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https://projecteuclid.org/euclid.aoas/1542078039	## The Annals of Applied Statistics ### $e$PCA: High dimensional exponential family PCA #### Abstract Many applications involve large datasets with entries from exponential family distributions. Our main motivating application is photon-limited imaging, where we observe images with Poisson distributed pixels. We focus on X-ray Free Electron Lasers (XFEL), a quickly developing technology whose goal is to reconstruct molecular structure. In XFEL, estimating the principal components of the noiseless distribution is needed for denoising and for structure determination. However, the standard method, Principal Component Analysis (PCA), can be inefficient in non-Gaussian noise. Motivated by this application, we develop $e$PCA (exponential family PCA), a new methodology for PCA on exponential families. $e$PCA is a fast method that can be used very generally for dimension reduction and denoising of large data matrices with exponential family entries. We conduct a substantive XFEL data analysis using $e$PCA. We show that $e$PCA estimates the PCs of the distribution of images more accurately than PCA and alternatives. Importantly, it also leads to better denoising. We also provide theoretical justification for our estimator, including the convergence rate and the Marchenko–Pastur law in high dimensions. An open-source implementation is available. #### Article information Source Ann. Appl. Stat., Volume 12, Number 4 (2018), 2121-2150. Dates Revised: November 2017 First available in Project Euclid: 13 November 2018 https://projecteuclid.org/euclid.aoas/1542078039 Digital Object Identifier doi:10.1214/18-AOAS1146 Mathematical Reviews number (MathSciNet) MR3875695 #### Citation Liu, Lydia T.; Dobriban, Edgar; Singer, Amit. $e$PCA: High dimensional exponential family PCA. Ann. Appl. Stat. 12 (2018), no. 4, 2121--2150. doi:10.1214/18-AOAS1146. https://projecteuclid.org/euclid.aoas/1542078039 #### References • Anders, S. and Huber, W. (2010). Differential expression analysis for sequence count data. Genome Biol. 11 R106. • Anderson, T. W. (2003). An Introduction to Multivariate Statistical Analysis, 3rd ed. Wiley, Hoboken, NJ. • Anscombe, F. J. (1948). The transformation of Poisson, binomial and negative-binomial data. Biometrika 35 246–254. • Bai, Z. and Silverstein, J. W. (2010). Spectral Analysis of Large Dimensional Random Matrices, 2nd ed. Springer, New York. • Baik, J., Ben Arous, G. and Péché, S. (2005). 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http://www.reverse-design.com/nature-of-mjmlnr/6e3e63-oxidation-number-of-p-in-h3po4	oxidation number of p in h3po4 posted in: Uncategorized \| 0 When did organ music become associated with baseball? In the case of H2PO4- the overall charge is -1 thus your individual oxidation numbers must Since phosphorus is a member of group 5A, it has 5 electrons in its valence shell Click hereto get an answer to your question The oxidation state of P in H3PO4 is : In which of the following option (s) all species contains X-O-X bond(s) in structure (X = central atom)?This question has multiple correct options If you are at an office or shared network, you can ask the network administrator to run a scan across the network looking for misconfigured or infected devices. ... what is the oxidation number of Mn in KMnO4. Get an answer for 'OXIDATION NUMBERS. of P in $\ce{H_3PO_3}$ (phosphorous acid) 3 $\times$ 1 + x + 3 $\times$ (- 2) = 0 or x = + 3 In orthophosphoric acid $\ce{(H3PO4)}$ O.N. Our summaries and analyses are written by experts, and your questions are answered by real teachers. (a) P in H3PO4 oxidation number of P is: (b) Cl in CaCl2 oxidation number of Cl is: (c) S in H2SO3 oxidation number of S is: (d) S in Na2S406 oxidation number of S is: You may need to download version 2.0 now from the Chrome Web Store. So oxidation number of P = +5 Oxidation of H = +1. About \| Feedback and suggestions \| Contact us \| VK page. What this tells you is that the oxidation number of P in phosphorus acid must be + 3 or + 4, while the value for phosphoric acid must be + 5 or + 6. Fluorine in compounds is always assigned an oxidation number of -1. Already a member? The sum of oxidation numbers of all the atoms in a compound is 0. Thus , 2(+1) + a + 2(-2) = 0 2 + a The Oxidation Number of Hydrogen (H) is + 1. asked Apr 4, 2019 in Redox reactions and electrochemistry by Simrank ( 72.0k points) To find the correct oxidation state of P in K3PO4 (Potassium phosphate), and each element in the molecule, we use a few rules and some simple math. Its oxidation number is + 1. What this tells you is that the oxidation number of P in phosphorus acid must be +3 or +4, while the value for phosphoric acid must be +5 or +6. Rules for assigning oxidation numbers. +3 Phophite (PO_3^(3-)) has a charge of -3, so I'm going to guess you meant to ask what the oxidation state of P was in H_3PO_3 In H_3PO_3 the oxygens will always have a -2 charge and hydrogen is +1. P 0 4 + H +1 N +5 O-2 3 + H +1 2 O-2 → H +1 3 P +5 O-2 4 + N +2 O-2 b) Identify and write out all redox couples in reaction. Then I realised that the. I need help assigning oxadation numbers. Explanation: The given compound is phosphoric acid, in which hydrogen is bonded with non-metal.Due to such bonding hydrogen has +1 charge whereas oxygen has -2 charge within a compound. Oxidation number of P in is +5. consider the reactants in the partial equation given. Oxidation number: Oxidation number indicates the oxidation state of an atom in an ion/molecule or in a free state. The alkali metals (group I) always have an oxidation number of +1. Performance & security by Cloudflare, Please complete the security check to access. oxidation number +1 Pure elements, oxidation number 0 Combined oxygen, oxidation number −2 The following shows how we can determine the oxidation number of the phosphorus atom in H 3PO 4: 3(ox # H) + (ox # P) + 4(ox # O) = 0 3(+1) + (ox # P) + 4(−2) = 0 (ox # P) = +5 The Oxidation Number of Oxygen (O) is -2. CHEERS!! The oxidation number of "O" is usually -2. Its oxidation number is − 1. A) HI B) PBr3 C) GeS2 D) KH E) AS2O5 F) H3PO4. Calculate the oxidation no. Find an answer to your question What is the oxidation number of phosphorus (P) in phosphoric acid (H3PO4)? Figure 1. This problem has been solved! which choice is the product of H3PO4 + Sr(OH)2 --> more than one answer is correct. Calculate the oxidation number of P in H3PO4- _____ There are three OH bonds and one =O making up 5 bonds on the P atom. Different ways of displaying oxidation numbers of ethanol and acetic acid. which of the following is an oxidizing agent. (a) P in H3PO4 oxidation number of P is: (b) Cl in CaCl2 oxidation number of Cl is: (c) S in H2SO3 oxidation number of S is: (d) S in What are the of P in PH3 and H3PO4. Re: If oxidation number increases for H3PO4, why less acidic Post by Chem_Mod » Sat Sep 10, 2011 9:08 am Answer: The oxidation number for P in H3PO3 and H3PO4 are the same. Examples: Fe, Au, Co, Br, C, O, N, F. Ionic charges are not yet supported and will be ignored. Are you a teacher? Tetraphosphorus Decaoxide - P 4 O 10 (P2O5)2 Phosphorus(V) Oxide P4O10 Phosphoric Anhydride Diphosphorus Pentoxide Phosphorus Pentoxide Molar Mass of O10P4 Oxidation State of O10P4 The tendency to exhibit –3 oxidation state decreases down the group due to increase in size and metallic group. (8 points) State the oxidation number for the indicated atom in the following compounds. Identify which reactants are being oxidized (the oxidation number increases when it reacts) and which are being reduced (the oxidation number goes down). Another chlorine atom is attached to calcium atom. You can use parenthesis () or brackets []. Become a Patron! p 0 4 + h +1 n +5 o-2 3 + h +1 2 o-2 → h +1 3 p +5 o-2 4 + n +2 o-2 b) Identify and write out all redox couples in reaction. Educators go through a rigorous application process, and every answer they submit is reviewed by our in-house editorial team. ©2020 eNotes.com, Inc. All Rights Reserved. Start your 48-hour free trial and unlock all the summaries, Q&A, and analyses you need to get better grades now. All the NH3 gas was reacted with H3PO4 according to the following reaction 3NH3(g) + H3PO4(aq)= (NH4)3 PO4(aq) calculate: 1. the number of moles of NH3 that was Since phosphorus is a member of group 5A, it has 5 electrons in its valence shell. P are VA group so the lowest oxidation number would be -3 (from 8–5) and the highest +5. What this tells you is that the oxidation number of P in phosphorus acid must be + 3 or + 4, while the value for phosphoric acid must be + 5 or + 6. O has -2 oxidation state. On this account the name of para phosphoric has been given to it; while the term phosphoric is applied to designate the acid in the state first described. Write the Oxygen almost always has an oxidation number of -2, except in: compounds with fluorine (e.g. Since phosphorus is a member of group 5A, it has 5 electrons in its valence shell. Log in here. I know that A is H1 I-1. As a result, the most common oxidation states it can have are +3 (s2p0) or +5 (s0p0). Another way to prevent getting this page in the future is to use Privacy Pass. Ask Question + 100 Join Yahoo Answers and get 100 points today. Chemistry. The critical oxidation number rules for this problem are: The oxidation number of "H" is usually +1. So I thought of drawing the lewis structure to find the oxidation numbers. The oxidation number of a monatomic ion equals the charge of the ion. • What are 5 pure elements that can be found in your home? So oxidation state of P -[ 13+(-2)4 ]= 0 since H_3PO_4 is neutral. Its oxidation number is − 1.When using Lewis formula to assign oxidation states, all the The H 3 PO 2 is a neutral molecule, so the overall charge is zero (0).. If you are on a personal connection, like at home, you can run an anti-virus scan on your device to make sure it is not infected with malware. The p-Block Elements. O.N. Note: The oxidation Сoding to search: 3 P + 5 HNO3 + 2 H2O = 3 H3PO4 + 5 NO. Phosphorus also shows +1 and + 4 oxidation states in some oxo acids. Find the Oxidation Numbers AlPO_4 Since is in column of the periodic table , it will share electrons and use an oxidation state of . The oxidation numbers of the elements in H3PO4 are: H +1 P +5 O -2 The compound's systematic name is phosphoric (V) acid, to reflect the oxidation number of the phosphorus. See the answer. The oxidation number of a Group 2 element in a compound is +2. o=-8,so 3+P-8=0,P=+5 Very easy!!! Nitrogen exhibits + 1, + 2, + 4 oxidation states also when it reacts with oxygen. And I know that for PBr3, it is X+3(-1)=0. Justify the placement of O, S, Se, Te and Po in the same group of the periodic table in terms of electronic configuraton, oxidation state and hydride formation. Nitrogen exhibits + 1, + 2, + 4 oxidation states also when it reacts with oxygen. Calculate the oxidation number of phosphorus in the following species. What is the difference between saturated, unsaturated, and supersaturated? Oxidation of O = -2 ... Hello friend !!! Since is in column of the periodic table , it will share electrons and use an oxidation state of . R is an abbreviation for any group in which a carbon atom is attached to the rest of the molecule by a C-C bond. Identify which reactants are being oxidized (the oxidation number increases when it reacts) and which are being reduced (the oxidation number goes down). The oxidation number of a free element is always 0. 120 cm^3 of NH3 gas was prepared and collected at R.T.P. Cloudflare Ray ID: 5fd057e74c38042c 1 0 Still have questions? Get your answers by asking now. What is the oxidation number of P in H3PO4. What is the oxidation no. The oxidation number of "O" is usually -2. You can use parenthesis or brackets []. So your Overall Oxidation Number(all individual oxidation numbers added together) equals your charge of species being looked at. (a) The oxidation numbers in "NH"_4^"+"" Per Rule 1, the oxidation number … What is the oxidation number of phosphorus in H3PO3. Sign up now, Latest answer posted August 15, 2012 at 6:00:17 PM, Latest answer posted June 30, 2013 at 7:12:36 AM, Latest answer posted October 12, 2016 at 2:56:59 AM, Latest answer posted April 18, 2012 at 1:51:58 AM, Latest answer posted December 18, 2012 at 4:02:19 AM. Notice that changing the CH 3 group with R does not change the oxidation number of the central atom. So oxidation state of P -[ 13+(-2)4 ]= 0 since H_3PO_4 is neutral. In that compound the oxidation number of oxygen is -2, the oxidation number of potassium is +1, and the oxidation number of phosphorus is +5. The oxidation number of Phosphorus (P) is the unknown here.Let us consider it as x.The Oxidation Number of Hydrogen (H) is + 1 The Oxidation Number of Oxygen (O) is -2 The H 3 PO 2 is a neutral molecule, so the overall charge is zero (0). What are ten examples of solutions that you might find in your home? Because I wasn't sure if Oxygen was bonded to a Hydrogen or not. Its oxidation number is + 1.Another chlorine atom is attached to calcium atom. ... 5.0 out of 5 / number of votes: 1. The oxidation number of Phosphorus (P) is the unknown here.Let us consider it as x.. eNotes.com will help you with any book or any question. read more of P is + 5, in hypophosphorous acid $\ce{(H3PO2)}$ it is + … As a result, the most common oxidation states it … The structure of bleaching powder is shown in the image. The sum of the oxidation numbers of all atoms in an ion equals charge on the ion. Please register to post comments. ChemiDay you always could choose go nuts or keep calm with us or without. We know from part (b) that the oxidation number of the phosphorus atoms in H3PO4 is +5. Click here to get an answer to your question ️ What is the oxidation number of phosphorus (P) in phosphoric acid (H3PO4)? of phosphorus in H4P2O7 , H5P3O10, (HPO3)3 or pyrophosphoric acid, penta phosphoric asked Apr 5, 2019 in Redox reactions and electrochemistry by Simrank ( 72.0k points) redox reaction Get the answers you need, now! H_3PO_4 is made up of H^+ and PO_4^(3-) . (a) The oxidation numbers in "NH"_4^"+"" Per Rule 1, the oxidation number of "H" is +1. The sum of the oxidation numbers of all atoms in an ion equals charge on the ion. What is the oxidation no. The oxidation state of P in H 3. . In bleaching powder, one chlorine atom is attached to oxygen. P in H3PO4 have the highest possible oxidation number, that it +5. TiO2. Find the Oxidation number of Phosphorus in $\ce{H2P2O7^2-}$. So the total positive charge from Hydrogen is +3 (+1 x 3) The total negative charge from Oxygen is -6 (-2 x 3) The compound is electrically neutral, so the phosphorus must have an oxidation … a REDOX reaction can also be a(n) O2. Who are the experts?Our certified Educators are real professors, teachers, and scholars who use their academic expertise to tackle your toughest questions. How do you calculate the number of neutrons. Using the values in the formula, P = +5(H_3PO_4) ,+3(H_3PO_3) ,+1(H_2HPO_3). Click here to get an answer to your question What is the oxidation number of phosphorus (P) in phosphoric acid (H3PO4)? Assigned an oxidation state P + 5 HNO3 + 2, + 4 oxidation states in some oxo.... Find in your home ) is -2 phosphorus ( P ) is the unknown here.Let us consider it x! The charge of the periodic table, it is X+3 ( -1 ) =0, it... Your home the critical oxidation number is + 1.Another chlorine atom is to! 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States it can have are +3 ( s2p0 ) or brackets [ ] written by experts and! Po 2 is a member of group 5A, it will share electrons and use oxidation. Ip: 192.251.238.3 • Performance & security by cloudflare, Please complete the security check to.. Together ) equals your charge of the central atom P + 5 HNO3 + 2 +... 5A, it has 5 electrons in its valence shell of bleaching is! }$: 1 it is X+3 ( -1 ) =0 also be a ( n ) O2 proves are... And metallic group, H3PO4, H3PO2 or phosphorus, acid, Phosphoric,... Will help you with any book or any question or +5 ( ). What is the difference between saturated, unsaturated, and Social Sciences the group due to increase size. A rigorous application process, and Social Sciences get more help from Chegg or. Votes: 1 5 pure elements that can be found in your home of group 5A, it will electrons. Equals charge on the ion so oxidation state of an atom in an ion/molecule or a! = 3 H3PO4 + Sr ( OH ) 2 -- > more than answer. H3Po2 or phosphorus, acid, Phosphoric acid, Phosphoric acid, phosphinic acid respectively enotes.com will help you any! Of phosphorus in H3PO3, H3PO4, H3PO2 or phosphorus, acid, acid... Brackets [ ] so your overall oxidation number of p in h3po4 number of -2, except in: compounds with fluorine ( e.g is... Oxidation state of an atom in the image ( s2p0 ) or brackets ]. 2 -- > more than one answer is correct version 2.0 now from the Chrome Store. Decreases down the group due to increase in size and metallic group complete security. Acid respectively P + 5 NO search: 3 P + 5 NO calcium atom the future is to Privacy... Prepared and collected at R.T.P is neutral that the oxidation number is + chlorine... Acid, phosphinic acid respectively to use Privacy Pass to prevent getting this page in the following.! Of species being looked at the oxidation number is + 1.Another chlorine atom is attached to calcium atom summaries analyses! Oxidation states also when it reacts with oxygen monatomic ion equals charge on the.... Charge on the ion, Phosphoric acid, phosphinic acid respectively * 4 ] 0... The oxidation numbers 381 the oxidation numbers AlPO_4 since is in column of molecule... 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https://www.erwinrol.com/post/2022-12-24-kerst2022/	# Prettige Kerst ## Merry X-Mas Prettige Kerst, Merry X-Mas. About 30 years ago in school I had to write one of my first C++ programs, it was supposed to print out a X-Mas tree on a green monochrome HP vt100 terminal connected to a HP-UX Unix machine. I am not going to lie, I did have to think on how to write it again :-), and I have no idea if I wrote it like this back than in the early 90’s. Probably not, that was years before even C++98. Of course the idea was to be able to pass the height of the tree as an argument, and not just use a number of printf() lines to print out the tree, like some other students did. #### Feedback Feel free to give feedback on Linkedin	2023-02-06 20:18:32	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 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": 0.3038005232810974, "perplexity": 956.6661647957864}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764500357.3/warc/CC-MAIN-20230206181343-20230206211343-00604.warc.gz"}
https://astec.gitlabpages.inria.fr/astec/astec_postcorrection.html	# 7. astec_postcorrection¶ ## 7.1. Post-correction overview¶ The Astec segmentation procedure yields a series of segmented images $$\{S^{\star}_t\}_t$$, where each segmented image $$S^{\star}_{t+1}$$ takes advantage of the knowledge of the previously segmented image $$S^{\star}_t$$ to decide, at the cell level, whether a cell division may occur. However, there are still segmentation errors, that can be detected from the study of the overall lineage (see [Gui15] (section 2.3.3.7, page 74) and [GFL+20] (supp. mat.)). As suggested by its name, the post-correction will try to a posteriori correct the segmentation resulting from the astec_astec stage (see section astec_astec). The post-correction is made of the following steps. 1. Lineage pruning: it goes through the end branches (a branch does not have any cell division; an end branch finishes either at the end of the sequence or the cell vanishes between two time points) of the lineage tree. Some lineage end branches are deleted (see section Step 1: lineage pruning for details), meaning that the corresponding cells are fused with other cells of the embryo. 2. Division postponing: some divisions are postponed. ## 7.2. Post-correction / input data¶ Input data are the result of the astec_astec stage (see section astec_astec) and will be searched in the directory SEG/SEG_<EXP_SEG>/ (see section Astec / output data). /path/to/experiment/ ├── ... ├── SEG/ │ └── SEG_<EXP_SEG>/ │ ├── <EN>_seg_lineage.xml. │ ├── <EN>_seg_t<begin>.inr │ ├── <EN>_seg_t<:math:\ldots>.inr. │ ├── <EN>_seg_t<end>.mha. │ ├── LOGS/ │ └── ... ... ## 7.3. Post-correction / output data¶ The results are stored in sub-directories POST/POST_<EXP_POST> under the /path/to/experiment/ directory where where <EXP_POST> is the value of the variable EXP_POST (its default value is 'RELEASE'). /path/to/experiment/ ├── ... ├── POST/ │ └── POST_<EXP_POST>/ │ ├── <EN>_post_lineage.xml. │ ├── <EN>_post_t<begin>.inr │ ├── <EN>_post_t<:math:\ldots>.inr. │ ├── <EN>_post_t<end>.mha. │ ├── LOGS/ │ └── ... ... The image format to be used (here mha) is given by the variable result_image_suffix, while the lineage format to be used (here xml) is given by the variable result_lineage_suffix. ## 7.4. Step 1: lineage pruning¶ Bifurcations of the lineage tree correspond to cell division, while branches (between two bifurcations or between a bifurcation and a leaf) corresponds to the lifespan of a cell. The purpose of this step is to detect suspicious end branches (terminating by a leaf) that may correspond to an over-segmentation error. An end branch is candidate for deletion if • either it terminates before the last time point (it corresponds then to a cell without daughter cell in the next time point), • or the volume of its last cell is too small (threshold given by the variable postcorrection_volume_minimal_value). An end branch candidate for deletion is deleted if • either it is too short (threshold given by the variable postcorrection_lifespan_minimal_value), • or (if the variable postcorrection_test_early_division is set to True) either its sister branch (which may not be an end branch) or its mother branch is too short, meaning that there are two divisions too close, (thresholds still given by the variable postcorrection_lifespan_minimal_value), • or if the Pearson correlation coefficient between the volumes of the candidate end branch and its sister branch is less than -postcorrection_correlation_threshold, meaning that the volumes are anti-correlated (typically the volumes of the candidate end branch are decreasing while the ones of the sister branch are increasing, indicating a fake division detection). ## 7.5. Step 2: division postponing¶ • postcorrection_volume_minimal_value branch ending with leaf cell below this value are candidate for deletion. Expressed in voxel unit. • postcorrection_lifespan_minimal_value • postcorrection_test_early_division • postcorrection_test_volume_correlation • postcorrection_correlation_threshold • postcorrection_lineage_diagnosis performs a kind of diagnosis on the lineage before and after the post-correction.	2023-01-28 20:59:26	{"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": 0, "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": 0.6286511421203613, "perplexity": 4820.646041563801}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764499654.54/warc/CC-MAIN-20230128184907-20230128214907-00070.warc.gz"}
https://toph.co/p/lucky-number-seven	# Lucky Number Seven Replay of Replay of Natio... Limits 1s, 512 MB People often consider the number $7$ as lucky. “But why?” - you might ask. Well, consider the fact that if we convert the number $7$ into base-2, we get $111$. Again, if we convert it to base-6, we get $11$. That means, on at least two bases ($b>1$), the number $7$ can be represented only using $1$. Now you must be thinking - “Then why stop at $7$? Why not other numbers as well?” To that, I say, “Sure! Why not?”. If a positive integer can be represented only using $1$ in at least two bases ($b>1$), we will call it lucky. In this problem, your task is to find the sum of all the lucky numbers less than $N$. ## Input The input consists of a single integer $\left(1 < N \leq 10^{12}\right)$. $N$ is in base-10. ## Output Print a single integer denoting the sum. As the answer can be large, print the answer modulo $10^9+7$. ## Sample InputOutput 8 8 There are two numbers below 8 that are lucky: $\{1, 7\}$. So the result is $(1 + 7) \mod (10^9 + 7) = 8$.	2022-08-11 17:23:19	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 16, "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": 0.9270643591880798, "perplexity": 622.4646010908982}, "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/1659882571483.70/warc/CC-MAIN-20220811164257-20220811194257-00229.warc.gz"}
https://www.clutchprep.com/chemistry/practice-problems/110134/in-the-lewis-structure-shown-here-a-d-e-q-x-and-z-represent-elements-in-the-firs-3	# Problem: In the Lewis structure shown here, A, D, E, Q, X, and Z represent elements in the first two rows of the periodic table (H - Ne).Identify Q so that its formal charge is zero. ###### FREE Expert Solution Formal Charge is the charge assigned to an atom in a molecule, assuming that electrons in all chemical bonds are shared equally between atoms, regardless of relative electronegativity. The steps that have to be taken to solve this problem are: 1. Use the formal charge equation to find the group number. 2. Note that the formal charge around Q is zero. 88% (398 ratings) ###### Problem Details In the Lewis structure shown here, A, D, E, Q, X, and Z represent elements in the first two rows of the periodic table (H - Ne). Identify Q so that its formal charge is zero. What scientific concept do you need to know in order to solve this problem? Our tutors have indicated that to solve this problem you will need to apply the Formal Charge concept. You can view video lessons to learn Formal Charge. Or if you need more Formal Charge practice, you can also practice Formal Charge practice problems.	2021-04-17 08:57:17	{"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": 0.8082510828971863, "perplexity": 803.5636859733618}, "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-2021-17/segments/1618038118762.49/warc/CC-MAIN-20210417071833-20210417101833-00560.warc.gz"}
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https://en.wikipedia.org/wiki/Method_of_quantum_characteristics	# Method of quantum characteristics Quantum characteristics are phase-space trajectories that arise in the phase space formulation of quantum mechanics through the Wigner transform of Heisenberg operators of canonical coordinates and momenta. These trajectories obey the Hamilton equations in quantum form and play the role of characteristics in terms of which time-dependent Weyl's symbols of quantum operators can be expressed. In the classical limit, quantum characteristics reduce to classical trajectories. The knowledge of quantum characteristics is equivalent to the knowledge of quantum dynamics. ## Weyl-Wigner association rule In Hamiltonian dynamics, classical systems with ${\displaystyle n}$ degrees of freedom are described by ${\displaystyle 2n}$ canonical coordinates and momenta ${\displaystyle ^{\;}\xi ^{i}=(x^{1},...,x^{n},p_{1},...,p_{n})\in \mathbb {R} ^{2n},}$ that form a coordinate system in the phase space. These variables satisfy the Poisson bracket relations ${\displaystyle ^{\;}\{\xi ^{k},\xi ^{l}\}=-I^{kl}.}$ The skew-symmetric matrix ${\displaystyle ^{\;}I^{kl}}$, ${\displaystyle \left\\|I\right\\|=\left\\|{\begin{array}{ll}0&-E_{n}\\E_{n}&0\end{array}}\right\\|,}$ where ${\displaystyle ^{\;}E_{n}}$ is the ${\displaystyle n\times n}$ identity matrix, defines nondegenerate 2-form in the phase space. The phase space acquires thereby the structure of a symplectic manifold. The phase space is not metric space, so distance between two points is not defined. The Poisson bracket of two functions can be interpreted as the oriented area of a parallelogram whose adjacent sides are gradients of these functions. Rotations in Euclidean space leave the distance between two points invariant. Canonical transformations in symplectic manifold leave the areas invariant. In quantum mechanics, the canonical variables ${\displaystyle ^{\;}\xi }$ are associated to operators of canonical coordinates and momenta ${\displaystyle {\hat {\xi }}^{i}=({\hat {x}}^{1},...,{\hat {x}}^{n},{\hat {p}}_{1},...,{\hat {p}}_{n})\in \operatorname {Op} (L^{2}(\mathbb {R} ^{n})).}$ These operators act in Hilbert space and obey commutation relations ${\displaystyle [{\hat {\xi }}^{k},{\hat {\xi }}^{l}]=-i\hbar I^{kl}.}$ Weyl’s association rule[1] extends the correspondence ${\displaystyle \xi ^{i}\rightarrow {\hat {\xi }}^{i}}$ to arbitrary phase-space functions and operators. ### Taylor expansion A one-sided association rule ${\displaystyle f(\xi )\to {\hat {f}}}$ was formulated by Weyl initially with the help of Taylor expansion of functions of operators of the canonical variables ${\displaystyle {\hat {f}}=f({\hat {\xi }})\equiv \sum _{s=0}^{\infty }{\frac {1}{s!}}{\frac {\partial ^{s}f(0)}{\partial \xi ^{i_{1}}...\partial \xi ^{i_{s}}}}{\hat {\xi }}^{i_{1}}...{\hat {\xi }}^{i_{s}}.}$ The operators ${\displaystyle {\hat {\xi }}}$ do not commute, so the Taylor expansion is not defined uniquely. The above prescription uses the symmetrized products of the operators. The real functions correspond to the Hermitian operators. The function ${\displaystyle ^{\;}f(\xi )}$ is called Weyl's symbol of operator ${\displaystyle {\hat {f}}}$. Under the reverse association ${\displaystyle f(\xi )\leftarrow {\hat {f}}}$, the density matrix turns to the Wigner function.[2] Wigner functions have numerous applications in quantum many-body physics, kinetic theory, collision theory, quantum chemistry. A refined version of the Weyl-Wigner association rule is proposed by Groenewold[3] and Stratonovich.[4] ### Groenewold-Stratonovich basis The set of operators acting in the Hilbert space is closed under multiplication of operators by ${\displaystyle c}$-numbers and summation. Such a set constitutes a vector space ${\displaystyle \mathbb {V} }$. The association rule formulated with the use of the Taylor expansion preserves operations on the operators. The correspondence can be illustrated with the following diagram: ${\displaystyle \left.{\begin{array}{c}{\begin{array}{c}\left.{\begin{array}{ccc}f(\xi )&\longleftrightarrow &{\hat {f}}\\g(\xi )&\longleftrightarrow &{\hat {g}}\\c\times f(\xi )&\longleftrightarrow &c\times {\hat {f}}\\f(\xi )+g(\xi )&\longleftrightarrow &{\hat {f}}+{\hat {g}}\end{array}}\right\}\;{\text{vector space}}\;\;\mathbb {V} \end{array}}\\{\begin{array}{ccc}{f(\xi )\star g(\xi )}&{\longleftrightarrow }&\;\;{{\hat {f}}{\hat {g}}}\end{array}}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\end{array}}\right\}{\text{algebra}}}$ Here, ${\displaystyle ^{\;}f(\xi )}$ and ${\displaystyle ^{\;}g(\xi )}$ are functions and ${\displaystyle {\hat {f}}}$ and ${\displaystyle {\hat {g}}}$ are the associated operators. The elements of basis of ${\displaystyle ^{\;}\mathbb {V} }$ are labelled by canonical variables ${\displaystyle ^{\;}\xi _{i}}$. The commonly used Stratonovich basis looks like ${\displaystyle {\hat {B}}(\xi )=\int {\frac {d^{2n}\eta }{(2\pi \hbar )^{n}}}\exp(-{\frac {i}{\hbar }}\eta _{k}(\xi -{\hat {\xi }})^{k})\in \mathbb {V} .}$ The Weyl-Wigner two-sided association rule for function ${\displaystyle ^{\;}f(\xi )}$ and operator ${\displaystyle {\hat {f}}}$ has the form ${\displaystyle f(\xi )=\operatorname {Tr} [{\hat {B}}(\xi ){\hat {f}}],}$ ${\displaystyle {\hat {f}}=\int {\frac {d^{2n}\xi }{(2\pi \hbar )^{n}}}f(\xi ){\hat {B}}(\xi ).}$ The function ${\displaystyle ^{\;}f(\xi )}$ provides coordinates of the operator ${\displaystyle {\hat {f}}}$ in the basis ${\displaystyle {\hat {B}}(\xi )}$. The basis is complete and orthogonal: ${\displaystyle \int {\frac {d^{2n}\xi }{(2\pi \hbar )^{n}}}{\hat {B}}(\xi )\operatorname {Tr} [{\hat {B}}(\xi ){\hat {f}}]={\hat {f}},}$ ${\displaystyle \operatorname {Tr} [{\hat {B}}(\xi ){\hat {B}}(\xi ^{\prime })]=(2\pi \hbar )^{n}\delta ^{2n}(\xi -\xi ^{\prime }).}$ Alternative operator bases are discussed also.[5] The freedom in choice of the operator basis is better known as the operator ordering problem. ## Star-product The set of operators ${\displaystyle Op(L^{2}(\mathbb {R} ^{n}))}$ is closed under the multiplication of operators. The vector space ${\displaystyle \mathbb {V} }$ is endowed thereby with an associative algebra structure. Given two functions ${\displaystyle f(\xi )=Tr[{\hat {B}}(\xi ){\hat {f}}]~~\mathrm {and} ~~g(\xi )=Tr[{\hat {B}}(\xi ){\hat {g}}],}$ one can construct a third function ${\displaystyle f(\xi )\star g(\xi )=Tr[{\hat {B}}(\xi ){\hat {f}}{\hat {g}}]}$ called ${\displaystyle \star }$-product [3] or Moyal product. It is given explicitly by ${\displaystyle f(\xi )\star g(\xi )=f(\xi )\exp({\frac {i\hbar }{2}}{\mathcal {P}})g(\xi ).}$ where ${\displaystyle {\mathcal {P}}=-{I}^{kl}{\overleftarrow {\frac {\partial }{\partial \xi ^{k}}}}{\overrightarrow {\frac {\partial }{\partial \xi ^{l}}}}}$ is the Poisson operator. The ${\displaystyle \star }$-product splits into symmetric and skew-symmetric parts ${\displaystyle f\star g=f\circ g+{\frac {i\hbar }{2}}f\wedge g.}$ The ${\displaystyle \circ }$-product is not associative. In the classical limit ${\displaystyle \circ }$-product becomes the dot-product. The skew-symmetric part ${\displaystyle f\wedge g}$ is known under the name of Moyal bracket. This is the Weyl's symbol of commutator. In the classical limit Moyal bracket becomes Poisson bracket. Moyal bracket is quantum deformation of Poisson bracket. ## Quantum characteristics The correspondence ${\displaystyle \xi \leftrightarrow {\hat {\xi }}}$ shows that coordinate transformations in the phase space are accompanied by transformations of operators of the canonical coordinates and momenta and vice versa. Let ${\displaystyle \mathbf {\hat {U}} }$ be the evolution operator, ${\displaystyle {\hat {U}}=\exp {\Bigl (}-{\frac {i}{\hbar }}{\hat {H}}\tau {\Bigr )},}$ and ${\displaystyle {\hat {H}}}$ is Hamiltonian. Consider the following scheme: ${\displaystyle \xi {\stackrel {q}{\longrightarrow }}{\acute {\xi }}}$ ${\displaystyle \updownarrow \;\;\;\;\;\;\updownarrow }$ ${\displaystyle {\hat {\xi }}{\stackrel {\hat {U}}{\longrightarrow }}{\acute {\hat {\xi }}},}$ Quantum evolution transforms vectors in the Hilbert space and, upon the Wigner association rule, coordinates in the phase space. In Heisenberg representation, the operators of the canonical variables are transformed as ${\displaystyle {\hat {\xi }}^{i}\rightarrow {\acute {{\hat {\xi }}^{i}}}={\hat {U}}^{+}{\hat {\xi }}^{i}{\hat {U}}.}$ The phase-space coordinates ${\displaystyle {\acute {\xi }}^{i}}$ that correspond to new operators ${\displaystyle {\acute {{\hat {\xi }}^{i}}}}$ in the old basis ${\displaystyle {\hat {B}}(\xi )}$ are given by ${\displaystyle \xi ^{i}\rightarrow {\acute {\xi }}^{i}=q^{i}(\xi ,\tau )=Tr[{\hat {B}}(\xi ){\hat {U}}^{+}{\hat {\xi }}^{i}{\hat {U}}],}$ with the initial conditions ${\displaystyle ^{\;}q^{i}(\xi ,0)=\xi ^{i}.}$ The functions ${\displaystyle ^{\;}q^{i}(\xi ,\tau )}$ define quantum phase flow. In the general case, it is canonical to first order in ${\displaystyle \tau }$.[6] ### Star-function The set of operators of canonical variables is complete in the sense that any operator can be represented as a function of operators ${\displaystyle {\hat {\xi }}}$. Transformations ${\displaystyle {\hat {f}}\rightarrow {\acute {\hat {f}}}={\hat {U}}^{+}{\hat {f}}{\hat {U}}}$ induce under the Wigner association rule transformations of phase-space functions: ${\displaystyle f(\xi ){\stackrel {q}{\longrightarrow }}{\acute {f}}(\xi )=Tr[{\hat {B}}(\xi ){\hat {U}}^{+}{\hat {f}}{\hat {U}}]}$ ${\displaystyle \updownarrow \;\;\;\;\;\;\;\;\;\;\,\updownarrow }$ ${\displaystyle {\hat {f}}\;\;\;\;{\stackrel {\hat {U}}{\longrightarrow }}\,{\acute {\hat {f}}}\;\;\;\;\;={\hat {U}}^{+}{\hat {f}}{\hat {U}}}$ Using the Taylor expansion, the transformation of function ${\displaystyle ^{\;}f(\xi )}$ under the evolution can be found to be ${\displaystyle f(\xi )\rightarrow {\acute {f}}(\xi )\equiv Tr[{\hat {B}}(\xi ){\hat {U^{+}}}f({\hat {\xi }}){\hat {U}}]=\sum _{s=0}^{\infty }{\frac {1}{s!}}{\frac {\partial ^{s}f(0)}{\partial \xi ^{i_{1}}...\partial \xi ^{i_{s}}}}q^{i_{1}}(\xi ,\tau )\star ...\star q^{i_{s}}(\xi ,\tau )\equiv f(\star q(\xi ,\tau )).}$ Composite function defined in such a way is called ${\displaystyle \star }$-function. The composition law differs from the classical one. However, semiclassical expansion of ${\displaystyle f(\star q(\xi ,\tau ))}$ around ${\displaystyle f(q(\xi ,\tau ))^{\;}}$ is formally well defined and involves even powers of ${\displaystyle \hbar }$ only. This equation shows that, given quantum characteristics are constructed, physical observables can be found without further addressing to Hamiltonian. The functions ${\displaystyle ^{\;}q^{i}(\xi ,\tau )}$ play the role of characteristics[7] similarly to classical characteristics used to solve classical Liouville equation. ### Quantum Liouville equation The Wigner transform of the evolution equation for the density matrix in the Schrödinger representation leads to a quantum Liouville equation for the Wigner function. The Wigner transform of the evolution equation for operators in the Heisenberg representation, ${\displaystyle {\frac {\partial }{\partial \tau }}{\hat {f}}=-{\frac {i}{\hbar }}[{\hat {f}},{\hat {H}}],}$ ${\displaystyle {\frac {\partial }{\partial \tau }}f(\xi ,\tau )=f(\xi ,\tau )\wedge H(\xi ).}$ ${\displaystyle \star }$-function solves this equation in terms of quantum characteristics: ${\displaystyle f(\xi ,\tau )=f(\star q(\xi ,\tau ),0).}$ Similarly, the evolution of the Wigner function in the Schrödinger representation is given by ${\displaystyle W(\xi ,\tau )=W(\star q(\xi ,-\tau ),0).}$ The Liouville theorem of classical mechanics fails, to the extent that, locally, the "probability" density in phase space is not preserved in time. ### Quantum Hamilton's equations Quantum Hamilton's equations can be obtained applying the Wigner transform to the evolution equations for Heisenberg operators of canonical coordinates and momenta ${\displaystyle {\frac {\partial }{\partial \tau }}q^{i}(\xi ,\tau )=\{\zeta ^{i},H(\zeta )\}\|_{\zeta =\star q(\xi ,\tau )}.}$ The right-hand side is calculated like in the classical mechanics. The composite function is, however, ${\displaystyle \star }$-function. The ${\displaystyle \star }$-product violates canonicity of the phase flow beyond the first order in ${\displaystyle \tau }$. ### Conservation of Moyal bracket The antisymmetrized products of even number of operators of canonical variables are c-numbers as a consequence of the commutation relations. These products are left invariant by unitary transformations and, in particular, ${\displaystyle q^{i}(\xi ,\tau )\wedge q^{j}(\xi ,\tau )=\xi ^{i}\wedge \xi ^{j}=-{I}^{ij}.}$ Phase-space transformations induced by the evolution operator preserve the Moyal bracket and do not preserve the Poisson bracket, so the evolution map ${\displaystyle \xi \rightarrow {\acute {\xi }}=q(\xi ,\tau ),}$ is not canonical.[7] Transformation properties of canonical variables and phase-space functions under unitary transformations in the Hilbert space have important distinctions from the case of canonical transformations in the phase space: ### Composition law Quantum characteristics can hardly be treated visually as trajectories along which physical particles move. The reason lies in the star-composition law ${\displaystyle q(\xi ,\tau _{1}+\tau _{2})=q(\star q(\xi ,\tau _{1}),\tau _{2}),}$ which is non-local and is distinct from the dot-composition law of classical mechanics. ### Energy conservation The energy conservation implies ${\displaystyle H(\xi )=H(\star q(\xi ,\tau ))}$, where ${\displaystyle H(\xi )=Tr[{\hat {B}}(\xi ){\hat {H}}]}$ is Hamilton's function. In the usual geometric sense, ${\displaystyle ^{\;}H(\xi )}$ is not conserved along quantum characteristics. ## Summary The origin of the method of characteristics can be traced back to Heisenberg’s matrix mechanics. Suppose that we have solved in the matrix mechanics the evolution equations for the operators of the canonical coordinates and momenta in the Heisenberg representation. These operators evolve according to ${\displaystyle {\hat {\xi }}^{i}\rightarrow {\hat {\xi }}^{i}(\tau )={\hat {U}}^{+}{\hat {\xi }}^{i}{\hat {U}}.}$ It is known that for any operator ${\displaystyle {\hat {f}}}$ one can find a function f(ξ) through which ${\displaystyle {\hat {f}}}$ is represented in the form ${\displaystyle f({\hat {\xi }})}$. The same operator ${\displaystyle {\hat {f}}}$ at time τ is equal to ${\displaystyle {\hat {f}}(\tau )=U^{+}{\hat {f}}U=U^{+}f({\hat {\xi }})U=f(U^{+}{\hat {\xi }}U)=f({\hat {\xi }}(\tau )).}$ This equation shows that ${\displaystyle {\hat {\xi }}(\tau )}$ are characteristics that determine the evolution for all of the operators in Op(L2(Rn)). This property is fully transferred to the phase space upon deformation quantization and, in the limit of ħ → 0, to the classical mechanics. CLASSICAL DYNAMICS QUANTUM DYNAMICS Liouville equation Finite-order PDE Infinite-order PDE ${\displaystyle {\frac {\partial }{\partial \tau }}\rho (\xi ,\tau )=-\{\rho (\xi ,\tau ),{\mathcal {H}}(\xi )\}}$ ${\displaystyle {\frac {\partial }{\partial \tau }}W(\xi ,\tau )=-W(\xi ,\tau )\wedge H(\xi )}$ Hamilton's equations Finite-order ODE Infinite-order PDE ${\displaystyle {\frac {\partial }{\partial \tau }}c^{i}(\xi ,\tau )=\{\zeta ^{i},{\mathcal {H}}(\zeta )\}\|_{\zeta =c(\xi ,\tau )}}$ ${\displaystyle {\frac {\partial }{\partial \tau }}q^{i}(\xi ,\tau )=\{\zeta ^{i},H(\zeta )\}\|_{\zeta =\star q(\xi ,\tau )}}$ Initial conditions Initial conditions ${\displaystyle ^{\;}c^{i}(\xi ,0)=\xi ^{i}}$ ${\displaystyle ^{\;}q^{i}(\xi ,0)=\xi ^{i}}$ Composition law ${\displaystyle \star }$-composition law ${\displaystyle ^{\;}c(\xi ,\tau _{1}+\tau _{2})=c(c(\xi ,\tau _{1}),\tau _{2})}$ ${\displaystyle q(\xi ,\tau _{1}+\tau _{2})=q(\star q(\xi ,\tau _{1}),\tau _{2})}$ Conservation of Poisson bracket Conservation of Moyal bracket ${\displaystyle ^{\;}\{c^{i}(\xi ,\tau ),c^{j}(\xi ,\tau )\}=\{\xi ^{i},\xi ^{j}\}}$ ${\displaystyle q^{i}(\xi ,\tau )\wedge q^{j}(\xi ,\tau )=\xi ^{i}\wedge \xi ^{j}}$ Energy conservation Energy conservation ${\displaystyle ^{\;}H(\xi )=H(c(\xi ,\tau ))}$ ${\displaystyle ^{\;}H(\xi )=H(\star q(\xi ,\tau ))}$ Solutions to Liouville equation ${\displaystyle ^{\;}\rho (\xi ,\tau )=\rho (c(\xi ,-\tau ),0)}$ ${\displaystyle ^{\;}W(\xi ,\tau )=W(\star q(\xi ,-\tau ),0)}$ Table compares properties of characteristics in classical and quantum mechanics. PDE and ODE are partial differential equations and ordinary differential equations, respectively. The quantum Liouville equation is the Weyl-Wigner transform of the von Neumann evolution equation for the density matrix in Schrödinger representation. The quantum Hamilton equations are the Weyl-Wigner transforms of the evolution equations for operators of the canonical coordinates and momenta in Heisenberg representation. In classical systems, characteristics ${\displaystyle ^{\;}c^{i}(\xi ,\tau )}$ satisfy usually first-order ODE, e.g., classical Hamilton's equations, and solve first-order PDE, e.g., classical Liouville equation. Functions ${\displaystyle ^{\;}q^{i}(\xi ,\tau )}$ are characteristics also, despite both ${\displaystyle ^{\;}q^{i}(\xi ,\tau )}$ and ${\displaystyle ^{\;}f(\xi ,\tau )}$ obey infinite-order PDE. The quantum phase flow contains entire information on the quantum evolution. Semiclassical expansion of quantum characteristics and ${\displaystyle \star }$-functions of quantum characteristics in power series in ${\displaystyle \hbar }$ allows calculation of the average values of time-dependent physical observables by solving a finite-order coupled system of ODE for phase space trajectories and Jacobi fields.[8][9] The order of the system of ODE depends on truncation of the power series. The tunneling effect is nonperturbative in ${\displaystyle \hbar }$ and is not captured by the expansion. The density of the quantum probability fluid is not preserved in phase-space, and there is no unambiguously defined notion of trajectories for quantum systems, as the quantum fluid "diffuses".[10] Quantum characteristics must therefore be distinguished from both the trajectories of the de Broglie - Bohm theory [11] and the trajectories of the path-integral method in phase space for the amplitudes [12] and the Wigner function. [13][14] So far, only a few quantum systems have been explicitly solved using the method of quantum characteristics. [15] ## References 1. ^ H. Weyl, Z. Phys. 46, 1 (1927). 2. ^ E. P. Wigner, On the quantum correction for thermodynamic equilibrium, Phys. Rev. 40, 749 (1932). 3. ^ a b H. J. Groenewold, On the Principles of elementary quantum mechanics, Physica, 12, 405 (1946). 4. ^ R. L. Stratonovich, Sov. Phys. JETP 4, 891 (1957). 5. ^ C. L. Mehta, J. Math. Phys. 5, 677 (1964). 6. ^ P. A. M. Dirac, The Principles of Quantum Mechanics, First Edition (Oxford: Clarendon Press, 1930). 7. ^ a b M. I. Krivoruchenko, A. Faessler, Weyl's symbols of Heisenberg operators of canonical coordinates and momenta as quantum characteristics, J. Math. Phys. 48, 052107 (2007). 8. ^ M. I. Krivoruchenko, C. Fuchs, A. Faessler, Semiclassical expansion of quantum characteristics for many-body potential scattering problem, Annalen der Physik 16, 587 (2007). 9. ^ S. Maximov, On a special picture of dynamical evolution of nonlinear quantum systems in the phase-space representation, Physica D238, 1937 (2009). 10. ^ J. E. Moyal, Quantum mechanics as a statistical theory, Proceedings of the Cambridge Philosophical Society, 45, 99 (1949). 11. ^ P. R. Holland, The quantum theory of motion, (Cambridge University Press, 1993). 12. ^ F. A. Berezin, Feynman path integrals in a phase space, Sov. Phys. Usp. 23, 763 (1980). 13. ^ M. S. Marinov, A new type of phase-space path integral, Phys. Lett. A 153, 5 (1991). 14. ^ Wong, C. Y. (2003). "Explicit solution of the time evolution of the Wigner function". Journal of Optics B: Quantum and Semiclassical Optics 5 (3): S420. doi:10.1088/1464-4266/5/3/381. 15. ^ G. Braunss, Quantum dynamics in phase space: Moyal trajectories 2, J. Math. Phys. 54, 012105 (2013). doi:10.1063/1.4773229 ## Textbooks • H. Weyl, The Theory of Groups and Quantum Mechanics, (Dover Publications, New York Inc., 1931). • V. I. Arnold, Mathematical Methods of Classical Mechanics, (2-nd ed. Springer-Verlag, New York Inc., 1989). • M. V. Karasev and V. P. Maslov, Nonlinear Poisson brackets. Geometry and quantization. Translations of Mathematical Monographs, 119. (American Mathematical Society, Providence, RI, 1993). • Some Applications of Quantum Mechanics, Ed. M. R. Pahlavani, (InTech, Zagreb, 2012).	2016-07-26 21:30:20	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 120, "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": 0.8636900186538696, "perplexity": 681.0543628407096}, "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/1469257825124.22/warc/CC-MAIN-20160723071025-00278-ip-10-185-27-174.ec2.internal.warc.gz"}
https://math.stackexchange.com/questions/544169/median-for-continuous-distribution	# Median for continuous distribution Consider a continuous random variable X with probability density function given by $f(x)=cx$ for $1 \le x \le 5$, zero otherwise. Find the median. First I calculate the CDF: $F(x)=cx^2/2$ for $1 \le x \le 5$, zero otherwise. Now we have to solve for constant c by using the definition of PDF, namely: $\int\limits_{-\infty}^{\infty}f(x)dx=1 \implies (cx^2/2)_1^5=1 \implies c=1/12$ Then to calculate the median, we set the CDF = 0.5: $0.5=(1/12)(1/2)x^2 \implies x=\sqrt{12}$ But the book solution is $\sqrt{13}$. Can someone tell me what I am doing wrong? Thank you. • @Eupraxis1981: I don't think so. Please check my answer. Oct 29, 2013 at 13:58 • The function $F(x)=cx^2/2$ is not the correct CDF here: its value at $1$ should be $0$. Oct 29, 2013 at 14:03 • "First I calculate the CDF" When I do that, I find $F(x)=\frac12c(x^2-1)$ for $x$ in $(1,5)$, $F(x)=0$ for $x\leqslant1$, and $F(x)=1$ for $x\geqslant5$, not what you wrote. – Did Oct 29, 2013 at 14:03 Recall that there is an integration constant when finding the CDF $F(x)$. Also recall that the CDF should take on the value ZERO when $x$ is from minus infinity to $x=1$ and it must take on the value ONE from $x=5$ to plus infinity. (i.e. the CDF is a non-decreasing function on the support of the density $f(x)$). If you counter verify, you will see that the above paragraph does not hold for the CDF you found above, in your question. Reworking on the problem, you should find an appropriate CDF. Simply put: the CDF should be $$F(x) = \frac{x^2}{24} - \frac{1}{24}$$ We see that $F(1) = 0$ and that $F(5) = 1$ indeed. Finally, $$0.5 = \frac{x^2}{24} - \frac{1}{24}$$ Solving for $x$ yields that the median equals $\sqrt{13}$ You forgot the integration constant. Calculating the CDF gives $F(x)=cx^2/2+d$ on $1\leq x\leq 5$, $F(x)=0$ for $x<1$ and $F(x)=1$ for $x>1$. Setting $F(1)=0$ and $F(5)=1$, we get $c=1/12$ and $d=-1/24$. The solution for $F(x)=1/2$ is then indeed $\sqrt{13}$. • I thought you only need integration constant if you are taking an indefinite integral. My integral is definite: from 1-5. Oct 29, 2013 at 13:59 • @Eupraxis1981: Please see my comment below the question to see why the constant matters. Oct 29, 2013 at 14:04 • user1527227: I've removed my comment, since Rasmus is correct. My oversight (I was moving too quickly) is not carrying through the lower bound on the definite integral. However, its not a "constant of integration" but merely the lower value of the definite integral, i.e., $F(1) = \frac{1}{24}$, so $F(x) = \frac{x^2}{24} - \frac{1}{24}$ thats where the additional term comes from. – user76844 Oct 29, 2013 at 14:05 • Yeah thanks guys! I also see I could have just changed my limits of integration from 1 to x and that would also work! Oct 29, 2013 at 14:13 • @Eupraxis1981 You may call it however you like. In any case, it's this constant in the integral that one needs to work things out. Oct 29, 2013 at 14:18 For median $$m$$: $$\int\limits_{-\infty}^{m} f(x) dx = \int\limits_{m}^{+\infty} f(x) dx\\ \int\limits_{1}^{m} f(x) dx = \int\limits_{m}^{5} f(x) dx\\ c \frac{x^2}{2}\Biggr\|_{1}^{m} = c \frac{x^2}{2}\Biggr\|_{m}^{5}\\ x^2\Biggr\|_{1}^{m} = x^2\Biggr\|_{m}^{5}\\ m = \sqrt{\frac{1^2 + 5^2}{2}}=\sqrt{13}$$	2022-05-17 02:30:05	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 2, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8904027342796326, "perplexity": 285.2204594017114}, "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/1652662515466.5/warc/CC-MAIN-20220516235937-20220517025937-00162.warc.gz"}
http://www.physicsforums.com/showthread.php?s=74b509a044c98253724b773ecefa3012&p=4012894	## Conservation of angular momentum 1. The problem statement, all variables and given/known data A 0.5-kg particle is located at the point r = 3m i + 4m k and is moving with a velocity v = 5m/s i -2m/s k . What is the angular momentum of this particle about the origin? A) 13 kgm^2/s k B) 13 kgm^2/s j C) 26 kgm^2/s k D) 26m^2/s j E) 13 kgm^2/s i 2. Relevant equations L = RxP where L is angular momentum, R is the distance from the origin and P is the momentum and the x stands for the cross product P = mv 3. The attempt at a solution P = 0.5(5m/s i - 2m/s k) = 5/2 kgm^2/s i - 1kgm^2/s L = (3i + 4i) x (5/2 kgm^2/s i - 1kgm^2/s) = -3j - 10j = -13j I'm fairly certain that the answer is B because I believe that the cross product should give me a value that is perpendicular to the two vectors, which in this case are i and k, thus the cross product should, I think, include j. Also, I believe that I can eliminate D because the units are incorrect. What I don't understand is why my answer is negative. PhysOrg.com science news on PhysOrg.com >> New language discovery reveals linguistic insights>> US official: Solar plane to help ground energy use (Update)>> Four microphones, computer algorithm enough to produce 3-D model of simple, convex room $L=m\cdot r\times v=\frac{1}{2}\cdot \det\begin{bmatrix}i&j&k\\3&0&4\\5&0&-2\end{bmatrix}=13j$ Similar discussions for: Conservation of angular momentum Thread Forum Replies Classical Physics 2 General Physics 16 Introductory Physics Homework 3 Introductory Physics Homework 2 General Engineering 0	2013-06-18 22:27:00	{"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": 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": 0.5520215034484863, "perplexity": 1800.8948583544689}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2013-20/segments/1368707435344/warc/CC-MAIN-20130516123035-00032-ip-10-60-113-184.ec2.internal.warc.gz"}
https://www.physicsforums.com/threads/we-call-l-the-limit-of-f-x-as-x.79517/	# We call L the limit of f(x) as x #### cscott Definition We call L the limit of f(x) as x approaches $\infty$ if for every number ε > 0 there exists a δ; such that whenever x > δ we have $$\left\| f(x) - L \right\| < \epsilon$$ When this holds we write $$\lim_{x \to \infty} f(x) = L$$ or $$f(x) \to L \quad as \quad x \to \infty.$$ Similarly, we call L the limit of f(x) as x approaches $-\infty$ if for every number ε > 0, there exists a number δ such that whenever x < δ we have $$\left\| f(x) - L \right\| < \epsilon$$ When this holds we write $$\lim_{x \to -\infty} f(x) = L$$ or $$f(x) \to L \quad as \quad x \to -\infty.$$ Notice the difference in these two definitions. For the limit of f(x) as x approaches $\infty$ we are interested in those x such that x > δ. For the limit of f(x) as x approaches $-\infty$ we are interested in those x such that x < δ. What is δ that they keep refering to in this definition? They first bring it up as if it's some quantity I'm supposed to know... maybe I'm just crazy :) #### whozum They dont stand for antghing in particular in this situation, they are just variables (which could be replaced by any other ones) with the loose definitions given to them above. When I first did limits the epsilons and deltas threw me off too. Do you understand how they're used in the limit definition? #### Galileo Homework Helper Suppose the limit exists, what it says is: Given $\epsilon>0$, there exists SOME number, such that whenever x is greater than that number, you have$\|f(x)-L\|<\epsilon$. It's just convenient to name that number and they chose to name it $\delta$. Example: $\lim_{x\to \infty} 1/x=0$ since given $\epsilon>0$ we can find a number $\delta$, such that $x>\delta \Rightarrow \|1/x\|<\epsilon$. Pick $\delta=1/\epsilon$, then for any $x>\delta=1/\epsilon$ we have $\|1/x\|<\|1/\delta\|=\epsilon$. Last edited: #### Hurkyl Staff Emeritus Gold Member For any real number N, there exists a real number M such that M > N. Do you understand the meaning of N and M in this? #### JoshHolloway Differentiation Problem I accidently posted in the wrong place, please forgive me. Last edited: #### SteveRives cscott said: What is δ that they keep refering to in this definition? They first bring it up as if it's some quantity I'm supposed to know... maybe I'm just crazy :) Short answer: δ is the "run" portion of slope. Limits are simple once you think of them as rise over run. Epsilon is rise, delta (δ) is run. The classic assingments are usually of the form that the book will give you an epsilon, and you have to find the corresponding delta (at some given point on a curve). Given what I have just said, you can now think of it this way: they give you a rise, you give the corresponding run. They give you an e, you give them a δ -- where δ is "run". I don't know of an easier way to think of it! #### dishuwen Epsilon-delta and limits I have finally begun to understand the relationship between epsilon and delta. I can pretty much now use the delta-epsilon method to prove limits. The (ironic) problem, for me, now is that I'm having trouble finidng the limits! Can anyone offer a bit of advice or, maybe, direct me to a site on which I can practice some problems? If anyone is having problems understanding the epsilon-delta relationship, I'll bge happy to try to explain my way of understanding it, if that would help. #### cscott Thanks for all your help, I understand it better now. #### SteveRives dishuwen said: The (ironic) problem, for me, now is that I'm having trouble finidng the limits! There are many things to say in answer to this; I will alert you to one of my favorites: L'Hopital's Rule. Learn the mechanics of it here: http://www.math.hmc.edu/calculus/tutorials/lhopital/ Given what you have asked so far, I suspect the problems you are working on now will not require this knowledge, but there you have it... ### Physics Forums Values We Value Quality • Topics based on mainstream science • Proper English grammar and spelling We Value Civility • Positive and compassionate attitudes • Patience while debating We Value Productivity • Disciplined to remain on-topic • Recognition of own weaknesses • Solo and co-op problem solving	2019-10-19 06:47:26	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7889307737350464, "perplexity": 734.8495079850156}, "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/1570986692126.27/warc/CC-MAIN-20191019063516-20191019091016-00196.warc.gz"}
https://forum.allaboutcircuits.com/threads/how-can-i-convert-24v-to-12v.120870/	# how can i convert 24v to 12v? #### CornyCorn Joined Feb 13, 2016 12 Hello, this is my first post so please bear with me. I am 12 years old and i am working on a prototype for an invention i have. I will need to turn 24vdc into 12vdc, but i want to use something that: -will not generate heat -that is small in size (about an inch) Thanks! #### panic mode Joined Oct 10, 2011 1,809 you can make voltage divider using two same value resistors. but you will quickly find drawbacks to this if you need significant current. in that case switching regulator is what you need. #### ScottWang Joined Aug 23, 2012 6,860 How much current does the 12V needed? The size is 1 inch=2.54cm? If just a light current then you can using one resistor and one 12V zener or LM2576-12V 3A, LM2596-12V 3A. #### dl324 Joined Mar 30, 2015 9,581 Welcome to AAC! I will need to turn 24vdc into 12vdc, but i want to use something that: -will not generate heat Don't want to discourage you, but what you want to do is not possible. Even the most efficient regulator will generate heat. #### Lestraveled Joined May 19, 2014 1,946 ...... Even the most efficient regulator will generate heat. Perhaps you are being too critical. He is 12 years old and not a crusty old fart like you. @CornyCorn What is your definition of "will not generate heat"? #### CornyCorn Joined Feb 13, 2016 12 Perhaps you are being too critical. He is 12 years old and not a crusty old fart like you. @CornyCorn What is your definition of "will not generate heat"? I apologize for being vague. i already have a transformer to 24v. I need to step down 24 volts to 12, power a few components, and then step it down to 5 to power some more components. Also should I have the voltage stepped down to 12v then 5v or should I have 24v stepped down to 5v then stepped up to 12? Thanks #### hp1729 Joined Nov 23, 2015 2,304 I apologize for being vague. i already have a transformer to 24v. I need to step down 24 volts to 12, power a few components, and then step it down to 5 to power some more components. Also should I have the voltage stepped down to 12v then 5v or should I have 24v stepped down to 5v then stepped up to 12? Thanks 24 Volts, 12 Volts and 5 Volts at how much current. Power (E x I) will determine the method you use to a big extent.. #### dl324 Joined Mar 30, 2015 9,581 I apologize for being vague. i already have a transformer to 24v. I need to step down 24 volts to 12, power a few components, and then step it down to 5 to power some more components. Also should I have the voltage stepped down to 12v then 5v or should I have 24v stepped down to 5v then stepped up to 12? If you specify your current requirement(s), you'll get more appropriate suggestions. #### Alec_t Joined Sep 17, 2013 10,713 If you don't know the current requirements, at least tell us what the 'few components' are that you will be powering at 12V and at 5V. Incidentally, 1" is just one dimension. What are the other dimensions? #### Evanguy Joined Dec 21, 2014 81 A simple buck converter should work for you. there are some circuits on google or they can be had on ebay for about 2$i assume you will be drawing less then a few amps. some even with led screens to show output as they can be varied #### Dodgydave Joined Jun 22, 2012 8,677 Lm7812, Lm7805, Lm2596, Lm317, all of these regulators will work, but until you give your current limit, we just keep guessing. #### AnalogKid Joined Aug 1, 2013 8,251 Where are you located, and what is the general nature of your invention? ak Thread Starter #### CornyCorn Joined Feb 13, 2016 12 Lm7812, Lm7805, Lm2596, Lm317, all of these regulators will work, but until you give your current limit, we just keep guessing. The current draw will be around 1-2 amps. Thread Starter #### CornyCorn Joined Feb 13, 2016 12 A simple buck converter should work for you. there are some circuits on google or they can be had on ebay for about 2$ i assume you will be drawing less then a few amps. some even with led screens to show output as they can be varied Are there any models that you recommend? I prefer to have ones with PCB pins. Last edited: #### hp1729 Joined Nov 23, 2015 2,304 Are there any models that you recommend? I prefer to have ones with PCB pins. 7812 for 12 V. Same circuit but 7805 for 5 V. Actually only the regulator is needed. The caps and stuff are not mandatory, depending on the rest of the circuit. Even with a small heat sink you are looking at about one cubic inch. #### GopherT Joined Nov 23, 2012 8,012 7812 for 12 V. Same circuit but 7805 for 5 V. Actually only the regulator is needed. The caps and stuff are not mandatory, depending on the rest of the circuit. Even with a small heat sink you are looking at about one cubic inch. He said no (little) heat. Now you recommend a part that works as a voltage regulator by purposely converting the excess (VxI) as heat? That poor little 7812 will dissipate 24-12volts of voltage drop = 12V and then 12V x 2 amps = 24 watts of heat. Try again. #### dl324 Joined Mar 30, 2015 9,581 The current draw will be around 1-2 amps. A 5V linear regulator would dissipate P = IV = 2A*19V = 38W. You should use a buck (step down switching) regulator to minimize power dissipation. They sell switching regulators that are drop-in replacements (PCB hole pattern only) for LM78xx regulators, but they're relatively expensive and probably won't handle more than 1A. #### ScottWang Joined Aug 23, 2012 6,860 7812 for 12 V. Same circuit but 7805 for 5 V. Actually only the regulator is needed. The caps and stuff are not mandatory, depending on the rest of the circuit. Even with a small heat sink you are looking at about one cubic inch. Did you see the current he needed -- around 1-2 amps? #### hp1729 Joined Nov 23, 2015 2,304 He said no (little) heat. Now you recommend a part that works as a voltage regulator by purposely converting the excess (VxI) as heat? That poor little 7812 will dissipate 24-12volts of voltage drop = 12V and then 12V x 2 amps = 24 watts of heat. Try again. Good point! He could possibly put together a switching regulator. One of the "Simple Switchers". You trade complexity and cost for heat and cheap. #### Alec_t Joined Sep 17, 2013 10,713 The current draw will be around 1-2 amps. For that current, the voltage converter is likely to be bigger than your "1 inch" you hope for.	2020-02-22 11:19:06	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.30983301997184753, "perplexity": 2503.989808138319}, "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-2020-10/segments/1581875145657.46/warc/CC-MAIN-20200222085018-20200222115018-00514.warc.gz"}
http://openstudy.com/updates/55ab6264e4b0ce105659a1f1	## yamyam70 one year ago I need help with these solve the inequality 0<3x + 2 <1 1. insa u can make two pairs out of that \|dw:1437296021976:dw\| 2. insa can u solve it now? 3. insa @yamyam70 4. yamyam70 -2/3 < x < -1/3 5. yamyam70 THANKS! 6. UsukiDoll another method $0<3x + 2 <1$ subtract 2 all over the inequality $0-2<3x + 2-2 <1 -2$ $-2<3x <-1$ and then divide 3 all over the inequality $\frac{-2}{3}<\frac{3}{3}x <\frac{-1}{3}$ $\frac{-2}{3}<x <\frac{-1}{3}$ 7. UsukiDoll @yamyam70 just for reference/information purposes. It's good to learn more than one valid method for solving a math problem as long as it doesn't break math rules :) 8. yamyam70 \|dw:1437297811873:dw\| 9. yamyam70 is that correct? 10. UsukiDoll hmmm that inequality we had earlier shows that it's between -2/3 < x < -1/3 and it's an open circle. looks similar to your drawing, but let's try using a number line \|dw:1437298046045:dw\| 11. yamyam70 considering -2/3 = -0.666 -1/3 = 0.3333 hey thanks maybe ill just put my values for X the way you wrote it :) TY! 12. UsukiDoll you're welcome :)	2016-10-25 01:38:29	{"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": 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": 0.7026988863945007, "perplexity": 2221.2695924358854}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-44/segments/1476988719843.44/warc/CC-MAIN-20161020183839-00031-ip-10-171-6-4.ec2.internal.warc.gz"}
http://www.lastfm.es/user/kzsc/library/music/James+Brown/_/I+Don't+Want+Nobody+To+Give+Me+Nothing+(Open+Up+The+Door+I'll+Get+It+Myself)?setlang=es	# Colección Música » James Brown » ## I Don't Want Nobody To Give Me Nothing (Open Up The Door I'll Get It Myself) 9 scrobblings \| Ir a la página del tema Temas (9) Tema Álbum Duración Fecha I Don't Want Nobody To Give Me Nothing (Open Up The Door I'll Get It Myself) 9:42 15 Ene 2012, 10:32 I Don't Want Nobody To Give Me Nothing (Open Up The Door I'll Get It Myself) 9:42 11 Dic 2011, 0:46 I Don't Want Nobody To Give Me Nothing (Open Up The Door I'll Get It Myself) 9:42 10 Dic 2011, 7:10 I Don't Want Nobody To Give Me Nothing (Open Up The Door I'll Get It Myself) 9:42 18 Ago 2011, 8:21 I Don't Want Nobody To Give Me Nothing (Open Up The Door I'll Get It Myself) 9:42 1 Ago 2011, 4:26 I Don't Want Nobody To Give Me Nothing (Open Up The Door I'll Get It Myself) 9:42 22 Jul 2011, 12:08 I Don't Want Nobody To Give Me Nothing (Open Up The Door I'll Get It Myself) 9:42 31 Dic 2010, 7:19 I Don't Want Nobody To Give Me Nothing (Open Up The Door I'll Get It Myself) 9:42 26 Ago 2010, 18:15 I Don't Want Nobody To Give Me Nothing (Open Up The Door I'll Get It Myself) 9:42 16 Jun 2010, 18:58	2014-07-14 04:45:06	{"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": 0.8601768612861633, "perplexity": 9442.266751352909}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-23/segments/1404776439916.87/warc/CC-MAIN-20140707234039-00094-ip-10-180-212-248.ec2.internal.warc.gz"}
https://koelnerkeyladen.de/en/how-many-digits-do-you-need-to-write-before-decimal-point-in-scientific-notation/	How many digits do you need to write before decimal point in scientific notation? Scientific notation of a number just writes the significant digits followed by an appropriate power of ten. The most common form of scientific notation inserts a decimal point after the first significant digit, follows the significant digits with times, “x”, and then 10 to a power. How many numbers can be before the decimal in scientific notation? Scientific notation is a system for writing very large and very small numbers that makes them easier to work with. Every number can be written in scientific notation as the product of two numbers (two numbers multiplied together): A decimal greater than or equal to 1 and less than 10. How do you write decimals in scientific notation? Quote from video: So let's see how far we had to go our decimal point started here and we moved one two three places to get five point one two. So that means that our exponent is negative three. How many digits should be to the left of the decimal when written in scientific notation? one digit When written in standard form there must be one digit, and only one digit to the left of the decimal point in the number N. Does there have to be a decimal in scientific notation? The proper format for scientific notation is a x 10^b where a is a number or decimal number such that the absolute value of a is greater than or equal to one and less than ten or, 1 ≤ \|a\| < 10. How do you write 0.000345 in scientific notation? Convert from Real Number to Scientific Notation: 0.000345 = 3.45 × 10. How do you write 0.00001 in scientific notation? Hence, the scientific notation of $0.00001$ is $1 \times {10^{ – 5}}$ . So, the correct answer is “ $1 \times {10^{ – 5}}$ ”. Note: Scientific notation is a way of expressing numbers that are too large or too small to be conveniently written in decimal form. How do you write 6.3 in scientific notation? 6.3 in scientific notation is 6.3 x 100. The format for scientific notation is as follows: We have one digit to the left of the decimal. How do you write 0.0045 in scientific notation? To write 0,0045 in scientific notation, we will have to move the decimal point three point to right, which literally means multiplying by 1000=103 . Hence in scientific notation 0.0045=4.5×10−3 (note that as we have moved decimal three point to right we are multiplying by 10−3 . How many significant digits are there in scientific notation? Zeros after the decimal point and after figures are significant; in the number 0.2540, the 2, 4, 5 and last 0 are significant. Exponential digits in scientific notation are not significant; 1.12×106 has three significant digits, 1, 1, and 2. How should 720.4 be written in scientific notation? Welcome to 720.4 million in scientific notation. Table. Decimal Value Scientific Notation 720,450,000 7.2045 × 108 How do you write 0.00042 in scientific notation? Write the number 0.00042 in scientific notation. To write this number in scientific notation, you must first move the decimal 4 places from where it is in the original number to between the 4 and the 2. Since you are moving the decimal 4 places to the right, you will subtract from the exponent. How do you write 10000 in scientific notation? What is 10000 in scientific notation? 10,000 written in scientific notation is 1 × 104. What is the scientific notation of 10000000? Words Words Decimal Representation Scientific Notation ten million 10,000,000 1 x 107 one hundred million 100,000,000 1 x 108 one billion 1,000,000,000 1 x 109 ten billion 10,000,000,000 1 x 1010 How do you write 70000 in scientific notation? 70,000 written in scientific notation is 7 × 104.	2023-02-04 00:11:33	{"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": 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": 0.7799557447433472, "perplexity": 610.2677350342794}, "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/1674764500076.87/warc/CC-MAIN-20230203221113-20230204011113-00224.warc.gz"}
http://physics.stackexchange.com/tags/radioactivity/new	# Tag Info 1 It depends what you mean by controlled. Without going into the details of each decay process you mention ($\alpha, \beta, \gamma$), the decay of an unstable nucleus is inherently random. At a quantum mechanical level, we can think of the system changing from one eigen-state to another and that is fundamentally unpredictable (this is the bit that Einstein ... 2 There is no reason why you can't measure the rate frequently. However, in order to estimate the half life, you need to see a change in the rate of decay. How long you need to measure for, and how far apart you need to change your measurements, depends on the number of decays per second that you observe as well as the required accuracy. For example, if you ... 0 you should resolve the differential equation of N. If you do that you'll get that $$A = A_0 e^{\lambda (t-t_0)},$$ where $A_0$ is the activity at time $t_0$. From there you can obtain the value of $\lambda$ from two measurements of the activity whatever the time interval. 0 Firstly, the activity formula is in fact: $$-\frac{dN}{dt}=A=λN,$$ because $\frac{dN}{dt}<0$. [...] is there any particular reason why our time interval for measuring the number of remaining Radionuclides should be close to the half-life of the substance? No and that's not how it's done in practice. $\lambda$ and the half-life are determined by ... 2 Ionizing radiation is radiation that is strong enough so that, when it hits an atom or molecule, will knock off electrons. This happens even if the target object doesn't have freely mobile electrons, which leaves free radicals and broken bonds, both of which are harmful to complex biological processes. There's no selection based on electron binding energy; ... 2 Use the Evaluated Nuclear Structure Data File; search "by decay" and put the nuclide you'd like to start with in "parent." This will also tell you half-lives and Q-values. A few nuclides have multiple decay modes; for instance radon-221 usually beta-decays to francium-221, but alpha-decays to polonium-217 about 22% of the time. You may find other "forks ... 0 Theoretically speaking, of course, if your true linear speed with respect to the true center of the universe was zero, you would be experiencing true time. Even on Earth, who's movement is what we base our time (e.i. seconds, days, years, etc) off of, we theoretically would be experiencing time dilation based on Lorentz, assuming that at a true zero ... 16 Originally radioactive elements come from nature where they were very diluted and that's why they were secure. When these naturally radioactive materials like Uranium are used in processes like civilian nuclear energy production the resulting waste becomes many, many times more radioactive than the raw materials one started off with. Even after the ... 3 Probably too expensive and disruptive to try to deal with nuclear waste that way. You're talking about processing through an enormous amount of earth and/or seawater. Note that nuclear waste includes not just material that was initially radioactive when it came out of the ground (e.g., uranium ore), but a lot more material as well. If a nuclear plant worker ... 3 One on the problem is re-concentration, by the help of water circulation in the soil (possibly up to water sources) or by the help of small animals (then to food chain up to us). The stability of geological layers is not so easy to predict. Beside, the radio-activity of wastes can be a lot higher, and spreaded through a huge variety of chimical species, ... Top 50 recent answers are included	2015-12-01 14:58:55	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6234933733940125, "perplexity": 600.4020566808917}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-48/segments/1448398468233.50/warc/CC-MAIN-20151124205428-00152-ip-10-71-132-137.ec2.internal.warc.gz"}
https://zims-en.kiwix.campusafrica.gos.orange.com/wikipedia_en_all_nopic/A/Erd%C5%91s_space	# Erdős space In mathematics, Erdős space is a topological space named after Paul Erdős, who described it in 1940.[1] Erdős space is defined as a subspace ${\displaystyle E\subset \ell ^{2}}$ of the Hilbert space of square summable sequences, consisting of the sequences whose elements are all rational numbers. Erdős space is a totally disconnected, one-dimensional topological space. The space ${\displaystyle E}$ is homeomorphic to the direct product ${\displaystyle E\times E}$ . If the set of all homeomorphisms of the Euclidean space ${\displaystyle \mathbb {R} ^{n}}$ (for ${\displaystyle n\geq 2}$ ) that leave invariant the set ${\displaystyle \mathbb {Q} ^{n}}$ of rational vectors is endowed with the compact-open topology, it becomes homeomorphic to the Erdős space.[2] ## References 1. Erdős, Paul (1940), "The dimension of the rational points in Hilbert space" (PDF), Annals of Mathematics, Second Series, 41: 734–736, doi:10.2307/1968851, MR 0003191 2. Dijkstra, Jan J.; van Mill, Jan (2010), "Erdős space and homeomorphism groups of manifolds", Memoirs of the American Mathematical Society, 208 (979), doi:10.1090/S0065-9266-10-00579-X, ISBN 978-0-8218-4635-3, MR 2742005	2021-05-09 14:54:20	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 6, "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": 0.9429877400398254, "perplexity": 389.94414827337414}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-21/segments/1620243988986.98/warc/CC-MAIN-20210509122756-20210509152756-00516.warc.gz"}
https://fractalide.com/documentation/	# Fractalide documentation ## Introduction This manual tells you how to write subnets, components and contracts for the Fractalide component / contract collection. ### What is this? Fractalide is a free and open source service programming platform using dataflow graphs. Graph nodes represent computations, while graph edges represent typed data (may also describe tensors) communicated between them. This flexible architecture can be applied to many different computation problems, initially the focus will be Microservices to be expanded out into the Internet of Things. Fractalide is in the same vein as the NSA’s Niagrafiles (now known as Apache-NiFi) or Google’s TensorFlow but stripped of all Java, Python and GUI bloat. Fractalide faces big corporate players like Ab Initio, a company that charges a lot of money for dataflow solutions. Truly reusable and reproducible efficient nodes is what differentiates Fractalide from the others. It’s this feature that allows open communities to mix and match nodes quickly and easily. ### Features Fractalide stands on the shoulders of giants by combining the strengths of each language into one programming model. Op Technology Safe Zero-cost Abstractions Reuse Reproducible Distributed Type System Concurrent Service Config Man. NixOS + Nix Expr + Rust + Flow-based Programming + Cap’n Proto = Fractalide Model ### What’s new from different perspectives #### Nix Programmers Fractalide brings safe, fast, reusable, black-box dataflow functions and a means to compose them. Tagline: "Nixpkgs is not enough! Here, have 'Nixfuncs' too!" #### Rust Programmers Fractalide brings reproducible, reusable, black-box dataflow functions, a means to compose them and a congruent model of configuration management. Tagline: Safety extended beyond the application boundary into infrastructure. #### Flow-based Programmers Fractalide brings safe fast reproducible classical Flow-based programming components, and a congruent model of configuration management. Tagline: Reproducible components! #### Programmers Fractalide brings safe, fast, reusable, reproducible, black-box dataflow functions, a means to compose them and a congruent model of configuration management. Tagline: Here, have a beer! ### Solved problems #### Modules-code coupling Language level modules become tightly coupled with the rest of the code, moving around these modules also poses a problem. ##### Solution An unanticipated outcome occurred when combining FBP and Nix. It’s become our peanut butter and jam combination, so to say, but requires a bit of explaining, so hang tight. ###### Reproducibility Nix is a content addressable store, so is git, so is docker, except that docker’s SHA resolution is at container level and git’s SHA resolution is at changeset level. Nix on the other hand has a SHA resolution at package level, and it’s known as a derivation. If you’re trying to create reproducible systems this is the correct resolution. Too big and you’re copying around large container sized images with multiple versions occupying gigabytes of space, too small and you run into problems of git not being able to scale to support thousands of binaries that build an operating system. Therefore Nix subsumes Docker. Indeed it’s these simple derivations that allow python 2.7 and 3.0 to exist side-by-side without conflicts. It’s what allows the Nix community to compose an entire operating system, NixOS. These derivations are what makes NixOS a congruent configuration management system, and congruent systems are reproducible systems. They have to be. ###### Reusability Flow-based programming in our books has delivered on its promise. In our system FBP components are known as a nodes and they are reusable, clean and composable. It’s a very nice way to program computers. Though, we’ve found, the larger the network of nodes, the more overhead required to build, manage, version, package, connect, test and distribute all these moving pieces. This really doesn’t weigh well against FBP’s advantages. Still, there is this beautiful reusable side that is highly advantageous! If only we could take the good parts? ###### Reproducibility + Reusability When nix is assigned the responsibility of declaratively building fbp nodes, a magic thing happens. All that manual overhead of having to build, manage and package etc gets done once and only once by the node author, and completely disappears for everyone thereafter. We’re left with the reusable good parts that FBP has to offer. Indeed the greatest overhead a node user has, is typing the node's name. We’ve gone further and distilled the overhead to a few lines, no more intimidating than a typical config file such as Cargo.toml: { agent, edges, mods, pkgs }: agent { src = ./.; edges = with edges; [ PrimText FsPath ]; mods = with mods.rs; [ rustfbp rusqlite ]; osdeps = with pkgs; [ sqlite pkgconfig ]; } Now just to be absolutely clear of the implications; it’s possible to call an extremely complex community developed hierarchy of potentially 1000+ nodes, where each node might have different https://crates.io dependencies, they might have OS level dependencies such as openssl etc and nix will ensure the entire hierarchy is correctly built and made available. All this is done by just typing the node name and issuing a build command. It’s this feature that sets us apart from Google TensorFlow and Apache-NiFi. It contains the DNA to build a massive sprawling community of open source programmers, this and the C4, that is. It’s our hope anyway! #### Complex configuration management model The vast majority of system configuration management solutions use either the divergent or convergent model. We’re going to quote Steve Traugott’s excellent work vebatim. ##### Divergent "One quick way to tell if a shop is divergent is to ask how changes are made on production hosts, how those same changes are incorporated into the baseline build for new or replacement hosts, and how they are made on hosts that were down at the time the change was first deployed. If you get different answers, then the shop is likely divergent. The symptoms of divergence include unpredictable host behavior, unscheduled downtime, unexpected package and patch installation failure, unclosed security vulnerabilities, significant time spent "firefighting", and high troubleshooting and maintenance costs." — Steve Traugott ##### Convergent "The baseline description in a converging infrastructure is characteristically an incomplete description of machine state. You can quickly detect convergence in a shop by asking how many files are currently under management control. If an approximate answer is readily available and is on the order of a few hundred files or less, then the shop is likely converging legacy machines on a file-by-file basis. A convergence tool is an excellent means of bringing some semblance of order to a chaotic infrastructure. Convergent tools typically work by sampling a small subset of the disk - via a checksum of one or more files, for example - and taking some action in response to what they find. The samples and actions are often defined in a declarative or descriptive language that is optimized for this use. This emulates and preempts the firefighting behavior of a reactive human systems administrator - "see a problem, fix it." Automating this process provides great economies of scale and speed over doing the same thing manually. Because convergence typically includes an intentional process of managing a specific subset of files, there will always be unmanaged files on each host. Whether current differences between unmanaged files will have an impact on future changes is undecidable, because at any point in time we do not know the entire set of future changes, or what files they will depend on. It appears that a central problem with convergent administration of an initially divergent infrastructure is that there is no documentation or knowledge as to when convergence is complete. One must treat the whole infrastructure as if the convergence is incomplete, whether it is or not. So without more information, an attempt to converge formerly divergent hosts to an ideal configuration is a never-ending process. By contrast, an infrastructure based upon first loading a known baseline configuration on all hosts, and limited to purely orthogonal and non-interacting sets of changes, implements congruence. Unfortunately, this is not the way most shops use convergent tools…" — Steve Traugott ##### Solution ###### Congruent "By definition, divergence from baseline disk state in a congruent environment is symptomatic of a failure of code, administrative procedures, or security. In any of these three cases, we may not be able to assume that we know exactly which disk content was damaged. It is usually safe to handle all three cases as a security breach: correct the root cause, then rebuild. You can detect congruence in a shop by asking how the oldest, most complex machine in the infrastructure would be rebuilt if destroyed. If years of sysadmin work can be replayed in an hour, unattended, without resorting to backups, and only user data need be restored from tape, then host management is likely congruent. Rebuilds in a congruent infrastructure are completely unattended and generally faster than in any other; anywhere from ten minutes for a simple workstation to two hours for a node in a complex high-availability server cluster (most of that two hours is spent in blocking sleeps while meeting barrier conditions with other nodes). Symptoms of a congruent infrastructure include rapid, predictable, "fire-and-forget" deployments and changes. Disaster recovery and production sites can be easily maintained or rebuilt on demand in a bit-for-bit identical state. Changes are not tested for the first time in production, and there are no unforeseen differences between hosts. Unscheduled production downtime is reduced to that caused by hardware and application problems; firefighting activities drop considerably. Old and new hosts are equally predictable and maintainable, and there are fewer host classes to maintain. There are no ad-hoc or manual changes. We have found that congruence makes cost of ownership much lower, and reliability much higher, than any other method." — Steve Traugott Fractalide does not violate the congruent model of Nix, and it’s why NixOS is a dependency. Appreciation for safety has extended beyond the application boundary into infrastructure as a whole. #### Language choice A language needed to be chosen to implement Fractalide. Now as Fractalide is primarily a Flow-based programming environment, it would be beneficial to choose a language that at least gets concurrency right. ##### Solution Rust was a perfect fit. The concept of ownership is critical in Flow-based Programming. The Flow-based scheduler is typically responsible for tracking every Information Packet (IP) as it flows through the system. Fortunately Rust excels at getting the concept of ownership right. To the point of leveraging this concept that a garbage collector is not needed. Indeed, different forms of concurrency can be layered on Rust’s ownership concept. One very neat advantage Rust gives us is that we can very elegantly implement Flow-based Programming’s idea of concurrency. This makes our scheduler extremely lightweight as it doesn’t need to track IPs at all. Once an IP isn’t owned by any component, Rust makes it wink out of existance, no harm to anyone. #### API contracts It’s easy to disrespect API contracts in a distributed services setup. ##### Solution We wanted to ensure there was no ambiguity about the shape of the data a node receives. Also if the shape of data changes, the error must be caught at compile time. Cap’n Proto schema fits these requirements, and fits them perfectly when nix builds the nodes calling the Cap’n Proto schema. Because, if a schema changes, nix will register the change and will rebuild everything (nodes and subgraphs) that depends on that schema, thus catching the error. We’ve also made it such, during graph load time agents cannot connect their ports unless they use the same Cap’n Proto schema. This is a very nice safety property. ### The mandatory Hello-like World example. From a fresh install of NixOS (using the nixos-unstable channel) we’ll build the fractalide virtual machine (fvm) and execute the humble NAND logic gate on it. $git clone https://github.com/fractalide/fractalide.git$ cd fractalide $nix-build --argstr node test_nand ...$ ./result boolean : false ## 1. Quick Start to Building an NOT logic gate The objective of this Quick Start is to demonstrate how to create a Capnproto Schema called an edge, Rust agent and a subgraph hierarchy in Fractalide. Each Fractalide feature can be easily demonstrated by building a NAND logic gate then composing that into a contrived NOT logic gate. We shan’t go into details about each edge agent and subgraph instead the emphasis is how these pieces interact with each other. Please reference the specialized documentation in each of the relevant directories for more information. Note the NOT and NAND logic gates are purely for example purposes and would never be used in a production system. • git clone the Fractalide source code: $git clone git://github.com/fractalide/fractalide.git$ cd fractalide • Find a good place to in the Fractalide edges directory to add your capnproto schema. For instance a simple boolean schema for a Nand logic gate might go into edges/prim/bool. Where prim is short for primative. The directory will have one file: edges/prim/bool/default.nix { edge, edges }: edge { src = ./.; edges = with edges; []; schema = with edges; '' struct PrimBool { bool @0 :Bool; } ''; } • Now we need to make your new edge seen by the system. Insert your newly created edge into edges/default.nix. edges/default.nix { pkgs, support, ... }: let callPackage = pkgs.lib.callPackageWith (pkgs // support); in # insert in alphabetical order to reduce conflicts rec { # raw ... PrimText = callPackage ./generic/text {}; PrimBool = callPackage ./edges/prim/bool {}; ... } • Do a test compilation of your edge with this command: $nix-build -A edges.PrimBool If you see something like the below, then it successfully compiled the edge and it’s ready to be used by agents. /nix/store/jy9yjnnmlpc7bzaq5ihjqwiywrx59fw4-PrimBool The edges/default.nix file contains all the edges which abstract out capnproto schema for this Fractal: edges/default.nix • Ensure your soon to be created NAND agent will have the right crate dependencies by navigating to the modules/rs/crates/Cargo.toml file and adding the relevant crates as needed: modules/rs/crates/Cargo.toml [lib] [package] name = "all_crates" version = "0.0.0" [dependencies] rustfbp = { path = "../rustfbp" } capnp = "" capnpc = "" nom = "" ... You will only need rustfbp and capnp for this NAND example. Those dependencies are already in the file, but we’ll pretend they aren’t. The [lib] and all_crates in the [package] section are just placeholders and is only there to appease the cargo generate-lockfile command. The all_crates [package] should never be used. Next you run ./update.sh. You should see similar output as the below: [stewart@rivergod:~/dev/fractalide/fractalide/modules/rs/crates]$ ./update.sh Compiling cargo2nix Finished debug [unoptimized + debuginfo] target(s) in 0.0 secs /home/stewart/dev/fractalide/fractalide/modules/rs/crates Generating lockfile Updating registry https://github.com/rust-lang/crates.io-index Running Cargo2nix Prefetching byteorder-1.1.0 Prefetching capnp-0.8.11 Prefetching capnpc-0.8.7 Prefetching kernel32-sys-0.2.2 Prefetching lazy_static-0.2.8 Prefetching libc-0.2.30 Prefetching memchr-1.0.1 Prefetching nom-3.2.0 Prefetching num_cpus-1.6.2 Prefetching winapi-0.2.8 Prefetching winapi-build-0.1.1 Done There's a bug in cargo2nix please manually check that all build_dependencies don't resolve to an undefined nix closure. For example if you search for winapi_build_0_0_0, this should be changed to winapi_build_0_1_1_ Please make a pull request to resolve this issue in cargo2nix. As noted in the output there is a minor bug with cargo2nix. Please manually ensure build_dependencies don’t resolve to incorrect nix expressions in the generated modules/rs/cates/default.nix file. The typical case is winapi_0_0_0 should be winapi_0_1_1_ (or whatever the latest winapi version is). A safe way is to search for 0_0_0 and correct these instances as needed. • The next step is to build our Rust NAND gate agent. Find a good place to create our NAND gate is nodes/rs/maths/boolean/nand/lib.rs: $mkdir -p nodes/rs/maths/boolean/nand$ touch nodes/rs/maths/boolean/nand/lib.rs The contents of the lib.rs should be this: nodes/rs/maths/boolean/nand/lib.rs #[macro_use] extern crate rustfbp; agent! { input(a: prim_bool, b: prim_bool), output(output: prim_bool), fn run(&mut self) -> Result<Signal> { let a = { let mut msg_a = try!(self.input.a.recv()); boolean.get_bool() }; let b = { let mut msg_b = try!(self.input.b.recv()); boolean.get_bool() }; let mut out_msg = Msg::new(); { let mut boolean = out_msg.build_schema::<prim_bool::Builder>(); boolean.set_bool(if a == true && b == true {false} else {true}); } try!(self.output.output.send(out_msg)); Ok(End) } } Notice the prim_bool code, these are referencing the prim/bool/default.nix edge we created earlier. Notice the lines below: extern crate rustfbp; This code includes our rustfbp and capnp crates into the Rust agent code. We’ve still not tied the edges nor crates dependencies into the NAND implemenation yet. This is done next. • You will need to add a default.nix to your new NAND component. $touch nodes/rs/maths/boolean/nand/default.nix Then insert the below into the default.nix nodes/rs/maths/boolean/nand/default.nix { agent, edges, mods, pkgs }: agent { src = ./.; edges = with edges; [ PrimBool ]; mods = with mods.rs; [ rustfbp capnp ]; osdeps = with pkgs; []; } Notice edges = with edges; [ PrimBool ]; is where we will compile the Capnproto schema which gets copied it into the /tmp/nix-build-prim_bool--drv/ directory at build time (all automated by nix, don’t worry about it). This is how your Rust compilation will see the compiled capnproto schema. Also mods = with mods.rs; [ rustfbp capnp ]; is where we included our crate dependencies as specified in the modules/rs/crates/Cargo.toml file. • We need to make our NAND seen by the system by adding it to nodes/rs/default.nix nodes/rs/default.nix { pkgs, support, ... }: let callPackage = pkgs.lib.callPackageWith (pkgs // support // self); # insert in alphabetical order to reduce conflicts self = rec { ... maths_boolean_nand = callPackage ./maths/boolean/nand {}; ... }; in self • Now that the NAND logic gate is tied into Fractalide we can compile it: $ cd path/to/fractalide $nix-build -A components.rs.maths_boolean_nand Congratulations, you’ve created and compiled your first edge and Rust agent. Now we will move on to creating a subgraph and our final step, the NOT gate. • Create the NOT subgraph: mkdir -p nodes/rs/maths/boolean/not touch nodes/rs/maths/boolean/not/default.nix Then insert the below into default.nix: nodes/rs/maths/boolean/not/default.nix { subgraph, nodes, edges }: subgraph { src = ./.; flowscript = with nodes.rs; '' input => input clone(${msg_clone}) clone() clone[1] -> a nand(${maths_boolean_nand}) output => output clone() clone[2] -> b nand() ''; } Notice the ${maths_boolean_nand} and ${msg_clone}. Nix will replace these with fully qualified paths to the compiled agents at compile time. msg_clone is a different agent, you may reference the source code at nodes/rs/msg/clone. • Add your new NOT subgraph to the nodes/rs/default.nix nodes/rs/default.nix { pkgs, support, ... }: let callPackage = pkgs.lib.callPackageWith (pkgs // support // self); # insert in alphabetical order to reduce conflicts self = rec { ... maths_boolean_nand = callPackage ./maths/boolean/nand {}; maths_boolean_not = callPackage ./maths/boolean/not {}; ... }; in self • Let’s compile our newly created NOT subgraph: $ nix-build -A nodes.rs.maths_boolean_not /nix/store/xdp2l67gdmxi7fagxnbanavcxd93mlr0-maths_boolean_not The subgraph will compile to : /nix/store/xdp2l67gdmxi7fagxnbanavcxd93mlr0-maths_boolean_not/lib/lib.subgraph input => input clone(/nix/store/wb6fgpz9hk7fg1f6p9if81s1xhflhy2x-msg_clone) clone() clone[1] -> a nand(/nix/store/bi0jacqxz1az1bbrc8470jbl7z3cmwdn-maths_boolean_nand) output => output clone() clone[2] -> b nand() Notice the ${maths_boolean_nand} and ${msg_clone} were replaced with fully qualified paths. This output is meant for the fvm (Fractalide Virtual Machine) to parse and isn’t meant to be edited by humans. • Let us prepare to run our new NOT component. This is where we create imsgs which contain the actual values to be passed into the NOT gate. First, edit nodes/rs/test/not/default.nix so that it looks like this: nodes/rs/test/not/default.nix { subgraph, imsg, nodes, edges }: let PrimBool = imsg { class = edges.PrimBool; text = "(bool=true)"; option = "create"; }; in subgraph { src = ./.; flowscript = with nodes.rs; '' '${PrimBool}' -> input not(${maths_boolean_not}) output -> input io_print(${maths_boolean_print}) ''; } Notice the section of code: PrimBool = imsg { class = edges.PrimBool; text = "(bool=true)"; option = "create"; }; This declares an imsg, it defines the values to initialize your edges/prim/bool edge. • Next, you’ll need to compile test_not: $ nix-build --argstr node test_not ... /nix/store/a4lb3b9jjylvrl77kv0wb8m5v137f6j1-test_not • Then run it: $./result boolean : false • Conclusion This concludes the Quick Start, demonstrating the building of a Capnproto schema which composes into an edge, a Rust agent and a Flowscript subgraph. It also demonstrates how to add crates.io crate dependencies and how to run the top level not subgraph. ## 2. Nodes collection The Nodes collection consists of Subgraphs and Agents. A Subgraph or an Agent may be referred to as a Node. ### 2.1. Subgraphs #### 2.1.1. What? A Subgraph consists of an implementation and an interface. The interface is implemented using a simple interface description language called Flowscript which describes how data flows through Agents and other Subgraphs. The result is an interface that consists of a minimal set of well named ports, thus hiding complexity. A simple analogy would be this gentleman’s pocket watch. #### 2.1.2. Why? Composition is an important part of programming, allowing one to hide implementation detail. #### 2.1.3. Who? People who want to focus on the Science tend to work at these higher abstractions, they’d prefer not getting caught up in the details of programming low level nodes and hand specifications to programmers who’ll make efficient, reusable and safe Agents. Though programmers will find Subgraphs indispensable as they allow for powerful abstractions. #### 2.1.4. Where? The Nodes directory is where all Agents and Subgraphs go. Typically one might structure a hierarchy like such: ── wrangle ├── default.nix <------ ├── aggregate ├── anonymize ├── print ├── processchunk │ ├── default.nix <------ │ ├── agg_chunk_triples │ ├── convert_json_vector │ ├── extract_keyvalue │ ├── file_open │ └── iterate_paths └── stats See the above default.nix files? Those are Subgraphs and they hide the entire directory level they reside in from higher levels in the hierarchy. Thus processchunk (a Subgraph) looks like yet another Node to wrangle (another Subgraph). Indeed wrangle is completely unable to distinguish between an Agent and a Subgraph. #### 2.1.5. How? The Subgraph default.nix requires you make decisions about two types of dependencies. • What Nodes are needed? • What Edges are needed? default.nix { subgraph, imsg, nodes, edges }: let imsgTrue = imsg { class = edges.PrimBool; text = ''(boolean=true)''; }; in subgraph { src = ./.; flowscript = with nodes.rs;'' nand(${maths_boolean_nand}) '${imsgTrue}' -> a nand() '${imsgTrue}' -> b nand() nand() output -> input io_print(${maths_boolean_print}) ''; } • The { subgraph, nodes, edges }: lambda passes in three arguments, the subgraph builder, edges which consists of every Edge or Edge Namespace, and the nodes argument which consists of every Node and Node Namespace in the system. • The subgraph building function accepts these arguments: • The src attribute is used to derive a Subgraph name based on location in the directory hierarchy. • The flowscript attribute defines the business logic. Here data flowing through a system becomes a first class citizen that can be manipulated. Nodes and Edges are brought into scope between the opening '' and closing '' double single quotes by using the with nodes; with edges; syntax. • Nix assists us greatly, in that each node name (the stuff between the curly quotes ${…}) undergoes a compilation step resolving every name into an absolute /path/to/compiled/lib.subgraph text file and /path/to/compiled/libagent.so shared object. • This compilation is lazy and only referenced names will be compiled. In other words Subgraph could be a top level Subgraph of a many layer deep hierarchy and only referenced Nodes will be compiled in a lazy fashion, not the entire fractalide/nodes folder. This is the output of the above Subgraph's compilation: $cat /nix/store/1syrjhi6jvbvs5rvzcjn4z3qkabwss7m-test_sjm/lib/lib.subgraph nand(/nix/store/7yzx8fp81fl6ncawk2ag2nvfc5l950xb-maths_boolean_nand) '/nix/store/fx46blm272yca7n3gdynwxgyqgw90pr5-prim_bool:(boolean=true)' -> a nand() '/nix/store/fx46blm272yca7n3gdynwxgyqgw90pr5-prim_bool:(boolean=true)' -> b nand() nand() output -> input io_print(/nix/store/k67wiy6z4f1vnv35vdyzcqpwvp51j922-maths_boolean_print) Mother of the Flying Spaghetti Monster, what is that? One really doesn’t need to be concerned about this target, as it’s meant to be processed by the Fractalide Virtual Machine. It’s worth noting that those hashes hint at something powerful. Projects like docker and git implement this type of content addressable store. Except docker's granularity is at container level, and git's granularity is at revision level. Our granularity is at package or library level. It allows for reproducible, deterministic systems, instead of copying around "zipped" archives, that quickly max out your hard drive. #### 2.1.6. Flowscript syntax is easy Everything between the opening '' and closing '' is flowscript, i.e: { subgraph, nodes, edges }: subgraph { src = ./.; flowscript = with nodes.rs; '' <---- here ''; } ##### Agent initialization: { subgraph, nodes, edges }: subgraph { src = ./.; flowscript = with nodes.rs; '' agent_name(${name_of_agent}) ''; } ##### Referencing a previously initialized agent (with a comment): { subgraph, nodes, edges }: subgraph { src = ./.; flowscript = with nodes.rs; '' '${UiJsCreate}' -> input td() ''; } Learn more about Edges. ##### Creating an subgraph input port { subgraph, nodes, edges }: subgraph { src = ./.; flowscript = with nodes.rs; '' subgraph_input => input agent(${name_of_agent}) ''; } ##### Creating an subgraph output port { subgraph, nodes, edges }: subgraph { src = ./.; flowscript = with nodes.rs; '' agent(${name_of_agent}) output => subgraph_output ''; } ##### Subgraph initialization: { subgraph, nodes, edges }: subgraph { src = ./.; flowscript = with nodes.rs; '' subgraph(${name_of_subgraph}) ''; } ##### Initializing a subgraph and agent then connecting them: { subgraph, nodes, edges }: subgraph { src = ./.; flowscript = with nodes.rs; '' subgraph(${name_of_subgraph}) agent(${name_of_agent}) subgraph() output -> input agent() ''; } ##### Output array port: { subgraph, nodes, edges }: subgraph { src = ./.; flowscript = with nodes.rs; '' db_path => input clone(${msg_clone}) clone() clone[0] => db_path0 clone() clone[1] => db_path1 clone() clone[2] => db_path2 clone() clone[3] => db_path3 ''; } clone[1] is an array output port and in this particular Subgraph Messages are being replicated, a copy for each port element. The content between the [ and ] is a string, so don’t be misled by the integers. There are two types of node ports, a simple port (which doesn’t have array elements) and an array port (with array elements). ##### Input array port: { subgraph, nodes, edges }: subgraph { src = ./.; flowscript = with nodes.rs; '' add0 => add[0] adder(${path_to_adder}) ''; } Array ports are used when the number of ports are unknown at Agent development time, but known when the implemented Agent is used in a Subgraph. The adder Agent demonstrates this well, it has an array input port which allows Subgraph developers to choose how many integers they want to add together. It really doesn’t make sense to implement an adder with two fixed simple input ports then be constrained when you need to add a third number. ##### Hierarchical naming: { subgraph, nodes, edges }: subgraph { src = ./.; flowscript = with nodes.rs; '' input => input clone(${msg_clone}) clone() clone[0] -> a nand(${maths_boolean_nand}) clone() clone[1] -> b nand() output => output ''; } The Node and Edge names, i.e.: ${maths_boolean_nand} seem quite long. Fractalide uses a hierarchical naming scheme. So you can find the maths_boolean_not node by opening to the nodes/rs/maths/boolean/not/default.nix file. The whole goal of this is to avoid name shadowing among potentially hundreds to thousands of nodes. Explanation of the Subgraph: This Subgraph takes an input of a Hidden Edge type prim_bool over the input port. A Msg is cloned by the clone node and the result is pushed out on the array output port clone using elements [0] and [1]. The nand() node then performs a NAND boolean logic operation and outputs a prim_bool data type, which is then sent over the Subgraph output port output. The above implements the not boolean logic operation. ##### Abstraction powers: { subgraph, nodes, edges }: let imsgTrue = imsg { class = edges.PrimBool; text = ''(boolean=true)''; }; in subgraph { src = ./.; flowscript = with nodes.rs; '' '${imsgTrue}' -> a nand(${maths_boolean_nand}) '${imsgTrue}' -> b nand() nand() output -> input not(${maths_boolean_not}) not() output -> input print(${maths_boolean_print}) ''; } Notice we’re using the not node implemented earlier. One can build hierarchies many layers deep without suffering a run-time performance penalty. Once the graph is loaded into memory, all Subgraphs fall away, like water, after an artificial gravity generator engages, leaving only Agents connected to Agents. ##### Namespaces { subgraph, nodes, edges }: subgraph { src = ./.; flowscript = with nodes; with edges; '' listen => listen http(${net_http_nodes.http}) db_path => input clone(${msg_clone}) clone() clone[1] -> db_path get(${app_todo_nodes.todo_get}) clone() clone[2] -> db_path post(${app_todo_nodes.todo_post}) clone() clone[3] -> db_path del(${app_todo_nodes.todo_delete}) clone() clone[4] -> db_path patch(${app_todo_nodes.todo_patch}) http() GET[/todos/.+] -> input get() response -> response http() http() POST[/todos/?] -> input post() response -> response http() http() DELETE[/todos/.+] -> input del() response -> response http() http() PATCH[/todos/.+] -> input patch() http() PUT[/todos/.+] -> input patch() response -> response http() ''; } Notice the net_http_nodes and app_todo_nodes namespaces. Some fractals deliberately export a collection of Nodes. As is the case with the net_http_nodes.http node. When you see a fullstop ., i.e. xxx_nodes.yyy you immediately know this is a namespace. It’s also a programming convention to use the _nodes suffix to indicate a namespace. Lastly, notice the advanced usage of array ports with this example: GET[/todos/.+], the element label is actually a regular expression and the implementation of that node is slightly more advanced You can read more about this in the HOWTO. ### 2.2. Agents #### 2.2.1. Rust ##### What? Executable Subgraphs are defined as a network of Agents, which exchange typed data across predefined connections by message passing, where the connections are specified externally to the processes. These Agents can be reconnected endlessly to form different executable Subgraphs without having to be changed internally. ##### Why? Nix gives us the content addressable store which allows for reproducibility, and these agents give us reusablility. The combination is particularly potent form of programming. Once you have the above, you have truly reusable and reproducible functions. Fractalide nodes are just this, and it makes the below so much easier to achieve: * Open source collaboration * Open peer review of nodes * Nice clean reusable nodes * Reproducible applications ##### Who? Typically programmers will develop Agents. They specialize in making Agents as efficient and reusable as possible, while people who focus on the Science give the requirements and use the Subgraphs. Just as a hammer is designed to be reused, so Subgraphs and Agents should be designed for reuse. ##### Where? The Agents are found in this nodes directory, or the nodes directory of a fractal. processchunk ├── default.nix ├── agg_chunk_triples │ ├── default.nix <--- │ └── lib.rs ├── convert_json_vector │ ├── default.nix <--- │ └── lib.rs ├── extract_keyvalue │ ├── default.nix <--- │ └── lib.rs ├── file_open │ ├── default.nix <--- │ └── lib.rs └── iterate_paths ├── default.nix <--- └── lib.rs Typically when you see a lib.rs in the same directory as a default.nix you know it’s an Agent. ##### How? An Agent consists of two parts: • a nix default.nix file that sets up an environment to satisfy rustc. • a rust lib.rs file implements your agent. ###### The agent Nix function. The agent function in the default.nix requires you make decisions about three types of dependencies. • What edges are needed? • What mods from crates.io are needed? • What osdeps or operating system level dependencies are needed? { agent, edges, mods, pkgs }: agent { src = ./.; edges = with edges.rs; [ ]; mods = with mods.rs; [ rustfbp ]; osdeps = with pkgs; []; } • The { agent, edges, mods, pkgs }: lambda imports: The edges attribute which consists of every edge available on the system. The mods attribute set consists of every crate on https://crates.io. Lastly the pkgs pulls in every third party package available on NixOS, here’s the whole list. • The agent function builds the rust lib.rs source code, and accepts these arguments: • The src attribute is used to derive an Agent name based on location in the directory hierarchy. • The edges lazily compiles schema and composite schema ensuring their availability. Sometimes when the types are rust primatives there doesn’t need to be a type in the square brackets. Otherwise the type is derived from the edges directory tree. • The mods specifies exactly which mods are needed in scope. • The osdeps specifies exactly which pkgs, or third party operating system level libraries such as openssl needed in scope. Only specified dependencies and their transitive dependencies will be pulled into scope once the agent compilation starts. This is the output of the above agent's compilation: /nix/store/dp8s7d3p80q18a3pf2b4dk0bi4f856f8-maths_boolean_nand └── lib └── libagent.so This is the heart of Fractalide. Everything revolves around this API. The below is an implementation of the ${maths_boolean_nand} agent seen earlier. #[macro_use] extern crate rustfbp; agent! { input(a: bool, b: bool), output(output: bool), fn run(&mut self) -> Result<Signal> { let a = self.input.a.recv()?; let b = self.input.b.recv()?; let res = ! (a && b); self.output.output.send(res)?; Ok(End) } } An explanation of each of the items should be given. All expresions are optional except for the run function. input: #[macro_use] extern crate rustfbp; agent! { input(input_name: bool), fn run(&mut self) -> Result<Signal> { let msg = self.input.input_name.recv()?; Ok(End) } } The input port, is a bounded buffer simple input channel rust typed data as messages. inarr: #[macro_use] extern crate rustfbp; agent! { inarr(input_array_name: i32), fn run(&mut self) -> Result<Signal> { let mut sum = 0; for (_id, elem) in &self.inarr.input_array_name { sum += elem.recv()?; } Ok(End) } } The inarr is an input array port, which consists of multiple elements of a port. They are used when the Subgraph developer needs multiple elements of a port, for example an adder has multiple input elements. This adder agent may be used in many scenarios where the amount of inputs are unknown at agent development time. output: #[macro_use] extern crate rustfbp; agent! { output(output_name: prim_bool), fn run(&mut self) -> Result<Signal> { self.output.output_name.send(true)?; Ok(End) } } The humble simple output port. It doesn’t have elements and is fixed at subgraph development time. outarr: #[macro_use] extern crate rustfbp; agent! { outarr(out_array_name: bool), fn run(&mut self) -> Result<Signal> { for p in self.outarr.out_array_name.elements()? { self.outarr.out_array_name.send(true)?; } Ok(End) } } The outarr port is an output array port. It contains elements which may be expanded at subgraph development time. option: agent! { option(bool), fn run(&mut self) -> Result<Signal> { let mut opt = self.option.recv(); // use opt to configure something Ok(End) } } The option port gives the subgraph developer a way to send in parameters such as a connection string and the message will not be consumed and thrown away, that message may be read on every function run. Whereas other ports will consume and throw away the message. accumulator: agent! { accumulator(prim_bool), fn run(&mut self) -> Result<Signal> { let acc = self.ports.accumulator.recv()?; // use the accumulator to start accumulating something. Ok(End) } } The accumulator gives the subgraph developer a way to start counting at a certain number. It’s a way of passing in initial state. run: This function does the actual processing and is the only mandatory expression of this macro. You’ve seen many examples already. #### 2.2.2. Idris ##### What? Executable Subgraphs are defined as a network of Agents, which exchange typed data across predefined connections by message passing, where the connections are specified externally to the processes. These Agents can be reconnected endlessly to form different executable Subgraphs without having to be changed internally. ##### Why? Functions in a programming language should be placed in a content addressable store, this is the horizontal plane. The vertical plane should be constructed using unique addresses into this content addressable store, critically each address should solve a single problem, and may do so by referencing multiple other unique addresses in the content addressable store. Users must not have knowledge of these unique addresses but a translation process should occur from a human readable name to a universally unique address. Read more about the problem. Nix gives us the content addressable store which allows for reproducibility, and these agents give us reusablility. The combination is particularly potent form of programming. Once you have the above, you have truly reusable and reproducible functions. Fractalide nodes are just this, and it makes the below so much easier to achieve: * Open source collaboration * Open peer review of nodes * Nice clean reusable nodes * Reproducible applications ##### Who? Typically programmers will develop Agents. They specialize in making Agents as efficient and reusable as possible, while people who focus on the Science give the requirements and use the Subgraphs. Just as a hammer is designed to be reused, so Subgraphs and Agents should be designed for reuse. ##### Where? The Agents are found in this nodes directory, or the nodes directory of a fractal. nodes └── idr └── file └── open ├── File <--- contains the file_open idris agent code │ └── .idr ├── Tests <--- contains the file_open idris agent tests │ └── .idr ├── default.nix <--- configures the environment to compile the agent └── agent.ipkg <--- idris uses the deps made available by default.nix When you see an agent.ipkg in the same directory as a default.nix you know it’s an Agent. ##### How? An Agent consists of three parts: • a nix default.nix file that sets up an environment to satisfy the idris repl/compiler. • an idris agent.ipkg tells idris how to compile the source code in the directories. • the idris .idr source code within capitalized directories. ###### The agent Nix function. The agent function in the default.nix requires you make decisions about three types of dependencies. • What edges inter-agent idris types are to be pulled in? • What mods which are idris library dependencies available online. • What osdeps or operating system level dependencies are needed? { agent, edges, mods, pkgs }: agent { src = ./.; edges = with edges.idr; [ TestVect ]; mods = with mods.idr; [ contrib ]; osdeps = with pkgs; [ openssl ]; } • The { agent, edges, mods, pkgs }: lambda imports: The edges attribute which consists of every edge available on the system. The mods attribute set consists of every mod in the modules/idr directory. Lastly the pkgs pulls in every third party package available on NixOS, here’s the whole list. • The agent function builds the idris agent.ipkg and associated source code, and accepts these arguments: • The src attribute is used to derive the source code and Agent name based on location in the nodes directory hierarchy. • The edges makes inter-agent idris types available just before build time. • The mods specifies exactly which modules are needed for the agent to build. • The osdeps specifies exactly which pkgs, or third party operating system level libraries such as openssl needed in scope. Only specified dependencies and their transitive dependencies will be pulled into scope once the agent compilation starts, or when you run a development shell. This is the output of the above agent's compilation: /nix/store/dp8s7d3p80q18a3pf2b4dk0bi4f856f8-file_open └── lib └── libagent.js ###### nix-shell It’s convenient to use a REPL, idris-mode with your editor while developing an agent. Here are the steps to setup a development environment. In the root directory of fractalide, issue these commands: • $ nix-shell -A mods.idr.idrisfbp ← or whatever the attribute path of the idris agent you want • $cd modules/idr/idrisfbpcd to your chosen agent in the nodes hierachy • $ source $setup • $ build • $run emacs . For the run command to work you need to have installed your editor via your system level configuration.nix file, as it references the /run/current-system/sw/bin/ path. You may also start a REPL and use your integrated idris-modes in your editor. Note all the dependencies have been alias’ed to idris. $ type idris idris is aliased to idris -i /nix/store/0ijgdwdb9bfwwkgcxac25p2mxl161ljb-base-1.1.0/lib/1.1.0-git:PRE/base -i /nix/store/12cs9pgsdq4rhnzxjdk5hqv5rc9v60pb-prelude-1.1.0/lib/1.1.0-git:PRE/prelude -i /nix/store/506r1qqayw6j2nb4dsfvb716n9x8ndmj-contrib-1.1.0/lib/1.1.0-git:PRE/contrib -i /home/stewart/dev/fractalide/fractalide/modules/idr/idrisfbp/idris_libs' Each idris command in each agent development environment will contain a different set of associated alias’ed paths. It might be a good idea to create a throw away agent that contains all the useful dependencies, create a development shell using that agent then navigate around the nodes hierarchy without closing your editer each time. ## 3. Edge Collection ### 3.1. What? An Edge is a bounded buffer message passing channel between two ports belonging to their respective agents. Terms: • A contract is defined as: The Cap’n Proto schema on the output port of the upstream agent MUST be the same as the Cap’n Proto schema on the input port of the downstream agent. If the two are the same, the contract is said to be satisfied, otherwise it is unsatisfied. There are three phases building up to a successful Edge formation: • During agent development time an agent's port is assigned a Cap’n Proto schema. • During subgraph development time the syntax -> or => is used to instruct an upstream agent's port to connect to a downstream agent's port later at run-time. It represents a connection between two nodes. Though it is not yet an edge because the contract might not be satisfied. • Lastly, the graph is successfully loaded into the virtual machine without errors. This act means all agent contracts were satisfied, and all subgraph connections are now classified as edges. Once an edge is formed, it becomes a bounded buffer message passing channel, which can only contain messages with data in the shape of whatever the Cap’n Proto schema is. So despite you seeing only Cap’n Proto schema in this directory, the concept of an Edge is more profound. Hence we would prefer naming this concept after it’s grandest manifestation, and in the process, the name encapsulates all of the above information. The name should also tie in with the concept of a node in graph theory, as such we use nodes and edges to construct subgraphs. #### 3.1.1. Exposed Edge or iMsg When developing a subgraph there comes a time when the developer wants to inject data into an agent or another subgraph. One needs to use an exposed edge or an imsg (initial message) which has this syntax: { subgraph, nodes, edges }: subgraph { src = ./.; flowscript = with nodes; with edges; '' '${prim_bool}:(boolean=true)' -> INPUT_PORT NAME() ''; } Exposed edges or imsgs kick start agents into action, otherwise they won’t start. Due to the dataflow nature of agents they will politely wait for data before they start doing anything. This is the equivalent of passing the path of a data source to some executable on the command line. Without the argument the program would just sit or fail. Another way of looking at it might be a pipeline that has come to the surface to accept some form of input. We use the name exposed edge to differentiate between a hidden edge, but by far the most common usage is imsg and edge. #### 3.1.2. Hidden Edge or Edge Hidden edges are represented with this syntax -> and =>, and are used to control the direction flowing data. Hence the process of programming a subgraph is essentially digging ditches and laying pipelines between buildings. Examples of hidden edges • From one agent to another agent: { subgraph, nodes, edges }: subgraph { src = ./.; flowscript = with nodes; with edges; '' agent1() output -> input agent2() ''; } • Into a subgraph: { subgraph, nodes, edges }: subgraph { src = ./.; flowscript = with nodes; with edges; '' output => input subgraph() ''; } • Out of a subgraph: { subgraph, nodes, edges }: subgraph { src = ./.; flowscript = with nodes; with edges; '' agent() output => output ''; } ### 3.2. Why? • Contracts between components are critical for creating living systems. • Schema ensure we do not need to parse strangely formatted stdin data. • A node does one and only one thing, and the schema represents a language the node speaks. If you want a node to do something, you have to speak it’s language. ### 3.3. Who? Typically subgraph developers will be interested in hidden and exposed edges. ### 3.4. Where? The edges directory is where all the schema go: ├── key │ └── value │ └── default.nix ├── list │ ├── command │ │ └── default.nix │ ├── text │ │ └── default.nix │ ├── triple │ │ └── default.nix │ └── tuple │ └── default.nix ├── maths │ ├── boolean │ │ └── default.nix │ └── number │ └── default.nix ### 3.5. How? Edges may depend on other edges. The { edge, edges }: lambda passes in two arguments, the edge builder and edges which consists of every Edge or Edge Namespace in the system. The edge building function accepts these arguments: • The src attribute is used to derive an Edge name based on location in the directory hierarchy. • The edges attribute resolve transitive dependencies and ensures your agent has all the needed files to type check. • a Cap 'n Proto schema. This is the heart of the contract, this is where you may create potentially complex deep hierarchies of structured data. Please read more about the schema language. { edge, edges }: edge { src = ./.; edges = with edges; [ command ]; schema = with edges; '' @0xf61e7fcd2b18d862; using Command = import "${command}/src/edge.capnp"; struct ListCommand { commands @0 :List(Command.Command); } ''; } #### 3.5.1. Naming of Cap’n Proto structs and enums Please use CamelCase for struct and enum names. The naming should reflect this manner: • if the schema is in folder /edges/maths/boolean then the struct should have name MathsBoolean. • if the schema is in fractal /edges/net/http/response then the struct should have name NetHttpResponse. The same naming applies for Cap’n Proto enums and interfaces. It’s crucial this naming is adopted. #### 3.5.2. One struct per Fractalide schema We prefer composition of schema, and the schema must have fully qualified struct names. Hence, this is example shouldn’t be used: { edge, edges }: edge { src = ./.; edges = with edges; [ command ]; schema = with edges; '' @0xf61e7fcd2b18d862; struct Person { name @0 :Text; birthdate @3 :Date; email @1 :Text; phones @2 :List(PhoneNumber); struct PhoneNumber { number @0 :Text; type @1 :Type; enum Type { mobile @0; home @1; work @2; } } } struct Date { year @0 :Int16; month @1 :UInt8; day @2 :UInt8; } ''; } The Date name can collide! Schema are pulled into agent's scope just before compile time, now we are unable to predict what combinations will happen. So if we have two schema that have struct Date … then a name collision will take place. Therefore to avoid this scenario please put struct Date … into it’s own schema and import it via this mechanism. #### 3.5.3. Cap’n Proto import Fractalide resolves transitive dependencies for you but you have to use this method: { edge, edges }: edge { src = ./.; edges = with edges; [ command ]; schema = with edges; '' @0xf61e7fcd2b18d862; using CommandInstanceName = import "${command}/src/edge.capnp"; struct ListCommand { commands @0 :List(CommandInstanceName.Command); } ''; } You must pull explicitly mention the edge you want to import via the edges = with edges; [ command ]; Then you must import it via this mechanism: using CommandInstanceName = import "${command}/src/edge.capnp"; and lastly use it … commands @0 :List(CommandInstanceName.Command); … Out of curiosity what does the output of the above list_command contract function look like? $cat /nix/store/3s25icpbf1chayvrxwbyxr9qckn7x669-list_command/src/edge.capnp @0xf61e7fcd2b18d862; using CommandInstanceName = import "/nix/store/bgh37035cbr49r7mracmdwwjx9sbf4nr-command/src/edge.capnp"; struct ListCommand { commands @0 :List(CommandInstanceName.Command); } The generated Rust code consists of the list_command, command and tuple contract concatenated together. ## 4. Services ### 4.1. What? Services are reusable and consist of agents, subgraphs and edges. They may be connected to other services on the same system or on remote machines in a high performance cluster quite easily. ### 4.2. Why? Services allow programmers to collaborate and distill common configurable patterns that solve general problems. Essentially a service in this context is no different from a top level subgraph, albeit the interface has been nix’ified. The nix interface is required because this is where nixos takes over and declaratively handles conguent configuration management. It does things like sets up dependencies such as database and other services that might need to be run for the subgraph to operate properly. This layer of software has been well tested by the nix community and ties in with every other service created by the nixos community. This abstraction allows you to pull in legacy services, plug a Cap’n Proto functionality and start talking to Fractalide services. ### 4.3. Who? Anyone who has completed a high quality, documented hierarchy of components, subnets and contracts, and where the hierarchy makes sense to expose as a service. We use the [C4](../CONTRIBUTING.md) so your services will be merged in quickly. ### 4.4. Where? Service implementations by convention exist in a fractal’s root folder. It doesn’t make sense to expose every fractal as a service. Though if the fractal is dependent on other services such as databases, then it might make sense. Nevertheless, the fractals which have a service.nix may have their components and contracts imported into higher level subnets via the Fractals importing mechanism. This higher level subnet may then be exposed as a service. ### 4.5. How? • First create a fractal • Update the service.nix file to reflect what options you want exposed on the service interface. • Ensure the subnet you want is correctly configured. • Add a line the fractalide/services/default.nix where you make your service visible to the rest of the fractalide community. #### 4.5.1. Two ways to run a service ##### From your Fractal This approach is when you have a service you don’t consider generic and is not worth sharing with the community. in your configuration.nix put these lines: configuration.nix { config, pkgs, ... }: let fractalide = import fetchFromGitHub { owner = "fractalide"; repo = "fractalide"; rev = "b9590b6{ git revision you want }f130ed79432c722"; sha256 = "0jz58mmax5{ correct sha256 generated by nix }yycmzr26xwfqa"; } {}; in { require = fractalide.services ++ [ "/path/to/private/dev/fractals/fractal_your_fractal/service.nix" ]; services.workbench = { enable = true; bindAddress = "192.168.0.14"; port = 8000; }; services.your_service = { enable = true; bindAddress = "192.168.0.15"; port = 8001; }; # ... other options you want } ##### From Fractalide in your configuration.nix put these lines: configuration.nix let fractalide = import fetchFromGitHub { owner = "fractalide"; repo = "fractalide"; rev = "b9590b6{ git revision you want }f130ed79432c722"; sha256 = "0jz58mmax5{ correct sha256 generated by nix }yycmzr26xwfqa"; } {}; in { require = fractalide.services; services.workbench = { enable = true; bindAddress = "127.0.0.1"; port = 8000; }; # ... other options you want • At this point, you may either choose to build nixops infrastructure scripts and deploy or simply run $ nixos-rebuild {test, boot , switch} to set up the services on your own system. • Do help out building generic services for programmers to speed up their work! ## 5. Fractals Fractals are third party Fractalide libraries. ### 5.1. What? Fractals are importable 3rd party sets of nodes, subgraphs and edges, this folder hierarchy is the single source of truth regarding where to find each community fractal. ### 5.2. Why? Fractals allow the community to develop their own projects in their own git repository, and once ready, may plug into dev/fractalide/fractals folder hierarchy which represents the spine of Fractalide. Once inserted your fractal is available for use by everyone. ### 5.3. Who? Anyone making high quality, documented nodes, subgraphs and edges may plug into this folder hierarchy. We use the C4 so your fractals will be merged in quickly. ### 5.4. Where? Each fractal needs to have it’s own hierarchical folder. For example, HTTP would be placed in the fractalide/fractals/net/http folder. ### 5.5. How? Say you wish to create a http project, these are the exact steps involved: • Ensure you use this directory structure convention: dev ├── fractalide │ └── fractals │ └── net │ └── http │ └── default.nix └── fractals ├── fractal_net_http └── ... more community fractals cloned $cd <your/development/directory>$ git clone \https://gitlab.com/fractalide/fractalide.git $NIX_PATH="nixpkgs=https://github.com/NixOS/nixpkgs/archive/125ffff089b6bd360c82cf986d8cc9b17fc2e8ac.tar.gz:fractalide=/path/to/dev/fractalide" && export NIX_PATH (1)$ mkdir dev/fractals && cd dev/fractals $git clone git://github.com/fractalide/fractal_workbench.git fractal_net_http (2)$ git remote set-url origin git://new.url.you.control.here $mkdir -p dev/fractalide/fractals/net/http/ (3) 1 Take note when setting the NIX_PATH environment variable, it must include the path to your newly cloned fractalide repo i.e.: NIX_PATH=…:fractalide=/path/to/dev/fractalide. Should you start a new shell, type -r then type 125ff this will search your command history for the above command, or just persist the command in your ~/.bashrc file. 2 The fractal_workbench repo provides a minimum correct structure for your fractal. Keep the repo naming convention fractal_* for your repo as it’ll be easy for the community to see this is a fractalide related project. 3 Create your needed directory hierarchy dev/fractalide/fractals/net/http/default.nix. Insert the below code into a file called default.nix which sits in the above folder. dev/fractalide/fractals/net/http/default.nix { pkgs , support , edges , nodes , fetchFromGitHub , ...}: let fractal = fetchFromGitHub { owner = "fractalide"; repo = "fractal_net_http"; rev = "66ad3bf74b04627edc71227b3e5b944561854367"; sha256 = "1vs1d3d9lbxnyilx8g45pb01z5cl2z3gy4035h24p28p9v94jx1b"; }; /fractal = ../../../../fractals/fractal_net_http;/ in import fractal {inherit pkgs support edges nodes; fractalide = null;} The pkgs, support, nodes, edges and fetchFromGitHub are arguments passed into this closure. The let expression contains a fetchFromGitHub expression describing the location of the fractal. owner of the git repository i.e.: github.com/fractalide repo is the git repository in question i.e.: github.com/fractalide/fractal_net_http rev indicates the git revision you want to import i.e.: https://github.com/fractalide/fractal_net_http/commit/66ad3bf74b04627edc71227b3e5b944561854367 sha256 is a neat nix mechanism to assist in deterministic builds. To obtain the correct sha256 try build your project with an incorrect sha256 (change the first alpha-numeric character in 1vs1d3d9lbxnyilx8g45pb01z5cl2z3gy4035h24p28p9v94jx1b to a 2). Nix will download the repository and check that the actual sha256 matches against what you incorrectly inserted. Nix will tell you what the correct sha256 is. Copy it and insert the correct sha256, replacing 2vs1d3d9lbxnyilx8g45pb01z5cl2z3gy4035h24p28p9v94jx1b. Please notice the line: /fractal = ../../../../fractals/fractal_net_http;/ This line allows you to tell nix not to refer to the remote repo but your local clone of fractal_net_http. Comment out the fetchFromGitHub expression and uncomment the above local repo clone path. When you publish your fractal upstream, ensure this line is commented out! Please use a relative path compatible with the above directory structure convention as it will work for everyone and we don’t have to hunt for the correct folder. Just un/comment and go! The line: import fractal {inherit pkgs support edges nodes; fractalide = null;} is where we import your closures into fractalide’s set of closures. Thus making your nodes available to everyone. The last step is to expose the exact nodes / edges to dev/fractalide/nodes/default.nix and dev/fractalide/edges/default.nix. This is done in this manner: • Open dev/fractalide/nodes/default.nix and seek out the net_http_nodes attribute. This is what it looks like: net_http_nodes = fractals.net_http.nodes; In this case there is no specific top level node you’d wish to expose so the convention _nodes is used to indicate this. Whereas if you have a specific node you’d wish to expose then you’d name it as such: net_http = fractals.net_http.nodes.http. Why the .http? well that’s what the node is named in the namespace here. Please notice the lack of the _nodes when exporting a single node. • Regarding importing edges, typically you don’t need to import edges, but there are times when you need a special edge which must operate on the public side of the fractal and thus usable across a number of fractals, say the net_http_edges.request You’d use a similar mechanism as above when exposing your edges. ### 5.6. Incremental Builds Incremental Builds speed up the development process, so that one doesn’t have to compile the entire crate from scratch each time you make a change to the source code. Fractalide expands the nix-build system for incremental builds. The Incremental Builds only work when debug is enabled. They also need the path to a cache folder. The cache folder can be created from an old result by the buildCache.sh script. Per default the cache folder is saved in the /tmp folder of your system. Incremental Builds permit you to compile a crate without having to recompiled the crate dependency tree. Here is an example how you can build with the Incremental Build System: $ cd dev/fractalide $nix-build --argstr debug true --argstr cache$(./support/buildCache.sh) --argstr subgraph workbench If you’re using NixOS, please ensure you have not set nix.useSandbox = true;, otherwise Incremental Compilation will fail. ### 5.8. Two ways to execute your fractal #### 5.8.1. Executing from within Fractalide $cd dev/fractalide$ nix-build --argstr rs workbench $./result • advantages • Incremental recompilation needed for development • disadvantages • wetware needed to plug into dev/fractalide/fractals to get incremental recompilation • long build command #### 5.8.2. Executing from with the Fractal $ cd /dev/fractals/fractal_workbench $nix-build$ ./result • faster to test by just issuing the nix-build command • convenient for CI & CD of you specific subgraph • don’t have to plug it into dev/fractalide/fractals • no incremental recompilation ## 6. HOWTO ### 6.1. Steps #### 6.1.1. Fractalide installation ##### Building the Fractalide Virtual Machine (FVM) Once logged into your virtualbox guest issue these commands: $git clone https://github.com/fractalide/fractalide.git$ cd fractalide $nix-build Let us inspect the content of the newly created symlink called result. $ readlink result /nix/store/ymfqavzrgmj3q3aljgwvh769fq9dszp2-fvm $tree result result └── bin └── fvm$ file result/bin/fvm result/bin/fvm: ELF 64-bit LSB shared object, x86-64, version 1 (SYSV), dynamically linked, interpreter /nix/store/8lbpq1vmajrbnc96xhv84r87fa4wvfds-glibc-2.24/lib/ld-linux-x86-64.so.2, for GNU/Linux 2.6.32, not stripped ##### Peek under the hood You shouldn’t need to care too much about this during your everyday programming, but it’s pleasant deviation from most normal workflows and thus should be explained. Let’s build a subgraph that runs a contrived maths_boolean_nand agent. $nix-build --argstr node test_nand This replaces the result symlink with a new symlink pointing to a generated file. $ readlink result /nix/store/zld4d7zc80wh38qhn00jqgc6lybd2cdi-test_nand Let’s investigate the contents of this executable file. $cat result /nix/store/ymfqavzrgmj3q3aljgwvh769fq9dszp2-fvm/bin/fvm /nix/store/jk5ibldrvi6cai5aj1j00p8rgi3zw4l7-test_nand Notice that we’re passing the path of the actual test_nand subgraph into the fvm. What does the contents of the actual /nix/store/jk5ibldrvi6cai5aj1j00p8rgi3zw4l7-test_nand file look like (the argument to fvm)? $ cat /nix/store/jk5ibldrvi6cai5aj1j00p8rgi3zw4l7-test_nand/lib/lib.subgraph '/nix/store/ynm9ipggdvxhzi5l2kkz9cgiqgvq2g87-prim_bool:(bool=true)' -> a nand(/nix/store/y919fp98qw33w0cs2wn5wzwgwpwgbchs-maths_boolean_nand) output -> input io_print(/nix/store/4fnk9dmky6jni4f4sbrzl1xsj50m3mb0-maths_boolean_print) '/nix/store/ynm9ipggdvxhzi5l2kkz9cgiqgvq2g87-prim_bool:(bool=true)' -> b nand() $file /nix/store/jk5ibldrvi6cai5aj1j00p8rgi3zw4l7-test_nand/lib/lib.subgraph /nix/store/jk5ibldrvi6cai5aj1j00p8rgi3zw4l7-test_nand/lib/lib.subgraph: ASCII text The --argstr node xxx are arguments passed into the nix-build executable. Specifically $ man nix-build ... --argstr name value This option is like --arg, only the value is not a Nix expression but a string. So instead of --arg system \"i686-linux\" (the outer quotes are to keep the shell happy) you can say --argstr system i686-linux. ... The name node refers to the top level graph to be executed by the fvm. nix compiles each of the agents and inserts their paths into subgraphs. The fvm knows how how to recursively load the entire hierarchy of subgraphs which contain fully qualified paths to their composed agents. ##### Quick feel of the system ###### A = (Graph setup + tear down): $nix-build --argstr node bench_load /nix/store/ij8jri0z1k5n447f9s0x5yfx5p9iqnnf-bench_load$ sudo nice -n -20 perf stat -r 10 -d ./result ... 3.684139058 seconds time elapsed ( +- 0.56% ) ###### B = (Graph setup + tear down + message pass + increment): $nix-build --argstr node bench /nix/store/mfl206ccv86wvyi2ra5296l8n1bks24x-bench$ sudo nice -n -20 perf stat -r 10 -d ./result Performance counter stats for './result' (10 runs): 6638.755996 task-clock (msec) # 1.443 CPUs utilized ( +- 0.57% ) 268,864 context-switches # 0.040 M/sec ( +- 0.47% ) 3,047 cpu-migrations # 0.459 K/sec ( +- 10.08% ) 82,417 page-faults # 0.012 M/sec ( +- 0.03% ) 18,012,749,608 cycles # 2.713 GHz ( +- 0.66% ) (50.10%) <not supported> stalled-cycles-frontend <not supported> stalled-cycles-backend 18,396,303,772 instructions # 1.02 insns per cycle ( +- 0.10% ) (62.48%) 3,008,536,908 branches # 453.178 M/sec ( +- 0.06% ) (73.97%) 13,396,472 branch-misses # 0.45% of all branches ( +- 1.01% ) (74.08%) 6,955,828,023 L1-dcache-loads # 1047.761 M/sec ( +- 0.50% ) (63.04%) 184,998,022 L1-dcache-load-misses # 2.66% of all L1-dcache hits ( +- 0.81% ) (29.73%) 49,018,759 LLC-loads # 7.384 M/sec ( +- 0.99% ) (26.13%) 3,032,354 LLC-load-misses # 6.19% of all LL-cache hits ( +- 1.56% ) (37.74%) 4.601455409 seconds time elapsed ( +- 0.66% ) ###### (Message Passing + Increment) = B - A: >>> 4.601455409 - 3.684139058 0.9173163509999998 This just gives you a feel for the system: • 3.7 secs to setup 10,000 [rust agents](./nodes/bench/inc/lib.rs) + teardown 10,000 agents. • 4.6 sces to setup 10,000 agents + message pass 10,000 times + increment 10,000 times + teardown 10,000 agents. • 0.9 sec to message pass 10,000 times + increment 10,000 times. ##### A Todo backend We will design an http server backend that’ll host a set of todos. It will provide the following HTTP features : GET, POST, PATCH/PUT, DELETE. The actual todos will be saved in a sqlite database. The client will use json to communicate with the server. A todo had the following fields : • id : a unique integer id, that is used to retrieve, delete and patch the todos. • title : a string, that represents the goal of the todo and will be displayed. • completed : a boolean, to remember if the todo has been completed or not. • order : a positive integer, used to display the todos in a certain order. The http server responds to these requests: • GET The request looks like GET http://localhost:8000/todos/1. The server, after it receives a "GET" request along with a numeric id, will respond with the corresponding todo in the database, otherwise it will return a 404. • POST The request looks like POST http://localhost:8000/todos. The content of the request must be json that correspond to a todo. The id field is ignored. e.g. : { "title": "Create a todo http server", "order": 1 } • PATCH or PUT The request looks like PUT http://localhost:8000/todos/1. The content of the request is the fields to update. ex : { "completed": true } • Delete The request looks like DELETE http://localhost:8000/todos/1. This will delete the todo with the id 1. ##### The Big Picture The centre of gravity revolves around the http agent. It receives requests from users and dispatches them to four other subgraphs, one subgraph for each HTTP feature. Each subgraph processes the request and provide a response. Before we approach the HTTP feature subgraphs let’s take a look at the http agent. ###### The HTTP Agent The implementation code can be found here. The http agent has one array output port for each HTTP method, and the elements of each array output ports is actually a fast rust regex. For example, http() GET[^/news/?$] will match the request with method GET and url http://../news or http://../news/. A Msg is sent on the output port of http with the schema net_http_request. We will just use the fields id, url, content. The id is the unique id for the request. It must be provided in the response corresponding to this request. The url is the url given by the user. The content is the content of the request, or the data given by the user. The http agent expects a Msg with the schema net_http_response. A response has an id, which corresponds to the request id. It also has a status_code, which is the response code of the request. By default, it’s 200 (OK). The content is the data that is sent back to the user. The http agent must be started with an iMsg of type net_http_address. It specifies the address and port on which the server listens: ###### The GET Subgraph { subgraph, nodes, edges }: subgraph { src = ./.; flowscript = with nodes; with edges; '' db_path => db_path get_sql() input => input id(${todo_get_id}) id -> get get_sql(${sqlite_local_get}) get_sql() id -> id todo_build_json(${todo_build_json}) get_sql() response -> todo todo_build_json() id() req_id -> id todo_add_req_id(${todo_add_req_id}) todo_build_json() json -> playload build_resp(${todo_build_response}) get_sql() error -> error build_resp() build_resp() response -> response todo_add_req_id() response => response ''; } A request follows this path: • Enters the subgraph via the virtual port request • Then enters the agent get_id. This agent has two output ports : req_id and id. The req_id is the id of the http request, given by the http agent. The id is todo id retrieved from the url (ie: given the url http://../todos/2, the number 2 will be sent over the id port). • The url id enters the sql_get agent, that retrieve a Msg from a database corresponding to the id. • If the id exists, a Msg is send to build_json that contains the json of the todo. • If the id doesn’t exist in the database, a Msg is send on the error port. • The build_request will receive Msg on one of its two input ports (error or playload). If there is an error, it will send a 404 response, or otherwise, it will send a 200 repsonse with the json as data. • This new response now goes into the add_req_id agent, which retrieves the req_id from the request, and sets it in the new response. • The response now leaves the subgraph. Now we can connect the http agent to the get subgraph, to retrieve all the GET http request. http() GET[^/todos/.+$] -> request get() get() response -> response http() Please understand how the code maps to the above diagram, as these particular diagrams shall not be repeated. ###### The POST Subgraph { subgraph, nodes, edges }: subgraph { src = ./.; flowscript = with nodes; with edges; '' db_path => db_path insert_todo() input => input todo_get_todo(${todo_get_todo}) todo -> input cl_todo(${msg_clone}) cl_todo() clone[0] -> insert insert_todo(${sqlite_local_insert}) cl_todo() clone[1] -> todo todo_build_json(${todo_build_json}) insert_todo() response -> id todo_build_json() todo_get_todo() req_id -> id todo_add_req_id(${todo_add_req_id}) todo_build_json() json -> playload todo_build_response(${todo_build_response}) todo_build_response() response -> response todo_add_req_id() response => response ''; } A request will follow this path : • Enters the subgraph by the virtual port request • Enters the agentget_todo. get_todo sends req_id and the content, which is converted from json into a new schema app_todo. • The todo schema is then cloned and sent to two agents. • One clone goes to sql_insert, which sends out the url id of the todo found in the database. This id is send in build_json. • The build_json receives the database id and the todo, and merges them together in json format. • This approach allows the building of a response with json as the content. • add_req_id then add the req_id in the reponse • The response is sent out The post subgraph is then connected to the http output port : http() POST[/todos/?$] -> request post() post() response -> response http() ###### The DELETE Subgraph { subgraph, nodes, edges }: subgraph { src = ./.; flowscript = with nodes; with edges; '' input => input id(${todo_get_id}) db_path => db_path delete_sql() id() id -> delete delete_sql(${sqlite_local_delete}) delete_sql() response -> playload build_resp(${todo_build_response}) id() req_id -> id todo_add_req_id(${todo_add_req_id}) build_resp() response -> response todo_add_req_id() response => response ''; } This subgraph is easier than the two before, hence nearly self-explainatory. • The req_id and the id are obtained in get_id. • The id is send to sql_delete, which returns the id to build_response. • build_response simply fill the http response with the id • add_req_id add the http id The delete subgraph is connect to the http output port : http() DELETE[/todos/.+] -> request delete() delete() response -> response http() ###### The PATCH Subgraph The patch subgraph is a little more complicated, because of the synch agent. Let first see what happend without it : The idea of the stream is this: • Get the new "todos" values in the request • In parallel, retrieve the old value of the todo from the database. • Then, send the old and the new values to a merge agent, which builds the resulting todo Now this graph has a problem; if there the todo is new then an old todo cannot be found in the database. In this case, the new edge between get_todo and merge and the error edge between sql_get and build_respone are completely concurrent, thus an issue will arise if a Msg is sent over the error edge when sql_get cannot find a todo in the database. At the same time get_todo will have recognized that it’s a new todo and will have sent a Msg over the new edge. This will insert 2 Msgs into the old input port, where the first Msg is incorrect. A solution is to add a synch agent which has outgoing edges old/new and error. If an error is received, it’s immediately communicated to build_respone and discards the old/new Msg. If it receives a new Msg, it forwards the new and old Msgs to merge. This ensures all Msgs are well taken care of. To simplify the graph a little, we’ve not mentioned the edge from synch to patch_sql. A Msg is send from the former with the todo id, whichs need to be updated. But all the logic, with synch, is exactly the same. The complete figure is: { subgraph, nodes, edges }: subgraph { src = ./.; flowscript = with nodes; with edges; '' input => input todo_get_todo(${todo_get_todo}) db_path => db_path patch_sql() todo_get_todo() id -> get get_sql(${sqlite_local_get}) synch(${todo_patch_synch}) get_sql() response -> todo synch() todo -> old merge(${todo_patch_json}) todo_get_todo() raw_todo -> raw_todo synch() raw_todo -> new merge() get_sql() id -> id synch() id -> id patch_sql(${sqlite_local_patch}) merge() todo -> msg patch_sql() patch_sql() response -> playload build_resp(${todo_build_response}) get_sql() error -> error synch() error -> error build_resp() ### 6.2. Tokio- We’re waiting patiently for the much anticipated https://github.com/tokio-rs/ code to land. That’s when we’ll get services talking to other services and http clients via tokio.	2018-12-15 02:16:38	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 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": 0.20185740292072296, "perplexity": 8344.785611752879}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-51/segments/1544376826686.8/warc/CC-MAIN-20181215014028-20181215040028-00168.warc.gz"}
https://www.physicsforums.com/threads/integration-of-rational-functions.15811/	Integration of rational functions 1. Mar 6, 2004 Kuja How do I solve the integration of a rational function such as: x^2 - 6x - 2 (x^2 + 2)^2 If possible, please list the general rule of solving, I DO NOT want the answer, I simply want to know the way of solving it. So far I got to the part where A = 1, B = -6 and C = -4, what can I do next? 2. Mar 6, 2004 Hurkyl Staff Emeritus Well, I guess that's one good thing about standardized notation across calc texts; when the student gives absolutely no indication about what they're doing, you can still tell from the variables they used! I'm assuming what you've done already is partial fractions. Anyways, the integrals you're left with should looke like the type of integral you've done in previous sections. (you spent an entire section on them!) You need to make a substitution... In general, you might need to complete the square in the quadratic factors in the denominator to get them in the right form. 3. Mar 6, 2004 Damned charming :) I am new here and I don't how to do fractions so ignore the ............ I need it for the space You have shown x^2 - 6x - 2 (x^2 + 2)^2 = 1 ................ -6x-4 ------- +.. ---------- (x^2+2) .... (x^2+2)^2 The first fraction integral is a standard integral ( inverse tan) and a hint is the first fraction can be rewritten as -3(2x) ............... -4 ----------- + .. -------- (x^2+2)^2 ..... (x^2+2)^2 Use the subsitution u =x^2+2 for the first fraction and I think the substitution u = square root of 2 tan u for the second fraction. Last edited: Mar 6, 2004 4. Mar 6, 2004 Kuja The problem is,I am in grade 12 and I have never done integration before (In fact, we are doing derivative right now) I am trying to check the derivative: x^2 - 6x - 2 (x^2 + 2)^2 that I got from: -x + 3 x^2 + 2 So I figure the anti-derivative (Integral) might be able to help me check, who would have thought it is so much work? I should have pick graphing instead. I haven't got a clue how to make the derivative looks like original function shown above by integration, and althought i have not done any integration, I would like to still give it a try. Any help would be welcome, thanks! Last edited: Mar 6, 2004 5. Mar 6, 2004 Hurkyl Staff Emeritus There are two things you need to look for in your text: "Substitution" and "Trigonometric substitution". Everything you need to integrate that (now that you've done "Partial Fractions") will be spelled out for you, probably much more thoroughly than you can get from a post on a message board. That being said, there are other ways you can check a derivative. You can try plotting your original function, and then check to make sure your derivative is positive whenever the original function is increasing. Rolle's theorem says that if a function is zero at two different points, the derivative has to be zero someplace in-between; so you could look for the zeroes of both. You could try making a differential approximation. Remember that: $$f(x + \delta x) \approxeq f(x) + f'(x) \delta x$$ is a good approximation when δx is small, so you could plug in some values and check. (The error is eventually much smaller than δx) Last edited: Mar 6, 2004 6. Mar 6, 2004 Damned charming :) Integration can be much harder than differentiation. One reason is that you do not have something as easy to use the product rule, quotient rule etc. (you can work backwards using integration by substitution but) to make this clear even though though 1 ---- 1+x^2 is a rational function it integrated to inverse tan x which is not one polynomial divided by another. 7. Mar 6, 2004 Kuja I got something like: 6x + 4 + In \|x^2 + 2\| + C x^2 + 2 which is nothing like: -x + 3 x^2 + 2 8. Mar 7, 2004 Hurkyl Staff Emeritus And in all truth, I think that antidifferentiation here is far more error prone that differentiation. You could probably compute the derivative several times (to make sure there are no arithmetic mistakes!) in the time it takes to perform the antidifferentiation, and you probably have smaller odds of making a mistake too. 9. Mar 7, 2004 NateTG I think Hurkyl is right on the money with his notion of numeric approximation of the derivative. It's a reasonable way to do a santiy check on the work. That said, when i differentiate: $$\frac{d}{dx} \frac{-x + 3}{x^2+2}$$ I get $$\frac{-1}{x^2+2} - 2x \frac{-x+3}{x^4+4x^2+4}$$ or $$\frac{x^2-6x-2}{x^4+4x^2+4}$$	2018-11-19 19:57:59	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8262037634849548, "perplexity": 873.4077845534053}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-47/segments/1542039746110.52/warc/CC-MAIN-20181119192035-20181119214035-00496.warc.gz"}
https://dsp.stackexchange.com/questions	All Questions 22,394 questions Filter by Sorted by Tagged with 16 views Help with literature pointers for improving estimation of underlying mean for signal with periodic noise I am in the starting phase of a master thesis and are having some trouble finding a good starting point in the literature for my thesis. The aim of my thesis is to provide as good of an estimate as ... 17 views Suppose, I sample a signal $z=\mathrm{e}^{\mathrm{i}\omega t}$ alternatingly in quadrature components $I=\mathrm{Re}(z(t))$ and $Q=\mathrm{Re}(\mathrm{e}^{\mathrm{i}\alpha}z(t))$. 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At this moment, my constellation in the best case looks like in the picture below: and the corresponding eye-pattern: Eyes are completely closed, which ... 19 views In OFDM, does the parameter time offset is important if there is only one mobile in the cell? I am learning about OFDM, and in my reading, I have noticed that (mentioned everywhere) synchronization is a very critical aspect of OFDM. Here, I am writing in the context of time synchronization (... 28 views What is the difference between sampling over $-4\pi$ to $4\pi$ and 0 to $8\pi$ I see no difference in magnitude, but the phase plots vary significantly while using NumPy and MatPlotLib to plot. Looking at the phase plots, I feel like I'm missing something important. Also, I'm a ... 45 views Analysis of the ratio of two moving average filters I have created a basic segmentation algorithm for 1-D signals (e.g. audio) based on a ratio of energy averages. 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I've been trying to get DFT from an impulse response, such as $(n+1)[u[n]-u[n-5]]$, but I don't have any clue of where I should start from Thanks	2021-05-06 13:46:46	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.807502031326294, "perplexity": 1357.1525794688941}, "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-2021-21/segments/1620243988753.97/warc/CC-MAIN-20210506114045-20210506144045-00319.warc.gz"}
https://community.wolfram.com/groups/-/m/t/1753418?sortMsg=Likes	# BoolEval package for fast vectorized evaluation of comparisons Posted 3 years ago 8690 Views \| \| 5 Total Likes \| It's been out for quite a while now, but for those who don't know about it, I would like to announce the BoolEval package.Note: You can try a preview of the main function from the Wolfram Function Repository.The package provides a fast vectorized way to evaluate array comparisons (equalities or inequalities, possibly joined with logical operations).Suppose we want to select the numbers greater than 3 from the list {1, 2, ..., 10}. In[1]:= Needs["BoolEval"] In[2]:= mask = BoolEval[Range[10] > 3] Out[2]= {0, 0, 0, 1, 1, 1, 1, 1, 1, 1} BoolEval returns an array which contains 1s where the condition is true and 0 elsewhere.We can use this result for various things, such as masking the array: mask*Range[10]; counting the number of elements that satisfy the condition: Total[mask]; or Picking them out. BoolEval includes convenient helper functions for this: In[3]:= arr = Range[10]; BoolPick[arr, arr > 3] Out[4]= {4, 5, 6, 7, 8, 9, 10} In[5]:= BoolCount[arr > 3] Out[5]= 7 Those who use basically any other scientific computing system (such as MATLAB, numpy, R, Julia, etc.) will immediately recognize this way of working. Indeed, the original idea for BoolEval came from a StackExchange thread asking for this feature back in 2012: Does Mathematica have advanced indexing?Of course, the very same operations can be performed with Select, Cases and Count. Why do we need BoolEval then? The answer: BoolEval is much, much faster. In[6]:= arr = RandomReal[1, 1000000]; In[7]:= res1 = Select[arr, # < 0.3 &]; // RepeatedTiming Out[7]= {0.490, Null} In[8]:= res2 = Cases[arr, x_ /; x < 0.3]; // RepeatedTiming Out[8]= {0.4638, Null} In[9]:= res3 = BoolPick[arr, arr < 0.3]; // RepeatedTiming Out[9]= {0.017, Null} In this case, it's faster by a factor of 30. It achieved this amazing speed by converting the comparisons to arithmetic. E.g. arr >= 0 can be computed as UnitStep[arr]. To see how BoolEval converts a complex expression into arithmetic, just pass a symbolic expression to it: In[11]:= BoolEval[3 < a < 5 \|\| a^2 < 1] Out[11]= Unitize[1 + (1 - UnitStep[3 - a]) (1 - UnitStep[-5 + a]) - UnitStep[-1 + a^2]] Advanced Mathematica users will often use the UnitStep trick to get fast results, but as you can see, the complexity of the arithmetic transcription will quickly get out of hand ... It's difficult not to make a mistake. BoolEval does the conversion automatically, and behind the scenes. It basically allows one to use a convenient, human-readable notation while still getting the performance boost.BoolEval works with any expression that Mathematica can do arithmetic on, including images. E.g., we can select image levels between 0.75 and 0.85 intensity: im = ColorConvert[ExampleData[{"AerialImage", "Oakland2"}], "Grayscale"] BoolEval[0.75 < im < 0.85] ` I am preparing to release a new version of BoolEval in the coming days. I made this post to collect feedback. Please let me know if you have any suggestion, especially if you have any ideas about how to improve the documentation or what cool examples to include. Please be sure to read the tutorial included in the documentation.	2022-05-26 05:28:59	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.3210658133029938, "perplexity": 2244.762410058258}, "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/1652662601401.72/warc/CC-MAIN-20220526035036-20220526065036-00293.warc.gz"}
https://courses.cs.washington.edu/courses/cse552/19sp/ProjectIdeas.html	## CSE 552 Spring 2019 Project Ideas Spiciness levels: ★★ An easy project. You must be very busy with your other classes. ★★★ A solid project. You’ll be busy. ★★★★ A very serious project. You’re shooting for a conference paper. Every project should have an evaluation component. Feel free to propose your own idea; these ideas should give you a reference for the kind of work we expect. Benchmark EPaxos ★★ A “benchmark” project is practice at writing a good Evaluation section of a paper. Not the design and implementation, although by the time you’ve evaluated, you may be inspired. First, drive EPaxos from a single client offering a back-to-back causally-dependent workload with varying "think time" delays between requests. Plot a latency-throughput curve. Then explore the same curve as the number of clients goes up to saturation. Inject failures. How does performance change in the presence of failure? Are all failure costs momentary, or are there interesting “brown failures” that are particularly bad? EPaxos can isolate commutative operations. What happens to performance when the client workloads have non-commutative overlaps of {0%, 5%, 50%, 100%}? What happens with different kinds of overlap? For example, • Read-Modify-Write vs. Read-Modify-Write. Here, the operations are reading and writing the same cell of the stored state. • Blind-Write vs. Blind-Write. Two operations emit writes that don’t depend on reading any state, hence they can occur in any order. They shouldn’t conflict. • Blind increments. Each operation increments a counter. The increment operation is not conditional on the prior state (hence "blind"), but if 6 increment operations complete, a later read should show the state value to be 6. • Each operation decrements a value (allocates a resource). If the prior value was zero, the operation fails and returns a failure code. Enhance EPaxos ★★★ EPaxos’s observation that two blind writes are conflict-free is a special case of the general idea of Conflict-Freedom. Consider the family of CRDTs (Conflict-free Replicated Data Types). Are there ways to map CRDTs into EPaxos operations that would expose commutativity in ways that applications might find useful (as evidenced by the fact that applications can use CRDTs)? Which ones get a performance boost “for free” in EPaxos? For those that don’t, can you change EPaxos to exploit their type of commutativity? The project should evaluate the performance improvement of introducing the CRDTs into EPaxos versus the baseline implementation using EPaxos’s existing primitives. Benchmark Corfu/Tango ★★ Drive Tango from a single client offering a back-to-back causally-dependent workload with varying "think time" delays between requests. Plot a latency-throughput curve. Explore the same curve as the number of clients goes up to saturation. Break costs down into sequencer vs. storage. What happens when you use (or simulate) different types of storage? What costs are associated with failure and recovery? Evaluate costs of TCP vs. UDP + protocol-level reliability ★★★ Some systems, like Mencius, use TCP instead of UDP and then design their protocol to take advantage of the ordering guarantees it provides. Others, like EPaxos and IronFleet, use UDP and take care of reliability via protocol design. What are the trade-offs here? Evaluate this question by updating various systems that use TCP to use UDP and vice-versa. Make TrInc-BFT more efficient ★★★★ The TrInc paper describes one way to implement a Byzantine-fault-tolerant state machine that needs only $2f+1$ nodes, each with a trinket: by using TrInc to implement A2M, and A2M-PBFT-EA to implement PBFT. This is likely far more inefficient than necessary. For instance, the presence of a trinket on each machine means that one probably doesn’t need all the phases (pre-prepare, prepare, and commit) that PBFT uses. Implement BFT directly on TrInc, using a streamlined protocol, prove it safe and live, and evaluate its improved performance resulting from its increased efficiency. Reconfigure blockchain peer-to-peer network using the blockchain ★★★★ Bitcoin uses a peer-to-peer network to disseminate transaction proposals and new blocks. This network relies on fixed seeds in the Bitcoin client code and other centralized mechanisms to bootstrap the process of joining. Devise a way to eliminate, or at least reduce the importance of, this centralized mechanism by having the blockchain itself contain the seeds, or perhaps servers, used to disseminate information. This will be trickier than in Paxos, where there’s an assumption of a central authority that can be used to approve proper reconfigurations; you need to have an argument for why miners and other potential malefactors can’t subvert the reconfiguration process to undermine the security properties required for dissemination. Implement your approach so you can demonstrate, by performance evaluation, some quantitative benefit of it. For instance, if you replace the peer-to-peer network with a set of well-provisioned servers, you may save a lot of network bandwidth and reduce the latency of information dissemination, and thereby perhaps reduce the probability of forks. Use programmable switches to implement network spies ★★★★ In the paper, “Detecting failures in distributed systems with the FALCON spy network”, the authors describe a system in which spies exist at multiple levels, but stop at the network switch. That’s because it has no way of detecting the failure of switches themselves. Performing network failure localization with spies is also tricky because there may be multiple routes from a failed node and they must all be shut down to truly kill a switch (or even a multi-homed machine). Leverage software-defined networking (SDN) to program all the switches in a network to implement a network of spies. Evaluate the overhead of your implementation, as well as the resulting improvement in failure detection time. Verify a Chain Replication implementation ★★★ Implement Chain Replication in a verified language that provides a runnable implementation, either via explicit code (Dafny) or via extraction (Ivy, Coq). Verify that the system provides serializable semantics. Evaluate the resulting implementation. Compare it with a non-verified implementation (ideally one you didn’t write). Is the performance comparable? Where it isn’t, why isn’t it? Do those reasons relate to limitations of the verification framework? This isn’t a verification class, but verification is a good way to be sure you really understand how invariants hold a safety property together. It’s also useful to understand why refinement is a good way to rigorously state behavioral specification.	2019-05-25 03:30:12	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.37873679399490356, "perplexity": 2942.543741673139}, "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-2019-22/segments/1558232257847.56/warc/CC-MAIN-20190525024710-20190525050710-00248.warc.gz"}
https://www.physicsforums.com/threads/gravitational-force-why-the-negative-sign.179893/	# Gravitational Force(Why the negative sign?) 1. Aug 9, 2007 ### Gunman 1. The problem statement, all variables and given/known data How come it is said that attractive forces such as gravitational force have negative sign? And how come the sign is ignored in the case of gravitational force? And also, how come gravitational potential energy carries a negative sign? Is it because the displacement of the object is always below the reference point which is infinity? 2. Relevant equations 3. The attempt at a solution My notes state that the negative sign is omitted because gravitational repulsion doesnot exist. I don't exactly comprehend that. What is meant by gravitational repulsion? Thanks for any help given to help me clarify my doubts. =) 2. Aug 9, 2007 ### malawi_glenn Force is a vector, it has a direction. Signs of the forces depends on which coordinate system you use. Gravitionial repulsion is that masses dont repel, only attract. As in electromagnic force, you have positive and negative charges. Opposite sign attract, and same sign repel. 3. Aug 9, 2007 ### Dick Forces are vectors - they are neither negative nor positive. A COMPONENT of a force can be negative or positive depending on whether it goes in the plus direction of your coordinate system or minus. There are NOT two different classes of forces called 'negative' and 'positive'. 4. Aug 9, 2007 ### neutrino The negative sign is just a convention. When we use a coordinate system in which the unit vector points in a direction opposite to that of the force, then we need a negative sign to show that. In the case of gravity (gravitostatics), it's usually the spherical polar coordinates with the "source mass" at the origin. On the other hand, the magnitude of the force does not carry a negative sign, since magnitude by definition is positive. Did you mean to ask something else? The case for the potential is a result of defining the force with a negative sign. The potential is defined as the negative of line integral of the force, integrated from a reference point to a point where you need to find the potential. If you're not familiar with vector calculus, I'll put it another way: It's the work you need to do against the gravitational force to bring a "test mass" from a reference point (usually infinity) to a particular point in the field. 5. Aug 9, 2007 ### Gunman Ohh. Okay thanks people. =) The replies helpes greatly. 6. Aug 9, 2007 ### learningphysics The formulas for gravitational forces and electric forces use the convential that the unit vector is radially outward (direction of repulsion)... So a positive value in the formula means the force is acting radially outward (repulsive)... when you get a negative value the force is radially inward (attractive). But for potential energy this convention doesn't matter you'll get the same result either way... because energy is a scalar. The potential energy when two masses are to a distance of R2 - the potential energy when they are at R1 = $$\deltaE = - Gm_1m_2/R_2 - (-Gm_1m_2/R_1)$$ Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook Have something to add?	2016-12-08 02:24:05	{"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": 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": 0.826481819152832, "perplexity": 522.7560794146958}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-50/segments/1480698542323.80/warc/CC-MAIN-20161202170902-00044-ip-10-31-129-80.ec2.internal.warc.gz"}
https://physics.stackexchange.com/questions/150746/red-shifted-photons-lost-energy-in-which-form?noredirect=1	# Red shifted photons lost energy in which form? Red shifted photons lost energy in which form? Photons which have experienced a change in frequency (red shift) due to gravity(or other red shifting affects), have necessarily lost energy, total energy is conserved. Red shifts happen because of various causes. there also exist blue shifts: Conversely, a decrease in wavelength is called blueshift and is generally seen when a light-emitting object moves toward an observer or when electromagnetic radiation moves into a gravitational field. Now on redshifts: Some redshifts are an example of the Doppler effect, familiar in the change in the apparent pitches of sirens and frequency of the sound waves emitted by speeding vehicles. A redshift occurs whenever a light source moves away from an observer. Energy is conserved by the motion of the source. Motion =kinetic energy . the red shifted adds to the kinetic energy of the source seen in the rest frame of the obsrver, and the blue adds to the energy of the photon again seen in the rest frame of the observer. Another kind of redshift is cosmological redshift, which is due to the expansion of the universe, and sufficiently distant light sources (generally more than a few million light years away) show redshift corresponding to the rate of increase in their distance from Earth. Again the motion takes up the energy balance Finally, gravitational redshift is a relativistic effect observed in electromagnetic radiation moving out of gravitational fields. The gravitational field picks up the balance of energy, again in the rest frame of the observer. • Anna, I was going to ask a similar question, and this one popped up. So considering $E=mc^2$ and $E=h\nu$ fully governs the energy conservation, then for a redshifted gamma photon from the big bang, $\nu$ decreases and so does $m$ since $c$ is constant. Right? Or are there more physics involved? – docscience Apr 14 '17 at 16:04 • @docscience E=mc^2 is misleading and we no longer use it in particle physics. This m is just a mathematical description of the extra inertia and confuses things. We only work with the rest mass, ( the "length" of the four vector) and the rest mass of the photon is always zero. So when the energy decreases nu becomes smaller, equivalent to the acoustic doppler shift – anna v Apr 14 '17 at 16:21 • Thanks! So then where did the energy go if it didn't change the mass? Considering say one photon by itself as a 'system'. Was it lost to 'space-time'? I'm still researching, but can the redshift be one source of the dark energy? – docscience Apr 14 '17 at 16:28 • @docscience For conservation of energy one has to consider the relative velocity of the observer to the photon source,. – anna v Apr 14 '17 at 17:00 • But considering the source is a dead star that was formed shortly after the big bang, how could the source matter any more today? Is it because, for the photon, time is 'stopped'? – docscience Apr 14 '17 at 17:45 It seems contradictory that red shifted light has lost energy yet total energy is conserved (where did the energy go?). The trick to understanding this is knowing that the energy measured depends on the frame of reference you are measuring from. Consider a ball flying towards you quite fast and hitting you in the head. From your perspective it has a lot of kinetic energy (and it hurt when it hit). Now consider yourself flying along at the same speed as the ball. It's stationary compared to you and it has no kinetic energy (compared to you). It can't hit you in the head and it can't hurt you because its not moving towards you any more. In both cases the ball is doing the same thing (and the energy in the system hasn't changed), its just your frame of reference that has changed and the same is happening with red shifted light. If you move away from light (at a fast enough speed) it will appear red shifted (less energy) and conversely if you move towards it (at a fast enough speed), it will appear blue shifted (appear to have gained energy). Going back to the ball example, if you drive towards a ball that is flying towards you, it will hurt more (have more energy) if you drive away from it, it will hurt less (have less energy). Nothing has changed in the ball, or it's energy, it's you that has changed (e.g. you've used energy to accelerate towards or away from it). • Technically, you can't "move away from light." You can move away from a light source, in which case, the spectrum of light that you receive from the source will be shifted toward the red from the spectrum that you expected to see based on your knowledge of the process that produced the light. – Solomon Slow Jul 27 '16 at 17:09 Energy is most definitely conserved in the case of gravitation-ally red shifted (GRS) photons. The sun is 4.6 billion years old and has energy output equal to 3.8x10+26 watts. If 1% of the energy output is lost to GRS, an enoumous quantity of energy is missing. If the lost energy were simply hanging out in the surrounding gravitational field, it should be somehow observable by now since the energy has been building up and stored for billions of years. However, that is not the case. Electromagnetic energy can not be "trapped" and somehow stored, but it can be converted to increasingly smaller frequency. All photon energy escapes EVERY gravitational field, but is red shifted. Said differently, a blue photon released from the sun is converted to many red photons as it escapes the gravitational field. Conservation of energy dictates this result. Imagine a transverse wave traveling down a stretched rope. The rope suddenly divides into two ropes. The initial wave is transformed into two smaller waves each of lower energy.	2019-10-19 14:29:04	{"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": 0.6026197075843811, "perplexity": 397.8025831557082}, "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/1570986696339.42/warc/CC-MAIN-20191019141654-20191019165154-00310.warc.gz"}
http://abstract.pugetsound.edu/aata/section-irreducible-poly.html	A nonconstant polynomial $f(x) \in F[x]$ is irreducible over a field $F$ if $f(x)$ cannot be expressed as a product of two polynomials $g(x)$ and $h(x)$ in $F[x]$, where the degrees of $g(x)$ and $h(x)$ are both smaller than the degree of $f(x)$. Irreducible polynomials function as the “prime numbers” of polynomial rings. ##### Example17.11 The polynomial $x^2 - 2 \in {\mathbb Q}[x]$ is irreducible since it cannot be factored any further over the rational numbers. Similarly, $x^2 + 1$ is irreducible over the real numbers. ##### Example17.12 The polynomial $p(x) = x^3 + x^2 + 2$ is irreducible over ${\mathbb Z}_3[x]$. Suppose that this polynomial was reducible over ${\mathbb Z}_3[x]$. By the division algorithm there would have to be a factor of the form $x - a$, where $a$ is some element in ${\mathbb Z}_3[x]$. Hence, it would have to be true that $p(a) = 0$. However, \begin{align} p(0) & = 2\\ p(1) & = 1\\ p(2) & = 2. \end{align} Therefore, $p(x)$ has no zeros in ${\mathbb Z}_3$ and must be irreducible. ##### Example17.16 Let $p(x) = x^4 - 2 x^3 + x + 1$. We shall show that $p(x)$ is irreducible over ${\mathbb Q}[x]$. Assume that $p(x)$ is reducible. Then either $p(x)$ has a linear factor, say $p(x) = (x - \alpha) q(x)$, where $q(x)$ is a polynomial of degree three, or $p(x)$ has two quadratic factors. If $p(x)$ has a linear factor in ${\mathbb Q}[x]$, then it has a zero in ${\mathbb Z}$. By Corollary 17.15, any zero must divide 1 and therefore must be $\pm 1$; however, $p(1) = 1$ and $p(-1)= 3$. Consequently, we have eliminated the possibility that $p(x)$ has any linear factors. Therefore, if $p(x)$ is reducible it must factor into two quadratic polynomials, say \begin{align} p(x) & = (x^2 + ax + b )( x^2 + cx + d )\\ & = x^4 + (a + c)x^3 + (ac + b + d)x^2 + (ad + bc)x + bd, \end{align} where each factor is in ${\mathbb Z}[x]$ by Gauss's Lemma. Hence, \begin{align} a + c & = - 2\\ ac + b + d & = 0\\ ad + bc & = 1\\ bd & = 1. \end{align} Since $bd = 1$, either $b = d = 1$ or $b = d = -1$. In either case $b = d$ and so \begin{equation}ad + bc = b( a + c ) = 1.\end{equation} Since $a + c = -2$, we know that $-2b = 1$. This is impossible since $b$ is an integer. Therefore, $p(x)$ must be irreducible over ${\mathbb Q}$. ##### Example17.18 The polynomial \begin{equation}f(x) = 16 x^5 - 9 x^4 + 3x^2 + 6 x - 21\end{equation} is easily seen to be irreducible over ${\mathbb Q}$ by Eisenstein's Criterion if we let $p = 3$. Eisenstein's Criterion is more useful in constructing irreducible polynomials of a certain degree over ${\mathbb Q}$ than in determining the irreducibility of an arbitrary polynomial in ${\mathbb Q}[x]$: given an arbitrary polynomial, it is not very likely that we can apply Eisenstein's Criterion. The real value of Theorem 17.17 is that we now have an easy method of generating irreducible polynomials of any degree. # SubsectionIdeals in $F\lbrack x \rbrack$¶ permalink Let $F$ be a field. Recall that a principal ideal in $F[x]$ is an ideal $\langle p(x) \rangle$ generated by some polynomial $p(x)$; that is, \begin{equation}\langle p(x) \rangle = \{ p(x) q(x) : q(x) \in F[x] \}.\end{equation} ##### Example17.19 The polynomial $x^2$ in $F[x]$ generates the ideal $\langle x^2 \rangle$ consisting of all polynomials with no constant term or term of degree 1. ##### Example17.21 It is not the case that every ideal in the ring $F[x,y]$ is a principal ideal. Consider the ideal of $F[x, y]$ generated by the polynomials $x$ and $y$. This is the ideal of $F[x, y]$ consisting of all polynomials with no constant term. Since both $x$ and $y$ are in the ideal, no single polynomial can generate the entire ideal. ##### Proof Throughout history, the solution of polynomial equations has been a challenging problem. The Babylonians knew how to solve the equation $ax^2 + bx + c = 0$. Omar Khayyam (1048–1131) devised methods of solving cubic equations through the use of geometric constructions and conic sections. The algebraic solution of the general cubic equation $ax^3 + bx^2 + cx + d = 0$ was not discovered until the sixteenth century. An Italian mathematician, Luca Pacioli (ca. 1445–1509), wrote in Summa de Arithmetica that the solution of the cubic was impossible. This was taken as a challenge by the rest of the mathematical community. Scipione del Ferro (1465–1526), of the University of Bologna, solved the “depressed cubic,” \begin{equation}ax^3 + cx + d = 0.\end{equation} He kept his solution an absolute secret. This may seem surprising today, when mathematicians are usually very eager to publish their results, but in the days of the Italian Renaissance secrecy was customary. Academic appointments were not easy to secure and depended on the ability to prevail in public contests. Such challenges could be issued at any time. Consequently, any major new discovery was a valuable weapon in such a contest. If an opponent presented a list of problems to be solved, del Ferro could in turn present a list of depressed cubics. He kept the secret of his discovery throughout his life, passing it on only on his deathbed to his student Antonio Fior (ca. 1506–?). Although Fior was not the equal of his teacher, he immediately issued a challenge to Niccolo Fontana (1499–1557). Fontana was known as Tartaglia (the Stammerer). As a youth he had suffered a blow from the sword of a French soldier during an attack on his village. He survived the savage wound, but his speech was permanently impaired. Tartaglia sent Fior a list of 30 various mathematical problems; Fior countered by sending Tartaglia a list of 30 depressed cubics. Tartaglia would either solve all 30 of the problems or absolutely fail. After much effort Tartaglia finally succeeded in solving the depressed cubic and defeated Fior, who faded into obscurity. At this point another mathematician, Gerolamo Cardano (1501–1576), entered the story. Cardano wrote to Tartaglia, begging him for the solution to the depressed cubic. Tartaglia refused several of his requests, then finally revealed the solution to Cardano after the latter swore an oath not to publish the secret or to pass it on to anyone else. Using the knowledge that he had obtained from Tartaglia, Cardano eventually solved the general cubic \begin{equation}a x^3 + bx^2 +cx +d = 0.\end{equation} Cardano shared the secret with his student, Ludovico Ferrari (1522–1565), who solved the general quartic equation, \begin{equation}a x^4 + b x^3 + cx^2 + d x + e = 0.\end{equation} In 1543, Cardano and Ferrari examined del Ferro's papers and discovered that he had also solved the depressed cubic. Cardano felt that this relieved him of his obligation to Tartaglia, so he proceeded to publish the solutions in Ars Magna (1545), in which he gave credit to del Ferro for solving the special case of the cubic. This resulted in a bitter dispute between Cardano and Tartaglia, who published the story of the oath a year later.	2017-03-26 22:53:07	{"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": 0.8747646808624268, "perplexity": 424.30277415263794}, "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-2017-13/segments/1490218189313.82/warc/CC-MAIN-20170322212949-00170-ip-10-233-31-227.ec2.internal.warc.gz"}
https://cs.stackexchange.com/questions/93297/will-hardware-implementation-affect-the-time-space-complexity-of-algorithms	# Will hardware/implementation affect the time/space complexity of algorithms? I’m not even a CS student, so this might be a stupid question, but please bear with me... In the pre-computer era, we can only implement an array data structure with something like an array of drawers. Since one have to locate the drawer with corresponding index before extracting the value from it, the time complexity of array lookup is $O(log(n))$, assuming binary search. However, the invention of computers made a big difference. Modern computers can read from their RAM so fast that we now consider the time complexity of array lookup to be $O(1)$ (even it’s technically not the case, because it takes more time to move the register over a greater distance, etc) Another example is Python dictionaries. While one might get a dictionary access complexity of $O(n)$ with an ill-written overloaded __hash__ magic method (or ridiculously bad luck, i.e. keys having lots of hash collisions), it’s usually presumed to be $O(1)$. In this case, time complexity depends on both the hash table implementation of Python dictionaries, and the keys’ implementation of the hash functions. Does this imply that hardware/implementation can affect the time complexity of algorithms? (While both examples are about data structures instead of algorithms, the latter are built on the former, and I've never heard of time complexity of data structures, so I'm using the term "algorithms" here) To me, algorithms are abstract and conceptual, whose properties like time/space complexity shouldn’t be affected by whether they’re implemented in a specific way, but are they? • Comments are not for extended discussion; this conversation has been moved to chat. – Gilles Jun 24 '18 at 9:58 Sure. Certainly. Here's how to reconcile your discomfort. When we analyze the running time of algorithms, we do it with respect to a particular model of computation. The model of computation specifies things like the time it takes to perform each basic operation (is an array lookup $O(\log n)$ time or $O(1)$ time?). The running time of the algorithm might depend on the model of computation. Once you've picked a model of computation, the analysis of the algorithm is a purely abstract, conceptual, mathematical exercise that no longer depends on hardware. However, in practice we usually want to pick a model of computation that reflects the reality of our hardware -- at least to a reasonable degree. So, if hardware changes, we might decide to analyze our algorithms under a different model of computation that is more appropriate to the new hardware. That is how the hardware can affect the running time. The reason this is non-obvious is because, in introductory classes, we often don't talk about the model of computation. We just implicitly make some assumptions, without ever making them explicit. That's reasonable, for pedagogical purposes, but it has a cost -- it hides away this aspect of the analysis. Now you know. • As you said, we use the random access model as model of computation but when we use GPU for certain calculations the time complexity for some algorithms changes as it uses the SIMD instructions. – Deep Joshi Jun 21 '18 at 6:04 • Also note that O() notation is an upper bound. Even if you use the drawer analogy finding a drawer in a limited size (real memory is limited in size) building takes O(1) time. Even if it takes you 20 minutes to reach the furthest drawer (all cache misses and you even have to load the data from swap) that is still O(1) time because 20 minutes will be your hidden constant for accessing memory. – Goswin von Brederlow Jun 21 '18 at 14:51 • For a real life example of what @GoswinvonBrederlow is saying, consider that searching for a value with a hash map is effectively $O(1)$. Searching for a value in an unordered array is $O(n)$. However, in a real application I wrote recently, using an array was faster than using a hash map. It was faster because I had a small number of values to search (less than 10), and the overhead of doing the hash and the hashmap algorithm was actually greater than the cost of just brute forcing the array-search! For practical sized problems, I sped my code up by a factor of 3 this way. – Cort Ammon Jun 21 '18 at 15:16 • @CortAmmon: Even on a large array, using linear search may be faster than using a hash map if all but a few of the elements on is searching for are very near the start. For example, if 50% of the elements match the first element, 25% match the second, 12.5% match the third, etc. except that one oddball element will match something that might be anywhere in the array, the expected number of comparisons to perform M lookups on a list of size N would be 2M+N. – supercat Jun 21 '18 at 18:42 • @DeepJoshi SIMD instructions don't change the complexity of algorithms. They only change the multiplicative constant. – Gilles Jun 21 '18 at 19:49 I think there's a fundamental misunderstanding in the question. You compare a person finding an object in a sorted list (e.g., a specific page in a book, given its number) with a computer looking up an item from an array. The reason that the former takes time $O(\log n)$ and the latter takes time $O(1)$ is not that the computer is so fast that it can do the binary search in the blink of an eye. Rather, it's because the computer doesn't use binary search at all. The computer has a mechanism to directly retrieve items from the array without searching. To retrieve the contents of an array cell, the computer just tells the memory controller the analog of "Give me page seventeen", the memory controller sets the voltages on the address wires to the binary representation of seventeen and the data comes back. So, yes, the hardware (i.e., the model of computation) does affect the running time of algorithms, as D.W. explains, but that's not what your array access example seems to be based on. • To be fair, you skipped all the pieces in between "the memory controller sets the voltages on the address wires to the binary representation of seventeen" and "the data comes back". One of those pieces almost certainly is a binary search tree of the kind described by the OP; but it nonetheless executes in constant time because log n is approximately 64, for all n. – Quuxplusone Jun 22 '18 at 2:01 • @Quuxplusone What part of the memory uses binary search? The address lines directly select the memory cells. – David Richerby Jun 22 '18 at 14:23 • We're operating far far outside my area of expertise, but what I was trying to imply is that an address decoder will be implemented in terms of a tree of demuxers. (Assuming that we are directly hitting physical memory, ignoring any extra complication that comes with caching.) Again, all this extra complication adds only O(lg size-of-memory), i.e., negligible — but that's exactly the bit OP was asking about! – Quuxplusone Jun 22 '18 at 18:10 No, the hardware doesn't affect the complexity of algorithms. But, it does affect the choice of algorithm, and it can affect the usefulness of complexity analysis to a point where the analysis becomes pretty much meaningless (or merely of academic interest). Finding the right drawer (as accessing an array element) uses the "open Nth element directly by index" algorithm, not the "search linearly" or "do binary search" algorithm. The algorithms are not changed, but the choice. On the other hand side, complexity analysis itself, or rather its meaningfulness, is affected greatly by hardware. Many algorithms that are stellar by their complexity analysis are poor performers or even useless in practice because the insignificant constant factor is not at all insignificant, but dominating. Or, because assumptions that were once true (or mostly true) no longer hold. Such as, for example, every operation is mostly the same (only small constant differences that don't matter), or it doesn't make a difference which memory locations you access in which order. By complexity analysis, you may conclude that some algorithm is vastly superior because it only needs so and so many operations. In practice, you may find that each operation causes a guaranteed cache miss (or worse yet, page fault), which introduces a k that's so huge that it is no longer insignificant, but dominating everything. If algorithm A takes 500 operations for processing a dataset of a given size and algorithm B takes only 5, but B causes 5 faults which burn twenty million cycles each, then despite what anaylsis or common sense may tell you, A is better. This has lead to funny surprises such as e.g. in Cuckoo Hashing a few years ago. Which was vastly superior because [long list of benefits]. After the hype cooled, it turned out that it was vastly inferior because it guaranteed two cache misses (faults, for larger data sets) on every access. Similar has happened to identifying and processing subsets of data. Often, the correct solution nowadays is: "just do it all", i.e. instead of figuring out what you need to proess and do that, process the complete dataset linearly even if you maybe only need half of it. Because, believe it or not, that's faster due to no branch mispredictions, no cache misses, no page faults. Need to read the first 8kB and the last 3kB of a 3MB file? Well, read the complete file, and throw away what you don't want, because seeking in between will be ten times slower than just reading the complete thing. Use a map because it has logarithmic complexity? Or a hash table, which has constant access time? Constant sounds awesome. Well, for anything with fewer than a thousand or so things (depending on hardware, data size, and access pattern), a linear search may be just as good or better. Surprise. So, it's not the algorithms per se that are affected, but their usefulness, and choice.	2019-03-18 14:36:57	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 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": 0.3735349774360657, "perplexity": 845.7481268902553}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-13/segments/1552912201329.40/warc/CC-MAIN-20190318132220-20190318154220-00117.warc.gz"}
https://nvda.groups.io/g/nvda/message/38011?p=,,,20,0,0,0::Created,,posterid%3A59943,20,2,0,16658351	#### Re: Maths? Michael Damien, The material below is from the appendix to a calculus book for blind students I have written. Its position in the market place has not yet been decided. I hope it helps you. Appendix 2 How To Modify The Speech Dictionary In NVDA NVDA is a screen reading program produced by NV Access(R). It speaks the information about various parts of the active window in response to keystrokes pressed by the user. If the keystroke is not an NVDA command, NVDA merely echoes the keystroke. But, if it is an NVDA command keystroke such as NVDA-key + t, NVDA tells the user something about the active window. In the case of NVDA-Key + t, NVDA announces the title of the active window. NVDA commands take the form of the NVDA-key followed by a letter. The INSERT key is by default the NVDA-key. However, NVDA allows the user to change the NVDA-key to the CapsLock key if desired. The user can control how NVDA announces a string of characters. If the user wants NVDA to speak a particular string of characters a certain way, he or she can make an entry in its speech dictionary for that particular string of characters. After this entry is saved into the speech dictionary, whenever that particular string of characters is encountered, NVDA speaks it according to the entry just made in the speech dictionary. For example, NVDA ordinarily speaks the word tortilla with an "L" sound rather than a "y" sound. To force NVDA to pronounce it with the "y" sound, there must be an entry in the speech dictionary to instruct NVDA to speak the string of characters "torteeya" as desired. To make such an entry do the following. STEP 1: Put focus on the main NVDA menu with the keystroke NVDA-key + n. The user can arrow down through a list of submenus. Or, press the letter "p" to go to the preferences menu. The first menu under "Preferences" is 'General'. The user may arrow down the the Speech Dictionary or type 'd" and NVDA will jump into the default speech dictionary, There are three entries. They are respectively "Default", "voice", and "temporary". I suggest putting my changes into the default dictionary. But before entering the default dictionnary, a word about the other two choices is in order. The user can arrow once to the voice dictionary which is the dictionary belonging to the synthesizer currently running. For instance, if the Microsoft Speech Platform synthesizer is running, arrowing to the Voice Dictionary opens the dictionary whose entries govern the speech under that synthesizer. The entries in this dictionary apply to the Microsoft Speech Platform session and not to any other synthesizer that has been installed. Whatever synthesizer is running, the entries in the default speech dictionary apply. The tempory dictionary is like the voice dictionary, but it is erased when you exit the NVDA session. Entering the default dictionary. A list of entries will appear. The user may add a new entry or arrow down to an entry he or she wishes to change. To add a new entry Tab once. To change an entry, arrow down the the entry in question and tab twice. Or to remove the entry in question, tab three times. But the object of this appendix is to add entries in support of the calculus book. So, tab only once to add an entry. Focus will be on the "Add" button. Press the ENTER key to activate it. NVDA says "add a dictionary entry dialogue, pattern edit". Focus is now in an edit box. The string of characters to be entered here is the string of characters as they appear in the text, that is, the actual character string encountered by the reading cursor. As an example, type in the word tortilla. Press the TAB key ". again, and the focus will be in another edit box. NVDA will say "replacement pattern edit". The replacement string to be typed will force NVDA to pronounce the word tortilla with a 'y' sound rather than an 'l' sound. That pattern is "tortee ya". Tab again to put focus in a comment edit box. Tabbing again puts focus on a check box called "Case sensitive". Its default value is "not checked". To make the replacement happen only when tortilla is upper case (either its first letter or all of them), press the space bar to check it. Tabbing again puts focus on a combo box whose entries are respectively "anywhere, "forward", and "regular expression". The "anywhere" choice means that the pattern being replaced may appear either by itself or anywhere in a longer string of non-blank characters. The "forward" choice means that the pattern being replaced must appear at the very front of a longer string of characters. Of course, it may appear by itself and not part of a longer string of characters. If the "forward" choice is picked, and the pattern being replaced appears following one or more non-blank characters, NVDA will not speak using the replacement pattern. Regular expressions are required to address certain situations where more flexibility is needed, and they will be discussed later. For now, suffice it to say, that in the case of a regular expression, the pattern to be replaced and the replacement pattern will contain special characters that will cause a more suffisticated replacement to take place. Now, the user is ready to implement the changes specified below. However, since NVDA supports regular expressions, some of them will not be entered into the NVDA speech dictionary. The speech dictionary dialog may ask you if the Actual Pattern should be case sensitive. If you check "no", then any characters in the Actual Pattern string may be either upper or lower case. If however, you check "yes", then the characters in the Actual Pattern string must exactly match the case of the characters encountered in the text before the Replacement Pattern will be triggorred. The changes listed below do not need case sensitivity unless otherwise stated. SECTION0.1 Modifications For Chapter 1 Number Systems Chapter 1 discusses number systems which uses * for multiplication, / for division, + for addition, and - for subtraction. One change for chapter 1 is that for section headings, "SECTION". Actual Pattern="SECTION", Replacement Pattern="section" The other change is that Subscripts in this book are indicated by the suffix underscore plus an integer. An example is x_0 for x sub zero. Actual Pattern=_", Replacement Pattern="sub" Some readers may wish to hear x^2 read as x squared rather than x raised to exponent 2 and also x^3 read as x cubed. . If the reader is willing to treat these two cases as special cases, the preferred reading can be supported as follows. Use two carot symbols instead of one, and preceed the carots with a space. Without that space, the variable name may not be spoken clearly. For example, write x cubed as x ^^3. This will require the user to make the following changes to the speech dictionary. Actual Pattern ^^2 and Replacement Pattern squared. The change for x cubed is similar. SECTION0.2 Modifications for Chapter 2 Functions Chapter 2 discusses functions of one and of two variables. I introduce notation for exponents, denominators, and absolute value. Below is a list of changes I propose to make to the speech dictionary for chapter 2. It would be nice if the characters "f(x)" is spoken as "f of x". But, we also want "g(t)" to be read as "g of t". IN other words, we want the function name to be any character upper or lower case and to have a subscript. We also want flexibility in the number of independent variables and also flexibility of variable names. Three regular expressions working together accomplish this goal. Regular expression recognizing the function name and opening parenthesis.: Actual Pattern: "([a-zA-Z])$$" Replacement Pattern: "\1 of open parenthesis " Regular expression recognizing all arguments which are followed by a comma: Actual Pattern: "([a-zA-Z]?)(\_?\d)([/\.]?\d) ([-]\|[+]?) ([a-zA-Z]?)(\_?\d)([/\.]?\d) [\,]{1}" Replacement Pattern: "\1 \2 \3 \4 \5 \6 \7 , " Regular expression recognizing an argument not followed by a comma: Actual Pattern: "([a-zA-Z])(\_?\d)([/\.]?\d) ([-]\|[+]?) ([a-zA-Z]?)(\_?\d)([/\.]?\d) $$" Replacement Pattern: "\1 \2 \3 \4 \5\6 \7 close parenthesis " H(x0) will be read as "h of x". The function name (f, g, h, or any other letter) does not matter. The following changes do not involve regular expressions. actualPattern=:R->R, Replacement Pattern= maps R into R Actual Pattern="+/-", Replacement Pattern="plus or minus" Actual Pattern="<=", Replacement Pattern="less than or equal to" Actual Pattern=">=", Replacement Pattern="greater than or equal to" ActualPattern=!, Replacement Pattern=factorial ActualPattern=_, Replacement Pattern=sub Actual Pattern="(/", Replacement Pattern="open parenthesis begin denominator" Actual Pattern="/)", Replacement Pattern="end of denominator close parenthesis" Actual Pattern="(^", Replacement Pattern="open parenthesis begin exponent" Actual Pattern="^)", Replacement Pattern="end of exponent close parenthesis" Actual Pattern="(\|", Replacement Pattern="open parenthesis begin absolute value" Actual Pattern="\|)", Replacement Pattern="end of absolute value close parenthesis" Actual Pattern="}/", Replacement Pattern="open brace begin denominator" Actual Pattern="/}", Replacement Pattern="end of denominator close brace" Actual Pattern="{^", Replacement Pattern="open brace begin exponent" Actual Pattern="^}", Replacement Pattern="end of exponent close brace" Actual Pattern="{\|", Replacement Pattern="open brace begin absolute value" Actual Pattern="\|}", Replacement Pattern="end of absolute value close brace" Actual Pattern="[/", Replacement Pattern="open bracket begin denominator" Actual Pattern="/]", Replacement Pattern="end of denominator close bracket" Actual Pattern="[^", Replacement Pattern="open bracket begin exponent" Actual Pattern="^]", Replacement Pattern="end of exponent close bracket" Actual Pattern="[\|", Replacement Pattern="open bracket begin absolute value" Actual Pattern="\|]", Replacement Pattern="end of absolute value close bracket" NOTE: In the text, I only use parentheses to specify begin and end of exponents, denominators, and absolute value. It is up to you whether to use braces and brackets as well. But, if you do decide to use braces and brackets n addition to parentheses, I strongly suggest that the Replacement Pattern mentions the brace or bracket . unless you do, the speech will not help you keep track of opening and closing braces and brackets. If documents you send to your professor do not have an equal number of opening and closing braces/brackets/parentheses, he or she will think SECTION0.3Modifications For Chapters 3, 4, and 5 The chapter on limits only needs two changes to the speech dictionary. No changes are needed for chapters 4 and 5. Actual Pattern="lim:", Replacement Pattern="limit as" Actual Pattern="->", Replacement Pattern="approaches" SECTION0.4Modifications For Chapters 6, 7, and 8 Regular expression for what variable a derivative is taken with respect to: Actual Pattern: "(\/)d([a-zA-Z])(\))" Replacement Pattern: "with respect to \2" Actual Pattern= (d0 Replacement Pattern="Open parenthesis ze rowth derivative" Actual Pattern="(d1" Replacement Pattern="Open parenthesis first derivative" Actual Pattern= "(d2" Replacement Pattern=" Open parenthesis second derivative" Actual Pattern= "(d3" Replacement Pattern="Open parenthesis third derivative" Ordinary derivative whose differentiation level is specified by an integer: Fourth derivative: Actual Pattern: "(d4" Replacement Pattern: "open parenthesis fourth derivative" Eighth derivative: Actual Pattern: "(d8" Replacement Pattern: "Open parenthesis eighthth derivative " The first, second, third, fourth, and eighth derivatives are entries that are not regular expressions. They are more understandable when not handled by a regular expression. However, the fifth, sixth, seventh, and ninth derivatives are all handled by a regular expression. Their replacement patterns are very understandable. Regular expression recognizing fifth, sixth, seventh, and ninth derivatives: Actual Pattern: "($$)d([5679])" Replacement Pattern: "open parenthesis \2 th derivative" Ordinary derivatives whose differentiation level is specified by an index variable: This regular expression comes into play in a future chapter, but it fits here. Actual Pattern: "(\()d([a-zA-Z])([-]\|[+]?)([1-9]?)" Replacement Pattern: "open parenthesis \2 th \3 \4 derivative" SECTION0.5 Modifications For Chapter 9, Rieman Integration This chapter introduces notations for summation and for integrals. I leave it to the reader whether or not to make this group case sensitive. I mean demanding that the characters "int" should be upper case to prevent collisions in the text. Actual Pattern="INT:", Replacement Pattern="integral" Actual Pattern="INT:{", Replacement Pattern="integral lower bound" Actual Pattern="}{", Replacement Pattern="and upper bound" Actual Pattern="}:", Replacement Pattern="of the integrand" Actual Pattern="INT{}", Replacement Pattern="indefinite integral" Actual Pattern="INT{}:", Replacement Pattern="indefinite integral of the integrand" Actual Pattern="INT:{}", Replacement Pattern="integral" I found myself entering the above integral notations using brackets instead of braces, so I recommend including the following just to make the entering text more forgiving. Actual pattern="INT [", replacement pattern="integral lower bound" actual pattern ="][", replacement pattern ="and upper bound" actual pattern ="]:", replacement pattern ="of the integrand" actual pattern ="INT[]", replacement pattern ="indefinite integral" actual pattern ="INT[]:", replacement pattern ="indefinite integral of the integrand" Your professor might prefer the representation of an upper bound used in latex which is "\to". Actual Pattern="\to", Replacement Pattern="and upper bound" Actual Pattern="INTI:", Replacement Pattern="inner integral" Actual Pattern="INTM:", Replacement Pattern="middle integral" Actual Pattern="INTO:", Replacement Pattern="outer integral" SECTION0.6 Modifications for Chapter 12 Vectors actual pattern={\|\|x+y\|\|}, Replacement Pattern=open brace begin norm x+y end norm close brace Actual pattern=DET2, Replacement Pattern=second order determinant Actual pattern=DET3, Replacement Pattern=third order determinant Actual pattern=(.), Replacement Pattern=dot product Actual pattern=(), Replacement Pattern=cross product SECTION0.7 Modifications For Chapter 13 Partial derivative whose differentiation level is specified by an integer: Fourth partial derivative: Actual Pattern: "(pd4" Replacement Pattern: "open parenthesis fourth partial derivative" Eighth partial derivative: Actual Pattern: "(pd8" Replacement Pattern: "open parenthesis eighth partial derivative" Regular expression recognizing fifth, sixth, seventh, and ninth partial derivatives: Actual Pattern: "(\()pd([5679] )" Replacement Pattern: "open parenthesis \2 th partial derivative Partial derivative whose differentation level is specified by an index variable: Actual Pattern: "(\()pd([a-zA-Z])([-]\|[+]?)([1-9]?)" Replacement Pattern: open parenthesis \2 th \3 \4 partial derivative SECTION0.8 Modifications For Chapters 14, 15, 16, and 17 actualWord=:R->Rn Replacement Pattern=maps R into R n actualWord=:R2->R2 Replacement Pattern=maps R2 into R2 actualWord=:R3->R3 Replacement Pattern=maps R3 into R3 actualWord=:Rn->Rn Replacement Pattern=maps R n into R n actualWord=:R2->R Replacement Pattern=maps R2 into R actualWord=:R3->R Replacement Pattern=maps R3 into R actualWord=:RN->R Replacement Pattern=maps RN into R This group is case sensitive. Actual Pattern="DINT::", Replacement Pattern="double integral" Actual Pattern="LINT::", Replacement Pattern="line integral over curve" Actual Pattern="SINT::", Replacement Pattern="surface integral" Actual Pattern="TINT::", Replacement Pattern="triple integral" There is a set of notations for derivatives for which I made no entries into the speech dictionary. It is not uncommon to represent a derivative by following the function name with an apostrophe. For example, f'(x) represents the first derivative of f(x). Likewise, f''(x) is the second derivative of f(x). You may occasionally encounter three apostrophes for third derivative such as for example f'''(x). However, it is unlikely that the student will ever see more than three apostrophes for differentiation. In fact, the use of apostrophe for differentiation is more prevalent in differential equations books than in a calculus text. So, if the student hears y' or y'', he or she should understand that differentiation is being represented. Just get used to it. Testing regular expressions involving carrot and parenthesis exposed a problem in two of the seven synthesizers I exercised. Those two are Microsoft Speech Platform and Microsoft SAPI5. The other five synthesizers passed the test. I tested regular expressions for beginning and ending exponential expressions. My syntax for these exponential expressions is that the expressions are bracketed between (^ and ^). Regular expression to recognize beginning of exponential expression: Actual Pattern: [\(][\^] Replacement Pattern: Begin exponent Regular expression to recognize end of exponential expression: Actual Pattern: [\^][$$] Replacement Pattern: end of exponent The regular expression for the end of an exponential expression fails to recognize the end. The regular expression for the beginning exponential expression works as expected. The two regular expressions are very similar. Why does one work and not the other? Here are some test cases: (^ ^) (^w+3^) I then wrote a regular expression to recognize the pair "()", and NVDA sees the closing parenthesis but not the opening one. Puzzling! The following synthesizers execute these regular expressions correctly. Eloquence, ESpeak NG, Soft Voice, Speech Player ESpeak, SVox Pico Synthesizer, The following synthesizers did not execute these regular expressions correctly. Microsoft Speech API Version 5, Microsoft Speech Platform For these last two synthesizers, I input the regular expressions into their Voice Dictionaries. (^ w + 3^) Begin exponent is recognized but not the end of exponent. This is not a fix but what I did to expose the problem. NOTE: The number of spaces specified in the replacement patterns for regular expressions should not be ignored. Some parts of the replacement string may be spoken so quickly, that you, the listener, may hear them as a nonintelligible blip. C Michael R. Cross -----Original Message----- From: nvda@nvda.groups.io [mailto:nvda@nvda.groups.io] On Behalf Of Damien Garwood Sent: Sunday, April 01, 2018 8:29 AM To: nvda@nvda.groups.io Subject: Re: [nvda] Maths? Hi, You can find extensive examples of problem 3 at: https://en.wikipedia.org/wiki/Scientific_notation Problem 2 can be seen at: https://en.wikipedia.org/wiki/Normalized_number Can't find one containing problem 1 off the top of my head, though like I said, problem 2 is becoming more common now. Cheers. Damien. -----Original Message----- From: Antony Stone Sent: Sunday, April 01, 2018 2:23 PM To: nvda@nvda.groups.io Subject: Re: [nvda] Maths? Please point us at an example of such an article so we can see for ourselves what it seems to contain. Otherwise we're just guessing, based on your guesses, about what might be going on. Antony. On Sunday 01 April 2018 at 14:31:53, Damien Garwood wrote: Hi, Not sure whether it's NVDA, or me. When I attempt to read articles that explain mathematical operations, explanations of scientific notation being a great example, I generally experience one or more of the following: 1. Graphics which, I assume would seem to have some sort of formula in it, but using strange symbols like \| and _ where I might expect to see operators like +, -, * or /. Though I'm not seeing these as often as I used to now. 2. Spaces, sometimes contained in lists, without anything in them, and with text following it which attempts to explain something which just doesn't seem to exist. This seems to have replaced the "graphical formula" strategy. 3. A formula written out in plaintext, but with some of the operators missing, making NVDA announce it as a big number (mainly happens when dealing with exponentiation), for example 102 instead of 10^2. Is there something I should be doing differently here? else...How wrong I am. Cheers. Damien. -- Most people have more than the average number of legs. list; me. Join nvda@nvda.groups.io to automatically receive all group messages. More	2021-01-20 17:55:23	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.1947038173675537, "perplexity": 6691.631881275576}, "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/1610703521139.30/warc/CC-MAIN-20210120151257-20210120181257-00497.warc.gz"}
https://newbedev.com/how-to-prevent-water-droplets-becoming-larger-on-the-tip-of-a-nozzle	# How to prevent water droplets becoming larger on the tip of a nozzle? The problem you have is surface tension. The drop will continue to grow until the weight of the drop is large enough that the "cost" of increasing the drop's surface by A (the contact area of the drop with the orifice) is less than the gain in energy from falling. This is described in some detail in this wikipedia entry. This describes how drop size in the pendant drop test can be used as a convenient measure of surface tension. This means that you need to do two things to make your drops smaller: 1. Make the total size of the nozzle smaller (not just the inner diameter, but the outer diameter) 2. Reduce the energy associated with this surface: this can be done either by changing the outer diameter of the nozzle, or by coating it with a hydrophobic coating, or by reducing the surface tension of the liquid. The math is a little bit complicated, since the area of a drop that is in the process of detaching actually increases before it decreases - but the extent to which it does so definitely depends on the outer diameter of the contact surface. See this image (from http://ej.iop.org/images/1367-2630/5/1/359/Full/img9.gif): As you can see, the area of the drop initially gets larger as it starts to fall - and it will do so when the gravitational energy associated with the lowering of the center of mass is greater than the energy involved in increasing the surface. Incidentally you might consider adding a controlled source of vibration to your setup: if you create a small mechanical "tap", this can shake the drop loose when it reaches the required size (basically by increasing the energy associated with separating the drop from the nozzle). And if you wonder "how small can you go", visit http://www.bnl.gov/sms/news/news.asp?a=617&t=pr where they show a zeptoliter pipette. That's $10^{-21}l$ if you were keeping score (yeah, I had never heard of it either...) UPDATE a bit of math. Clean water has a surface tension $\gamma$ of around 72 mN/m (at 25 °C) - it is a weak function of temperature. If your nozzle has (outer) diameter $d$, then the force with which it holds the water is $$F = \pi d \gamma$$ When this exceeds the weight of the drop, it will release. This gives us an approximate relationship between drop size and nozzle diameter: $$d = \frac{4 r^3 \rho g}{3 \gamma}$$ Plotting this for a range of values of drop diameter: You can see that releasing a small drop under gravity alone is quite challenging - which is why piezo actuators actually give the drop a "kick", literally shooting the water out of the hole in such a way that it will remain coherent, but not relying on gravity to pull the water out. It is clear that your nozzle with 0.2 mm diameter will never allow you to release a water drop under gravity - even if you coat the front surface with a hydrophobic coating, the area of the hole is too large. So you will have to use the piezo actuator - but note that when you set it up to produce a pulsed jet of water, this water will have quite a lot of kinetic energy and so will "shoot" out of the orifice. Basically, you need the momentum of the water to take the place of gravity - and from this I think you can estimate the velocity needed to escape. I have never tried this calculation before... but here goes with a rough estimation. If we assume that the escaping drop (velocity $v$) is increasing surface area at a rate $\Delta A = \pi d v \Delta t$ and force is given by $F = \pi d \gamma$ then the drop needs to have enough momentum to get away by say the radius of the nozzle before losing velocity. If we want the final drop to have a certain diameter $D$, volume $\frac16 \pi D^3$, then this represents a column of water in the nozzle of length $$l = \frac{\frac16 \pi D^3}{\frac14 \pi d^2}=\frac{2D^3}{3d^2}$$ This column of water will experience a force of surface tension as it tries to exit the nozzle, so the work done by the jet as it escapes would be $$W = F_{st} l = \pi d \gamma \frac{2D^3}{3d^2}$$ If we say that the kinetic energy of the water in the jet has to be sufficient to overcome this, we end up with an expression that relates the velocity needed to the size of the nozzle and the diameter of the target drop, with $$\frac12 m v^2 = W\\ \frac12 \frac16 \pi D^3 \rho v^2 = \pi \gamma \frac{2D^3}{3d}\\ v^2 = \frac{8\gamma}{\rho d}\\ v = \sqrt{\frac{8\gamma}{\rho d}}$$ Dimensionally that is correct... and it makes sense that the velocity has to be higher when the diameter of the nozzle is smaller, or the surface tension is higher. What is even more interesting is that the result does not depend on the size of the drop you want to create - although I suppose that we are assuming that the only liquid in the nozzle is the liquid we are trying to eject (there is no further momentum behind it pushing the liquid along). Anyway - for a nozzle diameter of 0.2 mm and a surface tension of 72 mN/m, the above gives a velocity of about 50 m/s. That seems high - but then the surface tension of the water is quite high. Still... it made me want to dig a little deeper and I found a fascinating paper on the subject of inkjet physics, that included this diagram: This states a jet velocity of 5 - 10 m/s for a much smaller nozzle: this suggests that some of my assumptions above were flawed (probably the fact that I was assuming all the energy had to be in the liquid being expelled, instead of assuming that the liquid behind it was available as a continuous force - like a plunger). However, a jet that is expelled in this way has a problem as shown in the next picture: And that leads me back to the original question - is there some other way to get the drop to gain the energy it needs without first becoming a long thin jet (which has the risk of not becoming a single coherent drop after release)? I wonder if it could be done electrostatically... If we could "increase gravity", then our original dropper would work just fine. Maybe an electric field could be used. If you put a charged plate with a hole close to the nozzle, then you would charge and attract the drop - and this would result in a force that can pull the water away from the nozzle. Again, the above paper gives some examples of how this is used. I don't have the time right now to do the calculation to see how feasible this is, but my hunch says that it ought to work for smaller drops. another update - this one inspired by an old drinking game. After the bottle is empty, bet your buddies you can still get 40 drops out of it. Then cut a small wedge of paper, and put it in the mouth of the bottle. Let the bottle sit on its side for a while, then tip it. The liquid that was stuck to the walls has accumulated in the bottom, and gets turned into tiny drops by the wedge. See this film clip. Picture of the setup (yes this is a tiny vodka bottle in keeping with the inspiration...) It looks like you get drops of about 1.4 mm or so with this method - sorry this was a crude late-night kitchen counter experiment. I suspect that using something other than a paper wedge may improve the experiment significantly - and if you were playing a nice loud noise on a loudspeaker, you could probably shake the drop off while it's still smaller. But the family is asleep and I'm not going to wake them up in the name of science. I think it would work though. Guitar amp, signal generator, boom. Adding a little surfactant (soap, alcohol) would probably help too. I am not sure the drop size is defined by the inner diameter of the tube, I would suspect that if you have a tube with a smaller outer diameter, you might get smaller droplets. You want to get the droplet to separate from the tip sooner than it otherwise would. I can think of some ways: • blast it off with a puff of air. • scrape it off with a hydrophobic knife-edge. • increase gravity (centrifuge). • eject no more liquid than you want, then transfer it to the desired surface (without forming a droplet). • create a large drop, and then dry it out, down to the desired size.	2022-06-25 20:39:56	{"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": 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": 0.5413401126861572, "perplexity": 329.512130720013}, "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/1656103036099.6/warc/CC-MAIN-20220625190306-20220625220306-00365.warc.gz"}
https://puzzling.stackexchange.com/questions/49324/travel-agency-a-zebra-puzzle-einsteins-riddle-variation	# Travel Agency - A Zebra Puzzle (Einstein's Riddle) variation You can play it online here: https://www.brainzilla.com/logic/zebra/travel-agency/ \| \| Woman #1 \| Woman #2 \| Woman #3 \| Woman #4 \| Woman #5 \| \|------------\|----------\|----------\|----------\|----------\|----------\| \| purse \| \| \| \| \| \| \| name \| \| \| \| \| \| \| age \| \| \| \| \| \| \| profession \| \| \| \| \| \| \| country \| \| \| \| \| \| \| duration \| \| \| \| \| \| Five women are side by side booking a travel in a travel agency. Each one is traveling to an different country. Follow the clues to discover where each one is going. • purse: blue, green, red, white, yellow • name: Ana, Glenda, Jessie, Lara, Rose • age: 24, 26, 28, 30, 32 • profession: biologist, hostess, judge, nurse, singer • country: China, Egypt, Italy, Mexico, Peru • duration: 5 days, 10 days, 15 days, 20 days, 25 days Clues: 1. The Singer is at the third position. 2. The woman traveling for 20 days is somewhere between the woman who is going to Peru and the owner of the Blue purse, in that order. 3. Ana is exactly to the left of the Biologist. 4. The 32 years old is going to see the Sahara. 5. The owner of the White purse is exactly to the right of the woman traveling to visit Machu Picchu. 6. Glenda is somewhere to the right of the woman who has the Green purse. 7. The person wearing the White purse is somewhere between the 30 years old woman and the owner of the Blue purse, in that order. 8. The 24 years old woman is going to visit an Aztec pyramid. 9. The woman wearing the White purse is somewhere to the left of the yougest woman. 10. The traveler going to Italy is exactly to the right of the woman traveling for 20 days. 11. The person who is going to travel for 25 days has the Red purse. 12. The Judge is in the first position. 13. The Nurse is exactly to the right of the woman who is going to travel for 20 days. 14. The Hostess is somewhere between Lara and the woman who has the Blue purse, in that order. 15. In the second position is the woman that is going to travel for 15 days. 16. Rose has the Green purse. 17. The woman who is traveling for less than a week is exactly to the left of the 32 years old woman. 18. The person traveling for 5 days is 28. 19. The Blue purse owner is somewhere between the 30 years old woman and the owner of the Yellow purse, in that order. Edit: fixed the 5th clue from "The owner of the White purse is going to visit Machu Picchu" to "The owner of the White purse is exactly to the right of the woman traveling to visit Machu Picchu". • I'd like to revert to the original puzzle, which was solvable and which I solved. – pb8330 Feb 22 '17 at 14:00 • @pb8330 this last had two solutions though – Beastly Gerbil Feb 22 '17 at 16:39 • But... why did Jamie change the fifth clue ? @Beast – pb8330 Feb 22 '17 at 16:50 From 1, 12 & 15 From 18 & 17, women of age 28 & 32 are in adjacent positions in that order. From 9 there are at-least two women on left of youngest woman and one of them is of age 30. Hence youngest woman is in position 3, women of age 28 & 32 are in position 4 & 5 respectively, woman at position 4 is traveling for 5 days and woman at position 2 has white purse and is of age 26. From 4 , 5 & 8 it also follows that woman at position 5 is visiting Sahara, woman at position 1 is visiting Machu Picchu & woman at position 3 is visiting Aztec Pyramid. From 10 & 2 woman traveling for 20 days is not at extreme left or extreme right. So woman traveling for 20 days is at position 3. From 10 & 13 woman at position 4 is a nurse and is going to Italy. From 14 it follows that hostess is not on extreme right and is therefore at position 2 & the name of the woman at position 1 is Lara. Which leaves position 5 to biologist. From 19 & 2 it follows that woman with blue purse is in position 4 and woman in position 5 has a yellow purse. From 16 it follows that the name of the woman at position 3 is Rose and she has a green purse. which leaves the color of the purse of woman at position 1 as red. From 11 it follows that woman 1 is traveling for 25 days and thus woman 5 is traveling for 10 days. Also from 2 it follows that Ana is at position 4, following which Glenda is in position 5, leaving position 2 to Jessie. Replacing the travel destinations by the country name and assigning the remaining country name as the destination of woman at position 2 the final result is. • I fixed the image for you – Beastly Gerbil Feb 25 '17 at 13:37 Using the offline version of this Java app, I filled in the following logic grid: And found that The first woman is going to China, the second to Peru, the third to Mexico, the fourth to Italy and the fifth to Egypt. Here is the website-confirmed answer: CSV: \| \| Woman #1 \| Woman #2 \| Woman #3 \| Woman #4 \| Woman #5 \| \|------------\|----------\|----------\|----------\|----------\|----------\| \| purse \| red \| white \| green \| blue \| yellow \| \| name \| Lara \| Jessie \| Rose \| Ana \| Glenda \| \| age \| 30 \| 26 \| 24 \| 28 \| 32 \| \| profession \| judge \| hostess \| singer \| nurse \|biologist \| \| country \| Peru \| China \| Mexico \| Italy \| Egypt \| \| duration \| 25 \| 15 \| 20 \| 5 \| 10 \| HOWEVER, @dcfyj points out in the comments that Machu Picchu isn't in China, it's in Peru. So therefore the person with the white purse should be going to Peru. However that doesn't work, so I think the website has made a mistake. I was timed and it took me 13 minutes. Adding Step by Step now STEP 1: Back to Basics First we should find the basic clues, the definites. The basic clues here are: 1, 12 and 15 GRID: \| \| Woman #1 \| Woman #2 \| Woman #3 \| Woman #4 \| Woman #5 \| \|------------\|----------\|----------\|----------\|----------\|----------\| \| purse \| \| \| \| \| \| \| name \| \| \| \| \| \| \| age \| \| \| \| \| \| \| profession \| judge \| \| singer \| \| \| \| country \| \| \| \| \| \| \| duration \| \| 15 \| \| \| \| STEP 2: Infermiera We know from the common statement in 10 and 13 that the Nurse is going to Italy. We also therefore know from statement 2 she is either person 4 or 5. STEP 3: Queen Ana IV From Step 2 we can work out that Ana is the fourth person. We know she is left of the biologist, meaning she can't be in fifth. We also know that the hostess is between someone from statement 14, so she can't be 5th. So now there are two possibilities: • judge - biologist - singer - hostess - nurse • judge - hostess - singer - nurse - biologist If we go with the first option, then that means that the fifth person has to have a blue purse from rule 14, but that contradicts rule 19 so the correct way is the second: GRID: \| \| Woman #1 \| Woman #2 \| Woman #3 \| Woman #4 \| Woman #5 \| \|------------\|----------\|----------\|----------\|----------\|----------\| \| purse \| \| \| \| \| \| \| name \| \| \| \| Ana \| \| \| age \| \| \| \| \| \| \| profession \| judge \| hostess \| singer \| nurse \|biologist \| \| country \| \| \| \| \| \| \| duration \| \| 15 \| \| \| \| STEP 4: Lara Judges a Nurse in the Blue Lara is person 1 from rule 14, and blue purse must be person 4 from 2 and 19. Rules 2 and 13 also mean that person 3 travels for 20 days. WILL HAVE TO FINISH TOMORROW, SLEEP CALLS US ALL, EVEN GERBILS... • We appear to have different conclusions ! I'll recheck mine later. – pb8330 Feb 21 '17 at 19:39 • @pb8330 Unfortunately for you I know mine is correct, as the website told me. Sorry :) – Beastly Gerbil Feb 21 '17 at 19:40 • Sorry about that mistake. Just fixed the 5th clue. Thanks for pointing it out @pb8330. – Jamie Feb 21 '17 at 20:05 • Hold on, why can't the original clues (i.e. white purse visiting Machu Picchu) work ? My answer seems to fit. (Am I being dippy ?) – pb8330 Feb 21 '17 at 20:48 • Will look into this tomorrow, @pb8330. Thanks again for your time pointing out mistakes. – Jamie Feb 21 '17 at 21:36	2019-12-08 04:07:03	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.18875093758106232, "perplexity": 1953.4429919955135}, "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-2019-51/segments/1575540504338.31/warc/CC-MAIN-20191208021121-20191208045121-00193.warc.gz"}
https://www.mathway.com/examples/precalculus/functions/determining-odd-and-even-functions?id=1063	# Precalculus Examples Determine if Odd, Even, or Neither Determine if the function is even. To check if a function is even, substitute in for and see if the resulting function is the same as the original. In other words, . Simplify each term. Apply the product rule to . Raise to the power of to get . Multiply by to get . The function is not even because the resulting function (after substituting in ) is not the same as the original. is not even is not even Determine if the function is odd. To check if the function is odd, substitute in for and check if the resulting function is the opposite of original function. In other words, determine if . Simplify each term. Apply the product rule to . Raise to the power of to get . Multiply by to get . The function is odd because is the opposite of . In other words, . The function is odd The function is odd Enter YOUR Problem Enter the email address associated with your Mathway account below and we'll send you a link to reset your password. Please enter an email address Please enter a valid email address The email address you entered was not found in our system The email address you entered is associated with a Facebook user We're sorry, we were unable to process your request at this time ### Mathway Premium Step-by-step work + explanations • Step-by-step work • Detailed explanations • No advertisements • Access anywhere Access the steps on both the Mathway website and mobile apps $--.--/month$--.--/year (--%) ### Mathway Premium Visa and MasterCard security codes are located on the back of card and are typically a separate group of 3 digits to the right of the signature strip. American Express security codes are 4 digits located on the front of the card and usually towards the right. This option is required to subscribe. Go Back Step-by-step upgrade complete! Mathway requires javascript and a modern browser.	2018-02-21 10:48:36	{"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": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.2257007658481598, "perplexity": 2065.217021071218}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-09/segments/1518891813608.70/warc/CC-MAIN-20180221103712-20180221123712-00504.warc.gz"}
https://www.shaalaa.com/question-bank-solutions/division-line-segment-construct-triangle-sides_7184	# Solution - Division of a Line Segment Account Register Share Books Shortlist ConceptDivision of a Line Segment #### Question Construct a triangle of sides 4 cm, 5cm and 6cm and then a triangle similar to it whose sides are 2/3 of the corresponding sides of the first triangle. Give the justification of the construction. #### Solution You need to to view the solution Is there an error in this question or solution? #### Reference Material Solution for concept: Division of a Line Segment. For the course CBSE S	2017-12-12 13:56:12	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5849015712738037, "perplexity": 1617.0340059938426}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-51/segments/1512948517181.32/warc/CC-MAIN-20171212134318-20171212154318-00746.warc.gz"}
https://brilliant.org/wiki/perfect-squares/	Perfect Squares, Cubes, and Powers Contents Cite as: Perfect Squares, Cubes, and Powers. Brilliant.org. Retrieved from https://brilliant.org/wiki/perfect-squares/ ×	2023-01-27 04:53:13	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 131, "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": 0.6583951711654663, "perplexity": 12150.758052147501}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764494936.89/warc/CC-MAIN-20230127033656-20230127063656-00358.warc.gz"}
https://physics.stackexchange.com/questions/linked/166095	17 questions linked to/from Do tachyons move faster than light? 303 views ### Tomorrowland movie and tachyons? [duplicate] Would be possible if tachyonic existed to send info to the past or see the future? since the tachyons according to special relativity could travel to the past? i have searched in wikipedia and the ... 596 views ### Can one classify partial differential equations according to the causality properties of their solutions (and if yes, then how)? Recently, I bumped into this interesting comment by Valter Moretti which made me wonder about the following, more general question (to which I suspect the answer is affirmative): Can we easily tell, ... 916 views ### What if a faster-than-light particle is found? What will be the consequence (severe ones) on laws of physics if a particle that travels faster than light is discovered? I am looking for a more general answer so that a high school student would be ... 865 views ### Why don't people use Hamilton's equations for a relativistic free charged particle? A charged relativistic free particle has the Hamiltonian in general: $$\mathcal{H} = \sqrt{{\bf p}^2c^2+m^2c^4}.$$ I read somewhere that says, it is possible to go further and say that the EoM are ... 1k views ### Would a tachyon be able to escape a black hole? Or at least escape from a portion of the hole inside the photon horizon? 797 views ### How do we know tachyons don't exist? As I mentioned yesterday, Hollywood screenwriter working on a TV pilot about physics trying to get the details right. What empirical evidence is there that tachyons do not exist? I understand that ... 359 views ### If Tachyons exists, can they escape event horizon? I was just wondering that, Tachyons travel faster than the speed of light. So in theory, if Tachyons exists can they escape event horizon? 577 views ### Does negative mass reverse the arrow of time? General relativity predicts that normal mass (positive mass) results in the curvature of spacetime which in return leads to gravitation. Since space and time are bonded together, any change on the ... 950 views ### Could dark matter/dark energy turn out to be tachyonic? Today, I was watching the Inexplicable Universe with Neil deGrasse Tyson. He spoke about Dark Matter, Dark Energy, and then he spoke about Tachyons. Naively, I thought what if the reason we can't ... 335 views ### Can strings go faster than light? I have been wondering if strings in string theory can go faster than light, as it is one dimensional when it does not vibrate? So do the same limits apply to it, I googled it but could not find a ... 185 views ### Is there any theory in physics that might support the existence of tachyons? According to Einstein, we all know that light is the fastest thing and it's impossible to beat it's speed. But isn't there a way to go around this? I read somewhere that tachyons gain speed per the ... 227 views ### Does the commutator of two quantum fields $\phi(x)$ and $\phi(y)$ vanish for $(x-y)^2<0$ and $m^2<0$? The commutator of two free scalar fields at two spacetime points $x$ and $y$ given by $$[\phi(x),\phi(y)]=\int\frac{d^3p}{(2\pi)^32E_{\textbf{p}}}[e^{-ip\cdot (x-y)}-e^{ip\cdot (x-y)}]$$ is Lorentz-... 214 views ### Can the momentum of a system be spacelike? For a particle, we always have $Eu=p$ where $u$ is the speed of the particle and correspondingly the momentum is always timelike. For a system of particles, if the potential energy is not considered ...	2020-02-26 01:04:22	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 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": 0.821029782295227, "perplexity": 576.5360790599057}, "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/1581875146176.73/warc/CC-MAIN-20200225233214-20200226023214-00551.warc.gz"}
http://math.eretrandre.org/tetrationforum/showthread.php?tid=828&page=2	• 0 Vote(s) - 0 Average • 1 • 2 • 3 • 4 • 5 Green Eggs and HAM: Tetration for ALL bases, real and complex, now possible? sheldonison Long Time Fellow Posts: 576 Threads: 22 Joined: Oct 2008 11/23/2013, 12:58 PM (11/22/2013, 11:30 PM)mike3 Wrote: .... where $b_n$ are properly-chosen "basis functions". I have thought about the Kneser-mapping solution (i.e. regular iteration warped with theta mapping) as a possible set of basis functions... but the problem is this only covers half of the plane (as given, the upper half-plane), and the Cauchy equations require both halves of the plane. .... Do you, perhaps, have any ideas as to how this could be done? The form of solution need not converge on the entire plane, only on and perhaps near the imaginary axis. You could try mapping the two unit infinite strip to the unit circle using, $f(z) = \frac{4}{\pi}\tan^{-1}(z)$ Here, f(z) maps the unit circle to an infinite strip, from -1 to +1. The left side of the unit circle is mapped to -1+iz, and the right side of the circle is mapped to 1+iz, where iz varies from $+/-\Im\infty$ I don't know if that would help or not, or whether the singularity at +/-I would be fairly mild, or not, given that the tetration solution also converges to a fixed point at $+/-\Im\infty$. - Sheldon mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 11/23/2013, 10:53 PM (This post was last modified: 11/23/2013, 11:14 PM by mike3.) (11/23/2013, 12:58 PM)sheldonison Wrote: (11/22/2013, 11:30 PM)mike3 Wrote: .... where $b_n$ are properly-chosen "basis functions". I have thought about the Kneser-mapping solution (i.e. regular iteration warped with theta mapping) as a possible set of basis functions... but the problem is this only covers half of the plane (as given, the upper half-plane), and the Cauchy equations require both halves of the plane. .... Do you, perhaps, have any ideas as to how this could be done? The form of solution need not converge on the entire plane, only on and perhaps near the imaginary axis. You could try mapping the two unit infinite strip to the unit circle using, $f(z) = \frac{4}{\pi}\tan^{-1}(z)$ Here, f(z) maps the unit circle to an infinite strip, from -1 to +1. The left side of the unit circle is mapped to -1+iz, and the right side of the circle is mapped to 1+iz, where iz varies from $+/-\Im\infty$ I don't know if that would help or not, or whether the singularity at +/-I would be fairly mild, or not, given that the tetration solution also converges to a fixed point at $+/-\Im\infty$. - Sheldon I'm not sure how that'd be useful, since what I need is a representation of the tetrational at the imaginary axis, that is, at $it$ where $t$ goes between $\pm \infty$, not $-1 + it$ or $1 + it$ (these are obtained by exp/log of the function's values at the imaginary axis within the integral equation under the integral sign, while on the other side of the equation (without the integral) is the function on the imaginary axis). So I don't see how this makes for a set of basis functions for the function on the imaginary axis. Edit: Although, I suppose that could work since you could just drop an "exp" or a "log" on the side of the integral equation that wants values at the imaginary axis. However, now that side of the integral equation can only be valid for input to the function in half the range of what the right side uses, namely the half of the unit circle we use to retrieve the values at the imaginary axis. So I'm not sure this'll work, as the integral equation, as specified, is for the whole function. mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 11/23/2013, 11:37 PM (This post was last modified: 11/23/2013, 11:49 PM by mike3.) However, the rescaled function $f(z) = \frac{\frac{4}{\pi} \tan^{-1}(z) + 1}{2} = \frac{2}{\pi} \tan^{-1}(z) + \frac{1}{2}$ might work. This maps the right half of the u.c. to $1 + ix$ and the left half to $ix$, thus allowing us to create a basis for the function at the imaginary axis. mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 11/24/2013, 01:14 AM (This post was last modified: 11/24/2013, 01:15 AM by mike3.) This shows $\mathrm{tet}(f(e^{it}))$ for your mapping (base-$e$ tetration), with $t$ going from $-\pi$ to $\pi$: There is a small corner in the real part (red curve). This may slow down convergence of a Fourier series. The rescaled mapping has a much nastier spike, however, and so doesn't seem very useful sheldonison Long Time Fellow Posts: 576 Threads: 22 Joined: Oct 2008 11/24/2013, 03:24 PM (11/24/2013, 01:14 AM)mike3 Wrote: .... There is a small corner in the real part (red curve). This may slow down convergence of a Fourier series. The rescaled mapping has a much nastier spike, however, and so doesn't seem very useful Yeah, the discontinuity in the derivative means that very large numbers of terms would be required in the fourier series. It doesn't seem like a terribly useful idea. With 8,192 terms, I was able to get accuracy to 1E-7 or so. Mapping an infinite strip to a unit circle seemed like a good idea, but in practice, not usable. - Sheldon mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 11/28/2013, 06:53 AM I've been experimenting with some other functions and mappings for this purpose. There do exist more complicated mappings which map the unit circle to the imaginary axis twice, with strong divergence to imaginary infinity at their singularities, so that the wrapped function will more quickly approach the fixed point and so be smoother. One such function is: $f(z) = \tanh\left(\frac{\pi}{4} (x - 1/x)\right)$ (unit circle -> imag axis). The inverse, giving the basis functions as powers of this, is $b(z) = f^{-1}(z) = \frac{1}{2} \left( \frac{4}{\pi} \tanh^{-1}(z) + \sqrt{\frac{16}{\pi^2} \tanh^{-1}(z)^2 + 4} \right)$. And the basis functions are $b_n(z) = b(z)^n$. The resulting mapping of the tetrational to base $e$, $\mathrm{tet}(f(e^{it}))$, looks like (red real, green imag): We can see this is much smoother and has no nasty corners. When we use this to get the Fourier series, however, we find that both positive and negative degree terms are required (i.e. the sum in terms of $b_n(z)$ I gave must go from $-\infty$ to $\infty$ instead of from $0$ to $\infty$), but it is significantly more accurate than the original mapping. In particular, with 203 terms (that's 101 positive and 101 negative degree terms plus the constant term), I get an error on the order of somewhat more than 10^-11 for the tetrational for imaginary-axis inputs close to 0. With 461 terms (230 positive, 230 negative), I get accuracy of around 10^-18 for close to 0, 10^-17 for inputs around $20i$. Clearly, this is very much improved, however I'm left wondering if it's possible a still better set of basis functions exists. While this is no use for getting an analytic approximation out of the HAM since it cannot be integrated exactly (the HAM was originally conceived, actually, as a method to get analytic, i.e. as formulas, approximations to the solutions of nonlinear problems, even though it can be operated as a numerical algorithm as well), we are interested in numerical approximations, and so this should still suffice for that. But it would be interesting if a simple enough basis function existed, that enabled the obtaining of an analytic approximation, since it might help in the finding of a series formula for tetration. sheldonison Long Time Fellow Posts: 576 Threads: 22 Joined: Oct 2008 06/24/2014, 10:57 PM (This post was last modified: 06/24/2014, 11:02 PM by sheldonison.) How about looking for a function that maps L to 1, and L* to -1? Call this first function "t", since its just a temporary step. $t(z) = \frac{(z-\Re(L))}{i\Im(L)}$ Now, if t(z)=+/-1, then 1-t(z)^2=0, so 1-t(L)^2=0 and 1-t(L)^2=0. So this might be your function, which maps L to 0 and L to 0, and does so in a nice analytic function. $f(z) = 1-t(z)^2 = (\frac{(z-\Re(L))}{i\Im(L)})^2-1$ $f(z) \approx 0.5592217758698x^2 - 0.3558121306015x - 1.056597524339$ Here is a graph of f(sexp(z)) at z=0, from -5i to +5i For all bases with a pair of complex fixed points, the function should exponentially decay to 0 as z goes to $\pm \Im \infty$. I would think f(z) would have an infinite convolution/Fourier representation. This does not work for real bases, b<=exp(1/e). - Sheldon mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 06/25/2014, 06:10 AM (This post was last modified: 06/25/2014, 06:10 AM by mike3.) (06/24/2014, 10:57 PM)sheldonison Wrote: How about looking for a function that maps L to 1, and L* to -1? Call this first function "t", since its just a temporary step. $t(z) = \frac{(z-\Re(L))}{i\Im(L)}$ Now, if t(z)=+/-1, then 1-t(z)^2=0, so 1-t(L)^2=0 and 1-t(L)^2=0. So this might be your function, which maps L to 0 and L to 0, and does so in a nice analytic function. $f(z) = 1-t(z)^2 = (\frac{(z-\Re(L))}{i\Im(L)})^2-1$ $f(z) \approx 0.5592217758698x^2 - 0.3558121306015x - 1.056597524339$ Here is a graph of f(sexp(z)) at z=0, from -5i to +5i For all bases with a pair of complex fixed points, the function should exponentially decay to 0 as z goes to $\pm \Im \infty$. I would think f(z) would have an infinite convolution/Fourier representation. This does not work for real bases, b<=exp(1/e). Hmm. It might work. However the function will not be complex-periodic, and therefore will not have a Fourier/exponential series expansion. It needs the periodicity in order to work. sheldonison Long Time Fellow Posts: 576 Threads: 22 Joined: Oct 2008 06/25/2014, 08:21 AM (This post was last modified: 06/25/2014, 08:25 AM by sheldonison.) (06/25/2014, 06:10 AM)mike3 Wrote: Hmm. It might work. However the function will not be complex-periodic, and therefore will not have a Fourier/exponential series expansion. It needs the periodicity in order to work.This one seems to converge pretty poorly; much slower than ideal. If you take $f(\text{sexp}(18\pi i))<\approx 10^{-32{$, and then wrap it around a unit circle, and then represent it with a Laurent series, then a 1000 term series from z^-500 to z^500 will be accurate to a little better than 10^-25. That's reasonably accurate, but it is also a very large number of Taylor/Laurent series terms. For comparison, the fast converging functions are like exp(-x^2), which is a perfect Gaussian. If you wrap it around a unit circle, so that f(-1)~-10^-32, then such a perfect Gaussian would require a little less than a 100 term Laurent series to be accurate to 32 decimal digits. So this function, needs 10x more terms and even then, it is less accurate -- not at all promising. Some background: I'm experimenting with Fourier/Laurent convolutions to calculate arbitrarily large Tetration Taylor series coefficients to arbitrary accuracy, and it works really well. But now I realize that it works best when the envelope function behaves like a Gaussian, which is often the case. But a Gaussian envelope converges to zero faster than this f(sexp(zI)) function. - Sheldon mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 06/29/2014, 12:03 AM (This post was last modified: 06/29/2014, 02:25 AM by mike3.) (06/25/2014, 08:21 AM)sheldonison Wrote: (06/25/2014, 06:10 AM)mike3 Wrote: Hmm. It might work. However the function will not be complex-periodic, and therefore will not have a Fourier/exponential series expansion. It needs the periodicity in order to work.This one seems to converge pretty poorly; much slower than ideal. If you take $f(\text{sexp}(18\pi i))<\approx 10^{-32{$, and then wrap it around a unit circle, and then represent it with a Laurent series, then a 1000 term series from z^-500 to z^500 will be accurate to a little better than 10^-25. That's reasonably accurate, but it is also a very large number of Taylor/Laurent series terms. For comparison, the fast converging functions are like exp(-x^2), which is a perfect Gaussian. If you wrap it around a unit circle, so that f(-1)~-10^-32, then such a perfect Gaussian would require a little less than a 100 term Laurent series to be accurate to 32 decimal digits. So this function, needs 10x more terms and even then, it is less accurate -- not at all promising. Some background: I'm experimenting with Fourier/Laurent convolutions to calculate arbitrarily large Tetration Taylor series coefficients to arbitrary accuracy, and it works really well. But now I realize that it works best when the envelope function behaves like a Gaussian, which is often the case. But a Gaussian envelope converges to zero faster than this f(sexp(zI)) function. Which makes one wonder: what about modifying the Cauchy integral equation so as to compute $e^{z^2} \mathrm{tet}(z)$ instead of $\mathrm{tet}(z)$ directly? Note that this function will still approach a limiting value at $\pm i\infty$, in particular, 0 at both ends! If we let $G(z) = e^{z^2} \mathrm{tet}(z)$, then $G(z+1) = e^{(z+1)^2} \mathrm{tet}(z+1) = e^{z^2 + 2z + 1} \exp(\mathrm{tet}(z)) = e^{z^2 + 2z + 1} \exp(e^{-z^2} G(z))$ $G(z-1) = e^{(z-1)^2} \mathrm{tet}(z-1) = e^{z^2 - 2z + 1} \log(\mathrm{tet}(z)) = e^{z^2 - 2z + 1} \log(e^{z^2} G(z))$. Then the Cauchy integral equation looks like $G_A(z) = \frac{1}{2\pi} \int_{-A}^{A} \frac{e^{-p^2 + 2ip + 1} \exp(e^{p^2} G(ip))}{1 + ip - z} dp - \frac{1}{2\pi} \int_{-A}^{A} \frac{e^{-p^2 - 2ip + 1} \log(e^{-p^2} G(ip))}{-1 + ip - z} dp$. where $G(z) = \lim_{A \rightarrow \infty} G_A(z)$. Note that unlike Kouznetsov's original Cauchy integral equation, there is no residual term $\mathcal{K}(z)$ because this function approaches 0 at $\pm i\infty$. This is just a theory, just a guess -- I have no idea if this will work or not. I am also concerned about the possibility of ambiguity because the fixed points are not specified in this equation and so you might wind up with the alternate fixed point solution instead. I suppose one could, perhaps, avoid that by adding a residual which is the Cauchy integral of the fixed point times the Gaussian, but that integral doesn't appear to be representable in terms of any standard functions. Or perhaps by trying to restrict it by some kind of restriction involving the branches of the $\log$ that appears there? Though the right initial guess should probably allow it to converge to the correct function, but finding a good initial guess for tetration seems tricky. But thought I'd toss it out there anyways... « Next Oldest \| Next Newest » Possibly Related Threads... Thread Author Replies Views Last Post Natural complex tetration program + video MorgothV8 1 179 04/27/2018, 07:54 PM Last Post: MorgothV8 complex base tetration program sheldonison 23 25,424 10/26/2016, 10:02 AM Last Post: Gottfried C++ program for generatin complex map in EPS format MorgothV8 0 1,529 09/17/2014, 04:14 PM Last Post: MorgothV8 "Kneser"/Riemann mapping method code for complex bases mike3 2 4,726 08/15/2011, 03:14 PM Last Post: Gottfried tiny q: superroots of real numbers x>e Gottfried 5 5,306 02/03/2009, 12:46 PM Last Post: bo198214 fractional iteration with complex bases/a bit of progress Gottfried 1 2,792 07/21/2008, 10:58 PM Last Post: Gottfried Tracing real values of x^(1/x) Gottfried 0 2,363 09/06/2007, 04:33 PM Last Post: Gottfried Users browsing this thread: 1 Guest(s)	2018-07-19 05:46:02	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 53, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8877679705619812, "perplexity": 893.2805829433648}, "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-2018-30/segments/1531676590559.95/warc/CC-MAIN-20180719051224-20180719071224-00356.warc.gz"}
https://www.physicsforums.com/threads/polytropic-processes.545848/	# Polytropic processes 1. Oct 31, 2011 ### jason.bourne polytropic processes are characterized by pvn = constant. are they valid for both reversible as well as irreversible processes? 2. Oct 31, 2011 ### Andrew Mason It depends. The relationship PV= constant (ie n = 1) for an ideal gas applies for all isothermal processes, reversible or non-reversible (ie. PV=nRT) However, the adiabatic condition for an ideal gas: $PV^\gamma = K$ applies only to a reversible adiabatic change. AM 3. Oct 31, 2011 ### jason.bourne thanks Andrew. suppose if there is a process in which heat flow is happening and the temperature of the system is not constant, lets assume that polytropic index is in the range 1 < n < γ. if the process was reversible, then it is reversible polytropic process and we can characterize it by pvn = constant. but what if there was friction or some other sort of irreversibility? how do we take that into account? how can we characterize such polytropic processes?	2017-09-21 12:29:15	{"extraction_info": {"found_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": 0.3936798870563507, "perplexity": 1473.2548186208453}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-39/segments/1505818687766.41/warc/CC-MAIN-20170921115822-20170921135822-00464.warc.gz"}
https://en.wikipedia.org/wiki/Soliton_(optics)	# Soliton (optics) In optics, the term soliton is used to refer to any optical field that does not change during propagation because of a delicate balance between nonlinear and linear effects in the medium.[1] There are two main kinds of solitons: • spatial solitons: the nonlinear effect can balance the diffraction. The electromagnetic field can change the refractive index of the medium while propagating, thus creating a structure similar to a graded-index fiber.[2] If the field is also a propagating mode of the guide it has created, then it will remain confined and it will propagate without changing its shape • temporal solitons: if the electromagnetic field is already spatially confined, it is possible to send pulses that will not change their shape because the nonlinear effects will balance the dispersion. Those solitons were discovered first and they are often simply referred as "solitons" in optics. ## Spatial solitons In order to understand how a spatial soliton can exist, we have to make some considerations about a simple convex lens. As shown in the picture on the right, an optical field approaches the lens and then it is focused. The effect of the lens is to introduce a non-uniform phase change that causes focusing. This phase change is a function of the space and can be represented with ${\displaystyle \varphi (x)}$, whose shape is approximately represented in the picture. The phase change can be expressed as the product of the phase constant and the width of the path the field has covered. We can write it as: ${\displaystyle \varphi (x)=k_{0}nL(x)}$ where ${\displaystyle L(x)}$ is the width of the lens, changing in each point with a shape that is the same of ${\displaystyle \varphi (x)}$ because ${\displaystyle k_{0}}$ and n are constants. In other words, in order to get a focusing effect we just have to introduce a phase change of such a shape, but we are not obliged to change the width. If we leave the width L fixed in each point, but we change the value of the refractive index ${\displaystyle n(x)}$ we will get exactly the same effect, but with a completely different approach. This has application in graded-index fibers: the change in the refractive index introduces a focusing effect that can balance the natural diffraction of the field. If the two effects balance each other perfectly, then we have a confined field propagating within the fiber. Spatial solitons are based on the same principle: the Kerr effect introduces a self-phase modulation that changes the refractive index according to the intensity: ${\displaystyle \varphi (x)=k_{0}n(x)L=k_{0}L[n+n_{2}I(x)]}$ if ${\displaystyle I(x)}$ has a shape similar to the one shown in the figure, then we have created the phase behavior we wanted and the field will show a self-focusing effect. In other words, the field creates a fiber-like guiding structure while propagating. If the field creates a fiber and it is the mode of such a fiber at the same time, it means that the focusing nonlinear and diffractive linear effects are perfectly balanced and the field will propagate forever without changing its shape (as long as the medium does not change and if we can neglect losses, obviously). In order to have a self-focusing effect, we must have a positive ${\displaystyle n_{2}}$, otherwise we will get the opposite effect and we will not notice any nonlinear behavior. The optical waveguide the soliton creates while propagating is not only a mathematical model, but it actually exists and can be used to guide other waves at different frequencies[citation needed]. This way it is possible to let light interact with light at different frequencies (this is impossible in linear media). ### Proof An electric field is propagating in a medium showing optical Kerr effect, so the refractive index is given by: ${\displaystyle n(I)=n+n_{2}I}$ We recall that the relationship between irradiance and electric field is (in the complex representation) ${\displaystyle I={\frac {\|E\|^{2}}{2\eta }}}$ where ${\displaystyle \eta =\eta _{0}/n}$ and ${\displaystyle \eta _{0}}$ is the impedance of free space, given by ${\displaystyle \eta _{0}={\sqrt {\frac {\mu _{0}}{\varepsilon _{0}}}}\approx 377\Omega .}$ The field is propagating in the ${\displaystyle z}$ direction with a phase constant ${\displaystyle k_{0}n}$. About now, we will ignore any dependence on the y axis, assuming that it is infinite in that direction. Then the field can be expressed as: ${\displaystyle E(x,z,t)=A_{m}a(x,z)e^{i(k_{0}nz-\omega t)}}$ where ${\displaystyle A_{m}}$ is the maximum amplitude of the field and ${\displaystyle a(x,z)}$ is a dimensionless normalized function (so that its maximum value is 1) that represents the shape of the electric field among the x axis. In general it depends on z because fields change their shape while propagating. Now we have to solve the Helmholtz equation: ${\displaystyle \nabla ^{2}E+k_{0}^{2}n^{2}(I)E=0}$ where it was pointed out clearly that the refractive index (thus the phase constant) depends on intensity. If we replace the expression of the electric field in the equation, assuming that the envelope ${\displaystyle a(x,z)}$ changes slowly while propagating, i.e. ${\displaystyle \left\|{\frac {\partial ^{2}a(x,z)}{\partial z^{2}}}\right\|\ll \left\|k_{0}{\frac {\partial a(x,z)}{\partial z}}\right\|}$ the equation becomes: ${\displaystyle {\frac {\partial ^{2}a}{\partial x^{2}}}+i2k_{0}n{\frac {\partial a}{\partial z}}+k_{0}^{2}[n^{2}(I)-n^{2}]a=0.}$ Let us introduce an approximation that is valid because the nonlinear effects are always much smaller than the linear ones: ${\displaystyle [n^{2}(I)-n^{2}]=[n(I)-n][n(I)+n]=n_{2}I(2n+n_{2}I)\approx 2nn_{2}I}$ now we express the intensity in terms of the electric field: ${\displaystyle [n^{2}(I)-n^{2}]\approx 2nn_{2}{\frac {\|A_{m}\|^{2}\|a(x,z)\|^{2}}{2\eta _{0}/n}}=n^{2}n_{2}{\frac {\|A_{m}\|^{2}\|a(x,z)\|^{2}}{\eta _{0}}}}$ the equation becomes: ${\displaystyle {\frac {1}{2k_{0}n}}{\frac {\partial ^{2}a}{\partial x^{2}}}+i{\frac {\partial a}{\partial z}}+{\frac {k_{0}nn_{2}\|A_{m}\|^{2}}{2\eta _{0}}}\|a\|^{2}a=0.}$ We will now assume ${\displaystyle n_{2}>0}$ so that the nonlinear effect will cause self focusing. In order to make this evident, we will write in the equation ${\displaystyle n_{2}=\|n_{2}\|}$ Let us now define some parameters and replace them in the equation: • ${\displaystyle \xi ={\frac {x}{X_{0}}}}$, so we can express the dependence on the x axis with a dimensionless parameter; ${\displaystyle X_{0}}$ is a length, whose physical meaning will be clearer later. • ${\displaystyle L_{d}=X_{0}^{2}k_{0}n}$, after the electric field has propagated across z for this length, the linear effects of diffraction can not be neglected anymore. • ${\displaystyle \zeta ={\frac {z}{L_{d}}}}$, for studying the z-dependence with a dimensionless variable. • ${\displaystyle L_{n\ell }={\frac {2\eta _{0}}{k_{0}n\|n_{2}\|\cdot \|A_{m}\|^{2}}}}$, after the electric field has propagated across z for this length, the nonlinear effects can not be neglected anymore. This parameter depends upon the intensity of the electric field, that's typical for nonlinear parameters. • ${\displaystyle N^{2}={\frac {L_{d}}{L_{n\ell }}}}$ The equation becomes: ${\displaystyle {\frac {1}{2}}{\frac {\partial ^{2}a}{\partial \xi ^{2}}}+i{\frac {\partial a}{\partial \zeta }}+N^{2}\|a\|^{2}a=0}$ this is a common equation known as nonlinear Schrödinger equation. From this form, we can understand the physical meaning of the parameter N: • if ${\displaystyle N\ll 1}$, then we can neglect the nonlinear part of the equation. It means ${\displaystyle L_{d}\ll L_{n\ell }}$, then the field will be affected by the linear effect (diffraction) much earlier than the nonlinear effect, it will just diffract without any nonlinear behavior. • if ${\displaystyle N\gg 1}$, then the nonlinear effect will be more evident than diffraction and, because of self phase modulation, the field will tend to focus. • if ${\displaystyle N\approx 1}$, then the two effects balance each other and we have to solve the equation. For ${\displaystyle N=1}$ the solution of the equation is simple and it is the fundamental soliton: ${\displaystyle a(\xi ,\zeta )=\operatorname {sech} (\xi )e^{i\zeta /2}}$ where sech is the hyperbolic secant. It still depends on z, but only in phase, so the shape of the field will not change during propagation. For ${\displaystyle N=2}$ it is still possible to express the solution in a closed form, but it has a more complicated form:[3] ${\displaystyle a(\xi ,\zeta )={\frac {4[\cosh(3\xi )+3e^{4i\zeta }\cosh(\xi )]e^{i\zeta /2}}{\cosh(4\xi )+4\cosh(2\xi )+3\cos(4\zeta )}}.}$ It does change its shape during propagation, but it is a periodic function of z with period ${\displaystyle \zeta =\pi /2}$. Soliton's shape while propagating with N = 1, it does not change its shape Soliton's shape while propagating with N = 2, it changes its shape periodically For soliton solutions, N must be an integer and it is said to be the order or the soliton. For ${\displaystyle N=3}$ an exact closed form solution also exists;[4] it has an even more complicated form, but the same periodicity occurs. In fact, all solitons with ${\displaystyle N\geq 2}$ have the period ${\displaystyle \zeta =\pi /2}$.[5] Their shape can easily be expressed only immediately after generation: ${\displaystyle a(\xi ,\zeta =0)=N\operatorname {sech} (\xi )}$ on the right there is the plot of the second order soliton: at the beginning it has a shape of a sech, then the maximum amplitude increases and then comes back to the sech shape. Since high intensity is necessary to generate solitons, if the field increases its intensity even further the medium could be damaged. The condition to be solved if we want to generate a fundamental soliton is obtained expressing N in terms of all the known parameters and then putting ${\displaystyle N=1}$: ${\displaystyle 1=N={\frac {L_{d}}{L_{n\ell }}}={\frac {X_{0}^{2}k_{0}^{2}n^{2}\|n_{2}\|\|A_{m}\|^{2}}{2\eta _{0}}}}$ that, in terms of maximum irradiance value becomes: ${\displaystyle I_{\max }={\frac {\|A_{m}\|^{2}}{2\eta _{0}/n}}={\frac {1}{X_{0}^{2}k_{0}^{2}n\|n_{2}\|}}.}$ In most of the cases, the two variables that can be changed are the maximum intensity ${\displaystyle I_{\max }}$ and the pulse width ${\displaystyle X_{0}}$. propagation of various higher-order optical solitons (image series: low power (no soliton), then n1-n7) Curiously, higher-order solitons can attain complicated shapes before returning exactly to their initial shape at the end of the soliton period. In the picture of various solitons, the spectrum (left) and time domain (right) are shown at varying distances of propagation (vertical axis) in an idealized nonlinear medium. This shows how a laser pulse might behave as it travels in a medium with the properties necessary to support fundamental solitons. In practice, in order to reach the very high peak intensity needed to achieve nonlinear effects, laser pulses may be coupled into optical fibers such as photonic-crystal fiber with highly confined propagating modes. Those fibers have more complicated dispersion and other characteristics which depart from the analytical soliton parameters. ### Generation of spatial solitons The first experiment on spatial optical solitons was reported in 1974 by Ashkin and Bjorkholm[6] in a cell filled with sodium vapor. The field was then revisited in experiments at Limoges University[7] in liquid carbon disulphide and expanded in the early '90s with the first observation of solitons in photorefractive crystals,[8][9] glass, semiconductors[10] and polymers. During the last decades numerous findings have been reported in various materials, for solitons of different dimensionality, shape, spiralling, colliding, fusing, splitting, in homogeneous media, periodic systems, and waveguides.[11] Spatials solitons are also referred to as self-trapped optical beams and their formation is normally also accompanied by a self-written waveguide. In nematic liquid crystals,[12] spatial solitons are also referred to as nematicons. ### Transverse-mode-locking solitons Localized excitations in lasers may appear due to synchronization of transverse modes. Confocal ${\displaystyle 2F}$ laser cavity with nonlinear gain and absorber slices in Fourier-conjugated planes In confocal ${\displaystyle 2F}$ laser cavity the degenerate transverse modes with single longitudinal mode at wavelength ${\displaystyle \lambda }$ mixed in nonlinear gain disc ${\displaystyle G}$ (located at ${\displaystyle z=0}$) and saturable absorber disc ${\displaystyle \alpha }$ (located at ${\displaystyle z=2F}$) of diameter ${\displaystyle D}$ are capable to produce spatial solitons of hyperbolic ${\displaystyle sech}$ form:[13] ${\displaystyle E(x,z=0)\sim {\rm {sech}}\left(\!{\frac {\pi xD}{2\lambda F}}{\sqrt {\frac {1-\alpha G}{G}}}\,\right)}$ ${\displaystyle E(x,z=2F)\sim {\rm {sech}}\left(\!{\frac {2\pi x}{D}}{\sqrt {\frac {G}{1-\alpha G}}}\,\right)}$ in Fourier-conjugated planes ${\displaystyle z=0}$ and ${\displaystyle z=2F}$ .[14] ## Temporal solitons The main problem that limits transmission bit rate in optical fibres is group velocity dispersion. It is because generated impulses have a non-zero bandwidth and the medium they are propagating through has a refractive index that depends on frequency (or wavelength). This effect is represented by the group delay dispersion parameter D; using it, it is possible to calculate exactly how much the pulse will widen: ${\displaystyle \Delta \tau \approx DL\,\Delta \lambda }$ where L is the length of the fibre and ${\displaystyle \Delta \lambda }$ is the bandwidth in terms of wavelength. The approach in modern communication systems is to balance such a dispersion with other fibers having D with different signs in different parts of the fibre: this way the pulses keep on broadening and shrinking while propagating. With temporal solitons it is possible to remove such a problem completely. linear and nonlinear effects on Gaussian pulses Consider the picture on the right. On the left there is a standard Gaussian pulse, that's the envelope of the field oscillating at a defined frequency. We assume that the frequency remains perfectly constant during the pulse. Now we let this pulse propagate through a fibre with ${\displaystyle D>0}$, it will be affected by group velocity dispersion. For this sign of D, the dispersion is anomalous, so that the higher frequency components will propagate a little bit faster than the lower frequencies, thus arriving before at the end of the fiber. The overall signal we get is a wider chirped pulse, shown in the upper right of the picture. effect of self-phase modulation on frequency Now let us assume we have a medium that shows only nonlinear Kerr effect but its refractive index does not depend on frequency: such a medium does not exist, but it's worth considering it to understand the different effects. The phase of the field is given by: ${\displaystyle \varphi (t)=\omega _{0}t-kz=\omega _{0}t-k_{0}z[n+n_{2}I(t)]}$ the frequency (according to its definition) is given by: ${\displaystyle \omega (t)={\frac {\partial \varphi (t)}{\partial t}}=\omega _{0}-k_{0}zn_{2}{\frac {\partial I(t)}{\partial t}}}$ this situation is represented in the picture on the left. At the beginning of the pulse the frequency is lower, at the end it's higher. After the propagation through our ideal medium, we will get a chirped pulse with no broadening because we have neglected dispersion. Coming back to the first picture, we see that the two effects introduce a change in frequency in two different opposite directions. It is possible to make a pulse so that the two effects will balance each other. Considering higher frequencies, linear dispersion will tend to let them propagate faster, while nonlinear Kerr effect will slow them down. The overall effect will be that the pulse does not change while propagating: such pulses are called temporal solitons. ### History of temporal solitons In 1973, Akira Hasegawa and Fred Tappert of AT&T Bell Labs were the first to suggest that solitons could exist in optical fibres, due to a balance between self-phase modulation and anomalous dispersion.[15] [16] Also in 1973 Robin Bullough made the first mathematical report of the existence of optical solitons. He also proposed the idea of a soliton-based transmission system to increase performance of optical telecommunications. Solitons in a fibre optic system are described by the Manakov equations. In 1987, P. Emplit, J.P. Hamaide, F. Reynaud, C. Froehly and A. Barthelemy, from the Universities of Brussels and Limoges, made the first experimental observation of the propagation of a dark soliton, in an optical fiber. In 1988, Linn Mollenauer and his team transmitted soliton pulses over 4,000 kilometres using a phenomenon called the Raman effect, named for the Indian scientist Sir C. V. Raman who first described it in the 1920s, to provide optical gain in the fibre. In 1991, a Bell Labs research team transmitted solitons error-free at 2.5 gigabits over more than 14,000 kilometres, using erbium optical fibre amplifiers (spliced-in segments of optical fibre containing the rare earth element erbium). Pump lasers, coupled to the optical amplifiers, activate the erbium, which energizes the light pulses[citation needed]. In 1998, Thierry Georges and his team at France Télécom R&D Centre, combining optical solitons of different wavelengths (wavelength division multiplexing), demonstrated a data transmission of 1 terabit per second (1,000,000,000,000 units of information per second)[citation needed]. In 2020, Optics Communications reported a Japanese team from MEXT, optical circuit switching with bandwidth of up to 90 Tbps (terabits per second), Optics Communications, Volume 466, 1 July 2020, 125677. ### Proof for temporal solitons An electric field is propagating in a medium showing optical Kerr effect through a guiding structure (such as an optical fibre) that limits the power on the xy plane. If the field is propagating towards z with a phase constant ${\displaystyle \beta _{0}}$, then it can be expressed in the following form: ${\displaystyle E(\mathbf {r} ,t)=A_{m}a(t,z)f(x,y)e^{i(\beta _{0}z-\omega _{0}t)}}$ where ${\displaystyle A_{m}}$ is the maximum amplitude of the field, ${\displaystyle a(t,z)}$ is the envelope that shapes the impulse in the time domain; in general it depends on z because the impulse can change its shape while propagating; ${\displaystyle f(x,y)}$ represents the shape of the field on the xy plane, and it does not change during propagation because we have assumed the field is guided. Both a and f are normalized dimensionless functions whose maximum value is 1, so that ${\displaystyle A_{m}}$ really represents the field amplitude. Since in the medium there is a dispersion we can not neglect, the relationship between the electric field and its polarization is given by a convolution integral. Anyway, using a representation in the Fourier domain, we can replace the convolution with a simple product, thus using standard relationships that are valid in simpler media. We Fourier-transform the electric field using the following definition: ${\displaystyle {\tilde {E}}(\mathbf {r} ,\omega -\omega _{0})=\int \limits _{-\infty }^{\infty }E(\mathbf {r} ,t)e^{-i(\omega -\omega _{0})t}\,dt}$ Using this definition, a derivative in the time domain corresponds to a product in the Fourier domain: ${\displaystyle {\frac {\partial }{\partial t}}E\Longleftrightarrow i(\omega -\omega _{0}){\tilde {E}}}$ the complete expression of the field in the frequency domain is: ${\displaystyle {\tilde {E}}(\mathbf {r} ,\omega -\omega _{0})=A_{m}{\tilde {a}}(\omega ,z)f(x,y)e^{i\beta _{0}z}}$ Now we can solve Helmholtz equation in the frequency domain: ${\displaystyle \nabla ^{2}{\tilde {E}}+n^{2}(\omega )k_{0}^{2}{\tilde {E}}=0}$ we decide to express the phase constant with the following notation: {\displaystyle {\begin{aligned}n(\omega )k_{0}=\beta (\omega )&=\overbrace {\beta _{0}} ^{\text{linear non-dispersive}}+\overbrace {\beta _{\ell }(\omega )} ^{\text{linear dispersive}}+\overbrace {\beta _{n\ell }} ^{\text{non-linear}}\\[8pt]&=\beta _{0}+\Delta \beta (\omega )\end{aligned}}} where we assume that ${\displaystyle \Delta \beta }$ (the sum of the linear dispersive component and the non-linear part) is a small perturbation, i.e. ${\displaystyle \|\beta _{0}\|\gg \|\Delta \beta (\omega )\|}$. The phase constant can have any complicated behaviour, but we can represent it with a Taylor series centred on ${\displaystyle \omega _{0}}$: ${\displaystyle \beta (\omega )\approx \beta _{0}+(\omega -\omega _{0})\beta _{1}+{\frac {(\omega -\omega _{0})^{2}}{2}}\beta _{2}+\beta _{n\ell }}$ where, as known: ${\displaystyle \beta _{u}=\left.{\frac {d^{u}\beta (\omega )}{d\omega ^{u}}}\right\|_{\omega =\omega _{0}}}$ we put the expression of the electric field in the equation and make some calculations. If we assume the slowly varying envelope approximation: ${\displaystyle \left\|{\frac {\partial ^{2}{\tilde {a}}}{\partial z^{2}}}\right\|\ll \left\|\beta _{0}{\frac {\partial {\tilde {a}}}{\partial z}}\right\|}$ we get: ${\displaystyle 2i\beta _{0}{\frac {\partial {\tilde {a}}}{\partial z}}+[\beta ^{2}(\omega )-\beta _{0}^{2}]{\tilde {a}}=0}$ we are ignoring the behavior in the xy plane, because it is already known and given by ${\displaystyle f(x,y)}$. We make a small approximation, as we did for the spatial soliton: {\displaystyle {\begin{aligned}\beta ^{2}(\omega )-\beta _{0}^{2}&=[\beta (\omega )-\beta _{0}][\beta (\omega )+\beta _{0}]\\[6pt]&=[\beta _{0}+\Delta \beta (\omega )-\beta _{0}][2\beta _{0}+\Delta \beta (\omega )]\approx 2\beta _{0}\,\Delta \beta (\omega )\end{aligned}}} replacing this in the equation we get simply: ${\displaystyle i{\frac {\partial {\tilde {a}}}{\partial z}}+\Delta \beta (\omega ){\tilde {a}}=0}$. Now we want to come back in the time domain. Expressing the products by derivatives we get the duality: ${\displaystyle \Delta \beta (\omega )\Longleftrightarrow i\beta _{1}{\frac {\partial }{\partial t}}-{\frac {\beta _{2}}{2}}{\frac {\partial ^{2}}{\partial t^{2}}}+\beta _{n\ell }}$ we can write the non-linear component in terms of the irradiance or amplitude of the field: ${\displaystyle \beta _{n\ell }=k_{0}n_{2}I=k_{0}n_{2}{\frac {\|E\|^{2}}{2\eta _{0}/n}}=k_{0}n_{2}n{\frac {\|A_{m}\|^{2}}{2\eta _{0}}}\|a\|^{2}}$ for duality with the spatial soliton, we define: ${\displaystyle L_{n\ell }={\frac {2\eta _{0}}{k_{0}nn_{2}\|A_{m}\|^{2}}}}$ and this symbol has the same meaning of the previous case, even if the context is different. The equation becomes: ${\displaystyle i{\frac {\partial a}{\partial z}}+i\beta _{1}{\frac {\partial a}{\partial t}}-{\frac {\beta _{2}}{2}}{\frac {\partial ^{2}a}{\partial t^{2}}}+{\frac {1}{L_{n\ell }}}\|a\|^{2}a=0}$ We know that the impulse is propagating along the z axis with a group velocity given by ${\displaystyle v_{g}=1/\beta _{1}}$, so we are not interested in it because we just want to know how the pulse changes its shape while propagating. We decide to study the impulse shape, i.e. the envelope function a(·) using a reference that is moving with the field at the same velocity. Thus we make the substitution ${\displaystyle T=t-\beta _{1}z}$ and the equation becomes: ${\displaystyle i{\frac {\partial a}{\partial z}}-{\frac {\beta _{2}}{2}}{\frac {\partial ^{2}a}{\partial T^{2}}}+{\frac {1}{L_{n\ell }}}\|a\|^{2}a=0}$ We now further assume that the medium where the field is propagating in shows anomalous dispersion, i.e. ${\displaystyle \beta _{2}<0}$ or in terms of the group delay dispersion parameter ${\displaystyle D={\frac {-2\pi c}{\lambda ^{2}}}\beta _{2}>0}$. We make this more evident replacing in the equation ${\displaystyle \beta _{2}=-\|\beta _{2}\|}$. Let us define now the following parameters (the duality with the previous case is evident): ${\displaystyle L_{d}={\frac {T_{0}^{2}}{\|\beta _{2}\|}};\qquad \tau ={\frac {T}{T_{0}}};\qquad \zeta ={\frac {z}{L_{d}}};\qquad N^{2}={\frac {L_{d}}{L_{n\ell }}}}$ replacing those in the equation we get: ${\displaystyle {\frac {1}{2}}{\frac {\partial ^{2}a}{\partial \tau ^{2}}}+i{\frac {\partial a}{\partial \zeta }}+N^{2}\|a\|^{2}a=0}$ that is exactly the same equation we have obtained in the previous case. The first order soliton is given by: ${\displaystyle a(\tau ,\zeta )=\operatorname {sech} (\tau )e^{i\zeta /2}}$ the same considerations we have made are valid in this case. The condition N = 1 becomes a condition on the amplitude of the electric field: ${\displaystyle \|A_{m}\|^{2}={\frac {2\eta _{0}\|\beta _{2}\|}{T_{0}^{2}n_{2}k_{0}n}}}$ ${\displaystyle I_{\max }={\frac {\|A_{m}\|^{2}}{2\eta _{0}/n}}={\frac {\|\beta _{2}\|}{T_{0}^{2}n_{2}k_{0}}}}$ or we can express it in terms of power if we introduce an effective area ${\displaystyle A_{\text{eff}}}$ defined so that ${\displaystyle P=IA_{\text{eff}}}$: ${\displaystyle P={\frac {\|\beta _{2}\|A_{\text{eff}}}{T_{0}^{2}n_{2}k_{0}}}}$ ## Stability of solitons We have described what optical solitons are and, using mathematics, we have seen that, if we want to create them, we have to create a field with a particular shape (just sech for the first order) with a particular power related to the duration of the impulse. But what if we are a bit wrong in creating such impulses? Adding small perturbations to the equations and solving them numerically, it is possible to show that mono-dimensional solitons are stable. They are often referred as (1 + 1) D solitons, meaning that they are limited in one dimension (x or t, as we have seen) and propagate in another one (z). If we create such a soliton using slightly wrong power or shape, then it will adjust itself until it reaches the standard sech shape with the right power. Unfortunately this is achieved at the expense of some power loss, that can cause problems because it can generate another non-soliton field propagating together with the field we want. Mono-dimensional solitons are very stable: for example, if ${\displaystyle 0.5 we will generate a first order soliton anyway; if N is greater we'll generate a higher order soliton, but the focusing it does while propagating may cause high power peaks damaging the media. The only way to create a (1 + 1) D spatial soliton is to limit the field on the y axis using a dielectric slab, then limiting the field on x using the soliton. On the other hand, (2 + 1) D spatial solitons are unstable, so any small perturbation (due to noise, for example) can cause the soliton to diffract as a field in a linear medium or to collapse, thus damaging the material. It is possible to create stable (2 + 1) D spatial solitons using saturating nonlinear media, where the Kerr relationship ${\displaystyle n(I)=n+n_{2}I}$ is valid until it reaches a maximum value. Working close to this saturation level makes it possible to create a stable soliton in a three-dimensional space. If we consider the propagation of shorter (temporal) light pulses or over a longer distance, we need to consider higher-order corrections and therefore the pulse carrier envelope is governed by the higher-order nonlinear Schrödinger equation (HONSE) for which there are some specialized (analytical) soliton solutions.[17] ## Effect of power losses As we have seen, in order to create a soliton it is necessary to have the right power when it is generated. If there are no losses in the medium, then we know that the soliton will keep on propagating forever without changing shape (1st order) or changing its shape periodically (higher orders). Unfortunately any medium introduces losses, so the actual behaviour of power will be in the form: ${\displaystyle P(z)=P_{0}e^{-\alpha z}}$ this is a serious problem for temporal solitons propagating in fibers for several kilometers. Consider what happens for the temporal soliton, generalization to the spatial ones is immediate. We have proved that the relationship between power ${\displaystyle P_{0}}$ and impulse length ${\displaystyle T_{0}}$ is: ${\displaystyle P={\frac {\|\beta _{2}\|A_{\text{eff}}}{T_{0}^{2}n_{2}k_{0}}}}$ if the power changes, the only thing that can change in the second part of the relationship is ${\displaystyle T_{0}}$. if we add losses to the power and solve the relationship in terms of ${\displaystyle T_{0}}$ we get: ${\displaystyle T(z)=T_{0}e^{(\alpha /2)z}}$ the width of the impulse grows exponentially to balance the losses! this relationship is true as long as the soliton exists, i.e. until this perturbation is small, so it must be ${\displaystyle \alpha z\ll 1}$ otherwise we can not use the equations for solitons and we have to study standard linear dispersion. If we want to create a transmission system using optical fibres and solitons, we have to add optical amplifiers in order to limit the loss of power. ### Generation of soliton pulse Experiments have been carried out to analyse the effect of high frequency (20 MHz-1 GHz) external magnetic field induced nonlinear Kerr effect on Single mode optical fibre of considerable length (50–100 m) to compensate group velocity dispersion (GVD) and subsequent evolution of soliton pulse ( peak energy, narrow, secant hyperbolic pulse).[18] Generation of soliton pulse in fibre is an obvious conclusion as self phase modulation due to high energy of pulse offset GVD, whereas the evolution length is 2000 km. (the laser wavelength chosen greater than 1.3 micrometers). Moreover, peak soliton pulse is of period 1–3 ps so that it is safely accommodated in the optical bandwidth. Once soliton pulse is generated it is least dispersed over thousands of kilometres length of fibre limiting the number of repeater stations. ## Dark solitons In the analysis of both types of solitons we have assumed particular conditions about the medium: • in spatial solitons, ${\displaystyle n_{2}>0}$, that means the self-phase modulation causes self-focusing • in temporal solitons, ${\displaystyle \beta _{2}<0}$ or ${\displaystyle D>0}$, anomalous dispersion Is it possible to obtain solitons if those conditions are not verified? if we assume ${\displaystyle n_{2}<0}$ or ${\displaystyle \beta _{2}>0}$, we get the following differential equation (it has the same form in both cases, we will use only the notation of the temporal soliton): ${\displaystyle {\frac {-1}{2}}{\frac {\partial ^{2}a}{\partial \tau ^{2}}}+i{\frac {\partial a}{\partial \zeta }}+N^{2}\|a\|^{2}a=0.}$ This equation has soliton-like solutions. For the first order (N = 1): ${\displaystyle a(\tau ,\zeta )=\tanh(\tau )e^{i\zeta }.\ }$ power of a dark soliton The plot of ${\displaystyle \|a(\tau ,\zeta )\|^{2}}$ is shown in the picture on the right. For higher order solitons (${\displaystyle N>1}$) we can use the following closed form expression: ${\displaystyle a(\tau ,\zeta =0)=N\tanh(\tau ).\ }$ It is a soliton, in the sense that it propagates without changing its shape, but it is not made by a normal pulse; rather, it is a lack of energy in a continuous time beam. The intensity is constant, but for a short time during which it jumps to zero and back again, thus generating a "dark pulse"'. Those solitons can actually be generated introducing short dark pulses in much longer standard pulses. Dark solitons are more difficult to handle than standard solitons, but they have shown to be more stable and robust to losses. ## References 1. ^ Taylo, James Roy (1992). Optical solitons : theory and experiment. Cambridge: Cambridge University Press. ISBN 9780521405485. OCLC 23975147. CS1 maint: discouraged parameter (link) 2. ^ Rashidian Vaziri, M R (2013). "Describing the propagation of intense laser pulses in nonlinear Kerr media using the ducting model". Laser Physics. 23 (10): 105401. Bibcode:2013LaPhy..23j5401R. doi:10.1088/1054-660X/23/10/105401. 3. ^ Chen, Chin-Lin (2006-09-11). Foundations for Guided-Wave Optics. John Wiley & Sons. ISBN 9780470042212. 4. ^ Chen, Chin-Lin (2006-09-11). Foundations for Guided-Wave Optics. John Wiley & Sons. ISBN 9780470042212. 5. ^ Agrawal, Govind P. (2007). Nonlinear Fiber Optics. Academic Press. ISBN 9780123695161. 6. ^ J.E. Bjorkholm; A. Ashkin (1974). "cw Self-Focusing and Self-Trapping of Light in Sodium Vapor". Phys. Rev. Lett. 32 (4): 129. Bibcode:1974PhRvL..32..129B. doi:10.1103/PhysRevLett.32.129. 7. ^ A. Barthelemy, S. Maneuf & C. Froehly (1985). "Propagation soliton et auto-confinement de faisceaux laser par non linearité optique de kerr". Opt. Commun. 55 (3): 201. Bibcode:1985OptCo..55..201B. doi:10.1016/0030-4018(85)90047-1. 8. ^ M. Segev; et al. (1992). "Spatial solitons in photorefractive media". Phys. Rev. Lett. 68 (7): 923–926. Bibcode:1992PhRvL..68..923S. doi:10.1103/PhysRevLett.68.923. PMID 10046033. 9. ^ E. DelRe & M. Segev (2009). "Self-Focusing and Solitons in Photorefractive Media". Self-focusing: Past and Present. Topics in Applied Physics. 114. pp. 547–572. Bibcode:2009sfpp.book..547D. doi:10.1007/978-0-387-34727-1_23. ISBN 978-0-387-32147-9. 10. ^ J.S. Aitchison; et al. (1992). "Observation of spatial solitons in AlGaAs waveguides". Electron. Lett. 28 (20): 1879. doi:10.1049/el:19921203. 11. ^ G.I. Stegeman & M. Segev (1999). "Optical Spatial Solitons and Their Interactions: Universality and Diversity". Science. 286 (5444): 1518–1523. doi:10.1126/science.286.5444.1518. PMID 10567250. 12. ^ J. Beeckman; K. Neyts; X. Hutsebaut; C. Cambournac; M. Haelterman (2004). "Simulations and Experiments on Self-focusing Conditions in Nematic Liquid-crystal Planar Cells". Opt. Express. 12 (6): 1011–1018. Bibcode:2004OExpr..12.1011B. doi:10.1364/OPEX.12.001011. PMID 19474916. [1][2] 13. ^ Okulov, A Yu (2000). "Spatial soliton laser: geometry and stability". Optics and Spectroscopy. 89 (1): 145–147. Bibcode:2000OptSp..89..131O. doi:10.1134/BF03356001. S2CID 122790937. 14. ^ Okulov, A Yu (2020). "Structured light entities, chaos and nonlocal maps". Chaos,Solitons&Fractals. 133 (4): 109638. arXiv:1901.09274. doi:10.1016/j.chaos.2020.109638. 15. ^ 16. ^ 17. ^ M. Gedalin, T.C. Scott, and Y.B. Band, "Optical Solitons in the Higher Order Nonlinear Schrödinger Equation", Phys. Rev. Lett. 78: 448–451 (1997) [3][4]. 18. ^ S.Chakraborty, "Report of soliton pulse generation within 50 m length of SM fibre by high frequency induced nonlinear intelligent feedback method" , Proceedings, IEEE National Conference on Applications of Intelligent System, Sonepat, India, pp.91–94, 2008, ISBN 978-81-906531-0-7.[verification needed] ## Bibliography • Saleh, B. E. A.; Teich, M. C. (1991). Fundamentals of Photonics. New York: John Wiley & sons, inc. ISBN 978-0-471-83965-1. • Agrawal, Govind P. (1995). Nonlinear fiber optics (2nd ed.). San Diego (California): Academic Press. ISBN 978-0-12-045142-5.	2021-04-16 22:33:48	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 130, "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": 0.7757784724235535, "perplexity": 1044.6219292346932}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-17/segments/1618038092961.47/warc/CC-MAIN-20210416221552-20210417011552-00584.warc.gz"}
https://www.studyadda.com/index.php?/notes/jee-main-advanced/mathematics/trigonometric-equations/area-of-triangle/9426	# JEE Main & Advanced Mathematics Trigonometric Equations Area of Triangle ## Area of Triangle Category : JEE Main & Advanced Let three angles of $\Delta ABC$ are denoted by $A,\,\,B,\,\,C$ and the sides opposite to these angles by letters $a,\,\,b,\,\,c$ respectively. (1) When two sides and the included angle be given : The area of triangle ABC is given by, $\Delta =\frac{1}{2}bc\sin A=\frac{1}{2}ca\sin B=\frac{1}{2}ab\sin C$ i.e., $\Delta =\frac{1}{2}$ (Product of two sides) $\times$ sine of included angle (2) When three sides are given : Area of $\Delta ABC=\,\Delta =\sqrt{s(s-a)(s-b)(s-c)}$ where semiperimeter of triangle $s=\frac{a+b+c}{2}$ (3) When three sides and the circum-radius be given : Area of triangle$\Delta =\frac{abc}{4R}$, where R be the circum-radius of the triangle. (4) When two angles and included side be given : $\Delta =\frac{1}{2}{{a}^{2}}\frac{\sin B\sin C}{\sin (B+C)}=\frac{1}{2}{{b}^{2}}\frac{\sin A\sin C}{\sin (A+C)}=\frac{1}{2}{{c}^{2}}\frac{\sin A\sin B}{\sin (A+B)}$ LIMITED OFFER HURRY UP! OFFER AVAILABLE ON ALL MATERIAL TILL TODAY ONLY! You need to login to perform this action. You will be redirected in 3 sec	2018-07-20 22:08:50	{"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": 0.7568246126174927, "perplexity": 2123.1478387980046}, "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/1531676591837.34/warc/CC-MAIN-20180720213434-20180720233434-00138.warc.gz"}
https://stats.stackexchange.com/questions/351314/r-how-to-find-cluster-centroids-with-tsclust	# R: how to find cluster centroids with tsclust? I have a time-series dataset and I am required to find similar clusters in the data. Based on my current knowledge and the requirements of my application, I used SBD measure (shape based distance) to calculate the dissimilarity matrix for my dataset and applied hierarchical clustering on it (using tsclust). The R command used is: library(dtwclust) hclust=tsclust(mydata,type="h", distance = "sbd") I also used cvi for cluster validation (cvi(hclust)) and was able to get a value of 0.508 for Silhouette width (which I believe is good enough). The problem is that I don't know at which point to cut this cluster tree - for how many clusters (value of k) or at what height (value of h) to get the Silhouette width of 0.5? Moreover, once I know this value of k or h, how do I find the centroids (time-series data) that represent these clusters? The answer was pretty straight-forward - thanks to Alexis who suggested me to read Appendix-A of the documentation, and @Haroon who suggested writing an email to Alexis. Here are the answers to my questions: 1) How to identify the value of k or h for which the Silhouette value was 0.5? I ran a loop for different number of clusters and cut the dendrogram for these many clusters. Then, I computed the silhouette value for each of the cluster sets and identified the value for k where the value was 0.5. The code for this is: library(dtwclust) hclust=tsclust(mydata,type='h',distance='sbd') #running the loop now numberofclusters=c(2:100) silValues=c(1:length(numberofclusters)) for(size in numberofclusters){ sbd_cluster=cutree(hclust,k=size) index=which(numberofclusters==size) x<-silhouette(sbd_cluster,dist=sbddist) silSBDclust[index]=mean(x[,3]) } plot(clustersizes,silSBDclust,xlab="Number of clusters",ylab="Average silhouette width") 2) How to identify cluster centroids? Once we have identified the number of clusters, we need to find the centroids for these clusters. Please note that the object returned by tsclust() is of type S4 (refer to the documentation of TSClusters-Class) and to access its formal elements, we need @ operator unlike the $ operator often used in R. When we know the value of k for our dendrogram, we can use this value in tsclust() again for better clustering. hclust=tsclust(mydata,k=k,type='h',distance='sbd') View(hclust@centroids) #gives you a list of centroids Hope it helps someone! • I tried this but I get "Error in View : trying to get slot "centroids" from an object of a basic class ("list") with no slots" Any idea on this? – Thusitha Thilina Dayaratne Jun 3 at 8:05 Just as a suggestion, what is being done by BajajG can be done directly with compare_clusterings: cfg <- compare_clusterings_configs( "h", k = 2L:100L, controls = list( hierarchical = hierarchical_control(method = "average", symmetric = TRUE) ), distances = pdc_configs("d", sbd = list()) ) evaluator <- cvi_evaluators("Sil") comparison <- compare_clusterings(mydata, "h", cfg, score.clus = evaluator$score, pick.clus = evaluator$pick) best <- repeat_clustering(mydata, comparison, comparison$pick$config_id) plot(best) Check also the results in comparison$results\$hierarchical. EDIT: and also set method = "all" in hierarchical_control if you want to test other linkage methods.	2019-08-20 01:05:16	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 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": 0.3507678508758545, "perplexity": 1199.3470923540438}, "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/1566027315174.57/warc/CC-MAIN-20190820003509-20190820025509-00404.warc.gz"}
https://astro.paperswithcode.com/paper/the-role-of-the-parker-instability-in	# The Role of the Parker Instability in Structuring the Interstellar Medium 18 Mar 2020 Heintz Evan Bustard Chad Zweibel Ellen The Parker instability, a Rayleigh-Taylor like instability of thermal gas supported against gravity by magnetic fields and cosmic rays, is thought to be dynamically important for galaxy evolution, possibly promoting molecular cloud formation and the galactic dynamo. In previous work, we examined the effect of three different cosmic ray transport models on the Parker instability: decoupled ($\gamma_c = 0$), locked to the thermal gas ($\gamma_c = 4/3$) and coupled to the gas with streaming by self-confinement... (read more) PDF Abstract # Code Add Remove Mark official No code implementations yet. Submit your code now # Categories • ASTROPHYSICS OF GALAXIES	2021-02-26 00:31:56	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.3433839976787567, "perplexity": 8001.089650800179}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-10/segments/1614178355944.41/warc/CC-MAIN-20210226001221-20210226031221-00615.warc.gz"}
https://www.techwhiff.com/issue/do-you-think-it-s-better-for-a-young-designer-to-use--446524	# Do you think it’s better for a young designer to use free, open-source art programs or to pay for commercial programs? Explain your answer, but come up with at least one counterargument that forces you to defend your position when you explain why you think one option is better than the other. ###### Question: Do you think it’s better for a young designer to use free, open-source art programs or to pay for commercial programs? Explain your answer, but come up with at least one counterargument that forces you to defend your position when you explain why you think one option is better than the other. ### Potential energy formally called the energy of Potential energy formally called the energy of... ### Somebody pls help I’ll brainlest Somebody pls help I’ll brainlest... ### What is the answer to number 5 What is the answer to number 5... ### Josh is 6 years older than 4 times Kim's age. The sum of their ages is less than 31. What is the oldest Kim could be? Josh is 6 years older than 4 times Kim's age. The sum of their ages is less than 31. What is the oldest Kim could be?... ### Gravity is needed to determine an object's _____. A. mass only B. weight only C. mass and weight Gravity is needed to determine an object's _____. A. mass only B. weight only C. mass and weight... ### Which calculations could be used to solve this problem? There were 24 dogs and cats that were adopted from the animal shelter last week. Three times as many dogs as cats were adopted. How many dogs were adopted? Choose all answers that are correct. A. Divide 24 ÷ 4. Subtract the quotient from 24. B. Divide 24 ÷ 4. Multiply the quotient by 3. C. Add 24 + 3. Divide the sum by 4. D. Multiply 24 × 3. Subtract 24 from the product. Which calculations could be used to solve this problem? There were 24 dogs and cats that were adopted from the animal shelter last week. Three times as many dogs as cats were adopted. How many dogs were adopted? Choose all answers that are correct. A. Divide 24 ÷ 4. Subtract the quotient fr... ### What is the ratio of 3x/7 to 5x/7 what is the ratio of 3x/7 to 5x/7... ### Geomorphic importance of earth materials (rocks and minerals) Geomorphic importance of earth materials (rocks and minerals)... ### 2) How is the agricultural revolution characterized? 2) How is the agricultural revolution characterized?... ### What major greek city fought off the Persians at a tiny seaport in marathon? Why? what major greek city fought off the Persians at a tiny seaport in marathon? Why?... ### Why was the declaration of indepence a necessary document for the founding of the new nation? why was the declaration of indepence a necessary document for the founding of the new nation?... ### The experession 4x(x-2)-3(5x+1) simplifies to (1) 4x^2 + 7x-3 (2) 4x^2 - 16x-3 (3) 4x^2-23x+3 (4)4x^2 -23x -3 ^2 is the power of 2 The experession 4x(x-2)-3(5x+1) simplifies to (1) 4x^2 + 7x-3 (2) 4x^2 - 16x-3 (3) 4x^2-23x+3 (4)4x^2 -23x -3 ^2 is the power of 2... ### (I is confusion)What fraction of an hour is 35 minutes?give your answer in simplest form (I is confusion)What fraction of an hour is 35 minutes?give your answer in simplest form... ### Compare and contrast the place of Irish immigrants in American society before and after the Great Famine of 1845 to 1850 Compare and contrast the place of Irish immigrants in American society before and after the Great Famine of 1845 to 1850... ### Simplify x^3+3x^2-x+2x^2+6x-2 simplify x^3+3x^2-x+2x^2+6x-2... ### Solve the exponential equation: 6x-3 = 216x-3 Solve the exponential equation: 6x-3 = 216x-3... ### Which us president had to retake the oath of office because of a mistake? Which us president had to retake the oath of office because of a mistake?...	2022-08-18 12:54:44	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 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": 0.34776729345321655, "perplexity": 2762.080476574718}, "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/1659882573197.34/warc/CC-MAIN-20220818124424-20220818154424-00510.warc.gz"}
https://www.nature.com/articles/s41396-020-00858-x?utm_campaign=related_content&utm_source=MICROBIO&utm_medium=communities&error=cookies_not_supported&code=e61afaff-c57b-496f-9671-d918148e85c6	Thank you for visiting nature.com. You are using a browser version with limited support for CSS. To obtain the best experience, we recommend you use a more up to date browser (or turn off compatibility mode in Internet Explorer). In the meantime, to ensure continued support, we are displaying the site without styles and JavaScript. # Selfishness driving reductive evolution shapes interdependent patterns in spatially structured microbial communities ## Abstract Microbes release a wide variety of metabolites to the environment that benefit the whole population, called public goods. Public goods sharing drives adaptive function loss, and allows the rise of metabolic cross-feeding. However, how public goods sharing governs the succession of communities over evolutionary time scales remains unclear. To resolve this issue, we constructed an individual-based model, where an autonomous population that possessed functions to produce three essential public goods, was allowed to randomly lose functions. Simulations revealed that function loss genotypes could evolve from the autonomous ancestor, driven by the selfish public production trade-off at the individual level. These genotypes could then automatically develop to three possible types of interdependent patterns: complete functional division, one-way dependency, and asymmetric functional complementation, which were influenced by function cost and function redundancy. In addition, we found random evolutionary events, i.e., the priority and the relative spatial positioning of genotype emergence, are also important in governing community assembly. Moreover, communities occupied by interdependent patterns exhibited better resistance to environmental perturbation, suggesting such patterns are selectively favored. Our work integrates ecological interactions with evolution dynamics, providing a new perspective to explain how reductive evolution shapes microbial interdependencies and governs the succession of communities. ## Introduction Microbes rarely live in isolated niches naturally, but interact with other individuals to form complex communities. It is widely believed that microbial interactions are central to the maintenance, stability, and productivity of these communities [1,2,3]. Although research has been focused on the evolution of different forms of microbial interactions and their impact on the fitness of the individuals, how the interplay among different individuals governs the succession of microbial communities in the process of long-term evolution has received less attention [4, 5]. Due to the wide range of uncertainty involved in evolutionary dynamics, with the given initial biotic and abiotic components, it is challenging to predict what kinds of genotypes could evolve and which type of interaction pattern could organize in the future community [6, 7]. Resolving this problem is not only important for understanding the formation and maintenance mechanism of microbial diversity, but also has implications for the evolutionary responses of the community to novel environments [8, 9]. One important form of microbial interaction is cooperation related to the production and exchange of so-called ‘public goods’. Public goods are products that, while costly to produce, provide a benefit to all the members of a community, especially to neighbors of the producer [10]. Many secretions released by microorganisms can be considered public goods, such as degradative enzymes [11, 12], siderophores [13,14,15], detoxification agents [16], and amino acids [17, 18]. Public goods sharing creates an opportunity for the evolution of cooperative interactions. The recently proposed Black Queen Hypothesis (BQH) explains how public goods dynamics drive the origin of dependencies over an evolutionary timescale, predicting that when an individual loses a costly, leaky function, it will receive a selective advantage and expand in the community until the production of public goods is just sufficient to support the equilibrium community [19]. According to the BQH, the origin of cooperative interactions may be based on the selfish trade-off of public goods production by individuals [19, 20]. The BQH has been widely applied to explain the evolution of metabolic dependencies through adaptive functions loss, both for free-living [16, 21,22,23,24] and host-associated [13, 21, 25] organisms. However, in complex ecosystems, microbes can exchange a variety of public goods, so multiple functions may be lost through reductive evolution, resulting in diverse ecological outcomes. Could diverse types of interaction patterns, especially the cross-feeding interdependent pattern, emerge from originally autonomous genotypes who can produce more than one public goods through Black Queen evolution? What decides the formation of different interaction patterns? Resolving these questions can help us understand how public goods exchange interactions govern the assembly and succession of microbial communities. Because of the difficulties in simulating long-term evolution and complicated natural conditions in experimental systems, some studies have applied mathematical modeling approaches to test the potential for cooperation driven by Black Queen evolution. Simulations initialized with diverse cooperative genotypes suggest that the emergence of microbial interdependencies only occurs under specific conditions [26, 27], and in some cases, cooperative interactions are associated with reduced productivity of communities and are, therefore, not selectively favored [26]. However, these models rarely considered stochastic events during evolution, for example, the randomness of the time order and positions for the emergence of different genotypes. A recent study by Mas et al. (2016) applied an agent-based model to test the BQH, and successfully simulated the invasion of a loss of function (LOF) genotype to its autonomously ancestral population [28]. These kinds of model (agent-based model or individual-based model, IBM) focused on spatially-structured environments, thus well-described the randomness of spatial positioning of different mutants. When numerous public goods can be secreted, this randomness will add the uncertainly to the evolution of an interdependent pattern because the emerging positions of LOF genotypes are unpredictable. In addition, IBMs focus on the behavior of individuals, thus can well model the occurrence of random mutations at the single-cell level, displaying good predictive power for the effect of random emergence order of different genotypes. Therefore, we used this type of model framework to investigate the complex evolutionary processes driven by public goods sharing and how it shapes microbial community diversity. Therefore, we proposed that different types of cooperative interdependent patterns could potentially evolve via Black Queen evolution, and constructed an individual-based model to test this idea. We simulated the spatiotemporal dynamics of one type of reductive evolution, starting with an ancestral population that could produce multiple fitness-promoting public goods, who was subsequently allowed to randomly mutate to lose those functions. We focused on investigating the conditions favoring the evolution of different patterns, considering both deterministic factors, such as function cost and functional redundancy, and stochastic factors, such as the randomness of the time order and positions, in our model. We also conducted numerous replicated simulations and genotype lineage tracing to capture the diversity and clustering of evolutionary paths. In sum, we built a new mathematical framework integrating the ecological interactions with evolution dynamics, providing a new perspective to explain how microbial interactions govern the succession of the communities over an evolutionary time scale. ## Results ### The logic of the model Our spatially-resolved model was simulated in discrete grid boxes of a 100 × 100 array, which included four basic assumptions: (1) Initial individuals were assumed to secrete three public goods but may randomly mutate to lose any of those functions with a certain probability; (2) Secreting a public good created a corresponding metabolic burden, therefore in losing a function the individual would gain a benefit; (3) All public goods were essential for growth. The net growth rates of individuals were dependent on the local concentrations of public goods; (4) Substrate and public goods diffused between two grid boxes at rates proportional to the concentration gradient. For the 1st assumption, we included three functions because it is the minimal unit and tersest design to simulate complex communities, allows for the emergence of three categories of interaction patterns, and a single cooperative LOF genotype might evolve from differential evolutionary paths (Fig. 1A, B). The genotypes were described by bit strings containing 1 and 0 which indicated the genotype could produce the corresponding public good or not, respectively. Eight genotypes could emerge during the simulations, which were the initial autonomous producer [1, 1, 1], three one-function loss genotypes (OFLGs, i.e., [1, 1, 0], [1, 0, 1], and [0, 1, 1]), three two-function loss genotypes (TFLGs, i.e., [1, 0, 0], [0, 1, 0], and [0, 0, 1]), and a nonproducing cheater [0, 0, 0] (Fig. 1A). The 2nd and 3rd assumptions were developed from the basic mathematical assumption of the BQH [19], and defined individual growth by integrating the benefit and cost of function loss (Fig. 1C). To conceptualize the cost of performing a function, we supposed a parameter (α) which is the fraction of biomass used to produce a public good per unit time of an individual. In addition, we defined a second parameter (β) as the ratio of the amount of public goods required during each step to account for the produced public goods. Therefore the redundant fraction of public goods production was 1−βj, and lower βj reflected a higher amount of redundant public goods that could be gained from the producers by the LOF genotypes, resulting in decreased risk in association with function loss (see Supporting Information S1 for more details). During the model simulation, spatiotemporal dynamic variables, i.e., positioning of genotypes and the time points at which genotypes evolved, would be collected. We initiated the simulations by randomly distributing 100 ancestor cells [1, 1, 1] into the grid boxes and iterated for at least 1,500,000 time steps. During each time step, individuals grew, decayed, reproduced, and mutated according to the previously mentioned assumptions (Fig. 1C). We paid attention to whether stable communities with various interdependent patterns could be formed after a specified number of iterations, as well as recorded the spatiotemporal dynamics of the communities. ### Diverse interdependent patterns emerged with high level of function cost and varied level of functional redundancy For model simulations, the function cost (parameter α) and functional redundancy (parameter β) were assigned to 0.0001, 0.0005, 0.001, and 0.4, 0.6, 0.8, respectively. A total of 2891 independent simulations with 9 parameter sets displayed different community structures (Fig. 2A). When the function cost was assigned to a low level, i.e., 0.0001, the autonomous ancestor dominated the community. When function costs were assigned to higher levels, 0.0005 and 0.001, new genotypes evolved and later interacted to form three distinct types of interdependent patterns even within the same α and β combination, i.e., asymmetric functional complementation (AFC), complete functional division pattern, and one-way dependency, with the relative amounts of 1677/2891, 143/2891, and 48/2891, respectively. In addition, higher functional redundancies favored the loss of more functions, increasing complexity of the community structures. Among the three possible kinds of interactions, the AFC pattern was the most widespread, which was the combination of a two-function-loss genotype (TFLG) and its complementary one-function-loss genotype (OFLG). For example, [0, 0, 1], which produced a single essential public good, depended on its functional complement one-function-loss partner [1, 1, 0], for the other two public goods. Specifically, three types of the asymmetric functional complementary pairs (AFCPs), that is, [0, 0, 1] coupled with [1, 1, 0], [0, 1, 0] coupled with [1, 0, 1], and [1, 0, 0] coupled with [0, 1, 1], colonized most of the grid with a similar frequency of emergence. Interestingly, under the condition of high level of cost, the emergence of AFC patterns was accompanied by some nonproducing cheaters, whose relative abundance rose with the increase in functional redundancy (Fig. 2A top row). The addition of cheaters significantly reduced the total biomass of the communities, suggesting that high functional redundancy favors the evolution of cheaters which may decrease the community productivity. In addition, function loss happened more easily with high function cost. As the function cost parameter α increased from 0.0005 to 0.001, relative abundance of TFLGs increased approximately from 55 to 70% (Fig. 2A). Besides the AFC patterns, two additional types of interdependent patterns evolved at a relatively lower frequency. The complete functional division pattern, that is, coexistence of [0, 0, 1], [0, 1, 0], and [1, 0, 0], only evolved when both factors were at high levels (α = 0.001, β = 0.4) with a frequency of approximately 45% (143 of 319 simulations, Fig. 2A, top right), which described a scenario with high benefit and low cost of function loss, favoring the loss of more functions and consequently more likely to maintain the evolution of TFLPs. Another form of interactions that emerged was one-way dependency, where one partner performs all functions and other none (i.e., coexistence of [1, 1, 1] and [0, 0, 0]). This form emerged at a low frequency (48 out of all 2891 simulations shown in Fig. 2A), but evolved with a higher probability under the condition of a mid-level function cost and low level of functional redundancy (α = 0.0005, β = 0.6, Fig. 2A, middle left), where the extinction of [1, 1, 1] was ~2.5 times slower than in other scenarios (Supplementary Fig. 1), leading to a higher potential for the spatial proximity between [1, 1, 1] and [0, 0, 0] during evolution. Taken together, these phenomena demonstrated that the mutualistic exchange of complementary functions happened only when function cost was high. The emergence of different interdependent interaction patterns was related to the function cost and function redundancy, especially for the complete functional division and one-way dependency pattern, which only emerged within a limited parameter range. However, even for a given combination of α and β, it still remained possible for the evolution of distinct interaction patterns, suggesting that stochastic processes may play a role. ### Same interdependent patterns might evolve via different modes Because the evolution of three kinds of AFCPs were the most common scenarios in our simulations, we then focused on the role of stochastic processes, i.e., the key random events, in deciding the winning complementary pair among the three similar but different AFCPs. As a first step, we traced the variation in the spatiotemporal dynamics, trying to cluster the numerous evolutionary dynamics into limited modes and divide the complex evolutionary courses into several stages. These simplifications would facilitate the search for key random events. Therefore, we analyzed the dynamics of 296 simulations with a typical parameter set (α = 0.001, β = 0.8), because under this condition, only the three types of AFCPs evolved, with a similar frequency of emergence (Fig. 2A, Top left), in order to avoid interference from the other interaction patterns. As described above, any of the three types of AFCPs could potentially take over the final community under this condition (Fig. 2B; Supplementary video 13). Using the emergence of AFCP [0, 0, 1] & [1, 1, 0] as an example, three categories of dynamic modes could give rise to its final domination. (1) After pair [0, 0, 1] & [1, 1, 0] emerged and formed a spatial aggregation, it rapidly expanded and took over the entire grid (Fig. 2C, first line; Supplementary video 3). (2) In addition to the pair [0, 0, 1] & [1, 1, 0], spatial aggregations of another AFCP also emerged (e.g., pair [0, 1, 0] & [1, 0, 1] in Fig. 2C, second line and Supplementary video 4). In this scenario, a special spatial pattern was established in a short period after the evolution of both AFCPs e, where pairs of two complementary members exhibited strong spatial mixing, while the two different AFCPs were totally segregated. Community succession was then governed by spatial competition between the two AFCPs. If pair [0, 0, 1] & [1, 1, 0] won the competition, it would dominate the final community. (3) Spatial aggregations of all three AFCPs emerged, and then pair [0, 0, 1] & [1, 1, 0] dominated the community after outcompeting the other two AFCPs (Fig. 2C, third line; Supplementary video 5). The clustering of these three possible modes of AFC patterns was also shown by the temporal dynamics of the α-diversity across different parameter sets (Supplementary Fig. 2), where the evolution modes of the AFC patterns were clearly clustered into three possible categories, suggesting that this clustering is independent of the determined factors α and β. In sum, the succession of interdependent patterns could be divided into two stages: (1) the emergence of spatial aggregations composed of two interdependent members with strong connections; (2) spatial competition among different aggregations drive the community to evolve to the final state, composed of only one type of interdependent interactions. Of course, if only one type of AFCP emerged, the spatial competition stage would be unnecessary during succession. ### Evolutionary random events play important roles in deciding the dominant AFCP in equilibrium communities The presence of two evolutionary stages lead us to hypothesize that the random events affecting ecological outcomes should arise from two aspects. First, in the initial evolutionary stage, the emergence of interdependent spatial aggregations should be related to the order in which new genotypes emerge. Second, the outcome of the spatial competition should be also influenced by the initial positioning of the new genotypes. The fact that each TFLP had two possible evolutionary paths (e.g., [1, 0, 0] could inherit its function from [1, 1, 0] or [1, 0, 1]), suggested that the effects of the random order of emergence for different genotypes were highly correlated with the evolutionary lineage. Therefore, to investigate the effects of this, we analyzed the evolutionary lineage of emergence, colonization, and loss of every genotype within the 296 simulations with the typical parameter set (α = 0.001, β = 0.8). In total, there were 24 evolutionary branches leading to the evolution of the three forms of AFC patterns (8 for each, Fig. 3). Among all these branches, we summarized two key random events (Fig. 3, red and blue boxes). The first event occurred after two types of OFLGs emerged. After this evolutionary time point, all three public functions were included in OFLGs. With the benefit of the function loss, these two OFLGs would expand and gradually outcompete the autonomous genotype [1, 1, 1]. Thus, the first key event was whether all three OFLGs could emerge before the autonomous genotype entirely disappeared (Fig. 3, blue box). If not, the third type of AFCP would never evolve; if so, all three types of AFCPs would still have a chance to dominate the final community. In the 296 simulations, the frequencies of these two scenarios were nearly same, that is, 147 simulations were clustered to the former, while 149 simulations were clustered to the latter. The 147 simulations, where the third type of AFCP never evolved, could be then divided into three categories with similar frequencies, where two of the three OFLGs occupied the whole space and excluded the ancestral population. The second key evolutionary event was the emergence of TFLGs (Fig. 3, red box). After the two or three types of OFLGs successfully colonized, whose functional complementary TFLGs first to emerge in the next evolutionary time would lead to the prior formation of the spatial aggregation of the AFCP. It is obvious that if no other AFCP aggregations formed later, this AFCP would dominate the final community (Fig. 3, red arrow indicated branches). Alternatively, if other AFCP aggregations formed during the expansion process, the spatial competition between different AFCPs would decide the dominant AFCP in the equilibrium communities (Fig. 3, blue arrow indicated branches). In our analysis, the chance of only one AFCP evolving reached 64.7% (198 of the 296 simulations). If only two OFLGs evolved after the first event, the frequency of only one AFCP evolving reached 79.6% (121 of the 152 simulations). In contrast, if three OFLGs evolved after the first event, there could be a relative higher possibility of two or three AFCPs evolving (47.4%), meaning that spatial competition could then be an important process. What decided the winner of the competition? We observed that after the segregated interdependent spatial pattern was newly established, the relative region sizes occupied by different AFCPs were the key to determining the winner (Fig. 2C, the second and third lines; Supplementary video 4 and 5). We analyzed the time gaps between the emergence of the two AFCPs in the second categories of succession modes and the size of the regions they occupied (Fig. 4A). The result indicated a significantly positive correlation between the length of the time gaps and the region size the prior AFCP occupied (t-test, p < 0.05; Fig. 4B); if the time gap was greater than ~17,000 time steps, or the prior occupied space was larger than 129 grid boxes, the first emerged AFCP would win the competition (Fig. 4C, D). Together with the lineage analysis, these results confirmed that the order of emergence by the different genotypes would largely decide the formation of different interdependent patterns in the final communities, i.e., the earlier an AFCP emerged, the greater the chance for it to dominate the equilibrium communities. To further prove this idea, we conducted additional simulations. Four genotypes, [0, 0, 1], [1, 1, 0], [0, 1, 0], and [1, 0, 1], were inoculated with the same initial number. We randomly distributed the genotypes in the 100 × 100 grids to obtain diverse initial spatial distribution (see Methods; Supplementary Fig. 3B). We then selected a series of distributions with a gradient of relative association degrees of AFCPs (PAI001:010), and ran simulations initialized with those values. The spatial self-organization also occurred in these cases, accompanied by the increase of PAD001 and PAD010 (Fig. 5B; Supplementary Fig. 4). For a given initial distribution, we conducted 100 replicate simulations to calculate the frequency of winning. A strong positive relationship between initial association degree of an AFCP and its winning frequency was found (Fig. 5B, C, P < 0.01), which further confirmed the important role of relative positioning on deciding the ecological outcome. One might ask, in absence of emergence order and relative positioning, whether other random events impacted the outcomes of spatial competition? To address this issue, we simulated communities starting from a symmetric initial distribution of cells, and two types of AFCPs, [0, 0, 1] & [1, 1, 0] and [0, 1, 0] & [1, 0, 1], were equally inoculated (Supplementary Fig. 5A). Eliminating initial spatial asymmetry and distinct space occupancy, the segregated interdependent spatial pattern still emerged, and one of the AFCPs dominated the final community in each run (Supplementary Fig. 5B). However, the frequencies of winning for the two AFCPs were nearly identical in the 100 repeated simulations (Supplementary Fig. 5C), suggesting other random events did not change the relative winning probability of the two AFCPs. ### Interdependent pattern of community is associated with increased stability Do the communities dominated by functional complementary LOF genotypes exhibit distinct community properties, e.g., different community stability? Our mathematical framework allowed us to make comparisons among communities at different time points. To investigate whether community stability changes with the succession, we imposed nutrient disturbances on the communities at initial and final time points. We stopped the nutrient supply for 12 h, and after restoring it, explored community response to the disturbance. Although all communities declined after nutrient depletion, the resistance of the evolved communities to disturbances was better than the original community composed of only the autonomous population, suggesting the evolved communities containing interdependent patterns were more stable (Fig. 6A). The improved community stability was attributed to the more effective resource allocation of the LOF genotypes. As shown in Fig. 6B, although the average available substrate for each cell was similar for both scenarios during the disturbance (solid line in the shaded region), the amount of resources required for producing public goods significantly decreased when an interdependent pattern formed (dashed line), indicating more resources can be assigned to growth or maintenance when the community is facing environment perturbations. This suggests that after the interdependent pattern evolved, the resources that were originally wastefully allocated to produce redundant public goods, were saved to fight against the harsh environmental change. This result can explain why the interdependent pattern is selectively favored at the community level. ## Discussion In this study, we tested how selfish public goods produced trade-offs that drove the reductive evolution of different interdependent patterns and governed the succession of microbial communities. We found the three classes of interaction patterns that emerged in the steady state communities correlated with different functional traits, including complete functional division, AFC, and one-way dependency. We highlighted the importance of random evolutionary events on the formation of different patterns, where the priority and the relative spatial position of emergence of the different genotypes decide which pair of functional complementary genotypes would dominate the final communities. In addition, communities with these interdependent patterns exhibit a more stable response to disturbance, attributed to the more effective pattern of resource allocation in these communities. In his work ‘The Origin of species’ [30], Charles Darwin clearly emphasized that natural selection favors the individuals who are selfish to get greatest personal reproductive success, so it was confusing for a long time how cooperation evolved, where an organism is selected to be selfless to enhance the fitness of others [31]. Here, by expanding the BQH framework, our model gives a new insight, that cooperation between auxotrophic genotypes can evolve automatically from the natural selection for selfishness. At the individual-level, LOF is a purely selfish trait, where the individual no longer contributes a public resource to the social group but saves the energy to increase its own reproductive success. However, at the community level, when several functionally complementary LOF genotypes evolve under the right conditions (with proper functional traits), at the right time (emerge simultaneously in the community), and in the right place (spatial proximity to each other), cooperation arises and the newly evolved community possesses the greater selective advantage of better resistance to environmental perturbation, benefiting all members. Thus, in our simulations, the evolution of reciprocal community traits is actually driven by the selection for selfish trait at individual level. This seemingly contradictory idea still follows Darwin’s rule that evolution gives fitness benefit to all the individuals, and gives a possible solution to the paradox of the evolution of cooperation. Do the interdependent patterns that we observed in our model ever evolve through reductive evolution in nature? A survey based on large-scale metabolic modelling has shown that metabolically interdependent groups are ubiquitous in microbial communities across diverse habitats [32], while further in silico experiments predict that interdependent patterns can arise associated with the production of specific public goods, such as several types of amino acids [33], and the cost of these secretions is a key driver of the mutualistic interactions [34]. Several examples are also reported in experimental systems, one of which is the production of vitamins in marine environments. Although vitamins are indispensable for growth, many plankton species lack a subset of the biosynthetic pathways [23, 35, 36]. This observation is also related to the genome streamlining often seen among oceanic microorganisms [37], because vitamin synthase genes are always energetic costly [38]. A recent study reported that B12 and B1 auxotrophy of the alga Ostreococcus tauri could be alleviated by co-culturing with a heterotrophic bacterial partner Dinoroseobacter shibae, which in turn relies on the alga to satisfy its requirements of three other B vitamins niacin (B3), biotin (B7), and p-aminobenzoic acid (a precursor for folate, B9) [39]. If we conceptualize these vitamins as five leaky public goods, the two members could be defined as genotype [1, 1, 0, 0, 0] and [0, 0, 1, 1, 1], forming an AFC interdependent pattern as we presented here. It is reasonable to speculate that both genotypes might evolve from their autonomous ancestral populations, respectively, and the selfish vitamin production trade-off drove the reductive evolution. The evolution of interdependent patterns has also been observed in experimental evolution systems. One direct piece of evidence arose from an innovative study by means of a system containing metabolic auxotrophic populations of Saccharomyces cerevisiae [40]. The authors found that the strains, originally prototrophic for four metabolites (histidine, leucine, uracil, and methionine) might gradually lose these public functions, when communities with cooperative metabolite exchange began to self-establish. This experimental evolution process exactly matched our simulation results. If we could conduct similar research repeatedly with a longer evolution time, we may experimentally identify the key roles of the function traits, evolutionary randomness, as well as the spatial structure governing the self-establishment of these communities. Our work also provides a potential explanation for microbial diversity. Leakiness of public goods is widespread within the microbial world, thus co-evolution driven by the sharing of public goods may be common in nature. Considering the complex effects of functional traits and evolutionary randomness, starting from a ‘super ancestor’ who contains multiple leaky functions, diverse LOF species can evolve automatically. Once cooperative interdependent patterns form, genotypic diversity will increase. These strains will share a pan-genome as a public genomic resource, which is identical with the genome of their common “super ancestor” [41]. Since most of these evolved strains are auxotrophic, this idea may also explain why many species are unable to be isolated as pure cultures in the laboratory [42,43,44,45]. Different from the previous related model, our model introduced the initial redundancy of public goods production (β) as an important parameter to describe functional traits. This assumption allows individuals to overproduce public goods, and should be the prerequisite for the evolution of the LOF genotypes, since they must rely on those redundant public goods for survive. As predicted, we observed that a higher level of the initial degree of functional redundancy was important for the evolution of those interdependent patterns (Fig. 2A). However, accompanied with function loss, the redundancy degree of the whole community decreased. The loss of functional redundancy, or functional distinction, is known as ‘niche complementarity’, where community members occupy complementary niches, buffering environmental perturbation, and conferring gains in productivity and efficiency [46]. This prediction matches with the increased community size (Fig. 2A), as well as increased community resistance (Fig. 6) after the interdependent pattern formed. However, we also found cheaters, who did not contribute to the community, were highly selected for at high levels of redundancy in public goods production, which reduced the community size, i.e., the productivity of the community (Fig. 2B, Top right). Thus, maintaining a proper level of function redundancy to prevent invasion from cheaters may be involved in the design of efficient synthetic communities. Despite these encouraging insights, our model still has some limitations, which should be addressed in future studies. Firstly, for the sake of simplicity, we assumed that all public functions have equal traits, i.e., equal function costs, degrees of redundancy, and essentiality. However, different public secretions will possess varying function traits. For instance, biosynthesis of different amino acids requires different function costs [18]. The heterogenetic traits of the public functions can potentially lead to divergent community compositions among different interaction patterns. For example, we tested scenarios where the three public functions had different functional redundancy. Although the three types of AFCPs showed similar probability of emergence, the relative abundances among different LOF genotypes, especially the fraction of cheaters ([0, 0, 0]), varied across different patterns (Supplementary Fig. 6). Therefore, we believe that heterogeneity in functional traits should play a role in shaping community structure, and requires further investigation. Secondly, we did not consider the inherent private benefit of the public goods producer. It has been reported that microbial cells could partially privatize some metabolites, allowing just a fraction of products to leak into the environment, resulting in unequal access to public goods between producer and non-producer [27, 47]. Research has also shown that mutual interdependency can be selectively favored at intermediate levels of privatization, while its absence will lead to the collapse of interaction [27]. Nevertheless, benefiting from our individual-based model, we considered spatial structure (limited mass diffusion), which is also thought to be a vital way for production privatization [29, 48]. Indeed, we observed that a clear concentration gradient of public goods was formed around the related producer cells (Supplementary Fig. 7), suggesting their private benefits. When we removed the spatial structure to perform a simulation in a well-mixed system, we observed that the unlimited expansion of nonproducer genotypes drove community collapse (Supplementary Fig. 8). We also performed simulations including the spatial structure but with varied diffusion coefficients. While the interdependent patterns evolved in a wide range of diffusion rate, increasing the rate of public goods diffusion favored the growth of the non-production cheaters (Supplementary Fig. 9), suggesting higher diffusion level weakened the private benefit of the producers. Our simulations also indicate that lower level of diffusion opposed the evolution of interdependent patterns, because in this scenario, the producers largely privatized the public goods, inhibiting the interactions dependent on public goods sharing. Therefore, private benefits of producers can be derived from both inherent privatization or limited mass diffusion in spatially structured environments, but further studies are still required to discuss the relative contribution of these two factors. Thirdly, we did not include active cell motility or cell movement via other physical factors (e.g., water flow). In absence of such movement, a daughter cell will be located near its mother cell, resulting in the formation of cell clusters (e.g., microcolony). However, cell movement will break this gathering, where the daughter cell may be soon separated from its mother. Thus, if we added this assumption to our simulations, it may challenge the formation of the segregated interdependent spatial pattern, instead, a pattern with high intermixing of different genotypes may be developed. We believe it would be very interesting to understand the effect of cell movement on the spatial organization of interdependent pattern by combining a new mathematical framework and experimental investigation in future work. ## Methods ### Individual-based model We used spatial simulations of 2D lattices with periodic boundaries, these were built based on previous studies [11, 12, 49]. A detailed model description is given in Supplementary information S1. Briefly, we used a 100 × 100 array to simulate a spatially structured environment. One microbial individual was allowed to occupy a specific spatial grid box, and could only divide into directly adjacent boxes. We used bit strings to describe the genotypes of microbes as described above (Fig. 1). Public goods and nutrient diffusion were computed using a second-order approximation. Microbial growth is assumed based on the mathematical assumption of the BQH, following the general form: $$\frac{{dX_i}}{{dt}} \,=\, \left[ {\left( {g_{max,i} - C_i} \right) - \left( {d_{max} - G_i} \right)} \right]X_i,$$ where Xi was the biomass (evaluated by biomass carbon) of the ith individual; gmax,i was the maximum growth rate, restricted by a limiting nutrient; Ci was the total cost paid by the individual that performed all functions it carried; dmax was the maximum death rate; Gi was the benefit from the local public goods, as a function of the public goods concentration of the grid box. Public goods diffiusion across the grids was caluculated following a second-order approximation for the 2D diffusion lattice as previously described [49]. Simulations were initialized with ancestral populations [1, 1, 1], and the initial biomass of each individual was set as X0 = 150 fg. Microbes reproduce when biomass reached a upper threshold 2X0 + ε, and died when biomass dropped below a lower threshold 0.2X0 + ε, where ε represents uniform random noise in the cell cycle. In addtion, after splitting, the daughter cell is allowed to randomly mutate to lose functions, thus the element ‘1’ in the bit string for the related function may turn into ‘0’ with a certain probability. To charaterize the dynamics of long-term evolution, time-lapse numerical simulations lasted for at least one million time steps, and for a given parameter set, more than 300 repeated simulations were conducted to capture the randomness during the evolution. At each time step, the computation order of the grid boxes is randomized to alleviate the effect of calculation order. All the variables used in the model are listed in Table S1, while all the parameters are provided in Table S2. The model was implemented by C++ language, and the source code are available on (https://github.com/RoyWang1991/Roy-Wang). ### Evolutionary trajectory analysis During each simulation of evolution dynamics, we specifically made notes of all ‘birth events’ (the first emergence of genotypes), mutation events, and extinction events (the extinction of genotypes). For a given event, the time point, spatial position of occurrence, and the related genotype, as well the genotype of its mother cell, were recorded in detail. These events were regarded as imporant nodes during evolution. Based on these nodes, we established the evolutionary affinities among different genotypes, primarily including what each genotype evolved from and when it evolved or became extinct, forming a tree-like diagram of evolutionary relationships. After trees of all the replicates were built, according to their simlarity, we clustered them manually to draw an overall map of evolutionary trajectories of all the interdependent repeated evolution dynamics (Fig. 3). This analysis was implemented by Wolfram Mathematica software (version 10.4). ### Simulations initialized with premixing two AFCPs To test if the relative PAD at key evolutionary time points had influence on the spatial competition, we initialized communities with four genotypes, [0, 0, 1], [1, 1, 0], [0, 1, 0], and [1, 0, 1], containing two types of AFCPs. We randomly premixed cells of these 4 genotypes at a constant cell number, 3000 cells for the TFLGs ([0, 0, 1] and [0, 1, 0]) and 1500 cells for the OFLGs ([1, 1, 0] and [1, 0, 1]), so that the relative proportion of OFLG and TFLG in an AFCP was roughly in line with the proportion in the final communities of the previous simulations (α = 0.001, β = 0.8), while the relative abundance of two AFCPs were equal to prevent the effect from differential initial group size. To obtain diverse communities with different relative PADs, 30,000 communities were initialized. The PAI001:010 of these communities ranged from 0.93 to 1.07 (Supplementary Fig. 3B), and could be approximately categorized into 15 groups by differential PAI values (i.e., 0.93, 0.94, 0.95…1.07). We then randomly picked a minimum of 40 communities from each group and ran initial simulations with these communities (α = 0.001, β = 0.8) to steady state (~120,000 time-steps). For a given initial distribution, we conducted 100 replicate simulations to calculate the frequency of winning. The community dynamics, accompanied by the dynamics of PAD values, were analyzed to find the relationship between initial relative PADs and the succession of the communities. These analyses were implemented by Wolfram Mathematica software (version 10.4), except for the simulations, which were conducted by modifying the previous C++ code. ### Simulations in well-mixed system To remove the effect of restricted diffusion of public goods, we conducted simulations in a well-mixed system, where at each time step, public goods and nutrient were rapidly and evenly distributed across the entire array. To realize this setting, at the beginning of each time step, the amount of each substance were summed up and equally spread among the 104 grid boxes, so when calculating microbial growth, there is no concentration difference of public goods and nutrient among the grid boxes. 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Monteverde DR, Gomez-Consarnau L, Suffridge C, Sanudo-Wilhelmy SA. Life’s utilization of B vitamins on early Earth. Geobiology. 2017;15:3–18. 37. 37. Giovannoni SJ, Thrash JC, Temperton B. Implications of streamlining theory for microbial ecology. Isme J. 2014;8:1553–65. 38. 38. Helliwell KE. The roles of B vitamins in phytoplankton nutrition: new perspectives and prospects. N Phytol. 2017;216:62–8. https://doi.org/10.1111/nph.14669. 39. 39. Cooper MB, Kazamia E, Helliwell KE, Kudahl UJ, Sayer A, Wheeler GL, et al. Cross-exchange of B-vitamins underpins a mutualistic interaction between Ostreococcus tauri and Dinoroseobacter shibae. Isme J. 2019;13:334–45. https://doi.org/10.1038/s41396-018-0274-y. 40. 40. Campbell K, Vowinckel J, Mulleder M, Malmsheimer S, Lawrence N, Calvani E, et al. Self-establishing communities enable cooperative metabolite exchange in a eukaryote. Elife. 2015;4. doi: ARTN e09943, https://doi.org/10.7554/eLife.09943. 41. 41. Fullmer MS, Soucy SM, Gogarten JP. 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Pande S, Kaftan F, Lang S, Svatos A, Germerodt S, Kost C. Privatization of cooperative benefits stabilizes mutualistic cross-feeding interactions in spatially structured environments. Isme J. 2016;10:1413–23. https://doi.org/10.1038/ismej.2015.212. 49. 49. Kreft JU, Booth G, Wimpenny JW. BacSim, a simulator for individual-based modelling of bacterial colony growth. Microbiology. 1998;144(Pt 12):3275–87. https://doi.org/10.1099/00221287-144-12-3275. ## Acknowledgements This work was supported by National Key R&D Program of China (2018YFA0902100 and 2018YFA0902103), and National Natural Science Foundation of China (91951204, 31761133006, 31770120, and 31770118). ## Author information Authors ### Corresponding authors Correspondence to Yong Nie or Xiao-Lei Wu. ## Ethics declarations ### Conflict of interest The authors declare that they have no conflict of interest. Publisher’s note Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. ## Rights and permissions Reprints and Permissions Wang, M., Liu, X., Nie, Y. et al. Selfishness driving reductive evolution shapes interdependent patterns in spatially structured microbial communities. ISME J 15, 1387–1401 (2021). https://doi.org/10.1038/s41396-020-00858-x	2021-10-27 10:42:51	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.5161746740341187, "perplexity": 3776.0743594501782}, "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/1634323588113.25/warc/CC-MAIN-20211027084718-20211027114718-00375.warc.gz"}
https://trac-hacks.org/ticket/6743	Opened 7 years ago Closed 6 years ago # [[FootNote]] macro should display Nothing when no footnotes are used Reported by: Owned by: Ryan J Ollos Ryan J Ollos lowest FootNoteMacro normal 0.11 ### Description I have inserted the following into my PageTemplates: {{{ #!comment Do not edit below this line. The FootNote macro will insert a resulting list of FootNotes if you use the FootNote macro within this document, and if you do not use it, then nothing will be added to the document. }}} [[FootNote]] A very minor issue is that a small horizontal line is inserted at the bottom of the page, even if no footnotes are generated. ### comment:1 Changed 6 years ago by Ryan J Ollos Resolution: → fixed new → closed Fixes in [8450]. We may want to revisit the change that was made in [8450] at some point to improve coding style. I'm very new to Python. ### Modify Ticket Action as closed The owner will remain Ryan J Ollos. The resolution will be deleted. Next status will be 'reopened'.	2017-02-20 08:42:29	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.3510124683380127, "perplexity": 3758.5228710873716}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-09/segments/1487501170434.7/warc/CC-MAIN-20170219104610-00319-ip-10-171-10-108.ec2.internal.warc.gz"}
https://bsc.rbms.info/DCRMR/series/	# 8 — Series ## 8.01.1 Sources of information 8.01.1.1 The sources of information for series elements are the series title page, monograph title page, cover (if issued by the publisher), dust jacket, and the rest of the manifestation, in that order of preference (see Data provenance, 0.1.5. For multipart monographs, prefer a source in the first volume. If the manifestation has both main series and subseries titles, prefer a source containing both titles. Series-like statements present on covers not issued by the publisher usually represent binders’ titles and should be treated as item-level information if considered important (see Title of item, 1.27.31.1). 8.01.1.2 If the Series statement, or any of its elements, is taken from a source other than the series title page, always make a Note on series statement (see 8.29.31.1). 8.01.1.3 If the Series statement appears on both the series title page and the monograph title page, indicate this in a Note on series statement if considered important (see 8.29.31.1). Transcribe the text of the latter statement as a second Series statement if the two differ. 8.01.1.4 If the Series statement appears as a stamp or on a label, transcribe it as found and make a Note on series statement to indicate the presence of the stamp or label (see 8.29.31.1). 8.01.1.5 If a Series statement is not present in the manifestation, but bibliographic or reference sources provide evidence that the book was issued as part of a publisher’s series, do not supply a Series statement. Rather, provide the series information in a Note on series statement if considered important (see 8.29.31.1). ## 8.01.2 Element order 8.01.2.1 General element order, punctuated and capitalized according to ISBD: Title of series : other title information of series / statement of responsibility relating to series ; numbering within sequence 8.01.2.2 If both a main series title and a subseries title appear on the resource, record the title of the main series first, followed by the title of the subseries. Title of series. Title of subseries 8.01.2.3 If there are parallel series titles, transcribe each subseries after the series title to which it relates. Title of series. Title of subseries = Parallel title of series. Parallel title of subseries 8.01.2.4 Transcribe Other title information of series relating to the series following the Title of series. If there are parallel series titles, transcribe the other title information after the series title to which it relates. Title of series : other title information of series = Parallel title of series Title of series : other title information of series = Parallel title of series : parallel other title information of series 8.01.2.5 Transcribe a Statement of responsibility relating to series following the Title of series. Title of series : other title information of series / statement of responsibility relating to series 8.01.2.6 If there are parallel series titles but the Statement of responsibility relating to series appears in only one language or script, transcribe the statement of responsibility after the last parallel title, following any other title information associated with the title. Title of series : other title information of series = Parallel title of series : parallel other title information of series / statement of responsibility relating to series. 8.01.2.7 If the statement of responsibility appears in more than one language or script, transcribe each statement after the Title of series (or other title information) to which it relates. Title of series : other title information of series / statement of responsibility relating to series = Parallel title of series : parallel other title information of series / parallel statement of responsibility relating to series 8.01.2.8 If series numbering is present, transcribe it as the last element in the Series statement. Title of series : other title information of series / statement of responsibility relating to series ; numbering within sequence 8.01.2.9 If there are parallel series titles and the series numbering also appears in more than one language or script, transcribe each number after the series title to which it relates (following any other title information or any statement of responsibility associated with the title). Title of series : other title information of series / statement of responsibility relating to series ; numbering within sequence = Parallel title of series : parallel other title information of series / parallel statement of responsibility relating to series ; numbering within sequence [relating to Parallel title of series] 8.01.2.91 If the series numbering appears only once, transcribe it after the series title to which it relates. Title of series ; numbering within sequence [relating to Title within series] = Parallel title of series 8.01.2.92 However, if the numbering relates to all, more than one, or none of the series titles, transcribe it at the end of the Series statement. Title of series = Parallel title of series ; numbering within sequence [relating to both the Title of series and the Parallel title of series] ## 8.01.3 Form and order of information 8.01.3.1 Transcribe Series information in the form and order in which it is presented on the preferred source of information, unless specifically instructed otherwise (see Transcription, 0.4.94.05). ## Contents: 8.2 — Series statement 8.21 — Title of series 8.215 — Parallel title of series 8.23 — Other title information of series 8.235 — Parallel other title information of series 8.25 — Statement of responsibility relating to series 8.255 — Parallel statement of responsibility relating to series 8.27 — Numbering within sequence 8.29 — Note on series statement	2021-09-22 21:34:01	{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8431812524795532, "perplexity": 3170.8204520169033}, "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-2021-39/segments/1631780057388.12/warc/CC-MAIN-20210922193630-20210922223630-00284.warc.gz"}
https://gitter.im/thunder-project/thunder/archives/2016/02/23	These are chat archives for thunder-project/thunder 23rd Feb 2016 Jeremy Freeman @freeman-lab Feb 23 2016 18:50 @jwittenbach @sofroniew continuing our chat here... i think a function for generating a movie from an images object would great Jason Wittenbach @jwittenbach Feb 23 2016 18:53 Are you thinking something that just does time/space downsampling via reduceByKey (time) and map (space)? another option would be to make the movie function generic — it takes whatever Images object you has and makes a full-res movie and then provide some other functions to let the user do the preprocessing necessary to get that Images object down to a size where you can make a reasonable movie out of it a lot of the Series preprocessing that you might want to do before making a movie (rolling avg for down-sampling, computing $\Delta F/F$, etc) are best doen of Blocks so I might be nice to provide some of these as methods on Blocks…but then we get into this problem where a lot of them will be things that we already have on Series Jeremy Freeman @freeman-lab Feb 23 2016 18:57 cool yeah, so i kinda think this should be a standalone thunder-movie module, it's getting very specific and makes assumptions beyond thunder's core objects (e.g. an images object isn't neccessarily temporal) i'd think of it as a function that takes an images object and returns a numpy array, with a few options to control different kinds of downsampling Jason Wittenbach @jwittenbach Feb 23 2016 18:59 makes sense; that’s nice and contained and easily defined Jeremy Freeman @freeman-lab Feb 23 2016 19:00 yeah eaxctly Jason Wittenbach @jwittenbach Feb 23 2016 19:00 any fancy would just have to be done elsewhere the downside is that, the representation that you need to do some of those preprocessing steps is perfect for doing things like temporal downsampling so you loose a little efficiency by needing to come back to Images each time well, maybe more than a little Jeremy Freeman @freeman-lab Feb 23 2016 19:02 i thought you would do it all in blobks Jason Wittenbach @jwittenbach Feb 23 2016 19:03 right Jeremy Freeman @freeman-lab Feb 23 2016 19:03 so thunder-movie would take an images object, go to blocks, do suff, then go to images Jason Wittenbach @jwittenbach Feb 23 2016 19:03 but what if I want to do $\Delta F/F$ too? and not just downsample Jeremy Freeman @freeman-lab Feb 23 2016 19:03 well, can either just write those methods to work on blocks within thunder-movie Jason Wittenbach @jwittenbach Feb 23 2016 19:04 true Jeremy Freeman @freeman-lab Feb 23 2016 19:04 or factor their implementation in thunder.series to be exportable like from thunder.series import detrend and then use that in the movie code Jason Wittenbach @jwittenbach Feb 23 2016 19:04 it might be nice to have thunder-movie allow you pass in an optional function that will be applied across all time series within each block Jeremy Freeman @freeman-lab Feb 23 2016 19:05 yeah definitely Jason Wittenbach @jwittenbach Feb 23 2016 19:05 and yeah, if we could get some of the Series methods exported, those would be great candidates for it Jeremy Freeman @freeman-lab Feb 23 2016 19:05 yup not exactly sure the best way to do that though want to create a repo for thunder-movie and sketch out the API in a readme? Jason Wittenbach @jwittenbach Feb 23 2016 19:06 yeah, will do Jeremy Freeman @freeman-lab Feb 23 2016 19:06 sweet! Jason Wittenbach @jwittenbach Feb 23 2016 19:07 I have some notebooks that do a lot of this in a disjointed way already, so I’ll throw one of those in there for reference Jason Wittenbach @jwittenbach Feb 23 2016 19:46 @freeman-lab would it be possible to up my permissions in thunder-project so that I’m able to create the repo? or you could create it and then I’ll commit to it. whichever works best. Jeremy Freeman @freeman-lab Feb 23 2016 20:14 oh totally, on it done Jason Wittenbach @jwittenbach Feb 23 2016 21:44 sweet, thanks! Davis Bennett @d-v-b Feb 23 2016 21:50 this sounds great	2019-06-25 19:40:01	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 2, "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": 0.28649255633354187, "perplexity": 4115.2273542216435}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-26/segments/1560627999946.25/warc/CC-MAIN-20190625192953-20190625214953-00298.warc.gz"}
https://quizlet.com/387082159/astr-110-exam-2-flash-cards/	# ASTR 110 Exam #2 4.0 (2 reviews) Understand the concept of the EM spectrum, and similarities & differences types of radiation Click the card to flip 👆 1 / 30 Terms in this set (30) COMPLETE SPECTURM OF LIGHT WAVELENGTHS - Sunlight passes through prism (rainbow color appears), prism separates light into component colors (light of each color is described by wavelength) - ROY G BIV (longer to shorter wavelengths) - Infrared waves: longer wavelength light energy, invisible to humans (can be sensed by thermometer - Ultraviolet: Beyond blue side of spectrum, special instruments can detect shorter wavelengths light energy - Longest wavelength (radio waves) to shortest wavelength (gamma rays) - Visible light: portion of EM spectrum our eyes are sensitive to (small fraction of EM spectrum) HOT, OPAQUE GASSES, LIQUIDS, & SOLIDS ALL EMIT BLACKBODY SPECTRA Will absorb all wavelengths & reflect one (the one we see) - Peak of blackbody spectrum is temp. of object determines the wavelength where spectrum has greatest brightness - Peak of wavelength in blackbody spectrum tells astronomers TEMP OF STAR Also, can tell temp by looking at color that's most INTENSE Continuous Spectrum: 1. Black-and-white photograph of a continuous spectrum is completely white b/c light energy at all wavelengths exposes the film Bright line or Emission spectrum: 1. Lines have same color as corresponding position in a continuous spectrum 2. Dark areas correspond to low brightness on graph whereas lines correspond to peaks in graph (high brightness) Absorption Spectrum 1. Shows a white background with a series of black vertical lines corresponding to wavelengths with NO LIGHT ENERGY o Graph of the brightness vs. wavelength of an absorption spectrum shows familiar blackbody curve interrupted by series of downward spikes associated with absorption lines o Wavelengths of absorption lines correspond exactly to wavelengths of bright lines produced by hot gasses o Absorption spectrum results from continuous spectrum created by star having particular wavelengths removed as continuous spectrum of light passes through thin outer atmosphere of the star o Wavelengths that are removed by gasses in outer atmosphere are exactly those that same gasses would produce if heated and observed directly Reflecting: largest telescopes, use mirrors instead of lenses, image is not as good as glass lens, mirrors are lighter, easier to make, easier to hold in place, and less expensive - Newtonian reflector used curved mirror in back of telescope to focus light onto a second, smaller mirror near the front telescope, the secondary mirror aims light to a hole in the side of the tube where the eyepiece is located EXAMPLE: SALT 11 m (South African large telescope) Refracting: uses only glass lens for magnification, largest lens is at end of telescope closest to object being viewed and it's called objective lens. Second lens is what astronomer looks through and it's the eyepiece lens - Give clearest and most crisp images Three main concerns 1. b/c different colors of light pass through glass lenses differently, stars viewed through a refractor can have circular rainbows around them - Adding a special coating on objective lens can reduce this chromatic aberration effect 2. Weight- hard to hold and heavy 3. Expense- big glass lenses have no flaws, difficult to make and expensive EXAMPLE: Chabot Space & Science Center in Oakland Cali (20-inch telescope) Similarities: both concentrate as much light as possible & allow you to see things far Differences: Reflector- easily make w/ large radius thus making better solution Refractor- uses a lens as an objective to form an image	2023-03-31 06:47:34	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 14, "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": 0.484886109828949, "perplexity": 5075.933428219481}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296949573.84/warc/CC-MAIN-20230331051439-20230331081439-00048.warc.gz"}
https://quizplus.com/quiz/154654-quiz-9-applications-of-cost-theory	# Managerial Economics Study Set 8 ## Quiz 9 :Applications of Cost Theory Question Type In making an entry/exit decision, if competitive pressure is projected to force the price down to $300, what is the break-even unit sales volume this company should have projected as part of its business plan before entering this market and should reconsider each time it considers leaving (exiting) this business altogether Free Essay Answer: Answer: Following table lists the cost of flying a charter flight from Baltimore to Las Vegas (and return next day) on a seven year old Boeing 737-8000 with 120 occupied seats. The break-point occurs when total revenue equals total costs. Break-even unit sales are calculated as …… (1) Where, FC are the fixed cost for the round-trip, P is the price per seat and VC are the variable cost for a round trip. In case of charter flight, variable cost would vary with the number of occupied seat, that is, food service with seat-by-seat purchase and JIT delivery at each departure. The number of seats occupied is 120 and the variable cost for 120 passengers is$2,400. Thus, the variable cost of sending one more person aboard a charter flight is each way, or $40 for a round-trip. All the costs except food service are fixed costs. Thus, the fixed cost of a round-trip is Substitute , and in (1) Thus, the break-even unit sales are . Tags Choose question tag What type of cost-output relationship (linear, quadratic, cubic) is suggested by these statistical results Free Essay Answer: Answer: The following regression equation is constructed on the basis of the statistical data on 109 selected high schools: Here, is the operating expenditure per student, is the students' average daily attendance, is the average teacher salary, is the number of credit units offered, is the average number of courses taught per teacher, is the change in enrollment between 1957 and 1960, and is the percentage of classrooms built after 1950. An asterisk over t-values in parenthesis indicates that the result is statistically significant at 0.01 level. It is evident from the information given that the term is statistically significant at 0.01 level. Thus, a quadratic average cost-output relationship, that is, a cubic total cost-output relationship is suggested by these statistical results. Tags Choose question tag McKee Corporation has annual fixed costs of$12 million. Its variable cost ratio is.60. a. Determine the company's break-even dollar sales volume. b. Determine the dollar sales volume required to earn a target profit of $3 million. Free Essay Answer: Answer: a) Company's breakeven dollar sales volume. McKee Corporation has annual fixed cost of$12,000,000 The ratio of variable cost is 0.60; hence, the price is $1. Required amount is . b) Sale volume required to earn the target profit of$3 million. Target volume is . Tags Choose question tag What are the variable costs for the decision to send one more person aboard a charter flight that is already 80 percent booked Essay Tags Choose question tag A study of the costs of electricity generation for a sample of 56 British firms in 1946-1947 yielded the following long-run cost function: 16 AVC = 1.24 +.0033Q +.0000029Q 2 .000046QZ .026Z +.00018Z 2 where AVC = average variable cost (i.e., working costs of generation), measured in pence per kilowatt-hour (kWh). (A pence was a British monetary unit equal, at that time, to 2 cents U.S.) Q = output; measured in millions of kWh per year Z = plant size; measured in thousands of kilowatts a. Determine the long-run variable cost function for electricity generation. b. Determine the long-run marginal cost function for electricity generation. c. Holding plant size constant at 150,000 kilowatts, determine the short-run average variable cost and marginal cost functions for electricity generation. d. For a plant size equal to 150,000 kilowatts, determine the output level that minimizes short-run average variable costs. e. Determine the short-run average variable cost and marginal cost at the output level obtained in Part (d). 16 Johnston, Statistical Cost Analysis, Chapter 4, op. cit. Essay Tags Choose question tag A study of 86 savings and loan associations in six northwestern states yielded the following cost function: 15 where C = average operating expense ratio, expressed as a percentage and defined as total operating expense ($million) divided by total assets ($ million) times 100 percent Q = output; measured by total assets ($million) X 1 = ratio of the number of branches to total assets ($million) Note: The number in parentheses below each coefficient is its respective t-statistic. a. Which variable(s) is(are) statistically significant in explaining variations in the average operating expense ratio b. What type of cost-output relationship (e.g., linear, quadratic, or cubic) is suggested by these statistical results c. Based on these results, what can we conclude about the existence of economies or diseconomies of scale in savings and loan associations in the Northwest 15 Holton Wilson, "A Note on Scale Economies in the Savings and Loan Industry," Business Economics (January 1981), pp. 45-49. Essay Tags Choose question tag Based on the results of this study, what can we conclude about the existence of economies or diseconomies in operating a public high school Essay Tags Choose question tag Based on the scatter diagram in Question, what kind of mathematical relationship would appear to exist between enrollment and operating expenditures per student In other words, do operating expenditures per student appear to (i) be constant (and independent of enrollment), (ii) follow a linear relationship as enrollment increases, or (iii) follow some sort of nonlinear U-shape (possibly quadratic) relationship as enrollment increases As part of this study, the following cost function was developed: C = f(Q, X 1 , X 2 , X 3 , X 4 , X 5 ) where C = operating expenditures per student in average daily attendance (measured in dollars) Q = enrollment (number of students in average daily attendance) X 1 = average teacher salary X 2 = number of credit units ("courses") offered X 3 = average number of courses taught per teacher X 4 = change in enrollment between 1957 and 1960 X 5 = percentage of classrooms built after 1950 Variables X 1 , X 2 , and X 3 were considered measures of teacher qualifications, breadth of curriculum, and the degree of specialization in instruction, respectively. Variable X 4 measured changes in demand for school services that could cause some lagging adjustments in cost. Variable X 5 was used to reflect any differentials in the costs of maintenance and operation due to the varying ages of school properties. Statistical data on 109 selected high schools yielded the following regression equation: Notes: The numbers in parentheses are the t-scores of each of the respective (b) coefficients. An asterisk (*) indicates that the result is statistically significant at the 0.01 level. Question Plot the data in columns B and C in an output (enrollment-) cost graph and sketch a smooth curve that would appear to provide a good fit to the data. Essay Tags Choose question tag Holding constant the effects of the other variables (X 1 through X 5 ), determine the enrollment level (Q) at which average operating expenditures per student are minimized. (Hint: Find the value of Q that minimizes the ( C/ Q function.) Essay Tags Choose question tag Charter contracts are negotiable, and charter carriers receive many contract offers that do not promise $300 prices or 80-percent-full planes. Should the airline accept a charter flight proposal from a group that offers to guarantee the sale of 90 seats at$250 Why or why not Essay Tags Choose question tag Referring to Exercise again: a. Holding constant the effects of branching (X 1 ), determine the level of total assets that minimizes the average operating expense ratio. b. Determine the average operating expense ratio for a savings and loan association with the level of total assets determined in Part (a) and 1 branch. Same question for 10 branches. Exercise A study of 86 savings and loan associations in six northwestern states yielded the following cost function: 15 where C = average operating expense ratio, expressed as a percentage and defined as total operating expense ($million) divided by total assets ($ million) times 100 percent Q = output; measured by total assets ($million) X 1 = ratio of the number of branches to total assets ($million) Note: The number in parentheses below each coefficient is its respective t-statistic. a. Which variable(s) is(are) statistically significant in explaining variations in the average operating expense ratio b. What type of cost-output relationship (e.g., linear, quadratic, or cubic) is suggested by these statistical results c. Based on these results, what can we conclude about the existence of economies or diseconomies of scale in savings and loan associations in the Northwest 15 Holton Wilson, "A Note on Scale Economies in the Savings and Loan Industry," Business Economics (January 1981), pp. 45-49. Essay Tags Choose question tag Assuming that all other factors remain unchanged, determine how a firm's breakeven point is affected by each of the following: a. The firm finds it necessary to reduce the price per unit because of increased foreign competition. b. The firm's direct labor costs are increased as the result of a new labor contract. c. The Occupational Safety and Health Administration (OSHA) requires the firm to install new ventilating equipment in its plant. (Assume that this action has no effect on worker productivity.) Essay Tags Choose question tag Again, holding constant the effects of the other variables, use the C/ Q function to determine, for a school with 500 students, the reduction in per-student operating expenditures that will occur as the result of adding one more student. Essay Tags Choose question tag If one were trying to decide whether to operate (fly) or not fly an unscheduled round-trip charter flight, what would be the total direct fixed costs and variable costs of the flight Essay Tags Choose question tag What variables (other than enrollment) would appear to be most important in explaining variations in operating expenditures per student Essay Tags Choose question tag What are the total contributions of the charter flight with 90 seats at \$250 per seat Essay Tags Choose question tag Again, holding the other variables constant, what would be the saving in perstudent operating expenditures of an increase in enrollment from 500 to 1,000 Essay Tags Choose question tag Plot the data in columns B and C in an output (enrollment-) cost graph and sketch a smooth curve that would appear to provide a good fit to the data. Essay	2022-08-18 08:50:05	{"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": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5099771618843079, "perplexity": 2847.5996023927555}, "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/1659882573172.64/warc/CC-MAIN-20220818063910-20220818093910-00152.warc.gz"}
http://accessemergencymedicine.mhmedical.com/content.aspx?bookid=454&sectionid=40199487	Chapter 81 Inhalant abuse is defined as the deliberate inhalation of vapors for the purpose of changing one's consciousness or becoming "high." It is also referred to as volatile substance abuse and was first described in 1951.32 Inhalants are appealing to adolescents because they are inexpensive, readily available, and sold legally. Initially, inhalant abuse was viewed as physically harmless, but reports of "sudden sniffing death" began to appear in the 1960s.9 Shortly thereafter, evidence surfaced of other significant morbidities, including organic brain syndromes, peripheral neuropathy, and withdrawal. The demographics of inhalant abuse differ markedly from those of other traditional substances of abuse. The 2003 National Survey on Drug Use and Health found that more than 2.6 million youths between the ages of 12 and 17 years used inhalants at least once in their lifetime.122 Inhalants were the most frequently reported illicit xenobiotics used in the past year by 12- and 13-year-old adolescents, and youths aged 12 to 17 years continued to report higher rates of past-year use than did adults aged 18 years and older. The lifetime prevalence of inhalant use peaked among 8th graders at 15.6%. The median age of first use is 13 years.7 A worrisome trend reported by the 2007 Monitoring the Future Study is that perceived risk of even one-time use of an inhalant has fallen steadily since 2001.89 Although long considered to be a problem among boys, there has been a steady increase of inhalant abuse among girls, and their lifetime prevalence now equals that of boys.11 In the United States, the problem is greatest among children of lower socioeconomic groups, and non-Hispanic white adolescents are the most likely and black adolescents the least likely to use inhalants.85 Although inhalant use is a problem in both urban and rural communities, its prevalence is higher in rural settings.116 This may relate to the easier access that teens in urban areas have to other drugs of abuse. Inhalant abuse includes the practices of sniffing, huffing, and bagging. Sniffing entails the inhalation of a volatile substance directly from a container, as occurs with airplane glue or rubber cement. Huffing involves pouring a volatile liquid onto fabric, such as a rag or sock, and placing it over the mouth and/or nose while inhaling and is the method used by more than 60% of volatile-substance abusers.85Bagging refers to instilling a solvent into a plastic or paper bag and rebreathing from the bag several times; spray paint is among the xenobiotics commonly used with this method. A newly reported form of abuse, "dusting," refers to the inhalation of compressed air cleaners containing halogenated hydrocarbons (eg, CRC Dust Off™), marketed for cleaning computers and electronics equipment. Reports of such exposures, including death and ventricular dysrhythmias, doubled in 2005 and tripled in 2006, yet dusting is not perceived to be harmful by users and, surprisingly, many users do not consider it a ... 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Subscription Options AccessEmergency Medicine Full Site: One-Year Subscription Connect to the full suite of AccessEmergency Medicine content and resources including advanced 8th edition chapters of Tintinalli’s, high-quality procedural videos and images, interactive board review, an integrated drug database, and more.	2017-03-27 06:44:39	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.19449609518051147, "perplexity": 4584.496854899018}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-13/segments/1490218189462.47/warc/CC-MAIN-20170322212949-00648-ip-10-233-31-227.ec2.internal.warc.gz"}
https://aliquote.org/micro/2018-10-02-08-22-00/	The ESS package somewhat sucks with Stata and font-lock. I have enabled the ado-mode again and it is so much better. I wish it could be merged into ESS directly. #emacs #stata	2018-10-18 20:25:58	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 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": 0.4532897174358368, "perplexity": 5277.466203698136}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-43/segments/1539583512014.61/warc/CC-MAIN-20181018194005-20181018215505-00408.warc.gz"}
https://www.txcorp.co.uk/images/docs/vsim/latest/html/VSimUserGuide/troubleshooting_es_sim.html	# Troubleshooting Electrostatic Simulations ## Simulation Crashes at Startup with PEC Dirichlet Boundaries PEC objects in the simulation must be entirely inside of the simulation grid in order to ensure proper problem setup. ## The Simulation Does Not Finish Properly The most common cause of crashes is improperly set up particle boundaries. The particle boundaries must completely surround the space in which particles are loaded. Otherwise particles can drift out of the grid and try to reference fields that do not exist. This leads to a Vorpal segmentation fault. Another possible reason for an electrostatic simulation not finishing properly is that a particle has crossed more than one cell in a time step. This could allow the particle to pass through a particle sink without being absorbed. • One solution is to reduce the duration of the time step. • Another solution is to limit the number of cells a particle can cross in one time step by artificially reducing the velocity of high speed particles. See VSimReferenceManual: Text Setup: Species: maxcellxing It could be that the definition of the Particle Species is incorrect. The following Species input block is not defined correctly: <Species electrons> kind = nonRelBoris emField = myZeroField ... </Species> • The problem: The input block does not specify mass and charge. • The result: The simulation runs normally with no complaints. The default mass and charge are those of a positron. • The solution: Include the mass and charge of your species every time they are defined. ## The Electrostatic Solver Does Not Converge If the electrostatic solver does not converge, this often indicates a problem with the setup. The matrix can be singular in a fully periodic system due to the failure to specify the value of the potential at one point.	2019-10-19 10:54:32	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 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": 0.7771904468536377, "perplexity": 955.381323836909}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-43/segments/1570986692723.54/warc/CC-MAIN-20191019090937-20191019114437-00066.warc.gz"}
https://socratic.org/questions/is-the-following-sequence-arithmetic-if-so-identify-the-common-difference-2-9-2-	Is the following sequence arithmetic? If so, identify the common difference. 2.9, 2.7, 2.5, 2.3, . . . yes, 2 yes, -0.2 no yes, 0.3 Apr 14, 2017 It is an arithmetic sequence with common difference $- 0.2$. Explanation: In an arithmetic sequence, difference between a term and it's immediately preceding term is always constant and is known as common difference. Here, $2.7 - 2.9 = - 0.2$ $2.5 - 2.7 = - 0.2$ and $2.3 - 2.5 = - 0.2$ As the difference is constantly $- 0.2$, It is an arithmetic sequence with common difference $- 0.2$.	2020-08-06 22:26:59	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 6, "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": 0.6659448742866516, "perplexity": 677.28782980597}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-34/segments/1596439737039.58/warc/CC-MAIN-20200806210649-20200807000649-00365.warc.gz"}
https://www.gradesaver.com/textbooks/math/precalculus/precalculus-10th-edition/chapter-6-trigonometric-functions-6-3-properties-of-the-trigonometric-functions-6-3-assess-your-understanding-page-395/100	## Precalculus (10th Edition) $\theta=k\pi$, where k is an integer Since $\csc(\theta)=\frac{1}{\sin\theta}$, then it is undefined when the denominator (which is $\sin{\theta}$) is $0$. Recall that $\sin\theta=0$ if $\theta=k\pi$, where $k$ is an integer. Therefore, the function is undefined when $\theta=k\pi$, where k is an integer.	2021-11-27 21:02:00	{"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": 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": 0.9897059202194214, "perplexity": 124.41807718639484}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-49/segments/1637964358233.7/warc/CC-MAIN-20211127193525-20211127223525-00575.warc.gz"}
https://codeforces.com/blog/entry/9145	### Igor_Kudryashov's blog By Igor_Kudryashov, 7 years ago, translation, , 353A - Domino Let's denote the sum of numbers on the upper halves of pieces as s 1, and the sum on the lower halves — s 2. If this sums are even, than the answer is obviously 0. Note, that if the numbers on both halves of piece have the same parity, than parity of s 1 and s 2 won't change after rotation this piece. If the numbers on halves have different parity, than parities of both s 1 and s 2 will change after rotation. Therefore, if s 1 and s 2 have different parities, than the answer is - 1. If both s 1 and s 2 are odd, than we should check, if there is a piece with numbers of different parities. If so, the answer is 1, otherwise, the answer is - 1. 353B - Two Heaps Let's say for shortness, that we put numbers, that are painted on cubes, in piles, instead of cubes themselves. Note, that the answer is the product of c 1... c 2, where c i is the number of different numbers in the i-th pile. Let's consider that all numbers are different. In this case the answer is n 2. Now, let's suppose that we have two equal numbers and all the other are different. Then, if we put them in different piles, the answer will be n 2, but if we put them in one — n·(n - 1). Obviously, the first case give greater product. Thinking in similar manner, you can conclude, that we should do the following. Take numbers, that appear more than once, and put one of them in the first pile, one of them in the second pile and the other put aside. After that, divide the numbers, that appears once, in two equal part and put the first part in the first pile and second part in the second pile. Finally, take the numbers, that we put aside, and separate them in two pile in any kind. 353C - Find Maximum Let's see on the highest bit of m. If it equals to zero, than for any there is a zero on the (n - 1)-th position, so a n - 1 doesn't affect the answer and we can put it aside and find the answer for smaller number of elements. If the highest bit of m equals to 1, than a n - 1 for some x will present in f(x), but for some will not. Let's consider such x, that a{n - 1} will present in f(x) with zero coefficient. It is obvious that . In this case f(x) will have maximum value when x = 2 n - 1 - 1. Try to update the answer by this value. Now we should analyze the case, when , find the maximum value of f(x) for all such x and try to update the answer by this value. Let's note, that in all such x there is 1 in (n - 1)-th position. Therefore we can find the maximum value of f(y) for all and add a n - 1 to it. 353D - Queue Note that if there are some girls in the begining of the line, they will never move. So let's remove them and will consider that the first schoolchildren in the line is a boy. Also note, the relative order of the girls doesn't change. Let's calculate for each girl such moment of time t i, that after it she won't move forever. Note, that for i-th girl t i ≥ t i - 1. Let's calculate t i in order from left to right. Let's denote y i is the position in the line where i-th girl will stop, ans x i is her current position. Therefore it is needed x i - y i second for girl to reach her finish position. So if x i - y i > t i - 1, then t i = x i - y i. Let's manage the case when x i - y i ≤ t i. The girl with number (i - 1) will be on y i-th position by t i - 1-th second, so t i ≥ t i - 1 + 1. Let's consider such moment of time p, when i-th girl stand right after (i - 1)-th, but not on y i-th position. After that, in (p + 1)-th moment of time (i - 1)-th girl and the boy standing in front of her will swap their positions, but i-th girl will save her position. Then since p + 2-th second till t i - 1 both girls will change their positions. Finally, at (t i - 1 + 1) - th second i-th girl will occupy her position. Therefore, t i = t i - 1 + 1 in this case. 353E - Antichain Let's divide our graph on chains. Denote chain as the maximal sequence of the edges, which have the same direction. If there are more than 2 edges in each chain, then the answer is the number of such chains. If there is a chain containing only one edge (u, v), then brute vertex, which we will take in maximal antichain (also consider the case, when we take none of them). Let's suppose we brute the vertex v. After that we put aside this vertex and all vertices, which is comparable with v. In received graph we can find the size of the maximal antichain by uning dynamic programming with O(n) time complexity. Let's show how to do this. Write all remaining vertices in line and renumerate them from left to right by the numbers from 1 to k. After that we are going to calculate d i — the size of the maximal antichain among vertices with numbers j ≥ i. So if i > k, then d i = 0. In other case we can skip i-th vertex and try to update d i by the value of d i + 1. Also we can try to take i-th vertex to the answer. In this case we should skip all vertices, that are reachable from i-th, or the vertices, from which we can reach i-th, take the answer from the next vertex, add 1 to it and try to update the answer. • +32 » 7 years ago, # \| +9 translation to English? • » » 7 years ago, # ^ \| +2 it will come soon) » 7 years ago, # \| 0 In fact we have the strict inequality: for every ti > ti - 1 » 7 years ago, # \| ← Rev. 2 → +1 problem B can someone tell me why this code gives wrong answer ? the expected answer is so close !! http://codeforces.com/contest/353/submission/4736607 • » » 7 years ago, # ^ \| 0 Read the comments here and you'll understand your mistake. » 7 years ago, # \| 0 Good alanysis for problem D » 7 years ago, # \| +4 I solve the problem E using Matching • » » 7 years ago, # ^ \| +1 That's too slow. There is no (not even closely) linear algorithm for matching, so you'd TLE. But in this case, the graph we're given is composed purely of a cycle of cliques connected in some vertices, on which a greedy idea works well. • » » » 7 years ago, # ^ \| ← Rev. 2 → +1 But he did. Look at this nice solution.It didn't TLE probably because of special structure of the graph. • » » » 7 years ago, # ^ \| ← Rev. 2 → 0 During the contest, I did it using matching too, via Hopcroft-Karp algorithm, and some friend of mine did the same. Here are our submissions:47350104735373My question is, why this doesn't TLE? Why this graph particular structure allow that? In Wikipedia I could only find that "for random graphs it runs in near-linear time.". Are the tests weak? » 7 years ago, # \| ← Rev. 3 → +6 I actually really like this round especially Problem D. It took me a before I could build up all the ideas to solve it. My track of thoughts is like: 1. The answer is the number of seconds needed to move the last girl to the finish place. 2. The number of seconds needed for girl i is at least the number of seconds needed for girl i - 1 + 1. 3. The number of seconds needed for girl i is at least the distance between the initial position and the finish position. 4. Each girl i will move as soon as she can, so she will be right at the back of the girl i - 1 or take at most dist(i) seconds to follow the girl i - 1. so we can calculate the T(i) = max(T(i-1) + 1, dist(i)) and the answer is just T(lastGirl). • » » 7 years ago, # ^ \| 0 how you computed dist[i]? • » » » 7 years ago, # ^ \| 0 dist[i] is just the distance between the initial position and the finish position (You already know the finish position of each girl). » 7 years ago, # \| +6 E can be solved by greedy, also in O(N). Let's count the lengths (as number of vertices) of individual chains (counted cyclically).Every chain of length 3 adds 1 to the answer. That's because we could always choose a vertex from it that doesn't belong to any other chain instead of one of the side vertices that do belong to one, or instead of not choosing any vertex at all.Now, we're left with either just a cyclic sequence of chains of length 2 (input 01010101...01 or reverse), for which the answer is the (length of the input string)/2, or some sequences of chains of length 2, each bordered by chains of length 3 on both sides. Greedy approach works here, too: no 2 chosen vertices from any such sequence can be adjacent, and we can't choose from the border. That means we can choose at most (length of such sequence-1)/2. » 7 years ago, # \| ← Rev. 2 → 0 Was beaten hardly by problem B!! :( A great round though, I have learnt a lot!I use a similar approach in probem D, i try to move the boys though: for each boy, we know his initial and final position, thats the minimum time for this boy to move Actually it's the # of girls initially behind this boy for each boy, if in the moving progress he has to stand and wait next boy to move, actually he can start in a later time so that he won's stop until arrive his final position, I call this the start time of the boy For step 2 we can greedily precompute, and the ans is max(start time + # of girls behind) among all boysHere's my code: 4742830 • » » 7 years ago, # ^ \| 0 I use same basic idea, the difference is only on implementation, I didn't store that much information. here is my solution: 4735037 » 7 years ago, # \| ← Rev. 2 → 0 http://codeforces.com/contest/353/submission/4745339 this code is wa for "011001" but it is accepted. » 7 years ago, # \| -11 Get it: GOOGLE TRANSLATE SUCKS. Russian editorials discriminate against non Russian-speaking people and go against what is written here, which I quote: "Codeforces' aim is to offer you a convenient platform to organize, run and discuss programming contests." If this competition was also available in English, an English editorial should also be here. » 7 years ago, # \| 0 I get wrong answer at Div2 D — Queue with this source, but I can't figure out why because of the large input. Could somebody give me a hint? » 6 years ago, # \| 0 Can anyone explain problem E in more simple words.Thanks in advance . » 4 months ago, # \| 0 for C why can we not just take prefix sum of all elements behind the right most index having a set bit • » » 4 months ago, # ^ \| ← Rev. 5 → 0 consider this case:n=4, m=1011, arr= (17 9 2 50 )	2020-08-15 08:00:26	{"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": 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": 0.5230836272239685, "perplexity": 709.2621728773212}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-34/segments/1596439740733.1/warc/CC-MAIN-20200815065105-20200815095105-00540.warc.gz"}
https://www.azdictionary.com/definition/gnomonology	# gnomonology definition • noun: • A treatise on gnomonics. • A treatise on gnomonics. • A treatise on gnomonics. • A treatise on gnomonics. • A treatise on dialing. • A treatise on dialing. • A treatise on gnomonics. • A treatise on gnomonics. • A treatise on dialing. • A treatise on gnomonics. • A treatise on gnomonics. • A treatise on dialing. • A treatise on gnomonics. • A treatise on gnomonics. • A treatise on dialing. • A treatise on gnomonics. • A treatise on gnomonics. • A treatise on gnomonics. • A treatise on gnomonics. • A treatise on gnomonics. • A treatise on gnomonics. • A treatise on gnomonics. • A treatise on gnomonics. • A treatise on dialing. • A treatise on dialing. • A treatise on gnomonics. • A treatise on gnomonics. • A treatise on dialing. • A treatise on gnomonics. • A treatise on gnomonics. • A treatise on dialing. • A treatise on dialing. • A treatise on dialing.	2017-05-29 13:26:39	{"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": 0.9310033321380615, "perplexity": 9813.673640915582}, "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/1495463612327.8/warc/CC-MAIN-20170529130450-20170529150450-00426.warc.gz"}
https://scicomp.stackexchange.com/tags/electromagnetism/hot	# Tag Info ## Hot answers tagged electromagnetism 10 Here I have an example: x = linspace(-5,5,100); y = linspace(-5,5,100); z = linspace(-5,5,100); [X, Y, Z] = meshgrid(x, y, z); Ex = sin(2pi/5Z); Ey = 0X; Ez = 0X; [Bx, By, Bz, V] = curl(X, Y, Z, Ex, Ey, Ez); Eplot = 0x; Bplot = 0x; for i=1:100 %% Integration-like procedure Eplot(i) = mean(mean(Ex(:,:,i),1),2); Bplot(i) = mean(mean(By(:,:,... 6 There are basically two methods: You can disrectize the angular part via grid points, or you can discretize it via basis expansion. I will focus on spherical symmetry here, the cylindrical case is quite similar. In the basis expansion approach, one applies the ansatz $$v(x,t) = \sum_{klm} a_{klm}(t)\,R_{klm}(r)\, Y_{lm}(\theta,\phi)$$ This is inserted ... 5 If you have simple, grid-aligned interior objects or accuracy is not crucial then stick with the method you have described. If you need to accurately represent arbitrary boundary shapes then you're probably best off moving to a (more complicated) finite element approach on an unstructured grid. You have described the simplest approach to solving this ... 5 I've found that if I reduce the radial domain to $8 \leq r \leq 20$, the condition number drops to ~10,000. This makes me think I need to scale my problem. I'm not sure how to do this, however, and I need to do it right. Nondimensionalization is partly repeated application of the chain rule, and partly art. The goal is to make as many quantities in your ... 5 A classic paper for evaluation of the integrals commonly present in computational electromagnetics (EM) is: D. R. Wilton, S. M. Rao, A. W. Glisson, D. H. Schaubert, O. M. Al-Bundak, and C. M. Butler, "Potential integrals for uniform and linear source distributions of polygonal and polyhedral domains," IEEE Trans. Antennas Propag., vol. AP-32, no. 3, pp. 276-... 5 The Maxwell system is a wave equation at heart, so your ansatz (the space where you seek solutions, the combination of your mesh and basis functions) must be able to faithfully represent waves. The Nyquist criterion sets an absolute lower limit to the "sample rate" of your mesh: two points per wavelength. In practice, you must upsample by a considerable ... 5 As a preamble, I would not expect that splitting $E$ into real/imaginary parts is very profitable. Normally, block 2x2 systems are motivated because one block of unknowns is "easier" to solve than the other in some sense (better conditioned? smaller in cardinality? etc). This is not the case for time-harmonic Maxwell, I think you'd be better off ... 4 As for any other domain, your mesh needs to be fine enough to resolve the features you have. This means that the mesh has to be finer than the geometric details of your unit cell, and it needs to be finer than the wavelengths of the waves you consider. Beyond this, the question of tets vs hexes is a minor issue. Hexes are generally more accurate, but if the ... 4 So Is it neccesary to use Poisson - Boltzmann equation if I only need to build electrostatic potential from a PQR file? No. You can use Poisson. Since you know the positions of each point charge, you know the charge distribution $\rho$, which is a sum of delta functions. You can thus solve numerically the Poisson's equation that links the charge ... 4 When you call Lapack's zgtsv, it doesn't just solve a tridiagonal system $Ax=b$. What it does first is perform an LU factorization (zgttrf) $A = LU$, where $L,U$ are lower- and upper-tridiagonal matrices, and only then proceeds to solve $LUx=b$. When you give it the lower, main, and upper diagonals of the matrix $A$, those diagonals are overwritten by the ... 4 ANSYS Maxwell is a Finite Element solver for Electromagnetism. So, I assume that you are looking for a Finite Element package that has a Python interface (or is written in Python). There are some popular options like: SfePy; FEniCS; and Agros2D. The last one provides a Python interface and a (nice) graphic interface, and is based in Hermes2D (so it ... 4 Interesting question. I would expect the Yee scheme to be indifferent to static (bias) fields induced by constant potentials. In the electrostatic case, if you have a constant electric potential $\phi$, it induces a field $\vec E = \nabla \phi$. The update equation is (modulo constants) $\frac{d}{dt}\vec B = \nabla \times \vec E$. But since $\vec E$ is a ... 3 Your questions suggest that you are new to implementing solvers for PDEs, and that you are not familiar with the usual data structures and algorithms used in this field. It would probably be very useful for you to see how other people write codes like the one you are trying to write, because you will see how they arrange data, how they represent geometry and ... 3 It's an interesting question - how does a dipole actually move? It seems to me you're not entirely sure what to expect, so we need to get a solid test case where we understand what's actually going on. On a related note, do try to add your imports and initial conditions the next time :) For now, let's assume both masses equal and opposite charges. We can ... 3 The paper: I. Anjam, J. Valdman, "Fast MATLAB assembly of FEM matrices in 2D and 3D: Edge elements", Applied Mathematics and Computation, 267, (2015), 252–263; states the following. http://www.sciencedirect.com/science/article/pii/S0096300315004191 "A finite element discretization is done in terms of edge elements, typically Raviart–Thomas elements [12] for ... 3 One issue (and this is mentioned by Mur in the first paper you linked in the comments above) is the fact that, while these edge functions provide tangential field continuity across interfaces and zero divergence within each element, they also allow for discontinuities in normal fields across interfaces. This behavior is non-physical, giving rise to ... 3 From what I understand, you'd like to see which numerical method best simulate the real physics relevant for a particular problem. MHD spans a wide scale of phenomena -- plasma physics (on length scales of ions and electrons) to ideal MHD (on the length scale relevant to accretion disks around blackhole or other compact objects). In this case, it's advisable ... 3 For this purpose you need to use a simulation software. One of the most common methods in Electromagnetics would be Finite element method, but you can also find Boundary Element Methods or Finite Difference Methods. Some common software in EM are Ansys Maxwell; and CST Studio A lot of people is also using COMSOL Multiphysics. But I would say that this ... 3 The Van der Walls radius - last column in the output - is calculated from the force field. Probably pdb2pqr and editconf uses different force fields, hence different radius. I don't use pdb2pqr, but it seems (1) AMBER99 is the default force field, though CHARMM, PARSE and TYL06 are supported. Gromacs' editconf reads force field from topology file generated ... 3 You're looking for waveguide port boundary conditions. I think the most accessible treatment is within Jin & Riley's Finite Element Analysis of Antennas and Arrays, Chapter 5. It's available on Amazon, see https://www.amazon.com/dp/0470401281/. A lot of these formulations were first introduced by Jin-Fa Lee, you can find his works in IEEE Microwave ... 3 Regarding simulation: There are several commercial solvers that might be helpful. A very popular FDTD Maxwell equation solvers for nanophotonics is Lumerical. It features opto-thermal and liquid crystal (which I would say is another keyword you should consider) simulations that should be very relevant. Another go-to multiphysics solver would be COMSOL. ... 3 I received a solution to this question from MATLAB's community. Essentially, I need to specify which contour lines to plot using the 'levels' spot on the 'contour()' command. Levels allows you to not only choose how many but which lines to plot. If you define a vector such as vlevel = linspace(20, 65, 10); and then place it in the 'levels' spot of ... 3 You don't need symbolic variables to compute the approximated potential for your Riemann sums. You can just use meshgrid to evaluate the potential in each point of interest. For the electric field, you can just compute the derivative analytically and then repeat the process for each component or compute a numerical gradient with gradient. There are better ... 3 Every iterative solver -- Jacobi, SSOR, CG, etc -- starts with an initial approximation. One often just uses the zero vector, but there is nothing wrong with using the solution of the previous time step. In fact, extrapolating from previous time steps to the current one is an even better idea -- one the authors apparently missed! For some iterative solvers, ... 3 It sounds like you are interested in a finite-element analysis, which is out of my area of expertise. But I can hopefully provide some insight from the perspective of finite-difference methods which may still have some relevance to your problem (since it is also used to solve wave equations). In general, a good rule of thumb is that specification of a ... 3 Using the typical expansion functions (1-forms/edge-elements for E, and 2-forms/facet-elements for B) the formulations are basically the same after spatial discretization and you'd expect more or less the same accuracy. I do think they express slightly different opinions about time integration. The mixed E/B formulation nudges you in the direction of ... 3 Your particle is a rounded proton (mass m = 2e-27 kg instead of 1.672e-27 kg). The equation of motion is $$\dot x=v,~~~ m\dot v = q\,v\times B,$$ where $B=(0,0,B_z)$ with $B_z=4T=4N/(m\,A)$ and $q=1e=1.602·10^{-19} C$, $C=A\,s$ This then gives for the acceleration m=2e-27 e_charge = 1.6e-19 q=+1e_charge Bz = 4 ax = q/mvyBz; ay = -q/mvx*Bz; az = 0 For ... 2 From browsing some forum posts, it appears that the numerical solution to these equations is not trivial to obtain. See, for example, this discussion (admittedly dated now). You indicate in your post that you are relatively new to numerical PDE solvers, and so the following references may be more involved than you were hoping for. In particular, they require ... 2 The main advantage of the tetrahedral vs hexahedral is the mesh generation, there are automatic mesh generators for tetrahedral meshes that give "good" elements. This process is not that easy for the hexahedral case and you will need to do a bigger effort to get a nice mesh. Nevertheless, is not a good idea to use linear tetrahedrals since they have linear ... Only top voted, non community-wiki answers of a minimum length are eligible	2021-12-07 09:04:17	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 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": 0.7028691172599792, "perplexity": 831.2423454740435}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "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-2021-49/segments/1637964363337.27/warc/CC-MAIN-20211207075308-20211207105308-00538.warc.gz"}
https://proofwiki.org/wiki/Definition:Hyperbolic_Tangent/Definition_3	# Definition:Hyperbolic Tangent/Definition 3 ## Definition The hyperbolic tangent function is defined on the complex numbers as: $\tanh: X \to \C$: $\forall z \in X: \tanh z := \dfrac {e^{2 z} - 1} {e^{2 z} + 1}$ where: $X = \set {z: z \in \C, \ e^{2 z} + 1 \ne 0}$	2019-11-19 01:05:04	{"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": 0.914191722869873, "perplexity": 8298.82820872997}, "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/1573496669868.3/warc/CC-MAIN-20191118232526-20191119020526-00182.warc.gz"}
http://openstudy.com/updates/55896cdee4b0757141181926	• anonymous In a radio station contest, the first prize is $2,000 and the second prize is$500. The probability of winning the first prize is 0.005 and the probability of winning the second prize is 0.008. The station runs this contest many, many times to increase its listening audience. What is the expected payout for each time the station offers the opportunity to win the prizes? Mathematics Looking for something else? Not the answer you are looking for? Search for more explanations.	2017-03-30 10:54:55	{"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": 0.5491654276847839, "perplexity": 558.0524859945318}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-13/segments/1490218193716.70/warc/CC-MAIN-20170322212953-00446-ip-10-233-31-227.ec2.internal.warc.gz"}
http://openstudy.com/updates/4fbb3b81e4b0556534303bdf	## Ishaan94 Group Title Integrate. $\large\int\limits_{-.9}^{.9}\lfloor x^2\rfloor\;dx$ 2 years ago 2 years ago 1. Ishaan94 Group Title It's Zero. Why didn't I use my head? :( 2. apoorvk Group Title It - is - not - zero. it is DEFINITELY zero. choose different limits. 3. Ishaan94 Group Title It's Zero because of the floor function and the limits. :/ I feel like kicking myslef, -1 in AIEEE now :( I had perfect 104 it's 99 now :(	2014-07-30 09:22:07	{"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": 0.6063859462738037, "perplexity": 9656.362953078935}, "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/1406510270313.12/warc/CC-MAIN-20140728011750-00375-ip-10-146-231-18.ec2.internal.warc.gz"}
https://clownhypothesis.com/category/platoons/	# Do Platoon Splits Mess Up Projections? Quick summary: I test the ZiPS and Marcel projection systems to see if their errors are larger for players with larger platoon splits. A first check says that they are not, though a more nuanced examination of the system remains to be conducted. First, a couple housekeeping notes: • I will be giving a short talk at Saberseminar, which is a baseball research conference held in Boston in 10 days! If you’re there, you should go—I’ll be talking about how the strike zone changes depending on where and when games are played. Right now I’m scheduled for late Sunday afternoon. • Sorry for the lengthy gap between updates; work obligations plus some other commitments plus working on my talk have cut into my blogging time. After the A’s went on their trading sprees last week at the trading deadline, there was much discussion about how they were going to intelligently deploy the rest of their roster to cover for the departure of Yoenis Cespedes. This is part of a larger pattern with the A’s as they continue to be very successful with their platoons and wringing lots of value out of their depth. Obviously, when people have tried to determine the impact of this trade, they’ve been relying on projections for each of the individual players involved. What prompted my specific question is that Jonny Gomes is one of those helping to fill Cespedes’s shoes, and Gomes has very large platoon splits. (His career OPS is .874 against left-handed pitchers and .723 against righties.) The question is what proportion of Gomes’s plate appearances the projection systems assume will be against right handers; one might expect that if he is deployed more often against lefties than the system projects, he might beat the projections substantially. Since Jonny Gomes in the second half of 2014 constitutes an extremely small sample, I decided to look at a bigger pool of players from the last few years and see if platoon splits correlated at all with a player beating (or missing) preseason projections. Specifically, I used the 2010, 2012, and 2013 ZiPS and Marcel projections (via the Baseball Projection Project, which doesn’t have 2011 ZiPS numbers). A bit of background: ZiPS is the projection system developed by Dan Szymborski, and it’s one of the more widely used ones, if only because it’s available at FanGraphs and relatively easy to find there. Marcel is a very simple projection system developed by Tangotiger (it’s named after the monkey from Friends) that is sometimes used as a baseline for other projection systems. (More information on the two systems is available here.) So, once I had the projections, I needed to come up with a measure of platoon tendencies. Since the available ZiPS projections only included one rate stat, batting average, I decided to use that as my measure of batting success. I computed platoon severity by taking the larger of a player’s BA against left-handers and BA against right-handers and dividing by the smaller of those two numbers. (As an example, Gomes’s BA against RHP is .222 and against LHP is .279, so his ratio is .279/.222 = 1.26.) My source for those data is FanGraphs. I computed that severity for players with at least 500 PA against both left-handers and right-handers going into the season for which they were projected; for instance, for 2010 I would have used career data stopping at 2009. I then looked at their actual BA in the projected year, computed the deviation between that BA and the projected BA, and saw if there was any correlation between the deviation and the platoon ratio. (I actually used the absolute value of the deviation, so that magnitude was taken into account without worrying about direction.) Taking into account the availability of projections and requiring that players have at least 150 PA in the season where the deviation is measured, we have a sample size of 556 player seasons. As it turns out, there isn’t any correlation between the two parameters. My hypothesis was that there’d be a positive correlation, but the correlation is -0.026 for Marcel projections and -0.047 for ZiPS projections, neither of which is practically or statistically significantly different from 0. The scatter plots for the two projection systems are below: Now, there are a number of shortcomings to the approach I’ve taken: • It only looks at two projection systems; it’s possible this problem arises for other systems. • It only looks at batting average due to data availability issues, when wOBA, OPS, and wRC+ are better, less luck-dependent measures of offensive productivity. • Perhaps most substantially, we would expect the projection to be wrong if the player has a large platoon split and faces a different percentage of LHP/RHP during the season in question than he has in his career previously. I didn’t filter on that (I was having issues collecting those data in an efficient format), but I intend to come back to it. So, if you’re looking for a takeaway, it’s that large platoon-split players on the whole do not appear to be poorly projected (for BA by ZiPS and Marcel), but it’s still possible that those with a large change in circumstances might differ from their projections. # Valuing Goalie Shootout Performance (Again) I wrote this article a few months ago about goalie shootout performance and concluded two things: • Goalies are not interchangeable with respect to the shootout, i.e. there is skill involved in goalie performance. • An extra percentage point in shootout save percentage is worth about 0.002 standings points per game. This is based on some admittedly sketchy calculations based on long term NHL performance, and not something I think is necessarily super accurate. I’m bringing this up because a couple of other articles have been written about this recently: one by Tom Tango and one much longer one by Michael Lopez. One of the comments over there, from Eric T., mentioned wanting a better sense of the practical significance of the differences in skill, given that Lopez offers an estimate that the difference between the best and worst goalies is worth about 3 standings points per year. That’s something I was trying to do in the previous post up above, and the comment prompted me to try to redo it. I made some simple assumptions that align with the one’s Lopez did in his followup post: • Each shot has equal probability of being saved (i.e. shooter quality doesn’t matter, only goalie quality). This probably reduces the volatility in my estimates, but since a goalie should end up facing a representative sample of shooters, I’m not too concerned. • The goalie’s team has an equal probability of converting each shot. This, again, probably reduces the variance, but it makes modelling things much simpler, and I think it makes it easier to isolate the effect that goalie performance has on shootout winning percentage. Given these assumptions, we can compute an exact probability that one team wins given team 1’s save percentage $p_1$ and team 2’s $p_2$. If you don’t care about the math, skip ahead to the table. Let’s call $P_{i,j}$ the probability that team $i$ scores $j$ times in the first three rounds of the shootout: $P_{i,j} = {3 \choose j} p_i^j(1-p_i)^{3-j}$ $P(\text{Team 1 Wins } \| \text{ } p_1, p_2) = \sum_{j=1}^3 \sum_{k=0}^{j-1} P_{1,j} \cdot P_{2,k} + \left ( \sum_{j=1}^3 P_{1,j}\cdot P_{2,j} \right ) \frac{p_1(1-p_2)} {1-(p_1p_2+(1-p_1)(1-p_2))}$ The first term on the right side is just the sum of the probabilities of the ways that team 1 can win the first three rounds, e.g. 2 goals for and 1 allowed or 3 goals for and none allowed. The term on the right is the sum of all the ways they can win if the first three rounds end in a tie, which can be expressed easily as the sum of a geometric series. Ultimately, we don’t really care about the formula so much as the results, so here’s a table and a plot showing the performance of a goalies who are a given percentage below or above league average when facing a league average goalie: Percentage Points Above/Below League Average Winning Percentage -20 26.12 -19 27.14 -18 28.18 -17 29.24 -16 30.31 -15 31.41 -14 32.52 -13 33.66 -12 34.81 -11 35.98 -10 37.17 -9 38.37 -8 39.60 -7 40.84 -6 42.10 -5 43.37 -4 44.67 -3 45.98 -2 47.30 -1 48.64 0 50.00 1 51.37 2 52.76 3 54.16 4 55.58 5 57.01 6 58.45 7 59.91 8 61.38 9 62.86 10 64.35 11 65.85 12 67.37 13 68.89 14 70.42 15 71.96 16 73.51 17 75.06 18 76.62 19 78.19 20 79.76 We would expect most of these figures to be close to league average, so if we borrow Tom Tango’s results (see the link above) we figure the most and least talented goalies are going to be roughly 6 percentage points away from the mean. The difference between +0.06 and -0.06 is about 0.16 in the simulation output, meaning the best goalies are likely to win sixteen shootouts per hundred more than the worst goalies assuming both play average competition. Multiplying this by 13.2%, the past frequency of shootouts, and we get an estimated benefit of only about 0.02 standings points / game from switching from the worst shootout goalie to the best. For a goalie making 50 starts, that’s only about 1 point added to the team, and that’s assuming maximal possible impact. Similarly, moving up this curve by one percentage point appears to be worth about 1.35 wins per hundred; multiplying that by 13.2% gives a value of 0.0018 standings points / game, which is almost exactly what I got when I did this empirically in the prior post, which leads me to believe that that estimate is a lot stronger than I initially thought. There’s obviously a lot of assumptions in play here, including the assumptions going into my probabilities and Tango’s estimates of true performance, and I’m open to the idea that one or another of those is suppressing the importance of this skill. Overall, though, I’m largely inclined to hew to my prior conclusions saying that for a difference in shootout performance to be enough to make one goalie better overall than another, it has to be a fairly substantial one, and the difference in actual save percentage has to be correspondingly fairly small. # Is a Goalie’s Shootout Performance Meaningful? One of the bizarre things about hockey is that the current standings system gives teams extra points for winning shootouts, which is something almost entirely orthogonal to, you know, actually being a good hockey team. I can’t think of another comparable situation in sports. Penalty shootouts in soccer are sort of similar, but they only apply in knockout situations, whereas shootouts in hockey only occur in the regular season. Is this stupid? Yes, and a quick Google will bring up a fair amount of others’ justified ire about shootouts and their effect on standings. I think the best solution is something along the lines of a 10 minute overtime (loser gets no points), and if it’s tied after 70 then it stays a tie. Since North Americans hate ties, though, I can’t imagine that change being made, though. What makes it so interesting to me, though, is that it opens up a new set of metrics for evaluating both skaters and goalies. Skaters, even fourth liners, can contribute a very large amount through succeeding in the shootout, given that it’s approximately six events and someone gets an extra point out of it. Measuring shooting and save percentage in shootouts is pretty easy, and there’s very little or no adjustment needed to see how good a particular player is. The first question we’d like to address is: is it even reasonable to say that certain players are consistently better or worse in shootouts, or is this something that’s fundamentally random (as overall shooting percentage is generally thought to be in hockey)? We’ll start this from the goalie side of things; in a later post, I’ll move onto the skaters. Since the shootout was introduced after the 2004-05 lockout, goalies have saved 67.1% of all shot attempts. (Some data notes: I thought about including penalty shots as well, but those are likely to have a much lower success rate and don’t occur all that frequently, so I’ve omitted them. All data come from NHL or ESPN and are current as of the end of the 2012-13 season. UPDATE: I thought I remembered confirming that penalty shots have a lower success rate, but some investigations reveal that they are pretty comparable to shootout attempts, which is a little interesting. Just goes to show what happens when you assume things.) Assessing randomness here is pretty tricky; the goalie in my data who has seen the most shootout attempts is Henrik Lundqvist, with 287. That might seem like a lot, but he’s seen a little over 14,000 shots in open play, which is a bit less than 50 times as many. This means that things are likely to be intensely volatile, at least from season to season. This intuition is correct, as looking at the year-over-year correlation between shootout save percentages (with each year required to have at least 20 attempts against) gets us a correlation of practically 0 (-0.02, with a wide confidence interval). Given that there are only 73 pairs of seasons in that sample, and the threshold is only 20 attempts, we are talking about a very low power test, though. However, there’s a different, and arguably better, way to do this: look at how many extreme values we see in the distribution. This is tricky when modelling certain things, as you have to have a strong sense of what the theoretical distribution really is. Thankfully, given that there are only two outcomes here, if there is really no goaltender effect, we would expect to see a nice neat binomial distribution (analogous to a weighted coin). (There’s one source of heterogeneity I know I’m omitting, and that’s shooter quality. I can’t be certain that doesn’t contaminate these data, but I see no reason it would introduce bias rather than just error.) We can test this by noting that if all goalies are equally good at shootouts, they should all have a true save percentage of 67% (the league rate). We can then calculate the probability that a given goalie would have the number of saves they do if they performed league average, and if we get lots of extreme values we can sense that there is something non-random lurking. There have been 60 goalies with at least 50 shootout attempts against, and 14 of them have had results that would fall in the most extreme 5% relative to the mean if they in fact performed at a league average rate. (This is true even if we attempt to account for survivorship bias by only looking at the average rate for goalies that have that many attempts.) The probability that at least that many extreme values occur in a sample of this size is on the order of 1 in 5 million. (The conclusion doesn’t change if you look at other cutoffs for extreme values.) To me, this indicates that the lack of year over year correlation is largely a function of the lack of power and there is indeed something going on here. The tables below shows some figures for the best and worst shootout goalies. Goalies are marked as significant if the probability they would get that percentage if they were actually league average is less than 5%. Player Attempts Saves Percentage Significant 1 Semyon Varlamov, G 71 55 77.46 Yes 2 Brent Johnson, G 55 42 76.36 Yes 3 Henrik Lundqvist, G 287 219 76.31 Yes 4 Marc-Andre Fleury, G 177 135 76.27 Yes 5 Antti Niemi, G 133 101 75.94 Yes 6 Mathieu Garon, G 109 82 75.23 Yes 7 Johan Hedberg, G 129 97 75.19 Yes 8 Manny Fernandez, G 63 46 73.02 No 9 Rick DiPietro, G 126 92 73.02 No 10 Josh Harding, G 55 40 72.73 No Player Attempts Saves Percentage Significant 1 Vesa Toskala, G 63 33 52.38 Yes 2 Ty Conklin, G 55 29 52.73 Yes 3 Martin Biron, G 76 41 53.95 Yes 4 Jason LaBarbera, G 77 43 55.84 Yes 5 Curtis Sanford, G 50 28 56.00 No 6 Niklas Backstrom, G 176 99 56.25 Yes 7 Jean-Sebastien Giguere, G 155 93 60.00 Yes 8 Miikka Kiprusoff, G 185 112 60.54 Yes 9 Sergei Bobrovsky, G 51 31 60.78 No 10 Chris Osgood, G 67 41 61.19 No So, some goalies are actually good (or bad) at shootouts. This might seem obvious, but it’s a good thing to clear up. Another question: are these the same goalies that are better at all times? Not really, as it turns out; the correlation between raw save percentage (my source didn’t have even strength save percentage, unfortunately) and shootout save percentage is about 0.27, which is statistically significant but only somewhat practically significant—using the R squared from regressing one on the other, we figure that goalie save percentage only predicts about 5% of the variation in shootout save percentage. You may be asking: what does all of this mean? Well, it means it might not be fruitless to attempt to incorporate shooutout prowess into our estimates of goalie worth. After all, loser points are a thing, and it’s good to get more of them. To do this, we should estimate what the relationship between a shootout goal and winning the shootout (i.e., collecting the extra point) is. To do this, I followed the basic technique laid in this Tom Tango post. Since shootouts per season are so small, I used lifetime data for each of the 30 franchises to come up with an estimate for the number of points that one shootout goal is worth. Regressing goal difference per game on winning percentage, we get a coefficient of 0.368. In other words, one shootout goal is worth about 0.368 shootout wins (that is, points). Two quick asides about this: one is that there’s an endemic flaw in this estimator even beyond sample size issues, and that’s that the skipping of an attempt when a team is up 2-0 (or 3-1) means that we are deprived of some potentially informative events simply due to the construction of the shootout. Another is that while this is not a perfect estimate, it does a pretty good job predicting things (R squared of 0.9362, representing the fraction of the variance explained by the goal difference). Now that we can convert shootout goals to wins, we can weigh the relative meaning of a goaltender’s performance in shootouts and in actual play. This research says that each goal is worth about 0.1457 wins, or 0.291 points, meaning that a shootout goal is worth about 26% more than a goal in open play. However, shootouts occur infrequently, so obviously a change of 1% in shootout save percentage is worth much less than a change of 1% in overall save percentage. How much less? To get this figure, we’re going to assume that we have two goalies facing basically identical, average conditions. The first parameter we need is the frequency of shootouts occurring, which since their establishment has been about 13.2% of games. The next is the number of shots per shootout, which is about 3.5 per team (and thus per goalie). Multiplying this out gets a figure of 0.46 shootout shots per game, a save on which is worth 0.368 points, meaning that a 1% increase in shootout save percentage is worth about 0.0017 points per game. To compute the comparable figure for regular save percentage, I’ll use the league average figure for shots in a game last year, which is about 29.75. Each save is worth about 0.29 points, so a 1% change in regular save percentage is worth about 0.087 points per game. This is, unsurprisingly, much much more than the shootout figure; it suggests that a goalie would have to be 51 percentage points better in shootouts to make up for 1 percentage point of difference in open play. (For purposes of this calculation, let’s assume that overall save percentage is equal to a goalie’s even strength save percentage plus an error term that is entirely due to his team, just to make all of our comparisons apples to apples. We’re also assuming that the marginal impact of a one percentage point change on a team’s likelihood of winning is constant, which isn’t too true.) Is it plausible that this could ever come into play? Yes, somewhat surprisingly. The biggest observed gap between two goalies in terms of shootout performance is in the 20-25% range (depends on whether you want to include goalies with 50+ attempts or only 100+). A 20% gap equates to a 0.39% change in overall save percentage, and that’s not a meaningless gap given how tightly clustered goalie performances can be. If you place the goalie on a team that allows fewer shots, it’s easier to make up the gap—a 15% gap in shootout performance is equivalent to a 0.32% change in save percentage for a team that gives up 27 shots a game. (Similarly, a team with a higher probability of ending up in a shootout has more use for the shootout goalie.) Is this particularly actionable? That’s less clear, given how small these effects are and how much uncertainty there is in both outcomes (will this goalie actually face a shootout every 7 times out?) and measurement (what are the real underlying save percentages?). (With respect to the measurement question, I’d be curious to know how frequently NHL teams do shootout drills, how much they record about the results, and if those track at all with in-game performance.) Still, it seems reasonable to say that this is something that should be at least on the table when evaluating goalies, especially for teams looking for a backup to a durable and reliable #1 (the case that means that a backup will be least likely to have to carry a team in the playoffs, when being good at a shootout is pretty meaningless). Moreover, you could maximize the effect of a backup goalie that was exceptionally strong at shootouts by inserting him in for a shootout regardless of whether or not he was the starter. That would require a coach to have a) enough temerity to get second-guessed by the press, b) a good enough rapport with the starter that it wouldn’t be a vote of no confidence, and c) confidence that the backup could perform up to par without any real warmup. This older article discusses the tactic and the fact that it hasn’t worked in a small number of cases, but I suspect you’d have to try this for a while to really gauge whether or not it’s worthwhile. For whatever it’s worth, the goalie pulled in the article, Vesa Toskala, has the worst shootout save percentage of any goalie with at least 50 attempts against (52.4%). I still think the shootout should be abolished, but as long as it’s around it’s clear to me that on the goalie end of things this is something to consider when evaluating players. (As it seems that it is when evaluating skaters, which I’ll take a look at eventually.) However, without a lot more study it’s not clear to me that it rises to the level of the much-beloved “market inefficiency.” EDIT: I found a old post that concludes that shootouts are, in fact, random, though it’s three years old and using slightly different methods than I am. The three years old portion is pretty important, because that means that the pool of data has increased by a substantial margin since then. Food for thought, however. # Tim McCarver and Going the Other Way During the Tigers-Red Sox game last night, Tim McCarver said he thought it was a little odd that the Tigers would bring in lefty Drew Smyly to face David Ortiz while also leaving the shift on, since lefties are more likely to go to the opposite field against a left-handed pitcher. (At the very least, I know he said this last part. Memory is a tricky thing, and I’m now not sure whether he said this about Ortiz or someone else, possibly Alex Avila.) Being Tim McCarver, he didn’t say why this might be true, nor did he cite a source for this information, putting this firmly in the realm of obnoxious hypotheses. The first question is whether or not this is true. For that, there are these handy aggregated spray charts, courtesy Brooks Baseball. Based on these data, I have to say it seems like McCarver’s assertion is true: they are slightly more likely to go to left against a left-handed pitcher. I don’t have enough information to say if the differences are either statistically significant (I’d guess it is, given the number of balls these guys have put into play in the last 5-7 years) or practically significant (I kinda doubt it). Regardless of the answer, though, the fact remains that the appropriate thing to do is to bring in the lefty and shift slightly less drastically, so who knows why McCarver brought this up to begin with. After all, Ortiz hits drastically worse against lefties (his OPS against lefties is 24% smaller than his lifetime rate, via baseball-reference), as does Avila (36%). There’s also the question of why this might be true, and in fairness to McCarver, there are some pretty plausible mechanisms for what he was saying. One is that a breaking pitch from a left-hander is more likely to be on the outer part of the plate for a left-handed batter than a similar pitch from a right-handed batter, and outside pitches are more likely to get hit the other way. Another is that left-handed batters can’t pick up a pitch as easily against a left-handed pitcher, so they are more likely to make late contact, which is in turn more likely to go to the opposite field. I can’t necessarily confirm either of these mechanisms empirically, though looking at Brooks splits for Avila and Ortiz suggests that the fraction of outside pitches they see against left-handers is about 3 percentage points larger than the fraction against righties. So, what McCarver said was true (though not terribly helpful), and there are seemingly good reasons for it to be true. I still posted something, though, because this is a great example of something that pisses me off about sports commentators–a tendency to toss out suppositions and not bother with supporting or explaining them. (Another good example of this is Hawk Harrelson.) That tendency, along with their love of throwing out hypotheses that are totally unfalsifiable (McCarver asserting that the pitching coach coming out to the mound is valuable, e.g.), is one of the things I plan to deal with pretty regularly in this space. (Happy first post, everyone.)	2022-01-26 18:26:00	{"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": 7, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5580019354820251, "perplexity": 1350.1709638846944}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320304959.80/warc/CC-MAIN-20220126162115-20220126192115-00333.warc.gz"}
http://math.stackexchange.com/questions/426167/frac-x-yx-can-it-be-simplified	# $\frac {(x-y)}{x}$ Can it be simplified? I appreciate anyone taking a look at this. It's been ages since I've been in algebra/calculus and need to figure out if $\dfrac {(x-y)}{x}$ can be simplified or would it be $\left(1 - \dfrac yx\right)$? Thank you, Josh Thanks to all the very speedy responses. I guess my algebra isn't as rusty as I thought. This question can be marked as solved/closed. - you already write its possibility to solve by $1-\frac yx$.Can you do more clear your ques? – iostream007 Jun 21 '13 at 13:29 @iostream007 I have changed the formatting of the title so as to make it take up less vertical space -- this is a policy to ensure that the scarce space on the main page is distributed evenly over the questions. See here for more information. Please take this into consideration for future questions. Thanks in advance. – Lord_Farin Jun 21 '13 at 13:41 @Lord_Farin I'll remember it – iostream007 Jun 21 '13 at 13:42 $1-\frac yx$ is the best you will do for most purposes. Sometimes the original will be better, depending on the rest of the expression. - Here's how to remember: it's always easy to add or subtract fractions with the same denominator. (If you cut a pie into eighths, and you select seven eighths of it, and take away four eigths, you're still left with three eighths.) So $$\frac{x - y}{z} = \frac{x}{z} - \frac{y}{z}$$ In your case $x = z$ and the first fraction simplifies to $1$ (eight eighths of a pie minus three eighths of a pie is five eighths of a pie). If you try to do the same reasoning with denominators of a fraction, you'll get junk (four eighths of a pie minus four sevenths of a pie $\neq$ four negative oneths of a pie). So don't try to simplify something like $$\frac{x}{x - y}.$$ - You can turn it to $\left(1-\dfrac yx\right)$ as you said . -	2014-08-31 06:25:47	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7692188620567322, "perplexity": 788.2719094201047}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-35/segments/1408500836108.12/warc/CC-MAIN-20140820021356-00274-ip-10-180-136-8.ec2.internal.warc.gz"}
http://cascadehikers.com/latex-homework-class.html	## Latex homework class Posted by \| in December 18, 2018 Homework Assignments class notes & Honework solutions on Sakai. Presentation days. Homework rewrites must be typeset in LaTeX. Hello, /r/math! I originally latex homework class this to /r/latex yesterday, but I thought this could be relevant to some users here as well. Homework 1 \| PDF \| LaTeX\| Hardcopy due at the start ain resume cover letter class on. I wrote a combination of LATEX commands for a new LATEX class implemented in file homework.cls, and Python latex homework class (in latex homework class file called homework.py) that make it. To conclude the Level 1 Safer Choices classes, students reviewed homework assignments, summarizing what. Before the (spring 2010) term started, I talked with some alumni about their experiences in the undergraduate seminar classes at M.I.T., and they suggested that I. Gallery — Homework Assignment. Here we provide a selection of homework assignments templates and. ### Uol audit essay Homework help expectations - Cooperate with our scholars to receive the top-notch essay. See Course Projects for more about the class project and some suggested topics. Jun 2015. The distribution is as follows: latex homework class 20, in-class participation 20%, midterms 20% and. View homewlrk 9 Bibliographies templates ». Note: For assignments that require LaTeX responses, a Preview in LaTeX option. Heres what parents can get started with your class not ask for students to help and. Answering a Writing Problem Attaching Files (Including Images) LaTeX and. This forum site is equipped with MathJax which supports LaTex, a wonderful. Oct 2016. Instead of class assignments being limited to the classroom space, or to. C. The Ut tyler cover letter of a LATEX Document The Text. The IRF may be useful for assignments in the mandatory classes in Latex homework class. ### How do you write a thesis statement for an expository essay No other LaTeX environments are recognized. Latex homework class 4: PREPARATION (HOMEWORK) Only conduct activity for products that. If you have a question on your homework or a general math question, you may. Starting off an essay with a quote example includes a clear title on the first page for the course. LATEX has a reputation for having a steep learning curve. Children with peanut or latex allergies may need more extensive plans that. Note 1: I strongly encourage you to typeset your answers (in Word, or Latex. See the instructions here, starter code and data here, and the LaTeX source here. Prior to discovering the exam class, I used the hmcpset latex homework class class from the. Slides of first class (power point,postscript,html). ### Application letter for atm card to bank manager Jan 2015. There are LaTeX document classes for typesetting books, articles, exams, presentations, and more. Your homwwork homework grade will be dropped and will not affect your final grade in the. Kudu Course. A complete textbook solution including lecture videos, text and homework sets. Learn to use LaTeX. See lates section on Homework below. Social class essay introduction myimaths hack 2018 writing assignments for punishment. Nov 26, 2018 \| Latex homework template stanford \| 0 comments. Note that in Matlab you can also use publish in clqss mode to create an m-file. Download: Latex Source sample strategic business plan format, full PDF notes, zip, the homework assignment or. LaTeX latex homework class free software, latex homework class is used by professional mathematicians to type mathematics. Grading: Your grade depends on your performance on homework, exams, and.	2019-02-23 03:04:23	{"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": 0.8704464435577393, "perplexity": 6193.200440767743}, "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-2019-09/segments/1550249434065.81/warc/CC-MAIN-20190223021219-20190223043219-00230.warc.gz"}
https://www.degruyter.com/view/j/eng.2017.7.issue-1/eng-2017-0022/eng-2017-0022.xml	Show Summary Details More options … # Open Engineering ### formerly Central European Journal of Engineering Editor-in-Chief: Ritter, William 1 Issue per year CiteScore 2017: 0.70 SCImago Journal Rank (SJR) 2017: 0.211 Source Normalized Impact per Paper (SNIP) 2017: 0.787 Open Access Online ISSN 2391-5439 See all formats and pricing More options … Volume 7, Issue 1 # An exponential-related function for decision-making in engineering and management Daniel O. Aikhuele / Faiz Mohd Turan Published Online: 2017-06-09 \| DOI: https://doi.org/10.1515/eng-2017-0022 ## Abstract An intuitionistic fuzzy TOPSIS model, which is based on an exponential-related function (IF-TOPSIS) and a fuzzy entropy method, has been proposed in this study. The exponential-related function, which represents the aggregated effect of positive and negative evaluations in the performance ratings of the alternatives, based on the intuitionistic fuzzy set (IFS) data. Serves, as a computational tool for measuring the separation distance of decision alternatives from the intuitionistic fuzzy positive and negative ideal solution to determine the relative closeness coefficient. The main advantage of this new approach is that (1) it uses a subjective and objective based approach for the computation of the criteria weight and (2) its simplicity both in its concept and computational procedures. The proposed method has successfully been implemented for the evaluation of some engineering designs related problems including the selection of a preferred floppy disk from a group of design alternatives, the selection of the best concept design for a new air-conditions system and finally, the selection of a preferred mouse from a group of alternatives as a reference for a new design. Also, for each of the three case studies, the method has been compared with some similar computational approaches. ## 1 Introduction The intuitionistic fuzzy set (IFS), which is an expansion of the traditional fuzzy set (FSs) theory was first proposed by Atanassov in 1986 [1]. It comprises of a membership and a non-membership function, which are used for the management of vagueness and uncertainty. As indicated by Wan and Li [2] and Aikhuele & Turan [3], the IFS are more adaptable, functional and capable than the traditional FS theory at handling uncertainty and vagueness in practices. The advantages of applying the IFS have been reported in [5] to include: (1) Its ability to model unknown information using hesitation degree, when the Decision-makers (DMs) are unsure about the preferences of an assessment. (2) It represents three grades of membership function, which include membership degree, non-membership degree, and hesitancy degree. Hence, the IFS can be said to consider opinions from three sides to arrive at the preferred one. (3) All the fuzzy numbers in the IFS theory can all be used to represent vagueness of “agreement”, although, they cannot be used to depict the “disagreement” of the DMs. As a mathematical tool, the IFS has demonstrated the ability to deal with fuzziness and uncertainty in information and data in a real-life situation and this has resulted in its many applications in diverse fields of study mostly for solving multiple criteria decision making (MCDM) problems [3, 6-10]. However, among the numerous applications of IFS for MCDM, the technique for order preference by similarity to the ideal solution (TOPSIS) by Hwang and Yoon [11] has remained the most extensively used method. TOPSIS method is based on the concept that the most appropriate alternative should have the shortest distance from the positive ideal solution and the farthest distance from the negative ideal solution and has remained one of the most reliable and practical decision-making tools which depend on preference information provided by the DMs [12, 13]. In matching up the preference information given by the DMs which are expressed in IFS, some metric methods were introduced, that is the score and accuracy functions as described in [14-17] and applied for solving MCDM problems. However, a recent investigation by Wu [17] suggests that the results obtained using the score and accuracy functions are not always consistent, while they also produce a negative priority vector in their applications. In addressing this issue, Wu [17], introduced the exponential score function. Although, the exponential score function appears to address these shortcomings, the function is only effective for determining priority weight that involves pairwise comparison. In this study, the exponential score function which have been extended in [18] (exponential-related function) is adopted in the intuitionistic fuzzy TOPSIS (IF-TOPSIS) model with intuitionistic fuzzy entropy method, for determining the criteria weight when the performance ratings are expressed in intuitionistic fuzzy number (IFN). The adoption of the new exponential-related function (ER) in the intuitionistic fuzzy decision-making method is undertaken to provide a flexible and a whole new approach to solving MCDM problems. In computing the weight of the criteria, the intuitionistic fuzzy entropy (IFE) originally proposed by Ye, [19] was adopted. The main contribution and advantages of the new method and approach lies in the use of an objective approach for the computation of the criteria weight, which allows for complete assessment of the actual performance and value of each of the criteria. The application of the matrix method (i.e. the exponential-related function), which represent the aggregated effect of the positive and negative evaluations in the performance ratings of the alternatives based on the intuitionistic fuzzy set (IFS) data. The integration of the exponential-related function and the intuitionistic fuzzy entropy into the traditional intuitionistic fuzzy TOPSIS model, introduction of the MCDM method, which can be described as simple both in its concept and computational procedures, compared to other existing methods and finally. The exponential-related function, which serves as a parameter and a better alternative to the Euclidian distance that often has correlation issues, in the computation of the separation measures of each alternative from the intuitionistic fuzzy positive and negative ideal solution which is used in the determination of the relative closeness coefficient. The rest of the paper is organized as follows; in section 2, the concept of the IFS, the intuitionistic fuzzy entropy, and the exponential-related function are presented. The algorithm of the Intuitionistic Fuzzy TOPSIS model based on the exponential-related function (IF-TOPSIS) and the intuitionistic fuzzy entropy (IFE) method are presented in section 3. In section 4, a numerical case study is presented to demonstrate the effectiveness of the model. While some come conclusions are presented in section 5. ## 2 The Basic Concept of IFS and the Exponential-Related Function This section presents, the fundamental definitions and concepts of the IFS theory as described in [1] and the proposed exponential-related function with the IFE. ## 2.1 Intuitionistic Fuzzy Set #### Definition 1 If the IFS A in X = {x is defined fully in the form A = {〈x, µA (x), vA (x), A (x)〉 \|xX, where µA : X → [0, 1], vA : X → [0, 1] and A : X → [0, 1]. The different relations and operations for the IFS are shown in Eq. (1) to (4). $AB={〈x,μA(x).μB(x),vA(x)+vB(x)−vA(x).vB(x)〉\|x∈X$(1) $A+B={〈x,μA(x)+μB(x)−μA(x).μB(x),vA(x).vB(x)〉\|x∈X$(2) $λA={x,1−1−μA(x)λ,(vA(x))λ\|x∈X},,λ>0.$(3) $Aλ={x,(μA(x))λ,1−(1−vA(x))λ\|x∈X},λ>0$(4) In the proceeding definition, comparisons between the IFS are presented, by introducing the score and accuracy functions as described in [14-16]. #### Definition 2 Let A = (µ, v) be an intuitionistic fuzzy number, a score function S and an accuracy function H of an intuitionistic fuzzy value can be represented as follow. $S(A)=(μ−v),whereS(A)∈[−1,+1]$(5)$H(A)=(μ+v),whereH(A)∈[0,1]$(6) #### Definition 3 Let A = (µ, v) be the intuitionistic fuzzy number, according to Wu (2015) the exponential score function Se of the intuitionistic fuzzy number can be represented as: $Se(A)=e(μ−v)whereSe(A)∈[1/e,e]$(7) ## 2.2 The Exponential Related Function (ER) #### Definition 4 [18] Let A = (µ, v) be the intuitionistic fuzzy number. The new exponential-related function ER of the intuitionistic fuzzy number can be defined as: $ER(A)=e1−μ2−v23,whereER(A)∈[1/e,e]$(8) #### Theorem 1 Let A = (µ, v) and B = (µ1, v1) be two intuitionistic fuzzy set, if AB then ER (A) ≤ ER (B). #### Proof Assume that A = (µ, v) and B = (µ1, v1) are two comparable alternatives with intuitionistic fuzzy numbers based on some criteria ci such that AB without loss of generality, let assume that ${\mu }_{1}^{2}\phantom{\rule{thinmathspace}{0ex}}\le \phantom{\rule{thinmathspace}{0ex}}{\mu }^{2},\phantom{\rule{thinmathspace}{0ex}}\mathrm{a}\mathrm{n}\mathrm{d}\phantom{\rule{thinmathspace}{0ex}}{v}^{2}\phantom{\rule{thinmathspace}{0ex}}\ge \phantom{\rule{thinmathspace}{0ex}}{v}_{1}^{2}$ such that ER (A) ≤ ER (B) By Definition 4, we have that: $ER(A)=e1−μ2−v23$ and $ER(B)=e1−μ12−v123$ Then $ER(B)−ER(A)=e1−μ12−v123−e1−μ2−v23=e1−μ12−v123−1−μ2−v23=e1−μ12−v12−1+μ2+v23=e(μ2−μ12+v2−v123)$ This can be rewritten as: $=e(μ2−μ123+v2−v123)$ Let assume the power of the exponential is multiply by 3, and then we have; $=e(μ2−μ12+v2−v12)$ Since, AB, ${\mu }_{1}^{2}\phantom{\rule{thinmathspace}{0ex}}\le \phantom{\rule{thinmathspace}{0ex}}{\mu }^{2},\phantom{\rule{thinmathspace}{0ex}}\mathrm{a}\mathrm{n}\mathrm{d}\phantom{\rule{thinmathspace}{0ex}}{v}^{2}\phantom{\rule{thinmathspace}{0ex}}\ge \phantom{\rule{thinmathspace}{0ex}}{v}_{1}^{2}$. Hence $\left({\mu }^{2}-{\mu }_{1}^{2}\right)\phantom{\rule{thinmathspace}{0ex}}\ge \phantom{\rule{thinmathspace}{0ex}}0,\phantom{\rule{thinmathspace}{0ex}}\mathrm{a}\mathrm{n}\mathrm{d}\phantom{\rule{thinmathspace}{0ex}}\left({v}^{2}-{v}_{1}^{2}\right)\phantom{\rule{thinmathspace}{0ex}}\ge \phantom{\rule{thinmathspace}{0ex}}0.$ Then it follows that ER (B) − ER (A) ≥ 0. □ #### Theorem 2 Let A = (µ, v) and B = (µ1, v1) be two intuitionistic fuzzy set, from the above theorem (1), we can conclude: 1. ER (B) > ER (A), if and only if B > A 2. ER (B) > ER (A), if and only if ${\mu }^{2}-{v}^{2}\phantom{\rule{thinmathspace}{0ex}}>\phantom{\rule{thinmathspace}{0ex}}{\mu }_{1}^{2}-{v}_{1}^{2}$ ## 2.3 The intuitionistic fuzzy entropy (IFE) Following the operations of the IFS, let us consider an intuitionistic fuzzy set A in the universe of discourse X = {x1, x2 x3,..., xn.The intuitionistic fuzzy set A is transformed into a fuzzy set to structure an entropy measure of the intuitionistic fuzzy set by means of µĀ (xi) = (µA (xi) + 1 - vA (xi))/2. Based on the definition of fuzzy information entropy Ye (2010) proposes the intuitionistic fuzzy entropy as follows: $E(A)=1n∑i=1nsinπ⋆[1+μA(xi)−vA(xi)]4+sinπ⋆[1−μA(xi)+vA(xi)]4−1⋆12−1$(9) When the criteria weights are completely unknown, we can use the IFE to determine the weights. The criteria weight is given as: $Wj=1−Hjn−∑j=0nHj$(10) where Wj ∈ [0, 1], ${\sum }_{j=1}^{n}{W}_{j}=1,\phantom{\rule{thinmathspace}{0ex}}{H}_{j}=\frac{1}{m}E\phantom{\rule{thinmathspace}{0ex}}\left({A}_{j}\right)$ and 0 ≤ Hj ≤ 1 for (j = 1, 2, 3,..., n). ## 3 Algorithm of the IF-TOPSIS and Intuitionistic Fuzzy Entropy (IFE) Method In this section, the algorithm for the IF-TOPSIS and the IFE Method is concisely expressed using the stepwise procedure. The schematic diagram of the proposed model is shown in Fig. 1. Figure 1 The schematic diagram of the proposed model #### Step 1 Set up a group of Decision Makers (DMs) and aggregate their evaluations using Intuitionistic Fuzzy Weighted Geometric (IFWG) operator [20]. Once the DMs has given their judgment using linguistic variables, the variables are then converted to the intuitionistic fuzzy number (IFNs), as shown in Table 1. The weight vector λ = (λ1, λ2, λ3, .., λl)T is used to aggregate all the DMs individual assessment matrices Dk (k = 1, 2, 3,..., l) into the group assessment matrix (i.e. intuitionistic fuzzy decision matrix) Ryxz(xij). $IFWG(d1d2d3,...,dn)=∏i=1nμijwj,1−∏i=1n1−vijwj$(11) Table 1 Fuzzy numbers for approximating the linguistic variable $Rmxn(aij)= [(μ11, v11)(μ12, v12)…(μ1n, v1n)(μ21, v21)(μ22, v22)⋯(μ2n, v2n)⋮⋮⋱⋮⋮⋮⋱⋮(μm1, vm1)(μm2, vm2)⋯(μmn, vmn)]$(12) #### Step 2 Determine the weight of each of the evaluating criteria wj using the IFE method. #### Step 3 Using the exponential related function ER (i.e. equation 8) convert the intuitionistic fuzzy decision matrix Ryxz(xij) to form the exponential related matrix EMyxz (ERij (aij)), which represents the aggregated effect of the positive and negative evaluations in the performance ratings of the alternatives based on the intuitionistic fuzzy set (IFS) data. $EMyxz (Eij (aij))= [ER11(x11)ER12(x12)⋯ER1n(x1z)ER21(x21)ER22(x22)⋯ER2n(x2z)⋮⋮⋱⋮⋮⋮⋱⋮ERy1(xy1)ERy2(xy2)⋯ERyz(xyz)]$(13) #### Step 4 Define the IFPIS A+ = (µj, vj) and IFNIS A = (µj, vj) for the alternatives. $A+=Cj,1,1\|Cj∈C,A−=Cj,0,0\|Cj∈C,j=1,2,3,....,z$ #### Step 5 Compute the exponential-related function-based separation measures in intuitionistic fuzzy environment $\left({d}_{i}^{+}\left({A}^{+},\phantom{\rule{thinmathspace}{0ex}}{A}_{i}\right)\right\phantom{\rule{thinmathspace}{0ex}}\mathrm{a}\mathrm{n}\mathrm{d}\phantom{\rule{thinmathspace}{0ex}}\left({d}_{i}^{-}\left({A}^{-},\phantom{\rule{thinmathspace}{0ex}}{A}_{i}\right)\right$ for each alternative for the IFPIS and IFNIS. $di+(A+,Ai)=∑i=1nwj(1−EMyxz(aij)2$(14)$di−(A−,Ai)=∑i=1nwjEMyxz(aij)2$(15) where wj is the weight of the criteria. #### Step 6 Compute the relative closeness coefficient, (CCi), which is defined to rank all possible alternatives with respect to the positive ideal solution A+. The general formula is given as; $CCi=di−A−,Aidi−A−,Ai+di+A+,Ai$(16) where CCi (i = 1, 2, ..n) is the relative closeness coefficient of Ai with respect to the positive ideal solution A+ and 0 ≤ CCi ≤ 1. #### Step 7 Rank the alternatives in the descending order. ## 4 Illustrative Examples #### Case 1 Let’s consider a practical decision-making problem originally reported in [21]. In this case, the original problem is modified to make a new example, however, using the same decision matrices while the attributes weight are derived using the intuitionistic fuzzy entropy method. Suppose a product manufacturing company want to select a preferred floppy disk from a group of candidates; S1, S2, and S3 as a reference disk for a new design. A group of three experts with the following weights values λ = {0.35, 0.36, 0.28} respectively, are to make a decision about the floppy disk with respect to the following criteria: Performance (C1), Appearance (C2) and Cost (C3). The experts’ preference judgments are given as shown in Table 2. Using the algorithm of the IF-TOPSIS and the IFE as given in section 3, the best floppy disk design from the three design alternatives with respect to the three criteria is selected. Table 2 The expert’s individual preference judgments In Steps 1&2, the individual expert’s assessments for the three designs with respect to the criteria are aggregated using the IFWG operator. The final comprehensive group assessment matrix for the expert’s assessment, called the intuitionistic fuzzy decision matrix R3x3(xij), is given in Table 3. The criteria weight is calculated from the intuitionistic fuzzy matrix using the IFE method which can be calculated by inputting the formula in a Microsoft excel program. The final result is given as w = {0.29, 0.23, 0.47 respectively. Table 3 Intuitionistic fuzzy decision matrix In step 3–5, using the exponential-related function, the intuitionistic fuzzy decision matrix R3x3(xij) is converted to form the exponential related matrix EM3x3 (ERij (aij)), while the exponential related function-based separation measures $\left({d}_{i}^{+}\left({A}^{+},\phantom{\rule{thinmathspace}{0ex}}{A}_{i}\right)\right\phantom{\rule{thinmathspace}{0ex}}\mathrm{a}\mathrm{n}\mathrm{d}\phantom{\rule{thinmathspace}{0ex}}\left({d}_{i}^{-}\left({A}^{-},\phantom{\rule{thinmathspace}{0ex}}{A}_{i}\right)\right$ (i = 1, 2, 3) is calculated using equation (14) and (15). In step 6–7, the relative closeness coefficient CCi, (i = 1, 2,3) to the ideal solution is calculated using equation (16), the relative closeness coefficients for each of the alternatives are ranked in the descending order. The results are given in Table 4. Table 4 The relative closeness coefficients for the three design alternatives From the ranking result of the three floppy design alternatives, we can conclude therefore that the design concept S2 is the best design based on the three evaluating criteria provided by the three Expert’s preference judgments. Table 5 shows that the result is totally in agreement with the result in [21]. This proves the effectiveness and feasibility of the proposed model at handling uncertainty and for decision-making. Table 5 Comparison of ranking results for the case 1 #### Case 2 Let’s consider another decision-making problem originally reported by Joshi & Kumar [22]. In this case, the problem has been modified to make a new example using the same decision matrix, while the attributes weights are derived using the intuitionistic fuzzy entropy method. Suppose a design company wants to select the best concept design for a new air-conditions system from the following alternatives S1, S2, S3, and S4. The DMs are to evaluate and select the best concept design with respect to Safety (C1), Attractive design (C2) and Reliability criteria (C3) design cost (C4) and compatibility design (C5). The aggregated DMs preference judgments are presented in Table 6 (i.e. Intuitionistic fuzzy decision matrix). From these the best concept design for the new air-conditions system can be selected based on the IF-TOPSIS and IFE method. Table 6 Intuitionistic fuzzy decision matrix Using the IF-TOPSIS algorithm we select the best concept design for an air conditions system, where the criteria weight is calculated from the intuitionistic fuzzy matrix using the IFE method. The result of the evaluation is given as: $w={0.161269,0.144649,0.14052,0.40608,0.147482},$ respectively. Using the exponential-related function, the intuitionistic fuzzy decision matrix R4x5(xij) is converted to form the exponential related matrix EM4x5 (ERij (aij)), while the exponential related function-based separation measures $\left({d}_{i}^{+}\left({S}^{+},\phantom{\rule{thinmathspace}{0ex}}{S}_{i}\right)\right\phantom{\rule{thinmathspace}{0ex}}\mathrm{a}\mathrm{n}\mathrm{d}\phantom{\rule{thinmathspace}{0ex}}\left({d}_{i}^{-}\right\left({S}^{-},\phantom{\rule{thinmathspace}{0ex}}{S}_{i}\right)$ (i = 1, 2,..., 4) is calculated using equation (14) and (15). In step 6–7, the relative closeness coefficient CCi, (i = 1, 2,..., 4) to the ideal solution is calculated using equation (16), the overall computational results as well as the ranking of the relative closeness coefficients for each of the alternatives are given in Table 7. Table 7 The relative closeness coefficients for the four design alternatives From the ranking result of the four air-conditions system design alternatives, we conclude that the S3 is the best design with respect to the five evaluating criteria. The result is totally in agreement with the result in [22] (Table 8). Table 8 Comparison of ranking results for the case 2 #### Case 3 Finally, Let us consider a practical MCDM problem originally reported by Ye [23] and adopted by Liu & Ren [24]. In this case, the original problem has been modified to make a new example, however, using the same decision matrix. Suppose a computer manufacturing company wants to select a preferred mouse from a group of candidates; A1, A2, A3 and A4 as a reference mouse for a new design. Again, a group of experts is asked to make a decision with respect to Performance (C1), Cost (C2) and Appearance (C3). The experts aggregated evaluations are given in Table 9. We select the preferred mouse using the IF-TOPSIS method. Table 9 Intuitionistic fuzzy decision matrix Using the IFE method, the criteria weight is calculated from the intuitionistic fuzzy matrix and the result is given as w = {0.377, 0.311, 0.313 respectively. Using the exponential-related function, just as in case 1&2, the intuitionistic fuzzy decision matrix R4x3 (xij) is converted to form the exponential related matrix and the exponential related function-based separation measures $\left({d}_{i}^{+}\left({A}^{+},\phantom{\rule{thinmathspace}{0ex}}{A}_{i}\right)\right\phantom{\rule{thinmathspace}{0ex}}\mathrm{a}\mathrm{n}\mathrm{d}\phantom{\rule{thinmathspace}{0ex}}\left({d}_{i}^{-}\left({A}^{-},\phantom{\rule{thinmathspace}{0ex}}{A}_{i}\right)\right$ (i = 1, 2,..., 4) is calculated for each of the alternative, while the relative closeness coefficient CCi, (i = 1, 2,..., 4) to the ideal solution is calculated using equation (14). The final results are shown in Table 10. Table 10 The relative closeness coefficients of the four candidates From the ranking result of the four alternative mouse designs, we conclude that the A2 is the best design with respect to the three evaluating criteria, and the ranking result is in agreement with the result in [23, 24] as shown in Table 11. Table 11 Comparison of ranking results for the case 3 ## 5 Conclusion In this paper, we have proposed a new matrix method (i.e. the exponential-related function (ER)) for comparing intuitionistic fuzzy sets, and as a replacement for the traditional exponential score function originally proposed by Wu [17], which have been found ineffective for solving some MCDM problems. The new exponential-related function (ER), which has been developed and adopted in the intuitionistic fuzzy TOPSIS model and intuitionistic fuzzy entropy is used for solving MCDM problems in which the weight of the evaluating criteria are completely unknown and the performance ratings of the alternatives are expressed in IFN. The criteria weight here, have been calculated using the intuitionistic fuzzy entropy method originally proposed by Ye [19]. The main advantage and contribution of the new method and approach is that (1) it uses an objective approach for the computation of the criteria weight, which allows for complete assessment of the actual performance of each of the criteria by assisting in the identification of the difference between the present situation (which is considered to be ideal) and the level of performance it intended to achieved in the future. (2) Simplicity in the MCDM method both in its concept and computational procedures as compared to other existing methods. (3) The application of the exponential-related function, which stands to represent the aggregated effect of the positive and negative evaluations in the performance ratings of the alternatives based on the intuitionistic fuzzy set (IFS) data and (4) finally, it serves as a parameter and a better alternative to the Euclidian distance that often has correlation issues, in the computation of the separation measures of each alternative from the intuitionistic fuzzy positive and negative ideal solution which is used in the determination of the relative closeness coefficient. To validate the feasibility and effectiveness of the method, the IF-TOPSIS, model has been applied for the assessment of some engineering designs related problems including selection of a preferred floppy disk from a group of design alternatives, the selection of the best concept design for a new air-conditions system and finally, for the selection of a preferred mouse from a group of alternatives as a reference for a new design. In the future, we will continue working on the application of the proposed method in other domain, specifically for problems with more criteria and alternatives and to make some comparisons with the adaptive fuzzy control of strict-feedback nonlinear time-delay systems, which have recently found applications in the intuitionistic fuzzy environment. ## References • [1] Atanassov K. T., Intuitionistic fuzzy sets, Fuzzy Sets Syst., 1986, 20 (1), 87-96 • [2] Wan S. P., Li D. 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Res. 2016, Univ. Malaysia Pahang, 2016, 374-384 Google Scholar • [14] Hong D. H., Choi C.H., Multi-criteria fuzzy decision making problems based on vague set theory, Fuzzy Sets Syst., 2000, 114 (1), 103-113 • [15] Chen S.M., Tan J.M., Handling multicriteria fuzzy decisionmaking problems based on vague set theory, Fuzzy Sets Syst., 1994, 67 (2), 163-172 • [16] Xu Z., Intuitionistic preference relations and their application in group decision making, Inf. Sci. (Ny)., 2007, 177 (11), 2363-2379 • [17] Wu J., Consistency in MCGDM Problems with Intuitionistic Fuzzy Preference Relations Based on an Exponential Score Function, Gr. Decis. Negot., 2015, 25 (2), 399-420 • [18] Aikhuele D. O., Turan F. M., Extended TOPSIS model for solving multi-attribute decision making problems in engineering, Decis. Sci. Lett., vol. 6, pp. 365-376, 2017 Google Scholar • [19] Ye J., Two effective measures of intuitionistic fuzzy entropy, Comput. (Vienna/New York), 2010, 87 (2), 55-62 • [20] Xu Z., Yager R. R., Some geometric aggregation operators based on intuitionistic fuzzy sets, Int. J. Gen. Syst., 2006, 35 (4), 417-433 • [21] Yue Z., An extended TOPSIS for determining weights of decision makers with interval numbers, Knowledge-Based Syst., 2011, 24 (1), 146-153 • [22] Joshi D., Kumar S., Intuitionistic fuzzy entropy and distance measure based TOPSIS method for multi-criteria decision making, Egypt. Informatics J., 2014, 15 (2), 97-104 • [23] Ye J., Fuzzy decision-making method based on the weighted correlation coefficient under intuitionistic fuzzy environment, Eur. J. Oper. Res., 2010, 205 (1), 202-204 • [24] Liu M., Ren H., A New Intuitionistic Fuzzy Entropy and Application in Multi-Attribute Decision Making, Information, 2014, 5 (4), 587-601 Accepted: 2017-05-16 Published Online: 2017-06-09 Citation Information: Open Engineering, Volume 7, Issue 1, Pages 153–160, ISSN (Online) 2391-5439, Export Citation	2018-07-18 16:48:24	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 32, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6447772979736328, "perplexity": 1948.0305176909624}, "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/1531676590295.61/warc/CC-MAIN-20180718154631-20180718174631-00104.warc.gz"}
http://www.progressingeography.com/CN/10.11820/dlkxjz.2011.11.004	• 水文过程 • 沂河流域水文特征变化及其驱动因素 1. 中国矿业大学资源与地球科学学院,徐州 221116 • 收稿日期:2010-11-01 修回日期:2011-03-01 出版日期:2011-11-25 发布日期:2011-11-25 • 作者简介:薛丽芳(1975-),女,博士,副教授,研究方向为城市与区域规划、流域规划。E-mail: xuel76@163.com • 基金资助: 国家自然科学基金项目(40371113);中国矿业大学青年科研基金项目(2008A027)。 Variations of the Hydrological Characteristics and Driving Factors in the Yihe River Basin XUE Lifang, TAN Haiqiao 1. School of Resource and Geo-Science, China University of Mining & Technology, Xuzhou 221116, China • Received:2010-11-01 Revised:2011-03-01 Online:2011-11-25 Published:2011-11-25 Abstract: Taking the Yihe River basin as a study region, this paper analyzes the variations of hydrological characteristics such as precipitation, runoff and peak discharge, during the period 1951-2002. Time series analysis methods, such as the Kendall method and orderly cluster analysis are used to test the change trend and the mutation of precipitation-runoff. The influences of climate change and human activities on runoff were examined quantitatively based on the comparison between the observed runoff and regression simulation data of the natural runoff at Linyi hydrological station. Some conclusions are drawn as follows. (1) During recent 50 years, the Kendall test value of the annual precipitation is -1.028 at Linyi hydrological station, showing that the precipitation decreased slightly. However, the value of annual runoff is -3.689, decreased evidently, and monthly runoff shows the same trend. That is the responses of runoff to precipitation tend to be weak. But the change trend of rainstorm runoff shows that single storm-runoff keeps strong consistency with rainfall. (2) Based on the runoff mutation, the annual precipitation-runoff process can be divided into three periods, 1951-1964, 1965-1975, and 1976-2002. In the 10-year scale, the observed runoff shows a relatively consistent trend with rainfall. To some extent, precipitation controls the evolution of runoff. (3) Since the mid-1960s, the mean annual runoff at Linyi hydrological station decreased by 148.8 mm, accounting for 51.6% of the mean annual natural runoff. The contribution rate of climate change and human activities to runoff reduction is 39.3% and 60.7%, respectively. The human activities, such as hydraulic engineering construction, land use and land cover change and water resources development, have a profound impact on runoff. The river basin sustainable development requires reasonable constraints to human activities to comply with the nature law of hydrological cycle.	2021-04-16 07:30:31	{"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": 0.18704567849636078, "perplexity": 6646.060054638211}, "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-2021-17/segments/1618038088731.42/warc/CC-MAIN-20210416065116-20210416095116-00427.warc.gz"}
https://blog.csdn.net/seek97/article/details/108664796	# 二、分类工作流程 ## 1.导入数据 letter = readtable("J.txt"); letter将成为一个类似excel的表格 plot(letter.X,letter.Y) 结果： axis equal ## 2.处理数据 letter = readtable("M.txt") letter.X = 1.5letter.X; plot(letter.X,letter.Y) axis equal letter.Time = letter.Time - letter.Time(1) letter.Time = letter.Time/1000 plot(letter.Time,letter.X) plot(letter.Time,letter.Y) ## 3.提取特征 ### Calculating Features What aspects of these letters could be used to distinguish a J from an M or a V? Instead of using the raw signals, the goal is to compute values that distill the entire signal into simple, useful units of information known as features. • For the letters J and M, a simple feature might be the aspect ratio (the height of the letter relative to the width). A J is likely to be tall and narrow, whereas an M is likely to be more square. • Compared to J and M, a V is quick to write, so the duration of the signal might also be a distinguishing feature. letter = readtable("M.txt"); letter.X = letter.X1.5; letter.Time = (letter.Time - letter.Time(1))/1000 plot(letter.X,letter.Y) axis equal dur = letter.Time(end)%%取最后一个时间 aratio = range(letter.Y)/range(letter.X) The range function returns the range of values in an array. That is, range(x) is equivalent to max(x)-min(x). ### Viewing Features load featuredata.mat features scatter(features.AspectRatio,features.Duration) gscatter(features.AspectRatio,features.Duration,features.Character) ## 4.Build a Model ### What is a Classification Model? A classification model is a partitioning of the space of predictor variables into regions. Each region is assigned one of the output classes. In this simple example with two predictor variables, you can visualize these regions in the plane. There is no single absolute “correct” way to partition the plane into the classes J, M, and V. Different classification algorithms result in different partitions. ### 下面开始训练模型：准备训练集和测试数据 load featuredata.mat features testdata ### 选用knn模型对训练集进行训练： knnmodel = fitcknn(features,"Character") ### Making Predictions Having built a model from the data, you can use it to classify new observations. This just requires calculating the features of the new observations and determining which region of the predictor space they are in. predictions = predict(knnmodel,testdata) The predict function determines the predicted class of new observations. predClass = predict(model,newdata) The inputs are the trained model and a table of observations, with the same predictor variables as were used to train the model. The output is a categorical array of the predicted class for each observation in newdata. The file featuredata.mat contains a table testdata that has the same variables as features. However, the observations in testdata are not included in features. Note that testdata contains observations for which the correct class is known (stored in the Character variable). This gives a way to test your model by comparing the classes predicted by the model with the true classes. The predict function will ignore the Character variable when making predictions from the model. ### Algorithm Options By default, fitcknn fits a kNN model with k = 1. That is, the model uses just the single closest known example to classify a given observation. This makes the model sensitive to any outliers in the training data, such as those highlighted in the image above. New observations near the outliers are likely to be misclassified. You can make the model less sensitive to the specific observations in the training data by increasing the value of k (that is, use the most common class of several neighbors). Often this will improve the model's performance in general. However, how a model performs on any particular test set depends on the specific observations in that set. knnmodel = fitcknn(features,"Character","NumNeighbors",5) predictions = predict(knnmodel,testdata) ### 最后，对比一下预测结果和测试样本： [predictions,testdata.Character] ## 6.Evaluate the Model load featuredata.mat testdata knnmodel = fitcknn(features,"Character","NumNeighbors",5); predictions = predict(knnmodel,testdata) iscorrect = predictions == testdata.Character accuracy = sum(iscorrect)/numel(predictions) 结果： iswrong = predictions ~= testdata.Character misclassrate = sum(iswrong)/numel(predictions) [testdata.Character(iswrong) predictions(iswrong)] 结果： Accuracy and misclassification rate give a single value for the overall performance of the model, but it can be useful to see a more detailed breakdown of which classes the model confuses. A confusion matrix shows the number of observations for each combination of true and predicted class. A confusion matrix is commonly visualized by shading the elements according to their value. Often the diagonal elements (the correct classifications) are shaded in one color and the other elements (the incorrect classifications) in another color. You can visualize a confusion matrix by using the confusionchart function. confusionchart(ytrue,ypred); where ytrue is a vector of the known classes and ypred is a vector of the predicted classes. confusionchart(testdata.Character,predictions); ### 实例 load featuredata13letters.mat features testdata gscatter(features.AspectRatio,features.Duration,features.Character) xlim([0 10]) knnmodel = fitcknn(features,"Character","NumNeighbors",5); predictions = predict(knnmodel,testdata); misclass = sum(predictions ~= testdata.Character)/numel(predictions) confusionchart(testdata.Character,predictions); # 三、导入和预处理数据 ## 1.Creating Datastores ### Handwriting Sample Files Samples of each letter were collected from many different volunteers. Some provided more than one sample of each letter. Each sample was saved in a separate file and all the files were stored in one folder. The file names have the form user003_B_2.txt This file would contain the second sample of the letter B written by the volunteer designated “user003”. Use the datastore function to make a datastore to all files containing the letter M. These files have _M_ in their name and a .txt extension. Store the datastore in a variable called letterds ### 通过匹配文件名来打开一系列文件，代表任意字符 letterds = datastore("_M_.txt") You can use the read function to import the data from a file in the datastore. Using the read function the first time will import the data from the first file. Using it a second time will import the data from the second file, and so on. data = read(letterds) plot(data.X,data.Y) Calling the read function again imports the data from the next file in the datastore. data = read(letterds) plot(data.X,data.Y) The readall function imports the data from all the files in the datastore into a single variable. data = readall(letterds) plot(data.X,data.Y) ## 2.Adding a Data Transformantion ### Custom Preprocessing Functions Typically you will want to apply a series of preprocessing operations to each sample of your raw data. The first step to automating this procedure is to make a custom function that applies your specific preprocessing operations. ### Transformed Currently, you still need to call your function manually. To automate your data importing and preprocessing, you want your datastore to apply this function whenever the data is read. You can do this with a transformed datastore. The transform function takes a datastore and a function as inputs. It returns a new datastore as output. This transformed datastore applies the given function whenever it imports data. ### Normalizing Data The location of a letter is not important for classifying it. What matters is the shape. A common preprocessing step for many machine learning problems is to normalize the data. Typical normalizations include shifting by the mean (so that the mean of the shifted data is 0) or shifting and scaling the data into a fixed range (such as [-1, 1]). In the case of the handwritten letters, shifting both the x and y data to have 0 mean will ensure that all the letters are centered around the same point. letterds = datastore("_M_.txt"); data = scale(data); plot(data.X,data.Y) axis equal plot(data.Time,data.Y) ylabel("Vertical position") xlabel("Time") function data = scale(data) data.Time = (data.Time - data.Time(1))/1000; data.X = 1.5data.X; data.X = data.X - mean(data.X,"omitnan"); data.Y = data.Y - mean(data.Y,"omitnan"); end preprocds = transform(letterds,@scale) plot(data.Time,data.Y) # 四、特征工程 ## Statistical Functions ### Measures of Central Tendency FunctionDescription meanArithmetic mean medianMedian (middle) value modeMost frequent value trimmeanTrimmed mean (mean, excluding outliers) geomeanGeometric mean harmeanHarmonic mean ### Measures of Spread FunctionDescription rangeRange of values (largest – smallest) stdStandard deviation varVariance madMean absolute deviation iqrInterquartile range (75th percentile minus 25th percentile) ### Measures of Shape FunctionDescription skewnessSkewness (third central moment) kurtosisKurtosis (fourth central moment) momentCentral moment of arbitrary order ## 1. Calculating Summary Statistic(Quantifying Letter Shapes) ### Descriptive Statistics The handwriting samples have all been shifted so they have zero mean in both horizontal and vertical position. What other statistics could provide information about the shape of the letters? Different letters will have different distributions of points. Statistical measures that describe the shape of these distributions could be useful features. load sampleletters.mat plot(b1.Time,b1.X) hold on plot(b2.Time,b2.X) hold off plot(b1.Time,b1.Y) hold on plot(b2.Time,b2.Y) hold off aratiob = range(b1.Y)/range(b1.X) medxb = median(b1.X,"omitnan") medyb = median(b1.Y,"omitnan") devyb = mad(b1.Y) aratiov = range(v1.Y)/range(v1.X) medxd = median(d1.X,"omitnan") medyd = median(d1.Y,"omitnan") devym = mad(m1.Y) plot(b1.X,b1.Y,b2.X,b2.Y) axis([-1 1 -1 1]) axis equal plot(d1.X,d1.Y,d2.X,d2.Y) axis([-1 1 -1 1]) axis equal ## 2.Finding Peaks Local minima and maxima are often important features of a signal. The islocalmin and islocalmax functions take a signal as input and return a logical array the same length as the signal. idx = islocalmin(x); The value of idx is true whenever the corresponding value in the signal is a local minimum. load sampleletters.mat plot(m1.Time,m1.X) idxmin = islocalmin(m1.X) idxmax = islocalmax(m1.X) plot(m1.Time,m1.X) hold on plot(m1.Time(idxmin),m1.X(idxmin),"o") plot(m1.Time(idxmax),m1.X(idxmax),"s") hold off Local minima and maxima are defined by computing the prominence of each value in the signal. The prominence is a measure of how a value compares to the other values around it. You can obtain the prominence value of each point in a signal by obtaining a second output from islocalmin or islocalmax. [idx,p] = islocalmin(x); [idx,prom] = islocalmin(m1.X); plot(m1.Time,prom) By default, islocalmin and islocalmax find points with any prominence value above 0. This means that a maximum is defined as any point that is larger than the two values on either side of it. For noisy signals you might want to consider only minima and maxima that have a prominence value above a given threshold. idx = islocalmin(x,"MinProminence",threshvalue) When choosing a threshold value, note that prominence values can range from 0 to range(x). idxmin = islocalmin(m1.X,"MinProminence",0.1); idxmax = islocalmax(m1.X,"MinProminence",0.1); nnz(idxmin) sum(idxmin) ## 3.Computing Derivative ### Approximating Velocity An important aspect of detecting letters written on a tablet is that there is useful information in the rhythm and flow of how the letters are written. To describe the shape of the signals through time, it can be useful to know the velocity of the pen, or, equivalently, the slope of the graph of position through time. The raw data recorded from the tablet has only position (not velocity) through time, so velocity must be calculated from the raw data. With discrete data points, this means estimating the velocity by using a finite difference approximation v=Δx/Δt load sampleletters.mat plot(m2.Time,m2.X) grid dX = diff(m2.X); dT = diff(m2.Time); dXdT = dX./dT; plot(m2.Time(1:end-1),dXdT) maxdx = max(dXdT) dYdT = diff(m2.Y)./dT; maxdy = max(dYdT) Due to limits on the resolution of the data collection procedure, the data contains some repeated values. If the position and the time are both repeated, then the differences are both 0, resulting in a derivative of 0/0 = NaN. However, if the position values are very slightly different, then the derivative will be Inf (nonzero divided by 0). Note that max ignores NaN but not Inf because Inf is larger than any finite value. However, for this application, both NaN and Inf can be ignored, as they represent repeated data. You can use the standardizeMissing function to convert a set of values to NaN (or the appropriate missing value for nonnumeric data types). xclean = standardizeMissing(x,0); Here, xclean will be the same as x (including any NaNs), but will have NaN wherever x had the value 0 dYdT = standardizeMissing(dYdT,Inf); maxdy = max(dYdT) Try calculating the derivatives of different sample letters. Note that a negative value divided by zero will result in -Inf. You can pass a vector of values to standardizeMissing to deal with multiple missing values at once. xclean = standardizeMissing(x,[-Inf 0 Inf]); dYdT = standardizeMissing(dYdT,[-Inf 0 Inf]); maxdy = max(dYdT) ## 4.Calculating Correlations ### Measuring Similarity The pair of signals on the left have a significantly different shape to the pair of signals on the right. However, the relationship between the two signals in each pair is similar in both cases: in the blue regions, the upper signal is increasing while the lower signal is decreasing, and vice versa in the yellow regions. Correlation attempts to measure this similarity, regardless of the shape of the signal. load sampleletters.mat plot(v2.X,v2.Y,"o-") For the first half of the letter V, the horizontal and vertical positions have a strong negative linear correlation: when the horizontal position increases, the vertical position decreases proportionally. Similarly, for the second half, the positions have a strong positive correlation: when the horizontal position increases, the vertical position also increases proportionally. The corr function calculates the linear correlation between variables. C = corr(x,y); C = corr(v2.X,v2.Y) C = NaN Because both variables contain missing data, C is NaN. You can use the "Rows" option to specify how to avoid missing values. C = corr(x,y,"Rows","complete"); C = corr(v2.X,v2.Y,"Rows","complete") C = 0.6493 The correlation coefficient is always between -1 and +1. • A coefficient of -1 indicates a perfect negative linear correlation • A coefficient of +1 indicates a perfect positive linear correlation • A coefficient of 0 indicates no linear correlation In this case, there is only a moderate correlation because the calculation has been performed on the entire signal. It may be more informative to consider the two halves of the signal separately. M = [v2.X(1:11) v2.Y(1:11) v2.X(12:22) v2.Y(12:22)] To calculate the correlation between each pair of several variables, you can pass a matrix to the corr function, where each variable is a column of the matrix. M = [x y z]; C = corr(M); Cmat = corr(M,"Rows","complete") The output Cmat is a 4-by-4 matrix of the coefficients of correlation between each pairwise combination of the columns of M. That is, Cmat(j,k) is the correlation of M(:,j) and M(:,k). The matrix is symmetric because the correlation between x and y is the same as the correlation between y and x. The diagonal elements are always equal to 1, because a variable is always perfectly correlated with itself. ## 5.Automating Feature Extraction ### Custom Preprocessing Functions Once you have determined the features you want to extract, you will need to apply the appropriate calculations to every sample in your data set. The first step to automating this procedure is to make a custom function that takes the data as input and returns an array of features as output. ### Creating a Feature Extraction Function Currently the script calculates six features for a given letter (stored in the variable letter). The six features are stored in six separate variables. You can use the table function to combine separate variables into a table. T = table(x,y,z); load sampleletters.mat letter = b1; aratio = range(letter.Y)/range(letter.X) idxmin = islocalmin(letter.X,"MinProminence",0.1); numXmin = nnz(idxmin) idxmax = islocalmax(letter.Y,"MinProminence",0.1); numYmax = nnz(idxmax) dT = diff(letter.Time); dXdT = diff(letter.X)./dT; dYdT = diff(letter.Y)./dT; avgdX = mean(dXdT,"omitnan") avgdY = mean(dYdT,"omitnan") corrXY = corr(letter.X,letter.Y,"rows","complete") featurenames = ["AspectRatio","NumMinX","NumMinY","AvgU","AvgV","CorrXY"]; By default, the table constructed with the table function has default variable names. To make a table with more useful names, use the 'VariableNames' option. T = table(x,y,z,'VariableNames',["X","Y","Z"]); Typically you can use either single or double quotes to specify option names. However, because strings can represent data for your table, you need to use single quotes when specifying the 'VariableNames' option. feat = table(aratio,numXmin,numYmax,avgdX,avgdY,corrXY) feat = table(aratio,numXmin,numYmax,avgdX,avgdY,corrXY,'VariableNames',featurenames) At the end of the script, add a local function called extract that takes a single variable, letter, as input and returns a table of features, feat, as output. Copy the code from the beginning of the script and from task 2 to make the body of the function. Test your function by calling it with b2 as input. Store the result in a variable called featB2 featB2 = extract(b2) function feat = extract(letter) aratio = range(letter.Y)/range(letter.X); idxmin = islocalmin(letter.X,"MinProminence",0.1); numXmin = nnz(idxmin); idxmax = islocalmax(letter.Y,"MinProminence",0.1); numYmax = nnz(idxmax); dT = diff(letter.Time); dXdT = diff(letter.X)./dT; dYdT = diff(letter.Y)./dT; avgdX = mean(dXdT,"omitnan"); avgdY = mean(dYdT,"omitnan"); corrXY = corr(letter.X,letter.Y,"rows","complete"); featurenames = ["AspectRatio","NumMinX","NumMinY","AvgU","AvgV","CorrXY"]; feat = table(aratio,numXmin,numYmax,avgdX,avgdY,corrXY,'VariableNames',featurenames); end ### Transformed Datastores To automate your feature extraction, you want your datastore to apply your extraction function whenever the data is read. As with preprocessing, you can do this with a transformed datastore. From the raw data, you will typically need to apply both preprocessing and feature extraction functions. You can apply the transform function repeatedly to add any number of transformations to the datastore to the raw data. The script currently applies the scale function to the files in the datastore letterds. The transformed datastore is stored in the variable preprocds. letterds = datastore(".txt"); preprocds = transform(letterds,@scale) featds = transform(preprocds,@extract) function data = scale(data) % Normalize time [0 1] data.Time = (data.Time - data.Time(1))/(data.Time(end) - data.Time(1)); % Fix aspect ratio data.X = 1.5data.X; % Center X & Y at (0,0) data.X = data.X - mean(data.X,"omitnan"); data.Y = data.Y - mean(data.Y,"omitnan"); % Scale to have bounding box area = 1 scl = 1/sqrt(range(data.X)range(data.Y)); data.X = scldata.X; data.Y = scl*data.Y; end function feat = extract(letter) % Aspect ratio aratio = range(letter.Y)/range(letter.X); % Local max/mins idxmin = islocalmin(letter.X,"MinProminence",0.1); numXmin = nnz(idxmin); idxmax = islocalmax(letter.Y,"MinProminence",0.1); numYmax = nnz(idxmax); % Velocity dT = diff(letter.Time); dXdT = diff(letter.X)./dT; dYdT = diff(letter.Y)./dT; avgdX = mean(dXdT,"omitnan"); avgdY = mean(dYdT,"omitnan"); % Correlation corrXY = corr(letter.X,letter.Y,"rows","complete"); % Put it all together into a table featurenames = ["AspectRatio","NumMinX","NumMinY","AvgU","AvgV","CorrXY"]; feat = table(aratio,numXmin,numYmax,avgdX,avgdY,corrXY,'VariableNames',featurenames); end Use the readall function to read, preprocess, and extract features from all the data files. Store the result in a variable called data. There are 12 files and the extract function calculates six features for each. Hence, data should be a 12-by-6 table. Visualize the imported data by making a scatter plot of AspectRatio on the x-axis and CorrXY on the y-axis. data = readall(featds) scatter(data.AspectRatio,data.CorrXY) The letters that the data represents are given in the data file names, which are of the form usernnn_X_n.txt. Note that the letter name appears between underscore characters (_X_). You can use the extractBetween function to extract text that occurs between given strings. extractedtxt = extractBetween(txt,"abc","xyz") If txt is the string array ["hello abc 123 xyz","abcxyz","xyzabchelloxyzabc"], then extractedtxt will be [" 123 ","","hello"]. knownchar = extractBetween(letterds.Files,"_","_") For classification problems, you typically want to represent the known label as a categorical variable. You can use the categorical function to convert an array to categorical type. xcat = categorical(x) By default, the unique values in x will be used to define the set of categories. knownchar = categorical(knownchar) It is convenient to have the known classes associated with the training data. Recall that you can create new variables in a table by assigning to a variable using dot notation. T.newvar = workspacevar data.Character = knownchar gscatter(data.AspectRatio,data.CorrXY,data.Character) # 五、分类模型 ## 1.Training a Model ### Handwriting Features The MAT-file letterdata.mat contains the table traindata which represents feature data for 2906 samples of individual letters. There are 25 features, including statistical measures, correlations, and maxima/minima for the position, velocity, and pressure of the pen. load letterdata.mat traindata histogram(traindata.Character) A boxplot is a simple way to visualize multiple distributions. boxplot(x,c) This creates a plot where the boxes represent the distribution of the values of x for each of the classes in c. If the values of x are typically significantly different for one class than another, then x is a feature that can distinguish between those classes. The more features you have that can distinguish different classes, the more likely you are to be able to build an accurate classification model from the full data set. boxplot(traindata.MADX,traindata.Character) Use the command classificationLearner to open the Classification Learner app. • Select traindata as the data to use. • The app should correctly detect Character as the response variable to predict. • Choose the default validation option. • Select a model and click the Train button. Try a few of the standard models with default options. See if you can achieve at least 80% accuracy. Note that SVMs work on binary classification problems (i.e. where there are only two classes). To make SVMs work on this problem, the app is fitting many SVMs. These models will therefore be slow to train. Similarly, ensemble methods work by fitting multiple models. These will also be slow to train. ## 2.Making Predictions The MAT-file letterdata.mat contains traindata, the table of data used to train the model knnmodel. It also contains testdata which is a table of data (with the same features as traindata) that the model has never seen before. Recall that you can use the predict function to obtain a model's predictions for new data. preds = predict(model,newdata) load letterdata.mat traindata knnmodel = fitcknn(traindata,"Character","NumNeighbors",5,"Standardize",true,"DistanceWeight","squaredinverse"); testdata predLetter = predict(knnmodel,testdata) misclassrate = sum(predLetter ~= testdata.Character)/numel(predLetter) The response classes are not always equally distributed in either the training or test data. Loss is a fairer measure of misclassification that incorporates the probability of each class (based on the distribution in the data). loss(model,testdata) testloss = loss(knnmodel,testdata) ## 3.Investigating Misclassifications（Identifying Common Misclassifications） ### The Confusion Matrix For any response class X, you can divide a machine learning model's predictions into four groups: • True positives (green) – predicted to be X and was actually X • True negatives (blue) – predicted to be not X and was actually not X • False positives (yellow) – predicted to be X but was actually not X • False negatives (orange) – predicted to be not X but was actually X ### False Negatives With 26 letters, you will need to enlarge the confusion chart to make the values visible. If you open the plot in a separate figure, you can resize it as large as you like. The row summary shows the false negative rate for each class. This shows which letters the kNN model has the most difficulty identifying (i.e., the letters the model most often thinks are som This model has particular difficulty with the letter U, most often mistaking it for M, N, or V. Some confusions seem reasonable, such as U/V or H/N. Others are more surprising, such as U/K. Having identified misclassifications of interest, you will probably want to look at some the specific data samples to understand what is causing the misclassification. ### Identifying Common Misclassifications When making a confusion chart, you can add information about the false negative and false positive rate for each class by adding row or column summaries, respectively. confusionchart(...,"RowSummary","row-normalized"); load letterdata.mat testdata predLetter confusionchart(testdata.Character,predLetter); confusionchart(testdata.Character,predLetter,"RowSummary","row-normalized"); Recall that the Files property of a datastore contains the file names of the original data. Hence, when you import the data and extract the features, you can keep a record of which data file is associated with each observation. The string array testfiles contains the file names for the test data. Use the logical array falseneg as an index into testfiles to determine the file names of the observations that were incorrectly classified as the letter U. Store the result in a variable called fnfiles. Similarly, use falseneg as an index into predLetter to determine the associated predicted letters. Store the result in a variable called fnpred. falseneg = (testdata.Character == "U") & (predLetter ~= "U"); fnfiles = testfiles(falseneg) fnpred = predLetter(falseneg) Use the readtable function to import the data in the fourth element of fnfiles into a table called badU. Visualize the letter by plotting Y against X. badU = readtable(fnfiles(4)); title("Prediction: "+string(fnpred(4))) ## 4.Investigating Misclassifications（Investigating Features） load letterdata.mat traindata testdata predLetter idx = (traindata.Character == "N") \| (traindata.Character == "U"); UorN = traindata(idx,:) idx = (testdata.Character == "U") & (predLetter ~= "U"); fnU = testdata(idx,:) Categorical variables maintain the full list of possible classes, even when only a subset are present in the data. When examining a subset, it can be useful to redefine the set of possible classes to only those that are in the data. The removecats function removes unused categories. cmin = removecats(cfull) UorN.Character = removecats(UorN.Character); You can use curly braces ({ }) to extract data from a table into an array of a single type. datamatrix = datatable{1:10,4:6} This extracts the first 10 elements of variables 4, 5, and 6. If these variables are numeric, datamatrix will be a 10-by-3 double array. UorNfeat = UorN{:,1:end-1}; fnUfeat = fnU{:,1:end-1}; parallel coordinates plot shows the value of the features (or “coordinates”) for each observation as a line. parallelcoords(data) To compare the feature values of different classes, use the "Group" option. parallelcoords(data,"Group",classes) parallelcoords(UorNfeat,"Group",UorN.Character) Because a parallel coordinates plot is just a line plot, you can add individual observations using the regular plot function. hold on plot(fnUfeat(4,:),"k") hold off Use the zoom tool to explore the plot. Note that N and U have similar values for many features. Are there any features that help distinguish these letters from each other? A kNN model uses the distance between observations, where the distance is calculated over all the features. Does this explain why N and U are hard to distinguish, even if there are some features that separate them? When plotting multiple observations by groups, it can be helpful to view the median and a range for each group, rather than every individual observation. You can use the "Quantile" option to do this. parallelcoords(...,"Quantile",0.2) parallelcoords(UorNfeat,"Group",UorN.Character,"Quantile",0.2) ## 5. Improving the Model Even if your model works well, you will typically want to look for improvements before deploying it for use. Theoretically, you could try to improve your results at any part of the workflow. However, collecting data is typically the most difficult step of the process, which means you often have to work with the data you have. If you have the option of collecting more data, you can use the insights you have gained so far to inform what new data you need to collect. In the handwriting example, volunteers were instructed only to write lower-case letters “naturally”. Investigating the data set reveals that there are often discrete groups within a particular letter, such as a block-letter style and a cursive style. This means that two quite different sets of features can represent the same letter. One way to improve the model would be treat these variants as separate classes. However, this would mean having many more than 26 classes. To train such a model, you would need to collect more samples, instruct the volunteers to write both block and cursive style, and label the collected data accordingly. If you have the option of collecting more data, you can use the insights you have gained so far to inform what new data you need to collect. In the handwriting example, volunteers were instructed only to write lower-case letters “naturally”. Investigating the data set reveals that there are often discrete groups within a particular letter, such as a block-letter style and a cursive style. This means that two quite different sets of features can represent the same letter. One way to improve the model would be treat these variants as separate classes. However, this would mean having many more than 26 classes. To train such a model, you would need to collect more samples, instruct the volunteers to write both block and cursive style, and label the collected data accordingly. Low accuracy in both your training and testing sets is an indication that your features do not provide enough information to distinguish the different classes. In particular, you might want to look at the data for classes that are frequently confused, to see if there are characteristics that you can capture as new features. However, too many features can also be a problem. Redundant or irrelevant features often lead to low accuracy and increase the chance of overfitting – when your model is learning the details of the training rather than the broad patterns. A common sign of overfitting is that your model performs well on the training set but not on new data. You can use a feature selection technique to find and remove features that do not significantly add to the performance of your model. You can also use feature transformation to perform a change of coordinates on your features. With a technique such as Principal Component Analysis (PCA), the transformed features are chosen to minimize redundancy and ordered by how much information they contain. Low accuracy in both your training and testing sets is an indication that your features do not provide enough information to distinguish the different classes. In particular, you might want to look at the data for classes that are frequently confused, to see if there are characteristics that you can capture as new features. However, too many features can also be a problem. Redundant or irrelevant features often lead to low accuracy and increase the chance of overfitting – when your model is learning the details of the training rather than the broad patterns. A common sign of overfitting is that your model performs well on the training set but not on new data. You can use a feature selection technique to find and remove features that do not significantly add to the performance of your model. You can also use feature transformation to perform a change of coordinates on your features. With a technique such as Principal Component Analysis (PCA), the transformed features are chosen to minimize redundancy and ordered by how much information they contain. The Classification Learner app provides an easy way to experiment with different models. You can also try different options. For example, for kNN models, you can vary the number of neighbors, the weighting of the neighbors based on distance, and the way that distance is defined. Some classification methods are highly sensitive to the training data, which means you might get very different predictions from different models trained on different subsets of the data. This can be harnessed as a strength by making an ensemble – training a large number of these so-called weak learners on different permutations of the training data and using the distribution of individual predictions to make the final prediction. For the handwriting example, some pairs of letters (such as N and V) have many similar features and are distinguished by only one or two key features. This means that a distance-based method such as kNN may have difficulty with these pairs. An alternative approach is to use an ensemble approach known as Error-Correcting Output Coding (ECOC) which use multiple models to distinguish between different binary pairs of classes. Hence, one model can distinguish between N and V, while another can distinguish between N and E, and another between E and V, and so on. When trying to evaluate different models, it is important to have an accurate measure of a model's performance. The simplest, and computationally cheapest, way to do validation is holdout – randomly divide your data into a training set and a testing set. This works for large data sets. However, for many problems, holdout validation can result in the test accuracy being dependent on the specific choice of test data. You can use k-fold cross-validation to get a more accurate estimate of performance. In this approach, multiple models are trained and tested, each on a different division of the data. The reported accuracy is the average from the different models. Accuracy is only one simple measure of the model's performance. It is also important to consider the confusion matrix, false negative rates, and false positive rates. Furthermore, the practical impact of a false negative may be significantly different to that of a false positive. For example, a false positive medical diagnosis may cause stress and expense, but a false negative could be fatal. In these cases, you can incorporate a cost matrix into the calculation of a model's loss. • 点赞 • 评论 • 分享 x 海报分享扫一扫，分享海报 • 收藏 3 • 手机看分享到微信朋友圈 x 扫一扫，手机阅读 • 打赏打赏南叔先生你的鼓励将是我创作的最大动力 C币余额 2C币 4C币 6C币 10C币 20C币 50C币 • 一键三连点赞Mark关注该博主, 随时了解TA的最新博文 04-23 619 06-10 6686 05-08 04-01 08-26 07-08 7337 ©️2020 CSDN 皮肤主题: 精致技术设计师:CSDN官方博客	2020-10-22 19:37:58	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 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": 0.5598230361938477, "perplexity": 1135.0258603549942}, "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-2020-45/segments/1603107880014.26/warc/CC-MAIN-20201022170349-20201022200349-00628.warc.gz"}
https://socratic.org/questions/what-is-the-name-of-the-binary-compound-that-has-the-formula-cao	# What is the name of the binary compound that has the formula CaO? $\text{Calcium oxide}$ This is also known as $\text{quick lime}$ or $\text{burnt lime}$.	2022-01-18 19:14:50	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 3, "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": 0.762566089630127, "perplexity": 261.13726161426285}, "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/1642320300997.67/warc/CC-MAIN-20220118182855-20220118212855-00276.warc.gz"}
https://hpmuseum.org/forum/showthread.php?tid=4047&pid=37473&mode=threaded	[WP 34s] Trapezoidal approximation of area under curve 06-16-2015, 10:47 AM (This post was last modified: 06-16-2015 10:49 AM by Marcio.) Post: #7 Marcio Senior Member Posts: 438 Joined: Feb 2015 RE: [WP 34s] Trapezoidal approximation of area under curve Hello again, Does anyone know how to do $$RCL+ Z$$ on the 35s? From what I saw in the manual, one has to use the EQN inside the program in order to recall the $$z$$-register, which is somewhat dangerous. Thanks. « Next Oldest \| Next Newest » Messages In This Thread [WP 34s] Trapezoidal approximation of area under curve - Marcio - 06-01-2015, 01:53 PM RE: [WP 34s] Trapezoidal approximation of area under curve - Marcus von Cube - 06-01-2015, 03:54 PM RE: [WP 34s] Trapezoidal approximation of area under curve - Dave Britten - 06-01-2015, 05:36 PM RE: [WP 34s] Trapezoidal approximation of area under curve - Thomas Klemm - 06-01-2015, 06:56 PM RE: [WP 34s] Trapezoidal approximation of area under curve - Dieter - 06-02-2015, 09:23 PM RE: [WP 34s] Trapezoidal approximation of area under curve - Marcio - 06-02-2015, 02:46 AM RE: [WP 34s] Trapezoidal approximation of area under curve - Marcio - 06-16-2015 10:47 AM RE: [WP 34s] Trapezoidal approximation of area under curve - Dieter - 06-16-2015, 11:31 AM RE: [WP 34s] Trapezoidal approximation of area under curve - Thomas Klemm - 06-16-2015, 12:38 PM RE: [WP 34s] Trapezoidal approximation of area under curve - Marcio - 06-16-2015, 12:57 PM User(s) browsing this thread: 1 Guest(s)	2022-05-25 16:43:57	{"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": 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": 0.5711573958396912, "perplexity": 14598.404218465806}, "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-2022-21/segments/1652662588661.65/warc/CC-MAIN-20220525151311-20220525181311-00015.warc.gz"}
http://www.physicsforums.com/showthread.php?p=3507219	# Will the earth and sun ever be tidally locked? by ARAVIND113122 Tags: earth, locked, tidally P: 54 suppose there are two bodies,one revolving in an orbit around the other[like the earth moon system]Differences in orbital and axial rotation of a small body results in a torque applied on it by the larger body. This results in the smaller body being tidally locked. THEN WHY ISN'T THE EARTH TIDALLY LOCKED WITH THE SUN?WILL IT EVER BE? Emeritus Sci Advisor PF Gold P: 2,361 There hasn't been enough time. Even the Moon, which has ~twice the tidal effect on the Earth as the Sun does hasn't had enough time to tidally lock the Earth it. P: 80 Quote by Janus There hasn't been enough time. Even the Moon, which has ~twice the tidal effect on the Earth as the Sun does hasn't had enough time to tidally lock the Earth it. what's this TIME got to do with the above question? really I can't understand it and could you please explain me in detail. P: 1,815 Will the earth and sun ever be tidally locked? As Janus says, the moon has ~twice the tidal effect as the sun, so it would seem unlikely until the moon's orbit moves far enough away from earth that the sun has a greater effect or that the period of lunar orbit equals one earth year. P: 269 Quote by Astro.padma what's this TIME got to do with the above question? really I can't understand it and could you please explain me in detail. As time progresses the Earth's rotation is slowing. The Earth and the moon might eventually be tidally locked if the moon doesn't drift away far enough to let the suns tidal forces on the Earth over come it's tidal forces on the Earth. At which point the Earth will become tidally locked with the sun. Emeritus PF Gold P: 2,361 Quote by Astro.padma what's this TIME got to do with the above question? really I can't understand it and could you please explain me in detail. You can estimate the time it would take for one body to tidal lock to another by the formula: $$\frac{\omega a^6 I Q}{3GM^2k_2 R^5}$$ Tidal locking takes time to occur. The factors include the initial rotation speed of the body, its distance from the other body, Its moment of Inertia, the Mass and radius of the body it is orbiting, plus a couple of coupling factors. I mentioned the Moon because its tidal effect on the Earth is larger than the Sun's, so if not enough time has passed for the Moon to slow the Earth's rotation to match its orbit, then definitely not enough time has passed for the Sun the tidally lock the Earth to it. P: 841 Ummm...The moon is tidally locked. The above equation is a good one though. P: 54 Thank you very much!!! Mentor P: 15,170 Quote by Travis_King Ummm...The moon is tidally locked. You missed the point. While the Moon is tidally locked to the Earth, the Earth is not tidally locked to the Moon. P: 841 Not to sound rude, but: so? The earth-moon and earth-sun systems are independent (barring the rotational effects the moon has on the earth). There's no sense in comparing the two. Besides, the question is whether the Earth will become tidally locked to the Sun. In this case, the earth is the satellite and the sun is the primary. In the earth-moon, the earth is the primary and the moon is the satellite. The OP asked whether or not the earth will be tidally locked to the Sun, not the other way around. P: 80 Quote by Janus You can estimate the time it would take for one body to tidal lock to another by the formula: $$\frac{\omega a^6 I Q}{3GM^2k_2 R^5}$$ Tidal locking takes time to occur. The factors include the initial rotation speed of the body, its distance from the other body, Its moment of Inertia, the Mass and radius of the body it is orbiting, plus a couple of coupling factors. I mentioned the Moon because its tidal effect on the Earth is larger than the Sun's, so if not enough time has passed for the Moon to slow the Earth's rotation to match its orbit, then definitely not enough time has passed for the Sun the tidally lock the Earth to it. K...Thanks for the reply but is this what you meant by the above? : The Moon's tidal effect on the Earth is larger than that of the Sun's. So only at that point of time, when the Moon's effect gets decreased, the Sun could tidally lock the Earth? P: 80 Quote by Janus $$\frac{\omega a^6 I Q}{3GM^2k_2 R^5}$$ I mentioned the Moon because its tidal effect on the Earth is larger than the Sun's, so if not enough time has passed for the Moon to slow the Earth's rotation to match its orbit, then definitely not enough time has passed for the Sun the tidally lock the Earth to it. Sir..Why in this context is the Moon's effect superior to that of the sun? Why do you think that the time to be taken by the Sun to tidally lock would be longer than the time taken by the Moon?? Emeritus PF Gold P: 2,361 Quote by Astro.padma Sir..Why in this context is the Moon's effect superior to that of the sun? Why do you think that the time to be taken by the Sun to tidally lock would be longer than the time taken by the Moon?? Tidal force is proportional to the mass exerting the force and inversely proportional to its distance. The moon is 1/27210884 the mass of the Sun, but it is 400 times closer. So the Sun's tidal force on the Earth is 27210884/400^3 = 0.4252 times that of the Moon. It is This tidal force acting on the Earth which would cause it to lock with either the Earth or Moon. Since the Moon exerts the greater tidal force on the Earth, It would be the first to tidally lock the Earth to it. Actually, if you look at the formula I gave for the time for tidal locking to occur, you will note that it increases by the distance between the bodies (a) to the power of 6, and decreases by the mass of the acting body by only the square of the mass. So 400^6/27210884^2 = 5.53, meaning that it would take ~5.5 times longer for the Sun to lock the Earth to it than it would for the Moon to lock the Earth. P: 841 I see what you are saying. I stand corrected. P: 80 Quote by Janus So 400^6/27210884^2 = 5.53, meaning that it would take ~5.5 times longer for the Sun to lock the Earth to it than it would for the Moon to lock the Earth. Oh K...now I got it :) but I really wonder why couldn't the Moon yet tidally lock the Earth?? Not Enough time or anything else?? If it is a matter of time, on what assumptions was the equation given by you framed? Ive tried to google it but couldn't find the answer. P: 841 The equation above describes the time required in those conditions to attain tidal locking. It's not a matter of assumptions, really. It's a matter of physics, and I'm sure the equations were derived painstakingly and are very long. P: 80 Quote by Travis_King The equation above describes the time required in those conditions to attain tidal locking. If am not bugging you people, here what does it mean by "those conditions" ? Related Discussions Introductory Physics Homework 1 Astronomy & Astrophysics 7 General Discussion 9 Astronomy & Astrophysics 1 General Discussion 15	2014-09-01 23:50:49	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4277592599391937, "perplexity": 522.2517845510438}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-35/segments/1409535920849.16/warc/CC-MAIN-20140909050316-00265-ip-10-180-136-8.ec2.internal.warc.gz"}
https://crypto.stackexchange.com/questions/76620/a5-2-session-key-derivation	# A5/2 session key derivation Let talk about Implementation and performance analysis of Barkan, Biham and Keller’s attack on A5/2. In this paper, mentioned that we need to brute force attack on register 4 ($$R_4$$) to get keystream from below: $$HP^{-1}C=HP^{-1}K. \tag{1}$$ Where $$HP^{-1}$$ is parity check and $$C$$ is ciphertext and $$K$$ is keystream. On the other hand, the equations on the key stream must be adapted into equations on the variables of the $$LFSRs$$. So we have the following linear system: $$Sr=K. \tag{2}$$ where $$S$$ is multiplication matrix of size $$1368*656$$ and $$K$$ is the concatenation of $$3$$ unknown key streams $$k_1$$, $$k_2$$, and $$k_3$$, and $$r$$ is the vector of unknowns representing the state of the $$LFSRs$$. 1. Where is $$R_4$$ influence in the first equation? 2. How we can construct the second equation(Please with details)? The dependencies between the LFSRs and the keystream vary greatly with the initial value of $$R_4$$ and since the variables of $$R_4$$ play no role in the value of the keystream, so it is not possible to find them. So, later on they suggest the values of $$R_4$$ must be bruteforced, as a precomputation. • Thanks, but it is not clear for me how multiplication matrix($S$) created? If it is scalar, how we can build it? – R. Jalaei Salahi Dec 29 '19 at 12:59	2021-04-16 08:39:47	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 20, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7093300819396973, "perplexity": 449.71801210888583}, "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-2021-17/segments/1618038088731.42/warc/CC-MAIN-20210416065116-20210416095116-00550.warc.gz"}
https://www.catalyzex.com/paper/arxiv:2210.13537	Get our free extension to see links to code for papers anywhere online! # Private Online Prediction from Experts: Separations and Faster Rates Oct 24, 2022 Hilal Asi, Vitaly Feldman, Tomer Koren, Kunal Talwar Online prediction from experts is a fundamental problem in machine learning and several works have studied this problem under privacy constraints. We propose and analyze new algorithms for this problem that improve over the regret bounds of the best existing algorithms for non-adaptive adversaries. For approximate differential privacy, our algorithms achieve regret bounds of $\tilde{O}(\sqrt{T \log d} + \log d/\varepsilon)$ for the stochastic setting and $\tilde O(\sqrt{T \log d} + T^{1/3} \log d/\varepsilon)$ for oblivious adversaries (where $d$ is the number of experts). For pure DP, our algorithms are the first to obtain sub-linear regret for oblivious adversaries in the high-dimensional regime $d \ge T$. Moreover, we prove new lower bounds for adaptive adversaries. Our results imply that unlike the non-private setting, there is a strong separation between the optimal regret for adaptive and non-adaptive adversaries for this problem. Our lower bounds also show a separation between pure and approximate differential privacy for adaptive adversaries where the latter is necessary to achieve the non-private $O(\sqrt{T})$ regret.	2023-01-28 00:58:16	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.644630491733551, "perplexity": 543.8090976199866}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764499468.22/warc/CC-MAIN-20230127231443-20230128021443-00383.warc.gz"}
http://cbse-notes.blogspot.com/2012/09/cbse-class-10-maths-ch1-real-numbers.html	Wednesday, September 5, 2012 CBSE Class 10 - Maths - Ch1 - Real Numbers (MCQs) Real Numbers Symbol to represent set of Real Numbers (MCQs) Q1: Which of the following has a non-terminating decimal expansion? (a) 77/210 (b) 23/8 (c) 17/8 (d) 35/50 Q2(CBSE 2010): The decimal expansion of 141/120 will terminate after how many places of decimals ? (a) 1 (b) 2 (c) 3 (d) will not terminate Q3: HCF of 84 and 270 is (a) 8 (b) 6 (c) 4 (d) 2 Q4(CBSE 2010): If p, q are two consecutive natural numbers, then HCF(p, q) is: (a) q (b) p (c) 1 (d) pq Q5: If HCF if 60 and 168 is 12, what is the LCM (a) 480 (b) 240 (c) 420 (d) 840 Q6(CBSE 2010): How many prime factors are there in prime factorisation of 5005 ? (a) 2 (b) 4 (c) 6 (d) 7 Q7(CBSE 2010): A rational number can be expressed as a terminating decimal if the denominator has factors (a) 2, 3 or 5 (b) 2 or 3 (c) 3 or 5 (d) 2 or 5 Q8(CBSE 2010): If p, q are two prime numbers, then LCM(p, q) is : (a) 1 (b) p (c) q (d) pq Q9: Let x = p/q be a rational number such that prime factorization of q is NOT in the form of 2n5m, where m and n are non-negative integers. Then x has a decimal representation which is: (a) terminating (b) Non-terminating, repeating (c) Non-terminating, non-repeating (d) None of these Q10(CBSE 2010): Euclid’s division lemma states that for any two positive integer ‘a’ and ‘b’ there exists unique integers q and r such that a=bq+r where r must satisfy: (a) 1 ≤ r < b (b) 0 < r ≤ b (c) 0 ≤ r < b (d) 0 < r < b Q11(CBSE 2010): Which of the following is not an irrational number ? (a) 5 - √3 (b) 5 + √3 (c) 4 + √2 (d) 5 + √9 Q12(CBSE 2011): Which of the following numbers has terminating decimal expansion? (a) 37/45 (b) 21/(2356) (c) 17/49 (d) 89/(2232) Q13: The mathematician who gave the term 'algorithm'? (a) Euclid (b) Gold Bach (c) Khwarizmi (d) Gauss Q14: The decimal expansion of π (a) is terminating. (b) is non-terminating and repeating (c) is non-terminating and non-repeating (d) None of these 1: (a) 77/210 [Hint: For a rational number in the form p/q, such that the prime factors of q are not of the form 2n5m, where n and m are non negative integers. Then the rational number has a decimal expansion which is non terminating repeating (recurring).] 2: (c) 3 [Hint: Rational form of number p/q, where q = 2352] 3: (b) 6 4: (c) 1 5: (d) 840 [Hint: LCM × HCF = Product of two numbers] 6: (b) 4 [Hint: 5005 = 5×7×11×13] 7: (d) 2 or 5 8: (d) pq 9: (b) Non-terminating, repeating 10: (c) 0 ≤ r < b 11:(d) 5 + √9 [Hint: √9 = 3, 5 + √9 = 8, a rational number] 12: (b) 21/(2356) [Hint: for rational number x = p/q, if factors are in the form of 2n5m, where n and m are non negative integers, the rational number has a decimal expansion of terminating type.] 13: (c) Khwarizmi 14: (c) is non-terminating and non-repeating	2015-01-25 18:15:29	{"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": 0.8214913010597229, "perplexity": 2849.318129468632}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-06/segments/1422115861305.18/warc/CC-MAIN-20150124161101-00078-ip-10-180-212-252.ec2.internal.warc.gz"}
https://cstheory.stackexchange.com/questions/27833/does-np-hardness-of-c-approximation-for-some-c1-imply-apx-hardness	# Does $NP$-hardness of $c$-approximation (for some $c>1$) imply $APX$-hardness? Assume that for a given minimization problem with only integer solutions, it is $NP$-hard to decide if the optimal solution is 5 or 6. I.e., a polynomial-time algorithm with an approximation ratio better than 6/5 would imply $P=NP$. 1) Does this imply that the problem is $APX$-hard as well? 2) Is there a common way of stating this inapproximability fact, besides stating that "it is $NP$-hard to approximate with an approximation ratio strictly better than 6/5"? Thank you! The answer for (1) is "unlikely". It is simple to show (reduce from $Partition$) there exists no $\alpha$-approximation for Bin Packing, for any $\alpha<\frac{3}{2}$, unless $P=NP$. That said, Crescenzi et al. have shown that unless the polynomial hierarchy collapses, Bin Packing is not APX-Hard. As for (2), perhaps you could phrase it as "Does not admit $PTAS$ unless $P=NP$". • @cs_student_273 - you are welcome. – R B Dec 15 '14 at 11:04	2021-10-24 18:11:04	{"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": 0.9606181979179382, "perplexity": 519.5323432233826}, "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-2021-43/segments/1634323587593.0/warc/CC-MAIN-20211024173743-20211024203743-00094.warc.gz"}
http://inperc.com/wiki/index.php?title=Restriction	This site contains: mathematics courses and book; covers: image analysis, data analysis, and discrete modelling; provides: image analysis software. Created and run by Peter Saveliev. # Restriction Suppose we have sets $X$ and $Y$, a function $f: X → Y$, and a subset $A$ of $X$. Then the restriction function $f\|_A: A → Y$ of $f$ is given by $$f\|_A(x) = f(x)$$ for all $x∈A$. In that case, the original function $f$ is an extension of $f\|_A$.	2013-12-11 22:10:42	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7534527778625488, "perplexity": 330.6480352756449}, "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-2013-48/segments/1386164050279/warc/CC-MAIN-20131204133410-00061-ip-10-33-133-15.ec2.internal.warc.gz"}
http://nrich.maths.org/public/leg.php?code=-68&cl=2&cldcmpid=4950	# Search by Topic #### Resources tagged with Visualising similar to Weekly Problem 26 - 2006: Filter by: Content type: Stage: Challenge level: ##### Other tags that relate to Weekly Problem 26 - 2006 Right angled triangles. Similarity. Circles. Squares. Circle theorems. Visualising. Area. Mathematical reasoning & proof. Pythagoras' theorem. ### There are 253 results Broad Topics > Using, Applying and Reasoning about Mathematics > Visualising ### Tied Up ##### Stage: 3 Challenge Level: In a right angled triangular field, three animals are tethered to posts at the midpoint of each side. Each rope is just long enough to allow the animal to reach two adjacent vertices. Only one animal. . . . ### Dissect ##### Stage: 3 Challenge Level: It is possible to dissect any square into smaller squares. What is the minimum number of squares a 13 by 13 square can be dissected into? ### Baravelle ##### Stage: 2, 3 and 4 Challenge Level: What can you see? What do you notice? What questions can you ask? ### Trice ##### Stage: 3 Challenge Level: ABCDEFGH is a 3 by 3 by 3 cube. Point P is 1/3 along AB (that is AP : PB = 1 : 2), point Q is 1/3 along GH and point R is 1/3 along ED. What is the area of the triangle PQR? ### Rolling Around ##### Stage: 3 Challenge Level: A circle rolls around the outside edge of a square so that its circumference always touches the edge of the square. Can you describe the locus of the centre of the circle? ### An Unusual Shape ##### Stage: 3 Challenge Level: Can you maximise the area available to a grazing goat? ### Inside Seven Squares ##### Stage: 2 Challenge Level: What is the total area of the four outside triangles which are outlined in red in this arrangement of squares inside each other? ### Framed ##### Stage: 3 Challenge Level: Seven small rectangular pictures have one inch wide frames. The frames are removed and the pictures are fitted together like a jigsaw to make a rectangle of length 12 inches. Find the dimensions of. . . . ### Intersecting Circles ##### Stage: 3 Challenge Level: Three circles have a maximum of six intersections with each other. What is the maximum number of intersections that a hundred circles could have? ### Like a Circle in a Spiral ##### Stage: 2, 3 and 4 Challenge Level: A cheap and simple toy with lots of mathematics. Can you interpret the images that are produced? Can you predict the pattern that will be produced using different wheels? ### Muggles Magic ##### Stage: 3 Challenge Level: You can move the 4 pieces of the jigsaw and fit them into both outlines. Explain what has happened to the missing one unit of area. ### Take Ten ##### Stage: 3 Challenge Level: Is it possible to remove ten unit cubes from a 3 by 3 by 3 cube made from 27 unit cubes so that the surface area of the remaining solid is the same as the surface area of the original 3 by 3 by 3. . . . ### Pattern Power ##### Stage: 1, 2 and 3 Mathematics is the study of patterns. Studying pattern is an opportunity to observe, hypothesise, experiment, discover and create. ### Buses ##### Stage: 3 Challenge Level: A bus route has a total duration of 40 minutes. Every 10 minutes, two buses set out, one from each end. How many buses will one bus meet on its way from one end to the other end? ### Convex Polygons ##### Stage: 3 Challenge Level: Show that among the interior angles of a convex polygon there cannot be more than three acute angles. ### Coordinate Patterns ##### Stage: 3 Challenge Level: Charlie and Alison have been drawing patterns on coordinate grids. Can you picture where the patterns lead? ### Christmas Boxes ##### Stage: 3 Challenge Level: Find all the ways to cut out a 'net' of six squares that can be folded into a cube. ### Square It ##### Stage: 1, 2, 3 and 4 Challenge Level: Players take it in turns to choose a dot on the grid. The winner is the first to have four dots that can be joined to form a square. ### Marbles in a Box ##### Stage: 3 and 4 Challenge Level: In a three-dimensional version of noughts and crosses, how many winning lines can you make? ### LOGO Challenge - Circles as Animals ##### Stage: 3 and 4 Challenge Level: See if you can anticipate successive 'generations' of the two animals shown here. ### Keep Your Distance ##### Stage: 3 Challenge Level: Can you mark 4 points on a flat surface so that there are only two different distances between them? ### Sprouts ##### Stage: 2, 3, 4 and 5 Challenge Level: A game for 2 people. Take turns joining two dots, until your opponent is unable to move. ### Square Coordinates ##### Stage: 3 Challenge Level: A tilted square is a square with no horizontal sides. Can you devise a general instruction for the construction of a square when you are given just one of its sides? ### Cubist Cuts ##### Stage: 3 Challenge Level: A 3x3x3 cube may be reduced to unit cubes in six saw cuts. If after every cut you can rearrange the pieces before cutting straight through, can you do it in fewer? ### Conway's Chequerboard Army ##### Stage: 3 Challenge Level: Here is a solitaire type environment for you to experiment with. Which targets can you reach? ### Khun Phaen Escapes to Freedom ##### Stage: 3 Challenge Level: Slide the pieces to move Khun Phaen past all the guards into the position on the right from which he can escape to freedom. ### Weighty Problem ##### Stage: 3 Challenge Level: The diagram shows a very heavy kitchen cabinet. It cannot be lifted but it can be pivoted around a corner. The task is to move it, without sliding, in a series of turns about the corners so that it. . . . ### Sea Defences ##### Stage: 2 and 3 Challenge Level: These are pictures of the sea defences at New Brighton. Can you work out what a basic shape might be in both images of the sea wall and work out a way they might fit together? ### The Old Goats ##### Stage: 3 Challenge Level: A rectangular field has two posts with a ring on top of each post. There are two quarrelsome goats and plenty of ropes which you can tie to their collars. How can you secure them so they can't. . . . ### Isosceles Triangles ##### Stage: 3 Challenge Level: Draw some isosceles triangles with an area of $9$cm$^2$ and a vertex at (20,20). If all the vertices must have whole number coordinates, how many is it possible to draw? ### All in the Mind ##### Stage: 3 Challenge Level: Imagine you are suspending a cube from one vertex (corner) and allowing it to hang freely. Now imagine you are lowering it into water until it is exactly half submerged. What shape does the surface. . . . ### Concrete Wheel ##### Stage: 3 Challenge Level: A huge wheel is rolling past your window. What do you see? ### Counting Triangles ##### Stage: 3 Challenge Level: Triangles are formed by joining the vertices of a skeletal cube. How many different types of triangle are there? How many triangles altogether? ### Coloured Edges ##### Stage: 3 Challenge Level: The whole set of tiles is used to make a square. This has a green and blue border. There are no green or blue tiles anywhere in the square except on this border. How many tiles are there in the set? ### Tetra Square ##### Stage: 3 Challenge Level: ABCD is a regular tetrahedron and the points P, Q, R and S are the midpoints of the edges AB, BD, CD and CA. Prove that PQRS is a square. ### Zooming in on the Squares ##### Stage: 2 and 3 Start with a large square, join the midpoints of its sides, you'll see four right angled triangles. Remove these triangles, a second square is left. Repeat the operation. What happens? ### On the Edge ##### Stage: 3 Challenge Level: Here are four tiles. They can be arranged in a 2 by 2 square so that this large square has a green edge. If the tiles are moved around, we can make a 2 by 2 square with a blue edge... Now try to. . . . ### Diagonal Dodge ##### Stage: 2 and 3 Challenge Level: A game for 2 players. Can be played online. One player has 1 red counter, the other has 4 blue. The red counter needs to reach the other side, and the blue needs to trap the red. ### Cutting a Cube ##### Stage: 3 Challenge Level: A half-cube is cut into two pieces by a plane through the long diagonal and at right angles to it. Can you draw a net of these pieces? Are they identical? ### Bands and Bridges: Bringing Topology Back ##### Stage: 2 and 3 Lyndon Baker describes how the Mobius strip and Euler's law can introduce pupils to the idea of topology. ### Rati-o ##### Stage: 3 Challenge Level: Points P, Q, R and S each divide the sides AB, BC, CD and DA respectively in the ratio of 2 : 1. Join the points. What is the area of the parallelogram PQRS in relation to the original rectangle? ### Chess ##### Stage: 3 Challenge Level: What would be the smallest number of moves needed to move a Knight from a chess set from one corner to the opposite corner of a 99 by 99 square board? ### A Square in a Circle ##### Stage: 2 Challenge Level: What shape has Harry drawn on this clock face? Can you find its area? What is the largest number of square tiles that could cover this area? ### Squares in Rectangles ##### Stage: 3 Challenge Level: A 2 by 3 rectangle contains 8 squares and a 3 by 4 rectangle contains 20 squares. What size rectangle(s) contain(s) exactly 100 squares? Can you find them all? ### Crossing the Atlantic ##### Stage: 3 Challenge Level: Every day at noon a boat leaves Le Havre for New York while another boat leaves New York for Le Havre. The ocean crossing takes seven days. How many boats will each boat cross during their journey? ### 3D Stacks ##### Stage: 2 and 3 Challenge Level: Can you find a way of representing these arrangements of balls? ### There and Back Again ##### Stage: 3 Challenge Level: Bilbo goes on an adventure, before arriving back home. Using the information given about his journey, can you work out where Bilbo lives? ### Rotating Triangle ##### Stage: 3 and 4 Challenge Level: What happens to the perimeter of triangle ABC as the two smaller circles change size and roll around inside the bigger circle? ### Troublesome Dice ##### Stage: 3 Challenge Level: When dice land edge-up, we usually roll again. But what if we didn't...? ### Triangles in the Middle ##### Stage: 3, 4 and 5 Challenge Level: This task depends on groups working collaboratively, discussing and reasoning to agree a final product.	2014-08-31 02:32:39	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 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": 0.3964742124080658, "perplexity": 2010.5176672743307}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-35/segments/1408500835872.63/warc/CC-MAIN-20140820021355-00428-ip-10-180-136-8.ec2.internal.warc.gz"}
https://akishore.in/2011/04/23/the-difficulty-with-science/	# The Difficulty with Science One of the first things that struck me when I started teaching is that science comes across as a difficult subject to a majority of the students. There are some to whom it seems to come naturally, and we label them “science-persons” while the others who seem to struggle, as “arts-persons” or “humanities-persons”. While there is no denying that there are variations in children’s ability, aptitude and interests, I’m gripped by the question whether everyone can be given a meaningful and enriching, and not painful science education. And if so, what would that be like. One of the approaches that could be taken, perhaps, is to study how science is normally taught and what are the sources of the difficulties which children face. How is scientific knowledge organised in our brains and what are the factors which make this process so difficult for so many? It might require some thought about the very nature of science and scientific knowledge. While I’m not an expert in any of these areas, some of the problems which I myself have faced as a student provides some hints. The problem may not be such a big mystery if you look a bit closer. Much of science is abstract, counter intuitive, and removed from real life experience. Take, for example, the kinetic theory of matter, which says that the particles which make up matter are in continuous motion and that the temperature of a body is a measure of the average kinetic energy of its particles. Most text books don’t explain why people came to believe that the particles are in motion, or why a body is hotter if the particles are moving faster. There is a lot that ends up having to be taken for granted. This is not an isolated example- far from it. Every theory or topic that children have to learn has some gap like this, and after a while instead of the concepts building up nicely into a jigsaw puzzle- an understanding of the way the world works, they end up being isolated and unrelated fragments that have to be “mugged up”. Related to this is the fact that scientific knowledge is often presented as absolute truth. Very little, if any importance is given to the process of discovery and the evolution of scientific knowledge. Text books do mention some history, but it is often lost within the vast sea of facts which children have to memorise. And in the process, only the end product, and not the evolution of the idea or the real life phenomenon that led to it, is given importance. Another difficulty which students often face is in forming mental pictures or representations of concepts or processes. If you are unable to visualise it, it quickly becomes abstract, meaningless information. For example, consider the case of common salt dissolving in water. I, the teacher, have a vivid picture of differently sized Na+ and Cl- ions packed tightly in the solid crystal, and water molecules floating (or flowing!) around with their partially positive H and partially negative O ends. And the moment you put NaCl in water, the partial charges of water molecules arranging themselves around the ions in the salt and pulling them apart. It is a complex mental representation, built up over time, with connections to many other concepts like kinetic theory, chemical bonding etc. How can a teacher help a student build her own mental picture? Some students do this effortlessly, but can the others do it with some help and more importantly, can having such coherent mental pictures help them learn science more easily and connect with it better? Most text books would introduce this concept with a sentence like “Sodium chloride dissolves well in water because it is a polar solvent.” And then go on to beat around the bush with all kinds of irrelevant information. How does the student visualise the term “polar solvent”? Does she think about it at all, or switch off completely? Or does she memorise the term without any clue as to what it means and move on and write the correct answer during the exam? It’s important to note that the problems I have mentioned are not solved by just doing “more practical work than theory”. The same problems arise in lab sessions also, because doing an experiment is one thing, and understanding what happens behind it is another. Of course, being usually more engaging than theory classes, there is a greater chance that the student will apply herself better and get the concept. But one cannot assume that the student has understood it just by carrying out the experiment. Again, let’s take an example, say precipitation reactions where solutions of two soluble salts are mixed together to form an insoluble salt. $Na_2CO_3 + CaCl_2 longrightarrow 2NaCl + CaCO_3 downarrow$ Some children are immediately able to connect the above equation to their understanding of ionic compounds and solutions. They will immediately conclude that calcium and carbonate ions can’t stay together in the solution and that’s why they precipitate, and if they had come across calcium carbonate in earlier experiments, they would already know that it is an insoluble salt, which all fits in nicely into a perfect mosaic of knowledge. They would already be predicting which other pairs of solutions would give precipitates. But for most, the equation wouldn’t mean anything deeper than what it literally states, and it needs to be made explicit to them what all information it represents and a visualisation of the process that is taking place. Otherwise, it is not unlikely that many would do this particular reaction, learn the equation by heart, do the next one, again learn the equation by heart and so on. It’s fortunate that I have had these problems myself as a student, especially during engineering days, so that I can connect with the difficulties which students face. It’s not as if only “non-science-persons” face these difficulties, and “science-persons” don’t face them at all. I, whom many would classify as a “science-person”, have had experiences of having gone through entire courses without being able to make out head or tail of what was taught, and having to mug up to pass the exams. The moment you are unable to form coherent mental pictures of a concept, it is going to become more and more meaningless. ## 7 thoughts on “The Difficulty with Science” 1. This post does give a great insight about the science education followed in most of the schools and how students learn science. I hope you would be able find better ways to teach science to students to make it more interesting and easy to learn for them. Best wishes with your teaching. 2. You can find on the net lots of essays and Youtube videos of the best science teacher who ever lived – Richard Feynman. Studying them would give you some idea as to how to approach the teaching process. I would also recommend Prof.Walter Lewin’s OCW lectures. You should also check out Tony Kuphaldt’s “Socratic Electronics”. Go through some of the worksheets and try to create something of that sort for chemistry! http://www.ibiblio.org/kuphaldt/socratic/ 3. AVG Warrier says: May be this has some relation to your theme. There is a drastic difference between water color painting and oil painting. In the first one takes the whiteness of the paper as the given and the forms are carved out of this whitenss with shades and shadows. In the second one takes the darkness of the background as the given and the forms are built on this by the play of light of several hues. If one apply the techniques of water color with oil paint the result is labored and unpleasant. The same happens if one uses the techniques of oil color with water color paints. Does an artist takes light for granted and is looking for patches of imperfections that brings out forms. Then the path of abstraction may be alien to him. Normally the artist and scientist live side by side in everyone. Is not cultivating both sides in a balanced way is also a role of an educator? 4. sabupaul says: It’s nice to see that you have invested so much thought into improving your teaching. I personally don’t have much experience with it thought I too enjoy explaining things. When I think of the best teachers that I have had, what comes to mind first is their ability to start with explaining the basics and ultimately describe how everything fits together and describe graphically the “big picture”. Even if everyone doesn’t get everything, they will at the very least see the beauty, power and importance of the concept to the overall scheme of things. I remember being asked when I was in the first year to explain what the significance of the Laplace transform was. Starting from the equ. you can finally arrive at a point where you can see the frequency components(real and imaginary) of any signal. It’s applications in everything from mechanical design to electronics to sound engineering can motivate even the most hardened science/math hater to at least acknowledge it’s importance. It is true that a lot of people especially children don’t naturally seek out the big picture. But, that’s the challenge, right!? It is up to the teacher to actively build up that kind of holistic thinking even in children who are not naturally inclined to do it. This should be done with greater zeal in the lower grades. With that kind of conditioning, kids will, I guess try and think at a higher level automatically when they reach the higher grades. I also remember seeing some study suggesting that some kids especially the girls tend to not use their hands or bodies or for that matter other physical objects to try and visualize things or as an aid in abstract thought processes. The researchers after observing this conducted a small session telling everyone to use their hands etc. to study maths and voila, they started performing at the same levels as their peers in maths and science. This is maybe one small point. But, I have always a noticed that a lot of people don’t like to “give shape” to their thoughts. Of course its just a matter of being introduced to it. I think if students are encouraged to think in physical terms it might improve their performance. 1. Sabu, “I have always a noticed that a lot of people don’t like to “give shape” to their thoughts. Of course its just a matter of being introduced to it. I think if students are encouraged to think in physical terms it might improve their performance.” Interesting thought … if you watch the Feynman video I have linked above (jiggling atoms), I think he too is trying to sort of give “shape” to ideas using his body language and expressions! 5. Very well expressed. I personally have been through this experience while helping out college-mates in undergrad. And I am surprised, for I was just about to mention Feynman lectures. 😛 😛 😛 Ramuettan beat me to it though. He was indeed the best, and I consider him an idol for his efforts at trying to serve the true purpose of education: to train the mind to think. Indian education is something to be permanently brooded over, and it is very difficult to do something about it unless you are in a like minded atmosphere. Starting from evolution to the basic concepts of Newtonian mechanics to Aristotle’s Physics (by the way, read it if physics is your forte. Mine is :D) to the brilliance of radiant chemistry. When I first read The Fountain Head, from a philosophical point of view, I found exactly what I believed in. The joy of understanding and uncovering knowledge is in itself a reward, and nothing else matters. But then that is never a notion these days, as marks and competitive progress and jobs take up the main reasons. I believe, in future, to live and teach by these principles where knowledge and its conceptual application and understanding take first priority. And you seem to be doing a great job already. Keep up with it. 🙂 And Ramuettan, Socratic electronics was brilliant. 🙂 One must read The Origin of Species, Godel Escher Bach, Aristotle’s Physics, Principia Mathematica and most importantly the works of Bertrand Russel. These are eye openers for teachers, and I feel sorry that hardly 0.00001% of the teachers do it.	2021-01-25 14:14:24	{"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": 1, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.35004082322120667, "perplexity": 755.8142909745973}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-04/segments/1610703581888.64/warc/CC-MAIN-20210125123120-20210125153120-00310.warc.gz"}
https://ankplanet.com/maths/application-of-derivatives/maxima-and-minima/	# Maxima and Minima To find the maxima and minima (maximum and minimum) value of a function $ƒ$ in an interval $(a,b)$, it is necessary to know the nature of the curve in the neighborhood of the point where the maximum or minimum value of the function occurs. The nature of the curve at a particular point is known by studying the slopes of the tangents at the point in its neighborhood. The slope or the gradient of a tangent line at a point of the graph of a function $ƒ$ may be positive, negative or zero. If the slope of $ƒ(x)$ is positive $\left(\text{i.e. } ƒ'(x)>0\right)$, the inclination $\theta$ of the tangent line is positive and the tangent line slopes upwards from left to right as shown in above figure. In the figures, within the interval $(a,b)$, where the slope of the tangent line at any point is positive, $x_2>x_1\Rightarrow ƒ(x_2)>ƒ(x_1)$ i.e. the function is increasing in its neighborhood. If the slope of $ƒ(x)$ is negative $\left(\text{i.e. } ƒ'(x)<0\right)$, the inclination $\theta$ of the tangent line is negative and the tangent line slopes downwards from left to right as shown in above figure. In the figures, within the interval $(a,b)$, where the slope of the tangent line at any point is negative, $x_2>x_1\Rightarrow ƒ(x_2)<ƒ(x_1)$ i.e. the function is decreasing in its neighborhood. If the slope of $ƒ(x)$ is zero $\left(\text{i.e. }ƒ'(x)=0\right)$, the tangent line neither slopes upwards nor downwards i.e. it remains horizontal or stationary. The points at which $ƒ'(x)=0$ or $ƒ'(x)$ is not defined are called critical (or stationary) points. $ƒ^{\prime\prime}(x)$ is the derivative of $ƒ'(x)$. If $ƒ^{\prime\prime}(x)>0$, $ƒ'(x)$ is increasing i.e. the slope of the tangent line is increasing. In the figure $\text{(a)}$, the slope $ƒ'(x)$ is increasing from a smaller positive value to a larger positive value and in the figure $\text{(b)}$, the slope is increasing from a larger negative value to a smaller negative value. The curve in the figure $\text{(c)}$ is just a combination of the figures $\text{(a)}$ and $\text{(b)}$. These curves are said to be concave upwards. Hence, if $ƒ^{\prime\prime}(x)>0$ in an interval $(a,b)$, the graph of $ƒ$ is concave upwards in the interval. If $ƒ^{\prime\prime}(x)<0$, $ƒ'(x)$ is decreasing i.e. the slope of the tangent line is decreasing. In the figure $\text{(a’)}$, the slope $ƒ'(x)$ is decreasing from a larger positive value to a smaller positive value, while in the figure $\text{(b’)}$, the slope is decreasing from a smaller negative value to a larger negative value. The curve in the figure $\text{(c’)}$ is just a combination of the figures $\text{(a’)}$ and $\text{(b’)}$. These curves are said to be concave downwards. Hence, if $ƒ^{\prime\prime}(x)<0$ in an interval $(a,b)$, the graph of $ƒ$ is concave downwards in the interval. Consider a function $ƒ$ with a portion of curve concave downwards in the interval $(a,b)$ as shown in figure. Hence, $ƒ^{\prime\prime}(x)<0$ so $ƒ'(x)$ decreases throughout the portion of the curve in the interval $(a,b)$. But in the interval $(a,c)$, $ƒ'(x)>0$ and in the interval $(c,b)$, $ƒ'(x)<0$. As the curve is continuous, there must exist a point $x_o$ such that $ƒ'(x_o)=0$ and at that point $x_o$, the function $ƒ$ has a local maximum. So, the conditions for a function $ƒ$ to have a local maximum at a point $x_o$ are $\text{1. } ƒ'(x_o)=0$ $\text{2. } ƒ^{\prime\prime}(x_o)<0$ Similarly, consider a function $ƒ$ with a portion of curve concave upwards in the interval $(a,b)$ as shown in figure. Hence, $ƒ^{\prime\prime}(x)>0$ so $ƒ'(x)$ increases throughout the portion of the curve in the interval $(a,b)$. But in the interval $(a,c)$, $ƒ'(x)<0$ and in the interval $(c,b)$, $ƒ'(x)>0$. As the curve is continuous, there must exist a point $x_o$ such that $ƒ'(x_o)=0$ and at that point $x_o$, the function $ƒ$ has a local minimum. So, the conditions for a function $ƒ$ to have a local minimum at a point $x_o$ are $\text{1. } ƒ'(x_o)=0$ $\text{2. } ƒ^{\prime\prime}(x_o)>0$ Again, consider the graph of a function $ƒ$ as shown in the figure. The portion of the graph in the interval $(a,c)$ is concave downwards so $ƒ^{\prime\prime}(x)<0$, while the portion of the graph in the interval $(c,b)$ is concave upwards so $ƒ^{\prime\prime}(x)>0$. Hence, there must exist a point in the interval at which $ƒ^{\prime\prime}(x)=0$. This point of the graph is called a point of inflection and it separates the portion of the graph which is concave downwards from the portion which is concave upwards. The local maxima and local minima are not necessarily the greatest and the least values of the function but simply the maximum and minimum values on the neighborhood of the point $x=a$. A function may have several maximum and minimum values which occur alternatively. But besides these maximum and minimum values, there may be the values greater than the maximum value and less than the minimum value i.e. there may be the greatest and the least values of the function which are termed as absolute maximum value and the absolute minimum value of the function. Hence, we have the following definitions for the following terms: ## Absolute Maxima A function $y=ƒ(x)$ is said to have the absolute maximum value or absolute maxima at $x=a$ if $ƒ(a)$ is the greatest of all of its values for all $x$ belonging to the domain of the function. The absolute maximum value is also known as the global maximum value. In other words, $ƒ(a)$ is the absolute maximum value of $ƒ(x)$ if $ƒ(a)≥ƒ(x)$ for all $x\in D(ƒ)$. In the figure, $ƒ(a)$ is the absolute maximum value of $ƒ(x)$ at $x=a$. ## Absolute Minima A function $y=ƒ(x)$ is said to have the absolute minimum value or absolute minima at $x=a$ if $ƒ(a)$ is the smallest of all of its values for all $x$ belonging to the domain of the function. The absolute minimum value is also known as the global minimum value. In other words, $ƒ(a)$ is the absolute minimum value of $ƒ(x)$ if $ƒ(a)≤ƒ(x)$ for all $x\in D(ƒ)$. In the figure, $ƒ(a)$ is the absolute minimum value of $ƒ(x)$ at $x=a$. ## Stationary Point A point on the graph of the function $y=ƒ(x)$ where the tangent is parallel to the $\text{X-axis}$ is known as the stationary point or critical point. At the stationary point, $\frac{dy}{dx}=ƒ'(x)=0$ ## Local Maxima A function $y=ƒ(x)$ is said to have the local maximum value or local maxima at $x=a$ if $ƒ(a)>ƒ(a\pm h)$ for sufficiently small positive value of $h$. The local maxima is also known as the relative maxima of the function. ## Local Minima A function $y=ƒ(x)$ is said to have the local minimum value or local minima at $x=a$ if $ƒ(a)<ƒ(a\pm h)$ for sufficiently small positive value of $h$. The local minima is also known as the relative minima of the function. ## Concavity and Convexity of Curves The graph of the function $y=ƒ(x)$ defined in an interval $(a,b)$ is concave upward (or convex downward) if each point (except the point of contact) on the graph (i.e. curve) lies above any tangent to it in that interval. But if each point on the graph (i.e. curve) lies below any tangent to the curve in the interval, then the graph of the function is said to be concave downward (or convex upward). The graph of the function $y=ƒ(x)$ will be concave upward if $\frac{d^2y}{dx^2}=ƒ^{\prime\prime}(x)>0$ and it will be concave downward if $\frac{d^2y}{dx^2}=ƒ^{\prime\prime}(x)<0$. ## Point of Inflection The point which divides the graph of the function from the shape of upward concavity (or downward concavity) to the downward concavity (or upward concavity) is known as the point of inflection. If $y=ƒ(x)$ represents a continuous curve, then the point where $\frac{d^2y}{dx^2}=ƒ^{\prime\prime}(x)=0$ and $\frac{d^3y}{dx^3}=ƒ^{\prime\prime\prime}(x)≠0$ is said to be the point of inflection. ### Find the absolute maximum and the absolute minimum values of the function $ƒ(x)=2x^3-15x^2+36x+10$ on $[1,4]$. Here, $ƒ(x)=2x^3-15x^2+36x+10$ $\therefore ƒ'(x)=6x^2-30x+36$ For stationary point, $ƒ'(x)=0$ $6x^2-30x+36=0$ $x^2-5x+6=0$ $(x-2)(x-3)=0$ $\text{either, }x=2 \text{ or, } x=3$ $\therefore ƒ(2)=2×2^3-15×2^2+36×2+10=38$ $ƒ(3)=2×3^3-15×3^2+36×3+10=37$ $ƒ(1)=2×1^3-15×1^2+36×1+10=33$ $ƒ(4)=2×4^3-15×4^2+36×4+10=42$ $\therefore \text{Absolute Maximum}=42 \text{ at }x=4$ $\text{Absolute Minimum}=33\text{ at }x=1$ ### Find the local maxima, the local minima and the point of inflection of the function $ƒ=x^3-6x^2+3$. Here, $ƒ(x)=x^3-6x^2+3$ $\therefore ƒ'(x)=3x^2-12x$ $ƒ^{\prime\prime}(x)=6x-12$ For stationary point, $ƒ'(x)=0$ $3x^2-12x=0$ $x(x-4)=0$ $\therefore x=0,4$ Now, $ƒ^{\prime\prime}(0)=0-12=-12<0$ $\therefore ƒ(x) \text{ is max. at } x=0.$ $\therefore \text{max. value }(ƒ_{\text{max}})=0-0+3=3$ And, $ƒ^{\prime\prime}(4)=6×4-12=12>0$ $\therefore ƒ(x) \text{ is mini. at } x=4.$ $\therefore \text{mini. value } (ƒ_{\text{mini}})=4^3-6×4^2+3$$=-29$ For point of inflection, $ƒ^{\prime\prime}(x)=0$ $6x-12=0$ $\therefore x=2$ Thus, the point of inflection is at $x=2$. ### Find the local maxima, the local minima and the point of inflection of the function $ƒ(x)=3x^2-6x+3$. Here, $ƒ(x)=3x^2-6x+3$ $ƒ'(x)=6x-6$ $ƒ^{\prime\prime}(x)=6$ For stationary point, $ƒ'(x)=0$ $6x-6=0$ $\therefore x=1$ Now, $ƒ^{\prime\prime}(1)=6>0$ $\therefore ƒ(x) \text{ is mini at } x=1.$ $\text{mini. value }(ƒ_{\text{mini}})=3×1-6×1+3=0$ For point of inflection, $ƒ^{\prime\prime}(x)=0$ $6=0 \text{ [No solution]}$ Hence, point of inflection does not exist. ### Show that the function $ƒ(x)=x^3-6x^2+12x-3$ has neither maximum nor minimum value. Here, $ƒ(x)=x^3-6x^2+12x-3$ $\therefore ƒ'(x)=3x^2-12x+12$ $ƒ^{\prime\prime}(x)=6x-12$ For stationary point, $ƒ'(x)=0$ $3x^2-12x+12=0$ $x^2-4x+4=0$ $(x-2)^2=0$ $x=2$ Now, $ƒ^{\prime\prime}(2)=6×2-12=0$ Hence, $ƒ(x)$ has neither maximum nor minimum value. ### Determine where the graph is concave upwards and where it is concave downwards of the function $ƒ(x)=x^4-8x^3+18x^2-24$. Here, $ƒ(x)=x^4-8x^3+18x^2-24$ $\therefore ƒ'(x)=4x^3-24x^2+36x$ $ƒ^{\prime\prime}(x)=12x^2-48x+36$ Now, curve will be concave downwards if $ƒ^{\prime\prime}(x)<0$ $12x^2-48x+36<0$ $x^2-4x+3<0$ $(x-1)(x-3)<0$ $\text{either, } (x-1)>0 \text{ and } (x-3)<0$ $x>1 \text{ and } x<3$ $\text{or, } (x-1)<0 \text{ and } (x-3)>0$ $x<1 \text{ and } x>3$ Hence, curve is concave downwards on $(1,3)$ and concave upwards on $(-\infty,1)\cup(3,\infty)$. ### A window is in the form of a rectangle surmounted by a semi-circle. If the total perimeter is $9$ metres, find the radius of the semi-circle for the greatest window area Let $ABCD$ be the rectangular part of the window. $AB=x=CD=\text{diameter}$ $BC=y=DA$ Then, $\text{Radius }(r)=\frac{x}{2}$ Now, $AB+BC+\overset{\huge\frown}{CD}+DA=9$ $x+y+\frac{1}{2}(2πr)+y=9$ $x+y+\frac{πx}{2}+y=9$ $2y+\left(1+\frac{π}{2}\right)x=9$ $2y=9-\left(\frac{π+2}{2}\right)x$ $y=\frac{9}{2}-\left(\frac{π+2}{4}\right)x$ And, $\text{Area }(A)=xy+\frac{1}{2}πr^2$ $A=x\left[\frac{9}{2}-\frac{(π+2)x}{4}\right]+\frac{1}{2}π\frac{x^2}{4}$ $A=\frac{9x}{2}-\frac{(π+2)x^2}{4}+\frac{π}{8}x^2$ $A=\frac{9x}{2}-\left(\frac{π+4}{8}\right)x^2$ $\therefore A(x)=\frac{9x}{2}-\left(\frac{π+4}{8}\right)x^2$ $A'(x)=\frac{9}{2}-\left(\frac{π+4}{4}\right)x$ $A^{\prime\prime}(x)=-\left(\frac{π+4}{4}\right)$ For stationary point, $A'(x)=0$ $\frac{9}{2}-\left(\frac{π+4}{4}\right)x=0$ $\frac{9}{2}=\left(\frac{π+4}{4}\right)x$ $18=(π+4)x$ $\therefore x=\frac{18}{π+4}$ Now, $A^{\prime\prime}\left(\frac{18}{π+4}\right)=-\left(\frac{π+4}{4}\right)<0$ Therefore, $A(x)$ is max. at $x=\frac{18}{π+4}$. $\therefore r=\frac{1}{2}\left(\frac{18}{π+4}\right)=\frac{9}{π+4}$ ### Find two numbers whose sum is 10 and the sum of whose squares is minimum. Let those two numbers be $x$ and $y$. Then, $x+y=10$ $y=10-x$ And, $S=x^2+y^2$ $S=x^2+(10-x)^2$ $S=x^2+100-20x+x^2$ $\therefore S(x)=2x^2-20x+100$ $S'(x)=4x-20$ $S^{\prime\prime}(x)=4$ For stationary point, $S'(x)=0$ $4x-20=0$ $x=5$ Now, $S^{\prime\prime}(x)=4>0$ Therefore, $S(x)$ is mini at $x=5$. And, $y=10-5=5$ Hence, the required two numbers are 5 and 5. © 2023 AnkPlanet - All Rights Reserved ·	2023-03-25 07:22:06	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.8782316446304321, "perplexity": 107.46082774631918}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296945317.85/warc/CC-MAIN-20230325064253-20230325094253-00156.warc.gz"}
https://physics.stackexchange.com/questions/403971/highly-conductive-lossless-medium	# Highly conductive lossless medium I've came across the term "highly conductive lossless medium" in the context of electromagnetic waves travelling in materials. I'm wondering how to make sense of that statement? I thought "highly conductive" implies there's energy through ohmic heating. However how can that then be a "lossless medium"? A good conductor is one for which $\frac{\sigma}{\omega\epsilon}\gg 1$ but a material is (nearly) lossless if the decay constant $$\alpha=\sqrt{\frac{\omega^2\mu\epsilon}{2}} \left[\sqrt{1+\frac{\sigma^2}{\omega^2\epsilon^2}}-1\right]^{1/2}$$ is small. Ultra-low frequency electromagnetic waves propagating in rock have $\sigma\sim 10^{-3}S/m$ and $\epsilon\sim 10\epsilon_0$, yielding $$\frac{\sigma}{\omega\epsilon}\sim 2\times 10^7$$ and thus a good conductor while $\alpha\sim 2\times 10^{-5}$, so rock is nearly lossless at that frequency.	2020-12-04 18:45:07	{"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": 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": 0.8414890170097351, "perplexity": 1122.0486915882052}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-50/segments/1606141740670.93/warc/CC-MAIN-20201204162500-20201204192500-00066.warc.gz"}
https://phys.libretexts.org/Bookshelves/Quantum_Mechanics/Advanced_Quantum_Mechanics_(Kok)/16%3A_Matter/16.3%3A_Molecules	# 16.3: Molecules Atoms can bond together. Sometimes, if one atom is able to completely steal an electron from another atom (as is the case with Chlorine and Sodium atoms, where a Sodium atom will donate an electron to a Chlorine atom), the resulting ions will then stick together as a result of the electrostatic attraction between their opposite net charge. More common, however, are molecules made from what is called covalent bonds. The electrons in the outer (unfilled) shell of an atom are known as “valence” electrons. Depending on the electronic configuration of an atom, it will have one or more effective valence electrons. In a molecule, the valence electrons are no longer associated with a single atom, but instead are shared between the electrons. In terms of the quantum mechanics involved, you wouldn’t find a solution to Schrödinger’s Equation for just the potential of one atom. Rather, you create a joint potential for the two atoms (including the effects of inner-shell electrons), and determine a solution for the system as a whole. The result is an electron wave function that indicates the electron probability cloud is shared between two or (for more complicated molecules) more of the atoms that composes the molecule. Just as nuclei have a binding energy, molecules have a binding energy, meaning that it is a lower energy state for these atoms to bind together and share an electron than it is for them to stay separate. Although this binding energy is typically a billionth of the mass energy of atoms, it is enough to create the vast majority of energy producing processes (e.g. burning gas to power a car) that we are familiar with in our everyday lives. Finding these solutions to multi-atom potentials is an extremely difficult problem, and cannot be solved analytically (as the Hydrogen atom may be). Describing the quantum mechanical state and electron orbitals of any molecule more complicated than something like $$H_{2}$$ generally involves both heavy-duty numerical calculations on computers and heavy-duty quantum chemists. This page titled 16.3: Molecules is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Pieter Kok via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.	2023-03-24 05:22:04	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 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": 0.7947749495506287, "perplexity": 493.244107483894}, "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/1679296945248.28/warc/CC-MAIN-20230324051147-20230324081147-00765.warc.gz"}
http://cmj.math.cas.cz/cmj48-4/8.html	Czechoslovak Mathematical Journal, Vol. 48, No. 4, pp. 701-710, 1998 # An algebraic characterization of geodetic graphs Abstract: We say that a binary operation $$ is associated with a (finite undirected) graph $G$ (without loops and multiple edges) if $$ is defined on $V(G)$ and $uv\in E(G)$ if and only if $u\not= v$, $u * v=v$ and $v*u=u$ for any $u$, $v\in V(G)$. In the paper it is proved that a connected graph $G$ is geodetic if and only if there exists a binary operation associated with $G$ which fulfils a certain set of four axioms. (This characterization is obtained as an immediate consequence of a stronger result proved in the paper).	2018-12-10 12:01:51	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8340822458267212, "perplexity": 66.5806870592795}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-51/segments/1544376823322.49/warc/CC-MAIN-20181210101954-20181210123454-00317.warc.gz"}
https://bibli.cirm-math.fr/listRecord.htm?list=link&xRecord=19253178146910713509	m • E F Nous contacter 0 # Documents 47A16 \| enregistrements trouvés : 8 O P Q Déposez votre fichier ici pour le déplacer vers cet enregistrement. ## A universal hypercyclic representation Glasner, Eli \| CIRM H Post-edited Research talks;Analysis and its Applications;Dynamical Systems and Ordinary Differential Equations For any countable group, and also for any locally compact second countable, compactly generated topological group, $G$, there exists a "universal" hypercyclic representation on a Hilbert space, in the sense that it simultaneously models every possible ergodic probability measure preserving free action of $G$. I will discuss the original proof of this theorem (a joint work with Benjy Weiss) and then, at the end of the talk, say some words about the development of this idea and its applications as expounded in a subsequent work of Sophie Grivaux. For any countable group, and also for any locally compact second countable, compactly generated topological group, $G$, there exists a "universal" hypercyclic representation on a Hilbert space, in the sense that it simultaneously models every possible ergodic probability measure preserving free action of $G$. I will discuss the original proof of this theorem (a joint work with Benjy Weiss) and then, at the end of the talk, say some words about ... Déposez votre fichier ici pour le déplacer vers cet enregistrement. ## Ergodic theory and related fields :2004-2006 chapel hill workshops on probability and ergodic theory#Feb. 15-18 Assani, Idris \| Amercian Mathematical Society 2007 Congrès - 145 p. ISBN 978-0-8218-3869-3 Contemporary mathematics , 0430 Localisation : Collection 1er étage théorie ergodique # système dynamique @ transformation conservant la mesure # équivalence d'orbite # transformation de Hilbert Déposez votre fichier ici pour le déplacer vers cet enregistrement. ## Recent progress on operator theory and approximation in spaces of analytic functions.Conference on completeness problems, Carleson measures, and spaces of analytic functions.Djursholm # June 29-July 3, 2015 Bénéteau, Catherine ; Condori, Alberto A. ; Liaw, Constanze ; Ross, William T. ; Sola, Alan A. \| American Mathematical Society 2016 Congrès - ix; 217 p. ISBN 978-1-4704-2305-6 Contemporary mathematics , 0679 Localisation : Collection 1er étage théorie des opérateurs # espace analytique # fonction analytique Déposez votre fichier ici pour le déplacer vers cet enregistrement. ## Some remarks regarding ergodic operators Matheron, Etienne \| CIRM H Multi angle Research talks;Dynamical Systems and Ordinary Differential Equations Let us say that a continuous linear operator $T$ acting on some Polish topological vector space is ergodic if it admits an ergodic probability measure with full support. This talk will be centred in the following question: how can we see that an operator is or is not ergodic? More precisely, I will try (if I’m able to manage my time) to talk about two “positive" results and one “negative" result. The first positive result says that if the operator $T$ acts on a reflexive Banach space and satisfies a strong form of frequent hypercyclicity, then $T$ is ergodic. The second positive result is the well-known criterion for ergodicity relying on the perfect spanning property for unimodular eigenvectors, of which I will outline a “soft" Baire category proof. The negative result will be stated in terms of a parameter measuring the maximal frequency with which (generically) the orbit of a hypercyclic vector for $T$ can visit a ball centred at 0. The talk is based on joint work with Sophie Grivaux. Let us say that a continuous linear operator $T$ acting on some Polish topological vector space is ergodic if it admits an ergodic probability measure with full support. This talk will be centred in the following question: how can we see that an operator is or is not ergodic? More precisely, I will try (if I’m able to manage my time) to talk about two “positive" results and one “negative" result. The first positive result says that if the ... Déposez votre fichier ici pour le déplacer vers cet enregistrement. ## The role of the spectrum in the cyclic behavior of composition operators Gallardo-Gutiérrez, Eva ; Montes-Rodriguez, Alfonso \| American Mathematical Society 2004 Ouvrage - 81 p. ISBN 978-0-8218-3432-9 Memoirs of the american mathematical society , 0791 Localisation : Collection 1er étage espace de fonction # opérateur linéaire # fonction de variable complexe # fonction hypergéométrique # opérateur de composistion # opérateur cyclique # opérateur supercyclique # opérateur hypercyclique # spectre # transformation de Fourier # mesure de Haar # zéro de fonction holomorphe # polynôme de Laguerre Déposez votre fichier ici pour le déplacer vers cet enregistrement. ## Dynamics of linear operators Bayart, Frédéric ; Matheron, Etienne \| Cambridge University Press 2009 Ouvrage - xiv; 337 p. ISBN 978-0-521-51496-5 Cambridge tracts in mathematics , 0179 Localisation : Collection 1er étage opérateur théorie # vecteur cyclique # théorie ergodique # dynamique topologique Déposez votre fichier ici pour le déplacer vers cet enregistrement. ## Generalized semigroups and cosine functions Kostic, Marko \| Matematicki Institut SANU 2011 Ouvrage - vi; 352 p. ISBN 978-86-80593-45-6 Posebna izdanja , 0023 Localisation : Ouvrage RdC (KOST) théorie des opérateurs # fonction cosinus # problème mal posés # problème de Cauchy Déposez votre fichier ici pour le déplacer vers cet enregistrement. ## Linear chaos Grosse-Erdmann, Karl-G. ; Peris Manguillot, Alfred \| Springer 2011 Ouvrage - xi; 386 p. ISBN 978-1-4471-2169-5 Universitext Localisation : Ouvrage RdC (GROS) chaos # système dynamique linéaire # opérateur hypercyclique # opérateur chaotique # dynamique des semi-groupes #### Filtrer ##### Codes MSC Ressources Electroniques (Depuis le CIRM) Books & Print journals Recherche avancée 0 Z	2019-09-20 13:03:48	{"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": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6316675543785095, "perplexity": 6622.93848006651}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-39/segments/1568514574018.53/warc/CC-MAIN-20190920113425-20190920135425-00031.warc.gz"}
http://www.solutioninn.com/as-a-wire-is-stretched-out-so-that-its-length	# Question As a wire is stretched out so that its length increases, its cross-sectional area decreases, while the total volume of the wire remains constant. (a) Will the resistance after the stretch be (1) greater than, (2) the same as, or (3) less than that before the stretch? (b) A 1.0-m length of copper wire with a 2.0-mm diameter is stretched out; its length increases by 25% while its cross-sectional area decreases, but remains uniform. Compute the resistance ratio (final to initial). Sales0 Views42	2016-10-24 21:02:15	{"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": 0.8417006731033325, "perplexity": 1635.7168015105544}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-44/segments/1476988719754.86/warc/CC-MAIN-20161020183839-00155-ip-10-171-6-4.ec2.internal.warc.gz"}
http://mathhelpforum.com/algebra/35715-arithmetic-sequences-print.html	# Arithmetic Sequences • Apr 23rd 2008, 05:03 PM KittyOfDoom Arithmetic Sequences I have been having trouble with the following problem: Write the next term in the sequence. Then write a rule for the nth term. 7,-8,9,-10... I found out that the next number in the sequence is 11, but I have yet to figure out the rule for the nth term. I have been trying to solve this problem for a while and cannot. Can you help? • Apr 23rd 2008, 05:22 PM Mathstud28 Quote: Originally Posted by KittyOfDoom I have been having trouble with the following problem: Write the next term in the sequence. Then write a rule for the nth term. 7,-8,9,-10... I found out that the next number in the sequence is 11, but I have yet to figure out the rule for the nth term. I have been trying to solve this problem for a while and cannot. Can you help? $a_n=(n+6)(-1)^{n+1}$ • Apr 23rd 2008, 05:22 PM Soroban Hello, KittyOfDoom! Quote: Write the next term in the sequence. Then write a rule for the nth term. . . . $7,\:-8,\:9,\:-10,\:\hdots$ Ignoring the signs, we have: . $7,\:8,\:9,\;10,\:\hdots$ . . The rule (so far) is: . $f(n) \:=\:n+6$ Now, how do we make the signs alternate? . . If you've never seen it done, this can be baffling. Consider $(-1)^n$ . . . negative-one raised to consecutive powers. We get: . . . . . . . $\begin{array}{ccc}(-1)^1 &=& -1 \\ (-1)^2 &=& +1 \\ (-1)^3 &=& -1 \\ (-1)^4 &=& +1 \\ (-1)^5 &=&-1 \\ \vdots & & \vdots \end{array}\quad\hdots$ . See? .Alternating signs! If we want to start with "minus", we use: . $(-1)^n$ If we want to start with "plus", we use: . $(-1)^{n+1}$ For this problem, the rule is: . $\boxed{f(n) \;=\;(-1)^{n+1}(n+6)}$ . . Plug in $n = 1,2,3,\hdots$ and check it out. • Apr 23rd 2008, 05:25 PM KittyOfDoom Thank you for the help ^.^	2017-02-27 23:13:28	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 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": 0.8976008296012878, "perplexity": 517.4995283926174}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-09/segments/1487501173866.98/warc/CC-MAIN-20170219104613-00030-ip-10-171-10-108.ec2.internal.warc.gz"}
https://www.numerade.com/questions/compute-the-work-done-by-the-force-field-mathbff-along-the-curve-c-mathbffx-ylangle-y-xrangle-c-is-t/	Vectors Vector Calculus ### Discussion You must be signed in to discuss. ##### Lily A. Johns Hopkins University ##### Kristen K. University of Michigan - Ann Arbor Lectures Join Bootcamp ### Video Transcript given a full six. Former right? Is it going to my ankle? My ex sees a square Wrong zero comma Zito to one comma Zero Do you want to go more? One, 20 Ghana one 201 was here. So they're dis combine. This is the first line segment we get from here. Excellent physical duel zero plus one minus zero Milstein and by two is equal to zero for the crown legal. The X one is it Will do. I didn t want From next line I get X two is equal dough one by two is equal to zero plus one minus zero That steam The X two is equal to zero. Do you like to do is beating too Next line segment we get X three years ago to one minus dean by trees equal to violent. So the X three years ago nine iss DT I mean Andi Divide trees equal to zero. In the last segment, I get exported. Zero explain it is one minus by Horace One last e So my i d x scores zero my dealer for is the default so function of X comma by and see the given by by the eggs less exterior. There s a multitude. So one of the function will be from 0 to 1. This is people Divide the X one plus x d by one blessed by the X to bless X dear I that's why the x three. Methinks the eggs the right tree classified the export plus x the X device Sexual devalues zero do one. So we have the Expo Nous de Dion and by zero So this becomes zero. Exit is be on divide to a zero. So again zero here rise D on the extra zero. So zero plus here X is one and you guys to is DT toe So I get one expression everyone GT two. Yes, Why is one minus d d three? So I get one into minus off GT three Does thanks. Is one minus Tabor Devise, you know. So I give you bless. Why is one plus T d. X or zero? So zero another rescue zero. Forget little best. My final expression is 0 to 1 one dd two last 0 to 1 bond and do d D three d 0 to 1 minus one in 20 to 1. So it isn't good to go. One minus one. It goes Other Schools #### Topics Vectors Vector Calculus ##### Lily A. Johns Hopkins University ##### Kristen K. University of Michigan - Ann Arbor Lectures Join Bootcamp 00:59 01:15 02:56	2021-04-13 05:31:40	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5286481380462646, "perplexity": 3530.830393471317}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-17/segments/1618038072082.26/warc/CC-MAIN-20210413031741-20210413061741-00277.warc.gz"}
https://electronics.stackexchange.com/questions/156745/3-axis-magnetometer-issue-in-one-axis	# 3 axis magnetometer issue in one axis I have several 3-axis magnetometers mounted on breadboards, and they all read correctly in the x and z directions. The Y direction is very strange; when turned to read 0, then turned 90 degrees counter clockwise, it reads 0 again. Then the values gradually get larger, peaking at about SE. What could cause this behaviour? I thought I had determined the breadboard itself was magnetic but I can't reproduce that. • I was going to say some magnetic material nearby. Can you add some leads and move the sensor around? – George Herold Feb 24 '15 at 2:12 • Same issue with leads. If it was something nearby I would expect it to affect every axis. Thought it could be the breadboard because that was stationary relative to the board. – evandentremont Feb 24 '15 at 2:15 • Yeah I wanted to get it off the bread board? How sensitive is it? Nickel is magnetic ... though Nickel plating is thin. – George Herold Feb 24 '15 at 2:15 • Yep. It's on 10" leads now and same behaviour. – evandentremont Feb 24 '15 at 2:16 • It definetly isn't something in the area, when tipping the magnetometer on it's side the Z axis works as expected, and the Y (now Z) shows grossly different numbers when facing up vs facing down. – evandentremont Feb 24 '15 at 2:22	2019-08-17 23:14:46	{"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": 0.34808549284935, "perplexity": 2245.6409056055095}, "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/1566027313501.0/warc/CC-MAIN-20190817222907-20190818004907-00230.warc.gz"}
https://search.datacite.org/members/ethz	Switzerland ### L’astuce dans l’escroquerie à l’assurance privée Katia Villard Le 19 juillet 2017, le Tribunal fédéral a confirmé la condamnation pour tentative d’escroquerie d’un assuré qui avait effectué, par téléphone, une fausse annonce de sinistre à son assurance casco. Ce faisant, notre Cour suprême admet qu’en matière d’escroquerie à l’assurance privée un simple mensonge oral peut déjà être constitutif de tromperie astucieuse au sens de l’art. 146 du code pénal suisse. La présente contribution a pour objectif de discuter du raisonnement du Tribunal fédéral... ### Swiss glaciers: An eternal world of ice? Fabian Schneider, Meda Hotea & Roman Walt ### Schweizer Gletscher: Eine ewige Eiswelt? Meda Hotea, Roman Walt & Fabian Schneider ### Unsupervised Learning of Phase-Change-Based Neuromorphic Systems Stanislaw Andrzej Wozniak Neuromorphic systems provide brain-inspired methods of computing. In a neuromorphic architecture, inputs are processed by a network of neurons receiving operands through synaptic interconnections, tuned in the process of learning. Neurons act simultaneously as asynchronous computational and memory units, which leads to a high degree of parallelism. Furthermore, owing to developments in novel materials, memristive devices were proposed for area- and energy-efficient mixed digital-analog implementation of neurons and synapses. In this dissertation, we propose neuromorphic... ### From 2D to 3D Characterization of Materials Subjected to Extreme Pressure and Temperature Conditions Farhang Nabiei Planetesimal were the first planetary objects to form in the solar system, which later grew to make the proto-planets. Most of these bodies were differentiated as a result of internal heating. Several differentiated bodies have, then, been accreted following the giant impacts to create the terrestrial planets. As a result of these impacts, the newly formed Earth was molten and completely differentiated. Subsequent crystallization has given rise to Earth's current structure. In order to bring... ### Advances in Vanadium-Based Catalyst Research for the Selective Catalytic Reduction of NOx by NH3 The most efficient after-treatment technology for reducing harmful NOx emissions from stationary and mobile sources of diesel exhaust is the selective catalytic reduction (SCR) of NOx with ammonia (NH3). Vanadium-based SCR catalysts reduce NOx selectively between ca. 200 and 500°C. Low temperature activity and high temperature stability are important for the automotive sector because of the large temperature fluctuations in the exhaust gas. 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In this thesis, we study a corpus of 4 million press articles covering a period of 200 years. The thesis tries to measure the evolution of written French on this period at the level of words and expressions, but also in a more global way by attempting to define integrated... ### Scanning tunneling microscopy on large bio-molecular systems on surfaces Sebastian Koslowski The ever growing demand for the development of new technologies and materials has led to extensive studies inspired by biological functional complexes. Peptides with their outstanding ability to efficiently self-assemble can express a broad spectrum of intriguing functionalities. A key question in mimicking their assembly and function is the precise understanding of the specific interactions between the contained amino acids on the level of a sub-molecular length scale. 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The work is motivated by a concrete problem which typically arises during sailing regattas, namely finding the best tacking strategy to reach the upwind buoy as quickly as possible. We assume that there is no loss of time when tacking. We first guess... ### Integrated Daylighting and Artificial Lighting Control based on High Dynamic Range Vision Sensors Ali Motamed One fifth of the electricity consumption of Swiss buildings is due to electric lighting. Integrated control of sun shading and artificial lighting can mitigate this demand while maintaining user comfort. However, the drawback of existing building control approaches is that they do not consider one of the main aspects of human-centric lighting: visual comfort. 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Due to their involvement in numerous essential cellular activity, dysfunctional mitochondria have been implicated in a wide range of human diseases. Confocal microscopy using fluorophores for molecular specificity remains the gold standard of intracellular imaging. However, fluorescent labels can be toxic to the cell upon prolonged exposure and still... ### Polynomial models in finance Damien Edouard Ackerer This thesis presents new flexible dynamic stochastic models for the evolution of market prices and new methods for the valuation of derivatives. These models and methods build on the recently characterized class of polynomial jump-diffusion processes for which the conditional moments are analytic. The first half of this thesis is concerned with modelling the fluctuations in the volatility of stock prices, and with the valuation of options on the stock. A new stochastic volatility model... ### From retinotopic processing to nonretinotopic representation Marc Michael Lauffs Encoding of visual information in the brain is retinotopic: Neighboring points in the visual field are mapped onto neighboring photoreceptors in the retina, and these neighborhood relations are maintained in the early stages of cortical processing. However, perception is nonretinotopic. First, vision is stable and continuous although the retinal image is in constant flux and frequently interrupted by eye blinks. Second, object parts are perceived relative to the object, rather than in retinal coordinates. For... ### Multi Level Monte Carlo Methods for Uncertainty Quantification and Robust Design Optimization in Aerodynamics Michele Pisaroni The vast majority of problems that arise in aircraft production and operation require decisions to be made in the presence of uncertainty. An effective and accurate quantification and control of the level of uncertainty introduced in the design phase and during the manufacturing and operation of aircraft vehicles is imperative in order to design robust and risk tolerant systems. Indeed, the geometrical and operational parameters, that characterize aerodynamic systems, are naturally affected by aleatory uncertainties... ### On some algebraic and extremal problems in discrete geometry In the present thesis, we delve into different extremal and algebraic problems arising from combinatorial geometry. Specifically, we consider the following problems. For any integer $n\ge 3$, we define $e(n)$ to be the minimum positive integer such that any set of $e(n)$ points in general position in the plane contains $n$ points in convex position. In 1935, Erd\H{o}s and Szekeres proved that $e(n) \le {2n-4 \choose n-2}+1$ and later in 1961, they obtained the lower... ### Fighting preterm birth with novel surgical tools and biomaterials engineering Yannick Robert Devaud With advances in fetal diagnosis and therapy, fetoscopy has become a good option to treat a series of life threatening diseases like twin-to-twin transfusion syndrome or severe congenital diaphragmatic hernia. The efficacy of fetoscopy for those types of disorders is not debatable; however, it comes with a daunting limitation: iatrogenic preterm premature rupture of the fetal membrane (iPPROM). The injury created by the instruments to access the amniotic cavity is the reason for fetal membrane... ### Design and applications of a clamp for Green Fluorescent Protein with picomolar affinity Patrick Ernst, Andreas Plückthun, Alexander Koch, Jakob C Stüber, Daniel Bojar, Simon Hansen & Alexander Batyuk ### Digitale Medien und das Kartellrecht Andreas Heinemann #### Resource Types • Text 536,172 • Image 472,880 • Dataset 228,765 • Physical object 2,502 • Event 321 • Workflow 246 • Other 85 • Software 80 • Model 79 • Audiovisual 40 • Collection 21 • Interactive resource 1 • 2016 17,175 • 2015 18,632 • 2014 244,056 • 2013 16,759 • 2012 12,890 • 2011 12,851 • 2010 12,873 • 2009 12,462 • 2008 12,937 • 1980 14,924 • 1970 12,537 • 1964 13,535 • 1950 12,350 • 1930 15,551 • 0000 34,048 • 2017 216,920 • 2016 399,606 • 2015 205,756 • 2014 105,103 • 2013 156,230 • 2012 165,483 • 2011 87,153 • 2010 293,578 • 2009 22,794 #### Data Centers • 027.7 - Zeitschrift für Bibliothekskultur 0 • 4TU.Centre for Research Data 0 • Aalborg University Library 0 • Acta Biologica Plantarum Agriensis 0 • Acta Regionis Rurum 0 0 0 • AERIS - Pôle français de données et services pour l'atmosphère 0 • Africa Health Research Institute 0 • African Population and Health Research Centre 0 0 • Agriculture Victoria 0 0 • Akademie der bildenden Künste Wien 0 • Aktionsbündnis Patientensicherheit e.V. 0 • Alexander von Humboldt Institut für Internet und Gesellschaft 0 • Alfred-Wegener-Institut 0 • Alkalmazott Nyelvtudomány 0 • Alkalmazott Pszichológia 0 • Állatorvostudományi Egyetem 0 • Alliance for Crops, Soils, and Environmental Science Societies 0 • ALT Proceedings / UB Bern 0 • altrelettere 0 • Alveo 0 • AMA Service GmbH 0	2017-12-16 20:50:36	{"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": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.3477173447608948, "perplexity": 6314.416451821085}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-51/segments/1512948589177.70/warc/CC-MAIN-20171216201436-20171216223436-00409.warc.gz"}
https://pos.sissa.it/336/111/	Volume 336 - XIII Quark Confinement and the Hadron Spectrum (Confinement2018) - C: Heavy quarks EFT determination of the heavy-hybrid spin potential W.K. Lai Full text: pdf Pre-published on: September 12, 2019 Published on: September 26, 2019 Abstract We study the spin splitting in the heavy quarkonium hybrid spectrum within the framework of an nonrelativistic effective field theory. We derive for the first time the spin-dependent part of the heavy-quark-antiquark potential for heavy quarkonium hybrids to order $1/m^2$ in the heavy-quark-mass expansion. We find that several operators that are not found in standard quarkonia appear, most remarkably an operator suppressed by only one power of the heavy-quark mass. By matching the weakly-coupled pNRQCD to the effective field theory in the regime of short heavy-quark-antiquark distances, we work out the matching coefficients of the spin-dependent operators, which are factorized into a perturbative and a nonperturbative part. The nonperturbative part can be expressed in terms of purely gluonic correlators. We fit the nonperturbative parts of the matching coefficients to lattice data of the charmonium hybrid spectrum and obtain results that respect the the power counting. Using the obtained nonperturbative pieces, we compute the bottomonium hybrid spectrum with the spin-dependent potential, for which results from the lattice are still sparse. DOI: https://doi.org/10.22323/1.336.0111 How to cite Metadata are provided both in "article" format (very similar to INSPIRE) as this helps creating very compact bibliographies which can be beneficial to authors and readers, and in "proceeding" format which is more detailed and complete. Open Access	2020-12-05 00:28:54	{"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": 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": 0.5433503985404968, "perplexity": 2378.8736493470874}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-50/segments/1606141745780.85/warc/CC-MAIN-20201204223450-20201205013450-00238.warc.gz"}
https://stats.stackexchange.com/questions/15160/poisson-regression-in-a-survival-setting-on-a-simulated-set	# Poisson regression in a survival setting on a simulated set I have a large dataset with patients and I'm studying a rare outcome (~ 2%) and death is a competing risk (mean age ~69 years). I've used the R "cmprsk" package for my statistics and it seems that competing risks and the Cox regression are performing similarly although the competing risk analysis is more conservative giving hazard ratios closer to 1. I've been suggested to do a Poisson regression on the data but the results don't make any sense and I would be really grateful to get some input on the benefits of doing this kind of analysis on survival data. I've created this simulation for creating a dataset with similar risk factors: library("cmprsk") # The time for the study accrual_time <- 10 followup_time <- 1 base_risk <- list("event" = .015, "cmprsk" = .1) risk_factors <- list(list("frequency"=.1, "event" = base_risk$event.5, "cmprsk" = base_risk$cmprsk2), list("frequency"=.05, "event" = base_risk$event1, "cmprsk" = base_risk$cmprsk1), list("frequency"=.05, "event" = base_risk$event-.5, "cmprsk" = base_risk$cmprsk0)) # Number of subjects n <- 5000 # Create base time, sequential inclusion time_in_study <- rep(c(1:n)/naccrual_time + followup_time, 1) set.seed(100) # Create empty sets x <- matrix(0, ncol=length(risk_factors), nrow=n) time_2_event <- rep(0, n) time_2_comprsk <- rep(0, n) # Create each studied observation and outcome for(i in 1:n){ # Set base risk event_risk <- base_risk$event comp_risk <- base_risk$cmprsk for(j in 1:length(risk_factors)){ x[i, j] <- rbinom(1, 1, risk_factors[[j]]$frequency)[1] # If there is a risk factor defined if (x[i, j] > 0){ event_risk <- event_risk + risk_factors[[j]]$event comp_risk <- comp_risk + risk_factors[[j]]$cmprsk } } # Time 2 event/risk is 1/rate meaning that higher number -> shorter time time_2_event[i] <- rexp(1, rate=event_risk)[1] time_2_comprsk[i] <- rexp(1, rate=comp_risk)[1] } cn <- c() for(i in 1:length(risk_factors)){ ev_rsk <- risk_factors[[i]]$event/base_risk$event+1 cmp_rsk <- risk_factors[[i]]$cmprsk/base_risk$cmprsk+1 name <- paste("Risk factor no: ", i, "\n ev=", ev_rsk, " cr=", cmp_rsk, " ", sep="") cn <- c(cn, name) } colnames(x) <- cn # Select the event that happens first: study ends, evenent occurs, a competing event occurs time <- apply(cbind(time_in_study, time_2_event, time_2_comprsk), 1, min) # Outcome identifiers event <- (time_2_event == time) + 0 comprsk <- (time_2_comprsk == time) + 0 cens <- event+2(event==0 & comprsk==1) out.cox_ev <- coxph(Surv(time, event)~x) summary(out.cox_ev) out.crr_ev <- crr(time, cens, x, failcode=1) summary(out.crr_ev) out.cox_cmprsk <- coxph(Surv(time, comprsk)~x) summary(out.cox_cmprsk) out.crr_cmprsk <- crr(time, cens, x, failcode=2) summary(out.crr_cmprsk) The output makes sense but when I do a: out.glm_pr <- glm(event ~ x, family="poisson") summary(out.glm_pr) It gives estimates of: • RF 1 ~ .14 • RF 2 ~ .41 • RF 3 ~ -.23 My questions: • Is the glm() code correct or should I somehow transform my data? • Does the Poisson output make any sense and how should if so interpret it? • What are the benefits/pitfalls in using Poisson regression for survival data? Thanks! ## UPDATE After adding exp(out.glm_pr$coefficients) the results are almost identical to the competing risk regression, here's a forest plot that compares the three: The x-axis is perhaps not entirely valid (should be "incident rate ratios" for the Poisson regression) but why are the outcomes for CRR & poisson almost identical? As for testing over-dispersion I've found these two methods: > library(qcc) > qcc.overdispersion.test(event) Overdispersion test Obs.Var/Theor.Var Statistic p-value poisson data 0.9391878 4695 0.99902 > > library(pscl) > out.glm_nb <- glm.nb(event ~ x) Warning messages: 1: In theta.ml(Y, mu, sum(w), w, limit = control$maxit, trace = control$trace > : iteration limit reached 2: In theta.ml(Y, mu, sum(w), w, limit = control$maxit, trace = control$trace > : iteration limit reached > odTest(out.glm_nb) Likelihood ratio test of H0: Poisson, as restricted NB model: n.b., the distribution of the test-statistic under H0 is non-standard e.g., see help(odTest) for details/references Critical value of test statistic at the alpha= 0.05 level: 2.7055 Chi-Square Test Statistic = -0.0139 p-value = 0.5 I conclude that there isn't any evidence of over-dispersion or are there other methods better suited for testing over-dispersion in this kind of survival data? The quasipoisson analysis gives similar values: > out.glm_quasi_pr <- glm(event ~ x, family=quasipoisson(link="log")) > round(exp(out.glm_quasi_pr\$coefficients), 3) (Intercept) xRF 1 xRF 2 xRF 3 0.059 1.152 1.509 0.794 Answering two of your three questions, because I'm not comfortable enough in R to diagnose coding errors - though your code looks right to me. For your estimates, as you didn't specify a link function, R is using the default log link function. In order for your estimates to make sense, you need to exp(estimate) - this will give you 1.15, 1.51 and 2.94 respectively. These should be close to the HRs coming off your Cox model. These numbers are "incident rate ratios" - the name is pretty suggestive of what they are. They can be interpreted very similarly to hazard ratios, and indeed should equal hazard ratios under certain assumption. As for the benefits (and drawbacks) of Poisson regression. Poisson survival analysis is a fully parametric, maximum likelihood method of estimating differences in the survival between groups, which has some nice properties for some uses. It estimates the baseline hazard, which if you intend to use the baseline hazard in further analysis (I often do), is something a Cox model expressly does not estimate. The incidence density (# of cases / time at risk) is also vastly more intuitive than the hazard. Now the drawbacks, of which there are many. The Poisson model is vulnerable to overdispersion - I'd rerun your model using quasipoisson or a negative binomial model to check and see if your results are sensitive to overdispersion. More important, IMO, is the assumptions the poisson model makes about the underlying survival function. The Cox Proportional Hazards model, as the name suggests, assumes the hazard function is proportional between the two groups over time. The Poisson model assumes not only are the hazards proportional, but constant. This is often a pretty major assumption, and should be checked. Adding a term or several in the model for time can help relax this assumption somewhat. As for the similarity of your results: I'm not sure what kind of model your competing risk package is assuming, but I'm guessing it's estimating a parametric model of the survival function - the Cox model, which is "semi-parametric", doesn't estimate this baseline hazard. If the estimated parametric survival function is close to a exponential model (the distribution of the survival function assumed with a Poisson model), you may get very similar results if your data isn't sensitive to the assumption of no competing risks. Your results don't seem terribly vulnerable to this assumption. What you do seem to be sensitive to is the violation of a constant hazard assumption, hence the dramatic difference in your estimates between Poisson and Cox models. If the hazard function was perfectly constant, the two models should actually produce the same estimate. I would try one of two things: 1. Add a time term or several to the Poisson model. Something like time and time*time, and see if your Poisson estimate moves closer to the Cox result. 2. You should be able to visualize the hazard function. The most common way to do this is to plot the log(-log(survival function)) versus the log of survival time for each of your variables, stratified by the groups. For the Cox model to be valid, they should be parallel. For the Poisson model to be valid, they should be straight and parallel. The survival functions in R must be able to do this, though I don't know how. I'd find it very odd to see this in a simulated data set introduced essentially by accident, but I'd try looking at it anyway. • Thank you for your excellent answer! Could you please clarify by what you mean by "Adding a term or several"? I've also added some over-dispersion analysis - is it performed correctly? Do you mind commenting on the similarity between CRR and the Poisson regression? – Max Gordon Sep 4 '11 at 11:00 • @MaxG What I mean by adding a term or several is essentially adding a time variable - one that makes sense for your study - or several in a more sophisticated form (polynomials of time for example) - which allows the constant hazard assumed by the poisson model to "flex" a bit. I would have used a likelihood ratio test to check for over-dispersion as well, so you should be good, though it seems your model may be having some issues converging (see the warnings about iteration limits reached). Comments on the similarity between CRR and Poisson in my answer. – Fomite Sep 5 '11 at 1:25	2019-08-18 09:29:29	{"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": 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": 0.7645904421806335, "perplexity": 1608.3070314356416}, "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/1566027313747.38/warc/CC-MAIN-20190818083417-20190818105417-00225.warc.gz"}
http://math.stackexchange.com/questions/127415/real-vs-complex-for-integrals-int-0-infty-fracdx1-x3	# Real VS Complex for integrals: $\int_0^\infty \frac{dx}{1 + x^3}$ The integral $$\int_0^\infty \frac{dx}{1 + x^4} = \frac{\pi}{2\sqrt2}$$ can be evaluated both by a complex method (residues) and by a real method (partial fraction decomposition). The complex method works also for the integral $$\int_0^\infty \frac{dx}{1 + x^3} = \frac{2\pi}{3\sqrt3}$$ but partial fraction decomposition does not give convergent integrals. I would like to know if there is some real method for evaluating this last integral. - Make the substitution $x = \frac{1}{t}$ and you get $$\int_{0}^{\infty} \frac{t}{1+t^3} \text{d}t$$ Write the one you want as $$\int_{0}^{\infty} \frac{1}{1+t^3} \text{d}t$$ Now you can add both and cancel that pesky $1+t$ factor. btw, a straightforward approach using partial fractions also works. You consider $$F(x) = \int_{0}^{x} \frac{1}{1+t^3} \text{d}t$$ Using partial fractions you can find that (I used Wolfram Alpha, I admit) $$F(x) = \frac{1}{6}\left(2\log(x+1) - \log(x^2 - x -1) + 2\sqrt{3} \arctan\left(\frac{2x-1}{\sqrt{3}}\right)\right) + \frac{\pi}{6\sqrt{3}}$$ Now as $x \to \infty$, we have that $2\log(x+1) - \log(x^2 - x + 1) \to 0$ . - See this answer: math.stackexchange.com/questions/34351/… to see that you can also evaluate the integral in that question to find your answer here, and then apply the other answers to that question (in particular, Eric's answer(s)). – Aryabhata Apr 2 '12 at 22:27 Thank you Aryabhata. This is a very elegant way of evaluation by reducing it to the $\int_0^\infty \frac{dx}{x^2 - x + 1}$. – Martin Apr 2 '12 at 22:45 @Martin: You are welcome. I have added another method, which does use partial fractions. – Aryabhata Apr 2 '12 at 22:47 Note that for $a > 0$, $$\int_0^N \frac{1}{x+a}\ dx = \ln(N+a) - \ln(a) = \ln(N) - \ln(a) + o(1)\ \text{as} \ N \to \infty$$ while \eqalign{\int_0^N \frac{x+a}{(x+a)^2 + b^2}\ dx &= \frac{1}{2} \left(\ln((N+a)^2+b^2) - \ln(a^2+b^2)\right)\cr &= \ln(N) - \ln(a^2+b^2) + o(1) \ \text{as} \ N \to \infty\cr} and (if $b > 0$) \eqalign{\int_0^N \frac{1}{(x+a)^2+b^2}\ dx = \frac{\arctan\left(\frac{N+a}{b}\right) - \arctan\left(\frac{a}{b}\right)}{b} = \frac{\pi}{2b} - \frac{\arctan\left(\frac{a}{b}\right)}{b} + o(1) \ \text{as} \ N \to \infty\cr} In particular, from the partial fraction decomposition $$\frac{1}{1+x^3} = \frac{1/3}{x+1} + \frac{(2-x)/3}{x^2 - x + 1} = \frac{1/3}{x+1} + \frac{1/2}{(x-1/2)^2+3/4} - \frac{(x-1/2)/3}{(x-1/2)^2 + 3/4}$$ you get $$\int_0^N \frac{1}{1+x^3} \ dx = \frac{\ln(N) - \ln(1))}{3} + \frac{\pi/2 + \arctan(1/\sqrt{3})}{\sqrt{3}} - \frac{\ln(N) - \ln((1/2)^2 + 3/4)}{3} + o(1)$$ i.e. $$\int_0^\infty \frac{1}{1+x^3} \ dx = \frac{\pi}{\sqrt{3}} + \frac{\arctan(1/\sqrt{3})}{\sqrt{3}} = \frac{2 \pi}{3 \sqrt{3}}$$ - Thank you Robert. Nice solution by taking asymptotics. Little typo: in the last integral the upper bound is $\infty$ – Martin Apr 3 '12 at 18:39 Thanks for spotting that, fixed it. – Robert Israel Apr 3 '12 at 21:01 For what is worth: Your integral evaluates in terms of the sine function: $$\int\limits_0^\infty \frac{1}{1+x^a}=\frac{\pi}{a}\sec\frac{\pi}{a}$$ refer to this question and the link in it. - I would like to know if there is some real method for evaluating this last integral. Actually, all integrals of the form $\displaystyle\int_0^\infty\frac{x^n}{1+x^m}dx$ can be solved by substituting $t=\dfrac1{1+x^m}$ , and then recognizing the expression of the beta function in the new integral, which can be written as a product of gamma functions. Then we use the reflection formula in order to finally arrive at the desired result, $I=\dfrac\pi m\cdot\csc\left[(n+1)\dfrac\pi m\right]$ — See my answer here for more information. -	2016-06-30 18:05:20	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9861369132995605, "perplexity": 439.7053541509207}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-26/segments/1466783399106.96/warc/CC-MAIN-20160624154959-00110-ip-10-164-35-72.ec2.internal.warc.gz"}
https://learn.careers360.com/ncert/question-find-the-rate-of-change-of-the-area-of-a-circle-with-respect-to-its-radius-r-when-r-4-cm/	Filters Q&A - Ask Doubts and Get Answers Q Find the rate of change of the area of a circle with respect to its radius r when r = 4 cm 1. b) Find the rate of change of the area of a circle with respect to its radius r when r = 4 cm Views Area of the circle (A) = $\pi r^{2}$ Rate of change of the area of a circle with respect to its radius r = $\frac{dA}{dr}$ = $\frac{d(\pi r^{2})}{dr}$ = $2 \pi r$ So, when r = 4, Rate of change of the area of a circle = $2 \pi (4)$ = $8 \pi$ Hence, Rate of change of the area of a circle with respect to its radius r when r = 4 is $8 \pi$ Exams Articles Questions	2020-02-19 04:18:38	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 7, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8518288731575012, "perplexity": 128.71956752481972}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-10/segments/1581875144027.33/warc/CC-MAIN-20200219030731-20200219060731-00446.warc.gz"}
https://blogs.mathworks.com/pick/2015/12/04/a-nice-easy-video-tutorial/?s_tid=blogs_rc_1	# A nice, easy video tutorial ### Contents Brett's Pick this week is Video Processing Tutorial, by prolific File Exchange author Image Analyst. For various reasons, I've been spending a lot of time working with videos in MATLAB lately--whether helping a customer work through some challenges, or developing new tools for detecting, tracking, and labeling regions of interest in individual frames. Getting started with video analysis can be a bit daunting--there are several options to confound the newer user, and it's easy to take mis-steps en route. I particularly like Image Analyst's tutorial precisely because (as one recent commenter put it), it's: "a nice starting place for video processing. Shows how easy it is for beginners to get into. Great job!" The tutorial is at just the right level to show video playback, and to demonstrate how to incorporate a simple frame-wise processing algorithm. Using a demo video ('rhinos.avi') that ships with the Image Processing Toolbox, Image Analyst shows how to: • Extract the folder information for the target video: folder = fileparts(which('rhinos.avi')); • Select a video from that directory: movieFullFileName = fullfile(folder, 'rhinos.avi'); videoObject = VideoReader(movieFullFileName) • And use it to read and display frames. (I show the code with slight modifications here): fontSize = 12; numberOfFrames = videoObject.NumberOfFrames; for frame = 1 : numberOfFrames image(thisFrame); caption = sprintf('Frame %4d of %d.', frame, numberOfFrames); title(caption, 'FontSize', fontSize); end There's much more.... Image Analyst shows the user the how to calculate and plot live statistics for the frame, and to perform some basic image analyses during the read/display process, and to visualize the results while the video plays: Optionally, you are prompted to write individual frames as images to a directory, and to recall them for subsequent playback. Very useful stuff! I have a few suggestions for consideration: Image Analyst's video player uses "frame-based" reading. In R2014b, we introduced "time-based" frame reading that can be more efficient. To modify the code to use time-based reading, consider commenting out the call to videoObject.NumberOfFrames. (That triggers frame-based reading, and causes an error if you subsequently try to read using time-based modalities.) Instead, you can calculate the number of frames using numberOfFrames = round(videoObject.FrameRate * videoObject.Duration);, if you need to. for frame = 1:numberOfFrames you can use a while loop: while hasFrame(videoObject) #### imshow and cdata instead of image Also, I really like imshow for displaying images. It recognizes that the input matrix is an image, and maintains the aspect ratio. (image-displayed images can be stretched.) It also suppresses axes tick marks automatically, and has some other nice behaviors. Moreover, once you've created an "image object," you can re-use it quite easily by modifying the "CData" of the object. There's no need to create a new image with each frame read: #### So... videoObject = VideoReader(movieFullFileName); img = imshow(frame); while hasFrame(videoObject) img.CData = thisFrame; caption = sprintf('Frame %4d of %d.', ... round(videoObject.CurrentTime*videoObject.FrameRate), numberOfFrames); title(caption, 'FontSize', fontSize); end #### And finally: Image Analyst uses a struct to store the frame information. The new "image datastore" functionality (introduced in R2015b) facilitates easy reference to, and visualization of, individual images. (Image Analyst shows how to create a new video from the extracted frames, but you can alternatively just display them in a loop, using the readimage method of the datastore object. (I'll leave it to you to peruse the documentation for those new capabilities.)	2020-07-15 08:03:25	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.234782412648201, "perplexity": 4583.709523738555}, "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-2020-29/segments/1593657163613.94/warc/CC-MAIN-20200715070409-20200715100409-00573.warc.gz"}
https://davidlowryduda.com/tag/automorphic-forms/	# Tag Archives: automorphic forms ## Notes from a Talk at Building Bridges 4 On 18 July 2018 I gave a talk at the 4th Building Bridges Automorphic Forms Workshop, which is hosted at the Renyi Institute in Budapest, Hungary this year. In this talk, I spoke about counting points on hyperboloids, with a certain focus on counting points on the three dimensional hyperboloid $$$$X^2 + Y^2 = Z^2 + h$$$$ for any fixed integer $h$. I gave a similar talk at the 32nd Automorphic Forms Workshop in Tufts in March. I don’t say this during my talk, but a big reason for giving these talks is to continue to inspire me to finish the corresponding paper. (There are still a couple of rough edges that need some attention). The methodology for the result relies on the spectral expansion of half-integral weight modular forms. This is unfriendly to those unfamiliar with the subject, and particularly mysterious to students. But there is a nice connection to a topic discussed by Arpad Toth during the previous week’s associated summer school. Arpad sketched a proof of the spectral decomposition of holomorphic modular cusp forms on $\Gamma = \mathrm{SL}(2, \mathbb{Z})$. He showed that $$L^2(\Gamma \backslash \mathcal{H}) = \textrm{cuspidal} \oplus \textrm{Eisenstein}, \tag{1}$$ where the cuspidal contribution comes from Maass forms and the Eisenstein contribution comes from line integrals against Eisenstein series. The typical Eisenstein series $$$$E(z, s) = \sum_{\gamma \in \Gamma_\infty \backslash \Gamma} \textrm{Im}(\gamma z)^s$$$$ only converges for $\mathrm{Re}(s) > 1$, and the initial decomposition in $(1)$ implicitly has $s$ in this range. To write down the integrals appearing in the Eisenstein spectrum explicitly, one normally shifts the line of integration to $1/2$. As Arpad explained, classically this produces a pole at $s = 1$ (which is the constant function). In half-integral weight, the Eisenstein series has a pole at $s = 3/4$, with the standard theta function $$$$\theta(z) = \sum_{n \in \mathbb{Z}} e^{2 \pi i n^2 z}$$$$ as the residue. (More precisely, it’s a constant times $y^{1/4} \theta(z)$, or a related theta function for $\Gamma_0(N)$). I refer to this portion of the spectrum as the residual spectrum, since it comes from often-forgotten residues of Eisenstein series. Thus the spectral decomposition for half-integral weight objects is a bit more complicated than the normal case. When giving talks involving half-integral weight spectral expansions to audiences including non-experts, I usually omit description of this. But for those who attended the summer school, it’s possible to at least recognize where these additional terms come from. The slides for this talk are available here. ## Paper: The Second Moments of Sums of Fourier Coefficients of Cusp Forms This is joint work with Thomas Hulse, Chan Ieong Kuan, and Alex Walker. We have just uploaded a paper to the arXiv on the second moment of sums of Fourier coefficients of cusp forms. This is the first in a trio of papers that we will be uploading and submitting in the near future. Suppose ${f(z)}$ and ${g(z)}$ are weight ${k}$ holomorphic cusp forms on ${\text{GL}_2}$ with Fourier expansions \begin{align} f(z) &= \sum_{n \geq 1} a(n) e(nz) \\ g(z) &= \sum_{n \geq 1} b(n) e(nz). \end{align} Denote the sum of the first ${n}$ coefficients of a cusp form ${f}$ by $$S_f(n) := \sum_{m \leq n} a(m). \tag{1}$$ We consider upper bounds for the second moment of ${S_f(n)}$. The famous Ramanujan-Petersson conjecture gives us that ${a(n)\ll n^{\frac{k-1}{2} + \epsilon}}$. So one might assume ${S_f(X) \ll X^{\frac{k-1}{2} + 1 + \epsilon}}$. However, we expect the better bound $$S_f(X) \ll X^{\frac{k-1}{2} + \frac{1}{4} + \epsilon}, \tag{2}$$ which we refer to as the “Classical Conjecture,” echoing Hafner and Ivić [HI]. Chandrasekharan and Narasimhan [CN] proved that the Classical Conjecture is true on average by showing that $$\sum_{n \leq X} \lvert S_f(n) \rvert^2 = CX^{k- 1 + \frac{3}{2}} + B(X), \tag{3}$$ where ${B(x)}$ is an error term, $$B(X) = \begin{cases} O(X^{k}\log^2(X)) \ \Omega\left(X^{k – \frac{1}{4}}\frac{(\log \log \log X)^3}{\log X}\right), \end{cases} \tag{4}$$ and ${C}$ is the constant, $$C = \frac{1}{(4k + 2)\pi^2} \sum_{n \geq 1}\frac{\lvert a(n) \rvert^2}{n^{k + \frac{1}{2}}}. \tag{5}$$ A application of the Cauchy-Schwarz inequality to~(3) leads to the on-average statement that $$\frac{1}{X} \sum_{n \leq X} \|S_f(n)\| \ll X^{\frac{k-1}{2} + \frac{1}{4}}. \tag{6}$$ From this, [HI] were able to show in some cases that $$S_f(X) \ll X^{\frac{k-1}{2} + \frac{1}{3}}. \tag{7}$$ Better lower bounds are known for ${B(X)}$. In the same work [HI] improved the lower bound of [CN] for full-integral weight forms of level one and showed that $$B(X) = \Omega\left(X^{k – \frac{1}{4}}\exp\left(D \tfrac{(\log \log x )^{1/4}}{(\log \log \log x)^{3/4}}\right)\right), \tag{8}$$ for a particular constant ${D}$. The question of better understanding ${B(X)}$ is analogous to understanding the error term in the circle problem or divisor problem. In our paper, we introduce the Dirichlet series $$D(s, S_f \times S_g) := \sum_{n \geq 1} \frac{S_f(n) \overline{S_g(n)}}{n^{s + k – 1}}$$ D(s, S_f \times \overline{S_g}) &:= \sum_{n \geq 1} \frac{S_f(n)S_g(n)}{n^{s + k – 1}} and provide their meromorphic continuations. From our review of the literature, these Dirichlet series and their meromorphic continuations are new and provide new approaches to the classical problems related to ${S_f(n)}$. Our primary result is the meromorphic continuation of ${D(s, S_f \times S_g)}$. As a first application, we prove a smoothed generalization to~(3). Theorem 1 Suppose either that ${f = g}$ is a Hecke eigenform or that ${f}$ and ${g}$ have real coefficients. \begin{equation} \frac{1}{X} \sum_{n \geq 1}\frac{S_f(n)\overline{S_g(n)}}{n^{k – 1}}e^{-n/X} = CX^{\frac{1}{2}} + O_{f,g,\epsilon}(X^{-\frac{1}{2} + \theta + \epsilon}) \end{equation} where \begin{equation} C = \frac{\Gamma(\tfrac{3}{2}) }{4\pi^2} \frac{L(\frac{3}{2}, f\times g)}{\zeta(3)}= \frac{\Gamma(\tfrac{3}{2})}{4\pi ^2} \sum_{n \geq 1} \frac{a(n)\overline{b(n)}}{n^{k + \frac{1}{2}}}, \end{equation} and ${\theta}$ denotes progress towards Selberg’s Eigenvalue Conjecture. Similarly, \begin{equation} \frac{1}{X} \sum_{n \geq 1}\frac{S_f(n)S_g(n)}{n^{k – 1}}e^{-n/X} = C’X^{\frac{1}{2}} + O_{f,g,\epsilon}(X^{-\frac{1}{2} + \theta + \epsilon}), \end{equation} where \begin{equation} C’ = \frac{\Gamma(\tfrac{3}{2})}{4\pi^2} \frac{L(\frac{3}{2}, f\times \overline{g})}{\zeta(3)} = \frac{\Gamma(\tfrac{3}{2})}{4\pi ^2} \sum_{n \geq 1} \frac{a(n)b(n)}{n^{k + \frac{1}{2}}}.\end{equation} We have a complete meromorphic continuation, and it would not be hard to give additional terms in the asymptotic. But the next terms come from zeroes of the zeta function and are complicated to nail down exactly. Choosing ${f = g}$, we recover a proof of the Classical Conjecture on Average. More interestingly, we show that the secondary growth terms do not arise from a pole, nor are there prescribed polar reasons for growth. The secondary growth in the classical result comes from choosing a sharp cutoff instead of the nicely behaving and natural smooth cutoffs. We prove analogous results for sums of normalized Fourier coefficients $$S_f^\alpha(n) := \sum_{m \leq n} \frac{a(m)}{m^\alpha} \tag{9}$$ for ${0 \leq \alpha < k}$. In the path to proving these results, we explicitly demonstrate remarkable cancellation between Rankin-Selberg convolution ${L}$-functions ${L(s, f\times g)}$ and shifted convolution sums $$Z(s, 0; f,g) := \sum_{n, h} \frac{a(n)\overline{b(n-h)}}{n^{s + k – 1}}. \tag{10}$$ Comparing our results and methodologies with the main results of [CN] guarantees similar cancellation for general level and general weight, including half-integral weight forms. We provide additional applications of the meromorphic continuation of ${D(s, S_f \times S_g)}$ in forthcoming works, which will be uploaded to the arXiv and described briefly here soon. For exact references, see the paper. Posted in Math.NT, Mathematics, Uncategorized \| \| 2 Comments	2021-09-19 05:14:19	{"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": 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": 3, "x-ck12": 0, "texerror": 0, "math_score": 0.9833970665931702, "perplexity": 775.2540446218933}, "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/1631780056711.62/warc/CC-MAIN-20210919035453-20210919065453-00406.warc.gz"}
https://en.academic.ru/dic.nsf/enwiki/256890	# Siphon Siphon Siphon principle A siphon used for homebrewing beer The word siphon (Greek: Σίφων, also spelled syphon) is sometimes used to refer to a wide variety of devices that involve the flow of liquids through tubes. But in the English language today, the word siphon usually refers to a tube in an inverted U shape which causes a liquid to flow uphill, above the surface of the reservoir, without pumps, powered by the fall of the liquid as it flows down the tube under the pull of gravity, and is discharged at a level lower than the surface of the reservoir. In practical siphons, atmospheric pressure pushes the liquid up the tube into the region of reduced pressure at the top of the tube. The reduced pressure is caused by liquid falling on the exit side. In the laboratory, some siphons have been demonstrated to work in a vacuum,[1][2][3][4][5] indicating the tensile strength of the liquid is contributing to the operation of siphons at very low pressures. ## History Egyptian reliefs from 1500 BC depict siphons used to extract liquids from large storage jars.[6] We have physical evidence for the use of siphons by Greek engineers in the 3rd century BC at Pergamon.[7] Hero of Alexandria wrote extensively about siphons in the treatise Pneumatica.[8] In the 9th century, the Banu Musa brothers invented a double-concentric siphon, which they described in their Book of Ingenious Devices.[9] ## Operation ### Theory A siphon works because gravity pulling down on the taller column of liquid causes reduced pressure at the top of the siphon. This reduced pressure means gravity pulling down on the shorter column of liquid is not sufficient to keep the liquid stationary so it flows from the upper reservoir, up and over the top of the siphon.[10][11] When the column of liquid is allowed to fall from C down to D, liquid in the upper reservoir will flow up to B and over the top.[10][11] No liquid tensile strength is needed. Even the falling lighter lower leg from C to D can cause the liquid of the heavier upper leg to flow up and over into the lower reservoir[12] An occasional misunderstanding of siphons is that they rely on the tensile strength of the liquid to pull the liquid up and over the rise.[10][11] While water has been found to have a great deal of tensile strength in some experiments (such as with the z-tube[13]), and siphons in vacuum rely on such cohesion, common siphons can easily be demonstrated to need no liquid tensile strength at all to function.[4][10][11] Furthermore, since common siphons operate at positive pressures throughout the siphon, there is no contribution from liquid tensile strength, because the molecules are actually repelling each other in order to resist the pressure, rather than pulling on each other.[4] To demonstrate, the longer lower leg of a common siphon can be plugged at the bottom and filled almost to the crest with liquid, leaving the top and the shorter upper leg completely dry and containing only air. When the plug is removed and the liquid in the longer lower leg is allowed to fall, the liquid in the upper reservoir will then typically sweep the air bubble down and out of the tube. The apparatus will then continue to operate as a siphon. As there is no contact between the liquid on either side of the siphon at the beginning of this experiment, there can be no cohesion between the liquid molecules to pull the liquid over the rise. Another simple demonstration that liquid tensile strength isn't needed in the siphon is to simply introduce a bubble into the siphon during operation. The bubble can be large enough to entirely disconnect the liquids in the tube before and after it, defeating any liquid tensile strength, and yet if the bubble isn't too big, the siphon will continue to operate with little change. The uphill flow of water in a siphon doesn’t violate the principle of continuity because the mass of water entering the tube and flowing upwards is equal to the mass of water flowing downwards and leaving the tube. A siphon doesn't violate the principle of conservation of energy because the loss of gravitational potential energy as liquid flows from the upper reservoir to the lower reservoir equals the work done in overcoming fluid friction as the liquid flows through the tube.[14] Once started, a siphon requires no additional energy to keep the liquid flowing up and out of the reservoir. The siphon will draw liquid out of the reservoir until the level falls below the intake, allowing air or other surrounding gas to break the siphon, or until the outlet of the siphon equals the level of the reservoir, whichever comes first. The maximum height of the crest is limited by atmospheric pressure, the density of the liquid, and its vapour pressure. When the pressure within the liquid drops to below the liquid's vapor pressure, tiny vapor bubbles can begin to form at the high point and the siphon effect will end. This effect depends on how efficiently the liquid can nucleate bubbles; in the absence of impurities or rough surfaces to act as easy nucleation sites for bubbles, siphons can temporarily exceed their standard maximum height during the extended time it takes bubbles to nucleate. For water at standard atmospheric pressure, the maximum siphon height is approximately 10 m (32 feet); for mercury it is 76 cm (30 inches). ### Analogy The chain model is a useful but flawed analogy to the operation of a siphon A simplified conceptual model of a siphon is that it is like a chain hanging over a pulley with one end of the chain piled on a higher surface than the other. Since the length of chain on the shorter side is lighter than the length of chain on the taller side, the chain will move up around the pulley and down towards the lower surface.[15] There are two problems with the chain model of a siphon. The first is that under most practical circumstances, dissolved gases, vapor pressure, and (sometimes) lack of adhesion with tube walls, conspire to render the tensile strength within the liquid ineffective for siphoning. Thus, unlike a chain which has significant tensile strength, liquids usually have little tensile strength under typical siphon conditions, and therefore the liquid on the rising side cannot be pulled up, in the way the chain is pulled up on the rising side.[11][4] The second problem with the chain model of the siphon is that the weight of liquid on the up side of the siphon can be greater than the liquid on the down side, yet the siphon can still function. For example, if the tube from the upper reservoir to the top of the siphon has a much larger diameter than the section of tube from the lower reservoir to the top of the siphon, the shorter upper section of the siphon may have a much larger weight of liquid in it, yet the siphon can function normally.[12] ## Practical requirements A plain tube can be used as a siphon. An external pump has to be applied to start the liquid flowing and prime the siphon. This can be a human mouth. This is sometimes done with any leak-free hose to siphon gasoline from a motor vehicle's gasoline tank to an external tank. (Siphoning gasoline by mouth often results in the accidental swallowing of gasoline, which is quite poisonous, or aspirating it into the lungs, which can cause death or lung damage.[16]) If the tube is flooded with liquid before part of the tube is raised over the intermediate high point and care is taken to keep the tube flooded while it is being raised, no pump is required. Devices sold as siphons come with a siphon pump to start the siphon process. When applying a siphon to any application it is important that the piping be as closely sized to the requirement as possible. Using piping of too great a diameter and then throttling the flow using valves or constrictive piping appears to increase the effect of previously cited concerns over gases or vapor collecting in the crest which serve to break the vacuum. Once the vacuum is reduced the siphon effect is lost. Reducing the size of pipe used closer to requirements appears to reduce this effect and creates a more functional siphon that does not require constant re-priming and restarting. In this respect, where the requirement is to match a flow into a container with a flow out of said container (to maintain a constant level in a pond fed by a stream, for example) it would be preferable to utilize two or three smaller separate parallel pipes that can be started as required rather than attempting to use a single large pipe and attempting to throttle it. ## Applications Siphoning the beer after a first fermentation When certain liquids needs to be purified, siphoning can help prevent either the bottom (dregs) or the top (foam and floaties) from being transferred out of one container into a new container. Siphoning is thus useful in the fermentation of wine and beer for this reason, since it can keep unwanted impurities out of the new container. Self-constructed siphons, made of pipes or tubes, can be used to evacuate water from cellars after floodings. Between the flooded cellar and a deeper place outside a connection is built, using a tube or some pipes. They are filled with water through an intake valve (at the highest end of the construction). When the ends are opened, the water flows through the pipe into the sewer or the river. Siphoning is common in irrigated fields to transfer a controlled amount of water from a ditch, over the ditch wall, into furrows. Large siphons may be used in municipal waterworks and industry. Their size requires control via valves at the intake, outlet and crest of the siphon. The siphon may be primed by closing the intake and outlets and filling the siphon at the crest. If intakes and outlets are submerged, a vacuum pump may be applied at the crest to prime the siphon. Alternatively the siphon may be primed by a pump at either the intake or outlet. Gas in the liquid is a concern in large siphons.[17] The gas tends to accumulate at the crest and if enough accumulates to break the flow of liquid, the siphon stops working. The siphon itself will exacerbate the problem because as the liquid is raised through the siphon, the pressure drops, causing dissolved gases within the liquid to come out of solution. Higher temperature accelerates the release of gas from liquids so maintaining a constant, low temperature helps. The longer the liquid is in the siphon, the more gas is released, so a shorter siphon overall helps. Local high points will trap gas so the intake and outlet legs should have continuous slopes without intermediate high points. The flow of the liquid moves bubbles thus the intake leg can have a shallow slope as the flow will push the gas bubbles to the crest. Conversely, the outlet leg needs to have a steep slope to allow the bubbles to move against the liquid flow; though other designs call for a shallow slope in the outlet leg as well to allow the bubbles to be carried out of the siphon. At the crest the gas can be trapped in a chamber above the crest. The chamber needs to be occasionally primed again with liquid to remove the gas. ## Siphon terminology ### Bowl siphon Bowl siphons are part of flush toilets. Siphon action in the bowl siphon siphons out the contents of the toilet bowl and creates the characteristic toilet "sucking" sound. Some toilets also use the siphon principle to obtain the actual flush from the cistern. The flush is triggered by a lever or handle that operates a simple diaphragm-like piston pump that lifts enough water to the crest of the siphon to start the flow of water which then completely empties the contents of the cistern into the toilet bowl. The advantage of this system was that no water would leak from the cistern excepting when flushed. Early urinals incorporated a siphon in the cistern which would flush automatically on a regular cycle because there was a constant trickle of clean water being fed to the cistern by a slightly open valve. Trap under a sink which functions as a siphon ### Inverted siphon An inverted siphon is not a siphon but a term applied to pipes that must dip below an obstruction to form a "U" shaped flow path. Inverted siphons are commonly called traps for their function in preventing smelly sewer gases from coming back out of drains and sometimes making dense objects like rings and electronic components retrievable after falling into a drain.[citation needed] Liquid flowing in one end simply forces liquid up and out the other end, but solids like sand will accumulate. This is especially important in sewage systems or culverts which must be routed under rivers or other deep obstructions where the better term is "depressed sewer". Large inverted siphons are used to convey water being carried in canals or flumes across valleys, for irrigation or gold mining. ### Back siphonage Back siphonage is a plumbing term applied to clean water pipes that connect directly into a reservoir without an air gap. As water is delivered to other areas of the plumbing system at a lower level, the siphon effect will tend to siphon water back out of the reservoir. This may result in contamination of the water in the pipes. Back siphonage is not to be confused with backflow. Back siphonage is a result of liquids at a lower level drawing water from a higher level. Backflow is driven entirely by pressure in the reservoir itself. Backflow cannot occur through an intermediate high-point. Back siphonage can flow through an intermediate high-point and is thus much more difficult to guard against. ### Anti-siphon valve Anti-siphon valves[18] are required in such designs. Building codes often contain specific sections on back siphonage and especially for external faucets. (See sample building code below.) The reason is that external faucets may be attached to hoses which may be immersed in an external body of water, such as a garden pond, swimming pool, aquarium or washing machine. Should the pressure within the water supply system fall, the external water may be siphoned back into the drinking water system through the faucet. Another possible contamination point is the water intake in the toilet tank. An anti-siphon valve is also required here to prevent pressure drops in the water supply line from siphoning water out of the toilet tank (which may contain additives such as "toilet blue") and contaminating the water system. Anti-siphon valves function as a one-direction check valve. Anti-siphon valves are also used medically. Hydrocephalus, or excess fluid in the brain, may be treated with a shunt which drains cerebrospinal fluid from the brain. All shunts have a valve to relieve excess pressure in the brain. The shunt may lead into the abdominal cavity such that the shunt outlet is significantly lower than the shunt intake when the patient is standing. Thus a siphon effect may take place and instead of simply relieving excess pressure, the shunt may act as a siphon, completely draining cerebrospinal fluid from the brain. The valve in the shunt may be designed to prevent this siphon action so that negative pressure on the drain of the shunt does not result in excess drainage. Only excess positive pressure from within the brain should result in drainage.[19][20][21] Note that the anti-siphon valve in medical shunts is preventing excess forward flow of liquid. In plumbing systems, the anti-siphon valve is preventing backflow. ### Other anti-siphoning devices Along with anti-siphon valves, anti-siphoning devices also exist. The two are unrelated in application. Siphoning can be used to remove fuel from tanks. With the cost of fuel increasing, it has been linked in several countries globally to the rise in fuel theft. Trucks, with their large fuel tanks, are most vulnerable. The anti-siphon device prevents thieves from inserting a tube into the fuel tank. ### Siphon barometer A siphon barometer is the term sometimes applied to the simplest of mercury barometers. A continuous U-shaped tube of the same diameter throughout is sealed on one end and filled with mercury. When placed into the upright position, mercury will flow away from the sealed end, forming a partial vacuum, until balanced by atmospheric pressure on the other end. The term "siphon" is used because the same principle of atmospheric pressure acting on a fluid is applied. The difference in height of the fluid between the two arms of the U-shaped tube is the same as the maximum intermediate height of a siphon. When used to measure pressures other than atmospheric pressure, a siphon barometer is sometimes called a siphon gauge and not to be confused with a siphon rain gauge. Siphon pressure gauges are rarely used today. ### Siphon bottle Siphon bottles A siphon bottle (also called a soda syphon or, archaically, a siphoid[22]) is a pressurized bottle with a vent and a valve. Pressure within the bottle drives the liquid up and out a tube. It is a siphon in the sense that pressure drives the liquid through a tube. A special form was the gasogene. ### Siphon cup A siphon cup is the (hanging) reservoir of paint attached to a spray gun. This is to distinguish it from gravity-fed reservoirs. An archaic use of the term is a cup of oil in which the oil is siphoned out of the cup via a cotton wick or tube to a surface to be lubricated. ### Siphon rain gauge A siphon rain gauge is a rain gauge that can record rainfall over an extended period. A siphon is used to automatically empty the gauge. It is often simply called a "siphon gauge" and is not to be confused with a siphon pressure gauge. ### Heron's siphon Heron's siphon is a siphon that works on positive air pressure and at first glance appears to be a perpetual motion machine. In a slightly differently configuration, it is also known as Heron's fountain.[23] ### Venturi Siphon A venturi siphon, also known as an eductor, is essentially a venturi which is designed to greatly speed up the fluid flowing in a pipe such that an inlet port located at the throat of the venturi can be used to siphon another fluid. See pressure head. The low pressure at the throat of the venturi is called a siphon when a second fluid is introduced, or an aspirator when the fluid is air. ### Siphonic roof drainage Siphonic roof drainage makes use of the siphoning principle to carry water horizontally from multiple roof drains to a single downpipe and to increase flow velocity.[24] Air baffles at the roof drain inlets reduce the injection of air which causes embolisms in siphons. One benefit to this drainage technique is the reduction in required pipe diameter to drain a given roof surface area, up to half the size. Another benefit is the elimination of pipe pitch or gradient required for conventional roof drainage piping. ### Siphon spillway A siphon spillway in a dam uses the siphon effect to increase the flow rate. A normal spillway flow is pressurized by the height of the reservoir above the spillway whereas a siphon flow rate is governed by the difference in height of the inlet and outlet. (See math section below.) ## Sample building code regulations regarding back siphonage From Ontario's building code:[25] 7.6.2.3.Back Siphonage 1. Every potable water system that supplies a fixture or tank that is not subject to pressures above atmospheric shall be protected against back-siphonage by a backflow preventer. 2. Where a potable water supply is connected to a boiler, tank, cooling jacket, lawn sprinkler system or other device where a non-potable fluid may be under pressure that is above atmospheric or the water outlet may be submerged in the non-potable fluid, the water supply shall be protected against backflow by a backflow preventer. 3. Where a hose bibb is installed outside a building, inside a garage, or where there is an identifiable risk of contamination, the potable water system shall be protected against backflow by a backflow preventer. ## Self-siphons The term self-siphon is used in a number of ways. Liquids that are composed of long polymers can "self-siphon"[26][27] and these liquids do not depend on atmospheric pressure. Self-siphoning polymer liquids work the same as the siphon-chain model where the lower part of the chain pulls the rest of the chain up and over the crest. This phenomenon is also called a tubeless siphon.[28] "Self-siphon" is also often used in sales literature by siphon manufacturers to describe portable siphons that contain a pump. With the pump, no external suction (e.g. from a person's mouth/lungs) is required to start the siphon and thus the product is described as a "self-siphon". If the upper reservoir is such that the liquid there can rise above the height of the siphon crest, the rising liquid in the reservoir can "self-prime" the siphon and the whole apparatus be described as a "self-siphon".[29] Once primed, such a siphon will continue to operate until the level of the upper reservoir falls below the intake of the siphon. Such self-priming siphons are useful in some rain gauges and dams. Capillary action can be used in self-priming siphons. In these, water soaks upwards (into a cotton-filled hose) and below the crest to begin the siphon gradually, and as weight is added to the down stream, this kind of siphon will speed up, but it will never be as fast as the same diameter of open hose. ## Siphons in nature ### Anatomy The term "siphon" is used for a number of structures in human and animal anatomy, either because flowing liquids are involved or because the structure is shaped like a siphon, but in which no actual siphon effect is occurring: see Siphon (biology). Biologists debate whether the siphon mechanism plays a role in blood circulation.[30] It is theorized that veins form a continuous loop with arteries such that blood flowing down veins help siphon blood up the arteries, especially in giraffes and snakes.[31] Some have concluded that the siphon mechanism aids blood circulation in giraffes.[32] Many others dispute this[33][34] and experiments show no siphon effects in human circulation.[35] Some cite negative pressure in the brain as supporting the role of the siphon effect in the brain.[36] ### Geology In speleology, a siphon is that part of a cave passage that lies under water and through which cavers have to dive to progress along it. ## Explanation using Bernoulli's equation Bernoulli's equation may be applied to a siphon to derive the flow rate and maximum height of the siphon. Example of a siphon with annotations to describe Bernoulli's equation Let the surface of the upper reservoir be the reference elevation. Let point A be the start point of siphon, immersed within the higher reservoir and at a depth −d below the surface of the upper reservoir. Let point B be the intermediate high point on the siphon tube at height +hB above the surface of the upper reservoir. Let point C be the drain point of the siphon at height −hC below the surface of the upper reservoir. Bernoulli's equation: ${v^2 \over 2}+gy+{P \over \rho}=\mathrm{constant}$ $v \;$ = fluid velocity along the streamline $g \;$ = gravitational acceleration downwards $y \;$ = elevation in gravity field $P \;$ = pressure along the streamline $\rho \;$ = fluid density Apply Bernoulli's equation to the surface of the upper reservoir. The surface is technically falling as the upper reservoir is being drained. However, for this example we will assume the reservoir to be infinite and the velocity of the surface may be set to zero. Furthermore, the pressure at both the surface and the exit point C is atmospheric pressure. Thus: ${0^2 \over 2}+g(0)+{P_\mathrm{atm} \over \rho}=\mathrm{constant}$ (Equation 1.) Apply Bernoulli's equation to point A at the start of the siphon tube in the upper reservoir where P = PA, v = vA and y = −d ${v_A^2 \over 2}-gd+{P_A \over \rho}=\mathrm{constant}$ (Equation 2.) Apply Bernoulli's equation to point B at the intermediate high point of the siphon tube where P = PB, v = vB and y = hB ${v_B^2 \over 2}+gh_B+{P_B \over \rho}=\mathrm{constant}$ (Equation 3.) Apply Bernoulli's equation to point C where the siphon empties. Where v = vC and y = −hC. Furthermore, the pressure at the exit point is atmospheric pressure. Thus: ${v_C^2 \over 2}-gh_C+{P_\mathrm{atm} \over \rho}=\mathrm{constant}$ (Equation 4.) ### Velocity As the siphon is a single system, the constant in all four equations is the same. Setting equations 1 and 4 equal to each other gives: ${0^2 \over 2}+g(0)+{P_\mathrm{atm} \over \rho}={v_C^2 \over 2}-gh_C+{P_\mathrm{atm} \over \rho}$ Solving for vC: Velocity of siphon: $v_C=\sqrt{2gh_C}$ The velocity of the siphon is thus driven solely by the height difference between the surface of the upper reservoir and the drain point. The height of the intermediate high point, hB, does not affect the velocity of the siphon. However, as the siphon is a single system, vB = vC and the intermediate high point does limit the maximum velocity. The drain point cannot be lowered indefinitely to increase the velocity. Equation 3 will limit the velocity to a positive pressure at the intermediate high point to prevent cavitation. The maximum velocity may be calculated by combining equations 1 and 3: ${0^2 \over 2}+g(0)+{P_\mathrm{atm} \over \rho}={v_B^2 \over 2}+gh_B+{P_B \over \rho}$ Setting PB = 0 and solving for vmax: Maximum velocity of siphon: $v_\mathrm{max}=\sqrt{2\left({P_\mathrm{atm} \over \rho}-gh_B\right)}$ The depth, −d, of the initial entry point of the siphon in the upper reservoir, does not affect the velocity of the siphon. No limit to the depth of the siphon start point is implied by Equation 2 as pressure PA increases with depth d. Both these facts imply the operator of the siphon may bottom skim or top skim the upper reservoir without impacting the siphon's performance. Note that this equation for the velocity is the same as that of any object falling height hC. Note also that this equation assumes PC is atmospheric pressure. If the end of the siphon is below the surface, the height to the end of the siphon cannot be used; rather the height difference between the reservoirs should be used. ### Maximum height Setting equations 1 and 3 equal to each other gives: ${0^2 \over 2}+g(0)+{P_\mathrm{atm} \over \rho}={v_B^2 \over 2}+gh_B+{P_B \over \rho}$ Maximum height of the intermediate high point occurs when it is so high that the pressure at the intermediate high point is zero; in typical scenarios this will cause the liquid to form bubbles and if the bubbles enlarge to fill the pipe then the siphon will 'break'. Setting PB = 0: ${P_\mathrm{atm} \over {\rho}}={v_B^2 \over 2}+gh_B$ Solving for hB: General height of siphon: $h_B={P_\mathrm{atm} \over \rho g} - {v_B^2 \over 2g}.$ This means that the height of the intermediate high point is limited by velocity of the siphon. Faster siphons result in lower heights. Height is maximized when the siphon is very slow and vB = 0: Maximum height of siphon: $h_{B\mathrm{,max}}={P_\mathrm{atm} \over \rho g}$ This is the maximum height that a siphon will work. It is simply when the weight of the column of liquid to the intermediate high point equates to atmospheric pressure. Substituting values will give approximately 10 metres for water and 0.76 metres for mercury. ## Vacuum siphons Experiments have shown that siphons can operate in a vacuum, provided that the liquids are pure and degassed and surfaces are very clean.[1][2][3][4][5][37][38] ## Oxford English Dictionary The Oxford English Dictionary (OED) entry on siphon, published in 1911, states that a siphon works by atmospheric pressure. Stephen Hughes of Queensland University of Technology criticised this in a 2010 article[15] which was widely reported in the media.[39][40][41][42] The OED editors stated, "there is continuing debate among scientists as to which view is correct. ... We would expect to reflect this debate in the fully updated entry for siphon, due to be published later this year."[43] Dr. Hughes continued to defend his view of the siphon in a late September post at the Oxford blog.[44] A set of experiments was recently published, seriously questioning Hughes's hypothesis. [10] ## References 1. ^ a b 2. ^ a b Minor, Ralph Smith (1914). "Would a Siphon Flow in a Vacuum! Experimental Answers". School Science and Mathematics. 14,2: 152. 3. ^ a b Duane1902 4. ^ a b c d e http://www.mindspring.com/~rwramette/nokes.pdf Nokes M C 1948 "Vacuum siphons" School Science Review 29 233 5. ^ a b Siphon Concepts 6. ^ Usher, Abbott Payson (April 1, 1988). A History of Mechanical Inventions (Revised Edition). Dover Publications. p. 461. ISBN 978-0486255934. 7. ^ Dora P. Crouch (1993). "Water management in ancient Greek cities". Oxford University Press US. p. 119. ISBN 0-19-507280-4 8. ^ "THE PNEUMATICS OF HERO OF ALEXANDRIA". www.history.rochester.edu. Retrieved 2010-05-11. 9. ^ Banu Musa (authors), Donald Routledge Hill (translator) (1979). The book of ingenious devices (Kitāb al-ḥiyal). Springer. p. 21. ISBN 9027708339. 10. ^ a b c d e http://hawaii.edu/news/article.php?aId=4128 11. ^ a b c d e "The Great Siphon Definition Debate". Retrieved 2010-05-31. [unreliable source?] 12. ^ a b 13. ^ Smith, Andrew M. (1991). "Negative Pressure Generated by Octopus Suckers: A Study of the Tensile Strength of Water in Nature". Journal of Experimental Biology 157 (1): 257–271. 14. ^ Streeter, Victor L., Fluid Mechanics, Section 10.2 (4th edition), McGraw-Hill, New York, USA. Library of Congress Catalog Card No. 66-15605 15. ^ a b Hughes, Stephen W (2010). "A practical example of a siphon at work". Physics Education 45 (2): 162–166. Bibcode 2010PhyEd..45..162H. doi:10.1088/0031-9120/45/2/006. 16. ^ 17. ^ "Siphons for Geosiphon Treatment Systems". sti.srs.gov. Retrieved 2010-05-11. 18. ^ "Toiletology ... Anti-siphon needs an explanation". www.toiletology.com. Retrieved 2010-05-11. 19. ^ Tokoro, Kazuhiko; Chiba, Yasuhiro; Abe, Hiroyuki; Tanaka, Nobumasa; Yamataki, Akira; Kanno, Hiroshi (1994). "Importance of anti-siphon devices in the treatment of pediatric hydrocephalus". Child's Nervous System 10 (4): 236–8. doi:10.1007/BF00301160. 20. ^ "Hydrocephalus and Shunts in the Person with Spina Bifida" (Press release). Spina Bifida Association of America. 2009. Retrieved November 9, 2010. 21. ^ Zemack, Göran; Romner, Bertil (1999). "Seven-year clinical experience with the Codman Hakim programmable valve: a retrospective study of 583 patients". Neurosurgical FOCUS 7 (4): 941–8. doi:10.3171/foc.1999.7.4.11. 22. ^ http://voronoi.ics.uci.edu/cgi-bin/Dict?Form=Dict2&Database=*&Query=siphoid 23. ^ Kezerashvili; Sapozhnikov (2003). "Magic Fountain". arXiv:physics/0310039v1 [physics.ed-ph]. 24. ^ Arthur, S; Wright, G (2007). "Siphonic roof drainage systems—priming focused design". Building and Environment 42 (6): 2421–31. doi:10.1016/j.buildenv.2006.08.021. 25. ^ [1] 26. ^ "Physics Demonstrations - Light". sprott.physics.wisc.edu. Retrieved 2010-05-11. 27. ^ School of Chemistry. Chem.soton.ac.uk. Retrieved on 2010-11-11. 28. ^ Tubeless Siphon and Die Swell Demonstration, Christopher W. MacMinn & Gareth H. McKinley, September 26, 2004 29. ^ Siphon. Grow.arizona.edu. Retrieved on 2010-11-11. 30. ^ [2] 31. ^ Seymour, RS; Arndt, JO (2004). "Independent effects of heart-head distance and caudal blood pooling on blood pressure regulation in aquatic and terrestrial snakes". The Journal of experimental biology 207 (Pt 8): 1305–11. doi:10.1242/jeb.00882. PMID 15010481. 32. ^ Hicks, JW; Badeer, HS (1989). "Siphon mechanism in collapsible tubes: application to circulation of the giraffe head". The American journal of physiology 256 (2 Pt 2): R567–71. PMID 2916707. 33. ^ Seymour, RS; Johansen, K (1987). "Blood flow uphill and downhill: does a siphon facilitate circulation above the heart?". Comparative biochemistry and physiology. A, Comparative physiology 88 (2): 167–70. doi:10.1016/0300-9629(87)90465-8. PMID 2890463. 34. ^ Seymour, RS; Hargens, AR; Pedley, TJ (1993). "The heart works against gravity". The American journal of physiology 265 (4 Pt 2): R715–20. PMID 8238437. 35. ^ Dawson, E. A. (2004). "Standing up to the challenge of standing: a siphon does not support cerebral blood flow in humans". AJP: Regulatory, Integrative and Comparative Physiology 287 (4): R911–4. doi:10.1152/ajpregu.00196.2004. 36. ^ Hicks, J. W. (2005). "The siphon controversy counterpoint: the brain need not be "baffling"". AJP: Regulatory, Integrative and Comparative Physiology 289 (2): R629–32. doi:10.1152/ajpregu.00810.2004. 37. ^ Ganci, S; Yegorenkov, V (2008). "Historical and pedagogical aspects of a humble instrument". European Journal of Physics 29 (3): 421–430. Bibcode 2008EJPh...29..421G. doi:10.1088/0143-0807/29/3/003. 38. ^ Nokes M. C. (1948). "Vacuum siphons". Am. J. Phys. 16: 254. 39. ^ QUT physicist corrects Oxford English Dictionary 40. ^ AOL News, For 99 Years, Oxford English Dictionary Got It Wrong 41. ^ Calligeros, Marissa, Dictionary mistake goes unnoticed for 99 years, Brisbane Times, May 10, 2010 42. ^ Malkin, Bonnie, Physicist spots 99-year-old mistake in Oxford English Dictionary, The Daily Telegraph (London), 11 May 2010 43. ^ "On The Definition of “Siphon”". OUPblog. Oxford University Press. 21 May 2010. Retrieved 23 May 2010. 44. ^ http://blog.oup.com/2010/05/siphon/comment-page-1/#comment-177572 Wikimedia Foundation. 2010. ### Look at other dictionaries: • siphon — [ sifɔ̃ ] n. m. • 1639; « tuyau pour tirer du vin » 1546; sifon « trombe » v. 1320; lat. sipho, gr. siphôn 1 ♦ Tube courbé utilisé pour transférer un liquide d un niveau donné à un niveau inférieur, en passant par un niveau supérieur aux deux… … Encyclopédie Universelle • Siphon — Si phon, n. [F. siphon, L. sipho, onis, fr. Gr. ??? a siphon, tube, pipe.] 1. A device, consisting of a pipe or tube bent so as to form two branches or legs of unequal length, by which a liquid can be transferred to a lower level, as from one… … The Collaborative International Dictionary of English • Siphon — [ˈziːfɔ̃, ziˈfɔ̃ː, ziˈfoːn] (von griech. σίφων [ˈsipʰɔːn] „Heber“) bedeutet: Siphon (Geruchsverschluss), ein U förmiger Gas oder Geruchsverschluss, hauptsächlich bei Kanalisationanschlüssen Herunter und wieder herauf geführte Rohrleitung, um eine … Deutsch Wikipedia • siphon — si‧phon [ˈsaɪfn] also syphon verb [transitive] to dishonestly take money from a business, account etc and use it for a purpose for which it was not intended: siphon something from/off something • I later found she had siphoned thousands of… … Financial and business terms • siphon — SIPHON. s. m. Tuyau recourbé dont les deux branches sont inégales. Siphon de verre. siphon de fer blanc. on se sert du siphon pour les experiences sur la pesanteur des liqueurs. Siphon, En terme de Marine, & principalement sur la Mediterranée,… … Dictionnaire de l'Académie française • siphon — [sī′fən] n. [Fr < L sipho (gen. siphonis) < Gr siphōn, tube, siphon] 1. a bent tube used for carrying liquid from a reservoir over the top edge of its container to a point below the surface of the reservoir: the tube must be filled, as by… … English World dictionary • Siphon — (grch., »Röhre, Heber«), mit Ausflußhahn versehene Flasche kohlensauren Wassers, bei welcher der Kohlensäuredruck die Flüssigkeit in einer Steigröhre bis zum Hahn empordrückt; auch geschlossene Wasserleitung von oder – förmiger Gestalt, u … Kleines Konversations-Lexikon • Siphon — Sm Gerät zur Erzeugung kohlesäurehaltiger Getränke; Geruchsverschluß per. Wortschatz fach. (19. Jh.) Entlehnung. Entlehnt aus frz. siphon, dieses aus l. sīpho ( ōnis) Spritze, Röhre, Heber , aus gr. sī ̌phōn. Ebenso nndl. sifon, ne. siphon,… … Etymologisches Wörterbuch der deutschen sprache • siphon — (n.) 1650s, from Fr. siphon (early 17c.), from L. sipho (gen. siphonis), from Gk. siphon pipe, tube, of unknown origin. The verb is attested from 1859; figurative sense of to draw off, divert is recorded from 1940. Related: Siphoned; siphoning … Etymology dictionary • Siphon — Si phon, v. t. (Chem.) To convey, or draw off, by means of a siphon, as a liquid from one vessel to another at a lower level. [1913 Webster] … The Collaborative International Dictionary of English • Siphon — (griech.), Saugröhre, Heber; Ausflußhahn an Flaschen mit moussierenden Getränken, auch eine mit solchem Hahn versehene Flasche selbst (s. Heronsball und Mineralwässer, S. 870); einförmig gebogenes Abflußrohr, das einen hydraulischen Verschluß… … Meyers Großes Konversations-Lexikon	2019-11-20 16:56:31	{"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": 0, "img_math": 18, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5231034755706787, "perplexity": 3199.0401550833017}, "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/1573496670597.74/warc/CC-MAIN-20191120162215-20191120190215-00234.warc.gz"}
https://unmethours.com/question/25480/why-my-heating-load-is-too-high-than-cooling-load/	Question-and-Answer Resource for the Building Energy Modeling Community Get started with the Help page Ask Your Question # why my heating load is too high than cooling load house is two storied house with basement on the ground and slope roof on the top .I divided each room of house in zones so I have 10 thermal zones.I used HVACTemplate:Zone:IdealLoadsAirSystem objects. to calculate zone wise sensible cooling and heating load hourly for 8760 hours. All I want is annual hourly based heating and cooling loads,somebody suggested me to use HVACTemplate:Zone:IdealLoadsAirSystem objects. in the same group to calculate hourly heating and cooling load. I am getting heating load too high than cooling load,the weather file I am using is of Dayton Ohio USA. please somebody suggest me? thermal load.PNG edit retag close merge delete ## Comments There could be many reasons for this. You can search the forum for reasons why heating load might be high. You could also share your model so other people can investigate. For a house, look at your assumptions for heating temperature setpoint, infiltration, and wall/roof/window U-values. You may also be providing way more ventilation than is typical. ( 2017-07-13 20:43:57 -0600 )edit ## 2 Answers Sort by » oldest newest most voted Hi @ben Using "HVACTemplate:Zone:IdealLoadsAirSystem" would be a right way to assess the heating and cooling loads. I just looked at the weather file "USA_OH_Dayton.Intl.AP.724290_TMY3". The ratio of HDD to CDD appears quite different for the two baseline temperatures which might partly explain your issue. - 1794 annual (wthr file) cooling degree-days (10°C baseline) - 1532 annual (wthr file) heating degree-days (10°C baseline) - 538 annual (wthr file) cooling degree-days (18°C baseline) - 3197 annual (wthr file) heating degree-days (18°C baseline) I never worked on a project in this area but if have experience in this area and if you think the ratio of heating loads to cooling loads is too skewed, you should possibly find a way to fix that based on the thermostat set points and thermostat schedules. I suggest taking a closer look at your thermostat set points and schedules and see if they are correctly modeled. For similar issues, I typically look through the .Zsz file to look at the peak heating and cooling loads (last row in the .zsz file) for each thermal zone to identify if the issue is caused by any particular zone. Hope that helps! Raghu more an answer - why cant the heating load be larger than the cooling load.Its dependent on the people living inside and the location.I suggest you take it as it is.But its quite a large difference. more ## Your Answer Please start posting anonymously - your entry will be published after you log in or create a new account. Add Answer ## Stats Asked: 2017-07-13 19:57:08 -0600 Seen: 705 times Last updated: Jul 14 '17	2022-01-21 01:46:54	{"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": 0.17548617720603943, "perplexity": 4974.21999066133}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320302715.38/warc/CC-MAIN-20220121010736-20220121040736-00423.warc.gz"}
https://brilliant.org/problems/hello-2/	# Hello!!! Calculus Level pending This is going to be a mixture between my 2 previous problems: Find the value for $$z$$ with $$(x , y)$$ = $$(5 , 12)$$ in the paraboloid tangent to $$x^{2} + y^{2} + z^{2} = 25$$ in its intersection points with $$z = -3$$ ×	2018-03-20 14:00:48	{"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": 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": 0.3900933265686035, "perplexity": 707.7388123946082}, "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-2018-13/segments/1521257647475.66/warc/CC-MAIN-20180320130952-20180320150952-00065.warc.gz"}
https://dmtcs.episciences.org/5245	## Dragan, Feodor F. and Mohammed, Abdulhakeem - Slimness of graphs dmtcs:4288 - Discrete Mathematics & Theoretical Computer Science, March 4, 2019, Vol. 21 no. 3 Slimness of graphs Authors: Dragan, Feodor F. and Mohammed, Abdulhakeem Slimness of a graph measures the local deviation of its metric from a tree metric. In a graph $G=(V,E)$, a geodesic triangle $\bigtriangleup(x,y,z)$ with $x, y, z\in V$ is the union $P(x,y) \cup P(x,z) \cup P(y,z)$ of three shortest paths connecting these vertices. A geodesic triangle $\bigtriangleup(x,y,z)$ is called $\delta$-slim if for any vertex $u\in V$ on any side $P(x,y)$ the distance from $u$ to $P(x,z) \cup P(y,z)$ is at most $\delta$, i.e. each path is contained in the union of the $\delta$-neighborhoods of two others. A graph $G$ is called $\delta$-slim, if all geodesic triangles in $G$ are $\delta$-slim. The smallest value $\delta$ for which $G$ is $\delta$-slim is called the slimness of $G$. In this paper, using the layering partition technique, we obtain sharp bounds on slimness of such families of graphs as (1) graphs with cluster-diameter $\Delta(G)$ of a layering partition of $G$, (2) graphs with tree-length $\lambda$, (3) graphs with tree-breadth $\rho$, (4) $k$-chordal graphs, AT-free graphs and HHD-free graphs. Additionally, we show that the slimness of every 4-chordal graph is at most 2 and characterize those 4-chordal graphs for which the slimness of every of its induced subgraph is at most 1. Source : oai:arXiv.org:1705.09797 Volume: Vol. 21 no. 3 Section: Graph Theory Published on: March 4, 2019 Submitted on: February 15, 2018 Keywords: Computer Science - Discrete Mathematics,Mathematics - Combinatorics	2019-06-17 13:35:35	{"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": 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": 0.5384811758995056, "perplexity": 1140.9394011239317}, "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-2019-26/segments/1560627998475.92/warc/CC-MAIN-20190617123027-20190617145027-00365.warc.gz"}
https://www.lmfdb.org/L/32?Submit=magma&download=1&query=%7B'degree':%2032%7D	## Results (1-50 of at least 1000) Next Label $\alpha$ $A$ $d$ $N$ $\chi$ $\mu$ $\nu$ $w$ prim arith $\mathbb{Q}$ self-dual $\operatorname{Arg}(\epsilon)$ $r$ First zero Origin 32-309e16-1.1-c0e16-0-0 $0.392$ $1.02\times 10^{-13}$ $32$ $3^{16} \cdot 103^{16}$ 1.1 $$[0.0]^{16} 0 0 0 0.726416 Modular form 309.1.l.a 32-31e16-1.1-c1e16-0-0 0.497 1.98\times 10^{-10} 32 31^{16} 1.1$$ $[1.0]^{16}$ $1$ $0$ $0$ $1.25116$ Modular form 31.2.g.a 32-548e16-1.1-c0e16-0-1 $0.522$ $9.79\times 10^{-10}$ $32$ $2^{32} \cdot 137^{16}$ 1.1 $$[0.0]^{16} 0 0 0 0.792563 Modular form 548.1.k.a 32-548e16-1.1-c0e16-0-0 0.522 9.79\times 10^{-10} 32 2^{32} \cdot 137^{16} 1.1$$ $[0.0]^{16}$ $0$ $0$ $0$ $0.0319393$ Modular form 548.1.j.a 32-579e16-1.1-c0e16-0-0 $0.537$ $2.36\times 10^{-9}$ $32$ $3^{16} \cdot 193^{16}$ 1.1 $$[0.0]^{16} 0 0 0 0.298340 Modular form 579.1.s.a 32-693e16-1.1-c0e16-0-0 0.588 4.19\times 10^{-8} 32 3^{32} \cdot 7^{16} \cdot 11^{16} 1.1$$ $[0.0]^{16}$ $0$ $0$ $0$ $0.635455$ Modular form 693.1.bp.a 32-44e16-1.1-c1e16-0-0 $0.592$ $5.39\times 10^{-8}$ $32$ $2^{32} \cdot 11^{16}$ 1.1 $$[1.0]^{16} 1 0 0 1.30057 Modular form 44.2.g.a 32-45e16-1.1-c1e16-0-0 0.599 7.72\times 10^{-8} 32 3^{32} \cdot 5^{16} 1.1$$ $[1.0]^{16}$ $1$ $0$ $0$ $1.05222$ Modular form 45.2.l.a 32-723e16-1.1-c0e16-0-0 $0.600$ $8.25\times 10^{-8}$ $32$ $3^{16} \cdot 241^{16}$ 1.1 $$[0.0]^{16} 0 0 0 0.448728 Modular form 723.1.bc.a 32-765e16-1.1-c0e16-0-0 0.617 2.03\times 10^{-7} 32 3^{32} \cdot 5^{16} \cdot 17^{16} 1.1$$ $[0.0]^{16}$ $0$ $0$ $0$ $0.816155$ Modular form 765.1.bz.a 32-772e16-1.1-c0e16-0-0 $0.620$ $2.35\times 10^{-7}$ $32$ $2^{32} \cdot 193^{16}$ 1.1 $$[0.0]^{16} 0 0 0 0.555024 Modular form 772.1.u.a 32-776e16-1.1-c0e16-0-0 0.622 2.56\times 10^{-7} 32 2^{48} \cdot 97^{16} 1.1$$ $[0.0]^{16}$ $0$ $0$ $0$ $0.870998$ Modular form 776.1.bp.a 32-799e16-1.1-c0e16-0-0 $0.631$ $4.08\times 10^{-7}$ $32$ $17^{16} \cdot 47^{16}$ 1.1 $$[0.0]^{16} 0 0 0 0.338365 Modular form 799.1.h.b 32-820e16-1.1-c0e16-0-0 0.639 6.18\times 10^{-7} 32 2^{32} \cdot 5^{16} \cdot 41^{16} 1.1$$ $[0.0]^{16}$ $0$ $0$ $0$ $0.526304$ Modular form 820.1.by.a 32-820e16-1.1-c0e16-0-1 $0.639$ $6.18\times 10^{-7}$ $32$ $2^{32} \cdot 5^{16} \cdot 41^{16}$ 1.1 $$[0.0]^{16} 0 0 0 0.664075 Modular form 820.1.bz.a 32-52e16-1.1-c1e16-0-0 0.644 7.80\times 10^{-7} 32 2^{32} \cdot 13^{16} 1.1$$ $[1.0]^{16}$ $1$ $0$ $0$ $0.828728$ Modular form 52.2.l.b 32-873e16-1.1-c0e16-0-0 $0.660$ $1.68\times 10^{-6}$ $32$ $3^{32} \cdot 97^{16}$ 1.1 $$[0.0]^{16} 0 0 0 0.203547 Modular form 873.1.br.a 32-55e16-1.1-c1e16-0-0 0.662 1.91\times 10^{-6} 32 5^{16} \cdot 11^{16} 1.1$$ $[1.0]^{16}$ $1$ $0$ $0$ $1.36077$ Modular form 55.2.j.a 32-884e16-1.1-c0e16-0-0 $0.664$ $2.05\times 10^{-6}$ $32$ $2^{32} \cdot 13^{16} \cdot 17^{16}$ 1.1 $$[0.0]^{16} 0 0 0 0.0953233 Modular form 884.1.ct.a 32-884e16-1.1-c0e16-0-1 0.664 2.05\times 10^{-6} 32 2^{32} \cdot 13^{16} \cdot 17^{16} 1.1$$ $[0.0]^{16}$ $0$ $0$ $0$ $0.676581$ Modular form 884.1.cn.a 32-896e16-1.1-c0e16-0-0 $0.668$ $2.55\times 10^{-6}$ $32$ $2^{112} \cdot 7^{16}$ 1.1 $$[0.0]^{16} 0 0 0 0.437125 Modular form 896.1.bm.a 32-30e32-1.1-c0e16-0-0 0.670 2.74\times 10^{-6} 32 2^{32} \cdot 3^{32} \cdot 5^{32} 1.1$$ $[0.0]^{16}$ $0$ $0$ $0$ $0.800737$ Modular form 900.1.bh.a 32-921e16-1.1-c0e16-0-0 $0.677$ $3.96\times 10^{-6}$ $32$ $3^{16} \cdot 307^{16}$ 1.1 $$[0.0]^{16} 0 0 0 0.840905 Modular form 921.1.p.a 32-927e16-1.1-c0e16-0-0 0.680 4.40\times 10^{-6} 32 3^{32} \cdot 103^{16} 1.1$$ $[0.0]^{16}$ $0$ $0$ $0$ $0.593173$ Modular form 927.1.v.a 32-935e16-1.1-c0e16-0-0 $0.683$ $5.05\times 10^{-6}$ $32$ $5^{16} \cdot 11^{16} \cdot 17^{16}$ 1.1 $$[0.0]^{16} 0 0 0 0.214447 Modular form 935.1.y.a 32-943e16-1.1-c0e16-0-0 0.686 5.79\times 10^{-6} 32 23^{16} \cdot 41^{16} 1.1$$ $[0.0]^{16}$ $0$ $0$ $0$ $0.383543$ Modular form 943.1.n.b 32-959e16-1.1-c0e16-0-1 $0.691$ $7.57\times 10^{-6}$ $32$ $7^{16} \cdot 137^{16}$ 1.1 $$[0.0]^{16} 0 0 0 0.732272 Modular form 959.1.t.a 32-959e16-1.1-c0e16-0-0 0.691 7.57\times 10^{-6} 32 7^{16} \cdot 137^{16} 1.1$$ $[0.0]^{16}$ $0$ $0$ $0$ $0.659919$ Modular form 959.1.r.a 32-960e16-1.1-c0e16-0-0 $0.692$ $7.70\times 10^{-6}$ $32$ $2^{96} \cdot 3^{16} \cdot 5^{16}$ 1.1 $$[0.0]^{16} 0 0 0 0.468476 Modular form 960.1.cl.a 32-964e16-1.1-c0e16-0-0 0.693 8.23\times 10^{-6} 32 2^{32} \cdot 241^{16} 1.1$$ $[0.0]^{16}$ $0$ $0$ $0$ $0.426260$ Modular form 964.1.bh.a 32-975e16-1.1-c0e16-0-0 $0.697$ $9.87\times 10^{-6}$ $32$ $3^{16} \cdot 5^{32} \cdot 13^{16}$ 1.1 $$[0.0]^{16} 0 0 0 0.352611 Modular form 975.1.bi.a 32-975e16-1.1-c0e16-0-1 0.697 9.87\times 10^{-6} 32 3^{16} \cdot 5^{32} \cdot 13^{16} 1.1$$ $[0.0]^{16}$ $0$ $0$ $0$ $0.596732$ Modular form 975.1.cj.a 32-61e16-1.1-c1e16-0-0 $0.697$ $1.00\times 10^{-5}$ $32$ $61^{16}$ 1.1 $$[1.0]^{16} 1 0 0 1.59992 Modular form 61.2.g.a 32-980e16-1.1-c0e16-0-0 0.699 1.07\times 10^{-5} 32 2^{32} \cdot 5^{16} \cdot 7^{32} 1.1$$ $[0.0]^{16}$ $0$ $0$ $0$ $0.252241$ Modular form 980.1.y.a 32-984e16-1.1-c0e16-0-1 $0.700$ $1.14\times 10^{-5}$ $32$ $2^{48} \cdot 3^{16} \cdot 41^{16}$ 1.1 $$[0.0]^{16} 0 0 0 0.416497 Modular form 984.1.cj.a 32-984e16-1.1-c0e16-0-0 0.700 1.14\times 10^{-5} 32 2^{48} \cdot 3^{16} \cdot 41^{16} 1.1$$ $[0.0]^{16}$ $0$ $0$ $0$ $0.0548250$ Modular form 984.1.cj.b 32-1020e16-1.1-c0e16-0-1 $0.713$ $2.03\times 10^{-5}$ $32$ $2^{32} \cdot 3^{16} \cdot 5^{16} \cdot 17^{16}$ 1.1 $$[0.0]^{16} 0 0 0 0.637329 Modular form 1020.1.cl.b 32-1020e16-1.1-c0e16-0-0 0.713 2.03\times 10^{-5} 32 2^{32} \cdot 3^{16} \cdot 5^{16} \cdot 17^{16} 1.1$$ $[0.0]^{16}$ $0$ $0$ $0$ $0.216696$ Modular form 1020.1.cl.a 32-1028e16-1.1-c0e16-0-0 $0.716$ $2.30\times 10^{-5}$ $32$ $2^{32} \cdot 257^{16}$ 1.1 $$[0.0]^{16} 0 0 0 0.125974 Modular form 1028.1.l.a 32-1045e16-1.1-c0e16-0-0 0.722 2.99\times 10^{-5} 32 5^{16} \cdot 11^{16} \cdot 19^{16} 1.1$$ $[0.0]^{16}$ $0$ $0$ $0$ $0.812487$ Modular form 1045.1.w.f 32-1067e16-1.1-c0e16-0-0 $0.729$ $4.17\times 10^{-5}$ $32$ $11^{16} \cdot 97^{16}$ 1.1 $$[0.0]^{16} 0 0 0 0.487225 Modular form 1067.1.bg.a 32-33e32-1.1-c0e16-0-0 0.737 5.79\times 10^{-5} 32 3^{32} \cdot 11^{32} 1.1$$ $[0.0]^{16}$ $0$ $0$ $0$ $0.474083$ Modular form 1089.1.s.b 32-1096e16-1.1-c0e16-0-1 $0.739$ $6.41\times 10^{-5}$ $32$ $2^{48} \cdot 137^{16}$ 1.1 $$[0.0]^{16} 0 0 0 0.591173 Modular form 1096.1.u.a 32-1096e16-1.1-c0e16-0-0 0.739 6.41\times 10^{-5} 32 2^{48} \cdot 137^{16} 1.1$$ $[0.0]^{16}$ $0$ $0$ $0$ $0.268622$ Modular form 1096.1.s.a 32-70e16-1.1-c1e16-0-0 $0.747$ $9.07\times 10^{-5}$ $32$ $2^{16} \cdot 5^{16} \cdot 7^{16}$ 1.1 $$[1.0]^{16} 1 0 0 0.603066 Modular form 70.2.k.a 32-1133e16-1.1-c0e16-0-0 0.751 0.000109 32 11^{16} \cdot 103^{16} 1.1$$ $[0.0]^{16}$ $0$ $0$ $0$ $0.325359$ Modular form 1133.1.s.a 32-72e16-1.1-c1e16-0-0 $0.758$ $0.000142$ $32$ $2^{48} \cdot 3^{32}$ 1.1 $$[1.0]^{16} 1 0 0 0.735497 Modular form 72.2.l.b 32-72e16-1.1-c1e16-0-1 0.758 0.000142 32 2^{48} \cdot 3^{32} 1.1$$ $[1.0]^{16}$ $1$ $0$ $0$ $1.39370$ Modular form 72.2.n.b 32-34e32-1.1-c0e16-0-1 $0.759$ $0.000150$ $32$ $2^{32} \cdot 17^{32}$ 1.1 $$[0.0]^{16} 0 0 0 0.850730 Modular form 1156.1.m.a 32-34e32-1.1-c0e16-0-0 0.759 0.000150 32 2^{32} \cdot 17^{32} 1.1$$ $[0.0]^{16}$ $0$ $0$ $0$ $0.362388$ Modular form 1156.1.l.a	2021-06-16 17:53:43	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 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": 0.9639120697975159, "perplexity": 308.2236647456301}, "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-2021-25/segments/1623487625967.33/warc/CC-MAIN-20210616155529-20210616185529-00577.warc.gz"}
https://preprint.impa.br/visualizar?id=5662	Preprint C117/2010 Singular Levi-flat hypersurfaces - An approach through holomorphic foliations Arturo Ulises Fernández Pérez Keywords: Levi-flat hypersurfaces \| Holomorphic foliations. The aim of this Thesis is to investigate normal forms of germs of real analytic Levi-flat hypersurfaces with singularities in complex spaces. Anexos:	2023-02-01 13:03:56	{"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": 0.8085708618164062, "perplexity": 5142.809300993456}, "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/1674764499934.48/warc/CC-MAIN-20230201112816-20230201142816-00012.warc.gz"}
https://17calculus.com/conics/ellipses/	You CAN Ace Calculus ### 17Calculus Subjects Listed Alphabetically Single Variable Calculus Absolute Convergence Alternating Series Arc Length Area Under Curves Chain Rule Concavity Conics Conics in Polar Form Conditional Convergence Continuity & Discontinuities Convolution, Laplace Transforms Cosine/Sine Integration Critical Points Cylinder-Shell Method - Volume Integrals Definite Integrals Derivatives Differentials Direct Comparison Test Divergence (nth-Term) Test Ellipses (Rectangular Conics) Epsilon-Delta Limit Definition Exponential Derivatives Exponential Growth/Decay Finite Limits First Derivative First Derivative Test Formal Limit Definition Fourier Series Geometric Series Graphing Higher Order Derivatives Hyperbolas (Rectangular Conics) Hyperbolic Derivatives Implicit Differentiation Improper Integrals Indeterminate Forms Infinite Limits Infinite Series Infinite Series Table Infinite Series Study Techniques Infinite Series, Choosing a Test Infinite Series Exam Preparation Infinite Series Exam A Inflection Points Initial Value Problems, Laplace Transforms Integral Test Integrals Integration by Partial Fractions Integration By Parts Integration By Substitution Intermediate Value Theorem Interval of Convergence Inverse Function Derivatives Inverse Hyperbolic Derivatives Inverse Trig Derivatives Laplace Transforms L'Hôpital's Rule Limit Comparison Test Limits Linear Motion Logarithm Derivatives Logarithmic Differentiation Moments, Center of Mass Mean Value Theorem Normal Lines One-Sided Limits Optimization p-Series Parabolas (Rectangular Conics) Parabolas (Polar Conics) Parametric Equations Parametric Curves Parametric Surfaces Pinching Theorem Polar Coordinates Plane Regions, Describing Power Rule Power Series Product Rule Quotient Rule Radius of Convergence Ratio Test Related Rates Related Rates Areas Related Rates Distances Related Rates Volumes Remainder & Error Bounds Root Test Secant/Tangent Integration Second Derivative Second Derivative Test Shifting Theorems Sine/Cosine Integration Slope and Tangent Lines Square Wave Surface Area Tangent/Secant Integration Taylor/Maclaurin Series Telescoping Series Trig Derivatives Trig Integration Trig Limits Trig Substitution Unit Step Function Unit Impulse Function Volume Integrals Washer-Disc Method - Volume Integrals Work Multi-Variable Calculus Acceleration Vector Arc Length (Vector Functions) Arc Length Function Arc Length Parameter Conservative Vector Fields Cross Product Curl Curvature Cylindrical Coordinates Directional Derivatives Divergence (Vector Fields) Divergence Theorem Dot Product Double Integrals - Area & Volume Double Integrals - Polar Coordinates Double Integrals - Rectangular Gradients Green's Theorem Lagrange Multipliers Line Integrals Partial Derivatives Partial Integrals Path Integrals Potential Functions Principal Unit Normal Vector Spherical Coordinates Stokes' Theorem Surface Integrals Tangent Planes Triple Integrals - Cylindrical Triple Integrals - Rectangular Triple Integrals - Spherical Unit Tangent Vector Unit Vectors Vector Fields Vectors Vector Functions Vector Functions Equations Differential Equations Boundary Value Problems Bernoulli Equation Cauchy-Euler Equation Chebyshev's Equation Chemical Concentration Classify Differential Equations Differential Equations Euler's Method Exact Equations Existence and Uniqueness Exponential Growth/Decay First Order, Linear Fluids, Mixing Fourier Series Inhomogeneous ODE's Integrating Factors, Exact Integrating Factors, Linear Laplace Transforms, Solve Initial Value Problems Linear, First Order Linear, Second Order Linear Systems Partial Differential Equations Polynomial Coefficients Population Dynamics Projectile Motion Reduction of Order Resonance Second Order, Linear Separation of Variables Slope Fields Stability Substitution Undetermined Coefficients Variation of Parameters Vibration Wronskian ### Search Practice Problems Do you have a practice problem number but do not know on which page it is found? If so, enter the number below and click 'page' to go to the page on which it is found or click 'practice' to be taken to the practice problem. An ellipse is formed when a plane crosses that is not parallel to one of sides of the cone and not parallel to the axis of the cone. A circle is a special form of an ellipse where the plane is perpendicular to the axis of the cone. On this page, we discuss ellipses in rectangular coordinates. For a discussion of ellipses in polar form, see this separate page. Ellipse The standard equation for an ellipse is $$\displaystyle{ \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 }$$. Figure 2 contains more information than we need right now but it will suffice. The longer axis is called the major axis (in this plot it is horizontal). The shorter axis is called the minor axis. The vertices are located on the ellipse where it crosses the major axis. The foci are also on the major axis, labeled F1 and F2 on this plot. The major axis is determined by the denominators, $$a^2$$ and $$b^2$$. The larger value is in the denominator of the major axis, i.e. if $$a > b$$ then the major axis is parallel to the x-axis. We need to define a value c where $$c^2=\abs{a^2-b^2}$$ which will help us determine the location of the foci. These tables contain the main attributes of an ellipse. We assume here that $$a > b$$. Similar equations exist for $$a < b$$. $$\displaystyle{ \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 }$$ center $$(h,k)$$ major axis $$y=k$$ vertices $$(h \pm a, k),$$ $$(h, k \pm b)$$ foci $$(h \pm c,k)$$ $$c^2=\abs{a^2-b^2}$$ eccentricity $$e=c/a$$ $$\displaystyle{ \frac{(x-h)^2}{b^2} + \frac{(y-k)^2}{a^2} = 1 }$$ center $$(h,k)$$ major axis $$x=h$$ vertices $$(h, k \pm a),$$ $$(h \pm b, k)$$ foci $$(h,k \pm c)$$ $$c^2=\abs{a^2-b^2}$$ eccentricity $$e=c/a$$ Notes 1. Since the foci are closer to the center than the vertices, it follows that $$c < a$$ and therefore $$0 < e < 1$$. 2. Notice in the standard form of the equation, both terms are positive. This is how you know the graph is an ellipse and not a hyperbola. 3. In the general form of the equation, $$Ax^2+Bxy+Cy^2+$$ $$Dx+Ey+F=0$$, $$A > 0$$ and $$C > 0$$. 4. The eccentricity e is not the same as the irrational constant $$e \approx 2.72$$. Okay, time for some fun videos about ellipses. Here are a couple of videos about playing pool on an elliptical table. They clearly show the relationship between the foci and demonstrate some fun physics at the same time. ### Numberphile - Elliptical Pool Table (1) [3min-39secs] video by Numberphile ### Numberphile - Elliptical Pool Table (2) [5min-52secs] video by Numberphile ### Practice Write the standard form of the equation for an ellipse, centered at the origin, vertical major axis of length 8 and minor axis of length 2. Problem Statement Write the standard form of the equation for an ellipse, centered at the origin, vertical major axis of length 8 and minor axis of length 2. Solution ### 1587 solution video video by PatrickJMT Write the standard form of the equation for an ellipse, centered at the origin, with x-intercepts at $$\pm 12$$ and foci at $$(0,\pm 5)$$. Problem Statement Write the standard form of the equation for an ellipse, centered at the origin, with x-intercepts at $$\pm 12$$ and foci at $$(0,\pm 5)$$. Solution ### 1588 solution video video by PatrickJMT Write the standard form of the equation for an ellipse, centered at the origin, with minor axis of length 6 and foci at $$(\pm 8, 0)$$. Problem Statement Write the standard form of the equation for an ellipse, centered at the origin, with minor axis of length 6 and foci at $$(\pm 8, 0)$$. Solution ### 1589 solution video video by PatrickJMT Find the intercepts of the ellipse $$\displaystyle{ \frac{y^2}{100} + \frac{x^2}{121} = 1 }$$. Problem Statement Find the intercepts of the ellipse $$\displaystyle{ \frac{y^2}{100} + \frac{x^2}{121} = 1 }$$. Solution ### 1590 solution video video by PatrickJMT Graph the ellipse $$\displaystyle{ 1 - \frac{y^2}{16} = x^2 }$$. Problem Statement Graph the ellipse $$\displaystyle{ 1 - \frac{y^2}{16} = x^2 }$$. Solution ### 1591 solution video video by PatrickJMT Find the center and the radius of the circle $$x^2+2x+y^2=4$$. Problem Statement Find the center and the radius of the circle $$x^2+2x+y^2=4$$. Solution ### 1594 solution video video by Krista King Math Sketch the circle $$x^2 + y^2 = 4x$$. Problem Statement Sketch the circle $$x^2 + y^2 = 4x$$. Solution ### 1595 solution video video by Krista King Math Sketch the circle $$x^2 + y^2 + 6y = 0$$. Problem Statement Sketch the circle $$x^2 + y^2 + 6y = 0$$. Solution ### 1596 solution video video by Krista King Math Sketch the circle $$x^2 + y^2 + 2x + 2y = 2$$. Problem Statement Sketch the circle $$x^2 + y^2 + 2x + 2y = 2$$. Solution ### 1597 solution video video by Krista King Math Sketch the circle $$x^2 + y^2 + 10x - 20y + 100 = 0$$. Problem Statement Sketch the circle $$x^2 + y^2 + 10x - 20y + 100 = 0$$. Solution ### 1598 solution video video by Krista King Math Sketch the circle $$2x^2 + 2y^2 + 2x - 2y = 1$$. Problem Statement Sketch the circle $$2x^2 + 2y^2 + 2x - 2y = 1$$. Solution ### 1599 solution video video by Krista King Math Sketch the circle $$9x^2 + 9y^2 - 6x - 12y = 11$$. Problem Statement Sketch the circle $$9x^2 + 9y^2 - 6x - 12y = 11$$. Solution ### 1600 solution video video by Krista King Math Graph $$\displaystyle{ \frac{x^2}{9} + \frac{y^2}{5} = 1 }$$. Problem Statement Graph $$\displaystyle{ \frac{x^2}{9} + \frac{y^2}{5} = 1 }$$. Solution ### 1601 solution video video by PatrickJMT	2018-01-22 06:16:04	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 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": 0.9147965312004089, "perplexity": 1630.548242328999}, "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/1516084891105.83/warc/CC-MAIN-20180122054202-20180122074202-00030.warc.gz"}
https://stats.stackexchange.com/questions/351444/catch-more-information-in-neighbouring-the-1-and-1-labels/359198	I have labels I feed to a LSTM model. I noticed that there were to few 1s and -1s compared to the number of 0s. I have at least 99.9% of 0s and the rest are 1s and -1s. I considered using weighted classes where I give more weight on 1 and -1 labels and a lot less weight on the 0 labels. Is it a good practice to put kinda neighbourhood of 1s and -1s around each 1 and -1 in my dataset? For instance, suppose I have: ..., 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, -1, 0, 0, 0, 0, 0, 0, ... I would like to create a function that will transform that to: ..., 0, 0, 1, 1, 1, 1, 1, 0, 0, 0, 0, -1, -1, -1, -1, -1, 0, 0, 0, 0, ... when k=2. So that way we catch the information around the initial labels.	2019-08-26 04:52:04	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6342939734458923, "perplexity": 153.74070211948316}, "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/1566027330968.54/warc/CC-MAIN-20190826042816-20190826064816-00000.warc.gz"}
https://www.calculus-online.com/exercise/1262	# Derivative by Definition – A cotan function – Exercise 1262 Exercise Find by definition the derivative of the function $$f(x)=\cot x$$ $$f'(x)=-\frac{1}{\sin^2 x}$$	2020-06-03 18:30:12	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 2, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9656738638877869, "perplexity": 1790.6628465214435}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-24/segments/1590347435987.85/warc/CC-MAIN-20200603175139-20200603205139-00575.warc.gz"}
https://python-forum.io/Thread-Calculus-in-python	Bottom Page • 1 Vote(s) - 3 Average • 1 • 2 • 3 • 4 • 5 Calculus in python pythonforumrocks Programmer named Tim Posts: 7 Threads: 1 Joined: Mar 2017 Reputation: 0 Likes received: 1 #1 Mar-19-2017, 11:39 PM (This post was last modified: Mar-19-2017, 11:39 PM by pythonforumrocks.) I don't have a great understanding of calculus (or python) but I need to find the average value of a curve for a project. I've been using a widget from wolframalpha.com. Can I duplicate that in python? Here's the widget: http://www.wolframalpha.com/widgets/view...dcbe645198 ichabod801 Number Four Posts: 4,092 Threads: 90 Joined: Sep 2016 Reputation: 254 Likes received: 1222 #2 Mar-20-2017, 12:45 AM The best I could find with a quick Google for the symbolic solving of integrals was sympy (http://www.scipy-lectures.org/advanced/sympy.html). You could of course do a simulation of the limit (dividing the area under the curve into a bunch of small rectangles, and calculating the total area of the rectangles), but that may not suit your needs. Mekire likes this post Craig "Ichabod" O'Brien - xenomind.com I wish you happiness. Recommended Tutorials: BBCode, functions, classes, text adventures zivoni Minister of Silly Walks Posts: 331 Threads: 3 Joined: Feb 2017 Reputation: 65 Likes received: 192 #3 Mar-20-2017, 12:46 AM sympy has support for both indefinite and definite integrals: Output:In [1]: from sympy import integrate, Symbol, init_printing In [2]: init_printing(use_unicode=True) In [3]: x = Symbol('x') In [4]: integrate(x2, x) Out[4]: 3 x ── 3 In [5]: integrate(x2, (x, 1, 2)) Out[5]: 7/3 Or you can use numerical integration provided by SciPy. micseydel, metulburr, buran like this post pythonforumrocks Programmer named Tim Posts: 7 Threads: 1 Joined: Mar 2017 Reputation: 0 Likes received: 1 #4 Mar-20-2017, 12:15 PM I don't see how the range ("from" and "to" on the wolframalpha.com widget) should be specified. Here's my formula: f(x) = ( a * b ) / ( c * 1.05^x ) from = 1 to = b http://www.wolframalpha.com/widgets/view...dcbe645198 I will have values for a, b, and c. zivoni Minister of Silly Walks Posts: 331 Threads: 3 Joined: Feb 2017 Reputation: 65 Likes received: 192 #5 Mar-20-2017, 02:18 PM (This post was last modified: Mar-20-2017, 02:20 PM by zivoni. Edited 1 time in total.) You can simplify your function as k * 1.05^(-x), where k = a * b / c. ( mathjax support would be handy here) Lets suppose that k = 5 and you want to integrate over interval [1,5]. You can use scipy's quad() - its arguments are function's definition (as a python function) and lower and upper limit: Output:In [1]: from scipy import integrate In [2]: k, a, b = 5, 1, 5 In [3]: integrate.quad(lambda x: k * 1.05*(-x), a, b) Out[3]: (17.30418300357802, 1.9211502392899616e-13) output is the result and an estimation of error. You can use sympy's integrate(), where first argument is a function definition and second is a tuple, with elements being variable you integrate by, lower limit, upper limit. With sympy you can write your function almost "naturally", but you need to specify that x has a special meaning (a symbol). If you are interested only in numerical result, scipy's quad is probably easier to use. Output:In [4]: from sympy import integrate, Symbol In [5]: x = Symbol('x') In [6]: integrate(k 1.05*(-x), (x, a, b)) Out[6]: 17.3041830035780 And finally, you can use a piece of paper (and a brain) and realize that the primitive function of k 1.05^(-x) is -k / log(1.05) * 1.05^(-x). After that its just Output:In [7]: from math import log In [8]: -k / log(1.05) * ( 1.05-b - 1.05-a ) Out[8]: 17.304183003578018 As python is not a domain-specific language for math/symbol manipulation, using python for calculus would require atleast basic understanding of python and basic understanding of a calculus, you likely need to spend some effort and study both a little. pythonforumrocks Programmer named Tim Posts: 7 Threads: 1 Joined: Mar 2017 Reputation: 0 Likes received: 1 #6 Mar-20-2017, 03:49 PM I'm getting a different output from the wolframalpha.com widget compared to your functions above. With: k, a, b = 5, 1, 5 and: k = a * b / c I get 4.32605 from the widget by using: f(x) = ( a * b ) / ( c * 1.05^x ) from = 1 to = b Could you point out my mistake and/or misunderstanding? zivoni Minister of Silly Walks Posts: 331 Threads: 3 Joined: Feb 2017 Reputation: 65 Likes received: 192 #7 Mar-20-2017, 04:34 PM (This post was last modified: Mar-20-2017, 05:30 PM by zivoni. Edited 2 times in total. Edit Reason: missing - ) Hmm, I was little unclear with notation. After first line i just stopped to care about original values of a, b, c - they were used only to compute k and forgotten (with k = a * b / c function k * 1.05^(-x) is same as (ab)/c 1.05^(-x) ) And rest was just integration of function k * 1.05^(-x) on interval [a, b], whatever values of k, a, b are. And for integral of 5 * 1.05^(-x) on interval [1, 5] wolfram alpha gives same result (17.3042) - just fill there 5 * 1.05^(-x), from 1, to 5. pythonforumrocks Programmer named Tim Posts: 7 Threads: 1 Joined: Mar 2017 Reputation: 0 Likes received: 1 #8 Mar-20-2017, 06:25 PM I have to confess I'm a bit lost. What would you say is the simplest python expression of what I've been doing in the Wolfram Alpha widget? zivoni Minister of Silly Walks Posts: 331 Threads: 3 Joined: Feb 2017 Reputation: 65 Likes received: 192 #9 Mar-20-2017, 10:35 PM I think that something like integrate.quad(lambda x: k * 1.05*(-x), a, b) is probably the simplest way possible - to compute a definite integral you need to have a function to integrate, a lower bound, an upper bound. And you must express that you want to compute the definite integral from these three things. So I am not sure if it even can be significantly simpler ... Anyway, I can try to rewrite what I think you tried to compute - just "verbatim" rewrite without any changes from scipy import integrate a, b, c = 2, 5, 2 # constants def my_func(x): # "defining" your function f_x = a b / ( c * 1.05*x ) # value of f(x) return f_x print( integrate.quad(my_func, 1, b) ) # you want to integrate function my_func on interval [1, b] and print result And output of it is: Output:(17.30418300357802, 1.9211502392899616e-13) I hope that it helps. And if not, perhaps someone else would volunteer ... Larz60+ likes this post pythonforumrocks Programmer named Tim Posts: 7 Threads: 1 Joined: Mar 2017 Reputation: 0 Likes received: 1 #10 Mar-21-2017, 03:26 PM Well, my goal is to get the same output from a python expression that I get from Wolfram Alpha but I'm getting different output. With: a, b, c = 2, 5, 2 I get ~17.30 from this (a one-liner is better for my application): integrate.quad(lambda x: (ab/c) * 1.05*(-x), 1, b) I get ~4.32 from Wolfram Alpha by using these inputs on the page (with a, b, and c replaced with their values above): f(x) = ( a b ) / ( c * 1.05^x ) from = 1 to = b http://www.wolframalpha.com/widgets/view...dcbe645198 Your function seems to be giving me exactly 4x what Wolfram Alpha is giving me. What am I missing? BTW, I think your code should read the following with 1 as the lower bound instead of a: integrate.quad(lambda x: k * 1.05**(-x), 1, b) « Next Oldest \| Next Newest » Top Page Possibly Related Threads... Thread Author Replies Views Last Post Some help with some calculus and fish agritheory 4 608 Aug-17-2018, 07:40 PM Last Post: agritheory Forum Jump: Users browsing this thread: 1 Guest(s)	2019-12-12 06:42:33	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6722391843795776, "perplexity": 3717.162228075466}, "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-2019-51/segments/1575540537212.96/warc/CC-MAIN-20191212051311-20191212075311-00137.warc.gz"}
https://www.squirr3l.com/posts/dactyl-manuform-guide-part-1/	#### Dactyl Manuform Keyboard supplemental build guide: Part 1 - Supplies ##### July 3, 2018 Mechanical Keyboard Ergonomic Guide # Step 1: Read the original guide First, start by looking at tshort’s guide. He is the designer of this keyboard and his readme does technically cover everything needed to build the keyboard. That said, it does leave a bit up to interpretation and glosses over some bits. The goal of this supplement and future ones is to fill in those gaps to make this easier for anyone to build. https://github.com/tshort/dactyl-keyboard # Step 2: Buy some supplies tshort’s list is accurate, but most of the specific links were dead and there are supplies for multiple different styles of the build included. I’m just going to list the supplies for my prefered way to simplify matters. I added examples of all the needed supplies to a pinterest board you can view here. Buy from wherever though, these links are just representations. ## Components You can find all of this on amazon, but if you have time to wait, buying from somewhere like AliExpress will make this MUCH cheaper. #### Required components: • Wire • Solid core hook up wire, heavier gauge is easier (I used 22AWG). Space is not an issue like with regular dactyl (non manuform keyboard). • I tried with uninsulated magnet wire like in tshorts’ guide, but I didn’t like it because it was too thin and thus difficult to work with, and uninsulated, so positioning was very important to not accidentally short anything out. If you have the right wire strippers, the insulated wire doesn’t really take that much more work. • 1N4148 Diodes (qty:100) • These are very cheap, buy extras because you will lose some. • Arduino Pro Micro (qty:2) • Cheap clones work fine. They cost about $3 each on AliExpress if you have the time to wait. • Cherry MX key switches (qty:number of keys you are using) • Personal preference here, but for easy of assembly, stick with cherry mx switches as the holes and tabs in the model are designed for this. If you go with a different style switch, then the keys won’t mount securely without some modifications. It doesn’t matter if you get PCB mounted switches or not as they all just snap in the same way. • Key caps • DSA or SA Row 3 profile preferably • Need to be Cherry MX compatible • Otherwise, all up to your personal preference. • These sets are expensive and hard to find. You also can’t get away with a set the exact size you want (unless you use blanks) because most of the non letter keys on this keyboard are non standard size. I ended up ordering a$50 144 piece set from Aliexpress and it had keycaps that worked for all the keys except space. • RJ-9 female connectors (qty:2) • Not really easy to find these days. I did see builds that used headphone style (TRRS) jacks instead, but the printed model is made for RJ-9 telephone connectors, and they snap in securely and are nice when finished. Just to be clear, these are just old school telephone jack connectors. • RJ-9 cord (telephone cord). • Rubber feet, or plastidip to keep it from sliding around. • The 3D Printed shells. • Do it yourself if you have a printer, use a service like shapeways, or bribe a friend with a printer. • Keep in mind these take a signifigant amount of time to print for most printers (but not much material). On my Prusa i3 Mk3, it took around 20 hours per side at 200um and used a couple dollars worth of PLA. #### Optional components: • Button or switch to wire up to the reset pin. Makes flashing new keymappings much easier. • Heat set inserts and screws. Only required if you want to screw on a flat piece of acrylic or something as a bottom piece. I just left mine open for now. Will probably come back to this. • Arcylic sheet to cut to fit the bottom. Could also 3d print something, but I like the idea of clear acryllic so everyone can see the mess. • Female/Female jumper wires. Makes it easy to make connections to the arduino without soldering directly to it, and you can just cut them in half and strip the end without the connector to connect to the columns / rows. ## Tools #### Required Tools: • Soldering Iron • Normal size tip is good for most work • A fine tip is helpful for soldering to the Arduino, if you choose to. • Solder (Regular works better than lead-free), just don’t eat it. • Needlenose pliers • Wire strippers. Specifically the Irwin vise grip wire strippers if you don’t mind spending the money as they really make moving the insulation around easy. #### Nice-to-have Tools: • 3D Printer • Very fine grip tweezers or needlenose pliers • Round nose pliers for making loops in wire. These are mostly used for making jewelery. I’ll make another post soon that supplements the actual build instructions.	2018-12-10 20:53:27	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.19964714348316193, "perplexity": 4453.785303054275}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-51/segments/1544376823442.17/warc/CC-MAIN-20181210191406-20181210212906-00415.warc.gz"}