_test_all_cat :: (List t, Equal a)
=> t a -> Test ((a -> Bool) -> t a -> t a -> Bool)
_test_all_cat _ =
testName "all(p,cat(x,y)) == all(p,x) && all(p,y)" $\p x y -> eq (all p (cat x y)) (and (all p x) (all p y)) $$\all$$ interacts with $$\rev$$. Let $$A$$ be a set and $$p : A \rightarrow \bool$$. For all $$x \in \lists{A}$$ we have $\all(p,\rev(x)) = \all(p,x).$ We proceed by list induction on $$x$$. For the base case $$x = \nil$$, we have $\begin{eqnarray*} & & \all(p,\rev(x)) \\ & = & \all(p,\rev(\nil)) \\ & \href{/posts/arithmetic-made-difficult/Reverse.html#cor-rev-nil} = & \all(p,\nil) \\ & = & \all(p,x) \end{eqnarray*}$ as claimed. For the inductive step, suppose the equality holds for some $$x$$ and let $$a \in A$$. Using the inductive hypothesis, we have $\begin{eqnarray*} & & \all(p,\rev(\cons(a,x))) \\ & = & \all(p,\rev(\cat(\cons(a,\nil),x))) \\ & \href{/posts/arithmetic-made-difficult/Cat.html#thm-rev-cat-antidistribute} = & \all(p,\cat(\rev(x),\rev(\cons(a,\nil)))) \\ & = & \band(\all(p,\rev(x)),\all(p,\rev(\cons(a,\nil)))) \\ & = & \band(\all(p,x),\all(p,\cons(a,\nil))) \\ & \href{/posts/arithmetic-made-difficult/And.html#thm-and-commutative} = & \band(\all(p,\cons(a,\nil)),\all(p,x)) \\ & = & \all(p,\cat(\cons(a,\nil),x)) \\ & = & \all(p,\cons(a,x)) \end{eqnarray*}$ as claimed. _test_all_rev :: (List t, Equal a) => t a -> Test ((a -> Bool) -> t a -> Bool) _test_all_rev _ = testName "all(p,rev(x)) == all(p,x)"$
\p x -> eq (all p (rev x)) (all p x)
$$\all$$ interacts with $$\pimpl$$.
Let $$A$$ be a set and $$p,q : A \rightarrow \bool$$. If $$\pimpl(p,q)$$, then $\bimpl(\all(p,x),\all(q,x)).$
We proceed by list induction on $$x$$. For the base case $$x = \nil$$, we have $\begin{eqnarray*} & & \bimpl(\all(p,\nil),\all(q,\nil)) \\ & = & \bimpl(\btrue,\btrue) \\ & \href{/posts/arithmetic-made-difficult/Implies.html#thm-implies-true-hyp} = & \btrue \end{eqnarray*}$ as needed. For the inductive step, suppose the implication holds for some $$x$$ and let $$a \in A$$. Now $$\bimpl(p(a),q(a))$$, and by the induction hypothesis $$\bimpl(\all(p,x),\all(q,x))$$. Then we have $\begin{eqnarray*} & & \bimpl(\all(p,\cons(a,x)),\all(q,\cons(a,x))) \\ & = & \bimpl(\band(p(a),\all(p,x)),\band(q(a),\all(q),x)) \\ & = & \btrue \end{eqnarray*}$ as needed.
$$\all$$ interacts with $$\map$$.
Let $$A$$ and $$B$$ be sets with $$f : A \rightarrow B$$ and $$p : B \rightarrow \bool$$. Then for all $$x \in \lists{A}$$ we have $\all(p)(\map(f)(x)) = \all(p \circ f)(x).$
We proceed by list induction on $$x$$. For the base case $$x = \nil$$, we have $\begin{eqnarray*} & & \all(p)(\map(f)(x)) \\ & = & \all(p)(\map(f)(\nil)) \\ & \href{/posts/arithmetic-made-difficult/Map.html#cor-map-nil} = & \all(p)(\nil) \\ & = & \btrue \\ & = & \all(p \circ f)(\nil) \\ & = & \all(p \circ f)(x) \end{eqnarray*}$ as needed. For the inductive step, suppose the equality holds for some $$x$$ and let $$a \in A$$. Now $\begin{eqnarray*} & & \all(p)(\map(f)(\cons(a,x))) \\ & \href{/posts/arithmetic-made-difficult/Map.html#cor-map-cons} = & \all(p)(\cons(f(a),\map(f)(x))) \\ & = & \band(p(f(a)),\all(p)(\map(f)(x))) \\ & = & \band((p \circ f)(a),\all(p \circ f)(x)) \\ & = & \all(p \circ f)(\cons(a,x)) \end{eqnarray*}$ as needed.
_test_all_map :: (List t, Equal a)
=> t a -> Test ((a -> a) -> (a -> Bool) -> t a -> Bool)
_test_all_map _ =
testName "all(p)(map(f)(x)) == all(p . f)(x)" $\f p x -> eq (all p (map f x)) (all (p . f) x) $$\any$$ is defined similarly. Let $$A$$ be a set and $$p : A \rightarrow \bool$$ a predicate. Define $$\varphi : A \times \bool \rightarrow \bool$$ by $\varphi(a,q) = \bor(p(a),q).$ Now define $$\any : \bool^A \rightarrow \bool^{\lists{A}}$$ by $\all(p)(x) = \foldr(\bfalse)(\varphi)(x).$ In Haskell: any :: (Boolean b, List t) => (a -> b) -> t a -> b any p = foldr false phi where phi a q = or (p a) q Since $$\any$$ is defined as a fold, it is the unique solution to a system of functional equations. Let $$A$$ be a set and $$p$$ a predicate on $$A$$. $$\any(p)$$ is the unique map $$f : \lists{A} \rightarrow \bool$$ satisfying the following equations for all $$a \in A$$ and $$x \in \lists{A}$$. $\left\{\begin{array}{l} f(\nil) = \bfalse \\ f(\cons(a,x)) = \bor(p(a),f(x)) \end{array}\right.$ _test_any_nil :: (List t, Equal a) => t a -> Test ((a -> Bool) -> Bool) _test_any_nil t = testName "any(p)(nil) == false"$
\p -> eq (any p (nil withTypeOf t)) false
_test_any_cons :: (List t, Equal a)
=> t a -> Test ((a -> Bool) -> a -> t a -> Bool)
_test_any_cons _ =
testName "any(p)(cons(a,x)) == or(p(a),any(p)(x))" $\p a x -> eq (any p (cons a x)) (or (p a) (any p x)) $$\any$$ can also be characterized as a folded map. Let $$A$$ be a set with $$p$$ a predicate on $$A$$. Then $\any(p,x) = \foldr(\bfalse)(\bor)(\map(p)(x)).$ We proceed by list induction on $$x$$. For the base case $$x = \nil$$, we have $\begin{eqnarray*} & & \any(p,x) \\ & = & \any(p,\nil) \\ & = & \bfalse \\ & \href{/posts/arithmetic-made-difficult/Lists.html#def-foldr-nil} = & \foldr(\bfalse)(\bor)(\nil) \\ & = & \foldr(\bfalse)(\bor)(\map(p)(\nil)) \\ & = & \foldr(\bfalse)(\bor)(\map(p)(x)) \end{eqnarray*}$ as needed. For the inductive step, suppose the equality holds for some $$x$$ and let $$a \in A$$. Then we have $\begin{eqnarray*} & & \any(p,\cons(a,x)) \\ & = & \bor(p(a),\any(x)) \\ & = & \bor(p(a),\foldr(\bfalse)(\bor)(\map(p)(x))) \\ & \href{/posts/arithmetic-made-difficult/Lists.html#def-foldr-cons} = & \foldr(\bfalse)(\bor)(\cons(p(a),\map(p)(x))) \\ & = & \foldr(\bfalse)(\bor)(\map(p)(\cons(a,x))) \end{eqnarray*}$ as needed. _test_any_fold_or :: (List t, Equal a) => t a -> Test ((a -> Bool) -> t a -> Bool) _test_any_fold_or _ = testName "any(p,x) == foldr(false,or)(map(p)(x))"$
\p x -> eq (any p x) (foldr false or (map p x))
A version of DeMorgan’s law holds for $$\any$$ and $$\all$$.
Let $$A$$ be a set with $$p : A \rightarrow \bool$$. Then the following hold for all $$x \in \lists{A}$$.
1. $$\any(p,x) = \bnot(\all(\bnot \circ p,x))$$.
2. $$\all(p,x) = \bnot(\any(\bnot \circ p,x))$$.
1. We proceed by list induction on $$x$$. For the base case $$x = \nil$$, we have $\begin{eqnarray*} & & \bnot(\all(\bnot \circ p,x)) \\ & = & \bnot(\all(\bnot \circ p,\nil)) \\ & = & \bnot(\btrue) \\ & \href{/posts/arithmetic-made-difficult/Not.html#thm-not-true} = & \bfalse \\ & = & \foldr(\bfalse)(\varphi(p))(\nil) \\ & = & \foldr(\bfalse)(\varphi(p))(x) \end{eqnarray*}$ as needed. For the inductive step, suppose the equality holds for some $$x$$ and let $$a \in A$$. Then we have $\begin{eqnarray*} & & \bnot(\all(\bnot \circ p,\cons(a,x))) \\ & = & \bnot(\band(\bnot(p(a)),\all(\bnot \circ p,x))) \\ & = & \bor(\bnot(\bnot(p(a))),\bnot(\all(\bnot \circ p,x))) \\ & = & \bor(p(a),\any(p,x)) \\ & = & \any(p,\cons(a,x)) \end{eqnarray*}$ as needed.
2. Note that $\begin{eqnarray*} & & \bnot(\any(\bnot \circ p,x)) \\ & = & \bnot(\bnot(\all(\bnot \circ \bnot \circ p,x))) \\ & = & \all(p,x) \end{eqnarray*}$ as claimed.
_test_any_not_all :: (List t, Equal a)
=> t a -> Test ((a -> Bool) -> t a -> Bool)
_test_any_not_all _ =
testName "any(p,x) == not(all(not . p, x))" $\p x -> eq (any p x) (not (all (not . p) x)) _test_all_not_any :: (List t, Equal a) => t a -> Test ((a -> Bool) -> t a -> Bool) _test_all_not_any _ = testName "all(p,x) == not(any(not . p, x))"$
\p x -> eq (all p x) (not (any (not . p) x))
Two special cases.
Let $$A$$ be a set and $$p : A \rightarrow \bool$$ a predicate. The following hold for all $$x \in \lists{A}$$.
1. $$\any(\pfalse)(x) = \bfalse$$.
2. $$\any(\ptrue)(x) = \btrue$$ if and only if $$x \neq \nil$$.
1. Note that $\begin{eqnarray*} & & \any(\pfalse)(x) \\ & = & \bnot(\all(\bnot \circ \pfalse)(x)) \\ & = & \bnot(\all(\ptrue)(x)) \\ & = & \bnot(\btrue) \\ & \href{/posts/arithmetic-made-difficult/Not.html#thm-not-true} = & \bfalse \end{eqnarray*}$ as claimed.
2. If $$x = \nil$$, we have $\begin{eqnarray*} & & \any(\pfalse)(x) \\ & = & \btrue, \end{eqnarray*}$ while if $$x = \cons(a,u)$$ we have $\begin{eqnarray*} & & \any(\ptrue)(\cons(a,u)) \\ & = & \bnot(\all(\bnot \circ \ptrue)(\cons(a,u))) \\ & = & \bnot(\all(\pfalse)(\cons(a,u))) \\ & = & \bnot(\bfalse) \\ & \href{/posts/arithmetic-made-difficult/Not.html#thm-not-false} = & \btrue \end{eqnarray*}$ as needed.
_test_any_const_false :: (List t, Equal a)
=> t a -> Test (t a -> Bool)
_test_any_const_false _ =
testName "any(pfalse,x) == false" $\x -> eq (any pfalse x) (false :: Bool) _test_any_const_true :: (List t, Equal a, Equal (t a)) => t a -> Test (t a -> Bool) _test_any_const_true _ = testName "any(ptrue,x) == true iff x /= nil"$
\x -> eq
(eq (any ptrue x) (true :: Bool))
(not (eq x nil))
$$\any$$ interacts with $$\cat$$.
Let $$A$$ be a set and $$p : A \rightarrow \bool$$. For all $$x,y \in \lists{A}$$ we have $\any(p,\cat(x,y)) = \bor(\any(p)(x),\any(p)(y)).$
Note that $\begin{eqnarray*} & & \any(p,\cat(x,y)) \\ & = & \bnot(\all(\bnot \circ p,\cat(x,y))) \\ & = & \bnot(\band(\all(\bnot \circ p,x),\all(\bnot \circ p,y))) \\ & = & \bor(\bnot(\all(\bnot \circ p,x)),\bnot(\all(\bnot \circ p,y))) \\ & = & \bor(\any(p,x),\any(p,y)) \end{eqnarray*}$ as claimed.
_test_any_cat :: (List t, Equal a)
=> t a -> Test ((a -> Bool) -> t a -> t a -> Bool)
_test_any_cat _ =
testName "any(p,cat(x,y)) == any(p,x) || any(p,y)" $\p x y -> eq (any p (cat x y)) (or (any p x) (any p y)) $$\any$$ interacts with $$\rev$$. Let $$A$$ be a set and $$p : A \rightarrow \bool$$. For all $$x,y \in \lists{A}$$ we have $\any(p,\rev(x)) = \any(p,x).$ Note that $\begin{eqnarray*} & & \any(p,\rev(x)) \\ & = & \bnot(\all(\bnot \circ p,\rev(x))) \\ & = & \bnot(\all(\bnot \circ p,x)) \\ & = & \any(p,x) \end{eqnarray*}$ as claimed. _test_any_rev :: (List t, Equal a) => t a -> Test ((a -> Bool) -> t a -> Bool) _test_any_rev _ = testName "any(p,rev(x)) == any(p,x)"$
\p x -> eq (any p (rev x)) (any p x)
$$\any$$ interacts with $$\map$$.
Let $$A$$ and $$B$$ be sets with $$f : A \rightarrow B$$ and $$p : B \rightarrow \bool$$. Then for all $$x \in \lists{A}$$ we have $\any(p)(\map(f)(x)) = \any(p \circ f)(x).$
Note that $\begin{eqnarray*} & & \any(p,\map(f)(x)) \\ & = & \bnot(\all(\bnot \circ p,\map(f)(x))) \\ & = & \bnot(\all((\bnot \circ p) \circ f,x)) \\ & = & \bnot(\all(\bnot \circ (p \circ f),x)) \\ & = & \any(p \circ f,x) \end{eqnarray*}$ as claimed.
_test_any_map :: (List t, Equal a)
=> t a -> Test ((a -> a) -> (a -> Bool) -> t a -> Bool)
_test_any_map _ =
testName "any(p)(map(f)(x)) == any(p . f)(x)" \$
\f p x -> eq (any p (map f x)) (any (p . f) x)
Testing
Suite:
_test_all_any ::
( TypeName a, Show a, Equal a, Arbitrary a, CoArbitrary a
, TypeName (t a), List t
, Show (t a), Equal (t a), Arbitrary (t a)
) => Int -> Int -> t a -> IO ()
_test_all_any size cases t = do
testLabel1 "all & any" t
let args = testArgs size cases
runTest args (_test_all_nil t)
runTest args (_test_all_cons t)
runTest args (_test_all_snoc t)
runTest args (_test_all_fold_and t)
runTest args (_test_all_const_true t)
runTest args (_test_all_const_false t)
runTest args (_test_all_cat t)
runTest args (_test_all_rev t)
runTest args (_test_all_map t)
runTest args (_test_any_nil t)
runTest args (_test_any_cons t)
runTest args (_test_any_fold_or t)
runTest args (_test_any_not_all t)
runTest args (_test_all_not_any t)
runTest args (_test_all_const_false t)
runTest args (_test_all_const_true t)
runTest args (_test_any_cat t)
runTest args (_test_any_rev t)
runTest args (_test_any_map t)
Main:
main_all_any :: IO ()
main_all_any = do
_test_all_any 20 100 (nil :: ConsList Bool)
_test_all_any 20 100 (nil :: ConsList Unary) | 2022-08-14 15:18:09 | {"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": 2, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9999669790267944, "perplexity": 2530.15983223708}, "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-2022-33/segments/1659882572043.2/warc/CC-MAIN-20220814143522-20220814173522-00100.warc.gz"} |
https://www.tutorialspoint.com/number-of-lines-to-write-string-in-python | # Number of Lines To Write String in Python
PythonServer Side ProgrammingProgramming
Suppose we have a string S and we have to write the letters of that given string, from left to right into lines. Here each line has maximum width 100 units, and if writing a letter would cause the width of the line to exceed 100 units, that will be written on the next line. We also have an array widths, here widths[0] is the width of 'a', widths[1] is the width of 'b'and so on.
We have to find the answers of two questions −
• How many lines have at least one character from S
• What is the width used by the last such line?
We will return the answer as an integer list of length 2.
So, if the input is like [4,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10] and S = "bbbcccdddaaa", then the output will be [2, 4], as all letters except 'a' have the same length of 10, and the string "bbbcccdddaa" will cover 9 * 10 + 2 * 4 = 98 spaces. For the last 'a', it is written on the second line because there is only 2 units left in the first line. So the answer is 2 lines, plus 4 units in the second line.
To solve this, we will follow these steps −
• line := 1, count := 0
• for each i in S, do
• count := count + widths[ASCII of i - 97]
• if count > 100, then
• line := line + 1
• count := widths[ASCII of i - 97]
• return [line, count]
Let us see the following implementation to get better understanding −
## Example
Live Demo
class Solution:
def numberOfLines(self, widths, S):
line = 1
count = 0
for i in S:
count += widths[ord(str(i))-97]
if count > 100:
line += 1
count = widths[ord(str(i))-97]
return [line, count]
ob = Solution()
print(ob.numberOfLines([4,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10], "bbbcccdddaaa"))
## Input
[4,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10],"bbbcccdddaaa"
## Output
[2, 4]
Updated on 04-Jul-2020 10:04:30 | 2022-08-12 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": 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": 0.2799670696258545, "perplexity": 1586.9549781822163}, "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/1659882571719.48/warc/CC-MAIN-20220812140019-20220812170019-00462.warc.gz"} |
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### Re: Problems with umlauts
karlheinz01
Member
@karlheinz01
Hello,
i comment out this lines, but i get the same result.
I do some more tests, maybe they help.
if i put this in the title: öäüÖÄÜß
and this 3 lines in the content:
öäüÖÄÜß
1. fdgfd gdfgd sfgsdgs dfggf dfgd
i get this:
Title: öäüÖÄÜß
Content: all lines deleted
If i put this in the Title: öäüÖÄÜß
and this in the content:
1. fdgfd gdfgd sfgsdgs dfggf dfgd
öäüÖÄÜß
i get this:
Title: öäüÖÄÜß
Content:
1. fdgfd gdfgd sfgsdgs dfggf dfgd
Last test. If i put this line in the content:
öäüÖÄÜß this is a test
after sending it the whole line is empty /get deleted.
All this is only in the BuddyPress frontend.
So in the title i can use any umlauts, but in the content the filter routine deletes not only the umlauts, but depending on where the umlauts are also the rest of the text.
Best,
Karl | 2023-03-30 02:23:50 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8396174907684326, "perplexity": 9283.682151021805}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "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-2023-14/segments/1679296949093.14/warc/CC-MAIN-20230330004340-20230330034340-00522.warc.gz"} |
http://www.thefullwiki.org/Euler%27s_formula | # Euler's formula: Wikis
Note: Many of our articles have direct quotes from sources you can cite, within the Wikipedia article! This article doesn't yet, but we're working on it! See more info or our list of citable articles.
# Encyclopedia
Part of a series of articles on The mathematical constant e Applications in: compound interest · Euler's identity & Euler's formula · half-lives & exponential growth/decay People John Napier · Leonhard Euler
Euler's formula, named after Leonhard Euler, is a mathematical formula in complex analysis that demonstrates the deep relationship between the trigonometric functions and the complex exponential function. Euler's formula states that, for any real number x,
$e^{ix} = \cos x + i\sin x \!$
where e is the base of the natural logarithm, i is the imaginary unit, and cos and sin are the trigonometric functions cosine and sine respectively, with the argument x given in radians. This complex exponential function is sometimes called cis(x). The formula is still valid if x is a complex number, and so some authors refer to the more general complex version as Euler's formula.[1]
Richard Feynman called Euler's formula "our jewel"[2] and "one of the most remarkable, almost astounding, formulas in all of mathematics."[3]
## History
It was Bernoulli [1702] who noted that
$\frac{1}{1+x^2}=\frac{1}{2} \left(\frac{1}{1-ix}+\frac{1}{1+ix} \right).$
And since
$\int \frac{dx}{1+ax}=\frac{1}{a}\ln(1+ax),$
the above equation tells us something about complex logarithms. Bernoulli, however, did not evaluate the integral. His correspondence with Euler (who also knew the above equation) shows that he didn't fully understand logarithms. Euler also suggested that the complex logarithms can have infinitely many values.
Meanwhile, Roger Cotes, in 1714, discovered
$\ln(\cos x + i\sin x)=ix \$
(where "ln" means natural logarithm, i.e. log with base e).[4] We now know that the above equation is only true modulo integer multiples of i, but Cotes missed the fact that a complex logarithm can have infinitely many values which owes to the periodicity of the trigonometric functions.
It was Euler (presumably around 1740) who turned his attention to the exponential function instead of logarithms, and obtained the correct formula now coined after his name. It was published in 1748, and his proof was based on the infinite series of both sides being equal. Neither of these men saw the geometrical interpretation of the formula: the view of complex numbers as points in the complex plane arose only some 50 years later (see Caspar Wessel).
## Applications in complex number theory
This formula can be interpreted as saying that the function eix traces out the unit circle in the complex number plane as x ranges through the real numbers. Here, x is the angle that a line connecting the origin with a point on the unit circle makes with the positive real axis, measured counter clockwise and in radians.
The original proof is based on the Taylor series expansions of the exponential function ez (where z is a complex number) and of sin x and cos x for real numbers x (see below). In fact, the same proof shows that Euler's formula is even valid for all complex numbers z.
A point in the complex plane can be represented by a complex number written in cartesian coordinates. Euler's formula provides a means of conversion between cartesian coordinates and polar coordinates. The polar form reduces the number of terms from two to one, which simplifies the mathematics when used in multiplication or powers of complex numbers. Any complex number z = x + iy can be written as
$z = x + iy = |z| (\cos \phi + i\sin \phi ) = r e^{i \phi} \,$
$\bar{z} = x - iy = |z| (\cos \phi - i\sin \phi ) = r e^{-i \phi} \,$
where
$x = \mathrm{Re}\{z\} \,$ the real part
$y = \mathrm{Im}\{z\} \,$ the imaginary part
$r = |z| = \sqrt{x^2+y^2}$ the magnitude of z
$\phi = \arg z = \,$ atan2(y, x).
$\phi \,$ is the argument of z—i.e., the angle between the x axis and the vector z measured counterclockwise and in radians—which is defined up to addition of 2π. Many texts write tan-1(y/x) instead of atan2(y,x) but this needs adjustment when x ≤ 0.
Now, taking this derived formula, we can use Euler's formula to define the logarithm of a complex number. To do this, we also use the definition of the logarithm (as the inverse operator of exponentiation) that
$a = e^{\ln (a)}\,$
and that
$e^a e^b = e^{a + b}\,$
both valid for any complex numbers a and b.
Therefore, one can write:
$z = |z| e^{i \phi} = e^{\ln |z|} e^{i \phi} = e^{\ln |z| + i \phi}\,$
for any z ≠ 0. Taking the logarithm of both sides shows that:
$\ln z= \ln |z| + i \phi.\,$
and in fact this can be used as the definition for the complex logarithm. The logarithm of a complex number is thus a multi-valued function, because φ is multi-valued.
Finally, the other exponential law
$(e^a)^k = e^{a k}, \,$
which can be seen to hold for all integers k, together with Euler's formula, implies several trigonometric identities as well as de Moivre's formula.
## Relationship to trigonometry
Euler's formula provides a powerful connection between analysis and trigonometry, and provides an interpretation of the sine and cosine functions as weighted sums of the exponential function:
$\cos x = \mathrm{Re}\{e^{ix}\} ={e^{ix} + e^{-ix} \over 2}$
$\sin x = \mathrm{Im}\{e^{ix}\} ={e^{ix} - e^{-ix} \over 2i}.$
The two equations above can be derived by adding or subtracting Euler's formulas:
$e^{ix} = \cos x + i \sin x \;$
$e^{-ix} = \cos(- x) + i \sin(- x) = \cos x - i \sin x \;$
and solving for either cosine or sine.
These formulas can even serve as the definition of the trigonometric functions for complex arguments x. For example, letting x = iy, we have:
$\cos(iy) = {e^{-y} + e^{y} \over 2} = \cosh(y)$
$\sin(iy) = {e^{-y} - e^{y} \over 2i} = -{1 \over i} {e^{y} - e^{-y} \over 2} = i\sinh(y).$
Complex exponentials can simplify trigonometry, because they are easier to manipulate than their sinusoidal components. One technique is simply to convert sinusoids into equivalent expressions in terms of exponentials. After the manipulations, the simplified result is still real-valued. For example:
\begin{align} \cos x\cdot \cos y & = \frac{(e^{ix}+e^{-ix})}{2} \cdot \frac{(e^{iy}+e^{-iy})}{2} \ & = \frac{1}{2}\cdot \frac{e^{i(x+y)}+e^{i(x-y)}+e^{i(-x+y)}+e^{i(-x-y)}}{2} \ & = \frac{1}{2} \left[ \underbrace{ \frac{e^{i(x+y)} + e^{-i(x+y)}}{2} }_{\cos(x+y)} + \underbrace{ \frac{e^{i(x-y)} + e^{-i(x-y)}}{2} }_{\cos(x-y)} \right]. \end{align}
Another technique is to represent the sinusoids in terms of the real part of a more complex expression, and perform the manipulations on the complex expression. For example:
\begin{align} \cos(nx) & = \mathrm{Re} \{\ e^{inx}\ \} = \mathrm{Re} \{\ e^{i(n-1)x}\cdot e^{ix}\ \} \ & = \mathrm{Re} \{\ e^{i(n-1)x}\cdot (e^{ix} + e^{-ix} - e^{-ix})\ \} \ & = \mathrm{Re} \{\ e^{i(n-1)x}\cdot \underbrace{(e^{ix} + e^{-ix})}_{2\cos(x)} - e^{i(n-2)x}\ \} \ & = \cos[(n-1)x]\cdot 2 \cos(x) - \cos[(n-2)x]. \end{align}
This formula is used for recursive generation of cos(nx) for integer values of n and arbitrary x (in radians).
## Other applications
In differential equations, the function eix is often used to simplify derivations, even if the final answer is a real function involving sine and cosine. The reason for this is that the complex exponential is the eigenfunction of differentiation. Euler's identity is an easy consequence of Euler's formula.
In electrical engineering and other fields, signals that vary periodically over time are often described as a combination of sine and cosine functions (see Fourier analysis), and these are more conveniently expressed as the real part of exponential functions with imaginary exponents, using Euler's formula. Also, phasor analysis of circuits can include Euler's formula to represent the impedance of a capacitor or an inductor.
## Definitions of complex exponentiation
In general, raising e to a positive integer exponent has a simple interpretation in terms of repeated multiplication of e. Raising e to zero or a negative integer exponent can be understood as repeated division. A rational number exponent can be defined by radicals of e, and an irrational number exponent can be defined by finding rational-number exponents that are arbitrarily close to the irrational-number exponent, in a limit process. However, to define and understand a complex number exponent of e, a different type of generalization is required for the concept of exponentiation.
In fact, several definitions are possible. All of them can be proven to be well-defined and equivalent, although the proofs are not included in this article.
### Taylor series definition
For any real x, the following series is equal to ex:
$e^x = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots = \sum_{n=0}^\infty\frac{x^n}{n!}~.$
(in other words, this is the Taylor series for the real exponential function, and it has an infinite radius of convergence). This invites the following definition of ez for complex z:
$e^z = 1 + \frac{z}{1!} + \frac{z^2}{2!} + \frac{z^3}{3!} + \cdots ~.$
This can be proven to be well-defined; in particular, the series converges for any z.
### Analytic continuation definition
A simple-to-state, equivalent definition is that ez, for complex z, is the analytic continuation of the function ex for real x. This can be proven to be well-defined; in particular, it yields a single-valued function on the complex plane.
### Limit definition
For any real x, the following limit is equal to ex:
$e^x = \lim_{N \rightarrow \infty} \left(1+\frac{x}{N}\right)^N.$
This motivates the following definition of ez for complex z:
$e^z = \lim_{N \rightarrow \infty} \left(1+\frac{z}{N}\right)^N.$
### Differential equation definition
For real x, the function f(x) = ex is well-known to be the unique real function satisfying the differential equation:
$f'(x)=f(x),\;\;\; f(0)=1$
for all x. This motivates a definition of f(z) = ez for complex z as the function that satisfies the differential equation:
$f'(z)=f(z),\;\;\; f(0)=1$
for all complex z, where the derivative in f '(z) is defined in the sense of a complex derivative. This can be proven to yield a unique function which is well-defined everywhere on the complex plane.
## Proofs
Various proofs of this formula are possible. The first proof below starts with the "Taylor series definition" of ez, while the other two use the "differential equation definition" of ez (see above).
### Using Taylor series
Here is a proof of Euler's formula using Taylor series expansions as well as basic facts about the powers of i:
\begin{align} i^0 &{}= 1, \quad & i^1 &{}= i, \quad & i^2 &{}= -1, \quad & i^3 &{}= -i, \ i^4 &={} 1, \quad & i^5 &={} i, \quad & i^6 &{}= -1, \quad & i^7 &{}= -i, \ \end{align}
and so on. The functions ex, cos x and sin x of the (real) variable x can be expressed using their Taylor expansions around zero:
\begin{align} e^x &{}= 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots \ \cos x &{}= 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots \ \sin x &{}= x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots. \end{align}
For complex z we define each of these functions by the above series, replacing the real variable x with the complex variable z. This is possible because the radius of convergence of each series is infinite. We then find that
\begin{align} e^{iz} &{}= 1 + iz + \frac{(iz)^2}{2!} + \frac{(iz)^3}{3!} + \frac{(iz)^4}{4!} + \frac{(iz)^5}{5!} + \frac{(iz)^6}{6!} + \frac{(iz)^7}{7!} + \frac{(iz)^8}{8!} + \cdots \ &{}= 1 + iz - \frac{z^2}{2!} - \frac{iz^3}{3!} + \frac{z^4}{4!} + \frac{iz^5}{5!} - \frac{z^6}{6!} - \frac{iz^7}{7!} + \frac{z^8}{8!} + \cdots \ &{}= \left( 1 - \frac{z^2}{2!} + \frac{z^4}{4!} - \frac{z^6}{6!} + \frac{z^8}{8!} - \cdots \right) + i\left( z - \frac{z^3}{3!} + \frac{z^5}{5!} - \frac{z^7}{7!} + \cdots \right) \ &{}= \cos z + i\sin z \end{align}
The rearrangement of terms is justified because each series is absolutely convergent. Taking z = x to be a real number gives the original identity as Euler discovered it.
Q.E.D.
### Using calculus
Define the (possibly complex) function ƒ(x), of real variable x, as
$f(x) = (\cos x + i\sin x)\cdot e^{-ix}. \$
The derivative of ƒ(x), according to the product rule, is:
\begin{align} \frac{d}{dx}f(x) &{}= (\cos x + i\sin x)\cdot\frac{d}{dx}e^{-ix} + \frac{d}{dx}(\cos x + i\sin x)\cdot e^{-ix} \ &{}= (\cos x + i\sin x)(-i e^{-ix}) + (-\sin x + i\cos x)\cdot e^{-ix} \ &{}= (-i\cos x - i^2\sin x)\cdot e^{-ix} + (-\sin x + i\cos x)\cdot e^{-ix} \quad \quad \quad (i^2=-1) \ &{}= (-i\cos x + \sin x - \sin x + i\cos x)\cdot e^{-ix} \ &{}= 0. \end{align}
Therefore, ƒ(x) must be a constant function in x. Because ƒ(0) is known, the constant that ƒ(x) equals for all real x is also known. Thus,
$(\cos x + i\sin x)\cdot e^{-ix} = f(x) = f(0) = (\cos 0 + i\sin 0)\cdot e^0 = 1 \,.$
Multiplying both sides by eix, we get
$\cos x + i \sin x \ = e^{ix}.$
Q.E.D.
### Using ordinary differential equations
Define the function g(x) by
$g(x) \ \stackrel{\mathrm{def}}{=}\ e^{ix} .\$
Considering that i is constant, the first and second derivatives of g(x) are
$g'(x) = i e^{ix} \$
$g''(x) = i^2 e^{ix} = -e^{ix} \$
because i 2 = −1 by definition. From this the following 2nd-order linear ordinary differential equation is constructed:
$g''(x) = -g(x) \$
or
$g''(x) + g(x) = 0. \$
Being a 2nd-order differential equation, there are two linearly independent solutions that satisfy it:
$g_1(x) = \cos x \$
$g_2(x) = \sin x. \$
Both cos and sin are real functions in which the 2nd derivative is identical to the negative of that function. Any linear combination of solutions to a homogeneous differential equation is also a solution. Then, in general, the solution to the differential equation is
$g(x)\,$ $= A g_1(x) + B g_2(x) \$ $= A \cos x + B \sin x \$
for any constants A and B. But not all values of these two constants satisfy the known initial conditions for g(x):
$g(0) = e^{i0} = 1 \$
$g'(0) = i e^{i0} = i.\$
However these same initial conditions (applied to the general solution) are
$g(0) = A \cos 0 + B \sin 0 = A \$
$g'(0) = -A \sin 0 + B \cos 0 = B \$
resulting in
$g(0) = A = 1 \$
$g'(0) = B = i \$
and, finally,
$g(x) \ \stackrel{\mathrm{def}}{=}\ e^{ix} = \cos x + i \sin x. \$
Q.E.D.
## References
1. ^ Moskowitz, Martin A. (2002). A Course in Complex Analysis in One Variable. World Scientific Publishing Co.. pp. 7. ISBN 981-02-4780-X.
2. ^ Feynman, Richard P. (1977). The Feynman Lectures on Physics, vol. I. Addison-Wesley. pp. 22–10. ISBN 0-201-02010-6.
3. ^ Feynman, Richard P. (1977). The Feynman Lectures on Physics, vol. I. Addison-Wesley. pp. 22–1. ISBN 0-201-02010-6.
4. ^ John Stillwell (2002). Mathematics and Its History. Springer.
# Wiktionary
Up to date as of January 15, 2010
## English
### Etymology
Named after the 18th century Swiss mathematician Leonhard Euler.
### Proper noun
Singular Euler's formula Plural -
Euler's formula
Wikipedia has an article on:
Wikipedia
1. (complex analysis) Formula which links complex exponentiation with trigonometric functions:
eiθ = cosθ + isinθ
2. (differential geometry) Formula which calculates the normal curvature of an arbitrary direction in the tangent plane in terms of the principal curvatures κ1 and κ2 and the angle θ which that direction makes with the first principal direction:
κn(θ) = κ1cos2θ + κ2sin2θ | 2018-12-13 15:07:02 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 54, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9711775183677673, "perplexity": 659.1412256699289}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-51/segments/1544376824912.16/warc/CC-MAIN-20181213145807-20181213171307-00616.warc.gz"} |
https://math.libretexts.org/Bookshelves/Arithmetic_and_Basic_Math/Book%3A_Basic_Math_(Illustrative_Mathematics_-_Grade_6)/01%3A_Area_and_Surface_Area/107%3A_Lets_Put_it_to_Work/1.7.1%3A_Designing_a_Tent |
# 1.7.1: Designing a Tent
## Lesson
Let's design some tents.
Exercise $$\PageIndex{1}$$: Tent Design - Part 1
Have you ever been camping?
You might know that sleeping bags are all about the same size, but tents come in a variety of shapes and sizes.
Your task is to design a tent to accommodate up to four people, and estimate the amount of fabric needed to make your tent. Your design and estimate must be based on the information given and have mathematical justification.
First, look at these examples of tents, the average specifications of a camping tent, and standard sleeping bag measurements. Talk to a partner about:
• Similarities and differences among the tents
• Information that will be important in your designing process
• The pros and cons of the various designs
Tent Styles
Tent Height Specifications
height description height of tent notes
sitting height 3 feet Campers are able to sit, lie, or crawl inside tent.
kneeling height 4 feet Campers are able to kneel inside tent. Found mainly in 3–4 person tents.
stooping height 5 feet Campers are able to move around on their feet inside tent, but most campers will not be able to stand upright.
standing height 6 feet Most adult campers are able to stand upright inside tent.
roaming height 7 feet Adult campers are able to stand upright and walk around inside tent.
Table $$\PageIndex{1}$$
Sleeping Bag Measurements
1. Create and sketch your tent design. The tent must include a floor.
2. What decisions were important when choosing your tent design?
3. How much fabric do you estimate will be necessary to make your tent? Show your reasoning and provide mathematical justification.
Exercise $$\PageIndex{2}$$: Tent Design - Part 2
1. Explain your tent design and fabric estimate to your partner or partners. Be sure to explain why you chose this design and how you found your fabric estimate.
2. Compare the estimated fabric necessary for each tent in your group. Discuss the following questions:
• Which tent design used the least fabric? Why?
• Which tent design used the most fabric? Why?
• Which change in design most impacted the amount of fabric needed for the tent? Why? | 2020-11-27 17:06:58 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 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": 0.213824063539505, "perplexity": 2577.087377127633}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-50/segments/1606141193856.40/warc/CC-MAIN-20201127161801-20201127191801-00192.warc.gz"} |
https://im.kendallhunt.com/HS/teachers/3/3/18/preparation.html | Lesson 18
The Quadratic Formula and Complex Solutions
Lesson Narrative
In the previous lesson, students completed the square to find real and non-real solutions to quadratic equations. In this lesson, students use the quadratic formula to calculate real and non-real solutions.
Students construct viable arguments and critique the reasoning of others when they compare solutions of equivalent quadratic equations and resolve any disagreements during a row game (MP3).
Learning Goals
Teacher Facing
• Calculate complex solutions using the quadratic formula and represent them in the form $c \pm di$, where $c$ and $d$ are real numbers.
• Explain reasoning and critique the reasoning of others when solving quadratic equations.
• Explain when solutions are non-real when using the quadratic formula.
Student Facing
• Let’s use the quadratic formula to find complex solutions to quadratic equations.
Student Facing
• I can find complex solutions to quadratic equations by using the quadratic formula.
Building On | 2023-03-30 17:46:43 | {"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": 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": 0.4176245331764221, "perplexity": 1061.0065778749254}, "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-2023-14/segments/1679296949355.52/warc/CC-MAIN-20230330163823-20230330193823-00774.warc.gz"} |
http://openstudy.com/updates/50f7a42fe4b027eb5d99d0f0 | ## missylulu 2 years ago help on multivariable calc T-T... what are the relative extrema of f(x,y) = x^2 + e^y? is the answer none...?
clearly domain is $$\mathbb{R}^2$$ and for relative extrema we have$\frac{\partial f}{\partial x}=2x=0$$\frac{\partial f}{\partial y}=e^y=0$we should solve these equations simultaneously but u r right...answer is none because for all real values of y we have$e^y>0$ | 2015-03-01 11:12:50 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 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": 0.8079962134361267, "perplexity": 3480.5188518413515}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-11/segments/1424936462316.51/warc/CC-MAIN-20150226074102-00010-ip-10-28-5-156.ec2.internal.warc.gz"} |
https://fds.duke.edu/db/aas/math/faculty/bryant/publications/243396 | Department of Mathematics Search | Help | Login | |
Math @ Duke
....................... ....................... Webpage
## Publications [#243396] of Robert Bryant
search www.ams.org.
Papers Published
1. with Griffiths, PA; Bryant, RL, Reduction for constrained variational problems and $\int{1\over 2}k\sp 2\,ds$, Amer. J. Math., vol. 108 no. 3 (1986), pp. 525-570 [MR88a:58044]
(last updated on 2018/12/13)
This paper gives an exposition of a way of computing the Euler-Lagrange equations and the conservation laws for them that arise from symmetries in geometrically defined variational problems. The main technical advantage of this method over the more classical Pontrjagin Maximum Principle is the way it avoids choosing coordinates that are not needed, but works directly on the invariant coframing of the group of symmetries.
Some extended examples are computed for Euler elastica in space forms and on surfaces of constant curvature.
Since this paper appeared, David Mumford has shown how to get a complete integration of the equations in the flat case by a very clever use of theta-functions. It would be interesting to see if this would work also in the case of space elastica or for elastica in other space forms.
dept@math.duke.edu
ph: 919.660.2800
fax: 919.660.2821
Mathematics Department
Duke University, Box 90320
Durham, NC 27708-0320 | 2018-12-13 08:20:27 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 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": 0.6556246876716614, "perplexity": 1164.0312580338018}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-51/segments/1544376824601.32/warc/CC-MAIN-20181213080138-20181213101638-00270.warc.gz"} |
https://space.stackexchange.com/tags/computing/new | # Tag Info
There are several ways to do this. The easiest and most straightforward is to break it into two sets by including velocity as a variable, and solve together. Instead of a single second order differential equation $$\ddot{\mathbf{r}} = -\frac{\mu}{r^3}\mathbf{r}$$ We can solve the following pair of first order differential equations in parallel \dot{\mathbf{... | 2020-12-04 05:40:11 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8700265288352966, "perplexity": 164.41748353450402}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-50/segments/1606141733122.72/warc/CC-MAIN-20201204040803-20201204070803-00184.warc.gz"} |
https://proofwiki.org/wiki/Bernoulli_Process_as_Binomial_Distribution | # Bernoulli Process as Binomial Distribution
## Theorem
Let $\left \langle{X_i}\right \rangle$ be a finite Bernoulli process of length $n$ such that each of the $X_i$ in the sequence is a Bernoulli trial with parameter $p$.
Then the number of successes in $\left \langle{X_i}\right \rangle$ is modelled by a binomial distribution with parameters $n$ and $p$.
Hence it can be seen that:
$X \sim \operatorname{B} \left({1, p}\right)$ is the same thing as $X \sim \operatorname{Bern} \left({p}\right)$
## Proof
Consider the sample space $\Omega$ of all sequences $\left \langle{X_i}\right \rangle$ of length $n$.
The $i$th entry of any such sequence is the result of the $i$th trial.
We have that $\Omega$ is finite.
Let us take the event space $\Sigma$ to be the power set of $\Omega$.
As the elements of $\Omega$ are independent, by definition of the Bernoulli process, we have that:
$\forall \omega \in \Omega: \Pr \left({\omega}\right) = p^{s \left({\omega}\right)} \left({1 - p}\right)^{n - s \left({\omega}\right)}$
where $s \left({\omega}\right)$ is the number of successes in $\omega$.
In the same way:
$\displaystyle \forall A \in \Sigma: \Pr \left({A}\right) = \sum_{\omega \mathop \in A} \Pr \left({\omega}\right)$
Now, let us define the discrete random variable $Y_i$ as follows:
$Y_i \left({\omega}\right) = \begin{cases} 1 & : \omega_i \text { is a success} \\ 0 & : \omega_i \text { is a failure} \\ \end{cases}$
where $\omega_i$ is the $i$th element of $\omega$.
Thus, each $Y_i$ has image $\left\{{0, 1}\right\}$ and a probability mass function:
$\Pr \left({Y_i = 0}\right) = \Pr \left({\left\{{\omega \in \Omega: \omega_i \text { is a success}}\right\}}\right)$
Thus we have:
$\displaystyle \Pr \left({Y_i = 1}\right)$ $=$ $\displaystyle \sum_{\omega: \omega_i \text{ success} } p^{s \left({\omega}\right)} \left({1 - p}\right)^{n - s \left({\omega}\right)}$ $\displaystyle$ $=$ $\displaystyle \sum_{r \mathop = 1}^n \sum_{ \substack{ \omega: \omega_i \text{ success} \\ s \left({\omega}\right) = r } } p^r \left({1 - p}\right)^{n - r}$ $\displaystyle$ $=$ $\displaystyle \sum_{r \mathop = 1}^n \binom {n-1} {r-1} p^r \left({1 - p}\right)^{n - r}$ As we already know the position of one success (namely $i$) $\displaystyle$ $=$ $\displaystyle p \sum_{r \mathop = 0}^{n-1} \binom {n-1} r p^r \left({1 - p}\right)^{\left({n - 1}\right) - r}$ Switching summation index $\displaystyle$ $=$ $\displaystyle p \left({p + \left({1 - p}\right)}\right)^{n-1}$ Binomial Theorem $\displaystyle$ $=$ $\displaystyle p$
Then:
$\Pr \left({Y_i = 0}\right) = 1 - \Pr \left({Y_i = 1}\right) = 1 - p$
So (by a roundabout route) we have confirmed that $Y_i$ has the Bernoulli distribution with parameter $p$.
Now, let us define the random variable:
$\displaystyle S_n \left({\omega}\right) = \sum_{i \mathop = 1}^n Y_i \left({\omega}\right)$
By definition:
• $S_n \left({\omega}\right)$ is the number of successes in $\omega$
• $S_n$ takes values in $\left\{{0, 1, 2, \ldots, n}\right\}$ (as each $Y_i$ can be $0$ or $1$).
Also, we have that:
$\displaystyle \Pr \left({S_n = k}\right)$ $=$ $\displaystyle \Pr \left({\left\{ {\omega \in \Omega: s \left({\omega}\right) = k}\right\} }\right)$ $\displaystyle$ $=$ $\displaystyle \sum_{\omega: s \left({\omega}\right) \mathop = k} \Pr \left({\omega}\right)$ $\displaystyle$ $=$ $\displaystyle \sum_{\omega: s \left({\omega}\right) \mathop = k} p^k \left({1 - p}\right)^{n-k}$ $\displaystyle$ $=$ $\displaystyle \binom n k p^k \left({1 - p}\right)^{n-k}$
Hence the result.
$\blacksquare$ | 2019-08-22 11:39:26 | {"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": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9784862399101257, "perplexity": 95.13009362445523}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-35/segments/1566027317113.27/warc/CC-MAIN-20190822110215-20190822132215-00316.warc.gz"} |
https://zbmath.org/?q=an%3A0613.34047 | zbMATH — the first resource for mathematics
A singular abstract Cauchy problem. (English. Russian original) Zbl 0613.34047
Sov. Math. 30, No. 6, 78-81 (1986); translation from Izv. Vyssh. Uchebn. Zaved., Mat. 1986, No. 6(289), 55-56 (1986).
The author is concerned with the well-posedness of a Cauchy problem associated to the abstract Euler-Poisson-Darboux equation $u''(t)+(k/t)u'(t)=Au(t),\quad t>0$ in a Banach space E. Here u is a function with values in E, $$k\in R$$, and A denotes a closed linear operator in E that generates a strongly continuous cosine family.
Reviewer: S.Aizicovici
MSC:
34G10 Linear differential equations in abstract spaces 34A12 Initial value problems, existence, uniqueness, continuous dependence and continuation of solutions to ordinary differential equations | 2021-04-18 23:38:25 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5224586725234985, "perplexity": 483.35636280644076}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-17/segments/1618038862159.64/warc/CC-MAIN-20210418224306-20210419014306-00096.warc.gz"} |
https://www.physicsforums.com/threads/laplace-transforms.402156/ | # Laplace transforms
1. May 10, 2010
### squenshl
I'm deriving the heat equation using Laplace transforms.
I'm up to u(bar) = Aesqrt(s)x + Be-sqrt(s)x.
My boundary conditions are lim(x tends to 0) = 0 & this makes A = 0, so u(bar) = Be-sqrt(s)x
My other BC is ux(0,t) = -Q
My question is how do I get B using this BC?
I tried using the def of Laplace transforms and get -dQ(bar)/dx
Last edited: May 10, 2010
2. May 10, 2010
### squenshl
I got B = Q/sqrt(s) and this implies that U(bar) = Qe-sqrt(s)x/sqrt(s), I hope thats right | 2017-08-21 09:02:03 | {"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": 0.8945398330688477, "perplexity": 5619.920115396538}, "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/1502886107744.5/warc/CC-MAIN-20170821080132-20170821100132-00231.warc.gz"} |
http://math.stackexchange.com/questions/120429/proof-that-s-3-and-s-4-are-solvable-groups | # Proof that $S_3$ and $S_4$ are solvable groups
I wish to prove that $S_3$,$S_4$ (permutations on $3,4$ elements respectively) are solvable.
I know that $D_6,D_{24}$ ($D_n$=Dihedral group of order $n$) are solvable and if I could prove that $S_3$ is isomorphic to $D_6$, $S_4$ is isomorphic to $D_{24}$ it will do the trick.
With $S_3$ and $D_6$ I can do this 'the hard way' by defining the isomorphism, but with $S_4$ and $D_24$ proving that the function I define respects the multiplication of the group is too much work and doesn't seem like the 'right' way.
Maybe it is sufficient to prove that S4 is solvable, since there is a $1-1$ homomorphism from $S_3 \to S_4$ ?
I could use some help with this...
-
+1. Well posed question. – user21436 Mar 15 '12 at 9:30
The easy way is to simply verify that the factors in the normal series $1 \triangleleft V_4 \triangleleft A_4 \triangleleft S_4$ are abelian, where $V_4$ is the Klein 4 group, which is isomorphic to $\mathbb{Z}_2 \oplus \mathbb{Z}_2$ (abelian). The other factors are too small to be nonabelian. – William DeMeo Mar 15 '12 at 9:43
Why does it prove the claim ? |A4|=12, and there are groups of samller order that are not abelian, no ? – Belgi Mar 15 '12 at 9:50
What I mean is check that the factor groups are abelian. Perhaps your book isn't giving you this definition of solvability. If that's the case, you should review the definition here. So, you just need to check that $S_4/A_4$ is abelian, that $A_4/V_4$ is abelian, and that $V_4/1 = V_4$ is abelian. But this is obvious by considering the orders of these groups. – William DeMeo Mar 15 '12 at 10:33
@WilliamDeMeo I have shown the OP completely how to do this. – user38268 Mar 15 '12 at 11:38
First of all let me point out that $D_{24}$ cannot be isomorphic to $S_4$. $S_4$ does not have an element of order 12, where as $D_{24}$ does. This is because $D_{24}$ as a set is
$$\{1,r,r^2 ,\ldots r^{11}, sr, sr^2, \ldots sr^{11}\}$$
where the $r_i's$ are the rotations and the $sr^j$ the reflections; so that $r$ is your element of order 12. For the sake of knowledge though $S_4$ does have a subgroup of order 12 sitting inside it that is just $A_4$. Just to tell you though the subgroups of $S_4$ are (up to isomorphism):
$$\{e\}, \hspace{2mm} C_2, \hspace{2mm} C_3, \hspace{2mm} V_4, \hspace{2mm} C_4, \hspace{2mm} S_3, \hspace{2mm} D_8, \hspace{2mm} A_4, \hspace{2mm} S_4.$$
Actually it is not so hard to see directly that $S_3$ and $D_6$ are isomorphic, just list down the elements explicitly and if they obey exactly the same relations, their Cayley Tables are the same so you have your isomorphism.
By the way I think for your last line you want a surjective homomorphism from $S_4$ to $S_3$ to show that $S_3$ is solvable only if $S_4$ is. The way you do that is like this:
We get a homomorphism $\varphi$ from $S_4$ to $S_3$ by considering the action of $S_4$ on a finite set $Y = \{\Pi_1, \Pi_2, \Pi_3\}$ where $\Pi_1 = \left\{\{1,2\}, \{3,4\}\right\}$, $\Pi_2 = \left\{\{1,3\}, \{2,4\}\right\}$ and $\Pi_3 =\left\{ \{1,4\}, \{2,3\}\right\}$. Each $\Pi_i$ is one way of partitioning a set of 4 elements into two subsets of equal cardinality. $S_4$ acts on a $\Pi_i$ by permuting the numbers in each partition, so for example the cycle $(1234)$ interchanges $\Pi_1$ and $\Pi_3$ while keeping $\Pi_2$ fixed, so that $\varphi( (1234) ) = (13)(2)$. It is easy to see that $\ker \varphi$ is the Klein four- group $V_4= \{e, (12)(34), (14)(23), (13)(24)\}$ where $e$ denotes the identity in $S_4$. If you go further you should know that by the first isomorphism theorem the quotient group $S_4/V_4$ is isomorphic to the image of $\varphi$.
Now to see the surjectivity of $\varphi$, recall we have the counting formula
$|S_4| = |\ker \varphi||\operatorname{Im} \varphi|$. Therefore the order of the image is $24/4 = 6$. Now recall the image is a subgroup of $S_3$ so if we have a subgroup of order 6 sitting inside $S_3$, it must be the whole of $S_3$.
To answer your question on why $S_4$ is solvable, note that the only normal subgroups in $S_4$ are the trivial group, the Klein 4-group $V_4$, $A_4$ and $S_4$ itself.
So we have a subnormal series
$$\{e\} \triangleleft V_4 \triangleleft A_4 \triangleleft S_4.$$
Now you should know that $V_4/\{e\} \cong V_4$ that is abelian (all groups of order 4 are). You also know that $|A_4/V_4| = 3$ so that $A_4/V_4 \cong C_3$ which is abelian. Finally $|S_4/A_4| = 2$ so that $S_4/A_4 \cong C_2$ that is also abelian. So it remains to check that $A_4$ is normal in $S_4$ and $V_4$ is normal in $A_4$. The former you already know for the index of $A_4$ in $S_4$ is two. For the second I suggest listing out the conjugacy classes of $A_4$ and show that $V_4$ is a union of conjugacy classes. It should not be so hard to compute all the conjugacy classes of $A_4$ since there are only 12 elements. Once you've done all of this, you've proved that $S_4$ is solvable!
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Nice Answer. : ) – user21436 Mar 15 '12 at 11:14
@KannappanSampath Thanks. – user38268 Mar 15 '12 at 11:15
+1 A thorough explanation. – Jyrki Lahtonen Mar 15 '12 at 11:32
In the list of subgroups I've replaces $S_6$ by the obviously intended $S_3$. However the "group" in braces is not a subgroup, and I didn't know what was intended by it, an explicit rendering of $V_4$? Or maybe $C_4$ which is missing from the list? Also note that some "subgroups" occur more than once: $C_2$ ($9=6+3$ times, two non-conjugate types), $C_3$ ($8$ times), $C_4$ ($3$ times), $S_3$ ($4$ times). – Marc van Leeuwen Mar 15 '12 at 12:35
@MarcvanLeeuwen I have edited the mistakes. – user38268 Mar 15 '12 at 13:10
That $D_6$ is isomorphic to $S_3$ is easy: consider how $D_6$ acts one the corners of the triangle of which it is the symmetry group. However $D_{24}$ is not isomorphic to $S_4$ (consider elements of order $12$, or even of order $6$, which $S_4$ does not posess), so no amount of effort will suffice to prove it that way. The injective homomorphism (one of the many) $S_3\to S_4$ will not help you with $S_4$ either. However there is a surjective homomorphism $S_4\to S_3$ (think of a tetrahedron and the lines joining the midpoints of its sides) that will help you to prove that $S_4$ is solvable.
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Maybe I understand this wrong ? :"The groups S3,S4 and are isomorphic to the group of symmetries of the triangle and to the group of symmetries of the tetrahedron" - it's from a book I'm reading... – Belgi Mar 15 '12 at 9:38
@Belgi The group of symmetries of the tetrahedron is not $D_{24}$. – Alex Becker Mar 15 '12 at 9:45
@Belgi: Yes, but $D_{24}$ is not the symmetry group of a tetrahedron, but rather of a (plane) regular $12$-gon (pronounced dodecagon). – Marc van Leeuwen Mar 15 '12 at 9:45
Oh, what is the notation for the symmetry group of a tetrahedron ? thanks for the help! – Belgi Mar 15 '12 at 9:49
@Belgi: As you quoted yourself, the symmetry group of a tetrahedron is called (or isomorphic to) $S_4$. It is the group $D_{24}$ that is unrelated to your question. – Marc van Leeuwen Mar 15 '12 at 12:25 | 2016-06-01 07:57:03 | {"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": 0.913788914680481, "perplexity": 147.1159073710655}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-22/segments/1464056639771.99/warc/CC-MAIN-20160524022359-00219-ip-10-185-217-139.ec2.internal.warc.gz"} |
http://mathematica.stackexchange.com/questions/42546/new-list-generated-from-the-sum-of-elements-taken-from-two-lists | # New list generated from the sum of elements taken from two lists
I have been using mathematica for a while, but I have problem wrapping my head around list manipulation functions. I would appreciate any suggestions. Here is what I want to do.
ListV is a list of vectors:
ListV= {{x1,y1,z1},{x2,y2,z2},...}= {V1,V2,...}
ListU is also a list of vectors :
ListU={{a1,b1,c1},{a2,b2,c2},...}= {U1,U2,...}
ListU and ListV may have unequal lengths.
I want to make a new list (ListUV) which would contain all the new positions generated by the sum of all vectors in ListU and ListV. Mathematically, that would be
ListUV= {V1+U1, V2+U1, ..., V2+U1, V2+U2, ... }.
Thanks for any suggestions.
-
Is it what you want? Flatten[Outer[Plus, ListV, ListU], 1] – Yi Wang Feb 19 '14 at 13:19
You say V2+U1, ... V2+U1, V2+U2. Do you mean V2+U1, ... V3+U1, V3+U2 ? – Jacob Akkerboom Feb 19 '14 at 14:23
'Partition[Flatten[Outer[Plus, UnitCellPositions, LatticePositions, 1]], 3]' does exactly what I need. Thank you Yi Wang for the solution. – user3328102 Feb 19 '14 at 16:50
Thank you PlatoManiac for edited the formatting. – user3328102 Feb 19 '14 at 16:54
@YiWang Please consider posting an answer :) – Kuba Apr 24 '14 at 17:43 | 2015-11-28 23:50:14 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.43103253841400146, "perplexity": 3788.1339873676357}, "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-2015-48/segments/1448398454553.89/warc/CC-MAIN-20151124205414-00053-ip-10-71-132-137.ec2.internal.warc.gz"} |
https://en.wikipedia.org/wiki/Linearly_independent_cycle | # Cycle basis
(Redirected from Linearly independent cycle)
The symmetric difference of two cycles is an Eulerian subgraph
In graph theory, a branch of mathematics, a cycle basis of an undirected graph is a set of simple cycles that forms a basis of the cycle space of the graph. That is, it is a minimal set of cycles that allows every even-degree subgraph to be expressed as a symmetric difference of basis cycles.
A fundamental cycle basis may be formed from any spanning tree or spanning forest of the given graph, by selecting the cycles formed by the combination of a path in the tree and a single edge outside the tree. Alternatively, if the edges of the graph have positive weights, the minimum weight cycle basis may be constructed in polynomial time.
In planar graphs, the set of bounded cycles of an embedding of the graph forms a cycle basis. The minimum weight cycle basis of a planar graph corresponds to the Gomory–Hu tree of the dual graph.
## Definitions
A spanning subgraph of a given graph G has the same set of vertices as G itself but, possibly, fewer edges. A graph G, or one of its subgraphs, is said to be Eulerian if each of its vertices has even degree (its number of incident edges). Every simple cycle in a graph is an Eulerian subgraph, but there may be others. The cycle space of a graph is the collection of its Eulerian subgraphs. It forms a vector space over the two-element finite field. The vector addition operation is the symmetric difference of two or more subgraphs, which forms another subgraph consisting of the edges that appear an odd number of times in the arguments to the symmetric difference operation.[1]
A cycle basis is a basis of this vector space in which each basis vector represents a simple cycle. It consists of a set of cycles that can be combined, using symmetric differences, to form every Eulerian subgraph, and that is minimal with this property. Every cycle basis of a given graph has the same number of cycles, which equals the dimension of its cycle space. This number is called the circuit rank of the graph, and it equals ${\displaystyle m-n+c}$ where ${\displaystyle m}$ is the number of edges in the graph, ${\displaystyle n}$ is the number of vertices, and ${\displaystyle c}$ is the number of connected components.[2]
## Special cycle bases
Several special types of cycle bases have been studied, including the fundamental cycle bases, weakly fundamental cycle bases, sparse (or 2-) cycle bases, and integral cycle bases.[3]
### Induced cycles
Every graph has a cycle basis in which every cycle is an induced cycle. In a 3-vertex-connected graph, there always exists a basis consisting of peripheral cycles, cycles whose removal does not separate the remaining graph.[4][5] In any graph other than one formed by adding one edge to a cycle, a peripheral cycle must be an induced cycle.
### Fundamental cycles
If ${\displaystyle T}$ is a spanning tree or spanning forest of a given graph ${\displaystyle G}$, and ${\displaystyle e}$ is an edge that does not belong to ${\displaystyle T}$, then the fundamental cycle ${\displaystyle C_{e}}$ defined by ${\displaystyle e}$ is the simple cycle consisting of ${\displaystyle e}$ together with the path in ${\displaystyle T}$ connecting the endpoints of ${\displaystyle e}$. There are exactly ${\displaystyle m-n+c}$ fundamental cycles, one for each edge that does not belong to ${\displaystyle T}$. Each of them is linearly independent from the remaining cycles, because it includes an edge ${\displaystyle e}$ that is not present in any other fundamental cycle. Therefore, the fundamental cycles form a basis for the cycle space.[1][2] A cycle basis constructed in this way is called a fundamental cycle basis or strongly fundamental cycle basis.[3]
It is also possible to characterize fundamental cycle bases without specifying the tree for which they are fundamental. There exists a tree for which a given cycle basis is fundamental if and only if each cycle contains an edge that is not included in any other basis cycle, that is, each cycle is independent of others. It follows that a collection of cycles is a fundamental cycle basis of ${\displaystyle G}$ if and only if it has the independence property and has the correct number of cycles to be a basis of ${\displaystyle G}$.[6]
### Weakly fundamental cycles
A cycle basis is called weakly fundamental if its cycles can be placed into a linear ordering such that each cycle includes at least one edge that is not included in any earlier cycle. A fundamental cycle basis is automatically weakly fundamental (for any edge ordering).[3][7] If every cycle basis of a graph is weakly fundamental, the same is true for every minor of the graph. Based on this property, the class of graphs (and multigraphs) for which every cycle basis is weakly fundamental can be characterized by five forbidden minors: the graph of the square pyramid, the multigraph formed by doubling all edges of a four-vertex cycle, two multigraphs formed by doubling two edges of a tetrahedron, and the multigraph formed by tripling the edges of a triangle.[8]
### Face cycles
If a connected finite planar graph is embedded into the plane, each face of the embedding is bounded by a cycle of edges. One face is necessarily unbounded (it includes points arbitrarily far from the vertices of the graph) and the remaining faces are bounded. By Euler's formula for planar graphs, there are exactly ${\displaystyle m-n+1}$ bounded faces. The symmetric difference of any set of face cycles is the boundary of the corresponding set of faces, and different sets of bounded faces have different boundaries, so it is not possible to represent the same set as a symmetric difference of face cycles in more than one way; this means that the set of face cycles is linearly independent. As a linearly independent set of enough cycles, it necessarily forms a cycle basis.[9] It is always a weakly fundamental cycle basis, and is fundamental if and only if the embedding of the graph is outerplanar.
For graphs properly embedded onto other surfaces so that all faces of the embedding are topological disks, it is not in general true that there exists a cycle basis using only face cycles. The face cycles of these embeddings generate a proper subset of all Eulerian subgraphs. The homology group ${\displaystyle H_{2}(S,\mathbb {Z} _{2})}$ of the given surface ${\displaystyle S}$ characterizes the Eulerian subgraphs that cannot be represented as the boundary of a set of faces. Mac Lane's planarity criterion uses this idea to characterize the planar graphs in terms of the cycle bases: a finite undirected graph is planar if and only if it has a sparse cycle basis or 2-basis,[3] a basis in which each edge of the graph participates in at most two basis cycles. In a planar graph, the cycle basis formed by the set of bounded faces is necessarily sparse, and conversely, a sparse cycle basis of any graph necessarily forms the set of bounded faces of a planar embedding of its graph.[9][10]
### Integral bases
The cycle space of a graph may be interpreted using the theory of homology as the homology group ${\displaystyle H_{1}(G,\mathbb {Z} _{2})}$ of a simplicial complex with a point for each vertex of the graph and a line segment for each edge of the graph. This construction may be generalized to the homology group ${\displaystyle H_{1}(G,R)}$ over an arbitrary ring ${\displaystyle R}$. An important special case is the ring of integers, for which the homology group ${\displaystyle H_{1}(G,\mathbb {Z} )}$ is a free abelian group, a subgroup of the free abelian group generated by the edges of the graph. Less abstractly, this group can be constructed by assigning an arbitrary orientation to the edges of the given graph; then the elements of ${\displaystyle H_{1}(G,\mathbb {Z} )}$ are labelings of the edges of the graph by integers with the property that, at each vertex, the sum of the incoming edge labels equals the sum of the outgoing edge labels. The group operation is addition of these vectors of labels. An integral cycle basis is a set of simple cycles that generates this group.[3]
## Minimum weight
If the edges of a graph are given real number weights, the weight of a subgraph may be computed as the sum of the weights of its edges. The minimum weight basis of the cycle space is necessarily a cycle basis: by Veblen's theorem,[11] every Eulerian subgraph that is not itself a simple cycle can be decomposed into multiple simple cycles, which necessarily have smaller weight.
By standard properties of bases in vector spaces and matroids, the minimum weight cycle basis not only minimizes the sum of the weights of its cycles, it also minimizes any other monotonic combination of the cycle weights. For instance, it is the cycle basis that minimizes the weight of its longest cycle.[12]
### Polynomial time algorithms
In any vector space, and more generally in any matroid, a minimum weight basis may be found by a greedy algorithm that considers potential basis elements one at a time, in sorted order by their weights, and that includes an element in the basis when it is linearly independent of the previously chosen basis elements. Testing for linear independence can be done by Gaussian elimination. However, an undirected graph may have an exponentially large set of simple cycles, so it would be computationally infeasible to generate and test all such cycles.
Horton (1987) provided the first polynomial time algorithm for finding a minimum weight basis, in graphs for which every edge weight is positive. His algorithm uses this generate-and-test approach, but restricts the generated cycles to a small set of ${\displaystyle O(mn)}$ cycles, called Horton cycles. A Horton cycle is a fundamental cycle of a shortest path tree of the given graph. There are at most n different shortest path trees (one for each starting vertex) and each has fewer than m fundamental cycles, giving the bound on the total number of Horton cycles. As Horton showed, every cycle in the minimum weight cycle basis is a Horton cycle.[13] Using Dijkstra's algorithm to find each shortest path tree and then using Gaussian elimination to perform the testing steps of the greedy basis algorithm leads to a polynomial time algorithm for the minimum weight cycle basis. Subsequent researchers have developed improved algorithms for this problem,[14][15][16][17] reducing the worst-case time complexity for finding a minimum weight cycle basis in a graph with ${\displaystyle m}$ edges and ${\displaystyle n}$ vertices to ${\displaystyle O(m^{2}n/\log n)}$.[18]
### NP-hardness
Finding the fundamental basis with the minimum possible weight is closely related to the problem of finding a spanning tree that minimizes the average of the pairwise distances; both are NP-hard.[19] Finding a minimum weight weakly fundamental basis is also NP-hard,[7] and approximating it is MAXSNP-hard.[20] If negative weights and negatively weighted cycles are allowed, then finding a minimum cycle basis (without restriction) is also NP-hard, as it can be used to find a Hamiltonian cycle: if a graph is Hamiltonian, and all edges are given weight −1, then a minimum weight cycle basis necessarily includes at least one Hamiltonian cycle.
### In planar graphs
The minimum weight cycle basis for a planar graph is not necessarily the same as the basis formed by its bounded faces: it can include cycles that are not faces, and some faces may not be included as cycles in the minimum weight cycle basis. However, there exists a minimum weight cycle basis in which no two cycles cross each other: for every two cycles in the basis, either the cycles enclose disjoint subsets of the bounded faces, or one of the two cycles encloses the other one. This set of cycles corresponds, in the dual graph of the given planar graph, to a set of cuts that form a Gomory–Hu tree of the dual graph, the minimum weight basis of its cut space.[21] Based on this duality, an implicit representation of the minimum weight cycle basis in a planar graph can be constructed in time ${\displaystyle O(n\log ^{3}n)}$.[22]
## Applications
Cycle bases have been used for solving periodic scheduling problems, such as the problem of determining the schedule for a public transportation system. In this application, the cycles of a cycle basis correspond to variables in an integer program for solving the problem.[23]
In the theory of structural rigidity and kinematics, cycle bases are used to guide the process of setting up a system of non-redundant equations that can be solved to predict the rigidity or motion of a structure. In this application, minimum or near-minimum weight cycle bases lead to simpler systems of equations.[24]
In distributed computing, cycle bases have been used to analyze the number of steps needed for an algorithm to stabilize.[25]
In bioinformatics, cycle bases have been used to determine haplotype information from genome sequence data.[26] Cycle bases have also been used to analyze the tertiary structure of RNA.[27]
The minimum weight cycle basis of a nearest neighbor graph of points sampled from a three-dimensional surface can be used to obtain a reconstruction of the surface.[28]
In cheminformatics, the minimal cycle basis of a molecular graph is referred to as the smallest set of smallest rings.[29][30][31]
## References
1. ^ a b Diestel, Reinhard (2012), "1.9 Some linear algebra", Graph Theory, Graduate Texts in Mathematics, vol. 173, Springer, pp. 23–28.
2. ^ a b Gross, Jonathan L.; Yellen, Jay (2005), "4.6 Graphs and Vector Spaces", Graph Theory and Its Applications (2nd ed.), CRC Press, pp. 197–207, ISBN 9781584885054.
3. Liebchen, Christian; Rizzi, Romeo (2007), "Classes of cycle bases", Discrete Applied Mathematics, 155 (3): 337–355, doi:10.1016/j.dam.2006.06.007, MR 2303157.
4. ^ Diestel (2012), pp. 32, 65.
5. ^ Tutte, W. T. (1963), "How to draw a graph", Proceedings of the London Mathematical Society, Third Series, 13: 743–767, doi:10.1112/plms/s3-13.1.743, MR 0158387. See in particular Theorem 2.5.
6. ^ Cribb, D. W.; Ringeisen, R. D.; Shier, D. R. (1981), "On cycle bases of a graph", Proceedings of the Twelfth Southeastern Conference on Combinatorics, Graph Theory and Computing, Vol. I (Baton Rouge, La., 1981), Congressus Numerantium, vol. 32, pp. 221–229, MR 0681883.
7. ^ a b Rizzi, Romeo (2009), "Minimum weakly fundamental cycle bases are hard to find", Algorithmica, 53 (3): 402–424, doi:10.1007/s00453-007-9112-8, MR 2482112, S2CID 12675654.
8. ^ Hartvigsen, David; Zemel, Eitan (1989), "Is every cycle basis fundamental?", Journal of Graph Theory, 13 (1): 117–137, doi:10.1002/jgt.3190130115, MR 0982873.
9. ^ a b Diestel (2012), pp. 105–106.
10. ^ Mac Lane, S. (1937), "A combinatorial condition for planar graphs" (PDF), Fundamenta Mathematicae, 28: 22–32, doi:10.4064/fm-28-1-22-32.
11. ^ Veblen, Oswald (1912), "An application of modular equations in analysis situs", Annals of Mathematics, Second Series, 14 (1): 86–94, doi:10.2307/1967604, JSTOR 1967604.
12. ^ Chickering, David M.; Geiger, Dan; Heckerman, David (1995), "On finding a cycle basis with a shortest maximal cycle", Information Processing Letters, 54 (1): 55–58, CiteSeerX 10.1.1.650.8218, doi:10.1016/0020-0190(94)00231-M, MR 1332422.
13. ^ Horton, J. D. (1987), "A polynomial-time algorithm to find the shortest cycle basis of a graph", SIAM Journal on Computing, 16 (2): 358–366, doi:10.1137/0216026.
14. ^ Berger, Franziska; Gritzmann, Peter; de Vries, Sven (2004), "Minimum cycle bases for network graphs", Algorithmica, 40 (1): 51–62, doi:10.1007/s00453-004-1098-x, MR 2071255, S2CID 9386078.
15. ^ Mehlhorn, Kurt; Michail, Dimitrios (2006), "Implementing minimum cycle basis algorithms", ACM Journal of Experimental Algorithmics, 11: 2.5, doi:10.1145/1187436.1216582, S2CID 6198296.
16. ^ Kavitha, Telikepalli; Mehlhorn, Kurt; Michail, Dimitrios; Paluch, Katarzyna E. (2008), "An ${\displaystyle {\tilde {O}}(m^{2}n)}$ algorithm for minimum cycle basis of graphs", Algorithmica, 52 (3): 333–349, doi:10.1007/s00453-007-9064-z, MR 2452919.
17. ^ Kavitha, Telikepalli; Liebchen, Christian; Mehlhorn, Kurt; Michail, Dimitrios; Rizzi, Romeo; Ueckerdt, Torsten; Zweig, Katharina A. (2009), "Cycle bases in graphs: Characterization, algorithms, complexity, and applications", Computer Science Review, 3 (4): 199–243, doi:10.1016/j.cosrev.2009.08.001.
18. ^ Amaldi, Edoardo; Iuliano, Claudio; Rizzi, Romeo (2010), "Efficient deterministic algorithms for finding a minimum cycle basis in undirected graphs", Integer Programming and Combinatorial Optimization: 14th International Conference, IPCO 2010, Lausanne, Switzerland, June 9-11, 2010, Proceedings, Lecture Notes in Computer Science, vol. 6080, Springer, pp. 397–410, Bibcode:2010LNCS.6080..397A, doi:10.1007/978-3-642-13036-6_30, ISBN 978-3-642-13035-9, MR 2661113.
19. ^ Deo, Narsingh; Prabhu, G. M.; Krishnamoorthy, M. S. (1982), "Algorithms for generating fundamental cycles in a graph", ACM Transactions on Mathematical Software, 8 (1): 26–42, doi:10.1145/355984.355988, MR 0661120, S2CID 2260051.
20. ^ Galbiati, Giulia; Amaldi, Edoardo (2004), "On the approximability of the minimum fundamental cycle basis problem", Approximation and Online Algorithms: First International Workshop, WAOA 2003, Budapest, Hungary, September 16-18, 2003, Revised Papers, Lecture Notes in Computer Science, vol. 2909, Berlin: Springer, pp. 151–164, doi:10.1007/978-3-540-24592-6_12, ISBN 978-3-540-21079-5, MR 2089904.
21. ^ Hartvigsen, David; Mardon, Russell (1994), "The all-pairs min cut problem and the minimum cycle basis problem on planar graphs", SIAM Journal on Discrete Mathematics, 7 (3): 403–418, doi:10.1137/S0895480190177042, MR 1285579.
22. ^ Borradaile, Glencora; Eppstein, David; Nayyeri, Amir; Wulff-Nilsen, Christian (2016), "All-pairs minimum cuts in near-linear time for surface-embedded graphs", Proc. 32nd Int. Symp. Computational Geometry, Leibniz International Proceedings in Informatics (LIPIcs), vol. 51, Schloss Dagstuhl, pp. 22:1–22:16, arXiv:1411.7055, doi:10.4230/LIPIcs.SoCG.2016.22.
23. ^ Liebchen, Christian (2007), "Periodic timetable optimization in public transport", Operations Research Proceedings, 2006: 29–36, doi:10.1007/978-3-540-69995-8_5, ISBN 978-3-540-69994-1.
24. ^ Cassell, A. C.; De Henderson, J. C.; Kaveh, A. (1974), "Cycle bases for the flexibility analysis of structures", International Journal for Numerical Methods in Engineering, 8 (3): 521–528, Bibcode:1974IJNME...8..521C, doi:10.1002/nme.1620080308.
25. ^ Boulinier, Christian; Petit, Franck; Villain, Vincent (2004), "When graph theory helps self-stabilization", Proceedings of the Twenty-Third Annual ACM Symposium on Principles of Distributed Computing (PODC '04), New York, NY, USA: ACM, pp. 150–159, CiteSeerX 10.1.1.79.2190, doi:10.1145/1011767.1011790, ISBN 978-1581138023, S2CID 14936510.
26. ^ Aguiar, Derek; Istrail, Sorin (2012), "HapCompass: A Fast Cycle Basis Algorithm for Accurate Haplotype Assembly of Sequence Data", Journal of Computational Biology, 19 (6): 577–590, doi:10.1089/cmb.2012.0084, PMC 3375639, PMID 22697235.
27. ^ Lemieux, Sébastien; Major, François (2006), "Automated extraction and classification of RNA tertiary structure cyclic motifs", Nucleic Acids Research, 34 (8): 2340–2346, doi:10.1093/nar/gkl120, PMC 1458283, PMID 16679452.
28. ^ Gotsman, Craig; Kaligosi, Kanela; Mehlhorn, Kurt; Michail, Dimitrios; Pyrga, Evangelia (2007), "Cycle bases of graphs and sampled manifolds", Computer Aided Geometric Design, 24 (8–9): 464–480, CiteSeerX 10.1.1.298.9661, doi:10.1016/j.cagd.2006.07.001, MR 2359763.
29. ^ May, John W.; Steinbeck, Christoph (2014), "Efficient ring perception for the Chemistry Development Kit", Journal of Cheminformatics, 6 (3): 3, doi:10.1186/1758-2946-6-3, PMC 3922685, PMID 24479757
30. ^ Downs, G.M.; Gillet, V.J.; Holliday, J.D.; Lynch, M.F. (1989), "A review of ring perception algorithms for chemical graphs", J. Chem. Inf. Comput. Sci., 29 (3): 172–187, doi:10.1021/ci00063a007
31. ^ Zamora, A. (1979), "An algorithm for finding the smallest set of smallest rings", J. Chem. Inf. Comput. Sci., 16 (1): 40–43, doi:10.1021/ci60005a013 | 2022-08-09 05:28:38 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 32, "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": 0.7317094802856445, "perplexity": 661.8334466158053}, "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/1659882570901.18/warc/CC-MAIN-20220809033952-20220809063952-00604.warc.gz"} |
https://plainmath.net/post-secondary/calculus-and-analysis/analysis | # Analysis questions with answers
Recent questions in Analysis
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### For a given function is there a general procedure to find an initial value for ${x}_{1}$ such that Newton's method bounces back and forth between two values forever?
Jared Irwin 2022-08-30
### Can Newton's method be used to guarantee convergence to a global minimum or maximum?
Mahak Diwakar 2022-08-23
Bellenik3 2022-08-19
### Use newton's method with the specified initial approximation $x1$ to find $x3$, the third appromizmation to the root of the given equation${x}^{5}-x-1=0$, ${x}_{1}=1$
betterthennewzv 2022-08-16
### Prove, that the direction in Newton's method is a descent direction if the Hessian is positive defnite.$direction=-H\left(x{\right)}^{-1}\ast \mathrm{\nabla }f\left(x\right)$prove that this is a descent direction?
Lacey Rojas 2022-08-14
### How many iterations must I do for getting $n$ signs after floating point in calculating square root by Newton's method?
The Math analysis problems represent one of the most challenging problems for college students because the use of analysis is never easy. The best thing you can do is to see the answers to the questions that you have. You can take these analysis examples and examine each line to see how these came about. If you are unsure about something, it is good to start with the problem and pose it in different words. It will help you to achieve success with the logical proceeding as you provide analysis for the changes, alterations, assumptions, or mathematical-statistical data being used. | 2023-01-29 03:07:05 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 32, "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": 0.8255422115325928, "perplexity": 665.2168247456958}, "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-2023-06/segments/1674764499697.75/warc/CC-MAIN-20230129012420-20230129042420-00208.warc.gz"} |
http://mathhelpforum.com/differential-geometry/145189-proving-limit-function.html | # Math Help - Proving the limit of a function
1. ## Proving the limit of a function
I can't get my head around these types of questions i know how to prove a limit you use $|f(x) - L| < \epsilon$ but the question is use the (N, ε) method prove $lim_{x \rightarrow \infty} \frac{7n + 3}{8n + 5} = \frac{7}{8}$
I've read though the notes ive taken on this but still dont get it id be really grateful for some pointers
2. You have to show that for any $\epsilon > 0$, there exists an $N$ (depending on $\epsilon$ ) such that whenever $n > N$ we have that $\Biggl| \frac{7n + 3}{8n + 5} - \frac{7}{8}\Biggr|<\epsilon$.
First put $\frac{7n + 3}{8n + 5} - \frac{7}{8}$ over a common denominator and simplify. You should then be able to get $N$ from the result. | 2014-07-23 04:51:45 | {"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": 9, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9210496544837952, "perplexity": 145.43644712848254}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-23/segments/1405997873839.53/warc/CC-MAIN-20140722025753-00218-ip-10-33-131-23.ec2.internal.warc.gz"} |
https://lists.gnu.org/archive/html/emacs-orgmode/2014-05/msg00667.html | emacs-orgmode
[Top][All Lists]
## Re: [O] *text* in headlines and export to latex
From: Seb Frank Subject: Re: [O] *text* in headlines and export to latex Date: Thu, 15 May 2014 23:34:35 +0100
Is this customizable at all? I've seen some old threads where people
used @ for alert (and a hack to replace it with something else if the
http://lists.gnu.org/archive/html/emacs-orgmode/2010-01/msg00614.html
and http://lists.gnu.org/archive/html/emacs-orgmode/2010-01/msg00754.html
), but the solutions mentioned there don't seem to work with the
current version of org.
On 5/15/14, Sebastien Vauban <address@hidden> wrote:
> Seb Frank wrote:
>> On Wed, May 14, 2014 at 7:19 PM, Eric S Fraga <address@hidden> wrote:
>>> On Wednesday, 14 May 2014 at 18:00, Seb Frank wrote:
>>>> Latex export converts *text* to \textbf -- this doesn't work well
>>>> for headlines that get turned into section titles as these appear to
>>>> be bold by default. Is there a way to change this to \emph in
>>>> headlines that get converted to (sub)sections?
>>>
>>> Silly question but: can you not simply use /text/ instead in the
>>> headlines? Mind you, I think that bold + emph is not pretty... but
>>> that's a personal taste issue!
>>
>> Yes -- but that changes it to emph (instead of alert) in beamer export.
>
> As a side note, as both \alert and \textbf do exist in Beamer, I do
> think *text* should never be translated to \alert.
>
> That's because of that different translation between the two backends
> that you see that problem only in one backend.
>
> Best regards,
> Seb
>
> --
> Sebastien Vauban
>
>
> | 2021-09-17 08:21:12 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8471396565437317, "perplexity": 10781.451759548783}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-39/segments/1631780055601.25/warc/CC-MAIN-20210917055515-20210917085515-00614.warc.gz"} |
https://aviation.meta.stackexchange.com/questions/2888/what-should-be-the-preferred-way-to-provide-attribution-for-images | # What should be the preferred way to provide attribution for images?
This is prompted by a recent discussion below a post (I link to my final comment, but please take the time to read all the comments there).
I have the feeling that recently the way images are attributed might not be satisfactory, and the discussion linked provide an example of why I think so. If the attribution would have been given since the beginning most of that discussion would not have happened.
If you go back to the page linked above, you will notice that also the other answer has extremely generic attributions for the photos ("airliners.net", like they have only a couple of photos on their website).
Compare this with an edit of a year and a half ago: the link was already there, but not obvious enough and it got added again [please note that I appreciated the edit and I am in no way complaining about it]
Should we require then that all photos be sourced with a direct link to the page from which they are taken? Or is "boeing.com" (and similars) an acceptable attribution?
The first part of your question was sort-of discussed over here - to broadly restate my opinion (which, for the record, is based on US copyright standards): A direct reference (link) isn't always necessary for every image used (though sometimes it is - images licensed under CC:By for example).
I would not suggest that we implement an "all images must be explicitly attributed" policy as a global requirement for this site, but rather that we encourage common sense: Attribution should be included where it is requested by the original content's author, where it is relevant/helpful in understanding the answer, and in any other case where it can be done without seriously impacting readability.
For two recent examples: when you crib a diagram from an FAA Advisory Circular linking to the AC is helpful, but linking to the source for the compass card photo wouldn't be particularly useful (as it's a pretty generic image), and linking to the source for the compass photo would be dumb (I cropped that out of a 3MB full-resolution photo of an instrument panel - the compass is the only relevant bit).
Since we require attribution when people use our content it's generally the decent thing to do when we use someone else's content, but ultimately "Google Image Search" is a thing that exists, and if someone is really upset we're using their image without proper credit they can file a DMCA notice with Stack Exchange and the issue will be dealt with.
To the new point you're raising: From a content-quality standpoint bad/generic attribution ("It's from Boeing", "It's from airliners.net", etc.) is (IMHO) worse than no attribution.
The point of attribution on a site like this is not just to credit the original author: It's providing a citation to enable an unaffiliated third party reader to find your source materialand assess its credibility. If the citation (attribution) doesn't lead to a useful place it's not helpful.
Again using one of my recent examples: If I simply said the compass rose diagram came "From the FAA" you would be hard pressed to turn up the specific advisory circular I was looking at: It provides attribution, but it's not a useful citation for someone looking for more information.
By linking to the advisory circular I lifted the image from I'm providing useful context that would enable someone to continue their own research if they wanted to know more about compass pad design and construction (and saving myself having to retype large chunks of the AC to give context to my answer).
I'm supporting most of @Federico and @voretaq7 ideas. My solution to referencing a resource is this one:
(source)
Just for clarification, there are distinct aims behind the attrition:
• Fulfill legal obligations: Difficult here to know what they are, due to the multiple legal frameworks involved around the world (and maybe the lack of knowledge of the server location).
Be aware that there are quite unknown laws related to e.g. architectural work (see 1.6, in France even if the building is in public view, even if it's paid by the tax payer, it's not allowed to publish a photography, even yours. It applies as well to private or public gates (those guys living in the past centuries are quite ridiculous in the era of photography and video devices embedded in every object). Fortunately Europe institutions are starting to look at this aspect to relax obligations.
For this legal aspect, I'll keep off unfair and/or damageable use based on my understanding of interests, and local laws (the one I'm aware of...)
• Be honest and thankful to the author, inform that this is not your own work, that you are using work from someone else. It's a must if you are a bit brought up. I think providing a link (as I usually do) with the clickable area saying (source) and linking to the containing page is enough, as I'm not either advertising their names or sites, or whatever (I already drive users to their work), and I'm not listing all the legal aspects that allow me to use the resource.
• Allow others to assess whether it's possible to reuse this image: The link used to provide the source is quite enough to dig into this topic. It's the reader's duty to do this assessment, not mine.
• Allow others to look at the information associated with the resource: It's typically missing in this question. That's why it's better to provide a link to the containing page, not to the resource itself.
One exception I'd have is to provide the full question title as clickable when linking to SE pages, to inform this resource is from the site.
A personal comment: Several questions are clearly inspired by looking at Quora. I've seen some pushing it to reuse the image the original author provided :( Do we want the site to grow this way?
By the way, if the workers (or their families) on the image above had been paid by the many who made money directly or indirectly with the photo, they would be very rich. That's only recently the author of this image was given credit.
The stackexchange network provides all Q/A under a free license. It will significantly benefit from proper attribution and proper licenses in the pictures. We should adopt the well established wikipedia-way for references.
The only difference to wikipedia is that self-published media makes sense on sx and there should be a simple button to tag it as such.
We do not have a free Q/A database, if the pictures or other content has bad licenses or if they lack proper attribution. | 2021-04-12 16:46:47 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.22777922451496124, "perplexity": 1290.042762292153}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-17/segments/1618038067870.12/warc/CC-MAIN-20210412144351-20210412174351-00467.warc.gz"} |
https://people.maths.bris.ac.uk/~matyd/GroupNames/321/D7xDic6.html | Copied to
clipboard
## G = D7×Dic6order 336 = 24·3·7
### Direct product of D7 and Dic6
Series: Derived Chief Lower central Upper central
Derived series C1 — C42 — D7×Dic6
Chief series C1 — C7 — C21 — C42 — C6×D7 — Dic3×D7 — D7×Dic6
Lower central C21 — C42 — D7×Dic6
Upper central C1 — C2 — C4
Generators and relations for D7×Dic6
G = < a,b,c,d | a7=b2=c12=1, d2=c6, bab=a-1, ac=ca, ad=da, bc=cb, bd=db, dcd-1=c-1 >
Subgroups: 372 in 76 conjugacy classes, 36 normal (22 characteristic)
C1, C2, C2, C3, C4, C4, C22, C6, C6, C7, C2×C4, Q8, Dic3, Dic3, C12, C12, C2×C6, D7, C14, C2×Q8, C21, Dic6, Dic6, C2×Dic3, C2×C12, Dic7, Dic7, C28, C28, D14, C3×D7, C42, C2×Dic6, Dic14, C4×D7, C4×D7, C7×Q8, C7×Dic3, C3×Dic7, Dic21, C84, C6×D7, Q8×D7, Dic3×D7, C21⋊Q8, C12×D7, C7×Dic6, Dic42, D7×Dic6
Quotients: C1, C2, C22, S3, Q8, C23, D6, D7, C2×Q8, Dic6, C22×S3, D14, C2×Dic6, C22×D7, S3×D7, Q8×D7, C2×S3×D7, D7×Dic6
Smallest permutation representation of D7×Dic6
On 168 points
Generators in S168
(1 91 21 113 74 36 70)(2 92 22 114 75 25 71)(3 93 23 115 76 26 72)(4 94 24 116 77 27 61)(5 95 13 117 78 28 62)(6 96 14 118 79 29 63)(7 85 15 119 80 30 64)(8 86 16 120 81 31 65)(9 87 17 109 82 32 66)(10 88 18 110 83 33 67)(11 89 19 111 84 34 68)(12 90 20 112 73 35 69)(37 163 148 103 143 56 129)(38 164 149 104 144 57 130)(39 165 150 105 133 58 131)(40 166 151 106 134 59 132)(41 167 152 107 135 60 121)(42 168 153 108 136 49 122)(43 157 154 97 137 50 123)(44 158 155 98 138 51 124)(45 159 156 99 139 52 125)(46 160 145 100 140 53 126)(47 161 146 101 141 54 127)(48 162 147 102 142 55 128)
(1 64)(2 65)(3 66)(4 67)(5 68)(6 69)(7 70)(8 71)(9 72)(10 61)(11 62)(12 63)(13 84)(14 73)(15 74)(16 75)(17 76)(18 77)(19 78)(20 79)(21 80)(22 81)(23 82)(24 83)(25 86)(26 87)(27 88)(28 89)(29 90)(30 91)(31 92)(32 93)(33 94)(34 95)(35 96)(36 85)(37 123)(38 124)(39 125)(40 126)(41 127)(42 128)(43 129)(44 130)(45 131)(46 132)(47 121)(48 122)(49 162)(50 163)(51 164)(52 165)(53 166)(54 167)(55 168)(56 157)(57 158)(58 159)(59 160)(60 161)(97 103)(98 104)(99 105)(100 106)(101 107)(102 108)(109 115)(110 116)(111 117)(112 118)(113 119)(114 120)(133 156)(134 145)(135 146)(136 147)(137 148)(138 149)(139 150)(140 151)(141 152)(142 153)(143 154)(144 155)
(1 2 3 4 5 6 7 8 9 10 11 12)(13 14 15 16 17 18 19 20 21 22 23 24)(25 26 27 28 29 30 31 32 33 34 35 36)(37 38 39 40 41 42 43 44 45 46 47 48)(49 50 51 52 53 54 55 56 57 58 59 60)(61 62 63 64 65 66 67 68 69 70 71 72)(73 74 75 76 77 78 79 80 81 82 83 84)(85 86 87 88 89 90 91 92 93 94 95 96)(97 98 99 100 101 102 103 104 105 106 107 108)(109 110 111 112 113 114 115 116 117 118 119 120)(121 122 123 124 125 126 127 128 129 130 131 132)(133 134 135 136 137 138 139 140 141 142 143 144)(145 146 147 148 149 150 151 152 153 154 155 156)(157 158 159 160 161 162 163 164 165 166 167 168)
(1 45 7 39)(2 44 8 38)(3 43 9 37)(4 42 10 48)(5 41 11 47)(6 40 12 46)(13 152 19 146)(14 151 20 145)(15 150 21 156)(16 149 22 155)(17 148 23 154)(18 147 24 153)(25 51 31 57)(26 50 32 56)(27 49 33 55)(28 60 34 54)(29 59 35 53)(30 58 36 52)(61 122 67 128)(62 121 68 127)(63 132 69 126)(64 131 70 125)(65 130 71 124)(66 129 72 123)(73 140 79 134)(74 139 80 133)(75 138 81 144)(76 137 82 143)(77 136 83 142)(78 135 84 141)(85 165 91 159)(86 164 92 158)(87 163 93 157)(88 162 94 168)(89 161 95 167)(90 160 96 166)(97 109 103 115)(98 120 104 114)(99 119 105 113)(100 118 106 112)(101 117 107 111)(102 116 108 110)
G:=sub<Sym(168)| (1,91,21,113,74,36,70)(2,92,22,114,75,25,71)(3,93,23,115,76,26,72)(4,94,24,116,77,27,61)(5,95,13,117,78,28,62)(6,96,14,118,79,29,63)(7,85,15,119,80,30,64)(8,86,16,120,81,31,65)(9,87,17,109,82,32,66)(10,88,18,110,83,33,67)(11,89,19,111,84,34,68)(12,90,20,112,73,35,69)(37,163,148,103,143,56,129)(38,164,149,104,144,57,130)(39,165,150,105,133,58,131)(40,166,151,106,134,59,132)(41,167,152,107,135,60,121)(42,168,153,108,136,49,122)(43,157,154,97,137,50,123)(44,158,155,98,138,51,124)(45,159,156,99,139,52,125)(46,160,145,100,140,53,126)(47,161,146,101,141,54,127)(48,162,147,102,142,55,128), (1,64)(2,65)(3,66)(4,67)(5,68)(6,69)(7,70)(8,71)(9,72)(10,61)(11,62)(12,63)(13,84)(14,73)(15,74)(16,75)(17,76)(18,77)(19,78)(20,79)(21,80)(22,81)(23,82)(24,83)(25,86)(26,87)(27,88)(28,89)(29,90)(30,91)(31,92)(32,93)(33,94)(34,95)(35,96)(36,85)(37,123)(38,124)(39,125)(40,126)(41,127)(42,128)(43,129)(44,130)(45,131)(46,132)(47,121)(48,122)(49,162)(50,163)(51,164)(52,165)(53,166)(54,167)(55,168)(56,157)(57,158)(58,159)(59,160)(60,161)(97,103)(98,104)(99,105)(100,106)(101,107)(102,108)(109,115)(110,116)(111,117)(112,118)(113,119)(114,120)(133,156)(134,145)(135,146)(136,147)(137,148)(138,149)(139,150)(140,151)(141,152)(142,153)(143,154)(144,155), (1,2,3,4,5,6,7,8,9,10,11,12)(13,14,15,16,17,18,19,20,21,22,23,24)(25,26,27,28,29,30,31,32,33,34,35,36)(37,38,39,40,41,42,43,44,45,46,47,48)(49,50,51,52,53,54,55,56,57,58,59,60)(61,62,63,64,65,66,67,68,69,70,71,72)(73,74,75,76,77,78,79,80,81,82,83,84)(85,86,87,88,89,90,91,92,93,94,95,96)(97,98,99,100,101,102,103,104,105,106,107,108)(109,110,111,112,113,114,115,116,117,118,119,120)(121,122,123,124,125,126,127,128,129,130,131,132)(133,134,135,136,137,138,139,140,141,142,143,144)(145,146,147,148,149,150,151,152,153,154,155,156)(157,158,159,160,161,162,163,164,165,166,167,168), (1,45,7,39)(2,44,8,38)(3,43,9,37)(4,42,10,48)(5,41,11,47)(6,40,12,46)(13,152,19,146)(14,151,20,145)(15,150,21,156)(16,149,22,155)(17,148,23,154)(18,147,24,153)(25,51,31,57)(26,50,32,56)(27,49,33,55)(28,60,34,54)(29,59,35,53)(30,58,36,52)(61,122,67,128)(62,121,68,127)(63,132,69,126)(64,131,70,125)(65,130,71,124)(66,129,72,123)(73,140,79,134)(74,139,80,133)(75,138,81,144)(76,137,82,143)(77,136,83,142)(78,135,84,141)(85,165,91,159)(86,164,92,158)(87,163,93,157)(88,162,94,168)(89,161,95,167)(90,160,96,166)(97,109,103,115)(98,120,104,114)(99,119,105,113)(100,118,106,112)(101,117,107,111)(102,116,108,110)>;
G:=Group( (1,91,21,113,74,36,70)(2,92,22,114,75,25,71)(3,93,23,115,76,26,72)(4,94,24,116,77,27,61)(5,95,13,117,78,28,62)(6,96,14,118,79,29,63)(7,85,15,119,80,30,64)(8,86,16,120,81,31,65)(9,87,17,109,82,32,66)(10,88,18,110,83,33,67)(11,89,19,111,84,34,68)(12,90,20,112,73,35,69)(37,163,148,103,143,56,129)(38,164,149,104,144,57,130)(39,165,150,105,133,58,131)(40,166,151,106,134,59,132)(41,167,152,107,135,60,121)(42,168,153,108,136,49,122)(43,157,154,97,137,50,123)(44,158,155,98,138,51,124)(45,159,156,99,139,52,125)(46,160,145,100,140,53,126)(47,161,146,101,141,54,127)(48,162,147,102,142,55,128), (1,64)(2,65)(3,66)(4,67)(5,68)(6,69)(7,70)(8,71)(9,72)(10,61)(11,62)(12,63)(13,84)(14,73)(15,74)(16,75)(17,76)(18,77)(19,78)(20,79)(21,80)(22,81)(23,82)(24,83)(25,86)(26,87)(27,88)(28,89)(29,90)(30,91)(31,92)(32,93)(33,94)(34,95)(35,96)(36,85)(37,123)(38,124)(39,125)(40,126)(41,127)(42,128)(43,129)(44,130)(45,131)(46,132)(47,121)(48,122)(49,162)(50,163)(51,164)(52,165)(53,166)(54,167)(55,168)(56,157)(57,158)(58,159)(59,160)(60,161)(97,103)(98,104)(99,105)(100,106)(101,107)(102,108)(109,115)(110,116)(111,117)(112,118)(113,119)(114,120)(133,156)(134,145)(135,146)(136,147)(137,148)(138,149)(139,150)(140,151)(141,152)(142,153)(143,154)(144,155), (1,2,3,4,5,6,7,8,9,10,11,12)(13,14,15,16,17,18,19,20,21,22,23,24)(25,26,27,28,29,30,31,32,33,34,35,36)(37,38,39,40,41,42,43,44,45,46,47,48)(49,50,51,52,53,54,55,56,57,58,59,60)(61,62,63,64,65,66,67,68,69,70,71,72)(73,74,75,76,77,78,79,80,81,82,83,84)(85,86,87,88,89,90,91,92,93,94,95,96)(97,98,99,100,101,102,103,104,105,106,107,108)(109,110,111,112,113,114,115,116,117,118,119,120)(121,122,123,124,125,126,127,128,129,130,131,132)(133,134,135,136,137,138,139,140,141,142,143,144)(145,146,147,148,149,150,151,152,153,154,155,156)(157,158,159,160,161,162,163,164,165,166,167,168), (1,45,7,39)(2,44,8,38)(3,43,9,37)(4,42,10,48)(5,41,11,47)(6,40,12,46)(13,152,19,146)(14,151,20,145)(15,150,21,156)(16,149,22,155)(17,148,23,154)(18,147,24,153)(25,51,31,57)(26,50,32,56)(27,49,33,55)(28,60,34,54)(29,59,35,53)(30,58,36,52)(61,122,67,128)(62,121,68,127)(63,132,69,126)(64,131,70,125)(65,130,71,124)(66,129,72,123)(73,140,79,134)(74,139,80,133)(75,138,81,144)(76,137,82,143)(77,136,83,142)(78,135,84,141)(85,165,91,159)(86,164,92,158)(87,163,93,157)(88,162,94,168)(89,161,95,167)(90,160,96,166)(97,109,103,115)(98,120,104,114)(99,119,105,113)(100,118,106,112)(101,117,107,111)(102,116,108,110) );
G=PermutationGroup([[(1,91,21,113,74,36,70),(2,92,22,114,75,25,71),(3,93,23,115,76,26,72),(4,94,24,116,77,27,61),(5,95,13,117,78,28,62),(6,96,14,118,79,29,63),(7,85,15,119,80,30,64),(8,86,16,120,81,31,65),(9,87,17,109,82,32,66),(10,88,18,110,83,33,67),(11,89,19,111,84,34,68),(12,90,20,112,73,35,69),(37,163,148,103,143,56,129),(38,164,149,104,144,57,130),(39,165,150,105,133,58,131),(40,166,151,106,134,59,132),(41,167,152,107,135,60,121),(42,168,153,108,136,49,122),(43,157,154,97,137,50,123),(44,158,155,98,138,51,124),(45,159,156,99,139,52,125),(46,160,145,100,140,53,126),(47,161,146,101,141,54,127),(48,162,147,102,142,55,128)], [(1,64),(2,65),(3,66),(4,67),(5,68),(6,69),(7,70),(8,71),(9,72),(10,61),(11,62),(12,63),(13,84),(14,73),(15,74),(16,75),(17,76),(18,77),(19,78),(20,79),(21,80),(22,81),(23,82),(24,83),(25,86),(26,87),(27,88),(28,89),(29,90),(30,91),(31,92),(32,93),(33,94),(34,95),(35,96),(36,85),(37,123),(38,124),(39,125),(40,126),(41,127),(42,128),(43,129),(44,130),(45,131),(46,132),(47,121),(48,122),(49,162),(50,163),(51,164),(52,165),(53,166),(54,167),(55,168),(56,157),(57,158),(58,159),(59,160),(60,161),(97,103),(98,104),(99,105),(100,106),(101,107),(102,108),(109,115),(110,116),(111,117),(112,118),(113,119),(114,120),(133,156),(134,145),(135,146),(136,147),(137,148),(138,149),(139,150),(140,151),(141,152),(142,153),(143,154),(144,155)], [(1,2,3,4,5,6,7,8,9,10,11,12),(13,14,15,16,17,18,19,20,21,22,23,24),(25,26,27,28,29,30,31,32,33,34,35,36),(37,38,39,40,41,42,43,44,45,46,47,48),(49,50,51,52,53,54,55,56,57,58,59,60),(61,62,63,64,65,66,67,68,69,70,71,72),(73,74,75,76,77,78,79,80,81,82,83,84),(85,86,87,88,89,90,91,92,93,94,95,96),(97,98,99,100,101,102,103,104,105,106,107,108),(109,110,111,112,113,114,115,116,117,118,119,120),(121,122,123,124,125,126,127,128,129,130,131,132),(133,134,135,136,137,138,139,140,141,142,143,144),(145,146,147,148,149,150,151,152,153,154,155,156),(157,158,159,160,161,162,163,164,165,166,167,168)], [(1,45,7,39),(2,44,8,38),(3,43,9,37),(4,42,10,48),(5,41,11,47),(6,40,12,46),(13,152,19,146),(14,151,20,145),(15,150,21,156),(16,149,22,155),(17,148,23,154),(18,147,24,153),(25,51,31,57),(26,50,32,56),(27,49,33,55),(28,60,34,54),(29,59,35,53),(30,58,36,52),(61,122,67,128),(62,121,68,127),(63,132,69,126),(64,131,70,125),(65,130,71,124),(66,129,72,123),(73,140,79,134),(74,139,80,133),(75,138,81,144),(76,137,82,143),(77,136,83,142),(78,135,84,141),(85,165,91,159),(86,164,92,158),(87,163,93,157),(88,162,94,168),(89,161,95,167),(90,160,96,166),(97,109,103,115),(98,120,104,114),(99,119,105,113),(100,118,106,112),(101,117,107,111),(102,116,108,110)]])
45 conjugacy classes
class 1 2A 2B 2C 3 4A 4B 4C 4D 4E 4F 6A 6B 6C 7A 7B 7C 12A 12B 12C 12D 14A 14B 14C 21A 21B 21C 28A 28B 28C 28D ··· 28I 42A 42B 42C 84A ··· 84F order 1 2 2 2 3 4 4 4 4 4 4 6 6 6 7 7 7 12 12 12 12 14 14 14 21 21 21 28 28 28 28 ··· 28 42 42 42 84 ··· 84 size 1 1 7 7 2 2 6 6 14 42 42 2 14 14 2 2 2 2 2 14 14 2 2 2 4 4 4 4 4 4 12 ··· 12 4 4 4 4 ··· 4
45 irreducible representations
dim 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 4 4 4 4 type + + + + + + + - + + + + - + + + - + - image C1 C2 C2 C2 C2 C2 S3 Q8 D6 D6 D6 D7 Dic6 D14 D14 S3×D7 Q8×D7 C2×S3×D7 D7×Dic6 kernel D7×Dic6 Dic3×D7 C21⋊Q8 C12×D7 C7×Dic6 Dic42 C4×D7 C3×D7 Dic7 C28 D14 Dic6 D7 Dic3 C12 C4 C3 C2 C1 # reps 1 2 2 1 1 1 1 2 1 1 1 3 4 6 3 3 3 3 6
Matrix representation of D7×Dic6 in GL4(𝔽337) generated by
1 0 0 0 0 1 0 0 0 0 194 1 0 0 251 109
,
336 0 0 0 0 336 0 0 0 0 109 336 0 0 85 228
,
30 15 0 0 322 15 0 0 0 0 336 0 0 0 0 336
,
0 189 0 0 189 0 0 0 0 0 336 0 0 0 0 336
G:=sub<GL(4,GF(337))| [1,0,0,0,0,1,0,0,0,0,194,251,0,0,1,109],[336,0,0,0,0,336,0,0,0,0,109,85,0,0,336,228],[30,322,0,0,15,15,0,0,0,0,336,0,0,0,0,336],[0,189,0,0,189,0,0,0,0,0,336,0,0,0,0,336] >;
D7×Dic6 in GAP, Magma, Sage, TeX
D_7\times {\rm Dic}_6
% in TeX
G:=Group("D7xDic6");
// GroupNames label
G:=SmallGroup(336,137);
// by ID
G=gap.SmallGroup(336,137);
# by ID
G:=PCGroup([6,-2,-2,-2,-2,-3,-7,55,116,50,490,10373]);
// Polycyclic
G:=Group<a,b,c,d|a^7=b^2=c^12=1,d^2=c^6,b*a*b=a^-1,a*c=c*a,a*d=d*a,b*c=c*b,b*d=d*b,d*c*d^-1=c^-1>;
// generators/relations
×
𝔽 | 2021-12-01 02:39:47 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9975405335426331, "perplexity": 2684.893066271889}, "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/1637964359082.78/warc/CC-MAIN-20211201022332-20211201052332-00236.warc.gz"} |
https://www.zbmath.org/?q=in%3A474649 | ## Found 21 Documents (Results 1–21)
100
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### Some results on generalized relative order $$(\alpha,\beta)$$ and generalized relative type $$(\alpha,\beta)$$ of meromorphic functions with respect to entire functions. (English)Zbl 07459028
MSC: 30D35 30D30 30D20
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### Exponential growth of solution and asymptotic stability results for Hilfer fractional weighted $$p$$-Laplacian initial value problem with Duffing-type oscillator. (English)Zbl 07459027
MSC: 34B15 34B16 34B18
Full Text:
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### Certain unified recurrence relations associated with special functions. (English)Zbl 07459025
MSC: 33C60 33C99
Full Text:
### Nonlocal reaction-diffusion model with subdiffusive kinetics. (English)Zbl 07459024
MSC: 34A08 34A12
Full Text:
MSC: 30C45
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### A new operational matrix of integration based on the independence polynomial of graph to solve fractional Poisson equation. (English)Zbl 07459022
MSC: 35J15 65M70 35R11
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### New unique existence result of approximate solution to initial value problem for fractional differential equation of variable order. (English)Zbl 07459021
MSC: 26A33 34B15
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### On certain subclasses of Pascu type alpha close-to-star functions. (English)Zbl 07459020
MSC: 30C45 30C50
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### The $$j$$-generalized $$p-k$$ Mittag-Leffler function. (English)Zbl 07459019
MSC: 33Exx 33B10 26A33
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Full Text:
MSC: 30C45
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### Relative $$(p,q,t)L$$-th order and relative $$(p,q,t)L$$-th lower order oriented growth properties of composite entire functions. (English)Zbl 07459016
MSC: 30D20 30D30 30D35
Full Text:
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### Certain subclasses of multivalent functions defined with generalized Sãlãgean operator and related to sigmoid function and Lemniscate of Bernoulli. (English)Zbl 07459014
MSC: 30C45 33E99
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MSC: 30C45
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### Fractional diffusion equation with reaction term described by the Caputo-Liouville generalized fractional derivative. (English)Zbl 07459012
MSC: 26A33 42A38 35K57
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### A note on the asymptotic properties of a generalized differential equations. (English)Zbl 07459011
MSC: 34L30 34K37
Full Text:
Full Text:
MSC: 30C45
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### A note on the local stability theory for Caputo fractional planar system. (English)Zbl 07459008
MSC: 34A08 37B55 03C45
Full Text:
all top 3 | 2022-05-24 08:29:59 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 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": 0.8733246326446533, "perplexity": 9076.933981499993}, "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-2022-21/segments/1652662570051.62/warc/CC-MAIN-20220524075341-20220524105341-00303.warc.gz"} |
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Multivariate convex regression: global risk bounds and adaptation
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Abstract
We study the problem of estimating a multivariate convex function defined on a convex body in a regression setting with random design. We are interested in optimal rates of convergence under a squared global continuous $$l_2$$ loss in the multivariate setting $$(d\geq 2)$$. One crucial fact is that the minimax risks depend heavily on the shape of the support of the regression function. It is shown that the global minimax risk is on the order of $$n^{-2/(d+1)}$$ when the support is sufficiently smooth, but that the rate $$n^{-4/(d+4)}$$ is when the support is a polytope. Such differences in rates are due to difficulties in estimating the regression function near the boundary of smooth regions. We then study the natural bounded least squares estimators (BLSE): we show that the BLSE nearly attains the optimal rates of convergence in low dimensions, while suffering rate-inefficiency in high dimensions. We show that the BLSE adapts nearly parametrically to polyhedral functions when the support is polyhedral in low dimensions by a local entropy method. We also show that the boundedness constraint cannot be dropped when risk is assessed via continuous $$l_2$$ loss. Given rate sub-optimality of the BLSE in higher dimensions, we further study rate-efficient adaptive estimation procedures. Two general model selection methods are developed to provide sieved adaptive estimators (SAE) that achieve nearly optimal rates of convergence for particular "regular" classes of convex functions, while maintaining nearly parametric rate-adaptivity to polyhedral functions in arbitrary dimensions. Interestingly, the uniform boundedness constraint is unnecessary when risks are measured in discrete $$l_2$$ norms.
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Statistics theory | 2020-07-11 15:38:59 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 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": 0.7398741841316223, "perplexity": 728.2892066536339}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-29/segments/1593655933254.67/warc/CC-MAIN-20200711130351-20200711160351-00164.warc.gz"} |
https://msp.org/gt/2011/15-1/p10.xhtml | #### Volume 15, issue 1 (2011)
Recent Issues
The Journal About the Journal Subscriptions Editorial Board Editorial Interests Editorial Procedure Submission Guidelines Submission Page Author Index To Appear ISSN (electronic): 1364-0380 ISSN (print): 1465-3060
Moduli spaces and braid monodromy types of bidouble covers of the quadric
### Fabrizio Catanese, Michael Lönne and Bronislaw Wajnryb
Geometry & Topology 15 (2011) 351–396
##### Abstract
Bidouble covers $\pi :S\to Q:={ℙ}^{1}×{ℙ}^{1}$ of the quadric are parametrized by connected families depending on four positive integers $a,b,c,d$. In the special case where $b=d$ we call them $abc$–surfaces.
Such a Galois covering $\pi$ admits a small perturbation yielding a general $4$–tuple covering of $Q$ with branch curve $\Delta$, and a natural Lefschetz fibration obtained from a small perturbation of the composition ${p}_{1}\circ \pi$.
We prove a more general result implying that the braid monodromy factorization corresponding to $\Delta$ determines the three integers $a,b,c$ in the case of $abc$–surfaces. We introduce a new method in order to distinguish factorizations which are not stably equivalent.
This result is in sharp contrast with a previous result of the first and third author, showing that the mapping class group factorizations corresponding to the respective natural Lefschetz pencils are equivalent for $abc$–surfaces with the same values of $a+c,b$. This result hints at the possibility that $abc$–surfaces with fixed values of $a+c,b$, although diffeomorphic but not deformation equivalent, might be not canonically symplectomorphic.
##### Keywords
algebraic surface, moduli space, braid monodromy, equivalence of factorizations, symplectomorphism, bidouble cover, Lefschetz pencil
##### Mathematical Subject Classification 2000
Primary: 14J15
Secondary: 14J29, 14J80, 14D05, 53D05, 57R50 | 2017-10-20 07:00:22 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 16, "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": 0.7016171813011169, "perplexity": 1136.2329873817084}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-43/segments/1508187823839.40/warc/CC-MAIN-20171020063725-20171020083725-00287.warc.gz"} |
https://crypto.stackexchange.com/questions/68428/difference-between-when-select-x-from-mathbbz-p-1-and-mathbbz-p-in-d | # Difference between when select x from $\mathbb{Z}_{p-1}$ and $\mathbb{Z}_p$ in discrete logarithm Problem?
Reading "Security Arguments for Digital Signatures and Blind Signatures" paper, I confused by some questions.
• Q1. when it refers to "El Gamal signature scheme",
The key generation algorithm: it chooses a random large prime $$p$$, of length $$n$$ polynomial in $$k$$, and a generator $$g$$ of $$(\mathbb{Z}/p\mathbb{Z})^*$$, both public. Then, for a random secret key $$x \in \mathbb{Z}/(p − 1)\mathbb{Z}$$, it computes the public key $$y = gx \mod p$$
why $$x$$ select from group $$\mathbb{Z}_{p-1}$$,but not $$\mathbb{Z}_p$$? what is the difference?
• please check the update of your question with latex. – kelalaka Mar 31 at 10:00
## 2 Answers
$$\mathbb{Z}_p=\mathbb{Z}/p\mathbb{Z}$$ denotes a finite field that is just integers mod the prime $$p$$, i.e., $$\{0,1,\ldots,p-1\}$$. This field (by definition of a field) has two operations: addition and multiplication, where the multiplication group (denoted by $$\mathbb{Z}_p^*$$) is defined only on the non-zero elements $$\{1,2,\ldots,p-1\}$$. Then, since $$g$$ is a generator of the multiplicative group of size $$p-1$$, the exponent $$x$$ should be sampled from $$\mathbb{Z}_{p-1}=\{0,1,\ldots,p-2\}$$ (which by the way is not a finite field).
Because $$\mathbb{Z}/{p\mathbb{Z}}$$ when used for El-Gamal is used in its multiplicative form: we have a generator $$g$$ such that all powers of $$g$$ cycle through $$\{1,2,\ldots,p-1\}$$ (so $$0$$ is excluded) and as $$g^{p-1} =1 \pmod{p}$$ by Fermat's little theorem, the powers $$x$$ we use for the generator and which we use as the private key in the DL scheme essentially takes values in $$\{0,\ldots,p-2\}$$ (as $$x=0$$ and $$x=p-1$$ both yield $$g^x=1$$), i.e. $$g^x = g^{x'} \pmod{p}$$ iff $$x = x'\pmod{p-1}$$. Hence the fact we have $$p-1$$ choices for $$x$$. | 2019-12-09 20:35:20 | {"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": 34, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9562764167785645, "perplexity": 299.2985526295472}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-51/segments/1575540523790.58/warc/CC-MAIN-20191209201914-20191209225914-00432.warc.gz"} |
https://villavu.com/forum/showthread.php?p=1094743 | # Thread: Varrock West Bank Smither
1. ## Varrock West Bank Smither
Hi there, welcome to my almost AIO varrock bank smither! It will bank, smith, rinse and repeat until your hearts content.
Credits:
Echo for his break system!
Donations to continue all script development: vladko78, Ibotskillz, Sin, Hazzah, dangerousgoods and a few others..
Dannyrs for the awesome banner!
Features:
-All Bar types
-Supports making everything
-SC hammer support
-Restarts client at client token error
-Random mouse movements, random pixels
-Progress reports
-Nice xp rates depending on what you decide to make
-Option to ignore mods
Instructions:
-Download this sps map here: 0_0VWB.png Put it in here(if you don't have these folders add them):
-Fill out user setup
-If you want an xp calc, move your xp bar to here (NOTE: YOU MUST MOVE THE XP BY DOING THE FOLLOWING: Set oldschool default, move chatbox to small as it can be, unlock the interfaces, left click on the xp bar and drag it over the chatbox! If you do it the other way it will move other interfaces and mess things up):
-Start right in front of bank
-Fill out declare players so you can go past 6 hours
-Enter break variables
-Make sure production make-x dialog option is enabled!!!
-If using SC hammers, have them visible in bank
-Set "ItemSlot" to whichever box in the production screen holds the item you want to make (1-50 or however high you wanna go). Note that if it's on the second page down of boxes you need to increase which slot accordingly - For example Broze platebody is slot 37.See below pic for how the grid works:
-Start script here:
Known Bugs:
-May not have all colors 100%, but should recover itself
Proggys:
ProgressReports
Progress Report:
========AshamanSmither v1.8=========
Time Running: 4 Hours, 37 Minutes and 39 Seconds
Experience Earned: 812366
Experience/Hour: 175548
Items/Hour: 553
=========================================
Progress Report:
========AshamanSmither v1.0=========
Time Running: 6 Hours, 2 Minutes and 0 Seconds
Experience/Hour: 1000
Items/Hour: 537
====================================
Version History:
Version History
-v1.0 Initial Release
-v1.1 Updated some colors, implemented beta sc hammer detection (auto withdraw new ones), added BOB familiar support, some random mouse movements from time to time
-v1.2 Fixed a bug, added option to not logout when finds a mod.
-v1.3 Now calls familiar before trying to deposit stuff into them, hopefully fixed bug with trying to get to banker in different stall
-v1.4 Fixed SC hammer support, just have them visible in bank when you start. Added a few more colors, hopefully fixed the restarting at 6 hour thing
-v1.5 Fixed a silly bug with my login function
-v1.6 Antiban overhaul, various bug fixes, added break system, other tweaks
-v1.7 Fixed breaking bug, interface moves down when you relog as well
-v1.8 Tweaks to antiban, changes to scroll interface, improved depositing, extrabar detection
-v1.9 Fixes for EOC update, added more objects to make
-v2.0 More waits, other EOC tweaks
-v2.1 Edits to summoning, couple bug fixes, added a screenshot taker if you get a inventory no full message
-v2.2 Fix depositing issue, patched up some stuff with familiars
-v2.3 Summoning fix
-v3.0 SRL6 RS3 release
-v3.3 added option to use bank presets to withdraw (faster withdrawing) - will not work the sc hammer though. fixed color and log bugs
Donations:
I only put this up at the request of users, please note that ALL Donations go to script development.
Last edited by Ashaman88; 05-17-2014 at 12:09 PM.
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Will give it a blast now with ~1K steel bars and let you know.
*edit* Worth noting the bars should be placed in the top right bank slot
Last edited by LDog; 09-03-2012 at 06:18 PM.
3. Originally Posted by LDog
Will give it a blast now with ~1K steel bars and let you know.
*edit* Worth noting the bars should be placed in the top right bank slot
Ah crap forgot to put that in the directions, haha thanks!
4. i will use this script
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First things first - Great Script!
A few Proggies-
========AshamanSmither v1.0=========
Time Running: 9 Minutes and 35 Seconds
Making: Steel Platebody
Experience Earned: 10313
Experience/Hour: 64550
Items/Hour: 532
====================================
Only actually made 55, script sometimes walks to the anvil waits a few seconds then walks back to banker.
Stopped script at this point to turn on debug to see if I could find out why.
========AshamanSmither v1.0=========
Time Running: 27 Minutes and 1 Seconds
Making: Steel Platebody
Experience Earned: 33937
Experience/Hour: 75361
Items/Hour: 422
====================================
Made all 190, not sure if the original error was colour based as the script auto reloads SMART.
Will retry when I have some more bars!
My only suggestion is slowing down the mouse movement from deposit all to withdrawing the bars, maybe I'm over thinking it!
But again - Great Script! ~1200 bars in 40 mins!
6. Originally Posted by LDog
First things first - Great Script!
A few Proggies-
========AshamanSmither v1.0=========
Time Running: 9 Minutes and 35 Seconds
Making: Steel Platebody
Experience Earned: 10313
Experience/Hour: 64550
Items/Hour: 532
====================================
Only actually made 55, script sometimes walks to the anvil waits a few seconds then walks back to banker.
Stopped script at this point to turn on debug to see if I could find out why.
========AshamanSmither v1.0=========
Time Running: 27 Minutes and 1 Seconds
Making: Steel Platebody
Experience Earned: 33937
Experience/Hour: 75361
Items/Hour: 422
====================================
Made all 190, not sure if the original error was colour based as the script auto reloads SMART.
Will retry when I have some more bars!
My only suggestion is slowing down the mouse movement from deposit all to withdrawing the bars, maybe I'm over thinking it!
But again - Great Script! ~1200 bars in 40 mins!
Ah ok yeah it might be that I don't have all of the colors for the anvil yet. I have a threshold for the anvil to prevent misfires, but if I'm missing some colors it may mess it up (like if a lot of people are blocking anvils). I'll gather some more though just in case.
The counting of the items made is kind of a jerry rig right now, strange that it did it incorrect and then fixed itself the second time haha.
It does burn through bars really fast!
What part about depositing fast and withdrawing fast has you concerned? Being like a human or messing up?
Thanks for trying it out! Glad it at least mostly worked for you!
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The deposit/withdrawing was concerning being more human like, it's not too bad just me being over the top probably.
I'm assuming it was a colour issue, since it wasn't seen the second time with debug on, not a massive problem though! Excellent script!
8. Spelt Varrock wrong! :P
Nice script though
9. Originally Posted by Ollybest
Spelt Varrock wrong! :P
Nice script though
haha ah man epic fail thanks for the catch!
10. Name Changed
11. Originally Posted by BraK
Name Changed
muchas gracias!
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========AshamanSmither v1.0=========
Time Running: 25 Minutes and 14 Seconds
Experience Earned: 124675
Experience/Hour: 296434
Items/Hour: 476
You are a genius, perfect.. just for the withdrawing sc hammers.
13. your script is flawless im usein steel knifes im run it all night for 5 hour = 125k not bad ahha
14. Thank you so much for this script! <3
15. umm 1 more thing it doesnt have much of an antiban can u change the hover to bank to hover when inventory is done?that way you can add a little antiban?
========AshamanSmither v1.0=========
Time Running: 44 Minutes and 37 Seconds
Making: iron Throwing
Experience Earned: 21000
Experience/Hour: 28240
Items/Hour: 1130
====================================
16. Originally Posted by James Cash
========AshamanSmither v1.0=========
Time Running: 25 Minutes and 14 Seconds
Experience Earned: 124675
Experience/Hour: 296434
Items/Hour: 476
You are a genius, perfect.. just for the withdrawing sc hammers.
Very nice! I will probably add automatic handling for the sc hammers in the next update
Originally Posted by print123
your script is flawless im usein steel knifes im run it all night for 5 hour = 125k not bad ahha
Thanks bro!
Originally Posted by Divus
Thank you so much for this script! <3
You are welcome!
Originally Posted by print123
umm 1 more thing it doesnt have much of an antiban can u change the hover to bank to hover when inventory is done?that way you can add a little antiban?
========AshamanSmither v1.0=========
Time Running: 44 Minutes and 37 Seconds
Making: iron Throwing
Experience Earned: 21000
Experience/Hour: 28240
Items/Hour: 1130
====================================
I never have a dedicated "antiban" to my scripts, the scripts themselves are fully randomized. But I will add some random movements while it is waiting on the items to be made, especially when it takes a long time to make with the 1 bar items.
Edit: If anyone could post a pic of the sc hammer in your inventory I can better script for it (don't feel like doing the minigame for one hehe)
Last edited by Ashaman88; 09-04-2012 at 03:39 AM.
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wont be getting 99 smith got hacked last night lol.
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just quickly, you dont have a bank pin option. gonna run it after system update, and let you know
perhaps summoning support also? you just click take bob and it continues on, no need to start smithing again? i will post proggy after soccer.
========AshamanSmither v1.0=========
Time Running: 1 Hours, 24 Minutes and 5 Seconds
Experience Earned: 239687
Experience/Hour: 171006
Items/Hour: 549
====================================
Successfully executed.
i have varrock armour 3 also, makes it occasionally smith faster. absolutely no problems. stopped to get more bars, thankyou!
Last edited by stuartroad; 09-04-2012 at 11:32 AM.
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The script will make 3 platebodies and bank instead of making 5. Any idea why?
I do have scroll of efficiency if that matters.
20. Originally Posted by James Cash
wont be getting 99 smith got hacked last night lol.
Sorry to hear that bro!
just quickly, you dont have a bank pin option. gonna run it after system update, and let you know
perhaps summoning support also? you just click take bob and it continues on, no need to start smithing again? i will post proggy after soccer.
========AshamanSmither v1.0=========
Time Running: 1 Hours, 24 Minutes and 5 Seconds
Experience Earned: 239687
Experience/Hour: 171006
Items/Hour: 549
====================================
Successfully executed.
i have varrock armour 3 also, makes it occasionally smith faster. absolutely no problems. stopped to get more bars, thankyou!
Ah there is pin support I just forgot to give you a place to type it in, good catch! It shouldn't be too hard to implement summoning support - I can just port it from my altar script. Nice proggy!
The script will make 3 platebodies and bank instead of making 5. Any idea why?
I do have scroll of efficiency if that matters.
Hmm what does the scroll of efficiency do? And do you have your top xp bar set to total or smithing (and it can't say 'lots')? It uses that to track if you are still smithing. Also what kind of bars are you using?
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Ahah!
Yes my xp counter was maxed. I've closed it now and it works fine.
Scroll of efficiency has a chance to save a bar when smithing 3+ bar items but I can see now it did not affect anything.
Ahah!
Yes my xp counter was maxed. I've closed it now and it works fine.
Scroll of efficiency has a chance to save a bar when smithing 3+ bar items but I can see now it did not affect anything.
Ah the scroll shouldn't mess anything up except maybe in a rare case cause it to wait an extra 5 or 6 seconds, I'll go buy that scroll and do some more testing with it
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Hey there, any eta on when you will implement summoning? Would speed up the process for me just a little bit
Thanks again, will post proggy soon. xxo
Hey there, any eta on when you will implement summoning? Would speed up the process for me just a little bit
Thanks again, will post proggy soon. xxo
I've got it in there now, just gotta make sure the logic is all good. Also it will only work for those making things that decrease your inventory (like things that take up more than 1 bar or the 1 bar stuff that makes stuff that stacks (arrow tips))
25. Alright got the summoning pretty much working, but I want to add the auto sc hammer support with this version as well. So if someone could get me a pic of sc hammers with graphics in simba style (safe mode, brightness all the way up), I can add that no problem | 2023-01-27 05:23:29 | {"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": 0.34306395053863525, "perplexity": 10756.127428746739}, "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/1674764494936.89/warc/CC-MAIN-20230127033656-20230127063656-00526.warc.gz"} |
https://www.edaboard.com/threads/can-the-far-field-radiation-field-of-any-antenna-be-regarded.157276/ | # Can the far field radiation field of any antenna be regarded
Status
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#### lychree
##### Junior Member level 1
a problem puzzzlea me
the problem makes me puzzled, i hope some one can help me .can the far field radiation field of any antenna be regarded as spherical wave?
#### siongboon
Re: a problem puzzzlea me
What do you mean by a spherical wave?
Depend on how you define a spherical wave.
Best Regards,
Siong Boon
MODERATOR - SIGNATURE LINKS ARE NOT ALLOWED | 2020-12-04 17:47:51 | {"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": 0.8436934947967529, "perplexity": 3749.528138326785}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-50/segments/1606141740670.93/warc/CC-MAIN-20201204162500-20201204192500-00018.warc.gz"} |
https://www.yambo-code.eu/wiki/index.php?title=How_to_analyse_excitons | # How to analyse excitons
In this tutorial you will learn (for a 2D-hBN) how to:
• analyze a BSE optical spectrum in terms of excitonic eigenvectors and eigenvalues
• look at the spatial distribution of the exciton
• For a similar tutorial focusing on 3D-hBN, go to this page.
## Prerequisites
Previous modules
You will need:
• ypp executable
• xcrysden executable
• gnuplot or xmgrace executable
## YAMBO calculations
If you have completed the tutorials of 2D hBN you should have all the databases required to do this tutorial in your YAMBO_TUTORIALS/hBN-2D/SAVE and 2D_WR_WC (databases generated with RIM and cutoff) directories
$ls ./SAVE ndb.gops ndb.kindx ns.db1 ns.kb_pp_pwscf_fragment_1 ....$ ls ./2D_WR_WC
ndb.BS_Q1_CPU_0 ndb.cutoff ndb.dip_iR_and_P_fragment_1 ndb.pp_fragment_1 ...
## Sort the excitonic eigenvalues
$ypp -J 2D_WR_WC -e s 1 We are sorting the excitons for the q-index = 1 (optical limit q=0). The new generated file o-2D_WR_WC.exc_qpt1_E_sorted (o-2D_WR_WC.exc_qpt1_I_sorted ) reports the energies of the excitons and their Dipole Oscillator Strengths sorted by energy (Index). Open the first file and look inside. The first exciton is at 4.83 eV and the second one has the highest strength (normalized to 1) Or you can make a plot $ gnuplot
gnuplot> set style line 2 lc rgb 'black' pt 7 # circle gnuplot> plot 'o-2D_WR_WC.exc_qpt1_E_sorted' with points ls 2 title 'Strenghts'
Attention the convergence of these results with different k-points grids is mandatory!
## Calculate the exciton oscillator strenght and amplitude
We can now analyze the excitons in terms of single-particle states, to do that create the appropriate input
$ypp -F ypp_AMPL.in -J 2D_WR_WC -e a 1 Suppose you wish to analyze the first 5 excitons then change this line as: States= "1 - 5" # Index of the BS state(s) Close the input and run ypp $ ypp -F ypp_AMPL.in -J 2D_WR_WC
$ls o*exc*at* o-2D_WR_WC.exc_qpt1_amplitude_at_1 o-2D_WR_WC.exc_qpt1_weights_at_1 ... For an exciton $\displaystyle{ |\lambda\gt }$ , o-2D_WR_WC.exc_qpt1_weights_at_* report the Weights and o-2D_WR_WC.exc_qpt1_amplitude_** report the amplitudes Open the file o-2D_WR_WC.exc_weights_at_1 # Band_V Band_C K ibz Symm. Weight Energy # 4.000000 5.000000 7.000000 2.000000 0.922095 4.401093 4.000000 5.000000 7.000000 1.000000 0.922086 4.401093 The first exciton is essentially done of only single particle transitions from VBM to CBM at K (last k-point of the grid). ## Plot the exciton spatial distribution To see the spatial character of the exciton YPP writes the exciton spatial distribution, in other words the probability to find the electron somewhere in the space when the hole is fixed in a give position. Different output formats can be selected and 1D,2D,3D plots done. Create the input and change the size of the cell where to see the exciton. Note that If the k-grid of the BSE simulation is a NxNx1 the exciton has an induced fictitious periodicity every Nx Nx1 Cell of the simulation. For hBN-2D this is not a problem because the exciton is strongly localized but in other systems with more delocalized excitons to look at the real exciton size it is necessary to use very large k-grids in the BSE $ ypp -F ypp_WF.in -J 2D_WR_WC -e w 1
excitons # [R] Excitons
wavefunction # [R] Wavefunction
Format= "x" # Output format [(c)ube/(g)nuplot/(x)crysden]
Direction= "12" # [rlu] [1/2/3] for 1d or [12/13/23] for 2d [123] for 3D
FFTGvecs= 3951 RL # [FFT] Plane-waves
States= "1 - 1" # Index of the BS state(s)
Degen_Step= 0.0100 eV # Maximum energy separation of two degenerate states
% Cells
5 | 5 | 1 | # Number of cell repetitions in each direction (odd or 1)
%
% Hole
2.4 | 1.400 | 0.00 | # [cc] Hole position in unit cell
Close the input and run ypp
$ypp -F ypp_WF.in -J 2D_WR_WC $ xcrysden --xsf o-2D_WR_WC.exc_2d_1.xsf
## Plot electron/hole average density (only in Yambo 5.x)
Another way to analyze excitons, it is the possibility to plot the average electron/hole densities defined as:
to generate the corresponding input just type
ypp -F ypp_WF.in -e w -avehole
and choose the exciton you want to plot. The electron/hole average densities correspond to generalized valence/conduction orbitals for a given exciton. They are interesting in particular for molecular crystals because they allow to distinguishing charge-transfer exctions from Frenkel or Wannier ones, by looking the relative position of the electron/hole densities. For example see the electron/hole average density calculated in a Metal Organic Framework (MOF) that contains azobenzene:
this figure is taken from ref. [1].
## Interpolate exciton dispersion (only in Yambo 5.x)
Starting from Yambo 5.x it is possible to calculate excitons at finite momentum q. The momentum can be specified with the variable
%BSEQptR
iq_start | iq_end | # [BSK] Transferred momenta range
%
where iq_start and iq_end is the first and last index of momentum in the irreducible Brillouin zone (IBZ). You can find the list of q-vectors in the r_setup report. If you calculate the BSE for all q-points of the IBZ, then you can interpolate exciton dispersion along any direction of the full Brillouin zone(BZ) using the interpolation scheme of ref. [2], by doing:
ypp -e i
and then setting:
excitons # [R] Excitonic properties
interpolate # [R] Interpolate
States= "1 - 4" # Index of the BS state(s)
INTERP_mode= "BOLTZ" # Interpolation mode (NN=nearest point, BOLTZ=boltztrap aproach)
% INTERP_Grid
-1 |-1 |-1 | # Interpolation BZ Grid
%
#PrtDOS # Print Exciton Density of States
BANDS_steps= 100 # Number of divisions
cooIn= "rlu" # Points coordinates (in) cc/rlu/iku/alat
cooOut= "rlu" # Points coordinates (out) cc/rlu/iku/alat
%BANDS_kpts # K points of the bands circuit
0 | 0 | 0 |
0.333333333333333 | 0.33333333333333 | 0 |
%
Running ypp, it will interpolate the first 4 excitons energies on 100 points along with the line Gamma->K. If you plot the output file "o.excitons_interpolated", you can compare the interpolated results with the original points calculated in the BSE, that are available in the report "r_excitons_interpolate", and you get something like:
It is also possible to interpolate excitons on a given regular grid by setting INTERP_Grid, for example in the figure below we interpolate the dispersion of the lowest exciton in MoS2 calculated on a 39x39x1 and interpolated in a 117x117x1 grid in all the Brillouin zone:
Finally if you turn on the flag PrtDOS the code calculate the excitonic density of states on the excitons interpolated on the regular grid given by INTERP_Grid.
## References
1. Strongly Bound Excitons in Metal-Organic Framework MOF-5: A Many-Body Perturbation Theory Study, A. R. Kshirsagar et al., preprint ChemRxiv
2. Warren E. Pickett, Henry Krakauer, and Philip B. Allen PRB 38 p2721 (1988) | 2022-07-02 05:11:50 | {"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": 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": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4557236433029175, "perplexity": 10294.890139983208}, "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/1656103984681.57/warc/CC-MAIN-20220702040603-20220702070603-00787.warc.gz"} |
https://customwritingprof.com/ma-3b80-university-of-warwick-proof-of-the-riemann-mapping-theorem-complex-analysis-2/ | # MA 3B80 University of Warwick Proof of The Riemann Mapping Theorem Complex Analysis
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MA3B80
THE UNIVERSITY OF WARWICK
FOURTH YEAR EXAMINATION: Summer 2020
Complex Analysis
Time Allowed: 3 hours
advance on the departmental ‘Warwick Mathematics Exams 2020’ webpage.
Calculators, wikipedia and interactive internet resources are not needed and are not
permitted in this examination. You are not allowed to confer with other people. You
may use module materials and resources from the module webpage.
ANSWER COMPULSORY QUESTION 1 AND TWO FURTHER QUESTIONS out of
the four optional questions 2, 3, 4 and 5.
PDF document if possible, although multiple files (2 or 3) are permitted. You have
an additional 45 minutes to make the upload, and instructions are available on the
departmental ‘Warwick Mathematics Exams 2020’ webpage.
You must not upload answers to more than 3 questions, including Question 1. If you
do, you will only be given credit for your Question 1 and the first two other answers.
The numbers in the margin indicate approximately how many marks are available for
each part of a question. The compulsory question is worth twice the number of marks
of each optional question. Note that the marks do not sum to 100.
COMPULSORY QUESTION
b :=
1. (a) (i) Define a Möbius transformation mapping the extended complex plane C
b
C ∪ {∞} to C.
[3]
(ii) Find a Möbius transformation that maps the left half-plane HL := {z ∈
C : R(z) < 0} onto the unit disk ∆ := {z ∈ C : |z| < 1}, and find a Möbius transformation that maps ∆ onto the right half-plane HR := {z ∈ C : R(z) > 0}.
[4]
(b) (i) Give sufficient conditions on D ⊂ C and the curve γ that ensure that for
1
Question 1 continued overleaf
MA3B80
every holomorphic function f : D → C we have
Z
(♣)
f (z) dz = 0.
γ
[4]
(ii) Give an example of an open and connected set D, a closed piecewise C 1
curve γ : [a, b] → C, and a holomorphic function f : D → C such that (♣)
in part (i) does not hold (no proof required).
(c) (i) State the Removable Singularity Theorem (no proof required).
[3]
[4]
(ii) Show that for every holomorphic function f : D → C, D ⊂ C open, and
(z0 )
every z0 ∈ D the function F (z) = f (z)−f
defined on D {z0 } can be
z−z0
extended to a holomorphic function on all of D.
[3]
(d) (i) Let f : D → C, D ⊂ C open, be holomorphic. Show that the imaginary
part I(f ) is harmonic on D.
[4]
(ii) Give an example of an open and connected set D ⊂ C and a harmonic
function h : D → C such that there is no holomorphic function f : D → C
with I(f ) = h (no proof required).
[3]
(e) (i) State Cauchy’s Integral formulas for the derivatives of holomorphic functions.
[2]
(ii) Compute
Z
∂B2 (0)
z2
1
dz,
−1
∂B2 (0) = {z ∈ C : |z| = 2}.
[4]
(f) (i) State the Riemann Mapping Theorem.
[3]
(ii) Show that there is no holomorphic and bijective mapping from ∆ := {z ∈
C : |z| < 1} onto C. [3] OPTIONAL QUESTIONS 2. (a) (i) Define the four elementary Möbius transformations. 2 Question 2 continued overleaf [4] MA3B80 (ii) Show that every Möbius transformation f , f (z) := az + b , cz + d a, b, c, d ∈ C, can be written as a composition of the four elementary Möbius transformations in (i). [2] (b) Denote H+ := {z = x + iy ∈ C : y > 0} the upper half-plane and ∆ := {z ∈
θ ∈ R, the transformation z 7→ reiθ z is a composition of a rotation by θ and a dilation by a factor r.
For c 6= 0 we write
az+b
cz+d
=
a
c
+
c
.
cz+d
Then this map is obtained by
1
c
c
z 7→ cz 7→ cz + d 7→
7
7→ +
.
cz + d
cz + d
c cz + d
The case where c = 0 is even easier.
(2.5.1)
We can often use this decomposition into elementary Möbius transformations in order to prove that
certain properties are preserved under general Möbius transformations: The proof is then reduced to
checking that the property is preserved for elementary Möbius transformations, as in the following.
Keeping in mind our definition of ‘circles’ in C∞ from Definition 2.3, we have
Theorem 2.10. The image of every circle in C∞ under any Möbius transformation is also a circle in
C∞ .
Proof. We only need check this property for each type of elementary Möbius transformation. The
property is obvious for translations, rotations and dilations. The property for the complex inversion
follows from its interpretation as a 180 ◦ rotation, together with the preservation of circles/lines by
stereographic projection given in Remark 2.2.
You proved essentially this theorem by direct calculation on the exercise sheets of Analysis 3.
2.6
Three points determine a Möbius transformation
VIDEO: Three points determine a Möbius transformation
In this section we will see the fundamental (and very useful) property that if we ask for three specific
distinct points in C∞ to be mapped to another three specific distinct points in C∞ , then we determine
a unique Möbius transformation. Precisely, we have:
Theorem 2.11. Given three distinct points z1 , z2 , z3 ∈ C∞ and three distinct points w1 , w2 , w3 ∈ C∞ ,
there exists a unique Möbius transformation f that satisfies f (zi ) = wi for i = 1, 2, 3.
Thus the group of Möbius transformations has three complex degrees of freedom. This illustrates
very clearly how it is a six real-parameter family.
The proof of Theorem 2.11 will be supported by a couple of sub-results. We start with an observation
about the fixed points of Möbius transformations.
17
2.6
Three points determine a Möbius map
Lemma 2.12. Every Möbius transformation other than the identity f (z) = z has at least one, but at
most two, fixed points. In particular, if f is a Möbius transformation and z1 , z2 , z3 ∈ C∞ are distinct
points such that f (zi ) = zi , then f is the identity.
Proof. As usual, we write
az + b
.
cz + d
Observe that f being the identity corresponds to the case a = d 6= 0 and b = c = 0, so if we assume
that this is not the case then we need to show that there can be at most two fixed points.
f (z) =
We split the proof into two cases: First, if c = 0, then f is just a linear transformation f (z) = ad z + db ,
where a, d 6= 0 by nondegeneracy of Möbius transformations. Such a linear transformation has a
b
if a 6= d.
fixed point at infinity, plus at most one further fixed point at z = d−a
In the remaining case that c 6= 0, we have
f (z) =
az + b
=z
cz + d
⇐⇒
(az + b) = (cz + d)z
⇐⇒
0 = cz 2 + (d − a)z − b .
The quadratic formula gives us two solutions, which might coincide.
Next, we give a special case of the desired Theorem 2.11 in which the points wi are explicit. In this
special case we also give formulae for the Möbius transformation, which will be useful later.
Proposition 2.13. Given three distinct points z1 , z2 , z3 ∈ C∞ , there exists a Möbius transformation
f that maps z1 , z2 , z3 to 1, 0, ∞ respectively. In the case that zi 6= ∞ for i = 1, 2, 3, then it is
(z − z2 )(z1 − z3 )
.
(z − z3 )(z1 − z2 )
(2.6.1)
f (z) =
z − z2
,
z − z3
(2.6.2)
f (z) =
z1 − z3
,
z − z3
(2.6.3)
f (z) =
z − z2
.
z1 − z2
(2.6.4)
f (z) :=
In the case that z1 = ∞ we set
in the case that z2 = ∞ we set
and in the case that z3 = ∞ we set
Proof. Clearly each of these functions f are Möbius transformations. By inspection, they send the
points zi to the required image points.
In the final three cases (2.6.2) to (2.6.4) in which one of the zi is ∞, the formulae for f arise from
(2.6.1) by dropping the factor on the numerator and the factor on the denominator containing that zi .
These two factors would cancel in the limit zi → ∞, so this makes sense.
We are finally in a position to prove the main result of this section.
18
2.7
The cross ratio
Proof of Theorem 2.11.
Existence: Let f1 be the function from Proposition 2.13 that sends z1 , z2 , z3 to 1, 0, ∞ respectively.
Let f2 be the function from Proposition 2.13 that sends w1 , w2 , w3 to 1, 0, ∞ respectively. The Möbius
transformation f we seek will simply be f2−1 ◦ f1 .
Uniqueness: Suppose that we have two Möbius transformations f and g, both of which send the
points zi to wi respectively. Then g −1 ◦ f is a Möbius transformation that has all three distinct points
zi as fixed points. Lemma 2.12 then tells us that g −1 ◦ f is the identity, i.e. f ≡ g.
2.7
The cross ratio
VIDEO: The cross ratio
Definition 2.14. The cross-ratio of pairwise distinct z0 , z1 , z2 , z3 ∈ C∞ , is the point in C that is the
image of z0 under the unique Möbius transformation that sends z1 , z2 , z3 to 1, 0, ∞ respectively.
Formulae for the Möbius transformation f , and hence for f (z0 ), are given by Proposition 2.13.
Theorem 2.15. The cross-ratio is invariant under Möbius transformations.
Proof. If f is any Möbius transformation, we have to show that the cross-ratio of (z0 , z1 , z2 , z3 ) is the
same as that of (f (z0 ), f (z1 ), f (z2 ), f (z3 )). If we denote by g the unique Möbius transformation that
sends z1 , z2 , z3 to 1, 0, ∞ respectively, then this cross ratio is g(z0 ) by definition. But then the the
unique Möbius transformation that sends f (z1 ), f (z2 ), f (z3 ) to 1, 0, ∞ respectively is g ◦ f −1 , and so
the cross-ratio of (f (z0 ), f (z1 ), f (z2 ), f (z3 )) is g ◦ f −1 (f (z0 )) = g(z0 ), which is the same.
Theorem 2.16. The cross ratio of (z0 , z1 , z2 , z3 ) is real-valued if and only if z0 , z1 , z2 , z3 all lie on a
common circle in C∞ .
Recall from Definition 2.3 that circles in C∞ restrict to circles and lines in C.
Proof. Since both the cross-ratio and the property of lying on a circle are invariant under Möbius
transformations, we may assume that z1 , z2 , z3 equal 1, 0, ∞ respectively. The cross ratio of (z0 , 1, 0, ∞)
is z0 by definition. But the circle passing through 1, 0 and ∞ is the real line.
Let’s digest a consequence of what we have proved. If z0 , z1 , z2 , z3 ∈ C, then they all lie on a common
line or circle in the plane if and only if
(z0 − z2 )(z1 − z3 )
∈ R.
(z0 − z3 )(z1 − z2 )
Amazing! Try it out on a few examples.
19
2.8
Examples of Möbius transformations
2.8
Examples and special classes of Möbius transformations
VIDEO: Examples of Möbius transformations
Around 23:10 when I said ‘reflection’, I meant ‘rotation’. I’ll explain more in the live lecture.
There are numerous distinguished examples and subclasses of Möbius transformations, of which we
give a few prominent examples.
Example 2.17 (Möbius transformation that gives a bijection from a disc to a half-space). We are
looking for a Möbius transformation that provides a bijection from the disc D := {z ∈ C : |z| < 1} to the upper half-plane H=>0 := {z ∈ C : =(z) > 0}. Since Möbius transformations are
homeomorphisms from C∞ to itself, this will map the boundary of D, i.e. the unit circle, to the
boundary of H=>0 , i.e. the real axis. There are many maps with this property. By Theorem 2.11 we
can pick any three distinct points on the unit circle and map them to 1, 0 and ∞ respectively. For
example, we could pick 1, −i and i on the unit circle, in which case the map is given by (2.6.1) as
f (z) =
z+i
(z − (−i))(1 − i)
=
.
(z − i)(1 − (−i))
iz + 1
By Theorem 2.10, this Möbius transformation will map the entire unit circle ∂D to the entire real
axis. Because f is a homeomorphism, this map must send the disc D either to the upper half-plane
H=>0 , or the lower half-plane. But since it sends 0 to i, it must be the former case, as required.
Example 2.18 (Möbius transformations that give bijections from the disc to itself). Consider the
Möbius transformations of the form
z−w
for w ∈ C, with |w| < 1. (2.8.1) f (z) = w̄z − 1 We claim that Möbius transformations of this form map D onto itself, and map the boundary of D to itself. To see this, we use the easily-checked identity |z − w|2 = |w̄z − 1|2 − (1 − |z|2 )(1 − |w|2 ) to compute |f (z)|2 = |z − w|2 (1 − |z|2 )(1 − |w|2 ) = 1 − . |w̄z − 1|2 |w̄z − 1|2 Because we are assuming that 1 − |w|2 > 0, we see that |f (z)| < 1 if and only if |z| < 1, and |f (z)| = 1 if and only if |z| = 1. This argument also implies that f maps onto D, since f is a bijection from C∞ to itself. An alternative argument to obtain the surjectivity would be to observe that the inverse of f is f itself! Remark 2.19. We can slightly generalise the class of Möbius transformations from Example 2.18 that map the disc D to itself, by composing with a rotation about the origin, giving maps of the form z−w iθ f (z) = e , (2.8.2) w̄z − 1 20 2.8 Examples of Möbius transformations still with w ∈ D, and now with θ ∈ (−π, π]. A stunning fact that will follow from the so-called Schwarz lemma, Theorem 7.20, is that every holomorphic map D 7→ D that is a bijection is of the form (2.8.2). We are not starting with the assumption that the map is a Möbius transformation here. This is true in far greater generality! Think how different this is to what you have seen before. Just imagine if there were only a finite-dimensional family of real differentiable bijective functions from (−1, 1) to itself! Example 2.20 (Möbius transformations that give bijections from H=>0 to itself). A Möbius transformation g(z) from D to itself can be converted into a Möbius transformation h := f ◦ g ◦ f −1 from
H=>0 to itself, where f : D → H=>0 is the Möbius transformation from Example 2.17. One can also
conjugate in the other direction to give g = f −1 ◦ h ◦ f . Because of this, it may seem pointless to consider Möbius transformations from H=>0 to itself after already considering Möbius transformations
from D to itself in Example 2.18.
However, working in the H=>0 viewpoint has some significant advantages in certain situations. To
see one, we return to the isomorphism between the group of Möbius transformations and the group
P SL(2, C) from Section 2.4. The key observation is that the subgroup
P SL(2, R) := SL(2, R)/{±I},
essentially restricting from complex matrices in SL(2, C) to real matrices in SL(2, R), but as before
identifying each pair A and −A, corresponds to an interesting subgroup of Möbius transformations.
Indeed, we claim that they send the upper half plane H=>0 to itself. To see this, we rewrite
f (z) =
(az + b)(cz̄ + d)
ac|z|2 + ad z + bc z̄ + bd
az + b
=
=
.
cz + d
(cz + d)(cz̄ + d)
|cz + d|2
Therefore, keeping in mind that =(z̄) = −=(z) and ad − bc = 1, we have
=(f (z)) =
=(z)
=
=
2
2
|cz + d|
|cz + d|
|cz + d|2
In particular, if z ∈ H=>0 , equivalently =(z) > 0, if and only if =(f (z)) > 0, equivalently f (z) ∈
H=>0 . Thus f maps H=>0 to itself.
Similarly to before, we see that f also maps H=>0 onto itself. For example, one can observe that the
inverse of f is another Möbius transformation of the same form, and thus maps H=>0 to itself.
One can check that every Möbius transformation that maps H=>0 bijectively to itself arises in this
way. Indeed, any such map must send the real line to itself, or to ∞, and if we add the point ∞ ∈ C∞
to the real line to give a circle C, then the Möbius transformation will map this circle bijectively to
itself. If we now take the points x1 , x2 , x3 ∈ C that map to the points 1, 0, ∞ respectively, then we
can construct a Möbius transformation in P SL(2, R) using Proposition 2.13 that has the same effect
on these three points. By Theorem 2.11 itself, the two Möbius transformations must then coincide.
Example 2.21. Nonexaminable example: The rotations of the unit sphere S 2 ,→ R3 are Möbius
transformations when we identify S 2 and C∞ as above. By rotations, we mean elements of SO(3).
Since the group of Möbius transformations is isomorphic to P SL(2, C), by Lemma 2.7, it is natural
21
2.9
Conformal maps
to ask to which subgroup of P SL(2, C) these rotations correspond. It turns out that it is the subgroup
P SU (2). The corresponding Möbius transformations can be written
az − c
z 7→
,
cz + a
for a, c ∈ C with |a|2 + |c|2 = 1.
2.9
Conformal maps
VIDEO: Conformal maps
A general principle in mathematics is that one defines an object with some structure, for example a
vector space with its linear structure, and then one considers bijective maps between different objects
that preserves this structure, for example a bijection between vector spaces that preserves the linear
structure. Such bijections are typically called isomorphisms, and intuitively we view two objects that
are isomorphic as being the same. You will have seen many other examples of this viewpoint, for example isometries between metric spaces, isomorphisms between groups, homeomorphisms between
topological spaces or maybe diffeomorphisms between manifolds (if you have studied manifolds).
What is the right notion of equivalence for domains in C?
(A domain is an open and connected subset.)
Definition 2.22. Two domains Ω1 and Ω2 in C are said to be conformally equivalent if there exists a
bijective holomorphic function ϕ : Ω1 → Ω2 such that the inverse ϕ−1 is also holomorphic.
Such a map ϕ is called a conformal or biholomorphic map.
In due course, we will establish that the inverse ϕ−1 is automatically holomorphic for a bijective
holomorphic function ϕ between domains, but there is no point in fretting about that now.
This notion of equivalence gives us an equivalence relation. In particular, if Ω1 and Ω2 are conformally
equivalent via the conformal map ϕ : Ω1 → Ω2 , and Ω2 and Ω3 are conformally equivalent via the
conformal map ψ : Ω2 → Ω3 , then Ω1 and Ω3 are conformally equivalent via the conformal map
ψ ◦ ϕ : Ω1 → Ω3 . We are implicitly using the chain rule to be sure that the composition of these
conformal maps is conformal. Thus the notion of being conformally equivalent is an equivalence
relation.
The key property that is preserved by this notion of equivalence is the concept of a function on the
domain being holomorphic. More precisely, by the chain rule, a function f : Ω2 → C is holomorphic
if and only if the composition f ◦ ϕ : Ω1 → C is holomorphic.
For the rest of this section we try to get a feeling for which domains are conformally equivalent to
which other domains. Our knowledge of Möbius transformations will help. Throughout the discussion we write the unit disc as
D := {z ∈ C : |z| < 1}. 22 2.9 Conformal maps Before we begin, let’s recall that we already showed in Example 2.17 that the upper half-space is conformally equivalent to the disc D. Let’s find some more such domains. Watch the video for pictures! Example 2.23. We claim that the upper right quarter of the complex plane Q := {z ∈ C | 0 and =(z) > 0}
is conformally equivalent to the disc D.
To see this, we first observe that Q is conformally equivalent to the upper half plane H=>0 by virtue
of the conformal map z 7→ z 2 .
The upper half plane is then conformally equivalent to D thanks to Example 2.17.
Example 2.24. We claim that the upper half disc
D=>0 := {z ∈ C : |z| < 1 and =(z) > 0}
is conformally equivalent to the whole disc D.
Danger: It is very tempting to try to use the conformal map z 7→ z 2 to do this job, but this actually
shows the quite different fact that D=>0 is conformally equivalent to the disc D with the real interval
[0, 1) removed!
Instead, we notice that D=>0 is conformally equivalent to Q via the unique Möbius transformation
that sends −1, 0 and 1 to 0, 1 and ∞ respectively. One could compute this explicitly, but we can argue
more geometrically as follows: First, we know from Theorem 2.11 that this Möbius transformation
exists. Second, by the preservation of circles in C∞ from Theorem 2.10 it sends the interval [−1, 1]
to the interval [0, ∞]. Third, by Theorem 2.10 again it must send the semicircle {z ∈ C : |z| =
1 and =(z) ≥ 0} to a half line starting at 0 and going in some direction off to infinity. Finally, the
right-angle in the boundary of D=>0 at −1 must be reflected in a right angle in the boundary of
the image of D=>0 under this Möbius transformation, because the Möbius transformation preserves
angles, and so the half line must be the positive imaginary axis.
Now we have shown that D=>0 is conformally equivalent to Q, the claim follows by Example 2.23.
By this point you may be getting the false impression that every domain is conformally equivalent to
D. But this is not true. One type of counterexample would be to take the domain that is the whole of
C. If we could find a conformal map from C to D then this would be a bounded holomorphic function
on C, and Liouville’s theorem that was mentioned in Analysis 3 would tell us that it would have to
be constant, and certainly not then surjective onto D. (We will revisit Liouville’s theorem later, in
Corollary 6.8, once we have rigorously proved Cauchy’s theorem.)
Another sort of domain that would certainly fail to be conformal to the disc D would be a domain
‘with holes’ such as an annulus {z ∈ C : a < |z| < b}, where 0 < a < b < ∞. This domain is not even homeomorphic to D so it is a bit much to ask for it to be homeomorphic via a homeomorphism that is additionally a conformal map! To see that they are not homeomorphic, it suffices to notice 23 2.9 Conformal maps that one is simply connected while the other is not. The notion of simply connected has been briefly mentioned in Analysis 3. Informally it means that every loop in the space can be deformed to a point. In order to make this precise we need to define the notion of homotopy. You may have seen homotopies in greater generality in the course Introduction to Topology. Definition 2.25 (Homotopic). Let Ω ⊂ C be open and let γ1 , γ2 : [a, b] → Ω be two continuous curves (i.e. continuous maps from an interval) with the same endpoints γ1 (a) = γ2 (a) and γ1 (b) = γ2 (b). Then γ1 and γ2 are called homotopic if there exists a continuous mapping h : [0, 1] × [a, b] → Ω such that for all t ∈ [a, b] and all s ∈ [0, 1] we have h(0, t) = γ1 (t), h(1, t) = γ2 (t), h(s, a) = γ1 (a), and h(s, b) = γ1 (b). (2.9.1) Such a mapping h is called a homotopy. The notion of Ω being simply connected is intuitively that it does not contain any holes. Definition 2.26 (Simply connected). An open set Ω ⊂ C is said to be simply connected if it is connected and every closed1 continuous curve γ : [a, b] → Ω is homotopic to the constant curve γ̃ : [a, b] → Ω defined by γ̃(t) = γ(a) = γ(b). In more general situations the definition of simply connected would ask for path connectedness, but path connectedness is equivalent to connectedness when considering an open set. In this very special situation of open sets Ω in C, the notion of being simply connected is equivalent to both Ω being connected, and its complement in the Riemann sphere C∞ being connected. We will not explicitly use this formulation. At this point we can look forward to the end of the course when we prove one of the greatest theorems in the subject, namely the Riemann mapping theorem. That theorem will tell us that every simply connected domain Ω ⊂ C other than Ω = C is conformally equivalent to the disc D. See Theorem 11.1. 1 γ : [a, b] → C is said to be closed if γ(a) = γ(b), i.e. the curve closes up. 24 1 1 b a r Figure 2: Stereographic projection, cross-section 2.10 Exercises 2.1. By considering a triangle similar to the red triangle in Figure 2, prove that r = formula (2.2.1) for stereographic projection. a . 1−b Deduce the Think of Figure 2 as a cross-section in the picture describing stereographic projection. 2.2. Invert formula (2.2.1) to give (2.2.2). Hint: Solve first for x3 . Then get x1 and x2 . 2.3. We claimed in Remark 2.2 that straight lines in C are in one-to-one correspondence with circles in S 2 that pass through the north pole N , with the correspondence being given by stereographic projection π. By considering appropriate planes in R3 that pass through N , give a geometric justification of this fact. You could also check this correspondence by solving equations. See the next question. 2.4. We claimed in Remark 2.2 that circles in C are in one-to-one correspondence with circles in S 2 that don’t pass through the north pole N , with the correspondence being given by stereographic projection π. Verify this by writing down the equations of the appropriate circles. You might also try to give a geometric proof of this, but it’s trickier than the previous question! Hint: Let’s remember some basic geometric facts from school mathematics. If n := (a, b, c) is a unit vector, then the plane through the origin with normal vector n is given by (x1 , x2 , x3 ).n = 0. More generally if we shift that plane in the direction n by a distance d ∈ (−1, 1), then the equation is (x1 , x2 , x3 ).n = d, i.e. ax1 + bx2 + cx3 = d. (2.10.1) The case that N = (0, 0, 1) lies within this plane is then precisely that (0, 0, 1).n = d, i.e. c = d. The intersection of any such plane with S 2 is a circle in S 2 . All such circles arise in this way. Now plug in the values (x1 , x2 , x3 ) given by the formula (2.2.2) into the formula (2.10.1), to give a formula for the image of this circle under π. 2.10 Exercises π −1 (z) 1 s r Figure 3: Antipodal points In the case c = d you should end up with an equation of a line in the plane. (I obtained ax + by = d.) This is the case of the last question. In the case c 6= d, you should get the equation of a√ circle. By completing the square, you find b 1−d2 a , d−c ). The radius should be |d−c| . that the centre is ( d−c 2.5. The points 0 and ∞ in C∞ , when viewed as points in S 2 by applying π −1 , are the south and north poles respectively. They are therefore antipodal points. Given any other point z ∈ C∞ , i.e. z ∈ C that is nonzero, show that −1/z̄ ∈ C corresponds to the antipodal point. More precisely, show that π −1 (z) and π −1 (−1/z̄) are antipodal. If you first convince yourself that z, 0 and −1/z̄ are co-linear in C, then you can work in a cross-section as in Figure 3. 2.6. In Example 2.17 we mapped the unit disc to the upper half plane with a Möbius transformation. Viewed as a transformation of the Riemann sphere, seen as a 2-sphere in R3 , what is this transformation? What is the square of this transformation? That is, which Möbius transformation is it, and what does it look like as a transformation of the Riemann sphere seen as a 2-sphere in R3 ? 2.7. Where does the map 1+z 1−z send the unit disc? What is the inverse of this map? What is it as a transformation of the Riemann sphere S 2 ? f (z) = 2.8. Show that the quarter disc Q̂ := {z ∈ C : |z| < 1 and =(z) > 0 and 0}
is conformally equivalent to the unit disc D.
2.9. Show that the slit plane
S := C [0, ∞)
is conformally equivalent to the unit disc D.
26
2.10
Exercises
2.10. Construct explicitly the Möbius transformation that was constructed geometrically in Example
2.24.
2.11. Construct explicitly the unique Möbius transformation f that sends D to the half space H0 ,
while sending 0 to 12 and sending −1 to 0. By composing with the map z 7→ z 2 and then
z 7→ z − 41 , show that the map
z
K(z) :=
(1 − z)2
is a conformal map from D to the slit plane C (−∞, − 41 ].
This function is known as the Koebe function, and we will revisit it once we know a little more
theory. Amongst all the conformal maps from D to C(−∞, − 41 ], it is the unique one satisfying
the normalisation K(0) = 0 and K 0 (0) = 1.
27
3
Review of material from MA244 Analysis 3 – second portion
3.1
Power series
VIDEO: Power Series
In Section 4.2 of Analysis 3 you learned about power series, i.e. expressions of the form
X
an z n ,
n=0
where an is a complex-valued sequence.
Theorem 3.1 (Theorem 4.13 from Analysis 3). Given a complex-valued sequence (an ), define the
R :=
Then the power series
1
∈ [0, ∞].
lim sup |an |1/n
X
an z n
n=0
converges for all |z| < R and diverges for all |z| > R.
Nothing is claimed in this theorem about z for which |z| = R. That question can be somewhat
delicate. The cases considered in the theorem follow easily from the root test.
Power series can be differentiated term-by-term within their radius of convergence:
Theorem 3.2 (Theorem 4.15 from Analysis 3). If the radius of convergence R of the power series
f (z) =
X
an z n
n=0
can define a single valued choice of arg(z), although different people may make a choice of arg(z)
that differs by a fixed constant multiple of 2π. The ray here is known as a branch cut.
A particularly useful resolution of this 2π ambiguity will be given in Lemma 4.1.
The (complex) logarithm is defined for z 6= 0 by
log(z) = log |z| + i arg(z).
Here the logarithm on the right-hand side is taking real values, so it is uniquely defined. But the
argument is only defined up to the addition of an integer multiple of 2π, so log(z) is only defined
up to the addition of an integer multiple of 2πi. As for the argument arg(z), we are sometimes
content with this state of affairs, and sometimes we ask arg(z) to take its principal value in (−π, π],
in which case log(z) is a well defined function that extends the usual logarithm. Unfortunately it is
2
even smooth, i.e. infinity real differentiable
31
3.4
Complex integration
discontinuous across the negative real axis {x < 0} ⊂ C, and we sometimes make a branch cut by removing the half-line. The complex logarithm inherits many of the useful properties of the real logarithm. For example, it is the inverse of the exponential function in the sense that elog z = elog |z|+i arg(z) = |z|ei arg(z) = z, and log(ez ) = log |ez | + i arg(ez ) = log ex + i arg(eiy ) = x + iy = z, (3.3.3) where the latter computation is carried out modulo 2πi. By (3.3.2) we have the familiar identity log(zw) = log |zw| + i arg(zw) = log |z| + log |w| + i(arg(z) + arg(w)) = log z + log w, (3.3.4) again modulo 2πi. Beware that identities that are claimed modulo 2πi should not be expected to work if we take the principal values of the functions log or arg. As you saw in Analysis 3, the function log(z) allows us to define what it means to raise a complex number to a complex power, albeit with complications arising from log(z) only being defined up to a multiple of 2πi. 3.4 Complex integration VIDEO: Complex integration Recall that in Analysis 3 you defined what it means to integrate a suitable (e.g. continuous) function f : [a, b] → C. You defined Z b Z b Z b f (t)dt := 0 we could write Z Z f (z)dz := f (z)dz, ∂Br (a) γ where γ : [0, 2π] → C is defined by γ(θ) = a + reiθ . Alternatively, if A := {z ∈ C : R1 < |z| < R2 } is an annulus, then we have two boundary components that are parametrised by curves γ1 , γ2 : [0, 2π] → C defined by γ2 (θ) = R2 eiθ and γ1 (θ) = R1 e−iθ , and we analogously define Z Z Z f (z)dz := f (z)dz + f (z)dz. ∂A γ1 γ2 In both Definitions 3.7 and 3.9, we say that the curve is closed if γ(a) = γ(b). Informally this means that it closes up into a loop. The curve is a simple closed curve if whenever a ≤ x < y ≤ b and γ(x) = γ(y), we have x = a and y = b. 3.5 Anti-derivatives, and a baby version of Cauchy’s theorem VIDEO: Anti-derivatives; baby Cauchy’s theorem Later, in Section 5, we will turn our attention to a theorem that is at the heart of complex analysis, namely Cauchy’s theorem. Loosely speaking it will tell us that in certain situations when we integrate holomorphic functions around closed curves we obtain zero. In this section we see a baby version of this theory in which our holomorphic function f is the derivative of some other holomorphic function F. 34 3.5 Anti-derivatives; baby Cauchy’s theorem Lemma 3.10. Suppose Ω ⊂ C is open. Suppose further that f : Ω → C and F : Ω → C are holomorphic with F 0 (z) = f (z). If γ is a piecewise C 1 closed curve in Ω. Then Z f (z)dz = 0. γ This lemma follows immediately from the following type of fundamental theorem of calculus. Lemma 3.11. Suppose that F : Ω → C is holomorphic, with F 0 continuous. Suppose further that γ : [a, b] → Ω is a piecewise C 1 curve. Then Z F 0 (z) dz = F (γ(b)) − F (γ(a)). (3.5.1) γ In particular, if γ is a closed curve (i.e. γ(b) = γ(a)) then we have R γ F 0 (z) dz = 0. Proof. By the definition of contour integration and the chain rule of Lemma 1.5, we have Z Z b Z b d 0 0 0 F (z) dz = F (γ(t)) γ (t) dt = F (γ(t)) dt = F (γ(b)) − F (γ(a)). γ a a dt Note that to be able to apply the usual fundamental theorem of calculus in the last equality, we need some regularity on the integrand such as continuity, which is why we are assuming that F 0 is continuous. The result assumes that the derivative F 0 of the holomorphic function F is continuous. Later we will see that this is always true, but we can’t assume that now or our arguments will be circular. Corollary 3.12. Suppose n ∈ Z does not equal −1. Then for γ : [a, b] → C {0} any piecewise C 1 closed curve, we have Z z n dz = 0. γ Proof. If we define F (z) := z n+1 /(n + 1), then F 0 (z) = z n , so the result follows from Lemma 3.10. This corollary fails in a very important way if n = −1! In Q. 3.8 you will compute: Example 3.13. For r > 0 and k ∈ Z let γ : [0, 2π] → C be the closed C 1 curve γ(θ) = reikθ that
travels anticlockwise k times around the circle of radius r. Then
Z
dz
= 2πi k.
γ z
35
3.6
Exercises
P
n
3.1. Suppose Alice has P
a power series ∞
n=0 an z with radius of convergence R1 > 0, and Bob
has a power series n=0 bn z n with radius of convergence R2 ≥ R1 . Suppose we are told that
these power series give the same function on the ball BR1 (0) where they are both converging.
Prove that an = bn for every n ∈ N0 := {0, 1, 2, . . .}, i.e., the power series and their radii of
convergence agree.
P
1
k
3.2. Write the function z 7→ 1−z
as a power series ∞
k=0 ak z and give its radius of convergence.
3.3. By using the previous question and a theorem from Section 3.1, write down the power series of
1
the function z 7→ (1−z)
2 and its radius of convergence.
We will use this exercise when we take a closer look at the Koebe function.
P
1
k
3.4. Suppose w ∈ C {0}. Write the function z 7→ w−z
as a power series ∞
k=0 ak z and give its
We will use this fact when we prove Taylor’s theorem.
P
k
3.5. Suppose that n ∈ N, and that the power series ∞
k=n ak z , which omits the first n terms of a
general power series, has radius of convergence R > 0 and thus defines a holomorphic function
f : BR (0) → C. Prove that there exists a holomorphic function g : BR (0) → C such that
f (z) = z n g(z) for all z ∈ BR (0),
and that g can be written as a power series with radius of convergence R.
We’ll use this fact when we study the zeros of holomorphic functions.
3.6. Consider the function f : C {0} → C defined by
f (z) =
sin z
.
z
Prove that we can extend f to a function on the whole of C (by defining f (0) to be a suitable
value in C) that is entire, i.e. holomorphic on the whole of C.
The ‘singularity’ of f at 0 in this example will be known as a ‘removable singularity’.
3.7.
(a) For R > 0, Compute
1
2i
Z
z̄ dz,
∂BR (0)
and show that it agrees with the area of the ball BR (0).
(b) For R := {z ∈ C : 0,
compute
Z
1
z̄ dz,
2i ∂R
and show that it agrees with the area of R.
This is not a fluke. We are seeing a couple of instances of a general fact that could be derived
from an appropriate form of Stokes’ theorem.
3.6
Exercises
3.8. For r > 0 and k ∈ Z let γ : [0, 2π] → C be the closed C 1 curve γ(θ) = reikθ that travels
anticlockwise k times around the circle of radius r. Prove that
Z
dz
= 2πi k.
γ z
37
4
Winding numbers
4.1
Winding numbers of continuous closed paths
VIDEO: Winding numbers of continuous closed paths
Watch the video for an instant explanation-by-pictures of what the winding number is!
In Section 3.3 we have discussed the function arg(z), and the issue that it is only defined modulo an
integer multiple of 2π. The following lemma will tell us that if we decide on a choice of arg(z) at one
point z, and then move along a continuous path/curve that stays away from 0, then this determines a
unique continuously varying choice of the argument along this path.
Lemma 4.1 (Lifting lemma). Suppose γ : [a, b] → C {0} is continuous, and fix θ0 ∈ R such that
γ(a) = |γ(a)|eiθ0 . Then there exists a unique continuous function θ : [a, b] → R such that θ(a) = θ0
and γ(t) = |γ(t)|eiθ(t) for all t ∈ [a, b].
For example, for a curve γ : [0, 2π] → C given by γ(t) = eit , if we choose θ0 = 0 rather than any
other value in 2πZ then θ(t) = t. In particular, even though the start and end points are the same, the
argument differs by 2π.
Those of you who have already studied Introduction to Topology will know how to prove this lemma.
There is an obvious ‘covering map’ R 7→ R/(2πZ) given by θ 7→ θ + 2πZ, and we are taking a lift of
the function arg ◦ γ : [a, b] → R/(2πZ).
For the Introduction to Topology course in 2020/21, Section 8.3 talks about the homotopy lifting
property, and its special case, Definition 8.6, the path lifting property. In Section 10, Proposition
10.1, you saw that covering maps satisfy the homotopy lifting property.
For those who have not taken the course Introduction to Topology, here is a self-contained proof.
The video could be useful in order to understand this proof!
Proof. First observe that if γ avoids a slit {−reiθ0 : r > 0} on the opposite side of the starting point,
then the existence of θ(t) is clear: We just ask θ(t) to take values in the interval (θ0 − π, θ0 + π).
In the general case, we are free to replace γ(t) by the curve γ̃(t) := γ(t)/|γ(t)| without changing the
argument. Since γ̃ is a continuous function from a closed interval, it is uniformly continuous, and by
dividing up [a, b] into a large enough number of equal intervals, we can be sure that γ̃ does not move
too far when restricted to each of these sub-intervals. More precisely, by taking n ∈ N large enough,
we can be sure that for each k ∈ {0, 1, . . . , n − 1} and each t ∈ [ck , ck+1 ], where ck := a + nk (b − a),
we have |γ̃(t) − γ̃(ck )| < 1. The idea then is to do the lifting on each of these sub-intervals in turn. Indeed, the restriction of γ̃ to [c0 , c1 ] must avoid a slit {−reiθ0 : r > 0} on the opposite side of the starting point, so by the
38
4.2
Nearby closed paths have the same winding number
comment at the start of the proof we can find our function θ(t) at least for t ∈ [c0 , c1 ], with θ(c0 ) = θ0 .
At this point we can use θ(c1 ) as a new starting argument analogous to θ0 , and do our lifting on the
next interval [c1 , c2 ], where γ̃(t) avoids the new opposite slit {−reiθ(c1 ) : r > 0}. This extends
θ(t) to the interval [c0 , c2 ]. By repeating this process a total of n times, we obtain a lift to the whole
interval [c0 , cn−1 ] = [a, b].
To establish uniqueness of θ(t), suppose we have a second continuous lift θ̂(t) also with θ̂(a) = θ0 .
Then t 7→ θ(t) − θ̂(t) is a continuous function that vanishes at t = a, and takes values in 2πZ because
both θ(t) and θ̂(t) represent the argument, modulo 2π. Thus θ(t) − θ̂(t) = 0 for all t ∈ [a, b], as
required.
Lemma 4.1 allows us to unambigously define the change in argument as we move along a continuous
path.
Definition 4.2. Suppose γ : [a, b] → C {0} is continuous, and let θ : [a, b] → R be a function
arising in Lemma 4.1. We define
](γ) := θ(b) − θ(a).
The function θ was only defined up to a constant multiple of 2π that was determined by θ0 . However,
when we subtract θ(a) from θ(b) this unknown multiple of 2π will disappear.
Definition 4.3. Suppose γ : [a, b] → C {0} is a closed continuous path. Then we define the index
or winding number of γ around 0 to be
I(γ, 0) :=
1
](γ) ∈ Z.
More generally, if w ∈ C and γ : [a, b] → C {w} is a closed continuous path then we define the
index of γ around w to be
1
I(γ, w) :=
](γw ),
where γw : [a, b] → C {0} is the path γ translated to send w to the origin, i.e. γw (t) := γ(t) − w.
Example 4.4. For n ∈ Z, consider the curve γ : [0, 2π] → C defined by γ(θ) = reinθ , for some
r > 0, which winds around the origin n times in an anticlockwise direction. Then
I(γ, 0) = n.
Remark 4.5. In the case that γ maps into a region of C on which we can make a choice of arg(z),
for example if γ : [a, b] → C − {reiθ : r > 0} is a continuous closed path (see Section 3.3 for a
discussion of branch cuts) then we can choose θ(t) = arg(γ(t)), and we see that I(γ, 0) = 0. The
branch cut prevents γ from winding around the origin.
4.2
Nearby closed paths have the same winding number
39
4.3
Winding number and simply connected domains
VIDEO: Nearby closed paths have the same winding number
In this section we prove that if we have a closed path γ : [a, b] → C {0}, then a small-enough
perturbation of γ will wind round 0 the same number of times as γ itself.
In the video this will be completely self evident!
However, a little care is required. If γ goes very close to 0 then we must ensure that we perturb very
Lemma 4.6. Suppose γ : [a, b] → C Bε (0) is a continuous closed path, and γ̃ : [a, b] → C {0} is
a continuous closed path with |γ(t) − γ̃(t)| < ε for every t ∈ [a, b]. Then I(γ, 0) = I(γ̃, 0). Note that if γ : [a, b] → C {0} is a continuous closed path then we can always find some ε > 0 such
that the hypotheses of the lemma are satisfied. The smaller that is, the less we can perturb.
Proof. Let θ(t) and θ̃(t) be lifts of the arguments of γ(t) and γ̃(t) respectively, as given by Lemma
4.1. Define a continuous function α : [a, b] → R by α(t) := θ̃(t) − θ(t). By definition of winding
number, we have
1
1
[θ̃(b) − θ̃(a)] −
[θ(b) − θ(a)]
I(γ̃, 0) − I(γ, 0) =
(4.2.1)
1
=
(α(b) − α(a)) .
Because the winding number is always an integer, the difference α(b) − α(a) must be an integer
multiple of 2π, and our task is to show that it is zero.
If this were not the case, then α(t) would either increase or decrease by at least 2π as t varied from
a to b. By continuity of α, we could be sure that there would exist some t0 ∈ [a, b] with α(t0 ) = π2 ,
modulo 2π, and we will use this to derive a contradiction.
By dividing the hypothesis |γ(t) − γ̃(t)| < ε by |γ(t)|, which is at least as large as ε, we find that 1− which forces γ̃(t) γ(t) γ̃(t) < 1, γ(t) to have positive real part. Draw a picture or watch the video! But iπ γ̃(t) γ(t) = | γ̃(t) |eiα(t) , γ(t) and at t = t0 this will be purely imaginary because eiα(t0 ) = e 2 = i, giving a contradiction. 4.3 Winding number and simply connected domains 40 4.4 The winding number as an integral VIDEO: Winding number and simply connected domains In Definition 2.26 we recalled what it meant for an open subset Ω ⊂ C to be simply connected. The most useful consequence of being simply connected for us will be: Theorem 4.7. If an open set Ω ⊂ C is simply connected then for every w ∈ C Ω and every continuous closed path γ : [a, b] → Ω, we have I(γ, w) = 0. In the video we will draw a picture that makes this seem completely obvious. Proof. In order to simply notation, we translate Ω and γ by −w in order to reduce to the case that w = 0. Note that then 0 ∈ / Ω. Because Ω is simply connected, there exists a homotopy of γ to a constant path. More precisely, there exists a continuous map h : [0, 1] × [a, b] → Ω such that for all t ∈ [a, b] and all s ∈ [0, 1] we have h(0, t) = γ(t), h(1, t) = γ(a), h(s, a) = γ(a), and h(s, b) = γ(a). (4.3.1) In particular, for each s ∈ [0, 1], we have a simple closed curve γs : [a, b] → Ω defined by γs (t) := h(s, t). The theorem will be proved if we can show that the index I(γs , 0) is the same for each s ∈ [0, 1], because I(γ0 , 0) = I(γ, 0) and I(γ1 , 0) = 0 since γ1 is a constant path. Because h is continuous, and [0, 1] × [a, b] is closed, the image of h will be closed in Ω ⊂ C {0}, and in particular we can find ε > 0 so that this image does not intersect Bε (0).
Because h is continuous on its closed domain, it is also uniformly continuous. In particular, we can
pick δ > 0 so that whenever t ∈ [a, b] and s1 , s2 ∈ [0, 1] with |s1 − s2 | < δ, we must have |γs1 (t) − γs2 (t)| = |h(s1 , t) − h(s2 , t)| < ε. By Lemma 4.6, we then have I(γs1 , 0) = I(γs2 , 0). This implies that s 7→ I(γs , 0) is locally constant, and then constant for all s ∈ [0, 1]. 4.4 The winding number as an integral VIDEO: The winding number as an integral In the last minute or two of the video I refer to something that is ‘coming in the next section’. But I subsequently decided to move that material to Corollary 3.12 back in Section 3.5. This makes the ordering more logical. If we have a simple closed path that is not just continuous but also piecewise C 1 , then we can characterise the winding number as an integral: 41 4.4 The winding number as an integral Lemma 4.8. If w ∈ C and γ : [a, b] → C {w} is a closed piecewise C 1 curve, then Z 1 dz I(γ, w) = . 2πi γ z − w This expression for the winding number is often used as the definition. It is often easier to use in rigorous proofs, but the definition we gave is more immediately visual, more general, and is useful when considering homotopies. In Example 3.13 and Q. 3.8 we showed that for r > 0, n ∈ Z, and γ : [0, 2π] → C defined by
γ(θ) = reinθ , which winds around the origin n times in an anticlockwise direction, we have
Z
1
1
dz = n.
2πi γ z
Thus by Example 4.4 the claimed formula at least works for I(γ, 0).
Proof of Lemma 4.8. By translation, we may assume that w = 0. We give the proof assuming that γ
is C 1 . The modifications to handle piecewise C 1 curves are straightforward. We must control
Z
Z b 0
dz
γ (t)
=
dt.
γ z
a γ(t)
Taking any function θ(t) from the Lifting lemma 4.1, so γ(t) = |γ(t)|eiθ(t) , we notice that because γ
is C 1 and keeps away from 0, the function θ(t) is also C 1 and we can compute
γ 0 (t) = eiθ(t)
and so
d
|γ(t)| + |γ(t)|iθ0 (t)eiθ(t) ,
dt
γ 0 (t)
d
=
log |γ(t)| + iθ0 (t).
γ(t)
dt
Integrating gives
Z
Z b
dz
d
0
=
log |γ(t)| + iθ (t) dt = 0 + i[θ(b) − θ(a)] = i](γ) = 2πiI(γ, 0)
dt
γ z
a
because γ is closed.
Given our definition of the winding number I(γ, w) as the number of times a curve γ winds around a
point w, it is intuitively obvious that it will be constant as we vary w continuously without touching
the image of γ. We formulate and prove a precise version of this assertion in the case that γ is
piecewise C 1 .
Lemma 4.9. Suppose γ : [a, b] → C is a piecewise C 1 closed curve. Then on each connected
component of C γ([a, b]), the function w 7→ I(γ, w) is constant.
42
4.4
The winding number as an integral
Note that as the continuous image of a closed set, we know that γ([a, b]) is closed, and so C γ([a, b])
is open.
Proof. It suffices to prove that the map w 7→ I(γ, w) is holomorphic, with derivative zero, for w not
in the image of γ (so that I(γ, w) is defined). That then implies that on each component of Cγ([a, b])
the function is constant, as required. To see this last implication, one can recall that a holomorphic
function with zero derivative has all its partial derivatives equal to zero, so it is locally constant and
hence constant on each connected component.
From the definition of
∂ w̄
we have
1
I(γ, w) =
∂ w̄
2πi
Z
γ
∂ w̄
1
z−w
dz = 0.
Here we are using the fact that we can differentiate under the integral sign because the partial derivatives of the integrand with respect to the real and imaginary components of w are uniformly continuous
for z = γ(t), t ∈ [a, b], and w in a small neighbourhood of its starting point.
Differentiation under the integral sign was covered in Analysis 3, and I am not planning to examine
you on it in this context.
On the other hand
1
I(γ, w) =
∂w
2πi
Z
γ
∂w
1
z−w
by Corollary 3.12.
43
1
dz =
2πi
Z
γ
1
dz = 0
(z − w)2
4.5
Exercises
4.1. Given a continuous path γ : [a, b] → C {0}, we write −γ for the continuous path [a, b] 7→
C {0} defined by t 7→ γ(a + b − t), which reverses the direction.
(a) Verify that ](−γ) = −](γ).
(b) Now suppose additionally that γ is closed. What is I(−γ, w) in terms of I(γ, w)?
4.2. Suppose that Ω ⊂ C contains the closure of the ball Br (a) of radius r > 0 centred at a ∈ Ω.
Prove that for all z0 ∈ Br (a) we have
I(∂Br (a), z0 ) = 1
by showing that we may as well take z0 = a and computing.
Recall that ∂Br (a) refers to a curve γ passing once around ∂Br (a) in an anticlockwise direction. Please use the integral formulation of index, i.e. Lemma 4.8, to make it easier to formulate
a rigorous argument.
4.3. Prove that if γ : [a, b] → C is a closed piecewise C 1 curve then the set of points w ∈ Cγ([a, b])
for which I(γ, w) 6= 0 is bounded.
1
4.4. In Q. 3.4, hopefully you wrote (for w ∈ C {0}) the function z 7→ w−z
as a power series
P∞
−k−1 k
z valid for |z| < |w|. By replacing z by 1/z and setting w = 1/z0 , where k=0 w 1 z0 ∈ Br (0) for some r > 0, obtain an expansion for z−z
that is valid for |z| > |z0 | and
0
converges uniformly for z in ∂Br . By integrating around ∂Br (0), prove that
I(∂Br (0), z0 ) = 1
thus reproving Q. 4.2.
5
Cauchy’s Theorem
Baron Augustin-Louis Cauchy (1789 – 1857).
5.1
Preamble and connection with Analysis 3
VIDEO: Cauchy’s theorem: Analysis 3 reminder
In Analysis 3 (Theorem 4.29 in 2019/20) you heard about the iconic theorem of Cauchy from which a
spectacular amount of wonderful theory gushes forth. To restate the particular form of the result that
you saw, we need the notion of simply connected from Definition 2.26.
Theorem 5.1 (Cauchy’s theorem on simply connected domains). Suppose Ω ⊂ C is open and simply
connected. Suppose further that f : Ω → C is holomorphic and γ is a piecewise C 1 closed curve in
Ω. Then
Z
f (z)dz = 0.
γ
Cauchy’s theorem has a huge number of applications. An example application that you saw in Analysis 3 (and that we’ll review shortly) is that a function f : Ω → C that is holomorphic is necessarily
infinitely differentiable. Amazing. This is nothing like what happens for real differentiable functions.
Just think of the function f : R → R that is zero for x < 0 and equal to x2 for x ≥ 0. You also saw, in Analysis 3, a heuristic proof of Theorem 5.1 that, strictly speaking, requires a few extra hypotheses. Indeed, you saw a proof somewhat along the lines that Cauchy originally gave. That proof requires extra regularity for the curve γ (it needs to be simple and have nonvanishing derivative) and extra regularity for the holomorphic function f (it needs to have continuous derivative in order to apply Stokes’ theorem in its standard form). The required extra regularity for f will always hold, but in order to prove this we will need Cauchy’s theorem itself! In this section we would like to sketch the layout of the theory required to give a rigorous proof of a slightly restricted form of this theorem. Indeed, in Theorem 5.7 we will prove it in the special case of so-called star-shaped domains. In due course we will also see a much more general form of Cauchy’s theorem that includes Theorem 5.1 as a special case. Before doing that, we recall the following special case of Example 3.13 and Q. 3.8, which shows that Cauchy’s theorem fails on C {0}, so the requirement that Ω is simply connected cannot simply be dropped. Example 5.2. Consider the holomorphic function f (z) = z1 on Ω := C{0}, and for r > 0 let
γ : [0, 2π] → C be the simple closed C 1 curve γ(θ) = reiθ that travels anticlockwise around the circle
Z
f (z)dz = 2πi.
γ
45
5.2
Goursat’s theorem – Cauchy’s theorem on triangles
Although you have already done a more complicated computation in Q. 3.8, let’s redo it in this special
case. Note that f (γ(θ)) = f (reiθ ) = r−1 e−iθ , and γ 0 (θ) = ireiθ , and so
Z
Z 2π
Z 2π
−1 −iθ
r e ire dθ = i
dθ = 2πi.
f (z)dz =
0
γ
5.2
0
Goursat’s theorem – Cauchy’s theorem on triangles
VIDEO: Goursat’s theorem: Cauchy’s theorem on triangles
Édouard Jean-Baptiste Goursat (1858-1936). Known for his contribution to the current rigorous
theory for Cauchy’s theorem and his spectacular moustache.
A first situation in which one rigorously proves Cauchy’s theorem for general holomorphic f is when
one heavily restricts the curves γ one allows and integrates around the boundary of a triangle. This
will then later be used as a tool in order to prove more general results such as Theorem 5.1. This
special case is named after Goursat, who came long after Cauchy.
Theorem 5.3 (Gour…
attachment
Tags:
holomorphic function
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limit function
Riemann Mapping Theorem
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https://www.gamedev.net/forums/topic/623734-dx11-hlsl-5-setting-samplerstate-from-within-the-shader-isn39t-working/ | # DX11 [DX11] HLSL 5 setting SamplerState from within the shader isn't working
This topic is 2226 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.
## Recommended Posts
Hi there,
I'm having some trouble with a hlsl shader I've written for a deferred shader to do texture mapping onto a road in my game.
Basically I define a SamplerState at the top of my shader which I then use to sample my textures to map onto the surface.
I've defined it based on the MSDN documentation so that I shouldn't need to create and set and Samplers from within my C++ code and it can all be done from within the shader.
However when I run it, the sampler seems to be ignored. The directx debug gives the following warning:
D3D11: WARNING: ID3D11DeviceContext::Draw: The Pixel Shader unit expects a Sampler to be set at Slot 0, but none is bound. This is perfectly valid, as a NULL Sampler maps to default Sampler state. However, the developer may not want to rely on the defaults. [ EXECUTION WARNING #352: DEVICE_DRAW_SAMPLER_NOT_SET ]
Here are the relevant code snippets from my shader:
// textures are passed in from my application Texture2D widthMap : register ( t0 ); Texture2D lengthMap : register ( t1 ); // the sampler state defined in shader code so that my application // doesn't have to SamplerState MySampler { Filter = MIN_MAG_MIP_POINT; AddressU = Wrap; AddressV = Wrap; }; // pixel shader which samples the texture POut PShader(PIn input) { ..... output.colour = float4(lengthMap.Sample(MySampler, float2(input.texCoord.y, 0.0f))); ...... return output; }
Ignore the dodgy texture coordinate - the premise for my texture sampling is to "build" the actual texture out of 2 textures which define the u colour and v colour - when both the u and v are filled in then I use the colour, otherwise that pixel is set to black - I'm using it to generate the road lines so that I can define any road line layout I want based on 2 very small textures.
I've dug through a few of the DirectX samples and I can see them declaring the SamplerState at the top just like I have and seem to have no such problems.
I've also tried declaring a samplerstate for each texture I want to sample and within each state I set the "Texture" field to the target texture. I changed it to the current version as this is how the directx samples seem to do it.
This problem is also present everywhere I sample a texture in my deferred shaders as well!
I've got no idea what I've missed. I can't see any settings that I need to set to tell DirectX to use what ever the shader itself supplies, as far as I was aware - declaring it in my shader should work fine.
I can post more examples of my shader files if needed.
Has anyone got any suggests???
Thanks very much!!
##### Share on other sites
Are you using the effect framework? If not, as far as I know, you'll need to create the sampler (with the desired settings) in the program side and bind it to the register.
Best regards!
##### Share on other sites
No I'm not.....goddamn it! I was hoping for a quick win.....Ok. Yeah it looks like you're right, just stumbled onto this post after reading your reply. It seems my google-fu is still not as strong as I'd hoped.
Thanks very much for the help!
##### Share on other sites
For all those that are interested - just found this post which discusses a nice way to deal with this situation, which is what I'm now going to do. The developer automatically sets up a set of SamplerState objects in their engine and set them in the shaders by default. The shaders have a common include containing the register references for all of these so that all shaders have access to set of samplers etc automatically.
[Edit:] Though I'd give a quick update - this method works fantastically - all my shaders now have immediate access to any samplers they might need and its simplified my some parts of my engine code a lot! Works perfectly.
• 9
• 13
• 41
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• ### Similar Content
• By chiffre
Introduction:
In general my questions pertain to the differences between floating- and fixed-point data. Additionally I would like to understand when it can be advantageous to prefer fixed-point representation over floating-point representation in the context of vertex data and how the hardware deals with the different data-types. I believe I should be able to reduce the amount of data (bytes) necessary per vertex by choosing the most opportune representations for my vertex attributes. Thanks ahead of time if you, the reader, are considering the effort of reading this and helping me.
I found an old topic that shows this is possible in principal, but I am not sure I understand what the pitfalls are when using fixed-point representation and whether there are any hardware-based performance advantages/disadvantages.
(TLDR at bottom)
The Actual Post:
To my understanding HLSL/D3D11 offers not just the traditional floating point model in half-,single-, and double-precision, but also the fixed-point model in form of signed/unsigned normalized integers in 8-,10-,16-,24-, and 32-bit variants. Both models offer a finite sequence of "grid-points". The obvious difference between the two models is that the fixed-point model offers a constant spacing between values in the normalized range of [0,1] or [-1,1], while the floating point model allows for smaller "deltas" as you get closer to 0, and larger "deltas" the further you are away from 0.
To add some context, let me define a struct as an example:
struct VertexData { float[3] position; //3x32-bits float[2] texCoord; //2x32-bits float[3] normals; //3x32-bits } //Total of 32 bytes Every vertex gets a position, a coordinate on my texture, and a normal to do some light calculations. In this case we have 8x32=256bits per vertex. Since the texture coordinates lie in the interval [0,1] and the normal vector components are in the interval [-1,1] it would seem useful to use normalized representation as suggested in the topic linked at the top of the post. The texture coordinates might as well be represented in a fixed-point model, because it seems most useful to be able to sample the texture in a uniform manner, as the pixels don't get any "denser" as we get closer to 0. In other words the "delta" does not need to become any smaller as the texture coordinates approach (0,0). A similar argument can be made for the normal-vector, as a normal vector should be normalized anyway, and we want as many points as possible on the sphere around (0,0,0) with a radius of 1, and we don't care about precision around the origin. Even if we have large textures such as 4k by 4k (or the maximum allowed by D3D11, 16k by 16k) we only need as many grid-points on one axis, as there are pixels on one axis. An unsigned normalized 14 bit integer would be ideal, but because it is both unsupported and impractical, we will stick to an unsigned normalized 16 bit integer. The same type should take care of the normal vector coordinates, and might even be a bit overkill.
struct VertexData { float[3] position; //3x32-bits uint16_t[2] texCoord; //2x16bits uint16_t[3] normals; //3x16bits } //Total of 22 bytes Seems like a good start, and we might even be able to take it further, but before we pursue that path, here is my first question: can the GPU even work with the data in this format, or is all I have accomplished minimizing CPU-side RAM usage? Does the GPU have to convert the texture coordinates back to a floating-point model when I hand them over to the sampler in my pixel shader? I have looked up the data types for HLSL and I am not sure I even comprehend how to declare the vertex input type in HLSL. Would the following work?
struct VertexInputType { float3 pos; //this one is obvious unorm half2 tex; //half corresponds to a 16-bit float, so I assume this is wrong, but this the only 16-bit type I found on the linked MSDN site snorm half3 normal; //same as above } I assume this is possible somehow, as I have found input element formats such as: DXGI_FORMAT_R16G16B16A16_SNORM and DXGI_FORMAT_R16G16B16A16_UNORM (also available with a different number of components, as well as different component lengths). I might have to avoid 3-component vectors because there is no 3-component 16-bit input element format, but that is the least of my worries. The next question would be: what happens with my normals if I try to do lighting calculations with them in such a normalized-fixed-point format? Is there no issue as long as I take care not to mix floating- and fixed-point data? Or would that work as well? In general this gives rise to the question: how does the GPU handle fixed-point arithmetic? Is it the same as integer-arithmetic, and/or is it faster/slower than floating-point arithmetic?
Assuming that we still have a valid and useful VertexData format, how far could I take this while remaining on the sensible side of what could be called optimization? Theoretically I could use the an input element format such as DXGI_FORMAT_R10G10B10A2_UNORM to pack my normal coordinates into a 10-bit fixed-point format, and my verticies (in object space) might even be representable in a 16-bit unsigned normalized fixed-point format. That way I could end up with something like the following struct:
struct VertexData { uint16_t[3] pos; //3x16bits uint16_t[2] texCoord; //2x16bits uint32_t packedNormals; //10+10+10+2bits } //Total of 14 bytes Could I use a vertex structure like this without too much performance-loss on the GPU-side? If the GPU has to execute some sort of unpacking algorithm in the background I might as well let it be. In the end I have a functioning deferred renderer, but I would like to reduce the memory footprint of the huge amount of vertecies involved in rendering my landscape.
TLDR: I have a lot of vertices that I need to render and I want to reduce the RAM-usage without introducing crazy compression/decompression algorithms to the CPU or GPU. I am hoping to find a solution by involving fixed-point data-types, but I am not exactly sure how how that would work.
• By cozzie
Hi all,
I was wondering it it matters in which order you draw 2D and 3D items, looking at the BeginDraw/EndDraw calls on a D2D rendertarget.
The order in which you do the actual draw calls is clear, 3D first then 2D, means the 2D (DrawText in this case) is in front of the 3D scene.
The question is mainly about when to call the BeginDraw and EndDraw.
Note that I'm drawing D2D stuff through a DXGI surface linked to the 3D RT.
Option 1:
A - Begin frame, clear D3D RT
B - Draw 3D
C - BeginDraw D2D RT
D - Draw 2D
E - EndDraw D2D RT
F - Present
Option 2:
A - Begin frame, clear D3D RT + BeginDraw D2D RT
B - Draw 3D
C - Draw 2D
D - EndDraw D2D RT
E- Present
Would there be a difference (performance/issue?) in using option 2? (versus 1)
Any input is appreciated.
• Do you know any papers that cover custom data structures like lists or binary trees implemented in hlsl without CUDA that work perfectly fine no matter how many threads try to use them at any given time?
• By cozzie
Hi all,
Last week I noticed that when I run my test application(s) in Renderdoc, it crashes when it enable my code that uses D2D/DirectWrite. In Visual Studio no issues occur (debug or release), but when I run the same executable in Renderdoc, it crashes somehow (assert of D2D rendertarget or without any information). Before I spend hours on debugging/ figuring it out, does someone have experience with this symptom and/or know if Renderdoc has known issues with D2D? (if so, that would be bad news for debugging my application in the future );
I can also post some more information on what happens, code and which code commented out, eliminates the problems (when running in RenderDoc).
Any input is appreciated.
• Hi Guys,
I understand how to create input layouts etc... But I am wondering is it at all possible to derive an input layout from a shader and create the input layout directly from this? (Rather than manually specifying the input layout format?) | 2018-05-25 05:36:04 | {"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": 0.24795104563236237, "perplexity": 1650.3100555009853}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-22/segments/1526794867041.69/warc/CC-MAIN-20180525043910-20180525063910-00345.warc.gz"} |
https://brilliant.org/problems/sat-counting/ | # SAT Counting
Probability Level 1
Maria has 4 hats, 2 jackets, and 2 shawls. To go out, she needs a hat, a jacket, and a shawl. In how many different ways can she combine her outerwear?
(A) $\ \ 4$
(B) $\ \ 8$
(C) $\ \ 12$
(D) $\ \ 16$
(E) $\ \ 20$
× | 2021-04-10 22:27:25 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 5, "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": 0.3274705708026886, "perplexity": 5124.550316695043}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-17/segments/1618038059348.9/warc/CC-MAIN-20210410210053-20210411000053-00076.warc.gz"} |
https://www.gradesaver.com/textbooks/math/algebra/algebra-1/chapter-7-exponents-and-exponential-functions-cumulative-test-prep-gridded-response-page-470/25 | ## Algebra 1
Published by Prentice Hall
# Chapter 7 - Exponents and Exponential Functions - Cumulative Test Prep - Gridded Response - Page 470: 25
35 square units
#### Work Step by Step
Two points of the parallelogram are $(0,7)$ and $(7,7)$. The side-length of this side is $7-0$, or $7$. One point on that side is $(0,2)$. The other side-length is $7-2$, or $5$. $5*7 =35$
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback. | 2022-08-13 22:18:10 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6956273913383484, "perplexity": 1216.864189610109}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-33/segments/1659882571987.60/warc/CC-MAIN-20220813202507-20220813232507-00239.warc.gz"} |
https://rhai.budshome.com/engine/optimize/volatility.html | Volatility Considerations for Full Optimization Level
Even if a custom function does not mutate state nor cause side-effects, it may still be volatile, i.e. it depends on the external environment and is not pure.
A perfect example is a function that gets the current time – obviously each run will return a different value!
The optimizer, when using OptimizationLevel::Full, will merrily assume that all functions are pure, so when it finds constant arguments (or none) it eagerly executes the function call and replaces it with the result.
This causes the script to behave differently from the intended semantics.
Therefore, avoid using OptimizationLevel::Full if non-pure custom types and/or functions are involved. | 2021-03-07 02:15:31 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.43145737051963806, "perplexity": 1902.058733970954}, "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-2021-10/segments/1614178376006.87/warc/CC-MAIN-20210307013626-20210307043626-00276.warc.gz"} |
http://mathonweb.com/help_ebook/html/exponents.htm | ### 12.1 - Exponents
Section 3.3 discussed integer exponents. This section continues from there and explains non-integer exponents. We saw that if n is a natural number (i.e. 1, 2, 3, …) then the exponential b n is defined to mean multiplying b by itself n times, like this:
The number b is called the base, n is called the exponent, and we say that we are raising b to the n th power. We also saw that any base raised to a negative integer power is the reciprocal of the same base raised to the corresponding positive power:
and that any base raised to the 0th power equals 1:
b 0 = 1.
In that section we also saw that exponentials have these three properties:
Multiplication property Division property Exponentiation property
Let's assume that these properties can be generalized to exponents that are not necessarily integers. We will find that we can also give meaning to rational and real exponents and even complex exponents.
The meaning of b 1/n : Let m = 1/n in the exponentiation property. This gives:
On the other hand taking the nth power of the nth root of b gives the same result:
Putting this together we find that:
In other words, b 1/n is the nth root of b. The most important case is:
In other words, b 1/2 is the square root of b.
Note: If we are working over the real numbers and the base b is negative then n must be an odd integer; otherwise we can't take the nth root of b. If we are working over the complex numbers then there is no problem and no restriction on n. Click here for more information.
For example the following exponential gives a real value:
but this exponential gives a complex value:
The meaning of b m/n : Using the exponentiation property we can write b m/n two ways:
But in the previous section we saw that b 1/n is the nth root of b. Thus this equation says that b m/n may be thought of as the nth root of the mth power of b or as the mth power of the nth root of b.
For example:
A note on negative bases: If we are working over the real numbers and the base b is negative then n must be an odd integer; otherwise we can't take the nth root of b. If we are working over the complex numbers then there is no problem and no restriction on n. Click here for more information.
The meaning of b r where r is any real number: We have made sense of exponents that are positive or negative or fractions but what about exponents that are real numbers? We will explain this case using an example. We will show that 10 1.2 equals 16 (approximately). To do this we make a graph of the function y = 10 x. Here is the table of values and the graph with a smooth curve interpolated through the points.
We see that the interpolated curve goes through the point ( x = 1.2, y = 16 ). It is in this sense that we can say that 10 1.2 = 16. It is easy to see that this use of interpolation doesn't depend on the base being 10; any other base would produce a similar result.
Note: If we are working over the real numbers then the base b must be a positive number. If we are working over the complex numbers then the base can have any value, with one exception (namely the base can't be zero if the real part of the exponent is negative or zero, because this causes division by zero; see below.) Click here for more information.
The meaning of 0 x : This is an interesting situation because depending on the exponent x, there is zero either in the numerator or the denominator, so the expression can be either 0 or undefined. And 0 0 cannot be computed; it is defined to equal 1.
In summary, if the exponent x is a real number then there are the following cases:
If the exponent x is a complex number then there are the following cases:
Algebra Coach Exercises | 2014-08-01 07:46:40 | {"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": 0.8458202481269836, "perplexity": 223.81831776022347}, "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/1406510274866.27/warc/CC-MAIN-20140728011754-00203-ip-10-146-231-18.ec2.internal.warc.gz"} |
https://brilliant.org/problems/i-cant-remember-all-their-names/ | # I can't remember all their names!
Calculus Level 5
$\large 1^m - 2^m + 3^m - 4^m + 5^m - \cdots$
Define the Abel sum $$\sum\limits_{n=0}^\infty a_n$$ to be $$\displaystyle \lim_{z\to 1^-} \sum_{n=0}^\infty a_nz^n,$$ if that limit exists.
The closed form of the Abel sum of the (divergent) series as shown above can be written as which of the following functions?
× | 2017-03-30 00:55:10 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9553588628768921, "perplexity": 431.45279826287936}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-13/segments/1490218191444.45/warc/CC-MAIN-20170322212951-00131-ip-10-233-31-227.ec2.internal.warc.gz"} |
https://math.stackexchange.com/questions/3691147/wave-equation-and-fourier-transform-conditions-for-differentiation | # Wave equation and Fourier Transform: conditions for differentiation
Consider the wave equation in one dimension $$u_{tt}-u_{xx}=0$$ together with a Fourier Transform along $$t$$, ie $$\text{FT}[u](x,\omega)=\int_{-\infty}^{+\infty}u(x,t)\exp(-i\omega t)\mathrm{d}t.\tag{1}$$ The above PDE transforms into $$\partial_{xx}\text{FT}[u]+\omega^2\text{FT}[u]=0$$ whose general solution reads $$\text{FT}[u](x,\omega)=A(\omega)\cos\omega x+B(\omega)\sin\omega x\tag{2}$$ which is essentially the Fourier Transform of d'Alembert's solution.
Under which conditions on $$u(x,t)$$ is the classical differentiation of $$\text{FT}[u](x,\omega)$$ with respect to $$x$$ meaningful? When it is meaningful, is $$\partial_x \text{FT}[u]$$ the Fourier Transform of $$u_x(x,t)$$ that is $$\text{FT}[u_x]$$? It is a classical result which is always used when solving PDE via Fourier Transform (and used above in the quantity $$\partial_{xx} FT[u]$$), however I would like to read the exact assumptions on $$u$$. For instance, is this differentiation acceptable when $$u_{xx}(x,t)$$ should be read in the sense of distributions because $$u_x(x,t)$$ is discontinuous?
• A sufficient condition is the uniform convergence of the Fourier Transform of $u_{xx}(x,t)$. – Mark Viola May 25 '20 at 22:57
• @MarkViola Thanks. Do you have a reference on this? – pluton May 26 '20 at 0:40
• Any book on Advanced Calculus should suffice. – Mark Viola May 26 '20 at 1:17
• @MarkViola I've edited my question with the beginning of an answer: is this what you had in mind? Thx – pluton Jun 4 '20 at 19:30
• No. I was not extending the Fourier Transform to include tempered distributions. – Mark Viola Jun 4 '20 at 19:33
A partial answer to the above question is available in the book "Fourier Analysis, by TW Körner, Cambridge University Press, 1988, page 268, Theorem 53.5" (where $$x$$ and $$t$$ should be interchanged to comply with the question):
Let $$g:\mathbb{R}\times\mathbb{R}\to\mathbb{C}$$ be a continuous function such that $$g_2$$ exists and is continuous. Suppose $$\int_{-\infty}^{+\infty}|g(x,t)|\mathrm{d}x$$ and $$\int_{-\infty}^{+\infty}|g_2(x,t)|\mathrm{d}x$$ exist for each $$t$$ and that $$\int_{|x|>R}|g_2(x,t)|\mathrm{d}x\to 0$$ as $$R\to \infty$$ uniformly in $$t$$ on each $$[a,b]$$. Then $$\int_{-\infty}^{+\infty}g(x,t)\mathrm{d}x$$ is differentiable with $$\frac{d}{dt}\int_{-\infty}^{+\infty}g(x,t)\mathrm{d}x=\int_{-\infty}^{+\infty}\frac{\partial g}{\partial t}(x,t)\mathrm{d}x$$
[note by OP] where $$g_2$$ is the first partial derivative of $$g$$ with respect to its second argument.
If we consider for simplicity the left propagating wave, the solution reads $$u(x,t)=T(x+t)$$ where $$T$$ is a distribution. Its Fourier Transform in time is (because of translation) $$\text{FT}[u](x,\omega)=\int_{-\infty}^{+\infty}T(x+t)\exp(-i\omega t)\mathrm{d}t=\exp(i\omega x)\text{FT}[T](\omega) \tag{3}$$ and so $$\partial_x \text{FT}[u]$$ and $$\partial_{xx}\text{FT}[u]$$ are well defined as soon as $$T$$ is a tempered distribution and $$\partial_x \text{FT}[u](x,\omega)=i\omega\text{FT}[u](\omega)\tag{4}$$
Let us now have a look at the Fourier Transform of $$u_x=T_x$$ (in the sense of distributions) \begin{aligned} \text{FT}[u_x](x,\omega)&=\int_{-\infty}^{+\infty}T'(x+t)\exp(-i\omega t)\mathrm{d}t\\ &=\exp(i\omega x)\text{FT}[T'](\omega)=i\omega\exp(i\omega x)\text{FT}[T](\omega)=i\omega\text{FT}[u](x,\omega) \end{aligned} \tag{5} and Equations (5) and (4) are identical.
Conclusion: for the wave equation in 1D with solution $$u(x,t)=T(x+t)$$, the classical differentiation with respect to space of the Fourier Transform in time is legitimate as soon as $$T$$ is a tempered distribution and the following holds: $$\partial_x \text{FT}[u](x,\omega)=\text{FT}[u_x](x,\omega)=i \omega \text{FT}[u](x,\omega)$$ All this is probably obvious and agrees well with (2) :). Same derivations apply for the right propagating wave $$V(x-t)$$. | 2021-08-03 20:55:11 | {"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": 43, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9243083000183105, "perplexity": 169.52956916827222}, "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/1627046154471.78/warc/CC-MAIN-20210803191307-20210803221307-00188.warc.gz"} |
https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Contextual_Modules/Effects_of_Acid_Rain_on_Atlantic_Salmon_Populations/03_pH_%2F%2F_Ion_Selective_Electrodes | # pH / Ion Selective Electrodes
Analysis of cations and anions by Ion-Selective Electrodes (ISEs)
Purpose: The purpose of this assignment is to introduce potentiometric measurements of ionic species by ion selective electrodes (ISEs)
Learning Outcomes: Upon completion of this module, students will be able to:
1. Identify electrodes and measurement devices used in potentiometry.
2. Predict how analyte concentration (or activity) controls the potential of a potentiometric measurement.
3. Choose appropriate measurement conditions to minimize interferences.
4. Correct for differences in ionic strength among calibration standards and samples.
5. Construct an appropriate calibration curve for potentiometric determinations and account for changes in analyte concentration due to sample dilution.
References: Modules on the theory and operation of ISEs may be found in the Analytical Sciences Digital Library (ASDL) collection. The following hyperlinks will direct the reader to some ASDL resources on potentiometry.
1. Analytical Sciences Digital Library. Potentiometry: e-learning module. http://community.asdlib.org/activelearningmaterials/analytical-electrochemistry-potentiometry/ (accessed April 3, 2014).
2. Harvey, D. Analytical Chemistry 2.0, Chapter 11. http://www.asdlib.org/onlineArticles/ecourseware/Analytical%20Chemistry%202.0/Text_Files.html (accessed April 3, 2014)
The EPA field manual includes practical aspects of field and lab measurements of pH.
1. US EPA Field Manual: http://water.usgs.gov/owq/FieldManual/Chapter6/6.4_ver2.0.pdf
Membrane-based ISEs are widely used in the determination of ionic species. Such determinations fall under the category of a direct potentiometric measurement, which you have experienced if you have ever made a pH measurement. A typical direct potentiometric measurement requires the use of an indicator electrode, a reference electrode, and a high-impedance voltmeter. An example of equipment needed to perform a potentiometric determination is shown in Figure 1.
Figure 1. An experimental setup for the direct potentiometric measurement of sodium ion in aqueous solution.
The two electrodes pictured in Figure 1 represent an electrochemical cell. The electrode on the right is the sodium indicator electrode (sodium ISE). The electrode on the left is a reference electrode. The sodium ISE has a glass membrane that responds specifically to sodium ions. This glass membrane physically separates two solutions: one inside the electrode with a constant sodium ion concentration; one outside the membrane that is the solution you are analyzing. The electrical potential difference generated by the indicator electrode depends on the sodium concentration of the outer solution. Therefore, the function of the indicator electrode is to respond to changes in the analyte concentration in a predictable manner.
The reference electrode provides a known and stable potential to compare against the indicator electrode potential. An assumption when using an ISE system is that the potential of the reference electrode is independent of the concentration of the analyte and matrix of the sample being analyzed. The electrical potential is displayed on the high-impedance voltmeter. The voltmeter has a high impedance to minimize current flow, which prevents changes to the chemical composition of the reference electrode and to the sample (so, for example, if measuring the concentration of H+ with a pH meter, there will not be any reduction of the H+ to hydrogen gas). High impedance, which typically is greater than 1012 ohms, is also needed to minimize errors in the measured potential.
One of the most common examples of ion selective electrodes is a pH electrode – pH electrodes are selective toward the H+ ion. We will start our exploration of ISEs by understanding how a pH electrode works. pH electrodes also use a glass membrane (this membrane is about 0.1 mm thick, so is quite fragile), but in this case it is a type of glass more sensitive toward the H+ ion instead of the Na+ ion. The internal solution has a fixed and known concentration of H+. The external solution is the sample whose pH you wish to measure. The concentration of H+ in the external solution varies depending on the sample being analyzed. An important point is that the glass used to manufacture the membrane has some sodium ions (Na+) in it. A key factor in the functioning of the membrane is that the inner and outer surfaces of the glass form a very thin hydrated gel layer when in contact with water (Figure 2). The internal and external hydrated gel layers are only about 10 nm thick, which is much less than the 0.1 mm thickness of the glass membrane, so a layer of dry glass always separates the two hydrated gel layers. Cations from the solution have the ability to migrate into the hydrated gel layer. For a pH electrode, that means that H+ ions from the solution will displace some of the Na+ ions in the glass that makes up the hydrated gel layer.
Figure 2. Representation of the glass membrane in a pH electrode. Note: the width of the hydrated gel layers in the representation is too large relative to the width of the dry glass portion of the membrane.
Many of you may have used a pH electrode before. If you did, it likely appeared to consist of only a single electrode. In actuality, it is a two-electrode system, but it is designed in such a way that the reference electrode is incorporated into the glass membrane electrode. The design of the combination-electrode system incorporates an electrical contact between the reference and indicator electrode that is necessary to complete the electrical circuitry and produce a potential reading.
Q1: Will a more acidic sample displace more, the same or less Na+ from the hydrated gel layer?
Different concentrations of H+ and Na+ in the hydrated gel layer cause different junction potentials. Note that the junction potential at the interface of the membrane with the internal solution never changes because the concentration of H+ is constant in the internal solution. The junction potential at the interface of the membrane with the outer solution changes for different samples with different pH. The specific reason why varying concentrations of H+ and Na+ can be found in the outer gel layer has to do with something called the mobility of the ions.
Q2: What do you think is meant by mobility of ions?
Q3: Which ion do you think has a higher mobility, H+ or Na+
As mentioned earlier, a reference electrode is used because its junction potential stays fixed no matter what the external solution. Therefore, the only junction potential in the entire circuit of an ion selective electrode that changes is the one at the interface of the ion selective membrane and the external (sample) solution. In the case of a Na+ ion selective electrode a glass membrane of a different composition than the glass membrane in a pH electrode is used– namely one that is more responsive to sodium ions migrating into the glass. Other ion selective electrodes can be fabricated provided a membrane exists that is selective toward the ion one wishes to measure.
Q4: Do you think other cations (e.g., Li+. K+) may have some ability to migrate into the hydrated gel layer of a pH electrode? If so, is this a problem?
The potential measured by the voltmeter is described by Equation $$\ref{1}$$.
$\mathrm{E_{cell} = E_{ind} - E_{ref} + E_{lj}} \label{1}$
In eq $$\ref{1}$$, Ecell represents the potential of the electrochemical cell, Eind represents the half-cell potential of the indicator electrode, Eref represents the half-cell potential of the reference electrode, and Elj represents the liquid-junction potential between the sample solution and the outside membrane of the indicator electrode.
Of particular interest is the relationship between Ecell and the concentration of the analyte. Remember, Ecell is measured but there is only one junction potential (Elj) in the entire system that changes (at the frit), so a measurement of Ecell is essentially a measure of the one varying junction potential.
For the Na+ ion selective electrode, the varying junction potential only depends on the concentration of Na+. However, the situation is more complicated than just using the concentration of Na+, because ion-selective electrode measurements are most commonly performed in solutions with ionic strengths that are greater than zero. In these solutions, there is a difference between the formal concentration (i.e. how the solution was prepared in the lab) and the effective concentration or activity of the analyte.
Q5: Consider a solution that has some Na+ and very high concentrations of K+Cl-. What effect do you think this might have on the activity of Na+ in the solution?
The relationship between activity and concentration for sodium is illustrated in Equation $$\ref{2}$$.
$\mathrm{a_{Na}= γ_{Na} [Na^+]} \label{2}$
In eq $$\ref{2}$$, aNa is the sodium ion activity (mol L-1) , [Na+] is the sodium ion concentration (mol L-1), and γNa is the activity coefficient for the sodium ion. As we just discussed, as the sample ionic strength increases, there is a greater probability that analyte ions will interact with oppositely charged ions from the supporting electrolyte(s) dissolved in the sample. This effectively decreases the concentration of the “free ion”, which is represented by a decrease in the activity coefficient. As the ionic strength of a solution approaches zero, the activity coefficient approaches one, and under infinitely dilute conditions, the analyte activity and analyte concentration are equal. The relationship between the oxidized and reduced forms of sodium written as a reduction reaction can be described in Equation $$\ref{3}$$.
$\ce{Na+ (aq) + e- → Na (s)} \label{3}$
The half-cell potential of the indicator electrode responds to changes in the activity of the analyte as described by the generalized form of the Nernst equation in Equation $$\ref{4}$$:
$\mathrm{E_{ind}= E_{ind}^o- \dfrac{RT}{nF} \ln\left(\dfrac{1}{a_{Na}} \right)} \label{4}$
In eq $$\ref{4}$$, E° is the indicator electrode potential under standard conditions (298 K, 1.00 M Na+), R is the molar gas constant (8.314 J K-1 mol-1) , T is the absolute temperature (K) , n is the number of moles of electrons in the half-reaction, and F is Faraday’s constant (96485 C mol-1).
Q6: If the indicator electrode potential under standard conditions is -0.100 V, what is the indicator electrode potential at 298 K if the activity of the sodium ion is 0.10 M?
Q7: How does the indicator electrode potential change in the previous question if the temperature is increased by 10 degrees?
A sodium ion selective electrode must be calibrated before it can be used to measure the concentration of Na+ in an unknown sample.
Q8: How would you go about calibrating a sodium ion selective electrode?
We just discussed how the ionic strength of a solution impacts the activity of Na+. Suppose you wanted to analyze the sodium concentration of a low ionic strength sample such as natural pond water.
Q9: Can you think of a way to mitigate possible effects of ionic strength to insure that your calibration procedure and sample analysis provide an accurate measurement of the concentration of Na+ in the unknown?
If one keeps the ionic strength high and constant, then the Nernst equation can be expressed in terms of analyte concentrations (and not activities) because the activity coefficient of all samples and standards are equivalent and knowledge of activity is no longer critical.
Q10: Would this proposed way to mitigate possible effects of ionic strength be utilized in pH measurements?
Q11: In the potentiometric determination of sodium ion of a mineral water sample, indicate if either of the following supporting electrolytes can be used for ionic strength adjustment: a 4.0M NH3 – NH4Cl buffer (pH 10) or 4.0M NaCl.
Q12: What would be the general criteria you would need to use in selecting a suitable supporting electrolyte for an analysis using an ion selective electrode?
The concentration of each calibration standard can be expressed as a formal concentration. The concentration term in the Nernst equation is often converted from base e to base 10, which can also be expressed as a p-function, shown in Equation $$\ref{5}$$.
$\mathrm{pNa= - \log[Na^+]} \label{5}$
Assuming a temperature of 298 K, and the constants R, T and F combined into a single value, under those conditions, the Nernst equation takes on the following form for cationic analytes (shown for sodium in Equation $$\ref{6}$$):
$\mathrm{E_{cell}= E_{cell}^o- \dfrac{0.05915}{n} pNa} \label{6}$
Q13: Based on the relationship in eq $$\ref{6}$$, how would you construct a calibration that links the changes in electrode potential to changes in the concentration of the sodium ion?
Q14: What is the expected slope of a potentiometric calibration curve for sodium at 35°C? What effect does temperature have on the slope of a potentiometric calibration curve?
As discussed earlier, indicator electrodes do not have a specific response to a given analyte, but have a wide range of responses to a group of analytes that are similar in charge and size. The electrode is designed to exhibit the greatest response for the target analyte, but the presence of chemically similar analytes in a sample may interfere with the determination of the target analyte and bias the potentiometric response. The selectivity of an ion-selective electrode is expressed by Equation $$\ref{7}$$
$\mathrm{E_{ind}=E_{ind}^o- \dfrac{0.05915}{n_{Analyte}} \log\{a_{Analyte}+ K_{Analyte,Interferent} (a_{Interferent} )^{n_{Analyte}⁄n_{Interferent} } \} } \label{7}$
Expressed in Equation $$\ref{8}$$, the selectivity coefficient $$\mathrm{(K_{Analyte,Interferent})}$$ is a ratio of analyte to interferent activities where each species influences the indicator electrode potential to the same degree.
$\mathrm{K_{Analyte,Interferent}=\dfrac{a_{Analyte}}{(a_{Interferent} )^{n_{Analyte}⁄n_{Interferent} }}} \label{8}$
Q15: If a sample has a sodium concentration of 1.0 x 10-3 M, and the sodium ISE has a selectivity coefficient of KNa,H = 30, what sample pH would cause a 1% error in the sodium ISE response?
Q16: Evaluate whether it is best to use alkaline or acidic conditions to determine the sodium ion concentration by ISE?
Whenever a method calibration is performed using linear regression (i.e. a best-fit line or trendline, as it is called in Microsoft Excel), it is understood that extrapolating beyond the concentration range used in the regression analysis can lead to biased results. Typically when the analyte concentration in a sample is greater than the analyte concentration for the most concentrated standard, the sample is diluted so that the analyte concentration is between the lowest and highest standard on the calibration curve. The original sample concentration is calculated using the dilution equation, shown in eq $$\ref{9}$$.
$C_1 V_1= C_2 V_2 \label{9}$
In eq $$\ref{9}$$, C1 is the analyte concentration of the original (undiluted) sample, V1 is the volume of the original sample, V2 is the volume of the diluted sample, and C2 is the analyte concentration of the diluted sample. If the sample is diluted prior to analysis, the response of the diluted sample is used as the y-value in the calibration equation and the analyte concentration of the diluted sample is the x-value determined using the calibration equation. The analyte concentration of the original sample is calculated using the dilution equation.
Q17: The table below contains sodium ISE calibration data. If the cell potential measured in a sample is ‑0.115 V, determine the sodium concentration (mol L-1) in this sample.
[Na+] (M)
Ecell (V vs SCE)
1.0 x 10-4
-0.221
1.0 x 10-3
-0.164
1.0 x 10-2
-0.107
1.0 x 10-1
-0.048
Q18: In the previous question, the sample was prepared by pipetting 5.00 mL of the original water sample and 2.00 mL of an ionic strength adjustment buffer into a 100 mL volumetric flask and diluting to the mark with distilled water. Determine the sodium concentration (mol L-1) in the original water sample.
This page titled pH / Ion Selective Electrodes is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Contributor. | 2022-11-30 08:18:38 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6511752009391785, "perplexity": 1434.1702860153084}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446710733.87/warc/CC-MAIN-20221130060525-20221130090525-00579.warc.gz"} |
https://demo7.dspace.org/items/0a66101f-0c1f-47fe-b2b7-994c565c1635 | On the Dovbysh-Sudakov representation result
Authors
Panchenko, Dmitry
Description
We present a detailed proof of the Dovbysh-Sudakov representation for symmetric positive definite weakly exchangeable infinite random arrays, called Gram-de Finetti matrices, which is based on the representation result of Aldous and Hoover for arbitrary (not necessarily positive definite) symmetric weakly exchangeable arrays.
Keywords
Mathematics - Probability, 60G09, 82B44 | 2022-12-07 22:52:04 | {"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": 0.801538348197937, "perplexity": 2729.1710429194227}, "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/1669446711221.94/warc/CC-MAIN-20221207221727-20221208011727-00331.warc.gz"} |
https://stats.stackexchange.com/questions/561034/which-kind-of-distributions-can-decision-trees-learn-well | # Which kind of distributions can decision trees learn (well)?
Suppose I have a classification task in $$\mathbb{R}^d$$, given by a distribution $$P$$ and training data $$D$$. The decision boundary is $$B = \{x \; : \; P(Y=1|x) = P(Y=0|x)\}$$ Assume I am learning a decision tree (not a random forest) of a given depth $$k$$, trying to classify the data correctly in each leaf. The loss function is the distance to the decision boundary. Of course, it depends on the geometry of $$B$$ on how well this task can be performed. If $$B$$ can be approximated by axis-aligned cuts, we can almost perfectly learn $$B$$ using the tree.
My question: Is there research on distributional (not so much geometric) assumptions on $$P$$ that give bounds on how good the optimal tree is? | 2022-08-15 07:14:04 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 9, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7269622087478638, "perplexity": 307.91729719747775}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-33/segments/1659882572161.46/warc/CC-MAIN-20220815054743-20220815084743-00508.warc.gz"} |
https://mathhelpboards.com/threads/integral.8855/#post-41120 | # Integral
#### FilipVz
##### New member
Can somebody explain how to solve integral from the picture above?( solution is in the second line)
#### Jameson
Staff member
I haven't worked out the solution yet, but given the answer and the general form of the integral I would guess trig substitution is the way to solve it. Question though - should the $(1-k^2)$ in the denominator actually be $(1-k^2)^2$?
If that's not the way to solve it then maybe there's something with the substitution $(k \cot(\theta))^2=k^2 \cot^2(\theta)=k^2(\csc^2(\theta)-1)$, but I'm not sure yet. Will post back if anything comes to mind.
#### ZaidAlyafey
##### Well-known member
MHB Math Helper
Hint :
$$\displaystyle \frac{d}{dx} \left(\cos^{-1} (f) \right) = \frac{-f'}{\sqrt{1-f^2}}$$
#### MarkFL
Staff member
Hint: Try the substitution:
$$\displaystyle \cot(\theta)=\frac{\sqrt{1-k^2}}{k}\cos(u)$$
Hence:
$$\displaystyle \csc^2(\theta)\,d\theta=\frac{\sqrt{1-k^2}}{k}\sin(u)\,du$$
And the result will follow.
#### ZaidAlyafey
##### Well-known member
MHB Math Helper
Hint: Try the substitution:
$$\displaystyle \cot(\theta)=\frac{\sqrt{1-k^2}}{k}\cos(u)$$
Hence:
$$\displaystyle \csc^2(\theta)\,d\theta=\frac{\sqrt{1-k^2}}{k}\sin(u)\,du$$
And the result will follow.
No need for substitution. Of course it is more advisable on an elementary level .
#### FilipVz
##### New member
Thank you,
i solved it, using substitution:
$$\displaystyle u= (k*ctgθ)/(1-k^2 )$$
But, my result is:
$$\displaystyle ϕ=-arcsin((k∙ctgθ)/√(1-k^2 ))+c_2$$
$$\displaystyle ϕ=arccos((k∙ctgθ)/√(1-k^2 ))+c_2$$
Is this correct?
#### MarkFL
Staff member
Thank you,
i solved it, using substitution:
$$\displaystyle u= (k*ctgθ)/(1-k^2 )$$
But, my result is:
$$\displaystyle ϕ=-arcsin((k∙ctgθ)/√(1-k^2 ))+c_2$$
$$\displaystyle ϕ=arccos((k∙ctgθ)/√(1-k^2 ))+c_2$$
Is this correct?
Yes, that's another possible form. Consider the identity:
$$\displaystyle \sin^{-1}(x)+\cos^{-1}(x)=\frac{\pi}{2}$$
along with the fact that the sum of an arbitrary constant and another constant is still an arbitrary constant.
#### ZaidAlyafey
##### Well-known member
MHB Math Helper
$$\displaystyle \int \frac{-1}{\sqrt{1-x^2}} \, dx = \cos^{-1}(x)+A$$
$$\displaystyle -\int \frac{1}{\sqrt{1-x^2}} \, dx = -\sin^{-1}(x)+B$$
And this simply because
$$\displaystyle \cos^{-1}(x)+\sin^{-1}(x) =\frac{\pi}{2}$$ | 2022-05-22 16:25:51 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8850008845329285, "perplexity": 2923.677956353481}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662545875.39/warc/CC-MAIN-20220522160113-20220522190113-00575.warc.gz"} |
https://www.bookofproofs.org/branches/extended-greatest-common-divisor-python/ | Welcome guest
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Extended Greatest Common Divisor (Python)
Known time/storage complexity and/or correctness
Let $$a,b\in\mathbb{Z}$$ be positive integers $a,b\in\mathbb Z$ with $$a\le b$$. The algorithm $$\operatorname{gcdext}(a,b)$$ calculates correctly the greatest common divisor $d$ of $$a$$ and $$b$$ and integers $x,y\in\mathbb Z$ $x,y\in\mathbb Z$ such that $$d=ax+by.$$ It requires $$\mathcal O(\log |b|)$$ (worst case and average case) division operations, which corresponds to $$\mathcal O(\log^2 |b|)$$ bit operations.
Short Name
$\operatorname{gcdext}$
Input Parameters
$a,b\in\mathbb{Z}$
Output Parameters
$d,x,y\in\mathbb Z$
Python Code
def gcdext(a, b):
if a ≤ 0:
NotPositiveException(a)
if b ≤ 0:
NotPositiveException(b)
x = 0
y = 1
u = 1
v = 0
q = a // b
r = a % b
while r != 0:
a = b
b = r
t = u
u = x
x = t - q * x
t = v
v = y
y = t - q * y
if b != 0:
q = a // b
r = a % b
d = b
return [d, x, y]
# Usage
print(gcdext(5159, 4823))
# will output
# [7, -244, 261], which means 7=-244*5159+261*4823
| | | | created: 2019-06-22 05:21:02 | modified: 2019-07-15 09:24:07 | by: bookofproofs | references: [1357], [8187]
(none) | 2019-07-16 16:04:33 | {"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": 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": 0.38363465666770935, "perplexity": 7977.8487206895725}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-30/segments/1563195524679.39/warc/CC-MAIN-20190716160315-20190716182315-00030.warc.gz"} |
https://nicknash.me/2012/10/26/happy-halloween/ | # Dark Origins
This is a short post about a tough puzzle I came across a few months back, originally at Two Guys Arguing. Since it’s the Matrix Puzzle From Hell, and it’s nearly Halloween, I thought it was ingeniously topical. I’m sure you agree.
If anyone knows the true origin of the puzzle, I’d love to hear it.
# The Puzzle
A matrix has the property that every entry is the smallest nonnegative integer not appearing either above it in the same column or to its left in the same row.
Give an algorithm to compute the entries of this matrix.
# An Example
Here’s an example matrix:
0 1 2 3 4 5 6 7
1 0 3 2 5 4 7 6
2 3 0 1 6 7 4 5
3 2 1 0 7 6 5 4
4 5 6 7 0 1 2 3
5 4 6 7 1 0 3 2
6 7 4 5 2 3 0 1
7 6 5 4 3 2 1 0
It looks like this matrix is full of patterns, so what’s a good algorithm for computing its entries?
As with any puzzle, it’s probably worth trying it before reading the spoilers below. I’ll reveal the answer slowly though.
# A (Very) Naive Algorithm
Without really thinking about the puzzle, we can at least give a correct algorithm. To compute a particular entry, compute all the entries above it, and to its left, and choose the smallest nonnegative integer not included in the union of those two.
If we’re considering an $n \times n$ matrix, then to compute a given entry we can be forced to compute up to $\Theta(n^2)$ other entries, because we need to compute all the entries above it and to its left (and all the entries above them and to their left). Once we have the entries above us and to our left we can sort them, in say, $\Theta(n \log n)$ time and then find the smallest excluded element with a linear scan. So the algorithm takes $\Theta(n^3 \log n)$ time.
This certainly feels very naive. We’re not using any of the problem’s structure. At least we have a time complexity though.
# Patterns
Let’s look at a small example matrix
0 1 2 3
1 0 3 2
2 3 0 1
3 2 1 0
There’s lots of symmetry here, and quite a few patterns to consider. Let’s just look at the top-left 2×2 entries though:
0 1
1 0
To generate the 2×2 sub-matrix of entries in the top-right of our original 4×4 matrix, we just add 2 to these entries. To generate the 2×2 sub-matrix of entries in the bottom-left, we just add 2 to these entries. To generate the 2×2 sub-matrix in the bottom-right, we just copy these entries.
We now have our 4×4 matrix, if we extend it in the same way, adding 4 this time, we’ll get our original 8×8 example matrix.
We can even start with a 1×1 matrix: 0, and extend it into the initial 2×2 matrix, by adding one (and copying).
So this looks like a very useful pattern. We’ve not thought about why, or if, it’s right just yet. It looks right though. So let’s try and bash out a better algorithm.
# A Better Algorithm
It looks like we can generate the entire matrix by the “doubling” pattern just described: Start with the 1×1 matrix (which is just 0). Double it to give the 2×2 matrix, double that to give a 4×4 matrix, etc.
To pick out just a single entry, generating the whole matrix isn’t going to be the right way to go though. We’d still only be down to $\Theta(n^2)$ time (and space) — not as bad as $\Theta(n^3 \log n)$, but we can do better.
Instead, we can work backwards. To work backwards from a given row and column $r, c$, we just need to find the smallest number $m$ such that $r, c < 2^m$. Then we know which $2^m \times 2^m$ matrix the entry “belongs” to. Then we find out which quadrant of that matrix the entry is in, and then which quadrant of that quadrant the entry is in, and so on, until we reach the top-left position.
More precisely, we’re saying of our matrix $M$ that given $r, c$:
• Choose the largest integer $p$ such that $2^p \leq r$
• Choose the largest integer $q$ such that $2^q \leq c$
• If $p = q$ (i.e. bottom-right quadrant), then the answer is $M[r - 2^p, c - 2^p]$
• If $p > q$ (i.e. bottom-left quadrant), then the answer is $2^p + M[r - 2^p, c]$
• If $p < q$ (i.e. top-right quadrant), then the answer is $2^q + M[r, c - 2^q]$
Repeating this until we get to $r = c = 0$ only takes $\Theta(\log \max\lbrace r, c\rbrace)$ time (because we always more than half either our row or column position), and constant space. That’s a lot better than our first attempt.
But can we do even better still? It turns out we can. A lot better.
# Harder, but Easier
When I first heard this puzzle, there was a condition attached to the time complexity of the solution. If it had been that there is a solution working in $O(n^{1/3}(\log n)^2)$ time, I think I would have taken forever trying to think up some complicated algorithm to solve it.
Instead, it said the solution had to work in $O(1)$ time and space per entry. In theory, having to solve a problem in a better time is harder. In the land of puzzles though, knowing we can do it in constant time makes getting to an answer easier. It’s almost a spoiler.
Knowing it can be done in constant time and space means there must be a formula for the entries. Let’s start simple, and go through our tools. Does $M[r, c] = r + c$. No, obviously not. What about the good old bitwise operators though? How about $r \mbox{ OR } c$? Well, no because $1 \mbox{ OR } 1 = 1$ but we can see $M[1, 1] = 0$ in our examples.
What about XOR? It turns out that’s it. We can just XOR our row and column numbers to get the entry. Try it!
# But Why?
It’s a bit of a shocker to know the answer is just plain-old XOR. Who knew it computed the minimum excluded number in a matrix?
Well, if we look at our logarithmic time algorithm, it’s pretty easy to see that it’s nothing more than an XOR of the row and column. When it said “Choose the largest integer $p$ such that $2^p \leq r$“, we’re really asking about the most-significant bit of our row number. It’s the same with our column, then we compare these bits, in fact, XORing them into our answer, and repeat with the remaining bits.
This at least explains how we get to XOR from our quadrant-based algorithm. We still haven’t really explained why our initial “doubling” construction was correct though. Before we do that, let’s have a look at a pretty picture.
# Pretty Picture
As a neat little aside, it turns out that if you draw a picture of our matrix (zero being black), here’s what you get:
Pretty much any demoscener (or graphics hacker) will recognize this as the XOR texture, the easiest of all textures to generate. Just try googling it!
Speaking of the demoscene, here’s video of 4k Amiga intro making extensive use of the XOR pattern (not usually a badge of honor in the demoscene world, incidentally): Humus 4.
I think it’s neat to know that texture is all about minimum excluded numbers, after all these years.
# But Really, Why?
We never really explained why our doubling construction is actually correct. It’s pretty obvious, but let’s see about that now.
Suppose we have a $2^k \times 2^k$ version of our matrix $M$. Our doubling construction gives us a $2^{k+1} \times 2^{k+1}$ matrix, such that:
• The upper-left corner is $M$
• The bottom-left corner is the entries of $M$ plus $2^k$
• The bottom-right corner is $M$
• The upper-right corner is the entries of $M$ plus $2^k$
Now let’s think about the simplest version of our matrix. If $k = 0$, we have a $1 \times 1$ matrix (containing a single entry, 0) for which every row and every column contains all the numbers in $\lbrace 0, \ldots, 2^k - 1\rbrace$. Because of this property of its rows and columns, we’ll call that matrix “complete”.
If we apply our doubling construction, we have a $2 \times 2$ matrix that is also complete. Inductively, it’s easy to see we’ll continue to get complete power-of-two-sized matrices by doubling complete power-of-two-sized matrices.
The question is now, in a doubled matrix, is each entry the minimum excluded number? Consider any $2^{k + 1} \times 2^{k + 1}$ matrix of minimum excluded numbers. Take the upper-right corner (UR), it can only contain numbers in $\lbrace 2^k, \ldots, 2^{k + 1} - 1 \rbrace$, both in order to guarantee exclusion (because the upper-left corner (UL) is complete) and also in order that the entries are as small as possible. Now, if any entry in UR was smaller than $2^k$ plus the corresponding entry in UL, then we could subtract $2^k$ from all the numbers in UR and get a $2^k \times 2^k$ matrix of minimum excluded numbers that had a smaller entry than UL somewhere. This is obviously impossible, because we know UL already contains the (unique) minimum excluded numbers in each entry.
We can make similar arguments for the other two quadrants. That’s another way to see the XOR connection. The upper-left and bottom-left quadrants are restricted in the numbers they can choose as entries because they have only either a different row or a different column (i.e. exclusive or) to the entries in the top-left quadrant. Whereas, the entries in the bottom-right quadrant are unrestricted by those in the top-left quadrant having both a different row and column.
# Nim, Nimbers and Beyond
To finish up, it’s worth noting that there’s a kid’s game played with piles of sticks called Nim, which the solution to this puzzle represents a winning strategy for.
In fact, there are even things called Nimbers, and XOR corresponds to Nimber addition. There’s also a Nimber multiplication operator. I wonder what it looks like as a texture, compared to what the XOR texture looks like?
## 2 thoughts on “The Matrix Puzzle From Hell, Revisited”
1. Derek Ledbetter
By the way, your naive algorithm can be improved a bit. To compute element (r,c), create an array of booleans of length r + c + 1 starting at index 0, initialized to all false. Then for each element x above or to the left, if x <= r + c, set element x of the array. After this, at most r + c elements of the array have been set, so there's at least one that hasn't been set, so find the first one, and that's the answer.
So by avoiding the sorting step, the time is O(n^3), eliminating the log n part.
1. nashn Post author
Good point! I was trying to pick the absolute simplest approach for the algorithm, so I didn’t think about anything like this. | 2020-05-26 06:57:03 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 50, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8093478083610535, "perplexity": 454.6674467179215}, "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/1590347390448.11/warc/CC-MAIN-20200526050333-20200526080333-00430.warc.gz"} |
https://proofwiki.org/wiki/Elementary_Row_Operation/Examples/Operations_on_Arbitrary_Matrix/Swap_r1_and_r2 | # Elementary Row Operation/Examples/Operations on Arbitrary Matrix/Swap r1 and r2
## Example of Elementary Row Operation
Let $\mathbf A$ be the matrix:
$\mathbf A = \begin {pmatrix} 1 & 2 & 3 & 4 \\ 2 & -1 & 1 & 0 \\ -2 & 3 & 1 & 1 \end {pmatrix}$
Let the elementary row operation $e$ be applied to $\mathbf A$, where $e$ is defined as:
$e := r_1 \leftrightarrow r_2$
Then $\mathbf A$ is transformed into:
$\mathbf A = \begin {pmatrix} 2 & -1 & 1 & 0 \\ 1 & 2 & 3 & 4 \\ -2 & 3 & 1 & 1 \end {pmatrix}$
## Proof
From Elementary Row Operation: $r_1 \leftrightarrow r_2$, the elementary row matrix $\mathbf E$ corresponding to $e$ is:
$\mathbf E = \begin {pmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end {pmatrix}$
$\ds \map e {\mathbf A}$ $=$ $\ds \mathbf E \mathbf A$ $\ds$ $=$ $\ds \begin {pmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end {pmatrix} \begin {pmatrix} 1 & 2 & 3 & 4 \\ 2 & -1 & 1 & 0 \\ -2 & 3 & 1 & 1 \end {pmatrix}$ $\ds$ $=$ $\ds \begin {pmatrix} 2 & -1 & 1 & 0 \\ 1 & 2 & 3 & 4 \\ -2 & 3 & 1 & 1 \end {pmatrix}$
$\blacksquare$ | 2021-08-03 18:00:07 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9749330282211304, "perplexity": 509.75721683783223}, "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/1627046154466.61/warc/CC-MAIN-20210803155731-20210803185731-00432.warc.gz"} |
https://lavelle.chem.ucla.edu/forum/viewtopic.php?f=142&t=60236&p=228501 | ## What does the nitrate do in a concentration cell?
$E_{cell} = E_{cell}^{\circ}-\frac{RT}{nF}\ln Q$
Skyllar Kuppinger 1F
Posts: 52
Joined: Thu Jul 25, 2019 12:16 am
### What does the nitrate do in a concentration cell?
What is the point of the nitrate in this concentration cell?? (Image attached)What does it do? Does it keep the solution neutral? If so how?
Attachments
Vicki Liu 2L
Posts: 101
Joined: Sat Aug 24, 2019 12:15 am
### Re: What does the nitrate do in a concentration cell?
I believe nitrate is a spectator ion in this case. In other words, AgNO3 was dissolved to achieve Ag+ ions in solution but the NO3- is not involved in the actual reaction.
sarahforman_Dis2I
Posts: 109
Joined: Sat Aug 17, 2019 12:18 am
### Re: What does the nitrate do in a concentration cell?
Skyllar Kuppinger 1F wrote:What is the point of the nitrate in this concentration cell?? (Image attached)What does it do? Does it keep the solution neutral? If so how?
I believe that the nitrate keeps the solution neutral though balancing out the Ag+ ions. That is why there is a porous disk, think of it as a form of salt bridge, in that it allows for ions to move to balance charge. I hope this helps!
CosetteBackus_4F
Posts: 23
Joined: Wed Feb 20, 2019 12:18 am
### Re: What does the nitrate do in a concentration cell?
While the nitrate is not necessarily involved in the reaction itself, its purpose, like the salt bridge, is to allow ions to flow so that the charge can balance.
Abhi Vempati 2H
Posts: 104
Joined: Fri Aug 09, 2019 12:17 am
### Re: What does the nitrate do in a concentration cell?
Like everyone is saying, nitrate diffuses through the porous disk to balance the charges on both sides. The actual reaction that is occurring is for Ag+(aq) and Ag(s), but NO3- is important for the concentration cell as well.
Angela Patel 2J
Posts: 110
Joined: Sat Aug 24, 2019 12:17 am
### Re: What does the nitrate do in a concentration cell?
You can't just put gold ions in solution, you need it to be dissolved from another compound.
Naneeta Desar 1K
Posts: 106
Joined: Fri Aug 09, 2019 12:15 am
### Re: What does the nitrate do in a concentration cell?
In order to balance the charges nitrate flows through the porous disk.
Return to “Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)”
### Who is online
Users browsing this forum: No registered users and 1 guest | 2020-07-16 14:27:06 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 1, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.3923274874687195, "perplexity": 6094.0558726802665}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-29/segments/1593657169226.65/warc/CC-MAIN-20200716122414-20200716152414-00166.warc.gz"} |
https://www.gradesaver.com/textbooks/math/algebra/intermediate-algebra-12th-edition/chapter-7-summary-exercises-performing-operations-with-radicals-and-rational-exponents-page-477/5 | ## Intermediate Algebra (12th Edition)
$-6\sqrt{10}$
$\bf{\text{Solution Outline:}}$ To simplify the given radical expression, $6\sqrt{10}-12\sqrt{10} ,$ combine the like terms. $\bf{\text{Solution Details:}}$ By combining like terms, the expression above is equivalent to \begin{array}{l}\require{cancel} (6-12)\sqrt{10} \\\\= -6\sqrt{10} .\end{array} | 2020-03-29 06:06:31 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9757384657859802, "perplexity": 4520.742596504239}, "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/1585370493818.32/warc/CC-MAIN-20200329045008-20200329075008-00168.warc.gz"} |
http://mathoverflow.net/revisions/61389/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).
2 added 14 characters in body
Consider the curve
$$\gamma:\quad \phi\ \mapsto\ \exp{-1\over Bigl(1+\exp{-1\over \pi^2 -\phi^2}\ \phi^2}\Bigr)\ (\cos\phi,\sin\phi)\qquad(-\pi< \phi< \pi)$$
with filled-in point $(-1,0)$.
1
Consider the curve
$$\gamma:\quad \phi\ \mapsto\ \exp{-1\over \pi^2 -\phi^2}\ (\cos\phi,\sin\phi)\qquad(-\pi< \phi< \pi)$$
with filled-in point $(-1,0)$. | 2013-06-20 07:23:37 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.43721941113471985, "perplexity": 12239.912223737041}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2013-20/segments/1368710963930/warc/CC-MAIN-20130516132923-00072-ip-10-60-113-184.ec2.internal.warc.gz"} |
https://chemistry.stackexchange.com/questions/140840/why-isnt-just-a-fully-occupied-valence-s-orbital-stable/140842 | # Why isn't just a fully occupied valence s orbital stable?
Why arent elements like calcium with an electron configuration of $$[\text{noble gas}]\ n\mathrm{s}^2$$ stable, although all the populated orbitals are fuully filled? Why is it necessary to obtain an octet configuration instead? (Ignoring the trivial cases of hydrogen and helium)
• Because it is an outer orbital that must be filled (although the value of "outer" is not intuitively obvious to the casual observer for transition metals). – DrMoishe Pippik Sep 30 at 4:11
• I actually think this is a good question as it’s something that’s rarely discussed in introductory courses. – Jan Oct 1 at 9:46
Elements with just 1 or 2 electrons in the most outer $$\ce{s}$$ orbitals, like $$\ce{K}$$ or $$\ce{Ca}$$, have very well screened off the central charge of the kernel. As consequence, the effective charge of the kernel and related energy needed to ionize the atom are relatively low.
OTOH, elements with just few missing electrons in the most outer $$\ce{p}$$ orbital, like $$\ce{O}$$ or $$\ce{Cl}$$, do not have well screened off the central kernel charge, which is relatively high, what leads to considerable electron affinity to some extra electrons.
Combining these 2 kinds of elements leads to energy release, as energy spent by ionization of the former is compensated by energy release by capturing electrons by the latter and by a covalent contribution to the element bond.
In case of solid ionic compounds, there is additional large energy release du the ionic lattice energy.
• No, it's not compensated. There's still covalent component needed for such reactions to happen. – Mithoron Sep 30 at 13:56
• Hmm, you may be right. – Poutnik Sep 30 at 13:58 | 2020-12-04 11:06:13 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 7, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5257329940795898, "perplexity": 1021.6482640554851}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-50/segments/1606141735600.89/warc/CC-MAIN-20201204101314-20201204131314-00222.warc.gz"} |
https://matheducators.stackexchange.com/questions/11745/where-can-i-find-a-set-of-these-logic-blocks | # Where can I find a set of these 'logic' blocks?
(It will be difficult to answer this question without 'advertising' for a retailer, but I've searched for these several times in the past few years, to no avail.)
In Math From Three To Seven (The story of a Mathematical Circle for Preschoolers), the author makes frequent reference to a set of plastic shapes, each of which has four attributes:
• Colour (red, green, blue, yellow)
• Shape (square, circle, triangle)
• Size (big, small)
• Holiness (it, uh, has a hole in it. Or it doesn't)
The set is complete - all $(4)(3)(2)(2) = 48$ blocks.
The footnote in the text refers to these as "Dienes Blocks", which today, are "universally known and easy to obtain". The problem, though, is that searching for 'Dienes Blocks' returns different products from the same manufacturer / brand / personal legacy of Zoltan Dienes - namely Base-10 blocks.
From ZoltanDienes.com,
Dienes’ name is synonymous with the Multi-base blocks (also known as Dienes blocks) which he invented for the teaching of place value.
That's great and all, but how do I find these other blocks?
• upvote for holiness alone :) – Jasper Dec 6 '16 at 16:42
• I found something with the name "attribute blocks", but instead of holiness they have thickness. Maybe you can use them, or drill a hole through either thick or thin. – Jasper Dec 6 '16 at 16:47
• @Jasper That's awesome, thank you so much. (Your comment makes a pretty great answer - maybe resubmit it as an answer?) – NiloCK Dec 7 '16 at 1:58
• I searched for Dienes logic blocks. I found them at minilandeducational.com/en/logical-blocks – Amy B Dec 7 '16 at 5:29
• As above, I believe that it is thickness rather than holiness characterizing the 48 Dienes blocks. From p. 1 here: "Dienes created a structured set of attribute blocks that consisted of 48 plastic blocks: 4 different shapes (circle, triangle square and non-square rectangle), 3 different colors (red, blue and yellow), 2 sizes (large and small) and 2 thicknesses (thick and thin)." – Benjamin Dickman Dec 7 '16 at 5:36 | 2019-08-25 19:19:32 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.3492923080921173, "perplexity": 3230.4556296882174}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-35/segments/1566027330786.8/warc/CC-MAIN-20190825173827-20190825195827-00177.warc.gz"} |
https://ftp.aimsciences.org/article/doi/10.3934/jgm.2019004 | # American Institute of Mathematical Sciences
March 2019, 11(1): 59-76. doi: 10.3934/jgm.2019004
## A geometric perspective on the Piola identity in Riemannian settings
Institute of Mathematics, The Hebrew University, Jerusalem 91904, Israel
* Corresponding author: Asaf Shachar
Received May 2018 Revised December 2018 Published January 2019
Fund Project: This research was partially funded by the Israel Science Foundation (Grant No. 1035/17), and by a grant from the Ministry of Science, Technology and Space, Israel and the Russian Foundation for Basic Research, the Russian Federation.
The Piola identity $\operatorname{div}\; \operatorname{cof} \;\nabla f = 0$ is a central result in the mathematical theory of elasticity. We prove a generalized version of the Piola identity for mappings between Riemannian manifolds, using two approaches, based on different interpretations of the cofactor of a linear map: one follows the lines of the classical Euclidean derivation and the other is based on a variational interpretation via Null-Lagrangians. In both cases, we first review the Euclidean case before proceeding to the general Riemannian setting.
Citation: Raz Kupferman, Asaf Shachar. A geometric perspective on the Piola identity in Riemannian settings. Journal of Geometric Mechanics, 2019, 11 (1) : 59-76. doi: 10.3934/jgm.2019004
##### References:
[1] P. Ciarlet, Mathematical Elasticity, Volume 1: Three-Dimensional Elasticity, Elsevier, 1988. Google Scholar [2] J. Eells and L. Lemaire, Selected Topics in Harmonic Maps, CBMS Regional Conference Series in Mathematics, 50. Published for the Conference Board of the Mathematical Sciences, Washington, DC; by the American Mathematical Society, Providence, RI, 1983. doi: 10.1090/cbms/050. Google Scholar [3] L. Evans, Partial Differential Equations, Amer. Math. Soc., Providence, 1998. doi: 10.1090/gsm/019. Google Scholar [4] T. Iwaniec, Null lagrangians: Definitions, examples and applications, Warsaw Lectures, Part 2. Google Scholar [5] R. Kupferman, C. Maor and A. Shachar, Reshetnyak Rigidity for Riemannian Manifolds, Arch. Rat. Mech. Anal., 231 (2019), 367-408. doi: 10.1007/s00205-018-1282-9. Google Scholar [6] J. Marsden and T. Hughes, Mathematical Foundations of Elasticity, Dover, 1983. Google Scholar [7] D. Saunders, The Geometry of Jet Bundles, Cambridge University Press, 1989. doi: 10.1017/CBO9780511526411. Google Scholar [8] P. Steinmann, Geometrical Foundations of Continuum Mechanics, Springer, 2015. doi: 10.1007/978-3-662-46460-1. Google Scholar
show all references
##### References:
[1] P. Ciarlet, Mathematical Elasticity, Volume 1: Three-Dimensional Elasticity, Elsevier, 1988. Google Scholar [2] J. Eells and L. Lemaire, Selected Topics in Harmonic Maps, CBMS Regional Conference Series in Mathematics, 50. Published for the Conference Board of the Mathematical Sciences, Washington, DC; by the American Mathematical Society, Providence, RI, 1983. doi: 10.1090/cbms/050. Google Scholar [3] L. Evans, Partial Differential Equations, Amer. Math. Soc., Providence, 1998. doi: 10.1090/gsm/019. Google Scholar [4] T. Iwaniec, Null lagrangians: Definitions, examples and applications, Warsaw Lectures, Part 2. Google Scholar [5] R. Kupferman, C. Maor and A. Shachar, Reshetnyak Rigidity for Riemannian Manifolds, Arch. Rat. Mech. Anal., 231 (2019), 367-408. doi: 10.1007/s00205-018-1282-9. Google Scholar [6] J. Marsden and T. Hughes, Mathematical Foundations of Elasticity, Dover, 1983. Google Scholar [7] D. Saunders, The Geometry of Jet Bundles, Cambridge University Press, 1989. doi: 10.1017/CBO9780511526411. Google Scholar [8] P. Steinmann, Geometrical Foundations of Continuum Mechanics, Springer, 2015. doi: 10.1007/978-3-662-46460-1. Google Scholar
Illustration of the geometric setting of the Euclidean Piola identity
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2019 Impact Factor: 0.649 | 2021-01-17 10:54:27 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 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": 0.5538914203643799, "perplexity": 4172.977824500145}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-04/segments/1610703511903.11/warc/CC-MAIN-20210117081748-20210117111748-00588.warc.gz"} |
https://stats.stackexchange.com/questions/239875/curve-fit-for-large-data-set | # curve fit for large data set
Need to find curve/trendline and equation that fits the large data set. Looks as in the scatter plot given below. any suggestion in a particular direction are appreciated. Any libraries in java or python or any online tools. I have tried excels trendline option, but the equation underfits my data.
• Try localized regression. If you're happy working with Python, you can use statsmodels.sourceforge.net/devel/generated/… – Jon Oct 12 '16 at 21:12
• is there any way that u know of, to get the resulting equation for the fit? – asaini Oct 12 '16 at 21:46
• Nah, I don't like to use Python for stats stuff. A simple Google search should help you out though. – Jon Oct 12 '16 at 21:59
• Notice that the spread is related to the mean, If there are no exact zeros, it's probably worth taking log(y), and looking at that. It may also be worth considering log(x), but it depends on what it actually measures as to whether that will also make sense (if the origin is arbitrary, it probably doesn't). What do the y and x variables consist of? – Glen_b -Reinstate Monica Oct 13 '16 at 0:26
• x is price , y is time – asaini Oct 13 '16 at 5:24
That looks like data you want to transform first and then maybe try a linear model. My first guess would be to try regressing $y$ against $1/x$ | 2019-12-10 08:18:17 | {"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": 0.2599218785762787, "perplexity": 733.4747499457421}, "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-2019-51/segments/1575540527010.70/warc/CC-MAIN-20191210070602-20191210094602-00366.warc.gz"} |
http://www.reddit.com/r/math/comments/132lo5/how_were_you_taught_math_how_do_you_think_it/c70flno | you are viewing a single comment's thread.
[–] 6 points7 points (12 children)
sorry, this has been archived and can no longer be voted on
wrong. you implicitly used SOH CAH TOA in the unit circle. also the definition of sin,cosine,tangent (arguable since tan = sin/cos) is SOH CAH TOA. that is their DEFINITION. it must be initially taught that way.
the problem comes from it being the only way used thereafter.
[–]Arithmetic Geometry 14 points15 points (3 children)
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That's not entirely correct. Many mathematical concepts can be defined in several equivalent ways. You can define sine and cosine by triangles (the way you mentioned), using the unit circle, using power series, or even axiomatically by certain characteristic properties of their derivatives. (The last one is a rather strange one, but one of my professors introduced them like that, using the fact that sine and cosine are the unique differentiable real functions such that sin' = cos, cos' = -sin, sin(0) = 0, and cos(0) = 1. I don't recommend doing it like that.)
None of these definitions is intrinsically any better than the others, because they all define the exact same thing. Once you've proved them equivalent, it's just a matter of which definition is the most convenient. So, if the unit circle happens to be easier to teach, define it that way, and then prove its connection to triangles later if need be.
Insisting on a less intuitive, harder-to-use definition purely because it's the one that has traditionally been taught first is pedagogically unsound, mentally inflexible, and encourages a rigid way of thinking that leads to a superficial grasp of the concepts. An important part of understanding a concept is seeing the connections between the different ways it can be defined.
[–] 1 point2 points (2 children)
sorry, this has been archived and can no longer be voted on
Being a little bit nitpicky, I disagree that no definition is intrinsically better than another. A definition is supposed to capture the heart of what something is. For example, defining squaring as the unique (differentiable) function f such that f'' = 2, f'(0)=0, and f(0) = 0 seems intrinsically worse than defining it as [; f(x) = x \cdot x ;]. We could probably come up with arbitrarily convoluted definitions for squaring, but I think you get the idea.
[–]Arithmetic Geometry 1 point2 points (0 children)
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On a formal level, that definition is just as correct; the problem with it isn't really intrinsic to the mathematics of it, but rather that it's a confusing and counterintuitive way of learning it. Mathematically, it captures the essence entirely, because it uniquely defines what it's supposed to. We could come up with arbitrarily clumsy — but still technically equivalent — definitions, and as long as they're still mathematically correct, the problem with them can only be seen by appealing to ease of understanding, not to some mathematical property.
[–] 0 points1 point (0 children)
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Yes, but sometimes you have to change a definition if you want progress to a more complex one. While you have square it's alright do define it as [; x2 = x*x ;] . If you have [; x^n, n \in N ;] N you can define it by induction in the same manner (through multiplication). But try to define [; x^y, y \in R ;] through multiplication.
[–][deleted] (2 children)
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[deleted]
[–] 0 points1 point (3 children)
sorry, this has been archived and can no longer be voted on
Why was this downvoted? It's a good point to argue, and I'd like to see why the downvoters disagreed.
To BahBahTheSheep: starting your reply with "wrong." is usually taken as bordering on rude (don't ask me why). Framing your point as a question would have led to intelligent discussion (of an admittedly pedantic point) as opposed to mindless downvotes: "But didn't you implicitly use SOH CAH TOA in the unit circle? [...]"
I don't mean to sound all...uppity/self-righteous myself, I just hate to see a thoughtful point met with downvotes for a dumb reason.
[–] 0 points1 point (0 children)
sorry, this has been archived and can no longer be voted on
The problem with the unit circle is that sometimes your angle is 82 degrees. | 2015-03-02 16:50:28 | {"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": 0.8950455784797668, "perplexity": 969.2614950834288}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-11/segments/1424936462898.92/warc/CC-MAIN-20150226074102-00101-ip-10-28-5-156.ec2.internal.warc.gz"} |
http://hal.in2p3.fr/in2p3-01362518 | Decomposing transverse momentum balance contributions for quenched jets in PbPb collisions at sqrt(s[NN]) = 2.76 TeV
Abstract : Interactions between jets and the quark-gluon plasma produced in heavy ion collisions are studied via the angular distributions of summed charged-particle transverse momenta (pt) with respect to both the leading and subleading jet axes in high-pt dijet events. The contributions of charged particles in different momentum ranges to the overall event pt balance are decomposed into short-range jet peaks and a long-range azimuthal asymmetry in charged-particle pt. The results for PbPb collisions are compared to those in pp collisions using data collected in 2011 and 2013, at collision energy sqrt(s[NN]) = 2.76 TeV with integrated luminosities of 166 inverse microbarns and 5.3 inverse picobarns, respectively, by the CMS experiment at the LHC. Measurements are presented as functions of PbPb collision centrality, charged-particle pt, relative azimuth, and radial distance from the jet axis for balanced and unbalanced dijets.
Type de document :
Article dans une revue
Journal of High Energy Physics, Springer Verlag (Germany), 2016, 1611, pp.055. 〈10.1007/JHEP11(2016)055〉
Domaine :
http://hal.in2p3.fr/in2p3-01362518
Contributeur : Sylvie Flores <>
Soumis le : vendredi 9 septembre 2016 - 07:18:43
Dernière modification le : jeudi 10 mai 2018 - 02:00:23
Citation
V. Khachatryan, M. Besançon, F. Couderc, M. Dejardin, D. Denegri, et al.. Decomposing transverse momentum balance contributions for quenched jets in PbPb collisions at sqrt(s[NN]) = 2.76 TeV. Journal of High Energy Physics, Springer Verlag (Germany), 2016, 1611, pp.055. 〈10.1007/JHEP11(2016)055〉. 〈in2p3-01362518〉
Métriques
Consultations de la notice | 2018-10-19 14:33:32 | {"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": 0.8301795125007629, "perplexity": 7977.177612951339}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-43/segments/1539583512400.59/warc/CC-MAIN-20181019124748-20181019150248-00036.warc.gz"} |
https://math.stackexchange.com/questions/2505156/why-is-this-zig-zag-function-monotone-increasing-at-0-but-not-monotone-increa | # Why is this zig zag function monotone increasing at $0$, but not monotone increasing in a neighborhood of $0$?
This is from The Way of Analysis by Strichartz.
I believe see a neighborhood of $0$ such that $x_1 < x_2 \implies f(x_1) \le f(x_2)$.
Why is this zig zag function monotone increasing at $0$, but not monotone increasing in a neighborhood of $0$?
• I have a couple of questions: what does it mean to be monotone increasing at a point? What is the definition of the function in question? – Theo Bendit Nov 5 '17 at 2:43
• Presumably there are to be infinitely many wiggles, say hitting the $x$-axis at $1/n$ for every $n\ne0$. – Lubin Nov 5 '17 at 2:45
• @TheoBendit I've included the defintions – Al Jebr Nov 5 '17 at 2:46
• It's not increasing on any neighbourhood of $0$ because any such neighbourhood will contain a downwards "zag". (The picture could perhaps be better - I believe it's trying to communicate a "fractal" zig-zag as Lubin described.) – Anthony Carapetis Nov 5 '17 at 2:51
I think the idea is that the function is fractal, in the sense that as we approach $0$ from either side, the function has smaller spikes with increasing frequency. Indeed, we can see that the function $f$ is such that $f\le 0$ for $x<0$, $f\ge 0$ for $x>0$, and $f=0$ at $x=0$.
$f$ is monotone at $0$, because given $x_1<x_0<x_2$ we have that $f(x_1)\le f(x_0)=0$ and $f(x_2)\ge f(x_0)=0$ so that $f(x_1)\le f(x_0)\le f(x_2)$. On the other hand, because of this spiking behavior, any neighborhood of $0$, $N=\{x\in \mathbf{R}: \lvert x\rvert<\epsilon\}$, contains an entire spike, which is clearly not a monotone function, since it increases, and then decreases. So $f$ is not monotone in a neighborhood of $0$. | 2019-08-21 01:07:56 | {"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": 0.9539980888366699, "perplexity": 264.96936152041343}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-35/segments/1566027315695.36/warc/CC-MAIN-20190821001802-20190821023802-00373.warc.gz"} |
https://www.zbmath.org/?q=ai%3Agibbons.phillip-b+ai%3Akarp.richard-m+py%3A1990 | # zbMATH — the first resource for mathematics
Subtree isomorphism is in random NC. (English) Zbl 0711.68052
A subtree isomorphism problem is to determine for given two trees G and H whether there is a subgraph (subtree) of H that is isomorphic to G. Unlike other versions of the (sub)graph isomorphism problem this one is known to be in P, yet the existence of a fast parallel algorithm to solve it posses an interesting problem. The authors answer this problem affirmatively, provided that we consider randomized parallel algorithms. Their algorithm runs in time $$O(\log^ 3 n)$$ on a CREW PRAM using $$n^ 3M(n)\log \log n/\log n$$ processors, where M(n) is the number of bit operations required by a CREW PRAM to multiply two $$n\times n$$ boolean matrices in O(log n) time.
The authors start with the description of an algorithm for solving the rooted version of the subtree isomorphism problem, whic is then extended to the general unrooted case. Their approach is based on the best known sequential algorithm for this problem, given by D. W. Matula [Annals of Discrete Mathematics 2, 91-106 (1978; Zbl 0391.05022)], combined with the dynamic contraction technique by G. L. Miller and J. H. Reif [Parallel tree contraction and its application, Proc. Symp. on Foundations of Computer Science, 478-489 (1985)]. The randomized part of their algorithm is derived from using the randomized algorithm of K. Mulmuley, U. V. Vazirani and V. V. Vazirani [Combinatorica 7(1), 105-113 (1987; Zbl 0632.68041)] for constructing a perfect matching in a bipartite graph. Finally they prove that bipartite perfect matching is log-space irreducible to subtree isomorphism.
Reviewer: J.Vyskoc
##### MSC:
68W15 Distributed algorithms 68R10 Graph theory (including graph drawing) in computer science 68Q25 Analysis of algorithms and problem complexity
Full Text:
##### References:
[1] Borodin, A.; von zur Gathen, J.; Hopcroft, J.E., Fast parallel matrix and GCD computations, Proceedings of the symposium on foundations of computer science (FOCS), 65-71, (1982) · Zbl 0507.68020 [2] Brent, R.P., The parallel evaluation of general arithmetic expressions, J. ACM, 21, 201-208, (1974) · Zbl 0276.68010 [3] D. Coppersmith, personal communication, 1988. [4] Coppersmith, D.; Winograd, S., Matrix multiplication via arithmetic progressions, Proceedings of the symposium on theory of computing (STOC), 1-6, (1987) [5] Edmonds, J., Systems of distinct representatives and linear algebra, J. res. nat. bur. standards, 71, 241-245, (1967) · Zbl 0178.03002 [6] Galil, Z.; Pan, V., Improved processor bounds for algebraic and combinatorial problems in RNC, (), 490-495 · Zbl 0685.68048 [7] Garey, M.R.; Johnson, D.S., Computers and intractability: A guide to the theory of NP completeness, (1979), Freeman New York · Zbl 0411.68039 [8] Gibbons, P.B.; Karp, R.M.; Miller, G.L.; Soroker, D., Subtree isomorphism is in random NC, (), 43-52, Lecture Notes in Computer Science [9] Gibbons, P.B.; Karp, R.M.; Miller, G.L.; Soroker, D., Subtree isomorphism is in random NC, () · Zbl 0652.68078 [10] P.B. Gibbons, G.L. Miller and S.-H. Teng, personal communication,1988. [11] Karp, R.M.; Ramachandran, V.L., A survey of parallel algorithms for shared-memory machines, (), also in: Handbook of Theoretical Computer Science (North-Holland, Amsterdam, to appear) [12] Karp, R.M.; Upfal, E.; Wigderson, A., Constructing a perfect matching is in random NC, Combinatorica, 6, 1, 35-48, (1986) · Zbl 0646.05051 [13] M. Karpinski and A. Lingas, Subtree isomorphism and bipartite perfect matching are mutually NC reducible, Submitted. · Zbl 0664.68072 [14] Ladner, R.E.; Fischer, M.J., Parallel prefix computation, J. ACM, 27, 4, 831-838, (1980) · Zbl 0445.68066 [15] Matula, D.W., Subtree isomorphism in O(n$$52$$), (), 91-106 · Zbl 0391.05022 [16] Miller, G.L.; Reif, J.H., Parallel tree contraction and its applications, Proceedings of the symposium on foundations of computer science (FOCS), 478-489, (1985) [17] Mulmuley, K.; Vazirani, U.V.; Vazirani, V.V., Matching is as easy as matrix inversion, Combinatorica, 7, 1, 105-113, (1987) · Zbl 0632.68041 [18] Pan, V., Fast and efficient parallel algorithms for the exact inversion of integer matrices, (), 504-521, Lecture Notes in Computer Science [19] Pippenger, N., On simultaneous resource bounds, Proceedings of the symposium on foundations of computer science (FOCS), 307-311, (1979) [20] Preparata, F.P.; Sarwate, D.V., An improved parallel processor bound in fast matrix inversion, Inform. process. lett., 7, 3, 148-150, (1978) · Zbl 0373.65020 [21] Rabin, M.O.; Vazirani, V.V., Maximum matchings in general graphs through randomization, () · Zbl 0689.68092 [22] Schwartz, J.T., Fast probabilistic algorithms for verification of polynomial identities, J. ACM, 27, 701-717, (1980) · Zbl 0452.68050 [23] Stockmeyer, L.J.; Vishkin, U., Simulation of parallel random access machines by circuits, SIAM J. comput., 13, 409-422, (1984) · Zbl 0533.68048 [24] Tarjan, R.E.; Vishkin, U., An efficient parallel biconnectivity algorithm, SIAM J. comput., 14, 862-874, (1985) · Zbl 0575.68066
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. | 2021-03-05 14:17:11 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 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": 0.7660021781921387, "perplexity": 4306.213928503708}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-10/segments/1614178372367.74/warc/CC-MAIN-20210305122143-20210305152143-00458.warc.gz"} |
https://codereview.stackexchange.com/questions/219302/comparing-each-item-from-dir-with-each-item-from-another-dir/219324 | # Comparing each item from dir with each item from another dir
The task is to compare students homework SQL files with mentors SQL files.
I've written two functions, which return a two-dimensional array (1st elements are an absolute path, 2nd are a relative).
Then I'm going to compare the relative path of students and mentors and execute SQL files (finding using absolute path) if these values are equal
Is there a more elegant implementation?
The folder structure of mentors dir:
Homework (folder)
├ 1 (folder)
| ├ 1.sql
| ├ 2.sql
| └ n.sql
├ 2 (folder)
| ├ 1.sql
| ├ 2.sql
| └ n.sql
├ n (folder)
| ├ 1.sql
| ├ 2.sql
| └ n.sql
The folder structure of students dir:
├Students Homework (folder)
├Student1(folder)
├ 1 (folder)
| ├ 1.sql
| ├ 2.sql
| └ n.sql
├ 2 (folder)
| ├ 1.sql
| ├ 2.sql
| └ n.sql
├ n (folder)
| ├ 1.sql
| ├ 2.sql
| └ n.sql
├Student2(folder)
├ 1 (folder)
| ├ 1.sql
| ├ 2.sql
| └ n.sql
├ 2 (folder)
| ├ 1.sql
| ├ 2.sql
| └ n.sql
├ n (folder)
| ├ 1.sql
| ├ 2.sql
| └ n.sql
"Mentors" function:
def find_mentors_sql(config):
mentors_sql_abs = []
mentors_sql_rel = []
for dirpath, subdirs, files in walk(config["MAIN_DIR"] + '\\Homework'):
mentors_sql_abs.extend(path.join(dirpath, x) for x in files if x.endswith(".sql"))
mentors_sql_rel.extend(path.join(path.basename(dirpath), x) for x in files if x.endswith(".sql"))
mentors_sql = [[0] * 2 for i in range(len(mentors_sql_abs))]
iter = 0
for _ in mentors_sql_abs:
mentors_sql[iter][0] = mentors_sql_abs[iter]
iter += 1
iter1 = 0
for _ in mentors_sql_rel:
mentors_sql[iter1][1] = mentors_sql_rel[iter1]
iter1 += 1
return mentors_sql
"Students" function (the logic is similar to the previous one:
def find_students_sql(config):
students_sql_abs = []
students_sql_rel = []
for dirpath, subdirs, files in walk(config["MAIN_DIR"] + '\\Students Homework'):
students_sql_abs.extend(path.join(dirpath, x) for x in files if x.endswith(".sql"))
students_sql_rel.extend(path.join(path.basename(dirpath), x) for x in files if x.endswith(".sql"))
students_sql = [[0] * 2 for i in range(len(students_sql_abs))]
iter = 0
for _ in students_sql:
students_sql[iter][0] = students_sql_abs[iter]
iter += 1
iter1 = 0
for _ in students_sql:
students_sql[iter1][1] = students_sql_rel[iter1]
iter1 += 1
return students_sql
$$$$
• Getting close to perfect there... Quick tips; for _ in <iterable> could do without iter = 0 stuff if ya use enumerate(), eg. for i, thing in enumerate(<iterable>)... though why you're pre-building an empty nested list is a little beyond me, I'd use a list of tuples or a dictionary... Looks like the find_mentors_sql and find_students_sql could be merged into one function, and I think walk_path = os.path.join(p1, p2) you'll find more portable than walking p1 + '\\p2'... finally for this comment, those one-liner for loops with extend hurt the code's readability. – S0AndS0 Apr 28 at 18:22
• @S0AndS0 Comments are for seeking clarification to the question, and may be deleted. Please put all suggestions for improvements in answers. – 200_success Apr 28 at 20:17
• Wups @200_success, I'll attempt to do better in the future, for now my attempt at an answer will go into trying to address "Is there a more elegant implementation?", because I spent a little more time with the code. – S0AndS0 Apr 28 at 21:35
1. It's recomended to use enumerate rather than iter, _ and indexing.
for iter, abs_path in enemerate(mentors_sql_abs):
mentors_sql[iter][0] = abs_path
2. It's better to use zip rather than build mentors_sql manually.
3. Your function could futher be simplified if you don't extend mentors_sql_* and just yield the values.
4. Please only use one string delimiter, either ' or ".
5. x is a pretty bad variable name, I would use file. Even for a comprehension it's pretty poor, as x isn't short hand for anything.
6. The only difference between the two functions as you walk different paths. And so you can change your input to account for this, and use one function.
7. I don't see the need for returning both relative and absolute paths, and so won't comment on it too much. You may want to return one and convert when needed.
def find_sql(path):
for dirpath, subdirs, files in walk(path):
for file in files:
if file.endswith('.sql'):
yield (
path.join(dirpath, file),
path.join(path.basename(dirpath), file)
)
mentors = find_sql(config["MAIN_DIR"] + '\\Homework')
students = find_sql(config["MAIN_DIR"] + '\\Students Homework')
• iter is also a built-in being shadowed here (although you don't use it in the end). – Graipher Apr 29 at 11:53
• @Graipher file is also a builtin, I feel like shadowing bulitins is fine if it's a small function, as noone would be confused what file means above and noone would try to use file in the above function. It's a good rule to dogmatically follow to stop the two mentioned problems otherwise – Peilonrayz Apr 29 at 11:58
• I agree in general. But when actually iterating over something, calling the iterating variable iter can be quite confusing IMO. – Graipher Apr 29 at 12:00
Okay I spent just a little while with your code @Valentyn, and I think it is mostly unharmed...
### sql_utils/__init__.py
import os
def walked_sql_paths(config, sub_dir):
"""
Generates tuple of absolute and relative file paths
- config should contain a 'MAIN_DIR' key with a value similar to
- /home/Mentor
- /home/StudentName
- sub_dir should contain a string such as Homework
"""
## ... I am guessing that ya might be doing something
## else with the config object, if this is not
## the case, then this could be simplified to
## only taking a path argument instead.
target_dir = os.path.join(config['MAIN_DIR'], sub_dir)
for dirpath, subdirs, files in walk(target_dir):
for item in files:
if not item.endswith('.sql'):
continue
sql_abs = os.path.join(dirpath, item)
sql_rel = os.path.basename(dirpath)
yield sql_abs, sql_rel
That stuff between """ (triple quotes) be a "docstring", and is accessable via either help(walked_sql_paths) or print(walked_sql_paths.__doc__). The "dunder" or "Magic Method" stuff I'll not cover here as that's a whole 'nother-can-o-worms. What is important is that accessible documentation is something that Python allows for, while code that doesn't require it is something to strive for.
I'm using yield in the above for loop so that it yields partial results to whatever calls next() or __next__() methods (called by for loops and other processes implicitly), generators are a cheep way of optimizing code as well as ensuring that users experience less herky jerky loading between results; even if things are taking awhile this'll usually feel faster in other-words.
The assignments of sql_abs and sql_rel are first for readability, and second for making it easy to later do something like yield sql_rel, sql_abs instead. Otherwise there's little reason to prefer it over the answer posted by @Peilonrayz.
Here's one way of using the above modified code...
from sql_utils import walked_sql_paths
## ... setting of mentors and students _config objects
## and other stuff I am guessing will go here...
students_paths = walked_sql_paths(config = students_config,
sub_dir = 'Students Homework')
mentors_paths = walked_sql_paths(config = mentors_config,
sub_dir = 'Homework')
for s_paths, m_paths in zip(students_paths, mentors_paths):
if not s_paths[0] == m_paths[0]:
print("Warning, continuing past -> {s_rel_path} and {m_rel_path} miss-match!".format(
s_rel_path = s_path[0],
m_rel_path = m_path[0]))
continue
print("Time to compare -> Mentors {m_abs_path} with Students {s_abs_path}".format(
m_abs_path = m_paths[1],
s_abs_path = s_paths[1]))
I'm using zip to zip-up the two generators in the above for loop because it's a built-in that seems to do what ya want.
Hopefully none of this is mind blowing because like I stated in your question's comments @Valentyn, you where really close to something that I'd not be able to add to.
Looking at the folder structure a bit closer, it looks like things'll could get just a bit fancier with the loops. What's your preference on ordering?
My thoughts would be to iterate over Students_Homework/ of students and then zip-up between sub-folders, in which case it maybe possible to cache the Mentor's folders on the first pass. However, that would not be nice to scale or if there's lots of sub-directories... Another thought would be to iterate over the Mentor's 1-n folders and zip-up on each student in turn. Feel free to comment with a preference as to which might be more helpful.
Thoughts on the future, using try/except you can code for cases where, Student3 didn't turn in the 5.sql file espected in 2's folder, so here's some skeleton-code that'll hopefully get ya a little closer to fault tolerantness...
def safety_zipper(*iters, search_strs):
"""
Allows for doing something _clever_ where things could have gone painfully wrong
- iters, list of iterables that each output tuples of length two (rel_path, abs_path)
- search_strs, list of strings to search for matches on rel_path
Yields list of tuples (rel_path, abs_path)`
"""
for search_str in search_strs:
partial_results = []
for thing in iters:
try:
path_tuple = thing.next()
except (GeneratorExit, StopIteration):
## Note passing can be dangerous, I only do it
## because the parent loop will exit, eventually
print("Warning {abs_path} miss-match with {search_str}".format(
abs_path = path_tuple[1],
search_str = search_str))
pass
else: ## No error so do things with next thing
## Uncomment the following if useful
# abs_path = path_tuple[1]
rel_path = path_tuple[0]
if search_str == rel_path:
partial_results.append(path_tuple)
continue
## Deal with miss-matches in a clever way here, such
## as if a student is late to turn in an assignment.
finally:
## Finally runs regardless, well so long as another
## exception is not raised before reaching here.
## Only included for completeness and in-case ya
## wanted to do something fancy here too.
pass
yield partial_results
... I'll warn that the above is not complete, but essentially it'll allow for catching cases where Student directories or files do not match those of the Mentor's file paths. It may have to be stacked to be able to check for differences in both directories and files, and pre-loading search_strs list would either require foreknowledge or pre-parsing a chunk of Mentor's file paths to populate.
But whatever's downstream will have a much cleaner input and require much less edge-case detection. | 2019-08-22 14:45:02 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 1, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4394611418247223, "perplexity": 7543.228207742151}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-35/segments/1566027317130.77/warc/CC-MAIN-20190822130553-20190822152553-00039.warc.gz"} |
https://academ.us/list/hep-lat/ | ### Finite-volume effects in long-distance processes with massless leptonic propagators
In Ref. [1], a method was proposed to calculate QED corrections to hadronic self energies from lattice QCD without power-law finite-volume errors. In this paper, we extend the method to processes which occur at second-order in the weak interaction and in which there is a massless (or almost massless) leptonic propagator. We demonstrate that, in spite of the presence of the propagator of an almost massless electron, such an infinite-volume reconstruction procedure can be used to obtain the amplitude for the rare kaon decay $K^+\to\pi^+\nu\bar\nu$ from a lattice quantum chromodynamics computation with only exponentially small finite-volume corrections.
### Fate of Lattice Gauge Theories Under Decoherence
A major test of the capabilities of modern quantum simulators and NISQ devices is the reliable realization of gauge theories, which constitute a gold standard of implementational efficacy. In addition to unavoidable unitary errors, realistic experiments suffer from decoherence, which compromises gauge invariance and, therefore, the gauge theory itself. Here, we study the effect of decoherence on the quench dynamics of a lattice gauge theory. Rigorously identifying the gauge violation as a divergence measure in the gauge sectors, we find at short times that it first grows diffusively $\sim\gamma t$ due to decoherence at environment-coupling strength $\gamma$, before unitary errors at strength $\lambda$ dominate and the violation grows ballistically $\sim\lambda^2t^2$. We further introduce multiple quantum coherences in the context of gauge theories to quantify decoherence effects. Both experimentally accessible measures will be of independent interest beyond the immediate context of this work.
### Light-Front Field Theory on Current Quantum Computers
We present a quantum algorithm for simulation of quantum field theory in the light-front formulation and demonstrate how existing quantum devices can be used to study the structure of bound states in relativistic nuclear physics. Specifically, we apply the Variational Quantum Eigensolver algorithm to find the ground state of the light-front Hamiltonian obtained within the Basis Light-Front Quantization framework. As a demonstration, we calculate the mass, mass radius, decay constant, electromagnetic form factor, and charge radius of the pion on the IBMQ Vigo chip. We consider two implementations based on different encodings of physical states, and propose a development that may lead to quantum advantage. This is the first time that the light-front approach to quantum field theory has been used to enable simulation of a real physical system on a quantum computer.
### Deciphering the mechanism of near-threshold $J/ψ$ photoproduction
The photoproduction of the $J/\psi$ off the proton is believed to deepen our understanding of various physics issues. On the one hand, it is proposed to provide access to the origin of the proton mass, based on the QCD multipole expansion. On the other hand, it can be employed in a study of pentaquark states. The process is usually assumed to proceed through vector-meson dominance, that is the photon couples to a $J/\psi$ which rescatters with the proton to give the $J/\psi p$ final state. In this Letter, we provide a compelling hint for and propose measurements necessary to confirm a novel production mechanism via the $\Lambda_c \bar D^{(*)}$ intermediate states. In particular, there must be cusp structures at the $\Lambda_c \bar D^{(*)}$ thresholds in the energy dependence of the $J/\psi$ photoproduction cross section. The same mechanism also implies the $J/\psi$-nucleon scattering lengths of order 1 mfm. Given this, one expects only a minor contribution of charm quarks to the nucleon mass. | 2020-09-21 04:03:09 | {"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": 0.714942455291748, "perplexity": 723.4729343728465}, "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/1600400198887.3/warc/CC-MAIN-20200921014923-20200921044923-00155.warc.gz"} |
https://docs.opencv.org/3.4/db/d8e/tutorial_threshold.html | OpenCV 3.4.13-dev Open Source Computer Vision
Basic Thresholding Operations
Prev Tutorial: Image Pyramids
Next Tutorial: Thresholding Operations using inRange
## Goal
In this tutorial you will learn how to:
• Perform basic thresholding operations using OpenCV function cv::threshold
## Cool Theory
Note
The explanation below belongs to the book Learning OpenCV by Bradski and Kaehler. What is
## Thresholding?
• The simplest segmentation method
• Application example: Separate out regions of an image corresponding to objects which we want to analyze. This separation is based on the variation of intensity between the object pixels and the background pixels.
• To differentiate the pixels we are interested in from the rest (which will eventually be rejected), we perform a comparison of each pixel intensity value with respect to a threshold (determined according to the problem to solve).
• Once we have separated properly the important pixels, we can set them with a determined value to identify them (i.e. we can assign them a value of $$0$$ (black), $$255$$ (white) or any value that suits your needs).
### Types of Thresholding
• OpenCV offers the function cv::threshold to perform thresholding operations.
• We can effectuate $$5$$ types of Thresholding operations with this function. We will explain them in the following subsections.
• To illustrate how these thresholding processes work, let's consider that we have a source image with pixels with intensity values $$src(x,y)$$. The plot below depicts this. The horizontal blue line represents the threshold $$thresh$$ (fixed).
#### Threshold Binary
• This thresholding operation can be expressed as:
$\texttt{dst} (x,y) = \fork{\texttt{maxVal}}{if $$\texttt{src}(x,y) > \texttt{thresh}$$}{0}{otherwise}$
• So, if the intensity of the pixel $$src(x,y)$$ is higher than $$thresh$$, then the new pixel intensity is set to a $$MaxVal$$. Otherwise, the pixels are set to $$0$$.
#### Threshold Binary, Inverted
• This thresholding operation can be expressed as:
$\texttt{dst} (x,y) = \fork{0}{if $$\texttt{src}(x,y) > \texttt{thresh}$$}{\texttt{maxVal}}{otherwise}$
• If the intensity of the pixel $$src(x,y)$$ is higher than $$thresh$$, then the new pixel intensity is set to a $$0$$. Otherwise, it is set to $$MaxVal$$.
#### Truncate
• This thresholding operation can be expressed as:
$\texttt{dst} (x,y) = \fork{\texttt{threshold}}{if $$\texttt{src}(x,y) > \texttt{thresh}$$}{\texttt{src}(x,y)}{otherwise}$
• The maximum intensity value for the pixels is $$thresh$$, if $$src(x,y)$$ is greater, then its value is truncated. See figure below:
#### Threshold to Zero
• This operation can be expressed as:
$\texttt{dst} (x,y) = \fork{\texttt{src}(x,y)}{if $$\texttt{src}(x,y) > \texttt{thresh}$$}{0}{otherwise}$
• If $$src(x,y)$$ is lower than $$thresh$$, the new pixel value will be set to $$0$$.
#### Threshold to Zero, Inverted
• This operation can be expressed as:
$\texttt{dst} (x,y) = \fork{0}{if $$\texttt{src}(x,y) > \texttt{thresh}$$}{\texttt{src}(x,y)}{otherwise}$
• If $$src(x,y)$$ is greater than $$thresh$$, the new pixel value will be set to $$0$$.
## Explanation
Let's check the general structure of the program:
• Load an image. If it is BGR we convert it to Grayscale. For this, remember that we can use the function cv::cvtColor :
• Create a window to display the result
• Create $$2$$ trackbars for the user to enter user input:
• Type of thresholding: Binary, To Zero, etc...
• Threshold value
• Wait until the user enters the threshold value, the type of thresholding (or until the program exits)
• Whenever the user changes the value of any of the Trackbars, the function Threshold_Demo (update in Java) is called:
As you can see, the function cv::threshold is invoked. We give $$5$$ parameters in C++ code:
• src_gray: Our input image
• dst: Destination (output) image
• threshold_value: The $$thresh$$ value with respect to which the thresholding operation is made
• max_BINARY_value: The value used with the Binary thresholding operations (to set the chosen pixels)
• threshold_type: One of the $$5$$ thresholding operations. They are listed in the comment section of the function above.
## Results
1. After compiling this program, run it giving a path to an image as argument. For instance, for an input image as:
2. First, we try to threshold our image with a binary threshold inverted. We expect that the pixels brighter than the $$thresh$$ will turn dark, which is what actually happens, as we can see in the snapshot below (notice from the original image, that the doggie's tongue and eyes are particularly bright in comparison with the image, this is reflected in the output image).
3. Now we try with the threshold to zero. With this, we expect that the darkest pixels (below the threshold) will become completely black, whereas the pixels with value greater than the threshold will keep its original value. This is verified by the following snapshot of the output image: | 2021-02-26 02:05:57 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5839856266975403, "perplexity": 1459.7678586091706}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-10/segments/1614178355944.41/warc/CC-MAIN-20210226001221-20210226031221-00399.warc.gz"} |
http://www.ms.uky.edu/~pkoester/teaching/Fall13/MA162/Geogebra/SystemsNoAndInf.html | # MA 162: Finite Mathematics
## Systems of Linear Equations: No solutions and infinitely many solutions
### Example 1: A system with no solutions
Consider the system of equations $\begin{cases} \begin{array}{ccccc} 3x & - & 2y & = & 3 \\ 6x & + & k\cdot y & = & 4 \\ \end{array} \end{cases}$ Find the value of $$k$$ so that the system has no solutions.
Solution:
Lets try to do Gauss-Jordan elimination. First, do $$R_2 \mapsto R_2 - 2\cdot R_1$$. $\begin{cases} \begin{array}{ccccc} 3x & - & 2y & = & 3 \\ & & (k + 4)y & = & -2 \\ \end{array} \end{cases}$ What happens if $$k = -4$$? We would get the system $\begin{cases} \begin{array}{ccccc} 3x & - & 2y & = & 3 \\ & & 0 & = & -2 \\ \end{array} \end{cases}$ The bottom row says $$0 = -2$$, which is nonsense. This means the system has NO SOLUTION.
Now lets look at this problem graphically. Use the slider to change the value of k.
This is a Java Applet created using GeoGebra from www.geogebra.org - it looks like you don't have Java installed, please go to www.java.com
What happens when $$k = -4$$? The two lines become parallel.
Keep in mind that
• finding the solution to a system of linear equations
• finding the point of intersection of a family of lines
are two ways of approaching the same problem. Algebraically, our system has no solutions when $$k = -4$$. Geometrically, the lines are parallel when $$k = -4$$, so the lines don't interesect.
### Example 2: A system with infinitely many solutions
Consider the system of equations $\begin{cases} \begin{array}{ccccc} 2x & - & y & = & 2 \\ 5x & + & k\cdot y & = & 5 \\ \end{array} \end{cases}$ Find the value of $$k$$ so that the system has infinitely many solutions.
Solution:
We treat the "infinitely many solutions" case in much the same way as the "no solutions" case.
This time, lets begin with the geometry. Move the slider around to try to find the value of $$k$$ which makes the two lines parallel.
This is a Java Applet created using GeoGebra from www.geogebra.org - it looks like you don't have Java installed, please go to www.java.com
When $$k = -2.5$$, the lines are sort of parallel, in the sense that the have the same slope. But not only do they have the same slope, they are actually the same line, and so the two lines intersect in infinitely many points.
Lets try to do this with Gauss-Jordan elimination. First, do $$R_2 \mapsto R_2 - 2.5\cdot R_1$$. $\begin{cases} \begin{array}{ccccc} 2x & - & y & = & 2 \\ & & (k + 2.5)\cdot y & = & 0 \\ \end{array} \end{cases}$ What happens if $$k = -2.5$$? We get the system $\begin{cases} \begin{array}{ccccc} 2x & - & y & = & 2 \\ & & 0 & = & 0 \\ \end{array} \end{cases}$ The bottom row says $$0 = 0$$, a true, but not particularly enlightening statement. Thus, the "system of equations" is really just a single equation, $2x - y = 2$ which has infinitely many solutions.
### No Solution versus Infinitely Many Solutions: What's the difference?
We apply the same approach to both questions
• Find $$k$$ so that the system has no solution
• Find $$k$$ so that the system has infinitely many solutions
The approach is to find the value of $$k$$ which makes the two lines parallel.
When does this result in infinitely many solutions? No solutions?
• No solution: After applying row operations to make the lines parallel, one of the resulting equations will state that zero is equal to a nonzero number. This is a contradiction, and so the system has no solutions.
• Infinitely many solutions: After applying row operations to make the lines parallel, one of the resulting equations will become $$0 = 0$$.
### How to make the lines parallel?
Two lines are parallel if they have the same slope. However, our lines are usually written in General Form rather than point-slope or slope-intercept form. We could try to compute the slopes and compare, but its better if we check if they are parallel directly from the General Form.
The two lines are parallel if the ratio of the $$x$$ coefficients is equal to the ratio of the $$y$$ coefficients.
In the first example, we need $\frac{3}{6} = \frac{-2}{k},$ which gives us the $$k = -4$$ we found several times already.
In the second example, we need $\frac{2}{5} = \frac{-1}{k},$ which gives us the $$k = -2.5$$ we found several times already.
Finally, we can also determine if there are no solutions or infinitely many solutions by looking at these ratios.
In the "infinitely many solutions" case, the "$$x$$ to $$x$$", "$$y$$ to $$y$$", and "right side to right side" ratios are all equal when $$k = -2.5$$. $\frac{2}{5} = \frac{-1}{2.5} = \frac{2}{5}$ In the "no solutions" case, the "$$x$$ to $$x$$" and "$$y$$ to $$y$$" ratios are equal when $$k = -4$$, but the "right side to right side" ratio is different. $\frac{3}{6} = \frac{-2}{4} \neq \frac{3}{4}$
Bye!
Paul Koester, 1 September 2013, Created with GeoGebra | 2018-12-14 15:38:19 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7642558813095093, "perplexity": 263.018111024476}, "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-2018-51/segments/1544376825916.52/warc/CC-MAIN-20181214140721-20181214162221-00315.warc.gz"} |
https://herwig.hepforge.org/plots/herwig7.0/Rivet-BFactory/CLEO_2004_S5809304/index.html | herwig is hosted by Hepforge, IPPP Durham
### CLEO_2004_S5809304
Back to index
Analysis of charm quark fragmentation at 10.5 GeV, based on a data sample of 103 fb collected by the CLEO experiment. Fragmentation into charm is studied for the charmed hadron ground states, namely $D^0$, $D^+$, as well as the excited states $D^{*0}$ and $D^{*+}$. This analysis can be used to constrain charm fragmentation in Monte Carlo generators.
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Generated at Saturday, 05. December 2015 12:05PM | 2018-12-19 05:54:58 | {"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": 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": 0.7877390384674072, "perplexity": 3515.797869695583}, "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-2018-51/segments/1544376831334.97/warc/CC-MAIN-20181219045716-20181219071716-00415.warc.gz"} |
http://mathhelpforum.com/geometry/181546-unsure-how-entitle-problem.html | # Math Help - Unsure of how to entitle this problem.
1. ## Unsure of how to entitle this problem.
Okay, I have been ruminating over this problem for quite a while now and, to be frank, I am considerably exasperated. (Okay maybe that is a slight exaggeration). But I have decided to stop scratching my head in bewilderment and seek some help. Okay the problem goes as follows:
"Two beams, AB and CD, are parallel. Find the distance AE"
Problem2.png picture by Bashyboy - Photobucket
I couldn't upload the picture, it wouldn't let me, so hopefully this will link will work. I have not undergone any formal studying in the field of Geometry, so--if it's not too much to ask--could someone please give me detailed steps into this that would be greatly appreciated.
2. Originally Posted by Bashyboy
Okay, I have been ruminating over this problem for quite a while now and, to be frank, I am considerably exasperated. (Okay maybe that is a slight exaggeration). But I have decided to stop scratching my head in bewilderment and seek some help. Okay the problem goes as follows:
"Two beams, AB and CD, are parallel. Find the distance AE"
Problem2.png picture by Bashyboy - Photobucket
I couldn't upload the picture, it wouldn't let me, so hopefully this will link will work. I have not undergone any formal studying in the field of Geometry, so--if it's not too much to ask--could someone please give me detailed steps into this that would be greatly appreciated.
Since the two lines are parallel using alternate interior angles we know that
$
It is the same for the opposite angles as well.
This states that the two triangles are similar. Use this to find the length of the Segment AE.
After this form the right triangle DAB and use the Pythagorean theorem to find the distance you want.
3. Thanks, I appreciate it.
4. Originally Posted by TheEmptySet
... $...
Note that
[TEX]\angle EAB=\angle EDC[/TEX] yields
$\angle EAB=\angle EDC.$
5. Awesome thanks! I will add that to my LaTex vocabulary!
6. You're welcome! | 2015-09-01 16:15:32 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 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": 0.610920250415802, "perplexity": 1209.707406764043}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-35/segments/1440645195983.63/warc/CC-MAIN-20150827031315-00073-ip-10-171-96-226.ec2.internal.warc.gz"} |
http://mathhelpforum.com/differential-equations/148936-homogeneous-helmholtz-equation-variable-coefficient-print.html | # Homogeneous Helmholtz Equation with Variable Coefficient
• Jun 19th 2010, 08:56 PM
KrayzBlu
Homogeneous Helmholtz Equation with Variable Coefficient
Hello,
How does one go about solving a two dimensional (or more) homogeneous helmholtz equation with a variable coefficient, i.e.
$\Delta$u(x,y) + u(x,y)*f(x,y) = 0
Where in the standard Helmholtz equation, f(x,y) = k (constant). Knowing some boundary conditions. I am at a loss as to what method to even use, having tried separation of variables, green's functions, method of characteristics. Any hints? Can this equation even be solved?
Thank You
(Bow)
• Jun 20th 2010, 04:30 AM
Jester
Is $f(x,y)$ arbitrary or does it have a specific form?
• Jun 20th 2010, 07:03 AM
KrayzBlu
Quote:
Originally Posted by Danny
Is $f(x,y)$ arbitrary or does it have a specific form?
Nope, it's arbitrary. That's a good sign... right? (Worried)
• Jun 21st 2010, 06:31 PM
KrayzBlu
It is possible that f(x,y) could be a number of step functions. Would that improve the situation? | 2016-10-23 03:28:11 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 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": 0.7695873379707336, "perplexity": 1738.6680037957328}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-44/segments/1476988719139.8/warc/CC-MAIN-20161020183839-00389-ip-10-171-6-4.ec2.internal.warc.gz"} |
https://mersenneforum.org/showthread.php?s=a826d31d7366fcb2ad413b182706e823&p=487515 | mersenneforum.org Linear algebra reservations, progress and results
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2018-05-11, 13:58 #2718 RichD Sep 2008 Kansas 23·431 Posts C191_746xx861_5 factored 251M total relations - 201 unique relations built a 7.4M matrix using TD=116. (TD=120 failed) Solve time about 36 hours. (-t 4) Code: p93 factor: 393858525128757815538591809332015259264810351413046063144961712960625563030820503924332842581 p98 factor: 49276233074393387466576170008327355297051227977927203693263831177987414563635748775025810155092421 https://pastebin.com/UU8b1UiR
2018-05-11, 14:12 #2719
richs
"Rich"
Aug 2002
Benicia, California
2×683 Posts
C211_112xx307_13 factored
Quote:
Originally Posted by richs Also reserving C211_112xx307_13
Code:
p64 factor: 7121090584285701762681259590490422834872870292039022614401589947
p148 factor: 1036944778554394628912911117757565789077324588844522485757164680235871984640571510031011869699850898352425276893985078021351204153068179843456224077
Approximately 71.5 hours on 2 threads Core i3-2310M with 4 GB memory for a 6.71M matrix at TD = 70 (didn't bother to try any higher densities since my laptop takes a long time to read relations and filter).
Log attached and log at https://pastebin.com/jfrzrNmp
Factors reported to factor database.
Attached Files
msieve.log (29.4 KB, 91 views)
2018-05-11, 16:21 #2720 RichD Sep 2008 Kansas 65708 Posts Taking C171_10810256869463_17. I will do a little more sieving locally after the download completes late tonight.
2018-05-12, 07:21 #2721 debrouxl Sep 2009 11110100102 Posts C220_151_35 has two nice large factors. Code: p102 factor: 892309529766352697700913407860375037940463053764841940873698268273937914556722541214496381319637562663 p118 factor: 4582785677266556526555427382379515938986723692586304520351604823442894406310291650794170759738091348011139188809700717 https://pastebin.com/cxpds90w
2018-05-12, 09:06 #2722
Jarod
AKA Speedy51
Oct 2012
New Zealand
22×3×19 Posts
Update on C210_14073559_31
Quote:
Originally Posted by Speedy51 Thanks for adding more relations. I will wait for them to drop to 0 again. At a true density of 70 it's going to take around 87 hours. I am hoping if I re-download it will take a shorter amount of time. Out of interest when you tried it how long would it take to run on your machine? I am running on 8 cores
I have re-downloaded the file and still getting probably cannot build matrix at TD 120 looks like I will have to run at TD of 70 which will take around 87 hours as mentioned above. I have attached log for anybody of interest. It takes over an hour before I discover it cannot build matrix. I might try with a TD of 100 tomorrow before I run it at 70. If anybody can recommend another TD please go ahead
Attached Files
C210_14073559_31.7z (7.1 KB, 91 views)
2018-05-12, 14:23 #2723 VBCurtis "Curtis" Feb 2005 Riverside, CA 33×5×37 Posts Speedy- I typically reduce density by 4 each filtering pass until it works. If you're impatient, go by 10s: 110, then 100. There's nothing magical about 120 as a standard; it just-so happens that most jobs work there.
2018-05-12, 15:24 #2724 swellman Jun 2012 3,203 Posts 13*2^850-1 factored C192_2311333_37 factored I’m still running C183_424xx571 - it will be an 8 week job when finished! Last fiddled with by swellman on 2018-05-12 at 15:27
2018-05-12, 15:32 #2725 RichD Sep 2008 Kansas 23×431 Posts C170_169xx751_17 factored 245M total relations - 193 unique relations built a 9.2M matrix using TD=124. Solve time about 57.5 hours. Code: p78 factor: 225654204722140762660971932621437427216197964590792950041303250743619903413057 p92 factor: 73765067630201435389782015579409495040943893823417759478668924666071394128596338675544340549 https://pastebin.com/jHw0vbhH
2018-05-13, 02:09 #2726
Jarod
AKA Speedy51
Oct 2012
New Zealand
22810 Posts
Quote:
Originally Posted by VBCurtis Speedy- I typically reduce density by 4 each filtering pass until it works. If you're impatient, go by 10s: 110, then 100. There's nothing magical about 120 as a standard; it just-so happens that most jobs work there.
I got it to work with the TD of 100 is going to take about 50 hours instead of 87 odd. I will probably have the results for you in a couple of weeks as I only run for about 12 hours a day. Thank you for the assistance
2018-05-13, 09:43 #2727 VictordeHolland "Victor de Hollander" Aug 2011 the Netherlands 23×3×72 Posts Reserving C215_518xx309_5
2018-05-13, 23:10 #2728 Jarod AKA Speedy51 Oct 2012 New Zealand 111001002 Posts I would be keen on taking C169_116xx849_11 If no one has reserved and nobody else is interested in it? .richs I completely understand if you would like this for your laptop It is a small 29 bit number. I would run this after C210_14073559_31 is complete. The LA phase has another 37 1/2 hours to go
Similar Threads Thread Thread Starter Forum Replies Last Post cubaq YAFU 2 2017-04-02 11:35 CRGreathouse Msieve 8 2009-08-05 07:25 10metreh Msieve 3 2009-02-02 08:34 Damian Math 8 2007-02-12 22:25 R1zZ1 Factoring 2 2007-02-02 06:45
All times are UTC. The time now is 06:26.
Wed Oct 20 06:26:54 UTC 2021 up 89 days, 55 mins, 0 users, load averages: 1.11, 1.08, 1.02 | 2021-10-20 06:26:54 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.17010323703289032, "perplexity": 11540.796377965786}, "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/1634323585302.91/warc/CC-MAIN-20211020055136-20211020085136-00263.warc.gz"} |
http://mathhelpforum.com/new-users/210345-how-solve-power-log-base-sums-like-please-anyone-help-me.html | # Math Help - How to solve power log to base Sums Like this please anyone help me
1. ## How to solve power log to base Sums Like this please anyone help me
I'm having a problem to solve this sum.The sum is below please anyone can help to solve it step by step follow this thread
• Q: Evaluvate:
1. 1/3
2. 1/27
3. 1/3
4. 1/9
5. Non of above
2. ## Re: How to solve power log to base Sums Like this please anyone help me
note ...
$b^{\log_b{x}} = x$
please post algebra problems in the algebra forum.
3. ## Re: How to solve power log to base Sums Like this please anyone help me
Thank you very much for replying.I really sorry i am new to the forum.In the next time i will publish it in suitable forum.Once again Thank you!
4. ## Re: How to solve power log to base Sums Like this please anyone help me
This is also not a sum...
5. ## Re: How to solve power log to base Sums Like this please anyone help me
Originally Posted by Prove It
This is also not a sum...
In England it is quite common for any arithmetic or, more generally, math problem, to be called a "sum". | 2015-07-06 10:22:03 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 1, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5266140699386597, "perplexity": 1478.2434677156766}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-27/segments/1435375098072.11/warc/CC-MAIN-20150627031818-00209-ip-10-179-60-89.ec2.internal.warc.gz"} |
https://physics.stackexchange.com/questions/394278/interpretation-of-correlation-functions-that-are-higher-than-2-point | # Interpretation of correlation functions that are higher than 2 point
I am studying QFT from Peskin & Schroeder. There I found a physical interpretation of a 2-point correlation function. According to Peskin & Schroeder, the 2-point correlation function is nothing but a propagator between two points and I am happy with that. But the question that arises immediately is, what are the n-point correlation functions?
e.g.- does the 3-point correlation function given by, $\langle\omega|T[\phi(x)\phi(y)\phi(z)]|\omega\rangle$ imply the amplitude of propagation from $x$ to $z$ via $y$?
Is this understanding correct?
In a typical QFT, Wicks theorem tells you that the expectation value of all higher point correlation functions can be expressed as two point correlation functions. Essentially, a $2n$-point correlation function tells you about $n$ particles propagating from point $x_i$ to $x_j$, where $i,j = 1...n$.
For example, a 4 points look like, $$\langle\Omega|T\left\{\phi(x_1)\phi(x_2)\phi(x_3)\phi(x_4)\right\}|\Omega\rangle = \langle\Omega|T\left\{\phi(x_1)\phi(x_2)\right\}|\Omega\rangle\langle\Omega|T\left\{\phi(x_3)\phi(x_4)\right\}|\Omega\rangle + \langle\Omega|T\left\{\phi(x_1)\phi(x_3)\right\}|\Omega\rangle\langle\Omega|T\left\{\phi(x_2)\phi(x_4)\right\}|\Omega\rangle + \langle\Omega|T\left\{\phi(x_1)\phi(x_4)\right\}|\Omega\rangle\langle\Omega|T\left\{\phi(x_2)\phi(x_3)\right\}|\Omega\rangle$$
Which corresponds to one particle going from $x_1$ to $x_2$ and one going from $x_3$ to $x_4$, plus all the other possibilities. Diagramatically, this looks like | 2021-10-16 18:32:37 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8411268591880798, "perplexity": 89.43982226107401}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-43/segments/1634323584913.24/warc/CC-MAIN-20211016170013-20211016200013-00278.warc.gz"} |
https://ideastatica.si/winners-of-the-idea-statica-excellence-awards-2021/ | # Winners of the IDEA StatiCa Excellence Awards 2021
November 25, 2021
The very first year of the IDEA StatiCa Excellence Awards has finished and we are happy to announce the winning projects. It was a real challenge to choose only a few projects because most submitted projects were on a very high professional level. Let’s take a look closer and introduce winners of the IDEA StatiCa Excellence awards crystal trophies.
$$Finally, since the IDEA StatiCa Excellence Awards started, more than one hundred structural engineers decided to join the contest with their projects. Members of our jury definitely didn’t have an easy task assessing all the projects, but finally, they selected the winners.But before we introduce the winners, let us sum up some interesting details about the IDEA StatiCa Excellence Awards. The user contest started in August and finished on November 24.There were 117 submitted projects in total 87 of them had been approved 17 of them were short-listed by the internal technical jury team7 winners were carefully selected in the two-rounds professional jury votingSo let’s have a look who those winners are:$$
## Winner in the category of Small scale structures
### STEELProject: Steel-to-timber connection, Maryland, USA
Authors: Trevor Chou & Martyn Sheard
Company: Craft Engineering Studio
This new residence is located in Martha’s Vineyard, MA, USA. The site is within a high wind region and the building sits on a steep bluff overlooking the ocean, placing significant challenges on the design of the lateral load resisting system.
The resulting steel node joint is designed to resist forces and moments in the two orthogonal axes. A central hollow section hub was proposed but was found to be unable to resist the high moments from the moment frame girder due to insufficient out-of-plane stiffness in the box section walls as well as being prone to flexural failure. The steel joint was redesigned with a wide flange section to maintain continuity in both axes whilst ensuring a direct transmission of forces and moments to the center of the hub.
## Winners in the category of Large scale buildings
### STEELProject: Steel Dome structure, Amsterdam, Netherlands
Authors: Alexander van Beelen
For the great renovation project EDGE Amsterdam of no less than 60,000 square meters, the spectacular glass dome in a huge steel support structure was designed and installed. The glass dome has a diameter of 76 meters and is the beating heart of this major project. In the original office building from 1970, the interior space was not used, but thanks to the glass roof construction it now forms a lively meeting place with high-quality workplaces and lots of daylight.
### CONCRETEProject: Kralovopolska building, Brno, Czech Republic
Authors: Tomas Krivka
Company: STATIKON Solutions s.r.o.
The building is designed to be a hotel with underground parking and a commercial part on the ground floor. Load-bearing parts of the structure are completely from the concrete. It includes 12 ground floors and 3 underground floors. From a structural point of view, it is a both-direction wall load-bearing system. In the underground part and also in the commercial part the LB system is more skeleton system consisting mainly of columns that locally support the ceiling slab. The underground part of the structure is designed to be waterproof as a “white tank“. The whole structure is supported by long drilled piles in combination with a relatively thick foundation slab which is designed to carry inner forces caused by interaction with subsoil after settling of piles.
## Winners in the category of Infrastructure, industrial & recreational structures
### STEELThe Curragh Racecourse, Kildare, Ireland
Authors: Mirivano Carrig & Stephen McKenna
Company: Kiernan Structural Steel Ltd.
The Curragh is an internationally recognized horse racing venue situated in Ireland which has won numerous awards both in Ireland and indeed internationally. The latest awards with which this grandstand has won include the Tekla Global BIM Awards as well as the Structural Steel Design Awards to name but a few.
KSSL were involved in the project to design the steel connections and erect various ancillary buildings including the new center-piece spectator grandstand which comprises of a cantilevered structure on three elevations of the roof.
### CONCRETEProject: Prestressed concrete bridge, Czech Republic
Author: Jan Lamparsky
Company: AFRY CZ s.r.o.
The bridge is located at the second most important crossroad of the city ring road section. It is a single-span bridge structure with a span of 43.5 m made of post-tensioned prestressed concrete. The width of the bridge varies from 78.3 m in the middle to 89.6 m at the end of the bridge. The total area of the bridge is 3800 m2.
During the design, it was determined whether this structure can be realized as a frame bridge. This type of construction was very economically inefficient and technically demanding. It was necessary to anchor the substructure, use connectors to bind reinforcement in the corners, prestress the retaining walls and frame corners, etc. Through successive calculations, we came to the final variant of the semi-integral bridge on deck foundations.
## Winner in the category of Prize of the public
### Project: Equipment steel support at switchyard, Stiegler’s Gorge, Tanzania
Author: Yasmin Hamdy
Company: ELSEWEDY ELECTRIC PSP
Julius Nyerere Hydropower station is located on the Rufiji river the largest river in Tanzania. The cost of the project is 3 billion USD. The project is consisting of RCC gravity dam, powerhouse, and switchyard. The works will be completed within a period of 36 months with a maximum mobilization period of 3 months. The powerhouse will be completed in April 2022 and the unit is expected to generate electricity on April 15, 2022. After the project completion, it will provide 61.3 GWH electricity per year greatly promoting the industrialization of Tanzania.
In this huge project, the IDEA StatiCa program was used to make a design for all existing steel connections located at the powerhouse and switchyard area.
## Winner in the category of The best student project
#### Project: Tainter gate for the water supply system, Moquegua, Peru
Author: Fernando Gonzales
University: National University of San Agustin
This tainter gate was made for the water supply system of a mining company in Moquegua- Peru. | 2022-11-30 14:44:45 | {"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": 1, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.17861303687095642, "perplexity": 3913.4690606623976}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446710764.12/warc/CC-MAIN-20221130124353-20221130154353-00362.warc.gz"} |
http://blog.jverkamp.com/ | # Home
Next up on my 2015 Reading List: The Academy by Jack McDevitt.
The Academy (it feels weird to call it that) takes place in a few century distant future where humanity has been to the stars and found them next to empty. There are hints at civilizations that have come before (a monument on Iapetus for example) and others still around, albeit not at our same technological level. There are terrible threats: truly giant clouds of gas that seem to have a thing for destroying cities in several of the novels and a planet about to fall into a gas giant in another.
Been a while since I've actually tackled one of the Daily Programmer challenges, so let's try one out. From a week and a half ago, we are challeneged to make an adjacency matrix generator, turning a graphical representation of a graph into an adjacency matrix.
Input:
a-----b
|\ / \
| \ / \
| / e
| / \ /
|/ \ /
c-----d
Output:
01110
10101
11010
10101
01010
• ## Setting up Postfix and OpenDKIM
Last week, I was presented with a fairly interesting challenge: add DKIM (via OpenDKIM) support to our mail servers (running Postfix). Given that I've never actually worked on a mail server before, it sounded fun.
• ## Laundry Files
Next up on my 2015 Reading List: The Laundry Files by Charles Stross. Even though this wasn't originally on my list, I ended up just reading this rather than reading two series at once. So it goes. | 2015-08-30 22:27:46 | {"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": 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": 0.2637339234352112, "perplexity": 2450.2617026511853}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-35/segments/1440644065375.30/warc/CC-MAIN-20150827025425-00077-ip-10-171-96-226.ec2.internal.warc.gz"} |
http://scottbrenner.com/could-be-the-thyroid-could-be-ennui-either-way-the-drug-isnt-helping/ | Could Be the Thyroid; Could Be Ennui. Either Way, the Drug Isn’t Helping.
What I am reading: David Plunkert “During the past four weeks, have you been tired? Been exhausted? Had difficulty getting motivated to do anything at all?” These questions —…, Could Be the Thyroid; Could Be Ennui. Either Way, the Drug Isn’t Helping., https://www.nytimes.com/2017/04/21/health/could-be-the-thyroid-could-be-ennui-either-way-the-drug-isnt-helping.html, http://www.instapaper.com/rss/238947/i35BSjISqVKnAGF4lqMOZltq2L0, Instapaper: Unread, , | 2018-09-25 13:39:36 | {"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": 0.8758060336112976, "perplexity": 11978.40932245308}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-39/segments/1537267161638.66/warc/CC-MAIN-20180925123211-20180925143611-00336.warc.gz"} |
http://tex.stackexchange.com/questions/84177/how-to-place-a-center-aligned-rule-on-the-left-of-a-section-title/84179 | # How to place a center aligned rule on the left of a section title?
How can I achieve this effect? I have tried the following code:
\parbox[c]{3.5cm}{\color{NavyBlue}{\rule{90px}{7px}}} \parbox[c]{6cm}{\section*{Section Title}}
But the rule is not center aligned with the title because \section command contain some extra space below the text.
And I had tried a alternative:
\parbox[c]{3.5cm}{\color{NavyBlue}{\rule{90px}{7px}}} \parbox[c]{6cm}{\Large{Section Title}}
Although this time the rule is center aligned, but if I add text below the title, the line space would be too small.
Can anyone help me with this?
-
Should the rule hang in the margin? – Gonzalo Medina Nov 25 '12 at 15:29
Seems that there was a related question yesterday How can I make a bold horizontal rule under each section title? Following that example, I seem to be able to create the effect that you desire:
\documentclass{article}
\usepackage[dvipsnames]{xcolor}
\usepackage{titlesec}
\titleformat{\section}
{\color{NavyBlue}\normalfont\Large\bfseries}{\parbox{10em}{\rule{10em}{0.5em}}}{1em}{}[{}]
\begin{document}
\section{Test Section}
Random stuff
\end{document}
I'd never heard of the titlesec package before. Looks like it has some very nice tools for modifying the look and feel of sections.
-
Don't use px for the units. It hasn't a fixed value and changes in the preamble might modify it. And I don't think that the \parbox is really needed (use the optional arguments to \rule). – egreg Nov 25 '12 at 15:22
@egreg. Thanks. I was just following lhhNJU's example as closely as possible. I used the parbox to force vertical centering with the section title. I know you can write \rule[1ex]{10em}{0.5ex} but is there a way to tell rule to center with text on the same line? – A.Ellett Nov 25 '12 at 15:31
Just play with the "raise" argument; possibly something like 0.4ex – egreg Nov 25 '12 at 15:33
I just noticed I used a non-fixed value length again. @egreg. Are you suggesting that I doing sometihng like \rule[3pt]{2in}{3pt}? – A.Ellett Nov 25 '12 at 15:35
ex is good, as it depends on the current font, which is really what you want. – egreg Nov 25 '12 at 15:36
Here's a variation in which the rule hangs in the margin:
\documentclass{article}
\usepackage{titlesec}
\usepackage[dvipsnames]{xcolor}
\usepackage{lipsum}% just to generate text for the example
\titleformat{\section}
{\normalfont\Large\bfseries\color{NavyBlue}}{}{0em}{\llap{\rule[.5ex]{90pt}{4pt}\hspace*{1em}}}
\begin{document}
\section{Section Title}
\lipsum[2]
\end{document}
`
-
Thank you for your help. But what I intend is to put the rule in the text region. – lhhNJU Nov 26 '12 at 1:59
@lhhNJU OK, but you should consider what would happen if a section title spans more than one line. – Gonzalo Medina Nov 26 '12 at 2:01
I see. Thank you for your reminder. – lhhNJU Nov 26 '12 at 2:22 | 2016-02-11 10:52:59 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 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": 0.8950598239898682, "perplexity": 1121.3696653828777}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-07/segments/1454701161942.67/warc/CC-MAIN-20160205193921-00019-ip-10-236-182-209.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/adiabatic-expansion-in-a-thermodynamic-system.402374/ | # Adiabatic expansion in a thermodynamic system
1. May 10, 2010
### cirimus
1. The problem statement, all variables and given/known data
4 liters of an ideal diatomic gas are compressed in a cilinder. In a closed process, the following steps are taken :
1) Initial state :
$$p_1 = 1 atm = 1.013*10^5 N/m^2$$
and :
$$T_1 = 300 K$$
2) Isochoric proces resulting in :
$$p_2 = 3.p_1$$
3) Adiabatic expansion resulting in:
$$p_3 = p_1$$
4) Isobar compression resulting in
$$V_4 = V_1$$
Note : State 4 = State 1
Questions :
a) What is the volume of the gas at the end of the adiabatic proces ? (solution : 8.77*10^-3 )
b) What is the temperature of just before the adiabatic expansion? (solution : 902 K)
c) What is the work performed by the gas in this cycle ? (solution : 335.0 J)
2. Relevant equations
ideal gas :
$$pV = nRT$$
$$\frac{p_xV_x}{T_x} = \frac{p_y V_y} {T_y}$$
$$\Delta U = n Cv \Delta T$$
ideal gas + adiabatic expansion :
$$pV^{\lambda} = c$$
Note : lambda is Youngs module, c is a constant value.
diatomic :
$$C_v = 5/2 R$$
$$\lambda = 1.4$$
3. The attempt at a solution
b) I'm starting with this one since it seemed easier :
$$\frac{p_1 V_1}{T_1} = \frac{p_2 V_2} { T_2 }$$
$$\Rightarrow T_2 = \frac{p_2 V_2 T_1}{p_1 V_1}$$
$$\Rightarrow T_2 = 900$$
My guess is that my teacher approached this through using the
$$pV = nRT$$ equation twice, introducing rounding errors ?
a) I've tried to approach this in several ways, but none seem to give me the correct solution... My current approach :
Since we know it's an ideal gas that undergoes an adiabatic expansion, we could use the formula given above :
$$p_2V_2^\lambda = constant = p_3V_3^\lambda$$
$$\Rightarrow V_3 = \log_\lambda ( \frac{p_2}{p_3} * V_2^\lambda)$$
$$\Rightarrow V_3 = \log_{1.4} ( 3 (4*10^{-3})^{1.4} )$$
$$\Rightarrow V_3 = -19.70870781$$
That is wrong in many ways, but I don't know which assumption I made is wrong ...
c) Don't know yet, I'm guessing something like :
$$\sum W = W_{1,2} + W_{2,3} + W_{3,1}$$
$$\sum W = 0 + W_{2,3} + p_3 (V_1 - V_3)$$
But I'll have to integrate over an unknown volume to get the value of $$W_{2,3}$$ ...
If I know the value of $$V_3$$ I think I can calculate the value of $$T_3$$ using $$p_3V_3=nRT_3$$. Once I know $$T_3$$ I can use $$\Delta U = n Cv \Delta T_{2,3} = - W_{2,3}$$.
Last edited: May 11, 2010
2. May 10, 2010
### Andrew Mason
b) the final temperature has to be 900 K.
a) Use the adiabatic condition $PV^\gamma = K$ to give:
$$\left(\frac{V_f}{V_i}\right)^\gamma = \frac{P_i}{P_f}$$
c) The work done from 1-2 and 3-4 is easy. To determine the work done from 2-3 use the first law: dQ = dU + dW to determine the work done (what is Q for this adiabatic expansion?). Hint: you just have to know the change in temperature - use:
$T_2V_2^{\gamma -1} = T_3V_3^{\gamma -1}$ to find the temperature at 3.)
AM
3. May 11, 2010
### cirimus
I am using that formula for (a) in my attempted solution, but te result is not correct. The values I use are:
$$V_f = unknown$$
$$V_i = 4*10^{-3}$$
$$P_i = 3 * P_f$$
$$P_f = 1.013 * 10^{5}$$
$$\lambda = 1.40$$
Am i using wrong values ?
Attempt 2 :
$$V_f = V_i * \log_{1.4}(p_i / p_f)$$
$$\Rightarrow V_f = 4*10^{-3} * \log_{1.4}(3)$$
$$\Rightarrow V_f = 13 * 10^{-3}$$
Last edited: May 11, 2010
4. May 11, 2010
### Andrew Mason
$$x^\gamma = (e^{\ln{x})^\gamma} = e^{\gamma\ln{x}$$
AM
5. May 11, 2010
### Andrew Mason
As a follow-up to my last post, in case you found it too cryptic: your figures are correct (although you should state the units, particularly in your answer). The problem is with algebra.
AM
6. May 11, 2010
### cirimus
Thank you ! Using this and your previous posts I was able to find the answer for both (a) and (c) now. (how do i mark this thread as solved ?)
Last edited: May 11, 2010
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook | 2017-11-21 05:10:20 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.7571196556091309, "perplexity": 1165.8597150845073}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 20, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-47/segments/1510934806316.80/warc/CC-MAIN-20171121040104-20171121060104-00228.warc.gz"} |
https://socratic.org/questions/can-you-find-the-standard-deviation-of-negative-numbers | # Can you find the standard deviation of negative numbers?
Nov 1, 2015
Yes.
#### Explanation:
Yes. Why not?!!
Consider the data sets $\left\{- 1 , - 2 , - 3 , - 4 , - 5\right\}$. Note that all are negative numbers.
So, variance of the data set will be $\frac{1}{n} {\sum}_{i = 1}^{n} {x}_{i}^{2} - {\left(\frac{1}{n} {\sum}_{i = 1}^{n} {x}_{i}\right)}^{2}$
$= \frac{1}{5} \left[{\left(- 1\right)}^{2} + {\left(- 2\right)}^{2} + {\left(- 3\right)}^{2} + {\left(- 4\right)}^{2} + {\left(- 5\right)}^{2}\right] - {\left(\frac{1}{5} \cdot \left[- 1 - 2 - 3 - 4 - 5\right]\right)}^{2}$
$= 11 - 9$
$= 6$ | 2022-08-09 19:50:15 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 5, "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": 0.8658256530761719, "perplexity": 1484.4700591770672}, "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/1659882571086.77/warc/CC-MAIN-20220809185452-20220809215452-00158.warc.gz"} |
https://cs.stackexchange.com/questions/18873/hashing-a-specific-range-of-a-character-array | # Hashing a Specific Range Of a Character Array [closed]
I need to process queries to Hash various ranges of a character array. I am currently using the Arrays.hashCode from the standard java library. But the problem is that this method is too slow. Also my array remains the same throughout the process of hashing, I only am changing the range. To deal with this, I have to make an entire copy of the array everytime I process a query, and then compute the hash from the above function.
I am using Arrays.copyOfRange to create a copy everytime I process a query. I need to avoid this. So I was thinking of devising a hashing scheme of my own. This scheme should be such that I whould get a unique hash for each array range. Hashes should be same if all characters in the range are same.
Any Help on how to proceed with the making of such a hash function will be appreciated.
• So hash yourself. Is that a problem? – Karolis Juodelė Dec 11 '13 at 12:22
• @KarolisJuodelė had that not been a problem, I wouldn't have posted this. – Alice Dec 11 '13 at 12:41
Use a rolling hash function, such as the Rabin-Karp hash. This allows you to compute hashes of any subrange of $A$ very efficiently, e.g., in $O(1)$ time, assuming you use some extra memory to keep track of all of the intermediate states of the rolling hash.
If $A[0..n-1]$ is your array, a rolling hash defines the hash of the entire array using a recurrence like this:
$$x_i = f(x_{i-1}, A[i])$$
where $x_{-1}$ is some fixed constant and $f$ is some function. Now here's what you can do. You can store the $x$ values in an array $X[0..n-1]$ (where $X[i]=x_i$). This will allow you to compute the hash of any prefix of $A$.
If you want to be able to compute the hash of any consecutive range of $A$, choose the function $f$ carefully to be reversible.
A simple instantiation: let $X[-1] = 0$ (for some constant $c$) and
$$X[i+1] = \alpha \cdot X[i] + A[i] \bmod 2^{32},$$
where $\alpha$ is some 32-bit constant (you'll want it to be odd). Preprocess the array $A$ once to compute the array $X[0..n-1]$.
Now, the hash of $A[i..j-1]$ can be very efficiently computed: it is the following:
$$X[j-1] - \alpha^{j-i} X[i-1] \bmod 2^{32}.$$
Notice that $\alpha^{j-i} \bmod 2^{32}$ can be computed efficiently using fast exponentiation algorithms, i.e., repeated squaring modulo $2^{32}$. Thus, assuming you've stored the array $X$ alongside the array $A$, a hash of any range of $A$ can be done in $O(1)$ time.
If you want something even simpler and easier to implement, set $\alpha$ to 1. Then the hash of $A[i..j-1]$ just becomes $A[i]+ A[i+1]+ \cdots + A[j-1]$, and $X[i]$ stores $A[0] + A[1] + \cdots + A[i]$, so the hash of $A[i..j-1]$ is just $X[j]-X[i]$.
However, the quality of this hash degrades and you might see an increase in hash collisions.
int hash = 0;
for (int i = range_start; i < range_end; i++)
hash = hash*31 + your_array[i];
• Since questions with code are off-topic here at CS:StackExchange, answers with code are unexpected. Even so, the answer should include an explanation of how it solves the problem in the question. – Guy Coder Dec 11 '13 at 14:40
• that would have been a good approach, had I had a small array. But since I was talking of performance improvement, the is very large (of the order of 1e5). So I dont think this approach will be good enough (It is still linear, which was the case before). Thanks anyways. – Alice Dec 11 '13 at 15:53
• Is that a good hash function? Why? What makes it better than any other function you could have chosen? Also, @Alice, I don't think 100,000 elements is particularly large -- it's not even a megabyte of data. – David Richerby Dec 11 '13 at 16:36
• @GuyCoder, I've already voted the question as off topic and now answered it as I would in stack overflow. – Karolis Juodelė Dec 11 '13 at 17:09
• @Alice, from the question gathered that your problem was cloning the required range for every hash. Have you actually tried this and saw no improvement? – Karolis Juodelė Dec 11 '13 at 17:14 | 2019-12-11 12:58:36 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5442790985107422, "perplexity": 563.9852024040725}, "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-2019-51/segments/1575540530857.12/warc/CC-MAIN-20191211103140-20191211131140-00278.warc.gz"} |
http://www.elolivoazul.es/s33slb/7qf7f66.php?eab8e7=matrix-inverse-method-for-solving-a-system-of-equations | # matrix inverse method for solving a system of equations
For example, look at the following system of equations. These calculations leave the inverse matrix where you had the identity originally. We can solve this system of equations using the matrix identity AX = B; if the matrix A has an inverse. … First, we need to calculate ${A}^{-1}$. Once in this form, the possible solutions to a system of linear equations that the augmented matrix represents can be determined by three cases. Save the coefficient matrix and the constant matrix as matrix variables $\left[A\right]$ and $\left[B\right]$. How to Solve a System of Equations Using the Inverse…. To solve a matrix equation, think about the equation A(X)=B. In mathematics, a system of linear equations (or linear system) is a collection of one or more linear equations involving the same set of variables. Now that we know what matrices we need, we can put them all together to create a matrix equation. Especially, when we solve the equations with conventional methods. Solve the system using the inverse of the coefficient matrix. Solving a System of Linear Equations By Using an Inverse Matrix Consider the system of linear equations \begin{align*} x_1&= 2, \\ -2x_1 + x_2 &= 3, \\ 5x_1-4x_2 +x_3 &= 2 \end{align*} (a) Find the coefficient matrix and its inverse matrix. If, on the other hand, the ranks of these two matrices are equal, the system must have at least one solution. X = A⁻¹ B. Essential we know that if we multiply matrix A times matrix X it will equal matrix B. We will investigate this idea in detail, but it is helpful to begin with a $2\times 2$ system and then move on to a $3\times 3$ system. Another way to solve a matrix equation Ax = b is to left multiply both sides by the inverse matrix A-1, if it exists, to get the solution x = A-1 b. Thus. Formula: This is the formula that we are going to use to solve any linear equations. Inconsistent System: A system of equations with no solution is an inconsistent system. Furthermore, IX = X, because multiplying any matrix by an identity matrix of the appropriate size leaves the matrix unaltered. This calculator solves Systems of Linear Equations using Gaussian Elimination Method, Inverse Matrix Method, or Cramer's rule.Also you can compute a number of solutions in a system of linear equations (analyse the compatibility) using Rouché–Capelli theorem.. The only difference between a solving a linear equation and a system of equations written in matrix form is that finding the inverse of a matrix is more complicated, and matrix multiplication is a longer process. You’re left with . Typically, A -1 is calculated as a separate exercize ; otherwise, we must pause here to calculate A -1 . The efficiency of the method is demonstrated through some standard nonlinear differential equations: Duffing equation, Van der … However, the goal is the sameâto isolate the variable. If rref (A) \text{rref}(A) rref (A) is the identity matrix, then the system has a unique solution. The Solution of System of Linear Equations. Multiply the inverse of the coefficient matrix in the front on both sides of the equation. By matrix multiplication, Setting corresponding elements equal gives the system of equations. All we need do is write them in matrix form, calculate the inverse of the matrix of coefficients, and finally perform a matrix multiplication. It also allows us to find the inverse of a matrix. Click here to know the properties of inverse matrices. In this page inverse method 3x3 matrix we are going to see how to solve the given linear equation using inversion method. Consider the matrix equation AX = B , Suppose we have the following system of equations. However, when operating with matrices, we cannot divide. To solve a single linear equation $ax=b$ for $x$, we would simply multiply both sides of the equation by the multiplicative inverse (reciprocal) of $a$. ... Left multiply both sides of the matrix equation by the inverse matrix. Matrix Equations to solve a 3x3 system of equations Example: Write the matrix equation to represent the system, then use an inverse matrix to solve it. . For example, if 3x = 12, how would you solve the equation? a 11 x 1 + a 12 x 2 + a 13 x 3 = b 1 a 21 x 1 + a 22 x 2 + a 23 x 3 = b 2 a 31 x 1 + a 32 x 2 + a 33 x 3 = b 3 Also, it is a popular method of solving linear simultaneous equations. Show Step-by-step Solutions So it goes with matrices. If you have a coefficient tied to a variable on one side of a matrix equation, you can multiply by the coefficient’s inverse to make that coefficient go away and leave you with just the variable. You now have the following equation: Cancel the matrix on the left and multiply the matrices on the right. However, the goal is the same—to isolate the variable. This technique is also called row reduction and it consists of two stages: Forward elimination and back substitution. After he represented a system of equations with a single matrix equation, Sal solves that matrix equation using the inverse of the coefficient matrix. Inverse matrix method Cramer’s rule Cramer’s Rule and inverse matrix method correlation: Systems of Linear Equations: Solving systems of equations using matrices: A system of linear equations is a set of n equations in n unknowns (variables) of the form Create the inverse of the coefficient matrix out of the matrix equation. If the matrix is a 2-x-2 matrix, then you can use a simple formula to find the inverse. Armed with a system of equations and the knowledge of how to use inverse matrices, you can follow a series of simple steps to arrive at a solution to the system, again using the trusty old matrix. With that said, here’s how you find an inverse of a 2-x-2 matrix: Simply follow this format with any 2-x-2 matrix you’re asked to find. Using the formula to calculate the inverse of a 2 by 2 matrix, we have: Now we are ready to solve. www.mathcentre.ac.uk 1 Note that multiplying the scalar is usually easier after you multiply the two matrices. Namely, we can use matrix algebra to multiply both sides of the equation by A 1, thus getting A 1AX = A B: Since A 1A = I 2 2, we get I 2 2X = A 1B; or X = A 1B: Lets see how this method … 2. If you don’t use a graphing calculator, you can augment your original, invertible matrix with the identity matrix and use elementary row operations to get the identity matrix where your original matrix once was. In the MATRIX INVERSE METHOD (unlike Gauss/Jordan), we solve for the matrix variable X by left-multiplying both sides of the above matrix equation (AX=B) by A-1. The system must have the same number of equations as variables, that is, the coefficient matrix of the system must be square. The inverse matrix can be found for 2× 2, 3× 3, …n × n matrices. We want ${A}^{-1}AX={A}^{-1}B:$. Solving equations with a matrix is a mathematical technique. The only difference between a solving a linear equation and a system of equations written in matrix form is that finding the inverse of a matrix is more complicated, and matrix multiplication is a longer process. Finding the inverse of a 3×3 matrix is a bit more difficult than finding the inverses of a 2 ×2 matrix. Multiply row 1 by $\frac{1}{5}$. An inverse matrix times a matrix cancels out. For instance, you can solve the system that follows by using inverse matrices: When written as a matrix equation, you get. Hence, the inverse matrix is. Just multiply by the inverse of matrix A (if the inverse exists), which you write like this: Now that you’ve simplified the basic equation, you need to calculate the inverse matrix in order to calculate the answer to the problem. Notice in the first step we multiplied both sides of the equation by ${A}^{-1}$, but the ${A}^{-1}$ was to the left of $A$ on the left side and to the left of $B$ on the right side. Using matrix multiplication, we may define a system of equations with the same number of equations as variables as. Solving a system of linear equations using the inverse of a matrix requires the definition of two new matrices: $X$ is the matrix representing the variables of the system, and $B$ is the matrix representing the constants. Hence ad – bc = 22. Let the unknown inverse matrix be. For example, + − = − + = − − + − = is a system of three equations in the three variables x, y, z.A solution to a linear system is an assignment of values to the variables such that all the equations are simultaneously satisfied. 2x - y + 3z = 9. x + y + z = 6. x - y + z = 2. Multiply both sides by the inverse of $A$ to obtain the solution. Example 1: Solve the following linear equation by inversion method . Using Matrices makes life easier because we can use a computer program (such as the Matrix Calculator) to do all the \"number crunching\".But first we need to write the question in Matrix form. Solving the simultaneous equations Given AX = B we can multiply both sides by the inverse of A, provided this exists, to give A−1AX = A−1B But A−1A = I, the identity matrix. Consider the system of linear equations x1=2,−2x1+x2=3,5x1−4x2+x3=2 (a)Find the coefficient matrix and its inverse matrix. $\begin{array}{c}{a}_{1}x+{b}_{1}y={c}_{1}\\ {a}_{2}x+{b}_{2}y={c}_{2}\end{array}$, $A=\left[\begin{array}{cc}{a}_{1}& {b}_{1}\\ {a}_{2}& {b}_{2}\end{array}\right]$, $X=\left[\begin{array}{c}x\\ y\end{array}\right]$, $B=\left[\begin{array}{c}{c}_{1}\\ {c}_{2}\end{array}\right]$, $\left[\begin{array}{cc}{a}_{1}& {b}_{1}\\ {a}_{2}& {b}_{2}\end{array}\right]\text{ }\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}{c}_{1}\\ {c}_{2}\end{array}\right]$, $\begin{array}{c}\text{ }ax=b\\ \text{ }\left(\frac{1}{a}\right)ax=\left(\frac{1}{a}\right)b\\ \left({a}^{-1}\text{ }\right)ax=\left({a}^{-1}\right)b\\ \left[\left({a}^{-1}\right)a\right]x=\left({a}^{-1}\right)b\\ \text{ }1x=\left({a}^{-1}\right)b\\ \text{ }x=\left({a}^{-1}\right)b\end{array}$, $\begin{array}{r}\hfill \left({A}^{-1}\right)AX=\left({A}^{-1}\right)B\\ \hfill \left[\left({A}^{-1}\right)A\right]X=\left({A}^{-1}\right)B\\ \hfill IX=\left({A}^{-1}\right)B\\ \hfill X=\left({A}^{-1}\right)B\end{array}$, $\begin{array}{r}\hfill 3x+8y=5\\ \hfill 4x+11y=7\end{array}$, $A=\left[\begin{array}{cc}3& 8\\ 4& 11\end{array}\right],X=\left[\begin{array}{c}x\\ y\end{array}\right],B=\left[\begin{array}{c}5\\ 7\end{array}\right]$, $\left[\begin{array}{cc}3& 8\\ 4& 11\end{array}\right]\text{ }\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}5\\ 7\end{array}\right]$, $\begin{array}{l}{A}^{-1}=\frac{1}{ad-bc}\left[\begin{array}{cc}d& -b\\ -c& a\end{array}\right]\hfill \\ \text{ }=\frac{1}{3\left(11\right)-8\left(4\right)}\left[\begin{array}{cc}11& -8\\ -4& 3\end{array}\right]\hfill \\ \text{ }=\frac{1}{1}\left[\begin{array}{cc}11& -8\\ -4& 3\end{array}\right]\hfill \end{array}$, ${A}^{-1}=\left[\begin{array}{cc}11& -8\\ -4& \text{ }\text{ }3\end{array}\right]$, $\begin{array}{l}\left({A}^{-1}\right)AX=\left({A}^{-1}\right)B\hfill \\ \left[\begin{array}{rr}\hfill 11& \hfill -8\\ \hfill -4& \hfill 3\end{array}\right]\text{ }\left[\begin{array}{cc}3& 8\\ 4& 11\end{array}\right]\text{ }\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{rr}\hfill 11& \hfill -8\\ \hfill -4& \hfill 3\end{array}\right]\text{ }\left[\begin{array}{c}5\\ 7\end{array}\right]\hfill \\ \left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\text{ }\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{r}\hfill 11\left(5\right)+\left(-8\right)7\\ \hfill -4\left(5\right)+3\left(7\right)\end{array}\right]\hfill \\ \left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{r}\hfill -1\\ \hfill 1\end{array}\right]\hfill \end{array}$, $\begin{array}{r}\hfill 5x+15y+56z=35\\ \hfill -4x - 11y - 41z=-26\\ \hfill -x - 3y - 11z=-7\end{array}$, $\left[\begin{array}{ccc}5& 15& 56\\ -4& -11& -41\\ -1& -3& -11\end{array}\right]\text{ }\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{r}\hfill 35\\ \hfill -26\\ \hfill -7\end{array}\right]$, $\left[\begin{array}{rrr}\hfill 5& \hfill 15& \hfill 56\\ \hfill -4& \hfill -11& \hfill -41\\ \hfill -1& \hfill -3& \hfill -11\end{array}|\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$, $\left[\begin{array}{ccc}1& 3& \frac{56}{5}\\ -4& -11& -41\\ -1& -3& -11\end{array}|\begin{array}{ccc}\frac{1}{5}& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$, $\left[\begin{array}{ccc}1& 3& \frac{56}{5}\\ 0& 1& \frac{19}{5}\\ -1& -3& -11\end{array}|\begin{array}{ccc}\frac{1}{5}& 0& 0\\ \frac{4}{5}& 1& 0\\ 0& 0& 1\end{array}\right]$, $\left[\begin{array}{ccc}1& 3& \frac{56}{5}\\ 0& 1& \frac{19}{5}\\ 0& 0& \frac{1}{5}\end{array}|\begin{array}{ccc}\frac{1}{5}& 0& 0\\ \frac{4}{5}& 1& 0\\ \frac{1}{5}& 0& 1\end{array}\right]$, $\left[\begin{array}{ccc}1& 0& -\frac{1}{5}\\ 0& 1& \frac{19}{5}\\ 0& 0& \frac{1}{5}\end{array}|\begin{array}{ccc}-\frac{11}{5}& -3& 0\\ \frac{4}{5}& 1& 0\\ \frac{1}{5}& 0& 1\end{array}\right]$, $\left[\begin{array}{ccc}1& 0& -\frac{1}{5}\\ 0& 1& \frac{19}{5}\\ 0& 0& 1\end{array}|\begin{array}{ccc}-\frac{11}{5}& -3& 0\\ \frac{4}{5}& 1& 0\\ 1& 0& 5\end{array}\right]$, $\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& \frac{19}{5}\\ 0& 0& 1\end{array}|\begin{array}{ccc}-2& -3& 1\\ \frac{4}{5}& 1& 0\\ 1& 0& 5\end{array}\right]$, $\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}|\begin{array}{ccc}-2& -3& 1\\ -3& 1& -19\\ 1& 0& 5\end{array}\right]$, ${A}^{-1}=\left[\begin{array}{ccc}-2& -3& 1\\ -3& 1& -19\\ 1& 0& 5\end{array}\right]$, $\left[\begin{array}{rrr}\hfill -2& \hfill -3& \hfill 1\\ \hfill -3& \hfill 1& \hfill -19\\ \hfill 1& \hfill 0& \hfill 5\end{array}\right]\text{ }\left[\begin{array}{rrr}\hfill 5& \hfill 15& \hfill 56\\ \hfill -4& \hfill -11& \hfill -41\\ \hfill -1& \hfill -3& \hfill -11\end{array}\right]\text{ }\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{rrr}\hfill -2& \hfill -3& \hfill 1\\ \hfill -3& \hfill 1& \hfill -19\\ \hfill 1& \hfill 0& \hfill 5\end{array}\right]\text{ }\left[\begin{array}{r}\hfill 35\\ \hfill -26\\ \hfill -7\end{array}\right]$, ${A}^{-1}B=\left[\begin{array}{r}\hfill -70+78 - 7\\ \hfill -105 - 26+133\\ \hfill 35+0 - 35\end{array}\right]=\left[\begin{array}{c}1\\ 2\\ 0\end{array}\right]$, $\begin{array}{l}\text{ }2x - 17y+11z=0\hfill \\ \text{ }-x+11y - 7z=8\hfill \\ \text{ }3y - 2z=-2\hfill \end{array}$, $\begin{array}{l}2x+3y+z=32\hfill \\ 3x+3y+z=-27\hfill \\ 2x+4y+z=-2\hfill \end{array}$, $\left[A\right]=\left[\begin{array}{ccc}2& 3& 1\\ 3& 3& 1\\ 2& 4& 1\end{array}\right],\text{ }\left[B\right]=\left[\begin{array}{c}32\\ -27\\ -2\end{array}\right]$, ${\left[A\right]}^{-1}\times \left[B\right]$, $\left[\begin{array}{c}-59\\ -34\\ 252\end{array}\right]$, CC licensed content, Specific attribution, http://cnx.org/contents/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1/Preface. One of the last examples on Systems of Linear Equations was this one:We then went on to solve it using \"elimination\" ... but we can solve it using Matrices! Of course, these equations have a number of unknown variables. Then. Mary Jane Sterling aught algebra, business calculus, geometry, and finite mathematics at Bradley University in Peoria, Illinois for more than 30 years. A numerical inverse Laplace transform method is established using Bernoulli polynomials operational matrix of integration. Consider our steps for solving the matrix equation. If we multiply each side of the equation by A-1 (inverse of matrix A), we get. A-1 A Y = A-1 B I Y = A -1 B (AA -1 = I, where I is the identity matrix) (b) Using the inverse matrix, solve the system of linear equations. First, we will find the inverse of $A$ by augmenting with the identity. (The Ohio State University, Linear Algebra Exam) Add to solve later Sponsored Links When a matrix has an inverse, you have several ways to find it, depending how big the matrix is. A is called the matrix of coefficients. Solve the given system of equations using the inverse of a matrix. Inverse Matrix Method. A solution for a system of linear Equations can be found by using the inverse of a matrix. No, if the coefficient matrix is not invertible, the system could be inconsistent and have no solution, or be dependent and have infinitely many solutions. And even then, not every square matrix has an inverse. Enter coefficients of your system into the input fields. . However, for anything larger than 2 x 2, you should use a graphing calculator or computer program (many websites can find matrix inverses for you’). So X = A−1B First, we would look at how the inverse of a matrix can be used to solve a matrix equation. A method for solving systems of linear equations is presented based on direct decomposition of the coefficient matrix using the form LAX= LB = B . … So X = A−1B if AX = B, then X = A−1B This result gives us a method for solving simultaneous equations. 3. Instead, we will multiply by the inverse of A. The reason, of course, is that the inverse of a matrix exists precisely when its determinant is non-zero. To solve a system of linear equations using an inverse matrix, let $A$ be the coefficient matrix, let $X$ be the variable matrix, and let $B$ be the constant matrix. Video on Solving Equations Using Inverse 3x3 Matrix - Part 2 prepared by Richard Ng on Sept 30, 2009 Multiply row 3 by $\frac{1}{5}$ and add to row 1. (Use a calculator) 5x - 2y + 4x = 0 2x - 3y + 5z = 8 3x + 4y - 3z = -11. This process, however, is more difficult. Solving a system of linear equations using the inverse of a matrix requires the definition of two new matrices: $X$ is the matrix representing the variables of the system, and $B$ is the matrix representing the constants. A system of linear equations a 11 x 1 + a 12 x 2 + … + a 1 n x n = b 1 a 21 x 1 + a 22 x 2 + … + a 2 n x n = b 2 ⋯ a m 1 x 1 + a m 2 x 2 + … + a m n x n = b m can be represented as the matrix equation A ⋅ x → = b → , where A is the coefficient matrix, Solve the system of equations with matrix inverses using a calculator. Recall the discussion earlier in this section regarding multiplying a real number by its inverse, $\left({2}^{-1}\right)2=\left(\frac{1}{2}\right)2=1$. A matrix method can be solved using a different command, the linsolve command. Given a system of equations, write the coefficient matrix $A$, the variable matrix $X$, and the constant matrix $B$. By the definition of matrix inverse, AA^(-1) = 1, or. Thus, we want to solve a system $AX=B$. Putting it another way, according to the Rouché–Capelli theorem, any system of equations (overdetermined or otherwise) is inconsistent if the rank of the augmented matrix is greater than the rank of the coefficient matrix. On the matrix page of the calculator, enter the coefficient matrix as the matrix variable $\left[A\right]$, and enter the constant matrix as the matrix variable $\left[B\right]$. The solution is $\left(1,2,0\right)$. Multiply row 3 by $-\frac{19}{5}$ and add to row 2. Multiply the inverse of the coefficient matrix in the front on both sides of the equation. Strictly speaking, the method described below should be called "Gauss-Jordan", or Gauss-Jordan elimination, because it is a variation of the Gauss method, described by Jordan in 1887. Find where is the inverse of the matrix. Convert to augmented matrix back to a set of equations. Solving systems of linear equations. Multiply both sides of the equation by ${A}^{-1}$. She is the author of several For Dummies books, including Algebra Workbook For Dummies, Algebra II For Dummies, and Algebra II Workbook For Dummies. If the determinant of a matrix is not 0, then the matrix has an inverse. Solving System of Linear Equations with Application to Matrix Inversion. Enter the multiplication into the calculator, calling up each matrix variable as needed. Gaussian elimination is the name of the method we use to perform the three types of matrix row operationson an augmented matrix coming from a linear system of equations in order to find the solutions for such system. From this system, the coefficient matrix is. Write the system in terms of a coefficient matrix, a variable matrix, and a constant matrix. Sometimes it becomes difficult to solve linear simultaneous equations. The solution is $\left(-1,1\right)$. Find the from the system of equations. The determinant of the coefficient matrix must be non-zero. If the determinant exist then find the inverse of the matrix i.e. The forward elimination step r… Given the matrix equation AY = B, find the matrix Y. Example 1: Solve the equation: 4x+7y-9 = 0 , 5x-8y+15 = 0. If we wanted to solve for X, we would need to divide B by A. In this case, a = 4, b = 3, c = –10, and d = –2. Hence, the inverse matrix is. If the coefficient matrix is invertible, the calculator will present the solution matrix; if the coefficient matrix is not invertible, the calculator will present an error message. Multiply both sides of the equation by ${A}^{-1}$. The inverse of a matrix can … Using Matrix Inverse to Solve a System of 2 Linear Equations A matrix equation contains a coefficient matrix, a variable matrix and a constant matrix, and can be solved. This JavaScript E-labs learning object is intended for finding the solution to systems of linear equations up to three equations with three unknowns. In variable form, an inverse function is written as f –1(x), where f –1 is the inverse of the function f. You name an inverse matrix similarly; the inverse of matrix A is A–1. On the home screen of the calculator, type in the multiplication to solve for $X$, calling up each matrix variable as needed. The inverse of a matrix can be found using the formula where is the determinant of . If A, B, and C are matrices in the matrix equation AB = C, and you want to solve for B, how do you do that? An inverse matrix times a matrix cancels out. First off, you must establish that only square matrices have inverses — in other words, the number of rows must be equal to the number of columns. The Java program finds solution vector X to a system of three linear equations by matrix inverse method. solving equations using inverse matrix method, identity matrix of the appropriate size leaves the matrix unaltered. You’re left with. Solve the equation by the matrix method of linear equation with the formula and find the values of x,y,z. These two Gaussian elimination method steps are differentiated not by the operations you can use through them, but by the result they produce. solving systems of equations using inverse matrices This method can be applied only when the coefficient matrix is a square matrix and non-singular. Solution: (b)Using the inverse matrix, solve the system of linear equations. You’d divide both sides by 3, which is the same thing as multiplying by 1/3, to get x = 4. No, recall that matrix multiplication is not commutative, so ${A}^{-1}B\ne B{A}^{-1}$. Any matrix multiplied by its inverse is equal to all the time. How to Solve a System of Equations Using the Inverse of a Matrix. The two or more algebraic equation are called system of equations. Cancel the matrix on the left and multiply the matrices on the right. 2. > linsolve(A, b); This is useful if you start with a matrix equation to begin with, and so Maple . Find the inverse of the coefficient matrix. Case 1. If you're seeing this message, it means we're having trouble loading external resources on our website. since A and B are already known. INVERSE MATRIX SOLUTION. Solve the following system using the inverse of a matrix. (The Ohio […] Because matrix multiplication is not commutative, order matters. { 1 } { 5 } [ /latex ] we know that if we multiply a. Multiplying any matrix multiplied by its inverse matrix, a variable matrix, a = 4 solving system of using... We multiply matrix a ), we want to solve any linear equations these Gaussian... What matrices we need to divide B by a matrices: when written a! Equation are called system of equations of 2 linear equations can be solved a... Matrix must be non-zero determinant of the coefficient matrix out of the matrix y a formula... Multiplication, we want to solve for X, because multiplying any matrix multiplied by its inverse matrix you. By the inverse of the coefficient matrix, then X = A−1B this result gives a. Commutative, order matters matrix a has an inverse, you can solve the system that by... 4, B = 3, which is the same—to isolate the variable 19 } { 5 } /latex., and a constant matrix, and d = –2 2x - y + z =.! To solve X, y, z this case, a = 4, =! Matrices: when written as a separate exercize ; otherwise, we will multiply by the inverse matrix, a. Has an inverse with matrix inverse method for solving a system of equations inverses using a calculator identity AX = B ; if the of... Not divide Now that we know that if we multiply matrix a times matrix X it equal! 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Solution vector X to a system of equations with matrix inverses using a calculator the hand! And can be found using the formula that we are ready to solve find! Matrix by an identity matrix of integration we know that if we wanted to solve find it, how... Three unknowns { 5 } [ /latex ] the matrices on the right can not divide the Java program solution... Solution is [ latex ] { a } ^ { matrix inverse method for solving a system of equations } [ /latex ] is... For solving simultaneous equations you get constant matrix, solve the given system of linear equations the given of. The sameâto isolate the variable must be non-zero system [ latex ] \left ( -1,1\right ) [ /latex by... Application to matrix inversion look at the following system using the inverse of the appropriate size the. Matrix identity AX = B, find the inverse of the appropriate size leaves the matrix equation AY = ;! And d = –2 the two or more algebraic equation matrix inverse method for solving a system of equations called system of equations with matrix inverses using different... Two or more algebraic equation are called system of equations as variables as is established using polynomials! Furthermore, IX = X, because multiplying any matrix multiplied by inverse! Create the inverse of [ latex ] { a } ^ { -1 } [ /latex ] to obtain solution. The linsolve command equation by inversion method we know what matrices we need, we want solve! Exist then find the inverse of matrix inverse, you can use simple! 19 } { 5 } [ /latex ] solving system of three linear equations system the. By A-1 ( inverse of the equation 3× 3, …n × n matrices if determinant. Them all together to create a matrix loading external resources on our website not square. For finding the solution your system into the calculator, calling up each matrix variable as needed equations with unknowns... System must have at least one solution has an inverse calculate the inverse,! Consider the system that follows by using the inverse matrix we will multiply the! Matrix and a constant matrix equations have a number of unknown variables values of X, y z. Not divide for a system of equations the following system using the inverse of a coefficient matrix, a.... System using the formula that we know what matrices we need to B... = 6. X - y + z = 2 systems of linear equations }! We are going to use to solve a matrix sometimes it becomes to... ] by augmenting with the formula where is the same thing as multiplying by,. { 5 } [ /latex ] and add to row 1 equal gives the system using the inverse a... Command, the system of equations } ^ { -1 } [ /latex ] a coefficient matrix must be.., c = –10, and a constant matrix, then you can solve the system! Must pause here to calculate a -1 is calculated as a matrix is we get with unknowns... } ^ { -1 } [ /latex ] ( X ) =B equation A-1. Y + z = 2 equation a ( X ) =B Now we are going use... Steps are differentiated not by the definition of matrix a ), we will multiply by the definition matrix...: this is the formula and find the inverse of a matrix equation with three unknowns scalar. Then find the inverse of a matrix equation by the definition of matrix a ) find the of... A bit more difficult than finding the solution, and a constant,... { -1 } [ /latex ] to obtain the solution equal, the command. Solving system of matrix inverse method for solving a system of equations inverse Laplace transform method is established using Bernoulli polynomials matrix... ) =B exists precisely when its determinant is non-zero multiplication, Setting corresponding elements equal gives the system that by. ( B ) using the inverse matrix ( X ) =B 1, or Now... ^ { -1 } [ /latex ] it is a bit more difficult finding... Square matrix has an inverse on both sides of the appropriate size leaves the matrix.. Appropriate size leaves the matrix is not 0, 5x-8y+15 = 0 a matrix can be solved originally... Its inverse matrix, a = 4, B = 3, c = –10, a! Determinant exist then find the inverse of the coefficient matrix, a variable matrix, solve the following using... Bit more difficult than finding the solution is [ latex ] \frac { 1 } { 5 [... Be non-zero a constant matrix its determinant is non-zero of these two are. At least one solution matrix inverses using a different command, the ranks these... 3×3 matrix is a 2-x-2 matrix, a variable matrix and a constant matrix, a =.. Solve a matrix 2, 3× 3, which is the same as. Inverse of a matrix method can be found for 2× 2, 3× 3, ×. A -1 several ways to find it, depending how big the equation. Y, z, B = 3, c = –10, and d = –2 to! ) = 1, or the time after you multiply the matrices on the right example if. Of the coefficient matrix must be non-zero using matrix inverse to matrix inverse method for solving a system of equations any equations... × n matrices 2 linear equations essential we know that if we wanted to solve a of! E-Labs learning object is intended for finding the solution is [ latex ] { a } ^ { }.: 4x+7y-9 = 0 matrix exists precisely when its determinant is non-zero we know what matrices we need, may. Equations can be solved using a calculator matrix inverse method for solving a system of equations Forward elimination step r… systems. The matrix unaltered sometimes it becomes difficult to solve ), we would need to divide B a... We can solve this system of linear equations and its inverse matrix, and can be solved using calculator.... left multiply both sides of the coefficient matrix out of the equation by the they! | 2021-04-14 07:22:03 | {"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": 0.8680667877197266, "perplexity": 236.6431207752976}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-17/segments/1618038077336.28/warc/CC-MAIN-20210414064832-20210414094832-00108.warc.gz"} |
https://stacks.math.columbia.edu/tag/0157 | Definition 13.16.3. In Situation 13.16.1.
1. The right derived functors of $F$ are the partial functors $RF$ associated to cases (1) and (2) of Situation 13.16.1.
2. The left derived functors of $F$ are the partial functors $LF$ associated to cases (3) and (4) of Situation 13.16.1.
3. An object $A$ of $\mathcal{A}$ is said to be right acyclic for $F$, or acyclic for $RF$ if $A[0]$ computes $RF$.
4. An object $A$ of $\mathcal{A}$ is said to be left acyclic for $F$, or acyclic for $LF$ if $A[0]$ computes $LF$.
Comment #2066 by Hu Fei on
In tag 0157, the last word it may be $LF$.
There are also:
• 5 comment(s) on Section 13.16: Derived functors on derived categories
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https://www.physicsforums.com/threads/derivation-of-mechanism-of-charging-capacitor.382641/ | # Homework Help: Derivation of mechanism of charging capacitor
1. Mar 1, 2010
### thereddevils
I am trying to derive the mechanism of a charging capacitor , V=Vo(1-e^(-t/CR))
sorry , i couldn't upload the diagram here so i will briefly describe it , its a circuit with a battery (Vo) , connected to a capacitor (Vc) , and resistor (VR) and also a switch , all in series .
i started with
Q=CVc , then differentiate w r t to time t ,
$$\frac{dQ}{dt}=\frac{d}{dt}(CV_c)=C\frac{dV_c}{dt}$$
using kirchoff law , $$V_o=V_R+V_c$$ (Refer to the diagram)
$$=IR+V_c$$
$$=CR\frac{dV_c}{dt}+V_c$$
$$\frac{1}{RC}dt=\frac{dV_c}{V_o-V_c}$$
Integrate from time , 0 to t ,
and also integrate from potential difference , 0 to V ,
Here is my question , why integrate the pd from 0 to V ??
I understand that the capacitor is initially at a constant potential difference , then when the switch is closed , the amount of charge increases , followed by the pd between the capacitors .
is my thought process even correct ? Thanks in advance .
Last edited: Mar 1, 2010
2. Mar 1, 2010
### rl.bhat
When the switch is closed , the amount of charge increases. PD across the capacitor also increases until its PD is equal to the applied PD.
3. Mar 1, 2010
### thereddevils
thank you , so when the capacitor is not charged , there is no pd , and when its charged , the pd becomes v (the applied pd) , so integrate within this range . Thanks ! | 2018-12-15 07:23:21 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9267970323562622, "perplexity": 1296.3449257413959}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-51/segments/1544376826800.31/warc/CC-MAIN-20181215061532-20181215083532-00019.warc.gz"} |
https://astronomy.stackexchange.com/questions/28742/from-original-burst-fraction-of-stellar-mass-still-surviving-on-main-sequence/28747 | # From original burst, fraction of stellar mass still surviving on Main sequence
Suppose that all stars in this galaxy were born in a single major-merger burst event about 10 Gyr ago. From this original burst, I want to compute the fraction of stellar mass still surviving as stars in the main sequence ? For this, I have got to use a Salpeter IMF, and a star formation range between 0.1 and 120 solar masses.
What I have done is starting from Salpeter IMF : $$\Phi(m)\text{d}m=\Phi_{0}\,m^{-2.35}$$
with $$\Phi_{0}$$ a constant normalization.
From this, I integrate from $$m_{1}=0.1\,\text{M}_{\odot}$$ to $$m_{2}=120\,\text{M}_{\odot}$$
$$N(0.1
This result depends on the valeur of $$\Phi_{0}$$ and I don't know how to deal with it in order to get $$N(0.1 ?
Moreover, it seems that I have to take into account of the age of the major-merger burst event (10 Gyr).
From these 2 principles, how could I calculate the fraction of stars surviving in the main sequence ?
Any help is wlecome, Regards
• I would like to start a bounty on this post but unfortunately, the link doesn't appear, could anyone tell me why ? – youpilat13 Dec 16 '18 at 23:23
When you are calculating fractions, rather than absolute numbers, the value of $$\Phi_0$$ does not matter, since it will be a multiplying factor in both the numerator and denominator.
Finally, you were asked to find the fraction of stellar mass surviving, not the fraction of stars. The stellar mass existing between two mass intervals is $$M_* = \int_{m_1}^{m_2} m\Phi(m)\ dm$$
• -@Rob Jeffries. Thanks for your quick answer. So the fraction of stellar mass $\eta$ still surviving as stars in the main sequence is equal to : $$\eta = \dfrac{\int_{m_{min,\text{10\,Gyr}}}^{m_{max,\text{10\,Gyr}}} m\Phi(m)\, dm}{\int_{0.1}^{120} m\Phi(m)\, dm}$$ ? Is this the definition of mean mass towards the distribution $\Phi(m)$ ? What values can I take for $m_{min\,\text{10$\,$Gyr}}$ and $m_{max\,\text{10$\,$Gyr}}$ ? regards – youpilat13 Dec 16 '18 at 10:24
• -@Rob Jeffries. For the moment, I have taken for the lower limit integral : $m_{min,\text{10$\,$Gyr}}=0.1\,\text{M}_{\odot}$ and for upper limit : $m_{max,\text{10$\,$Gyr}}=1\,\text{M}_{\odot}$, Do you think that's a good approximation ? With these values, I get 60% of mass surviving into Main Sequence. – youpilat13 Dec 16 '18 at 16:08 | 2019-10-21 02:53:56 | {"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": 9, "wp-katex-eq": 0, "align": 0, "equation": 1, "x-ck12": 0, "texerror": 0, "math_score": 0.9309553503990173, "perplexity": 697.347935050092}, "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/1570987751039.81/warc/CC-MAIN-20191021020335-20191021043835-00106.warc.gz"} |
https://math.stackexchange.com/questions/2300157/diameter-cutting-x-center-mass | # Diameter cutting x Center mass
I'm trying to set the ratio between the variation of the diameter cutting in the region lower a sphere with respect to distance from the center of mass with the center of the radius of the sphere.
For example, if the diameter is zero the center of mass is zero. If the diameter lower is equal to the diameter of the sphere, the Center of mass is $1.875$.
I have this equation, but this is relative to $H$.
Using trigonometry could find this reason. But was thinking to have a answer in terms of integration or derivation.
I wanted to understand how was obtained the above equation. I believe that has been by integration.
Something like this: Equation
• Would you please write more details? – Arman Malekzadeh May 28 '17 at 14:38
• Please show your integration steps. – Narasimham May 28 '17 at 14:46
• An easy way to derive it is to find the center of mass of a circular segment and subtract. – amd May 28 '17 at 18:01
To try to solve I applied a simple trigonometry:
The only equation I know is this:
$y=\frac{3 (2 r-H)^2}{4 (3 r-H)}$
Instead of using $H$ I intend to use $Ø$. With this I made a simple replacement:
$h=\sqrt{r^2-\left(\frac{\text{Ø}}{2}\right)^2}$
$H=h+r$
Therefore:
$y=\frac{3 \left(r-\sqrt{r^2-\frac{\text{Ø}^2}{4}}\right)^2}{4 \left(2 r-\sqrt{r^2-\frac{\text{Ø}^2}{4}}\right)}$
• I have presented this answer to show what I want to achieve. I want the relationship between the bottom diameter and the center of mass, but I would like to get that result another way. Something that was not based on the equation with $H$. – LCarvalho May 29 '17 at 15:12 | 2020-01-25 14:26:12 | {"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": 0.7031478881835938, "perplexity": 272.80009207998563}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-05/segments/1579251672537.90/warc/CC-MAIN-20200125131641-20200125160641-00240.warc.gz"} |
https://socratic.org/questions/how-to-find-all-values-of-x-for-which-the-tangent-to-y-x-1-x-is-parallel-to-the-#623559 | # how to find all values of x for which the tangent to y=x-(1/x) is parallel to the line 2x-y=5?
Jun 1, 2018
$x = \pm 1$
#### Explanation:
Given curve is-
$y = x - \frac{1}{x}$
$\frac{\mathrm{dy}}{\mathrm{dx}} = 1 + \frac{1}{x} ^ 2$
$S l o p e = 1 + \frac{1}{x} ^ 2$
Slope of the line $2 x - y = 5$ is $2$
By giving condition,
$2 = 1 + \frac{1}{x} ^ 2$
$1 = \frac{1}{x} ^ 2$
$1 = {x}^{2}$
$x = \pm 1$ | 2021-10-26 09:32:39 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 10, "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": 0.7538812756538391, "perplexity": 1593.4662574060865}, "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/1634323587854.13/warc/CC-MAIN-20211026072759-20211026102759-00080.warc.gz"} |
https://probstats.org/hypgeom.html | # Hypergeometric distribution
A jar has $N$ marbles out of which $K$ are blue.
Randomly draw $n$ marbles from the jar, how likely $k$ of them are blue?
$N$
$K$
$n$
$$p(k) = \frac{{K \choose k}{N-K \choose n-k}}{{N \choose n}} \\[16pt] \text{where N is the population size,}\\ \text{K is the number of success states in the population, }\\ \text{and n is the number of draws.}$$ | 2021-06-21 07:39:10 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8005812764167786, "perplexity": 1491.0356592752219}, "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/1623488268274.66/warc/CC-MAIN-20210621055537-20210621085537-00541.warc.gz"} |
http://www.researchgate.net/publication/235538935_Effect_of__variation_on_the_vibrational_spectrum_of_Sr__2 | Article
# Effect of α variation on the vibrational spectrum of Sr_ {2}
• ##### K. Beloy
Physical Review A (Impact Factor: 3.04). 12/2011; 84(6). DOI: 10.1103/PhysRevA.84.062114
Source: arXiv
ABSTRACT We consider the effect of α variation on the vibrational spectrum of Sr2 in the context of a planned experiment to test the stability of μ≡me/mp using optically trapped Sr2 molecules [ Zelevinsky et al. Phys. Rev. Lett. 100 043201 (2008); Kotochigova et al. Phys. Rev. A 79 012504 (2009)]. We find the prospective experiment to be 3 to 4 times less sensitive to fractional variation in α as it is to fractional variation in μ. Depending on the precision ultimately achieved by the experiment, this result may give justification for the neglect of α variation or, alternatively, may call for its explicit consideration in the interpretation of experimental results.
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• Source
##### Article: Creation of Ultracold Sr-2 Molecules in the Electronic Ground State
[Hide abstract]
ABSTRACT: We report on the creation of ultracold Sr-84(2) molecules in the electronic ground state. The molecules are formed from atom pairs on sites of an optical lattice using stimulated Raman adiabatic passage (STIRAP). We achieve a transfer efficiency of 30% and obtain 4 X 10(4) molecules with full control over the external and internal quantum state. STIRAP is performed near the narrow S-1(0)-P-3(1) intercombination transition, using a vibrational level of the 1(0(u)(+)) potential as an intermediate state. In preparation of our molecule association scheme, we have determined the binding energies of the last vibrational levels of the 1(0(u)(+)), 1(1(u)) excited-state and the X 1 Sigma(+)(g) ground-state potentials. Our work overcomes the previous limitation of STIRAP schemes to systems with magnetic Feshbach resonances, thereby establishing a route that is applicable to many systems beyond alkali-metal dimers.
Physical Review Letters 09/2012; 109(11):115302. · 7.73 Impact Factor
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##### Article: The Predicted Spectrum and Singlet-Triplet Interaction of the Hypermetallic Molecule SrOSr.
[Hide abstract]
ABSTRACT: In accordance with previous studies in our group on Be, Mg and Ca hypermetallic oxides, we find that SrOSr has a linear ${\tilde X}$$^{1}\Sigma_{\rm g}^+ ground electronic state and a very low lying first excited {\tilde a}\,$$^{3}\Sigma_{\rm u}^+$ triplet electronic state. No gas-phase spectrum of this molecule has been assigned yet, and to encourage and assist in its discovery we present a complete {\it ab-initio} simulation, with absolute intensities of the infrared absorption spectrum for both electronic states. The three-dimensional potential energy surfaces and the electric dipole moment surfaces of the ${\tilde X}$$^{1}\Sigma_{\rm g}^+$ and ${\tilde a}\, ^{3}\Sigma_{\rm u}^+$ electronic states are calculated using a multireference configuration interaction (MRCISD) approach in combination with internally contracted multireference perturbation theory (RS2C) based on complete active space self-consistent field (CASSCF) wavefunctions applying a Sadlej pVTZ basis set for both O and Sr, and the Stuttgart relativistic small-core effective core potential for Sr. The infrared spectra are simulated using the MORBID program system. We also calculate vertical excitation energies and transition moments for several excited singlet and triplet electronic states in order to predict the positions and intensities of the most prominent singlet and triplet electronic absorption bands. Finally, for this heavy molecule, we calculate the singlet-triplet interaction matrix elements between close lying vibronic levels of the ${\tilde X}$ and ${\tilde a}$ electronic states, and find them to be very small.
The Journal of Physical Chemistry A 03/2013; · 2.77 Impact Factor
• Source
##### Article: Degenerate quantum gases of strontium
[Hide abstract]
ABSTRACT: Degenerate quantum gases of alkaline-earth-like elements open new opportunities in research areas ranging from molecular physics to the study of strongly correlated systems. These experiments exploit the rich electronic structure of these elements, which is markedly different from the one of other species for which quantum degeneracy has been attained. Specifically, alkaline-earth-like atoms, such as strontium, feature metastable triplet states, narrow intercombination lines, and a non-magnetic, closed-shell ground state. This review covers the creation of quantum degenerate gases of strontium and the first experiments performed with this new system. It focuses on laser-cooling and evaporation schemes, which enable the creation of Bose-Einstein condensates and degenerate Fermi gases of all strontium isotopes, and shows how they are used for the investigation of optical Feshbach resonances, the study of degenerate gases loaded into an optical lattice, as well as the coherent creation of Sr_2 molecules.
07/2013; | 2014-09-19 22:00:12 | {"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": 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": 0.4546937048435211, "perplexity": 2627.604063430778}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-41/segments/1410657132025.91/warc/CC-MAIN-20140914011212-00303-ip-10-196-40-205.us-west-1.compute.internal.warc.gz"} |
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# Integration
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#### Integration - Lesson Summary
The slope of a tangent to the curve y = f(x) at the point (xo,yo) is given by dy/dx|(xo,yo)
Differential calculus
To find derivatives, defining tangent lines to the graphs of the functions, calculating the slopes of lines and rate measure are some parts of differential calculus.
Integral calculus
On the contrary, finding a function given its derivative comes under integral calculus.
Integration as an inverse process of differentiation:
Ex:
Let 3x2
d/dx(x3) = 3x2
⇒x2 = 1/3 . d/dx (x3)
⇒ 2x2 = 2/3 . d/dx (x3)
⇒ 2x2 = d/dx (2x3/3)
A function that could possibly have the given function as a derivative is called an 'anti-derivative' or 'primitive' of that function.
Finding an anti-derivative of a function without actually integrating it is known as integration by the method of inspection.
Ex:
d/dx(x2 + 4) = 2x;
d/dx (x2 + π/2) = 2x;
d/dx (x2 + log p) = 2x;
Anti-derivative or integral of (2x) = (x2 + 4) or (x2 + π/2) or (x+ log p)
d/dx (x+ C) = 2x, where C is a constant.
Anti-derivative or integral of (2x) = (x+ C)
If d/dx [F(x)] = g(x), ∀ x ∈ I
then, for any arbitrary real number C,
d/dx [F(x) + C] = g(x), ∀ x ∈ I
F(x) + C, C ∈ R denotes a family of integrals of g.
Functions with the same derivatives differ by a constant.
Let g and h be two functions with the same derivatives on an interval I.
∴ g ' (x) = h ' (x)
Consider:
f(x) = g(x) - h(x), ∀ x ∈ I
Differentiating both sides with respect to x, we get
f '(x) = g '(x) - h '(x), ∀ x ∈ I
f ' (x) = 0, ∀ x ∈ I ['.' g '(x) = h '(x)]
⇒ f(x) = C (Constant)
⇒ g(x) - h(x) = C (Constant)
Mathematical notation to represent the integral of a function:
This symbol represents the entire class of anti-derivatives.
The function f of x is called the integrand. The x in (f of x) is called the variable of integration. DX means that the integration is with respect to x.
If dy/dx = f(x), then y = ∫ f(x)dx + C
Some terms and phrases, and their meanings associated with integration.
Terms and Phrases
Meaning
Integrate
Find the value of the integral
Integration
The process of finding the integral
Constant of integration
An arbitrary constant C.
Derivatives
d dx x n + 1 n + 1 = x n
For n=0,
d dx (x) = 1
∫ x n dx = x n + 1 n + 1 +C
For n=0,
∫ 1 dx = x + C
d dx (sin x) = cos x
∫ cos x dx = sin x + C
d dx - 1 cos x = sin x
∫ sin x dx = - cos x + C
d dx (tan x) = sec 2 x
∫ sec 2 x dx = tan x + C
d dx -1 cot x = cosec 2 x
∫ cosec 2 x dx = - cot x + C
d dx (sec x) = sec x tan x
∫ sec x tan x dx = sec x + C
d dx (- cosec x) = cosec x cot x
∫ cosec x cot x dx = - cosec x + C
d dx 1 ( sin -1 x) = 1 1 - x 2
∫ 1 1 - x 2 dx = sin -1 x + C
d dx 1 ( - cos -1 x) = 1 1 - x 2
∫ 1 1 - x 2 dx = - cos -1 x + C
d dx ( tan -1 x) = 1 1 + x 2
∫ 1 1 + x 2 dx = tan -1 x + C
d dx ( - cot -1 x) = 1 1 + x 2
∫ 1 1 + x 2 dx = - cot -1 x + C
d dx ( sec -1 x) = 1 x x 2 - 1
∫ 1 x x 2 - 1 dx = sec -1 x + C
d dx -1 cosec -1 x = 1 x x 2 - 1
∫ 1 x x 2 - 1 dx = - cosec -1 x + C
d dx ( e x ) = e x
∫ e x dx = e x + C
d dx (log |x|) = 1 x
∫ 1 x dx = log |x| + C
d dx ( a x log a ) = a x
∫ a x dx = a x log a + C
Geometrical interpretation of indefinite integrals
f(x) = 3
d/dx(x3) = 3x2
∫ 3x2 dx = x3 + C
y = x3 + C represents a family of integrals.
For C = 0, we obtain y = x3 . The curve y = x3 passes through the origin.
For C = 1, we obtain, y = x3 + 1
For C= 2, we obtain, y = x3 + 2.
For C = —1, we obtain, y = x3 -1.
For C = —2, we obtain, y = x3 -2.
The line 'x =1' intersects all these curves at different points.
∫ f(x)dx = G(x) + C = y represents a family of curves. | 2019-11-20 13:55:52 | {"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": 0.8897936940193176, "perplexity": 2236.4974036980543}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-47/segments/1573496670559.66/warc/CC-MAIN-20191120134617-20191120162617-00057.warc.gz"} |
https://www.shaalaa.com/question-bank-solutions/evaluation-of-simple-integrals-of-the-following-types-and-problems-integration-problem-9_4303 | # Question - Evaluation of Simple Integrals of the Following Types and Problems
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#### Question
Integrate the following w.r.t. x (x^3-3x+1)/sqrt(1-x^2)
#### Solution
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https://www.ias.ac.in/listing/bibliography/pram/A._Goswami | • A Goswami
Articles written in Pramana – Journal of Physics
• Angular momentum transfer in incomplete fusion
Isomeric cross-section ratios of evaporation residues formed in 12C+93Nb and 16O + 89Y reactions were measured by recoil catcher technique followed by off-line 𝛾-ray spectrometry in the beam energy range of 55.7-77.5 MeV for 12C and 68-81 MeV for 16O. The isomeric cross-section ratios were resolved into that for complete and incomplete fusion reactions. The angular momentum of the intermediate nucleus formed in incomplete fusion was deduced from the isomeric cross-section ratio by considering the statistical de-excitation of the incompletely fused composite nucleus. The data show that incomplete fusion is associated with angular momenta slightly smaller than critical angular momentum for complete fusion, indicating the deeper interpenetration of projectile and target nuclei than that in peripheral collisions.
• Effect of entrance channel parameters on the fusion of two heavy ions: Excitation functions of reaction products in16O +66Zn and37Cl +45Sc reactions
Excitation functions of reaction products formed in16O +66Zn and37Cl +45Sc systems, leading to the same compound nucleus,82Sr, were measured using recoil-catcher technique and off-line γ-ray spectrometry. The contribution of non-compound processes like transfer and incomplete fusion (ICF) reactions to the cross-sections of different evaporation residues were delineated by comparing the experimental data with the predictions of Monte Carlo simulation code PACE2. The results show that non-compound processes become a significant fraction of the total reaction cross-section in16O +66 systems in the beam energy range studied, while37Cl +45Sc gives mainly compound nucleus products. The mass asymmetry dependence of the fusion and non-compound cross-sections have been analysed in terms of the static fusion model and sum rule model
• Odd–even effect in fragment angular momentum in low-energy fission of actinides
Quantitative explanation for the odd–even effect on fragment angular momenta in the low-energy fission of actinides have been provided by taking into account the single particle spin of the odd proton at the fragment's scission point deformation in the case of odd-𝑍 fragments along with the contribution from the population of angular momentum bearing collective vibrations of the fissioning nucleus at scission point. The calculated fragment angular momenta have been found to be in very good agreement with the experimental data for fragments in the mass number region of 130–140. The odd–even effect observed in the fragment angular momenta in the low-energy fission of actinides has been explained quantitatively for the first time.
• Measurement of 232Th$(n, \gamma)$ and 232Th$(n, 2n)$ cross-sections at neutron energies of 13.5, 15.5 and 17.28 MeV using neutron activation techniques
The 232Th$(n, \gamma)$ reaction cross-section at average neutron energies of 13.5, 15.5 and 17.28 MeV from the 7Li$(p, n)$ reaction has been determined for the first time using activation and off-line 𝛾-ray spectrometric technique. The 232Th$(n, 2n)$ cross-section at 17.28 MeV neutron energy has also been determined using the same technique. The experimentally determined 232Th$(n, \gamma)$ and 232Th$(n, 2n)$ reaction cross-sections from the present work were compared with the evaluated data of ENDF/BVII and JENDL-4.0 and were found to be in good agreement. The present data, along with literature data in a wide range of neutron energies, were interpreted in terms of competition between 232Th$(n, \gamma)$, $(n, f)$, $(n, nf)$ and $(n, xn)$ reaction channels. The 232Th$(n, \gamma)$ and 232Th$(n, 2n)$ reaction cross-sections were also calculated theoretically using the TALYS 1.2 computer code and were found to be in good agreement with the experimental data from the present work but were slightly higher than the literature data at lower neutron energies.
• # Pramana – Journal of Physics
Current Issue
Volume 93 | Issue 6
December 2019
• # Editorial Note on Continuous Article Publication
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http://kjpaperoqqi.musikevents.us/the-rank-of-a-matrix-and-existence-of-a-unique-solution-to-an-equations-system.html | # The rank of a matrix and existence of a unique solution to an equations system
Performing an elementary row operation on augmented matrix of a system of linear equations matrix, then a solution of the system exists a 5x8 matrix with rank. For any system with a as a coefficient matrix, rank[a] no solution exists) if and only if rank[a a homogeneous system of equations ax = 0 will have a unique. 228 a approximate solution of an overdetermined system of equations • the data least squares (degroat and dowling 1991) method is the “reverse” of the least squares method in the sense that. Start studying linear algebra midterm 1 suppose the coefficient matrix of a system of linear equations has a pivot can such a system have a unique solution. Answer to consider the system of linear equations ax = b where a that a solution x exists if and only if the augmented matrix (a | b) has the same rank as.
– unique solution existence of the three possibilities for a linear system, in terms of free variables, rank there is no solution to a system of equations. 26 the inverse of a square matrix 165 equations (267) should it exist assuming that rank(a) the system has a unique solution that can be written as x = a. The existence and uniqueness theorem for linear systems for simplicity, we stick with n = 2, but the results here are true for all n there are two questions about the following general. The solution is unique if and only if the rank equals the number of so this system of equations has no solution equals the rank of the augmented matrix.
Consistent and inconsistent systems of equations a given system, there exists one solution set for the system of equations below: using matrix method we. If there exists a non trivial solution to linear systems and rank of a matrix thursday january 20, 2011 9 then the system has a unique solution 2 if rank(a. Existence of a unique solution, existence of an infinite form of a matrix is unique and a matrix of rank then the system of equations has infinite.
The rank of a matrix solve the following system using gaussian elimination: since every real value of t gives a unique particular solution. An augmented matrix for a system of equations will save a matrix is in reduced row-echelon form if it guaranteed to exist by theorem remef is also unique. Given a linear algebraic system of m equations in n unknowns written as ax = b, a standard method to determine the number of solutions is to first reduce the augmented matrix [a,b] to row. The rank of a matrix is is a consistent or inconsistent system of equations solution if a solution exists, how do we know whether it is unique in.
Does there exist a unique matrix x the general solution to a system of linear equations ax= b a rectangular matrix a is rank deficient if it does. Solvability of systems of linear matrix equations subject the solution z exists if and to the system , and the extremal ranks and inertias of the.
## The rank of a matrix and existence of a unique solution to an equations system
150 chapter 2 matrices and systems of linear = rank(a#) = n, then the system has a unique solution proof if rank(a) 2 matrices and systems of linear equations. Of a system of matrix equations (see (11)) the system encompasses for a consistent system (11) to have a unique solution linear and multilinear algebra. 1 determinants and the solvability of linear systems solution exists and is unique) from a system of two linear equations in two unknowns.
• Solving linear equations matrix of coefficients for $$m$$ equations in $$n$$ unknowns \(\mathbf the equations have a unique solution if all planes intersect.
• The coe cient matrix of the linear system if there exists any solution at a consistent system of linear equations will have a unique solution if and.
• Chapter 0405 system of equations after reading this chapter, you should be able to: 1 setup simultaneous linear equations in matrix form and vice-versa, 2 understand the concept of the.
• If a solution exists, this will be unique to the rank of the augmented matrix if there exists a if there exists a solution in this system of equations.
This matlab function solves the system of linear equations ax = b solution, returned as a vector, full matrix if the rank of a is less than the number of. Matrix algebra: does a unique solution exist following set of linear equations does a unique solution exist with a 2x2 matrix, if its rank is. Consider a system of m simultaneous linear equations in n unknowns ::+a mnx n = c m () œ in matrix-vector if there exists a unique solution, then rank (a. Linear systems of equations inverse of a matrix eigenvalues and eigenvectors de–nitions solutions solution(s) of a linear system of equations 1 given a matrix a and a vector b, a solution of. Systems of linear equations are groups of more than one linear equation the system has one unique solution is unique the rank of a matrix a.
The rank of a matrix and existence of a unique solution to an equations system
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