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It estimates ... # Viterbi: # If we have a word sequence, what is the best tag sequence? What would you like to do? The full code can be found at: Also, here are the list of all the articles in this series: Filed Under: Machine Learning Tagged With: Decoding Problem, Dynamic Programming, Hidden Markov Model, Implementation, Machine Learning, Python, R, step by step, Viterbi. The Viterbi decoder itself is the primary focus of this tutorial. Consist of a learning module that calculates transition and emission probabilities of the training set and applies this model on the test data set. Please click on the ‘Code’ Button to access the files in the github repository. Skip to content. This means that all observations have to be acquired before you can start running the Viterbi algorithm. We will be using a much more efficient algorithm named Viterbi Algorithm to solve the decoding problem. Therefore HMM the following components along with components of Markov chain model mentioned above: The problem of POS tagging is modeled by considering the tags as states and the words as observations. σ2I(where Iis the K×Kidentity matrix) and unknown σ, VT, or CEM, is equivalent to the k-means clustering [9, 10, 15, 43]. Instead, we can employ a dynamic programming approach to make the problem tractable; the module that I wrote includes an implementation of the Viterbi algorithm for this purpose. The underflow problem and how to solve it. T) \) to solve. The code has been implemented from scratch and commented for better understanding of the concept. The basic idea here is that for unknown words more probability mass should be given to tags that appear with a wider variety of low frequency words. We will start with the formal definition of the Decoding Problem, then go through the solution and finally implement it. This would be easy to do in Python by iterating over observations instead of slicing it. Re-run EM with restarts or a lower convergence threshold. * * Program follows example from Durbin et. HMM Training (part 4) 13:16. As stated earlier, we need to find out for every time step t and each hidden state what will be the most probable next hidden state. However, just like we have seen earlier, it will be an exponentially complex problem $$O(N^T . Join and get free content delivered automatically each time we publish, # This is our most probable state given previous state at time t (1), # This is the probability of the most probable state (2), # Find the most probable last hidden state, # Flip the path array since we were backtracking, # Convert numeric values to actual hidden states, # ((1x2) . In all these cases, current state is influenced by one or more previous states. Cut Go through the example below and then come back to read this part. Note, here \( S_1 = A$$ and $$S_2 = B$$. The Viterbi algorithm works like this: for each signal, calculate the probability vector p_state that the signal was emitted by state i (i in [0,num_states-1]). The final most probable path in this case is given in the below diagram, which is similar as defined in fig 1. This is the best tutorial out there as i find the example really easy, easiest tbh. Start with some initial values ψ (0)= (P(0),θ ) and (use the Viterbi algorithm to) ﬁnd a realization of. The Viterbi algorithm is a dynamic programming algorithm for finding the most likely sequence of hidden states—called the Viterbi path—that results in a sequence of observed events, especially in the context of Markov information sources and hidden Markov models (HMM).. But is there anyway for me to show the Probabilities of Sequence ? a QGIS-plugin for matching a trajectory with a network using a Hidden Markov Model and Viterbi algorithm. In this post, we introduced the application of hidden Markov models to a well-known problem in natural language processing called part-of-speech tagging, explained the Viterbi algorithm that reduces the time complexity of the trigram HMM tagger, and evaluated different trigram HMM-based taggers with deleted interpolation and unknown word treatments on the subset of the Brown corpus. I noticed that the comparison of the output with the HMM library at the end was done using R only. Do share this article if you find it useful. Viterbi algorithm for Hidden Markov Models (HMM) taken from wikipedia - Viterbi.py We can repeat the same process for all the remaining observations. Viterbi algorithm is a dynamic programming algorithm for finding the most likely sequence of hidden states—called the Viterbi path—that results in a sequence of observed events, especially in the context of Markov information sources and hidden Markov models.. Similar to the most probable state ( at each time step ), we will have another matrix of size 2 x 6 ( in general M x T ) for the corresponding probabilities (2). author: becxer created: 2015-10-15 11:58:11 apriori clustering crf … For example, already visited locations in the fox's search might be given a very low probability of being the next location on the grounds that the fox is smart enough not to repeat failed search locations… I want to ask about the data used. Viterbi Algorithm is an algorithm to find the optimal path (or most likely path, or minimal cost path, etc) through the graph. Hidden Markov model and sequence annotation In Chapter 3, the n-ary grammar model marks the binary connection in the full segmentation word network from the fluency of word continuity, and then uses Viterbi algorithm to solve the path with the maximum likelihood probability. All these can be solved via smoothing. - [Narrator] Using a representation of a hidden Markov model … that we created in model.py, … we can now make inferences using the Viterbi algorithm. baum-welch-algorithm bayesian hidden-markov-models hmm hmm-viterbi-algorithm python. nkt1546789 / viterbi.py. It acts like a discounting factor. POS tagging is a fundamental block for Named Entity Recognition(NER), Question Answering, Information Extraction and Word Sense Disambiguation[1]. D D D + + 1 1 1 1 1 1 0 1 G 0 G 1 G 2 G 3 1+D+D2+D3 1+D+D3 C j1 C j2 1 input 2 outputs Impulse responses are P ( D) = 1 + +2 3. sT = i, v1, v2…vT | θ) We can use the same approach as the Forward Algorithm to calculate ωi( + 1) ωi(t + 1) = maxi(ωi(t)aijbjkv ( t + 1)) Now to find the sequence of hidden states we need to identify the state that maximizes ωi(t) at each time step t. He was appointed by Gaia (Mother Earth) to guard the oracle of Delphi, known as Pytho. You may use various prepro-cesssing steps on the dataset (lowercasing the tokens, stemming etc.). One straightforward method would be the brute force method, i.e., to calculate probabilities of all possible combinations. Few characteristics of the dataset is as follows: There are 2x1x4x2x2=32 possible combinations. Assume when t = 2, the probability of transitioning to $$S_2(2)$$ from $$S_1(1)$$ is higher than transitioning to $$S_1(2)$$, so we keep track of this. Figure 1: An illustration of the Viterbi algorithm. You can also use various techniques for unknown words. In hard decision decoding, where we are given a sequence of … Description of the Algorithms (Part 2) Performing Viterbi Decoding. Here is the result. The first part of the assignment is to build an HMM from data. Perhaps the single most important concept to aid in understanding the Viterbi algorithm is the trellis diagram. We will see what Viterbi algorithm is. ωi(t) = maxs1, …, sT − 1p(s1, s2…. ... _sentence(tagger_data, sentence): apply the Viterbi algorithm retrace your steps return the list of tagged words Implement the Viterbi algorithm, which will take a list of words and output the most likely path through the HMM state space. Which makes your Viterbi searching absolutely wrong. POS tagging refers labelling the word corresponding to which POS best describes the use of the word in the given sentence. # # The method above lets us determine the probability for a … Baum-Welch Updates for Multiple Observations. At issue is how to predict the fox's next location. The file must contain a word: and its POS tag in each line, seperated by ' \t '. If we draw the trellis diagram, it will look like the fig 1. We need to predict the sequence of the hidden states for the visible symbols. Calculating probabilites for 32 combinations might sound possible but as the length of sentences increases, the computations increase exponentially. Derivation and implementation of Baum Welch Algorithm for Hidden Markov Model. Implementation using Python. Our objective is to find the sequence {t1 t2 t3…tn} that maximizes the probability defined in the above equation. We will start with Python first. Here the sentence for which ithe POS tagging is done is considered as a set of sequence of words and sequence of tags. Part-Of-Speech refers to the purpose of a word in a given sentence. The Viterbi Algorithm (part 2) 15:04. When observing the word "toqer", we can compute the most probable "true word" using Viterbi algorithm in the same way we used it earlier, and get the true word "tower". Unknown words of the test are given a fixed probability. In this article we will implement Viterbi Algorithm in Hidden Markov Model using Python and R. Viterbi Algorithm is dynamic programming and computationally very efficient. During these 3 days, he told you, that he feels Normal (1st day), Cold (2nd day), Dizzy (3r… This can be calculated with the help HMM. Use dynamic programming to find the most probable combination based on the word frequency. I hope it will definitely be more easy to understand once you have the intuition. The parameters which need to be calculated at each step has been shown above. Why is this interesting? In the brute force method, to find the probability for the tag sequence {VBD, TO, JJ, DT, NN} and {VBD, TO, RB, DT, NN} we calculate the probability of the smaller paths(VBD->TO) twice. The last one can be solved by an iterative Expectation-Maximization (EM) algorithm, known as the Baum-Welch algorithm. Here is the same link: HMM Training (part 3) 13:33. Given the state diagram and a sequence of N observations over time, we need to tell the state of the baby at the current point in time. Sign in Sign up Instantly share code, notes, and snippets. 9.2 The Viterbi Decoder The decoding algorithm uses two metrics: the branch metric (BM) and the path metric (PM).Thebranchmetricisameasureofthe“distance”betweenwhatwastransmittedand what was received, and is deﬁned for each arc in the trellis. The 3rd and final problem in Hidden Markov Model is the Decoding Problem. Your email address will not be published. Discrete HMM in Code. HMM Training (part 1) 04:40. Implementation using Python. Now, I am pretty slow at recursive functions, so it took me some time to reason this myself. You are a doctor in a little town. There are set of rules for some POS tags dictating what POS tag should follow or precede them in a sentence. I have one doubt, i use the Baum-Welch algorithm as you describe but i don’t get the same values for the A and B matrix, as a matter of fact the value a_11 is practically 0 with 100 iterations, so when is evaluated in the viterbi algorithm using log produce an error: “RuntimeWarning: divide by zero encountered in log”, It’s really important to use np.log? Here is the link for the GitHub gist for the above code. We can use the same approach as the Forward Algorithm to calculate $$\omega _i(+1)$$. One approach would be to use the entire search history P1, P2,…, C to predict the next location. Then I have a test data which also contains sentences where each word is tagged. Uses Viterbi algorithm to classify text with their respective parts of speech tags. We will explain its performance by using a Java Applet that runs it. Download this Python file, which contains some code you can start from. Markov chain models the problem by assuming that the probability of the current state is dependent only on the previous state. One way out of this is to make use of the context of occurence of a word. The programming language Python has not been created out of slime and mud but out of the programming language ABC. Assuming you can store or generate every word form with your dictionary, you can use an algorithm like the one described here (and … This is where the Viterbi algorithm comes to the rescue. POS tagging). One implementation trick is to use the log scale so that we dont get the underflow error. The decoding problem is similar to the Forward Algorithm. So the Laplace smoothing counts would become . The*Viterbi#algorithm is*a*dynamicalprogramming*algorithm*that* allows*us*tocomputethemost*probablepath. You can find them in the python code ( they are structurally the same ). Given below is the implementation of Viterbi algorithm in python. Everything what I said above may not make a lot of sense now. def words_and_tags_from_file (filename): """ Reads words and POS tags from a text file. [5]Francis, W. Nelson, and Henry Kucera. The states are the tags which are hidden and only the words are observable. The Viterbi algorithm is used for decoding, i.e. Viterbi Algorithm is an algorithm to find the optimal path (or most likely path, or minimal cost path, etc) through the graph. The descriptions and outputs of each are given below: ###Viterbi_POS_WSJ.py It uses the POS tags from the WSJ dataset as is. Viterbi algorithm is within the scope of WikiProject Robotics, which aims to build a comprehensive and detailed guide to Robotics on Wikipedia. The R code below does not have any comments. Using HMMs for tagging-The input to an HMM tagger is a sequence of words, w. The output is the most likely sequence of tags, t, for w. -For the underlying HMM model, w is a sequence of output symbols, and t is the most likely sequence of states (in the Markov chain) that generated w. Needleman-Wunsch) HMM : Viterbi algorithm - a toy example H Start A 0.2 C … In hard decision decoding, where we are given a sequence of digitized parity bits, the branch metric is the Hamming distance between the … Please post comment in case you need more clarification to any of the section. Python had been killed by the god Apollo at Delphi. Your email address will not be published. Assume, in this example, the last step is 1 ( A ), we add that to our empty path array. *Its*principleis*similar*to the*DPprograms*used*toalign*2sequences*(i.e.Needleman GWunsch) HMM#:#Viterbi#algorithm#1 atoyexample H Start A****0.2 C****0.3 G****0.3 T****0.2 L A****0.3 C****0.2 G****0.2 T****0.3 0.5 0.5 0.5 0.4 0.5 0.6 G G C A C T G A A Viterbi#algorithm… You will be given a transition matrix, an … How to Choose the Number of Hidden States. I am only having partial result here. See the ref listed below for further detailed information. Take a look, https://www.oreilly.com/library/view/hands-on-natural-language/9781789139495/d522f254-5b56-4e3b-88f2-6fcf8f827816.xhtml, https://en.wikipedia.org/wiki/Part-of-speech_tagging, https://www.freecodecamp.org/news/a-deep-dive-into-part-of-speech-tagging-using-viterbi-algorithm-17c8de32e8bc/, https://sites.google.com/a/iitgn.ac.in/nlp-autmn-2019/, Build a Reinforcement Learning Terran Agent with PySC2 2.0 framework, What We Learned by Serving Machine Learning Models Using AWS Lambda, 10x Machine Learning Productivity With Stellar Questionnaire, Random Forest Algorithm for Machine Learning, The actor-Critic Reinforcement Learning algorithm, How to Use Google Cloud and GPU Build Simple Deep Learning Environment, A Gaussian Approach to the Detection of Anomalous Behavior in Server Computers. The baseline algorithm uses the most frequent tag for the word. #!/usr/bin/env python: import argparse: import collections: import sys: def train_hmm (filename): """ Trains a Hidden Markov Model with data from a text file. This is a screenshot taken from the lecture slides, so credits are to Columbia university. But since observations may take time to acquire, it would be nice if the Viterbi algorithm could be interleaved with the acquisition of the observations. In this post, we introduced the application of hidden Markov models to a well-known problem in natural language processing called part-of-speech tagging, explained the Viterbi algorithm that reduces the time complexity of the trigram HMM tagger, and evaluated different trigram HMM-based taggers with deleted interpolation and unknown word treatments on the subset of the Brown corpus. For example, a word that occurs between an determiner and a noun should be an adjective. In this section, we are going to use Python to code a POS tagging model based on the HMM and Viterbi algorithm. Define a method , HMM.viterbi, that implements the Viterbi algorithm to find the best state sequence for the output sequence of a given observation. CS447: Natural Language Processing (J. Hockenmaier)! This is highlighted by the red arrow from $$S_1(1)$$ to $$S_2(2)$$ in the below diagram. Assume we have a sequence of 6 visible symbols and the model $$\theta$$. To simplify things a bit, the patient can be in one of 2 states: (Healthy, Fever) and he can tell you 3 feelings: (Normal, Cold, Dizzy). Thanks again. Given a sentence it is not feasible to try out every possible combinations and find the one that best matches the semantic of the sentence. HMM is an extension of Markov chain. Viterbi Algorithm. Based on a prefix dictionary structure to achieve efficient word graph scanning. Required fields are marked *. python hmm.py data/english_words.txt models/two-states-english.trained v If the separation is not what you expect, and your code is correct, perhaps you got stuck in low local maximum. various techniques for unknown words. A good example of the utility of HMMs is the annotation of genes in a genome, which is a very difficult problem in eukaryotic organisms. “Brown corpus.”. /** * Implementation of the viterbi algorithm for estimating the states of a * Hidden Markov Model given at least a sequence text file. it becomes zero if u assign log no this kinds of problem … p(w_1 w_2 w_3…w_n, t_1 t_2 t_3…t_n) is the probability that the w_i is assigned the tag t_i for all 1≤i≤n. We went through the Evaluation and Learning Problem in detail including implementation using Python and R in my previous article. The dataset that we used for the implementation is Brown Corpus[5]. However, I found the Viterbi algorithm usage in tokenization is very different. In case you want a refresh your memories, please refer my previous articles. 05:05. For my training data I have sentences that are already tagged by word that I assume I need to parse and store in some data structure. if you can explain why is that log helps to avoid underflow error and your thoughts about why i don’t get the same values for A and B, it would be much appreciated, why log? The POS tag of a word can vary depending on the context in which it is used. Returns two lists of same: length: one containing the words and one containing the tags. """ pytrain: Machine Learning library for python. This is the 4th part of the Introduction to Hidden Markov Model tutorial series. the forward-backward algorithm, and the Baum{Welch algorithm. So, before moving on to the Viterbi Algorithm, ... We get an unknown word in the test sentence, and we don’t have any training tags associated with it. The algorithm has found universal application in decoding the convolutional codes used in both CDMA and GSM digital cellular, dial-up modems, … In the Viterbi algorithm and the forward-backward algorithm, it is assumed that all of the parameters are known|in other words, the initial distribution ˇ, transition matrix T, and emission distributions "i are all known. Imagine a fox that is foraging for food and currently at location C (e.g., by a bush next to a stream). Most Viterbi algorithm examples come from its application with Hidden Markov Model (e.g. Isolated-Word Speech Recognition Using Hidden Markov Models 6.962 Week 10 Presentation Irina Medvedev Massachusetts Institute of Technology April 19, 2001 – A free PowerPoint PPT presentation (displayed as a Flash slide show) on PowerShow.com - id: 545de3-NGM0O Hidden Markov Model is a probabilistic sequence model, that computes probabilities of sequences based on a prior and selects the best possible sequence that has the maximum probability. Done in collaboration with Prateek Chennuri, Latest news from Analytics Vidhya on our Hackathons and some of our best articles! DATOL: Phylogenetic Marker Discovery Pipeline Utilizing Deep Sequencing Data. Hi, If you would like to participate, you can choose to , or visit the project page (), where you can join the project and see a list of open tasks. Embed. So, revise it and make it more clear please. POS tagging). Moreover, often we can observe the effect but not the underlying cause that remains hidden from the observer. Given a sequence of visible symbol $$V^T$$ and the model ( $$\theta \rightarrow \{ A, B \}$$ ) find the most probable sequence of hidden states $$S^T$$. We want to find out if Peter would be awake or asleep, or rather which state is more probable at time tN+1. Most Viterbi algorithm examples come from its application with Hidden Markov Model (e.g. The other path is in gray dashed line, which is not required now. In this post we will focus on the famous Viterbi algorithm, the theory behind it and also a step-by-step implementation of it in python. Hidden Markov Model (HMM) helps us figure out the most probable hidden state given an observation. In the next section, we are going to study a practical example of the Viterbi algorithm; the maximum-likelihood algorithm based on convolutional codes. This “Implement Viterbi Algorithm in Hidden Markov Model using Python and R” article was the last part of the Introduction to the Hidden Markov Model tutorial series. Next we find the last step by comparing the probabilities(2) of the T’th step in this matrix. Last active Mar 18, 2017. The corpus consists of 9580 ambiguous types having more than 1 tags and 40237 types having unambiguous tags. However, the ambiguous types occur more frequently when compared to that of the unambiguous types. Star 0 Fork 0; Code Revisions 3. Embed Embed this gist in your website. where can i get the data_python.csv? Share Copy sharable link for this gist. In this section, we are going to use Python to code a POS tagging model based on the HMM and Viterbi algorithm. Main Functions 1. implement the Viterbi algorithm for finding the most likely sequence of states through the HMM, given "evidence"; and; run your code on several datasets and explore its performance. (1x2))      *     (1), #                        (1)            *     (1), # Due to python indexing the actual loop will be T-2 to 0, # Equal Probabilities for the initial distribution. The Viterbi algorithm is a dynamic programming algorithm for finding the most likely sequence of hidden states—called the Viterbi path—that results in a sequence of observed events, especially in the context of Markov information sources and hidden Markov models. In that previous article, we had briefly modeled th… The code has comments and its following same intuition from the example. … But, before jumping into the Viterbi algorithm, … let's see how we would use the model … to implement the greedy algorithm … that just looks at each observation in isolation. like Log Probabilities of V. Morning, excuse me. I believe these articles will help anyone to understand HMM. Since your friends are Python developers, when they talk about work, they talk about Python 80% of the time.These probabilities are called the Emission probabilities. HMM Training (part 2) 10:21. The intuition behind the Viterbi algorithm is to use dynamic programming to reduce the number of computations by storing the calculations that are repeated. Later we will compare this with the HMM library. From the above figure, we can observe that as the length of the sentence (number of tokens), the computation time of the algorithm also increases. This would be easy to do in Python by iterating over observations instead of slicing it. 04:53. There is a patient, who visited you for 3 days in a row. In case any of this seems like Greek to you, go read the previous articleto brush up on the Markov Chain Model, Hidden Markov Models, and Part of Speech Tagging. It does not take into account of what was the weather day before yesterday. The method should set the state sequence of the observation to be this Viterbi state sequence. In this assignment, you will implement the Viterbi algorithm for inference in hidden Markov models. Word embeddings can be generated using various methods like neural networks, co … Thank you for the awesome tutorial. Assuming you can store or generate every word form with your dictionary, you can use an algorithm like the one described here (and already mentioned by @amp) to divide your input into a sequence of words. For the unknown words, the ‘NNP’ tag has been assigned. 11:53. Viterbi algorithm on Python. For unknown words, a HMM-based model is used with the Viterbi algorithm. So as we go through finding most probable state (1) for each time step, we will have an 2x5 matrix ( in general M x (T-1) ) as below: The first number 2 in above diagram indicates that current hidden step 1 (since it’s in 1st row) transitioned from previous hidden step 2. In the case of Viterbi, the time complexity is equal to O (s * s * n) where s is the number of states and n is the number of words in the input sequence. C This article has been rated as C-Class on the project's quality scale. This means that all observations have to be acquired before you can start running the Viterbi algorithm. The previous locations on the fox's search path are P1, P2, P3, and so on. Consequently the transition and emission probabilities are also modified as follows. # Hidden Markov Models in Python # Katrin Erk, March 2013 updated March 2016 # # This HMM addresses the problem of part-of-speech tagging. You only hear distinctively the words python or bear, and try to guess the context of the sentence. Of occurence of a word usage in tokenization is very different respective parts of speech tags length of sentences,. Major POS can be described as follows, Segmental K-Means algorithm & Baum-Welch re-Estimation algorithm by. ( step ) be viterbi algorithm for unknown words python adjective history P1, P2, P3, and try to guess the in. The probabilities of the unambiguous types to identify the maximum likelihood hidden.! Are P1, P2, P3, and the information of the assignment is to make use of decoding... Then we find the previous state a fox that is foraging for food and at. Are 4 possible states ( p_signal ) in sign up instantly share code, notes, so... Used with the HMM library at the end was done using R only make it more clear please on... Sense now part 1. Python was created out of slime and but... ( HMM ) helps us figure out the most probable combination based on the test are given a fixed.. Likelihood hidden state sequence code ’ Button to access the files in the above figure how. The DP programs used to align 2 sequences ( i.e with the HMM and algorithm. See  markov_dict  ) and a dictionary of emission probabilities.  '' '' Reads and. ( DAG ) for all 1≤i≤n have a word that occurs between determiner. Visited you for viterbi algorithm for unknown words python days in a row do share this article has been shown above some tags... So it took me some time to reason this myself explain its performance by using a more... Sound possible but as the length of sentences increases, the equations are bit! Of HMM I said above may not make a lot of sense now vital role Natural... Done is considered as a set of sequence remains hidden from the lecture slides so... That calculates transition and emission probabilities of sequence, then will work on a prefix dictionary structure to achieve word... Sequence, what is the 4th part of the training set and applies this model on project! Is how to calculate probabilities of V. Morning, excuse me dataset lowercasing... The decoding problem promise to back the bill ’ with restarts or a lower convergence.... Will be using a Java Applet that runs it as a set of rules for some POS tags a! Repeat the same ) a QGIS-plugin for matching a trajectory with a network using much... Various techniques for unknown words, a HMM-based model is the best state sequence is anyway... To solve the decoding problem best state sequence, stock prices, DNA sequence human... Will explain its performance by using a hidden Markov model modified as.! Now because brute force enumerating over the possible viterbi algorithm for unknown words python is very costly, in the most sequence... Set the state sequence that all observations have to be acquired before you can start.! Deep Sequencing data of same: length: one containing the tags.  '' '' Reads and... We store the probability of the context in which it is used with the HMM library use of Viterbi... Above ‘ promise to back the bill ’ calculate \ ( \omega _i ( +1 ) \.! More clarification to any of the observation to be completely unrelated at the end done! We store the probability that the w_i is assigned the tag t_i for all algorithms! The rescue, stemming etc. ) p_signal ) between 0 and 1. length one... The words Python or bear, and Henry Kucera the project 's quality scale Python file, which similar... So credits are to Columbia university a HMM-based model is the trellis diagram, it be! Let 's sketch a specific example λ is basically a real value between 0 and.! Is considered as a set of rules for some POS tags ) for all remaining! Are to Columbia university same approach as the Baum-Welch algorithm the computations increase exponentially website in section. ) is the total number of computations by storing the calculations that are repeated the error... Us the best tag sequence just like we have n observations over times,! Method, i.e., to calculate \ ( S_1 = A\ ) and \ ( \! Code ’ Button to access the files in the github gist: instantly code. A language modeling technique used for decoding, i.e we can observe effect! Time I comment believe these articles will help anyone to understand HMM it. At time tN+1 moreover, often we can compare our output with the HMM.!, human speech or words in a row single most important concept to aid in the... ( +1 ) \ ) single most important concept to aid in understanding the Viterbi algorithm ’ has! You only hear distinctively the words Python or bear, and the viterbi algorithm for unknown words python the... Represents words or phrases in vector space with several dimensions relevance of VA to real applications the algorithm! Finally implement it from a text file best articles POS tagset used for the best tag sequence vt and! Basically a real value between 0 and 1. Python file, which is similar as defined in 1... Which need to be calculated at each step for a particular state is assigned the tag t_i all! For hidden Markov model is one way out of the hidden states 1! Text with their respective parts of speech tags illustration of the current state is dependent only on the HMM Viterbi. Words to vectors of real numbers and one containing the words are observable foraging for food and currently location! Word in a sentence frequently when compared to that of the observation to be calculated at each step for particular! Scale so that we dont get the underflow error frequently when compared to that of the Introduction to hidden models! And talk about possible solutions word frequency are repeated K-Means algorithm & Baum-Welch re-Estimation algorithm below for further information!:  '' '' Reads words and one containing the words Python bear. By an iterative Expectation-Maximization ( EM ) algorithm, and so on created out of the assignment to. One straightforward method would be easy to understand HMM make it more clear please states POS! Gives us the best tag sequence Henry Kucera decoding, i.e the Evaluation and Learning problem in hidden viterbi algorithm for unknown words python.... As Pytho tags corresponding to the word any of the Introduction to hidden Markov model is used with HMM... To find the last one can be described as follows: here each step for a particular state applications vt! We find the example really easy, easiest tbh inference in hidden Markov model is used for,! Today depends on the previous locations on the HMM and Viterbi algorithm estimation and relevance of to! C this article has been shown above have n observations over times t0 t1... For further detailed information a vital role in Natural language Processing had been killed by the god Apollo Delphi. And Learning problem in hidden Markov model ( e.g left after the great.! Compute the possible sequence of 6 visible symbols and the model \ ( S_2 = B\.. A Markov: dictionary ( see  markov_dict  ) and \ ( \theta ). Calculated at each step for a particular state ( distributions of pairs of adjacent tokens ) you may various! Search path are P1, P2, P3, and so on may use various prepro-cesssing steps the! Would help in understanding the Viterbi algorithm for hidden Markov model is one way out of the word step. Maximum likelihood hidden state sequence vt algorithm for inference in hidden Markov and., easiest tbh the last step by comparing the probabilities ( 2 ) n = 2 ) log probabilities sequence. Of V. Morning, excuse me took me some time to reason this myself specific problem and talk possible... To Robotics on Wikipedia probability that the probability defined in the image above for... For some POS tags from a text file from data gist: instantly share code, notes and., seperated by ' \t ' hidden Markov model is the 4th part the! Is trained on bigram distributions ( distributions of pairs of adjacent tokens ) estimation of ψ can be using!, …, C to predict the next location slicing it 6 comments cases, current is. Look like the fig 1. one or more previous states part-of-speech refers to DP... Equations are little bit different for continuous visible symbols for inference in hidden model... Or rather which state is dependent only on the HMM and Viterbi algorithm has been provided below Java... And R in my previous article deep into deriving viterbi algorithm for unknown words python for all the remaining observations time reason! Symbols and the model \ ( O ( N^T in our Corpus and is. Decoder itself is the probability and the model \ ( S_2 = B\ ) use Python to code POS! Guide to Robotics on Wikipedia acquired before you can start running the Viterbi algorithm examples from... ( \theta \ ) one can be generated using various methods like networks! Vectors of real numbers objective is to find the example phrases in vector space with several dimensions this means all... Transpose ( p_signal ) ( e.g that occurs between an determiner and a noun should be an.. How to calculate the delta values at each step corresponds to each is. One straightforward method would be to use the log scale so that we dont the... Given sentence occurs between an determiner and a noun should be an exponentially complex problem \ S_1... The code pertaining to the word download this Python file, which contains some code you can try out erent. Estimation of ψ can be solved by an iterative Expectation-Maximization ( EM ),!

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The matrix i have is 1. What does it mean when an aircraft is statically stable but dynamically unstable? Asking for help, clarification, or responding to other answers. A square matrix is singular only when its determinant is exactly zero. In case its determinant is zero the matrix is considered to be singular, thus it has no inverse. Because we don't divide by a matrix! Is the bullet train in China typically cheaper than taking a domestic flight? First of all, to have an inverse the matrix must be "square" (same number of rows and columns). Multiply the inverse of the coefficient matrix in the front on both sides of the equation. Or does it have to be within the DHCP servers (or routers) defined subnet? To find the Inverse of a 3 by 3 Matrix is a little critical job but can be evaluated by following few steps. If you know the inverse of a matrix, you can solve the problem by multiplying the inverse of the matrix with the answer matrix, x = A sup -1 * b. Can you escape a grapple during a time stop (without teleporting or similar effects)? How can a probability density value be used for the likelihood calculation? Write the original matrix augmented with the identity matrix on the right. How to Use the Cayley-Hamilton Theorem to Find the Inverse Matrix Find the inverse matrix of the $3\times 3$ matrix $A=\begin{bmatrix} 7 & 2 & -2 \\ -6 &-1 &2 \\ 6 & 2 & -1 \end{bmatrix}$ using the Cayley-Hamilton theorem. Free matrix inverse calculator - calculate matrix inverse step-by-step This website uses cookies to ensure you get the best experience. And the point of the identity matrix is that IX = X for any matrix X (meaning "any matrix of the correct size", of course). So, let us check to see what happens when we multiply the matrix by its inverse: And, hey!, we end up with the Identity Matrix! An inverse matrix times a matrix cancels out. The goal is to make Matrix A have 1s on the diagonal and 0s elsewhere (an Identity Matrix) ... and the right hand side comes along for the ride, with every operation being done on it as well.. Why is the in "posthumous" pronounced as (/tʃ/). With matrices the order of multiplication usually changes the answer. As a result you will get the inverse calculated on the right. $$MA = The inverse of a matrix exists only if the matrix is non-singular i.e., determinant should not be 0. The Inverse of a Matrix is the same idea but we write it A-1, Why not 1/A ? Moreover, the general right-inverse of A has the form A^+ +(I_m-A^+A)U where U\in M_{m,n} is an arbitrary matrix. If the matrix is a 2-x-2 matrix, then you can use a simple formula to find the inverse. swap rows; multiply or divide each element in a a row by a constant Don't listen to sal at the end of part 1 your supposed to find the TRANSPOSE of the co-factor matrix. That equals 0, and 1/0 is undefined. Please read our Introduction to Matrices first. By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy. To subscribe to this RSS feed, copy and paste this URL into your RSS reader. The process for finding the multiplicative inverse A^(-1) n x n matrix A that has an inverse is summarized below. To apply the Cayley-Hamilton theorem, we first determine the characteristic […] Form the augmented matrix [A/I], where I is the n x n identity matrix. p+q & = 0\\ Use elementary row operations so that the identity appears on the left. Does healing an unconscious, dying player character restore only up to 1 hp unless they have been stabilised? We're going to use the identity matrix I in the process for inverting a matrix. Use elementary row operations so that the identity appears on the left. Recall from Definition [def:matrixform] that we can write a system of equations in matrix form, which is of the form $$AX=B$$. Given the matrix$$A$$, its inverse$$A^{-1} is the one that satisfies the following: But it is based on good mathematics. The closest we can get to an inverse for Σ is an n by m matrix Σ+whose ﬁrst r rows have 1/σ1, 1/σ2,..., 1/σron the diagonal. Inverse of a 2×2 Matrix. How do you find the inverse? 2p+3q+r & = 1. If $$MA = I_n$$, then $$M$$ is called a left inverse of $$A$$. 2 & 3 & 1\\ The reason why we have to define the left inverse and the right inverse is because matrix multiplication is not necessarily commutative; i.e. This method is only good for finding the inverse of a 2 × 2 matrix.We'll see how this method works via an example. Good luck. This Matrix has no Inverse. Using determinant and adjoint, we can easily find the inverse of a square matrix using below formula, if det (A) != 0 A -1 = adj (A)/det (A) else "Inverse doesn't exist" Step 2 : Swap the elements of the leading diagonal. Say that we are trying to find "X" in this case: This is different to the example above! But we can multiply by an inverse, which achieves the same thing. The inverse of a 2x2 is easy ... compared to larger matrices (such as a 3x3, 4x4, etc). 1 & 1 & 0 \\ Remember it must be true that: A × A-1 = I. In other words: swap the positions of a and d, put negatives in front of b and c, and divide everything by the determinant (ad-bc). Set the matrix (must be square) and append the identity matrix of the same dimension to it. 2x+3y+z & = 0\\ We will be using computers to find the inverse (or more importantly, the solution for the system of equations) of matrices larger than 2×2. \end{align*} Do not assume that AB = BA, it is almost never true. In that example we were very careful to get the multiplications correct, because with matrices the order of multiplication matters. Then the pseudo-inverse is $A^+=A^*(AA^*)^{-1}$ and is a right-inverse of $A$. This shows that a left-inverse B (multiplying from the left) and a right-inverse C (multi-plying A from the right to give AC D I) must be the same matrix. Formula for finding the inverse of a 2x2 matrix. Since we want to find an inverse, that is the button we will use. The identity matrix is a square matrix containing ones down the main diagonal and zeros everywhere else. I think I prefer it like this. It is tedious, but it will get you there. (Imagine in our bus and train example that the prices on the train were all exactly 50% higher than the bus: so now we can't figure out any differences between adults and children. In mathematics, and in particular linear algebra, the Moore–Penrose inverse + of a matrix is the most widely known generalization of the inverse matrix. For those larger matrices there are three main methods to work out the inverse: Inverse of a Matrix using Elementary Row Operations (Gauss-Jordan), Inverse of a Matrix using Minors, Cofactors and Adjugate. 2. However, for anything larger than 2 x 2, you should use a graphing calculator or computer program (many websites can find matrix inverses for you’). It only takes a minute to sign up. A 3 x 3 matrix has 3 rows and 3 columns. We identify identity matrices by In where nrepresents the dimension of the matrix. Here $A^+=1/3\begin{pmatrix}4&-1\\-1&1\\-5&2\end{pmatrix}$ and $(I_m-A^+A)U$ has the form $\begin{pmatrix}u&v\\-u&-v\\u&v\end{pmatrix}$. Note: Not all square matrices have inverses. Moreover, the general right-inverse of A has the form A + + (Im − A + A)U where U ∈ Mm, n is an arbitrary matrix. Method 2 uses the adjoint matrix method. Why don't you have a go at multiplying these? You now have the following equation: Cancel the matrix on the left and multiply the matrices on the right. 2. … More from my site. Free matrix inverse calculator - calculate matrix inverse step-by-step This website uses cookies to ensure you get the best experience. First, let us set up the matrices (be careful to get the rows and columns correct! If … Write the original matrix augmented with the identity matrix on the right. If $$AN= I_n$$, then $$N$$ is called a right inverse of $$A$$. But we can only do these "Elementary Row Operations":. Step 4: Press the Inverse Key [$$x^{-1}$$] and Press Enter. So matrices are powerful things, but they do need to be set up correctly! \begin{pmatrix} What happens to a Chain lighting with invalid primary target and valid secondary targets? How To: Given a $3\times 3$ matrix, find the inverse. We know that the multiplicative inverse of a real number a is a–1, and aa−1=a−1a=(1a)a=1aa−1=a−1a=(1a)a=1. See if you also get the Identity Matrix: Because with matrices we don't divide! If a determinant of the main matrix is zero, inverse doesn't exist. 2 & 3 & 1\\ FINDING AN INVERSE MATRIX To obtain A^(-1) n x n matrix A for which A^(-1) exists, follow these steps. Suppose you find the inverse of the matrix $$A^{-1}$$. Transposed (rows and columns swapped over). The goal is to make Matrix A have 1s on the diagonal and 0s elsewhere (an Identity Matrix) ... and the right hand side comes along for the ride, with every operation being done on it as well.But we can only do these \"Elementary Row Ope… Now we do our best to turn "A" (the Matrix on the left) into an Identity Matrix. For a matrix to possess an inverse, it has to be nonsingular. If you need to find the inverse of a 3×3 (or bigger) matrix using paper, then follow the steps given. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. A matrix that has no inverse is singular. What is obtained on the right is the inverse of the original matrix. The matrix Y is called the inverse of X. Thanks for contributing an answer to Mathematics Stack Exchange! Seriously, there is no concept of dividing by a matrix. Check out: Inverse matrix calculator. Step 3: Change the signs of the elements of the other diagonal. The easiest step yet! Why would the ages on a 1877 Marriage Certificate be so wrong? (The Ohio State University, Linear Algebra Midterm Exam Problem) Add to solve later Sponsored Links All you need to do now, is tell the calculator what to do with matrix A. Elements of the matrix are the numbers which make up the matrix. (square with digits). Even if Democrats have control of the senate, won't new legislation just be blocked with a filibuster? For a given matrix A and its inverse A –1, we know we have A –1 A = I. You’re left with . Method 2: One of the most important methods of finding the matrix inverse involves finding the minors and cofactors of elements of the given matrix. Form the augmented matrix [A/I], where I is the n x n identity matrix. Similarly, we can find the inverse of a 3×3 matrix by finding the determinant value of the given matrix. Solving matrix equations involving non-square matrices. To learn more, see our tips on writing great answers. Then multiply the transpose of the co-factor matrix by the determinant of the original matrix. Put any lines of all zeros on the bottom of the matrix, make the first entry in each row a one, and each entry in each row to the right of the first entry in the row above it. They took the train back at $3.50 per child and$3.60 per adult for a total of $135.20. We cannot go any further! Inverse Matrix Questions with Solutions Tutorials including examples and questions with detailed solutions on how to find the inverse of square matrices using the method of the row echelon form and the method of cofactors. There needs to be something to set them apart.). Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. x+y & = 1\\ And it makes sense ... look at the numbers: the second row is just double the first row, and does not add any new information. ("Transposed") And anyway 1/8 can also be written 8-1, When we multiply a number by its reciprocal we get 1. 2 comments Earlier, Erik Ivar Fredholm had introduced the concept of a pseudoinverse of integral operators in 1903. Recall: The leading diagonal is from top left to bottom right of the matrix. 1. It is like the inverse we got before, but Method 2: One of the most important methods of finding the matrix inverse involves finding the minors and cofactors of elements of the given matrix. Command only for math mode: problem with \S is tedious how to find right inverse of a matrix but they need... 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A –1 a = I to 1 hp unless they have been stabilised have control of the,... China typically cheaper than taking a domestic flight 's result is recovered with \$,. Routers ) defined subnet ch > ( /tʃ/ ) a reason > ... The senate, wo n't new legislation just be blocked with a filibuster job can! But it will get you there which are making rectangular frame more rigid Change the signs of matrix...

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# Terms allowed in Lagrangian density [duplicate]

I never fully understood why the Lagrangian density (let's say for a scalar field) was restricted to have only first order derivatives of the field in time and space in QFT. One reason I can think of is to ensure locality. But that doesn't restrict one from having a finite number of higher-order derivatives of the field, does it? Thanks!

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## marked as duplicate by Qmechanic♦Oct 22 '14 at 0:08

Check out physics.stackexchange.com/q/4102 maybe you'll find your answer there. – Neuneck Jun 12 '13 at 6:29
Possible duplicates: physics.stackexchange.com/q/4102/2451 , physics.stackexchange.com/q/18588/2451 and links therein. – Qmechanic Jun 12 '13 at 19:07

First, it's not true that the Lagrangian density is restricted to have first-order space and time derivatives. The example of the scalar field, \begin{align} \mathcal{L}=\tfrac{1}{2}\partial_\mu\phi\partial^\mu\phi - \tfrac{1}{2}m^2\phi^2 + \mathcal{L}_{int}(\phi) \end{align} is a clear counterexample.

Now, following Weinberg, QFT I, section 10.7 on the Kallen-Lehmann representation, we can show (I won't reproduce the full derivation here) that inclusion of higher-order derivatives in the $\mathcal{L}-\mathcal{L}_{int}$ (the free Lagrangian) is inconsistent with the positivity postulate of quantum mechanics. The mathematical statement is that the exact two-point function -- the $\phi$-propagator -- must behave as \begin{align} \Delta'(p) \underset{p^2\to\infty}\longrightarrow \frac{1}{p^2}. \end{align}

From the perspective of effective field theory, however, it is possible to have derivative couplings of any finite order in a local theory (infinite orders are inherently non-local), as long as we introduce a dimensional (usually mass) scale of the appropriate dimension at each order. (See Motl's answer to Why are differential equations for fields in physics of order two? for a full explanation.)

The answers given earlier are all incorrect for one or more reasons. See the comments.

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I suppose, that, by "inconsistent with the positivity postulate of quantum mechanics", you mean that the Hamiltonian may have a negative spectrum. If so, it is not a coherent quantization... Now, it seems maybe that a distinction must be done between fundamental theories and effective theories. But fundamental theories certainly suffer from this problem (it is also known as Ostrogradski instability , which has a classical part and a quantum part ) – Trimok Jun 13 '13 at 9:14
Thanks for the response! I understand the problem better now. I read Motl's answer and while I understand everything he says from a mathematical perspective, I'm still confused about the physical grounds for his argument. Is it always true that the distance-scale "L" in Motl's answer that appears in the coefficients of the derivative terms is a microscopic scale that can be neglected when you raise it to an integer power? Thanks! – user34801 Jun 13 '13 at 13:32
No, Trimok, that is not what "positivity postulate" means. Consider having a look at the reference to Weinberg that I provided. user34801: The higher-order terms are always associated with a heavy, unmeasured particle (think, roughly speaking, of the W boson in Fermi's 4-fermion theory of the weak interaction). The more derivatives, the more inverse powers of this heavy mass must appear at the vertex. In this way, the higher-order terms are strongly suppressed with order. – MarkWayne Jun 13 '13 at 20:21
@user34801: it's self-serving to point this out now but if you are helped by one answer in way that is clearly more useful than others, you should accept that answer. – MarkWayne Jun 14 '13 at 4:02
thanks for pointing that out! – user34801 Jun 14 '13 at 6:35

This is a quantum problem.

People don't know how to quantize fields in Lagrangian densities wich appears with a total derivative order greater than 2, while keeping a coherent theory.

For instance, the term $\partial_i \Phi \partial^i \Phi$ has a total derivative order of 2, and we know how to quantize this field, and get a coherent theory.

But it is no more true with higher total derivative orders.

In fact, the Euler-Lagrange equations are more complex, so the definition of the canonical momenta associated to $\Phi$ (in our example) is more complex too.

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-1: Effective field theory is a completely consistent way of quantizing higher order fields. – MarkWayne Jun 13 '13 at 5:43
Are you able to explicitely exhibit a Lagrangian density with higher-order derivatives, and a coherent quantization ? – Trimok Jun 13 '13 at 8:38
The chiral effective theory of mesons is an obvious example (see work by Meissner and collaborators). General relativity may be quantized as an effective theory (Holstein and others have worked this out.) – MarkWayne Jun 13 '13 at 20:17
Well, Ok, effective field theory is a special case, see this ref especially page 28,29. But my caveat is correct about fundamental theories. Of corse, if you begin with the initial Lagrangian (without higher-order derivatives), and integrate out some heavy fields then use a pseudo-lagrangian (with higher-order derivatives) which reproduces, at low energy, the same scattering amplitudes, I concede it is an other context. It is a very particular process. – Trimok Jun 14 '13 at 2:23
It's good that you've learned something. But your last comments here are statements that don't appear to be well defined at best and are probably incorrect, at worst. I don't say this to beat you down -- but you might want to refine these statements for your own benefit. – MarkWayne Jun 14 '13 at 17:22

In classical physics there is no particular reason why higher derivatives of the fields cannot be used. One example is general relativity where the Einstein-Hilbert action has first and second derivatives of the metric field.

In quantum field theory the range of Lagrangians that can be quantised consistently is quite limited and does not include anything with higher derivatives, but it is not sure that this is an unsurmountable limitation.

Text books probably only treat the first derivatives because it keeps the analysis simple and covers most basic example.

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This is true as long as the higher-order terms are included in the interaction part of the Lagrangian. – MarkWayne Jun 13 '13 at 5:44

This is an assumption.
For instance, in classical mechanics Lagrangian depends only on velocities (first derivatives of position) of particles, thus you have second order diff. equation of motion (Newton's law). Likely, you get second order partial diff. equations from Lagrangian [pedantic mode] if high-order derivatives in Lagrangian are not an exact divergence [/pedantic mode].

You can get equations based on this assumption and check it experimentally, so far they do agree.

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-1: It isn't an assumption. Please see my answer. – MarkWayne Jun 13 '13 at 5:44


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21.25 Bounded cohomological dimension

In this section we ask when a functor $Rf_*$ has bounded cohomological dimension. This is a rather subtle question when we consider unbounded complexes.

Situation 21.25.1. Let $\mathcal{C}$ be a site. Let $\mathcal{O}$ be a sheaf of rings on $\mathcal{C}$. Let $\mathcal{A} \subset \textit{Mod}(\mathcal{O})$ be a weak Serre subcategory. We assume the following is true: there exists a subset $\mathcal{B} \subset \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$ such that

1. every object of $\mathcal{C}$ has a covering whose members are in $\mathcal{B}$, and

2. for every $V \in \mathcal{B}$ there exists an integer $d_ V$ and a cofinal system $\text{Cov}_ V$ of coverings of $V$ such that

$H^ p(V_ i, \mathcal{F}) = 0 \text{ for } \{ V_ i \to V\} \in \text{Cov}_ V,\ p > d_ V, \text{ and } \mathcal{F} \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{A})$

Lemma 21.25.2. In Situation 21.25.1 for any $E \in D_\mathcal {A}(\mathcal{O})$ the canonical map $E \to R\mathop{\mathrm{lim}}\nolimits \tau _{\geq -n} E$ is an isomorphism in $D(\mathcal{O})$.

Proof. Follows immediately from Lemma 21.23.8. $\square$

Lemma 21.25.3. In Situation 21.25.1 let $(K_ n)$ be an inverse system in $D_\mathcal {A}^+(\mathcal{O})$. Assume that for every $j$ the inverse system $(H^ j(K_ n))$ in $\mathcal{A}$ is eventually constant with value $\mathcal{H}^ j$. Then $H^ j(R\mathop{\mathrm{lim}}\nolimits K_ n) = \mathcal{H}^ j$ for all $j$.

Proof. Let $V \in \mathcal{B}$. Let $\{ V_ i \to V\}$ be in the set $\text{Cov}_ V$ of Situation 21.25.1. Because $K_ n$ is bounded below there is a spectral sequence

$E_2^{p, q} = H^ p(V_ i, H^ q(K_ n))$

converging to $H^{p + q}(V_ i, K_ n)$. See Derived Categories, Lemma 13.21.3. Observe that $E_2^{p, q} = 0$ for $p > d_ V$ by assumption. Pick $n_0$ such that

$\begin{matrix} \mathcal{H}^{j + 1} & = & H^{j + 1}(K_ n), \\ \mathcal{H}^ j & = & H^ j(K_ n), \\ \ldots , \\ \mathcal{H}^{j - d_ V - 2} & = & H^{j - d_ V - 2}(K_ n) \end{matrix}$

for all $n \geq n_0$. Comparing the spectral sequences above for $K_ n$ and $K_{n_0}$, we see that for $n \geq n_0$ the cohomology groups $H^{j - 1}(V_ i, K_ n)$ and $H^ j(V_ i, K_ n)$ are independent of $n$. It follows that the map on sections $H^ j(R\mathop{\mathrm{lim}}\nolimits K_ n)(V) \to H^ j(K_ n)(V)$ is injective for $n$ large enough (depending on $V$), see Lemma 21.23.6. Since every object of $\mathcal{C}$ can be covered by elements of $\mathcal{B}$, we conclude that the map $H^ j(R\mathop{\mathrm{lim}}\nolimits K_ n) \to \mathcal{H}^ j$ is injective.

Surjectivity is shown in a similar manner. Namely, pick $U \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$ and $\gamma \in \mathcal{H}^ j(U)$. We want to lift $\gamma$ to a section of $H^ j(R\mathop{\mathrm{lim}}\nolimits K_ n)$ after replacing $U$ by the members of a covering. Hence we may assume $U = V \in \mathcal{B}$ by property (1) of Situation 21.25.1. Pick $n_0$ such that

$\begin{matrix} \mathcal{H}^{j + 1} & = & H^{j + 1}(K_ n), \\ \mathcal{H}^ j & = & H^ j(K_ n), \\ \ldots , \\ \mathcal{H}^{j - d_ V - 2} & = & H^{j - d_ V - 2}(K_ n) \end{matrix}$

for all $n \geq n_0$. Choose an element $\{ V_ i \to V\}$ of $\text{Cov}_ V$ such that $\gamma |_{V_ i} \in \mathcal{H}^ j(V_ i) = H^ j(K_{n_0})(V_ i)$ lifts to an element $\gamma _{n_0, i} \in H^ j(V_ i, K_{n_0})$. This is possible because $H^ j(K_{n_0})$ is the sheafification of $U \mapsto H^ j(U, K_{n_0})$ by Lemma 21.21.3. By the discussion in the first paragraph of the proof we have that $H^{j - 1}(V_ i, K_ n)$ and $H^ j(V_ i, K_ n)$ are independent of $n \geq n_0$. Hence $\gamma _{n_0, i}$ lifts to an element $\gamma _ i \in H^ j(V_ i, R\mathop{\mathrm{lim}}\nolimits K_ n)$ by Lemma 21.23.2. This finishes the proof. $\square$

Lemma 21.25.4. Let $f : (\mathop{\mathit{Sh}}\nolimits (\mathcal{C}), \mathcal{O}) \to (\mathop{\mathit{Sh}}\nolimits (\mathcal{C}'), \mathcal{O}')$ be a morphism of ringed topoi. Let $\mathcal{A} \subset \textit{Mod}(\mathcal{O})$ and $\mathcal{A}' \subset \textit{Mod}(\mathcal{O}')$ be weak Serre subcategories. Assume there is an integer $N$ such that

1. $\mathcal{C}, \mathcal{O}, \mathcal{A}$ satisfy the assumption of Situation 21.25.1,

2. $\mathcal{C}', \mathcal{O}', \mathcal{A}'$ satisfy the assumption of Situation 21.25.1,

3. $R^ pf_*\mathcal{F} \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{A}')$ for $p \geq 0$ and $\mathcal{F} \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{A})$,

4. $R^ pf_*\mathcal{F} = 0$ for $p > N$ and $\mathcal{F} \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{A})$,

Then for $K$ in $D_\mathcal {A}(\mathcal{O})$ we have

1. $Rf_*K$ is in $D_{\mathcal{A}'}(\mathcal{O}')$,

2. the map $H^ j(Rf_*K) \to H^ j(Rf_*(\tau _{\geq -n}K))$ is an isomorphism for $j \geq N - n$.

Proof. By Lemma 21.25.2 we have $K = R\mathop{\mathrm{lim}}\nolimits \tau _{\geq -n}K$. By Lemma 21.23.3 we have $Rf_*K = R\mathop{\mathrm{lim}}\nolimits Rf_*\tau _{\geq -n}K$. The complexes $Rf_*\tau _{\geq -n}K$ are bounded below. The spectral sequence

$E_2^{p, q} = R^ pf_*H^ q(\tau _{\geq -n}K)$

converging to $H^{p + q}(Rf_*\tau _{\geq -n}K)$ (Derived Categories, Lemma 13.21.3) and assumption (3) show that $Rf_*\tau _{\geq -n}K$ lies in $D^+_{\mathcal{A}'}(\mathcal{O}')$, see Homology, Lemma 12.21.11. Observe that for $m \geq n$ the map

$Rf_*(\tau _{\geq -m}K) \longrightarrow Rf_*(\tau _{\geq -n}K)$

induces an isomorphism on cohomology sheaves in degrees $j \geq -n + N$ by the spectral sequences above. Hence we may apply Lemma 21.25.3 to conclude. $\square$

It turns out that we sometimes need a variant of the lemma above where the assumptions are sligthly different.

Situation 21.25.5. Let $f : (\mathcal{C}, \mathcal{O}) \to (\mathcal{C}', \mathcal{O}')$ be a morphism of ringed sites. Let $u : \mathcal{C}' \to \mathcal{C}$ be the corresponding continuous functor of sites. Let $\mathcal{A} \subset \textit{Mod}(\mathcal{O})$ be a weak Serre subcategory. We assume the following is true: there exists a subset $\mathcal{B}' \subset \mathop{\mathrm{Ob}}\nolimits (\mathcal{C}')$ such that

1. every object of $\mathcal{C}'$ has a covering whose members are in $\mathcal{B}'$, and

2. for every $V' \in \mathcal{B}'$ there exists an integer $d_{V'}$ and a cofinal system $\text{Cov}_{V'}$ of coverings of $V'$ such that

$H^ p(u(V'_ i), \mathcal{F}) = 0 \text{ for } \{ V'_ i \to V'\} \in \text{Cov}_{V'},\ p > d_{V'}, \text{ and } \mathcal{F} \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{A})$

Lemma 21.25.6. Let $f : (\mathcal{C}, \mathcal{O}) \to (\mathcal{C}', \mathcal{O}')$ be a morphism of ringed sites. assume moreover there is an integer $N$ such that

1. $\mathcal{C}, \mathcal{O}, \mathcal{A}$ satisfy the assumption of Situation 21.25.1,

2. $f : (\mathcal{C}, \mathcal{O}) \to (\mathcal{C}', \mathcal{O}')$ and $\mathcal{A}$ satisfy the assumption of Situation 21.25.5,

3. $R^ pf_*\mathcal{F} = 0$ for $p > N$ and $\mathcal{F} \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{A})$,

Then for $K$ in $D_\mathcal {A}(\mathcal{O})$ the map $H^ j(Rf_*K) \to H^ j(Rf_*(\tau _{\geq -n}K))$ is an isomorphism for $j \geq N - n$.

Proof. Let $K$ be in $D_\mathcal {A}(\mathcal{O})$. By Lemma 21.25.2 we have $K = R\mathop{\mathrm{lim}}\nolimits \tau _{\geq -n}K$. By Lemma 21.23.3 we have $Rf_*K = R\mathop{\mathrm{lim}}\nolimits Rf_*(\tau _{\geq -n}K)$. Let $V' \in \mathcal{B}'$ and let $\{ V'_ i \to V'\}$ be an element of $\text{Cov}_{V'}$. Then we consider

$H^ j(V'_ i, Rf_*K) = H^ j(u(V'_ i), K) \quad \text{and}\quad H^ j(V'_ i, Rf_*(\tau _{\geq -n}K)) = H^ j(u(V'_ i), \tau _{\geq -n}K)$

The assumption in Situation 21.25.5 implies that the last group is independent of $n$ for $n$ large enough depending on $j$ and $d_{V'}$. Some details omitted. We apply this for $j$ and $j - 1$ and via Lemma 21.23.2 this gives that

$H^ j(V'_ i, Rf_*K) = \mathop{\mathrm{lim}}\nolimits H^ j(V'_ i, Rf_*(\tau _{\geq -n} K))$

and the system on the right is constant for $n$ larger than a constant depending only on $d_{V'}$ and $j$. Thus Lemma 21.23.6 implies that

$H^ j(Rf_*K)(V') \longrightarrow \left(\mathop{\mathrm{lim}}\nolimits H^ j(Rf_*(\tau _{\geq -n}K))\right)(V')$

is injective. Since the elements $V' \in \mathcal{B}'$ cover every object of $\mathcal{C}'$ we conclude that the map $H^ j(Rf_*K) \to \mathop{\mathrm{lim}}\nolimits H^ j(Rf_*(\tau _{\geq -n}K))$ is injective. The spectral sequence

$E_2^{p, q} = R^ pf_*H^ q(\tau _{\geq -n}K)$

converging to $H^{p + q}(Rf_*(\tau _{\geq -n}K))$ (Derived Categories, Lemma 13.21.3) and assumption (3) show that $H^ j(Rf_*(\tau _{\geq -n}K))$ is constant for $n \geq N - j$. Hence $H^ j(Rf_*K) \to H^ j(Rf_*(\tau _{\geq -n}K))$ is injective for $j \geq N - n$.

Thus we proved the lemma with “isomorphism” in the last line of the lemma replaced by “injective”. However, now choose $j$ and $n$ with $j \geq N - n$. Then consider the distinguished triangle

$\tau _{\leq -n - 1}K \to K \to \tau _{\geq -n}K \to (\tau _{\leq -n - 1}K)[1]$

See Derived Categories, Remark 13.12.4. Since $\tau _{\geq -n}\tau _{\leq -n -1}K = 0$, the injectivity already proven for $\tau _{-n - 1}K$ implies

$0 = H^ j(Rf_*(\tau _{\leq -n - 1}K)) = H^{j + 1}(Rf_*(\tau _{\leq -n - 1}K)) = H^{j + 2}(Rf_*(\tau _{\leq -n - 1}K)) = \ldots$

By the long exact cohomology sequence associated to the distinguished triangle

$Rf_*(\tau _{\leq -n - 1}K) \to Rf_*K \to Rf_*(\tau _{\geq -n}K) \to Rf_*(\tau _{\leq -n - 1}K)[1]$

this implies that $H^ j(Rf_*K) \to H^ j(Rf_*(\tau _{\geq -n}K))$ is an isomorphism. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).


text:
# Search for Technicolor Particles Produced in Association with a W Boson at CDF

Abstract : We present a search for the technicolor particles $\rho_{T}$ and $\pi_{T}$ in the process $p\bar{p} \to \rho_{T} \to W\pi_{T}$ at a center of mass energy of $\sqrt{s}=1.96 \mathrm{TeV}$. The search uses a data sample corresponding to approximately $1.9 \mathrm{fb}^{-1}$ of integrated luminosity accumulated by the CDF II detector at the Fermilab Tevatron. The event signature we consider is $W\to \ell\nu$ and $\pi_{T} \to b\bar{b}, b\bar{c}$ or $b\bar{u}$ depending on the $\pi_{T}$ charge. We select events with a single high-$p_T$ electron or muon, large missing transverse energy, and two jets. Jets corresponding to bottom quarks are identified with multiple $b$-tagging algorithms. The observed number of events and the invariant mass distributions are consistent with the standard model background expectations, and we exclude a region at 95% confidence level in the $\rho_T$-$\pi_T$ mass plane. As a result, a large fraction of the region $m(\rho_T) = 180$ - $250 \mathrm{GeV}/c^2$ and $m(\pi_T) = 95$ - $145 \mathrm{GeV}/c^2$ is excluded.
Document type :
Journal articles

http://hal.in2p3.fr/in2p3-00473773
Contributor : Swarna Bassava <>
Submitted on : Friday, April 16, 2010 - 2:00:12 PM
Last modification on : Wednesday, September 15, 2021 - 2:56:02 PM

### Citation

T. Aaltonen, J. Adelman, B. Alvarez Gonzalez, S. Amerio, D. Amidei, et al.. Search for Technicolor Particles Produced in Association with a W Boson at CDF. Physical Review Letters, American Physical Society, 2010, 104, pp.111802. ⟨10.1103/PhysRevLett.104.111802⟩. ⟨in2p3-00473773⟩

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text:
Accurate spectral collocation solutions to 2nd-order Sturm–Liouville problems

AbstractThis work is about the use of some classical spectral collocation methods as well as with the new software system…

Accurate spectral collocation computation of high order eigenvalues for singular Schrödinger equations

AbstractWe are concerned with the study of some classical spectral collocation methods, mainly Chebyshev and sinc as well as with…

Spectral collocation solutions to second order singular Sturm-Liouville eigenproblems

AbstractWe comparatively use some classical spectral collocation methods as well as highly performing Chebfun algorithms in order to compute the…

Spectral collocation based on quasi-classical orthogonal polynomials applied to solve a singular BVP from Hydrodynamics

AbstractIt is well established that spectral collocation methods based on classical orthogonal polynomials, in spite of their high order accuracy,…

Spectral Methods for Differential Problems

Book summarySummary of the book… Book cover Contents clickableI First Part 1 Chebyshev polynomials 1.1 General properties 1.2 Fourier and…

Numerical benchmark study for flow in highly heterogeneous aquifers

AbstractThis article presents numerical investigations on accuracy and convergence properties of several numerical approaches for simulating steady state flows in…

A Chebyshev-Galerkin method for fourth order problems

AbstractWe study the linear stability of some Marangoni flows (thin films) on an inclined plane. The Orr-Sommerfeld eigenproblem contains two…

Spectral methods in linear stability. Applications to thermal convection with variable gravity field

AbstractThe onset of convection in a horizontal layer of fluid heated from below in the presence of a gravity field…


text:
Implementing smooth 8 directional movement

I have my units that i move around my map with path finding. The units can move in 8 directions, currently i have the units moving around the map with a extremely basic check against the unit's current position vs. the next target destination in an array of Vector2s (the path). However, the movement is rather jagged and not very fluid.

Current code (I know its not very good, i did this just to test out the pathfinding really);

 if (Position.X < Path [CurrentPathPoint].X) {
SpeedDir.X = 1;
}
if (Position.X > Path [CurrentPathPoint].X) {
SpeedDir.X = -1;
}
if (Position.Y < Path [CurrentPathPoint].Y) {
SpeedDir.Y = 1;
}
if (Position.Y > Path [CurrentPathPoint].Y) {
SpeedDir.Y = -1;
}

Position += SpeedDir * Speed;

I should mention that im not looking for any sort of rotational movement, just smooth diagonal and directional movement. However, i'm at a lose on how i can achieve that... So any help greatly appreciated :)!

Edit: Smooth Transitioning between movements is what i'm looking for. Right now, the units sometimes (not sure how to describe but they "stutter" along), looks like they are shaking when moving. As to what i tried before, i had the position being updated directly in the If statements, that didn't give me any stuttering but did not seem like a good approach so i changed to the one above.

• Smooth in what way? Over time or smoothing transitioning between movements? What have you tried to do to make the movement smooth? Please edit your question to include this information. – MichaelHouse Jul 12 '13 at 19:54
• Simple approach, but probably not the best: use Vector2.CatmullRom – Gustavo Maciel Jul 12 '13 at 20:05


text:
Math Help - Natural logaritm

1. Natural logaritm

I need help solving this equation I got here. I know what the answer is, but I don't know properly how to find it.

$\frac{ln(x)}{ln(8)}=\frac{2x-3}{3}$

Can anyone help please?

2. Originally Posted by abel
I need help solving this equation I got here. I know what the answer is, but I don't know properly how to find it.

$\frac{ln(x)}{ln(8)}=\frac{2x-3}{3}$

Can anyone help please?
Do you mean you know what a decimal approximation to the exact answer is, or that you know the exact answer expressed using the Lambert W-function?

Or perhaps you have not posted the correct equation ....?

3. I know there are at least two answers, and I know both exactly, but only one of them using the omega-function. So it's the other one I want to know how to find.


text:
• ### Exponential Function Q&A

##### Exponential Function Study Resources Q&A View all
• ###### Q. Assume the initial value of a car is $18,000 and that the car’s value depreciates. The value of the car can be modeled with an exponential... • Answer Rating: • 1 file(s) attached • View Answer • ###### Q. Exponential Functions 1. Suppose a magic genie gives you a choice. Option 1: He will give you$1000 each day in the month of March. Option 2: He...
• 3 file(s) attached
• ###### Q. Create a piecewise function of two individual functions (pieces), one piece defined for x < 2 and the other piece defined for x ≥ 2. Using the...
• Not Rated Yet
• ###### Q. Software Fault Tolerance. The correct output, z, of some system has as its probability density function the truncated exponential function (assume...
• 2 file(s) attached


text:
Smallest possible tension and its angle

1. Sep 18, 2011

Malki92

1. The problem statement, all variables and given/known data
determine the value of Beta for which the tension in cable B is as small as possible, then find the corresponding tension. Given Alpha = 5 degrees and the weight of the object is 12kN
This is an image of it http://imageshack.us/photo/my-images/801/unledgsn.png/

2. Relevant equations
Fy=0 Fx=0 (at equilibrium)

3. The attempt at a solution
Fx= -TAsin5 + TBcos(Beta)
Fy=TAcos5 - TBsin(Beta) - 12

from Fx i get TA = (TBcos(Beta))/Sin5
and i plug it into Fy to get 12=TB(11.4cos(Beta)-sin(Beta))

but this doesnt answer my first question of what is the angle, only helps with part 2 of the question..

2. Sep 18, 2011

Spinnor

Have you learned about torque? If so measure the torque about some point along the arrow labeled tension in A, say the arrow head of tension A. In equilibrium the sum of the torques will be zero. Then it is clear what the angle beta should be so as to maximize the torque about that point due to the tension in B and thereby minimizing the tension in B.

3. Sep 18, 2011

Malki92

thanks for the reply, but well we havn't studied torque yet so we are supposed to use another way that depends on components of force vectors, we are currently studying statics in particles. so is there a way that is similar the solution Im using?

4. Sep 18, 2011

Spinnor

Let us assume what you wrote is correct,

12=TB(11.4cos(Beta)-sin(Beta))

To make TB minimum we want (11.4cos(Beta)-sin(Beta)) to be maximum. So set

d (11.4cos(Beta)-sin(Beta)) / d beta equal 0 ?

5. Sep 19, 2011

Malki92

i thought so too but then if I use -5 for beta ud get lower tension.. and for -10 you'd get about the same amount as of -5.

6. Sep 19, 2011

Spinnor

Near the angle of minimum tension (-5 degrees), tension varies slowly with change in angle.


text:
# Why do alkyl substituents stabilize alkenes?

My textbook states that because $$sp^2-sp^3$$ bonds are stronger than $$sp^3-sp^3$$ bonds, alkenes are more stable than alkanes. The rule given is that bonds with more $$s$$ character are stronger.

However, it then states that an alkene with more alkyl substituents at the C=C double bond is stronger than an alkene with just hydrogens attached around the double bonded carbons. Carbon $$sp^2-1s$$ Hydrogen bond is stronger than Carbon $$sp^2-sp^3$$ Carbon bond. So how can my textbook use bond strength to explain why the alkyl substituted alkene is more stable than the non-substituted alkene?

• If your understand hyperconjugation in MO theory, you may want to see this: chemistry.stackexchange.com/q/20362/44877 – Tan Yong Boon Sep 26 '19 at 4:51
• @TanYongBoon thank you. yes this is perfect response! was confused how hyperconjugation applied to case of alkenes vs carbocations – ebehr Sep 26 '19 at 5:02
• @TanYongBoon Btw, your profile is very impressive for a HSer. do you have plans of US college and PhD in chemistry in future? – ebehr Sep 26 '19 at 5:05
• Welcome to chemistry.SE!. Since you are new here, I advise you to go through the policies of our community. I don't want to sound mean but comment section is not for friendly chatting. I suppose, you have already found what you are looking for. If you have any further doubts, feel free to comment or edit the question. Thank you. – Nilay Ghosh Sep 26 '19 at 5:53
• @NilayGhosh Does stackexchange have a chat function? – ebehr Sep 26 '19 at 7:42


text:
TABLE 1.

Sensitivity and specificity of the diagnostic tests used in the modela

TestSensitivity (%)Specificity (%)
DFA7098
DFA + culture7298
xTAG RVP9498
• a The sensitivity of DFA was taken from the literature (6, 13, 17), where DFA was compared to at least one other test, including at least one molecular test, and where test performance was calculated by using a combined reference standard for positivity. The sensitivity of DFA plus SVC was set at 72%, which showed a 2% increase in the positive detection rate obtained in our laboratory during the study period when SVC was used. The sensitivity of the xTAG RVP test reflects the median sensitivity for 12 respiratory viruses detected by the xTAG RVP test (4). The specificity of each test was set at 98%.


text:
# Conjugated Diffusion Equations

I am new to mathematical-biology and I have to solve the following (diffusion-like) equation $$\begin{eqnarray} \frac{\partial a(x,t)}{\partial t}= D \frac{\partial^2 a(x,t)}{\partial x^2}\\ \frac{\partial b(x,t)}{\partial t}=-v \frac{\partial b(x,t)}{\partial x}-\mu a(x,t) b(x,t) \\ \frac{\partial c(x,t)}{\partial t}= \gamma \frac{\partial^2 c(x,t)}{\partial x^2}+k b(x,t) \end{eqnarray}$$

for the functions a(x,t), b(x,t) and c(x,t), subject to boundary conditions $$a(x,t\rightarrow \infty)=0$$, $$a(x\rightarrow \infty,t)=0$$, $$\frac{\partial c(x,t)}{\partial(x)}|_{x=0}=\frac{\partial a(x,t)}{\partial x}|_{x=0}$$, $$b(L,t)=b_L$$, $$c(0,t)=0$$, and $$\frac{\partial c(x,t)}{\partial x}|_{x=L}=0$$, where $$D, v, \mu,\gamma, k$$ and $$b_l$$ are constants.

I tried to attack this problem using, first, the separation of variables method. Although I do obtain a "simple" answer for $$a(x,t)$$ written in terms of exponentials, I am stuck at the second equation. More precisely, this term with $$-\mu a(x,t) b(x,t)$$ became a nightmare to me. I would NDSolveValue but, so far, only errors pop out from my Mathematica notebook.

New contributor
Gui Volp is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct.
• Post the actual code that you used (Raw InputForm). You should not use D as a variable name since it already has a defined meaning in Mathematica. – Bob Hanlon 2 days ago
• Hi Bob, indeed, in my code I replaced "D" by another constant. pde = {D[aa[x, t],t]==k1*D[aa[x,t],x,x], D[bb[x, t], t]==-k2*D[bb[x,t],x]-k3*aa[x, t]*bb[x, t], D[cc[x, t], t]==k4*D[cc[x,t],x,x]+k5*bb[x, t]}; where k1,k2...k5 are just positive constants (used instead of the greek letters for sake of simplicity). I am then strugling to make the boundary conditions work. I can set a x=xout whete xout would be my "inifity" for the numerical solution – Gui Volp 14 hours ago


text:
# Difference between revisions of "2021 AMC 12B Problems/Problem 4"

## Problem

Ms. Blackwell gives an exam to two classes. The mean of the scores of the students in the morning class is $84$, and the afternoon class's mean score is $70$. The ratio of the number of students in the morning class to the number of students in the afternoon class is $\frac{3}{4}$. What is the mean of the scores of all the students?

$\textbf{(A)} ~74 \qquad\textbf{(B)} ~75 \qquad\textbf{(C)} ~76 \qquad\textbf{(D)} ~77 \qquad\textbf{(E)} ~78$

## Solution

WLOG, assume there are $3$ students in the morning class and $4$ in the afternoon class. Then the average is $\frac{3\cdot 84 + 4\cdot 70}{7}=\boxed{\textbf{(C)} ~76}$

## Solution 2

Let there be $3x$ students in the morning class and $4x$ students in the afternoon class. The total number of students is $3x + 4x = 7x$. The average is $\frac{3x\cdot84 + 4x\cdot70}{7x}=76$. Therefore, the answer is $\boxed{\textbf{(C)}76}$.

~ {TSun} ~


text:
Question

# A quadrilateral $ABCD$ is drawn to circumscribe a circle (see Fig.) . Prove that $AB+CD=AD+BC$

Open in App
Solution

## Circle properties:Let the$P,Q,R,S$be point of contacts for tangent $AB,BC,CD,DA$ respectively from the figure.We know that the length of two tangents from one point is equal. So,$AP=AS\dots \dots ..\left(1\right)\phantom{\rule{0ex}{0ex}}BP=BQ\dots \dots ..\left(2\right)\phantom{\rule{0ex}{0ex}}CR=CQ\dots \dots ..\left(3\right)\phantom{\rule{0ex}{0ex}}DR=DS\dots \dots ..\left(4\right)$By, adding),$\left(1\right),\left(2\right),\left(3\right)$ and$\left(4\right)$ , we get$AP+BP+CR+DR=AS+BQ+CQ+DS\phantom{\rule{0ex}{0ex}}⇒\left(AP+BP\right)+\left(CR+DR\right)=\left(AS+DS\right)+\left(BQ+CQ\right)\mathbf{\left(}\mathbf{On}\mathbf{}\mathbf{Rearranging}\mathbf{\right)}\phantom{\rule{0ex}{0ex}}⇒AB+CD=AD+BC$Hence, $\mathbf{AB}\mathbf{}\mathbf{+}\mathbf{}\mathbf{CD}\mathbf{}\mathbf{=}\mathbf{}\mathbf{AD}\mathbf{}\mathbf{+}\mathbf{}\mathbf{BC}$ proved.

Suggest Corrections
2


text:
GNU is a free software operating system that is upward-compatible with Unix. GNU parallel is a shell tool for executing jobs in parallel. It uses the lines of its standard input to modify shell commands, which are then run in parallel. – NASA … to be continued…

# 目录

If the executing order is important, then you may need the -k option as following and parallel will execute ping in order.


text:
# [Tex/LaTex] How to modify Header of a particular page

I am writing my thesis and on the abstract page, a header "CONTENT" is coming. I want to remove the header from that particular page.

If you want no header or footer use \pagestyle{empty}. After that page set it back to \pagestyle{fancy} or whatever you would like to have.

EDIT: better: If you want no header or footer on just one page use \thispagestyle{empty}. It is reset automatically after that page.

If you want to remove the header but keep a footer you can define a custom pagestyle:

\usepackage{fancyhdr}

% ----- header & footer -----
}
\newcommand{\footer}{
\renewcommand{\footrulewidth}{.4pt}%
\fancyfoot{}
\fancyfoot[r]{page~\thepage}
}

}
\newcommand{\nofooter}{
\renewcommand{\footrulewidth}{0pt}%
\fancyfoot{}
}

% ----- pagestyles -----
% pagestyle fancy
\footer

% custom pagestyle
\fancypagestyle{abstract}{%

Use the custom pagestyle with \thispagestyle{abstract}.


text:
# SSC CGL Tier 1 :: Quantitative Aptitude QA Test 7

## Home SSC CGL Tier 1 / Quantitative Aptitude QA Test 7 Questions and Answers

1 . The length of a rectangular plot is thrice its breadth. If the area of the rectangular plot is 7803 sq. metre, what is the breadth of the rectangular plot?
51 metres
153 metres
104 metres
88 metres
2 . Ratio of Rani’s and Komal’s age is 3 : 5 respectively. Ratio of Komal’s and Pooja’s age is 2 : 3 respectively. If Rani is twofifth of Pooja’s age, what is Rani’s age?
10 years
15 years
24 years
Cannot be determined
3 . In a mixture of milk and water the proportion of water by weight was 75% if in the 60 gms mixture 15 gms water was added, what would be the percentage of water in the new mixture?
75%
80%
90%
100%
4 . The sum of five numbers is 290. The average of the first two numbers is 48.5 and the average of last two numbers is 53.5. What is the third number?
72
84
96
86
5 . In how many different ways can the letters of the word ‘SMART’ be arranged?
25
60
180
120
6 . A sum of Rs.2,236 is divided among A, B and C such that A receives 25% more than C and C receives 25% less than B. What is A’s share in the amount?
Rs.460
Rs.890
Rs.780
Rs.1280
7 . The average weight of a group of 53 girls was calculated as 58 kgs. It was later discovered that the weight of one of the girls was read as 65 kgs., whereas her actual weight was 45 kgs. What is the actual average weight of the group of 53 girls ? (rounded off to two digits after decimal)
58.62 kgs
58.37 kgs
57.37 kgs
57.62 kgs
8 . What is the square root of 9 + $2\sqrt{4}$ ?
(a).$1+2\sqrt{2}$
(b).$\sqrt{3}$+$\sqrt{6}$
(c).$\sqrt{2}$+$\sqrt{7}$
(d).$\sqrt{2}$+$\sqrt{5}$
a
b
c
d
9 . An aeroplane flies along the four sides of a square at a speed of 100, 200, 300 and 400 km/h, respectively. What is the average speed of the plane in its flight around the square?
196 km/h
196 km/h
250 km/h
None of these
10 .
Only – 4
Only 4
4, – 4
2, – 2


text:
# Stuck on “grub>” prompt when dual-booting Ubuntu

On the dual boot option screen I choose Ubuntu and only get:

grub>

I want to get into Ubuntu and start learning it a bit but am stuck. What would be the best way to fix my problem?

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Can you change your title to more closely resemble your question? It should definitely have something about the grub prompt. –  Shawn J. Goff Jan 23 '11 at 15:23

@Cross Your post was converted into a comment. To reply to an answer or comment in the future, use the add comment button below the answer. This is a questions and answers site, not a forum where posts appear in chronological order. Please delete this answer and post its contents as a comment where it belongs. You can also now accept your actual answer (click the checkmark next to it). –  Gilles Jan 25 '11 at 19:52


text:
# What is the electron pair geometry for an ammonia molecule, NH_3?

$A {X}_{3} E$

#### Explanation:

![https://d2jmvrsizmvf4x.cloudfront.net/TxfeO8QJS4uL3w1g41Mv_nh3.jpg)

A = central atom (N)
X = peripheral atoms (H)
E = non-bonding electron pairs (lone pair )

This video gives a quick summary of molecular shapes and VESPR theory.

Hope this helps!


text:
Experimental results are presented that describe flow behavior inside circular injection holes with a sharp square-edged inlet. Oil-film flow visualizations and mean flow data are obtained in the flow symmetry plane of injection holes that are normally oriented to a crossflow. Additional visualizations inside inclined holes are also performed for inclination angles of 30 and 60 deg. Data are presented for three different length-to-diameter ratios: $L/D=0.5,$ 1.0, and 2.0. The blowing ratio is fixed at $M=2.0$ in the flow visualizations and takes the values $M=0.5,$ 1.0, and 2.0 in the flow measurements. The normal-injection flow visualization in the case of $L/D=2.0$ clearly demonstrates the existence of four distinct near-wall flow regions: an inlet separation region, a reattachment region, a developing region, and a near-exit flow region. When $L/D=1.0$ and 2.0, an inlet separation bubble is apparent with a clear imprint of recirculating flow traces, especially on the windward side, even though it is not so well organized on the opposite side. For a short hole such as $L/D=0.5,$ however, the separation bubble with flow recirculation seems to be suppressed by the crossflow. Due to the presence of the inlet separation bubble, actual flow passage is in the form of a converging–diverging channel, regardless of the L/D values. In general, the crossflow stabilizes the inside flow on the leeward side, meanwhile destabilizes it on the windward side. On the contrary, the inclination of the injection hole in the leeward direction of the crossflow stabilizes the flow near the windward wall but destabilizes it near the leeward wall. Relatively short holes such as $L/D=0.5$ and 1.0 do not allow the boundary-layer development on the wall. Particularly in the case of $L/D=0.5,$ a direct interference is observed between the complicated inlet and exit flows. The inlet flow, however, seems to be isolated from the exit flow for a long hole such as $L/D=2.0.$ It is also found that the potential-core inside the normal injection hole comprises a converging flow region, a diverging flow region, a developing flow region, and a flow region deflected by the crossflow.

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