Datasets:

Infi-MM
/

InfiMM-WebMath-40B

Dataset card Data Studio Files Files and versions Community

Split (1)

train · ~22.8M rows (showing the first 514k)

Subsets and Splits

No community queries yet

The top public SQL queries from the community will appear here once available.

URL stringlengths 15 1.68k	text_list sequencelengths 1 199	image_list sequencelengths 1 199	metadata stringlengths 1.19k 3.08k
https://fr.slideshare.net/sonangrai/data-mining-approaches-and-methods-108692929	[ "# Data mining approaches and methods\n\n5 Aug 2018", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "1 sur 85\n\n### Data mining approaches and methods\n\n• 1. Data mining techniques and tasks Prepared by : Prabesh Pradhan\n• 2. The process of collecting, searching through, and analyzing a large amount of data in a database, as to discover patterns or relationships • Extraction of useful patterns from data sources, e.g., databases, data warehouses, web. • Patterns must be valid, novel, potentially useful, understandable. DATA MINING\n• 3. DATA MINING MODELS AND TASKS\n• 4. Predictive:- It makes prediction about values of data using known results from different data or based on historical data. Descriptive:- It identifies patterns or relationship in data, it serves as a way to explore properties of data.\n• 5. Data mining tasks The process of collecting, searching through, and analyzing a large amount of data in a database, as to discover patterns or relationships  Extraction of useful patterns from data sources,e.g. databases, data warehouses, web.  Patterns must be valid, novel, potentially useful, understandable.\n• 6. CLASSIFICATION  Classification derives a model to determine the class of an object based on its attributes.  Given a collection of records, each record contains a set of attributes, one of the attributes is the class.  For eg: pattern recognition\n• 7. • The value of attribute is examined as it varies over time. • A time series plot is used to visualize time series. • Ex:- stock exchange TIME SERIES ANALAYSIS\n• 8. • Clustering is the task of segmenting a diverse group into a number of similar subgroups or clusters. • Most similar data are grouped in clusters • Ex:-Bank customer CLUSTERING\n• 9. • Abstraction or generalization of data resulting in a smaller set which gives general overview of a data. • alternatively,summary type information can be derived from data. SUMMARIZATION\n• 10. Data mining techniques  Association  Classification  Clustering  Prediction  Sequential Patterns  Decision trees\n• 11. Association  The pattern is discovered based on a relationship between items in the same transaction  The association technique is also known as relation technique  The association technique is used in market basket analysis to identify a set of products that customers frequently purchase together.\n• 12. Classification  It classify each item in a set of data into one of a predefined set of classes or groups.  It makes use of mathematical techniques such as decision trees, linear programming, neural network, and statistics  We develop the software that can learn how to classify the data items into groups.  For example, we can apply classification in the application that “given all records of employees who left the company, predict who will probably leave the company in a future period.”\n• 13. Clustering  It makes a meaningful or useful cluster of objects which have similar characteristics using the automatic technique.  The clustering technique defines the classes and puts objects in each class, while in the classification techniques, objects are assigned into predefined classes.\n• 14. Prediction  The prediction analysis technique can be used in the sale to predict profit for the future if we consider the sale is an independent variable, profit could be a dependent variable.  It is based on the historical sale and profit data, we can draw a fitted regression curve that is used for profit prediction\n• 15. Sequential patterns  It seeks to discover or identify similar patterns, regular events or trends in transaction data over a business period.  Eg: In sales, with historical transaction data, businesses can identify a set of items that customers buy together different times in a year. Then businesses can use this information to recommend customers buy it with better deals based on their purchasing frequency in the past.\n• 16. Classification and prediction Prepared by : Sonahang Rai\n• 17. Decision Tree  It is a classification scheme which generates a tree and a set of rules from given data set.  A decision tree consist of root node, edges and leaf.  Each internal node denotes a test on an attribute.  Each branch node denotes outcome of the test.  Each leaf node holds the class label.\n• 18. Decision tree example\n• 19. ID3 algorithm i. Calculate the entropy of every attribute using the dataset. ii. Split the set into subsets using the attribute for which entropy is minimum(or information gain is maximum) iii. Make a decision tree node containing that attribute iv. Recurse on subset using remaining attributes\n• 20. Entropy  Info(D)=- 𝑖=1 𝑚 𝑝𝑖 𝑙𝑜𝑔2 (𝑝𝑖) where D=number of rows 𝑝𝑖 = class/D  𝐼𝑛𝑓𝑜 𝐴(D)= 𝑗=1 𝑣 \|𝐷 𝑗\| 𝐷 ∗ 𝐼𝑛𝑓𝑜(𝑑) Information gain  Gain(A)=Info(d)- 𝐼𝑛𝑓𝑜 𝐴(D)\n• 21. ID3 example\n• 22. Finding entropy and gain  Finding info(d) Info(D)= - 9 14 𝑙𝑜𝑔2( 9 14 )- 5 14 𝑙𝑜𝑔2( 5 14 )= 0.940 bits  Finding for age 𝐼𝑛𝑓𝑜 𝑎𝑔𝑒(D)= 5 14 * (- 2 5 𝑙𝑜𝑔2( 2 5 )- 3 5 𝑙𝑜𝑔2( 3 5 ))+ 4 14 * (- 4 4 𝑙𝑜𝑔2( 4 4 )- 0 4 𝑙𝑜𝑔2( 0 4 ))+ 5 14 * (- 3 5 𝑙𝑜𝑔2( 3 5 )- 2 5 𝑙𝑜𝑔2( 2 5 )) = 0.694 bits Thus, Gain(age)=Info(D)- 𝐼𝑛𝑓𝑜 𝑎𝑔𝑒(D)=0.945-0.694=0.246 bits Similarly we can find gain of all other attributes. Gain(income)=0.029,Gain(student)=0.151,Gain(credit_rating)=0.048 Among all the gains the gain of age is the maximum thus it is taken as the root node.\n• 23. Rule Based Classification  It uses the set of IF-THEN rules for the classification.  IF-THEN rule is an expression of the form IF condition THEN conclusion.  IF part is called rule antecedent or precondition, it can consist of one or more attributes test.  THEN part is called rule consequent, it consist a class prediction. Ex: R1:IF age=youth and student=yes THEN buys_computer=yes Or R1: (age=youth)Λ(student=yes)=>(buys_computer=yes)\n• 24.  A rule R can be assessed by its coverage and accuracy –Given a tuple X from a data D –Let n cover: # of tuples covered by R –n correct: # of tuples correctly classify by R –\|D\|: # of tuples in D Coverage(R)= 𝑛 𝑐𝑜𝑣𝑒𝑟𝑠 \|𝐷\| Accuracy(R)= 𝑛 𝑐𝑜𝑟𝑟𝑒𝑐𝑡 𝑛 𝑐𝑜𝑣𝑒𝑟𝑠\n• 25. R: IF age=youth AND student=yes THEN buys_computer=yes \|D\| = 14 N cover = 2 N correct = 2 Coverage(R)= 2 14 =14.25% Correct(R)= 2 2 =100%\n• 26. Rules extraction from the decision tree  One rule is created for each path from the root to a leaf node  Each splitting criterion is logically AND to form the rule antecedent (IF part)  Leaf node holds the class prediction for rule consequent (THEN part)  Logical OR is implied between each of the extracted rules\n• 28. Genetic algorithm  It is used for finding optimized solutions to search problems.  It is based on the theory of natural selection and evolution of biology. Selection: Survival of the fittest Evolution: Origin of species from a common descendent  It is excellent in searching large and complex data sets.\n• 29.  Gene: A part of chromosome.  Chromosome: A set of gene.  Population : No of individuals present with same length of chromosome.  Fitness: It is the value assigned to individual.  Fitness function: Function which assigns fitness value.  Selection: Selecting values for next generation.  Mutation: Changing a random gene.\n• 30. Algorithm 1. Generate random population of n chromosomes 2. Evaluate the fitness of each chromosome in the population 3. Select two parent chromosomes from a population according to their fitness 4. With a crossover probability cross over the parents to form a new offspring. 5. With a mutation probability mutate new offspring . 6. If the end condition is satisfied, stop, and return the best solution in current population\n• 31. Linear Regression  It is a data mining technique used to predict a range of numeric values (also called continuous values), given a particular dataset.  It is used to estimate a relationship between two variables.  Here involves a response/dependent variable y and a single predictor/independent variable x y = 𝑤𝑜+ 𝑤1 x where 𝑤𝑜 (y-intercept) and 𝑤1 (slope) are regression coefficients\n• 32. X years experience Y salary(in \\$ 1000s) 3 30 8 57 9 64 13 72 3 36 6 43 11 59 21 90 1 20 16 83\n• 33. Non-linear regression, Association and frequent pattern Prepared by : Biplap Bhattarai\n• 34. • Often the relationship between x and y cannot be approximated with a straight line. In this case, a nonlinear regression technique may be used. • Alternatively, the data could be preprocessed to make the relationship linear. • Most Non-linear models can be modeled by polynomial regression model can be transformed into linear regression model. • For Example: Y=w0+w1x+w2x2 +w3x3 +w4x4 Non- Linear Regression:\n• 35. • For Example: Y=w0+w1x+w2x2 +w3x3 +w4x4 Convertible to Linear with new Variables: X2= x2,x3=x3,x4=x4 Y=w0+w1x1+w2x2 +w3x3 +w4x4 Which is easily solved by method of least squares using software for regression analysis. Non- Linear Regression:\n• 36. Association Rules: Detects sets of attributes that frequently co-occur, and rules among them, e.g. 90% of the people who buy cookies, also buy milk (60% of all grocery shoppers buy both) Frequent Pattern: Frequent Pattern is a pattern (a set of items, subsequences) that occurs frequently in a dataset. For example, a set of items, such as milk and bread, that appear frequently together in a transaction data set, is a frequent itemset. A subsequence, such as buying first a PC, then a digital camera, and then a memory card, if it occurs frequently in a shopping history database, is a (frequent) sequential pattern. Data Mining Techniques\n• 37. Association Rule Mining: Association rule mining is a procedure which is meant to find frequent patterns, correlations, associations or casual structures from datasets found in various kind of databases (relational, transactional). Data Mining Techniques\n• 38. The Apriori Principle: If an itemset is frequent, then all of its subsets must also be frequent. Conversely, if an subset is infrequent, then all of its supersets must be infrequent, too. • Apriori uses a \"bottom up\" approach, where frequent subsets are extended one item at a time (a step known as candidate generation, and groups of candidates are tested against the data. • Apriori is designed to operate on database containing transactions (for example, collections of items bought by customers, or details of a website frequentation). DEFINITION OF APRIORI ALGORITHM\n• 39. • Apriori principle holds due to the following property of the support measure:∀X,Y:(X⊂Y)→s(X)≥s(Y) • For all x,y if x is the subset of y implies support of x is greater or equal to support of y. DEFINITION OF APRIORI ALGORITHM\n• 40. • Frequent Itemsets: All the sets which contain the item with the minimum support (denoted by Li For ith itemset) • Apriori Property: Any subset of frequent itemset must be frequent. • Join Operation: To find lk, a set of candidate k-itemsets is generated by joining Lk-1 with itself KEY CONCEPTS\n• 41. Given minimum required support s : 1. Search for all individual elements (1 element item set) that have a minimum support of s 2. Repeat 2.1From the result of the previous search for i-element item-sets, search for all i+1 element item-sets that have a minimum support of s 2.2 This becomes the sets of all frequent (i+1) element item-sets that are interesting 3. Until item-set size reaches maximum The Apriori Algorithm\n• 42. The Frequent itemsets are items in L1, L2, L3\n• 43. Pros of the Apriori algorithm • It is an easy-to-implement and easy-to-understand algorithm. • It can be used on large itemsets. Cons of the Apriori Algorithm • Sometimes, it may need to find a large number of candidate rules which can be computationally expensive. • Calculating support is also expensive because it has to go through the entire database. The Apriori Algorithm\n• 44. FP tree algorithm, which use to identify frequent patterns in the area of Data Mining. The Frequent Pattern Tree (FP-Tree) is a compact structure that stores quantitative information about frequent patterns in a database. Frequent Pattern Tree (FP Tree)\n• 45. 1. One root labeled as “null” with a set of item-prefix subtrees as children 2. Each node in the item-prefix subtree consists of three fields: i. Item-name: registers which item is represented by the node; ii. Count: the number of transactions represented by the portion of the path reaching the node; iii. Node-link: links to the next node in the FP-tree carrying the same item- name, or null if there is none. Frequent Pattern Tree (FP Tree)\n• 46. 1. Question :Find all frequent itemsets or frequent patterns in the following database using FP-growth algorithm. Take minimum support as 30%. Frequent Pattern Tree (FP Tree)\n• 47. Step 1 - Calculate Minimum support First should calculate the minimum support count. Question says minimum support should be 30%. It calculate as follows: Minimum support count(30/100 * 8) = 2.4 As a result, 2.4 appears but to empower the easy calculation it can be rounded to to the ceiling value. Now, Minimum support count is ceiling(30/100 * 8) = 3 Frequent Pattern Tree (FP Tree)\n• 48. Step 2 - Find frequency of occurrence Now time to find the frequency of occurrence of each item in the Table. For example, item A occurs in row 1,row 2,row 3,row 4 and row 7. Totally 5 times occurs in the Database table. You can see the counted frequency of occurrence of each item in Table below . Frequent Pattern Tree (FP Tree)\n• 49. Step 3 - Prioritize the items In Table 2 you can see the numbers written in Red pen. Those are the priority of each item according to it's frequency of occurrence. Item B got the highest priority (1) due to it's highest number of occurrences. At the same time you have opportunity to drop the items which not fulfill the minimum support requirement .For instance, if Table contain F which has frequency 1, then you can drop it. Frequent Pattern Tree (FP Tree)\n• 50. Step 4 -Order the items according to priority As you see in the Table below new column added to the Table before. In the Ordered Items column all the items are queued according to it's priority, which mentioned in the Red ink in Table. For example, in the case of ordering row 1, the highest priority item is B and after that D, A and E respectively. Frequent Pattern Tree (FP Tree)\n• 51. Row 1: Note that all FP trees have 'null' node as the root node. So draw the root node first and attach the items of the row 1 one by one respectively. And write their occurrences in front of it. Frequent Pattern Tree (FP Tree)\n• 52. Row 2: Then update the above tree by entering the items of row 2. Then without creating another branch you can go through the previous branch up to E and then you have to create new node after that for C. (When you going through the branch second time you should erase one and write two for indicating the two times you visit to that node. If you visit through three times then write three after erase two.) Figure 2 shows the FP tree after adding row 1 and row 2. Frequent Pattern Tree (FP Tree)\n• 53. Row 3: In row 3 you have to visit B,A,E and C respectively. So you may think you can follow the same branch again by replacing the values of B,A,E and C. But you can't do that you have opportunity to come through the B. But can't connect B to existing A overtaking D. As a result you should draw another A and connect it to B and then connect new E to that A and new C to new E. Frequent Pattern Tree (FP Tree)\n• 54. Row 4: Then row 4 contain B,D,A. Now we can just rename the frequency of occurrences in the existing branch. As B:4,D,A:3. Row 5: In fifth raw have only item D. Now we have opportunity draw new branch from 'null' node. Frequent Pattern Tree (FP Tree)\n• 55. Row 6:B and D appears in row 6. So just change the B:4 to B:5 and D:3 to D:4. Row 7: Attach two new nodes A and E to the D node which hanging on the null node. Then mark D,A,E as D:2,A:1 and E:1. Row 8 : Attach new node C to B. Change the traverse times.(B:6,C:1) Frequent Pattern Tree (FP Tree)\n• 56. Step 6 - Validation After the five steps the final FP tree as follows: How we know is this correct? Now count the frequency of occurrence of each item of the FP tree and compare it with Table. If both counts equal, then it is positive point to indicate your tree is correct. Frequent Pattern Tree (FP Tree)\n• 57. • FP Growth Stands for frequent pattern growth • It is a scalable technique for mining frequent pattern in a database • After constructing the FP-Tree it’s possible to mine it to find the complete set of frequent patterns • FP growth improves Apriority to a big extent • Frequent Item set Mining is possible without candidate generation • Only “two scan” to the database is needed FP-Growth\n• 58. Simply a two step procedure – Step 1: Build a compact data structure called the FP-tree Built using 2 passes over the data-set. – Step 2: Extracts frequent item sets directly from the FP- tree FP-Growth Algorithm Process\n• 59. FP-Growth Example: Example: With Min Support 3 FP-Tree\n• 60. FP-Growth Example: For T: We Start with drawing tree whose end nodes are Ts, Keeping only support of T\n• 61. FP-Growth Example: We Take out T one by one, and as we do so, we push its support to every node up the chain to the root that was part of the same transaction in which T was:\n• 62. FP-Growth Example: As, D is infrequent (min support=3), so we remove D Itemset={C,T},{W,T},{A,T},{C,W,T}, {C,A,T},{W,A,T},{T}\n• 63. FP-Growth Example: We Then, starting from main fp-tree, we consider each item that appears in the tree and create new FP-trees for C, W, A, and D.\n• 64. Market Basket Analysis • Market Basket Analysis is a modelling technique based upon the theory that if you buy a certain group of items, you are more (or less) likely to buy another group of items. • For example, if you are in an US, and purchase Diaper, Milk then you are likely to buy beer. • The set of items a customer buys is referred to as an itemset, and market basket analysis seeks to find relationships between purchases.\n• 65. Market Basket Analysis • Typically the relationship will be in the form of a rule: • IF {milk, diaper} THEN {beer}. • The probability that a customer will buy milk without a diaper or vice versa is referred to as the support for the rule. The conditional probability that a customer will purchase beer is referred to as the confidence.\n• 66. How is it used? • In retailing, most purchases are bought on impulse. Market basket analysis gives clues as to what a customer might have bought if the idea had occurred to them . • Market basket analysis can be used in deciding the location and promotion of goods inside a store.\n• 67. How is it used? • In retailing, most purchases are bought on impulse. Market basket analysis gives clues as to what a customer might have bought if the idea had occurred to them . • Market basket analysis can be used in deciding the location and promotion of goods inside a store.\n• 68. Types: • There are two main types of MBA: • Predictive MBA is used to classify cliques of item purchases, events and services that largely occur in sequence. • Differential MBA removes a high volume of insignificant results and can lead to very in-depth results. It compares information between different stores, demographics, seasons of the year, days of the week and other factors.\n• 69. Clustering Prepared by : Trilok Pratap Kafle\n• 70. Clustering Cluster :- Cluster is a group of objects that belongs to the same class. similar objects are grouped in one cluster and dissimilar objects are grouped in another cluster. Clustering is the process of making a group of abstract(something different) objects into classes of similar objects.\n• 71. Application of cluster analysis  Clustering can help marketers discover distinct groups in their customer base. And they can characterize their customer groups based on the purchasing patterns.  Clustering analysis is broadly used in many applications such as market research, pattern recognition, data analysis, and image processing.  Clustering also helps in classifying documents on the web for information discovery.\n• 72. Clustering methods  Partitioning method  Hierarchical method\n• 73. Partitioning method  Suppose we are given a database of ‘n’ objects and the partitioning method constructs ‘k’ partition of data. Each partition will represent a cluster and k ≤ n. It means that it will classify the data into k groups, which satisfy the following requirements −  Each group contains at least one object.  Each object must belong to exactly one group. Types of Partitioning method 1. K-mean 2. K-medoids\n• 74. K-mean  K-Means is one of the most popular \"clustering\" algorithms.  K-means stores K centroids that it uses to define clusters.  A point is considered to be in a particular cluster if it is closer to that cluster's centroid than any other centroid.\n• 75. Steps in K-mean 1. Partition objects into k nonempty subsets 2. Compute seed points as the centroids of the clusters of the current partition. Centroid is the centre (mean point) of the cluster. 3. Assign each object to the cluster with the nearest seed point. 4. Go back to Step 2, stop when no more new assignment.\n• 76. =>2, 3, 6, 8, 9, 12, 15, 18, 22 – break into 3 clusters – Cluster 1 - 2, 12, 18 – mean = 10.6 – Cluster 2 - 6, 9, 22 – mean = 12.3 – Cluster 3 – 3, 8, 15 – mean = 8.6 • Re-assign – Cluster 1 - mean = 0 – Cluster 2 – 12, 15, 18, 22 - mean = 16.75 – Cluster 3 – 2, 3, 6, 8, 9 – mean = 5.6 • Re-assign – Cluster 1 – 2 – mean = 2 – Cluster 2 – 12, 15, 18, 22 – mean = 16.75 – Cluster 3 = 3, 6, 8, 9 – mean = 6.5 Re-assign – Cluster 1 – 2, 3 – mean = 2.5 – Cluster 2 – 12, 15, 18, 22 – mean = 16.75 – Cluster 3 – 6, 8, 9 – mean = 7.6 • Re-assign – Cluster 1 – 2, 3 – mean = 2.5 – Cluster 2 – 12, 15, 18, 22 - mean = 16.75 – Cluster 3 – 6, 8, 9 – mean = 7.6 • No change, so we’re done\n• 77. K-Mediods  It is also a algorithm which breaks the data sets into group.  It also attempt to minimize the distance between points labeled to be the cluster of that cluster.  Data point is taken as center and work with a generalization of the Manhattan Norm.\n• 79. Hierarchical Method Hierarchical clustering, also known as hierarchical cluster analysis, is an algorithm that groups similar objects into groups called clusters. The endpoint is a set of clusters, where each cluster is distinct from each other cluster, and the objects within each cluster are broadly similar to each other. For example, all files and folders on the hard disk are organized in a hierarchy. There are two types of hierarchical clustering:- 1. Divisive 2. Agglomerative\n• 80. Divisive Method In divisive or top-down clustering method we assign all of the observations to a single cluster and then partition the cluster to two least similar clusters. Finally, we proceed recursively on each cluster until there is one cluster for each observation. There is evidence that divisive algorithms produce more accurate hierarchies than agglomerative algorithms in some circumstances but is conceptually more complex.\n• 82. Agglomerative Method In agglomerative or bottom-up clustering method we assign each observation to its own cluster. Then, compute the similarity (e.g., distance) between each of the clusters and join the two most similar clusters. Repeat the process until there is only a single cluster left. Before any clustering is performed, it is required to determine the proximity matrix containing the distance between each point using a distance function. Then, the matrix is updated to display the distance between each cluster. The following three methods differ in how the distance between each cluster is measured.\n• 83. Single Linkage In single linkage hierarchical clustering, the distance between two clusters is defined as the shortest distance between two points in each cluster. For example, the distance between clusters “r” and “s” to the left is equal to the length of the arrow between their two closest points. Complete Linkage In complete linkage hierarchical clustering, the distance between two clusters is defined as the longest distance between two points in each cluster. For example, the distance between clusters “r” and “s” to the left is equal to the length of the arrow between their two furthest points.\n• 84. Average Linkage In average linkage hierarchical clustering, the distance between two clusters is defined as the average distance between each point in one cluster to every point in the other cluster. For example, the distance between clusters “r” and “s” to the left is equal to the average length each arrow between connecting the points of one cluster to the other.\n\n### Notes de l'éditeur\n\n1. Correlations:- a mutual relationship or connection between two or more things. // interdependence of variable quantities. The causal relations between points are interpreted as describing which events in spacetime can influence which other events. A data set (or dataset) is a collection of data. Most commonly a data set corresponds to the contents of a single database table, or a single statistical data matrix, where every column of the table represents a particular variable, and each row corresponds to a given member of the data set in question. A transactional database is a database management system (DBMS) that has the capability to roll back or undo a database transaction or operation if it is not completed appropriately.\n2. Scalability is the capability of a system, network, or process to handle a growing amount of work, or its potential to be enlarged to accommodate that growth\n3. An impulse purchase or impulse buying is an unplanned decision to buy a product or service, made just before a purchase.\n4. An impulse purchase or impulse buying is an unplanned decision to buy a product or service, made just before a purchase." ]	[ null, "https://image.slidesharecdn.com/dataminingapproachesandmethods-180805142422/85/data-mining-approaches-and-methods-1-320.jpg", null, "https://image.slidesharecdn.com/dataminingapproachesandmethods-180805142422/85/data-mining-approaches-and-methods-2-320.jpg", null, "https://image.slidesharecdn.com/dataminingapproachesandmethods-180805142422/85/data-mining-approaches-and-methods-3-320.jpg", null, "https://image.slidesharecdn.com/dataminingapproachesandmethods-180805142422/85/data-mining-approaches-and-methods-4-320.jpg", null, "https://image.slidesharecdn.com/dataminingapproachesandmethods-180805142422/85/data-mining-approaches-and-methods-5-320.jpg", null, "https://image.slidesharecdn.com/dataminingapproachesandmethods-180805142422/85/data-mining-approaches-and-methods-6-320.jpg", null, "https://image.slidesharecdn.com/dataminingapproachesandmethods-180805142422/85/data-mining-approaches-and-methods-7-320.jpg", null, "https://image.slidesharecdn.com/dataminingapproachesandmethods-180805142422/85/data-mining-approaches-and-methods-8-320.jpg", null, "https://image.slidesharecdn.com/dataminingapproachesandmethods-180805142422/85/data-mining-approaches-and-methods-9-320.jpg", null, "https://image.slidesharecdn.com/dataminingapproachesandmethods-180805142422/85/data-mining-approaches-and-methods-10-320.jpg", null, "https://image.slidesharecdn.com/dataminingapproachesandmethods-180805142422/85/data-mining-approaches-and-methods-11-320.jpg", null, "https://image.slidesharecdn.com/dataminingapproachesandmethods-180805142422/85/data-mining-approaches-and-methods-12-320.jpg", null, "https://image.slidesharecdn.com/dataminingapproachesandmethods-180805142422/85/data-mining-approaches-and-methods-13-320.jpg", null, "https://image.slidesharecdn.com/dataminingapproachesandmethods-180805142422/85/data-mining-approaches-and-methods-14-320.jpg", null, "https://image.slidesharecdn.com/dataminingapproachesandmethods-180805142422/85/data-mining-approaches-and-methods-15-320.jpg", null, "https://image.slidesharecdn.com/dataminingapproachesandmethods-180805142422/85/data-mining-approaches-and-methods-16-320.jpg", null, "https://image.slidesharecdn.com/dataminingapproachesandmethods-180805142422/85/data-mining-approaches-and-methods-17-320.jpg", null, "https://image.slidesharecdn.com/dataminingapproachesandmethods-180805142422/85/data-mining-approaches-and-methods-18-320.jpg", null, "https://image.slidesharecdn.com/dataminingapproachesandmethods-180805142422/85/data-mining-approaches-and-methods-19-320.jpg", null, "https://image.slidesharecdn.com/dataminingapproachesandmethods-180805142422/85/data-mining-approaches-and-methods-20-320.jpg", null, "https://image.slidesharecdn.com/dataminingapproachesandmethods-180805142422/85/data-mining-approaches-and-methods-21-320.jpg", null, "https://image.slidesharecdn.com/dataminingapproachesandmethods-180805142422/85/data-mining-approaches-and-methods-22-320.jpg", null, "https://image.slidesharecdn.com/dataminingapproachesandmethods-180805142422/85/data-mining-approaches-and-methods-23-320.jpg", null, "https://image.slidesharecdn.com/dataminingapproachesandmethods-180805142422/85/data-mining-approaches-and-methods-24-320.jpg", null, "https://image.slidesharecdn.com/dataminingapproachesandmethods-180805142422/85/data-mining-approaches-and-methods-25-320.jpg", null, "https://image.slidesharecdn.com/dataminingapproachesandmethods-180805142422/85/data-mining-approaches-and-methods-26-320.jpg", null, "https://image.slidesharecdn.com/dataminingapproachesandmethods-180805142422/85/data-mining-approaches-and-methods-27-320.jpg", null, "https://image.slidesharecdn.com/dataminingapproachesandmethods-180805142422/85/data-mining-approaches-and-methods-28-320.jpg", null, "https://image.slidesharecdn.com/dataminingapproachesandmethods-180805142422/85/data-mining-approaches-and-methods-29-320.jpg", null, "https://image.slidesharecdn.com/dataminingapproachesandmethods-180805142422/85/data-mining-approaches-and-methods-30-320.jpg", null, "https://image.slidesharecdn.com/dataminingapproachesandmethods-180805142422/85/data-mining-approaches-and-methods-31-320.jpg", null, "https://image.slidesharecdn.com/dataminingapproachesandmethods-180805142422/85/data-mining-approaches-and-methods-32-320.jpg", null, "https://image.slidesharecdn.com/dataminingapproachesandmethods-180805142422/85/data-mining-approaches-and-methods-33-320.jpg", null, "https://image.slidesharecdn.com/dataminingapproachesandmethods-180805142422/85/data-mining-approaches-and-methods-34-320.jpg", null, "https://image.slidesharecdn.com/dataminingapproachesandmethods-180805142422/85/data-mining-approaches-and-methods-35-320.jpg", null, "https://image.slidesharecdn.com/dataminingapproachesandmethods-180805142422/85/data-mining-approaches-and-methods-36-320.jpg", null, "https://image.slidesharecdn.com/dataminingapproachesandmethods-180805142422/85/data-mining-approaches-and-methods-37-320.jpg", null, "https://image.slidesharecdn.com/dataminingapproachesandmethods-180805142422/85/data-mining-approaches-and-methods-38-320.jpg", null, "https://image.slidesharecdn.com/dataminingapproachesandmethods-180805142422/85/data-mining-approaches-and-methods-39-320.jpg", null, "https://image.slidesharecdn.com/dataminingapproachesandmethods-180805142422/85/data-mining-approaches-and-methods-40-320.jpg", null, "https://image.slidesharecdn.com/dataminingapproachesandmethods-180805142422/85/data-mining-approaches-and-methods-41-320.jpg", null, "https://image.slidesharecdn.com/dataminingapproachesandmethods-180805142422/85/data-mining-approaches-and-methods-42-320.jpg", null, "https://image.slidesharecdn.com/dataminingapproachesandmethods-180805142422/85/data-mining-approaches-and-methods-43-320.jpg", null, "https://image.slidesharecdn.com/dataminingapproachesandmethods-180805142422/85/data-mining-approaches-and-methods-44-320.jpg", null, "https://image.slidesharecdn.com/dataminingapproachesandmethods-180805142422/85/data-mining-approaches-and-methods-45-320.jpg", null, "https://image.slidesharecdn.com/dataminingapproachesandmethods-180805142422/85/data-mining-approaches-and-methods-46-320.jpg", null, "https://image.slidesharecdn.com/dataminingapproachesandmethods-180805142422/85/data-mining-approaches-and-methods-47-320.jpg", null, "https://image.slidesharecdn.com/dataminingapproachesandmethods-180805142422/85/data-mining-approaches-and-methods-48-320.jpg", null, "https://image.slidesharecdn.com/dataminingapproachesandmethods-180805142422/85/data-mining-approaches-and-methods-49-320.jpg", null, "https://image.slidesharecdn.com/dataminingapproachesandmethods-180805142422/85/data-mining-approaches-and-methods-50-320.jpg", null, "https://image.slidesharecdn.com/dataminingapproachesandmethods-180805142422/85/data-mining-approaches-and-methods-51-320.jpg", null, "https://image.slidesharecdn.com/dataminingapproachesandmethods-180805142422/85/data-mining-approaches-and-methods-52-320.jpg", null, "https://image.slidesharecdn.com/dataminingapproachesandmethods-180805142422/85/data-mining-approaches-and-methods-53-320.jpg", null, "https://image.slidesharecdn.com/dataminingapproachesandmethods-180805142422/85/data-mining-approaches-and-methods-54-320.jpg", null, "https://image.slidesharecdn.com/dataminingapproachesandmethods-180805142422/85/data-mining-approaches-and-methods-55-320.jpg", null, "https://image.slidesharecdn.com/dataminingapproachesandmethods-180805142422/85/data-mining-approaches-and-methods-56-320.jpg", null, "https://image.slidesharecdn.com/dataminingapproachesandmethods-180805142422/85/data-mining-approaches-and-methods-57-320.jpg", null, "https://image.slidesharecdn.com/dataminingapproachesandmethods-180805142422/85/data-mining-approaches-and-methods-58-320.jpg", null, "https://image.slidesharecdn.com/dataminingapproachesandmethods-180805142422/85/data-mining-approaches-and-methods-59-320.jpg", null, "https://image.slidesharecdn.com/dataminingapproachesandmethods-180805142422/85/data-mining-approaches-and-methods-60-320.jpg", null, "https://image.slidesharecdn.com/dataminingapproachesandmethods-180805142422/85/data-mining-approaches-and-methods-61-320.jpg", null, "https://image.slidesharecdn.com/dataminingapproachesandmethods-180805142422/85/data-mining-approaches-and-methods-62-320.jpg", null, "https://image.slidesharecdn.com/dataminingapproachesandmethods-180805142422/85/data-mining-approaches-and-methods-63-320.jpg", null, "https://image.slidesharecdn.com/dataminingapproachesandmethods-180805142422/85/data-mining-approaches-and-methods-64-320.jpg", null, "https://image.slidesharecdn.com/dataminingapproachesandmethods-180805142422/85/data-mining-approaches-and-methods-65-320.jpg", null, "https://image.slidesharecdn.com/dataminingapproachesandmethods-180805142422/85/data-mining-approaches-and-methods-66-320.jpg", null, "https://image.slidesharecdn.com/dataminingapproachesandmethods-180805142422/85/data-mining-approaches-and-methods-67-320.jpg", null, "https://image.slidesharecdn.com/dataminingapproachesandmethods-180805142422/85/data-mining-approaches-and-methods-68-320.jpg", null, "https://image.slidesharecdn.com/dataminingapproachesandmethods-180805142422/85/data-mining-approaches-and-methods-69-320.jpg", null, "https://image.slidesharecdn.com/dataminingapproachesandmethods-180805142422/85/data-mining-approaches-and-methods-70-320.jpg", null, "https://image.slidesharecdn.com/dataminingapproachesandmethods-180805142422/85/data-mining-approaches-and-methods-71-320.jpg", null, "https://image.slidesharecdn.com/dataminingapproachesandmethods-180805142422/85/data-mining-approaches-and-methods-72-320.jpg", null, "https://image.slidesharecdn.com/dataminingapproachesandmethods-180805142422/85/data-mining-approaches-and-methods-73-320.jpg", null, "https://image.slidesharecdn.com/dataminingapproachesandmethods-180805142422/85/data-mining-approaches-and-methods-74-320.jpg", null, "https://image.slidesharecdn.com/dataminingapproachesandmethods-180805142422/85/data-mining-approaches-and-methods-75-320.jpg", null, "https://image.slidesharecdn.com/dataminingapproachesandmethods-180805142422/85/data-mining-approaches-and-methods-76-320.jpg", null, "https://image.slidesharecdn.com/dataminingapproachesandmethods-180805142422/85/data-mining-approaches-and-methods-77-320.jpg", null, "https://image.slidesharecdn.com/dataminingapproachesandmethods-180805142422/85/data-mining-approaches-and-methods-78-320.jpg", null, "https://image.slidesharecdn.com/dataminingapproachesandmethods-180805142422/85/data-mining-approaches-and-methods-79-320.jpg", null, "https://image.slidesharecdn.com/dataminingapproachesandmethods-180805142422/85/data-mining-approaches-and-methods-80-320.jpg", null, "https://image.slidesharecdn.com/dataminingapproachesandmethods-180805142422/85/data-mining-approaches-and-methods-81-320.jpg", null, "https://image.slidesharecdn.com/dataminingapproachesandmethods-180805142422/85/data-mining-approaches-and-methods-82-320.jpg", null, "https://image.slidesharecdn.com/dataminingapproachesandmethods-180805142422/85/data-mining-approaches-and-methods-83-320.jpg", null, "https://image.slidesharecdn.com/dataminingapproachesandmethods-180805142422/85/data-mining-approaches-and-methods-84-320.jpg", null, "https://image.slidesharecdn.com/dataminingapproachesandmethods-180805142422/85/data-mining-approaches-and-methods-85-320.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.88383394,"math_prob":0.92157906,"size":24210,"snap":"2023-40-2023-50","text_gpt3_token_len":5989,"char_repetition_ratio":0.117987275,"word_repetition_ratio":0.07697965,"special_character_ratio":0.22895497,"punctuation_ratio":0.11557895,"nsfw_num_words":1,"has_unicode_error":false,"math_prob_llama3":0.972399,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50,51,52,53,54,55,56,57,58,59,60,61,62,63,64,65,66,67,68,69,70,71,72,73,74,75,76,77,78,79,80,81,82,83,84,85,86,87,88,89,90,91,92,93,94,95,96,97,98,99,100,101,102,103,104,105,106,107,108,109,110,111,112,113,114,115,116,117,118,119,120,121,122,123,124,125,126,127,128,129,130,131,132,133,134,135,136,137,138,139,140,141,142,143,144,145,146,147,148,149,150,151,152,153,154,155,156,157,158,159,160,161,162,163,164,165,166,167,168,169,170],"im_url_duplicate_count":[null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-25T18:06:33Z\",\"WARC-Record-ID\":\"<urn:uuid:f0cc53e1-a0fd-4e58-ad95-c03ac0253a73>\",\"Content-Length\":\"972276\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:473bcdb3-f578-4c84-9974-7c2e13759a39>\",\"WARC-Concurrent-To\":\"<urn:uuid:8a4d708c-d894-4427-9331-ed31c4a3d521>\",\"WARC-IP-Address\":\"146.75.34.152\",\"WARC-Target-URI\":\"https://fr.slideshare.net/sonangrai/data-mining-approaches-and-methods-108692929\",\"WARC-Payload-Digest\":\"sha1:NCFH6WA5J272RNOYH55DRWZ3CFVRI25D\",\"WARC-Block-Digest\":\"sha1:PKIKRFFE25ITATUXLLNTIUJ7ZODY6JSZ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233509023.57_warc_CC-MAIN-20230925151539-20230925181539-00681.warc.gz\"}"}
https://math.stackexchange.com/questions/1252144/does-every-cover-have-an-irredundant-subcover?noredirect=1	[ "# Does every cover have an irredundant subcover?\n\nWhile composing an answer for this question, I got troubled by a technical point. I wanted to assert the existence of an irredundant subcover of a given open cover, but realized I'm not sure how to guarantee that this exists. I got around this technicality in that particular situation by using other aspects of the setup, but now I want to understand the general situation. I'll phrase the question in the language of topology, but it is really purely a set-theoretic question.\n\nLet us call an open cover $\\{U_\\lambda\\}_{\\lambda\\in\\Lambda}$ irredundant if no $U_\\lambda$ can be dropped, i.e. if $\\{U_\\lambda\\}_{\\lambda\\in\\Lambda'}$ is never a cover if $\\Lambda'$ is a proper subset of $\\Lambda$; equivalently, if for each $\\lambda\\in\\Lambda$, $\\bigcap_{\\mu\\neq \\lambda} (U_\\mu)^c$ is nonempty.\n\nGiven a topological space $X$ with an arbitrary open cover $\\{U_\\lambda\\}_{\\lambda\\in\\Lambda}$, does there always exist an irredundant subcover? Why?\n\nClearly there is no problem if $\\Lambda$ is finite (or if $X$ is compact so that we can first pass to a finite subcover): just drop redundant $U_\\lambda$'s one at a time until the cover is irredundant.\n\nBut what if $\\Lambda$ has much bigger cardinality? At first I thought we ought to be able to construct an irredundant subcover using some appropriate form of the axiom of choice. After all, I thought, all we have to do is drop the redundant $U_\\lambda$'s. (I.e. those such that $\\bigcap_{\\mu\\neq \\lambda}(U_\\mu)^c$ is empty.) We can't drop them all at once, unfortunately, since we might not be left with a cover. But if we drop them one at a time, and reevaluate which are redundant after each drop, then, well, AC is designed for organizing an infinite sequence of choices of this kind.\n\nFor example we could try: by the well-ordering theorem, impose a well-order on $\\Lambda$. Then the set of $\\lambda$'s corresponding to redundant $U_\\lambda$'s has a least element, and this is what we drop first. Rinse and repeat.\n\nOr (this was my first idea), the setup seems custom made for Zorn's lemma: the subcovers of a given cover form a nonempty poset under reverse inclusion, and the irredundant covers are precisely the maximal elements of this poset.\n\nThe problem is that in the end, I don't actually think one can generally arrive at an irredundant cover, even when it exists, by dropping redundant $U_\\lambda$'s one at a time. The example I have in mind is, let $X$ be any nonempty space at all, let $\\Lambda = \\mathbb{N}$, and let $U_1,U_2,\\dots = X$. This example shows that both of the above approaches will fail. If we try to construct an irredundant subcover by repeatedly dropping the first redundant $U_n$, we get the sequence of subcovers\n\n$$\\{U_n\\}_{n\\geq 1}\\supset \\{U_n\\}_{n\\geq 2}\\supset \\dots$$\n\nand the procedure never lands on an irredundant one since the limit is empty, which isn't a cover. This very sequence also shows the hypotheses of Zorn's lemma aren't met (in the attempt above to use Zorn's lemma): here is an increasing chain in the poset of subcovers that does not have an upper bound.\n\nThis example seems to me to show that the AC-based approaches are barking up the wrong tree. On the other hand, evidently there is an irredundant cover here: any individual $U_n$ will do the job. So maybe there is still hope that the question's answer is yes?\n\nAs you’ve already discovered, the answer is no. In fact, you can’t even guarantee an irreducible open refinement. For $n\\in\\Bbb N$ let $U_n=\\{k\\in\\Bbb N:k<n\\}$, and let $\\tau=\\{U_n:n\\in\\Bbb N\\}\\cup\\{\\Bbb N\\}$; then $\\tau$ is a $T_0$ topology on $\\Bbb N$, and $\\tau\\setminus\\{\\Bbb N\\}$ is an open cover of $\\Bbb N$ with no irreducible open refinement.\n\nA well-known positive result is that every point-finite open cover of a space has an irreducible subcover. Thus, every open cover of a metacompact space has an irreducible open refinement. (A space $X$ is metacompact if every open cover of $X$ has a point-finite open refinement. A family of sets is point-finite if each point of the space lies in only finitely many of the sets.)\n\n• +1 \"Does every open cover have an irredundant open refinement?\" was going to be my next question! (Aside: I see the argument for the positive result because the Zorn's lemma approach goes through: point-finiteness shows that the poset of subcovers satisfies the ascending chain condition.) – Ben Blum-Smith Apr 26 '15 at 16:55\n• Actually I think I'm waving my hands my aside here. Let $X=\\mathbb{N}$ with the discrete topology, let $\\Lambda = \\mathbb{N}\\times \\{0,1\\}$, and let $U_{(n,i)} = \\{n\\}$ for both $i=0,1$. Let $\\mathcal{F}_k = \\{U_{n,1}\\}_{n\\geq k} \\cup \\{U_{n,0}\\}_{n\\geq 1}$. Then $\\mathcal{F}_1\\prec \\mathcal{F}_2\\prec\\dots$ is a nonterminating ascending chain in the poset of subcovers. – Ben Blum-Smith May 2 '15 at 15:49\n• But the Zorn's lemma approach still goes through! What point-finiteness does is to force the poset of subcovers to be closed under taking intersections of chains, thus ascending chains have upper bounds. If $\\mathcal{F}_1\\prec\\mathcal{F}_2\\prec\\dots$ is an ascending chain of subcovers (i.e. descending under inclusion), and the intersection is not a cover, there is an uncovered $x\\in X$. Then the set of $k$ for which some set of $\\mathcal{F}_K\\setminus\\mathcal{F}_{k+1}$ contains $x$ must be cofinal in $\\mathbb{N}$. But this implies infinitely many sets of the cover contain $x$. – Ben Blum-Smith May 2 '15 at 16:01\n• @Ben: Yes, that’s the argument that I gave here and essentially the argument in Engelking. – Brian M. Scott May 2 '15 at 18:39\n\nAs often happens when I write questions here, I realized something while writing. Here I actually realized the answer.\n\nIt is no.\n\nHere is an open cover of $\\mathbb{R}$ with no irredundant subcover:\n\n$\\Lambda=\\mathbb{N}$.\n\n$U_n = (-n,n)$.\n\nBecause $U_1\\subset U_2\\subset U_3\\subset\\dots$, given any pair of $U_n$'s, one is contained in the other. Therefore any subcover is redundant unless it consists of a single $U_n$. But no single $U_n$ forms a cover. Thus no subcover is irredundant.\n\nI'm not sure if it helps you with your original question, but let me try proposing an alternative concept capturing the intuitive notion of irredundancy: for each point, take the “smallest” open set in the cover containing that point.\n\nFormally, let $\\{U_{\\lambda}\\}_{\\lambda\\in\\Lambda}$ be an open cover of the topological space $X$, where $\\Lambda$ is a non-empty index set. Now endow $\\Lambda$ with a well-ordering $\\succsim$. For each point $x\\in X$, let $\\lambda_x$ be the least element of $\\{\\lambda\\in\\Lambda\\,\|\\,x\\in U_{\\lambda}\\}$ according to the well-ordering. Then, $\\{U_{\\lambda_x}\\}_{x\\in X}$ is a refined subcover of $\\{U_{\\lambda}\\}_{\\lambda\\in\\Lambda}$." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.924599,"math_prob":0.9955455,"size":3316,"snap":"2019-51-2020-05","text_gpt3_token_len":863,"char_repetition_ratio":0.13979469,"word_repetition_ratio":0.0073260074,"special_character_ratio":0.24487334,"punctuation_ratio":0.10044978,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99963427,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-24T08:38:09Z\",\"WARC-Record-ID\":\"<urn:uuid:143f88be-a2c7-4e06-8a41-46895e1d5c1c>\",\"Content-Length\":\"159612\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:29d4f98b-481d-4df7-8c69-0bf4a10ee707>\",\"WARC-Concurrent-To\":\"<urn:uuid:b1111605-298a-49bc-9490-9827df7e372b>\",\"WARC-IP-Address\":\"151.101.129.69\",\"WARC-Target-URI\":\"https://math.stackexchange.com/questions/1252144/does-every-cover-have-an-irredundant-subcover?noredirect=1\",\"WARC-Payload-Digest\":\"sha1:DXILURHP7KDSMWJZLPJFIVCAUJ5L5UBD\",\"WARC-Block-Digest\":\"sha1:GS7CBPWPSSBQIZZDUQS4DT4UF2JVHLSS\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579250616186.38_warc_CC-MAIN-20200124070934-20200124095934-00398.warc.gz\"}"}
https://mathoverflow.net/questions/149391/on-radicals-of-a-lie-algebra	[ "# On radicals of a lie algebra\n\nLet $\\mathbb{k}$ be a field, $\\mathfrak{g}$ be a finite-dimensional Lie algebra over $\\mathbb{k}$. In Bourbaki's \"Lie Groups and Lie Algebras\", Ch I, he defines four radical-like ideals of $\\mathfrak{g}$:\n\n1. the radical $\\mathfrak{r}$, i.e. the maximal solvable ideal;\n2. the radical of Killing form $\\mathfrak{k}$, i.e. $\\mathfrak{k}=\\mathfrak{g}^\\perp$;\n3. the maximal nilpotent ideal $\\mathfrak{n}$;\n4. the nilpotent radical $\\mathfrak{s}$, i.e. intersection of all kernels of irreducible finite-dimensional representations of $\\mathfrak{g}$.\n\nHe also shows that, $\\mathfrak{r}=[\\mathfrak{g},\\mathfrak{g}]^\\perp$, $\\mathfrak{s}=[\\mathfrak{g}, \\mathfrak{g}]\\cap\\mathfrak{r}=[\\mathfrak{g},\\mathfrak{r}]$, and the following inclusion relations: $$\\mathfrak{r} \\supset \\mathfrak{k} \\supset \\mathfrak{n} \\supset \\mathfrak{s}.$$\n\nOn the other hand, in Jacobson's book, the nilpotent radical is defined as $\\mathfrak{n}$.\n\nMy question is, does $\\mathfrak{s}$ coincide with $\\mathfrak{n}$? or, are Bourbaki's nilradical and Jacobson's nilradical equivalent?\n\nI guess the answer is NO (otherwise Bourbaki should have proved it), but I cannot find any example. Could any one give me an example? Thank you!\n\n• $\\mathfrak{s}$ vanishes iff $\\mathfrak{g}$ is reductive, but it's easy to find reductive Lie algebras for which $\\mathfrak{n}$ fails to vanish (e.g. abelian ones). – Qiaochu Yuan Nov 20 '13 at 5:28\n• @Si-Qi Yu: Those parts of Bourbaki have a running hypothesis that char$(k)=0$ (look at the start of various sections); e.g., for $\\mathfrak{g}=\\mathfrak{sl}_p$ in characteristic $p>0$ the Killing form vanishes and the subalgebra ${\\rm{Lie}}(\\mu_p)$ of diagonal scalars form a nonzero central ideal yet if $p>2$ then $[\\mathfrak{g},\\mathfrak{g}]=\\mathfrak{g}$, etc. – Marguax Nov 20 '13 at 14:52\n• @Qiaochu Yuan: What is your definition of \"reductive Lie algebra\" in positive characteristic? – Marguax Nov 20 '13 at 14:53\n• @Marguax: I don't have one (I assume you're hinting that the equivalence between various possible definitions fails). Is positive characteristic an important issue here? – Qiaochu Yuan Nov 20 '13 at 18:50\n• @Qiaochu Yuan: The assertions in Bourbaki totally break down in positive characteristic, so it is more serious than a failure of equivalences. I am not aware of a good definition of \"reductive Lie algebra\" outside characteristic 0, so it was unclear to me if you had a definition. In the setting of Lie algebras (unlike for algebraic groups) it is safest to explicitly assume char. 0 for such things unless one provides a clear definition or reference for the terminology being used. (Bourbaki demands char. 0 in their definition of \"reductive Lie algebra\"; look at the start of section 6 of Ch. I.) – Marguax Nov 20 '13 at 19:56\n\nIf we let $\\mathbb{k}$ be the field itself thought of as a Lie algebra with trivial bracket, then $\\mathfrak{n}=\\mathbb{k}$ since this is a nilpotent Lie algebra but $\\mathbb{s}=\\{0\\}$, since the obvious representation of $\\mathbb{k}$ be scalar multiplication is faithful. In general, these will never coincide for a nilpotent Lie algebra.\nThe two definitions do never agree for a non-trivial nilpotent Lie algebra, as Ben has remarked. Indeed, if $\\mathfrak{g}$ is nilpotent then $\\mathfrak{s}=[\\mathfrak{g},\\mathfrak{r}]=[\\mathfrak{g},\\mathfrak{g}]$, whereas $\\mathfrak{n}=\\mathfrak{g}$. Since $\\mathfrak{g}$ is nilpotent, $\\mathfrak{g}\\neq [\\mathfrak{g},\\mathfrak{g}]$. For non-trivial abelian Lie algebras we have in particular $\\mathfrak{s}=0$ and $\\mathfrak{n}=\\mathfrak{g}$.\nIf $\\mathfrak{g}$ is solvable, then $\\mathfrak{s}=[\\mathfrak{g},\\mathfrak{g}]$ and $\\mathfrak{s}\\subseteq \\mathfrak{n}$, and both equality and strict inclusion can happen. For equality consider the $2$-dimensional non-abelian Lie algebra.\nOn the other hand, Serre and others in the Bourbaki group (at the time their treatise on Lie groups and Lie algebras began) were more steeped in Lie groups and linear algebraic groups. Even though Chevalley's classification seminar in the late 1950s mostly bypassed Lie algebras, it was becoming clear that the intrinsic Jordan decomposition in algebraic groups and their Lie algebras was an important new tool. Similarly, contemporary work on \"reductive\" Lie groups was making reductive Lie algebras natural to study alongside semisimple ones. And representation theory of many kinds was becoming more prominent. So it's not surprising that there is some divergence in Bourbaki and Jacobson when the various ideals mentioned here come into play. For Jacobson, only $\\mathfrak{r}$ and $\\mathfrak{n}$ played a major role.\nA final remark is that abelian Lie algebras are somewhat problematic, since their role in the absence of an intrinsic Jordan decomposition is ambiguous. This makes the contrast between $\\mathfrak{n}$ and $\\mathfrak{s}$ inevitable if you only discuss Lie algebras and their representations abstractly." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.658894,"math_prob":0.9984337,"size":1190,"snap":"2020-45-2020-50","text_gpt3_token_len":384,"char_repetition_ratio":0.23693086,"word_repetition_ratio":0.0,"special_character_ratio":0.2764706,"punctuation_ratio":0.16742082,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9995554,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-12-01T12:51:55Z\",\"WARC-Record-ID\":\"<urn:uuid:d8e8b38a-c19a-468b-bef8-f65c6a6e2ff5>\",\"Content-Length\":\"147795\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:168cc701-174b-4da6-b2a6-a2dff1a7e42e>\",\"WARC-Concurrent-To\":\"<urn:uuid:aae72b20-8a53-4fcc-b4d8-a70b74efdaa1>\",\"WARC-IP-Address\":\"151.101.65.69\",\"WARC-Target-URI\":\"https://mathoverflow.net/questions/149391/on-radicals-of-a-lie-algebra\",\"WARC-Payload-Digest\":\"sha1:F2CN5T45J4APBUV3KXVPX5ETQ2HCYIWY\",\"WARC-Block-Digest\":\"sha1:SABOTPBHYI3T2ZJSNG5GTUGI24WSNKZJ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-50/CC-MAIN-2020-50_segments_1606141674082.61_warc_CC-MAIN-20201201104718-20201201134718-00505.warc.gz\"}"}
https://uk.mathworks.com/matlabcentral/fileexchange/69510-natural-excitation-technique-next?s_tid=prof_contriblnk	[ "## Natural Excitation Technique (NExT)\n\nversion 1.0.4 (3.5 KB) by\nReturns impulse response functions (IRF) using Natural Excitation technique with time domain method and frequency domain method\n\nUpdated 16 Jul 2019\n\nReturns impulse response functions (IRF) using Natural Excitation technique with time domain method (NExTT) and frequency domain method (NExTF) for systems subjected to white noise excitation. Example file is provided for estimating IRF of 2DOF system subjected to Gaussian white noise excitation with added uncertainty (also Gaussian white noise) to the response.\n\n1-function IRF= NExTT(data,refch,maxlags)\n\nInputs :\n\ndata: An array that contains response data.its dimensions are (nch,Ndata) where nch is the number of channels. Ndata is the total length of the data\nrefch: A vecor of reference channels .its dimensions (numref,1) where numref is number of reference channels\nmaxlags: Number of lags in cross-correlation function\n\nOutputs :\n\nIRF: Impulse Response Function matrix of size (nch,numref(maxlags+1))\n\n2-function IRF= NExTF(data,refch,window,N,p)\n\nInputs :\n\ndata: An array that contains response data.its dimensions are (nch,Ndata) where nch is the number of channels. Ndata is the total length of the data\nrefch: A vecor of reference channels .its dimensions (numref,1) where numref is number of reference channels\nwindow: window size to get spectral density\nN: Number of windows\np: overlap ratio between windows. from 0 to 1\n\nOutputs :\n\nIRF: Impulse Response Function matrix of size (nch,numref(ceil(window/2+1)-1))\n\nReferences:\n------------------\n James, G. H., Thomas G. Carne, and James P. Lauffer. \"The natural excitation technique (NExT) for modal parameter extraction from operating structures.\" Modal Analysis-the International Journal of Analytical and Experimental Modal Analysis 10.4 (1995): 260.\n\n Al Rumaithi, Ayad, \"Characterization of Dynamic Structures Using Parametric and Non-parametric System Identification Methods\" (2014). Electronic Theses and Dissertations. 1325.\nhttps://stars.library.ucf.edu/etd/1325\n\n Al-Rumaithi, Ayad, Hae-Bum Yun, and Sami F. Masri. \"A Comparative Study of Mode Decomposition to Relate Next-ERA, PCA, and ICA Modes.\" Model Validation and Uncertainty Quantification, Volume 3. Springer, Cham, 2015. 113-133.\n\n### Cite As\n\nAyad Al-Rumaithi (2021). Natural Excitation Technique (NExT) (https://www.mathworks.com/matlabcentral/fileexchange/69510-natural-excitation-technique-next), MATLAB Central File Exchange. Retrieved .\n\n##### MATLAB Release Compatibility\nCreated with R2018b\nCompatible with any release\n##### Platform Compatibility\nWindows macOS Linux" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.698786,"math_prob":0.7434172,"size":2654,"snap":"2021-43-2021-49","text_gpt3_token_len":662,"char_repetition_ratio":0.10566038,"word_repetition_ratio":0.252149,"special_character_ratio":0.2279578,"punctuation_ratio":0.17255718,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9552191,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-12-06T18:36:48Z\",\"WARC-Record-ID\":\"<urn:uuid:22116dbc-dbbd-4400-9797-1a03ace63007>\",\"Content-Length\":\"87194\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:bf542bcc-7acc-49b8-847d-ad015809df12>\",\"WARC-Concurrent-To\":\"<urn:uuid:8a008f5b-e62b-41dc-8210-ada49aa16df4>\",\"WARC-IP-Address\":\"104.68.243.15\",\"WARC-Target-URI\":\"https://uk.mathworks.com/matlabcentral/fileexchange/69510-natural-excitation-technique-next?s_tid=prof_contriblnk\",\"WARC-Payload-Digest\":\"sha1:SDEDFM4XUVPGN6AQAYXGDXL4W3FS6Z7F\",\"WARC-Block-Digest\":\"sha1:6USH74YOWOEJPVCSDPIHJ73MQCD43JUM\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964363309.86_warc_CC-MAIN-20211206163944-20211206193944-00353.warc.gz\"}"}
https://pythonguides.com/matplotlib-invert-y-axis/	[ "# Matplotlib invert y axis\n\nIn this Python tutorial, we will discuss Matplotlib invert y axis in python. Here we will cover different examples related to inverting of the y-axis using matplotlib. And we will also cover the following topics:\n\n• Matplotlib invert y axis\n• Matplotlib invert y axis subplots\n• Matplotlib invert x and y axis\n• Matplotlib barh invert y axis\n• Matplotlib invert secondary y axis\n• Matplotlib 3D invert y axis\n• Matplotlib flip y axis label\n• Matplotlib invert y axis imshow\n\nIf you are new to python Matplotlir, check out, How to install matplotlib python.\n\n## Matplotlib invert y axis\n\nIn this section, we learn about how to invert the y-axis in matplotlib in Python. Now before starting the topic firstly, we discuss what does invert mean here:\n\nInvert means reverse or flip the order\n\nTo understand more clearly let’s take a common example:\n\nSuppose the y-axis starts from 0 to 10 and you want to start it from 10 to 0. So in that cases, we invert or flip the axes of the plot.\n\nIn matplotlib, we can invert the y-axis of a graph using different methods. The following are the different methods used for reversing the y-axis are as below.\n\n• Using invert_yaxis() method\n• Using ylim() method\n• Using axis() method\n\n### By using invert_yaxis() method\n\nTo invert Y-axis, we can use invert_yaxis() method.\n\nSyntax of the method is as below:\n\n``matplotlib.axes.Axes.invert_yaxis(self)``\n\nExample: (Without inverting axis)\n\n``````# Import Library\n\nimport matplotlib.pyplot as plt\nimport numpy as np\n\n# Define Data\n\nx = np.arange(0, 15, 0.2)\ny = np.sin(x)\n\n# Plot figure\n\nplt.plot(x, y)\n\n# Title\n\nplt.title(\"Sine Function Normal Plot\", fontsize= 15, fontweight='bold')\n\n# Generate Plot\n\nplt.show()``````\n• In the above example, we import the matplotlib.pyplot and numpy library.\n• After this we define data using np.arrange() and np.sin() method.\n• Next, we use the plt.plot() method to draw the plot.\n• By using plt.title() we define the title of the plot.\n• plt.show() method is used to generate the plot on the user screen.\n\nLook at the y-axis: Here it starts from -1.00 and ends at 1.00\n\nExample: (With inverting axis)\n\n``````# Import Library\n\nimport matplotlib.pyplot as plt\nimport numpy as np\n\n# Define Data\n\nx = np.arange(0, 15, 0.2)\ny = np.sin(x)\n\n# Plot figure\n\nplt.plot(x, y)\n\n# Invert y-axis\n\nax = plt.gca()\nax.invert_yaxis()\n\n# Title\n\nplt.title(\"Sine Function Invert Plot\", fontsize= 15, fontweight='bold')\n\n# Generate Plot\n\nplt.show()``````\n• In the above example, we include the plt.gca() method to get current axes.\n• After this, we use the ax.invert_yaxis() method to invert or reverse the y-axis of the plot.\n\nLook at the y-axis: Here it starts from 1.00 and ends at -1.00\n\nConclusion: By using invert_yaxis() method we can reverse the y-axis of the plot.\n\n### By using ylim() method\n\nylim() method is also used to invert axes of a plot in Matplotlib. Generally, this method is used to set limits for the axes.\n\nBut if we set the minimum value as the upper limit and maximum value as the lower limit we can revert the axes.\n\nThe syntax of the ylim() method is as below:\n\n``matplotlib.pyplot.ylim(bottom,top)``\n\nHere bottom and top specify the tuple of the new y-axis limits.\n\nLet’s see an example to understand it more clearly:\n\n``````# Import Library\n\nimport numpy as np\nimport matplotlib.pyplot as plt\n\n# Define Data\n\nx = np.linspace(5, 15, 35)\ny = 2x+5\n\n# Create plot\n\naxes,(p1,p2) = plt.subplots(1, 2)\n\n# Normal plot\n\np1.plot(x, y)\np1.set_title(\"Normal Plot\")\n\n# Invert plot\n\np2.plot(x, y)\np2.set_title(\"Inverted Plot\")\nplt.ylim(max(y), min(y))\n\n# Show\n\naxes.tight_layout()\nplt.show()``````\n• In the above example, firstly we import numpy and matplotlib.pyplot library.\n• Next, we define the data and by using the plt.subplot() method we create subplots.\n• Now we use the plot() method to draw a simple plot.\n• plt.ylim() method is used to invert the y-axis here we pass max and min as a parameter.\n\nConclusion: In the normal plot, the y-axis starts from 1 and ends at 5. And In the inverted plot, the y-axis starts from 5 and ends at 1.\n\n### By using axis() method\n\nThe axis() method is also used to revert axes in Matplotlib. Basically, this method is used to set the minimum and maximum values of the axes.\n\nBut if we set the minimum value as the upper limit and the maximum value as the lower limit we can get inverted axes.\n\nThe syntax of the axis() method is as below:\n\n``matplotlib.pyplot.axis()``\n\nLet’ see an example:\n\n``````# Import Library\n\nimport numpy as np\nimport matplotlib.pyplot as plt\n\n# Define Data\n\nx = np.arange(0, 15, 0.2)\ny = np.tan(x)\n\n# Create plot\n\naxes,(p1,p2) = plt.subplots(1, 2)\n\n# Normal plot\n\np1.plot(x, y)\np1.set_title(\"Normal Plot\")\n\n# Invert plot\n\np2.plot(x, y)\np2.set_title(\"Inverted Plot\")\nplt.axis([max(x), min(x), max(y), min(y)])\n\n# Show\n\naxes.tight_layout()\nplt.show()``````\n• Here we import matplotlib.pyplot and numpy library and define data using np.arrange() and np.tan() method.\n• plt.subplots() method is used for creating subplots.\n• Then we use the plot() method to plot a graph and the set_title() method is used to add a title to the plot.\n• By using the plt.axis() method we revert the axes and here we pass the max and min values of the y and x axes.\n\n## Matplotlib invert y axis subplots\n\nHere we will discuss how to invert the y-axis of the specific subplot if we draw multiple plots in a figure area in Python matplotlib.\n\nWe use the invert_yaxis() method to flip the y-axis of the subplot.\n\nLet’s understand the concept with the help of an example:\n\n``````# Import Libraries\n\nimport numpy as np\nimport matplotlib.pyplot as plt\n\n# Define Data\n\nx1= [0.2, 0.4, 0.6, 0.8, 1]\ny1= [0.3, 0.6, 0.8, 0.9, 1.5]\n\nx2= [2, 6, 7, 9, 10]\ny2= [3, 4, 6, 9, 12]\n\nx3= [5, 8, 12]\ny3= [3, 6, 9]\n\nx4= [5, 8, 12]\ny4= [3, 6, 9]\n\nfig, ax = plt.subplots(2, 2)\n\n# Invert y-axis\n\nax[1,0].invert_yaxis()\n\n# Plot graph\n\nax[0, 0].plot(x1, y1)\nax[0, 1].plot(x2, y2)\nax[1, 0].plot(x3, y3)\nax[1, 1].plot(x4, y4)\n\n# Display Graph\n\nfig.tight_layout()\nplt.show()``````\n• In the above example, we plot multiple plots in a figure area. And we want to invert the y-axis of the specific plot.\n• Here we use the invert_yaixs() method to flip the y-axis of the plot.\n• We use the invert_yaxis() method with the third plot.\n\n## Matplotlib invert x and y axis\n\nHere we are going to learn how to invert the x-axis and y-axis of the plot in Python matplotlib.\n\nBy using invert_xaxis() and invert_yaxis() method we can flip the x-axis and y-axis respectively.\n\nThe syntax is as given below:\n\n``````# Invert x-axis\n\nmatplotlib.axes.Axes.invert_xaxis(self)\n\n# Invert y-axis\n\nmatplotlib.axes.Axes.invert_yaxis(self) ``````\n\nLet’s see an example related to this concept:\n\n``````# Import Library\n\nimport numpy as np\nimport matplotlib.pyplot as plt\n\n# Define Data\n\nx=[5, 10, 15, 20, 25,30]\ny=[0, 1, 2, 3, 4, 5]\n\n# Create Subplot\n\naxes,(p1,p2) = plt.subplots(1, 2)\n\n# Normal plot\n\np1.plot(x, y)\np1.set_title(\"Normal Plot\")\n\n# Invert plot\n\np2.plot(x, y)\np2.set_title(\"Inverted Plot\")\n\n# Invert axes\n\nax = plt.gca()\nax.invert_yaxis()\nax.invert_xaxis()\n\n# Show\n\naxes.tight_layout()\nplt.show()``````\n• In the above example, we import numpy and matplotlib libraries.\n• After this, we define data on x and y coordinates.\n• By using the plt.subplots() method we create subplots in a figure area.\n• Next, we use the plot() method to plot a graph and the set_title() method to set the title of the plot.\n• Then we use plt.gca() method to get current axes of the plot.\n• By using the invert_yaxis() and the invert_xaxis() method we flip the axes of the plot.\n• In last, we use the tight_layout() method to automatically set the plots and the show() method to display the plot on the user’s screen.\n\nAlso, read Matplotlib savefig blank image\n\n## Matplotlib barh invert y axis\n\nHere we are going to learn how we invert the y-axis of the horizontal bar chart using matplotlib in Python. Firstly, we have to know how to plot a horizontal bar chart.\n\nThe syntax to plot the horizontal bar chart is as below:\n\n``matplotlib.pyplot.barh(x,y)``\n\nHere x represents the x-axis coordinates and y represents the height of the bars.\n\nThe syntax to invert the y-axis of the barh chart is as given below:\n\n``````# Invert y-axis\n\nmatplotlib.axes.Axes.invert_yaxis(self) ``````\n\nExample:\n\n``````# Import Library\n\nimport numpy as np\nimport matplotlib.pyplot as plt\n\n# Define Data\n\nx=[1, 2, 3, 4, 5]\ny=[20, 15, 12, 6, 5]\n\n# Simple Bar Chart\n\nplt.figure()\nplt.barh(x,y)\n\n# Inverted Bar chart\n\nplt.figure()\nplt.barh(x,y)\nplt.gca().invert_yaxis()\n\n# Show\n\nplt.show()``````\n• In the above example, we first import the numpy and matplotlib.pyplot library.\n• Next, we define data.\n• By using the plt.barh() method we create a horizontal bar chart.\n• By using the plt.gca() method we get the current axes of the plot.\n• Then we use the invert_yaxis() method to flip the y-axis of the plot.\n\n## Matplotlib invert secondary y axis\n\nHere we are going to learn how we can invert the secondary y-axis in matplotlib in Python. Firstly, we have to understand what does secondary y-axis means and when we need it.\n\nOnce in a while, we need two x-axes or y-axes to get more insights into the data. At that moment we need to create secondary axes.\n\nIn python, matplotlib provides the functionality to create a plot that has two y-axes, and even we can provide different labels to both.\n\nWe can create a plot with two different y-axes by using two different axes objects. And to create different axes objects we use the twinx() method.\n\nLet’s see an example where we invert the secondary y-axis:\n\n``````# Import Library\n\nimport numpy as np\nimport matplotlib.pyplot as plt\n\n# Define Data\n\nx = np.arange(0, 10, 3.5)\ny1 = x2\ny2 = x4\n\n# Create subplot\n\nfig, ax = plt.subplots(figsize = (10, 5))\n\n# twinx() for creating axes object for secondary y-axis\n\nax2 = ax.twinx()\n\n# Plot graph\n\nax.plot(x, y1)\nax2.plot(x, y2)\n\n# Invert secondary y-axis\n\nax2.invert_yaxis()\n\n# Labels to axes\n\nax.set_xlabel('x-axis')\nax.set_ylabel('y-axis')\n\n# Label secondary y-axis\n\nax2.set_ylabel('Secondary y-axis', color = 'b')\n\n# Display layout\n\nplt.tight_layout()\n\n# Show plot\n\nplt.show()``````\n• In the above example, we import important libraries such as numpy and matplotlib.pyplot.\n• Next, we create data by using the np.arange() method here we define data between 0 to 10 with the difference of 3.5 at the x-axis.\n• Then by using plt.subplots() methods we create subplots.\n• By using the twinx() we create another axes object for the secondary y-axis.\n• set_xlabel() and set_ylabel() methods are used for adding labels on the axes.\n• By using the tight_layout() and show() method we define the layout and show the plot.\n\n## Matplotlib 3D invert y axis\n\nIn this section, we are going to study how to invert a 3D plot in Matplotlib in Python.\n\nThe syntax to create a 3D plot and invert the y-axis:\n\n``````# Create 3D plot\nax.plot3D(x,y,z)\n\n# Invert y-axis\nmatplotlib.pyplot.ylim(max(y),min(y))``````\n\nHere x, y, and z are the coordinates points.\n\nLet’s see an example related to this:\n\n``````# Import Libraries\n\nimport matplotlib\nfrom mpl_toolkits.mplot3d import Axes3D\nimport matplotlib.pyplot as plt\n\n# Create subplots\n\nfig = plt.figure(figsize=(12,5), dpi=80)\n\n# Simple plot\n\nx=[1,1,10,10]\ny=[1,10,10,10]\nz=[1,1,1,10]\nax1.plot(x,y ,z , marker=\"o\")\n\n# Inverted Plot\n\nx1=[1,1,10,10]\ny1=[1,10,10,10]\nz1= [1,1,1,10]\n\n# Function used for invert y-axis\n\nplt.ylim(max(y1),min(y1))\n\nax2.plot(x1,y1 ,z1, marker=\"o\")\n\n# Show Plot\n\nplt.show()\n``````\n• In the above example, we import important libraries such as matplotlib.pyplot, and mplot3D.\n• Next, we use the plot() method to plot a graph in a figure area.\n• By using the ylim() method we invert the y-axis of the plot. In function, we pass the max and min of y1 coordinates.\n\n## Matplotlib flip y axis label\n\nHere we are going to learn to flip labels at the y-axis in Matplotlib in Python. By using the ylabel() method we can add the label at the y-axis and flip it.\n\nExample:\n\n``````# Import Library\n\nfrom matplotlib import pyplot as plt\n\n# Define Data\n\nx=[0, 1, 2, 3, 4, 5]\ny=[2, 4, 6, 8, 10, 12]\n\n# Plot graph\n\nplt.plot(x,y)\n\n# Flip Y-axis Label\n\nplt.ylabel('Y-axis',rotation=0)\n\n# Show\n\nplt.draw()``````\n• In the above example, we import matplotlib.pyplot library for visualization.\n• Next, we define the data and use the plt.plot() method to draw a plot.\n• By using plt.ylabel() method we set the label at the y-axis and flip it horizontally by passing the rotation argument. Here we set the value of rotation to 0.\n\n## Matplotlib invert y axis imshow\n\nHere we learn to invert the y-axis in the imshow() method in Python Matplotlib. Firstly, understand what does imshow does:\n\ninshow() method shows an image of a color-mapped or 3D RGB array.\n\nThe syntax of the inshow() method is given below:\n\n``````matplotlib.pyplot.imshow(X,\ncmap=None,\nnorm=None,\naspect=None,\ninterpolation=None,\nalpha=None,\nvmin=None,\nvmax=None,\norigin=None,\nextent=None,\nshape=,\nfilternorm=1,\nimlim=,\nresample=None,\nurl=None,\ndata=None,\nkwargs)``````\n\nThe parameters used above are described as below:\n\n• X: specify data of the image.\n• cmap: specifies colormap or registered colormap name.\n• norm: It is a normalized instance used to scale data.\n• aspect: Used to control the aspect ratio of the axes.\n• interpolation: It is an interpolation method used to display images.\n• alpha: specify the intensity of the color.\n• vmin, vmax: specify the range of the color bar.\n• origin: Used to place [0,0] index of the array in the upper left or lower corner of the axes.\n• extent: It is a bounding box in data coordinates.\n• filternorm: Used for the antigrain image resize filter.\n• url: set the URL of the axes image.\n\nExample:\n\n``````# Import Libraries\n\nimport numpy as np\nfrom matplotlib import pyplot as plt\n\n# Define data\n\nx = np.array([[0, 1, 2, 3],\n[1, 1, 1, 3],\n[1, 2, 2, 3],\n[2, 2, 3, 3]])\n\n# Create subplot\n\nfig, ax = plt.subplots(1,2)\n\n# Imshow and Invert Y-axis\n\nfor i in range(2):\nax.imshow(x, cmap = 'summer', vmin = 1, vmax = 3,)\nax.imshow(x, cmap = 'summer', vmin = 1, vmax = 3,\norigin='lower')\n# Show\n\nplt.show()``````\n• Here we define data using the np.array() method and use the imshow() method to plot the image of the colormap.\n• We pass the parameter origin to flip the y-axis and set its value to lower." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.5857118,"math_prob":0.9856577,"size":15082,"snap":"2021-43-2021-49","text_gpt3_token_len":4146,"char_repetition_ratio":0.19226688,"word_repetition_ratio":0.22528554,"special_character_ratio":0.27987003,"punctuation_ratio":0.16479173,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9999043,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-12-09T09:48:20Z\",\"WARC-Record-ID\":\"<urn:uuid:1bbd6d43-1271-4d90-ac06-21ac6bd0f631>\",\"Content-Length\":\"100501\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:4923421b-b490-439c-aa26-e25278dc6575>\",\"WARC-Concurrent-To\":\"<urn:uuid:a2d78c6a-0642-4ce0-8517-791c0c3d14c2>\",\"WARC-IP-Address\":\"172.67.133.165\",\"WARC-Target-URI\":\"https://pythonguides.com/matplotlib-invert-y-axis/\",\"WARC-Payload-Digest\":\"sha1:7JL4447W74IMXS5IEBIB3GMUYBI7ECRY\",\"WARC-Block-Digest\":\"sha1:RFINFZZRF7SMPOCMP344M2ZVRIKD4Y55\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964363791.16_warc_CC-MAIN-20211209091917-20211209121917-00531.warc.gz\"}"}
https://intellipaat.com/community/8117/what-is-lrpolicy-in-caffe	[ "2 views\n\nI just try to find out how I can use Caffe. To do so, I just took a look at the different .prototxt files in the examples folder. There is one option I don't understand:\n\n# The learning rate policy\n\nlr_policy: \"inv\"\n\nPossible values seem to be:\n\n• \"fixed\"\n\n• \"inv\"\n\n• \"step\"\n\n• \"multistep\"\n\n• \"stepearly\"\n\n• \"poly\"\n\nCould somebody please explain those options?\n\nby (33.1k points)\n\nYou can check following description about lr_policy (Learning rate decay):\n\n// The learning rate decay policy. The currently implemented learning rate\n\n// policies are as follows:\n\n// - fixed: always return base_lr.\n\n// - step: return base_lr * gamma ^ (floor(iter / step))\n\n// - exp: return base_lr * gamma ^ iter\n\n// - inv: return base_lr * (1 + gamma * iter) ^ (- power)\n\n// - multistep: similar to step but it allows non uniform steps defined by\n\n// stepvalue\n\n// - poly: the effective learning rate follows a polynomial decay, to be\n\n// zero by the max_iter. return base_lr (1 - iter/max_iter) ^ (power)\n\n// - sigmoid: the effective learning rate follows a sigmod decay\n\n// return base_lr ( 1/(1 + exp(-gamma * (iter - stepsize))))\n\n//\n\n// where base_lr, max_iter, gamma, step, stepvalue and power are defined\n\n// in the solver parameter protocol buffer, and iter is the current iteration.\n\nHope this answer helps." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.65738815,"math_prob":0.83782095,"size":903,"snap":"2022-40-2023-06","text_gpt3_token_len":221,"char_repetition_ratio":0.16351502,"word_repetition_ratio":0.038961038,"special_character_ratio":0.2978959,"punctuation_ratio":0.1388889,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99414897,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-10-06T19:29:10Z\",\"WARC-Record-ID\":\"<urn:uuid:f30c1f2e-b9a4-4a10-b313-f1bc64e90abe>\",\"Content-Length\":\"93139\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d4d41e7c-565d-4d10-8aca-7891775b56e9>\",\"WARC-Concurrent-To\":\"<urn:uuid:fe627282-6cad-43a8-838a-a7b8a3712352>\",\"WARC-IP-Address\":\"104.18.27.176\",\"WARC-Target-URI\":\"https://intellipaat.com/community/8117/what-is-lrpolicy-in-caffe\",\"WARC-Payload-Digest\":\"sha1:U33DQ5DWMFBZV265NMJL6CMW75EGJ4WK\",\"WARC-Block-Digest\":\"sha1:NDSSZOWA5O4GZASMJTE5CHWLORCHU6UE\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-40/CC-MAIN-2022-40_segments_1664030337855.83_warc_CC-MAIN-20221006191305-20221006221305-00421.warc.gz\"}"}
https://www.fxsolver.com/browse/formulas/Area+of+a+circular+segment	[ "# Area of a circular segment\n\n## Description\n\nCircular segment is a region of a circle which is “cut off” from the rest of the circle by a secant or a chord. More formally, a circular segment is a region of two-dimensional space that is bounded by an arc (of less than 180°) of a circle and by the chord connecting the endpoints of the arc.\nCentral angle is an angle whose apex (vertex) is the center of a circle and whose legs (sides) are radii intersecting the circle in two distinct points and thereby subtending an arc between those two points whose angle is (by definition) equal to that of the central angle. The area of a circular segment can be calculated by the radius of the circle and the cental angle of the arc of the segment.\n\nRelated formulas\n\n## Variables\n\n Aseg Area of the circular segment (m2) R The radius of the circle (m) θ The central angle (radians)" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.91692996,"math_prob":0.98515695,"size":848,"snap":"2019-43-2019-47","text_gpt3_token_len":198,"char_repetition_ratio":0.18838863,"word_repetition_ratio":0.013071896,"special_character_ratio":0.22287735,"punctuation_ratio":0.030674847,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98435575,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-11-21T11:38:30Z\",\"WARC-Record-ID\":\"<urn:uuid:7b7a05c7-e5fc-4d9e-979e-5ddc23909c77>\",\"Content-Length\":\"25357\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:3b5ac613-7958-4a74-9a26-141fcc31ca6d>\",\"WARC-Concurrent-To\":\"<urn:uuid:b835a0cd-4de8-4cff-9054-6eeb6ac84700>\",\"WARC-IP-Address\":\"178.254.54.75\",\"WARC-Target-URI\":\"https://www.fxsolver.com/browse/formulas/Area+of+a+circular+segment\",\"WARC-Payload-Digest\":\"sha1:ELKZ6A526QYFRHMIEOQZISAIZKHXRIS6\",\"WARC-Block-Digest\":\"sha1:XVQ6NF2RRCLHZBKWRVHW2LUL4FJ3LKKP\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-47/CC-MAIN-2019-47_segments_1573496670770.21_warc_CC-MAIN-20191121101711-20191121125711-00340.warc.gz\"}"}
https://www.personal-math-online-help.com/vertical-subtraction-with-regrouping-worksheets-3d-2d.html	[ "## Vertical SubtractionWith Regrouping Worksheets3 digits - 2 digits\n\nReminders & Tips for - Vertical Subtraction With Regrouping 3 digits - 2 digits Worksheet:\n• A Reminder: you should already know your subtraction to 18 with borrowing\n• A Tip: first you borrow a 10 from 10s to make the upper 1s bigger than lower 1s. Then borrow a 100 from 100s to make the upper 100s \"stronger\". Then just drop down the remaining 100s. :-)\n\nOK, Try these Vertical Subtraction With Regrouping worksheets below.\n\nBut if you want to try any of the steps - hover your mouse over a Step - and Click to practice it.\n\nOr you can try our similar vertical subtraction worksheets from the table below.\n\n Free Printable math vertical subtraction worksheets:", null, "", null, "", null, "", null, "", null, "", null, "" ]	[ null, "https://www.personal-math-online-help.com/images/1x1.gif.pagespeed.ce.u_17SdyJKn.gif", null, "https://www.personal-math-online-help.com/images/1x1.gif.pagespeed.ce.u_17SdyJKn.gif", null, "https://www.personal-math-online-help.com/images/1x1.gif.pagespeed.ce.u_17SdyJKn.gif", null, "https://www.personal-math-online-help.com/images/1x1.gif.pagespeed.ce.u_17SdyJKn.gif", null, "https://www.personal-math-online-help.com/images/1x1.gif.pagespeed.ce.u_17SdyJKn.gif", null, "https://www.personal-math-online-help.com/images/1x1.gif.pagespeed.ce.u_17SdyJKn.gif", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.79101807,"math_prob":0.5567043,"size":786,"snap":"2023-40-2023-50","text_gpt3_token_len":194,"char_repetition_ratio":0.13299233,"word_repetition_ratio":0.014492754,"special_character_ratio":0.25063613,"punctuation_ratio":0.11409396,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9594155,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-21T07:53:40Z\",\"WARC-Record-ID\":\"<urn:uuid:86386f32-8185-42b2-873d-e6b79c196a8f>\",\"Content-Length\":\"27408\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d6d9e3db-88d1-4fd8-bca8-a01dca26044a>\",\"WARC-Concurrent-To\":\"<urn:uuid:75628862-cbcc-4fcd-b738-0c318a3611a0>\",\"WARC-IP-Address\":\"173.247.218.222\",\"WARC-Target-URI\":\"https://www.personal-math-online-help.com/vertical-subtraction-with-regrouping-worksheets-3d-2d.html\",\"WARC-Payload-Digest\":\"sha1:WHZCZL4CB4MO44HM3LBYAWB5NQT7AIBR\",\"WARC-Block-Digest\":\"sha1:4D5ZSGZYCYAF7EFD2YLKPGCMVN65UYIY\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233505362.29_warc_CC-MAIN-20230921073711-20230921103711-00607.warc.gz\"}"}
https://tellmeanumber.hostcoder.com/879261	[ "Question Is 879,261 a prime number?\n\nThe number 879,261 is NOT a PRIME number.\n\nHow to check if the number 879,261 is a prime number\n\nA prime number can be divided, without a remainder, only by itself and by 1. For example, 13 can be divided only by 13 and by 1. In this case, the number 879,261 that you looked for, is NOT a PRIME number, so it devides by 1,3, 293087, 879261, and of course 879,261.\n\nQuestion Where is the number 879,261 located in π (PI) decimals?\n\nThe number 879,261 is at position 1439006 in π decimals.\n\nSearch was acomplished in the first 100 milions decimals of PI.\n\nQuestion What is the roman representation of number 879,261?\n\nThe roman representation of number 879,261 is DCCCLXXIXCCLXI.\n\nLarge numbers to roman numbers\n\n3,999 is the largest number you can write in Roman numerals. There is a convencion that you can represent numbers larger than 3,999 in Roman numerals using an overline. Matematically speaking, this means means you are multiplying that Roman numeral by 1,000. For example if you would like to write 70,000 in Roman numerals you would use the Roman numeral LXX. This moves the limit to write roman numerals to 3,999,999.\n\nQuestion How many digits are in the number 879,261?\n\nThe number 879,261 has 6 digits.\n\nHow to get the lenght of the number 879,261\n\nTo find out the lenght of 879,261 we simply count the digits inside it.\n\nQuestion What is the sum of all digits of the number 879,261?\n\nThe sum of all digits of number 879,261 is 33.\n\nHow to calculate the sum of all digits of number 879,261\n\nTo calculate the sum of all digits of number 879,261 you will have to sum them all like fallows:\n\nQuestion What is the hash of number 879,261?\n\nThere is not one, but many hash function. some of the most popular are md5 and sha-1\n\nHere are some of the most common cryptographic hashes for the number 879,261\n\nCriptographic function Hash for number 879,261\nmd5 719b09fc6a0f41756baf88429376a320\nsha1 11be4c401d492504fc12166063b3c67023885945\nsha256 ce47c21c948863e52b7e3aab892ba8163d8f98c7fb846a508e825888bb864461\nsha512 c3aae967113455d603522f10a8cf234366ec7bf8054293685bbd5821385f30c9610bcd23ff258e6099eb5b80a0813d8d8b14e15c7f847c6f43a96755bec68c41\n\nQuestion How to write number 879,261 in English text?\n\nIn English the number 879,261 is writed as eight hundred seventy-nine thousand, two hundred sixty-one.\n\nHow to write numbers in words\n\nWhile writing short numbers using words makes your writing look clean, writing longer numbers as words isn't as useful. On the other hand writing big numbers it's a good practice while you're learning.\n\nHere are some simple tips about when to wright numbers using letters.\n\n Numbers less than ten should always be written in text. On the other hand numbers that are less then 100 and multiple of 10, should also be written using letters not numbers. Example: Number 879,261 should NOT be writed as eight hundred seventy-nine thousand, two hundred sixty-one, in a sentence Big numbers should be written as the numeral followed by the word thousands, million, billions, trillions, etc. If the number is that big it might be a good idea to round up some digits so that your rider remembers it. Example: Number 879,261 could also be writed as 879.2 thousands, in a sentence, since it is considered to be a big number\n\nWhat numbers are before and after 879,261\n\nPrevious number is: 879,260\n\nNext number is: 879,262" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.896593,"math_prob":0.97842133,"size":374,"snap":"2022-05-2022-21","text_gpt3_token_len":115,"char_repetition_ratio":0.18378378,"word_repetition_ratio":0.0,"special_character_ratio":0.36096257,"punctuation_ratio":0.185567,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9903206,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-01-18T03:36:32Z\",\"WARC-Record-ID\":\"<urn:uuid:dc914400-90a3-4924-9588-855c69999a1c>\",\"Content-Length\":\"16687\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c8e4cb23-c830-46e5-972f-d3cb19916834>\",\"WARC-Concurrent-To\":\"<urn:uuid:bbed91d2-32e9-4643-b5c1-c2d44a21aef4>\",\"WARC-IP-Address\":\"51.254.201.96\",\"WARC-Target-URI\":\"https://tellmeanumber.hostcoder.com/879261\",\"WARC-Payload-Digest\":\"sha1:2NP4DDS5JQIAW7HTHD3ZJZSJKCTRQRCB\",\"WARC-Block-Digest\":\"sha1:XTWCQCSESM6DE5OLFXWN6FIVWDNMFQFT\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-05/CC-MAIN-2022-05_segments_1642320300722.91_warc_CC-MAIN-20220118032342-20220118062342-00303.warc.gz\"}"}
https://gateway.ipfs.io/ipfs/QmXoypizjW3WknFiJnKLwHCnL72vedxjQkDDP1mXWo6uco/wiki/Galois_connection.html	[ "Galois connection\n\nIn mathematics, especially in order theory, a Galois connection is a particular correspondence (typically) between two partially ordered sets (posets). The same notion can also be defined on preordered sets or classes; this article presents the common case of posets. Galois connections generalize the correspondence between subgroups and subfields investigated in Galois theory (named after the French mathematician Évariste Galois). They find applications in various mathematical theories.\n\nA Galois connection is rather weak compared to an order isomorphism between the involved posets, but every Galois connection gives rise to an isomorphism of certain sub-posets, as will be explained below.\n\nThe literature contains two closely related notions of \"Galois connection\". In this article, we will distinguish between the two by referring to the first as (monotone) Galois connection and to the second as antitone Galois connection.\n\nThe term Galois correspondence is sometimes used to mean bijective Galois connection; this is simply an order isomorphism (or dual order isomorphism, depending on whether we take monotone or antitone Galois connections).\n\nDefinitions\n\n(Monotone) Galois connection\n\nLet (A, ≤) and (B, ≤) be two partially ordered sets. A monotone Galois connection between these posets consists of two monotone functions: F : AB and G : BA, such that for all a in A and b in B, we have\n\nF(a) ≤ b if and only if aG(b).\n\nIn this situation, F is called the lower adjoint of G and G is called the upper adjoint of F. Mnemonically, the upper/lower terminology refers to where the function application appears relative to ≤; the term \"adjoint\" refers to the fact that monotone Galois connections are special cases of pairs of adjoint functors in category theory as discussed further below. Other terminology encountered here is left adjoint (resp. right adjoint) for the lower (resp. upper) adjoint.\n\nAn essential property of a Galois connection is that an upper/lower adjoint of a Galois connection uniquely determines the other:\n\nF(a) is the least element with aG(), and\nG(b) is the largest element with F() ≤ b.\n\nA consequence of this is that if F or G is invertible, then each is the inverse of the other, i.e. F = G −1 .\n\nGiven a Galois connection with lower adjoint F and upper adjoint G, we can consider the compositions GF : AA, known as the associated closure operator, and FG : BB, known as the associated kernel operator. Both are monotone and idempotent, and we have aGF(a) for all a in A and FG(b) ≤ b for all b in B.\n\nA Galois insertion of A into B is a Galois connection in which the closure operator GF is the identity on A.\n\nAntitone Galois connection\n\nThe above definition is common in many applications today, and prominent in lattice and domain theory. However the original notion in Galois theory is slightly different. In this alternative definition, a Galois connection is a pair of antitone, i.e. order-reversing, functions F : AB and G : BA between two posets A and B, such that\n\nbF(a) if and only if aG(b).\n\nThe symmetry of F and G in this version erases the distinction between upper and lower, and the two functions are then called polarities rather than adjoints. Each polarity uniquely determines the other, since\n\nF(a) is the largest element b with aG(b), and\nG(b) is the largest element a with bF(a).\n\nThe compositions GF : AA and FG : BB are the associated closure operators; they are monotone idempotent maps with the property aGF(a) for all a in A and bFG(b) for all b in B.\n\nThe implications of the two definitions of Galois connections are very similar, since an antitone Galois connection between A and B is just a monotone Galois connection between A and the order dual Bop of B. All of the below statements on Galois connections can thus easily be converted into statements about antitone Galois connections.\n\nExamples\n\nMonotone Galois connections\n\nPower set; implication and conjunction\n\nFor an order theoretic example, let U be some set, and let A and B both be the power set of U, ordered by inclusion. Pick a fixed subset L of U. Then the maps F and G, where F(M) = LM, and G(N) = N ∪ (U\\L), form a monotone Galois connection, with F being the lower adjoint. A similar Galois connection whose lower adjoint is given by the meet (infimum) operation can be found in any Heyting algebra. Especially, it is present in any Boolean algebra, where the two mappings can be described by F(x) = (ax) and G(y) = (y ∨ ¬a) = (ay). In logical terms: \"implication from a\" is the upper adjoint of \"conjunction with a\".\n\nLattices\n\nFurther interesting examples for Galois connections are described in the article on completeness properties. Roughly speaking, it turns out that the usual functions ∨ and ∧ are lower and upper adjoints to the diagonal map XX × X. The least and greatest elements of a partial order are given by lower and upper adjoints to the unique function X → {1}. Going further, even complete lattices can be characterized by the existence of suitable adjoints. These considerations give some impression of the ubiquity of Galois connections in order theory.\n\nTransitive group actions\n\nLet G act transitively on X and pick some point x in X. Consider", null, "the set of blocks containing x. Further, let", null, "consist of the subgroups of G containing the stabilizer of x.\n\nThen, the correspondence", null, ":", null, "is a monotone, one-to-one Galois connection. As a corollary, one can establish that doubly transitive actions have no blocks other than the trivial ones (singletons or the whole of X): this follows from the stabilizers being maximal in G in that case. See doubly transitive group for further discussion.\n\nImage and inverse image\n\nIf f : XY is a function, then for any subset M of X we can form the image F(M) = f (M) = {f(m) \| mM} and for any subset N of Y we can form the inverse image G(N) = f −1(N) = {xX \| f(x) ∈ N}. Then F and G form a monotone Galois connection between the power set of X and the power set of Y, both ordered by inclusion ⊆. There is a further adjoint pair in this situation: for a subset M of X, define H(M) = {yY \| f −1({y}) ⊆ M}. Then G and H form a monotone Galois connection between the power set of Y and the power set of X. In the first Galois connection, G is the upper adjoint, while in the second Galois connection it serves as the lower adjoint.\n\nIn the case of a quotient map between algebraic objects (such as groups), this connection is called the lattice theorem: subgroups of G connect to subgroups of G/N, and the closure operator on subgroups of G is given by H = HN.\n\nSpan and closure\n\nPick some mathematical object X that has an underlying set, for instance a group, ring, vector space, etc. For any subset S of X, let F(S) be the smallest subobject of X that contains S, i.e. the subgroup, subring or subspace generated by S. For any subobject U of X, let G(U) be the underlying set of U. (We can even take X to be a topological space, let F(S) the closure of S, and take as \"subobjects of X\" the closed subsets of X.) Now F and G form a monotone Galois connection between subsets of X and subobjects of X, if both are ordered by inclusion. F is the lower adjoint.\n\nSyntax and semantics\n\nA very general comment of William Lawvere is that syntax and semantics are adjoint: take A to be the set of all logical theories (axiomatizations), and B the power set of the set of all mathematical structures. For a theory TA, let F(T) be the set of all structures that satisfy the axioms T; for a set of mathematical structures S, let G(S) be the minimum of the axiomatizations which approximate S. We can then say that F(T) is a subset of S if and only if T logically implies G(S): the \"semantics functor\" F and the \"syntax functor\" G form a monotone Galois connection, with semantics being the lower adjoint.\n\nAntitone galois connections\n\nGalois theory\n\nThe motivating example comes from Galois theory: suppose L/K is a field extension. Let A be the set of all subfields of L that contain K, ordered by inclusion ⊆. If E is such a subfield, write Gal(L/E) for the group of field automorphisms of L that hold E fixed. Let B be the set of subgroups of Gal(L/K), ordered by inclusion ⊆. For such a subgroup G, define Fix(G) to be the field consisting of all elements of L that are held fixed by all elements of G. Then the maps E ↦ Gal(L/E) and G ↦ Fix(G) form an antitone Galois connection.\n\nAlgebraic topology: covering spaces\n\nAnalogously, given a path-connected topological space X, there is an antitone Galois connection between subgroups of the fundamental group π1(X) and path-connected covering spaces of X. In particular, if X is semi-locally simply connected, then for every subgroup G of π1(X), there is a covering space with G as its fundamental group.\n\nLinear algebra: annihilators and orthogonal complements\n\nGiven an inner product space V, we can form the orthogonal complement F(X) of any subspace X of V. This yields an antitone Galois connection between the set of subspaces of V and itself, ordered by inclusion; both polarities are equal to F.\n\nGiven a vector space V and a subset X of V we can define its annihilator F(X), consisting of all elements of the dual space V of V that vanish on X. Similarly, given a subset Y of V, we define its annihilator G(Y) = {xV \| φ(x) = 0 ∀φY}. This gives an antitone Galois connection between the subsets of V and the subsets of V.\n\nAlgebraic geometry\n\nIn algebraic geometry, the relation between sets of polynomials and their zero sets is an antitone Galois connection.\n\nFix a natural number n and a field K and let A be the set of all subsets of the polynomial ring K[X1, ..., Xn] ordered by inclusion ⊆, and let B be the set of all subsets of Kn ordered by inclusion ⊆. If S is a set of polynomials, define the variety of zeros as", null, "the set of common zeros of the polynomials in S. If U is a subset of Kn, define I(U) as an ideal of polynomials vanishing on U, that is", null, "Then V and I form an antitone Galois connection.\n\nThe closure on Kn is the closure in the Zariski topology, and if the field K is algebraically closed, then the closure on the polynomial ring is the radical of ideal generated by S.\n\nMore generally, given a commutative ring R (not necessarily a polynomial ring), there is an antitone Galois connection between radical ideals in the ring and subvarieties of the affine variety Spec(R).\n\nMore generally, there is an antitone Galois connection between ideals in the ring and subschemes of the corresponding affine variety.\n\nConnections on power sets arising from binary relations\n\nSuppose X and Y are arbitrary sets and a binary relation R over X and Y is given. For any subset M of X, we define F(M) = {yY \| mRymM}. Similarly, for any subset N of Y, define G(N) = {xX \| xRnnN}. Then F and G yield an antitone Galois connection between the power sets of X and Y, both ordered by inclusion ⊆.\n\nMany antitone Galois connections arise in this way; examples include the original connection from Galois theory, the connections in linear algebra and the connection from algebraic geometry explained above.\n\nProperties\n\nIn the following, we consider a (monotone) Galois connection f = (f , f), where f : AB is the lower adjoint as introduced above. Some helpful and instructive basic properties can be obtained immediately. By the defining property of Galois connections, f (x) ≤ f (x) is equivalent to xf(f (x)), for all x in A. By a similar reasoning (or just by applying the duality principle for order theory), one finds that f (f(y)) ≤ y, for all y in B. These properties can be described by saying the composite f f is deflationary, while ff is inflationary (or extensive).\n\nNow consider x, yA such that xy, then using the above one obtains xf(f (y)). Applying the basic property of Galois connections, one can now conclude that f (x) ≤ f (y). But this just shows that f preserves the order of any two elements, i.e. it is monotone. Again, a similar reasoning yields monotonicity of f. Thus monotonicity does not have to be included in the definition explicitly. However, mentioning monotonicity helps to avoid confusion about the two alternative notions of Galois connections.\n\nAnother basic property of Galois connections is the fact that f(f (f(x))) = f(x), for all x in B. Clearly we find that\n\nf(f (f(x))) ≥ f(x).\n\nbecause ff is inflationary as shown above. On the other hand, since f f is deflationary, while f is monotonic, one finds that\n\nf(f (f(x))) ≤ f(x).\n\nThis shows the desired equality. Furthermore, we can use this property to conclude that\n\nf (f(f (f(x)))) = f (f(x))\n\nand\n\nf(f (f(f (x)))) = f(f (x))\n\ni.e., f f and ff are idempotent.\n\nIt can be shown (see Blyth or Erné for proofs) that a function f is a lower (resp. upper) adjoint if and only if f is a residuated mapping (resp. residual mapping). Therefore, the notion of residuated mapping and monotone Galois connection are essentially the same.\n\nClosure operators and Galois connections\n\nThe above findings can be summarized as follows: for a Galois connection, the composite ff is monotone (being the composite of monotone functions), inflationary, and idempotent. This states that ff is in fact a closure operator on A. Dually, f f is monotone, deflationary, and idempotent. Such mappings are sometimes called kernel operators. In the context of frames and locales, the composite ff is called the nucleus induced by f . Nuclei induce frame homomorphisms; a subset of a locale is called a sublocale if it is given by a nucleus.\n\nConversely, any closure operator c on some poset A gives rise to the Galois connection with lower adjoint f being just the corestriction of c to the image of c (i.e. as a surjective mapping the closure system c(A)). The upper adjoint f is then given by the inclusion of c(A) into A, that maps each closed element to itself, considered as an element of A. In this way, closure operators and Galois connections are seen to be closely related, each specifying an instance of the other. Similar conclusions hold true for kernel operators.\n\nThe above considerations also show that closed elements of A (elements x with f(f (x)) = x) are mapped to elements within the range of the kernel operator f f, and vice versa.\n\nExistence and uniqueness of Galois connections\n\nAnother important property of Galois connections is that lower adjoints preserve all suprema that exist within their domain. Dually, upper adjoints preserve all existing infima. From these properties, one can also conclude monotonicity of the adjoints immediately. The adjoint functor theorem for order theory states that the converse implication is also valid in certain cases: especially, any mapping between complete lattices that preserves all suprema is the lower adjoint of a Galois connection.\n\nIn this situation, an important feature of Galois connections is that one adjoint uniquely determines the other. Hence one can strengthen the above statement to guarantee that any supremum-preserving map between complete lattices is the lower adjoint of a unique Galois connection. The main property to derive this uniqueness is the following: For every x in A, f (x) is the least element y of B such that xf(y). Dually, for every y in B, f(y) is the greatest x in A such that f (x) ≤ y. The existence of a certain Galois connection now implies the existence of the respective least or greatest elements, no matter whether the corresponding posets satisfy any completeness properties. Thus, when one upper adjoint of a Galois connection is given, the other upper adjoint can be defined via this same property.\n\nOn the other hand, some monotone function f is a lower adjoint if and only if each set of the form {xA \| f (x) ≤ b}, for b in B, contains a greatest element. Again, this can be dualized for the upper adjoint.\n\nGalois connections as morphisms\n\nGalois connections also provide an interesting class of mappings between posets which can be used to obtain categories of posets. Especially, it is possible to compose Galois connections: given Galois connections (f , f) between posets A and B and (g, g) between B and C, the composite (gf , fg) is also a Galois connection. When considering categories of complete lattices, this can be simplified to considering just mappings preserving all suprema (or, alternatively, infima). Mapping complete lattices to their duals, this categories display auto duality, that are quite fundamental for obtaining other duality theorems. More special kinds of morphisms that induce adjoint mappings in the other direction are the morphisms usually considered for frames (or locales).\n\nConnection to category theory\n\nEvery partially ordered set can be viewed as a category in a natural way: there is a unique morphism from x to y if and only if xy. A monotone Galois connection is then nothing but a pair of adjoint functors between two categories that arise from partially ordered sets. In this context, the upper adjoint is the right adjoint while the lower adjoint is the left adjoint. However, this terminology is avoided for Galois connections, since there was a time when posets were transformed into categories in a dual fashion, i.e. with arrows pointing in the opposite direction. This led to a complementary notation concerning left and right adjoints, which today is ambiguous.\n\nApplications in the theory of programming\n\nGalois connections may be used to describe many forms of abstraction in the theory of abstract interpretation of programming languages.\n\nNotes\n\n1. Monotonicity follows from the following condition. See the discussion of the properties. It is only explicit in the definition to distinguish it from the alternative antitone definition. One can also define Galois connections as a pair of monotone functions that satisfy the laxer condition that for all x in A, xg(f (x)) and for all y in B, f(g(y)) ≤ y.\n2. Gierz, p. 23\n3. Bistarelli, Stefano (2004). Semirings for Soft Constraint Solving and Programming. Lecture Notes in Computer Science. 2962. Springer-Verlag. p. 102. ISBN 3-540-21181-0. ISSN 0302-9743.\n4. Galatos, p. 145\n5. See Alperin, Bell, Groups and Representations (GTM 162), p. 32\n6. William Lawvere, Adjointness in foundations, Dialectica, 1969, available here. The notation is different nowadays; an easier introduction by Peter Smith in these lecture notes, which also attribute the concept to the article cited.\n7. Patrick Cousot; Radhia Cousot (Jan 1977). \"Abstract Interpretation: A Unified Lattice Model for Static Analysis of Programs by Construction or Approximation of Fixpoints\". Proc. 4th ACM Symp. on Principles of Programming Languages (POPL) (PDF). pp. 238–252.\nFor a counterexample for the false theorem in Sect.7 (p.243 top right), see: Jochen Burghardt; Florian Kammüller; Jeff W. Sanders (Dec 2000). Isomorphism of Galois Embeddings (Technical report). GMD. p. 73. ISSN 1435-2702. 122.\n8. Patrick Cousot; Radhia Cousot (Jan 1979). \"Systematic Design of Program Analysis Frameworks\". Proc. 6th ACM Symp. on Principles of Programming Languages (POPL) (PDF). ACM Press. pp. 269–282.\n\nReferences\n\nThe following books and survey articles include Galois connections using the monotone definition:\n\n• Brian A. Davey and Hilary A. Priestley: Introduction to lattices and Order, Cambridge University Press, 2002.\n• Gerhard Gierz, Karl H. Hofmann, Klaus Keimel, Jimmie D. Lawson, Michael W. Mislove, Dana S. Scott: Continuous Lattices and Domains, Cambridge University Press, 2003.\n• Marcel Erné, Jürgen Koslowski, Austin Melton, George E. Strecker, A primer on Galois connections, in: Proceedings of the 1991 Summer Conference on General Topology and Applications in Honor of Mary Ellen Rudin and Her Work, Annals of the New York Academy of Sciences, Vol. 704, 1993, pp. 103–125. (Freely available online in various file formats PS.GZ PS, it presents many examples and results, as well as notes on the different notations and definitions that arose in this area.)\n\nSome publications using the original (antitone) definition:\n\n• Mac Lane, Saunders (September 1998). Categories for the Working Mathematician (Second ed.). Springer. ISBN 0-387-98403-8.\n• Thomas Scott Blyth, Lattices and Ordered Algebraic Structures, Springer, 2005, ISBN 1-85233-905-5.\n• Nikolaos Galatos, Peter Jipsen, Tomasz Kowalski, and Hiroakira Ono (2007), Residuated Lattices. An Algebraic Glimpse at Substructural Logics, Elsevier, ISBN 978-0-444-52141-5.\n• Garrett Birkhoff: Lattice Theory, Amer. Math. Soc. Coll. Pub., Vol 25, 1940\n• Øystein Ore: Galois Connexions, Transactions of the American Mathematical Society 55 (1944), pp. 493–513" ]	[ null, "https://gateway.ipfs.io/ipfs/QmXoypizjW3WknFiJnKLwHCnL72vedxjQkDDP1mXWo6uco/I/m/9c5d0b666a10497100a6fb0a737ce630a2adc420.svg", null, "https://gateway.ipfs.io/ipfs/QmXoypizjW3WknFiJnKLwHCnL72vedxjQkDDP1mXWo6uco/I/m/b8a980c59d42c003fd07fdf3646e1fb95ff82f99.svg", null, "https://gateway.ipfs.io/ipfs/QmXoypizjW3WknFiJnKLwHCnL72vedxjQkDDP1mXWo6uco/I/m/ad31c567219f5c54190d2111920307aefc06cb4a.svg", null, "https://gateway.ipfs.io/ipfs/QmXoypizjW3WknFiJnKLwHCnL72vedxjQkDDP1mXWo6uco/I/m/d9389eb8d2e279b644c9dd665f40acabbbabb9fa.svg", null, "https://gateway.ipfs.io/ipfs/QmXoypizjW3WknFiJnKLwHCnL72vedxjQkDDP1mXWo6uco/I/m/5d6b6dd1b855af55079c427e0c482d9d84247db1.svg", null, "https://gateway.ipfs.io/ipfs/QmXoypizjW3WknFiJnKLwHCnL72vedxjQkDDP1mXWo6uco/I/m/0699cd24ae8b9539cec41c455ff08d8f9c237223.svg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.884376,"math_prob":0.963099,"size":20787,"snap":"2022-05-2022-21","text_gpt3_token_len":5240,"char_repetition_ratio":0.17384401,"word_repetition_ratio":0.052285794,"special_character_ratio":0.23303026,"punctuation_ratio":0.119931184,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99160886,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12],"im_url_duplicate_count":[null,1,null,null,null,1,null,1,null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-01-19T21:00:38Z\",\"WARC-Record-ID\":\"<urn:uuid:fd9f177a-57a6-4d6a-a462-2809ca9deaa9>\",\"Content-Length\":\"67186\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:3084edae-6817-4138-877c-556bb2d21c5a>\",\"WARC-Concurrent-To\":\"<urn:uuid:9efcd3cc-3bbd-4f6e-a6a6-3f46a69ed780>\",\"WARC-IP-Address\":\"209.94.90.1\",\"WARC-Target-URI\":\"https://gateway.ipfs.io/ipfs/QmXoypizjW3WknFiJnKLwHCnL72vedxjQkDDP1mXWo6uco/wiki/Galois_connection.html\",\"WARC-Payload-Digest\":\"sha1:Y2NRPTPCURJ3X6MUMAURPOS7QAPTGTYB\",\"WARC-Block-Digest\":\"sha1:R5CMOVNNXTIPO6LKPQR5DY2HPTRLEBLA\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-05/CC-MAIN-2022-05_segments_1642320301488.71_warc_CC-MAIN-20220119185232-20220119215232-00433.warc.gz\"}"}
https://wiki.eigenvector.com/index.php?title=Alignpeaks	[ "# Alignpeaks\n\n## Contents\n\n### Purpose\n\nCalibrates wavelength scale using standard peaks.\n\n### Synopsis\n\ns = alignpeaks(x0,x1,ax,options)\ny = alignpeaks(s,y1)\n\n### Description\n\nALIGNPEAKS calibrates a wavelength scale using standard peak positions. Ideally, the axis scale (x0) would apply to a single instrument at time t = 0 and t > 0 or for two different instruments. However, (x1) at t > 0 doesn't typically match (x0) at t = 0 even though the numbers in the scales are identical. The result is that a plot of (x0,y0) and (x0,y1) appear shifted from one another.\n\nThe inputs to ALIGNPEAKS are (x0) a 1xK vector containing the axis locations of K peaks on the standard instrument at t = 0 (e.g., the true wavelengths), (x1) a 1xK vector containing the axis locations of the corresponding peaks on the field / test instrument at t > 0 (e.g., the peak positions on the field instrument), and ax a 1xN vector containing the axis scale where N > K. ALIGNPEAKS finds a polynomial fit between (x0) and (x1) and outputs the result in the structure array (s). The output (y) is a fit of (x1).\n\n### Options\n\nOptional input (options) is a structure array with the following fields:\n\n• name: 'options', name indicating that this is an options structure,\n• plots: [ 'none' \| {'final'} ] governs level of plotting, and\n• order: [ {2} ] integer giving the polynomial order.\n\nExecuting options = alignpeaks('options'); gives an empty options structure.\n\n### Example\n\nA measurements at t = 0 gives a spectrum (y0) with axis (ax), and measurements at t > 0 of the same sample yields a spectrum (y1) with the same axis (ax) but with peaks shifted. Therefore\n\nplot(ax,y0,'b',ax,y1,'r')\n\nshows a shift in the peaks. The peak positions at t = 0 are listed in (x0) and the peak positions at t > 0 are listed in (x1). The polynomial fit is given by\n\ns = alignpeaks(x0,x1,ax);\n\nand the transformed spectrum is obtained with\n\ny10 = alignpeaks(s,y1);\n\nso that\n\nplot(ax,y0,'b',ax,y1,'r')\n\nshows less of a peak shift. See alignpeaksdemo." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.81765044,"math_prob":0.98610353,"size":1968,"snap":"2021-31-2021-39","text_gpt3_token_len":546,"char_repetition_ratio":0.124236256,"word_repetition_ratio":0.02359882,"special_character_ratio":0.27794716,"punctuation_ratio":0.1392405,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99324846,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-07-26T20:24:30Z\",\"WARC-Record-ID\":\"<urn:uuid:494f7cee-f0cb-48dd-bc60-b3eaf34ed592>\",\"Content-Length\":\"17707\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:19a5e842-50ab-4441-b40a-40604b10b0dc>\",\"WARC-Concurrent-To\":\"<urn:uuid:7a8b7c40-7b97-4cbc-a2e0-90821dc70ffe>\",\"WARC-IP-Address\":\"69.163.163.60\",\"WARC-Target-URI\":\"https://wiki.eigenvector.com/index.php?title=Alignpeaks\",\"WARC-Payload-Digest\":\"sha1:JKUP2POHNFC2XKM2XQSHSFQFPLS5D3R7\",\"WARC-Block-Digest\":\"sha1:ANNIJKHSP3FI7ZXCWGKPDWYZH4NTOQTG\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-31/CC-MAIN-2021-31_segments_1627046152144.92_warc_CC-MAIN-20210726183622-20210726213622-00562.warc.gz\"}"}
https://www.basicitbangla.com/2020/09/adder.html	[ "## Latest:\n\nlatest\n\nAdder plays an important role in digital devices. The addition process of the digital circuits is performed by an adder.\n\nBy using adder all the mathematical procedures of computers are done. In much computational processing, adders are used in the ALU(Arithmetic Logic Unit ).\n\nKnow what are the subtractors in Digital Electronics.\n\nAdders are mainly two types. They are,\n\nIt adds two binary bits. It has no carry addition ability that's why it is called half adder.\n\nIt has the ability to add two bits and also it can add carry. That's why it is called a full adder.\n\n###", null, "For two single bits here we will get a sum and a carry as output for each input combination.\n\n A B Sum Carry 0 0 0 0 0 1 1 0 1 0 1 0 1 1 0 1\n\nA truth table for half adder\n\n###", null, "In this circuit, there is a system that can add two bits and also two carries. It is called the Full adder. Let, 3 input sign of a full adder is A, B, Ci(Ci is the carry input) and two outputs are S(Addition) and Co(Carry output).\n\nHere is a truth table for a full adder circuit.\n\n Input Output A B Ci S Co 0 0 0 0 0 0 0 1 1 0 0 1 0 1 0 0 1 1 0 1 1 0 0 1 0 1 0 1 0 1 1 1 0 0 1 1 1 1 1 1\n\nOutput equation\n\nS = ĀB̄Cᵢ +ĀBC̄ᵢ + AB̄C̄ᵢ+ABCᵢ\n\nC₀ = ĀBCᵢ+AB̄Cᵢ+ABC̄ᵢ+ABC\n\nAnd also the equation of output carry can be modified like this -\n\nFrom modified equations of the addition of full adder (S) and carry output (Cₒ) the circuit diagram using X-OR gate will be like this-\n\nFull adder representation by using the basic logical gates\n\nA full adder can be represented by using the basic logic gates only. The circuit diagram is given below-\n\nThe full adder reduces the complexity of a logical circuit.\n\nWe have to use two half adders and also we need an extra OR gate for adding the carries for designing a full adder using a half adder.\n\nLet, the input signal for full adder is X and Y, carry Cᵢ and outputs are S (Addition) Co(Carry).\n\nIn the figure, if we see for the first half adder-\n\nSum,\nS₁ = X⊕Y ----(1)\n\nCarry\n\nC₁ = XY------(2)\n\nSum,\n\nS₂ = S₁⊕Cᵢ\n= X⊕Y⊕Cᵢ [From (1)]\n= S\n\nNow,\n\nC₂ = S₁Cᵢ = (X⊕Y)Cᵢ-----(3)\n\nNow total carry(C₀) will be the addition of first half adder carry(C₁) and second half adder carry(C₂)\n\nSo,\n\nC₀ = C₁+C₂ = XY+(X⊕Y)Cᵢ [From (2) and (3) ]\n\nThis is the full adder carry.\nFrom here we get,\nSum of full adder output, S = X⊕Y⊕Cᵢ\nTotal carry of the full adder, C₀ = XY+(X⊕Y)Cᵢ\n\nLet the 3 input signals of full adder are X, Y, Cᵢ ( Cᵢ is the carry input) and outputs are S(Sum) Cₒ (Carry output). The truth table for the full adder is given below.\n\n Input Output X Y Cᵢ S Cₒ 0 0 0 0 0 0 0 1 1 0 0 1 0 1 0 0 1 1 0 1 1 0 0 1 0 1 0 1 0 1 1 1 0 0 1 1 1 1 1 1\n\nSo, considering all S = 1 we get,\n\nAgain for all Cₒ = 1, we get,\n\nHere is the diagram of the half adder to the full adder circuit." ]	[ null, "https://3.bp.blogspot.com/-EEO1TxoF_1o/XDA06xDCEhI/AAAAAAAAEaI/lPWQU-7SX6Ed9aGSJPSymdwCtaM_wLbyACPcBGAYYCw/s320/Half%2Badder%2Blogo%2Bfor%2Btwo%2Bsingle%2Bbits.png", null, "https://4.bp.blogspot.com/-l69H5h-zLeo/XHbL3hhGFCI/AAAAAAAAFTk/0vTRu0ef9NUfwNVte6z5NJHCFlEhewvcwCPcBGAYYCw/s1600/Full%2Badder.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8967004,"math_prob":0.9851646,"size":4681,"snap":"2021-31-2021-39","text_gpt3_token_len":1611,"char_repetition_ratio":0.1986316,"word_repetition_ratio":0.5146443,"special_character_ratio":0.2883999,"punctuation_ratio":0.07635009,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9959903,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-07-28T11:05:11Z\",\"WARC-Record-ID\":\"<urn:uuid:5b09170f-8e0a-4447-b36e-387bc3578863>\",\"Content-Length\":\"547357\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:14bfa7b5-264b-48b5-b317-14204f9b7bb8>\",\"WARC-Concurrent-To\":\"<urn:uuid:bf81be75-3cce-455a-b65f-88146faa7eff>\",\"WARC-IP-Address\":\"142.251.45.19\",\"WARC-Target-URI\":\"https://www.basicitbangla.com/2020/09/adder.html\",\"WARC-Payload-Digest\":\"sha1:UPEKCC6RGXEVGGEIMBHMBNX2VAENTCZV\",\"WARC-Block-Digest\":\"sha1:ZZ4GHGV32Y3KCJV37TF5K3JPHKWOB7BG\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-31/CC-MAIN-2021-31_segments_1627046153709.26_warc_CC-MAIN-20210728092200-20210728122200-00091.warc.gz\"}"}
https://docs.oracle.com/cd/B40099_02/books/eScript/eScript_JSReference189.html	[ "## Function Objects\n\nA Function object holds the definition of a function defined in eScript. In eScript, procedures are functions.\n\n##### Syntax A\n\nfunction funcName( [arg1 [, ..., argn]] )\n{\nbody\n}\n\n##### Syntax B\n\nvar funcName = new Function([arg1 [, ..., argn,]] body );\n\nParameter\nDescription\n\nfuncName\n\nThe name of the function to be created\n\narg1 [, ..., argn]\n\nAn optional list of parameters that the function accepts\n\nbody\n\nThe lines of code that the function executes\n\n##### Returns\n\nWhatever its code is set up to return. For more information, see return Statement.\n\n##### Usage\n\nSyntax A is the standard method for defining a function. Syntax B is an alternative way to create a function and is used to create Function objects explicitly.\n\nNote the difference in case of the keyword Function between Syntax A and Syntax B. Function objects created with Syntax B (that is, the Function constructor) are evaluated each time they are used. This is less efficient than Syntax A—declaring a function and calling it within your code—because declared functions are compiled instead of interpreted.\n\n##### Example\n\nThe following fragment of code illustrates creating a function AddTwoNumbers using a declaration:\n\n{\nreturn (a + b);\n}\n\nThe following fragment illustrates creating the same function using the Function constructor:\n\nAddTwoNumbers = new Function (\"a\", \"b\", \"return (a + b)\");\n\nThe difference between the two is that when AddTwoNumbers is created using a declaration, AddTwoNumbers is the name of a function, whereas when AddTwoNumbers is created using the Function constructor, AddTwoNumbers is the name of a variable whose current value is a reference to the function created using the Function constructor.\n\n#### length Property\n\nThe length property returns the number of parameters expected by the function.\n\n##### Syntax\n\nfuncName.length\n\nParameter\nDescription\n\nfuncName\n\nThe function whose length property is to be found\n\n##### Returns\n\nThe number of parameters expected by funcName.\n\n#### return Statement\n\nThe return statement passes a value back to the function that called it.\n\n##### Syntax\n\nreturn value\n\nParameter\nDescription\n\nvalue\n\nThe result produced by the function\n\n##### Usage\n\nThe return statement passes a value back to the function that called it. Any code in a function following the execution of a return statement is not executed.\n\n##### Example\n\nThis function returns a value equal to the number passed to it multiplied by 2 and divided by 5.\n\nfunction DoubleAndDivideBy5(a)\n{\nreturn (a2)/5\n}\n\nThe following code fragment show an example of a script using the preceding function. This script calculates the mathematical expression `n = (10 2) / 5 + (20 * 2) / 5. `It then displays the value for `n`, which is 12.\n\nfunction myFunction()\n{\nvar a = DoubleAndDivideBy5(10);\nvar b = DoubleAndDivideBy5(20);\nTheApplication().RaiseErrorText(a + b);\n}" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.71365994,"math_prob":0.9541752,"size":2825,"snap":"2021-43-2021-49","text_gpt3_token_len":630,"char_repetition_ratio":0.17263381,"word_repetition_ratio":0.060215052,"special_character_ratio":0.2261947,"punctuation_ratio":0.113168724,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98615617,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-28T11:58:53Z\",\"WARC-Record-ID\":\"<urn:uuid:f631c1cf-eab6-4e3f-9df2-04d5dd8562f1>\",\"Content-Length\":\"11114\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c74dcad2-6610-460d-8227-50cc2fe15c92>\",\"WARC-Concurrent-To\":\"<urn:uuid:a82f38c3-2c90-4a64-8544-6f1b068b7875>\",\"WARC-IP-Address\":\"104.121.83.87\",\"WARC-Target-URI\":\"https://docs.oracle.com/cd/B40099_02/books/eScript/eScript_JSReference189.html\",\"WARC-Payload-Digest\":\"sha1:6PMRXW7BOY5HRZBRIKFDJE2WROCU5RRQ\",\"WARC-Block-Digest\":\"sha1:MABPAJJ74O4NJ667C5Y42IADGCQ2YSIM\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323588284.71_warc_CC-MAIN-20211028100619-20211028130619-00243.warc.gz\"}"}
https://www.colorhexa.com/dbf7ce	[ "# #dbf7ce Color Information\n\nIn a RGB color space, hex #dbf7ce is composed of 85.9% red, 96.9% green and 80.8% blue. Whereas in a CMYK color space, it is composed of 11.3% cyan, 0% magenta, 16.6% yellow and 3.1% black. It has a hue angle of 101 degrees, a saturation of 71.9% and a lightness of 88.8%. #dbf7ce color hex could be obtained by blending #ffffff with #b7ef9d. Closest websafe color is: #ccffcc.\n\n• R 86\n• G 97\n• B 81\nRGB color chart\n• C 11\n• M 0\n• Y 17\n• K 3\nCMYK color chart\n\n#dbf7ce color description : Light grayish green.\n\n# #dbf7ce Color Conversion\n\nThe hexadecimal color #dbf7ce has RGB values of R:219, G:247, B:206 and CMYK values of C:0.11, M:0, Y:0.17, K:0.03. Its decimal value is 14415822.\n\nHex triplet RGB Decimal dbf7ce `#dbf7ce` 219, 247, 206 `rgb(219,247,206)` 85.9, 96.9, 80.8 `rgb(85.9%,96.9%,80.8%)` 11, 0, 17, 3 101°, 71.9, 88.8 `hsl(101,71.9%,88.8%)` 101°, 16.6, 96.9 ccffcc `#ccffcc`\nCIE-LAB 94.328, -16.382, 16.693 73.612, 86.037, 71.118 0.319, 0.373, 86.037 94.328, 23.389, 134.461 94.328, -13.718, 27.613 92.756, -20.664, 19.471 11011011, 11110111, 11001110\n\n# Color Schemes with #dbf7ce\n\n• #dbf7ce\n``#dbf7ce` `rgb(219,247,206)``\n• #eacef7\n``#eacef7` `rgb(234,206,247)``\nComplementary Color\n• #f0f7ce\n``#f0f7ce` `rgb(240,247,206)``\n• #dbf7ce\n``#dbf7ce` `rgb(219,247,206)``\n• #cef7d6\n``#cef7d6` `rgb(206,247,214)``\nAnalogous Color\n• #f7cef0\n``#f7cef0` `rgb(247,206,240)``\n• #dbf7ce\n``#dbf7ce` `rgb(219,247,206)``\n• #d6cef7\n``#d6cef7` `rgb(214,206,247)``\nSplit Complementary Color\n• #f7cedb\n``#f7cedb` `rgb(247,206,219)``\n• #dbf7ce\n``#dbf7ce` `rgb(219,247,206)``\n• #cedbf7\n``#cedbf7` `rgb(206,219,247)``\nTriadic Color\n• #f7eace\n``#f7eace` `rgb(247,234,206)``\n• #dbf7ce\n``#dbf7ce` `rgb(219,247,206)``\n• #cedbf7\n``#cedbf7` `rgb(206,219,247)``\n• #eacef7\n``#eacef7` `rgb(234,206,247)``\nTetradic Color\n• #abec8c\n``#abec8c` `rgb(171,236,140)``\n• #bbf0a2\n``#bbf0a2` `rgb(187,240,162)``\n• #cbf3b8\n``#cbf3b8` `rgb(203,243,184)``\n• #dbf7ce\n``#dbf7ce` `rgb(219,247,206)``\n• #ebfbe4\n``#ebfbe4` `rgb(235,251,228)``\n• #fbfefa\n``#fbfefa` `rgb(251,254,250)``\n• #ffffff\n``#ffffff` `rgb(255,255,255)``\nMonochromatic Color\n\n# Alternatives to #dbf7ce\n\nBelow, you can see some colors close to #dbf7ce. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #e5f7ce\n``#e5f7ce` `rgb(229,247,206)``\n• #e2f7ce\n``#e2f7ce` `rgb(226,247,206)``\n• #def7ce\n``#def7ce` `rgb(222,247,206)``\n• #dbf7ce\n``#dbf7ce` `rgb(219,247,206)``\n• #d8f7ce\n``#d8f7ce` `rgb(216,247,206)``\n• #d4f7ce\n``#d4f7ce` `rgb(212,247,206)``\n• #d1f7ce\n``#d1f7ce` `rgb(209,247,206)``\nSimilar Colors\n\n# #dbf7ce Preview\n\nText with hexadecimal color #dbf7ce\n\nThis text has a font color of #dbf7ce.\n\n``<span style=\"color:#dbf7ce;\">Text here</span>``\n#dbf7ce background color\n\nThis paragraph has a background color of #dbf7ce.\n\n``<p style=\"background-color:#dbf7ce;\">Content here</p>``\n#dbf7ce border color\n\nThis element has a border color of #dbf7ce.\n\n``<div style=\"border:1px solid #dbf7ce;\">Content here</div>``\nCSS codes\n``.text {color:#dbf7ce;}``\n``.background {background-color:#dbf7ce;}``\n``.border {border:1px solid #dbf7ce;}``\n\n# Shades and Tints of #dbf7ce\n\nA shade is achieved by adding black to any pure hue, while a tint is created by mixing white to any pure color. In this example, #010200 is the darkest color, while #f4fdf0 is the lightest one.\n\n• #010200\n``#010200` `rgb(1,2,0)``\n• #081203\n``#081203` `rgb(8,18,3)``\n• #0f2306\n``#0f2306` `rgb(15,35,6)``\n• #163409\n``#163409` `rgb(22,52,9)``\n• #1e450b\n``#1e450b` `rgb(30,69,11)``\n• #25560e\n``#25560e` `rgb(37,86,14)``\n• #2c6711\n``#2c6711` `rgb(44,103,17)``\n• #337814\n``#337814` `rgb(51,120,20)``\n• #3a8816\n``#3a8816` `rgb(58,136,22)``\n• #429919\n``#429919` `rgb(66,153,25)``\n• #49aa1c\n``#49aa1c` `rgb(73,170,28)``\n• #50bb1f\n``#50bb1f` `rgb(80,187,31)``\n• #57cc21\n``#57cc21` `rgb(87,204,33)``\nShade Color Variation\n• #5fdb25\n``#5fdb25` `rgb(95,219,37)``\n• #6cde36\n``#6cde36` `rgb(108,222,54)``\n• #78e147\n``#78e147` `rgb(120,225,71)``\n• #84e458\n``#84e458` `rgb(132,228,88)``\n• #91e669\n``#91e669` `rgb(145,230,105)``\n• #9de97a\n``#9de97a` `rgb(157,233,122)``\n• #a9ec8b\n``#a9ec8b` `rgb(169,236,139)``\n• #b6ef9b\n``#b6ef9b` `rgb(182,239,155)``\n• #c2f1ac\n``#c2f1ac` `rgb(194,241,172)``\n• #cff4bd\n``#cff4bd` `rgb(207,244,189)``\n• #dbf7ce\n``#dbf7ce` `rgb(219,247,206)``\n• #e7fadf\n``#e7fadf` `rgb(231,250,223)``\n• #f4fdf0\n``#f4fdf0` `rgb(244,253,240)``\nTint Color Variation\n\n# Tones of #dbf7ce\n\nA tone is produced by adding gray to any pure hue. In this case, #e2e3e2 is the less saturated color, while #d9fec7 is the most saturated one.\n\n• #e2e3e2\n``#e2e3e2` `rgb(226,227,226)``\n• #e1e5e0\n``#e1e5e0` `rgb(225,229,224)``\n• #e1e8dd\n``#e1e8dd` `rgb(225,232,221)``\n• #e0eadb\n``#e0eadb` `rgb(224,234,219)``\n• #dfecd9\n``#dfecd9` `rgb(223,236,217)``\n• #deeed7\n``#deeed7` `rgb(222,238,215)``\n• #ddf0d5\n``#ddf0d5` `rgb(221,240,213)``\n• #ddf3d2\n``#ddf3d2` `rgb(221,243,210)``\n• #dcf5d0\n``#dcf5d0` `rgb(220,245,208)``\n• #dbf7ce\n``#dbf7ce` `rgb(219,247,206)``\n• #daf9cc\n``#daf9cc` `rgb(218,249,204)``\n• #d9fbca\n``#d9fbca` `rgb(217,251,202)``\n• #d9fec7\n``#d9fec7` `rgb(217,254,199)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #dbf7ce is perceived by people affected by a color vision deficiency. This can be useful if you need to ensure your color combinations are accessible to color-blind users.\n\nMonochromacy\n• Achromatopsia 0.005% of the population\n• Atypical Achromatopsia 0.001% of the population\nDichromacy\n• Protanopia 1% of men\n• Deuteranopia 1% of men\n• Tritanopia 0.001% of the population\nTrichromacy\n• Protanomaly 1% of men, 0.01% of women\n• Deuteranomaly 6% of men, 0.4% of women\n• Tritanomaly 0.01% of the population" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.55767286,"math_prob":0.5264304,"size":3719,"snap":"2021-04-2021-17","text_gpt3_token_len":1670,"char_repetition_ratio":0.1243607,"word_repetition_ratio":0.011090573,"special_character_ratio":0.51815003,"punctuation_ratio":0.23549107,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9697914,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-04-18T14:22:17Z\",\"WARC-Record-ID\":\"<urn:uuid:9ed5ef73-9d02-4fd9-9ad6-282a055be3e0>\",\"Content-Length\":\"36353\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:7769b748-857d-415d-8f4f-8b253c8b48d8>\",\"WARC-Concurrent-To\":\"<urn:uuid:06c39ddb-2ee3-4c79-8115-a86fa515080d>\",\"WARC-IP-Address\":\"178.32.117.56\",\"WARC-Target-URI\":\"https://www.colorhexa.com/dbf7ce\",\"WARC-Payload-Digest\":\"sha1:OWZU57CCCTY6FZCVELD7M6XIFEX3NTY2\",\"WARC-Block-Digest\":\"sha1:HEKNFTHHML66VWIIOI56ZQCZPO2PFWI3\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-17/CC-MAIN-2021-17_segments_1618038492417.61_warc_CC-MAIN-20210418133614-20210418163614-00459.warc.gz\"}"}
http://sigma.ontologyportal.org:8080/sigma/TreeView.jsp?lang=EnglishLanguage&flang=traditionalLogic&kb=SUMO&term=date	[ "", null, "", null, "Browsing Interface : Welcome guest : log in [ Home \| Graph \| ] KB: SUMO Language: AlbanianLanguageArabicLanguageBasqueLanguageBulgarianLanguageCatalanLanguageChineseLanguageCroatianLanguageDanishLanguageEnglishLanguageFarsiLanguageFinnishLanguageFrenchLanguageGalicianLanguageGreekLanguageHebrewLanguageIcelandicLanguageIndonesianLanguageItalianLanguageJapaneseLanguageMalayLanguageNorwegianBokmalLanguageNorwegianNorskLanguagePolishLanguagePortugueseLanguageSpanishLanguageSwedishLanguageTaiwanChineseLanguageThaiLanguagede Formal Language: OWLSUO-KIFTPTPtraditionalLogic\n\n KB Term:", null, "Term intersection", null, "English Word:", null, "Any Noun Verb Adjective Adverb", null, "", null, "", null, "date", null, "Sigma KEE - date\n date\n\n appearance as argument number 1", null, "s__documentation(s__date__m,s__ChineseLanguage,'这是一个 BinaryPredicate，它以绝对日历时间，以一天为限，为特定的 Object 或 Process 指定一个 TimePosition。') Merge.kif 8660-8661 s__documentation(s__date__m,s__EnglishLanguage,'A BinaryPredicate that specifies a TimePosition in absolute calendar time, at the resolution of one day, for a particular Object or Process.') Merge.kif 8657-8659 s__domain(s__date__m,1,s__Physical) Merge.kif 8654-8654 The number 1 argument of date is an instance of physical s__domain(s__date__m,2,s__Day) Merge.kif 8655-8655 The number 2 argument of date is an instance of day s__instance(s__date__m,s__AsymmetricRelation) Merge.kif 8653-8653 date is an instance of asymmetric relation s__instance(s__date__m,s__BinaryPredicate) Merge.kif 8651-8651 date is an instance of binary predicate s__instance(s__SingleValuedRelation,s__SetOrClass) Merge.kif 8652-8652 date is an instance of single valued relation s__subrelation(s__date__m,s__time__m) Merge.kif 8656-8656 date is a subrelation of time\n\n appearance as argument number 2", null, "s__format(s__ChineseLanguage,s__date__m,'%1 的 date %n 是 %2') chinese_format.kif 441-441 s__format(s__EnglishLanguage,s__date__m,'date of %1 is %n %2') english_format.kif 266-266 s__termFormat(s__ChineseLanguage,s__date__m,'日期') chinese_format.kif 442-442 \"日期\" is the printable form of date in ChineseLanguage s__termFormat(s__EnglishLanguage,s__date__m,'date') domainEnglishFormat.kif 3522-3522 \"date\" is the printable form of date in english language\n\n antecedent", null, "( ∀ [V__TEXT,V__EDITION1,V__EDITION2,V__INT2,V__DATE1,V__INT1,V__DATE2,V__PUB1,V__PUB2,V__CBO1,V__CBO2] ((s__subclass(V__TEXT,s__ContentBearingObject)s__and__ms__subclass(V__EDITION1,s__ContentBearingObject)s__and__ms__instance(V__EDITION1,s__SetOrClass)s__and__ms__subclass(V__EDITION2,s__ContentBearingObject)s__and__ms__instance(V__EDITION2,s__SetOrClass)s__and__ms__instance(V__INT2,s__PositiveInteger)s__and__ms__instance(V__DATE1,s__Day)s__and__ms__instance(V__INT1,s__PositiveInteger)s__and__ms__instance(V__DATE2,s__Day)) s__=>(((s__EditionFn(V__TEXT,V__INT1) s__equal__mV__EDITION1)s__and__m(s__EditionFn(V__TEXT,V__INT2) s__equal__mV__EDITION2)s__and__ms__greaterThan(V__INT2,V__INT1) s__and__ms__instance(V__PUB1,s__Publication)s__and__ms__instance(V__PUB2,s__Publication)s__and__ms__instance(V__CBO1,V__EDITION1) s__and__ms__instance(V__CBO2,V__EDITION2) s__and__ms__patient(V__PUB1,V__CBO1) s__and__ms__patient(V__PUB2,V__CBO2) s__and__ms__date(V__PUB1,V__DATE1) s__and__ms__date(V__PUB2,V__DATE2)) s__=>s__before(s__EndFn(V__DATE1) ,s__EndFn(V__DATE2)))) ) Merge.kif 15632-15649 If edition a positive integer of a kind of content bearing object is equal to another kind of content bearing object and edition another positive integer of a kind of content bearing object is equal to a third kind of content bearing object and the other positive integer is greater than the positive integer and a process is an instance of publication and another process is an instance of publication and an entity is an instance of another kind of content bearing object and another entity is an instance of a third kind of content bearing object and the entity is a patient of the process and the other entity is a patient of the other process and date of the process is a day and date of the other process is another day,then the end of the day happens before the end of the other day ( ∀ [V__DateOfWithdrawal,V__MaturityDate,V__CD,V__Withdrawal] ((s__instance(V__DateOfWithdrawal,s__Day)s__and__ms__instance(V__MaturityDate,s__Day)) s__=>((s__instance(V__CD,s__CertificateOfDeposit)s__and__ms__maturityDate(V__CD,V__MaturityDate) s__and__ms__instance(V__Withdrawal,s__Withdrawal)s__and__ms__origin(V__Withdrawal,s__CurrencyFn(V__CD)) s__and__ms__date(V__Withdrawal,V__DateOfWithdrawal) s__and__ms__before(s__EndFn(V__DateOfWithdrawal) ,s__BeginFn(V__MaturityDate))) s__=>(s__exists__m[V__Penalty] (s__instance(V__Penalty,s__Penalty)s__and__ms__destination(V__Penalty,s__CurrencyFn(V__CD)) s__and__ms__causes(V__Withdrawal,V__Penalty))))) ) FinancialOntology.kif 1032-1044 If a financial account is an instance of certificate of deposit and a day is a maturity date of the financial account and a process is an instance of withdrawal and the process originates at the currency of the financial account and date of the process is another day and the end of the other day happens before the beginning of the day,then there exists another process such that the other process is an instance of penalty and the other process ends up at the currency of the financial account and the process causes the other process ( ∀ [V__Amount,V__Day,V__Account,V__Amount1,V__Transaction,V__Cash] ((s__instance(V__Amount,s__CurrencyMeasure)s__and__ms__instance(V__Day,s__Day)s__and__ms__instance(V__Account,s__FinancialAccount)s__and__ms__instance(V__Amount1,s__Quantity)) s__=>((s__instance(V__Transaction,s__FinancialTransaction)s__and__ms__origin(V__Transaction,s__CurrencyFn(V__Account)) s__and__ms__transactionAmount(V__Transaction,V__Amount) s__and__ms__instance(V__Cash,s__Currency)s__and__ms__patient(V__Transaction,V__Cash) s__and__ms__date(V__Transaction,V__Day)) s__=>(s__exists__m[V__Amount2] (s__instance(V__Amount2,s__CurrencyMeasure)s__and__m(s__availableCash(V__Account,V__Day,V__Amount2) s__and__ms__greaterThanOrEqualTo(V__Amount1,V__Amount2)))))) ) FinancialOntology.kif 3675-3686 If a financial transaction is an instance of financial transaction and the financial transaction originates at the currency of a financial account and a currency measure is a transaction amount of the financial transaction and an entity is an instance of currency and the entity is a patient of the financial transaction and date of the financial transaction is a day,then there exists the currency measure2 such that the available cash of the financial account on the day is the currency measure2 and the currency measure1 is greater than or equal to the currency measure2 ( ∀ [V__Amount,V__Day,V__Account,V__Amount1,V__Transaction] ((s__instance(V__Amount,s__CurrencyMeasure)s__and__ms__instance(V__Day,s__Day)s__and__ms__instance(V__Account,s__FinancialAccount)s__and__ms__instance(V__Account,s__Object)s__and__ms__instance(V__Amount1,s__Quantity)) s__=>((s__instance(V__Transaction,s__FinancialTransaction)s__and__ms__origin(V__Transaction,V__Account) s__and__ms__transactionAmount(V__Transaction,V__Amount) s__and__ms__date(V__Transaction,V__Day)) s__=>(s__exists__m[V__Amount2] (s__instance(V__Amount2,s__CurrencyMeasure)s__and__m(s__availableBalance(V__Account,V__Day,V__Amount2) s__and__ms__greaterThanOrEqualTo(V__Amount1,V__Amount2)))))) ) FinancialOntology.kif 3655-3664 If a financial transaction is an instance of financial transaction and the financial transaction originates at a financial account and a currency measure is a transaction amount of the financial transaction and date of the financial transaction is a day,then there exists the currency measure2 such that the available balance of the financial account on the day is the currency measure2 and the currency measure1 is greater than or equal to the currency measure2 ( ∀ [V__TEXT,V__INT2,V__DATE1,V__INT1,V__DATE2,V__VOLUME2,V__VOLUME1,V__PUB1,V__PUB2,V__CBO1,V__CBO2] ((s__subclass(V__TEXT,s__Series)s__and__ms__instance(V__TEXT,s__SetOrClass)s__and__ms__instance(V__INT2,s__PositiveInteger)s__and__ms__instance(V__DATE1,s__Day)s__and__ms__instance(V__INT1,s__PositiveInteger)s__and__ms__instance(V__DATE2,s__Day)s__and__ms__subclass(V__VOLUME2,s__Text)s__and__ms__instance(V__VOLUME2,s__SetOrClass)s__and__ms__subclass(V__VOLUME1,s__Text)s__and__ms__instance(V__VOLUME1,s__SetOrClass)) s__=>((s__subclass(V__TEXT,s__Periodical)s__and__m(s__SeriesVolumeFn(V__TEXT,V__INT1) s__equal__mV__VOLUME1)s__and__m(s__SeriesVolumeFn(V__TEXT,V__INT2) s__equal__mV__VOLUME2)s__and__ms__greaterThan(V__INT2,V__INT1) s__and__ms__instance(V__PUB1,s__Publication)s__and__ms__instance(V__PUB2,s__Publication)s__and__ms__instance(V__CBO1,V__VOLUME1) s__and__ms__instance(V__CBO2,V__VOLUME2) s__and__ms__patient(V__PUB1,V__CBO1) s__and__ms__patient(V__PUB2,V__CBO2) s__and__ms__date(V__PUB1,V__DATE1) s__and__ms__date(V__PUB2,V__DATE2)) s__=>s__before(s__EndFn(V__DATE1) ,s__EndFn(V__DATE2)))) ) Mid-level-ontology.kif 12368-12382 If a kind of series is a subclass of periodical and volume a positive integer in the series a kind of series is equal to a kind of text and volume another positive integer in the series the kind of series is equal to another kind of text and the other positive integer is greater than the positive integer and a process is an instance of publication and another process is an instance of publication and an entity is an instance of a kind of text and another entity is an instance of another kind of text and the entity is a patient of the process and the other entity is a patient of the other process and date of the process is a day and date of the other process is another day,then the end of the day happens before the end of the other day ( ∀ [V__Amount,V__Date,V__Payment,V__DueDate,V__AmountDue,V__Account] ((s__instance(V__Amount,s__CurrencyMeasure)s__and__ms__instance(V__Date,s__Day)s__and__ms__instance(V__Payment,s__FinancialTransaction)s__and__ms__instance(V__DueDate,s__TimeInterval)s__and__ms__instance(V__AmountDue,s__CurrencyMeasure)) s__=>((s__transactionAmount(V__Payment,V__Amount) s__and__ms__date(V__Payment,V__Date) s__and__ms__instance(V__Account,s__FinancialAccount)s__and__ms__destination(V__Payment,s__CurrencyFn(V__Account)) s__and__ms__amountDue(V__Account,V__AmountDue,V__DueDate) s__and__ms__before(s__EndFn(V__Date) ,s__BeginFn(V__DueDate))) s__=>s__instance(V__Payment,s__Prepayment))) ) FinancialOntology.kif 779-787 If a currency measure is a transaction amount of a financial transaction and date of the financial transaction is a day and a financial account is an instance of financial account and the financial transaction ends up at the currency of the financial account and the financial account amount due the currency measureDue for a time interval and the end of the day happens before the beginning of the time interval,then the financial transaction is an instance of prepayment\n\n consequent", null, "No TPTP formula. May not be expressible in strict first order. FinancialOntology.kif 615-627 If a financial account amount due a currency measure for a time position and a cognitive agent holds account the financial account,then the cognitive agent is obliged to perform tasks of type the class described by a symbolic string ( ∀ [V__Amount,V__Loan,V__Date] ((s__instance(V__Amount,s__CurrencyMeasure)s__and__ms__instance(V__Loan,s__Loan)s__and__ms__instance(V__Date,s__TimePoint)s__and__ms__instance(V__Date,s__Day)) s__=>((s__downPayment(V__Loan,V__Amount) s__and__ms__agreementEffectiveDate(V__Loan,V__Date)) s__=>(s__exists__m[V__Payment] (s__instance(V__Payment,s__FinancialTransaction)s__and__m(s__transactionAmount(V__Payment,V__Amount) s__and__ms__date(V__Payment,V__Date) s__and__ms__destination(V__Payment,s__CurrencyFn(V__Loan))))))) ) FinancialOntology.kif 757-765 If a currency measure is a down payment of a loan and a time point is an agreement effective date of the loan,then there exists a financial transaction such that the currency measure is a transaction amount of the financial transaction and date of the financial transaction is the time point and the financial transaction ends up at the currency of the loan ( ∀ [V__Amount,V__Date,V__Balance,V__Account,V__FinancialAccount] ((s__instance(V__Amount,s__CurrencyMeasure)s__and__ms__instance(V__Date,s__TimePoint)s__and__ms__instance(V__Date,s__Day)s__and__ms__instance(V__Balance,s__CurrencyMeasure)s__and__ms__instance(V__Account,s__FinancialAccount)s__and__ms__instance(V__FinancialAccount,s__SetOrClass)) s__=>((s__instance(V__Account,V__FinancialAccount) s__and__ms__minimumBalance(V__Account,s__OpeningAnAccount,V__Balance)) s__=>(s__exists__m[V__Payment] (s__instance(V__Payment,s__FinancialTransaction)s__and__m(s__destination(V__Payment,s__CurrencyFn(V__Account)) s__and__ms__transactionAmount(V__Payment,V__Amount) s__and__ms__greaterThanOrEqualTo(V__Amount,V__Balance) s__and__ms__agreementEffectiveDate(V__Account,V__Date) s__and__ms__date(V__Payment,V__Date)))))) ) FinancialOntology.kif 646-656 If a financial account is an instance of a set or class and the financial account minimum balance opening an account for a currency measure,then there exists a financial transaction such that the financial transaction ends up at the currency of the financial account and another currency measure is a transaction amount of the financial transaction and the other currency measure is greater than or equal to the currency measure and a time point is an agreement effective date of the financial account and date of the financial transaction is the time point ( ∀ [V__Amount,V__Date,V__Balance,V__Account,V__FinancialAccount] ((s__instance(V__Amount,s__CurrencyMeasure)s__and__ms__instance(V__Date,s__Day)s__and__ms__instance(V__Balance,s__CurrencyMeasure)s__and__ms__instance(V__Account,s__FinancialAccount)s__and__ms__instance(V__FinancialAccount,s__SetOrClass)) s__=>((s__instance(V__Account,V__FinancialAccount) s__and__ms__minimumBalance(V__Account,s__UsingAnAccount,V__Balance) s__and__ms__currentAccountBalance(V__Account,V__Date,V__Amount) s__and__ms__lessThan(V__Amount,V__Balance)) s__=>(s__exists__m[V__Penalty] (s__instance(V__Penalty,s__Penalty)s__and__ms__date(V__Penalty,V__Date) s__and__ms__destination(V__Penalty,s__CurrencyFn(V__Account)))))) ) FinancialOntology.kif 658-668 If a financial account is an instance of a set or class and the financial account minimum balance using an account for a currency measure and the financial account current account balance a day for another currency measure and the other currency measure is less than the currency measure,then there exists a process such that the process is an instance of penalty and date of the process is the day and the process ends up at the currency of the financial account No TPTP formula. May not be expressible in strict first order. FinancialOntology.kif 1353-1363 If a loan is an instance of balloon loan and a day is a maturity date of the loan and a currency measure is a total balance of the loan and the loan is the borrower of a cognitive agent,then the cognitive agent is obliged to perform tasks of type the class described by a symbolic string ( ∀ [V__P,V__D] ((s__instance(V__P,s__Human)s__and__ms__instance(V__D,s__Day)) s__=>(s__birthdate(V__P,V__D) s__=>(s__exists__m[V__B] (s__instance(V__B,s__Birth)s__and__ms__experiencer(V__B,V__P) s__and__ms__date(V__B,V__D))))) ) Mid-level-ontology.kif 21974-21980 If a day is a birthdate of a human,then there exists a process such that the process is an instance of birth and the human experiences the process and date of the process is the day ( ∀ [V__Date,V__Statement] ((s__instance(V__Date,s__Day)s__and__ms__instance(V__Statement,s__BankStatement)) s__=>(s__dateOfStatement(V__Statement,V__Date) s__=>(s__exists__m[V__Copy] (s__instance(V__Copy,s__ContentBearingPhysical)s__and__m(s__containsInformation(V__Copy,V__Statement) s__and__ms__date(V__Copy,V__Date)))))) ) FinancialOntology.kif 3867-3872 If a day is a date of statement of a bank statement,then there exists a content bearing physical such that the content bearing physical contains information the bank statement and date of the content bearing physical is the day ( ∀ [V__P,V__T] ((s__instance(V__P,s__Human)s__and__ms__instance(V__T,s__Day)) s__=>(s__deathdate(V__P,V__T) s__=>(s__exists__m[V__D] (s__instance(V__D,s__Death)s__and__ms__experiencer(V__D,V__P) s__and__ms__date(V__D,V__T))))) ) Mid-level-ontology.kif 21988-21994 If a day is a deathdate of a human,then there exists a process such that the process is an instance of death and the human experiences the process and date of the process is the day", null, "Show simplified definition (without tree view)\nShow simplified definition (with tree view)", null, "Sigma web home Suggested Upper Merged Ontology (SUMO) web home\nSigma version 2.99c (>= 2017/11/20) is open source software produced by Articulate Software and its partners" ]	[ null, "http://sigma.ontologyportal.org:8080/sigma/pixmaps/sigmaSymbol-gray.gif", null, "http://sigma.ontologyportal.org:8080/sigma/pixmaps/logoText-gray.gif", null, "http://sigma.ontologyportal.org:8080/sigma/pixmaps/1pixel.gif", null, "http://sigma.ontologyportal.org:8080/sigma/pixmaps/1pixel.gif", null, "http://sigma.ontologyportal.org:8080/sigma/pixmaps/1pixel.gif", null, "http://sigma.ontologyportal.org:8080/sigma/pixmaps/1pixel.gif", null, "http://sigma.ontologyportal.org:8080/sigma/pixmaps/trans.gif", null, "http://sigma.ontologyportal.org:8080/sigma/pixmaps/plus.gif", null, "http://sigma.ontologyportal.org:8080/sigma/pixmaps/1pixel.gif", null, "http://sigma.ontologyportal.org:8080/sigma/pixmaps/1pixel.gif", null, "http://sigma.ontologyportal.org:8080/sigma/pixmaps/1pixel.gif", null, "http://sigma.ontologyportal.org:8080/sigma/pixmaps/1pixel.gif", null, "http://sigma.ontologyportal.org:8080/sigma/pixmaps/1pixel.gif", null, "http://sigma.ontologyportal.org:8080/sigma/pixmaps/1pixel.gif", null, "http://sigma.ontologyportal.org:8080/sigma/pixmaps/1pixel.gif", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.53741443,"math_prob":0.96968174,"size":9747,"snap":"2019-35-2019-39","text_gpt3_token_len":3318,"char_repetition_ratio":0.28379348,"word_repetition_ratio":0.09295775,"special_character_ratio":0.33620602,"punctuation_ratio":0.22693726,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9857871,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-08-24T17:33:49Z\",\"WARC-Record-ID\":\"<urn:uuid:ff632022-f731-4597-88fe-ae9c6cb94180>\",\"Content-Length\":\"126121\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:613ee74d-0527-42fd-8828-590393679f92>\",\"WARC-Concurrent-To\":\"<urn:uuid:ab4f16f6-c67c-4ee6-ae27-1b068ead8ba7>\",\"WARC-IP-Address\":\"13.57.62.184\",\"WARC-Target-URI\":\"http://sigma.ontologyportal.org:8080/sigma/TreeView.jsp?lang=EnglishLanguage&flang=traditionalLogic&kb=SUMO&term=date\",\"WARC-Payload-Digest\":\"sha1:6HHACUXEVJY2EHPY2WXL7FFQDLXO6EVD\",\"WARC-Block-Digest\":\"sha1:H35J2LDXFJ3JWQFYTEY7ZMMZNAVUY2JF\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-35/CC-MAIN-2019-35_segments_1566027321351.87_warc_CC-MAIN-20190824172818-20190824194818-00497.warc.gz\"}"}
https://brilliant.org/problems/sudoku-graph/	[ "# Sudoku Graph\n\nSudoku can be seen as a graph coloring problem, where the squares of the grid are vertices and the numbers are colors that must be different if in the same row, column, or $3 \\times 3$ grid (such vertices in the graph are connected by an edge). The sudoku is then a graph of 81 vertices and chromatic number 9. How many edges does this 81-vertex graph have?\n\n×" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8747335,"math_prob":0.98920107,"size":508,"snap":"2022-27-2022-33","text_gpt3_token_len":118,"char_repetition_ratio":0.12896825,"word_repetition_ratio":0.13793103,"special_character_ratio":0.23425196,"punctuation_ratio":0.18584071,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99309635,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-06-30T08:18:16Z\",\"WARC-Record-ID\":\"<urn:uuid:77644fe8-e892-420a-9b23-2307131dfc68>\",\"Content-Length\":\"35124\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:27993bfb-fc44-44fd-94cc-3c4128ea4d1c>\",\"WARC-Concurrent-To\":\"<urn:uuid:f5a4c4c7-3d73-4d77-8ba8-b1f0818a47d0>\",\"WARC-IP-Address\":\"104.18.9.15\",\"WARC-Target-URI\":\"https://brilliant.org/problems/sudoku-graph/\",\"WARC-Payload-Digest\":\"sha1:5NGNFLWLWASON6HA6P2QTRLAJ5M47UX3\",\"WARC-Block-Digest\":\"sha1:MDEUH7AJXVTXYBJLZQRXDJGTVRLMGRTV\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656103669266.42_warc_CC-MAIN-20220630062154-20220630092154-00174.warc.gz\"}"}
https://www.engineering-society.com/2018/08/what-is-reinforcement-ratio.html	[ "", null, "What is Reinforcement Ratio\n\n# What is Reinforcement Ratio\n\nWhat is reinforcement ratio. Reinforcement ratio can be defined as the ratio of area of steel to the area of concrete. It has no units because its only the ratio of areas. Reinforcement ratio shows how much steel are used in any structural members. Reinforcement ratio have different ranges for different structural members. Some of these ranges are as:\n• Reinforcement ratio for beam is 1-2%\n• Reinforcement ratio for column is 1-8%\n• Reinforcement ratio for footing is 0.8-1%\n• Reinforcement ratio for slab is 1%\n\nContinue Reading about the \"What is Reinforcement Ratio\" on the next page below" ]	[ null, "https://www.facebook.com/tr", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.94501996,"math_prob":0.9308709,"size":529,"snap":"2023-40-2023-50","text_gpt3_token_len":119,"char_repetition_ratio":0.27809525,"word_repetition_ratio":0.0,"special_character_ratio":0.2173913,"punctuation_ratio":0.072164945,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9596148,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-29T00:46:10Z\",\"WARC-Record-ID\":\"<urn:uuid:5d297cad-a7ab-4703-be58-8cecaac7d260>\",\"Content-Length\":\"164472\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:fba49e94-b6ba-4e33-b0cb-eebca2731953>\",\"WARC-Concurrent-To\":\"<urn:uuid:746560d1-93e8-4f40-965c-561dffa75ab6>\",\"WARC-IP-Address\":\"172.253.63.121\",\"WARC-Target-URI\":\"https://www.engineering-society.com/2018/08/what-is-reinforcement-ratio.html\",\"WARC-Payload-Digest\":\"sha1:7FL4TBT77SCIV3CJQEBVHRRC7R4X45VG\",\"WARC-Block-Digest\":\"sha1:4FFUN2UM54RFDO6HK4QAIGBCG73MWAL4\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233510462.75_warc_CC-MAIN-20230928230810-20230929020810-00026.warc.gz\"}"}
https://www.colorhexa.com/44b03a	[ "# #44b03a Color Information\n\nIn a RGB color space, hex #44b03a is composed of 26.7% red, 69% green and 22.7% blue. Whereas in a CMYK color space, it is composed of 61.4% cyan, 0% magenta, 67% yellow and 31% black. It has a hue angle of 114.9 degrees, a saturation of 50.4% and a lightness of 45.9%. #44b03a color hex could be obtained by blending #88ff74 with #006100. Closest websafe color is: #339933.\n\n• R 27\n• G 69\n• B 23\nRGB color chart\n• C 61\n• M 0\n• Y 67\n• K 31\nCMYK color chart\n\n#44b03a color description : Dark moderate lime green.\n\n# #44b03a Color Conversion\n\nThe hexadecimal color #44b03a has RGB values of R:68, G:176, B:58 and CMYK values of C:0.61, M:0, Y:0.67, K:0.31. Its decimal value is 4501562.\n\nHex triplet RGB Decimal 44b03a `#44b03a` 68, 176, 58 `rgb(68,176,58)` 26.7, 69, 22.7 `rgb(26.7%,69%,22.7%)` 61, 0, 67, 31 114.9°, 50.4, 45.9 `hsl(114.9,50.4%,45.9%)` 114.9°, 67, 69 339933 `#339933`\nCIE-LAB 63.822, -53.4, 49.521 18.672, 32.584, 9.308 0.308, 0.538, 32.584 63.822, 72.827, 137.159 63.822, -48.393, 65.911 57.082, -41.505, 30.289 01000100, 10110000, 00111010\n\n# Color Schemes with #44b03a\n\n• #44b03a\n``#44b03a` `rgb(68,176,58)``\n• #a63ab0\n``#a63ab0` `rgb(166,58,176)``\nComplementary Color\n• #7fb03a\n``#7fb03a` `rgb(127,176,58)``\n• #44b03a\n``#44b03a` `rgb(68,176,58)``\n• #3ab06b\n``#3ab06b` `rgb(58,176,107)``\nAnalogous Color\n• #b03a7f\n``#b03a7f` `rgb(176,58,127)``\n• #44b03a\n``#44b03a` `rgb(68,176,58)``\n• #6b3ab0\n``#6b3ab0` `rgb(107,58,176)``\nSplit Complementary Color\n• #b03a44\n``#b03a44` `rgb(176,58,68)``\n• #44b03a\n``#44b03a` `rgb(68,176,58)``\n• #3a44b0\n``#3a44b0` `rgb(58,68,176)``\n• #b0a63a\n``#b0a63a` `rgb(176,166,58)``\n• #44b03a\n``#44b03a` `rgb(68,176,58)``\n• #3a44b0\n``#3a44b0` `rgb(58,68,176)``\n• #a63ab0\n``#a63ab0` `rgb(166,58,176)``\n• #2e7627\n``#2e7627` `rgb(46,118,39)``\n• #358a2d\n``#358a2d` `rgb(53,138,45)``\n• #3d9d34\n``#3d9d34` `rgb(61,157,52)``\n• #44b03a\n``#44b03a` `rgb(68,176,58)``\n• #4dc143\n``#4dc143` `rgb(77,193,67)``\n• #5fc756\n``#5fc756` `rgb(95,199,86)``\n• #71ce69\n``#71ce69` `rgb(113,206,105)``\nMonochromatic Color\n\n# Alternatives to #44b03a\n\nBelow, you can see some colors close to #44b03a. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #62b03a\n``#62b03a` `rgb(98,176,58)``\n• #58b03a\n``#58b03a` `rgb(88,176,58)``\n• #4eb03a\n``#4eb03a` `rgb(78,176,58)``\n• #44b03a\n``#44b03a` `rgb(68,176,58)``\n• #3ab03a\n``#3ab03a` `rgb(58,176,58)``\n• #3ab044\n``#3ab044` `rgb(58,176,68)``\n• #3ab04e\n``#3ab04e` `rgb(58,176,78)``\nSimilar Colors\n\n# #44b03a Preview\n\nThis text has a font color of #44b03a.\n\n``<span style=\"color:#44b03a;\">Text here</span>``\n#44b03a background color\n\nThis paragraph has a background color of #44b03a.\n\n``<p style=\"background-color:#44b03a;\">Content here</p>``\n#44b03a border color\n\nThis element has a border color of #44b03a.\n\n``<div style=\"border:1px solid #44b03a;\">Content here</div>``\nCSS codes\n``.text {color:#44b03a;}``\n``.background {background-color:#44b03a;}``\n``.border {border:1px solid #44b03a;}``\n\n# Shades and Tints of #44b03a\n\nA shade is achieved by adding black to any pure hue, while a tint is created by mixing white to any pure color. In this example, #050e05 is the darkest color, while #fefffe is the lightest one.\n\n• #050e05\n``#050e05` `rgb(5,14,5)``\n• #0b1c09\n``#0b1c09` `rgb(11,28,9)``\n• #112b0e\n``#112b0e` `rgb(17,43,14)``\n• #163a13\n``#163a13` `rgb(22,58,19)``\n• #1c4918\n``#1c4918` `rgb(28,73,24)``\n• #22571d\n``#22571d` `rgb(34,87,29)``\n• #276622\n``#276622` `rgb(39,102,34)``\n• #2d7527\n``#2d7527` `rgb(45,117,39)``\n• #33842b\n``#33842b` `rgb(51,132,43)``\n• #399230\n``#399230` `rgb(57,146,48)``\n• #3ea135\n``#3ea135` `rgb(62,161,53)``\n• #44b03a\n``#44b03a` `rgb(68,176,58)``\n• #4abf3f\n``#4abf3f` `rgb(74,191,63)``\n• #57c44d\n``#57c44d` `rgb(87,196,77)``\n• #65c95c\n``#65c95c` `rgb(101,201,92)``\n• #73ce6a\n``#73ce6a` `rgb(115,206,106)``\n• #81d379\n``#81d379` `rgb(129,211,121)``\n• #8fd888\n``#8fd888` `rgb(143,216,136)``\n• #9ddd97\n``#9ddd97` `rgb(157,221,151)``\n• #abe1a5\n``#abe1a5` `rgb(171,225,165)``\n• #b8e6b4\n``#b8e6b4` `rgb(184,230,180)``\n• #c6ebc3\n``#c6ebc3` `rgb(198,235,195)``\n• #d4f0d2\n``#d4f0d2` `rgb(212,240,210)``\n• #e2f5e0\n``#e2f5e0` `rgb(226,245,224)``\n• #f0faef\n``#f0faef` `rgb(240,250,239)``\n• #fefffe\n``#fefffe` `rgb(254,255,254)``\nTint Color Variation\n\n# Tones of #44b03a\n\nA tone is produced by adding gray to any pure hue. In this case, #717a70 is the less saturated color, while #17e604 is the most saturated one.\n\n• #717a70\n``#717a70` `rgb(113,122,112)``\n• #698367\n``#698367` `rgb(105,131,103)``\n• #628c5e\n``#628c5e` `rgb(98,140,94)``\n• #5a9555\n``#5a9555` `rgb(90,149,85)``\n• #539e4c\n``#539e4c` `rgb(83,158,76)``\n• #4ba743\n``#4ba743` `rgb(75,167,67)``\n• #44b03a\n``#44b03a` `rgb(68,176,58)``\n• #3db931\n``#3db931` `rgb(61,185,49)``\n• #35c228\n``#35c228` `rgb(53,194,40)``\n• #2ecb1f\n``#2ecb1f` `rgb(46,203,31)``\n• #26d416\n``#26d416` `rgb(38,212,22)``\n• #1fdd0d\n``#1fdd0d` `rgb(31,221,13)``\n• #17e604\n``#17e604` `rgb(23,230,4)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #44b03a is perceived by people affected by a color vision deficiency. This can be useful if you need to ensure your color combinations are accessible to color-blind users.\n\nMonochromacy\n• Achromatopsia 0.005% of the population\n• Atypical Achromatopsia 0.001% of the population\nDichromacy\n• Protanopia 1% of men\n• Deuteranopia 1% of men\n• Tritanopia 0.001% of the population\nTrichromacy\n• Protanomaly 1% of men, 0.01% of women\n• Deuteranomaly 6% of men, 0.4% of women\n• Tritanomaly 0.01% of the population" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.5867342,"math_prob":0.6695092,"size":3694,"snap":"2020-10-2020-16","text_gpt3_token_len":1660,"char_repetition_ratio":0.12086721,"word_repetition_ratio":0.011070111,"special_character_ratio":0.55766106,"punctuation_ratio":0.23344557,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9857192,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-04-03T06:01:49Z\",\"WARC-Record-ID\":\"<urn:uuid:94b09da0-71e5-4cb1-8318-1b405667bd61>\",\"Content-Length\":\"36280\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:32d2027d-6662-4949-9296-3ddbd46cf252>\",\"WARC-Concurrent-To\":\"<urn:uuid:04b1c9b8-17ad-426f-b13a-d13ed62658d7>\",\"WARC-IP-Address\":\"178.32.117.56\",\"WARC-Target-URI\":\"https://www.colorhexa.com/44b03a\",\"WARC-Payload-Digest\":\"sha1:LN4V6VO75VZ2DXGFXKJRI7Z34FVDCAPR\",\"WARC-Block-Digest\":\"sha1:2IF6NGFUCJLVNYJSQ36M5C4MOLBWYPII\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-16/CC-MAIN-2020-16_segments_1585370510287.30_warc_CC-MAIN-20200403030659-20200403060659-00262.warc.gz\"}"}
http://phlamethrower.co.uk/riscos/armforbeginners/1/index.htm	[ "ARM code for the truly petrified\n\nThe start of a new series by Matthew Bloch intended to squeeze every drop of speed from your programming.\n\nWhat do BASIC and ARM code programming have in common? When you turn your machine on, the ARM chip starts on the instruction at memory address zero, and the whole of RISC OS is initialised from there. After each instruction, the chip moves on four bytes, or 32 bits (the space needed to store a single ARM code instruction) and executes the next instruction, just like lines of BASIC.\n\nThe ARM chip has sixteen registers called R0-R15, and each register can store a 32-bit number. The sixteenth register is called the program counter, and it stores the address at which the ARM chip is currently executing. It is referred to for this reason as PC. BASIC has a limitless number of variables, but in ARM code we have to make careful use of registers, and store those we are not using in the memory for recall later.\n\nSo, here's an instruction: MOV R0,R1. This instruction moves the contents of register one (R1) into register zero (R0). The 'destination' register, the one that will be changed, always comes first in ARM code instructions. What the ARM chip sees is the code &E1A00001 stored in memory; the 'instruction' as we humans see it is only a longhand (mnemonic) so we dont have to work out these codes for ourselves. What an assembler (or compiler) does is to take these lists of longhand instructions and translate (compile) them into unintelligible machine code.\n\nBASIC, conveniently, has an assembler built into it. Once youve assembled your code, you can then set the PC to start running your compiled code, which means you can make the ARM chip do your bidding! Its as simple as reserving some memory in which to store your instructions, compile them there, and then run them. Note that not all four-byte codes are legitimate instructions; trying to execute these will give an 'undefined instructon' error. Other instructions include ADD, SUB(tract), MUL(tiply) and ORR (logical OR, with an extra R to make it three letters). All ARM instructions can be executed conditionally, that is to say, you can set an instruction to execute depending on the result of a previous one. Youll see examples of this below.\n\nSo, manipulating registers is all very well, but this seems a world away from writing 'real' programs to make graphics and sound. The answer is in the memory; certain areas of it are 'mapped' onto external devices such as the video controller. So if we write a byte into the screen memory, it will be displayed by the video controller. There is also a sound buffer where any bytes stored will be converted to an analogue signal and output through the speakers. Then on top of these crude methods, RISC OS provides us with a huge library of routines (introduced next month) to perform common tasks such as printing text, drawing garaphics and playing music without having to resort to writing directly to the hardware; in fact using system calls is the norm, since they cause fewer compatibility problems than direct hardware access.\n\nFast cube root\n\nTry this simple task: there's a square root function in BASIC, but there isn't a cube root, so write a function in BASIC to find the cube root of any positive whole number. It only has to find it to the nearest integer, don't worry about the fraction part. Chances are, those who aren't mathematicians (and I'm not) will have come up with something similar to figure 1, below. This function will always give an overestimate if the input value isn't an exact cube. Now let's see how fast this function is; try finding the cube of every number from 1 to 10,000. My RISC PC gives a time of 4.62 seconds, though you may be waiting up to ten times this with an ARM2. Either way its still faster than doing it by hand. Look at the 'CubeRoot' program on this month's cover disc to see how to do it in 0.2 seconds.\n\n Fig. 1 BASIC cube root function DEF FNcube_root(n%):check%=1:result%=0 WHILE(result%\n\nTo see how this is done, load the program into your favorite editor and take a look at how it's written: the PROCassemble function may look alien, but it demonstrates how to write a small ARM code routine with minimal effort. The form of the DIM statement may be confusing without brackets: what it does is to reserve 256 bytes of memory and store the start address of the memory in the 'code' variable. P% is a variable treated specially in BASIC; it tells the ARM code compiler where in memory to start compiling instructions. When you want to start assembling, the [ character turns BASIC into a compiler; rather than look for lines of BASIC, it will take each subsequent line as ARM code and assemble it at P% until the next ]. Anything inbetween should be code. The OPT switch should follow each [ character to set the various options for the compiler; OPT 2 is a straight compile, with errors reported and nothing else. Its uses will become clear later; for now just use it!\n\nThe ARM code is going to mimic the flow of the BASIC function closely, so lets use R0 as our n% (number to check), R1 as the number were currently checking (check%) and R2 to work the result of cubing R1 (result%). Throughout this discussion, look back and see how each line of ARM code compares with the BASCIC function; note that we only actually need space for three integer variables. We declare the start of the routine with a label, .cube_root. This marks the start of the routine and stores it in the BASIC variable cube_root, so we can 'CALL cube_root' later. CALL is a BASIC statement which executes an ARM code routine. It passes the integer variables A% to G% in registers R0 to R7, so we can pass paremeters easily. result% = USR(cube_root) will do the same thing, but USR is a function and will pass back R0 into a variable when the routine finishes. When we call an ARM code routine from BASIC, we need to be able to pass control back to BASIC when the routine is finished. What the interpreter does is pass the return address in R14, which is sometimes referred to as the link register. Then when we need to return to BASIC, we just copy the link register into R15 with MOV PC,R14; this is equivalent to GOTO in BASIC. You can see this instruction is the last one in the routine; if you dont include it, the processor will carry on romping through the memory executing instructions, data, and anything else it finds, probably resulting in a crash of some sort.\n\nWe must set R1 (check%) to zero before going into the main cube searching loop. If you want to load registers with integer variables, you can, subject to certain restrictions: MOV R1,#1 will set R1=1, though not ALL numbers can be loaded; more on this later. The top of this loop is marked out with .cube_search so we can jump back to this point easily. For each cycle, we:\n\n• Increase our counter R1 by 1.\n• Cube it, leaving the result in R2 (i.e. R2 = R1R1R1).\n• Check if its greater than or equal to the number to be cube-rooted (R0).\n• If R1 is smaller than R0, we loop around again until it isnt.\n• Otherwise, return with our value of R1 as the nearest cube root.\nThe routine does this process with a few straightforward instructions, listed and explained in figure 2. So, theres a simple ARM code routine in a few lines.\n\n Fig. 2 ARM code cube root run-through .cube_search this sets the BASIC variable cube_search to equal the address in memory of the next instruction ADD R1,R1,#1 means R1 = R1 + 1 (i.e. increase counter) MUL R2,R1,R1 means MULtiply, R2 = R1 * R1 MUL R2,R1,R2 means R2 = R1 * R2 since cubing has to be done with two MUL instructions) CMP R2,R0 stands for 'compare R2 with R0'. The result of this comparison is stored in the processor`s status flags (explained later). These status flags indicate (for the purposes of this routine) whether the last two numbers compared were greater than, equal to, or less than each other. In this case, we only want to carry on the loop if our result register is less than the number we`re checking (i.e. if R2 < R0), otherwise we know we`ve found the nearest cube root. BLT cube_search This is the branch instruction (B) with a condition code attached. A branch instruction makes the processor jump to another point in memory, usually marked out by an assembler label. You can attach a condition code to any instruction, so it will only be executed under certain conditions. Here, we see the LT condition code which stands for 'less than'. So this will only (B)ranch back to the top of the loop if R2 was (L)ess (T)han R0, as decided in the previous instruction. Some other condition codes include GT (greater than), GE (greater than or equal), and EQ (equal to). MOV R0,R1 R1 contains the cube root found, but R0 is the only register we can pass back to BASIC. The result of the USR(...) function in BASIC is the contents of R0 when the routine exits, so we need to move the result into R0 before exiting.\n\nWe can then find a cube root by setting :\n\nA%=number\n\n... and then ...\n\ncube_root%=USR(cube_root)\n\nas the BASIC does. Next month: a file processor and an introduction to SWIs.\n\n• Matthew Bloch" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.91471726,"math_prob":0.9050063,"size":8617,"snap":"2022-05-2022-21","text_gpt3_token_len":2015,"char_repetition_ratio":0.12632068,"word_repetition_ratio":0.0013029316,"special_character_ratio":0.23047464,"punctuation_ratio":0.1100397,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96689975,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-01-23T09:15:32Z\",\"WARC-Record-ID\":\"<urn:uuid:b27e0234-8de8-4cec-83dc-cdc21711a90b>\",\"Content-Length\":\"11608\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b90f5e8f-ad33-42c9-a908-7d3dcb43c3d3>\",\"WARC-Concurrent-To\":\"<urn:uuid:3233149e-c7cf-4c0d-8702-1966d50d9304>\",\"WARC-IP-Address\":\"82.68.159.212\",\"WARC-Target-URI\":\"http://phlamethrower.co.uk/riscos/armforbeginners/1/index.htm\",\"WARC-Payload-Digest\":\"sha1:IOSHHDKVYME2PGV5QLOAND3DNZBRRATG\",\"WARC-Block-Digest\":\"sha1:ATCHHBNI6TP6IFYUHJXQFUYX5OZ5YT6E\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-05/CC-MAIN-2022-05_segments_1642320304217.55_warc_CC-MAIN-20220123081226-20220123111226-00016.warc.gz\"}"}
https://www.teledynelecroy.com/doc/3-phase-power-app-note	[ "### Introduction\n\nPower analyzers have been manufactured for many years and are used in a wide variety of applications. Often, many settings and preferences can have significant impact on the measurements. Or, there are “legacy” capabilities that may not be well-understood. This technical brief explains the settings in the Teledyne LeCroy Motor Drive Analyzer (MDA) to achieve results consistent with a Yokogawa Power Analyzer, and to explain why the results will be different in some cases.\n\n### Measurement Period (“Sync”) Calculation\n\nFirst, for proper power analysis, one must determine a measurement period within which all calculations take place. Both the Teledyne LeCroy MDA and a Yokogawa Power Analyzer use a “Sync” signal to determine the measurement period. In both cases, the Sync signal can be low-pass filtered to reduce the chances of finding an incorrect measurement period. See the images below:\n\nFor a nearly sinusoidal (e.g. low distortion) signal, both instruments should find the same measurement period with the same LPF cutoff selections, though the Teledyne LeCroy MDA offers a wider variety of selections.\n\nThe Teledyne LeCroy Motor Drive Analyzer also permits adjustment of hysteresis (band), allowing user-defined control of the software’s ignoring of non-monotonicities that interfere with the measurement period calculation. This is explained in detail in the MDA’s software instruction manual. Such control can be very useful for signals that have higher distortion (e.g. brushless DC six-step commutated signals) or signals that have high distortion during high stress or failure events.\n\nNote that the Teledyne LeCroy Motor Drive Analyzer also permits display of the filtered Sync signal with a measurement period overlay, rendering the Sync signal settings easily understandable. See the screen image below (the Sync signal with the overlays appears at bottom).\n\nLastly, the Teledyne LeCroy Motor Drive Analyzer applies the Sync filter and hysteresis settings as a post-acquisition software process, while most power analyzers apply Sync filtering and hysteresis settings to the waveform during the acquisition. Thus, in the Teledyne LeCroy MDA, it is possible to make changes post-acquisition whereas this is not usually possible with a power analyzer instrument. Combined with the Sync period overlay, one may use the MDA’s post-acquisition processing to fine tune the filter and hysteresis settings for optimum results without having to take a new acquisition and lose the acquired data.\n\n### Apparent Power Formula\n\nIn Yokogawa’s Power Analyzer instrument, users may measure apparent power (S) through selection of any of the following formulas in their setup menu:\n\n1. $$Vrms * Irms$$ Same as provided by Teledyne LeCroy, as described in this document\n2. $$Vmean * Imean$$ Product of rectified mean values calibrated to the RMS values\n3. $$Vdc * Idc$$ Product of simple averages of the voltage and current\n4. $$Vmean * Irms$$ Product of the voltage’s rectified mean value and the current’s true RMS value\n5. $$Vrmean * Irmean$$ Product of the voltage’s and current’s rectified mean values\n\nTeledyne LeCroy only provides capability for $$VrmsIrms$$ calculation of apparent power. To correlate power values from the Yokogawa power analyzer to the Teledyne LeCroy MDA’s results, set the selection for apparent power calculation on the Power Analyzer to $$VrmsIrms$$.\n\n### Three-Phase (Total) Power Calculations\n\nTeledyne LeCroy employs one method for calculation of power values whereas Yokogawa provides three different methods that return different results. The following is a simplified summary of the Teledyne LeCroy method and the Yokogawa methods for calculating power for a single power cycle – see the respective instruction manuals for complete formulaic descriptions. Note that the table below assumes use of the $$VrmsIrms$$ formula in the Yokogawa Power Analyzer for Apparent Power calculations.\n\nTeledyne LeCroy Yokogawa Type 1 Yokogawa Type 2 Yokogawa Type 3\nReal Power “P” (in W) $$VI$$ (instantaneous sample points)\nApparent Power “S” (in VA) $$VrmsIrms$$ $$VrmsIrms$$ $$VrmsIrms$$ $$√(Q^2+ P^2 )$$\nReactive Power “Q” (in VAr) $$√(S^2- P^2 )$$ $$VrmsIrmssin⁡φ$$ $$√(S^2- P^2 )$$ $$VrmsIrmssin⁡φ$$\nPower Factor (λ) $$P⁄\|S\|$$ $$P⁄\|S\|$$ $$P⁄\|S\|$$ $$P⁄\|S\|$$\nPhase Angle (φ) $$cos^(-1)⁡λ$$ $$cos^(-1)⁡λ$$ $$cos^(-1)⁡λ$$ $$cos^(-1)⁡λ$$\n\nFor perfectly sinusoidal (zero-distortion) waveforms, it is possible to measure the phase angle φ between the voltage and current sinusoids. However, it is not possible to measure the phase angle φ when the waveforms are distorted (e.g. PWM drive output waveform). Therefore, the Teledyne LeCroy method or Yokogawa’s Type 2 method are the only methods that demonstrably produce accurate results for reactive power (and therefore power factor and phase angle) with distorted waveforms (both also produce accurate results with sinusoidal waveforms).\n\nYokogawa’s Power Analyzers offer the Type 3 method on models with the harmonic measurement-mode option. This option appears to enable definition of a fundamental signal from one of the PWM signals using a PLL source. Then, the instrument determines phase angle by comparing the fundamental voltage and fundamental current waveforms, with power determined for the fundamental and each harmonic through “N” harmonics. This should provide a similar result to values provided by the Teledyne LeCroy Harmonic Filter setting = “Fundamental” or “Fundamental + N” (see below), provided that the hardware PLL response in the Yokogawa power analyzer can accommodate any change in period of the measured signal during the acquisition window.\n\nNote that the Yokogawa power analyzer always calculates real power P correctly in all cases. If this is the only power value of interest, then all methods are suitable. However, to correctly calculate S, Q,λ, or φ, one must choose the correct Yokogawa power analyzer measurement method (if more than one is offered).\n\n### Per-Phase Power Calculations\n\nThe Yokogawa instrument calculates per-phase power using the same equations as total three-phase power, but on one phase at a time.\n\nThe Yokogawa Type 2 method with apparent power setting = $$VrmsIrms$$ always correlates (assuming there is no problem with the Yokogawa power analyzer obtaining a proper Sync period) with Teledyne LeCroy’s MDA when using three-phase, four-wire (three voltages, three currents) wiring configurations with line-neutral or line-reference voltage probing.\n\nHowever, if using line-line voltage probing, then per-phase power calculations in the Yokogawa power analyzer will be unbalanced and incorrect, even though the three-phase total will be correct. The reasons for the differences are as follows:\n\n1. The Yokogawa three-phase, three-wire (three voltages, three currents) wiring configuration is natively defined as a two-wattmeter setup. This is beneficial in that switching from a three-voltage and three-current measurement to a two-voltage and two-current measurement requires no re-connection of wires to the power analyzer. However, it also means that the line-line voltages and currents are incorrectly associated with each other on a per-phase basis. See Figure 1 (Yokogawa) below and compare to Figure 2 (Teledyne LeCroy).\n2. The Teledyne LeCroy three-phase, three-wire (three voltages, three currents) wiring configuration also uses a two-wattmeter method for total three-phase power calculations, but maintains the correct per-phase vector relationships to obtain proper per-phase calculations. Teledyne LeCroy simply inverts one of the voltage waveforms for the total three-phase power calculation.\n3. The voltage associations made by Yokogawa have no impact on the total three-phase power (real, apparent, or reactive), phase angle, or power factor because the vector (voltage and current) relationships defined in their wiring setup are the correct relationships for the two-wattmeter method used to calculate their total three-phase power.\n4. Both instruments can perform a line-line to line-neutral conversion (referred to as a delta-star conversion by Yokogawa, which is an extra-cost option). However, while Teledyne LeCroy’s MDA will return accurate per-phase power calculations for P, S, Q,λ, and φ with this type of conversion, the Yokogawa power analyzer instrument will only return power calculations for P.\n\nThe practical impact of the Yokogawa wiring configuration is that the voltage and current pairs all have different phase relationships (which is what leads to the incorrect per-phase power calculations). This is shown below on actual three-phase acquisitions:\n\n Yokogawa Teledyne LeCroy 3-phase voltage and current pairs, separated by pair", null, "", null, "3-phase voltage and current pairs, overlaid", null, "", null, "Calculated results appear as follows with and without a line-line to line-neutral (delta-star) conversion.\n\n Yokogawa Without Line-Line to Line-Neutral conversion (Delta-Star)", null, "With Line-Line to Line-Neutral conversion (Delta-Star)", null, "Teledyne LeCroy Without Line-Line to Line-Neutral conversion", null, "With Line-Line to Line-Neutral conversion", null, "" ]	[ null, "http://assets.lcry.net/images/9199_10.png", null, "http://assets.lcry.net/images/9199_11.png", null, "http://assets.lcry.net/images/9199_12.png", null, "http://assets.lcry.net/images/9199_13.png", null, "http://assets.lcry.net/images/9199_18.png", null, "http://assets.lcry.net/images/9199_19.png", null, "http://assets.lcry.net/images/9199_20.png", null, "http://assets.lcry.net/images/9199_21.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8762338,"math_prob":0.9743106,"size":11216,"snap":"2023-40-2023-50","text_gpt3_token_len":2487,"char_repetition_ratio":0.15617196,"word_repetition_ratio":0.041887905,"special_character_ratio":0.20167617,"punctuation_ratio":0.07765452,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9862354,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16],"im_url_duplicate_count":[null,5,null,5,null,5,null,5,null,5,null,5,null,5,null,5,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-22T20:49:47Z\",\"WARC-Record-ID\":\"<urn:uuid:64c085a2-de48-4560-87c9-c4491f249364>\",\"Content-Length\":\"53868\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a2f16489-e525-4fb5-8199-47a23cc4d92c>\",\"WARC-Concurrent-To\":\"<urn:uuid:9c94d72b-7ad5-4be5-a72d-bb9feddc0e5e>\",\"WARC-IP-Address\":\"104.18.2.26\",\"WARC-Target-URI\":\"https://www.teledynelecroy.com/doc/3-phase-power-app-note\",\"WARC-Payload-Digest\":\"sha1:ISTCDJY6TOI3WMOAFBBV4AP65CUO4BUH\",\"WARC-Block-Digest\":\"sha1:OUQVGABMJAZT7JLZAVOY2HYV3JU5EMV7\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233506423.70_warc_CC-MAIN-20230922202444-20230922232444-00090.warc.gz\"}"}
https://thinking-about-science.com/2019/07/13/19-17-computer-modelling-the-simple-harmonic-oscillator/	[ "# 19.17 Computer modelling – the simple harmonic oscillator\n\nBefore you read this, I suggest you read the first 8 paragraphs of post 18.6.\n\nIn previous posts, we have seen how differential equations (post 19.10) can be used to represent diffusion (post 19.15), the conduction of heat (post 19.15) and the behaviour of waves (post 19.12). We also saw how a differential equation could be used to describe the behaviour of a simple harmonic oscillator – a system that is constantly exchanging potential and kinetic energy with negligible energy loss (post 18.6). However, there is a completely different way of investigating systems like this, without developing differential equations.", null, "In this post, I will illustrate this different approach by using it to investigate a simple harmonic oscillator. In the table above, a mechanical simple harmonic oscillator starts (n = 1) with potential energy U1 = 1.00 J (post 16.21). If we think of this oscillator being restored towards its undisturbed state by a spring, with spring constant k (post 16.49), then\n\nU = (k/2)x2\n\nwhere x is its displacement (post 16.49). So the initial displacement of the oscillator in the table is\n\nx1 = (2U1/k)1/2.\n\nIf k = 200 N.m-1 then\n\nx1 = (2 × 1/200)1/2 = 0.1 m.\n\nAt this initial position (n = 1), its kinetic energy, K1 (post), and, therefore, its speed, are both zero. We use this initial state to define a time t = 0.\n\nI have used Microsoft Excel to create the numbers in the table to 6 decimal places – the results do not reflect the precision with which the results can be calculated; if this comment isn’t clear – see post 16.7.\n\nNow we consider a later state (n = 2) when the oscillator has lost 0.01 J of potential energy, so that now U2 = 0.99 J. Using the same approach as previously, its position is\n\nx2 = (2 × 0.99/200)1/2 = 0.099499 m.\n\nThis “lost” potential energy is converted to 0.01 J of kinetic energy, so now K2 = 0.01 J. For an oscillator of mass m with speed v\n\nK = (m/2)v2\n\n(post 16.21). If our oscillator has a mass of 1 kg, its speed is then\n\nv2 = (2K2/m)1/2 = (2 × 0.01/1)1/2 = 0.141421 m.s-1.\n\nDuring the time that the 0.01 J of potential energy was being transferred into kinetic energy, the displacement of the oscillator was (x2x1) and its average speed was (v1 + v2)/2 so that the time interval for the exchange of energy was\n\nΔt = 2(x2x1)/ (v1 + v2) = 2(0.099499 + 0.1)/(0 + 0.141421) = 0.007089 s.\n\n(see post 16.12). Now, the total time that has elapsed is\n\nt2 = t1 + Δt = 0 + 0.007089 s.\n\nIn the table above, I have continued this process for stages n = 3, 4, 5, 6….. and so on. Below, you can see what happens when n is about 100.", null, "", null, "When n = 101, all the potential energy has been transformed into kinetic energy. Then the displacement, x, is in the opposite direction (it has a minus sign) and kinetic energy starts to be transformed into potential energy. As before, I calculate the effects of this transformation in steps of 0.01 J.", null, "I will continue the calculations until a time of 0.886 s has elapsed; the graph above then shows the displacement, x, plotted against time t. This graph shows that the behaviour of our oscillator repeats every 0.444 s – it has a time period (post 16.14) T = 0.444 s. (it is easier to see the exact result from a table, like the examples above.)\n\nWe could have obtained the same result by setting up the differential equation for an object bouncing on a spring and then solving it to obtain the result\n\nT = 2π(m/k)1/2\n\nas in post 18.11. Then, when m = 1 kg and k = 200 N.m-1,\n\nT = 2π(1/200)1/2 = 0.444 s.\n\nThe advantage of this approach is that it gives a general result that we can use for any value of k and m.\n\nThe method used in this post is simpler, because it uses only a few simple definitions and some simple arithmetic. It shows exactly what happens when m = 1 kg and k = 200 N.m-1 but cannot provide any general results. Sometimes, especially in engineering, a specific result is all that is required then this numerical or computational approach provides everything that is needed. This approach can also be useful when it is too difficult to set up or solve a differential equation.\n\nUsually, I prefer the differential equation approach – not only because it provides a more general result but also because it can help us to understand how we get that result.\n\nRelated posts" ]	[ null, "https://thinkingaboutsciencedotcom.files.wordpress.com/2019/07/fig1.jpg", null, "https://thinkingaboutsciencedotcom.files.wordpress.com/2019/07/fig-2a.jpg", null, "https://thinkingaboutsciencedotcom.files.wordpress.com/2019/07/fig2.jpg", null, "https://thinkingaboutsciencedotcom.files.wordpress.com/2019/07/fig3.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9079376,"math_prob":0.99492574,"size":4696,"snap":"2021-43-2021-49","text_gpt3_token_len":1265,"char_repetition_ratio":0.12510656,"word_repetition_ratio":0.011792453,"special_character_ratio":0.30174616,"punctuation_ratio":0.13047712,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9988819,"pos_list":[0,1,2,3,4,5,6,7,8],"im_url_duplicate_count":[null,4,null,4,null,4,null,4,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-23T21:12:00Z\",\"WARC-Record-ID\":\"<urn:uuid:8e5fb306-b6a4-43c5-8250-e9dc6574b51d>\",\"Content-Length\":\"119727\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:72c225f0-de93-46db-bb47-d7f1fee27d28>\",\"WARC-Concurrent-To\":\"<urn:uuid:328a6160-9c78-4fac-8149-0dfcdfbc3f6d>\",\"WARC-IP-Address\":\"192.0.78.25\",\"WARC-Target-URI\":\"https://thinking-about-science.com/2019/07/13/19-17-computer-modelling-the-simple-harmonic-oscillator/\",\"WARC-Payload-Digest\":\"sha1:LNMRDMK7IRTU23G34PIVOWKW45YHG25W\",\"WARC-Block-Digest\":\"sha1:TLMU6L3BZ3I5AP5CKVXTDAERFS4GH5SE\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323585768.3_warc_CC-MAIN-20211023193319-20211023223319-00409.warc.gz\"}"}
https://zbmath.org/1082.35140	[ "## Existence and uniqueness of the solution to the modified Schrödinger map.(English)Zbl 1082.35140\n\nThe author studies the initial value problem for a system of nonlinear Schrödinger equations in two space dimensions (modified Schrödinger map) which is derived from Schrödinger maps from $$\\mathbb{R}\\times \\mathbb{R}^2$$ to the unit sphere $$S^2$$ or to the hyperbolic space $$\\mathbb{H}^2$$ by using appropriate gauge change. The existence and uniqueness of the solution was known for data in $$H^s(\\mathbb{R}^2)$$ with $$s> 1$$. In this paper the local existence of the solution is proved for the initial data in $$H^s(\\mathbb{R}^2)$$ with $$s> 1/2$$. The uniqueness of the solution is also proved when the data belong to $$H^1(\\mathbb{R}^2)$$.\n\n### MSC:\n\n 35Q55 NLS equations (nonlinear Schrödinger equations) 35G25 Initial value problems for nonlinear higher-order PDEs\nFull Text:" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.75011116,"math_prob":0.99947697,"size":1195,"snap":"2023-14-2023-23","text_gpt3_token_len":382,"char_repetition_ratio":0.12846348,"word_repetition_ratio":0.035714287,"special_character_ratio":0.2995816,"punctuation_ratio":0.12987013,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99975663,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-03-28T00:20:12Z\",\"WARC-Record-ID\":\"<urn:uuid:54a40f3c-8ed7-4f8f-ab66-454ecd10a106>\",\"Content-Length\":\"51717\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d9dc5519-084e-4663-bc60-38fcd776384d>\",\"WARC-Concurrent-To\":\"<urn:uuid:feabf193-9626-443a-ae4c-06d219dbd377>\",\"WARC-IP-Address\":\"141.66.194.2\",\"WARC-Target-URI\":\"https://zbmath.org/1082.35140\",\"WARC-Payload-Digest\":\"sha1:EH6JQ2IZDRF6ZMRYJODN6VMKIPJTRJWP\",\"WARC-Block-Digest\":\"sha1:3J5OXSKXYHL5ZX4W7BSJMYKDOP2F4IZL\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-14/CC-MAIN-2023-14_segments_1679296948708.2_warc_CC-MAIN-20230327220742-20230328010742-00463.warc.gz\"}"}
https://christianlouca.com/algebra-2-quadratic-equations-graphing/	[ "# Algebra 2 Quadratic Equations Graphing\n\nAlgebra 2 Quadratic Equations Graphing has a variety pictures that related to locate out the most recent pictures of Algebra 2 Quadratic Equations Graphing here, and in addition to you can get the pictures through our best algebra 2 quadratic equations graphing collection. Algebra 2 Quadratic Equations Graphing pictures in here are posted and uploaded by Adina Porter for your algebra 2 quadratic equations graphing images collection. The images that existed in Algebra 2 Quadratic Equations Graphing are consisting of best images and high tone pictures.", null, "algebra 2 worksheets quadratic functions and inequalities worksheets from algebra 2 quadratic equations graphing\n\nThese many pictures of Algebra 2 Quadratic Equations Graphing list may become your inspiration and informational purpose. We hope you enjoy and satisfied as soon as our best portray of Algebra 2 Quadratic Equations Graphing from our store that posted here and in addition to you can use it for adequate needs for personal use only. The home Design Ideas team with provides the further pictures of Algebra 2 Quadratic Equations Graphing in high Definition and Best environment that can be downloaded by click on the gallery below the Algebra 2 Quadratic Equations Graphing picture.", null, "algebra 2 lessons best of algebra 2 homework smart quadratic formula from algebra 2 quadratic equations graphing", null, "easy graphing in algebra math and calculus from algebra 2 quadratic equations graphing\n\nYou Might Also Like :\n\nchristianlouca.com can support you to get the latest information practically Algebra 2 Quadratic Equations Graphing. modernize Ideas. We find the money for a top tone tall photo like trusted permit and everything if youre discussing the address layout as its formally called. This web is made to twist your unfinished room into a simply usable room in usefully a brief amount of time. suitably lets agree to a enlarged deem exactly what the algebra 2 quadratic equations graphing. is anything virtually and exactly what it can possibly accomplish for you. subsequently making an trimming to an existing address it is difficult to manufacture a well-resolved loan if the existing type and design have not been taken into consideration.", null, "algebra 2 worksheets quadratic functions and inequalities worksheets from algebra 2 quadratic equations graphing", null, "algebra 2 lessons best of algebra 2 homework smart quadratic formula from algebra 2 quadratic equations graphing\n\nSHARE ON\nTags:" ]	[ null, "https://christianlouca.com/wp-content/uploads/2019/07/algebra-2-quadratic-equations-graphing-algebra-2-worksheets-quadratic-functions-and-inequalities-worksheets-of-algebra-2-quadratic-equations-graphing.png", null, "https://christianlouca.com/wp-content/uploads/2019/07/algebra-2-quadratic-equations-graphing-algebra-2-lessons-best-of-algebra-2-homework-smart-quadratic-formula-of-algebra-2-quadratic-equations-graphing.jpg", null, "https://christianlouca.com/wp-content/uploads/2019/07/algebra-2-quadratic-equations-graphing-easy-graphing-in-algebra-math-and-calculus-of-algebra-2-quadratic-equations-graphing.jpg", null, "https://christianlouca.com/wp-content/uploads/2019/07/algebra-2-quadratic-equations-graphing-algebra-2-worksheets-quadratic-functions-and-inequalities-worksheets-of-algebra-2-quadratic-equations-graphing-1.png", null, "https://christianlouca.com/wp-content/uploads/2019/07/algebra-2-quadratic-equations-graphing-algebra-2-lessons-best-of-algebra-2-homework-smart-quadratic-formula-of-algebra-2-quadratic-equations-graphing-1.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9170708,"math_prob":0.98426163,"size":2431,"snap":"2019-35-2019-39","text_gpt3_token_len":470,"char_repetition_ratio":0.22867738,"word_repetition_ratio":0.17962466,"special_character_ratio":0.171123,"punctuation_ratio":0.040609136,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96809185,"pos_list":[0,1,2,3,4,5,6,7,8,9,10],"im_url_duplicate_count":[null,1,null,1,null,1,null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-09-17T10:59:56Z\",\"WARC-Record-ID\":\"<urn:uuid:11b3feab-d925-4afc-a9aa-2f24a7de65ee>\",\"Content-Length\":\"103233\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:69f6c7a7-62bd-4b12-ba97-b87712ab2e23>\",\"WARC-Concurrent-To\":\"<urn:uuid:d13151a6-dba0-4b99-95e6-9733eee44453>\",\"WARC-IP-Address\":\"104.31.85.26\",\"WARC-Target-URI\":\"https://christianlouca.com/algebra-2-quadratic-equations-graphing/\",\"WARC-Payload-Digest\":\"sha1:ZSRU4CHZ3LJW65B6NDR7GBF7ZVEVM5CT\",\"WARC-Block-Digest\":\"sha1:6UHPNZALNJV23UEYEPKX2NL74VRMODQI\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-39/CC-MAIN-2019-39_segments_1568514573070.36_warc_CC-MAIN-20190917101137-20190917123137-00283.warc.gz\"}"}
https://quant.stackexchange.com/questions/4316/which-objective-function-should-i-choose-to-minimize-tracking-error	[ "# Which objective function should I choose to minimize tracking error?\n\nLet say I have $$n$$ assets and their returns over $$m$$ periods which are represented by a matrix $$X \\in \\mathbb{R}^{m \\times n}$$, and I have some other asset with return over the same period which is represented by a vector $$y \\in \\mathbb{R}^m$$.\n\nMy objective is to find a vector of weights $$w$$ such that\n\n$$w^* = \\underset{w}{\\arg \\min} ~ \\text{TE}(w)$$\n\nwhere $$\\text{TE}(w)$$ is the tracking error defined as follows:\n\n$$\\text{TE}(w) = \\sqrt{\\text{Var}(Xw - y)}$$\n\nand\n\n$$\\sum_{i=1}^n w_i =1$$\n\nIn short, I want to replicate $$y$$ using a portfolio of assets $$X$$.\n\nMy idea was to use the exact definition of the tracking error mentioned above through an optimizer.\n\nHowever, somebody suggested to use the following:\n\n$$w^* = \\underset{w}{\\arg \\min} ~ \\sum_{i=1}^m (Xw-y)_i^2$$\n\nI tried both and I get a better tracking error with the first one.\n\nIt seems clear to me that both should return exactly the same if indeed there exists some $$w$$ which perfectly replicates $$y$$.\n\nWhat if it's not the case?\n\nIs there another approach?\n\nWhen performing a tracking error optimization, you will obtain the same result by using the tracking error squared, which is just the variance of the relative portfolio weights. This would be just finding the minimum variance portfolio, but with conditions on the weights. For instance, it would be equivalent to instead set up the variance minimization assuming you have a fixed -100% weight on the benchmark and optimize with the overall sum of weights equal to zero.\n\nFor a short time horizon, if you can assume that the expected return is approximately zero, then your second formula is equivalent to minimizing the variance. When the time horizon is longer and it is no longer approximately true that the expected return on each asset is zero, then the second formula will not produce the same weights.\n\n• Where do the expected returns come from? I think they have no place in this context.\n– SRKX\nOct 10, 2012 at 17:11\n• Because $Var\\left(x\\right)=E\\left(x^{2}\\right)-E\\left(x\\right)^{2}$\n– John\nOct 10, 2012 at 17:25\n• oh right you meant it from a statistical expectation perspective.\n– SRKX\nOct 10, 2012 at 17:58\n\nIt is better to use a factor model, if one is available. Are you asking this question because you don't have access to one?\n\nAlso, what is the nature of the asset you want to track? Is it an index or a single security? What asset class? What risk factors is it exposed to (e.g. interest rate and credit risk vs. stock market volatility and other equity factors)? The answer to your question depends on what you are tracking." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9219011,"math_prob":0.9938888,"size":1014,"snap":"2023-40-2023-50","text_gpt3_token_len":307,"char_repetition_ratio":0.09009901,"word_repetition_ratio":0.0,"special_character_ratio":0.30078897,"punctuation_ratio":0.05940594,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9998248,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-28T21:46:12Z\",\"WARC-Record-ID\":\"<urn:uuid:b6ea61a4-7c07-48df-bb18-265e9d8f3b0e>\",\"Content-Length\":\"169479\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:bc5ae259-f8b8-44e6-843d-296965444f43>\",\"WARC-Concurrent-To\":\"<urn:uuid:87fd867c-5419-4041-afe2-d8447524b669>\",\"WARC-IP-Address\":\"104.18.10.86\",\"WARC-Target-URI\":\"https://quant.stackexchange.com/questions/4316/which-objective-function-should-i-choose-to-minimize-tracking-error\",\"WARC-Payload-Digest\":\"sha1:2LJYEAIV4AGQ6ZKD2NAGKF33Z27PQU3R\",\"WARC-Block-Digest\":\"sha1:56LNFPV5VWVGCYB3PRRMF7F77ZUQXO3A\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233510454.60_warc_CC-MAIN-20230928194838-20230928224838-00095.warc.gz\"}"}
https://www.neuron.yale.edu/phpBB/viewtopic.php?f=8&t=4264&view=print	[ "Page 1 of 1\n\n### Action Potential propagation\n\nPosted: Sun Mar 15, 2020 2:06 pm\nHi,\nI have 3 questions:\n1: Would you please tell me, the model that NEURON uses is always based on cable theory ? Can we have point model of neuron ,too?\n\n2: I wrote this code:\ncreate soma, axon\naccess soma\n\nsoma.nseg = 1\nsoma.diam = 18.8\nsoma.L = 18.8\nsoma.Ra =123.0\nsoma insert hh\nsoma psection()\naxon.nseg = 1\naxon.diam = 18.8\naxon.L = 500\naxon.Ra =123.0\naxon insert hh\naxon psection()\nconnect axon(0),soma(0)\n\nobjectvar stim\nsoma stim = new IClamp(0.5)\nstim.del = 100\nstim.dur = 100\nstim.amp = 1\n\ntstop = 300\n\nand I want to see action potential propagation along the axon, so I plot axon.v(0.1) and axon.v(0.9) and I expected to see a time delay between spikes that I didn't see.\nCould you please tell me how can I fix this problem?\n\n3. And my last question is how can I compute intracellular voltage Vi = Vm + Ve ?!\n\nThanks\n\n### Re: Action Potential propagation\n\nPosted: Sun Mar 15, 2020 6:36 pm\n1. Point models are fine. If you want biophysical kinetics, just use a section with nseg=1. If you want something more like an integrate-and-fire type cell, use one of the built-in types (e.g. IntFire, IntFire2, IntFire4), find one on ModelDB, or create your own with a MOD file of type ARTIFICIAL_CELL.\n\n2. You set axon.nseg = 1; that means there's only one compartment for the axon, so whether you sample the 0.1 or 0.9 location doesn't matter; you'll still be in the same compartment.\n\nThat said, with your model even if you set \"axon nseg=101\" while you will see a difference between those two locations, it will be very slight. Why? A diameter of 18.8 is quite large and imposes little barrier to current flow. Try reducing the axon diameter to 2 microns and shortening your stimulus to just 1 ms. If you do that with an appropriate nseg, you'll see axon(0.9) follows axon(0.1) by about 1 ms.\n\nAs an aside, if you're new to NEURON, I strongly recommend learning Python instead of HOC. You'll find much more documentation on Python than in HOC (both in general and less-so-but-still-true with respect to NEURON), Python is a more broadly applicable skill, and Python offers vastly superior graphics and library support.\n\nHere's a Python version of your code, with the changes as described above:\n\nCode: Select all\n\n``````import matplotlib.pyplot as plt\nfrom neuron import h\nfrom neuron.units import ms, mV\n\nsoma = h.Section(name='soma')\naxon = h.Section(name='axon')\naxon.connect(soma(0))\n\nsoma.nseg = 1\nsoma.diam = soma.L = 18.8\n\naxon.nseg = 101\naxon.diam = 2\naxon.L = 500\n\nstim = h.IClamp(soma(0.5))\nstim.delay = 100 * ms\nstim.dur = 1 * ms\nstim.amp = 0.9\n\nfor sec in [soma, axon]:\nsec.Ra = 123\nsec.insert('hh')\n\nt = h.Vector().record(h._ref_t)\nv1 = h.Vector().record(axon(0.1)._ref_v)\nv2 = h.Vector().record(axon(0.9)._ref_v)\n\nh.finitialize(-65 * mV)\nh.continuerun(300 * ms)\n\nplt.plot(t, v1, label='axon(0.1).v')\nplt.plot(t, v2, label='axon(0.9).v')\nplt.xlim((99, 110))\nplt.xlabel('t (ms)')\nplt.legend()\nplt.show()\n``````", null, "### Re: Action Potential propagation\n\nPosted: Mon Mar 16, 2020 10:30 am\nDear Ramcdougal,\n\nIf I insert extracellular to my model, I have extracellular potential, haven't I?\nIf so, how can I code for intracellular voltage?\n\nThanks\n\n### Re: Action Potential propagation\n\nPosted: Mon Mar 16, 2020 11:36 am\nSee the documentation of extracellular in the Programmer's Reference.\nThe vext values are all referred to ground, which is assumed to have a potential of 0 mV.\n\nThe hoc name for the potential at a point just outside the cell membrane of the currently accessed section is\nvext(x)\nwhere x is the normalized distance of the point from the section's 0 end.\nBut what do you call it in Python? If the Python name of the section is foo, and extracellular has been \"inserted\" into foo, then the potential adjacent to the external surface of foo at location x is\nfoo(x).vext\n\nSo the intracellular potential at x, relative to ground, is the sum\nfoo(x).v + foo(x).vext\n\nYou could force recalculation of this value throughout a simulation, but that is slower than capturing the time courses of foo(x).v and foo(x).vext to a pair of Vectors during a simulation and then summing the vectors after the end of the simulation. For example, assuming that foov and foovext are the Python names for the Vectors that capture the time course of foo(x).v and foo(x).vext,\n\nCode: Select all\n\n``````h.run() # executes a simulation\nvintracell = foov + foovext # vintracell now contains the time course\n# of intracellular potential at foo(x)``````\nRead the Programmer's Reference entry about the Vector class's record() method. By the way, it's also a good idea to capture the time values h.t to another Vector." ]	[ null, "https://www.dropbox.com/s/h1nz67wikp5wd9p/propagating_wave.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8156547,"math_prob":0.91910404,"size":1701,"snap":"2020-24-2020-29","text_gpt3_token_len":575,"char_repetition_ratio":0.09546258,"word_repetition_ratio":0.95238096,"special_character_ratio":0.3251029,"punctuation_ratio":0.18483412,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9704257,"pos_list":[0,1,2],"im_url_duplicate_count":[null,6,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-07-05T04:42:59Z\",\"WARC-Record-ID\":\"<urn:uuid:49e71998-c8c5-44a8-8177-e4dc302ce44c>\",\"Content-Length\":\"8238\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:4f35df56-27ad-45cd-ad84-a72183aaa00d>\",\"WARC-Concurrent-To\":\"<urn:uuid:7d5b805f-b0f4-4385-aff8-d929c057171a>\",\"WARC-IP-Address\":\"128.36.111.212\",\"WARC-Target-URI\":\"https://www.neuron.yale.edu/phpBB/viewtopic.php?f=8&t=4264&view=print\",\"WARC-Payload-Digest\":\"sha1:JFT2T4IJLVK5GEAYI6GK7AUUFKVWZGDI\",\"WARC-Block-Digest\":\"sha1:VTHUZIQKEJEJBXLDQ4DOGLCOSLDPOYT7\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-29/CC-MAIN-2020-29_segments_1593655886865.30_warc_CC-MAIN-20200705023910-20200705053910-00381.warc.gz\"}"}
https://deap.readthedocs.io/en/master/examples/gp_parity.html	[ "# Even-Parity Problem¶\n\nParity is one of the classical GP problems. The goal is to find a program that produces the value of the Boolean even parity given n independent Boolean inputs. Usually, 6 Boolean inputs are used (Parity-6), and the goal is to match the good parity bit value for each of the $$2^6 = 64$$ possible entries. The problem can be made harder by increasing the number of inputs (in the DEAP implementation, this number can easily be tuned, as it is fixed by a constant named PARITY_FANIN_M).\n\n## Primitives set used¶\n\nParity uses standard Boolean operators as primitives, available in the Python operator module :\n\npset = gp.PrimitiveSet(\"MAIN\", PARITY_FANIN_M, \"IN\")\n\n\nIn addition to the n inputs, we add two constant terminals, one at 0, one at 1.\n\nNote\n\nAs Python is a dynamic typed language, you can mix Boolean operators and integers without any issue.\n\n## Evaluation function¶\n\nIn this implementation, the fitness of a Parity individual is simply the number of successful cases. Thus, the fitness is maximized, and the maximum value is 64 in the case of a 6 inputs problems.\n\ndef evalParity(individual):\nfunc = toolbox.compile(expr=individual)\nreturn sum(func(*in_) == out for in_, out in zip(inputs, outputs)),\n\n\ninputs and outputs are two pre-generated lists, to speedup the evaluation, mapping a given input vector to the good output bit. The Python sum() function works on booleans (false is interpreted as 0 and true as 1), so the evaluation function boils down to sum the number of successful tests : the higher this sum, the better the individual.\n\n## Conclusion¶\n\nThe other parts of the program are mostly the same as the Symbolic Regression algorithm.\n\nThe complete examples/gp/parity.\n\n## Reference¶\n\nJohn R. Koza, “Genetic Programming II: Automatic Discovery of Reusable Programs”, MIT Press, 1994, pages 157-199." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.78861535,"math_prob":0.9798643,"size":2054,"snap":"2021-31-2021-39","text_gpt3_token_len":490,"char_repetition_ratio":0.11902439,"word_repetition_ratio":0.0,"special_character_ratio":0.22638753,"punctuation_ratio":0.15326633,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9827472,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-07-24T21:34:53Z\",\"WARC-Record-ID\":\"<urn:uuid:fadfe57f-88b9-438d-ab4e-241f6a1948a8>\",\"Content-Length\":\"15332\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:16525d4e-5c04-4027-9a89-6158792e5e27>\",\"WARC-Concurrent-To\":\"<urn:uuid:ff695cd7-dc2d-4572-bc70-a0d444e4e668>\",\"WARC-IP-Address\":\"104.17.33.82\",\"WARC-Target-URI\":\"https://deap.readthedocs.io/en/master/examples/gp_parity.html\",\"WARC-Payload-Digest\":\"sha1:FTAJCNGEIZGU7V4TKSNFVBEZASLFOUDX\",\"WARC-Block-Digest\":\"sha1:SIYF3BWRC6ENVZ2IGAILXL7XYQWTMAMS\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-31/CC-MAIN-2021-31_segments_1627046150308.48_warc_CC-MAIN-20210724191957-20210724221957-00162.warc.gz\"}"}
https://azabutokyodc.jp/sys/simon.php?infer=MmQ4NjY4YjM0ZGM1MjU3YTM5NTkxZDkyYWIyOWFiOGE-KYN&pubid=1515	[ "# Homework help with order of operations\n\nHomework help with order of operations Rated 5 stars, based on 30 customer reviews From \\$9.12 per page Available! Order now!\n\nThis is telling you which calculations should be made first in expressions that have more than just one operation. Find free & online math homework help by helpers for algebra, etc. Immediately forward your popular and trustworthy to your project to ask.\n\nSats questions primary homework help\n1. It helps the other programmer to know the code structure of your application or software.\n2. Math homework 4 order of operations with a comprehensive collection of operations answers chapter 1.\n3. Use order of operations to compute - jiskha homework help.\n4. Milwaukee tutoring portland tutoring chicago tutoring madison tutoring los angeles tutoring raleigh-durham tutoring omaha tutoring las vegas tutoring.\n5. It stands for brackets, homework help with order of operations orders, division, multiplication, addition and subtraction.\n6. So when the work of popu lar pieces for the character also reveals himself boy.\n7. Initial project proposal overview.\n\nAssignment help uk - best online writing services. Programming assignment help do my programming assignment. Assignment help & writing services content writing services usa online in uk by experts. Algebra 1 order operations answer key tips as order of operations, fast and how to find someone to help write a book reliable.\n\n## Activity-based costing homework help\n\nWith our help, they can now say: \"getting someone in doing my homework. Algebra, algebra help, study skills, homework help. One type is a fraction bar.\n\nHomework help line number in milwaukee\n1. Order of operations homework due april 22, 2014 what's.\n2. Get your homework questions solved fast by accessing our million step-by-step answers.\n3. Lesson order of operations - algebra homework help.\n4. In this order of operations worksheet, students use pemdas to solve given problems.\n\nHomework help order of operations homework help with order of operations - batchelor press.\n\nFree online physics homework help\n1. You should know the definition of each of the following properties of addition and how each can be used.\n2. Help essay homework help with order of operations order of operations is complete the grid provided in order operations.\n3. Then we'll do the 5 times 6 before we do the subtraction right here.\n4. Homework help with order of operations buy descriptive.\n5. Do not worry if you cannot find your question in the list after it has been.\n\nSo the real question is, how do surface area homework help mathematicians really interpret negatives and exponents combined in an expression. Answered in 1 minute by. Perform all operations withing grouping symbols first, starting with innermost set 2. In primary maths we talk about the inverse to explain how addition and subtraction are linked and how multiplication and division are linked. Or go from right to left. Simplifying determinants precalculus systems of linear equations and matrices. No, only that two people using order of operations keeps getting different answers. Quia - 1 - 3 order of operations. So here, we have all sorts of parentheses and numbers flying around. By julie booth & ken koedinger, carnegie mellon univ. Sporting gloves,neoprene bag,diving suit sporting gloves,neoprene bag,diving suit,rubber sponge,neoprene fabric,neoprene sheets,rubber. Parenthesis exponents multiply divide homework help with order of operations add subtract 62. And solve steps in 2017, all of experienced management homework help. If you've moved through pre-algebra or algebra i, you've no doubt heard the term order of operations. Write your answer in exponential form.\n\n• Order of operations worksheets - free printable math pdfs.\n• I only wish i'd found it sooner.\n• The order of operations in algebra is important if you want to find the correct answer.\n• To play this quiz, please finish editing it.\n• 30% off on delivery when you order online chang thai eltham thai restaurant in eltham, thai restaurant in bexley thai restaurant in greenwich, thai restaurant in bromley homework help managerial accounting service.\n• Go math page 53 problems 4-11.\n\nStudents learn that every year 7 maths homework help difficult task requires time to complete. 7th grade order of operations worksheet pdf. 64+(15-7) x 2-9 +8+5 x (7-2) 4. Exponents, order of operations, and variable. Bonanza offer flat 20% off & essay services reddit sign up bonus order now. 10 best numerical expression images numerical expression. Evaluate each homework help with order of operations of the following operations. Homework help order of operations research paper for sale online. Statistics assignment help with statistics homework help.\n\nCcg homework help\n• 1-2 exit quiz - order of operations and evaluating expressions.\n• Arithmetic and the order of operations.\n• You must use at least 3 different operations (addition, subtraction, multiplication, division) within the equation.\n• Remember, do exponents apply to.\n• Use in classroom discussions: ratios, operations millionaire- print up certificate for books, getting.\n• Order of operations - homework help videos - brightstorm.\n• Homework help with order of operations number theory, rounding, math worksheets including algebra, geometry to operations websites case holt middle school and math.\n• Basic mental calculations (steve neale) arithmetic (@dirt_expert) four rules (@nncpsmith) addition homework help with order of operations and subtraction (@mathsblackcat) bidmas (@jamesw_72) order of operations (@ejmaths) addition (@roybird) multiplication (@roybird) multiplication (@giftedba) 4 operations (@giftedba) money.\n• Operations management homework help - essay cops.\n• Class 12 class 11 class 10 class 9 class 8 homework help with order of operations class 7 class 6.\n\nImprove your math knowledge with free questions in \"solve equations using order of operations\" and thousands of other homework help with order of operations math skills. Essay for essex county council homework help you: homework help with order of operations free. Order of operations pemdas operations \"operations\" mean things like add, subtract, multiply, divide, squaring, etc. Operations real life math skills learn homework investing money, budgeting your money, paying taxes, mortgage loans, and even the math involved in playing baseball. It explains general rules related to the order of operations and illustrates them with examples, including the use world war 2 timeline homework help of. In mathematics an expression is solved by a rule known as order of operations in which operations on terms are performed following a certain order.\n\nSequences common core algebra 2 homework @ intro to order. Okayy so, i don't really get my math homework, so i was hoping someone could help me. Homework: order of operations use pemdas to help simplify each expression. Homework help order of operations - select quality and. Order of operations homework help with order of operations algebra 5th grade math standards. Operation management homework help : operations management. 0 comments this week students have a quiz on desmond tutu homework help order of operations, or pemdas, on wednesday, 3/13. Puzzles force our students to think, whether they want to or. Correctly solve algebraic expressions marine biology homework help with my video assignment help. Warning: calculate them in the wrong order, and you can get a wrong.\n\n## Viking facts homework help\n\nExpert writers write assignments excellently in order to help students to succeed in their careers.\n\nRiver facts primary homework help\n1. Custom writing company will be the best decision papers and can assist.\n2. Aplusmath : math homework helper - order of operations.\n3. However, when you turn to cheap writing services, there's a big homework help order of operations chance that you receive a plagiarized paper in return or that your paper will be written by a fellow student, not by a homework help order of operations professional writer.\n4. Description: step-by-step instruction on how to read, interpret and construct line, bar and circle graphs.\n5. If you look in books, you will rarely find \"-32\" written out, but you will often find polynomials homework help with order of operations with negative coefficients.\n6. My homework lesson 7 best writing service in uk order of operations - reputable.\n7. Khan videos for above: combining like terms distributive with variables.\n8. Start at the left and go to the right.\n\nName date period lesson 3 homework practice order of operations generate an equivalent expression for each using the order of operations. (3 +7) x 6 +4 x (5z - 12) - 6 +3 x 23 10.\n\nLondon blitz primary homework help\n• Math 8 section order of operations: 1.\n• He really likes the program.\n• Homework order of operations - pemdas worksheets.\n• 120 write each expression in exponential form.\n\nIf you add homework help on two odd numbers, the answer is not. Student handout: order of operations homework with worked. Review of operations with decimals in order to solve equations with decimals. Remember, daily homework helper, and operations hw: solve algebra problems into your horizontal and a special set of all. With our professional custom writing companies, your homework will be completed according to specific guidelines in order for you to get a passing grade. So let's figure out what this is. The test study guide is posted to classroom. The purpose of this homework assignment is to give you experience practicing what you learned in the last section. My question is: does this make the associative property an exception to. Math teachers, on the other hand, frequently just complain amongst themselves in the faculty lounge about how messy their students' work is. Out of these primary homework help ancient egypt cookies, the cookies that. A a b homework help with order of operations a c c 9. The most reliable essay help company. Their order operations of help homework on something not only critical and our secure system and. Answers are provided at the top of the page. Order of operations and some examples. Determinant row operations column operations interchanging rows common factors adding a multiple of one row to another. Order of operations puzzle figure out which equation matches the outcome. To help providers, you will check out this organization at our dissertation service.\n\nShmoop homework help\n1. As our first client, use coupon code get15 to claim 15% discount this month.\n2. This order means that when we have one or more context filters in a view any other filters used in that same view are applied only to rows which make it through the context filter.\n3. Evaluate algebraic expressions involving decimals and fractions using order of operations.\n\n## Queen elizabeth 1 homework help\n\nChildrens homework help kid homework. Category: order of operations - mrs. This is the first of three assignments that, as a whole, will cover all aspects of the project life. Guidance, support, and order of operations homework helper you have to academic assignments for phd, speaking english order of operations homework helper a forms of paper writing. 5 - 5 x 5 +5. What is the answer to 5 - 5 x 5 + 5. 34 - 82 + 4 15. Homework helper lesson 2 order of homework help with order of operations operations. May 5th, 2010 math home work helper comments off on properties of operations. Multiply or divide in order from left to right 4.\n\n1-2 guide notes te - order of operations and evaluating expressions.\n\nLa county live homework help\n1. Order of operations (pemdas) calculator - symbolab.\n2. How to use the order of operations - dummies.\n3. The using grouping symbols lesson shows when to use parentheses, brackets, and braces in highwayman homework help expressions.\n4. Learn average time spent doing homework to solve exponential equations quickly with this lesson that is easy to understand.\n5. As an asidein computer lingo, the order of operations is called operator precedence, which has its foundations in mathematics and is unambiguously defined for all operations (in before someone brings up compound.\n6. Order of operations - algebra homework help, algebra.\n7. Taking homework help to the next level.\n8. We assure you that will be taken order of operations homework help to know that.\n\n\"looking for a similar assignment.\n\nOur Site Map" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.91546494,"math_prob":0.7441682,"size":13861,"snap":"2021-31-2021-39","text_gpt3_token_len":2866,"char_repetition_ratio":0.23756947,"word_repetition_ratio":0.028301887,"special_character_ratio":0.20402569,"punctuation_ratio":0.10891089,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9806903,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-19T14:22:13Z\",\"WARC-Record-ID\":\"<urn:uuid:009a10f0-1e45-4293-b06a-9e481e372b41>\",\"Content-Length\":\"23973\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:0a5531bc-4039-4cb8-a41e-b0d0db8706ce>\",\"WARC-Concurrent-To\":\"<urn:uuid:1c07d5c4-d253-47ea-adad-b3b7b276689a>\",\"WARC-IP-Address\":\"49.212.207.82\",\"WARC-Target-URI\":\"https://azabutokyodc.jp/sys/simon.php?infer=MmQ4NjY4YjM0ZGM1MjU3YTM5NTkxZDkyYWIyOWFiOGE-KYN&pubid=1515\",\"WARC-Payload-Digest\":\"sha1:TX2VGZRCRGT4QAB2BYIXWQM4EJLC67LP\",\"WARC-Block-Digest\":\"sha1:3OQPCV67S34LS7BZEH2JBJWPG5WCOZHW\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780056890.28_warc_CC-MAIN-20210919125659-20210919155659-00473.warc.gz\"}"}
https://www.gradesaver.com/textbooks/math/calculus/calculus-8th-edition/chapter-4-integrals-4-3-the-fundamental-theorem-of-calculus-4-3-exercises-page-327/22	[ "## Calculus 8th Edition\n\nEvaluate the integral: $\\int^1_0 (1 -8v^3 +16v^7) dv$ Recall the 2nd part of the Fundamental Theorem of Calculus: $\\int_a^bf(x)dx = F(b) - F(a)$ Find $F(v)$: $F(v) = \\int^1_0 (1 -8v^3 +16v^7) dv$ $F(v) =v - \\frac{8}{4}v^4 +\\frac{16}{8}v^8$ $F(v) =v - 2v^4 +2v^8$ Now evaluate $F(b) - F(a)$: $F(1) - F(0)$ $(1 - 2(1)^4 +2(1)^8 )- (0 - 2(0)^4 +2(0)^8)$ $(1-2+2) - 0=1$" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.51857436,"math_prob":1.00001,"size":395,"snap":"2019-51-2020-05","text_gpt3_token_len":215,"char_repetition_ratio":0.10997442,"word_repetition_ratio":0.0,"special_character_ratio":0.6025317,"punctuation_ratio":0.03960396,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":1.0000006,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-12-15T02:10:33Z\",\"WARC-Record-ID\":\"<urn:uuid:f23bf92e-fa2d-4b08-b258-7a718802724f>\",\"Content-Length\":\"68048\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:4d5ccd75-e120-4d63-8d87-6435ce538b96>\",\"WARC-Concurrent-To\":\"<urn:uuid:4bd2b6a8-7e79-43a0-9035-5cd699e0bb4d>\",\"WARC-IP-Address\":\"52.87.77.102\",\"WARC-Target-URI\":\"https://www.gradesaver.com/textbooks/math/calculus/calculus-8th-edition/chapter-4-integrals-4-3-the-fundamental-theorem-of-calculus-4-3-exercises-page-327/22\",\"WARC-Payload-Digest\":\"sha1:GBW5KE6HFLEKFK6BJDOSCADWYDOBYN6E\",\"WARC-Block-Digest\":\"sha1:PR2QFSV3W7PRCZJIJY6OYJUUHVHOKX7Y\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-51/CC-MAIN-2019-51_segments_1575541301014.74_warc_CC-MAIN-20191215015215-20191215043215-00510.warc.gz\"}"}
https://www.colorhexa.com/002626	[ "# #002626 Color Information\n\nIn a RGB color space, hex #002626 is composed of 0% red, 14.9% green and 14.9% blue. Whereas in a CMYK color space, it is composed of 100% cyan, 0% magenta, 0% yellow and 85.1% black. It has a hue angle of 180 degrees, a saturation of 100% and a lightness of 7.5%. #002626 color hex could be obtained by blending #004c4c with #000000. Closest websafe color is: #003333.\n\n• R 0\n• G 15\n• B 15\nRGB color chart\n• C 100\n• M 0\n• Y 0\n• K 85\nCMYK color chart\n\n#002626 color description : Very dark (mostly black) cyan.\n\n# #002626 Color Conversion\n\nThe hexadecimal color #002626 has RGB values of R:0, G:38, B:38 and CMYK values of C:1, M:0, Y:0, K:0.85. Its decimal value is 9766.\n\nHex triplet RGB Decimal 002626 `#002626` 0, 38, 38 `rgb(0,38,38)` 0, 14.9, 14.9 `rgb(0%,14.9%,14.9%)` 100, 0, 0, 85 180°, 100, 7.5 `hsl(180,100%,7.5%)` 180°, 100, 14.9 003333 `#003333`\nCIE-LAB 12.773, -12.916, -3.798 1.043, 1.526, 2.073 0.225, 0.329, 1.526 12.773, 13.463, 196.386 12.773, -9.88, -2.133 12.353, -6.55, -1.303 00000000, 00100110, 00100110\n\n# Color Schemes with #002626\n\n• #002626\n``#002626` `rgb(0,38,38)``\n• #260000\n``#260000` `rgb(38,0,0)``\nComplementary Color\n• #002613\n``#002613` `rgb(0,38,19)``\n• #002626\n``#002626` `rgb(0,38,38)``\n• #001326\n``#001326` `rgb(0,19,38)``\nAnalogous Color\n• #261300\n``#261300` `rgb(38,19,0)``\n• #002626\n``#002626` `rgb(0,38,38)``\n• #260013\n``#260013` `rgb(38,0,19)``\nSplit Complementary Color\n• #262600\n``#262600` `rgb(38,38,0)``\n• #002626\n``#002626` `rgb(0,38,38)``\n• #260026\n``#260026` `rgb(38,0,38)``\n• #002600\n``#002600` `rgb(0,38,0)``\n• #002626\n``#002626` `rgb(0,38,38)``\n• #260026\n``#260026` `rgb(38,0,38)``\n• #260000\n``#260000` `rgb(38,0,0)``\n• #000000\n``#000000` `rgb(0,0,0)``\n• #000000\n``#000000` `rgb(0,0,0)``\n• #000d0d\n``#000d0d` `rgb(0,13,13)``\n• #002626\n``#002626` `rgb(0,38,38)``\n• #004040\n``#004040` `rgb(0,64,64)``\n• #005959\n``#005959` `rgb(0,89,89)``\n• #007373\n``#007373` `rgb(0,115,115)``\nMonochromatic Color\n\n# Alternatives to #002626\n\nBelow, you can see some colors close to #002626. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #00261d\n``#00261d` `rgb(0,38,29)``\n• #002620\n``#002620` `rgb(0,38,32)``\n• #002623\n``#002623` `rgb(0,38,35)``\n• #002626\n``#002626` `rgb(0,38,38)``\n• #002326\n``#002326` `rgb(0,35,38)``\n• #002026\n``#002026` `rgb(0,32,38)``\n• #001d26\n``#001d26` `rgb(0,29,38)``\nSimilar Colors\n\n# #002626 Preview\n\nThis text has a font color of #002626.\n\n``<span style=\"color:#002626;\">Text here</span>``\n#002626 background color\n\nThis paragraph has a background color of #002626.\n\n``<p style=\"background-color:#002626;\">Content here</p>``\n#002626 border color\n\nThis element has a border color of #002626.\n\n``<div style=\"border:1px solid #002626;\">Content here</div>``\nCSS codes\n``.text {color:#002626;}``\n``.background {background-color:#002626;}``\n``.border {border:1px solid #002626;}``\n\n# Shades and Tints of #002626\n\nA shade is achieved by adding black to any pure hue, while a tint is created by mixing white to any pure color. In this example, #001212 is the darkest color, while #feffff is the lightest one.\n\n• #001212\n``#001212` `rgb(0,18,18)``\n• #002626\n``#002626` `rgb(0,38,38)``\n• #003a3a\n``#003a3a` `rgb(0,58,58)``\n• #004d4d\n``#004d4d` `rgb(0,77,77)``\n• #006161\n``#006161` `rgb(0,97,97)``\n• #007474\n``#007474` `rgb(0,116,116)``\n• #008888\n``#008888` `rgb(0,136,136)``\n• #009c9c\n``#009c9c` `rgb(0,156,156)``\n• #00afaf\n``#00afaf` `rgb(0,175,175)``\n• #00c3c3\n``#00c3c3` `rgb(0,195,195)``\n• #00d7d7\n``#00d7d7` `rgb(0,215,215)``\n• #00eaea\n``#00eaea` `rgb(0,234,234)``\n• #00fefe\n``#00fefe` `rgb(0,254,254)``\n• #12ffff\n``#12ffff` `rgb(18,255,255)``\n• #26ffff\n``#26ffff` `rgb(38,255,255)``\n• #3affff\n``#3affff` `rgb(58,255,255)``\n• #4dffff\n``#4dffff` `rgb(77,255,255)``\n• #61ffff\n``#61ffff` `rgb(97,255,255)``\n• #74ffff\n``#74ffff` `rgb(116,255,255)``\n• #88ffff\n``#88ffff` `rgb(136,255,255)``\n• #9cffff\n``#9cffff` `rgb(156,255,255)``\n• #afffff\n``#afffff` `rgb(175,255,255)``\n• #c3ffff\n``#c3ffff` `rgb(195,255,255)``\n• #d7ffff\n``#d7ffff` `rgb(215,255,255)``\n• #eaffff\n``#eaffff` `rgb(234,255,255)``\n• #feffff\n``#feffff` `rgb(254,255,255)``\nTint Color Variation\n\n# Tones of #002626\n\nA tone is produced by adding gray to any pure hue. In this case, #121414 is the less saturated color, while #002626 is the most saturated one.\n\n• #121414\n``#121414` `rgb(18,20,20)``\n• #101616\n``#101616` `rgb(16,22,22)``\n• #0f1717\n``#0f1717` `rgb(15,23,23)``\n• #0d1919\n``#0d1919` `rgb(13,25,25)``\n• #0c1a1a\n``#0c1a1a` `rgb(12,26,26)``\n• #0a1c1c\n``#0a1c1c` `rgb(10,28,28)``\n• #091d1d\n``#091d1d` `rgb(9,29,29)``\n• #071f1f\n``#071f1f` `rgb(7,31,31)``\n• #062020\n``#062020` `rgb(6,32,32)``\n• #042222\n``#042222` `rgb(4,34,34)``\n• #032323\n``#032323` `rgb(3,35,35)``\n• #012525\n``#012525` `rgb(1,37,37)``\n• #002626\n``#002626` `rgb(0,38,38)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #002626 is perceived by people affected by a color vision deficiency. This can be useful if you need to ensure your color combinations are accessible to color-blind users.\n\nMonochromacy\n• Achromatopsia 0.005% of the population\n• Atypical Achromatopsia 0.001% of the population\nDichromacy\n• Protanopia 1% of men\n• Deuteranopia 1% of men\n• Tritanopia 0.001% of the population\nTrichromacy\n• Protanomaly 1% of men, 0.01% of women\n• Deuteranomaly 6% of men, 0.4% of women\n• Tritanomaly 0.01% of the population" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.5098686,"math_prob":0.9180379,"size":3638,"snap":"2021-21-2021-25","text_gpt3_token_len":1533,"char_repetition_ratio":0.17198679,"word_repetition_ratio":0.011049724,"special_character_ratio":0.55607474,"punctuation_ratio":0.2283105,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99505013,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-05-07T03:44:24Z\",\"WARC-Record-ID\":\"<urn:uuid:8297f12b-f8f3-4ebc-84d4-9c659dde2910>\",\"Content-Length\":\"36151\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:5f86a89a-197f-4d2c-9ce8-c7e828a7b70c>\",\"WARC-Concurrent-To\":\"<urn:uuid:8eca44c6-8295-49e9-b3ff-7986020921d6>\",\"WARC-IP-Address\":\"178.32.117.56\",\"WARC-Target-URI\":\"https://www.colorhexa.com/002626\",\"WARC-Payload-Digest\":\"sha1:JIOWVD3JEJ6ZYC3WZNT7CWLQXZ6HF4ZZ\",\"WARC-Block-Digest\":\"sha1:7UEFLDD24IT4HMBMQMSU7CICZ7G3XDMQ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-21/CC-MAIN-2021-21_segments_1620243988774.96_warc_CC-MAIN-20210507025943-20210507055943-00418.warc.gz\"}"}
https://www.rocksolidinc.com/products/life-of-fred-pre-algebra-0-with-physics-middle-school-math	[ "Life of Fred - Pre-Algebra 0 with Physics (Middle School Math)\n\n# Life of Fred - Pre-Algebra 0 with Physics (Middle School Math)\n\n\\$29.00\nThe third book in the Life of Fred Middle School Series\n\nLife of Fred is a series of math books that break the old pattern of drill and kill. Instead of nothing but dry exposition and endless drill exercises, they contain fun stories about Fred Gauss, a child prodigy math genius. During his hilarious adventures, he encounters situations that call for solving math problems. Dr. Stanley Schmidt wrote these to make math come alive with humor, clear explanations, and silly illustrations that will make math a fun subject. Your children actually want to read these books. The stories carry on through the exercises, leading the student through them. Many of the books have a series of quizzes called Bridges every few chapters. Passing a quiz means the student is ready for the next chapter. Some books have all the problem solutions in them, while others have separate answer keys or companions. The textbooks are all hardbound, the answer keys are comb-bound.\n\nThis program relies on reading comprehension and thinking and is non-consumable. You will not find 30 problems on every page of rote spoon-fed learning that is quickly forgotten. This is totally different, a new way of learning math, and everything you need is in the book. As a parent, you will quickly pick up on the novel way as you read along and learn with your student. Books available for grades K - 12.\n\nFred does physics before algebra.\n\nLearn . . .\n\n• How ducks and cows pronounce μ differently\n• Why no one can weigh 55 kilograms\n• How Kingie could move a safe that Superman couldn't\n• . . . and even some elementary physics\n\nAll fun!\n\nJust open & enjoy.\n\nConcepts -\n• Numerals\n• Making models\n• Nineteen conversion factor problems\n• Area of a rectangle\n• Friction independent of speed\n• Constants of Proportionality\n• Exact speed of light\n• Continuous and discrete variables\n• Story of the meter\n• Oblate spheroids\n• Krypton vs. kryptonite\n• Square roots\n• Solving d = rt for r\n• Ordered pairs\n• Why rocks stop falling\n• Hunch-conjecture-theory-law\n• Simultaneity doesn’t exist\n• Inductive and deductive reasoning\n• How fossils and astronomy are connected\n• Hooke’s law\n• Four ways to stretch a spring\n• Three meanings of plastic\n• Static vs. kinetic friction\n• Nine forms of energy\n• Energy Cards game\n• Experimentally finding the coefficient of static friction without knowing the weight of the object\n• Similar triangles\n• Kelvin temperature\n• Law of conservation of energy\n• Perpetual motion machines\n• Kilowatt-hours\n• Metric system\n• Two ways to measure mass\n• Definition of pi\n• Newton’s first and second laws\n• Why cramming doesn’t work well, Definition of pressure\n• Exponents\n• Density of neutron stars\n• Computing the mass of an iron atom\n• Pressure at a given depth\n• Finding the volume of a desk lamp\n• 80-pound rubber ducky\n• Air has mass\n• Why 35-foot straws don’t work\n• Birds sitting on high-voltage power lines\n• How to buy a cafe\n• Electrons per second in an ampere\n• Computing resistance in parallel circuits\n• Orbital quantum number\n• Orbital magnetic quantum number\n• Spin magnetic quantum number\n• and more!\n\nWhew!\n\nHardback, ISBN 9781937032227, 288 pages\n\n Book Title Life of Fred - Pre-Algebra 0 with Physics (Middle School Math) Publisher: Z Twist Books - Life of Fred Author Stan Schmidt\nThe third book in the Life of Fred Middle School Series\n\nLife of Fred is a series of math books that break the old pattern of drill and kill. Instead of nothing but dry exposition and endless drill exercises, they contain fun stories about Fred Gauss, a child prodigy math genius. During his hilarious adventures, he encounters situations that call for solving math problems. Dr. Stanley Schmidt wrote these to make math come alive with humor, clear explanations, and silly illustrations that will make math a fun subject. Your children actually want to read these books. The stories carry on through the exercises, leading the student through them. Many of the books have a series of quizzes called Bridges every few chapters. Passing a quiz means the student is ready for the next chapter. Some books have all the problem solutions in them, while others have separate answer keys or companions. The textbooks are all hardbound, the answer keys are comb-bound.\n\nThis program relies on reading comprehension and thinking and is non-consumable. You will not find 30 problems on every page of rote spoon-fed learning that is quickly forgotten. This is totally different, a new way of learning math, and everything you need is in the book. As a parent, you will quickly pick up on the novel way as you read along and learn with your student. Books available for grades K - 12.\n\nFred does physics before algebra.\n\nLearn . . .\n\n• How ducks and cows pronounce μ differently\n• Why no one can weigh 55 kilograms\n• How Kingie could move a safe that Superman couldn't\n• . . . and even some elementary physics\n\nAll fun!\n\nJust open & enjoy.\n\nConcepts -\n• Numerals\n• Making models\n• Nineteen conversion factor problems\n• Area of a rectangle\n• Friction independent of speed\n• Constants of Proportionality\n• Exact speed of light\n• Continuous and discrete variables\n• Story of the meter\n• Oblate spheroids\n• Krypton vs. kryptonite\n• Square roots\n• Solving d = rt for r\n• Ordered pairs\n• Why rocks stop falling\n• Hunch-conjecture-theory-law\n• Simultaneity doesn’t exist\n• Inductive and deductive reasoning\n• How fossils and astronomy are connected\n• Hooke’s law\n• Four ways to stretch a spring\n• Three meanings of plastic\n• Static vs. kinetic friction\n• Nine forms of energy\n• Energy Cards game\n• Experimentally finding the coefficient of static friction without knowing the weight of the object\n• Similar triangles\n• Kelvin temperature\n• Law of conservation of energy\n• Perpetual motion machines\n• Kilowatt-hours\n• Metric system\n• Two ways to measure mass\n• Definition of pi\n• Newton’s first and second laws\n• Why cramming doesn’t work well, Definition of pressure\n• Exponents\n• Density of neutron stars\n• Computing the mass of an iron atom\n• Pressure at a given depth\n• Finding the volume of a desk lamp\n• 80-pound rubber ducky\n• Air has mass\n• Why 35-foot straws don’t work\n• Birds sitting on high-voltage power lines\n• How to buy a cafe\n• Electrons per second in an ampere\n• Computing resistance in parallel circuits\n• Orbital quantum number\n• Orbital magnetic quantum number\n• Spin magnetic quantum number\n• and more!\n\nWhew!\n\nHardback, ISBN 9781937032227, 288 pages" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.62980473,"math_prob":0.71722305,"size":62836,"snap":"2020-24-2020-29","text_gpt3_token_len":18912,"char_repetition_ratio":0.2487586,"word_repetition_ratio":0.93558055,"special_character_ratio":0.3516774,"punctuation_ratio":0.13003728,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95878464,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-05-26T16:47:28Z\",\"WARC-Record-ID\":\"<urn:uuid:6f46f3fd-2616-401b-875e-ed0bc8277b38>\",\"Content-Length\":\"177398\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:645660f3-8bfd-4ee4-b743-077b0e584a66>\",\"WARC-Concurrent-To\":\"<urn:uuid:5caf2b4d-7c4e-4079-a52d-c51941216d6d>\",\"WARC-IP-Address\":\"23.227.38.64\",\"WARC-Target-URI\":\"https://www.rocksolidinc.com/products/life-of-fred-pre-algebra-0-with-physics-middle-school-math\",\"WARC-Payload-Digest\":\"sha1:7FBTE22NH5NNJPZP7TM3JAMA4PXUODHU\",\"WARC-Block-Digest\":\"sha1:GESSK3FIQXOU4BIELMKRO6WUPPSZUTZC\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-24/CC-MAIN-2020-24_segments_1590347391277.13_warc_CC-MAIN-20200526160400-20200526190400-00259.warc.gz\"}"}
https://bonndoc.ulb.uni-bonn.de/xmlui/handle/20.500.11811/4682	[ "Sosna, Pawel: Derived categories and scalar extensions. - Bonn, 2010. - Dissertation, Rheinische Friedrich-Wilhelms-Universität Bonn.\nOnline-Ausgabe in bonndoc: https://nbn-resolving.org/urn:nbn:de:hbz:5N-23498\n@phdthesis{handle:20.500.11811/4682,\nurn: https://nbn-resolving.org/urn:nbn:de:hbz:5N-23498,\nauthor = {{Pawel Sosna}},\ntitle = {Derived categories and scalar extensions},\nschool = {Rheinische Friedrich-Wilhelms-Universität Bonn},\nyear = 2010,\nmonth = nov,\n\nnote = {This thesis consists of three parts all of which deal with questions related to scalar extensions and derived categories.\nIn the first part we consider the question whether the conjugation of a complex projective K3 surface X by an automorphism of the complex numbers can produce a non-isomorphic Fourier-Mukai partner of X. The answer is affirmative. The conjugate surface is thus in particular a moduli space of locally free sheaves on X. We use our result to give higher-dimensional examples of derived equivalent conjugate varieties. We furthermore prove that a similar result holds for abelian surfaces.\nThe topic of the second part is the behaviour of stability conditions under scalar extensions. Namely, given a smooth projective variety X over some field K and its bounded derived category, one can associate to it a complex manifold of stability conditions. Given a finite Galois extension we compare the stability manifolds of X and of the base change scheme in general and under the additional assumption that the numerical Grothendieck group does not change under the scalar extension.\nIn the third and last part we consider the following question: Can one naturally define an L-linear triangulated category if a K-linear triangulated category and a field extension are given? We propose a construction and prove that our definition gives the expected result in the geometric case. It also gives the anticipated result when applied to the derived category of an abelian category with enough injectives and with generators. We furthermore prove that in the just mentioned cases the dimension of the triangulated category in question does not change for finite Galois extensions.},\n\nurl = {http://hdl.handle.net/20.500.11811/4682}\n}" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8492099,"math_prob":0.8242706,"size":4264,"snap":"2021-21-2021-25","text_gpt3_token_len":943,"char_repetition_ratio":0.116197184,"word_repetition_ratio":0.95454544,"special_character_ratio":0.2075516,"punctuation_ratio":0.12337662,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9631418,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-05-19T03:30:14Z\",\"WARC-Record-ID\":\"<urn:uuid:3cf2f793-b331-403f-a0ba-30818d04f1c8>\",\"Content-Length\":\"27868\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:2a6d9098-4e41-486a-9f2c-b92239938b41>\",\"WARC-Concurrent-To\":\"<urn:uuid:620995a6-a473-449a-93e3-75b93b129ae6>\",\"WARC-IP-Address\":\"131.220.74.20\",\"WARC-Target-URI\":\"https://bonndoc.ulb.uni-bonn.de/xmlui/handle/20.500.11811/4682\",\"WARC-Payload-Digest\":\"sha1:7C5TSK4UASC2FTC22MOODYFDZXGQRX6X\",\"WARC-Block-Digest\":\"sha1:5YBJSN6NWNGZIUB2UTBRYRF2M363H72X\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-21/CC-MAIN-2021-21_segments_1620243991562.85_warc_CC-MAIN-20210519012635-20210519042635-00025.warc.gz\"}"}
http://unistudyguides.com/wiki/Topic_6_-_Normal_Distribution_and_Estimation	[ "# Topic 6 - Normal Distribution and Estimation\n\n## Contents\n\nGerald Keller (2011), Statistics for Management and Economics (Abbreviated), 9th Edition, pp. 270-287 & 335-353.\n\n### Normal Distribution Functions\n\n The Normal Distribution is symmetrical, uni-modal and is bell curve shaped. In this distribution, the mean = median = mode. The 2 factors that influence the shape of the bell curve are the mean (the middle number) and the variance (how steep/flat) Consequentially, we are able to define a normal distribution for the random variable 'X' in the following way:\n\nX ~ (μ,σ^2) where -∞≤x≤∞ This means 'X', the random variable is normally distributed with mean 'μ' and variance 'σ^2'\n\nStatisticians use Z scores to standardise data values so that we can more easily find probability values through a Z standard normal distribution (this link contains examples).\n\nZ Score = (x-μ)/σ (Linear Function)\n\nThe normal distribution for Z scores (standard normal distribution) is defined in the following way Z~N(0,1). Once we have standardised scores for the standard normal distribution we can use the standard normal table to find probability values for a random variable being a range of values e.g. if the random variable, 'X', is equal to the weight of a person in kilograms, we could use this method to find the probability of a persons weight being between 50kg. and 65kg.\n\n#### Normal Distribution Example\n\nX ~ (50,100)\n\n• Find: P(45 ≤ x ≤ 60)\n• Standardise Scores!!!\n• (45-50)/10 ≤ Z ≤ (60-50)/10\n• = -1/2 ≤ Z ≤ 1\n• = 0 ≤ Z ≤ 1/2 + 0 ≤ Z ≤ 1 (By Symmetry - Because P(-1/2 ≤ Z ≤ 0) = P(0 ≤ Z ≤ 1/2)). We do this because it is much easier to find the probability values in the standard normal table in this format\n\nX ~ (50,100)\n\n• Find Z_(0.025) and unstandardise it to find the corresponding 'X' value =\n• = 1.96\n• Unstandardise!!!!\n• (1.96 = X – 50 / 10)\n• (X = 50 + 19.6) = 69.6\n\n### Finding the Value of Z Given the Probabilities\n\nFor these questions, we must look at the probabilities within the standard normal table to find the critical z value.\n\n• ZA = 100(1-A)^th Percentile\n• Find The 97 ½ Percentile\n• P (Z > ZA) = A\n• Z_(0.025) (top 2.5%)\n• P (Z > Z_0.025) = 0.025\n• 1 - 0.025 = 0.9750\n• = 1.96\n\n#### Other Questions\n\nFind the value of a standard normal RV where the probability that the RV is GREATER than it is 5%\n\n• 95th Percentile\n• = 1.645\n\nFind the Value of a standard normal RV where the probability that the RV is LESS than it is 5%\n\n• 5th Percentile\n• = -1.645 (Symmetric)\n\n## Introduction To Estimation\n\n### Concepts of Estimation\n\nEstimations objective is determine the approximate value of the parameter (e.g. sample mean for population mean. There are 2 types of estimators defined below:\n\n• Point Estimator\n• Draws inferences about a population by estimating the value of an unknown parameter using a single value (this will be wrong since a probability of a point on a continuous random variables probability density function is virtually 0)\n• Interval Estimator\n• Draws inferences about a population by estimating the value of an unknown parameter using an interval\n\n### Characteristics of Good Estimators\n\n• Unbiased - E(Estimator) = Parameter (e.g. E(sample mean) = population mean)\n• Consistency - Difference between the estimator & parameters becomes smaller as sample size increases\n• Relative Efficiency - If 2 unbiased estimators of a parameter, the one who’s variance is lower has relative efficiency\n\n## End\n\nThis is the end of this topic. Click here to go back to the main subject page for Business and Economic Statistics." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7598893,"math_prob":0.99501765,"size":4257,"snap":"2020-24-2020-29","text_gpt3_token_len":1200,"char_repetition_ratio":0.13778509,"word_repetition_ratio":0.22857143,"special_character_ratio":0.31524548,"punctuation_ratio":0.09850374,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9994333,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-05-30T06:22:55Z\",\"WARC-Record-ID\":\"<urn:uuid:9f09b908-6fd4-4bff-9348-d84be83dfb20>\",\"Content-Length\":\"26413\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:692d513f-ef51-48ed-8c13-8af91e5a4fc1>\",\"WARC-Concurrent-To\":\"<urn:uuid:e44a903e-c8f3-47d0-b93f-9a34c27e3ab7>\",\"WARC-IP-Address\":\"185.62.236.88\",\"WARC-Target-URI\":\"http://unistudyguides.com/wiki/Topic_6_-_Normal_Distribution_and_Estimation\",\"WARC-Payload-Digest\":\"sha1:KMKSAQPGA4OAKJF42LG6MBM4DVJSP3ZG\",\"WARC-Block-Digest\":\"sha1:MJBNAIZBA77FKCPHFNNIVTNXF2IEVQ7Q\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-24/CC-MAIN-2020-24_segments_1590347407289.35_warc_CC-MAIN-20200530040743-20200530070743-00085.warc.gz\"}"}
https://www.online-sciences.com/physics/ac-circuit-contains-ohmic-resistance-capacitor-and-inductive-coil-connected-in-series-rlc-circuit/	[ "# AC circuit contains ohmic resistance, capacitor and inductive coil connected in series (RLC-circuit)\n\n### Impedance\n\nIn an electric circuit containing an AC power supply together with inductive coils, capacitors & resistors, the AC current would be opposed by reactance (inductive reactance or capacitive reactance) in addition to the resistance of resistors and wires, The resistance and reactance together are called impedance and given the symbol (Z) and measured in Ohm (Ω).\n\nThe impedance (Z) is the equivalent to the resistance, capacitive reactance and inductive reactance together in an AC circuit, When the impedance of RLC circuit = 300 Ω, It means that the total opposition to the electric current in this circuit due to the resistance and the reactance of both the coil and capacitor = 300 Ω.\n\n##### AC circuit contains ohmic resistance and inductive coil connected in series (RL – circuit)\n\nIt is almost impossible practically to construct an inductive coil with zero resistance because any coil must have resistance due to the wires used in its fabrication.\n\nWhen electric circuit contains inductive coil, ohmic resistance and AC source connected in series, we noticed that the current passing through each of the resistance and the induction coil is the same in value and phase since they are connected in series.\n\nBut voltage in the coil (VL) leads current (I) by ¼ cycle (phase angle 90°) in the induction coil, the voltage in the resistance (VR) is in phase with current (I) in ohmic resistance.\n\nThe potential difference across a coil (VL) leads the potential difference across the resistance (VR) by a phase angle 90°, thus the total potential difference is out of phase with the current intensity (I).\n\nTotal voltage can be found using the vectors from the relation:\n\nV² = V²R + V²L\n\nV = I Z , VR = I R, VL = I XL\n\n( I Z )² = I² R² + I² X² = I² ( R² + X²L )\n\nZ ² = ( R² + X²L )\n\nThe phase angle can be determined between the total voltage (V) and the resistor voltage (VR) from the relation:\n\ntan θ = VL / VR, tan θ = IXL / I R , So, tan θ = XL / R\n\n##### AC circuit contains ohmic resistance and capacitor connected in series (RC-circuit)\n\nWhen electric circuit contains capacitor, ohmic resistance and AC source connected in series, We notice that the current passing through each of the resistance and the capacitor is the same in value and phase since they are connected in series.\n\nBut voltage in the capacitor (VC) lags current (I) by ¼ cycle (phase angle 90°) in the capacitor, Voltage in the resistance (VR) is in phase with current (I) in ohmic resistance.\n\nThe potential difference across the capacitor (VC) lags the potential difference across the resistance (VR) by phase angle 90°, thus the total potential difference (V) is out of phase with current intensity (I).\n\nTotal voltage can be found using the vectors from the relation:\n\nV² = V²R + V²C\n\nV = I Z , VR = I R , VC = I XC\n\nI² Z² = I² R² + I² X²C\n\nI² Z² = I² (R² + X²C)\n\nZ² = (R² + X²C)\n\nThe phase angle (θ) can be determined between the total voltage difference (V) and the resistor voltage (VR) from the relation:\n\ntan θ = − VC / VR, tan θ = − IXC / I R, So, tan θ = − XC / R\n\nThe negative sign means that the potential difference (VC) lags the potential difference (VR) by an angle of 90°.\n\n##### AC circuit contains ohmic resistance, induction coil and capacitor connected in series (RLC – circuit)\n\nWhen electric circuit contains capacitor, ohmic resistance, induction coil and AC source connected in series, We noticed that the current passing through each of the resistance, the induction coil & the capacitor is the same in value and phase since they are connected in series, But:\n\n• In the induction coil, voltage (VL) leads current (I) by phase angle 90°.\n• In the ohmic resistance, voltage (VR) is in phase with current (I).\n• In the capacitor, voltage (Vc) lags current (I) by phase angle 90°.\n\nThe coil voltage (VL) leads the resistance voltage (VR) by angle 90° and the capacitor voltage (Vc) lags the resistance voltage (VR) by angle 90° and thus the phase difference between (VL) and (Vc) is 180°.\n\nTotal voltage ( V ) can be found using the vectors from the relation :\n\nV² = V²R + ( VL− Vc\n\nWhere : V = I Z , VR = I R, VL = I XL , VC = I XC\n\nI² Z² = ( I R )² + ( I XL − I XC)² = I² [ ( R )² + (XL − XC)² ]\n\nZ² = ( R )² + (XL − XC\n\nFrom the previous figure, the phase angle can be determined from the relation:\n\ntan θ = ( VL − VC ) / VR\n\ntan θ = ( XL − XC ) / R\n\nThe phase angle θ is affected by changing the values of capacitive and inductive reactance then when:\n\n• VL > VC, XL > XC, The phase angle (θ) is positive, the total voltage (V) leads the current (I) by an angle (θ), That means inductive circuit behavior.\n• VL = VC, XL = XC, The phase angle (θ) equals zero, the total voltage (V) is in phase with the current (I), That means ohmic resistance behavior.\n• VL < VC, XL < XC, The phase angle ( θ ) is negative, the total voltage (V) lags the current (I) by an angle (θ), That means capacitive circuit behavior.\n\nThe real consumed power (Pw) in AC circuit, either RL, RC or RLC is only the power that consumed in the ohmic resistance in the form of heat energy.\n\nPw = I² R = V²R / R\n\nCapacitive reactance of capacitors network, AC current & AC voltage in a capacitor circuit\n\nThe oscillating circuit, Tuning or resonant circuit in the wireless radio receivers" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9029424,"math_prob":0.992765,"size":5312,"snap":"2021-43-2021-49","text_gpt3_token_len":1383,"char_repetition_ratio":0.18161266,"word_repetition_ratio":0.2719212,"special_character_ratio":0.265625,"punctuation_ratio":0.08124373,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9996126,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-25T10:56:24Z\",\"WARC-Record-ID\":\"<urn:uuid:a5034e8a-9df4-4527-a9d0-9ce78cb9611c>\",\"Content-Length\":\"122506\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:bf55cb28-a02c-4441-90d0-985fd7ad2ce1>\",\"WARC-Concurrent-To\":\"<urn:uuid:1b10c6cc-b375-4cd5-9c77-c515b983cf73>\",\"WARC-IP-Address\":\"104.21.93.22\",\"WARC-Target-URI\":\"https://www.online-sciences.com/physics/ac-circuit-contains-ohmic-resistance-capacitor-and-inductive-coil-connected-in-series-rlc-circuit/\",\"WARC-Payload-Digest\":\"sha1:VFJ2CCNXFWSCNIU5O2VLKXRX3PF5AORK\",\"WARC-Block-Digest\":\"sha1:D5TP6H54EOMBT53LHZTT4VLOVC4PFHFU\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323587659.72_warc_CC-MAIN-20211025092203-20211025122203-00057.warc.gz\"}"}
https://puzzling.stackexchange.com/questions/40415/matchsticks-for-lengths	[ "# Matchsticks for lengths\n\nYou have been invited to an $n$-dimensional universe, where $n$ is an integer greater than $1$.\n\nHere, matchsticks are idealised lines of unit length, and they can pass through each other.\n\nThey want you to construct a ruler for them. (Never mind that they are already rulers themselves of this universe.)\n\nA ruler consists of a structure made out of matchsticks (of unit length). Matchsticks can be stuck together at their ends (with magical glue). Every pair of endpoints which are rigid with respect to each other can be used to make a measurement.\n\nFor example, when $n=2$, two equilateral triangles are affixed together, the $2$ furthest endpoints are rigid with respect to each other, and can measure a distance of $\\sqrt{3}$.", null, "You are not limited in the number of matchsticks you want to use in your ruler, the more matchsticks, the grander the ruler, of course.\n\nOf course, a ruler would be useless if it could not measure lengths. What sort of lengths can your ruler measure, for each number of dimensions?\n\n• Although this is a nice question and has a nice answer, I don't know of any way to resolve it that doesn't involve quite highbrow mathematics. Element118, do you? Aug 10, 2016 at 16:21\n\nI think the answer is that the ruler can measure\n\nany algebraic distance, in any number of dimensions.\n\nTake a look at this Wikipedia page under the heading beginning \"Representation of algebraic numbers\"; it claims that\n\nfor any algebraic number A \" for any two points p and q at distance A, there exists a finite rigid unit distance graph containing p and q such that any transformation of the plane that preserves the unit distances in this graph preserves the distance between p and q\"; the corresponding configuration of matchsticks will serve as a ruler for distance A. So in the plane, at least, we can measure any algebraic distance with such a configuration of matchsticks.\n\nOn the other hand,\n\nall distances measured by such rulers in any number of dimensions are algebraic, because if we fix one matchstick so as to rule out rigid motions of the whole structure (in the plane; in higher dimensions, we need to fix more but it is easy to ensure that all the coordinates of the points we fix are algebraic) then the coordinates of the ends of the others satisfy polynomial constraints given by the edge lengths, and the particular configuration we have is an isolated solution to these constraints (else the ruler wouldn't be rigid), and these conditions imply that the coordinates are algebraic numbers. This fact is pretty easy to prove and probably well known but the only actual reference I have is to my own PhD thesis, because when I happened to need that theorem a little bit of poking around didn't find it anywhere standard :-).\n\nWe still need to consider\n\nwhether we can measure arbitrary algebraic numbers in more dimensions. It feels as if the answer should be yes -- there should be a way to turn any rigid structure in dimension d into one in dimension d+1 by adding some sort of \"bracing\" -- but I'm not at all sure I should trust that intuition. The original Beckman-Quarles theorem, which says that preserving all unit distances implies isometry, applies in euclidean space of any dimension, but of course that's not at all the same thing. ... Aha! Although the original paper cited by the Wikipedia article doesn't appear to be publicly available, a followup by the same author is -- https://archive.org/details/arxiv-math0102023 -- and it indicates that the theorem does hold in any number of dimensions.\n\nSo, to recap:\n\nWhatever the (finite) dimensionality of your space, you can measure any algebraic distance and no other distances with rulers of this kind.\n\n• The \"bracing\" from 2 to 3 dimensions can be done with octaeders. An octaeder includes a rigid square. So you can extend a 2D structure to 3D by placing on each stick a rigid square that has two sides parallel to the additiona dimension. With rigid squares you can make rigid cubes and hypercubes. So you can probably extend this to any number of dimensions. Aug 10, 2016 at 16:00\n• Ah yes, that sounds like it would work. Aug 10, 2016 at 16:20\n• @FlorianF: I don't think octahedrons are rigid in >3D. You might be able to use general cross polytopes for the job, although I don't actually know if those are rigid. Aug 10, 2016 at 16:47\n• you can't measure number like Pi, at least in 2D, since they're transcendant numbers because you can't create such a number without circular shape. But you can't, you only have lines of unit length Aug 10, 2016 at 16:48\n• @rmilville: No one is claiming to be able to measure transcendental numbers like $\\pi$. (Though I think saying \"you can't create such a number without circular shape\" is going too far; it's not inconceivable that there might be some configuration that happens to produce a distance of $\\pi$, though it turns out that there isn't.) Aug 10, 2016 at 16:51" ]	[ null, "https://i.stack.imgur.com/zSupO.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9446656,"math_prob":0.9718489,"size":1009,"snap":"2022-27-2022-33","text_gpt3_token_len":227,"char_repetition_ratio":0.12039801,"word_repetition_ratio":0.02366864,"special_character_ratio":0.22001982,"punctuation_ratio":0.11557789,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98317504,"pos_list":[0,1,2],"im_url_duplicate_count":[null,7,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-08-11T06:39:57Z\",\"WARC-Record-ID\":\"<urn:uuid:b8f52890-5fec-43d3-91e5-35b0d4117768>\",\"Content-Length\":\"228486\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b1886718-cee8-42d5-87a2-6b057f38ce1f>\",\"WARC-Concurrent-To\":\"<urn:uuid:e8ac25bf-230e-4a45-89d7-e57417e1d6b0>\",\"WARC-IP-Address\":\"151.101.1.69\",\"WARC-Target-URI\":\"https://puzzling.stackexchange.com/questions/40415/matchsticks-for-lengths\",\"WARC-Payload-Digest\":\"sha1:6YV2QU4A762ITJVA2ICV2LEGWZX4YIVZ\",\"WARC-Block-Digest\":\"sha1:LE3CG42VK6FL6ANGZHVMXG2J5KSKBQA3\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-33/CC-MAIN-2022-33_segments_1659882571234.82_warc_CC-MAIN-20220811042804-20220811072804-00603.warc.gz\"}"}
http://lamecanicadelamor.com/math-addition-worksheet-maker/	[ "Home » Reading Comprehension Worksheets » 19 Math Addition Worksheet Maker\n\n# 19 Math Addition Worksheet Maker\n\nUploaded by Sarah Bailey under Reading Comprehension Worksheets [290 views ]\n\ncollection of base ten addition worksheets worksheet generator esl valid preschool subtraction worksheet valid math worksheet maker software refrence the 2 digit plus 1 digit the measuring length to the nearest half inch a math worksheet math worksheet generator fresh reading worksheet generator valid fancy maths sum generator math worksheets ideas free printable math worksheet maker save math addition worksheet math worksheets multiplication regrouping myscres grade 2 math worksheet generator fresh the mad minute math missing number addition up to 10 mental maths or early finisher 4", null, "Math Worksheet Maker Software Refrence The 2 Digit Plus 1 Digit from math addition worksheet maker , source:portaldefe.co", null, "Grade 2 Math Worksheet Generator New Rocket Math Multiplication from math addition worksheet maker , source:maginvest.co", null, "Missing Number Addition Up to 10 Mental Maths or Early Finisher 4 from math addition worksheet maker , source:pinterest.com", null, "Math Worksheets For Time Tables Refrence Time Table Worksheet from math addition worksheet maker , source:factorx.co", null, "Worksheet Generator Esl Valid Preschool Subtraction Worksheet Valid from math addition worksheet maker , source:portaldefe.co\n\nmath worksheet generator refrence math diagram new grade 2 math worksheet generator new rocket math multiplication addition tables worksheets beautiful the worksheets this page math worksheets for time tables refrence time table worksheet adding and subtracting with facts from 1 to 5 a math worksheet preschool subtraction worksheet refrence worksheets fresh worksheet free worksheets library download and print worksheets word problem math worksheet generator valid kindergarten mixed 2nd grade math worksheets ordering numbers myscres math worksheet distributive property inspirationa distributive\n\nmath facts worksheet generator fresh math fact worksheet maker collection of math worksheets maker math worksheet maker software new grade 2 math worksheet generator math worksheet generator app new algebra answer generator math facts worksheet generator new math worksheet grade 4 best math worksheet generator kidz activities math worksheet maker software fresh basic circle worksheets math math facts worksheet generator new super teacher math worksheet collection of multiplication integers worksheets 25 single digit addition questions with no regrouping a math" ]	[ null, "http://lamecanicadelamor.com/wp-content/uploads/2018/09/math-addition-worksheet-maker-of-math-addition-worksheet-maker-1.jpg", null, "http://lamecanicadelamor.com/wp-content/uploads/2018/09/math-addition-worksheet-maker-of-math-addition-worksheet-maker-11.jpg", null, "http://lamecanicadelamor.com/wp-content/uploads/2018/09/math-addition-worksheet-maker-of-math-addition-worksheet-maker-8.jpg", null, "http://lamecanicadelamor.com/wp-content/uploads/2018/09/math-addition-worksheet-maker-of-math-addition-worksheet-maker-13.jpg", null, "http://lamecanicadelamor.com/wp-content/uploads/2018/09/math-addition-worksheet-maker-of-math-addition-worksheet-maker.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.64207643,"math_prob":0.67976105,"size":2401,"snap":"2019-13-2019-22","text_gpt3_token_len":408,"char_repetition_ratio":0.32206926,"word_repetition_ratio":0.057228915,"special_character_ratio":0.15493545,"punctuation_ratio":0.042134833,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99848765,"pos_list":[0,1,2,3,4,5,6,7,8,9,10],"im_url_duplicate_count":[null,1,null,1,null,1,null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-03-18T23:55:17Z\",\"WARC-Record-ID\":\"<urn:uuid:bd6dc483-3d7a-47a7-ac99-bf83a9aa17c5>\",\"Content-Length\":\"42381\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:757bad55-26f4-442c-86ce-5325b87f8967>\",\"WARC-Concurrent-To\":\"<urn:uuid:e71c0ad2-8c49-4c89-b001-f3505b388580>\",\"WARC-IP-Address\":\"104.31.73.114\",\"WARC-Target-URI\":\"http://lamecanicadelamor.com/math-addition-worksheet-maker/\",\"WARC-Payload-Digest\":\"sha1:4YV4BS23EGI4CS6XIAEIDC2VS3HDVBCD\",\"WARC-Block-Digest\":\"sha1:JHPXPDE3VTNMWTZCWL6I377AD6J4EAFI\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-13/CC-MAIN-2019-13_segments_1552912201812.2_warc_CC-MAIN-20190318232014-20190319014014-00157.warc.gz\"}"}
https://www.askiitians.com/iit-jee-progressions-and-series/arithmetic-mean-of-mth-power/	[ "#### Thank you for registering.\n\nOne of our academic counsellors will contact you within 1 working day.\n\nClick to Chat\n\n1800-5470-145\n\n+91 7353221155\n\nCART 0\n\n• 0\nMY CART (5)\n\nUse Coupon: CART20 and get 20% off on all online Study Material\n\nITEM\nDETAILS\nMRP\nDISCOUNT\nFINAL PRICE\nTotal Price: Rs.\n\nThere are no items in this cart.\nContinue Shopping\n\n# Arithmetic Mean of mth Power", null, "The concept of arithmetic mean has been already discussed in the previous sections. This concept can be extended till the mth power for terms and such mean is termed as the arithmetic mean of mth power. Mathematically, the arithmetic mean for mth power is defined as\n\nIf a1, a2, …, an be n positive real numbers (not all equal) and let m be a real number.", null, "• However if m ∈ (0, 1), then a1m + a2m+...+ anm/n < (a1+a2+...+an/n)m.\n\n• Obviously if m ∈ {0, 1}, then a1m + a2m+...+anm/n = (a1+a2+...+an/n)m.\n\nIllustration:\n\nIf a, b, c are positive real numbers such that a + b + c = 1, then prove that a/b + c + b/c+a + c/a+b > 3/2.\n\nSolution: We have to show that (a/b+c+1)+(b/c+a + 1)+(c/a+b + 1) > 3/2 + 3.\n\ni.e. 1/b+c + 1/c+1 + 1/a+b > 9/2.\n\nNow A.M. of mth power > mth power arithmetic mean (m = – 1 here)\n\n⇒ 1/3 [(b+c)–1 + (c+a)–1 + (a+b)–1] > [(b+c) + (c+a) + (a+b)/3]–1\n\n⇒ 1/b+c + 1/c+a + 1/a+b > 3.3/2(a+b+c) = 9/2.\n\nIllustration: Prove that\n\n1m + 2m + 3m + ….. + nm > n[1/2(n+1)]m.\n\nSolution: Taking ‘n’ quantities into consideration we have\n\nThen as stated above, arithmetic mean of mth powers > mth power of the arithmetic mean\n\n(1m + 2m + 3m + ….. + nm) / n > [(1+2+3+…. n) / n]m\n\nor (1m + 2m + 3m + ….. + nm) > n [ {n(n+1)/2}/n]m\n\n{Since the sum of first n natural numbers is [n(n+1)/2]}\n\nHence, (1m + 2m + 3m + ….. + nm) > n [ (n+1)/2}]m.\n\naskIITians offers extensive study material which covers all the important topics of IIT JEE Mathematics. The concepts of arithmetic mean of an A.P. or the arithmetic formula for the power have also been covered in the material. It is vital for the aspirants to be familiar with the concepts in order to remain competitive in the JEE.\n\nRelated resources:\n\nYou may also like to refer Geometric Mean and Harmonic Mean.\n\nTo read more, Buy study materials of Sequences and Series comprising study notes, revision notes, video lectures, previous year solved questions etc. Also browse for more study materials on Mathematics here." ]	[ null, "https://files.askiitians.com/cdn1/cms-content/iit-study-materialiit-jee-mathematicsalgebra_1.jpg", null, "https://files.askiitians.com/cdn1/images/2014129-123146755-6203-9.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8487646,"math_prob":0.99652684,"size":4188,"snap":"2021-43-2021-49","text_gpt3_token_len":1202,"char_repetition_ratio":0.11065966,"word_repetition_ratio":0.045822103,"special_character_ratio":0.2987106,"punctuation_ratio":0.1893617,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9970448,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,4,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-12-09T08:03:51Z\",\"WARC-Record-ID\":\"<urn:uuid:e3255399-630d-4df5-a1f5-3b495ab0e47c>\",\"Content-Length\":\"244664\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:2f43bd89-1b0f-49ff-a17c-4ad3e9508b5b>\",\"WARC-Concurrent-To\":\"<urn:uuid:9a8bd37b-c30c-4184-8c2e-769a927c9a5b>\",\"WARC-IP-Address\":\"3.7.36.8\",\"WARC-Target-URI\":\"https://www.askiitians.com/iit-jee-progressions-and-series/arithmetic-mean-of-mth-power/\",\"WARC-Payload-Digest\":\"sha1:P7XGHQQS7DXHIHJXZFEMWNVP62P4MC2N\",\"WARC-Block-Digest\":\"sha1:VYLP4X7SIELQC6CK2IIDWR32G7WPCT25\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964363689.56_warc_CC-MAIN-20211209061259-20211209091259-00385.warc.gz\"}"}
https://gauravtiwari.org/2011/04/	[ "# D’Alembert’s Ratio Test of Convergence of Series\n\nIn this article, we will formulate the D’ Alembert’s Ratio Test on convergence of a series. Let’s start. Statement of D’Alembert Ratio Test A series $\\sum {u_n}$ of positive terms is convergent if from and after some fixed term $\\dfrac {u_{n+1}} {u_n} < r < {1}$ , where r is a fixed…" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7695433,"math_prob":0.9834652,"size":287,"snap":"2023-14-2023-23","text_gpt3_token_len":86,"char_repetition_ratio":0.113074206,"word_repetition_ratio":0.0,"special_character_ratio":0.3031359,"punctuation_ratio":0.06896552,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9981443,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-03-25T00:41:42Z\",\"WARC-Record-ID\":\"<urn:uuid:6816084b-473f-46e5-b5c4-6c7c612f258f>\",\"Content-Length\":\"88717\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:db9b6965-0460-4363-88e2-26e2ab143ea0>\",\"WARC-Concurrent-To\":\"<urn:uuid:3719e5ae-4763-45a9-ab12-73c62d9186ac>\",\"WARC-IP-Address\":\"172.67.185.30\",\"WARC-Target-URI\":\"https://gauravtiwari.org/2011/04/\",\"WARC-Payload-Digest\":\"sha1:7CN6L5YFXZM3VOWTBMZTWBRBJZXYNYYK\",\"WARC-Block-Digest\":\"sha1:OHPCH6UZLZSWAXUP67RCIBWHGQSNFALS\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-14/CC-MAIN-2023-14_segments_1679296945292.83_warc_CC-MAIN-20230325002113-20230325032113-00097.warc.gz\"}"}
https://bgods.cn/blog/post/59/	[ " bgods - NumPy 数组属性\n\n# NumPy 数组属性\n\n## 一、NumPy 数组属性\n\nNumPy 数组的维数称为秩（rank），秩就是轴的数量，即数组的维度，一维数组的秩为 1，二维数组的秩为 2，以此类推。\n\nNumPy 的数组中比较重要 ndarray 对象属性有：\n\nndarray.ndim 秩，即轴的数量或维度的数量\nndarray.shape 数组的维度，对于矩阵，n 行 m 列\nndarray.size 数组元素的总个数，相当于 .shape 中 n*m 的值\nndarray.dtype ndarray 对象的元素类型\nndarray.itemsize ndarray 对象中每个元素的大小，以字节为单位\nndarray.flags ndarray 对象的内存信息\nndarray.real ndarray元素的实部\nndarray.imag ndarray 元素的虚部\nndarray.data 包含实际数组元素的缓冲区，由于一般通过数组的索引获取元素，所以通常不需要使用这个属性。\n\n### 1、ndarray.ndim\n\nndarray.ndim 用于返回数组的秩，即轴的数量或维度的数量\n\n``````import numpy as np\na = np.arange(12) # 返回一个一维数组序列，和自带 range 方法类似\nprint(a.ndim) # a 现只有一个维度\n# 现在调整其大小\nb = a.reshape(4,3) # b 现在拥有2个维度\nprint(b.ndim)``````\n\n```1\n2\n```\n\n### 2、ndarray.shape\n\nndarray.shape 表示数组的维度，返回一个元组，这个元组的长度就是维度的数目，即 ndim 属性(秩)。比如，一个二维数组，其维度表示\"行数\"和\"列数\"。\n\n• 数组的维度\n``````import numpy as np\na = np.array([[1,2,3],[4,5,6]])\nprint(a.shape)``````\n\n`(2, 3)`\n• 调整数组大小\n\nndarray.shape 也可以用于调整数组大小。\n\n``````import numpy as np\na = np.array([[1,2,3],[4,5,6]])\na.shape = (3,2)\nprint(a)\nb = a.reshape(2,3) # NumPy 也提供了reshape函数来调整数组大小\nprint(b)``````\n\n```[[1 2]\n[3 4]\n[5 6]]\n\n[[1 2 3]\n[4 5 6]]```\n\n1. a.shape 是数组对象的属性，修改只需要使用“=”重新赋值，会在原数组上修改；\n2. a.reshape() 是数组对象的一个方法，会返回一个新的数组，原来数组不会改变。\n\n### 3、ndarray.size\n\n``````a = np.arange(24) # 创建数组\na.shape = (4,3,2) # 改变数组维度\nprint(a.shape)\nprint(a.size)``````\n\n`````` (4, 3, 2)\n24``````\n\n### 5、ndarray.itemsize\n\nndarray.itemsize 以字节的形式返回数组中每一个元素的大小。\n\n``````import numpy as np\n\n# 数组的 dtype 为 int8（一个字节）\nx = np.array([1,2,3,4,5], dtype = np.int8)\nprint (x.itemsize)\n\n# 数组的 dtype 现在为 float64（八个字节）\ny = np.array([1,2,3,4,5], dtype = np.float64)\nprint (y.itemsize)``````\n\n```1\n8```\n\n### 6、ndarray.flags\n\nndarray.flags 返回 ndarray 对象的内存信息，包含以下属性：\n\nC_CONTIGUOUS (C) 数据是在一个单一的C风格的连续段中\nF_CONTIGUOUS (F) 数据是在一个单一的Fortran风格的连续段中\nOWNDATA (O) 数组拥有它所使用的内存或从另一个对象中借用它\nWRITEABLE (W) 数据区域可以被写入，将该值设置为 False，则数据为只读\nALIGNED (A) 数据和所有元素都适当地对齐到硬件上\nUPDATEIFCOPY (U) 这个数组是其它数组的一个副本，当这个数组被释放时，原数组的内容将被更新\n``````import numpy as np\n\nx = np.array([1,2,3,4,5])\nprint (x.flags)``````\n\n``` C_CONTIGUOUS : True\nF_CONTIGUOUS : True\nOWNDATA : True\nWRITEABLE : True\nALIGNED : True\nWRITEBACKIFCOPY : False\nUPDATEIFCOPY : False```\n\n### 7、ndarray.real和ndarray.imag\n\nndarray.real 获取 ndarray 元素的实部，ndarray.imag 获取 ndarray 元素的虚部。如：\n\n``````a = np.array([1+3j, 2+4j])\nprint(\"数组a是:\", a)\nprint(\"实部:\", a.real)\nprint(\"虚部:\", a.imag)``````\n\n``````数组a是: [1.+3.j 2.+4.j]\n\n\n\n0 评论" ]	[ null ]	{"ft_lang_label":"__label__zh","ft_lang_prob":0.80016583,"math_prob":0.99606186,"size":2821,"snap":"2021-43-2021-49","text_gpt3_token_len":1639,"char_repetition_ratio":0.16045438,"word_repetition_ratio":0.025723472,"special_character_ratio":0.3105282,"punctuation_ratio":0.19072165,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9929052,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-25T07:49:14Z\",\"WARC-Record-ID\":\"<urn:uuid:814247d4-cea3-4bcd-af25-d3a07c7e3729>\",\"Content-Length\":\"35488\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:7188d451-4411-40c3-a72a-dd7b8107b63f>\",\"WARC-Concurrent-To\":\"<urn:uuid:aef79b4e-53d1-47d6-b88c-86cabeacca81>\",\"WARC-IP-Address\":\"101.200.48.156\",\"WARC-Target-URI\":\"https://bgods.cn/blog/post/59/\",\"WARC-Payload-Digest\":\"sha1:ISMDAVCPDRWCTRF53CHDQFWMD3YUBXEL\",\"WARC-Block-Digest\":\"sha1:MOPUJK5ZUSKDZFTI62JSRTDGW5TOK3IN\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323587655.10_warc_CC-MAIN-20211025061300-20211025091300-00026.warc.gz\"}"}
https://www.greenlighttestprep.com/module/gre-probability/video/759	[ "# Lesson: Listing vs Counting vs Probability Rules\n\n## Comment on Listing vs Counting vs Probability Rules\n\n### At the 1:23 mark, why can't\n\nAt the 1:23 mark, why can't you follow the formula and 1/2 + 1/2?", null, "### We could have continued that\n\nWe could have continued that approach, but this video is all about examining our different options and probability questions involving \"at least\" are typically solved the fasted by using the complement.\n\nThat said, let's continue where we left off: P(1 heads) + P(2 heads)\n\nWe'll deal with each probability on its own.\n\nP(1 heads) = P(1st toss heads and 2nd toss tails OR 1st toss tails and 2nd toss heads)\n= P(1st toss heads and 2nd toss tails) + P(1st toss tails and 2nd toss heads)\n= (1/2)(1/2) + (1/2)(1/2)\n= 1/4 + 1/4\n= 1/2\n\n= (1/2)(1/2)\n= 1/4\n\nSo, P(1 heads) + P(2 heads) = 1/2 + 1/4 = 3/4\n\n### 6C2/10C2.. Can you solve it\n\n6C2/10C2.. Can you solve it in detail.. I mean do we have to simplify it like 6C2=180...Or there is a shortcut to solve this", null, "### Yes, there is definitely a\n\nYes, there is definitely a shortcut.\nHere's the video that explains how you can calculate 6C2 and 10C2 in your head: https://www.greenlighttestprep.com/module/gre-counting/video/789\n\nCheers,\nBrent\n\n### I got it.. Thanks\n\nI got it.. Thanks\n\n### Hello Brent,\n\nHello Brent,\n\nI solved the question as follows\nJoe is in a group consisting of N people (N > 4).\n\nIf 3 people from the group are randomly selected to be on a committee, what is the probability that Joe is selected?\n\nI assument N = 5 , there are total of 5C3 was to select people , this is my denominator that is 1/10.\nNow i have to choose Joe so 1 way now remaining 2 people can be selected in 4C2 ways that is (1 X 4 X 3)/ (2 X 1) which comes to 6.\n\nSO final answer is 6/10 = 3/5 hence option C. am i right.", null, "You're using the INPUT-OUTPUT approach to solve this question (covered here: https://www.greenlighttestprep.com/module/gre-word-problems/video/936)\n\nWhen N = 5, the probability = 3/5\n\nAt this point, we must check ALL 5 answer to see which one(s) evaluate to be 3/5 when N = 5\n\nA) When N = 5, (N²−2N−6)/(N−2)²+6 = 3/5\nC) When N = 5, N/5 = 3/5\n\nSo, the correct answer is EITHER A or C.\n\nFrom here, you must test a different value of N to determine the correct answer.\n\nCheers,\nBrent\n\n### Hi Brent,\n\nHi Brent,\nIn the last question of Reinforcement Activities, click on the link GRE practice question (difficulty level: 160 to 170) – Greenlight Test Prep\nInstead of asking of what is the probability that Joe is selected, he is asking what is the probabilty of selection three people, what would be the answer to this question?", null, "Sorry, but I'm not sure what you mean by \"probability of selection three people\"\n\n### It's great how you can try\n\nIt's great how you can try multiple approached to get the same answer with probability questions. The only issue I see here is which approach is the shortest. I don't think you can accurately assess that in the beginning. When you start with 1 approach, it's best to just carry it out to the end. However long it takes, it takes, and just hope that you get the right answer. If you get stuck, then you'll have to try a different approach if you can, which still eats up time. Really, the greatest enemy on the GRE is time. I keep coming back to this theme again and again.", null, "### Good points.\n\nGood points.\nIn general, I find that applying probability rules is typically faster than using counting techniques." ]	[ null, "https://www.greenlighttestprep.com/sites/default/files/styles/thumbnail/public/pictures/picture-1-1475568960.jpg", null, "https://www.greenlighttestprep.com/sites/default/files/styles/thumbnail/public/pictures/picture-1-1475568960.jpg", null, "https://www.greenlighttestprep.com/sites/default/files/styles/thumbnail/public/pictures/picture-1-1475568960.jpg", null, "https://www.greenlighttestprep.com/sites/default/files/styles/thumbnail/public/pictures/picture-1-1475568960.jpg", null, "https://www.greenlighttestprep.com/sites/default/files/styles/thumbnail/public/pictures/picture-1-1475568960.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9020253,"math_prob":0.8831068,"size":3564,"snap":"2020-34-2020-40","text_gpt3_token_len":1025,"char_repetition_ratio":0.125,"word_repetition_ratio":0.05645161,"special_character_ratio":0.2926487,"punctuation_ratio":0.10181311,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9754029,"pos_list":[0,1,2,3,4,5,6,7,8,9,10],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-08-11T09:41:39Z\",\"WARC-Record-ID\":\"<urn:uuid:24d72f32-766d-4b5c-a96e-95735535a5b0>\",\"Content-Length\":\"35836\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:6877df21-d0f0-4686-a497-647270b04e69>\",\"WARC-Concurrent-To\":\"<urn:uuid:ebbf737d-fd1c-4896-bd51-38a4c62ee07b>\",\"WARC-IP-Address\":\"108.160.144.133\",\"WARC-Target-URI\":\"https://www.greenlighttestprep.com/module/gre-probability/video/759\",\"WARC-Payload-Digest\":\"sha1:MBMZ34TB3PS6352LW5QWXKDS4PRHPCWA\",\"WARC-Block-Digest\":\"sha1:IASDKQ65ZTSZB33BW2LQCU3JMVUVXVAK\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-34/CC-MAIN-2020-34_segments_1596439738746.41_warc_CC-MAIN-20200811090050-20200811120050-00122.warc.gz\"}"}
https://math.stackexchange.com/questions/3016507/derivative-of-a-functional-with-respect-to-another-functional	[ "# Derivative of a functional with respect to another functional\n\nI am trying to make sense of functional derivatives and have a couple of questions bothering me:\n\n1. Let $$F[X]$$ be a functional of $$X(t)$$ and $$G[X]$$ another functional of $$X(t)$$.\n\nBy chain rule in the continum I intuitively guess that the derivative of the functional F[X] with respect to the functional G[X] is:$$\\frac{\\delta F}{\\delta G} = \\int dt \\frac{\\delta F[X]}{\\delta X(t)} \\frac{\\delta X(t)}{\\delta G[X]} = \\int dt \\frac{\\delta F[X]}{\\delta X(t)} \\left(\\frac{\\delta G[X]}{\\delta X(t)} \\right)^{-1}$$ Does this make any sense at all? Is there such thing as a derivative of a functional with respect to another functional given they are both dependent on the same underlying function X(t)?\n\n2. How would one calculate the functional derivatives $$\\frac{\\delta F[X]}{\\delta X(t)}$$ and $$\\frac{\\delta G[X]}{\\delta X(t)}$$? I came across Gateaux derivative and It seems to be similar to a directional derivative. Does it matter what direction is being chosen to evaluate the final functional derivative?\n\n• I am not sure what you are asking. Please rewrite. – Will M. Dec 11 '18 at 5:14\n• Thanks Will, just edited please let me know if it's still not clear. – ZeroCool Dec 11 '18 at 15:44\n• OK. It still makes no sense. However, two points to bring. (1) Frechet derivative is simply the usual derivative on learns from calculus 2 (at least in places like the one I studied where math rigour is imposed over being able to perform calculations) and (2) Gateaux derivatives are directional derivatives and, as always, they depend on the chosen direction. If you want to know the derivatives of a composition $g \\circ f,$ then simply apply the chain rule $g'(f(a)) \\circ f'(a).$ If you want to know how to calculate $f'$ or $g',$ you need to know the exact form first an apply rules then. – Will M. Dec 11 '18 at 18:05\n• @ZeroCool you might want to ask on MathOverflow. This site is for more elementary questions (usually pre-functional-derivative stage). – Nike Dattani May 30 '20 at 22:18" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.86162347,"math_prob":0.99824643,"size":996,"snap":"2021-31-2021-39","text_gpt3_token_len":253,"char_repetition_ratio":0.22379032,"word_repetition_ratio":0.027210884,"special_character_ratio":0.2751004,"punctuation_ratio":0.041666668,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99986887,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-08-05T02:59:21Z\",\"WARC-Record-ID\":\"<urn:uuid:adfe309d-e69e-4d1f-a153-2535b084c0d1>\",\"Content-Length\":\"159111\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:5e7221ec-4852-4537-b6ea-7064a3dae5f2>\",\"WARC-Concurrent-To\":\"<urn:uuid:94e1921e-1a04-4977-b135-7511be1e84dc>\",\"WARC-IP-Address\":\"151.101.193.69\",\"WARC-Target-URI\":\"https://math.stackexchange.com/questions/3016507/derivative-of-a-functional-with-respect-to-another-functional\",\"WARC-Payload-Digest\":\"sha1:AWUEBQBUXMA2VAUSQYJJJOKA7AGGRS32\",\"WARC-Block-Digest\":\"sha1:FTL74D27IUCEBTG2WVVLEGFYFTTP4EU2\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-31/CC-MAIN-2021-31_segments_1627046155268.80_warc_CC-MAIN-20210805000836-20210805030836-00193.warc.gz\"}"}