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https://www.studyadda.com/ncert-solution/11th-chemistry-redox-reactions_q74/497/38534	• # question_answer 74) On the basis of standard electrode potential values, suggest which of the following reactions would take place?(Consult the book for ${{E}^{O-}}$ value).(i) $Cu+Z{{n}^{2+}}\to C{{u}^{2+}}+Zn$(ii) $Mg+F{{e}^{2+}}\to M{{g}^{2+}}+Fe$(iii) $B{{r}_{2}}+2C{{l}^{-}}\to C{{l}_{2}}+2B{{r}^{-}}$(iv)$Fe+C{{d}^{2+}}\to Cd+F{{e}^{2+}}$ (i) $Cu+Z{{n}^{2+}}\to C{{u}^{2+}}+Zn$ $E_{\operatorname{Re}dox\,process}^{{}^\circ }=E_{\operatorname{Re}duced\,species}^{{}^\circ }-E_{Oxidised\,species}^{{}^\circ }$ $=E_{Z{{n}^{2+}}/Zn}^{{}^\circ }-E_{C{{u}^{2+}}/Cu}^{{}^\circ }$ $=(-0.76)-(+0.34)=-1.10V$ Negative value shows that the redox process is not feasible. (ii) $Mg+F{{e}^{2+}}\to M{{g}^{2+}}+Fe$ $E_{\operatorname{Re}dox\,process}^{{}^\circ }=E_{\operatorname{Re}duced\,species}^{{}^\circ }-E_{Oxidised\,species}^{{}^\circ }$ $=E_{F{{e}^{2+}}/Fe}^{{}^\circ }-E_{M{{g}^{2+}}/Mg}^{{}^\circ }$ $=-0.44-(-2.36)=+1.92V$ Positive value shows that the redox process is feasible. (iii) $B{{r}_{2}}+2C{{l}^{-}}\to C{{l}_{2}}+2B{{r}^{-}}$ $E_{\operatorname{Re}dox\,process}^{{}^\circ }=E_{\operatorname{Re}duced\,species}^{{}^\circ }-E_{Oxidised\,species}^{{}^\circ }$ $=E_{B{{r}_{2}}/Br}^{{}^\circ }-E_{C{{l}_{2}}/C{{l}^{-}}}^{{}^\circ }$ $=+1.09-1.36=-ve$ Negative value shows that the redox process is not feasible. (iv) $Fe+C{{d}^{2+}}\to Cd+F{{e}^{2+}}$ $E_{\operatorname{Re}dox\,process}^{{}^\circ }=E_{\operatorname{Re}duced\,species}^{{}^\circ }-E_{Oxidised\,species}^{{}^\circ }$ $=E_{C{{d}^{2}}/Cd}^{{}^\circ }-E_{F{{e}^{2+}}Fe}^{{}^\circ }$ $=(-0.40)-(-0.44)=+0.04V$ Positive value shows that the redox process is feasible.	2020-09-24 20:14:04	{"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": 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": 0.9280741810798645, "perplexity": 10343.679410601526}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-40/segments/1600400220495.39/warc/CC-MAIN-20200924194925-20200924224925-00394.warc.gz"}
http://imomath.com/index.php?options=785	# The 54th International Mathematical Olympiad: Problems and Solutions ## Day 1 (July 23th, 2013) Problem 1 Assume that $$k$$ and $$n$$ are two positive integers. Prove that there exist positive integers $$m_1$$, $$\dots$$, $$m_k$$ such that $1+\frac{2^k-1}{n}=\left(1+\frac1{m_1}\right)\cdots \left(1+\frac1{m_k}\right).$ Problem 2 Given $$2013$$ red and $$2014$$ blue points in the plane, assume that no three of the given points lie on a line. A partition of the plane is called perfect if no region contains points of different colors. Determine the smallest number $$k$$ for which a perfect partition can always be achieved by drawing $$k$$ straight lines. Problem 3 Let $$A_1$$, $$B_1$$, and $$C_1$$ be the points at which the excircles touch the sides $$BC$$, $$CA$$, and $$AB$$ of the triangle $$ABC$$. Prove that if the circumcenter of $$\triangle A_1B_1C_1$$ belongs to the circumcircle of $$\triangle ABC$$, then one of the angles of $$\triangle ABC$$ is $$90^{\circ}$$. ## Day 2 (July 24th, 2013) Problem 4 Let $$ABC$$ be an acute triangle with orthocenter $$H$$, and let $$W$$ be a point on the side $$BC$$ between $$B$$ and $$C$$. The points $$M$$ and $$N$$ are the feet of perpendiculars from $$B$$ and $$C$$, respectively. Let $$\omega_1$$ be the circumcircle of $$\triangle BWN$$, and let $$X$$ be the point such that $$WX$$ is a diameter of $$\omega_1$$. Let $$\omega_2$$ be the circumcircle of $$\triangle CWM$$, and let $$Y$$ be the point such that $$WY$$ is a diameter of $$\omega_2$$. Prove that the points $$X$$, $$Y$$, and $$H$$ are collinear. Problem 5 Let $$\mathbb Q_+$$ be the set of all positive rational numbers. Let $$f:\mathbb Q_+\to\mathbb R$$ be a function that satisfies the following three conditions: • (i) $$f(x)f(y)\geq f(xy)$$ for all $$x,y\in\mathbb Q_+$$, • (ii) $$f(x+y)\geq f(x)+f(y)$$ for all $$x,y\in\mathbb Q_+$$, • (iii) There exists a rational number $$a> 1$$ such that $$f(a)=a$$. Prove that $$f(x)=x$$ for all $$x\in\mathbb Q_+$$. Problem 6 Given an integer $$n\geq 3$$, assume that $$n+1$$ equally spaced points are marked on a circle. Consider all labelings of these points with the numbers $$0$$, $$1$$, $$\dots$$, $$n$$ such that each label is used exactly once. Two labelings are considered the same if one can be obtained from the other by a rotation of the circle. A labeling is beautiful if, for any four labels $$a< b< c< d$$ with $$a+d=b+c$$, the chord joining points labeled $$a$$ and $$d$$ does not intersect the chord joining points labeled $$b$$ and $$c$$. Let $$M$$ be the number of beautiful labelings and let $$N$$ be the number of ordered pairs $$(x,y)$$ of positive integers such that $$x+y\leq n$$ and $$\mbox{gcd }(x,y)=1$$. Prove that $$M=N+1$$. 2005-2017 IMOmath.com \| imomath"at"gmail.com \| Math rendered by MathJax	2017-07-25 20:36:11	{"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": 0.8047477602958679, "perplexity": 74.34965991993133}, "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-2017-30/segments/1500549425381.3/warc/CC-MAIN-20170725202416-20170725222416-00648.warc.gz"}
http://byu.apmonitor.com/wiki/index.php/Main/OptionVdvl?action=diff&source=n&minor=y	Main ## Main.OptionVdvl History June 08, 2017, at 10:49 PM by 10.5.113.159 - Changed lines 5-6 from: to: June 01, 2017, at 06:39 AM by 45.56.3.173 - (:title VDVL - APMonitor Option:) (:keywords VDVL, Optimization, Estimation, Option, Configure, Default, Description:) (:description Delta validity limit:) Type: Floating Point, Input Default Value: 1.0e20 Description: Delta validity limit VDVL is the maximum change of a measured value before it is considered an unrealistic change. The change in measured values is recorded at every cycle of the application and compared to the VDVL limit. Validity limits are placed to catch instrument errors that may otherwise create bad inputs to the application. If a delta validity limit is exceeded, the action is to either freeze the measured value at the last good value (VLACTION=0) or change the measured value by a maximum of VDVL in the direction of the bad measurement (VLACTION=1). Another way to minimize the impact of unrealistic changes in measurements is to set FSTATUS between 0 and 1 with values closer to 0 becoming insensitive to measurement changes.	2018-05-26 00:09:08	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.6117925643920898, "perplexity": 4360.1691919076875}, "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-2018-22/segments/1526794867254.84/warc/CC-MAIN-20180525235049-20180526015049-00289.warc.gz"}
https://byjus.com/questions/what-are-the-two-types-of-consumer-motives/	# What Are the Two Types of Consumer Motives? The two types of consumer motives are: Stay connected with BYJU’S for more such questions and answers on various commerce topics.	2022-01-16 09:46:40	{"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": 0.8387859463691711, "perplexity": 3096.1255880222775}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320299852.23/warc/CC-MAIN-20220116093137-20220116123137-00281.warc.gz"}
https://opendatagroup.github.io/data%20science/2019/08/15/Machine-Learning-Model-Interpretation.html	# Machine Learning Model Interpretation on Thu, Aug 15, 2019 To either a model-driven company or a company catching up with the rapid adoption of AI in the industry, machine learning model interpretation has become a key factor that helps to make decisions towards promoting models into business. This is not an easy task – imagine trying to explain a mathematical theory to your parents. Yet business owners should always be curious about these models, and some questions easily raise: • How do machine learning models work? • How reliable are the predictions made from these models? • How carefully do we have to monitor these models? As people understand more and more about the need and importance of model interpretation, data scientists need to build up a model interpretation strategy at the beginning of model building process that helps them to answer the questions above, for many possible reasons such as auditing. Such a strategy requires a structure aligned with the scope for interpretability: • model algorithm and model hyperparameter, • the model training process, • model interpretability metrics, • human-friendly explanations. At this point, we are talking about a model as an asset in business. We need to understand that producing predictions is only the basics of a model. We also need to make sure that the results of the model can be understood in a business sense. A good model interpretation requires a thorough understanding of the model life cycle as well as a clear and intuitive demonstration in each step. ## Interpretable Models One of the easier ways to handle model interpretation is to start with an interpretable model, such as linear regression, logistic regression, decision tree, CNN, etc. High interpretability and efficiency are part of the main reasons that these models are widely used across different industries. They have a relative easy algorithm and training process, which are easy to monitor and adjust as a result. Feature importance is a very intuitive score that is implemented in many packages. A feature with higher importance not only contributes more to the prediction made by the model but also accounts for more in the prediction error. As a numerical score, importance can be used to create easy visualizations to present model training results. Feature importance not only provides a straightforward understanding of features but also provides a reliable metric for the dimension reduction process. • Filter methods select a subset of features and compare model performance, and they generate metrics like linear discriminant analysis, ANOVA, Chi-Square, etc. • Wrapper methods train a model using a selected subset of features, and feature importance serves as a criterion depending on the selection or elimination method used. ## Neural Networks People have seen how powerful some of the neural network models can be. Cool visualizations illustrating how these models work help a lot for people to understand them more. Examples include illustrating cues to find the “focus” of deep neural networks in image recognition. Because of the linear nature of some neural networks, it is not hard to trace back several layers of neurons and figure out where the model is focusing on when it is making a prediction. Source: Open AI However, even though sometimes a nice neural network demo looks nice and informative, it may not show you the feature importance information, i.e. which factors drive the model to make such a prediction. The classic example of tensorflow MNIST model using Convolutional Neural Network has an easy trace-back of which pixels are driving the prediction: Source: Not another MNIST tutorial with TensorFlow Recently, tensorflow also introduced a new tf-explain package that helps to generate such trace-back for image recognition as well. from tf_explain.callbacks.occlusion_sensitivity import OcclusionSensitivityCallback callbacks = [ validation_data=(x_val, y_val), layer_name="activation_1", class_index=0, output_dir=output_dir, ) ] model.fit(x_train, y_train, batch_size=2, epochs=2, callbacks=callbacks) !tensorboard --logdir logs Source: TensorFlow Explain We have to keep in mind that each example above uses methods designed for the specific type of models or problems. There are a lot of exploratory works and hyperparameter tuning to do before such a nice demonstration can be made. ## LIME and Shapley Values Over the years, data scientists and researchers have developed tools to help with model interpretation, and Local Interpretable Model-Agnostic Explanations (LIME) is one of the tools based on the paper “Why Should I Trust You?”: Explaining the Predictions of Any Classifier. The goal is obvious from the title of the paper. LIME does a good job of giving a meaningful estimate of feature importance for a given test data using the idea of Shapley Values, which is a game theory method of assigning weights to features depending on their contribution to the final prediction. However, in contrast to a global surrogate model, which is an approximated interpretable model of a black-box model, being local is one of the biggest disadvantages. Seeing LIME results for a particular data point, which could be quite an abnormal one, does not tell enough story of the whole model. The correct interpretation of “local” is hard to answer as well. To have a better understanding of the model, we would need to run tests on a large number of data points. To go from a local observation to a global picture could cost us a lot of computation power. Also, creating a local surrogate model works fine when the black box model consists of a smooth function, but not so well when the black box model contains many discrete (categorical) variables. ## Concluding Remarks Even with all these data science tools, model interpretability is still one of the main challenges in machine learning as our models are also growing in complexity. It is one of the most important attributes of a model asset from a business perspective. Researchers are trying their best to reveal the mysterious yet fascinating “black-box” of AI models. Only after we understand our AI models from a deeper level can we improve our AI models to a whole new level. ## Resources Interpretable Machine Learning is a very good online book about this topic by Christoph Molnar for people interested in more details.	2020-09-25 16:54:37	{"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": 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": 0.3442588150501251, "perplexity": 721.1591194251404}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-40/segments/1600400227524.63/warc/CC-MAIN-20200925150904-20200925180904-00638.warc.gz"}
https://www.coursehero.com/file/7460209/Suppose-the-population-proportion-of-passengers-who-carry-on/	hw.prop.CI.Hypo.pVal.STS.KEY # Suppose the population proportion of passengers who This preview shows page 1. Sign up to view the full content. This is the end of the preview. Sign up to access the rest of the document. Unformatted text preview: on board. Suppose the population proportion of passengers who carry on their bags is 0.70 . We intend to take a sample of 25. What is the question being asked of passengers? KEY: are you checking your bag? What is the symbol for the population proportion? KEY: p What is the symbol for the sample proportion? KEY: ^ p Show working in deciding whether it is OK to use the Normal distribution as an estimate. KEY: OK to use Z if np>5 and n(1-p)>5: 25(0.7)=17.5 > 5, 25(0.3)=7.5 > 5 so OK to use Z Compute the probability that the sample proportion will be greater than 0.55 . When you use a formula, show the raw formula with symbols (no numbers) first. Draw a graph and label everything. ^ p-p p(1- p) n 0.55 - 0.70 -0.15 0.55 - 0.70 z= = = 0.09 = Z = -1.67 0.09 0.70(0.30) 25 Area for Z=-1.67 is 0.4525 ^ P(p>0.55) = 0.4525 + 0.5000 = 0.9425 ^ KEY: P(p>0.55): Z = 6. The population proportion of airline passengers checking baggage is unknown. A sample of 25 showed that 16 intended to check their baggage. We'll construct a 95% confidence interval. 6a. This is a CI for what (symbol)? KEY: p p(1 - p) 6b. Write out the standard deviation (in symbols) we would prefer to use: KEY: n 6c. Show working in deciding whether it is OK to use the Normal distribution as an estimate. KEY: We do NOT have p, but must use ^ instead , and ^ = 16/25 = 0.64 p p ^>5 and n(1-p)>5: 25(0.64)=16 > 5, 25(0.36)=9 > 5 so OK to use Z ^ OK to use Z if np ^ (1 - ^) p p 6d. Write out the standard deviation (in symbols) that we must use KEY: n 6e. Construct the 95% confidence interval. First show f... View Full Document ## This note was uploaded on 01/27/2013 for the course STAT 385 taught by Professor Szatrowski during the Spring '08 term at Rutgers. Ask a homework question - tutors are online	2017-05-26 14:30:16	{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8989997506141663, "perplexity": 1432.955333187914}, "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-2017-22/segments/1495463608665.33/warc/CC-MAIN-20170526124958-20170526144958-00470.warc.gz"}
http://mathoverflow.net/revisions/3910/list	3 added 199 characters in body I am quite curious about the definition and applications of the primary decomposition for modules. 1. The definition of a primary submodule. (Let's assume we work over a commutative noetherian ring R $R$ and an R-module M) $R$-module $M$) When I first worked on Atiyah-Macdonald I used this definition: A submodule N $N$ of M $M$ is primary if any zero divisor on M/N $M/N$ is nilpotent. But recently I saw the definition in Matsumura's commutative algebra, which is slightly different: A submodule N $N$ of M $M$ is primary if any zero divisor on M/N $M/N$ is locally nilpotent, i.e. if a $a$ is a zero divisor, then for any $x \in M/NM/N$, there exists n $n$ possibliy depending on x $x$ such that $a^n x = 00$. Of course, these two definitions agree when M $M$ is a finite R-module. $R$-module. (which I guess is the most interesting case) But what should be the "right" definition in the general situation? 2 1. The application of this. Is this generality of any use? If M is finite, then I know it admits a filtration with quotients being R/p_i $R/{\mathfrak{p}_i}$ where p_i $\mathfrak{p}_i$ are associated primes. This seems to be quite useful in some proofs. But what about the case where M is infinite? 3 Geometric meaning. Primary decomposition of an ideal I in R is related to the irreducible components of Spec R/I. $\mathrm{Spec}(R/I)$. Is there something similar for the module case? Thanks very much! Edit: As there still does not seem to be a clear consensus of answers, it would be great if experts could weigh in. 2 retag 1 # Primary decomposition for modules I am quite curious about the definition and applications of the primary decomposition for modules. 1. The definition of a primary submodule. (Let's assume we work over a commutative noetherian ring R and an R-module M) When I first worked on Atiyah-Macdonald I used this definition: A submodule N of M is primary if any zero divisor on M/N is nilpotent. But recently I saw the definition in Matsumura's commutative algebra, which is slightly different: A submodule N of M is primary if any zero divisor on M/N is locally nilpotent, i.e. if a is a zero divisor, then for any x in M/N, there exists n possibliy depending on x such that a^n x = 0. Of course, these two definitions agree when M is a finite R-module. (which I guess is the most interesting case) But what should be the "right" definition in the general situation? 2 The application of this. Is this generality of any use? If M is finite, then I know it admits a filtration with quotients being R/p_i where p_i are associated primes. This seems to be quite useful in some proofs. But what about the case where M is infinite? 3 Geometric meaning. Primary decomposition of an ideal I in R is related to the irreducible components of Spec R/I. Is there something similar for the module case? Thanks very much!	2013-05-24 12:56:58	{"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": 0.7749160528182983, "perplexity": 316.1863629320272}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2013-20/segments/1368704658856/warc/CC-MAIN-20130516114418-00044-ip-10-60-113-184.ec2.internal.warc.gz"}
https://mathematica.stackexchange.com/questions/69567/cant-install-workbench-on-surface-pro-3?noredirect=1	# Can't install Workbench on Surface Pro 3 I've been trying to install Wolfram Workbench 2 for a few days on a Surface Pro 3 i5 256 Mb, Windows 8.1, Mathematica 10.0.1, Java 8, without success, I'm getting a bit desperate. For the Eclipse version I've tried Kepler and Luna. The strange thing is that it used to work on a similar configuration but with a Surface Pro 2. I've tried also the solution from Rolf Mertig on this site. Has somebody had this problem and managed to get it running ? The error I get is An internal error occurred during: "Launching test.nb (1)". at com.wolfram.eclipse.MEET.launch.MathematicaConsoleLaunch.launchOperation(MathematicaConsoleLaunch.java:374) at com.wolfram.eclipse.MEET.launch.MathematicaConsoleLaunch.launch(MathematicaConsoleLaunch.java:215) at org.eclipse.debug.internal.core.LaunchConfiguration.launch(LaunchConfiguration.java:858) at org.eclipse.debug.internal.core.LaunchConfiguration.launch(LaunchConfiguration.java:707) at org.eclipse.debug.internal.ui.DebugUIPlugin.buildAndLaunch(DebugUIPlugin.java:1018) at org.eclipse.debug.internal.ui.DebugUIPlugin\$8.run(DebugUIPlugin.java:1222) at org.eclipse.core.internal.jobs.Worker.run(Worker.java:53) • Were you using MMA 10.0.1 on your Surface Pro 2? If not I personally wouldn't bother with WB2 for MMA 10.0.x, as they don't seem to play nice together (at least for my purpose/syntax). Might be worth trying with MMA 9.x if you haven't already. – Gordon Coale Dec 23 '14 at 10:43 • Yes I was, and I'm using mm10's Association/Dataset, so I need mm10. – faysou Dec 23 '14 at 10:56 • @faysou you could try contacting support and asking for the Workbench 3.0 beta. This issue should be fixed in WB3.0. – Stefan R Dec 23 '14 at 21:38	2019-07-22 06:32:11	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.3177380859851837, "perplexity": 4203.2191908479}, "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/1563195527531.84/warc/CC-MAIN-20190722051628-20190722073628-00171.warc.gz"}
http://mymathforum.com/geometry/338015-vector-how-find-force.html	My Math Forum (Vector)How to find the force? User Name Remember Me? Password Geometry Geometry Math Forum December 11th, 2016, 06:27 AM #1 Newbie Joined: Dec 2016 From: HK Posts: 2 Thanks: 0 (Vector)How to find the force? Hi everyone, I need help~ how to find the force ? thank you very much~ December 11th, 2016, 06:47 AM #2 Newbie Joined: Dec 2016 From: Burswood Posts: 1 Thanks: 0 These formulas may help. Sent from my iPhone using Tapatalk December 11th, 2016, 10:04 AM #3 Math Team Joined: Jul 2011 From: Texas Posts: 2,207 Thanks: 1050 (i) $\hat{e} = \dfrac{\vec{e}}{\|\vec{e}\|}$ (ii) $\vec{F} = \|\vec{F}\|\hat{e}$ (iii) $\vec{r} = \vec{B} - \vec{A}$ (iv) work ... if $\vec{F} = ai+bj+ck$ and $\vec{r} = xi+yj+zk$, then $\vec{F} \cdot \vec{r} = ax + by + cz$ Thanks from topsquark December 11th, 2016, 11:36 PM #4 Newbie Joined: Dec 2016 From: HK Posts: 2 Thanks: 0 Quote: Originally Posted by skeeter (i) $\hat{e} = \dfrac{\vec{e}}{\|\vec{e}\|}$ (ii) $\vec{F} = \|\vec{F}\|\hat{e}$ (iii) $\vec{r} = \vec{B} - \vec{A}$ (iv) work ... if $\vec{F} = ai+bj+ck$ and $\vec{r} = xi+yj+zk$, then $\vec{F} \cdot \vec{r} = ax + by + cz$ thank you ~~ i got it Tags find, force, vectorhow Thread Tools Display Modes Linear Mode Similar Threads Thread Thread Starter Forum Replies Last Post leo255 Calculus 1 December 16th, 2014 08:48 PM girlbadatmath Physics 9 July 17th, 2014 01:24 AM OriaG Calculus 1 May 22nd, 2013 09:14 AM yoyosuper8 Calculus 6 April 26th, 2009 08:26 PM Contact - Home - Forums - Top	2017-02-22 19:38:14	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6703540682792664, "perplexity": 9938.333454649952}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-09/segments/1487501171043.28/warc/CC-MAIN-20170219104611-00599-ip-10-171-10-108.ec2.internal.warc.gz"}
http://gamr.shop/another-way-cia/fca-emergency-grants-covid-19-fund-a40e6a	The inverse exponent of the index number is equivalent to the radical itself. 4) You may add or subtract like radicals only Example More examples on how to Add Radical Expressions. For problems 5 â 7 evaluate the radical. CCSS.Math: HSN.CN.A.1. When radicals, itâs improper grammar to have a root on the bottom in a fraction â in the denominator. Therefore, we have â1 = 1, â4 = 2, â9= 3, etc. Just as the square root undoes squaring, so also the cube root undoes cubing, the fourth root undoes raising things to the fourth power, et cetera. The only difference is that this time around both of the radicals has binomial expressions. 7. To solve the equation properly (that is, algebraically), I'll start by squaring each side of the original equation: x â 1 â£ = x â 7. Email. Section 1-3 : Radicals. And take care to write neatly, because "katex.render("5\\,\\sqrt{3\\,}", rad017);" is not the same as "katex.render("\\sqrt[5]{3\\,}", rad018);". Watch how the next two problems are solved. Reminder: From earlier algebra, you will recall the difference of squares formula: In other words, we can use the fact that radicals can be manipulated similarly to powers: There are various ways I can approach this simplification. For example , given x + 2 = 5. I was using the "times" to help me keep things straight in my work. The approach is also to square both sides since the radicals are on one side, and simplify. Variables with exponents also count as perfect powers if the exponent is a multiple of the index. Google Classroom Facebook Twitter. The expression " katex.render("\\sqrt{9\\,}", rad001); " is read as "root nine", "radical nine", or "the square root of nine". Follow the same steps to solve these, but pay attention to a critical pointâsquare both sides of an equation, not individual terms. That is, by applying the opposite. The imaginary unit i. . Then: katex.render("\\sqrt{144\\,} = \\mathbf{\\color{purple}{ 12 }}", typed01);12. Example 1: $\sqrt{x} = 2$ (We solve this simply by raising to a power both sides, the power is equal to the index of a radical) $\sqrt{x} = 2 ^{2}$ $x = 4$ Example 2: $\sqrt{x + 2} = 4 /^{2}$ $\ x + 2 = 16$ $\ x = 14$ Example 3: $\frac{4}{\sqrt{x + 1}} = 5, x \neq 1$ Again, here you need to watch out for that variable $x$, he canât be ($-1)$ because if he could be, weâd be dividing by $0$. Is the 5 included in the square root, or not? There are certain rules that you follow when you simplify expressions in math. I used regular formatting for my hand-in answer. The radical symbol is used to write the most common radical expression the square root. 4â81 81 4 Solution. Algebra radicals lessons with lots of worked examples and practice problems. Sometimes, we may want to simplify the radicals. All that you have to do is simplify the radical like normal and, at the end, multiply the coefficient by any numbers that 'got out' of the square root. Not only is "katex.render("\\sqrt{3}5", rad014);" non-standard, it is very hard to read, especially when hand-written. Neither of 24 and 6 is a square, but what happens if I multiply them inside one radical? For example Division of Radicals (Rationalizing the Denominator) This process is also called "rationalising the denominator" since we remove all irrational numbers in the denominator of the fraction. When writing an expression containing radicals, it is proper form to put the radical at the end of the expression. In this section we will define radical notation and relate radicals to rational exponents. There is no nice neat number that squares to 3, so katex.render("\\sqrt{3\\,}", rad03B); cannot be simplified as a nice whole number. A radical. Then my answer is: This answer is pronounced as "five, times root three", "five, times the square root of three", or, most commonly, just "five, root three". I could continue factoring, but I know that 9 and 100 are squares, while 5 isn't, so I've gone as far as I need to. Examples of radicals include (square root of 4), which equals 2 because 2 x 2 = 4, and (cube root of 8), which also equals 2 because 2 x 2 x 2 = 8. Constructive Media, LLC. Rules for Radicals. ( x â 1 â£) 2 = ( x â 7) 2. x + 2 = 5. x = 5 â 2. x = 3. Some radicals do not have exact values. Web Design by. 3) Quotient (Division) formula of radicals with equal indices is given by More examples on how to Divide Radical Expressions. =xâ7. Intro to the imaginary numbers. Some radicals have exact values. The most common type of radical that you'll use in geometry is the square root. One would be by factoring and then taking two different square roots. open radical â © close radical â ¬ â radical sign without vinculum â â © Explanation. In other words, since 2 squared is 4, radical 4 is 2. Oftentimes the argument of a radical is not a perfect square, but it may "contain" a square amongst its factors. The simplest case is when the radicand is a perfect power, meaning that itâs equal to the nth power of a whole number. If the radicand is 1, then the answer will be 1, no matter what the root is. $\ 4 = 5\sqrt{x + 1}$ $\ 5\sqrt{x + 1} = 4 /: 5$ \$\sqrt{x + 1} = \frac{4}{5â¦ This is the currently selected item. To simplify this sort of radical, we need to factor the argument (that is, factor whatever is inside the radical symbol) and "take out" one copy of anything that is a square. Therefore we can write. Property 3 : If we have radical with the index "n", the reciprocal of "n", (That is, 1/n) can be written as exponent. You can solve it by undoing the addition of 2. In the second case, we're looking for any and all values what will make the original equation true. Since I have two copies of 5, I can take 5 out front. Radicals are the undoing of exponents. (Technically, just the "check mark" part of the symbol is the radical; the line across the top is called the "vinculum".) âw2v3 w 2 v 3 Solution. (In our case here, it's not.). In the same way, we can take the cube root of a number, the fourth root, the 100th root, and so forth. The radical of a radical can be calculated by multiplying the indexes, and placing the radicand under the appropriate radical sign. In math, a radical is the root of a number. Learn about radicals using our free math solver with step-by-step solutions. In elementary algebra, the quadratic formula is a formula that provides the solution(s) to a quadratic equation. . In the first case, we're simplifying to find the one defined value for an expression. Perhaps because most of radicals you will see will be square roots, the index is not included on square roots. How to Simplify Radicals with Coefficients. Dr. Ron Licht 2 www.structuredindependentlearning.com L1â5 Mixed and entire radicals. 35 5 7 5 7 . But we need to perform the second application of squaring to fully get rid of the square root symbol. This is because 1 times itself is always 1. â¦ You don't want your handwriting to cause the reader to think you mean something other than what you'd intended. You can accept or reject cookies on our website by clicking one of the buttons below. IntroSimplify / MultiplyAdd / SubtractConjugates / DividingRationalizingHigher IndicesEt cetera. \small { \sqrt {x - 1\phantom {\big\|}} = x - 7 } xâ1â£â£â£. Now I do have something with squares in it, so I can simplify as before: The argument of this radical, 75, factors as: This factorization gives me two copies of the factor 5, but only one copy of the factor 3. When doing this, it can be helpful to use the fact that we can switch between the multiplication of roots and the root of a multiplication. URL: https://www.purplemath.com/modules/radicals.htm, Page 1Page 2Page 3Page 4Page 5Page 6Page 7, © 2020 Purplemath. Math Worksheets What are radicals? Pre-Algebra > Intro to Radicals > Rules for Radicals Page 1 of 3. How to simplify radicals? The number under the root symbol is called radicand. Examples of Radical, , etc. Property 2 : Whenever we have two or more radical terms which are dividing with same index, then we can put only one radical and divide the terms inside the radical. You could put a "times" symbol between the two radicals, but this isn't standard. Solve Practice Download. In particular, I'll start by factoring the argument, 144, into a product of squares: Each of 9 and 16 is a square, so each of these can have its square root pulled out of the radical. a square (second) root is written as: katex.render("\\sqrt{\\color{white}{..}\\,}", rad17A); a cube (third) root is written as: katex.render("\\sqrt[{\\scriptstyle 3}]{\\color{white}{..}\\,}", rad16); a fourth root is written as: katex.render("\\sqrt[{\\scriptstyle 4}]{\\color{white}{..}\\,}", rad18); a fifth root is written as: katex.render("\\sqrt[{\\scriptstyle 5}]{\\color{white}{..}\\,}", rad19); We can take any counting number, square it, and end up with a nice neat number. We can deal with katex.render("\\sqrt{3\\,}", rad03C); in either of two ways: If we are doing a word problem and are trying to find, say, the rate of speed, then we would grab our calculators and find the decimal approximation of katex.render("\\sqrt{3\\,}", rad03D);: Then we'd round the above value to an appropriate number of decimal places and use a real-world unit or label, like "1.7 ft/sec". can be multiplied like other quantities. are some of the examples of radical. Radical equationsare equations in which the unknown is inside a radical. The radical can be any root, maybe square root, cube root. Rationalizing Radicals. In the example above, only the variable x was underneath the radical. Lesson 6.5: Radicals Symbols. In general, if aand bare real numbers and nis a natural number, n n n n nab a b a b . is the indicated root of a quantity. You don't have to factor the radicand all the way down to prime numbers when simplifying. Since most of what you'll be dealing with will be square roots (that is, second roots), most of this lesson will deal with them specifically. For example, in the equation âx = 4, the radical is canceled out by raising both sides to the second power: (âx) 2 = (4) 2 or x = 16. I'm ready to evaluate the square root: Yes, I used "times" in my work above. For instance, relating cubing and cube-rooting, we have: The "3" in the radical above is called the "index" of the radical (the plural being "indices", pronounced "INN-duh-seez"); the "64" is "the argument of the radical", also called "the radicand". (Other roots, such as –2, can be defined using graduate-school topics like "complex analysis" and "branch functions", but you won't need that for years, if ever.). 8+9) â 5 = â (25) â 5 = 5 â 5 = 0. Rationalizing Denominators with Radicals Cruncher. For example the perfect squares are: 1, 4, 9, 16, 25, 36, etc., because 1 = 12, 4 = 22, 9 = 32, 16 = 42, 25 = 52, 36 = 62, and so on. 5) You may rewrite expressions without radicals (to rationalize denominators) as follows A) Example 1: B) Example 2: For instance, if we square 2 , we get 4 , and if we "take the square root of 4 ", we get 2 ; if we square 3 , we get 9 , and if we "take the square root of 9 ", we get 3 . All Rights Reserved. Solve Practice. If you believe that your own copyrighted content is on our Site without your permission, please follow this Copyright Infringement Notice procedure. The radical sign, , is used to indicate âthe rootâ of the number beneath it. (a) 2â7 â 5â7 + â7 Answer (b) 65+465â265\displaystyle{\sqrt[{{5}}]{{6}}}+{4}{\sqrt[{{5}}]{{6}}}-{2}{\sqrt[{{5}}]{{6}}}56â+456ââ256â Answer (c) 5+23â55\displaystyle\sqrt{{5}}+{2}\sqrt{{3}}-{5}\sqrt{{5}}5â+23ââ55â Answer Property 1 : Whenever we have two or more radical terms which are multiplied with same index, then we can put only one radical and multiply the terms inside the radical. But the process doesn't always work nicely when going backwards. For example . Download the free radicals worksheet and solve the radicals. On the other hand, we may be solving a plain old math exercise, something having no "practical" application. In the opposite sense, if the index is the same for both radicals, we can combine two radicals into one radical. On a side note, let me emphasize that "evaluating" an expression (to find its one value) and "solving" an equation (to find its one or more, or no, solutions) are two very different things. All right reserved. For example . Then they would almost certainly want us to give the "exact" value, so we'd write our answer as being simply "katex.render("\\sqrt{3\\,}", rad03E);". Free math problem solver answers your algebra, geometry, trigonometry, calculus, and statistics homework questions with step-by-step explanations, just like a math tutor. Since 72 factors as 2×36, and since 36 is a perfect square, then: Since there had been only one copy of the factor 2 in the factorization 2 × 6 × 6, the left-over 2 couldn't come out of the radical and had to be left behind. If the radical sign has no number written in its leading crook (like this , indicating cube root), then it â¦ For example, the fraction 4/8 isn't considered simplified because 4 and 8 both have a common factor of 4. And also, whenever we have exponent to the exponent, we can multiplâ¦ The expression is read as "a radical n" or "the n th root of a" The expression is read as "ath root of b raised to the c power. In mathematics, an expression containing the radical symbol is known as a radical expression. Before we work example, letâs talk about rationalizing radical fractions. \small { \left (\sqrt {x - 1\phantom {\big\|}}\right)^2 = (x - 7)^2 } ( xâ1â£â£â£. There are other ways of solving a quadratic equation instead of using the quadratic formula, such as factoring (direct factoring, grouping, AC method), completing the square, graphing and others. When doing your work, use whatever notation works well for you. So, , and so on. For instance, consider katex.render("\\sqrt{3\\,}", rad03A);, the square root of three. Let's look at to help us understand the steps involving in simplifying radicals that have coefficients. For instance, [cube root of the square root of 64]= [sixth roâ¦ 3âx2 x 2 3 Solution. To understand more about how we and our advertising partners use cookies or to change your preference and browser settings, please see our Global Privacy Policy. Radicals can be eliminated from equations using the exponent version of the index number. Khan Academy is a 501(c)(3) nonprofit organization. More About Radical. "Roots" (or "radicals") are the "opposite" operation of applying exponents; we can "undo" a power with a radical, and we can "undo" a radical with a power. 3ââ512 â 512 3 Solution. These worksheets will help you improve your radical solving skills before you do any sort of operations on radicals like addition, subtraction, multiplication or division. For instance, 4 is the square of 2, so the square root of 4 contains two copies of the factor 2; thus, we can take a 2 out front, leaving nothing (but an understood 1) inside the radical, which we then drop: Similarly, 49 is the square of 7, so it contains two copies of the factor 7: And 225 is the square of 15, so it contains two copies of the factor 15, so: Note that the value of the simplified radical is positive. For example, which is equal to 3 × 5 = ×. We use first party cookies on our website to enhance your browsing experience, and third party cookies to provide advertising that may be of interest to you. That is, we find anything of which we've got a pair inside the radical, and we move one copy of it out front. The product of two radicals with same index n can be found by multiplying the radicands and placing the result under the same radical. Basic Radicals Math Worksheets. 7ây y 7 Solution. Intro to the imaginary numbers. Similarly, the multiplication n 1/3 with y 1/2 is written as h 1/3 y 1/2. 4 4 49 11 9 11 994 . 6âab a b 6 Solution. is also written as But my steps above show how you can switch back and forth between the different formats (multiplication inside one radical, versus multiplication of two radicals) to help in the simplification process. For problems 1 â 4 write the expression in exponential form. Free math problem solver answers your algebra, geometry, trigonometry, calculus, and statistics homework questions with step-by-step explanations, just like a math tutor. In math, sometimes we have to worry about âproper grammarâ. While either of +2 and –2 might have been squared to get 4, "the square root of four" is defined to be only the positive option, +2. Microsoft Math Solver. But when we are just simplifying the expression katex.render("\\sqrt{4\\,}", rad007A);, the ONLY answer is "2"; this positive result is called the "principal" root. Rejecting cookies may impair some of our website’s functionality. The radical sign is the symbol . Here's the rule for multiplying radicals: * Note that the types of root, n, have to match! Radicals and rational exponents â Harder example Our mission is to provide a free, world-class education to anyone, anywhere. Generally, you solve equations by isolating the variable by undoing what has been done to it. Radicals quantities such as square, square roots, cube root etc. In mathematical notation, the previous sentence means the following: The " katex.render("\\sqrt{\\color{white}{..}\\,}", rad17); " symbol used above is called the "radical"symbol. Perfect cubes include: 1, 8, 27, 64, etc. As soon as you see that you have a pair of factors or a perfect square, and that whatever remains will have nothing that can be pulled out of the radical, you've gone far enough. For example. © 2019 Coolmath.com LLC. For example, the multiplication of âa with âb, is written as âa x âb. For instance, x2 is a â¦ We will also give the properties of radicals and some of the common mistakes students often make with radicals. Sometimes you will need to solve an equation that contains multiple terms underneath a radical. This tucked-in number corresponds to the root that you're taking. We can raise numbers to powers other than just 2; we can cube things (being raising things to the third power, or "to the power 3"), raise them to the fourth power (or "to the power 4"), raise them to the 100th power, and so forth. You probably already knew that 122 = 144, so obviously the square root of 144 must be 12. For example, â9 is the same as 9 1/2. Practice solving radicals with these basic radicals worksheets. For instance, if we square 2, we get 4, and if we "take the square root of 4", we get 2; if we square 3, we get 9, and if we "take the square root of 9", we get 3. The multiplication of radicals involves writing factors of one another with or without multiplication sign between quantities. That is, the definition of the square root says that the square root will spit out only the positive root. Very easy to understand! Rejecting cookies may impair some of our website’s functionality. Since I have only the one copy of 3, it'll have to stay behind in the radical. We will also define simplified radical form and show how to rationalize the denominator. So, for instance, when we solve the equation x2 = 4, we are trying to find all possible values that might have been squared to get 4. Here are a few examples of multiplying radicals: Pop these into your calculator to check! That one worked perfectly. Another way to do the above simplification would be to remember our squares. The square root of 9 is 3 and the square root of 16 is 4. While " katex.render("\\sqrt[2]{\\color{white}{..}\\,}", rad003); " would be technically correct, I've never seen it used. To simplify a term containing a square root, we "take out" anything that is a "perfect square"; that is, we factor inside the radical symbol and then we take out in front of that symbol anything that has two copies of the same factor. No, you wouldn't include a "times" symbol in the final answer. In case you're wondering, products of radicals are customarily written as shown above, using "multiplication by juxtaposition", meaning "they're put right next to one another, which we're using to mean that they're multiplied against each other". For example, -3 * -3 * -3 = -27. This is important later when we come across Complex Numbers. Learn about the imaginary unit i, about the imaginary numbers, and about square roots of negative numbers. To indicate some root other than a square root when writing, we use the same radical symbol as for the square root, but we insert a number into the front of the radical, writing the number small and tucking it into the "check mark" part of the radical symbol. Similarly, radicals with the same index sign can be divided by placing the quotient of the radicands under the same radical, then taking the appropriate root. Sometimes radical expressions can be simplified. This problem is very similar to example 4. 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Understand the steps involving in simplifying radicals that have Coefficients 's not. ) â write!	2021-06-24 08:49:21	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8152711391448975, "perplexity": 755.9699573158882}, "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/1623488552937.93/warc/CC-MAIN-20210624075940-20210624105940-00376.warc.gz"}
https://www.physicsforums.com/threads/quantum-mechanics-problem-quick-help-kindly-requested.201083/	# Quantum Mechanics Problem: Quick help kindly requested 1. Nov 28, 2007 ### bktennis2006 1. The problem statement, all variables and given/known data 1. Suppose we have a potential V(x) which is zero everywhere except 0<x<a, where it is equal to –Vo, where Vo is positive (and thus –Vo is negative). For the case where E > 0, it asks a. to give the form of the wavefunctions in all three regions, defining k1, k2 and k3 properly. b. for scattering off the potential by particles coming in from the left, determine what wavefunction amplitudes can be set to zero. c. derive appropriate boundary conditions relating wave function amplitudes at the boundaries between regions. 2. It then asks to answer the same question using the case where E < 0, and also for b. it asks to determine which wavefunctions should be set to zero for bound states. 3. Solve explicitly for the transmission coefficient in Problem 1, and in what cases is the coefficient exactly 1? 2. Relevant equations The time-independent Shrodinger equation, of course. 3. The attempt at a solution Unfortunately, I was only able to find the wavefunctions in #1, and I don't even know if they're right. I did it by solving for $$\psi$$ using the Shrodinger equation, but does it also want $$\Psi$$? The second part of both #1 and #2 confuse me, as I don't even know what they're asking for. Any help would be greatly appreciated, but unfortunately the problems are due today, so a quick response would be greatly appreciated...thanks! 2. Nov 28, 2007 ### robphy So, what did you get for #1? Once you have that... what does it mean to have "particles coming in from the left"?	2018-01-18 16:49:35	{"extraction_info": {"found_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": 0.7148897647857666, "perplexity": 581.6279401876802}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-05/segments/1516084887423.43/warc/CC-MAIN-20180118151122-20180118171122-00617.warc.gz"}
http://www.gamedev.net/topic/645539-issue-with-memory-mapped-files/	• Create Account # Issue with memory mapped files Old topic! Guest, the last post of this topic is over 60 days old and at this point you may not reply in this topic. If you wish to continue this conversation start a new topic. 10 replies to this topic ### #1Chris_F Members - Reputation: 2207 Like 1Likes Like Posted 16 July 2013 - 04:47 PM I'm trying to work with the memory mapped file API in Windows, and am having some pretty serious issues. I'm trying to generate a 16GB file so I started with a little test to see how things would go. Here is the code: HANDLE file = CreateFile("output.raw", GENERIC_READ \| GENERIC_WRITE, 0, NULL, CREATE_NEW, FILE_ATTRIBUTE_NORMAL, NULL); HANDLE filemap = CreateFileMapping(file, NULL, PAGE_READWRITE, 0x4, 0, NULL); unsigned char* data = (unsigned char)MapViewOfFile(filemap, FILE_MAP_ALL_ACCESS, 0, 0, 0); As far as I know I set this up right. This should create a 16GB file and give me a pointer to the full 16GB of data. To make sure it works I output a magic number every 4 million bytes or so, for the entire length of the file, and then I opened it up in a hex editor to make sure it had worked, which it had. Then I tried writing a gigabytes worth of data, again spread out over the entire 16GB. My computer promptly froze and 10 minutes later, unable to move my mouse or get any other kind of feedback, I pulled the plug. What am I doing wrong? ### #2Chris_F Members - Reputation: 2207 Like 0Likes Like Posted 16 July 2013 - 05:40 PM OK, so I just noticed that if I right 4GB of data to the first 4GB of the file, my memory usage jumps 4GB in task manager. It doesn't say that my program is using the 4GB of memory, but presumably the OS is using it to cache the data while it writes it to disk? How the heck do I stop this? The whole reason I'm even trying to use this is because I need to work with an array that is much larger than my memory. ### #3Khatharr Crossbones+ - Reputation: 2939 Like 0Likes Like Posted 16 July 2013 - 05:56 PM There are limits to how much you can have paged at one time. You need to use 'views' of the map to isolate the section that you're wanting to work with. It's described in the MSDN docs on the subject. Please check those out and keep us posted. Edited by Khatharr, 16 July 2013 - 05:58 PM. void hurrrrrrrr() {__asm sub [ebp+4],5;} There are ten kinds of people in this world: those who understand binary and those who don't. ### #4Chris_F Members - Reputation: 2207 Like 0Likes Like Posted 16 July 2013 - 06:03 PM There are limits to how much you can have paged at one time. You need to use 'views' of the map to isolate the section that you're wanting to work with. It's described in the MSDN docs on the subject. Please check those out and keep us posted. I need to be able to address the entire 16GB (very random access pattern) and I don't understand why I shouldn't be able too on a computer that has an address space that is 16 exabytes. ### #5Bacterius Crossbones+ - Reputation: 8324 Like 1Likes Like Posted 16 July 2013 - 09:36 PM Have you tried passing the FILE_FLAG_RANDOM_ACCESS flag to CreateFile? It advises the OS you will be doing random access on that file so it may reclaim the virtual memory pages more quickly. Ideal behaviour in your case would be caching the last few megabytes read in case you suddenly start reading sequentially, but no further to avoid exhausting physical memory, and I would expect the OS to do just that. So, yeah, try the flag to see if it does anything. Either way Windows isn't going to be running a statistical analysis on your memory access patterns to figure out how best to manage your huge mapping, though I admit it is kind of dumb just letting the system run out of memory like that. The slowsort algorithm is a perfect illustration of the multiply and surrender paradigm, which is perhaps the single most important paradigm in the development of reluctant algorithms. The basic multiply and surrender strategy consists in replacing the problem at hand by two or more subproblems, each slightly simpler than the original, and continue multiplying subproblems and subsubproblems recursively in this fashion as long as possible. At some point the subproblems will all become so simple that their solution can no longer be postponed, and we will have to surrender. Experience shows that, in most cases, by the time this point is reached the total work will be substantially higher than what could have been wasted by a more direct approach. - Pessimal Algorithms and Simplexity Analysis ### #6achild Crossbones+ - Reputation: 1631 Like 0Likes Like Posted 16 July 2013 - 10:46 PM If it's going to be entirely random and your computer has less than 20 GB of RAM, this is going to be painful no matter what. Does this at least target SSD drives? Or is it targeting normal hard disk drives? The best flags for performance will be quite different between these two... Also if it is so random, can you do this on multiple threads at once? Or does the algorithm/whatever require serial processing? Why not just say it all huh? Okay, so worst case scenario is that you are using a normal hdd. In this case, you have two options. 1. Use default options, hint that it will use random access. Typically, windows is going to keep caching as more and more pages of memory get read (since you are not releasing anything and have to map it all at once). It is possible the random access flag will help this somewhat, but don't count on much here (I'd love to be wrong). You will typically get decent performance until memory fills up, followed by a very long time of "frozen" computer. Windows is simply paging a bunch of stuff in and out of memory. No need to pull the plug usually. Just go eat some lunch. Possibly take a nap. Seriously. 2. For consistent time without the crazy hiccups, use the write-through flag. Now you can skip the cache. There may be another flag... perhaps direct buffer when you open the file or something. I can look tomorrow at our code and see. Single threaded for both of these options is the way to go. Perhaps it is best to put all your I/O on its own single thread though... On the other end of the spectrum would be a computer with 20+ GB RAM, and 2-4 SSDs in a RAID 0 configuration (sure it's slightly risky but not nearly as flimsy as the internet would have you believe) 1. No need for write-through, it will actually slow you down. 2-4 threads writing is the sweet spot. Test this though. All our tests were for maximum sequential throughput in high speed video acquisition. Your usage is very different. There's really not much more to say here... you'll notice 2 different limits at different stages of testing - 1 your RAM bus speed, and 2 your total SSD(s) speed. Again - writing such small bits randomly will probably prove this entirely wrong. Remember you'll actually be reading/writing chunks based on 1 memory page size and 2 sector size - not much difference between an 8 byte and 4096 byte transaction. I'm guessing your configuration is with a normal hdd given your frozen computer. I'd say shoot for something in between these 2 situations. With any luck you can at least switch it with a known stable SSD. You'll see a dramatic difference. Is it okay to ask what the application for this is? Edited by achild, 16 July 2013 - 11:04 PM. ### #7RIGHT_THEN Members - Reputation: 109 Like 0Likes Like Posted 17 July 2013 - 12:20 AM Chirs_F I am not an expert but with such large RAM. the file is still not read at once. then could it be a paging problem on the system? the ram could be occupied with other processes on the system and therefore the file may not have adequate space as desired. and maximum size allowed on your system maybe less than the demand your applications are putting on the system. if at all it has anything to do with paging then at least on windows 7 this is i am sure you must know about paging but just for the benefit of doubt paging ->( when computer uses hard disk space to assist RAM because of the lack of it. ) i am sorry if my opinion is not related to your problem Thanks Regards ### #8samoth Crossbones+ - Reputation: 4660 Like 0Likes Like Posted 17 July 2013 - 01:38 AM There are several things you need to consider. If your goal is mainly to create a file of this length, then it is not necessary to map the view. Creating the file mapping (assuming proper access rights and protection mode) will create a file to the desired length already. Mapping the view is another story. As you've pointed out, your computer should be able to address 16 exabytes. Yes, but only if your program is compiled as 64-bit binary (on a 64-bit computer running a 64-bit operating system). You didn't specify what compiler you're using, but let's assume that -- like most people -- you use a 32-bit compiler, mapping such a large memory area will simply fail, even if your OS is 64 bits. Note that creating a mapping also consumes considerable amounts of memory and is non-neglegible work, so one wouldn't want to actually map such huge amounts of memory unless really necessary. 16 GiB of memory corresponds to somewhat over 4 million pages table entries, which consume 64 MiB of physical memory. Mapping 16 GiB when your address space allows for it but you do not have the physical RAM (you'll probably need upwards of 20 GiB) is yet another story, and of course there's the working set size limits. The maximum working set is limited to ridiculously small sizes by default, unless you change it. Which means that pages will be added to and removed from your working set all the time with such a huge dataset. This means literally millions of page faults (although they are "soft" faults as long as the OS does not run low on zero pages) Lastly, the amount of zero memory pages is not unlimited. Normally you never notice that, because the idle task always clears unused pages and you normally don't consume that many, so there's always spare ones. However, asking for 16 GiB in one go (and also touching it!) may make you feel it. The OS is required to zero all pages, both in the file on the disk and pages that are mapped into your address space. Since both are "opaque" to your application, the OS cheats as much as it can to hide the overhead. It will for example allocate a file without initializing it and "remember" that anything that is to be read from those sectors is "zero". The same goes for the pages in your address space, it merely "remembers" that these pages are new pages that you've never accessed. However, eventually, the OS has to write all those zero sectors to disk. Also, eventually, it has to provide a phyically present zeroed memory page (at the very least, the moment you try to access one -- but possibly sooner). At some point, the OS cannot cheat any more, but has to do the actual work. This is when you start to feel it. Unluckily, Windows is not particularly intelligent when it comes to dirty page writeback either (though I don't know if other operating systems are much smarter). I've experienced this when I wrote a quick-and-dirty free-sector-eraser for my wife's computer (for doing a "kind of security wipe" when getting a computer upgrade with the requirement to turn the old machine in, but with a still intact Windows installation -- so the plan was to first delete the documents folder using Explorer, and then overwrite the now free sectors on the disk 5-6 times simply by writing huge files until the disk was full, thus overwriting all available disk space with random values). My little tool would map a 1 GiB file, fill the memory with random, close the mapping, and create/map the next 1 GiB file. The idea was that this memory mapping was probably the most efficient way of writing huge amounts of "data" to the disk. Pages that you unmap become "free" as soon as they are no longer "dirty". New zero pages that you allocate do not need to physically exist unless you touch them, at which point they fault. So, yes, there is a bit of racing for physical pages, but the worst thing to happen is the writer thread is stalled by a page fault which needs to wait for a page to become free and zeroed. That's OK, because all we want is the disk to keep writing with maximum possible speed, we don't actually care about the application's performance. This theory proved true until the machine ran out of physical memory. At that time, Windows started copying dirty pages which it should just flush to disk and be done with them to the page file (no, I'm not joking!) to make phyical pages available for newly touched pages, then page the dirty pages in again (paging out the now freshly touched pages), and finally write them to disk. Result: Writing at ca. 110 MiB/s the first 2 seconds (pretty much the disk drive's theoretical maximum), then drop to ca. 2 MiB/s. Now, for the funny part, the "fix" was simply inserting a Sleep(5000); after unmapping each file (cough*). This gave the disk just about enough time to flush enough pages to disk and throw them away before the application was asking for a new pages. What a crap solution, but it worked... ### #9Chris_F Members - Reputation: 2207 Like 0Likes Like Posted 17 July 2013 - 12:31 PM Is it okay to ask what the application for this is? Nothing serious. I was just trying to render a large (128K x 128K) Sacks spiral. I was being lazy about it so I though if I simply memory mapped the entire file, I could render it naively and not have to waste any thought optimizing the problem. I knew it would be show as heck with a mechanical hard drive, but I though Windows would handle it more gracefully, i.e. not grind to a halt. ### #10achild Crossbones+ - Reputation: 1631 Like 0Likes Like Posted 17 July 2013 - 12:58 PM I see. Well, best suggestion is to not use any file flags such as random_access. To keep the system from hanging, you will want to judiciously use FlushViewOfFile followed by UnmapViewOfFile. On the other hand, that will be very slow if you do it every time you plot a pixel. Hmmm... It would possibly be much, much, much faster to map, say, 1 or 2 GB at a time. Go through all the values you are plotting, and if they are within the current "mapped" region of the output image, plot them. Otherwise ignore. Then do the next 1 or 2 GB, and go through all values again. The calculations are going to be tremendously quicker than mapping and flushing+unmapping 65kb pages every time you access a pixel. ### #11Waterlimon Crossbones+ - Reputation: 2441 Like 0Likes Like Posted 17 July 2013 - 02:06 PM I guess you could have a list of buckets representing mappable areas of the files. Then, when you create your plottable values, you throw them in the appropriate bucket (what portion of the file they would be based on x,y), including x,y,value. After you have a lot of them, you go through the buckets one by one: 1.Load chunk of file corresponding to bucket 2.Write all the values to that file chunk (using the x,y coord you stored along with the value) 4.Repeat for next bucket This would make the writing to file more sequential instead of having to repeatedly load and unload different parts which would happen with completely random access. o3o Old topic! Guest, the last post of this topic is over 60 days old and at this point you may not reply in this topic. If you wish to continue this conversation start a new topic. PARTNERS	2014-09-01 11:34:37	{"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": 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": 0.20396706461906433, "perplexity": 1269.64283806927}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-35/segments/1409535917663.12/warc/CC-MAIN-20140901014517-00393-ip-10-180-136-8.ec2.internal.warc.gz"}
https://en.m.wikisource.org/wiki/Royal_Naval_Biography/Inglefield,_John_Nicholson	# Royal Naval Biography/Inglefield, John Nicholson JOHN N. INGLEFIELD, Esq [Retired Captain.] This officer attained the rank of Lieutenant about the year 1768; and served as such in the Courageux and the Robust 74’s, commanded by the brothers Captains Samuel and Alexander Hood, both of whom were afterwards advanced to the peerage. He was made a Post-Captain October 11, 1780; and soon after appointed to the command of the Barfleur, a second rate, bearing the flag of Rear-Admiral Sir Samuel Hood, whom he accompanied to the West Indies, with a reinforcement for the squadron in that quarter, under the orders of Sir George B. Rodney. On the 29th April, 1781, Sir Samuel Hood having been detached with eighteen sail of the line to cruise off Martinique, fell in and had a partial action with the French fleet under the Count de Grasse, consisting of twenty-four ships of the line and two of 50 guns. In this affair the Barfleur had 5 men killed. The total loss sustained by the British was 41 slain and 130 wounded. Amongst the former were Captain Nott, of the Centaur, and Mr. Plowden, his first Lieutenant, two brave and excellent officers. The next day, the enemy’s van and centre being at some distance from their rear, Sir Samuel Hood, by a bold manoeuvre, notwithstanding his inferiority, attempted to cut them off; but having failed in this resolute enterprise, he was obliged, from the bad condition of many of his ships, to bear away for Antigua. On the 31st July following, he assumed the chief command of the fleet, Sir George B. Rodney having sailed for England. In the ensuing month Captain Inglefield, who had been removed into the Centaur on the death of her late commander, accompanied Sir Samuel Hood to the coast of America, in pursuit of M. de Grasse. He returned to Barbadoes with the same officer, after the surrender of Earl Cornwallis to the combined armies of France and America[1]. During the subsequent operations at the island of St. Christopher, Captain Inglefield was several times sent with flags of truce to the Marquis de Bouille and the Count de Grasse. He was also employed in the hazardous service of establishing signals between the fleet and the garrison of Brimstone Hill, by means of personal communication with the officer commanding there[2]. The melancholy fate, of the Centaur was still more deplorable. The squall had laid her so much on her beam ends, that the water burst through from the hold between decks; she lay motionless, and seemed irrecoverably overset. Her masts falling overboard, she in some degree righted, with the loss of her rudder, and such extreme violence as caused unspeakable mischief and confusion. The guns broke loose, the shot were thrown out of the lockers, and the water that came from the hold swept away every thing between decks, as effectually as the waves had from the upper. The officers, when the ship overset, ran up from their beds naked; neither could they get at a single article of clothes to put on in the morning, nor receive any assistance from those who were upon deck, they themselves having no other but what they had on. On the 16th day after their departure from the ship, the last ration of bread and water was distributed, and all hope vanished. The Almighty, however, who had conducted these unfortunate people through so many perils, still favored them with his divine protection; and on the same day, to their inexpressible joy, land was discovered, for which they instantly steered, and before night arrived safely in the harbour of Fayal, where they met with every humane attention, and from whence they soon after proceeded to England[5]. On the 25th Jan. 1783, Captain Inglefield and the other survivors of the Centaur, were tried by a court-martial at Portsmouth, for the loss of that vessel, and fully acquitted of all blame on account thereof[6]. Immediately after his trial, Captain Inglefield was appointed to the Scipio of 64 guns, stationed as a guard-ship in the river Medway. His next appointment was in the autumn of 1788, to the Adventure of 44 guns; in which ship he went to the coast of Africa, and returned from thence in Aug. 1789. He afterwards made three successive voyages to the same station, in the Medusa of 50 guns. The Medusa, coming up Channel in Sept. 1792, passed the frigate in which our late venerable monarch was making his usual marine excursion from Weymouth. After saluting the royal standard, Captain Inglefield followed her to the anchorage, and on the following morning was received by the King with marked distinction and approbation, and honored with a long conference on the esplanade. On the Medusa’s arrival at Chatham, she was ordered to be put out of commission; and Captain Inglefield soon after obtained the command of l’Aigle frigate, in which ship we find him serving at the reduction of Corsica, under the orders of Lord Hood, by whom he was appointed, conjointly with Vice-Admiral Goodall, Captain James Young, and his Lordship’s Secretary, Mr. M‘Arthur, to draw up the articles of the capitulation, by which Bastia was surrendered to the British arms. In the spring of 1794, our officer was appointed to succeed the late Sir Hyde Parker, as Captain of the Mediterranean fleet; and towards the close of the same year, he returned to England with Lord Hood, in the Victory of 100 guns. From this period until the summer of 1811, he appears to have been employed as a resident Commissioner of the Navy, successively, at Corsica, Malta, Gibraltar, and Halifax. Preferring the retention of his civil appointment to a flag, he was placed on the retired list of Post-Captains in Feb. 1799. Captain Inglefield is the reputed author of “A View of the Naval Force of Great Britain,” published in 1791. His son, Samuel Hood Inglefield, obtained post rank in 1807, and his daughter is the lady of that excellent officer, Vice-Admiral Sir Benjamin Hallowell, K.C.B. Agent.– William Marsh, Esq. 1. After the partial action off the Chesapeake, September 5, 1781, which we have already noticed in our first volume, p. 133, the British fleet, commanded by the Rear-Admirals Graves, Hood, and Drake, returned to Sandy Hook, and took on board 7000 troops under Sir Henry Clinton, destined for the relief of Earl Cornwallis, who was closely invested at York and Gloucester, by the French and rebel armies. On the 24th October the armament arrived off the Chesapeake, when the British commanders had the mortification to find that his Lordship, owing to the exhausted and sickly state of his army, and being without any hopes of relief, had entered into a capitulation for the surrender of those important posts on the 17th. By this unfortunate event 6000 British troops, and 1500 seamen, fell into the hands of the enemy. 2. Sir Samuel Hood, after his return from America, remained in Carlisle Bay, with his fleet moored in order of battle, in daily expectation of a visit from the French, till January 14, 1782, when he received intelligence that the Count de Grasse had relinquished his plan of attacking Barbadoes, and gone to St. Christopher’s; on his arrival at which island the Marquis de Bouille was landed with 8000 troops, and the British garrison consisting of only 600 men, under Brigadier-General Fraser, obliged to retire into the fort at Brimstone Hill. The Rear-Admiral, notwithstanding the superiority of the enemy, determined on a measure of unusual boldness, for the preservation of that valuable island. Instead of waiting their approach, he resolved to confound the enemy by an immediate attack, and to engage them as they lay at their anchors. For this purpose he immediately put to sea from Carlisle Bay, embarked General Prescott and the few troops that could be spared from Antigua, and proceeded without loss of time to attack the enemy in Basseterre Road. At day-break on the 24th the signal was made to form the line of battle, for the purpose of bearing down to the attack; but the untoward accident of the Alfred’s running foul of the Nymph, arrested the prosecution of this well-concerted design, and obliged the fleet to bring to whilst the former vessel repaired her damages. Towards the evening of the same day the Count de Grasse quitted his anchorage and put to sea, that his ships might have full room to act, and thus secure the advantages of their superiority in point of number. At day-light on the 25th, the enemy’s fleet was observed about three leagues to leeward, formed in order of battle, and consisting of twenty-nine sail of the line. Sir Samuel Hood, who had only twenty-two line-of-battle ships, instantly perceived the great advantages to be derived from this movement, and carried on every appearance of an immediate and determined attack, which drove the enemy farther to leeward, whilst he himself pushed for Basseterre, and anchored his fleet in line of battle a-head, in Frigate Bay. The Count de Grasse, astonished at this excellent manoeuvre, and apprehensive that all communication with the army might be cut off, made a most furious attack upon the rear of the British fleet, commanded by Commodore Affleck; but that gallant officer made so noble a defence, and was so ably supported by his seconds, the Hon. William Cornwallis in the Canada, and Lord Robert Manners in the Resolution, who kept up an incessant fire, covering the other ships of the division while they brought up in their stations, particularly the Prudent, whose wheel was shot away, and the rudder choked by a shot which had lodged between it and the stem-post, that the enemy, finding they could not make any impression on the resolute firmness of the British, bore up and stood to sea. The next morning, at 8 o’clock, the French fleet stood in, as if determined to force the British line, which they attacked with great violence from van to rear, without making the least visible impression on it; they then wore and stood to sea. Sir Samuel Hood, having observed that the rear of his fleet was too much exposed, took this opportunity to change the position thereof, and directed the Alfred, Canada, Prudent, Resolution, Belliqueux, Centaur, and Monarch, to extend themselves in a line towards the town of Basseterre, forming an obtuse angle, by which means no one part of the fleet could suffer a partial attack. The Count de Grasse, not yet discouraged, renewed the engagement in the afternoon, directing his attack principally against the centre and rear divisions; he was again repulsed, and suffered more material damage than in the preceding battle. The Ville de Paris, bearing de Grasse’s flag, was upon the heel all the next day, covering her shot-holes; and according to information which Sir Samuel Hood subsequently received from the shore, upwards of 1,000 wounded Frenchmen were sent to St. Eustatius. The loss sustained by the British, in all the attacks, amounted to 72 killed, and 244 wounded. On the 28th, part of the 13th regiment, and the whole of the 28th and 69th, were landed under cover of four frigates. After a smart skirmish with a detachment of French troops, which were beaten, and obliged to retreat with great loss into Basseterre, General Prescott took post upon a commanding hill. The following morning, the Marquis de Bouille arrived with 4,000 troops from Sandy Point; but finding the British General’s position to be too strong to venture an attack, he proceeded to the siege of Brimstone Hill. As no object could be gained by General Prescott remaining on shore, he re-embarked the same evening. Soon after the arrival of the fleet, Captain Inglefield of the Centaur, was sent to Brigadier-General Fraser with a message of importance, and returned in safety, after establishing signals between the fort and the squadron. The vigilance of the enemy cut off all further communication. Many attempts were afterwards made to throw succours into the garrison, all of which proved ineffectual; and several officers sent with messages to the Brigadier, were detected and taken prisoners. The enemy prosecuted the siege with unabating vigour till the 13th Feb., when a practicable breach was made in the works, and BrigadierGeneral Fraser and the Governor, having given up all hope of succour, reluctantly consented to capitulate. On the morning of the 14th, the French fleet, reinforced by five ships of the line, anchored off Nevis; and it being no longer necessary for the British to continue in its present situation, which was useless and dangerous, not only from the vast superiority of the enemy’s fleet, but that they were preparing to erect gun and mortar-batteries on a hill commanding the anchorage, Sir Samuel Hood issued orders to the respective Captains to slip or cut their cables without signal, at 11 P.M., the sternmost and leewardmost ships first, and so on in succession, then to proceed under an easy sail until directed otherwise by signal. That this order might be punctually obeyed, the Captains were directed to set their watches by Sir Samuel’s time-piece . This was performed with the utmost order and regularity, without being molested or pursued by the French fleet; which was lying within five miles, and must have witnessed the manoeuvre. The British fleet anchored at Antigua on the 19th, and a few days after was joined by Sir George B. Rodney, with a reinforcement from, England. 3. Mr. Renny was afterwards made a Lieutenant, and appointed to the command of a cutter, which foundered on her passage to Gibraltar with despatches, and all on board perished. 4. Mr. Baylis died a Lieutenant of the Mercury frigate, at St. John’s, Newfoundland, Sept. 1., 1799. 5. Thomas Matthews, a quarter-master, died in the boat the day before land was discovered. Those who escaped from the ill-fated Centaur, in addition to Captain Inglefield, the Master, and Midshipman mentioned above, were Mr. James Clark, Surgeon’s Mate; Timothy Sullivan, the Captain’s coxswain; John Gregory, a Quarter-Master; and five seamen. 6. The following is a list of the ships of war which sailed from Jamaica under the orders of Rear-Admiral Graves; and will show how they were disposed of: Ramillies 74 ${\displaystyle \scriptstyle {\left\{{\begin{matrix}\ \\\ \end{matrix}}\right.}}$ Rear-Admiral T. GravesCaptain S. Monarty. ${\displaystyle \scriptstyle {\left.{\begin{matrix}\ \\\ \end{matrix}}\right\}\,}}$ Abandoned after being set on fire. * Ville de Paris 110 Captain A. Wilkinson. ${\displaystyle \scriptstyle {\left.{\begin{matrix}\ \\\ \end{matrix}}\right\}\,}}$ Foundered, and their crews perished. * Glorieux 74 Captain Hon. T. Cadogan. Canada 74 Captain Hon. W. Cornwallis. Arrived in England, with the loss of her mizen-mast. Centaur 74 Captain J. N. Inglefield. Foundered, only 11 of her crew preserved. * Hector 74 Captain J. Bouchier. Foundered, crew saved by a letter of marque. † Jason 64 Captain John Aylmer. Arrived in England. † Caton 64 Captain T. Fisher. Arrived at Halifax. * Ardent 64 Captain R. Lucas. Returned to Jamaica. Pallas 36 Captain C. Parker. Went to Halifax very leaky, and afterwards lost on one of the Western Islands; crew saved. * Taken by Sir George B. Rodney, April 12, 1782. † Taken by Sir Samuel Hood, in the Mona Passage, April 19, 1782.	2019-05-24 20:19:30	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 3, "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": 0.3923065960407257, "perplexity": 8893.018301803411}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-22/segments/1558232257731.70/warc/CC-MAIN-20190524184553-20190524210553-00543.warc.gz"}
https://owly.wiki/en/Pohlig%E2%80%93Hellman_algorithm/	# Pohlig–Hellman algorithm Steps of the Pohlig–Hellman algorithm. In group theory, the Pohlig–Hellman algorithm, sometimes credited as the Silver–Pohlig–Hellman algorithm,[1] is a special-purpose algorithm for computing discrete logarithms in a finite abelian group whose order is a smooth integer. The algorithm was introduced by Roland Silver, but first published by Stephen Pohlig and Martin Hellman (independent of Silver). ### Groups of prime-power order As an important special case, which is used as a subroutine in the general algorithm (see below), the Pohlig–Hellman algorithm applies to groups whose order is a prime power. The basic idea of this algorithm is to iteratively compute the -adic digits of the logarithm by repeatedly "shifting out" all but one unknown digit in the exponent, and computing that digit by elementary methods. (Note that for readability, the algorithm is stated for cyclic groups — in general, must be replaced by the subgroup generated by , which is always cyclic.) Input. A cyclic group of order with generator and an element . Output. The unique integer such that . 1. Initialize 2. Compute . By Lagrange's theorem, this element has order . 3. For all , do: 1. Compute . By construction, the order of this element must divide , hence . 2. Using the baby-step giant-step algorithm, compute such that . It takes time . 3. Set . 4. Return . Assuming is much smaller than , the algorithm computes discrete logarithms in time complexity , far better than the baby-step giant-step algorithm's . ### The general algorithm In this section, we present the general case of the Pohlig–Hellman algorithm. The core ingredients are the algorithm from the previous section (to compute a logarithm modulo each prime power in the group order) and the Chinese remainder theorem (to combine these to a logarithm in the full group). (Again, we assume the group to be cyclic, with the understanding that a non-cyclic group must be replaced by the subgroup generated by the logarithm's base element.) Input. A cyclic group of order with generator , an element , and a prime factorization . Output. The unique integer such that . 1. For each , do: 1. Compute . By Lagrange's theorem, this element has order . 2. Compute . By construction, . 3. Using the algorithm above in the group , compute such that . 2. Solve the simultaneous congruence The Chinese remainder theorem guarantees there exists a unique solution . 3. Return . The correctness of this algorithm can be verified via the classification of finite abelian groups: Raising and to the power of can be understood as the projection to the factor group of order . ### Complexity The worst-case input for the Pohlig–Hellman algorithm is a group of prime order: In that case, it degrades to the baby-step giant-step algorithm, hence the worst-case time complexity is . However, it is much more efficient if the order is smooth: Specifically, if is the prime factorization of , then the algorithm's complexity is group operations.[2] ### Notes 1. ^ Mollin 2006, pg. 344 2. ^ Menezes, et. al 1997, pg. 108 ### References • Mollin, Richard (2006-09-18). An Introduction To Cryptography (2nd ed.). Chapman and Hall/CRC. p. 344. ISBN 978-1-58488-618-1. • S. Pohlig and M. Hellman (1978). "An Improved Algorithm for Computing Logarithms over GF(p) and its Cryptographic Significance" (PDF). IEEE Transactions on Information Theory (24): 106–110.CS1 maint: uses authors parameter (link) • Menezes, Alfred J.; van Oorschot, Paul C.; Vanstone, Scott A. (1997). "Number-Theoretic Reference Problems" (PDF). Handbook of Applied Cryptography. CRC Press. pp. 107–109. ISBN 0-8493-8523-7. Original content from Wikipedia, shared with licence Creative Commons By-Sa - Pohlig–Hellman algorithm	2021-08-03 03:43:49	{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9523478746414185, "perplexity": 1115.1354761956015}, "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/1627046154420.77/warc/CC-MAIN-20210803030201-20210803060201-00629.warc.gz"}
http://hoorah2heroes.org/6mhrn1k/aaf9d8-dot-product-latex	\lessdot \boldsymbol{\pi} \mathbf{A}_{\boldsymbol{0} }. \leftharpoonup \lg \precnapprox \Longleftrightarrow \breve{x} That is, they are simple symbols, in class 0. We've documented hundreds of LaTeX macros, packages, and configuration options, and we're adding more content every week! No installation, real-time collaboration, version control, hundreds of LaTeX templates, and more. > \gcd Find the dot product of two vectors. To combine the dot product and transpose in the definition itself is confusing. Blackboard bold (no lowercase) is used to represent standard sets of numbers, e.g. \leftleftarrows \ddagger Therefore, special environments have been declared for this purpose. ; The Comprehensive LaTeX Symbol List. The dot product, or any inner product, is generally considered to take two vectors in the same vector space to yield a scalar. \coprod. = Obtain this by typing the fraction and pressing space: 1/2 1 2 {\displaystyle {\frac {1}{2}}} Linear fraction (resp. The geometric definition of the dot product says that the dot product between two vectors $\vc{a}$ and $\vc{b}$ is $$\vc{a} \cdot \vc{b} = \\|\vc{a}\\| \\|\vc{b}\\| \cos \theta,$$ where $\theta$ is the angle between vectors $\vc{a}$ and $\vc{b}$. 1.3 Using physics in your LATEX document To use the physics package, simply insert \usepackage{physics} in the preamble of your document, before \smile The dot product is a negative number when $90^\circ \lt \phi \le 180^\circ$ and is a positive number when $0^\circ\le \phi \lt 90^\circ$. \varlimsup \varrho We've documented hundreds of LaTeX macros, packages, and configuration options, and we're adding more content every week! The dot product can help us understand the angle between two vectors. \overbrace{xxx} \vdash \exp \lneqq \varsupsetneqq \lnapprox \succneqq LaTeX The LaTeX command that creates the icon. However, \mathbf cannot be applied to Greek symbols, for instance. 18mu equals 1em. This Wikipedia article has more details on dot products. \nshortmid \vartriangleleft The following example illustrates the difference between \\prod and \\Pi. Export (png, jpg, gif, svg, pdf) and save & share" \looparrowleft Inner product spaces generalize Euclidean spaces (in which the inner product is the dot product, also known as the scalar product) to vector spaces of any (possibly infinite) dimension, and are studied in functional analysis. Many script-languages use backslash \"\\" to denote special commands. \varsigma \gvertneqq \widehat{xxx} \curlyeqprec \simeq Dot Product â Let we have given two vector A = a1 * i + a2 * j + a3 * k and B = b1 * i + b2 * j + b3 * k. Where i, j and â¦ \tau Vector Products [ edit ] The dot product (inner product) can be displayed using the centered dot symbol "\cdot" e.g. The \pmb command is not supported by the Wikia's LaTeX parser. \twoheadrightarrow \lfloor \rfloor, \vert or \| \arg \succnapprox \nVdash Features & Benefits. \supsetneq \ngtr \times \ggg or \gggtr \hookleftarrow 1 \sdiv 2) and pressing space. \sqsupseteq \emptyset \rightharpoondown \blacktriangle \in \nleqq \subseteqq LaTeX symbols have either names (denoted by backslash) or special characters. \triangleq, \circlearrowleft \Vvdash, ( ) \perp This formula gives a clear picture on the properties of the dot product. \doublebarwedge \overset{}{X} But it's fun to take and it's interesting because it results-- so this is a1, a2, all the way down to a n. That vector dot my b vector: b1, b2, all the way down to bn is going to be equal to the product of each â¦ Dot product is also known as scalar product and cross product also known as vector product. The first command in a powerdotpresentation must be after that the usual data (author, title and date) can be included in the preamble. The \smallint command is not supported by the Wikia's LaTeX parser. \varsupsetneq \rightarrowtail \nRightarrow Within the document, the commands \maketitle and \sectionwill create a new slide to display the corresponding information. \upharpoonright or \parallel In very simple terms dot product is a way of finding the product of the summation of two vectors and the output will be a single vector. The default is vertically aligned as illustrated below. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.. Visit Stack Exchange Dot product is also known as scalar product and cross product also known as vector product. \underset{}{X} Simple symbols (class 0) have no spaces around them, Operators (class 1) have thin spaces around them, thicker 6mu space provided by backslash followed by blank. \thickapprox In linear algebra, a dot product is the result of multiplying the individual numerical values in two or more vectors. Most of the stock math commands are written for typesetting math or computer science papers for academic journals, so you might need to dig deeper into LaTeX commands to get the vector notation styles that are common in physics textbooks and articles. \flat \measuredangle Find the dot product of A and B, treating the rows as vectors. Defined matrix operations. \succsim \nexists \varprojlim \nsupseteq dot treats the columns of A and B as vectors and calculates the dot product of corresponding columns. \ntrianglerighteq ˙, ¨) can be obtained by following a letter variable with "\dot" for a first derivative and "\ddot" for a second derivative. \varsubsetneq Dot Product – Let we have given two vector A = a1 * i + a2 * j + a3 * k and B = b1 * i + b2 * j + b3 * k. Where i, j and k are the unit vector along the x, y and z directions. \blacksquare Provide details and share your research! Learn about the conditions for matrix multiplication to be defined, and about the dimensions of the product of two matrices. Where Î¸ is the angle between vectors $\vec{a}$ and $\vec{b}$. The difference between this and \\Pi, which generates the capital letter , is that \\product appears larger, and that it supports the limits to be displayed below and above the symbol. \leqq \arctan \wedge or \land \circlearrowright \dagger In powerdot, to create a new slide the corresponding text and images must be enclosed in a special environment. Now, if two vectors are orthogonal then we know that the angle between them is 90 degrees. \upuparrows, \backepsilon While you can also do this by right-clicking on the equation and clicâ¦ \pi \circleddash \limsup \geqslant \lVert \rVert, \langle \rangle \Longleftarrow The following formula should make it clear where and are vectors. The symbol \ ( \times\ ) \cdot '' e.g with the commands \maketitle and \sectionwill create a new the. Installation, real-time collaboration, version control, hundreds of LaTeX macros, packages, and more • Page of. Number when â¦ an online LaTeX editor that 's easy to use is using! 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'S LaTeX parser 0 ( Ord ) symbols: simple / ordinary â¦ Thanks for contributing an to!	2021-04-14 14:09:33	{"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": 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": 0.9626045227050781, "perplexity": 2108.4615700363574}, "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/1618038077818.23/warc/CC-MAIN-20210414125133-20210414155133-00051.warc.gz"}
https://acm.sdut.edu.cn/onlinejudge2/index.php/Home/Index/problemdetail/pid/3316.html	### The Great Pan Time Limit: 1000 ms Memory Limit: 65536 KiB #### Problem Description As a programming contest addict, Waybl is always happy to take part in various competitive programming contests. One day, he was competing at a regional contest of Inventing Crappy Problems Contest(ICPC). He tried really hard to solve a "geometry" task without success. After the contest, he found that the problem statement is ambiguous! He immediately complained to jury. But problem setter, the Great Pan, told him "There are only four possibilities, why don\'t you just try all of them and get Accepted?". Waybl was really shocked. It is the first time he learned that enumerating problem statement is as useful as trying to solve some ternary search problem by enumerating a subset of possible angle! Three years later, while chatting with Ceybl, Waybl was told that some problem "setters" (yeah, other than the Great Pan) could even change the whole problem 30 minutes before the contest end! He was again shocked. Now, for a given problem statement, Waybl wants to know how many ways there are to understand it. A problem statement contains only newlines and printable ASCII characters (32 ≤ their ASCII code ≤ 127) except \'{\', \'}\', \'\|\' and \'′.Wayblhasalreadymarkedallambiguityinthefollowingtwoformats:1.A\|B\|C\|D\|...indicatesthispartcouldbeunderstandasAorBorCorDor....2.blah blah$indicates this part is printed in proportional fonts, it is impossible to determine how many space characters there are. Note that A, B, C, D won\'t be duplicate, but could be empty. (indicate evil problem setters addedclarified it later.) Also note that N consecutive spaces lead to N+1 different ways of understanding, not 2N ways. It is impossible to escape from "$$" and "{}" markups even with newlines. There won\'t be nested markups, i.e. something like "${A\|B}$" or "{$A$\|B}" or "{{A\|B}\|C}" is prohibited. All markups will be properly matched. #### Input Input contains several test cases, please process till EOF. For each test case, the first line contains an integer n, indicating the line count of this statement. Next n lines is the problem statement. ≤ ≤ 1000, size of the input file will not exceed 1024KB. #### Output For each test case print the number of ways to understand this statement, or "doge" if your answer is more than 105. #### Sample Input 9 I\'ll shoot the magic arrow several times on the ground, and of course the arrow will leave some holes on the ground. When you connect three holes with three line segments, you may get a triangle. {\|It is hole! Common sense!\| No Response, Read Problem Statement\|don\'t you know what a triangle is?} 1 Case$1: = >$5$/This is my code printed in proportional font, isn\'t it cool?/ printf(\"Definitely it is cooooooool \\ %d\\n\",4 * 4 * 4 * 4 * 4 * 4 * 4 * 4 * 4 * 4 * 4 * 4 * 4 * 4 * 4 * 4 * 4 * 4);$2$Two space\$ and {blue\| red} color! #### Sample Output 4 4 doge 6	2020-01-23 01:55:30	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 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": 0.2941007614135742, "perplexity": 2800.0153695211566}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-05/segments/1579250608062.57/warc/CC-MAIN-20200123011418-20200123040418-00535.warc.gz"}
https://math.stackexchange.com/questions/4612166/stochastic-lorenz-model	# Stochastic Lorenz model Consider Lorenz model \begin{align} \frac{dx}{dt}&=\sigma(y-x)\\ \frac{dy}{dt}&=\rho x-y-xz\\ \frac{dz}{dt}&=xy-\beta z \end{align} with $$\sigma=10$$, $$\rho=28$$ and $$\beta=\frac{8}{3}$$. I want to embed this model in stochastic environment by adding exponentially colored noise process to parameter $$\sigma$$: $$d\widetilde{\sigma_t}=-\alpha\widetilde{\sigma_t}dt+\gamma dW_t\text{, }\sigma_t=10+\widetilde{\sigma_t}$$ where $$\alpha$$ and $$\gamma$$ are values I will do experiments with. I came up with $$d\begin{bmatrix}x_t\\y_t\\z_t\\\widetilde{\sigma_t}\end{bmatrix}=\begin{bmatrix}(10+\widetilde{\sigma_t})(y-x)\\\rho x-y-xz\\xy-\beta z\\-\alpha\widetilde{\sigma_t}\end{bmatrix}dt+\begin{bmatrix}0\\0\\0\\\gamma\end{bmatrix}dW_t$$ 1. Is stochastic differential equation above correct? I want to apply Euler scheme to approximate given model numerically. I know that Euler scheme requires equation to be in Ito form. 1. How can I determine if equation above is in Ito form? Finally, I want to analyze order of convergence through experiments. 1. Does equation above have exact solution? If not, what should I use as reference solution to calculate error against? • Your system seems correct, but as is, it is not linear, which quite annoying. However, $\sigma_t$ is decoupled from the other variables and its equation of motion can be solved on its own (use Ito's lemma on the function $f(\sigma_t,t) = e^{\alpha t}\sigma_t$). The remaining system is linear with respect to $x,y,z$. Jan 5 at 16:29 • @Abezhiko Applying Ito's lemma on $f(\sigma_t,t)$, I get $df=\alpha e^{\alpha t}\sigma_tdt+e^{\alpha t}d\sigma_t$. Substituting $-\alpha\sigma_tdt+\gamma dW_t$ for $d\sigma_t$, I get $\gamma e^{\alpha t}dW_t$. I am confused how this implies that $e^{\alpha t}\sigma_t$ is exact solution for $d\sigma_t$. Did I apply Ito's lemma incorrectly? Jan 6 at 12:24 • That's correct, then you find $e^{\alpha t}\sigma_t - e^{\alpha t_0}\sigma_{t_0} = \int_{t_0}^te^{\alpha s}\,\mathrm{d}W_s$ by integration, hence $\sigma_t = e^{-\alpha(t-t_0)}\sigma_{t_0} + e^{-\alpha t}\int_{t_0}^te^{\alpha s}\,\mathrm{d}W_s$ and it cannot be simplified further. Nevertheless, note that $X_t := \int_{t_0}^te^{\alpha s}\,\mathrm{d}W_s$ follows a zero-mean normal law, whose variance is $\mathrm{Var}[X_t] = \int_{t_0}^te^{2\alpha s}\,\mathrm{d}t = \frac{e^{2\alpha t}-e^{2\alpha t_0}}{2\alpha}$ by Itô's isometry. Jan 6 at 12:44 • Oops, I forgot the last part : [...] and thus $\displaystyle\tilde{\sigma}_t\sim\mathcal{N}\left(\tilde{\sigma}_{t_0}e^{-\alpha(t-t_0)},\frac{1-e^{-2\alpha(t-t_0)}}{2\alpha}\right)$. Jan 6 at 14:25 • @Abezhiko I haven't run any experiments yet but from your answers, it seems like $\gamma$ has no effect. Is it true? Also, should I interpret the resulting stochastic differential equation as Ito integral or do I need to convert it from Stratonovich integral to Ito integral in order to apply Euler scheme? Jan 9 at 18:33	2023-03-27 23:16:20	{"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": 9, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9563499093055725, "perplexity": 418.40898965893484}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296948708.2/warc/CC-MAIN-20230327220742-20230328010742-00376.warc.gz"}
https://hal-mines-paristech.archives-ouvertes.fr/hal-00789164v3	Riemannian L p Averaging on Lie Group of Nonzero Quaternions - Archive ouverte HAL Accéder directement au contenu Article Dans Une Revue Advances in Applied Clifford Algebras Année : 2014 ## Riemannian L p Averaging on Lie Group of Nonzero Quaternions Jesus Angulo #### Résumé This paper discusses quaternion $L^p$ geometric weighting averaging working on the multiplicative Lie group of nonzero quaternions $\mathbb{H}^{}$, endowed with its natural bi-invariant Riemannian metric. Algorithms for computing the Riemannian $L^p$ center of mass of a set of points, with $1 \leq p \leq \infty$ (i.e., median, mean, $L^p$ barycenter and minimax center), are particularized to the case of $\mathbb{H}^{}$. Two different approaches are considered. The first formulation is based on computing the logarithm of quaternions which maps them to the Euclidean tangent space at the identity $\mathbf{1}$, associated to the Lie algebra of $\mathbb{H}^{*}$. In the tangent space, Euclidean algorithms for $L^p$ center of mass can be naturally applied. The second formulation is a family of methods based on gradient descent algorithms aiming at minimizing the sum of quaternion geodesic distances raised to power $p$. These algorithms converges to the quaternion Fr\'{e}chet-Karcher barycenter ($p=2$), the quaternion Fermat-Weber point ($p=1$) and the quaternion Riemannian 1-center ($p=+\infty$). Besides giving explicit forms of these algorithms, their application for quaternion image processing is shown by introducing the notion of quaternion bilateral filtering. #### Domaines Informatique [cs] Traitement des images [eess.IV] Loading... ### Dates et versions hal-00789164 , version 1 (16-02-2013) hal-00789164 , version 2 (26-10-2013) hal-00789164 , version 3 (21-01-2015) ### Identifiants • HAL Id : hal-00789164 , version 3 • DOI : ### Citer Jesus Angulo. Riemannian L p Averaging on Lie Group of Nonzero Quaternions. Advances in Applied Clifford Algebras, 2014, 24 (2), pp.355-382. ⟨10.1007/s00006-013-0432-2⟩. ⟨hal-00789164v3⟩ ### Exporter BibTeX TEI Dublin Core DC Terms EndNote Datacite ### Collections 252 Consultations 1343 Téléchargements ### Partager Gmail Facebook Twitter LinkedIn More	2023-03-30 21:20:08	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.42013129591941833, "perplexity": 2718.3019007541525}, "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-2023-14/segments/1679296949387.98/warc/CC-MAIN-20230330194843-20230330224843-00384.warc.gz"}
http://mathhelpforum.com/calculus/120996-implicit-differentiation.html	Math Help - Implicit differentiation 1. Implicit differentiation could you tell me how to attack this one 2. Originally Posted by Da Freak thanks a lot man. could you tell me how to attack this one too implicit differentiation 3. Originally Posted by skeeter implicit differentiation how do you do that, i had a huge amount of trouble with that previously and got destroyed big time on that part of a test earlier in the year. 4. Originally Posted by Da Freak how do you do that, i had a huge amount of trouble with that previously and got destroyed big time on that part of a test earlier in the year. $\frac{d}{dx} (x^2y^2+2y=x^3)$ $x^2 \cdot 2y \cdot \frac{dy}{dx} + y^2 \cdot 2x + 2 \cdot \frac{dy}{dx} = 3x^2$ $2x^2y \cdot \frac{dy}{dx} + 2 \cdot \frac{dy}{dx} = 3x^2 - 2xy^2$ $\frac{dy}{dx}(2x^2y + 2) = 3x^2 - 2xy^2$ $\frac{dy}{dx} = \frac{3x^2 - 2xy^2}{2x^2y + 2}$ 5. Originally Posted by skeeter $\frac{d}{dx} (x^2y^2+2y=x^3)$ $x^2 \cdot 2y \cdot \frac{dy}{dx} + y^2 \cdot 2x + 2 \cdot \frac{dy}{dx} = 3x^2$ $2x^2y \cdot \frac{dy}{dx} + 2 \cdot \frac{dy}{dx} = 3x^2 - 2xy^2$ $\frac{dy}{dx}(2x^2y + 2) = 3x^2 - 2xy^2$ $\frac{dy}{dx} = \frac{3x^2 - 2xy^2}{2x^2y + 2}$ thanks a lot man, i really appreciate the help.	2014-12-20 16:36:24	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 10, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5668327212333679, "perplexity": 1291.0707175469768}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-52/segments/1418802770043.48/warc/CC-MAIN-20141217075250-00134-ip-10-231-17-201.ec2.internal.warc.gz"}
https://erwinjr2.readthedocs.io/en/stable/manual/intro.html	# Introduction¶ ErwinJr2 is a cross-platform software with a combination of: 1. a C library for solving 1D quantum problem and the related thermal and electrical problems; 2. a Python interface for the C library; 3. a set of Python modules for loading, saving, organizing, solving quantum eigen-states in semiconductor quantum wells or super lattices, and calculating relevant physics parameters for these states, especially for quantum cascade laser (QCL) design purposes; 4. a GUI front-end interface for the above Python modules. The C library is based on ANSI C and is tested using GCC (under Linux and MacOS) and Visual Studio (on Windows). It also has optional OpenMP support for parallel computing if the environment supports. The C library Python interface is based on ctypes in standard Python library and numpy. The Python module for simulation adds scipy.constants requirement for scientific constants, and uses json for saving and loading super-lattices (or quantum wells) information. The GUI interface is based on PyQt5 and matplotlib. Quantum cascade lasers (QCLs) are semiconductor lasers that emit light through inter-subband transitions. These lasers consist of periodic series of thin layers of various semiconductor materials which creates a one-dimensional multiple-quantum-well confinement. Compare to conventional semiconductor lasers which use single material, QCLs have the advantage of both a higher output efficiency due to possible quantum cascades across different quantum wells, and an improved flexibility in tuning the frequencies. ## Models and Formulas¶ The physics model and formulas used in the software are discussed in Physics Model and Formulas Todo List references and validations	2021-01-16 17:33:09	{"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": 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": 0.17575116455554962, "perplexity": 3820.9902185287024}, "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-2021-04/segments/1610703506832.21/warc/CC-MAIN-20210116165621-20210116195621-00378.warc.gz"}
http://physics.aps.org/synopsis-for/10.1103/PhysRevLett.107.101801	Synopsis: Controlling for the “look-elsewhere effect” When searching for new physics at the Large Hadron Collider, researchers have to keep track of how many places they look. One of the primary purposes of the Large Hadron Collider (LHC) is to search for the unexpected. Usually this means looking for a “bump” in the data—an excess of events over a known background level. Such an excess could be due to new physics. However, experimentalists must account for the fact that if they look at enough areas of parameter space, they are certain to see statistical fluctuations. To control for this “look-elsewhere effect,” the data must be normalized for the number of places searched in which a fluctuation could be observed. The Compact Muon Solenoid (CMS) collaboration at the LHC has now performed many searches for new physics, and seen very few potential bumps. But, in a paper appearing in Physical Review Letters, in which they analyze data taken in 2010, CMS reports a small excess of events that could correspond to a pair of new $390$ giga-electron-volt particles, each decaying into three hadronic jets. According to theory, supersymmetric partners of the gluon, called gluinos, can produce such events if a symmetry known as $R$ parity is violated, but at a rate lower than what CMS has seen. After taking into account the “look-elsewhere effect,” the statistical significance of the bump is only $1.9$ standard deviations. As of now, the collaboration has taken $30$ times more data than they did last year. Analyzing these new data should show whether CMS is indeed just seeing a fluctuation. – Robert Garisto Announcements More Announcements » Subject Areas Particles and Fields Previous Synopsis Atomic and Molecular Physics Graphene Related Articles Particles and Fields Viewpoint: Ghostly Neutrino Comes into Sharper Focus The first results from the NOvA experiment set new constraints on charge-parity violation in neutrinos and on the ordering of neutrino masses. Read More » Particles and Fields Synopsis: Explaining a 750 GeV Bump Theorists try to explain data from the LHC that could be hinting at the existence of new particles. Read More » Particles and Fields Synopsis: The Heavy Limit of Dark Matter A theoretical investigation of super-heavy dark matter particles finds that their existence might be discerned in the cosmic microwave background. Read More »	2016-04-30 13:10:11	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 8, "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": 0.6476897597312927, "perplexity": 1276.941464986287}, "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-2016-18/segments/1461860111865.15/warc/CC-MAIN-20160428161511-00002-ip-10-239-7-51.ec2.internal.warc.gz"}
http://xrpp.iucr.org/A1a/ch2o1v0001/sec2o1o3o3/	International Tables for Crystallography Volume A1 Symmetry relations between space groups Edited by Hans Wondratschek and Ulrich Müller International Tables for Crystallography (2006). Vol. A1, ch. 2.1, pp. 46-48 \| 1 \| 2 \| Section 2.1.3.3. Basis transformation and origin shift Hans Wondratscheka* and Mois I. Aroyob aInstitut für Kristallographie, Universität, D-76128 Karlsruhe, Germany, and bDepartamento de Física de la Materia Condensada, Facultad de Ciencias, Universidad del País Vasco, Apartado 644, E-48080 Bilbao, Spain Correspondence e-mail: [email protected] 2.1.3.3. Basis transformation and origin shift \| top \| pdf \| Each t-subgroup is defined by its representatives, listed under sequence' by numbers each of which designates an element of . These elements form the general position of . They are taken from the general position of and, therefore, are referred to the coordinate system of . In the general position of , however, its elements are referred to the coordinate system of . In order to allow the transfer of the data from the coordinate system of to that of , the tools for this transformation are provided in the columns matrix' and shift' of the subgroup tables. The designation of the quantities is that of IT A Part 5 and is repeated here for convenience. In the following, columns and rows are designated by boldface italic lower-case letters. Point coordinates , translation parts of the symmetry operations and shifts are represented by columns. The sets of basis vectors and are represented by rows [indicated by , which means transposed']. The quantities with unprimed symbols are referred to the coordinate system of , those with primes are referred to the coordinate system of . The following columns will be used ( is analogous to w): The matrices W and of the symmetry operations, as well as the matrix P for a change of basis and its inverse , are designated by boldface italic upper-case letters ( is analogous to W): Let be the row of basis vectors of and the basis of , then the basis is expressed in the basis by the system of equations 2or In matrix notation, this is The column p of coordinates of the origin of is referred to the coordinate system of and is called the origin shift. The matrix–column pair (P, p) describes the transformation from the coordinate system of to that of , for details, cf. IT A, Part 5 . Therefore, P and p are chosen in the subgroup tables in the columns matrix' and shift', cf. Section 2.1.3.2. The column matrix' is empty if there is no change of basis, i.e. if P is the unit matrix I. The column shift' is empty if there is no origin shift, i.e. if p is the column o consisting of zeroes only. A change of the coordinate system, described by the matrix–column pair (P, p), changes the point coordinates from the column x to the column . The formulae for this change do not contain the pair (P, p) itself, but the related pair : Not only the point coordinates but also the matrix–column pairs for the symmetry operations are changed by a change of the coordinate system. A symmetry operation is described in the coordinate system of by the system of equations i.e. by the matrix–column pair (W, w). The symmetry operation will be described in the coordinate system of the subgroup by the equation and thus by the pair . This pair can be calculated from the pair by solving the equations and Example 2.1.3.3.1 Consider the data listed for the t-subgroups of , No. 31:This means that the matrices and origin shifts are (1) (2) (3) The first subgroup is monoclinic, the symmetry direction is the b axis, which is standard. However, the glide direction is nonconventional. Therefore, the basis of is transformed to a basis of the subgroup such that the b axis is retained, the glide direction becomes the axis and the axis is chosen such that the basis is a right-handed one, the angle and the transformation matrix P is simple. This is done by the chosen matrix . The origin shift is the o column. With equations (2.1.3.8) and (2.1.3.9), one obtains for the glide reflection , which is after standardization by . (4) For the second monoclinic subgroup, the symmetry direction is the (nonconventional) a axis. The rules of Section 2.1.2.5 require a change to the setting unique axis b'. A cyclic permutation of the basis vectors is the simplest way to achieve this. The reflection is now described by . Again there is no origin shift. (5) The third monoclinic subgroup is in the conventional setting unique axis c', but the origin must be shifted onto the screw axis. This is achieved by applying equation (2.1.3.9) with , which changes of to of . Example 2.1.3.3.2 Evaluation of the t-subgroup data of the space group , No. 151, started in Example 2.1.3.2.4. The evaluation is now continued with the columns sequence', matrix' and `shift'. They are used for the transformation of the elements of to their conventional form. Only the monoclinic t-subgroups are of interest here because the trigonal subgroup is already in the standard setting. One takes from the tables of subgroups in Chapter 2.3 Designating the three matrices by , , , one obtains with the corresponding inverse matrices and the origin shifts For the three new bases this means All these bases span ortho-hexagonal cells with twice the volume of the original hexagonal cell because for the matrices holds. In the general position of , No.151, one findsThese entries represent the matrix–column pairs : Application of equations (2.1.3.8) on the matrices and (2.1.3.9) on the columns of the matrix–column pairs results in All translation vectors of are retained in the subgroups but the volume of the cells is doubled. Therefore, there must be centring-translation vectors in the new cells. For example, the application of equation (2.1.3.9) with to the translation of with the vector , i.e. , results in the column , i.e. the centring translation of the subgroup. Either by calculation or, more easily, from a small sketch one sees that the vectors for , for (and for ) correspond to the cell-centring translation vectors of the subgroup cells.	2019-05-19 17:39:54	{"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": 0.8752470016479492, "perplexity": 792.2299127743227}, "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-2019-22/segments/1558232255071.27/warc/CC-MAIN-20190519161546-20190519183546-00246.warc.gz"}
http://cnx.org/content/m44934/latest/	# Connexions You are here: Home » Content » Total Variation Denoising (An MM Algorithm) ### Recently Viewed This feature requires Javascript to be enabled. # Total Variation Denoising (An MM Algorithm) Module by: Ivan Selesnick. E-mail the author Summary: Total variation denoising (TVD) is an approach for noise reduction developed so as to preserve sharp edges in the underlying signal. Unlike a conventional low-pass filter, TV denoising is defined in terms of an optimization problem. This module describes an algorithm for TV denoising derived using the majorization-minimization (MM) approach, developed by Figueiredo et al. [ICIP 2006]. To keep it simple, this module addresses TV denoising of 1-D signals only. For computational efficiency, the algorithm may use a solver for sparse banded systems. ## Introduction Total variation denoising (TVD) is an approach for noise reduction developed so as to preserve sharp edges in the underlying signal [12]. Unlike a conventional low-pass filter, TV denoising is defined in terms of an optimization problem. The output of the TV denoising filter' is obtained by minimizing a particular cost function. Any algorithm that solves the optimization problem can be used to implement TV denoising. However, it is not trivial because the TVD cost function is non-differentiable. Numerous algorithms have been developed to solve the TVD problem, e.g. [5], [7], [4], [15], [16]. Total variation is used not just for denoising, but for more general signal restoration problems, including deconvolution, interpolation, in-painting, compressed sensing, etc. [2]. In addition, the concept of total variation has been generalized and extended in various ways [13], [10], [3]. These notes describe an algorithm for TV denoising derived using the majorization-minimization (MM) approach, developed by Figueiredo et al. [9]. To keep it simple, these notes address TV denoising of 1-D signals only (ref. [9] considers 2D TV denoising for images). Interestingly, it is possible to obtain the exact solution to the TV denoising problem (for the 1-D case) without optimization, but instead using a direct algorithm based on a characterization of the solution. Recently, a fast algorithm has been developed and C code made available [6]. Total variation denoising assumes that the noisy data y(n)y(n) is of the form y ( n ) = x ( n ) + w ( n ) , n = 0 , , N - 1 y ( n ) = x ( n ) + w ( n ) , n = 0 , , N - 1 (1) where x(n)x(n) is a (approximately) piecewise constant signal and w(n)w(n) is white Gaussian noise. TV denoising estimates the signal x(n)x(n) by solving the optimization problem: arg min x n = 0 N - 1 \| y ( n ) - x ( n ) \| 2 + λ n = 1 N - 1 \| x ( n ) - x ( n - 1 ) \| . arg min x n = 0 N - 1 \| y ( n ) - x ( n ) \| 2 + λ n = 1 N - 1 \| x ( n ) - x ( n - 1 ) \| . (2) The regularization parameter λ>0λ>0 controls the degree of smoothing. Increasing λλ gives more weight to the second term which measures the fluctuation of the signal x(n)x(n). MATLAB software is available online at: http://eeweb.poly.edu/iselesni/lecture_notes/TVDmm/index.html ### Notation The NN-point signal xx is represented by the vector x = [ x ( 0 ) , , x ( N - 1 ) ] t . x = [ x ( 0 ) , , x ( N - 1 ) ] t . (3) The 11 norm of a vector vv is defined as v 1 = n \| v ( n ) \| . v 1 = n \| v ( n ) \| . (4) The 22 norm of a vector vv is defined as v 2 = n \| v ( n ) \| 2 1 2 . v 2 = n \| v ( n ) \| 2 1 2 . (5) The matrix DD is defined as D = - 1 1 - 1 1 - 1 1 . D = - 1 1 - 1 1 - 1 1 . (6) The first-order difference of an NN-point signal xx is given by DxDx where DD is of size (N-1)×N(N-1)×N. Note, for later, that DDtDDt is a tridiagonal matrix of the form: D D t = 2 - 1 - 1 2 - 1 - 1 2 - 1 - 1 2 . D D t = 2 - 1 - 1 2 - 1 - 1 2 - 1 - 1 2 . (7) The total variation of the NN-point signal x(n)x(n) is given by TV ( x ) : = D x 1 = n = 1 N - 1 \| x ( n ) - x ( n - 1 ) \| . TV ( x ) : = D x 1 = n = 1 N - 1 \| x ( n ) - x ( n - 1 ) \| . (8) With this notation, the TV denoising problem Equation 2 can be written compactly as arg min x R N y - x 2 2 + λ D x 1 . arg min x R N y - x 2 2 + λ D x 1 . (9) ## Majorization-Minimization Majorization-minimization (MM) is an approach to solve optimization problems that are too difficult to solve directly. Instead of minimizing the cost function F(x)F(x) directly, the MM approach solves a sequence of optimization problems, Gk(x)Gk(x), k=0,1,2,.k=0,1,2,. The idea is that each Gk(x)Gk(x) is easier to solve that F(x)F(x). The MM approach produces a sequence xkxk, each being obtained by minimizing Gk-1(x)Gk-1(x). To use MM, one must specify the functions Gk(x)Gk(x). The trick is to choose the Gk(x)Gk(x) so that they are easy to solve, but they should also each approximate F(x)F(x). The MM approach requires that each function Gk(x)Gk(x) is a majorizor of F(x)F(x), i.e., G k ( x ) F ( x ) , x G k ( x ) F ( x ) , x (10) and that it agrees with F(x)F(x) at xkxk, G k ( x k ) = F ( x k ) . G k ( x k ) = F ( x k ) . (11) In addition, Gk(x)Gk(x) should be convex functions. The MM approach then obtains xk+1xk+1 by minimizing Gk(x)Gk(x). Figure 1 illustrates the MM procedure with a simple example. For clarity, the figure illustrates the minimization of a univariate function. However, the MM procedure works in the same way for the minimization of multivariate functions, and it is in the multivariate case where the MM procedure is especially useful. The majorization-minimization approach to minimize the function F(x)F(x) can be summarized as: 1. Set k=0k=0. Initialize x0x0. 2. Choose Gk(x)Gk(x) such that 1. Gk(x)F(x)Gk(x)F(x) for all xx 2. Gk(xk)=F(xk)Gk(xk)=F(xk) 3. Set xk+1xk+1 as the minimizer of Gk(x)Gk(x). xk+1= arg minxGk(x)xk+1= arg minxGk(x) (12) 4. Set k=k+1k=k+1 and go to step (2.) When F(x)F(x) is convex, then under mild conditions, the sequence xkxk produced by MM converges to the minimizer of F(x)F(x). More details about the majorization-minimization procedure can be found in [8] and references therein. Example majorizor. An upper bound (majorizor) of f(t)=\|t\|f(t)=\|t\| that agrees with f(t)f(t) at t=tkt=tk is g ( t ) = 1 2 \| t k \| t 2 + 1 2 \| t k \| g ( t ) = 1 2 \| t k \| t 2 + 1 2 \| t k \| (13) as illustrated in Figure 2. The figure makes clear that g ( t ) f ( t ) , t g ( t ) f ( t ) , t (14) g ( t k ) = f ( t k ) g ( t k ) = f ( t k ) (15) The derivation of the majorizor in Equation 13 is left as exercise Item 11. It is convenient to use second-order polynomials as majorizors because they are easy to minimize. Setting the derivatives to zero gives linear equations. A higher order polynomial could be used to give a closer fit to the function f(t)f(t) to be minimized, however, then the minimization will be more difficult (involving polynomial root finding, etc.) ## TV Denoising Algorithm One way to apply MM to TV denoising is to majorize TV (x) TV (x) by a quadratic function of xx, as described in ref. [9]. Then the TVD cost function F(x)F(x) can be majorized by a quadratic function, which can in turn be minimized by solving a system of linear equations. To that end, using Equation 13, we can write 1 2 \| t k \| t 2 + 1 2 \| t k \| \| t \| t R 1 2 \| t k \| t 2 + 1 2 \| t k \| \| t \| t R (16) Using v(n)v(n) for tt and summing over nn gives n 1 2 \| v k ( n ) \| v 2 ( n ) + 1 2 \| v k ( n ) \| n \| v ( n ) \| n 1 2 \| v k ( n ) \| v 2 ( n ) + 1 2 \| v k ( n ) \| n \| v ( n ) \| (17) which can be written compactly as 1 2 v t Λ k - 1 v + 1 2 v k 1 v 1 1 2 v t Λ k - 1 v + 1 2 v k 1 v 1 (18) where ΛkΛk is the diagonal matrix Λ k : = \| v k ( 1 ) \| \| v k ( 2 ) \| \| v k ( N ) \| = diag ( \| v k \| ) . Λ k : = \| v k ( 1 ) \| \| v k ( 2 ) \| \| v k ( N ) \| = diag ( \| v k \| ) . (19) In the notation, diag (\|v\|) diag (\|v\|), the absolute value is applied element-wise to the vector vv. Using DxDx for vv, we can write 1 2 x t D t Λ k - 1 D x + 1 2 D x k 1 D x 1 1 2 x t D t Λ k - 1 D x + 1 2 D x k 1 D x 1 (20) where Λ k : = diag ( \| D x k \| ) . Λ k : = diag ( \| D x k \| ) . (21) Note in Equation 20 that the majorizor of Dx1Dx1 is a quadratic function of xx. Also note that the term Dxk1Dxk1 in Equation 20 should be considered a constant — it is fixed as xkxk is the value of xx at the previous iteration. Similarly, ΛkΛk in Equation 20 is also not a function of xx. A majorizor of the TV cost function Equation 9 can be obtained from Equation 20 by adding y-x22y-x22 to both sides, y - x 2 2 + λ 1 2 x t D t Λ k - 1 D x + λ 1 2 D x k 1 y - x 2 2 + λ D x 1 . y - x 2 2 + λ 1 2 x t D t Λ k - 1 D x + λ 1 2 D x k 1 y - x 2 2 + λ D x 1 . (22) Therefore a majorizor Gk(x)Gk(x) for the TV cost function is given by G k ( x ) = y - x 2 2 + λ 1 2 x t D t Λ k - 1 D x + λ 1 2 D x k 1 , Λ k = diag ( \| D x k \| ) G k ( x ) = y - x 2 2 + λ 1 2 x t D t Λ k - 1 D x + λ 1 2 D x k 1 , Λ k = diag ( \| D x k \| ) (23) which satisfies Gk(xk)=F(xk)Gk(xk)=F(xk) by design. Using MM, we obtain xkxk by minimizing Gk(x)Gk(x), x k + 1 = arg min x y - x 2 2 + λ 1 2 x t D t Λ k - 1 D x + λ 1 2 D x k 1 . x k + 1 = arg min x y - x 2 2 + λ 1 2 x t D t Λ k - 1 D x + λ 1 2 D x k 1 . (24) An explicit solution to Equation 24 is given by x k + 1 = I + λ 2 D t Λ k - 1 D - 1 y . x k + 1 = I + λ 2 D t Λ k - 1 D - 1 y . (25) A problem with update Equation 25 is that as the iterations progress, some values of DxkDxk will generally go to zero, and therefore some entries of Λk-1Λk-1 in Equation 25 will go to infinity. This issue is addressed in Ref. [9] by rewriting the equation using the matrix inverse lemma. By the matrix inverse lemma, I + λ 2 D t Λ k - 1 D - 1 = I - D t 2 λ Λ k + D D t - 1 D I + λ 2 D t Λ k - 1 D - 1 = I - D t 2 λ Λ k + D D t - 1 D (26) where Λ k = diag ( \| D x k \| ) . Λ k = diag ( \| D x k \| ) . (27) Now the update equation Equation 25 becomes x k + 1 = y - D t 2 λ diag ( \| D x k \| ) + D D t - 1 D y . x k + 1 = y - D t 2 λ diag ( \| D x k \| ) + D D t - 1 D y . (28) Observe that even if some elements of DxkDxk are zero, no division by zero arises in Equation 28. The update Equation 28 calls for the solution to a linear system of equations. In general, it is desirable to avoid such a computation in an iterative filtering algorithm due to the high computational cost of solving linear systems (especially when the signal yy is very long and the system is very large). However, the matrix [2λ diag (\|Dxk\|)+DDt][2λ diag (\|Dxk\|)+DDt] in Equation 28 is a sparse banded matrix; it consists of only three diagonals — the main diagonal, one upper diagonal, and one lower diagonal. This is because DDtDDt is tridiagonal as shown in Equation 7. Therefore, the linear system in Equation 28 can be solved very efficiently [11]. Further, the whole matrix need not be stored in memory, only the three diagonals. The MATLAB function TVD_mm implements TV denoising based on the update Equation 28. The function uses the sparse matrix structure in MATLAB so as to avoid high memory requirements and so as to invoke sparse banded system solvers. MATLAB uses LAPACK [1] to solve the sparse banded system in the program TVD_mm. The algorithm used by MATLAB to solve a sparse linear system can be monitored using the command spparms('spumoni',3). function [x, cost] = TVD_mm(y, lam, Nit) % [x, cost] = TVD_mm(y, lam, Nit) % Total variation denoising using majorization-minimization % and a banded linear systems. % % INPUT % y - noisy signal % lam - regularization parameter % Nit - number of iterations % % OUTPUT % x - denoised signal % cost - cost function history % % Reference % 'On total-variation denoising: A new majorization-minimization % algorithm and an experimental comparison with wavalet denoising.' % M. Figueiredo, J. Bioucas-Dias, J. P. Oliveira, and R. D. Nowak. % Proc. IEEE Int. Conf. Image Processing, 2006. % Ivan Selesnick, [email protected], 2011 y = y(:); % Ensure column vector cost = zeros(1, Nit); % Cost function history N = length(y); e = ones(N-1, 1); DDT = spdiags([-e 2e -e], [-1 0 1], N-1, N-1); % DD' (sparse matrix) D = @(x) diff(x); % D (operator) DT = @(x) [-x(1); -diff(x); x(end)]; % D' x = y; % Initialization Dx = D(x); for k = 1:Nit F = 2/lam * spdiags(abs(Dx), 0, N-1, N-1) + DDT; % F : Sparse matrix structure % F = 2/lam * diag(abs(D(x))) + DDT; % Not stored as sparse matrix x = y - DT(F\D(y)); % Solve sparse linear system Dx = D(x); cost(k) = sum(abs(x-y).^2) + lam * sum(abs(Dx)); % Save cost function history end An example of TV denoising is shown in Figure 3. The history of the cost function through the progression of the algorithm is shown in the figure. It can be seen that after 20 iterations the cost function has leveled out, suggesting that the algorithm has practically converged. Another algorithm for 1-D TV denoising is Chambolle's algorithm [5], a variant of which is the iterative clipping' algorithm [14]. This algorithm is computationally simpler than the MM algorithm because it does not call for the solution to a linear system at each iteration. However, it may converge slowly. For the denoising problem illustrated in Figure 3, the convergence of both the iterative clipping and MM algorithms are shown in Figure 4. It can be seen that the MM algorithm converges in fewer iterations. ## Optimality Condition It turns out that the solution to the TV denoising problem can be concisely characterized [6]. Suppose the noisy data is yy and the regularization parameter is λλ. If xx is the solution to the TV denoising problem, then it must satisfy \| s ( n ) \| λ 2 , n = 0 , , N - 1 \| s ( n ) \| λ 2 , n = 0 , , N - 1 (29) where s(n)s(n) is the cumulative sum' of the residual, i.e. s ( n ) : = k = 0 n y ( k ) - x ( k ) . s ( n ) : = k = 0 n y ( k ) - x ( k ) . (30) The condition Equation 29 is illustrated in Figure 5a for the TV denoising example of Figure 3. The condition Equation 29 by itself is not sufficient for x(n)x(n) to be the solution to the TV denoising problem. It is further necessary that x(n)x(n) satisfy d ( n ) > 0 , s ( n ) = λ / 2 d ( n ) < 0 , s ( n ) = - λ / 2 d ( n ) = 0 , \| s ( n ) \| < λ / 2 d ( n ) > 0 , s ( n ) = λ / 2 d ( n ) < 0 , s ( n ) = - λ / 2 d ( n ) = 0 , \| s ( n ) \| < λ / 2 (31) where d(n)d(n) is the first-order difference function of x(n)x(n), i.e. d ( n ) = x ( n + 1 ) - x ( n ) . d ( n ) = x ( n + 1 ) - x ( n ) . (32) The condition Equation 31 is illustrated in Figure 5b. The figure shows (d(n),s(n))(d(n),s(n)) as a scatter plot. It can be seen that this condition requires the points to lie on a curve consisting of three line segments (a double-L' shape). Notice in the figure that d(n)d(n) is mostly zero, reflecting the sparsity of the derivative of x(n)x(n). ## Conclusion Total variation (TV) denoising is a method to smooth signals based on a sparse-deriviative signal model. TV denoising is formulated as the minimization of a non-differentiable cost function. Unlike a conventional low-pass filter, the output of the TV denoising filter' can only be obtained through a numerical algorithm. Total variation denoising is most appropriate for piecewise constant signals, however, it has been modified and extended so as to be effective for more general signals. ## Exercises 1. Reproduce figures like those of the example (using a blocky' signal). Try different values of λλ. How does the solution change as λλ is increased or decreased? 2. Compare TV denoising with low-pass filtering (e.g. a Butterworth or FIR filter, etc). Apply each method to the same signal. Plot the denoised/filtered signals using each method and discuss the differences you observe. 3. Perform TV denoising on a signal that is not blocky' (which has slopes or oscillatory behavior). You should see stair-case' artifacts in the denoised signal. Show these artifacts in a figure and explain why they arise. 4. Is TV denoising linear? (Conventional low-pass filters are linear, e.g. Butterworth filter.) Illustrate that TV denoising satisfies (or does not) the superposition property by performing TV denoising on each of two signals and their sum. 5. Find a majorizor of the function f(t)=\|t\|f(t)=\|t\| of the form g(t)=at2+bt+c,g(t)=at2+bt+c, (33) that coincides with f(t)f(t) at t=tkt=tk. As illustrated in Figure 2, the function g(t)g(t) should satisfy g(t)f(t)tR,g(t)f(t)tR, (34) g(tk)=f(tk).g(tk)=f(tk). (35) 6. Show the solution to problem Equation 24 is given by Equation 25. 7. For denoising a noisy signal using TV denoising, devise a method or formula to set the regularization parameter λλ. You can assume that the variance σ2σ2 of the noise is known. Show examples of your method. 8. Explain why DtDt can be implemented in MATLAB by the command DT = @(x) [-x(1); -diff(x); x(end)]; 9. Modify the TV denoising MATLAB program so that the matrix FF is not sparse. Measure the run-times of the original and modified programs. Is the sparse version faster? Use a long signal and many iterations to see the difference more clearly. 10. Second-order TV denoising is based the second-order difference instead of the first-order difference. Modify the algorithm and MATLAB program so that it performs second-order TV denoising. Compare first and second order TV denoising using blocky' and non-blocky' signals, and comment on your observations. ## References 1. Anderson, E. and Bai, Z. and Bischof, C. and Demmel, J. and Dongarra, J. and Croz, J. Du and Greenbaum, A. and Hammarling, S. and McKenney, A. and Ostrouchov, S. and Sorensen, D. (1992). LAPACK's user's guide. [http://www.netlib.org/lapack]. SIAM. 2. Bect, J. and Blanc-Feaud, L. and Aubert, G. and Chambolle, A. (2004). A L1-Unified Variational Framework for Image Restoration. In European Conference on Computer Vision, Lecture Notes in Computer Sciences. (Vol. 3024, pp. 1-13). 3. Bredies, K. and Kunisch, K. and Pock, T. (2010). Total generalized variation. SIAM J. Imag. Sci., 3(3), 492-526. 4. Chambolle, A. (2004). An algorithm for total variation minimization and applications. J. of Math. Imaging and Vision, 20, 89-97. 5. Chambolle, A. and Lions, P.-L. (1997). Image recovery via total variation minimization and related problems. Numerische Mathematik, 76, 167–188. 6. Condat, L. (2012). A direct algorithm for 1D total variation denoising. [http://hal.archives-ouvertes.fr/]. Technical report. Hal-00675043. 7. Chan, T. F. and Osher, S. and Shen, J. (2001, February). The Digital TV Filter and Nonlinear Denoising. IEEE Trans. Image Process., 10(2), 231-241. 8. Figueiredo, M. and Bioucas-Dias, J. and Nowak, R. (2007, December). Majorization-minimization algorithms for wavelet-based image restoration. IEEE Trans. Image Process., 16(12), 2980-2991. 9. Figueiredo, M. and Bioucas-Dias, J. and Oliveira, J. P. and Nowak, R. D. (2006). On total-variation denoising: A new majorization-minimization algorithm and an experimental comparison with wavalet denoising. In Proc. IEEE Int. Conf. Image Processing. 10. Hu, Y. and Jacob, M. (2012, May). Higher Degree Total Variation (HDTV) Regularization for Image Recovery. IEEE Trans. Image Process., 21(5), 2559-2571. 11. Press, W. H. and Teukolsky, S. A. and Vetterling, W. T. and Flannery, B. P. (1992). Numerical recipes in C: the art of scientific computing (2nd ed.). Cambridge University Press. 12. Rudin, L. and Osher, S. and Fatemi, E. (1992). Nonlinear total variation based noise removal algorithms. Physica D, 60, 259-268. 13. Rodriguez, P. and Wohlberg, B. (2009, February). Efficient Minimization Method for a Generalized Total Variation Functional. IEEE Trans. Image Process., 18(2), 322-332. 14. Selesnick, I. and Bayram, I. (2009). Total Variation Filtering. [http://cnx.org/content/m31292/1.1/]. Connexions Web site. 15. Wang, Y. and Yang, J. and Yin, W. and Zhang, Y. (2008). A new alternating minimization algorithm for total variation image reconstruction. SIAM J. on Imaging Sciences, 1(3), 248-272. 16. Zhu, M. and Wright, S. J. and Chan, T. F. (2008). Duality-based algorithms for total-variation-regularized image restoration. J. Comp. Optimization and Applications. ## Content actions PDF \| EPUB (?) ### What is an EPUB file? EPUB is an electronic book format that can be read on a variety of mobile devices. 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https://wiki.documentfoundation.org/Documentation/Calc_Functions/BINOMDIST	# Documentation/Calc Functions/BINOMDIST BINOMDIST ## Category: Statistical Analysis ## Summary: Calculates binomial distribution probabilities from either the probability mass function or the cumulative distribution function. The binomial distribution is a discrete probability distribution that is used to analyze data in many domains. ## Syntax: BINOMDIST(X; Trials; SP; C) ## Returns: Returns a real number in the range [0, 1], which is the binomial distribution probability for the given arguments. ## Arguments: X is a non-negative integer, or a reference to a cell containing that integer, that is the number of trial successes for which the probability is required. Trials is a non-negative integer, or a reference to a cell containing that integer, that is the total number of independent trials. SP is a real number (expressed as a percentage, such as 2.5%, or a decimal fraction, such as 0.025), or a reference to a cell containing that number, that is the probability of a successful outcome on each trial. As a probability, SP lies in the range [0, 1] (or equivalently 0% ≤ SP ≤ 100%). C is a logical value, or a reference to a cell containing that value, that determines whether the required probability is taken from the probability mass function or the cumulative distribution function. If C is set to 0 or FALSE, a value from the probability mass function is calculated. For any other values of C, a value from the cumulative distribution function is calculated. • If any of X, Trials, SP, and C is non-numeric, then BINOMDIST reports a #VALUE! error. • If either of X or Trials is a non-integer value, then BINOMDIST truncates it to an integer value. • If SP is less than 0.0 or greater than 1.0, then BINOMDIST reports an invalid argument error (Err:502). • BINOMDIST checks (after any truncation) that Trials ≥ 0, X ≥ 0, and TrialsX. If any of these checks fail, then BINOMDIST reports an invalid argument error (Err:502). • Calc's BINOMDIST and BINOM.DIST functions perform the same calculations. The requirements for BINOMDIST are specified in ODF 1.2; BINOM.DIST is provided for interoperability with Microsoft Excel. • The binomial distribution should be used in its intended circumstances and this is when there are a fixed number of independent trials, each having two possible outcomes (success or failure), with a constant probability of achieving a successful outcome for each trial. • The formula for BINOMDIST when C is set to 0 or FALSE (probability mass function) is: $\displaystyle{ \text{BINOMDIST}(x;\:n;\:p;\:0)~=~\frac{n!}{x!(n-x)!}\:\times\:p^{x}\:\times\:(1-p)^{n-x} }$ The $\displaystyle{ \frac{n!}{x!(n-x)!} }$ term in the above equation is often written as $\displaystyle{ \binom{n}{x} }$ and referred to as "n choose x" or the binomial coefficient. • The formula for BINOMDIST when C is set to any value other than 0 and FALSE (cumulative density function) is: $\displaystyle{ \text{BINOMDIST}(x;\:n;\:p;\:TRUE)~=~\sum_{j=0}^{x} \frac{n!}{j!(n-j)!}\:\times\:p^{j}\:\times\:(1-p)^{n-j} }$ • For valid input arguments c, n, p, x, and y, the following relations arise: • $\displaystyle{ \text{BINOMDIST}(x;\:n;\:p;\:c)~=~\text{BINOM.DIST}(x;\:n;\:p;\:c) }$ • $\displaystyle{ \text{B}(n;\:p;\:x;\:y)~=~\text{BINOMDIST}(y;\:n;\:p;\:TRUE) - \text{BINOMDIST}(x-1;\:n;\:p;\:TRUE) }$ • $\displaystyle{ \text{B}(n;\:p;\:0;\:y)~=~\text{BINOMDIST}(y;\:n;\:p;\:TRUE) }$ • $\displaystyle{ \text{B}(n;\:p;\:x) = \text{BINOMDIST}(x;\:n;\:p;\:FALSE) }$. The second of these is only valid when x > 0. For the final three relations, BINOMDIST could be replaced with calls to the functionally equivalent BINOM.DIST. ## Examples: Formula Description Returns =BINOMDIST(A1; A2; A3; A4) where cell A1 contains the number 2, cell A2 contains the number 10, cell A3 contains the formula =1/6, and cell A4 contains the formula =FALSE(). Suppose that we roll a fair (unbiased) die 10 times. For each roll, the result will be either 1, 2, 3, 4, 5, or 6, and each of these will occur with an equal probability of 1/6. Here the function calculates the probability that a specific number (it doesn't matter which) will come up exactly twice in the 10 rolls, using the probability mass function. 0.290710049201722 =BINOMDIST(3; 10; 1/6; 0) Assuming the same scenario as in the previous example, here the function calculates the probability that a specific number (it doesn't matter which) will come up exactly three times in the 10 rolls, using the probability mass function. 0.155045359574252 =BINOMDIST(2; 10; 1/6; 0) + BINOMDIST(3; 10; 1/6; 0) Assuming the same scenario as in the previous examples, here the function calculates the probability that a specific number (it doesn't matter which) will come up exactly two or exactly three times in the 10 rolls, using the probability mass function. Note that using the cumulative distribution function, the formula =BINOMDIST(3; 10; 1/6; 1) - BINOMDIST(1; 10; 1/6; 1) performs exactly the same calculation. 0.445755408775974 =BINOMDIST(3; 10; 1/6; 1) Assuming the same scenario as in the previous examples, here the function calculates the probability that a specific number (it doesn't matter which) will come up exactly zero times, or exactly one time, or exactly two times, or exactly three times in the 10 rolls. The number returned in this example is a value from the cumulative distribution function. Note that the formula =BINOMDIST(0; 10; 1/6; 0) + BINOMDIST(1; 10; 1/6; 0) + BINOMDIST(2; 10; 1/6; 0) + BINOMDIST(3; 10; 1/6; 0) performs exactly the same calculation. 0.930272157445512 =BINOMDIST(7; 15; 50%; 0) Suppose that we toss a fair (unbiased) coin 15 times. For each toss, the result will be either a head or a tail, and each of these will occur with an equal probability of 0.5 or 50%. Here the function calculates the probability that heads will come up exactly seven times in the 15 tosses, using the probability mass function. Note that the formula =BINOMDIST(8; 15; 50%; 0) returns the same result because the binomial distribution is symmetrical when the probability of success is equal to 0.5. 0.196380615234375 =1 - BINOMDIST(7, 20, 25%, TRUE()) Suppose that we are to sit an important Quantum Cosmology examination but, unfortunately, we have not attended any lectures and know nothing at all about the subject. We discover that the exam comprises 20 multiple-choice questions, each with four possible options. Our strategy will be to randomly guess the answer to each question from these four options. Here the function calculates the probability that we will achieve a mark of a least 40% using this strategy, using the cumulative distribution function. 0.101811856922723 BINOMDIST	2023-03-26 18:38:48	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 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": 0.7890625, "perplexity": 545.0645054295061}, "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/1679296946445.46/warc/CC-MAIN-20230326173112-20230326203112-00255.warc.gz"}
https://planetmath.org/LocallyFree	# locally free A sheaf of $\mathcal{O}_{X}$-modules $\mathcal{F}$ on a ringed space $X$ is called locally free if for each point $p\in X$, there is an open neighborhood (http://planetmath.org/Neighborhood) $U$ of $x$ such that $\mathcal{F}\|_{U}$ is free (http://planetmath.org/FreeModule) as an $\mathcal{O}_{X}\|_{U}$-module, or equivalently, $\mathcal{F}_{p}$, the stalk of $\mathcal{F}$ at $p$, is free as a $(\mathcal{O}_{X})_{p}$-module. If $\mathcal{F}_{p}$ is of finite rank (http://planetmath.org/ModuleOfFiniteRank) $n$, then $\mathcal{F}$ is said to be of rank $n$. Title locally free LocallyFree 2013-03-22 13:52:31 2013-03-22 13:52:31 mps (409) mps (409) 13 mps (409) Definition msc 14A99	2018-11-21 01:38:00	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 16, "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": 0.9064717292785645, "perplexity": 453.7962238864622}, "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-2018-47/segments/1542039746926.93/warc/CC-MAIN-20181121011923-20181121033923-00539.warc.gz"}
http://docs.scipy.org/doc/scipy/reference/generated/scipy.signal.morlet.html	# scipy.signal.morlet¶ scipy.signal.morlet(M, w=5.0, s=1.0, complete=True)[source] Complex Morlet wavelet. Parameters: M : int Length of the wavelet. w : float Omega0. Default is 5 s : float Scaling factor, windowed from -s2pi to +s2pi. Default is 1. complete : bool Whether to use the complete or the standard version. morlet : (M,) ndarray Notes The standard version: pi*-0.25 exp(1jwx) * exp(-0.5(x2)) This commonly used wavelet is often referred to simply as the Morlet wavelet. Note that this simplified version can cause admissibility problems at low values of w. The complete version: pi-0.25 (exp(1jwx) - exp(-0.5(w2))) exp(-0.5(x2)) The complete version of the Morlet wavelet, with a correction term to improve admissibility. For w greater than 5, the correction term is negligible. Note that the energy of the return wavelet is not normalised according to s. The fundamental frequency of this wavelet in Hz is given by f = 2swr / M where r is the sampling rate. #### Previous topic scipy.signal.daub scipy.signal.qmf	2015-07-04 20:47:31	{"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": 0.2562618851661682, "perplexity": 4719.225466020299}, "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/1435375096944.75/warc/CC-MAIN-20150627031816-00251-ip-10-179-60-89.ec2.internal.warc.gz"}
https://www.physicsforums.com/threads/rutherfords-atomic-model.192987/	# Rutherford's atomic model 1. Oct 21, 2007 ### GuitarDean Ok, I just posted this in the advanced forum, but looking at some of the topics on there my question might belong in here instead.. I understand Rutherford proposed that electrons orbit around a central nucleus. However, since accelerating charges produce electromagnetic radiation, the orbiting electron should lose energy via E&M and spiral into the nucleus. But my question is: How do I calculate the time it takes for the electron to spiral into the nucleus, given the rate of energy loss (as a function of acceleration) and the initial electron-nucleus distance? The power loss equation is: P = (e^2 a^2 ) / (6 pi epsilon c^3) So far I've thought of calculating the initial energy of the system and integrating the power, and then equating the lost energy to the initial energy; however the final energy is negative inifinity, so this doesn't seem to work. Algebraic manipulation of circular motion equations didn't get me anywhere either; I'm not really sure how else to proceed now.	2018-03-17 22:51:06	{"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": 0.8478384613990784, "perplexity": 500.83396146556476}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-13/segments/1521257645362.1/warc/CC-MAIN-20180317214032-20180317234032-00163.warc.gz"}
http://mathoverflow.net/questions/53930/time-integral-of-a-smooth-vector-valued-function-of-a-planar-brownian-bridge	# Time-integral of a smooth, vector-valued function of a planar Brownian bridge I'm looking for information on how to compute the distribution of the random vector $$Z = \int_0^t f(B_s) ds$$ where $t>0$ is fixed, $B_s$ is a 2D Brownian bridge with $B_0 = 0$, $B_t=b \in \mathbb{R}^2$, and $f : \mathbb{R}^2 \rightarrow \mathbb{R}^K$ has components $f_k : \mathbb{R}^2 \rightarrow \mathbb{R}$ that are positive, bounded, and in $C^\infty$. Here's a reference to the local time of a 1D Brownian bridge, that gives an explicit density for the local-time $L_t(x)$ so a time-integral can be computed as $\int_\mathbb{R} f(x)L_t(x) dx$. But it doesn't appear to work for processes in the plane. Any pointers to relevant papers or textbook chapters would be appreciated. The motivation is a problem in experimental biophysics, which I'm happy to outline if others are interested. This is my first venture into stochastic integration. Edit: By request, a summary of the motivating problem: The overall problem is optical tracking of a microscopic biological process consisting of many component parts, some diffusing and others not. It requires estimating the size and rate of diffusion of an ensemble of particles in a membrane, under unfavorable conditions: particle sizes smaller than the optical resolution, other objects in the field, unknown per-particle brightness, and most significantly, a hidden process that adds new particles to the ensemble at random times. The random vector $Z$ is one piece of our probability model of this (rather complicated) biological process. It represents the light due to a single particle, collected for a brief exposure $[0,t)$ by a digital camera attached to our microscope. Each of the scalar components $f_k$ is the parameter of a Poisson random variable modeling the number of photons detected by pixel $k$ during the exposure. A simple model of the optics has $$f_k(x) = \int_{A_k} g(x - y) dy$$ where $A_k \subset \mathbb{R}^2$ is the rectangular area "seen" by pixel $k$ (so the $A_k$ are disjoint) and $g$ is a Gaussian approximating the microscopes' point spread function. The fixed end-point $B_t=b$ arises from the way we move from this continuous-time model to our discrete time experimental data. We use a Hidden Markov Model in which each discrete time $i \in \mathbb{N}$ corresponds to an instant $t_i$ between camera exposures; and the HMM emissions $Z$ are conditioned upon a transition $b_i \mapsto b_j$. Then we can evaluate the likelihood of an experimental observation, a complete digital movie containing $n$ exposures, in only $O(n)$ operations by using the forward algorithm to iterate over the exposures. I welcome comments and criticism of this approach. - I'd love to hear about the problem. – weakstar Feb 1 '11 at 0:09 To compute the distribution of $Z$ for a general function $f$ might be difficult. To evaluate it, one could rewrite the process $(B_s)_{0\le s\le t}$ as $B_s=W_s+(b-W_t)(s/t)$ for every $0\le s\le t$, where $(W_s)_{s\ge0}$ is a standard 2D Brownian motion starting from $W_0=0$. By "evaluate it" do you mean evaluate the probability density function of $Z$ at arbitrary points in $\mathbb{R}^K$? If so, could you elaborate on how to proceed after re-writing $B_s$ in terms of $W_s$? – Gabriel Feb 1 '11 at 16:33 Sorry if this was unclear. To evaluate the distribution of $Z$, one can simulate a large number of i.i.d. copies. To simulate one copy of $Z$, one can discretize the integral defining $Z$ and compute $f(W_s+(b-W_t)(s/t))$ at every point $s=it/N$ with $i\le N$. If $s=it/N$, $W_s-(s/t)W_t$ is $\sqrt{t/N}$ times $(1-k/N)(X_1+\cdots+X_i)+(X_{i+1}+\cdots+X_N)$, where $(X_k)_{1\le k\le N}$ is an i.i.d. sample of 2D normal random variables. For $K$ copies of $1/N$ discretizations of $Z$, one needs $2KN$ i.i.d. standard normal random variables. – Did Feb 1 '11 at 16:59 Ah, yes of course. I was hoping there might be some further analysis, taking advantage of the structure of $f$, that could eliminate or reduce the need for Monte Carlo simulation. But I suppose this may be the best approach. Upvoted. Will wait a few days and then mark accepted if my silver bullet hasn't arrived. – Gabriel Feb 1 '11 at 17:35 You are right that I did not use the structure of $f$. But even taking it into account, simply to compute $E(Z_1)$ is not that easy. Assuming that $g$ is the density of a standard Gaussian random variable, centered and with variance $I_2$, (I think that) $E(Z_1)$ is the integral from $0$ to $t$ of $g((A_1-b(s/t))/\sigma(s))$ with $\sigma^2(s)=1+(s/t)(t-s)$ (where $g(B)$ is the integral of $g$ over $B$). Sure, if the rectangle $A_1$ is parallel to the axes, each $g((A_1-b(s/t))/\sigma(s))$ factors into the product of two one-dimensional Gaussian integrals... but I do not see how to go further. – Did Feb 1 '11 at 18:04	2014-10-24 21:01:00	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8427233695983887, "perplexity": 235.06815240224972}, "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-2014-42/segments/1414119646518.17/warc/CC-MAIN-20141024030046-00224-ip-10-16-133-185.ec2.internal.warc.gz"}
https://mathoverflow.net/questions/238514/obstruction-to-rationality-of-del-pezzo-surfaces-of-degree-4/238553#238553	# Obstruction to rationality of del Pezzo surfaces of degree 4 Let $X$ be a del Pezzo surface over a number field $k$. (A del Pezzo surface over $k$ is a smooth, projective, geometrically connected surface whose anti-canonical class $K_X$ is ample.) Let $d := K_X^2$ be the degree of $X$. It is well-known that $d$ satisfies $1 \leq d \leq 9$. If $d \geq 5$, and $X(k) \neq \emptyset$, then $X$ is birationally equivalent to $\mathbb{P}^2_k$. Moreover, if $d=5$ or $d=7$, the condition on $X(k)$ is automatically satisfied. (This is Theorem 9.4.7 in Bjorn Poonen's Rational Points on Varieties.) In other words, for del Pezzo surfaces of degree $5$ and higher, the only obstruction to $k$-rationality is the possible lack of $k$-rational points. My question concerns the "first" (i.e. highest-degree) non-trivial case, as far as rationality is concerned, namely the case where $d=4$. In this case, there are examples of $X$ with $X(k)\neq \emptyset$ but where $X$ is non-rational. An obstruction to rationality is given by the (non-trivial part of the) Brauer group $\operatorname{Br}(X)$ of $X$: that is, if $\operatorname{Br}(X)/\operatorname{Br}(k) \neq 0$, then $X$ is not $k$-rational. (To be sure, such $X$ exist, even among those $X$ that have rational points. This is where the $d=4$ case differs from the higher degree cases.) Indeed, this simply follows from the fact that $\operatorname{Br}$ is a birational invariant of smooth, projective, geometrically connected varieties, and from the fact that $\operatorname{Br}(\mathbb{P}^2_k) = \operatorname{Br}(k)$. My question is: Is the converse true? That is, if $X$ is a del Pezzo surface of degree $4$ over a number field $k$ with $X(k)\neq\emptyset$, and $\operatorname{Br}(X) = \operatorname{Br}(k)$, does it follow that $X$ is a $k$-rational surface? • In the rather special but related case of diagonal cubic surfaces, what you ask follows from the calculations of Colliot-Thélène, Kanevsky and Sansuc: see Proposition 1 and the following Lemme 1 in their article. May 11 '16 at 7:38 • You can assume that your $X$ is minimal, since otherwise blowing down reduces to the case of higher degree. For dp4s, minimal is equivalent to Picard rank 1. So you asking whether it's possible to have a surface with $\mathrm{H}^1(k,\mathrm{Pic}\,\bar{X})=0$ and $\mathrm{H}^0(k,\mathrm{Pic}\,\bar{X})=\mathbb{Z}$, and with a rational point. May 11 '16 at 8:25 • Daniel Loughran already pointed out that the answer is no. But if you want at least a partial result in this direction, see theorem 3.36 in Wittenberg's book Intersections de deux quadriques et pinceaux de courbes de genre 1. May 11 '16 at 8:43 • Hi Martin. It is not quite true that for a DP4, minimal is equivalent to Picard rank $1$. You can have minimal DP4s with a conic bundle structure (the example I construct in my answer is of this type). For cubic surfaces, however, of course minimal is equivalent to Picard rank $1$. May 11 '16 at 8:44 • @Gro-Tsen: The result you mention in Wittenberg's book concerns the Hasse principle. How does this give applications to rationality? May 11 '16 at 8:50 Interesting question. But, alas, the answer is no. The issue is that you have missed an extra non-rationality criterion. Namely, it is possible that such a surface $X$ has $\mathrm{Br}(X) = \mathrm{Br}(k)$, yet we have $\mathrm{Br}(X_K) \neq \mathrm{Br}(K)$ for some finite extension $K/k$. Problems of this type for cubic surfaces over finite fields are considered in the PhD thesis of Shuijing Li: http://arxiv.org/abs/0904.3555 Though she only works over finite fields, the analysis for surfaces over number fields with a cyclic splitting field is more-or-less identical. Here is how the counter-example goes. Take your DP4 $X$ and blow it up in a rational point not on a line to obtain a cubic surface $S$. We then choose this surface so that it has conjugacy class $C_7$ in the notation of Manin's table (p176 of his book); such a surface exists over $\mathbb{Q}$ by the recent resolution of the inverse Galois problem for cubic surfaces by Elsenhans and Jahnel. Its splitting field is cyclic of order $6$. You see from the table that the Brauer group of $S$ is constant. However this surface is non-rational. This is because the Brauer group (modulo constants) becomes $(\mathbb{Z}/2\mathbb{Z})^2$ after a cubic extension, as it becomes the class $C_3$ from Manin's table. I don't know whether this more general "extra non-rationality criterion" I mention is sufficient over number fields. That it is sufficient over finite fields is proven in Theorem 5.3.7 in Li's thesis. To approach this general problem over number fields, I guess one could just enumerate all conjugacy classes of subgroups of the appropriate Weyl group in magma, and study which Brauer groups arise as well as the configurations of lines, and use the non-rationality criterion given in Theorem 5.3.1 in Li. • Thanks Daniel! The "extra non-rationality criterion" that you mention looks promising, maybe I will look into it later. It should be a finite check, as you suggest. – RP_ May 11 '16 at 12:23 • @René: If you want to discuss this approach more, feel free to contact me via email. May 11 '16 at 12:32 • In Tschinkel's talk maths.ed.ac.uk/cheltsov/edge2017/pdf/yura.pdf, on slide 15 he says that it is a conjecture of Colliot-The'le`ne that existence of a point plus vanishing of Br(X_K)/Br(K) for all field extensions K/k implies stable rationality for del Pezzo surfaces. Mar 21 '20 at 10:16 I will supplement Dan's nice answer by claiming that the answer is almost always no. Specifically, almost every degree 4 del Pezzo surface over the rational numbers with a rational point is non-rational but has trivial Brauer group. To support this, I wrote a little Magma program that picked two random quadrics in $\mathbb{P}^4$ containing the point $(1:0:0:0:0:0)$ and with integer coefficients in $[-5,5]$. Let $X$ be the intersection of these two quadrics. I computed the Fano scheme of lines on $X$. If the Fano scheme is irreducible over $\mathbb{Q}$, that is, the lines are all Galois-conjugate, then a calculation shows that the Brauer group is trivial and the surface is not rational. I ran the program 100 times and, apart from 1 time when $X$ was singular, this was always the case. I'm sure it would be straightforward to show that the Fano scheme is always irreducible outside a thin set of the space parametrising such pairs of quadrics. • Hi Martin. I like your answer, but can you provide more details how you show non-rationality of such a surface? I mean a DP5 can have irreducible Fano scheme of lines, yet is still rational. May 11 '16 at 9:29 • OK, maybe I need a little more. I just computed the Galois group of the field of definition of the 16 lines in one random example, and it was the whole Weyl group $W(D_5)$. I hope that's enough to guarantee non-rationality. And if that happens in one example, then I think it has to be the generic behaviour: specialising can only make the Galois action smaller. May 11 '16 at 9:43 • I have realised that you can prove non-rationality in this case using a result of Iskovskih on conic bundles. Your DP4 has Picard number $1$. The blow-up in a point off a line is a cubic surface of Picard number $2$ with a line. Such cubic surfaces are non-rational by Corollary 2.6 of "Iskovskih - Rational surfaces with a sheaf of rational curves and with a positive square of canonical class". May 11 '16 at 11:09	2021-10-24 01:19:50	{"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": 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": 0.8751471638679504, "perplexity": 261.0976901703468}, "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/1634323585828.15/warc/CC-MAIN-20211023224247-20211024014247-00597.warc.gz"}
http://mathoverflow.net/questions/130492/higher-dimensional-convex-hull	# Higher dimensional convex hull Let $CH(S)$ be a convex hull of a finite set $S$ and denote the set of all the vertices of $CH(S)$ as $Vert(S)$. For a vertex $v \in Vert(S)$, it has an associated set $E(v)$ which is defined as $E(v)=$ { $x \in Vert(S)$ \| $xv$ is an edge of $CH(S)$ }. My question is best stated using an example in 2-D: In this example, $S =${ $v_1, v_2, v_3, v_4, v_5$} and $v_1, v_2, v_3, v_4, v_5$ in clockwise order constitutes $Vert(S)$. There exists a line $L$ between $v_1$ and $CH(S -${ $v_1$}$)$. If we link $v_1$ with every other 4 points, we will obtain 4 new points $p_2, p_3, p_4, p_5$ on $L$, where $p_i$ is the intersection of $L$ and the line $v_1v_i$. You will find that { $p_2, p_5$ } $= Vert($ { $p_2, p_3, p_4, p_5$ } $)$. Here comes my problem. Given a finite set $S$ in higher dimensional (i.e. >= 3) vector space, we can find $Vert(S) =$ { $v_1, v_2, \cdots, v_h$ }. Now consider a point $v \in Vert(S)$. There exists a hyperplane between $v$ and $CH(S -${$v$}$)$. If we link $v$ with every other points in $Vert(S)$, we will get a set $P$ containing $h - 1$ points on the hyperplane. Will the points in $P$ corresponding to $E(v)$ constitute $Vert(P)$? Or is there some related theorems? - For your first paragraph, specify that $S$ is a finite set. Then vertex and edge make sense. – Gerald Edgar May 13 '13 at 16:59 I think the answer is yes. First observe that $CH(P) \subset CH(S)\cap H$: if $x\in CH(P)$ then $x$ is written as a convex combination of things which are convex combinations of vertices of $CH(S)$, so is a convex combination of vertices of $CH(S)$. Then we note that $CH(S)\cap H$ has as its vertices the points in $P$ corresponding to $E(v)$ (in general this is going to be all of $E(v)$ are, which are all contained in $P$ of course so $CH(S)\cap H \subset CH(P)$ and the two sets are equal, and we know the vertices of $CH(S)\cap H$ To see that $CH(S) \cap H$ has the correct vertices, the faces of $CH(S) \cap H$ correspond to faces of $CH(S)$ with their dimension dropped by at most 1 (if the face happens to be parallel to the hyperplane, then it will not drop in dimension). The edges in $E(v)$ cannot be parallel to the hyperplane because that means the hyperplane does not separate $v$ and the corresponding vertex on the other side of the edge - I'm just a beginner. Could you explain the second and the third paragraphs more clearly? Or what theorems that is your statement based on? – echo May 13 '13 at 17:40 If you think of CH(S) as being a polyhedron: A polyhedron P of dimension n has a face F of dimension d if there are n-d linearly independent inequalities for P which are active (have an equality at that point). So if we take P=CH(S), the edges are the places were n-1 inequalities are active. If we restrict to $CH(S)\cap H$, inequalities for $P$ correspond to inequalities for $CH(S)\cap H$ inside of $H$. The edge which originally had n-1 active inequalities still has n-1 active inequalities inside of $H$ as long as $H\cap edge$ is not equal to the whole edge – David Benson-Putnins May 13 '13 at 17:51 The operation that you are describing is known as the "vertex figure". I suggest that you have a look at Günter Ziegler's book "Lectures on Polytopes". This is well explained at Chapter 2. I guess that the particular statement that you are looking for is Proposition 2.4 that states a bijection between (k-1)-faces of the vertex figure at v with k-faces of the original polytope that contain v. -	2015-03-06 11:05:53	{"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": 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": 0.8301313519477844, "perplexity": 178.9257799094121}, "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/1424936468536.58/warc/CC-MAIN-20150226074108-00330-ip-10-28-5-156.ec2.internal.warc.gz"}
http://math.stackexchange.com/questions/249587/linear-algebra-small-problem	# Linear Algebra (small problem) I would like to count the below : $(a^{2}I_{n} + b^{2}I_{n})^{-1}$ * $(aI_{n} bI_{n})$ = ? Any idea? Note the second bracket is a matrix (1x2) and $I_{n}$ is an identity matrix. - How come the second bracket is $1\times2$? The identity matrix, by definition, is a square matrix. So, your two bracketed terms, as well as their products, are square matrices. – user1551 Dec 2 '12 at 23:45 I think $(aI_nbI_n)$ is meant to be $(aI_n\ bI_n)$, a matrix with $n$ rows and $2n$ columns. – Gerry Myerson Dec 3 '12 at 10:28 I see. That makes sense, but we still need the OP to confirm it. – user1551 Dec 3 '12 at 12:16 Assuming that $a,b$ are scalars, it is just $$\frac{ab}{a^2+b^2}I_n$$ The question is not entirely clear, but here goes. First, $$a^2I_n+b^2I_n=(a^2+b^2)I_n$$ It follows that $$(a^2I_n+b^2I_n)^{-1}=(a^2+b^2)^{-1}I_n$$ Now I'm going to assume that $(aI_nbI_n)$ is actually the $n$-by-$2n$ matrix $(aI_n\ \ bI_n)$. Then $$(a^2I_n+b^2I_n)^{-1}(aI_n\ \ bI_n)=(a^2+b^2)^{-1}I_n(aI_n\ \ bI_n)=(a^2+b^2)^{-1}(aI_n\ \ bI_n)=\left({a\over a^2+b^2}I_n\ \ {b\over a^2+b^2}I_n\right)$$	2016-04-30 12:04:01	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.9398369789123535, "perplexity": 311.22463977244007}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-18/segments/1461860111838.20/warc/CC-MAIN-20160428161511-00202-ip-10-239-7-51.ec2.internal.warc.gz"}
https://ftp.aimsciences.org/article/doi/10.3934/dcdss.2022041	# American Institute of Mathematical Sciences doi: 10.3934/dcdss.2022041 Online First Online First articles are published articles within a journal that have not yet been assigned to a formal issue. This means they do not yet have a volume number, issue number, or page numbers assigned to them, however, they can still be found and cited using their DOI (Digital Object Identifier). Online First publication benefits the research community by making new scientific discoveries known as quickly as possible. Readers can access Online First articles via the “Online First” tab for the selected journal. ## On some qualitative aspects for doubly nonlocal equations Dipartimento di Matematica, Università degli Studi di Bari Aldo Moro, Via E. Orabona 4, 70125 Bari, Italy Corresponding author: Silvia Cingolani Dedicated to the memory of Professor Rosella Mininni, a brilliant mathematician, and a very nice person Received October 2021 Early access February 2022 In this paper we investigate some qualitative properties of the solutions to the following doubly nonlocal equation $$$\label{eq_abstract} (- \Delta)^s u + \mu u = (I_\alphaF(u))F'(u) \quad \text{in } \mathbb{R}^N \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{(P)}$$$ where $N \geq 2$ , $s\in (0, 1)$ , $\alpha \in (0, N)$ , $\mu>0$ is fixed, $(-\Delta)^s$ denotes the fractional Laplacian and $I_{\alpha}$ is the Riesz potential. Here $F \in C^1(\mathbb{R})$ stands for a general nonlinearity of Berestycki-Lions type. We obtain first some regularity result for the solutions of (P). Then, by assuming $F$ odd or even and positive on the half-line, we get constant sign and radial symmetry of the Pohozaev ground state solutions related to equation (P). In particular, we extend some results contained in [23]. Similar qualitative properties of the ground states are obtained in the limiting case $s = 1$ , generalizing some results by Moroz and Van Schaftingen in [52] when $F$ is odd. Citation: Silvia Cingolani, Marco Gallo. On some qualitative aspects for doubly nonlocal equations. Discrete and Continuous Dynamical Systems - S, doi: 10.3934/dcdss.2022041 ##### References: [1] C. Argaez and M. Melgaard, Solutions to quasi-relativistic multi-configurative Hartree-Fock equations in quantum chemistry, Nonlinear Anal., 75 (2012), 384-404. doi: 10.1016/j.na.2011.08.038. [2] T. Bartsch, Y. Liu and Z. Liu, Normalized solutions for a class of nonlinear Choquard equations, SN Partial Differ. Equ. Appl., 1 (2020), Paper No. 34, 25 pp. doi: 10.1007/s42985-020-00036-w. [3] W. Beckner, Sobolev inequalities, the Poisson semigroup, and analysis on the sphere $S^n$, Proc. Natl. Acad. Sci. USA, 89 (1992), 4816-4819. doi: 10.1073/pnas.89.11.4816. [4] J. 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[62] P. Tod and I.M. Moroz, An analytical approach to the Schrödinger-Newton equations, Nonlinearity, 12 (1999), 201-216. doi: 10.1088/0951-7715/12/2/002. [63] M. Willem, Minimax Theorems, Progress in Nonlinear Differential Equations and Their Applications, 24, Birkäuser, 1996. doi: 10.1007/978-1-4612-4146-1. [1] Patricio Felmer, César Torres. Radial symmetry of ground states for a regional fractional Nonlinear Schrödinger Equation. Communications on Pure and Applied Analysis, 2014, 13 (6) : 2395-2406. doi: 10.3934/cpaa.2014.13.2395 [2] Jincai Kang, Chunlei Tang. Ground state radial sign-changing solutions for a gauged nonlinear Schrödinger equation involving critical growth. Communications on Pure and Applied Analysis, 2020, 19 (11) : 5239-5252. doi: 10.3934/cpaa.2020235 [3] Zupei Shen, Zhiqing Han, Qinqin Zhang. Ground states of nonlinear Schrödinger equations with fractional Laplacians. 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Communications on Pure and Applied Analysis, 2016, 15 (3) : 991-1008. doi: 10.3934/cpaa.2016.15.991 2021 Impact Factor: 1.865	2022-08-18 16:38:01	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 1, "x-ck12": 0, "texerror": 0, "math_score": 0.5352596044540405, "perplexity": 2923.308325582608}, "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/1659882573242.55/warc/CC-MAIN-20220818154820-20220818184820-00738.warc.gz"}
http://www.jessesadler.com/tags/dh-2.0/	## Introduction to Network Analysis with R ### Creating static and interactive network graphs Over a wide range of fields network analysis has become an increasingly popular tool for scholars to deal with the complexity of the interrelationships between actors of all sorts. The promise of network analysis is the placement of significance on the relationships between actors, rather than seeing actors as isolated entities. The emphasis on complexity, along with the creation of a variety of algorithms to measure various aspects of networks, makes network analysis a central tool for digital humanities.1 This post will provide an introduction to working with networks in R, using the example of the network of cities in the correspondence of Daniel van der Meulen in 1585. There are a number of applications designed for network analysis and the creation of network graphs such as gephi and cytoscape. Though not specifically designed for it, R has developed into a powerful tool for network analysis. The strength of R in comparison to stand-alone network analysis software is three fold. In the first place, R enables reproducible research that is not possible with GUI applications. Secondly, the data analysis power of R provides robust tools for manipulating data to prepare it for network analysis. Finally, there is an ever growing range of packages designed to make R a complete network analysis tool. Significant network analysis packages for R include the statnet suite of packages and igraph. In addition, Thomas Lin Pedersen has recently released the tidygraph and ggraph packages that leverage the power of igraph in a manner consistent with the tidyverse workflow. R can also be used to make interactive network graphs with the htmlwidgets framework that translates R code to JavaScript. This post begins with a short introduction to the basic vocabulary of network analysis, followed by a discussion of the process for getting data into the proper structure for network analysis. The network analysis packages have all implemented their own object classes. In this post, I will show how to create the specific object classes for the statnet suite of packages with the network package, as well as for igraph and tidygraph, which is based on the igraph implementation. Finally, I will turn to the creation of interactive graphs with the vizNetwork and networkD3 packages. ## Geocoding with R ### Using ggmap to geocode and map historical data In the previous post I discussed some reasons to use R instead of Excel to analyze and visualize data and provided a brief introduction to the R programming language. That post used an example of letters sent to the sixteenth-century merchant Daniel van der Meulen in 1585. One aspect missing from the analysis was the visualization of the geographical aspects of the data. This post will provide an introduction to geocoding and mapping location data using the ggmap package for R, which enables the creation of maps with ggplot. There are a number of websites that can help geocode location data and even create maps.1 You could also use a full-scale geographic information systems (GIS) application such as QGIS or ESRI. However, an active developer community has made it possible to complete a full range of geographic analysis from geocoding data to the creation of publication-ready maps with R.2 Geocoding and mapping data with R instead of a web or GIS application brings the general advantages of using a programming language in analyzing and visualizing data. With R, you can write the code once and use it over and over, while also providing a record of all your steps in the creation of a map.3 This post will merely scratch the surface of the mapping capabilities of R and will not enter into the domain of the more complex specific geographic packages available for R.4 Instead, it will build on the dplyr and ggplot skills discussed in my brief introduction to R. The example of geocoding and mapping with R will also provide another opportunity to show the advantages of coding. In particular, geocoding is a good example of how code can simplify the workflow for entering data. Instead of dealing with separate spreadsheets, one containing information about the letters and the other with geographic information, with code the geographic information can be created directly from the contents of the letters data. This has the added advantage that the code to find the longitude and latitude of locations can be saved as a R script and rerun if new data is added to ensure that the information is always kept up to date. ## Excel vs R: A Brief Introduction to R ### With examples using dplyr and ggplot Quantitative research often begins with the humble process of counting. Historical documents are never as plentiful as a historian would wish, but counting words, material objects, court cases, etc. can lead to a better understanding of the sources and the subject under study. When beginning the process of counting, the first instinct is to open a spreadsheet. The end result might be the production of tables and charts created in the very same spreadsheet document. In this post, I want to show why this spreadsheet-centric workflow is problematic and recommend the use of a programming language such as R as an alternative for both analyzing and visualizing data. There is no doubt that the learning curve for R is much steeper than producing one or two charts in a spreadsheet. However, there are real long-term advantages to learning a dedicated data analysis tool like R. Such advice to learn a programming language can seem both daunting and vague, especially if you do not really understand what it means to code. For this reason, after discussing why it is preferable to analyze data with R instead of a spreadsheet program, this post provides a brief introduction to R, as well as an example of analysis and visualization of historical data with R.1 ## My Approach to Digital Humanities Digital humanities holds the promise of increasing the means by which scholars are able to analyze and present data. Though some sentiments about the significance of digital humanities might be overblown, there is no doubt that the more ways we have to analyze sources the better. Learning a variety of the tools that make up the rather nebulous universe of digital humanities is like learning a new language. It opens up new possibilities that were previously closed or necessitated the expertise of others. This frames digital humanities as a collection of skills rather than a means to a predetermined end. I have adopted this perspective in learning about the possibilities opened by digital humanities and working on a digital humanities project. I am hardly the first person to take these steps, but I hope that by explaining my thought process I can set a basis for future posts on digital humanities. If there has been one guiding force in my approach to digital humanities, it is to learn skills and tools in the process of production. In a way, this is simply the application of critical thinking to how I create my scholarly work. Instead of doing things in what seems the de facto manner, I have sought to question if there is either a more efficient way or a way in which I could gain or improve my competency. The goal of efficiency was particularly significant in thinking about how to organize my research and writing (DH 1.0), while learning new skills has been more important in the production of digital humanities projects (DH 2.0). This may not be the easiest way to complete a project in the short run, but by doing things the hard way, I am looking to open up new opportunities for future projects. With this theoretical approach in mind, let me now discuss a few concrete principles of my digital humanities practices concerning text, applications, and producing digital humanities projects.	2018-01-21 22:19:02	{"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": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.2564595341682434, "perplexity": 715.493565667181}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-05/segments/1516084890893.58/warc/CC-MAIN-20180121214857-20180121234857-00261.warc.gz"}
http://connection.ebscohost.com/c/articles/91942090/vertex-graceful-labeling-some-path-related-graphs	TITLE # Vertex Graceful Labeling-Some Path Related Graphs AUTHOR(S) Selvaraju, P.; Balaganesan, P.; Renuka, J. PUB. DATE September 2013 SOURCE International Journal of Mathematical Combinatorics;Sep2013, Vol. 3, p44 SOURCE TYPE DOC. TYPE Article ABSTRACT In this article, we show that an algorithm for VG of a caterpillar and proved that A(mj, âˆª P3 is vertex graceful if mj is monotonically increasing, 2 â‰¤ j â‰¤ n, when n is odd, 1 â‰¤ m2 â‰¤ 3 and m1 â‰¤ m2, (mj, n) âˆª P3 is vertex graceful if mj is monotonically increasing, 2 â‰¤ j â‰¤ n, when n is odd, 1 â‰¤ m2 = 3, m1 < m2 and Cn âˆª Cn+1 is vertex graceful if and only if n â‰¥ 4. ACCESSION # 91942090 ## Related Articles • The Cycle Discrepancy of Three-Regular Graphs. Abbasi, Sarmad; Aslam, Laeeq // Graphs & Combinatorics;Jan2011, Vol. 27 Issue 1, p27 Let G = ( V, E) be an undirected graph and $${{\mathcal C}(G)}$$ denote the set of all cycles in G. We introduce a graph invariant cycle discrepancy, which we define as We show that, if G is a three-regular graph with n vertices, then This bound is best possible and is achieved by very simple... • On Mean Graphs. Vasuki, R.; Arockiaraj, S. // International Journal of Mathematical Combinatorics;Sep2013, Vol. 3, p22 Let G(V,E) be a graph with p vertices and q edges. For every assignment f : V (G) â†’ {0, 1, 2, 3, ..., q}, an induced edge labeling f* : E(G) â†’ {1, 2, 3, ..., q} is Â… if f(u) and f(v) are of the same parity otherwise for every edge uv âˆˆ E(G). If f*(E) = {1, 2, ..., q}, then... • On Mean Cordial Graphs. Ponraj, R.; Sivakumar, M. // International Journal of Mathematical Combinatorics;Sep2013, Vol. 3, p78 Let f be a function from the vertex set V (G) to {0, 1, 2}. For each edge uv assign the label [f(u) + f(v)/2]. f is called a mean cordial labeling if \|vf(i) - vf(j)\| â‰¤ 1 2 and \|ef(i) - ef(j)\| â‰¤ 1, i, j âˆˆ {0, 1, 2}, where vf(x) and ef(x) respectively are denote the number of... • Solution of a Conjecture on Skolem Mean Graph of Stars K1,l Ï… K1,m Ï… K1,n. Balaji, V. // International Journal of Mathematical Combinatorics;Dec2011, Vol. 4, p115 In this paper, we prove a conjecture that the three stars K1,l Ï… K1,m Ï… K1,n is a skolem mean graph if \|m - n\| < 4 + l for integers l, m â‰¤ 1 and l â‰¥ m < n. • H-magic covering on some classes of graphs. Roswitha, Mania; Baskoro, Edy Tri // AIP Conference Proceedings;5/22/2012, Vol. 1450 Issue 1, p135 For a graph G(V,E), an edge-covering of G is a family of different subgraphs H1,...Hk such that any edge of E belongs to at least one of the subgraphs Hi, 1 â‰¤ i â‰¤ k. If every Hi is isomorphic to a given graph H, then G admits an H-covering. Graph G is said to be H-magic if G has an... • Vertex-antimagic labelings of regular graphs. Ahmad, Ali; Ali, Kashif; Bača, Martin; Kovář, Petr; Semaničová-Feňovčíková, Andrea // Acta Mathematica Sinica;Sep2012, Vol. 28 Issue 9, p1865 Let G = ( V,E) be a finite, simple and undirected graph with p vertices and q edges. An ( a, d)-vertex-antimagic total labeling of G is a bijection f from V ( G) âˆª E( G) onto the set of consecutive integers 1, 2, ..., p + q, such that the vertex-weights form an arithmetic progression with... • Degree Splitting Graph on Graceful, Felicitous and Elegant Labeling. Selvaraju, P.; Balaganesan, P.; Renuka, J.; Balaj, V. // International Journal of Mathematical Combinatorics;Apr2012, Vol. 2, p96 We show that the degree splitting graphs of Bn,n; Pn; Km,n; n(k4 -3e)I; n(k4 - 3e)II(b); n(k4 - e)II and n(k4 - 2e)II(a) are graceful [3]. We prove C3 ï¿½K1,n is graceful, felicitous and elegant [2], Also we prove K2,n is felicitous and elegant. • Fibonacci and Super Fibonacci Graceful Labelings of Some Cycle Related Graphs. Vaidya, S. K.; Prajapati, U. M. // International Journal of Mathematical Combinatorics;Dec2011, Vol. 4, p59 No abstract available. • UNDERLYING GRAPHS OF 3-QUASI-TRANSITIVE DIGRAPHS AND 3-TRANSITIVE DIGRAPHS RUIXIA WANG, SHIYING WANG. RUIXIA WANG; SHIYING WANG // Discussiones Mathematicae: Graph Theory;2013, Vol. 33 Issue 2, p429 A digraph is 3-quasi-transitive (resp. 3-transitive), if for any path x0x1 x2x3 of length 3, x0 and x3 are adjacent (resp. x0 dominates x3). CÃ©sar HernÃ¡ndez-Cruz conjectured that if D is a 3-quasi-transitive digraph, then the underlying graph of D, UG(D), admits a 3-transitive orientation.... Share	2018-07-21 09:34:30	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.6899552941322327, "perplexity": 6537.361844876874}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-30/segments/1531676592475.84/warc/CC-MAIN-20180721090529-20180721110529-00420.warc.gz"}
http://survey.aperiodical.com/	# Exactly how bad is the 13 times table? Let’s recite the $13$ times table. Pay attention to the first digit of each number: \begin{array}{l} \color{blue}13, \\ \color{blue}26, \\ \color{blue}39, \\ \color{blue}52 \end{array} What happened to $\color{blue}4$‽ A while ago I was working through the $13$ times table for some boring reason, and I was in the kind of mood to find it really quite vexing… Read more » # The OEIS now contains 300,000 integer sequences The Online Encyclopedia of Integer Sequences just keeps on growing: at the end of last month it added its 300,000th entry. Especially round entry numbers are set aside for particularly nice sequences to mark the passing of major milestones in the encyclopedia’s size; this time, we have four nice sequences starting at A300000. These were… Read more » # Proof by sedition $n > 2$ An unexpected bit of controversy involving mathematical notation hit the internet last week, when China's government briefly blocked all Chinese internet users from viewing any page or message containing the letter n. Apparently, those in charge of the Great Firewall feared that those who disapproved of Xi Jinping removing the two-term limit on his presidency of China would use the letter n to refer to the now-arbitrary number of terms for which he can remain in power. There's some more context in a post by Victor Mair on Language Log, and in the Guardian. # LMS Meeting in honour of Maryam Mirzakhani The London Mathematical Society are organising an event later this month in honour of the late Fields Medalist Maryam Mirzakhani. It's at the University of Warwick on 22nd March, and will include talks outlining some of Mirzakhani's work, followed by a drinks reception and dinner. The event is part of a larger EPSRC symposium on Teichmüller dynamics. # Faces of Women in Mathematics This is nice for International Women's Day. Filmmaker Irina Linke and mathematician Eugénie Hunsicker have put together this montage of women in maths from all around the world. # πkm Running Challenge: 7-day update Katie’s running πkm per day for charity, every day in March. Seven days in, here’s how she’s getting on. # Carnival of Mathematics 155 The next issue of the Carnival of Mathematics, rounding up blog posts from the month of February, and compiled by Ben, is now online at Math Off The Grid. The Carnival rounds up maths blog posts from all over the internet, including some from our own Aperiodical. See our Carnival of Mathematics page for more information.	2018-03-21 03:19:31	{"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": 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": 0.6786243915557861, "perplexity": 4029.0536434634787}, "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-2018-13/segments/1521257647567.36/warc/CC-MAIN-20180321023951-20180321043951-00504.warc.gz"}
https://mathoverflow.net/questions/249068/is-there-a-concept-of-uniform-hausdorff-dimension	# Is there a concept of uniform Hausdorff dimension? Let $M$ be a metric space and let $U \subset M$ be open. Then the Hausdorff dimension of $U$ is defined in the usual way. If there is a single dimension number $d$ that is the Hausdorff dimension of every open set $U$ in $M$ I say the metric space $M$ has uniform Hausdorff dimension $d$. What I want to know is whether some concept like this already exists, possible under a different name. I want to use this as a property that 'nice' metric spaces have. A simple example: Let $I$ denote the unit interval and $D$ the unit ball in $\mathbb{R}^2$. Then both $I$ and $D$ have a uniform Hausdorff dimension but the disjoint union of $I$ and $D$ does not. • Well, the unit interval $I$ is not open as a subset of $\mathbb{R}^2$, so it doesn't invalidate the "uniform Hausdorff dimension" of the disjoint union of $I$ and $D$ being 2 according to your definition. – N Unnikrishnan Sep 6 '16 at 7:01 There is a significantly stronger property that is commonly used and does have a name: A metric space is called Ahlfors $n$-regular if there is a constant $K$ such that for each closed ball $B(x,r)$ in the space ($0<r\leq\text{diam}(X)$), $$K^{-1} r^n \leq H^n(B(x,r)) \leq K r^n$$ where $H^n$ is $n$-dimensional Hausdorff measure. In your example, $I$ and $D$ are of course Ahlfors $1$-regular and Ahlfors $2$-regular, respectively, while the union is not Ahlfors regular for any $n$.	2020-02-27 11:58:28	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.9291249513626099, "perplexity": 74.3968989739346}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-10/segments/1581875146681.47/warc/CC-MAIN-20200227094720-20200227124720-00347.warc.gz"}
https://tex.stackexchange.com/questions/175410/how-to-format-multiple-authors-separated-by-and	How to format multiple authors separated by \and? I have a document class for theses which, up until now, only accepted one author name. But now, due to a change in the university's regulation, the document will have to accept more than one author name. Currently, I am using the standard \author{} command to get the author name . My question is: how can I get the separate author names in order to apply the correct formatting to them? Are there different variable names for each one? I know multiple authors are separated by \and and, after I am able to get each author name, I would like to put them one below the other, in a centered position (just like the example below, but with each additional author name below the previous one). Here I have the code of my \maketitle command, which is the only one that uses the author names. In the image below, I show the resulting title page. \renewcommand{\maketitle}{ \clearpage % clears pages just to be sure \begin{center} %text is centered \textbf{ % and bold \MakeUppercase{\@institution}\\[\baselineskip] % institution name \MakeUppercase{\@author} % author name \vfill \MakeUppercase{\@title}\ifthenelse{\isundefined{\@subtitle}}{}{:\@subtitle}} % title and optional subtitle \vfill \@city\\ % city \number\year %year \end{center} } Clarifying Update I believe I wasn't clear before, but now I have additional information regarding the issue. Following the recent answer, I decided to forget about \and and try to separate author names with \\ so they could each go in a separate line, but it seems that \MakeUppercase{} does not allow for line breaks to be inserted inside it. Any advice? So you see, my issue is not related to the \and command in itself, but rather with a way to break a string in many lines while keeping it in uppercase. I guess an example of what I'm trying to do, but which is not working, is the following: \documentclass{report} \author{Doug \\ Lou} \begin{document} \MakeUppercase{\@author} \end{document} End result update Just for posterity, I will add here the end result, after using the chosen solution from this post. \def\and{\\} % redefine so it is compatible with other classes \renewcommand{\maketitle}{ \clearpage % clears pages just to be sure \begin{center} %text is centered \textbf{ % and bold \MakeUppercase{\@institution}\\[\baselineskip] % institution name \uppercase\expandafter{\@author} % uppercase author names!! \vfill \MakeUppercase{\@title}\ifthenelse{\isundefined{\@subtitle}}{}{:\@subtitle}} % title and optional subtitle \vfill \@city % city \number\year % year \end{center} } • For a more specific answer, you would have to include a compilable MWE that recreates your situation. As it is, I cannot make your code snippet work in action, as it has too many user-defined portions. – cslstr May 5 '14 at 18:12 • @cslstr I just expanded my question based on your answer. – Douglas De Rizzo Meneghetti May 5 '14 at 20:25 The report.cls defines the author portion of \maketitle to be a tabular array, that is {\large \lineskip .75em% \begin{tabular}[t]{c}% \@author \end{tabular}\par}% The default definition of \and from latex.ltx is used to end the tabular environment, add a bit of horizontal space, and then start a new tabular environment: \def\and{% % \begin{tabular} \end{tabular}% \hskip 1em \@plus.17fil% \begin{tabular}[t]{c}}% % \end{tabular} So, when you write \author{First \and Second}, the result is that both names get typeset side-by-side in individual tabular environments. You could just as easily write \author{First \\ Second} to have the two names typeset in a column (that is, in the single centered column of the first tabular environment). The same could be done by redefining \and as \def\and{\\}; only through chance does simply writing \author{First & Second} do the same thing, as the "extra" alignment character, &, is interpreted as \\ automatically. But this last approach will only work for you if you include something along the lines of the original author portion of \maketitle, that defines the tabular array in your own \maketitle. Updated from OP Edit: You can use \uppercase to do the uppercase translation in the following manner. As noted, I suggest you use the entire structure (with tabular) as originally defined in report.cls for the author portion in your own \maketitle. This keeps the original format with \and working, along with the other methods I mention above. \documentclass{report} \author{Doug \\ Lou} \begin{document} \makeatletter \begin{center} % Suggest using the tabular in your \maketitle definition {\large \lineskip .75em% \begin{tabular}[t]{c}% \uppercase\expandafter{\@author} \end{tabular}\par}% \end{center} \makeatother \end{document} • That is very enlightening, but it did not solve my problem, unfortunately. I just expanded my question to see if I could clarify some things. – Douglas De Rizzo Meneghetti May 5 '14 at 20:26 • @DouglasDeRizzoMeneghetti I have updated my answer. Also note that you must wrap any code using the @ character in \makeatletter ... \makeatother for it to compile correctly. – cslstr May 5 '14 at 20:48 • That's exact what I needed, that line with \uppercase\expandafter did the trick. Since I will be using this inside a cls file, there is no need to use \makeatletter and \makeatother. I also added \def\and{\\} so that it would be compatible with my class style. Thank you very much. – Douglas De Rizzo Meneghetti May 6 '14 at 1:33 Addressing only the matter that's mentioned in the "update" part of your posting: Whenever a macro -- in the present case, \@author -- contains one or more @ ("at") symbols, one has to encase it (the macro) in a pair of \makeatletter and \makeatother directives. \documentclass{report} \author{Doug \\ Lou} \begin{document} \makeatletter % new \noindent % new, for more consistent formatting \MakeUppercase{\@author} \makeatother % new \end{document} • As I explained above, my plan is to use this code inside a cls file, so I will not need to encase it between \makeatletter and \makeatother. I only did that in the question so I could give the minimal working example. – Douglas De Rizzo Meneghetti May 6 '14 at 1:47	2019-11-12 23:13:23	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 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": 0.9753080010414124, "perplexity": 1584.9795535863977}, "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-2019-47/segments/1573496665809.73/warc/CC-MAIN-20191112230002-20191113014002-00141.warc.gz"}
http://nklein.com/math/clifford-algebras	## “Visualizing Quaternions” by Andrew J. HansonJuly 10th, 2009 Patrick Stein Sunday night, I finished reading Visualizing Quaternions by Andrew J. Hanson. Unfortunately, the events of the week have kept me from writing this summary sooner. I feel like it would have been longer had I written it Monday. Alas, let’s play the cards in front of me. This is a wonderful book. In addition to having gobs and gobs of useful content, it is a fine specimen of book design. The layout and illustrations are tremendous. It tends to go from well-written, conversational passages that explain things in great detail to terse passages that mean nothing if you’re not a tensor-analysis whiz. But, even if you skim the parts that assume you know things you don’t, there’s still lots of book here to read. There is even a nice appendix about Clifford algebras which folds in nicely with the complex number, quaternions, Clifford algebra posts that I’ve made here recently. If you do three-dimensional simulations or you like a good mathematical read, you should give this book a look. ## Clifford Algebras for Rotating, Scaling, and Translating SpaceJuly 6th, 2009 Patrick Stein In (very much) earlier articles, I described: Today, it is time to tackle rotating, translating, and scaling three-dimensional space using Clifford algebras. ### Three dimensions now instead of two Back when we used Clifford algebras to rotate, translate, and scale the plane, we were using the two-dimesional Clifford algebra. With the two-dimensional Clifford algebra, we represented two-dimensional coordinates $(x,y)$ as $xe_1 + ye_2$. It shouldn’t surprise you then to find we’re going to represent three-dimensional coordinates $(x,y,z)$ as $xe_1 + ye_2 + ze_3$. As before, we will have $e_1e_1 = 1$ and $e_2e_2 = 1$. Similarly, we will have $e_3e_3 = 1$. In the two-dimesional case, we showed that $e_1e_2 = -e_2e_1$. By the same logic as the two-dimensional case, we also find that $e_1e_3 = -e_3e_1$ and $e_2e_3 = - e_3e_2$. We could potentially also end up multiplying $e_1$, $e_2$, and $e_3$ all together. This isn’t going to be equal to any combination of the other things we’ve seen so we’ll just leave it written $e_1e_2e_3$. ## Quaternions for Rotating, Scaling, and Translating SpaceJune 11th, 2009 Patrick Stein In earlier posts, I described how complex numbers can be used to rotate, scale, and translate the plane, how Clifford algebras can be used to rotate, scale, and translate the plane, and why I resorted to an awkward trick for the Clifford algebra rotations of the plane. In this post, I am going to explain what the quaternions are and describe how they can be used to represent a rotation in three-dimensional space. ### What are the quaternions Okay, remember how we got the complex numbers? We needed something that was the square root of negative one. Now, imagine that you are Sir William Rowan Hamilton. The year is 1843. It is springtime. You know how to use the complex numbers to represent points in the plane. And, you know that when you do that, you can use complex numbers to rotate, scale, and translate the points. That’s all well and good, but you don’t live in a two-dimensional world. How are you going to do the same sort of thing with three-dimensional space? How are you going to multiply triples? You spend months on this. If only you could say, How about I let there be another number that is different from $i$ (and from $-i$) that has the same property that its square is negative one? You fight with this for months. You try to represent a point with coordinates $(x,y,z)$ as $x + yi + zj$. But, nothing you come up with makes any sense. Your kids are harassing you, Daddy, did you figure out how to multiply triples yet? You have to answer them every morning with a polite, No, not yet. Then, you’re walking along the Royal Canal in Dublin. It’s mid-October already. My, how the year has flown by. Bam, it hits you. If you add a third number like $i$ and $j$ which is equal to $i\cdot j$, everything works out. You get so excited, that you carve your equations into a stone bridge over the canal: $i^2 = j^2 = k^2 = ijk = -1$ ## What Was Up With That Rotation Trick?June 10th, 2009 Patrick Stein In my prior post about using Clifford algebras to do plane rotations, I finished with a non-intuitive step at the end. Rather than multiplying on the right by an element representing a rotation of angle $\theta$, I multiplied on the left by an element representing a rotation of angle $\frac{\theta}{2}$ and multiplied on the right by an element representing a rotation of angle $-\frac{\theta}{2}$. Why did I do this? Well, I mentioned it would be awkward for the two-dimensional case, but that it will be important when we get to three or more dimensions. Well, work for a moment with $\frac{\theta}{2}$ being a quarter rotation (ninety degrees, $\frac{\pi}{2}$ radians). This means our total rotation is going to be a half turn (180 degrees, $\pi$ radians). For that $\frac{\theta}{2}$, $r = e_1e_2$ and so $\overline{r} = -e_1e_2$. Let’s just look at what it does to our unit vectors $e_1$ and $e_2$ to multiply on the left by $r$ and on the right by $\overline{r}$. For $e_1$, we get $-e_1e_2e_1e_1e_2 = -e_1e_2e_2 = -e_1$. Similarly, for $e_2$, we get $-e_1e_2e_2e_1e_2 = -e_2$. So far, we were only working in two dimensions. As such, there wasn’t any $e_3$ to worry about. But, what if there were? What happens to the $z$-coordinate of something if you rotate things parallel to the $xy$-plane? It remains unchanged. Well, what would happen if we multiplied $e_3$ on the right by $\cos\theta + \sin\theta e_1e_2$? We would end up with $\cos\theta e_3 + \sin\theta e_3e_1e_2 = \cos\theta e_3 + \sin\theta e_1e_2e_3$. We’ve ended up scaling $e_3$ and adding in a trivector $e_1e_2e_3$. We’ve made a mess. Let’s try it instead with our trick. We’re going to start with $-e_1e_2e_3e_1e_2$. Every time we transpose elements with different subscripts, we flip the sign. Every time we get two elements next to each other with the same subscript, they cancel out. So, switching the $e_3$ with the second $e_1$, we get $e_1e_2e_1e_3e_2$. From there, we can switch the first two elements to get $-e_2e_1e_1e_3e_2$ which is just $-e_2e_3e_2$. We can switch the $e_3$ with the second $e_2$ to get: $e_2e_2e_3$ which is just $e_3$. So, our trick leaves $e_3$ unchanged. In the above, there is nothing special about the subscript three. It would work for any subscript except one or two. So, the trick allows us to break the rotation up into two parts that still do what we want with $e_1$ and $e_2$ but leave our other directions unchanged (or, maybe it’s easier to think of them as changing them and then changing them right back). ## Clifford Algebras for Rotating, Scaling, and Translating the PlaneJune 8th, 2009 Patrick Stein In my previous post, I reviewed how the complex numbers can be used to represent coordinates in the plane and how, once you’ve done that, complex arithmetic leads naturally to rotations, scalings, and translations of the plane. Today, we’re going to do the same with the Clifford algebra $\mathcal{C}\ell_2$. ### What are Clifford Algebras In our previous post, we used two different ways to represent coordinates in the plane. We used an ordered pair of real numbers like $(3,5)$ and we used the real and imaginary parts of a complex number like $3 + 5i$. Another way we could have written the coordinates in the plane is as a vector. Typically, to express a vector, we pick an axis (say, the x-axis) and then pick a second axis perpendicular to it (say, the y-axis). Most physics books would call these $\hat{i}$ and $\hat{j}$. I am going to use $e_1$ and $e_2$ instead so that there is no confusion with $i = \sqrt{-1}$ and because it is a notation that we will continue throughout Clifford algebras. We could then express $(3,5)$ as $3e_1 + 5e_2$. Here, $e_1$ and $e_2$ are called unit vectors. We assume they have length one so that when we take three of them, we get something with length three (for arbitrary values of three). Having done that, we can easily multiply any vector by a real number using the normal distributive law: $s \cdot \left(xe_1 + ye_2\right) = (sx)e_1 + (sy)e_2$. And, we can add vectors just like we added complex numbers: summing like parts. So, $(ae_1 + be_2) + (ce_1 + de_2) = (a+c)e_1 + (b+d)e_2$. Of course, subtraction goes the same way: $(ae_1 + be_2) - (ce_1 + de_2) = (a-c)e_1 + (b-d)e_2$.	2013-05-25 01:58:57	{"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": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 75, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7185661792755127, "perplexity": 336.3651265827977}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2013-20/segments/1368705318091/warc/CC-MAIN-20130516115518-00016-ip-10-60-113-184.ec2.internal.warc.gz"}
https://scirate.com/arxiv/1703.06132	# Implications of the AdS/CFT Correspondence on Spacetime and Worldsheet OPE Coefficients We explore the connection between the operator product expansion (OPE) in the boundary and worldsheet conformal field theories in the context of AdS$_{d+1}$/CFT$_d$ correspondence. Considering single trace scalar operators in the boundary theory and using the saddle point analysis of the worldsheet OPE [1], we derive an explicit relation between OPE coefficients in the boundary and worldsheet theories for the contribution of single trace spin $\ell$ operators to the OPE. We also consider external vector operators and obtain the relation between OPE coefficients for the exchange of single trace scalar operators in the OPE. We revisit the relationship between the bulk cubic couplings in the Supergravity approximation and the OPE coefficients in the dual boundary theory. Our results match with the known examples from the case of AdS$_3$/CFT$_2$. For the operators whose two and three point correlators enjoy a non renormalization theorem, this gives a set of three way relations between the bulk cubic couplings in supergravity and the OPE coefficients in the boundary and worldsheet theories. Submitted 17 Mar 2017 to High Energy Physics - Theory [hep-th] Published 20 Mar 2017 Updated 25 Apr 2017 Author comments: 34 pages, 1 figure http://arxiv.org/abs/1703.06132 http://arxiv.org/pdf/1703.06132.pdf	2017-04-26 21:25:41	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.7317471504211426, "perplexity": 653.6584250476255}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-17/segments/1492917121665.69/warc/CC-MAIN-20170423031201-00027-ip-10-145-167-34.ec2.internal.warc.gz"}
https://www.physicsforums.com/threads/the-origin.803265/	# The origin 1. Mar 15, 2015 ### LagrangeEuler Almost always in xy plane we take that origin is $(0,0)$. Is it possible to take that origin is in the point $(1,1)$, or some other point? 2. Mar 15, 2015 ### Staff: Mentor The point (0,0) is called origin. That is just a definition. It does not make sense to give another point the same name. 3. Mar 15, 2015 ### LagrangeEuler If I want to use some translation in x-axis I need more then one coordinate system, for example. So origin of first system for instance is $(0,0)$ and for second is $(2,0)$? 4. Mar 15, 2015 ### DrewD The origin of the second system is at $(2,0)$ in the coordinates of the first system, but using the coordinates of the second system the origin, by definition, is at $(0,0)$. As mfb said, it doesn't make sense to use the term to apply to some other point in the coordinate system.	2017-10-23 06:34:07	{"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": 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": 0.6487755179405212, "perplexity": 538.7922067576079}, "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/1508187825700.38/warc/CC-MAIN-20171023054654-20171023074654-00813.warc.gz"}
https://mathoverflow.net/questions/331644/generalizing-polynomial-identities-for-rings	# Generalizing polynomial identities for rings For a ring $$R$$, a polynomial identity of $$R$$ is a polynomial (in non-commuting variables) $$f(x_1,\ldots,x_n)\in \mathbb{Z}[x_1,\ldots, x_n]$$ such that for any choice of $$a_i\in R$$, $$f(a_1,\ldots, a_n)=0$$. For example, for all $$n$$, the standard (or symmetric) polynomial $$S_{2n}(x_1,\ldots,x_{2n})=\sum_{\sigma\in S_{2n}}$$sign$$(\sigma)x_{\sigma(1)}\cdots x_{\sigma(2n)}$$ is a polynomial identity for the ring $$R=M_n(\mathbb{C})$$. I would like to understand how this idea might extend to considering, rather than polynomials in the $$x_i$$, polynomials in the $$x_i$$ and their inverses, i.e., an element of $$\mathbb{Z}\langle x_1,\ldots,x_n\rangle$$. Obviously, to even be defined, one would need to restrict any 'generalized' polynomial identity to the units of $$R$$. An easy example is: Example: If $$R^\times$$ is abelian, then $$f(x,y) = 1 - xyx^{-1}y^{-1}$$ is such a generalized polynomial identity. Similar statements can be made if $$R^\times$$ is nilpotent, solvable, &c. I'm primarily interested in the situation of $$R=M_n(\mathbb{C})$$, with $$R^\times=\textrm{GL}_n(\mathbb{C})$$. Is anyone aware of previous work on this subject? Are there any 'obvious' such identities? More optimistically, I wonder if anything might be said by taking the (skew) field of fractions of $$\mathbb{Z}[x_1,\ldots,x_n]$$ and asking what identities arise here. For example, something of the form $$f(x,y)=x(x-y)^{-1}y$$. In this case, the domain of the identity $$f$$ would need to be restricted to those $$x,y\in R$$ so that $$x-y\in R^\times$$. More generally, the domain might become even stranger (consider $$f(x)=(1+(1+x)^{-1})^{-1}$$, $$f(x)=(1+(1+(1+x)^{-1})^{-1})^{-1}$$, ...). I'll add that I'm not particularly worried about how nasty this domain of definition may be for a particular identity, just whether such an identity exists with nonempty domain; again, I'm interested in $$R=\textrm{GL}_n(\mathbb{C})$$, and here (I believe) for any fixed such $$f$$, the domain will be nonempty unless some inverted term is itself an identity. New contributor M. Nichols is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct.	2019-05-20 19:37:03	{"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": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 27, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9047136902809143, "perplexity": 331.60700327913486}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-22/segments/1558232256100.64/warc/CC-MAIN-20190520182057-20190520204057-00085.warc.gz"}
http://odin-osiris.usask.ca/ReST/SasktranIF/climatology/osiris_ecmwf.html	# OSIRIS_ECMWF¶ A climatology class that provides the background atmospheric state used in Odin-OSIRIS retrievals. The atmospheric state parameters are derived from the ECMWF climatology interpolated to individual OSIRIS scan. The advantage of the class is that a much smaller file size can store the ECMWF information for one scan and it is much quicker to read in the small file than the entire ECMWF climatology. The class only considers mjd when updating the cache and completely ignores any latitude and longitude parameters provided by the user. The class is very useful for providing the atmospheric state of OSIRIS scans but has limited value outside this purpose: mjd = 52393.3792987115; climate = ISKClimatology('OSIRIS_ECMWF') climate.UpdateCache( [0,0,0,mjd]) [ok, value] = climate.GetParameter( SKCLIMATOLOGY_AIRNUMBERDENSITY_CM3(), ... [0.0, 0.0, 25000, mjd]); [ok, value] = climate.GetParameter( SKCLIMATOLOGY_PRESSURE_PA(), [0.0, 0.0, 25000, mjd]); [ok, value] = climate.GetParameter( SKCLIMATOLOGY_TEMPERATURE_K(), [0.0, 0.0, 25000, mjd]); ## Supported Species¶ • SKCLIMATOLOGY_AIRNUMBERDENSITY_CM3 • SKCLIMATOLOGY_TEMPERATURE_K • SKCLIMATOLOGY_PRESSURE_PA ## Software Package¶ This climatology is installed as part of the OSIRIS Level 1 Services (Version 2.48 or higher) and your system has access to the OSIRIS Level 1 data products.	2017-03-23 20:21:10	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.17536279559135437, "perplexity": 7614.534980284443}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-13/segments/1490218187206.64/warc/CC-MAIN-20170322212947-00193-ip-10-233-31-227.ec2.internal.warc.gz"}
https://physics.stackexchange.com/questions/439187/2d-isotropic-quantum-harmonic-oscillator-polar-coordinates?noredirect=1	# 2D isotropic quantum harmonic oscillator: polar coordinates This may come a bit elemental, what I was working on a direct way to find the eigenfunctions and eigenvalues of the isotropic two-dimensional quantum harmonic oscillator but using polar coordinates: $$H=-\frac{\hbar}{2M}\left(\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}\right)+\frac{M\omega^2}{2}\left(x^2+y^2\right).$$ I can easily solve the 2-dimensional case in cartesian coordinates as we can separate the hamiltonian in independent oscillators for each coordinate. For the polar case in two dimensions, we can rewrite, $$H=-\frac{\hbar}{2M}\left(\frac{\partial^2}{\partial r^2}+\frac{1}{r}\frac{\partial}{\partial r}+\frac{1}{r^2}\frac{\partial^2}{\partial \phi^2}\right)+\frac{\omega^2}{2}r^2.$$ With $$r^2=x^2+y^2$$ and $$\phi=\arctan(y/x)$$. Using separation of variables $$\psi(r,\phi)=R(r)\Phi(\phi)$$ and plugging into the Schrödinger equation, we can easily solve for the angular part $$\Phi=e^{im\phi}$$, where $$m\in \mathbb{Z}$$. Plugging back into the Schrodinger equation, for the radial part, we get: $$r^2R''+rR'+(r^2E-m^2-M\omega^2r^4)R=0.$$ While I have an idea for the solution by making an analogy with the 3D case (where we get Laguerre polynomials), I'm not sure how to correctly proceed from here. I appreciate any input or even useful references* (all of the references I've found deal with the 3D case, which I have no problem solving). *I read online this problem is treated in the book of "Wave Mechanics" from Pauli, but unfortunately it isn't available on neither of my campus libraries nor online (it's only available for purchase and I lack the funds to buy it). • Thanks for the typo, I indeed forgot the last $R$. On the other hand, I indeed looked at those articles but they have specific solutions, and I was looking to construct the general one in terms of special functions. – Charlie Nov 6 '18 at 18:14 Indeed, as suggested by phase-space quantization, most of these equations are reducible to generalized Laguerre's, the cousins of Hermite. As universally customary, I absorb $$\hbar$$, M and ω into r,E. Note your E is twice the energy. Since $$r\geq 0$$ you don't lose negative values, and you may may redefine $$r^2\equiv x$$, so that $$r\partial_r = 2x \partial_x \qquad \Longrightarrow r\partial_r (r\partial_r)= r^2\partial_r^2+ r\partial_r=4(x^2\partial_x^2+x\partial_x),$$ hence your radial equation reduces to $$\left ( \partial_x^2+ \frac{1}{x}\partial_x +\frac{E-x}{4x} -\frac{m^2}{4x^2} \right ) R(m,E)=0 ~.$$ Now, further define $$R(m,E)\equiv x^{\|m\|/2} e^{-x/2} ~ \rho(m,E),$$ to get $$\partial_x R(m,E)= x^{\|m\|/2} e^{-x/2} \left (-1/2 +\frac{\|m\|}{2x} + \partial_x \right )~ \rho(m,E) \\ \partial_x^2 R(m,E)= x^{\|m\|/2} e^{x/2} \left (-1/2 +\frac{\|m\|}{2x} + \partial_x \right )^2~ \rho(m,E),$$ whence the generalized Laguerre equation, $$x \partial_x^2\rho(m,E) +\left({m+1} -x\right )\partial_x \rho(m,E)+\frac{1}{2}(E/2-m-1) \rho(m,E)=0~.$$ This equation has well-behaved solutions for non-negative integer $$k=(E/2-m-1)/2\geq 0 ~,$$ to wit, generalized Laguerre (Sonine) polynomials $$L^{(m)}_k (x)=x^{-m}(\partial_x -1)^k x^{k+m}/k!$$. Plugging into the factorized solution and the above substitutions nets your eigen-wavefunctions. The ground state is $$k=0=m$$, ($$E=2$$ in your conventions), so a radially symmetric Gaussian, $$e^{-r^2/2}$$. Again, in your idiosyncratic convention, the degeneracy is E/2. So, degeneracy 2 for $$E=4$$ : $$m=1$$, $$k=0$$; you may check this is just $$r e^{-r^2/2 +i\phi}$$. You may choose the $$\cos \phi$$ and $$\sin \phi$$ solutions, if you wish, constituting a doublet of the underlying degeneracy group SU(2). • I see, so indeed we should get Laguerre polynomials. I'll repeat the calculations according to what you put so I grasp the solution for this case and the general one. By the way, what do you mean by idiosincratic convention? If you refer to the case that m=1 I think someone badly edited my OP since I was actually working in the general case with arbitrary $m$ and $\omega$. I don't know why they removed the $m$ and $\omega$ (they didn't even leave it in natural units). – Charlie Nov 6 '18 at 18:13 • the energy is E /2 as things stand.... – Cosmas Zachos Nov 6 '18 at 20:59	2019-12-06 15:55:05	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 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": 27, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9182183146476746, "perplexity": 464.3185672039309}, "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/1575540488870.33/warc/CC-MAIN-20191206145958-20191206173958-00286.warc.gz"}
https://cstheory.stackexchange.com/tags/graph-colouring/new	# Tag Info 2 This result is included in Ore's book The Four-Colour Problem (see Theorem 7.4.3). I saw a paper that states this as a folklore result and cites Ore. Interestingly, the book gives a different proof for (ii)$\implies$(i). It seems that at that time, it wasn't known that the mapping $f\longmapsto f^*$ is a bijection. Sorry to disappoint; but that's the best I ... Top 50 recent answers are included	2020-07-14 15:39:26	{"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": 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": 0.8935021162033081, "perplexity": 811.4534496013476}, "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-2020-29/segments/1593655897168.4/warc/CC-MAIN-20200714145953-20200714175953-00132.warc.gz"}
https://www.gradesaver.com/textbooks/science/physics/physics-principles-with-applications-7th-edition/chapter-10-fluids-problems-page-286/16	## Physics: Principles with Applications (7th Edition) (a) The absolute pressure at the bottom of the pool is $1.19\times 10^5~Pa$ The total force on the bottom of the pool is 500 N (b) Since the pressure acts in all directions, the pressure against the side of the pool near the bottom is also $1.19\times 10^5~Pa$ (a) We can find the absolute pressure at the bottom of the pool. $P = 1~atm+\rho~g~h$ $P = 1.01\times 10^5~Pa +(1.00\times 10^3~kg/m^3)(9.80~m/s^2)(1.8~m)$ $P = 1.19\times 10^5~Pa$ The absolute pressure at the bottom of the pool is $1.19\times 10^5~Pa$ We can find the total force on the bottom of the pool. $F = PA$ $F = {(1.19\times 10^5~Pa)}{(28.0~m)(8.5~m)}$ $F = 2.810^7~N$ The total force on the bottom of the pool is 2.8*10^7 N. (b) Since the pressure acts in all directions, the pressure against the side of the pool near the bottom is also $1.19\times 10^5~Pa$.	2018-06-23 04:47:07	{"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": 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": 0.8690074682235718, "perplexity": 94.17509477014603}, "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/1529267864940.31/warc/CC-MAIN-20180623035301-20180623055301-00521.warc.gz"}
https://www.nag.com/numeric/nl/nagdoc_27.1/adhtml/s/sconts.html	# NAG AD LibraryS (Specfun)Approximations of Special Functions Settings help S (Specfun) Chapter Introduction (FL Interface) – A description of the Chapter and an overview of the algorithms available. Routine Mark of Introduction Purpose s01ba_a1w_f 26.2 nagf_specfun_log_shifted_a1w $\mathrm{ln}\left(1+x\right)$ s15ab_a1w_f 27.1 nagf_specfun_cdf_normal_a1w Cumulative Normal distribution function $P\left(x\right)$ Complement of error function $\mathrm{erfc}\left(x\right)$ Modified Bessel functions ${I}_{\nu +a}\left(z\right)$, real $a\ge 0$, complex $z$, $\nu =0,1,2,\dots$	2021-06-24 13:24:03	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 7, "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": 0.8590342402458191, "perplexity": 12928.32865279567}, "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/1623488553635.87/warc/CC-MAIN-20210624110458-20210624140458-00570.warc.gz"}
http://hal.in2p3.fr/view_by_stamp.php?label=LPNHE&langue=en&action_todo=view&id=in2p3-00704430&version=1	362 articles – 2926 references [version française] HAL: in2p3-00704430, version 1 arXiv: 1112.4358 Physical Review D 85 (2012) 072001 Search for standard model Higgs boson production in association with a W boson using a matrix element technique at CDF in pp̅ collisions at √s=1.96 TeV For the CDF collaboration(s) (2012-04) This paper presents a search for standard model Higgs boson production in association with a $W$ boson using events recorded by the CDF experiment in a dataset corresponding to an integrated luminosity of 5.6 fb-1. The search is performed using a matrix element technique in which the signal and background hypotheses are used to create a powerful discriminator. The discriminant output distributions for signal and background are fit to the observed events using a binned likelihood approach to search for the Higgs boson signal. We find no evidence for a Higgs boson, and 95% confidence level (C.L.) upper limits are set on the Higgs boson production rate. The observed limits range from 3.5 to 37.6 relative to the standard model expectation for Higgs boson masses between 100 and 150 GeV. The 95% C.L. expected limit is estimated from the median of an ensemble of simulated experiments and varies between 2.9 and 32.7 relative to the production rate predicted by the standard model over the Higgs boson mass range studied. Research team: APC - AHE Subject(s) : Physics/High Energy Physics - ExperimentSciences of the Universe/Astrophysics/High Energy Astrophysical PhenomenaPhysics/Astrophysics/High Energy Astrophysical Phenomena	2014-07-23 23:41:09	{"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": 0.6000896692276001, "perplexity": 623.4116042197055}, "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/1405997883898.33/warc/CC-MAIN-20140722025803-00003-ip-10-33-131-23.ec2.internal.warc.gz"}
https://curriculum.illustrativemathematics.org/k5/teachers/grade-5/unit-8/lesson-6/lesson.html	# Lesson 6 Revisit Volume ## Warm-up: Estimation Exploration: Sugar Cubes (10 minutes) ### Narrative The purpose of an Estimation Exploration is for students to practice the skill of estimating a reasonable answer based on experience and known information. In this activity, students estimate the number of sugar cubes in the bowl. During the synthesis, revisit what students know about rectangular prisms and volume and connect it to the image of the sugar cubes. ### Launch • Groups of 2 • Display the image. • “What is an estimate that’s too high?” “Too low?” “About right?” • 1 minute: quiet think time ### Activity • 1 minute: partner discussion • Record responses. ### Student Facing How many cubes are in the bowl? Record an estimate that is: too low about right too high $$\phantom{\hspace{2.5cm} \\ \hspace{2.5cm}}$$ $$\phantom{\hspace{2.5cm} \\ \hspace{2.5cm}}$$ $$\phantom{\hspace{2.5cm} \\ \hspace{2.5cm}}$$ ### Student Response For access, consult one of our IM Certified Partners. ### Activity Synthesis • “What would make it easier to find the exact number of cubes?” (If the cubes were organized in a way that we could count groups. If we could see them all.) ## Activity 1: 126 Cubes (15 minutes) ### Narrative The purpose of this activity is for students to review the concept that volume is the number of unit cubes required to fill a space without gaps or overlaps. Students are asked to find all of the different ways that they can arrange 126 sugar cubes to create a rectangular prism. This provides practice with factoring since the side lengths will be factors of 126. If students struggle to find factors of 126, it may be worthwhile to pause early on in the task and discuss the different strategies students are using to find factors of 126. A variety of different suggestions for how to pack the cubes should be anticipated and encouraged with the focus on how students decide on a particular shape of rectangular prism. In practice, many concerns influence the actual choice such as the amount of packaging material needed and how the package fits on a store shelf. When students interpret the meaning of the numbers in the context, they reason abstractly and quantitatively (MP2). The goal of the synthesis is to share ideas about predictions for how the cubes are packaged and how students decided they should be packaged. MLR7 Compare and Connect. Synthesis: Invite partners to prepare a visual display that shows the strategy they used to pack the sugar cubes. Encourage students to include details that will help others interpret their thinking. For example, using different colors, shading, arrows, labels, notes, diagrams or drawings. Give students time to investigate each others’ work. During the whole-class discussion, ask students, “Did anyone solve the problem the same way, but would explain it differently?” Action and Expression: Develop Expression and Communication. Synthesis: Give students access to cubes to build smaller rectangular prisms and invite students to make connections to the task. Supports accessibility for: Visual-Spatial Processing, Attention, Organization ### Launch • Groups of 2 • Display the image from the student workbook: • “These are sugar cubes. They are used to sweeten coffee and tea. Has anyone ever seen or tasted a sugar cube?” • “You are going to investigate different ways you can arrange 126 cubes to make a rectangular prism and then look at a particular example.” ### Activity • 2 minutes: independent work time • 5–7 minutes: partner work time ### Student Facing A company packages 126 sugar cubes in each box. The box is a rectangular prism. 1. What are some possible ways they could pack the cubes? 2. How would you choose to pack the cubes? Explain or show your reasoning. 3. The side lengths of the box are about $$1\frac{7}{8}$$ inches by $$3\frac{3}{4}$$ inches by $$4\frac{3}{8}$$ inches. What can we say about how the sugar cubes are packed? ### Student Response For access, consult one of our IM Certified Partners. ### Activity Synthesis • Invite students to share different ways they suggested packing the cubes and their reasoning for the choice. • “How did you find the different side lengths?” (I divided 126 by different numbers and used multiplication facts I knew to find other combinations.) • “Why is a 1 by 1 by 126 arrangement not useful for packaging the sugar cubes?” (It would be too long. It would break easily. It would be difficult to handle.) ## Activity 2: Colossal Structures Old and New (15 minutes) ### Narrative The purpose of this activity is for students to solve problems about the volume of different buildings. While students can find products of the given numbers, those products do not represent the volume of the structure. In both cases, the Great Pyramid of Egypt and the Empire State Building, neither structure is a rectangular prism. The pyramid steadily decreases in size as it gets taller while the Empire State Building also decreases in size at higher levels but not in the same regular way as the pyramid. With not enough information to make a definitive conclusion, students can see that both structures are enormous and that their volumes are roughly comparable, close enough that more studying would be needed for a definitive conclusion (MP1). ### Launch • Groups of 2 • To help understand how large the Great Pyramid and the Empire State Building are, consider estimating the size of the classroom. Estimates will vary but should be a few hundred cubic meters (versus several million for these huge structures). ### Activity • 5 minutes: independent work time • 5 minutes: partner work time • Monitor for students who • use the standard algorithm for multiplication • make estimates rather than using the standard algorithm for multiplication • identify that it is not possible with the given information to find the exact volume of either structure ### Student Facing 1. The base of the Great Pyramid of Egypt is a square. One side length of the base is 230 meters. The pyramid is 140 meters tall. If the pyramid was shaped like a rectangular prism, what would the volume of the prism be? 2. The Empire State Building is in New York City. The base is 129 meters by 59 meters. The building is 373 meters tall. Estimate the volume of the Empire State Building. 3. Which do you think is larger, the Great Pyramid or the Empire State Building? Explain or show your reasoning. ### Student Response For access, consult one of our IM Certified Partners. ### Activity Synthesis • Invite students to share their calculations for the volumes of the 2 structures. • “Why is it hard to find the exact volume of the Great Pyramid?” It’s not a rectangular prism. It has slanted sides.) • “Is the product of the area of the base and the height larger than the volume of the pyramid or smaller? How do you know?” (Larger because the pyramid does not fill all of that space. It gets more and more narrow toward the top.) • “Why is it hard to find the exact volume of the Empire State Building?” (It’s also not a rectangular prism. It also gets narrower toward the top.) • “Which do you think has greater volume?” (I think it’s too close to tell. I think the Great Pyramid is bigger because it looks like the base of the Empire State Building does not go up very far. It gets a lot narrower quickly. The Great Pyramid gets narrower more gradually.) ## Lesson Synthesis ### Lesson Synthesis “We started the year by exploring volume. What do you remember about the work we did in unit 1?” (We used cubes. We learned formulas.) “How did you apply what you learned in unit 1 in today’s lesson?” (I knew that the volume of a rectangular prism is the product of length, width, and height and that it is the product of the area of a base and the height. I used those formulas to calculate prism volumes.) “The blocks used in the Great Pyramid in Egypt have a volume of about 1 cubic meter. About how many blocks were used to build the Great Pyramid?” (There are probably more than one million. There are about 2 million.) ## Cool-down: Reflection: Volume (5 minutes) ### Cool-Down For access, consult one of our IM Certified Partners.	2021-10-20 01:59:31	{"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": 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": 0.30776673555374146, "perplexity": 1329.9784012752098}, "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/1634323585290.83/warc/CC-MAIN-20211019233130-20211020023130-00004.warc.gz"}
https://newbedev.com/do-strong-acids-actually-dissociate-completely	# Chemistry - Do strong acids actually dissociate completely? ## Solution 1: Yes, for sufficiently strong acids, in sufficiently dilute conditions, in sufficiently basic solvents. However, things are hazier than you might expect, and depending on your definition, "clear-cut" examples of complete dissociation only become possible when you start swatting flies with nuclear bombs. The problem is that chemistry works on equilibria, and at some point it becomes difficult to measure tiny concentrations - in a haystack of one billion straws, how can you be sure there isn't at least one needle? And in one trillion? And what about in a septillion? First, let us perform a qualitative investigation by considering the generic acid dissociation equilibrium in water, as follows: $$\ce{HA(aq)->H+(aq) + A-(aq)} \quad \quad \quad \mathrm{K_a=\frac{a_{H^+(aq)}a_{A^-(aq)}}{a_{HA(aq)}}=10^{-pK_a}}$$ To simplify the following discussion, we will make the very crude approximation where we equate the thermodynamic activity of a species with its molar concentration. In other words: $$\mathrm{K_a=\frac{a_{H^+(aq)}a_{A^-(aq)}}{a_{HA(aq)}} \approx \frac{[H^+(aq)][A^-(aq)]}{[HA(aq)]}}$$ Our objective now is to virtually eliminate the presence of undissociated $$\ce{HA}$$ molecules. For clarity, we rearrange the above equation: $$\mathrm{[HA(aq)] \approx \frac{[H^+(aq)][A^-(aq)]}{K_a}}$$ Clearly, to make $$\ce{[HA]}$$ as small as possible, the numerator must be very small, and the denominator must be very large. Let us set a goal of making $$\mathrm{[HA]<10^{-27}\ mol\ L^{-1}}$$, which implies less than $$\mathrm{0.001}$$ molecules of undissociated $$\ce{HA}$$ in a litre of solution. At this point, I should mention that aqueous solutions of acids in ambient conditions must always have $$\mathrm{[H^+(aq)]>10^{-7}\ mol\ L^{-1}}$$, thanks to the self-ionization of water. We take the best-case scenario (which physically corresponds to very dilute solutions): $$\mathrm{10^{-27} \gtrsim \frac{10^{-7}\ [A^-(aq)]}{K_a} \quad \longrightarrow \quad 10^{-20} \gtrsim \frac{[A^-(aq)]}{K_a}}$$ We now take a typical strong acid, hydrogen chloride ($$\ce{HCl}$$) which has an estimated pKa of -5.9 in water. In other words: $$\mathrm{10^{-20} \gtrsim \frac{[Cl^-(aq)]}{10^{5.9}}=\frac{[Cl^-(aq)]}{790\ 000} \quad \longrightarrow \quad [Cl^-(aq)] \lesssim 8\times 10^{-15}\ mol\ L^{-1}}$$ We see therefore that meeting the criterion of full dissociation requires preparing an aqueous $$\ce{HCl}$$ solution with femtomolar concentration. Essentially every solution used in a chemistry laboratory will have a higher concentration, so every $$\ce{HCl}$$ solution anyone has ever touched will have undissociated $$\ce{HCl}$$ molecules. Strictly speaking, you could say the dissociation is never complete. However, if you have a mind to make a fully dissociated solution of $$\ce{HCl}$$, it is physically possible. Just diluting acid solutions into oblivion until every molecule dissociates is kind of boring. The more exciting prospect is, of course, to increase the strength of the acid. The acidity scale goes far, far beyond aqueous solutions of $$\ce{HCl}$$. Interestingly, for sufficiently strong acids (typically $$\mathrm{pK_a<-10}$$, meaning acids at least 10 000 times stronger than aqueous $$\ce{HCl}$$), it is possible to isolate hydronium salts - when the strong acid is mixed with water in a 1:1 molecular ratio, a solid salt is formed, where the cation is $$\ce{H3O+}$$ and the anion is supplied by the conjugate base of the acid. The stoichiometry can be made exact. Surely this represents full dissociation of the acid? Well, again, it depends on your strictness. If you actually look at the crystal structures of many of these salts, it's clear that there is a strong, close interaction between the $$\ce{H3O+}$$ cation and the anion, and you could argue this still does not count as complete dissociation. That said, you can go crazy and consider some of the strongest acids we can put in a bottle, such as carborane superacids. The corresponding hydronium salt of these superacids can have extremely weak interactions between the hydronium cation and the anion. For example, when the hydronium salt $$\ce{(H3O+)(CHB11Cl11^-)}$$ is prepared in a benzene solution, the solid that forms actually contains three molecules of benzene surrounding the hydronium cation, which is completely detached from the anion. If this doesn't imply "complete dissociation" of the carborane superacid in water, I don't know what would. The acidity scale goes further still than even carborane superacids. There are a few known acids which will practically always fully dissociate, no matter what. The representatives of this class are rather exotic things such as $$\ce{H_3^+}$$, $$\ce{HN2^+}$$, $$\ce{HNe^+}$$, $$\ce{HHe^+}$$, and so forth. These acids can only be detected in the gas phase, as they fully dissociate by protonating virtually anything they come into contact with (including any anion, which is why they cannot be isolated as a neutral compound). The last way to ensure full dissociation of an acid in a solution is to change the solvent in which it is dissolved. All things considered, water is a pretty weak base. If, for example, ammonia (am) were used as the solvent, the relevant equilibrium for $$\ce{HCl}$$ would be: $$\ce{HCl(am)->H+(am) + Cl-(am)} \quad \quad \quad \mathrm{K_a(am)=\frac{a_{H^+(am)}a_{Cl^-(am)}}{a_{HCl(am)}}=10^{-pK_a(am)}}$$ If I were to give a crude guess, I'd imagine that the $$\mathrm{pK_a}$$ for $$\ce{HCl}$$ in liquid ammonia would very roughly be somewhere around -15; that is, dissociation of $$\ce{HCl}$$ in liquid ammonia is about a billion times easier compared to water. This is because ammonia is a stronger base than water. You know how we were talking about hydronium salts as if they were some unusual thing? Well, the equivalent concept in ammonia would be the formation of ammonium salts. And indeed, $$\ce{HCl}$$ does easily form a salt with ammonia in a exact 1:1 molecular ratio, namely ammonium chloride, $$\ce{NH4Cl}$$. Not exactly an unusual compound. There's nothing stopping you from using an extremely strong base as a solvent. For example, in pure liquid phosphazene superbases, even very weak acids may be fully dissociated in ambient conditions, even in the most rigorous sense. ## Solution 2: Fluoroantimonic acid is so strongly disassociating that molecules of $$\ce{HSbF6}$$ apparently have never been isolated. Only compounds in which a proton has been transferred to form ionic species are seen. (Replacing the fluorine with chlorine or bromine also gives an isolable compound, which may or may not be molecular depending on if water of hydration accepts a proton.) In its usual application as a solution in hydrofluoric acid solvent (the latter, by the way, is itself close to being a superacid, even without the fluoroantimonic acid solute), fluoroantimonic acid consists overwhelmingly of solvated $$\ce{H_{n+1}F_n^+}$$ cations and $$\ce{Sb_nF_{5n+1}^-}$$ anions, which equilibrate with $$\ce{HF + SbF5}$$ rather than with a molecular $$\ce{HSbF6}$$ species. When a "pure" fluoroantimonic acid compound is obtained from the solution, we still do not get association into neutral molecules: Two related products have been crystallized from HF-SbF5 mixtures, and both have been analyzed by single crystal X-ray crystallography. These salts have the formulas $$\ce{[H2F+][Sb2F_{11}−]}$$ and $$\ce{[H3F2+][Sb2F_{11}−]}$$. In both salts, the anion is $$\ce{Sb2F_{11}−}$$.[1] As mentioned above, $$\ce{SbF_6−}$$ is weakly basic; the larger anion $$\ce{Sb2F_{11}−}$$ is expected to be still weaker. Cited reference: 1. Mootz, Dietrich; Bartmann, Klemens (March 1988). "The Fluoronium Ions $$\ce{H2F+}$$ and $$\ce{H3F2+}$$: Characterization by Crystal Structure Analysis". Angewandte Chemie International Edition. 27 (3): 391–392. Link	2022-06-29 19:08:19	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 45, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8213945627212524, "perplexity": 1493.60327507846}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-27/segments/1656103642979.38/warc/CC-MAIN-20220629180939-20220629210939-00452.warc.gz"}
https://es.mathworks.com/help/dsp/ref/dsp.channelsynthesizer-system-object.html	# dsp.ChannelSynthesizer Polyphase FFT synthesis filter bank ## Description The dsp.ChannelSynthesizer System object™ merges multiple narrowband signals into a broadband signal by using an FFT based synthesis filter bank. The filter bank uses a prototype lowpass filter and is implemented using a polyphase structure. You can specify the filter coefficients directly or through design parameters. To merge multiple narrowband signals into a broadband signal: 1. Create the dsp.ChannelSynthesizer object and set its properties. 2. Call the object with arguments, as if it were a function. ## Creation ### Description example synthesizer = dsp.ChannelSynthesizer creates a synthesizer object, using the default properties. synthesizer = dsp.ChannelSynthesizer(Name,Value) specifies additional properties using Name,Value pairs. Unspecified properties have default values. Example: synthesizer = dsp.ChannelSynthesizer('NumTapsPerBand',20,'StopbandAttenuation',140) ## Properties expand all Unless otherwise indicated, properties are nontunable, which means you cannot change their values after calling the object. Objects lock when you call them, and the release function unlocks them. If a property is tunable, you can change its value at any time. Filter design parameters or filter coefficients, specified as one of these options: • "Number of taps per band and stopband attenuation" — Specify the filter design parameters through the NumTapsPerBand and StopbandAttenuation properties. • "Coefficients" — Specify the filter coefficients directly using LowpassCoefficients. Number of filter coefficients each polyphase branch uses, specified as a positive integer. The number of polyphase branches matches the number of frequency bands. The total number of filter coefficients for the prototype lowpass filter is given by product of the number of frequency bands and NumTapsPerBand. For a given stopband attenuation, increasing the number of taps per band narrows the transition width of the filter. As a result, there is more usable bandwidth for each frequency band at the expense of increased computation. #### Dependencies This property applies when you set Specification to "Number of taps per band and stopband attenuation". Data Types: single \| double \| int8 \| int16 \| int32 \| int64 \| uint8 \| uint16 \| uint32 \| uint64 Stopband attenuation of the lowpass filter, specified as a positive real scalar in dB. This value controls the maximum amount of aliasing from one frequency band to the next. Larger is the stopband attenuation, smaller is the passband ripple. #### Dependencies This property applies when you set Specification to "Number of taps per band and stopband attenuation". Data Types: single \| double Coefficients of the prototype lowpass filter, specified as a row vector. The default vector of coefficients is obtained using rcosdesign(0.25,6,8,"sqrt"). There must be at least one coefficient per frequency band. If the length of the lowpass filter is less than the number of frequency bands, the object zero-pads the coefficients. If you specify complex coefficients, the object designs a prototype filter that is centered at a nonzero frequency, also known as a bandpass filter. The modulated versions of the prototype bandpass filter appear with respect to the prototype filter and are wrapped around the frequency range [−Fs Fs]. Tunable: Yes #### Dependencies This property applies when you set Specification to "Coefficients". Data Types: single \| double Complex Number Support: Yes ## Usage ### Description example synthOut = synthesizer(input) merges the narrowband input signals contained as columns in input into broadband signal, synthOut. ### Input Arguments expand all Narrowband signals, specified as a matrix or a 3-D array. Each narrowband signal is stored as a column in the input signal. The number of columns in input corresponds to the number of frequency bands of the filter bank. If input is three-dimensional, each matrix corresponds to a separate channel. If M is the number of frequency bands, and input is an L-by-M matrix, then the output signal, synthOut, has dimensions L×M-by-1. If input has more than one channel, that is, it has dimensions L-by-M-by-N, then synthOut has dimensions L×M-by-N. This object also accepts variable-size inputs. That is, once the object is locked, you can change the size of each input channel. The number of channels cannot change. Data Types: single \| double Complex Number Support: Yes ### Output Arguments expand all Merged broadband signal, returned as a matrix or a 3-D array. If M is the number of frequency bands, and input is an L-by-M matrix, then the output signal, synthOut, has dimensions L×M-by-1. If input has more than one channel, that is, it has dimensions L-by-M-by-N, then synthOut has dimensions L×M-by-N. Data Types: single \| double Complex Number Support: Yes ## Object Functions To use an object function, specify the System object as the first input argument. For example, to release system resources of a System object named obj, use this syntax: release(obj) expand all coeffs Coefficients of prototype lowpass filter tf Return transfer function of overall prototype lowpass filter polyphase Return polyphase matrix step Run System object algorithm release Release resources and allow changes to System object property values and input characteristics reset Reset internal states of System object ## Examples collapse all Channelize and synthesize a sine wave signal with multiple frequencies using an M -channel filter bank. The M -channel filter bank contains an analysis filter bank section and a synthesis filter bank section. The dsp.Channelizer object implements the analysis filter bank section. The dsp.ChannelSynthesizer object implements the synthesis filter bank section. These objects use an efficient polyphase structure to implement the filter bank. For more details, see Polyphase Implementation under Algorithms on the object reference pages. Initialization Initialize the dsp.Channelizer and dsp.ChannelSynthesizer System objects. Each object is set up with 8 frequency bands, 8 polyphase branches in each filter, 12 coefficients per polyphase branch, and a stopband attenuation of 140 dB. Use a sine wave with multiple frequencies as the input signal. View the input spectrum and the output spectrum using a spectrum analyzer. offsets = [-40,-30,-20,10,15,25,35,-15]; sinewave = dsp.SineWave('ComplexOutput',true,'Frequency',... offsets+(-375:125:500),'SamplesPerFrame',800); channelizer = dsp.Channelizer('StopbandAttenuation',140); synthesizer = dsp.ChannelSynthesizer('StopbandAttenuation',140); spectrumAnalyzer = spectrumAnalyzer('ShowLegend',true,... 'SampleRate',sinewave.SampleRate,... 'ChannelNames',{'Input','Output'},... 'Title',"Input and Output Spectra"); Streaming Use the channelizer to split the broadband input signal into multiple narrow bands. Then pass the multiple narrowband signals into the synthesizer, which merges these signals to form the broadband signal. Compare the spectra of the input and output signals. The input and output spectra match very closely. for i = 1:5000 x = sum(sinewave(),2); y = channelizer(x); v = synthesizer(y); spectrumAnalyzer(x,v) end expand all expand all ## References [1] Harris, Fredric J, Multirate Signal Processing for Communication Systems, Prentice Hall PTR, 2004. [2] Harris, F.J., Chris Dick, Michael Rice. "Digital Receivers and Transmitters Using Polyphase Filter Banks for Wireless Communications." IEEE Transactions on microwave theory and techniques. Vol. 51, Number 4, April 2003. ## Version History Introduced in R2016b	2022-05-17 17:12:03	{"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": 0.551845133304596, "perplexity": 3473.697778123116}, "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-2022-21/segments/1652662519037.11/warc/CC-MAIN-20220517162558-20220517192558-00721.warc.gz"}
http://math.stackexchange.com/tags/graph-theory/hot	# Tag Info A partial answer : We have $a \sim b$ if and only if there exist a sequence of integers $a_1, \ldots, a_n$ such that $a \ \mathcal{R} \ a_1 \ \mathcal{R} \ \cdots \ \mathcal{R} \ a_n \ \mathcal{R} \ b$. The relation $ab-1 = c (a + b)$ can be written as $(ac)(bc)=c^2+1$ and can be solves $a=c+d$ and $b=c+ \dfrac{c ^ 2 + 1} d$ where d is a divisor of $c^ 2 ... 2 No, you cannot map an each edge in$G_1$to an arbitrary one in$G_2$. For example the 1-4 edge in your$G_1$connects a degree-3 vertex with a degree-2 vertex, and cannot map isomorphically to the 1-4 edge in$G_2$which goes bewteen the two degree-3 vertices. Rather than having two isomorphic graphs, it seems to be easier to think in terms of how many ... 2 Remember the "first theorem of graph theory:" $$\sum_{v \in V(G)} \deg v =2\|E(G)\|,$$ where$V(G)$denotes the vertex set and$E(G)$the edge set. If$\deg v$is odd for each vertex, what does the above say about the parity (even or odd) of$\|V(G)\|$? Once you've figured this out, you still need to translate this into a statement about the size of the graph, ... 2 (This is too long for a comment and not part of the answer I gave a year ago, so I'm posting it as a new answer under community wiki.) I made an exhaustive computer search of the solutions for eight people per side and found out that there is really only one unique solution up to symmetries. Stating this fact and showing the structure of the solution might ... 1 Try working it backwards: can you draw$K_{3,3}$, then add the missing edges to get$K_6$, with the missing edges forming a$6$-cycle? Answer: no, because the missing edges must form two$3$-cycles. 1 For the converse I would proceed by contrapositive with the following ideas. (1) Assume$x$and$y$are vertices for which neither is an ancestor of the other. (2) Argue that there is a lowest common ancestor (I am picturing the tree drawn with the root at the top and going downwards) of$x$and$y$, say$a$. (3) Argue that one of$x$and$y$, say$x$is ... 1$\Longleftarrow$: Assume that (a)$x$stands before$y$in the pre-order traversal of$B$and that (b)$x$stands after$y$in the post-order traversal of$B$. Now assume that (c)$x$is not an ancestor of$y$(a proof by contradiction will follow). There are two cases:$y$is an ancestor of$x$: a node$u$isn't visited by pre-order traversal until all ... 1 The correct way, it seems to me, would be writing it as follows: $$\sum_{\{i,j\}\in\cal E}i+j$$ The index set is all the edges, and for each edge we sum its two vertices. Of course, this means that the vertices belong to some structure which includes addition. To your second question$\{1,2\}$is the set with two elements,$1$and$2$. The set ... 1 Is this enough of a proof? I would have to say no. If every face boundary is a cycle of even length, every face has an even degree. This is essentially saying "If X, then X." The catch is that there might be cycles that are not face boundaries. There are no cycles of odd degree and the graph must be bipartite. This is what we need to prove (and we need ... 1 Each vertex has$d$edges and no two vertices in$A$(or$B$) are connected. So in total,$A$has$\|A\|d$edges coming out of it and$B$has$\|B\|d$edges going into it. Can those be different? 1 I wanted to vote up the Karolis's answer and add a comment for$d=0$but I don't have enough reputation, so I write an answer. If$d=0$then your graph has no edges. Any partitioning into disjoint sets$A$and$B$would render a legal bipartition, since there are no edges among the vertices in either of the sets. In particular different number of vertices ... 1 Here is an example of two non-isomorphic graphs$G$and$H$with isomorphic 2-step graphs$G^{(2)}$and$H^{(2)}$. 1 Some useful properties : http://en.wikipedia.org/wiki/Bipartite_graph#Properties A graph is bipartite if and only if it has no odd cycle. Also, a graph is bipartite if and only if it is 2-colorable. The rest is up to you :) 1 The eccentricity of every other vertex can be$4$, as looking at a path of length$5$will tell you (there, the third element has eccentricity of$2$, while the edges have an eccentricity of$4$). Can you show that the eccentricity cannot be more than$4$? 1 The problem you describe is a specialization of the 1-center problem on graphs. This problem can be mathematically formulated as follows: Given a graph$G = (V, E)$with node set$V$and edge set$E \subseteq V \times V$with distances$d: E \rightarrow \mathbb{R}$, find the point$xin the graph minimizing \begin{align} \max_{v \in V} d(v, x) \end{align} ... 1 This should work: LetA'$be all vertices with in-degree 0, and let$A''$be their neighbors Add vertices in$A'$to$A$Add vertices in$A''$to$V \setminus A$Remove$A' \cup A''$from$G$and repeat steps 1-3 until the graph is empty Correctness: Since there are no edges between vertices with in-degree 0, property (i) holds for$A'\$. As all ...	2014-07-26 11:24:59	{"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": 1, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 1.0000026226043701, "perplexity": 1040.819146800573}, "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/1405997901076.42/warc/CC-MAIN-20140722025821-00202-ip-10-33-131-23.ec2.internal.warc.gz"}
http://math.stackexchange.com/questions/86864/subspaces-in-linear-algebra	# Subspaces in Linear Algebra Find the $\operatorname{Proj}_wv$ for the given vector $v$ and subspace $W$. Let $V$ be the Euclidean space $\mathbb{R}^4$, and $W$ the subspace with basis $[1, 1, 0, 1], [0, 1, 1, 0], [-1, 0, 0, 1]$ (a) $v = [2,1,3,0]$ ans should be - $[7/5,11/5,9/5,-3/5]$ My attempt at the solution was basically we can find the basis perpendicular to $W$ as $[ 1,-2,2, 1]$ then, $[2, 1, 3, 0] = a[1, 1, 0, 1] + b[0, -1, 1, 0] + c[0 ,2, 0,3] + d[1,-2,2,1]$ We solve for $a,b,c,d$ and get $a = 16/3,b=29/3,c=-2/3,d=-10/3$ now the problem is what do I do from here? - To type $\mathbb{R}^4$, type $\mathbb{R}^4$. – Arturo Magidin Nov 29 '11 at 22:54 If you find a basis for $W^\perp$ that is orthogonal to $W$, say with Gram-Schmidt, finding the projecting is much easier. – Joe Johnson 126 Nov 29 '11 at 23:10 Or, finding an orthonormal basis for $W$ itself makes finding the projection very easy. – Arturo Magidin Nov 29 '11 at 23:15 You can do it that way (though you must have an arithmetical error somewhere; the denominator of $3$ cannot be right), and the remaining piece is then simply to take $a[1, 1, 0, 1] + b[0, -1, 1, 0] + c[0 ,2, 0,3]$, forgetting the part perpendicular to $W$. However, it is much easier to normalize your $[1,-2,2,1]$ to $n=\frac{1}{\sqrt{10}}[1,-2,2,1]$ -- then the projection map is simply $v\mapsto v - (v\cdot n)n$. (If you write that out fully, the square root even disappears). I did that and I end up getting $(9/5,3/5,9/5,0)$. Just redid it here to see if I had made an error - bit.ly/s9wIOw on WolframAlpha – eWizardII Nov 29 '11 at 23:33	2016-05-25 09:48:56	{"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": 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": 0.9444782137870789, "perplexity": 274.14951039910517}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-22/segments/1464049274324.89/warc/CC-MAIN-20160524002114-00153-ip-10-185-217-139.ec2.internal.warc.gz"}
https://math.stackexchange.com/questions/1715324/what-is-the-difference-between-algebraic-sets-and-algebraic-varieties/1715335	# What is the difference between algebraic sets and algebraic varieties? I am wondering what is the difference between algebraic sets and algebraic varieties in complex projective space. It seems that both are zero sets of polynomials, so what is the difference? If we give $\mathbb A^n$ (resp. $\mathbb P^n$) their Zariski topology, an algebraic set is just a closed subset $V\subset \mathbb A^n$ (resp. $V\subset \mathbb P^n$). Every algebraic set, which a priori is a topological subspace, can be endowed with the structure of algebraic variety: the supplementary datum consists of decreeing which functions on open subsets $U\subset V$ are considered acceptable, thus obtaining the ring $\mathcal O_V(U)$ of "regular" functions on $U$. However there are many algebraic varieties which are neither affine nor projective, and thus are completely different from algebraic subsets: the simplest example is the plane with the origin removed, $V=\mathbb A^2\setminus \{(0,0)\}$.	2021-03-06 02:42:13	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 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": 0.8562032580375671, "perplexity": 238.5981386721982}, "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/1614178374217.78/warc/CC-MAIN-20210306004859-20210306034859-00290.warc.gz"}
https://researchportal.bath.ac.uk/en/publications/petascale-solvers-for-anisotropic-pdes-in-atmospheric-modelling-o	# Petascale solvers for anisotropic PDEs in atmospheric modelling on GPU clusters Research output: Contribution to journalArticle 6 Citations (Scopus) ### Abstract Memory bound applications such as solvers for large sparse systems of equations remain a challenge for GPUs. Fast solvers should be based on numerically efficient algorithms and implemented such that global memory access is minimised. To solve systems with trillions ($\order(10^{12})$) unknowns the code has to make efficient use of several million individual processor cores on large GPU clusters. We describe the multi-GPU implementation of two algorithmically optimal iterative solvers for anisotropic PDEs which are encountered in (semi-) implicit time stepping procedures in atmospheric modelling. In this application the condition number is large but independent of the grid resolution and both methods are asymptotically optimal, albeit with different absolute performance. In particular, an important constant in the discretisation is the CFL number; only the multigrid solver is robust to changes in this constant. We parallelise the solvers and adapt them to the specific features of GPU architectures, paying particular attention to efficient global memory access. We achieve a performance of up to 0.78 PFLOPs when solving an equation with $0.55\cdot 10^{12}$ unknowns on 16384 GPUs; this corresponds to about $3\%$ of the theoretical peak performance of the machine and we use more than $40\%$ of the peak memory bandwidth with a Conjugate Gradient (CG) solver. Although the other solver, a geometric multigrid algorithm, has a slightly worse performance in terms of FLOPs per second, overall it is faster as it needs less iterations to converge; the multigrid algorithm can solve a linear PDE with half a trillion unknowns in about one second. Original language English 53-69 20 Parallel Computing 50 28 Oct 2015 https://doi.org/10.1016/j.parco.2015.10.007 Published - 1 Dec 2015 ### Fingerprint Data storage equipment Modeling Unknown Fast Solvers Iterative Solvers Semi-implicit Time Stepping Asymptotically Optimal Condition number System of equations Efficient Algorithms Discretization Bandwidth Grid Converge Iteration Graphics processing unit Architecture ### Keywords • iterative solver • multigrid • Graphics Processing Unit • massively parallel • atmospheric modelling ### Cite this In: Parallel Computing, Vol. 50, 01.12.2015, p. 53-69. Research output: Contribution to journalArticle @article{40d7aba0676e475791bafd936c9c027a, title = "Petascale solvers for anisotropic PDEs in atmospheric modelling on GPU clusters", abstract = "Memory bound applications such as solvers for large sparse systems of equations remain a challenge for GPUs. Fast solvers should be based on numerically efficient algorithms and implemented such that global memory access is minimised. To solve systems with trillions ($\order(10^{12})$) unknowns the code has to make efficient use of several million individual processor cores on large GPU clusters.We describe the multi-GPU implementation of two algorithmically optimal iterative solvers for anisotropic PDEs which are encountered in (semi-) implicit time stepping procedures in atmospheric modelling. In this application the condition number is large but independent of the grid resolution and both methods are asymptotically optimal, albeit with different absolute performance. In particular, an important constant in the discretisation is the CFL number; only the multigrid solver is robust to changes in this constant. We parallelise the solvers and adapt them to the specific features of GPU architectures, paying particular attention to efficient global memory access. We achieve a performance of up to 0.78 PFLOPs when solving an equation with $0.55\cdot 10^{12}$ unknowns on 16384 GPUs; this corresponds to about $3\{\%}$ of the theoretical peak performance of the machine and we use more than $40\{\%}$ of the peak memory bandwidth with a Conjugate Gradient (CG) solver. Although the other solver, a geometric multigrid algorithm, has a slightly worse performance in terms of FLOPs per second, overall it is faster as it needs less iterations to converge; the multigrid algorithm can solve a linear PDE with half a trillion unknowns in about one second.", keywords = "iterative solver, multigrid, Graphics Processing Unit, massively parallel, atmospheric modelling", author = "Eike Mueller and Robert Scheichl and Eero Vainikko", year = "2015", month = "12", day = "1", doi = "10.1016/j.parco.2015.10.007", language = "English", volume = "50", pages = "53--69", journal = "Parallel Computing", issn = "0167-8191", publisher = "Elsevier", } TY - JOUR T1 - Petascale solvers for anisotropic PDEs in atmospheric modelling on GPU clusters AU - Mueller, Eike AU - Scheichl, Robert AU - Vainikko, Eero PY - 2015/12/1 Y1 - 2015/12/1 N2 - Memory bound applications such as solvers for large sparse systems of equations remain a challenge for GPUs. Fast solvers should be based on numerically efficient algorithms and implemented such that global memory access is minimised. To solve systems with trillions ($\order(10^{12})$) unknowns the code has to make efficient use of several million individual processor cores on large GPU clusters.We describe the multi-GPU implementation of two algorithmically optimal iterative solvers for anisotropic PDEs which are encountered in (semi-) implicit time stepping procedures in atmospheric modelling. In this application the condition number is large but independent of the grid resolution and both methods are asymptotically optimal, albeit with different absolute performance. In particular, an important constant in the discretisation is the CFL number; only the multigrid solver is robust to changes in this constant. We parallelise the solvers and adapt them to the specific features of GPU architectures, paying particular attention to efficient global memory access. We achieve a performance of up to 0.78 PFLOPs when solving an equation with $0.55\cdot 10^{12}$ unknowns on 16384 GPUs; this corresponds to about $3\%$ of the theoretical peak performance of the machine and we use more than $40\%$ of the peak memory bandwidth with a Conjugate Gradient (CG) solver. Although the other solver, a geometric multigrid algorithm, has a slightly worse performance in terms of FLOPs per second, overall it is faster as it needs less iterations to converge; the multigrid algorithm can solve a linear PDE with half a trillion unknowns in about one second. AB - Memory bound applications such as solvers for large sparse systems of equations remain a challenge for GPUs. Fast solvers should be based on numerically efficient algorithms and implemented such that global memory access is minimised. To solve systems with trillions ($\order(10^{12})$) unknowns the code has to make efficient use of several million individual processor cores on large GPU clusters.We describe the multi-GPU implementation of two algorithmically optimal iterative solvers for anisotropic PDEs which are encountered in (semi-) implicit time stepping procedures in atmospheric modelling. In this application the condition number is large but independent of the grid resolution and both methods are asymptotically optimal, albeit with different absolute performance. In particular, an important constant in the discretisation is the CFL number; only the multigrid solver is robust to changes in this constant. We parallelise the solvers and adapt them to the specific features of GPU architectures, paying particular attention to efficient global memory access. We achieve a performance of up to 0.78 PFLOPs when solving an equation with $0.55\cdot 10^{12}$ unknowns on 16384 GPUs; this corresponds to about $3\%$ of the theoretical peak performance of the machine and we use more than $40\%$ of the peak memory bandwidth with a Conjugate Gradient (CG) solver. Although the other solver, a geometric multigrid algorithm, has a slightly worse performance in terms of FLOPs per second, overall it is faster as it needs less iterations to converge; the multigrid algorithm can solve a linear PDE with half a trillion unknowns in about one second. KW - iterative solver KW - multigrid KW - Graphics Processing Unit KW - massively parallel KW - atmospheric modelling U2 - 10.1016/j.parco.2015.10.007 DO - 10.1016/j.parco.2015.10.007 M3 - Article VL - 50 SP - 53 EP - 69 JO - Parallel Computing JF - Parallel Computing SN - 0167-8191 ER -	2019-11-21 08:06:18	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.3793918788433075, "perplexity": 1166.879577393047}, "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/1573496670743.44/warc/CC-MAIN-20191121074016-20191121102016-00303.warc.gz"}
https://serverfault.com/questions/877712/how-can-i-pass-a-variable-parameter-to-a-windows-remoteapp	# How Can I Pass a Variable Parameter to A Windows RemoteApp? I have published a remote app, MyApp.exe, from Server X. MyApp.exe can be launched with a variable optional parameter. That parameter can have a larger number of different values. I have downloaded the RDP file from the gateway website that Server X publishes. I named the rdp file MyApp.rdp. Is there anyway to pass a parameter to MyApp.exe running on Server X when I launch the RDP file? Something like this would be the most natural: MyApp.RDP XYZ This would run MyApp.exe on Server X passing it parameter XYZ. I've looked all around the town and I can't find a clear yes or no or how-to for this question. 3. Use the /REMOTECMDLINE parameter of mstsc Ex: mstsc /REMOTECMDLINE:foo path\to\app.rdp will result in app.exe foo on the server You must remove the remoteapplicationcmdline:s: parameter from the rdp file lest you get the error message "This RDP file has settings that cannot be overridden by command line." You must also have "Allow any command-line parameters" selected on the app configuration on the server. 4. Use the /REMOTEFILE parameter of mstsc	2021-01-16 20:53:23	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 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": 0.2546582520008087, "perplexity": 4604.062813173036}, "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/1610703507045.10/warc/CC-MAIN-20210116195918-20210116225918-00158.warc.gz"}
https://itwissen.info/en/characteristic-sound-pressure-129279.html	# characteristic sound pressure The characteristic sound pressure is a characteristic value of loudspeakers. It is the sound pressure measured when applying a power of 1 W at a distance of 1 m. The characteristic sound pressure can also refer to the applied voltage, which is 2.83 V for a power of 1 W and a nominal impedance of 8 ohms. The characteristic sound pressure is given in dB/W/m or dB/2.83 V/m. The impedance of loudspeakers is strongly frequency-dependent, therefore the power specification of 1 W or the application of a constant voltage of 2.83 V is recommended. The characteristic sound pressure is determined in the frequency range between 125 Hz and 4 kHz and, depending on the loudspeaker, ranges from about 80 dB/W to over 100 dB/W for loudspeaker cabinets. The efficiency of the loudspeakers can be calculated from the characteristic sound pressure. Informations: Englisch: characteristic sound pressure Updated at: 12.09.2017 #Words: 133 Links: sound, pressure, power (P), voltage, impedance (Z) Translations: DE Sharing:	2022-06-29 14:15:49	{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8787315487861633, "perplexity": 2287.034115054989}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-27/segments/1656103639050.36/warc/CC-MAIN-20220629115352-20220629145352-00232.warc.gz"}
https://www.queryhome.com/puzzle/10510/poles-height-and-situated-such-that-sun-rays-through-the-pole	# Two poles of height 12 m and 4 m are situated in such a way that the sun rays through the top of a pole... 351 views Two poles of height 12 m. and 4 m. are situated in such a way that the sun rays through the top of a pole also passes through the top of other. What is the distance between the poles? It is given that the sun rays make an angle of 45 degree with the horizon posted Oct 16, 2015 +1 vote A 45 degree angle means for every meter height there must be one meter distance. The height difference between the two poles is 8 meter, so they must be 8 meters apart. Similar Puzzles A body is projected at velocity of 40 m/s, after 2 sec it crosses a vertical pole of height 20.4 m. What is angle of projection and horizontal range? At a particular time the shadow of a men of 170 cm height is 204 cm on the ground and at the same time and on the same ground the shadow of a pole is 600 cm. Find the height of the pole?	2021-09-28 20:43:36	{"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": 0.8969838619232178, "perplexity": 346.348124686471}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-39/segments/1631780060882.17/warc/CC-MAIN-20210928184203-20210928214203-00118.warc.gz"}
https://www.groundai.com/project/resummation-of-soft-and-coulomb-corrections-for-tbarth-production-at-the-lhc/	Resummation of soft and Coulomb corrections for t\bar{t}h production at the LHC # Resummation of soft and Coulomb corrections for t¯th production at the LHC Wan-li Ju School of Physics and State Key Laboratory of Nuclear Physics and Technology, Peking University, Beijing 100871, ChinaCollaborative Innovation Center of Quantum Matter, Beijing, ChinaCenter for High Energy Physics, Peking University, Beijing 100871, China and Li Lin Yang School of Physics and State Key Laboratory of Nuclear Physics and Technology, Peking University, Beijing 100871, ChinaCollaborative Innovation Center of Quantum Matter, Beijing, ChinaCenter for High Energy Physics, Peking University, Beijing 100871, China ###### Abstract In this paper, a combined resummation of soft and Coulomb corrections is performed for the associated production of the Higgs boson with a top quark pair at the LHC. We illustrate the similarities and critical differences between this process and the production process. We show that up to the next-to-leading power, the total cross section for production admits a similar factorization formula in the threshold limit as that for production. This fact, however, is not expected to hold at higher powers. Based on the factorization formula, we perform the resummation at the improved next-to-leading logarithmic accuracy, and match to the next-to-leading order result. This allows us to give NLL+NLO predictions for the total cross sections at the LHC. We find that the resummation effects enhance the NLO cross sections by about 6%, and significantly reduce the scale dependence of the theoretical predictions. ## 1 Introduction After the discovery of the Higgs boson at the Large Hadron Collider (LHC) in 2012 Aad:2012tfa (); Chatrchyan:2012xdj (), a main task of particle physics is to investigate its properties. One particularly important property is the Yukawa coupling between the top quark and the Higgs boson, which is crucial to understand the origin of the large top quark mass. Precise knowledge of the top quark Yukawa coupling will also help us to constrain new physics effects in other couplings such as the Higgs boson self-coupling Goertz:2014qta (); Azatov:2015oxa (); Cao:2015oaa (); Shen:2015pha (). At the LHC, the top-Higgs Yukawa coupling can be probed by measuring the cross section for Higgs boson production in association with a top quark pair ( production). The observation of this process has been established very recently by the ATLAS Aaboud:2018urx () and CMS Sirunyan:2018hoz () collaborations. The measured cross sections are in agreement with theoretical predictions, although there are still large experimental uncertainties due to limited statistics. In the future, more precise measurements will be carried out at the upgraded LHC and the High Luminosity phase of the LHC (HL-LHC) Apollinari:2015bam (). For this reason, it is necessary to improve the theoretical understanding of this process. At the leading order (LO), production can be initialized by or from the colliding protons, which has been studied many years ago Ng:1983jm (); Kunszt:1984ri (); Dicus:1988cx (). The next-to-leading order (NLO) quantum chromodynamics (QCD) corrections were calculated in Beenakker:2001rj (); Reina:2001bc (); Reina:2001sf (); Beenakker:2002nc (); Dawson:2002tg (); Dawson:2003zu (), while the NLO electroweak contributions were calculated in Yu:2014cka (); Frixione:2014qaa (); Frixione:2015zaa (). Given the complicated structure of this scattering process, it will be very challenging to perform an exact calculation of the next-to-next-to-leading order corrections. Therefore, a lot of efforts have been devoted to approximations of the higher order QCD corrections in various kinematic limits Kulesza:2015vda (); Broggio:2015lya (); Broggio:2016lfj (); Kulesza:2017ukk (). The derivation of an approximation often involves a factorization formula, which can be used to resum a class of large corrections to all orders in perturbation theory. For example, Refs. Broggio:2015lya (); Broggio:2016lfj (); Kulesza:2017ukk () have investigated the limit , where is the partonic center-of-mass energy and is the invariant mass of the system. In this limit, large logarithmic corrections are present at each order in perturbation theory. These corrections are resummed to all orders in the strong coupling up to the next-to-next-to-leading logarithmic (NNLL) accuracy Broggio:2016lfj (); Kulesza:2017ukk (). In this paper, we consider a different kinematic limit , where and are the masses of the top quark and the Higgs boson, respectively. The limit under consideration is a bit different from the limit taken in Broggio:2016lfj (); Kulesza:2017ukk (). There are power-like corrections arising from exchanges of Coulomb gluons, besides logarithmic corrections coming from soft gluon emissions. This limit has been studied in Kulesza:2015vda (), where the soft gluon contributions are resummed, but the Coulomb gluon contributions are only incorporated at fixed-order. In this paper, we will derive a factorization formula which can resum simultaneously both kinds of higher-order corrections. The framework presented in this paper closely resembles that in production Beneke:2009ye (); Beneke:2010da (); Beneke:2011mq (). However, it should be emphasized that production is more complicated than production. In particular, in the threshold limit, the pair has a non-vanishing transverse momentum given by the recoil against the extra Higgs boson. This will lead to more involved interplay between the soft and Coulomb gluons, as will be clear later. The derivation of the factorization formula therefore requires new analyses other than those in Beneke:2009ye (); Beneke:2010da (). It is also much more difficult to calculate the hard function describing the contributions from hard gluons with typical momentum scale around . The paper is organized as follows. In Sec. 2, we use the analytic form of the LO partonic cross sections to analyze their behavior in the threshold limit. In Sec. 3, we present the derivation of the factorization and resummation formulas using effective field theory methods. In Sec. 4, we show the numeric results based on our resummation formula. We conclude in Sec. 5. ## 2 Analyses of leading order results The total cross section for inclusive production at hadron colliders can be expressed as Collins:1989gx () σ(s,mt,mh)=∑i,j∫1τmindτ^σij(τ,mt,mh,μf)ffij(τ,μf), (1) where the sums are over all the partons within the colliding hadrons, i.e, ; is the center-of-mass energy of the collider; and are the top quark mass and the Higgs mass, respectively; is the factorization scale; and τmin=(2mt+mh)2s. (2) The above factorization formula involves the partonic cross section and the effective parton luminosity function . The definition of the latter is ffij(τ,μf)=∫1τdξξfi/N1(ξ,μf)fj/N2(τ/ξ,μf), (3) where is the non-perturbative parton distribution function (PDF) of the parton in the hadron . It is universal and can be extracted from experimental data. The partonic cross section can be calculated in perturbative QCD. In this work, we are interested in its behavior near the threshold limit , or . Here is the partonic center-of-mass energy defined by . To analyze this region, we define . The parameter goes to zero in the threshold limit, and represents the typical momenta of final state particles. To see this, consider the energy conservation condition in the partonic center-of-mass frame √^s=√m2t+→p2t+√m2t+→p2¯t+√m2h+→p2h+EX, (4) where and are the 3-momenta of the (anti-)top quark and the Higgs boson, respectively, and is the total energy of other emitted particles in the final state. In the limit , the 3-momenta of the (anti-)top quark and the Higgs boson becomes much smaller than their rest mass. The right side of the above equation can then be expanded in the small momenta and we obtain the following relation: →p2t2mt+→p2¯t2mt+→p2h2mh+EX∼√^s−2mt−mh∼√^s2β2. (5) We therefore have the power-counting √^s∼mt∼mh,\|→pt,¯t,h\|∼√^sβ,EX∼√^sβ2. (6) This will be important for establishing the effective field theory description later in the next section. For the moment, we are going to investigate the behavior of the partonic cross sections in the limit . The Born-level partonic cross section can be written as (7) To the first order in , one can approximate the -functions in the above formula as δ(√^s−2mt−mh−→p2t2mt−→p2¯t2mt−→p2h2mh)δ3(→pt+→p¯t+→ph), (8) and the squared-amplitudes can be expanded according to the counting in eq. (6). The integrals over the 3-momenta can then be carried out, and we arrive at approximate expressions of the partonic cross sections in the following form ^σ(0)Appq¯q→t¯th=2β4m5tα2s9v2m3/2h(mh+2mt)7/2,^σ(0)Appgg→t¯th=β4mtα2s(2m4h−7m2hm2t+14m4t)192v2m3/2h(mh+2mt)7/2, (9) where is the vacuum expectation value of the Higgs field and is the strong coupling constant. A crucial feature of eq. (9) is that the leading order partonic cross sections are proportional to in the threshold limit. This should be contrasted to the case of a similar process, production, where the threshold behavior is given by Czakon:2008cx (), σ(0)Appq¯q→t¯t=πα2sβ9m2t,σ(0)Appgg→t¯t=7πα2sβ192m2t. (10) That is, the Born partonic cross sections are linear in in the threshold limit. The different behaviors imply that the threshold region is less important in the case of production than for production. In order to study these different behaviors more precisely, we will numerically compare the approximate results to the exact ones in the following. For our numerical computations, we set , and Tanabashi:2018oca (). The strong coupling constant is evolved from the initial condition to the renormalization scale , where is the mass of the boson. In Fig. 1, we numerically compare the approximate Born partonic cross sections for production with the exact ones, i.e, . The approximate results are obtained from Eq. (9). To calculate , we first employ the programs FeynArts Hahn:2000kx () and FeynCalc Shtabovenko:2016sxi (); Mertig:1990an () to generate the transition amplitudes and then use the Cuba library Hahn:2004fe (); Hahn:2014fua () to perform the phase space integration. Our numerical results are checked against the automatic program MadGraph5_aMC@NLO (MG5) Alwall:2014hca (). As shown in Fig. 1, the approximate and exact partonic cross sections approach each other as becomes small. They are both highly suppressed in the threshold limit, as can be expected. When grows larger, the approximate results start to overestimate the exact ones, indicating that there are important negative power corrections to the approximate formula. In Fig. 2, we show a similar comparison for the production process. The exact results are obtained from Moch:2008qy (), while the approximate results are computed using Eq. (10). The first impression is that the small- region is less suppressed compared to the case, which is clear from the vs. behaviors. This means that the reliability of small- resummation can be quite different in these two processes, a fact not often mentioned in the literatures. Beside this, here we also observe that the approximate result in the channel overestimates the exact one as grows. Interestingly, for the subprocess, the approximate result underestimates the exact one for most values of , contrary to the case shown in Fig. 1. It is therefore possible that, for the sum of the two channels, the approximate result stays closer to the exact one than in the case of individual channels. This should be investigated at the hadron level as we are going to do below. The partonic cross sections need to be convoluted with the parton distribution functions (PDFs) to arrive at the hadronic cross sections. It is therefore interesting to see how the approximate and exact results compare at the hadron level. We show in Fig. 3 the hadronic differential cross sections for production. From the left plot, we see again that the approximate results in both the and the channels overshoot a lot over the exact ones for large . It is also clear that the small- region is highly suppressed due to the behavior. In the right plot, we show the ratio between the approximate results and the exact ones as a function of . One can see that in the small- region, the approximate results are in good agreements with the exact ones. As goes above , the approximation quickly fails. On the other hand, we show in Fig. 4 the results for production. Again, we find that the small- region is more important in this case compared to production due to the behavior. Besides, one can also see the accidental cancellation between the power corrections in the and channel mentioned in the last paragraph. The above two facts lead to the observation that the small- expansion provides a reasonable approximation to the hadronic total cross section for production Bonciani:1998vc (); Moch:2008qy (); Beneke:2009ye (); Beneke:2010da (); Beneke:2011mq (); Piclum:2018ndt (). This is clearly not the case for production. While the above analyses show that the small- region for production is not as important as that for production, it is still interesting to study the small- behavior of the cross section at higher orders in QCD. First of all, theoretically, production is similar but slightly different from production. The pair is recoiled against the Higgs boson and therefore has a non-vanishing transverse momentum already at the lowest order in QCD. The interplay between the soft gluons and the Coulomb gluons can therefore be a bit different from the case of production. This poses a question of how to properly factorize these contributions to all orders in perturbation theory, which was not addressed in the literature. Secondly, this process is closely related to the process at future electron-positron colliders, which receives important QED corrections, especially in the threshold region Denner:2003zp (). The investigation of soft and Coulomb gluons in the process can therefore be applied straightforwardly to the soft and Coulomb photons in the process. Finally, while the small- limit is not significant for the total cross section, it could be important if one specifically wants to study certain kinematic configurations sensitive to the threshold region by, e.g., vetoing additional jets. In this work, we are going to study the threshold region by applying a cut on the variable. While this is not a physical cut (since cannot be measured), it simplifies the theoretical considerations. We define the following quantity Moch:2008qy () σcut(s,mt,mh,βcut) =∑ij∫1τmindτ^σij(τ,mt,mh,μf)θ(βcut−β)ffij(τ,μf). (11) Apparently, for sufficiently small , should be well approximated by the leading power expression of the threshold expansion. It is also obvious that as , approaches the total cross section defined in Eq. (1). The constrained cross section is the main object we are going to study in the rest of the paper. Before entering the technical details, we first investigate its leading order behavior against the variation of . We again choose to work with the LHC. We take the exact results and the approximate results for the leading order partonic cross sections and , and plug them into Eq. (11). We denote the results as and , and plot their ratio in Fig. 5 as a function of . From the figure, we find that up to , provides a rather good approximation to the exact result. This fact should be kept in mind when we later combine the small- resummation and the fixed-order calculation. At higher orders in perturbation theory, exchanges of Coulomb gluons and soft gluons lead to threshold-enhanced terms such as and . For small , these terms represent the dominant contributions to the constrained cross section . The rest of the paper will be devoted to deriving a factorization formula for the partonic cross section in the threshold limit, and resumming these enhanced terms to all orders in perturbation theory. ## 3 Factorization and resummation in the threshold limit ### 3.1 Higher order QCD corrections in the threshold limit Beyond the Born level, the cross sections receive contributions from exchange of virtual gluons and emission of real gluons. We will investigate these contributions as a power expansion in in the threshold limit . In this limit, there will be -enhanced terms and -enhanced terms at higher orders in . Schematically, we are going to consider corrections of the form ^σNLL′ij∼α0s{1,β}+αs{ln2β,lnβ,1,1β,βln2β,βlnβ}+α2s{ln4β,ln3β,ln2β,1β2,1β,ln2ββ,lnββ,βln4β,βln3β}+⋯. (12) The collection of these terms are referred to as the improved next-to-leading logarithmic (NLL) corrections. Note that due to the presence of two kinds of terms, one needs to insist on a consistent logarithmic counting for both of them, which we take as . Using this counting, it can be seen that the NLL corrections include terms up to order . It is also clear from this counting that one needs to include formally next-to-leading power (NLP) terms besides the leading power (LP) ones in the power expansion. This greatly complicates the analysis of factorization, as will be clear below. The behavior of higher order corrections in the threshold limit can be studied using the method of regions Beneke:1997zp (); Jantzen:2011nz (). We work in the partonic center-of-mass frame where the momenta of the two incoming partons are given by pμ1=√^s2nμ,pμ2=√^s2¯nμ, (13) where and are two light-like vectors satisfying and . For a given momentum , we perform the light-cone decomposition as kμ=k+2nμ+k−2¯nμ+kμ⊥, (14) with and . We identify the following momentum regions relevant to our problem: hard:kμ∼√^s,soft:kμ∼√^sβ,potential:k0∼√^sβ2,→k∼√^sβ,ultrasoft:kμ∼√^sβ2,collinear:(k+,k−,k⊥)∼√^s(1,β2,β),anticollinear:(k+,k−,k⊥)∼√^s(β2,1,β). (15) These serve as the basis for constructing the effective field theoretic description of the process, and for deriving the factorization formula for the cross sections. At this point, it should be noted that there is a subtle difference between production here and production discussed in Beneke:2009ye (); Beneke:2009rj (); Beneke:2010da (); Beneke:2011mq (); Piclum:2018ndt (). In production, the 3-momentum of the pair is of the ultrasoft scale . This means that the rest frame is formally equivalent to the partonic center-of-mass frame. On the other hand, in production, the pair is recoiled by the Higgs boson and has a 3-momentum of the potential scale . Therefore, an ultrasoft mode in the partonic center-of-mass frame will become a potential mode in the rest frame. The impact of this difference on the factorization and resummation will be discussed in this section. ### 3.2 Effective field theories In order to derive the factorization and resummation formulas in the threshold limit, it is useful to employ the language of effective field theories (EFTs). According to the momentum regions in Eq. (15), the relevant EFTs are the soft-collinear effective theory (SCET) and the non-relativistic QCD (NRQCD). SCET Bauer:2000ew (); Bauer:2000yr (); Bauer:2001yt (); Beneke:2002ph (); Beneke:2002ni () describes the interactions among collinear, anticollinear and ultrasoft modes. At leading power and next-to-leading power in , the effective Lagrangians are given by L0SCET −12Tr{FμνusFusμν}, (16) L1aSCET =¯ξn(xμ⊥nν¯¯¯¯¯¯WngsFusμν¯¯¯¯¯¯W†n)⧸¯n2ξn+(n↔¯n), (17) L1bSCET =Tr{nμFnμν¯¯¯¯¯¯Wni[xρ⊥¯nρFusρσ,¯¯¯¯¯¯W†n(iDνn⊥¯¯¯¯¯¯Wn)]¯¯¯¯¯¯W†n}−Tr{nμFμν⊥n¯¯¯¯¯¯Wn¯nρFusρν⊥¯¯¯¯¯¯W†n} +(n↔¯n), (18) L1cSCET =¯ξni⧸Dn⊥¯¯¯¯¯¯Wnqus+h.c.+(n↔¯n), (19) where and denote the collinear and ultrasoft quark fields; and represent the collinear (ultrasoft) gluon fields, with their field strength tensors; is the collinear Wilson line. To describe the interactions among the potential, soft and ultrasoft modes, we employ the potential non-relativistic QCD (pNRQCD) Pineda:1997bj (); Brambilla:1999xf (); Beneke:1999zr (); Beneke:1999qg (). The leading power and next-to-leading power effective Lagrangians can be written as Beneke:1999zr (); Beneke:1999qg (); Kniehl:2002br () L0pNRQCD(x) =ψ†⎛⎝iD0us+→∂22mt⎞⎠ψ+χ†⎛⎝iD0us−→∂22mt⎞⎠χ −∫d3→rψ†Taψ(x0,→x+→r)(αsr)χ†Taχ(x0,→x), (20) L1apNRQCD(x) =−ψ†(x)gs→x\textperiodcentered→Eus(x0,→0)ψ(x)−χ†(x)gs→x\textperiodcentered→Eus(x0,→0)χ(x), (21) L1bpNRQCD(x) =−∫d3→rψ†Taψ(x0,→x+→r)α2s4πr[a1+2β0ln(eγEμr)]χ†Taχ(x0,→x), (22) where and are Pauli spinor fields annihilating the top quark and creating the anti-top quark, respectively; are the chromoelectric components of the ultrasoft field strength tensor. The coefficient was calculated in Fischler:1977yf (); Billoire:1979ih () and is given by . The one-loop coefficient of the QCD -function is given in the Appendix. Note that in the pNRQCD power counting, and in Eq. (20) and (22) are considered as order . Worthy of particular attention is that in Refs. Beneke:1999zr (); Beneke:1999qg (); Kniehl:2002br (), the pNRQCD Lagrangian is derived in the rest frame of the quarkonium, where the heavy quark pair is recoiled by ultrasoft momenta. This is different from our case of production, where the top quark pair is recoiled by the Higgs boson. However, since the LO and NLO potentials in Eqs. (20)-(22) only involve the relative momentum between the heavy quark and anti-quark, in our case the LP and NLP Lagrangians take the same form. It should be stressed that this fact is not expected to hold beyond NLP. For example, as shown in Appendix B, new structures depending on the recoil momentum appear in the NNLP Lagrangian. To derive the factorization formula, we first match the QCD amplitudes onto an effective Hamiltonian constructed out of the SCET and pNRQCD fields. Generically, we write H≡HLP+HNLP+⋯≡∑I,mCImLPOImLP+∑I,mCImNLPOImNLP+⋯, (23) where labels different color structures and for Lorentz structures, and are leading power and next-to-leading power effective operators describing the scattering process under consideration, while and are their Wilson coefficients arising from the hard region contributions. Note that the NLP effective Hamiltonian actually does not contribute to the cross section at next-to-leading power. The reason is that such a contribution would be given by the interferences between and , which vanish due to angular momentum conservation. To obtain the Wilson coefficients , we need to calculate on-shell scattering amplitudes in the limit using both QCD and the effective Hamiltonian. In this limit, the loop integrals in the effective theories are scaleless and vanish in dimensional regularization. Therefore, we only need to calculate the QCD amplitudes up to NLO. For this calculation we employ the program packages FeynArts Hahn:2000kx (), FeynCalc Mertig:1990an (); Shtabovenko:2016sxi () and FIRE5 Smirnov:2014hma () to generate the amplitudes and perform the reduction to master integrals. The resulting master integrals can be evaluated to analytic expressions using Package-X Patel:2015tea (); Patel:2016fam (). The interface connecting Package-X, FIRE5 and FeynCalc is provided by FeynHelpers Shtabovenko:2016whf (). We renormalize the top quark mass in the on-shell scheme, and the strong coupling in the scheme. The Wilson coefficients can then be extracted after renormalizing the effective operators (which is equivalent to subtracting the infrared poles from the QCD amplitudes). For the purpose of this paper, we don’t need the explicit forms of individual Wilson coefficients and effective operators, but the combinations of them entering the cross section for production. These combinations will be given in the next subsection as the “hard functions”. The discussion in the last subsection tells us that for the cross section up to NLP, we only need to consider the amplitudes of . At leading power, we use together with the LP Lagrangians and to calculate the cross section (24) where labels the initial state, and represents phase-space integration over the momentum of the final state particle . In the LP Lagrangians (16) and (20), the interactions of the ultrasoft gluon with the collinear fields and heavy quark fields are encoded in the covariant derivative . Such interactions can be removed by the decoupling transformations Bauer:2001yt (); Beneke:2010da () ξn(¯n)(x)→Sqn(¯n)(x)ξn(¯n)(x),An(¯n)(x)→Sgn(¯n)(x)An(¯n)(x), ψ(x)→Sv(x)ψ(x),χ(x)→Sv(x)χ(x), (25) where and are ultrasoft Wilson lines in the fundamental representation along the directions implied by the subscripts, while are ultrasoft Wilson lines in the adjoint representation. After the decoupling, the factorization of the LP cross section follows along the same line of arguments as the production Beneke:2010da (). The only difference comes from the appearance of the Higgs momentum . The factorization formula therefore reads ^σLPij=12^s∫dΦhdωHJIij(μ)JαLP(EJ−ω2,→pJ)SαIJij(ω,μ), (26) where EJ=√^s−2mt−mh−\|→ph\|22mh,→pJ=−→ph. (27) The potential function is given by (28) where and are the projectors to the singlet-octet color states of the system, which are given by P(1){a}=13δa1a2δa3a4,P(8){a}=2Tca1a2Tca4a3. (29) Note that since the interactions in the LP Lagrangian are spin-independent, the two fields in the definition of the potential function share the same polarization index , and similar for the two fields. It should be stressed that the potential function here is different from that in production, due to the presence of the recoil momentum . The soft function in Eq. (26) is defined as (30) where the color basis for the channel is given by C(1)q¯q,{a}=13δa2a1δa3a4,C(2)q¯q,{a}=1√2Tca2a1Tca3a4, (31) and that for the channel is C(1)gg,{a}=12√6δa2a1δa3a4,C(2)gg,{a}=12√3Fca2a1Tca3a4,C(3)gg,{a}=12√35Dca2a1Tca3a4, (32) where and . This soft function is the same as that for production in Beneke:2010da (), since it does not feel the presence of the recoil momentum. It is diagonal in the color basis we have chosen. In practice, it is more convenient to quote its Laplace transform ~sαIJij(L,μ)=∫∞0dωexp(−Nω2mt+mh)SαIJij(ω,μ), (33) where L=2lnμNeγE2mt+mh. (34) Up to the NLO, the result reads Beneke:2010da () ~sαIIij(L,μ)=1+αs4π[(Ci+Cj)(L2+π26)+2Cα(L+2)], (35) where for , for , for (singlet), and for (octet). Note that the soft function actually does not depend on the index . Finally, the hard function is defined as the product of LP Wilson coefficients projected onto the -wave spin structure determined by the potential function. In general, the hard function is not diagonal. However, since the soft function is diagonal, only the diagonal entries of the hard function contribute to the cross section. Furthermore, since the soft function does not depend on the index , we can project the diagonal entries of the hard function onto the singlet-octet basis as H(1)ij=H11ij,H(8)q¯q=H22q¯q,H(8)gg=H22gg+H33gg. (36) We have calculated these entries at NLO explicitly using the method described in the last subsection, and write the result as Hαij=4π2α2sm2tv2m2h(mh+2mt)4HLO,αij×[1+αs4π[−4(Ci+Cj)L2H+(4β0+γH,αij,0)LH+HNLO,αij]], (37) where , and the LO coefficients are given by HLO,(1)q¯q =0, HLO,(8)q¯q =329, HLO,(1)gg =CF(m2h−4m2t)264m4t, HLO,(8)gg =−2HLO,(1)gg+C2ACF(m4h−4m2hm2t+8m4t)64m4t. (38) The one-loop hard anomalous dimensions are given by γH,(8)q¯q,0=−12CF−4CA,γH,(1)gg,0=−4β0,γH,(8)gg,0=−4β0−4CA. (39) The analytic expressions for the NLO coefficients are too tedious to be shown. To get some impression of their sizes, we quote the numeric values here with and : HNLO,(8)q¯q≈4.93,HNLO,(1)gg≈37.0,HNLO,(8)gg≈23.8. (40) To achieve NLL accuracy for the resummation, we need to further consider next-to-leading power corrections to the cross section. These amount to contributions from the NLP interactions in and to the squared-amplitude in Eq. (24). In analogy to the arguments in Beneke:2009ye (); Beneke:2010da () for production, it can be shown that single insertions of give vanishing results due to angular momentum conservation, while does not contribute due to baryon number conservation. The terms in involve subleading potentials between the top and anti-top quarks. These contributions can be incorporated by upgrading the potential function to the NLO, which we will discuss in the next subsection. Finally, we need to consider the corrections induced by . The terms in involve extra interactions between ultrasoft gluons and potential modes, which are not removed by the decoupling transform (25). In Beneke:2010da (), it was proved that these interactions do not contribute to the cross section at NLP. However, the arguments there rely on the fact that the partonic center-of-mass frame and the rest frame are the same, and therefore the potential function does not depend on an external 3-momentum. However, for production, the recoil momentum from the extra Higgs boson spoils the proof, and we need to reinvestigate the contributions from here. We begin with an explicit diagram depicted in Fig. 6. Its contribution to the partonic cross section in the threshold limit can be written as (up to overall factors due to coupling constants, color factors, Wilson coefficients, etc.) Δ^σ∝∫dΦhA(→ph), (41) where is given by A(→ph)=∫dΦtdΦ¯tdΦgδ(EJ−Et−E¯t−p0g)δ(3)(→ph+→pt+→p¯t)∫d4k(2π)4M, (42) where is defined in Eq. (27), Et=\|→pt\|22mt,E¯t=\|→p¯t\|22mt, (43) and M=1−\|→k−→pt\|21k0−p0g−\|→k\|2/(2mt)1Et+E¯t+p0g−k0−\|→k−→pt−→p¯t\|2/(2mt)×(∂∂→k1k0−\|→k\|2/(2mt))\textperiodcentered(v% \textperiodcenteredn)→pg−(v\textperiodcenteredpg)→nn\textperiodcenteredpg. (44) Note that we have suppressed the imaginary part in the propagators. We now observe that the last factor in the above expression does not depend on , , and , while the other factors do not depend on . Together with the fact that , we can conclude that after integrating over , , and , the function must be proportional to (multiplied by a function of and other scalar quantities). As a result, after performing the integration over as in Eq. (41), the contribution of this diagram to the partonic total cross section must vanish. The argument above can be generalized to all contributions from a single insertion of in a more formal way. The cross section induced by can be written as (45) where denotes time-ordered product. We can perform the usual decoupling transforms (25) to remove the leading power interaction between ultrasoft and potential modes. The remaining interaction is of the form from . As a result, we can write the cross section as ^σ1aij =12^s∫dΦhdωHij(μ)∫d4k(2π)4J1a(EJ−ω2,→pJ,k)∫d4ze−ik\textperiodcenteredzi→z% \textperiodcentered→S1aij(ω,z0,μ)+h.c. =12^s∫dΦhdωdz0dk02πe−ik0z0Hij(μ)→j1	2021-03-09 11:07:42	{"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": 0.843542218208313, "perplexity": 839.627899351651}, "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/1614178389798.91/warc/CC-MAIN-20210309092230-20210309122230-00528.warc.gz"}
http://taggedwiki.zubiaga.org/new_content/8e37bc020b5c0f10c5e56e5fbbc18f78	# Sharpe ratio The Sharpe ratio or Sharpe index or Sharpe measure or reward-to-variability ratio is a measure of the excess return (or Risk Premium) per unit of risk in an investment asset or a trading strategy, named after William Forsyth Sharpe. Since its revision made by the original author in 1994, it is defined as: $S = \frac{R-R_f}{\sigma} = \frac{E[R-R_f]}{\sqrt{\mathrm{var}[R-R_f]}},$ where R is the asset return, Rf is the return on a benchmark asset, such as the risk free rate of return, E[RRf] is the expected value of the excess of the asset return over the benchmark return, and σ is the standard deviation of the asset excess return.[1] Note, if Rf is a constant risk free return throughout the period, $\sqrt{\mathrm{var}[R-R_f]}=\sqrt{\mathrm{var}[R]}.$ The Sharpe ratio is used to characterize how well the return of an asset compensates the investor for the risk taken. When comparing two assets each with the expected return E[R] against the same benchmark with return Rf, the asset with the higher Sharpe ratio gives more return for the same risk. Investors are often advised to pick investments with high Sharpe ratios. However like any mathematical model it relies on the data being correct. Pyramid schemes with a long duration of operation would typically provide a high Sharpe ratio when derived from reported returns but the inputs are false. When examining the investment performance of assets with smoothing of returns (such as With profits funds) the Sharpe ratio should be derived from the performance of the underlying assets rather than the fund returns. Sharpe ratios, along with Treynor ratios and Jensen's alphas, are often used to rank the performance of portfolio or mutual fund managers. ## History This ratio was developed by William Forsyth Sharpe in 1966.[2] Sharpe originally called it the "reward-to-variability" ratio in before it began being called the Sharpe Ratio by later academics and financial professionals. Sharpe's 1994 revision acknowledged that the risk free rate changes with time. Prior to this revision the definition was $S = \frac{E[R]-R_f}{\sigma}$ assuming a constant Rf . Recently, the (original) Sharpe ratio has often been challenged with regard to its appropriateness as a fund performance measure during evaluation periods of declining markets.[3] ## Examples Suppose the asset has an expected return of 15% in excess of the risk free rate. We typically do not know the asset will have this return; suppose we assess the risk of the asset, defined as standard deviation of the asset's excess return, as 10%. The risk-free return is constant. Then the Sharpe ratio (using a new definition) will be 1.5 (RRf = 0.15 and σ = 0.10). As a guide post, one could substitute in the longer term return of the S&P500 as 10%. Assume the risk-free return is 3.5%. And the average standard deviation of the S&P500 is about 16%. Doing the math, we get that the average, long-term Sharpe ratio of the US market is about 0.40625 ((10%-3.5%)/16%). But we should note that if one were to calculate the ratio over, for example, three-year rolling periods, then the Sharpe ratio would vary dramatically. ## Strengths and Weaknesses The Sharpe ratio has as its principle advantage that it is directly computable from any observed series of returns without need for additional information surrounding the source of profitability. Unfortunately, some authors are carelessly drawn to refer to the ratio as giving the level of 'risk adjusted returns' when the ratio gives only the volatility adjusted returns when interpreted properly. Other ratios such as the Bias ratio (finance) have recently been introduced into the literature to handle cases where the observed volatility may be an especially poor proxy for the risk inherent in a time-series of observed returns. ## References 1. ^ Sharpe, W. F. (1994). "The Sharpe Ratio". Journal of Portfolio Management 21 (1): 49–58. 2. ^ Sharpe, W. F. (1966). "Mutual Fund Performance". Journal of Business 39 (S1): 119–138. doi:10.1086/294846. 3. ^ Scholz, Hendrik (2007). "Refinements to the Sharpe ratio: Comparing alternatives for bear markets". Journal of Asset Management 7 (5): 347–357. doi:10.1057/palgrave.jam.2250040.	2019-10-23 21:42:00	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 3, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8182829022407532, "perplexity": 2042.4666080300942}, "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/1570987836295.98/warc/CC-MAIN-20191023201520-20191023225020-00142.warc.gz"}
http://ckunte.net/2015/ph	Purging history Many of my repositories, including this site, contain binary files. This results in repos getting heavier over time, with version after version, every copy ever pushed is still around for me to dig back and recover from — the general light and magic of git. That said, most repos I store tend to be in forward motion, and rarely do I find the need to revert back to an old copy, especially after initial few days of adding something new. Now, I do not know if this is the best way to manage old baggage, but once in a while, I find purging git history handy — after a particular repo has stabilized in fixes and updates, especially those with obsolete binary files. It makes for faster pulls, leaner on disk space, and it results in generally faster responses from server. To automate this, I wrote a generic script. Here it is. #!/usr/bin/env python # -- coding: UTF-8 -- # ph.py -- purge (git) history, 2015 ckunte import os def main(): fullpath = os.getcwd() foldername = os.path.basename(os.path.normpath(fullpath)) repo = "[email protected]:ckunte/" + foldername + ".git" cmd1 = 'git reset --hard; rm -rf .git/; git init; git add .; git commit -m "first commit.";' cmd2 = 'git remote add origin ' + repo + '; git push --force origin master;' os.system(cmd1 + cmd2) pass if __name__ == '__main__': main() It needs to be run from the repo’s root. For example, if the repo is at ~/Projects/hcalc, then I run the script from within the hcalc folder. Another way to do this is as follows: #!/usr/bin/env zsh git checkout --orphan newBranch	2020-08-04 16:31:48	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.18148267269134521, "perplexity": 8306.591466045566}, "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-2020-34/segments/1596439735881.90/warc/CC-MAIN-20200804161521-20200804191521-00328.warc.gz"}
http://tex.stackexchange.com/questions/146663/pgfplots-fails-to-generate-boxplots-for-some-data-set	# Pgfplots fails to generate boxplots for some data set I use pgfplots to generate boxplots, however with some small data sets as the following, I get an error message. speed 2 3 1 2 2 11 Here is the error message : Sorry, an internal routine of the floating point unit got an ill-floating point number '. The unreadable part was near '' To avoid this problem, I need to add white noise to my data set when generating it, but it is only a dirty workaroud. Do you know how to avoid the pgfplots error message ? You can for example try the following code : \documentclass{article} \usepackage{pgfplots,filecontents} \pgfplotsset{compat=1.9} \usepgfplotslibrary{statistics} \begin{filecontents}{test.dat} speed 2 3 1 2 2 11 \end{filecontents} \begin{document} \begin{tikzpicture} \begin{axis} `	2014-09-30 16:43:27	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8231160640716553, "perplexity": 1541.8107478553327}, "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/1412037663036.27/warc/CC-MAIN-20140930004103-00298-ip-10-234-18-248.ec2.internal.warc.gz"}
https://machinelearningmastery.com/logistic-regression-with-maximum-likelihood-estimation/	# A Gentle Introduction to Logistic Regression With Maximum Likelihood Estimation Last Updated on October 28, 2019 Logistic regression is a model for binary classification predictive modeling. The parameters of a logistic regression model can be estimated by the probabilistic framework called maximum likelihood estimation. Under this framework, a probability distribution for the target variable (class label) must be assumed and then a likelihood function defined that calculates the probability of observing the outcome given the input data and the model. This function can then be optimized to find the set of parameters that results in the largest sum likelihood over the training dataset. The maximum likelihood approach to fitting a logistic regression model both aids in better understanding the form of the logistic regression model and provides a template that can be used for fitting classification models more generally. This is particularly true as the negative of the log-likelihood function used in the procedure can be shown to be equivalent to cross-entropy loss function. In this post, you will discover logistic regression with maximum likelihood estimation. After reading this post, you will know: • Logistic regression is a linear model for binary classification predictive modeling. • The linear part of the model predicts the log-odds of an example belonging to class 1, which is converted to a probability via the logistic function. • The parameters of the model can be estimated by maximizing a likelihood function that predicts the mean of a Bernoulli distribution for each example. Kick-start your project with my new book Probability for Machine Learning, including step-by-step tutorials and the Python source code files for all examples. Let’s get started. A Gentle Introduction to Logistic Regression With Maximum Likelihood Estimation Photo by Samuel John, some rights reserved. ## Overview This tutorial is divided into four parts; they are: 1. Logistic Regression 2. Logistic Regression and Log-Odds 3. Maximum Likelihood Estimation 4. Logistic Regression as Maximum Likelihood ## Logistic Regression Logistic regression is a classical linear method for binary classification. Classification predictive modeling problems are those that require the prediction of a class label (e.g. ‘red‘, ‘green‘, ‘blue‘) for a given set of input variables. Binary classification refers to those classification problems that have two class labels, e.g. true/false or 0/1. Logistic regression has a lot in common with linear regression, although linear regression is a technique for predicting a numerical value, not for classification problems. Both techniques model the target variable with a line (or hyperplane, depending on the number of dimensions of input. Linear regression fits the line to the data, which can be used to predict a new quantity, whereas logistic regression fits a line to best separate the two classes. The input data is denoted as X with n examples and the output is denoted y with one output for each input. The prediction of the model for a given input is denoted as yhat. • yhat = model(X) The model is defined in terms of parameters called coefficients (beta), where there is one coefficient per input and an additional coefficient that provides the intercept or bias. For example, a problem with inputs X with m variables x1, x2, …, xm will have coefficients beta1, beta2, …, betam, and beta0. A given input is predicted as the weighted sum of the inputs for the example and the coefficients. • yhat = beta0 + beta1 * x1 + beta2 * x2 + … + betam * xm The model can also be described using linear algebra, with a vector for the coefficients (Beta) and a matrix for the input data (X) and a vector for the output (y). • y = X * Beta So far, this is identical to linear regression and is insufficient as the output will be a real value instead of a class label. Instead, the model squashes the output of this weighted sum using a nonlinear function to ensure the outputs are a value between 0 and 1. The logistic function (also called the sigmoid) is used, which is defined as: • f(x) = 1 / (1 + exp(-x)) Where x is the input value to the function. In the case of logistic regression, x is replaced with the weighted sum. For example: • yhat = 1 / (1 + exp(-(X * Beta))) The output is interpreted as a probability from a Binomial probability distribution function for the class labeled 1, if the two classes in the problem are labeled 0 and 1. Notice that the output, being a number between 0 and 1, can be interpreted as a probability of belonging to the class labeled 1. — Page 726, Artificial Intelligence: A Modern Approach, 3rd edition, 2009. The examples in the training dataset are drawn from a broader population and as such, this sample is known to be incomplete. Additionally, there is expected to be measurement error or statistical noise in the observations. The parameters of the model (beta) must be estimated from the sample of observations drawn from the domain. There are many ways to estimate the parameters. There are two frameworks that are the most common; they are: Both are optimization procedures that involve searching for different model parameters. Maximum Likelihood Estimation is a frequentist probabilistic framework that seeks a set of parameters for the model that maximizes a likelihood function. We will take a closer look at this second approach in the subsequent sections. ### Want to Learn Probability for Machine Learning Take my free 7-day email crash course now (with sample code). Click to sign-up and also get a free PDF Ebook version of the course. ## Logistic Regression and Log-Odds Before we dive into how the parameters of the model are estimated from data, we need to understand what logistic regression is calculating exactly. This might be the most confusing part of logistic regression, so we will go over it slowly. The linear part of the model (the weighted sum of the inputs) calculates the log-odds of a successful event, specifically, the log-odds that a sample belongs to class 1. • log-odds = beta0 + beta1 * x1 + beta2 * x2 + … + betam * xm In effect, the model estimates the log-odds for class 1 for the input variables at each level (all observed values). What are odds and log-odds? Odds may be familiar from the field of gambling. Odds are often stated as wins to losses (wins : losses), e.g. a one to ten chance or ratio of winning is stated as 1 : 10. Given the probability of success (p) predicted by the logistic regression model, we can convert it to odds of success as the probability of success divided by the probability of not success: • odds of success = p / (1 – p) The logarithm of the odds is calculated, specifically log base-e or the natural logarithm. This quantity is referred to as the log-odds and may be referred to as the logit (logistic unit), a unit of measure. • log-odds = log(p / (1 – p) Recall that this is what the linear part of the logistic regression is calculating: • log-odds = beta0 + beta1 * x1 + beta2 * x2 + … + betam * xm The log-odds of success can be converted back into an odds of success by calculating the exponential of the log-odds. • odds = exp(log-odds) Or • odds = exp(beta0 + beta1 * x1 + beta2 * x2 + … + betam * xm) The odds of success can be converted back into a probability of success as follows: • p = odds / (odds + 1) And this is close to the form of our logistic regression model, except we want to convert log-odds to odds as part of the calculation. We can do this and simplify the calculation as follows: • p = 1 / (1 + exp(-log-odds)) This shows how we go from log-odds to odds, to a probability of class 1 with the logistic regression model, and that this final functional form matches the logistic function, ensuring that the probability is between 0 and 1. We can make these calculations of converting between probability, odds and log-odds concrete with some small examples in Python. First, let’s define the probability of success at 80%, or 0.8, and convert it to odds then back to a probability again. The complete example is listed below. Running the example shows that 0.8 is converted to the odds of success 4, and back to the correct probability again. Let’s extend this example and convert the odds to log-odds and then convert the log-odds back into the original probability. This final conversion is effectively the form of the logistic regression model, or the logistic function. The complete example is listed below. Running the example, we can see that our odds are converted into the log odds of about 1.4 and then correctly converted back into the 0.8 probability of success. Now that we have a handle on the probability calculated by logistic regression, let’s look at maximum likelihood estimation. ## Maximum Likelihood Estimation Maximum Likelihood Estimation, or MLE for short, is a probabilistic framework for estimating the parameters of a model. In Maximum Likelihood Estimation, we wish to maximize the conditional probability of observing the data (X) given a specific probability distribution and its parameters (theta), stated formally as: • P(X ; theta) Where X is, in fact, the joint probability distribution of all observations from the problem domain from 1 to n. • P(x1, x2, x3, …, xn ; theta) This resulting conditional probability is referred to as the likelihood of observing the data given the model parameters and written using the notation L() to denote the likelihood function. For example: • L(X ; theta) The joint probability distribution can be restated as the multiplication of the conditional probability for observing each example given the distribution parameters. Multiplying many small probabilities together can be unstable; as such, it is common to restate this problem as the sum of the log conditional probability. • sum i to n log(P(xi ; theta)) Given the frequent use of log in the likelihood function, it is referred to as a log-likelihood function. It is common in optimization problems to prefer to minimize the cost function rather than to maximize it. Therefore, the negative of the log-likelihood function is used, referred to generally as a Negative Log-Likelihood (NLL) function. • minimize -sum i to n log(P(xi ; theta)) The Maximum Likelihood Estimation framework can be used as a basis for estimating the parameters of many different machine learning models for regression and classification predictive modeling. This includes the logistic regression model. ## Logistic Regression as Maximum Likelihood We can frame the problem of fitting a machine learning model as the problem of probability density estimation. Specifically, the choice of model and model parameters is referred to as a modeling hypothesis h, and the problem involves finding h that best explains the data X. We can, therefore, find the modeling hypothesis that maximizes the likelihood function. • maximize sum i to n log(P(xi ; h)) Supervised learning can be framed as a conditional probability problem of predicting the probability of the output given the input: • P(y \| X) As such, we can define conditional maximum likelihood estimation for supervised machine learning as follows: • maximize sum i to n log(P(yi\|xi ; h)) Now we can replace h with our logistic regression model. In order to use maximum likelihood, we need to assume a probability distribution. In the case of logistic regression, a Binomial probability distribution is assumed for the data sample, where each example is one outcome of a Bernoulli trial. The Bernoulli distribution has a single parameter: the probability of a successful outcome (p). • P(y=1) = p • P(y=0) = 1 – p The probability distribution that is most often used when there are two classes is the binomial distribution.5 This distribution has a single parameter, p, that is the probability of an event or a specific class. — Page 283, Applied Predictive Modeling, 2013. The expected value (mean) of the Bernoulli distribution can be calculated as follows: • mean = P(y=1) * 1 + P(y=0) * 0 Or, given p: • mean = p * 1 + (1 – p) * 0 This calculation may seem redundant, but it provides the basis for the likelihood function for a specific input, where the probability is given by the model (yhat) and the actual label is given from the dataset. • likelihood = yhat * y + (1 – yhat) * (1 – y) This function will always return a large probability when the model is close to the matching class value, and a small value when it is far away, for both y=0 and y=1 cases. We can demonstrate this with a small worked example for both outcomes and small and large probabilities predicted for each. The complete example is listed below. Running the example prints the class labels (y) and predicted probabilities (yhat) for cases with close and far probabilities for each case. We can see that the likelihood function is consistent in returning a probability for how well the model achieves the desired outcome. We can update the likelihood function using the log to transform it into a log-likelihood function: • log-likelihood = log(yhat) * y + log(1 – yhat) * (1 – y) Finally, we can sum the likelihood function across all examples in the dataset to maximize the likelihood: • maximize sum i to n log(yhat_i) * y_i + log(1 – yhat_i) * (1 – y_i) It is common practice to minimize a cost function for optimization problems; therefore, we can invert the function so that we minimize the negative log-likelihood: • minimize sum i to n -(log(yhat_i) * y_i + log(1 – yhat_i) * (1 – y_i)) Calculating the negative of the log-likelihood function for the Bernoulli distribution is equivalent to calculating the cross-entropy function for the Bernoulli distribution, where p() represents the probability of class 0 or class 1, and q() represents the estimation of the probability distribution, in this case by our logistic regression model. • cross entropy = -(log(q(class0)) * p(class0) + log(q(class1)) * p(class1)) Unlike linear regression, there is not an analytical solution to solving this optimization problem. As such, an iterative optimization algorithm must be used. Unlike linear regression, we can no longer write down the MLE in closed form. Instead, we need to use an optimization algorithm to compute it. For this, we need to derive the gradient and Hessian. — Page 246, Machine Learning: A Probabilistic Perspective, 2012. The function does provide some information to aid in the optimization (specifically a Hessian matrix can be calculated), meaning that efficient search procedures that exploit this information can be used, such as the BFGS algorithm (and variants). This section provides more resources on the topic if you are looking to go deeper. ## Summary In this post, you discovered logistic regression with maximum likelihood estimation. Specifically, you learned: • Logistic regression is a linear model for binary classification predictive modeling. • The linear part of the model predicts the log-odds of an example belonging to class 1, which is converted to a probability via the logistic function. • The parameters of the model can be estimated by maximizing a likelihood function that predicts the mean of a Bernoulli distribution for each example. Do you have any questions? ## Get a Handle on Probability for Machine Learning! #### Develop Your Understanding of Probability ...with just a few lines of python code Discover how in my new Ebook: Probability for Machine Learning It provides self-study tutorials and end-to-end projects on: Bayes Theorem, Bayesian Optimization, Distributions, Maximum Likelihood, Cross-Entropy, Calibrating Models and much more... ### 38 Responses to A Gentle Introduction to Logistic Regression With Maximum Likelihood Estimation 1. Norman November 1, 2019 at 4:18 pm # Hi, Jason One suggestion. Could you consider to use the notation “yhat ^ y * (1 – yhat) ^(1 – y)” to define the likelihood? This one has the same result as the original one for Bernoulli distribution. But this one is easier for calculating log-likelihood by math. These links below for your reference. • Jason Brownlee November 2, 2019 at 6:39 am # Thanks for the suggestion Norman. • Mateusz Serocki November 2, 2020 at 9:41 am # I totaly agree with that. 2. Anthony The Koala November 2, 2019 at 3:32 am # Dear Dr Jason, I would like clarification please on today’s topic: * odds: odds = p/(1-p). If have a die, probability of “1” face up = p = 1/6 The odds = p/(1-p) = 0.2 What is the interpretation of 0.2? * In the section on Logistic Regression and MLE What is the interpretation of Thank you, Anthony of Sydney • Jason Brownlee November 2, 2019 at 7:09 am # Good question. If you have a probability for of 1/6 (prob) and for reference, a probability against of 5/6. Then the odds in favor of rolling a “1” are: prob / (1 – prob) = 0.2 The odds against (e.g. rolling a not 1) are (1 – prob) / prob = 5 which is 1:25. For more on odds vs probs see: https://en.wikipedia.org/wiki/Odds#Mathematical_relations The second part of your question is about the likelihood function for a Bernoulli trial. E.g. we are reporting a probability of matching the positive outcome. It’s redundant, but it provides the basis for the log likelihood that follows – in that section I want to to see that it’s not doing anything exciting. 3. Adam December 13, 2019 at 2:50 pm # Does yhat_i = P(y_i\|x_i ; h)? • Jason Brownlee December 14, 2019 at 6:05 am # Yes. • Robert August 17, 2020 at 10:51 pm # Hmm… isn’t P(y_i\|x_i;h) = yhat_iy_i ? Or what am I missing in the last example at the very end when the (log-)likelihood is calculated going from the expression maximize sum i to n log(P(yi\|xi ; h)), introduced earlier to the expression maximize sum i to n log(yhat_i) y_i + log(1 – yhat_i) * (1 – y_i) 4. Omar December 21, 2019 at 9:00 pm # I have some conceptual questions to ask. Mostly referring to log-odds with natural logarithm is written as ln( prob_event / (1 – prob_event) ) = b_0 + b_1 * X_1 + … + b_n * X_n 1. When dealing with real data, how do I know the probability of the event? 2. Suppose that I have no idea about the probability of the event. I assumed we can calculate the log-odds by fitting multiple linear regression (please correct me if I am wrong) since the right hand side of the equation above is a multiple linear regression. What should I do to calculate the beta (b_0, b_1, … b_n) ? I understand how to get the values using python, but have no idea to calculate them manually. I also know that we can fit the logistic regression using Maximum Likelihood Estimation but I don’t know how to do it manually. 3. Also suppose that I have a dataset with 100 rows, divided into 20 windows with each window containing 5 rows to do classification with labels corresponding to the window. Logistic Regression from sklearn can classify them. Do you have any intuition to guide me how can sklearn’s logistic regression do this? What I understand is that after we have the beta, we can easily plug the data into X, but I don’t know what actually happens if the value we want to plug to X is not a single row (in this case, 5 rows). Thank you very much for your assistance. • Jason Brownlee December 22, 2019 at 6:11 am # Good questions! We estimate the probability of an event from historical observations (Frequency), or we use domain expertise to define our belief about the likelihood of an event (Bayesian). In a model, we can assume a likelihood distribution over events, and guess at the probability of new events. This is what we do in logistic regression. Coefficients are optimized to minmize log loss on a training dataset. No need to worry about the coefficients for a single observation. A prediction is made by multipling input by the coefficients. • Omar December 22, 2019 at 6:32 pm # Ah I see, thanks! Please let me rephrase the third question. Assume we have Y = b_0 + b_1X_1 (a logistic regression model with only one predictor). Let’s say that my data is only 20 samples with 20 target variable, with each sample contain 5 rows (so that the total rows is 100). My question is, what is the math behind fitting/predicting samples with multiple rows inside? For example, if b_0 = 2, b_1 = 0.5, and X_1 = 4, I can calculate Y = 2 + 0.5 * 4. But if X_1 is a list [2,3,4], I don’t know the math to predict it. Plugging more than one row as a sample in sklearn seems fine (no error or warning shown) • Jason Brownlee December 23, 2019 at 6:47 am # Sorry, I don’t follow. Logistic regression assumes you have one target variable with 2 classes and some number of input variables. I don’t know what “20 samples with 20 target variable, with each sample contain 5 rows” means. • Omar December 23, 2019 at 10:06 pm # Please refer to this image: https://imgur.com/VHikbUn Given 6 observations, I want to make “windows of observations” which each window contains 3 observations. In this case, I have 2 samples, the blue and orange one. The blue window will be assigned class 0 (not fault), and orange with class 1 (fault). This means I will have 2 samples (blue and orange). My goal is whenever I pass 3 observation, my model will see them as “one window” and will predict whether it is a fault or not. Using fit method in sklearn Logistic Regression, this means X has two samples (blue and orange), and y also two samples (0 and 1). I succesfully fitted them. My question may be answered from two perspective: 1. From python or scikit-learn perspective, how is it not throwing any error? 2. From statistics perspective, how does one estimate beta parameters and also do prediction with X similar with my case? This is my best effort to explain the case. Very much appreciated your help Jason. • Jason Brownlee December 24, 2019 at 6:42 am # Thanks for elaborating. We would call the 3 numbers a sample, not a window. This is a standard prediction problem, you pass samples of 3 number and predict the class. You can do this manually or via sklearn. The parameters will be estimated from training data. 5. Omar December 24, 2019 at 12:28 am # Dear Jason, please ignore my last question, looks like I made an error on the code that somehow makes it able to be processed and misunderstood the complete context of the problem. 6. Omar December 25, 2019 at 2:22 am # Dear Jason, I think now I have a bit of insights about my case above. Please refer again to this image: https://imgur.com/VHikbUn I was able to construct the data to be 3-dimensional array (can be plugged into LSTM models), with shape (2,3,1) for (samples, timesteps, feature). To make sure I can pass the sample (with three numbers inside like blue or orange) to sklearn classifier, I reshaped the data with this code: nsamples, nx, ny = sample_array.shape new_sample_array = sample_array.reshape((nsamples,nxny)) #could also be done using .reshape((nsamples, -1)) In my understanding, this will cause my new_sample_array having shape of (2,3). It seems that the three rows inside my sample turned into three columns. I assumed that the columns mean first sample with first time steps, first sample with second time steps, and so on. (Please also refer to this image for the reshaping reference: https://imgur.com/E3G4rLb) Therefore, I’d like to ask two questions: 1. Is my reshaping method correct? 2. If I pass the reshaped data into Logistic Regression (let’s say the classifier is clf) and do clf.coef_, I got an array with three values. Does this corresponds to b_1, b_2, and b_3? Thank you very much for your help 7. Grzegorz Kępisty April 21, 2020 at 4:06 pm # Good morning Jason! Thank you for the article. Two questions on the topic: 1) In this blog entry you describe that the model can be optimized by maximizing the likelihood function for given input data. On the other hand there (“How To Implement Logistic Regression From Scratch in Python”) you show that we can optimize a model by minimizing error of predictions. When and how do we choose between those 2 aim functions? 2) What would be the difference between those models optimized in two different ways (maximum likelihood or minimizing the error)? If error is minimial in the second case why shall we use likelihood? Thank you in advance for your responce and have a pleasant day! • Jason Brownlee April 22, 2020 at 5:50 am # You’re welcome. They both solve the problem the same general way, just with different optimization algorithms. Use the solver if the data fits in ram, use SGD if it doesn’t. 8. Sabine Katzdobler July 20, 2020 at 10:12 pm # Good day, currently, I started to rethink my usage of the (ordinal) logit distribution and wondered if maximum-likelihood-distribution might be better suited. Thus, I am asking: Would you recommend maximum likelihood estimation (with Bernoulli-distribution), when the dependent variable is ordinal involving 5 levels? • Jason Brownlee July 21, 2020 at 6:03 am # The distribution does not matter for the framework. I believe for a binomial distribution, you will arrive at a cross-entropy loss. 9. Tristan October 11, 2020 at 9:58 pm # Hi Jason, Firstly, thanks a lot for the insightful post. I have one question which I am trying to find an answer to and that no searches have provided Insight on. You stated “ Recall that this is what the linear part of the logistic regression is calculating: log-odds = beta0 + beta1 x1 + beta2 * x2 + … + betam * xm” Where is the step that shows we can state that the log odds is equal to linear equation? I have gone through 5 derivations and they all do the same thing as you have done. I am curious to understand how that statement is derived. Kind regards, Tristan • Jason Brownlee October 12, 2020 at 6:43 am # It is a sum of weighted terms which by definition is a linear equation. • Tarun Gupta January 19, 2021 at 8:50 pm # Same question !! How does bthese tow become equal. “sum of weighted terms which by definition is a linear equation” true but how they became equal ? • Jason Brownlee January 20, 2021 at 5:42 am # We are defining a term, it is the log odds defined as a weighted sum. 10. Chew Jing Wei October 29, 2020 at 2:38 pm # Hello! Thank you for the post, your explanations are very clear. However, I think there might be a mistake in this equation: likelihood = yhat * y + (1 – yhat) * (1 – y) Based on the lecture notes here: http://web.stanford.edu/class/archive/cs/cs109/cs109.1178/lectureHandouts/220-logistic-regression.pdf, the likelihood based on a single observation should be yhat ** y * (1 – yhat) ** (1 – y) instead. • Jason Brownlee October 30, 2020 at 6:48 am # I believe it is correct, I recommend the references at the end of the tutorial. 11. Harrison Bohl November 5, 2020 at 12:52 pm # Please use latex to write your maths equations, it’s really hard to understand what is happening and also it looks bad. Thanks 12. Jon March 14, 2021 at 9:36 am # How do you use the above ( Logistic Regression Likelihood Function ) to calculate AIC or BIC 13. Paul September 30, 2021 at 8:19 pm # Is the this method an alternative to Gradient Descent? A bit confused by the difference between the two. Thank you! • Adrian Tam October 1, 2021 at 12:43 pm # If you mean logistic regression and gradient descent, the answer is no. Logistic regression is to take input and predict output, but not in a linear model. Gradient descent is an algorithm to do optimization. In this case, we optimize for the likelihood score by comparing the logistic regression prediction and the real output data. 14. John December 20, 2021 at 1:21 am # How do I calculate the intercept and coefficients of logistic regression?	2022-09-30 06:30:46	{"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": 0.7406259179115295, "perplexity": 720.762363431091}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030335444.58/warc/CC-MAIN-20220930051717-20220930081717-00322.warc.gz"}
https://zbmath.org/?q=an:1126.35067	# zbMATH — the first resource for mathematics Blowup theory for the critical nonlinear Schrödinger equations revisited. (English) Zbl 1126.35067 Consider the nonlinear Schrödinger equation $i\partial_{t}u+\bigtriangleup{u}+\| u\| ^{\tfrac{4}{d}}u=0,\quad x\in\mathbb{R}^d, \;t>0.$ The authors prove the following theorem: Let $$\{\nu_{n}\}_{n=1}^{\infty}$$ be a bounded family of functions in $$\text{H}^1(\mathbb{R}^d)$$ such that $\limsup_{n\to\infty}\| \| \bigtriangledown\nu_{n}\| \| _{\text{L}^2}\leq{M},\quad \limsup_{n\to\infty}\| \| \nu_{n}\| \| _{\text{L}^{\tfrac{4}{d}+2}}\geq{m}.$ Then, there exists $$\{x_{n}\}_{n=1}^{\infty}\subset\mathbb{R}^d$$ such that, up to a subsequence, $\nu_{n}(\cdot\;+x_{n})\rightharpoonup\text{V}\;\text{weakly,}$ with $$\|\text{V}\\| _{\text{L}^2}\geq{(\tfrac{d}{d+2})^{\tfrac{d}{4}}}(\frac{m^{\tfrac{d}{2}+1}} {M^{\tfrac{d}{2}}})\\|Q\\|_{\text{L}^2},$$ with $$\bigtriangleup{Q}-Q+\| Q\| ^{\tfrac{4}{d}}Q=0$$. ##### MSC: 35Q55 NLS equations (nonlinear Schrödinger equations) 35B40 Asymptotic behavior of solutions to PDEs ##### Keywords: Cauchy problem; Sobolev space; blow up Full Text:	2021-10-26 11:35:32	{"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": 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": 0.5322172045707703, "perplexity": 801.652044922501}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-43/segments/1634323587877.85/warc/CC-MAIN-20211026103840-20211026133840-00416.warc.gz"}
https://openstax.org/books/calculus-volume-3/pages/7-chapter-review-exercises	Calculus Volume 3 # Chapter Review Exercises Calculus Volume 3Chapter Review Exercises True or False? Justify your answer with a proof or a counterexample. 117. If $yy$ and $zz$ are both solutions to $y″+2y′+y=0,y″+2y′+y=0,$ then $y+zy+z$ is also a solution. 118. The following system of algebraic equations has a unique solution: $6z1+3z2=84z1+2z2=4.6z1+3z2=84z1+2z2=4.$ 119. $y=excos(3x)+exsin(2x)y=excos(3x)+exsin(2x)$ is a solution to the second-order differential equation $y″+2y′+10=0.y″+2y′+10=0.$ 120. To find the particular solution to a second-order differential equation, you need one initial condition. Classify the differential equation. Determine the order, whether it is linear and, if linear, whether the differential equation is homogeneous or nonhomogeneous. If the equation is second-order homogeneous and linear, find the characteristic equation. 121. $y″−2y=0y″−2y=0$ 122. $y″−3y+2y=cos(t)y″−3y+2y=cos(t)$ 123. $(dydt)2+yy′=1(dydt)2+yy′=1$ 124. $d2ydt2+tdydt+sin2(t)y=etd2ydt2+tdydt+sin2(t)y=et$ For the following problems, find the general solution. 125. $y″+9y=0y″+9y=0$ 126. $y″+2y′+y=0y″+2y′+y=0$ 127. $y″−2y′+10y=4xy″−2y′+10y=4x$ 128. $y″=cos(x)+2y′+yy″=cos(x)+2y′+y$ 129. $y″+5y+y=x+e2xy″+5y+y=x+e2x$ 130. $y″=3y′+xe−xy″=3y′+xe−x$ 131. $y″−x2=−3y′−94y+3xy″−x2=−3y′−94y+3x$ 132. $y″=2cosx+y′−yy″=2cosx+y′−y$ For the following problems, find the solution to the initial-value problem, if possible. 133. $y″+4y′+6y=0,y″+4y′+6y=0,$ $y(0)=0,y(0)=0,$ $y′(0)=2y′(0)=2$ 134. $y″=3y−cos(x),y″=3y−cos(x),$ $y(0)=94,y(0)=94,$ $y′(0)=0y′(0)=0$ For the following problems, find the solution to the boundary-value problem. 135. $4y′=−6y+2y″,4y′=−6y+2y″,$ $y(0)=0,y(0)=0,$ $y(1)=1y(1)=1$ 136. $y″=3x−y−y′,y″=3x−y−y′,$ $y(0)=−3,y(0)=−3,$ $y(1)=0y(1)=0$ For the following problem, set up and solve the differential equation. 137. The motion of a swinging pendulum for small angles $θθ$ can be approximated by $d2θdt2+gLθ=0,d2θdt2+gLθ=0,$ where $θθ$ is the angle the pendulum makes with respect to a vertical line, g is the acceleration resulting from gravity, and L is the length of the pendulum. Find the equation describing the angle of the pendulum at time $t,t,$ assuming an initial displacement of $θ0θ0$ and an initial velocity of zero. The following problems consider the “beats” that occur when the forcing term of a differential equation causes “slow” and “fast” amplitudes. Consider the general differential equation$ay″+by=cos(ωt)ay″+by=cos(ωt)$ that governs undamped motion. Assume that $ba≠ω.ba≠ω.$ 138. Find the general solution to this equation (Hint: call $ω0=b/aω0=b/a$). 139. Assuming the system starts from rest, show that the particular solution can be written as $y=2a(ω02−ω2)sin(ω0−ωt2)sin(ω0+ωt2).y=2a(ω02−ω2)sin(ω0−ωt2)sin(ω0+ωt2).$ 140. [T] Using your solutions derived earlier, plot the solution to the system $2y″+9y=cos(2t)2y″+9y=cos(2t)$ over the interval $t=[−50,50].t=[−50,50].$ Find, analytically, the period of the fast and slow amplitudes. For the following problem, set up and solve the differential equations. 141. An opera singer is attempting to shatter a glass by singing a particular note. The vibrations of the glass can be modeled by $y″+ay=cos(bt),y″+ay=cos(bt),$ where $y″+ay=0y″+ay=0$ represents the natural frequency of the glass and the singer is forcing the vibrations at $cos(bt).cos(bt).$ For what value $bb$ would the singer be able to break that glass? (Note: in order for the glass to break, the oscillations would need to get higher and higher.)	2020-06-05 07:42:38	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 46, "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": 0.8643633723258972, "perplexity": 380.8431177995648}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-24/segments/1590348493151.92/warc/CC-MAIN-20200605045722-20200605075722-00527.warc.gz"}
https://optimization-online.org/2012/10/3649/	# Variational Analysis of the Spectral Abscissa at a Matrix with a Nongeneric Multiple Eigenvalue The spectral abscissa is a fundamental map from the set of complex matrices to the real numbers. Denoted $\alpha$ and defined as the maximum of the real parts of the eigenvalues of a matrix $X$, it has many applications in stability analysis of dynamical systems. The function $\alpha$ is nonconvex and is non-Lipschitz near matrices with multiple eigenvalues. Variational analysis of this function was presented in \cite{BurOveMatrix}, including a complete characterization of its regular subgradients and necessary conditions which must be satisfied by all its subgradients. A complete characterization of all subgradients of $\alpha$ at a matrix $X$ was also given for the case that all active eigenvalues of $X$ (those whose real part equals $\alpha(X)$) are nonderogatory (their geometric multiplicity is one) and also for the case that they are all nondefective (their geometric multiplicity equals their algebraic multiplicity). However, necessary and sufficient conditions for all subgradients in all cases remain unknown. In this paper we present necessary and sufficient conditions for the simplest example of a matrix $X$ with a derogatory, defective multiple eigenvalue. ## Citation MPI Magdeburg, Sandtorstr 1. 39106 Magdeburg, Germany,10/2012 ## Article Download View Variational Analysis of the Spectral Abscissa at a Matrix with a Nongeneric Multiple Eigenvalue	2023-03-27 19:24:55	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 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": 0.6747564077377319, "perplexity": 306.19973564148546}, "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-2023-14/segments/1679296948684.19/warc/CC-MAIN-20230327185741-20230327215741-00209.warc.gz"}
https://engdic.org/20-other-ways-to-say-sorry/	# 20 Other Ways to Say Sorry 20 Other Ways to Say Sorry • I’m sorry. • I’m sorry for the way I acted. • I’m sorry for what I said. • I’m sorry for hurting you. • I’m sorry for making you upset. • I’m sorry if I made you feel uncomfortable. • I’m sorry if I offended you. • Can you please forgive me? • Will you forgive me? • Let’s put this behind us and move on. • This is water under the bridge. • This is in the past. • I’ve learned my lesson, and I promise it won’t happen again. • I’m sorry for what I did, and I’ll make it up to you. • I know I messed up, and I’ll do whatever it takes to make things right. • Please give me another chance. • I’m begging you, please forgive me!	2022-10-01 17:03:29	{"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": 0.8108189105987549, "perplexity": 1912.4523495259439}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030336880.89/warc/CC-MAIN-20221001163826-20221001193826-00064.warc.gz"}
https://www.socratease.in/content/12/newton-laws-motion-1/58/multiple-forces-4	menu Genius! We need to calculate all the forces along the y-direction. $$N$$ and $$m g$$ are already acting along the y-direction. For the y-component of $$F$$, we take the $$\sin$$ component. We sum them all up and set it equal to $$0$$.	2018-11-20 00:10:22	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8667037487030029, "perplexity": 152.4045429619759}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-47/segments/1542039746171.27/warc/CC-MAIN-20181119233342-20181120015342-00266.warc.gz"}
https://math.libretexts.org/Bookshelves/Applied_Mathematics/Math_in_Society_(Lippman)/02%3A_Voting_Theory/2.16%3A_Exercises	# 2.16: Exercises 1. To decide on a new website design, the designer asks people to rank three designs that have been created (labeled A, B, and C). The individual ballots are shown below. Create a preference table. ABC, ABC, ACB, BAC, BCA, BCA, ACB, CAB, CAB, BCA, ACB, ABC 1. To decide on a movie to watch, a group of friends all vote for one of the choices (labeled A, B, and C). The individual ballots are shown below. Create a preference table. CAB, CBA, BAC, BCA, CBA, ABC, ABC, CBA, BCA, CAB, CAB, BAC 1. The planning committee for a renewable energy trade show is trying to decide what city to hold their next show in. The votes are shown below. $$\begin{array}{\|c\|c\|c\|c\|c\|} \hline \textbf { Number of voters } & \textbf { 9 } & \textbf { 19 } & \textbf { 11 } & \textbf { 8 } \\ \hline \textbf { 1st choice } & \text { Buffalo } & \text { Atlanta } & \text { Chicago } & \text { Buffalo } \\ \hline \textbf { 2nd choice } & \text { Atlanta } & \text { Buffalo } & \text { Buffalo } & \text { Chicago } \\ \hline \textbf { 3rd choice } & \text { Chicago } & \text { Chicago } & \text { Atlanta } & \text { Atlanta } \\ \hline \end{array}$$ 1. How many voters voted in this election? 2. How many votes are needed for a majority? A plurality? 3. Find the winner under the plurality method. 4. Find the winner under the Borda Count Method. 5. Find the winner under the Instant Runoff Voting method. 6. Find the winner under Copeland’s method. 1. A non-profit agency is electing a new chair of the board. The votes are shown below. $$\begin{array}{\|c\|c\|c\|c\|c\|} \hline \textbf { Number of voters } & \mathbf{1 1} & \mathbf{5} & \mathbf{1 0} & \mathbf{3} \\ \hline \textbf { 1st choice } & \text { Atkins } & \text { Cortez } & \text { Burke } & \text { Atkins } \\ \hline \textbf { 2nd choice } & \text { Cortez } & \text { Burke } & \text { Cortez } & \text { Burke } \\ \hline \textbf { 3rd choice } & \text { Burke } & \text { Atkins } & \text { Atkins } & \text { Cortez } \\ \hline \end{array}$$ 1. How many voters voted in this election? 2. How many votes are needed for a majority? A plurality? 3. Find the winner under the plurality method. 4. Find the winner under the Borda Count Method. 5. Find the winner under the Instant Runoff Voting method. 6. Find the winner under Copeland’s method. 1. The student government is holding elections for president. There are four candidates (labeled A, B, C, and D for convenience). The preference schedule for the election is: $$\begin{array}{\|c\|c\|c\|c\|c\|c\|c\|} \hline \textbf {Number of voters} & \mathbf{1 2 0} & \mathbf{5 0} & \mathbf{4 0} & \mathbf{9 0} & \mathbf{6 0} & \mathbf{1 0 0} \\ \hline \textbf {1st choice } & \mathrm{C} & \mathrm{B} & \mathrm{D} & \mathrm{A} & \mathrm{A} & \mathrm{D} \\ \hline \textbf{2nd choice } & \mathrm{D} & \mathrm{C} & \mathrm{A} & \mathrm{C} & \mathrm{D} & \mathrm{B} \\ \hline \textbf{3rd choice } & \mathrm{B} & \mathrm{A} & \mathrm{B} & \mathrm{B} & \mathrm{C} & \mathrm{A} \\ \hline \textbf{4th choice } & \mathrm{A} & \mathrm{D} & \mathrm{C} & \mathrm{D} & \mathrm{B} & \mathrm{C} \\ \hline \end{array}$$ 1. How many voters voted in this election? 2. How many votes are needed for a majority? A plurality? 3. Find the winner under the plurality method. 4. Find the winner under the Borda Count Method. 5. Find the winner under the Instant Runoff Voting method. 6. Find the winner under Copeland’s method. 1. The homeowners association is deciding a new set of neighborhood standards for architecture, yard maintenance, etc. Four options have been proposed. The votes are: $$\begin{array}{\|c\|c\|c\|c\|c\|c\|c\|} \hline \textbf { Number of voters } & \mathbf{8} & \mathbf{9} & \mathbf{1 1} & \mathbf{7} & \mathbf{7} & \mathbf{5} \\ \hline \textbf { 1st choice } & \text { B } & \text { A } & \text { D } & \text { A } & \text { B } & \text { C } \\ \hline \textbf { 2nd choice } & \text { C } & \text { D } & \text { B } & \text { B } & \text { A } & \text { D } \\ \hline \textbf { 3rd choice } & \text { A } & \text { C } & \text { C } & \text { D } & \text { C } & \text { A } \\ \hline \textbf { 4th choice } & \text { D } & \text { B } & \text { A } & \text { C } & \text { D } & \text { B } \\ \hline \end{array}$$ 1. How many voters voted in this election? 2. How many votes are needed for a majority? A plurality? 3. Find the winner under the plurality method. 4. Find the winner under the Borda Count Method. 5. Find the winner under the Instant Runoff Voting method. 6. Find the winner under Copeland’s method. 1. Consider an election with 129 votes. 1. If there are 4 candidates, what is the smallest number of votes that a plurality candidate could have? 2. If there are 8 candidates, what is the smallest number of votes that a plurality candidate could have? 1. Consider an election with 953 votes. 1. If there are 7 candidates, what is the smallest number of votes that a plurality candidate could have? 2. If there are 8 candidates, what is the smallest number of votes that a plurality candidate could have? 1. Does this voting system having a Condorcet Candidate? If so, find it. $$\begin{array}{\|c\|c\|c\|c\|} \hline \textbf { Number of voters } & \mathbf{1 4} & \mathbf{1 5} & \mathbf{2} \\ \hline \textbf { 1st choice } & \mathrm{A} & \mathrm{C} & \mathrm{B} \\ \hline \textbf { 2nd choice } & \mathrm{B} & \mathrm{B} & \mathrm{C} \\ \hline \textbf { 3rd choice } & \mathrm{C} & \mathrm{A} & \mathrm{A} \\ \hline \end{array}$$ 1. Does this voting system having a Condorcet Candidate? If so, find it. 1. The marketing committee at a company decides to vote on a new company logo. They decide to use approval voting. Their results are tallied below. Each column shows the number of voters with the particular approval vote. Which logo wins under approval voting? $$\begin{array}{\|c\|c\|c\|c\|} \hline \textbf { Number of voters } & \mathbf{8} & \mathbf{7} & \mathbf{6} \\ \hline \textbf { 1st choice } & \mathrm{A} & \mathrm{C} & \mathrm{B} \\ \hline \textbf { 2nd choice } & \mathrm{B} & \mathrm{B} & \mathrm{C} \\ \hline \textbf { 3rd choice } & \mathrm{C} & \mathrm{A} & \mathrm{A} \\ \hline \end{array}$$ 1. The downtown business association is electing a new chairperson, and decides to use approval voting. The tally is below, where each column shows the number of voters with the particular approval vote. Which candidate wins under approval voting? $$\begin{array}{\|c\|c\|c\|c\|c\|c\|c\|c\|} \hline \textbf { Number of voters } & \mathbf{8} & \mathbf{7} & \mathbf{6} & \mathbf{3} & \mathbf{4} & \mathbf{2} & \mathbf{5} \\ \hline \mathbf{A} & \mathrm{X} & \mathrm{X} & & & \mathrm{X} & & \mathrm{X} \\ \hline \mathbf{B} & \mathrm{X} & & \mathrm{X} & \mathrm{X} & & & \mathrm{X} \\ \hline \mathbf{C} & & \mathrm{X} & \mathrm{X} & \mathrm{X} & & \mathrm{X} & \\ \hline \mathbf{D} & \mathrm{X} & & \mathrm{X} & & \mathrm{X} & \mathrm{X} & \\ \hline \end{array}$$ 2.16: Exercises is shared under a CC BY-SA 3.0 license and was authored, remixed, and/or curated by David Lippman via source content that was edited to conform to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.	2022-05-23 04:27:22	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.25707781314849854, "perplexity": 2333.197478548557}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662555558.23/warc/CC-MAIN-20220523041156-20220523071156-00569.warc.gz"}
https://tex.stackexchange.com/questions/386030/how-to-draw-orbital-elements/553482#553482	# How to draw orbital elements I'm trying to draw a diagram representing the orbital elements (only the angles) to obtain something like the following: At the moment I have only drawn the Line of Nodes and the right ascension of the ascending node $\Omega$. I'm having problems in drawing the actual orbit (the circle in solid line) because I can't figure out how to set tikz-3dplot's rotated coordinate system. Any suggestion on how to go about that? Here is my code and the result so far: \documentclass[border=5pt]{standalone} \usepackage{tikz,tikz-3dplot} \begin{document} \tdplotsetmaincoords{70}{110} \begin{tikzpicture}[tdplot_main_coords,scale=4] \pgfmathsetmacro{\r}{.8} \pgfmathsetmacro{\O}{45} % right ascension of ascending node [deg] \pgfmathsetmacro{\i}{30} % inclination [deg] \draw[->] (0,0,0) -- (1,0,0) node[anchor=north east]{$x$}; \draw[->] (0,0,0) -- (0,1,0) node[anchor=north west]{$y$}; \draw[->] (0,0,0) -- (0,0,1) node[anchor=south]{$z$}; \tdplotdrawarc[dashed]{(0,0,0)}{\r}{0}{360}{}{} \tdplotsetrotatedcoords{\O}{0}{0} \draw [tdplot_rotated_coords] (-1,0,0) -- (1,0,0) node [below] {Line of Nodes}; \tdplotdrawarc[->]{(0,0,0)}{.5\r}{0}{\O}{anchor=north}{$\Omega$} \end{tikzpicture} \end{document} • If you want to change your plane to, say 30^\circ, then replace \tdplotsetrotatedcoords{\O}{0}{0} with \tdplotsetrotatedcoords{\O}{30}{0}. Aug 12, 2017 at 14:43 • @JohnKormylo Thanks for your comment. However, the rotation you suggest is made around the unrotated y-axis, while I would like to rotate around the transformed x-axis. In short, my transformed frame is should be obtained by a first rotation around the z-axis and then another rotation around the new x-axis. I could achieve the first one but not the second one, sorry if I wasn't clear enough in the question. Aug 12, 2017 at 14:58 • The rotaions are applied sequentially, and the third argument is in the same direction as the first. For example \tdplotsetrotatedcoords{90}{90}{-90} IIRC will swap the y and z axes. Aug 12, 2017 at 15:04 • This question seems to be something similar: tex.stackexchange.com/questions/119871/… Aug 12, 2017 at 20:46 Thanks to @JohnKormylo's suggestions I could reproduce the figure to a satisfactory degree, even if it is not 100% accurate. \documentclass[border=5pt]{standalone} \usepackage{tikz,tikz-3dplot} \begin{document} \tdplotsetmaincoords{70}{110} \begin{tikzpicture}[tdplot_main_coords,scale=5] \pgfmathsetmacro{\r}{.8} \pgfmathsetmacro{\O}{45} % right ascension of ascending node [deg] \pgfmathsetmacro{\i}{30} % inclination [deg] \pgfmathsetmacro{\f}{35} % true anomaly [deg] \coordinate (O) at (0,0,0); \draw [->] (O) -- (2,0,0) node[anchor=north east] {$x$}; \draw [->] (O) -- (0,1,0) node[anchor=north west] {$y$}; \draw [->] (O) -- (0,0,1) node[anchor=south] {$z$}; \node at (0,-\r,0) [left,text width=4em] {Ecliptic Plane}; \tdplotdrawarc[dashed]{(O)}{\r}{0}{360}{}{} \tdplotsetrotatedcoords{\O}{0}{0} \draw [tdplot_rotated_coords] (-1,0,0) -- (1,0,0) node [below right] {Line of Nodes}; \tdplotdrawarc[->]{(O)}{.33\r}{0}{\O}{anchor=north}{$\Omega$} \tdplotsetrotatedcoords{-\O}{\i}{0} \tdplotdrawarc[tdplot_rotated_coords]{(O)}{\r}{0}{360}{}{} \begin{scope}[tdplot_rotated_coords] % \draw[->] (O) -- (1,0,0) node [above] {$x'$}; % \draw[->] (O) -- (0,1,0) node [above] {$y'$}; \draw[->] (O) -- (0,0,1) node [above] {$\hat{h}$}; \draw (1,0,0) -- (-1,0,0); \tdplotdrawarc[->]{(O)}{.33\r}{90}{180}{anchor=west}{$\omega$} \coordinate (P) at (180+\f:\r); \draw (O) -- (P); \tdplotdrawarc[->]{(O)}{.33\r}{180}{180+\f}{anchor=south west}{$\nu$} \end{scope} \tdplotsetrotatedcoords{-\O+\f}{\i}{0} \tdplotsetrotatedcoordsorigin{(P)} \begin{scope}[tdplot_rotated_coords,scale=.2,thick] \draw [->] (P) -- (-1,0,0) node [right] {$\hat{r}$}; \draw [->] (P) -- (0,-1,0) node [above] {$\hat{\theta}$}; \draw [->] (P) -- (0,0,1) node [above] {$\hat{k}$}; \fill (P) circle (.33ex); \end{scope} \tdplotsetthetaplanecoords{-\f} \tdplotdrawarc[tdplot_rotated_coords,->]{(O)}{.75\r}{0}{\i}{anchor=south}{$i$} % not accurate :( \end{tikzpicture} \end{document} Visit this Overleaf project to find a Latex Beamer example. \documentclass[compress,9pt]{beamer} \usepackage{pgfpages} \usepackage{tikz} %TikZ is required for this to work. Make sure this exists before the next line \usepackage{tikz-3dplot} %requires 3dplot.sty to be in same directory, or in your LaTeX installation \begin{document} % Orbital elements or Keplerian elements \begin{frame}[fragile] % \begin{figure}[H] \centering \def\r{3.5} \pgfmathsetmacro{\inclination}{35} \pgfmathsetmacro{\nuSatellite}{55} \pgfmathsetmacro{\gammaAngle}{290} \tdplotsetmaincoords{70}{165} \begin{tikzpicture}[tdplot_main_coords] \onslide<1->{ \fill (0,0) coordinate (O) circle (5pt) node[left =7pt] {$M_\oplus$}; % Draw equatorial ellipse %\tdplotdrawarc[thin]{(0,0,0)}{\r}{-90}{205}{label={[xshift=-3.7cm, yshift=0.9cm]Equatorial plane}}{} %\tdplotdrawarc[dotted]{(0,0,0)}{\r}{205}{270}{}{} % Draw equatorial plane \draw[] (0,-\r,0) -- (\r,-\r,0) node[below]{Equatorial plane} -- (\r,\r,0) -- (-\r,\r,0) -- (-\r,-0.65\r,0); \draw[dotted] (-\r,-0.65\r,0) -- (-\r,-\r,0) -- (0,-\r,0); % Draw ellipses intersection. Line of nodes \draw[dashed] (0,-1.3\r,0) -- (0,1.3\r,0) node[right] {Line of nodes}; % Draw gamma direction } \onslide<2->{ % Set gamma direction \tdplotsetcoord{Pg}{1.3\r}{90}{\gammaAngle} \draw[->] (0,0,0) -- (Pg) node[anchor=east] {Reference direction $\boldsymbol{\gamma}$}; } \onslide<1->{ % Create a new rotated system in the center \tdplotsetrotatedcoords{0}{\inclination}{90} % Draw orbital ellipse \tdplotdrawarc[tdplot_rotated_coords,thin,blue]{(0,0,0)}{\r}{-125}{180}{label={[xshift=-5.7cm, yshift=-2.2cm]Orbital plane}}{} \tdplotdrawarc[tdplot_rotated_coords,dotted,blue]{(0,0,0)}{\r}{180}{235}{}{} % Define m position \pgfmathsetmacro{\omegaSatellite}{90} \pgfmathsetmacro{\xmRot}{\rcos(\omegaSatellite+\nuSatellite)} \pgfmathsetmacro{\ymRot}{\rsin(\omegaSatellite+\nuSatellite)} \pgfmathsetmacro{\zmRot}{0} % Draw a vector to m \draw[tdplot_rotated_coords,thin,->,blue] (0,0,0) -- (\xmRot,\ymRot,\zmRot); % Draw a mass \filldraw[tdplot_rotated_coords, blue] (\xmRot,\ymRot,\zmRot) circle (2pt) node[above left] {$m$}; } \onslide<5->{ % Draw periapsis line \draw[dashed,tdplot_rotated_coords,blue] (0,0,0) -- (0,\r,0) node[anchor=south west] {Periapsis}; } \onslide<5->{ % Draw omega angle \tdplotdrawarc[tdplot_rotated_coords,thick,-stealth,blue]{(0,0,0)}{0.4\r}{0}{\omegaSatellite}{anchor=south west}{$\omega$} % Draw nu angle \tdplotdrawarc[tdplot_rotated_coords,thick,-stealth,blue]{(0,0,0)}{0.4\r}{\omegaSatellite}{\omegaSatellite+\nuSatellite}{anchor=south west}{$\nu$} } \onslide<3->{ % Create rotated shifted system at (0,\r,0) \tdplotresetrotatedcoordsorigin \tdplotsetrotatedcoords{0}{0}{180} % Draw \Omega % Hidden part of the arc %% \tdplotdrawarc[tdplot_rotated_coords,dashed,thick,brown]{(0,0,0)}{0.4\r}{0}{90}{anchor=south}{}%{$\Omega$} % Visible part of the arc \tdplotdrawarc[tdplot_rotated_coords,thick,-stealth,brown]{(0,0,0)}{0.4\r}{\gammaAngle-180}{270}{anchor=north east}{$\Omega$} % Shift the rotated coordinates \coordinate (Shift) at (0,\r,0); \tdplotsetrotatedcoordsorigin{(Shift)} % \draw[thick,tdplot_rotated_coords,->,blue] (0,0,0) -- (.5,0,0) node[anchor=north west]{$x_2$}; % \draw[thick,tdplot_rotated_coords,->,blue] (0,0,0) -- (0,.5,0) node[anchor=north]{$y_2$}; % \draw[thick,tdplot_rotated_coords,->,blue] (0,0,0) -- (0,0,.5) node[anchor=south west]{$z_2$}; } \onslide<4->{ % Draw inclination angle \tdplotsetrotatedthetaplanecoords{0} \tdplotdrawarc[tdplot_rotated_coords,thick,-stealth,brown]{(Shift)}{0.3\r}{90}{90-\inclination}{anchor=west}{$i$} } \end{tikzpicture} \caption{Orbital elements or Keplerian elements}\label{fig:elipseNodos2} \end{figure} \end{frame} \end{document} • Good Idea. Thank you! Jul 16, 2020 at 9:35 Following code show my try without tikz-3dplot. Just for fun:) \documentclass[tikz, border=1cm]{standalone} \usetikzlibrary{3d, calc} \makeatletter \tikzset{ plane/.code args={#1and#2}{ \tikz@scan@one@point\pgf@process#1 \edef\temp@a{(\the\pgf@x, \the\pgf@y)}; \tikz@scan@one@point\pgf@process#2 \edef\temp@b{(\the\pgf@x, \the\pgf@y)}; \pgfkeysalso{ plane x={\temp@a}, plane y={\temp@b}, canvas is plane, } }, } \makeatother \begin{document} \begin{tikzpicture} \draw[thick, red!20] (135:4cm) -- (-45:4cm); \foreach \i [evaluate=\i as \c using \i/170100] in {0, 10, ..., 170} { \draw[plane={(-45:0.5cm) and (\i:1cm)}, fill=red!\c!violet, opacity=.1] (0, 0) circle (2cm); } % \fill[plane={(-45:0.5cm) and (0:1cm)}, red, opacity=.5] (-3, -3) rectangle (3, 3); \end{tikzpicture} \end{document}	2022-08-12 00:08:20	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.8816470503807068, "perplexity": 7687.298372645564}, "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/1659882571536.89/warc/CC-MAIN-20220811224716-20220812014716-00000.warc.gz"}
https://cs.stackexchange.com/questions/128257/consistent-theory-based-on-l-and-nota-a-is-a-theorem	Consistent theory based on L and not(A->A) is a theorem I am working on this problem in which I have a theory $$T$$ based on language $$\mathcal{L}$$ and the only information we have is that T is consistent and $$\vdash \lnot(A \rightarrow A)$$. Given this information, how can I know if this theory is sound, complete and/or decidable? My only guess is that I can say that $$T$$ is sound because since $$T$$ is consistent and we can derive $$\lnot(A \rightarrow A)$$ from axioms and inference rules, $$\lnot(A \rightarrow A)$$ is a theorem and because of that we can assume that $$T$$ is sound (because the premises and conclusions are true). Thank you! • Your problem statement makes no sense because in first-order classical logic $\lnot (A \to A)$ is false, and therefore it can only be a theorem in inconsistent theory. Please double check that you transcribed it correctly. – Andrej Bauer Jul 21 '20 at 20:59 The first thing to point out is that $$\lnot(A \rightarrow A)$$ leads to a contradiction: ($$A \land \lnot A$$). Since propositional logic itself is consistent, it is impossible to derive a contradiction, so $$\vdash \lnot(A \rightarrow A)$$ is impossible. So perhaps your intention was $$T \vdash \lnot(A \rightarrow A)$$. This can only happen if $$T$$ is itself inconsistent. Given that $$T$$ is inconsistent, $$T$$ is vacuously complete (since an inconsistent theory can prove any formula) and decidable (for the same reason). As for soundness, it's a property of the logical system, not any particular theory like $$T$$. Propositional logic is sound.	2021-01-26 10:58:34	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 16, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8134570121765137, "perplexity": 172.68781621758296}, "config": {"markdown_headings": false, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-04/segments/1610704799741.85/warc/CC-MAIN-20210126104721-20210126134721-00235.warc.gz"}
https://www.nag.com/numeric/py/nagdoc_latest/naginterfaces.library.stat.pdf_normal.html	# naginterfaces.library.stat.pdf_normal¶ naginterfaces.library.stat.pdf_normal(x, xmean, xstd)[source] pdf_normal returns the value of the probability density function (PDF) for the Normal (Gaussian) distribution with mean and variance at a point . For full information please refer to the NAG Library document for g01ka https://www.nag.com/numeric/nl/nagdoc_29/flhtml/g01/g01kaf.html Parameters xfloat , the value at which the PDF is to be evaluated. xmeanfloat , the mean of the Normal distribution. xstdfloat , the standard deviation of the Normal distribution. Returns pdffloat The value of the PDF. Raises NagValueError (errno ) On entry, . Constraint: , where is the safe range parameter as defined by machine.real_safe. (errno ) Computation abandoned owing to underflow of . (errno ) Computation abandoned owing to an internal calculation overflowing. Notes The Normal distribution has probability density function (PDF)	2023-03-26 21:07:57	{"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": 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": 0.9155614376068115, "perplexity": 1414.8789954174138}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296946535.82/warc/CC-MAIN-20230326204136-20230326234136-00768.warc.gz"}
http://farside.ph.utexas.edu/teaching/336k/lectures/node59.html	Next: Exercises Up: Rotating Reference Frames Previous: Coriolis Force # Foucault Pendulum Consider a pendulum consisting of a compact mass suspended from a light cable of length in such a manner that the pendulum is free to oscillate in any plane whose normal is parallel to the Earth's surface. The mass is subject to three forces: first, the force of gravity , which is directed vertically downward (we are again ignoring centrifugal acceleration); second, the tension in the cable, which is directed upward along the cable; and, third, the Coriolis force. It follows that the apparent equation of motion of the mass, in a frame of reference which co-rotates with the Earth, is [see Equation (414)] (438) Let us define our usual Cartesian coordinates (,,), and let the origin of our coordinate system correspond to the equilibrium position of the mass. If the pendulum cable is deflected from the downward vertical by a small angle then it is easily seen that , , and . In other words, the change in height of the mass, , is negligible compared to its horizontal displacement. Hence, we can write , provided that . The tension has the vertical component , and the horizontal component , since --see Figure 27. Hence, the Cartesian equations of motion of the mass are written [cf., Equations (424)-(426)] (439) (440) (441) To lowest order in (i.e., neglecting ), the final equation, which is just vertical force balance, yields . Hence, Equations (439) and (440) reduce to (442) (443) Let (444) Equations (442) and (443) can be combined to give a single complex equation for : (445) Let us look for a sinusoidally oscillating solution of the form (446) Here, is the (real) angular frequency of oscillation, and is an arbitrary complex constant. Equations (445) and (446) yield the following quadratic equation for : (447) The solutions are approximately (448) where we have neglected terms involving . Hence, the general solution of (446) takes the form (449) where and are two arbitrary complex constants. Making the specific choice , where is real, the above solution reduces to (450) Now, it is clear from Equation (444) that and are the real and imaginary parts of , respectively. Thus, it follows from the above that (451) (452) These equations describe sinusoidal oscillations, in a plane whose normal is parallel to the Earth's surface, at the standard pendulum frequency . The Coriolis force, however, causes the plane of oscillation to slowly precess at the angular frequency . The period of the precession is (453) For example, according to the above equations, the pendulum oscillates in the -direction (i.e., north/south) at , in the -direction (i.e., east/west) at , in the -direction again at , etc. The precession is clockwise (looking from above) in the northern hemisphere, and counter-clockwise in the southern hemisphere. The precession of the plane of oscillation of a pendulum, due to the Coriolis force, is used in many museums and observatories to demonstrate that the Earth is rotating. This method of making the Earth's rotation manifest was first devised by Foucault in 1851. Next: Exercises Up: Rotating Reference Frames Previous: Coriolis Force Richard Fitzpatrick 2011-03-31	2018-04-27 04:45:09	{"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": 0.8050744533538818, "perplexity": 700.0997524691024}, "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-2018-17/segments/1524125949036.99/warc/CC-MAIN-20180427041028-20180427061028-00175.warc.gz"}
https://planetcalc.com/908/	homechevron_rightStudychevron_rightMathchevron_rightGeometry Side length of a triangle This calculator computes side length of a triangle given two sides and angle between them (cosine law) Articles that describe this calculator Side length of a triangle Digits after the decimal point: 2 Side C Calculators used by this calculator PLANETCALC, Side length of a triangle	2020-07-07 10:32:03	{"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": 0.8620986938476562, "perplexity": 2952.4007679161273}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-29/segments/1593655891884.11/warc/CC-MAIN-20200707080206-20200707110206-00117.warc.gz"}
https://aviation.stackexchange.com/questions/90462/once-pressure-distribution-is-established-why-soft-wing-at-zero-aoa-without-bat	# Once pressure distribution is established, why soft wing at zero AoA without battens can't maintain shape? Imagine we have soft wing with curved battens installed in wind tunnel set at zero AoA.Wing has camber,so it produce lift. If we remove battens from wing during wind tunnel test when wing produce lift, why wing will lose shape if it seems that pressure distribution must hold the shape,high pressure under, low above?When I say maintain shape,I dont mean stricly follow batten shape,but at least wing has some belly and dont flutter. Reality is, wing cloth will start to flutters like flag in the wind. I know that lack of support at leading edge where stagnation pressure push will deform this part but peak pressure drop and all low pressure above and high pressure under has tendency to lift cloth up and make some belly,at least theoreticaly. (I ask this purely out of curiosity,normally it makes no sense to use soft wing without battens.) • You certainly can have a soft wing without battens—fore-and-aft sails work like wings and most don't have battens. But they have to be positioned to generate certain minimum lift to maintain the shape, so not at zero AoA. Nov 27, 2021 at 23:36 • With sails, really a tale of two winds. The wind (from any direction) fills the sail. Any close approximation to wings would be sailing into the wind. Forward motion is created from the suction peak of the cambered sail (and the hull). This is why thermal riding birds have heavily cambered wings. Without the updraft, many "slipper" gliders reduce camber and go for maximum distance by reducing drag and increasing airspeed. Nov 28, 2021 at 9:50 • Re "Reality is, wing cloth will start to flutters like flag in the wind." -- this question would benefit from an explanation of how you know this to be true. Also, you could suggest that as an alternative thought experiment, rather than removing the battens while the wing was in the wind tunnel with the fan on, you might simply start with a battenless wing at a higher angle-of-attack and slowly reduce the aoa to zero and see if the wing flutters. Dec 29, 2021 at 13:29 • Anyway, battenless hang gliders were the norm back in the early days, so such aircraft do indeed exist, and I'd need further evidence than has been provided in the question so far, to accept as a "given" that such aircraft always automatically experienced "luffing" at zero degrees a-o-a, if that condition was approached from a previous higher a-o-a. Now, at some negative a-o-a, sure the wing would "luff". And maybe even at zero a-o-a -- but I'm not sure readers should be asked to just accept that as a "given". Can you provide some links to additional content that supports this assertion? Dec 29, 2021 at 13:37 We might be able to start by trying to fathom why a flag flutters in a breeze. If one considers a soft (non-battened) airfoil they may do well to consider the material it is made of. Many sail fabrics stretch and contract with wind drag, resulting in a disruption of the airflow that holds the soft sail shape. Woven titanium may have less of a tendency to do this (wind tunnel study). Sails, the obvious application, are not designed at all to function at low angles of attack as sail boats can only point so far into the wind before they lose forward propulsion. The "luff" or flutter is a good sign to increase the angle to the wind and allow it to "fill" the sail. This might lead one to believe bottom pressure holds the sail shape, and indeed, at higher angles of attack, the suction peak moves forward, helping propel the boat. At lower angles of attack (with a soft thin undercambered airfoil), inconsistencies and turbulence under the wing will cause changes in its shape, which affects the airflow above. Local pressure fluctuations then cause the "luff". Battens$$^1$$ help prevent this. One may imagine sails at higher angles of attack to be "stalled", but we must remember they are not required to produce "lift" (which would force the boat sideways), only thrust from the forward suction peak. Aircraft airfoils simply are different in that they must support the aircraft weight with as little drag as possible, and any "luffing" would be disastrous. Therefor aircraft airfoils must be supported to avoid sudden collapse at low or negative AOA. Although non supported airfoils may be possible for ultralight aircraft, they would carry higher risks best avoided, particularly with two separate wings as compared with a parachute, delta, or Rogallo wing. $$^1$$ see performance of cambered battened sails In simple terms: Without battens in place to hold the shape, at zero AOA the airflow on the bottom will not “fill” the concave area underneath. Instead it will simply rush on by, “sucking out” existing high pressure air along with it. This will negate any pressure differential that might hold the shape and cause it to flap.	2022-08-14 12:42:37	{"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": 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": 2, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5009765625, "perplexity": 2802.1888238757433}, "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/1659882572033.91/warc/CC-MAIN-20220814113403-20220814143403-00694.warc.gz"}
http://www.physicsforums.com/showthread.php?t=104854	## Simple Hydralic Lift help I have this project for school and it is all about Pascal's Principle. Well Ive decided I would like to make a hydraulic lift to go along with my presentation and just need some help with the design. I figure I can make it out of PVC or Copper tubing, either way I dont mind buying most of the materials. I just need a sort of "U" shape lift that I think would work, my problem is that I dont know what I could use for the "piston" portion. If it came down to it I guess I could just use two different size seringes(sp?) but I dont want to settle for less. Any ideas? Not clear enough? PhysOrg.com science news on PhysOrg.com >> Leading 3-D printer firms to merge in $403M deal (Update)>> LA to give every student an iPad;$30M order>> CIA faulted for choosing Amazon over IBM on cloud contract Recognitions: Gold Member I guess this doesn't count as doing your homework for you. Until I read down to where you discounted it, I was actually going to recommend using syringes. I got well over 100 lbs. of lift from a 50cc unit using compressed air. If you want something bigger and/or classier, you could use an old style bicycle pump or modify a hatch-back gas strut. In the event that you want to make your own, go ahead and use ABS or PVC pipe. You can make a piston simply by filing a groove around the circumference of a puck-shaped item (in fact, why not a puck?) and installing an O-ring for sealing. Screw a piece of theaded rod or just a big bolt into the centre and attach your clevis or whatever to the other end. Ok thanks that helps a lot! Blog Entries: 7 Recognitions: Gold Member Homework Help ## Simple Hydralic Lift help Another idea for a piston would be to take any cylindrical object such as a jar or hocky puck as suggested by Danger, and wrap tape around it to build it up so the OD fits just right into the ID of the pipe. If you wanted to use the O-ring idea, it might be easier to layer on some tape and put two more layers above and below the O-ring to center it. Hmmm... tape again. Must tell ya something about my mentality. Recognitions: Gold Member Quote by Q_Goest Hmmm... tape again. Must tell ya something about my mentality. You have got to be a Red Green fan.	2013-06-20 08:16:45	{"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": 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": 0.22086642682552338, "perplexity": 1844.8860350325087}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2013-20/segments/1368711005723/warc/CC-MAIN-20130516133005-00034-ip-10-60-113-184.ec2.internal.warc.gz"}
https://math.stackexchange.com/questions/601136/reweight-a-graph-to-give-it-a-small-max-cut	# Reweight a graph to give it a small max cut Let $G = (V, E)$ be an undirected, unweighted graph. I wish to assign weights (possibly negative, not all zero) to the edges to minimize the value of: $$\frac{m}{\\|w\\|}$$ where $m$ is the value of the maximum magnitude cut on the new weighted graph, and $w$ is the $\|E\|$-length vector containing the new edge weights. What is the smallest I can make $\frac{m}{\\|w\\|}$? Can I make it $o(1)$ (little-oh)? Is there a polynomial time procedure for generating some $w$ that comes within a constant factor of the best value for $\frac{m}{\\|w\\|}$?	2019-08-22 13:36:02	{"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": 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": 0.4776686728000641, "perplexity": 239.87243451760503}, "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/1566027317130.77/warc/CC-MAIN-20190822130553-20190822152553-00451.warc.gz"}
https://socratic.org/questions/if-you-measured-the-volume-of-something-that-had-a-length-of-5-0-m-and-a-width-o	# If you measured the volume of something that had a length of 5.0 m and a width of 3.456 m, how many significant figures would be in your answer? Sep 27, 2017 Your final answer should consist of two significant figures to reflect the number of significant figures in your least accurate given data. #### Explanation: In this example, the length was not measured with the same degree of accuracy as the width, possibly due to different measuring equipment, different method, or that other person being sloppy. Regardless, the resulting answer can only be returned with precision to two significant figures. To maintain the accuracy as much as possible, calculations should use all the significant figures given, with the rounding-off occurring only at the answer. Example: $V = l w h = 5.0 m \times 3.456 m \times h m = 17.28 \left(h\right) {m}^{3}$ This would need to round-off to $17 \left(h\right) {m}^{3} \to$ [ANS] So volume in this case could not be $17.2 \left(h\right) {m}^{3}$ nor $17.28 \left(h\right) {m}^{3}$	2021-06-24 15:02:16	{"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": 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": 0.7568532228469849, "perplexity": 458.4268347485795}, "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/1623488556133.92/warc/CC-MAIN-20210624141035-20210624171035-00540.warc.gz"}
http://sci-gems.math.bas.bg/jspui/handle/10525/83	Please use this identifier to cite or link to this item: http://hdl.handle.net/10525/83 Title: Semantic Modelling for Product Line Engineering Authors: Roshchin, MikhailGraubmann, PeterKamaev, Valery Keywords: Variability ModelingSemantic ModelingProduct Line EngineeringMDA Issue Date: 2008 Publisher: Institute of Information Theories and Applications FOI ITHEA Abstract: The aim of our work is to present solutions and a methodical support for automated techniques and procedures in domain engineering, in particular for variability modeling. Our approach is based upon Semantic Modeling concepts, for which semantic description, representation patterns and inference mechanisms are defined. Thus, model-driven techniques enriched with semantics will allow flexibility and variability in representation means, reasoning power and the required analysis depth for the identification, interpretation and adaptation of artifact properties and qualities. URI: http://hdl.handle.net/10525/83 ISSN: 1313-0463 Appears in Collections: Volume 15 Number 4 Files in This Item: File Description SizeFormat	2022-09-30 22:31:49	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 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": 0.2818484306335449, "perplexity": 11190.180729596927}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030335504.37/warc/CC-MAIN-20220930212504-20221001002504-00661.warc.gz"}
http://www.zora.uzh.ch/7211/	Quick Search: Browse by: Zurich Open Repository and Archive ## Maintenance: Tuesday, July the 26th 2016, 07:00-10:00 ZORA's new graphical user interface will be relaunched (For further infos watch out slideshow ZORA: Neues Look & Feel). There will be short interrupts on ZORA Service between 07:00am and 10:00 am. Please be patient. Permanent URL to this publication: http://dx.doi.org/10.5167/uzh-7211 # Kappeler, T; Serier, F; Topalov, P (2008). On the symplectic phase space of KdV. Proceedings of the American Mathematical Society, 136(5):1691-1698. PDF (Verlags-PDF) - Registered users only 1MB View at publisher Preview Accepted Version PDF 163kB ## Abstract We prove that the Birkhoff map $\Omega$ for KdV constructed on $H^{-1}_0(\mathbb{T})$ can be interpolated between $H^{-1}_0(\mathbb{T})$ and $L^2_0(\mathbb{T})$. In particular, the symplectic phase space $H^{1/2}_0(\mathbb{T})$ can be described in terms of Birkhoff coordinates. ## Citations 2 citations in Web of Science® 2 citations in Scopus®	2016-07-24 01:35:53	{"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": 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": 0.3019658923149109, "perplexity": 8374.069785117565}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-30/segments/1469257823805.20/warc/CC-MAIN-20160723071023-00048-ip-10-185-27-174.ec2.internal.warc.gz"}
https://dsp.stackexchange.com/questions/34069/where-to-get-transtion-matrix-for-kalman-filter	# Where to get transtion matrix for Kalman filter? I am trying to understand the Kalman filter, and I am struggling to find out how to choose the transition matrix (denoted as $\mathbf F(k)$ or $\mathbf A(k)$. This matrix is used to update the weight error covariance matrix as follows $$\mathbf P(k+1) = \mathbf F(k+1) \mathbf P(k) \mathbf F^\mathrm T (k) + \mathbf Q$$ In the paper On the Intrinsic Relationship Between the Least Mean Square and Kalman Filters it is description of "The Kalman filter for deterministic states.", what does not use the part with prediction, so it is without the transtion matrix. But there is also the general Kalman filter, with the prediction according to transtion matrix. I have not understand how this matrix should be obtained. I quess, that this transition matrix is from state space formulation of the model. But what should I do if I do not have a model? Is this mean that the general Kalman filter is usable only If I have exact model of the system? Simple: if you do not have a model, you cannot apply the Kalman filter. Or you could and make up a model, but you cannot expect any of the optimality properties of the filter to hold. Based on that paper, it seems they assume that $\mathbf{x}_k$ is known or at least directly measureable...	2022-10-07 02:48:54	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 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": 0.770809531211853, "perplexity": 241.7486394427896}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030337906.7/warc/CC-MAIN-20221007014029-20221007044029-00696.warc.gz"}
https://math.stackexchange.com/questions/1604822/counting-how-many-possible-images-in-an-800x5003color-image	# Counting how many possible images in an 800x500(3color) image. I want to compare what takes more space, 1,200,000 byte 800x500 image with 3 colors, or space required to store the number of all possible images that can be exist in this space(eg, if there was a hypothetical database of all possible images with a unique identifier and the address of the image, would identifying the image by the unique identifier take less space than the image takes up by itself). For example. a 1x1 image in this format could be stored using 3 bytes (excluding the width/height at start of file which takes 4 bytes each). But this 1x1 space would contain 16,581,375(255^3) possible images(as each color pixel is a unique image), and 16,581,375 can be stored in 3 bytes as well, as 3 bytes represent 0 through 16,777,215. How do I determine how many possible images exist in such a space beyond a 1x1 image? They take up the same space. The number of images is $$256^{800 \times 500 \times 3} = 256^{1.2 \times 10^6},$$ which can be expressed in a number $1.2$ million bytes long. • Actually, you need $1\,200\,001$ bytes to represent that number: The first byte is $1$, all others are $0$. – celtschk Jan 8 '16 at 20:45 • i suspected something like this, but I wasn't sure. I typed ln(256^(8005003))/ln(256) in wolframalpha and it confirmed it is near 1,200,000 bytes... Although it didn't give a complete number. Thanks. – Dmitry Jan 8 '16 at 20:46 • Well, that expression indeed gives exactly $1\,200\,000$. However, consider the number of sequences of 3 decimal digits. There are $10^3=1000$ such sequences, but the number $1000$ already has 4 digits (but $\ln 1000/\ln 10=3$). – celtschk Jan 8 '16 at 20:52 • Looking again at the three-digit example, interpreting the digits as number gives numbers from $0$ to $999$. That are $1000$ numbers, but the number $1000$ itself is not among them; indeed, it's the first one that's too large to fit in three digits. Another way to see it is: If it were only the numbers $1$ to $999$, you'd have $999$ numbers. But additionally you've got the $0$, and thus one more. But $999$ is already the largest 3-digit number, so you get a four-digit number. BTW, to write a positive integer $n$ in base $b$, you always need $\lfloor\log n/\log b\rfloor + 1$ digits. – celtschk Jan 8 '16 at 21:08 • Yes, as I wrote above: The most significant byte would be 1, and the 1,200,000 other bytes would be 0. However note that you only need 1,2000,000 bytes to represent $256^{8005003}$ different values (which would, obviously, not include the value $256^{8005003}$; the highest representable value in 1,200,000 bytes is $256^{8005003}-1$). – celtschk Jan 8 '16 at 21:27 Well, each byte can be one of $256$ things and may be chosen independently of the other bytes in an image - note that you've used the incorrect value of $255$ in your calculations, which possibly ignores that bytes can be $0$. In particular, this means that the number of possible images is: $$256^{1200000}=9.09\times 10^{2889887}$$ which is a very large number. It is, however, fundamentally impossible to assign every image a shorter description than its representation as an array of pixels, since one needs to get the right number of strings, and there just aren't enough shorter strings. That said, compression algorithms operate by assigning shorter strings to some images at the expense of assigning longer strings to others. One should note that the vast majority of this space does not contain meaningful images (i.e. it's mostly noise) and the human eye cannot discern images which are close together (e.g. if you allow for each color channel to vary by $1$ from the truth, then there are typically $3^{3\times 800\times 500}=3.20\times 10^{572545}$ indistinguishable images from a given one). Compression can take advantage of both of these facts.	2019-06-20 23:55:59	{"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": 0.527583122253418, "perplexity": 514.0468116406169}, "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-2019-26/segments/1560627999291.1/warc/CC-MAIN-20190620230326-20190621012326-00386.warc.gz"}
https://cs.stackexchange.com/questions/10977/finding-recurrence-when-master-theorem-fails	# Finding recurrence when Master Theorem fails Following method is explained by my senior. I want to know whether I can use it in all cases or not. When I solve it manually, I come to same answer. $T(n)= 4T(n/2) + \frac{n^2}{\lg n}$ In above recurrence master theorem fails. But he gave me this solution, when for $T(n) = aT(n/b) + \Theta(n^d \lg^kn)$ if $d = \log_b a$ if $k\geq0$ then $T(n)=\Theta(n^d \lg^{k+1})$ if $k=-1$ then $T(n)=\Theta(n^d\lg\lg n)$ if $k<-1$ then $T(n)=\Theta(n^{\log_ba})$ using above formulae, the recurrence is solved to $\Theta(n^2\lg\lg n)$. When I solved manually, I come up with same answer. If it is some standard method, what it is called ? • See also our reference question for solving recurrences. In particular, the first case you have been given is covered by the master theorem. But then, even the Akra-Bazzi method does not cover your example. Oh well. By manually, do you mean using recursion trees? – Raphael Apr 2, 2013 at 19:16 • ^yes. Basically I meant without using Master Theorem or Akra-Bazi method. Here's one solution : chuck.ferzle.com/Notes/Notes/DiscreteMath/… – avi Apr 3, 2013 at 12:38 • I see; that would be guess & proof, then. Legit, but arduous: you need to deal with lower and upper bound separately and perform induction proofs for both. – Raphael Apr 3, 2013 at 14:08 OK, try Akra-Bazzi (even if Raphael thinks it doesn't apply...) $$T(n) = 4 T(n / 2) + n^2 / \lg n$$ We have $g(n) = n^2 / \ln n = O(n^2)$, check. We have that there is a single $a_1 = 4$, $b_1 = 1 / 2$, which checks out. Assuming that the $n / 2$ is really $\lfloor n / 2 \rfloor$ and/or $\lceil n / 2 \rceil$, the implied $h_i(n)$ also check out. So we need: $$a_1 b_1^p = 4 \cdot (1 / 2)^p = 1$$ Thus $p = 2$, and: $$T(n) = \Theta\left(n^2 \left( 1 + \int_2^n \frac{u^2 du}{u^3 \ln u} \right) \right) = \Theta\left(n^2 \left( 1 + \int_2^n \frac{du}{u \ln u} \right) \right) = \Theta(n^2 \ln \ln n)$$ (The integral as given with lower limit 1 diverges, but the lower limit should be the $n_0$ for which the recurrence starts being valid, the difference will usually just be a constant, so using 1 or $n_0$ won't make a difference; check the original paper.) • Ah, so you are allowed/supposed to change the lower boundary of the integral -- that was my problem exactly! Your explanation does not make a lot of sense to me, though: the integral does not converge on $[1,2]$, so the difference it not a constant! I guess I'll have to look at the paper at some point... if only it was readily available.	2022-07-03 11:45:41	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.864338219165802, "perplexity": 428.1329767279024}, "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/1656104240553.67/warc/CC-MAIN-20220703104037-20220703134037-00457.warc.gz"}
https://necromuralist.github.io/index-9.html	# Packet Capturing Bibliography ## Packet Capturing • [R3] Schulman A, Levin D, Spring N. On the fidelity of 802.11 packet traces. InInternational Conference on Passive and Active Network Measurement 2008 Apr 29 (pp. 132-141). Springer, Berlin, Heidelberg. (PDF Link) • [R4] Kashyap A, Paul U, Das SR. Deconstructing interference relations in WiFi networks. InSensor Mesh and Ad Hoc Communications and Networks (SECON), 2010 7th Annual IEEE Communications Society Conference on 2010 Jun 21 (pp. 1-9). IEEE. (PDF Link) • [R5] Dujovne D, Turletti T, Filali F. A taxonomy of IEEE 802.11 wireless parameters and open source measurement tools. IEEE Communications Surveys & Tutorials. 2010 Apr 1;12(2):249-62. (PDF Link) • [R6] Serrano P, Zink M, Kurose J. Assessing the fidelity of COTS 802.11 sniffers. InINFOCOM 2009, IEEE 2009 Apr 19 (pp. 1089-1097). IEEE. (PDF Link) ## Key • [R<number>] - Research Paper # Networking Tools ## Packet Capturing ### Gulp Gulp purports to be better at capturing packets than tcpdump (although they can work together). There is more than one version out there: - This one says it applied a patch to it five years ago. • This one says it is the original but hasn't been updated in six years. • This blog post has updated versions of it including one in 2017 that says it has a major bug fix (but I don't know if it's a gulp bug or not) ### tcpdump More easily obtainable and better documentation available (although still not enough). ### ssldump Captures packets and decodes SSL/TLS packets. ## Packet Examining ### Compressed PCAP Packet Indexing Prograpm (cppip) This adds indexing to bgzip compressed LibPCAP files which then lets you extract them while the original files are still compressed. ### tcpslice This lets you extract part of or combine files created by tcpdump when using file rotation. ### ngrep Describes itself as like GNU grep but for packets. ### pylibpcap, pypcap Python code to work with libpcap. ### WireShark These are installed when you install wireshark. #### tshark/wireshark Packet capturing and examining (better documented than most of the other programs) #### reordercap Reorders the packets by timestamp. #### capinfos This prints summary information about packe files (works with gzipped files). #### mergecap Merges multiple packet files together. Mergecap will try to keep timestamps in order when merging, but it assumes each individual file to merge is already in order. ## Packet Flows ### tcpick Track, reassemble, reorder TCP streams. ### tcptrace Gives connection information taken from a capture file. ### tcpflow Separates out TCP flows into separate files. ## Capture Summarizing ### ipsumdump Summarizes packet information in ASCII format ### tcpdstat Gives summary statistics for a pcap file ## Network Monitoring ### ntop Like top but for the network. Part of iproute2 ## Miscellaneous ### tcpreplay A suite of programs to edit and replay pcap files. ### netcat Read from and write to TCP/UDP network connections. ### socat Route data between byte streams. # The Flask Debug Server ## Background This is documented on the Flask site, but I was trying to help someone debug some old server code that I'd written and couldn't remember how to debug it, so I'm documenting it here as I go through remembering it again so I'll have a single reference to use the next time. Some of the settings look different from what I remember using so I think that Flask has changed a little over time, but since I didn't document it the first time I don't have a record to compare against (well, I probably have some notes in a notebook but it's hard to refer to that when sitting at someone else's desk). ## Setup The Flask Quickstart tells you what to do, but for some reason when we googled it, the instructions were different, I think it might have lead us to an older form of the documentation. This is the current version (May 20, 2018.) ### The Environment Variables #### The Flask App First you have to tell flask which file contains your flask app by setting the FLASK_APP environment variable. In my case I'm using connexion, an oddly-named adapter that adds support for swagger/OpenApi to Flask. So the file that has the app has this line in it. application = connexion.FlaskApp(__name__) In this case that's a file named api.py which we'll say is in the server folder (it isn't, but that's okay) so we need to set our environment accordingly. I use the fish shell so the syntax is slightly different from the Quick Start example. Also - and this caused me a lot of trouble - when I didn't pass in the name of my FlaskApp instance I got this error: Error: Failed to find application in module "server.api". Are you sure it contains a Flask application? Maybe you wrapped it in a WSGI middleware or you are using a factory function. So I had to specifically tell flask the name of my app by appending it to the end of the setting (perhaps it is looking for app specifically, but I called mine application). set -x FLASK_APP server.api:application #### Development Mode If you want the server to automatically reload when it detects changes then you should to set the FLASK_ENV environment variable to development. This is similar to using FLASK_DEBUG but I think it adds the reloading. Anyway, it does more than just set debug mode. set -x FLASK_ENV development ## Run It ### The Development server This is the output of the help string for the development server, note that it uses -h for host so you need to pass in --help to see this output or you will get an error. flask run --help ### Public Server The default server runs on localhost, but since I'm hosting the code on a raspberry pi sitting on the network somewhere but debugging it remotely, I need to run it on a public address. flask run --host=0.0.0.0 # Monitor Mode With airmon-ng ## Introduction I'm looking into setting up a wireless (WiFi and bluetooth) monitoring station to collect data that correlates with how my network is performing and what the state of the network is and I thought that, as a first step, I'd get some packet capturing logs going. I'm primarily a python programmer who's kept my toe in the Linux command-line world but it's been a little while since I really dove into the wireless networking world. I had some vague notion about doing it with iwconfig or iw, but then I found airmon-ng and realized that it was what I was really looking for. Why is it better? Well, to be honest, I'm not informed enough to say that it's better, but when I tried to use iw it failed without really telling me why, while airmon-ng not only didn't fail, but it told me that there were other processes already using my wireless interface which is likely why iw failed and it told me how to fix it. On the one hand, since it's hiding so much from you airmon-ng lets you be a little ignorant and still do stuff, on the other - what's wrong with that? ## Setting Up I'm using Ubuntu 18.04 (Bionic Beaver) - which seems to have both fixed and broken a surprising amount of stuff (nice that you let me log in with Dvorak now, but maybe you should let me know the keyboard layout has changed ahead of time) - so these instructions are based on that. First, airmon-ng is part of the aircrack-ng package so you need to install it to get what we want. sudo apt install aircrack-ng Once you do this you'll see that airmon-ng is installed. which airmon-ng Interestingly, if you check it out, you'll see that all it is is a bash script. file which airmon-ng The file is kind of long. wc -l which airmon-ng So I won't list it here - you can check it out if you're interested. It's actually very informative if you want to learn how to do this kind of stuff, but for this case, we just need to know it works. ## Monitor Mode ### Starting Up Monitor Mode #### Finding your interface In the good old days you could be pretty sure that your wireless interface was wlan0 (assuming you only had one) but then ubuntu/freedesktop went and changed things so now you should probably check what your interface name is using iw. iw dev So it looks like we have a wireless interface named wlp2s0 that we want to change from managed to monitor mode. ### Okay, now monitor it The syntax to start monitor mode is airmon-ng start <interface>. sudo airmon-ng start wlp2s0 Found 5 processes that could cause trouble. If airodump-ng, aireplay-ng or airtun-ng stops working after a short period of time, you may want to run 'airmon-ng check kill' PID Name 1505 wpa_supplicant 1524 NetworkManager 1541 avahi-daemon 1748 avahi-daemon 2298 dhclient PHY Interface Driver Chipset phy0 wlp2s0 iwlwifi Intel Corporation Wireless 7260 (rev 73) (mac80211 monitor mode vif enabled for [phy0]wlp2s0 on [phy0]wlp2s0mon) (mac80211 station mode vif disabled for [phy0]wlp2s0) #+END_SRC The first thing you should notice is that there are five potentially interfering processes. This is probably what interferes with the =iw= method, but we'll leave it alone and see if it works. Why don't we check on the interface. #+BEGIN_SRC bash :results raw iw dev #+END_SRC #+RESULTS: phy#0 Interface wlp2s0mon ifindex 5 wdev 0x3 addr 7c:5c:f8:f7:f5:c6 type monitor channel 10 (2457 MHz), width: 20 MHz (no HT), center1: 2457 MHz txpower 0.00 dBm So you can see that running =airmon-ng start= killed our original =wlp2s0= interface and replaced it with =wlp2s0mo= which is in monitor mode on channel 10. Unforturnately I wanted channel 6 but forgot to specify it. Let's try that again. The first thing we have to do is to turn off monitor mode. #+BEGIN_EXAMPLE sudo airmon-ng stop wlp2s0mon Note that we are stopping the new monitor-mode interface, not our original wireless interface. Now we can start the monitor-mode interface set to channel 6. The syntax is airmon-ng start <interface> <channel>. sudo airmon-ng start wlp2s0 6 #+END_SRC There's some output from the command, but we want to know what =iw= thinks is going on. #+BEGIN_SRC bash :results raw iw dev #+END_SRC #+RESULTS: phy#0 Interface wlp2s0mon ifindex 7 wdev 0x6 addr 7c:5c:f8:f7:f5:c6 type monitor channel 6 (2437 MHz), width: 20 MHz (no HT), center1: 2437 MHz txpower 0.00 dBm So now we have an interface (=wlp2s0mon=) on channel 6 in monitor mode. We can make sure that it's working using [[https://tcpdump.org][tcpdump]]. #+BEGIN_EXAMPLE sudo tcpdump -i wlp2s0mon -n Note that we need to use the new interface name. Also, if it wasn't obvious up to now, putting the interface into monitor mode will break any networking capabilities for that interface on your computer (so if it was your internet connection, don't expect to access the web when it's in monitor mode). ## Cleaning Up We already got a preview of turning off monitor mode earlier. The syntax is airmon-ng stop <interface>. sudo airmon-ng stop wlp2s0mon This will bring back the original wireless interface, but it won't (likely) re-establish your connection to your wireless access point. To get back onto the network you will probably need to open network manager and go through the setup process again. ## Summary These were my notes on setting up monitor mode using airmon-ng. The main point I wanted to get across is how easy it is to do using airmon-ng as opposed to the other methods. I didn't actually show how much harder it is to use iwconfig, but if you have tried you might know what it entails. In any case, hopefully these notes will help me in the future as I keep watching the packets. # TCP Dump Notes These are notes I made while surfing the web looking into TCP Dump. You will most likely need to use sudo to run most of the commands, but I'm leaving it off to make it shorter. ## About TCP Dump • It has more filtering capabilities and can filter while capturing packets, but it doesn't have the analytical tools that something like wireshark has1. ## Some Examples ### Listing interfaces You can ask tcpdump which interfaces it is able to listen to2. tcpdump -D ### Capture packets on an interface To capture packets on an interface you pass its name to the -i flag2 (here the interface I'll use is eno1). tcpdump -i eno1 #### Save the packet capture to a file The default behavior is for tcpdump to send the output to standard output, to have it save the packets to a files use the -w flag2 (you can call it anything, I'll call it dump.pcap). tcpdump -i eno1 -w dump.pcap #### Increase the verbosity of the capture To increase the amount if information that's captured, pass multiple v arguments2 (in this case I'll use -vvv). tcpdump -i eno1 -vvv -w dump.pcap ### Filtering #### By IP address You can get all the packets being sent or received by a host using the host argument3. tcpdump host 192.168.1.12 #### By Sender IP Address You can filter out all the packets except those that are being sent by a host using the src host argument2. tcpdump -i eno1 src host 192.168.1.12 You can leave off the host argument and just use src3 #### By Target IP Address To filter out all the packets except those that are going to a specific target use the dst host argument2. tcpdump -i eno1 dst host 192.168.1.1 #### Sender and Target IP Addresses You can combine parameters using the logical operators and, or, and not3. tcpdump 'src 192.168.1.1 and dst 192.168.1.12' The single quotes are optional and are just used to group the arguments together. #### By Subnet You can grab all the packets on a network or subnet using the net argument and CIDR notation3. This example grabs all the packets on the 192.168.1.* subnet. tcpdump net 192.168.1.0/24 #### By port and/or protocol If you want to only catch activity on a certain port and by a certain protocol then you use the port argument and the name of the protocol (e.g. udp)3. This would catch all the tcp traffic over SSH. tcpdump tcp port 22 You can use tcp, udp, or icmp for the protocols and add multiple ports using a comma4. tcpdump tcp port 22,80 #### Turn off hostname and port translation The default behavior for tcpdump is to translate the hostnames and ports to something human-readable if possible. To turn this off you pass in the -n argument3. Since this stops having to look things up it will reduce the amount of overhead needed by tcpdump. tcpdump -n -i eno1 port 22 ## Footnotes: 1 Diogenes, Y. & Ozkaya, E. (2018). Cybersecurity, Attack and Defense Strategies : infrastructure security with Red Team and Blue Team tactics. Birmingham, UK: Packt Publishing.] 2 Johansen, G. (2017). Digital forensics and incident response : an intelligent way to respond to attacks. Birmingham, UK: Packt Publishing. 3 Beltrame, J. (2017). Penetration testing bootcamp : quickly get up and running with pentesting techniques. Birmingham, UK: Packt Publishing. 4 McPhee. & Beltrame, J. (2016). Penetration testing with Raspberry Pi : learn the art of building a low-cost, portable hacking arsenal using Raspberry Pi 3 and Kali Linux 2. Birmingham, UK: Packt Publishing. 5 Baxter, J., Orzach, Y. & Mishra, C. (2017). Wireshark revealed : essential skills for IT professionals : get up and running with Wireshark to analyze your network effectively. Birmingham, UK: Packt Publishing. # MNIST Digits With Keras ## Imports These are the parts that will make up the model. ### The Sequential Model The Keras Sequential Model is a stack of layers that will make up the neural network. from keras.models import Sequential ### The Dense Layers The Keras Dense layer is a densely-connected layer within our neural network. from keras.layers.core import Activation ### Activation The Activation represents the activation function for each layer (e.g. relu). from keras.layers.core import Activation ### Adam To tune the model to the data we'll use the Adam optimizer from keras.optimizers import Adam ### Categorical Converter Finally, since our problem is a classification problem (identify which of 10 digits an image represents) I'll import the Keras to_categorical function to enable classification of our data. from keras.utils import np_utils The MNIST dataset is made up of human-classified hand-written digits. Keras includes it as part of their installation so we can load it directly from keras. from keras.datasets import mnist We're going to use numpy to reshape the data. import numpy To make our output the same every time, I'll set the random seed to April 28, 2018 as a string of digits. numpy.random.seed(4282018) # bokeh org-mode ## Introduction This is an illustration of how to use bokeh with org-mode in nikola. There is a more extensive and flexible explanation of how to do this in this post on cherian.net but I made these notes to understand how it works and to have a simpler example to refer to. I was interested in doing this because I'm trying to re-work some of what I did for the Coursera Data Science With Python specialization by changing the data-sets and building them as blog posts. I might convert the posts to restructured text at some point, but while I'm working with them I'm using org-mode. Also, while most of the time I use matplotlib for plotting since this is going to be a blog-first approach I decided to go with bokeh. I had previously written about how to get bokeh into Nikola using restructured text, but as an intermediate step I want to do the work in org-mode and still be able to see the output as I'm working. The magic mix for this seems to be to use: ## Creating the Bokeh Plot ### Imports These are the dependencies. It's really all bokeh, numpy is just there to generate the data-values. # from pypi from bokeh.models import HoverTool from bokeh.plotting import figure, ColumnDataSource from bokeh.embed import autoload_static, file_html import bokeh.resources import numpy I probably should save bokeh to this repository to keep the post from breaking in the future, but I'm lazy so I'm just going to import it from a CDN. bokeh = bokeh.resources.CDN ### The Data To get a simple example going I'm just going to use some random outputs generated by numpy. X = numpy.arange(10) Y = numpy.random.randint(0, 10, 10) In order to create a data-structure that bokeh can use (similar to a pandas dataframe) you need to use a ColumnDataSource. source = ColumnDataSource(data=dict( x=X, y=Y, desc=["a", "b", "c", "d", "e", "f", "g", "h", "i", "j"], )) The keys in the data-dict are essentially the column headers. Key Description x the x-axis values y the y-axis values desc The labels for the tooltip Now to get some tool-tips to pop up when you hover over the plot, I'll create a HoverTool. hover = HoverTool(tooltips=[ ('index', '$index'), ('(x, y)', '($x, $y)'), ('desc', '@desc'), ]) The tooltips list maps the labels that will show up in the tooltip (the first argument to each tuple) to variables in the ColumnDataSource (if preceded by an @) or generated value (if preceded by a $) The index value is the index in the array where the data point sits (so for the first point it will be 0, the second will be 1, etc.). The (x, y) values are the coordinate locations of your pointer when you hover over the data points, and the desc will be replaced by the label I set in the ColumnDataSource for a particular data-point. ### The Plot Now I'll create the actual plot (figure). fig = figure(title="Random Example", x_axis_label="x", y_axis_label="y", tools=[hover]) fig.line('x', 'y', source=source) fig.circle('x', 'y', size=10, source=source) ## Getting the Bokeh Plot Into The Post Finally I'll save the javascript and HTML files needed and then output the blob needed to embed the plot into this post. The autoload_static function takes the bokeh plot object (fig), the bokeh javascript that I loaded earlier (bokeh), and the name of the javascript file that you want it to creat (test.js) and returns the javascript to save (javascript) and the HTML fragment that will include the javascript (source). Note that because of the way nikola structures things I have to create a folder named files/posts/bokeh-org-mode and save the files there. Since nikola will automatically look in this folder the name you pass into autoload_static should just be the filename without the path, but when you save the javascript file you will save it there so you need to add the relative path. If my explanation seems a little convoluted, just look at the code below, it's fairly simple. First I'll create a variable to hold the path to the folder to save the files in. All files for nikola posts go into sub-folders of files/posts/ and since the source file for this post is called bokeh-org-mode.org, the files to include in it go into the folder files/posts/bokeh-org-mode (files/posts/ plus the slug for the post). FOLDER_PATH = "../files/posts/bokeh-org-mode/" ### The Javascript Now, I'll create the javascript source for the plot. FILE_NAME = "test.js" javascript, source = autoload_static(fig, bokeh, FILE_NAME) with open(FOLDER_PATH + FILE_NAME, "w") as writer: writer.write(javascript) The javascript variable holds the actual javascript source code (which then gets saved) while the source variable holds the string with the HTML to embed the javascript into this post (which I show at the end of this post). ### Embedding the Plot Finally, we need to print out the string that is stored in the source variable which then tells org-mode to embed the files into this post. I'll output the full org-block so you can see the header arguments. #+BEGIN_SRC python :session bokeh :results output raw :exports results print('''#+BEGIN_EXPORT html {} #+END_EXPORT'''.format(source)) #+END_SRC And there you have it. I don't have a lot to say about it, other than that if you hover over the data with your cursor and then look up above at the ColumnDataSource above, you'll see that the variables match the inputs ## Summary To get a bokeh figure into an org-mode document in nikola: 1. Create the bokeh plot. 2. Create a folder in the files/posts/ folder that matches the slug for the post. 3. Use autoload_static to convert the bokeh object to javascript and create the HTML tag to embed it. 4. Save the javascript in the files/posts/<slug>/ folder that you created 5. Print the HTML fragment in an org-mode #+BEGIN_EXPORT html block. # Data Visualization Bibliography MW.1. Playing With The. (2018) Drawing animated GIFs with matplotlib - Eli Bendersky's website. Retrieved February 27, 2018, from https://eli.thegreenplace.net/2016/drawing-animated-gifs-with-matplotlib/ This shows how to save the animation as a GIF using ImageMagick. MW.2. Repeatedly Calling A. (2018) animation module — Matplotlib 2.1.2 documentation. Retrieved February 27, 2018, from https://matplotlib.org/api/animation_api.html This is the matplotlib documentation for creating animations.	2018-08-21 00:01:36	{"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": 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": 0.21032394468784332, "perplexity": 4109.508213419068}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-34/segments/1534221217901.91/warc/CC-MAIN-20180820234831-20180821014831-00265.warc.gz"}
https://www.scalarl.com/course/tabular/n_step_bootstrapping.html	# WARNING! Note - this is VERY EARLY DAYS! All of the files in the course with this warning are the raw, totally unprocessed notes that I generated during my first reading of “Reinforcement Learning: An Introduction”. I’ll be converting these into proper course sections with wonderful embedded code that you can try out. I’m not there yet, but I wanted to develop all of this in the open. Maybe you’ll find these notes interesting, but don’t expect anything special… yet! # Notes Really, this is the main thing we should be coding. And then we can specialize back to get everything else. Build up, as in the book, but I need to know where I’m building toward! These let us shift smoothly between the TD(0) method and the full-on, infinite step Monte Carlo methods. We hear another mention of eligibility traces… but I don’t know what that is, so let’s move on! n-step methods allow bootstrapping to occur very fast, to get online behavior, but also if we have time to start extending updates out into the future, to get the benefit of the long view. • Prediction first, • then control. We start with n-step Sarsa… but then we come up with some other methods that are in fact extensions of a grand unifying thing! That I want to go into. We sort of have to solve for THIS stuff… and then all of the previous work becomes a generalization. Well, sort of. Dynamic programming becomes a… is it a special case, still? I guess not since we don’t have the dynamics at play. But if we do then we can use them. That is what comes back in chapter 8. ## Chapter Sketch The individual chapters. What is up? ### n-step TD Prediction Nice picture of the backup diagrams, which I still don’t really get / like. What do we do if we want to extend beyond that first step? We need to keep playing a game, generating an episode, but as we play we can start passing info back. For example, a two-step reward would be based on the first two rewards and the estimated value of the state two steps later. (p. 142) The methods are still TD methods because they still change an earlier estimate based on how it differs from a later estimate. (p. 142) This still is a constant-alpha method; BUT NOTE that we are sneaking in an idea with the constant alpha thing!!! That is definitely NOT the only way to aggregate this stuff! That is just a way to privilege recent games. You can totally have different methods of aggregation. Then we look at the random walk task about show how different settings of alpha or n cause different overshoots. Thing to note is that an intermediate value of n worked nicely, halfway between monte carlo and TD. This is a state-value updater because the policy already decides what to do… it has some action, presumably, that it is going to take for everything. Or some chance of taking each action that it based on an action-value function that you assume someone else gave you. It definitely has knowledge of the actions, in any case. ### n-step Sarsa Remember, to come up with a policy we need to switch from state-value to action-value. That way we can use the collection of action-values to modify the policy, potentially. We still need the guarantee that we’re possibly going to explore all states, so for on policy we need an $\epsilon\text{-greedy}$ method. Keep the action-value around then update. Expected sarsa is easy… just use the expected approximate values at the end. ### n-step Off-Policy Learning Woah, generalize again! Obviously I have to code this first, and then specialize it back to all of the previous stuff. Show how single state gives you bandits, for example. Note that our importance sampling ratio starts and ends one step later than if we were estimating the action-value. off-policy n-step expected sarsa too, here. just showing off. ### Per-decision methods with Control Variates Research topic alert! Look into this more in the exercises, but this should just be a different aggregator. This is a way of saying, okay, my update is actually a weighted average of the training policy’s thing, what it learned, and then what the target policy ALREADY has. This is using importance sampling in place of alpha, I think? Think of a way to describe what is happening. But show again that it is just another aggregator. ### Off policy learning WITHOUT importance sampling - the n-step tree backup algo Is there a non-importance-sampling method that we can use for off-policy learning? We did this for the one-step case before… can we do this for the good stuff, the multi-step? You can get some budget and go chase promising stuff, unfolding as you go, if you have time. You’re sort of learning from your model. I like it, easy. ### n-step $Q(\sigma)$ Can we unify all of the algos we’ve already seen? yeah, this is what to actually code, since we can… show that the $\sigma$ parameter can be set anywhere from 0 to 1. ## Programming Exercises • Note that when we do the constant alpha thing, that is just a particular way to aggregate. If we used $1 / n$ then we’d be doing a straight-up average. But you can weight different games differently, even, instead of going by time. The reason to weight games played more recently is that you’ve got more experience that went into choosing that trajectory. • Maybe when you average in the monoid you do a different thing - you weight by how many games you’ve played, instead of by alpha. • exercise 7.2… plug in the “sum of TD errors” thing from 7.1! That is actually a DIFFERENT way of propagating the information around. Implement this as an aggregator as well. • implement n-step sarsa • expected n-step sarsa • q-learning • Can we do a sliding window product? I think so right? Since products are just repeated addition? Well, that’s a nice extension. • exercise 7.10, implement the control variates thing. • off-policy without importance sampling… tree backup algo! • n-step q-sigma	2019-11-18 19:18:04	{"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": 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": 0.5903640389442444, "perplexity": 1243.7738421957201}, "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/1573496669813.71/warc/CC-MAIN-20191118182116-20191118210116-00200.warc.gz"}
http://www2.macaulay2.com/Macaulay2/doc/Macaulay2-1.19/share/doc/Macaulay2/Polyhedra/html/___Working_spwith_spfans.html	Working with fans We start by constructing a fan, which consists of a single cone and all of its faces: i1 : C = coneFromVData matrix {{1,0,0},{0,1,0},{0,0,1}} o1 = C o1 : Cone i2 : F = fan C o2 = F o2 : Fan By this, we have already constructed the fan consisting of the positive orthant and all of its faces. The package saves the generating cones of the fan, which can be accessed by: i3 : maxCones F o3 = {{0, 1, 2}} o3 : List Now we could expand the fan by adding more cones, for example the following: i4 : C1 = coneFromVData matrix {{1,0,0},{1,1,0},{0,0,-1}} o4 = C1 o4 : Cone But in this case we can not, because the two cones are not compatible, i.e. their intersection is not a face of each. So, when one tries to add a cone to a fan that is not compatible with one of the generating cones of the fan, the function addCone gives an error. For two cones one can check if their intersection is a common face by using commonFace: i5 : commonFace(C,C1) o5 = false Since the intersection of both is already computed in this function there is a different function, which also returns the intersection, to save computation time when one needs the intersection afterward anyway: i6 : (b,C2) = areCompatible(C,C1) o6 = (false, C2) o6 : Sequence i7 : rays C2 o7 = \| 0 1 \| \| 1 1 \| \| 0 0 \| 3 2 o7 : Matrix ZZ <--- ZZ So we can make the cone compatible and add it to the fan. i8 : C1 = coneFromVData matrix {{1,0,0},{0,1,0},{0,0,-1}} o8 = C1 o8 : Cone i9 : F = addCone(C1,F) o9 = F o9 : Fan Instead of creating a fan with one cone and then adding more cones, we can also make a fan out of a list of cones: i10 : C2 = coneFromVData matrix {{-1,0,0},{0,1,0},{0,0,1}}; i11 : C3 = coneFromVData matrix {{-1,0,0},{0,1,0},{0,0,-1}}; i12 : C4 = coneFromVData matrix {{-1,0,0},{0,-1,0},{0,0,1}}; i13 : C5 = coneFromVData matrix {{-1,0,0},{0,-1,0},{0,0,-1}}; i14 : F1 = fan {C2,C3,C4,C5} o14 = F1 o14 : Fan Furthermore, we could add a list of cones to an existing fan: i15 : C6 = coneFromVData matrix {{1,0,0},{0,-1,0},{0,0,1}}; i16 : C7 = coneFromVData matrix {{1,0,0},{0,-1,0},{0,0,-1}}; i17 : F1 = addCone( {C6,C7}, F1) o17 = F1 o17 : Fan So, fan and addCone are the methods to construct fans ''from scratch'', but there are also methods to get fans directly, for example normalFan, which constructs the inner normal fan of a polytope. i18 : P = hypercube 4 o18 = P o18 : Polyhedron i19 : F2 = normalFan P o19 = F2 o19 : Fan Now we have seen how to construct fans, so we turn to functions on fans, for example the direct product (directProduct: i20 : F3 = fan {coneFromVData matrix {{1}},coneFromVData matrix {{-1}}} o20 = F3 o20 : Fan i21 : F1 = F3 * F1 o21 = F1 o21 : Fan The result is in the direct product of the ambient spaces. i22 : ambDim F1 o22 = 4 Of course, we can check if two fans are the same: i23 : F1 == F2 o23 = false A bit more on fans can be found in part 2: Working with fans - Part 2.	2023-02-01 21:41:24	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 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": 0.4628206193447113, "perplexity": 1842.4199495684172}, "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-2023-06/segments/1674764499953.47/warc/CC-MAIN-20230201211725-20230202001725-00607.warc.gz"}
https://plainmath.net/82735/where-are-radio-waves-located-on-the-ele	# Where are radio waves located on the electromagnetic spectrum? Where are radio waves located on the electromagnetic spectrum? You can still ask an expert for help • Questions are typically answered in as fast as 30 minutes Solve your problem for the price of one coffee • Math expert for every subject • Pay only if we can solve it Kendrick Jacobs Radio waves are a type of electromagnetic radiation, forming board part of spectrum with wavelengths in the electromagnetic spectrum longer than infrared light. The frequencies lies between 300 GHz to as low as 3 kHz, and corresponding wavelengths from 1 millimeter to 100 kilometers, and radio waves travel at the speed of light.	2022-08-13 03:35:21	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 4, "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": 0.41786015033721924, "perplexity": 1014.3848844005893}, "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/1659882571869.23/warc/CC-MAIN-20220813021048-20220813051048-00705.warc.gz"}
https://hamnvag.no/y1xbvmr/egyptian-fractions-worksheet-941655	For K-12 kids, teachers and parents. Stories For Comprehension. Fractions really aren't that difficult to master especially with the support of our wide selection of worksheets. An egyptian number is any number equal which can be expressed as the sum of an integer plus the sum of an Egyptian fraction. Yes, it’s a unit fraction. Egyptian fraction expansion. You don't have to be an ancient Egyptian to decipher fractions in this activity that focuses on adding fractions with unlike denominators and developing fraction number sense. Lesson materials located below the video overview. Old Egyptian Math cats never repeated the same fraction when adding. These grade 1 worksheets introduce students to fractions. Some of the worksheets displayed are Egyptian mathematics, Egyptian fractions work pdf, Egyptian fractions, Egyptian unit fractions, Egyptian fractions what are known and unknown, Egyptian mathematics, Egyptian mathematics, Mega fun fractions. Reading Comprehension Worksheets For Free. Egyptian Fractions Lesson Plans & Worksheets Reviewed by Teachers You can & download or print using the browser document reader options. Fractions of shapes Skill: Learning equal parts. The representation of rational numbers as sums of unit fractions dates back to the time of ancient Egypt. The single exception was having a symbol for 2/3. The people of ancient Egypt represented fractions as sums of unit fractions (vulgar fractions with the numerator equal to 1). Egyptian Fractions: The Answer Sheet; To Be Continued… Read all the posts from the January/February 1999 issue of my Mathematical Adventures of Alexandria Jones newsletter. In this unit we want to explore that situation. Square We would slice each cake into 4 … Egyptian fractions A wonderfully different way to practise fraction addition. Worksheet will open in a new window. In this first lesson we have a look at the sum of two Egyptian Fractions to see if we can get another Egyptian Fraction. In this unit we want to explore that situation. Articles that describe this calculator. Yes, it’s a unit fraction. Some of the worksheets for this concept are Egyptian mathematics, Egyptian mathematics, Egyptian fractions representations as sums of unit fractions, Egyptian mathematics, A complete overview for all those who never, Count like an egyptian, Second edition teachers guide 4, Math mammoth fractions 1 worktext. Showing top 8 worksheets in the category - Egyptian Fractions. Next. These fractions worksheets are great for practicing how to add fractional inch measurements that you would find on a tape measure. The file includes a student worksheet for Egyptian fractions from 1/2 to 9/10. As a result, any fraction with numerator > 1 must be written as a combination of some set of Egyptian fractions. Identify a proper fraction, improper fraction, mixed number, unit fraction, like fractions, unlike fractions like a pro with these printable types of fractions worksheets. Egyptians represented fractions differently than we do. The lesson is split into two parts. Basic Reading Comprehension Worksheets Free. The Ancient Egyptians used a 'canon of proportions' which dictated how the figure would be drawn. Some sort of function that finds the standard fractional notation version of the given Egyptian fraction. The range of fraction lessons includes hands-on explorations and activities that invoke problem solving, reasoning and proving, communicating, connecting, and representing fractions. Worksheets > Math > Grade 2 > Fractions. This website and its content is subject to our Terms and We start you off with the obvious: modeling fractions. This calculator allows you to calculate an Egyptian fraction using … Worksheets > Math > Grade 6 > Fractions vs. Decimals > Converting decimals to fractions. Reduce the proper fraction, improper fraction and … Enjoy the videos and music you love, upload original content, and share it all with friends, family, and the world on YouTube. This problem set is based on this one from Illustrative_Mathematics. Old Egyptian Math cats never repeated the same fraction when adding. Fraction worksheets for children from kindergarten to 7th grades. Grade 2 fraction worksheets. Learn how the ancient Egyptians used unit fraction sums to write different fractions… Print the PDF: mixed operations and compare the fractions This test or worksheet provides fraction problems involving mixed operations, requiring adding, subtracting, multiplying, and dividingIf you are using this printable as a test, you'll find out whether students understand when they need to find a common denominator before working the fraction … Learn fraction in a fun way by coloring or shading the pictures. The calculator transforms common fraction into sum of unit fractions. But to make fractions like 3/4, they had to add pieces of pies like 1/2 + 1/4 = 3/4. Adding And Subtracting Scientific Notation Worksheet With Answer Key. Egyptian Fractions: The Answer Sheet. Unit fractions are written additively: 1/4 1/26 means 1/4 + 1/26. Worksheets > Math > Grade 1 > Fractions. The Egyptian fractions were particularly useful when dividing a number of objects equally for more number of people. Many of the fractions were easy. Worksheet 1. Now subtract 1/4 from 3/10 to see if we have an Egyptian Fraction or not. 12 fun Ancient Egyptian themed maths worksheets, covering a wide range of topics from addition with 2-digit numbers to finding equivalent fractions of 2-digit numbers, all tailored for 2nd Grade classes. What Egyptian Fraction is smaller than 0.3 but closest to it? Egyptian Fractions - Displaying top 8 worksheets found for this concept. Egyptian You can start by exploring unit fractions at Keep it Simple In this problem we are going to start by considering how the Egyptians might have written fractions with a numerator of 2 (i.e. These fractions worksheets will use 1/2's, 1/4's, 1/8's. Lesson 5 Homework Practice Solve Multi Step Equations Answers. Students draw a 'modern' representation of an Egyptian figure, using the sample and grid First few slides are a fun intro. All Egyptian fractions share this same property: there are always endless ways to write an Egyptian fraction. Egyptian Art is…. Equivalent fraction worksheets contain step-by-step solving process, identifying missing numbers, finding the value of the variables, completing the chain of equivalent fractions, writing equivalent fractions represented by pie models and fraction bars and representing the visual graphics in fractions. Ask them if they can think of any reason why the Egyptians were hooked on fractions with a one in the numerator. This bundle contains 12 worksheets on 12 different topics all listed in detail below. We look at how fractions can be represented in terms of Egyptian Fractions. The answers are now posted. (1/4) So start with 1/4 as the closest Egyptian Fraction to 3/10. Adding Decimals Practice. These worksheets are pdf files.. 107, (2000), pages 62-63 and 1/4 + 1/28 = our 2/7. Egyptian fractions You are encouraged to solve this task according to the task description, using any language you may know. These are unit fractions - fractions whose numerator is one. My art worksheets show aspects of Egyptian Art that are unusual, unconventional and relatively unknown. Reading Comprehension Past Simple Regular Verbs PDF Preposition Worksheet … Reading Comprehension Worksheets By Grade Level. Introduce the idea of Egyptian Fractions to the class. All ancient Egyptian fractions, with the exception of 2/3, are unit fractions, that is fractions with numerator 1. What type of fraction is 1/4? Fractions worksheets: addition with like denominators. She knew that… Therefore, as soon as she figured out one fraction, she had the answer to all of its equivalents. Showing top 8 worksheets in the category - Egyptian Fractions. An egyptian fraction is an expression of the sum of unit fractions 1/a + 1/b + 1/c +..., where the denominators a,b,c,... are increasing. Some of the worksheets displayed are Egyptian mathematics, Egyptian fractions work pdf, Egyptian fractions, Egyptian unit fractions, Egyptian fractions what are known and unknown, Egyptian mathematics, Egyptian mathematics, Mega fun fractions. Accept a… This is one of our more popular pages most likely because learning fractions is incredibly important in a person's life and it is a math topic that many approach with trepidation due to its bad rap over the years. This bundle contains 12 worksheets on 12 different topics all listed in detail below. But to make fractions like 3/4, they had to add pieces of pies like 1/2 + 1/4 = 3/4. Subtracting Fractions Worksheets. Old Egyptian Math Cats knew fractions like 1/2 or 1/4 (one piece of a pie). Addition And Subtraction For Class 4. Simplifying Fraction. Egyptian fractions for 4/n and the Erdös-Straus Conjecture Every fraction of the form 3/n where n is not a multiple of 3 and odd can be written as 1/a + 1/b + 1/c for distinct odd a, b and c. For a proof see A Proof of a Conjecture on Egyptian Fractions T. R. Hagedorn The American Mathematical Monthly, Vol. There are over 100 free fraction worksheets in PDFs below to support the many concepts encountered with fractions. This page includes Fractions worksheets for understanding fractions including modeling, comparing, ordering, simplifying and converting fractions and operations with fractions. Course Home. When starting with fractions, begin by focusing on 1/2 and then a 1/4 before moving to equivalent fractions and using the 4 operations with fractions … Created: Sep 22, 2013\| Updated: Feb 22, 2018, Pupils can learn about fractions by doing divisions the Egyptian way! The reason the Egyptians chose this method for representing fractions is not clear, although André Weil characterized the decision as "a wrong turn" … Reading Comp Worksheet. Old Egyptian Math Cats knew fractions like 1/2 or 1/4 (one piece of a pie). Old Egyptian Math cats never repeated the same fraction when adding. Instead of 3/4, the Ancient Egyptians would use the unit fractions 1/2 + 1/4. Just a part of the whole. Previous. This resource is designed for UK teachers. Below are six versions of our grade 6 math worksheet on converting decimal numbers to proper fractions. Fractions notation as we know it was not in use generally until after 1500 AD. Old Egyptian Math Cats knew fractions like 1/2 or 1/4 (one piece of a pie). They are often referred to as unit fractions. Your third grader will shade in parts of rectangles and circles in this coloring math worksheet to match a given fraction amount. Egyptian fractions A wonderfully different way to practise fraction addition. For example, 23 can be represented as \$${1 \over 2} +{1 \over 6} \$$. These worksheets will generate 10 fractional inch problems per worksheet. Egyptian Fractions Displaying top 8 worksheets found for - Egyptian Fractions . Egyptian Fractions. 3rd grade. To download/print, click on pop-out icon or print icon to worksheet to print or download. In this download packet you will find 12 pages of Egyptian Art and one Teacher's Guide. Egyptian fractions; Published: 20/04/2012 KS3 2 pages. 3000 BCE) were very practical in their approaches to mathematics and always sought answers to problems that would be of most convenience to the people involved. The number of terms in the Egyptian fraction representation of x/y is the sum of the odd terms after the first in the continued fraction list, which is at most x. Quickly find that inspire student learning. For some reason that is not clear, Ancient Egyptians only used fractions with a numerator of 1, with one exception (2/3). Ask them if they can think of any reason why the Egyptians were hooked on fractions with a one in the numerator. Tes Global Ltd is registered in England (Company No 02017289) with its registered office … (intro) 2. Egyptian fraction notation was developed in the Middle Kingdom of Egypt, altering the Old Kingdom's Eye of Horus numeration system. Any uneven distribution of food ration among the labors could easily kindle dispute and disrupt their work process. Subtraction Question For Class 1. For a more exciting powerpoint, please check out cbarclay99’s at this link: ... Fractions of Amounts Differentiated Worksheet … Think of slicing 3 cakes among 4 workers. you can see that fractions take part of this. Math explained in easy language, plus puzzles, games, quizzes, videos and worksheets. The Egyptians insisted on writing all their fractions as sums of fraction with numerators equal to 1. 1. This is the android version of the Mathcats Old Egyptian Fractions: The app is about: proper fractions and their equivalent egyptian fractions (which in this case, are the addition of 2 or 3 unit fractions). Kindergarten Worksheet Examples; Newspaper Worksheet Examples in PDF; When you purchase goods, when you eat, when you clean your house, when you count your money, when you cook or bake, when you see carpenters or architects making a house or a building, when kids play, even in computer games! Subtraction Problems Year 5. As a result of this mathematical quirk, Egyptian fractions are a great way to test student understanding of adding and combining fractions with different denominators (grade 5-6), and for understanding the relationship between fractions with different denominators (grade 5). EGYPTIAN NUMERATION - FRACTIONS For reasons unknown, the ancient Egyptians worked only with unit fractions, that is, fractions with a numerator of 1. Find egyptian fractions lesson plans and teaching resources. Grade 1 fractions worksheets. Generally, Egyptian Art spans 5,000 years. Pupils can learn about fractions by doing divisions the Egyptian way! For some reason that is not clear, Ancient Egyptians only used fractions with a numerator of 1, with one exception (2/3). Egyptian Art is… Cats 3. These fraction sheets are printable PDF pages and can be used by a big or small group at home or in school. A UNIT fraction is a fraction with a numerator one(1). What type of fraction is 1/4? Today this subject survives mainly as a source of mathematical puzzles and problems in abstract number theory, but the subject is also of historical and anthropological 1/16's and there is an option to select 1/32's and 1/64's. person_outlineAntonschedule 2019-10-29 20:02:56. London WC1R 4HQ. Egyptian Fractions. So tables were provided to help with this task. Includes Guided Reading Handouts. They did this because fractions of the form 1 N have a good intuitive feel to them. The fact that the sum of the fractions is the original input number is a straightforward but tedious exercise in algebraic manipulation. These fractions worksheets will use 1/2's, 1/4's, 1/8's. Students are shown the method of dividing the numerator by the denominator to find the first fraction and then dividing the remainder(s) to find the other fractions. Eating fractions is a great way to advance kids' understanding of fractions. The representation of rational numbers as sums of unit fractions dates back to the time of ancient Egypt. Mathematics / Number / Fractions, decimals, percentage, equivalence, 2020 KS2 SATs Revision Ultimate 15-in-1 Maths Organiser, Fractions 2 - Equivalent Fractions & Simplifying (+ worksheet). PDF (2.44 MB) 12 fun Ancient Egyptian themed maths worksheets, covering a wide range of topics from addition with 2-digit numbers to finding equivalent fractions of 2-digit numbers, all tailored for 2nd Grade classes. Egyptian fractions; Egyptian fraction expansion. D. Russell. Students should simplify all fractions. Paragraphs With Comprehension Questions. Math Grade 3 Year 3 Free Printable Math Worksheets Worksheet … The unit fraction is made by writing the number with a “mouth” symbol over the top – $$\dfrac{1}{2}=$$ Egyptians did not like repeating fractions, therefore, each fraction must be unique. After this brief introduction, hand out Egyptian Fractions, and have the students begin working either individually or in small groups to solve the problems. Similar: She had the most trouble with the 7ths and 9ths. The answer is 1/20. 2 Step Subtraction Word Problems Year 4. As a result, any fraction with numerator > 1 must be written as a combination of some set of Egyptian fractions. Egyptian Fractions Ancient Egyptians used unit fractions, such as $\frac{1}{2}$ and $\frac{1}{3}$, to represent all fractions. Sixth graders take a brief look at what is known about Egyptian Fractions. For example 1/2, 1/7, 1/34. The Secret of Egyptian Fractions. Learn how the ancient Egyptians used unit fraction sums to write different fractions. Kindergarten Worksheet Examples; Newspaper Worksheet Examples in PDF; When you purchase goods, when you eat, when you clean your house, when you count your money, when you cook or bake, when you see carpenters or architects making a house or a building, when kids play, even in computer games! registered in England (Company No 02017289) with its registered office at 26 Red Lion Get Free Access See Review Illness Management Recovery Guide In Spanish. The focus is on gaining a conceptual understanding of what a fraction means; this is largely approached by comparing fractions to simple shapes which have been divided into equal parts.Fractions of a group or set are also introduced. Egyptian Fractions. Egyptian Fractions - Displaying top 8 worksheets found for this concept.. An Egyptian fraction is a sum of positive (usually) distinct unit fractions.The famous Rhind papyrus, dated to around 1650 BC contains a table of representations of as Egyptian fractions for odd between 5 and 101. 1. This was practically important because many of the Egyptian structures required massive labor work. (Be sure to use the words numerator and denominator.) 3 Digit Subtraction With Regrouping Worksheets. View US version . of the form $\frac{2}{n}$). Acce… This website and its content is subject to our Terms and Conditions. Conditions. There are several NRICH problems based on Egyptian fractions. All other fractions were written as a sum of unit fractions: 6 ; 5 8 5 6 <. Tes Global Ltd is For a more exciting powerpoint, please check out cbarclay99âs at this link: http://www.tes.co.uk/teaching-resource/Egyptian-Fractions-6255731/ If you like this resource then please check out my other stuff on here! But to make fractions like 3/4, they had to add pieces of pies like 1/2 + 1/4 = 3/4. Simplifying Fraction This Egyptian Fractions Lesson Plan is suitable for 5th Grade. In this fraction addition worksheet, 5th graders review the term unit fraction and then find the sums for the fraction equations and study how the Egyptians wrote the fractions. These worksheets will generate 10 fractional inch problems per worksheet. Welcome to the fractions worksheets page at Math-Drills.com where the cup is half full! Reading Passages For Children. (Be sure to use the words numerator and denominator.) Common fraction. Egyptian Unit Fractions . Math. Explore some of these worksheets for free. LaTex does not handle Egyptian hieroglyphs — or at least, I don’t know how to make it do so. Some of the worksheets for this concept are Egyptian mathematics, Egyptian fractions work pdf, Egyptian fractions, Egyptian unit fractions, Egyptian fractions what are known and unknown, Egyptian mathematics, Egyptian mathematics, Mega fun fractions. This worksheet asks kids to solve fraction word problems about food. Old Egyptian Math Cats knew fractions like 1/2 or 1/4 (one piece of a pie). you can see that fractions take part of this. They never wrote: 1/4 + 1/4 + 1/4 = 3/4 Found worksheet you are looking for? Introduce the idea of Egyptian Fractions to the class. Egyptian fractions are the reciprocals of the positive integers where the numerator is always one. In this fraction addition worksheet, ... Egyptian Fractions For Teachers 6th. Some of the worksheets for this concept are Egyptian mathematics, Egyptian mathematics, Egyptian fractions representations as sums of unit fractions, Egyptian mathematics, A complete overview for all those who never, Count like an egyptian, Second edition teachers guide 4, Math mammoth fractions 1 worktext. Egyptian Fraction. This worksheet is designed as a short exercise in drawing human figures in the style of the Ancient Egyptians. Math worksheets: Converting decimals to proper fractions. But to make fractions like 3/4, they had to add pieces of pies like 1/2 + 1/4 = 3/4. In this first lesson we have a look at the sum of two Egyptian Fractions to see if we can get another Egyptian Fraction. This means that our Egyptian Fraction representation for 4/5 is 4/5 = 1/2 + 1/4 + 1/20; Scholars of ancient Egypt (ca. This math worksheet will give your child practice identifying fractions of shapes and filling in the missing numbers in fractions. 1. Cross-curricular activities link fractions to language arts, music, science, art, and social studies. 4.1 Egyptian Fractions. This set features one-step addition and subtraction inequalities such as “5 + x > 7″ and “x – 3″ 21″. (Maybe ¾, too. Types of Fractions. Alex made a poster of Egyptian-style fractions, from 1/2 to 9/10. :). As they work, circulate around the room offering encouragement, quietly assessing progress, and answering questions. Finally we suggest a … Identify a proper fraction, improper fraction, mixed number, unit fraction, like fractions, unlike fractions like a pro with these printable types of fractions worksheets. These fractions worksheets are great for practicing how to add fractional inch measurements that you would find on a tape measure. after each example, there is an exercise consisting of 5 questions. 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Eye of Horus numeration system Grade 1 > fractions the Egyptians were hooked on fractions with numerator....	2022-08-19 17:22:28	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.4083396792411804, "perplexity": 2142.7190751652697}, "config": {"markdown_headings": true, "markdown_code": false, "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/1659882573744.90/warc/CC-MAIN-20220819161440-20220819191440-00421.warc.gz"}
https://www.rosettacommons.org/node/10118	# ERROR: Assertion Fdensity.u2 and Fdensity.u3 failed 14 posts / 0 new ERROR: Assertion Fdensity.u2 and Fdensity.u3 failed #1 I'm using "rosetta_src_2017.36.59679_bundle". I'm trying to fit some strucutres to an EM density map. My script exited with the below error. Please help me troubleshoot it and let me know if you need further details. This is the script that's I'm running: <ROSETTASCRIPTS> <SCOREFXNS> <ScoreFunction name="cen" weights="score4_smooth_cart"> <Reweight scoretype="elec_dens_fast" weight="20"/> </ScoreFunction> <ScoreFunction name="dens_soft" weights="beta_soft"> <Reweight scoretype="cart_bonded" weight="0.5"/> <Reweight scoretype="pro_close" weight="0.0"/> <Reweight scoretype="elec_dens_fast" weight="%%denswt%%"/> </ScoreFunction> <ScoreFunction name="dens" weights="beta_cart"> <Reweight scoretype="elec_dens_fast" weight="%%denswt%%"/> <Set scale_sc_dens_byres="R:0.76,K:0.76,E:0.76,D:0.76,M:0.76,C:0.81,Q:0.81,H:0.81,N:0.81,T:0.81,S:0.81,Y:0.88,W:0.88,A:0.88,F:0.88,P:0.88,I:0.88,L:0.88,V:0.88"/> </ScoreFunction> </SCOREFXNS> <MOVERS> <SetupForDensityScoring name="setupdens"/> <SwitchResidueTypeSetMover name="tocen" set="centroid"/> <MinMover name="cenmin" scorefxn="cen" type="lbfgs_armijo_nonmonotone" max_iter="200" tolerance="0.00001" bb="1" chi="1" jump="ALL"/> <CartesianSampler name="cen5_50" automode_scorecut="-0.5" scorefxn="cen" mcscorefxn="cen" fascorefxn="dens_soft" strategy="auto" fragbias="density" rms="%%rms%%" ncycles="200" fullatom="0" bbmove="1" nminsteps="25" temp="4" fraglens="7" nfrags="25"/> <CartesianSampler name="cen5_60" automode_scorecut="-0.3" scorefxn="cen" mcscorefxn="cen" fascorefxn="dens_soft" strategy="auto" fragbias="density" rms="%%rms%%" ncycles="200" fullatom="0" bbmove="1" nminsteps="25" temp="4" fraglens="7" nfrags="25"/> <CartesianSampler name="cen5_70" automode_scorecut="-0.1" scorefxn="cen" mcscorefxn="cen" fascorefxn="dens_soft" strategy="auto" fragbias="density" rms="%%rms%%" ncycles="200" fullatom="0" bbmove="1" nminsteps="25" temp="4" fraglens="7" nfrags="25"/> <CartesianSampler name="cen5_80" automode_scorecut="0.0" scorefxn="cen" mcscorefxn="cen" fascorefxn="dens_soft" strategy="auto" fragbias="density" rms="%%rms%%" ncycles="200" fullatom="0" bbmove="1" nminsteps="25" temp="4" fraglens="7" nfrags="25"/> <ReportFSC name="report" testmap="%%testmap%%" res_low="10.0" res_high="%%reso%%"/> <BfactorFitting name="fit_bs" max_iter="50" wt_adp="0.0005" init="1" exact="1"/> <FastRelax name="relaxcart" scorefxn="dens" repeats="1" cartesian="1"/> </MOVERS> <PROTOCOLS> </PROTOCOLS> <OUTPUT scorefxn="dens"/> </ROSETTASCRIPTS> This is the error I'm getting: protocols.rosetta_scripts.ParsedProtocol: =======================BEGIN MOVER ReportFSC - report======================= rho_calc 6807 of 6807 core.scoring.electron_density.ElectronDensity: Forcing apix to 1.375,1.375,1.375 protocols.rosetta_scripts.ParsedProtocol: [ ERROR ] Exception while processing procotol: protocols.jd2.JobDistributor: [ ERROR ] [ERROR] Exception caught by JobDistributor for job tubulin_PF_dep_noligand_0001 [ ERROR ] EXCN_utility_exit has been thrown from: src/core/scoring/electron_density/ElectronDensity.cc line: 1053 ERROR: Assertion Fdensity.u1()==Fdensity2.u1() && Fdensity.u2()==Fdensity2.u2() && Fdensity.u3()==Fdensity2.u3() failed. protocols.jd2.JobDistributor: [ WARNING ] tubulin_PF_dep_noligand_0001 reported failure and will NOT retry protocols.jd2.JobDistributor: no more batches to process... protocols.jd2.JobDistributor: 1 jobs considered, 1 jobs attempted in 30641 seconds protocols::checkpoint: Deleting checkpoints of FastRelax protocols::checkpoint: Deleting checkpoints of FastRelax protocols::checkpoint: Deleting checkpoints of FastRelax protocols::checkpoint: Deleting checkpoints of FastRelax protocols::checkpoint: Deleting checkpoints of FastRelax protocols::checkpoint: Deleting checkpoints of FastRelax Error: [ ERROR ] Exception caught by rosetta_scripts application:1 jobs failed; check output for error messages Error: [ ERROR ] ERROR: Assertion Fdensity.u1()==Fdensity2.u1() && Fdensity.u2()==Fdensity2.u2() && Fdensity.u3()==Fdensity2.u3() failed. ERROR:: Exit from: src/core/scoring/electron_density/ElectronDensity.cc line: 1053 BACKTRACE: /home/labusr/rosetta/main/source/build/src/release/linux/3.10/64/x86/gcc/4.8/default/libutility.so(utility::exit(std::string const&, int, std::string const&, int)+0x27f) [0x7f02d445448f] /home/labusr/rosetta/main/source/build/src/release/linux/3.10/64/x86/gcc/4.8/default/libcore.3.so(core::scoring::electron_density::ElectronDensity::getFSC(ObjexxFCL::FArray3D<std::complex<double> > const&, ObjexxFCL::FArray3D<std::complex<double> > const&, unsigned long, double, double, utility::vector1<double, std::allocator<double> >&, bool)+0x213) [0x7f02d6a6b6e3] /home/labusr/rosetta/main/source/build/src/release/linux/3.10/64/x86/gcc/4.8/default/libprotocols.3.so(protocols::electron_density::ReportFSC::apply(core::pose::Pose&)+0x11fd) [0x7f02da4f9a6d] /home/labusr/rosetta/main/source/build/src/release/linux/3.10/64/x86/gcc/4.8/default/libprotocols.1.so(protocols::rosetta_scripts::ParsedProtocol::apply_mover(core::pose::Pose&, protocols::rosetta_scripts::ParsedProtocol::MoverFilterPair const&)+0x22c) [0x7f02d924f30c] /home/labusr/rosetta/main/source/build/src/release/linux/3.10/64/x86/gcc/4.8/default/libprotocols.1.so(protocols::rosetta_scripts::ParsedProtocol::sequence_protocol(core::pose::Pose&, __gnu_cxx::__normal_iterator<protocols::rosetta_scripts::ParsedProtocol::MoverFilterPair const*, std::vector<protocols::rosetta_scripts::ParsedProtocol::MoverFilterPair, std::allocator<protocols::rosetta_scripts::ParsedProtocol::MoverFilterPair> > >)+0x3b) [0x7f02d924ff3b] /home/labusr/rosetta/main/source/build/src/release/linux/3.10/64/x86/gcc/4.8/default/libprotocols.1.so(protocols::rosetta_scripts::ParsedProtocol::apply(core::pose::Pose&)+0x1a7) [0x7f02d9250707] /home/labusr/rosetta/main/source/build/src/release/linux/3.10/64/x86/gcc/4.8/default/libprotocols.1.so(protocols::jd2::JobDistributor::run_one_job(std::shared_ptr<protocols::moves::Mover>&, long, std::string&, std::string&, unsigned long&, unsigned long&, bool)+0xb78) [0x7f02d9313528] /home/labusr/rosetta/main/source/build/src/release/linux/3.10/64/x86/gcc/4.8/default/libprotocols.1.so(protocols::jd2::JobDistributor::go_main(std::shared_ptr<protocols::moves::Mover>)+0xc1) [0x7f02d9314f71] /home/labusr/rosetta/main/source/build/src/release/linux/3.10/64/x86/gcc/4.8/default/libprotocols.1.so(protocols::jd2::FileSystemJobDistributor::go(std::shared_ptr<protocols::moves::Mover>)+0x4a) [0x7f02d92ec6ba] /home/labusr/rosetta/main//source/bin/rosetta_scripts.linuxgccrelease() [0x403408] /lib64/libc.so.6(__libc_start_main+0xf5) [0x7f02d2449c05] /home/labusr/rosetta/main//source/bin/rosetta_scripts.linuxgccrelease() [0x403d4b] ./Tublin_exercise_refine.sh: line 8: -ignore_unrecognized_res: command not found Post Situation: Wed, 2017-12-06 07:38 I wish someone would help with this. Thanks a lot. Mon, 2017-12-11 08:30 Hey sorry for the delay. It looks like you are trying to do an FSC calculation using two maps of different volume. You'll need to make both maps the same size to get it to run. This should be doable in chimera without too much trouble. -Brandon Mon, 2017-12-11 09:49 brandon.frenz Thanks a lot. I'm still new to chimera, so I wodon't mind at all if you tell me how to do it, or refer me to a tutorial. Thanks a lot. Mon, 2017-12-11 12:40 I'm not actually sure I haven't had to do it myself. I would check out this to see if there's something you can use on there. https://www.cgl.ucsf.edu/chimera/docs/UsersGuide/midas/vop.html Wed, 2017-12-13 18:51 brandon.frenz Thanks. I actually have been trying a few stuff like using molmap and masking then fitting the two maps, however, I'm not entire sure what you mean by having the same volume. Could you please elaborate on that? I'm still getting that eror unless I use the exact same map. Mon, 2017-12-18 11:51 The maps are basically a 3 dimensional grid where every point has a value associated with it. The grid sizes need to be the same to run a FSC calculation. So if the size of your grid for the first map is 10x10x10 the size of the grid for the second map has to be the same. I believe the resample command is the one you want to use https://www.cgl.ucsf.edu/chimera/docs/UsersGuide/midas/vop.html#resample Fri, 2017-12-22 14:12 brandon.frenz Thanks a lot. I looked into that, the two maps actually are taken from an original map by doing a 10 angstrom molmap and then masking. Because of that, they should have the same grid size. I tried 1) fitting in the two maps together 2) changing the origin index from map viewer->features_>coordinates to make them match, I still keep getting this error. I attached the two maps here, and I would appreciate if you take a look at them: https://drive.google.com/drive/folders/1Cfw3IB2DHezZ5j2zdnlY_J33dcUWVuGD?usp=sharing Sun, 2018-01-21 21:01 Any help would be appreciated. Sun, 2018-01-28 09:48 You ran it through with Rosetta using the maps with the same coordinates after aligning? I am not sure what the issue is but I will ask the author of the code to take a look. Have you tried running the chimera command vop resample #1 onGrid #2? Tue, 2018-01-30 11:04 brandon.frenz I just tried vop resample, I think it worked, but let me try it one more time to verify it. Tue, 2018-02-27 09:00 Yep, to confirm this works. However, let me ask you, what is the FSC being done "between" exactly? The two map volumes? If so, why? It would make more sense to do some kind of other cross correlation to validate the model generated in the other map, not the map volumes themselves. Am I assuming something wrong here? Regards. Mon, 2018-03-05 10:50	2019-10-23 18:45:46	{"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": 0.6463028192520142, "perplexity": 11489.279168934534}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-43/segments/1570987835748.66/warc/CC-MAIN-20191023173708-20191023201208-00232.warc.gz"}
https://physics.stackexchange.com/questions/276742/demonstration-of-general-relativity?noredirect=1	Demonstration of General relativity I've been trying to at least roughly understand gravitation in general relativity, but so far I haven't found a demonstration which I'd find satisfactory. The most common demonstration is description via objects with various weights on a trampoline, demonstrated with this image: I just cannot accept this demonstration as they try to explain gravitation via gravitation (of the weights of the objects on the trampoline). I found this demonstration which I find quite plausible, but I haven't found any references to it nor whether it is a correct (source is here: http://spark.sciencemag.org/generalrelativity/): They explain gravitation via warping time part of the spacetime, however I find it a bit too simplified. Is there some better demonstrations or some other relevant ones? Can you confirm the quality of the second demonstration? • All pictures, demonstrations, illustrations etc assume an inherently false notion that somehow 4 D spacetime can be demonstrated by 3 D pictures drawn usually on a 2 D surface. Ultimately, imo its much easier to take the GR equation and spend a little while trying to to understand that, seriously, than any incorrect picture or diagram. And you get the true idea. – user108787 Aug 27 '16 at 19:56 • The second picture is actually pretty good as far as simplifications go. This question reminds me of xkcd #895. – EL_DON Aug 27 '16 at 23:13 • – Qmechanic Aug 27 '16 at 23:24 The second diagram is a hundred percent correct. One of the great principles of general relativity - the equivalence principle - can be used to reduce the diagram to one in special relativity. We care about a person in freefall, jumping from the roof of a house and hitting the ground. The equivalence principle states that this is the same as someone standing in a spaceship which is accelerating at 9.81 meters per second squared. So we can consider an equivalent problem: The spaceship is accelerating at 9.81 meters per second squared, and passes a freefalling person, who "falls" from the nose of the spaceship to its tail. In the frame of the person (the free-falling observer), the spaceship (or ground) is accelerating towards him, and he is in flat spacetime! The equivalence principle states that "the ground rushing up to meet you" is a 100% accurate phrase. Let's work this problem out completely. First we have to solve for what path the spaceship traces out. To make the units nice, I assume the rear of the spaceship is accelerating at $9\cdot 10^{16}\mbox{ meters per second}^2$ - a big number to compensate for the axes being "meters" in my graph. It turns out the $(ct,x)$ coordinates of the rear of the spaceship trace out the path $(\sinh(k),\cosh(k)-1)$ (measured in meters), and the front of the spaceship traces out the path $(ct_2,x_2)=((1+\ell)\sinh(k),(1+\ell)\cosh(k)-1)$, where $\ell$ is the proper length of the spaceship in meters. I plug in $\ell=0.5\mbox{ meters}$. This looks like the following: Note that the coordinate position of the nose of the spacecraft slowly approaches the rear of the spacecraft due to length contraction. This is all described in flat, Minkowski spacetime. A freefalling observer feels none of the acceleration forces, and so simply moves in a straight line on this graph: This is your original picture! Note that the reason the acceleration is so high on this graph is that one tick mark on the horizontal axis (the time axis) is only 3.3 nanoseconds! The real version, for our 9.81 meters per second squared, would just be a less exaggerated version of this graph. Due to length contraction, the tip of the spacecraft is always moving a tiny bit slower than the rear of the spacecraft. Time dilation states that time ticks by more slowly for faster moving objects. Therefore, time ticks by on the tip of the spacecraft a tiny bit faster than on the rear of the spacecraft, according to our freefalling observer. This is exactly as stated in your example, where time ticks by faster on the roof than on the ground. We've now solved the problem in the frame of our freefalling observer, but we could move back to the frame of the spaceship using the equivalence principle. Once we do so, our coordinates are warped into Rindler coordinates. In our warped Rindler coordinates, free-falling objects no longer follow straight lines, but instead satisfy the geodesic equation. Weirdly enough, all of this has nothing to do with the warping of spacetime. This example takes place completely in flat spacetime. While we can describe an elevator this way, we can't describe the whole spherical Earth this way. The whole surface of the Earth can't be "accelerating outwards" all at the same time... unless spacetime is curved. Explanation of the formulas. In special relativity, if the back of the spaceship with a constant proper acceleration $g$, then it travels in a spacetime hyperbola. $(ct,x)=(\frac{c^2}{g}\sinh(\frac{g \tau}{c}),\frac {c^{2}}{g}\left(\cosh {\frac {g\ \tau }{c}}-1\right))$, where $\tau$ is the proper time at the back of the spaceship. We don't actually care about the proper time at the back of the spaceship, so we can just define $k=\frac{g \tau}{c}$ and $a=\frac {c^{2}}{g}$. I'll choose $a=1\mbox{ meter}$, representing an acceleration of $g=\frac{(3\cdot 10^8\mbox{ meters per second})^2}{1\mbox{ meter}}=9\cdot 10^{16}\mbox{ meters per second}^2$. This is a lot more than Earth's 9.81 meters per second squared, but it makes the math a lot nicer! So, where the units are in meters, the back of the spaceship traces out the path: $(ct_1,x_1)=(\sinh(k),\cosh(k)-1)$ Once we know what a constantly accelerating observer looks like, we can come up with Rindler coordinates. We conclude that the front of the spaceship traces out the path $(ct_2,x_2)=((1+\ell)\sinh(k),(1+\ell)\cosh(k)-1)$. This is another spacetime hyperbola with a slightly smaller acceleration. It corresponds to constant coordinates in Rindler coordinates, separated by a distance $\ell$. I initially incorrectly reasoned as follows: I want the spaceship to always have proper length $\ell$, so I'll say that the front of the spaceship is $(ct,x+\frac{\ell}{\gamma})$, where $\gamma=\frac{1}{\sqrt{1-v^2/c^2}}$ is the length contraction factor and $v$ is the velocity of the spaceship. We have $\frac{v}{c}=\frac{dx}{c dt}=\frac{\sinh(k)dk}{\cosh(k)dk}=\tanh{k}$. Plugging this in and using hyperbolic trig identities, we find $\gamma=\cosh{k}$, and that the front of the spaceship traces out the path $(\sinh(k),\cosh(k)-1+\frac{\ell}{\cosh(k)})$. This is inconsistent because it implies the tip is moving more slowly than the rear. I can't derive the length contraction factor by assuming the whole body of the spaceship is moving at the same velocity, and then conclude that the tip is moving slower than the rear! (with respect to our inertial observer) • This is so interesting. So is it accurate (though strange) to say that I am pulled downwards on earth because time moves slower for my feet than my head (assuming I am standing up)? – killdash9 Sep 3 at 17:46 Here is the best picture of all to get an idea of General Relativity. If you are prepared to put a bit of time into understanding the above picture, you will understand GR much better than any other method, without distortion and using the original source material. In general terms it's saying that the curvature of 4 dimensional space on the left hand side is balanced by the amount on mass energy on the right hand side. So $R_{v\mu}$ is the part of the equation that tells you how much spacetime curves in the presence of mass/energy. The next term $\frac {1}{2}Rg_{v\mu}$ is a term that, amongst other things involves differentiation (no offence intended but I am guessing you know the basic idea of differentiation, if a function A = function B, then their derivatives should also equal each other, so Einstein needed to include this term). The last term $\Lambda g_{v\mu }$ is called the cosmological constant, Einstein had to include this term, otherwise his equation would have predicted that gravitation would have pulled the galaxies back together a long time ago. He needed a term to act as a repulsive force to balance gravitational pull inwards as, at the time of his equation, he believed the universe was static. This is the term that is now linked with the Dark Energy concept. On the right hand side is $T_{v\mu }$ (the numbers in front of it you probably know already are Newton's Gravitational constant and c, the speed of light to the power of 4), which are used to balance both sides of the equation. The term $T_{v\mu }$ is an indication of how much mass/energy exists in a given region, this region could include the area around the Earth, all the way outwards, right up to the size of the visible universe. Hyperphysics or Wikipedia tells you much more about this equation, which is not really that complicated, and in my opinion is certainly not any more complicated that trying to figure out what an artist is trying to describe with , say trampoline shaped space. The only way we 3 D creatures can truly understand 4D spacetime, is through math.	2019-10-17 06:04:55	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 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": 0.8356039524078369, "perplexity": 332.4780682939284}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-43/segments/1570986672723.50/warc/CC-MAIN-20191017045957-20191017073457-00463.warc.gz"}
https://testbook.com/question-answer/the-frequency-of-a-sinusoidal-signal-is-50-hz-wha--60b1d70a120210989d1d01d1	The frequency of a sinusoidal signal is 50 Hz. What will be the period of the signal? Free Practice With Testbook Mock Tests Options: 1. 20 ms 2. 30 ms 3. 50 ms 4. 10 ms Correct Answer: Option 1 (Solution Below) This question was previously asked in PGCIL Diploma Trainee EE Official Paper (Held on 17 Dec 2020) Solution: Concept: Frequency: The number of waves that pass a fixed point in unit time; also, the number of cycles or vibrations undergone during one unit of time by a body in periodic motion is called frequency. The frequency of a signal, given the time period, is given by: $$f=\frac{1}{T}$$ The time period is, therefore: $$T=\frac{1}{f}$$ This is explained with the help of the following sine wave diagram: That means the frequency is inversely proportional to the time period. For example: Let, the time period of the wave = 10 μsec, i.e. T = 10 × 10-6 sec ∴ The frequency will be: $$f=\frac{1}{T}=\frac{1}{10× 10^{-6}}$$ f = 0.1 × 106 Hz f = 100 × 103 Hz Since 103 Hz = 1 kHz, the above frequency becomes: f = 100 kHz Calculation: In a periodic waveform time period (T) is inverse of frequency (f) T = 1/f Given, f = 50 Hz T= 1/50 T= 0.02 s = 20 ms	2021-07-31 14:16:29	{"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": 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": 0.7541629076004028, "perplexity": 2182.8176330264196}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-31/segments/1627046154089.68/warc/CC-MAIN-20210731141123-20210731171123-00015.warc.gz"}
https://openastronomy.org/rcsc18/chapters/05-writing-effective-tests/03-testing-your-code	# Tests and Exceptions In this lesson we will look at how to make your code more reliable by writing tests. Tests when used cleverly can give you a lot of confidence in your code and therefore your results! def square(x): return x*3 Tests can take many forms, they can compare your code against known results i.e. ones in a published paper, or they can just test that the result of some function returns the type of object you expect or even just check that your code results always stays the same, so you know if something breaks. A simple test for our square function we defined above might look like: def test_square(): assert square(10) == 1010 ## The assert statement As we will see later, the way to make a test fail is to raise an error. Therefore the assert statement in Python is a useful shortcut when writing tests. The assert statement will raise an error if a condition is not satisfied. The general form of the assert statement is: assert condition, "message" i.e. assert 5 == 6, "Five is not equal to six" We can run our test function the same way that we run any other function. Although this dosent scale very well to thousands of tests! test_square() --------------------------------------------------------------------------- AssertionError Traceback (most recent call last) <ipython-input-3-5351b87f150c> in <module>() ----> 1 test_square() <ipython-input-2-0fd7f2671b21> in test_square() 1 def test_square(): ----> 2 assert square(10) == 1010 AssertionError: ## Writing Tests The following function has bugs in it. Write some tests below the function to find all the bugs. def quadratic(a, b, c): return ((-1 b) * np.sqrt(b*2 - 4ac) / (2 a)) import numpy as np def test_quadratic(): Once you have a few test functions written, you will probably start getting bored with typing out their names and running them one-by-one. There are a few different modules to help you write and run tests. The one we will be using is called pytest. For the next section of this session we will be using the two Python (.py) files named askdl.py and lsadkj.py.	2022-01-21 07:43:01	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 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": 0.19191613793373108, "perplexity": 1440.3642731555267}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320302740.94/warc/CC-MAIN-20220121071203-20220121101203-00660.warc.gz"}
https://www.physicsforums.com/threads/how-do-i-check-my-solution.813101/	# How do I check my solution? 1. May 10, 2015 ### randy_1981 Would somebody be so kind as to tell how I can check my solution to this problem(s) below: Find parametric equations for the line through the points P=(-2,0,3) and Q=(3,5,2). More specifically, I feel that I should be able to substitute some value for t or (x,y,z) to confirm my solutions. Thanks! 2. May 10, 2015 ### Staff: Mentor What did you get as solution? You should be able to plug in some values of t into your solution to get your two points. 3. May 11, 2015 ### HallsofIvy Staff Emeritus If you have parametric equations, take any two values for the parameter and check if the points you get are on the same lie as the two given points. Depending upon how you find the parametric equations, it should be easy to choose the parameters so you get the given points. Of course, a line is determined by two points so you only need to check two. 4. May 11, 2015 ### randy_1981 Ok, I'm about to reveal my ignorance but how do I determine what the parameter values are? 5. May 11, 2015 ### randy_1981 My solution is r(t)=< 5t-2, 5t , 3-5t > 6. May 11, 2015 ### Staff: Mentor Is there some value of the parameter t so that r(t) = (-2, 0, 3)? Is there another value of t so that r(t) = (3, 5, 2)? If the answers to these questions aren't obvious by inspection, set up an equation with <5t - 2, 5t, 3 - 5t> on one side, and either of the given points on the other side, and solve for t. 7. May 13, 2015 ### randy_1981 Gottca, thanks for the help! I'm undergoing a self study program and some things that should be obvious aren't at times. 8. May 13, 2015 ### Ray Vickson Basically, that is what "through" means in this type of problem When you say the line passes through the points P=(-2,0,3) and Q=(3,5,2), this means that for some value of t you will have <5t - 2, 5t, 3 - 5t> = <-2,0,3>, while for some other value of t you will have <5t - 2, 5t, 3 - 5t> = <3,5,2>. Componentwise, <a,b,c> = <d,e,f> means a = d, b = e and c = f. BTW: another (easier?) way is to note that the line through $P = \langle -2,0,3 \rangle$ and $Q = \langle 3,5,2 \rangle$ has the form $$(1-t) P + t Q = (1-t) \langle -2,0,3 \rangle + t \langle 3,5,2 \rangle$$ For $0 \leq t \leq 1$ the points are on the line-segment joining P and Q (with point = P when t = 0 and point = Q when t = 1). For t < 0 the points are on the line "before" P (that is, on the side away from Q), while for t > 1 they are on the line "after" Q (that is, on the side away from P). 9. May 17, 2015 ### HallsofIvy Staff Emeritus If you are referring to something like "$3x- 7\le 2$ so $3x\le 9$, $x\le 3$, you can't do that for functions that are not "one to one". 10. May 17, 2015 ### Staff: Mentor You can always simplify $3x- 7\le 2$ to $x \le 3$ (assuming real numbers). And all the functions that might come up here are injective.	2017-08-21 20:16:34	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5107524394989014, "perplexity": 671.9161756308266}, "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/1502886109525.95/warc/CC-MAIN-20170821191703-20170821211703-00117.warc.gz"}
https://ai.stackexchange.com/questions/27027/predicting-the-probability-of-a-periodically-happening-event-occurring-at-a-give	# Predicting the probability of a periodically happening event occurring at a given time I have encountered this problem on how to predict the probability of a periodically happening event occurring at a given time. For example, we have an event called being_an_undergrad. There are many data points: bob is an undergrad from (1999 - 2003), Bill is an undergrad from (1900 - 1903), Alice is an undergrad from (1900 - 1905), and there are many other data points such as (2010 - 2015), (2011 - 2013) .... There are many events(data points) of being_an_undergrad. The lasting interval varies, it might be 1 year, 2 years, 3 years, .... or even 10 years. But the majority is around 4 years. However, I am wondering given all the data points above. If I now know that Jason starts college in 2021, and how can I calculate/predict the probability that he will still be an undergrad in 2022? and 2023? and 2024 .... 2028, etc. My current dataset consists of 10000 tuples representing events of different relations. The relations are all continuous relations similar to the example above. There are about 10 continuous relations in total in this dataset, such as isMarriedTo, beingUndergrad, livesIn, etc. For each relation, there are about 1000 data points(1000 durations) about this relation, for example, <Leo, isUndergrad, Harvard, 2010 - 2011>, <Leo, isUndergrad, Stanford, 2013 - 2016>..... <Jason, livesIn, US, 1990 - 2021>, <Richard, livesIn, UK, 1899- 1995> ... My problem now is that I want to get a confidence level(probability) when I want to predict one event happening at a specific time point. For example, I want to predict the probability that event <Jason, livesIn, US, 2068> happens, given: 1.the above datasets which includes info about the relation: livesIn 2.the starting time when Mike lives in US, say he started to live in US since 2030. I have used normal distribution to simulate, but I am wondering if there are any other better AI / ML / Stats approaches. Thanks a lot! • If the only data you have is samples from population of period lengths, and you want to predict probability of a new randomly sampled one (from the same population), then this looks like a basic question in statistics, which is not really an AI question. If you still think this is an AI question, could you explain where AI is involved? For instance, if there was a lot more contextual data about the time periods, then this might be a suitable problem for machine learning. If that is the case, please use edit to explain roughly what contextual data you have. Mar 27 at 11:06 • Hi @NeilSlater! Thanks a lot for your comment. Yup, initially I thought of using statistics approaches such as survival function or poisson distribution. But their accuracy is not that good. Thus, I am thinking of using AI/ML to do the prediction. Actually from my point of view, AI is also like a way of modelling(just like physics/math). For example, we can also use ML/DL to learn the Newton's rules right haha(by inputting lots of experiment data to train the Newton's rule's model). Thus, I was wondering if there are similar time-series approaches in ML/DL that could tackle this problem. Mar 27 at 12:05 • Unless there is some context that you have not explained in your data, a simple statistics approach will get the best accuracy, most ML methods should get close to it but will not be better, and can easily be worse. ML cannot improve on basic probabilities when the prediction is essentially a guessing game. Mar 27 at 13:45 • Advanced AI might even be able to use information such as the student's name, because that may be correlated with demographics that do affect length of the course and likelihood to complete it. It is still not clear from your question that this is what you are asking though. Mar 27 at 13:49 • Hi @NeilSlater! Thank you so much for your very detailed explanation : ) Actually in my context, I have a dataset of continuous relations similar to the example above. There are about 10 relations in total, such as isMarriedTo, beingUndergrad, livesIn, etc. For each relation, there are about 1000 data points about this relation(1000 durations), such as <Leo, isUndergrad, 2010 - 2011> and <Leo, isUndergrad, 2013 - 2016> ........ Mar 27 at 14:40	2021-12-07 13:12:38	{"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": 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": 0.6029223203659058, "perplexity": 833.8908506471048}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-49/segments/1637964363376.49/warc/CC-MAIN-20211207105847-20211207135847-00190.warc.gz"}
https://learn.careers360.com/ncert/question-solve-the-following-pairs-of-equations-by-reducing-them-to-a-pair-of-linear-equations-v-7x-minus-2y-by-xy-equals-5-8x-plsu-7y-by-xy-equals-15/	Q # Solve the following pairs of equations by reducing them to a pair of linear equations: (v) 7x - 2y / xy = 5 8x + 7y / xy = 15 Q1. Solve the following pairs of equations by reducing them to a pair of linear equations: (v) $\\\frac{7x - 2y}{xy} = 5\\ \frac{8x + 7y}{xy} = 15$ Views Given Equations, $\\\frac{7x - 2y}{xy} = 5\\\\\Rightarrow\frac{7}{y} -\frac{2}{x}=5\\ \frac{8x + 7y}{xy} = 15\\\Rightarrow \frac{8}{y}+\frac{7}{x}=15$ Let, $\frac{1}{x}=p\:and\:\frac{1}{y}=q$ Now, our equation becomes $7q-2p=5........(1)$ And $8q+7p=15..........(2)$ By Cross Multiplication method, $\frac{q}{b_1c_2-b_2c_1}=\frac{p}{c_1a_2-c_2a_1}=\frac{1}{a_1b_2-a_2b_1}$ $\frac{q}{(-2)(-15)-(7)(-5)}=\frac{p}{(-5)(8)-(-15)(7)}=\frac{1}{(7)(7)-(8)(-2)}$ $\frac{q}{30+35}=\frac{p}{-40+105}=\frac{1}{49 +16}$ $\frac{q}{65}=\frac{p}{65}=\frac{1}{65}$ $p=1,\:and\:q=1$ And Hence, $x=1\:and\:y=1.$ Exams Articles Questions	2020-02-21 22:33:07	{"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": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 11, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7584539651870728, "perplexity": 927.1086355543937}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-10/segments/1581875145538.32/warc/CC-MAIN-20200221203000-20200221233000-00024.warc.gz"}
https://www.physicsforums.com/threads/change-of-basis.15697/	# Change of basis 1. ### Norman 922 Hello all, I need some help... If I know the form of a wavefunction in the $S_z$ basis, say it is spin up, how do I convert that to a wavefunction expressed in the $S_x$ basis? Is there a very simple way to do this? Thanks 2. ### Norman 922 Anyone??? Do I just rotate the spinor using a rotation operator? Help? 3. ### Tom Mattson 5,538 Staff Emeritus You've basically got it. To convert from Sz to Sx, you rotate the state vector p/2 rad about the y-axis.	2015-08-30 09:56:45	{"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": 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": 0.6193105578422546, "perplexity": 1477.2679413690742}, "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-2015-35/segments/1440644065241.14/warc/CC-MAIN-20150827025425-00052-ip-10-171-96-226.ec2.internal.warc.gz"}
https://byjus.com/question-answer/time-taken-by-an-electron-to-complete-one-revolution-in-the-bohr-orbit-of-hydrogen/	Question # Time taken by an electron to complete one revolution in the Bohr orbit of hydrogen atom in terms of its radius r and quantum number n is: A 4π2mr2nh B 2πrnh C π2r28πr2 D nh28πr Solution ## The correct option is B 4π2mr2nhWe need to find time taken by an e− to complete one revolution in Bohr orbit of H-atom in terms of radius ′r′ and quantum no. ′n′. We have two relations; Tn=2πrnvn ...(i) mvnrn=nh2π ...(ii) where, Tn=Time period rn=radius vn=velocity n=Orbit number From equation (ii), vn=nh2πmrn ...(iii) Putting this value in equation (i), Tn=2πrnnh2πmrn=4π2mr2nhChemistry Suggest Corrections 1 Similar questions View More People also searched for View More	2022-01-23 15:20:22	{"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": 0.847915530204773, "perplexity": 6542.498458232335}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320304287.0/warc/CC-MAIN-20220123141754-20220123171754-00137.warc.gz"}
https://doc.sagemath.org/html/en/reference/groups/sage/groups/raag.html	Right-Angled Artin Groups¶ A right-angled Artin group (often abbreviated as RAAG) is a group which has a presentation whose only relations are commutators between generators. These are also known as graph groups, since they are (uniquely) encoded by (simple) graphs, or partially commutative groups. AUTHORS: • Travis Scrimshaw (2013-09-01): Initial version • Travis Scrimshaw (2018-02-05): Made compatible with ArtinGroup class sage.groups.raag.CohomologyRAAG(R, A) The cohomology ring of a right-angled Artin group. The cohomology ring of a right-angled Artin group $$A$$, defined by the graph $$G$$, with coefficients in a field $$F$$ is isomorphic to the exterior algebra of $$F^N$$, where $$N$$ is the number of vertices in $$G$$, modulo the quadratic relations $$e_i \wedge e_j = 0$$ if and only if $$(i, j)$$ is an edge in $$G$$. This algebra is sometimes also known as the Cartier-Foata algebra. REFERENCES: class Element An element in the cohomology ring of a right-angled Artin group. algebra_generators() Return the algebra generators of self. EXAMPLES: sage: C4 = graphs.CycleGraph(4) sage: A = groups.misc.RightAngledArtin(C4) sage: H = A.cohomology() sage: H.algebra_generators() Finite family {0: e0, 1: e1, 2: e2, 3: e3} degree_on_basis(I) Return the degree on the basis element clique. EXAMPLES: sage: C4 = graphs.CycleGraph(4) sage: A = groups.misc.RightAngledArtin(C4) sage: H = A.cohomology() sage: sorted([H.degree_on_basis(I) for I in H.basis().keys()]) [0, 1, 1, 1, 1, 2, 2] gen(i) Return the i-th standard generator of the algebra self. This corresponds to the i-th vertex in the graph (under a fixed ordering of the vertices). EXAMPLES: sage: C4 = graphs.CycleGraph(4) sage: A = groups.misc.RightAngledArtin(C4) sage: H = A.cohomology() sage: H.gen(0) e0 sage: H.gen(1) e1 gens() Return the generators of self (as an algebra). EXAMPLES: sage: C4 = graphs.CycleGraph(4) sage: A = groups.misc.RightAngledArtin(C4) sage: H = A.cohomology() sage: H.gens() (e0, e1, e2, e3) ngens() Return the number of algebra generators of self. EXAMPLES: sage: C4 = graphs.CycleGraph(4) sage: A = groups.misc.RightAngledArtin(C4) sage: H = A.cohomology() sage: H.ngens() 4 one_basis() Return the basis element indexing $$1$$ of self. EXAMPLES: sage: C4 = graphs.CycleGraph(4) sage: A = groups.misc.RightAngledArtin(C4) sage: H = A.cohomology() sage: H.one_basis() () class sage.groups.raag.RightAngledArtinGroup(G, names) Bases: sage.groups.artin.ArtinGroup The right-angled Artin group defined by a graph $$G$$. Let $$\Gamma = \{V(\Gamma), E(\Gamma)\}$$ be a simple graph. A right-angled Artin group (commonly abbreviated as RAAG) is the group $A_{\Gamma} = \langle g_v : v \in V(\Gamma) \mid [g_u, g_v] \text{ if } \{u, v\} \notin E(\Gamma) \rangle.$ These are sometimes known as graph groups or partially commutative groups. This RAAG’s contains both free groups, given by the complete graphs, and free abelian groups, given by disjoint vertices. Warning This is the opposite convention of some papers. Right-angled Artin groups contain many remarkable properties and have a very rich structure despite their simple presentation. Here are some known facts: • The word problem is solvable. • They are known to be rigid; that is for any finite simple graphs $$\Delta$$ and $$\Gamma$$, we have $$A_{\Delta} \cong A_{\Gamma}$$ if and only if $$\Delta \cong \Gamma$$ [Dro1987]. • They embed as a finite index subgroup of a right-angled Coxeter group (which is the same definition as above except with the additional relations $$g_v^2 = 1$$ for all $$v \in V(\Gamma)$$). • In [BB1997], it was shown they contain subgroups that satisfy the property $$FP_2$$ but are not finitely presented by considering the kernel of $$\phi : A_{\Gamma} \to \ZZ$$ by $$g_v \mapsto 1$$ (i.e. words of exponent sum 0). • $$A_{\Gamma}$$ has a finite $$K(\pi, 1)$$ space. • $$A_{\Gamma}$$ acts freely and cocompactly on a finite dimensional $$CAT(0)$$ space, and so it is biautomatic. • Given an Artin group $$B$$ with generators $$s_i$$, then any subgroup generated by a collection of $$v_i = s_i^{k_i}$$ where $$k_i \geq 2$$ is a RAAG where $$[v_i, v_j] = 1$$ if and only if $$[s_i, s_j] = 1$$ [CP2001]. The normal forms for RAAG’s in Sage are those described in [VW1994] and gathers commuting groups together. INPUT: • G – a graph • names – a string or a list of generator names EXAMPLES: sage: Gamma = Graph(4) sage: G = RightAngledArtinGroup(Gamma) sage: a,b,c,d = G.gens() sage: acd^4a^-3b v0^-2v1v2v3^4 sage: Gamma = graphs.CompleteGraph(4) sage: G = RightAngledArtinGroup(Gamma) sage: a,b,c,d = G.gens() sage: acd^4a^-3b v0v2v3^4v0^-3v1 sage: Gamma = graphs.CycleGraph(5) sage: G = RightAngledArtinGroup(Gamma) sage: G Right-angled Artin group of Cycle graph sage: a,b,c,d,e = G.gens() sage: dbad v1v3^2v0 sage: e^-1cbeb^-1c^-4 v2^-3 We create the previous example but with different variable names: sage: G.<a,b,c,d,e> = RightAngledArtinGroup(Gamma) sage: G Right-angled Artin group of Cycle graph sage: dbad bd^2a sage: e^-1cbeb^-1*c^-4 c^-3 REFERENCES: class Element(parent, lst) Bases: sage.groups.artin.ArtinGroupElement An element of a right-angled Artin group (RAAG). Elements of RAAGs are modeled as lists of pairs [i, p] where i is the index of a vertex in the defining graph (with some fixed order of the vertices) and p is the power. cohomology(F=None) Return the cohomology ring of self over the field F. EXAMPLES: sage: C4 = graphs.CycleGraph(4) sage: A = groups.misc.RightAngledArtin(C4) sage: A.cohomology() Cohomology ring of Right-angled Artin group of Cycle graph with coefficients in Rational Field gen(i) Return the i-th generator of self. EXAMPLES: sage: Gamma = graphs.CycleGraph(5) sage: G = RightAngledArtinGroup(Gamma) sage: G.gen(2) v2 gens() Return the generators of self. EXAMPLES: sage: Gamma = graphs.CycleGraph(5) sage: G = RightAngledArtinGroup(Gamma) sage: G.gens() (v0, v1, v2, v3, v4) sage: Gamma = Graph([('x', 'y'), ('y', 'zeta')]) sage: G = RightAngledArtinGroup(Gamma) sage: G.gens() (vx, vy, vzeta) graph() Return the defining graph of self. EXAMPLES: sage: Gamma = graphs.CycleGraph(5) sage: G = RightAngledArtinGroup(Gamma) sage: G.graph() Cycle graph: Graph on 5 vertices ngens() Return the number of generators of self. EXAMPLES: sage: Gamma = graphs.CycleGraph(5) sage: G = RightAngledArtinGroup(Gamma) sage: G.ngens() 5 one() Return the identity element $$1$$. EXAMPLES: sage: Gamma = graphs.CycleGraph(5) sage: G = RightAngledArtinGroup(Gamma) sage: G.one() 1 one_element() Return the identity element $$1$$. EXAMPLES: sage: Gamma = graphs.CycleGraph(5) sage: G = RightAngledArtinGroup(Gamma) sage: G.one() 1	2021-02-24 19:52:36	{"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": 0.7593479156494141, "perplexity": 1841.2633689176557}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-10/segments/1614178347321.0/warc/CC-MAIN-20210224194337-20210224224337-00246.warc.gz"}
https://dsp.stackexchange.com/questions/28294/implementation-of-dct-for-mfcc	# Implementation of DCT for MFCC I am in the process of implementing MFCC and I'm unsure about the DCT part. This is how I implemented it (using DCT II). //some coefficients let k = Math.PI/numFilters; let w1 = 1.0/Math.sqrt(2); let w2 = Math.sqrt(2.0/numFilters); let numCoeffs = numFilters; let dctMatrix = new Float32Array(numCoeffsnumFilters); // compute the dct matrix for(let i = 0; i < numCoeffs; i++){ for (let j = 0; j < numFilters; j++) { let idx = i + (jnumCoeffs); if(i === 0){ dctMatrix[idx] = w1 * Math.cos(k * (i+1) * (j+0.5)); } else{ dctMatrix[idx] = w2 * Math.cos(k * (i+1) * (j+0.5)); } } } // apply it on the logged energies. let mfccs = new Float32Array(numCoeffs); for (let k = 0; k < numCoeffs; k++) { let v = 0; for (let n = 0; n < numFilters; n++) { let idx = k + (nnumCoeffs); v += (dctMatrix[idx] loggedMelBands[n]); } mfccs[k] = v; } I'm getting 26 values (which is the number of filters I defined), both negative and positive values, where the first value is always has highest magnitude and the rest are in the range of -1 and 1. Does this implementation seem correct to you? • why are you implementing a very common algorithm yourself? – Marcus Müller Jan 19 '16 at 10:16 • Why not ? :D I want to learn. – nevos Jan 20 '16 at 15:41 • fair point, really! I think the best way to verify it would be checking it against a reference implementation. Get your octave ready :) – Marcus Müller Jan 20 '16 at 16:15	2020-01-27 09:16:45	{"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": 0.272409051656723, "perplexity": 3320.917443733794}, "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/1579251696046.73/warc/CC-MAIN-20200127081933-20200127111933-00344.warc.gz"}
http://www.askiitians.com/forums/Analytical-Geometry/24/44462/surface-area-and-volume.htm	A closed cone is placed with water half of its height as shown below in fig 1 . If it will completely revolved then water goes down and height will change . So at what height from the base water will stay? Find X. 2 years ago Share More Questions On Analytical Geometry Post Question Vouchers To Win!!! The equation of a straight line passing through the point (-5,4) and which cuts an intercept of root(2) units between the lines x+y+1=0 and x+y-1 is? Hello Student. Please find the solutionx+y+1=0 and x+y-1 are two parallel lines and distance between them is mod\|\frac{1-(-1)}{\sqrt{2}} \|So distabce between them is 2/root2 = root2So our... Nishant Vora one month ago Find the equation of the line perpendicular to the line x-7y+5=0 and having x-intercept 3 7x+y-21=0 is the eqn perpendicular and having x-intercept to be 3 VINAYREDDY one year ago 7x+y-21=0 Akhil sai one year ago Tangents are drawn from the point (α,β) to the hyperbola 3x 2 -2y 2 =6 and are inclined at angles θ and φ to the x-axis. If tanθ.tanφ =2, prove that β 2... Sunil Raikwar 9 months ago Hi Pranjal, There is slight technical issue. Please post these questions again in analytical Geometry. We will upload the answers for the same. askIITians Faculty sunil raikwar 9 months ago Given equation is Equations of tangents are the roots of this equation is therefore Thanks &Regards, Sunil Raikwar, askIITians faculty. sunil raikwar 9 months ago R=(7+4root3)^{2n}=I+f find R(1-f) Hello Student, G=(7-4root3)^{2n}=Even integer. I+f+G=an even integer. f+G=1=>G=1-f =>R(1-f)=RG=1 Thanks & Regards Arun Kumar Btech, IIT Delhi Askiitians Faculty Arun Kumar 4 months ago What is the Intregration of 1/sin x + cos x dx ?	2014-11-23 12:10:40	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 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": 0.6818310618400574, "perplexity": 6772.903469450067}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-49/segments/1416400379466.35/warc/CC-MAIN-20141119123259-00108-ip-10-235-23-156.ec2.internal.warc.gz"}
https://ch.mathworks.com/help/symbolic/meijerg.html	# meijerG Meijer G-function ## Description example meijerG(a,b,c,d,z) returns the Meijer G-Function. meijerG is element-wise in z. The input parameters a, b, c, and d are vectors that can be empty, as in meijerG([], [], 3.2, [], 1). ## Examples collapse all syms x meijerG(3, [], [], 2, 5) ans = 25 Call meijerG when z is an array. meijerG acts element-wise. a = 2; z = [1 2 3]; meijerG(a, [], [], [], z) ans = 0.3679 1.2131 2.1496 Convert numeric input to symbolic form using sym, and find the Meijer G-function. For certain symbolic inputs, meijerG returns exact symbolic output using other functions. meijerG(sym(2), [], [], [], sym(3)) ans = 3exp(-1/3) meijerG(sym(2/5), [], sym(1/2), [], sym(3)) ans = (2^(4/5)3^(1/2)gamma(1/10))/80 For symbolic variables or expressions, meijerG returns an output in terms of simple or special functions. syms a b c d z f = meijerG(a,b,c,d,z) f = (gamma(c - a + 1)(1/z)^(1 - a)hypergeom([c - a + 1, d - a + 1],... b - a + 1, 1/z))/(gamma(b - a + 1)gamma(a - d)) Substitute values for the variables by using subs, and convert values to double by using double. fVal = subs(f, [a b c d z], [1.2 3 5 7 9]) fVal = (2669^(1/5)hypergeom([24/5, 34/5], 14/5, 1/9))/(25gamma(-29/5)) double(fVal) ans = 5.7586e+03 Calculate fVal to higher precision using vpa. vpa(fVal) ans = 5758.5946416377834597597497022199 Show relations between meijerG and simpler functions for given parameter values. Show that when a, b, and d are empty, and c = 0, then meijerG reduces to exp(-z). syms z meijerG([], [], 0, [], z) ans = exp(-z) Show that when a, b, and d are empty, and c = [1/2 -1/2], then meijerG reduces to 2Kv(1,2z1/2). meijerG([], [], [1/2 -1/2], [], z) ans = 2besselk(1, 2*z^(1/2)) Plot the real and imaginary values of the Meijer G-function for values of b and z, where a = [-2 2] and c and d are empty. Fill the contours by setting Fill to on. syms b z f = meijerG([-2 2], b, [], [], z); subplot(2,2,1) fcontour(real(f),'Fill','on') title('Real Values of Meijer G') xlabel('b') ylabel('z') subplot(2,2,2) fcontour(imag(f),'Fill','on') title('Imag. Values of Meijer G') xlabel('b') ylabel('z') ## Input Arguments collapse all Input, specified as a number or vector, or a symbolic number, variable, vector, function, or expression. Input, specified as a number or vector, or a symbolic number, variable, vector, function, or expression. Input, specified as a number or vector, or a symbolic number, variable, vector, function, or expression. Input, specified as a number or vector, or a symbolic number, variable, vector, function, or expression. Input, specified as a number or vector, or a symbolic number, variable, vector, function, or expression. collapse all ### Meijer G-Function The Meijer G-function meijerG([a1,…,an], [an + 1,…,ap], [b1,…,bm], [bm + 1,…,bq], z) is a general function that includes other special functions as particular cases, and is defined as ${G}_{p,q}^{m,n}\left(\begin{array}{c}{a}_{1},\dots ,{a}_{p}\\ {b}_{1},\dots ,{b}_{q}\end{array}\|z\right)=\frac{1}{2\pi \text{i}}\int \frac{\left(\prod _{j=1}^{m}\Gamma \left({b}_{j}-s\right)\right)\left(\prod _{j=1}^{n}\Gamma \left(1-{a}_{j}+s\right)\right)}{\left(\prod _{j=m+1}^{q}\Gamma \left(1-{b}_{j}+s\right)\right)\left(\prod _{j=n+1}^{p}\Gamma \left({a}_{j}-s\right)\right)}{z}^{s}ds.$ ## Algorithms For the Meijer G-function meijerG([a1,…,an], [an + 1,…,ap], [b1,…,bm], [bm + 1,…,bq], z), for ai ∊ (a1,…,an) and bj ∊ (b1,…,bm), no pair of parameters ai − bj should differ by a positive integer. The Meijer G-function involves a complex contour integral with one of the following types of integration paths: • The contour goes from - i ∞ to i ∞ so that all poles of $\Gamma \left({b}_{j}-s\right)$, j = 1, …, m lie to the right of the path, and all poles of $\Gamma \left(1-{a}_{k}+s\right)$, k = 1, …, n lie to the left of the path. The integral converges if $c=m+n-\frac{p+q}{2}>0$, \|arg(z)\| < c π. If \|arg(z)\| = c π, c ≥ 0, the integral converges absolutely when p = q and (ψ) < - 1, where $\Psi =\left(\sum _{j=1}^{q}{b}_{j}\right)-\left(\sum _{i=1}^{p}{a}_{i}\right)$. When pq, the integral converges if you choose the contour so that the contour points near i ∞ and - i ∞ have a real part σ satisfying $\left(q-p\right)\sigma >\Re \left(\psi \right)+1-\frac{q-p}{2}$. • The contour is a loop beginning and ending at and encircling all poles of $\Gamma \left({b}_{j}-s\right)$, j = 1, …, m moving in the negative direction, but none of the poles of $\Gamma \left(1-{a}_{k}+s\right)$, k = 1, …, n. The integral converges if q ≥ 1 and either p < q or p = q and \|z\| < 1. • The contour is a loop beginning and ending at - ∞ and encircling all poles of $\Gamma \left(1-{a}_{k}+s\right)$, k = 1, …, n moving in the positive direction, but none of the poles of $\Gamma \left({b}_{j}+s\right)$, j = 1, …, m. The integral converges if p ≥ 1 and either p > q or p = q and \|z\| > 1. The integral represents an inverse Laplace transform or, more specifically, a Mellin-Barnes type of integral. For a given set of parameters, the contour chosen in the definition of the Meijer G-function is the one for which the integral converges. If the integral converges for several contours, all contours lead to the same function. The Meijer G-function satisfies a differential equation of order max(p, q) with respect to a variable z: $\left({\left(-1\right)\right)}^{m+n-p}z\left(\prod _{i=1}^{p}\left(z\frac{d}{dz}-{a}_{i}-1\right)\right)-\prod _{j=1}^{q}\left(z\frac{d}{dz}-{b}_{j}\right)\right){G}_{p,q}^{m,n}\left(\begin{array}{c}{a}_{1},\dots ,{a}_{p}\\ {b}_{1},\dots ,{b}_{p}\end{array}\|z\right)=0.$ If p < q, this differential equation has a regular singularity at z = 0 and an irregular singularity at z = ∞. If p = q, the points z = 0 and z = ∞ are regular singularities, and there is an additional regular singularity at z = (−1)m + n - p. The Meijer G-function represents an analytic continuation of the hypergeometric function [1]. For particular choices of parameters, you can express the Meijer G-function through the hypergeometric function. For example, if no two of the bh terms, h = 1, …, m, differ by an integer or zero and all poles are simple, then ${G}_{p,q}^{m,n}\left(\begin{array}{c}{a}_{1},\dots ,{a}_{p}\\ {b}_{1},\dots ,{b}_{p}\end{array}\|z\right)=\sum _{h=1}^{m}\frac{\left(\prod _{j=1\dots m,j\ne h}\Gamma \left({b}_{j}-{b}_{h}\right)\right)\left(\prod _{j=1}^{n}\Gamma \left(1+{b}_{h}-{a}_{j}\right)\right)}{\left(\prod _{j=m+1}^{q}\Gamma \left(1+{b}_{h}-{b}_{j}\right)\right)\left(\prod _{j=n+1}^{p}\Gamma \left({a}_{j}-{b}_{h}\right)\right)}{z}^{b}{h}_{p}{F}_{q-1}\left({A}_{h};{B}_{h};{\left(-1\right)}^{p-m-n}z\right).$ Here p < q or p = q and \|z\| < 1. Ah denotes ${A}_{h}=1+{b}_{h}-{a}_{1},\dots ,1+{b}_{h}-{a}_{p}.$ Bh denotes ${B}_{h}=1+{b}_{h}-{b}_{1},\dots ,1+{b}_{h}-{b}_{\left(h-1\right)},1+{b}_{h}-{b}_{h+1},\dots ,1+{b}_{h}-{b}_{q}.$ Meijer G-functions with different parameters can represent the same function. • The Meijer G-function is symmetric with respect to the parameters. Changing the order inside each of the following lists of vectors does not change the resulting Meijer G-function: [a1, …, an], [an + 1, …, ap], [b1, …, bm], [bm + 1, …, bq]. • If z is not a negative real number and z ≠ 0, the function satisfies the following identity: ${G}_{p,q}^{m,n}\left(\begin{array}{c}{a}_{1},\dots ,{a}_{p}\\ {b}_{1},\dots ,{b}_{q}\end{array}\|z\right)={G}_{q,p}^{n,m}\left(\begin{array}{c}1-{b}_{1},\dots ,1-{b}_{p}\\ 1-{a}_{1},\dots ,1-{a}_{p}\end{array}\|\frac{1}{z}\right).$ . • If 0 < n < p and r = a1 - ap is an integer, the function satisfies the following identity: ${G}_{p,q}^{m,n}\left(\begin{array}{c}{a}_{1},{a}_{2},\dots ,{a}_{p-1},{a}_{p}\\ {b}_{1},{b}_{2},\dots ,{b}_{q-1},{b}_{q}\end{array}\|z\right)={G}_{p,q}^{m,n}\left(\begin{array}{c}{a}_{p},{a}_{2},\dots ,{a}_{p-1},{a}_{1}\\ {b}_{1},{b}_{2},\dots ,{b}_{q-1},{b}_{q}\end{array}\|z\right).$ . • If 0 < m < q and r = b1 - bq is an integer, the function satisfies the following identity: ${G}_{p,q}^{m,n}\left(\begin{array}{c}{a}_{1},{a}_{2},\dots ,{a}_{p-1},{a}_{p}\\ {b}_{1},{b}_{2},\dots ,{b}_{q-1},{b}_{q}\end{array}\|z\right)={\left(-1\right)}^{\gamma }{G}_{p,q}^{m,n}\left(\begin{array}{c}{a}_{1},{a}_{2},\dots ,{a}_{p-1},{a}_{p}\\ {b}_{q},{b}_{2},\dots ,{b}_{q-1},{b}_{1}\end{array}\|z\right).$ . According to these rules, the meijerG function call can return meijerG with modified input parameters. ## References [1] Luke, Y. L., The Special Functions and Their Approximations. Vol. 1. New York: Academic Press, 1969. [2] Prudnikov, A. P., Yu. A. Brychkov, and O. I. Marichev, Integrals and Series. Vol 3: More Special Functions. Gordon and Breach, 1990. [3] Abramowitz, M., I. A. Stegun, Handbook of Mathematical Functions. 9th printing. 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